Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Can Angles A and B In A Trapezium Be Solved Using Basic Geometry?
Can angles A and B be solved? Neither the area nor the perimeter was given. Thank you very much if you can help! :)
| If I'm not mistaken, I think the followings tell us that the angles $A,B$ are not determined. Set
$$A(0,0),B(9,0),C(9+7\cos(180^\circ-B),7\sin(180^\circ-B)),D(4\cos A,4\sin A).$$
Then, we have
$$CD^2=5.9^2=\{9+7\cos(180^\circ-B)-4\cos A\}^2+\{7\sin(180^\circ-B)-4\sin A\}^2$$
$$\iff 5.9^2=9^2+7^2+4^2-126\cos B-72\cos A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Given any positive real numbers $a,b,c$, we have $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$ I have a beautiful inequality, but I can only prove part of cases. Given any positive real numbers $a,b,c$, we have
$$(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$$
How can we prove this statement?
| Hint: you can use Cauchy Schwarz inequality to prove stronger inequality
$$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\ge 9(ab+bc+ac)$$
Use Cauchy-Schwarz inequality we have
$$(b^2+1)(1+c^2)\ge (b+c)^2, (b^2+c^2)(1+1)\ge (b+c)^2$$
so
$$(b^2+2)(c^2+2)=(b^2+1)(1+c^2)+b^2+c^2+3\ge (b+c)^2+\dfrac{1}{2}(b+c)^2+3=3[1+\dfrac{(b+c)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
| After two applications of L'Hôpital's rule: $$ \displaystyle \begin{aligned} \lim_{x \to 0} \left( \dfrac{1-\cos^3 (x)}{x\sin (2x)} \right) & \ = \ \lim_{x \to 0} \left( \dfrac{3\sin (x) \cos^2 (x)}{\sin (2x) + 2x\cos (2x)} \right) \\ & \ = \ \lim_{x \to 0} \left( \dfrac{3\cos^3 (x) - 6\cos (x) \sin (x)}{4\cos (2x) - 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888081",
"timestamp": "2023-03-29T00:00:00",
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Solve trigonometric function $x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0$ I need to solve a trigonometric function similar to the following one for $\alpha$.
$$
x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0
$$
I found a solution to a very similar probl... | Using the complex representation, we have $z:=e^{i\alpha}$, $\cos\alpha=\frac{z+z^{-1}}2$, $\sin\alpha=\frac{z-z^{-1}}{2i}$, and $\cos2\alpha=\frac{z^2+z^{-2}}2$, $\sin2\alpha=\frac{z^2-z^{-2}}{2i}$.
$$x_1\frac{z^2-z^{-2}}{2i}+x_2\frac{z^2+z^{-2}}2-x_3\frac{z-z^{-1}}{2i}-x_4\frac{z+z^{-1}}2=0,$$
or
$$x_1\frac{z^4-1}{2i... | {
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"timestamp": "2023-03-29T00:00:00",
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Find a closed form for this infinite sum: $ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+ \dots$ How to find a closed form for the expression??
$$ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+\frac{1\times 2\times 3\times 4}{2 \times 5\t... | First of all, let us note that
$$\int_0^1(1-x)^{n-1}x^{-1/3}\,dx=\mathrm{B}\left(n,\frac23\right)=\frac{\Gamma(n)\Gamma(\frac23)}{\Gamma(n+\frac23)}=\frac{(n-1)!\,\Gamma(\frac23)}{(n-\frac13)(n-\frac43)\ldots\cdot \frac23 \Gamma(\frac23)}=\frac{3^n (n-1)!}{2\cdot 5\cdot\ldots\cdot(3n-1)},$$
which is almost the main ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/889117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Evaluate a limit using Taylor series Let $$\lim\limits_{x\to 0}\frac{({\ln(1+x) -x +\frac{x^2}{2})^4}}{(\cos(x)-1+\frac{x^2}{2})^3}$$
Now, I know that I should utilize Taylor polynomial.
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...$
$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ...$
Plugin it into the limit:
... | You might rearrange and factor some parts of the fraction appropriately to use $$\frac{\ln(1+x)}{x}\to 1,~~~\frac{\cos x-1}{x}\to 0$$ while $x\to 0$. Note that when $x\to 0$ then, we can simplify something like $\frac{x^4}{x^2}$ and get $x^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How do you find the modular inverse of $5\pmod{\!11}$ I need to find out the modular inverse of 5(mod 11), I know the answer is 9 and got the following so far and don't understand how to than get the answer. I know how to get the answer for a larger one such as 27(mod 392) but am stuck because they are both low numbers... | In finding a modular inverse, you are trying to solve the modular equation
$$
ax\equiv 1\pmod n.
$$
Ordinarily, you use the Extended Euclidean Algorithm for this to solve the equation $ax+ny=1$. If the numbers $a$, and $n$ are small, then simple trial and error is probably just as fast or faster.
For your example, we h... | {
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How to solve recursion? I have tried to solve some recursion:
$$f_n = \frac{2n-1}{n}f_{n-1} - \frac{n-1}{n} f_{n-2} + 1, \quad f_0 = 0, f_1 = 1$$
I would like to use a generating function:
$$F(x) = \sum_{n=0}^{\infty}f_nx^n$$
Then:
$F(x) = f_0 + f_1x + \sum_{n=2}^{\infty} \frac{2n-1}{n}f_{n-1}x^n - \sum_{n=2}^{\inft... | You have:
$$ x\, F'(x) = \sum_{n=1}^{+\infty} n\,f_n\, x^n,$$
$$ x\, F(x) = \sum_{n=1}^{+\infty} f_{n-1} x^n,$$
$$ x^2\, F'(x) + x F(x) = \sum_{n=1}^{+\infty}n\, f_{n-1}\, x^n,$$
$$ x^2 F(x) = \sum_{n=2}^{+\infty} f_{n-2} x^n,$$
$$ 2x^2 F(x) + x^3 F'(x) = \sum_{n=2}^{+\infty} n f_{n-2} x^n.$$
The recursion now gives:
... | {
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"timestamp": "2023-03-29T00:00:00",
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Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\... | A short and direct answer to your question, distilling the wisdom in answers given by others here earlier:
Both $S_n$ and $s_n$ are of the form $n(An+B)$.
However when you take the ratio, $n$ gets cancelled out.
Therefore when you take the ratio of the $\underline{\text{difference}}$, $n$ (and also $n-1$) should be r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894079",
"timestamp": "2023-03-29T00:00:00",
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Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
| This is another way of solving this problem using Completing the Square method.
$x^2+2xy-y^2=??$
First we have,
$x^2+2xy-y^2=0$
$x^2+2xy=y^2,$
By adding both sides by $y^2$ to make it perfect square then,
$x^2+2xy+y^2=y^2+y^2$
$(x+y)^2=2y^2$
$x+y=[2^(1/2)]y$
$x=(+ or -)[2^(1/2)]y-y$
$x=[2^(1/2)]y-y ; x=-[2^(1/2)]y-y $... | {
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Ranges in trigonometry How to find the range of the sum or difference of two trigonometric functions?
$2\sin x-3\cos x$
Before this whenever the question of range i have solved they were either single trigonometric function or if they were in pair then they were in form of squires (eg. $2\cos^2x + \sin^2x$) so i used ... | Let $\theta$ be an angle which satisfies $\cos \theta = \dfrac{2}{\sqrt{13}}$ and $\sin \theta = -\dfrac{3}{\sqrt{13}}$. Then we have:
$2\sin x - 3\cos x$
$= \sqrt{13}\left(\dfrac{2}{\sqrt{13}}\sin x - \dfrac{3}{\sqrt{13}}\cos x\right)$
$= \sqrt{13}\left(cos\theta\sin x + \sin\theta\cos x\right)$
$= \sqrt{13}\sin(x+\th... | {
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Distinct balls into distinct boxes with a minimal number of balls in each box
Find the number of ways to distribute $8$ distinct balls into $3$ distinct boxes if each box must hold at least $2$ balls.
