Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove $\sin^2 \theta +\cos^4 \theta =\cos^2 \theta +\sin^4 \theta $ Prove $$\sin^2(\theta)+\cos^4(\theta)=\cos^2(\theta)+\sin^4(\theta)$$
I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and fac... | I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $\sin^2\theta + \cos^2\theta = 1 $ rearranged into $\sin^2\theta = 1 - \cos^2\theta $ and $\cos^2\theta = 1 - \sin^2\theta $
We can see that: $\cos^4\theta = \cos^2\theta\cos^2\theta = (1-\sin^2\theta)(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 1
} |
Range of a Rational Function How to find the Range of function $$f(x)= \frac{x^2-3x-4}{x^2 - 3x +4}$$
I tried to equate the expression to $y$, then cross multiplied
$$ y= \frac{x^2-3x-4}{x^2 - 3x +4}$$
$$ y(x^2 - 3x +4)= x^2-3x-4 $$
bought the terms to one side so it becomes a quadratic and made Discriminant to zero ... | $$\implies x^2(y-1)+x(3-3y)+4y+4=0$$
The discriminant
$$=(3-3y)^2-4(y-1)(4y+4)=-(y-1)(7y+25)$$ which needs to be $\ge0$
Now $(x-a)(x-b)\le0, a\le b\implies a\le x\le b$
Alternatively,
$$\dfrac{x^2-3x-4}{x^2-3x+4}=1+\dfrac{x^2-3x-4}{x^2-3x+4}-1=1-\dfrac8{x^2-3x+4}$$
Now $x^2-3x+4=\dfrac{4x^2-12x+16}4=\dfrac{(2x-3)^2+7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the integral $\int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds$ I am attempting to evaluate the integral (where $t \rightarrow \infty$)
$$I(t) = \int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds$$
which occurs in the calculation of the second term in the a... | Given the integral
$$I(t) = \int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds$$
then by integration by parts
$$I = \left[ -\frac{(s-1)^2}{3}e^{-t(s-1)^2/2}\right]_0^{\infty} + \frac{2}{3}\int_0^{\infty}(s-1)e^{-t(s-1)^2/2} ds.$$
The remaining integral is $t^{-1}$ times the derivative of the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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I'm stuck in this one of trig substitution for fuctions. I got this:
$$\int\frac{dx}{\sqrt{(4x^2-9)^3}}.$$
I know that the answer is:
$$\frac{x}{9*\sqrt{4x^2-9}}+c.$$
And with the steps that I know about this type of substitution, I came up here, but.. I don´t know how to continue to the answer:
$$\frac{3}{2}\int \frac... | Let
$$I = \int\frac{dx}{\sqrt{(4x^2-9)^3}}$$
and make the substitution $x = \frac{3}{2} \, \operatorname{sec}\theta$ to obtain
\begin{align}
I &= \frac{3}{2} \, \int \frac{\operatorname{sec}\theta \, \tan\theta \, d\theta}{ 27 \, \tan^{3}\theta} \\
&= \frac{1}{18} \, \int \frac{\operatorname{sec}\theta}{\tan^{2}\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How do you factor $\frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}=$? \begin{align}
& \frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}= \frac{2x^2-x-1}{(x-3)(x+3)} \cdot \frac{x+3}{2x+1} \\[10pt]
= {} & \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)(2x+1)}
\en... | $2x^2 -x -1$ has $2$ factors: $x-1$ and $x+1/2$ . Hence:
$$
2x^2 -2x -1 = (2x+1)(x-1)
$$
You have done the factorization wrong.
Hence, final answer will be :
$$
(x-1)/(x-3)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Problem of Integration by Parts involving algebraic and exponential functions Can anyone please help me in solving this integration problem $\int \frac{e^x}{1+ x^2}dx \, $?
Actually, I am getting stuck at one point while solving this problem via integration by parts.
| This is an alternative to the other answer I have provided. I have decided to add it as another answer because I think it uses a sufficiently different approach, and nobody else seems to have hinted at it. We begin just as we begun in my other answer to this same question, and continue until we reach $$\int e^{\tan{u}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Answer Clarification Let $d_n$ be the number of ordered sequences of die rolls (i.e., sequences of integers from $1$ to $6$) that add up to $n$. For example, $d_4=8$, because a total of $4$ can be rolled in $8$ ways:
$$\begin{array}{*4c} 4 & 3+1 & 2+2 & 1+3 \\ \\ ~2+1+1~ & ~1+2+1~ & ~1+1+2~ & ~1+1+1+1~ \end{array}$$
an... | If you "know for sure that $\frac{1}{D(x)}$ is referring to $\frac{1}{1-(x+x^2+x^3+x^4+x^5+x^6)}$" you should invert both sides and get $D(x)=1-(x+x^2+x^3+x^4+x^5+x^6)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the initial Value Problem Solve the initial value problem:
$x'=\frac{-1}{1+t}x+2$, $x(0)=1$
What I have done so far:
$\frac{dx}{dt}= \frac{-1}{1+t}x$
$\frac{dx}{dt}-2= \frac{-1}{1+t}x$
$dx-2dt= \frac{-dt}{1+t}x$
$dx=\frac{-dt}{1+t}x +2dt$
| $$x'=\frac { -1 }{ 1+t } x+2\\ { x }^{ \prime }+\frac { 1 }{ 1+t } x=0\\ \frac { dx }{ dt } =-\frac { x }{ 1+t } \\ \int { \frac { dx }{ x } } =-\int { \frac { dt }{ 1+t } } \\ \ln { \left| x \right| } =-\ln { C\left| 1+t \right| =\ln { \frac { C }{ \left| 1+t \right| } } } \\ x=\frac { C }{ 1+t } \\ { x }^{ \pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove $\sum_{k=0}^n {n\choose k}^2 {k\choose {n-m}} = {n\choose m} {{m+n}\choose m}$? $$\sum_{k=0}^n {n\choose k}^2 {k\choose {n-m}} = {n\choose m} {{m+n}\choose m}$$
I am struggling with this identity, with no progress. Can someone show me the proof?
| Suppose we seek to verify that
$$\sum_{k=0}^{n} {n\choose k}^2 {k\choose n-m}
= {n\choose m} {m+n\choose m}.$$
where presumably $n\ge m.$
Introduce
$${n\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{k+1}} \; dz.$$
and
$${k\choose n-m}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding x in an Olympiad simultaneous equation I have been practicing for a an upcoming intermediate math olympiad and I came across the following question:
Let $x$ and $y$ be positive integers that satisfy the equations $$\begin{cases} xy = 2048 \\ \frac{x}{y}-\frac{y}{x}= 7.875.\end{cases}$$ Find x.
My approach... | Your solution is incorrect because it was stated that x is a positive integer, and $\sqrt{2048}$ is not an integer, therefore it can't be a solution.
Hints to solve the problem:
$$2048 = 2^{11}$$
$$ 7.875 = 8 - \frac{1}{8} $$
(Spoiler alert) My solution:
Because $2048 = xy = 2^{11}$, we can say that $x = 2^a$ and $y = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 2
} |
How to find the sum $\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+..+\frac{2015}{2015^4+2015^2+1}$? How to find the sum $\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+..+\frac{2015}{2015^4+2015^2+1}$ ?
I'm not being able to approach the problem.Hints please!
| Notice that:
$$ \frac{n}{n^4+n^2+1}=\frac{n}{(n^2+n+1)(n^2-n+1)} = \frac{1}{2}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)$$
and that:
$$ \frac{1}{(n+1)^2-(n+1)+1}=\frac{1}{n^2+n+1}.$$
Together they give:
$$\begin{eqnarray*} \sum_{n=1}^{2015}\frac{n}{n^4+n^2+1}&=&\frac{1}{2}\sum_{n=1}^{2015}\left(\frac{1}{n^2-n+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$
Evaluation of $$\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \int\frac{\sin 2x}{(3+4\cos x)^3}dx = 2\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^3 x.$
So we get $$\displaystyle... | $$\left( -\frac{1}{3(3\sec x+4)^2}\right)'=\frac23\dfrac{3\sin x}{\cos^2x}\frac1{(3\sec x+4)^3}=\frac{2\sin x\cos x}{(3+4\cos x)^3},$$
you did nothing wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Derivation of series of $\ln(x)$ for $x > 0$ How is the following expansion obtained?
$$
\ln(z) = 2 \left[\frac{z-1}{z+1} + \frac{1}{3} \left( \frac{z-1}{z+1} \right)^3 + \frac{1}{5} \left( \frac{z-1}{z+1} \right)^5 + \frac{1}{7} \left( \frac{z-1}{z+1} \right)^7 + \cdots \right]
$$
As far as I understand from http://... | Hint: For $x\in (-1,1), \ln (1+x) = x-x^2/2 + x^3/3 - \cdots .$ So for such $x,$
$$\ln \left (\frac{1+x}{1-x}\right ) = \dots\ ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$ So my work,
Squaring both sides $$(\sin\theta+\cos\theta)^2=1$$
$$1+2\sin\theta\cos\theta=1\ \ \ \ \ \text{-------(i)}$$
$$\sin\theta\cos\theta=0 \ \ \ \ \ \text{------(ii)}$$
So reverting back to $(i)$,
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\t... | $\cos \theta + \sin \theta=1$ is easily solved for $\theta$ in a graphical way, since it describes the intersection between a line and the goniometric circle:
$$
\begin{cases}
X+Y=1\\
X^2+Y^2=1,
\end{cases}
$$
where $X=\cos\theta$, $Y=\sin\theta.$
So either $\cos\theta =0$, $\sin\theta=1$, or viceversa. Plugging this i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove: $n \in Z^{\geq 2}$, $f_nf_{n+1} - f_{n-1}f_{n+2} = (-1)^{n+1}$ $n \in Z^{\geq 2}$, $f_nf_{n+1} - f_{n-1}f_{n+2} = (-1)^{n+1}$.
