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Claculate limit $\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$ I have a problem to calculte this limit:
$$\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$$
I used Taylor expansion for $\sin(x), \cos(x)$ and considered also $1-\cos(\alpha)=2\sin^2(\frac{\alpha}{2})$ and $\alpha=2-2\sqrt{\frac{\sin(x)}{x}}$ (I have no clue, what to do next with it), but with Taylor and ended up with:
$$\lim_{x\to 0}\frac{\sqrt{1-\frac{x^2}{6}+o(x^2)}+o(\sqrt{1-\frac{x^2}{6}+o(x^2)})}{x^4} $$
which tends to infinity
|
Hint
Compose Taylor series one piece at the time
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\sqrt{\frac{\sin (x)}{x}}=1-\frac{x^2}{12}+\frac{x^4}{1440}+O\left(x^6\right)$$
$$\cos \left(1-\sqrt{\frac{\sin (x)}{x}}\right)=1-\frac{x^4}{288}+\frac{x^6}{17280}+O\left(x^8\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
HINT: $a^2+b^2=c^2$ cannot hold for $a,b$ odd and $c$ even. (using congruences) Here is what I have: Suppose the contrary. Thus,
(2k+1)(2k+1) + (2j+1)(2j+1) = (2p)(2p)
Take mod 2 of both sides
[1][1]+[1][1]=[0]
[2]=[0]
[0]=[0]
No contradiction ... am I approaching this correctly? I want to figure this out myself but would like a small hint
|
With $a$, $b$ odd we may write
$a = 2m + 1, \tag 1$
$b = 2n + 1; \tag 2$
then
$a^2 = 4m^2 + 4m + 1, \tag 3$
$b^2 = 4n^2 + 4n + 1, \tag 4$
$a^2 + b^2 = 4(m^2 + n^2 + m + n) + 2; \tag 5$
also,
$c = 2p, \tag 6$
whence
$c^2 = 4p^2; \tag 7$
if
$a^2 + b^2 = c^2, \tag 8$
then substituting in (3), (4) and (7) we find
$4(m^2 + n^2 + m + n) + 2 = 4p^2; \tag 9$
we reduce this modulo $4$ and obtain
$2 \equiv 0 \mod 4, \tag{10}$
a contradiction which shows there are no such $a$, $b$, and $c$.
|
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|
Summation of $n$th partial products of the square of even numbers diverges, but for odd numbers they converge in this series I'm looking at. Why? So I have the two following series:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}$$
I figured out the $n$th partial products:
$$\prod_{k=1}^n(2k)^2=4^n(n!)^2$$
$$\prod_{k=0}^n (2k+1)^2=\frac{((2n+1)!)^2}{4^n(n!)^2}$$
So putting these back into my series they become the following:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}=\sum_{n=1}^\infty\frac{4^n(n!)^2}{(2n+2)!}$$
Now this diverges as expected by the limit test test. However when I look at my other series:
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}=\sum_{n=0}^\infty\frac{(
(2n+1)!)^2}{4^n(n!)^2(2n+3)!}$$
By the limit test maybe diverges or maybe doesn't, and the ratio test is inconclusive. Since I wasn't sure what to use for the a comparison test I threw this into wolfram alpha and it told me it converges which is baffling to me since both series are very similar if we write them out:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}=\frac{2^2}{4!}+\frac{2^24^2}{6!}+\frac{2^24^26^2}{8!}\cdot\cdot\cdot\cdot$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}=\frac{1^2}{3!}+\frac{1^23^2}{5!}+\frac{1^23^25^2}{7!}+\cdot\cdot\cdot$$
They both have the nth parial product of the even/odd integers squared in the numerator, and are over a factorial that is two greater than $n$, so I'm not sure why one is diverging and the other is converging. Is wolframalpha wrong, as it can be at times? Or is there someething here that I am missing?
