Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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 factoring.
After some help I found that you move the identity around, so:
$\sin^2(\theta)-\cos^2(\theta)=\sin^4(\theta)-\cos^4(\theta)$
Then,
$\sin^2(\theta)-\cos^2(\theta)=(\sin^2(\theta)+\cos^2(\theta))(\sin^2(\theta)-\cos^2(\theta))$
the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?
| 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-\sin^2\theta) = 1-2\sin^2\theta + \sin^4\theta $
$\sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta $
$\sin^2\theta + (1-2\sin^2\theta + \sin^4\theta) = \cos^2\theta + \sin^4\theta $
$\sin^2\theta + 1-2\sin^2\theta + \sin^4\theta = \cos^2\theta + \sin^4\theta $
$\sin^4\theta-\sin^2\theta+1= \cos^2\theta + \sin^4\theta $
$\sin^4\theta-(1-\cos^2\theta)+1=\cos^2\theta + \sin^4\theta $
$\sin^4\theta-1+\cos^2\theta+1=\cos^2\theta + \sin^4\theta $
$\sin^4\theta+\cos^2\theta=\cos^2\theta + \sin^4\theta $
| {
"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 , but i cant seem to reach anywhere.. Any suggestions?
| $$\implies x^2(y-1)+x(3-3y)+4y+4=0$$
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 $x^2-3x+4=\dfrac{4x^2-12x+16}4=\dfrac{(2x-3)^2+7}4$
Now $0\le(2x-3)^2\le\infty\iff\dfrac74\le x^2-3x+4\le\infty\iff\dfrac47\ge \dfrac1{x^2-3x+4}\ge0$
Can you take it from here?
| {
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"url": "https://math.stackexchange.com/questions/1414298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 asymptotic form of the gamma function. I believe the answer should be $\frac{1}{12t}\sqrt{\frac{2\pi}{t}}$.
Edit: Apparently this integral evaluates to zero. As mentioned I was attempting to calculate the asymptotics of the Gamma function using
$$\Gamma(t+1) \sim t^{t+1}e^{-t}\int_0^{\infty}e^{-t(s-1)^2/2}\left(1 + \frac{t(s-1)^3}{3} + \left( \frac{t(s-1)^3}{3} \right)^2 + \dots \right)^2 $$
Hmm...
| 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 exponential and leads to
\begin{align}
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 \\
&= \frac{1}{3} \, e^{-t/2} - \frac{2}{3 \, t} \, \int_{0}^{\infty} \frac{d}{ds} \left( e^{- \frac{t (s-1)^{2}}{2}} \right) \, ds \\
&= \frac{1}{3} \, e^{- t/2} + \frac{2}{3 \, t} \, e^{-t/2}
\end{align}
From this
\begin{align}
\int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds = \frac{1}{3} \, \left( 1 + \frac{2}{t} \right) \, e^{-t/2}.
\end{align}
In the case $t \to \infty$ the value of the integral tends to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$$\frac{3}{2}\int \frac{\tan\theta \sec\theta}{(9\tan^2\theta)(3\tan\theta)} \,d\theta.$$
| 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} \, d\theta = \frac{1}{18} \, \int \frac{\cos\theta}{\sin^{2}\theta} \, d\theta \\
&= - \frac{1}{18} \, \frac{1}{\sin\theta} + c_{0}
\end{align}
Now,
\begin{align}
\frac{2 \, x}{3} &= \frac{1}{\cos\theta} \to \cos\theta = \frac{3}{2 \, x} \\
\theta &= \cos^{-1} \left(\frac{3}{2 \, x}\right) \\
\sin\theta &= \frac{1}{\sqrt{1 - \left(\frac{3}{2 \, x}\right)^{2}}} = \frac{2 \, x}{\sqrt{4 \, x^2 - 9}}
\end{align}
leading to
$$I = - \frac{1}{9} \, \frac{x}{\sqrt{4 \, x^2 - 9}} + c_{0}.$$
| {
"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)}
\end{align}
Then I use quadratic formula on numerator to factor it :
$a=2,b=-1,c=-1$
$$=\frac{2(x+2)(x-\frac{5}{2})}{(x-3)(2x+1)}$$
But apparently this can be factored further. What else can I do?
| $2x^2 -x -1$ 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 $$\int e^{\tan{u}}\,\mathrm{d}u.$$ From here, the two answers diverge radically, and the content of this post will be entirely concerned with evaluating this integral, which in turn answers the question.
In the other answer, we expanded $e^{\tan{u}}$ as a power series in $\tan{u}$. Now, instead, we shall first expand $\tan{u}$ as a power series in terms of $u$. Then $e^{\tan{u}}$ is an infinite product of exponentials, each of which can be expanded as a power series. From this sequence of expansions, we produce a power series for $e^{\tan{u}}$ which can be integrated term-by-term.
The power series for $\tan{u}$, valid for $|u|<\pi/2$, is $$\tan{u} = \sum_{n=1}^{\infty} {B_{2n}(-4)^{n}(1-4^{n}) \over (2n)!}u^{2n-1} = u+\frac{u^{3}}{3}+\frac{2u^{5}}{15}+\ldots.$$
Hence, for $|u|<\pi/2$, we have
\begin{eqnarray*}
\exp{\tan{u}} & = & \exp{\left(\sum_{n=1}^{\infty} {B_{2n}(-4)^{n}(1-4^{n}) \over (2n)!}u^{2n-1}\right)}\\
& = & \prod_{n=1}^{\infty}\exp{\left( {B_{2n}(-4)^{n}(1-4^{n}) \over (2n)!}u^{2n-1}\right)}\\
& = & \exp{u}\cdot\exp{\frac{u^{3}}{3}}\cdot\exp{\frac{2u^{5}}{15}}\cdot\ldots\\
& = & \sum_{k=1}^{\infty} \left({B_{2}^{k}(-4)^{k}(1-4)^{k} \over k!2!^{k}}u^{k}\right) \cdot \sum_{k=1}^{\infty} \left({B_{4}^{k}(-4)^{2k}(1-4^{2})^{k} \over k!4!^{k}}u^{3k}\right)\cdot\ldots\\
& = & \left( 1+\frac{u}{1!}+\frac{u^{2}}{2!}+\ldots \right)\left( 1+\frac{u^{3}}{3^{1}\cdot1!}+\frac{u^{6}}{3^{2}\cdot2!}+\ldots \right)\left( 1+\frac{2u^{5}}{15\cdot1!}+\frac{2^{2}u^{10}}{15^{2}\cdot2!}+\ldots \right)\\
& = & 1 + u + \frac{u^{2}}{2!} + \left(\frac{u^{3}}{3!}+\frac{u^{3}}{3\cdot1!}\right) + \left(\frac{u^{4}}{4!}+\frac{u}{1!}\cdot\frac{u^{3}}{3\cdot1!}\right)\\
&& \;\; + \left(\frac{u^{5}}{5!} +\frac{u^{2}}{2!}\cdot\frac{u^{3}}{3\cdot1!}+ \frac{2u^{5}}{15}\right)+\ldots\\
& = & 1 + u + \frac{u^{2}}{2} + \frac{u^{3}}{2}+\frac{3u^{4}}{8}+\frac{37u^{5}}{120}+\ldots.
\end{eqnarray*}
This gives us a series that I think we should be able to integrate term-by-term, to get
\begin{eqnarray*}
\int e^{\tan{u}}\,\mathrm{d}u & = & \int(1 + u + \frac{u^{2}}{2} + \frac{u^{3}}{2}+\frac{3u^{4}}{8}+\frac{37u^{5}}{120}+\ldots)\,\mathrm{d}u\\
& = & \text{constant} + u + \frac{u^{2}}{2} + \frac{u^{3}}{6}+\frac{u^{4}}{8}+\frac{3u^{5}}{40}+\frac{37u^{6}}{720}+\ldots.
\end{eqnarray*}
Therefore, assuming that our interval of integration is within $|\tan^{-1}{x}|<\pi/2$, we have
\begin{eqnarray*}
\int\frac{e^{x}}{1+x^{2}}\,\mathrm{d}x = \text{constant} + \tan^{-1}{x} + \frac{(\tan^{-1}{x})^{2}}{2}+ \frac{(\tan^{-1}{x})^{3}}{6}+\frac{(\tan^{-1}{x})^{4}}{8}+\frac{3(\tan^{-1}{x})^{5}}{40}+\frac{37(\tan^{-1}{x})^{6}}{720}+\ldots,
\end{eqnarray*}
and I'm pretty sure we can adjust this to other intervals of integration using the fact that $\tan{(u+\pi)}=\tan{u}$ for all $u\in\mathbb{R}$ for which $\tan{u}$ is defined.
| {
"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:
$$\begin{array}{*4c} 4 & 3+1 & 2+2 & 1+3 \\ \\ ~2+1+1~ & ~1+2+1~ & ~1+1+2~ & ~1+1+1+1~ \end{array}$$
and $d_0=1$, since $0$ can be rolled in one way (roll no dice).
Let $D(x)$ be the generating function
$$D(x) = d_0 + d_1x + d_2x^2 + d_3x^3 + \cdots .$$
Then $\frac 1{D(x)}$ is a polynomial. What polynomial is it?
I know for sure that $\frac{1}{D(x)}$ is referring to $\frac{1}{1-(x+x^2+x^3+x^4+x^5+x^6)}$ the geometric function. But I don't understand what to put as the answer?
I understand this is a duplicate question, but I don't understand how to translate and clarify.
| 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
} |
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:
$\frac{dx}{dt}= \frac{-1}{1+t}x$
$\frac{dx}{dt}-2= \frac{-1}{1+t}x$
$dx-2dt= \frac{-dt}{1+t}x$
$dx=\frac{-dt}{1+t}x +2dt$
| $$x'=\frac { -1 }{ 1+t } x+2\\ { x }^{ \prime }+\frac { 1 }{ 1+t } x=0\\ \frac { dx }{ dt } =-\frac { x }{ 1+t } \\ \int { \frac { dx }{ x } } =-\int { \frac { dt }{ 1+t } } \\ \ln { \left| x \right| } =-\ln { C\left| 1+t \right| =\ln { \frac { C }{ \left| 1+t \right| } } } \\ x=\frac { C }{ 1+t } \\ { x }^{ \prime }=\frac { { C }^{ \prime }\left( t \right) \left( 1+t \right) -C\left( t \right) }{ { \left( 1+t \right) }^{ 2 } } \\ \frac { { C }^{ \prime }\left( t \right) \left( 1+t \right) -C\left( t \right) }{ { \left( 1+t \right) }^{ 2 } } +\frac { 1 }{ 1+t } \frac { C }{ 1+t } =2\\ \frac { { C }^{ \prime }\left( t \right) }{ { 1+t } } =2\\ C\left( t \right) =2\int { \left( 1+t \right) dt } =2t+{ t }^{ 2 }+C_{ 1 }\\ x\left( t \right) =\frac { 2t+{ t }^{ 2 }+C_{ 1 } }{ 1+t } \\ x\left( 0 \right) =\frac { 2\cdot 0+{ 0 }^{ 2 }+C_{ 1 } }{ 1+0 } =1\Rightarrow { C }_{ 1 }=1\\ $$
$$x\left( t \right) =\frac { 2t+{ t }^{ 2 }+1 }{ 1+t } =\frac { { \left( 1+t \right) }^{ 2 } }{ 1+t } =1+t\\ \\ \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$${k\choose n-m}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w^{n-m+1}} \; dw.$$
This gives for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}}
\sum_{k=0}^n {n\choose k} \frac{(1+w)^k}{z^k}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}}
\left(1+\frac{1+w}{z}\right)^n
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}}
(1+w+z)^n
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}}
{n\choose n-m} (1+z)^{n-(n-m)}
\; dz
\\ = {n\choose m} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+m}}{z^{n+1}} \; dz
\\ = {n\choose m} {n+m\choose n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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
$$xy=2048$$
$$x=\frac{2048}{y}$$
$$\frac{\frac{2048}{y}}{y} - \frac{y}{\frac{2048}{y}} = 7.875$$
$$\frac{2048}{y^2} -\frac{y^2}{2048} = 7.875$$
$$2048^2 -y^4 =7.875\cdot 2048 y^2$$
If we compare the the first equation (x/y) - (y/x) = 7.875 to the euqtion that we just found ((sqrt. 2048)/y) - (y/(sqrt. 2048)) = 7.875 , x is in the same place so, x must be equal to the sqrt. of 2048.
Is this solution correct or are there any better solutions?
Thank you :)
| 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 $y = 2^b$, where $a+b=11$ (equation 1).
Now, we see that:
$$\frac{2^a}{2^b} - \frac{2^b}{2^a} = 7.875$$
$$\frac{2^a}{2^b} - \frac{2^b}{2^a} = 8 - 1/8$$
Therefore:
$$\frac{2^a}{2^b} = 8$$
Which can be rewritten as:
$$2^{a-b} = 2^3$$
$$a-b = 3 (Equation 2)$$
Now, adding equation 1 and 2:
$$2a = 14$$
$$a = 7$$
Therefore, $x = 2^{7} = 128$
| {
"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:
$$\begin{eqnarray*} \sum_{n=1}^{2015}\frac{n}{n^4+n^2+1}&=&\frac{1}{2}\sum_{n=1}^{2015}\left(\frac{1}{n^2-n+1}-\frac{1}{(n+1)^2-(n+1)+1}\right)\\&=&\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2016^2-2016+1}\right)=\color{blue}{\frac{2031120}{4062241}}.\end{eqnarray*}$$
| {
"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 $$\displaystyle I = 2\int\frac{\sec x\cdot \tan x}{(3\sec x+4)^3}dx$$
Now Put $\displaystyle (3\sec x+4) = t\;,$ Then $3\sec x\cdot \tan xdx = dt$
So we get $$\displaystyle I = \frac{2}{3}\int \frac{1}{t^3}dt = -\frac{1}{3t^2}+\mathcal{C} = -\frac{1}{3(3\sec x+4)^2}+\mathcal{C}$$
But answer is http://www.wolframalpha.com/input/?i=INTEGRATION+OF+%28sin+2x%29%2F%283%2B4cos+x%29%5E3
Where I gave Done Wrong,
Thanks
| $$\left( -\frac{1}{3(3\sec x+4)^2}\right)'=\frac23\dfrac{3\sin x}{\cos^2x}\frac1{(3\sec x+4)^3}=\frac{2\sin x\cos x}{(3+4\cos x)^3},$$
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://www.math.com/tables/expansion/log.htm it is valid for $z > 0$?
| 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",
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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 $(i)$,
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-4\sin\theta\cos\theta=1-4\sin\theta\cos\theta$$
$$(\cos\theta-\sin\theta)^2=1-4\sin\theta\cos\theta$$
$$\cos\theta-\sin\theta=\pm1$$
But my teacher says that there is a shorter solution than that, so please can someone help me find that?
| $\cos \theta + \sin \theta=1$ 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 into $\cos\theta-\sin\theta$ you either get $+1$ or $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$$
\begin{pmatrix}f_{n}&f_{n+2}\\f_{n-1}&f_{n+1}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{pmatrix}
\begin{pmatrix}f_{n-1}&f_{n+1}\\f_{n-2}&f_{n}\end{pmatrix}
$$
and so, by induction,
$$
\begin{pmatrix}f_{n}&f_{n+2}\\f_{n-1}&f_{n+1}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1}
\begin{pmatrix}f_{1}&f_{3}\\f_{0}&f_{2}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1}
\begin{pmatrix}1&2\\0&1\end{pmatrix}
$$
Now take determinants, noting that $(-1)^{n-1}=(-1)^{n+1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 using the table function on a scientific calculator.
By checking whole number choices for $a$, with $-5 \le a \le 5$, we see that $a=-1$ work.
Since $\mathrm{f}(-1) = 0$, the Factor Theorem tells us that $(x-(-1)) = (x+1)$ divides $\mathrm{f}(x)$. Hence:
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x^2+px+q)$$
where $p$ and $q$ are numbers that you need to find. You can do one of two things:
*
*Expand the right-hand side and then compare coefficients,
*Use polynomial long division.
If we expand, we see that $(x+1)(x^2+px+q) \equiv x^3+(p+1)x^2+(p+q)x+q$.
