Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Computing $\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \frac{(x+2)^{2017}}{(x-2)^{2015}}$ I'm studying for an exam, but I have trouble with computing the following limit:
$$\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \f... | This is another case of intimidation via use of large numbers. The part $(x + 2)^{2017}/(x - 2)^{2015}$ is a rational function which is defined for $x = 0$ and hence its limit as $x \to 0$ is same as its value at $x = 0$ and thus the limit of this part is $-4$. We can thus proceed as follows
\begin{align}
L &= \lim_{x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help m... | Consider the following picture. We see that the triangle with vertices $(\cos A, \sin A), (\cos(A+120^\circ), \sin(A+120^\circ)), (\cos(A+240^\circ), \sin(A+240^\circ))$ are the vertices of an equilateral triangle with centriod at the origin.
Thus if we set $a = \cos A, b = \cos(A+120^\circ), c = \cos(A+240^\circ)$, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
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Epsilon-delta limit definition I got $\lim_{x \rightarrow 3} \frac{x^2-2x+1}{x-2} = 4$
I need to prove that by delta epsilon. I came to delta ={1/2epsilon,1/2}
Is that right? If not can you explain me how?
| |$\frac {(x-3)(x-3)}{x-2}|$< $\epsilon$
$\delta$ <= $\frac{1}{2}$
|x-3|<$\delta$ <=$\frac{1}{2}$
-$\frac{1}{2}$ < x-3 < $\frac{1}{2}$
$\frac{1}{2}$ < x-2 < $\frac{3}{2}$
2>$\frac {1}{x-2}$>$\frac{2}{3}$
$\delta$ = min{$\frac{1}{2}$$\epsilon$, $\frac{1}{2}$}
|$\frac {(x-3)(x-3)}{x-2}|$ < $\delta$*2 = $\epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find function $f(x)$, such that $f(x+ \frac{1}{x}) = x^2 + \frac{1}{x^2}$. The question is,
Find function $f(x)$, if $f(x+ \frac{1}{x}) = x^2 + \frac{1}{x^2}$.
What does this mean?
Do I have to find $x$ in $x+ \frac{1}{x} = x^2 + \frac{1}{x^2}$?
In this case (not counting solutions in the complex plane), $x = 1; f(x)... | Observe that
$$\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2$$
so in fact
$$f(x):=x^2-2\;\;\;\text{gives}\;\;\;f\left(x+\frac1x\right)=\left(x+\frac1x\right)^2-2=x^2+\frac1{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Distributing 4 distinct balls between 3 people
In how many ways can you distribute 4 distinct balls between 3 people such that none of them gets exactly 2 balls?
This is what I did (by the inclusion–exclusion principle) and I'm not sure, would appreciate your feedback:
$$3^4-\binom{3}{1}\binom{4}{2}\binom{2}{1}^2+\bi... | Method 1: If no person receives exactly two balls, there are two possible cases. One person receives all four balls or one person receives three balls and another person receives one ball.
Case 1: One person receives all four balls.
There are three ways of selecting the person who receives all the balls.
Case 2: Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
$P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous. I need to prove $P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous.
I just proved that the sum is continuous, but I'm lost at how to manipulate the inequalities in this case.
My attempt:
Let $\epsilon>0$.
$d((x,y),(a,b)) = \sqrt{(x-a)^2 + (y-b)^2} $.... | Fix $a,b\in\mathbb{R}$, then we have the following:
Scratch work: If $d((x,y),(a,b))=|x-a|+|y-b|<\delta$ then $|x-a|<\delta$ and $|y-b|<\delta$. So we have:
\begin{align*}
x-a&<\delta\\
x+a&<\delta+2a\\
\end{align*}
and so we have:
\begin{align*}
|(x+a)(y-b)|&=|xy-ab + (ay-bx)|\geq |xy-ab|\\
&\Downarrow\\
|xy-ab|&\leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove inequality $2e^x>x^3+x^2$ If $x \in \Bbb R$, show that
$$2e^x>x^3+x^2$$
This inequality is right, see
Own ideas: If $x\in \Bbb R$,
$$f(x)=2e^x-x^3-x^2$$
$$f'(x)=2e^x-3x^2-2x$$
$$f''(x)=2e^x-6x-2$$
$$f'''(x)=2e^x-6$$
$$f''''(x)=2e^x>0$$
Like the symbol cannot judge $f$ sign.
So how can we show this $$f(x)>0 \text... | *
*For $x \leq 1$, notice that
$$ x^3 + x^2 \leq x + 1 \leq e^x \leq 2e^x. $$
*For $x \geq 1$, it suffices to prove that $f(x) := \log(2e^x) - \log(x^3 + x^2) > 0$. Differentiating twice,
$$ f'(x) = 1 - \frac{2}{x} - \frac{1}{x+1}, \qquad f''(x) = \frac{2}{x^2} + \frac{1}{(x+1)^2} $$
This shows that $f$ is strictly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A problem about quadratic function and inequality Let $f(x) = ax^2 + bx + c(c \neq 0)$. If
\begin{equation*}
|f(0)| \leqslant 1, |f(1)| \leqslant 1, |f(-1)| \leqslant 1,
\end{equation*} try to show that
\begin{equation*}
\text{for each } |x| \leqslant 1, \text{ we have }|f(x)| \leqslant \frac{5}{4}.
\end{equation*}
A... | We have that
$$f(x)=f(0)(1-x)(1+x)+\frac{f(1)}{2}x(1+x)-\frac{f(-1)}{2}x(1-x)$$
then for $x\in [-1,1]$,
$$|f(x)|\leq |f(0)||1-x^2|+\frac{|f(1)|}{2}|x(1+x)|+\frac{|f(-1)|}{2}|x(1-x)|\\
\leq |1-x^2|+\frac{1}{2}|x(1+x)|+\frac{1}{2}|x(1-x)|\\
=(1-x^2)+\frac{1}{2}|x|(1+x)+\frac{1}{2}|x|(1-x)$$
Now the RHS is an even functio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Convergence/divergence of $\sum_{k=1}^\infty\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}$ A problem asks me to determine if the series
$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$$
converges or div... | The series is divergent since $a_k>1$ for every $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{... | A similar approach to the one suggested by Jack to obtain the loose inequality: the function $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}$ is convex in $a,b,c$ being the sum of three convex functions (they're the restriction of the Euclidean norm to the planes $\{a=0\}$ etc.). The domain of the function is
$$\{a,b,c\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
not able to prove a trig identity For the identity,
$$\dfrac{\cos A}{1 - \tan A} + \dfrac{\sin A}{1 - \cot A}=\sin A + \cos A$$
What I have been able to perform,
$$
\dfrac{\cos A}{\dfrac{\cos A}{\cos A}-\dfrac{\sin A}{\cos A}} + \dfrac{\sin A}{\dfrac{\sin A}{\sin A} - \dfrac{\cos A}{\sin A}}$$
$$
\dfrac{\cos^2 A}{\cos ... | Okay, so we want to verify that $$\frac{\cos{A}}{1 - \tan{A}} + \frac{\sin{A}}{1 - \cot{A}} = \sin{A} + \cos{A}.$$
Staring with the left hand side, we have (based on what you've already done):
\begin{split} \frac{\cos{A}}{1 - \tan{A}} + \frac{\sin{A}}{1 - \cot{A}} &= \frac{\cos{A}}{\left(\frac{\cos{A} - \sin{A}}{\cos{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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limit of $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞ $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞
Both limits tend to infinity, but +∞ −(+∞) doesn't make sense. How would I get around to solving this?
| Note that:
$$
a_n=\sqrt{n^2+2} - \sqrt{n^2+1}=\frac{\big(\sqrt{n^2+2}-\sqrt{n^2+1}\big)\big(\sqrt{n^2+2}+\sqrt{n^2+1}\big)}{\sqrt{n^2+2} + \sqrt{n^2+1}}= \\
=\frac{n^2+2-n^2-1}{\sqrt{n^2+2} + \sqrt{n^2+1}}=\frac{1}{\sqrt{n^2+2} + \sqrt{n^2+1}}
$$
Thus:
$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}\big(\sqrt{n^2+2} - \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the equation of two ...
Find the single equation of two straight lines that pass through the point $(2,3)$ and parallel to the line $x^2 - 6xy + 8y^2 = 0$.
My Attempt:
Let, $a_1x+b_1y=0$ and $a_2x+b_2y=0$ be the two lines represented by $x^2-6xy+8y^2=0$.
then,
$$(a_1x+b_1y)(a_2x+b_2y)=0$$
$$(a_1a_2)x^2+(a_1b_2+b... | Note that $x^2-6xy+8y^2=(x-2y)(x-4y)$, so the lines represented by the equation $x^2-6xy+8y^2=0$ are
$$
x-2y=0 \qquad \text{and} \qquad x - 4y = 0.
$$
The line parallel to $x-2y=0$ and passing through $(2,3)$ is
$$
(x-2) - 2(y-3) =0,
$$
which in expanded form is $x - 2y + 4 = 0$. Similarly, the other line is $x-4y+10=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polar to cartesian form of r=sin(4θ)? The Polar to cartesian form of $ r = \sin(2\theta)$ is fairly simple.