The stars and bars approach would not work because the balls are non-identical. Stirling Numbers of the second kind w... | Associate a different colour to each of the three boxes; then you are looking for the number of ways to colour the balls such that the frequency vectors of colours used is one of
$$
(2,2,4), (2,3,3), (2,4,2), (3,2,3), (3,3,2), (4,2,2)
$$
The number for each of these colourings is given by a trinomial coefficient, and... | {
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Sum of an unorthodox infinite series $ \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots $
This is a pretty unorthodox problem, and I'm not quite sure how to simplify it. Could I get a solution? Thanks.
| Another approach is to use summation by parts. This method works as follows. If
$$S_N = \sum_{n=1}^{N} a_n b_n$$
then we can define
$$B_n = \sum_{k=1}^{n} b_k$$
Then
$$S_N = a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n)$$
In this problem, we can set $a_n = 2n-1$ and $b_n = 1/2^n$. Then
$$B_n = \sum_{k=1}^{n} \frac{1}... | {
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"timestamp": "2023-03-29T00:00:00",
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How prove this triangulation with indentity
let $x,y,z\in (0,\pi)$, prove or disprove
$$\sin{(x+y)}\cdot\sin{(y+z)}\cdot\sin{(x+z)}\cdot\sin{(x+y+z)}
=[\sin{(x+y+z)}\cdot\sin{x}+\sin{y}\cdot\sin{z}]\times[\sin{(x+2y+z)}\cdot\sin{z}+\sin{(x+y)}\cdot\sin{y}]$$
if this is ture, we can use this
$$2\sin{x}\sin{y}=\cos{(x-... | HINT:
From the RHS,
Using Werner Formula
$$2\sin(x+2y+z)\sin z=\cos(x+2y)-\cos(x+2y+2z)$$
and $$2\sin(x+y)\sin y=\cos x-\cos(x+2y)$$
Using Prosthaphaeresis formula,
$$2\sin(x+2y+z)\sin z+2\sin(x+y)\sin y=\cos x-\cos(x+2y+2z)=2\sin(y+z)\sin(x+y+z)$$
Can you apply the same method on $$[\sin(x+y+z)\sin x]+[\sin y\sin z]?$... | {
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Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$ I am trying to find out the sum of this
$$1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\frac{1\cdot 2\cdot 3\cdot 4}{2\cdot 5\cdot 8\cdot 11}+\cdots$$.
I tried with binomial theorem with ration... | Since $$S=1+\sum_{k=0}^{+\infty}\prod_{j=0}^k\frac{j+1}{3j+2}=1+\sum_{k=1}^{+\infty}k\, 3^{-k} B(2/3,k),$$
where $B(2/3,k)$ is the Euler Beta function:
$$ B(2/3,k)=\int_{0}^{1}x^{-1/3}(1-x)^{k-1}$$
we have: $$S=1+3\int_{0}^{1}\frac{dx}{(1-x)^{1/3}(3-x)^2}.$$
Now the last integral can be computed explicitly, and leads ... | {
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Squared binomial paradox? When you square this $$(5-2)^2$$ you will get 49 $$ 5^2 - 2 * 5 * (-2) + (-2)^2$$
$$25 + 20 + 4 = 49$$ but if you do it like this (5-2) * (5-2) you will get 9 $$ 5(5-2) - 2(5-2)$$
$$25-10-10+4$$
$$25-20+4$$
$$5+4$$
$$9$$
Why do I always get different results if I'm doing the same thing?Am I do... | The binomial formular is $(a + b)^2 = a^2 + 2ab + b^2$
If you apply it:
$$(5 - 2)^2 = 5^2 + 2\cdot5\cdot(-2) + 2^2 = 25-20 + 4 = 9$$
The second binomial formular is $(a - b)^2 = a^2 - 2ab + b^2$
If you apply it:
$$(5 - 2)^2 = 5^2 - 2\cdot5\cdot 2 + 2^2 = 25-20 + 4 = 9$$
Your mistake was, that you used the second binom... | {
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Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
| I find this easier to visualize if I write this problem in terms of powers:
$$\dfrac{n^{1/2}}{\left[n + \left(n + n^{1/2}\right)^{1/2}\right]^{1/2}}\text{.} $$
Division by $n^{1/2}$ for both the numerator and denominator turns the numerator into $1$ and the denominator into
$$\begin{align*}
\dfrac{\left[n + \left(n + n... | {
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A hard square root question This is my first question on StackExchange. So my question is:
If $$x = \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{\sqrt{\sqrt5 + 1}} + \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{2\sqrt{\sqrt5 + 1}} - \sqrt{3-2\sqrt2}$$
What is the value of $x^2$?
If someone can also tell me how to input mathemat... | Letting
$$\frac{\sqrt{\sqrt 5+2}+\sqrt{\sqrt 5-2}}{\sqrt{\sqrt 5+1}}=\alpha,$$
we have
$$\alpha^2=\frac{(\sqrt 5+2)+(\sqrt 5-2)+2\sqrt{(\sqrt 5+2)(\sqrt 5-2)}}{\sqrt 5+1}=\frac{2(\sqrt 5+1)}{\sqrt 5+1}=2\Rightarrow \alpha=\sqrt 2.$$
Hence, we have
$$x=\alpha+\frac{\alpha}{2}-\sqrt{(\sqrt 2-1)^2}=\frac{3}{2}\alpha-(\sqr... | {
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Minimum eigenvalue of product of two matrices Abstract description:
Let $\mathbf{A}$ and $\mathbf{B}$ be two $n \times n$ real matrices. Let $\sigma( \mathbf{A B} )$ denote the spectrum of $\mathbf{A B}$.
Assume that
(A1) $\mathbf{A}$ is symmetric and positive definite but not diagonal.
(A2) $\mathbf{B}$ is not symmetr... | According to the new statement, $A=\begin{pmatrix}20&-4&0\\-4&4&0\\0&0&1\end{pmatrix},B=\begin{pmatrix}1&0&0\\2&2&0\\0&0&0\end{pmatrix}$.
When the next change ?
| {
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Exotic proofs of $\sum_{j=0}^{n-1}\binom{p+j}{p}=\binom{p+n}{p+1}$ Let $p,n$ be positive integers.
The following identity $\displaystyle \sum_{j=0}^{n-1}\binom{p+j}{p}=\binom{p+n}{p+1}$ may be proved by induction or by successive uses of Pascal's rule (both techniques are demonstrated here).
Question
Is there a combina... | First Proof:
\begin{align}
\sum^{n-1}_{j=0}\binom{p+j}{p}
&=\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^p}{z^{p+1}}\sum^{n-1}_{j=0}(1+z)^j{\rm d}z\\
&=\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^{n+p}-(1+z)^p}{z^{p+2}}{\rm d}z\\
&=\frac{1}{(p+1)!}\lim_{z \to 0}\frac{{\rm d}^{p+1}}{{\rm d}z^{p+1}}\left[(1+z)^{n+p}-(1+z)^p\rig... | {
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"source": "stackexchange",
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Is my solution to this differential equation correct?
My answer is: $[(1+x^2)^3]y = \dfrac{(1+x^2)^3}3+C$
But this option is not given, so is it correct?
Thanks
| $$(1+x^2)\frac{dy}{dx}+6xy=2x \Rightarrow (1+x^2)\frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2)\frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{1}{1-3y}dy=\frac{2x}{1+x^2}dx \Rightarrow \int \frac{1}{1-3y}dy=\int \frac{2x}{1+x^2}dx \\ \Rightarrow \frac{-1}{3}\ln{|1-3y|}=\ln{|1+x^2|}+c\Rightarrow \ln{|1-3y|}^{\frac{-1}{3}}=\ln{|1+x^2... | {
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How to simplify $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1}$? This is the original problem:
$\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = x$.
I'm really confused about how to solve this problem, I come as far as saying this: $\sqrt[4]{5} + \sqrt{1}\cdot \sqrt[4]{5}-\sqrt{1}$.
| Hint: Start with $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = \sqrt{(\sqrt{5}+1) \cdot (\sqrt{5}-1)}$, and then use the difference of squares identity, i.e. $(a+b)(a-b) = a^2-b^2$.
| {
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If the equation $|x^2+bx+c|=k$ has four real roots .....
If the equation $|x^2+bx+c|=k$ has four real roots then which of the following option is correct :
(a) $b^2-4c >0$ and $0<k<\frac{4c-b^2}{4}$
(b) $b^2-4c <0$ and $0<k<\frac{4c-b^2}{4}$
(c) $b^2-4c >0$ and $0<k>\frac{4c-b^2}{4}$
Please suggest how to proceed i... | Hint: Clearly $k$ is positive. Also, both
$$ x^2+bx + (c-k) = 0 $$
and $$ x^2+bx + (c+k) = 0 $$
have real roots (that is, both equation discriminants are positive).
(I have been assuming everywhere that there are four distinct real roots.)
