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Differential equations of first order but of higher degree Solve $(x-py)(px-y)=2p$ where $p=\frac{dy}{dx}$
My attempt:
I started by making it a quadratic in $p$ but after a point a could not proceed further:
$$p^2-\frac{p(x^2+y^2-2)}{xy} + 1 = 0$$
$$p = (x^2+y^2-2)\pm\frac{(x^2-y^2)^2-4(x^2+y^2-1)^{1/2}}{2xy}$$
Now I don't know how to proceed further.
|
We are trying to solve
$$\tag 1 (px-y)(x-py)=2p , ~\text{where}~p=\frac{dy}{dx}$$
Continuing on the path used leads to something unsolvable, so let's try another approach.
Let $x^2 = X \implies 2 x dx = dX, y^2 = Y \implies 2 y dy = dY$, so
$$P = \dfrac{dY}{dX} = \dfrac{2 y dy}{2 x dx} = \dfrac{y}{x} p \implies p = \dfrac{x}{y}P$$
Substituting into $(1)$
$$\left(\left(\dfrac{x}{y} P\right) x - y\right)\left(x - \left(\dfrac{x}{y} P\right) y \right) = 2 \dfrac{x}{y}P$$
Simplifying
$$\left( \dfrac{x^2}{y}P - y\right) (x(1-P))= 2 \dfrac{x}{y}P$$
Solving for $y$
$$y^2 = P x^2 - 2 \dfrac{P}{1-P} = P X - 2 \dfrac{P}{1-P}$$
This is a Clairaut's Equation of the form $y = px + f(p)$, so we can write the solution as
$$y^2 = c x^2 -2 \dfrac{c}{1-c}$$
|
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|
Closed form for $f(x)=\ _3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$
I am seeking a closed form for the function $$f(x)=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$$
I expect there to be one, because of this post and Wolfram. The Wolfram link produces closed forms involving $\mathrm{Li}_2$ for any value of $x$ that I've tried so far, so I can only assume that a general closed form exists.
I've started my attempts by noticing that
$$f(x)=\frac12\int_0^1 \frac{_2F_1(\tfrac12,\tfrac12;\tfrac32;xt)}{\sqrt{t}}dt,$$
because
$$\frac12\int_0^1 \frac{(xt)^n}{\sqrt{t}}dt=\frac{x^n}{2n+1}$$
which would introduce another factor of $$\frac{n+1/2}{n+3/2}$$
when computing the ratio of the terms. Similarly,
$$_2F_1\left(\tfrac12,\tfrac12;\tfrac32;x\right)=\frac12\int_0^1 \frac{_1F_0(\tfrac12;;xt)}{\sqrt{t}}dt.$$
The last hypergeometric I was able to recognize as $$_1F_0\left(\tfrac12;;xt\right)=\frac1{\sqrt{1-xt}}.$$
So, all in all,
$$f(x)=\frac14\int_0^1\int_0^1 \frac{1}{\sqrt{vu}\sqrt{1-xvu}}dvdu,$$
which looks like the Beta function's evil cousin.
I do not know how to turn this integral into something containing $\mathrm{Li}_2$ and I need some help. Thanks!
|
Starting from the integral representation, we let $v = w/u$ and change the order of integration to find
\begin{align}
f(x) &= \frac{1}{4} \int \limits_0^1 \int \limits_0^u \frac{1}{u \sqrt{w(1-xw)}} \, \mathrm{d} w \, \mathrm{d} u = \frac{1}{4} \int \limits_0^1 \frac{\mathrm{d} w}{\sqrt{w(1-xw)}} \int \limits_w^1 \frac{\mathrm{d} u}{u} \\
&= \frac{1}{4} \int \limits_0^1 \frac{-\log(w)}{\sqrt{w(1-xw)}} \, \mathrm{d} w \, .
\end{align}
The combination of the next few substitutions can be written as $w = \sin^2(t/2)/x$, which yields
\begin{align} f(x) &= \frac{1}{2 \sqrt{x}} \int \limits_0^{2\arcsin(\sqrt{x})} - \log\left(\frac{\sin\left(\frac{t}{2}\right)}{\sqrt{x}}\right) \, \mathrm{d} t \\
&= \frac{\arcsin(\sqrt{x})}{\sqrt{x}} \log(2 \sqrt{x}) + \frac{1}{2\sqrt{x}} \int \limits_0^{2\arcsin(\sqrt{x})} - \log\left(2 \sin\left(\frac{t}{2}\right)\right) \, \mathrm{d} t \\
&= \frac{1}{\sqrt{x}} \left[\arcsin(\sqrt{x}) \log(2\sqrt{x}) + \frac{1}{2} \operatorname{Cl}_2(2\arcsin(\sqrt{x}))\right]
\end{align}
for $x \in (0,1]$, while $f(0) = 1$. The Clausen function is of course related to the dilogarithm. Interesting special values include $f(1) = \frac{\pi}{2} \log(2)$, $f\left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}} \left[\frac{\pi}{4} \log(2) + \mathrm{G}\right]$ and $f\left(\frac{1}{4}\right) = \operatorname{Cl}_2\left(\frac{\pi}{3}\right)$ (see this question).
For $y > 0$ similar steps lead to
\begin{align}
f(-y) &= \frac{\operatorname{arsinh}(\sqrt{y})}{\sqrt{y}} \log(2 \sqrt{y}) + \frac{1}{2\sqrt{y}} \int \limits_0^{2\operatorname{arsinh}(\sqrt{y})} - \log\left(2 \sinh\left(\frac{t}{2}\right)\right) \, \mathrm{d} t \\
&= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{1}{2} \int \limits_0^{2\operatorname{arsinh}(\sqrt{y})} - \log\left(1 - \mathrm{e}^{-t}\right) \, \mathrm{d} t \right] \\
&= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{\pi^2}{12} - \frac{1}{2} \operatorname{Li}_2\left(\mathrm{e}^{-2\operatorname{arsinh}(\sqrt{y})}\right) \right] \\
&= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{\pi^2}{12} - \frac{1}{2} \operatorname{Li}_2\left[\left(\sqrt{1+y} - \sqrt{y}\right)^2\right] \right] ,
\end{align}
which closely resembles Claude Leibovici's result (some dilogarithm identities should do the trick). Here $f\left(-\frac{1}{4}\right) = \frac{\pi^2}{10}$ looks rather nice, as does $f\left(-\frac{1}{8}\right) = \frac{\pi^2 - 3 \log^2(2)}{6 \sqrt{2}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing real integral using residues Let's consider the following integral
$$\int \limits_{-\infty}^{\infty} \frac{\cos2x}{(x^2+4)^2} \, dx. \tag{1}$$
I would like to compute $(1)$ using residue theory. Let's consider a complex function
$$f(z) = \frac{\cos2z}{(z^2+4)^2} = \frac{\cos2x}{(z-2i)^2(z+2i)^2}.$$
Of course $\text{Im}(-2i) < 0$ thus I am to calculate residue only in the point $z_0 = 2i$.
Noting that $z_0$ is a double pole we have
$$R = \text{res}_{z_0}f(z) = \lim_{z \to 2i} \frac{d}{dz} \bigg((z-2i)^2 \frac{\cos(2z)}{(z-2i)^2(z+2i)^2} \bigg).$$
After some calculations we get
$$R = \frac{i \big(-5 + 3 e^8 \big)}{64 e^4}.$$
That implies
$$\int \limits_{-\infty}^{\infty} \frac{\cos2x}{(x^2+4)^2} \, dx = 2 \pi i \frac{i \big(-5 + 3 e^8 \big)}{64 e^4} = -\frac{(-5 + 3 e^8) \pi}{32 e^4}.$$
According to WolframAlpha $(1)$ is equal to
$$\frac{5 \pi}{16 e^4}.$$
What am I doing wrong?
|
I want to write more details, may helps someone. Consider
$$\int_C \frac{e^{2iz}}{(z^2+4)^2} \, dz$$
where $C$ is the contour, the semicircle in upper half plane (as you have gotten I think), then the integrand has poles at $z=\pm 2i$, which it's reside at $z=2i$ is
$$\operatorname{Res}_{f}(2i) = \lim_{z\to2i} \frac{d}{dz} \bigg((z-2i)^2 \frac{e^{2iz}}{(z-2i)^2(z+2i)^2} \bigg) = \lim_{z\to2i} e^{2iz}\dfrac{2iz-6}{(z+2i)^3}=-i\dfrac{5}{32}e^{-4}$$
then
$$\int_{-R}^{R}\dfrac{\cos2z+i\sin2z}{(z^2+4)^2} \, dz+\int_\gamma \frac{e^{2iz}}{(z^2+4)^2} \, dz=2\pi i\times-i\dfrac{5}{32}e^{-4}=\dfrac{5\pi}{16}e^{-4}$$
where $\gamma$ is upper semicircle $|z|=1$. Finally take the limit $R\to\infty$ and $find the result.
|
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|
If $a^b$ has $b$ digits, what is the greatest value of $b$?
For natural numbers $a$ and $b$, what is the greatest value of $b$ so that $a^b$ has $b$ digits?
I knew that the greatest value of $b$ is $21$, where $9^{21}=\underset{21 \text{ digits}}{\underbrace{109,418,989,131,512,359,209}}$, but I am not sure how to prove that there is no greater value of $b$.
|
If $a^b$ has $b$ digits then
$10^{b-1} \le a^b < 10^b$.
$\log 10^{b-1} \le \log a^b < \log 10^b$ (Notation: $\log$ means $\log_{10}$)
$b-1 \le b\log a < b$
$1 - \frac 1b \le \log a <1$. so $a < 10$
The largest we can allow $b$ to be, the closer to $1$ that $\log a$ will be.
Smaller values of $a$ will result is smaller options for $b$ and large options of $a$ will allow for larger options for $b$. And if $a <a'$ and $b < b'$ we have $a^b < a'^{b'}$ we clearly will get larger values of $a^b$ if $a$ is as large as possible. So we can make $b$ largest (and therefore $a^b$ largest) when $a = 9$.
so $1-\frac 1b \le \log 9< 1$
which means $b \le \frac 1{1-\log 9}$
Welp.... calculator time... $b \le 21.85$ so $9^{21}$ is indeed the largest such number.
|
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|
Prove that $a_n^2 - 2b_n^2 = 1$ if $a_n+b_n\sqrt{2}=(a_{n-1}+b_{n-1}\sqrt{2})^2$
Two sequences of positive integers $a_n$ and $b_n$ are defined by $a_1 = b_1 = 1$ and $$a_n+\sqrt{2}b_n=(a_{n-1}+\sqrt{2}b_{n-1})^2$$ for $n \ge 2$. Prove that $(a_n)^2 - 2(b_n)^2 = 1$.
I found that
$$a_n+b_n\sqrt{2}=(1+\sqrt{2})^{2^{n-1}}$$
and I want to prove
$$a_n-b_n\sqrt{2}=(a_{n-1}-b_{n-1}\sqrt{2})^2$$
but it didn’t work. Please give me some hints. Thanks for your advances.
|
If $a_n+\sqrt{2}b_n=(a_{n-1}+\sqrt{2}b_{n-1})^2=a_{n-1}^2+2b_{n-1}^2+2\sqrt2a_{n-1}b_{n-1}$
then $a_{n}=a_{n-1}^2+2b_{n-1}^2$ and $b_n=2a_{n-1}b_{n-1},$
so you can prove that $a_n^2-2b_n^2=1$ by induction because
$a_n^2-2b_n^2=(a_{n-1}^2+2b_{n-1}^2)^2-2(2a_{n-1}b_{n-1})^2=a_{n-1}^4+4b_{n-1}^4+4a_{n-1}^2b_{n-1}^2-8a_{n-1}^2b_{n-1}^2$
$=a_{n-1}^4+4b_{n-1}^4-4a_{n-1}^2b_{n-1}^2 =(a_{n-1}^2-2b_{n-1}^2)^2.$
|
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"timestamp": "2023-03-29T00:00:00",
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Calculating complex roots "Find all the complex roots of the following polynomials
A) $S(x)=135x^4 -324x^3 +234x^2 -68x+7$, knowing that all its real roots belong to the interval $(0.25;1.75)$
B)$M(x)=(x^3 -1+i)(5x^3 +27x^2 -28x+6)$
"
Well, in A) I don't know how to use the given information about real roots. I mean, I know that I can apply Bolzano but I don't think that's very useful. To find the complex roots I should have some information about a complex root in particular so that I could use Ruffini, but this is not the case.
And in B) I know that $(x^3 -1+i)$ is giving me some information related to a complex root, but that "^3" bothers me. If it wasn't there, I would know that $1-i$ is a root...
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Since $\frac{1}{3}$ is a root of $S$, we obtain:
$$S=135x^4-324x^3+234x^2-68x+7=$$
$$=135x^4-45x^3-279x^3+93x^2+141x^2-47x-21x+7=$$
$$=(3x-1)(45x^3-93x^2+47x-7)=$$
$$=(3x-1)(45x^3-15x^2-78x^2+26x+21x-7)=$$
$$=(3x-1)^2(15x^2-26x+7)=(3x-1)^2(15x^2-5x-21x+7)=(3x-1)^3(5x-7).$$
Since $\frac{3}{5}$ is a root of $5x^3+27x^2-28x+6$, we obtain:
$$5x^3+27x^2-28x+6=5x^3-3x^2+30x^2-18x-10x+6=(5x-3)(x^2+6x-2)=$$
$$=(5x-3)((x+3)^2-11)=(5x-3)(x+3-\sqrt{11})(x+3+\sqrt{11}).$$
Also, $$\sqrt[3]{1-i}=\sqrt[6]2\sqrt[3]{\cos315^{\circ}+i\sin315^{\circ}}=$$
$$=\sqrt[6]2(\cos(105^{\circ}+120^{\circ}k)+i\sin(105^{\circ}+120^{\circ}k)),$$
where $k\in\{0,1,2\}$.
|
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How to prove $\int^{\pi}_0 \cos(x)\log(\tan(\frac{x}{4}))\,\mathrm dx = -2$ How to prove $\int^{\pi}_0 \cos(x)\log(\tan(\frac{x}{4}))\,\mathrm dx = -2$?
Or $\int^1_0 \log(u)\frac{(u^4-6u^2+1)}{(u^2+1)^3} \,\mathrm du = -1/2$.
(I substituted $\tan(x/4) = u$.)
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\begin{align*}
\int_0^\pi\cos x\log\left(\tan \frac x4\right)dx&=\left[\sin x\log\left(\tan\frac x4\right)\right]_0^\pi-\int_0^\pi\sin x\cdot\frac{\frac14\sec^2\frac x4}{\tan\frac x4}dx\\
&=0-\int_0^\pi\frac{\sin x}{4\sin\frac x4\cos\frac x4}dx\\
&=-\int_0^\pi\frac{\sin x}{2\sin\frac x2}dx\\
&=-\int_0^\pi\cos\frac x2dx\\
&=-\left[2\sin\frac x2\right]_0^\pi\\
&=-2
\end{align*}
Note that
\begin{align*}
\lim_{x\to0^+}\sin x\log\left(\tan\frac x4\right)&=\lim_{x\to0^+}\frac{\log\left(\tan\frac x4\right)}{\csc x}\\
&=\lim_{x\to0^+}\frac{\frac{\sec^2\frac x4}{4\tan\frac x4}}{-\cot x\csc x}\\
&=\lim_{x\to0^+}\left(-\sin x\tan x\cos\frac x2\right)\\
&=0
\end{align*}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find the polynomial equation when I know the roots A polynomial of minimum degree has rational coefficients and has the roots: $x_1=-1-\sqrt5;x_2=1+2i$ so there are $x_3=-1+\sqrt5$ and $x_4=1-2i$. I need to find the polynomial equation.
I tried to use $(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ but the calculations are too "heavy" and too long.There is an easier method to solve this?
Right answer: $x^4-3x^2+18x-20$
|
Just to give a different approach, from $x_1+x_2+x_3+x_4=0$ and $x_1x_2x_3x_4=-20$, we know the answer has the form
$$P(x)=x^4+ax^2+bx-20$$
Now
$$P(1)=(2-\sqrt5)(2+\sqrt5)(-2i)(2i)=(4-5)(4)=-4$$
and
$$P(-1)=(-\sqrt5)(\sqrt5)(-2-2i)(-2+2i)=(-5)(4+4)=-40$$
so $1+a+b-20=-4$ and $1+a-b-20=-40$, or
$$\begin{align}
a+b&=15\\
a-b&=-21
\end{align}$$
from which we find $2a=-6$ and $2b=36$, so $a=-3$ and $b=18$.
Whether this is "easier" than computing the two quadratic factors of $P(x)$ and simply multiplying them together is unclear.
|
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|
Prove $\sum _{n \geq 1} \left[\frac{1}{n}-\frac{2}{n+1}{_2F_1}(1,\frac{n+1}{2};\frac{n+3}{2};-1)\right]=\frac{\pi}{4}-\frac{\log 2}{2}$
Can we prove the following closed form by directly using hypergeometric series?
$$S=\sum _{n=1}^{\infty} \left[\frac{1}{n}-\frac{2 }{n+1}\, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-1\right)\right]=\frac{\pi}{4}-\frac{\log 2}{2}$$
I have come to this result by considering the integral:
$$I = \int_0^\infty \frac{1-\frac{1}{\cosh x}}{e^x-1} dx$$
We can expand the denominator as geometric series and then find the integrals for each term.
$$\int_0^\infty \frac{2 e^{-n x}dx}{e^x+e^{-x}}=2 \int_0^\infty \frac{t^n dt}{1+t^2}=\int_0^\infty \frac{u^{(n-1)/2} du}{1+u}= \\ = B \left(1,\frac{n+1}{2} \right) \, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-1\right)$$
As for the integral as a whole, it's easy enough to rewrite it:
$$I= \int_0^\infty \frac{e^x+e^{-x}-2}{(e^x-1)(e^x+e^{-x})}dx= \\ =\int_0^\infty \frac{dx}{e^x+e^{-x}}-\int_0^\infty \frac{dx}{e^{2x}+1}= \\ =\int_1^\infty \frac{dt}{t^2+1}-\int_1^\infty \frac{dt}{t(t^2+1)}= \frac{\pi}{4}-\frac{\log 2}{2}$$
|
It's easy enough to work with the hypergeometric function directly, since $(n+1)/2$ and $(n+3)/2$ differ by 1. This means we have
\begin{align}_2F_1\left(1,\frac{n+1}2;\frac{n+3}2;-1\right)&=(n+1)\sum_{k=0}^\infty\frac{(-1)^k}{2k+n+1}\\&=(n+1)\sum_{k=0}^\infty\int_0^1(-x^2)^kx^n~\mathrm dx\\&=(n+1)\int_0^1\sum_{k=0}^\infty(-x^2)^kx^n~\mathrm dx\\&=(n+1)\int_0^1\frac{x^n}{1+x^2}~\mathrm dx\end{align}
Shoving this into your sum gives
\begin{align}S&=\sum_{n=1}^\infty\frac1n-2\int_0^1\frac{x^n}{1+x^2}~\mathrm dx\\&=\sum_{n=1}^\infty\int_0^1x^{n-1}-\frac{2x^n}{1+x^2}~\mathrm dx\\&=\int_0^1\sum_{n=1}^\infty x^{n-1}-\frac{2x^n}{1+x^2}~\mathrm dx\\&=\int_0^1\frac1{1-x}-\frac{2x}{(1+x^2)(1-x)}~\mathrm dx\end{align}
which is easy to integrate with partial fractions and gives the result smoothly.
Hopefully this is what you wanted.
|
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If $x,y,z$ are real numbers satisfying$x/(y+z) +y/(z+x) +z/(x+y) =1$ then $x^2/(y+z) +y^2/(z+x)+z^2/(x+y)=$ I have tried it a few times but I am not making any progress. Please help.
|
$$\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}$$
$$=\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z-(x+y+z)$$
$$=(x+y+z)\left(\dfrac x{y+z}+\cdots\right)-(x+y+z)$$
$$=?$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the following inequality $a,b,c\in R_{+}^{*}$ and $2(a+b+c)=3$ Prove :
$\sum_{cyc}\frac{4a}{b^2+2(b+1)}≤3$
Where :
$a,b,c\in R_{+}^{*}$ , $2(a+b+c)$=3$
I think we use Cauchy inequality :
$\sum_{cyc}\frac{4a}{b^2+2(b+1)}≤4\sqrt{\sum_{cyc}\frac{a^2}{(b^2+2(b+1))^2}}$
Then I use :
$a^{2}+b^{2}+c^{2}≤(a+b+c)^{2}=\frac{9}{4}$
But what about the rest ?
I don't if my idea help me or no ?
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As $b^2+2b+2\geq 2$, we have $$\sum_{cyc}\frac{4a}{b^2+2b+2}\leq \sum_{cyc} \frac{4a}{2}=3$$
Equality holds at $a=b=0$ and $c=\frac32$.
|
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|
Taylor series at infinity I'm required to make a Taylor series expansion of a function $f(x) = \arctan(x)$ at $x = +\infty$. In order to do this I introduce new variable $z = \frac{1}{x}$, so that $x \to +\infty$ is the same as $z \to +0$. Thus I can expand $f(z)$ at $z = 0$: $$f(z) = z - \frac{z^3}{3}+\frac{z^5}{5}-...$$
Then I try to reverse the substitution but this is either incomplete or incorrect: $$f\bigg(\frac{1}{x}\bigg) = \frac{1}{x} - \frac{1}{3x^3} + \frac{1}{5x^5} - ...$$
|
We could use the Taylor expansion of $\arctan(x)$ when $x \rightarrow 0$ and that
$$
\forall x \in \mathbb{R}^*_+, \arctan(x) + \arctan\bigg(\frac{1}{x}\bigg) = \frac{\pi}{2}
$$
and find the correct result.
But as @janosch points out, it's faster to use
$$
\arctan'\bigg(\frac{1}{x}\bigg)= -\frac{1}{x^2 +1}
$$
and fix the integrating constant to
$\lim_{x \rightarrow 0} \arctan(1/x) = \frac{\pi}{2}$ and finally get
$$
\arctan(x) =_{x \rightarrow + \infty} \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + o\bigg(\frac{1}{x^5}\bigg).
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$ Prove the following for all real $x$
i. $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$
ii. $⌊x⌋-2⌊x/2⌋$ is equal to either $0$ or $1$
For ($ii$.) I attempted to split it into cases of whether the fraction part {$x$} is $≥.5$ or $<5$ but that ended up being too tedious and I know there must be a more elegant, simpler method.
For ($i$) I tried cases like in part ($ii$) but since there are 2 variables that would lead to 4 cases.
An elegant and easy solution will be much appreciated
|
For (i), define $\{x\} = x - \lfloor x \rfloor$. Then
\begin{align*}
\lfloor 2x \rfloor + \lfloor 2y \rfloor = 2\lfloor x\rfloor + 2\lfloor y \rfloor + \lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor
\end{align*}
and
\begin{align*}
\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x + y \rfloor = 2\lfloor x\rfloor + 2\lfloor y \rfloor + \lfloor \{x\} + \{y\} \rfloor
\end{align*}
So the inequality to be proven is equivalent to
\begin{align*}
\lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor \ge \lfloor \{x\} + \{y\} \rfloor
\end{align*}
But this is true, since
\begin{align*}
\lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor \ge \lfloor 2\max(\{x\}, \{y\})\rfloor \ge \lfloor \{x\} + \{y\} \rfloor
\end{align*}
|
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|
Solve $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$ for $x\in\mathbb{R}.$ Solve $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$ for $x\in\mathbb{R}.$ There is a brute force method that relies on breaking each exponential up into $2^x$’s and $3^x$’s and substituting, but I am looking for a more elegant one line or so proof (not brute force). Any suggestions?
|
Here's the way I did it: $$\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x)^3+(3^x)^3}{(3\cdot2^2)^x+(2\cdot3^2)^x}=\frac{(2^x)^3+(3^x)^3}{3^x\cdot(2^x)^2+2^x\cdot(3^x)^2}\tag1$$
Let $a=2^x, b=3^x$ so that $(1)$ is equivalent to $$\frac{a^3+b^3}{ba^2+ab^2}=\frac{(a+b)(a^2-ab+b^2)}{(a+b)(ab)}=\frac{7}{6}\tag2$$
Then cancelling the $a+b$ term and cross multiplying (we are allowed to do so since $a\neq -
b \text{ for all } x)$ yields $$6a^2-6ab+6b^2=7ab\iff6a^2-13ab+6b^2=0\tag3$$ Then $(3)$ yields the nice simplifications $$a=\frac{2b}{3}\text{ or } a=\frac{3b}{2}\implies 2^x=\frac{2\cdot3^x}{3}\text { or } 2^x=\frac{3\cdot 3^x}{2}\tag4$$
Thus $(4)$ gives $$2^{x-1}=3^{x-1}\text{ or }\space\space 2^{x+1}=3^{x+1}\iff \boxed{x=1\text{ or } x=-1}$$
|
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|
Finding the length of a side and comparing two areas from a given figure The following figure came in two national-exam paper (different question in each paper).
First Question:
What is the length of $AE$?
