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Fibonacci sum for $\pi$: $\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}$ Wikipedia's "List of formulae involving $\pi$" entry states that
$$\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}.$$
Why is this true?
If $\varphi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2$, then we can rewrite this identity as
$$F(\varphi)-F(\psi)=\frac{4\pi^2}{25}\quad\text{ where }F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}.$$
$F(x)$ looks similar to the power series for arcsin.
|
In this answer, it is shown that
$$
\frac{\arcsin(x)}{\sqrt{1-x^2}}
=\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag1
$$
Integrating $(1)$ and multiplying by $4$ yields
$$
\begin{align}
2\arcsin^2(x)
&=\sum_{k=0}^\infty\frac1{(2k+1)(2k+2)}\frac{4^{k+1}}{\binom{2k}{k}}x^{2k+2}\\
&=\sum_{k=0}^\infty\frac{4^{k+1}x^{2k+2}}{(k+1)^2\binom{2k+2}{k+1}}\\
&=\sum_{k=1}^\infty\frac{4^kx^{2k}}{k^2\binom{2k}{k}}\tag2
\end{align}
$$
Applying $F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}
&=\frac2{\sqrt5}\left(\arcsin^2\left(\frac\phi2\right)-\arcsin^2\left(\frac{1-\phi}2\right)\right)\\
&=\frac2{\sqrt5}\left(\frac{9\pi^2}{100}-\frac{\pi^2}{100}\right)\\
&=\frac{4\pi^2}{25\sqrt5}\tag3
\end{align}
$$
Addendum: Trigonometric Values Used Above
Note that
$$
\begin{align}
4\cos^2(2\pi/5)+2\cos(2\pi/5)-1
&=\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)^2+\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)-1\\[6pt]
&=e^{4\pi i/5}+e^{2\pi i/5}+1+e^{-2\pi i/5}+e^{-4\pi i/5}\\
&=\frac{e^{10\pi i/5}-1}{e^{2\pi i/5}-1}\,e^{-4\pi i/5}\\
&=0\tag4
\end{align}
$$
Solving for the positive root of $(4)$ gives
$$
\begin{align}
\cos(2\pi/5)
&=\frac{-1+\sqrt{5}}4\\
&=\frac{\phi-1}2\tag5
\end{align}
$$
Since $\cos(2\pi/5)=2\cos^2(\pi/5)-1$ and $\cos(\pi/5)$ is positive, apply $\phi^2=\phi+1$ to get
$$
\cos(\pi/5)=\frac\phi2\tag6
$$
Since $\cos(x)=\sin(\pi/2-x)$, $(5)$ and $(6)$ become
$$
\begin{align}
\sin(\pi/10)&=\frac{\phi-1}2\tag7\\
\sin(3\pi/10)&=\frac\phi2\tag8
\end{align}
$$
|
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|
Contradict the uniqueness of least-squares solution
Find a least-squares solution of $Ax=b$ for
$A=\begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 \\
\end{pmatrix}
,b=\begin{pmatrix}
-3\\
-1\\
0\\
2\\
5\\
1
\end{pmatrix}$
My approach:
$$A^{T}A=\begin{pmatrix}
6 & 2 & 2 & 2 \\
2 & 2 & 0 & 0 \\
2 & 0 & 2 & 0 \\
2 & 0 & 0 & 2
\end{pmatrix}$$
$$A^{T}b=\begin{pmatrix}
4\\
-4\\
2\\
6
\end{pmatrix}$$
Seems $A^{T}A$ is not invertible. The aumented matrix for $A^{T}A\hat{x}=A^{T}b$ is $$\begin{pmatrix}
6 & 2 & 2 & 2 & 4\\
2 & 2 & 0 & 0 & -4\\
2 & 0 & 2 & 0 & 2\\
2 & 0 & 0 & 2 &6
\end{pmatrix}\sim\begin{pmatrix}
1 & 0 & 0 & 1 & 3\\
0 & 1 & 0 & -1 & -5\\
0 & 0 & 1 & -1 & -2\\
0 & 0 & 0 & 0 & 0
\end{pmatrix}$$ Hence the general least-squares solution of $Ax=b$ has the form
$$\hat{x}=\begin{pmatrix}3\\-5\\-2\\0\end{pmatrix}+t\begin{pmatrix}-1\\1\\1\\1\end{pmatrix}$$
But the orthogonal projection $\hat{x}$ is always unique then why that free variable come$?$ Eventually I can't interpret my outcome$!$ Can anyone explain this outcome Geometrically as well as why it contradict uniqueness please.
Update: So the columns of $A$ are not linearly independent hence free variable arrive but then $\color{red}{why}$ the projection of $b$ on the column space are so many$?$
|
Matrix $A$ is not full rank column (since $C_1=C_2+C_3+C_4$) and therefore we have one free parameter for the solution which is not unique.
Indeed recall that least square method corresponds to find solutons for $Ax=p$ where $p$ is the projection of $b$ onto $Col(A)$. Therefore since $A$ is not full rank column the result follows.
|
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|
What is the two-parametric solution of the biquadratic Diophantine equation? On the following biquadratic Diophantine equation, Leonard Eugene Dickson mentioned a parameter equation in his works (HOTTON Vol-II), but did not give it in detail
$$a^4+6a^2b^2+b^4=c^4+6c^2d^2+d^4$$
A. Gérardin noted that, to find two right triangles having the same sum of squares of the hypotenuse and one leg, we have to solve
$$(x^2+y^2)^2+4x^2y^2=(\alpha^2+\beta^2)^2+4\alpha^2\beta^2$$
and gave a solution in which x,y,α,β are functions of the seventh degree of two parameters.
[HOTTON Vol-II] "History Of The Theory Of Numbers Vol-II" by Leonard Eugene Dickson
Chapter 4, p188
https://archive.org/details/HistoryOfTheTheoryOfNumbersVolII/page/n214
What is the two-parametric solution of the biquadratic Diophantine equation obtained by André Gérardin?
%%%%%%%%%%
Thanks for Piezas' results
https://sites.google.com/site/tpiezas/018
\begin{align*}
\begin{cases}
a=3 t^7+7 t^6+3 t^5+19 t^4-15 t^3+37 t^2+9 t+1\\
b=t^7-9 t^6+37 t^5+15 t^4+19 t^3-3 t^2+7 t-3\\
c=-3 t^7+7 t^6-3 t^5+19 t^4+15 t^3+37 t^2-9 t+1\\
d=t^7+9 t^6+37 t^5-15 t^4+19 t^3+3 t^2+7 t+3
\end{cases}
\end{align*}
Although it can be transformed into a two parameter solution by $t=u/v$, I'm not sure that Gérardin got this result.
[157]» Sphinx-Oedipe, 5, 1910, 187.
https://books.google.co.uk/books?id=cfgSAQAAMAAJ
(I found it in Google Books, but I don't seem to be able to access the 187th page)
|
I am deeply indebted to a French friend for his help.
\begin{align*}
\begin{cases}
x=3 t^7+7 t^6+\,\,\,3 t^5+19 t^4-15 t^3+37 t^2+9 t+1\\
y=\,\,\,t^7-9 t^6+37 t^5+15 t^4+19 t^3-\,\,\,3 t^2+7 t-3\\
\alpha=\,\,\,t^7+9 t^6+37 t^5-15 t^4+19 t^3+\,\,\,3 t^2+7 t+3\\
\beta=3 t^7-7 t^6+\,\,\,3 t^5-19 t^4-15 t^3-37 t^2+9 t-1
\end{cases}
\end{align*}
(a^2 + b^2)^2 + (2 a*b)^2 - (c^2 + d^2)^2 - (2 c*d)^2 /. {
a -> 3 m^7 + 7 m^6 n + 3 m^5 n^2 + 19 m^4 n^3
- 15 m^3 n^4 +37 m^2 n^5 + 9 m*n^6 + n^7,
b -> m^7 - 9 m^6 n + 37 m^5 n^2 + 15 m^4 n^3
+ 19 m^3 n^4 - 3 m^2 n^5 + 7 m*n^6 - 3 n^7,
c -> m^7 + 9 m^6 n + 37 m^5 n^2 - 15 m^4 n^3
+ 19 m^3 n^4 + 3 m^2 n^5 + 7 m*n^6 + 3 n^7,
d -> 3 m^7 - 7 m^6 n + 3 m^5 n^2 - 19 m^4 n^3
- 15 m^3 n^4 - 37 m^2 n^5 + 9 m*n^6 - n^7} // FullSimplify
|
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|
Is there a closed form for $3\cdot\frac{3}{\sqrt{6}}\cdot\frac{3}{\sqrt{6+\sqrt{6}}}\cdot\frac{3}{\sqrt{6+\sqrt{6+\sqrt{6}}}}\cdots$? Inspired by Vieta's Formula for $\pi$,
$$\pi=2\cdot\frac{2}{\sqrt{2}}\cdot\frac{2}{\sqrt{2+\sqrt{2}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$
I became interested in a more generalized case for Vieta's Formula.
For $m=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}}$, what is the closed form of
$$m\cdot\frac{m}{\sqrt{a}}\cdot\frac{m}{\sqrt{a+\sqrt{a}}}\cdot\frac{m}{\sqrt{a+\sqrt{a+\sqrt{a}}}}\cdots$$?
To find the value of $m$, we can solve the following equation
$$m=\sqrt{a+m}$$,
which gives
$$m=\frac{1+\sqrt{1+4a}}{2}$$
One example is that
$$3=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}}$$
And I want to find the exact value of
$$3\cdot\frac{3}{\sqrt{6}}\cdot\frac{3}{\sqrt{6+\sqrt{6}}}\cdot\frac{3}{\sqrt{6+\sqrt{6+\sqrt{6}}}}\cdots\approx3.815$$
but to no avail. Could anyone provide any insight on the possible closed form for this infinite product?
|
Just a comment, as the comment section is really getting long, you may really want to know the connection why it is $2$ in the radical that gives this amazing result. It is due to it's connection to trigonometric functions. More specifically, a quick research into this can give that a result, (link)
$$\frac{\sin x}{x}=\cos \frac{x}{2}\cdot\cos \frac{x}{4}\cdot\cos \frac{x}{8}...$$
And also the fact that
$$\cos\frac{x}{2^n}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+...(n-1) times}}}$$
or more specifically speaking;
$$A=\frac{\sin A}{\cos\frac{A}{2}\cdot \cos\frac{A}{4}\cdot \cos\frac{A}{8}...}$$
So a generalized version of this would be doubtful. And also one can obviously make a conjecture that that your stated problem will result in a transendental number which couldn't be expressed in terms of closed form algebraic numbers or functions.
A very good paper, if you're really interested in square roots will be this, by Dixon J. Jones
|
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|
Solve system of equations $\frac{x}{y}+\frac{y}{x}=\frac{10}{3}$, $x^2-y^2=8$
Solve the system:
$$\begin{array}{l}\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{10}{3} \\
x^2-y^2=8\end{array}$$
First, we have $x \ne 0$ and $y \ne 0$. We can rewrite the first equation as $$\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}$$
What should I do next?
|
Denote: $\frac xy=a$. Then from the first equation:
$$a+\frac1a=\frac{10}{3} \Rightarrow 3a^2-10a+1=0 \Rightarrow a=\frac13,3 \Rightarrow \\
x=3y,y=3x$$
Plug these into the second equation:
$$(3y)^2-y^2=8\Rightarrow y=\pm 1 \Rightarrow x=\pm 3\\
x^2-(3x)^2=8 \Rightarrow x^2=-1 \Rightarrow \emptyset.$$
|
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|
Calculate the sum: $C_{2n}^n+2C_{2n-1}^n+4C_{2n-2}^n+...+2^nC_n^n$ Calculate this sum:$$C_{2n}^n+2C_{2n-1}^n+4C_{2n-2}^n+...+2^nC_n^n.$$
What I tried:
$$ C^n_{2n}=\frac{(2n)!}{(n!)^2}$$
$$ 2C^n_{2n-1}=\frac{2(2n-1)!}{n!(n-1)!}=\frac{2n(2n)!}{n!n!(2n)}=\frac{(2n)!}{(n!)^2}$$
$$ 4C^n_{2n-2}=\frac{4(2n-2)!}{n!(n-2)!}=\frac{4(2n)!(n)(n-1)}{(2n)(2n-1)(n!)^2}=\frac{2(2n)!(n-1)}{(2n-1)(n!)^2}$$
$$ 2^nC_n^n=2^n$$
So our original sum is equal to this:
$$C_{2n}^n\left( 1+1+\frac{2(n-1)}{(2n-1)}+\frac{2^2(n-1)(n-2)}{(2n-1)(2n-2)}+...+\frac{2^n}{C_{2n}^n} \right)$$
$$=C_{2n}^n\sum_{k=1}^n \frac{2^k(n-1)!(2n-k)!}{(n-k)!(2n-1)!}$$
$$=C_{2n}^n\sum_{k=1}^n \frac{2n}{n-1}=C_{2n}^n2n\sum_{k=1}^n\frac{1}{n-1}=C_{2n}^n \left( \frac{2n(n-1)}{n-1} \right) = 2nC_{2n}^n$$
So... how do I go from here? Also, I'm not sure if the last 2 steps are correct.
Edit: I was starting the sum at $k=0$, fixed it to $k=1$, and then changed the rest after that.
Edit: I found the answer on a book, it's $2^{2n}$, so my answer is wrong... Still, I don't know what I did wrong, or how to correctly solve this.
|
Let
\begin{align}
f(n)
&=\sum_{k=0}^n2^k\binom{2n-k}{n}\\
&=\sum_{h=0}^n2^{n-h}\frac{(n+h)!}{n!h!}
\end{align}
Then
\begin{align}
f(n+1)
&=\sum_{h=0}^{n+1}2^{n+1-h}\frac{(n+1+h)!}{(n+1)!h!}\\
&=\sum_{h=0}^{n+1}2\frac{n+1+h}{n+1}2^{n-h}\frac{(n+h)!}{n!h!}\\
&=2\sum_{h=0}^{n+1}2^{n-h}\frac{(n+h)!}{n!h!}+2\sum_{h=0}^{n+1}\frac{h}{n+1}2^{n-h}\frac{(n+h)!}{n!h!}\\
&=2f(n)+\frac{(2n+1)!}{n!(n+1)!}+2\sum_{h=1}^{n+1}2^{n-h}\frac{(n+h)!}{(n+1)!(h-1)!}\\
&=2f(n)+\frac{(2n+1)!}{n!(n+1)!}+\frac 12\sum_{h=1}^{n+1}2^{(n+1)-(h-1)}\frac{((n+1)+(h-1))!}{(n+1)!(h-1)!}\\
&=2f(n)+\frac{(2n+1)!}{n!(n+1)!}+\frac 12\sum_{u=0}^{n}2^{(n+1)-u}\frac{((n+1)+u)!}{(n+1)!u!}\\
&=2f(n)+\frac{(2n+1)!}{n!(n+1)!}+\frac 12f(n+1)-\frac 12\frac{(2n+2)!}{(n+1)!(n+1)!}\\
\end{align}
from which
\begin{align}
\frac 12f(n+1)
&=2f(n)+\frac{(2n+1)!}{n!(n+1)!}-\frac 12\frac{(2n+2)!}{(n+1)!(n+1)!}\\
&=2f(n)+\frac{(2n+1)!}{n!(n+1)!}\left(1-\frac 12\frac{2n+2}{n+1}\right)\\
&=2f(n)
\end{align}
from which $f(n+1)=4f(n)$, hence $f(n)=4^n$.
|
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|
Calculate $ f [f (x)] $ Where $ f (x) = \frac {1} {1 + x ^ 2} $, $ f [f (x)] $ is equal to:
a
I was factoring and found $f(f(x))= \frac{1}{1+(\frac{1}{1+x^2})^2} = \frac{1}{\frac{x^4+2x^2+2}{x^4+2x^2+1}}= \frac{x^4+2x^2+1}{x^4+2x^2+2}$, but the answer is x. I am wrong?
|
Your calculation of
$$f(f(x))=\frac{x^4+2x^2+1}{x^4+2x^2+2}$$
is correct. Both the numerator and denominator are fourth degree polynomials with a coefficient of $1$ before the $x^4$ term. Therefore, there is a horizontal asymptote at $y=1$.
So, the answer cannot be $f(f(x))=x$. As the graph of $f(f(x))$ is bounded above by $y=1$, it cannot be equal to the value of $x$ when $x>1$.
|
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|
Value of $x$ in floor Inequality: $\lfloor \sin^{-1}(x)\rfloor >\lfloor \cos^{-1}(x)\rfloor$
Find value of $x$ for which $\displaystyle \lfloor \sin^{-1}(x)\rfloor >\lfloor \cos^{-1}(x)\rfloor,$ is
What I try $\displaystyle \sin^{-1}(x)\in \bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$ and $\displaystyle \cos^{-1}(x)\in [0,\pi]$
$\displaystyle \lfloor \sin^{-1}(x)\rfloor \in \{-2,-1,0,1,2\}$ and $\lfloor \cos^{-1}(x)\rfloor \in \{0,1,2,3\}$
$\bullet\; $ If $\lfloor \sin^{-1}(x)\rfloor =2\Rightarrow 2\leq \sin^{-1}(x)<3\Rightarrow \sin 2<x<\sin 3$
Then $\lfloor \cos^{-1}(x)\rfloor =0\Rightarrow 0\leq \cos^{-1}(x)<1\Rightarrow \cos 1<x\leq 1$
$\bullet\ \bullet $ If $\lfloor \sin^{-1}(x)\rfloor =2\Rightarrow 2\leq \sin^{-1}(x)<3\Rightarrow \sin 2<x<\sin 3$
Then $\lfloor \cos^{-1}(x)\rfloor =1\Rightarrow 1\leq \cos^{-1}(x)<2\Rightarrow \cos 2<x\leq \cos 1$
$\bullet\;\bullet\bullet $ If $\lfloor \sin^{-1}(x)\rfloor =1\Rightarrow 1\leq \sin^{-1}(x)<2\Rightarrow \sin 1<x<\sin 2$
Then $\lfloor \cos^{-1}(x)\rfloor =0\Rightarrow 0\leq \cos^{-1}(x)<1\Rightarrow \cos 1<x\leq 1$
How do I find common solution Help me please
|
$$\lfloor\arcsin x\rfloor>\lfloor\arccos x\rfloor\iff\arcsin x\ge\lfloor\arccos x\rfloor+1.$$
As the arc sine cannot exceed $\dfrac\pi2$ and $\arccos x\ge0$, we have the only case
$$\lfloor\arccos x\rfloor=0\to x>\cos1\land \arcsin x\ge 1\implies x\ge\sin 1.$$
Graphically:
|
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|
Trapezoid Volume and Surface Area
Given $AB = 6$, $DC = 12$, $CG = 8$. Find the surface area and volume of the object.
I tried solving for h first. By inspection, you can see that the height is equal to the side of the 45°.
Using $h=(6-h)(\tan 60°)$. I got $h=9-3 \sqrt{3}$ then I plug it in to solve the volume.
$$V = \left[\frac12(12+6)(9-3 \sqrt{3)})\right]\times 8$$
$$V = 273.877$$
Then for the surface area, I just added the areas of each faces.
$$SA = 2(9*(9-\sqrt{3}) + 2(8(9-\sqrt{3}) + 18 + 96 $$
$$SA = 298.756$$
I am unsure whether this is correct
|
The volume is completely correct. Since this is a prism, its volume is simply the area of trapezium $ABCD$ (or $EFGH$) multiplied by $CG$. So the volume is $\frac 12(12+6)(9-3\sqrt 3) \times 8 \approx 273.88$
For the area, just sum the areas of the $4$ rectangles and $2$ (identical) trapeziums.
$ABEF$ has area $8 \times 6 =48$
$DCGH$ area = $8\times 12 = 96$
Trapeziums $ABCD$ and $EFGH$ are congruent, each with area $\frac 12(12+6)(9-3\sqrt 3)$, so total area of these two is $18(9-3\sqrt 3)$
Finally, "slanted" rectangles $BCGF$ and $ADHE$
$BC = \frac h{\sin 60^{\circ}} = 6(\sqrt 3-1)$
Area $BCGF = 8 \times BC = 48(\sqrt 3-1)$
Similarly, $AD = h\sqrt 2 = 9\sqrt 2 - 3\sqrt 6$
Area $ADHE = 8 \times (9\sqrt 2 - 3\sqrt 6) = 72\sqrt 2 - 24\sqrt 6$
Adding it all up, total surface area $\approx 290.64$
|
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|
Proof by mathematical induction proof verification If $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
The base case where $n=1$ is clearly true: $\frac{1}{2!}=\frac{1}{2}$ and $1-\frac{1}{(1+1)!}=\frac{1}{2}$.
Now to show that $S_k\Rightarrow S_{k+1}$, we assume that $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!}=1-\frac{1}{(k+1)!}$ and observe the following equalitites:
$\begin{align*}\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k+1}{((k+1)+1)!}&=\\
(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!})+\Big(\frac{(k+1)}{((k+1)+1)!}\Big)&=\\
1-\frac{1}{(k+1)!}+\frac{k+1}{((k+1)+1)!}&=\frac{(k+2)(k+1)!-(k+2)+(k+1)}{(k+2)(k+1)!}\\
&=\frac{(k+2)(k+1)!-1}{(k+2)(k+1)!}\\
&=1-\frac{1}{(k+2)!}\\
&=1-\frac{1}{((k+1)+1)!}\end{align*}$
Therefore we conclude that if $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
I know this is a relatively straightforward example but I'm very new to the concept of induction and I'm wondering if I'm on the right track with this proof. It seems correct to me but I'm worried I may have overlooked something. Any advice is helpful, thanks!
|
Your proof is correct. Very well done. However, I would strongly recommend familiarising yourself with using sums as an easier way to express your work.
We are proving:
$$\sum_{r=1}^n\frac{r}{(r+1)!}=1-\frac{1}{(n+1)!} \tag A$$
1) Your base case was done well.
2) Assumption step, assume true for $n=k$
$$\sum_{r=1}^k\frac{r}{(r+1)!}=1-\frac{1}{(k+1)!}\tag B $$
3) Use to show for $n=k+1$. From $(B)$ we get:
$$\sum_{r=1}^{k+1}\frac{r}{r+1}=1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}$$
While from $(A)$ we get:
$$\sum_{r=1}^{k+1}\frac{r}{r+1}=1-\frac{1}{(k+2)!}$$
We show these are identical, which implies:
$$\frac{1}{(k+2)!}\equiv\frac{1}{(k+1)!}-\frac{k+1}{(k+2)!}$$
$$RHS\to \frac{(k+2)!-{(k+1)!}(k+1)}{(k+1)!(k+2)!}\equiv\frac{(k+2)-(k+1)}{(k+2)!}\equiv\frac{1}{(k+2)!} a.r. $$
I use "a.r." for "as required". Just a bit of flair.
You should then conclude in the manner of: "I have shown (statement) to be true for $n=1$ and true for $n=k+1$ when $n=k$ has been assumed, hence it is true for all $n\in\Bbb N$"
|
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value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
$$\lim_{x\rightarrow 0}\frac{x^{n+1}\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
How do I solve it? Help me please.
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The denominator factors as
$$(x-\sin x)\sum_{k=0}^{n-1} x^{n-k-1}\sin^kx\sim ax^3x^{n-1}$$ while the numerator is $\sim bx^{2n}$.
So $2n=n+2$.
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Prove that $R$ is reflexive, symmetric, and transitive. Define a relation $R$ on $\Bbb Z$ by declaring that $xRy$ if and only if $x^2\equiv y^2\pmod{4}$. Prove that $R$ is reflexive, symmetric, and transitive.
Suppose $x\in\Bbb Z$. Then $x^2\equiv x^2\pmod {4}$ means that $4\mid (x^2-x^2)$, so $x^2-x^2=4a$ where $a=0\in\Bbb Z$. Therefore $R$ is reflexive.
Now suppose $x^2\equiv y^2\pmod {4}$. This means $4\mid (x^2-y^2)$ and so $x^2-y^2=4a$, for some $a\in\Bbb Z$. Multiplying by $-1$ we have $-1(x^2-y^2=4a)\\\rightarrow -x^2+y^2=-4a\\\rightarrow y^2-x^2=4(-a)$
so $4\mid(y^2-x^2)$ and $y^2\equiv x^2\pmod{4}$. This shows that $R$ is symmetric.
Now we assume that $x^2\equiv y^2\pmod{4}$ and $y^2\equiv z^2\pmod{4}$. This means $4\mid(x^2-y^2)$ and $4\mid(y^2-z^2)$. Then we have $x^2-y^2=4a$ and $y^2-z^2=4b$ for some $a,b\in\Bbb Z$. Rearranging we get $x^2=4a+y^2$ and $z^2=y^2-4b$.
Then
$\begin{align*}x^2-z^2&=(4a+y^2)-(y^2-4b)\\&=4a+4b\\&=4(a+b)\end{align*}$
This shows that $4\mid(x^2-z^2)$ and $x^2\equiv z^2\pmod{4}$, therefore $R$ is transitive.
