Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $\cos 25^\circ + \sin 25^\circ = k,$ then what is $\cos 20^\circ$? Question:
If $$\cos 25^\circ + \sin 25^\circ = k,$$ then what is $\cos 20^\circ$?
What I did:
I tried to square both sides, and obtained that $\sin 50 = k^2 -1$, however, this didn't get me anywhere. Then I tried splitting 25 into 20 + 5 but that didn't get me anywhere either. Can someone just point me in the right direction?
| Then we have
$$k^2=(\cos 25^\circ + \sin 25^\circ)^2=1+2\cos 25^\circ \sin 25^\circ=1+\sin 50^\circ.$$
Note that $\cos 40^\circ=\cos(90^\circ-50^\circ)=\sin 50^\circ$. Hence, we have
$$\tag{1}\cos 40^\circ=k^2-1.$$
Now
$$\tag{2}\cos 40^\circ=2\cos^2 20^\circ-1.$$
Now combining $(1)$ and $(2)$ gives you the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/857986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Express $3.72444\ldots$ as a fraction using the formula for geometric progressions I did the following:
$$3.72\overline{4} = 3.724+\left(\left(\frac{4}{10^4}\right)+\left(\frac{4}{10^5}\right)+\left(\frac{4}{10^6}\right)+\cdots\right)$$
where $a=3.724$ and $r=\dfrac{1}{10}$
Using $S=\dfrac{a}{1-r}$ I get $\dfrac{37.24}{9}\neq3.72\overline{4}$
The answer given by the text is $\dfrac{838}{225}$
I've been staring at this for an hour and can't figure out where I'm going wrong.
| Only part of this is geometric:
$$
3.72\bar{4}= 3.72 +0.00\bar 4=\frac{372}{100}+\frac{4}{1000}\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\cdots\right)
$$
$$
=\frac{372}{100}+\frac{4}{1000}\cdot\frac{1}{1-\frac{1}{10}} = \frac{372}{100}+\frac{4}{1000}\cdot\frac{10}{9} =\frac{372}{100}+\frac{4}{900}
$$
$$
=\frac{4\cdot93}{4\cdot25}+\frac{4\cdot1}{4\cdot225}=\frac{9\cdot93}{9\cdot25} +\frac{1}{225}=\frac{837}{225}+\frac{1}{225}=\boxed{\dfrac{838}{225}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving a trigonometric limit $\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$ First off, please excuse my n00bishness I have only just begun learning about algebraic manipulation of limits so this is probably a really dumb or obvious question.
I'm trying to solve the following limit:
$$
\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}
$$
This limit is $0/0$ if evaluated directly, so I tried multiplying by the conjugate of the denominator:
$$
\begin{align}
& = \lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)}\\
& = \lim_{x\to\pi/6}\frac{(2\sin{x} - 1)(2\sqrt{3}\cos{x} - 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{3} + 3)} \\
& = \lim_{x\to\pi/6}\frac{2\sin{x} - 1}{2\sqrt{3}\cos{x} + 3}\\
& = \frac{2(1/2) - 1}{2\sqrt{3}\frac{\sqrt{3}}{2} + 3}\\
& = \frac{0}{6}\\
& = 0
\end{align}
$$
But according to WolframAlpha this is incorrect, and the limit should be 1. What have I done wrong?
Also, as I have only just begun I am unfamiliar with L'Hopital's rule.
| $$
\begin{align}
\lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)}
\end{align}
$$
You need to multiply out the denominator (and possibly the numerator):
$$
\begin{align}
\lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{12\cos^2{x} - 9}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
$12\frac{\sin 45^\circ}{\sin 60^\circ}$ Need help breaking this down. Otherwise known as $12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}$
How do you simplify this multi level fractional radical expression into $4\sqrt{6}$.
| $$
12\cdot \frac{1}{\sqrt{2}}\cdot\frac{2}{\sqrt{3}} = 12\cdot \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}\sqrt{2}}{\sqrt{3}} = 12\cdot\frac{\sqrt{2}}{\sqrt{3}} = 12\cdot\frac{\sqrt{2}\sqrt{3}}{\sqrt{3}\sqrt{3}}= 12\cdot\frac{\sqrt{6}}{3} =\cdots\cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Proving Cauchy inequality involving four expression Show that $$(a^2 + b^2 + c^2) (a^2b^2 +b^2c^2 +c^2a^2) \geq (a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2)$$
i should prove this inequality by making it a Cauchy form inequality(as teacher stated).
my problem : this inequality is involving 4 parentheses but in other question i proved previously there was one or two in left side and one in right side.
things i have done so far:
$(a^2 + b^2 + c^2) (a^2b^2 +b^2c^2 +c^2a^2) \geq (a^2b + b^2c + c^2a)^2$
| It's $$\sum_{cyc}(a^4b^2+a^4c^2-a^3b^3-a^4bc)\geq0,$$
which is just Muirhead:
$$\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3)+\sum_{cyc}(a^4b^2+a^4c^2-2a^4bc)\geq0,$$
which is true because $(4,2,0)\succ(3,3,0)$ and $(4,2,0)\succ(4,1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove trigonometry identity for $\sec x -\sin x$ I'm trying to prove this equality but I' stuck at the second step.
Please give me some hints or other ways to proceed.
\begin{gather}\frac{\tan^2x + \cos^2x}{\sin x+ \sec x} \equiv \sec x - \sin x \\
\sin x = 0 \\
\cos x = y \\
\frac{\frac{x^2}{y^2}+ \frac{y^4}{y^2}}{\frac{xy}{y} + \frac{1}{y}} \equiv \frac{1}{y} - x = \frac{1-xy}{y} \tag{1} \\
\frac{ \frac{x^2+y^4}{y^2} }{ \frac{xy+1}{y} }\equiv \tag{2} \\
\frac{x^2+y^4}{y(xy+1)} \equiv\tag{3}
\end{gather}
| $$\cos^2(x)-1=-\sin^2(x)$$
$$\tan^2(x)=\sec^2(x)-1$$
$$\frac{\tan^2(x)+\cos^2(x)}{\sin(x)+\sec(x)}=\frac{\left(\sec^2(x)-1 \right)+\cos^2(x)}{\sin(x)+\sec(x)}$$
$$=\frac{\sec^2(x)-\sin^2(x)}{\sin(x)+\sec(x)}=\frac{(\sin(x)+\sec(x))(\sin(x)-\sec(x))}{\sin(x)+\sec(x)}$$
I think you can do the rest from here. Though try not to substitute a function for the same variable (e.g. use $t=\cos(x)$ not $x=\cos(x)$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/867145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Global Max and Min Problem I'm working on a problem which asks me to find local and global extrema of the following function.
$$f(x,y) = x^2y^2e^{(-x^2 - 2y^2)}$$
I went through and found all of the relevant partial derivatives.
\begin{align*}
f_x &= (2xy^2)(e^{(-x^2 - 2y^2)}) + (x^2y^2)(e^{(-x^2 - 2y^2)})(-2x)\\
f_x &= (e^{(-x^2-2y^2)})(2xy^2 -2x^3y^2)\\
\\
f_y & = (2x^2y)(e^{(-x^2-2y^2)}) + (x^2y^2)(e^{(-x^2-2y^2)})(-4y)\\
f_y &= (e^{(-x^2-2y^2)})(2x^2y-4x^2y^3)\\
\\
f_{xx} &= (e^{(-x^2-2y^2)})(-2x)(2xy^2 -2x^3y^2) + (e^{(-x^2-2y^2)})(2y^2 -6x^2y^2)\\
f_{xx} &= (e^{(-x^2-2y^2)})(-10x^2y^2 + 4x^4y^2 + 2y^2)\\
\\
f_{yy} &= (e^{(-x^2-2y^2)})(-4y)(2x^2y-4x^2y^3) + (e^{(-x^2-2y^2)})(2x^2 - 12x^2y^2)\\
f_{yy} &= (e^{(-x^2-2y^2)})(-20x^2y^2 + 16x^2y^4 + 2x^2)\\
\\
f_{xy} &= (e^{(-x^2-2y^2)})(-4y)(2xy^2 -2x^3y^2) + (e^{(-x^2-2y^2)})(4xy-4x^3y)\\
f_{xy} &= (e^{(-x^2-2y^2)})(-8xy^3 + 8x^3y^3 +4xy - 4x^3y)\\
\end{align*}
However, I'm not sure what to do after this. I thought I was supposed to set $f_x$ and $f_y$ equal to 0 but I don't know how to solve the equations that I get. Can someone please help me? Did I make a mistake while I was determining my partial derivatives?
EDIT: I made a mistake calculating the partial derivatives and I edited that
| Another route is to change coordinates. If the exponent were $x^2+y^2$ then polar coordinates would be obvious; in lieu of this, we use the parameterization $$(x,y)=\left(r\cos\theta,\frac{1}{\sqrt{2}} r\sin\theta\right)$$ and so obtain $$f(r,\theta) = \frac{1}{2}r^4 e^{-r^2} \cos^2\theta\sin^2\theta=\frac{1}{8}r^4 e^{-r^2}\sin^22\theta.$$ This separates $f(r,\theta)$ into angular and radial parts which are easy to maximize/minimize.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/868349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Prove this trig identity $$\sin^2(x) - \cos^2(x) - \tan^2(x) = \frac{2\sin^2(x) - 2\sin^4(x) - 1 }{ 1-\sin^2(x)}.$$
I tried this but I can't figure out how they got $-2\sin^4(x)$.
| To get you started:
$$\begin{align}\sin^2 x - \cos^2 x - \tan^2 x & = \sin^2 x- \cos^2 x - \frac{ \sin^2 x }{\cos^2 x} \\ & = \frac{\sin^2 x\cos^2 - \cos^4 x - \sin^2 x}{\cos^2 x} \\ & = \frac{\sin^2 x (1-\sin^2 x)+(1-\sin^2 x)^2 - \sin^2 x}{1-\sin^2 x} & \\ & = \ldots \end{align}$$
Just expand and simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/870527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Inequality question $$a,b,c,d\ge 0$$
$$a\le 1$$
$$a+b\le 5$$
$$a+b+c\le 14$$
$$a+b+c+d\le 30$$
Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$.
We can subtract inequalities to get the answer, but that is wrong... I can't think of any another method... Any hints or suggestions will be helpful.
| Using AM-QM and the assumptions we get
\begin{align*}
\sqrt a + \sqrt b + \sqrt c + \sqrt d &
= \sqrt a + 2\sqrt{\frac b4} + 3\sqrt{\frac c9} + 4 \sqrt{\frac d{16}} \\
& \le 10 \sqrt{\frac{a+2\cdot \frac b4 + 3 \cdot \frac c9 + 4 \cdot \frac d{16}}{10}} \\
& = \sqrt{10a+5b+\frac{10}3 c + \frac 52 d} \\
& = \sqrt{5a+\frac 53 (a+b) + \frac 56 (a+b+c) + \frac 52(a+b+c+d)} \\
& \le \sqrt{5 \cdot 1 + \frac 53 \cdot 5 + \frac 56 \cdot 14 + \frac 52 \cdot 30}\\
& = 10.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Write the given expression as an algebraic expression in $x$: $\tan(2 \cos^{−1} x)$. OK I first assigned $\theta = \cos^{−1} x$
and so $\cos(\theta) = x$
drew a triangle where adj = $x$ hyp = $1$ and opp = $\sqrt{1-x^2}$ by pyth theorem.
Then $\tan(2\cos^{−1} x) = \tan(2\theta)$
and I used the double angle formula for tangent so
$\tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$
Then used the triangle: $\tan(\theta) = \frac{\sqrt{1-x^2}}{x}$
but my answer is
The computer states I am wrong. Am I setting the problem up wrong?
| You set the problem up fine:
Let me retrace your work for you.
So you are correct with all your steps.
$$tan(\theta) = \frac{\sqrt{1-x^2}}{x} $$
Now,
$$\tan{2\theta} = \frac{2*\frac{\sqrt{1-x^2}}{x}}{1 - (\frac{\sqrt{1-x^2}}{x})^2}$$
$$ = \frac{\frac{2\sqrt{1-x^2}}{x}}{1 - \frac{1-x^2}{x^2}}$$
$$= \frac{\frac{2\sqrt{1-x^2}}{x}}{\frac{2x^2 - 1}{x^2}}$$
$$= \frac{2\sqrt{1-x^2}}{x}*\frac{x^2}{2x^2 -1}$$
$$= \frac{2x\sqrt{1-x^2}}{2x^2 -1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of Sequence $a_n=1+\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3n-2}$ Apply Cauchy's principle of convergence to prove that the sequence $\langle a_n\rangle$ defined by $$a_n=1+\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3n-2}$$ is not convergent
My attempt :
consider,
\begin{align*}
|a_{2m}-a_{m}|&=|1+1/4+1/7+\ldots+\frac{1}{3m-2}+\frac{1}{3m+1}+\ldots+\frac{1}{6m-2}\\
&-(1+1/4+1/7+\ldots+\frac{1}{3m-2})| \\
&=|\frac{1}{3m+1}+\ldots+\frac{1}{6m-2}|\end{align*}
What do i do next ? i get confused after this step
| Note that
$$a_n=\sum_{k=1}^n\frac{1}{3k-2}>\sum_{k=1}^n\frac{1}{3k}=\frac{1}{3}\sum_{k=1}^n\frac{1}{k}.$$
Now, since $\displaystyle\sum_{k=1}^n\frac{1}{k}\to\infty$, when $n\to\infty$, we have
$$a_n\to\infty,\ \mbox{when}\ n\to\infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/873733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum\limits_{cyc}\frac{a}{b(3+a-b)}\ge 1$ Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 3$. Prove that
$$\sum_{cyc}\frac{a}{b(3+a-b)}\ge1$$
I tried applying the Cauchy-Schwarz inequality by doing:
$$\sum_{cyc}\frac{a}{b(3+a-b)}=\sum_{cyc}\frac{a^2}{ab(3+a-b)}\ge \frac{(a+b+c)^2}{\sum_{cyc}ab(3+a-b)}=\frac{9}{\sum_{cyc}ab(3+a-b)}.$$
We need therefore to prove that
$$\sum_{cyc}ab(3+a-b)\le 9.$$
Thanks for any help.
| Your way works!
By C-S
$$\sum_{cyc}\frac{a}{b(3+a-b)}=\sum_{cyc}\frac{a}{b(2a+c)}=\sum_{cyc}\frac{a^2}{ab(2a+c)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}ab(2a+c)}.$$
Thus, it remains to prove that
$$(a+b+c)^3\geq3\sum\limits_{cyc}ab(2a+c)$$ or
$$\sum_{cyc}(a^3-3a^2b+3a^2c-abc)\geq0,$$
which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/874455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find a function $\phi(x)$ such that $\sqrt{1+y^2} - \sqrt{1+x^2} \geq \phi(x) (y-x)$ for each $x,y\in \mathbb{R}$ How to find a function $\phi(x)$ such that $\sqrt{1+y^2} - \sqrt{1+x^2} \geq \phi(x) (y-x)$ for each $x,y\in \mathbb{R}$.
Here are some of my ideas:
Also by applying Mean Value theorem, we know that for every $x,y\in \mathbb{R}$ there exists a constant $c$ between $x,y$ such that
$$\sqrt{1+y^2} - \sqrt{1+x^2} = \frac{c}{\sqrt{1+c^2}}(y-x).$$
I know that $\phi(0) = 0$, by taking $x=0$; $y= \phi(0)$ and $y=-\phi(0)$. Also that $|\phi|$ is bounded above by $1$.
Could anyone give me a hint on how to continue? Thanks!
| Using: $$\sqrt{A}-\sqrt{B} = \frac{A-B}{\sqrt{A}+\sqrt{B}}$$
So $$\sqrt{1+y^2}-\sqrt{1+x^2} = \frac{y^2-x^2}{\sqrt{1+y^2}+\sqrt{1+x^2}} = (y-x)\frac{y+x}{\sqrt{1+y^2}+\sqrt{1+x^2}}$$
So you need a $\phi(x)$ so that for $y>x$, $$\phi(x)\leq \frac{y+x}{\sqrt{1+y^2}+\sqrt{1+x^2}}$$ and for $y<x$
$$\phi(x)\geq\frac{y+x}{\sqrt{1+y^2}+\sqrt{1+x^2}}$$
So $$\sup_{y<x} \frac{y+x}{\sqrt{1+x^2}+\sqrt{1+y^2}}\leq \phi(x)\leq \inf_{y>x} \frac{y+x}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$
Looks like $\phi(x)=\frac{x}{\sqrt{x^2+1}}$ is the only option, since when $y$ gets close to $x$ from either direction, we see that $\phi(x)$ must be both $\leq$ and $\geq$ this function.
That doesn't prove this function works.
It's not hard to prove that $g_x(y)=\frac{x+y}{\sqrt{1+x^2}+\sqrt{1+y^2}}$ is increasing, which would finish the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/874800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Let $a_n$ be the $nth$ term of the sequence $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \dots$ Question:Let $a_n$ be the $nth$ term of the sequence $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6\dots$, constructed by including the integer $k$ exactly $k$ times. Show that $a_n = \lfloor \sqrt{2n} + \dfrac{1}{2}\rfloor$.
Attempt:
Try to prove $k = a_n=\lfloor \sqrt{2n} + \dfrac{1}{2}\rfloor$ then $k = a_{n+k-1}=\lfloor \sqrt{2(n+k-1)} + \dfrac{1}{2}\rfloor$.
The inequality of $k = a_n=\lfloor \sqrt{2n} + \dfrac{1}{2}\rfloor$, is the following,
$$k \leq \sqrt{2n} + \dfrac{1}{2} < k + 1$$
$$k-\dfrac{1}{2} \leq \sqrt{2n} < k + \dfrac{1}{2}$$
$$\Big( k-\dfrac{1}{2} \Big)^2 \leq 2n < \Big(k +\dfrac{1}{2}\Big)^2$$
$$\dfrac{1}{2}\Big( k-\dfrac{1}{2} \Big)^2 \leq n < \dfrac{1}{2}\Big(k +\dfrac{1}{2}\Big)^2$$
$$\dfrac{1}{2}\Big( k-\dfrac{1}{2} \Big)^2 + k - 1 \leq n + k -1 < \dfrac{1}{2}\Big(k +\dfrac{1}{2}\Big)^2 + k - 1$$
$$\Big( k-\dfrac{1}{2} \Big)^2 + 2k - 2 \leq 2(n + k -1) < \Big(k +\dfrac{1}{2}\Big)^2 + 2k - 2$$
$$k^2 - k + \dfrac{1}{4} + 2k - 2 \leq 2(n+k - 1) < k^2 + k + \dfrac{1}{4} + 2k - 2$$
$$k^2 + k + \dfrac{1}{4} - 2 \leq 2(n+k - 1) < k^2 + 3k + \dfrac{9}{4} - \dfrac{9}{4} + \dfrac{1}{4} - 2$$
$$\Big( k+\dfrac{1}{2} \Big)^2 - 2 \leq 2(n+k - 1) < \Big( k+\dfrac{3}{2} \Big)^2 - 4$$
From here $2(n+k - 1) < \Big( k+\dfrac{3}{2} \Big)^2 - 4 \implies 2(n+k - 1) < \Big( k+\dfrac{3}{2} \Big)^2$ so I can take the square root of both sides. The problem is the left hand side, $\Big( k+\dfrac{1}{2} \Big)^2 - 2 \leq 2(n+k - 1)$ does not imply $\Big( k+\dfrac{1}{2} \Big)^2 \nleq 2(n+k - 1)$, thus we can't really get rid of the square.
It is easy to see that if we have $k = a_{n+k}=\lfloor \sqrt{2(n+k)} + \dfrac{1}{2}\rfloor$, the problems we encountered are avoided, but $k \neq a_{n+k}$.
| I am trying this question in a different way.
As is evident from series interger 1 occurs first time at 1 st position
integer 2 occurs first time after first term (1 + 0)
integer 3 occurs for the first time after (0 +1 + 2 ) terms
integer 4 occurs after (0 + 1 + 2 + 3 ) terms
so integer k will be present after (k(k-1)/2) terms, which means at [k(k-1)/2 + 1]th position
so if [k(k + 1)/2 + 1]th term is k
then for nth term n=k(k+1)/2 + 1.
Just express k in terms of n.
I did not try it so i don't know if this is correct
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
number of solution to the given equation. a,b,c, are all non-negative integers such that
a + b + c=100
and
1000a + 300b + 50c = 10000
How many such triplets are possible?
i have tried to reduce the equations to a relation between a b and c.
18a + 4b = c
but this would satisfy many non negative integers. I think i am wrong somewhere.
| You begain with these two equations:
$$
\begin{align}
a + b + c &= 100\\
1000 a + 300 b + 50 c &= 10\ 000
\end{align}
$$
First divide your second equation by 50, so that now we have:
$$
\begin{align}
a + b + c &= 100\\
20 a + 6 b + c &= 200
\end{align}
$$
Subtracting these gives:
$$
19 a + 5b = 100
$$
Since b is non-negative, we have that
$$
\begin{align}
19 a &\leq 100\\
a &\leq 100/19 \text{ which is more than 5}\\
a &\leq 5 \text{ since it is an integer}
\end{align}
$$
But from the above equation:
$$
19 a = 100 - 5b = 5(20-b)
$$
So a must be a multiple of 5.
Hence $a=0$ or $a=5$, since it is less than or equal to 5.
If $a=0$, we get $5b = 100$, so $b = 20$. Subbing this into your first equation gives so $c = 80$.
If $a=5$ we get $95 + 5b = 100$, so $b=1$. Subbing this into your first equation gives $c = 94$.
So there are two solutions: $a = 0$, $b = 20$, $c = 80$ and $a = 5$, $b = 1$, $c = 94$.
| {
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"url": "https://math.stackexchange.com/questions/878478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use the binomial theorem to give a formula for positive integers $x_{k}$ and $y_{k}$ such that $(3 + 2\sqrt{2})^{^{x}} = x_{k} + y_{k}\sqrt{2}$. Use the binomial theorem to give a formula for positive integers $x_{k}$ and $y_{k}$ such that
$$(3 + 2\sqrt{2})^{^{x}} = x_{k} + y_{k}\sqrt{2}.$$
Is this simply just applying the binomial theorem?
I get something like
$$(3 + 2\sqrt{2})^{^{k}} = (1 + (2 + 2\sqrt{2}))^{^{k}} $$
$$= \sum_{j=0}^{k} \binom{k}{j} 2^{^{j}} (1 + \sqrt{2})^{^{j}}$$
but I am struggling to proceed from here. I don't know how to arrive to the last part of the equality, $x_{k} + y_{k}\sqrt{2}$. Any help would be greatly appreciated.
Edit: I guess I'm rather confused about what the question is asking me to do exactly. It's saying that I should find one formula for $x_{k}$ and one formula for $y_{k}$ individually, right?
| $$(3+2\sqrt{2})^k= \sum_{j=0}^{k} \binom{k}{j} 3^{^{k-j}} (2\sqrt{2})^{^{j}}$$
Now split the sum in terms with $j$ odd and $j$ even. The part where all $j$ are even is an integer, while the art where $j$ is odd has the form integer times $\sqrt{2}$.