How do you do the inductive step of this proof, every time I do it I cannot find a way to use the definition of a Fibonacci sequence to simplify the right side of the equation enough.
| Here is an approach to how you might discover this result.
Notice that
$$
f_nf_{n+1} - f_{n-1}f_{n+2} =
\det \begin{pmatrix}f_{n}&f_{n+2}\\f_{n-1}&f_{n+1}\end{pmatrix}
$$
Using the definition of the Fibonacci sequence, we have
$$
\begin{pmatrix}f_{n}&f_{n+2}\\f_{n-1}&f_{n+1}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Express $y= x^{3} - x^{2} - 5x - 3$ in its fully factorised form Don't know how to do this, please help. I have never done factorising cubic polynomials and don't know how to go about this
| The usual way to do this is to use the Factor Theorem.
A polynomial $\mathrm{f}(x)$ is divisible by $(x-a)$ if $\mathrm{f}(a)=0$.
In your case, you have $\mathrm{f}(x) = x^3 - x^2 - 5x - 3$, and you need to find an $a$ for which
$$\mathrm{f}(a) = a^3 - a^2 - 5a - 3 =0 $$
This can often be done by trial-and-error, or by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$. Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$.
When i solved this question,i got the answer $x<\frac{-28}{9}$ but the answer given in the book is $x\leq-3$.
This is how i solved.
$\sqrt{x^2-5x-24}-x-2>0\Rightarrow \sqrt{x^2-5x-24}>x+2$
$$\therefore x^2-5x-24>x^2+4x+4\Rig... | First note $\sqrt{x^2-5x-24}$ is a real number iff $x\in(-\infty, -3]\cup[8,\infty)$
Notice that
$$x\ge 8\,\,\text{ and }\,\,\sqrt{x^2-5x-24}>x+2\quad \implies\quad x^2-5x-24>(x+2)^2\iff 9x+28<0$$
which is imposible.
On the other hand
$$x\le -3 \implies\quad \sqrt{x^2-5x-24}\ge 0 \quad\text{ and }\,\,\,-1 = -3 +2 \ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Find a $3\times 3$ matrix whose minimal polynomial is $x^2$. Find a $3\times 3$ matrix whose minimal polynomial is $x^2$.
My try:
Since a characteristic polynomial and a minimal polynomial have the same roots ,so the characteristic polynomial must be $x^3$ since $0$ is the only characteristic value of multiplicity $2... | I think this would work:
Let A be a $3\times3$ matrix whose minimal polynomial is $x^{2}$ (the only root of which is x=0).
Since the minimal polynomial and the characteristic polynomial have same roots, the characteristic polynomial of A must then be $x^{3}$.
So we start by considering A as a triangular matrix (say low... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \... | $$\begin{align}
3! \cdot 5! \cdot 7! &= 6 \cdot 120 \cdot 7! \\
&= 6 \cdot 15 \cdot 8! \\
&= 2 \cdot 5 \cdot 9! \\
&= 10 \cdot 9! \\
&= 10!
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 6
} |
Show that the range of the function $f(x)=(3x+2)/(x^2+5)$ is bounded
Show that the set $\{\frac{3x+2}{x^2+5}|x\in \mathbb{R}\}$ is bounded.
A friend suggested that I should you Cauchy inequality to prove this. But I want to find a simpler way. Please help.
| $$f(x)=\frac{3x+2}{x^2+5}$$
is a continuous function on $\mathbb{R}$ for which
$$ \lim_{x\to \pm\infty}f(x) = 0,$$
hence $f$ is bounded. By computing $f'(x)$, we may check that the stationary points of $f(x)$ occur at $x=-3$ and $x=\frac{5}{3}$. If we compute the values of $f(x)$ at such points, we get:
$$-\frac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$\int_{0}^{\pi/2}\log(\sin^2\theta+k^2\cos^2\theta)d\theta=\pi\log\frac{1+k}{2},k\geq0$ Prove that $$\int_{0}^{\pi/2}\log(\sin^2\theta+k^2\cos^2\theta)d\theta=\pi\log\frac{1+k}{2},k\geq0$$
I tried but stuck in between.
Let $$I=\int_{0}^{\pi/2}\log(\sin^2\theta+k^2\cos^2\theta)d\theta...........(1)$$
$$I=\int_{0}^{\pi/2... | Let $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln\left(\sin^2 \theta +k^2\cos^2 \theta\right)d\theta = \int_{0}^{\frac{\pi}{2}}\ln\left(\cos^2 \theta +k^2\sin^2 \theta\right)d\theta $$
above we used $$\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$$
Now Using Differentiation under Integral Sign.
Now $$\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
$\bf{My\; try::}$ We can write $\displaystyle 1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots$ as
$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldo... | Hint:
$$
\begin{align}\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx&=
1-{1\over4}+{1\over6}-{1\over9}+{1\over11}-{1\over14}+\ldots\\&=
\sum_{k=0}^\infty{3\over(5k+1)(5k+4)}\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Prove $\prod\limits_{\mathrm{cyc}}(2-x^2+x) \le 4 +x+y+z+xyz$ for $x^2+y^2+z^2=3$
Let $x,y,z \in \mathbb{R}$ and $x^2+y^2+z^2=3$. Prove that
$$ (2-x^2+x)(2-y^2+y)(2-z^2+z) \leqslant 4 +x+y+z+xyz.$$
Observations
*
*At first, I think this inequality is easy. I applied AM-GM to the term $(2-x^2+x) \text{ }$,$(2-y^2+y)... | hint:
$3(x^2+y^2+z^2) \ge (x+y+z)^2 \implies 3\ge x+y+z \ge -3,|xyz|^3 \le \sqrt{\dfrac{x^2+y^2+z^2}{3}}=1 \implies -1 \le xyz \le 1 \implies RHS \ge 0$
In Case LHS $<0$, the inequality is trivial true.
so you only need to consider LHS$ \ge 0$
there is two cases:
case 1: $2-x^2+x,2-y^2+y,2-z^2+z$ all positive. so yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Complex numbers - how to solve $(\sqrt{3}-i)z^6+16=0$ When $z = x + yi$ (or $a + bi$), I need to solve:
$$(\sqrt{3}-i)z^6+16=0$$
Here is how I started:
$(\sqrt{3}-i)z^6=-16$
$z^6=\frac{-16}{\sqrt{3}-i}$
$z=\sqrt[6]\frac{-16}{\sqrt{3}-i}$
In other cases I get a normal complex number under the root sign in the right side... | $$\frac{-16}{\sqrt3-i}=\frac{(-16)}{(\sqrt3-i)}.\frac{(\sqrt3+i)}{(\sqrt3+i)}=\frac{-16\sqrt3-16i}{4}=-4\sqrt3-4i$$
Now by $x=r\cos\theta;y=r\sin\theta$ and $r=\sqrt{x^2+y^2}$
$$r=\sqrt{48+16}=8$$
So $$-4\sqrt3=8\cos\theta$$
$$\frac{-\sqrt3}{2}=\cos\theta$$
So $\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$
So we get that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proving that $\alpha:\mathbb{R}\to\mathbb{R}$ where $\alpha(x)=\frac{x^{3}}{x^{2}+1}$ is bijective Using the usual method that I was taught, in order to find that something is injective, I assume $f(x)=f(y)$ for some $x,y\in\mathbb{R}$, and then I demonstrate that $x=y$. This is not working out so well for me, using th... | You're almost done proving that $f$ is injective.
If $f(x)=f(y)$, then $x$ and $y$ have the same sign, because the denominators are always positive. Hence, $xy\ge0$.
Now,
$
x^2+xy+y^2=-x^2y^2
$
implies
$
x^2+xy+y^2+x^2y^2=0
$.