Note that $k>0$ and positivity of the second discriminant, $b^2-4(c+k)>0$ imp... | {
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Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$ if $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$
if $a,b,c,d$ are positive real numbers and $$a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$$
Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$
Things i have done: from assumption $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$ I can conclude that $$a^2+b^2-ab=c^2+d^2-cd$$
Powe... | Observe that:
$$
a^4+b^4+(a-b)^4-c^4-d^4-(c-d)^4\\
=2(a^2-ab+b^2-c^2+cd-d^2)(a^2-ab+b^2+c^2-cd+d^2)
$$
so the result follows because you have proved the first term on the RHS is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 1
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Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we ha... | $(x+y)^2 \geq xy$
If one of $x, y$ is negative, this is obvious due to sign. Otherwise, WLOG, assume both $x$ and $y$ are positive. Since $x+y \geq x$ and $x+y \geq y$, then $(x+y)^2 \geq xy$.
| {
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"answer_id": 5
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Complex analysis: Rewrite $\cos^{-1}{i}$ in algebraic form I'm stuck in this problem (complex analysis), my answer is not the one reported in the book:
Rewrite $\cos^{-1}{i}$ in the algebraic form. A: $k\pi + i \frac{\ln{2}}{2}\ \forall\ k \in \mathbb{Z}$
So I tried this approach in order to solve it:
*
*As $\cos^... | Let me obtain an answer similar to yours by different means. Rather than use the algebraic form of the inverse cosine, I'll invert to solve for $x=e^{iz}$ and then invert again to get $z$. Carrying this out gives
\begin{align}
z=\cos^{-1} i
&\underset{\text{invert}}{\implies} i=\cos z=\frac{1}{2}(x+x^{-1})\implies x^2... | {
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Proving a trigonometric inequality $(1-\sin a)x^2 -2x\cos a + 1+ \sin a \ge 0$ $(1-\sin (a))x^2 -2x\cos(a) + 1+ \sin( a) \ge 0$, where $a,x$ are any two real constants.
Any suggestions on how to prove this ? I tried playing with it, but nothing useful came out.
| $\cos^2 a = 1-\sin^2a=(1+\sin a)(1-\sin a)$
$1+\sin a = \frac {\cos^2 a}{1-\sin a}$
Substituting this value in the equation --
$(1-\sin a)x^2 -2x\cos(a) + [1+ \sin( a)] $
=$(1-\sin a)x^2 -2x\cos(a) + \frac {\cos^2 a}{1-\sin a} $
=$[x \root \of {1-\sin a} - \frac {\cos a}{\root \of {1-\sin a}} ]^2$
$\ge 0$
| {
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"question_score": "3",
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Linear algebra system of equations Given the following linear system:
$x - y + 2z + 3w = 12$
$ x - 2y + z + w = 8$
$4x + y + 2z - w = 3$
1) find all solutions of the system. Write the solution in Vector Form
2) find all solutions of the corresponding homogeneous system
I have turned this into a matrix and wo... | Oh of course there is more that can be done!
So starting from where you left, I would first multiple $R_2$ with the scalar $(-1)$ and then add the new $R_2$ to $R_1$ and replace it. Then I get,
$$
\left({\begin{matrix}
1 & 0 & 3 & 5 & 16 \\
0 & 1 & 1 & 2 & 4 \\
0 & 0 & -11 & -23 & -65 ... | {
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"answer_count": 1,
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} |
Find the generating function of the sequence $a_n = \sum\limits_{k=0}^n k(k-1)$
Find the generating function of the sequence $ a_n =\sum\limits_{k=0}^n k(k-1)$
My try:
Let's assume $k(k-1)$ is genereated by $F(x)$ then $a_n$ is generated by $\frac{F(x)}{1-x}$ (that's a common trick).
So have we reduced the problem to... | from your initial conditions, we have
$$a_n=\sum_{k=0}^nk^2-\sum_{k=0}^nk\\
=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\\
=\frac{n(n+1)(n-1)}{3}$$
thus the generating function is
$$G(a_n,x)=\frac{1}{3}\sum_{n=0}^\infty(n^3-n)x^n\\
=\frac{1}{3}\left[\sum_{n=0}^\infty n^3x^n-\sum_{n=0}^\infty nx^n\right]$$
other hand, we ha... | {
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Find $m_1 , m_2,m_3,m_4\in\mathbb{Q}$ s.t. $\forall a_k,b_k\in\mathbb Z,\,m_1(a_1^2+a_2^2)+m_2(a_3^2+a_4^2)\neq m_3(b_1^2+b_2^2)+m_4(b_3^2+b_4^2)$ Let us assume that $a_1 , a_2 , a_3 ,a_4,b_1,b_2,b_3,b_4\in\mathbb{Z}$.
If $m_1 , m_2,m_3,m_4\in\mathbb{Q}$, then how can I choose $m_1,m_2,m_3,m_4$, such that the following... | As I said, for 8 unknown parameters and the formula goes bulky.
$$a(z_1^2+z_2^2)+b(z_3^2+z_4^2)=c(z_5^2+z_6^2)+j(z_7^2+z_8^2)$$
3 - the formula looks like this: Solutions to $ax^2 + by^2 = cz^2$
Will consider here the special case when: $a+b=c+j$ $(1)$
Then the solutions are of the form:
$$z_1=js^2+jt^2+ck^2+cp^2-bq^2... | {
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Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$ I have a problem with the following integral:
$$
\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\,
{{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}
$$
The first idea was to use the integration by parts because
$$
\int{{\rm d}x \over \,\sqrt{x\... | We have:
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx = 2\pi\log\frac{2+\sqrt{3}}{2}.\tag{1}$$
This happens because:
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=2\int_{0}^{1}\frac{\log(3+x^2)}{\sqrt{1-x^2}}\,dx =\int_{-\pi/2}^{\pi/2}\log(3+\cos^2\theta)\,d\theta,$$
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-... | {
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"answer_count": 3,
"answer_id": 2
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Find $ \int \frac{dx}{x\sqrt{1-x^4}}$ Find $\displaystyle \int \dfrac{dx}{x\sqrt{1-x^4}}$
I cannot figure out how start this problem, can anyone explain
| The substitution $u=\sqrt{1-x^{4}}$ converts the integral into a rational
function of $u$. Since $x^{4}=1-u^{2}$ and $du=-\frac{2x^{3}}{u}dx$, we get
\begin{equation*}
I=\int \frac{dx}{x\sqrt{1-x^{4}}}=\int \frac{1}{xu}\left( -\frac{u}{2x^{3}}%
\right) du=\frac{1}{2}\int \frac{1}{u^{2}-1}\,du.
\end{equation*}
Expanding... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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find the slope of the tangent line Sorry I couldn't figure out how to use the editor right now and I really need help with this question.
Let $f(x)=\frac{1}{\sqrt[3]{x}}$. find the slope of the tangent line at $(8,\frac{1}{2})$
So far this is what I have I just don't know what to do from here
$$
\begin{array}{lll}
m&=... | From where you're at, do this trick:
\begin{align*}
&\lim_{h \to 0} \frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}=\lim_{h \to 0} \frac{\left(2-\sqrt[3]{8+h}\right)}{2h\sqrt[3]{8+h}}\frac{\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}{\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}=\\
&\lim_{h\to 0}\frac{-h}{2h(8+h)^{1/3}\left((8+... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that the square root of 3 is irrational I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if $x$ is divisible by $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of $6$?
| suppose $\sqrt{3}$ is rational, then $\sqrt{3}=\frac{a}{b} $ for some $(a,b)
$
suppose we have $a/b$ in simplest form.
\begin{align}
\sqrt{3}&=\frac{a}{b}\\
a^2&=3b^2
\end{align}
if $b$ is even, then a is also even in which case $a/b$ is not in simplest form.
if $b$ is odd then $a$ is also odd.
Therefore:
\begin{align}... | {
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} |
Integrate $\int\frac{\cos^2x}{1+\tan x}dx$ Integrate $$I=\int\frac{\cos^2x}{1+\tan x}dx$$
$$I=\int\frac{\cos^3xdx}{\cos x+\sin x}=\int\frac{\cos^3x(\cos x-\sin x)dx}{\cos^2x-\sin^2x}=\int\frac{\cos^4xdx}{1-2\sin^2x}-\int\frac{\cos^3x\sin xdx}{2\cos^2x-1}$$
Let $t=\sin x,u=\cos x,dt=\cos xdx,du=-\sin xdx$
$$I=\underbra... | The integral is equivalent to $ \displaystyle{\int\frac{\cos^3 x}{\sin x + \cos x}\,\mathrm{d}x}. $ Now, consider
$$ \mathcal{I}_{1} = \int\frac{\cos^3 x}{\sin x + \cos x}\,\mathrm{d}x \qquad\text{and}\qquad\mathcal{I}_{2}=\int\frac{\sin^3 x}{\sin x + \cos x}\,\mathrm{d}x. $$
Observe that
$$
\begin{aligned}
\mathcal{I}... | {
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"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$\frac{\partial \coth ^{-1}(x)}{\partial x}$ I am asked to find
$\dfrac{d\coth ^{-1}(x)}{dx}$
I rewrite it to become
$x=\dfrac{1}{\tan(y)}$
$\dfrac{\text{dx}}{\text{dy}}=-\dfrac{1}{\frac{\sec ^2(x)}{\tan ^2(x)}}=-\sin^2(x)$
However the answer should be
$\dfrac{1}{1-x^2}$
What am I doing wrong?
| First note that
$$ \coth^{-1} x=\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right), \mbox{for}\ |x|\gt 1$$
Then
$$ \frac{d}{dx}[\coth^{-1} x]=\frac{1}{2}\frac{d}{dx}\left[\ln\left(\frac{x+1}{x-1}\right)\right]$$
$$ =\frac{1}{2}\left[\frac{d}{dx}[\ln(x+1)]-\frac{d}{dx}[\ln(x-1)]\right] $$
$$ =\frac{1}{2}\left[\frac{1}{x+1}-\fra... | {
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"source": "stackexchange",
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Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align... | The expression under limit can be written as $$\frac{y^{1/3}-1}{x}=\frac{y^{1/3}-1}{y-1}\cdot\frac{y-1}{x}\tag{1}$$ where $$y=\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}=\frac{x}{1+\sqrt{1-x}}\cdot\frac{\sqrt{1+x}+1}{x}=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}\tag{2}$$ Since $y\to 1$ as $x\to 0$ we can see that the first fraction in $(... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Multiplying both sides of an equation when there's a limit on one side?