$\blacksquare$
Please forgive my rushed formatting, just wondering if my arguments here work and if this is a valid proof. Any feedback is appreciated, thanks!
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Your solution is ok. I’d just like to point out that your problem is in fact just a specific case of a much more general phenomenon:
Let $A$, $B$ be sets, $f:A\to B$ a function, and $R\subseteq B\times B$ an equivalence relation. Define a relation $S\subseteq A\times A$ such that $$(x,y)\in S\Leftrightarrow (f(x),f(y))\in R.$$ Then, $S$ is also an equivalence relation.
This is straightforward to prove:
*
*$S$ is reflexive since $f(x)Rf(x)$ implies $xSx$.
*$S$ is symmetric since, if $xSy$, then $f(x)Rf(y)$, $f(y)Rf(x)$, and $ySx$.
*$S$ is transitive since, if $xSy$ and $ySz$, then $f(x)Rf(y)$ and $f(y)Rf(z)$, so that $f(x)Rf(z)$ and $xSz$.
Your result is now immediate: take $A=B=\mathbb Z$, $R$ to be equivalence $\text{mod }4$, and $f(x)=x^2$ to be the squaring function.
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|
Evaluate $\int \frac{\cos^2x}{1+\tan x}dx$ Evaluate $\int \dfrac{\cos^2x}{1+\tan x}dx$
Here are my various unsuccessful attempts:-
Attempt $1$:
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
$$\ln(1+t)=y$$
$$\dfrac{1}{1+t}=\dfrac{dy}{dt}$$
$$\int \dfrac{dy}{\left(1+(e^y-1)^2\right)^2}$$
$$\int \dfrac{dy}{(1+e^{2y}+1-2e^y)^2}$$
$$\int \dfrac{dy}{(e^{2y}+2-2e^y)^2}$$
$$\int \dfrac{dy}{e^4y+4+4e^{2y}+4e^{2y}-4e^y(e^{2y}+2)}$$
$$\int \dfrac{dy}{e^4y-4e^{3y}+8e^{2y}-8e^y+4}$$
From here I gave up on this method.
Attempt $2$:
$$\int \dfrac{\cos^3x}{\cos x+\sin x}$$
$$\int\dfrac{3\cos x+\cos 3x}{4(\cos x+\sin x)}$$
$$\dfrac{1}{4}\left(3\cdot\int\dfrac{\cos x}{\cos x+\sin x}+\int \dfrac{\cos 3x}{\cos x+\sin x}\right)$$
Solving the first integral
$$\cos x=A(\cos x+\sin x)+B(-\sin x+\cos x)+C$$
$$A+B=1$$
$$A-B=0$$
$$A=\dfrac{1}{2},B=\dfrac{1}{2}$$
$$\dfrac{1}{4}\left(\dfrac{3}{2}\left(x+\ln|\sin x+\cos x|\right)+\int \dfrac{\cos 3x}{\cos x+\sin x}\right)$$
I was not understanding how to proceed for second integration.
Attempt $3$:
For @mvpq
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
Trying to write expression as
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A}{1+t}+\dfrac{B}{1+t^2}+\dfrac{C}{(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A(1+t^2)^2+B(1+t^2)+C(1+t)}{(1+t)(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=(A+B+C)+Ct+(B+2A)t^2+At^4$$
$$A+B+C=1\tag{1}$$
$$A=0$$
$$A+2B=0$$
$$B=0$$
$$C=0$$
But $A+B+C=1$, hence no solution, so cannot be solved by partial fractions.
Attempt $4$:
For @heropup
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
Trying to write expression as
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A}{1+t}+\dfrac{Bt+C}{1+t^2}+\dfrac{Dt+E}{(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A(1+t^2)^2+(Bt+C)(1+t^2)(1+t)+(Dt+E)(1+t)}{(1+t)(1+t^2)^2}$$
$$1=A(1+t^2)^2+(Bt+C)(1+t^2)(1+t)+(Dt+E)(1+t)$$
Placing $t=-1$ in the equation
$$4A=1, A=\dfrac{1}{4}$$
$$1=(Bt+C)(1+t+t^2+t^3)+(Dt+Dt^2+E+Et)$$
$$1=Bt+Bt^2+Bt^3+Bt^4+C+Ct+Ct^2+Ct^3+Dt+Dt^2+E+Et$$
$$1=Bt^4+(B+C)t^3+(B+C+D)t^2+(B+C+D+E)t+(C+E)$$
$$B=0$$
$$B+C=0$$
$$C=0$$
$$B+C+D=0$$
$$D=0$$
$$B+C+D+E=0$$
$$E=0$$
$$C+E=1$$
$$E=1$$
This is contradiction right?
Any hints?
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Your partial fraction decomposition fails because the appropriate choice for numerators are not constants, but instead $$\frac{1}{(1+t^2)^2(1+t)} = \frac{A}{1+t} + \frac{Bt + C}{1+t^2} + \frac{Dt + E}{(1+t^2)^2}.$$ When the denominator is an integer power of a quadratic polynomial, the numerator is a linear function of $t$.
In order to solve for the coefficients in a correct manner, you need to first collect all of the terms on the RHS over a common denominator; this you have done:
$$\frac{1}{(1+t^2)^2(1+t)} = \frac{A(1+t^2)^2 + (Bt+C)(1+t)(1+t^2) + (Dt+E)(1+t)}{(1+t^2)^2(1+t)}.$$
Next, you need to equate the numerator on the LHS with the numerator on the RHS:
$$1 = A(1+t^2)^2 + (Bt+C)(1+t)(1+t^2) + (Dt+E)(1+t).$$
Here is where you make a mistake: You need to expand the RHS as a polynomial in $t$ and then collect like terms in $t$:
$$1 = (A+B) t^4 + (B+C) t^3 + (2A + B + C + D) t^2 + (B + C + D + E)t + (A + C + E).$$
After this, you equate corresponding coefficients to obtain the system
$$\begin{cases} 1 &= A + C + E \\
0 &= B + C + D + E \\
0 &= 2A + B + C + D \\
0 &= B + C \\
0 &= A + B \end{cases}$$
Moreover, your substitution trick was performed incorrectly. When $t = -1$, you would get $$1 = A(1 + (-1)^2)^2 = 4A,$$ hence $A = 1/4$. I do not understand how you got $A = 0$.
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|
Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the determinant into that form, but I am not seeing it. And directly computing the determinant by the cofactor expansion looks very nasty. So is there a simpler way to compute this determinant?
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Let $b=ax,c=axy$. Then:
$$\small\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} =\begin{vmatrix}a^2x^2(1+y^2)&a^2x&a^2xy\\a^2x&a^2(1+x^2y^2)&a^2x^2y\\a^2xy&a^2x^2y&a^2(1+x^2)\end{vmatrix} =a^6x^2\begin{vmatrix}1+y^2&1&y\\1&1+x^2y^2&x^2y\\y&x^2y&1+x^2\end{vmatrix} =\\
\small\stackrel{R_1-R_2\to R_1}=
a^6x^2\begin{vmatrix}y^2&-x^2y^2&y-x^2y\\1&1+x^2y^2&x^2y\\y&x^2y&1+x^2\end{vmatrix} =a^6x^2y\begin{vmatrix}y&-x^2y&1-x^2\\1&1+x^2y^2&x^2y\\y&x^2y&1+x^2\end{vmatrix} \stackrel{R_3-R_1\to R_3}=\\
\small a^6x^2y\begin{vmatrix}y&-x^2y&1-x^2\\1&1+x^2y^2&x^2y\\0&2x^2y&2x^2\end{vmatrix}=
2a^6x^4y\begin{vmatrix}y&-x^2y&1-x^2\\1&1+x^2y^2&x^2y\\0&y&1\end{vmatrix}=\\
\small =2a^6x^4y\cdot [y(1+x^2y^2-x^2y^2)-(-x^2y-y(1-x^2)]=4a^6x^4y^2=4a^2b^2c^2.$$
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|
Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$
In this new video Dr. Loh claims to discover a new way of solving the quadratic equation! How? It is the same as the above formula, by using the quadratic formula, the only thing I see different, is he rewrite it in the above form!
Can someone please explain to me how this is a new way?
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For $b^2-4ac\geq0$ and $a=1$ they are the same:
$$\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}+\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-c}$$ and
$$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}-\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-c}.$$
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|
How many ways can we get a number by addition if each part of the addition has to be smaller or equal to a set value? For example, if we need to get 5 with the largest number we can use being 3, we can use:
*
*3 + 2
*3 + 1 + 1
*2 + 2 + 1
*2 + 1 + 1 + 1
*1 + 1 + 1 + 1 + 1
Is there any way to find out the solution with any two numbers without calculating every one of them?
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We can calculate the number of ways with the help of generating functions. We encode the usage of
*
*zero or more $1$s as $1+x+x^2+x^3+\cdots=\frac{1}{1-x}$
*zero or more $2$s as $1+x^2+x^4+x^6+\cdots=\frac{1}{1-x^2}$
*zero of more $3$s as $1+x^3+x^6+x^9+\cdots=\frac{1}{1-x^3}$
and we look for the coefficient of $x^5$ denoted with $[x^5]$ of the product $\frac{1}{(1-x)(1-x^2)(1-x^5)}$.
This needs a little algebra, but we can keep it simple since we can skip powers greater than $5$ when multiplying out.
We obtain
\begin{align*}
\color{blue}{[x^5]}&\color{blue}{\frac{1}{(1-x)(1-x^2)(1-x^3)}}\\
&=[x^5]\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x^2+x^4\right)\left(1+x^3\right)\tag{1}\\
&=[x^5]\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x^2+x^3+x^4+x^5\right)\tag{2}\\
&=[x^5]\left(1\cdot x^5+x\cdot x^4+x^2\cdot x^3+x^4\cdot x+x^5\cdot 1\right)\tag{3}\\
&=[x^5]5x^5\\
&\,\,\color{blue}{=5}
\end{align*}
Comment:
*
*In (1) we expand $\frac{1}{1-x}$, $\frac{1}{1-x^2}$ and $\frac{1}{1-x^3}$ but only up to powers raised to $x^5$ since higher powers do not contribute to $[x^5]$.
*In (2) we multiply out the two right-most terms again skipping powers greater than $5$.
*In (3) we multiply out skipping all factors which do not give $x^5$.
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$\lim_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}$
$$\lim_{n\to\infty}\frac{e^{1/n^2}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}, \quad n \in \mathbb{N}$$
I'm stuck trying to evaluate this limit. I have tried applying l'Hôpital's rule to the corresponding function in $\mathbb{R}$, but even after differentiating all that stuff I still end up with a $\frac{0}{0}$ form. There has to be something I'm missing. Any hints?
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The best way to solve limits like these is to split them in to standard limits.
$\lim\limits_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos {\frac{1}{n}}}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}$
Hence, we can split the term in the numerator to apply standard logarithmic and trigonometric limits.
$\lim\limits_{n\to\infty}\frac{e^{\frac{1}{n^2}}-1+1-\cos {\frac{1}{n}}}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}$
$\Rightarrow\lim\limits_{n\to\infty}\frac{e^{\frac{1}{n^2}}-1}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}\cdot\frac{n^2}{n^2}+\lim\limits_{n\to\infty}\frac{2\sin^2 (\frac{1}{2n})}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}\cdot\frac{4n^2}{4n^2}$
$$\color{blue}{\lim\limits_{h\to0}\frac{e^h-1}{h}=1}, \color{red}{\lim\limits_{h\to0}\frac{\sin h}{h}=1} $$
$\Rightarrow\lim\limits_{n\to\infty}\color{blue}{\frac{e^{\frac{1}{n^2}}-1}{\frac{1}{n^2}}}\cdot\frac{1}{n\log(\frac{n+1}{n})-\frac{\sin^2\frac{2}{n}}{\frac{1}{n^2}}}+\lim\limits_{n\to\infty}\color{red}{\frac{2\sin^2(\frac{1}{2n})}{(\frac{1}{2n})^2}}\cdot\frac{1}{n\log(\frac{n+1}{n})-\frac{\sin^2\frac{2}{n}}{\frac{1}{n^2}}}\cdot\frac{1}{4}$
$\Rightarrow\lim\limits_{n\to\infty}\frac{\color{blue}{1}+\color{red}{\frac{1}{2}}}{n\log(\frac{n+1}{n})-\frac{\sin^2\frac{2}{n}}{\frac{1}{n^2}}}$
$\Rightarrow\lim\limits_{n\to\infty}\frac{1.5}{\frac{\log(1+\frac{1}{n})}{\frac{1}{n}}-4\frac{\sin^2(\frac{2}{n})}{\frac{4}{n^2}}}$.
Hence the limits is $\frac{-1}{2}$
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Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2-x^2] $$
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Note $x^2$ is the only $x$-term,
$$1 - y-x^2-y^2-yx^2+y^3$$
$$=1-y-y^2+y^3 -(1+y)x^2$$
$$=(1-y) -y^2(1-y) -(1+y)x^2$$
$$=(1-y)(1-y^2)-(1+y)x^2 =(1+y)(1-y)^2-(1+y)x^2 $$
$$= ( 1+y)[(1-y)^2-x^2] $$
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Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, please.
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Apply CS inequality: $ LHS \ge \dfrac14\cdot ( 1- a + a - b + b - c + c )^2 = \dfrac14 = RHS $ with equality occurs when $1 - a = a - b = b - c = c$. From this you can find $a, b, c, d$. Specifically, $ b = 2c, a = 2b - c = 4c - c = 3c \implies 1 = 2a - b = 6c - 2c = 4c \implies c = \dfrac14, b = 2c = \dfrac12, a = 3c = \dfrac34$ .
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evaluate the integral $\int x^3(5x^2+3)^5 dx$ How to evaluate for the integral:
$$\int x^3(5x^2+3)^5 dx$$
What I've tried:
For example,
(1) $u = x^2$ whence $du = 2x$ $dx$
(2)$u = ax^2+b$ whence $du = 2ax$ $dx$
Therefore,
(3) $u = 5x^2+3$ whence $du = 10x$ $dx$
$$I = \int \frac{du}{10}u^5 = \frac{1}{10} \int \frac{u^6}{6} du= \frac{1}{10}\frac{1}{6}u^6$$
However, this does not lead me to the answer:
$$\frac{(10x^2-1)(5x^2+3)^6}{700}$$
Where did I go wrong and how do I correct this?
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There are a few small mistakes in your reasoning. First you forgot the $dx$. Your initial expression should be
$$I=\int x^3(5x^2+3)^5\ dx$$
Applying the change of variables
$$u=5x^2+3 \qquad du=10xdx$$
Then
$$I=\int x^3(5x^2+3)^5\ dx=\frac1{10}\int x^2(5x^2+3)^5\ 10xdx$$
We will need to get rid of $x^2$ as well.
$$u=5x^2+3 \implies x^2=\frac{u-3}{5}$$
Now we are able to proceed
$$I=\frac1{10}\int \frac{u-3}{5}u^5\ du$$
Can you take it from here?
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Is this solution correct for equation $3x^2-4y^2=13$? Prove equation $3x^2-4y^2=13$ has no integer solution.
Solution: Suppose (y, 3)=1, We have:
$y≡( 1, 2) \mod (3)$
⇒ $4y^2≡( 1, 2) \ mod(3)$
$3x^2≡0 \mod (3)$
The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:
⇒ $3x^2-4y^2≡ 0 \ mod(3)$
But, $13≡1 \mod (3)$
Hence this equation has no integer solution. Similar result comes out when we consider remainder of nomials on 4.
Is this solution correct?
|
If $3 \mid y$, then $3 \mid -4y^2$ and since $3 \mid 3x^2$, you would have $3 \mid 13$, which is not true. Thus, as you stated, $3 \not\mid y$, which means $y^2 \equiv 1 \pmod 3$.
As such, $3x^2 - 4y^2 \equiv -y^2 \equiv 2 \pmod 3$. Note, you made a mistake when you stated "common remainder between $3x^2$ and $4y^2$ is $0$" (I'm not quite sure how you determined this) giving $3x^2 - 4y^2 \equiv 0 \pmod 3$. Also, $4y^2 \equiv 1 \pmod 3$ only, i.e., $4y^2 \not\equiv 2 \pmod 3$, which you indicated is a possible solution. Regardless, you have $13 \equiv 1 \pmod 3$. Since this doesn't match the left hand side, it means there are no integer solutions to $3x^2 - 4y^2 = 13$. This is basically what J. W. Tanner's question comment states.
As you state, you can also prove there are no solutions by using modulo $4$ instead. In this case, $4 \not\mid x$ (since, otherwise, you would need $4 \mid 13$ which is not true), so $x^2 \equiv 1 \pmod 4 \implies 3x^2 \equiv 3 \pmod 4$, so $3x^2 - 4y^2 \equiv 3 \pmod 4$, but $13 \equiv 1 \pmod 4$. As these congruence values don't match, it means there's no integer solution.
|
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|
$50\cos^2 x + 5\cos x = 6\sin^2 x$, find $\tan x$
$50\cos^2 x + 5\cos x = 6\sin^2 x$
Find $\tan x$
I used $\cos^2 x + \sin^2 x = 1$ to get the equation $$56\cos^2 x + 5\cos x -6 = 0$$
I then solve this to get $\cos x = \dfrac27, -\dfrac38$
Then I used generic trig ratios to get $\tan x = \pm\dfrac{3\sqrt5}{2}, \pm\dfrac{\sqrt{55}}{3}$
Are these $\pm$ signs correct? My reasoning came for a CAST diagram, can you confirm if this is correct and if not, why?
|
It's correct. The signs are necessary because $\cos x=\frac 27$ doesn't tell us which quadrant (either 1st or 4th) the argument is in. You can also use the identity
$$1+\tan^2x=\frac{1}{\cos^2 x}\Rightarrow \tan x=\pm\sqrt{\frac{1}{\cos^2 x}-1} $$
to directly compute $\tan x$ from $\cos x$.
|
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|
Summation to n terms of series \begin{gather}
In\ a\ problem\ given\ to\ us\ by\ our\ teacher,\ one\ part\ of\ it\ requires\ the\ summation\ of\ the\ \notag\\
following\ sequence\ to\ a\ finite\ number\ of\ terms\ ( n) , \notag\\
t_{r\ } \ =\ \frac{1}{r\times 2^{r}} \notag\\
\notag\\
S\ =\ \frac{1}{1\times 2} +\frac{1}{2\times 2^{2}} +\frac{1}{3\times 2^{3}} +....\ \frac{1}{r\times 2^{r}} \notag\\
\notag\\
I\ tried\ the\ following\ approach\ :\ \notag\\
\notag\\
\ S\ =\ \frac{1}{1\times 2} +\frac{1}{2\times 2^{2}} +\frac{1}{3\times 2^{3}} +....\ \frac{1}{r\times 2^{r}} \ \ \ ( i) \notag\\
\ \frac{S}{2} \ =\ \frac{1}{1\times 2^{2}} +\frac{1}{2\times 2^{3}} +\frac{1}{3\times 2^{4}} +....\ \frac{1}{( r-1) \times 2^{r}} \ +\ \frac{1}{r\times 2^{r+1}} \ \ \ \ ( ii) \notag\\
( i) \ -\ ( ii) \ \notag\\
\frac{S}{2} \ =\ \frac{1}{1\times 2} \ -\left( \ \left( 1-\frac{1}{2}\right) 2^{-1} \ +\ \left(\frac{1}{2} \ -\ \frac{1}{3}\right) 2^{-2} \ +...\left(\frac{1}{r-1} -\frac{1}{r}\right) 2^{-r}\right) \ -\ \frac{2^{-r}}{r}\\
\notag\\
\frac{S}{2} \ =\ \frac{1}{1\times 2} \ -\left( \ \left(\frac{1}{2}\right) 2^{-1} \ +\ \left(\frac{1}{6}\right) 2^{-2} \ +...\left(\frac{1}{r( r-1)}\right) 2^{-r}\right) \ -\ \frac{2^{-r}}{r} \notag\\
Here\ I\ got\ stuck\ ,\ as\ the\ series\ generated\ above\ again\ generates\ another\ similar\ sequence.\ \notag\\
\notag\\
\notag\\
\notag
\end{gather}
I don't think the sum telescopes as well. Can calculus be used in some way to calculate the sum?
I'll just add the actual problem, in case it offers some help :
\begin{gather*}
t_{r} \ =\ \frac{r+2}{r( r+1)} \ 2^{-r}\\
\\
\\
\end{gather*}
|
Considering the actual probem
$$t_{r} =\frac{r+2}{r( r+1)} x^r=2\frac{x^r}{r}-\frac{x^r}{r+1}=2\frac{x^r}{r}-\frac 1x\frac{x^{r+1}}{r+1}$$
$$\sum_{r=1}^n t_r=2\sum_{r=1}^n \frac{x^r}{r}-\frac 1x\sum_{r=1}^n\frac{x^{r+1}}{r+1}$$ Differentiate both sides with respect to $x$
$$\left(\sum_{r=1}^n t_r \right)'=2\sum_{r=1}^n x^{r-1}-\frac 1x\sum_{r=1}^n{x^{r}}$$
I am sure that you can take it from here.
At the end, make $x=\frac 12$ and $n\to \infty$
|
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|
Intuitive way to prove that $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots = 0$ I’m trying to come up with an intuitive way to prove that $$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots = 0$$ That is, I want to prove that $$\prod_{n=1}^\infty \frac {(1)^n}{2}=0$$
I thought of this way. $$S= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots$$ Multiply both sides by 2 to give $$2S= 1 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots$$ But $$2S=S$$ So $S=0$. Is this proof valid?
|
Let $p_n = \prod_{i=1}^n \frac12 = \left(\frac12\right)^n$. Then $\frac{p_{n+1}}{p_n} = \frac12<1$ so $\{p_n\}$ is decreasing, and the product of positive numbers is a positive number, so $p_n>0$ for all $n$. Since $p_n$ is decreasing and bounded below, it converges to its infimum. Now, given $\varepsilon>0$, we may choose a positive integer $N$ such that $N>\frac{\log \varepsilon}{\log\frac12}$. It follows that for $n\geqslant N$ we have
$$
p_n= \left(\frac12\right)^n <\varepsilon,
$$
so that $\lim_{n\to\infty} p_n=0$.
|
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|
Factoring $3n^3 - 39n^2 + 360n + 20$ I am wondering how to factor
$$f(n) = 3n^3 - 39n^2 + 360n + 20$$ the right way. I think the factors are equal to
$$(n - 39.9762)(n - 12.0791)(n + 0.055248)$$
|
Set
$$
P(x)=3x^3-39x^2+360 x+20.
$$
Then set $x=y+\frac{13}{3}$, then
$$
\frac{1}{3}P(x)=P_1(y):=y^3+\frac{191}{3}y+\frac{9826}{27}
$$
If $\rho_0=\sqrt[3]{A}+\sqrt[3]{B}$ is the real root of $P_1(y)=0$, then
$$
\rho_0^3=A+B+3(AB)^{1/3}(\sqrt[3]{A}+\sqrt[3]{B})=s+3p^{1/3}\rho_0.
$$
Hence $s=A+B=-\frac{9826}{27}$, $p=AB=-\left(\frac{191}{9}\right)^3$ and the equation
$$
X^2+\frac{9826}{27}X-\left(\frac{191}{9}\right)^3=0,
$$
have roots
$$
A=\frac{1}{27}(-4913-12\sqrt{216010})\textrm{ , }B=\frac{1}{27}(-4913+12\sqrt{216010}).
$$
Hence we find
$$
\rho_0=\sqrt[3]{\frac{1}{27}(-4913+12\sqrt{216010})}-\sqrt[3]{\frac{1}{27}(4913+12\sqrt{216010})}.
$$
Now $P_1(y)$ have the other two roots such (Vieta)
$$
\rho_1+\rho_2=-\rho_0\textrm{ and }\rho_1\rho_2=\frac{9826}{27}\rho_0^{-1}
$$
Solving
$$
X^2+\rho_0X+\frac{9826}{27}\rho_0^{-1}=0
$$
we get the other two roots of $P_1(y)=0$.
By this way every third degree polynomial equation reduced solving only two degree equations.
NOTE. We have used $\sqrt[3]{-|a|^3}=-|a|$, since the equation $x^3+|a|^3=0$, have solution $x=-|a|$.
|
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|
Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there some clever way to show this other than to add the fractions together by brute-force? For example, is there some way to group terms together and say something like "These terms sum to more than $\frac{1}{3}$, these terms sum to more than $\frac{1}{2}$, and these terms sum to larger than $\frac{1}{6}$, so the whole thing sums to more than $1$"?
|
By C-S we obtain:
$$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}=$$
$$=\frac{17}{60}+\frac{17}{66}+\frac{17}{70}+\frac{17}{72}\geq\frac{17\cdot4^2}{60+66+70+72}=\frac{68}{67}>1.$$
|
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|
JBMO TST macedonia Number Theory
Find all $x,y$ positive integers $$x + y^2 + (\gcd(x, y))^2 = xy \cdot
\gcd(x, y)$$
I tried supposing $\gcd(x,y)=d$ and letting $x=ad$, $y=bd$ but didn't get anything the closest thing I think is usefull that $(b+1)^2+4a^2b$ is a perfect square.
|
Your substitutions give the equation
$$ad+b^2d^2+d^2=abd^3$$
Modulo $bd^2-1$ we obtain $$ad+b+d^2\equiv ad.$$ Then $bd^2-1$ is a factor of $b+d^2$ and so $bd^2-1\le b+d^2$. Either $d=1$ or $d\ge2$ and $$b\le \frac {d^2+1}{d^2-1}=1+ \frac {2}{d^2-1}<2.$$
IF $b=1.$
Then $a+2d=ad^2$. Therefore $a|2d$ and $d|a$ and so $a=d$ or $2d$. Neither possibility gives a solution.