The LHS is a sum of positive terms and so each term must be zero. This means that $x=y=0$.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Having trouble find sum of this equation which has exponents. Can someone please help me solve this My question is:
Find the value of
$$\sum\limits_{j=1}^{100}(3^j+3\cdot 2^j)$$
Leave answer as powers of $2$ and $3$.
I've really tried to think of a way to solve this but cant seem to find one. Could someone plea... | Since ${\sum_{j=0}^{n}ar^j=a\left(\frac{1-r^{n+1}}{1-r}\right)}\quad r\ne 1$, it follows
\begin{align}
\sum_{j=1}^{100}\left(3^j+3\cdot 2^j\right)&=\sum_{j=1}^{100}3^j+3\cdot\sum_{j=1}^{100}2^j\\
&=3\cdot\sum_{j=0}^{99}3^j+3\cdot2\cdot\sum_{j=0}^{99}2^j\\
&=3\left(\frac{1-3^{100}}{1-3}\right)+3\cdot 2\cdot\left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fundamental theorem of Calculus: $\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt$ Possible textbook mistake using the FTC we are supposed to evaluate
$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt
$$
and the answer, the textbook says, should be a constant. When I evaluate that derivative I get
$$
\frac{d}{dx} \int... | Your answer is correct. Note if the textbook was correct then the integral would be independent of $x$: clearly this is false because the function is always positive so increasing $x$ will add more to the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Chinese Remainder Theorem: solving $x^2 \equiv 1 \pmod {91}$. I am trying to solve the following problem: find all solutions to the congruence $x^2 \equiv 1 \pmod {91}$.
I have solved already the congruence $x^2 \equiv 1 \pmod 7$ and $\!\!\pmod {13}$, and I am trying to use the Chinese Remainder Theorem. However, I am ... | If $x\equiv 1\pmod 7$ and $x\equiv 1\pmod {13}$ then $x\equiv 1\pmod {91}$.
If $x\equiv -1\pmod 7$ and $x\equiv -1\pmod {13}$ then $x\equiv -1\pmod {91}$.
If $x\equiv 1\pmod 7$ and $x\equiv -1\pmod {13}$ then $x\equiv 64\pmod {91}$.
If $x\equiv -1\pmod 7$ and $x\equiv 1\pmod {13}$ then $x\equiv 27\pmod {91}$.
In each c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the cube roots of $-11-2i$. How do I find the roots of $\sqrt[3]{ - 11 - 2i}$ ?
Tried to use Moivre's theorem, but can not find the solutions by using the polar form:
$z_k=\sqrt{5}[\cos(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})+i.\sin(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})]$, for $k=0, 1$ and $2$.
Also resorted to ... | Here is an elementary way of resolving (without any guesswork) the given system
$$a^3-3ab^2=-11, b^3-3a^2b=2$$
From the first equation, we obtain
$b^2=\frac{a^3+11}{3a}$.
Hence, the second equation yields
$$b=\frac{2}{b^2-3a^2}=\frac{2}{\frac{a^3+11}{3a}-3a^2}=\frac{6a}{11-8a^3}$$
Plugging this into the first equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$ without L'Hopital Without L'Hopital:
$$\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$$
Rationalize:
$$\frac{x}{\sqrt{1-3x}-1}\cdot \frac{\sqrt{1-3x}+1}{\sqrt{1-3x}+1}$$
$$\frac{x\cdot(\sqrt{1-3x}+1)}{(1-3x)-1}$$
This will still yield $\frac{0}{0}$. Maybe I should now try variable sub... | In the beginning of your post, you have
$$\frac{x}{\sqrt{1-3x}-1}\cdot \frac{\sqrt{1-3x}+1}{\sqrt{1-3x}+1}=\frac{x\cdot(\sqrt{1-3x}+1)}{(1-3x)-1}=\frac{\sqrt{1-3x}+1}{-3}$$
for all $x\leq 1/3$.
Hence
$$\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}=\lim_{x\to0}\frac{\sqrt{1-3x}+1}{-3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
A... | How is $\;\;x(\sqrt{3+x}+\sqrt{3-x}) = 2\sqrt{3}x$? Instead of simplifying any further you could also solve it by canceling and then taking the limit:
\begin{align}
\lim_{x\rightarrow 0} \frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\rightarrow 0} \frac{(\sqrt{3+x}+\sqrt{3-x})}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find the solution of $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$ Is anyone able to help me with the following equation concerned the floor function $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$
I don't know how to deal with the floor terms properly.
| $$\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$$
\begin{matrix}
x-1 &\lt &\lfloor x\rfloor &\lt &x+1\\
x^2-1 &\lt &\lfloor x^2\rfloor &\lt &x^2+1\\
\\
3x-1 &\lt &\lfloor 3x\rfloor &\lt &3x+1\\
-3x-1 &\lt &-\lfloor 3x\rfloor &\lt &-3x+1\\
\\
x^2-3x-2 &\lt &\lfloor x^2 \rfloor-\lfloor 3x\rfloor &\lt &x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
$\frac{2n\choose n}{n+2}\not\in\mathbb N$ and $n\neq3k+1$ and $n\neq4k+2$
Are there any natural numbers $n\not\equiv1\bmod3$, and $n\not\equiv2\bmod4$, so that $~\dfrac{\displaystyle{2n\choose n}}{n+2}\not\in\mathbb N$ ?
Since $C_n=\dfrac{\displaystyle{2n\choose n}}{n+1}\in\mathbb N$ for all n, this is equivalent to ... | Note that Catalan Numbers are $${2n\choose n}-{2n\choose n+1}=\frac{(2n)!}{n!(n+1)!}(n+1-n)$$
Try a similar thing for these numbers:
$$A{2n\choose n}+B{2n\choose n+1}+C{2n\choose n+2}\\
=\frac{(2n)!}{n!(n+2)!}\left[A(n+2)(n+1)+Bn(n+2)+Cn(n-1)\right]\\
A+B+C=0,3A+2B-C=1,2A=1\\
A=1/2,B=-1/3,C=-1/6\\
D_n=\frac1{n+2}{2n\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \fra... | This is a well known double integral proof by Beukers, Kolk, and Calabi. First consider the double integral:
$$\int_{0}^{1}\int_{0}^{1} \frac{1}{1-x^2y^2} dydx.$$
Since $0<x,y<1$, rewrite the integrand as a geometric series:
$$\frac{1}{1-x^2y^2}=\sum_{n=0}^{\infty}(xy)^{2n}.$$
Now notice:
$$\int_{0}^{1}\int_{0}^{1}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Find all conditions of a and b for when matrix has determinant zero Find all conditions on a and b for which A =
\begin{bmatrix} a & 0 &a^2 \\ 2 & a & 3a \\ a & -1 & ba+b\end{bmatrix}
has determinant 0.
| Let $A$ be the matrix$$\begin{pmatrix} a & 0 & a^2 \\ 2 & a & 3a \\ a & -1 & ab + b \end{pmatrix}.$$
If $a=0$, then
$$\det\begin{pmatrix} a & 0 & a^2 \\ 2 & a & 3a \\ a & -1 & ab + b \end{pmatrix} = \det\begin{pmatrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & -1 & b \end{pmatrix},$$
which is obviously 0 from the first row. So, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Area under curves , two given curves and finding function of 3rd curve from relationship between their area . Let $C_1,C_2$ be the graphs of $y=x^{2} , y=2x, 0<x<1$ respectively. Let $C_3$ be the graph of an unknown function $y=f(x), 0<c<1$ and $f(0)=0$ for a point $P$ on $C_1$ let the line pass through $P$, parallel t... | Let the x-coordinate of $P$ be $x$. Then the points are: $P(x,x^2), Q(\frac{1}{2}x^2,x^2),R(x,f(x))$. Let $S$ be the projection of point $P$ onto the x-axis. and $T$ the projection of points $P,Q$ onto the y-axis.
Then
$$\begin{align}
Area(OPQ)&=Area(OSPT)-Area(OSP)-Area(OQT)\\[1em]
&=x\cdot x^2-\frac{1}{3}x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin... | Notice, $$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}$$
$$=\lim_{x\to 0}\frac{\sin x\frac{(\cos x-1)}{\cos x}}{2x^2\sin x\cos x}$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\cos x-1}{x^2\cos^2 x}$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\cos x-1}{x^2}\cdot \lim_{x\to 0}\frac{1}{\cos^2x}$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\left(1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Determinant of big matrix Let:
$$M_n:=\left(\begin{array}{ccccc} 1 & 2 & 3 & \dots & n \\ 1^3 & 2^3 & 3^3 & \dots & n^3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1^{2n-1} & 2^{2n-1} & 3^{2n-1} & \dots & n^{2n-1} \end{array}\right),$$
for $n\in\mathbb N$. How could I compute $\det M_n$? I tried computing the firs... | $$M_n:=\left(\begin{array}{ccccc} 1 & 2 & 3 & \dots & n \\ 1^3 & 2^3 & 3^3 & \dots & n^3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1^{2n-1} & 2^{2n-1} & 3^{2n-1} & \dots & n^{2n-1} \end{array}\right) = \left(\begin{array}{ccccc} 1 \times 1^0 & 2 \times 2^0 & 3 \times 3^0 & \dots & n \times n^0 \\ 1 \times 1^2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the maximum of $\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$ for nonnegative $a$, $b$, $c$.