What is the Cartesian form of the polar equation r=sin(4θ)?
[edit]
$$r=4sin(θ)cos(θ)(cos(θ)^2-sin(θ)^2)$$, so $$r^5=4rsin(θ)rcos(θ)(r^2cos(θ)^2-r^2sin(θ)^2)$$,so $$r^5=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5/2}=4xy(x^... | Hint.
$$\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$$
Then again you may use
$$\sin(2\theta) = 2\sin\theta\ \cos\theta$$
and
$$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$
Can you go on?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of polynomial times rational function of trig function over multiple periods. $$\int_0^{12\pi} \frac{x}{6+\cos 8x} dx$$
I tried all kinds of stuff but end up with $\arctan\frac{5^{1/2} \tan(x/2)}{7^{1/2}}$
Please help :)
| It looks like the OP is trying to compute the antiderivative and use the fundamental theorem of calculus. With multiple periods, that approach is paved with all sorts of difficulty that is really an artifice related to the functional form of the antiderivative. In truth, there should be no such difficulty. A better ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Does $\sum \frac{n^2 + 1}{n^{3.5} -2}$ converge? Does $\sum \frac{n^2 + 1}{n^{3.5} -2}$ converge? I think it does. But I cannot find a series to compare. I tried to compare it with $\sum \frac{1}{n^{1.5}}$, but I do not know how.
$$\sum \frac{n^2 + 1}{n^{3.5} -2}?\sum\frac{n^2}{n^{3.5}-2}?\sum\frac{n^2}{n^{3.5}}$$
| Beside the good and simple solutions you already received, setting $x=\frac 1 n$ and using Taylor series around $x=0$, you could show that $$\frac{n^2+1}{n^{7/2}-2}=x^{3/2}+x^{7/2}+2 x^5+2 x^7+4 x^{17/2}+O\left(x^{21/2}\right)$$ Back to $n$ $$\frac{n^2+1}{n^{7/2}-2}=\frac{1}{n^{3/2}}+\frac{1}{n^{7/2}}+\frac{2}{n^5}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
5 odd-numbered taxis out of 9 to 3 airports
A fleet of 9 taxis must be dispatched to 3 airports: three to airport A, five to B and one to C. If the cabs are numbered 1 to 9, what is the probability that all odd-numbered cabs are sent to airport B?
What I have have come up with so far:
*
*Probability the cabs are o... | If I understand your question correctly, you can rephrase it as, "What's the probability of selecting all odd numbers when sampling uniformly from the collection $\{1,2, \ldots, 9\}$?" Well, the probability of selecting an odd number out of these $9$ possible numbers is $P(1) = \frac{5}{9}$. Having selected this firs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Recurrence relation of type $a_{n+1} = a^2_{n}-2a_{n}+2$
A sequence $\{a_{n}\}$ is defined by $a_{n+1} = a^2_{n}-2a_{n}+2\forall n\geq 0$ and $a_{0} =4$
And another sequence $\{b_{n}\}$ defined by the formula $\displaystyle b_{n} = \frac{2b_{0}b_{1}b_{2}..........b_{n-1}}{a_{n}}\forall n\geq 1$ and $\displaystyle b_{... | For the first problem, as LaloVelasco answered, you have $$c_{n+1}=c_{n}^{2}$$ Take logarithms $$\log(c_{n+1})=2\log(c_n)$$ Define $d_n=\log(c_n)$ which makes $$d_{n+1}=2d_n\implies d_n=k\, 2^{n-1}\implies c_n=e^{k\, 2^{n-1}}\implies a_n=1+e^{k\, 2^{n-1}}$$ and $k=2 \log (a_0-1)$ which makes $$a_n=(a_0-1)^{2^n}+1$$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-... | One must have $$a^2+3a+3=\pm1\\a^2-3a+3=\pm1$$ From these we get, discarding the constant term $4$, only two equations
$$a^2+3a+2=0=(a+1)(a+2)\\a^2-3a+2=0=(a-1)(a-2)$$ which give
both the primes $7$ and $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do t... | Hint:
\begin{align}
f(x) = \log|x| - \frac{1}{2}\log|x^2+1|.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Why $\forall n\gt2\in\Bbb N:\lfloor\sqrt{n\cdot\sqrt{n\cdot\lfloor\sqrt{...\lfloor\sqrt{n\cdot\lfloor\sqrt{n}\rfloor}}\rfloor\rfloor}}\rfloor=n-2$? I have found this just by chance, but can not understand why the limit turns out to be $n-2, \forall n\gt 2 $.
Def:
$a_0 = \lfloor \sqrt{n} \rfloor$
$a_1 = \lfloor \sqrt{n... | All $n\gt 2$ converge to $n-2$ as you surmise.
The easiest way to see it is to look what happens if $a_n=n-3$. Then $a_{n+1}=\lfloor \sqrt {n(n-3)} \rfloor = \lfloor \sqrt {n^2-3n} \rfloor $. If $n \ge 4, n^2-3n \ge n^2-4n+4 = (n-2)^2$ so you will climb from $n-3$ to $n-2$ As you have pointed out, once you get to $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving a Coupled Second-Order Differential Equation I'm trying to solve the real-valued differential equation
$$xy\ddot{y}+2y\dot{x}\dot{y}+\alpha\ddot{x}+x\dot{y}^2=0$$
where $\alpha\in\mathbb{R}$ and $\alpha>0$ for both $x(t)$ and $y(t)$. How would I even approach this problem?
EDIT: Sorry, I forgot a conservation ... | Is this a homework problem or is it some sort of practice question? If not, any possible symmetries coming from the formulation of the problem? If not, I don't know how far one can go, but let's see. It may help if you tell us how you derived the equation, weather it is an Euler-Lagrange equation of some sort. Maybe i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving system of $9$ linear equations in $9$ variables I have a system of $9$ linear equations in $9$ variables:
\begin{array}{rl}
-c_{1}x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} - c_{2}x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} - c_{3}... | If $T = x_1 + x_2 + \ldots + x_9$, you can write this as
$$ T - (1+c_1) x_1 = T - (1 + c_2) x_2 = \ldots = T - (1+c_9) x_9 = 0$$
If any $c_i = -1$, then $T = 0$, and any $x_j$ for which $c_j \ne -1$ must be $0$, while those for which $c_j = -1$ must add to $0$.
On the other hand, if all $c_i \ne -1$,
then each $x_i = T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$.
Let $x,y$ be real numbers s.t.
$a + b = 6$,
$ax + by = 10$,
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$.
I am new to problem solving exercises like this and therefo... | I prefer to add another answer for the most general case.
Consider the four equations which I shall write $$ax^{i-1}+b y^{i-1}=c_i \qquad (i=1,2,3,4)$$ Manipulating them, we can show that $x y=\lambda$ and $a b=\mu$ using $$\lambda=\frac{c_3^2-c_2 c_4}{c_2^2-c_1 c_3}$$
$$\mu=-\frac{\left(c_2^2-c_1 c_3\right)^3}{(4 c_4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Lower bound of product. We have the product :
$$\prod{(1-\frac{1}{2^i})}$$ I know that product has upper bound and possibly has lower bound.
My question is : how can I get the lower bound of this product?
My idea was to estimate this product with other product , but I didn't get the product with non-zero limit.
| Observe for $n\geq 4$ we have
\begin{align}
2^n \geq n^2
\end{align}
then it follows
\begin{align}
\prod^\infty_{n=4}\left(1-\frac{1}{n^2}\right) \leq \prod^\infty_{n=4}\left( 1- \frac{1}{2^n}\right)
\end{align}
which means
\begin{align}
\prod^\infty_{n=2}\left(1-\frac{1}{n^2}\right) \leq
\frac{\left(1-\frac{1}{2^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $a,b,c>0$ and $a^3+b^3=c^3$ then prove that $a^2+b^2-c^2>6(c-a)(c-b)$. If $a,b,c>0$ and $a^3+b^3=c^3$ then prove that $$a^2+b^2-c^2>6(c-a)(c-b)$$.
I tried factoring but things got more complicated. I dont think standard AM-Gm can be applied directly. Can somebody help me to proceed? Thanks a lot.
| We need to prove that
$$c(a^2+b^2)-(a^3+b^3)>6c(c-a)(c-b)$$ or
$$\frac{(c^3-b^3)(c-a)}{a}+\frac{(c^3-a^3)(c-b)}{b}>6c(c-a)(c-b)$$ or
$$\frac{b^2+bc+c^2}{a}+\frac{a^2+ac+c^2}{b}>6c$$
which is AM-GM.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integration by completing the square I need to complete the square on the following integral. Once this is done apparently I will be able to use on of the integration tables in the back of my book.