Choices:
A) $\frac{5}{4}$ cm
B) $\frac{\sqrt{41}}{5}$ cm
C) $\frac{7}{5}$ cm
D) $\frac{15}{8}$ cm
Second Question
Compare:
First value: area of $\triangle ABC$
Second value: area of $\triangle CDE$
Choices:
A) First value > Second value
B) First value < Second value
C) First value = Second value
D) Given information is not enough
My Attempt:
Finding the equation of $AC$ and $BD$ in order to find the coordinates of E:
We can assume that $C$ is the origin $(0,0)$ and $A$ is $(3,4)$, then the equation of $AC$ is $y=\frac{4}{3}x$, and the equation of $BD$ is $y=-\frac{4}{5}x+4$.
Next, solve the system of equations for $x$ and $y$, we end up with the coordinates of $E(\frac{15}{8},\frac{5}{2})$.
To solve the first question, we can use the distance between two points formula,
therefore $AE=\sqrt{(3-\frac{15}{8})^2+(4-\frac{5}{2})^2}=\frac{15}{8}$ cm.
Hence D is the correct choice.
To solve the second question, we can use the area of triangle formula,
area of $\triangle ABC=\frac{1}{2}\times 3\times 4=6$ cm$^2$.
area of $\triangle CDE=\frac{1}{2}\times 5\times \frac{5}{2}=6.25$ cm$^2$.
Hence B is the correct choice.
In the exam, calculators are not allowed. It is simple to determine to coordinates of $E$ without a calculator, but it will take time!
For the first question, I think there is a good way to solve, maybe $BD$ divides $AC$ into two parts ($AE$ and $CE$) in a ratio, this ratio can be calculated somehow.
For the second question, I think also a good way to think, like moving $D$ will make $\triangle CDE$ bigger or smaller, while $\triangle ABC$ will be unchanged. But this has a limitation since the areas $6$ and $6.25$ cm$^2$ are close, it will not be obvious to us.
If we have $75$ seconds (on average) to solve each of these questions, then how can we solve them? [remember that in some questions in the national exam, we do not need to be accurate $100$%].
Any help will be appreciated. Thanks.
|
Triangles $ABE$ and $CDE$ are similar, with ratio $3:5$.
Hence
$$AE=\frac{3}{8}AC=\frac{3}{8}\cdot 5=\frac{15}{8}$$
By the same similarity, if $h$ is the height of triangle $CDE$ from $E$ to $CD$, then
$$h=\frac{5}{8}BC=\frac{5}{8}\cdot 4=\frac{5}{2}$$
hence the area of triangle $CDE$ is
$$\frac{1}{2}\cdot CD\cdot h=\frac{1}{2}\cdot 5\cdot \frac{5}{2}=\frac{25}{4}=6.25$$
whereas the area of triangle $ABC$ is $6$.
|
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|
Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ A = \begin{pmatrix}
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t X^t
\end{matrix}$$
Let $B$ be the standard basis for $\mathbb{R}^{2 \times 2}$ :
$$B =\left\{ \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
1 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
0 & 1\\
\end{pmatrix} \right\}$$
Calculate $\textsf{M}_B(\varphi)$ we come to $$\textsf{M}_B(\varphi) = \begin{pmatrix}
0 & 0 & 0 & 0\\
-1 & 1 & -1 & 1 \\
1 & -1 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$$
We calculate a basis for the kernel like this:
If
$$X:= \begin{pmatrix}
a & b\\
c & d\\
\end{pmatrix}$$
then $$\varphi(X) = \begin{pmatrix}
a-b & -a+b\\
c-d & -c+d\\
\end{pmatrix}+\begin{pmatrix}
a-b & c-d\\
-a+b & -c+d\\
\end{pmatrix} = \begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
Now we have to look, for what values $$\begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
by definition, the kernel of a linear transformation is $\varphi(X) = 0$, therefore our basis for $\ker(\varphi)$ should be
$$\left\{ \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \right\}$$
Now here comes the part where I'm confused. How do I calculate a basis for $\operatorname{im}(\varphi)$ ?
$\textsf{M}_B(\varphi)$ is the transformation matrix. I've read that you'd just transpose the matrix $\textsf{M}_B(\varphi)$ and row reduce to calculate a basis. I just don't get it.
The solution according to the solution it should be the basis $$ \left\{\begin{pmatrix}
0 & 1 \\
1 & -2 \end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & -1 \end{pmatrix} \right\}$$
Also little side questions. I do know about
$$\dim(A) = \dim(\operatorname{im} A) + \dim(\ker A)$$
but how exactly do you know the dimension of the Kernel/Image?
|
We can represent a $2\times 2$ matrix as a $1\times 4$ vector.
i.e. $\pmatrix{c_1\\c_2\\c_3\\c_4} = c_1\pmatrix{1&0\\0&0}+ c_2\pmatrix{0&1\\0&0}+c_3\pmatrix{0&0\\1&0}+c_4\pmatrix{0&0\\0&1}$
$\phi(X) = \pmatrix{2&-2&0&0\\-1&1&1&-1\\-1&1&1&-1\\0&0&2&-2}$
The linearly independent columns of the matrix above will give the image of the transformation.
|
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|
$n$-th term of a sequence from generating function $\frac{1}{(1-x^2)(1-x^3)(1-x^7)}$ I have a generating function of a sequence such that
$$\frac{1}{(1-x^2)(1-x^3)(1-x^7)}.$$
They are the product of three geometric series with coefficient 1.
Now I want to transform this into infinite sum form so that I can get nth term explicitly. However, I can not refactor denominator to get power of $(1-x)$ or partial fractions. Is there any identity I must know or is this completely wrong way of solving?
|
Use partial fractions, and pry out the coefficients of interest:
$\begin{align*}
\frac{1}{(1 - x^2) (1 - x^3) (1 - x^7)}
&= \frac{x^4 + 2 x^3 + 2 x^2 + 2}
{7 (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)}
+ \frac{x + 2}{9 (x^2 + x + 1)}
+ \frac{1}{8 (x + 1)}
- \frac{17}{72 (x - 1)}
+ \frac{3}{28 (x - 1)^2}
- \frac{1}{42 (x - 1)^3} \\
&= \frac{(x^4 + 2 x^3 + 2 x^2 + 2)(1 - x)}
{7 (1 - x^7)}
+ \frac{(x + 2) (1 - x)}{9 (1 - x^3)}
+ \frac{1}{8 (x + 1)}
- \frac{17}{72 (x - 1)}
+ \frac{3}{28 (x - 1)^2}
- \frac{1}{42 (x - 1)^3}
\end{align*}$
Multiply out the numerators, and you are left with picking off the coefficients of powers of $x$ in geometric series and series of the form:
$\begin{align*}
(1 - x)^m
&= \sum_{k \ge 0} (-1)^k \binom{-m}{k} x^k \\
&= \sum_{k \ge 0} \binom{k + m - 1}{m - 1} x^k
\end{align*}$
|
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|
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?
|
Starting with what you have written we quickly see that $$0=x^3-18x^2+81x-8=(x-8)(x^2-10x+1)$$
It is easy to see that $x\neq 8$ so we must have $\boxed {x^2-10x+1=0}$
Now, go back to the original equation. Multiply by $x$ to get $$x^2-\sqrt {8x}=9x\implies x-\sqrt {8x}=10x-x^2$$
But the boxed equation tells us that $10x-x^2=1$ so $$\boxed {x-\sqrt {8x}=1}$$
|
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|
Finding values of $\tan^{-1} (2i)$. I am trying to find all solutions of $\tan^{-1} (2i)$. I don't see anything that I have done wrong, my answer doesn't match the one in the textbook. Here is what I have. (The convention in Brown and Churchill is to use $\log$ for a complex number and $\ln$ for a real number.)
\begin{align*}
\tan^{-1} (2i) & = \frac{i}{2} \log \frac{i + 2i}{i - 2i} \\
& = \frac{i}{2} \log \frac{3i}{-i} \\
& = \frac{i}{2} \log (-3) \\
& = \frac{i}{2} \left(\ln 3 + (2n + 1) \pi i \right) \\
& = \frac{i}{2} \ln 3 - \frac{2n + 1}{2} \pi \\
& = \frac{i}{2} \ln 3 - \left(n + \frac{1}{2}\right) \pi.
\end{align*}
The answer in the textbook, however, is:
\begin{align*}
\frac{i}{2} \ln 3 + \left(n + \frac{1}{2}\right)\pi.
\end{align*}
This leads me to believe that I have misplaced a sign somewhere, but I cannot see where. Might there just be a typo in the book?
Thanks in advance.
|
Since $\tan(x)=2i$
Using Euler's formula
$\tan(x)={\frac {e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}}$ witch must equal $2i$
Manipulating a bit we get $3e^{ix}+e^{-ix}=0$ and if $e^{ix}=y$, $3y+\frac{1}{y}=0$ thus $e^{ix}=y=±i \frac{\sqrt 3}{3}$
This, of course, means that $\ln\left(±i \frac{\sqrt 3}{3}\right)=ix$
|
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|
Simplify a rational identity
Simplify:
$$\frac{\dfrac{a}{b}-\dfrac{b}{a}}{1+\dfrac{b}{a}}$$
I have a feeling the solution has to do with factoring, but I'm really not sure, and would appreciate any help.
|
Well, in view of addition the rule is
$$\frac{a}{b}+\frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad+bc}{bd}$$
i.e., the fractions need to be extended first to obtain the same denominator and then the addition can be performed (same for subtraction).
In view of multiplication the rule is
$$\frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}$$
i.e., numerator and denominator are multiplied separately.
In view of division the rule is
$$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b}:\frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}$$
i.e., in the fraction (denominator) $c/d$ the numerator and denominator are flipped $d/c$ and then multiplied with $a/b$.
You can easily use these rules to simplify your fraction:
$$\frac{\frac{a}{b}-\frac{b}{a}}{1+\frac{b}{a}}
= \frac{\frac{a^2}{ab}-\frac{b^2}{ab}}{\frac{a}{a}+\frac{b}{a}}
= \frac{\frac{a^2-b^2}{ab}}{\frac{a+b}{a}}
= \frac{a^2-b^2}{ab}\cdot \frac{a}{a+b}
= \frac{(a^2-b^2)a}{ab(a+b)}
= \frac{a^2-b^2}{b(a+b)}
=\frac{(a-b)(a+b)}{b(a+b)}
=\frac{a-b}{b}.$$
|
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|
Find a condition on real numbers $a$ and $b$ such that $\left(\frac{1+iz}{1-iz}\right)^n = a+ib$ has only real solutions I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that
$$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$
has only real solutions
This is what I got till now.
$$\left(\frac{1+i(a+ib)}{1-i(a+ib)}\right)^n = a+ib$$
$$\left(\frac{(1-b)+ia}{(1+b)-ia}\right)^n = a+ib$$
$$\frac{(1-b)+ia}{(1+b)-ia}.\frac{(1+b)+ia}{(1+b)+ia} = \frac{1-b^2+2ia-a^2}{1+2b+b^2+a^2}$$
then
$$\left(\frac{1-b^2-a^2}{1+2b+b^2+a^2}+\frac{2ia}{1+2b+b^2+a^2}\right)^n = a+ib$$
where
$$a=0 \text{; & } 1+2b+b^2\neq0$$
|
that is an awesome answer given by Martin R.
alternatively, let
$$
a+ib = \sqrt{a^2+b^2} e^{i \theta}
$$
$c$ be the real solution for satisfied $a,b$. One would obtain
$$
\label{eq}
\sqrt[n]{a^2+b^2} e^{ \frac{i (\theta + 2k\pi )}{n} } = \frac{1-c^2 +i 2c}{1+c^2} \tag1
$$
note that
$$
( \frac{1-c^2 }{1+c^2} ) ^2 + ( \frac{2c}{1+c^2} ) ^2 = 1
$$
which means $\sqrt[n]{a^2+b^2} =1$ equivalently
$$a^2+b^2 =1 $$
And for each $e^{ \frac{i (\theta + 2k\pi )}{n} } $ there exists only one $c \in \mathbb{R}$ satisfy $\eqref{eq}$
|
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|
Exercises to see if certain series converge.
*
*Does the series $\sum \frac{2^k+1}{3^k}$ converge?
*Does the series $\sum (-1)^k\frac{1}{k^2}$ converge?
*Does the series $\sum \frac{2^k+2^{2k}}{4^k}$ converge?
*Does the series $\sum \frac{k!}{(2k)!}$ converge?
For the first question, do we use the comparison test comparing it to $\left(\frac{2}{3}\right)^k$???
For the second question, this series does converge. To prove this observe that this converges absolutely because
\begin{equation*}
\sum_{k=1}^{\infty} \left|(-1)^k\frac{1}{k^2}\right| = \sum_{k=1}^{\infty} \frac{1}{k^2}
\end{equation*}
converges. To see that this does converge, we can use the integral test.
The function $f:[1,\infty) \mapsto \mathbb{R}$ defined by $f(x) := \frac{1}{x^2}$ is continuous, non-negative and decreasing by examining the derivative $f'(x) = -\frac{2}{x^3} < 0$ for all $x\geq 1$. So we have
\begin{equation*}
\begin{split}
\int_{1}^{\infty} \frac{1}{x^2} \; dx &= \left[-\frac{1}{x}\right]_{1}^{\infty} \\
&= \lim_{a\to\infty} \left(-\frac{1}{a}+1\right) \\
&= 1
\end{split}
\end{equation*}
This integral converges so by the integral test $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges as well and hence, $\sum_{k=1}^{\infty} (-1)^k\frac{1}{k^2}$ converges.
For the third question, this series does NOT converge. To show that this series diverges, we just need to show that
\begin{equation*}
\lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} \neq 0
\end{equation*}
or doesn't exist. Well, for $k\geq 0$,
\begin{equation*}
\frac{2^k+2^{2k}}{4^k} = \frac{2^k+4^k}{4^k} = \left(\frac{1}{2}\right)^k+1.
\end{equation*}
Thus,
\begin{equation*}
\begin{split}
\lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} &= \lim_{k\to\infty} \left(\frac{1}{2}\right)^k+1 \\
&= 0+1 \\
&= 1 \neq 0.
\end{split}
\end{equation*}
For the fourth question, if we let $a_k = \frac{k!}{(2k)!}$ then using the ratio test we get
\begin{equation*}
\begin{split}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty} \left|\frac{\frac{(k+1)!}{(2k+1)!}}{\frac{k!}{(2k)!}}\right| \\
&= \lim_{k\to\infty} \left|\frac{(k+1)!}{(2k+1)!}\times \frac{(2k)!}{k!}\right| \\
&= \lim_{k\to\infty} \left|\frac{(k+1)\times k\times (k-1)\times\ldots\times (2k)\times (2k-1)\times (2k-2)\times\ldots}{(2k+1)\times 2k\times (2k-1)\times\ldots\times k\times (k-1)\times (k-2)\times\ldots}\right| \\
&= \lim_{k\to\infty} \frac{k+1}{2k+1} \\
&= \lim_{k\to\infty} \frac{1+\frac{1}{k}}{2+\frac{1}{k}} \\
&= \frac{1}{2} < 1.
\end{split}
\end{equation*}
Hence, $\sum_{k=0}^{\infty} \frac{k!}{(2k)!}$ converges.
|
As regards the last one, this is an alternative way: for $k\geq 1$,
$$\frac{k!}{(2k)!}=\frac{1}{(2k)\cdot(2k-1)\cdots (k+1)}\leq \frac{1}{(2)\cdot(2)\cdots (2)}=\frac{1}{2^k}$$
and therefore the non-negative series $\sum \frac{k!}{(2k)!}$ converges by the comparison test.
Your answers for 2) and 3) are correct. For the first one,
$$\frac{2^k+1}{3^k}\leq \frac{2^k+2^k}{3^k}=2\left(\frac{2}{3}\right)^k$$
and use the comparison test, or you may note that
$$\sum \frac{2^k+1}{3^k}=\sum \left(\frac{2}{3}\right)^k+\sum \frac{1}{3^k}$$
which are both convergent.
|
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|
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$
Firstly is this right? Secondly how to complete?
|
$$(9)^{ \displaystyle\frac{a}{b} } + (16)^{ \displaystyle\frac{b}{a} } = \left(3^{\displaystyle \color{magenta}{2}}\right)^{ \displaystyle\frac{\color{red}{a}}{b} } + \left(4^{\displaystyle \color{brown}{2}}\right)^{ \displaystyle\frac{\color{blue}{b}}{a} } =\left(3^{\displaystyle \color{red}a}\right)^{ \displaystyle\frac{\color{magenta}2}{b} } + \left(4^{\displaystyle \color{blue}b}\right)^{ \displaystyle\frac{\color{brown}2}{a} } $$
Now if $3^{\displaystyle a} = 4^{\displaystyle b}$, then
$$\left(3^{\displaystyle a}\right)^{ \displaystyle \frac{2}{b} } + \left(4^{\displaystyle b}\right)^{ \displaystyle \frac{2}{a} } = \left(4^{\displaystyle b}\right)^{ \displaystyle \frac{2}{b} } + \left(3^{\displaystyle a}\right)^{ \displaystyle \frac{2}{a} } = 25 $$
The rules are:
*
*If you have three numbers $x,y$ and $z$, then $(x^y)^z = x^{y~\cdot ~z}= x^{z~\cdot~ y}=(x^z)^y$.
*If you have three numbers $x,y$ and $z\neq 0$, then $\frac{x~\cdot ~y}{z} = x~\cdot~ \frac{y}{z}= y~\cdot~ \frac{x}{z} =\frac{y~\cdot~ x}{z}$.
|
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"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding lower bounds for a function Given a function $w\left(x\right)=\sum_{n=1}^x\lfloor\frac{x}{n}\rfloor$.
And $f(x) = w(6x+1)-w(6x-2)$, prove that for $x>3; f(x) \geq 12$
EDIT: This doesn't really come from anywhere, I was just messing around with discrete functions and wondered if I could show that there are an infinite amount of instances where f(x) is equal to 12. What I tried was manually counting the trivial values. For example for n = 1 we would already have 3 and by nature $w(6x+1)$ will be 3 more than $w(6x-2)$.
|
You have the $2$ functions
$$w(x) = \sum_{n=1}^x \left\lfloor \frac{x}{n} \right\rfloor \tag{1}\label{eq1}$$
\begin{align}
f(x) & = w(6x + 1) - w(6x - 2) \\
& = \sum_{n=1}^{6x+1} \left\lfloor \frac{6x+1}{n} \right\rfloor - \sum_{n=1}^{6x-2} \left\lfloor \frac{6x-2}{n} \right\rfloor \\
& = \left\lfloor \frac{6x+1}{6x+1} \right\rfloor + \left\lfloor \frac{6x+1}{6x} \right\rfloor + \left\lfloor \frac{6x+1}{6x-1} \right\rfloor + \sum_{n=1}^{6x-2} \left(\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor\right) \\
& = 3 + \sum_{n=1}^{6x-2} \left(\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor\right) \tag{2}\label{eq2}
\end{align}
Consider any integer $x \gt 3$. In the summation part of \eqref{eq2}, each term $\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor \ge 0$. Also, among these terms, for $n = 1$, you get $(6x+2) - (6x-1) = 3$; for $n = 2$, you get $(3x) - (3x-1) = 1$; for $n = 3$, you get $(2x) - (2x-1) = 1$; for $n = 6$, you get $x - (x-1) = 1$; for $n = x$, you get $6 - 5 = 1$; for $n = 2x$, you get $3 - 2 = 1$; and for $n = 3x$, you get $2 - 1 = 1$. Note since $x \gt 3$, each of these values of $n$ being used (i.e., $1,2,3,6,x,2x,3x$) are distinct, except for where $x = 6$. In this case, you can replace the $n$ value of $x$ with $n = 4$ instead since it has $9 - 8 = 1$.
Adding these parts together gives $3 + 1 + 1 + 1 + 1 + 1 + 1 = 9$. Thus, along with the initial $3$ in \eqref{eq2} and that each term in the sum is non-negative, this shows the overall value must be at least $12$, i.e., $f(x) \ge 12$, for all $x \gt 3$.
|
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"url": "https://math.stackexchange.com/questions/3309911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Matrix rearrangement I have a matrix A in this form:
$$
A=
\left[
\begin{array}{cccc}
x_1 & x_2 & 0 & 0 \\
0 & 0& x_1 & x_2
\end{array} \right]
$$
Here, $x_1$ and $x_2$ are variables, and A is a $2 \times 4$ matrix. I
would like to rearrange the matrix product so that
$$
A^TA=BC
$$
$A^T$ is the transpose of A. $B$ and $C$ are matrices to be found. $B$ is a $4 \times 2$ matrix that only contains constants, and C is a $2 \times 4$ matrix that contains constants and variables ($x_1,x_2$ etc). Is this possible to have such $B$ and $C$?
|
We have $$A^T A = M =
\left[
\begin{array}{cccc}
x_1^2 & x_1 x_2 & 0 & 0 \\
x_1 x_2 & x_2^2 & 0 & 0 \\
0 & 0 &x_1^2 & x_1 x_2 \\
0 & 0 &x_1 x_2 & x_2^2 \\
\end{array}
\right]$$
We will try to write this as $M = CB$, where $B$ is a $2*4$ matrix of constants, and $C$ is a $4*2$ matrix of constants and variables. I applied a transpose to the original problem in order to switch $B$ and $C$, because it's easier to have the right matrix be the matrix of constants. (ie. you have $ M = M^{t} = B^{t}C^{t}$, where $B^{t}$ would be your original $B$ . However, I work with its transpose).
Now, because $B$ is a $2*4$ matrix of constants, it has a constant kernel of dimension at least $2$. There exists at least $2$ constant linearly independent vectors $v_{1}$, $v_{2}$, such that $B{v_1} = 0$ and $B{v_2} = 0$ . Eg. if $B=
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1& 0 & 0
\end{array} \right]$, an example of such vectors would be $v_1 =
\left[
\begin{array}{c}
0 \\
0 \\
1 \\
0 \\
\end{array} \right]$, $v_2 =
\left[
\begin{array}{c}
0 \\
0 \\
0 \\
1 \\
\end{array} \right]$. Regardless of $B$, we can always find such vectors. Now, if $M= CB$, that means $M$ "inherits" the kernel-spanning constant vectors from $B$: there must exist constant linear independent vectors $v_1,v_2$ such that $Mv_1 = 0$ and $Mv_2 = 0$. The kernel of $M$ must always contain this constant subspace of dimension $2$ spanned by $v_1, v_2$. For $x_1 = 1 $, $x_2 = 0$, then $M$ is
$$ M =
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 &1 & 0 \\
0 & 0 &0 & 0 \\
\end{array}
\right]$$
and its kernel is spanned by $\left[
\begin{array}{c}
0 \\
0 \\
0 \\
1 \\
\end{array} \right],
\left[
\begin{array}{c}
0 \\
1 \\
0 \\
0 \\
\end{array} \right]$ . On the other hand, for $x_1 = 0$, $x_2 = 1$ gives
$$ M =
\left[
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 &0 & 0 \\
0 & 0 &0 & 1 \\
\end{array}
\right]$$ and its kernel is spanned by $\left[ \begin{array}{c}
0 \\
0 \\
1 \\
0 \\
\end{array} \right],
\left[
\begin{array}{c}
1 \\
0 \\
0 \\
0 \\
\end{array} \right]$. However, these two subspaces are disjoint (except the null space). They cannot both contain linearly independent constant vectors $v_1,v_2$ that span the kernel of $B$ in $M = CB$. Therefore, no decomposition of the form $M=CB$ with $B$ constant $2*4$ matrix exists. The kernel of $M$ "moves too much" for that to happen.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the maximum and minimum values of $a^2\sin^2\theta + b^2\csc^2\theta$
How do I find the maximum and minimum values of the following?
$$a^2\sin^2\theta + b^2\csc^2\theta$$
Is the max value $\infty$?
I tried to find the minimum value by using A.M$\geq$G.M. inequality (is there any other way?) and got it as 2|a||b| which seem to work for all the positive values when a>b (I checked it on Desmos by putting different values of a and b) but when a$\le$b it just don't match in the graph. So, what am I doing wrong here? And what should be the correct answer?
|
The function is periodic with period $\pi$ so let us consider $ 0\le x \le \pi.$
Case-1: When $a>b$ in $$f(x)=a^2\sin^2 x+ b^2 \csc^2 x~~~(1)$$ then
$$f'(x)=2a^2 \sin x \cos x-2b^2 \cot x \csc^2 x.= 2 \cos x(a^2 \sin^4 -b^2)/\sin^2 x ~~~(2)$$
$f'(x)=0,$ gives $\cos x=0$ or $\sin^4 x = b^2/a^2 \Rightarrow \sin^2 x= b/a \Rightarrow \sin x = \pm \sqrt{b/a} \Rightarrow x_1 = \pi/2, ~ \mbox{or} ~ x_{2,3}=\pm \sin^{-1} \sqrt{b/a}$.
