Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequa... | The sum can be written as
\begin{align}
\frac{1}{n+1} + \frac{1}{n+3} + \ldots + \frac{1}{3n - 1} & = \sum_{i=1}^n \frac{1}{n + 2i - 1}.
\end{align}
Now recall the AM-HM inequality:
$$
\frac 1n\sum_{i=1}^n(n + 2i - 1) > \frac{n}{\sum_{i=1}^n \frac{1}{n + 2i - 1}}.
$$
(The requirement that $n > 1$ guarantees that the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1... | You are on the right road.... Continuing calculation,
\begin{align}
\frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2 B}}&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -1+2\cos B-\cos^2 B }{\sin^2 B}}\\
&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -\sin^2 B -\cos^2 B+2\cos B-\cos^2 B }{\sin^2 B}}\\
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to derive the equation of tangent to an arbitrarily point on a ellipse?
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$
I've tried implicit differentiation $\to \frac{2x... | To show that the given line is tangent to the ellipse, one can show that the system of equations
\begin{align}
\frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \tag{I} \\
\frac{xx_0}{a^2} + \frac{yy_0}{b^2} &= \tag{II} 1
\end{align}
with $x$, $y$ variables has exactly one solution—namely $(x_0, y_0)$. We further assume that $(x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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Numerical evaluation of first and second derivative We start with the following function $g: (0,\infty)\rightarrow [0,\infty)$,
$$ g(x)=x+2x^{-\frac{1}{2}}-3.$$ From this function we need a 'smooth' square-root. Thus, we check $g(1)=0$,
$$g'(x)=1-x^{-\frac{3}{2}},$$ $g'(1)=0$ and
$$g''(x)=\frac{3}{2}x^{-\frac{5}{2}}\... | For the first question, since $g(x)=\frac{(\sqrt{x}-1)^2(\sqrt{x}+2)}{\sqrt{x}}$, then
$$f(x)=sign(x-1)\sqrt{\frac{(\sqrt{x}-1)^2(\sqrt{x}+2)}{\sqrt{x}}}$$
The $sign(x-1)$ part is necessary for $f$ to be smooth, as $\lim_{x\rightarrow1^-}f'(x)\neq\lim_{x\rightarrow1^+}f'(x)$.
Indeed, if we write
$$f'(x)=\frac{1}{2}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f (x+xy+f(y) )= (f(x)+ \frac 12 )\ (f(y)+ \frac 12 \ ).$ Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that
$$f\left (x+xy+f(y) \right )=\left (f(x)+ \frac 12 \right )\left (f(y)+ \frac 12 \right ).$$
for every $x,y \in \mathbb R$.
My work... | From $f(-1) = -\frac{1}{2}$ we obtain
$$ f \left( - \frac{1}{2} \right) = f (f (-1)) = 0.$$
So by choosing $y= - \frac{1}{2}$ we obtain
$$f\left( \frac{1}{2} x \right) = f\left(x - \frac{1}{2} x + f\left( - \frac{1}{2} \right) \right) = \frac{1}{2} f(x) + \frac{1}{4}.$$
So $f(0) = \frac{1}{2}$.
Choosing $y=0$ gives
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Complex Addition: $(i+\sqrt3/2)^{200} + (i-\sqrt3/2)^{200}$ Solving a question, I need to find the value of following in between the solution.
$$\left(\frac{i+\sqrt3}{2}\right)^{200} + \left(\frac{i-\sqrt3}{2}\right)^{200}$$
The only useful thing I got was
$$\left(\frac{i+\sqrt3}{2}\right)^{100}\left(\frac{i-\sqrt3}{... | If $a=\dfrac{i+\sqrt3}2, b=\dfrac{\sqrt3-i}2$
$a+b=\sqrt3,ab=1\implies a^2+b^2=(a+b)^2-2ab=1$
So, $a,b$ are the roots of $t^2-\sqrt3t+1=0$
$\implies(t^2+1)^2=(\sqrt3t)^2$
$\iff t^4-t^2+1=0\implies(t^2+1)(t^4-t^2+1)=0\iff t^6=-1$
$\implies t^{2+6(2n+1)}=(t^6)^{2n+1}\cdot t^2=-t^2$
$\implies a^{12n+8}+b^{12n+8}=-(a^2+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Let A be a square matrix such that $A^3 = 2I$ Let $A$ be a square matrix such that $A^3 = 2I$
i) Prove that $A - I$ is invertible and find its inverse
ii) Prove that $A + 2I$ is invertible and find its inverse
iii) Using (i) and (ii) or otherwise, prove that $A^2 - 2A + 2I$ is invertible and find its inverse as a polyn... | For parts (i) and (ii):
\begin{align}
(A-I)^{-1} & = A^2+A+I \\
(A+2 I)^{-1} & = \frac{1}{10}(A^2-2A+4).
\end{align}
For (iii):
$$
A^2-2A+2I=A^2-2A+A^3=A(A+2I)(A-I)
$$
The inverse of $A$ is $\frac{1}{2}A^2$. So,
\begin{align}
(A^2-2A+2I)^{-1}&=(A-I)^{-1}(A+2I)^{-1}A^{-1}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Strange behavior of infinite products $\prod^{\infty}_{n=1} \ln (1+ \frac{1}{n} )^n$ and $\prod^{\infty}_{n=1} \ln (1+ \frac{1}{n} )^{n+1}$ There are two expressions marking the lower and upper bounds for number $e$:
$$\left(1+\frac{1}{n} \right)^n \leq e \leq \left(1+\frac{1}{n} \right)^{n+1}$$
Naturally, I wanted to ... | Note $$\log\left(1+\frac 1n\right)=\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac1{n^2}\right)$$
So $$\log\left(1+\frac 1n\right)^n=n\log\left(1+\frac 1n\right)= 1-\frac{1}{2n}+o\left(\frac1n\right).$$
Now, since $\prod \left(1-\frac{1}{2n}\right)$ does not converge to a positive value, neither does the left side product.
You... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Prove $\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$ is always divisible by $6$ when $n$ is an integer.
Prove $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$$ is always divisible by $6$ when $n$ is an integer.
I have done a similar proof that $\binom{2n}{n}$ is divisible by $2$ by showing that $$\binom{2n}{n}=\binom{2n-1}{n-1}+... | $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!} = \binom{3n}{n}\binom{2n}{n}$$
$$\binom{3n}{n} = \binom{3n -1}{n-1} +\binom{3n-1}{n} = \frac{(3n - 1)!}{(n-1)!(3n -1 -(n-1))!} + \frac{(3n - 1)!}{n! ((3n -1 -n)!}$$
$$=\frac{(3n-1)!)}{(n-1)!(2n-1)!}(\frac{1}{2n} + \frac{1}{n})$$
$$=\frac{(3n-1)!)}{(n-1)!(2n-1)!} \frac{3}{2n}$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
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Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers... Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers. Prove that $7$, $1 + 2\sqrt {5}i$, and $1 - 2\sqrt {5} i$ are all prime in $C$.
-I am really lost in this qu... | Hint: Use the norm: $N(a+b\sqrt 5i)= a^2+5b^2$.
Suppose $7=ab$. This implies $N(u)N(v) = N(7)=49$.
$N(u)=1, N(v)=49$, write $u=a+i\sqrt5 b$, $N(u) =a^2+5b^2=1$ implies $b=0, a^2=1$.
$N(u)=7=N(v)$. Write $u=a+ i\sqrt5 b$, $N(u) =a^2+5b^2=7$ we deduce that $\mid b\mid \leq 1$. Suppose that $\mid b\mid =1$, $a^2 +5=7$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trouble proving through induction after establishing a basecase What is the following sum?
$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + ... + \frac{1}{(n-1)n}$
Experiment, conjecture the value, and then prove it by induction.
I found the sum to be $\frac{n-1}{n}$ and established the base of n=2 being... | By induction assume that:$\frac 1 2 +\frac 1 6 +\frac {1}{12}+\dots \frac{1}{n(n-1)}=\frac{n-1}{n}$
For $n+1:$
$\underbrace{\frac 1 2 +\frac 1 6 +\frac {1}{12}+\dots \frac{1}{n(n-1)}}_{=\frac{n-1}{n}}+\frac{1}{n(n+1)}\stackrel{\color{red}?}{=}\frac{n}{n+1}$
$$\frac{1}{n(n+1)}+\frac{n-1}{n}=\frac{1+(n-1)(n-1)}{n(n+1)}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1655882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the sum of all $abc$
Let $T$ be the set of all triplets $(a,b,c)$ of integers such that $1 \leq a < b < c \leq 6$. For each triplet $(a,b,c)$ in $T$ , take number $a \cdot b \cdot c$. Add all these numbers corresponding to all the triplets in $T$. Prove that the answer is divisible by $7$.
Attempt
Let $S = \disp... | For any such triplet $(a, b, c)$ with $a<b<c$,
we have a corresponding triplet $(7-c, 7-b, 7-a)$ with $7-c<7-b<7-a$.