$\int x \sqrt{x^2 + 6x +3} dx $
This is what I have come up with so far:
$\int x \sqrt{(x+3)^2 -6} $
I really am at a l... | $$ I= \int x \sqrt {x^2 +6x +3} dx $$
This is best done by substitution.
$$I= \int x \sqrt {(x+3)^2-6} dx \\ x+3= \sqrt 6 \sec \theta \implies dx=\sqrt 6 \sec \theta \tan \theta d \theta$$
$$ \therefore I= \sqrt 6 \int \sec \theta \tan \theta ( \sqrt 6 \sec \theta -3) \sqrt {6( \sec^2 \theta -1)} \\ I=6 \int \sec \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculating $ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} $. $ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} = η(8t)η(16t)$.
(Kac-Peterson)
I try to confirm the formula, however I don't know the left.
(The right is $q -q^9 -2q^{17} + q^{25} + 2q^{41} + \cdots $)
Define $a(k)$ as the following.
$ \sum_{n \g... | $(2n + 1)^2 - 32m^2 = 9$ and $n \ge 3|m|⇔ (n, m) = (1, 0)$.
So $a(9) = (-1)^1 = -1.$
$(2n + 1)^2 - 32m^2 = 17$ and $n \ge 3|m|⇔ (n, m) = (3, -1), (3, 1)$.
So $a(17) = (-1)^3 + (-1)^3 = -2.$
The other $a(k)$ is calculated by the same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the Maclaurin Series for $\frac{x}{x^2 + 1}$ How to find the Maclaurin Series for $$\frac{x}{x^2 + 1}?$$
I know that the Maclaurin Series is given by:
$$f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dots + \dfrac{f^{(k)}(0)}{k!}x^k.$$
So from repeatedly deriving $f(x)$ until there was a pattern. I have found th... | $${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-1)^nx^{n}$$
let $x\rightarrow x^2$
$${\frac {1}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n}$$
then multiply by $x$
$${\frac {x}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the real solutions of the equations:$x+y=2$, $xy-z^2=1$ Can anyone help with this? I will show you what I did.
Find the real solutions of the equation:$$x+y=2\tag{1}$$ $$xy-z^2=1 \tag{2}$$
From equation (2), I have $$xy=1+z^2 \tag{3}$$ From equations (1) and (3), I have a new equation $$r^2-2r+(1+z^2)=0$$ The ro... | Given $x+y=2$ and $xy-z^2=1$
So $xy=1+z^2\geq 1>0$.
So $x,y>0$
Using $\bf{A.M\geq G.M},$ we get $$\frac{x+y}{2}\geq \sqrt{xy}\Rightarrow (x+y)^2\geq 4xy$$ and equality hold when $x=y$
So $4\geq 4xy\Rightarrow xy\leq 1\Rightarrow 1+z^2\leq 1\Rightarrow z^2\leq 0\Rightarrow z=0$
and $x=y=1.$ So we get $(x,y,z) = (1,1,0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Square root of a $2 \times 2$ matrix
How can I find the square root of the following non-diagonalisable matrix?
$$\begin{pmatrix}
5 & -1 \\
4 & 1
\end{pmatrix}$$
I have shown that
$$A^{n}=3^{n-1}\begin{pmatrix}
2n+3 & -n \\
4n & 3-2n
\end{pmatrix}$$
though I'm not sure how to use this fact.
| The Jordan decomposition $A=SJS^{-1}$ is given by the matrices $$S = \begin{pmatrix}1&\frac12\\2&0 \end{pmatrix},\quad J = \begin{pmatrix}3&1\\0&3 \end{pmatrix}. $$
Let $\lambda = 3$ and write $J = \lambda (I+K)$, where $$K=\begin{pmatrix}0&\frac13\\0&0 \end{pmatrix}.$$ Using the Mercator series, we compute the logarit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$
Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$
I don't know how to begin to solve this problem
| Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.
Hence, $(a+b)(b+c)(c+d)(d+a)-\prod\limits_{cyc}(a+b+c-d)=$
$$4(u^2+v^2+w^2-uv-vw)a^2+(4(u^3+v^3+w^3)-2u^2v-2v^2u-2v^2w-2w^2v)a+$$
$$+\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)+u^2w^2$$
Since $\sum\limits_{cyc}(2u^3-u^2v-u^2w)=\sum\limits_{cyc}(u+v)(u-v)^2\geq0$,
it remai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer
Given a sequence $(a_n)$, with $a_1 = 4$ and $a_{n+1} = a_n^2-2$ for all $n \in \mathbb{N}$, prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer.
We must ... | You have proven that $$S_n = \frac{1}{4}\sqrt{3a_n^2(a_n^2-4)}$$
Suppose $S_n$ is an integer.
$$S_n^2 = \frac{3}{16}a_n^2(a_n^2-4)$$
We have
\begin{align}
S_{n+1}^2 &= \frac{3}{16}a_{n+1}^2(a_{n+1}^2-4)\\
& = \frac{3}{16}(a_{n+1}^2-4) a_{n+1}^2\\
& = \frac{3}{16}((a_{n}^2-2)^2-4) a_{n+1}^2\\
&= \frac{3}{16}(a_n^4-4a_n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that point $P $ and $\Delta ABC$ satisfy the given condition
Let ABC be a triangle and $P$ be any point in complex plane. Then show that
$$1.BC^2+CA^2+AB^2 \le 3 (PA^2+PB^2+PC^2)$$
$$2.BC^2+CA^2+AB^2 = 3 (PA^2+PB^2+PC^2) \iff \text {P is centroid of} \space\Delta ABC$$
I tried to solve it by assuming point... | Your expression is wrong. At least, you must point to what is the LHS and RHS. Let me guess. You mean LHS is $AB^2 + BC^2 + CA^2$ and RHS is $3(PA^2 + PB^2 + PC^2)$, but we have $$LHS = AB^2 + BC^2 + CA^2 = 2\sum x_i^2 + 2\sum y_i^2 -\sum 2x_ix_j - 2\sum y_iy_j,$$
$$RHS = 3(PA^2 + PB^2 + PC^2) = 3[3(x^2+y^2) + \sum x_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limits at infinity by rationalizing I am trying to evaluate this limit for an assignment.
$$\lim_{x \to \infty} \sqrt{x^2-6x +1}-x$$
I have tried to rationalize the function:
$$=\lim_{x \to \infty} \frac{(\sqrt{x^2-6x +1}-x)(\sqrt{x^2-6x +1}+x)}{\sqrt{x^2-6x +1}+x}$$
$$=\lim_{x \to \infty} \frac{-6x+1}{\sqrt{x^2-6x +1... | it should be $$\lim _{ x\to \infty } \frac { -6x+1 }{ \sqrt { x^{ 2 }-6x+1 } +x } =\lim _{ x\to \infty } \frac { x\left( -6+\frac { 1 }{ x } \right) }{ x\left( \sqrt { 1-\frac { 6 }{ x } +\frac { 1 }{ { x }^{ 2 } } } +1 \right) } =\frac { -6 }{ 2 } =-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\arccos(\cos\frac{15\pi }{4})$ I have a problem with understanding of this exercise:
$\arccos(\cos\frac{15\pi }{4})= ?$
$\cos(\frac{15\pi}{4}-2\pi)=\cos(\frac{7\pi}{4})$
$\cos(\frac{7\pi}{4}+\pi)= -\cos(\frac{3\pi}{4})$
then $\arccos(-\cos\frac{3\pi}{4})$
All above I understand pretty clearly. We did it at ... | Since $\arccos x: [-1, 1] \to [0, \pi]$, we need to find the angle in $[0, \pi]$ that has the same cosine as $15\pi/4$. Since coterminal angles have the same cosine and
$$\frac{15\pi}{4} = -\frac{\pi}{4} + 4\pi = -\frac{\pi}{4} + 2 \cdot 2\pi$$
we have
$$\cos\left(\frac{15\pi}{4}\right) = \cos\left(-\frac{\pi}{4}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions of $\sin2x-\sin x>0$ with $x\in[0,2\pi]$ What are the solutions of this equation with $x\in[0,2\pi]$?
$$\sin2x-\sin x>0$$
I took this to
$$(\sin x)(2\cos x-1)>0$$
Now I need both terms to be the same sign. Can you please help me solve this?
| $$\begin{align}\sin(2x)-\sin(x)&=2\sin(x)\cos(x)-\sin(x)\\
&=2\sin(x)(\cos(x)-\frac{1}{2})\tag{1}
\end{align}$$
We examine the two factors:
$$\cos(x)=\frac{1}{2}\Rightarrow x=\frac{\pi}{3},\frac{5\pi}{3}$$
And $$\cos(x)-\frac{1}{2}>0,\, 0<x<\frac{\pi}{3} \text{ and } \frac{5\pi}{3}<x<2\pi$$
$$\cos(x)-\frac{1}{2}<0,\, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to set up equation to find the unknown values if a limit exist.
