Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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General Inequality Problem. The question goes like this :
Let $a,b,c,d$ be positive reals and given that $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$
My approach goes like this:
I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$
Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$
Although ... | By Chebyshev's inequality,
$\frac{a^2+b^2+c^2+d^2}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a+b+c+d}{4}$
$\implies a^2+b^2+c^2+d^2 \geq \frac{1}{4}$
Similarly,
$\frac{a^3+b^3+c^3+d^3}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a^2+b^2+c^2+d^2}{4}$
$\implies 4(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2$
$6(a^3+b^3+c^3+d^3) \geq a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution? The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:
$$\int\frac{dx... | $$
\begin{aligned}
\int \frac{d x}{\sqrt{x^{2}+4}} &\stackrel{y=\frac{1}{x} }{=}-\int \frac{d y}{y \sqrt{1+4 y^{2}}} \\
&=-\frac{1}{4} \int \frac{d (\sqrt{1+4 y^{2}})}{y^{2}} \\
&=-\int \frac{d (\sqrt{1+4 y^{2}})}{\left(\sqrt{1+4 y^{2}}\right)^{2}-1} \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{1+4 y^{2}}+1}{\sqrt{1+4 y^{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Limiting Sum including Stirling numbers I would like to simplify the following summation including $S(m,n)$, Stirling numbers of the Second kind.
$\begin{gather*}\\
&&
\frac{1}{n^{n+1}}\sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,n-k\right)
\left( n-k\right) !k\left( 1-\left( \frac{n-2}{n-1}\right) ^{k}\right)
&&
\end{gathe... | We get for the sum term
$$\sum_{k=1}^{n-1} {n\choose k} {n\brace n-k} (n-k)!
\times k x^k
\\ = n \sum_{k=1}^n
{n-1\choose k-1} {n\brace n-k} (n-k)! \times x^k
\\ = n x \sum_{k=0}^{n-1}
{n-1\choose k} {n\brace n-1-k} (n-1-k)! \times x^k
\\ = n x \times n! [z^n] \sum_{k=0}^{n-1}
{n-1\choose k} (\exp(z)-1)^{n-1-k} x^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Error with Differential Forms I'm learning a little bit about differential forms in my Complex Analysis course, and I'm not sure if I'm making an error in what follows. I wonder if anyone could offer any insight into this. Thank you.
So, the question is to show that $\alpha = \frac{xdy - ydx}{x^2 + y^2}$ is closed. To ... | If $\phi$ is a $p$-form and $\psi$ is a $q$-form, then $d(\phi\wedge\psi) = d\phi\wedge\psi + (-1)^p\phi\wedge d\psi$. In the second case, $f$ is a one-form and $g$ is a $0$-form, so $d\alpha = d(f\wedge g) = df\wedge g + (-1)^1f\wedge dg = df\wedge g \color{red}{-} f\wedge dg$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Angle bisected in a parallelogram
$ABCD$ is a parallelogram and $ID =JB$.
the point $P$ is the intersection of $DJ$ and $IB$.
Prove that the angle $X=Y$
$a,b,α,β$ are angles.
What I realized:
$a=b$.
and if we sum up these angles.
$$180=180-(x+α)+180-(y+β)+a $$
$$\iff 180=x+y+α+β-a \iff 360+2a=2(x+y)+2(α+β)$$
Please He... | It is straightforward by sine law,
$\angle PID = 180^0 - \beta, \angle PJB = 180^0 - \alpha, \angle DPI = \angle BPJ = \gamma$
So in $\triangle PID, \frac{\sin \beta}{PD} = \frac{\sin \gamma}{ID}$
Similarly in $\triangle PBJ, \frac{\sin \alpha}{PB} = \frac{\sin \gamma}{BJ}$
As $BJ = ID$, $ PD \sin \alpha = PB \sin \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inequality Induction Proof, how should I proceed? I have been trying to prove inequalities using induction to no avail.
For example,
Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$
Base Case:
$\frac{1}{2^{1}} = 2$
$\frac{1}{2}<2\;$ Which is true.
We assume, for n=k, that
$\... | You may first prove the following equality:
Prove that $\frac 1 {2^1} + \frac 2 {2^2} + \dots + \frac k {2^k} = 2 - \frac {k + 2} {2^k}$.
This can be proved by induction on $k$ (easy exercise).
Once it is proved, the original inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$ Assume that all logarithms are natural.
Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\rig... | Here's a proof
by comparing the power series.
Eliding some computation,
$\log\left( \frac{1+t}{1-t}\right)
=2\sum_{n=1}^{\infty} \dfrac{t^{2n-1}}{2n-1}
$.
$\sqrt{1+t}-\sqrt{1-t}
=2\sum_{n=1}^{\infty} \binom{1/2}{2n-1}t^{2n-1}
=2\sum_{n=1}^{\infty} \dfrac{1}{2^{4n-3}(2n-1)}\binom{4n-4}{2n-2}t^{2n-1}
$
So we want
$\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $
$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$
I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.
| We are given $I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)}$
Here we have $7x-10-x^2=(x-2)(5-x)$,
therefore put $\sqrt{7x-10-x^2} =(x-2)t$
on solving and factoring,
we get $x=\frac{\left(2t^2+5\right)}{\left(t^2+1\right)}\ $ therefore
$dx=\frac{-6t}{(t^2+1)^2}$ and $x-2=$ $\frac{(2t^2+5)}{(t^2+1)}-2 =$$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$ Question:
$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $
Answer key:
$\implies \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \tex... | $$(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$$
Let $x = \cos \theta$ and $y = \sin \theta$. Then we get
\begin{align}
(y - x)(2+xy) &\leq 2 \\
x^2 + y^2 &= 1
\end{align}
We can now argue
\begin{align}
-(y-x)^2 &= -1 + 2xy \\
5 - (y-x)^2 &= 2(2 + xy) \\
\hline
(y-x)(5 - (y-x)^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
Equality involving cosine and binomial coefficients - check if true for every $n$ While doing some computations with complex numbers I think I proved that
$$(\cos{ \frac{n \cdot \pi}{4} )\cdot 2^{n/2}} = \binom{n}{0} - \binom{n}{2} + \binom{n}{4} - \binom{n}{6} + ... $$
How? Well, these are two different forms of
$$\fr... | Without complex numbers, by induction, assume that for some integer $n\ge 0$,
$$\begin{align*}
2^{n/2}\cos\frac{n\pi}{4} = \binom{n}{0}-\binom n2+\binom{n}{4}-\binom n6+\cdots\\
2^{n/2}\sin\frac{n\pi}{4} = \binom{n}{1}-\binom n3+\binom{n}{5}-\binom n7+\cdots\\
\end{align*}$$
Then by compound angle formulae,
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Question about simplifying $(a+b+c)(ab+bc+ac)=abc$ To simplify this expression, according to math110's answer, we can write LHS as:
$$(a+b+c)(ab+bc+ac)=(a+b)(b+c)(a+c)+abc$$
then $abc$ cancels and we get: $(a+b)(b+c)(a+c)=0$
But my question is, how we can recognize that we have this equality? I have no clue about it. i... | The original equation is
$$\frac1a + \frac1b+ \frac1c = \frac1{a+b+c}$$
We see that $a=-b, b=-c, c=-a$ are trivial solutions to the equation. Thus
$$abc(a+b+c)\left(\frac1a + \frac1b+ \frac1c - \frac1{a+b+c}\right)$$
must contain $(a+b)(b+c)(c+a)$ as a factor; indeed it is equal to $(a+b)(b+c)(c+a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The center of the circumcircle lies on a side of a triangle Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.
$$AB=\dfrac{144\sqrt5}{5}$$
By Triangle-Angle-Bisector Theorem $$\dfrac... | $\angle POC = \angle A$, so $\triangle COP \sim \triangle CAB$
Hence $\frac{R}{AB} = \frac{20}{36} = \frac{5}{9}$ ...(i)
$\angle APB = \frac{1}{2} \angle AOB = 90^0 - \angle A$
$AB = 2R \sin \angle APB = 2R \cos A$
So from (i), $\frac{R}{2R \cos A} = \frac{5}{9}$
$\cos A = \frac{9}{10} = 1 - 2 \sin^2{\frac{A}{2}} \impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric system with different coefficients I have a trigonometric system as below and I want to solve it for $x$.
\begin{gather}
A\cos(x)+B\cos(2x)+C\sin(2x) = D \\
-C\cos(2x)-A\sin(x)+B\sin(2x) = F
\end{gather}
Can anyone suggest a solution?
Regards.
| Rewrite the two equations as
$$\sqrt{B^2+C^2}\cos(2x-a) = D-A\cos x $$
$$\sqrt{B^2+C^2}\sin(2x-a) = F+A\sin x $$
with $a = \tan^{-1} \frac CB$. Square and add the two equations
$$B^2+C^2= D^2+F^2 +A^2 +2A(F \sin x - D\cos x)\\
= D^2+F^2 +A^2 +2A\sqrt{D^2+F^2} \sin (x-b)$$
with $b= \tan^{-1} \frac DF$. Now, it should be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding area based on only the sides I came across this simple looking question:
The perimeter of a triangle is $42$ cm. One side of a triangle is $8$ cm longer than the smallest side and the third side is $1$ cm less than $3$ times the smallest side. Find the area of the triangle.
