Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
probability that $(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = 1$
If $\omega$ is a complex cube root of unity with $0<\arg(\omega)<2\pi. $ A fair die is thrown three times
If $a,b,c$ are the numbers obtained on the dice , Then probability that
$(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = 1$
$\bf{Attempt:... | You are looking for the probability of
$$(a-b)^2+(b-c)^2+(c-a)^2 = 2, \qquad \text{ when }a,b,c \in \{1,2,3,4,5,6\}$$
For the sum of the three squares to be $2$, the only way it can happen is when two of them are $1$ and the third is a $0$. For example, $a=b$, then $c=a \pm 1$. So you are looking for triples of the fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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primality of $\frac{(2^{19}+1)}{3}$ and $\frac{(2^{23}+1)}{3}$ This is from an exercise in Burton's Number theory book. I have a hard time solving this. How could we prove that $\displaystyle\frac{(2^{19}+1)}{3}$ and $\displaystyle\frac{(2^{23}+1)}{3}$ are primes. I know that the primes dividing the numbers are of the ... | In general, if $q>3$ is prime and $p$ a prime divisor of $N=\frac{2^q+1}{3}$, then, as you argued, $p=2qk+1$ for some $k$.
So: $2^{(p-1)/2}\equiv (-1)^k\pmod p$.
If $k$ is even, then $2$ is a square modulo $p$ which means that $p\equiv \pm 1\pmod{8}$ or $2qk\equiv 0,6\pmod{8}$ or $qk\equiv 0,3\pmod{4}$. But since $k$ ... | {
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"url": "https://math.stackexchange.com/questions/2341855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the correct exponential generating function for word counting problem
Find the number of five-letter words that use an odd number of $A$s, at least two $B$s, and at most two $C$s.
The given answer: $65.$
I use the expression $(A + \frac{A^3}{3!})(\frac{B^2}{2!} + \frac{B^3}{3!} + \frac{B^4}{4!})(1 + C + \frac... |
We are looking for
\begin{align*}
120[z^5]\left(z+\frac{z^3}{3!}+\frac{z^5}{5!}\right)\left(\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}\right)
\left(1+z+\frac{z^2}{2!}\right)
\end{align*}
... can you figure out why?
As you correctly noted, the term $\frac{A^5}{5!}$ does not contribute to the result.... | {
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"timestamp": "2023-03-29T00:00:00",
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Probability of a random selection of $2$ out of $5$ shoes
From a set of five pairs of shoes, two of the shoes are selected at random. Find the probability of each of the following:
a) Both are from the same pair.
My answer: $\dfrac{\binom{10}{1}\binom{1}{1}}{\binom{10}{2}}$
b) One left shoe and one right shoe are se... | The numerators in your solutions are incorrect. Considering the two different questions:
*
*Both shoes must be from the same pair. Using probability on the first and second draw, we find: $$\frac{10}{10} \cdot \frac{1}{9} = \frac{1}{9}$$ With your suggested approach, however, we find: $$\frac{10 \cdot 1}{45} = \frac... | {
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Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$ Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$.
Here it is how I proceeded:
\begin{align*}
(x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\
& = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\
& = (x^2 + 8x + 7)[(x^2 + 8x +... | One way to solve this problem would be to use the Rational Roots Theorem to find the roots $-2$ and $-6$, then use polynomial division to get a quadratic which is easily solved. However this method will not work in general, as a polynomial does not need to have any rational roots at all
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| Consider the monic polynomial with roots $a$, $b$, $c$
$$P(x)=(x-a)(x-b)(x-c)$$
Expanding out:
$$P(x)=x^{3}+x^{2}(-a-b-c)+x(ab+bc+ac)-abc$$
(This is essentially Vieta's formula).
Then plugging in your values this polynomial equals:
$$P(x)=x^{3}-2x^{2}+x-abc$$
Further, since $a$, $b$, $c$ are roots
$$P(a)=a(a^{2}-2a+1-b... | {
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Calculate the area of the triangular region ABC How to calculate the area of the triangle from the figure
| Let $O$ be the centre of the circle and $r$ its radius.
Then $AB=2\sqrt{r^2-(r-1)^2}=2\sqrt{2r-1}$, $BC=2\sqrt{r^2-(r-2)^2}=2\sqrt{4r-4}$, $AC=2\sqrt{r^2-(r-3)^2}=2\sqrt{6r-9}$.
Now depending on what edge we take as base, the area of $ABC$ can be expressed in three forms:
$$\sqrt{2r-1}\cdot (2r-1)=\sqrt{4r-4}\cdot (2r... | {
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Two ways to calculate the primitive of $f(x) =\frac{x^2}{x^4-1}$ but two different results, why? When I search primitives of $f(x)= \dfrac{x^2}{x^4-1}$, using partial fraction decomposition the solutions are
$\boxed {F(x) =\dfrac{1}{4} \ln|x-1| + \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
But if I use an o... | $\boxed {F(x) =\dfrac{1}{4} \ln|x-1| - \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
F is defined over $\mathbb{R}\backslash \{-1,1\}$
Regarding your result, Your function is defined only over $]-1,1[$, why?
Now for the message from Bernard, thank you, I realised I must used the u-substitution. So :
$\display... | {
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Finding a Closed Form for $\int_0^\infty A\sin(\frac{2\pi}{T}x) \exp(-bx)dx$ I'm struggling in finding a closed form for,
$$ \int_0^\infty Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right) \,dx $$
My Attempt: Let $H = \int Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx$. First I integrate by parts, splitting $f=e^{-bx}$ and $d... | This is the Laplace Transform of $\sin \left(\dfrac{2 \pi x}{T}\right)$
Its value is $\dfrac{2 \pi A T}{b^2 T^2+4 \pi ^2}$
I just looked on the Laplace transform table
| {
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Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is ev... | The solution to the equation $x^2+x+1=0$ is $e^{\frac{2\pi}{3}}$ and $e^{\frac{4\pi}{3}}$ .Having $1990=3\times 663 +1$ and $200=3\times 66 +2 $, we can evaluate the second equation as
if $ x= e^{\frac{2\pi}{3}} $ then $x^{1990}=x^{3\times 663 +1}=x,x^{200}=x^{3\times 66 +2}=x^2$, so $e^{\frac{2\pi}{3}}$ is root of ... | {
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If a and b are two distinct real values such that $F (x)= x^2+ax+b $ And given that $F(a)=F(b)$ ; find the value of $F(2)$
If $a$ and$ b$ are two distinct real values such that
$$F (x)= x^2+ax+b $$
And given that $F(a)=F(b)$ ; find the value of $F(2)$
My try:
$$F(a)=a^2+a^2+b= 2a^2+b,\quad
F(b)=b^2+ab+b $$
$F... | $F(2) = 2^2+2a+b = 4+2a+b$. From $2a^2 = b^2+ab \implies a^2-b^2 = ab - a^2 \implies (a-b)(a+b) = a(b-a) \implies (a-b)(2a+b) = 0\implies a = b$ or $2a+b = 0$. Since $a \neq b, 2a+b = 0 \implies F(2) = 4+0 = 4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Degrees of the irreducible factors of $X^{13}-1$ I need some hints to determine the degrees of irreducible factors of $X^{13}-1$ in $\Bbb F_5[X]$. I know the first factor is $X-1$ and it is a single root. There are not other roots. I also checked from Mathematica that $1+X+ \dots X^{12}$ is not irreducible in $\Bbb F_5... | We know that $X^{5^n} - X$ is the product of all irreducible polynomials over $\mathbb{F}_5$ whose degree divides $n$. You can compute
$$\mathrm{gcd}(X^{13} - 1, X^5 - X) = X - 1,$$
$$\mathrm{gcd}(X^{13} - 1, X^{25} - X) = X - 1,$$
$$\mathrm{gcd}(X^{13} - 1, X^{625} - X) = X^{13} - 1$$ with the Euclidean algorithm and ... | {
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proof that $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ Let $n$ be a positive integer number.