Comparing coefficients gives $p+1=-1$, $p+q=-5$ and $q=-3$. Clearly $p=-2$ and $q=-3$. Hence
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x^2-2x-3)$$
Now we need only factorise the quadratic $x^2-2x-3$, which gives $(x-3)(x+1)$. Hence
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x+1)(x-3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435580",
"timestamp": "2023-03-29T00:00:00",
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} |
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.
$\sqrt{x^2-5x-24}-x-2>0\Rightarrow \sqrt{x^2-5x-24}>x+2$
$$\therefore x^2-5x-24>x^2+4x+4\Rightarrow9x<-28\Rightarrow x<\frac{-28}{9} \tag 1$$
and
$$x^2-5x-24\geq0\Rightarrow x\leq -3 \text{ or } x\geq 8 \tag 2$$
When taking the intersection of (1) and (2),we get $x<\frac{-28}{9}$. Where have i gone wrong? Please help me.
| 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
$$x\le -3 \implies\quad \sqrt{x^2-5x-24}\ge 0 \quad\text{ and }\,\,\,-1 = -3 +2 \ge x+2$$
Then, $$\sqrt{x^2-5x-24}-x-2\ge 1>0$$
So, $\ln(\sqrt{x^2-5x-24}-x-2)$ is a real number iff $x\le -3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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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 $2$.
So the matrix $A$ must be of the form \begin{bmatrix} 0 & 0 & 0\\ b & 0 & 0 \\ c & a & 0 \end{bmatrix}
Since the minimal polynomial is $x^2$ so rank $A=2$,so we must have a non-zero minor of order $2$ .Hence we should have $a\neq 0,b\neq 0;a,b,c\in \mathbb R$ .
Is the solution correct?Please suggest edits if required.
| 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 lower triangular) with all diagonal entries as zero.
(Why? Since then the characteristic polynomial = determinant of the triangular matrix (xI-A) = product of diagonal entries (x-0),(x-0),(x-0) = $x^{3}$)
Thus, A as of now, is of the form
$$ A = \left[\begin{array}\\0&0&0\\a&0&0\\b&c&0\end{array}\right]$$
where a,b,c$\in \Bbb R$
Now we want $A\neq 0$ and $A^2=0$ so that the polynomial $x^2$ is the minimum degree polynomial that annihilates A, i.e.we need,
$$ A = \left[\begin{array}\\0&0&0\\a&0&0\\b&c&0\end{array}\right]\neq 0,\space \space \space and \space A^2 = \left[\begin{array}\\0&0&0\\0&0&0\\ac&0&0\end{array}\right] = 0$$,
$\Rightarrow$a,b,c cannot all be zero, and both a,c cannot be non-zero.
Hence, $$ \left[\begin{array}\\0&0&0\\0&0&0\\9&0&0\end{array}\right] , \left[\begin{array}\\0&0&0\\0&0&0\\-1&1&0\end{array}\right], \left[\begin{array}\\0&0&4\\0&0&13\\0&0&0\end{array}\right], \left[\begin{array}\\0&-7&4\\0&0&0\\0&0&0\end{array}\right] $$ are all examples of $3\times3$ matrices whose minimal polynomial is $x^2$.
Thank You!
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\
\end{array}$$
I really didn't see this helping me.
I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.
Any ideas?
| $$\begin{align}
3! \cdot 5! \cdot 7! &= 6 \cdot 120 \cdot 7! \\
&= 6 \cdot 15 \cdot 8! \\
&= 2 \cdot 5 \cdot 9! \\
&= 10 \cdot 9! \\
&= 10!
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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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:
$$-\frac{1}{2}\leq f(x) \leq \frac{9}{10}.$$
Anyway, the Cauchy-Schwarz inequality is a pretty fast way to go:
$$ \left|3x+2\right| = \left|3x+\frac{2}{\sqrt{5}}\cdot\sqrt{5}\right|\leq \sqrt{9+\frac{4}{5}}\sqrt{x^2+5} $$
hence:
$$ \left| f(x)\right |\leq \frac{7}{\sqrt{5}\sqrt{x^2+5}}\leq \frac{7}{5}.$$
The AM-GM plus the triangular inequality work pretty well, too:
$$\left|\frac{3x+2}{x^2+5}\right|\leq \frac{2}{5}+\frac{3|x|}{x^2+5} \leq \frac{2}{5}+\frac{3}{2\sqrt{5}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441267",
"timestamp": "2023-03-29T00:00:00",
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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 $$I=\int_{0}^{\pi/2}\log(\sin^2\theta+k^2\cos^2\theta)d\theta...........(1)$$
$$I=\int_{0}^{\pi/2}\log(\cos^2\theta+k^2\sin^2\theta)d\theta............(2)$$
Adding $(1)$ and $(2),$ we get
$$2I=\int_{0}^{\pi/2}\log(\cos^2\theta+k^2\sin^2\theta)(\sin^2\theta+k^2\cos^2\theta)d\theta$$
$$2I=\int_{0}^{\pi/2}\log[(1+k^4)\cos^2\theta\sin^2\theta+k^2(\sin^4\theta+\cos^4\theta)]d\theta$$
But i am unable to solve further.Please help me.
| 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 $$\displaystyle \frac{dI}{dk} = \int_{0}^{\frac{\pi}{2}}\frac{2k\sin^2 \theta}{\cos^2 \theta+k^2\sin^2 \theta}d\theta = 2k\int_{0}^{\frac{\pi}{2}}\frac{2k\tan^2 \theta\cdot \sec^2 \theta}{(1+ +k^2\tan^2 \theta)\cdot (1+\tan^2 \theta)}d\theta$$
Now Put $\tan \theta = t\;,$ Then $\sec^2 \theta d\theta = dt$ and changing limit, We get
$$\displaystyle \frac{dI}{dk} = 2k\int_{0}^{\infty}\frac{t^2}{(1+k^2t^2)\cdot (1+t^2)}dt = \frac{2k}{k^2-1}\int_{0}^{\infty}\frac{(1+k^2t^2)-(1+t^2)}{(1+k^2t^2)\cdot (1+t^2)}dt$$
So $$\displaystyle \frac{dI}{dk} = \frac{2k}{k^2-1}\int_{0}^{\infty}\left[\frac{1}{1+t^2}-\frac{1}{1+k^2t^2}\right]dt$$
So we get $$\displaystyle \frac{dI}{dk} = \frac{2k}{k^2-1}\int_{0}^{\infty}\left[\frac{\pi}{2}-\frac{1}{k}\cdot \left(\tan^{-1}(tk)\right)_{0}^{\infty}\right] = \frac{2k}{k^2-1}\left[\frac{\pi}{2}-\frac{\pi}{2k}\right]$$
So we get $$\displaystyle \frac{dI}{dk} =\frac{2k}{k^2-1}\cdot \frac{\pi}{2}\cdot \left(\frac{k-1}{k}\right) =\frac{\pi}{k+1}$$
Now $$\displaystyle \int \frac{dI}{dk}dk = \pi\int \frac{1}{k+1}dk$$
So $$\displaystyle I = \pi\cdot \ln(k+1)+\mathcal{C}...............(1)$$
Now If $k=1\;,$ Then $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln\left(\cos^2 \theta +\sin^2 \theta\right)d\theta = 0$$
Now put $k=1$ in equation...............$(1)$
So we get $$\displaystyle 0 = \frac{\pi}{2}+\mathcal{C}\Rightarrow \mathcal{C} = -\frac{\pi}{2}$$
So put $\displaystyle \mathcal{C} = -\frac{\pi}{2}$ in equation .. ...........$(1)$
So we get $$\displaystyle I = \pi\cdot \ln\left(\frac{1+k}{2}\right)\;,$$ Where $k\geq 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldots ) = \frac{(1-x^3)(1)}{1-x^5}$$
So we can write it as $\displaystyle \frac{(1-x)(x^2+x+1)}{(1-x)(x^4+x^3+x^2+x+1)}$
So our Integral Convert into $\displaystyle \int_{0}^{1}\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$
Now How can I solve it, Help me
Thanks
| Hint:
$$
\begin{align}\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx&=
1-{1\over4}+{1\over6}-{1\over9}+{1\over11}-{1\over14}+\ldots\\&=
\sum_{k=0}^\infty{3\over(5k+1)(5k+4)}\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $(2-x^2+x) \text{ }$,$(2-y^2+y) \text{ }$,$(2-z^2+z) \text{ }$ then try to rewrite the new inequality in term of $u=x+y+z \text{ }$,$w=xyz \text{ }$,$\text{ } v=xy+yz+zx$. Then I realized that the term $2-x^2+x$ can be negative. So the AM-GM approach fail.
*I then try to homogenize the inequality, and hope that after expand every thing. I can obtain some obvious inequality. Well, I fail to homogenize it.
*I try to assume $x\leqslant y \leqslant z$ and come up with some estimations but I get stuck.
I appreciate if anyone can attempt to solve it.
| 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 you can use what you want.(but it is not easy also)
case2: two of them negtive, suppose $x<-1,y<-1$, then let $a=-x>1,b=-y>1,2<a+b \le \sqrt{2(3-z^2)} ,ab>1,|z|<1$
then RHS$ > f(z)>$LHS,try to find $f(z)$
another approach is expand LHS and replace with $uvw$ at both sides, then you will have to prove $f(w) \ge 0 \implies \Delta \le 0$, with $u^2-2v=3,|v|\le 3,|u|\le 3$, you should be able to get $\Delta \le 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 (in the form of $x+yi$) and then I represent this number in its trigonometric form and apply De Moivre's formula. But this case seems different... What should I do next?
| $$\frac{-16}{\sqrt3-i}=\frac{(-16)}{(\sqrt3-i)}.\frac{(\sqrt3+i)}{(\sqrt3+i)}=\frac{-16\sqrt3-16i}{4}=-4\sqrt3-4i$$
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 $$-4\sqrt3-4i=8(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$$
Now use De Moivre's Theorem or Euler's Formula as per your comfort.
$$\text{De Movire's Theorem}$$
$$(\cos\theta+i\sin\theta)^{\frac1n}=(\cos\frac{\theta}{n}+i\sin\frac{\theta}{n})$$
$$\text{Euler's Formula}$$
$$e^{i\theta}=\cos\theta+i\sin\theta$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 the following steps:
\begin{align}
\frac{x^{3}}{x^{2}+1}&=\frac{y^{3}}{y^{2}+1}\\
x^3(y^2+1)&=y^3(x^2+1)\\
x^3y^2+x^3&=x^2y^3+y^3\\
x^3-y^3&=x^2y^3-x^3y^2\\
(x-y)(x^2+xy+y^2)&=x^2y^2(y-x)\\
(x^2+xy+y^2)&=-x^2y^2\\
x^2+y^2&=-x^2y^2-xy\\
x^2+y^2&=-xy(xy+1)
\end{align}
And here is where I'm stuck. Am I assuming correctly that I should find that $x=y$ somehow?
| 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 $x=y=0$.
This proves that $f$ is injective.
(Note that when you deduced $x^2+xy+y^2)=-x^2y^2$ from $(x-y)(x^2+xy+y^2)=x^2y^2(y-x)$ by canceling $x-y$, you assumed that $x-y\ne0$, that is, that $x\ne y$.)
To see that $f$ is surjective, take $a\in\mathbb R$. Then the equation $f(x)=a$ is a cubic equation in $x$ and so has a real solution.
| {
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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 please show the steps to how they would solve this.
| 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{1-2^{100}}{1-2}\right)\\
&=\frac{3}{2}\left(3^{100}-1\right)+3\cdot 2\left(2^{100}-1\right)
\end{align}
| {
"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
$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt = \frac{d}{dx} \int_{-x}^{0} \frac{1}{3+t^2} \ dt + \frac{d}{dx} \int_{0}^{x} \frac{1}{3+t^2} \ dt\\
= - \frac{d}{dx} \int_{0}^{-x} \frac{1}{3+t^2} \ dt + \frac{d}{dx} \int_{0}^{x} \frac{1}{3+t^2} \ dt\\
= - \frac{1}{3+(-x)^2} \cdot (-1) + \frac{1}{3+x^2}\\
= \frac{1}{3+x^2} + \frac{1}{3+x^2}\\
= \frac{2}{3+x^2}
$$
but I suppose that the textbook is expecting the answer $0$ (which for me is not correct since the chain rule, on the first term, makes the fraction not cancel). Is there really a mistake?
Thank you.
| 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",
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"source": "stackexchange",
"question_score": "1",
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} |
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 puzzled by how exactly to use it in this case. I do know that $x \equiv \pm 1 \pmod 7$ and $\!\!\pmod {13}$, since $7$ and $13$ are both prime.
Any help is appreciated here.
| 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 case, the chinese remainder theorem guarantees that the solution you found (by trial-and-error) $\pmod{91}$ is the only one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
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 $(a+bi)^3=-11-2i$ and did not have success, because I could not solve
$a³-3ab² = -11$ and $3a²b - b³ = -2$.
Solutions at the end of the book are $1+2i$ and $-\frac{1+2\sqrt{3}}{2}+\frac{\sqrt{3}-2}{2}i$ and $-\frac{1-2\sqrt{3}}{2}-\frac{\sqrt{3}-2}{2}i$.
| 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, we obtain
$$a^3-3a\frac{36a^2}{(11-8a^3)^2}=-11$$
Setting $a^3=x$ and simplifying, this yields
$$x-\frac{108x}{(11-8x)^2}=-11$$
$$(x+11)(11-8x)^2=108x$$
$$64x^3+528x^2-1923x+1331=0$$
Noticing that $x=1$ is a trivial solution of this one, we can proceed to the following:
$$(x-1)(64x^2+592x-1331)=0$$
So we obtain the solution $a^3=x=1$ which gives $a=1$ and hence $b=2$.
So $z=a+bi=1+2i$ is a solution.
The others can be obtained by computing $\omega \cdot z$ and $\omega^2 \cdot z$ where $\omega$ is the third root of unity.
Remark: These are also the solutions which you obtain when resolving the quadratic in $x$ and then compute the third root of these solutions. But the way using the roots of unity is much nicer..
| {
"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 substitution to eliminate the root. Let
$$w^2 = 1-3x$$
So
$$w\to 1$$
Therefore
$$\lim_{w\to1}\frac{x\cdot(\sqrt{w^2}+1)}{(1-3x)-1} = \lim_{w\to1}\frac{x\cdot(w+1)}{(1-3x)-1}$$
We find that
$$x = \frac{1-w^2}{3}$$
So
$$\lim_{w\to1}\frac{\left( \frac{1-w^2}{3}\right)\cdot(w+1)}{\left(1-3\left( \frac{1-w^2}{3}\right)\right)-1}$$
$$\lim_{w\to1}\frac{\left( \frac{1-w^2}{3}\right)\cdot(w+1)}{\left(1-3+3w^2\right)-1}$$
This will still evaluate to $\frac{0}{0}$.
I just usted the two methods that I always use to calculate limits. Rationalize/factorize and variable substitution. I'm not supposed to use L'Hopital.
What did I do wrong, and how should I have done it?
| 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",
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"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:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
And now:
$$\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x} = \infty$$
What I'm doing wrong? I know that answer is $\sqrt{3}$, but where is my mistake?
| 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{align}
| {
"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^2-3x+2\\
x^2-3x-2 &\lt &-2 &\lt &x^2-3x+2\\
-2 &\lt &-x^2+3x-2 &\lt &2\\
-2 &\lt &x^2-3x+2 &\lt &2\\
0 &\lt &x &\lt &3
\end{matrix}
The places where $\lfloor x^2\rfloor$ changes value are
$\{1, \sqrt 2, \sqrt 3, 2, \sqrt 5, \sqrt 6, \sqrt 7, \sqrt 8, 3 \}$
The places where $\lfloor 3x\rfloor$ changes value are
$\{\frac 13, \frac 23, 1, \frac 43, \frac 53, 2, \frac 73, \frac 83, 3 \}$
The sorted list of the union of these two sets is
$$\left\{\frac 13, \frac 23, 1, \frac 43, \sqrt 2, \frac 53,\sqrt 3 , 2,
\sqrt 5, \frac 73, \sqrt 6, \sqrt 7, \frac 83, \sqrt 8, 3 \right\}$$
\begin{matrix}
\text{breakpoint} & \lfloor x^2\rfloor &\lfloor 3x\rfloor & \lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2\\
\frac 13 & 0 & 1 & 1 &\\
\frac 23 & 0 & 2 & 0 & \checkmark\\
1 & 1 & 3 & 0 & \checkmark\\
\frac 43 & 1 & 4 & -1 &\\
\sqrt 2 & 2 & 4 & 0 & \checkmark\\
\frac 53 & 2 & 5 & -1 &\\
\sqrt 3 & 3 & 5 & 0 & \checkmark\\
2 & 4 & 6 & 0 & \checkmark\\
\sqrt 5 & 5 & 6 & 1 &\\
\frac 73 & 5 & 7 & 0 &\checkmark\\
\sqrt 6 & 6 & 7 & 1 &\\
\sqrt 7 & 7 & 7 & 2 &\\
\frac 83 & 7 & 8 & 1 &\\
\sqrt 8 & 8 & 8 & 2 &\\
3 & 9 & 9 & 2 &\\
\end{matrix}
Hence the solution set is
$$ x \in
\left[ 2/3, 4/3 \right) \cup
\left[ \sqrt 2, 5/3 \right) \cup
\left[ \sqrt 3, \sqrt 5 \right) \cup
\left[ 7/3, \sqrt 6 \right)
$$
| {
"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 asking whether there are any Catalan
numbers not divisible by $n+2$, with n neither of the form $3k+1$, nor of the form $4k+2$.