IF $d=1,b\ge2.$
Then $b-1$ is a factor of $b+1$. Therefore $b+1\ge2(b-1)$ and $b=2$ or $3$.
The solutions are $x=5,y=2$ and $x=5,y=3.$
|
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|
Determine all six exact values if $\sqrt[6]{\frac{2i}{1+i}}$. Determine all six exact values if $\sqrt[6]{\frac{2i}{1+i}}$, expressing each in simplified, exact rectangular form.
What I have done:
I set $\sqrt[6]{\frac{2i}{1+i}}=z$ and put the both to the sixth power to get ${\frac{2i}{1+i}}=z^6$. I then multiplied by the conjugate to get $z^6(1-i)=\frac{2i-2i^2}{2}$ and simplified to $z^6(1-i)=i-i^2$ and simplified farther to get $z^6-z^6i=i+1$. I then moved the equation to one side to set it equal to zero and got $z^6-z^6i-i-1=0$.
This is where I got stuck, and I do not know if I even did this right.
Can anyone help me, please?
|
Use polar coordinates and De Moivre Theorem
Let $w = \frac {2i}{1+i}=\frac {2i(1-i)}{(1+i)(1-i)} =\frac {2+ 2i}{2} = 1 + i$
$w = re^{i(\theta + 2\pi k)} = r(\cos \theta + i\sin \theta) = 1 + i$ where
$r = \sqrt {1^2 + 1^2} =\sqrt 2$ and $\theta = \arctan \frac 11 = \frac \pi 4$.
So $w = \sqrt 2*e^{i(\frac \pi 4+ 2\pi k)}$.
So $z = w^{\frac 16} = \sqrt[6]{\sqrt 2}e^{i(\frac {\frac \pi 4+ 2\pi k}6)}$ where $k=0,...5$.
$= \sqrt[12]{2} (\cos \phi + i \sin \phi)$ where $\phi = \frac \pi {24}+ \frac 13\pi k$ where $k = 0... 5$ are the six sixth roots.
|
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|
Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagrange multiplier method has not gotten me further either.
|
You can show it just by squaring, using GM-AM and applying the two facts
*
*$(1)$: $a+b = ab$ and
*$(2)$: $ab \geq 4$
LHS:
$$\left(\sqrt{a^2+4}+\sqrt{a^2+4}\right)^2 =a^2+b^2+8 + 2\sqrt{(a^2+4)(b^2+4)}\stackrel{GM-AM}{\leq}2(a^2+b^2+8)$$
RHS:
$$\left(\frac{\sqrt 2}{4}(a+b)^2\right)^2 \stackrel{(1)}{=} \frac 18 (a+b)^2(ab)^2 \stackrel{(2)}{\geq}2(a^2+b^2+2ab)$$
$$ \stackrel{(2)}{\geq}2(a^2+b^2+8)$$
|
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|
Evaluate a triple integral with spherical coordinates $$\iiint_{Q}z\ dV$$ Where $Q$ is the common region of the spheres $x^{2}+y^{2}+z^{2}\leq 1$ and $x^{2}+y^{2}+(z-1)^{2}\leq 1$
I have tried nothing.
|
We have to use spherical coordinates $(x=r\sin\theta\cos\varphi, \quad y=r\sin\theta\sin\varphi, z=r\cos\theta)$, so we will have $$x^2+y^2+z^2 \le 1 \Leftrightarrow r^2 \le 1 \Leftrightarrow r \le 1$$ and $$x^2+y^2+(z-1)^2 \le 1 \Leftrightarrow x^2+y^2+z^2 \le 2z \Leftrightarrow r^2 \le 2r\cos\theta \Leftrightarrow r \le 2\cos\theta.$$
Next, since both spheres have their centers on $z$-axis, one center is $P_1=(0,0,0)$ and another one is $P_2=(0,0,1)$ we know that it will hold: $$0 \le \varphi\le 2\pi,$$ and since both spheres have radius $1$ it is not hard to see that it will hold: $$0 \le \theta \le \frac{\pi}{2}.$$
Finally, your integral will be equal to $$\int_{0}^{2\pi}d\varphi \int_{0}^{\frac{\pi}{2}}\sin\theta \cos\theta d\theta \int_{2\cos\theta}^{1}r^3 dr=2\pi\int_{0}^{\frac{\pi}{2}}\sin\theta \cos\theta(\frac{1}{4} - 4\cos^4\theta)d\theta =...= \frac{-13\pi}{12}.$$
|
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|
How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0}^{1}\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\left[\int\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b\right]\mathrm{d}b
\end{align}
Based on it, the first summand is given by
\begin{align}\label{eq8}
\int_{0}^{1}\frac{b^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \int_{0}^{1}\frac{(1 - 2b + b^{2}) - (1 - 2b)}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\frac{(2-2b) - 1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - 2\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b + \int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
The first of the last three integrals is given by
\begin{align}
\int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \frac{1}{\theta}\int_{0}^{1}\frac{\theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta}\int_{0}^{1}\frac{1 - \theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta} + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
According to the substitution $u = \sqrt{\theta}(1-b)$, it results that
\begin{align}\label{eq10}
\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\sqrt{\theta}}\int_{0}^{\sqrt{\theta}}\frac{1}{1 - u^{2}}\mathrm{d}u = \frac{1}{2\sqrt{\theta}}\ln\left|\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right|
\end{align}
Finally, based on the same substitution, we have
\begin{align}\label{eq11}
\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\theta}\int_{0}^{\sqrt{\theta}}\frac{u}{1-u^{2}}\mathrm{d}u = -\frac{1}{\theta}\ln|1-\theta^{2}|
\end{align}
Combining all these results, we get the first summand from the integration by parts method. The problem arises when I try to determine the second part.
|
In a similar way, we can make this:
$b^3=\dfrac{-1}{\theta}(b-\theta b+2\theta b^2-\theta b^3-b+\theta b-2\theta b^2)$
$\dfrac{b^3}{1-\theta (1-b)^2}=\dfrac{-1}{\theta}(b+\dfrac{-b+\theta b-2\theta b^2}{1-\theta (1-b)^2})$
We can write the last term as $2\dfrac{\frac{-b}{2}+\frac{b\theta}{2}-\theta b^2}{1-\theta (1-b)^2}=2 \dfrac{1-\theta +2\theta b-\theta b^2-1+\theta-2\theta b -\frac{b}{2}+\frac{b\theta}{2}}{1-\theta (1-b)^2}$
$=2(1+\dfrac{-1+\theta-2\theta b -\frac{b}{2}+\frac{b\theta}{2}}{1-\theta (1-b)^2})$
For the last numerator follows $-1+\theta+\dfrac{-(3\theta+1)}{2}b=\dfrac{1}{2\theta}(-2\theta+2\theta^2+\dfrac{-(3\theta+1)}{2}2\theta b)$
$=\dfrac{(3\theta+1)}{4\theta}(\dfrac{-4\theta +4\theta^2}{3\theta+1}-2\theta b+2\theta -2\theta)$
So, we have, finally:
$\dfrac{-1}{\theta}\int_{0}^{1}(b+2+\dfrac{(3\theta+1)}{2\theta}\dfrac{2\theta-2\theta b}{1-\theta (1-b)^2}-\dfrac{(3\theta+1)}{2\theta}(\dfrac{4\theta -4\theta^2}{3\theta+1}+2\theta)\dfrac{1}{1-\theta (1-b)^2})db$
$=\dfrac{-1}{\theta}(\dfrac{1}{2}+2-\dfrac{(3\theta+1)}{2\theta}\ln|1-\theta|-\dfrac{\theta+3}{\sqrt \theta}(\ln|1+\sqrt \theta|-\ln|\sqrt {1-\theta}|))$
|
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|
Square numbers and divisibility I'm trying to prove the following statement:
If $kn+1$ and $(k+1)n+1$ are both perfect squares, then $2k+1$ divides n, where k and n are positive integers.
Case $k=1$:
Put $n+1=x^2$, $2n+1=y^2$. Quadratic residues can only be $0, 1 \mod 3$.
Hence
$n \equiv 1 \mod 3 \Rightarrow x^2 = n+1 \equiv 2 \mod 3$
$n \equiv 2 \mod 3 \Rightarrow y^2 = 2n+1 \equiv 5 \equiv 2 \mod 3$
Both of them are contradictions.
Case $k=2$:
Put $2n+1=x^2$, $3n+1=y^2$. Quadratic residues can only be $0, 1, 4 \mod 5$. Hence
$n \equiv 1 \mod 5 \Rightarrow x^2 = 2n+1 \equiv 3 \mod 5$
$n \equiv 2 \mod 5 \Rightarrow y^2 = 3n+1 \equiv 2 \mod 5$
$n \equiv 3 \mod 5 \Rightarrow x^2 = 2n+1 \equiv 2 \mod 5$
$n \equiv 4 \mod 5 \Rightarrow y^2 = 3n+1 \equiv 3 \mod 5$
All of them are contradictions.
Case $k=3$:
See If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
It might be possible to prove further individual cases for different k.
However I'm not sure if the statement is true for all positive k and if yes, how to prove it.
Some observations:
If $kn+1 = x^2$ and $(k+1)n+1 = y^2$ we get
$y^2-x^2=n$
$y^2+x^2=(2k+1)n+2$
$(k+1)x^2-ky^2=1$
but I have no good idea how to proceed. Any hint/help is appreciated.
|
This answer explains how your question relates to the Pell's equation and points you to various online sources which can help you to solve your problem.
The equation you have, where I've switched the $x$ and $y$ around, of
$$(k+1)y^2 - kx^2 = 1 \implies kx^2 - (k+1)y^2 = -1 \tag{1}\label{eq1A}$$
is a form of a generalized Pell equation. There's an online Solving a generalized Pell equation page where it allows you to check the solubility of equations of the form
$$ax^2 - by^2 = \pm 1 \tag{2}\label{eq2A}$$
where $1 \lt a \lt b$, $\gcd(a,b) = 1$ and $ab$ is not a perfect square. Your cases are satisfied with the right hand side of \eqref{eq1A} for $k \ge 2$ since $a = k$ and $b = k + 1$ satisfies $1 \lt k \lt k + 1$, $\gcd(k, k + 1) = 1$ and $k(k+1)$ not being a perfect square. To see the last point, note $k^2 \lt k(k+1) \lt (k+1)^2$.
The online solution calculator has a link to an explanation of how the algorithm works at A Midpoint Criterion for the Diophantine Equation $ax^2 - by^2 = \pm 1$. Also, it gives a link to a GNU BC program source code implementation. Using this, if there are no solutions, then you're done for that $k$. Otherwise, you can check each solution to see if $\frac{x^2 - 1}{k + 1} = \frac{y^2 - 1}{k} = n$ where $n$ is an integer and, if so, then whether or not $2k + 1 \mid n$.
Alternatively, take the left hand side of \eqref{eq1A}, switch $x$ and $y$ around, and multiply both sides by $k + 1$, to get
$$\begin{equation}\begin{aligned}
(k+1)^2 x^2 - k(k+1)y^2 & = k + 1 \\
(x')^2 - Dy^2 & = N
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
where $x' = (k+1)x$. This is now in the more usual form of the generalized Pell equation. There's various sites with information about how to solve this, such as at Solving the generalized Pell equation, plus multiple posts on this site, e.g., at Generalized Pell's equation.
If you use this method, then if there are any solutions at all, for each one, first confirm $x = \frac{x'}{k+1}$ is an integer. If so, then do the other previous checks mentioned (but with $x$ and $y$ switched around) for a resulting integer $n$ and, if so, then that $2k + 1 \mid n$.
|
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|
Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore, $\frac{d}{du} (-\frac{1}{u-2})= \frac{1}{x^2}$ and
$$\frac{d}{du} (-\frac{1}{u-2})= \frac{d}{du} (-\frac{1}{-2(1-\frac{u}{2})})=\frac{d}{du}(\frac{1}{2} \frac{1}{1-\frac{u}{2}})=\frac{d}{du} \Bigg( \frac{1}{2} \sum_{n=0}^\infty \bigg(\frac{u}{2}\bigg)^n\Bigg)$$
$$= \frac{d}{du} \Bigg(\sum_{n=0}^\infty \frac{u^n}{2^{n+1}}\Bigg)= \frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=$$
$$\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
From this we can conclude that
$$f(x)=\frac{1}{x^2}=\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
Is this solution correct?
|
Narrowly on your question: no, it is not correct. But it is almost correct.
The error is that just before the end you forgot to get the range for $n$ correct. When you differentiate a term in $x^0$ you get 0, not a term in $x^{-1}$.
As a quite separate point there are other ways of getting a power series for $\frac{1}{(u-2)^2}$, but you did not ask for alternative ways.
|
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|
Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$. So i had to prove:
Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$.
Here is my proof:
So we have
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx=1+2(xy+yz+zx)=0\Rightarrow2(xy+yz+xz)=-1$
Let's assume that all of $xy,zx,yz$ are bigger than $-\frac{1}{3}$
$xy>-\frac{1}{3}$
$yz>-\frac{1}{3}$
$zx>-\frac{1}{3}$
Adding that we have $xy+yz+zx>-1$
$2(xy+yz+zx)>xy+yz+zx>-1\Rightarrow 2(xy+yz+zx)>-1$
Which is contradiction because $2(xy+yz+xz)=-1$, so one of $xy,yz,zx$ must be less or equal $-\frac{1}{3}$
Is my proof valid?
|
I found another, easier answer, so I thought it might be worth posting it.
Without loss of generality we may assume that $x \leqslant y \leqslant z$. From our assumptions we see that $x<0$ and $0<z<1$. If $0 \,\leqslant y$, then
$$1=x^2+y^2+z^2=(-y-z)^2+y^2+z^2=2y^2+2z^2+2yz \geqslant 2y^2+2y^2+2y^2=6y^2$$
If $y<0%$:
$$1=x^2+y^2+z^2=x^2+y^2+(-x-y)^2=2x^2+2y^2+2xy \geqslant 2y^2+2y^2+2y^2=6y^2$$
In both cases we have: $\; y^2\leqslant \frac16\ $, and also $1=x^2+(-x-z)^2+z^2=2x^2+2z^2+2xz \,$,which means that:
$$xz= \frac12-(x^2+z^2)=\frac12-(1-y^2)=y^2-\frac12\leqslant\frac16-\frac12=-\frac13 $$
|
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|
Edwards Calculus chapter 4 problem 81 I can see that the terms arise from the differentiation of $\log(1+x^{2^r} )$ and tried to do it but couldn't figure it out in the end.
Show that
$$
\frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} + \ldots = \sum_{n=0}^{\infty} \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}} = \frac{1}{1-x} \qquad \text{when}~|x|<1
$$
|
Let's assume
$$S=\dfrac{1}{1+x}+\dfrac{2x}{1+x^2}+\dfrac{4x^3}{1+x^4}+...$$
The common term of the series, as seen from the pattern, is $\displaystyle \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}}$ for $n=0,1,2,3,...$ and the sum is then $\displaystyle\sum_{n=0}^{\infty} \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}}$.
Now, noting that the numerator in each case is the derivative of the denominator, as you've already done, we integrate $$\int_0^x Sdx=\left[\ln(1+x)+\ln(1+x^2)+\ln(1+x^4)+...\right]^x_0$$
$$=\ln[(1+x)(1+x^2)(1+x^4)...]$$
$$=\ln(1+x+x^2+x^3...)$$
Here, $\ln$ denotes the natural logarithm and I've used the facts that for the lower limit, $\ln1=0$, then $\ln a+\ln b= \ln{ab}$ and expanded the multiplication in the last step.
Note that the series $1+x+x^2+x^3+...$ is the series expansion of $\dfrac{1}{1-x}$ and this is valid only when $|x|<1$. So the term inside the logarithm converges provided that $|x|<1$ and the sum is $\dfrac{1}{1-x}$.
Thus, $\displaystyle\int_0^x Sdx=\ln {\dfrac{1}{1-x}}$ and so,
$$ S=\dfrac{d}{dx} \ln\left( \dfrac{1}{1-x} \right)=\dfrac{1}{1-x}$$
|
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|
Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) = \dfrac{\sin^2(2x)}{4}$$
$$\sin^2(2x) - 4 \sin(2x) -4 = 0$$
Here we can use the notation $t = \sin(2x)$ with the condition that $t \in [-1,1]$.
$$t^2-4t-4=0$$
Solving this quadratic equation we get the solutions:
$$t_1 = 2+ 2\sqrt{2} \hspace{3cm} t_2 = 2 - 2\sqrt{2}$$
I managed to prove that $t_1 \notin [-1, 1]$ and that $t_2 \in [-1, 1]$. So the only solution is $t_2 = 2 - \sqrt{2}$. So we have:
$$\sin(2x) = 2 - 2\sqrt{2}$$
From this, we get:
$$2x = \arcsin(2-2\sqrt{2}) + 2 k \pi \hspace{3cm} 2x = \pi - \arcsin(2-2\sqrt{2}) + 2 k \pi$$
$$x = \dfrac{1}{2} \arcsin(2-2\sqrt{2}) + k \pi \hspace{3cm} x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin(2 - 2\sqrt{2}) + k \pi$$
Is this solution correct? It's such an ungly answer, that I kind of feel like it can't be right. Did I do something wrong?
|
Using auxiliary angle, we first change the equation
$$
\sin x+\cos x=\sin x \cos x \tag*{(*)}
$$
equivalently to
$$
\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)= \frac{\sin 2 x}{2}
$$
Putting $y=x-\dfrac{\pi}{4} $ gives a quadratic equation in $\cos y $
$$
\sqrt{2} \cos y=\frac{1}{2}\left(2 \cos ^{2} y-1\right)
$$
Solving the quadratic equation yields
$$
\cos y=\frac{\sqrt{2}-2}{2} \text { or } \frac{\sqrt{2}+2}{2} \text { (rejected) }
$$
Plugging the general solution for $(*)$ is
$$
x=\frac{(8 n+1) \pi}{4} \pm \cos ^{-1}\left(\frac{\sqrt{2}-2}{2}\right),
$$
where $n\in Z.$
|
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|
Combination of problem having identical objects without using generating function approach
A person goes in for an examination in which there are $4$ papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is
My Attempt
Number of ways of partitioning $n$ identical objects into $r$ distinct groups, if empty groups are allowed is:${}^{n+r-1}C_{r-1}$.
Here, $n=2m$ and $r=4$
Req. number of ways = ${}^{n+r-1}C_{r-1}-$[#{more than m marks for paper 1}+#{more than m marks for paper 2}+#{more than m marks for paper 3}+#{more than m marks for paper 4}]=$${}^{2m+4-1}C_{4-1}-[{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}]={}^{2m+4-1}C_{4-1}-4.{}^{m+4-1}C_{4-1}\\
={}^{2m+3}C_{3}-4.{}^{m+3}C_{3}=\frac{(2m+3)(2m+2)(2m+1)}{3.2}-4\frac{(m+3)(m+2)(m+1)}{3.2}\\
=\frac{m+1}{3}.[4m^2+8m+3-2m^2-10m-12]=\frac{(m+1)(2m^2-2m-9)}{3}
$$
But my reference gives the solution $\dfrac{(m+1)(2m^2+4m+3)}{3}$, so what is going wrong here ?
Note: I am trying to solve this problem without using the generating function approach
|
You should consider $\#\{\text{strictly more than $m$ marks for paper $j$}\}$. So we start with $m+1$ marks in paper $j$ and then distribute the remaining $m-1$ marks among the 4 papers. Thus we get $$\frac{(2m+3)(2m+2)(2m+1)}{3\cdot2}-4\frac{(m+2)(m+1)m}{3\cdot2}$$
$$=\frac{m+1}{3}(4m^2+8m+3-2m^2-4m)$$
$$=\frac{(m+1)(2m^2+4m+3)}{3}$$
|
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|
Solve $T(n) = 2^nT(n/2) + n^n$ Let $$T(n) = 2^nT(n/2) + n^n$$ Can't be solved using the master theorem, because the equation doesn't satisfy $$T(n) = aT(n/b) + f(n)$$
How would you approach this to find the time complexity?
|
If
$T(n) =
2^nT(n/2) + n^n
$,
then,
putting $2^m$ for $n$,
$T(2^m)
= 2^{2^m}T(2^{m-1}) + (2^m)^{2^m}
= 2^{2^m}T(2^{m-1}) + 2^{m2^m}
$.
Let
$U(m) = T(2^m)$.
then
$U(m)
= 2^{2^m}U(m-1) + 2^{m2^m}
$.
Since
$\sum 2^m = 2^{m+1}$,
divide by
$2^{2^{m+1}}$
to get
$\dfrac{U(m)}{2^{2^{m+1}}}
= \dfrac{2^{2^m}U(m-1)}{2^{2^{m+1}}} + \dfrac{2^{m2^m}}{2^{2^{m+1}}}\\
= \dfrac{U(m-1)}{2^{2^{m+1}-2^m}} + 2^{m2^m-m-1}\\
= \dfrac{U(m-1)}{2^{2^m}} + 2^{m2^m-m-1}
$
Let
$V(m)
=\dfrac{U(m)}{2^{2^{m+1}}}
$.
This becomes
$V(m)
= V(m-1) + 2^{m2^m-m-1}
$
and this reduces to computing
$\sum 2^{m2^m-m-1}
$.
Another way is
$\begin{array}\\
T(n)
&=2^nT(n/2) + n^n\\
&=2^n(2^{n/2}T(n/4) + (n/2)^{n/2}) + n^n\\
&=2^{n(3/2)}T(n/4) + 2^{3n/2}(n/2)^{n/2} + n^n\\
&=2^{n(3/2)}T(n/4) + (2^{3/2}n/2)^{n/2} + n^n\\
&=2^{n(3/2)}T(n/4) + (2^{1/2}n)^{n/2} + n^n\\
&=2^{n(3/2)}(2^{n/4}T(n/8) + (n/4)^{n/4}) + (2^{1/2}n)^{n/2} + n^n\\
&=2^{n(7/4)}T(n/8) + (n/4)^{n/4}) + (2^{1/2}n)^{n/2} + n^n\\
\end{array}
$
Not sure how to
nicely express this,
so I'll stop here.
|
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|
Algebraic and Trigonometric expression is $>0$ for all real $x$
Prove that $$2x^2\sin x+2x\cos x+2x^2+1$$ is always positive for all real $x$.
From completing the square method
Write $1$ as $\sin^2 x+\cos^2 x$
$$x^4+2x^2\sin x+\sin^2 x+x^2+2x\cos x+\cos^2 x+x^2-x^4$$
$$(x^2+\sin x)^2+(x+\cos x)^2+x^2(1-x^2)$$
$\bullet\; $ In $|x|\leq 1$ our expression is $>0$.
Is there is any way to prove expression $>0$ for $|x|>1$?
Thanks in advance.
|
For $\sin{x}+1>0$ we obtain: $$2x^2\sin x+2x\cos x+2x^2+1=2(1+\sin{x})x^2+2x\cos{x}+1\geq0$$ because
$$\frac{\Delta}{4}=\cos^2x-2(1+\sin{x})=-(1+\sin{x})^2<0.$$
If $\sin{x}+1=0$ so $$2x^2\sin x+2x\cos x+2x^2+1=1>0.$$
|
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|
Integrate $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$
Evaluate
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I have tried substitution of $\sin x$ as well as $\cos x$ but it is not giving an answer.
Do not understand if there is a formula for this or not.
|
Recall that $$\begin{align}\sin(x) &= \frac{\tan(x)}{\sec(x)} \\ \cos(x) &= \frac1{\sec(x)} \\ \sec^2(x) &= 1 + \tan^2(x)\end{align}$$
Then,
$$\int\frac{\sqrt{\sin(x)}}{\sqrt{\sin(x)} + \sqrt{\cos(x)}}\,\mathrm dx\equiv\int\sec^2(x)\cdot\frac{\sqrt{\tan(x)}}{\sqrt{\tan(x)} \left(1 + \tan^2(x)\right) + 1 + \tan^2(x)}\,\mathrm dx$$
Now, let $u = \sqrt{\tan(x)}\implies2\cdot\mathrm du = \dfrac{\sec^2(x)}{\sqrt{\tan(x)}}\,\mathrm dx$. So,
$$\begin{align}\int\sec^2(x)\cdot\frac{\sqrt{\tan(x)}}{\sqrt{\tan(x)} \left(1 + \tan^2(x)\right) + 1 + \tan^2(x)}\,\mathrm dx&\equiv\int2\cdot\frac{u^2}{u\left(1 + u^4\right) + 1 + u^4}\,\mathrm du \\ &= 2\int\frac{u^2}{(1 + u)(1 + u^4)}\,\mathrm du\end{align}$$
From here on, partial fractions and linearity are your friends.
|
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|
Find $f(x) = \lim_{n\to\infty}\frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n}$ I found this cool problem in a textbook. I googled it and used MSE's search tool to check if it has been asked before or not, but it seems that it hasn't been asked before.