For what values of the constants $a$ and $b$ does the following limit exist?
$$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$
for this question, $$f(x) = \frac{-(x+3)(\sqrt{ax+b}-2)}x,x<-3$$
$$f(x) = \frac{(x+3)(\sqrt{ax+b}-2)}x,x>=-3$$
Firstly, I foun... | $\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x
$
Since
$\lim_{x\to0} |x+3|
=3$,
that doesn't matter.
What is left is
$\lim_{x\to0}\frac{(\sqrt{ax+b}-2)}x
$.
For this limit to exist,
we must have
$\lim_{x\to0}\sqrt{ax+b}-2
=0
$
or
$\lim_{x\to0}\sqrt{ax+b}
=2
$.
Since
$\lim_{x\to0}ax
=0
$,
we must have
$b=4
$.
The answer is,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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if $abc=1$,show that $a^3+b^3+c^3+\frac{256}{(a+1)(b+1)(c+1)}\ge 35$
Let $a,b,c>0,abc=1$ show that
$$a^3+b^3+c^3+\dfrac{256}{(a+1)(b+1)(c+1)}\ge 35\tag{1}$$
iff $a=b=c=1$
I know use AM-GM inequality
$$a^3+b^3+c^3\ge 3abc=3$$
and
$$(a+1)(b+1)(c+1)\ge 2\sqrt{a}\cdot 2\sqrt{b}\cdot 2\sqrt{c}=8$$
In this way, will ... | Alternative proof
WLOG, assume $c \in (0, 1]$.
Denote the inequality by $f(a, b, c) \ge 0$. We have
\begin{align}
&f(a, b, c) - f(\sqrt{ab}, \sqrt{ab}, c)\\
=\ & (a + b - 2\sqrt{ab})\Big[(a+b)^2 + (a+b)\cdot 2\sqrt{ab} + (2\sqrt{ab})^2 - 3ab\Big]\\
&\qquad - \frac{256(a + b - 2\sqrt{ab})}{(c+1)(ab + a+b+1)(ab + 2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Probability that a number is divisible by 11 The digits $1, 2, \cdots, 9$ are written in random order to form a nine digit number. Then, the probability that the number is divisible by $11$ is $\ldots$
I know the condition for divisibility by $11$ but I couldn't guess how to apply it here.
Please help me in this regard... | since $10 = -1 \pmod {11}$, a number $abcdefghi$ is a multiple of $11$ if and only if $(a+c+e+g+i)-(b+d+f+h)$ is a multiple of $11$.
Since $(a+c+e+g+i)+(b+d+f+h) = 45 = 1 \pmod {11}$, this is equivalent to $1-2(b+d+f+h) = 0 \pmod {11}$, and to $(b+d+f+h) = 6 \pmod {11}$.
So we want to know, when is the sum of $4$ numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
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A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\t... | Subtract the first column from other columns to reduce $n^2$-s, then subtract twice the second column from the third one to reduce $n$:
$$\begin{vmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\
(n+1)^2 & (n+2)^2 & (n+3)^2 \\
(n+2)^2 & (n+3)^2 & (n+4)^2
\end{vmatrix}
=
\begin{vmatrix}
n^2 & 2n+1 & 4n+4 \\
(n+1)^2 & 2n+3 & 4n+8 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
find $\lim_{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+y^4}$ if $$f(x,y)=\frac{x^2y^2}{x^4+y^4}$$ is a 2 variable function, find $$\lim_{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+y^4}$$
I really don't understand 2 variable limits. I understand that if limits from 2 or more paths aren't the same, the limit doesn't exist, but I don't kno... | Approach #1: $\epsilon$-$\delta$ proof...probably too much work for the given problem.
Approach #2: Convert to polar/spherical
Recalling that $x=r \cos \theta$ and $y= r \sin \theta$, we note that $(x,y) \to (0,0) \iff r \to 0$. As such,
$$\lim_{(x,y) \to (0,0) } \frac{ x^2 y^2}{x^4 + y^4} = \lim_{r \to 0} \frac{r^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the extreme values of the given function, subject to constraints $f(x,y) = y^2 - x^2 ; x^2 + y^2 = 36$
Find the extreme values of the function, subject to constraints
$$f(x,y) = y^2 - x^2 ; x^2 + y^2 = 36$$
Here is what I have done:
Using the substitution method we can say:
$$y = \sqrt{-x^2 + 36}$$
Then plugging... | Let $x = 6\cos \theta, y = 6\sin \theta \implies F(x,y) = F(\theta) = 36\cos (2\theta)\le 36$, and the max is $36$. The min is $-36$ and this occurs when $\cos (2\theta) = -1$ which corresponds to $x^2 - y^2 = -36$. Since $x^2+y^2 = 36$, we have: $x = 0, y = \pm 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.
Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$
I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
M... | By the induction hypothesis
$$
4^3+4^4+\ldots+4^n=\frac{4(4^n-16)}{3}=\frac{4^{n+1}-64}{3}
$$
so
$$
4^3+4^4+\ldots+4^n+4^{n+1}=\frac{4^{n+1}-64}{3}+4^{n+1}=\frac{4^{n+1}-64+3\cdot4^{n+1}}{3}\\
=\frac{4\cdot 4^{n+1}-64}{3}=\frac{4^{n+2}-64}{3}=\frac{4(4^{n+1}-16)}{3}
$$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$
Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$.
I have to prove this inequality using induction, I proved it for $n=1$ and now I have to prove it for $n+1$ assuming $n$ as hypothesis, but this seems impossible t... | Define $\displaystyle{S_n=\sum_{i=1}^n \dfrac{1}{n+i}}$. Then we have for every $n$ that:
\begin{align*}
S_{n+1}-S_n&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i}-\sum_{i=1}^n \dfrac{1}{n+i}\\
&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\sum_{i=2}^n \dfrac{1}{n+i}\right)-\dfrac{1}{n+1}\\
&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Can we show that all $2 \times 2$ matrices are sums of matrices with determinant 1? I came across a paper on the Sums of 2-by-2 Matrices with Determinant One. In the paper, which I have conveniently indicated here for reference, the author claims, but without proof, that a $2 \times 2$ is a sum of elements of the speci... | It is sufficient to show that the matrices $\left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & 0 \\ c & 0 \end{matrix}\right)$, and $\left(\begin{matrix} 0 & 0 \\ 0 & d \end{matrix}\right)$ can be written as the sum of matrices ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ and $ S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i}$ . Find $S_1 + S_2 $ I know to find sum of series using method of difference. I tried sum of write the term as (6n-1)(6n+1). i don't know how to proceed further.
| \begin{align*}
S_1+S_2&=\sum_{n\geq 1}\frac{1}{36n^2-1}+\frac{2}{(36n^2-1)^2}\\
&=\sum_{n\geq 1}\frac{1}{2}\frac{1}{(6n-1)^2}+\frac{1}{2}\frac{1}{(6n+1)^2}\\
&=\frac{1}{2}\sum_{\substack{n\geq 5\\n\equiv \pm1\,\!\!\!\mod 6}}\frac{1}{n^2}\\
&=\frac{1}{2}\left[\sum_{n\geq 1}\frac{1}{n^2}-\sum_{\substack{n\geq 1\\n\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve the system of inequalities Solve the system:
$$|3x+2|\geq4|x-1|$$
$$\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$$
So for $|3x+2|\geq4|x-1|$ I got the solution $x=6$ and $x=2/7$ and Wolframalpha agrees with me. I'm having troubles writing the final solution for $\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$.
