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Calculating $\lim_{x \rightarrow 1}(\frac{23}{1-x^{23}} - \frac{31}{1-x^{31}})$ How to calculate following limit?
$$\lim_{x \rightarrow 1}\left(\frac{23}{1-x^{23}} - \frac{31}{1-x^{31}}\right)$$
| Alternatively, applying L'Hopital's rule, we get:
\begin{align}
\lim_{x \to 1}{\frac{23}{1-x^{23}}-\frac{31}{1-x^{31}}}& =\lim_{x \to 1}{\frac{23(x^{23}+x^{24}+ \ldots +x^{30})-8(1+x+\ldots +x^{22})}{(1-x^{31})(1+x+ \ldots +x^{22})}} \\
& =\lim_{x \to 1}{\frac{23(23x^{22}+24x^{23}+ \ldots +30x^{29})-8(1+ 2x+\ldots +22x^{21})}{-31x^{30}(1+x+ \ldots +x^{22})+(1-x^{31})(1+2x+\ldots+22x^{21})}} \\
& =\frac{23(22(8)+\frac{8(9)}{2})-8(\frac{22(23)}{2})}{-31(23)} \\
& =-4
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/335423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.
Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
Here's my idea:
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
And I'm stuck here.
I need to prove that:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$
In the first case using Cauchy-Schwarz Inequality I prove that:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$3(a^2 + b^2 + c^2) \ge (a+b+c)^2$
Now I need to prove that:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$
I need I don't know how to continue.
In the second case I tried proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and
$a^2 + b^2 + c^2 \ge a+b+c$
Using Cauchy-Schwarz Inequality I proved:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$
$a^2 + b^2 + c^2 \ge a+b+c$
But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$
So please help me with this problem.
P.S
My initial idea, which is proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
maybe isn't the right way to prove this inequality.
| Since $ab+ac+bc=\frac{(a+b+c)^2-a^2-b^2-c^2}{2}=\frac{9-a^2-b^2-c^2}{2}$, we need to prove that
$$\sum_{cyc}\left(\sqrt{a}+\frac{a^2}{2}-\frac{3}{2}\right)\geq0$$ or
$$\sum_{cyc}(a^2+2\sqrt{a}-3)\geq0$$ or
$$\sum_{cyc}(\sqrt{a}-1)(a\sqrt{a}+a+\sqrt{a}+3)\geq0$$ or
$$\sum_{cyc}\left((\sqrt{a}-1)(a\sqrt{a}+a+\sqrt{a}+3)-3(a-1)\right)\geq0$$ or
$$\sum_{cyc}(\sqrt{a}-1)^2\sqrt{a}(\sqrt{a}+2)\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 1
} |
Surface integral over ellipsoid I've problem with this surface integral:
$$
\iint\limits_S {\sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}{dS}
$$, where
$$
S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\}
$$
| What an interesting integral. I had to resort to referring to the first differential form of the spherical parametrization, but doing that, I am amazed at how this turns out.
We parametrize in the usual way:
$$x=a\sin{u} \cos{v}$$
$$y=b\sin{u} \sin{v}$$
$$z=c \cos{u}$$
where $u \in [0,\pi)$ and $v \in [0,2 \pi)$. The coefficients of the first differential form are
$$E=(a^2 \sin^2{v}+b^2 \cos^2{v}) \sin^2{u}$$
$$F=(b^2-a^2) \sin{u} \cos{u} \sin{v} \cos{v}$$
$$G=(a^2 \cos^2{v}+b^2 \sin^2{v}) \cos^2{u}+c^2 \sin^2{u}$$
The stated integral is equal to
$$\int_0^{\pi} du \: \int_0^{2 \pi} dv \: \sqrt{E G-F^2} \sqrt{\frac{\sin^2{u} \cos^2{v}}{a^2} + \frac{\sin^2{u} \sin^2{v}}{b^2} + \frac{\cos^2{u}}{c^2}}$$
There is an enormous amount of algebra involved in simplifying the integrand. Miraculously, it simplifies a lot, and the integral is equal to
$$\frac{1}{a b c} \int_0^{\pi} du \: \sin{u} \int_0^{2 \pi} dv\: (a^2 b^2 \cos^2{u} + b^2 c^2 \sin^2{u} \cos^2{v} + a^2 c^2 \sin^2{u} \sin^2{v})$$
I really couldn't believe this myself at first, but it does check out for the case $a=b=c$. In any case, these integrals are much easier than one would expect from first seeing this problem, and the reader should have no trouble evaluating them by hand. The result is
$$\frac{4 \pi}{3} \left ( \frac{a\, b}{c} + \frac{a\, c}{b} + \frac{b\, c}{a} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
If $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}}(x≥1) $ then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y$ How do I prove following?
If $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}},(x≥1)$, then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y$
| Complete the square:
$2x = y^\frac{1}{m} + y^\frac{-1}{m}$
$2xy^\frac{1}{m} = y^\frac{2}{m} + 1$
$0 = y^\frac{2}{m} - 2xy^\frac{1}{m} + 1$
$0 = (y^\frac{2}{m} - 2xy^\frac{1}{m} + x^2) - x^2 + 1$
$0 = (y^\frac{1}{m} - x)^2 - x^2 + 1$
$(y^\frac{1}{m} - x)^2 = x^2 - 1$
$y^\frac{1}{m} - x = \sqrt{x^2 - 1}$
$y^\frac{1}{m} = x + \sqrt{x^2 - 1}$
$y = (x+\sqrt{x^2 - 1})^m$
Differentiate from there and plug in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find a closed form of the series $\sum_{n=0}^{\infty} n^2x^n$ The question I've been given is this:
Using both sides of this equation:
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$
Find an expression for $$\sum_{n=0}^{\infty} n^2x^n$$
Then use that to find an expression for
$$\sum_{n=0}^{\infty}\frac{n^2}{2^n}$$
This is as close as I've gotten:
\begin{align*}
\frac{1}{1-x} & = \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \frac{d^2}{dx^2} \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \sum_{n=2}^{\infty} n(n-1)x^{n-2} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n(n-1)\frac{x^n}{x}(x+1) \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} (n^2x + n^2 - nx - n)\frac{x^n}{x} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n^2x^n + n^2\frac{x^n}{x} - nx^n - n\frac{x^n}{x} \\
\end{align*}
Any help is appreciated, thanks :)
| Hint: $n^2=n(n-1)+n$ and $x^2 x^{n-2}=x^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Solving linear recurrence relation Solve the following linear recurrence relation:
$$h_n=4h_{n-1}-4h_{n-2}+n^2 2^n$$
for $n\geq2$ and $h_0=h_1=1$
| Use generating functions. Define $H(z) = \sum_{n \ge 0} h_n z^n$, and write the recurrence as:
$$
h_{n + 2} = 4 h_{n + 1} - 4 h_n + 4 (n + 2)^2 2^n
$$
By the properties of ordinary generating functions (see e.g. Wilf's "generatingfunctionology") with the operator $z D = z \dfrac{d}{d z}$
:
$$
\begin{align*}
\frac{H(z) - 1 - z}{z^2} &= 4 \frac{H(z) - 1}{z} - 4 H(z)
+ 4 (z D + 2)^2 \frac{1}{1 - 2 z} \\
H(z) &= \frac{2}{(1 - 2 z)^5}
- \frac{3}{(1 - 2 z)^4}
- \frac{3}{(1 - 2 z)^3}
+ \frac{21}{2 (1 - 2 z)^2}
- \frac{19}{2 (1 - 2 z)}
+ 4
\end{align*}
$$
Using $\binom{-r}{s} = (-1)^r \binom{r + s - 1}{r - 1}$ the powers give polynomials in $n$ multiplying $2^n$, a computer algebra system like maxima
reduces this mess to a simpler expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to solve $5^{2x}-3\cdot2^{2y}+5^x\cdot2^{y-1}-2^{y-1}-2\cdot5^x+1=0$ in $\mathbb{Z}$ how to solve in $\Bbb Z$:
$$5^{2x}-3\cdot2^{2y}+5^x\cdot2^{y-1}-2^{y-1}-2\cdot5^x+1=0$$
| Note that $5^{2x}-2\cdot5^x+1=(5^x-1)^2$, and $5^x\cdot2^{y-1}-3\cdot2^{2y}-2^{y-1}=(5^x-1)2^{y-1}-3\cdot2^{2y}$, so the original equation can be written
$$(5^x-1)(5^x-1+2^{y-1})-3\cdot2^{2y}=0\;.\tag{1}$$
The recurring element $5^x-1$ suggests making a substitution $u=5^x-1$, and since $3\cdot2^{2y}$ can be expressed simply in terms of $2^{y-1}$, we might also try substituting $v=2^{y-1}$ and rewriting $(1)$ as
$$u(u+v)-12v^2=u^2+uv-12v^2=(u-3v)(u+4v)=0\;.$$
Thus, either $u=3v$, or $u=-4v$. That is, $5^x-1=3\cdot2^{y-1}$, or $5^x-1=-4\cdot2^{y-1}=-2^{y+1}$. It seems preferable to write these as $5^x=3\cdot2^{y-1}+1$ and $5^x=1-2^{y+1}$. Clearly the latter produces no solutions, so we need only find the solutions to $5^x=3\cdot2^{y-1}+1$. By inspection $x=2,y=4$ is a solution; it’s not immediately clear to me whether there are any others.
Added: As Ivan Loh points out in the comments, this is the unique solution. First, it’s clear that $x$ and $y$ must be positive. since $5^x\equiv(-1)^x\pmod3$, we must have $x$ even; say $x=2a$, with $a\ge 1$. Then we have $5^{2a}-1=3\cdot2^{y-1}$, and we can factor the lefthand side to get
$$(5^a-1)(5^a+1)=3\cdot2^{y-1}\;.$$
Clearly $5^a+1\equiv2\pmod 4$. On the other hand, $5^a+1$ must be either $2^k$ or $3\cdot2^k$ for some $k\le y-1$, so we must have $k=1$ and $5^a$ either $2$ or $3\cdot2=6$. Then $a=0$ or $a=1$. If $a=0$, then $5^a-1=0$, which is impossible, since $3\cdot2^{y-1}\ne0$., so $a=1$, $5^1-1=4$, and we have the solution $x=2,y=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/341536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ $$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$
I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.
The way I did it:
$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
| For a slightly different solution,
$2\cos^2 A-2\sin^2 B$
$=2\cos^2 A-1+1-2\sin^2 B$
$=\cos 2A+\cos 2B$
$=2\cos (A+B) \cos (A-B)$
and halve each side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Integrating $\int_0^\infty \sin(1/x^2) \, \operatorname{d}\!x$ How would one compute the following improper integral:
$$\int_0^\infty \sin\left(\frac{1}{x^2}\right) \, \operatorname{d}\!x$$
without any knowledge of Fresnel equations?
I was thinking of using the comparison theorem since $\sin x$ is never greater than 1. However, I can't find a function always greater than 1 between 0 and infinity such that its integral from 0 to infinity converges.
Thanks.
| I prefer to evaluate the more general integral below
$$\displaystyle I(a):=\int_{-\infty}^{\infty} \sin \left(\frac{a}{x^2} \right)dx, \tag*{}$$
where $a\geq 0.$
Letting $y=\frac{1}{x^2}$ transforms the integral into
$\displaystyle \begin{aligned} I(a)&=2 \int_{\infty}^0 \sin (a y)\left(-\frac{1}{2 y^{\frac{3}{2}}} d y\right) = \int_0^{\infty} \frac{\sin (a y)}{y^{\frac{3}{2}}} d y \end{aligned} \tag*{} $
Differentiating $I(a)$ w.r.t. $a$, we have
$\displaystyle \begin{aligned} I^{\prime}(a) &=\int_0^{\infty} \frac{\cos (a y)}{y^{\frac{1}{2}}} d y\stackrel{u^2=ay}{=}\frac{2}{\sqrt{a}} \int_0^{\infty} \cos u^2 d u \end{aligned} \tag*{} $
Using the result(refer to footnote for details):$ $
$$\displaystyle \int_0^{\infty} \cos x^2 d x=\frac{1}{2}\int_{-\infty}^{\infty} \cos x^2 d x =\sqrt{\frac{\pi}{8}}\tag*{} $$
we obtain
$$\displaystyle I^{\prime}(a)=\sqrt{\frac{\pi}{2}} a^{-\frac{1}{2}} \tag*{} $$
Integrating back yields
$$\boxed{\displaystyle \int_{-\infty}^{\infty} \sin \left(\frac{a}{x^2} \right)dx=\sqrt{\frac{\pi}{2}} \int_0^a t^{-\frac{1}{2}} d t=\sqrt{2 a \pi}} \tag*{} $$
Putting $a=1$ yields our integral
$\boxed{\displaystyle \int_0^{\infty} \sin \left(\frac{1}{x^2} \right)dx=\sqrt{2 \pi}} \tag*{} $
Furthermore, for $a<0$, we have
$\displaystyle \int_{-\infty}^{\infty} \sin \left(\frac{a}{x^2} \right)dx= -\sqrt{-2 a\pi}\tag*{}$
Footnote:
$\displaystyle \begin{aligned} \int_{-\infty}^{\infty} \cos x^2 d x-i \int_{-\infty}^{\infty} \sin x^2 d x =& \int_{-\infty}^{\infty} e^{-x^2 i} d x \\ =&\frac{\sqrt{2}}{1+i} \int_{-\infty}^{\infty} e^{-\left[\left(\frac{1+i}{\sqrt{2}}\right)x\right]^2} d\left(\frac{1+i}{\sqrt{2}} x\right) \\ =& (1-i)\sqrt{\frac{\pi}{2} } \end{aligned} \tag*{} $
By comparing the real and imaginary parts, we have
$\displaystyle \int_{-\infty}^{\infty} \cos x^2 d x=\int_{-\infty}^{\infty} \sin x^2 d x=\sqrt{\frac{\pi}{2}} \tag*{} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $a,b$, and $c$ are reals satisfying $ \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=6$, calculate $ x = \frac{(a+b+c)^3}{a^3+b^3+abc}$ $$ \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=6$$
$$ x = \frac{(a+b+c)^3}{a^3+b^3+abc}$$
As a trivial solution, I found $$a=b=c$$
then, $x$ will always be $9$.
Despite this, my algebra teacher told me there was another solution that I haven't found yet. Is it possible inside the reals? If so, how?
| You can also have $a=b=c=-2$. In that case, $x$ is still $9$. Given the wording of the problem, the value of $x$ should not depend upon what solution you find for $a,b,c$, so you should be able to find one, evaluate $x$ from it, and quit. If they want yo to prove it independent of $a,b,c$ they should ask.
Added: $a=b=c$ is a solution for any non-zero $a$, which gives $9$ for $x$ You can probably use the AM-GM inequality to justify that $a=b=c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Stuck finding a recursive recurrence relation. I am analyzing the following algorithm:
QUANT(n):
if n == 0 or n == 1:
return 1
else
return (n-1)*QUANT(n-1) + n
I need to find the recurrence relation of this algorithm and prove it using mathematical induction. Here is what I have tried so far:
$$a_0 = 1, a_1 = 1, a_n = (n-1)*a_{n-1} + n$$
$$a_0 = 1$$
$$a_1 = 1$$
$$a_2 = 1 + 2 = 3$$
$$a_3 = (2)(3) + 3 = 9$$
$$a_4 = (3)(9) + 4 = 31$$
$$a_5 = (4)(31) + 5 = 129$$
Also tried backwards substitution:
$$a_0 = 1$$
$$a_1 = 1$$
$$a_2 = (n-1) + n = 2n - 1$$
$$a_3 = (n-1)(2(n-1)-1) + n = 2n^2 -4n + 3$$
$$a_4 = (n-1)(2(n-1)^2-2(n-1)+1)+n=2n^3-10n^2+10n-1$$
$$a_5 = (n-1)(2(n-1)^3-4(n-1)^2+4(n-1)-1)+n=2n^4-18n^3+52n^2-58n+23$$
I am having trouble finding a pattern. For the backwards one I can detect some things like the first term is always $2n^{n-1}$, all the powers of $n$ are expressed. Please point out what concept of series I am missing to find the pattern, if there is one.
| You have the recurrence:
$$
a_{n + 1} = n a_n + n + 1 \quad a_0 = 1
$$
This is a non-homogeneous linear first order recurrence. If you have:
$$
x_{n + 1} - u_n x_n = f_n
$$
Dividing by $u_n \ldots u_0$ gives:
$$
\frac{x_{n + 1}}{u_n \ldots u_0} - \frac{x_n}{u_{n - 1} \ldots u_0}
= \frac{f_n}{u_n \ldots u_0}
$$
This is easy to sum:
$$
\frac{x_n}{u_{n - 1} \ldots u_0} = x_0
+ \sum_{0 \le k \le n - 1} \frac{f_k}{u_k \ldots u_0}
$$
In our case $u_n = n$, $f_n = n + 1$. To avoid division by 0, better start at index 1. Then:
$$
\begin{align*}
\frac{a_n}{(n - 1)!} &= \frac{a_1}{0!}
+ \sum_{1 \le k \le n - 1} \frac{k + 1}{(k + 1)!} \\
a_n &= (n - 1)! + (n - 1)! \sum_{1 \le k \le n - 1} \frac{1}{k!} \\
&= (n - 1)! \sum_{0 \le k \le n - 1} \frac{1}{k!}
\end{align*}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The question is whether the two values of $I_2$ as obtained below are same? Integrate, $\int \frac{\sin^{2}{2\theta}}{2\cos{2\theta}}d\theta$
Let, $\sin2\theta=t$
then, ${2\cos2\theta} d\theta=dt$
$dt=\frac{1}{2}\frac{1}{\sqrt{1-t^2}}$
Integral =$\frac{1}{4}\int {\frac{t^2}{1-t^2}} dt$
Integral =$\frac{1}{4}\int {\frac{t^2}{1-t^2}} dt$
Integral =$-\frac{1}{4}\int {\frac{t^2-1+1}{t^2-1}} dt$
Integral =$-\frac{1}{4}\int ({1+\frac{1}{t^2-1}} )dt$
Integral =$-\frac{1}{4} (t+\int{\frac{1}{t^2-1}} dt)$
Integral =$-\frac{1}{4} (t+I_2)$
$I_2=\int \frac{1}{t^2-1}dt$
Now, $I_2=\frac{1}{2}\int (\frac{1}{t-1}-\frac{1}{t+1})dt$
Now, $I_2=\frac{1}{2}log(\frac{t-1}{t+1})+c_2dt$ ---(I)
we can also have the following,
$I_2=-\int \frac{1}{1-t^2}dt$
Now, $I_{2}=-\frac{1}{2}\int (\frac{1}{1+t}+\frac{1}{1-t})dt$
Now, $I_{2}=-\frac{1}{2}log(\frac{1+t}{1-t})+c_{22}$ ---(II)
The question is whether these two values of $I_2$ are same ? If, yes how do we show that both the values of $I_2$ as obtained in (I)&(II) are same ?
| A few things to point out:
*
*$\int\frac{1}{x}dx=\ln|x|+C$.
*$\ln\frac{a}{b}=-\ln\frac{b}{a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Determining series formula Is there any procedure to follow when determining the function of a series? This seems simple but for I can't figure it out.
$$ \frac15 + \frac18 + \frac1{11} +\frac1{14} + \frac1{17}+\cdots$$
| Generally speaking, when I see a series like this, I try to construct a generating function of some sort. In this case:
$$f(x) = \sum_{k=1}^{\infty} \frac{x^{3 k+2}}{3 k+2}$$
You can easily show that
$$f'(x) = \frac{x^4}{1-x^3}$$
So we integrate and use $f(0)=0$ to evaluate $f(1)$. In this case, it should be clear that this value will be infinite, but it may be instructive to see how the sum diverges.
We evaluate the integral by partial fractions:
$$\frac{x^4}{1-x^3} = x \left ( \frac{1}{1-x^3} - 1\right ) = \frac{1}{3} \left ( \frac{1}{1-x} + \frac{x+1/2}{(x+1/2)^2 + 3/4} - \frac{3/2}{(x+1/2)^2 + 3/4}\right)-x$$
Doing the integration and applying the initial condition reveals
$$f(x) = \frac{1}{6} \log{(1+x+x^2)} - \frac{1}{\sqrt{3}} \arctan{\left[\frac{2 x+1}{\sqrt{3}}\right]} - \frac{1}{3} \log{(1-x)} - \frac{1}{2} x^2 + \frac{\pi}{6 \sqrt{3}}$$
So that, as $x \rightarrow 1^-$, $f(x)$ diverges logarithmically as one may expect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/354236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I generate my linear transformation matrix in this example? Let $T$ be the linear transformation of space of polynomials $P^3$ given by
$$T(a + bx + cx^2) = a + b(x + 1) + c(x + 1)^2$$
Find all eigenvalues and eigenvectors of $T$.
| Hints: $a+bx+cx^2$ is equivalent to the vector $[a,b,c]^{T}$. Expanding the RHS
\begin{align}
T=(a+b+c)+(b+2c)x+cx^2
\end{align}
In terms of matrices, $T$ should be equivalent to
\begin{align}
\begin{bmatrix}
a+b+c \\ b+2c \\ c
\end{bmatrix}=
\begin{bmatrix}
1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
a \\ b \\ c
\end{bmatrix}
\end{align}
Now try to find the eigenvectors.
| {
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"source": "stackexchange",
"question_score": "2",
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Finding the maximum of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$ If $a,b,c,d$ are distinct real numbers such that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}=4$ and $ac=bd$.
Then how would we calculate the maximum value of $$\dfrac{a}{c}+\dfrac{b}{d}+\dfrac{c}{a}+\dfrac{d}{b}.$$
I was unable to proceed due to the 'distinct'.
| Substituting $d=\frac{ac}{b}$ the constraint becomes
$$\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)=4$$
while the function to be maximized becomes
$$\frac{a}{c}+\frac{b^2}{ac}+\frac{c}{a}+\frac{ac}{b^2}=\frac{b}{c}\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{c}{b}\left(\frac{a}{b}+\frac{b}{a}\right)=\left(\frac{a}{b}+\frac{b}{a}\right)\left(\frac bc+\frac cb\right).$$
Now set
$$x=\frac{a}{b}+\frac{b}{a}\qquad \mathrm{and}\qquad y=\frac bc+\frac cb$$
so that $x+y=4$ and we want to maximize $xy$. Observe that if $x$ is positive then $a$ and $b$ have the same sign and thus
$$x=\frac ab+\frac ba=\frac{(a-b)^2}{ab}+2>2$$
because $a$ and $b$ are distinct. The same clearly also holds for $y$.