How do we show that the irreducible polynomial $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ for all $n$?
| $$P \equiv n^4-2n^3-n^2+2n \equiv (n-2)\cdot(n-1)\cdot n\cdot(n+1)\equiv 0\mod 24 $$
The last one is because four consecutive numbers must include one number that is divisible by $4$, another number that is divisible by $2$, and one number that is divisible by $3$.
| {
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"url": "https://math.stackexchange.com/questions/2356982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $-\int_0^1\frac{1-x}{(1-x+x^2)\log x}\,dx$ I was trying do variations of an integral representation for $\log\frac{\pi}{2}$ due to Jonathan Sondow, when I am wondering about if it is possible to evaluate
$$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx,\tag{1}$$
Wolfram Alpha online calculator provide me a clos... | Let
$$I=\int_0^1\frac{1+x}{1+x^3}\frac{x-1}{\ln(x)}dx$$
and
$$I(a)=\int_0^1\frac{1+x}{1+x^3}\frac{x^a-1}{\ln(x)}dx$$
and notice that $I(1)=I$ and $I(0)=0.$
$$I'(a)=\int_0^1\frac{x^a+x^{a+1}}{1+x^3}dx\overset{x=t^{1/3}}{=}\frac13\int_0^1\frac{t^{\frac{a-2}{3}}+t^{\frac{a-1}{3}}}{1+t}dt$$
By using
$$\int_0^1\frac{x^n}{1+... | {
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Determine if these functions are injective
Determine if the following functions are injective.
$$f(x) = \frac{x}{1+x^2}$$
$$g(x) = \frac{x^2}{1+x^2}$$
My answer:
$f(x) = f(y)$
$$\implies \frac{x}{1+x^2} = \frac{y}{1+y^2}$$
$$\implies x+xy^2 =y+yx^2$$
$$\implies x=y$$
Hence $f(x)$ is injective
$g(x) = g(y)$
$$\implies... | Where you had $x+xy^2=y+yx^2,$ you canceled $xy^2$ from one side and $yx^2$ from the other side, but those are not the same.
But the equation can be written as $yx^2 - (y^2+1) x + y = 0,$ and that is $ax^2+bx+c=0,$ where $a=y,$ $b=-(y^2+1),$ and $c=y.$
The solution for $x$ of the equation $ax^2+bx+c=0$ is $x = \dfrac{-... | {
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Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \thet... | Let $x = r^2 \implies 1 \le x \le 49 $, and consider $f(x) = \dfrac{x}{16} + \dfrac{36}{x}\implies f'(x) = \dfrac{1}{16}- \dfrac{36}{x^2}\implies f'(x) = 0 \iff x = 24$, and $f(1) = 36+\dfrac{1}{16} = 36.0625, f(49) = \dfrac{49}{16}+\dfrac{36}{49}=3.8, f(24) = 3.$ $3\cos(2\theta)$ has a min of $-3$, and a max of $3$. P... | {
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On understanding the use of binomial theorem to find asymptotes of a real valued function. I have the function
$$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$
Wolfram Alpha says that it has a non linear asymptote at
$y=\frac{x^2}{2} - \frac{3x}{16}+ \frac{251}{256}$.
I have been told to expand using the binomial theorem to o... | Consider first the denominator of $$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$ $${\sqrt{4x^2+3x+2}}=2x{\sqrt{1+\frac{3x}{4x^2}+\frac2{4x^2}}}=2x{\sqrt{1+\frac{3}{4x}+\frac1{2x^2}}}$$ Now apply the binomial therorem or Taylor series to get
$${\sqrt{1+\frac{3}{4x}+\frac1{2x^2}}}=1+\frac{3}{8 x}+\frac{23}{128 x^2}+O\left(\frac... | {
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | HINT:
Set $1/x=h$ to get $$F=\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1}h$$
Now $\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1=\tan^{-1}\left(\dfrac{\dfrac{h+1}{4h+1}-1}{1+\dfrac{h+1}{4h+1}}\right)=\tan^{-1}\dfrac{-3h}{5h+2}$
$$\implies F=\lim_{h\to0^+}\dfrac{\tan^{-1}\left(\dfrac{-3h}{5h+2}\right)}{\left(\dfra... | {
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If $A+B=225^\circ$ then $\frac{\cot A}{1+\cot A}\cdot\frac{\cot B}{1+\cot B}=\ldots$
If $A+B=225^\circ$ then $\frac{\cot A}{1+\cot A}\cdot\frac{\cot B}{1+\cot B}=\ldots$
I tried using $\cot(A+B)$ formula but failed to get a proper answer.
Options are:(A) $1$; (B) $-1$; (C) $0$; (D) $\frac12$.
| $$\frac{\cot{A}}{1+\cot{A}}\cdot\frac{\cot{B}}{1+\cot{B}}=\frac{\cot{A}}{1+\cot{A}}\cdot\frac{\cot\left(45^{\circ}-A\right)}{1+\cot\left(45^{\circ}-A\right)}=$$
$$=\frac{\cos{A}}{\sin{A}+\cos{A}}\cdot\frac{\cos{A}+\sin{A}}{\cos{A}-\sin{A}+\cos{A}+\sin{A}}=\frac{1}{2}$$
| {
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How do you calculate the expected value of geometric distribution without diffrentiation? Is there any way I can calculate the expected value of geometric distribution without diffrentiation? All other ways I saw here have diffrentiation in them.
Thanks in advance!
| I know at least two ways off hand and there are probably others.
First I'll show you a concrete way to do it. After that I'll show you how to express the same thing exactly. (Together these make up only one of those "two ways". The other one now appears in the answer posted by "Henry".)
$$
\begin{array}{ccccccccccccccc... | {
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Inequality in sum of fractions
Is there a constant $C\ge 0$ such that for any $a,b,c,d>0$ the inequality holds:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge C+\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}
$$
My attempt:
$a=b=c=d$ gives $1\ge C$.
Now, take $b=c=d$; this leads to inequality with two vari... | Using scaling of $a,b,c,d$ you can set $d=1$ without loss of generality.
This gives the function
$$
f(a,b,c) \equiv \left( \frac{a}{b} + \frac{b}{c} + c + \frac{1}{a} \right) - \left( \frac{a+b}{b+c} + \frac{b+c}{c+1} + \frac{c+1}{1+a}\right)
$$
to minimize.
Taking the derivative $\frac{\partial f(a,b,c)}{\partial b}... | {
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"timestamp": "2023-03-29T00:00:00",
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is there a formula for this binomial sum? Are there any formulas to calculate the below sum?
$$\sum_{n=1}^{1918}n\binom{2017-n}{99}$$
Or, more generally,
$$\sum_{n=1}^{1918}n\binom{2017-n}{k}$$
| A general solution where $p = 2017$ and $k = 99$ in your case:
First rewrite the sum to look nicer:
\begin{align*}
\sum_{n=0}^{p-k}n\binom{p-n}k &= \sum_{n=k}^p(p-n)\binom{n}k = p\sum_{n=k}^p\binom{n}k - \sum_{n=k}^p n\binom{n}k \\
&= p\sum_{n=0}^{p-k}\binom{n+k}k - \sum_{n=0}^{p-k}(n+k)\binom{n+k}k \\
&= (p-k)\sum_{n=... | {
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Intersection of rectangular and polar equations Find the points of intersection for the following relations:
$x^2+y^2-20 = 0$ and $\theta = -3\pi / 4$
So the first one is obviously a circle, and the second one, converting into rectangular form, is y/x = 1, and thus y=x. Substituting y as x in the first equation, I then... | $x = r\cos\theta, \ y= r\sin\theta$.
$\theta = - 3\pi/4$.
$r = (\sqrt{20})$.
Combining:
$x = (\sqrt{20})\cos(-3\pi/4)$ ;
$y = (\sqrt{20})\sin(-3\pi/4)$ .
$\cos(-3\pi/4) = \cos(3\pi/4) =$
$- (1/2)\sqrt{2}$.
$\sin(-3\pi/4) = - \sin(3\pi/4) =$
$ -(1/2)\sqrt{2}$.
Finally:
$x = -(\sqrt{20})/(\sqrt{2}) = - (\sqrt{10})$;
... | {
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There exists rational number $a_n$, $(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$ Let $n\in \mathbb{N}$. Prove that there exists a rational number $a_n$ for which $$(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$$
My attempt :
I try $n=2$,
$(x^2+\frac{1}{2}x+1)(x^2-\frac{1}{2}x+1)=x^4+\frac{7}{4}x^2+1$
$a_2 = \frac{7}{4}$
| just an idea
The roots of the left polynom are
$$x_1=\frac{-1+i\sqrt {15} }{4}=e^{it}$$
$$x_2=\frac {-1-i\sqrt {15} }{4}=e^{-it} $$
they are also roots of the right one
$$e^{2int}+a_ne^{int}+1=0$$
$$e^{-2int}+a_ne^{-int}+1=0$$
thus
$$a_n=-\frac {\sin (2nt)}{\sin (nt)} $$
$$=-2\cos (nt) $$
For example,
$$a_2=-2\cos (2t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Convert a equation with fractions into whole numbers So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$
$$36a=a^2$$
$$36=a$$
However, I wa... | Method 1: $\frac ab x + \frac cd x = gy$
Multiple by both $b$ and $d$. Multiplying by $bd$ will always eleminate everything. so $bd\frac abx + bd \frac cd x = adx + bc x = bdgy$.