|
Convergence
Using the asymptotic approximation given in inequality $(9)$ of this answer, we get
$$
\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag1
$$
Therefore,
$$
\begin{align}
\frac{\prod\limits_{k=1}^n(2k)^2}{(2n+2)!}
&=\frac{4^nn!^2}{(2n)!(2n+1)(2n+2)}\\
&=\frac{\color{#090}{4^n}}{\color{#090}{\binom{2n}{n}}\color{#C00}{(2n+1)(2n+2)}}\\
&\sim\frac{\color{#090}{\sqrt{\pi n}}}{\color{#C00}{4n^2}}\\
&=\frac{\sqrt\pi}{4}\frac1{n^{3/2}}\tag2
\end{align}
$$
and
$$
\begin{align}
\frac{\prod\limits_{k=0}^n(2k+1)^2}{(2n+3)!}
&=\frac{(2n+1)!^2}{4^nn!^2(2n+3)!}\\
&=\frac{\color{#090}{\binom{2n}{n}}\color{#C00}{(2n+1)}}{\color{#090}{4^n}\color{#C00}{(2n+2)(2n+3)}}\\
&\sim\frac1{\color{#090}{\sqrt{\pi n}}\,\color{#C00}{2n}}\\
&=\frac1{2\sqrt\pi}\frac1{n^{3/2}}\tag3
\end{align}
$$
The sums of both $(2)$ and $(3)$ converge by comparison to a $p$-series with $p=3/2$.
Evaluation
In this answer, it is shown that
$$
\begin{align}
\arcsin^2(x)
&=\sum_{k=1}^\infty\frac{4^kx^{2k}}{2k^2\binom{2k}{k}}\\
&=\sum_{k=1}^\infty\frac{4^k}{\binom{2k}{k}}\frac{x^{2k}}{2k^2}\\
&=\sum_{k=0}^\infty\frac{4^k}{\binom{2k}{k}}\frac{2x^{2k+2}}{(2k+1)(2k+2)}\tag4\\
\end{align}
$$
and in this answer, it is shown that
$$
\begin{align}
\arcsin(x)
&=\sum_{k=0}^\infty\frac2{2k+1}\binom{2k}{k}\left(\frac{x}{2}\right)^{2k+1}\\
&=\sum_{k=0}^\infty\frac{\binom{2k}{k}}{4^k}\frac{x^{2k+1}}{2k+1}\\
&=x+\sum_{k=0}^\infty\frac{\binom{2k}{k}}{4^k}\frac{(2k+1)x^{2k+3}}{(2k+2)(2k+3)}\tag5
\end{align}
$$
Applying $(4)$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{\prod\limits_{k=1}^n(2k)^2}{(2n+2)!}
&=\sum_{n=1}^\infty\frac{4^n}{\binom{2n}{n}(2n+1)(2n+2)}\\
&=\frac12\arcsin(1)^2-\frac12\\
&=\frac{\pi^2}8-\frac12\tag6
\end{align}
$$
Applying $(5)$, we get
$$
\begin{align}
\sum_{n=0}^\infty\frac{\prod\limits_{k=0}^n(2k+1)^2}{(2n+3)!}
&=\sum_{n=0}^\infty\frac{\binom{2n}{n}(2n+1)}{4^n(2n+2)(2n+3)}\\
&=\arcsin(1)-1\\[6pt]
&=\frac\pi2-1\tag7
\end{align}
$$
|
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|
Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And we know that $\zeta_n=e^{\frac{2\pi i}{n}}$
So the roots are $2^\frac{1}{3}, -(2^\frac{1}{3}) ~and~ 2^\frac{1}{3}\frac{1}{2}(\sqrt{3}i-1).$
Is this correct?
Note: I don't like factoring method. I studied in schools time. So don't help with that method.
Thanks!
|
The best way to do it is to use the methods that other responses have detailed. I am going to show you an inferior method only because understanding the relationship between this inferior method and the better method (already described by others) will stretch your intuition.
Let $a$ denote $2^{(1/3)}.$
Given $0 = (x^3 - 2) = (x^3 - a^3) = (x - a)(x^2 + ax + a^2)$, then
one of the roots will be given by the factor $(x - a)$.
The other two roots will come from applying the quadratic eqn against $(x^2 + ax + a^2) = 0$.
This gives $$x = \frac{1}{2} \times \left[-a \pm\sqrt{a^2 - 4a^2}\right]$$
$$= \frac{1}{2} \times \left[-a \pm i \times \sqrt{3} \times a\right]$$
$$= \frac{1}{2} \times a \times \left[-1 \pm i \times \sqrt{3}\right]$$
$$= 2^{(1/3)} \times \frac{1}{2} \times \left[-1 \pm i \times \sqrt{3}\right].$$
|
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|
Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ in the plane $x+y+z=4$.
Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ with respect to plane $x+y+z=4$.
my method:
Any point in the given line will be $(3t,2t,t+1)$.
Thus the line meets the plane in $(3/2,1,3/2)$.
We now take a random point $(0,0,1)$ on this line. Clearly its image will lie on lines image wrt to this plane.
We have to find image of $(0,0,1)$.
Since the perpendicular to the plane passing through $(0,0,1)$ is $$x=y=z-1$$. It will meet the plane in $(1,1,2)$. i.e image is $(2,2,3)$.
Thus the image line which passes through $(3/2,1,3/2)$ ad $(2,2,3)$ is $$\frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}$$
I am however looking for alternative methods
|
The line $r:\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ is the intersection of the planes $\alpha:\frac{x}{3}=\frac{y}{2};\;\beta:\frac{x}{3}=z-1$.
Any linear combination $\mathcal{P}$ of the two planes $\alpha:2x-3y=0$ and $\beta:x-3z+3=0$ passes through the line $r$
$$\mathcal{P}:h(2x-3y)+k(x-3z+3)=0;\;\forall h,k\in\mathbb{R}\land(h^2+k^2\ne 0)$$
Reorder the equation of $\mathcal{P}$
$$\mathcal{P}:x(2h+k)-3hy-3kz+3k=0$$
We have the plane passing through $r$ perpendicular to the given plane $\pi:x+y+z=4$ if the normal vectors $p=(2h+k,-3h,-3k)$ and $u=(1,1,1)$ are orthogonal, that is $p\cdot u=0$, which means
$$2h+k-3h-3k=0\to h=-2k$$
Plug in the equation of $\mathcal{P}$
$$\mathcal{P*}:x(-4k+k)+6ky-3kz+3k=0\to -3k(x-2y+z-1)=0$$
The plane perpendicular to $\pi$ passing through $r$ has equation $\gamma:x-2y+z=1$.
Thus the projection $r_{\pi}$ of $r$ on $\pi$ is the line
$$\begin{cases}
x+y+z=4\\
x-2y+z=1\\
\end{cases}
$$
that is $$r_{\pi}:\left( 3-t,1,t\right)$$
or
$$
\begin{cases}
x=3-z\\
y=1\\
\end{cases}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Partial fraction decomposition $\frac{1}{(x-y)^2}\frac{1}{x^2}$ I want to integrate $$\frac{1}{(x-y)^2}\frac{1}{x^2}$$ with respect to $x$.
I know that I have to apply partial fraction decomposition but which ansatz do I have to make to arrive at
$$\frac{1}{(x-y)^2}\frac{1}{x^2}=-\frac{2}{y^3(x-y)}+\frac{1}{y^2(x-y)^2}+\frac{2}{y^3x}+\frac{1}{y^2x^2}$$
|
Here is the (comparatively) fast way to determine the partial fraction decomposition, using the poles of the fraction. To make the computation clearer, I'll denote temporarily $y$ as the constant $a$.
The decomposition has the form
$$\frac 1{x^2(x-a)^2}=\frac Ax+\frac B{x^2}+\frac C{x-a}+\frac D{(x-a)^2}.\tag 1$$
Multiplying both sides by $x^2(x-a)^2$, you obtain
$$1=Ax(x-a)^2+B(x-a)^2+Cx^2(x-a)+Dx^2.$$
Setting $x=0$ yields instantly $\;B=\dfrac1{a^2}$. Similarly, setting $x=a$ yields $D=\dfrac1{a^2}$.
To obtain the last two coefficients, multiply both sides of $(1)$ by $x$ and let $x\to\infty$: you obtain the relation $\:A+C=0$.