Find the function $f(x)$ that the following limit defines: $$f(x) = \lim_{n\to\infty}\frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n}$$
I have already solved it and I have shared my solution as an answer. Other solutions are welcome too.
|
Solution: We know that $x-1<\lfloor x \rfloor \leq x$, hence
$$x + \cdots + x^n - n < \lfloor x \rfloor + \cdots + \lfloor x^n \rfloor \leq x + \cdots + x^n$$
Dividing both sides by $x^n$
$$\frac{x + \cdots + x^n}{x^n} - \frac{n}{x^n} < \frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n} \leq \frac{x + \cdots + x^n}{x^n}$$
Using the identity $x+\cdots+x^n = \frac{x-x^{n+1}}{1-x}$, we have
$$\frac{x}{x-1}\cdot \frac{x^n-x}{x^n} - \frac{n}{x^n} < \frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n} \leq \frac{x}{x-1}\cdot \frac{x^n-x}{x^n}$$
If $|x|>1$, then $\lim_{n\to\infty}\frac{n}{x^n} = 0$. To prove it, first ssume $x>1$, we can write it as $x=1+r$ where $r>0$, hence
$$x^n=(1+r)^n > 1+nr+\frac{n(n-1)}{2}r^2$$
$$0 \leq \lim_{n\to\infty}\frac{n}{x^n} \leq \lim_{n\to\infty}\frac{n}{1+nr+\frac{n(n-1)}{2}r^2}=0$$
For the case $x<-1$, just replace $x$ with $(-x)>1$ and notice that $\lim_{n\to\infty} (-1)^n\frac{n}{(-x)^n} = 0$.
Hence, for $|x| > 1$, the squeeze theorem proves that
$$\lim_{n\to\infty}\frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n} = \frac{x}{x-1}$$
For $0<x<1$, $f(x)=0$ and $f(x)$ is undefined on $[-1,0] \cup \{1\}$. Q.E.D.
|
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|
About the sequence: $1, 3, 8, 24, 29, 87, 92, ?...$ find $a_n$ This is a classic and curious recurrence sequence used in logic tests. Your rule can be determined as follows:
for $n$ even: $a_n = 3a_{n-1}$
for $n$ odd: $a_n = a_{n-1} + 5$
That is, the ratio alternates with each term.
Will there be a single formula for $a_n$ in terms of $n$?
|
For an even value
\begin{eqnarray*}
a_{2n}=3a_{2n-1}=5+3 a_{2n-2}.
\end{eqnarray*}
Iterate this
\begin{eqnarray*}
a_{2n}=5+3\times 5 +3^2 \times 5 + \cdots 3^{n-1} \times 5 +3^n.
\end{eqnarray*}
Sum the geometric series
\begin{eqnarray*}
a_{2n}=5\frac{3^{n}-1}{3-1}+3^n = \frac{7 \times 3^n -5}{2}.
\end{eqnarray*}
|
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|
Deriving $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ using Fourier Series for $f(x)=|x|$ One can find that $$f(x) = \frac{\pi}{2} + \sum_{n=1}^{\infty}\frac{2(-1+(-1)^n)}{\pi n^2}\cos(nx)$$
Now look at the case for $x=0$. We can find that $\sum_{n \text{ odd}, n \geq 1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{8}$. However, I am having a hard time proving the original equality since all the even terms would vanish. Thus it seems like the total sum should be equivalent to the sum of odd terms, but I know this is not the case. What nuance am I missing?
|
Rearranging gives
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1+(-1)^{n+1}}{n^2} =\frac{\pi^2}{4}.
\end{eqnarray*}
Now let
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{n^2} =x.
\end{eqnarray*}
So
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{(2n)^2} =\frac{x}{4} \\
\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} =\frac{3x}{4} \\
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} =\frac{x}{2}. \\
\end{eqnarray*}
Do some linear algebra and we have the usual result
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{\pi^2}{6}.
\end{eqnarray*}
|
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|
Prove that $a^{b} \equiv 3 \,( \text{mod}\, 4)$ implies $a,b$ odd. So far I've shown that
$a^{b} \equiv 3 \,( \text{mod} \, 4) \implies a^{b} \,\text{odd}\implies a \, \text{odd}$. I also know that since $a^{b} \equiv 3 \, (\text{mod} \, 4)$, there exists prime $p \mid a^{b}$ such that $p \equiv 3 \, (\text{mod} \, 4)$.
Any help would be appreciated, thanks!
|
As you've shown, $a$ must be odd as any even value can't be congruent to $3$ modulo $4$. Thus, either $a \equiv 1 \pmod 4$, which doesn't work as $a^b \equiv 1 \pmod 4$ for any $b$, or $a \equiv 3 \pmod 4$. For the second case, note $a^2 \equiv 3^2 \equiv 1 \pmod 4$. Thus, for any even $b = 2c$ for some integer $c$, you have $a^b \equiv (3^2)^c \equiv 1^c \equiv 1 \pmod 4$.
This means $b$ must be odd. Note this gives that $b = 2d + 1$ for some $d$, so $a^b \equiv (3^2)^d(3) \equiv 3 \pmod 4$.
|
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|
Interval of convergence of a series How to get the interval of convergence for the given function,
$$f(x) = \frac{1}{2+x-x^2}$$
I have computed the Maclaurin series and the generalized power series as follows however I am unable to proceed with the valid interval of convergence. Kindly help.
$$\frac{1}{2+x-x^2} = \sum_{n=0}^{\alpha }\frac{1}{6}x^n(2(-1)^n+2^{-n}) \\
\frac{1}{2+x-x^2}=\frac{1}{2}-\frac{1}{4}x+\frac{3}{4}\frac{x^{2}}{2!}+ \ldots$$
|
We can also calculate the partial fraction decomposition
\begin{align*}
\frac{1}{2+x-x^2}=\frac{1}{3(x+1)}-\frac{1}{3(x-2)}\tag{1}
\end{align*}
and obtain this way geometric series. We see the radius of convergence of the geometric series at $x=0$ is $1$ resp. $2$. So, the radius of convergence of the left-hand side of (1) is the minimum distance from $x=0$ to the nearest singularity at $x=-1$ and the interval of convergence is $(-1,1)$.
|
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|
recursive sequence such as $x_{n+2}-2*x_{n+1}+4*x_{n}=0$ I have the recursive sequence such as $x_{n+2}-2*x_{n+1}+4*x_{n}=0$
We are looking for some solutions in the form $x_n=A* \lambda^n $
The equation becomes $A* \lambda ^{n+2}-2A \lambda ^{n+1}+4* A\lambda ^n=0
\longrightarrow P( \lambda)=\lambda ^2 - 2\lambda +4=0\longrightarrow \lambda_ {1/2}=1 \pm i\sqrt3=2 (\frac{1}{2} \pm i\frac{\sqrt3}{2})=2 (\cos\frac{\pi}{3} \pm i\sin\frac{\pi}{3})$.
$x_n=A \lambda_1^n+B \lambda_2^n=A 2^n(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3})+B2^n(\cos\frac{n\pi}{3} - i\sin\frac{n\pi}{3})=2^n (A+B)(\cos\frac{n\pi}{3} )+ 2^n (A-B)i\sin\frac{n\pi}{3}=2^n (C)(\cos\frac{n\pi}{3} )+ 2^n (D)i\sin\frac{n\pi}{3} \forall C,D \in R$
I don't understand why in the solution in my book there is
$x_n=2^n (C)(\cos\frac{n\pi}{3} )+ 2^n (D)\sin\frac{n\pi}{3} \forall C,D \in R$ without the $i$
|
The sine term's coefficient needs to be real for $x_n$ to be real. If $x_n$ is allowed to be non-real complex, so are the coefficients.
|
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|
$m=\sum_{k=1}^{2017}k^3\binom{n}{k}$. Find m. $$m=\sum_{k=1}^{2017}k^3\binom{n}{k}$$
Find m.
I was given this question for a math class I'm taking and I don't really know how to start or what identities to use. Could someone give me a hint?
|
The problem seems to mix up $n$ and $2017$ somehow. We can show
\begin{align*}
\sum_{k=1}^nk^3\binom{n}{k}=2^{n-3}n^2(n+3)\tag{1}
\end{align*}
and a special case is evaluating (1) at $n=2\,017$.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^n}&\color{blue}{k^3\binom{n}{k}}=n\sum_{k=1}^nk^2\binom{n-1}{k-1}\tag{2}\\
&=n\sum_{k=1}^{n}\left((k-1)(k-2)+3(k-1)+1\right)\binom{n-1}{k-1}\tag{3}\\
&=n\sum_{k=3}^{n}(k-1)(k-2)\binom{n-1}{k-1}
+3n\sum_{k=2}^{n}(k-1)\binom{n-1}{k-1}+n\sum_{k=1}^{n}\binom{n-1}{k-1}\tag{4}\\
&=n(n-1)(n-2)\sum_{k=3}^{n}\binom{n-3}{k-3}
+3n(n-1)\sum_{k=2}^{n}\binom{n-2}{k-2}+n\sum_{k=1}^{n}\binom{n-1}{k-1}\tag{5}\\
&=n(n-1)(n-2)\sum_{k=0}^{n-3}\binom{n-3}{k}
+3n(n-1)\sum_{k=0}^{n-2}\binom{n-2}{k}+n\sum_{k=0}^{n-1}\binom{n-1}{k}\tag{6}\\
&=n(n-1)(n-2)2^{n-3}+3n(n-1)2^{n-2}+n2^{n-1}\tag{7}\\
&\,\,\color{blue}{=2^{n-3}n^2(n+3)}
\end{align*}
and the claim (1) follows.
Comment:
*
*In (2) we use the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$.
*In (3) we represent $k^2$ as polynomial in $k-1$ and $k-2$ as preparation for the next steps.
*In (4) we multiply out and set the lower limits accordingly skipping zero terms.
*In (5) we repeatedly apply the binomial identity from (2) again.
*In (6) we shift the indices to start with $k=0$.
*In (7) we use the binomial identity $(1+1)^q=2^q$ and simplify the expression in the last step.
|
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|
minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\displaystyle x+\frac{1}{x}=t,$ then $\displaystyle x^3+\frac{1}{x^3}=t^3-3t$
$$f(t)=\frac{(t-1)^3}{t^3-3t}$$
How do i solve it without derivatives
Help me please
|
Here is a 'non-calculus' solution.
We want to minimize the function $f:(-\infty,-2] \cup [2,\infty) \to \mathbb{R}$
$$f(t) = \frac{(t-1)^3}{t^3-3t-1}$$
Let's pretend we don't know calculus and that we're really clever to spot that for $t=1+\sqrt{3}$, we have $f(t)=2\sqrt{3}-3$. So, we will prove:
$$\frac{(t-1)^3}{t^3-3t-1}\geq 2\sqrt{3}-3$$
If $t \leq -2$, we have $6t-3t^2<0$ and $t^3-3t-1 < 0$, so:
$$\frac{(t-1)^3}{t^3-3t-1} = 1+\frac{6t-3t^2}{t^3-3t-1} > 1 > 2\sqrt{3}-3$$
If $t \geq 2$, we have $t^3-3t-1 > 0$, so it will be enough to prove:
$$(t-1)^3 \geq (2\sqrt{3}-3)(t^3-3t-1)$$
Since we know $t=1+\sqrt{3}$ is the equality case, this is easy to factor into:
$$(t-1-\sqrt{3})^2\left[(4-2\sqrt{3})t+4\sqrt{3}-7\right]\geq 0$$
and this is obviously true, because
$$(4-2\sqrt{3})t+4\sqrt{3}-7 \geq 2(4-2\sqrt{3})+4\sqrt{3}-7=1 > 0$$
completing the proof.
|
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|
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
$$a+b=0;a-b+c=1; a-c=0 $$
$$a=c=\frac{1}3 \qquad b=-\frac{1}3$$
Then I wrote:
$$\frac{1}3\int \frac{1}{x-1}\,\mathrm dx-\frac{1}3\int\frac{x-1}{x^2+x+1} \, \mathrm dx$$
so the first one is just $\frac{1}{3}\ln|x-1|$. Which makes my calculations already wrong, most likely.
With the second one I tried a few different things ( involving u-substitution mostly) and got stuck.
I know I'm supposed to get this:
$$\frac{1}6\ln \frac{(x-1)^3}{x^2+x+1}+\frac{1}{\sqrt{3}} \arctan\frac{2x+1}{\sqrt{3}}$$
What have I already done wrong? What am I supposed to do?
|
You can write that
$$
x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=\frac{3}{4}\left(1+\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)^2\right)
$$
and you can then put your fraction under the form
$$
\frac{u'}{1+u^2}
$$
that integrated to an arctan.
|
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|
Prove that $6 | (a+b+c)$ if and only if $6 | (a^3 + b^3 +c^3).$ I have tried the question but not sure if my solution is correct or not...
My try..
\begin{align}a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)\end{align}
So , if
\begin{align}6 |(a^3 + b^3 + c^3)\end{align}
Then,
\begin{align}6 | [(a+b+c)^3 - 3(a+b)(b+c)(c+a)]\end{align}
So, also
\begin{align}6 | (a+b+c)^3\end{align}
\begin{align}6 | (a+b+c)(a+b+c)(a+b+c)\end{align}
So,
\begin{align}6|(a+b+c)\end{align}
Is my solution correct ?
|
The reason for your incorrect solution has been pointed out by Iris in the comments.
\begin{align}
&(a^3+b^3+c^3) - (a+b+c) = (a^3 - a) + (b^3-b) + (c^3-c) \\ &\quad \quad \ = (a-1)a(a+1) + (b-1)b(b+1) + (c-1)c(c+1)
\end{align}
The set of terms in $(a-1)a(a+1)$ contains at least one multiple of $2$ and one multiple of $3$ by the pigeonhole principle. By coprimality of $2,3$, each such number is a multiple of $6$. So $(a^3+b^3+c^3) - (a+b+c)$ is a multiple of $6$ : if one is divisible by $6$, so is the other.
|
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|
A simple equation involving trigonometry If $$\cos\theta+ \sin\theta + \tan\theta=\sqrt2+1$$
then what is the maximum value of$$\sin\theta,\sin^2\theta?$$
|
Let $a=\sin \theta,\ b=\cos \theta$. Then $a^2+b^2=1$ and
$$a+b+\frac{a}{b}=\sqrt{2}+1$$
Notice that:
$$a+b\leq |a+b| \leq \sqrt{2(a^2+b^2)} = \sqrt{2}$$
and thus we deduce that $\dfrac{a}{b}\geq 1$. So $a,b$ have the same sign. Since we want the maximum of $a$, we are only interested in the case when $a$ and $b$ are positive. In this case, using AM-GM:
$$\sqrt{2}+1=a+b+\frac{1}{\sqrt{2}} \frac{a}{b}+\left(1-\frac{1}{\sqrt{2}}\right)\frac{a}{b}\geq 3\sqrt[3]{\frac{a^2}{\sqrt{2}}}+1-\frac{1}{\sqrt{2}}$$
and from here, it's easy to find that $a\leq \dfrac{1}{\sqrt{2}}$.
|
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|
sum of binomial series with alternate terms
Evaluation of series $\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}$
what I tried:
from Binomial Identity
$$\binom{n+k}{2k}=\binom{n+k-1}{2k}+\binom{n+k-1}{2k-1}$$
series is $$\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k}+\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k-1}$$
let $\displaystyle S_{1}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{0}-4\binom{n}{2}+4^2\binom{n+1}{4}+\cdots +(-4)^n\binom{2n-1}{2n}$
let $\displaystyle S_{2}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{-1}-4\binom{n}{1}+4^2\binom{n+1}{3}+\cdots +(-4)^n\binom{2n-1}{2n-1}$
Help me please
did not know to simplify $S_{1}$ and $S_{2}$
|
Let
$$S_n=\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}\tag{*}$$
This issue can be connected in a natural way to Chebyshev polynomials of the second kind, with explicit expression see Wikipedia article.
$$U_n(x)=\sum_{k=0}^n(-2)^k\binom{n+k+1}{2k+1}(1-x)^k\tag{1}$$
If we take $x=-1$ in (1),
$$U_n(-1)=\sum_{k=0}^n(-4)^k\binom{n+k+1}{2k+1}=(-1)^n (n+1),\tag{2}$$
the second expression for $U_n(-1)$ being a known property (see Remark below).
Now, let us write the same expression as (2), but for $n-1$ instead of $n$ :
$$U_{n-1}(-1)=\sum_{k=0}^{n-1}(-4)^k\binom{n+k}{2k+1}=(-1)^{n-1} (n)\tag{3}$$
Subtracting (3) from (2),
$$\sum_{k=0}^{n-1}(-4)^k\underbrace{\left(\binom{n+k+1}{2k+1}-\binom{n+k}{2k+1}\right)}_{\binom{n+k}{2k}}+(-4)^n=\underbrace{(-1)^n (n+1) - (-1)^{n-1} (n)}_{(-1)^n (2n+1)}\tag{4}$$
(we have used the classical recurrence relationship between binomial coefficients)
It is finished there because (4) expresses the fact that $S_n$ (given by (*)) is :
$$S_n=(-1)^n (2n+1)$$
Remark : Why do we have $U_n(-1)=(-1)^n (n+1)$ ?
It can be easily seen by setting $x=-1$ into the generating function of the $U_n$s :
$$\sum_{n=0}^{\infty}U_n(x)t^n=\dfrac{1}{1-2tx+t^2}$$
giving
$$\dfrac{1}{(1+t)^2}=1-2t+3t^2-4t^3+\cdots$$
|
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|
Find all the relative extrema of $f(x)=x^4-4x^3$ Find all the relative extrema of $f(x)=x^4-4x^3$
$Solution:$
Step 1: Solve $f'(x)=0$.
$f'(x)=4x^3-12x^2=0$
$\rightarrow$ $4x^2(x-3)=0$
$\rightarrow$ $x=0$ and $x=3$
Step 2: Draw a number line and evaluate the sign of the derivative on each section (I don't know how to draw a number line on the computer but I'll do what I can).
Lets pick a number in the region $(-\infty,0)$, how about $x=-1$:
$f'(-1)=4(-1)^2(-4)=-16<0$
Now lets pick a number in the region $(0,3)$, how about $x=2$:
$f'(2)=4(2)^2(2-3)=-1<0$
Now lets pick a number in the region $(3,\infty)$, how about $x=4$:
$f'(4)=4(4)^2(4-3)=64>0$
So our function is decreasing as $x$ increases towards $0$, then our function decreases as $x$ increases towards $3$, then our function increases as $x$ increases towards $\infty$. Therefore we have a relative minimum when $x=3$ and no relative maximum. Lets solve for the corresponding $y$ value:
$f(3)=3^4-4(3)^3 = 81-4(27)=-27$
So we have a relative minimum at $(3,-27)$ and our relative maximum DNE
|
Solution without derivative.
By AM-GM $$x^4-4x^3=3\cdot\frac{x^4}{3}+27-4x^3-27\geq4\sqrt[4]{\left(\frac{x^4}{3}\right)^3\cdot27}-4x^3-27=$$
$$=4|x^3|-4x^3-27\geq-27.$$
The equality occurs for $x=3,$ which says that we got a minimal value.
The maximum does not exist because $$\lim\limits_{x\rightarrow+\infty}(x^4-4x^3)=\lim_{x\rightarrow+\infty}x^4\left(1-\frac{4}{x}\right)=+\infty.$$
|
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|
Taylor's Polynomial Evaluation Hi guys now started off doing some taylor expansion wanted some help with the following question:
$$ f:[0,\infty)\to \mathbb{R}$$
$$ f(x) = 2e^{-x/2} + e^{-x} $$
Determine the nth degree taylor's polynomila $T_n$ of f about c = 0
My attempt to this question can be seen below;
$$P(x) = f(x) + f'(x)(x-c) + \frac{f''(x)(x-c)^2}{2!} + \frac{f'''(x)(x-c)^3}{3!} + \frac{f''''(x)(x-c)^4}{4!}$$
Evaluation of the function up to the fourth derivative
$$f'(x) = -e^{-x/2} - e^{-x}$$
$$f''(x) = \frac{1}{2}e^{-x/2} + e^{-x}$$
$$f'''(x) = -\frac{1}{4}e^{-x/2} - e^{-x}$$
$$f''''(x) = \frac{1}{8}e^{-x/2} + e^{-x}$$
substituting x = 0 and we obtain the following
$$f'(x) = -2 $$
$$f''(x) = \frac{3}{2} $$
$$f'''(x) = \frac{-5}{4}$$
$$f''''(x) = \frac{9}{8} $$
substituting c = 0 and the derivatives into the taylor's polynomial we obtain the following:-
$$P(x) = 3 - 2x + \frac{\frac{3}{2}x^2}{2!} - \frac{\frac{5}{4}x^3}{3!} + \frac{\frac{9}{8}x^3}{4!} $$
I am not sure if my evaluation is correct and if so I am having difficulty putting into the form as required below hoping someone can help.
$$T_n (x) = \sum_{k=o}^\infty \frac{f^{k}(c)(x-c)^k}{k!}$$
|
If you start with the usual
$$e^{-t}=\sum_{n=0}^\infty \frac{(-1)^n}{n!} t^n$$ and apply it twice
$$f(x)=2 e^{-x/2}+e^{-x}=2\sum_{n=0}^\infty \frac{(-1)^n}{2^n\,n!}x^n+\sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(1+2^{-n+1} \right)x^n$$
|
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|
Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$. So, I tried to prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}\ge -1$$
Using Wolfram, I found this is a square:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} + 1 = \frac{(a^2+b^2+ab-a-b)^2}{(a+b)^2} $$
so it is positive.
My question is, can we prove this with more traditional and natural solution, maybe with Cauchy-Schwarz?
|
If you see this as a map $\mathbb{R}^2 \to \mathbb{R}$, it is differentiable on $\mathbb{R}^2 \backslash \{a=-b\}$. You can calculate the critical points and the hessian to determine if it is a minimum.
This only works to find a local minimum, so you should also look at the behaviour around the line $a=-b$
|
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|
Evaluate $\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy$ How to prove
$$\small\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy=
-4 \pi ^2 \left(\frac{\pi }{\sqrt{3}}+\log (2)-\frac{\psi ^{(1)}\left(\frac{1}{6}\right)}{2 \sqrt{3} \pi }\right)$$
Where $\psi^{(1)}$ denotes Trigamma? This identity arises from J. Borwein's review on experimental mathematics (which refer this formula to V. Adamchik) with no related source offered. Any help will be appreciated!
|
This is easily tackled by a series of substitution, let $I$ be your integral, then
$$\begin{aligned}I &= \int_0^{2\pi} \int_0^{2\pi} \log(3-2\cos \frac y2 \cos(x+\frac y2)-\cos y) dxdy\\
&= \int_0^{2\pi} \int_0^{2\pi} \log(3-2\cos \frac y2 \cos x-\cos y) dxdy\\
&= 4\pi\int_0^{\pi} \log\left[ \frac{1}{2} \left(-\cos \frac{y}{2}+2 \sec \frac{y}{2}+\sqrt{\cos ^2\frac{y}{2}+4 \sec ^2\frac{y}{2}-5}\right)\right] dy\\
&= 8\pi\int_0^1 \log\left(\frac{2-u^2+\sqrt{u^4-5u^2+4}}{2u}\right) \frac{1}{\sqrt{1-u^2}} du \\
&= 64\pi \int_{\pi/6}^{\pi/2} \log(2\sin t) \frac{2-\cos 2 t}{5-4 \cos 2 t} dt
\end{aligned}$$
where $4\sin^2 t = 2-u^2+\sqrt{u^4-5u^2+4}$, and $\int_0^1 \frac{\log(2u)}{\sqrt{1-u^2}}du = 0$ is used.
The last integral is straightforward by writing
$$\log(2\sin t) = -\sum_{k=1}^\infty \frac{\cos 2kt}{k}\qquad \frac{2-\cos 2 t}{5-4 \cos 2 t} = \frac{1}{2}+\frac{1}{2}\sum_{k=1}^\infty\frac{\cos 2kt}{2^k}$$
Therefore $$I = \sum_{r\geq 0, s>0} \frac{a_{r,s}}{2^r s^2}$$
where $a_{r,s}$ only depends on $r+s \pmod{3}$, the expression of $I$ in terms of polygamma follows immidiately.
|
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|
Taylor approximation is different from the actual value The problem is to use the $n$th Taylor polynomial to approximate the value of $F(x)=x^{2}\cos{x}$ with $n=3,x_{0}=-2,x=-1.09$.