Determine the value of $a\in\mathbb{R}$, such that $\displaystyle\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$
My attempt:
\begin{align*}
&\lim_{x\to 1} \dfrac{x^2+(3-a)x+3a}{x-1}=7\\
&\implies\lim_{x\to 1} x^2+(3-a)x+3a=7x-7\\
&\implies\lim_{x\to ... | $$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=\lim_{x\to 1}7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}-\lim_{x\to 1}7=0$$
$$\lim_{x\to 1}\left(\dfrac{x^2+(3-a)x+3a}{x-1}-7\right)=0$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a-7x+7}{x-1}=0$$
$$\lim_{x\to 1}\dfrac{x^2+(-4-a)x+3a+7}{x-... | {
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"question_score": "3",
"answer_count": 11,
"answer_id": 4
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Are all those numbers coprime? The values of $4m^2+1$ and $4m^2+4m+5$ for $m\geq{1}$ are (resp.) 5,17,37,... and 13,29,53,... Those numbers seem to be all coprime : how to prove it if it is true, please ?
| Suppose there is a common divisor $d$. Then $d|4m^2+1$ and $d|4m^2+4m+5$ implies that $d|4m^2+4m+5-4m^2-1$, which is $d|4m+4$. Since $4m^2+1$ is an odd number, so is $d$. Therefore $d|m+1$. Also, $d|4(m+1)^2$, so $d|4(m+1)^2-(4m^2+4m+5)$ and then $d|4m-1$.
Now combine $d|4m-1$ and $d|4m+4$. This gives us $d|5$.
| {
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If $\alpha$, $\beta$ are two values of $\theta$ satisfying the equation $\cos\theta/a+\sin\theta/b=1/c$, prove that $\cot ((\alpha+\beta)/2) = b/a$ What I did was
$$b\ \cos (\theta) + a \sin (\theta) = \dfrac{ab}{c} \\
b\ \cos (\theta) = \frac{ab}{c} - a\ \sin (\theta) $$
Square both sides and using sum of roots and p... | Hint. Write
$$\theta=\frac{\alpha+\beta}{2}\ ,\quad \phi=\frac{\alpha-\beta}{2}\ .$$
You need to find $\cot\theta$, and you are given that
$$\frac{\cos(\theta+\phi)}{a}+\frac{\sin(\theta+\phi)}{b}
=\frac{\cos(\theta-\phi)}{a}+\frac{\sin(\theta-\phi)}{b}\ .$$
See if you can take it from here.
| {
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Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdo... | $$\begin{align}
y &= \sum_{k=1} \frac{\prod_{j=2}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{A}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
\frac{\left(-\frac 12\right)\prod_{j=0}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{B}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
\frac{\left(-\frac 13\right)^{k+1}\left(-\frac 12\right)\prod... | {
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"url": "https://math.stackexchange.com/questions/936146",
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"question_score": "2",
"answer_count": 3,
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Sum of the series $\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\dots$ I am recently struck upon this question that asks to find the sum until infinite terms
$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+.....∞$$
I tried my best to get something telescoping or somethin... | $$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}$$
Rewrite the sum as
\begin{align}
\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots
&=\sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!}\\
&=\sum^\infty_{n=0}\frac{(2n+1)!}{2^{2n+2}n!(n+2)!}\\
&=\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}\... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Stuck on this intergral $\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{\tan^2x}{x-\tan x} dx $ calculus I $$\int^{\pi/3}_{\pi/4} \frac{\tan^2x}{x-\tan x} dx $$
this is that I have tried
$$\int^{\pi/3}_{\pi/4} \frac{\frac{\sin^2x}{\cos^2 x}}{x-\frac{\sin x}{\cos x}} dx $$
$$\int^{\pi/3}_{\pi/4} \frac{\sin^2(x)}{x \cos^2(x)-... | Hint: For the integral
$$\int \dfrac{\tan^2(x)}{x-\tan(x)}\, dx$$
try the substitution $u = x-\tan(x)$, $du = 1-\frac{1}{\cos(x)^2} \, dx = -\tan^2(x)\, dx$
As requested here is the elaboration:
With the substitution $u = x-\tan(x)$ we get
\begin{align*}du &= \dfrac{d}{dx}\left(x-\frac{\sin(x)}{\cos(x)}\right)dx=1-\fr... | {
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Closed form of $\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2}$ I'm new in the area of the series involving Bessel function of the first kind. What are
the usual tools you would recommend me for computing such a series? Thanks.
$$\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2}$$
| Here is a route.
Recall that
$$
J_0(2n)=\frac 1\pi \int_0^\pi \cos (2n \sin x)\:{\rm d}x \tag1
$$
and that
$$\sum_{n=1}^\infty\frac{\cos nt}{n^2}=\frac{\pi^2}{6}-\frac{\pi t}{2}+\frac{t^2}{4},\quad 0\leq t\leq 2\pi. \tag2 $$
Then, due to normal convergence of the series on $[0,2\pi] $, we may write
$$
\begin{align}
\su... | {
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"question_score": "19",
"answer_count": 2,
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Please explain the Quotient Rule I am currently working on an equation but I'm having a hard time understanding how to get the answer.
the answer is ${(x^2-4)(x^2+4)(2x+8)-(x^2+8x-4)(4x^3)\over (x^2-4)^2(x^2+4)^2}$
The equation is $f(x)= {x\over x^2-4}-{x-1\over x^2+4}$
When I apply the quotient rule i get $f'(x)= {(1)... | $$\frac{a}{b}-\frac{a}{c} \neq \frac{a-a}{b}$$ but $$\frac{a}{b}-\frac{a}{c}=\frac{ac-ab}{bc}$$
Notice that you have different denominators.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculate $\pi$ By Hand? All over the internet the only hand equation i found was
$$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$
But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?
| I skimmed the other answers. I believe this one is different in that it gives a way to rapidly calculate the digits of $\pi$ and proves from scratch that that number in fact is $\pi$.
It's obvious that $\pi = 6 \times \sin^{-1}(\frac{1}{2})$. It turns out that the derivative of $\sin^{-1}$ is elementary so we can take ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 7
} |
What is the closed form of $\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(n\right)} ^2}{{n}^4}}$? Using Maple I am obtaining the numerical approximation
$$0.5902373619$$
Please, let me know what is the closed form. Many thanks.
| Hint.
Observe that
$$
J_0(n)=\frac 1\pi \int_0^\pi \cos (n \sin x)\:{\rm d}x \tag1
$$
and that
$$\sum_{n=1}^\infty\frac{\cos nt}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2 t^2}{12}+\frac{\pi t^3}{12}-\frac{t^4}{48},\quad 0\leq t\leq 2\pi. \tag2 $$
Then, due to normal convergence of the series on $[0,2\pi] $, we may write
$$
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left... | I played a little bit with Jack's answer. One can expand with the Newton formula
$$ (1 + \sqrt{3} t)^n = \sum_{k=0}^n \binom{n}{k} (\sqrt{3}t)^k $$
then the integrals can be performed using
$$
\int_0^\infty dt\, t^k e^{- t} \cos(s t) = \frac{T_{1+k}\bigl(1/\sqrt{1 + s^2}\bigr) k!}{(1+s^2)^{(k+1)/2}} $$
where $T_a(x)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Solving quadratic complex equation I'm trying to solve this equation:
$z^2+2iz+11-16i=0$
where $z=x+iy$ with $ x,y \in \mathbb{R}$.
I tried it with the quadratic formula $z_{1/2}=- \frac p2 \pm \sqrt{\left(\frac p2\right)^2 -q}$ and get:
$z_{1/2}=-i \pm \sqrt{i^2-11+16i} \\ \Leftrightarrow z_{1/2}=-i \pm \sqrt{12i^2+... | Completing the square might be simpler:
$z^2+2iz+i^2 = -11+16i+i^2$
$(z+i)^2 = -12+16i$
$z = -i \pm \sqrt{-12+16i}$
$z = -i \pm 2\sqrt{-3+4i}$ (factor out a $4$ from under the squareroot).