$1)$ $x^{2}+x-2\leq0... | If $x\ge 1$ we have $$|3x+2|\geq4|x-1|\iff 3x+2\geq 4x-4 \iff x\le 6.$$ So $[1,6]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have
$$|3x+2|\geq4|x-1|\iff 3x+2\geq 4-4x \iff x\ge 2/7.$$ So $[2/7,1]$ is solution of the system. Finally, if $x\le -2/3$ then
$$|3x+2|\geq4|x-1|\iff -3x-2\geq 4-4x \iff ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show $\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$
Positive real numbers $a,b,c$ satisfy $abc=1$. Prove
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
I tried AM–GM, Cauchy–Schwarz and Jensen's but they all failed.
| Following stewbasic's hint in the comments:
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
$$\iff(b^2+b+1)(c^2+c+1)+(c^2+c+1)(a^2+a+1)+(a^2+a+1)(b^2+b+1)\ge(a^2+a+1)(b^2+b+1)(c^2+c+1)$$
The LHS expands as
$$a^2b^2+a^2b+ab^2+\color{blue}{a^2}+b^2+ab+a+b+1\\
+b^2c^2+b^2c+bc^2+\color{blue}{b^2}+c^2+bc+b+c+1\\
+c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
value of $a$ for which $25^x+(a+2)5^x-(a+3)<0$for at least one real $x$
Find the values of $a$ for which the inequality is satisfied for $25^x+(a+2)5^x-(a+3)<0$
for at least one real value of $x$
$\bf{My\; Try::}$ We can write it as $a(5^x-1)<-\left[25^x+2\cdot 5^x-3\right]$
So $$a < -\left(\frac{25^x-5^x+3\cdot 5^x-... | $z=5^x \to $ $z^2+(a+2)z-(a+3) \lt 0$
now $ \\ \Delta = \to (a+2)^2-4(1)(-(a+3)) = \\a^2+4+4a+4a+12=a^2+8a+16=(a+4)^2 \geq0 \\$
now $$z=\frac{-(a+2)\pm \sqrt{(a+4)^2}}{2}=\\z=1 ,z=-(a+3)\\z=1 \to 5^x=1 \to x=0\\ (5^x-1)(5^x+a+3)<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
| We can say:
\begin{align}P(x)-1&=(x-1)^3(ax^2+bx+c)\\
P(x)&=x^3(dx^2+ex+f)
\end{align}
Therefore, we can say that \begin{align}(x-1)^3(ax^2+bx+c)+1&=x^3(dx^2+ex+f)\\
(x^3-3x^2+3x-1)(ax^2+bx+c)+1&=dx^5+ex^4+fx^3\\
ax^5+(b-3a)x^4+(c-3b+3a)x^3+(3b-3c-a)x^2+(3c-b)x+(1-c)&=dx^5+ex^4+fx^3
\end{align}
We can then equate coeff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
If $ f(x) = \frac{b+2x-x^2}{ax-1}\forall x \in \mathbb{R}$ and $\min(f(x)) = -5,$ Then $a+b$
If $\displaystyle f(x) = \frac{b+2x-x^2}{ax-1}\forall x \in \mathbb{R}$ and $\min(f(x)) = -5,$ Then $a+b=$
Let $$y = \frac{b+2x-x^2}{ax-1}\Rightarrow axy-y=b+2x-x^2$$
So $x^2+(ay-2)x-(y+b) = 0$
Now for real values of $y,$ Equ... | As $x \rightarrow \infty$, $f(x)$ will be dominated by the $x^2$ term in the numerator. Hence if $a > 0$, $f(x) \rightarrow -\infty$. Again, if $a < 0$, $f(x) \rightarrow -\infty$ as $x \rightarrow -\infty$. Thus $a=0$.
Now
$$f(x) = x^2-2x-b = (x-1)^2 -b-1 $$ and if the minimum of $f(x) = -5$, it follows that $b=4$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $
Prove
$$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.
My... | Without Induction
We can werite series as $$\sum^{n}_{r=1}\frac{1}{\sin 2^{r}x} = \sum^{n}_{r=1}\frac{\sin(2^{r}x-2^{r-1}x)}{\sin 2^{r-1}x\cdot \sin 2^{r}x} = \sum^{n}_{r=1}(\cot 2^{r-1}x-\cot 2^{r}x )$$
Now use Telescopic Sum
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Summation using Euler's formula I must verify the following formula:
$$
\sum _{i=1}^n \sin ^2 (2k-1) \theta = -\frac{n}{2} - \frac{\sin 4n \theta}{4 \sin 2 \theta}
$$
I believe that I must do this by using the Euler's formula,
$$
e^{i \theta} = \cos \theta + i \sin \theta
$$
and taking only the imaginary parts of it.
H... | \begin{align*}
\sum _{i=1}^n \sin ^2 (2k-1) \theta &= \frac{1}{2}\sum _{k=1}^n (1-\cos 2(2k-1)\theta)\\
&= \frac{n}{2} - \frac{1}{2}\sum _{k=1}^n \cos (4k - 2) \theta\\
&= \frac{n}{2} - {\text {Real part of }}\frac{1}{2}\sum _{k=1}^n e^{(4k - 2)i\theta}\\
&=\frac{n}{2} - {\text {Real part of }}\frac{1}{2}e^{2i\theta}(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$ where $0\leq x\leq 360$ Solve
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot \sin 2x \cdot \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x \cdot \cos x ). (4 \sin x \cdot \cos x \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I pr... | \begin{align}
\sin 3x
&= \sin(2x + x) \\
&= \sin 2x \; \cos x + \cos 2x \; \sin x \\
&= 2 \sin x \; \cos^2 x + \cos 2x \; \sin x \\
&= \sin x \; (2 \cos^2 x + \cos 2x) \\
&= \sin x \; (2 \cos 2x + 1)
\end{align}
\begin{align}
4 \sin 2x \; \sin 4x
&= 2(\cos 4x \; \cos 2x + \sin 4x \; \sin 2x)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Using induction to prove formula I am revising for my test from Discrete math. I have come to this problem.
I am to prove by using mathematical induction that
$6\times7^{n} - 2 \times 3^{n}$ is divisible by 4. for $n \ge 1$ ;
I created basic step :
$6\times7^{1} - 2\times3^{1} = 36 $
and induction step
$\forall n\ge ... | First, show that this is true for $n=1$:
$6\cdot7^{1}-2\cdot3^{1}=4\cdot9$
Second, assume that this is true for $n$:
$6\cdot7^{n}-2\cdot3^{n}=4\cdot{k}$
Third, prove that this is true for $n+1$:
$6\cdot7^{n+1}-2\cdot3^{n+1}=$
$6\cdot7\cdot7^{n}-2\cdot3\cdot3^{n}=$
$7\cdot6\cdot7^{n}-3\cdot2\cdot3^{n}=$
$7\cdot6\cdot7^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Determine using the comparison test whether the series $\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^2 + 1}}$ diverges How can I determine whether or not the following series converges using the comparison test?
$$\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}}$$
As $n$ goes to infinity, the sum is roughly equal to $\sum_{... | Note that $n^2+1 \le 2 n^2$ so you have
${1 \over \sqrt[3]{n^2+1}} \ge {1 \over \sqrt[3]{2}} {1 \over \sqrt[3]{n^2}} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$. I'm trying to figure out how to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$.
Hints?
| $$\begin{align}6x^2-20x+23&=\frac{2}{3}\left(\frac326x^2-\frac3220x+\frac3223\right)\\
&=\frac23\left(9x^2-15x+\frac{69}2\right)\\
&=\frac23\left(9x^2-15x+\frac{50}2+\frac{19}{2}\right)\\
&=\frac23\left(9x^2-15x+25+\frac{19}{2}\right)\\
&=\frac23\left((3x-5)^2+\frac{19}{2}\right)\\
&=\frac23\left(3x-5\right)^2+\frac23\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int_0^\pi \frac{\cos^4(x)}{1+\sin^2(x)}dx$ using contour integrals How do I go about calculating $\int_0^\pi \frac{\cos^4(x)}{1+\sin^2(x)}dx$ using contour integrals?
| Suppose we seek to evaluate
$$J = \int_0^{\pi} \frac{\cos^4 x}{1+\sin^2 x} dx
= \frac{1}{2} \int_0^{2\pi} \frac{\cos^4 x}{1+\sin^2 x} dx.$$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) dx$ and hence
$\frac{dz}{iz} = dx$ to obtain
$$\frac{1}{2} \int_{|z|=1}
\frac{(z+1/z)^4/2^4}{1+(z-1/z)^2/(2i)^2} \frac{dz}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim \sup, \lim \inf$ for sequence $s_n$ defined by $s_1=0$, $s_{2m}=\frac{s_{2m-1}}{2}$ $s_{2m+1}=\frac{1}{2}+s_{2m}$ I want to find $\lim \sup, \lim \inf$ for the sequence $s_n$ defined by $s_1=0$, $s_{2m}=\frac{s_{2m-1}}{2}$ $s_{2m+1}=\frac{1}{2}+s_{2m}$. So $s_n$ depends on $s_{n-1}$, but the dependence is... | You just have to note the following pattern in your sequence.
$$
s_n = \begin{cases}
1- \dfrac 1{2^{\frac{n-1}{2} }} & n \neq 1 \text{ odd} \\
\dfrac 12 - \dfrac 1{2^{\frac{n}{2} }} & n \text{ even}
\end{cases}
$$
To see this, the base cases are easy to see: Put $n = 3$ and $s_3 = \frac{1}{2} = 1 - \frac 1{2^1}$, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find the volume of the region bounded by the paraboloid $z=x^2+y^2$ and the cylinder $x^2+y^2=9$ My attempt:
$\int_{x=-3}^3 \int_{y=-3}^3 x^2+y^2 dy dx$
$\int_{-3}^3 6x^2 + 18 dx$
$= 216?$
| The limits of your solution actually define the squared cylinder $[-3,3]\times[-3,3]$. You could either try by changing the limits to $-\sqrt{9 - x^2} \le y \le \sqrt{9 - x^2} $ and $-3 \le x \le 3$ or by changing your coordinate system to $x = r\cos\theta$, $y = r sin\theta$
$$
\int\int_{x^2 + y^2 < 9} dxdy\; (x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Squared magnitude of two complex numbers I believe one easy questions for maths experts. If I have two complex numbers x=a+ib and y=c+di is the squared magnitude of their sum equal to:
\begin{equation}
|x+y|^2=|x|^2+2|xy|+|y|^2
\end{equation}
| Your LHS equals
$$
(a+c)^2+(b+d)^2=(a^2+b^2)+2(ac+bd)+(c^2+d^2)=|x|^2+2(ac+bd)+|y|^2
$$
so it's almost the RHS, except for the fact that
\begin{aligned}
2|xy|&=2|(ab-cd)+i(ac+bd)|=2\sqrt{(ab-cd)^2+(ac+bd)^2}\\
&=2\sqrt{a^2b^2+c^2d^2+a^2c^2+b^2d^2}\\
&\neq 2(ac+bd).