At first it seemed easy enough. The s... | Here is a proof for Heron's formula, as requested.
We begin with the Cosine rule:
$$a^2=b^2+c^2-2bc\cos A\iff\cos A=\frac{b^2+c^2-a^2}{2bc}$$
Recall the identity $$\sin^2x+\cos^2x\equiv 1\iff\sin^2x\equiv1-\cos^2x$$
This means that $$\begin{align}
\sin^2A&=1-\cos^2A\\
&=1-\frac{(b^2+c^2-a^2)^2}{(2bc)^2}\\
&=\frac{(2bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$? This is a multiple choice question, where using calculators is not allowed. Candidates have, on average, two minutes to solve the problem.
PROBLEM:
How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?
CHOI... | The function is periodic with period $2\pi$ and achieves local extrema
$$y' = \cos x - 2\sin 2x = 0 \implies \cos x (1-4\sin x) = 0$$
at
$$\pm\sqrt{1-\cos^2x}+2\cos^2x-1\Bigr|_{\cos x = 0} = 0,-2$$
and
$$\sin x + 1-2\sin^2x\Bigr|_{\sin x = \frac{1}{4}} = \frac{9}{8}$$
which means the only interval we can expect interse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to show that $\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$ How to show that $$\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$$
I know that
$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x\right|... | Easier: if $(x,y) = (r\cos\theta,r\sin\theta)$,
$$
\left|\frac{x^2 y^2}{|x|^3+|y|^3}\right|
= \frac{r^4\cos^2\theta\sin^2\theta}{r^3(|\cos\theta|^3 + |\sin\theta|^3)}\le\frac{r}{m} = \frac{\sqrt{x^2 + y^2}}{m},
$$
with
$$m = \min_{\theta\in[0,2\pi]}(|\cos\theta|^3 + |\sin\theta|^3) > 0.$$
(Why $> 0$)?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the expectation and variance of X A pdf is defined as
\begin{equation}
f(x)=
\begin{cases}
C(x+\frac{3}{2}),\quad0<x<2\\
0,\quad\quad\quad\quad\text{otherwise}
\end{cases}
\end{equation}
*
*Find the value of C.
*Find the expectation and variance of X.
*Find the expectation of random variable $Z=\frac{X}{2X+3}$... | points 1 is ok
the expectation is ok
As variance is concerned
$$E[X^2]=\int_0^2 x^2f(x)dx=\frac{8}{5}$$
thus
$$V[X]=\frac{8}{5}-\left(\frac{17}{15}\right)^2=\frac{71}{225}$$
point 3, easy but incorrect
$$E[Z]=\int_0^2 \frac{x}{2x+3}f(x)dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is there a mistake to the answer given to this problem: Find the matrix of the reflection through the line y=−2x/3. This is the answer given:
I think this answer is not correct because the line given by y=-2x/3 makes an angle that is below the x axis, so the order is incorrect right? We need to start with a countercl... | Here the slope of the line = $m = tan \gamma = -\frac{2}{3}$.
$\therefore sin \gamma = \frac{m}{\sqrt{1+m^2}},\; cos \gamma = \frac{1}{\sqrt{1+m^2}}$.
The transformation matrix for reflection around the line is given by the product of 3 matrices (rotate by angle -$\gamma$ to make it horizontal, reflect w.r.t the horizo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that the least upper bound of the sequence $\left\{ \frac{2^n - 1}{2^{n-1}} \right\}$ is $2$ It can be easily verified that the sequence $\left\{ \dfrac{2^n -1} {2^{n-1}} \right\}$ is strictly positive, the task now is to prove that 2 is an upper bound.
Proof:
Suppose for the sake of contradiction that 2 is not t... | Your answer is okay but it's ludicrously complicated.
Why not just say $\frac {2^n -1}{2^{n-1}} < \frac {2^n}{2^{n-1}} = 2$?
That's it. You are done.
But is that all of the question?
Are you sure you weren't also being as to prove that $2$ is the least upper bound.
In that case if $x < 2$ you must prove there exists a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4042388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$ Evaluate the following expression:$$\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$$
I tried combining the whole into a single fraction and using double-angle identity, product/sum to sum/product, but it didn... | Per the identity
$\sin3t =3\sin t -4\sin^3t$,
it is straightforward to verify that $\sin20$, $\sin40$, $-\sin80$ are the roots of
$4x^3-3x+\frac{\sqrt3}2=0$, which leads to
$$\sin20\sin40\sin80=\frac{\sqrt3}8$$
Then
\begin{align}
& \frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ\\
=& \frac{64}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find $ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta$ Evaluate $$ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left[\frac{1+2r^2}{1-r^2\cos\left(2\theta\right)}\right]^{1/3}{\rm d}\theta $$
Question- Can I take the limit inside the integral?
My try-
$$I= \lim_{r \to 1^{-}} \int_{-\pi}^{\... | This is not an unswer, just a hint and too long for a comment: The integral above may be transformed with $\theta = \frac{\arccos(y)}{2}$
$$2\, \left(1+2\, r^2\right)^{1/3} \int_{-1}^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dy$$
Next we can split the integral in two parts with $p=-y$ in the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove thi... | $$a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}\implies \log(a_n)=\log \left(\Gamma \left(\frac{n}{2}+\frac{1}{4}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}+\frac{3}{4}\right)\right)$$ Now, using Stirling approximation
$$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \log \left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Calculate $2 C_{n}^{k-1} + 3 C_{n}^{k-2} + 4 C_{n}^{k-3} + ... + (k + 1)C_{n}^{0}$ Are there any easy ways to calculate this?
$$ \sum_{i=1}^{k} (i + 1) \cdot C_{2k}^{k-i} = 2 C_{2k}^{k-1} + 3 C_{2k}^{k-2} + 4 C_{2k}^{k-3} + ... + (k + 1) C_{2k}^{0} $$
I tried to "turn" the expression:
$$ \sum_{i=1}^{k} (i + 1) \cdot C_... | Consider the formula provided by this answer:
$\sum_{0\leq j\leq n/2}\binom{n}{j}x^j = (x+1)^n - \binom{n}{\left\lfloor\frac{n}{2}\right\rfloor + 1} x^{\left\lfloor\frac{n}{2}\right\rfloor + 1} {_2F_1}\left(1,\left\lfloor\frac{n}{2}\right\rfloor - n + 1;\left\lfloor \frac{n}{2}\right\rfloor + 2;-x\right),$
where $_2F_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$
Suppose $p$ is an odd prime. Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$, where $1$ is the identity.
I know that the conjugate of power is the power of conjugate, so $yx^{k}y^{-1} = (x^{p+1})^{k}$, and used thi... | $yx^k = x^{k(p+1)}y$, so $y^a x^k = x^{k(p+1)^a} y^a$, so:
$$\begin{array}{rcl}
(yx^{-1})^p
&=& y x^{-1} y x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)}y^2 x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)-(p+1)^2} y^3 x^{-1} \cdots yx^{-1} \\
&=& \cdots \\
&=& x^{-(p+1)-(p+1)^2-\cdots-(p+1)^{p-1}} y^p \\
&=& x^{-(p+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplifying the sum of a product of multinomial coefficients From the multinomial theorem the following holds
$$
\sum_{k_1 + k_2 + \ldots + k_m=n} {n \choose k_1, k_2, \ldots, k_m} = m^n
$$
I have the following sum
$$
\sum_{\substack{k_1 + k_2 + \ldots + k_m &=& B \\ g_1 + g_2 + \ldots + g_m &=& W}} {B \choose k_1, ... | We assume positive integers $m,n\geq 2$. Since $m\mid n$ we know that $n=Nm$ is a multiple of $m$ with $N\geq 1$. This way we can write $W=Nm-B$ and $x=N$. We show by induction of $m$ that for all $N\geq 1, m\leq B\leq Nm$ the following is valid:
\begin{align*}
\color{blue}{\sum_{\substack{k_1+\cdots+k_m=B\\k_1,\ldots,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4048704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to so... | The question asks how to prove a polynomial
$\,f(n)\,$ takes only integer values for any
integer $\,n\,.$ If the polynomial is of
degree $0$, then it is a constant and if
that constant is an integer we are done.
If the polynomial is of degree $1$, then if
any two consecutive values,
$\,f(k), f(k+1)\,$ are integers,
the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{... | To determine the remainder of some computation, $C, \pmod{70}$, simply determine the remainder of $C, \pmod{2}, \pmod{5}$, and $\pmod{7}.$ Then employ Number Theory methods to solve the resultant congruencies.