Added:
$$ \sum_{j=0}^{k} \binom{k}{j} 3^{^{k-j}} (2\sqrt{2})^{^{j}}=\\
=\sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} (2\sqrt{2})^{^{2j}}+\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} (2\sqrt{2})^{^{2j+1}}\\
=\sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} 8^{^{j}}+\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} 8^j (2\sqrt{2})\\
=\sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} 8^{^{j}}+2\sqrt{2}\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} 8^j \\$$
Thus
$$x_k= \sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} 8^{^{j}}\\
y_k=2\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} 8^j $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that there is an angle $\theta$ such that $a=\cos\theta$ and $b=\sin\theta$ My problem is from Israel Gelfand's Trigonometry textbook.
Page 50. Exercise 3: Suppose that $\alpha$ is some angle. If $a=4\cos^3\alpha-3\cos\alpha$ and $b=3\sin\alpha-4\sin^3\alpha$, show that there is an angle $\theta$ such that $a=\cos\theta$ and $b=\sin\theta$
The attempt at a solution:
In order to show that, I understand that I have to show that $a^2+b^2=1$, now i have expanded $(4\cos^3\alpha-3\cos\alpha)^2+(3\sin\alpha-4\sin^3\alpha)^2$, but all I get is messy expression. I would appreciate some hints, thank you in advance.
| By De Moivre's formula, we have that
$$\begin{align}
\cos3x+i\sin3x&=(\cos x+i\sin x)^3
\\&=\cos^3x+3i\cos^2x\sin x-3\sin^2x\cos x-i\sin^3 x
\\&=\cos^3x-3(1-\cos^2x)\cos x+i[-\sin^3x+3(1-\sin^2x)\sin x]
\\&=4\cos^3x-3\cos x+i[3\sin x-4\sin^3 x]
\end{align}$$
which implies
$$\cos3x=4\cos^3x-3\cos x\,\,\,\,\,\,\,\,\,\,\,\,\text{ and }\,\,\,\,\,\,\,\,\,\,\,\,\sin 3x=3\sin x-4\sin^3x$$
This just about solves it (let $\theta=3\alpha$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int x^2 \sqrt{x^2-1} dx$ How do I evaluate the following indefinite integral?
$$\int x^2 \sqrt{x^2-1} dx$$
Through integration of parts, I have obtained
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int (x^2-1)^{3/2} dx $$
I've attempted evaluating the second term through substitution, where
$$ x = \sec(u)$$
However, I am stuck with
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int \tan^4(u) \sec (u) du $$
What would be my next step?
| Used a different version of the secant reduction formula. I really enjoyed this problem!
Let $x=\sec(u)$ and $dx=\sec(u)\tan(u)\,dx$ so that:
$$\int x^2\sqrt{x^2-1}\,dx=\int\sec^2(u)\sqrt{\sec^2(u)-1}\sec(u)\tan(u)\,du$$
Now simplify, noting that $\sec^2(u)-1=\tan^2(u)$:
$$=\int\sec^3(u)\tan^2(u)\,du$$
And again, using $\sec^2(u)-1=\tan^2(u)$ and simplifying:
$$\int(\sec^5(u)-\sec^3(u))\,du$$
Apply the version of the secant reduction formula containing tangent, which is:
$$\int\sec^n(u)\,du=\frac{\tan(u)\sec^{n-2}(u)}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}(u)\,du$$
To get:
$$\int(\sec^5(u)-\sec^3(u))\,du=\frac{\tan(u)\sec^3(u)}{4}+\frac{3}{4}\bigg(\frac{\tan(u)\sec(u)}{2}+\frac{1}{2}\ln|\sec(u)+\tan(u)|\bigg)$$
$$-\bigg(\frac{\tan(u)\sec(u)}{2}+\frac{1}{2}\ln|\sec(u)+\tan(u)|\bigg)$$
Next apply the substitutions $x=\sec^{-1}(u)$ and $\tan(\sec^{-1}(u)=x\sqrt{1-\frac{1}{x^2}}$ and simplify:
$$\int x^2\sqrt{x^2-1}\,dx=\frac{1}{4}x^4\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}x^2\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}\ln\bigg|x+\sqrt{1-\frac{1}{x^2}}\bigg|+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x \to 2} \frac{e^x - e^2 - (x^2 - 4x + 4)\sin (x - 2)}{\sin [(x - 2)^2] + \sin (x-2)}$ I make a mistake somewhere but I cannot find where. The answer is supposed to be $e^2$. I think it can be solved with l'Hopital rule, but that is tedious and error-prone. I was looking for a faster way.
$$\begin{align}
\lim_{x \to 2} \frac{e^x - e^2 - (x^2 - 4x + 4)\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} &= \lim_{x \to 2} \frac{e^x - e^2 - (x - 2)^2\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} =\\
&=\left [ t = x - 2, x \to 2 \implies t \to 0 \right ] =\\
&=\lim_{t \to 0} \frac{e^{t + 2} - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\
&=\lim_{t \to 0} \frac{e^2 - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\
&=\lim_{t \to 0} \frac{- t^2(t + o(t))}{t^2 + o(t^2) + t + o(t)} = \tag{1} \\
&=\lim_{t \to 0} \frac{-t^3}{t} = 0
\end{align}
$$
I think I may be making a mistake in $(1)$ by either simplifying away the two $e$'s or rewriting the expression with the little-o notation.
| Or, expand in powers of t
$$
\lim_{t->0} \frac{e^2(e^t-1)-t^2 \sin t}{\sin t^2 + \sin t} \\
=
\lim_{t->0} \frac{e^2(t + t^2/2 + \ldots) - (t^3 - t^4/6 + \ldots)}{t^2 - t^6/6 + \ldots + t-t^3/6+\ldots} \\
= e^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality with $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}$ Inspired by this recent question, I suggest this.
Let $n=2,3,4, \ldots .$ Then
$$
\frac{7}{12} < \cfrac 1 {1 + \cfrac {1^2} {1 + \cfrac {2^2} {\ddots + \cfrac \vdots { 1 + \, {n^2} \,}}}} \leq \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \tag1
$$
Could you prove $(1)$?
| Here is a brute force method to prove the first inequality. We will see the calculations for $n=2,3,4,5$ and that will be sufficient to prove for general $n$. Here goes:
$n=2$
$$\frac{7}{12} \leq \cfrac{1}{1+\cfrac{1^2}{1 + 2^2}} \iff \frac{12}{7} \geq 1+\cfrac{1^2}{1 + 2^2} \iff \frac{5}{7} \geq \cfrac{1^2}{1 + 2^2}$$
which is true. For $n=3$, we'll take the calculations from here to get
$$ \frac{5}{7} \geq \cfrac{1^2}{1 + \cfrac{2^2}{1+3^2}} \iff \frac{7}{5} \leq 1 + \cfrac{2^2}{1+3^2} \iff \frac{2}{5} \leq \cfrac{2^2}{1+3^2} $$ which is again true. Carrying on for $n=4$
$$ \frac{2}{5} \leq \cfrac{2^2}{1+ \cfrac{3^2}{1+4^2}} \iff 10 \geq 1+ \cfrac{3^2}{1+4^2} \iff 1 \geq \cfrac{1}{1+4^2} $$ which is true. Carrying on for $n=5$, we have
$$ 1 \geq \cfrac{1}{1+ \cfrac{4^2}{1+5^2}} \iff 1 \leq 1+ \frac{4^2}{1+5^2} \iff 0 \leq \frac{4^2}{1+5^2} $$ which is true.
For $n\geq 6$, the equivalent condition will just be
$$ 0 \leq \cfrac{4^2}{1+ \cfrac{5^2}{1+ \cfrac{6^2}{1+ ...}}} $$
which is of course true. These calculations are not revealing, but I sometimes find them cute.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A Trigonometric Question If
$$\frac{\cos\left(x\right)}{\cos\left(\theta\right)}+\frac{\sin\left(x\right)}{\sin\left(\theta\right)}\:=1=\:\frac{\cos\left(y\right)}{\cos\left(\theta\right)}+\frac{\sin\left(y\right)}{\sin\left(\theta\right)}\:\:$$
Then what's the value of :
$$\frac{\cos\left(x\right)\:\cos\left(y\right)}{\cos^2\left(\theta\right)}+\frac{\sin\left(x\right)\:\sin\left(y\right)}{\sin^2\left(\theta\right)}$$
Options:
(a)2 (b)0 (c)1 (d)-1
**Solution given in my guide:**
From Hypothesis:
$sin\left(x+θ\right)=sin\left(y+θ\right)$
Therefore
$2cos\left(\frac{x+y+2θ}{2}\right)sin\left(\frac{x-y}{2}\right)=0$
Now
$x-y\ne 2n\pi \:\:⇒\:sin\left(\frac{x-y}{2}\right)\ne \:0$ //How does this step work?
Hence
$\frac{x+y+2θ}{2}=\left(2n+1\right)\left(\frac{\pi }{2}\right)$
$x+y=\left(2n+1\right)\pi -2θ$
And proceeds to the final answer as -1.
| $$\frac{\cos x}{\cos\theta}+\frac{\sin x}{\sin\theta}=1\iff\sin\theta\cos\theta=\cos x\sin\theta+\sin x\cos\theta$$
$$\iff\sin x\cos\theta=\sin\theta(\cos\theta-\cos x)$$
Squaring we get, $$(1-\cos^2x)\cos^2\theta=\sin^2\theta(\cos^2\theta+\cos^2x-2\cos\theta\cos x)$$
$$\iff\cos^2x-2\sin^2\theta\cos\theta\cos x+(\sin^2\theta-1)\cos^2\theta=0\ \ \ \ (1)$$
If we start with $\displaystyle\frac{\cos y}{\cos\theta}+\frac{\sin y}{\sin\theta}=1;$ we shall reach at the same equation.
So, the roots of $(1)$ are $\displaystyle\cos x,\cos y\implies\cos x\cos y=\frac{(\sin^2\theta-1)\cos^2\theta}1=-\cos^4\theta$
Similarly if we start with $$\frac{\cos x}{\cos\theta}+\frac{\sin x}{\sin\theta}=1\iff\cos\theta(\sin\theta-\sin x)=\cos x\sin\theta,$$
we shall reach at $\displaystyle\sin x\sin y=-\sin^4\theta$
You should take it home from here
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$ $$\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$$
The square root really got me confused here. I've tried using the standard trick with $x = x + \sqrt{x^2+4}$ but failed. That might still be the best approach.
What's the best way to solve this and how to do it?
| For real $x,$
$$\sqrt{4x^4+16x^2}=\sqrt{4x^2(x^2+4)}=2|x|\sqrt{x^2+4}$$
As $x\ge0,$ $$\sqrt{4x^4+16x^2}=2x\sqrt{x^2+4}$$
Set $x^2+4=u$ or $x^2+4=u^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given that $w = e^{(2\pi/5)i}$, show that $w + \overline{w}$ is a root > of $z^2 + z - 1 = 0$ Hoping to get some help with the following problem
Given that $w = e^{(2\pi/5)i}$, show that $w + \overline{w}$ is a root
of $z^2 + z - 1 = 0$.
I've noticed that $w + \overline{w} = 2Re(w)$, but I'm not sure on how to proceed from there. $2\pi/5$ is a weird angle, so substituting a cosine doesn't seem to make much sense.
| $w + \overline{w} = 2\cos (2\pi/5)$, and $(2\cos (2\pi/5))^2 + 2\cos (2\pi/5) - 1 = 4\left(\dfrac{1+\cos (4\pi/5)}{2}\right) + 2\left(2\cos^2(\pi/5) - 1\right) - 1 = 2\left(1-\cos (\pi/5)\right) + 4\cos^2(\pi/5) - 2 - 1 = 4\cos^2(\pi/5) - 2\cos (\pi/5) - 1 = S$.
We have: $\cos 36^\circ = \sin 54^\circ$ gives: $1-2x^2 = 3x - 4x^3$ with $x = \sin 18^\circ = \sin (\pi/10)$. Thus: $4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$. Since $0 < x < 1$, $4x^2 + 2x - 1 = 0$. Thus : $\cos (\pi/5) = 1 - 2x^2 = 1 - \dfrac{1-2x}{2} = \dfrac{1+2x}{2}$, and $\cos^2(\pi/5) = \left(\dfrac{1+2x}{2}\right)^2 = \dfrac{1+4x+4x^2}{4} = \dfrac{1+4x+1-2x}{4} = \dfrac{1+x}{2}$. Thus: $S = 4\left(\dfrac{1+x}{2}\right) - 2\left(\dfrac{1+2x}{2}\right) - 1 = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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For a triangle $ABC$, $a^2+b^2+c^2=8R^2$ then it is a right triangle? $ABC$ is a triangle, $a^2+b^2+c^2=8R^2$ then how do we prove it is a right triangle?
| We use the well-known formula $R = \frac{abc}{4A}$, where $A$ is the area of the triangle.
Thus $8R^2 = \frac{8a^2b^2c^2}{16A^2}$, and by Heron's formula
$$16A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c).$$
Now assume that $a^2 + b^2 + c^2 = 8R^2$. Then we have
$$8a^2b^2c^2 - (a^2 + b^2 + c^2)(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = 0.$$
Expanding this gives
$$a^6 + b^6 + c^6 - a^2b^4 - b^2c^4 - c^2a^4 - a^4b^2 - b^4c^2 - c^4a^2 + 2 a^2 b^2 c^2 = 0,$$
which can be factored as
$$(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(c^2 + a^2 - b^2) = 0.$$
Thus the claim follows from Pythagorean theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two and Three Variable Limit Questions Find the following limits, if they exist.
$$\lim_{x,y\rightarrow 0,0}\frac{x^2 + \sin^2 y}{\sqrt{x^2+y^2}}$$
I believe we're suppose to use the squeeze theorem on this first one above. Possibly utilizing the fact that sin(y) is always between -1 and 1? I know the end result is suppose to be zero, but I'm having a hard time getting there.
$$\lim_{x,y,z\rightarrow 0,0,0}\frac{x^2 yz}{x^8 + y^4 + z^2}$$
I know the end result is suppose to be a DNE. I attempted to set multiple variables equal to zero to see what it would come out to, and got differing values. So, when $y,z = 0$, the limit $= 0$, when $y,z = x$, the limit $= \infty$. Since these values are different -> DNE. However I'm quite sure I'm not getting the full picture here either.
If someone could go over the process and logic associated with these problems, I'd greatly appreciate it. This is NOT homework, this is test prep.
| (1) For an application of the squeeze theorem, note that for $y \in [-\pi/2,\pi/2]$
$$\frac{2|y|}{\pi} \leq |\sin y| \leq |y|$$
and
$$(4/\pi^2)\sqrt{x^2+y^2} \leq\frac{x^2 +(4/\pi^2)y^2}{\sqrt{x^2+y^2}} \leq\frac{x^2 +\sin^2 y}{\sqrt{x^2+y^2}} \leq \frac{x^2 +y^2}{\sqrt{x^2+y^2}} = \sqrt{x^2+y^2}.$$
Since $\lim_{(x,y) \rightarrow (0,0)}\sqrt{x^2+y^2}=0,$ the squeeze theorem implies that your limit is $0$.
(2) For the second limit, consider a path where $y = x^2$ and $z= x^4$, then
$$\lim_{(x,y,z)\rightarrow (0,0,0), y = x^2, z=x^4}\frac{x^2 yz}{x^8 + y^4 + z^2}=\lim_{x\rightarrow 0}\frac{x^8}{3x^8}=\frac1{3}\neq 0.$$
Next consider a path where $y = x$ and $z= x$, then
$$\lim_{(x,y,z)\rightarrow (0,0,0), y = z=x}\frac{x^2 yz}{x^8 + y^4 + z^2}=\lim_{x\rightarrow 0}\frac{x^4}{x^8+x^4+x^2}=\lim_{x\rightarrow 0}\frac{x^2}{x^6+x^2+1}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\forall x>0:\ln(1+x) > x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$ I'm reading a proof which aim to show that:
$$\forall x>0:\ln(1+x) > x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$$
the Taylor expansion of $\ln(1+x)$ is (not by chance):
$$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}...$$
Now, the proof claims that the remainder, starting from $\frac{x^5}{5}$ is positive.
But, calculating the remainder using the Lagrange's form gives:
$$\frac{\ln(1+x)^{(6)}(z)}{6!}x^6 < 0$$
Because $$\ln(1+x)^{(6)} = -\frac{120}{(x+1)^6} < 0$$
Where is the mistake?
| A much simpler proof is as follows. Let $t > 0$ and then we have $1 - t^{4} < 1$ so that on division by $(1 + t)$ we get $$(1 - t)(1 + t^{2})< \frac{1}{1 + t}$$ or $$1 - t + t^{2} - t^{3} < \frac{1}{1 + t}$$ If $x > 0$ then we can integrate the above inequality to get $$\int_{0}^{x} \left(1 - t + t^{2} - t^{3}\right)\,dt < \int_{0}^{x}\frac{dt}{1 + t}$$ and we get $$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} < \log(1 + x)$$ for $x > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying a trigonometric identity Simplify $1 + \tan^2x$
My attempt:
$$\begin{align}1 + \tan^2x&\\
&= \frac{1}{1} + \frac{\sin^2x}{\cos^2x}\\
&= \frac{1(\cos^2x)}{1(\cos^2x)} +\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x + \sin^2x}{\cos^2x\cos^2x}\\
&= \frac{\sin^2x}{\cos^2x}\\
&= \tan^2x\end{align}$$
The correct answer, however..is $sec^2x$ Wherever I went wrong, please show.
| $${\cos^2x\over\color{red}{\cos^2x}}+{\sin^2x\over\color{red}{\cos^2x}}={\cos^2x+\sin^2x\over\color{red}{\cos^2x}}={1\over\cos^2x}=\sec^2x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892065",
"timestamp": "2023-03-29T00:00:00",
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any other method for evaluating $\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx$? I tried below and its getting tedious :
$\begin{align}\\ \int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx &= \int\limits \frac{(2x-1)+ x^2-x+2 }{ \sqrt{x^2-x+1} } dx \\~\\ &= \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ (x^2-x+1 )dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ dx }{ \sqrt{x^2-x+1} } \\~\\&\cdots\\~\\ \end{align}$
wolfram shows very much simplified answer :
http://www.wolframalpha.com/input/?i=%5Cint+%28x%5E2%2Bx%2B1%29%2F%28sqrt%28x%5E2-x%2B1%29%29
I'm wondering if there is any nice way to work this
| $$
\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx = \frac{2}{\sqrt{3}}\int \frac{x^2+x+1 }{\sqrt{\frac{4}{3}\left(x-\frac{1}{2}\right)^2 + 1} }
$$
set $t= \frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)$
we obtain
$$
\frac{1}{4}\int\frac{3t^2+4\sqrt{3}t + 7}{\sqrt{t^2+1}}
$$
This is the same way as yours but a little cleaner to solve :).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to calculate the integral of $\sum_{n=1}^\infty (1/r)^{n+1} r^2$? How to calculate this integral?
$$\int\limits_0^1 {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{r}} \right)} } ^{n + 1} r^2 dx$$
Here $r$ is a real number
| I'll assume that what you wrote was intended to be written
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2 \right)dr
$$
Notice that the sum is really
$$
\sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2=\frac{1}{r^2}r^2+\frac{1}{r^3}r^2+\frac{1}{r^4}r^2+\cdots=1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\cdots
$$
which is clearly a geometric series which converges if $|r|>1$ (which means we could say that we are done right here! Why?). Then we have
$$
\sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2=\sum_{n=0}^\infty \left(\frac{1}{r}\right)^n=\frac{1}{1-\frac{1}{r}}=\frac{r}{r-1}
$$
Therefore, we have
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2 \right)dr=\int_0^1 \frac{r}{r-1} dr
$$
which clearly does not converge over $[0,1]$ as you can easily check.
If you meant
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}\right) r^2 dr
$$
this doesn't matter as
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2 \right)dr=\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1} \right)r^2 dr
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $t^3+t^2-2t-2$ given that $t$ is greater than or equal to 2 The original question was to find the range of the function f defined by:
$$f(x)=\frac {(1+x+x^2)(1+x^4)}{x^3}$$ for $x>0$
Evidently, differentiating is not very helpful. So I wrote $f(x)$ as:
$$f(x)=t^3+t^2-2t-2=(t^2-2)(t+1)$$
Where $t=x+\frac 1x$
The maximum value clearly approaches $+\infty$.
How do I find the minimum value, where $t >=2$? Here, too differentiating would not help, unless I want to plot a graph, which would again be somewhat cumbersome.
| Expanding $f(x)$ and breaking the fraction,
$$f(x)=\frac{(1+x+x^2)(1+x^4)}{x^3}=\frac{1+x+x^2+x^4+x^5+x^6}{x^3}=\frac 1{x^3}+\frac 1{x^2}+\frac 1x+x+x^2+x^3.$$
Can you differentiate $f(x)$ now? Only the power rule is required for the last expression.
| {
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"url": "https://math.stackexchange.com/questions/897427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Find all complex number $z\in\Bbb{C}$ such that $\vert z\vert=\vert z^{-1}\vert=\vert z-1\vert$
Find all complex number $z\in\Bbb{C}$ such that $$\vert z\vert=\vert z^{-1}\vert=\vert z-1\vert$$
I tried to write $z=a+ib$, clearly $z=1$ is not a solution. I have to solve
$$\left\{
\begin{array}{l}
a^2+b^2=1 \\
1=\sqrt{(a-1)^2-b^2}\sqrt{a^2+b^2}
\end{array}
\right.$$
By multiplying $\sqrt{a^2+b^2}$, (I hope it's correct).
Which it implies to solve
$$
\left\{
\begin{array}{l}
a^2+b^2=1 \\
1=(a-b)^2\bigl((a^2-b^2)-2a\bigr)
\end{array}
\right.
$$
Here I am stuck, thanks in advance for your help.
| If $z=re^{i2\theta}$ where $r\ge0,\theta$ are real
$|z^{-1}|=\dfrac1r\implies r=\dfrac1r\iff r^2=1\implies r=1$
$z-1=e^{i2\theta}-1=\cos2\theta+i\sin2\theta-1=2i\sin\theta(\cos\theta+i\sin\theta)$
$\implies |z-1|=2|\sin\theta|$
$|z|=|z-1|\implies 2|\sin\theta|=1\iff\sin\theta=\pm\frac12\implies\cos2\theta=1-2\sin^2\theta=\dfrac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $y''-3y'+2y=x^2$ Solve $$y''-3y'+2y=x^2$$
My approach:
Homogen solution:
$$y = Ae^x +Be^{2x}$$
Particular solution:
$$ y_p = x(Ax^2+Bx+C) = Ax^3+Bx^2+Cx $$
$$ y_p' = 3Ax^2 + 2Bx + C$$
$$y_p'' = 6Ax + 2B$$
Put his into the initial equartion to get A, B and C gives me:
$A=0, B=1/2, C=3/2$
This leads me to the answer:
$$y = Ae^x +Be^{2x} + x^2/2 + 3x/2$$
However the correct answer is
$$y = Ae^x +Be^{2x} + x^2/2 + 3x/2+ 7/4$$
Where's my miss? Where comes the last term from?
| Another way is to use the operator D:
$$
y^{\prime \prime} -3y^{\prime} + 2y = x^2 \quad \Rightarrow \quad (D^2-3D + 2)y = x^2 \quad \Rightarrow
$$
$$
(D-1)(D-2)y = 0 + x^2 \quad \Rightarrow \quad y(x) = \dfrac{1}{(D-1)(D-2)}\cdot 0 + \dfrac{1}{2 - 3D + D^2}\cdot x^2
$$
Hence,
$$
y(x) = C_1e^x + C_2e^{2x} + \biggl[\dfrac{1}{2} + \dfrac{3D}{4} + \dfrac{7D^2}{8} + \dfrac{15D^3}{8} + \ldots\biggr]x^2
$$
$$
y(x) = C_1e^x + C_2e^{2x} + \dfrac{x^2}{2} + \dfrac{3x}{2} + \dfrac{7}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Checking for convergence of series To check the convergence of the series
$\displaystyle \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..\infty$
Attempt 1: Term $\displaystyle u_n= \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$
So the series becomes ...