We need a last equation to determine $A$ & $C$. The best way seems to let $x=\frac a2$ (the arithmetic mean of the poles), which yields the relation
$$A-C=\frac 4{a^3},\quad\text{whence}\quad A=\frac 2{a^3},\enspace C=-\frac 2{a^3}.$$
Note:
If the fraction has a pole of higher order, it is generaly still faster to use division of polynomials by increasing degrees.
|
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|
Integral $\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$ vanishes Come across
$$I=\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$$
and break the integrand as
$$\frac{1-x(2-\sqrt x)}{1-x^3}=\frac{1-x}{1-x^3}- \frac{x(1-\sqrt x)}{1-x^3}
$$
The first term simplifies and the second term transforms with $\sqrt x\to x$. Then, the integral becomes
$$I=\int_0^\infty \frac{1}{1+x+x^2}dx
- 2\int_0^\infty \frac{x^3}{1+x+x^2+x^3+x^4+x^5}dx
$$
The first integration is straightforward $\frac{2\pi}{3\sqrt3}$ and the second is carried out via partial fractional decomposition. After some tedious and lengthy evaluation, it surprisingly yields the same value $\frac{2\pi}{3\sqrt3}$.
Given its vanishing value, there may exist a shorter and more direct derivation, without evaluation of any explicit intermediate values.
|
I'm not sure if this is something that is of value to you since it uses your approach, but once you transform $\sqrt{x} \to x$ for the second integral you can use PFD to cancel out the first integral and be left with two simple integrals, instead of expressing it as you did:
$$-\int_0^{\infty} \frac{2x^3(1-x)}{1-x^6} \; \mathrm{d}x=-\int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\frac{1}{3}\int_0^{\infty} \frac{2x-1}{x^2-x+1} \; \mathrm{d}x+\frac{2}{3} \int_0^{\infty} \frac{\mathrm{d}x}{x+1} $$
And so you have,
\begin{align*}
I&= \int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\frac{1}{3}\int_0^{\infty} \frac{2x-1}{x^2-x+1} \; \mathrm{d}x+\frac{2}{3} \int_0^{\infty} \frac{\mathrm{d}x}{x+1} \\
&= \bigg [-\frac{1}{3} \ln(x^2-x+1)+\frac{2}{3}\ln(x+1) \bigg]\bigg \rvert_0^{\infty} \\
&=0
\end{align*}
|
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|
Trigonometric equation: $3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$
Solve the following equation:
$$3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$$
My attempt:
$$3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$$
$$\implies 3\sin x = -\tan x \cdot \cot 2x - \cot x$$
$$\implies 3\sin x + \tan x \cdot \cot 2x + \cot x = 0$$
Since $\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x} \implies \cot 2x = \dfrac{1}{\tan 2x} = \dfrac{1 - \tan^2 x}{2 \tan x}$, so:
$$3\sin x + \dfrac{1 - \tan^2 x}{2} + \cot x = 0$$
I got stuck from here. How can I proceed this?
|
I think that after writing that tou want to find the zeros of
$$f(x)=3\sin (x) + \dfrac{1 - \tan^2 (x)}{2} + \cot (x) $$ the problem becomes purely numerical.
If you plot the function, it is quite awful because of the discontinuities due to the tangent and cotangent. It would be better to remove them multipluing everythng by $\sin(x)$ (because of the cotangent) and by $\cos^2(x)$ (because of the tangent). Doing it and using a few trigonometric identities, you could instead try to find the zeros of function
$$g(x)=3-2 \sin (x)+2 \sin (3 x)+6 \cos (x)+2 \cos (3 x)-3 \cos (4 x)$$
Ploting the funtion $g(x)$, we see positive solutions close to $1.3$, $2.0$, $2.6$, $3.8$, $7.5$, $10.0$, $\dots$ and negative solutions close to $-11.3$, $-10.6$, $-10.0$, $-8.8$, $-2.5$.
Using any of these as a starting point, using Newton method will converge quite fast. For example for the fourth positive root, the iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 3.80000 \\
1 & 3.74477 \\
2 & 3.74671 \\
3 & 3.74671
\end{array}
\right)$$ As said in comments, there is more than likely a problem with the problem itself.
As @am301 commented, making $y=\sin(x)$ leads to a monster.
Using the tangent half-angle substitution would require to solve for $t$
$$\frac {t^8-t^7-14 t^6+5 t^5+24 t^4+5 t^3-10 t^2-t-1 } {2 t\left(t^2-1\right)^2 \left(t^2+1\right) }=0$$ the real solutions of which being
$$\{ -3.20369, 0.69728, 1.50324, 3.8625\}$$
|
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|
How to find a useful variable change for this integral I would like to find the area of the following region
$$
D=\left \{(x,y): -\sqrt{1+y^2}\leq x\leq \sqrt{1+y^2}; -1\leq y\leq (x+1)/2\right \}.
$$
I try to calculate the double integral brute force, but following this path
I came across some very unpleasant integrals. So I think that maybe a variable
change may be an appropriate approach here.