Next $$f''(x)= b^2(4 \cot^2x \csc^2 x + 2 \csc^2 x)+2a^2(\cos^2 x- \sin^2 x)~~~~~~(3).$$ Get $f''(x_1)=-2(a^2-b^2)< 0,$ so local max. at $x=\pi/2$. Use $\sin^2 x=b/a, \cos^2 x=(a-b)/a$ etc. in (3) to get $f''(x_{2,3})=8a(a-b)>0,$ so two local minima at $x=x_{2,3}.$
Hence when $a>b$, the local max. and min. are given as
$$f_{max}=f(\pi/2)=a^2+b^2, f_{min}f(x_{2,3})=2ab.$$
Case-2: When $a<b$, then $x_{2,3}$ do not exist and $f''(\pi/2)=-2(a^2-b^2)>0,$ so there exists only one local min. and $f_{min}=f(\pi/2)=a^2+b^2$
In both the cases the function $f(x)$ is positive and unbounded which can becomes $\infty$ at $x=0, \pi.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $ \sin a +\sin c =2 \sin b $, show that $ \tan\frac{a+b}{2}+\tan\frac{b+c}{2} = 2 \tan\frac{c+a}{2}$ Found the question in the textbook.
I tried many methods of manipulating the identity to be proved but I did not even see a clue of using the given condition(of different forms like
$$
\sin\frac{a+c}{2} \cos\frac{a-c}{2} = \sin \frac{b}{2} \cos \frac{b}{2}
$$
and
$$
\cos \frac{a+b}{2} \sin \frac{a-b}{2} = \cos \frac{b+c}{2} \sin \frac{b-c}{2}
$$
|
We need to prove that
$$\frac{\sin\left(\frac{a+b}{2}-\frac{c+a}{2}\right)}{\cos\frac{a+b}{2}\cos\frac{a+c}{2}}=\frac{\sin\left(\frac{c+a}{2}-\frac{b+c}{2}\right)}{\cos\frac{c+a}{2}\cos\frac{b+c}{2}}$$ or
$$\frac{\sin\frac{b-c}{2}}{\cos\frac{a+b}{2}}=\frac{\sin\frac{a-b}{2}}{\cos\frac{b+c}{2}}$$ or
$$\sin{b}-\sin{c}=\sin{a}-\sin{b},$$ which is given.
|
{
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"url": "https://math.stackexchange.com/questions/3311279",
"timestamp": "2023-03-29T00:00:00",
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|
Find function $f(x)$ knowing that $[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$ is satisfied for $\forall x \in \mathbb R$.
Find function $f(x)$ knowing that $$\large [f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$$ is satisfied for $\forall x \in \mathbb R$ $(f(x) \ne c$ with $c$ being a constant$)$.
Okay, where to start. $f(x) = x$ doesn't work as $$(x^2 - x + 1)(x^3 + 1) - (x^4 - 2x^3 + x + 1)(x + 1) = 3(x - 1)x(x + 1)$$. Furthermore, if $f(x) = x^2 - x$ then $f(x^2 - x) = x^4 - 2x^3 + x$. However, at this case,
$$[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1) = (x - 1)x(x + 1)(x^6 - 3x^4 + 2x^3 + 1)$$
|
The function
$$f(x) = -\frac 1 2 x$$
satisfies the above property.
This is how I found it. First, notice that the equation $x^2 - x = x^4 - 2x^3 + x$ has solutions $x = -1, 0, 1, 2$. If you let $x = -1, 0, 1$ in the equality you get an identity, because $-1, 0, 1$ are also solutions of the equation $x^3 + 1 = x + 1$. If you let $x = 2$, though, you obtain
$$[f(2) + 1]f(9) = [f(2) + 1]f(3)$$
which means that either $f(2) = -1$ or $f(9) = f(3)$.
The simplest non-constant function such that $f(2) = -1$ is $f(x) = -\frac 1 2 x$. I checked if such function satisfied the property for all $x \in \mathbb R$, and indeed it does.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Here is my attempt:
$x = 0$ and $y = 0$, these both line go through the $x$ and $y$ axes. And also the circle touches those two lines. So the center will be $C(p,p)$. That means $r = k = h = p$. The circle also touches the $x = a$ line. That means $r = a/2$. From that, I determined the center as $C(a/2, a/2)$.
The equation:
$$\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2$$
$$x^2 + y^2 - 2ax - 2ay + \frac{a^2}{4} = 0$$
which is not the correct answer.
Now can anyone tell me what's wrong with this attempt?
|
Your error is very simple: you have wronged expanding the square, in fact $\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2$ is not $x^2 + y^2 - 2ax - 2ay + \frac{a^2}{4} = 0$, but: $$x^2-ax+\frac{a^2}{4}+y^2+ay=0$$ In other words: $$4x^2+4y^2-4ax-4ay+a^2=0$$
Also, by simmetry, I get $\left(x - \frac{a}{2}\right)^2 + \left(y + \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2$: $$4x^2+4y^2-4ax+4ay+a^2=0$$
To clarify, see this graph where $a=6$:
|
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|
Can the integral $\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$ be solved analytically? I'm trying to solve the following integrals:
$$I_1(a)=\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$$
and
$$I_2(a)=\int_0^\infty\ \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx.$$
They seem similar to the two integrals solved in this post, so I wonder if there is something in the Bessel function family that will do the trick. But so far, substitutions like $x=a \sinh(t)$ and $u=\sqrt{x^2+a^2}$ are not giving me integrals that can be manipulated into one of the Bessel function forms on Wikipedia or Wolfram.
|
Assume $a\neq0$ for the key case.
For $\int_0^\infty\dfrac{1}{\sqrt{x^2+a^2}\left(1+e^\sqrt{x^2+a^2}\right)}~dx$ ,
$\int_0^\infty\dfrac{1}{\sqrt{x^2+a^2}\left(1+e^\sqrt{x^2+a^2}\right)}~dx$
$=\int_0^\infty\dfrac{1}{\sqrt{a^2\sinh^2t+a^2}\left(1+e^\sqrt{a^2\sinh^2t+a^2}\right)}~d(|a|\sinh t)$
$=\int_0^\infty\dfrac{1}{1+e^{|a|\cosh t}}~dt$
$=\int_0^\infty\dfrac{e^{-|a|\cosh t}}{1+e^{-|a|\cosh t}}~dt$
$=\int_0^\infty\sum\limits_{n=0}^\infty(-1)^ne^{-|a|(n+1)\cosh t}~dt$
$=\sum\limits_{n=0}^\infty(-1)^nK_0(|a|(n+1))$
For $\int_0^\infty\dfrac{1}{1+e^\sqrt{x^2+a^2}}~dx$ ,
$\int_0^\infty\dfrac{1}{1+e^\sqrt{x^2+a^2}}~dx$
$=\int_0^\infty\dfrac{1}{1+e^\sqrt{a^2\sinh^2t+a^2}}~d(|a|\sinh t)$
$=\int_0^\infty\dfrac{|a|\cosh t}{1+e^{|a|\cosh t}}~dt$
$=\int_0^\infty\dfrac{|a|e^{-|a|\cosh t}\cosh t}{1+e^{-|a|\cosh t}}~dt$
$=\int_0^\infty\sum\limits_{n=0}^\infty|a|(-1)^ne^{-|a|(n+1)\cosh t}\cosh t~dt$
$=\sum\limits_{n=0}^\infty|a|(-1)^nK_1(|a|(n+1))$
|
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|
A disc is cut into 12 sectors with areas in arithmetic progression. The largest angle is twice the smallest. Find the smallest angle. I was given a question which states -
A circular disc is cut into twelve sectors whose areas are in an
arithmetic sequence. The angle of the largest sector is twice the
angle of the smallest sector.
Find the size of the angle of the smallest sector.
Here's what I have done so far,
$\Rightarrow U_1 = w$
$\Rightarrow U_n = 2w$
$\Rightarrow w + (n - 1)d = 2w$
$\Rightarrow n = 12$
$\therefore\ d = \frac{w}{11}$
And now I am clueless about where to go from here...
|
You got $d = a$1$/11$.
($a$1 is angle of the smallest sector)
We can write: $11d = a$1 ........(1)
Now, we know:
Area of sector = ${\frac{1}2}.{r^2}.\theta$
And,
Sum of area of all sectors = Area of circle
(${\frac{1}2}.{r^2}.a$1) + (${\frac{1}2}.{r^2}.a$2) + ..... + (${\frac{1}2}.{r^2}.a$12) = $\pi.r^2$
${\frac{1}2}.r^2$.{ $a$1 + $a$2 + ..... + $a$12 } = $\pi.r^2$
$a$1 + $a$2 + ..... + $a$12 = 2$\pi$
Applying the formula for the sum of $n$ terms of an AP:
${\frac{12}2}.${ $2a$1 + $(12-1).d$ } = 2$\pi$
$6$.{ $2a$1 + $11d$ } = 2$\pi$
Putting value of $11d$ from (1):
$6$.{ $2a$1 + $a$1 } = 2$\pi$
$a$1 = $\frac{\pi}9$
|
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|
Show that $7a^{2} = 13b^{2}$ Here is another question which states that -
The first and the third terms of an arithmetic sequence are
a and b respectively. The sum of the first n terms is denoted by $S_n$.
(a) Find $S_4$ in terms of a and b.
(b) Given that $S_4$, $S_5$,
$S_7$ are consecutive terms of a geometric sequence, show that
$$7a^{2} = 13b^{2}$$
My approach to part (a) -
$\Rightarrow\ U_1 = a$
$\Rightarrow\ U_3 = b$
$\Rightarrow\ a + 2d = b$
$\therefore\ d = \frac{b-a}{2}$
$\Rightarrow\ U_4 = b + \frac{b - a}{2}$
$\therefore\ U_4 = \frac{3b - a}{2}$
$\Rightarrow\ S_n = \frac{n}{2}(U_1 + U_4)$
$\Rightarrow\ S_4 = \frac{4}{2}(a + \frac{3b - a}{2})$
$\therefore\ S_4 = a+3b$ (Answer)
I do not know how to approach part (b) without the need for tedious calculations.
|
I know that:
$$\left\{\begin{matrix}
c_1=a+3b
\\c_1\cdot q=5b
\\ c_1\cdot q^2=\frac{7}{2}(3b-a)
\end{matrix}\right.$$
For simplicity, I put the first term $c_1$ equal to $S_4$.
From this I have $q=\frac{5b}{c_1}$. Substituting in the third equation: $$(a+3b)\cdot \frac{25b^2}{(a+3b)^2}=\frac{7}{2}(3b-a)$$
In other words: $50b^2=7(3b-a)(3b+a)$; so $-13b^2=-7a^2$ and $13b^2=7a^2$
|
{
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|
How to maximize area of a square inscribed in a equilateral triangle? We have an equilateral triangle and want to inscribe a square, in such way that maximizes the area of the square.
I sketched two possible ways, not to scale and not perfect.
Note I am not sure if the second way will really have all square corners touching the triangle sides.
The second case appears to have bigger side-lengths of the square, so bigger area. But I do not know how to determine the angles involved. How to solve this?
|
Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below.
With the angles in the diagram,
$$d_1=\frac{1}{2\tan 30}a_1=\frac{\sqrt{3}}{2}a_1$$
$$d_2=\frac{\sin 15}{\sin 30}a_2=\frac{1}{2\cos 15}a_2$$
Assume both equilateral triangles have unit height.
$$1=a_1+d_1=\left(1+\frac{\sqrt{3}}{2}\right)a_1=\frac{1}{2}(2+\sqrt{3})a_1$$
$$1=\sqrt{2}a_2+d_2=\left(\sqrt{2}+\frac{1}{2\cos 15}\right)a_2=\frac{1}{2}(\sqrt{2}+\sqrt{6})a_2$$
So, their ratio is
$$\frac{a_1}{a_2}= \frac{\sqrt{2}+\sqrt{6}}{2+\sqrt{3}} =\left(\frac{8+4\sqrt{3}}{7+4\sqrt{3}}\right)^{\frac{1}{2}} > 1$$
|
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|
Prove inequality for a sum to $2^{n+1}$ Problem:I realize I made a mistake on one of the last questions I posted.Here is the problem:
Prove for all $n \in \mathbb{N}$
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k} > \frac{n}{2}$
Attempt:
I'm having trouble recognizing the pattern for the general term of the inequality, can someone help me?
trying cases
for $n=1$ then $\sum\limits_{k=1}^{2^{1+1}} \frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}$
for $n=2$ then $\sum\limits_{k=1}^{2^{2+1}}\frac
{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+ \frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}$
Can someone tell me if this is the correct pattern to use in the proof?Would this be a proof by induction or a direct proof?
Direct proof attempt
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k}>\frac{n2^{n}}{2^{n+1}}=\frac{n}{2}$
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Note that\begin{align}\left(\sum_{k=1}^{2^{n+2}}\frac1k\right)-\left(\sum_{k=1}^{2^{n+1}}\frac1k\right)&=\sum_{k=2^{n+1}+1}^{2^{n+2}}\frac1k\\&>\overbrace{\frac1{2^{n+2}}+\frac1{2^{n+2}}+\cdots+\frac1{2^{n+2}}}^{2^{n+1}\text{ times}}\\&=\frac{2^{n+1}}{2^{n+2}}\\&=\frac12.\end{align}Can you take it from here?
|
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|
Least value of $x+y+z$ where $ax=by=cz$ The following question is a generalization of the case $a=3$, $b=4$, $c=5$ from a MindYourDecisions YouTube video (which I am not going to actually link here).
Given positive integers $a$, $b$, and $c$, what is the smallest possible value of $x+y+z$ where $x$, $y$, and $z$ are positive integers with $ax=by=cz$?
If $a$, $b$, and $c$ are pairwise coprime, then the answer is just $bc+ac+ab$, because $lcm(a,b,c)=abc$.
The case $a=3$, $b=4$, $c=7$ was asked 6 years ago in Least Value Of $x+y+z$.
|
Let $N = ax = by = cz$. Since $N$ is a multiple of $a, b, c$ it must be a multiple of $\mathrm{lcm}(a, b, c)$. Thus, we should take
$$
\begin{align*}
x &= N/a \\
y &= N/b \\
z &= N/c
\end{align*}
$$
The sum will be minimal when $N = \mathrm{lcm}(a, b, c)$.
|
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|
Show that the general value of $\theta$ satisfying $\sin\theta=\sin\alpha$ and $\cos\theta = \cos\alpha$ is given by $\theta = 2n\pi + \alpha$
The general value of $\theta$ simultaneously satisfying equations, $$\sin\theta = \sin\alpha \quad\text{and}\quad \cos\theta = \cos\alpha$$
is given by
$$\theta = 2n\pi + \alpha \ \forall \ n \in \mathbb{Z} \\$$
My attempt:
Adding the two equations,
$$\sin\theta + \cos\theta = \sin\alpha + \cos\alpha$$
$$\sin\theta - \sin\alpha = \cos\alpha - \cos\theta$$
$$2\cos\left( \frac{\theta + \alpha}{2} \right)\sin\left( \frac{\theta - \alpha}{2} \right)= 2\sin\left( \frac{\theta + \alpha}{2} \right)\sin\left( \frac{\theta - \alpha}{2} \right)$$
$$\cos^2\left( \frac{\theta + \alpha}{2} \right) = \sin^2\left( \frac{\theta + \alpha}{2} \right)$$
$$\cos(\theta + \alpha) = 0$$
$$\therefore \theta + \alpha = (2n+1)\frac{\pi}{2},\ n \in \mathbb{Z}$$
$$\theta = (2n+1)\frac{\pi}{2} - \alpha, \ n \in \mathbb{Z} $$
Why doesn't my solution match with the correct solution?
Please help!
|
One problem is you divided by $\sin\left(\frac{\theta - \alpha}{2}\right)$ when going from the third to fourth lines. However, with $\theta = 2n\pi + \alpha$, then $\frac{\theta - \alpha}{2} = n\pi$, but $\sin(n\pi) = 0$. When you divide by $0$, basically anything can occur from it, including incorrect results such as what you got.
|
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|
Why cant I prove this trigonometric equation straight down? The question is as follows:
Given that $\tan^2a - 2 \tan^2b = 1$. Show that $\cos2a + \sin^2b = 0$.
After a few attempts, I successfully came up with a solution as follows:
$$\tan^2a - 2 \tan^2b = 1$$
$$(\sec^2a - 1) - 2(\sec^2b - 1) = 1$$
$$\sec^2a - 2sec^2b = 0$$
$$\frac{1}{\cos^2a} - \frac{2}{\cos^2b} = 0$$
$$\cos^2b - 2\cos^2a= 0$$
$$(1-\sin^2b) - (1 - \sin^2a) - \cos^2a = 0$$
$$\sin^2a - \cos^2a + \sin^2b = 0$$
$$\cos2a + \sin^2b = 0$$
And the proof is finished. However, before I found this solution, I came up with something like this:
$$\tan^2a - 2 \tan^2b = 1$$
$$\frac{\sin^2a}{\cos^2a} - 2\frac{\sin^2b}{\cos^2b} = 1$$
$$\sin^2a\cos^2b - 2\sin^2b\cos^2a - \cos^2a\cos^2b = 0$$
I wanted to extract the $\cos2a $ here. So I do the following
$$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0$$
$$\cos2a + \frac{2\sin^2b\cos^2a}{\cos^2b} = 0$$
It suffices to show that
$$\frac{2\sin^2b\cos^2a}{\cos^2b} = \sin^2b$$
which is equalvalent as saying
$$\frac{\cos^2a}{\cos^2b} = \frac{1}{2}$$
Therefore, I looked at the original equation and proved that it is true like this:
$$\tan^2a - 2 \tan^2b = 1$$
$$(\sec^2a - 1) - 2(\sec^2b - 1) = 1$$
$$\sec^2a = 2sec^2b$$
$$\frac{\cos^2a}{\cos^2b} = \frac{1}{2}$$
My confusion is the reason why I need to look at the orginal equation again to finish the proof. $\tan^2a - 2 \tan^2b = 1$ is identical to $\cos2a + \sin^2b = 0$. And I didn't lose information during my proof. Shouldn't I be able to finish the proof directly like my first proof?
It may sounds like a strange question. But I hope that someone can explain it to me because it sometimes happens and is confusing to me.
|
Starting from this line continue as follows (instead of dividing by $\cos ^2b$, express it as $1-\sin^2b$):
$$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0 \\
(\sin^2a - \cos^2a) (1-\sin^2b) - 2\sin^2b\cos^2a = 0\\
-\cos 2a-\sin^2a\sin^2b+\sin^2b\cos^2a-2\sin^2b\cos^2a=0 \\
-\cos2a-\sin^2a\sin^2b-\sin^2b\cos^2a=0\\
\cos2a+\sin^2b=0.$$
|
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|
Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!
|
Why did things get compilicated?
If you replace $x$ with $y^2$ and $\sqrt x$ with $y$ you should get
$y^2 - 8y +7 =0$ and that should not be complicated.
========
Factor or use the quadratic formula so
$(y - 7)(y-1) = 0$ so $y = 7$ or $y=1$.
Or $y = \frac {-(-8) \pm \sqrt {64-4*7*1}}2 = \frac {8\pm {36}}2 = \frac {8\pm 6}2=4\pm 3=1,7$; so $y=7$ or $y =1$.
So $x = y^2$ so $x = 7^2 = 49$ or $x = 1$. Thus are the two solutions.
Check:
If $x = 49$ then $x - 8\sqrt x +7 = 49 - 8*7 + 7 = 49- 56 + 7 = 0$.
And if $x = 1$ then $x - 8\sqrt x + 7 = 1-8*1 + 7 = 0$.
|
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|
If $x>4$, what is the minimum value of $\frac{x^4}{(x-4)^2}$. If $x>4$, what is the minimum value of $\frac {x^4}{(x-4)^2}$ ?
I have tried using AM-GM Inequality here by letting $y=x-4$ and ended up getting $224$ but that does not seem to be the correct answer. I find out by trial and error that the minimum value is when $x=8$ which is $256$. Are there any better ways to solve for the minimum value other than trial and error?
|
Using AM-GM:
$$\frac {x^4}{(x-4)^2}=\left(\frac{x^2-16+16}{x-4}\right)^2=\left(\color{red}{x-4+\frac{16}{x-4}}+8\right)^2\ge (\color{red}8+8)^2=256,$$
equality occurs for $x=8$.
|
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|
Given two points endpoints of circle chord find the locus of midpoint
$A(2 \cos \theta_1 , 2 \sin \theta_1)$ and $B(2 \cos \theta_2 , 2 \sin \theta_2)$ are two end points of a variable chord $AB$ of the circle, and $M$ is the midpoint of the chord. Suppose that the slope of the chord $AB$ is $\textbf{always}$ equal to $m$. Find the equation of the locus of $M$.
While trying to solve the problem I found that:
*
*Equation of the circle is $x^2 + y^2 = 4$
*Coordinates of chord midpoint $M$ are $\left( \cos \theta_1 + \cos \theta_2 , \sin \theta_1 + \sin \theta_2 \right)$
*Slope of chord AB is $m =
\dfrac{\sin \theta_2 - \sin \theta_1}{ \cos \theta_2 - \cos \theta_1}$
What should I do to get the solution: $x + my = 0$ ?
|
The slope of the line of midpoints must be $-\dfrac 1m$ and the line must contain the origin. So the equation is $y = -\dfrac 1mx$ or $x+my=0$
You know that
\begin{align}
M_{1,2}
&= \left(\cos\theta_1 + \cos\theta_2 , \
\sin\theta_1 + \sin\theta_2 \right) \\
&= \left(2\cos\dfrac{\theta_1+\theta_2}{2}
\cos\dfrac{\theta_1-\theta_2}{2} , \
2\sin\dfrac{\theta_1+\theta_2}{2}
\cos\dfrac{\theta_1-\theta_2}{2} \right)
\end{align}
We also know that $M_0 =(0,0)$ is the midpoint of the line with slope $m$ that passes through the origin. The slope of the line through $M_0$ and $M_{1,2}$ is
$m' = \dfrac
{2\sin\dfrac{\theta_1+\theta_2}{2}\cos\dfrac{\theta_1-\theta_2}{2}}
{2\cos\dfrac{\theta_1+\theta_2}{2}\cos\dfrac{\theta_1-\theta_2}{2}}
= \tan \dfrac{\theta_1+\theta_2}{2}
$
You also know that
\begin{align}
m
&= \dfrac{\sin \theta_2 - \sin \theta_1}
{ \cos \theta_2 - \cos \theta_1} \\
&= \dfrac{ 2 \sin \dfrac{\theta_2 - \theta_1}{2}
\cos \dfrac{\theta_2 + \theta_1}{2}}
{-2 \sin \dfrac{\theta_2 - \theta_1}{2}
\sin \dfrac{\theta_2 + \theta_1}{2}} \\
&= -\cot \dfrac{\theta_2 + \theta_1}{2}
\end{align}
So, for all $\theta_1$ and $\theta_2$ that represent the angles corresponding to the endpoints of a line with slope $m$ intersecting the circle $x^2+y^2 = 4$, $m \cdot m' = -1$.
In other words, the locus of the bisectors is a line with slope $-\dfrac 1m$.
|
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|
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means the maximum must be $4$, but the answer key says it's $16$. How come?
|
You've observed that $x\in[-1,2]$, but your mistake is that the maximum need not be at one of the endpoints. Once you know that
$$4(x+1)(x-2) + (y+1)^2=0,$$
you see that any solution $(x,y)$ must have $-1\leq x\leq 2$. Then
$$y=-1\pm2\sqrt{-(x+1)(x-2)},$$
and so you want to find the maximum of
$$5x+6y=5x-1\pm2\sqrt{-(x+1)(x-2)},$$
for $x\in[-1,2]$, which I'm sure you can do by standard methods.
|
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|
Integer $N$, in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b<16$, then what is the successor of $b$?
An integer $N$, expressed in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b$ is less than $16$, then what is the successor of $b$?
I couldn't develop anything more than
$$N_b = \left(\;n\cdot (b-1)\;\right)_b = 6789_b$$
(with subscripts indicating "in base $b$").
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If $N$ is a multiple of $b-1$ then by the rule of nines[1] the sum of the digits is a multiple of $b-1$ as well.
So $b-1| 6+ 7 + 8 + 9 = 30$. So $ b-1 = 1,2,3,5,6,10,15,30$
Bu $9$ is a digit so $b\ge 10$. And $b < 16$ so $b-1 < 15$.
... so $b-1 < 15$ and so $b-1 =10$ and $b =11$ and then successor of $b$ is $12$.
[1] Rule of nines: $\sum_{k=0}^n a_k b^k \equiv \sum_{k=0}^n a_k \pmod {(b-1)}$.
Pf: $\sum_{k=0}^n a_k b^k - \sum_{k=0}^n a_k = \sum_{k= 0}^n a_k (b^k-1) $
Now each $b^0 - 1 = 0$ and $b^1 - 1 =b-1$ and for $k > 1$ then $b^k -1 = (b-1)(b^{k-1} + b^{k-1} + ...... + 1)$ so $b-1|b^k-1$. So $b-1|\sum_{k= 0}^n a_k (b^k-1)$
And so $\sum_{k=0}^n a_k b^k - \sum_{k=0}^n a_k\equiv 0 \pmod {(b-1)}$.
In other words, $N =\sum_{k=0}^n a_k b^k$ and the sum of the digits of $N$ (in base $b$) have the same remainder when divided by $b-1$.
And therefor $b-1|N$ if and only if $b-1|$ the sum of the digits if $N$ (in base $b$)
|
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|
The maximum of real function with 4 prime parameters and $\lfloor \ \rfloor$ Let $a$,$b$,$c$ and $d$ be prime numbers such that $a>b>c>d$. Let $x$ be an integer greater than $a$.