Inspired by the afore-mentioned Catalan numbers, I began investigating integers with the property
that $~\dfrac{\displaystyle{2n\choose n}}{n+2}\in\mathbb N$, and then soon afterwards arrived at the conclusion that there were too many of
them for such a quest to be even remotely interesting, so I negated the above property, and, after a
rather short while, I immediately started noticing that the new results were either of the form $n=$
$=3k+1$, or of the form $n=4k+2$. Updating the search parameters so as to eliminate these two
classes as well, I eventually came to the realization that there were no solutions for $n\le10^5$.
| 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:
$$A{2n\choose n}+B{2n\choose n+1}+C{2n\choose n+2}\\
=\frac{(2n)!}{n!(n+2)!}\left[A(n+2)(n+1)+Bn(n+2)+Cn(n-1)\right]\\
A+B+C=0,3A+2B-C=1,2A=1\\
A=1/2,B=-1/3,C=-1/6\\
D_n=\frac1{n+2}{2n\choose n}=\frac12{2n\choose n}-\frac13{2n\choose n+1}-\frac16{2n\choose n+2}\\
=\frac12{2n\choose n}-\frac13\left[{2n\choose n+1}-{2n\choose n+2}\right]-\frac12{2n\choose n+2}$$
${2n\choose n}=2{2n-1\choose n}$ so the first term is an integer.
$$(n-1)\left[{2n\choose n+1}-{2n\choose n+2}\right]=\frac{3(2n)!}{(n-2)!(n+2)!}
$$
which is a multiple of $3$, so when you divide by $n-1$ it is still a multiple of $3$ unless $n=1\mod 3$.
That leaves
$$\frac12{2n\choose n+2}=\frac n{n+2}{2n-1\choose n+1}$$
which only has an extra 2 in the denominator if $n=2\mod 4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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 :
$$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$
Now how can we prove that:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$
| 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:
$$\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}(xy)^{2n}dydx$$
is the same as:
$$\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}(xy)^{2n}dydx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}.$$
Make the change of variables:
$$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(u)}.$$
The Jacobian Determinant is:
$$\det\frac{\partial (x,y)}{\partial(u,v)}=\begin {vmatrix} \frac{\cos(u)}{\cos(v)}&&\frac{\sin(u)\sin(v)}{(\cos(v))^2} \\ \frac{\sin(v)\sin(u)}{(\cos(u))^2}&&\frac{\cos(v)}{\cos(u)}\\\end{vmatrix}=1-x^2y^2,$$
which cancels with the integrand, and the region of integration is the open iscosceles triangle formed by the inequalities: $$0<u+v<\frac{\pi}{2},0<u,v<\frac{\pi}{2}.$$
Using either geometry or evaluating the double integral:
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}-v}1dudv,$$
the area of the isosceles triangle is $\frac{\pi^2}{8}$. So the result is that:
$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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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, assume that $a\ne 0$ and compute via row-reduction as follows:
$$\det \begin{pmatrix} a & 0 & a^2 \\ 2 & a & 3a \\ a & -1 & ab + b \end{pmatrix}=\det \begin{pmatrix} a & 0 & a^2 \\ 0 & a & a \\ 0 & -1 & ab + b - a^2 \end{pmatrix} = \det \begin{pmatrix} a & 0 & a^2 \\ 0 & a & a \\ 0 & 0 & -a^2 + ab + b +1 \end{pmatrix}.$$
Therefore, we have
$$\det(A)=0 \iff a=0 \mbox{ or }a^2(-a^2+ab+b+1)=0 \iff a=0 \mbox{ or } -a^2+ab+b+1=0.$$
Doing a bit of algebra, we have the following solutions:
$$\det(A)=0 \iff a=0, a=-1 \mbox{ or } b=a-1 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 to the axes, meet $C_2$ and $C_3$ at $Q$ and $R$ respectively . If every position of $P ($$on $ $C_1)$ the areas of the shaded region $OPQ $ and $ORP$ are equal then find $f(x)$ ?
diagram :
| Let the x-coordinate of $P$ be $x$. Then the points are: $P(x,x^2), Q(\frac{1}{2}x^2,x^2),R(x,f(x))$. Let $S$ be the projection of point $P$ onto the x-axis. and $T$ the projection of points $P,Q$ onto the y-axis.
Then
$$\begin{align}
Area(OPQ)&=Area(OSPT)-Area(OSP)-Area(OQT)\\[1em]
&=x\cdot x^2-\frac{1}{3}x^3-\frac{1}{2}(\frac{1}{2}x^2)(x^2) \\[1em]
&=\frac{2}{3}x^3-\frac{1}{4}x^4 \\[1em]
&=Area(ORP)
\end{align}$$
So $$\int_0^X{x^2-f(x)\,dx}=\frac{2}{3}X^3-\frac{1}{4}X^4$$
By the fundamental theorem of calculus, we then have
$$X^2-f(X)=\frac{d}{dX}\left[\frac{2}{3}X^3-\frac{1}{4}X^4\right]=2X^2-X^3$$
so
$$\boxed{f(x)=x^3-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\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 x}{x^2\cdot\sin 2x\cdot \cos x}$$
Split that:
$$\frac{\sin x\cdot \cos x}{x^2\cdot\sin 2x\cdot \cos x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
In the left side, we can cancel the $\cos x$ and also apply $\frac{\sin x}{x} = 1$ once:
$$\frac{1}{x\cdot\sin 2x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
That was probably a bad idea, since $x \cdot \sin2x$ will definitely be $0$... But anyway, let's keep going with the right side. There, we can apply the identity $\frac{\sin x}{x} = 1$ again:
$$\frac{1}{x\cdot\sin 2x} - \frac{1}{x\cdot\sin 2x\cdot \cos x}$$
Hey, I could get rid of the $\sin 2x$ on the left side if I multiply and divide by $2x$... the same on the right side:
$$\frac{1}{2x^2} - \frac{1}{2x^2\cdot \cos x}$$
Looking pretty, but sadly that's not going anywhere. What can I do?
| Notice, $$\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-\frac{x^2}{2!}+O(x^2)\right)-1}{x^2}\cdot 1$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\left(-\frac{x^2}{2!}+O(x^2)\right)}{x^2}$$
$$=\frac{1}{2}\lim_{x\to 0}\left(-\frac{1}{2!}+O(1)\right)$$
$$=\frac{1}{2}\left(-\frac{1}{2}+0\right)=\color{red}{-\frac{1}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant of big matrix Let:
$$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),$$
for $n\in\mathbb N$. How could I compute $\det M_n$? I tried computing the first few and seeing if I saw an apparent pattern to try inducting, but the third one was already too much of a mess and no pattern was evident. This was an exercise in an exam, so there must be a smart way of doing it. Is there? Which is it?
| $$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 & 2 \times 2^2 & 3 \times 3^2 & \dots & n \times n^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 \times 1^{2n-2} & 2 \times 2^{2n-2} & 3 \times 3^{2n-2} & \dots & n \times n^{2n-2} \end{array}\right)$$
Factor out $n!$ (in the form of $j$ from the $j$th column):
$$\text{det} = n! \left \vert \left(\begin{array}{ccccc} 1^0 & 2^0 & 3^0 & \dots & n^0 \\ 1^2 & 2^2 & 3^2 & \dots & n^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1^{2n-2} & 2^{2n-2} & 3^{2n-2} & \dots & n^{2n-2} \end{array}\right) \right \vert$$
This is a Vandermonde matrix with $a_i = i^2$; the determinant of that matrix is $$\prod_{1 \leq i < j \leq n} (i^2 - j^2)$$
Therefore $$\text{det}(M_n) = n! \prod_{1 \leq i < j \leq n} (i+j)(i-j)$$
If you want to "simplify" it a bit more, you can do the following to get a feel for it, but it doesn't yield a nice closed form that I can see.
First, let $n$ be odd.
In the $i+j$ terms, $3$ and $4$ turn up once; $5$ and $6$ turn up twice; $7$ and $8$ three times; and so on up to $n$ and $n+1$, which occurs $\frac{n-1}{2}$ times. Then $n+2$ turns up $\frac{n-1}{2}$ times, $n+3$ and $n+4$ turn up $\frac{n-3}{2}$ times, and so on down to $2n-2$ and $2n-1$ which happen once. Therefore $$\prod_{1 \leq i < j \leq n} (i+j) = 3 \times 4 \times 5^2 \times 6^2 \times \dots \times n^{(n-1)/2} \times (n+1)^{(n-1)/2} \times (n+2)^{(n-1)/2} \times (n+3)^{(n-1)/2-1} \times \dots \times (2n-2) \times (2n-1)$$
You can do similar things if $n$ is even, and then analyse the $i-j$ terms in a similar way. (The $i-j$ terms lead to a product of factorials.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum of $\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$ for nonnegative $a$, $b$, $c$.
Show that the maximum of
$$\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 + \dots + a_n \right)^n - n^n \, a_1 \dots a_n
}
{
a_1^n + \dots + a_n^n - n \, a_1 \dots a_n
}
\le
\left( n-1 \right)^{n-1},
$$
for nonnegative $a_1, \dots, a_n$? In other words,
$$
r_n(a_1, \dots, a_n)
\equiv
\frac{
\dfrac{a_1^n+\dots+a_n^n}{n} - \,a_1 \dots a_n
}
{
\left(\dfrac{ a_1+\dots+a_n }{n} \right)^n - a_1 \dots a_n
}
\ge
\left( \frac{n}{n-1} \right)^{n-1}.
$$
This would imply that
$$
\lim_{n \rightarrow \infty} r_n(a_1, \dots, a_n)
\ge e.
$$
| Here is an analytical proof for the general case, although I suspect better solutions exist. The $n = 2$ case is obvious, and we shall assume $n \ge 3$.
Without loss of generality, we can assume $a_1 \le a_2 \le \dots \le a_n$. First, since the function $f(x) = x^n$ is convex, we have
$$
a_2^n + \dots + a_n^n
\ge
(n - 1)
\left(
\frac{ a_2 + \dots + a_n } { n - 1 }
\right)^n,
$$
with the equality achieved at $a_2 = \dots = a_n$.
1) If $a_1 = 0$, then $a_1 \cdots a_n = 0$, we have the
\begin{align}
R
&\equiv
\frac{ \left(
a_1 + a_2 + \dots + a_n
\right)^n - n^n \, a_1 \cdots a_n
}{
a_1^n + a_2^n + \dots + a_n^n
- n \, a_1 \cdots a_n
} \\
&= \frac{ \left(
a_2 + \dots + a_n
\right)^n
}{
a_2^n + \dots + a_n^n
}
\le (n - 1)^{n - 1}.
\end{align}
2) If $a_1 > 0$, all other $a_k > 0$. The inequality makes sense only if at least two $a_k$ are different. Thus, we can assume $a_n > a_1$. Now suppose for some $m$ ($1 < m < n$), we have $a_1 \le a_m < a_n$. Consider the following infinitesimal adjustment
\begin{align}
a_1 &\rightarrow a_1 - (a_n - a_m) \, a_1 \, dt \\
a_m &\rightarrow a_m + (a_n - a_1) \, a_m \, dt \\
a_n &\rightarrow a_n - (a_m - a_1) \, a_n \, dt.
\end{align}
It leaves $a_1 + \dots + a_n$ and $a_1 \cdots a_n$ invariant, and
changes $a_1^n + \dots + a_n^n$ by
\begin{align}
d (a_1^n + \dots + a_n^n)
&=
d a_1^n + d a_m^n + d a_n^n \\
&=
n \, \left[
-(a_n - a_m) \, a_1^n
+(a_n - a_1) \, a_m^n
-(a_m - a_1) \, a_n^n
\right] \, dt \\
&=
n \, \left[
(a_n - a_m) (a_m^n - a_1^n)
-(a_m - a_1) (a_n^n - a_m^n)
\right] \, dt \\
&=
\big[
(a_m^{n-1} + a_m^{n-2} a_1 + \cdots + a_1^{n-1}) \\
&\;
-(a_m^{n-1} + a_m^{n-2} a_n + \cdots + a_n^{n-1})
\big] \,
n \, (a_n - a_m) (a_m - a_1)\, dt \\
&< 0.
\end{align}
The last expression is less than zero, because $a_1 < a_n$. This means that our adjustment always increases the target function $R$, and since it decreases $a_1$, we should set $a_1$ to the smallest possible value, which is $0$. This is the case 1).
3) Last, if no such $m$ exists, then for $m > 1$, $a_m = a_n$. Then, we can choose $m = 2$ and use the same transform, which will create some difference between $a_n$ and $a_m$, and return to case 2). This completes the proof.
| {
"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^2/2} + \int (y^2 + y^{-2})e^{-y^2/2} dy$, but I'm not sure what to do with the integral.
| 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^{\left( -\frac{x^{2}}{2}\right) }+C.
\end{equation*}
Its proof maybe found at
Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$
So we are done if we find a function $f(x)$ such that
\begin{equation*}
h(x)=f^{\prime }(x)+g^{\prime }(x)f(x).
\end{equation*}
In what follows, I will show that $f(x)=x^{-3}-x^{-1},$ and therefore
\begin{equation*}
\int \left( 1-\frac{3}{x^{4}}\right) e^{\left( -\frac{x^{2}}{2}\right)
}dx=\left( x^{-3}-x^{-1}\right) e^{\left( -\frac{x^{2}}{2}\right) }+C.
\end{equation*}
$\color{red}{\bf Problem:}$ We want to write $1-\frac{3}{x^{4}}$ as $f^{\prime
}(x)+g^{\prime }(x)f(x)$ where $g(x)=-\frac{x^{2}}{2},$ $g^{\prime }(x)=-x$
and $f(x)$ is to be determined.
First, it is easy to see that
\begin{equation*}
-\frac{3}{x^{4}}=-3x^{-4}=(x^{-3})^{\prime }.
\end{equation*}
It follows that if we put $f_{1}(x)=x^{-3},$ then
\begin{equation*}
f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)=\left( -\frac{3}{x^{4}}\right)
+\left( -x\right) \left( x^{-3}\right) =\left( -\frac{3}{x^{4}}-\frac{1}{%
x^{2}}\right)
\end{equation*}
This suggests to add and to substract the term $-\frac{1}{x^{2}}$ as follows
\begin{equation*}
\left( 1-\frac{3}{x^{4}}\right) =\left( 1-\frac{3}{x^{4}}\right) -\frac{1}{
x^{2}}+\frac{1}{x^{2}}=\left( -\frac{3}{x^{4}}-\frac{1}{x^{2}}\right)
+\left( \frac{1}{x^{2}}+1\right) ,
\end{equation*}
Now let us find $f_{2}(x)$ such that :
\begin{equation*}
\left( \frac{1}{x^{2}}+1\right) =f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x).
\end{equation*}
It is easy to see that
\begin{equation*}
\frac{1}{x^{2}}=x^{-2}=(-x^{-1})^{\prime }.
\end{equation*}
Then if we put $f_{2}(x)=-x^{-1},$ it follows that
\begin{equation*}
f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)=\left( -x^{-1}\right) ^{\prime
}-x(-x^{-1})=\frac{1}{x^{2}}-x\left( -x^{-1}\right) =\left( \frac{1}{x^{2}}
+1\right) .