Let $C = 16^3 + 17^3 + 18^3 + 19^3.$
Since the second and fourth terms of $C$ (only) are odd, it is immediate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\binom{80}{40}\bmod 2000$
Find $\displaystyle\binom{80}{40}\bmod 2000$.
So far, I've found that $\displaystyle\binom{80}{40}$ is divisible by $2^2$ and $5^1$, so the answer isn't $0$. Usually, with smaller numbers, I would split the $\bmod 2000$ into $2^4$ and $5^3$, find the answer through brute force for each... | Since $${80 \choose 40} = \frac{80\cdot 79 \cdot 78 \cdot 77 \cdot ...\cdot 41}{40 \cdot 39 \cdot 38 \cdot 37 \cdot ... \cdot 1}$$ we can cancel the terms in the denominator with terms in the numerator leaving a factor of $2$, so $2^{40}$ is a factor of the answer. It only remains to see how many factors of 5 remain. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given area of parallelogram, find smallest distance from origin to furthest point (complex numbers)
On the complex plane, the parallelogram formed by the points 0, $z,$
$\frac{1}{z},$ and $z + \frac{1}{z}$ has area $\frac{35}{37}.$ If the
real part of $z$ is positive, let $d$ be the smallest possible value
of $\left| ... | Let $a$ and $b$ two $2$-dimensional vectors satisfying $||a||\cdot||b||=1$. Then (via Gram) the square of area of the parallelogram spanned by these vectors is $(35/37)^2=||a||^2\cdot||b||^2-\langle a,b\rangle^2=1-\langle a,b\rangle^2$ from where $\langle a,b\rangle=\pm12/37$.
Now
$$\begin{align}
d^2=||a+b||^2&=||a||^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Binomial coefficients identity : $\sum_{k=1}^{n-m+1} k\binom{n-k+1}{m}=\binom{n+2}{m+2}$ For any positive integer m&n.$n\ge m$ , let $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) = {}^n{C_m}$. Prove that $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
{n - 1}... | Starting from
$$\sum_{k=1}^{n-m+1} k {n-k+1\choose m}$$
we re-write as
$$\sum_{k=1}^{n-m+1} k {n-k+1\choose n-m+1-k}
= [z^{n-m+1}] (1+z)^{n+1}
\sum_{k=1}^{n-m+1} k \frac{z^k}{(1+z)^k}$$
Here the coefficient extractor enforces the upper limit of the sum and
we obtain
$$[z^{n-m+1}] (1+z)^{n+1}
\sum_{k\ge 1} k \frac{z^k}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find angle at centroid of triangle by its edge lengths? I need to write a program that takes length of a triangle's edges and calculates the angle $\angle APB$ ($P$ is the centroid of the triangle). Thanks for any help or clue.
| $$
PA^2=\frac {2b^2+2c^2-a^2}9,\quad PB^2=\frac {2a^2+2c^2-b^2}9\\
\implies \cos(\angle APB)=\frac {PA^2+PB^2-c^2}{2\cdot PA \cdot PB}
=\frac {a^2+b^2-5c^2}{2\sqrt{(2b^2+2c^2-a^2)(2a^2+2c^2-b^2)}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the intergal of $\int {\dfrac{1}{(x^2+9)^2}}\,dx$ with recursive formula? I used partial integration and I got this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2}{(x^2+9)^3}}\,dx$
According to the recursice formula I should continue like this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}... | You have found the integral of $(x^2+9)^{-3}$ in terms of the integral of $(x^2+9)^{-2}$. So, instead, integrate $(x^2+9)^{-1}$ by parts
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$p_1,p_2$ roots of $x^2+3x+5=0$, then the value of the expression $M=\frac{p_1^2+5p_1+7}{p_1^2+7p_1+5}+\frac{p_2^2+5p_2+7}{p_2^2+7p_2+5}$ If $p_1,p_2$ are the roots of the equation $x^2+3x+5=0$, then the value of the expression $M=\frac{p_1^2+5p_1+7}{p_1^2+7p_1+5}+\frac{p_2^2+5p_2+7}{p_2^2+7p_2+5}$
I solved the questio... | Since $p_1,p_2$ are roots of $x^2+3x+5=0$, then $p_1^2+3p_1+5=p_2^2+3p2+5=0$ and we can simplify the expression to
$$
\frac{2p_1+2}{4p_1}+\frac{2p_2+2}{4p_2}=\frac{p_1+1}{2p_1}+\frac{p_2+1}{2p_2}=\frac{p_1p_2+p_2+p_1p_2+p_1}{2p_1p_2}=1+\frac{p_1+p_2}{2p_1p_2}
$$
Now $p_1+p_2=-3$ and $p_1p_2=5$ so $M=\frac{7}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculating the limit of a geometrical construction Suppose I have a rectangle ABCD, with AD larger than AB. I trace its diagonal AC, then the perpendicular to AC that goes through B. It intersects AC in $B_1$. I then trace the perpendicular to AD that goes through $B_1$, which intersects AD in $A_1$, and the perpendic... | The lengths of the next step you obtained are correct.
Now consider the ratio between the initial height and width, i.e. write $b = ka, k>1$.
The new lengths are given by:
$$\frac {a^3}{a^2+ b^2} = \frac {a^3}{a^2 + k^2a^2} = \frac a{1+k^2}$$
$$\frac {b^3}{a^2+b^2} = \frac {k^3a^3}{a^2+b^2}$$
hence the new ratio of hei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent. Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
... | Alternative way using Taylor series:
$$\ln\left(\cos \frac{1}{n}\right) = \ln\left(1-O\left(\frac{1}{n^2}\right) \right) = O\left(\frac{1}{n^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine the value of K so the function have only one point of one horizontal point of tangency
Consider the polynomial function define by
$ f(x) = x^3 + x^2 + kx - 1 $
Determine the value of k so f has only on point of horizontal
tangency.
So I know that $ f^{'}(x) = 3x^2 + 2x + k $
And I should consider $ f... | Continuing where you left off:
We know that the horizontal tangent line is found when $$f'(x) = 3x^2 + 2x + k =0$$
So we need to find the value of $k$ which only gives us $1$ root for this polynomial. Let's us use the quadratic formula to solve for the roots.
$$x=\frac{-2\pm \sqrt{2^2-4(3)(k)}}{2(3)}\\
x=\frac{-2\pm \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding a specific polynomial function, with conditions.
Find a polynomial such that $f(x) = a + bx + cx^2 +dx^3 + ex^4$ such
that $f(1)=1, f(2) = -1, f(-1)=5, f(3) = -59, f(-2) = -29$
Any hints on how to approach this? I was thinking about plugging in the f value and then the result in order to build a matrix like t... | $f(1)=1, f(2)=-1,f(-1)=5,f(3)=-59,f(-2)=-29$
Provided a linear $5\times 5$ sysytem of equation as $MV=U$
where $$M=\begin{pmatrix} 1 & 1 &1 &1 &1 \\ 1 & 2 & 4 & 8 & 16 \\ 1 & -1 & 1 &-1 & 1\\ 1 & 3 & 9 & 27 & 81 \\ 1 & -2 & 4 & -8 & 16 \end{pmatrix}, V=\begin{pmatrix} a \\ b\\ c\\ d\\ e \end{pmatrix}, U =\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate the Limit of Double Sum Compute
\begin{equation}
L=\lim _{n \rightarrow \infty}\frac{1}{n} \sum_{a=1}^n \sum_{b=1}^n \frac{a}{a^2+b^2 }.
\end{equation}
My attempt:
Define \begin{equation} f(n,m)= \frac{1}{n} \sum_{a=1}^n \frac{1}{m}\sum_{b=1}^m \frac{\frac{a}{n}}{(\frac{a}{n})^2+(\frac{b}{m})^2 } = \frac{1}{m... | While this approach arrives at the correct value for the limit it is not easily justified.
At first glance,
$$f(n,n) = \frac{1}{n}\sum_{a=1}^n\sum_{b=1}^n\frac{a}{a^2 + b^2} = \frac{1}{n^2}\sum_{a=1}^n\sum_{b=1}^n\frac{a/n}{(a/n)^2 + (b/n)^2} $$
is a Riemann sum for the integral of $g:(x,y) \mapsto \frac{x}{x^2+ y^2}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Sketch the curve $\frac{ax + b}{cx + d}$ I am working through a pure maths book out of interest. I am stuck on this problem:
If $a$, $b$, $c$ and $d$ are positive, sketch the curve $y = \frac{ax + b}{cx + d}$ when
i) $ad - bc \lt 0$
ii) $ad - bc = 0$
iii) $ad - bc \gt 0$
I can work out the basics. Thus
$x$-intercept:... | From your last observation, $$\frac{ax+b}{cx+d} =\frac 1c \left( a +\frac{bc-ad}{cx+d} \right)$$
You can think of this as a series of transformations applied to the graph of $y=\frac 1x$.