I found that the $3$rd Taylor polynomial with remainder term is \begin{align*}
x^{2}\cos{x} =& 4\cos(2)+[-4\cos(2)+4\sin(2)][x+2]+[-\cos(2)-4\sin(2)][x+2]^{2}+ \\
& [2\cos(2)+ \dfrac{1}{3}\sin(2)][x+2]^{3} + [\dfrac{(x+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))]
\end{align*}
approximating $x=-1.09$ using the Taylor polynomial,
\begin{align*}
(-1.09)^{2}\cos(-1.09) =& 4\cos(2)+[-4\cos(2)+4\sin(2)][(-1.09)+2]+[-\cos(2)-4\sin(2)][(-1.09)+2]^{2}+ \\
& [2\cos(2)+ \dfrac{1}{3}\sin(2)][(-1.09)+2]^{3} + [\dfrac{((-1.09)+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))] \\
=& 0.0938985 + [\dfrac{((0.91)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))]
\end{align*}
However, $(-1.09)^{2}\cos(-1.09)=0.549479$ which is a lot of decimals different from Taylor approximation of $0.0938985$.
Is there something that I miss?
|
Expand just the cosine series around $x=-2$ and you will have:
$$x^2cos x = x^2\left(cos(2) + sin(2)[x+2] - \frac{cos(2)[x+2]^2}{2} - \frac{sin(2)[x+2]^3}{6}\right )$$
$$\\$$
(Expanding the $x^2$ in a Taylor polynomial and multiplying everything by brute force also works, but the Taylor series for $x^2$ will just reduce back to $x^2$ anyway, so it is not necessary)
$$\\$$
Evaluated at $x=-1.09$:
$$ (-1.09)^2cos(-1.09) \approx 0.549479$$
$$\\$$
$$(-1.09)^2\left(cos(2) + sin(2)[0.91] - \frac{cos(2)[0.91]^2}{2} - \frac{sin(2)[0.91]^3}{6}\right )\approx 0.557713$$
|
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|
Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$
I started with:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\sqrt{x+6})}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}=\lim_{x\to 3} \frac{3-x}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}$$
$3+\sqrt{x+6} \to 6$ is determinate, so I only need to calculate:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$$
but I could only do it with l'Hospital:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{-1}{6\cdot \frac{\pi}{18}\cos \frac{\pi x}{18}-1}=\frac{-1}{\frac{\pi\sqrt{3}}{6}-1}$$
Can I get some help without l'Hospital?
|
I assume you meant elementary limits are allowed. I'll use:
$$\lim_{x\to 0} \frac{\sin x}{x}=1$$
Instead of the fraction, it's easier to look at its inverse:
$$
\begin{aligned}
\lim_{x\to 3} \frac{6\sin \frac{\pi x}{18}-x}{3-x}&=1+\lim_{x\to 3}\frac{6\sin \frac{\pi x}{18}-3}{3-x}\\
&= 1+ 6\lim_{x\to 3}\frac{\sin \frac{\pi x}{18}-\sin \frac{\pi}{6}}{3-x}\\
&= 1+ 12\lim_{x\to 3}\frac{\sin \left(\frac{\pi x}{36}-\frac{\pi}{12}\right)\cos \left(\frac{\pi x}{36}+\frac{\pi}{12}\right)}{3-x}\\
&= 1+6\sqrt{3}\lim_{x\to 3}\frac{\sin \left(\frac{x-3}{36}\pi\right)}{3-x}\\
&= 1-6\sqrt{3}\lim_{x\to 3}\frac{\sin \left(\frac{\pi}{36}(x-3)\right)}{\frac{\pi}{36}(x-3)} \cdot \frac{\pi}{36} \\
&=1-\frac{\pi\sqrt{3}}{6}
\end{aligned}
$$
Therefore your limit equals $\dfrac{6}{6-\pi\sqrt{3}}$.
|
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|
Prove or Disprove this statement . I can't find a counter-example to the following statement :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b\geq c$ and $13a^2+5b^2\geq 13b^2+5c^2\geq 13c^2+5a^2$ then $\exists n>1$ such that :$$\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\geq \frac{13a^2+5b^2}{54}$$
$$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)}{54^2}$$
$$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\Big(\frac{c^3}{n}+ \frac{(n-1)(13c^2+5a^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)(13c^2+5a^2)}{54^3}$$
Pari-Gp have run and there is nothing against this statement .But I have a doubt .The first line is obvious .So my question concern only the two others .
If someone could prove or disprove this it will be cool.
Thanks a lot to your time .
Edit : If it's works we can add to my reasoning the Buffalo's way like here.
|
Put $N=54$, $m=n-1>0$, $x=13b^2+5c^2$, $y=13c^2+5a^2$, and $z=13a^2+5b^2$. Then the second inequality transforms to $rm+s\ge 0$, where $r=N(a^3x+b^3z)-2xz$ and $s=N^2a^3b^3-xz$. The third inequality transforms to
$um^2+vm+w\ge 0$, where $u=N(a^3xy+b^3yz+c^3xz)-3xyz$, $v=N^2(a^3b^3y+a^3c^3x+b^3c^3z)-3xyz$, and $w=N^3a^3b^3c^3-xyz$.
Now it’s time to follow the Buffalo way. Put $b=c+p$, $a=c+p+q$, and replace $xz$ by $(a+b+c)xz$ to make the total degrees of all
monomials equal. When we open the brackets in the expressions for $r$ and $u$ and simply (I did this with Mathcad), we obtain long sums of products of non-negative numbers with positive coefficients. Each of the sums equal to zero iff $p=q=0$. Thus if $p=q=0$ then $a=b=c$, $x=y=z=18a^2$, so $r=s=u=v=w=0$ and both inequalities become equalities. Otherwise $r,u>0$ and both inequalities are satisfied for sufficiently big $m$.
|
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|
Finding pattern in a sequence of polynomials So I have a set of polynomials, with variables $n_0, n_1, n_2$. I would like to figure out the general formula for these polynomials given the number.
$$f(2) = n_{0} + 2 n_{1} + n_{2}$$
$$f(3) = n_{0}^{3} + 3 n_{0} n_{1}^{2} + 3 n_{1}^{2} n_{2} + n_{2}^{3}$$
$$f(4) = n_{0}^{6} + 4 n_{0}^{3} n_{1}^{3} + 6 n_{0} n_{1}^{4} n_{2} + 4 n_{1}^{3} n_{2}^{3} + n_{2}^{6}$$
$$f(5) = n_{0}^{10} + 5 n_{0}^{6} n_{1}^{4} + 10 n_{0}^{3} n_{1}^{6} n_{2} + 10 n_{ 0} n_{1}^{6} n_{2}^{3} + 5 n_{1}^{4} n_{2}^{6} + n_{2}^{10}$$
$$f(6) = n_{0}^{15} + 6 n_{0}^{10} n_{1}^{5} + 15 n_{0}^{6} n_{1}^{8} n_{2} + 20 n _{0}^{3} n_{1}^{9} n_{2}^{3} + 15 n_{0} n_{1}^{8} n_{2}^{6} + 6 n_{1}^{5} n_{2}^{10} + n_{2}^{15}$$
$$f(7) = n_{0}^{21} + 7 n_{0}^{15} n_{1}^{6} + 21 n_{0}^{10} n_{1}^{10} n_{2} + 35 n_{0}^{6} n_{1}^{12} n_{2}^{3} + 35 n_{0}^{3} n_{1}^{12} n_{2}^{6} + 21 n_{0} n_{1}^{10} n_{2} ^{10} + 7 n_{1}^{6} n_{2}^{15} + n_{2}^{21}$$
As you can see that the coefficients are given by binomial expansion but the variables multiply and have degrees in a non-trivial manner. Is there a pattern to degree and combinations of variables, can a closed-form in the number of constants (i.e $1,2,3,4 ..$) be achieved ?? I can post a larger list of polynomials if needs be.
|
I think it is
$$
f(k)=\sum\limits_{j = 0}^{k} \binom{k}{j}n_0^{\frac{(k - j)(k - j - 1)}{2}} n_1^{(k-j)j} n_2^{\frac{j(j - 1)}{2}}
$$
for $k\geq 2$.
|
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|
Evaluate the summation $\sum_{n=1}^{10} n \left( \frac{1^2}{1 + n} + \frac{2^2}{2 + n} + ....+\frac{10^2}{10 + n}\right)$ $$P = \sum_{n=1}^{10} n \left( \frac{1^2}{1 + n} + \frac{2^2}{2 + n} + ....+\frac{10^2}{10 + n}\right)$$
My attempt: Get the constant square terms out and evaluate the summations left in $n$
$$P = 1^2\sum_{n=1}^{10}\frac{n}{1 + n} + 2^2\sum_{n=1}^{10}\frac{n}{2 + n} + ... + 10^2\sum_{n=1}^{10}\frac{n}{10 + n}$$
for the summation $\sum_{n=1}^{10}\frac{n}{k + n}$:
$$S(k) = \sum_{n=1}^{10}1-\frac{k}{k + n} \\ = 10 - k\sum_{n=1}^{10}\frac1{k+n}$$
Substituting $S(k)$ back, the sum reduces to
$$(1^2+2^2+...+10^2)10 - \sum_{k=1}^{10}\sum_{n=1}^{10}\frac{k^3}{k+n}$$
If the double summation on the right can be evaluated, the sum would be solved. Have I made a complex problem even more complex via this approach? Are there any easier ways to solve this problem?
P.S: The answer mentioned is $P = \frac{(1+2+3+....+10)^2}{2}$.
Edit: On checking with Maple, the correct answer is $55$ and not the above answer. I still do not know how maple obtained 55 (apparently $\sum n$ till 10) from this expression.
|
$P=S(10)$ where $$S(m)=\sum_{n=1}^m\sum_{k=1}^m\frac{nk^2}{k+n}=\frac{1}{2}\sum_{n=1}^m\sum_{k=1}^m\frac{nk^2+kn^2}{k+n}=\frac{1}{2}\sum_{n=1}^m\sum_{k=1}^m nk=\frac{1}{2}\left(\sum_{n=1}^m n\right)^2\color{LightGray}{=\ldots}$$
|
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|
Prove $\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$ If $a,b,c>0$, prove that:
$$\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
My try: I used $a^4+b^4 \geq ab(a^2+b^2)$ to get:
$$(a^4+b^4)(a^4+c^4)(b^4+c^4) \geq a^2b^2c^2(a^2+b^2)(a^2+c^2)(b^2+c^2)$$
I am stuck with this inequality:
$$\sqrt[3]{(a^2+b^2)(a^2+c^2)(b^2+c^2)}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
I am not sure if it is true or not.
|
Raising to the $6$th power, the inequality is equivalent with:
$$(1+abc)^3(a^4+b^4)^2(b^4+c^4)^2(c^4+a^4)^2 \geq a^4b^4c^4(a^3+bc)^3(b^3+ca)^3(c^3+ab)^3$$
This is wrong for large variables as Michael Rozenberg stated. However, I found that
$$(1+abc)^3(a^4+b^4)^2(b^4+c^4)^2(c^4+a^4)^2 \geq a^2b^2c^2(a^3+bc)^3(b^3+ca)^3(c^3+ab)^3$$
is true. Notice that, using Holder's inequality, we have:
$$(a^4+b^4)(a^4+c^4)(abc+1)\geq (a^3\sqrt[3]{bc}+bc\sqrt[3]{bc})^3=bc(a^3+bc)^3$$
and now multiply with the other two similar inequalities to complete it.
|
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|
Integrate $\int \frac{1}{1+x+x^4}dx$
Find $$\int \frac{{\rm d}x}{1+x+x^4}.$$
WA gives a result, which introduces complex numbers. But according to the algebraic fundamental theorem, any rational fraction can be integrated over the real number field.
How to do this?
|
Factorize
$$x^4 +x+ 1= \left(x^2+ax+\frac{a^3-1}{2a}\right)\left(x^2-ax+\frac{a^3+1}{2a}\right)$$
where $a$ satisfies $a^6-4a^2-1=0$, or
$$a=\frac2{\sqrt[4]3}\sqrt{\cos\left(
\frac13\cos^{-1}\frac{3\sqrt3}{16} \right)}
$$
Then, decompose the integrand
$$\frac1{1+x+x^4}
= \frac a{2a^6+1} \bigg(\frac{2a^2x+2a^3+1}{x^2+ax+\frac{a^3-1}{2a}}
-\frac{2a^2x-2a^3+1}{x^2-ax+\frac{a^3+1}{2a}} \bigg)$$
and integrate
\begin{align}
&\int \frac{dx}{1+x+x^4}\\
=&\ \frac a{2a^6+1} \int \bigg(\frac{a^2(2x+a)+ a^3+1}{x^2+ax+\frac{a^3-1}{2a}}
-\frac{a^2(2x-a)-a^3+1}{x^2-ax+\frac{a^3+1}{2a}} \bigg)dx\\
=&\ \frac {a^{3/2}}{a^6+\frac12} \bigg(
\frac {a^3+1}{\sqrt{a^3-2}}\tan^{-1}\frac{\sqrt a(2x+a)}{\sqrt{a^3-2}}
+\frac {a^3-1}{\sqrt{a^3+2}}\tan^{-1}\frac{\sqrt a(2x-a)}{\sqrt{a^3+2}}\\
&\hspace{20mm} + \frac{a^{3/2}}2\ln \frac{x^2+ax+\frac{a^3-1}{2a}}{x^2-ax+\frac{a^3+1}{2a}} \bigg)+C
\end{align}
|
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|
Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$ Problem: Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$.
I manage to make progress, but then end up always getting a $\sum\frac{n}{2^{n}}$. For example, $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}$ = $\sum_{n=0}^{\infty}\frac{n}{2^{n+2}}$ = $\frac{1}{4}\sum_{n=0}^{\infty}\frac{n}{2^{n}}$, which is the sum we're not allowed to use.
Any help on how to proceed would be appreciated!
|
Render
$S=\dfrac{0}{2^1}+\dfrac{1}{2^2}+\dfrac{2}{2^3}+...$
Divide by $2$:
$S/2=\dfrac{0}{2^2}+\dfrac{1}{2^3}+\dfrac{2}{2^4}+...$
Now take the difference between terms with like denominators, e.g. $1/2^2$ from the first equation minus $0/2^2$ from the second. This gives
$S-S/2=\dfrac{0}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...$
where all nonzero terms on the right side have $1$ as numerator. The left side is $S/2$ and the right side may be summed as a geometric series, from which we then extract $S$.
This is essentially the way to prove the sum of a series where an arithmetic sequence is multiplied by a geometric one.
|
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|
Using the necessary and sufficient conditions compute all minima and maxima of a given function
Using the necessary and sufficient conditions compute all minima and maxima of
$$f(x,y) = x^2 + 8x\cos(y) − 2x + 10\cos(y) − 8\cos^2(y) + 8\cos^3(y)$$
for $(x,y) \in \mathbb{R}^2$.
My attempt:
Take the partial derivatives on both variables w.r.t $\,x$ and $y$, this gives $$2x+8\cos(y)-2$$
and
$$\sin(y)(-8x-10+16\cos(y)-24\cos^2(y)),$$
respectively.
Equating each of these two equations to $0$, we get either $\sin(y)=0$ giving $y=2k\pi$ and $x=-3$, or $y=2k\pi+\pi$ and $x=5$ for $k\in\mathbb{Z}$.
My problem is when $(-8x-10+16\cos y-24\cos^2y)=0,$ what are the values of $x$ and $y$?
|
If $2x+8\cos(y)−2=0$ then
$$\cos(y)=\frac{1-x}{4}.$$
Moreover $-8x-10+16\cos(y)-24\cos^2(y)=0$ implies
$$4x=-5+8\cos(y)-12\cos^2(y)=-5+2(1-x)-(3/4)(1-x)^2$$
that is
$$x^2+6x+5=0$$
which has two real solutions: $x=-1$ and $x=-5$.
Hence
1) if $x=-1$ then $\cos(y)=1/2$ and we have further critical points:
$$(-1,\pi/3+2k\pi),\quad (-1,-\pi/3+2k\pi)\quad\text{for $k\in\mathbb{Z}$.}$$
2) If $x=-5$ then $\cos(y)=3/2$: no real solutions.
|
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|
For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$ A 11th grade inequality problem:
Let $a,b,c$ be non-negative real numbers. Prove that
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
Do you have any hints to solve this inequality? Any hints would be fine.
I tried this:
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
By Cauchy's inequality, $$a^2+1\ge(2a)$$ and I did the same to $b$ and $c$ and applied it to the problem but the results are $$2abc\ge1+a^2b^2c^2$$ and this is wrong.
|
The Macavity's beautiful idea we can release also by C-S:
$$2(1-a+a^2)(1-b+b^2)\geq2\sqrt{\frac{1+a^4}{2}}\cdot\sqrt{\frac{1+b^4}{2}}\geq1+a^2b^2.$$
|
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|
Integrating odd Legendre polynomials using generating function I must show using generating function of Legendre polynomials, that
\begin{align}
\int_0^1 P_{2n+1}(x)\phantom{1}dx = (-1)^n\frac{(2n)!}{2^{2n+1}n!(n+1)!}
\end{align}
My attempt is to change the generating function $\Phi(u,x)=(1-2ux+u^2)^{-1/2}=\sum_{n=0}^{\infty}u^nP_n(x)$ into infinite series using Newton's generalized binomial theorem
\begin{align}
(1-2ux+u^2)^{-1/2} = \sum_{n=0}^{\infty}(-1)^n\frac{(2n)!}{2^{2n}(n!)^2}(-2ux+u^2)^n=\sum_{n=0}^{\infty}u^nP_n(x)
\end{align}
and then integrating from $0$ to $1$ with respect to $x$. But it seems unsuccessfull. Can you give me some hint to answer this? Or, I must starting from where?
|
Expand the result of the integral of the generating function with Taylor series, using
$$\sqrt{1+z} = \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}z^n = 1+\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}z^{n+1}}{2^{2n+1}(n+1)}$$
As follows:
\begin{equation}
\begin{split}
I &= \int_0^1 (1-2ux+u^2)^{-1/2}dx\\
&= \left[-\frac{1}{u}\sqrt{1-2ux+u^2}\right]_{x=0}^{x=1}\\
&= -\frac{1}{u}(\sqrt{1-2u+u^2}-\sqrt{1+u^2})\\
&= -\frac{1}{u}(1-u-\sqrt{1+u^2})\\
&= -\frac{1}{u}(1-u-\left(1+\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}u^{2n+2}}{2^{2n+1}(n+1)}\right))\\
&= 1+\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}u^{2n+1}}{2^{2n+1}(n+1)}
\end{split}
\end{equation}
Comparing with $I = \sum_{m=0}^\infty u^m \int_0^1 P_m(x)dx$, we conclude that
$$ \int_0^1 P_{2n+1}(x)dx = \frac{(-1)^n\binom{2n}{n}}{2^{2n+1}(n+1)}=\frac{(-1)^n(2n)!}{2^{2n+1}(n+1)(n!)^2} = \frac{(-1)^n(2n)!}{2^{2n+1}n!(n+1)!}$$
|
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"url": "https://math.stackexchange.com/questions/3586831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove $\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$ (part of sum) I'm looking for a specific clarification for a part of the solution in proving the following identity.
$\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$
Here I'm taking, $\theta=\tan^{-1}m;$ $-\pi/2<\theta<\pi/2$
so I get $\tan\theta=m$---(1)
I need to find $\sin \theta $and $\cos \theta$ in terms of m
By Trigonometric identity I can easily derive, $\cos \theta$
$\tan^2\theta+1=\sec^2\theta$
$\cos^2\theta=\frac{1}{m^2+1}$
$\cos\theta=+\sqrt\frac{1}{m^2+1}$ ( here only plus due to range of $\theta$)
Now if I deduce $\sin\theta$ from,
$\sin^2\theta+\cos^2\theta=1$
I get, $\sin\theta=\pm\sqrt\frac{m^2}{m^2+1}$ ( I have to take $\pm$ because of range of $\theta$)
But if I deduce $\sin\theta$ from (1)
I get, $\sin\theta=\frac{m}{\sqrt {m^2+1}}$
Which of the following method is correct to find $\sin\theta$? Please help me. Thank you!
P.S. I'm not interested in the solution. What I need to know is how to find $\sin\theta$
|
I don't have means to put up the image for my plot so bear with me.
Take a right-angled triangle $ABC$ such that AB is the height $=m$ and BC is the base $=1$. $\widehat{ABC}=90^\circ$ and $\widehat{ACB}=\tan^{-1} m$
Let $AC=y$. Draw a line $AE$ perpendicular to $AC$ at $A$ equal to $ny$. $\widehat{EAC}=90^\circ$ and $\widehat{ACE}=\tan^{-1} \dfrac{ny}{y}=\tan^{-1} n$. Additionally, let $EC=z$
Extend $AB$ past $A$. Draw a perpendicular from E to that line and name that intersection $D$. $\widehat{EAD}=\widehat{ACB}$. This is because of the right angle $\widehat{EAC}$. Hence $\triangle ABC \sim \triangle EDA$. This equality follows:
$\dfrac{AB}{AC}=\dfrac{ED}{EA}$
$\dfrac{m}{y}=\dfrac{ED}{ny}$
$ED=mn$
Draw a perpendicular to $BC$ at $C$ and extend $DE$ past $E$. Label the point where these lines meet as $F$. We have created a rectangle in $DBCF$, thus $DF=BC=1$:
$EF+DE=1$
$EF=1-mn$
Additionally, $DF \parallel BC$ hence $\widehat{ECB}=\widehat{CEF} \text{ Alternate angles}$
$\widehat{ECB}=\widehat{ACB}+\widehat{ECA}=\tan^{-1} m+\tan^{-1} n$
$\widehat{CEF}=\cos^{-1} \dfrac{EF}{EC}$
$\dfrac{EF}{EC}=\dfrac{1-mn}{z}$
Remember now all the right-angled triangles we drew.
$z^2=(ny)^2+y^2=y^2(n^2+1)$
$y^2=m^2+1$
$z^2=(n^2+1)(m^2+1)$
$z=\sqrt{(n^2+1)(m^2+1)}$
$\dfrac{EF}{EC}=\dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}$
$\widehat{CEF}=\cos^{-1} \dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}$
But $\widehat{ECB}=\widehat{CEF}$, hence:
$\boxed{\tan^{-1} m+\tan^{-1} n=\cos^{-1} \dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}}$
Edit: the formula kinda breaks down when $m$ and $n$ don't have the same sign i.e $n \lt 0 \lt m$ for example. But they won't be that different. For example, using $-\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ yields $-30^\circ$ and $30^\circ$ on the left and right respectively. But they do have the same cosine value meaning they're not too distinct.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving inequality by induction produces an unclear answer I need to prove following assertion by induction: $∀n ∈ N , ∀ a,b >0: a^{n-1}b \leq (\frac{(n-1)a+b}{n})^n$
After successfully proving the assertion for n = 1, I establish the following assumption:
$a^{k-1}b \leq (\frac{(k-1)a+b}{k})^k$
Now when I try to prove for k + 1, which yields $a^{k}b \leq (\frac{((k+1)-1)a+b}{k+1})^{k+1}$,
I get following inequality:
$a^{k-1}a b \leq ((\frac{(k-1)a+b}{k})^{n} ) a $.
and this is where I have some troubles. I am unable to proof that $((\frac{(k-1)a+b}{k})^{n} ) a \leq (\frac{((k+1)-1)a+b}{k+1})^{k+1}$, which ultimately hinders me from finding a solution.
|
The inequality is
$a^{n-1}b
\leq (\frac{(n-1)a+b}{n})^n
= (a+\frac{b-a}{n})^n
= a^n (1+\frac{b/a-1}{n})^n
$
or
$b/a
\le
(1+\frac{b/a-1}{n})^n
$
or,
letting $r = b/a$,
$r
\le (1+(r-1)/n)^n
$
or,
letting $r = 1+ns$,
$1+ns
\le (1+s)^n
$.
This is Bernoulli's inequality
which is easily proved
by induction.
The induction step is
$\begin{array}\\
(1+s)^{n+1}
&=(1+s)^n(1+s)\\
&\ge (1+ns)(1+s)\\
&= 1+(n+1)s+ns^2\\
&\ge 1+(n+1)s\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Inequality from Israel TST Let $a, b, c, d$ be nonnegative numbers such that $a+b+c+d=18.$
Prove that:
$$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$
These are my attempts to solve the Problem:
There is an equality case in $(a, b, c, d) = (8, 1, 8, 1)$ and similar cyclic permutations.
To preserve the equality case we'll use Cauchy that way:
$$\sum_{cyc}\sqrt{\frac{a}{b+6}}=\sum_{cyc}\sqrt{\frac{a}{a+6}\cdot\frac{a+6}{b+6}}\leq\sqrt{\sum_{cyc}\frac{a}{a+6}\sum_{cyc}\frac{a+6}{b+6}}$$
Which preserves the equality case. I don't know how to proceed now.
Any ideas?
|
Let $\{a,b,c,d\}=\{x,y,z,t\}$ such that $x\geq y\geq z\geq t$.