Show that the maximum of
$$\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$$
is $4$ for nonnegative $a$, $b$, $c$.
An elegant elementary solution is preferred.
Generally, is there an easy way to show that
$$
\frac{
\left( a_1 + \... | Here is an analytical proof for the general case, although I suspect better solutions exist. The $n = 2$ case is obvious, and we shall assume $n \ge 3$.
Without loss of generality, we can assume $a_1 \le a_2 \le \dots \le a_n$. First, since the function $f(x) = x^n$ is convex, we have
$$
a_2^n + \dots + a_n^n
\ge
(n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Computing $\int (1 - \frac{3}{x^4})\exp(-\frac{x^{2}}{2}) dx$ How does one compute
$$\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$
Mathematica gives $(x^{-3} - x^{-1})\exp(-x^2/2)$ which indeed the correct answer, but how does one get there?
Integration by parts gives $(y + y^{-3})e^{-y^2/... | The integral is of the form
\begin{equation*}
\int \left( 1-\frac{3}{x^{4}}\right) e^{\left( -\frac{x^{2}}{2}\right)
}dx=\int h(x)e^{g(x)}dx.
\end{equation*}
This form recalls the well-known formula
\begin{equation*}
\int \left( f^{\prime }(x)+g^{\prime }(x)f(x)\right) e^{\left( -\frac{x^{2}}{
2}\right) }dx=f(x)e^{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Homogenous Linear Equations in the form of Determinants In Arfken's "Mathematical Methods for Physicists"
How did he arrive to:
$x_1/x_3 = \frac{(a_2b_3-a_3b_2)}{(a_1b_2-a_2b_1)}$
Starting from:-
$$
a_1x_1+a_2x_2+a_3x_3=0 ; \\
b_1x_1+b_2x_2+b_3x_3=0 ; \\
c_1x_1+c_2x_2+c_3x_3=0
$$
Given that C is a linear combination o... | Since $C$ is linearly dependent on $A$ and $B$, we need not consider it, it does not add any information.
If you look at the system
$$\begin{align}
a_1 x_1 + a_3x_3 = -a_2x_2 \\
b_1 x_1 + b_3x_3 = -b_2x_2
\end{align}$$
you can see, by Cramer's rule, that
$$x_1 = \frac{\begin{vmatrix} -a_2x_2 & a_3 \\ -b_2x_2 & b_3 \\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer
Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:
*
*the denominator has to be one,
*the numerator has to be a multiple of the denominator or
*the numerator and denominator have to be the same.... | Presumably you want $a,b$ to be integers. If $x = a/b$, solve the equation
$$x + \dfrac{1}{x} = n$$
to get $$x = \dfrac{n}{2} \pm \dfrac{\sqrt{n^2-4}}{2} $$
The question is whether $\sqrt{n^2-4}/2$ is ever rational. This would require $\sqrt{n^2-4}$ to be an integer (because the square root of an integer is only rat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
square root of $\frac{2+\sqrt{3}}{4}$
Find the square root of $\frac{2+\sqrt{3}}{4}$.
Attempt:
I'm thinking of equating it to $a + bi$. And then finding the root. Is that right?
| For convenience we will work with $4(2+\sqrt3)$ to avoid fractions.
Let us search a solution of the form $a+b\sqrt3$, giving
$$(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt3=8+4\sqrt3.$$
By identifying $2ab$ and $4$, we have $b=\dfrac2a$ and
$$a^2+\frac{12}{a^2}+4\sqrt3=8+4\sqrt3.$$
By inspection (or resolution of a quadratic equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Shorter way to calculate this limit I want to calculate this (without L'Hospital) :
$$\lim_{x \to 0} \frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3}$$
I already solved it in a looooong way:
\begin{align}
\frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3} &= \frac{\tan x - \sin x}{x^3} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ ... | Your proof looks correct (but you slightly abused notation at the end - you should say $\to 1 \cdot 1^2 \cdots$, not $= 1 \cdot 1^2 \cdots$). You can shorten it if you use $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$.
$$\frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Let $m_1$ and $m_2$ are the slopes of the tangents drawn to circle $x^2+y^2-4x-8y-5=0$ from the point $P(-1,-2)$,and $|m_1+m_2|=\frac{p}{q}$ Let $m_1$ and $m_2$ are the slopes of the tangents drawn to circle $x^2+y^2-4x-8y-5=0$ from the point $P(-1,-2)$,and $|m_1+m_2|=\frac{p}{q}$,where $p$and $q$ are relatively prime ... | Note that $OP=3\sqrt 5$ and that the radius is $5$. So, you can have
$$\tan\theta=\frac{5}{\sqrt{(3\sqrt 5)^2-5^2}}=\frac{\sqrt 5}{2}.$$
So, you have
$$\frac{|m-2|}{|1+2m|}=\frac{\sqrt 5}{2}.$$
Squaring the both sides gives
$$16m^2+36m-11=0.$$
So, by Vieta's formulas,
$$|m_1+m_2|=\left|-\frac{36}{16}\right|=\frac 94.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Drawing without replacement - prob. for an Ace followed by an Ace? Given a standard 52-cards deck:
You are extracting cards from the deck without replacement, until you get an "Ace" for the first time. What is the probability that the next card will be "Ace" too?
I've already seen the following Q&A:
Probability of draw... | We use a fairly crude counting approach, in order to rely minimally on intuition. There are $\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the "favourables."
Maybe the first two Aces are in positions 1 and 2. That leaves $\binom{50}{2}$ for the rest.
Maybe they are in posit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?
The options are:
*
*$$\sqrt{\frac{1-x^2}{1-y^2}}$$
*$$\sqrt{\frac{1-y^2}{1-x^2}}$$
*$$\frac{1-x^2}{1-y^2}$$
*$$\frac{1-y^2}{1-x^2}$$
I trie... |
As stated in the comments this is actually a wee-bit lengthier and
hence not recommended.
Since i always try to resort to trigonometric substitutions whenever fractional powers are involved I'm gonna attempt this by substituting $x=sinA$ and $y=sinB$
$cosA +cosB=a(sinA-sinB)$
$2cos\frac{(A+B)}{2} \times cos\frac{(A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to take the derivative of $Y=\log(x+\sqrt{a^2+x^2})$?
$$Y=\log(x+\sqrt{a^2+x^2})$$ Find $\dfrac {dY}{dx}$.
My answer:
$$\dfrac 1{x+\sqrt{a^2+x^2}}\cdot\frac{d}{dx}\sqrt{a^2+x^2})$$
Which again goes to a chain rule as $\frac{2}{3} \sqrt{a^2+x^2}\cdot \frac{d}{dx} (a^2+x^2)$
Is there an alternative way other than t... | I think you missed some terms $$Y=\log(x+\sqrt{a^2+x^2})$$ Let $f=x+\sqrt{a^2+x^2}$ which makes $Y=\log(f)$. So $$\frac{dY}{dx}=\frac{dY}{df} \times \frac{df}{dx}=\frac 1 f\times \left(1+\frac{x}{\sqrt{a^2+x^2}}\right)=\frac 1 f\times \left(\frac{x+\sqrt{a^2+x^2}}{\sqrt{a^2+x^2}}\right)$$ Replacing $f$ by its expressio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that this sum is an integer. I have to show that
$$g\left(\frac{1}{2015}\right) + g\left(\frac{2}{2015}\right) +\cdots + g\left(\frac{2014}{2015}\right) $$
is an integer. Here $g(t)=\dfrac{3^t}{3^t+3^{1/2}}$.
I tried to solve it using power series, but I can not finish with any convincing argument to said that th... | By a rigid translation you can see that $g$ is an odd function around the point $(\frac 1 2, \frac 1 2)$:
$$
\begin{align}
f(x) &= g(x+\frac 1 2) - \frac 1 2 \\
&= \frac{3^{x + 1/2}}{3^{x + 1/2} + 3^{1/2}} - \frac 1 2 \\
&= \frac{3^x 3^{1/2}}{3^x 3^{1/2} + 3^{1/2}} - \frac 1 2 \\
&= \frac{3^x 3^{1/2}}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
} |
The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})?$ What is the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots (1+x^{100})?$
I manually expanded $(1+x)(1+x^2)(1+x^3)...(1+x^{10})$ and calculated the coefficient of $x^9$ as $8$ but i dont know how to solve it withou... | The coefficient of $x^9$ is the number of partitions of $9$ into distinct parts, i.e., the number of ways of writing $9$ as the sum of distinct positive integers when the order of the summands doesn’t matter. The sequence of these numbers is OEIS A000009, and as you can see there, there is no nice formula. However, for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Solve the recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n $
Solve the recurrence relation
$$a_n = 4a_{n-1} - 3a_{n-2} + 2^n $$
With initial conditions:
$a_1 = 1$
$a_2 = 11$
I have done similar recurrence relation problems to this, but none that were a non-homogeneous recurrence relation such as this one.