Let $f(x) = \left(\dfrac{x}{a}\right) – \left(\left(\dfrac{x}{ab}\right) + \left(\dfrac{x}{ac}\right) + \left(\dfrac{x}{ad}\right)\right) + \left(\left(\dfrac{x}{abc}\right) + \left(\dfrac{x}{abd}\right) + \left(\dfrac{x}{acd}\right)\right) - \left(\dfrac{x}{abcd}\right) $
Let $g(x) =
\left\lfloor \dfrac{x}{a} \right\rfloor –
\left(\left\lfloor \dfrac{x}{ab} \right\rfloor +
\left\lfloor \dfrac{x}{ac} \right\rfloor +
\left\lfloor \dfrac{x}{ad} \right\rfloor \right) +
\left(\left\lfloor \dfrac{x}{abc} \right\rfloor +
\left\lfloor \dfrac{x}{abd} \right\rfloor +
\left\lfloor \dfrac{x}{acd} \right\rfloor \right) -
\left\lfloor \dfrac{x}{abcd} \right\rfloor -1$
What value of $x$ results in the maximum difference between $f(x)$ and $g(x)$ (i.e. $f(x) - g(x)$) and what is the value of the difference in terms of $a,b,c$ and $d$?
Edit: I changed the square brackets to parentheses since they were causing confusion.
There are multiple values of $x$ where the difference between $f(x)$ and $g(x)$ reaches a maximum. If you graph the difference between $f(x)$ and $g(x)$, you will notice that the curve is cyclical with a period equal to $a\times b\times c\times d$. In other words $f(z) - g(z) = f(z + abcd) - g(z + abcd)$ for any integer $z$.
For example, $a=7$, $b=5$, $c=3$, $d=2$.
By plotting the difference between $f(x)$ and $g(x)$, I have found that
the maximal difference occurs at $x=76 + 210n$ where $n$ is any integer.
The value of the maximal difference is $\frac{398}{210}$.
Why does the maximum occur at $x=76$? Why is the maximal difference equal to $\frac{398}{210}$?
Edit2: Notice that the difference between $f(x)$ and $g(x)$ reaches a peak at $x=an-1$ where $n$ is an integer. This is because the difference between the first terms of $f(x)$ and $g(x)$ is at a maximum.
$\left(\dfrac{an-1}{a}\right) - \left\lfloor \dfrac{an-1}{a} \right\rfloor = \dfrac{a-1}{a}$.
In my example where $a=7$, the peaks occur at $x = 48,76,90,118,132,160,202,$etc. All these values are in the form $x=7n-1$ and the largest peak occurs at $x=76$.
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Note: Most of the work was already done by OP. This answer is a kind of summary.
In order to find the maximum difference $\max_{x\in\mathbb{N}}\{f(x)-g(x)\}$ the constant term $-1$ of $g(x)$ is not relevant and will be neglected. We denote with $\{x\}$ the fractional part of a real number $x\geq 0$.
\begin{align*}
x=\lfloor x \rfloor+\{x\}\qquad\qquad 0\leq \{x\}<1
\end{align*}
Taking primes $a>b>c>d$ we can write the difference as function $h:\mathbb{N}\to\mathbb{R}$ with
\begin{align*}
h(x)&=\frac{x}{a}-\frac{x}{ab}-\frac{x}{ac}-\frac{x}{ad}+\frac{x}{abc}+\frac{x}{abd}+\frac{x}{acd}-\frac{x}{abcd}\\
&\qquad-\left\lfloor\frac{x}{a}\right\rfloor+\left\lfloor\frac{x}{ab}\right\rfloor+\left\lfloor\frac{x}{ac}\right\rfloor+\left\lfloor\frac{x}{ad}\right\rfloor\\
&\qquad-\left\lfloor\frac{x}{abc}\right\rfloor-\left\lfloor\frac{x}{abd}\right\rfloor-\left\lfloor\frac{x}{acd}\right\rfloor+\left\lfloor\frac{x}{abcd}\right\rfloor\\
&=\left\{\frac{x}{a}\right\}-\left\{\frac{x}{ab}\right\}-\left\{\frac{x}{ac}\right\}-\left\{\frac{x}{ad}\right\}\\
&\qquad+\left\{\frac{x}{abc}\right\}+\left\{\frac{x}{abd}\right\}+\left\{\frac{x}{acd}\right\}-\left\{\frac{x}{abcd}\right\}\tag{1}
\end{align*}
In order to better see what's going on we look at OP's example $a=7,b=5,c=3,d=2$ and consider according to (1):
\begin{align*}
&\left\{\frac{x}{7}\right\}-\left\{\frac{x}{35}\right\}-\left\{\frac{x}{21}\right\}-\left\{\frac{x}{14}\right\}\\
&\qquad+\left\{\frac{x}{105}\right\}+\left\{\frac{x}{70}\right\}+\left\{\frac{x}{42}\right\}-\left\{\frac{x}{210}\right\}\to \max\tag{2}
\end{align*}
We deduce from (2):
*
*The dominant part is $\left\{\frac{x}{7}\right\}$ which is maximal whenever $x=7n-1$, $n\geq 1$ and zero whenever $x$ is a multiple of $7$.
*Since each of the denominators is a divisor of $210$, we consider WLOG $1\leq x=7n-1 \leq 210$, $n=1,\ldots,30$.
*The first term $\left\{\frac{x}{7}\right\}$ evaluated at $x=7n-1$ is the constant $\frac{6}{7}$.
We conclude from the three aspects above:
The maximum of $h$ is given as
\begin{align*}
&\max_{1\leq x\leq 210}h(x)\\
&\qquad=\max_{1\leq x\leq 210}\left(\left\{\frac{x}{7}\right\}-\left\{\frac{x}{35}\right\}-\left\{\frac{x}{21}\right\}-\left\{\frac{x}{14}\right\}\right.\\
&\qquad\qquad\qquad\qquad\left.+\left\{\frac{x}{105}\right\}+\left\{\frac{x}{70}\right\}+\left\{\frac{x}{42}\right\}-\left\{\frac{x}{210}\right\}\right)\\
&\qquad=\max_{1\leq n\leq 30}\left(\frac{6}{7}
-\left(\frac{(7n-1)\mathrm{mod}(35)}{35}\right\}-\left\{\frac{(7n-1)\mathrm{mod}(21)}{21}\right\}\right.\\
&\qquad\qquad\qquad\qquad-\left\{\frac{(7n-1)\mathrm{mod}(14)}{14}\right\}+\left\{\frac{(7n-1)\mathrm{mod}(105)}{105}\right\}\\
&\qquad\qquad\qquad\qquad+\left\{\frac{(7n-1)\mathrm{mod}(70)}{70}\right\}
+\left\{\frac{(7n-1)\mathrm{mod}(42)}{42}\right\}\\
&\qquad\qquad\qquad\qquad\left.-\left\{\frac{(7n-1)\mathrm{mod}(210)}{210}\right\}\right)\tag{3}
\end{align*}
Calculation gives a maximum value for $n=11$ resulting in
\begin{align*}
h(x)=h(7n-1)=\color{blue}{h(76)=\frac{94}{105}\doteq 0,8952}
\end{align*}
In general we obtain as in (3) for primes $a>b>c>d$ the maximum of $h$ as:
\begin{align*}
&\max_{1\leq x\leq abcd}h(x)\\
&\qquad=\max_{1\leq n\leq bcd}\left(1-\frac{1}{a}
-\left\{\frac{(an-1)\mathrm{mod}(ab)}{ab}\right\}-\left\{\frac{(an-1)\mathrm{mod}(ac)}{ac}\right\}\right.\\
&\qquad\qquad\qquad\qquad-\left\{\frac{(an-1)\mathrm{mod}(ad)}{ad}\right\}+\left\{\frac{(an-1)\mathrm{mod}(abc)}{abc}\right\}\\
&\qquad\qquad\qquad\qquad+\left\{\frac{(an-1)\mathrm{mod}(abd)}{abd}\right\}+\left\{\frac{(an-1)\mathrm{mod}(acd)}{acd}\right\}\\
&\qquad\qquad\qquad\qquad\left.-\left\{\frac{(an-1)\mathrm{mod}(abcd)}{abcd}\right\}\right)
\end{align*}
|
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|
Find the solution to $y' = \frac{1+y}{x^2+x}$ that satisfies $y(1) = -1$ I'm having trouble when conditions like domain on variables are present in ODE's.
I am asked to find the solution to $y' = \dfrac{1+y}{x^2+x}$ that satisfies
$(a) \quad y(-2) = 1$
$(b) \quad y(1) = -1$
$(c) \quad y(1) = -2$
$(d) \quad y(-\frac{1}{2}) = 0$
and to state for which $x$ the solution exist.
In particular I have a problem with $(b)$.
I reasoned like this:
To begin with, $\dfrac{1+y}{x^2+x} = \dfrac{1+y}{x(x+1)}$ so $x \neq 0, -1$
If we assume $y \neq -1$ we can rewrite the equation
$y' = \dfrac{1+y}{x^2+x} "\iff" \dfrac{dy}{1+y} = \dfrac{dx}{x(x+1)} "\iff" \int \dfrac{1}{1+y} dy = \int \dfrac{1}{x(x+1)} dx, \quad (1.)$
I write $"\iff"$ because that how my teacher writes, since we haven't really defined what it means to multiply or divide with $dy, dx$.
Since $\int \dfrac{1}{x(x+1)} dx = \int \dfrac{1}{x} - \dfrac{1}{x+1} dx = C + \ln(|x|) - \ln(|x+1|) = C + \ln\left(\dfrac{|x|}{|x+1|} \right), C\in\mathbb{R} $
$\int \dfrac{1}{1+y} dy = C + \ln(|1+y|), C\in\mathbb{R}$
It follows from $(1.)$ that
$\ln(|1+y|) = C + \ln\left(\dfrac{|x|}{|x+1|} \right), \quad C\in\mathbb{R} \iff $
$\exp(\ln|1+y|) = \exp\left(C + \ln\left(\dfrac{|x|}{|x+1|} \right)\right), \quad C\in\mathbb{R} \iff$
$|1+y| = e^C \dfrac{|x|}{|x+1|} = C_2 \dfrac{|x|}{|x+1|}, \quad C_2 > 0 \quad (2.)$
Now we consider different cases based on the value of $y$
Case 1 : If $y > -1$ then it follows from $(2.)$ that $y = C_2 \dfrac{|x|}{|x+1|} - 1, \quad C_2 > 0$
Case 2 : If $y < -1$ then it follows from $(2.)$ that $y = -C_2 \dfrac{|x|}{|x+1|} - 1, \quad C_2 > 0$
Next we consider different cases based on the value of $y$ and $x$
Case 1.1 : $y > -1$ and $x > 0$ implies that
$y = C_2 \dfrac{|x|}{|x+1|} - 1 = C_2 \dfrac{x}{x+1} - 1 = \dfrac{C_2x - x - 1}{x+1} = \dfrac{(C_2-1)x - 1}{x+1} = \\ \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$
Case 1.2 $y > -1$ and $-1 < x < 0$ implies that
$y = C_2 \dfrac{|x|}{|x+1|} - 1 = C_2 \dfrac{-x}{x+1} - 1 = \dfrac{-C_2x - x - 1}{x+1} = \dfrac{-(C_2+1)x - 1}{x+1} = \\ \dfrac{C_4x - 1}{x+1}, \quad C_4 < -1$
Similar reasoning shows that for the rest of the cases the following is true:
Case 1.3 $y > -1$ and $x < -1$ implies that $y = \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$
Case 2.1 $y < -1$ and $x > 0$ implies that $y = \dfrac{C_4x - 1}{x+1}, \quad C_4 < -1$
Case 2.2 $y < -1$ and $-1 < x < 0$ implies that $y = \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$
Case 2.3 $y < -1$ and $x < -1$ implies that $y = \dfrac{C_4x - 1}{x+1}, \quad C_4 < -1$
Now, to solve $(a)$ we note that in this case $x = -2 < -1$ and $y = 1 > -1$ so this fits into case 1.3 where $y = \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$. Plugging in $x = -2$ and $y = 1$ we get $C_3 = 0 > -1$ so that $y = -\dfrac{1}{x+1}, \quad x < -1$
I reasoned similarly in $(c)$ and $(d)$ and the answers seem to be correct.
However, in $(b)$, $y = -1$ which I assumed to be false when revriting the equation to $(1.)$. So I figured I have to try a different way than using any of the derived formulas for $y$. Just by inspecting the original equation $y' = \dfrac{1+y}{x^2+x}$ I noticed that $y(x) = -1$ solves the equation for all $x \neq 0, -1$, but in the answer they state that $y(x) = -1, \quad x > 0$ is the correct solution. So I wonder, why is the condition $x > 0$ necessary? Couldn't it just as well be $x < -1$ or $-1 < x < 0$ ?
|
If you consider $x < 0$ too, there will not be a unique solution to the IVP. Indeed, any solution to the IVP on the three intervals $(-\infty, -1), (-1, 0), (0, \infty)$ can be spliced together to form a solution to the differential equation. So long as the function is constantly $-1$ on $(0, \infty)$ (and whatever other solution on the other two intervals) it will satisfy the IVP.
For example, the following satisfies the IVP:
$$y(x) = \begin{cases} \frac{-3x}{x + 1} - 1 & \text{if } x < -1 \\ \frac{2x}{x + 1} - 1 & \text{if } -1 < x < 0 \\ -1 & \text{if } x > 0. \end{cases}$$
The question seems to want the unique solution locally around $x = 1$ and the largest domain on which it's unique. The question doesn't sound like it makes this clear.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the 94th term of this sequence? $1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$
What is the 94th term of the following sequence?
$$1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$$
*
*8
*9
*10
*11
My Attempt: I found that the answer is 3rd option i.e. 94th term is 10. As every number is written 2n: n is natural number. Here 94 = 2(47) so sum of first few natural numbers should be greater than or equal to 47. Since $$1+2+3+4+5+6+7+8+9 = 45 < 47$$ so options 1,2 are not possible and $$1+2+3+4+5+6+7+8+9+10 = 55 >47$$ But this is a lengthy process.
Please tell me easiest way to approach the answer.
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To detremine the $94$-th term, I can use the rules of the aritmethic progression, in particular: $$S_n=\frac{n}{2}(2a_0+(n-1)d)$$ Substituing the numbers, I have: $$S_n=\frac{n}{2}(2+2n)$$ Imposig $S_n=94$, I obtain: $n^2+n-94=0$ and so $n=10$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\{e_1,e_2,e_3\}$ and $\{v_1,v_2,v_3\}$ are two different basis, $[v]_B\neq [v]_V$. So what is wrong in my argument What's wrong here ? Let $B=\{e_1,e_2,e_3\}$ the canonical basis of $\mathbb R^3$ and let $V\{v_1,v_2,v_3\}$ an other basis of $\mathbb R^3$. I denote $$\left[\begin{pmatrix}x\\ y\\ z\end{pmatrix}\right]_B=xe_1+ye_2+ze_3.$$
Let $v=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\in\mathbb R^3$.
I know that $$[e_1]_B=[v_1]_V=\begin{pmatrix}1\\0\\0\end{pmatrix},\quad [e_2]_B=[v_2]_V=\begin{pmatrix}0\\1\\0\end{pmatrix}\quad \text{and}\quad [e_3]_B=[v_3]_V=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$
So $$[v]_B=\left[\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\right]_B=1\left[e_1\right]_B+2\left[e_2\right]_B+3\left[e_3\right]_B=1\begin{pmatrix}1\\0\\0\end{pmatrix}+2\begin{pmatrix}0\\1\\0\end{pmatrix} +3\begin{pmatrix}0\\0\\1\end{pmatrix}=1[v_1]_V+2[v_2]_V+3[v_3]_V=[v]_V.$$
What is wrong in my argument ? (I know it's wrong since the writing of $v$ in $B$ and in $V$ should be different).
|
I see two problematic parts: the "red" and "blue" parts. I highlighted them from your argument. I hope you are not colorblind.
$\color{red}{\text{Let }v=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\in\mathbb R^3}$.
I know that $$[e_1]={\color{red}{[v_1]_V}}=\begin{pmatrix}1\\0\\0\end{pmatrix},\quad [e_2]_B={\color{red}{[v_2]_V}}=\begin{pmatrix}0\\1\\0\end{pmatrix}\quad \text{and}\quad [e_3]_B={\color{red}{[v_3]_V}}=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$
So $$[v]_B=\left[\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\right]_B=1\left[e_1\right]_B+2\left[e_2\right]_B+3\left[e_3\right]_B=1\begin{pmatrix}1\\0\\0\end{pmatrix}+2\begin{pmatrix}0\\1\\0\end{pmatrix} +3\begin{pmatrix}0\\0\\1\end{pmatrix}=1{{[v_1]_V}}+2{{[v_2]_V}}+3{{[v_3]_V}}{\color{blue}{{}=[v]_V}}.$$
The blue equality you claim may not necessarily be true.
$v_1$, $v_2$, and $v_3$ have conflicting meaning. They are introduced as a basis; $v_i$ are then vectors. Then when $v:=(1,2,3)$ one could interpret $v_i$ to be the $i$-th projection of $(1,2,3)$; $v_i$ are then scalars.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrating $\int\ln x \arccos\left( 7x^2-\sqrt{49x^4-50x^2+1}\right) dx$ This is a problem that my calculus professor gave to his students many years ago.
$$\int\ln x \arccos\left( 7x^2-\sqrt{49x^4-50x^2+1}\right) dx$$
Wolfram doesn't find any solution in terms of standard mathematical functions. I'm sure that this integral has a solution otherwise my professor wouldn't have assigned it.
Could someone help me?
|
Cross-posted on AoPS here. Answered by ysharifi. I will just put it here in length.
That is much neater than your other integral. Assuming you want to stay in real numbers, first we need to have $x > 0$ because of the presence of $\ln x.$ Secondly we also need to have
$$49x^4-50x^2+1=(1-x^2)(1-49x^2) \ge 0,$$
and so we have either $0 < x \le \frac{1}{7}$ or $x > 1.$
Thirdly, we also need to have $-1 \le 7x^2-\sqrt{49x^4-50x^2+1} \le 1$ and so $7x^2-1 \le \sqrt{49x^4-50x^2+1},$ which does not hold for $x > 1.$ So the domain of the function you want to integrate is $0 < x \le \frac{1}{7}.$ Now see that
$$\cos^{-1}(7x^2-\sqrt { 49x^4-50x^2+1})=\cos^{-1}(7x^2-\sqrt{1-x^2}\sqrt{1-49x^2})$$
$$=\cos^{-1}(\cos(\cos^{-1}x)\cos(\cos^{-1}(7x))-\sin(\cos^{-1}x)\sin(\cos^{-1}(7x)))$$
$$=\cos^{-1}(\cos(\cos^{-1}x+\cos^{-1}(7x)))=\cos^{-1}x+\cos^{-1}(7x).$$
So
$$I:=\int \ln x \cos^{-1}(7x^2-\sqrt{49x^4-50x^2+1})\ \mathrm dx=\int \ln x \ (\cos^{-1}x+\cos^{-1}(7x))\ \mathrm dx.$$
The rest of the solution is straightforward. Use integration by parts with $\ln x=u$ and $(\cos^{-1}x+\cos^{-1}(7x)) \ \mathrm dx=\mathrm dv.$ Then
$$\mathrm du=\frac{\mathrm dx}{x}, \ \ \ \ \ \ v=x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}$$
and hence
$$I=\ln x \left(x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}\right)-\int \left(\cos^{-1}x+\cos^{-1}(7x)-\frac{\sqrt{1-x^2}}{x}-\frac{\sqrt{1-49x^2}}{7x}\right)\ \mathrm dx$$
$$=(\ln x-1) \left(x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}\right)+\int \left(\frac{\sqrt{1-x^2}}{x}+\frac{\sqrt{1-49x^2}}{7x}\right)\mathrm dx. \ \ \ \ \ \ \ \ \ \ (1)$$
Also, given $a > 0,$ the substitution $1-a^2x^2=t^2$ gives
$$\int \frac{\sqrt{1-a^2x^2}}{x} \ \mathrm dx=\sqrt{1-a^2x^2}+\ln x-\ln(1+\sqrt{1-a^2x^2})+ \text{constant}. \ \ \ \ \ \ \ \ \ \ \ (2)$$
Now $(1),(2)$ together complete the solution.
Brilliantly done; as always by him.
|
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|
Differentiate $\sqrt{\frac{1 +\sin x}{1 -\sin x}}$ I have tried a lot of ways to solve this question but I am unable to get the answer as same as my textbook.
The text book answer is as follow: $$\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)$$
The steps which I took is as follows:
$$\sqrt{\frac{1+ \sin x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}}$$
Then Secondly
$$\sqrt{\frac{\left(1+\sin x\right)^2}{1-\sin^2 x}}$$
Then I got
$$\dfrac{1+\sin x}{\cos x}$$
When I differentiated this I got the following
$$\frac{\cos ^2\left(x\right)+\sin \left(x\right)\left(1+\sin \left(x\right)\right)}{\cos ^2\left(x\right)}$$
Can Anyone tell me what I am doing wrong?
I also know that $$\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)=\frac{2}{\left(\cos \frac{x}{2}-\sin\frac{x}{2}\right)^2}$$
Thank you for the help!
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Alternatively, using the product rule:
$$\begin{align}\left(\sqrt{\dfrac{1 +\sin x}{1 -\sin x}}\right)'
&=(\sqrt{1+\sin x})'\cdot (1-\sin x)^{-1/2}+\sqrt{1+\sin x}\cdot ((1-\sin x)^{-1/2})'=\\
&=\frac{\cos x}{2\sqrt{1+\sin x}}\cdot \frac1{\sqrt{1-\sin x}}+\sqrt{1+\sin x}\cdot \frac{\cos x}{2(1-\sin x)\sqrt{1-\sin x}}=\\
&=\frac{\cos x}{2\sqrt{\cos ^2x}}+\frac{\cos x\sqrt{(1+\sin x)^2}}{2(1-\sin x)\sqrt{1-\sin ^2x}}=\\
&=\frac12+\frac{1+\sin x}{2(1-\sin x)}=\\
&=\frac1{1-\sin x}=\cdots =\\
&=\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)\end{align}$$
Can you show the equality of the last two expressions using what you stated you know?
Answer (see the hidden area):
$$\frac1{1-\sin x}=\frac{1}{\sin^2x+\cos^2x-2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{1}{(\sin \frac x2-\cos \frac x2)^2}=\\=\frac{1}{2(\frac{1}{\sqrt{2}}\sin \frac x2-\frac{1}{\sqrt{2}}\cos \frac x2)^2}=\frac{1}{2\cos^2(\frac{\pi}{4}+\frac x2)}=\frac12\sec^2(\frac{\pi}{4}+\frac x2).$$
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $a^2+u^2+d^2-b^2-c^2-v^2>-4w^2$ Let $ABCD$ be cyclic quadrilateral of the circle $O$ with: $$R=w\text{ is radius };AB=a;BC=b;CD=c;DA=d;AC=u;BD=v$$. Prove that $$a^2+u^2+d^2-b^2-c^2-v^2>-4w^2$$
We have $$u=\sqrt{\frac{\left(ac+bd\right)\left(ad+bc\right)}{ab+cd}};v=\sqrt{\frac{\left(ac+bd\right)\left(ab+cd\right)}{ad+bc}}$$ and $$R=\frac{1}{4}\sqrt{\frac{\left(ab+cd\right)\left(ac+bd\right)\left(ad+bc\right)}{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}} \text{for
} s=\frac{a+b+c+d}{2}$$
Then by BW and computer we're done. But it's very ugly.I have no idea to solve it without computer. Help me.
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Let us denote by $x,y$ the two angles in $A$ delimited by the diagonal $AC$ and the sides $AD$, respectively $AB$. Let us also denote by $s,t$ the two angles in $C$ delimited by the same diagonal $CA$ and the sides $CD$, respectively $CB$. We have
$$
x+y+s+t=\pi\ .
$$
Then we can express all the data in the inequality in terms of $R$ and sine functions in (sums of some of) the variables $x,y,s,t$, for instance, $a/2=R\sin t$, $b/2=R\sin y$, $v/2=R\sin(x+s)=R\sin(y+t)$.
Then $a^2=4R^2\sin^2t=2R^2(1-\cos(2t))$, and similarily for the other squares, so it is useful to introduce $X,Y,S,T$ equal to respectively $2x,2y,2s,2t$ to lower the degrees of the appearing trigonometric functions,
$$
X+Y+S+T=2\pi\ .$$
Then we have to show equivalently, step by step:
$$
\begin{aligned}
4R^2 + a^2 + u^2 + d^2 &> c^2 + v^2 + b^2\ ,
\\
1 + \sin^2t + \sin^2(x+s) + \sin^2s &>
\sin^2x + \sin^2(x+y) + \sin^2y\ ,
\\
2
-\cos T-\cos(X+S)-\cos(S) &>
-\cos X-\cos(X+Y)-\cos(Y) \ ,
\\
2
+\cos X+\cos(X+Y)+\cos(Y)
&>
\cos T+\cos(X+S)+\cos(S)\ ,
\\
2
+\cos X+\cos(X+Y)+\cos(Y)
&>
\cos (X+Y+S)+\cos(X+S)+\cos S\ ,
\\
2
+\cos X+\cos(X+Y)+\cos(Y)
&>
\cos S\Big[\ \cos(X+Y)+\cos X+1\ \Big]
\\
&\qquad-\sin S\Big[\ \sin(X+Y)+\sin X\ \Big] =:E(S,X,Y)\ .