But then $x$ and $y$ must have opposite signs: they can not be both negative because $x+y=4$, and they can not be both positive because the sum would be strictly greater than $4$. Suppose for example that $x<0$, then
$$x=\frac ab+\frac ba=\frac{(a+b)^2}{ab}-2\leq -2$$
because $a$ and $b$ have opposite signs. Therefore $y=4-x\geq 6$ which implies $xy\leq -12$. Putting wlog $b=1$ and solving for the equality cases we obtain the equations
$$a+\frac 1a=-2\qquad\mathrm{and}\qquad c+\frac1c=6$$
which have solutions and so the upper bound $-12$ is attained.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Laurent series $z=i$ To find the Laurent series expansion for $\frac{1}{1+z^2}$, centered at $z=i$ would using partial fraction decomposition be the right idea? So, $\frac{1}{1+z^2}$=$\frac{1}{(z+i)(z-i)}$=$\frac{\frac{-1}{2i}}{z+i} +\frac{\frac{1}{2i}}{z-i}$?
| First:
$$\frac{1}{z^2+1}=\frac{1}{(z-i)(z+i)}=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)$$
Now
$$|z-i|<1\implies\frac{1}{z+i}=\frac{1}{z-i+2i}=\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}=\frac{1}{2i}\left(1-\frac{z-i}{2i}+\frac{(z-i)^2}{-4\cdot 2!}+\ldots\right)\implies$$
$$\frac{1}{z^2+1}=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{2i}\left(1-\frac{z-i}{2i}-\frac{(z-i)^2}{8}+\ldots\right)\right)=$$
$$\frac{1}{2i(z-i)}+\frac{1}{4}-\frac{z-i}{8i}-\frac{(z-i)^2}{32}+\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The right way to calculate the volume obtained by rotating the area between 2 graphs around the x axis If i have 2 graphs: $f(x)=x\cdot \frac{\sqrt{1-x^{2}}}{2},\:g(x)=\frac{\sqrt{1-x^{2}}}{2}$
And need to calculate the volume obtained by rotating the area between $f(x)$ and $g(x)$
Around the $x$-axis.
I know i need to compute the integral: $\pi \int_{-1}^{1}(g^{2}(x)-f^{2}(x))$
But can i do that direct or do i need to calculate: $\pi( \int_{-1}^{0}(g^{2}(x)-f^{2}(x))+\int_{0}^{1}(g^{2}(x)-f^{2}(x)))$?
Because it doesn't give the same result.
Thanks.
| Since you have concerns, we do the calculation. We have $g^2(x)=\frac{1-x^2}{4}$ and $f^2(x)=\frac{x^2(1-x^2)}{4}$, and therefore the difference is $\frac{(1-x^2)^2}{4}$, which is $\frac{1}{4}(1-2x^2+x^4)$. Integrate from $0$ to $1$. An antiderivative is $\frac{1}{4}(x-\frac{2}{3}x^3+\frac{1}{5}x^5)$.
Do the substitution. We get $\frac{1}{4}\left(1-\frac{2}{3}+\frac{1}{5}\right)$, which should be $\frac{2}{15}$.
To integrate from $-1$ to $0$, we again need to substitute. We get $-\frac{1}{4}\left(-1+\frac{2}{3}-\frac{1}{5}\right)$, the same number.
To integrate if we integrate directly from $-1$ to $1$, we get
$$\frac{1}{4}\left(-1+\frac{2}{3}-\frac{1}{5}\right)-\frac{1}{4}\left(-1+\frac{2}{3}-\frac{1}{5}\right),$$
which is exactly what we get by integrating separately and adding.
But there was no need to do the calculation for negative $x$. For our integrand is an even function. The integral over any interval $[-a,a]$ is twice the integral from $0$ to $a$.
Remark: I would almost automatically note the symmetry, integrate from $0$ to $1$, and double. But I am exceptionally bad at handling negative numbers, and really don't like them. They are so $\dots$ negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\tan(75^\circ) = 2 + \sqrt{3}$ My (very simple) question to a friend was how do I prove the following using basic trig principles:
$\tan75^\circ = 2 + \sqrt{3}$
He gave this proof (via a text message!)
$1. \tan75^\circ$
$2. = \tan(60^\circ + (30/2)^\circ)$
$3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ) $
$4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$
Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ?
EDIT
Using the answers given I'm almost there:
*
*$\tan75^\circ$
*$\tan(45^\circ + 30^\circ)$
*$\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$
*$(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$
*$\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$
*$\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$
*multiply throughout by $(\sqrt{3}) + 1)$
Another alternative approach:
*
*$\tan75^\circ$
*$\tan(45^\circ + 30^\circ)$
*$\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$
*$\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$
*at point 6 in above alternative
| The formula you want to see is: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ for any degrees $x$ and $y$.
On the other hand, proving this tangent equality from the formulas $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$ and $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ will be a good exercise for a beginner.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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A sufficient condition to ensure $\alpha=\beta$ Let $\alpha$, $\beta$ be acute angles satisfying
$$
\frac{\sin 2\alpha}{\sin(2\alpha+\beta)}=\frac{\sin2\beta}{\sin(2\beta+\alpha)}
$$
Show then $\alpha=\beta$.
| I got a solution for this problem. Here we have to use the following identities
\begin{eqnarray*}
\sin A\sin B&=&-\frac{1}{2}[\cos(A+B)-\cos(A-B)],\\
\cos A-\cos B&=&-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},\\
\sin 3A&=&3\sin A-4\sin^3 A,\\
\sin 5A&=&16\sin^5 A-20\sin^3 A+5\sin A.
\end{eqnarray*}
We obtain
\begin{eqnarray*}
&&\frac{\sin 2\alpha}{\sin(2\alpha+\beta)}=\frac{\sin 2\beta}{\sin(2\beta+\alpha)}\\
&\Leftrightarrow&\sin 2\alpha\sin(2\beta+\alpha)=\sin 2\beta\sin(2\alpha+\beta) \\
&\Leftrightarrow&-\frac{1}{2}[\cos(3\alpha+2\beta)-\cos(\alpha-2\beta)]=-\frac{1}{2}[\cos(3\beta+2\alpha)-\cos(\beta-2\alpha)]\\
&\Leftrightarrow&\cos(3\alpha+2\beta)-\cos(3\beta+2\alpha)=\cos(\alpha-2\beta)-\cos(\beta-2\alpha)\\
&\Leftrightarrow&-2\sin\frac{5(\alpha+\beta)}{2}\sin\frac{\alpha-\beta}{2}=-2\sin\frac{3(\alpha+\beta)}{2}\sin\frac{-\alpha-\beta}{2}\\
&\Leftrightarrow&\sin\frac{5(\alpha+\beta)}{2}\sin\frac{\alpha-\beta}{2}=-\sin\frac{3(\alpha-\beta)}{2}\sin\frac{\alpha+\beta}{2}\\
&\Leftrightarrow&\sin\frac{\alpha-\beta}{2}\left[16\sin^5 \frac{\alpha+\beta}{2}-20\sin^3 \frac{\alpha+\beta}{2}+5\sin \frac{\alpha+\beta}{2}\right]\\
&&=-\sin\frac{\alpha-\beta}{2}\left[3-4\sin^2\frac{\alpha-\beta}{2}\right]\sin\frac{\alpha+\beta}{2}.
\end{eqnarray*}
Thus we have
Case 1: $\sin\frac{\alpha-\beta}{2}=0$. Since $\alpha,\beta$ are acute angles, so $\alpha=\beta$.
Case 2: $16\sin^5 \frac{\alpha+\beta}{2}-20\sin^3 \frac{\alpha+\beta}{2}+5\sin \frac{\alpha+\beta}{2}=-\left[3-4\sin^2\frac{\alpha-\beta}{2}\right]\sin\frac{\alpha+\beta}{2}$. Since $\sin\frac{\alpha+\beta}{2}\neq 0$, we obtain
$$ 4\sin^4 \frac{\alpha+\beta}{2}-5\sin^2 \frac{\alpha+\beta}{2}+2=\sin^2\frac{\alpha-\beta}{2}.$$
Letting $x=\sin^2\frac{\alpha+\beta}{2}$, we have
$$ 4x^2-5x+1=-1+\sin^2\frac{\alpha-\beta}{2}$$
or
$$ (1-x)(1-4x)=-1+\sin^2\frac{\alpha-\beta}{2}. $$
Note $x\in[0,1]$. If $\frac{1}{4}\le x\le 1$, then $1-4x\le 0$ and hence
$$ (1-x)(1-4x)=-(1-x)(4x-1)=-\frac{1}{4}(4-4x)(4x-1)\ge -\frac{1}{4}\frac{9}{4}=-\frac{9}{16} $$
from which, we have $-1+\sin^2\frac{\alpha-\beta}{2}\ge -\frac{9}{16}$ or $\sin^2\frac{\alpha-\beta}{2}\ge \frac{7}{16}>0$. This implies $\alpha\neq \beta$.
If $0\le x<\frac{1}{4}$, we have $(1-x)(1-4x)> 0$ and $-1+\sin^2\frac{\alpha-\beta}{2}\le 0$ and hence this will never occur.
Thus only Case 1 holds or $\alpha=\beta$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimal Distance between two curves What is the minimal distance between curves?
*
*$y = |x| + 1$
*$y = \arctan(2x)$
I need to set a point with $\cos(t), \sin(t)$?
| Let $(a,|a|+1)$ be a point on the first curve and let $(b,\arctan(2b))$ be a point on the second curve.
Half the distance between the two points squared is $$\frac{1}{2}d^2 = \frac{1}{2}(a-b)^2+\frac{1}{2}(|a|+1-\arctan(2b))^2.$$
To find the minimum of this expression we set the partial derivatives to zero:
$$\frac{1}{2}\frac{\partial d^2}{\partial a} = (a-b)+(|a|+1-\arctan(2b))\frac{a}{|a|} = 0$$
and
$$\frac{1}{2}\frac{\partial d^2}{\partial b} = (b-a)+(|a|+1-\arctan(2b))\frac{-2}{1+4b^2}=0.$$
Adding these two equations gives
$$ (|a|+1-\arctan(2b))\left(\frac{a}{|a|}-\frac{2}{1+4b^2}\right)=0.$$
If the first term is to be zero, then $\frac{1}{2}\frac{\partial d^2}{a}=0$ implies $a=b$, there is no solution for $1+|a|=\arctan(2a)$ however.
If the second term is to be zero, we have $a>0$ and $b=\pm\frac{1}{2}$.
$\frac{1}{2}\frac{\partial d^2}{a}=0$ then reduces to $$ 0=a\mp\frac{1}{2}+(a+1-\arctan(\pm 1)) = 2a+1\mp\left(\frac{1}{2}+\frac{\pi}{4}\right).$$
Since $a>0$ we need to pick $b=\frac{1}{2}$ and thus $a = \frac{\pi-2}{8}$.
Hence the closest points are $(\frac{\pi-2}{8},\frac{\pi+6}{8})$ and $(\frac{1}{2},\frac{\pi}{4})$ and their distance is $$\sqrt{\left(\frac{\pi-2}{8}-\frac{1}{2}\right)^2+\left(\frac{\pi+6}{8}-\frac{\pi}{4}\right)^2} = \frac{\sqrt{2}(6-\pi)}{8}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Pythagorean triples and perfect squares This problem is giving me difficulty:
Show that in any Pythagorean triple there exist at most a single perfect square
So far I've been working with the equations for primitive Pythagorean Triples (ie $x = m^2 - n^2$, $y = 2mn$, $z = m^2 + n^2$) but that hasn't really worked out and the problem doesn't necessarily require the triples to be primitive, so I'm stuck here. I will appreciate any help.
| Are you familiar with the following?
The area of a Pythagorean triangle is never square.
For a proof, let $H$ be the set of $h>0$ such that there exists a
primitive Pythagorean triangle of hypotenuse $h$ whose area is square. Suppose that
$H\ne\emptyset$, then we can take $h_0=\min\left(H\right)$, and write
$h_0^2=x^2+y^2$, for some $x,y>0$. One of $x$, $y$ must be even, so assume that $2\mid x$.
The area of the Pythagorean triangle is given by $A=\frac{1}{2}xy$, and since the triple is primitive, we may write:
\begin{equation*}
A = \frac{1}{2}xy= \frac{1}{2}\cdot 2pq\left(p^2-q^2\right)
= pq\left(p+q\right)\left(p-q\right).
\end{equation*}
But $p$, $q$, $p+q$, $p-q$ are pairwise coprime, and their product is a
square, so we may write:
\begin{align*}
p &= r^2, & q &= s^2, & p+q &= t^2, & p-q &= u^2,
\end{align*}
for some $r,s,t,u>0$. Putting these together we get that:
\begin{equation*}
t^2=u^2+2s^2\Longleftrightarrow 2s^2=\left(t+u\right)\left(t-u\right),
\end{equation*}
so $2\mid \left(t+u\right)\left(t-u\right)$. The product of two integers is
even if and only if at least one of the integers is even. But if $2\mid\left(t+u\right)$
or $2\mid\left(t-u\right)$, then $t$ and $u$ must have the same parity, and it follows that
$2\mid\left(t\pm u\right)$.
Since $\left(t^2,u^2\right)=1$, we must have
$\left(t+u,t-u\right)=2$, and so we may write:
\begin{align*}
t+u &= 2v_0, & t-u &= 2w_0,
\end{align*}
for some $v_0,w_0>0$ with $\left(v_0,w_0\right)=1$. Recall that:
\begin{align*}
2s^2 &= \left(t+u\right)\left(t-u\right)= 4v_0w_0, \\
s^2 &= 2v_0w_0.
\end{align*}
So one of $v_0$, $w_0$ must be a square, and the other must be twice a
square. Hence $v_0=v^2$, $w_0=2w^2$, and:
\begin{align*}
t+u &= 2v^2, & t-u &= 4w^2,
\end{align*}
or vice versa. Note that in either case, we get that $s=2vw$. Now adding and subtracting the above equations we see that:
\begin{align*}
t &= v^2+2w^2, & \pm u &= v^2-2w^2.
\end{align*}
Consequently:
\begin{equation*}
r^2=p=\frac{1}{2}\left(t^2+u^2\right)=v^4+4w^4,
\end{equation*}
and $\left(v^2,2w^2,r\right)$ forms a Pythagorean triangle with area
$\left(vw\right)^2$, so $r=\sqrt{p}\in H$. But $h_0=p^2+q^2$, and clearly
$\sqrt{p}<p^2+q^2$, so $r<h_0$, contradicting the minimality of $h_0$.
Any Pythagorean triple with more than one perfect square would satisfy the Diophantine equation $x^4+y^4=z^2$, or $x^4+y^2=z^4$, and you can find Pythagorean triangles with square areas if either of these equations have integer solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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$f: \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$ The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.
| Let $x-1=y$, then,
$$(y-1)f(y+1)-(y+2)f(y)=3$$
$$\implies f(y+1)=\frac{3+(y+2)f(y)}{y-1} \;\;\;\;\;(1)$$
Lemma: $\forall \; n \geq 2 \in \mathbb{N}$, $f(n)=n(n-1)(n+1)-1$.
Base Case: If $n=2$ then $f(n)=1 \cdot 2 \cdot 3 -1=5$ which is true by information provided in the question.
Inductive Step: Assume for $n=k$ that $f(k)=k(k-1)(k+1)-1$.
Then by equation $(1)$, $$f(k+1)=\frac{3+(k+2)(k(k-1)(k+1)-1)}{k-1}$$
$$=\frac{k^4+2k^3-k^2-3k+1}{k-1}$$
$$=\frac{(k-1)(k^3+3k^2+2k-1)}{k-1}$$
$$=k^3+3k^2+2k-1$$
$$=k(k^2+3k+2)-1$$
$$=k(k+1)(k+2)-1$$
Thus completing the induction.
Hence our Lemma is true and $f(n)=n(n-1)(n+1)-1 \; \forall \; n \geq 2 \in \mathbb{N}$
In particular, if $n=2013$, $f(2013)=2013 \cdot 2012 \cdot 2014-1=8157014183$ (I confess I used a calculator).
P.S I found the lemma by trying small cases.
| {
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Evaluating a trigonometric integral using residues Finding the trigonometric integral using the method for residues:
$$\int_0^{2\pi} \frac{d\theta}{ a^2\sin^2 \theta + b^2\cos^2 \theta} = \frac{2\pi}{ab}$$ where $a, b > 0$.
I can't seem to factor this question
I got up to $4/i (z) / ((b^2)(z^2 + 1)^2 - a^2(z^2 - 1)^2 $
I think I should be pulling out $a^2$ and $b^2$ out earlier but not too sure how.
| Without residues and same dirty trigonometric trick as before:
$$\frac{1}{a^2\sin^2t+b^2\cos^2t}=\frac{1}{b^2\cos^2t}\frac{1}{1+\frac{a^2}{b^2}\tan^2t}=\frac{1}{ab}\frac{\frac{a}{b\cos^2t}}{1+\left(\frac{a}{b}\tan t\right)^2}$$
and since
$$\frac{a}{b\cos^2t}=\left(\frac{a}{b}\tan t\right)'$$
we finally get:
$$\int\limits_0^{2\pi}\frac{dt}{a^2\sin^2t+b^2\cos^2t}=2\int\limits_{-\pi/2}^{\pi/2}\frac{dt}{a^2\sin^2t+b^2\cos^2t}=\left.\frac{2}{ab}\arctan\frac{a}{b}\tan t\right|_{-\pi/2}^{\pi/2}=\frac{2\pi}{ab}$$
| {
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"url": "https://math.stackexchange.com/questions/367798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Symmetry properties of $\sin$ and $\cos$. Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$? For a question such as:
If $\sin(x) = 0.34$, find the value of $\cos\left(\frac{3\pi}{2} - x\right)$.
The solution says that:
\begin{align*}
\cos\left(\frac{3\pi}{2} - x\right) &= \cos\left(-\frac{\pi}{2} - x\right)\\
&= \cos\left(\frac{\pi}{2} + x\right)\\
&= -\sin(x)
\end{align*}
Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$?
Similarly, if $\cos(x) = 0.6$, find $\sin\left(\frac{3\pi}{2} + x\right)$, the solution says $\sin\left(\frac{3\pi}{2} + x\right)$ is equal to $\cos(\pi+x)$. How did they get $\cos(\pi+x)$?
| For your first question, recall that $-\frac{\pi}{2}$ is just a different "angle name" for the same direction as $\frac{3\pi}{2}$. The first is obtained by turning clockwise from the positive x-axis ($\theta = 0$) by $90^{\circ}$ or $\frac{\pi}{2}$ radians, and the second by turning counter-clockwise from $\theta = 0$ by $270^{\circ}$ or $\frac{3\pi}{2}$ radians.
As for your second question, the solution is using what are called the "co-relations", which include
$$\sin \theta = \cos (\frac{\pi}{2} - \theta) , \cos \theta = \sin (\frac{\pi}{2} - \theta) , $$
$$\cos \theta = \sin (\frac{\pi}{2} + \theta) , -\sin \theta = \cos (\frac{\pi}{2} + \theta) , \text{etc.} $$
So we can write $\sin (\frac{3 \pi}{2} + x ) = \cos(\frac{\pi}{2} - [\frac{3\pi}{2} + x ]) = \cos(-\pi - x) = \cos( - [\pi + x ])$ . But we also know that $\cos(-\theta) = \cos\theta$ , so $\sin (\frac{3 \pi}{2} + x ) = \cos( - [\pi + x ]) = \cos( \pi + x )$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
What is going on with this constrained optimization? I'd like to figure out what is going on when trying to maximize a function (below $a_i$ are real numbers)
$F = a_1a_2 + a_2a_3 + \cdots + a_{n - 1}a_n + a_na_1;$
When we have active constraints
$h_1 = a_1 + a_2 + \cdots + a_n = 0;$
$h_2 = a_1^2 + a_2^2 + \cdots +a_n^2 = 1;$
So my gradients
$\nabla F = (a_2+a_n,a_3+a_1, \ldots , a_{n-2}+a_n,a_{n-1}+a_1);$
$\nabla h_1 = (1, 1, \ldots , 1);$
$\nabla h_2 = (2a_1,2a_2,\ldots , 2a_n);$
Kuhn-Tucker should provide necessary conditions in this case which is I guess pretty much the same as the method of Lagrange multipliers:
$$\begin{cases}
a_2+a_n = \lambda_1 + 2\lambda_2a_1; \\
a_3+a_1 = \lambda_1 + 2\lambda_2a_2; \\
a_4+a_2 = \lambda_1 + 2\lambda_2a_3; \\
\ldots \\
a_{n-1} + a_1 = \lambda_1 + 2\lambda_2 a_n;
\end{cases}$$
In a matrix form this is
$$\begin{pmatrix}
2\lambda_2 & -1 & 0 & 0 & \ldots & 0 & -1 \\
-1 & 2\lambda_2 & -1 & 0 & \ldots & 0 & 0 \\
0 & -1 & 2\lambda_2 & -1 & \ldots &0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 &0 &0 & 0 & \ldots & 2\lambda_2 & -1 \\
-1 &0 &0 & 0 & \ldots & -1 & 2\lambda_2
\end{pmatrix}
\begin{pmatrix}
a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_{n-1} \\ a_n
\end{pmatrix} =
- \lambda_1\begin{pmatrix}
1 \\ 1 \\ 1 \\ \vdots \\ 1 \\ 1
\end{pmatrix}
$$
How should I proceed?
| Well, on adding up the n equations you have, you get
$$2\sum_{i=1}^{n}a_i = n\lambda_1 + 2\lambda_2\sum_{i=1}^{n}a_i$$
which gives, $\lambda_1 = 0$ as $\sum_{i=1}^{n}a_i=0$.