$\displaystyle \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+..\infty$
This is an alternating series. So the series is said to be oscillatory or non-convergent.
Attempt 2: We have $\displaystyle u_n= \frac{1}{n^2}\frac{1}{1+1/n}$
and $\displaystyle v_n= \frac{1}{n^2}$
Then $\displaystyle \lim_{n \to\infty}\frac{u_n}{v_n}=\lim_{n \to\infty}\frac{1}{1+1/n}=1$
Therefore, both $u_n$ and $v_n$ converge or diverge together. As $\displaystyle \sum v_n=\sum \frac{1}{n^2}$ converges, $\sum u_n$ must also converge.
Now my question is that which of this approaches is correct. The given answer says "convergent". So Attempt 2 should be correct.
But, how do I know which method to choose for solving unknown problems. Please advise.
| To check the convergence:
$$n(n+1)=n^2+n\geq n^2$$
then $$\frac{1}{n(n+1)}\leq \frac{1}{n^2}$$
and so $$\sum_{n=1}^\infty \frac{1}{n(n+1)}$$
converge.
To calculate the limit,
$$\sum_{n=1}^N\frac{1}{n(n+1)}=\sum_{n=1}^N\frac{1}{n}-\sum_{n=1}^N\frac{1}{n+1}=\sum_{n=1}^N\frac{1}{n}-\sum_{n=2}^{N+1}\frac{1}{n}=1+\sum_{n=2}^N\frac{1}{n}-\sum_{n=2}^N\frac{1}{n}-\frac{1}{N+1}=1+\frac{1}{N+1}$$
When you make tighten $N\to\infty $, you finally obtain
$$\sum_{n=1}^\infty \frac{1}{n(n+1)}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Evaluation of $\int \sec^3 (x)dx$ How Can I evaluate $\displaystyle \int \sec^3 (x)dx$
(Without Using Weierstrass Substution or Integration by parts.)
$\bf{My\; Try::}$ Let $\displaystyle I = \int\sec^3(x)dx = \int \frac{1}{\cos^3(x)}dx = \int \frac{1}{\sin ^3\left(\frac{\pi}{2}-x\right)}dx$
Now Let $\displaystyle \left(\frac{\pi}{2}-x\right) = t\;,$ Then $\displaystyle dx = -dt$. So $\displaystyle I = -\int \frac{1}{\sin^3 t}dt = -\int\frac{1}{2\sin^3\left(\frac{t}{2}\right)\cdot \cos^3 \left(\frac{t}{2}\right)}dt$
Now How can I solve after that
Help me
Thanks
| Hint:
Substitite
$$\sec (x)= \cosh (u) $$
$$\tan(x) = \sinh(u)$$
$$\sec^2 (x) \, \mathrm dx = \cosh (u) \,\mathrm du$$
$$\mathrm dx = \frac1{\cosh(u)} \, \mathrm du$$
Edit: Why was this downvoted?
Here is the entire process
$$\begin{align}
\int \sec^3 x \, \mathrm dx &{}= \int \cosh^2 u\,\mathrm du \\
&= \frac{1}{2}\int ( \cosh 2u +1) \,\mathrm du \\
&= \frac{1}{2} \left( \frac{1}{2}\sinh2u + u\right) + C\\
&= \frac{1}{2} ( \sinh u \cosh u + u ) + C \\
&= \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| + C
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/903306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
I need compute a rational limit that involves roots I need compute the result of this limit without l'hopital's rule, I tried different techniques but I did not get the limit, which is 1/32, I would appreciate if somebody help me. Thanks.
$$\lim_{y\to32}\frac{\sqrt[5]{y^2} - 3\sqrt[5]{y} + 2}{y - 4\sqrt[5]{y^3}}$$
| $$\lim_{y\to32}\frac{\sqrt[5]{y^2} - 3\sqrt[5]{y} + 2}{y - 4\sqrt[5]{y^3}}$$
taking $\sqrt[5]{y}=x$
we have that
$$\lim_{x\to2}\frac{x^2-3x+2}{x^5-4x^3}=\lim_{x\to2}\frac{x^2-x-2x+2}{x^3(x^2-4)}=$$
$$=\lim_{x\to2}\frac{x(x-1)-2(x-1)}{x^3(x-2)(x+2)}=\lim_{x\to2}\frac{(x-1)(x-2)}{x^3(x-2)(x+2)}$$
$$=\lim_{x\to2}\frac{(x-1)}{x^3(x+2)}=1/32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/903492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How solve this equation $\sin x\cdot \sin20=2\sin(110-x) (\sin10)^2$ let
$0<x<90$, and such
$$\sin x\cdot \sin20=2\sin{(110-x)}(\sin10)^2$$
find the $x$
my idea: since
$$\sin x\cdot 2\sin10\cos10=2\sin(70+x)(\sin10)^2$$
so
$$\cot10=\dfrac{\sin(70+x)}{\sin x}$$
then How find it?
| $$\sin20^\circ=2\sin10^\circ\cos10^\circ$$
$$\implies\sin x\cos10^\circ=\sin(70^\circ+x)\sin10^\circ$$
$$\implies2\sin x\cos10^\circ=2\sin(70^\circ+x)\cos80^\circ$$
Using Werner’s Formula,
$$\sin(x-10^\circ)+\sin(x+10^\circ)=\sin(x+150^\circ)+\sin(x-10^\circ)$$
$$\implies x+10^\circ=180^\circ n+(-1)^n(x+150^\circ)$$ where $n$ is any integer
If $n$ is even $=2m$(say), $$x+10^\circ=180^\circ(2m)+(x+150^\circ)\iff 360^\circ m+140^\circ=0\text{ (Is it possible?)}$$
If $n$ is odd, $=2m+1$(say), $$x+10^\circ=180^\circ(2m+1)-(x+150^\circ)$$
$$\iff x=90^\circ(2m+1)-70^\circ=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the limit of the following sum $$\lim_{n\to\infty}\sum_{k=1}^n \ln\Big(1+\frac{k}{n^2}\Big)$$
According to me, the answer is $0$. I'm curious as to what answers might others come up with, as well as the method of reasoning.
| Rather boring answer:
We have $x-{x^2 \over 2} \le \log(1+x) \le x$ for all $x >0$.
We have
$\sum_{k=1}^n \log(1+ { k \over n^2}) \le \sum_{k=1}^n { k \over n^2} = {1 \over n^2} (1+2 + \cdots + n) = {1 \over n^2} {1 \over 2} n (n+1)$.
We also have $\sum_{k=1}^n \log(1+ { k \over n^2}) \ge \sum_{k=1}^n ( { k \over n^2} -{ 1\over 2} ({ k \over n^2})^2 ) = {1 \over n^2} {1 \over 2} n (n+1) - { 1\over 2} {1 \over n^4} (1^2+2^2+\cdots + n^2)$.
Since $1^2+2^2+\cdots + n^2 = {1 \over 3} n^3 + {1 \over 2} n^2 +{1 \over 6} n$, we
see that the limit of the sum is ${ 1\over 2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Evaluate $\int\frac{8x+20}{5x^2+25x+20}dx$ I tried to solve it and got $\frac{4}{5} \ln(4+5 x+x^2)+C$ as an answer, but my online homework program says it's incorrect. What did I do wrong?
I pulled out $\frac{4}{5}$ as a constant and saw that the numerator was the derivative of the denominator. So I put the denominator in a natural log.
| Here is a general method for integrals of this form.
$$ I = \int \frac{mx±n}{ax^2+bx+d}dx$$
$$I = \int \frac{\frac{m}{2a}(2ax+b) + (n - \frac{mb}{2a})}{(x+ \frac{b}{2a})^2 + (c -\frac{b^2}{4a^2})}dx = \frac{m}{2a}ln|ax^2 + bx + c| + (n- \frac{mb}{2a}) \int \frac{dx}{ax^2+bx+c}dx$$
depending on the quadratic in the bottom, the second integral will be in the form of either $\frac{1}{a}\arctan(\frac{u}{a})$ + C or $ \frac{1}{2a}ln|\frac{u-a}{u+a}| + C $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Finding the limit of $\left(\frac{(1+2x)^{1/x}}{e^2}\right)^{1/x}$ at $x=0$ I can't seem to find a solution to this.
$$\lim_{x\to0} \left(\frac{(1+2x)^{1/x}}{e^2}\right)^{1/x}$$
i tried to manipulate to apply Lhopitals rule but i can't see to do it
| There have been many nice solutions provided by others earlier.
This is an alternative approach, without using logs or l'Hopital:
$$\begin{align}
\dfrac {\left(1+\dfrac kn \right) ^n}{e^k}
&= {\left(1+\dfrac kn \right) ^n}e^{-k}\\
&={\left(1+\dfrac kn \right) ^n}\left(1-k+\frac{k^2}{2!}-\frac{k^3}{3!}+... \right)\\
&=\left[ 1+n\left(\frac kn\right)+\dfrac{n(n-1)}{2!}\left(\frac kn \right)^2+\cdots \right] \left[1-k+\frac{k^2}{2!}-\frac{k^3}{3!}+\cdots \right]\\
&\approx \left[1+k+\frac{n-1}{2n}k^2 \right]\left[ 1-k+\frac{k^2}2\right]\\
&\approx 1+k-k+k^2 \left(\frac{n-1}{2n}-1+\frac 12 \right) \\
&\approx 1-\frac{k^2}{2n}\\
\lim\limits_{n\to\infty} \left(\dfrac{\left(1+ \dfrac kn\right)^n}{e^k}\right)^n
&=\lim\limits_{n\to\infty} \left(1-\dfrac{k^2}{2n}\right)^n\\
&=e^{-k^2/2}
\end{align}$$
Put $k=2$ and $n=\dfrac 1x$:
$$\begin{align}
\lim\limits_{x\to 0}
\left(\dfrac{\left(1+ 2x \right)^{\frac 1x}}{e^2}\right)^{\frac 1x}
&=e^{-2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How can I distribute 15 pennies (1 cent) and 17 nickels (5 cents)? How can I distribute 15 pennies (1 cent) and 17 nickels (5 cents), between four children, with the following restriction:
*
*A child receives at leat 1 penny and 3 nickels
*The children 2,3 and 4, receive at least 3 pennies but not more than 9 nickels
I developed the following generating function and I would like to know if it is correct.
$f(x)=[(x+x^2+x^3+...)(x^5+x^{10}+x^{15}+...)][(x^3+x^4+x^5+...)(x^5+x^{10}+...+x^{15})]^3$
$f(x)=[x(1+x+x^2+...)x^5(1+x+x^2+...)][x^3(1+x+x^2+...)(x^5+x^{10}+...+x^{45})]^3$
$f(x)=[x(1+x+x^2+...)x^5(1+x+x^2+...)][x^3(1+x+x^2+...)x^5(1+x+...+x^{9})]^3$
$f(x)=[x^6(1+x+x^2+...)^2][x^8(1+x+x^2+...)(1+x+...+x^{9})]^3$
$f(x)=x^6[(1+x+x^2+...)^2]x^{24}[(1+x+x^2+...)(1+x+...+x^{9})]^3$
$f(x)=x^{30}(1+x+x^2+...)^2(1+x+x^2+...)^3[(1+x+...+x^{9})]^3$
$f(x)=x^{30}\frac{1}{(1-x)^5}(\frac{1-x^{10}}{1-x})^3$
Then the answer is the coefficient of $x^{32}$, but I cant find it...
| I am getting:
$\left(x+x^2+x^3+x^4+...+x^{15}\right)\left(x^{15}+x^{20}+x^{25}+...+x^{85}\right)$
for the first kid.
$(\left(x^3+x^4+x^5+...+x^{15}\right) \left(x^{15}+x^{20}+x^{25}+...+x^{45}\right))^3$
for the next 3 kids. Then multiply both of them and check the coefficient of $x^{100}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $ \dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)} $, then what type of triangle is $\triangle ABC $? In $\triangle ABC$
$
\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}
$
then what type of triangle is $\triangle ABC $ ?
My try :
By componendo and dividendo
$\dfrac{a^2}{b^2} = \tan A \cot B$
Not able to conclude, any help ?
| $$\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}$$
$$\dfrac{a^2-b^2}{a^2+b^2} = \frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\cos A\sin B}$$
$$\dfrac{a^2-b^2+a^2+b^2}{a^2-b^2-a^2-b^2} = \frac{\sin A\cos B-\cos A\sin B+\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B-\sin A\cos B-\cos A\sin B}$$
$$\dfrac{a^2}{b^2} = \frac{\sin A\cos B}{\cos A\sin B}$$
$$\dfrac{a^2}{b^2} = \frac{ak.\cos B}{\cos A.bk}$$
Because,
$$\text{sine rule}\implies\frac{\sin A}a=\frac{\sin B}b=k$$
$$\dfrac{a}{b} = \frac{\cos B}{\cos A}$$
$$a\cos A = b\cos B$$
$$\sin A\cos A=\sin B\cos B$$
$$\sin 2A=\sin 2B$$
Or, A=B.
So, the triangle is an isoceles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the limit of a trigonometric function $\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$ Could someone please clarify whether my calculation on the following limit problem is correct?
Determine the following limit:
$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$
$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$ = $\lim_{x \to \frac{\pi}{2}} \sin x\frac{\sin x-1}{\sin x-1} = \lim_{x \to \frac{\pi}{2}}\sin x \frac{1}{1} = \lim_{x \to \frac{\pi}{2}} \sin x = 1$
Thank you.
| Using $a^2-b^2=(a+b)(a-b)$,
$$\frac{\sin^2 (x) -1}{\sin (x)-1}=\sin (x)+1$$ The rest is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$
Let $x,y,z>0$. Prove that
$$\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}.$$
I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
| $a=x+2y,b=y+2z,c=z+2x \implies x=\dfrac{a-2b+4c}{9},y=\dfrac{b-2c+4a}{9},z=\dfrac{c-2a+4b}{9}$
the inequality become:
$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \ge \dfrac{27}{5(ab+bc+ac)-2(a^2+b^2+c^2)}$
now use UVW method:
$3u=a+b+c,3v^2=ab+bc+ac,w^3=abc \implies u\ge v\ge w$
$\iff \dfrac{(3v^2)^2-6uw^3}{w^6}\ge \dfrac{3}{3v^2-2u^2} \iff w^6+2u(3v^2-2u^2)w^3-3v^4(3v^2-u^2) \le 0$
let $w^3=x,f(x)=x^2+2u(3v^2-2u^2)x-3v^4(3v^2-2u^2)$
$2u(3v^2-2u^2) \ge 0 $
$f_{max}(x)=f(w^3|w=v)=f(v^3)$, when $w=v \implies u=v=w \implies f(v^3)=0 \implies f(x) \le 0 $
when $u=v=w \implies a=b=c \implies x=y=z$
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Why does this simplifies like that?
I don't understand how do we simplify this fraction ? How does 3^n becomes 3^(n+1) and how does 1-(1/3) becomes 2 ???
Is there a general rule for this ?
| The following is an explicit path from your statement to it's simplification:
$$\frac{3^n}{ (1 - \frac{1}{3}) }
= \frac{3^n}{(\frac{3}{3} - \frac{1}{3})}
= \frac{3^n}{(\frac{3-1}{3})}
=\frac{3^n}{(\frac{2}{3})}
= 3^n\cdot (\frac{2}{3})^{-1}
= 3^n\cdot\frac{3}{2}
= \frac{3^n \cdot 3^1}{2}
= \frac{3^{n+1}}{2} $$
or generally, you could try the following:
$$1\cdot\frac1{1-\frac1{x}} = \frac{x}{x}\cdot\frac1{1-\frac1{x}} = \frac{x}{x(1 -\frac{1}{x})} = \frac{x}{x -1}$$
It doesn't matter how you simplify as long as you do it right and understand how to get the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to multiply the binomials $(2x^3 - x)\left(\sqrt{x} + \frac{2}{x}\right)$ I am sorry if the numbers are not formatted, I have searched but found nothing on how. I am trying to multiply $$(2x^3 - x)\left(\sqrt{x} + \frac {2}{x}\right)$$ together and I arrive at a different answer that does not match the one given. The approach I take is to assign the numbers powers then multiply them by the lattice method. So $$(2x^3) (X^{\frac {1}{2}}) + (2x^3) (2x^{-1}) + (x^{\frac {1}{2}}) (-x)) + (2x^{-1}) (-x)$$. I am not asking for a solution but rather an explanation on how to go about multiplying the terms with negative and fractional exponents.
| We distribute just as we would distribute $$(a + b)(c + d) = a(c+d) + b(c + d) = ac + ad + bc + bd.$$
$$(2x^3 - x)(\color{blue}{\sqrt x + 2x^{-1}}) = 2x^3(\color{blue}{\sqrt x + 2x^{-1}}) - x(\color{blue}{\sqrt x + 2x^{-1}})\\ =2x^3\sqrt x + 4x^3x^{-1} - x\sqrt x - 2xx^{-1}\\ = 2x^{6/2}x^{1/2} + \frac {4x^3}{x} - x^{2/2}x^{1/2} - \frac {2x}{x}\\
= 2x^{6/2 + 1/2} + 4x^2 - x^{2/2 + 1/2} - 2 \\
= 2x^{7/2} + 4x^2 - x^{3/2} - 2$$
We are using the fact that $$a^ba^c = a^{b+c}$$
This holds for negative exponents as well. So when we approach this simply by looking at products of terms with the same base, each raised to an exponent, we can see that $4x^3x^{-1} = 4x^{3+ (-1)} = 4x^2$.
Note that this is consistent with what I wrote, e.g., $$4x^3x^{-1} = \frac{4x^3}{x}.$$ We can see that we can cancel the common factor of $x$ in the numerator and denominator: $\dfrac{4x^3}x = 4x^2$ and in doing so, we are, essentially, subtracting the exponent in the denominator from the exponent in the numerator. We can generalize this into a handy "rule":
$$\frac {a^b}{a^c} = a^{b-c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Help Evaluating $\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\frac{\pi}{2}-x}-\tan {x}\right)$ Does anyone know how to evaluate the following limit?
$$\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\frac{\pi}{2}-x}-\tan {x}\right)$$
The answer is 0 , but I want to see a step by step solution if possible.
| Here are the steps
$$
\lim_{x\to \frac{\pi}{2}} \left[\frac{1}{\frac{\pi}{2}-x}-\tan x\right]= \lim_{x\to \frac{\pi}{2}} \left[\frac{2}{\pi-2x}-\frac{\sin x}{\cos x}\right]
$$
$$
= \lim_{x\to \frac{\pi}{2}} \left[\frac{2\cos x-(\pi-2x)\sin x}{(\pi-2x)\cos x}\right]= \lim_{x\to \frac{\pi}{2}} \left[\frac{\frac{d}{dx}[2\cos x-(\pi-2x)\sin x]}{\frac{d}{dx}[(\pi-2x)\cos x]}\right]
$$
$$
= \lim_{x\to \frac{\pi}{2}} \left[\frac{-2\sin x-(\pi-2x)\cos x+2\sin x}{-(\pi-2x)\sin x-2\cos x}\right] = \lim_{x\to \frac{\pi}{2}} \left[\frac{\frac{d}{dx}[(\pi-2x)\cos x]}{\frac{d}{dx}[(\pi-2x)\sin x+2\cos x]}\right]
$$
$$
= \lim_{x\to \frac{\pi}{2}} \left[\frac{-(\pi-2x)\sin x-2\cos x}{(\pi-2x)\cos x-2\sin x-2\sin x}\right] = \lim_{x\to \frac{\pi}{2}} \left[\frac{(\pi-2x)\sin x+2\cos x}{4\sin x -(\pi-2x)\cos x}\right]
$$
$$
= \frac{0\cdot 1+2\cdot 0}{4\cdot 1 -0\cdot 0}= \frac{0}{4}=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
matrix-finding determinant of adj of inverse matirx if A is a $3$x$3$ matrix and let A=$2$,then what will be the value of det(adj(adj(adj($A^{-1}$)))?
1.$\dfrac{1}{512}$
2.$\dfrac{1}{1024}$
3.$\dfrac{1}{128}$
4.$\dfrac{1}{256}$
| Assume you meant $det(A)=2$. Then $det(A^{-1})=\frac{1}{2}$. Work the problem outward:
\begin{align*}
adj(A^{-1})&=det(A^{-1})A=\frac{1}{2}A,\\
adj(\frac{1}{2}A)&=(1/2)^2 adj(A)=\frac{1}{4}det(A)A^{-1}=\frac{1}{2}A^{-1},\\
adj(\frac{1}{2}A^{-1})&=(1/2)^2adj(A^{-1})=\frac{1}{4}det(A^{-1})A=\frac{1}{8}A,\\
det(\frac{1}{8}A)&=(1/8)^3det(A)=\frac{1}{512}2=\frac{1}{256}.
\end{align*}
For justifications of the steps, you can wiki "adjugate matrix" and "determinant."
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Factoring $(x+1)(x+2)(x+3)(x+4)+1$ I have to factor this polynomial:
$$(x+1)(x+2)(x+3)(x+4)+1$$
WolframAlpha gives
$$(x^2+5x+5)^2$$
How can I prove it without expanding the result?
Thanks!
| I answered a similar question here. You can do it like this:
$(x+2)(x+3) = x^2+5x+5+1$, $(x+1)(x+4) = x^2+5x+5-1$, so
$$
(x+1)(x+2)(x+3)(x+4) = (x^2+5x+5+1)(x^2+5x+5-1) = (x^2+5x+5)^2-1^2.
$$
This gives the answer $(x+1)(x+2)(x+3)(x+4)+1 =(x^2+5x+5)^2 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Integral of $1/(1+x \tan(x))^2$ How would you solve the following integral?
$$\int \frac{1}{(1+x\tan(x))^2} dx$$
Any help would be appreciated.
| $\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{(1+x\tan x)^2}dx = \int\frac{\cos ^2x}{(x\sin x+\cos x)^2}dx$
So $\displaystyle I = \int \frac{\cos x}{x}\cdot \left\{\frac{x\cos x}{(x\sin x+\cos x)^2}\right\}dx$
Now Using Integration by Parts....