Can someone suggest me a useful variable change to calculate the area of this region?
|
See the diagram. You need to find the area of the region ABCO. If you integrate along $y$ axis taking strips of thickness $dy$ parallel to $x$ axis, you can see that from $y = -1$ to $y = 0$, both the left and right ends are bound by the hyperbola but for strips at $0 \leq y \leq \frac{4}{3}$, the left is bound by the line $2y = x+1$ and the right is bound by the hyperbola. So we divide our integral in two parts.
Now to find the intersection of the line $2y = x+1$ and the hyperbola $x^2 - y^2 = 1$,
$x^2 - y^2 = 1$
At intersection, $x^2 - \frac{(x + 1)^2}{4} = 1$. That gives us $x = \frac{5}{3}, y = \frac{4}{3}$
So, $A = \displaystyle \int_{-1}^0 ({x_r - x_l}) \, dy \, \, + \int_{0}^{4/3} ({x_r - x_l}) \, dy$
$A = \displaystyle \int_{-1}^0 (\sqrt{1+y^2} - (-\sqrt{1+y^2}) \, dy \, \, + \int_{0}^{4/3} (\sqrt{1+y^2} - (2y-1)) \, dy$
$A = \displaystyle 2\int_{-1}^0 \sqrt{1+y^2} \, dy \, \, + \int_{0}^{4/3} (\sqrt{1+y^2} - 2y + 1) \, dy$
To integrate $\sqrt{1+y^2}$, one of the ways is to substitute $y = \tan \theta$.
Integral of $\sqrt{1+y^2}$ is given by $\frac{y}{2} \sqrt{1+y^2} + \frac{1}{2} \ln ({y + \sqrt{1+y^2}})$.
You can check WolframAlpha for the same.
|
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|
(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
The first few terms of this series shows me that summation $\lfloor \log_2 n \rfloor$ for $n=1$ to $n=10$ give $2^{n +1}$.
|
Denote the sum by $S_{n}$. Note that for any $k\in\mathbb{N}$ there are $2^k$ positive integers $x$ for which $[\log_{2}(x)]=k$, and those are $x=2^{k},2^{k}+1,\ldots,2^{k+1}-1$. Thus we have $$S_{2^{k}-1}=0 + (1+1) + (2+2+2+2+) + \cdots + \bigl((k-1)+(k-1)+\cdots + (k-1)\bigr)$$ where there number of $(k-1)$ terms is $2^{k-1}$. It follows that $$S_{2^{k}-1} = (k-2)2^{k}+2$$ Putting $k=8$ we see that $S_{255}=1538<1994$ and putting $k=9$ we see that $S_{511}=3586>1994$. Thus it's clear that our $n$ should satisfy $2^{8}-1<n<2^{9}-1$. Now we have $$1994=S_{n}=S_{255}+(n-255)8=8n-502$$ which gives $n=312$.
|
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|
Limit multivarible How do I solve this limit?
$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz} $$
This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$
The first one is a standard limit with value one, but I'n not sure about the other term.
|
Yes your idea is very good to put away the $\sin$ term, now we have that, for example by spherical coordinates
$$\frac{xyz}{x^2+y^2+z^2} =r \cos \theta\sin \theta\cos^2 \phi\sin \phi $$
or as an alternative by AM-GM we can use that
$$x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2y^2z^2} \implies \left|\frac{xyz}{x^2+y^2+z^2}\right| \le \frac{|xyz|^\frac13}{3}$$
concluding in both cases by squeeze theorem.
|
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"url": "https://math.stackexchange.com/questions/3887468",
"timestamp": "2023-03-29T00:00:00",
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|
Find the product of all values of $(1+i\sqrt 3)^{\frac{3}{4}}$. Find the product of all values of $(1+i\sqrt 3)^{\frac{3}{4}}$.
My try:
$(1+i\sqrt 3)^{\frac{3}{4}}=\exp (\frac{3}{4}(Log(1+i\sqrt 3)))=\exp(\frac{3}{4}(\log2+i\frac{\pi}{3}+2n\pi i))$.
I am kinda stuck on how to find the product of all values of the above expression.