\end{aligned}
In fact,
$$
(ac+bd)^2=a^2c^2+b^2d^2+2(ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rig... | An integer is divisible by $3^k$ if and only if the sum of the digits is divisible by $3^k$. In this case, note the sum of the digits is always $9$!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Fibonacci Identity with Binomial Coefficients A friend showed me this cool trick: Take any row of Pascal's triangle (say, $n = 7$):
$$1, 7, 21, 35, 35, 21, 7, 1$$
Leave out every other number, starting with the first one:
$$7, 35, 21, 1$$
Then these are backwards base-5 "digits", so calculate:
$$7 + 35 \cdot 5 + 21 \cd... | Suppose we seek to show that
$$\frac{1}{2^{n-1}} \sum_{k=0}^{\lfloor n/2\rfloor}
5^k {n\choose 2k+1} = F_n$$
a Fibonacci number. Introduce
$${n\choose 2k+1} = {n\choose n-2k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2k}} (1+z)^n \; dz.$$
Observe that the largest index $k$ producing a non-zero value ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
What is $(-1)^{\frac{1}{7}} + (-1)^{\frac{3}{7}} + (-1)^{\frac{5}{7}} + (-1)^{\frac{9}{7}} + (-1)^{\frac{11}{7}}+ (-1)^{\frac{13}{7}}$? The question is as given in the title. According to WolframAlpha, the answer is 1, but I am curious as to how one gets that. I tried simplifying the above into $$6(-1)^{\frac{1}{7}}$$ ... | Notice:
$$-1=\left|-1\right|e^{\left(\arg\left(-1\right)+2\pi\text{k}\right)i}=e^{\pi\left(1+2\text{k}\right)i}$$
Where $\text{k}\in\mathbb{Z}$
So, when $\text{n}\in\mathbb{R}$:
$$\left(-1\right)^\text{n}=e^{\text{n}\pi\left(1+2\text{k}\right)i}=\cos\left(\text{n}\pi\left(1+2\text{k}\right)\right)+\sin\left(\text{n}\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Integration of Erf function Let $f(x) = \int^\sqrt{x}_1 e^{-t^2}dt$. Find $\int^1_0 \frac{f(x)}{
\sqrt{x}}dx$.
Could anyone give me any hint how to start? The Erf function $f(x)$ seems not to be easily integrated.
| Brevan Ellefsen's solution is the most straightforward and efficient. Here is an alternative solution.
\begin{equation}
f(x) = \int\limits_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt
\label{eq:161108-1}
\tag{1}
\end{equation}
\begin{equation}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
\label{eq:161108-2}
\tag{2}
\end{equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?
| Note that
\begin{aligned}
\sum_{i=1}^{n}i^5 &= \sum_{i=0}^{n-1}(i+1)^5 \\
& = \sum_{i=0}^{n-1}\left(i^5 + 5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
&= \sum_{i=0}^{n-1}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
&= -n^5 + \sum_{\bbox[yellow]{i=1}}^{\bbox[yellow]{n}}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Sum of the digits of $N=5^{2012}$ The sum of the digits of $N=5^{2012}$ is computed.
The sum of the digits of the resulting sum is then computed.
The process of computing the sum is repeated until a single digit number is obtained.
What is this single digit number?
| You want to know the value of $5^{2012} \pmod 9$.
Since $\varphi(9) = 3^2 - 3 = 6$ and $\gcd(5,9) = 1$, then, by Euler's theorem, $5^6 \equiv 1 \pmod 9$.
Since $2012 = 335 \times 6 + 2$,
$$5^{2012} \equiv (5^6)^{335} \times 5^2
\equiv 1^{335} \times 25 \equiv 7 \pmod 9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show this equality? This is a equation from a book of control theory:
$$c(sI-A)^{-1}b=\frac{\det(sI-A)-\det(sI-A-bc)}{\det(sI-A)}$$
$I$ is identity matrix, $A$ is $n\times n$, $b$ is $n \times 1$ vector, $c$ is $1 \times n$ vector.
I was trying to use induction, but seems not work. I would very appreciate some a... | Note that $bc$ is a rank-1-matrix. We will find a formula for $(B - bc)^{-1}$, where $B$ is any $n\times n$-matrix ($sI - A$ in your case). We will prove
$$ \det (B - bc) = (1 - cB^{-1}b)\det(B) $$
To prove that, note
\begin{align*}
\det(B- bc) &= \det B\det(I - B^{-1}bc)\\
&= \det B \det\begin{pmatrix} I- B^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does the following series converges or diverges? $\sum_{n=1}^{+\infty}\left[n \log\left ( \frac{2n+1}{2n-1} \right )-1\right]$ ,converges or not
My attempt
$a_n=n \log\left ( \frac{2n+1}{2n-1} \right )-1 \\ \\
\log \left ( \frac{2n+1}{2n-1} \right )=\log \left ( 1+\frac{2}{2n-1} \right ) \sim \frac{2}{2n-1} \ (n\righta... | Your mistake was assuming this: If $x_n \sim y_n,$ then $x_n - 1 \sim y_n -1.$ A counterexample is $x_n =1+ 1/n, y_n = 1+1/n^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find a limit without l'Hospital: $\lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ Find $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$.
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ =
$\displaystyle \lim_{x\to -\infty}\left(\frac{\l... | We have
$$
\ln(1+t)=t+o(t^2).
$$
Thus, since $2^x\to0$ and $3^x\to0$ when $x\to-\infty$,
$$
\frac{\ln(1+2^x)}{\ln(1+3^x)}=\frac{2^x+o(2^{2x})}{3^x+o(3^{2x})}
=\left(\frac23\right)^x\,\frac{1+o(2^x)}{1+o(3^x)}.
$$
The second fraction goes to $1$ when $x\to-\infty$. So the limit will be equal to
$$
\lim_{x\to-\infty}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$
Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$ $(x,y,z>0)$
Can I apply Jensen here, sum of the squareroots is greater then squareroot of the sum, so I make the RHS smaller then show also then the inequality ... | Recall $\|(x,y,z)\| = \sqrt{x^2+y^2+z^2}$.
By C-S
$$x+y+ z= (x,y,z)\cdot (1,1,1) \le \|(x,y,z)\|\|(1,1,1)\|=\|(x,y,z)\|\sqrt{3}.$$
But by triangle inequality (follows from C-S):
$$ \|(x,y,z)\| = \frac 12 \|(x,y,0) +(0,y,z)+ (x,0,z)\|\le \frac 12 \left(\|(x,y,0)\|+\|(0,y,z)\|+\|(x,0,z)\|\right).$$
Summarizing:
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A proof of fraction inequality by induction Prove that $$\left(\frac{a+b}{2}\right)^n\le\frac{a^{n}+b^{n}}{2}$$ I have checked the base case and assumed the inequality is valid and continued like this $$\left(\frac{a+b}{2}\right)^{n+1}\le\frac{a^{n+1}+b^{n+1}}{2}\Rightarrow \left(\frac{a+b}{2}\right)^n\left(\frac{a+b}{... | Work:
$\left( \dfrac{a+b}{2}\right)^{n+1}=\left(\dfrac{a+b}{2}\right)^n\left(\dfrac{a+b}{2}\right)$
$ \hspace{2.6 cm}\leq \dfrac{a^n+b^n}{2}\cdot \left(\dfrac{a+b}{2}\right)$
$\hspace{2.6 cm} = \dfrac{a^{n+1}+a^nb+ab^n+b^{n+1}}{4}$
Now write what you want to obtain and subtract extras:
$\hspace{2.6 cm} = \dfrac{a^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $k$ be an odd positive integer. Prove that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$. Problem:Let $k$ be an odd positive integer. Prove that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$.
My Attempt: Proof (By Induction on $k$): Let $P(k)$ be the proposition that $(1+2+... | Your identity is wrong so your solution is wrong.