Ex: $\frac 23 a^2 - \frac 49 a^2 = 8a$
$3*9\frac 23 a^2 -3*9\frac 49 a^2 = 3*9*8a$
$18a^2 - 12a^2 = 216 a$
But notice that was over kill. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If... | We have $3$ consecutive numbers so one of these will give a factor of $3$.
If $n$ is even then $n$ will give a factor of $16$. If $n$ is odd then $(n-1)$ and $(n+1)$ will both be even and if $(n+1)^2$ only gives $2$ factors of $2$ then $(n-1)$ will be divisible by $4$, so $(n-1)(n+1)^2$ will be divisible by $16$.
So $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Definite integral for a 4 degree function
The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$
I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?
| One way to solve this problem is to do integration by parts.
$$\int{x^3 \frac{x}{(x^2+a^2)^4}dx}=-\frac{x^3}{6(x^2+a^2)^3}+\frac{1}{2}\int{x\frac{x}{(x^2+a^2)^3} dx}.$$
Continuing further
$$\int{x\frac{x}{(x^2+a^2)^3} dx} = -\frac{1}{4} \frac{x}{(x^2+a^2)^2}+\frac{1}{4}\int{\frac{1}{(x^2+a^2)^2}dx}.$$
Now you can subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Inequality $\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$
Let $a,b\in \mathbb{R+}$ and $n\in \mathbb{N}$. Prove that:
$$\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$$
I have a solution using induction, but without induction I can not ... | For positives $a$ and $b$ By C-S we obtain:
$$\left(\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb}\right)^2$$
$$\leq n\left(\frac{1}{(a+b)^2}+\frac{1}{(a+2b)^2}+\cdots+\frac{1}{(a+nb)^2}\right)$$
$$<n\left(\frac{1}{(a+\frac{b}{2})(a+\frac{3b}{2})}+\frac{1}{(a+\frac{3b}{2})(a+\frac{5b}{2})}+\cdots+\frac{1}{(a+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\s... | The function is increasing in $\left(0, \dfrac{\pi}{2}\right)$ which can be seen easily by differentiating and seeing that the derivative is positive (in the given interval).
It is easy to see that $x = \dfrac{\pi}{4}$ is a root and so is the only root in $\left(0, \dfrac{\pi}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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If $\frac{\sin x}{\sin y}= {3}$ and $\frac{\cos x}{\cos y}= \frac{1}{2}$, then find $\frac{\sin2x}{\sin2y}+\frac{\cos2x}{\cos2y}$
Let $x$ and $y$ be real numbers such that $\dfrac{\sin x}{\sin y}= {3}$ and $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$. Find a value of $$\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}.$$
My a... | hint
$$\sin (y)=\frac {1}{3}\sin(x) $$
$$\cos (y )=2\cos (x) $$
$$\implies \sin^2(x )+36\cos^2 (x)=9$$
$$\implies 35\cos^2 (x)=8$$
$$\implies \cos^2 (x)=\frac {8}{35} $$
the second expression is
$$\frac {16-35}{64-35}=-19/29$$
the final result is
$$3/2-19/29=49/58$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Maximize linear function over disk of radius $2$
Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.
$f(x, y)$ has no CP's so thats something gone.
I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt... |
Max $z=x+2y$ subject to $x^2+y^2\le4$.
The contour lines: $y=-\frac{1}{2}x+\frac{1}{2}z$, the largest value of $z$ occurs when $y$ is tangent to the constraint circle $x^2+y^2=4$ or $y=\sqrt{4-x^2}$.
Thus the slope of tangent line must be equal to the slope of $y=-\frac{1}{2}x+\frac12z:$
$$y'=\frac{-2x}{2\sqrt{4-x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rationalize $\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$ I am having trouble rationalizing the denominator of $$\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$$
I tried grouping the denominator as $(2 + 2\sqrt[6]{2}) + 2\sqrt[3]{2}$ and multiplying top and bottom by $(2 + 2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\s... | Let $a=\sqrt[6]2$. Then your expression is$$\frac1{2+2a+2a^2}=\frac{1-a}{2(1+a+a^2)(1-a)}=\frac{1-a}{2(1-a^3)}=\frac{1-\sqrt[6]2}{2\bigl(1-\sqrt2\bigr)}=\frac{\bigl(1-\sqrt[6]2\bigr)\bigl(1+\sqrt2\bigr)}{-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\be... | $$
\begin{aligned}
& \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\\=& \frac{(1-\cos \theta)+\sin \theta}{(1+\cos \theta)+\sin \theta}\\
=& \frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)+2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)+2 \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Calculation of a partial derivative I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.
1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\f... | Note:
$$\frac{\partial}{\partial x}\left( \left(\sqrt{x^2+y^2}\right)^{-1}\right)=(-1)\left(\sqrt{x^2+y^2}\right)^{-2}\frac{\partial}{\partial x}\sqrt{x^2+y^2}$$
$$\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)=\frac{0-1\cdot \frac{\partial}{\partial x}\sqrt{x^2+y^2}}{\left(\sqrt{x^2+y^2}\right)^2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question related to the existence of a prime Let $I_m=[m!+2,m!+m]\cap\mathbb N$ an "interval" of $\mathbb N$; it can obviously be as long as we want and it is easy to prove $I_m$ does not contain any prime. Prove the following:
$$\text { if }n^2+(n+1)^2\in I_m\text{ then }4n^2+1\notin I_M$$
Note that if in a large in... | Remark(I):
At first notice that:
$\dfrac{3+\sqrt{9+4}}{2} \leq \dfrac{4+4}{2}=4$ ,
so for $4 \leq n$ we have:
$$0 \leq n^2-3n+1 \ \ \Longrightarrow \ \
3n^2+3n+2 \leq 4n^2+1 \ \ \Longrightarrow \ \
\\
\dfrac{3}{2}\Big(2n^2+2n+1 \Big) < 4n^2+1 \ \ \Longrightarrow \ \
\dfrac{3}{2}\Big(n^2+(n+1)^2 \Big) < 4n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the determinant of the following $4 \times 4$ matrix
Use a cofactor expansion across a row or column to find the determinant of the following matrix
$$B=\begin{pmatrix}1 &c&0&0\\0&1&c&0\\0&0&1&c\\c&0&0&1\end{pmatrix}$$
Clearly indicate the steps you take.
I have tried
$$\det B = 1 \det \begin{pmatrix}1 &c&0\\0&1... | Expanding $\det B$ on first column we have that
$$\det B=1\det\begin{pmatrix}1 &c&0\\0&1&c\\0&0&1\end{pmatrix}-c\begin{pmatrix}c &0&0\\1&c&0\\0&1&c\end{pmatrix}=1-c^4$$
The determinant of triangular matrix is the product of the diagonal entries
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$
Let $a$, $b$ and $c$ be positive real numbers. Prove that:
$$\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}+\frac{\sqrt{b^3a}}{2\sqrt{c^3b}+3ca}+\frac{\sqrt{c^3b}}{2\sqrt{a^3c}+3ab}\geq \frac{3}{5}$$
My attempt :
Substitute $a=x^2, b=y^2, c=z^2$, ... | I think your start is good.
Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Hence, by C-S
$$\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}=\sum_{cyc}\frac{x^3z}{y^2(3z^2+2xy)}=\frac{1}{x^2y^2z^2}\sum_{cyc}\frac{x^5z^3}{3z^2+2xy}=$$
$$=\frac{1}{x^2y^2z^2}\sum_{cyc}\frac{x^6z^4}{xz(3z^2+2xy)}\geq\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove the the general equation of the family of circles through the intersection points of two circles? Let $ \mathscr{C_{1}} $ and $ \mathscr C_{2} $ be two intersecting circles determined by the equations $$ x^2 + y^2 + A_{1}x + B_{1}y + C_{1} = 0 $$ and $$ x^2 + y^2 + A_{2}x + B_{2}y + C_{2} = 0. $$ For any n... | Let the two points of the intersection be A and B and consider the straight line AB then, since the equation $x^2 + y^2 + \frac{A_{1}+kA_{2}}{1+k}x + \frac{B_{1}+kB_{2}}{1+k}y + \frac{C_{1}+kC_{2}}{1+k} = 0$ is the equation of a circle of centre $(\frac{A_{1}+kA_{2}}{2(1+k)}, \frac{B_{1}+kB_{2}}{2(1+k)})$ pasing throu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right)<1 $ $$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right) $$
Which is the solution of$\ f(x)<1 $ ? I did:
$$\ \frac{2x+3}{7-x}>0 $$
$\ x<7 $ results in $\ x> \frac{-3}{2} $ whereas $\ x>7 $ results in $\ x< \frac{-3}{2} $ so x=($\ \frac{-3}{2} $,7) and also
$\ x-2>0 $ so $\ ... | $$\log _{(x-2)}\left(\frac{2x+3}{7-x}\right)<1$$
Hint
$1)$ $x-2>0$ and $x-2\ne 1$
$2)$ $\frac{2x+3}{7-x}>0$
$2)$ If $x-2>1$ then
$$\frac{2x+3}{7-x}<x-2$$
$3)$ If $0<x-2<1$ then
$$\frac{2x+3}{7-x}>x-2$$
Now you can solve it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $1^n+2^n+ \ldots +n^n=k!$ over positive integers If $k,n \in \mathbb{N}^*$, solve the following equation
$1^n+2^n+ \ldots +n^n=k!$, where $k! $ denotes $1 \cdot 2 \cdot 3 \cdots k$.
| I will answer for $n$ odd, because barto above answered the question for $n$ even.