Since we can pair up all the triplets in this manner,
and $abc+(7-c)(7-b)(7-a)\equiv 0\pmod{7}$,
the sum of the products of each of the triplets is divisible by 7.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve $3 = -x^2+4x$ by factoring I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$.
\begin{align*}
3 & = -x^2 + 4x\\
x^2-4x+3 & = 0
\end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but... | $$ 3 = -x^2 +4x \implies \ x^2 -4x+3 = 0$$
$$ x^2 - 3x - x +3 = 0$$
$$ x(x-3)-1(x-3)$$
$$ (x-3)(x-1) = 0$$
$$x = 1, x = 3$$
For more on the methods, visit this link.
Alternatively, you could use the quadratic formula:
$$x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
This was obtained from the quadratic equation:
$$ax^2 + bx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$ If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$
Ideas for solution include factorizing the expression into a multiple of $x+y$ and expressing the left hand side as a sum of some perfect square expressions.
| Let $s=x+y$ and $d=x-y$. Then $x=(s+d)/2$ and $y=(s-d)/2$. Making this substitution, we find that
$$
x^2+y^2-xy-x-y+1=\frac{3d^2}{4}+\frac{(s-2)^2}{4}.
$$
Hence, being a sum of squares, if the original expression is zero, $d=0$ and $s-2=0$. This means that $s=x+y=2$, with $x=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: L... | For the inductive step, it suffices to show $\sqrt{n+1} - \sqrt{n} < \frac {1}{\sqrt{n+1}} = \frac{\sqrt{n+1}}{n+1}$ since that implies the inequality gets stronger as $n$ increases. This is clear since
$$\sqrt{n+1} - \sqrt{n} = \frac{\sqrt{n+1}}{n+1} + \frac{(-\sqrt{n}\sqrt{n+1})(\sqrt{n+1} - \sqrt{n})}{n+1}$$
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are,... | $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)$$
The key here is the identity (for all $n\in\mathbb Z_{\ge 3}$):
$$a^n+b^n+c^n=(a+b+c)\left(a^{n-1}+b^{n-1}+c^{n-1}\right)-$$
$$-(ab+bc+ca)\left(a^{n-2}+b^{n-2}+c^{n-2}\right)+abc\left(a^{n-3}+b^{n-3}+c^{n-3}\right)$$
Therefore: $$a^3+b^3+c^3=3abc\\ a^4+b^4+c^4=2(ab+bc+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 2
} |
Integrate $I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$
$$I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$$
My Endeavour :
\begin{align}I&= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x\\ &= \int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x - \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x\end{align}
\begin{align}\te... | Here is another approach.
Let $x^2=\tan(\phi)$
$$
\begin{align}
\int\frac{x^2-1}{x\sqrt{1+x^4}}\,\mathrm{d}x
&=\frac12\int\frac{x^2-1}{x^2\sqrt{1+x^4}}\,\mathrm{d}x^2\\
&=\frac12\int\frac{\tan(\phi)-1}{\tan(\phi)\sec(\phi)}\,\mathrm{d}\tan(\phi)\\
&=\frac12\int(\sec(\phi)-\csc(\phi))\,\mathrm{d}\phi\\
&=\frac12\log(\se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
I know how to find equation with numbers, but got really confused with this one. If anybody could break it dow... | The function $y = \sqrt{x} = x^{\frac{1}{2}}$ has derivative
$$y' = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$$
as you found. The tangent line to the graph of $y = \sqrt{x}$ at the point $(a, \sqrt{a})$ has slope
$$y'(a) = \frac{1}{2\sqrt{a}}$$
which is defined for $a > 0$.
You made a mistake when you ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\righ... | Let us shift the variable by $3$ and get
$$\lim_{x\to\infty}\sqrt[5]{(x-2)(x-1)x(x+1)(x+2)}-x+3=\lim_{x\to\infty}x\left(\sqrt[5]{1-\frac5{x^2}+\frac4{x^4}}-1\right)+3.$$
Then by L'Hospital,
$$\lim_{t\to0}\frac{\sqrt[5]{1-5t^2+4t^4}-1}t=\lim_{t\to0}\frac{-10t+16t^3}{5\sqrt[5]{1-5t^2+4t^4}}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 0
} |
Different constants of integration on different intervals I tried to prove that on the interval $(-1,\infty)$ that: $$\arctan\left(\frac{x-1}{x+1}\right)=\arctan(x)-\frac{\pi}{4}$$
So I defined: $$f(x)=\arctan\left(\frac{x-1}{x+1}\right)-\arctan(x)$$
Then apparently: $$f'(x)=0$$
Therefore $f(x)$ is constant thus:
$$f(x... | From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),
$$\arctan x+\arctan y=\begin{cases} \arctan\frac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\frac{x+y}{1-xy} & \mbox{if } xy>1 \\ \text{sgn}(x)\dfrac\pi2 & \mbox{if } xy=1 \end{cases} $$
As $\arctan(-u)=-\arctan u,$
Setting $y=-1,$
$$\arctan x-\dfrac\pi4=\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Geometric problem about triangle [Edited]
Let $|BC|=|AD|=1, \quad\angle BCA=\frac{\pi}{2}$ and $\angle ACD=\frac{\pi}{6}$. Find $|AB|$.
Can anyone show how to solve it?
| Put $|BC|=:a$, $|AB|=:c$, $|AC|=:b$, and denote the angle at $D$ by $\beta$. Two applications of the sine law then give
$${a+c\over\sin 120^\circ}={a\over\sin\beta},\qquad {a\over\sin30^\circ}={b\over\sin\beta}\ .$$
Dividing the first of these by the second we obtain
$${a+c\over a}\cdot{1/2\over\sqrt{3}/2}={a\over b}={... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Number of $4\times3$ matrices of rank 3 over a field with 3 elements. I am finding number of $4\times3$ matrices of rank 3 over a field with 3 elements. If i count it as number of linearly independent columns i.e $3$ then its answer is $(3^{4}-1)(3^{4}-3)(3^{4}-3^{2}).$ But when i like to obtain the same formula as num... | Start with the empty set and successively add the rows. The dimension of the space spanned by the set must increase from $0$ to $3$, and it can increase by at most $1$ in each step, so there must be exactly $3$ steps where it's incremented and one step where it stagnates. The incrementing steps always have the same cou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the absolute value and argument for $(3+4i)^{-1}$ When I have to find the argument and absolute value of $z=(3+4i)^{-1}$ I thought that I had to find the inverse of $z$ by applying the rule $z^{-1}=\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right)$ but the answer suggests that $\lvert(3+4i)^{-1}\rvert$=$\frac{1}{\... | The inverse of a complex number $z=a+bi$ is given by:
$$z^{-1} = \frac{1}{a+bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a-bi}{a^2+b^2}.$$
Thus, its absolute value is
$$|z^{-1}| = \sqrt{\frac{a^2}{(a^2+b^2)^2} + \frac{b^2}{(a^2+b^2)^2}} = \frac{\sqrt{a^2+b^2}}{a^2+b^2} = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{|z|}.$$
Using the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2})$ - Unique factorization? Personal question : We know that $5 + \sqrt{2}$, $2-\sqrt{2}$, $11-7\sqrt{2}$ and $2+\sqrt{2}$ are irreductible in $\mathbb{Z}[\sqrt{2}]$ and that $$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2}).$$ Why this fact doesn't contradict t... | This is because the factorisation is unique up to units, and $\Bbb Z[\sqrt 2]$ has units, for example $(\sqrt 2+1)$.
In $\Bbb Z$, you had $15 = 3 \times 5 = (-3) \times (-5)$. This doesn't contradict the unique factorisation theorem either because each factor of one factorisation differs from a factor in the other by a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+C... | One may observe that
$$
\int \frac{2t}{\left(1+t^2\right)^2} \, dt=\int \frac{\left(1+t^2\right)'}{\left(1+t^2\right)^2} \, dt=-\frac{1}{ 1+t^2}
$$ then, using an integration by parts, one has
$$
\int \frac{2t^2}{\left(1+t^2\right)^2} \, dt=t \times \left(-\frac{1}{ 1+t^2} \right)+\int \frac1{\left(1+t^2\right)} \, dt=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
row exchange results in wrong answer? This is probably a very simple, but why is it that when I do row exchanges I end up with the wrong answer. I have the matrix equation $Ax = B$, as:
$$
\underbrace{
\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -2 \\ -1 & -1 & 1 \end{bmatrix}
}_A
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatr... | $$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
0 & 1 & -2 & 1 \\
-1 & -1 & 1 & -6 \\
\end{array}\right]$$
Exchange row 2 and 3 :
$$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
-1 & -1 & 1 & -6 \\
0 & 1 & -2 & 1 \\
\end{array}\right]$$
Add row 1 to row 2:
$$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this pattern occur? I first saw this pattern when I was trying to factor quadratics.
let $a + b = k$ where $k$ is any constant,
Now let $(a - b)/2 = x$
It appears to be that $(k^2/4) - x^2 = a*b$
For example, let's say $k = 8$
then,
$4*4 = 16$
$3*5 = 15$
$2*6 = 12$
$1*7 = 7$
$0*8 = 0$
$-1*9 = -9$
If we subtr... | In the original post, $x=a-b$ and was changed to $x=\frac{a-b}{2}$. This answer addresses both formulae.