\end{equation*}
It follows that
\begin{eqnarray*}
\left( 1-\frac{3}{x^{4}}\right) &=&\left( -\frac{3}{x^{4}}-\frac{1}{x^{2}}
\right) +\left( \frac{1}{x^{2}}+1\right) \\
&& \\
&=&\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +\left(
f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) \\
&& \\
&=&\left( (f_{1}(x)+f_{2}(x))^{\prime }+g^{\prime
}(x)(f_{1}(x)+f_{2}(x))\right)
\end{eqnarray*}
then, it suffices to take
\begin{equation*}
f(x)=f_{1}(x)+f_{2}(x)=x^{-3}-x^{-1}.\ \ \
\color{red}
\blacksquare
\end{equation*}
| {
"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
} |
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
$$
Given that C is a linear combination of A and B.
| Since $C$ is linearly dependent on $A$ and $B$, we need not consider it, it does not add any information.
If you look at the system
$$\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 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
= -x_2 \frac{\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
$$
and also that
$$x_3 = \frac{\begin{vmatrix} a_1 & -a_2x_2 \\ b_1 & -b_2x_2 \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
= -x_2 \frac{\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
$$
and so you will have:
$$\frac{x_1}{x_3} = -x_2 \frac{\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}} \Bigg/ -x_2 \frac{\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}} =
\frac{\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}}
= \frac{a_2b_3 - a_3b_2}{a_1b_2 - a_2b_1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer
Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:
*
*the denominator has to be one,
*the numerator has to be a multiple of the denominator or
*the numerator and denominator have to be the same.
Obviously, with the given conditions case 3 is eliminated since $a \neq b$.
For case 1, if $b=1$, then $a=1$ which is a contradiction to the given condition that says $a \neq b$.
For case 2, I would think of examples. For example if $b=2$ then a multiple is $4$, so $a=4$. Then we have:
$$\frac{a}{b}+\frac{b}{a}=\frac{4}{2}+\frac{2}{4}=\frac{10}{4}$$
which is not an integer, but how would i proceed to show this case for all integers.
| Presumably you want $a,b$ to be integers. If $x = a/b$, solve the equation
$$x + \dfrac{1}{x} = n$$
to get $$x = \dfrac{n}{2} \pm \dfrac{\sqrt{n^2-4}}{2} $$
The question is whether $\sqrt{n^2-4}/2$ is ever rational. This would require $\sqrt{n^2-4}$ to be an integer (because the square root of an integer is only rational if it is an integer). But then if $m = \sqrt{n^2-4}$, we would have $n^2 - m^2 = 4$. The only squares of integers that differ by $4$ are $2^2$ and $0^2$, but that corresponds to $x = 1$, i.e. $a = b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
square root of $\frac{2+\sqrt{3}}{4}$
Find the square root of $\frac{2+\sqrt{3}}{4}$.
Attempt:
I'm thinking of equating it to $a + bi$. And then finding the root. Is that right?
| For convenience we will work with $4(2+\sqrt3)$ to avoid fractions.
Let us search a solution of the form $a+b\sqrt3$, giving
$$(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.$$
By inspection (or resolution of a quadratic equation), $a^2=2$, so that
$$\left(\sqrt2+\frac2{\sqrt2}\sqrt3\right)^2=8+4\sqrt3,$$
$$\left(\frac{\sqrt2+\sqrt6}4\right)^2=\frac{2+\sqrt3}4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Shorter way to calculate this limit I want to calculate this (without L'Hospital) :
$$\lim_{x \to 0} \frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3}$$
I already solved it in a looooong way:
\begin{align}
\frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3} &= \frac{\tan x - \sin x}{x^3} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &=
\frac{\sin x (1- \cos x )}{x^3 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &=
\frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &=
\frac{\sin x}{x} \frac{(1- \cos x)(1+\cos x) }{x^2 \cos x \ (1+\cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &=
\frac{\sin x}{x} \frac{1- \cos^2 x }{x^2} \frac{1}{\cos x \ (1 + \cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &=
\frac{\sin x}{x} \frac{\sin^2 x }{x^2} \frac{1}{\cos x \ (1 + \cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &=
1 \cdot 1^2 \cdot \frac{1}{1 \cdot (1 +1)} \frac{1}{\sqrt{1+0}+\sqrt{1+0}} \tag{as $x \to 0$} \\ \\ &=
\frac 1 4
\end{align}
Is there a shorter way?
EDIT: I am wondering whether there is a trick like $x= \arctan u^4$ etc.
| Your proof looks correct (but you slightly abused notation at the end - you should say $\to 1 \cdot 1^2 \cdots$, not $= 1 \cdot 1^2 \cdots$). You can shorten it if you use $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$.
$$\frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}$$
$$=\frac{\sin x}{x}\frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x\left(\sqrt{1+\tan x}+\sqrt{1+\sin x}\right)}$$
$$\stackrel{x\to 0}\to 1\cdot \frac{1}{2}\cdot \frac{1}{1\cdot (\sqrt{1+0}+\sqrt{1+0})}=\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Let $m_1$ and $m_2$ are the slopes of the tangents drawn to circle $x^2+y^2-4x-8y-5=0$ from the point $P(-1,-2)$,and $|m_1+m_2|=\frac{p}{q}$ Let $m_1$ and $m_2$ are the slopes of the tangents drawn to circle $x^2+y^2-4x-8y-5=0$ from the point $P(-1,-2)$,and $|m_1+m_2|=\frac{p}{q}$,where $p$and $q$ are relatively prime natural numbers,then find $p+q.$
Slope of $OP=m=\frac{2}{1}$
$\tan\theta=\frac{|m_1-2|}{|1+2m_1|}$ and $\tan\theta=\frac{|m_2-2|}{|1+2m_2|}$
$\frac{|m_1-2|}{|1+2m_1|}=\frac{|m_2-2|}{|1+2m_2|}$
$|m_1-2||1+2m_2|=|m_2-2||1+2m_1|$
$|(m_1-2)(1+2m_2)|=|(m_2-2)(1+2m_1)|$
Taking plus and minus signs,we get
either $m_1=m_2$ or $4-4m_1m_2+3m_1+3m_2=0$
But i am not able to find the value of $|m_1+m_2|$.Please help me.
| Note that $OP=3\sqrt 5$ and that the radius is $5$. So, you can have
$$\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.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471756",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Drawing without replacement - prob. for an Ace followed by an Ace? Given a standard 52-cards deck:
You are extracting cards from the deck without replacement, until you get an "Ace" for the first time. What is the probability that the next card will be "Ace" too?
I've already seen the following Q&A:
Probability of drawing an Ace: before and after
According to that thread, the answer should be:
$$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$
But the official solution says that the answer is:
$$\frac{4}{52}$$
which doesn't make sense IMHO. They solved it only with intuition or "mind trick", as they wrote..
My calculation:
Assuming that the 1st card is Ace, then:
$$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$
Assuming that the 2nd card is Ace, then:
$$\frac{(52-4) \cdot 4 \cdot 3}{52 \cdot 51 \cdot 50} = \frac{24}{5525}$$
We notice a pattern here.
Having the 1st Ace at the $k$'th draw, then the probability (for a second Ace after that) is:
$$
p_1 = \frac{ {_{52-4}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }
$$
We need to consider the least-possible scenario - we draw 48 non-Ace cards, then:
$$
p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} } = \frac{1}{270725}
$$
So, the required probability is:
$$\begin{align}
p &= p_2 + \sum\limits^{48}_{k=1} p_1 \\
&= p_2 + \sum\limits^{48}_{k=1} \frac{ {_{48}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }\\
&= \frac{1}{270725} + \frac{1696}{20825}\\
&= \frac{1297}{15925}\\
&\cong 0.081444
\end{align}
$$
But my answer is far from either the official solution and from the answer in that thread.
| We use a fairly crude counting approach, in order to rely minimally on intuition. There are $\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the "favourables."
Maybe the first two Aces are in positions 1 and 2. That leaves $\binom{50}{2}$ for the rest.
Maybe they are in positions 2 and 3. That leaves $\binom{49}{2}$ for the rest.
Maybe they are in positions 3 and 4. That leaves $\binom{48}{2}$ for the rest.
And so on, up to positions 49 and 50, leaving $\binom{2}{2}$ for the rest. That gives probability
$$\frac{\binom{50}{2}+\binom{49}{2}+\cdots+\binom{2}{2}}{\binom{52}{4}}.$$
The numerator can be simplified in various ways. My favourite is to count the number of ways to choose $3$ numbers from the first $51$. This can be done in $\binom{51}{3}$ ways. But the smallest number can be $1$, leaving $\binom{50}{2}$ ways to choose the other two. Or the smallest can be $2$, and so on. We get the number of favourables obtained above.
Our probability is therefore $\frac{\binom{51}{3}}{\binom{52}{4}}$. This simplifies to $\frac{1}{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?
The options are:
*
*$$\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}$$
I tried differentiating the entire expression:
$$\begin{align}
\frac{-2x}{2\sqrt{1-x^2}}+\frac{-2y}{2\sqrt{1-y^2}}\frac{dy}{dx} &= a(1-\frac{dy}{dx}) \\
a + \frac{x}{\sqrt{1-x^2}} &= \frac{dy}{dx}(a-\frac{y}{\sqrt{1-y^2}}) \\
\frac{dy}{dx} &= (a+\frac{x}{\sqrt{1-x^2}})(a-\frac{y}{\sqrt{1-y^2}})^{-1}
\end{align}$$
I am not sure how to proceed now. Is there a better approach? Is it advisable to use my approach?
|
As stated in the comments this is actually a wee-bit lengthier and
hence not recommended.
Since i always try to resort to trigonometric substitutions whenever fractional powers are involved I'm gonna attempt this by substituting $x=sinA$ and $y=sinB$
$cosA +cosB=a(sinA-sinB)$
$2cos\frac{(A+B)}{2} \times cos\frac{(A-B)}{2} =a\times 2Cos\frac{(A+B)}{2}\times Sin\frac{(A-B)}{2}$
$cot\frac{(A-B)}{2} =a$
$A-B=2cot^{-1}a$
Magically reducing it to
$sin^{-1}x-sin^{-1}y =2cot^{-1}a$
Now differentiate…
$\frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-y^{2}}} \times \frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$
If you are preparing for JEE ADV i would recommend practising differential calculus book by cengage , and for theory i would recommend Differential Calculus by AmitMAgarwal
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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How to take the derivative of $Y=\log(x+\sqrt{a^2+x^2})$?
$$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})$$
Which again goes to a chain rule as $\frac{2}{3} \sqrt{a^2+x^2}\cdot \frac{d}{dx} (a^2+x^2)$
Is there an alternative way other than this? Hints, please.
| I think you missed some terms $$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)$$ Replacing $f$ by its expression finally leads to $$\frac{dY}{dx}=\frac{1}{\sqrt{a^2+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that this sum is an integer. I have to show that
$$g\left(\frac{1}{2015}\right) + g\left(\frac{2}{2015}\right) +\cdots + g\left(\frac{2014}{2015}\right) $$
is an integer. Here $g(t)=\dfrac{3^t}{3^t+3^{1/2}}$.
I tried to solve it using power series, but I can not finish with any convincing argument to said that the sum is an integer. (Also the use of a computer isn't allowed.)
| By a rigid translation you can see that $g$ is an odd function around the point $(\frac 1 2, \frac 1 2)$:
$$
\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}}{(3^x+1) 3^{1/2}} - \frac 1 2 \\
&= \frac{3^x}{3^x+1} - \frac 1 2 \\
&= \frac{2(3^x)-(3^x+1)}{2(3^x+1)} \\
&= \frac{1}{2}\frac{3^x-1}{3^x+1} \\
&= \frac{1}{2}\frac{e^{x\log(3)}-1}{e^{x\log(3)}+1} \\
&= c_1 \tanh(c_2x) \ \ \text{ with $c_1 = \frac 1 2$ and $c_2 = \log(3)$ }
\end{align}
$$
The last formula is the hyperbolic tangent of a multiple of $x$, so it's symmetric with respect to the origin. From this, it follows that the $f(x) + f(-x) = 0$, which, after the transformation, means $g(x) + g(1-x) = 1$.
From this the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
} |
The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})?$ What is the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots (1+x^{100})?$
I manually expanded $(1+x)(1+x^2)(1+x^3)...(1+x^{10})$ and calculated the coefficient of $x^9$ as $8$ but i dont know how to solve it without expanding.Please help me.
| The coefficient of $x^9$ is the number of partitions of $9$ into distinct parts, i.e., the number of ways of writing $9$ as the sum of distinct positive integers when the order of the summands doesn’t matter. The sequence of these numbers is OEIS A000009, and as you can see there, there is no nice formula. However, for small $n$ it’s not hard to write out the partitions by hand: $9$, $8+1$, $7+2$, $6+3$, $5+4$, $6+2+1$, $5+3+1$, and $4+3+2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve the recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n $
Solve the recurrence relation
$$a_n = 4a_{n-1} - 3a_{n-2} + 2^n $$
With initial conditions:
$a_1 = 1$
$a_2 = 11$
I have done similar recurrence relation problems to this, but none that were a non-homogeneous recurrence relation such as this one.
So far I have:
$$r^n = 4^{n-1} - 3^{n-2} $$
Divide both sides by $$\frac{1}{r^{n-2}}$$
Giving me this as my Auxiliary Equation:
$$ r^n - 4r + 3 = 0 $$
I then solved for the $r$ values and got $r = -4$ and $r = 1$
I am stumped from here as to where the non-homogeneous piece comes into play, any help is appreciated.
| A general technique is to use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write your recurrence as:
$$
a_{n + 2}
= 4 a_{n + 1} - 3 a_n + 4 \cdot 2^n
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= 4 \sum_{n \ge 0} a_{n + 1} z^n - 3 \sum_{n \ge 0} a_n z^n
+ 4 \sum_{n \ge 0} 2^n z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
&= 4 \frac{A(z) - a_0}{z} - 3 A(z) + 4 \frac{1}{1 - 2 z} \\
\frac{A(z) - 1 - 11 z}{z^2}
&= 4 \frac{A(z) - 1}{z} - 3 A(z) + 4 \frac{1}{1 - 2 z}
\end{align*}$
Solve for $A(z)$, as partial fractions:
$$
A(z)
= \frac{7}{1 - 3 z} - \frac{4}{1 - 2 z} - \frac{2}{1 - z}
$$
This is just geometric series. We want:
$$ a_n = [z^n] A(z) = 7 \cdot 3^n - 4 \cdot 2^n - 2 $$
| {
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"url": "https://math.stackexchange.com/questions/1475469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
How many numbers less than $1000$ with digit sum to $11$ and divisible by $11$ How many positive (integers) numbers less than $1000$ with digit sum to $11$ and divisible by $11$?
There are $\lfloor 1000/11 \rfloor = 90$ numbers less than $1000$ divisible by $11$.
$N = 100a + 10b + c$ where $a + b + c = 11$ and $0 \le a, b, c \le 9$
I got $\binom{13}{2} - 9 = 69$ solutions.
| Digitsum is related to the modulo 9 operation. A weakening of the conditions given is that you are counting how many $0\leq n\leq 1000$ satisfy the coungruencies:
$\begin{array}{} n\equiv 2\pmod{9}\\
n\equiv 0\pmod{11}\end{array}$
By the chinese remainder theorem, we get that
$n\equiv 11\pmod{99}$
So, we can look at the possible solutions and trim the ones that don't meet the stronger requirement that the digit sum be $11$ (as opposed to $2$ or $20$ or $29$ or $37$)
We have the list then $\{11,110,209,308,407,506,605,704,803,902\}$
All but the first two have digitsum 11 (whereas the first two have only digit sum equaling 2).
The answer is then $8$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with solving recurrence relations using iterative substitution I need help solving these two recurrences with iterative substitution. I've looked at examples, and tried to follow them, but I just don't understand the whole plugging the recurrence into itself. I tried them out, not sure if they are correct, but if someone can point me in the right direction I would be very greatful!!!
For both assume $T(n) = \theta(1)$ for $n \leq 1$.