Here is one way to attempt the problem. We know $1+2+ \cdots +n=\dfrac{n(n+1)}{2}$ and $\gcd (n,n+1)=1$. We consider following two cases:
Case 1. If $n$ is even then there are $n/2$ pairs of $i^k,(n-i+1)^k$ and see that $n+1 \mid i^k+(n-i+1)^k$ since $k$ is odd. Thus, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$
Problem Statement:-
Solve the equation $$x^2+\dfrac{9x^2}{(x+3)^2}=27$$
I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value.
These are the things that I have... | HINT:
Using $a^2+b^2=(a-b)^2+2ab,$
$$x^2+\left(\dfrac{3x}{x+3}\right)^2=\left(x-\dfrac{3x}{x+3}\right)^2+2\cdot x\cdot\dfrac{3x}{x+3}=\left(\dfrac{x^2}{x+3}\right)^2+6\cdot\dfrac{x^2}{x+3}$$
Generalization :
For $a^2+b^2=k$
If $\dfrac{ab}{a+b}=c$ where $c$ is a non-zero finite constant,
$$\implies k=(a+b)^2-2ab=(a+b)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to p... | Let us conjugate freak_warrior's tip with a telescopic sum to find the closed formula.
Let us set $a_n = n(n+1)(n+2)(n+3) $
You want to find
$$\sum_{i = 1}^{k} a_i $$
Rewrite
$$a_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5} - \frac{(n-1)n(n+1)(n+2)(n+3)}{5} $$
One can prove this is true by factoring out $n(n+1)(n+2)(n+3)$.
Set... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 5
} |
Find all primes $p$ such that $p^3-4p+9=x^2$, where $x$ is a positive integer. Problem:Find all primes $p$ such that $p^3-4p+9=x^2$, where $x$ is a positive integer.
My Attempt: I have made the following observations while trying tot solve this problem:
*
*$p=2$ yields $x=3$.
*For $p>2$ the following is true: $x^2... | If $p^3-4p+9=n^2$ then $n^2\equiv 9 \pmod{p}$ and $\pm n \equiv 3 \pmod{p}$ since $n^2-9\equiv 0$ can only have two roots modulo a prime.
We then have
$$
\begin{array}{ll}
n^2=(-n)^2 &= (kp+3)^2 \\
p^3-4p+9 &= k^2p^2+6kp+9 \\
0 &= p^2-k^2p-(6k+4) \\
p & = \frac{k^2\pm \sqrt{k^4+24k+16}}{2}
\end{array}
$$where we used $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\lim _\limits{n\to \infty }\left(4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n\right)$ How can I calculate this limit?
$\lim_\limits{n\to \infty }\left(4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n\right)\:$
Can I apply the rule $a_n+1/a_n$ to proove its convergence?
| Hint. Note that
$$a_n:=4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n=4n^2\left(\frac{1+3/n+3/n^3}{2+1/n^2-1/n^3}\right)^n.$$
Can we say that $a_n\sim\frac{4n^2}{2^n}$?
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$... | Hint
It's easier to begin by the end
$$\left(\frac{(n+1)(n+2)}{2}\right)^2-\frac{n(n+1)}{2}^2=$$
$$\left(\frac{n+1}{2}\right)^2\left((n+2)^2-n^2\right)=$$
$$\left(\frac{n+1}{2}\right)^2(4(n+1))=$$
$$(n+1)^3.$$
qed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions of the complex equation: $(z+1)^3 = 36(z^*+1)$ today in an exam a friend had this equation with complex number:
$$(z+1)^3 = 36(z^*+1)$$
where with z* I mean conjugate z; can someone solve this?
We tried and there is probably a way to solve it doing a big system with everyterm with i = 0 and everyterm without... | Set $z + 1 = w$, then $\overline{z} + 1 = \overline{w} + 1$ as well, and you want to solve
$$
w^3 = 36\overline{w}.
$$
Multiplying both sides by $w$, we find
$$
w^4 = 36\left|w\right|.
$$
If $w = re^{i\theta}$, then $r^4e^{i4\theta} = 36 r^2$. It follows that either $r = 0$ (in which case $w = 0$), or $r^2 = 36$, so $r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Need help conjugating I need some help conjugating because I feel like I am overlooking a step:
$$\lim_{x \rightarrow 4} \frac{3(x-4)(\sqrt{x+5})}{3 - \sqrt{x+5}}$$
I've done the conjugation by multiplying by $(3 + \sqrt{x+5})$ to clear out the bottom, but how would the top look like once multiplied?
| $$\lim _{ x\rightarrow 4 } \frac { 3(x-4)(\sqrt { x+5 } ) }{ 3-\sqrt { x+5 } } =\lim _{ x\rightarrow 4 } \frac { 3(x-4)(\sqrt { x+5 } )\left( 3+\sqrt { x+5 } \right) }{ \left( 3-\sqrt { x+5 } \right) \left( 3+\sqrt { x+5 } \right) } =\\ =\lim _{ x\rightarrow 4 } \frac { 3(x-4)(\sqrt { x+5 } )\left( 3+\sqrt { x+5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ABC$ is an acute angle triangle, then point $P((\cos B-\sin A),(\sin B-\cos A))$ lies in which quadrant triangle $ABC$ is an acute angle triangle, then point $P((\cos B-\sin A),(\sin B-\cos A))$ lies in which quadrant $.........$
$\cos B-\sin A= \sin(\frac{\pi}{2}-B)-\sin A = 2\cos(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2... |
$\sin(\frac{\pi}{2}-B)-\sin A = 2\cos(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
$\cos(\frac{\pi}{2}-B)-\cos A=-2\sin(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
These are incorrect. They should be
$$\sin\left(\frac{\pi}{2}-B\right)-\sin A=2\cos\left(\frac{\frac{\pi}{2}+A-B}{\color{red}{2}}\right)\sin\left(\frac{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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summation of a infinite series i want to know if this infinite sum $\sum_{n=1}^{\infty}$ $\frac{a(a+1)(a+2)......(a+n-1)}{b(b+1)(b+2)......(b+n-1)}$ converges or diverges ? where a>0 and b>a+1.
if the sum converges what is the sum i.e where it will converge? i need a concrete explanation
i found some inequalities $\fra... | Since Hamou has answered the question of convergence, I'll answer the question about the sum of the series. For $b - a > 1$, the sum of the series is $$-1 + \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$
Indeed, Abel's continuity theorem gives
$$\sum_{n = 1}^\infty \frac{a(a+1)\cdots (a + n-1)}{b(b + 1)\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Regular non-orthodox semigroups Is there a resource somewhere with smallest finite and other examples of regular semigroups that are not orthodox? I want concrete examples for my private research. I once installed GAP and two semigroup packages too but I have no idea how difficult it would be to calculate such examples... | There is only one finite semigroup of order 4 whose idempotents don't form a subsemigroup (up to anti-isomorphism), and has the following Cayley table:
\begin{array}{l|llll}
& 1 & 2 & 3 & 4 \\ \hline
1 & 1 & 1 & 1 & 1 \\
2 & 1 & 1 & 1 & 2 \\
3 & 1 & 2 & 3 & 2 \\
4 & 1 & 1 & 1 & 4
\end{array}
Sadly it is not regular. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| Multiply by $2$, then we get:
$$8a^2+2b^2+2 \ge 4ab+4a+2b$$
and rearranging we get:
$$(4a^2-4ab+b^2)+(b^2-2b+1)+(4a^2-4a+1) \ge 0$$
and then
$$(2a-b)^2+(b-1)^2+(2a-1)^2 \ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Probability only 2 blue balls are selected out of 6 different colored balls? We don't have an answer key for this test questions and I just wanted to confirm my answers.
We have a bag with 6 balls, 3 blue, 2 green, and 1 red.
If we select 3 at random, what is the chance we select only 2 of 3 are blue balls?
So I have... | Yes, that is okay. $9/20$ and $6/20$ are the answers for the reasons you gave.
The probability of selecting exactly $2$ from $3$ blue ball (and $1$ from $3$ non-blue) when selecting $3$ from all $6$ balls is more simply $\dbinom 3 2\dbinom 31\big/\dbinom 6 3$.
But yes, $\dbinom 3 2\Big(\dbinom 2 1 +\dbinom 1 1\Big)\... | {
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"url": "https://math.stackexchange.com/questions/2036689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $x^2 + y^2 = z^2$, then at leeast one of x and y is a multiple of 3. Show that if $x^2 + y^2 = z^2$, then at least one of $x$ and $y$ is a multiple of 3.
Attempt:
Given $x,y,z$ is pyhagorean triple,
$$
x^2+y^2=z^2.
$$
Assume neither $x$ nor $y$ is divisible by $3$. Since $x^2+y^2=z^2$, $x^2+y^2 \equiv z^2 ... | Suppose none of $x,y,z$ is divisible by $3$.
then
$$x\equiv \pm 1\mod 3,$$
$$y\equiv \pm 1 \mod 3,$$
and
$$z\equiv \pm 1 \mod 3$$
$\implies$
$$x^2+y^2-z^2\equiv 1+1-1\mod 3$$
$$\implies x^2+y^2\neq z^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Calculate: $\lim\limits_{n\to\infty} \sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}$ Calculate: $$\lim\limits_{n\to\infty} \sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}$$
I thought a Riemann sum could lead to something, but couldn't find a suitable partition.