So far ... | A general technique is to use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write your recurrence as:
$$
a_{n + 2}
= 4 a_{n + 1} - 3 a_n + 4 \cdot 2^n
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= 4 \sum_{n \ge 0} a_{n + 1} z^n - 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How many numbers less than $1000$ with digit sum to $11$ and divisible by $11$ How many positive (integers) numbers less than $1000$ with digit sum to $11$ and divisible by $11$?
There are $\lfloor 1000/11 \rfloor = 90$ numbers less than $1000$ divisible by $11$.
$N = 100a + 10b + c$ where $a + b + c = 11$ and $0 \le ... | Digitsum is related to the modulo 9 operation. A weakening of the conditions given is that you are counting how many $0\leq n\leq 1000$ satisfy the coungruencies:
$\begin{array}{} n\equiv 2\pmod{9}\\
n\equiv 0\pmod{11}\end{array}$
By the chinese remainder theorem, we get that
$n\equiv 11\pmod{99}$
So, we can look at t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Help with solving recurrence relations using iterative substitution I need help solving these two recurrences with iterative substitution. I've looked at examples, and tried to follow them, but I just don't understand the whole plugging the recurrence into itself. I tried them out, not sure if they are correct, but if ... | Regarding the first recurrence:
\begin{align} T(n) &= T(n - 2) + 7 \\ &= T(n - 2 - 2) + 7 - 2
\end{align}
Should it not be
$$
T(n-2) \mapsto T(n-4)+7
$$
and
\begin{align}
T(n) &= T(n - 2) + 7 \\
&= (T((n - 2) - 2) + 7) + 7 \\
&= T(n + 2\cdot (-2)) + 2\cdot 7 \\
&= \vdots \\
&= T(n + k\cdot (-2)) + k\cdot 7
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the derivative for a product of two polynomial functions? In my problem, I am attempting to find $f'(x)$ when $f(x)=(5x^2-2x+8)(4x^2+7x-3)$. For my work I have:
\begin{align}
& \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt]
= {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt]
= {} & 40x^3+35x^2-16... | $4x^2 \cdot -2 = -8x^2$. You have written $-16x^2$. Purely an arithmetic mistake. Rest looks good to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Proof of trigonometric relationship of angle bisector I have to prove the following equation:
$$CD=\frac{2ab\cos\frac{C}{2}}{a+b}$$ where $CD$ is the angle bisector of $C$ in $\Delta ABC $.
My attempt:
I've started by rearranging the equation in terms of $\cos \frac{C}{2}$ ,then by squaring and subtracting $\sin^2 \fr... | Just use the law of sines; call $\gamma=2\delta$ the angle in $C$, $\alpha$ the angle in $A$, $\beta$ the angle in $B$. Also let $a=BC$, $b=AC$, $c=AB$. Finally, let $R$ be the radius of the circumscribed circle to $ABC$. The law of sines for $ABC$ says that
$$
a=2R\sin\alpha,\quad
b=2R\sin\beta,\quad
c=2R\sin\gamma
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find solutions of $2^m\cdot p^2+1=q^5$ $2^m\cdot p^2+1=q^5$
$p$ and $q$ are prime numbers
find $p$ and $q$
I think it will be useful to transfer $1$ to the other side of the equation
$2^m\cdot p^2=(q-1)(q^4+q^3+q^2+q+1)$
and we know $gcd(q-1,q^4+q^3+q^2+q+1)=1$ or $5$
we know if
I)$q-1|2^m \implies p^2|q^4+q^3+q^2+q+1... | Let $f(q) = q^4+q^3+q^2+q+1$. Since $f(q)$ is odd, it must be equal to $p$ or to $p^2$. The first is not possible because it would imply that $q-1=2^mp$, i.e. $q>p$ which contradicts with $f(q)=p$.
Thus $f(q)=p^2$ and $q-1=2^m$. Since $q\equiv 1 \pmod{2^m}$, we get $p^2=f(q)\equiv 5 \pmod{2^m}$. This is possible only f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\begin{align*}
(-8)^{4/3} &= (-8)^{1+1/3} \\
&= -8\times(-8)^{1/3} \\
&= -8\times (-1)^{1/3}\times 8^{1/3} \\
&= -2\times 8\times (-1)^{1/3} \\
... | They are both correct.
$$(-1)^{3} = -1$$
because it is
$$-1 \times -1 \times -1 $$
and a negative $\times$ a negative is a positive:
\begin{align*}
(-1 \times -1) \times -1 \\
= 1 \times -1 \\
= -1 \\
\end{align*}
Because of that, a solution to "What is the cube root of -1 ($\sqrt[3]{-1}$)" is $-1$.
This means that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve: $(1-x)^2y''-4xy'-(1+x^2)y=x$ To solve the differential equation:
$$(1-x)^2y''-4xy'-(1+x^2)y=x$$
I tried to solve the question by using the normal form of the equation but got stuck. Please help.
| There are no Liouvillian solutions (solutions that can be expressed using exp, algebraic functions, and integration).
There is an irregular point at $x=1$.
You might try a series solution in powers of $x$:
$$ y = \sum_{j=0}^\infty a_j x^j$$
where $a_j$ satisfy the recurrence
$$ \eqalign{-a_n &+ \left( {n}^{2}-n-7 \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more si... |
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
False.
It also shows the remaining four solutions are roots of $z^4 + z^2 + 1 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 3
} |
Fibonacci and Matrices Consider Matrix $$ A = \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix} $$
Investigate the sequence of powers of $A$
(i.e. $A^n$ for $n = 1, 2, 3, 4,\ldots$.
Verify that $$A^n = \begin{pmatrix}F_{n+1} &F_n \\ F_n & F_{n−1}\end{pmatrix}$$ for $n \geq 20$, where $F_n$ is the $n^{th}$ Fibonacci number.
I... | Hint:
$$A^{n+1} = A\cdot A^n = \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}\cdot\begin{pmatrix} F_{n+1} & F_{n}\\ F_{n} & F_{n-1} \end{pmatrix}$$$$$$
$$ = \begin{pmatrix} F_{n+1} + F_n & F_{n} + F_{n-1}\\ F_{n} + F_{n-1} & F_{n} \end{pmatrix} = \begin{pmatrix} F_{n+2} & F_{n+1}\\ F_{n+1} & F_{n} \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Karush-Kuhn-Tucker NLP
Consider the nonlinear program
Minimize: \begin{align}f(x,y) = \frac{1}{2}x^2 - 10xy + 10y^2\end{align}
Subject to: \begin{align}2x +y^2 &\le 5 \implies g_1(x,y)=2x + y^2 -5 \le0 \\ x^2 - ay &\le 2 \implies g_2(x,y) =x^2 -ay -2 \le 0\end{align}
a) Is the above problem a convex optimisation probl... | When $(x,y)=(1,1)$, we have $2x+y^2=3<5$. That is, the first constraint is strictly satisfied, and $2x+y^2-5\neq 0$. That and (3) implies $\lambda_1 = 0$.
Substituting $(x,y)=(1,1)$ and $\lambda_1=0$ into (1) yields $\lambda_2=-4.5$.
Since $\lambda_2$ is nonzero, (4) dictates that $x^2-ay-2=1-a-2=0$, or $a=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the shortest distance from the origin to the hyperbola $x^2+8xy+7y^2=225$
Find the shortest distance from the origin to the hyperbola $x^2+8xy+7y^2=225$
i know that
$$d(x_0, x) = \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$$
I also found this formula in my notes
$$ d(x_0,p) = \frac{|ax_0+by_0+cz_{0}-c|}{\sqrt{a^2+b^2+c... | To minimise the distance, the gradient at that point must be tangent to the circle centred at the origin (*):
$$2x + 8 \left(y + x \frac{dy}{dx} \right) + 14y \frac{dy}{dx} = 0$$
$$\implies (8x + 14y) \frac{dy}{dx} = -(2x+8y)$$
$$\implies \frac{dy}{dx} \frac{y}{x} = -1 \implies -\frac{x+4y}{4x+7y} \frac{y}{x} = -1 \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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2nd degree matrix equation Let $X$ be a matrix with 2 rows and 2 columns.