We have $1^n+2^n+ \ldots+n^n=1+(2^n+n^n)+(3^n+(n-1)^n) + \ldots \equiv 1 \pmod {n+2}$.
So, $k! \equiv 1 \pmod {n+2}$. If $k \geqslant n+2$, we have $k! \equiv 0 \pmod {n+2}$.
So, $n<k<n+2$, which yields $k=n+1$.
We have now $(n+1)! \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding the inverse of the function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$ Let $f$ be the function defined by $$f(x)=x^2-4x$$ for $x\ge 2$. Find the inverse function indicating its domain and the range.
When I try to make $x$ the subject I get $+$and $-$ two answers for the inverse function. Is it the proper ... | The function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$ has domain $\text{Dom}_f = [2, \infty)$. We can find its range by completing the square.
\begin{align*}
f(x) & = x^2 - 4x\\
& = (x^2 - 4x + 4) - 4\\
& = (x - 2)^2 - 4
\end{align*}
The graph of $y = (x - 2)^2 - 4$ is a parabola with vertex $(2, -4)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I sti... | Let $\sqrt{6 - \sqrt{20}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers. Squaring both sides of the equation
$$\sqrt{a} - \sqrt{b} = \sqrt{6 - \sqrt{20}}$$
yields
\begin{align*}
a - 2\sqrt{ab} + b & = 6 - \sqrt{20}\\
a - 2\sqrt{ab} + b & = 6 - 2\sqrt{5}
\end{align*}
Matching rational and irrational p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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A quicker approach to the integral $\int{dx\over{(x^3+1)^3}}$ Source: A question bank on challenging integral problems for high school students.
Problem:
Evaluate the indefinite integral
$$\int{dx\over{(x^3+1)^3}}$$
Seems pretty compact but upon closer look, no suitable substitution comes to mind. I can use partial fra... | In order to show the reduction formula, we use differentiation.
$$
\begin{aligned}
\frac{d}{d x}\left(\frac{x}{3 n\left(x^{3}+1\right)^{n}}\right) &=\frac{\left(x^{3}+1\right)^{n}-x n\left(x^{3}+1\right)^{n-1}\left(3 x^{2}\right)}{3 n\left(x^{3}+1\right)^{2 n}} \\
&=\frac{\left(x^{3}+1\right)^{n}-3 nx^{3}\left(x^{3}+1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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The quadratic equation with two unknowns: $ x^2 - 2axy + by^2 = 0 $ Please, could you help me with the next problem:
The problem:
Determine real parameters $a, b \in \Bbb{R}$, such that with:
$$ \langle x, y \rangle = x_1y_1 - ax_2y_1 - ax_1y_2 + bx_2y_2 $$
the inner product is defined in (or is it 'on') the vector s... | Here's an answer building on the comments.
\begin{align}
\langle x,x \rangle &= x_1^2 - 2ax_1x_2 + bx_2^2 \\
&= x_1^2 - 2ax_1x_2 + a^2x_2^2 + (b-a^2)x_2^2 \\
&= \underbrace{(x_1 - ax_2)^2}_{\geq 0} + \underbrace{(b-a^2)x_2^2}_{\geq 0 \text{ if }a^2\leq b}
\end{align}
So we are guaranteed that $\langle x,x \rangle\geq 0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three positive integers $a$, $b$, and $c$ satisfy $abc=8!$ where $aThree positive integers $a$, $b$, and $c$ satisfy $abc=8!$ where $a<b<c$. What is the smallest possible value of $c-a$?
I know that $40,320=8!=8*7*6*5*4*3*2*1=7*5*3^2*2^7*1$.
I'm trying to see how I can choose $a$, $b$, and $c$ such that $abc=8!$ and $a... | One has $8!=2^7\cdot 3^2\cdot 5\cdot 7$ and ${\root3\of 8!}\approx34.3$. We now should factor $8!$ into three factors as equal as possible, which means: as near to $34.3$ as possible.. It seems that $a=2^5=32$, $b=5\cdot 7=35$, $c=2^2\cdot3^2=36$ is the best we can do. Note that neither $33$ nor $34$ can be attained w... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positive integers $n > 1$ such that the polynomial $P(x)$ belongs to the ideal generated by the polynomial $x^2 +x +1$ in $\Bbb Z_n[x]$
Find all positive integers $n > 1$ such that the polynomial $x^4 + 3x^3 + x^2 + 6x + 10$ belong to the ideal generated by the polynomial $x^2 + x + 1$ in $\Bbb Z_n[x]$.
My a... | Hint $\ 6x\!+\!12 = 0\,$ in $\,\Bbb Z_n[x]\iff 6=0=12\in \Bbb Z_n\!\iff n\mid 6,12\iff n\mid (6,12) = 6.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rudin Chapter 3 exercise 14 For exercise 14c of Baby Rudin:
If $\{s_n\}$ is a complex sentence, define its arithmetic means $\sigma_n$ by
$$\sigma_n = \frac{s_0 + s_1 + \cdots + s_n}{n + 1},$$
where $n = 0, 1, 2, ...$
14c: Can it happen that $s_n>0$ for all $n$ and that $\limsup s_n = \infty$, although $\lim \sigma_n ... | Your example does not work since
$$\frac{1}{n+1} \sum_{k=1}^n \sqrt{k} \geqslant \frac{1}{n+1}\int_0^n \sqrt{x} \, dx = \frac{2n^{3/2}}{3(n+1)}= \frac{2 \sqrt{n}}{3 + 3/n} \to \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| As I explained in 2011, one easy way is to write the number in Horner form in the original base, then do the base conversion from the inside-out, e.g. below where $\rm\color{#c00}{red}$ means radix $9$ notation
$$\begin{align} 397
\,&=\, (3\cdot 10\, +\, 9)\,10 +7\\
&=\, (\color{#c00}{3\cdot 11+10})10+7\\
&=\qquad\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
How can I prove the following sequence converges Given the following n-th term sequence:
$$a_{n} = \sqrt[n]{1^2+2^2+...+n^2}$$
You're asked to evaluate the limit of the given sequence, justifying your operations.
What strategy should I take on this? I have considered taking some inequality in order to, eventually, be ... | Hint
Show by induction that
$$\sum_{k=1}^n k^2=\dfrac{n^3}{3}+\dfrac{n^2}{2}+\dfrac n6. $$ Thus
$$\sqrt[n]{1^2+2^2+...+n^2}\le \sqrt[n]{\dfrac{5n^3}6 }.$$
Can you conclude from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find $\iiint (x^2+y^2+z^2)~dV$, above the cone $z=\sqrt{3(x^2+y^2)}$, inside the sphere $x^2+y^2+z^2\leq a^2$.
Find
$$\iiint (x^2+y^2+z^2)~dV$$
above the cone $z=\sqrt{3(x^2+y^2)}$, and inside the sphere $x^2+y^2+z^2\leq a^2$.
I think that the solution for this problem is following:
Use cylindrical coordinates:
$... | $\mathbf{HINT}$: First we have $\sqrt{3(x^2+y^2)}\leq z \leq \sqrt{a^2-x^2-y^2} $.
Second solve system: $z=\sqrt{3(x^2+y^2)}$, $x^2 + y^2 + z^2 = a^2 $. Solution: $D:x^2+y^2 = \frac{a^2}{4}$.