So $\displaystyle I = -\frac{\cos x}{x}\cdot \frac{1}{(x\sin x+\cos x)}-\int \frac{(x\sin x+\cos x)}{x^2}\cdot \frac{1}{(x\sin x+\cos x)}dx$
So $\displaystyle I = -\frac{1}{x\cdot (1+x\tan x)}+\frac{1}{x}+\mathcal{C} = \frac{\tan x}{(1+x\tan x)}+\mathcal{C}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove or disprove $xyz+\frac{8}{27}\ge xy+yz+zx$ if $x+y+z=1$
if $x,y,z$ are positive and $x+y+z = 1$,Prove:$$xyz+\frac{8}{27}\ge xy+yz+zx$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done: I could rewrite question as:$$(x^2+y^2+z^2)+2xyz+\frac{16}{27}\ge 2(xy+yz+zx)+(x^2+y^2+z^2)=(x+y+z)^2=1$$
it remains to prove $(x^2+y^2+z^2)+2xyz\ge\frac{11}{27}$.by the well-known inequality that $3(x^2+y^2+z^2)\ge(x+y+z)^2=1$ So $x^2+y^2+z^2\ge\frac{1}{3}=\frac{9}{27}$.So it remains to Prove $xyz \ge \frac{1}{27}$.but obviously If we take $x,y$ small enough it turns to be wrong.So there are two possibilities. The question is wrong or I made a mistake on previous steps.
| Hint:
$$(1-x)(1-y)(1-z) = (x+y)(y+z)(z+x)=xy+yz+zx-xyz.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Use Taylor Series method to solve $y''-2xy+y=0$ I am doing some practice problems for solving second order ODEs, and I am a bit stuck on this one.
Here is what I have:
$y''-2xy'+y=0$
Let $y = \sum_{n=0}^{\infty} C_nx^n \implies y' = \sum_{n=0}^{\infty} nC_nx^{n-1} \implies y'' = \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2} $
Substituting this into the ODE, and I get:
$$ \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2} -2\sum_{n=0}^{\infty} nC_nx^{n}+ \sum_{n=0}^{\infty} C_nx^n = 0$$
Then getting each term to $x^n$ and starting each sum at $n=0$, I have:
$$ \sum_{n=0}^{\infty} [(n+2)(n+1)C_{n+2}-2 nC_n+ C_n]x^n = 0 $$
$$ \implies C_{n+2} = \frac{(2n-1)C_n}{(n+2)(n+1)}$$
I notice that this decouples into two series' for odd and even terms, but I am having trouble with determining the general formula for $C_n$ for each series:
For $n$ even:
When $n=0: C_2 = \frac{-C_0}{2} $
When $n=2: C_4 = \frac{3C_2}{4 \cdot 3} = \frac{-3C_0}{4!} $
When $n=4: C_6 = \frac{7C_4}{6 \cdot 5} = \frac{-7 \cdot 3C_0}{6!} $
When $n=6: C_8 = \frac{11C_6}{8 \cdot 7} = \frac{-11 \cdot 7 \cdot 3C_0}{8!} $
For $n$ odd:
When $n=1: C_3 = \frac{C_1}{3 \cdot 2} $
When $n=3: C_5 = \frac{5C_3}{5 \cdot 4} = \frac{5C_3}{5!} $
When $n=5: C_7 = \frac{9C_5}{7 \cdot 6} = \frac{9 \cdot 5C_1}{7!} $
When $n=7: C_9 = \frac{13C_7}{9 \cdot 8} = \frac{13 \cdot 9 \cdot 5C_1}{9!} $
I am mainly finding it difficult to determine the closed formula for the numerator in each series, so that I can calculate the radius of convergence of each one.
Thanks so much, any help is greatly appreciated.
| Well, may be I am wrong but it seems for me a general formula can be easily deduced from
your recurrent one:
$$
C_{2n}=\frac{(2n-2))!}{(2n)!} \prod_{i=1}^n (4i-5) C_0\\
C_{2n+1}=\frac{(2n-1))!}{(2n+1)!} \prod_{i=1}^n (4i-3) C_1\\
$$
The product also can expressed in factorials but I let it do for somebody else :). $C_0$ and $C_1$ are free constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Factoring the following polynomials
Factorize the following polynomial:
*
*$t^3 -9t +8$
*$t^6 -91t^2 +90$
| It's not so bad to be baffled... :)
Putting $t=1$ and and $t^2=1$ into the first and second expressions respectively will make them zero, so $(t-1)$ and $(t^2-1)$ are factors for the first and second expressions respectively.
You can work out the other factor(s) by equating coefficients of powers of $t$.
$$\begin{align}t^3-9t+8&=(t-1)(t^2+t-8)\\ \\
t^6-91t^2+90&=(t^2-1)(t^4+t^2-90)\\
&=\left[(t-1)(t+1)\right]\left[(t^2-9)(t^2+10)\right]\\
&=(t-1)(t+1)(t-3)(t+3)(t^2+10)\\
&=(t-3)(t-1)(t+3)(t+3)(t^2+10)\\\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can't figure out what's wrong with my solution to an indefinite integral I tried integrating $\int{\frac{xdx}{(Ax + B)^n}}$, and my solution was
$$
\int{\frac{xdx}{(Ax + B)^n}} = \frac{1}{A^2}\left[\frac{(Ax + B)^{-n}}{-n} - B\frac{(Ax + B)^{-n + 1}}{-n + 1}\right] + c
$$
which I got by substituting $u = Ax + B$ and then integrating $\frac{1}{A^2}\int{\frac{u-B}{u^n}du} = 1/A^2(\int{\frac{du}{u^{n-1}}} - B\int{\frac{du}{u^n}})$. WolframAlpha gives me a different answer, and I can't determine if the two answers are equivalent (which it doesn't look like they are). Is my answer the same as WolframAlpha's answer? And what algorithm or procedure might WolframAlpha have used to get a solution to this integral?
EDIT: Okay maybe I should post my solution, so that people get an idea of my thought process.
$$
\begin{align}
\int{\frac{xdx}{(Ax + B)^n}} & = \int{\frac{1/A(u - B)}{u^n}\cdot\frac{1}{A}du}, \ u = Ax + B \\
& = \frac{1}{A^2}\left(\int{\frac{udu}{u^n}} - B\int{\frac{du}{u^n}}\right) \\
& = \frac{1}{A^2}\left(\int{\frac{du}{u^{n-1}}} - B\int{\frac{du}{u^n}}\right) \\
& = \frac{1}{A^2}\left(\frac{u^{-n}}{-n} - B\frac{u^{-n+1}}{-n+1}\right) + c \\
& = \frac{1}{A^2}\left(\frac{(Ax + B)^{-n}}{-n} - B\frac{(Ax + B)^{-n+1}}{-n+1}\right) + c
\end{align}
$$
It seems alright to me, but I don't see how
$$
\frac{1}{A^2}\left(\frac{(Ax + B)^{-n}}{-n} - B\frac{(Ax + B)^{-n+1}}{-n+1}\right) = -\frac{(Ax + B)^{-n+1}(A(n-1)x+B)}{A^2(n-2)(n-1)}
$$
where the right hand side is what WolframAlpha is giving. So are the two expressions equivalent?
| $$\int \frac{x}{(ax+b)^n}dx=\\\frac{1}{a}\int \frac{ax}{(ax+b)^n}dx =\\\frac{1}{a}\int \frac{ax+b-b}{(ax+b)^n}dx =\\\frac{1}{a}\int (\frac{ax+b}{(ax+b)^n}dx +\frac{-b}{(ax+b)^n}dx)=\\\frac{1}{a}\int (\frac{1}{(ax+b)^{n-1}}dx +\frac{-b}{(ax+b)^n}dx)=\\\frac{1}{a}\int ((ax+b)^{-(n-1)}dx -b(ax+b)^{-n}dx)=\\\frac{1}{a}(\frac{(ax+b)^{-(n-1)+1}}{(-(n-1)+1)a} -b \frac{(ax+b)^{-(n)+1}}{(-(n)+1)a})=\\\frac{1}{a}(\frac{(ax+b)^{-(n-1)+1}}{(-(n-1)+1)a} -ab \frac{(ax+b)^{-(n)+1}}{(-(n)+1)a^2})=\\now\\factor\\\frac{1}{a^2}\\\frac{1}{a^2}(\frac{(ax+b)^{-(n-1)+1}}{(-(n-1)+1)} -ab \frac{(ax+b)^{-(n)+1}}{(-(n)+1)})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
A closed-form of product the gamma functions containing $\pi$ and $\phi$ Playing with gamma functions by randomly inputting numbers to Wolfram Alpha, I got the following beautiful result
\begin{equation}
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=\frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi}
\end{equation}
where $\phi$ is golden ratio.
Could anyone here please help me to prove it by hand? I mean without using table for the specific values of $\Gamma(x)$ except for $\Gamma\left(\frac{1}{2}\right)$. As usual, preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
| We will show that
$$
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=\frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi},
$$
where $\phi$ is the golden ratio.
Because of the Gauss's multiplication formula we know that
$$\Gamma(2z)=\frac{1}{\sqrt{\pi}}2^{2z-1}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right).$$
Because $\Gamma$ is nowhere zero, we can divide the formula by $\Gamma(z)$, and we get
$$\frac{\Gamma(2z)}{\Gamma(z)}=\frac{1}{\sqrt{\pi}}2^{2z-1}\Gamma\left(z+\frac{1}{2}\right).$$
We put $z:=2/10$ into the formula and get
$$\color{red}{\frac{\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}}=\frac{1}{\sqrt{\pi}}2^{\frac{4}{10}-1}\color{blue}{\Gamma\left(\frac{7}{10}\right)}.$$
Now take a look at the Euler's reflection formula.
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(z\pi)}.$$
With $z:=3/10$ we get
$$\color{green}{\Gamma\left(\frac{3}{10}\right)}\color{blue}{\Gamma\left(\frac{7}{10}\right)}=\frac{\pi}{\sin\left(\frac{3}{10}\pi\right)}.$$
Putting this all together, we get
$$
\color{red}{\frac{\color{green}{\Gamma\left(\frac{3}{10}\right)}\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}}=\frac{1}{\sqrt{\pi}}2^{\frac{4}{10}-1}\frac{\pi}{\sin\left(\frac{3}{10}\pi\right)}.$$
Now we need that
$$\sin\left(\frac{3}{10}\pi\right)=\frac{1+\sqrt{5}}{4},$$
and of course we also know that $\pi / \sqrt{\pi} = \sqrt{\pi}$.
Using this we get
$$
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=2^{-\frac{3}{5}} \cdot \sqrt{\pi} \cdot\frac{4}{1+\sqrt{5}} = \frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi},$$
and this completes the proof.
One last fun fact, that we can generalize the problem like this
$$\frac{\Gamma\left(\frac{1}{2}-z\right)\Gamma(2z)}{\Gamma(z)}=\frac{2^{2z-1}
\cdot \sqrt{\pi}}{\cos(z\pi)},$$
for all $z \notin -\mathbb{N}$ and $z \neq n-1/2, \ n \in \mathbb{Z}$. You can get your result by $z:=2/10.$
And really at last an other related formula is the following
$$\frac{\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)}{\Gamma\left(\frac{1}{10}\right)}=\frac{\sqrt[10]{3} \cdot \sqrt[5]{2} \cdot \sqrt{\pi}}{\phi}.$$
This is even more interesting, because with the same idea you have to use the multiplication formula twice for $3z$ and also for $2z$ and after that the reflection formula. Because the equation $6z=z+2/3$ only has one solution, and it is $2/15$, that's why $2/15$ has a very important role in the formula, and this problem does not have a generalization like I gave above. I could imagine other ways to generalize, but it would not be nice.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Find a power series expansion of $\frac{4x^2+2x}{1-3x-10x^2}$ about the point $x = \frac{1}{5}$ Find a power series expansion of $\frac{4x^2+2x}{1-3x-10x^2}$
Now I know that $\frac{1}{5}$ is a singularity of the $\frac{4x^2+2x}{1-3x-10x^2}$
and I know that $f(z) = f\left(\frac{1}{5}\right)+\frac{f'\left(\frac{1}{5}\right)}{1!}\left(x-\frac{1}{5}\right) + \ldots$
Now my question is this, how do I go about calculating $f\left(\frac{1}{5}\right)$, $f'\left(\frac{1}{5}\right)$, and so on.
| Factor your expression and you have
$$\begin{align}
\frac{2x(2x+1)}{(1-5x)(1+2x)}
&=\frac{2x}{1-5x}\\
&=-{\frac{1}{5}}\frac{2\left(x-\frac15\right)+\frac25}{x-\frac15}\\
&=-{\frac{2}{5}}-\frac2{25}\left(x-\frac15\right)^{-1}
\end{align}$$
Note that computing a true power series about $\frac15$ is impossible precisely because there is a singularity there. So this is a Laurent series with just two terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/917703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simple limit problem with squares I'm doing a refreshment course in math but I'm stuck with some problem. Although this problem doesn't look hard I don't know what I'm doing wrong.
$$\lim_{x \to 4} \frac{\sqrt{x-3}-1}{2\sqrt{2}-\sqrt{x^2-3x+4}}$$
I have tried to take the conjuct of both squares but I still got the indeterminate form of $\frac{0}{0}$.
Thanks in advance...
| Multiply numerator and denominator by $\displaystyle (2\sqrt{2}+\sqrt{x^2-3x+4})(\sqrt{x-3}+1)$:
$$\lim_{x \to 4} \frac{\sqrt{x-3}-1}{2\sqrt{2}-\sqrt{x^2-3x+4}}=\lim_{x \to 4} \frac{ (2\sqrt{2}+\sqrt{x^2-3x+4})(\sqrt{x-3}+1)(\sqrt{x-3}-1)}{(2\sqrt{2}-\sqrt{x^2-3x+4}) (2\sqrt{2}+\sqrt{x^2-3x+4})(\sqrt{x-3}+1)}= \\ = \lim_{x \to 4}\frac{(2\sqrt{2}+\sqrt{x^2-3x+4})(x-3-1)}{-(x^2-3x-4)(\sqrt{x-3}-1)}=\lim_{x \to 4}\frac{(x-4)(2\sqrt{2}+\sqrt{x^2-3x+4})}{-(x+1)(x-4)(\sqrt{x-3}+1)}=\\=\lim_{x \to 4}\frac{(2\sqrt{2}+\sqrt{x^2-3x+4})}{-(x+1)(\sqrt{x-3}+1)}=\frac{-(2\sqrt{2}+2\sqrt{2})}{5(\sqrt{2}+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/918064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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How to prove that a derivative of a formula equals to another formula. If $u= \ln(\tan x+\tan y+\tan z)$ prove $$\sin 2x \dfrac{du}{dx} + \sin 2y \dfrac{du}{dy} + \sin 2z \dfrac{du}{dz}=2 $$
My answwer was like this:
$$u' =\dfrac{ 1}{\tan x+\tan y+\tan z} \cdot( \sec^2 x +\sec^2 y +\sec^2 z )$$
$$ =\dfrac{ 1}{\dfrac{\sin x}{\cos x} + \dfrac{\sin y}{\cos y} + \dfrac{\sin z}{\cos z}}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
$$ =\dfrac{ \cos x+\cos y+\cos z}{\sin x+\sin y+\sin z}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
$$ = \sin 2x (\dfrac{\cos x}{\sin x})+ \sin 2y (\dfrac{\cos y}{\sin y}) + \sin 2z (\dfrac{\cos z}{\sin z})\cdot( \sec^2 x +\sec^2 y +\sec^2 z )$$
$$ =\dfrac{ 2\sin x\cos x}{\sin x} +\dfrac{ 2\sin y \cos y}{\sin x} +\dfrac{ 2\sin z\cos z}{\sin z}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
$$=\dfrac{ \sin 2x}{\sin x} + \dfrac{ \sin 2y}{\sin x} + \dfrac{ \sin 2z}{\sin z}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
I solved until here and I got stuck. Is my answer right until now? How could I finish it to prove that it equals to 2? Sorry for the wrong codes.
| We have that
$$u(x,y,z)=\ln(\tan x+\tan y+\tan z)$$
Then
$$\frac{\partial u}{\partial x}=\frac{\sec^2x}{\tan x+\tan y+\tan z}\\
\frac{\partial u}{\partial y}=\frac{\sec^2y}{\tan x+\tan y+\tan z}\\
\frac{\partial u}{\partial z}=\frac{\sec^2z}{\tan x+\tan y+\tan z}$$
Then
$$\sin(2x)\frac{\partial u}{\partial x}+\sin(2y)\frac{\partial u}{\partial y}+\sin(2z)\frac{\partial u}{\partial z}=\\
\frac{\sin(2x)\sec^2x+\sin(2y)\sec^2y+\sin(2z)\sec^2z}{\tan x+\tan y+\tan z}=\\
\frac{2\sin x\cos x\sec^2x+2\sin y\cos y\sec^2y+2\sin z\cos z\sec^2z}{\tan x+\tan y+\tan z}=\\
2\frac{\tan x+\tan y+\tan z}{\tan x+\tan y+\tan z}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/918897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How find this maximum $\sqrt{x^2+x^4}+\sqrt{(x-\frac{3}{2})^2+(x^2-\frac{9}{4})^2}$ Question:
Let $0<x<\dfrac{3}{2}$, then find the maximum of
$$\sqrt{x^2+x^4}+\sqrt{\left(x-\dfrac{3}{2}\right)^2+\left(x^2-\dfrac{9}{4}\right)^2}$$
I Found when $x=\dfrac{1}{2}$ it's maximum,see
the maximum .
This result seems interesting, so I think there exist simple methods.
My idea:
Let $$f(x)=\sqrt{x^2+x^4}+\sqrt{\left(x-\dfrac{3}{2}\right)^2+\left(x^2-\dfrac{9}{4}\right)^2}$$
then I found $f'(x)$ is very ugly,then I can't find this roots by hand.
| A geometric solution:
Note that the given expression equals sum of distances from $P = (x,x^2)$ to $(0,0)$ and $(\frac{3}{2},\frac{9}{4})$. These are points on the parabola $y = x^2$.
If you ignore the constraint that $P$ has to lie on parabola, the locus of points for which the sum is constant is a ellipse with foci $(0,0)$ and $(\frac{3}{2},\frac{9}{4})$. Maximising sum thus equals maximising radius of the ellipse such that the ellipse still intersects parabola. (Since we are working on closed interval, this maximum exists.)
But then ellipse has to be tangent to parabola (otherwise, essentially, radius of the ellipse could be increased a bit). Now, since tangent of ellipse reflects ray from the other focus to another (see Reflective Properties of Ellipse), tangent of parabola must also reflect ray from $(0,0)$ to $(\frac{3}{2},\frac{9}{4})$, which then again means that reflection of $(0,0)$ with respect to tangent line at $(x,x^2)$, $(x,x^2)$ itself, and $(\frac{3}{2},\frac{9}{4})$ are collinear.
Now, you can work out that the reflection point is
$$
\left( \frac{4x^3}{4x^2+1}, -\frac{2x^2}{4x^2+1} \right).
$$
Checking the collinearity is easy; slopes have to match. After some manipulation you are left out with
$$
\frac{3}{2} = 4x^3+2x
$$
for which the unique real root is $\frac{1}{2}$ which therefore give the maximum.
It requires some work, but it gives also the general solution when $\frac{3}{2}$ and $\frac{9}{4}$ are replaced by $a$ and $a^2$ (just replace $\frac{3}{2}$ with $a$ in the last equation) and an immediate answer for the search of the minimum for the original function (triangle inequality).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $3(\sin x-\cos x)^4 + 6(\sin x+ \cos x)^2 + 4(\sin^6 x + \cos^6 x) -13 = 0$
Q) Prove that $3(\sin \theta-\cos \theta)^4 + 6(\sin \theta+ \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) -13 = 0$
Source: Trigonometric Functions, Page 5.9, Mathematics XI - R.D. Sharma
My Attempt::
For writing convenience, let $\color{red}{s} = \sin \theta$ and $\color{blue}{c} = \cos \theta$
$
\begin{align}
\text{To Prove }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) - 13 = 0\\
\equiv \text{TP }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) = 13
\end{align}
$
$$
\require{cancel}
\begin{align}
\text{LHS }
&= 3\left[(\color{red}{s}-\color{blue}{c})^2\right]^2
+ 6(\color{red}{s} +\color{blue}{c})^2
+ 4(\color{red}{s^6} + \color{blue}{c^6})
\\
&= 3(\color{limegreen}{s^2 + c^2} - 2\color{red}{s}\color{blue}{c})^2
+ 6(\color{limegreen}{s^2 + c^2} + 2\color{red}{s}\color{blue}{c})
+ 4\left[(\color{red}{s^2})^3 + (\color{blue}{c^2})^3\right]
\\
&= 3(1 - 2\color{red}{s}\color{blue}{c})^2 + 6(1+2\color{red}{s}\color{blue}{c})
+ 4(\color{limegreen}{s^2 + c^2})(\color{red}{s^4}
- \color{red}{s^2}\color{blue}{c^2}
+ \color{blue}{c^4})
\\
&= 3(1 - 4\color{red}{s}\color{blue}{c} + 4\color{red}{s^2}\color{blue}{c^2}) + 6 (1+2\color{red}{s}\color{blue}{c}) + 4(\color{red}{s^4} - \color{red}{s^2} \color{blue}{c^2}+\color{blue}{c^4})
\\
&= 3
\cancel{- 12\color{red}{s}\color{blue}{c}}
+12\color{red}{s^2}\color{blue}{c^2}
+ 6
\cancel{+ 12\color{red}{s}\color{blue}{c}} + 4\color{red}{s^4}
-4\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4}\\
&= 4\color{red}{s^4} + 8\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4} + 9\\
&= 4(\color{red}{s^4} + 2\color{red}{s^2}\color{blue}{c^2} + \color{blue}{c^4}) + 9\\
&= 4(\color{limegreen}{s^2 + c^2})^2 + 9\\
&= 4 + 9 = 13 = \text{RHS} \tag{Q.E.D.}
\end{align}
$$
Thanks to @mathlove, I found and corrected the mistake in my attempt.
$\Huge\color{lightgrey}{☺}$ Although, a quicker alternate way will always be nice.
| *
*$$
\begin{align}
(\sin x+\cos x)^2 &=\sin^2+\cos^2 +2\sin x \cos x \\
&= 1+ 2 \sin x \cos x\\
&=1+\sin 2x
\end{align}$$
*$$
\begin{align}
(\sin x+\cos x)^4 &= (\sin^2+\cos^2 +2\sin x \cos x)2 \\
&=(1+ 2 \sin x \cos x)^2\\
&=(1+\sin 2x)^2
\end{align}
$$
*$$
\begin{align}
(\sin^6 x+\cos^6 x) &=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cdot\cos^2x)\\
&=1\left((\sin^2x+\cos^2x)^2-2\sin^2x\cdot\cos^2x-\sin^2x\cdot\cos^2x\right)\\
&=(1-3\sin^2x\cdot\cos^2x)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.
step 1 $\frac {2x^2}{x+2} < x-2$
step 2 $\frac {2x^2}{x+2} - (x-2) < 0$
step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$
step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} < 0$
step 5 $\frac {2x^2 - x^2 + 4}{x+2} < 0$
step 5 $\frac {x^2 + 4}{x+2} < 0$
step 6 I used character study to get the result x > 2. But this is incorrect.
Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.
Thanks!
| $\left(\frac{x^2+4}{ x+2}\right) < 0$ implies $\left(\frac{1}{x+2}\right) < 0$ because $x^2 + 4 >0$ which implies $x+2 < 0$ and then $x < -2$ Please note $\left(\frac{1}{x+2}\right) = 0$ can never occur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
$\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$ I am clueless about this homework question. Looking at it, I see I could use the compound angle and and sum formulas, and tried using them. Unfortunately, couldn't proceed beyond that. Help?
$$\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$$
| Using Prosthaphaeresis Formulas,
$$\sin2A+\sin2B=2\sin(A+B)\cos(A-B)$$
and $$\sin(2A+2B+2C)-\sin2C=2\sin(A+B)\cos(A+B+2C)$$
Finally,
$$\cos(A-B)-\cos(A+B+2C)=2\sin(A+C)\sin(B+C)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}This is a math problem from the German Math Olympiad, but in this case I do not know where to start, probably because I do not have enough intuition regarding inequalities.