First recurrence:
$$
\begin{equation}
\begin{split}
T(n) &= T(n - 2) + 7 \\
& = T(n - 2 - 2) + 7 - 2 \\
&= T(n - 4) + 5 \\
&= T(n - 2^i) + 5 - 2^i \newline
&=T(n) =\theta(n)
\end{split}
\end{equation}
$$
Second recurrence:
$$
\begin{equation}
\begin{split}
T(n) &= nT(n - 1) + 1 \\
& = nT(n - 1 - 1) + 1 \\
&= n^2T(n - 2) + 1 \\
&= n^iT(n - i) + 1 \\
&= \theta(n)
\end{split}
\end{equation}
$$
I'm pretty sure the last one is wrong, I'm guessing it is $\log (n)$ from when I use the master theorem on it instead. I'm not sure though.
| Regarding the first recurrence:
\begin{align} T(n) &= T(n - 2) + 7 \\ &= T(n - 2 - 2) + 7 - 2
\end{align}
Should it not be
$$
T(n-2) \mapsto T(n-4)+7
$$
and
\begin{align}
T(n) &= T(n - 2) + 7 \\
&= (T((n - 2) - 2) + 7) + 7 \\
&= T(n + 2\cdot (-2)) + 2\cdot 7 \\
&= \vdots \\
&= T(n + k\cdot (-2)) + k\cdot 7
\end{align}
instead?
Regarding the second recurrence:
\begin{align} T(n) &= nT(n - 1) + 1 \\ & = nT(n - 1 - 1) + 1
\end{align}
This should be
$$
T(n-1) \mapsto (n-1) T(n-2) + 1
$$
and
\begin{align}
T(n) &= n T(n - 1) + 1 \\
&= n ((n-1) T(n - 2) + 1) + 1 \\
&= n(n-1) T(n - 2) + n + 1
\end{align}
The resulting expression might be useful, if the argument of $T$ is reduced to some known initial value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the derivative for a product of two polynomial functions? In my problem, I am attempting to find $f'(x)$ when $f(x)=(5x^2-2x+8)(4x^2+7x-3)$. For my work I have:
\begin{align}
& \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt]
= {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt]
= {} & 40x^3+35x^2-16x^2-14x+64x+56+40x^3-16x^2+70x^2-14x-30x+6 \\[8pt]
= {} & 80x^3 + 73x^2+6x+62
\end{align}
But when I plugged my original equation [$f(x)=(5x^2-2x+8)(4x^2+7x-3)$] into an online derivative calculator to check my answer, it comes out as:
$$=80x^3+81x^2+6x+62\ldots\text{ ?}$$
Can anyone spot where I am going wrong (if I am)?
| $4x^2 \cdot -2 = -8x^2$. You have written $-16x^2$. Purely an arithmetic mistake. Rest looks good to me.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of trigonometric relationship of angle bisector I have to prove the following equation:
$$CD=\frac{2ab\cos\frac{C}{2}}{a+b}$$ where $CD$ is the angle bisector of $C$ in $\Delta ABC $.
My attempt:
I've started by rearranging the equation in terms of $\cos \frac{C}{2}$ ,then by squaring and subtracting $\sin^2 \frac{C}{2}$ from both sides, yielding thus:
$$\cos^2 \frac{C}{2}-sin^2\frac{C}{2} =\frac{(a+b)^2 \cdot CD^2}{(2ab)^2}- \ \sin^2\frac{C}{2}\tag{1}$$
$$\cos C =\frac{(a+b)^2 \cdot CD^2}{(2ab)^2} - \frac{(s-a)(s-b)}{ab}\tag{2}$$
Now i express $ \cos C $ in terms of $a,b,c$ ,then I simplify by multiplying for $ab$ both sides and I multiply out $(s-a)(s-b)$ , getting now:
$$2(b^2+a^2-c^2)=\frac{(a+b)^2 \cdot CD^2}{4ab}-(ab+bc-b^2+c^2-a^2)\tag{3}$$
By simplifying this and rearranging for $CD^2$ I get:
$$CD^2=\frac{(a^2+b^2-c^2 +ab+ bc)(ab)}{(a+b)^2}\tag{4}$$
Finally by replacing $CD^2=ab-(abc^2)/(a+b)^2\tag{5}$
and doing all the algebraic manipulation i get the final result that
$a+c=2a$ which is clearly wrong...And here i am asking for humble help on math.stackexchange
| Just use the law of sines; call $\gamma=2\delta$ the angle in $C$, $\alpha$ the angle in $A$, $\beta$ the angle in $B$. Also let $a=BC$, $b=AC$, $c=AB$. Finally, let $R$ be the radius of the circumscribed circle to $ABC$. The law of sines for $ABC$ says that
$$
a=2R\sin\alpha,\quad
b=2R\sin\beta,\quad
c=2R\sin\gamma
$$
whereas, calling $s$ the bisector to be found, the law of sines says
$$
\frac{s}{\sin\alpha}=\frac{AD}{\sin\delta},\quad
\frac{s}{\sin\beta}=\frac{DB}{\sin\delta}
$$
so
$$
c=AD+DB=
s\sin\delta\left(\frac{1}{\sin\alpha}+\frac{1}{\sin\beta}\right)=
s\sin\delta\left(\frac{2R}{a}+\frac{2R}{b}\right)
$$
Therefore
$$
2R\sin\gamma=s\sin\delta\left(\frac{2R}{a}+\frac{2R}{b}\right)
$$
and, since $\sin\gamma=2\sin\delta\cos\delta$,
$$
2\cos\delta=s\left(\frac{1}{a}+\frac{1}{b}\right)
$$
and the relation
$$
s=\frac{2ab\cos\delta}{a+b}
$$
follows immediately.
Note that the above relations with $AD$ and $DB$ also prove a theorem of elementary geometry
$$
\frac{AD}{DB}=\frac{\sin\beta}{\sin\alpha}=\frac{a}{b}
$$
(the second equality follows from the law of sines).
| {
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"timestamp": "2023-03-29T00:00:00",
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Find solutions of $2^m\cdot p^2+1=q^5$ $2^m\cdot p^2+1=q^5$
$p$ and $q$ are prime numbers
find $p$ and $q$
I think it will be useful to transfer $1$ to the other side of the equation
$2^m\cdot p^2=(q-1)(q^4+q^3+q^2+q+1)$
and we know $gcd(q-1,q^4+q^3+q^2+q+1)=1$ or $5$
we know if
I)$q-1|2^m \implies p^2|q^4+q^3+q^2+q+1$ or
II)$2^m|q-1 \implies q^4+q^3+q^2+q+1|p^2$
if $gcd(q-1,q^4+q^3+q^2+q+1)=5 \implies$
I)$5|2^m \implies$
Inconsistency
II)$5|p^2 \implies p=5$
$\implies gcd(q-1,q^4+q^3+q^2+q+1)=1$
But I went to this part of the problem and more of this I could not continue
| Let $f(q) = q^4+q^3+q^2+q+1$. Since $f(q)$ is odd, it must be equal to $p$ or to $p^2$. The first is not possible because it would imply that $q-1=2^mp$, i.e. $q>p$ which contradicts with $f(q)=p$.
Thus $f(q)=p^2$ and $q-1=2^m$. Since $q\equiv 1 \pmod{2^m}$, we get $p^2=f(q)\equiv 5 \pmod{2^m}$. This is possible only for $m \leq 2$ since all odd squares are $1 \pmod 8$.
Now $m=1$ gives $q=3$ and $p=11$, while $m=2$ giving $q=5$ does not yield a valid solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\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} \\
&= -16\times (-1)^{1/3}.
\end{align*}
$$
So, which is the correct?
| They are both correct.
$$(-1)^{3} = -1$$
because it is
$$-1 \times -1 \times -1 $$
and a negative $\times$ a negative is a positive:
\begin{align*}
(-1 \times -1) \times -1 \\
= 1 \times -1 \\
= -1 \\
\end{align*}
Because of that, a solution to "What is the cube root of -1 ($\sqrt[3]{-1}$)" is $-1$.
This means that $$-16 \times -1^{1/3} = 16$$
can also be written as
$$-16 \times -1 = 16$$
which is clearly true.
Also, it may be easier to solve
$$−8^{4/3}$$
with the following method:
\begin{align*}
-8^{4/3} \\
&=\sqrt[3]{-8}^4 \\
&=2^4 \\
&= 2 \times 2 \times 2 \times 2 \\
&=16 \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Solve: $(1-x)^2y''-4xy'-(1+x^2)y=x$ To solve the differential equation:
$$(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 \right) a_{n+2} +
\left( -2\,{n}^{2}-10\,n-12 \right) a_{n+3} + \left( {n}
^{2}+7\,n+12 \right) a_{ n+4} = 0\cr
a_0 &= y(0)\cr
a_1 &= y'(0)\cr
a_2 &= y''(0)/2 = a_0/2\cr
a_3 &= y'''(0)/6 = 1/6 - a_0/3 + 5 a_1/6\cr}$$
$$y \left( x \right) =a_{{0}}+a_{{1}}x+{\frac {a_{{0}}}{2}}{x}^{2}+
\left( {\frac {a_{{0}}}{3}}+{\frac {5\,a_{{1}}}{6}}+{\frac {1}{6}}
\right) {x}^{3}+ \left( {\frac {17\,a_{{0}}}{24}}+{\frac {5\,a_{{1}}
}{6}}+{\frac {1}{6}} \right) {x}^{4}+ \left( {\frac {29\,a_{{0}}}{30}}
+{\frac {161\,a_{{1}}}{120}}+{\frac {31}{120}} \right) {x}^{5}+
\left( {\frac {205\,a_{{0}}}{144}}+{\frac {347\,a_{{1}}}{180}}+{
\frac {67}{180}} \right) {x}^{6}+ \left( {\frac {5203\,a_{{0}}}{2520}}
+{\frac {14141\,a_{{1}}}{5040}}+{\frac {2731}{5040}} \right) {x}^{7}+
\left( {\frac {120257\,a_{{0}}}{40320}}+{\frac {2917\,a_{{1}}}{720}}+
{\frac {493}{630}} \right) {x}^{8}+ \left( {\frac {194149\,a_{{0}}}{
45360}}+{\frac {2109857\,a_{{1}}}{362880}}+{\frac {407527}{362880}}
\right) {x}^{9}
+\ldots
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
|
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
False.
It also shows the remaining four solutions are roots of $z^4 + z^2 + 1 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 3
} |
Fibonacci and Matrices Consider Matrix $$ 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}$$ for $n \geq 20$, where $F_n$ is the $n^{th}$ Fibonacci number.
I don't get it, please help. Thank you!
| Hint:
$$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}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Karush-Kuhn-Tucker NLP
Consider the nonlinear program
Minimize: \begin{align}f(x,y) = \frac{1}{2}x^2 - 10xy + 10y^2\end{align}
Subject to: \begin{align}2x +y^2 &\le 5 \implies g_1(x,y)=2x + y^2 -5 \le0 \\ x^2 - ay &\le 2 \implies g_2(x,y) =x^2 -ay -2 \le 0\end{align}
a) Is the above problem a convex optimisation problem?
b) For which value of $a$ is $(x,y)=(1,1)$ a KKT point?
My attempt:
a)
\begin{align}\nabla f(x,y) &= \begin{bmatrix}x - 10y \\ -10x + 20y\end{bmatrix}\end{align} and \begin{align}H_f(x,y) =\begin{bmatrix}1 & -10 \\ -10 & 20\end{bmatrix}\end{align} which is positive definite, hence $f$ is strictly convex.
\begin{align}\nabla g_1(x,y) &= \begin{bmatrix}2 \\ 2y\end{bmatrix} \implies H_{g_1}(x,y) = \underbrace{\begin{bmatrix}0 & 0 \\ 0 & 2\end{bmatrix}}_{\text{Positive Semidefinite}} \\
\nabla g_2(x,y) &= \begin{bmatrix}2x \\ -a\end{bmatrix} \implies H_{g_2}(x,y) = \underbrace{\begin{bmatrix}2 & 0 \\ 0 & 0\end{bmatrix}}_{\text{Positive Semidefinite}}\end{align}
Thus both $f$ and the set of constraints are convex. Hence this is, in fact, a convex optimization problem.
b) Let $$L(x,y,\lambda_1, \lambda_2) = \frac{1}{2}x^2 - 10xy + 10y^2 - \lambda_1( 2x +y^2 -5) - \lambda_2(x^2 -ay -2)$$
The Kuhn-Tucker conditions are given by
\begin{align}x - 10y -2 \lambda_1 -2 \lambda_2 x &= 0 \tag{$1$} \\
-10x + 20y - 2 \lambda_1 y + a \lambda-2 &=0 \tag{$2$} \\
\lambda_1(2x + y^2 -5) &=0 \tag{$3$} \\
\lambda_2( x^2 -ay -2) &=0 \tag{$4$}\\
2x + y^2 -5 &\le 0 \tag{$5$} \\
x^2 -ay -2 &\le 0 \tag{$6$} \\
\lambda_1, \lambda_2 &\le 0 \tag{$7,8$}\end{align}
This is where I am stuck. Can anyone please help guide me as to how I can go about solving the rest of the problem?
| When $(x,y)=(1,1)$, we have $2x+y^2=3<5$. That is, the first constraint is strictly satisfied, and $2x+y^2-5\neq 0$. That and (3) implies $\lambda_1 = 0$.
Substituting $(x,y)=(1,1)$ and $\lambda_1=0$ into (1) yields $\lambda_2=-4.5$.
Since $\lambda_2$ is nonzero, (4) dictates that $x^2-ay-2=1-a-2=0$, or $a=-1$.
| {
"language": "en",
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} |
Find the shortest distance from the origin to the hyperbola $x^2+8xy+7y^2=225$
Find the shortest distance from the origin to the hyperbola $x^2+8xy+7y^2=225$
i know that
$$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^2}}$$
I just haven't seen it been applied in class so i'm a bit confused.
This was in a week we were learning about Lagrange Multipliers and we don't seem to be given a constraint.
| To minimise the distance, the gradient at that point must be tangent to the circle centred at the origin (*):
$$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 \tag{*}$$
$$\implies xy + 4y^2 = 4x^2 + 7xy$$
$$\implies 2(2x^2 + 3xy - 2y^2) = 0 \implies (2x - y)(x + 2y) = 0 \implies y = 2x,-\frac{x}{2}$$
and factorising the condition, $\left(x+y\right)\left(x+7y\right)=225$, thus $(3x)(15x) = 225 \Rightarrow x = ±\sqrt{5}$, $y = ±2 \sqrt{5}$, or $\left(\frac{1}{2} x\right)\left(-\frac{5}{2} x \right) =225$ which is impossible since the LHS is always not positive.
Hence the minimum distance is $\sqrt{x^2+y^2} = \sqrt{5 + 4 \cdot 5} = \boxed{5}$.
| {
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2nd degree matrix equation Let $X$ be a matrix with 2 rows and 2 columns.
Solve the following equation:
$$ 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 = 3\\
ab + bd = 5\\
ac + cd = -5\\
d^2 + bc = 8
\end{matrix}\right. $$
At this point I got stucked.
If you know how to solve this please help me! Thank you!
| We have the following criteria which you already stated correctly, but you missed one more information $(5)$ - still, you can solve this root problem without this additional knowledge by plugging in recursively - which comes from the determinant, we get then
\begin{align}
a^2 + bc &= 3 \tag1\\
ab + bd &= 5\tag2 \\
ac + cd &= -5\tag3 \\
d^2 + bc &= 8\tag4 \\
\det(X)=ad-bc&=7=\sqrt{\det(M)} \tag{5a}
\end{align}
Remark remember that we have $\det(AB)=\det(A)\det(B)$
This gives us
\begin{align} (1)-(4)&=a^2-d^2=(a-d)(a+d)=-5\\ (2)&=b(a+d)=5 \\
(3)&=c(a+d)=-5 \\ (1)+(5a)&=a(a+d)=10 \end{align}
so we get from $(2)\wedge(3)$ $b=-c$ and further $a-d=c$ and $a=-2(a-d)\iff\frac32a=d$
therefore
\begin{align} a(a+d)=10=a(a+\frac32a)=\frac52a^2\iff4=a^2 \end{align}
and thus we get for $a=2$
\begin{align}
a=2,d=3,c=-1,b=1
\end{align}
so
$$
X_1=\begin{pmatrix}2 &1\\ -1&3\end{pmatrix}
$$
and for $a=-2$
\begin{align}
a=-2,d=-3,c=1,b=-1
\end{align}
so
$$
X_2=\begin{pmatrix}-2 &-1\\ 1&-3\end{pmatrix}=-X_1
$$
Remark due to Robert Israel:
Indeed we have to investigate the other possible determinant solution
\begin{align}
\det(X)=ad-bc&=-7 \tag{5b}
\end{align}
then we get
\begin{align}
(1)+(5b)&=a(a+d)=-4 \\
(1)-(4)&=a^2-d^2=(a-d)(a+d)=-5
\end{align}
which gives us $\frac54a=(a-d)\iff-\frac14a=d$ and therefore
\begin{align}
a(a+d)=-4=a(a-\frac14a)\iff-4=\frac34a^2
\end{align}
which leads, if we stay in the field of the real numbers, to a contradiction. However, one might find for example other complex solutions. For a more detailed discussion please check out the comment section.
| {
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"answer_id": 0
} |
Show $\frac{1}{n - 1} \geq e^\frac{1}{n} - 1, for~n \gt 1$ I am to show that
$\frac{1}{n - 1} \geq e^\frac{1}{n} - 1, for~n \in \mathbb{N}^+, n \gt 1$.