So now you need to find the nullspace of the matrix, which should hopefully be easier?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/370251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the polar form of $ z = 1- \sin (\alpha) + i \cos (\alpha) $? How do I change $ z = 1- \sin (\alpha) + i \cos (\alpha) $ to polar? I got $r = (2(1-\sin(\alpha))^{\frac{1}{2}} $. I have problems with the exponential part. What should I do now?
| $1-\sin\alpha\ge 0$
If $1-\sin\alpha=0,\cos\alpha=0, z=0+i\cdot0$
If $1-\sin\alpha> 0$
Let $1-\sin\alpha=r\cos t, \cos\alpha=r\sin t$ where $r> 0$
Squaring & adding we get
$r^2=2(1-\sin\alpha)=2(\cos\frac\alpha2-\sin\frac\alpha2)^2=4\cos^2(\frac\pi4+\frac\alpha2)$
On division, $\frac{r\sin t}{r\cos t}=\frac{\cos \alpha}{1-\sin \alpha}$
$\implies \tan t=\frac{\cos^2\frac\alpha2-\sin^2\frac\alpha2}{(\cos\frac\alpha2-\sin^2\frac\alpha2)^2}$
$=\frac{\cos\frac\alpha2+\sin\frac\alpha2}{\cos\frac\alpha2-\sin\frac\alpha2}$
$=\frac{1+\tan \frac\alpha2}{1-\tan\frac\alpha2}=\tan(\frac\pi4+\frac\alpha2)$
$t=n\pi+\frac\pi4+\frac\alpha2$ where $n$ is any integer
If $n$ is odd$=2m+1$(say), $t=(2m+1)\pi+\frac\pi4+\frac\alpha2,\sin t=\sin\{(2m+1)\pi+\frac\pi4+\frac\alpha2\}=-\sin(\frac\pi4+\frac\alpha2)$ and $\cos t=\cos\{(2m+1)\pi+\frac\pi4+\frac\alpha2\}=-\cos(\frac\pi4+\frac\alpha2)$
If $n$ is even$=2m$(say), $t=2m\pi+\frac\pi4+\frac\alpha2,\sin t=\sin(2m\pi+\frac\pi4+\frac\alpha2)=\sin(\frac\pi4+\frac\alpha2)$ and $\cos t=\cos(2m\pi+\frac\pi4+\frac\alpha2)=\cos(\frac\pi4+\frac\alpha2)$
Now, $\cos(\frac\pi4+\frac\alpha2)$ will be $<0$
if $2n\pi+\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac{3\pi}2\iff 4n\pi+\frac{\pi}2<\alpha<4n\pi+\frac{5\pi}2$
Then $r=-2\cos(\frac\pi4+\frac\alpha2)$ Clearly $t=(2m+1)\pi+\frac\pi4+\frac\alpha2\equiv \pi+\frac\pi4+\frac\alpha2\pmod{2\pi}$
Similarly, $\cos(\frac\pi4+\frac\alpha2)$ will be $>0$
if $2n\pi-\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac\pi2\iff 4n\pi-\frac{3\pi}2<\alpha<4n\pi+\frac\pi2$
Then, $r=2\cos(\frac\pi4+\frac\alpha2),$ Clearly $t=2m\pi+\frac\pi4+\frac\alpha2\equiv \frac\pi4+\frac\alpha2\pmod{2\pi}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Does one always use augmented matrices to solve systems of linear equations? The homework tag is to express that I am a student with no working knowledge of math.
I know how to use elimination to solve systems of linear equations. I set up the matrix, perform row operations until I can get the resulting matrix into row echelon form or reduced row echelon form, and then I back substitute and get the values for each of my variables.
Just some random equations: $a + 3b + c = 5 \\ a + 0 + 5c = -2$
My question is, don't you always want to be using an augmented matrix to solve systems of linear equations? By augmenting the matrix, you're performing row operations to both the left and right side of those equations. By not augmenting the matrix, aren't you missing out on performing row operations on the right side of the equations (the $5$ and $-2$)?
The sort of problems I'm talking about are homework/test-level questions (as opposed to real world harder data and more complex solving methods?) where they give you $Ax = b$ and ask you to solve it using matrix elimination.
Here is what I mean mathematically:
$[A] =
\begin{bmatrix}
1 & 3 & 1 \\
1 & 0 & 5 \\
\end{bmatrix}
$
$ [A|b] =
\left[\begin{array}{ccc|c}
1 & 3 & 1 & 5\\
1 & 0 & 5 & -2\\
\end{array}\right]
$
So, to properly solve the systems of equations above, you want to be using $[A|b]$ to perform Elementary Row Operations and you do not want to only use $[A]$ right?
The answer is: yes, if you use this method of matrices to solve systems of linear equations, you must use the augmented form $[A|b]$ in order to perform EROs to both $A$ and $b$.
| Augmenting the matrix is just notational shorthand.
When you do row operations, what you're really doing is multiplying both sides of an equation by some matrix $P_1$:
$$\begin{align*}
Ax &= b \\
P_1Ax &= P_1b \tag{$P_1$ represents some row operation} \\
P_2P_1Ax &= P_2 P_1 b \tag{$P_2$ represents some row operation} \\
&\vdots \\
Dx &= P_k \cdots P_2 P_1b \tag{$D$ is now a triangular matrix}
\end{align*}
$$
Doing in situ row operations obscures this fact, and augmenting it obscures it even further.
Here is an example of solving a system using augmented matrices:
$$\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}$$
becomes
$$
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 2 \\ 0 & 1 & 1 & 3 \end{array}\right)$$
Then you do the following operations:
$$
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\ 0 & -1 & 0 & 1 \\ 0 & 1 & 1 & 3
\end{array}\right) \tag{subtract row 1 from row 2}$$
$$
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\ 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 4
\end{array}\right) \tag{add row 2 to row 3}$$
And now you have an upper-triangular matrix that you can backsolve.
What you're really doing, however, is this:
$$\begin{align*}
\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} &= \begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix} \\
\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} &= \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix} \\
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}
\end{align*}$$
Multiply everything out, you get:
$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 4\end{pmatrix}$$
which is exactly what appears in your augmented form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/373667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$
Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$.
Please brief about the concept behind this to solve such problems. Thanks.
| The idea is to reach at $\equiv\pm1\pmod n$ for a given modulo integer $n$
In general we should utilize Fermat's little theorem for prime modulo
or Euler's Totient Theorem or Carmichael Function for non-prime modulo
unless we can reach $\pm1$ easily like below.
$2^3=8\equiv1\pmod 7\implies 2^{100}=2\cdot (2^3)^{33}\equiv2\cdot 1^{33}\pmod 7\equiv2$
$4^3=64\equiv1\pmod 7\implies 4^{100}=4\cdot (4^3)^{33}\equiv4\cdot1^{33}\pmod 7\equiv4$
$3^3=27\equiv-1\pmod 7\implies 3^{100}=3\cdot (3^3)^{33}\equiv3\cdot(-1)^{33}\pmod 7\equiv-3$
$5^3=125\equiv-1\pmod 7\implies 5^{100}=5\cdot (5^3)^{33}\equiv5\cdot(-1)^{33}\pmod 7\equiv-5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/377378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
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Discrete Time Fourier Transform example: $x = [1 \; 2 \; 3 \; 4]^T \; \rightarrow \; X=?$ How do I find the Discrete Fourier Transform of the sequence below?
$$ x = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$$
Show all steps.
| Apply Discrete Time Fourier Transformation on the vector
$$ x = \begin{bmatrix} x[0] \\ x[1] \\ x[2] \\ x[3] \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} $$
to find the frequency domain vector
$$ \mathcal{F} \left \{ x \right \} = X = \begin{bmatrix} X[0] \\ X[1] \\ X[2] \\ X[3] \end{bmatrix}. $$
Use the formula
$$ X[k] = \sum_{n=0}^{N-1} x[n] \cdot e^{-\dfrac{j 2 \pi k n}{N}}, $$
where $N$ is the number of elements in the time domain vector (i.e.; $\;N\!=\!4\;$ in our case). If we rewrite the sum in the open form:
$$ \begin{array}{llll}
X[k] & = 1e^{-j\dfrac{0\pi}{4}k} & + 2e^{-j\dfrac{2\pi}{4}k} & + 3e^{-j\dfrac{4\pi}{4}k} & + 4e^{-j\dfrac{6\pi}{4}k} \\
& = 1 & + 2e^{-j\dfrac{\pi}{2}k} & + 3e^{-j\pi k} & + 4e^{-j\dfrac{3\pi}{2}k} \\
\end{array} $$
Next, find the elements of $X$ one by one:
$$ \begin{array}{lllll}
X[0] & = 1 & + 2 & + 3 & + 4 & = \mathbf{+10} \\
\\
X[1] & = 1 & + 2e^{-j\dfrac{\pi}{2}} & + 3e^{-j\pi} & + 4e^{-j\dfrac{3\pi}{2}} \\
& = 1 & - j2 & - 3 & + j4e & = \mathbf{-2 + j2} \\
\\
X[2] & = 1 & + 2e^{-j\pi} & + 3e^{-j2\pi} & + 4e^{-j3\pi} \\
& = 1 & - 2 & + 3 & - 4 & = \mathbf{-2} \\
\\
X[3] & = 1 & + 2e^{-j\dfrac{3\pi}{2}} & + 3e^{-j3\pi} & + 4e^{-j\dfrac{9\pi}{2}k} \\
& = 1 & + j2 & - 3 & - j4 & = \mathbf{-2 - j2}
\end{array} $$
Hence, the Discrete Fourier Transform of $x$ is
$$ X = \mathcal{F} \left \{ x \right \} = \mathcal{F} \left \{ \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \right \} = \begin{bmatrix} 10 \\ −2+i2 \\ −2 \\ −2−i2 \end{bmatrix}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/380287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex roots of polynomial equations with real coefficients Consider the polynomial $x^5 +ax^4 +bx^3 +cx^2 +dx+4$ where $a, b, c, d$ are real
numbers. If $(1 + 2i)$ and $(3 - 2i)$ are two roots of this polynomial then what is the value of $a$ ?
| As the coefficients of the different powers of $x$ are real, all the complex roots must occur with conjugate pair.
So, the other two roots are $1-2i,3+2i$
So if $p$ is the fifth root,
using Vieta's Formulas $(1-2i)(3+2i)(1+2i)(3-2i)p=(-1)^5\frac41$
and $(1-2i)+(3+2i)+(1+2i)+(3-2i)+p=-\frac a1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/381114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by mathematical induction that $1 + 1/4 +\ldots + 1/4^n \to 4/3$ Please help. I haven't found any text on how to prove by induction this sort of problem:
$$
\lim_{n\to +\infty}1 + \frac{1}{4} + \frac{1}{4^2} + \cdots+ \frac{1}{4^n} = \frac{4}{3}
$$
I can't quite get how one can prove such. I can prove basic divisible "inductions" but not this. Thanks.
| I'm not really sure if it is possible to prove this using induction but we can do something else.
The sum $\displaystyle 1 + \frac{1}{4} + \frac{1}{4^2} + \cdots \frac{1}{4^k}$ can be written as $\displaystyle\sum_{n=0}^k \frac{1}{4^n}$. Now $\displaystyle\sum_{n=0}^k \displaystyle\frac{1}{4^n} = \frac{1-\frac{1}{4^{k+1}}}{1-\frac{1}{4}}$. Now letting $k \rightarrow \infty$ allows us to conclude that $\displaystyle 1 + \frac{1}{4} + \frac{1}{4^2} + \cdots \frac{1}{4^k} \rightarrow \displaystyle\frac{4}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/382295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to write this conic equation in standard form? $$x^2+y^2-16x-20y+100=0$$
Standard form? Circle or ellipse?
| $$
\begin{align}
x^2+y^2-16x-20y+100 & = 0 \\ \\
\color{blue}{\bf x^2 -16x} + \color{red}{\bf y^2 -20y } + 100 & = 0
\end{align}
$$
We complete the square: $$x^2+bx+(b/2)^2-(b/2)^2+c= (x+b/2)^2+c-(b/2)^2$$
$$\begin{align}
\color{blue}{\bf x^2 - 16x} + \underbrace{\bf \left(\frac{-16}{2}\right)^2}_{\color{green}{\bf \large +64}} + \color{red}{\bf y^2 - 20y} + \underbrace{\bf \left(\frac{-20}{2}\right)^2}_{\color{green}{\bf \large + 100}} + 100 {\color{green}{\bf - 64 -100}} & = 0 \\ \\
(x^2 - 16x + 64) + (y^2 - 20 y + 100) - 64 & = 0\\ \\
(x - 8)^2 + (y - 10)^2 - 64 & = 0\\ \\
(x - 8)^2 + (y - 10)^2 & = 64 = (8)^2 \\
\end{align}
$$
The equation of a circle with center $(a, b)$ and radius $r$ is given by $$(x - a)^2 + (y - b)^2 = r^2$$
In your case, we have a circle centered at $(8, 10)$ with radius $r = 8$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integrating a school homework question. Show that $$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \frac{a\sqrt{3}+b-\pi}{6},$$ where $a$ and $b$ are constants to be found.
Answer is: $$\frac{24\sqrt3-48-\pi}{6}$$
Thank you in advance!
| $$I=\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \int_0^1\frac{4x-5}{\sqrt{4-(x-1)^2}}dx$$
Let $x-1=2\sin\theta\implies dx=2\cos\theta d\theta$
If $x=0, \sin\theta=\frac12, \theta=\frac\pi6$
If $x=1, \sin\theta=0, \theta=0$
$$\text{So,}I=\int_{\frac\pi6}^0\frac{4(2\sin\theta+1)-5}{2\cos\theta}\cdot2\cos\theta d\theta$$
$$=8\int_{\frac\pi6}^0\sin\theta d\theta-\int_{\frac\pi6}^0\theta d\theta$$
$$=8[-\cos\theta]_{\frac\pi6}^0-\left(0-\frac\pi6\right)$$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{1}{4^n}\binom{1/2}{n} \stackrel{?}{=} \frac{1}{1+2n}\binom{n+1/2}{2n}$ - An identity for fractional binomial coefficients In trying to write an answer to this question:
calculate the roots of $z = 1 + z^{1/2}$ using Lagrange expansion
I have come across the identity
$$
\frac{1}{4^n}\binom{1/2}{n} = \frac{1}{1+2n}\binom{n+1/2}{2n}.
\tag{1}
$$
Could anyone help me prove this? The few identities I know for binomial coefficients aren't enough to get anywhere useful and I don't see a way to account for the large difference in the number of factors in their respective numerators and denominators.
| Simply compute ratios of each side. Introducing notation $\Theta_n f(n) = \frac{f(n+1)}{f(n)}$:
$$
\Theta_n \frac{1}{4^n} \binom{1/2}{n} = \frac{1}{4} \Theta_n \binom{1/2}{n} =\frac{1}{4} \frac{\Gamma(n+1) \Gamma\left(\frac{3}{2}-n\right)}{\Gamma(n+2) \Gamma\left(\frac{1}{2}-n\right)} = -\frac{2n-1}{8(n+1)}
$$
On another hand:
$$
\Theta_n \frac{1}{2n+1} \binom{n+1/2}{2n} = \frac{2n+1}{2n+3} \Theta_n \binom{n+1/2}{2n} = \frac{2n+1}{2n+3} \frac{\frac{\Gamma(n+5/2)}{\Gamma(2n+3) \Gamma(1/2-n)}}{\frac{\Gamma(n+3/2)}{\Gamma(2n+1) \Gamma(3/2-n)}} = \frac{2n+1}{2n+3} \frac{(n+3/2)(1/2-n)}{(2n+2)(2n+1)} = -\frac{1}{8} \frac{2n-1}{n+1}
$$
The identity is obviously true for $n=0$, which finishes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Find the following integral: $\int {{{1 + \sin x} \over {\cos x}}dx} $ My attempt:
$\int {{{1 + \sin x} \over {\cos x}}dx} $,
given : $u = \sin x$
I use the general rule:
$\eqalign{
& \int {f(x)dx = \int {f\left[ {g(u)} \right]{{dx} \over {du}}du} } \cr
& {{du} \over {dx}} = \cos x \cr
& {{dx} \over {du}} = {1 \over {\cos x}} \cr
& so: \cr
& \int {{{1 + \sin x} \over {\cos x}}dx = \int {{{1 + u} \over {\cos x}}{1 \over {\cos x}}du} } \cr
& = \int {{{1 + u} \over {{{\cos }^2}x}}du} \cr
& = \int {{{1 + u} \over {\sqrt {1 - {u^2}} }}du} \cr
& = \int {{{1 + u} \over {{{(1 - {u^2})}^{{1 \over 2}}}}}du} \cr
& = \int {(1 + u){{(1 - {u^2})}^{ - {1 \over 2}}}} du \cr
& = {(1 - u)^{ - {1 \over 2}}} + u{(1 - {u^2})^{ - {1 \over 2}}}du \cr
& = {1 \over {({1 \over 2})}}{(1 - u)^{{1 \over 2}}} + u - {1 \over {\left( {{1 \over 2}} \right)}}{(1 - {u^2})^{{1 \over 2}}} + C \cr
& = 2{(1 - u)^{{1 \over 2}}} - 2u{(1 - {u^2})^{{1 \over 2}}} + C \cr
& = 2{(1 - \sin x)^{{1 \over 2}}} - 2(\sin x){(1 - {\sin ^2}x)^{{1 \over 2}}} + C \cr
& = {(1 - \sin x)^{{1 \over 2}}}(2 - 2\sin x) + C \cr} $
This is wrong, the answer in the book is:
$y = - \ln |1 - \sin x| + C$
Could someone please explain where I integrated wrongly?
Thank you!
| Do this:
$ \int { \frac { 1+\sin { x } }{ \cos { x } } dx } =\int { \frac { \left( 1+\sin { x } \right) \left( 1-\sin { x } \right) }{ \cos { x } \left( 1-\sin { x } \right) } dx= } \int { \frac { \cos { x } }{ \left( 1-\sin { x } \right) } } dx$
Then do the substitution rule
$\left( 1-\sin { x } \right) =u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Finding the sum of the coefficients of polynomial of degree 21 Problem : Find the sum of the coefficients of the polynomial$ p(x) =(3x-2)^{17}(x+1)^4$
Solution :
$ p(x) =(3x-2)^{17}(x+1)^4 $
$= (a_0+a_1x+....a_{17}x^{17})(b_0+b_1x+....b_4x^4)$ for some $a_i; b_j$
= $(c_0+c_1x+...c_{21}x^{21})$ for some $c_k$
Sum of the coefficients of p(x) =$c_0+c_1+c_2+....c_{21}$
In solution it is given that the sum of the coefficients can be obtained by putting x =1 , my question is why x =1?
Please explain this concept.. else is clear in this problems... Thanks..
| If $P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{0}$, and we want to find $a_{n} + a_{n-1} + \cdots + a_{1} + a_{0}$, then we see that:
$P(1) = a_{n}(1)^{n} + a_{n-1}(1)^{n-1} + \cdots + a_{1}(1) + a_{0} = a_{n}(1) + a_{n-1}(1) + \cdots + a_{1}(1) + a_{0} = a_{n} + a_{n-1} + \cdots + a_{1} + a_{0}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify $\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $ How to simplify the following expression :
$$\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
| Write $1$ in the numerator as : $$\sec^2(\theta) - \tan^2(\theta)$$
$$\frac{(\sec\theta -\tan\theta)^2+\sec^2\theta - \tan^2\theta}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta -\tan\theta)^2+(\sec\theta - \tan\theta)(\sec\theta + \tan\theta)}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta - \tan\theta)(\sec\theta - \tan\theta + \sec\theta + \tan\theta)}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta - \tan\theta)(2 \sec\theta)}{\csc\theta(\sec\theta - \tan\theta)}$$
$$2 \tan\theta$$
Hence the simplified result is: $$2 \tan\theta$$
Hope the answer is clear !
| {
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What is the centroid of a hollow spherical cap? I have a unit hollow sphere which I cut along a diameter to generate two equivalent hollow hemispheres. I place one of these hemispheres on an (x,y) plane, letting it rest on the circular planar face where the cut occurred.
If the hemisphere was solid, we could write that its centroid in the above case would be given as $(0,0,\frac{3}{8})$. Given that the hemisphere is hollow, can we now write its centroid as $(0,0,\frac{1}{2})$?
| In Cartesian coordinates, the hemisphere can be represented by a rotation about the z-axis of a circle, for which we can use, say, $y^2 + z^2 = 1$ to produce the infinitesimal "belt" of surface area $ dS = 2 \pi y \ ds \ $ from
$$\frac{d}{dz} [y^2 + z^2] \ = \ \frac{d}{dz} [1] \ \Rightarrow \ \frac{dy}{dz} \ = \ -\frac{z}{y} $$
$$\Rightarrow \ ds \ = \ \sqrt{(\frac{dy}{dz})^2 + 1} \ \ dz \ = \ \sqrt{(-\frac{z}{y})^2 + 1} \ \ dz \ = \ \frac{\sqrt{y^2 + z^2}}{y} \ dz \ = \ \frac{1}{y} \ dz $$
$$\Rightarrow \ S \ = \ 2 \pi \int_a^b y \ ds \ = \ 2 \pi \int_0^1 y \ \cdot \frac{1}{y} \ dz \ = \ 2 \pi z \ |_0^1 \ = \ 2 \pi \ , $$
$$\Rightarrow \ M_{xy} \ = \ 2 \pi \int_a^b z \cdot y \ ds \ = \ 2 \pi \int_0^1 z \ dz \ = \ \pi z^2 \ |_0^1 \ = \ \pi \ , $$
giving the height of the centroid as $$\overline{z} = \frac{M_{xy}}{S} = \frac{1}{2} \ . $$
$$ $$
But the actual reason I decided to chime in here at all is that there is a variant method we can employ by extending it to one dimension higher than it is often used. Pappus' "centroid theorems" would give us the centroid of a semi-circular arc from
$$2 \pi \overline{x} \ \cdot \ \pi R \ = \ 4 \pi R^2 \ \Rightarrow \ \overline{x} \ = \frac{2}{\pi}R \ , $$
with $\pi R$ being the arclength of the curve and $4 \pi R^2$ the surface area of the sphere generated by revolution; by the same token, the centroid of a semi-circular region is given by
$$2 \pi \overline{x} \ \cdot \ \frac{1}{2}\pi R^2 \ = \ \frac{4}{3} \pi R^3 \ \Rightarrow \ \overline{x} \ = \ \frac{4}{3 \pi}R \ , $$
where $\frac{1}{2}\pi R^2$ is the area of the region and $\frac{4}{3} \pi R^3$ , the volume of the sphere of revolution. (These are probably familiar enough results.)
Here, we are working with a hemispherical shell and a hemispherical volume. It turns out that we can extend Pappus' theorems to deal with revolution generating the hypersurface area and hypervolume of a 4-sphere. Thus, for the shell,
$$2 \pi \overline{x} \ \cdot \ 2 \pi R^2 \ = \ 2 \pi^2 R^3 \ \Rightarrow \ \overline{x} \ = \frac{1}{2}R \ , $$
and for the solid hemisphere,
$$2 \pi \overline{x} \ \cdot \ \frac{2}{3}\pi R^3 \ = \ \frac{1}{2} \pi^2 R^4 \ \Rightarrow \ \overline{x} \ = \ \frac{3}{8}R \ , $$
from which your values for the unit sphere follow. This works as it does because, although Pappus would not have said it that way, his relations are connected with moment integrals.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form for $\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}$ Here is another infinite sum I need you help with:
$$\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}.$$
I was told it could be represented in terms of elementary functions and integers.
| Note that
$$\frac{2^{-n}}{2^{2^{-n}}-1}-\frac{2^{-(n-1)}}{2^{2^{-(n-1)}}-1} = \frac{2^{-n}}{2^{2^{-n}}+1} $$
Thus we have a telescoping sum. However, note that
$$\lim_{n \to \infty} \frac{2^{-n}}{2^{2^{-n}}-1} = \frac{1}{\log{2}}$$
Therefore the sum is
$$a_1-a_0 + a_2-a_1 + a_3-a_2 + \ldots + \frac{1}{\log{2}} = \frac{1}{\log{2}}- a_0$$
where
$$a_n = \frac{1}{2^n \left ( 2^{2^{-n}}-1\right)}$$
or
$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}= \frac{1}{\log{2}}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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Contour Integral and Complex Identity While studying I came across this problem:
(a) For $z=x+iy$, show that
$$|\cos \pi z|^2=\frac{1}{2}(\cos(2\pi x)+\cosh(2\pi y))$$
(b) For a positive integer let $\gamma N$ be the square connecting the points $\pm N\pm Ni$ oriented in positive direction. Show that
$$|\cos \pi z|\geq 1$$
for all $z$ on $\gamma N$.