Consider
$$
\frac{k^2}{4}-x^2=\frac{(a+b)^2}{4}-(a-b)^2=\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}-a^2+2ab-b^2=-\frac{3}{4}a^2+\frac{5}{2}ab-\frac{3}{4}b^2=-\frac{3}{4}(a-b)^2+ab
$$
This is not $ab$, except in fairly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum of the series $\sum_{k=1}^\infty ((\frac{x+1}{2})^2)^k$ Find the sum of the series for those values of $$\sum_{k=1}^\infty ((\frac{x+1}{2})^2)^k$$
I have found that $$a=\frac{x^2+2x+1}{4}$$ and $$r=\frac{x^2+2x+1}{4}$$
The sum of the series would than equal $$=\frac{a}{1-r}=\frac{(\frac{x^2+2x+1}{4})}{1-(\frac{x^2... | Note that $r=\left(\frac{x+1}{2}\right)^2$. We have
$$\left(\frac{x+1}{2}\right)^2\lt 1$$
if and only if
$$-1\lt \frac{x+1}{2}\lt 1.$$
The right-hand inequality holds if and only if $x\lt 1$.
The left-hand inequality holds if and only if $-3\lt x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving this $\gcd(2^n-1,3^n+2)=1$ for all postive integers $n$ I have found
$$\gcd(2-1,3+2)=\gcd(1,5)=1$$
$$\gcd(2^2-1,3^2+2)=\gcd(3,11)=1$$
$$\gcd(2^3-1,3^3+2)=\gcd(7,29)=1$$
$$\gcd(2^4-1,3^4+2)=\gcd(15,83)=1$$
$$\gcd(2^5-1,3^5+2)=\gcd(31,245)=1$$
$$\cdots\cdots$$
I conjecture $\gcd(2^n-1,3^n+2)=1$,I can't prove th... | I have checked with Sage that, for $n=176$, we have
\begin{align*}
2^n-1 &= 95780971304118053647396689196894323976171195136475135 \\
3^n+2 &= 940461086986004843694934910131056317906479029659199959555574885740211572136210345923,
\end{align*}
and $gcd(2^n-1,3^n+2) = 257$.
$n=176$ is the smallest. The next ones are: 432, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and... | \begin{align}
\int x^{13}\sqrt{x^7+1}\,dx&=\int x^6(x^7+1)\sqrt{x^7+1}\,dx-\int x^6\sqrt{x^7+1}\,dx\\
&=\int x^6(x^7+1)^{3/2}\,dx-\int x^6(x^7+1)^{1/2}\,dx
\end{align}
Then, use $u=x^7+1$ in order to evaluate these integrals.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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2,3,7 is the only triple $\geq 2$ that satisfies this property The property being that "taking the product of two of the numbers and adding one yields a number that is divisible by the third".
Clearly, this holds for 2,3,7 since
*
*$2\cdot 3 + 1 = 1\cdot 7$
*$2\cdot7 + 1 = 5\cdot 3$
*$3\cdot 7 + 1 = 11\cdot 2$
b... | HINT
Since $a|bc+1$, $b|ca+1$, $c|ab+1$, multiplying these together gives us that $$abc|ab+bc+ca+1$$
Note that $$a,b,c \ge 4 \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}<1 \Leftrightarrow abc>ab+bc+ca+1$$
This gives us that one of $a,b,c$ must be smaller than $4$.
I think you can continue to divide c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$ Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$
The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\... | Hint:
$$\cot\theta + \sec\theta = \frac{1}{\tan\theta} + \frac{1}{\cos\theta} = \frac{\cos\theta + \tan\theta}{\tan\theta\cos\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$ Then $\lfloor A \rfloor\;\;,$
If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$
My Try:: Using $\b... | Let $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}=\frac 1K$$
$$K=\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}<\frac{1}{1980}+\frac{1}{1980}+\frac{1}{1980}+........+\frac{1}{1980}=\frac{33}{1980}=\frac{1}{60}$$
$$A=\frac 1K >60$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Find all integer solutions for $x*y = 5x+5y$ For this equation $x*y = 5x + 5y$ find all possible pairs.
The way I did it was: $x=5y/(y-5)$
And for this I wrote a program to brute force a couple of solutions. If it helps, some possibilities are: [4,-20], [6, 30], [10, 10]
So my question is: What is a mathematically ... | $$5x+5y-xy=0$$
$$5x+5y-xy-25=-25$$
$$x(5-y)-5(5-y)=-25$$
$$(x-5)(y-5)=25$$
1) $x-5=1; y-5=25; \Rightarrow x=6; y=30$;
2) $x-5=-1; y-5=-25; \Rightarrow x=4; y=-20$;
3) $x-5=5; y-5=5; \Rightarrow x=10; y=10$
4) $x-5=-5; y-5=-5; \Rightarrow x=0; y=0$
5) $x-5=25; y-5=1; \Rightarrow x=30; y=6$
6) $x-5=-25; y-5=-1; \Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integrate square root of quotient polynomials I need to estimate this integral:
$$\displaystyle\int_a^\lambda \sqrt{\frac{(z-a)(z-b)}{z(z-1)}}dz$$
where $a$ and $b$ are closed to 1/2, and $z$ is complex.
Mathematica does not know how to do that. Any idea or reference ?
| Not quite what you asked but using Mathematica 10.2 with the input:
FullSimplify[Integrate[Sqrt[(z - a) (z - b)/(z (z - 1))], z]]
I got the following output:
$$\left(\sqrt{\frac{(-a+z)(-b+z)}{(-1+z)z}}\left((a-b)(b-z)\sqrt{\frac{(-1+b)(a-z)}{(a-b)(-1+z)}}z\sqrt{\frac{b-z}{b-bz}}+(a-b)b\sqrt{\frac{(-1+a)(b-z)}{(-a+b)(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Cauchy integral formula- Evaluate the integrals The contour $\Gamma$ is parameterised by $r:[-\pi,\pi] \rightarrow \mathbb{C}$ is given by $r(\theta)=3e^{i\theta}+1$.
(a) $\int_{\Gamma} \frac{\sin(z)}{z-1}\ dz = 2 \pi i \sin(1)$?
(b) $\int_{\Gamma} \frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-1)(z-2)}\ dz = \frac{\frac{\sin(... | The first one is correct, but you forgot to take into account all of the isolated singular points. Let's work through the problems one by one.
Throughout this post I will use the theorem that if we can write $f(z)=\frac{p(z)}{q(z)}$ where $p(z_0)\neq 0$, $q(z_0)=0$, and $q'(z_0)\neq0$ for some isolated singularity $z_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$ converge? $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$
and I need to check whether this sequence converges to a limit without finding the limit itself. I think about using... | $$\frac{1}{\sqrt{n^2+2n-1}}+\frac{1}{\sqrt{n^2+2n-1}}\cdots+\frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}\cdots+\frac{1}{\sqrt{n^2+n}}$$
$$\frac{n}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving for $z$ given 3 constraints: $\DeclareMathOperator{\Re}{Re}\Re[z^4]=1/2$ , $z\bar{z}+2|z|-3=0$, $\arg z \leq \frac{\pi}{4}.$
Let $z$ be a complex number satisfying
$$\DeclareMathOperator{\Re}{Re}\Re[z^4]=\frac{1}{2}$$
$$z\bar{z}+2|z|-3=0$$
$$\arg z \leq \frac{\pi}{4}.$$
Find $z$
As a side note this is the s... | The third condition gives you - Both $x$ and $y$ are positive and $x>y$
So, from your set of answers(assuming they are right), the answer would be
$x=\frac{\sqrt 6+\sqrt 2}{4}$
$y=\frac{\sqrt 6-\sqrt 2}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $4k^3 + 17k^2 - 228k -1116 = 0$ The equation given to me is $$4x^4 + 16x^3 - 17x^2 - 102x -45 = 0$$
I'm asked to find it's resolvent cubic which is not so difficult to find. But the problem is that the question further asks to find the solution of resolvent cubic.
I have found resolvent cubic using Ferrari's me... | Ferrari's method: If $a$, $b$, $c$, $d$ are the roots of the quartic, form the equation with roots
$$a b + c d, a c + b d, a d + b c$$
The coefficients of this equation are symmetric in $a$, $b$, $c$, $d$ ( since we have $a b + c d$ and all its symmetric transforms).