I tried substituting using $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$, which gives: $\frac{1}{n-1} \geq \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} ...$, and I also tried using the inequality $e^\frac{1}{n-1} \geq e^\frac{1}{n}$, but from there on I am stuck.
Can you give me a hint?
| You have said $e^{1/n} - 1 = \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} + \ldots$
You could also say $\frac{1}{n-1}= \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \ldots$, and then $n\gt 1$ will give you what you want since each term in the second series is greater than or equal to the corresponding term in the first series and both series are absolutely convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490942",
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"source": "stackexchange",
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Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge? Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge?
| $$\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{1}{3n+3}> \frac{1}{3n+1}$$
So no.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491522",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Is there a way to solve for $x$ and $y$ in this simultaneous equation. Is there a way to solve for x and y in this simultaneous equation?
$$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}{2} \\
8 = 5 \\
\end{cases}
$$
And $8\neq 5$ so no solutions!
| {
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"answer_id": 7
} |
Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule
Find the following limit
$$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
without using L'Hopital's rule.
I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x\right)^\frac1x=e,\lim_{x\to0}x^x=1$ and equivalent infinitesimals at $x\to0$ such as $x\sim\sin x,a^x\sim1+x\ln a,(1+x)^a\sim1+ax$. This is what I did so far:
$$\begin{align}\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\lim_{x\to0}\frac{\left(\frac{3-(1-\cos x)}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(\frac{3-\frac12x^2}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(1-\frac16x^2\right)^x-1}{x^3}\end{align}$$
I used fact that $x\sim\sin x$ at $x\to0$. After that, I tried to simplify $\left(1-\frac16x^2\right)^x$. Problem is because this is not indeterminate, so I cannot use infinitesimals here. Can this be solved algebraically using fundamental limits, or I need different approach?
| Another approach starts with using the half-angle formula for the sine function to establish the identity
$$\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 \frac14 x^2 \tag 3$$
Note that we can establish $(1)$ and $(2)$ using nothing more than the limit definition $e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$ of the exponential function, the fact that $\left(1-\frac xn\right)^n$ is a decreasing sequence See This Answer, and Bernoulli's Inequality. In addition, we can establish $(3)$ by appealing to the geometric interpretation of the sine and cosine functions.
Then, using $(1)-(3)$, for $x>0$ reveals
$$\begin{align}
\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\frac{1}{x^3}\left(e^{x\log\left(1-\frac23 \sin^2(x/2)\right)}-1\right)\\\\
&\le-\frac16 \frac{\cos^2(x/2)}{1+\frac16x^3\cos^2(x/2)}
\end{align}$$
and $$\begin{align}
\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\frac{1}{x^3}\left(e^{x\log\left(1-\frac23 \sin^2(x/2)\right)}-1\right)\\\\
&\ge-\frac16 \frac{1}{1-\frac23 \sin^2(x/2)}
\end{align}$$
Therefore, the right-sided limit
$$\lim_{x\to 0^+}\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)=-\frac16$$
A similar development for $x<0$ shows that the left-sided limit is also $-1/6$. Therefore, using only (i) standard inequalities that can be established without using calculus and (ii) the squeeze theorem yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)=-\frac16}$$
| {
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Solving this second order ODE with power series method Problem: Use the power series method to find a general solution for the following differential equation: $$ (x^2 - 3) y'' + 2x y' = 0 $$
Attempt: We see that $x = \pm \sqrt{3}$ is a singular point of this ODE. Hence we apply the power series method around $x = 0$, and look for a solution of the form $$ y(x) = \sum_{n=0}^{\infty} a_n x^n. $$ We have that \begin{align*} y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \qquad \qquad y''(x) = \sum_{n=2}^{\infty} (n-1) n a_n x^{n-2} \end{align*} Plugging this in the ODE and distributing gives \begin{align*} \sum_{n=2}^{\infty} (n-1) n a_n x^n - 3 \sum_{n=2}^{\infty} (n-1)n a_n x^{n-2} + 2 \sum_{n=1}^{\infty} n a_n x^n = 0 \end{align*} Then if I shift the indices to get the exponents equal I get \begin{align*} \sum_{n=2}^{\infty} (n-1) n a_n x^n - 3 \sum_{n=0}^{\infty} (n+1)(n+2) a_{n+2} x^n + 2 \sum_{n=1}^{\infty} n a_n x^n = 0 \end{align*} From the identity theorem we have the following recurrence relation for the coefficients: \begin{align*} (n-1) n a_n - 3 (n+1) (n+2) a_{n+2} + 2 n a_n = 0 \end{align*} We can rewrite this as \begin{align*} a_{n+2} = \frac{n}{3(n+2)} a_n \end{align*}
From what I understand, we need to find two solutions now (since the order of the differential equation is two), one for the even and one for the odd coefficients. A general solution is then a linear combination of the two. I tried solving the recurrence relation for the even coefficients, but I got stuck. For the even coefficients we have \begin{align*} \frac{2n}{3 (2n + 2)} a_{2n} = a_{2n + 2} \end{align*} Now I want to expand this, and write $a_{2n+2}$ in terms of $a_2$. I noticed that \begin{align*} a_{2n} = a_{2n-2} \frac{2 (n-1)}{3 \big( 2 (n-1) + 2 \big)} \end{align*} and \begin{align*} a_{2n-2} = a_{2n - 4} \frac{2}{3} \frac{n-2}{2(n-2) + 2} \end{align*} so that \begin{align*} a_{2n+2} = \frac{2}{3} \frac{n(n-1)(n-2)}{(2n+2)(2(n-1) +2 ) (2 (n-2) + 2) } a_{2n-4} \end{align*} So I figure that if I continue this until $a_2$ shows up at the RHS, I have to write $n!$ in the numerator. But how do I know with what terms I should stop the denominator? Also, is this the general method for solving a problem like this?
| Prior to substituting your power series, consider starting off by reducing the order of the ODE via the substitution $y'=z$, so you have
$$(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_{n-2}-3na_n)x^{n-1}-3a_1-6a_2x+2a_0x=0$$
So you have $a_1=0$ and $a_0=3a_2$ as initial conditions for the recurrence relation,
$$3a_n=a_{n-2}$$
which you might agree is a bit simpler than the one you have. This can be solved via the method of generating functions. Let $A(t)=\sum\limits_{n\ge0}a_nt^n$ be the generating function for $a_n$, then the recurrence relation can be written as
$$\begin{align*}
3\sum_{n\ge2}a_nt^n&=\sum_{n\ge2}a_{n-2}t^n\\[1ex]
3\left(\sum_{n\ge0}a_nt^n-a_0-a_1t\right)&=t^2\sum_{n\ge2}a_{n-2}t^n\\[1ex]
3A(t)-3a_0&=t^2\sum_{n\ge0}a_nt^n\\[1ex]
3A(t)-3a_0&=t^2A(t)\\[1ex]
A(t)&=\frac{a_0}{1-\frac{t^2}{3}}\\[1ex]
&=\frac{a_0}{2}\left(\frac{1}{1-\frac{t}{\sqrt3}}+\frac{1}{1+\frac{t}{\sqrt3}}\right)
\end{align*}$$
Recall the well-known power series,
$$\frac{1}{1-t}=\sum_{n\ge0}t^n\quad\text{for }|t|<1$$
This means the generating function can be written as
$$\begin{align*}
A(t)&=\frac{a_0}{2}\left(\sum_{n\ge0}\left(\frac{t}{\sqrt3}\right)^n+\sum_{n\ge0}\left(-\frac{t}{\sqrt3}\right)^n\right)\\[1ex]
&=\frac{a_0}{2}\sum_{n\ge0}3^{-n/2}\left(1+(-1)^n\right)t^n
\end{align*}$$
for $|t|<\sqrt3$. Comparing to the general proposed form of $A(t)$, it stands to reason that
$$a_n=\frac{a_0}{2}3^{-n/2}\left(1+(-1)^n\right)$$
All this to say that the solution to this first-order ODE is
$$y'(x)=z(x)=\frac{a_0}{2}\sum_{n\ge0}\frac{1+(-1)^n}{3^{n/2}}x^n$$
Integrating once, you get
$$y(x)=\frac{a_0}{2}\sum_{n\ge0}\frac{1+(-1)^n}{3^{n/2}(n+1)}x^{n+1}$$
(noting that the constant of integration disappears, since $y(0)=\sum\limits_{n\ge0}0=0$).
This might suggest only one solution, but in fact you can extract another. Since odd terms of $a_n$ disappear, upon setting $n=2k$ you have
$$y=a_0\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)3^k}$$
which is the power series for $a_0\sqrt3\text{arctanh}\dfrac{x}{\sqrt3}$, which can be expanded into two linearly independent logarithmic terms. See line (3).
| {
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Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$
$|(1+i\sqrt{3})(1+i)|=8$
Using Moivre's theorem on the LHS:
$$\sqrt{(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)}=2\sqrt{2}\left(\cos\frac{\phi+2k\pi}{2}+i\sin\frac{\phi+2k\pi}{2}\right),k=0,1$$
This is not correct (checked for $\phi=\pi/3$).
How to prove this equation?
| We have $$(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\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right).$$
| {
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min polynomial of $\sqrt{6+2\sqrt{5}}$ min polynomial of $\sqrt{6+2\sqrt{5}}$ over $\mathbb{Q}$
I found $\alpha^4 -12\alpha^2 +16 = 0$ but wolframalpha does not agree with me. furthermore, would anything change is it was over $\mathbb{R}$?
| You can try to factor $\alpha^4-12\alpha^2+16$. Assuming that $\sqrt{6+2\sqrt{5}}$ is not in $\mathbb{Q}$, then the best that you can hope for is that this factors into quadratics. We write the following:
$$
\alpha^4-12\alpha^2+16=(\alpha^2+r\alpha+s)(\alpha^2+t\alpha+u).
$$
Multiplying this out, one has
$$
\alpha^4+(r+t)\alpha^3+(s+rt+u)\alpha^2+(ru+st)\alpha+su.
$$
Since the coefficient $\alpha^3$ is zero, $t=-r$. This makes the coefficient of $\alpha$ $r(s-u)$. Assuming that $r\not=0$ (which can be checked separately), this means that $s=u$. Therefore, the factorization (if it exists) is
$$
\alpha^4-12\alpha^2+16=(\alpha^2+r\alpha+s)(\alpha^2-r\alpha+s).
$$
Since $s^2=16$, $s=\pm 4$ and since $2s-r^2=-12$, we know that the only solution in the integers is $s=-4$ and $r=\pm 2$.
Therefore,
$$
\alpha^4-12\alpha^2+16=(\alpha^2+2\alpha-4)(\alpha^2-2\alpha-4).
$$
We can check that this factorization is correct, and the roots of one of these polynomials is the given expression.
The roots of the first factor are $-1\pm\sqrt{5}$ and the roots of the second factor are $1\pm\sqrt{5}$. Since the given expression is positive, we can eliminate all negative possibilities, so the given expression is either $-1+\sqrt{5}$ or $1+\sqrt{5}$. Since $\sqrt{5}>2$, the expression in the radical is at least $10$, so its square root is at least $3$. Since $-1+\sqrt{5}$ is at most $2$ (since $5<9$), this means that the given expression is $1+\sqrt{5}$.
Therefore, $\alpha^2-2\alpha+4$ is the minimal polynomial for this expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
In order for both roots of the equation $x^2 + ( m + 1 ) x + 2m - 1 = 0$ to not be real, it must be ... In order for both roots of the equation $x^2 + ( m + 1 ) x + 2m - 1 = 0 $ to not be real, it must be the case that ...
A. $m > 1$
B. $1 < m < 5$
C. $1 \leq m \leq 5$
D. $m < 1$ or $m > 5$
E. $m \leq 1$ or $m \geq 5$
| In case you forget the discriminant like me: The equation
$$
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
$$
because then the square root of a negative number prevents real solutions $x$ to exist.
That right hand side is a quadratic function
$$
f(m) = \frac{1}{4} \left(m^2-6m+5\right)
$$
with positive factor for $m^2$, so its graph is an upwards open parabola.
The roots are
$$
m = 3 \pm \sqrt{4} = 3\pm 2\in \{ 1, 5\}
$$
so $f(m)$ is negative on the intervall $(1, 5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Make all the sixes. Fun Given the following it is possible to complete each case so that they are all true? (i.e. so that the equation .... = 6 is true)
You can add any mathematical operations and parentheses but you cannot add any numbers.
i.e. $\sqrt{4}$ is fine but $\sqrt[3]{8}$ is not.
So...
$$0\text{ }\text{ }\text{ }0\text{ }\text{ }\text{ }0 = 6$$
$$1\text{ }\text{ }\text{ }1\text{ }\text{ }\text{ }1 = 6$$
$$2\text{ }\text{ }\text{ }2\text{ }\text{ }\text{ }2 = 6$$
$$3\text{ }\text{ }\text{ }3\text{ }\text{ }\text{ }3 = 6$$
$$4\text{ }\text{ }\text{ }4\text{ }\text{ }\text{ }4 = 6$$
$$5\text{ }\text{ }\text{ }5\text{ }\text{ }\text{ }5 = 6$$
$$6\text{ }\text{ }\text{ }6\text{ }\text{ }\text{ }6 = 6$$
$$7\text{ }\text{ }\text{ }7\text{ }\text{ }\text{ }7 = 6$$
$$8\text{ }\text{ }\text{ }8\text{ }\text{ }\text{ }8 = 6$$
$$9\text{ }\text{ }\text{ }9\text{ }\text{ }\text{ }9 = 6$$
For example, the number 6 line could be...
$$6+6-6 = 6$$
There are no trick questions here and it doesn't require using the $\neq$ sign.
Can you find the correct mathematical operations and parentheses to complete each line?
| $$(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 }={ R }_{ 1 }+{ R }_{ 2 } = \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
2 & 1 & -1 & 1 \\
1 & -1 & 4 & 2\\
4 & 4 & 1 & 3
\end{array}\right]$$
$$\Rightarrow { R }_{ 2 }={ R }_{4}-2{ R }_{ 2 } = \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
0 & 2 & 3 & 1 \\
1 & -1 & 4 & 2\\
4 & 4 & 1 & 3
\end{array}\right]$$
$$ \Rightarrow { R }_{ 3 }=4{ R }_{3}-{ R }_{ 4 } = \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
0 & 2 & 3 & 1 \\
0 & -8 & 15 & 5\\
4 & 4 & 1 & 3
\end{array}\right]$$
At this point I have no idea what to do to get solve this system of equations. To be honest, I don't even know how to handle Gaussian Elimination with $4$ equations with $3$ unknowns. I would like an explanation as to what I need to do (in layman terms, please) and then point me in the right direction.
| Hint:if you want o solve the first threee equations you will get:
he system
$$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$$
aer simplifiation we get
$$y-z=-1$$
$$y-z=-1/2$$
this is a contradiction thus no solution exists
| {
"language": "en",
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Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$
Prove: $${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)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1$).
How to prove this equation?