(c) Making use of the contour integral
$$\int_{\gamma N} \frac{1}{z^3\cos \pi z}dz$$
evaluate the sum of the series
$$\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)^3}$$
I wish I could show more effort, but I honestly don't know where to begin or what the connection between parts is. ANY help appreciated.
| For (a), use the cosine addition formula: $\cos{(a+i b)} = \cos{a} \cos{i b} - \sin{a} \sin{i b}$. Note that $\cos{i b} = (e^{i (i b)} + e^{-i (i b)})/2 = (e^{-b}+e^b)/2 = \cosh{b}$. Similarly, you can show that $\sin{i b} = i \sinh{b}$. Then
$$\cos{(a+i b)} = \cos{a} \cosh{b} - i \sin{a} \sinh{b}$$
so that
$$\begin{align}|\cos{\pi z}|^2 &= \cos^2{\pi x} \, \cosh^2{\pi y} + \sin^2{\pi x} \, \sinh^2{\pi y}\\ &= \left (\frac{1+\cos{2 \pi x}}{2} \right )\left (\frac{1+\cosh{2 \pi y}}{2} \right )+\left (\frac{1-\cos{2 \pi x}}{2} \right )\left (\frac{\cosh{2 \pi y}-1}{2} \right )\\ &= \frac12 (\cos{2 \pi x} + \cosh{2 \pi y})\end{align}$$
Note that, on $\gamma_N$,
$$|\cos{\pi z}|^2 = \frac12 (1+\cosh{2 \pi N}) = \cosh^2{\pi N} \ge 1$$
Therefore, because $|\cos{\pi z}|\ge 1$:
$$\left | \int_{\gamma_N} \frac{dz}{z^3 \cos{\pi z}} \right | \le \frac{8 N}{N^3} = \frac{8}{N^2}$$
Therefore, the integral vanishes as $N \to \infty$. In that case, the sum of the residues of the poles of the integrand also vanish. There is a pole of order 3 at the origin, and poles at $z=(2 n+1)/2$ for $n \in \mathbb{Z}$. For the pole at the origin, the residue is
$$\frac12 \left[\frac{d^2}{dz^2} \sec{\pi z}\right ]_{z=0} = \frac{\pi^2}{2}$$
For the pole at $z=(2 n+1)/2$, the residue is
$$\frac{2^3}{(2 n+1)^3 (-\pi \sin{(n+1/2)\pi})} = -\frac{(-1)^{n} 8}{\pi (2 n+1)^3}$$
Summing over all $n$, we express that the sum of the residues is zero as
$$-\frac{8}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(2 n+1)^3} + \frac{\pi^2}{2} = 0 $$
Note that the sum is even in $n$; this the sum from $(-\infty,\infty)$ is twice the sum from $[0,\infty)$. Therefore
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)^3} = \frac{\pi^3}{32}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?
Thanks!
| For odd $n$ you can always write:
$$f_n(x) = \sum_{i=0}^n x^{2i} = (x^2+1)(x^{2n-2}+x^{2n-6}+x^{2n-10}+...)$$
For even $n$ this trick no longer works.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$
Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$
We have removed one square root.
Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?
Simplify: $2x-5 = 2\sqrt{(x-1)} + x$
Simplify more: $x-5 = 2\sqrt{(x-1)}$
Now do the "square root" thing again:
Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$
Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$
Square root removed
Thank you in advance for your help
| It is an identity:
$$(1+x)^2=1+2x+x^2$$
How? Well, consider this:
$$\begin{equation*}
\begin{split}
(1+x)^2&=(\color{blue}{1}+\color{red}{x})(1+x)\\
&=\color{blue}{1}(1+x)+\color{red}{x}(1+x)\\
&=1+x+x+x^2\\
&=1+2x+x^2\\
\end{split}
\end{equation*}$$
In general, $(a+b)^2=a^2+2ab+b^2$, which you can try to deduce yourself.
Hope this helps. Ask anything if not clear :)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to integrate $\int \sqrt{x^2+a^2}dx$ $a$ is a parameter. I have no idea where to start
| Since $\sinh^2(x)+1
=\cosh^2(x)
$,
this suggests
letting $x = a \sinh(y)$.
$dx = a \cosh(y)$
and $x^2+a^2 = a^2(\sinh^2(y)+1)
= a^2 \cosh^2(y)$,
so $a \cosh(y) = \sqrt{x^2+a^2}$.
$\begin{align}
\int{\sqrt{x^2+a^2}}dx
&=\int a^2 \cosh^2(y) dy\\
&= a^2 \int (e^{2y}+2+e^{-2y})\,dy/4\\
&= (a^2/4) (e^{2y}/2 + 2y - e^{-2y}/2)\\
&=(a^2/4)(\sinh(2y)+2y)\\
&=(a^2/4)(2\sinh(y)\cosh(y)+2y)\\
&=(a^2/2)((x/a) (1/a)\sqrt{x^2+a^2}+\sinh^{-1}(x/a))\\
&=(x \sqrt{x^2+a^2})/2+(a^2/2)\sinh^{-1}(x/a)
\end{align}
$
I'll leave it at this - you can find $\sinh^{-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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positive Integer value of $n$ for which $2005$ divides $n^2+n+1$
How Can I calculate positive Integer value of $n$ for which $2005$
divides $n^2+n+1$
My try:: $2005 = 5 \times 401$
means $n^2+n+1$ must be a multiple of $5$ or multiple of $401$
because $2005 = 5 \times 401$
now $n^2+n+1 = n(n+1)+1$
now $n(n+1)+1$ contain last digit $1$ or $3$ or $7$
$\bullet $ if last digit of $n(n+1)+1$ not contain $5$. So it is not divisible by $5$
Now how can I calculate it? please explain it to me.
| If a number is divisible by $2005=5\cdot401,$ it must be divisible by $5$ and by $401$
Now, as you have identified $n^2+n+1$ is not divisible by $5,$ it can not be divisible by any multiple of $5$ like $2005=5\cdot401$
Alternatively,
$$n^2+n+1\equiv0\pmod {2005}\implies n^2+n+1\equiv0\pmod 5$$
$$\implies 4(n^2+n+1)\equiv0\pmod 5\implies (2n+1)^2\equiv-3\equiv2$$
Now, $(\pm1)^2\equiv1\pmod 5,(\pm2)^2\equiv4$
So, there is no solution to $ (2n+1)^2\equiv-3\pmod 5$
| {
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Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$ Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?
$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$
Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?
| The answer is: YES, we can!
$$
\begin{align}
(\sqrt{1001}+\sqrt{999})^2&=2000+2\sqrt{1001\times 999}
\\
&=2000+2\sqrt{(1000+1)(1000-1)}
\\
&=2000+2\sqrt{1000^2-1}
\end{align}
$$
\begin{align}
\text{and that: }(2\sqrt{1000})^2&=4000
\\
&=2000+2\sqrt{1000^2}
\end{align}
Since $2000+2\sqrt{1000^2-1}<2000+2\sqrt{1000^2}$
$\therefore \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 11,
"answer_id": 8
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Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.
| Setting $a=1$ in
$$ \int_0^\infty\frac{\operatorname{Li}_a(-x)}{1+x^2}\mathrm{d}x=-2^{-a-1}\pi\, \eta(a)-a\beta(a+1)$$
we have
$$\int_0^\infty\frac{\operatorname{Li}_1(-x)}{1+x^2}\mathrm{d}x=-\int_0^\infty\frac{\ln(1+x)}{1+x^2}\mathrm{d}x=-\frac14\ln(2)-\beta(2)=-\frac14\ln(2)-G$$
Other results:
\begin{gather*}
\int_0^\infty\frac{\operatorname{Li}_2(-x)}{1+x^2}\mathrm{d}x=-\frac{\pi^3}{96}-2\beta(3);\\
\int_0^\infty\frac{\operatorname{Li}_3(-x)}{1+x^2}\mathrm{d}x=-\frac{3\pi}{64}\zeta(3)-3\beta(4);\\
\int_0^\infty\frac{\operatorname{Li}_4(-x)}{1+x^2}\mathrm{d}x=-\frac{7\pi^5}{23040}-4\beta(5);\\
\int_0^\infty\frac{\operatorname{Li}_5(-x)}{1+x^2}\mathrm{d}x=-\frac{15\pi}{1024}\zeta(5)-5\beta(6).
\end{gather*}
Proof: By writing the integral representation of $\operatorname{Li}_a(-x)$, we have
\begin{gather*}
\int_0^\infty\frac{\operatorname{Li}_a(-x)}{1+x^2}\mathrm{d}x=\int_0^\infty\frac{1}{1+x^2}\left(\frac{(-1)^a}{(a-1)!}\int_0^1\frac{x\ln^{a-1}(y)}{1+xy}\mathrm{d}y\right)\mathrm{d}x\\
\{\text{change the order of the integration}\}\\
=\frac{(-1)^a}{(a-1)!}\int_0^1\ln^{a-1}(y)\left(\int_0^\infty\frac{x}{(1+x^2)(1+xy)}\mathrm{d}x\right)\mathrm{d}y\\
\{\text{compute the inner integral by partial fraction decomposition}\}\\
=\frac{(-1)^a}{(a-1)!}\int_0^1\ln^{a-1}(y)\left(\frac{\pi}{2}\frac{y}{1+y^2}-\frac{\ln(y)}{1+y^2}\right)\mathrm{d}y\\
=\frac{(-1)^a\pi}{2(a-1)!}\underbrace{\int_0^1\frac{y\ln^{a-1}(y)}{1+y^2}\mathrm{d}y}_{y=\sqrt{x}}-\frac{(-1)^a}{(a-1)!}\int_0^1\frac{\ln^a(y)}{1+y^2}\mathrm{d}y\\
=\frac{(-1)^a\pi}{2^{a+1}(a-1)!}\int_0^1\frac{\ln^{a-1}(x)}{1+x}\mathrm{d}x-\frac{(-1)^a}{(a-1)!}\int_0^1\frac{\ln^a(y)}{1+y^2}\mathrm{d}y\\
=-2^{-a-1}\pi\, \eta(a)-a\beta(a).
\end{gather*}
| {
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"timestamp": "2023-03-29T00:00:00",
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A special curve $\left( \frac { x^{ 2 }+1 }{ 1-x^{ 2 } } \right)$. Let $a,x\in \mathbb{Q}$ and $n\in\mathbb{N}$
For $0<x<1$, $a>0$ and $n>1$, show that
$$\ \frac { x^{ 2 }+1 }{ 1-x^{ 2 } } \neq { a }^{ n }$$.
For example, for $x = 0.3$, then
$\frac {109 }{ 91 }$ can not be shown as $a^2$ or $a^3$ etc.
| Assume on the contrary that $\dfrac{1+x^2}{1-x^2}=a^n$ for some $x, a \in \mathbb{Q}, n \in \mathbb{N}, 0<x<1, a>0, n>1$. Write $x=\frac{s}{t}, s<t, s, t \in \mathbb{Z}^+, \gcd(s, t)=1$, so $\dfrac{t^2+s^2}{t^2-s^2}=a^n$
We consider 2 cases:
Case 1: $s, t$ have different parities.
Then $\gcd(t^2+s^2, t^2-s^2)=1$, so we have $t^2+s^2=b^n, t^2-s^2=c^n, a=\frac{b}{c}$ for some $b, c \in \mathbb{Z}^+$. Thus $c^n=(t-s)(t+s)$, so since $\gcd(t-s, t+s)=1$, we have $t-s=d^n, t+s=e^n, c=de$ for some $d, e \in \mathbb{Z}^+$. Thus $d^{2n}+e^{2n}=(t-s)^2+(t+s)^2=2(t^2+s^2)=2b^n$.
If $n \geq 3$, by results of Darmon and Merel, we have no non-trivial primitive solutions to the equation $x^n+y^n=2z^n$ (where trivial means $|xyz| \leq 1$) Clearly $\gcd(d, e)=1$, so $(x, y, z)=(d^2, e^2, b)$ is a primitive solution to the Fermat-like equation $x^n+y^n=2z^n$. It must thus be trivial, so $|d^2e^2b| \leq 1$, so $d=e=b=1$ since $d, e, b$ are positive integers. This implies that $2s=e^n-d^n=0$, a contradiction.
Therefore $n=2$. Now $t^2+s^2=b^2, t^2-s^2=c^2$, so $t^4-s^4=(t^2-s^2)(t^2+s^2)=(bc)^2$. We proceed via infinite descent (by comparing the size of $b$) to show that there are no positive integer solutions $(t, s, b, c)$ satisfying both equations simultaneously, where $t, s$ are of different parities and $\gcd(t, s)=1$.
Note that if $t^2=s^2+c^2 \not \equiv 0 \pmod{4}$, so $t$ cannot be even, so $s$ is even. Using the parameterization of primitive Pythagorean triplets we get $t^2=k^2+l^2, s=2kl, c^2=k^2-l^2$, where $k, l$ are of different parities and $\gcd(k, l)=1$. Then $(k, l, t, c)$ is also a solution of the 2 equations, and we have $t<b$. We have thus obtained a smaller positive integer solution, and this leads (via infinite descent) to a contradiction.
Case 2: $s, t$ are of the same parity.
Since $\gcd(s, t)=1$, $s, t$ are both odd. We have $\gcd(t^2+s^2, t^2-s^2)=2$, so $t^2+s^2=2b^n, t^2-s^2=2c^n$ for some $b, c \in \mathbb{Z}^+$. If $n$ is even, then we may take $n$ to be $2$, since we can replace $b, c$ with $b^{\frac{n}{2}}, c^{\frac{n}{2}}$ respectively. We thus have $t^2+s^2=2b^2, t^2-s^2=2c^2$, so $t^2=b^2+c^2, s^2=b^2-c^2$. Note that $\gcd(b, c)=1$ and $b, c$ are of different parities (since $s, t$ are odd), so applying the result from Case 1 (With a variable change $(t, s, b, c) \to (b, c, t, s)$), we get a contradiction.
Therefore $n$ is odd. Thus $2b^n=(t^2-s^2)+2s^2=2c^n+2s^2$, so $b^n+(-c)^n=s^2$. If $n \geq 5$, then again by results of Darmon and Merel, we have no non-trivial primitive solutions to the equation $x^n+y^n=z^2$. Clearly $\gcd(b, c)=1$, so $(x, y, z)=(b, -c, s)$ is a primitive solution to the equation $x^n+y^n=z^2$. It must thus be trivial, so $|bcs| \leq 1$, so $b=c=s=1$ since $b, c, s$ are positive integers. However this implies that $t^2=s^2+2c^n=3$, a contradiction.
Therefore $n=3$, and we have $t^2+s^2=2b^3, t^2-s^2=2c^3$. We have $2c^3=(t-s)(t+s)$, so we have $\{(t-s), (t+s)\}=\{2d^3, e^3\}$ for some $d, e \in \mathbb{Z}^+$. Notice that it does not matter which of $t+s, t-s$ is $2d^3$ and which is $e^3$, since we will only consider $4b^3=2(s^2+t^2)=(t-s)^2+(t+s)^2=4d^6+e^6$. Note that $\gcd(t-s, t+s)=2$, so $e$ is even and $d$ is odd. This also implies that $\gcd(d, e)=1$.
Write $e=2f, f \in \mathbb{Z}^+, \gcd(d, f)=1$. Then $4b^3=4d^6+64f^6$, so $b^3=d^6+16f^6$, so $b^3+(-d^2)^3=2(2f^2)^3$. Since $\gcd(-d^2, 2f^2)=1$ and $|2bd^2f^2|>1$, $(x, y, z)=(b, -d^2, 2f^2)$ is a non-trivial primitive solution to the equation $x^3+y^3=2z^2$, which again by results of Darmon and Merel (as in Case 1) leads to a contradiction.
Conclusion: We have obtained a contradiction in both cases, so the equation $\dfrac{1+x^2}{1-x^2}=a^n$ has no solutions for $x, a \in \mathbb{Q}, n \in \mathbb{N}, 0<x<1, a>0, n>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Non-linear first order differential equation I've found this particular equation rather tough, can you give me some hints on how to solve
$$\dot{y}+t\cos\frac{\pi y}{2}+1-t=y$$
Thanks a lot.
| $\dot{y}+t\cos\dfrac{\pi y}{2}+1-t=y$
$\dfrac{dy}{dt}=\left(1-\cos\dfrac{\pi y}{2}\right)t+y-1$
Let $u=y-1$ ,
Then $y=u+1$
$\dfrac{dy}{dt}=\dfrac{du}{dt}$
$\therefore\dfrac{du}{dt}=\biggl(1-\cos\dfrac{\pi(u+1)}{2}\biggr)t+u$
$\left(\left(1-\cos\left(\dfrac{\pi u}{2}+\dfrac{\pi}{2}\right)\right)t+u\right)\dfrac{dt}{du}=1$
$\left(\left(1+\sin\dfrac{\pi u}{2}\right)t+u\right)\dfrac{dt}{du}=1$
Let $v=t+\dfrac{u}{1+\sin\dfrac{\pi u}{2}}$ ,
Then $t=v-\dfrac{u}{1+\sin\dfrac{\pi u}{2}}$
$\dfrac{dt}{du}=\dfrac{dv}{du}+\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$
$\therefore\left(1+\sin\dfrac{\pi u}{2}\right)v\left(\dfrac{dv}{du}+\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}\right)=1$
$\left(1+\sin\dfrac{\pi u}{2}\right)v\dfrac{dv}{du}+\left(\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{1+\sin\dfrac{\pi u}{2}}-1\right)v=1$
$\left(1+\sin\dfrac{\pi u}{2}\right)v\dfrac{dv}{du}=\left(1-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{1+\sin\dfrac{\pi u}{2}}\right)v+1$
$v\dfrac{dv}{du}=\left(\dfrac{1}{1+\sin\dfrac{\pi u}{2}}-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}\right)v+\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $v=\dfrac{1}{w}$ ,
Then $\dfrac{dv}{du}=-\dfrac{1}{w^2}\dfrac{dw}{du}$
$\therefore-\dfrac{1}{w^3}\dfrac{dw}{du}=\left(\dfrac{1}{1+\sin\dfrac{\pi u}{2}}-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}\right)\dfrac{1}{w}+\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$
$\dfrac{dw}{du}=-\dfrac{w^3}{1+\sin\dfrac{\pi u}{2}}+\left(\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}\right)w^2$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
| {
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"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\sqrt{3x-2} +2-x=0$
Solve the equation: $$\sqrt{3x-2} +2-x=0$$
I squared both equations $$(\sqrt{3x-2})^2 (+2-x)^2= 0$$
I got $$3x-2 + 4 -4x + x^2$$
I then combined like terms $x^2 -1x +2$
However, that can not be right since I get a negative radicand when I use the quadratic equation.
$x = 1/2 \pm \sqrt{((-1)/2)^2 -2}$
The answer is 6
| What you did was not squaring both sides of the equation. If we have an equation
$$a+b=0,$$
then squaring both sides produces
$$\begin{align*}
(a+b)^2&=0^2\\a^2+2ab+b^2&=0.
\end{align*}$$
In your case, you have $a=\sqrt{3x-2}$ and $b=2-x$. But when you squared the left side of the equation, what you've written is instead $a^2+b^2$. But
$$(a+b)^2=a^2+2ab+b^2\neq a^2+b^2.$$
Thus, when correctly squaring both sides of the equation you've been given, the result is
$$(\sqrt{3x-2})^2 +2(\sqrt{3x-2})(2-x)+(2-x)^2=0.$$
However, this is not a productive place to start solving the question. A better approach would be to move the things not inside the radical to the other side, and then square, like this:
$$\begin{align*}
\sqrt{3x-2} +2-x&=0\\\\
\sqrt{3x-2} &=x-2\\\\
\qquad\qquad\qquad(\sqrt{3x-2})^2&=(x-2)^2\qquad \leftarrow \text{squared both sides}
\end{align*}$$
Can you take it from here?
| {
"language": "en",
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"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to solve this system of equation. $x^2-yz=a^2$
$y^2-zx=b^2$
$z^2-xy=c^2$
How to solve this equation for $x,y,z$. Use elementary methods to solve (elimination, substitution etc.).
Given answer is:$x=\pm\dfrac{a^4-b^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$, $y=\pm\dfrac{b^4-a^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$ ,$z=\pm\dfrac{c^4-b^2a^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}$
| Subtract them we get
$$(x^2-y^2)+xz-yz = a^2-b^2 \implies (x-y)(x+y+z) = (a-b)(a+b)$$
$$(y^2-z^2)+yx-zx = b^2-c^2 \implies (y-z)(x+y+z) = (b-c)(b+c)$$
$$(z^2-x^2)+zy-xy = c^2-a^2 \implies (z-x)(x+y+z) = (c-a)(c+a)$$
Now consider the cases $x+y+z=0$ and $x+y+z\neq 0$. I trust you can finish it off from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Listing subgroups of a group I made a program to list all the subgroups of any group and I came up with satisfactory result for $\operatorname{Symmetric Group}[3]$ as
$\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 2 & 3 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 3 & 2 \\
\end{array}
\right)\right]\right\}\right\}$
It excludes the whole set itself though it can be added seperately. But in case of $SymmetricGroup[4]$ I am getting following
$\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
2 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
3 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 3 \\
2 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 4 \\
2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 2 & 3 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 3 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 2 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 4 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 3 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
1 & 4 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
2 & 3 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
2 & 4 & 3 \\
\end{array}
\right)\right]\right\}\right\}$
The matrix form shows double transposition. Can someone please check for me if I am getting appropriate results? I doubt I am!!
| By using GAP, you can make them all so easier and visible.
gap> LoadPackage("sonata");;
gap> s4:=SymmetricGroup(IsPermGroup,4);;
gap> g:=Subgroups(s4);;
gap> List([1..Size(g)],k->g[k]);
| {
"language": "en",
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"answer_count": 2,
"answer_id": 0
} |
Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)
Is this the right way?
| First observe that since $5 \le n$, we have:
$$
4(n+1) =4n+4 < 4n+5 \le 4n+n = 5n \le (n)n = n^2
$$
Hence, since $4(n+1)<n^2 \Rightarrow \boxed{n+1<\dfrac{1}{4}n^2}$ (and $f(x)=\sqrt{x}$ is monotonically increasing and $n\ge0$), we have:
$$
\sqrt{n+1}+\sqrt{n} \le \sqrt{n+1}+\sqrt{n+1} = 2\sqrt{n+1}<2\sqrt{\dfrac{1}{4}n^2}=2\left(\dfrac{1}{2}n\right) = n
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
} |
If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod a$ and $z\equiv y\pmod b$
If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod{a}$ and $z\equiv y\pmod{b}$
What I have so far:
Let $z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$ and $
z \equiv y\pmod b
$. Then by the chinese remainder theorem there is a unique $z\pmod{\text{lcm}(a,b)}$ which satisfies this...