After we solve the cubic, the quadric is solved usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Presentation of the unit Let $N -$ number of fives $(a_1, a_2, a_3, a_4, a_5)$ positive integers,
satisfying the condition
$$\frac {1}{a_1}+\frac {1}{a_2}+\frac {1}{a_3}+\frac {1}{a_4}+\frac {1}{a_5}=1.$$
Find out even or odd number is $N$.
My work so far:
$$\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=... | HINT
You just have to determine if $N$ is even or odd.
If $a_1, a_2, a_3, a_4, a_5$ are all distinct, there are $120$ ways to arrange them.
Similarly, if two of them were the same there are $60$ ways to arrange them.
In both cases, there are an even number of ways to arrange them.
However, these cases, which are pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| You are better off using partial fractions, which you should know if you have been given this integral to evaluate. You have the following factors for the denominator:
$x^4+x^2-2
= (x^2-1)(x^2+2)
= (x+1)(x-1)(x^2+2)$
And then
$x^2/(x^4+x^2-2) = a/(x+1)+b/(x-1)+(cx+d)/(x^2+2)$
You are to find the condtants a, b, c, d t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Simultaneous Diagonalization of two bilinear forms I need to diagonalize this two bilinear forms in the same basis (such that $f=I$ and $g$=diagonal matrix):
$f(x,y,z)=x^2+y^2+z^2+xy-yz $
$g(x,y,z)=y^2-4xy+8xz+4yz$
I know that it is possible because f is positive-definite, but I don't know how can I do it
| As I commented, a more difficult version is at Congruence and diagonalizations
This one is easier, the matrix $C$ has three distinct eigenvalues that are integers.
$$
A =
\left(
\begin{array}{rrr}
1 & \frac{1}{2} & 0 \\
\frac{1}{2} & 1 & -\frac{1}{2} \\
0 & -\frac{1}{2} & 1
\end{array}
\right)
$$
$$
B =
\left(
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to expand $f(z)$ as a Laurent Series Given
$$\frac{z}{(z-1)(z+2i)}$$
expand $f(z)$ in the following regions: $|z|<1$, $1<|z|<2$, $|z|>2$
I'm preparing for an exam and Laurent Series are a weakness of mine. I would love advice regarding interpretation, understanding, and solution of the problem.
| Here is a somewhat detailed description of the Laurent expansion of $f(z)$ around $z=0$.
The function
\begin{align*}
f(z)&=\frac{z}{(z-1)(z+2i)}
=\frac{1-2i}{5}\cdot\frac{1}{z-1}+\frac{4+2i}{5}\cdot\frac{1}{z+2i}\\
\end{align*}
has two simple poles at $1$ and $-2i$.
Since we want to find a Laurent expansion with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 \cdot 1 $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 \cdot (\sin^2(x)+\cos^2(x)) $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5\sin^2(x)+5\cos^2(x) $$
$$ \sin^2(x)+\sin(x)\cos(x)-6\cos^2(x)=0$$
$$\tan^2(x)+\tan(x)-6=0$$
$\tan(x)=2$ or $\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
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Which two digit number when you find the product of the digits yields a number that is half the original? Which two digit number when you find the product of the digits yields a number that is half the original?
Let x=$ab$ be the $2$-digit number. So $x=10a+b$.
Then $ab=\frac{x}{2} \implies ab=\frac{10a+b}{2} \implies ... | From what you figured out so far, $ab=5a+\frac{b}{2}$. If there is an integer solution, $b$ is even and greater than $5$, so $b$ is either $6$ or $8$. Also (see the picture), $\frac{b}{2}$ is divisible by $b-5$, which rules out $b=8$. If $b=6$, then the rightmost rectangle below is $1$ unit across and has area $\frac{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove $x^2+y^4=1994$ Let $x$ and $y$ positive integers with $y>3$, and $$x^2+y^4=2(x-6)^2+2(y+1)^2$$
Prove that $x^2+y^4=1994$.
I've tried finding an upper bound on the value of $x$ or $y$, but without sucess. Can anyone help me prove this problem? Note that $x^2+y^4=1994$ is the result we are trying to prove, not an a... | Rearrange the equation given into
$$
(x-12)^2 = 71 + (y^2-1)^2 - 4y.
$$
$(y^2-1)^2$ is a square. The square before $(y^2-1)^2$ is $(y^2-2)^2$, which is $2y^2-1$ less. For $y > 2$, $2y^2-1 > 4y > 4y-71$. The square after $(y^2-1)^2$ is $y^4$, which is $2y^2+1$ greater. For $y\geq 6$, $2y^2+1 > 71 > 71-4y$.
You the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Unable to find solution for $a^2+b^2-ab$, given $a^2+b^2-ab$ is a prime number of form $3x+1$ I have a list of prime numbers which can be expressed in the form of $3x+1$. One such prime of form $3x+1$ satisfies the expression: $a^2+b^2-ab$.
Now I am having list of prime numbers of form $3x+1$ (i.e., $7,19 \ldots$). But... | Use the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$, and note that, by Little Fermat, $x^3\equiv x\mod 3$, so $a^3+b^3\equiv a+b \mod3$. Hence, if $a+b\not\equiv 0\mod 3$, necessarily $a^2-ab+b^2\equiv 1\mod3$.
Now suppose you've found $a$ and $b$ such that $a^3+b^3$ is the product of two primes. Then one of them will be con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$ If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$
Let $y=\frac{x^2+ax+b}{x^2+2x+3}$
$$x^2y+2xy+3y=x^2+ax+b$$
$$x^2(y-1)+x(2y-a)+3y-b=0$... | Begin with
$$g(y)=-8y^2+y(-4a+4b+12)+a^2-4b\geq 0,\quad\forall y\in[-5,4].$$
Then we wish to solve $g(-5)=g(4)=0$, that is,
\begin{align}
-200-5(-4a+4b+12)+a^2-4b=0,\\
-128+4(-4a+4b+12)+a^2-4b=0,
\end{align}
or
\begin{align}
a^2+20a-24b&=260,\\
a^2-16a+12b&=80.
\end{align}
Thus
\begin{align}
a-b=5\quad\mbox{and}\quad
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
solve $3^x +4^x = 5^x$ in $\Bbb R$ solve the following equation :
$$3^x +4^x = 5^x \text{ in } \Bbb R$$
the trivial answer is $x=2$ .
| $$\left( \frac{3}{5} \right)^x+\left( \frac{4}{5} \right)^x=1$$
Then $x=2$
$f(x)=\left( \frac{3}{5} \right)^x+\left( \frac{4}{5} \right)^x$decreasing function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the largest integer $n$ such that $n^2$ is the difference of two consecutive cubes and $2n +79$ is a perfect square.
Find the largest integer $n$ such that $n^2$ is the difference of two consecutive cubes and $2n +79$ is a perfect square.
This is an AIME problem. I have been trying and have been going round in c... | Let $$n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1$$
Note that $$(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2$$
Since $$\gcd(2n-1,2n+1)=1$$
Since their product is three times a square, one of them must be a square and the other three times a square. If we have $2n - 1=3a^2$, then $2n + 1=b^2=3a^2+2 \equiv 2 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What type of functional equation is this? I'm trying to solve the following functional equation
$f\left(x\right)=A\mbox{ exp}\left\{ \int\frac{1}{f\left(x\right)x^{2}+Bx}dx\right\}$
where $f\left(x\right):\mathbb{R}_{+}\rightarrow\mathbb{R}_{\geq0}$, A and B are constants in $\mathbb{R}$.
Does anybody recognize this ty... | First, there is no concept about "indefinite integral equation" , so you should modify the question as $f(x)=Ae^{\int_k^x\frac{1}{f(x)x^2+Bx}dx}$
$\ln\dfrac{f(x)}{A}=\int_k^x\dfrac{1}{f(x)x^2+Bx}dx$
$\dfrac{1}{f(x)}\dfrac{df(x)}{dx}=\dfrac{1}{f(x)x^2+Bx}$ with $f(k)=A$
$f\dfrac{dx}{df}=fx^2+Bx$ with $x(A)=k$
Case $1$: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Equation of sphere through variable points and origin Question : Find the equation of the sphere through four points $(0,0,0) , (-a,b,c) , (a,-b,c)$ and $(a,b,-c)$. Also find the centre and the radius of the sphere.
Now I know that the sphere passes through origin. Therefore, the constant term in the equation of the sp... | The four points form two triangles on the sphere. The planes on which these triangles lie, intersect the sphere in two circles, which are the circumscribed circles of the two triangles. The two lines perpendicular to the planes, through the center of the circles, intersect each other in the center of the sphere. The ce... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $F(\sqrt{a+b+2\sqrt{ab}}) = F(\sqrt{a},\sqrt{b})$? Let $F$ be a field of characteristic $\neq 2$. Let $a \neq b \in F$, and $F(\sqrt{a},\sqrt{b})$ is of degree 4 over $F$. I've shown that $F(\sqrt{a}+\sqrt{b}) = F(\sqrt{a},\sqrt{b})$.