So, $$\iiint (x^2+y^2+z^2) dV = \iint_{D} \int_{\sqrt{3(x^2+y^2)}}^{\sqrt{a^2-x^2-y^2}} (x^2+y^2+z^2) dz dxdy,$$
then we use cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the closed form of $\sum_{k=2}^{\infty}{\lambda(k)-1\over k}?$ Given that:
$$\sum_{k=2}^{\infty}{\lambda(k)-1\over k}\tag1$$
Where $\lambda(k)$ is Dirichlet Lambda Function
We are seeking to determine the closed form $(1)$ and came close to estimates it to $1-{\frac12}\left(\gamma+\ln{\pi}\right)$.
where $\... | You have found it. Interchange the order of summation, and a few rewrites plus Wallis's product yield the result:
\begin{align}
\sum_{k = 2}^{\infty} \frac{1}{k}\bigl(\lambda(k) - 1\bigr)
&= \sum_{k = 2}^{\infty} \frac{1}{k} \sum_{n = 1}^{\infty} \frac{1}{(2n+1)^k} \\
&= \sum_{n = 1}^{\infty} \sum_{k = 2}^{\infty} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove identity holds Just wondering if there is any other way to show that for each positive integer $n$ holds
$$2\left(\sqrt{n} - 1\right) < 1 + \frac{1}{\sqrt{2}} +\cdots+\frac{1}{\sqrt{n}} < 2\sqrt{n}$$
other than by mathematical induction~
| For the right inequality:
$$\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\right) = \frac{1}{n}\left(\sqrt{n}+\sqrt{\frac{n}{2}}+\cdots+\sqrt{\frac{n-1}{n}} + \sqrt{\frac{n}{n}}\right)$$
This is a Riemann sum of a function $f(x) = \frac{1}{\sqrt{x}}$ on the segment $[0,1]$ with the equidistant s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
How to prove $a+b\sqrt{2}=0$ iff $a=b=0$ I want to prove for $a,b \in \mathbb{Q}$
\begin{align}
a+b\sqrt{2}=0 \quad \Leftrightarrow \quad a=b=0
\end{align}
For $\Leftarrow$ is trivial, now I am in trouble of showing $\Rightarrow$
Simply I can say $\sqrt{2}$ is irrational and go proceed, but I want some formal mathemat... | Assume:($a\neq b\neq 0$)
$$a+b\sqrt{2}=0$$
$$(a+b\sqrt{2})(\frac{a-b\sqrt{2}}{a-b\sqrt{2}})=0$$
$$\frac{a^2-2b^2}{a-b\sqrt{2}}=0$$
$$\left\{\begin{matrix}
a^2-2b^2=0 & & \\
a-b\sqrt{2}\neq 0& &
\end{matrix}\right.$$
$$\left\{\begin{matrix}
\frac{a}{b}=\sqrt{2} & & \\
\frac{a}{b}\neq \sqrt{2}& &
\end{matrix}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ I was evaluating
$$\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$$
My work:
I see that in the denominator, there is the radical $\sqrt{x^2 + 3}$. This reminds me of the trigonometric substitution $\sqrt{u^2 + a^2}$ and letting
$ u = a \tan \theta$. With ... | By letting $x=1/t$ and by integration by parts, we get
\begin{align*}
\int_1^{\infty} \frac{1}{x^4 \sqrt{x^2 + 3} } dx
&=
\int_0^1\frac{t^3}{\sqrt{3t^2 + 1}}\,dt=
\frac{1}{3}\int_0^1t^2\frac{d(\sqrt{3t^2 + 1})}{dt}\,dt\\
&=\frac{1}{3}\left[t^2\sqrt{3t^2 + 1}\right]_0^1-
\frac{2}{3}\int_0^1 t\sqrt{3t^2 + 1} dt\\
&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that for any integer $n, n^2+4$ is not divisible by $7$. The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n... | The question boils down to 3 being a quadratic residue modulo 7 or non-residue
$$n^2+4 \equiv 0 \; (mod\;7)$$
$$n^2\equiv 3 \;(mod\;7)$$
For this we have Legendre symbol
$$\left ( \frac{a}{p} \right )=\begin{cases}
& 1\text{ if }a\text{ is a quadratic residue modulo }p\text{ and }p\text{ does not divide }a \\
& -1\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Distance between two lines in parametric form
a) Parametrize the line $L$ through $P = (2, 1, 2)$ that intersects the
line
$x = 1 + t$, $y = 1 − t$, $z = 2t$
perpendicularly.
b) Parametrize the $z$ axis.
c) What is the distance from this line $L$ to the $z$-axis?
My work For the part a), I got the equation was $x... | Let $B(1+t,1-t,2t)$ and $A(2,1,2)$.
Thus, $$\vec{AB}\perp\vec{(1,-1,2)}$$ or
$$\vec{(t-1,-t,2t-2)}\vec{(1,-1,2)}=0$$ or
$$t-1+t+4t-4=0$$ or
$$t=\frac{5}{6}$$ and since
$$\vec{\left(\frac{5}{6}-1,-\frac{5}{6},\frac{5}{3}-2\right)}=\vec{\left(-\frac{1}{6},-\frac{5}{6},-\frac{1}{3}\right)},$$
we obtain:
$$L:(2,1,2)+s(1,5,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding generating function for a given sequence The task is to find the generating function for the sequence $$d_n = (\sum_{k=0}^n \binom{n}{k})*(\sum_{k=0}^n \frac{(k-1)^2}{2^k})$$
let's call this function $D(x)$.
I marked $a_n = \frac{1}{2^n}$; $b_n = \frac{-2n}{2^n}$; $c_n = \frac{n^2}{2^n}$ with respective generat... | Hint: The situation is somewhat different.
Write
\begin{align*}
d_n &= \sum_{k=0}^n \binom{n}{k}\sum_{k=0}^n \frac{(k-1)^2}{2^k}\\
&=\sum_{k=0}^n(k-1)^22^{n-k}
\end{align*}
and consider the Cauchy-product $\sum_{k=0}^na_kb_{n-k}$ when multiplying generating functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1 $ is divisible by $ \ \large 2^{n+2} $ .
Answer:
For $ n=1 \ $ we have
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1.
For $ n=2 $ we have
$ \large 3^{2^2}-1=81-1=80 \ \... | We can write,
$$(3)^{2^n}-1$$
$$(4-1)^{2^n}-1$$
Using binomial,
$$=\color{red}{\binom{2^n}{0}}-\color{blue}{\binom{2^n}{1}4}+\color{green}{\binom{2^n}{2}4^2}-\cdots-\color{orange}{\binom{2^n}{2^n-1} 4^{2^n-1}}+\color{purple}{\binom{2^n}{2^n}4^{2^n}}-\color{red}{1}$$
$$$$
$$\binom{n}{r}=n (r+1) \binom{n-1}{r+1}$$
$$-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$
Solve the following inequality. $$\sqrt{5-2\sin x}\geq 6\sin x-1.$$
My tries:
As $5-2\sin x>0$ hence we do not need to worry about the domain.
Case-1: $6\sin x-1\leq0\implies \sin x\leq\dfrac{1}{6}\implies -1\leq\sin x\leq\dfrac{1}{6}\tag*{}$
Case-2:$6\sin x-1>0\implies \d... | solving the inequality $$18\sin(x)^2-5\sin(x)-2\le 0$$ we have
$$2 \pi c_1-\sin ^{-1}\left(\frac{2}{9}\right)\leq x\leq \frac{1}{6} \left(12 \pi c_1+\pi
\right)\lor \frac{1}{6} \left(12 \pi c_1+5 \pi \right)\leq x\leq 2 \pi c_1+\pi +\sin
^{-1}\left(\frac{2}{9}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\frac{3}{abc} \ge a + b + c$, prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c$ Let $a$, $b$ and $c$ be positive numbers such that $\frac{3}{abc} \ge a + b + c$. Prove that: $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c.$$
Any way I use - I get stuck after 2 or 3 steps...
| $$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\ge 3\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge abc(a+b+c)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=(a+b+c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the coordinates of the points on $y=-(x+1)^2+4$ that have a distance of $\sqrt {14}$ to $(-1,2)$ Create a function that gives the distance between the point $(-1,2)$ the graph of $$y=-(x+1)^2+4.$$ Find the coordinates of the points on the curve that have a distance of $\sqrt {14}$ units from the point $(-1,2)$. ... | You need to solve $$\sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}=\sqrt{14}.$$ Squaring both parts and substituting $u=(x+1)^2$ we get a quadratic equation $u^2-3u-10=0$ which has two roots, $u=5$ and $u=-2$. Only $(x+1)^2=5$ is possible for real numbers so we get $x=\sqrt5-1$ and $x=-\sqrt5-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
I want to compute this sum by computing one term at a time. It's clear that
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right tr... | I think of these sorts of things in terms of complex numbers. Let $\theta = \arctan 1/7$. Then $e^{i\theta} = \cos \theta + i \sin \theta$. We need $\sin \theta / \cos \theta = 1/7$, so this will be a multiple of $7 + i$, but it must have length 1, so $e^{i \theta} = (7+i)/\sqrt{50}$. Similarly let $\phi = \arctan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
distance between 2 lines in 3d
Calculate the distance between the lines
$$L_1:x=4+2t,y=3+2t,z=3+6t$$
$$L_2: x=-2+3s ,y=3+4s ,z=5+9s$$
I tried subtraction $L_1$ from $L_2$ then multiplying the resting vector by the $t$'s and $s$'s original values and trying to find value for $t$ to or $s,$ but I found $t=\frac{19}... | The distance between the lines
$L_1:x=4+2t,y=3+2t,z=3+6t$ and $L_2: x=-2+3s ,y=3+4s ,z=5+9s$ is
$d=2 \sqrt{10}$
Indeed consider the function which gives the distance between a generic point of the first line and a generic point of the second line
$f(t,s)=\sqrt{(2 t-4 s)^2+(-3 s+2 t+6)^2+(-9 s+6 t-2)^2}$
and set to zero... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Is sum of digits of $3^{1000}$ divisible by $7$? I'm working on a little exercise I found in my high school book (printed in 2007) which is pretty complicated.