Is this the right approach here? I can't figure out how to get from $$z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$$ what I need.
| Existence
Bezout's Identity says that we have $g_a,g_b$ so that
$$
g_aa+g_bb=\gcd(a,b)\tag{1}
$$
Thus,
$$
\begin{align}
g_bb&\equiv\gcd(a,b)&\pmod{a}\\
g_bb&\equiv0&\pmod{b}
\end{align}\tag{2}
$$
Since $x\equiv y\pmod{\gcd(a,b)}$, we have $\gcd(a,b)\mid x{-}y$. Multiplying $(2)$ by $\frac{x-y}{\gcd(a,b)}$ gives
$$
\begin{align}
\frac{x-y}{\gcd(a,b)}g_bb&\equiv\frac{x-y}{\gcd(a,b)}\gcd(a,b)=x-y&\pmod{a}\\[6pt]
\frac{x-y}{\gcd(a,b)}g_bb&\equiv\frac{x-y}{\gcd(a,b)}0=0&\pmod{b}
\end{align}\tag{3}
$$
Adding $y$ to $(3)$ and setting $z=\frac{x-y}{\gcd(a,b)}g_bb+y$, we have
$$
\begin{align}
z&\equiv x&\pmod{a}\\
z&\equiv y&\pmod{b}
\end{align}
$$
Uniquenes
Suppose that $z_1\equiv x\pmod{a}$ and $z_2\equiv x\pmod{a}$. Then $z_1\equiv z_2\pmod{a}$.
Suppose that $z_1\equiv y\pmod{b}$ and $z_2\equiv y\pmod{b}$. Then $z_1\equiv z_2\pmod{b}$.
Therefore, we have $z_a,z_b$ so that
$$
z_1-z_2=z_aa=z_bb\tag{4}
$$
Multiplying $(1)$ by $(4)$ yields
$$
\begin{align}
g_aa\color{#C00000}{z_bb}+g_bb\color{#C00000}{z_aa}&=\gcd(a,b)\color{#C00000}{(z_1-z_2)}\\
(g_az_b+g_bz_a)\frac{ab}{\gcd(a,b)}&=z_1-z_2
\end{align}
$$
Therefore, since $\gcd(a,b)\,\mathrm{lcm}(a,b)=ab$, we have
$$
z_1\equiv z_2\pmod{\mathrm{lcm}(a,b)}
$$
| {
"language": "en",
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"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$ I have the following integral:
$$\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$$
I think I may be to evaluate this with beta and gamma functions but I am not quite sure how. Any help?
| An alternative way, using elementary methods, which begins with the same substitution as sos440's solution:
$$\int_0^1 \frac{\ln x\ln (1-x)}{\sqrt{x(1-x)}}\,dx=8\int_0^{\frac{\pi}{2}}\ln \sin x\ln \cos x\,dx$$
Note that:
$$\begin{aligned}2\int_0^{\frac{\pi}{2}}\ln \sin x\ln \cos x\,dx &=\int_0^{\frac{\pi}{2}}(\ln \sin x+\ln \cos x)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\int_0^{\frac{\pi}{2}}(\ln \sin 2x-\ln 2)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\int_0^{\frac{\pi}{2}}(\ln \sin x-\ln 2)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\ln^2 2\int_0^{\frac{\pi}{2}}\,dx-2\ln 2\int_0^{\frac{\pi}{2}}\ln \sin x\,dx-\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\frac{3\pi\ln^2 2}{2}-\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\end{aligned} $$
Now we are left to evaluate the latter integral, for which a little preliminary work must be done:
$$ \sum_{k=1}^n \sin 2kz =\frac{1}{2\sin z} \Big( \cos z-\cos (2n+1)z\Big)$$ which upon integration with $0<x<\frac{\pi}{2}$ gives:
$$ \int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\sum_{k=1}^n \frac{\cos 2k x}{k}+\sum_{k=1}^n \frac{(-1)^{k+1}}{k}+\ln \sin x$$
Furthermore, $$\int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\frac{(-1)^{n+1}}{2n+1}+\frac{\sin (2n+1) x}{(2n+1)\sin x}- \int_x^{\frac{\pi}{2}} \frac{\sin (2n+1) z\cos z}{(2n+1)\sin^2 z}\,dz$$
Which $\to 0$ as $n\to\infty$. Hence we obtain the relation: $$ \ln \sin x= -\sum_{k=1}^{\infty} \frac{\cos 2k x}{k}-\ln 2$$
$$ \ln^2 \sin x= \ln^2 2+\left(\sum_{a=1}^{\infty} \frac{\cos 2a x}{a}\right)\left(\sum_{b=1}^{\infty} \frac{\cos 2b x}{b}\right)+2\ln 2\sum_{k=1}^{\infty} \frac{\cos 2k x}{k}$$
Integrating over $(0,\frac{\pi}{2})$ the latter sum clearly vanishes and all the paired cosines for which $a\neq b$ as well. Thus we have:
$$\begin{aligned}\int_0^{\frac{\pi}{2}}\ln^2 \sin x\,dx &=\ln^2 2\int_0^{\frac{\pi}{2}}dx+\sum_{a=1}^{\infty}\frac{1}{a^2}\int_0^{\frac{\pi}{2}}\cos^2 2a x\,dx\\&=\frac{\pi\ln^2 2}{2}+\sum_{a=1}^{\infty}\frac{\pi}{4a^2}\\&=\frac{\pi\ln^2 2}{2}+\frac{\pi^3}{24}\end{aligned}$$
Combining everything and multiplying by $4:$
$$\int_0^1 \frac{\ln x\ln (1-x)}{\sqrt{x(1-x)}}\,dx=4\pi\ln^2 2-\frac{\pi^3}{6}$$
| {
"language": "en",
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"answer_count": 2,
"answer_id": 1
} |
Prove inequality: $74 - 37\sqrt 2 \le a+b+6(c+d) \le 74 +37\sqrt 2$ without calculus
Let $a,b,c,d \in \mathbb R$ such that $a^2 + b^2 + 1 = 2(a+b),
c^2 + d^2 + 6^2 = 12(c+d)$, prove inequality without calculus (or
langrange multiplier): $$74 - 37\sqrt 2 \le a+b+6(c+d) \le 74
+37\sqrt 2$$
The original problem is find max and min of $a+b+6(c+d)$ where ...
Using some calculus, I found it, but could you solve it without calculus.
| First rewrite the first restriction to $(a-1)^2+(b-1)^2 = 1^2$ and the second restriction to $(c-6)^2+(d-6)^2=6^2$. Hence we may choose $\phi,\theta\in[0,2\pi)$ such that $a = 1+\sin(\phi)$, $b=1+\cos(\phi)$, $c=6+6\sin(\theta)$ and $d = 6+6\cos(\theta)$.
Now $\sin(x)+\cos(x) = \sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \in \left[-\sqrt{2},\sqrt{2}\right]$, hence $a+b = 2+\sin(\phi)+\cos(\phi)\in\left[2-\sqrt{2},2+\sqrt{2}\right]$ and $c+d = 12+6\sin(\theta)+6\cos(\theta)\in\left[12-6\sqrt{2},12+6\sqrt{2}\right]$.
Finally we get $a+b+6(c+d) \in\left[74-37\sqrt{2},74+37\sqrt{2}\right]$.
| {
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How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| Let $ a-b = x $ , $ b -c= y $ and $ c-a =z$ . So actually you want to know the value of $ x^3+y^3+z^3-3xyz$ . Here the technique resides .
$ x+y = (a-c) =-z $ ,
$x^3+y^3+z^3-3xyz = (x+y)^3-3xy(x+y) +z^3-3xyz
= -z^3 +z^3 +3xyz -3xyz = 0 $
Hence the value is $ 0 $
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability puzzle - the 3 cannons (Apologies if this is the wrong venue to ask such a question, but I don't understand how to arrive at a solution to this math puzzle).
Three cannons are fighting each other.
Cannon A hits 1/2 of the time. Cannon B hits 1/3 the time. Cannon C hits 1/6 of the time.
Each cannon fires at the current "best" cannon. So B and C will start shooting at A, while A will shoot at B, the next best. Cannons die when they get hit.
Which cannon has the highest probability of survival? Why?
Clarification: B and C will start shooting at A.
| I assume they all fire balls simultaneously. The probability for the different outcomes of the first volley are as then as follows (I will keep my eighteenths unsimplified for simplicity):
1) $A$ is hit: $\frac{1}{3} + \frac{1}{6} - \frac{1}{18} = \frac{8}{18}$
2) $B$ is hit: $\frac{1}{2}$.
And these two are independent, so we get the following odds for who's dead before round 2:
1) $A$ and $B$: $\frac{4}{18}$
2) Just $B$: $\frac{5}{18}$
3) Just $A$: $\frac{4}{18}$
4) None (back to square one): $\frac{5}{18}$.
This means that after the first volley where someone died, it is just as likely that just $A$ died as it is that both $A$ and $B$ died, so $C$ certainly has better odds than $B$ (since $C$ does have some probability of winning a duel between the two). We therefore need to look at $A$'s odds of survival given that $B$ just died.
In $4$ out of $9$ cases, $A$ will have died in the same volley that killed $B$, so $C$ is crowned winner. In $5$ out of the $9$, a duel breaks out. We can now analyze the probabilities of the outcomes of a single volley the same way as before:
1) $C$ dies with probability $\frac{1}{2}$
2) $A$ dies with probability $\frac{1}{6}$
This gives the following odds for the outcome:
1) They both die: $\frac{1}{12}$
2) They both survive: $\frac{5}{12}$
3) Only $A$ survives: $\frac{5}{12}$
4) Only $C$ survives: $\frac{1}{12}$
So, if $B$ is (among) the first to die (which has probability $\frac{9}{13}$), the probability is $\frac{4}{9}$ that $A$ was also killed, and $\frac{5}{9}$ that we have a duel between $A$ and $C$. In this duel, $A$ has $\frac{5}{7}$ chance to come out on top, and $\frac{1}{7}$ chance to come out on bottom. All in all, this gives survival chances of $\frac{9}{13}\cdot\frac{5}{9}\cdot \frac{5}{7} = \frac{25}{91}$ for $A$. For $C$, there is the $\frac{4}{13} = \frac{28}{91}$ chance of winning without any dueling, plus the $\frac{5}{91}$ for winning by duel against $A$, and similarily, $\frac{1}{13} = \frac{7}{91}$ for winning by duelling against $B$, and we clearly see who is most often the winner here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with a trig-substitution integral I'm in the chapter of trigonometric substitution for integrating different functions. I'm having a bit of trouble even starting this homework question:
$$\int \frac{(x^2+3x+4)\,dx} {\sqrt{x^2-4x}}$$
|
Make the substitution suggested by completing the square, $\sqrt{x^2-4x}=\sqrt{(x-2)^2-4}$, that is, $y=x-2$. Then, by the easily constructed triangle, we have
\begin{align}
\frac{y}{2}
& = \sec\theta\\
\frac{\sqrt{y^2-4}}{2}
& = \tan\theta\\
x^2+3x+4
& =(y+2)^2+3(y+2)+4\\
& = y^2+7y+14\\
\frac{dy}{2}
& = \sec\theta\tan\theta\ d\theta.
\end{align}
The integral becomes
\begin{align}
\int \frac{(x^2+3x+4)} {\sqrt{x^2-4x}}dx
& = \int\frac{y^2+7y+14}{\sqrt{y^2-4}}dy\\
& = \int\frac{4\sec^2\theta+14\sec\theta+14}{2\tan\theta}2\sec\theta\tan\theta\ d\theta\\
& = \int4\sec^3\theta+14\sec^2\theta+14\sec\theta\ d\theta.
\end{align}
Andre Nicolas suggests in another response that the powers of secant are unsavory, and that a hyperbolic trig sub would be more apt. (Claude Leibovici explores this.)
(I vaguely recall methods for the $\sec^3\theta$ integrand in Stewart's Calculus. The $\sec\theta$ is in most trig integral lists, and perhaps the $\sec^2\theta$ is amenable to some fortuitous trig identity.)
Here are some resources on those:
http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integrands_involving_only_secant
http://en.wikipedia.org/wiki/Integral_of_the_secant_function
http://en.wikipedia.org/wiki/Integral_of_secant_cubed
| {
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Convergence of $\sum_{n=1}^{\infty}\frac{1+\sin^{2}(n)}{3^n}$?
Using either the Direct or Limit Comparison Tests, determine if $\sum_{n=1}^{\infty}\frac{1+\sin^{2}(n)}{3^n}$ is convergent or divergent.
I seem to be completely stuck here.
I've chosen my series to be $\sum\frac{1}{3^n}$, which is clearly a convergent geometric series. But when I do either of the two tests, I get inconclusive answers.
\begin{align}
\lim_{n\to\infty}\frac{1+\sin^2(n)}{3^n}\times\frac{3^n}{1}&=\lim_{n\to\infty}1+\sin^2(n)\\
&=\infty
\end{align}
and
\begin{align}
\frac{1+\sin^2(n)}{3^n}\leq\frac{1}{3^n}\Longrightarrow 1+\sin^2(n)\leq 1
\end{align}
but this inequality is not true.
But I just had a thought now... Am I allowed to choose the series $\sum\frac{2^n}{3^n}$ to help solve this? Clearly $1+\sin^2(n)\leq 2^n,\forall n\geq 1$, and $\sum\frac{2^n}{3^n}$ is also a convergent geometric series.
| Since $1\le1+\sin^2(n)\le2$, we can compare to the geometric series
$$
\sum_{n=1}^\infty\frac1{3^n}=\frac12\quad\text{and}\quad\sum_{n=1}^\infty\frac2{3^n}=1
$$
In fact, this is actually the sum of three geometric series:
$$
\begin{align}
\sum_{n=1}^\infty\frac{1+\sin^2(n)}{3^n}
&=\sum_{n=1}^\infty\frac{1-\frac14(e^{2in}-2+e^{-2in})}{3^n}\\
&=\frac32\sum_{n=1}^\infty\frac1{3^n}-\frac12\mathrm{Re}\left(\sum_{n=1}^\infty\frac{e^{2in}}{3^n}\right)\\
&=\frac32\cdot\frac{1/3}{1-1/3}-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3}{1-e^{2i}/3}\right)\\
&=\frac34-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3}{1-e^{2i}/3}\cdot\frac{1-e^{-2i}/3}{1-e^{-2i}/3}\right)\\
&=\frac34-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3-1/9}{10/9-2\cos(2)/3}\right)\\
&=\frac34-\frac14\frac{3\cos(2)-1}{5-3\cos(2)}\\
&=\frac{4-3\cos(2)}{5-3\cos(2)}\\[9pt]
&\doteq0.83996006708282
\end{align}
$$
As estimated, the sum is between $\frac12$ and $1$.
| {
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Finding all the integer solutions for :$y^2=x^6+17$ Assume that $x,y$ are integers .How to find the solutions for:
$$y^2=x^6+17$$
| $y^2-x^6=17\Rightarrow (y-x^3)(y+x^3)=17$.And we know $17$ is a prime.
So the only possibility is
$y+x^3=17$ and $y-x^3=1$ .
or $y+x^3=-17$ and $y-x^3=-1$
or $y+x^3=1$ and $y-x^3=17$
(As $x,y$ are integers so $(y-x^3)(y+x^3)=17$ implies that the only factors of $17 $ are $y+x^3$ and $y-x^3$ )
$\Rightarrow y=9,x=2$
or $y=9,x=-2$
or $x=-9,y=2$
or $x=-9,y=-2$
| {
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How find this arithmetic sequence of $n$ if there exist positive integer sequence $a_{1},a_{2},a_{3},\cdots,a_{n}$,such that
$$a_{1}a_{2},a_{2}a_{3},a_{3}a_{4},\cdots,a_{n-1}a_{n},a_{n}a_{1}$$ is arithmetic sequence,and the common difference $d=a_{i+1}a_{i+2}-a_{i}a_{i+1}\neq 0,i=1,\cdots,n-1.a_{n+1}=a_{1}$
find the value $n$
I Think this problem is very nice,Thank you everyone.
my idea: first, we have $n\ge 3$,and
when $n=3$,then
$$2a_{2}a_{3}=a_{1}a_{2}+a_{3}a_{1}$$
$$\Longrightarrow \dfrac{2}{a_{1}}=\dfrac{1}{a_{2}}+\dfrac{1}{a_{3}}$$
and
$$\dfrac{2}{5}=\dfrac{1}{3}+\dfrac{1}{15}$$
and when $n=4,5,\cdots,$?
\
\
I think $n=2k-1$ is true,but I can't prove it
| This can be done for any odd $n$. Take $d=n!$ and define $a_1=1\cdot 3 \cdot 5 \cdots$ and $a_2=2 \cdot 4 \cdot 6 \cdots$, in both cases progressing through the odd or even integers up to $n$ until they run out. The sequence of products will be the arithmetic sequence $d,2d,3d,...,nd$ where $nd=n\cdot n!$ The values of the $a_k$ are then determined by the requirement that this be the sequence of products of the adjacent $a_k.$
For example if $n=5$ we put $a_1=1\cdot 3 \cdot 5=15$ and $a_2=2 \cdot 4=8$, obtaining for the sequence of five $a$ values
$$15,\ 8,\ 30,\ 12,\ 40$$
Note that for $n=6$ if we put $a_1=1 \cdot 3 \cdot 5=15$ and $a_2=2 \cdot 4 \cdot 6=48,$ the $a$ are
$$15,\ 48,\ 30,\ 72,\ 40,\ 90,\ 48,$$
but if as required in the OP we use $a_7=a_1=15$ then
$$a_7a_6-a_6a_5=15\cdot 48 - 48 \cdot 90=-3600.$$
It would come out $720$ had we used the
value $105=3\cdot 5 \cdot 7$ for $a_7$, which is closely related to $a_1$, and I believe in the even case if one uses $a_{n+1}=(n+1)a_1$ in the $n$ even case we have a solution to a slightly altered problem than asked by math110.
The relations to compute the $a_k$ come from the requirement that the sequence of products be a progression as stated with common difference $d=n!$. We have in general
$$\frac{a_k}{a_{k-2}}=\frac{k-1}{k-2}.$$
This gives each $a_k$ from the one two index values before, and also explains why they all remain integers, since e.g. for the odd indices $a_1$ starts out with the full sequence of odd factors it needs in order not to give a noninteger for $a_3,a_5,\cdots.$
Impossibility for even $n$ (with $d \neq 0$ as stated in OP).
First assume that $d>0$.
Since the $a_k>0$ we have from $a_{k+1}a_k-a_ka_{k-1}=0$ that $a_{k+1}>a_{k-1}$. Therefore
$a_1<a_3<\cdots a_{n-1},$ so that
$$a_1<a_{n-1}. \tag{1}$$ But then also from
$$a_{n+1}a_{n}-a_na_{n-1}=d$$ from which $a_{n+1}>a_{n-1}$ which contradicts $(1)$.
and since $a_1=a_{n+1}$ this means $a_1>a_{n-1}.$
If instead $d<0$ we can proceed in the same way, with inequalities reversed, and arrive at a similar contradiction.
Note on why it works for $n$ odd: In this case $a_1=1\cdot 3 \cdots n$ with $(n+1)/2$ odd factors, while $a_2=2 \cdot 4 \cdots (n-1)$ with $(n-1)/2$ even factors. In computing $a_{n+1}$ via the recursion, since $n+1$ is even we begin with $a_2$ and apply the recursion $a_{k+2}=a_k \cdot \frac{k+1}{k}$ a total of $(n-1)/2$ times. This gives
$$a_{n+1}=2\cdot 4 \cdots (n-1)\cdot \frac32 \frac54 \cdots \frac{n}{n-1},$$
i.e. $a_{n+1}=3 \cdot 5 \cdots n$, which happens to equal $a_1$ as desired since the first factor of $a_n$ is $1$ and the rest are $3,5,...,n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| $c=-a-b$, so
$$a^2+b^2+c^2=2a^2+2ab+2b^2=1$$
let $a=x+y,b=x-y$
then
$$3x^2+y^2=1/2$$
with the we only consider $ab\le 0$
and $c=-(a+b)=-2x$,
$$c^2=a^2+2ab+b^2\le a^2+b^2=1-c^2$$
so $$8x^2\le 1$$
so $$abc=ab(-a-b)=-(a+b)ab=-2x(x^2-y^2)=-x(8x^2-1)$$
then
$$a^2b^2c^2=x^2(8x^2-1)(8x^2-1)=\dfrac{1}{16}\cdot 16x^2(1-8x^2)(1-8x^2)\le \dfrac{1}{16}\dfrac{8}{27}=\dfrac{1}{54}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a formula to a given $\sum$ using generating functions Find a close formula to the sum $\sum_{k=0}^{n}k\cdot 5^k$
I tried using generating functions using the differences sequences with no luck.
| Set $g_k=\frac{1}{4}5^{k+1}$. Note that $\Delta g_k=g_{k+1}-g_k=\frac{1}{4}(5^{k+1}-5^k)=5^k$. Set $f_k=k$; we have $\Delta f_k=f_{k+1}-f_k=1$. The partial summation formula gives:
We have $$\sum_0^nk5^k=\sum_0^nf_k\Delta g_k=(f_{n+1}g_{n+1}-f_0g_0)-\sum_0^ng_{k+1}\Delta f_k$$
The RHS simplifies to $$(n+1)\frac{1}{4}5^{n+2}-0-\sum_{0}^n \frac{1}{4}5^{k+2}=\frac{1}{4}\left((n+1)5^{n+2}-5^2\frac{5^{n+1}-1}{5-1}\right)=\frac{25}{4}\left((n+1)5^n-\frac{1}{4}(5^{n+1}-1)\right)=\frac{25}{4}\left(\left(n-\frac{5}{4}\right)5^n+\frac{1}{4}\right)$$
| {
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Function problem Show that function $f(x) =\frac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $0 < c \leq 1$ Problem :
Show that function $f(x)=\dfrac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $0 < c \leq 1$
My approach :
Let the given function $f(x) =\dfrac{x^2+2x+c}{x^2+4x+3c} = t $ where $t$ is any arbitrary constant.
$\Rightarrow (t-1)x^2+2(2t-1)x+c(3t-1)=0$
The argument $x$ must be real, therefore $(2t-1)^2-(t-1)(3tc-c) \geq 0$.