Is it possible to use induction?
$${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1};$$
$$(2n+1)!!=\frac{(2n)!(2n+1)}{2^nn!}\Rightarrow \frac{n!2^n}{(2n+1)!!}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)};$$
$$\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)}$$
What now?
| ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}
=\frac{n!2^n}{(2n+1)!!}
$
Since
$(1+x)^n
=\sum_{k=0}^n \binom{n}{k} x^k
$,
$(1-x^2)^n
=\sum_{k=0}^n \binom{n}{k}(-1)^k x^{2k}
$.
Integrating from $0$ to $1$,
$\int_0^1 (1-x^2)^n\,dx
=\sum_{k=0}^n \binom{n}{k}(-1)^k \int_0^1 x^{2k}\,dx
=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{2k+1}
$.
$\begin{align*}
\int_0^1 (1-x^2)^n\,dx
&=\int_0^1 (1-x)^n (1+x)^n\,dx\\
&=\int_0^1 y^n (2-y)^n\,dy
\qquad(y = 1-x)\\
&=2\int_0^\frac12 (2z)^n (2-2z)^n\,dz
\qquad(y = 2z)\\
&=2^{2n+1}\int_0^\frac12 z^n (1-z)^n\,dz\\
&=2^{2n}\int_0^1 z^n (1-z)^n\,dz\\
&=2^{2n}B(n+1, n+1)
\qquad\text{(Beta function)}\\
&=2^{2n}\frac{(n!)^2}{(2n+1)!}\\
\end{align*}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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A club with 32 women and 10 men needs to form a committee of size 5 I want to make sure that I'm thinking of this correctly.
A club with 32 women and 10 men needs to form a committee of
size 5:
(a) How many committees are possible? - I said $$
\begin{pmatrix}
42 \\
5 \\
\end{pmatrix}
$$
(b) How many committees are possible if a committee must have 3
women and 2 men? - $$
\begin{pmatrix}
32 \\
3 \\
\end{pmatrix}
\begin{pmatrix}
10 \\
2 \\
\end{pmatrix}
$$
(c) How many committees are possible if a committee must consist
of all women or all men? $$
\begin{pmatrix}
32 \\
5 \\
\end{pmatrix} +
\begin{pmatrix}
10 \\
5 \\
\end{pmatrix}
$$
(d) How many committees are possible if a committee must have
exactly one women? - $$
\begin{pmatrix}
32 \\
1 \\
\end{pmatrix}
\begin{pmatrix}
10 \\
4 \\
\end{pmatrix}
$$
(e) How many committees of five different executive positions are
possible? (e.g., chair, treasurer, secretary, etc.) - (42)(41)(40)(39)(38)
Thanks for any help!
| In part (e), order matters. Selecting Angela to be chair and Barbara to be treasurer is different from selecting Barbara to be chair and Angela to be treasurer. Therefore, there are $42$ ways to fill the position of chair, $41$ ways to pick the treasurer, $40$ ways to select the secretary, and so forth. Hence, there are
$$42 \cdot 41 \cdot 40 \cdot 39 \cdot 38 = \frac{42!}{(42 - 5)!} = \frac{42!}{37!}$$
ways to select a committee consisting of five different executive positions. This is a permutation in which $5$ members are selected from a set with $42$ members. The number of ordered selections of $k$ elements from a set of $n$ elements is $$P(n, k) = \frac{n!}{(n - k)!}$$ since there are $n$ ways to make the first selection, $n - 1$ ways to make the second selection, and so forth until there are $n - k + 1$ to make the $k$th selection and
$$n(n - 1) \cdots (n - k + 1) = \frac{n(n - 1) \cdots (n - k + 1) \cdot (n - k)!}{(n - k)!} = \frac{n!}{(n - k)!}$$
Unless members of a committee are assigned to specific posts, the order in which the committee is selected does not matter. Selecting Alice, Benjamin, Charles, Diana, and Evelyn to serve on the committee produces the same committee as selecting Diana, Charles, Evelyn, Alice, and Benjamin. There are $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ orders in which we could select the same five people. Since there are $$42 \cdot 41 \cdot 40 \cdot 39 \cdot 38 = \frac{42!}{(42 - 5)!}$$ ways to make an ordered selection, the number of ways we can select a committee of $5$ people from the $42$ members of the club is $$\frac{42 \cdot 41 \cdot 40 \cdot 39 \cdot 38}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{42!}{(42 - 5)!5!} = \frac{42!}{37!5!}$$ ways to select a committee of five people from the club. This is a combination in which $5$ members are selected from a set with $42$ members. The number of unordered selections (subsets) of $k$ elements from a set with $n$ elements is $$\binom{n}{k} = C(n, k) = \frac{n!}{k!(n - k)!}$$
where we divide the number of ordered selections of $k$ elements by the $k!$ orders in which those elements could be selected.
See if you can revise your answers to parts (b), (c), and (d).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Square of a mod
Find $9^{16} \pmod{25}$ by separation.
The separation means, that $25 = 5^2$.
But $9^{16} \equiv 1 \pmod{5}$, so I how I derive that $9^{16} \equiv 16 \pmod{25}$?
| As $9=3^2,9^{16}=(3^2)^{16}=3^{32}$
Now $3^3\equiv2\pmod{25}\implies3^{32}=3^2(3^3)^{10}\equiv3^2\cdot2^{10}$
Now $2^5\equiv7\implies2^{10}=(2^5)^2\equiv7^2\equiv-1$
$\implies3^2\cdot2^{10}\equiv9\cdot-1\equiv16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$ How can I compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$?
| $$\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_{x\to 0} \frac{\sin^2(x)\cos(x)}{x^2}=$$
$$\frac{\left(\lim_{x\to 0}\cos(x)\right)\left(\lim_{x\to 0}\frac{\sin^2(x)}{x^2}\right)}{8}=$$
$$\frac{\left(\cos(0)\right)\left(\lim_{x\to 0}\frac{\sin^2(x)}{x^2}\right)}{8}=$$
$$\frac{\left(1\right)\left(\lim_{x\to 0}\frac{\sin^2(x)}{x^2}\right)}{8}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\sin^2(x)}{x^2}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin^2(x)}{\frac{\text{d}}{\text{d}x}x^2}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{2x}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\sin(x)\cos(x)}{x}=$$
$$\frac{\left(\lim_{x\to 0}\cos(x)\right)\left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)}{8}=$$
$$\frac{\left(\cos(0)\right)\left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)}{8}=$$
$$\frac{\left(1\right)\left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)}{8}=$$
$$\frac{\lim_{x\to 0}\frac{\sin(x)}{x}}{8}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin(x)}{\frac{\text{d}}{\text{d}x}x}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\cos(x)}{1}=$$
$$\frac{1}{8}\lim_{x\to 0}\cos(x)=$$
$$\frac{1}{8}\cdot\cos(0)=$$
$$\frac{1}{8}\cdot 1=\frac{1}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that for $n$ an integer larger than $0$, $\frac{n^2}{n+1}$ is never an integer. I have a couple questions.
$1)$ Is my proof is considered valid $2)$ If it is good may you tell me how to improve it or if you have a better proof $3)$ how may I prove the obvious assumption I made that an integer and a non integer may never add to an integer.
Sorry for all the questions, here is my proof:
Playing around with the equation:
$$\frac{n^2+2n+1}{n+1}=n+1$$
I get:
$$\frac{n^2}{n+1}+\frac{2n+1}{n+1}=n+1$$
Let $I$ represent an integer, $N$ a non integer. I recall that:
$N+I=I$ is not possible
So either $\frac{n^2}{n+1}$ and $\frac{2n+1}{n+1}$ are both integers or they are both non
integers.
$\frac{2n+1}{n+1}=f(n)$ is increasing for $n>0$ and is bounded by horizontal asymptote $y=2$. Thus if $f(n)$ was an integer larger than $0$ it could only be $1$. This would imply $n=0$ which does not satisfy our bound on $n$. Thus for $n$ an integer larger than $0$:
$\frac{2n+1}{n+1}$ is not an integer.
Recall:
$I+N=I$ is not possible
Thus it is not possible (with $n$ larger than zero) for:
$\frac{n^2}{n+1}$ to be an integer
| 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 not an integer,
so neither is
$\frac{n^2}{n+1}$.
| {
"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 as that of $E_1$. Initially $E_1$ and $E_2$ are touching each other at the end of their major axis. Keeping $E_1$ fixed $E_2$ is now rolled over $E_1$(without sliding). Find the locus of the centre of $E_2$. Please help me to solve this problem.
| 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\theta}{a}\cdot\frac{X}{2}+\frac{\sin\theta}{b}\cdot\frac Y2=1.$$
Solving these gives
$$X=\frac{2ab^2\cos\theta}{a^2\sin^2\theta+b^2\cos^2\theta},\quad Y=\frac{2a^2b\sin\theta}{a^2\sin^2\theta+b^2\cos^2\theta}$$
Here, let $\tan\varphi=\frac YX=\frac ab\tan\theta$. Then,
$$\begin{align}r^2&=\frac{4a^2b^2}{a^2\sin^2\theta+b^2\cos^2\theta}\\\\&=\frac{4a^2b^2(1+\tan^2\theta)}{a^2\tan^2\theta+b^2}\\\\&=\frac{4a^2b^2(1+(\frac ba\tan\varphi)^2)}{a^2(\frac ba\tan\varphi)^2+b^2}\\\\&=4(a^2\cos^2\varphi+b^2\sin^2\varphi)\end{align}$$
Hence, the answer is
$$\color{red}{r=2\sqrt{a^2\cos^2\varphi+b^2\sin^2\varphi}}$$
where $(X,Y)$ is the centre of $E_2$ and $\tan\varphi=\frac YX$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
If one has the result it is not difficult to verify it by induction, but how can I solve it without induction ?
| We have
$$\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{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Proportion with 3 variables - binomial coefficients So, if we're given something like this:
$$\binom{n}{k}:\binom{n+1}{k}:\binom{n+1}{k+1}=3:4:8$$
How do I rewrite this so I can manipulate it?
Edit: Is there a general procedure for n variables?
|
Note, that the equality of proportions
\begin{align*}
a:b:c=x:y:z
\end{align*}
is a compact notation for
\begin{align*}
\frac{a}{b}=\frac{x}{y},\qquad\frac{a}{c}=\frac{x}{z},\qquad\frac{b}{c}=\frac{y}{z}\tag{1}
\end{align*}
Each equality in (1) can be derived from the other two.
So, in order to check if we can solve
\begin{align*}
\binom{n}{k}:\binom{n+1}{k}:\binom{n+1}{k+1}=3:4:8
\end{align*}
we take two proportions and calculate $n$ and $k$.
We find by solving
\begin{align*}
\frac{\binom{n}{k}}{\binom{n+1}{k}}=\frac{3}{4}\qquad \frac{\binom{n+1}{k}}{\binom{n+1}{k+1}}=\frac{4}{8}
\end{align*}
the solutions $n=7,k=2$ and can finally conclude
\begin{align*}
\binom{7}{2}:\binom{8}{2}:\binom{8}{3}=21:28:56=3:4:8
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding real solutions to $\sqrt{x}+\sqrt[3]{x^2-1}+\sqrt[4]{x^3+15}=x^2+2$ I found that $x=1$ is a solution of
$$\sqrt{x}+\sqrt[3]{x^2-1}+\sqrt[4]{x^3+15}=x^2+2.$$
How would one find other real solutions?
| To find a polynomial with that root:
You have $A+B+C=x^2+2$. Consider $(x^2+2)^2=(A+B+C)^2= x+B^2+C^2+2AB+2AC+2BC$.
Take the first 25 powers of $x^2+2$, from $1=(x^2+2)^0$ to $(x^2+2)^{24}$. They will all be linear combinations of $A,B,C,B^2,C^2,C^3,AB,AB^2,AC,ABC,AB^2C,...$ with coefficients that are polynomials in $x$. But that is a 24-dimensional space, so there is a linear relation between the 25 formulas. This linear relation is a polynomial in $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the identity $\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$
Let $n,m \in \mathbb{N}$. Prove the identity $$\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$$
This seems very similar to Vandermonde identity, which states that for nonnegative integers we have $\sum^{m}_{k=0}\binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$. But, clearly this identity is somehow different from it. Any ideas?
| For $t=0,1,\cdots, n$, $\binom{m+t}{m}$ is the coefficient of $x^m$ in the expansion of $(1+x)^{m+t}$.
Hence,
$$\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}.$$
Here, we have
$$(1+x)^m+(1+x)^{m+1}+\cdots +(1+x)^{m+n}$$$$=\frac{(1+x)^m((1+x)^{n+1}-1)}{x}=\frac{(1+x)^{m+n+1}}{x}-\frac{(1+x)^m}{x}.$$
Since there is no term of $x^m$ in $\frac{(1+x)^m}{x}$, what we want is the coefficient of $x^{m+1}$ in $(1+x)^{m+n+1}$, i.e. $\binom{m+n+1}{m+1}=\binom{m+n+1}{n}$.
| {
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"url": "https://math.stackexchange.com/questions/1521147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $x^4 + x^2 + y^4 + y^2 = 2$ ?
Wolfram|Alpha plot
| 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 you have a challenging integral to compute, and you may need to resort to an approximation technique.
Note $$4r^3\left(\cos^4t+\sin^4t\right)\frac{dr}{dt}+r^4\cdot4\left(-\cos^3t\sin t+\sin^3t\cos t\right)+2r\frac{dr}{dt}=0$$
$$\frac{dr}{dt}=\frac{2r^3\sin t\cos t\left(\cos^2t-\sin^2t\right)}{2r^2\left(\cos^4t+\sin^4t\right)+1}=\frac{2r^3\sin t\cos t\left(\cos^2t-\sin^2t\right)}{\sqrt{1+8\left(\cos^4t+\sin^4t\right)}}$$
$$\left(\frac{dr}{dt}\right)^2=\frac{4r^6\sin^2 t\cos^2 t\left(\cos^2t-\sin^2t\right)^2}{1+8\left(\cos^4t+\sin^4t\right)}$$
You can either use this to try to conquer the integral, or you can use this to help with an approximation technique like say Simpson's Rule (where it would be wise to integrate from $0$ to $\frac{\pi}{4}$ and octuple the result.)