\\[3mm]
&\qquad\text { Here we break the chain of equivalences.}
\\
&\qquad\text { We take the maximum w.r.t. $S$ on the R.H.S. above.}
\\
&\qquad\text { Let us show first:}
\\
(2
+\cos X+\cos(X+Y)+\cos(Y))^2
&\ge
%(\cos^2 S+\sin^2 S)
%\Big[\
\Big(\cos(X+Y)+\cos X+1\Big)^2+\Big(\sin(X+Y)+\sin X\Big)^2
%\ \Big]
\\
&\qquad\text{ i.e. equivalently}
\\
(2
+\underbrace{\cos X+\cos(X+Y)+\cos(Y)}_{=:u})^2
&\ge
1+1+1+
\underbrace{2\cos X+2\cos(X+Y)+2\cos Y}_{=2u}\ .
\end{aligned}
$$
Above $u\in[-3/2,\ 3]$ (so the above quantity $2-u$ is indeed $>0$, and we could apply the square function in that inequality, obtaining an equivalent inequality,) is a substitute for the sum of cosine functions in $X,X+Y,Y$. The inequality $(2+u)^2\ge3+2u$ becomes $(1+u)^2\ge0$. This is clear. The strict inequality fails in the case $u=-1$, equivalently either $X=\pi$, or $Y=\pi$. By initial symmetry, we consider only $Y=\pi$. In this special case, we have to show $2+\cos X>\cos S$. The equality is possible only in case of $X=\pi$. This is a degenerated case with $A=B=D$...
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"timestamp": "2023-03-29T00:00:00",
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|
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
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As the degree of the divisor is $3$, so the degree of the remainder is at most $2$. Let the remainder $R\left(x\right)$ be $ax^2+bx+c$. As $x^3-x=x\left(x-1\right)\left(x+1\right)$, we'll find that: $$\begin{cases} R\left(-1\right)=a-b+c=\left(-1\right)^{81}+\left(-1\right)^{49}+\left(-1\right)^{25}+\left(-1\right)^9+\left(-1\right)=-5 \\ R\left(0\right)=c=0^{81}+0^{49}+0^{25}+0^9+0=0 \\ R\left(1\right)=a+b+c=1^{81}+1^{49}+1^{25}+1^9+1=5 \end{cases}$$ After some calculation, you'll get $a=0,b=5,c=0$, so the remainder is $\boxed{5x}$.
|
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"url": "https://math.stackexchange.com/questions/3352608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Closed form for series of multiple sums I have a series that goes like that:
\begin{align}
(n=1)\qquad &\sum_{i=0}^x\bigg(\sum_{j=1}^i 1\bigg)\\
(n=2)\qquad &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j 2\bigg)\bigg)\\
(n=3)\qquad &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j \bigg(2+\sum_{l=2}^k 3\bigg)\bigg)\bigg)\\
(n=4)\qquad &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j \bigg(2+\sum_{l=2}^k \bigg(3+\sum_{m=3}^l 4\bigg)\bigg)\bigg)\bigg)
\end{align}
and so on.
For each $n$, it is easy to write the sums in closed expressions:
\begin{align}
(n=1)\qquad &\frac{1}{2} (x+1)\,x \\
(n=2)\qquad &\frac{1}{6} (x+1)\,(x+2)\,(2x+3)\\
(n=3)\qquad &\frac{1}{24} (x+1)\,(x+2)\,(3x^2+5x+12)\\
(n=4)\qquad &\frac{1}{120} (x+1)\,(x+2)\,(4x^3+3x^2+33x+60)
\end{align}
What I would like to have is a closed or more compact expression that can give me the polynomials in $x$ for arbitrary $n$. I tried OEIS for the coefficients with little success, and also couldn't find if the polynomials (or parts thereof) are known.
Has someone seen such a construction for a sequence and has an idea how I could go about finding a closed or more compact expression?
|
Here we give a somewhat more compact expression. We look at the case $n=3$ again and derive from it a representation for the general case.
We have
\begin{align*}
\color{blue}{\sum_{j_0=0}^x}&\color{blue}{\left(\sum_{j_1=0}^{j_0}\left(1+\sum_{j_2=1}^{j_1}\left(2+\sum_{j_3=2}^{j_2}3\right)\right)\right)}\\
&=\sum_{j_0=0}^x\sum_{j_1=0}^{j_0}1+2\sum_{j_0=0}^x\sum_{j_1=0}^{j_0}\sum_{j_2=1}^{j_1}1+3\sum_{j_0=0}^x\sum_{j_1=0}^{j_0}\sum_{j_2=1}^{j_1}\sum_{j_3=2}^{j_2}1\tag{1}\\
&=\sum_{0\leq j_1\leq j_0\leq x}1+2\sum_{1\leq j_2\leq j_1\leq j_0\leq x}1+3\sum_{2\leq j_3\leq j_2\leq j_1\leq j_0\leq x}1\tag{2}\\
&\,\,\color{blue}{=\binom{x+2}{2}+2\binom{x+2}{3}+3\binom{x+2}{4}}\tag{3}\\
&=\frac{1}{2}(x+2)(x+1)+\frac{1}{3}(x+2)(x+1)x+\frac{1}{8}(x+2)(x+1)x(x-1)\\
&=\frac{1}{24}(x+2)(x+1)(3x^2+5x-3)
\end{align*}
Comment:
*
*In (1) we multiply out and factor out the constants.
*In (2) we use convenient representation to better see the index range.
*In (3) we note the number of summands given by the index range $2\leq j_3\leq j_2\leq j_1\leq j_0\leq x$ is the number of ordered $4$-tupel $(j_0,j_1,j_2,j_3)$ between $2$ and $x$ with repetition. This number is given by the binomial coefficient
\begin{align*}
\binom{4+(x-1)-1}{4}=\binom{x+2}{4}
\end{align*}
In general we have in the case $n=N$:
\begin{align*}
\color{blue}{\sum_{j_0=0}^x}&\color{blue}{\left(\sum_{j_1=0}^{j_0}\left(1+\sum_{j_2=1}^{j_1}\left(2+\cdots+\sum_{j_N=N-1}^{j_{N-1}}N\right)\cdots\right)\right)}\\
&=\sum_{0\leq j_1\leq j_0\leq x}1+2\sum_{1\leq j_2\leq j_1\leq j_0\leq x}1+\cdots+N\sum_{N-1\leq j_N\leq\cdots\leq j_2\leq j_1\leq j_0\leq x}1\\
&=\binom{x+2}{2}+2\binom{x+2}{3}+\cdots+N\binom{x+2}{N+1}\\
&\,\,\color{blue}{=\sum_{k=1}^N k\,\binom{x+2}{k+1}}\\
\end{align*}
|
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|
Prove that sum of the $k$ numbers in the $k$th group = ${\frac{1}{2}\left(k(k^2+1)\right)}$. Consider an arrangement of the positive integers, grouped as shown, so that the $k$th group has $k$ elements: $(1),(2,3),(4,5,6),(7,8,9,10), \ldots$.
The expression for the sum of the $k$ numbers in the $k$th group turns out to be ${\frac{1}{2}\left(k(k^2+1)\right)}$.
However, how would you prove this? I am assuming that you would have to proof by induction, but I can't seem to construct it as of now.
|
Let $a_{i,j}$ be the $j$'th term of the $i$'th group. The first element of the $k$'th group is the sum of the previous $1$ to $k - 1$ groups, plus $1$. Since each $i$'th group has $i$ elements, from Arithmetic progression, the first term of the $k$'th group is
$$a_{k,1} = 1 + \frac{(k-1)(1 + (k - 1))}{2} = 1 + \frac{k(k-1)}{2} \tag{1}\label{eq1}$$
Since the last term of the $k$'th group would be $1$ less than the first term of the $(k+1)$'th group, \eqref{eq1} gives it would be $\frac{k(k+1)}{2}$. Thus, the sum of the $k$'th group is
$$\begin{equation}\begin{aligned}
S_k & = \frac{k(a_{k,1} + a_{k,k})}{2} \\
& = \frac{1}{2}k\left(1 + \frac{k(k-1)}{2} + \frac{k(k+1)}{2}\right) \\
& = \frac{1}{4}k\left(2 + (k^2 - k) + (k^2 + k)\right) \\
& = \frac{1}{4}k\left(2 + 2k^2\right) \\
& = \frac{1}{2}(k(k^2 + 1))
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
|
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|
Solving critical points and classifying the maxima, minima and saddle points I need help to find critical points of the function:
$$f(x,y,z)={(x^2+2y^2+3z^2) e ^{-(x^2+y^2+z^2)}}$$
Then I have to classify these critical points as local maxima/minima or saddle points.
What I would do is to take the partial derivatives $\frac{\partial f}{\partial x}(x, y, z)$, $\frac{\partial f}{\partial y}(x, y, z)$ and $\frac{\partial f}{\partial z}(x, y, z)$.
But now I will get
$\frac{\partial f}{\partial x}(x, y, z)$ = $ {2x-6e^{-x^2-y^2-z^2}xz^2} $
$\frac{\partial f}{\partial y}(x, y, z)$ = $ {4y-6e^{-x^2-y^2-z^2}yz^2} $
$\frac{\partial f}{\partial z}(x, y, z)$ = $ 3(-2e^ {-x^2-y^2-z^2}z^3+2e^{-x^2-y^2-z^2}z) $
Normally I would put them in the matrix form and solve the system of equation for x, y and z.
But I don't know how to solve the system of equation with these partial derivatives. What I am doing wrong or what should be done in order to solve this problem?
|
Computing the partial derivatives, we then have
$$\frac{\partial f(x, y, z)}{\partial x}=-2 x \left(x^2+2 y^2+3 z^2-1\right)e^{-(x^2+y^2+z^2)}$$
$$\frac{\partial f(x, y, z)}{\partial y}=-2 y \left(x^2+2 y^2+3 z^2-2\right)e^{-(x^2+y^2+z^2)}$$
$$\frac{\partial f(x, y, z)}{\partial z}=-2 z \left(x^2+2 y^2+3 z^2-3\right)e^{-(x^2+y^2+z^2)}$$ which means that you need to solve
$$x \left(x^2+2 y^2+3 z^2-1\right)=0$$
$$y \left(x^2+2 y^2+3 z^2-2\right)=0$$
$$z \left(x^2+2 y^2+3 z^2-3\right)=0$$ for which the solutions are quite simple
$$\{\{ 0, 0, -1\}, \{0, 0, 1\}, \{0, -1,
0\}, \{0, 1, 0\}, \{-1, 0, 0\}, \{1, 0,
0\}, \{0, 0, 0\}\}$$
Now, your turn.
|
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|
Discussion about $\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)}$ May I have some further discussion on
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)} = \frac{\pi^3}{96} - \frac{\pi}{8}\ln^2\!2 \tag1$$
where $H_{n}$ is harmonic-number and $\psi(x)$ is digamma function.
Formerly, I deduce the result by this elementary approach, also the only solution I have due to the asset for me is quite insufficient.
Start with famous series
$$\frac1{2} \arctan x \ln (1+x^2) = \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n+1}}$$
by which somehow I find (1) actually equal to
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)} = \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x(1+x)} \mathrm{d}x}$$
and this integral is not impossible to solve.
Question 1: Any other possible access to summation as (1)?
For digamma function has strong relation with harmonic-number, here $\psi(n+1)-\psi(n+\tfrac1{2})$ can also be substitution for $H_{2n}-H_{n}$ with some other constants. So what if I want a part of (1) like
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}^{2}}{2n+1}} \quad \text{or} \quad \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}H_{n}}{2n+1}} \tag2$$
but, of course, I have no idea how to deal with such summation involving higher order of harmonic-number.
Question 2: How to find the closed form for summation in (2)?
|
Solution to question 1:
Since you managed to write your sum in (1) as $\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx$, so lets evaluate the integral.
From here we have
$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{3\pi^3}{96}\tag{1}$$
and from here we have
$$\int_0^1\frac{\arctan x}x\ln\left(\frac{(1+x^2)^3}{(1+x)^2}\right)dx=0\tag{2}$$
By combining (1) and (2) we get
$$\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^22$$
.
BONUS:
Since
$$\psi(n+1)-\psi(n+\tfrac12)=H_n-H_{n-\small{\frac12}}=2H_n-2H_{2n}-2\ln2$$
Then the quality in (1) can be written as
$$\small{\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^2}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}-\ln2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}=\frac12\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx}$$
Rearranging
$$\small{\sum_{n=0}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}=\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^2}{2n+1}-\ln2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}-\frac12\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx}$$
Lets evaluate the second sum:
\begin{align}
\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}&=\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\\
&=\Im\left\{\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n}-\sum_{n=1}^\infty\frac{(i)^n}{n^2}\right\}\\
&=\Im\left\{\frac12\ln^2(1-i)+\operatorname{Li}_2(i)-\operatorname{Li}_2(i)\right\}\\
&=\Im\left\{\frac12\ln^2(1-i)\right\}\\
&=-\frac{\pi}{8}\ln2
\end{align}
So by plugging this result, along with
$$\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}^2}{2n+1}=\frac{5}{96}\pi^3+\frac{3\pi}{16}\ln^22+\frac12\ln2\ G+2\Im\left\{\operatorname{Li_3}(1-i)\right\}$$
and
$$\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^22$$
we finally get
$$\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}H_n}{2n+1}=\frac{3}{64}\pi^3+\frac{3\pi}{8}\ln^22+\frac12\ln2\ G+2\Im\left\{\operatorname{Li_3}(1-i)\right\}$$
|
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|
What did I get wrong when solving $\int\frac{\sqrt{x^2-1}}{x^4}dx$? I'm not sure that this is the problem, but I think I may not know how to find the $\theta$ value when solving an integral problem with trigonometric substitution.
I got $\frac{\sin^3(\sec^{-1}(x))}{3}+C$ for the answer, but the answer should be, $\frac{1}{3}\frac{(x^2-1)^{3/2}}{x^3}+C$
$$\int\frac{\sqrt{x^2-1}}{x^4}dx$$
Let $x=\sec\theta$
Then $dx=\sec\theta\tan\theta d\theta$
$$\int\frac{\sqrt{\sec^2\theta-1}}{\sec^4\theta}\sec\theta\tan\theta d\theta$$
$$=\int\frac{\sec\theta}{\sec^4\theta}\sqrt{\tan^2\theta}\tan\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta} \tan^2\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta}\frac{\sec^2\theta}{\csc^2\theta}d\theta$$
$$=\int\frac{1}{\sec\theta}\frac{1}{\csc^2\theta}d\theta$$
$$=\int \cos\theta\sin^2\theta d \theta$$
Using $u$-substition, let $u=\sin\theta$
Then $du=\cos\theta d\theta$ and $dx = \frac{1}{\cos\theta}du$
$$\int\cos\theta u^2 \frac{1}{\cos\theta}du$$
$$=\int u^2 du$$
$$=\frac{u^3}{3}+C$$
$$=\frac{\sin^3\theta}{3}+C$$
Since $x=\sec\theta$, $\sec^{-1}(x)=\theta$
$$=\frac{\sin^3(\sec^{-1}(x))}{3}+C$$
What am I doing wrong?
|
Since $x=\sec\theta$, so $\cos\theta=\frac1x$ and hence $\sin\theta=\frac{\sqrt{x^2-1}}{x}$. Thus
$$\sin(\sec^{-1}(x))=\frac{(x^2-1)^{3/2}}{x^3}. $$
|
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|
Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/smarter way to do this?
|
Let $y=\frac1{x}$. Then,
\begin{align}
(x+\frac1{x})^4(x-\frac1{x})^2&=(x+y)^4(x-y)^2\\
&=(x+y)^2(x+y)^2(x-y)^2\\
&=(x+y)^2(x^2-y^2)^2\\
&=\big((x+y)(x^2-y^2)\big)^2\\
&=(x^3+x^2y-xy^2-y^3)^2\\
&=(x^3+x-\frac1{x}-\frac1{x^3})^2\\
&=\bigg(\frac{x^6+x^4-x^2-1}{x^3}\bigg)^2\\
&=\frac{x^{12} + 2x^{10} - x^8 - 4 x^6 - x^4 + 2 x^2 + 1}{x^6}
\end{align}
(or $x^6+2x^4-x^2-4-\frac1{x^2}+\frac2{x^4}+\frac1{x^6}$, if you'd like).
|
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|
A box with 3 red balls and 2 white balls. 3 balls taken randomly with replacement, A box containing 3 red balls and 2 white balls. A ball is taken randomly and with replacement. This is done 3 times.
*
*What is the probability that exactly one red ball is obtained?
*Probability at least 1 red ball is obtained?
Attempt
For the first part, the possibilities are: red ball at 1st draw, red ball at 2nd draw, or red ball at 3rd draw. The probabiity for each is $p = \frac{3}{5} \frac{2}{5} \frac{2}{5}$, so the answer is $3 \times p$.
For the second part, possibilities are: 1 red ball taken, 2 red balls taken, or 3 red balls taken. probability of 2 red balls taken is: $3 \times \frac{3}{5} \frac{3}{5} \frac{2}{5}$. So answer is:
$$ \left( 3 \times \frac{3}{5} \frac{2}{5} \frac{2}{5} \right) + \left( 3 \times \frac{3}{5} \frac{3}{5} \frac{2}{5} \right) + \frac{3}{5}\frac{3}{5}\frac{3}{5} $$
|
Both parts are correct. As regards the second one, it is easier to evaluate the complement: the probability that at least one red ball is obtained is $1$ minus the probability that all the three balls are white, namely
$$1-\left(\frac{2}{5}\right)^3.$$
|
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|
Solving $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$ $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$
So, getting rid of the denominators I got to:
$(x+1)(x+5) + x(x+5) < 2x(x+1)$
$\rightarrow x < \frac{-5}{9}$
And also keeping in mind we can't have $-1,-5$ in the solution set, I got that the solution set was:
$\left(-\infty ,-5\right)\cup \left(-5,-1\right)\cup \left(-1,-\frac{5}{9}\right)$
But this is not correct. Help appreciated!
|
When multiply we need to take into account the sign; in this case your step are not correct since you are assuming positive terms.
As an alternative to avoid multiplication, we have
$$\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5} \iff\frac{1}{x} + \frac{1}{x+1} - \frac{2}{x+5}<0 \iff \frac{9x+5}{x(x+1)(x+5)}<0$$
then study the sign of each term.
|
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|
Evaluating $\iint_R\big(x^2+y^2\big)\,dA$
Evaluate the following double integral:
$$\iint_R\big(x^2+y^2\big)\,dA,$$
where $R$ is the region given by plane $x^2+y^2\leq a^2$.
My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\
&=\int_{-a}^{a}\left(x^2y+\dfrac{y^3}{3}\right)_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dx\\
&=\dfrac{2}{3}\int_{-a}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx.
\end{align}
I can't go further from here, please help.
|
$$\iint_R x^2+y^2 dA=\int_0^{2\pi}\int_0^a r^3drd\theta=2\pi\int_0^a r^3 dr$$
|
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|
Solve Integration
$$I = \int\frac{2\sin x+\sin 2x}{(\cos x-1)\sqrt{\cos x+\cos^2x+\cos^3x}}dx$$
I have tried for this question to solve it first I tried to separate the numerator but that doesn't work then i tried for substituting the value in the denominator but still I am not able to convert the whole function in terms of the substituted variable function. So please help me out to solve this problem.
|
First, use the double angle identity $\sin2x = 2\sin x \cos x$ to get that
$$I = \int\frac{2\sin x + \sin 2x}{(\cos x - 1)\sqrt{\cos x + \cos^2 x + \cos^3 x}}dx$$ $$ = \int \frac{(1+\cos x)(-2\sin x)dx}{(1-\cos x)\sqrt{\cos x + \cos^2 x + \cos^3 x}}$$
Now the tricky part is deciding what substitution to choose. Using the substitution $$\frac{1+\cos x}{1-\cos x} = t\implies \cos x = 1 - \frac{2}{t+1}\implies -\sin x dx = \frac{2}{(t+1)^2}dt$$
we can simplify the integral. @ClaudeLeibovici caught my error in the original simplification of my substitution. The correct simplification is
$$I = \int \frac{4t}{\sqrt{3t^4-2t^2-1}}dt$$
Complete the square on the inside of the square root:
$$I =\frac{2}{\sqrt{3}} \int \frac{2t}{\sqrt{(t^2-\frac{1}{3})^2-\frac{4}{9}}}dt$$
Now let $$t^2 = \frac{2}{3}\cosh \tau +\frac{1}{3}\implies 2tdt = \frac{2}{3}\sinh \tau d\tau$$ which gives us the integral
$$\frac{2}{\sqrt{3}} \int \frac{\frac{2}{3}\sinh \tau}{\sqrt{\frac{4}{9}\sinh^2\tau}}d\tau = \frac{2}{\sqrt{3}} \int d\tau = \frac{2}{\sqrt{3}}\tau$$
leaving us with an answer of
$$I = \frac{2}{\sqrt{3}} \cosh^{-1}\left(\frac{3t^2-1}{2}\right)+C$$
Plugging back in for $x$ and utilizing $1-\cos x = 2\sin^2\left(\frac{x}{2}\right)$ we get our final answer:
$$I = \frac{2}{\sqrt{3}} \cosh^{-1}\left(\frac{1+4\cos(x)+\cos^2(x)}{4\sin^4\left(\frac{x}{2}\right)}\right)+C$$
|
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|
How many solutions $x_1+\dots+x_8=30$ has?
Enumerate the solutions of
$$x_1+\dots+x_8=30$$
where $2\leq x_i\leq 5$ for $i=1,\dots,6$,
and $x_7$ and $x_8$ must be or 5 or 10.
I am not sure with my process.
and I would like to ask you if you have a better way to solve this?
and if I am right.
$S_0 - S_1 + S_2 - S_3\dots$ etc. is the solution.
but i think im doing something wrong:
my solution
|
A possible way is using generating functions:
*
*First set $y_i = x_i-2$ for $i=1,\ldots , 6$ and $y_i = x_i -5$ for $i=7,8$. So, you look for the number of solutions of
$$\mbox{}y_1 + \cdots + y_6 + y_7 + y_8 = 8 \mbox{ with } 0\leq y_i \leq 3 \;(i=1,\ldots ,6) \mbox{ and } y_i \in \{0,5\}\;(i=7,8)$$
*So you look for the coefficient of $x^8$ of the generating function of this problem (look at the exponents): $$(1+x+x^2+x^3)^6(1+x^5)^2 = \left(\frac{1-x^4}{1-x}\right)^6(1+x^5)^2$$
*Using $\frac{1}{(1-x)^6} = \sum_{n=0}^{\infty}\binom{n+5}{5}x^n$ (which can easily be verified by repeated differentiation), you get
\begin{eqnarray*} [x^8]\left(\frac{1-x^4}{1-x}\right)^6(1+x^5)^2
& = & [x^8](1-6x^4+\binom{6}{2}x^8)(1+2x^5)\sum_{n=0}^{\infty}\binom{n+5}{5}x^n \\
& = & [x^8](1-6x^4+2x^5 + \binom{6}{2}x^8)\sum_{n=0}^{\infty}\binom{n+5}{5}x^n \\
& = & \binom{8+5}{5} - 6\binom{4+5}{5} + 2\binom{3+5}{5} + \binom{6}{2}\\
& = & \boxed{658}
\end{eqnarray*}
|
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|
How to find the probability of the event that exactly two of the three events occur? My professor posted some practice problems with answers (without explanation) for our midterm and this was one of them:
Three independent events have respective probabilities, $\frac{1}{3},\frac{2}{5},\frac{1}{4}$
a.) Find the probability of the event that exactly two of the three events occur
b.) Find the probability of the even that at least two of the three events occur.
Answer:
a.) $\frac{1}{3}\frac{2}{5}\frac{3}{4} + \frac{1}{3}\frac{3}{5}\frac{1}{4}+ \frac{2}{3}\frac{2}{5}\frac{1}{4}$
b.) $1 -(\frac{1}{3}\frac{3}{5}\frac{3}{4} + \frac{2}{3}\frac{2}{5}\frac{3}{4}+ \frac{2}{3}\frac{3}{5}\frac{1}{4}+ \frac{2}{3}\frac{3}{5}\frac{3}{4})$
Could someone please explain how he derived the answer (referencing any laws, theorems, properties, formulas, etc)
|
In a) he's simply adding up the probabilities. The first term is the probability that the first two events happen and the third one does not, for instance.
In b) he's getting the complement of the event that no more than one event occurs. The first three terms are the probabilities that exactly one event occurs (similar to a) and the last term is the probability that none of the events occurs.