Is the sum of digits of $3^{1000}$ a multiple of $7$?
Do you have any advice to solve this type of problem (without programming of course!)
PS :
We are a group ... | I am not answering the question but the post asks for clues so here it is a couple of ideas.
If $a_0,a_1,a_2, \cdots , a_{477} $ are the decimal digits of $3^{1000}$ then the numbers
$$b_i=a_{6i}+a_{6i+1}\cdot 10+a_{6i+2}\cdot 10^2+a_{6i+3}\cdot 10^3+a_{6i+4}\cdot 10^4+a_{6i+5}\cdot 10^5$$
for $i=0,1,2,\cdots, 79$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 1,
"answer_id": 0
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Given $a^3+b^3+c^3=(a+b+c)^3$, show that $a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}$
Assume that $\{a,b,c\} \subset \Bbb R$, $n\in\mathbb N$ and $a^3+b^3+c^3=(a+b+c)^3$.
Show that $a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}.$
This is from a list of problems used for training a team for a math olympics. I tried to use... | $$(a+b+c)^3-a^3-b^3-c^3=\sum_{cyc}(3a^2b+3a^2c+2abc)=3(a+b)(a+c)(b+c)$$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Is there any sequence $\{a_n:n\in \mathbb N\}$ with each non-zero terms s.t. $\displaystyle \sum_{n=1}^{\infty}a_n=0$
Is there any sequence $\{a_n:n\in \mathbb N\}$ with each non-zero terms s.t. $\displaystyle \sum_{n=1}^{\infty}a_n=0$?
I'm finding an infinite linearly dependent set which has no finite linearly depen... | Let convergence be understood in the $\|\cdot\|_\infty$ norm.
Consider the familiy of sequences $$\mathcal{F} = \left\{-\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right) : n \in \mathbb{N}\right\} \cup \left\{\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If four dice are thrown, what is the probability that the sum of numbers thrown up will be 15? How about 16?
If four dice are thrown, what is the probability that the sum of numbers thrown up will be 15? How about 16?
What's a good method of finding the answer? I know that overall there are 1296 possibilities.
| Quoting from this answer:
It is the coefficient of $x^{15}$ in the expansion of the generating function $$(x^6+x^5+x^4+x^3+x^2+x^1)^4=\left(\frac{x(1-x^6)}{1-x}\right)^4$$ which is $$1\cdot{{x}^{24}}+4\cdot {{x}^{23}}+10\cdot {{x}^{22}}+20\cdot {{x}^{21}}+35\cdot {{x}^{20}}+56\cdot {{x}^{19}}+80\cdot {{x}^{18}}\\+104\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A formula for $1^4+2^4+...+n^4$ I know that
$$\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)$$
and
$$\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.$$
Here is the question: is there a formula for
$$\sum^n_{i=1}i^4.$$
| We can get the formula by the following way.
$$(n+1)^5-1=\sum_{k=1}^n((k+1)^5-k^5)=\sum_{k=1}^n(5k^4+10k^3+10k^2+5k+1).$$
Thus,
$$\sum_{k=1}^nk^4=$$
$$=\frac{1}{5}\left((n+1)^5-1-10\cdot\frac{n^2(n+1)^2}{4}-10\cdot\frac{n(n+1)(2n+1)}{6}-5\cdot\frac{n(n+1)}{2}-n\right)=$$
$$=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the value of $P(X < KY)$ The random variables X and Y are independent and have the pdf's as follows:
$f_X(x) = \begin{cases}
xe^{-\frac{1}{2}x^2}& \text{if $x\geq0$} \\
0 & \text{otherwise}
\end{cases}$
and
$f_Y(y) = \begin{cases}
ye^{-\frac{1}{2}y^2}& \text{if $y\geq0$} \\
0 & \text{otherwise}
\end{cases}... | if $K \leq 0$, $P(X \leq KY ) = 0$.
If $K>0$, the integral should be
\begin{align*}
P(X \leq KY) &= \color{blue}{\int_0^\infty}\color{red}{\int_0^{Ky}}xe^{-\frac{1}{2}x^2}ye^{-\frac{1}{2}y^2}\, \color{red}{dx}\,\color{blue}{dy}
\end{align*}
Your final solution should be an expression that is depending only on $K$, it s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating limits without L'Hopital's I need to evaluate without using L'Hopital's Rule. They say you don't appreciate something until you lose it. Turns out to be true.
*
*$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}$
*$\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$
I have tried to "rationalise" Q1 by using the identi... | As regards 2), note that
$$\frac{\cos(3x) - \cos(x)}{x^2}=\frac{1 - \cos x}{x^2}-9\cdot\frac{1 - \cos (3x)}{(3x)^2}.$$
Hence
$$\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}=
\lim_{x \to 0}\frac{1 - \cos x}{x^2}-9\cdot\lim_{x \to 0}\frac{1 - \cos (3x)}{(3x)^2}\\=(1-9)\lim_{x \to 0}\frac{1 - \cos x}{x^2}
=-8\lim_{x \to 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Multivariable $\epsilon-\delta$ proof verification Prove that $\lim\limits_{(x,y) \to (1,1)} xy=1$
Of course, I am aware that this is "obvious", but I want to add some rigor to it. When I searched around for multivariable limits using $\epsilon-\delta$, most of the examples had $(x,y) \rightarrow (0,0)$, but in this ca... | Your proof is correct and nice.
Take $\delta<1$
By doing this a priori you will find the delta you need more easily and save some time from the calculations
You proved that $$|xy-1| \leq |x-1||y-1|+|x-1|+|y-1|$$
Thus $$|x-1|=\sqrt{(x-1)^2} \leq \sqrt{(x-1)^2+(y-1)^2} < \delta$$
$$|y-1|=\sqrt{(y-1)^2} \leq \sqrt{(x-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Using the Chinese Remainder theorem to solve the systems of congruence's. We got introduced to the Chinese Reminder Theorem, but I still haven't quite grasped it and I have a problem that asks:
Find $x \bmod 77$ if $x \equiv 2 \pmod 7$ and $x \equiv 4 \pmod {11}$.
My attempt was like so:
$$B = 7 * 77 * 11 = 5929$$
... | Let us use the Euclidean algorithm to express $\gcd(7,11)$ as a linear combination:
$$4 = 11 - 7\\
3 = 7 - 4 = 2 \cdot 7 - 11 \\
1 = 4 - 3 = 2 \cdot 11 - 3 \cdot 7.$$
Now, $x = (2 \cdot 11 - 3 \cdot 7) x = 2 (11x) - 3 (7x)$. By the given congruences, $11x \equiv 11 \cdot 2 = 22 \pmod{77}$ and $7x \equiv 7 \cdot 4 = 28... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solving for equation of a circle when tangents to it are given. Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.
To find: All the circles that are tangent to these three lines.
Outline of the method :
If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(... | Consider $\triangle ABC$,
$A=(2,5),B=(2,-1),C=(10,5)$
and its incircle and three excircles,
\begin{align}
a&=10,\quad
b=8,\quad
c=6,\quad
\\
r&=2,\quad
r_a=12,\quad
r_b=6,\quad
r_c=4
,\\
O_i&=(4,3),\quad
O_a=(14,-7),\quad
O_b=(8,11),\quad
O_c=(-2,1)
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why does the augmented matrix method for finding an inverse give different results for different orders of elementary row operations? Why does the augmented matrix method for finding an inverse give different results for different orders of elementary row operations?
Consider the example of elements from the Heisenberg... | $$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & c & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
I believe there is a typo in your working, I believe you mean subtract $c$ times the third row and add it to the second row
$$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & 0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\arcsin{(2x^2-1)}+2\arcsin{x} = -\frac{\pi}{2}$ First of, we have to restrict the domain of the equation by looking at the argument of the first term. The domain for $\arcsin$ is $[-1,1]$, so the inequality $ -1\leq 2x^2-1 \leq1$ has to hold. So $x\geq0$ and $-1\leq x\leq 1$. It follow that the domain is $[-1,1]... | Let $f(x)=\arcsin{(2x^2-1)}+2\arcsin{x}+\frac{\pi}{2}$.