Observe that $x=\sqrt{a} + \sqrt{b}$ and $x = \sqrt{a+b+2\sqrt{ab}}$ both sa... | As you and @D_S showed, $F(\sqrt{a+b+2\sqrt ab})=F(\sqrt a+\sqrt b)$ since they have the same minimal polynomial, so it suffices to show that $F(\sqrt a+\sqrt b)=F(\sqrt a, \sqrt b)$, which can be shown by showing that $\sqrt a \in F(\sqrt a+\sqrt b)$ (showing that $\sqrt b \in F(\sqrt a+\sqrt b)$ is very similar and s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a$,$b$ and $c$ are roots of the equation $x^3-x^2-x-1=0$ The roots of the equation $x^3-x^2-x-1=0$ are $a$,$b$ and $c$.
if $n \gt 21 $ and $n \in \mathbb{N}$ The find the possible values of $$E=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$ in $[0 \: 2]$ are?
Since $x^3=x^2+x+1$ from the graphs of $x^3$... | HINT: If
$E_n=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$
then, because of $x^3=x^2+x+1$ you have $x^n=x^{n-1}+x^{n-2}+x^{n-3}$ and:
$E_n=E_{n-1} + E_{n-2} + E_{n-3}$.
It is easy to find out the first values of this recurrence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha s... | We consider $$f(x) = x - 4x^2 - 8x^3 $$
Notice that for $x \ge \frac{1}{2}, \>\> f(x) < -1$ siince $f(\frac{1}{2}) = -\frac{3}{2} $ and $f'(\frac{1}{2}) < 0$ and $f''(x) < 0$ for all $x > -1$
Now it's only important to show that $\cos \frac{2 \pi}{7} \ge \frac{1}{2}$
It should be relatively easy to show $\cos \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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Riemann Sum to show convergence help? I have the function $f(x)=\sin\left(\dfrac{\pi x}{2}\right)$ on the partition of $[0,1]$ given by $$P_{n}: 0 < \frac{1}{n} < \frac{2}{n} < ... < \frac{n-1}{n} < 1$$
I have shown that $$L(f,P_{n})= \sum^{n-1}_{j=0} \frac{1}{n}\sin\left(\frac{j \pi}{2n}\right)$$
However i don't know ... | It appears you need to compute the limit of the lower Riemann sum using some analytical tools -- as opposed to declaring simply that it converges to the integral.
The following identity is useful
$$\sum_{j=1}^n \sin (jx) = \frac{\sin\left( \frac{nx}{2}\right)\sin\left( \frac{(n+1)x}{2}\right)}{\sin\left( \frac{x}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac... | $$(\sin(x) + \cos(x))^2 = 1/9$$
$$\sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) = 1/9$$
$$1 + \sin(2x) = 1/9$$
$$\sin(2x) = -8/9$$
$$x = \arcsin(-8/9) / 2$$
$$ x = -0.547457... $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
Show that the group is abelian Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$ with $x,y,z\in M$.
I have shown that all the elements of the center $Z(G)$ are the matrices of the form $\begin{pmatrix}
1 & 0 & \tilde{y} \\... | The multiplication is
$$
\begin{pmatrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & x+a & y+az+b \\
0 & 1 & z+c \\
0 & 0 & 1
\end{pmatrix}
$$
Similarly
$$
\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find last two digits of $33^{100} $ Find last two digits of $33^{100}$.
My try:
So I have to compute $33^{100}\mod 100$
Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$
So we have $33^{40}\equiv 1 \pmod{100}$
Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$
Since $100=2\cdot40+20$ so we have $33^{100}=1... | $33^{100} \equiv (33^2)^{50} \equiv 1^{50} \equiv 1 \bmod 4$, because $\varphi(4)=2$.
And $33^{100} \equiv (33^{20})^5 \equiv 1^5\equiv 1 \bmod 25$, because $\varphi(25)=20$.
So $33^{100} \equiv 1 \bmod 100$.
Similarly, $a^{100} \equiv 1 \bmod 100$ for all $a$ coprime to $100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Closed formula for finite product series I need to solve the recurrence:
$$
\begin{align*}
T(n) &= kT\left(\frac{n}{2}\right) + (k - 2)n^3 \\
&\textit{where}\; k \in \mathbb{Z}: k \geq 2 \\
&= k(kT\left(\frac{n}{4} + \frac{(k - 2)n^3}{2}\right) + (k - 2)n^3 \\
&= k^2T\left(\frac{n}{4}\right) + \f... | Here is a quick way to bound many kinds of functions defined by a recurrence if one is interested in the function's behaviour for large $n$.
Let $\lg x=\log_2 x$.
Suppose we are given a recurrence of the form $f(n)=cf(n/2)+p(n), f(1)=1$, where $p(n)$ is a function depending on $n$ and $c$ is a constant.
If $p(n)\ge 0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate the integral using the theory of residues: $\int_0^{2\pi} \frac{(\cos \theta)^2 d \theta}{3-\sin \theta}$ $$\int_0^{2\pi} \frac{(\cos \theta)^2 d \theta}{3-\sin \theta}$$
I''m having trouble simpliyfing this into a form that will allow me to use Residue Theorem. I got it to the point where the integrand looks ... | On the unit circle, Using the substitution $z=e^{i\theta}$, we get
$$
\begin{align}
\sin(\theta)&=\frac1{2i}\left(z-\frac1z\right)\\
\cos(\theta)&=\frac12\left(z+\frac1z\right)\\
\mathrm{d}\theta&=\frac{\mathrm{d}z}{iz}
\end{align}
$$
Therefore, on a counter-clockwise contour along the unit circle,
$$
\begin{align}
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate: $\int x\sqrt{\frac{a-x}{a+x}}\hspace{1mm}dx$
Evaluate: $$\int x\sqrt{\dfrac{a-x}{a+x}}\hspace{1mm}\mathrm{d}x$$
I don't know where to start. Hints/suggestions will be appreciated
| $$
\frac{a-x}{a+x} = u^2 \rightarrow x = a\frac{1-u^2}{1+u^2} \rightarrow dx = \frac{-4au}{(1+u^2)^2}du
$$
$$
\int x\sqrt{\dfrac{a-x}{a+x}}\hspace{1mm}\mathrm{d}x = \int a\frac{1-u^2}{1+u^2} u\hspace{1mm}\frac{-4au}{(1+u^2)^2}du = -4a^2 \int \frac{u^2(1-u^2)}{(1+u^2)^3}du =
$$
$$
-4a^2 \int \frac{u^2(1-u^2)}{(1+u^2)^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the number of words each consisting of $3$ consonants and $3$ vowels that can be formed from the letters of the word CIRCUMFERENCE. Find the number of words each consisting of $3$ consonants and $3$ vowels that can be formed from the letters of the word CIRCUMFERENCE.In how many of these $c$'s will be together?
T... | Arrangements of the vowels:
You correctly calculated that the number of distinguishable ways to arrange three E's is $1$ and that the number of ways to arrange three different vowels is $3!$. However, you calculated the number of arrangements with two E's incorrectly.
There are $\binom{3}{2}$ ways of selecting the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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An integral calculus I know that $\displaystyle \frac{1}{2i\pi}\int_0^{2\pi}\frac{ib \cos t - a \sin t}{a\cos t+ib\sin t} \, dt =1.$
I'm trying to use this to find the value of $\displaystyle \int_0^{2\pi}\frac{dt}{a^2\cos^2t+b^2\sin^2t}.$
Is it possible? Any hint would be much appreciated.
| Note that we can write
$$\begin{align}
\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}&=\frac{iab }{a^2\cos^2(t)+b^2\sin^2(t)}+\frac{\frac12(b^2-a^2)\sin(2t) }{a^2\cos^2(t)+b^2\sin^2(t)}\end{align} \tag 1$$
Therefore, integrating $(1)$ reveals
$$\frac{1}{2\pi i}\int_0^{2\pi}\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
definite integrals with trigonometric substitution How do you integrate $$\int_{-7}^{-5} \frac{2}{x^4\sqrt{x^2-25}}dx,$$ I seem to be getting my answer off by -1 all the time.
| Note that since the integrand is an even function, we can write
$$\int_{-7}^{-5}\frac{2}{x^4\sqrt{x^2-25}}\,dx=\int_5^7\frac{2}{x^4\sqrt{x^2-25}}\,dx$$
Now, enforcing the substitution $x\to 5\sec(x)$ reveals
$$\begin{align}
\int_5^7\frac{2}{x^4\sqrt{x^2-25}}\,dx&=\int_{0}^{\arccos(5/7)}\frac{2}{3125\sec^4(x)\tan(x)}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove with some AM-GM inequality? I have proved the following inequality:
Let $a,b,c>0$
$$\dfrac{(a+\sqrt{ab}+\sqrt[3]{abc})}{3}\le \sqrt[3]{a\cdot\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}}$$
My solution is:$$a\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3} =\dfrac{1}{3^3}(a+a+a)(a+\dfrac{a+b}{2}+b)(a+b+c)\ge\dfrac{1}{3^3}(a+a+a)(a+\s... | This is a proof by computer.