To simplify $\sqrt{-3+4i} = a+bi$, we need to find reals $a,b$ such that $(a+bi)^2 = (a^2-b^2)+(2ab)i = -3+4i$.
Since the number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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To prove triangle is equilateral> given one equation and one combined equation. show that the lines $ x^2 - 4xy +y^2 $ and $x +y=3 $ form an equilateral trinagle . Also Find Area.
Here is what I have tried:
$ l1 =x+y=3 $
The combined equation is:
$ x^2 - 4xy + y^2 $
compare it with general formula :
$ ax^2 +2hxy+y^2 $... | If the angle between $y=mx$ and $x+y=3\iff y=-x+3$ is $60^\circ,$
$$\left|\frac{m-(-1)}{1+m(-1)}\right|=\tan60^\circ=\sqrt3$$
$$\iff\frac{m+1}{m-1}=\pm\sqrt3\iff(m+1)^2=3(m-1)^2\iff m^2-4m+1=0$$
Setting $m=\dfrac yx\implies y^2-4xy+x^2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the integer solutions of the equation $3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14$ $
3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14
$ .
I already solved this using the Cauchy–Schwarz inequality and got $x=4$ and $y=5$. But I'm sure there is a prettier, simpler solution to this and I was won... | This is a trivial solution so that we get answer in integer
8 - x and 6 - y are under the root so x <8 and y <6 , for making 6 - y to be perfect square y=2 or 5. and similarly x=4 or 7.
Now x+y is also under the root so x+y has to be perfect square. So we get two pairs of (x, y)=(7,2) and (2,5) but (7,2) will not satis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^... | Let $$I = \int^{\frac{\pi}{4}}_{0}\frac{1}{\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x}dx = \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x(\tan ^2x+\cot^2 x-1)}dx$$
So $$I =\int^{\frac{\pi}{4}}_{0}\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+1}dx = \left[\tan^{-1}(\tan x-\cot x)\right]^{\frac{\pi}{4}}_{0}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Bivariate distribution with normal conditions Define the joint pdf of $(X,Y)$ as:
$$f(x,y)\propto \exp(-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]),$$
where $A,B,C,D$ are constants.
Show that the distribution of $X\mid Y=y$ is normal with mean $\frac{By+C}{Ay^2+1}$ and variance $\frac{1}{Ay^2+1}$. Derive a corresponding result f... | You would need to integrate out $y$, not $x$, to get a marginal density that is a function of $x$.
What you have inside of $\exp\left(\dfrac{-1}2\left(\bullet\bullet\bullet\right)\right)$ is
$$
Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy.
$$
Think of how this behaves as a function of $x$. You want
$$
\left(\frac{x-\mu}\sigma\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/954523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Bound for $\sum_{k=1}^\infty\left(\frac{1}{2^k+k^2}\right)$ I found for the series:
$$S=\sum_{k=1}^\infty\left(\dfrac{1}{2^k+k^2}\right)$$ a bound:
$$S\le\dfrac{\pi^2}{6+\pi^2}$$
which is in good agreement with the approximate value of $S$ calculated with Maple or Mathematica $(S=0.588239...)$, while the ratio: $\dfrac... | One idea:
$$\sum_{n=1}^{\infty} \frac{1}{n^2+2^n}=\sum_{n=1}^{k} \frac{1}{n^2+2^n}+\sum_{n=k+1}^{\infty} \frac{1}{n^2+2^n}$$
For some $k$
Now $\displaystyle \sum_{n=1}^{k} \frac{1}{n^2+2^n}$ is finite sum, so it's possible to calculate it directly.
In second series use inequality $\frac{1}{n^2+2^n} \leq \frac{1}{2^n}$,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Divisibility property of $(a+b)^n-a^n-b^n$ Let $n$ be a natural number of the form $n=6k+1$ (while $k$ is a positive integer).
Show that $(a^2+ab+b^2)^2$ divides $(a+b)^n-a^n-b^n$ for all integer numbers $a,b$ (such that $a^2+ab+b^2\ne0$).
| Note: This proof replaces an earlier less elementary version that used the unique factorization property of $\mathbf{Z}[x,y]$.
Let $c = a^2 + ab + b^2$. Since $(a-b)(a^2 + ab + b^2) = a^3 - b^3$, we have $a^3 \equiv b^3$ modulo $c$. It follows therefore that for any $k$ we have $a^{6k} \equiv b^{6k}$ modulo $c$, hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Two differentiation results of $\sin^{-1}(2x\sqrt{1-x^2})$ While trying to differentiate $\sin^{-1}(2x\sqrt{1-x^2})$, if we put
$x = \sin\theta$, we get,
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})\\
&= \sin^{-1}(2\sin\theta\cos\theta)\\
&= \sin^{-1}(\sin2\theta... | Remember that $\sin^{-1}$ is not a "true inverse". Here is a graph of $\sin^{-1}(\sin(2x))$.
I think it is clear that this is the most likely source for your sign error. As you note, you could have a different choice for arcsin which would give you the opposite sign when differentiating.
I think the easiest way to rea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is this the correct procedure for Integral Partial Fraction. $∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$
The first step I did is distributed the denominator so that I can find out if I should use synthetic division. Which after doing this I discovered that the denominator's exponent is greater therefore I do not use synthetic d... | You have to use what is called a coefficient matrix.
So as you said we have that:
$$∫ \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)}dx =∫ \frac{Ax +B}{x^2+1} + \frac{Cx+D}{x^2+3}dx$$
Then multiplying both sides we get that:
$$x^3+x^2+x+3 = (x^2+3)(Ax+B) + (x^2+1)(Cx+D) = Ax^3+Bx^2+3Ax+3B + Cx^3+Dx^2+Cx+D$$
Now we analyze the coeff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration Question: Completing the Square/Trig Sub yields a different answer than integral table. After the completing the square,
$$\int \frac{dx}{x^2 + 2x - 3}$$
becomes,
$$ \int \frac{dx}{(x+1)^2 - 4}$$
The integral table in my book says the antiderivative is,
$$\frac{1}{2a} ln \, \Biggl\lvert \frac{x-a}{x+a} \Big... | $$-\frac{1}{2} \ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert = -\frac12 \ln \left| \sqrt{\frac{(x+3)^2}{(x-1)(x+3)}} \right| = -\frac14 \ln \left| \frac{x+3}{x-1} \right| = \frac14 \ln \left|\frac{x-1}{x+3}\right|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Suddenly a wild Z-transform appears... I am attempting to complete the z transform of the following formula:
$$x(n) = \frac{1}{2}(n^2+n) u(n-1)$$
I got it into the summation form like so:
$$X(z) = \frac{3}{2} \sum_{n=1}^{\infty} n^2(\frac{1}{3} z^{-1})^2 + \frac{3}{2} \sum_{n=1}^{\infty} n(\frac{1}{3} z^{-1})^2$$
$$X(z... | The crucial idea here is to use the fact that
$$
\sum_{n=1}^\infty na(n)z^{-n}=-z\frac{d}{dz}\sum_{n=1}^\infty a(n)z^{-n}.
$$
The Z-transform of your $x(n)$ will by definition be
$$
X(z)
=\sum_{n=-\infty}^\infty \frac{1}{2}(n^2+n)u(n-1)z^{-n}
=\frac{1}{2}\sum_{n=1}^\infty n^2z^{-n}+\frac{1}{2}\sum_{n=1}^\infty nz^{-n},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt... | $$\lim_{x\rightarrow \infty} (\sqrt{x^2+2x} - \sqrt{x^2-7x}) $$$$=\lim_{x\rightarrow \infty} \bigg(\sqrt{x^2+2x+1} - \sqrt{x^2-7x+\frac{49}{4}}\bigg) $$ $$=\lim_{x\rightarrow \infty} \bigg( x+1 - \bigg(x-\frac{7}{2}\bigg)\bigg) =\frac{9}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
$\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers of $x$ and $y$ Give a convincing argument that $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers $x$ and $y$. Could someone please explain how to prove this? I attempted to say that the largest ... | Let $x = \lfloor x \rfloor + \epsilon_1$ and $y = \lfloor y \rfloor + \epsilon_2$.
Then $\lfloor x + y \rfloor = \lfloor\lfloor x \rfloor + \epsilon_1 + \lfloor y \rfloor + \epsilon_2\rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor \epsilon_1 + \epsilon_2 \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/963245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Equation of line passing through point. The straight line $3x + 4y + 5 = 0 $ and $4x - 3y - 10 = 0$ intersect at point $A$. Point $B$ on line $3x + 4y + 5 = 0 $ and point C on line $4x - 3y - 10 = 0$ are such that $d(A,B)=d(A,C)$.
Find the equation of line passing through line $\overline{BC}$ if it also passes through... | Intersection point $A$ has coordinates $(1, -2)$.
The first line equation can be written as $y = -\frac34x - \frac54$, the second: $y = \frac43x-\frac{10}{3}$, so $k_1 = -\frac34, k_2 = \frac43$.