However, I tried to apply the standard AM-GM equation which didn't help, and in general I could not find a good up-/ downward estimation for the root term.
I also looked at the function plots and it seems that those three terms are very close to each other which seems to make this problem even more difficult to prove.
Any tips on how to begin? Please do not post a complete solution here.
BTW usually those problems provide an elegant solution based on non-university level knowledge, so I do not like to use an approach like a series expansion of the root term!
EDIT: Alright, the right side should've been obvious:
$$0<\frac{1}{4x^2}\Leftrightarrow \sqrt{x^2+1}<\sqrt{x^2+1+\frac{1}{4x^2}}=\sqrt{x^2+2x\frac{1}{2x}+\frac{1}{4x^2}}=x+\frac{1}{2x}$$
EDIT 2: So, I noted that squaring the left inequality (considering $x>0$ as well as $LHS>0$) gives me an inequality equivalent to $x\gt\sqrt{\frac{1}{8}}$.
Therefore, I only have to show that the Assumption $LHS\gt 0$ is implying exactly this.
I thus noted that $0=x+\frac{1}{2x}-\frac{1}{8x^3}$ has only one positive solution $\frac{\sqrt{\sqrt{3}-1}}{2}\gt\sqrt{\frac{1}{8}}$ -- but how do I show this as an inequality? I always get very confused about inequations with quadratic polynomials, especially regarding the direction of the inequality sign...
EDIT 3: Got it! So, let me state the whole proof ;)
To prove: $L:=x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}=:R,\quad x>0$
a) Consider $L\leq0$. $x>0$ implies $R>0 \Rightarrow L<R$
b) Consider $L\gt0$. We get
$$
0<x+\frac{1}{2x}-\frac{1}{8x^3} \Rightarrow 0\lt x^4+\frac{1}{2}x^2-\frac{1}{8} = (x^2)^2 + 2\frac{1}{4}x^2+\frac{1}{4^2}-\frac{1}{4^2}-\frac{1}{8} = (x^2+\frac{1}{4})^2-\frac{3}{16}\\
\Leftrightarrow \frac{\sqrt{3}}{4}\lt x^2+\frac{1}{4} \Leftrightarrow x^2 \gt \frac{\sqrt{3}-1}{4}
\Leftrightarrow x \gt \sqrt{\frac{\sqrt{3}-1}{4}} \Rightarrow x\gt \sqrt{\frac{1}{8}}
$$
From that, we get
$$
\frac{1}{8x^2}\lt 1 \Leftrightarrow \frac{1}{8x^2}\frac{1}{8x^4}\lt \frac{1}{8x^4}
\Leftrightarrow (\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt0 \\
\Leftrightarrow x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt x^2+1
$$
Noticing that $L^2=x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}$ and $R^2=x^2+1$ gives you
$$
\Leftrightarrow L^2 \lt R^2 \Leftrightarrow L\lt R
$$
...quod erat demonstrandum.
Sorry it took so long, I am not used to write that much TeX ;)
BTW, can anyone leave a comment regarding the "beauty" of my proof? I tried to compose this final version as clean as possible!
| A start: Square everything. The inequality on the right will be obvious. The one on the left is a little more unpleasant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximum value of $a+b$ given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{20}$ What is the maximum value of $a+b$ given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{20}$ here $a,b \in \mathbb{Z^+}$?
What I have gotten so far:
From the above, $\frac{a+b}{ab} = \frac{1}{20} \implies 20(a+b) = ab $
I noticed that $(a-1)(b-1) = ab - a-b+1 = k > 0$ when $a,b > 1$, because when $a, b = 1$ is clearly not a solution. $$\therefore 20(a+b) = ab = a+b - 1 + k > 0 \implies 19(a+b) = (a-1)(b-1) - 1 $$
Using AM-GM I got that $\frac{a+b}{2} \geq \sqrt{ab} \implies ab = 2(a+b) \geq 40\sqrt{ab} \implies \sqrt{ab} \geq 40 \implies ab \geq 1600$.
When $ab = 1600$, one obvious answer is $a=b=40$.
Here, I am stuck... How do I find the maximum value of $a+b$ or perhaps prove that maximum occurs when $a=b=40$?
| From $20(a+b) = ab$, we have $b = \frac{20a}{a-20}$. This is an integer, so $a - 20 \mid 20a$.
Let $g = \gcd(a, 20)$. Then $a = ga'$ and $a' - \frac{20}{g} \mid 20a'$. But $\gcd(a'-\frac{20}{g}, 20a') = 1$ (since if not we have a contradiction with $g$ being the greatest divisor of $a$ and $20$), so we must have $a'-\frac{20}{g} = 1$.
Rearranging and substituting back in, $a = 20+g$ and $b=\frac{20(20+g)}{g} = \frac{400}{g} + 20$.
Therefore the problem is to select which of the divisors of $20$, that is $\{1,2,4,5,10,20\}$, maximises $40 + g + \frac{400}{g}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac 1{1+x+y^{-1}}+\frac1{1+y+z^{-1}}+\frac1{1+z+x^{-1}}=1$ if $xyz=1$ If $x.y.z=1$ show that $\dfrac 1{1+x+y^{-1}}+\dfrac1{1+y+z^{-1}}+\dfrac1{1+z+x^{-1}}=1$
My attempt - L.H.S$=\dfrac 1{1+x+y^{-1}}+\dfrac1{1+y+z^{-1}}+\dfrac1{1+z+x^{-1}}$
$=\dfrac y{y+xy+1}+\dfrac z{z+yz+1}+\dfrac x{x+zx+1}$
$=\dfrac{yz}{yz+xyz+z}+\dfrac z{z+yz+1}+\dfrac x{x+zx+1}$
What to do next?
| yours continue
$$\frac{yz}{yz+xyz+z}+\frac z{z+yz+1}+\frac x{x+zx+1}$$
$$=\frac{yz}{yz+1+z}+\frac{xz}{xz+xyz+x}+\frac{x}{x+zx+1}$$
$$=\frac{xyz}{xyz+x+zx}+\frac{xz}{xz+1+x}+\frac{x}{x+zx+1}$$
$$=\frac{1}{1+x+zx}+\frac{xz}{xz+1+x}+\frac{x}{x+zx+1}$$
$$=\frac{1+x+zx}{1+x+zx}$$
$$=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Cosine of the sum of two solutions of trigonometric equation $a\cos \theta + b\sin \theta = c$ Question:
If $\alpha$ and $\beta$ are the solutions of $a\cos \theta + b\sin \theta = c$, then show that:
$$\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}$$
No idea how to even approach the problem. I tried taking two equations, by substituting $\alpha$ and $\beta$ in place of $\theta$ in the equation and manipulating them, but that didn't get me anywhere. Please help!
| We have $\displaystyle a\cos\theta=c-b\sin\theta,$
Squaring we get,
$\displaystyle(c-b\sin\theta)^2=(a\cos\theta)^2=a^2(1-\sin^2\theta)$
$\displaystyle\iff (a^2+b^2)\sin^2\theta-2bc\sin\theta+c^2-a^2=0$
So, $\displaystyle\sin\alpha\sin\beta=\dfrac{c^2-a^2}{a^2+b^2}$
Similarly find $\displaystyle\cos\alpha\cos\beta$ by squaring $\displaystyle b\sin\theta=c-a\cos\theta$
Finally use $\displaystyle\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Quadratic Formula for complex variable with real coefficients I have been upto proving the following
$$(\forall x\in \mathbb C, ax^2 + b x + c = 0) \land(a\neq0)\Leftrightarrow \left(x = {\frac {-b \pm \sqrt{b^{2}-4ac}} {2a}}\right)$$
Due to equality we need to prove bi implication,i.e. $a\Rightarrow b$ and $a\Leftarrow b$.It involves just simple arithmetic to go from roots to quadratic equation, but the other way is bit tricky. This is what I have done so far
$$(\forall x\in \mathbb C, ax^2 + b x + c = 0) \Rightarrow \left(x = {\frac {-b \pm \sqrt{b^{2}-4ac}} {2a}}\right)$$
$$(\forall x\in \mathbb C, c = -ax^2 - b x ) \Rightarrow (2ax = {-b \pm \sqrt{b^{2}-4ac}} )$$
$$(\forall x\in \mathbb C, c = -ax^2 - b x ) \Rightarrow (2ax + b = {\pm \sqrt{b^{2}-4ac}} )$$
Using value of $c$ we get complete square, equation looks like this
$$(\forall x\in \mathbb C, c = -ax^2 - b x ) \Rightarrow ((2ax + b) = {\pm \sqrt{(2ax+b)^2} })$$
On this forum I have found discussion on forms like
$$(x = {\pm \sqrt{x^2} })$$
For complex numbers, I had an impression that it is not true. But it raises question on the goal I am trying to proof.So, if I summarize, I need comments regarding
1-is goal legal?
2-If legal then how to deal with resulting form?
3- If not (1), then looking for some alternative approach to handle it.
thanks in advance
| The trinomial
$$ax^2+bx+c=0$$
May be solved when $a,b,c$ are complex numbers, and it works fine with real $a,b,c$, since $\Bbb R\subset\Bbb C$.
However, we'll start with the real case.
Real resolution
We start with the equation
$$ax^2+bx+c=0$$
Where $a\neq0$, and $a,b,c\in\Bbb C$.
Then, rewrite the equation
$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0$$
Then, according to the sign of $\Delta=b^2-4ac$, there are three possibilities:
*
*If $\Delta=0$, the equation amounts to
$$\left(x+\frac{b}{2a}\right)^2=0$$
It has thus one double root, $x=-\frac{b}{2a}$.
But since $\Delta=0$, we can also write
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$
*
*If $\Delta>0$, then $\frac{b^2-4ac}{4a^2}$ is positive, so it's the square of number $\frac{\sqrt{\Delta}}{2a}$, and we can write
$$\left(x+\frac{b}{2a}\right)^2-\left(\frac{\sqrt{\Delta}}{2a}\right)^2=0$$
$$\left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)\left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right)=0$$
Hence
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$
*
*If $\Delta<0$, then the equation has no real solution, since it would amount to
$$\left(x+\frac{b}{2a}\right)^2=\frac{\Delta}{4a^2}$$
Where the left-hand side is $\geq0$, whereas the right-hand side is $<0$.
However, we can still write that $\Delta$ is the square of complex number $i\sqrt{4ac-b^2}$, hence we can write
$$\left(x+\frac{b}{2a}\right)^2-\left(\frac{i\sqrt{-\Delta}}{2a}\right)^2=0$$
$$\left(x+\frac{b}{2a}+\frac{i\sqrt{-\Delta}}{2a}\right)\left(x+\frac{b}{2a}-\frac{i\sqrt{-\Delta}}{2a}\right)=0$$
$$x=\frac{-b\pm i\sqrt{-\Delta}}{2a}$$
But, with the convention that $\sqrt{t}=i\sqrt{-t}$ when $t<0$, we can again write
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$
So, in all cases, the solutions are
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$
Hence your equivalence holds.
Complex resolution
Again, we start with the equation
$$ax^2+bx+c=0$$
Where $a\neq0$, and $a,b,c\in\Bbb C$.
Then, rewrite the equation
$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0$$
Now, a complex number $z$ has always a square root (two distinct ones if $z\neq0$, see for example here). We go on with a square root $\alpha$ of $b^2-4ac$, that is, $\alpha^2=b^2-4ac$, so that the equation can be written
$$\left(x+\frac{b}{2a}\right)^2-\frac{\alpha^2}{4a^2}=0$$
It's a difference of squares, and we know that $A^2-B^2=(A+B)(A-B)$, so
$$\left(x+\frac{b}{2a}+\frac{\alpha}{2a}\right)\left(x+\frac{b}{2a}-\frac{\alpha}{2a}\right)=0$$
So the roots are
$$x=-\frac{b}{2a}\pm\frac{\alpha}{2a}$$
Or if you prefer
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Important note: there is no canonical way to write that $\sqrt{t}$ is a specific square root of $t\in\Bbb C$, hence this symbol is meaningless. However, since it appears with a $\pm$, we will always work with both roots, so using the symbol here is harmless. Nevertheless, it should be avoided.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/928516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Polya urn scheme probability calculation Consider an urn with $b$ black and $r$ red balls. In each step we randomly choose a ball. Then we put it back in and put $c$ balls of the same color in the urn. Let us denote $B_m$ as the event that $m^{\text{th}}$ draw has resulted in black. Then $$\Pr(B_m\cap B_n)=\frac{b(b+c)}{(b+r)(b+r+c)},\forall m \text{ such that }\;m<n$$
How to do this? I tried induction. For $m=1$ this is trivial. Now let us assume it to be true for some $m$. How to extend it to $m+1$? Can someone help? Some other solution is also welcomed. I think induction is not the best way. I think it all depends on breaking it into two disjoint events and use conditional probability. But I just can't do it. Thanks.
| Firstly, $$P(B_m \cap B_n) = P(B_n \mid B_m)\,P(B_m).$$
Now use induction to establish the value of each factor on the RHS.
Our first claim is that for all $m \geq 1$: $$P(B_m) = \dfrac{b}{b+r}$$
Initial case, $m=1$: $$P(B_1) = \dfrac{b}{b+r}$$ is obviously true since there are $b$ chances to choose a black ball out of $b+r$ balls.
Now assume the claim is true for $m=k$ for some $k \geq 1$. Then, conditioning on $B_1$,
\begin{eqnarray*}
P(B_{k+1}) &=& P(B_{k+1} \mid B_1)\,P(B_1) \,+\, P(B_{k+1} \mid B_1^c)\,P(B_1^c).
\end{eqnarray*}
Now,
\begin{eqnarray*}
P(B_{k+1} \mid B_1) &=& P(B_k) \qquad\mbox{starting with $b+c$ black and $r$ red balls} \\
&=& \dfrac{b+c}{b+r+c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
Also,
\begin{eqnarray*}
P(B_{k+1} \mid B_1^c) &=& P(B_k) \qquad\mbox{starting with $b$ black and $r+c$ red balls} \\
&=& \dfrac{b}{b+r+c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
Therefore, we have,
\begin{eqnarray*}
P(B_{k+1}) &=& \dfrac{b+c}{b+r+c} \dfrac{b}{b+r} + \dfrac{b}{b+r+c} \dfrac{r}{b+r} \\
&=& \dfrac{b}{b+r}.
\end{eqnarray*}
This proves the case for $m=k+1$ and the inductive proof of the first claim is done.
Our second claim is that for all $n \gt 1$ and $m: 1 \leq m < n$: $$P(B_n \mid B_m) = \dfrac{b+c}{b+r+c}.$$
We use induction on $m$. Initial case is for any $n \gt 1$ and with $m=1$: In proving the first claim we have already shown that $$P(B_n \mid B_1) = \dfrac{b+c}{b+r+c}.$$
Now assume that our second claim is true for any $n \gt 1$ and some $k: 1 \leq k < n$. Conditioning on $B_1$,
$$P(B_n \mid B_{k+1}) = P(B_n \mid B_1 \cap B_{k+1})\,P(B_1 \mid B_{k+1}) \,+\, P(B_n \mid B_1^c \cap B_{k+1})\,P(B_1^c \mid B_{k+1}).$$
Evaluating the four probabilities on the RHS,
\begin{eqnarray*}
P(B_n \mid B_1 \cap B_{k+1}) &=& P(B_{n-1} \mid B_k) \qquad\mbox{starting with $b+c$ black and $r$ red balls} \\
&=& \dfrac{b+2c}{b+r+2c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
$\\$
\begin{eqnarray*}
P(B_n \mid B_1^c \cap B_{k+1}) &=& P(B_{n-1} \mid B_k) \qquad\mbox{starting with $b$ black and $r+c$ red balls} \\
&=& \dfrac{b+c}{b+r+2c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
$\\$
\begin{eqnarray*}
P(B_1 \mid B_{k+1}) &=& \dfrac{P(B_{k+1} \mid B_1) P(B_1)}{P(B_{k+1})} \\
&& \\
&=& \dfrac{b+c}{b+r+c} \dfrac{b}{b+r} \bigg/ \dfrac{b}{b+r} \\
&& \qquad\mbox{first factor is by our initial case, the second and third terms by our first claim} \\
&& \\
&=& \dfrac{b+c}{b+r+c}.
\end{eqnarray*}
$\\$
\begin{eqnarray*}
P(B_1^c \mid B_{k+1}) &=& \dfrac{P(B_{k+1} \mid B_1^c) P(B_1^c)}{P(B_{k+1})} \\
&& \\
&=& \dfrac{b}{b+r+c} \dfrac{r}{b+r} \bigg/ \dfrac{b}{b+r} \\
&& \qquad\mbox{by similar reasoning as previously} \\
&& \\
&=& \dfrac{r}{b+r+c}.
&&
\end{eqnarray*}
$\\$ $\\$
Putting these together, we have,
\begin{eqnarray*}
P(B_n \mid B_{k+1}) &=& \dfrac{b+2c}{b+r+2c} \dfrac{b+c}{b+r+c} + \dfrac{b+c}{b+r+2c} \dfrac{r}{b+r+c} \\
&& \\
&=& \dfrac{b+c}{b+r+c}.
\end{eqnarray*}
Thus the claim is true for $m=k+1$ and our induction is complete for our second claim.
Therefore, as required, we have $$P(B_m \cap B_n) = P(B_n \mid B_m)P(B_m) = \dfrac{b(b+c)}{(b+r)(b+r+c)}.$$
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
Least squares fitting using cosine function? Hello I am trying to fit a harmonic of the form $$y = b + c\cos(x)$$ to four data points (0,6.1) (.5,5.4) (1,3.9) (1.5,1.6) using least squares for homework. I know that the error $= Y_i - f(x_i)$ but am pretty confused with the partial derivatives and linear algebra part of this question. Apparently the professor wants only $y=b+c\cos(x)$ and not $y=b+c\cos(x)+d\sin(x)$
I get two equations from taking the derivative of $E = \sum_{i=0}^n [y_i - (b+c\cos(x_i))]^2$, with respects to $b$ and $c$.
$$ \frac{dE}{db} = bn + \sum_{i=0}^n c\cos(x_i) = \sum_{i=0}^n y_i $$ and
$$ \frac{dE}{dc} = b\cos(x_i) + \sum_{i=0}^n c\cos^2(x_i) = \sum_{i=0}^n \cos(x_i) y_i $$
Putting this into matrix form gives me:
$$ \left[ \begin{array}{cc|c}
n & \sum_{i=0}^n (x_i) & \sum_{i=0}^n (y_i) \\
\sum_{i=0}^n \cos(x_i) & \sum_{i=0}^n \cos^2(x_i) & \sum_{i=0}^n \cos(x_i)y_i \\
\end{array}\right] $$
which lands me with $$ \left[\begin{array}{cc|c}
4 & 3 & 17 \\
2.48862 & 2.067077 & 13.05931
\end{array}\right] $$
Sorry for the terrible formatting | is meant to make this an augmented matrix (the four data points are included at the top). Anyways
I'm not really sure how to solve this matrix so I assume I did something wrong with the partial derivatives. Is there anyone out there that could be so helpful as to let me know what I did wrong or any hints on how to solve this system if it is indeed solvable?
| The solution of @user147263 relies on calculus. Another option is to exploit linear algebra.
Form linear system
Start with a series of data points $(x_{k}, y_{k})_{k=1}^{m}$, and the trial function
$$
y(x) = c_{1} + c_{2} \cos x,
$$
We have the linear system
$$
\begin{align}
A c & = y \\
\left[
\begin{array}{cc}
1 & \cos x_{1} \\
1 & \cos x_{2} \\
\vdots & \vdots\\
1 & \cos x_{m}
\end{array}
\right]
\left[
\begin{array}{c}
c_{1} \\
c_{2}
\end{array}
\right]
& =
\left[
\begin{array}{c}
y_{1} \\
y_{2} \\
\vdots \\
y_{m}
\end{array}
\right]
\end{align}
$$
Find the solution vector $c$ which minimizes the sum of the squares of the residuals:
$$
r^{2}(c) = \lVert A c - y \rVert_{2}^{2} = \sum_{k=1}^{m} (y_{k} - c_{1} - c_{2} \cos x_{k})^2.
$$
Normal equations: Form the normal equations
$$
A^{T}A c = A^{T}y.
$$
Solve linear system
The solution vector is
$$
c = (A^{T}A)^{-1} A^{T}y.
$$
Defining the $m-$vector
${\mathbf{1}} =
\left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \right]$,
and the the $m-$vector
$\eta =
\left[ \begin{array}{c} \cos x_{1} \\ \cos x_{2} \\ \cos x_{3} \\ \cos x_{4} \end{array} \right]$, the system matrix has the column vector form
$A = \left[ \begin{array}{c} \mathbf{1} & \eta \end{array} \right]$, and the product matrix is
$$
A^{T} A =
\left[
\begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot \eta \\
\eta \cdot \mathbf{1} & \eta \cdot \eta
\end{array}
\right] =
%
\left[
\begin{array}{cl}
m & \sum_{k=1}^{m} \cos x_{k} \\
\sum_{k=1}^{m} \cos x_{k} & \sum_{k=1}^{m} \cos^{2} x_{k}
\end{array}
\right]
$$
with inverse
$$
( A^{T} A )^{-1} =
(\det (A^{T}A))^{-1}
\left[
\begin{array}{rc}
\eta \cdot \eta & -\mathbf{1} \cdot \eta \\
- \mathbf{1} \cdot \eta & m
\end{array}
\right]
$$
where the determinant is
$$
\det (A^{T}A) = m (\eta \cdot \eta) - (\mathbf{1}\cdot \eta)^{2}.
$$
For the data set above, $x =
\left[
\begin{array}{l}
0 \\
0.5 \\
1 \\
1.5
\end{array}
\right]$, $\ $
$\eta =
\left[
\begin{array}{l}
1 \\
0.877583 \\
0.540302 \\
0.0707372
\end{array}
\right]$, $\ $
$y =
\left[
\begin{array}{c}
6.1 \\
5.4 \\
3.9 \\
1.6
\end{array}
\right]$, $\ $
and $\mathbf{1} =
\left[
\begin{array}{c}
1 \\
1 \\
1 \\
1
\end{array}
\right].$
The inner products are
$$
\begin{align}
\mathbf{1} \cdot \mathbf{1} &= 4, \\
\mathbf{1} \cdot \eta = \eta \cdot \mathbf{1} &= 2.48862, \\
\eta \cdot \eta & = 2.06708.
\end{align}
$$
The matrices are
$$
A =
\left[
\begin{array}{rl}
1 & 1 \\
1 & 0.877583 \\
1 & 0.540302 \\
1 & 0.0707372
\end{array}
\right], \quad
%
A^{T}A =
\left[
\begin{array}{rl}
4 & 3 \\
3 & 3.5
\end{array}
\right], \quad
%
(A^{T}A)^{-1} =
\det (A^{T}A)^{-1}
\left[
\begin{array}{rr}
2.06708 & -2.48862 \\
-2.48862 & 4
\end{array}
\right].
$$
The determinant is $\det (A^{T}A) = 2.07509$. The solution vector is
$$
c =
\left[
\begin{array}{c}
c_{1} \\
c_{2}
\end{array}
\right]
=
\left[
\begin{array}{r}
1.27258 \\
4.78565
\end{array}
\right]
$$
Check result
The solution curve is plotted against the data points.