Substitute by $a=x^3, b=y^3, c=z^3$, the original inequality is equivalent to
show a polynomial in $x,y,z$ of degree $30$ is non-negative. A Cylindrical Decomposition shows this is true.
The code is the following
ieq1 = 3 (a + Sqrt[a b] + (a b c)^(1/3)) <= (7 + (4 Sqrt[a b])/(
a + b))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Is there an elegant way to solve $\int \frac{(\sin^2(x)\cdot \cos(x))}{\sin(x)+\cos(x)}dx$? The integral is:
$$\int \frac{(\sin^2(x)\cdot \cos(x))}{\sin(x)+\cos(x)}dx$$
I used weierstraß substitution
$$t:=\tan(\frac{x}{2})$$
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$dx=\frac{2}{1+t^2}dt$$
Got this... | \begin{align}
\int\frac{\sin^2 x\cos x}{\sin x+\cos x}\,\mathrm dx
&=
\int\frac{\sin^2 x\sin\left(\frac\pi2-x\right)}{\sin x+\sin\left(\frac\pi2-x\right)}\,\mathrm dx
\\
&=
\int\frac{\sin^2\left(\frac\pi4+u\right)\sin\left(\frac\pi4-u\right)}{\sin \left(\frac\pi4+u\right)+\sin\left(\frac\pi4-u\right)}\,\mathrm d\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Function that maps the "pureness" of a rational number? By pureness I mean a number that shows how much the numerator and denominator are small.
E.g. $\frac{1}{1}$ is purest, $\frac{1}{2}$ is less pure (but the same as $\frac{2}{1}$), $\frac{2}{3}$ is less pure than the previous examples, $\frac{53}{41}$ is worse, ....... | You can take,
$$\frac{a}{b}\mapsto \frac{a+b}{\gcd(a,b)},$$
Note that this is independent of the choice of representative since $\gcd(na,nb)=n\gcd(a,b)$ for non-negative integers $n$.
For your examples,
$$\frac{1}{1}\mapsto 2,\quad \frac{1}{2}\mapsto 3,\quad\frac{2}{3}\mapsto 5,\quad\frac{53}{41}\mapsto 94.$$
Another p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 9,
"answer_id": 4
} |
Area between two trigonometric curves I need to find the area between two curves:
$$\begin{cases} x=\sqrt { 2 } \cos { t } \\ y=4\sqrt { 2 } \sin { t } \end{cases}\\ y=4\quad (y\ge 4)$$
I came up with:
$$\frac { 1 }{ 4 } \left( 8\pi n +\pi \right) \le t\le \frac { 1 }{ 4 } \left( 8\pi n + 3\pi \right) $$
$$\int _{... | As I said in my comment, you need to restrict your angles, both in general, to have only one revolution around the origin, and when you want to meet your additional restriction on $y$. For the first case, it's obvious that $t \in [0, 2\pi)$. For the second one, solve an easy equation
$$
y = 4 \implies 4\sqrt 2 \sin t =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate the value of the following integral: $ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} $
The question is show that $$ \int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx $$
Hence or otherwise, calculate the value of the following integral
$$ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(... | So you have shown that:
$\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} dx= \int^{\frac{\pi}{2}}_0 \frac{\cos^n(x)}{\cos^n(x) + \sin^n(x)}dx$
$2\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)}dx =$$\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)}dx + \int^{\frac{\pi}{2}}_0 \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Characteristic polynomials of matrices related How to show that the characteristic polynomials of matrices A and B are $\lambda^{n-1}(\lambda ^2-\lambda -n)=0$ and $\lambda^{n-1}(\lambda^2+\lambda-n)=0$ respectively by applying elementary row or column operations.
$A=\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 ... | About first matrix.
Lets take the bottom line. As we know :
$$\det{A} = \sum{(-1)^{i+j}\cdot a_{i,j}\cdot M_{i,j}},$$ so we got:
$$S_{n+1} = a_{n+1,n+1}\cdot (-1)^{2n+2}\cdot S_{n} + (-1)^{n+2}a_{n+1,1}S'_{n},$$ where $S'_{n}= (-1)^{1 + n} \lambda^{n-1}(-1)^{n-1}$, because of :
$S'_{n} = \begin{bmatrix}
1 & 1 & \cdots... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Simplify this vector expression I have a question which I know the answer to but am having difficulty showing it. It's about simplifying the following vector equation (I'm aware I've grouped the terms in an arguably strange way):
$ (p^2 + q^2 |\mathbf{P}|^2 + (\mathbf{P} \bullet \mathbf{Q})^2)\mathbf{X} + (\mathbf{Q}... | Many thanks for the reply. After a few hours straight of trying to tackle this, I finally solved the problem. I'm going to use these two vector identities (which I've pulled from Wikipedia and trust to be correct).
$ (\mathbf{A} \times \mathbf{B}) \bullet (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \bullet \mathbf{C})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745224",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Testing whether the circumcenter of a cyclic quadrilateral lies inside it For a triangle with sides $a, b, c$ (where $c$ is the biggest side) there is a simple check to see whether it's circumcenter lies inside of it:
$$a^2 + b^2 < c^2$$
Is there such an inequality for a cyclic quadrilateral, given its side lengths $a,... | The circumcenter of a triangle is inside the triangle if and only if
$$
(a^2+b^2-c^2)(c^2+a^2-b^2)(b^2+c^2-a^2)\gt0\tag{1}
$$
The diagonal with sides $a$ and $b$ on one side and $c$ and $d$ on the other is
$$
e^2=\frac{\frac{a^2+b^2}{ab}+\frac{c^2+d^2}{cd}}{\frac1{ab}+\frac1{cd}}\tag{2}
$$
Then the circumcenter is in... | {
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"url": "https://math.stackexchange.com/questions/1745840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to compute taylor series for $f(x)=\frac{1}{1-x}$ about $a=3$ and $f(x)=\sin{x}$ about $a=\frac{\pi}{4}$? How to compute taylor series for $f(x)=\frac{1}{1-x}$ about $a=3$ and $f(x)=\sin{x}$ about $a=\frac{\pi}{4}$ ?
For the first one, using substitution, let $t=x-3$, then $x=t+3$. Then $f(x)=\frac{1}{1-(t+3)}=\s... | Hint. One may write
$$
f(x):=\frac{1}{1-x}=-\frac12\frac{1}{1+\frac{(x-3)}2}=-\frac12\sum_{n=0}^\infty(-1)^n\frac{(x-3)^n}{2^n}, \quad |x-3|<2,
$$ and one may write, for any real number $x$,
$$
\begin{align}
f(x)&:=\sin x
\\\\&=\frac{\sqrt{2}}2 \left(\sin\left(x-\frac{\pi}4\right)+ \cos\left(x-\frac{\pi}4\right)\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\sum^{\infty}_{n=2}\frac{(4n^2+8n+3)2^n}{n!}$? How to compute $\sum^{\infty}_{n=2}\frac{(4n^2+8n+3)2^n}{n!}$?
I am trying to connect the series to $e^x$
My try: $\sum^{\infty}_{n=2}\frac{(4n^2+8n+3)2^n}{n!}=\sum^{\infty}_{n=2}\frac{(2n+1)(2n+3)2^n}{n!}$
Let $x=\sqrt{2}$, then the series becomes $\sum^{\... | HINT:
Let $f(x)=e^x-(1+x)$. Then, note that
$$f(x)=e^x-(1+x)=\sum_{n=2}^\infty \frac{x^n }{n!}$$
$$xf'(x)=x(e^x-1)=\sum_{n=2}^\infty \frac{n\,x^{n} }{n!}$$
$$x(xf'(x))'=x(x+1)e^x-x=\sum_{n=2}^\infty \frac{n^2\,x^{n} }{n!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How can you prove this by strong induction? The sequence $b_1,b_2,...$ is defined recursively as:\begin{align} b_1&=0;\\ b_2&=1;\\ b_n&=2b_{n-1}-2b_{n-2}-1 \ \text{for} \ n\geq3. \end{align} Prove that this means: $$\forall n\geq1: b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$$
Edit:
I have tried to prove th... | Comment on the Approach in the Question
The inductive step needs to show
$$
\begin{align}
2b_{n-1}-2b_{n-2}-1
&=2\left(2^{(n-1)/2}\sin\left(\tfrac{(n-1)\pi}4\right)-1\right)-2\left(2^{(n-2)/2}\sin\left(\tfrac{(n-2)\pi}4\right)-1\right)-1\\
&=2^{(n+1)/2}\sin\left(\tfrac{(n-1)\pi}4\right)-2^{n/2}\sin\left(\tfrac{(n-2)\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to square both the sides of an equation? Question: $x^2 \sqrt{(x + 3)} = (x + 3)^{3/2}$
My solution: $x^4 (x + 3) = (x + 3)^3$
$=> (x + 3)^2 = x^4$
$=> (x + 3) = x^2$
$=> x^2 -x - 3 = 0$
$=> x = (1 \pm \sqrt{1 + 12})/2$
I understand that you can't really square on both the sides like I did in the first step, howe... | Why can't you square both sides like you did? You absolutely can. If you do this you just need to make sure you didn't introduce any extraneous solutions. You can make sure of this by checking each solution you get. Admittedly that may be a little difficult with answers like $x = (1 \pm \sqrt{13})/2$, but it is wha... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
| We use the fact that the 1-form
$$\eta = \frac{x dy - y dx}{x^2 + y^2}$$
has integral of $2 \pi$ over $\gamma_r(t) = (r \cos t, r \sin t)$.