As far as justification goes, the multiplications are okay because the events are independent. (Recall that if $A$ and $B$ are independent events, then $A$ and $B^c$ are also independent.) The additions are justified because all the events are mutually exclusive. These are two extremely important facts. Make sure you understand them.
|
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|
Show that $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $z=-\frac12(1\pm i\cot(k\pi/8))$; and follow-up questions
*
*(a) Show that the equation $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $-\frac12\left(1\pm i\cot\frac{k\pi}{8}\right)$, where $k=1,2,3$.
*(b) Hence show that
$$(z+1)^8-z^8=\tfrac18(2z+1)(2z^2+2z+1)\left(4z^2+4z+\csc^2\tfrac{\pi}{8}\right)\left(4z^2+4z+\csc^2\tfrac{3\pi}{8}\right)$$
*(c) By making a suitable substitution into this identity, deduce that
$$\cos^{16}\theta - \sin^{16}\theta = \tfrac1{16}\cos 2\theta(\cos^22\theta+1)\left(\cos^22\theta+\cot^2\tfrac{\pi}{8}\right)\left(\cos^22\theta+\cot^2\tfrac{3\pi}{8}\right) $$
original problem image
How do you reformat the polynomial to reach the answer required?
I have tried to turn the LHS into $[(z+1)/z]^8 = 0$ and solve the roots of unity, but I keep getting $\operatorname{cis}(2k\pi/8)$ instead of $k\pi/8$ which is required.
Thanks in advance
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By your method
$$(z+1)^8-z^8=0 \iff \left(1+\frac1z\right)^8=1$$
then let
$$w=1+\frac1z$$
and solve
$$w^8=1 \implies w=e^{\frac{2i\pi k}8}\quad k=0,1,2,3,4,5,6,7$$
but only $7$ solutions are valid since $k=0$ is not acceptable, indeed the original equation is of degree $7$.
The solutions are given by
$$z=\frac1{e^{\frac{2i\pi k}8}-1}=\frac{e^{\frac{-i\pi k}8}}{e^{\frac{i\pi k}8}-e^{\frac{-i\pi k}8}}=\frac{\cos\left(\frac{\pi k}8\right)-i\sin\left(\frac{\pi k}8\right)}{2i\sin\left(\frac{\pi k}8\right)}=-\frac12-\frac i2\cot \left(\frac{\pi k}8\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$
If $\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ is satisfied for all real $x>0$ then obtain the possible values of the parameter $a$.
My attempt is as follows:
$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0$$
First condition: If $\forall$ $x>0$, inequality is satisfied, it means $x=0$ must have been the root of the equation $\quad \{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}=0\quad$ at which given inequality would not be satisfied.
Second condition: If one root is real, then other root must be real as complex roots occur as conjugates if all quadratic equation coefficients are real. So $D>=0$
Third condition: $a>0$ as $\forall$ $x>0$, $\quad\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0\quad$
First condition:
$$x=0$$
$$a+\sqrt{2}=0$$
$$a=-\sqrt{2}$$
Second condition:
$$D>=0$$
$$4(a^2-2)^2-4(a-\sqrt{2})(a^2+a-3)(a+\sqrt{2})>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a^2-2-a^2-a+3)>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a-1)<=0$$
$$a\in \left(-\infty,-\sqrt{2}\right]\quad \cup \quad[1,\sqrt{2}]$$
Third condition:
$$a>0$$
$$(a-\sqrt{2})(a^2+a-3)>0$$
$$(a-\sqrt{2})\left(a-\left(\frac{-1+\sqrt{13}}{2}\right)\right)\left(a-\left(\frac{-1-\sqrt{13}}{2}\right)\right)>0$$
$$a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$
Taking intersection of all three conditions would give
$$a\in \{-\sqrt{2}\}\quad$$
But answer is $a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$
What am I missing here, I tried to think of it a lot but didn't any breakthroughs. Please help me in this.
|
Well I got the answer, thanks to @Robert Z
I) First lets see what should be the condition when $a>0 \text{\{a here means coefficient of $x^2$\}}$
So $$\text{parameter }a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$
Now there can be two sub-cases:
1) When $D<0$, then at all values of $x$, inequality will hold.
So for $D<0$ and $a>0$, $\text{parameter }a\in \left(-\sqrt{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right)$
2) When $D>=0$, then inequality will hold only when $x=0$ is the root of the given quadratic equation.
So for $a>0$ and $D>=0$ and $\text{$x=0$ is the root}$, we get parameter $a\in \{-\sqrt{2}\}$
II)When $a<0$, as in that case inequality cannot hold till $+\infty$, so no solutions.
III) When $a=0$, then parameter $a$ can be $\{\sqrt{2},\frac{-1+\sqrt{13}}{2},\frac{-1-\sqrt{13}}{2}\}$
If parameter $a=\sqrt{2}$, then $\sqrt{2}+\sqrt{2}>0$ which is always true, so $\sqrt{2}$ is the solution.
If parameter $a=\frac{-1+\sqrt{13}}{2}$, then we get $x<4.5$, so inequality will hold only for $x<4.5 $, but we want inequality to hold $\forall x>0$, so $a=\frac{-1+\sqrt{13}}{2}$ can't be the solution.
If parameter $a=\frac{-1-\sqrt{13}}{2}$, then we get $x>0.133$, so inequality will hold only for $x>0.133$, but we want inequality to hold $\forall x>0$, so $a=\frac{-1-\sqrt{13}}{2}$ can't be the solution.
So taking union of all cases I),II),III), we get $\text {parameter }a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$ which is the correct answer.
|
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|
Ramanujan, sum of two cubes - how was it discovered? I'm looking for motivation for, and hopefully a derivation of, Ramanujan's sum of cubes formula
$$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-4xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$
I can't see where one could start to justify this without actually expanding everything out.
Additionally, does it provide a parametrisation of the surface
$$X^3+Y^3=Z^3+1$$ in $\mathbb{Q}^3$
with
$$X=\frac{x^2+7xy-9y^2}{2x^2+10y^2}$$
$$Y=\frac{2x^2-4xy+12y^2}{2x^2+10y^2}$$
$$Z=\frac{x^2-9xy-y^2}{2x^2+10y^2}$$
|
"OP" inquired about finding parameterization of two cubes in two different way's.
Consider the below equation;
$s^3+t^3=u^3+v^3$ ----$(1)$
Tomita has given a method to parameterize equation $(1)$.
In his solution, take $(p,q,r,s)=(-1,9,10,12)$
then the below mentioned solution is arrived at:
$s=36x^2+22xy+y^2$
$t=40x^2-40xy+12y^2$
$u=48x^2-40xy+10y^2$
$v= 4x^2+22xy+9y^2$
For, $(x,y)=(0,1)$, we get,
$(s,t,u,v)=(1,12,10,9)$
His webpage link is given Here
|
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|
Is the function $f(x)=\int_0^{\infty} \left[ 1 - a^x \sin \left( \frac{1}{a^x} \right) \right] da$ writable in a nicer way? I was idly exploring properties of this strange function exploiting wolframalpha and I can't understand it. Some calculus seems to show that this function is defined only for $x>\frac{1}{2}$ (if $x \le \frac{1}{2}$ the integral diverges) and has a minimum near $1$ (maybe in $1$?). Maybe for bigger $x$ the function has limit $1$. For $x > 1$, $f(x)$ seems takes the form $\xi(x) \Gamma \left( 1 - \frac{1}{x} \right)$ (but $\Gamma$ function is surely present even in points smaller than 1, for example $f \left( \frac{3}{5} \right) = \frac{9\sqrt{3}}{32} \Gamma \left( \frac{1}{3} \right) \approx 1,3 $) where $\xi(x)$ satisfies
$\xi \left( \frac{4}{3} \right) = \frac{2}{7} \sqrt{2 - \sqrt{2}} $
$\xi \left( \frac{3}{2} \right) =\frac{3}{10}$
$\xi \left( 2 \right) = \frac{\sqrt{2}}{3} $
$\xi \left( 3 \right) = \frac{3\sqrt{3}}{8} $
$\xi \left( 4 \right) = \frac{2}{5} \sqrt{2 + \sqrt{2}} $
$\xi \left( 5 \right) = \frac{5}{12} \sqrt{\frac{5+\sqrt{5}}{2}} $
$\xi \left( 6 \right) = \frac{3}{7} \sqrt{2+\sqrt{3}} $
... I can't see a pattern and I have no idea of how wolframalpha was able to solve these integrals (but it can't solve the general integral, it solves only if I assign a value to $x$). In addiction $f(1)$ has the simple value $\frac{\pi}{4}$, this suggests an charming way of seeing $\pi$: it is the area under this pathological function
|
The substitution $u = a^{-x}$ yields
$$ f(x) = \int \limits_0^\infty [1 - a^x \sin(a^{-x})] \, \mathrm{d} a = \frac{1}{x} \int \limits_0^\infty \frac{u - \sin(u)}{u^{2 + \frac{1}{x}}} \, \mathrm{d} u \, .$$
The integrand behaves like $u^{1 - \frac{1}{x}}$ near $u = 0$, so $x > \frac{1}{2}$ is indeed necessary (and sufficient) for the integral to converge. Now we can integrate by parts twice to obtain
$$ f(x) = \frac{1}{1+x} \int \limits_0^\infty \frac{1 - \cos(u)}{u^{1+\frac{1}{x}}} \, \mathrm{d} u = \frac{x}{1+x} \int \limits_0^\infty \frac{\sin(u)}{u^{\frac{1}{x}}} \, \mathrm{d} u \, .$$
The remaining integral has already been calculated here and we find the dependence on the gamma function you have predicted:
$$ f(x) = \frac{x}{1+x} \cos\left(\frac{\pi}{2x}\right) \operatorname{\Gamma} \left(1-\frac{1}{x}\right) \, .$$
Euler's reflection formula and the recurrence relation allow us to rewrite the result as
$$ f(x) = \frac{\pi}{2 x \sin\left(\frac{\pi}{2x}\right) \operatorname{\Gamma} \left(2+\frac{1}{x}\right)} = \frac{1}{\operatorname{sinc}\left(\frac{\pi}{2x}\right) \operatorname{\Gamma} \left(2+\frac{1}{x}\right)} \, , $$
from which $f(1) = \frac{\pi}{4}$ is obvious. We also see that $\lim_{x \to \infty} f(x) = 1$ holds. The minimum is located at $x \simeq 1.084$.
|
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|
Using a line integral to find work For the life of me I can't see where I'm going wrong with this.
Given the vector field
$$ F(x, y) = \left[ xy, \frac{2y}{x} \right] $$
a particle travels on the hyperbola
$$ \frac{x^2}{4} - y^2 = 1 $$
from $ \left(2\sqrt{10}, -3 \right) $ to $ \left(2, 0 \right) $ how much work is done by the force over the path?
My take on this is to solve for x and parametrize $t = y^2 + 1$ such that
$$ x = 2\sqrt{t} $$ $$ y = \sqrt{t-1} $$
$$ t\in\left[10,1\right]$$
for which
$$ \int_C F(r(t)) \cdot r'(t) dt = \int_{10}^1 2\sqrt{t}\sqrt{t-1}\frac{1}{\sqrt{t}} + \frac{2\sqrt{t-1}}{2\sqrt{t}}\frac{1}{2\sqrt{t-1}} $$
$$ = \left. \left(\frac{4}{3}\left(t-1\right)^{\frac{3}{2}} + \sqrt{t} \right) \right|_{10}^1 = -35 - \sqrt{10} $$
would be grateful for any input.
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Since the $y$-coordinate of the trajectory is negative, you should parametrize the path as
$$ x = 2\sqrt{t}, \>\>\> y = -\sqrt{t-1},\>\>\> t\in\left[10,1\right]$$
to allow negative values for the $y$-variable. Then, the work integral becomes
$$ \int_C F(r(t)) \cdot r'(t) dt = \int_{10}^1 \left[2\sqrt{t}(-\sqrt{t-1})\frac{1}{\sqrt{t}} + \frac{2\sqrt{t-1}}{2\sqrt{t}}\frac{1}{2\sqrt{t-1}}\right]dt $$
$$ = \left. \left(-\frac{4}{3}\left(t-1\right)^{\frac{3}{2}} + \sqrt{t} \right) \right|_{10}^1 = 37 - \sqrt{10} $$
|
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|
Uniform convergence of $\sum\limits_{k = 1}^{\infty} \frac{\sin (\sqrt{x}/k)}{\sqrt{x^2 + k^2}}$ I need to check uniform convergence of $\sum\limits_{k = 1}^{\infty} \frac{\sin (\sqrt{x}/k)}{\sqrt{x^2 + k^2}}, x \in [0, +\infty)$. I've tried Weierstrass Test, but it didn't work.
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The convergence is not uniform for $x \in [0,\infty)$.
With $x_n = \frac{\pi^2n^2}{4}$ and $n < k \leqslant 2n$ we have
$$\frac{\pi}{4} = \frac{\pi n}{2} \frac{1}{2n} \leqslant \frac{\sqrt{x_n}}{k} < \frac{\pi n}{2} \frac{1}{n}=\frac{\pi}{2}$$
Thus, for all $n \in \mathbb{N}$
$$\left|\sum_{k=n+1}^{2n}\frac{\sin \frac{\sqrt{x_n}}{k}}{\sqrt{x_n^2 + k^2}} \right| =\sum_{k=n+1}^{2n}\frac{\sin \frac{\sqrt{x_n}}{k}}{\sqrt{x_n^2 + k^2}} \geqslant n \cdot\frac{ \frac{1}{\sqrt{2}}}{\sqrt{\frac{\pi^2}{4}n^2 + 4n^2}}\\ = \frac{1}{\sqrt{\frac{\pi^2}{2} + 8}}$$
This violates the Cauchy criterion for uniform convergence which requires that for any $\epsilon > 0$ there exists a positive integer $N$ such that for all $m > n > N$ and for all $x \in [0,\infty)$ we have
$$\left|\sum_{k=n+1}^{m}\frac{\sin \frac{\sqrt{x}}{k}}{\sqrt{x^2 + k^2}} \right| < \epsilon$$
|
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|
The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$
$\sqrt{4^0+\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\cdots}}}}=?$
I found this problem in a book. I tried to solve this but couldn't. Using calculator, I found the value close to $2$. But how can this problem be solved with proper procedure?
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Take $f(x,n)=x+2^n$. We can see that;
$$\begin{aligned}
f(x,n) &= \sqrt{2^{2n}+x\left(x+2^{n+1}\right)} \\
&= \sqrt{2^{2n}+xf(x,n+1)} \\
&= \sqrt{2^{2n}+x\sqrt{2^{2\left(n+1\right)}+x\sqrt{2^{2\left(n+2\right)}+x\sqrt{...}}}}\\
&=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\\
\end{aligned}$$
Taking $x=1,n=0$; we get;
$$2=\sqrt{4^{0}+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}$$
|
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|
Evaluate $ \int_0^\frac{\pi}{2}\frac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta. $
Evaluate
$$
\int_0^\frac{\pi}{2}\frac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta.\qquad\text{(1)}
$$
By letting $t=\tan\theta$, $(1)$ equals
$$
\int_0^\infty\frac{1}{(1+t)(1+t^2)}
\,dt,$$
and then?
By letting $t=\tan\frac{\theta}{2}$, (1) equals
$$
\int_0^1\frac{(1-t^2)\cdot2}{(1+t^2)\cdot(1-t^2+2t)}\,dt,
$$
and then?
All need to much effort.
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Use symmetry instead. Notice under the change of variable $\theta \mapsto \frac{\pi}{2}-\theta$
$$I = \int_0^{\frac{\pi}{2}} \frac{\sin\theta}{\cos\theta + \sin\theta}d\theta$$
by the trig identity $\cos\left(\frac{\pi}{2}-\theta\right) = \sin\theta$ and vice-versa. Then add the two integrals:
$$2I = \int_0^{\frac{\pi}{2}} \frac{\cos\theta}{\cos\theta + \sin\theta}d\theta + \int_0^{\frac{\pi}{2}} \frac{\sin\theta}{\cos\theta + \sin\theta}d\theta = \int_0^{\frac{\pi}{2}}d\theta = \frac{\pi}{2}$$
Therefore $I = \frac{\pi}{4}$
|
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|
Solve $x^3+x+57\equiv 0$ mod $125$
Solve $x^3+x+57\equiv 0$ mod $125$
So first I try to solve $x^3+x+2\equiv 0$ mod $5$
I have $f^\prime(x)=3x^2+1$
I found $f(4)$ to be the only solution.
And since $f^\prime(4)\equiv 4$ there is a unique solution mod $25$
So I then need to solve $f(4+t5)=f(4)+t\cdot 5\cdot f^\prime(4)\equiv 0$ mod $25$
$f(4)\equiv 20$ mod $25$ and $f^\prime(4)\equiv 24$ mod $25$
I'm not sure where to go from here,
I have that $20+t\cdot 5 \cdot 24\equiv 0$ mod $25$ and I want to solve for $t$
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$x^3+x+57\equiv 0\pmod {125}\Rightarrow x^3+x-68\equiv 0\pmod {125}\Rightarrow (x-4)(x^2+4x+17)\equiv 0\pmod {125}.$
How did I get $x-4?$ I simply used the fact that $x-4$ is a factor of the polynomial $x^3+x-68.$
Since the quadratic $x^2+4x+17$ has no solutions $\pmod {125},$ the only solution is $x=4\pmod {125}.$ You didn't need to start with $\pmod 5.$
|
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|
Where is the error in this apparent contradiction arising from the function $f(\pm n)=\left(\frac{n}{n+1}\right)^{\pm 1}$? We define a function $ f \left( \pm n \right) =\left( \frac{ n }{ n + 1 } \right) ^ { \pm 1 } $.
That is, $$ f \left( m \right) = \begin{cases}\frac{ m }{ m + 1 } &\text{ if } m \ge 0 \\ \frac{ -m + 1 }{ -m } & \text{ if }m \lt 0 \end{cases}. $$
From this, $ f \left( - a \right) = \frac{ a + 1 }{ a } = \frac{ a }{ a } + \frac{ 1 }{ a }= 1 + \frac{ 1 }{ a } $
However, we can also do this:
$ f \left( \frac{ \alpha }{ \beta } > 0 \right) = \frac{ \alpha }{ \beta } \div \left( \frac{ \alpha }{ \beta } + 1 \right) = \frac{ \alpha }{ \beta } \div \left( \frac{ \alpha }{ \beta } + \frac{ \beta }{ \beta } \right) = \\ = \frac{ \alpha }{ \beta } \div \frac{ \alpha + \beta }{ \beta } = \frac{ \alpha \beta }{ \alpha \beta + \beta ^ 2 } $
Since $ \frac{ \alpha \beta + \beta ^ 2 }{ \alpha \beta } = 1 + \frac{ \beta ^ 2 }{ \alpha \beta } = 1 + \frac{ \beta }{ \alpha } = \frac{ \alpha + \beta }{ \alpha } $, $ \frac{ \alpha \beta }{ \alpha \beta + \beta ^ 2 } = \frac{ \alpha }{ \alpha + \beta } $.
Since $ f \left( k \right) = \frac{ 1 }{ f \left( - k \right)} $, $ f \left( - \frac{ \gamma }{ \delta } \right) = \frac{ \gamma }{ \gamma + \delta } $.
But this conflicts with the previous definition!
It says that $f \left( -2 \right) = \frac{ 1 }{ 2 }$ AND $= 1 + \frac{ 1 }{ 2 }$!
Where is the problem?
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The following is wrong: $f\left(-\frac{\gamma}{\delta}\right)=\frac{\gamma}{\gamma+\delta}$. In fact:
$$\frac{\gamma}{\gamma+\delta}=f\left(\frac{\gamma}{\delta}\right)=\frac{1}{f\left(-\frac{\gamma}{\delta}\right)}$$
hence
$$f\left(-\frac{\gamma}{\delta}\right)=\frac{\gamma+\delta}{\gamma}.$$
So $$f(-2)=f\left(-\frac{2}{1}\right)=\frac{3}{2}=1+\frac{1}{2}.$$
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|
Find all matrices $A\in \mathbb{R}^{2\times2}$ such that $A^2=\bf{0}$ Attempt:
Let's consider $A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$.
$$\begin{align}
A^2 &=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\cdot
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \\&=
\begin{bmatrix}
a\cdot a+b\cdot c & a\cdot b + b\cdot d \\
c\cdot a + d\cdot c & c\cdot b + d\cdot d
\end{bmatrix}\\ &= \bf{0}
\end{align}$$
This gives us the system of equations:
$$\begin{align}
a\cdot a+b\cdot c &= 0 \tag{1}\\
a\cdot b + b\cdot d = b\cdot(a+d)&= 0 \tag{2}\\
c\cdot a + d\cdot c = c\cdot(a+d)&= 0 \tag{3}\\
c\cdot b + d\cdot d &= 0 \tag{4}
\end{align}$$
Now from equation $(2)$ and $(3)$, we have eight cases:
*
*$b = 0$
*$c = 0$
*$a+d = 0$
and 5 combinations of (1,2,3) which I won't bother listing.
Case 1 ($b=0$):
$b=0$ implies $a = 0$ in equation $(1)$ and $d = 0$ in equation $(4)$. This means that if $A = \begin{bmatrix}0&0\\c&0\end{bmatrix}$ then $A^2=\bf{0}$.
Case 2 ($c=0$):
From symmetry with $b$, $c=0 \implies A=\begin{bmatrix}0&b\\0&0\end{bmatrix}$.
So, we only consider cases where $b\neq0$ and $c\neq 0 $ which leaves us only with case 3 ($a+d=0$).
Case 3 ($a+d=0$):
In equation (1), $a+d=0 \implies a\cdot d - b\cdot c = 0$. So $A=\begin{bmatrix}a&b\\c&-a\end{bmatrix}$ is not invertible, $A^2 = \bf{0}$.
In summary, if $A$ has one of the following forms:
$$\begin{bmatrix}0&b\\0&0\end{bmatrix},
\begin{bmatrix}0&0\\c&0\end{bmatrix},
\begin{bmatrix}a&b\\c&-a\end{bmatrix} \text{ (and not invertible) }$$
then $A^2=\bf{0}$.
Questions:
*
*Is this a correct proof?
*What is the standard proof?
"Strange question":
*
*How can I know if there was only the 8 cases? As in, how do I know that only these 8 cases are relevant to $A^2 = \bf{0}$?
|
If $A^2=0$, then $A$ is singular. Hence $Au=0$ for some nonzero vector $u$. Pick any vector $v$ that is linearly independent of $u$. Then $Av=au+bv$ for some scalars $a$ and $b$. It follows from $A^2=0$ that
$$
0=A^2v=A(Av)=A(au+bv)=aAu+bAv=bAv=b(au+bv)=abu+b^2v.
$$
Thus $b$ must be zero. Hence $Av=au$ and $Au=0$. Conversely, if $Av=au$ and $Au=0$, it is straightforward to verify that $A^2$ maps both $u$ and $v$ to zero. Thus $A^2=0$.
In short, when $A$ is $2\times2$, $A^2=0$ iff there is some basis $\{u,v\}$ such that $Au=0$ and $Av=au$ for some scalar $a$. In terms of matrices, if $P$ denotes the augmented matrix $[u,v]$, we see that $A^2=0$ iff
$$
A=P\pmatrix{0&a\\ 0&0}P^{-1}
$$
for some nonsingular matrix $P$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that the number of different ways to write an integer X as a sum of positive integers is $2^z$, if the order matters. As an example for number 5, it is equal to $2^4$.
We can prove it by brute-forcing it, since it's a small number, as follows:
$1+1+1+1+1 $
$1+1+1+2 $
$1+1+2+1 $
$1+2+1+1 $
$2+1+1+1$
$1+1+3 $
$1+3+1$
$3+1+1$
$2+2+1$
$2+1+2$
$1+2+2$
$2+3$
$3+2$
$1+4$
$4+1$
$5$
What is a faster, more elegant way to do this?
|
Write down $1+1+\cdots+1=X$. Now group runs of adjacent ones by going from left to right and for each $+$ sign you encounter deciding whether or not a group boundary occurs there. Parenthesize each group to get a partition of $X$ into a sum of nonnegative integers. Each possible choice of group boundaries corresponds to a distinct partition if order matters (prove this!) and there are $X-1$ decision points, therefore there are $2^{X-1}$ such partitions of $X$.
|
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|
how to find $\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$ how to find
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$$ without L'hopital or taylor/Laurent series
I tried but did not get any answer:
$$\frac{\tan^{2}\left(x\right)\tan\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\sin\left(x\right)\cos^{2}\left(x\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\left(\sin\left(x\right)\left(1-\sin^{2}\left(x\right)\right)\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2}{1}\frac{-3\left(\sin(x)-\sin^{2}\left(x\right)\right)+\sin^{3}\left(x\right)}{\left(1-\sin^{2}\left(x\right)\right)(\cos(x))\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}$$
|
$$\begin{aligned}\frac{\tan^3 x-3\tan x}{\cos\left(x+\frac{\pi}{6}\right)}
&=\tan(x)\cdot\frac{\tan^2 x-3}{\cos x\cos\left(\frac{\pi}{6}\right)-\sin x\sin\left(\frac {\pi}{6}\right)},\quad\quad x\neq \frac{\pi}{3}
\\
&=\frac{2\sin x}{\cos^{3} x}\cdot\frac{\sin x^{2}-3\cos x^{2}}{\sqrt{3}\cos x-\sin x}\quad\quad\text{(1)}
\\
&=\frac{2\sin x}{\cos^{3} x}\cdot\frac{\left(\sin x^{2}-3\cos x^{2}\right)\left(\sqrt{3}\cos x+\sin x\right)}{\left(\sqrt{3}\cos x-\sin x\right)\left(\sqrt{3}\cos x+\sin x\right)}
\\
&=-\frac{2\sin x}{\cos^{3} x}\cdot\frac{4\cos^{2} x-1}{4\cos^{2} x-1}\cdot\left(\sqrt{3}\cos x+\sin x\right)\quad\quad\text{(2)}
\end{aligned}$$
Equation $(1)$ uses $\tan x=\frac{\sin x}{\cos x}$ and $(2)$ uses $\sin^2 x=1-\cos^2 x$. Hence, by substitution, the limit is $-24$.
|
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|
Prove $3|z-1|^2 = |z+1|^2 \iff |z-2|^2 = 3$ I'm struggling with this prove question. I tried starting with let $z = x + iy$, and substituting the three in the first eqn with $|z-2|^2$, with the intention of eventually equating the LHS with the RHS, and I ended up with something most likely incorrect.