Now how to proceed further? Please guide. Thanks.
| Note that from your quadratic equation in $x$:
$$
(t-1)x^2 + (4t-2)x^2 + (3tc-c) = 0
$$
its discriminant is:
$$
\Delta_x = (4t-2)^2-4(t-1)(3tc-c)=(-12c+16)t^2+(16 c-16)t+(-4c+4)
$$
Note that it suffices to show that $\Delta_x$ is always nonnegative. To do this, we consider its discriminant with respect to $t$:
$$
\Delta_t = (16c-16)^2-4(-12c+16)(-4c+4) = 64c(c-1)
$$
Now consider what happens when $0<c\le 1$. Observe that $\Delta_t\le0$, so the parabola $(-12c+16)t^2+(16 c-16)t+(-4c+4)$ either lies completely on one side of the $t$-axis or just barely touches the $t$-axis. But since $c\le1 \implies -12c+16\ge4$, we know that the parabola must be opening upwards so that $\Delta_x \ge 0$, as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Algebraic manipulation with square roots I have always had problems with the algebraic manipulation of square roots. For example, recently I encountered this in a problem I was working on:
$$\sqrt{\left(\dfrac{x-1}{2x}\right)^2 - \dfrac{y}{x}} = \dfrac{1}{2x} \sqrt{(x-1)^2 -4xy}$$
I still don't grasp why this is correct and in general, I have trouble knowing when you can factor out something when dealing with square roots. Can someone enlighten me.
| Observe that:
$$ \begin{align*}
\sqrt{\left(\dfrac{x-1}{2x}\right)^2 - \dfrac{y}{x}}
&= \sqrt{\dfrac{(x-1)^2}{(2x)^2} - \dfrac{y}{x} \cdot \dfrac{4x}{4x}} \\
&= \sqrt{\dfrac{(x-1)^2}{(2x)^2} - \dfrac{4xy}{4x^2}} \\
&= \sqrt{\dfrac{(x-1)^2}{(2x)^2} - \dfrac{4xy}{(2x)^2}} \\
&= \sqrt{\dfrac{(x-1)^2-4xy}{(2x)^2}} \\
&= \sqrt{\dfrac{1}{(2x)^2} \cdot \left( (x-1)^2-4xy \right)} \\
&= \sqrt{\left(\dfrac{1}{2x}\right)^2 \left((x-1)^2 -4xy\right)} \\
&= \sqrt{\left(\dfrac{1}{2x}\right)^2}\cdot \sqrt{(x-1)^2 -4xy} \\
&= \dfrac{1}{2x} \sqrt{(x-1)^2 -4xy}
\end{align*} $$
assuming that $x>0$.
| {
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Why $x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$?
Why $f(x) = x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$ (where $\zeta = \zeta_7$ is a primitive root of the unit) over $\mathbb{Q}$?
Of course it's irreducible by the Eisenstein criterion, however it apparently does not satisfies $1 + 2(\zeta + \zeta^2 + \zeta^4)$ as a root, I tried to calculate several times however I couldn't get $f(1 + 2(\zeta + \zeta^2 + \zeta^4))$ = 0$.
Thanks in advance.
| Slightly easier:
$(x-1)^2 = 4 ( \zeta^2 + \zeta^4 + \zeta^8 + 2 \zeta^3 + 2 \zeta^5 + 2 \zeta^6) = 4 ( - 2 - \zeta -\zeta^2 - \zeta^4) = 4 ( -2 - \frac{x-1}{2})$
Hence $ x^2 - 2x + 1 = -8 - 2x + 2$ or $x^2 + 7 = 0 $.
| {
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Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.
| Let $\displaystyle{\sin\theta=\frac{a}{c}}$ and $\displaystyle{\cos\theta=\frac{b}{c}}$ such that $\displaystyle{a^2 + b^2=c^2}$.
$$\displaystyle{\frac{sin\theta - \cos\theta + 1}{\sin\theta+\cos\theta -1}}$$ $$\displaystyle{=\frac{a-b+c}{a+b-c}}$$
Now $\displaystyle{\tan\theta=\frac{a}{b}}$ and $\displaystyle{\sec\theta=\frac{c}{b}}$. Hence, $$\displaystyle{\frac{1}{\sec\theta-\tan\theta}}$$ $$\displaystyle{=\frac{1}{\frac{c}{b}-\frac{a}{b}}}$$ $$\displaystyle{=\frac{b}{c-a}}$$
Let us assume that $$\displaystyle{=\frac{a-b+c}{a+b-c}\neq\frac{b}{c-a}}$$
This simplifies to $$\displaystyle{a^2 + b^2\neq c^2}$$
This is obviously false. Hence, $$\displaystyle{\frac{a-b+c}{a+b-c}=\frac{b}{c-a}}$$
QED
| {
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How to find the partial derivative of this complicated function? Find the partial derivative with respect to a (treat other variables as constants):
$\displaystyle f(a,b,c)=\frac {a}{\sqrt{a^2+8bc}}-\frac {a^r}{a^r+b^r+c^r}$
The article I'm reading says it should be:
$\displaystyle \frac {\sqrt{a^2+8bc}-\dfrac{a^2}{\sqrt{a^2+8bc}}}{a^2+8bc}-\frac{ra^{r-1}(a^r+b^r+c^r)-a^r \cdot ra^{r-1}}{(a^r+b^r+c^r)^2}$
but I don't know how to get there.
Thanks in advance!
| $$\dfrac{a}{\sqrt{a^2+8bc}}\ge\dfrac{a^r}{a^r+b^r+c^r}\cdots \cdots (1)$$
I guess you do this show that
$$\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ca}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge 1$$
so you can Substituting $b=c=1$ in (1)
we obtain the following condition on $r:f(a)\ge 0$ for all $a>0$,wher
$$f(a)=a(a^r+2)-a^r\sqrt{a^2+8}$$
Note that $f(1)=0,$
use Lemma 3:https://www.awesomemath.org/wp-content/uploads/reflections/2009_1/MR_1_2009_article1.
so $$f'(a)=(r+1)a^r+2-ra^{r-1}\sqrt{a^2+8}-a^r\cdot\dfrac{a}{\sqrt{a^2+8}}$$
so $$f'(1)=0\Longrightarrow r=\dfrac{4}{3}$$
| {
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"url": "https://math.stackexchange.com/questions/428742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A proposed proof by induction of $1+2+\ldots+n=\frac{n(n+1)}{2}$
Prove: $\displaystyle 1+2+\ldots+n=\frac{n(n+1)}{2}$.
Proof
When $n=1,1=\displaystyle \frac{1(1+1)}{2}$,equality holds.
Suppose when $n=k$, we have $1+2+\dots+k=\frac{k(k+1)}{2}$
When $n = k + 1$:
\begin{align}
1+2+\ldots+k+(k+1) &=\frac{k(k+1)}{2}+k+1 =\frac{k(k+1)+2k+2}{2}\\
&=\frac{k^2+3k+2}{2}\\
\text{[step]}&=\displaystyle\frac{(k+1)(k+2)}{2}=\displaystyle\frac{(k+1)((k+1)+1)}{2}
\end{align}
equality holds.
So by induction, the original equality holds $\forall n\in \mathbb{N}$.
Question 1: any problems in writing?
Question 2: Why [step] happen to equal? i.e., why does $k^2+3k+2=(k+1)(k+2)$ hold?
| No, all the steps look pretty good; as for the unknown $[\text{step}]$, that is just factoring - although, it is easier to notice that
$$
\frac{k(k+1) + 2k + 2}{2} = \frac{k(k+1) + 2 (k+1)}{2} = \frac{(k+1)(k+2)}{2} .
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Math Parlor Trick
A magician asks a person in the audience to think of a number $\overline {abc}$. He then asks them to sum up $\overline{acb}, \overline{bac}, \overline{bca}, \overline{cab}, \overline{cba}$ and reveal the result. Suppose it is $3194$. What was the original number?
The obvious approach was modular arithmetic.
$(100a + 10c + b) + (100b + 10a + c) + (100b + 10c + a) + (100c + 10a + b) + (100c + 10b + a) = 3194$
$122a + 212b + 221c = 3194$
Since $122, 212, 221 \equiv 5 (mod\space9)$ and $3194 \equiv 8 (mod\space9)$
$5(a + b + c) \equiv 8 (mod\space9)$
So, $a + b + c = 7$ or $16$ or $26$
Hit and trial produces the result $358$. Any other, more elegant method?
| The sum of all six combinations is $222(a+b+c)$
So, $3194+100a+10b+c=222(a+b+c)$
As $3194/222>14$
If $a+b+c=15, 100a+10b+c=222(15)-3194=136$
$\implies a+b+c=1+3+6=10\ne 15$
If $a+b+c=16, 100a+10b+c=222(16)-3194=358$
$\implies a+b+c=3+5+8=16$ as needed
| {
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How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac {-\sqrt{1 - (t-1)^2}}{t} = \int \frac {-\sqrt{t^2 + 2t}}{t} = \int \frac {-\sqrt t \cdot \sqrt t \cdot \sqrt{t +
2}}{\sqrt t \cdot t} $$
$$= \int \frac {- \sqrt{t + 2}}{\sqrt t } = \int - \sqrt{1 + \frac 2t} = ? $$
What next? I don’t know. Also, I’ve tried another “substitution”, namely $1 + \cos x = 2 \cos^2 \frac x2) $
$$ \int \frac {dx}{1+ \cos x} = \int \frac {dx}{2 \cos^2 \frac x2} = \int \frac 12 \cdot \sec^2 \frac x2 = ? $$
And failed again. Help me, please.
| $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$
Therefore, $$\int_0^{\pi/2}\dfrac{1}{1+\cos x}dx=\int_0^{\pi/2}\dfrac{1}{1+\sin x}dx$$
As you did $$\int_0^{\pi/2}\dfrac{1}{1+\cos x}dx=\frac{1}{2}\int_0^{\pi/2}\sec^2\frac{x}{2}dx$$
Substitute $x=2t$ and use $\int\sec^2 tdt=\tan t+C$
EDIT: Proof of $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$:
Let $I=\int_a^bf(x)dx$
Now, substitute $x=a+b-t\implies dx=-dt$
then $I=-\int_b^af(a+b-t)dt=\int_a^bf(a+b-t)dt=\int_a^bf(a+b-x)dx$$
| {
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"timestamp": "2023-03-29T00:00:00",
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simple limit but I forget how to prove it I have to calculate the following limit
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - x$$
it is in un undeterminated form.
I tried to rewrite it as follows:
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - \sqrt{|x|^2}$$
but seems a dead road.
Can anyone suggest a solution?
thanks for your help
| HINT:
Putting $y=-x,$
$$ \sqrt{x^2+2x+2} - x=\sqrt{y^2-2y+2} +y= \frac{(\sqrt{y^2-2y+2} +y)(\sqrt{y^2-2y+2} -y)}{(\sqrt{y^2-2y+2} -y)}$$
$$=\frac{y^2-2y+2-y^2}{\sqrt{y^2-2y+2} -y}=\frac{-2+\frac2y}{\sqrt{1-\frac2y+\frac2{y^2}}-1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Question about Permutations, and the distinct differences I have the following question regarding permutations of the sequence $(1,2,\cdots,n)$:
For what values of $n$ does there exist a permutation $(x_1,x_2,\cdots,x_n)$ of $(1,2,\cdots,n)$, such that the differences $|x_k-k|$ for each $k\in\{1,2,\cdots,n\}$ are all distinct?
I have shown that $n$ must not be congruent to 2 or 3 modulo 4. I have tried to construct a permutation for each value of $n$ congruent to 0 or 1 modulo 4, but I was not able to find a pattern, and as such I was unable to come up with a permutation for higher values of $n$.
Is there a nice way of actually constructing the desired permutations, or are there yet more values of $n$ for which such a permutation does not exist?
| I'm not sure, but I think I've found a construction for $n=0\mod 4$. First, some examples:
$$\begin{pmatrix}1 & 4 & 2 & 3\\
4 & 2 & 1 & 3
\end{pmatrix} \\
\begin{pmatrix}1 & 8 & 2 & 7 & 3 & 5 & 4 & 6\\
8 & 2 & 7 & 3 & 5 & 4 & 1 & 6
\end{pmatrix} \\
\begin{pmatrix}1 & 12 & 2 & 11 & 3 & 10 & 4 & 8 & 5 & 7 & 6 & 9\\
12 & 2 & 11 & 3 & 10 & 4 & 8 & 5 & 7 & 6 & 1 & 9
\end{pmatrix} \\
\begin{pmatrix}1 & 16 & 2 & 15 & 3 & 14 & 4 & 13 & 5 & 11 & 6 & 10 & 7 & 9 & 8 & 12\\
16 & 2 & 15 & 3 & 14 & 4 & 13 & 5 & 11 & 6 & 10 & 7 & 9 & 8 & 1 & 12
\end{pmatrix}
$$
Generally, if $n=4k$, the permutation has a single cycle that is composed from alternating numbers from $1,2,\dots$ and $n,n-1,\dots$ but skips $3k$.
Also for $n=1\mod 4$ consider the examples:
$$
\begin{pmatrix}1 & 5 & 3 & 4 & 2\\
5 & 3 & 4 & 1 & 2
\end{pmatrix}
\\
\begin{pmatrix}1 & 9 & 2 & 8 & 4 & 7 & 5 & 6 & 3\\
9 & 2 & 8 & 4 & 7 & 5 & 6 & 1 & 3
\end{pmatrix}
\\
\begin{pmatrix}1 & 13 & 2 & 12 & 3 & 11 & 5 & 10 & 6 & 9 & 7 & 8 & 4\\
13 & 2 & 12 & 3 & 11 & 5 & 10 & 6 & 9 & 7 & 8 & 1 & 4
\end{pmatrix}
$$
Generally, if $n=4k+1$, the permutation has a single cycle that is composed from alternating numbers from $1,2,\dots$ and $n,n-1,\dots$ but skips $k+1$.
| {
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How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists? I would like to investigate the convergence of
$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt\ldots}}}}$$
Or more precisely, let $$\begin{align}
a_1 & = \sqrt 1\\
a_2 & = \sqrt{1+\sqrt2}\\
a_3 & = \sqrt{1+\sqrt{2+\sqrt 3}}\\
a_4 & = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt 4}}}\\
&\vdots
\end{align}$$
Easy computer calculations suggest that this sequence converges rapidly to the value 1.75793275661800453265, so I handed this number to the all-seeing Google, which produced:
*
*OEIS A072449
* "Nested Radical Constant" from MathWorld
Henceforth let us write $\sqrt{r_1 + \sqrt{r_2 + \sqrt{\cdots + \sqrt{r_n}}}}$ as $[r_1, r_2, \ldots r_n]$ for short, in the manner of continued fractions.
Obviously we have $$a_n= [1,2,\ldots n] \le \underbrace{[n, n,\ldots, n]}_n$$
but as the right-hand side grows without bound (It's $O(\sqrt n)$) this is unhelpful. I thought maybe to do something like:
$$a_{n^2}\le [1, \underbrace{4, 4, 4}_3, \underbrace{9, 9, 9, 9, 9}_5, \ldots,
\underbrace{n^2,n^2,\ldots,n^2}_{2n-1}] $$
but I haven't been able to make it work.
I would like a proof that the limit $$\lim_{n\to\infty} a_n$$
exists. The methods I know are not getting me anywhere.
I originally planned to ask "and what the limit is", but OEIS says "No closed-form expression is known for this constant".
The references it cites are unavailable to me at present.
| The proofs previous to this one have all been non-constructive. They all state an upper bound, observe that the sequence is monotonically increasing, and then appeal to the Monotone Convergence Theorem. This type of argument cannot provide a computable error bound. The following does.
Call the expression $\sqrt{1 + \sqrt{2 + \sqrt{3+\dotsc}}}$ by the name $R$. Observe that if you partially expand $R$, and leave an $x$ for the unexpanded bit, you get something like $\sqrt{1 + \sqrt{2 + \sqrt{3 + x}}}$. Now consider replacing $x$ with some upper bound $U$, and some lower bower bound $L$. You get:
$$\sqrt{1 + \sqrt{2 + \sqrt{3 + L}}} \leq R \leq \sqrt{1 + \sqrt{2 + \sqrt{3 + U}}}.$$
The top answer for this question suggests to set $U = \sqrt 4 \phi$, where $\phi$ is the golden ratio. Why does this work? Because $\sqrt4\phi = \sqrt4\sqrt{1 + \sqrt{1 + \sqrt{1+\dotsc}}} = \sqrt{4^{2^0}+\sqrt{4^{2^1}+\sqrt{4^{2^2}+\dotsc}}} \geq \sqrt{4 + \sqrt{5 + \sqrt{6 + \dotsc}}}$.
We can set $L = \sqrt 4$.
We now estimate the difference between the upper bound and lower bound. We get:
$$\begin{align}
&\sqrt{1 + \sqrt{2 + \sqrt{3 + U}}} - \sqrt{1 + \sqrt{2 + \sqrt{3 + L}}}
\\&\leq \sqrt{0 + \sqrt{0 + \sqrt{0 + U}}} - \sqrt{0 + \sqrt{0 + \sqrt{0 + L}}} \tag{drive 1,2,3 to 0}
\\&=U^{2^{-3}} - L^{2^{-3}}
\\&=(\sqrt{4}\phi)^{2^{-3}} - (\sqrt{4})^{2^{-3}}
\\&=(\sqrt{4})^{2^{-3}}(\phi^{2^{-3}} - 1)
\\&\approx 0.06760864818
\end{align}$$
In general, you have that when $R$ is partially expanded to $\sqrt{1+\sqrt{2+\dotsc\sqrt{n+\sqrt{n+1}}}}$, the error is at most $(\sqrt{n+1})^{2^{-n}}(\phi^{2^{-n}} - 1)$. We demonstrated this in the special case where $n=3$. The error clearly converges to zero. The technique generalises to all similar problems.
| {
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convergence to a generalized Euler constant and relation to Zeta serie Let $0 \leq a \leq 1$ be a real number. I would like to know how to prove that the following sequence converges:
$$u_n(a)=\sum_{k=1}^n k^a- n^a \left(\frac{n}{1+a}+\frac{1}{2}\right)$$
For $a=1$:
$$u_n(1)=\sum\limits_{k=1}^{n} k- n \left(\frac{n}{1+1}+\frac{1}{2}\right)= \frac{n(n+1)}{2}-\frac{n(n+1)}{2}=0$$
so $u_n(1)$ converges to $0$.
for $a=0$:
$$u_n(0)=\sum\limits_{k=1}^{n} 1- \left(\frac{n}{1+0}+\frac{1}{2}\right) = n-n+\frac{1}{2}=\frac{1}{2}$$
so $u_n(0)$ converges to $1/2$.
In genaral, the only idea I have in mind is the Cauchy integral criterion but it does not work because $k^a$ is an increasing function,
Do the proof involves Zeta serie ?
| To avoid a black box formula
Since the function $x\mapsto x^a$ is increasing then we find easily the asymptotic expansion
$$\sum_{k=1}^n k^a\sim_\infty \int_1^n x^a dx\sim_\infty \frac{1}{a+1}n^{a+1}\tag{AE}$$
hence we have
$$\sum_{k=1}^n k^a=\frac{1}{a+1}n^{a+1}+O\left(n^{a}\right)\tag{1}$$
Now let's improve the equality $(1)$ so let
$$v_n=\sum_{k=1}^n k^a-\frac{1}{a+1}n^{a+1}$$
so we have
$$v_{n+1}-v_n=(n+1)^a-\frac{1}{a+1}(n+1)^{a+1}+\frac{1}{a+1}n^{a+1}$$
then we expand using the binomial formula we find
$$v_{n+1}-v_n\sim_\infty \frac{a}{2}n^{a-1}$$
and by telescoping and repeating the same method as in $(AE)$ we have:
$$v_n\sim_\infty \frac{a}{2}\int_1^n x^{a-1}dx\sim_\infty \frac{n^a}{2}$$
hence finaly we find
$$\sum_{k=1}^n k^a=\frac{1}{a+1}n^{a+1}+ \frac{n^a}{2}+O\left(n^{a-1}\right)$$
and we conclude.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum and maximum of $ \sin^2(\sin x) + \cos^2(\cos x) $ I want to find the maximum and minimum value of this expression:
$$ \sin^2(\sin x) + \cos^2(\cos x) $$
| Following George's hint,
Because $-1\le \sin x \le 1$, and $\sin x$ is strictly increasing on $-1\le x\le 1$, we see that $\sin (\sin x)$ (and hence $\sin^2(\sin x)$) is maximized when $\sin x=1$, e.g. at $x=\pi/2$.
On the other hand, $\cos x$ is maximized when $x=0$, so $\cos (\cos x)$ (and hence $\cos^2(\cos x)$ is maximized when $\cos x=0$, e.g. at $x=\pi/2$.
Hence the combined function is maximal at $\pi/2$, when it is $1+\sin^21\approx 1.7$.
For the other direction, $x=0$ gives $\sin^2(\sin x)=0$, clearly minimal. Because $-1\le \cos x\le 1$, and $\cos x$ is increasing on $[-1,0)$ and decreasing on $(0,1]$, we minimize $\cos x$ for $x=\pm 1$. Hence in particular $x=0$ minimizes $\cos(\cos x)$ and thus $\cos^2(\cos x)$. Combining, the minimum value is $0+\cos^21\approx 0.29$
| {
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Recursion Question - Trying to understand the concept Just trying to grasp this concept and was hoping someone could help me a bit. I am taking a discrete math class. Can someone please explain this equation to me a bit?
$f(0) = 3$
$f(n+1) = 2f(n) + 3$
$f(1) = 2f(0) + 3 = 2 \cdot 3 + 3 = 9$
$f(2) = 2f(1) + 3 = 2 \cdot 9 + 3 = 21$
$f(3) = 2f(2) + 3 = 2 \cdot 21 + 3 = 45$
$f(4) = 2f(3) + 3 = 2 \cdot 45 + 3 = 93$
I do not see how they get the numbers to the right of the equals sign. Please someone show me how $f(2) = 2f(1) + 3 = 2 \cdot 9 + 3$. I see they get "$2\cdot$" because of $2f$ but how and where does the $9$ come from? I also see why the $+3$ at the end of each equation but how and where does that number in the middle come from?
| The function $f$ is given by a starting condition $f(0)=3$, and by the condition $f(n+1)=2\cdot f(n)+3$. This means that we calculate the value of $f(2)$ from the value of $f(1)$, which itself is calculated from $f(0)$ -- which was given to us.
$$f(1)=9\implies f(2)=2\cdot f(1)+3=2\cdot 9+3$$
| {
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Discrete Math Question: arithmetic progression A lumberjack has $4n + 110$ logs in a pile consisting of n layers. Each layer has two more logs than the layer directly above it. If the top layer has six logs, how many layers are there? Write the steps to calculate the equation for the problem and state the number of layers.
It's given that the top layer has 6 logs and each layer below that gains 2 logs per layer.