Thus, $$f'(x)=\frac{2x}{|x|\sqrt{1-x^2}}+\frac{2}{\sqrt{1-x^2}}=0$$
for all $-1<x<0$.
Thus, $f(x)=conctant$ for $-1\leq x\leq0$ and since $f(0)=0$ we obtain $[-1,0]$.
In another hand $f'(x)>0$ for all $0<x<1$.
Thus, $f(x)>f(0)=0$ and on $[-1,0]$ we got all solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive numbers $a,b,c$ that $a+b+c=3$ prove that: For positive numbers $a,b,c$ that $a+b+c=3$ prove that:
$\frac{a(a+b-2c)}{ab+1}+\frac{b(b+c-2a)}{bc+1}+\frac{c(c+a-2b)}{ac+1} \ge 0$
This problem wad in the first part of the inequality book(AM-GM inequality)But it doesn't seem to have a solution by AM_GM inequali... | C0nsider $(1.2,1.2,0.6)$ one has $LHS = \frac{1.2^2}{1.2^2 + 1} + \frac{-0.6\times 1.2}{1.2\times 0.6 + 1} + \frac{- 0.6\times 0.6}{1.2\times 0.6 + 1} < 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About the property of a sequence of polynomials
There is a sequence of polynomials $P_n(x)$ such that $P_0(x) = x^3 - 4x$ and, for $n>0$,
$$P_n(x) = P_{n-1}(1+x)P_{n-1}(1-x) - 1.$$
Prove that $x^{2016}$ divides $P_{2016}(x)$.
I tried induction, but then I had to prove that $x^n \vert P_n(x)P_n(2 - x) + P_n(-x)P... |
We show by induction that $x^n(x-2)(x+2)$ divides $P_n(x)$ for any even integer $n$.
For $n=0$, $P_0=x(x+2)(x-2)$ and the property holds. Let $n>0$. Note that
$$P_{n+2}(x)=(P_{n}(2+x)P_{n}(-x)-1)(P_{n}(2-x)P_{n}(x)-1)-1.$$
If $x^n(x-2)(x+2)$ divides $P_n(x)$ then, for some polynomial $Q_n(x)$,
$$P_{n}(2+x)P_{n}(-x)=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find value of $\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$ Find value of
$$S=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$$
I started with $$S+\frac{1}{S}=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}+\frac{\sum... | Amazingly, it appears that
$$ \dfrac{\sum_{k=1}^{m^2-1} \sqrt{m + \sqrt{k}}}{\sum_{k=1}^{m^2-1} \sqrt{m - \sqrt{k}}} = 1 + \sqrt{2} $$
for all integers $m \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all points on which a function is discontinuous. $ f(x,y) = \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} &\quad\text{if} [x,y] \neq [0,0]\\[2ex] 0 &\quad\text{if}[x,y] = [0,0]\\ \end{cases} $
The only point it could be discontinuous in is [0,0]. How do I find the limit of the function for $(x,y) \rightarrow (0,0)$? $ \... | For $x,y \in \mathbb R \setminus \{0\}$ we have
$$\vert f(x,y)\vert = \left\vert\frac{x^3 + y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2+ y^2} \right\vert + \left\vert \frac{y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2} \right\vert + \left\vert \frac{y^3}{y^2}\right\vert = \vert x \vert + \vert ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
SAT II Geometry Find the missing side length
I'm thinking the answer is choice A but I want someone to back up my reasoning/check. So since DE and DF are perpendicular to sides AB and AC respectively that must make EDFA a rectangle. Therefore AF must be 4.5 and AE must be 7.5. Since they state AB = AC that must mean E... | Since $\triangle ABC$ is isosceles, $\angle ABC \cong \angle ACB$. Since $\angle BED$ and $\angle CFD$ are right angles, they are congruent. Hence, $\triangle BED \sim \triangle CFD$ by the Angle-Angle Similarity Theorem. Hence,
$$\frac{|CD|}{|BD|} = \frac{|CF|}{|BE|} = \frac{|FD|}{|ED|} = \frac{7.5}{4.5} = \frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x... | You want to prove that for every $\epsilon>0$ there exists $\delta>0$ such that
$$0<\sqrt{x^2+y^2}<\delta\implies\left|\frac{x^3+y^3}{x^2+y^2}\right|<\epsilon.$$
Let $\delta=\frac{\epsilon}{2}$. Then
$$\left|\frac{x^3+y^3}{x^2+y^2}\right|\leq\left|\frac{x^3}{x^2+y^2}\right|+\left|\frac{y^3}{x^2+y^2}\right|$$
$$=|x|\und... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$ How to show that
$${2\over (2-1)(2^2-1)(2^3-1)}+{2^2\over (2^2-1)(2^3-1)(2^4-1)}+{2^3\over (2^3-1)(2^4-1)(2^5-1)}+\cdots={1\over 9}?\tag1$$
We may rewrite $(1)$ as
$$\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\... | Well done so far. Now shift the indices on the last two sums:
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}=\\
{1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Prove $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$ Let $x,y,z$ be real numbers all greater than $1$, then prove that
$$\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$$
My Attempt:
I am trying to use $A.M>GM$ but... | To prove it, let's subtract LHS from RHS and call it $\Delta$:
\begin{align}
\Delta&=\frac{x-1}{y-1}-\frac{x+1}{y+1}+\frac{y-1}{z-1}-\frac{y+1}{z+1}+\frac{z-1}{x-1}-\frac{z+1}{x+1}\\
&= 2\left(\frac{x-y}{y^2-1}+\frac{y-z}{z^2-1}+\frac{z-x}{x^2-1}\right)\\
\end{align}
Assuming $1<x\leq y\leq z$, we can see:
$$\frac{z-x}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Determine $\textrm{sign}[x + \ln y]$ I need to determine the sign of
\begin{align}
f(a,b) = \textrm{sign}\left[\frac{(a-1)(b-1)^2}{ba^2(b(a-1)+1)} + \ln\left(\frac{b(a-1)+1}{ab}\right)\right] = ?
\end{align}
with $a \geq 1$, and $b > 0$.
We have $f(a,b=1) = 0$. I conjecture also that $f(a,b<1) > 0$ and $f(a,b>1) < 0$... | for $0<b<1$:
$$
[\ldots]=\left[\underbrace{\ldots}_{> 0}+\ln\Big(\underbrace{1+\frac{1-b}{ab}}_{> 1}\Big)\right]>0.
$$
for $b>1$: use the fact that $\ln(1+x)\le x$, for $x>-1$, i.e.
$$
[\ldots]\le\frac{a-1}{a-1+1/b}\cdot\Big(\frac{b-1}{ab}\Big)^2
-\frac{b-1}{ab}=\epsilon\cdot q^2-q<0
$$
since $0<\epsilon<1$ and $0<q<1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$ Evaluate the sum $${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$$
I have tried comparing this to the similar problem here.
I believe ... | $$\begin{align}
S&=1-\frac 12\binom n1+\frac 13 \binom n2-\cdots+(-1)^n\frac 1{n+1}\binom nn\\
\times (n+1):\hspace{1cm}\\
(n+1)S&=(n+1)-\frac {n+1}2\binom n1+\frac {n+1}3\binom n2-\cdots+(-1)^n\frac {n+1}{n+1}\binom nn\\
&=\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}\\
&=\color{blue}{\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Calculating residue of $f(z) = \frac{(z^2 - 1)^4}{z^5}$ at $z=0$
Calculate the residue of $$f(z) = \frac{(z^2 - 1)^4}{z^5}$$ in $z = 0$
I let $g(z) = (z^2 - 1)^4.$
I'm using a theorem which states:
Suppose $g$ is holomorphic around $z = \alpha$ and that $N$ is a
positive integer, then $$RES_{z=\alpha}\frac{g(z)}{... | $$
\frac{(z^2 - 1)^4}{z^5}
= \frac{z^8 - 4 z^6 + 6 z^4 - 4 z^2 + 1}{z^5}
= \cdots + \frac{6}{z} + \cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Epsilon- Delta Proof $$
\lim_{x\to 1} \frac{1}{x^2+2} = \frac{1}{3}
$$
I'm having a problem proving this using $\epsilon-\delta$ proofs. For some reason When I solve for $\epsilon$, I get a negative number. Since this value is supposed to equal $\delta$ and $\delta$ can't be negative I'm not sure how to move forward.