Solve the following equation:
$$ X^2 = \begin{pmatrix}
3 & 5\\
-5 & 8
\end{pmatrix} $$
Here is what I did:
Let $ X = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix} $. After multiplying I got the following system:
$$ \left\{\begin{matrix}
a^2 + bc... | We have the following criteria which you already stated correctly, but you missed one more information $(5)$ - still, you can solve this root problem without this additional knowledge by plugging in recursively - which comes from the determinant, we get then
\begin{align}
a^2 + bc &= 3 \tag1\\
ab + bd &= 5\tag2 \\
ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show $\frac{1}{n - 1} \geq e^\frac{1}{n} - 1, for~n \gt 1$ I am to show that
$\frac{1}{n - 1} \geq e^\frac{1}{n} - 1, for~n \in \mathbb{N}^+, n \gt 1$.
I tried substituting using $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$, which gives: $\frac{1}{n-1} \geq \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} ...$, and I also ... | You have said $e^{1/n} - 1 = \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} + \ldots$
You could also say $\frac{1}{n-1}= \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \ldots$, and then $n\gt 1$ will give you what you want since each term in the second series is greater than or equal to the corresponding term in the firs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge? Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge?
| $$\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{1}{3n+3}> \frac{1}{3n+1}$$
So no.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is there a way to solve for $x$ and $y$ in this simultaneous equation. Is there a way to solve for x and y in this simultaneous equation?
$$2x - 3y = 4 $$
$$4x - 6y = 5 $$
Attempt:
I tried solving it but $x$ and $y$ keeps eliminating.
| $$
\begin{cases}
2x - 3y = 4 \\
4x - 6y = 5 \\
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x = 2- \frac{3y}{2} \\
4x - 6y = 5 \\
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x = 2- \frac{3y}{2} \\
4\left(2- \frac{3y}{2}\right) - 6y = 5 \\
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x = 2- \frac{3y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
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Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule
Find the following limit
$$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
without using L'Hopital's rule.
I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x... | Another approach starts with using the half-angle formula for the sine function to establish the identity
$$\frac{2+\cos x}{3}=1-\frac23 \sin^2(x/2)$$
Then, recall the inequalities
$$\frac{-x}{1-x}\le \log(1-x)\le -x \tag 1$$
$$1-x\le e^{-x}\le \frac{1}{1+x} \tag 2$$
and
$$\frac14 x^2\cos^2 (x/2) \le \sin^2 (x/2)\le \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Solving this second order ODE with power series method Problem: Use the power series method to find a general solution for the following differential equation: $$ (x^2 - 3) y'' + 2x y' = 0 $$
Attempt: We see that $x = \pm \sqrt{3}$ is a singular point of this ODE. Hence we apply the power series method around $x = 0$, ... | Prior to substituting your power series, consider starting off by reducing the order of the ODE via the substitution $y'=z$, so you have
$$(x^2-3)z'+2xz=0$$
Now, plugging in terms of power series you have
$$(x^2-3)\sum_{n\ge1}na_nx^{n-1}+2x\sum_{n\ge0}a_nx^n=0$$
Distributing, rewriting, etc, you have
$$\sum_{n\ge3}(na_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$
$|(1+i\sqrt{3})(1+i)|=8$
U... | We have $$(1+i\sqrt{3})=2(\cos \frac{\pi}{3}+i\sin\frac{\pi}{3}),$$
$$(1+i)=\sqrt{2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4}).$$ Then the product of these is $2\sqrt{2}(\cos \frac{7\pi}{12}+i\sin\frac{7\pi}{12})$. So $$2\sqrt{2}(\cos \frac{7\pi}{12}+i\sin\frac{7\pi}{12})(\cos {\phi}+i\sin{\phi})=2\sqrt{2}\left(\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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min polynomial of $\sqrt{6+2\sqrt{5}}$ min polynomial of $\sqrt{6+2\sqrt{5}}$ over $\mathbb{Q}$
I found $\alpha^4 -12\alpha^2 +16 = 0$ but wolframalpha does not agree with me. furthermore, would anything change is it was over $\mathbb{R}$?
| You can try to factor $\alpha^4-12\alpha^2+16$. Assuming that $\sqrt{6+2\sqrt{5}}$ is not in $\mathbb{Q}$, then the best that you can hope for is that this factors into quadratics. We write the following:
$$
\alpha^4-12\alpha^2+16=(\alpha^2+r\alpha+s)(\alpha^2+t\alpha+u).
$$
Multiplying this out, one has
$$
\alpha^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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In order for both roots of the equation $x^2 + ( m + 1 ) x + 2m - 1 = 0$ to not be real, it must be ... In order for both roots of the equation $x^2 + ( m + 1 ) x + 2m - 1 = 0 $ to not be real, it must be the case that ...
A. $m > 1$
B. $1 < m < 5$
C. $1 \leq m \leq 5$
D. $m < 1$ or $m > 5$
E. $m \leq 1$ or $m \geq 5$
| In case you forget the discriminant like me: The equation
$$
x^2 + (m+1) x + 2m - 1 = 0
$$
leads to the condition
$$
\left(x + \frac{m+1}{2}\right)^2 = \left(\frac{m+1}{2}\right)^2 - 2m + 1 < 0
$$
because then the square root of a negative number prevents real solutions $x$ to exist.
That right hand side is a quadrati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Make all the sixes. Fun Given the following it is possible to complete each case so that they are all true? (i.e. so that the equation .... = 6 is true)
You can add any mathematical operations and parentheses but you cannot add any numbers.
i.e. $\sqrt{4}$ is fine but $\sqrt[3]{8}$ is not.
So...
$$0\text{ }\text{ }\tex... | $$(0! + 0! + 0!)! = 6$$
$$(1 + 1 + 1)! =6$$
$$2 + 2+ 2 = 6$$
$$3! +3 - 3 = 6$$
$$4 + 4 - \sqrt{4} = 6$$
$$5+5/5 = 6$$
$$7 - 7/7 = 6$$
$$\lfloor \sqrt{8} \rfloor +\lfloor \sqrt{8} \rfloor +\lfloor \sqrt{8} \rfloor =6$$
$$\sqrt{9} + (9 / \sqrt{9}) = 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Where am I going wrong when trying to solve this system of equations using Gaussian Elimination? $$3x-y+z=5 \\ 2x+y-z=1 \\ x-y+z=2 \\ 4x+4y+z=3$$
Steps I took:
$$\left[\begin{array}{rrr|r}
3 & -1 & 1 & 5 \\
2 & 1 & -1 & 1 \\
1 & -1 & 1 & 2\\
4 & 4 & 1 & 3
\end{array}\right]$$
$$\Rightarrow { R }_{ 1 }... | Hint:if you want o solve the first threee equations you will get:
he system
$$x-y+z=2$$
$$2x+y-z=1$$
$$3x-y+z=5$$
multplying the fist equation by (-2) ad adding this to the second one we get
$$3y-3z=-3$$
multiplying the irs equation by -3 and adding this to the third one we obtain
$$2y-2z=-1$$
aer simplifiation we get
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cd... | ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}
=\frac{n!2^n}{(2n+1)!!}
$
Since
$(1+x)^n
=\sum_{k=0}^n \binom{n}{k} x^k
$,
$(1-x^2)^n
=\sum_{k=0}^n \binom{n}{k}(-1)^k x^{2k}
$.
Integrating from $0$ to $1$,
$\int_0^1 (1-x^2)^n\,dx
=\sum_{k=0}^n \binom{n}{k}(-1)^k \int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
A club with 32 women and 10 men needs to form a committee of size 5 I want to make sure that I'm thinking of this correctly.
A club with 32 women and 10 men needs to form a committee of
size 5:
(a) How many committees are possible? - I said $$
\begin{pmatrix}
42 \\
5 \\
\end{pmatrix}
$$ ... | In part (e), order matters. Selecting Angela to be chair and Barbara to be treasurer is different from selecting Barbara to be chair and Angela to be treasurer. Therefore, there are $42$ ways to fill the position of chair, $41$ ways to pick the treasurer, $40$ ways to select the secretary, and so forth. Hence, there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Square of a mod
Find $9^{16} \pmod{25}$ by separation.
The separation means, that $25 = 5^2$.
But $9^{16} \equiv 1 \pmod{5}$, so I how I derive that $9^{16} \equiv 16 \pmod{25}$?
| As $9=3^2,9^{16}=(3^2)^{16}=3^{32}$
Now $3^3\equiv2\pmod{25}\implies3^{32}=3^2(3^3)^{10}\equiv3^2\cdot2^{10}$
Now $2^5\equiv7\implies2^{10}=(2^5)^2\equiv7^2\equiv-1$
$\implies3^2\cdot2^{10}\equiv9\cdot-1\equiv16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$ How can I compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$?
| $$\lim_{x\to 0} \frac{\sin^3(x)}{(2x)^3}=$$
$$\lim_{x\to 0} \frac{\sin^3(x)}{8x^3}=$$
$$\frac{1}{8}\lim_{x\to 0} \frac{\sin^3(x)}{x^3}=$$
$$\frac{1}{8}\lim_{x\to 0} \frac{\frac{\text{d}}{\text{d}x}\sin^3(x)}{\frac{\text{d}}{\text{d}x}x^3}=$$
$$\frac{1}{8}\lim_{x\to 0} \frac{3\sin^2(x)\cos(x)}{3x^2}=$$
$$\frac{1}{8}\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Prove that for $n$ an integer larger than $0$, $\frac{n^2}{n+1}$ is never an integer. I have a couple questions.