A contour plot showing $r^{2}$ as a function of the solution vector $c$ with a white dot marking the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proof of $\cos \theta+\cos 2\theta+\cos 3\theta+\cdots+\cos n\theta=\frac{\sin\frac12n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta$
State the sum of the series $z+z^2+z^3+\cdots+z^n$, for $z\neq1$.
By letting $z=\cos\theta+i\sin\theta$, show that
$$\cos \theta+\cos 2\theta+\cos 3\theta+\cdots+\cos n\theta=\frac{\sin\frac12n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta$$
Where $\sin\frac12\theta\neq0$.
I know the first part, The second part Im kind of stuck in showing that
My Attempt:
$$\begin{align}\Re{(\cos\theta+i\sin\theta)+(\cos\theta+i\sin\theta)^2+\ldots+(\cos\theta+i\sin\theta)^n}\end{align}$$
I realized that this is a Geometric Progression, so its in the form:
$a+ar+ar^2+....+ar^n$ , where $a=(\cos\theta+i\sin\theta)$ and $r=(\cos\theta+i\sin\theta)$
So I will apply the formula for the Sum of a G.P to my problem.
$$\begin{align}\Re \frac{(\cos\theta+i\sin\theta)(1-(\cos\theta+i\sin\theta)^n)}{1-(\cos\theta+i\sin\theta)}\end{align}$$
I applied the De Movire Theorem and simplified as follows:
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\frac{(1+(cos\theta+isin\theta))}{(1+(cos\theta+isin\theta))}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos^2\theta+2icos\theta sin\theta-sin^2\theta-cos\theta cos(n+1)\theta- isin \theta cos(n+1)\theta+i cos\theta sin(n+1)\theta-sin\theta sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(cos\theta+isin \theta)+(i cos\theta -sin \theta) sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta}{2sin^2\theta-2i cos\theta sin\theta}\end{align}$$
$$\begin{align}\Re \frac{[\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta(sin\theta-i cos\theta)} \frac{(sin\theta+i cos\theta)}{(sin\theta+i cos\theta)}\end{align}$$
$$\begin{align}\frac{[-cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta} \frac{(sin\theta+i cos\theta)}{1}\end{align}$$
$$\begin{align}\frac{[-cos(n+1)\theta(sin\theta+cos\theta sin\theta - cos\theta sin \theta)+(sin\theta-cos^2\theta -sin^2 \theta) sin(n+1)\theta]}{2sin\theta} \end{align}$$
$$\begin{align}\frac{[-cos(n+1)sin\theta+(sin\theta-1) sin(n+1)\theta]}{2sin\theta} \end{align}$$
I have no idea right now where I am taking this, I just dont know what the next step I should take. Please dont send me the solution (at least yet). Can anyone give me a hint (a little boost to my little mind) as to what I should do next, (make sure its a small hint, Don't give me the full next step) Just the help in order for me to construct the next step and carry on.
| Here are the main steps.
You may write
$$
\begin{align}
\sum_{k=1}^{n} \cos (k\theta)&=\Re \sum_{k=1}^{n} e^{ik\theta}\\\\
&=\Re\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\
&=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\
&=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\
&=\Re\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\Re\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos ((n+1)\theta/2).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Does $\sum_1^\infty \frac{n}{n^2 + 4}$ converge or diverge? Is my solution correct? Does $\sum_1^\infty \frac{n}{n^2 + 4}$ converge or diverge?
I am confused because my friend insists the series converges conditionally. I think the series diverges. Here is my process and solution:
$\sum_1^\infty \frac{n}{n^2 + 4}$
Limit Comparison Test:
Let $\sum_1^\infty \frac{n}{n^2 + 4} = \sum_1^\infty a_n$
Let $\sum_1^\infty b_n = \sum_1^\infty \frac{1}{n}$
Since $$\lim \limits_{n \to \infty} | \frac{a_n}{b_n}| = \lim \limits_{n \to \infty} | \frac{\frac{n}{n^2 + 4}}{\frac{1}{n}}| = \lim \limits_{n \to \infty} \frac{n^2}{n^2 + 4} = \lim \limits_{n \to \infty} \frac{n^2}{n^2(1 + \frac{4}{n^2})} = \lim \limits_{n \to \infty} \frac{1}{1 + \frac{4}{n^2}} = 1$$
and $1 > 0$ and $1$ is a finite number,
we can say that the behavior of $\sum_1^\infty a_n$ is the same as the behavior of $\sum_1^\infty b_n$.
Since $\sum_1^\infty b_n$ diverges ( $\sum_1^\infty \frac{1}{n}$ diverges by p-series, $p=1$), $\sum_1^\infty a_n$ diverges.
So by the Limit Comparison Test, $\sum_1^\infty \frac{n}{n^2 + 4}$ diverges.
Right?
Also, this problem was an exercise in the Ratio/Root Test section. Both tests, however seem to fail. Can the convergence be solved with the Root Test or the Ratio Test?
| Direct comparison is a little easier here:
$$\sum_{n=1}^\infty \frac{n}{n^2+4} \geq \sum_{n=2}^\infty \frac{n}{n^2+4} \geq \sum_{n=2}^\infty \frac{n}{n^2+n^2} = \frac{1}{2}\sum_{n=2}^\infty \frac{1}{n} = \infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Showing linear dependence in a matrix Let
$v_1=\begin{bmatrix}
1\\
-3\\
2\\
\end{bmatrix}
$
Let $V_2=\begin{bmatrix}
-3\\
9\\
-6\\
\end{bmatrix}
$
Let $V_3=\begin{bmatrix}
5\\
-7\\
h\\
\end{bmatrix}
$
For what value of h is the span {v1,v2,v3} linear dependent.
This be my work.
Let $A= \begin{bmatrix}
1 & -3 & 5& 0 \\
-3&9& -7& 0 \\
2& -6 & h& 0 \\
\end{bmatrix}
$
then doing 3r1+r3 and -2r1+r2
Let $A= \begin{bmatrix}
1 & -3 & 5& 0 \\
0&0& 8& 0 \\
0&0 & h-10& 0 \\
\end{bmatrix}
$
| You must have zeros in last row, so $h+30=0$ and $h=-30$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Equation of the line passing through the intersection of two lines and is parallel to another line. The Question is :
Find the equation of the line through the intersection of the lines $3x+2y−8=0,5x−11y+1=0$ and parallel to the line $6x+13y=25$
Here is how I did it..
$L_1 = 3x + 2y -8 = 0$
$L_2 = 5x -11y +1 = 0$
$L_3= ?$
$L_4 = 6x + 13y -25= 0$
I found the point of intersection : $(-2, -1)$
Using formula: $L_1 + kL_2$ = 0
$(3x + 2y -8) + k(5x -11y +1)=0$
$(3(2) + 2(1) -88) + k(5(2) -11(1) +1) = 0$
$(6 +2 -8) + k(10 -11 +1) =0$
$8-8 + k(11-11) =0$
$0 +k(0) = 0$
What's wrong ?
| What;'s wrong is that $(-2, -1)$ is not on the line
$$
5x - 11y + 1 = 0
$$
because
$$
5(-2) - 11(-1) + 1 = -10 + 11 + 1 = 2.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$ One of the possible ways of computing the series is to obtain the generating function, but
this might be a tedious, hard work, pretty hard to obtain. What would you propose then?
$$\sum_{n=1}^{\infty} \frac{H_{
n}}{2^nn^4}$$
| Here is a magical solution:
We proved here
\begin{align}
I&=\int_0^1\frac{\ln^2(1-x)}{1-x}\left(\ln^2(1+x)-\ln^2(2)\right)\ dx\\
&=\small{\boxed{\frac18\zeta(5)-\frac12\ln2\zeta(4)+2\ln^22\zeta(3)-\frac23\ln^32\zeta(2)-2\zeta(2)\zeta(3)+\frac1{10}\ln^52+4\operatorname{Li}_5\left(\frac12\right)\quad}}\tag{1}
\end{align}
On the other hand and by integration by parts, we have
\begin{align}
I&=\frac23\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{1+x}\ dx\overset{\color{red}{1-x\ \mapsto\ x}}{=}\frac13\int_0^1\frac{\ln^3x\ln(2-x)}{1-x/2}\ dx\\
&=\frac{\ln2}{3}\int_0^1\frac{\ln^3x}{1-x/2}\ dx+\frac13\int_0^1\frac{\ln^3x\ln(1-x/2)}{1-x/2}\ dx\\
&=\frac{\ln2}{3}\sum_{n=1}^\infty\frac{1}{2^{n-1}}\int_0^1x^{n-1}\ln^3x\ dx-\frac13\sum_{n=1}^\infty\frac{H_n}{2^n}\int_0^1x^n\ln^3x\ dx\\
&=\frac{\ln2}{3}\sum_{n=1}^\infty\frac{1}{2^{n-1}}\left(-\frac{6}{n^4}\right)-\frac13\sum_{n=1}^\infty\frac{H_n}{2^n}\left(-\frac{6}{(n+1)^4}\right)\\
&=-4\ln2\sum_{n=1}^\infty\frac{1}{n^42^n}+2\sum_{n=1}^\infty\frac{H_n}{(n+1)^42^n}\\
&=\boxed{-4\ln2\operatorname{Li}_4\left(\frac12\right)+4\sum_{n=1}^\infty\frac{H_n}{n^42^n}-4\operatorname{Li}_5\left(\frac12\right)}\tag{2}
\end{align}
From $(1)$ and $(2)$, we get
\begin{align}
\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2)
+\frac12\ln^22\zeta(3)\\
&\quad-\frac18\ln2\zeta(4)-
\frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52
\end{align}
Note: Full credit goes to Cornel for proposing such amazing problem in $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/944065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 1,
"answer_id": 0
} |
Proving the summation formula using induction: $\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$ I am trying to prove the summation formula using induction:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$
So far I have...
Base case:
*
*Let n=1 and test
$\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
$\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
$\frac{1}{2} = \frac{1}{2}$
*
*True for n=1
Induction Hypothesis:
*
*Assume the statement is true for the n-th case
$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
Inductive Step:
*
*Prove, using the Inductive Hypothesis as a premise, that
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$
To prove
$$ 1-\frac{1}{n+2} = \frac{n+1}{n+2} $$
Multiply both sides by $n+2$ to get an equivalent expression.
$$ (1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2) $$
$$ n+1=n+2−1 $$
Does this all make sense? How can this be improved upon?
| Assume for $n-1$ then you will be very clear.
I am giving the final step only...
$$\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n-1}\frac{1}{k(k+1)}+\frac 1 {n(n+1)}=1-\frac 1 n+\frac 1 {n(n+1)}=1+\frac {-n-1+1}{n(n+1)}\\=1-\frac {1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to show $\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \ dx$? $$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \ dx.$$
Someone please show that this equation is correct !?
| It is known that
$$
\arcsin(x)^2=\frac{1}{2}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\binom{2n}{n}}
$$
see Companion to Concrete Mathematics - Mathematical Techniques and Various Applications Z. A. Melzak p. 108
Then
$$
\begin{aligned}
\int_{0}^{1/2}\frac{4\arcsin(x)^2}{x}
&=\int_{0}^{1/2}\frac{2}{x}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\binom{2n}{n}}dx\\
&=\sum_{n=1}^\infty\frac{2^{2n+1}}{n^2\binom{2n}{n}} \int_{0}^{1/2}x^{2n-1}dx\\
&=\sum_{n=1}^\infty\frac{2^{2n+1}}{n^2\binom{2n}{n}} \frac{\left(\frac{1}{2}\right)^{2n}}{2n}\\
&=\sum_{n=1}^\infty\frac{1}{n^3\binom{2n}{n}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find the last non-zero digit of $50!$ A week ago i made a similar question but nobody help me, i´ve been trying but i still don't get it.
I want to know how to find the last non-zero digit of $50!$.
my try:
First i have to know how much Zeros $50!$ has so i did this:
$$E_{5}50 = \sum _{1\leq k <n} \Bigg[\frac{50}{5^{k}}\Bigg] =\Bigg[\frac{50}{5}\Bigg] + \Bigg[\frac{50 }{25}\Bigg] = 12$$
So $50!$ has $12$ zeros which means that the last digit of $\quad\frac{50!}{10^{12}}\quad$ is the number that i´m looking for.
so if $x = \frac{50!}{10^{12}}$ i need to find $x (mod 10)$ to get it but this is such a big number and i still don't know how to reduce it.
Thanks in advance!!
| My Solution:: calculation of last non zero digit in $50!$
Now first we will calculate number of zeros at the end of $50!$, which is obtained
when we multiplied $2$ and $5$. So we will calculate no. of $2$ and $5$ in $50!$
$\displaystyle 50! = \lfloor \frac{50}{2}\rfloor +\lfloor \frac{50}{2^2} \rfloor+\lfloor \frac{50}{2^3}\rfloor+\lfloor \frac{50}{2^4}\rfloor+\lfloor \frac{50}{2^5}\rfloor+........... = 25+12+6+3+1 = 47$
$\displaystyle 50! = \lfloor \frac{50}{5}\rfloor+\lfloor \frac{50}{5^2}\rfloor+..... = 10+2 = 12$
So Total no. of zeros at the end of $50! = 12$
Now $\displaystyle 50! = \lfloor \frac{50}{3}\rfloor+\lfloor\frac{50}{3^2}\rfloor+\frac{50}{3^3}\rfloor+..... = 16+5+1 = 22$
Now we can write $\displaystyle 50! = 1\times 3^{22}\times 7^8\times 11^4 \times 13^3 \times 17 \times 19\times 23\times 29\times 31 \times 17\times 37 \times 41 \times 43 \times 23 \times 47 \times 10^{12}\times 2^{35} = \bf{last digit\; (6)} = 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\int_0^\pi\frac{\cos x \cos 4x}{(2-\cos x)^2}dx=\frac{\pi}{9} (2160 - 1247\sqrt{3})$ Prove that
$$\int_0^\pi\frac{\cos x \cos 4x}{(2-\cos x)^2}dx=\frac{\pi}{9} (2160 - 1247\sqrt{3})$$
I tried to use Weierstrass substitution but the term $\cos 4x$ made horrible algebraic-forms since $\cos 4x = \sin^4 x + \cos^4 x - 6\sin^2 x \cos^2 x$. My friend suggests me use a contour integration method but I am not familiar with that method. Any idea? Any help would be appreciated. Thanks in advance.
| I would like to try to tackle the problem with elementary integration techniques though it is a bit tedious. First of all, using double angle formula, we get
$$\cos x\cos 4x= 8 \cos ^{5} x-8 \cos ^{3} x+\cos x$$ and hence
$$
I=\int_{0}^{\pi} \frac{8 \cos ^{5} x-8 \cos ^{3} x+\cos x}{(2-\cos x)^{2}} d x
$$
By division, we reduce the integrand to a proper fraction,
$$
\begin{aligned}
I \displaystyle &=\int_{0}^{\pi}\left[8 \cos ^{3} x+32 \cos ^{2} x+88 \cos x+224\right] d x + \displaystyle \int_{0}^{\pi} \frac{545 \cos x-896}{(\cos x-2)^2} d x \\&= 240 \pi+545\underbrace{\int_{0}^{\pi} \frac{d x}{\cos x-2}}_{J}+194 \underbrace{\int_{0}^{\pi} \frac{d x}{(\cos x-2)^{2}}}_{K}
\end{aligned}
$$
For any $a>1,$ let’s define a definite integral$$
\begin{aligned}
\displaystyle I(a):&=\int_{0}^{\pi} \frac{d x}{a-\cos x} \stackrel{x\mapsto \pi-x}{=} \int_{0}^{\pi} \frac{d x}{a+\cos x}\\
\\2 I(a)&=\int_{0}^{\pi}\left(\frac{1}{a-\cos x}+\frac{1}{a+\cos x}\right) d x\\
&=4 a \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2}-\cos ^{2} x} d x\\
&=4 a \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{a^{2} \sec ^{2} x-1} d x\\
&=4 a \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{a^{2} \tan ^{2} x+\left(a^{2}-1\right)}\\
&=\frac{4}{\sqrt{a^{2}-1}}\left[\tan ^{-1}\left(\frac{a \tan x}{\sqrt{a^{2}-1}}\right)\right]_{0}^{\frac{\pi}{2}}\\
&=\frac{2\pi}{\sqrt{a^{2}-1}}\\
\therefore \quad \int_{0}^{\pi} \frac{d x}{a-\cos x}&=\frac{\pi}{\sqrt{a^{2}-1}}
\end{aligned}
$$
In particular, $$J=-I(2)=-\frac{\pi}{\sqrt{3}}$$
For $K$, we just differentiate $I(a)$ w.r.t. $a$.
$$
\begin{aligned}
I^{\prime}(a)=-\frac{\pi}{\left(a^{2}-1\right)^{\frac{3}{2}}}
&\Rightarrow \int_0^{\pi}\frac{1}{(a-\cos x)^{2}} d x=\frac{\pi a}{\left(a^{2}-1\right)^{\frac{3}{2}}} \\
\therefore K=\int_{0}^{\pi} \frac{1}{(2-\cos x)^{2}} d x&=\frac{2 \pi}{3 \sqrt{3}} \\
\end{aligned}
$$
Now we can conclude that
$$I =240 \pi+545\left(-\frac{\pi}{\sqrt{3}}\right)+194\left(\frac{2 \pi}{3 \sqrt{3}}\right) =\frac{\pi}{9}(2160 - 1247 \sqrt{3})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
} |
(complex variables) Expand $\frac{2z+3}{1+z}$ in a power series of $z-1$ and comment on its convergence Question:
Expand $\frac{2z+3}{1+z}$ in a power series of $z-1$. What can we say about its convergence?
Attempt:
First, notice $ \frac{2z+3}{1+z} = \frac{2z+3}{1} \frac{1}{1+z}$.
Let $w = 1 -z$. Using substitution, I can get
$$ \frac{5}{2} - \sum_{n=1}^\infty (z-1)^n (-1)^n 2^{-n}.$$
So, it converges when $ \left| z - 1 \right| < 1$.
However, if I plug this into wolfram alpha, I get:
$$ \frac{5}{2} + \sum_{n=1}^\infty (z-1)^n (-1)^n 2^{-1-n}.$$
Notice that in the second case, the summation is added to the fraction and there is an additional $\frac{1}{2}$ in the summation. Can anyone tell me where I'm going wrong?
| Let $w=z-1$ so
$$\frac{2z+3}{1+z}=\frac{2w+5}{w+2}=2+\frac{1}{w+2}=2+\frac12.\frac{1}{1+\frac{w}{2}}=2+\frac12\sum_{n=0}^\infty(-1)^n\Big(\frac{w}{2}\Big)^n$$
and this series with convergence redius 2 expand as $$\frac52-\frac{w}{4}+\frac{w^2}{8}-\frac{w^3}{16}+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the eigenvalues of tridiagonal Toeplitz matrix? Assume the tridiagonal matrix $T$ is in this form:
$$
T = \begin{bmatrix}
a & c & & & &\\
b & a & c & & &\\
& b & a & c & &\\
& & &\ddots & &\\
& & & b & a & c\\
& & & & b & a\\
\end{bmatrix}
$$
we must show that its eigenvalues are of the form
$$a + 2 \sqrt{bc} \, \cos \left( \frac{k \pi}{n+1} \right)$$
where $$a=qh^2−1, ~~ b=1- \frac{ph}{2}, ~~ c =1+\frac{ph}{2} , ~~q \leq 0.$$
| Once you get a recursion formula there are at least three different ways to
solve the problem. Using the Z transform, using Chebyshev polynomials, and using the traditional (which I include here) method to solve difference equations.
This document illustrate this. From it, I will
rewrite here one of the methods.
The eigenvalues of $U$ are roots of the characteristic polynomial That is
\begin{eqnarray*}
p(\lambda) = \det( U - \lambda I) = 0.
\end{eqnarray*}
That is,
\begin{eqnarray*}
p_n(\lambda) = \det \begin{pmatrix}
a-\lambda & c & \\
b & a-\lambda & c \\
& b & a-\lambda & c \\
& & \ddots &\ddots &\ddots \\
& & & \ddots &\ddots &\ddots \\
& & & & b & a-\lambda & c \\
& & & & & b & a-\lambda
\end{pmatrix} = 0
\end{eqnarray*}
where $n$ is the number of rows (or columns) of $U$. At the moment
we assume $n \ge 2$. We expand the determinant through the first row to find
\begin{eqnarray*}
p(\lambda) = (a- \lambda) p_{n-1} - c \det B
\end{eqnarray*}
where $B$ is the matrix
\begin{eqnarray*}
\begin{pmatrix}
b & c & \\
& a-\lambda & c \\
& b & a-\lambda & c \\
& & \ddots &\ddots &\ddots \\
& & & \ddots &\ddots &\ddots \\
& & & & b & a-\lambda & c \\
& & & & & b & a-\lambda
\end{pmatrix}
\end{eqnarray*}
Then we expand the determinant of $B$ through the first column to find
\begin{eqnarray*}
p_{n}(\lambda) = (a- \lambda) p_{n-1} - bc p_{n-2} ,
\end{eqnarray*}
and
\begin{eqnarray}
p_{n}(\lambda) - (a- \lambda) p_{n-1} - bc p_{n-2} = 0.
(1)
\end{eqnarray}
This is a difference equation.
We want to find an analytic solution for the function $p_n(\lambda)$ which is
valid for all $n$. This equation needs two initial conditions due to the fact
that $p_n$ depends on two previous instances $p_{n-1}$ and $p_{n-2}$.
That is, for $n=1$, $p_1(\lambda)=a-\lambda$ and for $n=2$
\begin{eqnarray}
p_2(\lambda)= \det
\begin{pmatrix}
a - \lambda & c \\
b & a-\lambda
\end{pmatrix}
= (a-\lambda)^2 - bc,
(2)
\end{eqnarray}
This could be the initial conditions. All other values of $p_n(\lambda)$
could be found starting with these two initial conditions on the
recursive equation (1). However, and to simplify operations
we can state simpler initial conditions. Although there are no matrices
with $n=0$ rows we could define $p_0(\lambda)=1$, and keep the initial
condition $p_1(\lambda)=a - \lambda$. Using the recursion (1)
we find that
\begin{eqnarray*}
p_2(\lambda) = (a - \lambda)^2- bc
\end{eqnarray*}
which is exactly equation (2). The solution of $p_n(\lambda)$ for higher $n$
values will not change.
We then have the initial conditions
\begin{eqnarray}
p_0(\lambda)=1 \quad , \quad p_1(\lambda)=a -\lambda.
(3)
\end{eqnarray}
We apply the theory about difference equations to obtain the solution of our problem.
Returning to equation (1) and replacing $p_i(\lambda)$ by $t^i$
we find the characteristic equation
\begin{eqnarray*}
t^n - (a- \lambda) t^{n-1} - bc t^{n-2} = t^{n-2}[t^2 - (a - \lambda) t - bc] = 0.
\end{eqnarray*}
That is, we need to solve the quadratic equation
\begin{eqnarray}
t^2 - (a - \lambda) t - bc = 0.
(4)
\end{eqnarray}
The two solutions are
\begin{eqnarray}
t_{\pm} = \frac{(a - \lambda) \pm \sqrt{(a-\lambda)^2 - 4 bc}}{2}.