Thus, since $(\sqrt{x},\sqrt{y},\sqrt{z},\sqrt{t})$ and $\left(\frac{1}{\sqrt{t+6}},\frac{1}{\sqrt{z+6}},\frac{1}{\sqrt{y+6}},\frac{1}{\sqrt{x+6}}\right)$ are same ordered,
by Rearrangement, AM-GM and by Jensen for $f(x)=\sqrt{x+12}$ and for $g(x)=-\frac{1}{\sqrt{x+12}}$
we obtain:
$$\sum_{cyc}\sqrt{\frac{a}{b+6}}\leq\sqrt{\frac{x}{t+6}}+\sqrt{\frac{y}{z+6}}+\sqrt{\frac{z}{y+6}}+\sqrt{\frac{t}{x+6}}=$$
$$=\sum_{cyc,x\rightarrow y,t\rightarrow z}\left(\sqrt{\frac{x}{t+6}}+\sqrt{\frac{t}{x+6}}\right)=\sum_{cyc}\sqrt{\frac{x}{t+6}+\frac{t}{x+6}+2\sqrt{\frac{xt}{(x+6)(t+6)}}}=$$
$$=\sum_{cyc}\sqrt{\frac{x+t}{6}+\frac{x}{t+6}-\frac{x}{6}+\frac{t}{x+6}-\frac{t}{6}+2\sqrt{\frac{xt}{(x+6)(t+6)}}}=$$
$$=\sum_{cyc}\sqrt{\frac{x+t}{6}-\frac{xt(x+t+12)}{6(x+6)(t+6)}+2\sqrt{\frac{xt}{(x+6)(t+6)}}}=$$
$$=\frac{1}{\sqrt6}\sum_{cyc}\sqrt{x+t-\frac{xt(x+t+12)}{(x+6)(t+6)}+2\sqrt{\frac{36xt}{(x+6)(t+6)}}}\leq$$
$$\leq\frac{1}{\sqrt6}\sum_{cyc}\sqrt{x+t-\frac{xt(x+t+12)}{(x+6)(t+6)}+\frac{xt(x+t+12)}{(x+6)(t+6)}+\frac{36}{x+t+12}}=$$
$$=\frac{1}{\sqrt6}\sum_{cyc}\sqrt{\frac{(x+t)^2+12(x+t)+36}{x+t+12}}=\sum_{cyc}\frac{x+t+6}{\sqrt{6(x+t+12}}=$$
$$=\sum_{cyc}\frac{x+t+12-6}{\sqrt{6(x+t+12}}=\sum_{cyc}\left(\frac{1}{\sqrt6}\cdot\sqrt{x+t+12}-\sqrt6\cdot\frac{1}{\sqrt{x+y+12}}\right)\leq$$
$$\leq\frac{2}{\sqrt6}\sqrt{\frac{x+t+y+z}{2}+12}-\frac{2\sqrt6}{\sqrt{\frac{x+t+y+z}{2}+12}}=\frac{2\sqrt{21}}{\sqrt6}-\frac{2\sqrt6}{\sqrt{21}}=5\sqrt{\frac{2}{7}}.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3591099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Can someone confirm if my answer is right?( complex numbers) I'm fairly new to complex numbers and there wasn't much I could find online but could someone see if my answer is right? we need to find $z = a+bi$ given that $z^2 = -144+12i\sqrt{8}$
for LHS:
$$z^2 = (a+bi)^2 = a^2 + 2abi - b^2$$
$$a^2 +2abi - b^2 = -144+12i\sqrt{8}$$
equating real numbers : $$a^2 - b^2 = -144$$
equating imaginary numbers : $$2ab = 12\sqrt{8}$$
solving simultaneously gives
$$a = +1.4 \text{ or } -1.4 \\
b = +12.1 \text{ or } -12.1 $$
hence $z = 1.4 + 12.1i \text{ or } z = -1.4 - 12.1i$
|
Your calculation are correct, in fact $a,b$ are solutions to the following system:
$$\left\{\begin{matrix}
a^2-b^2=-144
\\ 2ab=12\sqrt{8}
\end{matrix}\right.$$
From here, it's very easy to see that the solutions are:
$$\left\{\begin{matrix}
a=\frac{7}{5}
\\ b=\frac{30\sqrt{8}}{7}
\end{matrix}\right.$$
and:
$$\left\{\begin{matrix}
a=-\frac{7}{5}
\\ b=-\frac{30\sqrt{8}}{7}
\end{matrix}\right.$$
As a general rule, I don't advise you to trunc the solutions, but write:
$$z_1=\frac{7}{5}+\frac{30\sqrt{8}}{7}i \: \vee \: z_2=-\frac{7}{5}-\frac{30\sqrt{8}}{7}i$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3591265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find a set $C(N)$ of $n \times n$ matrices that commute with $\it{N}$ I have a $n \times n$ matrix $N$, where $n \in \mathbb{N}$.
$N = \begin{bmatrix}
0 & 1 & 0 & \dots & 0 & 0 \\
0 & 0 & 1 & \dots & 0 & 0 \\
0 & 0 & 0 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 0 & 1 \\
0 & 0 & 0 & \dots & 0 & 0\end{bmatrix}$
Determine a set $C(N)$ that contains all the $n \times n$ matrices that commute with $N$
I have tried defining some matrix $A$ and taking its $j$ column $A_j$ and then tried to multiply it with vector $e_j = \begin{bmatrix} 0 & \dots & 0 & 1 & 0 & \dots & 0 \end{bmatrix}^T$, that has $1$ only on index $j$, so things would maybe simplify, but then I get completely lost. I would really need an explanation on this.
|
Consider a generic matrix $A=(a_{i,j})_{i,j}$ and impose $AN=NA$. You have to be very careful, but you obtain these condition:
\begin{equation}
AN=
\left[
\begin{matrix}
0 & a_{1,1} & a_{1,2} & \cdots & \cdots & a_{1,n-1}\\
0 & a_{2,1} & a_{2,2} & \cdots & \cdots & a_{2,n-1}\\
0 & a_{3,1} & a_{3,2} & \cdots & \cdots & a_{3,n-1}\\
\vdots & & & & & \vdots\\
\vdots & & & & & \vdots\\
0 & a_{n-1,1}& a_{n-1,2} & \cdots & \cdots & a_{n-1,n-1}\\
0 & a_{n,1} & a_{n,2} & \cdots & \cdots & a_{n,n-1}\\
\end{matrix}
\right]
=\left[
\begin{matrix}
a_{2,1} & a_{2,2} & a_{2,3} & \cdots & \cdots & a_{2,n}\\
a_{3,1} & a_{3,2} & a_{3,3} & \cdots & \cdots & a_{3,n}\\
a_{4,1} & a_{4,2} & a_{4,3} & \cdots & \cdots & a_{4,n}\\
\vdots & & & & & \vdots\\
\vdots & & & & & \vdots\\
a_{n,1} & a_{n,2} & a_{n,3} & \cdots & \cdots & a_{n,n-1}\\
0 & 0 & 0 & \cdots & \cdots & 0\\
\end{matrix}
\right] =NA
\end{equation}
Reading these conditions along the diagonals you can observe that $a_{i,j} = a_{i+1,j+1}$ for all $i,j$ admitted. In particular you have that $A$ is upper triangular of this form:
\begin{equation}
\left[
\begin{matrix}
b_1 & b_2 & b_3 & \cdots & \cdots & b_n\\
0 & b_1 & b_2 & \cdots & \cdots & b_{n-1}\\
0 & 0 & b_1 & \cdots & \cdots & b_{n-2}\\
\vdots & & & & & \vdots\\
\vdots & & & & & \vdots\\
0 & 0 & 0 & \cdots & \cdots & b_{2}\\
0 & 0 & 0 & \cdots & \cdots & b_1\\
\end{matrix}
\right]
\end{equation}
|
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"url": "https://math.stackexchange.com/questions/3591427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
A three-parameters identity involving Stirling numbers of both kinds Let $n, m, k $ be three natural numbers, ${n \brack k}$ and ${n \brace k}$ the Stirling numbers of first and second kind respectively.
We have:
$$ \tag{*} {n-1 \choose m}{n-m \brack k}= \sum_i (-1)^{i-m}{k-1+i
\choose k-1}{i \brace m}{n \brack i+k} $$
where the bounds for $i$ in the sum on the rhs don't need to be specified as there is only a finite number of values of $i$ whose corresponding summand is non-zero and the sum is understood over all such $i$.
This identity can be verified numerically and can be derived from another three parameters identity involving the second kind of Stirling numbers only- namely Eq. (6.28) in Concrete Mathematics Second Edition, R. L. Graham, D. E. Knuth, O. Patashnik)
$$ \tag{**} {\ell+m \choose \ell}{n \brace \ell+m}= \sum_k {k\brace \ell}{n-k \brace m}{n \choose k} $$
which is obtained rather easily via the exponential generating functions of ${n \brace l+m}$, ${n \brace m}$ and ${n \brace l}$ .
Indeed, if we replace $m$ by $-m$ and $n$ by $-n$ in (**), taking into account that ${-a \brace -b}$ = ${b \brack a}$ and ${-n \choose k}=(-1)^k{n+k-1\choose k}$, we obtain
\begin{align*} {\ell-m \choose \ell}{-n \brace \ell-m}&= \sum_k {k\brace \ell}{-n-k \brace -m}{-n \choose k} \\
(-1)^\ell{m-1 \choose \ell}{m- \ell \brack n}&= \sum_k {k\brace \ell}{m \brack n+k}(-1)^k{n+k-1 \choose k}\end{align*}
which is (*) after the appropriate change of notation.
But in Concrete Mathematics, the identity (**) is given under the condition $\ell,m,n \ge 0$, so I am note sure whether it is licit to do such negation of the indices.
Then my question is: how can we derive (*) directly, without resorting to (**). Maybe with generating functions, coefficient extractors or things like that?
|
We seek to verify that
$$\sum_{q=m}^{n-k} (-1)^{q-m} {k-1+q\choose k-1}
{q\brace m} {n\brack q+k}
= {n-1\choose m} {n-m\brack k}.$$
Using the standard EGFs the LHS becomes
$$\sum_{q=m}^{n-k} (-1)^{q-m} {k-1+q\choose k-1}
q! [z^q] \frac{(\exp(z)-1)^m}{m!}
n! [w^n] \frac{1}{(q+k)!} \left(\log\frac{1}{1-w}\right)^{q+k}
\\ = \frac{n!}{(k-1)! \times m!} [w^n] \sum_{q=m}^{n-k} (-1)^{q-m}
[z^q] (\exp(z)-1)^m
\frac{1}{q+k} \left(\log\frac{1}{1-w}\right)^{q+k}
\\ = \frac{(n-1)!}{(k-1)! \times m!} [w^{n-1}]
\sum_{q=m}^{n-k} (-1)^{q-m}
[z^q] (\exp(z)-1)^m
\left(\log\frac{1}{1-w}\right)^{q+k-1} \frac{1}{1-w}
\\ = \frac{(n-1)!}{(k-1)! \times m!} [w^{n-1}] \frac{1}{1-w}
\\ \times \sum_{q=m}^{n-k} (-1)^{q-m}
[z^{q+k-1}] z^{k-1} (\exp(z)-1)^m
\left(\log\frac{1}{1-w}\right)^{q+k-1}
\\ = \frac{(n-1)!}{(k-1)! \times m!} [w^{n-1}] \frac{1}{1-w}
\\ \times \sum_{q=m+k-1}^{n-1} (-1)^{q-(k-1)-m}
[z^{q}] z^{k-1} (\exp(z)-1)^m
\left(\log\frac{1}{1-w}\right)^{q}.$$
Now as $\log\frac{1}{1-w} = w + \cdots$ when $q\gt n-1$ there is
no contribution from the logarithmic power term due to the coefficient
extractor $[w^{n-1}]$ so we find
$$(-1)^{m+(k-1)} \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1}] \frac{1}{1-w}
\\ \times \sum_{q\ge m+k-1} (-1)^{q}
\left(\log\frac{1}{1-w}\right)^{q}
[z^{q}] z^{k-1} (\exp(z)-1)^m.$$
Note that $z^{k-1} (\exp(z)-1)^m = z^{m+k-1} + \cdots$ which
means that the remaining sum / coefficient etractor pair covers the
entire series and we get
$$(-1)^{m+(k-1)} \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1}] \frac{1}{1-w}
\\ \times (-1)^{k-1} \left(\log\frac{1}{1-w}\right)^{k-1}
\left(\exp\left(-\log\frac{1}{1-w}\right)-1\right)^m
\\ = (-1)^{m+(k-1)} \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1}] \frac{1}{1-w}
\\ \times (-1)^{k-1} \left(\log\frac{1}{1-w}\right)^{k-1} (-w)^m
\\ = \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1-m}] \frac{1}{1-w}
\left(\log\frac{1}{1-w}\right)^{k-1}
\\ = \frac{(n-1)!}{m!}
[w^{n-1-m}] \frac{1}{1-w}
\frac{1}{(k-1)!} \left(\log\frac{1}{1-w}\right)^{k-1}
\\ = \frac{(n-1)!}{m!}
(n-m) [w^{n-m}]
\frac{1}{k!} \left(\log\frac{1}{1-w}\right)^{k}
\\ = \frac{(n-1)!}{m! \times (n-1-m)!}
(n-m)! [w^{n-m}]
\frac{1}{k!} \left(\log\frac{1}{1-w}\right)^{k}
\\ = {n-1\choose m} {n-m\brack k}.$$
This is the claim.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Line Integral with Change of Variables I'm a bit rusty on my computational math, and genuinely can't solve this question which is frustrating me.
QUESTION:
$$\int_Csin(y)dx+xcos(y)dy$$ where C is the ellipse defined as follows: $x^2+xy+y^2=1$
MY ATTEMPT:
Define variable $u= \sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y}$
Define variable $v= \sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y$
Hence:
$u^2+v^2=(\sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y})^2+(\sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y)^2$
$=(\frac{3}{4}x^2+\frac{(2)(3)}{4}xy+\frac{3}{4}y^2)+(\frac{1}{4}x^2-\frac{2}{4}xy+\frac{1}{4}y^2)$
$=(\frac{3}{4}+\frac{3}{4})x^2+(\frac{3}{2}-\frac{1}{2})xy+(\frac{3}{4}+\frac{1}{4})y^2$
$=(x^2+xy+y^2) = C$
Therefore, we have Jacobian:
$$
\begin{bmatrix}
\frac{\partial{u}}{\partial{x}} & \frac{\partial{u}}{\partial{y}} \\
\frac{\partial{v}}{\partial{x}} & \frac{\partial{v}}{\partial{y}} \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
\sqrt\frac{3}{4} & \sqrt\frac{3}{4} \\
\sqrt\frac{1}{4} & -\sqrt\frac{1}{4} \\
\end{bmatrix}
$$
$$ = \frac{\sqrt{3}}{2} $$
Beyond this point, I must incorporate the original dot product with the differential $(dx,dy)$ but don't know how
|
The integral is path independent because
$$\nabla(x\sin y) = (\sin y, x\cos y)$$
So, the integral along the closed contour $C$ vanishes:
$$\int_C (\sin y\,dx + x\cos y\,dy) =0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$ Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$
My Attempt:
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}{2}\log_2 (x^2+7))=-2$$
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+1+\log_{1}{2}(\frac {1}{2} \log_2(x^2+7))=-1$$
$$\log_\frac {3}{4} (\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {3}{4}} (\dfrac {3}{4})=-(1+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-(\log_{\frac {1}{2}} (\frac {1}{2})+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-\log_{\frac {1}{2}} (\frac {1}{4} \log_2 (x^2+7))$$
|
I want to do some replacement but it's going to be tricky.
$\log_8 (x^2+7)=a$
$8^a=x^2+7$
$2^{3a}=x^2+7$
$2^{-3a}=(x^2+7)^{-1}$
$(2^{-2})^{\frac{3}{2}a}=(x^2+7)^{-1}$
$\left(\dfrac{1}{4} \right)^{\frac{3}{2}a}=(x^2+7)^{-1}$
$\log_{\frac{1}{4}} (x^2+7)^{-1}=\dfrac{3}{2}a$
The whole thing changes to:
$\log_{\frac{3}{4}} a+\log_{\frac{1}{2}} \dfrac{3}{2}a=-2$
And with the change of base formula (you'll realize the base doesn't matter) then:
$\dfrac{\log a}{\log \dfrac{3}{4}}+\dfrac{\log \dfrac{3}{2}+\log a}{\log \dfrac{1}{2}}=-2$
$\log a \left(\log \dfrac{1}{2}+\log \dfrac{3}{4} \right)+\log \dfrac{3}{4}\log \dfrac{3}{2}=-2 \cdot \log \dfrac{3}{4}\log \dfrac{1}{2}$
$\log a \log \dfrac{3}{8}=\log \dfrac{3}{4}\log \left(\dfrac{1}{2} \right)^{-2}-\log \dfrac{3}{4}\log \dfrac{3}{2}$
$=\log \dfrac{3}{4} \left(\log 4-\log \dfrac{3}{2} \right)=\log \dfrac{3}{4}\log \dfrac{8}{3}$
$\log a=\dfrac{\log \dfrac{3}{4}\log \dfrac{8}{3}}{\log \dfrac{3}{8}}=\dfrac{-1 \cdot \log \dfrac{3}{4}\log \dfrac{3}{8}}{\log \dfrac{3}{8}}$
$=-\log \dfrac{3}{4}=\log \dfrac{4}{3}$
$a=\dfrac{4}{3}$
$8^{\frac{4}{3}}=16=x^2+7$
$x=\pm 3$
I'm tired now. Let me log out.
|
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|
How to find the common factor I am currently stomped at this problem $(2 x^2-(y+z) (y+z-x))/(2 y^2-(z+x) (z+x-y))$ I am supposed to factor it and I can't find a way how to get to the answer.
It says that
$(2 x^2+x y-y^2+x z-2 y z-z^2)/(2 y^2-x^2+x y-2 x z+y z-z^2)$ has a common factor of x+y+z. Can anyone elaborate on how that is possible? What am I not seeing...
|
Consider the numerator $2x^2-(y+z)(y+z-x).$ The first thing to notice here is the sum $y+z.$ You may set $y+z=p$ if it makes the following more convenient to follow. Now expanding gives $$2x^2-(y+z)^2+x(y+z).$$ You almost have a difference of squares -- indeed you do already, but it doesn't help much, so we split into four terms so we can group in pairs, as follows $$x^2-(y+z)^2+x^2+x(y+z)=(x-y-z)(x+y+z)+x(x+y+z)=(x+y+z)(x-y-z+x),$$ or $(x+y+z)(2x-y-z).$
A similar procedure, or just performing the transformations $x\mapsto y$ and $y\mapsto x,$ gives the denominator in factored form as $(x+y+z)(2y-x-z).$
|
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|
Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(\frac{|x|^2+|y|^2}{\sqrt{x^2+y^2}})|\le |\cos(\frac{2\sqrt{x^2+y^2}^2}{\sqrt{x^2+y^2}})| \le |\cos(2\sqrt{x^2+y^2})|$$
Is this correct? I couldn't get Wolfram to compute this properly for some reason.
|
HINT
Consider the polar coordinates $x = r\cos(\theta)$ and $y = r\sin(\theta)$. Then we get
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\cos\left(\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}}}\right) = \lim_{r\rightarrow 0}\cos(r\cos(2\theta)) = \ldots
\end{align*}
Can you take it from here?
|
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|
Inequality involving the angle bisectors of a triangle
Let $l_a,l_b,l_c$ denote the lengths of angle bisectors of a triangle with sides $a,b,c$ and semiperimeter $s$. I am looking for the best constant $K>0$ such that
$$l_a^2+l_b^2+l_c^2> K s^2.$$
I found that $K=2/3$ works, but I suspect that best constant is $K=8/9>2/3$. Any proof or reference?
BTW it is known that $l_a^2+l_b^2+l_c^2\leq s^2$.
Proof for $K=2/3$. According to Cut-the-knot,
$$m_a l_a+m_b l_b+m_c l_c\ge s^{2}$$
where $m_a,m_b,m_c$ are the medians. Therefore, by Cauchy–Schwarz inequality,
$$(m_a^2+m_b^2+m_c^2)(l_a^2+l_b^2+l_c^2)\geq (m_a l_a+m_b l_b+m_c l_c)^2\geq s^4$$
which implies
$$l_a^2+l_b^2+l_c^2\geq \frac{s^4}{m_a^2+m_b^2+m_c^2}> \frac{2s^2}{3}$$
in view of
$$m_a^{2}+m_b^{2}+m_c^{2}=\frac{3(a^2+b^2+c^2)}{4}< \frac{3s^2}{2}.$$
EDIT. I found a reference that $K=8/9$ is the best constant. See 11.7. at p. 218 in Recent Advances in Geometric Inequalities by Mitrinovic et al.
No proof is given.
|
Let $a=b=1$ and $c\rightarrow2^-$.
Thus, $K<\frac{8}{9}.$
We'll prove that $\frac{8}{9}$ it's a best constant.
Indeed, we need to prove that:
$$\sum_{cyc}\left(\frac{2bc\cos\frac{\alpha}{2}}{b+c}\right)^2\geq\frac{8}{9}\cdot\frac{(a+b+c)^2}{4}$$ or
$$\sum_{cyc}\left(\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}{b+c}\right)^2\geq\frac{2(a+b+c)^2}{9}$$ or
$$\sum_{cyc}\frac{bc(b+c-a)}{(b+c)^2}\geq\frac{2(a+b+c)}{9}.$$
Now, let $a=x+u$, $b=x+v$ and $c=x+u+v,$ where $x>0$ and $u$ and $v$ are non-negatives.
Thus, we need to prove that:
$$48x^7+224(u+v)x^6+16(23u^2+61uv+23v^2)x^5+$$
$$+16(u+v)(16u^2+75uv+16v^2)x^4+$$
$$+(65u^4+894u^3v+1859u^2v^2+894uv^3+65v^4)x^3+$$
$$+(4u^5+185u^4v+900u^3v^2+900u^2v^3+185uv^4+4v^5)x^2+$$
$$+(5u^6-7u^5v+121u^4v^2+275u^3v^3+121u^2v^4-7uv^5+5v^6)x+$$
$$+2(u+v)^3(u-v)^4\geq0,$$
which is obvious.
|
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Find the exact value of integration $ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$ Can you help me find the exact value for integration with the given steps?
$$ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$$
Some of my attempts as indefinite Integral
$$
\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}} \, dx\approx \left(\sqrt{x+1}+\left(-\frac{1}{\sqrt{x+1}+1}-1\right) \sqrt{1-x}+\frac{1}{\sqrt{x+1}+1}-\frac{2 \left(0.707107 \sqrt{x+1}\right)}{\sin }\right)+C
$$
Is it considered Improper Integral?
|
Setting $x=\cos(2t)$, we have:
$1-x=2\sin^2t\;\;$ and $\;\;1+x=2\cos^2t$
|
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|
Find $\lim _{n\to \infty}\frac{n-1}{n-2}$ I'm taking a university real analysis course and I have been tasked with proving that the sequence $x_n = \frac{n-1}{n-2}$ converges using first principles.
First fix $\epsilon >0$.
Using the algebra of limits we find the limit in question.
$x_n = \frac{n-1}{n-2} = \frac{1-\frac{1}{n}}{1-\frac{2}{n}} \rightarrow x_n = 1 $ as $n\rightarrow\infty$
We observe a limit if for any $\epsilon > 0$ there is an $N \in \mathbb{N}$ such that
$n \geq N \implies \mid \frac{n-1}{n-2}- 1 \mid < \epsilon$
Provided $n \geq N$
For all $n>3$, we can remove absolute value.
$\frac{n-1}{n-2} - \frac{n-2}{n-2} = \frac{1}{n-2} \leq \frac{3}{n} <\epsilon$ $\iff \frac{3}{\epsilon}<n$
Thus any integer $N$ greater than $max(3, \frac{3}{\epsilon})$ has the required property
Any feedback would be much appreciated!
|
Your proof is correct, but I think you should explicitly specify a choice of $N$ as a function of $\epsilon$. Something like this: For any $\epsilon$ and for all $n' \geq N = \max\{\frac{4}{\epsilon}, 3\}$, then
$$
\left\lvert \frac{n'-1}{n'-2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{3}{N} = \frac{3}{4} \epsilon < \epsilon.
$$
Thus the sequence $(x_n) \rightarrow 1$.
Here's another bound that works: For any $\epsilon$ and for all $n' \geq N = \max\{3, \lceil 2 + \frac{2}{\epsilon} \rceil\}$, then
$$
\left\lvert \frac{n' - 1}{n' - 2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{1}{ 2 - 2 + \frac{2}{\epsilon}} = \frac{1}{2}\epsilon < \epsilon
$$
Thus the sequence $(x_n) \rightarrow 1$.
|
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|
Methods to solve $\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx$
Find $$\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx.$$
I'd like to know in what ways can one approach this integral that can be found here, since the post was about using feynman's trick to evaluate integrals i used the parameter,
$$I=\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx$$
$$I\left(a\right)=\int _0^{\infty }\frac{x^{\frac{4}{5}a}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx$$
$$I'\left(a\right)=\frac{4}{5}\int _0^{\infty }\frac{x^{\frac{4}{5}a}}{x^2+1}\:dx$$
where $I\left(a=1\right)=I$ and $I\left(a=\frac{5}{6}\right)=0$.