Hint, please?
| One may recognize a Riemann sum, by writing
$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}=\frac1n \cdot\sum_{k=0}^{n} \frac{2+\frac{k}n }{1+(2+\frac{k}n)^2},
$$ then letting $n \to \infty$, to obtain
$$
\frac1n \cdot\sum_{k=0}^{n} \frac{2+\frac{k}n}{1+(2+\frac{k}n)^2} \to \int_0^1 \frac{2+x }{1+(2+x)^2}\:dx.\tag1
$$
Add-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Showing $\Gamma\left(x+1\right) \ge x^{x/2}$, when $x\ge2$ It is easy to show that $(2n)! \ge (2n)^n$.
Indeed, if we have $ 1\le k\le n$, for integer $k,n$ , we can write:
$$
\begin{aligned}
0\lt n+\frac{1}{2}-k &\le n - \frac{1}{2} \\
\left(n+\frac{1}{2}-k\right)^2 &\le \left(n - \frac{1}{2}\right)^2 \\
k\cdot(2n-k+1)... | Let
$$f(x) = \log \Gamma(x + 1) - \frac{x \log(x)}{2} = \log(x) + \log \Gamma(x) - \frac{x \log(x)}{2}$$
We will prove that $f$ is a convex function with $f(2) = 0$ and $f'(2) > 0$. Using the subderivative property of convex functions we then get
$$f(x) \geq f(2) + (x - 2) f'(2) \ge 0 \qquad \text{for } x \ge 2,$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Number of occurrences of k consecutive 1's in a binary string of length n (containing only 1's and 0's) Say a sequence $\{X_1, X_2,\ldots ,X_n\}$ is given, where $X_p$ is either one or zero ($0 < p < n$). How can I determine the number of strings, which do contain at least one occurrence of consequent $1$'s of length $... | One systematic way(not very cool to do it by hand tho) to do it is using automatons and the Chomsky-Schûtzenberger theorem in the following way.
Case $k = 2$:
The automata that accepts your language(namely $F = \{x\in \{0,1\}^*:\underbrace{11\cdots 11}_{\text{$k$ times}}\in Sub(x)\}$) is described by the image below($S... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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How many solutions does $x^2 + 3x +1 \equiv 0\, \pmod{101}$ have? $x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,
$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$
$\therefore$ The equations have... | Since
$$
(x-49)^2-77\equiv x^2+3x+1\pmod{101}
$$
we are looking for solutions to $(x-49)^2\equiv77\pmod{101}$. You have verified that there is a root, so $77$ is a quadratic residue mod $101$, thus, there are two solutions for
$$
(x-49)^2\equiv77\pmod{101}
$$
Alternatively, working mod $101$, by squaring and multiplyin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Let $a,b,c$ be the length of sides of a triangle then prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Let $a,b,c$ be the length of sides of a triangle then prove that:
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Please help me!!!
| let
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)=0$ (1)
let
$x=-a+b+c; y=a-b+c; z=a+b-c$
$z,y,x$ are twice the length of the segments between the vertices and the touching point of the incircles. so
$a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}$
substitute them to (1) and multiply the inequality by 16
$(y+z)^2(z+x)(y-x)+(z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$, find the value of $a+b$
$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.
I am not able to proceed with solving this ... | To find $\sqrt{a+\sqrt{a+\cdots}} $, solve the equation
$x = \sqrt{a+x}$
The solution of $\sqrt{31+\sqrt{31+\cdots}}$ gives us $x = \frac{1\pm 5\sqrt{5}}{2}$.
The solution of $\sqrt{1+\sqrt{1+\cdots}}$ gives us $y = \frac{1\pm \sqrt{5}}{2}$.
Thus, we have $$\frac{x}{y} = 6-\sqrt{5}$$ Hence, $a=6, b=5 \Rightarrow a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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All fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$ Find all fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$
for some integers $k,l$.
Please check my answer and tell me is correct or not....
$$\frac{43}{31},\f... | $$\frac{7k-5}{5k-3}=\frac{6l-1}{4l-3}$$
$$28kl-20l-21k+15=30kl-18l-5k+3$$
$$2kl+2l+16k-12=0$$
$$kl+l+8k-6=0$$
either:
$$l(1+k)=2(3-4k)$$
so
$$l=2\frac{3-4k}{1+k}$$
or:
$$k(l+8)=6-l$$
so
$$k=\frac{6-l}{l+8}$$
Let's go with this second one to complement the other answer.
Then $$k=-\frac{l-6}{l+8}=-\left(\frac{l+8-14}{l+8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{3-x}-\sqrt{-x^2+8x+10}=1$ Consider the equation
$$
\sqrt{3-x}-\sqrt{-x^2+8x+10}=1.
$$
I have solved it in a dumb way by solving the equation of degree four. So, the
only real solution is $x = -1$. Can you please suggest, maybe there are better or easier ways to solve it?
| This is a simple way used when I was in high school.
Basically, all you have to do is draw out the minimal polynomial of the known solutions (which make this even useful with most algebraic solutions, not only integers).
Firstly, match each square root with its real value.
\begin{align} (\sqrt{3-x}-2)-(\sqrt{-x^2+8x+10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can we compute $\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3}$ and $\lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$? Could we compute the limits
$$\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3} \\ \lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$$
without using the l'Hospital rule and the Taylor expansion?
| The first one is equal to $1+\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}.$ We can apply to this limit the technique in this answer of mine.
Let $L=\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}$. This is not infinite if it exists because near $0$ one has $$-\frac12\leftarrow\frac{\cos x-1}{x^2}<\frac{\frac{\sin x}{x}-1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x... | Let $F(x,y) = \frac{1}2 \log(x^2+y^2)$. Then $f$ is harmonic everywhere away from the origin, and so are its derivatives $\partial_1 F$ and $\partial_2 F$. To prove the first claim, note that
$$\partial_1 F = \frac{x}{x^2+y^2}$$
$$\partial_{11} F = \frac{y^2-x^2}{(x^2+y^2)^2}$$
and by symmetry
$$\partial_{22} F = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Struggling with a strong induction problem involving recurrence relations The problem states:
Let $b_0$ = $12$ and $b_1$ = $29$, and for all integers $k ≥ 2$, let $b_k$ = $5b_{k-1}-6b_{k-2}$.
Prove that for all $n ≥ 0$, $b_n$ = $5\:\cdot \:3^n\:+\:7\:\cdot\:2^n$
I'm familiar with questions where I'm supposed... | I think you just have to show that $b_n = 5\cdot 3^n + 7\cdot 2^n$ fulfills the recurrence relation and initial conditions for every instance $n$.
As the relation involves up to two prior sequence elements I would use strong induction.
You start with direct proofs for $b_0$, then $b_1$ and the induction from $n=2$.
Upd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Hall and Knight- Higher Algebra Problem 23, Example V. b If $r < 1$ and positive and m is a positive integer, show that $(2m+1)r^m(1-r)<1-r^{2m+1}$. Hence show that $nr^n$ is indefinitely small when $n$ is indefinitely great.
I have tried by taking $\frac{1-r^{2m+1}}{1-r}$ as the sum of the series $1 + r + r^2 ... + r... | You have
$$\begin{aligned}
\frac{1 - r^{2m+1}}{1-r}&= 1+r+ \dots +r^{2m}\\
&= r^m(\frac{1}{r^m} + \frac{1}{r^{m-1}} + \dots + \frac{1}{r} + 1+ r +r^2 + \dots + r^{m-1}+r^m)\\
&= r^m[1 + (r^m+\frac{1}{r^m}) + (r^{m-1} + \frac{1}{r^{m-1}}) + \dots +(r + \frac{1}{r})]\\
&> r^m(1 + \underbrace{2 + \dots + 2}_{m \text{ time... | {
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"url": "https://math.stackexchange.com/questions/2064803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = \sqrt{2}$, and $s_{n+1} = \sqrt{2 + \sqrt{s_n}}$, what is the limit of this sequence? Here's Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2 + \sqrt{s_n}} \ \ (n = 1, 2, 3, \ldots),$$ ... | If the limit $s$ of $s_n$ exists, then
$$
s=\lim s_n=\lim s_{n+1}=\lim \sqrt{2+\sqrt{s_n}}=\lim \sqrt{2+\sqrt{s}}.
$$
Hence, the limit satisfies the equation
$$
s=\sqrt{2+\sqrt{s}}.
$$
Thus,
$$
s^2=2+\sqrt{s}\qquad\text{or equivalently}\qquad (s^2-2)^2=s.
$$
Thus the limit is the unique solution of $s^4-4s^2-s+4=0$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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.