(5)
\end{eqnarray}
The roots could be equal or different. We consider two cases;
1.
No repeated roots:
The general solution $p_n(\lambda)$ can be written as
\begin{eqnarray}
p_n(\lambda)= c_1 t^n_+ + c_2 t^n_-.
\label{pnl}
\end{eqnarray}
From the initial conditions (3) we have
\begin{eqnarray}
1 &=& c_1 + c_2 \nonumber \\
\quad \quad \quad (5a) \\
a - \lambda &=& c_1 t_+ + c_2 t_- . \nonumber
\end{eqnarray}
This system has solution if its determinant does not vanish. That is, if
\begin{eqnarray*}
\Delta = \det
\begin{pmatrix}
1 & 1 \\
t_+ & t_- \\
\end{pmatrix}
= t_- - t_+ = -\sqrt{(a-\lambda)^2 - 4 bc} \ne 0.
\end{eqnarray*}
Please observe that this is consistent with the fact that $t_+ \ne t_-$.
We have then that $(a-\lambda)^2 - 4 bc \ne 0$. To solve system (5a)
we observe that $c_2=1-c_1$ and
\begin{eqnarray*}
a-\lambda = c_1 t_+ + (1-c_1) t_- = c_1(t_+ - t_-) + t_-,
\end{eqnarray*}
That is,
\begin{eqnarray}
c_1 = \frac{(a-\lambda) - t_-}{t_+ - t_-}
\quad , \quad
c_2 = 1-c_1 = \frac{t_+ - (a - \lambda)}{t_+-t_-},
\label{ec1}
\end{eqnarray}
so that the solution for $p_n(\lambda$) is given by
\begin{eqnarray}
p_n(\lambda) &=& t^n_+ \frac{(a-\lambda) - t_-}{t_+ - t_-} +
t^n_- \frac{t_+ - (a - \lambda)}{t_+-t_-}.
\end{eqnarray}
Now, since the sum of the roots of equation (5)
is given by $t_+ + t_-=(a-\lambda)$ we write
\begin{eqnarray*}
p_n(\lambda) = t_+^n \frac{t_+ + t_- - t_-}{t_+ - t_-}
+ t_-^n \frac{t_+ - t_+ - t_-}{t_+ - t_-}
&=& \frac{t_+^{n+1} - t_-^{n+1}}{t_+ - t_-}.
\end{eqnarray*}
This is the analytic solution of (1) with initial
conditions (3) . Since we need to know $\lambda$ such
that $p_n(\lambda=0)$ we have that the eigenvalues $\lambda$ satisfy
\begin{eqnarray}
t_+^{n+1} = t_-^{n+1}.
(6)
\end{eqnarray}
Note that equation $t_+^{n+1} - t_-^{n+1}=0$ is of degree $n+1$ in $\lambda$ since
$t_+^{n+1} - t_-^{n+1} = (t_+ - t_-) p_n(\lambda)$. We introduced a new root
to the original problem. This root satisfy the equation $t_+ - t_-=0$.
We are assuming $t_+ - t_- \ne 0$, $\lambda$ can not be a root
of $\Delta = 0$. Please observe that $\Delta=0$ implies
$(a-\lambda)^2 - 4 bc =0$. That is, the roots
\begin{eqnarray}
\lambda = a \pm 2 \sqrt{bc}
\label{tworaices}
\end{eqnarray}
cannot be included in the set of solutions to the problem. Now,
\begin{eqnarray}
\frac{t_+}{t_-}
= \frac{t_+^2}{t_+ t_-}
= \frac{t_+^2}{bc}
= \left ( \frac{t_+}{\sqrt{bc}} \right)^2
(7)
\end{eqnarray}
where we used the product of the two roots of the quadratic equation (4)
$t_+ t_- =bc$. We find, using (6)
\begin{eqnarray*}
\left ( \frac{t_+}{\sqrt{bc}} \right)^{2n+2} = 1.
\end{eqnarray*}
The solution of this equation corresponds with $2n+2$ roots located uniformly
along the unit circle. These roots are represented by
\begin{eqnarray*}
\frac{(t_+)_k}{\sqrt{bc}} = \mathrm{e}^{\frac{ k \pi \mathrm{i}}{2n+2}}
\quad , \quad k=0, 1, \cdots , 2n-1.
\end{eqnarray*}
so that $(t_+)_k=\sqrt{bc} \, \mathrm{e}^{k \pi \mathrm{i}/(2n+2)}$.
From $t_+ t_- = bc$ we find $(t_-)_k=\sqrt{bc} \, \mathrm{e}^{-k \pi
\mathrm{i}/(2n+2)}$ and from $\mathrm{e}^{\theta} + \mathrm{e}^{-\theta}=2 \cos
\theta$,
\begin{eqnarray*}
(t_+)_k + (t_-)_k = 2 \sqrt{bc} \cos \frac{k \pi}{2n +2} .
\end{eqnarray*}
Now, since $(t_+)_k + (t_-)_k=a-\lambda$ we have that
\begin{eqnarray*}
a - \lambda_k = 2 \sqrt{bc} \cos \frac{k \pi }{2n +2}
\end{eqnarray*}
and
\begin{eqnarray*}
\lambda_k = a - 2 \sqrt{bc} \cos \frac{k \pi }{2n+2}
\quad , \quad k=0,1, \cdots , 2n+1.
\end{eqnarray*}
However $p_n(\lambda)=0$ only has $n$ roots and we found $2n+2$.
We already eliminated two roots, corresponding
to $k=0$ and $k=n+1$. We need to exclude $n$ other roots.
If we observe equation (7) we see that the function
$t_+$ was squared when substituting $t_-$ for $\sqrt{cd}/t_+$.
This introduced $n$ new roots, corresponding to $k=1,3,5, \cdots, 2n+1$,
which are solutions of the equation $(t_+/\sqrt{bc})^{2n+2}$ instead of the
corresponding equation $(t_+/\sqrt{bc})^{n+1}$ where $t_+$ is not squared.
With this, the solutions $\lambda_k$ are given by the set
\begin{eqnarray*}
\lambda_k = a - 2 \sqrt{bc} \cos \frac{k \pi }{2n+2}
\quad , \quad k=2,4, \cdots , 2n,
\end{eqnarray*}
or
\begin{eqnarray*}
\lambda_k = a - 2 \sqrt{bc} \cos \frac{k \pi }{n+1}
\quad , \quad k=1,2, \cdots , n.
\end{eqnarray*}
*
Repeated roots:
We have $t_+=t_-=(a-\lambda)/2$. That is, the quadratic equation is a
perfect square. The recursive equation (1) has a solution of the form
\begin{eqnarray*}
p_n(\lambda) = c_1 t_+^n + c_2 n t_+^n.
\end{eqnarray*}
Again, to find $c_1$ y $c_2$ we need to apply the initial conditions to
find a two-by-two linear system of equations.
\begin{eqnarray*}
p_0(\lambda) &=& 1 = c_1 \\
p_1(\lambda) &=& a-\lambda = (c_1 + c_2 )t_+
\end{eqnarray*}
and so
\begin{eqnarray*}
c_2 = \frac{a-\lambda}{t_+} -1 = 1.
\end{eqnarray*}
Then
\begin{eqnarray*}
p_n(\lambda) = t_+^n + n t_+^n
= \left ( \frac{a-\lambda}{2} \right )^n(1 + n).
\end{eqnarray*}
From here we find that all eigenvalues are equal.
Recall that $\Delta=0$ means $\lambda = a \pm 2 \sqrt{bc}$,
and given that $\lambda=a$ we have that $b=0$ or $c=0$. That is, the original matrix
is lower triangular($b=0$) or upper triangular($c=0$) where all its eigenvalues are sitting
along the diagonal.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there any polygon exists such that sum of length of square of two adjacent sides is equal to another side/diagonal? In Right angle triangle we have $ a^2 + b^2 = c^2$ where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$ $b^2 = (x_3-x_2)^2 + (y_3-y_2)^2 $and $c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$
And in Square we have
$ a^2 + b^2 = c^2$ $ d^2 + e^2 = c^2$ $ a^2 + d^2 = f^2$ $ b^2 + e^2 = f^2$ and $a=b=d=e , c=f$
where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$ $b^2 = (x_3-x_2)^2 + (y_3-y_2)^2 $ , $c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$ , $d^2= (x_1-x_4)^2+(y_1-y_4)^2$ ,$e^2= (x_3-x_4)^2+(y_3-y_4)^2$ ,and $f^2= (x_2-x_4)^2+(y_2-y_4)^2$
Is there any other polygon exists which posses this property (sum of length of square of two adjacent sides is equal to another side/diagonal ) ? It's a sufficient condition to say if this property is satisfied by a polygon then it's a right angle triangle or square ?
| Two adjacant sides and the corresponding diagonal fulfill this equality iff they form a right triangle. Some simple examples of such polygons are polyominoes, or any polygon with only horizontal and vertical edges. If you allow the "$c$" to be an unrelated diagonal or side, many many other polygons are possible.
| {
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Summing $\sum^n_{k=1} \frac i {2^k}$ Could someone help me sum following series as $n$ goes to infinity:
$$\sum^n_{i=1} \frac i {2^i} = \frac 1{2^1} + \frac 2{2^2} + \frac 3{2^3} +\cdots+ \frac n{2^n}$$
I don't need the answer so much as being pointed towards what I need to know to figure out the answer...I assume there are different ways to do it. I used to know this stuff a long time ago but don't know where to begin now. Thanks...
| Easier hint:
$$
\begin{array}{ccll}
\sum_1^\infty\frac{n}{x^n} &=& \frac{1}{x^1} + & \frac{1}{x^2} + &\frac{1}{x^3} + \ldots \\
&+& & \frac{1}{x^2} +& \frac{1}{x^3} + \ldots \\
&+& & & \frac{1}{x^3} + \ldots \\
&+& \cdots
\end{array}
$$
Each of these is a simple geometric series (for example, row 2 sums to $\frac{1}{x^2} \frac{1}{1-1/x}$).
But each of those series is $1/x$ times the one above it. So you have a second geometric sum to do.
| {
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Why does summation of infinite series end up in powers of pi? For instance, we have
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...=\frac{\pi}{4}$$
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\frac{\pi^2}{6}$$
$$1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+...=\frac{\pi^3}{32}$$
$$1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+...=\frac{\pi^4}{90}$$
All the $\pi$s come up for no apparent reason. Is there any reason why infinite series give rise to $\pi$, especially in the case when it ends up in powers of $\pi$?
I can think of Fourier transform and Reimann Zeta function as an approach, but I'm not a math guy so I have no idea how to explain this.
Edit: Further powers:
$$1-\frac{1}{3^5}+\frac{1}{5^5}-\frac{1}{7^5}+...=\frac{5\pi^5}{1536}$$
$$1+\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{4^6}+...=\frac{\pi^6}{945}$$
$$1-\frac{1}{3^7}+\frac{1}{5^7}-\frac{1}{7^7}+...=\frac{61\pi^7}{184320}$$
$$1+\frac{1}{2^8}+\frac{1}{3^8}+\frac{1}{4^8}+...=\frac{\pi^8}{9450}$$
| One way to arrive at some such series expressions involving $\pi$ is to start from $$\tag1\frac1\pi\sin \pi z = z\prod_n(1-\frac{z^2}{n^2})$$
Of course, first of all you have to justify $(1)$; you may at least notice that the zeroes are in the right places and that naive (but justifyable) differentiation produces the same derivative at $z=0$. Next, (again: naively, but this can be justified) develop into powers of $z$, to find
$$z-\frac{\pi^2}6z^3+\frac{\pi^4}{120}z^5\pm\ldots =z-z^3\sum\frac1{n^2}+z^5\prod_{n<m}\frac1{(nm)^2}\pm\ldots$$
and compare coefficients. This gives you $\sum_n \frac1{n^2}=\frac{\pi^2}6$ directly and then from
$$ \sum_n\frac1{n^4} = \left(\sum_n\frac1{n^2}\right)^2 - 2\sum_{n<m}\frac1{(nm)^2}=\frac{\pi^4}{6^2}-2\cdot\frac{\pi^4}{120}=\frac{\pi^4}{90}$$
and so on
| {
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In $\triangle ABC$, if $\sin^2{A}+\cos^2{C}=\cos^2{B}$,then $C=?$ In $\triangle ABC$, if $\sin^2{A}+\cos^2{C}=\cos^2{B}$,then $C=?$
We have $\sin^2{A}+\cos^2{C}=\cos^2{B}
\implies 2\sin^2{A}+2\cos^2{C}=2\cos^2{B}
\implies 1-\cos{2A}+\cos{2C}-1=\cos{2B}-1
\implies 1+\cos{2C}=\cos{2B}+\cos{2A}
\implies 1+\cos{2C}=2\cos(A+B)\cos(A-B)
\implies 1+\cos{2C}=2\cos(\pi-A)\cos(A-B)$, since $A+B+C=\pi$
$\implies 1+\cos{2C}=2\cos{A}\cos(A-B)$
How can I find $C$ ? Please help.
| Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin^2A=\cos^2B-\cos^2C=\sin^2C-\sin^2B=\sin(C+B)\sin(C-B)$$
$$\implies\sin A\sin(\pi-\overline{B+C})=\sin(\pi-A)\sin(C-B)$$
$$\implies\sin A[\sin(B+C)-\sin(C-B)]=0$$
$$\implies\sin A[2\sin B\cos C]=0$$
$$\implies C = \frac\pi 2$$(as $ A,B,C\ne 0 $)
Alternatively, $$\cos^2C=\cos^2B-\sin^2A=\cos(B+A)\cos(B-A)$$
Again, $\cos(A+B)=\cos(\pi-C)=-\cos C$
$$\implies\cos C[-\cos(A+B)]=-\cos C\cos(B-A)\iff\cos C[\cos(B-A)-\cos(B+A)]=0$$
$$\iff\cos C[2\sin B\sin A]=0$$
$$\implies C = \frac\pi 2$$(as $ A,B,C\ne 0 $)
| {
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How do I take the limit of this function? $$\lim_{x \to 2^+} \frac{3}{x-2} - \frac{6}{x^2-4}$$
I'm unable to simplify it, but when I put it into my calculator (i.ie i'll put 2.1, 2.0001, etc), I keep getting $15$. When I try to simplify it, I get nowhere at all. How can I make this answer be 15?
| $$\frac{3}{x-2} - \frac{6}{x^2-4} = \frac{3}{x-2} + \frac{3}{2}\frac{1}{x-2} -\frac{3}{2} \frac{1}{x+2} $$ $$= \frac{9}{2}\frac{1}{x-2} - \frac{3}{2}\frac{1}{x+2} $$
Since $$ \lim_{x\rightarrow 2^+} \frac{1}{x+2} = \frac{1}{4}$$
we have $$ \lim_{x\rightarrow 2^+} \bigg( \frac{9}{2}\frac{1}{x-2} - \frac{3}{2}\frac{1}{x+2} \bigg)=\infty$$
| {
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If $abc=1$, then $\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} $
Let $a$, $b$ and $c$ be positive real numbers with $abc=1$. Prove that
$$
\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n}
$$
for each integer $n$.
I have used Cauchy-Schwarz inequalities and Jensen inequality. But I am stuck. I need some idea and advice on this problem. Induction would be cruel.
| You can see that every term has value $\frac{1}{n}$ in case $a = b = c = 1$.
Try to prove that changing one of the variables (let $a > 1$, $b < 1$, $c < 1$ be the first case and $a > 1$, $b > 1$, $c < 1$ - the second one and $a > 1$, $b < 1$, $c > 1$ - the third one) increases the total sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the heat equation (PDE) I have the terms:
$$\left\{\begin{array}{l l} u_t=u_{xx}, 0<x<1, t>0\\u(0,t)=1, u(1,t)=3, t>0\\ u(x,0)=2x+1-\sin(2\pi x), 0<x<1\end{array}\right.\ $$
And I should determine the heat equation. I can begin like this: $$ u(x,t)=f(x)g(t)\\u_t = u_{xx} \Rightarrow f(x)g'(t)=f''(x)g(t)\\\frac{g'(t)}{g(t)}=\frac{f''(x)}{f(x)}=\lambda$$
But then I'm stuck. Where should I go from here? I should probably solve HL and VL separately but is there some obvious knowledge about solving those that I should know? It seems to me that all the examples I have looked is doing some magic here...
| Given ( slightly more general):
\begin{align}
u_{t} &= u_{xx} \hspace{10mm} 0 \leq x \leq L , t \geq 0 \\
u(0,t) &= a , \, \, u(L, t) = b \\
u(x, 0) &= \frac{1}{L} + \frac{2x}{L} - \sin\left(\frac{2 \pi x}{L}\right)
\end{align}
The equation $u_{t} = u_{xx}$ can be solved, as stated, by letting $u(x,t) = f(t) \, g(x)$ for which
\begin{align}
g \, f_{t} = f \, g_{xx} \rightarrow \frac{f_{t}}{f} = - \lambda^{2} = \frac{g_{xx}}{g}
\end{align}
This leads to the equations $f_{t} + \lambda^{2} f = 0$ and $g_{xx} + \lambda^{2} g = 0$. The solution for $f$ is seen to be $f(t) = e^{- \lambda^{2} t}$. The solution for $g$ is seen as $g(x) = A \cos(\lambda x) + B \sin(\lambda x)$.
Initial conditions: $u(0,t) = a$ and $u(L,t) = b$ lead to $g(0) = a$ and $g(L) = b$. From this it is seen that
\begin{align}
g(0) &= a = A \\
g(L) &= b = A \cos(\lambda L) + B \sin(\lambda L).
\end{align}
This leads to the equation to determine $\lambda$ as
\begin{align}
B \, \tan(\lambda L) = b \sec(\lambda L) - a.
\end{align}
Supposing this equation leads to solutions, let these values be given by $\lambda_{n}$, $0 \leq n \leq \infty$.
General solution:
The general solution is of the form
\begin{align}
u(x,t) = a \, e^{- \lambda_{0}^{2} t} + \sum_{n=1}^{\infty} e^{-\lambda_{n}^{2} t} \left( a \cos(\lambda_{n} x) + B_{n} \sin(\lambda_{n} x) \right)
\end{align}
Now, for $t=0$, $u(x, 0) = 1/L + \frac{2x}{L} - \sin\left(\frac{2 \pi x}{L}\right)$ which leads to
\begin{align}
\frac{1}{L} - a + \frac{2x}{L} - \sin\left(\frac{2 \pi x}{L}\right) = \sum_{n=1}^{\infty} \left( a \cos(\lambda_{n} x) + B_{n} \sin(\lambda_{n} x) \right).
\end{align}
Case for $a=b=0$
When $a=0$ the equations reduce to a more solvable case. From the initial conditions: $g(0) = 0$ and $g(L) = 0$ lead to $A = 0$ and $0 = B \sin(\lambda L)$. This leads to $\lambda L = n \pi$. The solution now becomes
\begin{align}
u(x,t) = \sum_{n=1}^{\infty} B_{n} \, \sin\left( \frac{n \pi x}{L} \right) \, e^{- \left( \frac{n \pi}{L} \right)^{2} \,t }.
\end{align}
Now, for $t=0$, $u(x, 0) = 1/L + \frac{2x}{L} - \sin\left(\frac{2 \pi x}{L}\right)$ which leads to
\begin{align}
\frac{1}{L} + \frac{2x}{L} - \sin\left(\frac{2 \pi x}{L}\right) = \sum_{n=1}^{\infty} B_{n} \sin\left(\frac{n \pi x}{L}\right).
\end{align}
The coefficients $B_{n}$ can be found by making use of the Fourier series integral
\begin{align}
B_{n} &= \frac{2}{L} \, \int_{0}^{L} \left[ \frac{1}{L} + \frac{2x}{L} - \sin\left(\frac{2 \pi x}{L}\right) \right] \, \sin\left(\frac{n \pi x}{L}\right) \\
&= \frac{2}{L} \left[ \frac{1 - (-1)^{n}}{n \pi} - \frac{2 L \, (-1)^{n}}{n \pi} - \frac{L}{2} \delta_{n,2} \right].
\end{align}
This leads to the solution
\begin{align}
u(x,t) &= \frac{ 4 (1 + L) }{\pi \, L} \, \sin\left( \frac{\pi x}{L} \right) \, e^{- \frac{\pi^{2} t}{L^{2}}} - \frac{(2+\pi)}{\pi} \, \sin\left( \frac{2 \pi x}{L} \right) \, e^{- \frac{4 \pi^{2} t}{L^{2}}} \\
& \hspace{5mm} + \frac{1}{\pi} \sum_{n=3}^{\infty} \left[ \frac{1 - (1 + 2 L) (-1)^{n}}{n} \right] \, \sin\left( \frac{n \pi x}{L} \right) \, e^{- \left( \frac{n \pi}{L} \right)^{2} \,t }.
\end{align}
| {
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$\displaystyle\lim_{x\to -8} \frac {8 − |x|} {8 + x} \ $ How do I solve? \begin{align*}
\lim_{x\to -8^+} \frac {8 − |x|} {8 + x} &=
\lim_{x\to -8^+} \frac {8 − x} {8 + x}\\
&= \frac {8 − 8} {8 + -8} \\ &= 0
\end{align*}
\begin{align*}
\lim_{x\to -8^-} \frac {8 − |x|} {8 + x} &=
\lim_{x\to -8^-} \frac {8 − (-x)} {8 + x} \\
&= \frac {8 + − 8} {8 + -8} \\ &= 0
\end{align*}
However, the answer key says that I am wrong. Where have I failed?
| Let $\alpha$ stand for a number infinitesimally close to 0, but not equal to 0.
Now, we have \begin{align*}
\lim_{x\to -8^+} \frac {8 − |x|} {8 + x} &
\end{align*}
becomes
\begin{align*}
\frac {8 − |-8+\alpha|} {8 + -8+\alpha} &
\end{align*}.