Furthermore, if $\Gamma(0) = \Gamma(2\pi)$ and if the intervals $[\gamma(t), \Gamma(t)]$ do not contain $\mathbf{0}$ for any $t \in [0, 2 \pi]$, then the integral over $\Gamma$ is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Integrating $\int \frac{\sqrt{x^2-x+1}}{x^2}dx$ Evaluate $$I=\int\frac{\sqrt{x^2-x+1}}{x^2}dx$$ I first Rationalized the numerator and got as
$$I=\int\frac{(x^2-x+1)dx}{x^2\sqrt{x^2-x+1}}$$ and splitting we get
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^2}-\frac{1}{x}}{\sqrt{x^2-x+1}}dx$$ i.e.,
$$I=\int\fr... | So the last step should be changing variable $t=\frac{2}{\sqrt3}(x-\frac{1}{2})$ and use the identity
$$\int \frac{1}{\sqrt{x^2+1}}dx=arcsinh(x)$$
so we get
$$I=-\frac{\sqrt{x^2-x+1}}{x}+\frac{3}{2}\int\frac{1}{\sqrt{x^2-x+1}}dx=
-\frac{\sqrt{x^2-x+1}}{x}+\frac{3}{2}arcsinh[\frac{2}{\sqrt3}(x-\frac{1}{2})]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find a limit for Doublet Stream function In fluid Mechanics,
The superimposed stream function of point source and sink is:
$\psi=-\frac{Qcos\theta_1}{4\pi}+\frac{Qcos\theta_2}{4\pi}$
Graphical image of the function
and for a sink - source doublet, we need to show that as: $l\rightarrow 0$
we get:
$\psi=\frac{m}{r}sin... | The positions of the sink and source are $(-l/2,0)$ and $(l/2,0),$ respectively.
Hence,
$$\psi = \frac{Q}{4 \pi}(\cos \theta_2 - \cos \theta_1) \\= \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{(x-l/2)^2 + y^2}}- \frac{x+l/2}{\sqrt{(x+l/2)^2 + y^2}}\right) \\ = \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{x^2 + y^2 -lx +l^2/4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760931",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Using the $\epsilon-\delta$ definition show that $f(x) = \frac 1 {x^2}$ is a continuous function at any $x_0 = a, a > 0$ Using the $\epsilon-\delta$ definition show that $f(x) = \frac 1 {x^2}$
is a continuous function at any
$x_0 = a, a > 0$
I have expressed in the form:
$$lim_{x\to a}\frac1{x^2}=\frac1{a^2}$$
and thu... | There were some (really) stupid errors in my other answer.
We want
$$|\frac{(x-a)(x+a)}{x^2a^2}|<\epsilon$$ We need to bound $x$ about $a$, that is of course what makes this problem harder.
If we set at least $\delta <a/2$ then
$$-\frac{a}{2}<x-a<\frac{a}{2}$$ and so $$\frac{a}{2}<x<\frac{3a}{2}$$
this gives
$$|\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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inequality involving heights and bisectors Let $a,b,c,a \le b \le c$ be the sides of the triangle $ABC$, $l_a,l_b,l_c$ the lengths of its bisectors and $h_a,h_b,h_c$ the lengths of its heights. Prove that:
$$\frac {h_a+h_c} {h_b} \ge \frac {l_a+l_c} {l_b}$$
| Here's a proof.
The formulae for the lenghts of the angular bisectors and the heights are known. (see e.g. https://en.wikipedia.org/wiki/Triangle ). Let $T$ be the area of the triangle. We have
$$ l_a = \sqrt{bc (1 - \frac{a^2}{(b+c)^2})}$$
and
$$h_a = \frac{2 T}{a}$$
and cyclic shifts of those.
With these formulae, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Bronstein Integral 21.42 Good morning.
I came across the following integral in some field theory calculation:
$\int_0^\pi dx\,\log\left(a^2+b^2-2ab\cos x\right)=2\pi\log\left(\max\lbrace a,b\rbrace\right)$
for $0<a,b\in\mathbb{R}$. The notation has been adapated to make contact with Bronstein's integral 21.42.
Despite ... | Suppose WLOG $b > a$. Then with $c = b/a$ we have
$$\begin{align} I &= \int_0^\pi \log(a^2 + b^2 - 2ab \cos x) \, dx \\ &=\int_0^\pi \log a^2 \, dx + \int_0^\pi \log(1 + (b/a)^2 - 2(b/a) \cos x) \, dx \\ &= 2\pi \log a + \int_0^\pi \log(1 + c^2 - 2c \cos x) \, dx \end{align}$$
Now we can evaluate the second integral... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find $\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$ Find $$\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$$
The general term is $\frac{1}{r}\sin\frac{r\pi}{3}$
Let $z=e^{i\frac{\pi}{3}}$
Then, $$\frac{1}{r}z^r=\frac{1}{r}e^{i\frac{r\pi}{3}}$$... | In your line
$$P=-\ln(1-e^{i\pi/3})$$
Now $$\ln(1-e^{i\pi/3})=\ln(-1)+\ln(e^{i\pi/6})+\ln(2i\sin\dfrac\pi6)\equiv i\pi+\dfrac{i\pi}6+\ln(i)\pmod{2\pi i}$$
$$\equiv i\left(\pi+\dfrac\pi6+\dfrac\pi2\right)\equiv-\dfrac{i\pi}3$$
So, the principal value of the imaginary part of $P$ is $$\dfrac\pi3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
System of equations that can be solved by inequalities: $(x^3+y^3)(y^3+z^3)(z^3+x^3)=8$ and $\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32$
S367. Solve in positive real numbers the system of equations:
\begin{gather*}
(x^3+y^3)(y^3+z^3)(z^3+x^3)=8,\\
\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32.... | The inequality $$\frac{a^2+b^2}{a+b}\geq\sqrt[3]{\frac{a^3+b^3}{2}}$$ we can prove also by AM-GM:
$$2(a^2+b^2)^3=\frac{2}{27}(3a^2+3b^2)^3=$$
$$=\frac{2}{27}\left(2(a^2-ab+b^2)+2\cdot\frac{(a+b)^2}{2}\right)^3\geq$$
$$\geq\frac{2}{27}\left(3\sqrt[3]{2(a^2-ab+b^2)\cdot\left(\frac{(a+b)^2}{2}\right)^2}\right)^3=$$
$$=(a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\sum_{k=1}^n \frac{1}{n+k} = \sum_{k=1}^{2n} \frac{1}{k}(-1)^{k-1}$ using induction I'm trying to prove (using induction) that:
$$\sum_{k=1}^n \frac{1}{n+k} = \sum_{k=1}^{2n} \frac{1}{k}(-1)^{k-1}.$$
I have found problems when I tried to establish an induction hypothesis and solving this because I've learne... | We can write
$$\begin{align}\sum_{k=1}^{n+1}\frac{1}{n+1+k}&=\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n+2}\\&=\left(-\frac{1}{n+1}+\frac{1}{n+1}\right)+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n+2}\\&=\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}\right)-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\\&=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that : $\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$
Prove inequality for positive numbers:
$$\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$$
My work so far:
Lemma: If $x>y>0, t>z>0$, then $$\frac{x}{x+z}>\frac y{y+t}.$$
Proof. Real... | First, let's write the left side of the inequality as:
$L(c)=\dfrac{c+A}{c+B}+\dfrac{c+D}{c+E}$,
the derivative of $L(c)$ is:
$L'(c)=\dfrac{B-A}{(c+B)^2}+\dfrac{E-D}{(c+E)^2}$,
which is always positive for $c\geq 0$, because $B>A$ and $E>D$. Hence, L(c) is cresent for $c\geq 0$ and so, $L(c)>L(0)$ for $c\geq 0$, i.e.
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Prove that exist a $a_n<0$ by the recursion $a_{n+2}-Pa_{n+1}+Qa_n = 0$ with the condition $\Delta:= P^2-4Q < 0$ Suppose $P, Q > 0$ such that $\Delta:= P^2-4Q < 0$. The sequence $\{a_n\}$ is defined by the recursion $a_{n+2}-Pa_{n+1}+Qa_n = 0$ where both $a_1$ and $a_2$ are real and at least one is non-zero.