1st layer: 6
2nd layer: 6 + 1(2) = 6 + 2 = 8
3rd layer: 6 + 2(2) = 6 + 4 = 10
4th layer: 6 + 3(2) = 6 + 6 = 12
5th layer : 6 + 4(2) = 6 + 8 = 14
and so on
the nth layer: $a_n = 6 + 2(n-1)$
1st term 6 and common difference 2 with n terms
1st term: $a_n = 6$
last term: $a_n = 6 + 2n + 4$
$S_n = (n/2)(a_1 + a_n)$
$$4n + 110 = (n/2)(6 + 6 + 2(n-1))$$
Can anyone help break this equation down to solve for n?
| As you observe, you have an arithmetic progression, and your work is fine to the point you left off; solving for $n$ is mostly a matter of algebra:
Picking up where you left off
$$\begin{align} 4n + 110 & = \dfrac n2\left(6 + 6 + 2(n-1)\right) \\ \\
& = \dfrac n2(12) + \dfrac {2n\cdot (n-1)}{2}\\ \\
& = 6n + n(n - 1) \\ \\
& = 6n + n^2 - n \\ \\
4n + 110 & = n^2 + 5n\end{align}$$
Now, we move everything to one side of the equation: $$\begin{align} n^2 + 5n - 4n - 110 & = 0 \\ \\ n^2 + n - 110 & = 0\\ \\(n-10)(n+11) &= 0\end{align}$$
So $n = 10, \; n=-11$ solve the quadratic, but we need the value where $n > 0$. Hence, $n = 10$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the derivative of $x\sin x$? Ok so I know the answer of $\frac{d}{dx}x\sin(x) = \sin(x)+ x\cos(x)$...but how exactly do you get there? I know $\frac{d}{dx} \sin{x} = \cos{x}$. But where does the additional $\sin(x)$ (in the answer) come in?
| Depending on how familiar you are with the chain rule and derivative of logarithmic functions, you could differentiate $x \sin{x}$ using this alternate method that doesn't directly use the product rule,
$$\begin{align} y &= x \sin{x} \\
\ln(y) &= \ln( x \sin{x}) \\
\ln(y) &= \ln(x) + \ln(\sin{x}) \\
\frac{1}{y} \frac{dy}{dx} &= \frac{1}{x} + \frac{\cos{x}}{\sin{x}} \\
\frac{dy}{dx} &= \left( \frac{1}{x} + \frac{\cos{x}}{\sin{x}} \right) \cdot y \\
\frac{dy}{dx} &= \left( \frac{1}{x} + \frac{\cos{x}}{\sin{x}} \right) \cdot x \sin{x}\\
\frac{dy}{dx} &= \sin{x} + x \cos{x}. \\ \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find y'' using implicit differentiation Pretty sure I messed up
$$
x^3 - 3xy + y^3 = 1
$$
$$
3x^2-3xy'-3y+3y^2y'=0
$$
$$
3y^2y'-3xy'=3y-3x^2
$$
$$
(3y^2-3x)y'=3y-3x^2
$$
$$
y' = \dfrac{y-x^2}{y^2-x}
$$
$$
y'' = \dfrac{(y^2-x)(y'-2x)-(y-x^2)(2yy'-1)}{(y^2-x)^2}
$$
$$
y'' = \dfrac{y^2y'-2xy^2-xy'+2x^2-2y^2y'+y+2x^2yy'-x^2}{(y^2-x)^2}
$$
$$
y''=\dfrac{x^2-2xy^2+y+(2x^2y-x-y^2)y'}{(y^2-x)^2}
$$
After subbing in $y'$, expanding and simplifying...
$$
y'' = \dfrac{x^2-2xy^2+y+3x^2y^2-2x^4y-xy+x^3-y^3}{(y^2-x)^3}
$$
The answer should be
$$
y'' = -\dfrac{4xy}{(y^2-x)^3}
$$
| As pointed out in the comments, the error occurred when plugging back in the value for $y'$. The part that reads:
$$x^2-2xy^2+y$$
Should read:
$$(x^2-2xy^2+y)(y^2-x)$$
This multiplies out and helps simplify further.
| {
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Are there $a,b>1$ with $a^4\equiv 1 \pmod{b^2}$ and $b^4\equiv1 \pmod{a^2}$? Are there solutions in integers $a,b>1$ to the following simultaneous congruences?
$$
a^4\equiv 1 \pmod{b^2} \quad \mathrm{and} \quad b^4\equiv1 \pmod{a^2}
$$
A brute-force search didn't turn up any small ones, but I also don't see how to rule them out.
| I would think this could be attacked as follows. Write $b = b_1b_2b_3$ for integers $b_1,b_2,b_3$, such that $b_1^2 \mid (a-1)$ and $b_2^2 \mid (a+1)$ and $b_3^2 \mid (a^2+1)$. Hence the first condition is satisfied, i.e.,
$$ a^4-1 = (a-1)(a+1)(a^2+1) \equiv 0 \pmod{b^2}.$$
Write $a-1 = c_1b_1^2$ and $a+1=c_2b_2^2$ and $a^2+1=c_3b_3^2$ for integers $c_1, c_2,c_3$.
From this point, I would think there would be a number of interesting ways to solve or contradict the second condition
$$ b^4-1 = (b-1)(b+1)(b^2+1) \equiv 0 \pmod{a^2}.$$
Good luck!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/446269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Big $\Omega$ question! Prove $(n-1)(n-2)(n-3)$ is $\Omega(n^3)$ Problem
Prove $(n-1)(n-2)(n-3)$ is $\Omega(n^3)$.
Attempt @ Solution
*
*$f(n) = n^3(1-6/n+11/n^2-6/n^3)$
*$g(n) = n^3$
*Show that there exists a $C > 0$ and $n_0$ such that $f(n) \ge Cg(n)$ for all $n > n_0$.
*I tried plugging in different numbers for $n$ that would make $f(n) > n^3$. I found that setting $n = 7$ makes sure that $f(n)$ is greater than $g(n)$. So, is that my answer? Evaluating the expression with $n=7$ to solve for $C$, and setting $n_0$ as $7$? Is that a sufficient proof? Also, Does my constant have to be a Natural number, or can it simply be a Rational number?
| Your claim is that $n_0=7$ and $C=1$ work: that if $n>7$, then $f(n)\ge g(n)$, but you haven’t actually proved it. And you cannot possible prove it, because
$$\frac{f(n)}{g(n)}=\frac{(n-1)(n-2)(n-3)}{n^3}=\frac{n-1}n\cdot\frac{n-2}n\cdot\frac{n-3}n<1^3=1\tag{1}$$
for all $n\ge 1$, and therefore $f(n)<g(n)$ for all $n\ge 1$.
You want to find $n_0$ and $C>0$ such that $f(n)\ge Cg(n)$ whenever $n>n_0$. Since $g(n)\ne 0$ when $n\ge 1$, this amounts to wanting a $C>0$ such that $\frac{f(n)}{g(n)}\ge C$ whenever $n>n_0$. In view of $(1)$, any such $C$ is going to have to be less than $1$, since $\frac{f(n)}{g(n)}$ is always less than $1$.
As you already computed, we do have
$$\frac{f(n)}{g(n)}=\frac{n^3-6n^2+11n-6}{n^3}=1-\frac6n+\frac{11}{n^2}-\frac6{n^3}\;.$$
Let’s find something smaller than this that is still positive, at least when $n$ is large enough. I’ll do it by progressively increasing the amount that I’m subtracting from $1$ on the righthand side, thereby making the difference smaller, and at the same time making the expression on the righthand side simpler:
$$\begin{align*}
\frac{f(n)}{g(n)}&=1-\frac6n+\frac{11}{n^2}-\frac6{n^3}\\\\
&>1-\frac6n-\frac{11}{n^2}-\frac6{n^3}\\\\
&>1-\frac{11}n-\frac{11}{n^2}-\frac{11}{n^3}\\\\
&=1-11\left(\frac1n+\frac1{n^2}+\frac1{n^3}\right)\\\\
&\ge1-11\left(\frac1n+\frac1n+\frac1n\right)\\\\
&=1-\frac{33}n
\end{align*}$$
whenever $n\ge 1$. And now I see that if $n\ge 34$, then
$$\frac{f(n)}{g(n)}\ge 1-\frac{33}{34}=\frac1{34}\;.$$
In other words, I’ve proved that if $n\ge 34$, then $f(n)\ge\frac1{34}g(n)$. This shows that $f(n)$ really is $O\big(g(n)\big)$.
(To answer the last question, $C$ can be any positive real number; it need not even be rational, let alone an integer.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$6$ people, $3$ rooms, $1$ opening door $6$ people spread in $3$ distinguishable rooms, every room needs one person who opens the door.
There are ${6 \choose 3}\cdot 3 \cdot 2$ options to choose the three door opener persons and let them open one certain room, so that every room is opened by one person.
Further, there are then $3$ options to have all left three guys in one of the rooms, $3\cdot 2$ options to have exactly one other guy in every room, and ${3 choose 2}\cdot 3 \cdot 2$ options to have one guy in one room, two in the other and none in the last room.
So in total there are ${6 \choose 3}\cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot 2 {3 \choose 2} \cdot 3 \cdot 2$ options to have all rooms opened by exactly one person and spread the others in these rooms.
If only there are 3 people who can open the doors (only they have keys for all the rooms). There are only $3 \cdot 2 \cdot 3 \cdot 3 \cdot 2 {3 \choose 2} \cdot 3 \cdot 2$ options left, right?
| Here's an easy way to look at the problem:
First, we make sure that we have a door opener in each room. There are $6$ choices for the first room, $5$ for the second, and $4$ for the third. Thus, we may select door openers in $6\times5\times4=120$ ways.
Now, each of the three remaining people can be put in any of the three rooms. So, we have $3\times3\times3=27$ ways of distributing the remaining people after the door openers are chosen.
Thus, the total number of possibilities is $120\times27=3240$
As far as I can tell, you've followed the same steps, but I think this is a bit less redundant. For example, instead of writing that there are $\binom{6}{3}\times3\times2$ ways of selecting door openers, you could have noticed that
$$
\begin{align}
\binom{6}{3}\times3\times2 &= \frac{6!}{3!3!}\times3!\\
&=\frac{6!}{3!}\\
&=\frac{6\times5\times4\times3\times2}{3\times2}\\
&=6\times5\times4
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A problem about the convergence of an improper integral Let $f:\mathbb R\longrightarrow\mathbb R$ be a function with
$$f(x)=\frac{1}{3}\int_x^{x+3} e^{-t^2}dt$$
and consider $g(x)=x^nf(x)$ where $n\in\mathbb Z$.
I have to discuss the convergence of the integral
$$\int_{-\infty}^{+\infty}g(x)dx$$ at the varying of $n\in\mathbb Z$.
Any idea? We have an integral of an "integral function", and I'm a bit confused!
Thanks in advance.
| I think I can give you expression for $\int_{-\infty}^{\infty}g(x)\ dx$. Then, we can think about how it should converge.
We start by noting that $$\large f(x)=\frac{\sqrt{\pi}}{3}\left(Q\left(\sqrt{2}x\right)-Q\left(\sqrt{2}(x+3)\right)\right)$$ and $$\large f(-x)=\frac{\sqrt{\pi}}{3}\left(-Q\left(\sqrt{2}x\right)+Q\left(\sqrt{2}(x-3)\right)\right)$$
Also we keep in mind that $$\large I=\int_{-\infty}^{\infty}g(x)\ dx=\int_{-\infty}^{\infty}x^nf(x)\ dx\\ \large =\int_{0}^{\infty}\left(x^nf(x)+(-x)^nf(-x)\right)\ dx $$
Case $1$: $n$ even
In this case,
\begin{equation}
\large
\begin{split}
I=&\int_{0}^{\infty}x^n\left(f(x)+f(-x)\right)\ dx\\
\ =& \frac{\sqrt{\pi}}{3}\int_{0}^{\infty}x^n\left(Q\left(\sqrt{2}(x-3)\right)-Q\left(\sqrt{2}(x+3)\right)\right)\ dx\\
\ =& \frac{\sqrt{\pi}}{3}\int_{0}^{\infty}\frac{x^n}{\sqrt{2\pi}}\int_{\sqrt{2}(x-3)}^{\sqrt{2}(x+3)}\exp\left({-\frac{u^2}{2}}\right)\ du \ dx\\
\ =&\frac{1}{3}\int_{0}^{\infty}x^n\int_{x-3}^{x+3}\exp\left({-\frac{u^2}{2}}\right)\ du \ dx\\
\end{split}
\end{equation}
Now, if the order of integration is changed then, as one can verify without much difficulty, we get the following two integrals $$\large I=I_1+I_2$$ where $$\large I_1=\frac{1}{3}\int_{-3}^{3}e^{-u^2}\int_{0}^{u+3}x^n\ dx\ du\\
\large =\frac{1}{3(n+1)}\int_{-3}^{3}e^{-u^2}(u+3)^{n+1}\ du\\ \large =\frac{1}{3(n+1)}\int_{0}^{3}e^{-u^2}\left\{(u+3)^{n+1}-(u-3)^{n+1}\right\} \ du\\ \large (\mbox{Since}\ n \ \mbox{is even})\\
\large I_2=\frac{1}{3}\int_{3}^{\infty}e^{-u^2}\int_{u-3}^{u+3}x^n\ dx\ du\\
\large =\frac{1}{3(n+1)}\int_{3}^{\infty}e^{-u^2}\left\{(u+3)^{n+1}-(u-3)^{n+1}\right\} \ du$$
Then,
\begin{equation}
\large
\begin{split}
I=& I_1+I_2\\
\ =& \frac{1}{3(n+1)}\int_{0}^{\infty}e^{-u^2}\left\{(u+3)^{n+1}-(u-3)^{n+1}\right\} \ du\\
\ =& \frac{1}{3(n+1)}\int_{-\infty}^{\infty}e^{-u^2}(u+3)^{n+1} \ du\ (\mbox{Since}\ n \ \mbox{is even})\\
\ =& \frac{1}{3(n+1)}\int_{-\infty}^{\infty}e^{-(u-3)^2}u^{n+1} \ du\\
\ =& \frac{\sqrt{\pi}}{3(n+1)}\mathbb{E}(X^{n+1})
\end{split}
\end{equation} where $\displaystyle X\sim \mathcal{N}\left(3,\frac{1}{2}\right)$.
Case $2$: $n$ odd
In this case,
\begin{equation}
\large
\begin{split}
I=& \int_{0}^{\infty}x^n(f(x)-f(-x))\ dx\\
\ =& \frac{\sqrt{\pi}}{3}\int_{0}^{\infty}x^n
\left[2Q\left(\sqrt{2}x\right)-Q\left(\sqrt{2}(x+3)\right)-Q\left(\sqrt{2}(x-3)\right)\right]\ dx\\
\ =& \frac{2I_2-(I_1+I_3)}{3}\\
\end{split}
\end{equation}
Now, \begin{equation}
\large
\begin{split}
I_1=&\sqrt{\pi}\int_{0}^{\infty}x^nQ\left(\sqrt{2}(x+3)\right)\ dx\\
\ =&\int_{0}^{\infty}x^n \int_{x+3}^{\infty}\exp\left(-u^2\right)\ du \ dx\\
\ =&\int_{3}^{\infty}\exp\left(-u^2\right) \int_{0}^{u-3}x^n\ dx \ du\ (\small \mbox{Interchanging order of integration})\\
\ =&\frac{1}{n+1}\int_{3}^{\infty}\exp\left(-u^2\right)(u-3)^{n+1} \ du\\
\ =&\frac{1}{n+1}\int_{3}^{\infty}\exp\left(-(u+3)^2\right)u^{n+1} \ du\\
\ =&\frac{1}{n+1}\int_{-\infty}^{0}\exp\left(-(u-3)^2\right)u^{n+1} \ du\ (\because\ n\ \mbox{is odd})\\
\end{split}
\end{equation}
Similarly, we will get $$\large I_3=\frac{1}{n+1}\int_{0}^{\infty}\exp\left(-(u-3)^2\right)u^{n+1} \ du$$
and \begin{equation}
\begin{split}
I_2=&\int_{0}^{\infty}\exp\left(-u^2\right) \int_{0}^{u}x^n\ dx \ du\\
\ =&\frac{1}{n+1}\int_{0}^{\infty}\exp\left(-u^2\right) u^{n+1} \ du\\
\ =&\frac{1}{2(n+1)}\Gamma\left(1+\frac{n}{2}\right)
\end{split}
\end{equation}
So, $$\large I=\frac{\Gamma\left(1+\frac{n}{2}\right)-\int_{-\infty}^{\infty}\exp\left(-(u-3)^2\right)u^{n+1} \ du}{3(n+1)}\\
\large =\frac{\sqrt{\pi}}{3(n+1)}\left(\frac{(2n-1)!!}{2^n}-\mathbb{E}\left(X^{n+1}\right)\right)\\
\large =\frac{\sqrt{\pi}}{3(n+1)}\left(\mathbb{E}(Y^{2n})-\mathbb{E}\left(X^{n+1}\right)\right)$$ where $\displaystyle X\sim \mathcal{N}\left(3,\frac{1}{2}\right),\ Y\sim \mathcal{N}\left(0,\frac{1}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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What is the number of real solutions of the following? $ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $ What is the number of real solutions of the following?
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
My solution:
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
$$ \implies \sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}-3)^2} = 1 $$
$$ \implies (\sqrt{x-1}-2) + (\sqrt{x-1}-3) = 1 $$
$$ \implies \sqrt{x-1} = 3$$
So, $ x = 10$ is the only solution.
But the answer key (and Wolfram alpha too) says there are infinite number of solutions to this equation. Where I am going wrong?
| It happens, for the choices of the arguments of the radicals in this problem, that we can make this a bit less of a headache to think about by noting that setting $ \ u = \sqrt{x-1} \ \Rightarrow \ u^2 = x - 1 \ $ reduces the original equation to
$$ \sqrt{(u^2 + 4) - 4u} \ + \ \sqrt{(u^2 + 9) - 6u} \ = \ 1 \ \Rightarrow \ | u - 2 | \ + \ | u - 3 | = 1 \ . $$
[To this point, this is similar to Adriano's argument.]
Since these terms must be positive or zero, we can choose, say, $ \ | u - 2 | = a \ $ and $ \ | u - 3 | = 1 - a \ , $ with $ \ 0 \le a \le 1 \ . $ Two of the possible equations, $ \ u - 2 = a \ $ and $ \ 3 - u = 1 - a \ $ produce $ \ u = 2 + a \ $ , while using $ \ 2 - u \ = a \ \Rightarrow \ u = 2 - a \ $ or $ \ u - 3 \ \Rightarrow \ 1 - a \ \Rightarrow \ u = 4 - a \ $ are not mutually consistent results.
So we have the single interval,
$$ 0 \ \le \ a \ \le \ 1 \ \Rightarrow \ 2 \ \le \ u = 2 + a \ \le \ 3 \ . $$
(Graphing $ \ | u - 2 | \ + \ | u - 3 | = 1 \ $ confirms this.) From this, we have
$$4 \ \le \ u^2 = x - 1 \ \le \ 9 \ \Rightarrow \ 5 \ \le \ x \ \le \ 10 \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $(2x^2-3x+1)(2x^2+5x+1)=9x^2$,then prove that the equation has real roots. If $(2x^2-3x+1)(2x^2+5x+1)=9x^2$,then prove that the equation has real roots.
MY attempt:
We can open and get a bi quadratic but that is two difficult to show that it has real roots.THere must be an easy way.!
| First divide both members of the equation by $x^2$:
$$(2x + \frac{1}{x} - 3)(2x + \frac{1}{x} + 5) = 9$$
With the notation $2x + \frac{1}{x} = y$ equation is obtained in $y$: $$y^2 + 2y - 24 =0$$ with roots $y_1 = -6$ and $y_2=4$.
The equation $$2x + \frac{1}{x} = -6$$ has roots $x_1,_2 = \frac{-3\pm\sqrt{7}}{2}$;
The equation $$2x + \frac{1}{x} = 4$$ has roots $x_3,_4 = \frac{2\pm\sqrt{2}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement. "Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$"
Resolution:
$$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)$$and$$8\mid n(n^{2}-1)(3n+2)?$$$$$$
Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you.
| We know from here or here, the product $n$ consecutive integers is divisible by $n!$ for integer $n>0$
As $24=4!$and we already have $n(n^2-1)=(n-1)n(n+1)$ the product of $3$ consecutive integers
So, if we can arrange the next or previous term as multiplier, we shall have the product of $4$ consecutive integers, hence divisible by $4!=24$
If we consider the previous term $(n-2),$
$3n+2=3(n-2)+4$
$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n-2)+4\}$$
$$=3\underbrace{(n+1)n(n-1)(n-2)}_{\text{ the product of } 4 \text{ consecutive integers }} +4 \underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$
If we consider the next term $(n+2),$
$3n+2=3(n+2)-4$
$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n+2)-4\}$$
$$=3\underbrace{(n+2)(n+1)n(n-1)}_{\text{ the product of } 4 \text{ consecutive integers }}-4\underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $83^{27} +1 $ a prime number? I'm having problems with exercises on proving whether or not a given number is prime. Is $83^{27} + 1$ prime?
| We have a chain of divisibilities, based on the fact that $(a-b)\mid(a^n-b^n)$,
$$
83^1-(-1)^1\mid83^3-(-1)^3\mid83^9-(-1)^9\mid83^{27}-(-1)^{27}=83^{27}+1
$$
Using this chain, we get, using $a^3-b^3=(a-b)(a^2+ab+b^2)$,
$$
\begin{align}
83^{27}+1
&=\frac{83^{27}+1}{83^9+1}\times\frac{83^9+1}{83^3+1}\times\frac{83^3+1}{83^1+1}\times\left(83^1+1\right)\\
&=\left(83^{18}-83^9+1\right)\times\left(83^6-83^3+1\right)\times\left(83^2-83^1+1\right)\times\left(83^1+1\right)\\[9pt]
&=34946659039493167203883141969862007\times326939801583\times6807\times84
\end{align}
$$
Thus, $83^{27}+1$ is not prime.
Note: none of these factors are guaranteed to be prime, just factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
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$(a^{n},b^{n})=(a,b)^{n}$ and $[a^{n},b^{n}]=[a,b]^{n}$? How to show that $$(a^{n},b^{n})=(a,b)^{n}$$ and $$[a^{n},b^{n}]=[a,b]^{n}$$ without using modular arithmetic? Seems to have very interesting applications.$$$$Try: $(a^{n},b^{n})=d\Longrightarrow d\mid a^{n}$ and $d\mid b^n$
| Bezout's Identity says that $\gcd(a,b)$ is the smallest positive element of $\{ax+by:x,y\in\mathbb{Z}\}$.