Starting from it, first shift to the left by $d$. That gives $\frac {1}{x+d}$. Then squish parallel to the $x$ axis by a factor of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Unusual Constant appearing for definite integral I’m learning integration and have come across a question in a textbook. I arrived to the same answer as the textbook itself, but I know it is not correct. It starts as part a) show that $$\frac{d}{dx} tan^{-1} \left(\frac{3tan(x)}{2} \right)= \frac{6}{5sin^2(x)+4}$$ Henc... | This happens because $\tan x$ is not defined at the points $x = \frac{\pi}{2}, \frac{3\pi}{2} $ which lie in $(0,7)$ so $\tan^{-1} \left(\frac{3\tan(x)}{2} \right)$ becomes discontinuous at these points.
This is why you have to split the interval into sub-intervals $(0,\frac{\pi}{2}), (\frac{\pi}{2}, \frac{3\pi}{2}), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4079263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculating limit of a square root function, in terms of N and epsilon I've been trying to solve this question:
Prove directly from the definition of a limit, in terms of $N$ and $\varepsilon$:
$$\lim_{n\to\infty} \sqrt{\frac{4n+1}{n}} =2$$
I'm not supposed to use any known properties and use the definition directly.... | Alternative approach
$\sqrt{4 + 1/n} < 2 + \epsilon = \sqrt{4 + 4\epsilon + \epsilon^2}.$
Therefore, you have to choose $n$ such that
$(4 + 1/n) < (4 + 4\epsilon + \epsilon^2)$.
This means that $(1/n) < 4\epsilon + \epsilon^2.$
This means that $n > \frac{1}{\epsilon} \times
\frac{1}{4 + \epsilon}.$
Since $\frac{1}{4 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4079417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots... | There is an argument using the Farey tree. Briefly, the Farey algorithm fills in rational numbers between $x = \frac{p}{q}$ and $y = \frac{r}{s}$ by
$\frac{p+r}{q+s}$ (and so on recursively) [What some have called the Bart Simpson addition of fractions). The basic fact is that if you start with $\frac12$ and $1,$ the n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 0
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General term for series $0,2,12,70, 408...$ Given the curves $y=\sqrt{2x^2+1}$ and $x=\sqrt{2y^2+1}$, the first line (line with smallest y coordinate) which intersects the curves with integer coordinates is $x+y=1.$ This line intersects the curves at $(0,1)$ and $(1,0)$. Along with this it intersects the curves at $(-2... | I would select the smaller absolute value with each pair of coordinates, thus $0,2,12,70,408,...$.
Suppose you have a line $x+y=a_n+b_n$ which intersects $y^2-2x^2=1$ at $(a_n,b_n)$ and thus intersects $x^2-2y^2=1$ at $(b_n,a_n)$. The other intersection with $y^2-2x^2=1$ is determined by
$(a_n+b_n-x)^2-2x^2=1$
$-x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Rolling a die twice in either order with not mutual exclusive events I am reading the following problem:
If a single die is rolled twice, find the probability of rolling an
odd number and a number greater than $4$ in any either order
My solution:
Probability of rolling an odd number = $$\frac{3}{6}$$
Probability of r... | The outcome potentially counted twice is $(5,5)$, not $5$.
The number of pairs $(a,b)$ where $a$ is odd and $b>4$ is $3\times2=6$. This includes $(5,5)$.
The number of pairs $(a,b)$ where $a>4$ and $b$ is odd is $2\times3=6$. This includes $(5,5)$.
The sum of these is $6+6=12$, but $(5,5)$ was counted twice; there are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Trigonometric integral $\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}$ The function
$f(x_1,x_2)=\frac{e^{-\sqrt{x^2_1 + x^2_2}}}{\sqrt{x^2_1+x^2_2}}$, $(x_1,x_2)\neq(0,0)$ is integrable in $\mathbb{R}^2\setminus\{(0,0\})$ as $f>0$ and
\begin{align}
\int_{\mathbb{R}^2}f(x_1,x_2)\,dx_1dx_2=2\pi\int^\infty_0e^{-r}\,dr
\end{... | Let $I=\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}=2\int_0^{\pi/2} \frac{d\theta}{1+a^2\sin^2\theta}$ Noting that $\csc^2\theta=1+\cot^2\theta$, divide numerator and denominator by $\sin^2\theta$ to get:
$\frac I2=\int_0^{\pi/2} \frac{\csc^2\theta d\theta}{1+\cot^2\theta+a^2}$. Now substitue $t=\cot \theta$ so that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that there is a simultaneously tangent line to both the curves $y = e^x$ and $y = \ln x$ I found the derivatives of both of the curves but I'm having trouble on how to move on from here:
$y = e^x\implies y' = e^x$
$y = \ln x\implies y' = \frac{1}{x}$
| Given \begin{align} y = e^x\implies y' = e^x$, $y = \ln x\implies y' = \frac{1}{x}\end{align}
Tangent line to $y = e^x$ at x=a
\begin{align} y=e^ax+e^a(1-a)\end{align}
Tangent line to $y=lnx$ at x=b\begin{align} y=\frac{1}{b}x+\ln b-1\end{align}
When the two lines become one line:
\begin{align} e^a=\frac{1}{b}, e^a(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
The equation ${4 \choose k}=6$ Find the solution of the equation $${4 \choose k}=6.$$
So $k$ must be a natural number $(n\in \mathbb{N})$ and we can find that when $k=1 \rightarrow {4\choose 1}=4$ and when $k=2\rightarrow {4\choose2}=6$, so the solution of the equation is actually $k=2$. Can we solve the problem withou... | You can write $n\choose{k}$ = $\frac{n!}{(n-k)!k!}$. Substituting $n = 4$ into there, we get $\frac{24}{(4-k)!k!}$ = 6. Multiplying both sides by $(4-k)!k!$, we get $24 = 6(4-k)!(k!)$. Dividing both sides by $6$, we get $4 = (4 - k)!(k!)$. Realizing $4 = 2^2$, we can assume that $k! = 2$ and $4 - k = 2$.
The only value... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How should I check that $6^{\log_{10} x} + 8^{\log_{10} x} = x $ does not have other solutions? The question is to solve the equation $6^{\log_{10} x} + 8^{\log_{10} x} = x $
I know one of the solutions is $x=100$ using Pythagorean triples but I can't show that this is the only solution.
I'm looking for an idea without... | Solving for $x$ in $6^{\log_{10} x} + 8^{\log_{10} x} = x$
is equivalent to solving for $y$ in $6^y + 8^y = 10^y$ where $y = \log_{10} x.$
I have an easier time visualizing this equation with $y$ than the one with $x,$ so I would prefer to solve for $y$.
You can also write this $6^{y-2} 6^2 + 8^{y-2} 8^2 = 10^{y-2} 10^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral including an Incomplete Gamma function Does anyone have an idea of solving following integral
$$ I =\int_{0}^{\infty}\frac{x\Gamma \left(a,b x\right)}{(x^2+1)}\,dx\,,$$
where $a,b>0$ are positive real values?
Mathematica gives an answer, but I do not know how it can be derived by using integral tables.
\begin{... | Assuming that $a$ is a positive integer and $b>0$, considering
$$I_a =\int_{0}^{\infty}\frac{x}{x^2+1}\,\Gamma \left(a,b x\right)\,dx$$ there is only one which is simple
$$I_1=\frac{\pi}{2} \sin (b)-(\text{Ci}(b) \cos (b)+\text{Si}(b) \sin (b))$$ where appear the since and cosine integral functions.
The problem is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Exact Diff. Equation $ \frac{d y}{d x}=\frac{x y^{2}-\cos x \sin x}{y\left(1-x^{2}\right)} $ Solve:$$
\frac{d y}{d x}=\frac{x y^{2}-\cos x \sin x}{y\left(1-x^{2}\right)}.$$
$${(xy^2 -\cos x\sin x)dx+y(1-x^2)dy=0}.$$
So the first term, $(xy^2 -\cos x\sin x)dx$ is $=M(x,y)$ and the second, $ y(1-x^{2} )dy$ term is $=N(x... | $$\frac{d y}{d x}=\frac{x y^{2}-\cos x \sin x}{y\left(1-x^{2}\right)}.$$
$$y(1-x^2){d y}=x y^{2}dx-\cos x \sin xdx$$
$$dy^2-2x^2yd y-2xy^2dx=2\cos x d\cos x$$
$$dy^2-(x^2d y^2+y^2dx^2)=2\cos x d\cos x$$
$$y^2-x^2 y^2-\cos^2 x =C$$
Your solution looks correct to me.
| {
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If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which... | You have a few mistakes:
*
*$(b+\frac1y)(y+\frac1a)(a+\frac1b)\neq a^2b^2y^2+\frac1{aby}$, as you write above
*$(b+\frac1y)(y+\frac1a)(a+\frac1b)\neq \frac{a^2b^2y^2+1}{aby}$, as you apply the formula you wrote
The actual solution is:
$$(b+\frac1y)(y+\frac1a)(a+\frac1b)=\frac1{aby}(yb+1)(ay+1)(ab+1)=\frac1{aby}(aby... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2} $ The formula we need to prove is
$$
\displaystyle \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d... | Integral on the left is the integration over the $\color{red}{red}$ circle, and the integral on the right is the integration over the $\color{blue}{blue}$ square (see the picture below).