$1)$ Is my proof is considered valid $2)$ If it is good may you tell me how to improve it or if you have a better proof $3)$ how may I prove the obvious assumption I made that an integer and a non integer ma... | Note that
$\frac{2n+1}{n+1}-2
=\frac{2n+1-2(n+1)}{n+1}
=\frac{-1}{n+1}
$.
Since
$\frac{-1}{n+1}$
is not an integer,
$\frac{2n+1}{n+1}$
is not an integer.
To handle
$\frac{n^2}{n+1}$,
use
$(n+1)(n-1)
=n^2-1
$.
$\frac{n^2}{n+1}-(n-1)
=\frac{n^2-(n-1)(n+1)}{n+1}
=\frac{n^2-(n^2-1)}{n+1}
=\frac{1}{n+1}
$.
Again,
this is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Rolling of one ellipse on another ellipse of same size when initially touching each other at the end of their major axis. I have a question related to conic section which i could not understood. The question is
$E_1$:$ \frac{x^2}{a^2} +\frac{y^2}{b^2}=1(a>b)$ is a given ellipse. Another ellipse $E_2$: is of same size ... | Let $O'(X,Y)$ be the centre of $E_2$.
Let $P(a\cos\theta,b\sin\theta)$ be the tangent point of $E_1,E_2$. Also, let $l$ be the common tangent at $P$. Then, note that $E_1,E_2$ are symmetric about $l$: $\frac{\cos\theta}{a}x+\frac{\sin\theta}{b}y=1$. We have
$$\frac YX=\frac{a\sin\theta}{b\cos\theta}$$
$$\frac{\cos\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
If one has the result it is not difficult to verify it by induction, but how can I solve it without induction ?
| We have
$$\begin{align}
1+\sum_{k=1}^{j-1}\frac{1\cdot 2 \cdot 3\cdots k}{3\cdot 4\cdot5\cdots k+2}&=1+\sum_{k=0}^{j-1}\frac{k!}{(k+2)!/2}\\\\
&=1+2\sum_{k=0}^{j-1}\frac{1}{(k+1)(k+2)}\\\\
&=1+2\sum_{k=0}^{j-1}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\\\\
&=1+2\left(1-\frac{1}{j+1}\right)\\\\
&=3-\frac{2}{j+1}
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proportion with 3 variables - binomial coefficients So, if we're given something like this:
$$\binom{n}{k}:\binom{n+1}{k}:\binom{n+1}{k+1}=3:4:8$$
How do I rewrite this so I can manipulate it?
Edit: Is there a general procedure for n variables?
|
Note, that the equality of proportions
\begin{align*}
a:b:c=x:y:z
\end{align*}
is a compact notation for
\begin{align*}
\frac{a}{b}=\frac{x}{y},\qquad\frac{a}{c}=\frac{x}{z},\qquad\frac{b}{c}=\frac{y}{z}\tag{1}
\end{align*}
Each equality in (1) can be derived from the other two.
So, in order to check if we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding real solutions to $\sqrt{x}+\sqrt[3]{x^2-1}+\sqrt[4]{x^3+15}=x^2+2$ I found that $x=1$ is a solution of
$$\sqrt{x}+\sqrt[3]{x^2-1}+\sqrt[4]{x^3+15}=x^2+2.$$
How would one find other real solutions?
| To find a polynomial with that root:
You have $A+B+C=x^2+2$. Consider $(x^2+2)^2=(A+B+C)^2= x+B^2+C^2+2AB+2AC+2BC$.
Take the first 25 powers of $x^2+2$, from $1=(x^2+2)^0$ to $(x^2+2)^{24}$. They will all be linear combinations of $A,B,C,B^2,C^2,C^3,AB,AB^2,AC,ABC,AB^2C,...$ with coefficients that are polynomials in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the identity $\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$
Let $n,m \in \mathbb{N}$. Prove the identity $$\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$$
This seems very similar to Vandermonde identity, which states that for nonnegative integers we have $\sum^{m}_{k=0}\binom{m}{k}\binom{n}{r-k} = \binom{m... | For $t=0,1,\cdots, n$, $\binom{m+t}{m}$ is the coefficient of $x^m$ in the expansion of $(1+x)^{m+t}$.
Hence,
$$\binom{m}{0}+\binom{m+1}{1}+\cdots +\binom{m+n}{n},$$
i.e.
$$\binom{m}{m}+\binom{m+1}{m}+\cdots +\binom{m+n}{m}$$
is the coefficient of $x^m$ in the expansion of
$$(1+x)^m+(1+x)^{m+1}+\cdots +(1+x)^{m+n}.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$? Consider
$$x^4 + x^2 + y^4 + y^2 = 2$$
It is a smooth non-intersecting circle like curve in the plane.
A bit like a Hyperellipse.
See https://en.m.wikipedia.org/wiki/Superellipse
What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$ ... | In polar coordinates: $$r^4\cos^4t+r^2\cos^2t+r^4\sin^4t+r^2\sin^2t=2$$
$$r^4\left(\cos^4t+\sin^4t\right)+r^2-2=0$$
$$r^2=\frac{-1+\sqrt{1+8\left(\cos^4t+\sin^4t\right)}}{2\left(\cos^4t+\sin^4t\right)}$$
Now the arc length for polar coordinates is $$\int_0^{2\pi}\sqrt{r^2+\left(\frac{dr}{dt}\right)^2}dt$$ So it appears... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Trigonometric problem? This may be really easy but is giving a hard time. I'm a beginner at trigonometry so hope I can understand it.
*
*If $\sin{\frac{x}{2}}+\cos{\frac{x}{2}}= 1.4$, then $\sin{x}=?$
*If $\sin{a}-\cos{a}=a$, then $\sin(2a)=?$
| The trick with these questions is finding the way to start. We'll need two identities
$$\sin^2 t + \cos^2 t = 1, \qquad\text{and}\qquad \sin(2t) = 2\sin t \cos t$$
All we have to do is square each side of the equation
$$\left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2 = (1.4)^2$$
A bit of algebra gives
$$\sin^2 \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1524752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $ \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x} $. I need to solve the following integral:
$$
I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.
$$
Wolfram Alpha gives the answer as $ \dfrac{\pi}{2 \sqrt{2}} $.
I think it’s achievable by complex analysis, but I really ... | This is the general antiderivative. Just take the limits and you're good.
$$\int{\frac{\sqrt{x-1}}{(x+1)^2}dx}$$
Substitute $u = \sqrt{x-1}$
$$= \int{\frac{u}{(u^2+2)^2}du}$$
$$= 2\int\left(\frac{u}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$
$$= \int\frac{u}{\frac{u^2}{2}+1}du-4\int\frac{2}{(u^2+2)^2}du$$
Substitute $s = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Find the limit of the sequence $\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) _{n\in N}$ My answer is as follows, but I'm not sure with this:
$\lim _{n\rightarrow \infty }\dfrac {\sqrt {2n^{2}+n}}{\sqrt {2n^{2}+2n}}=\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}$
$\lim _{n\ri... | If you know that for small $x$ we have that $\sqrt{1+x}$ behaves like $1+\frac{x}{2}$, then $\sqrt{2n^2+2an} = \sqrt{2}\cdot\sqrt{(n+a)^2-a^2}$ behaves like $\sqrt{2}(n+a)\left(1-\frac{a^2}{2(n+a)^2}\right)$, or $n\sqrt{2}+a\sqrt{2}+O\left(\frac{1}{n}\right)$, for large $n$. To compute the limit from that is straightfo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the coefficient of partial expansion $\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}$ I want to decompose the equation:
$$\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}=\frac{A}{1-x}+\frac{B}{1+0.5x+0.5x^2}$$
I found $A$ by multiple both side with $1-x$ and plug $x=1$. However, it is so difficult to find $B$. Could you help me to find $B$
| Starting from Mario G's suggestion in the comments...
$$\frac{x+\frac{3}{2}}{(1-x)(2+x+x^2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+2}$$
Multiply through by $(x-1)(2+x+x^2)$,
$$x+\frac{3}{2}=A(x^2+x+2)+(Bx+C)(x-1)$$
Collect powers of $x$ on the RHS.
$$x+\frac{3}{2}=(A+B)x^2+(A-B+C)x+(2A-C)$$
Now due to the linear independence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
convergence of series $ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$ $$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$$
determine the convergence of the series
$$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\} < {n}^{p} \cdot \frac{1}{\sqrt{n-1}\s... | Hint:
$$\begin{align}
\frac{1}{\sqrt{n-1}} - \frac{1}{\sqrt{n}}
&= \frac{1}{\sqrt{n}}\left( \frac{1}{\sqrt{1-\frac{1}{n}}} - 1\right)
= \frac{1}{\sqrt{n}}\left( 1+\frac{1}{2n} + o\left(\frac{1}{n}\right) - 1\right) \\
&= \frac{1}{2n^{3/2}} + o\left(\frac{1}{n^{3/2}}\right)
\end{align}$$
using Taylor expansions: $(1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_{\frac{-\pi}4}^{\frac{\pi}4}\ln(\sin x+\cos x)\mathrm{d}x$ $$\int_{\frac{-\pi}4}^{\frac{\pi}4} \ln(\sin x+\cos x)\mathrm{d}x $$
I just can't think of any technique to solve this question.