$AB$ would lie on the first line if $y_B - y_A = k_1(x_B - x_A)$.
$AC$ would lie on the second line if $y_C - y_A = k_2(x_C ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solving $2\cos2x-4\sin x\cos x=\sqrt{6}$ How I solve the following equation for $0 \le x \le 360$:
$$
2\cos2x-4\sin x\cos x=\sqrt{6}
$$
I tried different methods. The first was to get things in the form of $R\cos(x \mp \alpha)$:
$$
2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\
2\cos2x-2\sin2x=\sqrt{6}\\
R = \sqrt{4} = 2 \\
\alph... | Note that
$$2\cos(2x)-2\sin (2x)=\sqrt{2^2+2^2}\cos(2x+\alpha)$$
where
$$\cos\alpha=\sin\alpha=\frac{2}{2\sqrt 2}=\frac{1}{\sqrt 2}.$$
Hence, you'll have
$$2\sqrt 2\cos(2x+45^\circ)=\sqrt 6\Rightarrow\cos(2x+45^\circ)=\frac{\sqrt 6}{2\sqrt 2}=\frac{\sqrt 3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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how to solve $\int\frac{1}{1+x^4}dx$ i want find the answer and method of solve of $\int\frac{1}{1+x^4}dx$.
I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,
How I can use this to solve of that integration.
| $$\frac{1}{1+x^4}=\frac{Ax+B}{2\sqrt2 (-x^2+\sqrt 2 x-1)}+\frac{Cx+D}{2\sqrt2 (x^2+\sqrt 2 x+1)}=\dots A=C=1, D=-B=\sqrt 2$$
Simplify even further
$$\frac{x-\sqrt 2}{2\sqrt2 (-x^2+\sqrt 2 x-1)}= -\frac{\sqrt 2-2x}{2-x^2+\sqrt 2 x-1)}-\frac{1}{\sqrt 2(-x^2+\sqrt 2 x-1}$$
Substitute $u=-x^2+\sqrt 2 x-1$ then it's trivia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find $\int (x^3+1)(\cos(2x)) dx$ What I've tried for this problem is expanding it to $x^3\cos(2x) + \cos(2x)$ and then evaluating the respective functions as separate integrals. The first one uses tabular and the second one is simple u substitution. Is my procedure correct?
| what we want to do is eliminate x for the integral, thus do the integral by parts.
$$
\begin{align}
\int (x^3+1)\cos (2x))\mathrm dx
&=\frac{1}{2}(x^3+1)\sin2x - \int 3x^2\sin 2x\mathrm dx\\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{2}\int 3x\cos2x\mathrm dx\\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify: $\sin \frac{2\pi}{n} +\sin \frac{4\pi}{n} +\ldots +\sin \frac{2\pi(n-1)}{n}$. Can you help me solve this problem?
Simplify: $\sin \dfrac{2\pi}{n} +\sin \dfrac{4\pi}{n} +\ldots +\sin \dfrac{2\pi(n-1)}{n}$.
| Take the terms in opposite pairs, and note the change of sign,
$$\sin \dfrac{2k\pi}{n}+\sin \dfrac{2\pi(n-k)}{n}=\sin \dfrac{2k\pi}{n} +\sin(2\pi-\dfrac{2k\pi}{n})=\sin \dfrac{2k\pi}{n} -\sin\dfrac{2k\pi}{n}=0.$$
In case that $n$ is even, the central term remains, but $$\sin \dfrac{2n\pi}{2n}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Drawing a Right Triangle With Legs Not Parallel to x/y Axes? I have been presented with an interesting problem. How can I decide whether a right triangle with given side lengths can be placed (with integer coordinate vertices) on a Cartesian plane so that the legs are not parallel to the x or y axes?
I'm a little stump... | Here's an example of a right triangle with integer coordinates at every vertex
and legs that are not parallel to either coordinate axis:
The vertices of this triangle are $A=(6,4),$ $B=(-6,9),$ and $C=(0,0).$
Now to find out whether a right triangle with given side lengths
can be placed with all three vertices at inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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A problem on positive semi-definite quadratic forms/matrices Suppose $a+b+c=0$ and (without loss of generality) $a\leq b\leq 0\leq c$, $a^2+b^2+c^2=1$, is the following quadratic form positive semi-definite? Thank you very much.
\begin{equation*}
\begin{split}
I(x,y,z)=&(2a^2+5b^2+5c^2)a^2x^2+(5a^2+2b^2+5c^2)b^2y^2\\
+... | A sum-of-squares decomposition of the polynomial is possible (the constraints are not needed).
Here is an implementation in the MATLAB Toolbox YALMIP (developed by me). YALMIP (together with a semidefinite programming solver) easily derives a decomposition $I(a,b,c,x,y,z) = v(a,b,c,x,y,z)^TQv(a,b,c,x,y,z)$ where $Q$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that inequality holds How would you show that the following inequality holds? Could you please write your reasoning by solving this problem too?
$a^2 + b^2 + c^2 \ge ab + bc + ca$ for all positive integers a, b, c
I have tried:
$a^2 + b^2 + c^2 - ab - ab + ab \ge bc + ca $
$(a^2 -2ab + b^2) + c^2 + ab \ge bc + ca... | Hint: Consider $(a-b)^2+(b-c)^2+(c-a)^2$.
Remark: Using the above hint, you can show that the desired inequality holds for all real numbers $a$, $b$, and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Basic Math, exponents and algebra I have the equation
$$\frac{x_1^{-\frac{1}{2}}}{{x_2^{-\frac{1}{2}}}} = p_l/p_2$$
How do I get $x_2$ on its own?
I have
$$x_2^{-\frac{1}{2}} = \frac{p_2(x_1^{-\frac{1}{2}})}{p_1}$$
And if you have a useful link that reviews this info, it would be highly appreciated.
| By definition, if $x \neq 0$, then
$$x^{-n} = \frac{1}{x^n}$$
Thus,
$$x_1^{-\frac{1}{2}} = \frac{1}{x_1^{\frac{1}{2}}}$$
If we make the substitution $n = -m$ in the equation
$$x^{-n} = \frac{1}{x^n}$$
we obtain
$$x^{m} = \frac{1}{x^{-m}}$$
Hence,
$$\frac{1}{x_2^{-\frac{1}{2}}} = x_2^{\frac{1}{2}}$$
Therefore, we c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
For the series $S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2$...... Problem :
For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series.
We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1... | You don't need to bother with calculating $S_n - S_{n-1}$ here as the terms are explicitly given in the summation.
Here $T_n = \frac{(1+2+...+n)^2}{(1+3+5+...+(2n-1))}$
The numerator is the square of the first $n$ integers, so is equal to $(\frac{1}{2}n(n+1))^2$.
The denominator is the sum of the first $n$ odd integers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I u... | \begin{align}
\int_0^{\Large\frac{\pi}{4}} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx&=\int_0^1\frac{\sqrt{y(1-y)}}{1+y^2}\,dy\quad\Rightarrow\quad y=\tan x\\
&=\int_0^\infty\frac{\sqrt{t}}{(1+t)(1+2t+2t^2)}\,dt\quad\Rightarrow\quad t=\frac{y}{1-y}\\
&=\int_0^\infty\frac{2z^2}{(1+z^2)(1+2z^2+2z^4)}\,dz\quad\Rightarrow\quad z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 5
} |
How to Change Summation Expression $\sum_{i=1}^N \mathbf{X}_i^{\top}\mathbf{\Omega}^{-1}\mathbf{X}_i$ into Matrix Expression Let $\mathbf{X}_i$ be a $G \times K$ matrix, and suppose are $i=1,...,N$ of these matrices. Note that
\begin{align}
\sum_{i=1}^N \mathbf{X}_i^{\top}\mathbf{X}_i
&=
\begin{bmatrix}
\mathbf{X}_1^{\... | Let's try this.
\begin{align}
\sum_{i=1}^N \mathbf{X}_i^{\top}\mathbf{\Omega}^{-1}\mathbf{X}_i&=\mathbf{X}_1\mathbf{\Omega}^{-1}\mathbf{X}_1+\mathbf{X}_2\mathbf{\Omega}^{-1}\mathbf{X}_2+\cdots+\mathbf{X}_N\mathbf{\Omega}^{-1}\mathbf{X}_N \\
&=\begin{bmatrix}
\mathbf{X}_1^{\top} & \mathbf{X}_2^{\top} & \cdots & \mathbf{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+... | $$1+4+7+…+(3n-2) = \frac{n}{2}(3n-1)$$
Given identity is true For $n=1$
$$\frac{1\cdot(3-1)}{2}=1\tag{1}$$
Assume it's true for some integer $k$
$$1+4+7+…+(3k-2) = \frac{k}{2}(3k-1)\tag{2}$$
for $k+1$ .