We note that the area of the circle is $S_\circ = \pi\cdot 1^2 = \pi$ and the area of the square is $S_\square = (\sqrt \pi)^2 = \pi$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to expand this expression in series? From Poisson's first Memoire sur la distribution d'electricite (1812), $\S$39. Suppose $b$ is very small relative to $c-a$ where $a,b,c$ are positive real scalars. Consider the equations
\begin{align*}
af(x)+\frac{b^2}{c-ax} F\left(\frac{b}{c-ax}\right)&=h,\\
bF(x)+\frac{a^2}{c-... | The expansion is around $x=0$.
Then $$bF(x)=g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)$$
Expanding in Taylor series at $x=0$, the constant term is $$g-\frac{a^2}cf\left(\frac ac\right)$$
The linear term is given by the first derivative:
$$\left(g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)\right)'=\frac{a^2}{(c-bx)^... | {
"language": "en",
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"source": "stackexchange",
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Extraneous solutions without performing squaring operation
Solve in $[0, 2\pi]$ $$\sec x+\tan x=2\cos x$$
My mind boggled while solving the trigonometric equation in two different ways:
Method $1.$
Assuming $x\ne \frac{\pi}{2},\frac{3\pi}{2}$
We have:
$$\sec x+\tan x=2\cos x$$
$\implies$
$$\sec x-\tan x=\frac{1}{2\co... | Hint
Subtract the two equation to find $$2\tan x=2\cos x-\dfrac1{2\cos x}$$
As $\cos x\ne0,$
multiply out both sides by $2\cos x,$ to find $$4\sin x=4\cos^2x-1=4(1-\sin^2x)-1$$
Your solution needs to satisfy this equation as well
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Modulus function: finding set of values. I didn't learn this in my syllabus but it is in some past papers. I need a small explanation for how to find the possible values for p to give 0, 1 or 2 roots.
Here is an example question: $f(x)=6-|x+2|$ with roots (-8,0) and (4,0)
Find the set of values for p for which the equa... | I think what you mean it to find the values for which $6-|x+2| = px + 5$ have $0, 1$ or $2$ solutions. The solutions, not roots, are when $6-|x+2|$ and $px+5$ both equal the same number which might not and probably will not be $0$. $6-|x+2|$ has two roots; $x=4$ and $x = -8$ and $px +5$ will have one root if $p\ne 0$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
At $n=4$, $$5^4<4^5$$ which is indeed true.
By mathematical induction, we need to prove that $$(n+2)^{n+1} < (n+1)^{n+2}$$
$$\implies (n+2)^n\times (n+2) < (n+1)^n\times(n+1)^2 $$
I am not getting how to proceed further than th... | Here is a non-inductive proof that involves a nice application of the AM-GM inequality.
It suffices for us to prove the following inequality:
\begin{align}
\forall \ & n>3, \left(1+\dfrac{1}{n}\right)^n <4 \\
& \iff \left(\dfrac{n+1}{n}\right)^n <4 \\
& \iff \left(\dfrac{n}{n+1}\right)^n >\dfrac{1}{4} \\
& \iff \dfrac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing the generator and error polynomials of a $\left[15, 7, 5\right]$ BCH code My question: Let $\alpha$ be a root of $1+x+x^4 \in \text{F}_{2}\left[x\right]$ and let C be a narrow-sense BCH code with length 15, and designed distance 5.
Find the generator polynomial of $C$ and determine the error position of $1000... | After taking the comments above into consideration, I believe I have found the solution.
So, using the transpose of $H$,
$$H^{T} = \begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
| obten la raiz negativa de la siguiente funcion con 4 cifras decimales.
$2x^4-2x^3+x^2+3x-4=0 \quad\quad$ newton raphson
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $\int \sqrt{2me-mkr^2-\frac{1}{2}m br^4 - \frac{a^2}{r^2}} \,dr$ I was trying to find Hamiltons principle function, $S$, for the Hamiltonian:
$$ H = \frac{1}{2} m \left( P_{r}^2 + \frac{P_{\theta}^2}{r^2} \right) + \frac{1}{2}kr^2 + \frac{1}{4} b r^4$$
After considering the form $S = f(r, \theta) - Et = R(... | $$\sqrt{2mE-mkr^2-\frac12mbr^4-\frac{a^2}{r^2}}\,dr=\frac1r\sqrt{2mEr^2-mkr^4-\frac12mbr^6-a^2}\,dr$$
Now with the substitution $u=r^2\Rightarrow dr=\frac{du}{2r}$ so:
$$\frac{1}{2u}\sqrt{2mEu-mku^2-\frac12mbu^3-a^2}du$$
so we have reduced the order of the polynomial inside the square root. However, integrals of this f... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$ I am trying to solve the following integral:
$$\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$$
However the polynomial have imaginary roots:
$$-(-i A + B^2)^{1/2},\; (-i A + B^2)^{1/2},\; -(i A + B^2)^{1/2},\;(i A + B^2)^{1/2}$$
I am only interested on the solution from $... | Using the variable change $x\to Bx$, one can immediately reduce to the case $B=1$. If $A$ is real, we have:
\begin{align*}
\int_0^1\frac1{A^2+(1-x^2)^2}dx&=\frac1{2Ai}\int_0^1\left[\frac1{(x^2-1)-Ai}-\frac1{(x^2-1)+Ai}\right]dx\\
&=\frac1A\Im\left(\int_0^1\frac1{(x^2-1)-Ai}dx\right)\\
&=\frac1A\Im\left(\frac1{2\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the following converge to 0? Does the following converge to $0$ for $\theta>-1/2$?
$$\lim_{n\rightarrow\infty}\frac{\sum_{i=1}^ni^{3\theta}}{\left(\sum_{i=1}^ni^{2\theta}\right)^{3/2}}$$
I'd like to use the comparison test but no idea what to compare it to. If $\theta=1$, based on code I wrote it seems to be going... | Rearranging terms we have that
$$\frac{\sum_{i=1}^ni^{3\theta}}{\left(\sum_{i=1}^ni^{2\theta}\right)^{\frac{3}{2}}}\cdot\frac{\frac{1}{n^{3\theta}}}{\frac{1}{n^{3\theta}}}\cdot\frac{\frac{1}{n^{\frac{3}{2}}}}{\frac{1}{n^{\frac{3}{2}}}} = \frac{\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta}\frac{1}{n}}{\left(\sum_{i=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which of the following is not the correct quadratic equation? If $3/2$ and $4$ are the two roots of a quadratic equation, then which one of the following is not the correct quadratic equation?
(A) $2x^2-11x+6=0$,
(B) $6x^2-33x+18=0$,
(C) $-10x^2-55x-30=0$,
(D) $4x^2-41x+24=0$.
I tried using the analogy $(x-3/2)(x-4)$ b... | Your textbook is suggesting a uselessly complicated method.
The sum of the roots is $11/2$ and their product is $6$. Dividing by the leading coefficient should make the equation in the form
$$
x^2-\dfrac{11}{2}x+6=0
$$
and the form of this equation is unique. Now we have
(A) $x^2-\dfrac{11}{2}x+3=0$ (divide by $2$)
(B)... | {
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Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$? Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$?
If $\displaystyle I = \int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c$, then $\left\lfloor \dfrac{1}{g'(2)} \right\rfloor = \cdots$,
(where $\lf... | Your approach is correct.The formula given clearly indicates that $g(x)$ is the indefinite integral of the function:
$$ \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} $$
So differentiating $g(x)$ gives us back that function of $x$, and evaluating it at $x=2$ after differentiation is correct.
We should get $g'(2) = \frac{1}{3\sqrt{1... | {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int_{0}^{\infty}\frac{\cos(x)}{(1+x^2)^2}dx$ without the residue theorem? Consider the integral
$$\int_{0}^{\infty}\frac{\cos(x)}{(1+x^2)^2}dx$$
Are there any other ways to compute this integral besides using the residue theorem?
Edit: Thank you all, kind people, for the answers. I had this on an exam and... | Let us define integral with parameter:
\begin{align}
I(t) = \int_0 ^\infty \frac{\cos(xt)}{(1+x^2)^2}dt.