Can anyone help me with at least how to begin?
| Suppose one knows that,
1)$\displaystyle \int_0^1 \dfrac{\ln x}{1+x^2}dx=-G$
2) $\displaystyle \int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{2}-G$
Where $G$ is the Catalan constant.
$I=\displaystyle \int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\sin x+\cos x)dx$
$I=\displaystyle \int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
A nonlinear optimization problem with difficult Kuhn-Tucker system of equations I know about the sufficient optimality theorem Kuhn-Tucker, and this problem can use the Kuhn-Tucker theorem directly, but ridiculously, I got stuck on the system of equations to find one root for optimization value. Hope someone can help m... | $$\theta_{min}=\theta(x_0,y_0,z_0)=63-\frac{3 \sqrt{132685}}{35}$$
$${x_0}=1-\frac{54 \sqrt{132685}}{7805},\;{y_0}=\frac{61 \sqrt{132685}}{15610}-\frac{5}{2},\;{z_0}=\frac{3}{2}-\frac{3 \sqrt{132685}}{3122}$$
Used free CAS Maxima
http://maxima.sourceforge.net/
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $ \sum_{k=0}^n k4^k = \frac 49((3n-1)4^n + 1) $ by induction Prove that for every position integer $n$ that
$$ \sum_{k=0}^n k4^k = \frac 49((3n-1)4^n + 1) $$
Proof: Let $P(n)$ denote the above statement.
Base case: $n=1$ : Note that $$ \sum_{k=1}^1 k4^k = \frac 49((3(1)-1)4^{(1)} + 1) $$
$\frac 49((3(1)-1)4^{(... | So starting from where you ended we just have to manipulate properly:
$$\frac{4}{9}((3s - 1)4^s + 1) + (s + 1)4^{s + 1} $$
$$= \frac{4}{3}s4^s - \frac{4}{9}4^s + \frac{4}{9} + s4^{s + 1} + 4^{s + 1}$$
$$=\Big( \frac{s}{3} - \frac{1}{9} + s + 1 \Big)4^{s + 1} + \frac{4}{9}$$
$$=\frac{4}{9}\Big( (3s + 2)4^{s + 1} + 1 \Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)... | Take a look for $$x^2-4x+1=0 \\x_1.x_2=1 \\x^2+1=4x \\\div x
\\x+\frac{1}{x}=4 \\$$so we can write an equation with $x_1^2,x_2^2 ,roots$
$$s'=x_1^2+x_2^2=s^2-2p=16-4\\x_1^2.x_2^2=1^2 \\\to z^2-14z+1=0 \\z=x^2 \\\to x^4+\frac{1}{x^4}=14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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Find sum of infinite anharmonic(?) series I need help with this:
$$
\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot5\cdot6}\dots
$$
I don't know how to count sum of this series. It is similar to standard anharmonic series so it should have similar solution. However I can't work it out.... | What you're evaluating is the infinite series
$$\sum_{n = 1}^\infty \frac{1}{n(n+1)(n+2)(n+3)}.$$
Since
$$\frac{1}{n(n+3)} = \frac{1}{3n}-\frac{1}{3(n+3)},$$
then
$$\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{3n(n+1)(n+2)}-\frac{1}{3(n+1)(n+2)(n+3)}.$$
So your series telescopes to
$$\frac{1}{3(1)(2)(3)} = \frac{1}{18}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
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How to find matrix from general solution?
Find a $2 \times 3$ system (two equations with three unknowns) such that its general solution has form $\begin{pmatrix}1\\1\\0 \end{pmatrix}+s\begin {pmatrix}1\\2\\1\end{pmatrix},\ s \in \Bbb R.$
I tried thinking that $s\begin {pmatrix}1\\2\\1\end{pmatrix}$ is solution to ke... | Let $$A\mathbf{x}=\mathbf{b}\tag{1}$$ be the system in matrix form, where $A$ is the coefficient matrix for the asked system in matrix form. Where
$\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$
Notice that $\ker(A)=\{k\mathbf{v}|k\in\mathbb{R}\}$, where
$\mathbf{v}=\begin{pmatrix}1\\2\\1\end{pmatrix}$
You can t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is $a\sqrt[3]{2} + b\sqrt[3]{4}$ irrational? I need to prove that $$ a\sqrt[3]{2} + b\sqrt[3]{4}$$ is irrational, while $a$,$b $ are non zero rationals.
I know that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational and I also know how to prove it, but I can't think of any reasonable implication that would state: $a\sqrt[3]{2... | Assume $a\sqrt[3]2+b\sqrt[3]4$ is rational. This gives that $$
(a\sqrt[3]2+b\sqrt[3]4)^2=2b^2\sqrt[3]2+a^2\sqrt[3]4+4ab
$$is also rational. Subtracting $4ab$ doesn't change rationality. This means that
$$
a(2b^2\sqrt[3]2+a^2\sqrt[3]4)-2b^2(a\sqrt[3]2+b\sqrt[3]4)=(a^3-2b^3)\sqrt[3]4
$$
is rational. But $\sqrt[3]4$ is ir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| Before starting, let us note that the formula is not really $a$ nested
radicals formula, because it suffices to replace $\sqrt{B}$ by $B,$ and
replace $C^{2}=A^{2}-B,$ by $C^{2}=A^{2}-B^{2}.$ So let me show you how starting
from the l.h.s. written as
\begin{equation*}
\sqrt{A+B}
\end{equation*}
to arrive to the r.h.s w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
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Find all eigenvalues and eigenvectors of this $3\times3$ matrix I'm reviewing for a Differential Equations exam and one of the questions in the practice exam asks me to find all the eigenvectors of $3 \times 3$ matrix $\mathbf{A}$ given that $1$ is an eigenvalue of the matrix.
The matrix $\mathbf{A}$ is
$$
\left(
\begi... | Problem statement
Resolve the eigensystem for
$$
\mathbf{A} =
%
\left[
\begin{array}{crr}
1 & -1 & 1 \\
1 & 0 & -1 \\
0 & -1 & 2 \\
\end{array}
\right]
$$
Eigenvalues
The problem is succinctly formulated in the comment of @Bye_World. Start with
$$
\mathbf{A} - \lambda \mathbf{I}_{3} =
\left[
\begin{array}{ccc}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show this function of two variables $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at (0,0) Is the function $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at $(0, 0)$?
Trying to use the epsilon-delta definition of continuity to prove this but can't figure it out.
$\sqrt{x^2+y^2} < \delta$ implies $\left|\frac{2x^2y^3}{x... | **Hint:**As you approach $(0,0)$ along $y^3=mx^2$, you get $lim_{(x,y)→(0,0)}f(x,y)=2m/(1+m^2)$ which depends on arbitrary $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $f(n)$ , $f(n)=\frac{1}{2^n}+\frac{1}{2^{n+2}}\binom{n+1}{2}+ \frac{1}{2^{n+4}}\binom{n+3}{4}+\frac{1}{2^{n+6}}\binom{n+5}{6}+ \cdots$ I would appreciate if somebody could help me with the following problem
Q: Find $f(n)$? (n $\in \mathbb{N}$ )
$$f(n)=\frac{1}{2^n}+\frac{1}{2^{n+2}}\binom{n+1}{2}+ \frac{1}{2^{n+4}... | $(1+x)^{-n}=1+(-n)x+\dfrac{(-n)(-n-1)}{2!}x^2+\dfrac{(-n)(-n-1)(-n-2)}{3!}x^3+...$
$(1-x)^{-n}=1-(-n)x+\dfrac{(-n)(-n-1)}{2!}x^2-\dfrac{(-n)(-n-1)(-n-2)}{3!}x^3+...$
By adding
$\dfrac{1}{2}((1+x)^{-n}+(1-x)^{-n})=1+\dfrac{(-n)(-n-1)}{2!}x^2+\dfrac{(-n)(-n-1)(-n-2)(-n-3)}{4!}x^4+...$
$1+\binom{n+1}{2}x^2+\binom{n+3}{4}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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