Any help would be greatly appreciated, the question is as follows:
Prove $$ 3|z-1|^2 = |z+1|^2 \quad \iff \quad|z-2|^2 = 3 $$
|
Generalization:
Let $$|z-a|=b,z=a+b\cos t+ib\sin t$$ where $a,b,t$ are real
$$\left|\dfrac{z-c}{z+c}\right|=\sqrt{\dfrac{(a-c+b\cos t)^2+b^2\sin^2t}{(a+c+b\cos t)^2+b^2\sin^2t}}=\sqrt{\dfrac{(a-c)^2+b^2+2b(a-c)\cos t}{(a+c)^2+b^2+2b(a+c)\cos t}}$$
which will be $$\left|\dfrac{a-c}{a+c}\right|$$ if
$$\dfrac{2b(a-c)}{2b(a+c)}=\dfrac{(a-c)^2+b^2}{(a+c)^2+b^2}$$
$$\iff\dfrac{a-c}{a+c}=\dfrac{a^2+b^2+c^2-2ca}{a^2+b^2+c^2+2ca}$$
$$\iff\dfrac ac=\dfrac{a^2+b^2+c^2}{2ca}\iff a^2=b^2+c^2$$
Here $a=2, b=\sqrt3, c=1$
|
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|
Evaluate $\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2 t} dt$ \begin{align}
\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt
\end{align}
I am trying to evaluate above integral. The results is $2\pi$ according to Mathematica. I want to obtain this result by integrating properly
Can this integral be evaluated using simple trigometric identities?
Do I have to use complex analysis i.e., $\cos(\theta) = \frac{z+\frac{1}{z}}{2}$ and do residue calculus?
|
Rewrite,
\begin{align}
I=\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt =
2\int_0^{\pi} \frac{2-\cos t}{3\cos^2 t -4\cos t +2} dt
\end{align}
and then substitute $\cos t = \frac{1-x^2}{1+x^2}$, along with $dt = \frac{2dx}{1+x^2}$,
$$I= 4\int_0^{\infty} \frac{1+3x^2}{9x^4 -2x^2 +1} dx $$
$$= 4\int_0^{\infty} \frac{\frac{1}{x^2}+3}{9x^2 +\frac{1}{x^2}-2} dx $$
$$=4\int_0^{\infty} \frac{d\left(3x-\frac{1}{x}\right)} {\left(3x-\frac{1}{x}\right)^2+4}
$$
$$=2\tan^{-1}\left[\frac12\left(3x-\frac{1}{x}\right)\right]_0^{\infty}=2\pi$$
|
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|
Prove $\frac{1}{1^{5}\cosh(\frac{\pi}{2})}-\frac{1}{3^{5}\cosh(\frac{3\pi}{2})}+\frac{1}{5^{5}\cosh(\frac{5\pi}{2})}+\cdots=\frac{\pi^{5}}{768}$ This is an identity from Ramanujan's letter, I am just curious. How do you prove this. My math level knowledge is still very basic so a simplified proof is preferred:
$$\frac{1}{1^{5}\cosh(\frac{\pi}{2})}-\frac{1}{3^{5}\cosh(\frac{3\pi}{2})}+\frac{1}{5^{5}\cosh(\frac{5\pi}{2})}+\cdots=\frac{\pi^{5}}{768}$$
|
This sum can be found using the partial fraction expansion of secant, with a similar approach to the one found here.
Beginning with the formula
$$ \frac{\pi}{4} \frac{1}{\cosh \left(\tfrac{\pi}{2}x \right)} = \sum_{k=0}^{\infty} \frac{(-1)^k (2k+1)}{(2k+1)^2+x^2} $$
valid for all real $x$ (see this post)
set $x=2n+1$, divide through by $(2n+1)^5$, and sum both sides to get
$$ \frac{\pi}{4} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5 \cosh \left(\frac{(2n+1)\pi}{2} \right)} = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^n (-1)^k (2k+1)}{(2n+1)^5 ((2k+1)^2+(2n+1)^2)} $$
The fraction inside the double sum may be rewritten as
$$ \frac{(-1)^n (-1)^k}{(2n+1)^5(2k+1)} - \frac{(-1)^n (-1)^k}{(2n+1)^3(2k+1)^3}+\frac{(-1)^n(-1)^k}{(2n+1)(2k+1)^5}-\frac{(-1)^n(-1)^k(2n+1)}{(2k+1)^5((2n+1)^2+(2k+1)^2)} $$
Notice that the rightmost fraction is the same as the one we started with if we switch the dummy variables. Therefore we have
$$ \hspace{-10cm} 2\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^n (-1)^k (2k+1)}{(2n+1)^5 ((2k+1)^2+(2n+1)^2)} $$
$$ \hspace{5cm} = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \left(\frac{(-1)^n (-1)^k}{(2n+1)^5(2k+1)} - \frac{(-1)^n (-1)^k}{(2n+1)^3(2k+1)^3}+\frac{(-1)^n(-1)^k}{(2n+1)(2k+1)^5} \right) $$
Using the Dirichlet beta function values
$$ \beta(1) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4} $$
$$ \beta(3) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{32} $$
$$ \beta(5) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5} = \frac{5\pi^5}{1536} $$
we find that
$$ \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^n (-1)^k (2k+1)}{(2n+1)^5 ((2k+1)^2+(2n+1)^2)} = \beta(1)\beta(5)-\frac{1}{2}\beta^2(3) = \frac{\pi^6}{3072} $$ and so $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5 \cosh \left(\frac{(2n+1)\pi}{2} \right)} = \frac{\pi^5}{768} $$
|
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|
Help Understanding Proof for Inequality Problem #3 from Problem Solving Strategies The problem is the following:
Prove for $a$, $b$, $c$, $d$ that
$$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$
I understand the proof saying
$${\frac{abc+abd+acd+bcd}{4}}=\frac{(ab)(c+d)+(cd)(a+b)}{4}$$
Apply AM-GM to $ab$ and $cd$ to yield
$$\frac{\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)}{4}=\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{a+b+c+d}{4}$$
However it then states the following:
$$\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{a+b+c+d}{4}=\left(\frac{a+b+c+d}{4}\right)^3$$
I don't understand this step in the proof. Unless I'm missing something basic, the above expression does not factor as shown. It doesn't adduce a theorem to justify the equivalence either so I don't know how it was deduced.
|
Assuming $a,b,c,d\geq 0$ Maclaurin's inequality gives
$$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}\leq \frac{a+b+c+d}{4} $$
and by the AM-QM (i.e. Cauchy-Schwarz) inequality we have
$$ \frac{a+b+c+d}{4}\leq\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$$
|
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|
Evaluate $\displaystyle\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}$ I have been trying to evaluate
\begin{equation*}
\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}.
\end{equation*}
We have
\begin{equation*}
\frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x} = \frac{x^2 + x}{\ln(1 + x + x^2) - x} \cdot \frac{1}{\sqrt{x^2 + x + 4} + 2} \quad \text{for all}\ x \in \mathbb{R},
\end{equation*}
so I think my exercise boils down to
\begin{equation*}
\lim_{x \to 0^{-}} \frac{x^2 + x}{\ln(1 + x + x^2) - x},
\end{equation*}
and this limit equals $-\infty$ by De L'Hôpital's Theorem.
Can I evaluate this limit without De L'Hôpital's Theorem?
|
Define $f(y) = \ln(1+y)$. We know that, $f'(0) = 1$. Thus $$\lim_{y\to 0} \frac{\ln(1+y)}{y} = \lim_{y\to 0} \frac{f(y)-f(0)}{y-0} = f'(0) = 1 \hspace{1cm}(1)$$
Now, consider the limit $$\lim_{x\to 0^{-}}\frac{\ln(1+x+x^{2})-x}{x^{2}+x} \hspace{1cm} (2)$$ Note that $x^{2}+x = (x+\frac{1}{2})^{2}-\frac{1}{4}$. Take $u = x+\frac{1}{2}$. If $x \to 0^{-}$ than $u \to \frac{1}{2}^{-}$. Now, (2) becomes $$ \lim_{u\to \frac{1}{2}^{-}}\frac{\ln(1+u^{2}-\frac{1}{4})-\bigg{(}u-\frac{1}{2}\bigg{)}}{u^{2}-\frac{1}{4}} = \lim_{u\to \frac{1}{2}^{-}}\frac{\ln(1+(u+1/2)(u-1/2))-\bigg{(}u-\frac{1}{2}\bigg{)}}{(u+1/2)(u-1/2)}$$
We break into two limits. The second one is $$ \lim_{u\to \frac{1}{2}^{-}}\frac{-\bigg{(}u-\frac{1}{2}\bigg{)}}{\bigg{(}u+\frac{1}{2}\bigg{)}\bigg{(}u-\frac{1}{2}\bigg{)}} = -1$$
The first one is $$\lim_{u\to \frac{1}{2}^{-}} \frac{\ln(1+(u+1/2)(1-1/2))}{(u+1/2)(u-1/2)}$$
is equivalent to the limit given by (1). Thus, the limit (2) goes to zero from the left.
|
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|
Parabola equation in form of quadratic $ax^2+bx+c=y$ where $a+b+c$ is an integer Suppose a parabola has vertex $(\frac{1}{4},\frac{-9}{8})$ and equation $ax^2+bx+c=y$ where $a>0$ and $a+b+c $ is an integer. Find the minimum possible value of $a$ under the given condition.
My approach
$(x-\frac{1}{4})^2=4a'(y+\frac{9}{8})$
$\frac{x^2-\frac{x}{2}+\frac{1}{16}}{4a'}-\frac{9}{8}=y$
$\frac{1}{4a'}-\frac{1}{4a'}+\frac{1}{64a'}-\frac{9}{8}=Z$, where $Z$ is an integer
$\frac{9}{64a'}-\frac{9}{8}=Z$, from here I am confused
|
The equation of the parabola is:
$$a \left(x-\frac{1}{4}\right)^2 - \frac{9}{8}$$
$$=a \left(x^2-\frac{1}{2}x+\frac{1}{16} \right)- \frac{9}{8}$$
$$=ax^2-\frac{a}{2}x+ \left(\frac{a}{16}-\frac{9}{8} \right)$$
Then we find that: $$a+b+c = \frac{9}{16}(a-2).$$
When $a=0$, $a+b+c = -\frac{9}{8}$. Since the gradient of the line is positive, if we try $a+b+c=-1$, we will get a positive value of $a$.
Therefore we have that: $-1 = \frac{9}{16}(a-2) \Rightarrow a = \frac{2}{9}$.
|
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|
Solve $x_{n+1}=Ax_n+\frac{B}{x_n^5}+\frac{C}{x_n^9}$
For $$x_{n+1}=Ax_n+\frac{B}{x_n^5}+\frac{C}{x_n^9}$$
a. Find $A,B,C$ that will give an optimal approximation (Highest order) for $\sqrt{2}$
b. Solve the system using LU decomposition, what is the given order?
So we are looking to approximate the function $f(x)=x^2-2$
therefore we will demand that $x=\sqrt{2}$ is a fixed point of the iterative method $g(x)=Ax+\frac{B}{x^5}+\frac{C}{x^9}$ i.e:
$$\sqrt{2}=g(\sqrt{2})=\sqrt{2}A+\frac{B}{(\sqrt{2})^5}+\frac{C}{(\sqrt{2})^9}$$
$$\sqrt{2}=\sqrt{2}A+\frac{B}{4\sqrt{2}}+\frac{C}{16\sqrt{2}}$$
Therefore:
$$a)\sqrt{2}=\sqrt{2}(A+\frac{B}{4\cdot 2}+\frac{C}{16\cdot 2})\rightarrow 1=A+\frac{B}{8}+\frac{C}{32}$$
Now we will demand that $g'(\sqrt{2})=0$
$$0=g'(\sqrt{2})=A-\frac{5x^4B}{x^{10}}-\frac{9x^8C}{x^{18}}\Rightarrow b. 0=A-\frac{5B}{8}-\frac{9C}{32}$$
can we next demand that $g''(\sqrt{2})=0$? (the $x^2-2$ second derivative is non zero?)
If so we get:
$$0=g(\sqrt{2})=\frac{30B}{x^7}+\frac{90C}{x^11}\Rightarrow 0=\frac{30B}{8\sqrt{2}}+\frac{90C}{32\sqrt{2}}\Rightarrow c) 0=\frac{15B}{4\sqrt{2}}+\frac{45C}{16\sqrt{2}}$$
Now we have $3$ equations so we can find $A,B,C$:
a)-b):
$$1=\frac{6B}{8}+\frac{10C}{32}$$
From c) we get $$-B=\frac{3}{4}C$$
Plugin it back we get $C=-4, B=3, A=\frac{3}{4}$
Is this answer correct? How to use LU decomposition here if there is no matrices?
|
Your answer is just fine... Your linear system is
$$
\begin{cases}
g(\sqrt 2) = \sqrt 2\\
g'(\sqrt 2)=0\\
g''(\sqrt 2)=0
\end{cases}
$$
or, in terms of $A,B,C$:
$$
\begin{bmatrix}
\sqrt 2 & \frac{\sqrt 2}{8} & \frac{\sqrt 2}{32}\\
1 & -\frac 58 & -\frac{9}{32}\\
0 & \frac{15 \sqrt 2}{8} &\frac{45 \sqrt 2}{32}
\end{bmatrix}\begin{bmatrix}A \\ B \\ C\end{bmatrix}=\begin{bmatrix}\frac{\sqrt 2}{2} \\ 0 \\0\end{bmatrix}
$$
Now, you just need to solve this system using the LU decomposition. In the end you should obtain the values of $A, B, C$ you already have...
Edit:
The requirement for $g'(\sqrt 2) = g''(\sqrt 2)=0$ comes from the fact that a convergent fixed point iteration of the form $x_{n+1}=g(x_n)$ converges with at least order $p$ when $g'(z)=\cdots = g^{(p-1)}(z)=0$. This comes from Taylor's formula:
$$
g(x_n) = g(z) + g'(z)(x_n-z) + \cdots \frac{1}{(p-1)!}g^{(p-1)}(z)(x_n-z)^{p-1} + \frac{1}{p!} g^{(p)}(\xi)(x_n-z)^p
$$
If you substitute $g'(z)=\cdots = g^{(p-1)}(z)=0$ and note that $g(x_n)=x_{n+1}$, you have that
$$
x_{n+1} - z = \frac{1}{p!} g^{(p)}(\xi)(x_n-z)^p
$$
from where you can deduce that
$$
\lim \frac{|x_{n+1}-z|}{|x_n-z|^p} = \frac{g^{(p)}(z)}{p!}.
$$
|
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|
Find the values of a, b, c, and d from a matrix equation Here is the equation I'm working on:
\begin{align}
\begin{bmatrix}2a+b&a-2b\\5c-d&4c+3d\end{bmatrix} = \begin{bmatrix}4&-3\\11&24\end{bmatrix}\\
\end{align}
My solutions:
Row 1 column 1
\begin{align}
2a + b = 4\\
2a = 4 - b\\
a = 2 - \frac 12b\\
\end{align}
Row 1 column 2
\begin{align}
a - 2b = -3\\
2 - \frac 12b - 2b = -3\\
2 - \frac 52b = -3\\
2\left(-\frac 52b \right) = 2(-5)\\
-5b = -10\\
b = 2
\end{align}
Row 2 column 1
\begin{align}
5c - d = 11\\
5c = 11 + d\\
c = \frac {11}{5} + \frac {1}{5}d\\
\end{align}
Row 2 column 2
\begin{align}
4c + 3d = 24\\
4\left(\frac {11}{5} + \frac {1}{5}d \right) + 3d = 24\\
\frac {44}{5} + \frac {4}{5}d + 3d = 24\\
\frac {44}{5} + \frac {19}{5}d = 24\\
\frac {19}{5}d = 24 - \frac {44}{5}\\
\frac {19}{5}d = \frac {76}{5}\\
19d = 5\left(\frac {76}{5} \right)\\
d = \frac {76}{19}\\
d = 4\\
\end{align}
Based on b value which is 2, I've found the a value:
\begin{align}
2a + b = 4\\
2a + 2 = 4\\
2a = 2\\
a = 1\\
\end{align}
Likewise, found the c value based on d value which is 4:
\begin{align}
5c - 11d = 11\\
5c - 4 = 11\\
5c = 15\\
c = 3\\
\end{align}
Thus,
\begin{align}
a = 1, b = 2, c = 3, d = 4\\
\end{align}
I'm new to algebra and matrices. My question is: Is this the right approach to solving the problem?
|
Yes your approach is correct. As an alternative you could consider the $4$ by $5$ augmented matrix for the system in the unknowns $a$,$b$,$c$ and $d$
$$\left[\begin{array}{cccc|c}
2&1&0&0&4 \\
1&0&0&-2&-3 \\
0&0&5&-1&11 \\
0&0&4&3&24
\end{array}\right]$$
and solve by gaussian elimination.
|
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|
Problem regarding inconsistent solution
My attempt :
For First answer i try to used by elementary row operation $ A= \begin{bmatrix} 1&2&3\\2&4&8 \\3&6&7 \end{bmatrix}$ that give $ A= \begin{bmatrix} 1&2&3\\0&0&2 \\0&0&0 \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3 \end{bmatrix}=\begin{bmatrix} 1 \\0\\5\end{bmatrix}$
so i got $ x_1 + 2x_2 + 3x_3=1$ , $2x_2=0$, $x_3= 5$
Now i got $ x_1=-4, x_2=0 ,x_3 =5$
For the second answer it is very easy take $b= (0,0,0)$ that will produce no solution
Is its true ?
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That way to proceed is not correct, indeed we need to apply the RREF on the augmented matrix
$$\left[\begin{array}{ccc|c} 1&2&3&1\\2&4&8&0 \\3&6&7&5 \end{array}\right]$$
For the second question let consider
$$\left[\begin{array}{ccc|c} 1&2&3&b_1\\2&4&8&b_2 \\3&6&7&b_3 \end{array}\right]$$
and apply RREF again to obtain
$$\left[\begin{array}{ccc|c} 1&2&3&b_1\\2&4&8&b_2 \\3&6&7&b_3 \end{array}\right]\to \left[\begin{array}{ccc|c} 1&2&3&b_1\\0&0&2&b_2-2b_1 \\0&0&-2&b_3-3b_1 \end{array}\right]\to \left[\begin{array}{ccc|c} 1&2&3&b_1\\0&0&2&b_2-2b_1 \\0&0&0&b_2+b_3-5b_1 \end{array}\right]$$
then a vector which leads to an inconsisten system is for example $(1,1,1)$.
|
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|
For all real $x$ the expression$ \frac{x^2-2x+4}{x^2+2x+4}$ lies between $\frac 13$ and $3$.
The values between which the expression $$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$ is
A solution I saw says that the expression can be written as
$$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2.(3.3^x)+4}$$ which is analogous to the previous expression and hence would have the same interval. Why are they analogous? Clearly it has $x$ as a power, so something has to change right? Why are they the same?
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$$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$
$$\frac{(3)^2(3^x)^2+2(3)(3^x)+4}{(3)^2(3^x)^2-2(3)(3^x)+4}$$
$$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2(3.3^x)+4}$$
$$\frac{(z)^2+2(z)+4}{(z)^2-2(z)+4}$$
|
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Solve this differential equation $(x^2+1)y' = xy$ $$\begin{align}
(x^2+1)\frac{dy}{dx} &= xy \\
\frac{1}{y}\frac{dy}{dx} &= \frac{x}{(x^2+1)} \\
\int \frac{1}{y} dy &= \int \frac{x}{(x^2+1)} dx \\
\ln(y) &= \frac{1}{2\ln(x^2+1)} + C \\
y &= e^{\frac{1}{2\ln(x^2+1)} + C}
\end{align}$$
When I integrated $\frac{x}{x^2+1}$, I tried to use u substitution by setting $u = x^2+1$ and $ du = \frac{1}{2}x$.
The answer says $y = K\sqrt{x^2+1}$ and after combing through my work, I am not sure where I went wrong except possibly during the integration part and need some help, thanks
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Your hunch is right. Indeed there is a mistake at integration step:
$$\int \frac{x}{(x^2+1)} dx =\dfrac{1}{2}\ln|x^2+1|+C$$
Might help memorizing:
$$\int \dfrac{f'(x)}{f(x)}dx = \ln |f(x)|+C$$
|
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|
Problem with equation in complex numbers
I am supposed to calculate:
$x^{2}=5+i$
I used formula:
$\left | \cos \frac{\alpha }{2} \right |=\sqrt{\frac{1+\cos \alpha }{2}}$
$\left | \sin \frac{\alpha }{2} \right |=\sqrt{\frac{1-\cos \alpha }{2}}$
and I came to the point where:
x= $\sqrt[4]{26}\left ( \frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}\sqrt[4]{26}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}\sqrt[4]{26}} \right)$= $\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
I know that k=0,1 and also that $\sin x, \cos x $ are positive in the first quadrant, so my solution will simple be :
$x_{0}=\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
$x_{1}=-\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
It does not seem like the correct solution. Can anyone please tell me, where I made the mistake?
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You can also write $$5+i=a^2-b^2+2abi$$ so we will get $$5=a^2-b^2$$ and $$1=2ab$$
Plugging $$b=\frac{1}{2a}$$ in the first equation you will get
$$20a^2=4a^4-1$$
|
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|
Sums and Sequences I want some help, please, to solve this problem.
Let $$\forall n\in\mathbb{N}^*,\ S_n=\sum_{k=1}^n\frac{1}{\sqrt{k}}$$
I want to prove that $\forall n\in\mathbb{N}^*,\ S_n\leqslant\sqrt{n-1}+\sqrt{n}$
and that $\forall n\in\mathbb{N}^*,\ S_n\geqslant2\sqrt{n+1}-2$
I used the Mathematical Induction method, so for $n=1$ we have : $S_1=1$ and $\sqrt{1-1}+\sqrt{1}=1$
Then the statement in this case is satisfied. Next we assume another real non-null number $k/\ S_k\leqslant\sqrt{k-1}+\sqrt{k}$ , and we prove that $S_{k+1}\leqslant\sqrt{k}+\sqrt{k+1}$ , but I'm stuck.
So please help me if you can.
Thanks!
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We have
$$S_{k+1}=S_k+\frac1{\sqrt{k+1}}\stackrel{?}\le \sqrt{k}+\sqrt{k+1}$$
which is true indeed
$$S_k+\frac1{\sqrt{k+1}}\le \sqrt{k-1}+\sqrt{k}+\frac1{\sqrt{k+1}}\le \sqrt{k}+\sqrt{k+1}$$
since
$$\sqrt{k-1}+\sqrt{k}+\frac1{\sqrt{k+1}}\le \sqrt{k}+\sqrt{k+1}\iff \sqrt{k-1}+\frac1{\sqrt{k+1}}\le \sqrt{k+1}$$
$$\sqrt{k^2-1}+1\le k+1 \iff \sqrt{k^2-1}\le k \iff k^2-1 \le k^2 \iff -1\le 0$$
Edit
We also need to prove that
$$S_{k+1}=S_k+\frac1{\sqrt{k+1}}\stackrel{?}\ge 2\sqrt{k+2}-2$$
which is true indeed
$$S_k+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+1}-2+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+2}-2$$
since
$$2\sqrt{k+1}-2+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+2}-2\iff 2\sqrt{k+1}+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+2}$$
$$2(k+1)+1\ge 2\sqrt{(k+2)(k+1)} \iff 2k+3 \ge 2\sqrt{(k+2)(k+1)}$$
$$4k^2+12k+9 \ge 4k^2+12k+8 \iff 1\ge 0$$
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k.
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30?
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You can show divisiblity by $30$ by showing divisibilty by $2,3$ and $5$
So show $21\times5^{2k+3} - 5\times3^{2k+3} + 4\times30M$ is divisible by $2,3$ and by $5$
So
$21\times5^{2k+3} - 5\times3^{2k+3} + 4\times30M=$
$15(7\times 5^{2k+2} - 3^{2k+2} + 4\times 2M)$.
Obviously $15$ is divisible by $3$ and by $5$.