But that integral doesnt seem so simple to tackle. i'd appreciate any ideas or different approaches to the integral.
|
Consider
$$J(a)=\int_0^\infty\frac{x^a-1}{(x^2+1)\ln x}dx$$
where $a\in\Bbb C$ with $|\Re a|<1$.
Then
$$J'(a)=\int_0^\infty\frac{x^a}{x^2+1}dx.$$
Using the keyhole contour you can see that
$$J'(a)=\frac{2\pi i}{1-e^{2\pi a}}\left(\frac{e^{\frac{\pi a}{2}}}{2i}+\frac{e^{\frac{3\pi a}{2}}}{-2i}\right)=\frac{\pi}{2}\sec\frac{\pi a}{2}.$$
Alternatively with $y=x^2$ and $u=\frac{y}{y+1}$, note that
\begin{align}J'(a)&=\frac{1}{2}\int_0^\infty \frac{y^{\frac{a+1}{2}-1}}{y+1}dy=\frac12\int_0^1 u^{\frac{a+1}{2}-1}(1-u)^{\left(1-\frac{a+1}{2}\right)-1}du\\&=\frac12\mathrm{B}\left(\frac{a+1}{2},1-\frac{a+1}{2}\right)=\frac12\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(1-\frac{a+1}{2}\right).\end{align} From Euler's reflection formula, $J'(a)=\frac{\pi}{2}\operatorname{cosec}\left(\pi\frac{a+1}{2}\right)=\frac{\pi}{2}\sec\frac{\pi a}{2}$.
Therefore
$$J(a)=\int_0^a \frac{\pi}{2}\sec \frac{\pi t}{2} dt=\int_0^{\frac{\pi a}{2}} \sec\theta d\theta=\ln\left(\tan\frac{\pi a}{2}+\sec\frac{\pi a}{2}\right).$$
Therefore
$$I=J\left(\frac45\right)-J\left(\frac23\right)=\ln\frac{\tan\frac{2\pi}{5}+\sec \frac{2\pi}{5}}{\tan\frac{\pi}{3}+\sec{\frac{\pi}{3}}}=\ln\frac{\sqrt{5+2\sqrt{5}}+1+\sqrt{5}}{\sqrt3 +2}\approx 0.525772.$$
|
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|
For what $x$ and $y$ polynomial has maximum value?
For what $x,y\in\mathbb R$ does the polynomial
$$-5x^2-2xy-2y^2+14x+10y-1$$
attain a maximum?
My attempt:
I called $\alpha$ maximum value.
$$-5x^2-2xy-2y^2+14x+10y-1\leqslant\alpha$$
$$-5x^2-2xy-2y^2+14x+10y-1-\alpha\leqslant 0$$
$$5x^2+2xy+2y^2-14x-10y+1+\alpha\geqslant 0$$
$$(x+y)^2+(y-5)^2+3x^2+(x-7)^2-73+\alpha\geqslant0$$
$$\alpha\geqslant73$$
So the lowest maximum value turned out to be $73$, but after checking answers I was wrong-maximum is $16$, so my further plans to calculate from that $x$ and $y$ seemed purposless. I'd like to see solution using only high school knowledge.
Ans: $x=1$, $y=2$
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let $k$ be a maximal value.
Thus, for any $x$ and $y$ we have $$-5x^2-2xy-2y^2+14x+10y-1\leq k$$ or
$$5x^2+2(y-7)x+2y^2-10y+1+k\geq0,$$ which gives
$$(y-7)^2-5(2y^2-10y+1+k)\leq0$$ or
$$9y^2-36y-44+5k\geq0,$$ which gives $$18^2-9(-44+5k)\leq0,$$ which gives $$k\geq16.$$
The equality occurs for $$y=\frac{36}{2\cdot9}=2$$ and $$x=-\frac{2(2-7)}{2\cdot5}=1,$$ which says that a minimal value of $k$,
for which the first inequality is true for any values of $x$ and $y$, it's $16$.
|
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|
Row Echelon, Partial Fractions, and Numerator Coefficients I am trying to get the numerator values for the partial fraction decomposition of:
$$\dfrac{x^2+1}{x(x-1)(x+1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{D}{x-4}$$
I really started hitting speed bumps on this one, and was forced to take it slow so I decided to use a Augmented Matrix to solve the problem except that the matrix did not give the right solutions that Symbolab gives.
$$\begin{bmatrix}1&1&1&1&|&0\\-4&-3&-5&0&|&1\\ -1&-4&4&-1&|&0\\4&0&0&0&|&1\end{bmatrix}$$
This matrix does not give me the right values Symbolab gives, I would like to really learn to solve this problem using an augmented matrix it would help my skills, I also using row echelon form to solve it. If anybody has any hints, and suggestions it would greatly appreciated!
Edit
My matrix was constructed by the following technique I get a system of equations
This is $x^3$
$$A+B+C+D=0$$
This is $x^2$
$$-4A-3B-5C+0D=1$$
This is $x$
$$-A-4B+4C-D=0$$
This is constants
$$4A+0B+0C+0D=1$$
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Lets use Wolfram Alpha to get a final result of the original expression using partial fractions.
We see that we would get
$$A = \dfrac{1}{4}, B = -\dfrac{1}{3}, C = -\dfrac{1}{5} , D = \dfrac{17}{60}$$
We want to solve the partial fraction expansion of
$$\dfrac{x^2+1}{x(x-1)(x+1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{D}{x-4}$$
If we combine the RHS, we get the expression
$\dfrac{x^2+1}{x(x-1)(x+1)(x-4)} = \\ \dfrac{a x^3-4 a x^2-a x+4 a+b x^3-3 b x^2-4 b x+c x^3-5 c x^2+4 c x+d x^3-d x}{x(x-1)(x+1)(x-4)} $
Using this, we can setup four equations and four unknowns and get the augmented matrix (your matrix setup looks perfect),
$$\left[\begin{array}{rrrr|r}1&1&1&1&0\\-4&-3&-5&0&1\\ -1&-4&4&-1&0\\4&0&0&0&1\end{array}\right]$$
We perform the RREF and arrive at
$$\left[\begin{array}{rrrr|r}
1 & 0 & 0 & 0 & \dfrac{1}{4} \\
0 & 1 & 0 & 0 & -\dfrac{1}{3} \\
0 & 0 & 1 & 0 & -\dfrac{1}{5} \\
0 & 0 & 0 & 1 & \dfrac{17}{60} \\
\end{array}\right]$$
If you want to see the steps of those reductions, click this link.
That matches the Wolfram result.
Where is your result failing?
The limit approach is much cleaner.
|
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|
Sum of Beta functions. Given the summation
$$ S_{n} = \sum_{k=0}^{n-1} \binom{n+k-1}{k} \, ( B(a+k, b+n) + B(a+n, b+k) )$$
is it possible to show that it equals the suspected value of $B(a, b)$?
For the first two $n$ values it can be shown that
\begin{align}
S_{1} &= B(a, b+1) + B(a+1, b) = B(a,b) \\
S_{2} &= B(a, b+2) + B(a+2, b) + 2 \, ( B(a+1, b+2) + B(a+2, b+1) ) \\
&= \frac{\Gamma(a) \, \Gamma(b)}{\Gamma(a+b+2)} \, ((a+b)^2 + (a+b)) = B(a,b)
\end{align}
which leads to the suspected value that $S_{n} = B(a,b)$. Are there other ways that this can be shown to be true?
|
A solution using induction. We have $S_n=\int_0^1 x^{a-1}(1-x)^{b-1}Q_n(x)\,dx$, where \begin{align}
Q_n(x)&:=(1-x)^n P_n(x)+x^n P_n(1-x),\\
P_n(x)&:=\sum_{k=0}^{n-1}\binom{n+k-1}{k}x^k,\\
P_{n+1}(x)&=\sum_{k=0}^{n}\binom{n+k-1}{k}x^k+\sum_{k=1}^{n}\binom{n+k-1}{k-1}x^k
\\&=P_n(x)+\binom{2n-1}{n}x^n+x\left(P_{n+1}(x)-\binom{2n}{n}x^n\right),
\end{align}
hence $(1-x)P_{n+1}(x)=P_n(x)+\binom{2n}{n}\left(\frac{x^n}{2}-x^{n+1}\right)$ and $$Q_{n+1}(x)=Q_n(x)+\binom{2n}{n}x^n(1-x)^n\left(\frac12-x+\frac12-(1-x)\right)=Q_n(x).$$
Thus, $Q_n(x)=Q_1(x)\equiv 1$ and $S_n=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx=\mathrm{B}(a,b)$.
|
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|
Evaluating $\lim_{x\to1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$ I don't know how to solve these kind of limit problems. I was wondering if someone could help me about it to learn them. $$\lim_{x\to 1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$$ Thanks in advance.
|
Hint
You can always write $f(x)-g(x)$ as $$\frac{f(x)^2-g(x)^2}{f(x)+g(x)}.$$
So, taking $$f(x)=\sqrt{x+1}+\sqrt{x^2+1}\quad \text{and}\quad g(x)=\sqrt{x^3+1},$$ at the numerator and $$f(x)=\sqrt{x-1}+\sqrt{x^2+1}\quad \text{and}\quad g(x)=\sqrt{x^4+1},$$
at the denominator should do the work.
|
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How to prove this inequality: $\sum_{i=1}^{n}\sum_{j=1}^{n}\text{lcm}(i,j)\ge\frac{7}{8}n^3$?
Let $n$ be postive integers. Show that
$$\sum_{i=1}^{n}\sum_{j=1}^{n}[i,j]\ge\dfrac{7}{8}n^3\,,$$ where $[a,b]$ denote the least common multiple of $a$ and $b$.
For $n=1,2,3 $, it is clear. How to prove this inequality? Thanks!
|
Let $S_n:=\sum\limits_{i=1}^n\,\sum\limits_{j=1}^n\,\text{lcm}(i,j)$. We have $S_1=1\geq \dfrac{7}{8}\cdot 1^3$ and $S_2=7\geq \dfrac{7}{8}\cdot 2^3$. Assume that $n>2$ is an integer such that $$S_{n-1}\geq \dfrac{7}{8}\,(n-1)^3\,.$$
Then,
$$S_{n}-S_{n-1}=n+2\,\sum_{k=1}^{n-1}\,\text{lcm}(k,n)\,.$$
Since $n>3$,
$$S_n-S_{n-1}\geq n+2\,\big(\text{lcm}(n-1,n)+\text{lcm}(n-2,n)\big)\,.$$
We see that $\text{lcm}(n-1,n)=(n-1)n$ and $\text{lcm}(n-2,n)\geq \dfrac{(n-2)n}{2}$. Ergo,
$$S_n-S_{n-1}\geq n+2\,\left((n-1)n+\frac{(n-2)n}{2}\right)=3n^2-3n\,.$$
For $n>2$, we have
$$3n^2-3n\geq 3\cdot 3^2-3\cdot 3=18>7;$$
therefore,
$$3n^2-3n> \frac{7}{8}\,(3n^2-3n+1)=\frac{7}{8}\,\big(n^3-(n-1)^3\big)\,.$$
Consequently,
$$S_n>\frac{7}{8}\,\big(n^3-(n-1)^3\big)+S_{n-1}\geq \frac{7}{8}\,\big(n^3-(n-1)^3\big)+\frac{7}{8}\,(n-1)^3=\frac{7}{8}\,n^3\,.$$
By induction,
$$\sum_{i=1}^n\,\sum_{j=1}^n\,\text{lcm}(i,j)=S_n\geq \frac{7}{8}\,n^3$$
for every positive integer $n$. The equality holds if and only if $n=2$.
Let $H_n:=\sum\limits_{k=1}^n\,\dfrac1k$ for each positive integer $n$.
Apparently, we also have $$S_n\geq 2\,\sum_{k=1}^n\,k^2\,H_k-\frac{n(n+1)(4n-1)}{6}\,.$$
The inequality above becomes an equality if and only if $n=1$ and $n=2$.
If $\gamma$ is the Euler–Mascheroni constant, then using $H_n>\ln(n)+\gamma$ for every positive integer $n$ yields
$$\begin{align}S_n&\geq 2\,\int_0^n\,x^2\,\big(\ln(x)+\gamma)\,\text{d}x-\frac{n(n+1)(4n-1)}{6}
\\&=\dfrac{2n^3}{9}\,\big(3\ln(n)-1\big)+\frac{2\gamma n^3}{3}-\frac{n(n+1)(4n-1)}{6}\,.\end{align}$$
I am going to predict again as I have done in many threads that
$$S_n=\alpha n^3\ln(n)+\beta n^3+o(n^3)\,,$$
where $\alpha$ and $\beta$ are constants. (Clearly, $\alpha\geq \frac{2}{3}$.) However, I can be wrong about this. I have not tried to find an upper bound of $S_n$ that is of the form $\mathcal{O}\big(n^3\,\ln(n)\big)$. (The OP posted another question regarding an upper bound.)
|
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|
Expected Waiting Time in a Queuing System $(M | M | 2 | 5)$
In the queuing system $(M | M | 2 | 5)$, the input flow rate is $240$ requests per hour, the average service time for one request is $30$ seconds. Find the average waiting time for an application in the queue.
How do I find that?
|
Let $\pi_k$ denote the steady-state probability that there are $k$ requests. We have $\lambda=4\,\texttt{min}^{-1}$ and $\mu=2\,\texttt{min}^{-1}$. Write $\rho:=\dfrac{\lambda}{\mu}=2$. We have the balance equations
$$\lambda\,\pi_0=\mu\,\pi_1\,,$$
and
$$\lambda\,\pi_k=2\mu\,\pi_{k+1}$$
for $k=1,2,3,4$. That is,
$$\pi_1=\rho\,\pi_0\,\text{ and }\,\pi_{k+1}=\frac{\rho}{2}\,\pi_k$$
for $k=1,2,3,4$. If $p:=\pi_0$, then
$$\pi_1=\rho\,p\,\text{ and }\,\pi_k=\frac{\rho^k}{2^{k-1}}\,p$$
for $k=2,3,4,5$. Since $\sum\limits_{k=0}^5\,\pi_k=1$, we obtain
$$p=\frac{1}{1+\rho+\frac{\rho^2}{2}+\frac{\rho^3}{4}+\frac{\rho^4}{8}+\frac{\rho^5}{16}}\,.$$
The expected queue size is
$$L_q=0\,\pi_0+0\,\pi_1+0\,\pi_2+1\,\pi_3+2\,\pi_4+3\,\pi_5=\frac{\frac{\rho^3}{4}+\frac{2\rho^4}{8}+\frac{3\rho^5}{16}}{1+\rho+\frac{\rho^2}{2}+\frac{\rho^3}{4}+\frac{\rho^4}{8}+\frac{\rho^5}{16}}\,.$$
From Little's Law, the expected wait time $W_q$ is
$$W_q=\frac{L_q}{\lambda}=\frac{1}{\lambda}\,\left(\frac{\frac{\rho^3}{4}+\frac{2\rho^4}{8}+\frac{3\rho^5}{16}}{1+\rho+\frac{\rho^2}{2}+\frac{\rho^3}{4}+\frac{\rho^4}{8}+\frac{\rho^5}{16}}\right)\,.$$
For $\lambda=4\,\texttt{min}^{-1}$ and $\rho=2$, we get
$$W_q=\left(\frac{1}{4}\,\texttt{min}\right)\frac{2+4+6}{1+2+2+2+2+2}=\frac{3}{11}\,\texttt{min}\,.$$
|
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|
Find the curve of a family of curves with the following conditions
Let $\mathfrak{J}$ be the family of curves such that their normal line at each point is tangent to the parabola of equation $$y=kx^2$$ that passes through that point. Find the curve $C\in\mathfrak{J}$ that passes through $(0,1)$.
My work: $$\mathfrak{J}\colon y=kx^2\to k=\frac{y}{x^2},\quad y'=2kx=2\frac{y}{x^2}x=2\frac{y}{x}.$$ Moreover, $$\mathfrak{J}^\perp\colon -\frac{1}{y'}=2\frac{y}{x}\to-\frac x2=yy'\to\frac{y^2}2=-x^2+c\to y^2=-2x^2+C.$$ Since $y(0)=1$, we have that $1=C$, so the final answer is $$\boxed{y^2=-2x^2+1}$$
Is it correct?
The answer does not match with the book, the answer should be:
$x^2+2y^2=2$.
|
This line in your attempt is correct:
$$yy'=-\frac x 2$$
Then you have :
$$ 2yy'=-x \implies (y^2)'=-x$$
Integrate:
$$y^2=-\frac {x^2}2+c$$
$$2y^2+ {x^2}=2c$$
$$2y^2+ {x^2}=2$$
|
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|
A type of Combinatorial equality：$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$ When computing the Taylor series of the function $f(z)=e^z\cos z,$
I use two methods:
On the one hand, using Cauchy product,
\begin{align*}
e^z\cos z
&=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)
\left(\sum_{n=0}^{\infty}\frac{\cos\frac{n}{2}\pi}{n!}z^n\right)\\
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{1}{(n-k)!}
\frac{\cos\frac{k}{2}\pi}{k!}\right)z^n\ (\text{Cauchy Product})\\[3pt]
&=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\sum_{k=0}^{n}\binom{n}{k}\cos\frac{k}{2}\pi
\right)z^n,\ z\in\mathbb{C};
\end{align*}
On the other hand,
\begin{align*}
e^z\cos z
&=e^z\cdot\frac{e^{i z}+e^{-i z}}{2}\\
&=\frac{e^{(1+i)z}+e^{(1-i)z}}{2}\\[3pt]
&=\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{(1+i)^n}{n!}z^n
+\sum_{n=0}^{\infty}\frac{(1-i)^n}{n!}z^n\right)\\
&=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(1+i)^n+(1-i)^n}{n!}z^n\\
&=\frac{1}{2}\sum_{n=0}^{\infty}
\frac{2^{\frac{n}{2}}\left(e^{\frac{n}{4}\pi i}+e^{-\frac{n}{4}\pi i}\right)}{n!}z^n\\
&=\sum_{n=0}^{\infty}\left(\frac{2^{\frac{n}{2}}}{n!}\cos\frac{n}{4}\pi\right)z^n,\ z\in\mathbb{C}.
\end{align*}
So Compare the corresponding coefficients, we get the following Combinatorial equality:
$$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$$
What I want to konw: is there an elementary method or constructive method ( which is suitable for high school student!) to prove this Combinatorial equality?
Any help and hint will welcome!
|
There is a fundamental way to prove this.
I used a simple binomial expansion and Euler's identity to solve this...
Here's the solution:
$$
\begin{aligned}
&\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \cos \frac{k \pi}{2}=?\\
&\text { Let } T_{k}=\left(\begin{array}{l}
n \\
k
\end{array}\right)\left\{\cos \left(\frac{k \pi}{2}\right)+i \sin \left(\frac{k \pi}{2}\right)\right\}\\
&=\left(\begin{array}{l}
n \\
k
\end{array}\right) e^{\frac{i k \pi}{2}} \quad[\text { Euler's identity }]
\end{aligned}
$$
$$
\begin{aligned}
\sum_{k=0}^{n} T_{k} &=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) e^{\frac{i k \pi}{2}} \\
&=\left(e^{i \pi / 2}+1\right)^{n} \quad[\text { Binomial expansion }] \\
&=(1+i)^{n} \\
&=2^{n / 2}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{n} \\
&=2^{n / 2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{n} \\
S_{k} &=2^{n / 2} e^{\frac{i n \pi}{4}}
\end{aligned}
$$
$$
\text { But, } \operatorname{Im}\left(T_{k}\right)=\left(\begin{array}{l}
n \\
k
\end{array}\right) \cos \frac{k \pi}{2}
$$
$$
\operatorname{Im}\left(S_{k}\right)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \cos \frac{k \pi}{2}=2^{n / 2} \cos \frac{n \pi}{4}
$$
|
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|
How to determine direct solution of determinant? How to show that the determinant of the following $(2n+1)×(2n+1)$ matrix $A$?
\begin{equation}
\det A = \begin{array}{|cccccccccc|cc}
1 & -1 & 0 & \dots & 0 & 0 & 0 & \dots & 0 & 0 & & {\color{blue}{\text{row }1}}\\
-1 & 2 & -1 & \dots & 0 & 0 & 0 & \dots & 0 & 0& & {\color{blue}{\text{row }2}} \\
0 & -1 & 2 & \dots & 0 & 0 & 0 & \dots & 0 & 0& & {\color{blue}{\text{row }3}} \\
\vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots& \vdots & &{\color{blue}{\vdots}}\\
0 & 0 & 0 & \dots & 2 & -1 & 0 & \dots & 0 & 0 && {\color{blue}{\text{row }j-1}}\\
0 & 0 & 0 & \dots & -1 & 3 & -1 & \dots & 0 & 0 &{\color{blue}{\rightarrow}}& {\color{blue}{\text{row }j}}\\
0 & 0 & 0 & \dots & 0 & -1 & 2 & \dots & 0 & 0 && {\color{blue}{\text{row }j+1}} \\
\vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots& \vdots & &{\color{blue}{\vdots}} \\
0 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & 2 & -1 && {\color{blue}{\text{row }2n}} \\
0 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & -1 & 1 && {\color{blue}{\text{row }2n+1}}\\
\end{array}
\end{equation}
By a direct calculation of determinant. here $3$ is at the $j$-th row for some $2\leq j\leq 2n$.
Therefore, $\det (L_{A_{11}}) = (-1)^n 2^{n-1}.$
if $n+1\leq i\leq 3n+1$ we denote the resulting matrix by $A_{12}$ after
deleting the $i$-th row and column of $L_A$. Then, by a
similar computation, we have $\det (L_{A_{12}}) = (-1)^n 2^{n}.$
$ \det A = (-1)^n 2^{n-1}n + (-1)^n 2^{n}(2n+1)=(-1)^n 2^{n-1}(5n+2)$
equal to $(-1)^n 2^{n-1}(5n+2)$ ?
|
For each $X\in\text{Mat}_{m\times m}(\mathbb{R})$, $a,b\in\mathbb{R}$, and $i,j=1,2,\ldots,m$, the notation $$Y:=(R_i\leftleftarrows a\, R_i+b\, R_j)(X)$$
means $Y$ is obtained from $X$ by changing the $i$-th row of $X$ to $a$ times the $i$-th row of $X$ plus $b$ times the $j$-th row of $X$. Let $m:=2n+1$. Consider
$$A[1]:=(R_2\leftleftarrows R_2+R_1)\big(A\big)\,,$$
$$A[2]:=(R_3\leftleftarrows R_3+R_2)\big(A[1]\big)\,,$$
$$A[3]:=(R_4\leftleftarrows R_4+R_3)\big(A[2]\big)\,,$$
$$\vdots$$
$$A[j-1]:=(R_j\leftleftarrows R_j+R_{j-1})\big(A[j-2]\big)\,.$$
Then,
$$A[j]:=\left(R_{j+1}\leftleftarrows R_{j+1}+\frac{1}{2}\,R_j\right)\big(A[j-1]\big)\,,$$
$$A[j+1]:=\left(R_{j+2}\leftleftarrows R_{j+2}+\frac{2}{3}\,R_{j+1}\right)\big(A[j]\big)\,,$$
$$A[j+2]:=\left(R_{j+3}\leftleftarrows R_{j+3}+\frac{3}{4}\,R_{j+2}\right)\big(A[j+1]\big)\,,$$
$$\vdots$$
$$A[m-1]:=\left(R_m\leftleftarrows R_m+\frac{m-j+1}{m-j+2}\,R_{m-1}\right)\big(A[m-2]\big)\,.$$
The final matrix $A[m-1]$ is upper-triangular with diagonal entries
$$\underbrace{1\,,\,\,1\,,\,\,1\,,\,\,\ldots\,,\,\,1}_{j-1\text{ ones }}\,\,,\,\,2\,,\,\,\frac{3}{2}\,,\,\,\frac{4}{3}\,,\,\,\ldots\,,\,\,\frac{m-j+1}{m-j}\,,\,\,\frac{1}{m-j+1}\,.$$
Therefore,
$$\begin{align}\det\big(A\big)&=\det\big(A[1]\big)=\det\big(A[2]\big)=\ldots=\det\big(A[m-1]\big)\\&=1^{j-1}\cdot 2\cdot \frac{3}{2}\cdot\ldots \cdot\frac{m-j+1}{m-j+2}\cdot\frac{1}{m-j+1}=1\,.\end{align}$$
Let $x\in\mathbb{C}$. If we want to evaluate
\begin{equation}
\begin{array}{|cccccccccc|cc}
1 & -1 & 0 & \dots & 0 & 0 & 0 & \dots & 0 & 0 & & {\color{blue}{\text{row }1}}\\
-1 & 2 & -1 & \dots & 0 & 0 & 0 & \dots & 0 & 0& & {\color{blue}{\text{row }2}} \\
0 & -1 & 2 & \dots & 0 & 0 & 0 & \dots & 0 & 0& & {\color{blue}{\text{row }3}} \\
\vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots& \vdots & &{\color{blue}{\vdots}}\\
0 & 0 & 0 & \dots & 2 & -1 & 0 & \dots & 0 & 0 && {\color{blue}{\text{row }j-1}}\\
0 & 0 & 0 & \dots & -1 & 3 & -1 & \dots & 0 & 0 &{\color{blue}{\rightarrow}}& {\color{blue}{\text{row }j}}\\
0 & 0 & 0 & \dots & 0 & -1 & 2 & \dots & 0 & 0 && {\color{blue}{\text{row }j+1}} \\
\vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots& \vdots & &{\color{blue}{\vdots}} \\
0 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & 2 & -1 && {\color{blue}{\text{row }m-1}} \\
0 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & -1 & {\color{red}x}&& {\color{blue}{\text{row }m}}\\
\end{array}
\end{equation}
instead, the answer turns out to be
$$1^{j-1}\cdot 2\cdot \frac{3}{2}\cdot\ldots \cdot\frac{m-j+1}{m-j+2}\cdot\left((x-1)+\frac{1}{m-j+1}\right)=(m-j+1)\,(x-1)+1\,.$$
|
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|
Sequences and series that I cant partial fraction decomposition I have to calculate the partial sum for an equation. How can I calculate the sum for
$$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-5}$$
And I think that is not simple as $$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-3}.$$
TIA
|
For computing partial fractions, use the one you know how to work as a model for the other. The method is the same.