Using Simpson, with $n=2$ over $[0,\pi/4]$ and then octuple: $$8\cdot\frac{\pi}{24}\left(1+4\sqrt{\frac{-1+\sqrt{7}}{3/2}+\frac1{28}\left(\frac{-1+\sqrt7}{3/2}\right)^3}+\sqrt{-1+\sqrt{5}}\right)\approx6.6923\ldots$$ which is within $0.16\%$ of Robert's Maple decimal output.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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
$$\sin^2 \frac{x}{2} + \cos^2\frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1.96$$
$$1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1.96$$
$$2 \sin \frac{x}{2} \cos \frac{x}{2} = 0.96$$
And with the second identity,
$$\sin\left( 2\times\frac{x}{2}\right) = 0.96$$
Thus
$$\sin x = 0.96$$
The same principle applies to the second problem
| {
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"timestamp": "2023-03-29T00:00:00",
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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 have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?
| 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 $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}$
$$= \sqrt{2}\int\frac{1}{s^2+1}ds-4\int\frac{2}{(u^2+2)^2}du$$
$$= \sqrt{2}\arctan(s)-4\int\frac{2}{(u^2+2)^2}du$$
Substitute $u = \sqrt{2}\tan (p)\quad$ and $\quad du = \sqrt{2}\sec^2(p) dp\quad$ and $\quad(u^2 + 2)^2 = (2\tan^2(p)+2)^2 = 4\sec^4(p)\quad$ and $\quad p=\arctan\frac{u}{\sqrt{2}}$
$$= \sqrt{2}\arctan(s)-\sqrt{2}\int\cos^2(p)du$$
$$= \sqrt{2}\arctan(s)-\sqrt{2}\int\left(\frac{1}{2} \cos(2p) + \frac{1}{2}\right)du$$
$$= \sqrt{2}\arctan(s)-\frac{1}{\sqrt{2}}\int\cos(2p) - \frac{1}{\sqrt{2}}\int du$$
$$= \sqrt{2}\arctan(s)-\frac{p}{\sqrt{2}} - \frac{\sin(p)\cos(p)}{\sqrt{2}}$$
$$= \sqrt{2}\arctan(s)-\frac{\arctan\frac{u}{\sqrt{2}}}{\sqrt{2}} - \frac{\sin(\arctan\frac{u}{\sqrt{2}})\cos(\arctan\frac{u}{\sqrt{2}})}{\sqrt{2}}$$
Note that $\cos(\arctan(z)) = \frac{1}{\sqrt{z^2 + 1}}$ and $\sin(\arctan(z)) = \frac{z}{\sqrt{z^2+1}}$
$$= \frac{2\sqrt{2}(u^2+2)\arctan(s)+\sqrt{2}(u^2+2)\arctan\frac{u}{\sqrt{2}}+2u}{2(u^2+2)}$$
$$=\frac{\sqrt{2}(u^2+2)\arctan \frac{u}{\sqrt{2}}-2u}{2(u^2+2)}$$
$$=\frac{\sqrt{2}(x+1)\arctan \frac{\sqrt{x-1}}{\sqrt{2}}-2\sqrt{x-1}}{2(x+1)}$$
$$=\frac{\arctan \frac{\sqrt{x-1}}{\sqrt{2}}}{\sqrt{2}}-\frac{\sqrt{x-1}}{x+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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:
$\lim _{n\rightarrow \infty }\dfrac {\sqrt {2n^{2}+n}}{\sqrt {2n^{2}+2n}}=\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}$
$\lim _{n\rightarrow \infty }\dfrac {2n^{2}+n}{2n^{2}+2n}=\lim _{n\rightarrow \infty }\dfrac {2+\dfrac {1}{n}}{2+\dfrac {2}{n}}$
since $\lim _{n\rightarrow \infty }\dfrac {1}{n}=0$, $\lim _{n\rightarrow \infty }\dfrac {2n^{2}+n}{2n^{2}+2n}=1$
hence $\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}=\left( 1\right) ^{\dfrac {1}{2}}=1$ (by composite rule)
hence $\sqrt {2n^{2}+n}=\sqrt {2n^{2}+2n}$ as $n\rightarrow \infty $
so $\lim _{n\rightarrow \infty }\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) =0$
| 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 straightforward.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 of the powers of $x$, we can write three equations, matching the coefficients of each power.
$$\begin{align*}
0&=A+B \\
1&=A-B+C \\
\frac 32&=2A-C
\end{align*}$$
All that's left is to solve the above system. Adding all three equations gives:
$$\frac 52=4A \implies A= \frac 58 $$
Plug this result back into the first equation to get
$$B=-A = -\frac 58$$
And use the second equation to get
$$C=1+B-A = 1-\frac 54 = -\frac 14$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\} < {n}^{p} \cdot \frac{1}{\sqrt{n-1}\sqrt{n}} < \frac{{n}^{p}}{n-1} < \frac{1}{{n}^{1-p}-{n}^{-p}} < \frac{1}{{n}^{1-p}} $$
so converges when p<0 how about other cases?
| 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: $(1+x)^\alpha = 1+\alpha x + o(x)$ for any fixed $\alpha \in \mathbb{R}$, when $x\to 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.
$I=\displaystyle \int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\sin x+\cos x)dx$
$I=\displaystyle \int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln\left(\dfrac{\sin x+\cos x}{\cos x}\right)dx+\int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\cos x)dx$
Since $x\rightarrow \ln(\cos x)$ is an even function,
$\displaystyle I=\int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\tan x+1)dx+2\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx$
$\displaystyle I=\int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\tan x+1)dx-\int_{0}^{\tfrac{\pi}{4}}\ln((\sec x)^2)dx$
Perform the change of variable $y=\tan x$,
$\displaystyle I=\int_{-1}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
$\displaystyle I=\int_{-1}^0 \dfrac{\ln(1+x)}{1+x^2}dx+\int_{0}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
Perform the change of variable $y=-x$ in the first integral.
$\displaystyle I=\int_{0}^1 \dfrac{\ln(1-x)}{1+x^2}dx+\int_{0}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$ in the first integral:
$\displaystyle I=\int_{0}^1 \dfrac{\ln\left(\dfrac{2x}{1+x}\right)}{1+x^2}dx+\int_{0}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
Therefore,
$\displaystyle I=\dfrac{\pi\ln 2}{4}+\int_0^1 \dfrac{\ln x}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{4}-G+G-\dfrac{\pi\ln 2}{2}=-\dfrac{\pi\ln 2}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534784",
"timestamp": "2023-03-29T00:00:00",
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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 me or give me some hint to solve it. Thanks.
Find the minimum of $\theta(x,y,z) = (x + 3)^2 + (y-2)^2 +(z-4)^2 + 12$, with $x, y,z \in R$ which satisfies $g(x,y,z) = x^2 + y^2 + z^2 -3x + 2y -8z + 2 \le 0$ and $h(x,y,z) = x + 3y + 5z = 1$
Using Kuhn-Tucker theorem, I need to solve the system of equation and in equation:
$$
\left\{
\begin{array}{c}
\nabla \theta(x) + u\nabla g(x) + vB= 0 \\
g(x) \le 0 \\
Bx = d \\
ug(x) = 0 \\
u \ge 0
\end{array}
\right.
$$
Here $B = \left[ \begin{matrix}1& 3 &5\end{matrix} \right]$
Which is equivalent to:
$$
\left\{
\begin{array}{c}
(2x + 6) + u(2x - 3) + v = 0 (1)\\
(2y - 4) + u(2y + 2) + 3v = 0 (2)\\
(2z - 8) + u(2z - 8) + 5v = 0 (3)\\
x^2 + y^2 + z^2 -3x + 2y -8z + 2 \le 0 (4)\\
u(x^2 + y^2 + z^2 -3x + 2y -8z + 2) = 0 (5)\\
x + 3y + 5z = 1 (6)\\
u \ge 0 (7)
\end{array}
\right.
$$
I calculated $1 * (1) + 3 * (2) + 5 * (3)$, then use (6), I come to conclusion that $35v - 35u = 44$. From this I can conclude that $u \neq 0$, therefore, we have this system:
$$
\left\{
\begin{array}{c}
(2x + 6) + u(2x - 3) + v = 0 \\
(2y - 4) + u(2y + 2) + 3v = 0 \\
(2z - 8) + u(2z - 8) + 5v = 0 \\
x^2 + y^2 + z^2 -3x + 2y -8z + 2 = 0\\
x + 3y + 5z = 1 \\
u \ge 0
\end{array}
\right.
$$
And this is where I got stuck
| $$\theta_{min}=\theta(x_0,y_0,z_0)=63-\frac{3 \sqrt{132685}}{35}$$
$${x_0}=1-\frac{54 \sqrt{132685}}{7805},\;{y_0}=\frac{61 \sqrt{132685}}{15610}-\frac{5}{2},\;{z_0}=\frac{3}{2}-\frac{3 \sqrt{132685}}{3122}$$
Used free CAS Maxima
http://maxima.sourceforge.net/
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $$ \sum_{k=1}^1 k4^k = \frac 49((3(1)-1)4^{(1)} + 1) $$
$\frac 49((3(1)-1)4^{(1)} + 1) = \frac49((2)4+1) = \frac49(8+1) = \frac 49(9) = 4$
$k4^k = (1)4^{(1)} = 4$
So P(1) holds.
Inductive Step: Let $s\ge1$. Assume P(s), so
$$ \sum_{k=1}^s k4^k = \frac 49((3s-1)4^s + 1) $$
Note
$$ \sum_{k=1}^{s+1} k4^k = \sum_{k=1}^{s} k4^k + (s+1)4^{s+1} $$
and by inductive hypothesis:
**
$$ \frac 49((3s-1)4^s + 1) + (s+1)4^{s+1} $$
**
I'm afraid I'm stuck after this point. I know my endpoint needs to be:
$$ \sum_{k=1}^{s+1} k4^k = \frac 49((3(s+1)-1)4^{s+1} + 1) $$
but I don't know how to get from the asterisks to the above. Any help would be greatly appreciated.
| So starting from where you ended we just have to manipulate properly:
$$\frac{4}{9}((3s - 1)4^s + 1) + (s + 1)4^{s + 1} $$
$$= \frac{4}{3}s4^s - \frac{4}{9}4^s + \frac{4}{9} + s4^{s + 1} + 4^{s + 1}$$
$$=\Big( \frac{s}{3} - \frac{1}{9} + s + 1 \Big)4^{s + 1} + \frac{4}{9}$$
$$=\frac{4}{9}\Big( (3s + 2)4^{s + 1} + 1 \Big)$$
$$=\frac{4}{9}\Big( (3(s + 1) - 1)4^{s + 1} + 1 \Big)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)
| 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 $x_1^2,x_2^2 ,roots$
$$s'=x_1^2+x_2^2=s^2-2p=16-4\\x_1^2.x_2^2=1^2 \\\to z^2-14z+1=0 \\z=x^2 \\\to x^4+\frac{1}{x^4}=14$$
| {
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"answer_id": 5
} |
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
} |
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 kernel of asked matrix, and tried matrix $\begin {pmatrix}1&-1&1\\1&-1&1\end{pmatrix}$, but it also includes $\begin{pmatrix}1\\1\\0 \end{pmatrix}$ in its kernel!
| 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 take any matrix $A$ as above, for instance
$$A=\begin{pmatrix}2&-1&0\\0&1&-2\end{pmatrix}$$
since $\mathbf{x}=\begin{pmatrix}1\\1\\0\end{pmatrix}$ is a particular solution we have $\mathbf{b}=\begin{pmatrix}2&-1&0\\0&1&-2\end{pmatrix}\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}$. Hence the system
$$\begin{pmatrix}2&-1&0\\0&1&-2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}$$
satisfies the problem.
| {
"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
} |
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: $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational $\implies$ $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, so I could show it is a contradiction.
| 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 irrational, so that can only be true if $a^3=2b^3$, which, under the assumption that $b\neq0$, implies that $\frac ab=\sqrt[3]2$ is rational. This is clearly a contradiction, and we are done.
| {
"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
} |
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 written as
\begin{equation*}
\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ \ with\ \ \ C=\sqrt{%
A^{2}-B^{2}}.
\end{equation*}
Consider a rectangle triangle with sides $A$, $B$, and $C$ such that
(pythagor theorem)
\begin{equation*}
A^{2}=B^{2}+C^{2}\ \ \ \ \ \ (\ast )
\end{equation*}
then
\begin{eqnarray*}
B^{2} &=&A^{2}-C^{2} \\
B^{2} &=&(A+C)(A-C) \\
B^{2} &=&4\frac{(A+C)}{2}\frac{(A-C)}{2}
\end{eqnarray*}
then
\begin{equation*}
B=2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}
\end{equation*}
and by adding $A$ to both sides
\begin{equation*}
A+B=A+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}
\end{equation*}
However, note that
\begin{equation*}
A=\frac{A+C}{2}+\frac{A-C}{2}
\end{equation*}
then
\begin{eqnarray*}
A+B &=&A+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\
&=&\frac{A+C}{2}+\frac{A-C}{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\
&=&\left( \sqrt{\frac{A+C}{2}}\right) ^{2}+\left( \sqrt{\frac{A-C}{2}}%
\right) ^{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\
A+B &=&\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right) ^{2}
\end{eqnarray*}
and then
\begin{equation*}
\sqrt{A+B}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
\end{equation*}
where $C$ is given by (*), that is, $C=\sqrt{A^{2}-B^{2}}.$
| {
"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
} |
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
$$
\left(
\begin{array}{crr}
1 & -1 & 1 \\
1 & 0 & -1 \\
0 & -1 & 2 \end{array}
\right)$$
The solution mentions something about the eigenvectors satisfying
$$
\left( \mathbf{A} - \lambda_{k} \mathbf{I} \right) v_{k} = \mathbf{0}
$$
but I'm not sure where this came from or why it's true. Could someone please explain?
| 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
$$
\mathbf{A} - \lambda \mathbf{I}_{3} =
\left[
\begin{array}{ccc}
\boxed{1-\lambda} & \boxed{-1} & \boxed{1} \\
1 & -\lambda & -1 \\
0 & -1 & 2-\lambda \\
\end{array}
\right]
$$
Define characteristic equation
$$
p(\lambda) = \det \left( \mathbf{A} - \lambda \mathbf{I}_{3} \right)
$$
Construct determinant from minors
$$
%
\begin{align}
%
\det \left( \mathbf{A} - \lambda \mathbf{I}_{3} \right)
%
&= \boxed{\left( 1 - \lambda \right)}
\left|
\begin{array}{cc}
-\lambda & -1 \\
-1 & 2-\lambda \\
\end{array}
\right|
%
- \boxed{-1}
\left|
\begin{array}{cc}
1 & -1 \\
0 & 2-\lambda \\
\end{array}
\right|
%
+ \boxed{1}
\left|
\begin{array}{cc}
1 & -\lambda \\
0 & -1 \\
\end{array}
\right| \\[3pt]
%
&= -\lambda \left(\lambda ^2-3 \lambda +2\right) \\
%
&= -\lambda \left(\lambda - 2 \right)\left(\lambda - 1 \right) \\
%
\end{align}
%
$$
The roots $p\left( \lambda \right) = 0$ are the eigenvalues.
The eigenvalue spectrum is
$$
\lambda \left( \mathbf{A} \right) = \left\{ 2, 1, 0\right\}
$$
Eigenvectors
Solve $$ \left( \mathbf{A} -\lambda_{k} \mathbf{I}_{3} \right) u = 0$$
$ \lambda_{1} = 2$:
$$
%
\begin{align}
%
\left[
\begin{array}{c}
-u_{1} - u_{2} + u_{3} \\
u_{1} - 2 u_{2} - u_{3} \\
-u_{2} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
0 \\
0 \\
0 \\
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
v_{1} =
\left[
\begin{array}{c}
1 \\
0 \\
1 \\
\end{array}
\right]
%
\end{align}
%
$$
$ \lambda_{2} = 1$:
$$
%
\left[
\begin{array}{c}
- u_{2} + u_{3} \\
u_{1} - u_{2} - u_{3} \\
-u_{2} + u_{3} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
0 \\
0 \\
0 \\
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
v_{2} =
\left[
\begin{array}{c}
2 \\
1 \\
1 \\
\end{array}
\right]
%
$$
$ \lambda_{3} = 0$:
$$
%
\left[
\begin{array}{c}
u_{1} - u_{2} + u_{3} \\
u_{1} - u_{3} \\
-u_{2} + 2u_{3} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
0 \\
0 \\
0 \\
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
v_{3} =
\left[
\begin{array}{c}
1 \\
2 \\
1 \\
\end{array}
\right]
%
$$
| {
"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
} |
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 $\left|\frac{2x^2y^3}{x^4+y^6}\right| < \epsilon$.
Using polar coordinates $r < \delta$ implies
$$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| < \epsilon^4$$
$$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| \leq \left|\frac{2r}{\cos^4(\theta)+r^2\sin^6(\theta)}\right|$$
But from here I cant simplify the equation further.
| **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
} |
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 $\in \mathbb{N}$ )
$$f(n)=\frac{1}{2^n}+\frac{1}{2^{n+2}}\binom{n+1}{2}+ \frac{1}{2^{n+4}}\binom{n+3}{4}+\frac{1}{2^{n+6}}\binom{n+5}{6}+ \cdots$$
| $(1+x)^{-n}=1+(-n)x+\dfrac{(-n)(-n-1)}{2!}x^2+\dfrac{(-n)(-n-1)(-n-2)}{3!}x^3+...$
$(1-x)^{-n}=1-(-n)x+\dfrac{(-n)(-n-1)}{2!}x^2-\dfrac{(-n)(-n-1)(-n-2)}{3!}x^3+...$
By adding
$\dfrac{1}{2}((1+x)^{-n}+(1-x)^{-n})=1+\dfrac{(-n)(-n-1)}{2!}x^2+\dfrac{(-n)(-n-1)(-n-2)(-n-3)}{4!}x^4+...$
$1+\binom{n+1}{2}x^2+\binom{n+3}{4}x^4+...=\dfrac{1}{2}((1+x)^{-n}+(1-x)^{-n})$
substituting $x=\dfrac{1}{2}$,
$1+\binom{n+1}{2}\dfrac{1}{2^2}+\binom{n+3}{4}\dfrac{1}{2^4}+...=\dfrac{1}{2}((\dfrac{3}{2})^{-n}+(\dfrac{1}{2})^{-n})$
multiplying through $\dfrac{1}{2^n}$,
$f(n)=\dfrac{1}{2}({3^{-n}+1})$
| {
"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
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.