Since |-8+$\alpha$|=8, we thus end up with $\alpha$/$\alpha$=1, for the first limit. Similarly reasoning yields 1 for the second limit.
| {
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Partial fraction decomposition of a rational function with denominator $x^5+2x^4+x^3-x^2-2x-1$
Factorize the denominator completely and write $f(x)$ as a partial fraction given $$f(x) = \frac{2x^5+15x^4+15x^3+2x^2+2}{x^5+2x^4+x^3-x^2-2x-1}$$
Any ideas for this partial fraction question? Not a clue how to go about it, the normal methods Im used to dont work. Any tips for direction will be greatly appreciated
I tried the general methods of breaking down the denominator and placing the parts over A and B, as well as C - the 2 factorizations that I tried using were (x + 1)^2 (x^3 - 1) and (x+1)^2 ( x^2 + x + 1) (x-1). I tried long division of the polynomials as well but for some reason I wasnt able to get it to work. Thanks for the feed back!
| $$f(x) = \frac{2x^5+15x^4+15x^3+2x^2+2}{x^5+2x^4+x^3-x^2-2x-1}$$
First: By polynomial long division, we obtain $$f(x)= \frac{2x^5+15x^4+15x^3+2x^2+2}{x^5+2x^4+x^3-x^2-2x-1} = 2 + \frac{5x^4 + 13x^3+ 4x^2 + 4x+4}{x^5 + 2x^4 +x^3-x^2-2x - 1}$$
Now, we factor the denominator. Noting that both $x=-1$ and $x = 1$ are roots of the denominator, we know that $$x^5+2x^4 + x^3 - x^2 - 2x - 1 = (x-1)(x+2)g(x)$$ To solve for the third degree polynomial $g(x)$, we divide the denominator by $(x-1)(x+1) = x^2 - 1$ to obtain $$g(x) = x^3 + 2x^2 + 2x +1 = (x+1)(x^2 + x+1)$$
That gives us $$f(x) = 2 +\frac{5x^4+13x^3 + 4x^2 + 4x + 4}{(x-1)(x+1)^2(x^2 + x + 1)} = 2 + \frac{A}{x-1} + \frac{B}{x+1} + \frac C{(x+1)^2} + \frac{Dx + E}{x^2 + x + 1}$$
Can you take it from here? We need only solve for $A, B, C, D, E$, knowing that $$A(x+1)^2(x^2 + x+1) + B(x-1)(x+1)(x^2 + x + 1) + C(x-1)(x^2 + x + 1) + (Dx+E)(x-1)(x+1)^2 = 5x^4 + 13x^3 + 4x^2 + 4x + 4$$
| {
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How to solve $10^{x^2+x}+\log{x} = 10^{x+1}$? In one of my recent exam, I was ask to solve this:
$$
10^{x^2+x}+\log{x} = 10^{x+1}
$$
My attempt to solve it was:
$$
10^{x^2+x}+\log{x} = 10^{x+1} \\
\log{x}=10^{x+1}-10^{x^2+x} \\
\log{x} = 10^{x+1}(1-10^x) \\
\log(\log{x})=(x+1)+\log(1-10^x) \\
$$
At this point I got stuck because I don't know how to solve an equation with double logs.
| Note that $x=1$ is obviously a solution. We prove that this is the only solution.
Suppose that $x>1$. Then $x^2>1$, so $x^2+x>x+1$. Exponentiating both sides we get that:
$$10^{x^2+x}>10^{x+1}$$
since $\log(x)>0$ when $x>1$, then we also have:
$$10^{x^2+x}+\log(x)>10^{x+1}$$
so there are no solutions in this range.
Similarly, when $0<x<1$, then $x^2+x<x+1$, so
$$10^{x^2+x}<10^{x+1}$$
and because $\log(x)<0$ in this range, then:
$$10^{x^2+x}+\log(x)<10^{x+1}$$
so there are no solutions in this range either.
The equation is undefined when $x\leq0$, so this proves that $x=1$ is the only solution.
| {
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Proving by induction $5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$ is divisible by $4$ I want to prove the following twice. Once by induction then again by any other method.
$$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$$ is a multiple of 4 for all nonnegative integers n.
Let n=0 , since it is the first nonnegative integer
$$5^{3(0)}+2*5^{2(0)}-5^{0}-2 = 0 $$
Factoring gives us $(5-1)(5+1)(5+2)$
| Basis(n=1): $5^3+50-5-2=168 = 4*42$
Inductive step: assume that $5^{3n}+2*5^{2n}-5^n+1 = 4k, k\in \mathbb{Z}$. Then,
$$
5*(5^{3n})+10*{5^{2n}}-5*5^n + 1 = 5(5^{3n}+2*5^{2n}-5^n)+1 = 5(4k-1)+1=20k-4=4(5k-1).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this $\lim_{x\to 1} (\frac{x^x-x}{\ln{x}-x+1}+2)/( x-1)$ Find the limit
$$I=\lim_{x\to 1}\dfrac{\left(\dfrac{x^x-x}{\ln{x}-x+1}+2\right)}{x-1}$$
I want let $x-1=t$
$$I=\lim_{t\to 0}\dfrac{\dfrac{(t+1)^{t+1}-(t+1)}{\ln{(t+1)}-t}+2}{t}=\lim_{t\to 0}\dfrac{(t+1)^{t+1}-(t+1)+2\ln{(t+1)}-2t}{t(\ln{(t+1)}-t)}=\lim_{t\to 0}\dfrac{(t+1)^{t+1}+2\ln{(t+1)}-3t-1}{-\dfrac{t^3}{2}}$$
I have solve this result $-\dfrac{7}{3}$by hand ,is true ?
| Taylor series, built at $x=1$ could work $$\log(x)=(x-1)-\frac{1}{2} (x-1)^2+\frac{1}{3} (x-1)^3+O\left((x-1)^4\right)$$ $$\log(x)-x+1=-\frac{1}{2} (x-1)^2+\frac{1}{3} (x-1)^3+O\left((x-1)^4\right)$$ $$x^x-x=(x-1)^2+\frac{1}{2} (x-1)^3+O\left((x-1)^4\right)$$ Combining the pieces $$\frac{\left(\frac{x^x-x}{\ln{x}-x+1}+2\right)}{x-1}=-\frac{7}{3}-\frac{11 }{9}(x-1)-\frac{83}{135} (x-1)^2+O\left((x-1)^3\right)$$
| {
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} |
Prove that $\sum\limits_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$ This sum $\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} $showed up as I was computing the expected value of a random variable.
My calculator tells me that $\,\,\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$.
How can I show that?
I know how to find the value of the sum $\,\displaystyle \sum_{n=1}^\infty \frac{n^2}{2^n},\,$ but I can't deal with $\displaystyle \sum_{n=1}^\infty \frac{n^3}{2^n}$
| For $\lvert x\rvert<1$ we have that
$\displaystyle\sum_{n=0}^\infty x^n=\frac{1}{1-x},$
hence differentiating both sides we get
$$
\sum_{n=0}^\infty (n+1)x^{n}=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2},
$$
while differentiating once more
$$
\sum_{n=0}^\infty (n+2)(n+1)x^{n}=\frac{2}{(1-x)^3},
$$
and once again
$$
\sum_{n=0}^\infty (n+3)(n+2)(n+1)x^{n}=\frac{6}{(1-x)^4}.
$$
Next, we express $n^2(n-1)$ as a linear combination of
$(n+3)(n+2)(n+1)$, $(n+2)(n+1)$, $(n+1)$ and $1$:
$$
n^2(n-1)=(n+3)(n+2)(n+1)-7(n+2)(n+1)+10(n+1)-2,
$$
and hence
$$
\sum_{n=0}^\infty n^2(n-1)x^n=\frac{6}{(1-x)^4}-\frac{14}{(1-x)^3}+\frac{10}{(1-x)^2}-\frac{2}{1-x},
$$
and setting $x=1/2$ we obtain that
$$
\sum_{n=0}^\infty \frac{n^2(n-1)}{2^n}=6\cdot 16-4\cdot 8+10\cdot 4-4=20.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
How can I find a closed form for the summation (i^2)(-1^i+1) systematically? In one of my homeworks I was given the following sequence $1^2-2^2+3^2-4^2+\dots (-1)^{n+1}n^2$, and I'm supposed to find a closed form formula and prove that it works.
Rewriting this as a sum gives the following
$$\sum_{i=1}^{n}(-1)^{n+1}i^2$$
I have found the closed form, which is $-\frac{1}{2} (-1)^n n (n+1)$, and I also have the proof, which was trivial to get. The problem is that I do not know how to get to this formula in a systematic way, without a wild guess or a tool like Mathematica, which would somehow just spit out the answer.
I started reading Knuth's Concrete Mathematics, which seems to describe how to approach these problems in general, but I'm having a hard time understanding the text (I'm in first semester studying Computer Science.)
I tried comparing the data this sum generates ($1, -4, 9, -16, 25, -36, 49, -64$), and tried looking at the prefix sums ($1, -3, 6, -10, 15, -21, 28, -36, 45, -55$), but I just don't know how this is supposed to help me in finding this formula.
TL;DR: How can I systematically find a closed form formula for this particular summation, or any arbitrary one in general?
I tried looking through the other questions, but they tend to address more complicated formulars which I'm having hard trouble understanding.
|
A method to systematically find a closed formula for this and similar expressions is based upon formal power series. The idea is to start from the most basic sequence
\begin{align*}
(1)_{n\geq 0}=(1,1,1,\ldots)
\end{align*}
and iteratively apply operators to obtain the wanted sequence
\begin{align*}
\left(\sum_{j=0}^n (-1)^{j+1}j^2\right)_{n\geq 0}=(1,-3,6,-10,15,\ldots)
\end{align*}
Here we do it systematically in four steps
\begin{align*}
(1)_{n\geq 0}
\quad\rightarrow\quad (n)_{n\geq 0}
\quad\rightarrow\quad (n^2)_{n\geq 0}
\quad\rightarrow\quad ((-1)^{n+1}n^2)_{n\geq 0}
\quad\rightarrow\quad \left(\sum_{j=0}^{n}(-1)^{j+1}j^2\right)_{n\geq 0}
\end{align*}
We start with the constant sequence $(1)_{n\geq 0}=(1,1,1,\ldots)$ represented by the generating function
\begin{align*}
\frac{1}{1-x}&=\sum_{n=0}^{\infty}x^n=1+x+x^2+\cdots
\end{align*}
We next observe that applying differentiation and multiplication with $x$ of a formal power series $A(x)=\sum_{n=0}^{\infty}a_nx^n$ results in
\begin{align*}
(xD)A(x)&=(xD)\sum_{n=0}^{\infty}a_nx^n\\
&=x\sum_{n=0}^{\infty}na_nx^{n-1}\\
&=\sum_{n=0}^{\infty}na_nx^n
\end{align*}
We conclude that application of the operator $xD$ to a power series $A(x)=\sum_{n=0}^{\infty}a_nx^n$ transforms the sequence
\begin{align*}
(a_n)_{n\geq 0}\qquad \text{to}\qquad (na_n)_{n\geq 0}
\end{align*}
and applying the operator $xD$ to a power series $A(x)$ iteratively $k$ times, we obtain
\begin{align*}
(xD)^kA(x)=\sum_{n=0}^{\infty}n^ka_nx^n
\end{align*}
Applying the operator $xD$ to the geometric series $\frac{1}{1-x}$ we obtain
\begin{align*}
(xD)\frac{1}{1-x}&=\frac{x}{(1-x)^2}\\
&=\sum_{n=0}^\infty nx^n\\
\end{align*}
Applying the operator $xD$ twice, we obtain
\begin{align*}
(xD)^2\frac{1}{1-x}&=(xD)\frac{x}{(1-x)^2}\\
&=\frac{1+x}{(1-x)^3}\\
&=\sum_{n=0}^\infty n^2x^n
\end{align*}
Next we replace $x$ with $-x$ and multiply with $-1$ to obtain
\begin{align*}
(-1)\left.\left((xD)^2A(x)\right)\right|_{x=-x}&=-\frac{x(1-x)}{(1+x)^3}\\
&=\sum_{n=0}^{\infty}(-1)^{n+1}n^2x^n\tag{1}
\end{align*}
The last step is to transform a sequence
\begin{align*}
(a_n)_{n\geq 0}\qquad\text{to}\qquad\left(\sum_{j=0}^{n}a_j\right)_{n\geq 0}
\end{align*}
which can be done in terms of formal power series by multiplication with $\frac{1}{1-x}$
\begin{align*}
\frac{1}{1-x}A(x)&=\frac{1}{1-x}\sum_{n=0}^{\infty}a_nx^n\\
&=\sum_{n=0}^\infty\left(\sum_{j=0}^na_j\right)a_nx^n
\end{align*}
Applying this multiplication to (1) results in
\begin{align*}
\frac{-1}{1-x}\left.\left((xD)^2A(x)\right)\right|_{x=-x}&=
\frac{x}{(1+x)^3}\\
&=\sum_{n=0}^{\infty}\left(\sum_{j=0}^\infty(-1)^{j+1}j^2\right)x^n
\end{align*}
You may notice that with some routine it's a quick and easy job to conclude:
The generating function of the summation formula under consideration is
\begin{align*}
\frac{x}{(1+x)^3}=\sum_{n=0}^{\infty}\left(\sum_{j=0}^n(-1)^{j+1}j^2\right)x^n
\end{align*}
Now it is time to harvest. We use the binomial series expansion to find a closed formula. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We obtain for $n\geq 1$
\begin{align*}
\sum_{j=0}^n(-1)^{j+1}j^2&=[x^n]\frac{x}{(1+x)^3}\\
&=[x^{n-1}]\frac{1}{(1+x)^3}\tag{2}\\
&=[x^{n-1}]\sum_{j=0}^{\infty}\binom{-3}{j}x^j\tag{3}\\
&=\binom{-3}{n-1}\tag{4}\\
&=\binom{n+1}{n-1}(-1)^{n-1}\tag{5}\\
&=(-1)^{n-1}\frac{n(n+1)}{2}
\end{align*}
Comment:
*
*In (2) we use the rule $[x^n]x^k=[x^{n-k}]$
*In (3) we apply the binomial series expansion
*In (4) we extract the coefficient of $x^{n-1}$
*In (5) we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q=\binom{p+q-1}{p-1}(-1)^q
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
An Inequality; $a^2+b^2=1$ $a,b$ are tho real numbers such that $a^2+b^2=1$.
To prove that ;
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{3}{1+\cfrac{(a+b)^2}{4}}$$
When I first saw this question, I thought of applying Titu's Lemma, to get
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{9}{1+1+1+a^2+b^2+ab}=\dfrac{9}{3+(a+b)^2-ab}OR=\dfrac{9}{4+ab}$$
Now, from hereI am confused, how to proceed. Can anybody kindly help me over this problem ?
| We need to prove that
$$\frac{1}{2a^2+b^2}+\frac{1}{a^2+2b^2}+\frac{1}{a^2+ab+b^2}\geq\frac{12}{5a^2+2ab+5b^2}$$ or
$$(a-b)^2(a^4+3a^3b+7a^2b^2+3ab^3+b^4)\geq0,$$
which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding max and min in $f(x) = 5\sin x + 12\sin(x+\frac{\pi}{3})$ For the function $f(x) = 5\sin x + 12\sin(x+\frac{\pi}{3})$, find the max and min value the function can be.
Own thoughts
I first noted that the function had no constant, and so the max = |amplitude|, that also means that min = -|amplitude|.
I tried what I could rewriting the function, because I know that a trig function's coefficient is its amplitude; alas I did not succeed.
$5\sin x+12\sin(x+\frac{\pi}{3}) = 5\sin x + 12(\sin x\cos\frac{\pi}{3}+\cos x\sin\frac{\pi}{3}) =\\= 5\sin x+ 6\sin x + 6\sqrt3\cos x = 11\sin x + 6\sqrt3\cos x$
Still a $\cos$ term. How can I solve this?
| Hint: Once you have an expression in the form $a\sin x+b\cos x$, think of it as being in the form $$\sqrt{a^2+b^2}\left({a\over\sqrt{a^2+b^2}}\sin x+{b\over\sqrt{a^2+b^2}}\cos x \right)$$
Then think of $a/\sqrt{a^2+b^2}$ and $b/\sqrt{a^2+b^2}$ as the sine and cosine (or vice versa) of some angle $\phi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the series $\sum_{n=1}^\infty \frac{n+1}{n^3+10n}$ converge? Using the ratio test, we evaluate:
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty}\left| \frac{(n+1) + 1}{(n+1)^3 + 10(n+1)} \cdot \frac{n^3+10n}{n+1} \right| = \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{(n+1)^4 + 10(n+1)^2} \right|$$
$$ < \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{n^4 + 10n^2} \right| = 1$$
Hence $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ and the series converges.
Is this an appropriate solution?
| The series is bounded above by
$$\sum \frac{n + 1}{n^3} = \sum \frac{1}{n^2} + \sum \frac{1}{n^3},$$
so by the comparison test the series converges because both sums on the right-hand side are convergent $p$-series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$? Some time ago I asked How to find $\displaystyle{\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$.
Thanks to great effort of several MSE users, we now know that
\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln x}xdx=&\,\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)\\&\,-\frac{21}4\zeta(3)\ln^22-12\operatorname{Li}_4\!\left(\tfrac12\right)\ln2-12\operatorname{Li}_5\!\left(\tfrac12\right)\tag1
\end{align}
Now, a natural follow-up to that question is to bump the power of the logarithm and to ask:
Question: What is a closed form for the next integral?
$$I=\int_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx\tag2$$
I think it is likely that $I$ has a closed form, because there are several very similar integrals having known closed forms:
$$\int_0^1\frac{\ln^2(1+x)\,\ln^2x}xdx=\frac{\pi^2\,\zeta(3)}3-\frac{29\,\zeta(5)}8\tag3$$
$$\int_0^1\frac{\ln^3(1-x)\,\ln^2x}xdx=12\zeta^2(3)-\frac{23\pi^6}{1260}\tag4$$
\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln^2x}{x^2}dx=&\,\frac{3\zeta(3)}2+2\pi^2\zeta(3)+\frac{3\zeta(5)}2-\frac{21\zeta(3)}2\ln^22\\&\,-\frac{63\zeta(3)}2\ln2+\frac{23\pi^4}{60}-\frac{4\ln^52}5-\frac{3\ln^42}2\\&\,-4\ln^32+\frac{2\pi^2}3\ln^32+\frac{3\pi^2}2\ln^22-24\operatorname{Li}_5\!\left(\tfrac12\right)\\&\,-36\operatorname{Li}_4\!\left(\tfrac12\right)-24\operatorname{Li}_4\!\left(\tfrac12\right)\ln2\tag5
\end{align}
| This is not going to be a complete answer but since this kind of approach has not been presented here yet and since I believe that it can be brought to a successful completion given enough time under disposal( which I lack now) I present the approach now.
Denote:
\begin{eqnarray}
{\mathcal I}^{(2,3)} := \int\limits_0^1 \frac{\log(\xi)^2 \log(1+\xi)^3}{\xi} d\xi
\end{eqnarray}
Then we have:
\begin{eqnarray}
&&{\mathcal I}^{(2,3)} = \left. \frac{\partial^2}{\partial \theta_1^2} \frac{\partial^3}{\partial \theta_2^3} \int\limits_0^1 \xi^{\theta_1-1} (1+\xi)^{\theta_2} d\xi \right|_{\theta_1=\theta_2=0} \\
&&= \left. \frac{\partial^2}{\partial \theta_1^2} \frac{\partial^3}{\partial \theta_2^3}
\left[\sum\limits_{l=0}^\infty \frac{ (\theta_2)_{(l)} }{ \theta_1^{(l+1)} } \cdot 2^{\theta_2-l} (-1)^l \right]
\right|_{\theta_1=\theta_2=0}\\
&&= \sum\limits_{l=1}^\infty
\left(\log(2)^2 + \frac{\log(4)}{l} + \frac{2}{l^2} + [H_l]^2 - H_l^{(2)}-\frac{2}{l} H_l - 2 \log(2) H_l \right)\cdot \\
&&\left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l\cdot 2^l} \right)
\end{eqnarray}
The first line is straightforward. In the second line we computed the integral in question by integrating by parts. Finally in the last line we computed the partial derivatives using Higher order derivatives of the binomial factor and the chain rule. Now, the sums look scary but it appears that those sums have actually a much simpler integrals representation than the original integral we want to compute. As a matter of fact the following holds:
\begin{eqnarray}
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) x^l = -\int\limits_0^1 \frac{x}{1-\xi x} \cdot [\log(1-\xi)]^3 d\xi
\end{eqnarray}
Using the generation function above we compute the harmonic sums in question. We have:
\begin{eqnarray}
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) \cdot \frac{1}{2^l} &=& \frac{21}{4} \zeta(4)\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} &=& -\frac{3 \pi ^2 \zeta (3)}{8}+12 \zeta (5)-\frac{7}{120} \pi ^4 \log (2)\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^3} \right) \cdot \frac{1}{2^l} &=& -\int\limits_0^1 \frac{1}{\xi} Li_2(\frac{\xi}{2}) \log(1-\xi)^3 d\xi\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) \cdot \frac{1}{2^l} \cdot H_l &=& -\frac{7}{8} \pi^2 \zeta(3) + \frac{279}{16} \zeta(5)\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} \cdot H_l &=& \int\limits_{0}^1 \frac{Li_2(-\xi)}{\xi(1+\xi)} \cdot [\log(\frac{1-\xi}{1+\xi})]^3 d\xi\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} \cdot \left([H_l]^2-H_l^{(2)}\right) &=&
-12 \left( \zeta(-4,1,1)-\zeta(4,-1,1)\right)- \frac{1}{8} \left(\pi^4 \log(2) + 14 \pi^2 \zeta(3) - 279 \zeta(5)\right)
\end{eqnarray}
It is clear that the remaining sums are more complicated and more time is required to bring this thread to completion. We will finish this work as soon as possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 2,
"answer_id": 0
} |
Sum Representation of log(1 + x) $\log(1+x) = \sum_{k=1}^{\infty} \left(\dfrac{x}{1+x}\right)^{k} \dfrac{1}{k} = \sum_{k=1}^{\infty} \left(1 - \dfrac{1}{1+x}\right)^k \dfrac{1}{k}$
Why is this true? The most sum representation of $\log(1+x)$ (in this case a MacLaurin series) is $\sum_{k=1}^{\infty} (-1)^{k+1} \dfrac{x^k}{k}$
| From
\begin{align}
\ln(1+x) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \, x^{k}}{k}
\end{align}
let $x = -u$ for which
\begin{align}
\ln(1-u) = - \sum_{k=1}^{\infty} \frac{u^{k}}{k}
\end{align}
Now let $u = \frac{x}{1+x}$ to obtain
\begin{align}
\ln(1+x) = \sum_{k=1}^{\infty} \frac{1}{k} \left( \frac{x}{1+x} \right)^{k}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the intersection points of 2 circles with a given equation, center and radius Determine the points of intersection of the circle with equation $x^2 + y^2 - 12x - 4y + 30 = 0$ and the circle with center $(3,5)$ and radius $4$.
My attempt :
With given center and radius of the second circle, I make an equation:
$(x - a)^2 + (y-b)^2 = r^2$
$(x - 3)^2 + (y-5)^2 = 4^2$
$ x^2 -6x +9 + y^2 -10y +25 = 16 $
$ x^2 + y^2 - 6x -10y +18 = 0 $
And then I equate both equations:
$ x^2 + y^2 - 6x -10y +18 = x^2 + y^2 - 12x - 4y + 30 $
and at the end it gives me:
$-x + y = -2$
Am I doing it in the right way? what should I do next?
| So you have the points of intersection of the circles satisfy $y = x-2$. Using this in any one circle gives you the points. e.g.
$$(x-3)^2+(x-7)^2=16 \implies x = 3,7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Differentiate the following function: $y = \frac{2x^2 + 6\sqrt{x} }{9x}$ My answer is $- \dfrac{1}{9 \sqrt{x} }$, however, the program I am using states that I am wrong. Where have I went wrong?
| $$y = 2x^2 + \frac{6\sqrt x}{9x} = 2x^2 +\frac {2}{3\sqrt x} = 2x^2 + \frac 23\cdot x^{-1/2}$$
$$y' = 4x - \frac 12\cdot\frac 23 \cdot x^{-3/2} = 4x - \frac{1}{3x^{3/2}}$$
EDIT:
Given the OP's comments and attempted edit, it seems that the function of interest is intended to be $$y = \frac{2x^2 + 6\sqrt{x}}{9x}= \frac 29 x + \frac 23\cdot x^{-1/2}$$
If so, then
$$\begin{align} \frac{dx}{dy} & = \frac 29 - \frac 12\cdot\frac 23 \cdot x^{-3/2}\\ \\
& = \frac 29 - \frac{1}{3x^{3/2}}\\ \\
& = \frac{2x^{3/2} - 3}{9x^{3/2}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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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.