We have to show that
$$1+4+7+…+(3k-2)+(3(k+1)-2) = \frac{k+1}{2}(3(k+1)-1)$$
$$1+4+7+…+(3k-2)+(3k+1) = \frac{k+1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the bounds on a triple integral Problem: Find the volume enclosed by the cone $$x^2 + y^2 = z^2$$ and the plane $$2z - y -2 = 0$$
So I know that I need to do a triple integral over this region, and the integrand will be 1.
My problem is with finding the bounds for the integral. I set the $z$s equal to each oth... | The intersection is indeed an ellipse, but you seem to have made some algebra mistakes:
\begin{align*}
x^2 + y^2 &= (\tfrac{y + 2}{2})^2 \\
x^2 + y^2 &= \tfrac{1}{4}y^2 + y + 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y) &= 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y + \tfrac{4}{9} - \tfrac{4}{9}) &= 1 \\
x^2 + \tfrac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$? Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$?
| To compute this series let's rewrite it first as a polygamma function :
\begin{align}
\tag{1}S&=\sum_{k=0}^{\infty }\frac{1}{(4k+3)^3}\\
S&=\frac 1{4^3}\sum_{k=0}^{\infty }\frac{1}{\left(k+\frac 34\right)^3}\\
\tag{2}S&=-\frac 1{2!\,4^3}\psi^{(2)}\left(\frac 34\right)\\
\end{align}
Since (from the previous link) $$\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
prove that $a^3 - b(b+1) = (a-2)(c^2 + 1) + 2$ has infinitely many solutions How do I prove that $$a^3 - b(b+1) = (a-2)(c^2 + 1) + 2$$ has infinitely many solutions if a, b and c are natural numbers?
I have opening the brackets and moving all the terms to one side which gets rid of the constant 2. I have also tried sub... | Ad-hoc solution: note that when $a=2$, $c$ doesn't matter. Solving
$$
a^3-b(b+1)=2
$$
for $b$ (given that $a=2$) gives us $b=2$ as well. Then, any triple $(2,2,c)$ $(c\in\mathbb{N})$ is a solution. There are infinitely many such triples so the claim follows.
Better solution:
solving for $c$ gives you
$$
c=\sqrt{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How integrate $ \iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0I'm trying to resolve this integral
$$
\iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0<a<b
$$
I tried with polar coordinates:
$$
x = r\cos{\theta} \\
y = r\sin{\theta} \\
Ja... | if you are integrating over an annulus, then by symmetry
$$
\iint_{D} (\frac{x^2}{x^2+y^2})dA = \iint_{D} (\frac{y^2}{x^2+y^2})dA
$$
so by adding you have
$$
\iint_{D} (\frac{x^2+y^2}{x^2+y^2})dA = \iint_{D} dA = \pi(b^2-a^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I know if two vectors with $n$ components are parallel? How do I know if two vectors with $n$ components are parallel?
For example
$$\begin{pmatrix}5\\2\\1\\3\\4\end{pmatrix} \text{, and } \begin{pmatrix}4\\1\\2\\3\\6\end{pmatrix}.$$
| Put them together like: $$\begin{pmatrix} 5 & 4 \\ 2 & 1 \\ 1 & 2 \\ 3 & 3 \\ 4 & 6\end{pmatrix},$$
or in rows. If one of the $2 \times 2$ determinants there is non-zero, they're not parallel. They are not parallel because, for example, $\left|\begin{matrix} 5 & 4 \\ 2 & 1\end{matrix}\right| = -3 \neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the volume of the solid generated by revolving the region bounded by $y=x$ and $y=x^2$ about the line $y=x$ Find the volume of the solid generated by revolving the region bounded by $y=x$ and $y=x^2$ about the line $y=x$
I am confused, how do we approach such problems, where the rotation lines are not vertical/hor... | One way to do this is to rotate everything counter-clockwise about the angle $\theta = \pi/4$, so the line $y=x$ would line up with the $y$-axis
We can parametrize the curves:
$$y = x \rightarrow x_1 = t, \quad y_1 = t $$
$$y = x^2 \rightarrow x_2 = t, \quad y_2 = t^2 $$
Applying the rotation gives new coordinates
$$ x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
derivative of $y=x^3\sqrt{x}-\frac{1}{x^2\sqrt{x}}$ $y=x^3\sqrt{x}-\dfrac{1}{x^2\sqrt{x}}$
Both terms require the product rule, right? My try:
$x^3\dfrac{1}{2}x^{-1/2}+3x^2x^{1/2}-\dfrac{-1}{2}x^{-3/2}x^{-2}--2x^{-3}x^{-1/2}$
What am I doing wrong? The correct answer is: $y\;'=3.5x^2\sqrt{x}+\dfrac{2.5}{x^3\sqrt... | $$y=x^3x^{1/2} -x^{-2}x^{-1/2}$$
$$y'=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} -\left( -2 x^{-3}x^{-1/2} -\frac{1}{2}x^{-2}x^{-3/2}\right)$$
$$=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} +2 x^{-3}x^{-1/2} +\frac{1}{2}x^{-2}x^{-3/2}$$
$$=3x^{5/2} + \frac{1}{2}x^{5/2} + 2 x^{-7/2} + \frac{1}{2}x^{-7/2},$$
which matches the correct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}... | Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that
$$\sum_{cyc}\frac{\frac{x}{y}}{\sqrt{1+\frac{x}{y}}}\geq\frac{3}{\sqrt2}$$ or
$$\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\geq\frac{3}{\sqrt2}.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 0
} |
Proving Identity of Combination The lecturer had given two questions of proving that are
$$\binom{r}{r}+\binom{r+1}{r}+...+\binom{n}{r}=\binom{n+1}{r+1}\text{for }n\geq{r}\geq{1} $$
$$\binom{r}{0}+\binom{r+1}{1}+...+\binom{r+k}{k}=\binom{r+k+1}{k}\text{for }r,k\geq{1}$$
I tried to use the induction to prove these two... | $\bf{My\; Solution::}$ Given
$\displaystyle \binom{r}{0}+\binom{r+1}{1}+\binom{r+2}{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot +\binom{r+k}{k} = \binom{r+k+1}{k}\;,$ where $r,k\geq 1$
Now Using The formula of $\displaystyle \binom{n}{r} = \binom{n}{n-1}$ on $\bf{L.H.S\;}$
We can Write $\bf{L.H.S}$ as
$\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove $n!\leq(\frac{n+1}{2})^n$ Prove that for $n\in\mathbb{N}$ $$n!\leq(\frac{n+1}{2})^n.$$
I've solved base case for $n=1$
$$1\leq(\frac{1+1}{2})^1=1$$
The second step I've made was that I assumed that $n!\leq(\frac{n+1}{2})^n$ And then I have $(\frac{n+2}{2})^{n+1}$. What do I need to do now?
| The induction proof can be made to work; use the fact that
$$\begin{align*}
\frac{\left(\frac{n+2}2\right)^{n+1}}{\left(\frac{n+1}2\right)^n}&=\frac{n+2}2\left(\frac{n+2}{n+1}\right)^n\\
&=\frac{n+2}2\left(1+\frac1{n+1}\right)^n\\\\
&>\frac{n+2}2\left(1+\frac{n}{n+1}\right)\\\\
&=\frac{(n+2)(2n+1)}{2n+2}\\\\
&=\frac{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int \cos(x^2 + x)dx$ I need to evaluate following integral,
$$\int \left( x + \frac{1}{2} \right) \cos(x^2 + x)\,dx$$
but this one has got me pretty stumped! Any suggestions would be appreciated. Thanks!
| Complete the square $$x^2+x=(x+\frac12)^2-\frac14$$ and change variable $y=x+\frac12$. So, $$\cos(x^2 + x)=\cos(y^2-\frac14)=\cos(y^2)\cos(\frac14)+\sin(y^2)\sin(\frac14)$$ and then $$I=\int cos(x^2 + x)dx=\cos(\frac14)\int \cos(y^2)dy+\sin(\frac14)\int \sin(y^2)dy$$where you recognized Fresnel integrals $$\int \cos(y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Can two pythagoras triplet have a common number If I have a pythagoras triplet $(a,b,c)$ such that
$$a^2+b^2=c^2$$
then is there another triplet $(a,d,e)$ possible such that
$$a^2+d^2=e^2, \; b\neq d$$
| Each of $\quad A,B,C\quad$ will have as many primitive triples as its number of unique prime factors.