\end{align}
Taking the derivative of I:
\begin{align}
\frac{dI}{dt} = \frac{d}{dt} \int_0 ^\infty \frac{\cos(xt)}{(1+x^2)^2}dt
=\int_0 ^\infty \frac{\partial}{\partial t} \frac{\cos(xt)}{(1+x^2)^2}dt= - \int_0^\inf... | {
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"question_score": "2",
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Summation over $a+b+c=5$ Let $a,b,c$ be positive integers. Compute
$$\sum_{a+b+c=5} (6-a)(6-b)(6-c).$$
The first thing I notice is symmetry, so that I can permute $3!=6$ ways, but i'm not really sure how that works with the condition $a+b+c=10.$ The other method is to fix $a$, but that is reall time-consuming and unfe... | By using generating functions, we show that for any non-negative integer $n$,
$$\begin{align}\sum_{a+b+c=n} (6-a)(6-b)(6-c)
&=[x^n]\left(\sum_{k=1}^{\infty}(6-k)x^k\right)^3\\
&=[x^n]\frac{(x(5-6x))^3}{(1-x)^6}\\
&=[x^n]\frac{125x^3-450x^4+540x^5-216x^6}{(1-x)^6}\\
&=125\binom{n+2}{5}-450\binom{n+1}{5}+540\binom{n}{5}-... | {
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"timestamp": "2023-03-29T00:00:00",
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Study of $\sum_{n=1}^{\infty}n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)$ For $\alpha\in \mathbb{R}$ I want to study the behaviour of
$$\sum_{n=1}^{\infty}n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)$$
I have thought to apply the asymptotic test but in order to be able t... | We have $$
n^\alpha\left(\frac{1}{n^2+n}-\frac{2n+1}{2(n^2(n+1))}+o\left(\frac{1}{n^2}\right)\right)=\frac{n^\alpha}{n(n+1)}\left(-\frac1{2n}+o\left(\frac{1}{n^2}\right)\right)<0$$ for $n$ sufficiently large. It's just as good to have all the terms negative as positive, of course, so you can proceed.
Perhaps I ought... | {
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Calculate $\lim_{x\rightarrow0}\frac{(e^{\sin x}+ \sin x)^{\frac{1}{\sin x}}-(e^{\tan x}+ \tan x)^{\frac{1}{\tan x}}}{x^3}$ How to calculate the following limit?
$$\lim_{x\rightarrow0}\frac{(e^{\sin x}+ \sin x)^{\frac{1}{\sin x}}-(e^{\tan x}+ \tan x)^{\frac{1}{\tan x}}}{x^3}$$
I thought of L'Hopital's rule, Taylor expa... | Hint 1: Consider the Taylor series of $f(\sin(x)) - f(\tan(x))$ for a general function $f$. What information do you need about $(e^x + x)^{1/x}$ to solve the problem?
Hint 2:
The information can be gotten from the Taylor series for $\ln(e^x + x)/x$, which is easier to find.
Full solution:
You can verify by substit... | {
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The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
Let $a,b,c$ be there positive integers. So, $a+b+c=20$
Total number of solutions would be $^{19}C_2=171$
For... | Because in the case of a triangle's sides, order has no meaning. The set $\{7,8,5\}$, representing $a,b$ and $c$ forms the same triangle as $\{8,5,7\}$ and likewise. So, if $a,b,c$ are distinct, the $6$ possible ordered pairs give only $1$ combination, hence division by $6$. If two of $a,b,c$ are equal, then $3$ differ... | {
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I'm facing a problem in differentiating an infinite square root function. How can I differentiate this function?
$y$ = $\displaystyle\frac{x}{x+\frac{\sqrt{x}}{x+\frac{\sqrt{x}}{x+....}}}$
I think I should break this function into an implicit function:
$\displaystyle\frac{y}{\sqrt{x}}=\frac{\sqrt{x}}{x+\frac{y}{\sqrt{x... | You have to first solve for the function and then differentiate. You already know that
$\frac{y}{\sqrt{x}} = \frac{\sqrt{x}}{x + \frac{y}{\sqrt{x}}}$
so let $z = \frac{y}{\sqrt{x}}$, then we have
$z = \frac{\sqrt{x}}{x + z}$
Then $z^2 + xz = \sqrt{x}$. By the quadratic formula, $z = \frac{-x \pm \sqrt{x^2 + 4 \sqrt{x}}... | {
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Find b to make the three numbers in AP For a positive number '$b$', the value of '$b$' for which the numbers $3^x + 3^{-x} , b ,9^x+ 9^{-x}$ are in A.P. can be : (A) 1 (B) 2 (C) 3 (D) 5
Hello! This question has multiple correct answers and belongs to the topic Arithmetic Progression(Sequence and series). I was able to... | \begin{align}
\frac{3^x+3^{-x}+9^x + 9^{-x}}{2} &= \frac{3^x+3^{-x}+(3^x+3^{-x})^2-2}{2}\\
&= -1 + \frac{(3^x+3^{-x})+(3^x+3^{-x})^2}{2}
\end{align}
We know that the range of $3^{x}+3^{-x}$ is $[2, \infty)$
Hence the range of $(3^x+3^{-x})+(3^x+3^{-x})^2$ is $[6, \infty)$.
Hence $b$ can attain any number at least $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding expected total number of die rolls Question
Ann and Bob take turns to roll a fair six-sided die. The game ends after a six or three consecutive fives come up, with the winner being the last person who threw the die. Ann will go first.
$(a)\quad$ Find the probability that Ann will win.
$(b)\quad$ Find the expect... | If $x$ is the expected number of rolls, $y$ is the expected number of rolls from $S(5)$ and $z$ is the expected number of rolls from $S(55)$, then we have
$\left
\{\begin{array}
{l}x = 1 + \dfrac{1}{6} \cdot y + \dfrac{4}{6} \cdot x \\ y = 1 + \dfrac{4}{6} \cdot x + \dfrac{1}{6} \cdot z \\
z = 1 + \dfrac{4}{6} \cdot x
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$a^2\phi=2$ , what is the value of $2a(\phi+1)$? I reduced a problem to this:
We have $a^2\phi=2$ where $a>0$ what is the value of $2a(\phi+1)$ ?
$1)2\sqrt{2\sqrt5+4}\qquad\qquad2)2\sqrt{\sqrt5+4}\qquad\qquad3)2\sqrt{2\sqrt5+1}\qquad\qquad4)2\sqrt{\sqrt5+1}$
Where $\phi$ is golden ratio ($\frac{1+\sqrt5}2)$.
This is a ... | Just solve for $a$ and plug it in.
Assuming $a\ge 0$ we have
$a^2 \phi = 2$
$a^2 = \frac 2 {\phi}$
$a= \sqrt{\frac 2\phi } = \frac {\sqrt 2}{\sqrt \phi}$.
So
$2a(\phi +1)= 2\frac {\sqrt 2}{\sqrt \phi} (\phi+1)=2\sqrt 2(\sqrt \phi + \frac 1{\sqrt \phi})=....$
Okay here we can use $\phi^2 = \phi + 1$.
$2\sqrt 2\frac {\ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $ \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
Find $\displaystyle \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
I couldn't find my answer so I looked up the solution which is as follows
$$\displaystyle \begin{align}&\int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx\\&= \int \frac{2\cos x+1}{(2+\cos x)^2}\... | Analogous of how integration by parts comes from the product rule, from the quotient rule of differentiation:
$$\begin{align*}
\frac uv &= \int \frac {du}{v} - \int\frac{u\ dv}{v^2}\\
\int\frac{u\ dv}{v^2} &= -\frac uv + \int \frac {du}{v}\\
\end{align*}$$
Or directly deriving from integration by parts:
$$
\int \frac{u... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
What I Tried:- Given this expr... | Edit: This is similar Moko19's answer (written in paralel but posted too late...), but I keep it here as it describes process how to get the solutions for the $a^3+b^3=1729$.
Notice that $m^3+3m^2+3m=(m+1)^3-1$, and similarly $$64k^3 + 24k^2 + 3k=\frac{1}{8}((8k)^3+3(8k)^2+3(8k))=\frac{1}{8}((8k+1)^3-1).$$
And so
$$
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $f : \mathbb R\to\mathbb R$ such that $f(f(x+y)) = f(x+y) + f(x)f(y) -xy$ Find all $f : \mathbb R\to\mathbb R$ such that
$$f(f(x+y)) = f(x+y) + f(x)f(y) -xy$$
for all reals $x,y.$ (Belarusian Mathematical Olympiad-1995)
My answer:
Consider $f(0) = c. ...(i)$
Let $x,y = 0$ at first.
$$f(f(x+y)) = f(x+y) + f(x)f... | $$f(f(x+y))=f(x+y)+f(x)f(y)-xy$$
As mentioned above, I have assumed $f(0)=c$ for some real $c$.
*
*$(x,y)\equiv(c,-c)$
$$f(c)=c+f(c)f(-c)+c^2$$
*$(x,y)\equiv(0,0)$
$$f(c)=c+c^2$$
The above equations imply that $f(c)f(-c)=0$. Hence, there exists some real $k$ such that $f(k)=0$.