The smallest positive element of $\{acx+bcy:x,y\in\mathbb{Z}\}$ is $\gcd(ac,bc)$; it is also $c$ times the smallest positive element of $\{ax+by:x,y\in\mathbb{Z}\}$ which is $\gcd(a,b)$. Therefore
$$
\gcd(ac,bc)=c\gcd(a,b)\tag{1}
$$
Suppose $ax+by=1$ and $au+cv=1$, then
$$
\begin{align}
by\,cv&=(1-ax)(1-au)\\
&=1-a(x+u-axu)\\
a(x+u-axu)+bc\,vy&=1
\end{align}
$$
Thus,
$$
\gcd(a,b)=1\quad\text{and}\quad\gcd(a,c)=1\implies\gcd(a,bc)=1\tag{2}
$$
Using $(2)$ and induction, we get that
$$
\gcd(a,b)=1\implies\gcd\left(a^n,b^n\right)=1\tag{3}
$$
Using $(1)$ and $(3)$, we have
$$
\begin{align}
\gcd(a,b)\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)&=\gcd(a,b)&&(1)\\
\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)&=1&&\text{cancel}\\
\gcd\left(\frac{a^n}{\gcd(a,b)^n},\frac{b^n}{\gcd(a,b)^n}\right)&=1&&(3)\\
\gcd(a,b)^n\gcd\left(\frac{a^n}{\gcd(a,b)^n},\frac{b^n}{\gcd(a,b)^n}\right)&=\gcd(a,b)^n&&\text{multiply}\\[9pt]
\gcd\left(a^n,b^n\right)&=\gcd(a,b)^n&&(1)\tag{4}
\end{align}
$$
In this answer, it is shown that
$$
\gcd(a,b)\,\mathrm{lcm}(a,b)=ab\tag{5}
$$
Therefore,
$$
\begin{align}
\mathrm{lcm}(a,b)^n
&=\frac{(ab)^n}{\gcd(a,b)^n}&&(5)\\
&=\frac{a^n\,b^n}{\gcd(a^n,b^n)}&&(4)\\[4pt]
&=\mathrm{lcm}(a^n,b^n)&&(5)\tag{6}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Quadratic inequality with boundaries Here is a very old high school exam question I am trying to solve (purely for interest only):
If $a,b,c$ are real numbers such that $-1 \le ax^2+bx+c \le 1$ for $-1 \le x \le 1$ prove that $-4 \le 2ax+b \le4$ for $-1 \le x \le 1$ (Hint: Consider the functions at the end-points and at the mid-point of the interval).
I can see (graphically) that if $a>0$ then, as $2ax+b$ is the gradient function, the max and min gradients will occur when the parabola passes through the end-points $(-1,1)$ and $(1,1)$ and the mid-point $(0,-1)$. This gives 3 equations with 3 unknowns and is solved to give $a=2, b=0$ and $c=-1$. The required result follows easily from this. Due to symmetry, a<0 gives the same result.
Can someone please help turn my "partial" solution into a more convincing/algebraic solution.
| We can show the required inequlity in an entirely algebraic manner:
We first note that by setting $x=0$ in our initial inequality we get:
$$-1\le c\le1 \\ |c|\le1.$$
If we set $x=\pm1$ in our given inequality we will get:
$$
\text{(i)}\;\;-1\le a+b+c\le 1 \;\;\;\text{and}\;\;\;\text{(ii)}\;\;-1\le a-b+c\le 1.
$$
Adding these two inequalities together will give us:
$$
-2\le 2a+2c\le2 \\
-1\le a+c\le1 \\
-2\le-c-1\le a\le-c+1\le2 \\
|a|\le2.
$$
Thus, considering only $-1\le x\le1$, we see that if $a\gt 0$ then:
$$2ax+b \;\le\; 2a+b \;=\; a+b+c + (a-c)\;\le\;1 + (a-c) \;\le\;1+1+2\;=\;4,$$
and
$$-4 \;=\; -2-1 -1\;\le\; (-a+c) - 1\;\le\; (-a+c) - a+b-c\;\le\; -2a+b \;\le\;2ax+b.$$
Similarily, if $a\lt0$ then:
$$2ax+b \;\ge\; 2a+b \;=\; a+b+c + (a-c)\;\ge\;-1 + (a-c) \;\ge\;-1-1-2\;=\;-4,$$
and similarily:
$$4 \;=\; 2+1 +1\;\ge\; (-a+c) + 1\;\ge\; (-a+c) - a+b-c\;\ge\; -2a+b \;\ge\;2ax+b.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/460889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Positive semi-definite matrix problem If $A，B$ and $M$ are positive semi-definite matrices, and we have
$$ A+B \succeq M .$$
Do there always exist two positive semi-definite matrices $ M_{1}, M_{2} , $ such that
$$ A \succeq M_{1}, \quad B \succeq M_{2}, \quad M_{1}+M_{2}=M ?$$
| The statement is false even when we impose stronger conditions of positive definiteness on the matrices.
Let $p > 0$ be a number to be determined. Consider the following martices:
$$
A = \begin{pmatrix}2p + 1& 0\\0&1\end{pmatrix}, \quad
B = \begin{pmatrix}1 & 0\\0 & 2p + 1\end{pmatrix} \quad\text{ and }\quad
M = \begin{pmatrix}p+1 & p\\p & p+1\end{pmatrix}
$$
It is clear $A, B, M \succ 0$ and $A + B - M = \begin{pmatrix}p+1&-p\\-p&p+1\end{pmatrix} \succ 0 \implies A + B \succ M$
If $M_1 = \begin{pmatrix}a_1&b_1\\b_1&c_1\end{pmatrix} \succeq 0$ and $A \succeq M_1$, we must have:
$$0 \le a_1 \le 2p+1,\quad 0\le c_1 \le 1\quad\text{ and }\quad a_1 c_1 - b_1^2 \ge 0$$
This implies $|b_1| \le \sqrt{a_1c_1} \le \sqrt{2p+1}$. By a similar argument, if
$M_2 = \begin{pmatrix}a_2&b_2\\b_2&c_2\end{pmatrix} \succeq 0$ and $B \succeq M_2$, we will have $|b_2| \le \sqrt{2p+1}$.
If we choose $p$ such that $$2\sqrt{2p+1} < p\quad\iff\quad p > 4 + 2\sqrt{5}$$
then there is no way such a pair of $M_1, M_2$ can sum to $M$. The off diagonal element simply won't match.
| {
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"url": "https://math.stackexchange.com/questions/461694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I show that $\gcd(a^2, b^2) = 1$ when $\gcd(a,b)=1$? How do I show that $\gcd(a^2, b^2) = 1$ when $\gcd(a,b)=1$?
I can show that $\gcd(a,b)=1$ implies $\gcd(a^2,b)=1$ and $\gcd(a,b^2)=1$. But what do I do here?
| If $\gcd(a, b) = 1$, this means exactly that $a$ and $b$ do not have any prime factors in common. Any common factor would constitute a common divisor. If $a$ and $b$ have some prime factors in common, then the product of those common factors (including any repeated ones of higher multiplicity) is the greatest common divisor.
For instance $24 = 2 \cdot 3\cdot 4$ and $60 = 3\cdot 4\cdot 5$. The common factors are $3$ and $4$ and so $\gcd(24, 60) = 3\cdot 4 = 12$.
If we take an integer $a$ and square it, the resulting integer $a^2$ does not have any new prime factors which are not already present in $a$. It simply has the same factors, with double the multiplicity.
For instance $12 = 3\cdot 2\cdot 2$, and $144 = 12^2 = (3\cdot 2\cdot 2)(3\cdot 2 \cdot 2) = 3\cdot 3\cdot 2\cdot 2\cdot 2\cdot 2$.
So if $a$ and $b$ have no prime factors in common, then $a^2$ and $b^2$ have no prime factors in common either, and so $\gcd(a^2, b^2) = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the sum of this by telescopic method? $$\sum_{r=1}^\infty \frac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}$$
This is to be done by telescopic method. I've used many things like writing $r^4+r^2+1=(r^2+r+1)(r^2-r+1)$ but have failed. I think all it needs is the correct rearranging of terms.
| Hints:
*
*Since $r^3+(r^2+1)^2=r(r^2+r)+(r^4+r^2+1)$,
$$
\frac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}=\frac{r}{r^4+r^2+1}+\frac1{r^2+r}.
$$
*Since $r^4+r^2+1=(r^2+r+1)(r^2-r+1)$,
$$
\frac{2r}{r^4+r^2+1}=a_r-a_{r+1},\qquad a_r=\frac{1}{r(r-1)+1}.
$$
*Since $r^2+r=r(r+1)$,
$$
\frac1{r^2+r}=b_r-b_{r+1},\qquad b_r=\frac1r.
$$
And now, grab your telescope!
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\sec \theta + \tan \theta = 5$ , Find $\csc \theta + \cot \theta $. The question is to find the value of $ \csc \theta + \cot \theta $ if $\sec \theta + \tan \theta = 5$ .
Here is what I did :
$\sec \theta + \tan \theta = 5$
$\sec \theta = 5 - \tan \theta $
Squaring both sides ,
$$\sec^2 \theta = 25 + \tan^2 \theta -10\tan \theta$$
Substituting $1+\tan^2 \theta$ for $\sec^2 \theta$ ,
$$1+\tan^2 \theta = 25 + \tan^2 \theta -10\tan \theta$$
Thus , $$\tan \theta=24/10$$
So , $\cot \theta = 10/24 $ and $\csc \theta=26/24$
Thus $ \csc \theta + \cot \theta =3/2$ .
But I checked the answer sheet and the answer is not 3/2 but $(3+\sqrt5 )/2$ .
Where have I went wrong ? Please help.
| Too long for a comment:
...
THIS IS HOW I THINK THE ANSWER SHEET WRITER DID THE MISTAKE:
Normally the equation is
$$\sin\theta+1=5\cos\theta.$$
Squaring they "mistakenly" got $\sin^2\theta+2\sin\theta+1=5\cos^2\theta\implies$ $\sin^2\theta+2\sin\theta+1=5-5\sin^2\theta\implies$ $3\sin^2\theta+\sin\theta-2=0\implies$ $(3\sin\theta-2)(\sin\theta+1)=0$.
Thus, $\sin\theta=\frac{2}{3}$ is a solution. Finally, $\csc\theta=\frac{3}{2}$, $\cot\theta=\frac{\sqrt{5}}{2}$ and $\csc\theta+\cot\theta=\frac{3+\sqrt{5}}{2}$.
CORRECT SOLUTION:
But in fact squaring we get $(13\sin\theta-12)(\sin\theta+1)=0$.
Thus, $\sin\theta=\frac{12}{13}$ is a solution. Finally, $\cot\theta=\frac{5}{12}$, $\csc\theta=\frac{13}{12}$ and
$$\csc\theta+\cot\theta=\frac{13+5}{12}=\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solve differential equation using given substitution Solve the following equation by making substitution
$$y=xz^n \text{ or } z=\frac{y}{x^n}$$
and choosing a convenient value of n.
$$\frac{dy}{dx}= \frac{2y}{x} +\frac{x^3}{y} +x\tan\frac{y}{x^2}$$
I thought it can be solved in 2 ways
1.making it exact differential equation
2.making it linear differential equation
But I do not understand when i will do substitution.
Please help me solve this question.
| Since the ODE has the term $x\tan\dfrac{y}{x^2}$ present, so it is obviously to choose $n=2$ .
Let $z=\dfrac{y}{x^2}$ ,
Then $y=x^2z$
$\dfrac{dy}{dx}=x^2\dfrac{dz}{dx}+2xz$
$\therefore x^2\dfrac{dz}{dx}+2xz=2xz+\dfrac{x}{z}+x\tan z$
$x^2\dfrac{dz}{dx}=x\left(\dfrac{1}{z}+\tan z\right)$
$\dfrac{dx}{x}=\dfrac{z}{z\tan z+1}dz$
$\int\dfrac{dx}{x}=\int\dfrac{z}{z\tan z+1}dz$
$\int\dfrac{dx}{x}=\int\dfrac{z\cos z}{z\sin z+\cos z}dz$
$\int\dfrac{dx}{x}=\int\dfrac{d(z\sin z+\cos z)}{z\sin z+\cos z}$
$\ln x=\ln(z\sin z+\cos z)+c$
$x=C(z\sin z+\cos z)$
$x=C\left(\dfrac{y}{x^2}\sin\dfrac{y}{x^2}+\cos\dfrac{y}{x^2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Area ratio in triangle? Given: $\triangle ABC$. In the side $AB$, we choose point $D$. From this point $D$, we draw a line $DF$ such that intersect side $AC$ and line $DE$ such that intersect side $BC$. If $DF\parallel BC$, $DE\parallel AC$, and the area of $\triangle BDE = p$ times area of $\triangle$ADF, what is the ratio of area $\triangle CEF$ and $\triangle ABC$ ?
| The quadrilateral $CEDF$ is a parallelogram, and therefore $CE=DF=d$, $CF=DE=e$, and $\triangle CEF \equiv \triangle DEF$.
Suppose $h$ is the height of parallelogram $CEDF$ and $\ell$ is the height of $\triangle ADF$. Then, $h$ will also be the height of $\triangle BDE$ and $\triangle CEF$.
$$\therefore \ \text{Area of }\triangle CEF = \text{Area of }\triangle DEF = \frac12 \text{ Area of $CEDF$ parallelogram} = \frac12 dh$$
As shown in the diagram (use your knowledge on parallelograms), you may find $\triangle BDE$ and $\triangle ADF$ are similar. Thus, their corresponding side lengths have the same ratio:
$$ \frac{BE}{DF} = \frac{BD}{DA} = \frac{DE}{AF} = \frac{c}{d} = \frac{b}{a} = \frac{e}{f} \tag1$$
Suppose areas of $\triangle BDE$ and $\triangle ADF$ are $A_{\triangle BDE}$ and $A_{\triangle ADF}$, respectively. Given that:
$$\frac{A_{\triangle BDE}}{A_{\triangle ADF}} = p \ \Rightarrow \ A_{\triangle ADF} = \frac{1}{p} A_{\triangle BDE} \tag2$$
$$\therefore \ \frac{\frac12ch}{\frac12d\ell} = \frac{ch}{d\ell} = p \ \Rightarrow \ \frac{c}{d} = \frac{\ell}{h}p \tag 3$$
From the equations $(1)$ and $(2)$, $\frac{b}{a} = \frac{\ell}{h}p$ and $\frac{e}{f} = \frac{\ell}{h}p$, as well.
You may also find $\triangle CGF$ and $\triangle AFH$ are similar as well. Thus, their corresponding side lengths have the same ratio:
$$ \frac{AF}{CF} = \frac{AH}{FG} = \frac{f}{e} = \frac{\ell}{h} \ \Rightarrow \ \frac{e}{f} = \frac{h}{\ell} $$
However, we previously found $\frac{e}{f} = \frac{\ell}{h}p$
$$\therefore \frac{\ell}{h}p = \frac{h}{\ell} \ \Rightarrow \ \frac{h^2}{\ell^2} = p \ \Rightarrow \ \frac{h}{\ell} = \sqrt{p} \tag4$$
$$\frac{A_{\triangle BDE}}{A_{\triangle CEF}} = \frac{\frac12ch}{\frac12dh} = \frac{ch}{dh} = \frac{c}{d} = \frac{\ell}{h}p = \frac{p}{\sqrt{p}} \ \Rightarrow \ \therefore \ A_{\triangle BDE} = \frac{p}{\sqrt{p}} A_{\triangle CEF} \tag5$$
From the equation $(2)$:
$$A_{\triangle ADF} = \frac{1}{p} A_{\triangle BDE} = \frac{1}{p} \frac{p}{\sqrt{p}} A_{\triangle CEF} = \frac{1}{\sqrt{p}} A_{\triangle CEF} \tag6$$
If the area of $\triangle ABC$ is $A_{\triangle ABC}$, then:
$$A_{\triangle ABC} = A_{\triangle ADF} + A_{\triangle BDE} + 2 \times A_{\triangle CEF} = \frac{1}{\sqrt{p}} A_{\triangle CEF} + \frac{p}{\sqrt{p}} A_{\triangle CEF} + 2 \times A_{\triangle CEF} \\= \left(\frac{1}{\sqrt{p}} + \frac{p}{\sqrt{p}} + 2 \right)A_{\triangle CEF} $$
$$\therefore \ \frac{ A_{\triangle ABC}}{A_{\triangle CEF} } = \frac{\sqrt{p}}{p}(1+p) + 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the 12th term and the sum of the first 12 terms of a geometric sequence. A geometric series has a first term $\sqrt{2}$ and a second term $\sqrt{6}$ . Find the 12th term and the sum of the first 12 terms.
I can get to the answers as irrational numbers using a calculator but how can I can obtain the two answers in radical form $243 * \sqrt{6}$ and $364 \left(\sqrt{6}+\sqrt{2}\right)$ ?
The closest I get with the 12th term is $\sqrt{2} \left(\sqrt{6} \over \sqrt{2}\right)^{(12-1)}$ or $\sqrt{2} * 3^\left({11\over 2}\right)$
And for the sum ${\sqrt{2}-\sqrt{2}*(\sqrt{3})^{12} \over 1 - \sqrt{3}}$
| The general term of a geometric series is $a r^{n-1}$, where $r$ is the ratio, and $a$ is the first term. In your case, $a=\sqrt{2}$ and $r=\sqrt{3}$. The 12th term is then
$$\sqrt{2} (\sqrt{3})^{11} = 243 \sqrt{6}$$
The sum of the first 12 terms is
$$a \sum_{k=0}^{11} r^k = a \frac{r^{12}-1}{r-1} = \sqrt{2} \frac{728}{\sqrt{3}-1} = 364 (\sqrt{2}+\sqrt{6})$$
Note that I used $3^5 = 243$ and $3^6=729$ in the above.
EDIT
Note that
$$ \sqrt{2} \frac{728}{\sqrt{3}-1} = \sqrt{2} \frac{728}{\sqrt{3}-1} \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{2} (\sqrt{3}+1) 728}{(\sqrt{3})^2-1^2} = \frac{(\sqrt{6}+\sqrt{2}) 728}{3-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468331",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^{\infty} {y^2 \cos^2(\frac{\pi y}{2}) \over (y^2-1)^2} dy$ I´m having trouble with the following integral
$$
\int_0^{\infty} {y^2 \cos^2(\frac{\pi y}{2}) \over (y^2-1)^2} dy
$$
I have tried lots of approaches and nothing works. Mathematica says it does not converge but that is not true. It appears in a Physical problem (it is the energy of a system) and the answer should be (by conservation of energy): $\frac{\pi^2}{8}$ but I cannot show it.
| Let $I$ denote the integral. Then
\begin{align*}
I &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{y^2 \cos^{2} (\pi y/2)}{(y^{2} - 1)^{2}} \, dx \\
&= \frac{1}{4} \int_{-\infty}^{\infty} \frac{y^2 (1 + \cos \pi y)}{(y^{2} - 1)^{2}} \, dx \\
&= \frac{1}{4} \Re \mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{y^2 (1 + e^{i\pi y})}{(y^{2} - 1)^{2}} \, dx.
\end{align*}
Now considering an upper-semicircular contour with two vanishing upper-semicircular indents at $z = \pm 1$, it follows that
\begin{align*}
\mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{z^2 (1 + e^{i\pi z})}{(z^{2} - 1)^{2}} \, dz
&= \pi i \left( \mathrm{Res}_{z=-1} \frac{z^2 (1 + e^{i\pi z})}{(z^{2} - 1)^{2}} + \mathrm{Res}_{z=1} \frac{z^2 (1 + e^{i\pi z})}{(z^{2} - 1)^{2}} \right) \\
&= \pi i \left( -\frac{i\pi}{4} -\frac{i\pi}{4} \right) \\
&= \frac{\pi^{2}}{2}.
\end{align*}
Therefore the conclusion follows.
| {
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"url": "https://math.stackexchange.com/questions/468664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof: for a pretty nasty limit Let $$ f(x) = \lim_{n\to\infty} \dfrac {[(1x)^2]+[(2x)^2]+\ldots+[(nx)^2]} {n^3}$$.
Prove that f(x) is continuous function.
Edit: $[.] $ is the greatest integer function.
| $k^2x^2-1 < [(kx)^2] \le k^2x^2$
Summing the three expressions from $k=1$ to $n$ and dividing by $n^3$, we get
$\frac{x^2}{6}(1+\frac{1}{n})(2+\frac{1}{n}) - \frac{1}{n^2}< \frac{[(1x)^2] + \cdots + [(nx)^2]}{n^3} \le \frac{x^2}{6}(1+\frac{1}{n})(2+\frac{1}{n}) $.
Taking limits as $n \rightarrow \infty$, we get
$\frac{x^2}{3} \le f(x) \le \frac{x^2}{3}$ for every $x \in \mathbb{R}$. Thus $f(x)=\frac{x^2}{3}$ which is a continuous function.
(Note that the above, also proves that the limit always exists.)
| {
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"url": "https://math.stackexchange.com/questions/470247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving inequalities, simplifying radicals, and factoring. (Pre calculus) (Q.1) Solve for $x$ in $x^3 - 5x > 4x^2$
its a question in pre calculus for dummies workbook, chapter 2. The answer says:
then factor the quadratic: $x(x-5)(x+1)>0$. Set your factors equal to $0$ so you can find your key points.When you have them, put these points on a number line and plug in test numbers form each possible section to determine whether the factor would be positive or negative. Then, given that you're looking for positive solution, think about the possibilities: (+)(+)(+) = +,
++- = - , -+- = +, --- = -. Therefore, your solution is $-1 < x < 0$ or $x > 5$.
so i know how he got the $x>5$ but i don't get the $-1 < x < 0$ cuz it suppose to be $x > -1$ and what these possibilities have to do with the solution ? please explain to me in details.
(Q.2) solve for x in $x^\frac{5}{3} - 6x = x^\frac{4}{3}$
same book. The answer says: Next, factor out an $x$ from each term: $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$. The resulting expression is similar to $y^3(y^2 - y - 6)$, which factors into $y^3(y+2)(y-3)$.Similarly, you can factor $x(x^\frac{2}{3} - x^\frac{1}{3} - 6)$ into $x(x^\frac{1}{3} + 2)(x^\frac{1}{3} - 3) = 0$
I don't get how did he factor the main equation and $x^\frac{5}{3}$ became $x^\frac{2}{3}$ and $x^\frac{4}{3}$ became $x^\frac{1}{3}$. I know how to factor like this but with numbers not fractions. And also how this expression $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$ is similar to that $y^3(y^2 - y - 6)$. I'm solving for x so why he got the y into the answer now ?! (weird)
(Q.3) simplify $\frac{8}{4^\frac{2}{3}}$
same book. The answers says: change it to $\frac{8}{\sqrt[3]{4^2}}$ ( i get this one ) but then he said multiply the numerator and denominator by one more cube root of 4 . so how i can multiply $\frac{8}{\sqrt[3]{4^2}}$ to $\frac{8}{ \sqrt[3]{4}}$ and get $4$ ? isn't supposed to be $\frac{8}{\sqrt[3]{4^2}}*\frac{8}{\sqrt[3]{4^2}}$equals $4$ ?
| multiply the numerator AND denominator by one more cube root of 4.
$$\frac{8}{4^{\frac{2}{3}}} = \frac{8}{\sqrt[3]{4^2}}\cdot \frac{\sqrt[3]{4^1}}{\sqrt[3]{4^1}} = \frac{8\cdot\sqrt[3]4}{\sqrt[3]{4^3}}=\frac{8\cdot\sqrt[3]4}{4}=2\cdot\sqrt[3]4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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