The zeroes of $16n^2-8n-3$ are $\frac{8\pm\sqrt{256}}{32}$, so $\frac34$ and $-\frac14$, which means you can write the polynomial as $16(n-\frac34)(n+\frac14) = (4n-3)(4n+1)$. Your candidate for the partial fraction is then
$$\frac{A}{4n-3} + \frac{B}{4n+1}.$$
Use the same method for the other expression.
The zeroes of $16n^2-8n-5$ are $\frac{8\pm 8\sqrt{6}}{32}$, so $\frac14 + \frac14\sqrt{6}$ and $\frac14 - \frac14\sqrt{6}$, which means you can write the polynomial as
$16(n-\frac14 - \frac14\sqrt{6})(n - \frac14 + \frac14\sqrt{6}) = (4n - 1 - \sqrt{6})(4n - 1 + \sqrt{6})$. Your candidate for the partial fraction is then
$$\frac{A}{4n-1 - \sqrt{6}} + \frac{B}{4n - 1 + \sqrt{6}}.$$
Additional comment: This works for any quadratic, even if the zeroes of the polynomial are not real (!). For example,
$$x^2+1 = (x-i)(x+i)$$
so
$$\frac{1}{x^2+1} = \frac{A}{x-i}+\frac{B}{x+i}$$ would be the candidate for the partial fraction decomposition (what are $A$ and $B$ in this case?).
|
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|
Use mathematical induction to prove that for all integers $n \geq 3,\, 2n + 1 < 2^{n}$ This is what I've got so far.
Let $P(n)$ be the statement that $2n + 1 < 2^n$
Basis:
Let $n = 3$. Show that $P(3)$ is true.
$2(3) + 1 = 7$ and $2^3 = 8$.
Since $7 < 8$, $P(3)$ is true.
Inductive Hypothesis:
Suppose that $P(k)$ is true for an arbitrary integer $k\ge 3$.
This implies that $2k + 1 < 2^k$.
Show that $2(k + 1) + 1 < 2^{k + 1}$.
\begin{align*}
2(k + 1) + 1 &= 2k + 2 + 1\\
&= (2k + 1) + 2
\end{align*}
From the IH,
\begin{align*}
2k + 1 < 2^k &\Rightarrow (2k + 1) + 2 < 2^k + 2\\
&\Rightarrow (2k + 1) + 2 < 2^k + 2 \\
&\Rightarrow (2k + 2) + 1 < 2^k + 2 \\
&\Rightarrow 2(k + 1) + 1 < 2^k + 2
\end{align*}
Now we need to show that $2^k + 2 < 2^{k + 1}$.
$2^k < 2^{k + 1}\Rightarrow 2^k + 2 < 2^{k + 1} + 2$
This is where I'm stuck. Any help on how to proceed is much appreciated.
|
You're doing fine up until near end. Note that for $k \ge 3$, (actually, for any $k \gt 1$), you have $2 \lt 2^k$, so
$$2^k + 2 \lt 2^k + 2^k = 2^{k + 1} \tag{1}\label{eq1A}$$
with which you can then finish your inductive proof.
|
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|
Proving a complicated looking inequality in a simple way This is again a search for alternative proofs:
Let $0 <s \le 1$, and suppose that $0 <a,b $ satisfy
$$ ab=s,a+b=1+\sqrt{s}. \tag{1}$$
I have a proof for the assertion
$$ 2(1-\sqrt s)^3 \le |a-1|^3+|b-1|^3, \, \, \, \text{for every } \, s \ge \frac{1}{9}$$
but it is rather involved.
Actually, I am sure that the lower bound of $ \frac{1}{9}$, is not tight; the inequality holds for some $s > s^*$ where $s^* < \frac{1}{9}$.
Define $F(x,y):=|x-1|^3+|y-1|^3$. My proof is based on finding the global minimum $ \min_{xy=s} F(x,y)$.
However, here we need to show "only that" $F(\sqrt s, \sqrt s) \le F(a,b)$ for the specific $a,b$ described above in $(1)$.
Is there a way to prove this inequality "directly", without solving the harder global optimization problem?
Bonus: Is there a natural way to find the exact threshold $s^*$?
Edit:
There are now some very nice answers. I still wonder whether one can prove this without solving explicitly the quadratic described implicitly in $(1)$.
Here is an elementary proof for when $s \ge \frac{4}{9}$:
Suppose that $a \ge b$. The conditions on $a,b$ easily imply that $a \ge 1$, hence $s=ab \ge b$. Thus, we have
$$ b \le s \le \sqrt s \le 1 \le a.$$
So, replacing $b$ with $\sqrt s$ clearly lowers the value of $F$, since we get closer to $1$. Now it is beneficial to replace $a$ by $\sqrt s$ when
$$|\sqrt s -1|=1-\sqrt s \le a-1 \iff 2-\sqrt s \le a \iff 4-2\sqrt s \le 2a. \tag{2}$$
Solving explicitly the quadratic
$ a^2-(1+\sqrt s)a+s=0$, we get (assuming $a \ge b$) that
$$ a=\frac{1}{2}(1+\sqrt s+\sqrt{1+2\sqrt s-3s}).$$ Thus, inequality $(2)$ beceoms
$$
4-2\sqrt s \le 1+\sqrt s+\sqrt{1+2\sqrt s-3s},
$$
or $3-3\sqrt s \le \sqrt{1+2\sqrt s-3s}$. Squaring this and simplifying gives
$$
3s-5\sqrt s +2 \le 0,
$$
which holds exactly for $\frac{4}{9} \le s \le 1$.
|
Let $\sqrt{s}=t$.
Thus, $\frac{1}{3}\leq t\leq 1.$
Since $a$ and $b$ are roots of the equation $$x^2-(1+t)x+t^2=0,$$ we need to prove that
$$\left|\frac{1+t+\sqrt{(1+t)^2-4t^2}}{2}-1\right|^3+\left|\frac{1+t-\sqrt{(1+t)^2-4t^2}}{2}-1\right|^3\ge2(1-t)^3$$ or
$$\left|\frac{\sqrt{(1-t)(1+3t)}-(1-t)}{2}\right|^3+\left|\frac{\sqrt{(1-t)(1+3t)}+(1-t)}{2}\right|^3\ge2(1-t)^3$$ or
$$\left(\sqrt{1+3t}-\sqrt{1-t}\right)^3+\left(\sqrt{1+3t}+\sqrt{1-t}\right)^3\ge16\sqrt{(1-t)^3}$$ or
$$\sqrt{(1+3t)^3}+3(1-t)\sqrt{1+3t}\ge8\sqrt{(1-t)^3}$$ or
$$(1+3t)^3+6(1-t)(1+3t)^2+9(1-t)^2(1+3t)\ge64(1-t)^3$$ or
$$4t^3-12t^2+15t-3\ge0,$$ which is true even for $t\ge\frac{1}{4}:$
$$4t^3-12t^2+15t-3=4t^3-t^2-11t^2+\frac{11}{4}t+\frac{49}{4}t-\frac{49}{16}+\frac{1}{16}=$$
$$=(4t-1)\left(t^2-\frac{11}{4}t+\frac{49}{16}\right)+\frac{1}{16}>0.$$
|
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|
Find all polynomials $P(x)$ such that $2 P\left(2 x^{2}-1\right)=P(x)^{2}-2$ Question -
(Romania, 1990)Find all polynomials
$P(x)$ such that
$$
2 P\left(2 x^{2}-1\right)=P(x)^{2}-2
$$
for all $x \in \mathbb{R}$.
Solution
Putting $x=1,$ we get a quadratic equation in $P(1)$ and hence $P(1)=1 \pm \sqrt{3} .$ If $P(1)=1+\sqrt{3},$ we may write $P(x)=(x-1) Q(x)+1+\sqrt{3} .$ This gives a relation for $Q(x)$.
$$
4(x+1) Q\left(2 x^{2}-1\right)=(x-1) Q(x)^{2}+2(1+\sqrt{3}) Q(x)
$$
Taking $x=1,$ we see that $Q(1)=0 .$ Thus $(x-1)$ divides $Q(x) .$
We may use the induction to prove that $(x-1)^{n}$ divides $Q(x)$ for all positive integers $n .$ Thus $Q(x) \equiv 0$ and $P(x)=1+\sqrt{3}$ for all $x .$ Similarly, $P(1)=1-\sqrt{3}$ gives $P(x)=1-\sqrt{3}$ for all $x$
now i did not understand how to prove that $(x-1)^{n}$ divides $Q(x)$ for all positive integers $n .$ using induction....
Any help will be appreciated
thankyou
|
Firstly, when using induction, you need the initial/base case. We conclude that $(x-1) = (x-1)^1$ divides the polynomial $Q(x)$.
Note that I will be using the following notation; $Q(x) = (x-1) \cdot Q^{(1)}(x)$, and in general \begin{equation}Q^{(n)}(x) = (x-1) \cdot Q^{(n+1)}(x)\end{equation} or \begin{equation}Q(x) = (x-1)^n \cdot Q^{(n)}(x).\end{equation}
Now for induction-step we will assume that $(x-1)^n$ divides $Q(x)$. We get the following: \begin{equation}P(x)=(x−1)^n Q^{(n)}(x) + 1 \pm \sqrt3 \end{equation}
To answer one must prove "the inductive step". Prove that if we assume the statement holds for $n$, it also holds for $n+1$.
So using the above, assumed identity, and plugging it into the original equation, we get:
\begin{equation}2 \left(((2x^2-1)−1)^n Q^{(n)}(2x^2-1) +1\pm\sqrt3\right) = \left((x−1)^n Q^{(n)}(x) +1\pm\sqrt3\right)^2 - 2\end{equation}
You should work out the brackets to yield:
\begin{equation} 4(x-1)^n(x+1)^n Q(2x^2-1) + 2(1\pm \sqrt3) =\\ (x-1)^{2n} Q^{(n)}(x)^2 + (x-1)^n Q^{(n)}(x) + 1 \pm 2\sqrt3 +3 -2\end{equation}
The constant cancels, we divide by $(x-1)^n$ and we get:
\begin{equation}4(x+1)^n Q(2x^2-1) = (x-1)^n Q^{(n)}(x)^2 +Q^{(n)}(x) \end{equation}
Filling in $x=1$ gives:
\begin{equation} 2^{n+2} Q^{(n)}(1) = Q^{(n)}(1)\end{equation}
\begin{equation} (2^{n+2}-1) Q^{(n)}(1) = 0 \end{equation}
We conclude $Q^{(n)}(1) = 0$, thus $(x-1)$ is a divisor of $Q^{(n)}(x)$.
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|
Simplifying $\frac{ \cos^2 x - \sin^2 x }{\sin{2x}}$ Please consider the following problem and my answer to it:
Problem:
Simplify the following expression:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} $$
Answer:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{ \cos^2 x - \sin^2 x }{ 2 \sin x \cos x} $$
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{\cos x}{2 \sin x} - \frac{\sin x}{2 \cos x} $$
The result is not much simpler than what I started with. Is there an additional simplification that can be made? Am I missing something?
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$\cos^2x -\sin^2 x = \cos (2x)$. So the result can be written as $\cot(2x)$.
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|
Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$ Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$
My Attempt:
Given
$$y-3px+ayp^2=0$$
$$3px=y+ayp^2$$
$$x=\frac {1}{3} \cdot \frac {y}{p} + \frac {a}{3} \cdot yp$$
This is solvable for x. Differentiating both sides with respect to $y$
$$\frac {dx}{dy}=\frac {1}{3} \cdot \frac {p-y\frac {dp}{dy}}{p^2} + \frac {a}{3} (y\cdot \frac {dp}{dy} +p)$$
$$\frac {dx}{dy} = \frac {1}{3p} - \frac {y}{3p^2} \cdot \frac {dp}{dy} + \frac {ay}{3} \cdot \frac {dp}{dy} + \frac {ap}{3}$$
$$\frac {1}{p} - \frac {1}{3p} -\frac {ap}{3} = \frac {y}{3} \cdot \frac {dp}{dy}\cdot \frac {ap^2-1}{p^2}$$
$$\frac {3-1-ap^2}{3p} = \frac {y}{3} \cdot \frac {ap^2-1}{p^2} \cdot \frac {dp}{dy}$$
$$(2-ap^2) = \frac {y}{p} \cdot (ap^2-1) \cdot \frac {dp}{dy}$$
$$\frac {dy}{y} = \frac {ap^2-1}{p(2-ap^2)} dp$$
How do I integrate the RHS of above equation?
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From @Rebellos work $$\int \frac{dy}{y}=\int \frac{dp}{p}+\int \frac{dp}{p(2-ap^2)}$$
In the third integral let $p=1/t \implies -dt/t^2$, then $2t^2-a=u$, we get
$$\ln y= \ln p-\frac{1}{4}\ln u \implies y(p)=\frac{Cp\sqrt{p}}{(2-ap^2)^{1/4}}~~~~(1)$$
Inserting this in the given ODE, we get
$$x(p)=\frac{y+ayp^2}{3p}~~~~(2)$$
Eqs.(1) and (2) define parametric equation of family of curves when $p$ acts like a parametre, $C$ is the arbitrary constant of integration for a fixed value of $a$.
|
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|
How to prove that $a + b \neq 2^{n+1} (2c+1) $ with $ab = 4^n - 1$ and $a,b, c, n \in \mathbb{N}$? How to prove that $a + b \neq 2^{n+1} (2c+1) \quad \text{; with } ab = 4^n - 1 \text{ and } a,b, c, n \in \mathbb{N}$ (without zero)?
I already know that:
\begin{align}
a,b &\equiv 1 \pmod{2} \\
a + b &\equiv 0\pmod{4}
\end{align}
And because of the symmetry, I can define:
\begin{align}
a &\equiv 1 \pmod{4} \\
b &\equiv 3 \pmod{4}
\end{align}
For even $n$, I am already able to prove the statement by:
\begin{align}
2^{n+1} (2c+1) &\equiv 2 \pmod{4} \\
2^{n+1} (2c+1) &\not\equiv a + b \pmod{4}
\end{align}
For odd $n$, however, I am stuck. From the perspective of the congruence, the equation could hold for odd $n$ by $n+1 = 2m; \text{ with } m \in \mathbb{N}$ then:
\begin{align}
4^m (2c+1) &\equiv 0 \pmod{4} \\
4^m (2c+1) &\equiv a + b \pmod{4}
\end{align}
|
Define $v_2(n)$ to be the exponent of the highest power of $2$ that divides $n$, for example $v_2(24)=v_2(2^3\cdot 3)=3$ and $v_2(7)=0$.
Suppose that $a+b=2^{n+1}(2c+1)$. This gives $$v_2(a+b)=n+1.$$ It is not hard to see that $$n+1=v_2(4^n+a+b)=v_2((a+1)(b+1))=v_2(a+1)+v_2(b+1).$$ As seen above, WLOG $b\equiv_4 1$, so $v_2(b+1)=1$. This gives $v_2(a+1)=n,$ but because $v_2((a+1)+(b-1))=n+1$, we must also have $v_2(b-1)=n$. Thus, we have $a=2^n\cdot x-1$ and $b=2^n\cdot y+1$ for some odd $x$ and $y$. Plugging this in $ab=4^n-1$ we get $x=y=1$, so $c=0\not\in\mathbb{N}.$
|
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|
Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$ Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$
My Attmept:
$$x+yp=ap^2$$
$$x=ap^2-yp$$
This is solvable for $x$ so differentiating both sides w.r.t $y$
$$\frac {dx}{dy} = 2ap\cdot \frac {dp}{dy} - y\cdot \frac {dp}{dy} -p$$
$$\frac {1}{p} + p = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {1+p^2}{p} = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {dy}{dp} = \frac {2ap^2}{1+p^2} - \frac {yp}{1+p^2}$$
$$\frac {dy}{dp} + \frac {p}{1+p^2} \cdot y = \frac {2ap^2}{1+p^2}$$
$$\textrm {This is Linear so}$$
$$\textrm {Integrating Factor}=e^{\int \frac {p}{1+p^2} dp}=\sqrt {1+p^2}$$
So we have
$$y\sqrt {1+p^2} = \int \frac {2ap^2}{1+p^2} \times \sqrt {1+p^2} dp + c$$
$$y\sqrt {1+p^2} = 2a \int \frac {p^2}{\sqrt {1+p^2}} dp + c $$
I could not solve further from here.
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Usually you get fastest to the relevant equations if you start with the derivative by $p$, using $\dot y(p)=p\dot x(p)$. Then
$$
\dot x+p\dot y+y=2ap
$$
Multiply with $p$ and eliminate $x$
$$
(1+p^2)\dot y+py=2ap^2\implies \sqrt{1+p^2}y(p)=\int \frac{2ap^2}{\sqrt{1+p^2}}dp
$$
Now substitute $p=\sinh(u)$ in the last integral,
$$
=\int a(\cosh(2u)-1)du
=a\sinh(u)\cosh(u)-au+C
=ap\sqrt{1+p^2}-a\sinh^{-1}(p)+C.
$$
|
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|
Is there a $4$-by-$4$, rank $3$, positive semidefinite matrix with $a_{ii}=3$, $|a_{12}|\neq 1$, and principal minors having minimal eigenvalue $1$? Could anyone help me search for a positive semidefinite matrix $\left(a_{i,j}\right)_{4\times4}$ of rank 3 with $a_{i,i}=3$
and its all 3 by 3 principal minor matrices having minimal eigenvalue $\lambda_{\min}=1$, but $\left|a_{1,2}\right|\ne1$, or could anyone explain why such a matrix would not exist?
Thanks a lot.
|
Suppose $A$ is a rank-$3$ positive semidefinite $4\times4$ matrix such that the minimum eigenvalues of its principal $3\times3$ submatrices are equal to all $1$. We will prove that
$$
A=D\pmatrix{3&1&1&1\\ 1&3&-1&-1\\ 1&-1&3&-1\\ 1&-1&-1&3}D^\ast
$$
for some unitary diagonal matrix $D$ (and hence all of its off-diagonal entries have unit moduli and the answer to your question is negative). Note that
\begin{aligned}
A&:=\pmatrix{3&a&b&c\\ \overline{a}&3&d&e\\ \overline{b}&\overline{d}&3&f\\ \overline{c}&\overline{e}&\overline{f}&3}\\
&=\pmatrix{1&\frac a2&\frac b2&0\\ \frac{\overline{a}}{2}&1&\frac d2&0\\ \frac{\overline{b}}{2}&\frac{\overline{d}}{2}&1&0\\ 0&0&0&0}
+\pmatrix{1&\frac a2&0&\frac c2\\ \frac{\overline{a}}{2}&1&0&\frac e2\\ 0&0&0&0\\ \frac{\overline{c}}{2}&\frac{\overline{e}}{2}&0&1}
+\pmatrix{1&0&\frac b2&\frac c2\\ 0&0&0&0\\ \frac{\overline{b}}{2}&0&1&\frac f2\\ \frac{\overline{c}}{2}&0&\frac{\overline{f}}{2}&1}
+\pmatrix{0&0&0&0\\ 0&1&\frac d2&\frac e2\\ 0&\frac{\overline{d}}{2}&1&\frac f2\\ 0&\frac{\overline{e}}{2}&\frac{\overline{f}}{2}&1}\\
&=:X+Y+Z+W.
\end{aligned}
By assumption, $\lambda_\min\left(A(1:3,1:3)\right)=1$. Since $X(1:3,1:3)=\frac12\left(A(1:3,1:3)-I_3\right)$, we have $\lambda_\min\left(X(1:3,1:3)\right)=0$ and hence $X$ is positive semidefinite. Similarly, $Y,Z$ and $W$ are PSD too.
As $A$ is singular, it has an eigenvector $v$ in its null space. Since $X,Y,Z,W$ are PSD and
$$
0=v^\ast Av=v^\ast Xv+v^\ast Yv+v^\ast Zv+v^\ast Wv,
$$
we must have $v^\ast Xv=v^\ast Yv=v^\ast Zv=v^\ast Wv=0$ and in turn $Xv=Yv=Zv=Wv=0$. By relabelling the rows and columns of $A$ and by scaling $v$ if necessary, we may assume that $v=(x,y,z,1)^\top$. The matrix equations $Xv=0,Yv=0,Zv=0$ and $Wv=0$ can then be rewritten as four systems of linear equations
$$
\begin{array}{ll}
\begin{cases}
2x+ay+bz=0,\\
\overline{a}x+2y+dz=0,\\
\overline{b}x+\overline{d}y+2z=0,
\end{cases}
&
\begin{cases}
2x+ay+c=0,\\
\overline{a}x+2y+e=0,\\
\overline{c}x+\overline{e}y+2=0,
\end{cases}
\\
\\
\begin{cases}
2x+bz+c=0,\\
\overline{b}x+2z+f=0,\\
\overline{c}x+\overline{f}z+2=0,
\end{cases}
&
\begin{cases}
2y+dz+e=0,\\
\overline{d}y+2z+f=0,\\
\overline{e}y+\overline{f}z+2=0.
\end{cases}
\end{array}
$$
Rearrange these equations into four groups, namely, those equations with the constant term $2$, those with the term $2x$, those with the term $2y$ and those with the term $2z$:
\begin{align}
&\overline{c}x+\overline{e}y+2=\overline{c}x+\overline{f}z+2=\overline{e}y+\overline{f}z+2=0\\
&2x+ay+bz=2x+ay+c=2x+bz+c=0\\
&\overline{a}x+2y+dz=\overline{a}x+2y+e=2y+dz+e=0\\
&\overline{b}x+\overline{d}y+2z=\overline{b}x+2z+f=\overline{d}y+2z+f=0.
\end{align}
From each group of equations in the above, we obtain
\begin{align}
&\overline{c}x=\overline{e}y=\overline{f}z=-1,\tag{1}\\
&ay=bz=c=-x,\tag{2}\\
&\overline{a}x=dz=e=-y,\tag{3}\\
&\overline{b}x=\overline{d}y=f=-z.\tag{4}
\end{align}
Substitute $c=-x$ (from $(2)$) into $\overline{c}x=-1$ (from $(1)$), we get $|c|=|x|=1$. Similarly, we also have $|e|=|y|=1$ and $|f|=|z|=1$. It follows from $(2),(3)$ and $(4)$ that $|a|,|b|$ and $|c|$ are also equal to $1$.
It follows that if we replace $A$ by $D^\ast AD$ for some appropriate unitary diagonal matrix $D$, we may assume that $a=b=c=1$. Then $(2)$ implies that $(x,y,z)=(-1,1,1)$, and $(3),(4)$ give $d=e=f=-1$. Now we are done.
|
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|
How to solve exercises with polynomial with 2 parameters having all real roots? $f=2x^{4}+4x^{3}+3x^{2}+bx+c$ has all real roots, find b,c (b,c are from R).
Thanks a lot, I tried with substitution, I don't know, is there something with derivative? please help thanks
|
$f(x) = 2x^4+4x^3+3x^2+bx+c$
If $f(x)$ has all $4$ roots to be real and complex, then it's discriminate $$\Delta = (-72)b^2+176b^3-108b^4-864c-2304bc+1536b^2c+2304c^2-3072bc^2+2048c^3$$ must be real, $\Delta > 0$
Also if you divide $f(x)$ by 2, and make a translation $x = y-\frac{1}{2}$
$2x^4+4x^3+3x^2+bx+c = 0$
$x^4+2x^3+\frac{3/2}x^2+\frac{b/2}x+\frac{c/2} = 0$
Say $x = y-\frac{1}{2}$
The equation is depressed that suddenly appear to be
$y^4+(\frac{b}{2}-\frac{1}{2})y+\frac{c}{2}-\frac{b}{4}+\frac{3}{16} = 0$
Notice that the $y^2$ automatically vanishes
Defined $D = 512c-256b+192$
For the polynomial to therefore have real distinct root $D < 0$ and $\Delta > 0$
Nature of roots
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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.