*$(x,y)\equiv(k,0)$
$$f(f(k+0))=f(k+0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$
How can I prove this?
$$\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$$
Here... | Repeating your steps for the antiderivarive ( assuming $a>0$ and $b >0$)$$\int\frac{\log (1+x)}{a^2b^2(1+x)^2+({a+b})}\, dx=\frac 1{a^2\,b^2}\int\frac{\log (1+x)}{(1+x)^2+k^2}\, dx$$ with $k=\sqrt{\frac {a+b}{a^2\,b^2}}$ Now, let $r$ and $s$ be the complex roots of the denominator and use
$$\frac{1}{(1+x)^2+k^2}=\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $u-v=\sqrt5$ given that $u-v>0$ given that $u=b+b^4$ and $v=b^2+b^3$ We're also given that $b$ is a root of $z^5-1=0$ $b^4+b^3+b^2+b+1=0$
If $u=b+b^4$ and $v=b^2+b^3$, show that
i) $u+v=uv=-1$
ii) $u-v=\sqrt5$ given that $u-v>0$
I managed to do part i):
Plugging in $u$ and $v$: $(b+b^4)+(b^2+b^3)=-1$ (using $... | You can compute $(u-v)^2=(u+v)^2-4uv$ simply from what you already know.
You have the elementary symmetric polynomials and $(u-v)^2$ is symmetric, so it will have an expression in terms of $uv$ and $u+v$. The computation is equivalent to solving the quadratic with roots $u$ and $v$, but is more straightforward - you do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How can you quickly use the graph of $3x-6y=4$ to produce the graph of $3x-6y=10$? To start with, I am asked to graph the lines $3x-2y=k$ for $k=1,2,3$ and $4$.
I find that the equations produce different parallel lines.
For each line where $k=2,3$ and $4$, the coordinates points are $0.5k$ below the coordinates points... | Great question. Let's start with a couple of general lines here:
$$ ax + by = k_1 , $$
$$ ax + by = k_2 . $$
As you note, these lines are parallel because their slopes are the same:
$$ ax + by = k_1 \Rightarrow y = -\frac{a}{b}x + \frac{k_1}{b} , $$
$$ ax + by = k_1 \Rightarrow y = -\frac{a}{b}x + \frac{k_2}{b} . $$
An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove all $2\times2$ real matrices with eigenvalues $1$ and $-1$ can be represented as My exercise asks:
Prove that all $2\times2$ real matrices with eigenvalues $\lambda_1=1$ and $\lambda_2=-1$ can be represented as
\begin{equation}
\begin{bmatrix}
\cos\theta & a\sin\theta \\
\frac{1}{a}\sin\theta & -\cos\theta
\end{b... | $a,b,c,d$ are fixed numbers and we have to show that they can be represented as $\cos\theta, k\sin\theta, \frac1k \sin\theta, -\cos\theta$
Let $\cos \theta=a \implies d = -a = -\cos \theta$
This can be done because there always exists a $\theta \in \mathbb{C}$ such that $\cos\theta = a$
Now,
$$ a^2 + bc=1$$
$$ bc = 1-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplifying a summation with binomial coefficients Simplify the sum
$$
f(k,q)=\sum_{i=0}^k\binom{2i}{i+q}\binom{2k-2i}{k-i}
$$
where $k,q$ are given integers satisfying $0<q\le k$.
I tried to simplify it combinatorially. For a lattice path $p$ using steps (1,1) and (1,-1) and starts from the origin, define $g_q(p)$ to ... | Supposing we seek to simplify
$$\sum_{j=0}^k {2j\choose j+q} {2k-2j\choose k-j}.$$
where $0\le q\le k.$ This is
$$[z^k] (1+z)^{2k}
\sum_{j=0}^k {2j\choose j+q} \frac{z^j}{(1+z)^{2j}}.$$
Here the coefficient extractor enforces the upper limit of the sum
and we find
$$[z^k] (1+z)^{2k}
\sum_{j\ge 0} {2j\choose j+q} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?
$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$
First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neithe... | Simply square the constraint and you get
$4 \sin^2(x)+4\sin(x)\cos(x)+\cos^2(x)=3$
then double angle and trig identity
$3\sin^2(x)+2\sin(2x)=2$
furthermore this is equivalent to
$2\sin(2x) = 3\cos^2(x)-1$
and with your definition of the left hand side
$\tan^2(x)+4\tan(x)=\frac{\sin^2(x)+2\sin(2x)}{\cos^2(x)}$
we can in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does the triangle inequality seem to be false when rewritten based on $|x| = \sqrt{x^{2}}$? The triangle inequality is $$|x + y| \leq |x| + |y|.$$
Also, we know that $|x| = \sqrt{x^{2}}$. Then,
\begin{align*}
\sqrt{x + y} &\leq \sqrt{x} + \sqrt{y} \\
x + y &\leq x + y + 2\sqrt{xy} \\
0 &\leq 2\sqrt{xy} ... | The statement you have written using $|x|=\sqrt {x^2}$ is wrong. The correct statement would be: $$\sqrt {(x+y)^2}\leq \sqrt {x^2}+\sqrt {y^2}$$ which, indeed, does hold for all real $x,y$, as can be verified by squaring and simplifying.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My a... | Hint:
As the roots of $x^2+x+1=0$ are $w,w^2$ where $w$ is a complex cube root of unity,
the roots of $x^2-x+1=0$ are $-w,-w^2$
we can write $$f(x)$$
$$=p(x)(x^2+x+1)(x^2-x+1)+A(x-w)(x-w^2)(x+w)+B(x-w)(x-w^2)(x+w^2)+C(x-w^2)(x+w)(x+w^2)+D(x-w)(x+w)(x+w^2)$$
$$=p(x)(x^2+x+1)(x^2-x+1)+(x^2-x+1)(c(x-w^2)+d(x-w))+\cdots$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$ Let $k>0~$ be fixed. Find
$$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$$
over all $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$
==================
In case we have 2 positive numbers $a+b=ab$ results that $a, b > 1$ and $a= \dfrac{b}{b-1}$,
so the mi... | Sketch of my solution:
Set WLOG $a\leq b\leq c.$ We denote $bc=p^2$ and $b+c=2s.$ Clearly, $s\geq p\geq1.$ By AM-GM and CBS,we get:
$$\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\geq\sqrt{ka+1}+2\cdot\sqrt{\sqrt{kb+1}\cdot\sqrt{kc+1}}\geq\sqrt{ka+1}+2\cdot\sqrt{kp+1}.$$
Case 1: $p>2.$ We have:
$$E(a,b,c,k)\geq \sqrt{ka+1}+2\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help for $\int _0^1\int _0^z\int _0^y\frac{1}{\left(1-x^2\right)\left(1+y^2\right)\left(1+z^2\right)}\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z$ With the help of programs I have been able to conjecture
$$\int _0^1\int _0^z\int _0^y\frac{1}{\left(1-x^2\right)\left(1+y^2\right)\left(1+z^2\right)}\:\mathrm{d}x\:\mathrm{d}y\:... | The last integral, after the substitution $$x=\frac\pi4-\arctan y=\arctan\frac{1-y}{1+y},$$ becomes $\int_0^{\pi/4}x\ln\tan x\,dx$ which is evaluated here (or here).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Proving that $f(t)=\frac{n^2}{2}\cdot t^{n-4}(1-t^2)\left(t^2-\frac{n-3}{n}\right)$ is bounded above by $1$, for $n\geq6$ and $t\in[0,1]$ I have a problem that looks like a typical problem of maximizing functions in a compact interval. However, I am not being able to prove the bound I need.
Let $n\geq 6$ be an integer... | In case a less-clever estimate by separation of cases is useful:
Our goal is to show
$$
\frac{n^{2}}{2}\, t^{n-4}(1 - t^{2}) \left(t^{2} - \frac{n-3}{n}\right)
= \frac{n^{2}}{2}\, t^{n-4}(1 - t^{2}) \left(\frac{3}{n} - (1 - t^{2}\right)
\leq 1
$$
for $0 \leq t \leq 1$.
Write $n = 4 + m$ for $m \geq 2$ and $u = 1 - t^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $ab... | Let $a=x+2$, $b=y+2$ and $c=z+2$.
Thus, $x\geq0$, $y\geq0$, $z\geq0$ and $$(x+2)(y+2)(z+2)=x+y+z+8$$ or
$$xyz+2(xy+xz+yz)+3(x+y+z)=0,$$ which gives $x=y=z=0$ and $a=b=c=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
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Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + ... | $$\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}} =\frac{\sqrt{6+2\sqrt5}}{\sqrt{2}(\sqrt{2} + \sqrt {10})} = \frac{\sqrt{5}+1}{2+2\sqrt{5}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$.
Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$ and has no critical points.
For increasing function, $$f'(x)\gt0\\2\cos(2x) - 8(a... | You are almost done! Each of the two inequalities has to be true for all $x \in \mathbb{R}$.