... | The problem is that “solving for $\epsilon$” is too restrictive in general. Let's unpack @mfl's hint.
$$
\left| \frac{1}{x^2+2} - \frac{1}{3} \right|
= \left|\frac{1-x^2}{x^2+2}\right|
= \left|\frac{(x-1)(x+1)}{x^2+2}\right|
= |x-1|\cdot |x+1| \cdot \frac{1}{x^2+2}
$$
We want to make this product smal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Proving that: $ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$ This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$
It would be really great if somebody could come up with a solution to this problem.
| \begin{eqnarray*}
\color{blue}{a^2b^2(a-b)^2} \geq 0 \\
\color{red}{a^6} +\color{blue}{a^4b^2+a^2 b^4} +\color{red}{b^6} \geq \color{red}{a^6} +\color{blue}{2 a^3b^3} +\color{red}{b^6} \\
(a^2+b^2)(a^4+b^4) \geq (a^3+b^3)^2.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help with homework of 3rd order equation I can't solve this equation :
$$t^3-4t^2-5=0$$
I think it should be solved by cover it to $(a+b)^3$, but I can't figure it out
| Let $u=t-\frac{4}{3}$. One then has $$(u+\frac{4}{3})^3-4(u+\frac{4}{3})^2-5 = 0$$
$$u^3-\frac{16}{3}u - \frac{263}{27}=0.$$
Now, let $x+y = u$, then $x^3+y^3 = (x+y)^3-3xy(x+y) = u^3 - 3xyu$.
One can set $xy = \frac{16}{9}$, $x^3+y^3= \frac{263}{27}$. One can solve to find $x$, $y$ then $u= x+y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that $\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$
Prove that $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$$
I tried to rationalize the denominator but I always end up with a large fraction that won't cancel out. Is there something I'm m... | $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)+2i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-2i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
} |
How to differentiate $(x+1)^{3/2}$ using the limit definition of the derivative? I want to use the definition of the derivative to find $f'$ of $f(x)=(x+1)^{3/2}$.
I solved it using the chain rule. Would like to try to solve it using the definition of a derivative:
$$
\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.
$$
D... | Let $f(x)=(x+1)^{3/2}$, then $$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h} \cdot \frac {(x+1+h)^{3/2}+(x+1)^{3/2}}{(x+1+h)^{3/2}+(x+1)^{3/2}}= \lim_{h \to 0} \frac{(x+1+h)^3 -(x+1)^3}{h[(x+1+h)^{3/2}+(x+1)^{3/2}]} = \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Foot of perpendicular proof. I know there is few answer on Foot of perpendicular, but my doubt is different, so please don't mark this as the duplicate.
My book says:
Foot of the perpendicular from a point $(x_1,y_1)$ on the line $ax+by+c=0$ is $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-\dfrac{ax_1+by_1+c}{a^2+b^2}$$
My pr... | First the algebraic distance $r$ should be $\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$ depending on the position of $(x_1;y_1)$ for a given system of coordinates.
The correct formula is $$\dfrac{x-x_1}{a}=-\dfrac{ax_1+by_1+c}{a^2+b^2}.$$
To see this take the scalar product to be zero for $(x,-\dfrac{ax}{b}-\frac{c}{b})$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find a + b + c + bc. My math trainer friend ask my help to look for the shortest possible solution for this problem:
Let $a$, $b$ and $c$ be positive integers such that
$$\left \{\begin{matrix}a + b + ab = 15 \\
b + c + bc = 99 \\
c + a + ca = 399\end{matrix}\right. $$
Find $a + b + c + bc.$
I tried elimination but... | $(a+1)(b+1) = 16 $
$(b+1)(c+1) = 100$
$(c+1)(a+1) = 400$
Multiply all $3$ equations to get value of $(a+1)(b+1)(c+1) = 800$ then
$c+1 = 50,\quad a+1 = 8,\quad b+1 = 2$ , required is $a+(99) = 106$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find : $\lim_{x \to \pi/2}\frac{1-\sin x}{\sin x+\sin 3x}$ How to find :
$$\lim_{x \to \pi/2}\frac{1-\sin x}{\sin x+\sin 3x}$$
My Try :
$$x-\pi/2=t \to x=t+\frac{\pi}{2}$$
And:
$$ \sin (t+\frac{\pi}{2})=\cos t \\ \sin 3(t+\frac{\pi}{2})=-\cos 3t $$
So :
$$\lim_{t \to 0}\frac{1-\cos t}{\cos t-\cos 3t}=\\\lim_{t... | $$\\ \lim _{ t\to 0 } \frac { 1-\cos t }{ t^{ 2 } } \cdot \frac { t^{ 2 } }{ \cos t-\cos 3t } =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 } } }{ 2{ \left( \frac { t }{ 2 } \right) }^{ 2 } } \cdot \frac { t\cdot t }{ 2\sin { t } \sin { 2t } } =\\ =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to show that $\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?$ How can we show that $(1)$
$$\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?\tag1$$
$\sin^3(x)={3\over 4}\sin(x)-{1\over 4}\sin(3x)$
$1+\cos^2(x)=2-\sin^2(x)$
$$\int_{0}^{\pi}[{3\over 4}\sin(x)-{1\over 4}\sin(3x)]{\mathrm dx\ove... | For the integrand, use the trigonometric identitity $\sin^2(x)=1-\cos^2(x)$ and consider the two changes of variables \begin{align*}
u=\cos(x), &\,\,du=-\sin(x)\,dx\\
u=\tan(s), &\,\,du=\sec^2(s)\,ds
\end{align*}
to get\begin{align*}
\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}&=\int_0^\pi\frac{\sin(x)(1-\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Need Clarification on Trig Substitution $$\int\frac{x}{(1+x^2)^\frac{3}{2}}\ \mathrm dx$$
becomes
$$\sin\theta = \frac{x}{\sqrt{x^2+1}}$$
I am aware that we would use $x = a^2\tan^2\theta$. and that our answer before we plug in from the triangle becomes $\sin\theta$.
I just took a test and got the answer of
$$\sin\th... | It doesn't need any trigonometric substitution $$\int \frac { x }{ \left( 1+x^{ 2 } \right) ^{ \frac { 3 }{ 2 } } } dx=\frac { 1 }{ 2 } \int \frac { d\left( 1+{ x }^{ 2 } \right) }{ \left( 1+x^{ 2 } \right) ^{ \frac { 3 }{ 2 } } } =\frac { 1 }{ 2 } \int { { \left( 1+{ x }^{ 2 } \right) }^{ -\frac { 3 }{ 2 } } }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If and when a linear system has exactly three solutions
Does the following system have exactly three solutions?
$$\left\{ \begin{array}{l}
2x - y +3z = 1 \\
x + 4y - 2z = -7 \\
3x + y -z = 4 \\
\end{array} \right.$$
I marked this answer as True.
I proceeded to row reduce it and obtained the resultant matrix as -
... | Start with the augmented matrix:
\begin{eqnarray}
\left(\begin{array}{cccc} 2&-1&3 &: 1\\ 1&4&2&:-7\\ 3&1&-1&:4\end{array}\right)
\end{eqnarray}
First: Interchange rows 1 and 2 to get:
$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 2&-1&3 &: 1\\ 3&1&-1&:4\end{array}\right)$$
Second: Use the first element of the first row ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Maximum value of diagonals angle in parallelogram Let $ABCD$ be a parallelogram with sides a and b $(a>b)$ and let $r=a/b$ the ratio of the two sides.
The two diagonals form 4 angles which are by two congruent. Let's call the acute angle φ and the obtuse θ.
Find the maximum value that φ can take, in relation to r.
I ha... | Let $\alpha$ be a measure of the angle of our parallelogram.
Thus, $S_{ABCD}=ab\sin\alpha=rb^2\sin\alpha$.
In another hand, by law of cosines we obtain:
$$a^2=\frac{AC^2}{4}+\frac{BC^2}{4}-\frac{AC\cdot BC\cos(180^{\circ}-\varphi)}{2}$$ and
$$b^2=\frac{AC^2}{4}+\frac{BC^2}{4}-\frac{AC\cdot BC\cos\varphi}{2},$$
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove combinatoric identity How would you prove this identity using combinatorics? Any hints or advice?
For all positive integers $n>1$,
$\sum_{k=0}^{n} \frac{1}{k+1} {n\choose k} (-1)^{k+1}=\frac{-1}{n+1} $
|
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}(-1)^{k+1}}
&=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k+1}(-1)^{k+1}\tag{1}\\
&=\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}(-1)^{k}\tag{2}\\
&=\frac{1}{n+1}(1-1)^{n+1}-\frac{1}{n+1}\tag{3}\\
&\color{blue}{=-\frac{1}{n+1}}
\end{align*}
Comment:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadra... | For example:
$$z^2-2i=z^2-(1+i)^2=(z-1-i)(z+1+i).$$
I think the following a bit of better.
$$z^4+4=z^4+4z^2+4-4z^2=(z^2+2)^2-(2z)^2=$$
$$=(z^2-2z+2)(z^2+2z+2)=((z-1)^2-i^2)((z+1)^2-i^2)=$$
$$=(z-1-i)(z-1+i)(z+1-i)(z+1+i).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"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.