So just remains to show $ 7\times 5^{2k+2} - 3^{2k+2} + 4\times 2M$ is divisible by $2$ (or in other words even).
=======
But it's easier without induction.
$5|5^{2n+1}$ and so need to show $5|3^{2n+1}+ 2^{3n+1}$.
This is the "freshman dream". but if you know modular arithmetic.
$3^{2n+1} \equiv (-2)^{2n+1} \equiv -2^{2n+1} \pmod 5$ so $3^{2n+1}+2^{2n+1} \equiv 0 \pmod 5$.
But if you dont know modular arithmetic: $3^{2n+1}+ 2^{2n+1} = (3+2)(3^{2n}- 2*3^{2n-1} + ...... - 2^{2n-1}*3 +2^{2n})$ and $3+2 = 5$.
...
Likewise $3|3^{2n+1}$ so we need to show $3|5^{2n+1}-2^{2n+1}$.
That the same thing $5 \equiv 2 \pmod 3$ so $5^{2n+1}\equiv 2^{2n+1}\pmod 3$ so $3|5^{2n+1} -2^{2n+1}$.
Or $5^{2n+1} -2^{2n+1}= (5-2)(5^{2n} + 5^{2n-1}*2 + .... + 5*2^{2n-1}+ 2^{2n})$. And $5-2=3$.
....
And $2|2^{2n+1}$ so we have to show $2|5^{2n+1} - 3^{2n+1}$ which we can do the exact same ways as above, or we can not that ODD minus ODD is even.
So $2,3,5$ each divide $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ so $30$ does as well.
|
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Problem of Apollonius: Find the radius of the circle tangent to $3$ other circles $O_1$, $O_2$ and $O_3$ have radius of $a$, $b$ and $c$ Find the radius of the circle tangent to $3$ other circles $O_1$, $O_2$ and $O_3$ have radius of $a$, $b$ and $c$
The Wikipedia page about the Problem of Apollonius can be found here
I can't provide you an exact image because the construction is too complex. What have I done is draws tons of lines to attempt to use Pythagoras's theorem and trig to solve it.
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If you can endure the algebraic slog (a computer algebra system really helps), you can find some structure that simplifies the solution.
Let the given circles have radii $r_A$, $r_B$, $r_C$, and (non-collinear) centers $A:=(x_A,y_A)$, $B:=(x_B,y_B)$, $C:=(x_C,y_C)$; I specifically used
$$A = (0,0) \qquad B = (c,0) \qquad C = (b\cos A, b \sin A)$$
As Wikipedia's Problem of Apollonius entry notes, there are eight solutions that depend on whether each circle's tangency with the target circle is external or internal. The entry's coordinate discussion accommodates this by attaching sign variables to the radii; here, we can just assert that the radii $r_A$, $r_B$, $r_C$ are signed values.
Let the target circle have center $P:=(x_P,y_P)$ and radius $r$. We can express $P$ in barycentric coordinates:
$$P = \frac{\alpha A+\beta B+\gamma C}{\alpha+\beta+\gamma} \tag{1}$$
Now, we'll follow Wikipedia's lead and consider the equations
$$\begin{align}
(x_P-x_A)^2 + (y_P-y_A)^2=(r-r_A)^2 \\
(x_P-x_B)^2 + (y_P-y_B)^2=(r-r_B)^2 \\
(x_P-x_C)^2 + (y_P-y_C)^2=(r-r_C)^2
\end{align} \tag{2}$$
Upon subtracting the latter two equations from the first, we eliminate troublesome quadratic terms, leaving a linear homogeneous system of two equations in the three unknowns $\alpha$, $\beta$, $\gamma$. The reader can verify that we can write the solution as
$$\begin{align}
\alpha &= a^2 b c \cos A - a^2 r_A (r_A - 2 r) + a b \cos C\,r_B (r_B - 2 r) + c a \cos B\,r_C (r_C - 2 r) \\[4pt]
\beta &= a b^2 c \cos B - b^2 r_B (r_B - 2 r) + b c \cos A\,r_C (r_C - 2 r) + a b \cos C\,r_A (r_A - 2 r) \\[4pt]
\gamma &= a b c^2 \cos C - c^2 r_C (r_C - 2 r) + c a \cos B\,r_A (r_A - 2 r) + b c \cos A\,r_B (r_B - 2 r)
\end{align} \tag{3}$$
where $a$, $b$, $c$ are the sides of $\triangle ABC$. Conveniently, $\alpha+\beta+\gamma = 16 |\triangle ABC|^2$, by Heron's Formula.
Substituting $P$ with these $\alpha$, $\beta$, $\gamma$ values back into the first equation of $(1)$, we get this quadratic in $r$:
$$\begin{align}
0 &= 4 r^2 \left(\;-4|\triangle ABC|^2
+ \sum_{cyc} \left(a^2 r_A^2 - 2 b c \cos A\,r_B r_C \right) \;\right) \\
&-4r \left(\;
\sum_{cyc}\left( a^2 r_A^3 - b c \cos A (a^2 r_A + r_B^2 r_C + r_B r_C^2) \right)
\;\right) \\[4pt]
&+a^2 b^2 c^2 + \sum_{cyc} \left( a^2 r_A^4 - 2 b c \cos A (a^2 r_A^2 + r_B^2 r_C^2)\right)
\end{align} \tag{8}$$
Somewhat surprisingly, the discriminant simplifies nicely:
$$\triangle = 64 |\triangle ABC|^2 \left(a^2 - (r_B - r_C)^2\right)\left( b^2 - (r_C - r_A)^2\right) \left(c^2 - (r_A - r_B)^2\right) \tag{9}$$
The last three factors invite defining, say, $d$, $e$, $f$ such that
$$d^2 = a^2 - (r_B - r_C)^2 \qquad e^2 = b^2 - (r_C - r_A)^2 \qquad f^2 = c^2 - (r_A - r_B)^2 \tag{10}$$
Since the differences could be negative, we allow that $d$, $e$, $f$ could be imaginary; but $d^2e^2f^2$ must be non-negative (so, $def$ is real). As it turns out, it's helpful to define (possibly-imaginary) "angles" $D$, $E$, $F$ with
$$\cos D = \frac{-d^2+e^2+f^2}{2e f} \qquad \cos E = \frac{-e^2+f^2+d^2}{2fd} \qquad \cos F = \frac{-f^2+d^2+e^2}{2de} \tag{11}$$
and (possibly-imaginary) "area" $|\triangle DEF|$ with
$$|\triangle DEF|^2 = \frac1{16}(d+e+f)(-d+e+f)(d-e+f)(d+e-f) \tag{12}$$
How helpful? Well, helpful enough that $(8)$ shrinks to
$$\begin{align}
0 &= 16 r^2 |\triangle DEF|^2 - 4r d e f ( d r_A\cos D + er_B \cos E + fr_C \cos F) \\[4pt]
&-\left(d^2 e^2 f^2 + d^4 r_A^2 + e^4 r_B^2 + f^4 r_C^2 - 2 e^2 f^2 r_B r_C - 2 f^2 d^2 r_C r_A - 2 d^2 e^2 r_A r_B \right)
\end{align} \tag{8'}$$
(If we defined possibly imaginary $s_A$, $s_B$, $s_C$ with $s_A^2 = d^2 r_A$, etc, the constant term could be written $(s_A+s_B+s_C)(-s_A+s_B+s_C)(s_A-s_B+s_C)(s_A+s_B-s_C)-d^2 e^2 f^2$, but that isn't particularly helpful going forward.)
Solving $(8')$ yields
$$r = \frac{def}{8|\triangle DEF|^2} \left(\;
dr_A \cos D
+ e r_B \cos E
+ f r_C \cos F
\pm 2|\triangle ABC|
\;\right) \tag{13}$$
Note that, if we were to multiply-through the $def$ factor, and expand the cosines via $(11)$, the first three terms would only involve even powers $d$, $e$, $f$, making for a real (possibly-negative) value. We know that $def$ is real, so the $\pm 2def\,|\triangle ABC|$ term is also real. This guarantees that $r$ itself is real.
Since the signs of $r_A$, $r_B$, $r_C$ account for the eight possible tangency configurations, presumably one choice of "$\pm$" is always extraneous. Can we decide which one ahead of time? Maybe, but I'm running out of steam ...
Substituting $(13)$ into a $def$ version of $(7)$, and dividing-through by a common factor, yields a final-ish form for parameter $\alpha$:
$$\begin{align}
\alpha = d
\left(
\begin{array}{l}
\phantom{+} \cos D
\left(
4 |\triangle DEF|^2 + \sum_{cyc} \left( d^2 r_A^2 - 2 e f r_B r_C \cos D \right)
\right) \\
\pm 2 |\triangle ABC|\; (d r_A - e r_B \cos F - f r_C \cos E)
\end{array}
\right)
\end{align} \tag{14}$$
If we define $u := d r_A$, $v := e r_B$, $w := f r_C$, we reduce a bit of clutter in $(13)$ and $(14)$:
$$\begin{align}
r &= \frac{def}{8|\triangle DEF|^2}
\left(\;u \cos D + v \cos E + w \cos F\pm 2|\triangle ABC|\;\right) \\
\alpha &= d
\left(
\begin{array}{l}
\phantom{+} \cos D
\left(
4 |\triangle DEF|^2 + u^2 + v^2 + w^2 - 2 v w \cos D - 2 w u \cos E - 2 u v \cos F
\right) \\[6pt]
\pm 2 |\triangle ABC|\; (u - v \cos F - w \cos E)
\end{array}
\right)
\end{align} \tag{15}$$
As a final note, I'll mention that the values $d$, $e$, $f$ have geometric significance related to the power of certain points with respect to the given circles. It's a little tricky to describe, and I don't (yet) see how it might contribute to the discussion, so I won't bother including it.
|
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|
$ \int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx$ I am tring to obtain following formula [actually I obtained this resut via mathematica]
\begin{align}
\int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx
= \frac{\sqrt{2} \sinh(x)}{a^2 \sqrt{a^2-b^2 + (a^2+b^2) \cosh(2x)}} + C
\end{align}
Since the answer contains $\cosh(2x)$
My first trial was
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x) = a^2 \cosh(2x) + (b^2 -a^2) \cosh^2(x)
\end{align}
Seems not good..
My second trial is
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x) = (a^2+b^2) \sinh^2(x) + b^2
\end{align}
and this also does not good for integrand....
How one can obtain this integral?
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Ket us use $$\int \frac{dt}{(m^2+t^2)^{1]2}}=\ln (t+\sqrt{m^2+t^2}]+C_1~~~)1)$$
D. w.r.t $a$ we get $$\int \frac{dt}{(m^2+t^2)^{3/2}}=\frac{t}{m^2\sqrt{m^2+t^2}}+C_2~~~~(2) $$
Then $$I= \int \frac{\cosh x}{[(a^2+b^2) \sinh^2 x+b^2]^{3/2}} dx= \frac{1}{(a^2+b^2)^{3/2}}\int \frac{dt} {(m^2+t^2)^{3/2}}, m=b/\sqrt{a^2+b^2}.$$
Now (2) can be used.
|
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|
Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
|
$2a^2+b+2=2a^2+3-a=(\frac{a^2+1}{2}-a)+(\frac{3a^2+5}{2})\geq\frac{3a^2+5}{2}$.
On the other hand by QM-AM $\sqrt{\frac{a^2+a^2+a^2+1+1+1+1+1}{8}}\geq\frac{3a+5}{8}$, therefore $\sqrt{3a^2+5}\geq\frac{3a+5}{\sqrt{8}}$. Now we return back: $\sqrt{2a^2+b+2}\geq\sqrt{\frac{3a^2+5}{2}}\geq\frac{3a+5}{\sqrt{2\times8}}=\frac{3a+5}{4}$.
Therefore $\sqrt{2a^2+b+2}+\sqrt{2b^2+a+2}\geq\frac{(3a+5)+(3b+5)}{4}=\frac{3(a+b)+10}{4}=\frac{16}{4}=4$
|
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Limit with radicals, $\cos$, $\ln$ and powers $\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(\frac{x-1}{x})^{\frac{1}{x}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}\Big)}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}=0$
My answer is rather imprecise:$$\underset{x\rightarrow +\infty}\lim{\cos{\frac{1}{x^3}}}=1\implies\underset{x\rightarrow +\infty}\lim{\sqrt[6]{1-\cos{\frac{1}{x^3}}}}=0$$
$$\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}}=0?$$
I am aware of the mistake I have made by writting $0$ for an undefined term.
$$\underset{x\rightarrow +\infty}\lim{\Big(1-\frac{1}{x^3}\Big)}=1\implies \ln{\Big(1-\frac{1}{x}\Big)}<0\implies\underset{x\rightarrow +\infty}\lim{\Bigg(\frac{1}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}\Bigg)=-\infty}$$
The limit of the denumerator is $0$ $\&$ the limit of the whole expression is $0$.
How can I prove this concisely?
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Examine the expression you have,
$$\underset{x\rightarrow \infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}\Big)}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}$$
Note that in the limit $x\rightarrow \infty$,
$$ \sqrt[6]{1-\cos\frac 1{x^3}} \rightarrow
\sqrt[6]{1-(1-\frac12 \frac 1{x^6})}
=\frac1{\sqrt[6]2}\frac1x$$
$$\frac{3^{\frac1x}-2^{\frac1x}}{6^{\frac1x}}
\rightarrow (1+\ln3 \frac1x)-( 1+\ln2 \frac1x)= \ln\frac32\frac1x$$
$$\frac1x \ln(1-\frac1x) \rightarrow - \frac1{x^2}$$
Thus, the limiting value is
$$\lim_{x\rightarrow \infty}\frac {\left( \frac1{\sqrt[6]2}\frac1x\right) \left( \ln\frac32\frac1x\right)}{-\frac1{x^2}}
=- \frac1{\sqrt[6]2}\ln\frac32$$
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"url": "https://math.stackexchange.com/questions/3448285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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|
How to find the limit of this function using L'Hospital's rule?
$$\lim_{x\rightarrow 0} \,\,\left( \sqrt[3]{1+2x+x^3} - \frac{2x}{2x+3} \right) ^ {\frac1{x^3}} $$
I have already tried several options, but the only answer I have gotten so far is
$e^{\infty}$, which is incorrect. The correct answer is $e^\frac{43}{81}$, which is easy to get by using Taylor series, but our task was to get the same one by using L'Hospital's rule. Can you help me with it?
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The standard approach here is to take logarithm and then evaluate the limit. If $f(x) $ denotes the expression under limit and it tends to a limit $L$ then
\begin{align}
\log L&=\lim_{x\to 0}\frac{1}{x^3}\log\left(\sqrt[3]{1+2x+x^3}-\frac{2x}{2x+3}\right)\notag\\
&=\lim_{x\to 0}\frac{g(x)}{x^3}\cdot\frac{\log(1+g(x))}{g(x)}\notag\\
&=\lim_{x\to 0} \frac{g(x)} {x^3}\notag\\
&=\lim_{x\to 0}\frac{1}{x^3}\left(\sqrt[3]{1+2x+x^3}-1-\frac{2x}{2x+3}\right)\notag
\end{align}
The algebraic limit at the end can be evaluated using standard techniques. Here is one approach which requires some algebra. If $$A=\sqrt[3]{1+2x+x^3},B=\frac{4x+3}{2x+3}$$ then $A,B$ both tend to $1$ and $$A^3-B^3=x^3+\frac{(1+2x)(2x+3)^3-(4x+3)^3}{(2x+3)^3}=x^3+\frac{16x^3(1+x)} {(2x+3)^3}$$ so that $(A^3-B^3)/x^3\to 43/27$. And therefore $$\frac{A-B} {x^3}=\frac{A^3-B^3}{x^3}\cdot\frac{1}{A^2+AB+B^2}\to \frac{43}{81}$$ Thus the desired limit is $L=e^{43/81}$.
Advanced tools like L'Hospital's Rule and Taylor series are often unnecessary for simple limit problems and instead a little bit of algebra helps.
Incidentally this exercise helps us to know that $(3+2x)/(3+x)$ is a very good approximation to $\sqrt[3]{1+x}$ for small values of $x$. For example if $x=0.1$ then we have $$\sqrt[3]{1.1}=1.032280\dots,\frac{3.2}{3.1}=1.032258\dots$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/3448772",
"timestamp": "2023-03-29T00:00:00",
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Manipulation of Taylor expansion of $e^x$ I have an expression:
$$ f(x) = \sum\limits_{N=0}^\infty (N-x)^2 \frac{x^N}{N!}$$
I want to figure out whose Taylor Expansion this is.
I've found that I can separate out the above expression, and get the following:
$$f(x) = \sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} - 2 \sum\limits_{N=0}^\infty xN \frac{x^N}{N!} + \sum\limits_{N=0}^\infty x^2 \frac{x^N}{N!}$$
I figured out this much:
$$x^2 e^x = \sum\limits_{N=0}^\infty x^2 \frac{x^N}{N!} = \sum\limits_{N=0}^\infty xN \frac{x^N}{N!}$$
$$xe^x = \sum\limits_{N=0}^\infty N \frac{x^N}{N!}$$
But I cannot figure out who the first mysterious sum belongs to!
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Hint:
$$(x-n)^2=x^2-2nx+n^2=x^2-2nx+n(n-1)+n$$
so that
$$\sum_{n=0}^\infty(x-n)^2\frac{x^n}{n!}\\
=\sum_{n=0}^\infty x^2\frac{x^n}{n!}-2\sum_{n=0}^\infty nx\frac{x^n}{n!}+\sum_{n=0}^\infty n(n-1)\frac{x^n}{n!}+\sum_{n=0}^\infty n\frac{x^n}{n!}\\
=x^2\sum_{n=0}^\infty \frac{x^{n}}{n!}-2x\sum_{n=1}^\infty \frac{x^{n}}{(n-1)!}+\sum_{n=2}^\infty\frac{x^n}{(n-2)!}+\sum_{n=1}^\infty\frac{x^n}{(n-1)!}\\
=x^2\sum_{n=0}^\infty \frac{x^{n}}{n!}-2x\sum_{n=0}^\infty \frac{x^{n+1}}{n!}+\sum_{n=0}^\infty\frac{x^{n+2}}{n!}+\sum_{n=0}^\infty\frac{x^{n+1}}{n!}.$$
$$xe^x.$$
The key to the solution is to decompose $n^2$ in terms of $n$ and $n(n-1)$ and to simplify with the factorials at the denominators. This generalizes to any polynomial in $n$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3449582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\left(\dfrac{B}{2}\right)=\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{C}{2}\right)$$
$$2\sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$$
$$2\sqrt{\dfrac{(s-a)(s-c)(s-b)}{s(s-b)^2}}=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-a)^2}}+\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-c)^2}}$$
$$\dfrac{2}{s-b}=\dfrac{1}{s-a}+\dfrac{1}{s-c}$$
$$\dfrac{2}{s-b}=\dfrac{s-c+s-a}{(s-a)(s-c)}$$
$$\dfrac{2}{s-b}=\dfrac{b}{(s-a)(s-c)}$$
$$2\left(\dfrac{a+b+c}{2}-a\right)\left(\dfrac{a+b+c}{2}-c\right)=b\left(\dfrac{a+b+c}{2}-b\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$b^2-a^2-c^2+2ac=ba+bc-b^2$$
$$2b^2-a^2-c^2+2ac-ba-bc=0\tag{1}$$
$$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$$
$$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$$
$$\cos A+\cos C=\dfrac{ab^2+ac^2-a^3+a^2c+b^2c-c^3}{2abc}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac^2-a^3+a^2c-c^3}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac(a+c)-(a+c)(a^2+c^2-ac)}{b}}{2ac}$$
Using equation $(1)$, $2ac-a^2-c^2=ba+bc-2b^2$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{(a+c)(ba+bc-2b^2)}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+(a+c)(a+c-2b)}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+a^2+c^2+2ac-2ba-2bc}{2ac}$$
$$\cos A+\cos C=\dfrac{a^2+c^2+2ac-ab-bc}{2ac}$$
Using equation $(1)$, $2ac-ba-bc=a^2+c^2-2b^2$
$$\cos A+\cos C=\dfrac{a^2+c^2+a^2+c^2-2b^2}{2ac}$$
$$\cos A+\cos C=\dfrac{2a^2+2c^2-2b^2}{2ac}$$
$$\cos A+\cos C=2\cdot\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A+\cos C=2\cos B$$
Is there any nice way to solve this question, mine goes very long. I tried various methods but this was the only way I was able to prove the required result.
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Rewrite the equation as
$$2\frac{\sin\frac B2}{\cos\frac B2}
=\frac{\sin\frac A2}{\cos\frac A2}+\frac{\sin\frac C2}{\cos\frac C2}
=\frac{\sin\frac {A+C}2}{\cos\frac A2\cos\frac C2}$$
Then, with $A+C = \pi - B$,
$$2\sin\frac B2 \cos\frac A2 \cos\frac C2 = \cos\frac B2\sin\frac {\pi-B}2=\cos^2\frac B2$$
$$2\sin\frac B2(\cos\frac {A+C}2 + \cos\frac {A-C}2) = 1 + \cos B$$
$$2\sin^2\frac B2+ 2\sin\frac B2\cos\frac {A-C}2 = 1 + \cos B$$
$$2\cos\frac {A+C}2\cos\frac {A-C}2 = 2\cos B$$
$$\cos A + \cos C = 2\cos B$$
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"url": "https://math.stackexchange.com/questions/3450441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc} \sqrt{(a+b)^2-ab}$$
and now I want to apply Cauchy-Schwarz but it is the wrong direction.
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I’m not 100% sure but this is what I got.
$$
a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2>(a+b/2)^2
$$
From this we get
$$
a+b/2+b+c/2+c+a/2=9/2>\sqrt3
$$
Or
$$
a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2> \frac{3}{4}b^2
$$
From this we get
$$
\frac{\sqrt{3}}{2}(a+b+c)= \frac{3\sqrt{3}}{2}>\sqrt{3}
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3450882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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|
Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$ Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$.
By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.
Question from Jalil Hajimir
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If you want some calculus/analysis argument:
After establishing there must exist a global minimum, let $p$ be the global minimum. Then we must have that
$$f(x) = x^2\left((y^2+1)(z^2+1) - p\right) - 2xp(y+z) + (y^2+1)(z^2+1) - p(y+z)^2\geq 0$$
as a quadratic in $x.$ So the discriminant is non-positive:
$$D =4\left[p^2(y+z)^2 - (y^2+1)^2(z^2+1)^2 - p^2(y+z)^2+(y^2+1)(z^2+1)p(1+(y+z)^2)\right]\leq 0\iff $$
$$p\leq\min\dfrac{(y^2+1)(z^2+1)}{1+(y+z)^2}.$$
But
$$4(y^2+1)(z^2+1) - 3 - 3(y+z)^2 = 4y^2z^2+y^2+z^2-6yz+1 = (y-z)^2+(2yz-1)^2\geq 0.$$ So $p = \dfrac{3}{4}$ by continuity argument and it is achieved by $y = z = \dfrac{1}{\sqrt{2}},$ which in return easily tells us that that $x$ is also $\dfrac{1}{\sqrt{2}}$ for the minimum to be attained.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3454095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\int \frac{dx}{x^3+x^2\sqrt{x^2-1}-x}$ solve:
$$\int \frac{\mathrm dx}{x^3+x^2\sqrt{x^2-1}-x}$$
I tried:
$$\begin{align}\int \frac{\mathrm dx}{x(x^2-1)+x^2\sqrt{x^2-1}}&=\int \frac{\mathrm dx}{x(\sqrt{x^2-1}\sqrt{x^2-1})+x^2\sqrt{x^2-1}}\\&=\int \frac{\mathrm dx}{x\sqrt{x^2-1}(\sqrt{x^2-1}+x)}\end{align}$$
$x=\sin t$
$$\int \frac{\mathrm dt}{\cos t\sin t-\sin^2t}$$
And I can not continue from here.
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Hint: Put $x＝\dfrac{1}{\cos t}$, then
$$\int \frac{\cos^3 t}{\sin t(\sin t+1)}\frac{\sin t}{\cos^2 t}dt＝\int \frac{d\sin t}{\sin t+1}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3456281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Limit ${\lim\limits_{x \to \frac{\pi}{2}}\Big(\tan^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}\Big)\Big)$ $\displaystyle\lim_{x \to \frac{\pi}{2}}\left(\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)$
I wanted to use L' Hopital's rule so I wrote the term as:
$$\frac{\sin^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)}{\cos^2{x}}$$
and found the first derivative of the denumaerator & denominator:
$\frac{{\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\sin{x}\cos{x}}{2\cos{x}(-\sin{x})}$
$$\lim_{x\to \frac{\pi}{2}}{\left(-2\left({\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)\sec{x}\csc{x}\right)}$$
But I didn't know how to continue.
I drew a graph in Geogebra:
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Hint:
Set $\sin x=s$
Rationalize the numerator
$$s^2\cdot\dfrac{\sqrt{2s^2+3s+4}-\sqrt{s^2+6s+2}}{1-s^2}$$
$$=s^2\cdot\dfrac{(s^2-3s+2)}{1-s^2}\cdot\dfrac1{\cdots}$$
As $s\to1,s\ne1$ so cancel out $s-1$ and set $s\to1$
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"url": "https://math.stackexchange.com/questions/3457123",
"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.