Let us consider the first situation:
$$\text{cos} \ x<a+1-\sqrt{5} , \ (\forall) \ x \in \mathbb{R}$$
Since the maximum value of $\text{cos} \ x$ is $1$ we get that $$1<a+1-\sqrt{5} \iff \sqrt{5} <a$$
For the second case:
$$\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Patterns in indefinite integral (1) So far, I've found out this:
$\int \arcsin x \mathrm d x = x \arcsin x + \sqrt{1-x^2}+C$
$\int x\arcsin x \mathrm d x = \dfrac{x^2}{2}\arcsin x +\dfrac {x}{4}\sqrt{1-x^2} -\dfrac{1}{4}\arcsin x + C$
$\int x^2 \arcsin x \mathrm d x = \dfrac {x^3}{3}\arcsin x+\dfrac{x^2}{9}\sqrt{1-x^2}... | Note: please help with formatting
Here's what I found:
$\int x^n \arcsin x \mathrm d x$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \int \dfrac{x^{n+1}}{(n+1)\sqrt{1-x^2}}\mathrm d x$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1}\int \dfrac{x^{n+1}}{\sqrt{1-x^2}}\mathrm d x$
[ let x = sin u ]
$= \dfrac{x^{n+1}}{n+1}\arcsin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Point inside a right angled triangle A right angled triangle $ABC$ $(\measuredangle C=90^\circ)$ is given with $\measuredangle BAC=\alpha$. The point $O$ lies inside the triangle $ABC$ such that $\measuredangle OAB=\measuredangle OBC=\measuredangle OCA=\varphi$. Show that $\tan\varphi=\sin\alpha\cdot\cos\alpha.$
We can... | Let the hypotenuse $AB = 1$. Then $BC=\sin\alpha$.
Consider $\triangle OAB$. $\angle OAB = \varphi$ and $\angle ABO = (90^\circ-\alpha) - \varphi$, so $\angle BOA = 90^\circ + \alpha$. By the laws of sine,
$$\begin{align*}
\frac{OB}{\sin \angle OAB} &= \frac{AB}{\sin\angle BOA}\\
OB &= \frac{1\cdot\sin \varphi}{\sin (9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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A question involving the use of complex numbers.
Evaluate $2^{n-1}\left(\cos \theta -\cos(\frac{\pi}{n})\right)\left(\cos \theta -\cos(\frac{2\pi}{n})\right)...\left(\cos \theta -\cos(\frac{(n-1)\pi}{n})\right)$
In the above question the terms $\cos(\frac{\pi}{n}),\cos(\frac{2\pi}{n}),...$ are the real part of a comp... | Let $z^{2n}=1$
$$\implies z^{2n}-1=(z-z_1)(z-z_2)(z-z_3)....(z-z_{2n})$$
where $z_1,z_2...z_{2n}$ are the roots of unity
Since $z_k^{2n}=e^{2k\pi i}$,
$$z_k=e^{i(\frac{k\pi}{n})}=\cos\bigg(\frac{k\pi}{n}\bigg)+i\sin\bigg(\frac{k\pi}{n}\bigg)$$
Notice that
$$z_{2n-k}=\cos\bigg(\frac{(2n-k)\pi}{n}\bigg)+i\sin\bigg(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequa... | Note that the sum of any two sides is greater then the third side, which shows $$0 < (a+b-c)(a-b+c) = a^2 - (b-c)^2.$$
Similarly $$\begin{eqnarray}
a^2 &>& (b-c)^2\\
b^2 &>& (a-c)^2\\
c^2 &>& (a-b)^2
\end{eqnarray}.$$ Summing these lines:
$$a^2+b^2+c^2 > 2(a^2+b^2+c^2 - ab - ac - bc)\\ = 3(a^2+b^2+c^2) - (a+b+c)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$. I want to solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$.
I started by saying $\cos (z) = \frac {e^{iz} + e^{-iz}}{2}$.
I am therefore led to solve:
$$e^{iz} + e^{-iz} = 2 + 2i $$ this implies that $$e^{2iz}+1 =(2 + 2i) e^{iz}$$
By setting $x = e^{iz}$
Consider ... | Try to use the complex definition of the inverse cosine and its analytic extension to simplify:
$$\cos^{-1}(z)=\frac\pi 2+i \ln\left(iz+\sqrt{1+z}\sqrt{1-z}\right)+2k\pi\Bbb Z$$
$$\mathop \implies^\text{z=1+i}\cos^{-1}(1+i)=\frac\pi2+i\ln\left(i(1+i)+\sqrt{1-(1+i)^2}\right)=\frac\pi2+i\ln\left(i-1+\sqrt{1-2i}\right)$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $
$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $$
My Approach:
$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{(1-\sin x)(1+\sin x)} \biggl|$$... | Let me use $\ln$ in place of $\log$ and get from your third line
$$\frac{1}{\ln(\sin x)}+\frac{1}{(1-\sin x)} =\frac{1-\sin x+\ln(\sin x)}{(1-\sin x)\ln(\sin x)} = \\=
\frac{1-\sin x+\sin x -1+(\sin x-1)^2\frac{1}{2}-(\sin x - 1)^3\frac{1}{3}+\cdots}{-(\sin x - 1)^2}\to -\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the sum of the series $1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$ The following is a question which has been bugging me for quite a while,
Find the sum of the series
$$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$
Where $n$ is odd
I started by ... | The given sum is equal to
$$\begin{align}
1^2 + 2 × 2^2 + 3^2 + &2 × 4^2 + 5^2 + 2 × 6^2 + \dots + 2(n − 1)^2 + n^2\\
&=1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + \dots + (n − 1)^2 + n^2\\
&\qquad\;+2^2 + 4^2 + 6^2 + \dots + (n − 1)^2\\
&=\sum_{k=1}^n k^2+\sum_{k=1}^{m}(2k)^2=\sum_{k=1}^n k^2+4\sum_{k=1}^{m}k^2\\
&=\frac{n(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Are there infinitely many positive integer solutions to $(xz+1)(yz+1)=P(z)$? Let $P(z)\equiv 1($ mod $ \ z) $ be a polynomial of degree $n>3$ with integer coefficients. Are there infinitely many positive integers $x, y, z$ such that $(xz+1)(yz+1)=P(z)$?
If $P(z) = a_nz^n+1$, it has be proven that the Diophantine equati... | I figured out all solutions in positive integers $\ x, y , z \ $ to $(xz+1)(yz+1)=z^4+z^3+z^2+z+1$.
First, define some sequences as follows:
$r_m = m^2+m-1, \ \ \ m=1, 2, \ldots $
$q_m = (r_m+2)^2-2,\ \ \ m=1, 2, \ldots $
$a_m = m+1, \ \ \ m=1,2,\ldots $
$b_m = m^5 + 3m^4 + 5m^3 + 4m^2 +m, \ \ \ m=1,2,\ldots $
$c_m = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Trigonometric equality proof $ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$ Looking to prove
$x = A\cos(\omega t)\\
y = A\cos(\omega t + \delta)
\\
\\$
YIELDS
$x^2-2xy\cos(\delta)+y^2=A^2\sin^2(\delta)$
Specifically we're trying to express the equatio... | Alternative approach, similar to Deepak's answer.
$\underline{\text{intermediate results}}$
Lemma 1: $~~\cos(a+b) \cos(b-a) = \cos^2(a) - \sin^2(b).$
Proof:
$(\cos a \cos b - \sin a \sin b) \times (\cos a \cos b + \sin a \sin b)$
$= (\cos^2 a ~\cos^2 b) - (\sin^2 a ~\sin^2 b)$
$= [(\cos^2 a)(1 - \sin^2 b)] - [(1 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\int_0^\infty \ln(x)\left(e^{-(x+c)^2}+e^{-(x-c)^2}\right)dx$? Mathematica is telling me
$$\int_0^\infty \ln(x)\left(e^{-(x+c)^2}+e^{-(x-c)^2}\right)dx=-\frac{\sqrt\pi}2\left(\gamma+\ln4+f'(0)\right),$$
where
$$f(a)={}_1F_1(a;1/2;-c^2)$$
is the confluent hypergeometric function. I already know how to deri... | A ridiculous approach compared to the simple and elegant solution given by @WillG.
Hoping to tecognize some patterns, I used a series expansion of
$$f(x)=e^{-(x+c)^2}+e^{-(x-c)^2}$$ around $c=0$ which gives
$$f(x)=e ^{-x^2}\, \sum_{n=0}^\infty \frac{P_n(x^2)}{n!}c^{2n}$$ and
$$\int_0^\infty x^{2k}\,e ^{-x^2}\,\log(x)\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
I'm stuck in derivatives for finding Curvature of radius I know the formula How can we find the radius of curvature for this equation
$$(x^2+y^2)^2=a^2(x^2-y^2)$$
Here I know the formula but I just want to find y' and y''
I'm stuck in it
here is what know
$$\delta=\frac{(1+y'^2)^\frac{3}{2}}{y''}$$
i got y'
$$y'=\frac{... | It is easier to do this in polar coordinates,
$(x^2+y^2)^2=a^2(x^2-y^2)$
Using $x = r \cos\theta, y = r \sin\theta$,
$r^4 = a^2 r^2 \cos2\theta \implies r^2 = a^2 \cos2\theta$
$r_{\theta} = - \cfrac{a^2 \sin2\theta}{r} \implies r_{\theta}^2 = \cfrac{a^4 - r^4}{r^2}$
$r_{\theta \theta} = \cfrac{a^2 \sin2\theta}{r^2} r_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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