Triples are often generated with Euclid's formula
$$\text{where} \quad A=m^2-k^2 \quad B=2mk \quad C=m^2+k^2$$
$A=3\times5=15\longrightarrow
f(4,1)=(15,8,17)\quad
f(8,7)=(15,112,113)$
$B=2^2\times3=12\longrightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving Trigonometric Equation Problem im kinda stuck to solve the following problems below
Problem: 4cos²2x+sin2x=3 (0 < x <= π)
Steps:
4cos²2x+sin2x=3
2+2cos4x+sin2x=3
2cos4x+sin2x=1
2(1-2sin²2x)+sinx=1
2-4sin²2x+sinx=1
Bring every term to the right side
0= 1-2+4sin²2x-sinx
4sin²2x-sinx-1=0
Let M= sin2x, sin
4m²-m-... | You have:
$$4\cos^2(2x)+\sin(2x) = 3$$
Use $\cos^2(2x) = 1-\sin^2(2x)$ then:
$$4(1-\sin^2(2x))+\sin(2x) = 3\implies4-4\sin^2(2x) + \sin(2x) = 3\implies\\1+\sin(2x)-4\sin^2(2x) = 0$$
which has roots:
$$\sin(2x) = \frac{1}{8}(1-\sqrt{17})\\\sin(2x) = \frac{1}{8}(1+\sqrt{17})$$
For the first equations:
$$\sin(2x) = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of $\left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- 1/ {n^4}} {4+ n^7/ {3^n}}$ as $n$ tends to infinity I already took some steps to simplifying the original question and im stuck at this point:
$$ \lim_{n \to \infty} \left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- \frac {1} {n^4}} {4+ \frac {n^7} {3^... | I think you can get this by doing a bit of algebra. Since $1-\frac{1}{n^4}<4+\frac{n^7}{3^n}$ for all $n \in \Bbb{N}$ then $$ \left(\frac{2}{3}\right)^n \cdot n^4 \cdot \frac{1- \frac{1}{n^4}} {4+ \frac{n^7}{3^n}}< \left(\frac{2}{3}\right)^n \cdot n^4$$ Now let's use the fact that exponents grow faster than powers. We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$168... | Since $41^2 - 40^2 = (41-40)(41+40) = 81 = 9^2$, there exists an angle $\phi$ satisfying $$\sin \phi = \frac{40}{41}, \quad \cos \phi = \frac{9}{41},$$ hence $$\begin{align*} 1 &= \frac{9}{41} \sin \theta + \frac{40}{41} \cos \theta \\ &= \sin\theta \cos\phi + \cos\theta \sin\phi \\ &= \sin(\phi+\theta),\end{align*}$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Proving $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\ldots\left(1+\frac{1}{n^3}\right)<3$ for all positive integers $n$
Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)<3$ for all positive integers $n$
This problem is copied from Math Olympiad Tr... | If we need to prove the strong inequality and proving the weak inequalities for the simplest to imagine values is easy...
\begin{align*}
\left(3-\frac1n\right)\left(1+\frac1{(n+1)^3}\right)
&=
3+\frac3{(n+1)^3}-\frac1n-\frac1{n(n+1)^3}\\
&=3+\frac{3n-(n+1)^3-1}{n(n+1)^3}\\
&=3+\frac{-n^3-3n^2-2}{n(n+1)^3}\\
&=3+\frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 2
} |
Completing Cayley table for a group The task is to complete the following Cayley table
for a given group. $e$ is of course the identity element. Together with group axioms and the fact that every Cayley table of a group must be a latin square, I arrived at
Is it correct? I know there is only one possible table and ... | I agree with your answer.
Completing the Cayley table of order $6$ (if $pq=e$, it must be $qp=e$, etc.), you get a non commutative ($pr\ne rp$) magma, with identity element $e$.
Each element of it has unique inverse (each one is inverse of itself, but $p$ and $q$ mutually inverses).
Finally, if you relabel $e,p,q,r,s,t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
probability: continuous uniform distribution mean by symmetry I am trying to show that the mean of the uniform continuous distribution is $(b+a)/2$ by symmetry. The direct method is fairly simple but, for some reason, I cant get this one.
\begin{align}
E[X] &= \int_{-\infty}^{\infty}xp_X(x)dx\\
&= \int_{-\infty}^{\frac... | Sorry, I didn't read your question carefully enough.
The limit on the second of your integrals after doing the substitutions is wrong. Since $u=-x + (b+a)/2$ and $x$ runs from $(b+a)/2$ to $\infty$, you should have $u$ running from $0$ to $-\infty$. I think this is what has mislead you. The two integrals evaluate to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $\sin A+\sin B =a,\cos A+\cos B=b$, find $\cos(A+B),\cos(A-B),\sin(A+B)$ If $\sin A+\sin B =a,\cos A+\cos B=b$,
*
*find $\cos(A+B),\cos(A-B),\sin(A+B)$
*Prove that $\tan A+\tan B= 8ab/((a^2+b^2)^2-4a^2)$
| $$\begin{align*} \tan A + \tan B &= \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} \\
&= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} \\
&= \frac{\sin(A+B)}{\frac{\cos(A+B) + \cos(A-B)}{2}} \\
&= \frac{2\sin(A+B)}{\cos(A+B) + \cos(A-B)} \end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Give a combinatorial argument Give a combinatorial argument to show that $$\binom{6}{1} + 2 \binom{6}{2} + 3\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} + 6 \binom{6}{6} = 6\cdot2^5$$
Not quite where to starting proving this one. Thanks!
| $$2^5$$ is just a way to count the elements in a 5 member set. (For each member of the set you make a binary decision to include it or not)
Now let's say you have a six element set.
Let's look at $$ \binom{6}{1} $$
Fix any member of the 6 element set. Now you have 5 members left. So that is $$ \binom{5}{1} $$ But you c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Use limit definition to find derivative of $x+\sqrt x$ The function is $f(x) = x + \sqrt x$.
How would you use the limit definition of the derivative to find the derivative of that equation?
| We have
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
Let $$f(x) = x + \sqrt x$$
$$\begin{align}f'(x)&=\lim_{h\to0}\frac{(x+h + \sqrt {x+h})-(x + \sqrt x)}{h}\\
&=\lim_{h\to0}\frac{h + \sqrt {x+h} - \sqrt x}{h}\\
&=\lim_{h\to0}1+\frac{ \sqrt {x+h} - \sqrt x}{h}\\
&=\lim_{h\to0}1+\frac{ \sqrt {x+h} - \sqrt x}{h}\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find maximum/minimum for $\cos(2x) + \cos(y) + \cos(2x+y) $ I have not been able to find the critical points for $\cos(2x) + \cos(y) + \cos(2x+y) $
| When you set the partial derivatives equal to $0$, you obtain the two equations
\begin{align*}
\sin 2x + \sin (2x+y) &= 0 \\
\sin y + \sin (2x+y) &= 0\,.
\end{align*}
This means that we must have $\sin 2x = \sin y$, and so either $y=x+(2k)\pi$ or $y=(2k+1)\pi-x$ for some integer $k$. Substituting, for example, $y=2x$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified? I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?
| Hint: \begin{align}64x^4 + 16 x^2 + 4 &= (8x^2-2)^2 + 32x^2 + 16x^2
\\&= (8x^2-2)^2 + 48x^2
\\&= (8x^2-2)^2 - (4i\sqrt 3 x)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Volume of the solid cut by a plane.
I'd like to find the volume of the following solid.
The solid enclosed by the paraboloid $z=4-x^2-y^2$ and the plane $x+y+z=1$.
Actually original problem is the following (I made upper problem...)
The solid enclosed by the cylinder $x^2+y^2=1$ and the plane $x+y+z=1$ and $z=-5$
In ... | $$\begin{align}
z & = 4 - x^2 - y^2\\
1 - x - y & = 4 - x^2 - y^2\\
0 & = 3 - x^2 + x - y^2 - y\\
0 & = 3 - x^2 + x - \frac14 + \frac14 - y^2 + y - \frac14 + \frac14\\
0 & = 3 - (x - \frac12)^2 + \frac14 - (y - \frac12)^2 + \frac14\\
\frac72 & = (x - \frac12)^2 + (y - \frac12)^2
\end{align}$$
So the region of intersect... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x... | You can use l'Hospital as many times as needed as long as the indeterminate forms conditions are fulfilled. In this case, using Taylor series can be helpful, too:
$$\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \ldots = x - \frac{x^3}6 + \mathcal O(x^5)$$
$$\implies \frac{\sin x - x + \frac{x^3}6}{x^3} = \frac{\mathcal ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Manipulating roots of a cubic Given that $A,B,C$ are the roots of the equation $x^3-5x^2+x+1$, how do I find the value of $$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B}$$ I know the Vieta's formulas but I am not able to manipulate the above expression into something known. And taking the LCM doesn't help. Please help m... | Hint: $(A+B)(B+C)(C+A) = (AB+AC+B^2+BC)(C+A) = ABC+A^2B+AC^2+A^2C+B^2C+B^2A+BC^2+BCA = 2ABC + A^2(B+C) + B^2(C+A) + C^2(A+B) = 2\cdot 1 + A^2(5-A) + B^2(5-B) + C^2(5-C) = 2 + 5(A^2+B^2+C^2) - (A^3+B^3+C^3) = 2 + (A+B+C) + 3 = 2+5+3=10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
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