Then, whic... | As user1952009
said,
if
$a_{n+2}-Pa_{n+1}+Qa_n = 0
$,
by assuming that
$a_n
=r^n
$,
we get,
after dividing by
$r^n$,
$0
=r^2-Pr+Q
$
or
$r
=\frac12(P \pm \sqrt{D})
$
where
$D
= P^2-4Q
< 0
$.
By considering the two possible
values of $r$,
which I will call
$u$ and $v$,
we can find
$A$ and $B$ such that
$A+B = a_0$
and
$A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Random points on a sphere — expected angular distance Suppose we randomly select $n>1$ points on a sphere (all independent and uniformly distributed).
*
*What is the expected angular distance from a point to its closest neighbor?
*What is the expected angular distance from a point to its $m^{\text{th}}$ closest n... | Let
*
*$a = m -1$, $b = n - m - 1$, $d = a + b = n - 2$.
*$\theta_{ij}$, $1 \le i \ne j \le n$ be the angular separation between point $x_i$ and $x_j$.
*$\ell_m$ be the expected angular separation among a pair of $m^{th}$ nearest neighbors.
*$\mathcal{E}_m$ be the event that $x_2$ is the $m^{th}$ nearest neighbo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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numerical values of points in cantor set Let $C$ be the standard middle thirds Cantor set in the interval $[0,1]$. The "endpoints" of $C$ have very simple numerical values that can be listed off: $$0,1,1/3,2/3,1/9,2/9,6/9,7/9,1/27,2/27,7/27,8/27,... $$
What I am looking for is some numerical values of the "nonendpoints... | For example,
$$\frac 14 = \frac{0}{3} + \frac{2}{3^2} + \frac{0}{3^3} + \frac{2}{3^4} + \frac{0}{3^5} + \frac{2}{3^6} + \cdots$$
is in the Cantor Set.
Edit: Actually, elements of the Cantor Set are of the form
$$\sum_{n=1}^\infty \frac{a_n}{3^n}$$
where $a_n$ is any sequence consisting of only $0$ and $2$'s.
So, when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$
If $p(x) = ax^3+bx^2+cx+d$ and $|p(x)|\leq 1\forall |x|\leq 1$, what is the $\max$ value of $|a|+|b|+|c|+|d|$?
My try:
*
*Put $x=0$, we get $p(0)=d$,
*Similarly put $x=1$, we get $p(1)=a+b+c+d$,
*similarly put $x=-1$, we get ... | I'll assume that $x$ is real.
Consider system of four equations with four variables.
$$
a + b + c + d = p(1),
$$
$$
-a+b-c+d = p(-1),
$$
$$
\frac{a}{8}+\frac{b}{4} + \frac{c}{2} + d = p(1/2),
$$
$$
-\frac{a}{8}+\frac{b}{4} - \frac{c}{2} + d = p(-1/2).
$$
You can just solve it and find values of $a,b,c,d$.
From this eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use polar coordinates to evaluate function
To find:
$$\iint_R\frac{1}{1+x^2+y^2}\,dA$$
where R is first quadrant bounded by $y=0$, $y=x$, $x^2+y^2=4$
$$\iint_Rf(r\cos \theta,r\sin\theta)r\,dA$$
which is as far as I got -
$$\iint\frac{1}{1+r^2\cos^2\theta +r^2\sin^2\theta}\,dA$$
$$I=\iint \frac{1}{1+r^2}\,dr\,d\th... | So
$$\iint_R\frac{1}{1+x^2+y^2}\,dA$$
Since $r^2 = x^2 + y^2$ and $dA = r \,dr \,d\theta$
$$\iint_R\frac{1}{1+r^2}r \,dr \,d\theta$$
Now, $R$ is the region bounded by $r = [0, 2]$ and $\theta = [0, \pi/4]$. So
$$\int_{0}^{\pi/4} \int_{0}^{2} \frac{r}{1+r^2} \,dr \,d\theta$$
$$\int_{0}^{\pi/4} \,d\theta \times \int_{0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, ... | If you know about the Jordan Canonical Form you can use that.
Another method, probably more elementary, was mentioned in a comment. The comment was deleted; I don't know why. Note that $A=I+N$, where $N^3=0$. It follows that $$A^k=I+kN+\frac{k(k-1)}{2}N^2.$$You can use that to calculate $e^{At}=\sum t^kA^k/k!$.
Edit: O... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 1
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If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution: $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y... | Let $x\ge y\ge z$. Then
$$3^x +3^y +3^z=3^{26}$$
$$3^z(3^{x-z}+3^{y-z}+1)=3^{z}\cdot3^{26-z}$$
$$3^{x-z}+3^{y-z}+1=3^{26-z} \Leftrightarrow x=z=y=25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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I have a problem when I go to calculate $\lim_{x\to\infty}\left( \frac {2x+a}{2x+a-1}\right)^{x}.$ The limit:
$\lim_{x\rightarrow\infty}
\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$
I make this:
$\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$=${{\rm e}^{{\it x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$
Then... | $(\frac{2x+a}{2x+a-1})^x=(\frac{2x+a+1-1}{2x+a-1})^x=(1+\frac{1}{\alpha})^{\frac{\alpha}{2}+\frac{1-a}{2}}\longrightarrow e^{\frac{1}{2}}$. Where $\alpha=2x+a-1$ and $\alpha\rightarrow \infty,$ when $x\rightarrow\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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What is the probability of selecting five of the winning balls and one of the supplementary balls? So I'm just doing a bit of probability questions and wanted to make sure I got it right.
I have $50$ balls numbered $1-50$, and we pick $6$ winning balls and $2$ supplementary without replacement.
So the chance to get t... | I will do it another way. The Tax on the Poor Corporation chooses $6$ "main" numbers and $2$ supplementary numbers. There are $\binom{50}{6}$ ways to choose the main numbers. Presumably they are all equally likely.
We find the probability that you get all $6$ main numbers. There is only $1$ hand that will do the job, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the limit of $\lim_{n\to \infty}\frac{2^n n^2}{n!}$ I am trying to find the limit of the following, $$\lim_{n\to \infty}\frac{2^n n^2}{n!}$$
L'Hospital is not going to work. Hoping for a squeeze, by the observation that $2^n<n!$ for $n\geq 4$ does not help either as one side of the limit goes to $\infty$. How can ... | Note that:
$$n!\ge1\cdot 2\cdot3\cdot3\cdots\cdot3\cdot(n-2)\cdot(n-2)\cdot(n-2)=2\cdot3^{n-5}(n-2)^3$$
So for $n>5$:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot3^{n-5}(n-2)^3}=\frac{3^5}{2}\left(\frac{2}{3}\right)^n\left(\frac{n}{n-2}\right)^3\frac{1}{n}<\frac{3^5}{2}\left(\frac{2}{3}\right)^5\left(\frac{5}{3}\right)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Find the range of $y = \sqrt{x} + \sqrt{3 -x}$ I have the function $y = \sqrt{x} + \sqrt{3 -x}$. The range in wolfram is $y \in\mathbb R: \sqrt{3} \leq y \leq \sqrt{6}$
(solution after correction of @mathlove)
$\sqrt{x} + \sqrt{3 -x} = y$
$$
\begin{cases}
x \geq 0\\
x \leq 3
\end{cases}
$$
then
$(\sqrt{x} + \sqrt{3 ... | $f(x) = \sqrt{x} + \sqrt{3-x}$. From this you have correctly deduced that the domain is $[0,3]$.
\begin{align*}
f'(x) &= \frac{1}{2\sqrt x} - \frac{1}{2\sqrt{3-x}}\\[0.3cm]
&= \frac{\sqrt{3-x}}{2\sqrt x \sqrt{3-x}} - \frac{\sqrt{x}}{2\sqrt x\sqrt{3-x}}
\end{align*}
$f'(x) = 0$ when $\sqrt{3-x} = \sqrt{x}$, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+... | Your A, B and C are wrong. Assuming your A,B and C are right, the integraiotn is right
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Express last equation of system as sum of multiples of first two equations The question says to 'Express the last equation of each system as a sum of multiples of the first two equations."
System in question being:
$ x_1+x_2+x_3=1 $
$ 2x_1-x_2+3x_3=3 $
$ x_1-2x_2+2x_3=2 $
The question gives a hint saying "Label the e... | You can do row reduction on the transpose:
\begin{align}
\begin{bmatrix}
1 & 2 & 1\\
1 & -1 & -2\\
1 & 3 & 2\\
1 & 3 & 2\\
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 2 & 1\\
0 & -3 & -3\\
0 & 1 & 1\\
0 & 1 & 1\\
\end{bmatrix}
&&\begin{aligned}
R_2&\gets R_2-R_1\\
R_3&\gets R_3-R_1\\
R_4&\gets R_4-R_1
\end{aligned}
\\[6px]
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.