Can someone please help me out.
|
$$1+i\sqrt3=2e^{i\pi/3}$$
The fourth roots of this number have common magnitude $2^{1/4}$ and arguments $\pi/12,7\pi/12,-5\pi/12,-11\pi/12$. The $\frac34$-powers thus have common magnitude $2^{3/4}$ and arguments $\pi/4,7\pi/4\equiv-\pi/4,-5\pi/4\equiv3\pi/4,-11\pi/4\equiv-3\pi/4$. The producf of all the four possible values of $(1+i\sqrt3)^{3/4}$ therefore has magnitude equal to the product of the magnitudes of each possible value, which is $8$, and argument equal to the sum of the arguments of each possible value, which is $0$. The final answer is $8$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$ One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$
I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * $\frac{1}{(1+x)^5}$ * $\frac{1}{(1-x)^6}$
Then got $f(x) = x^3 * \sum_{k=0}^\infty \binom{n+4}{4}(-1)^kx^k * \sum_{k=0}^\infty \binom{n+5}{5}x^k$
I dont know what to do from here, can you tell me what is the next step?
|
$$
\begin{align}
\frac{x^3}{(1+x)^5 (1−x)^6}
&=\frac{x^3+x^4}{\left(1-x^2\right)^6}\tag1\\
&=\left(x^3+x^4\right)\sum_{k=0}^\infty(-1)^k\binom{-6}{k}x^{2k}\tag2\\
&=\left(x^3+x^4\right)\sum_{k=0}^\infty\binom{k+5}{5}x^{2k}\tag3\\
&=\sum_{k=0}^\infty\binom{k+5}{5}\left(x^{2k+3}+x^{2k+4}\right)\tag4\\
&=\sum_{k=2}^\infty\binom{k+3}{5}\left(x^{2k-1}+x^{2k}\right)\tag5
\end{align}
$$
Explanation:
$(1)$: multiply by $\frac{1+x}{1+x}$
$(2)$: use the series for $(1+x)^{-6}$
$(3)$: $\binom{-6}{k}=(-1)^k\binom{k+5}{5}$ when $k\ge0$
$\phantom{\text{(3):}}$ (negative binomial coefficients)
$(4)$: distribute the $x^3+x^4$
$(5)$: substitute $k\mapsto k-2$
Thus, if $n=2k-1$ or $n=2k$, then the coefficient of $x^n$ is $\binom{k+3}{5}$.
This can be stated as the coefficient of $x^n$ is $\binom{\left\lfloor\frac{n+7}2\right\rfloor}{5}$
|
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|
Finite sum which sums to $x e^x$ In my analysis class, we have to prove this and another two related sums that I think I could prove if I knew this. However, I don't know how to begin solving this; any hints would be appreciated.
$$\sum_{n=0}^{\infty} \left( e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac{x^n}{n!} \right)=xe^x$$
One method of proving this mentions summation by parts, and the only reference I found in relation to that is Abel's lemma which I tried to use but doesn't help me. What other solutions are there?
|
\begin{align}
& \sum_{n=0}^\infty \left( e^x-1-\frac x {1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac{x^n}{n!} \right) = \sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!} \\[8pt]
& \begin{array}{cccccccccc}
= & x & + & \dfrac{x^2} 2 & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & + & \dfrac{x^2} 2 & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & & & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & & & & & + & \dfrac{x^5}{120} & + & \cdots
\end{array} \\[10pt]
= {} & x + x^2 + \frac{x^3} 2 + \frac{x^4} 6 + \frac{x^5}{24} + \cdots \\[8pt]
= {} & x\left( 1 + x + \frac{x^2} 2 + \frac{x^3} 6 + \frac{x^4}{24} +\cdots \right) = xe^x.
\end{align}
Postscript: Maybe it is useful to express it in a way in which the general form is explicit rather than being indicated by three dots that mean "continue with the same pattern."
\begin{align}
& \sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!} \\[8pt]
= {} & \sum_{n,k\,:\,k\,\ge\,n+1\,\ge\,1} \frac{x^k}{k!} \\[8pt]
= & \sum_{k=1}^\infty \left( \sum_{n=0}^{k-1} \frac{x^k}{k!} \right)
\end{align}
But no $\text{“}n\text{”}$ appears in this sum in which $n$ goes from $0$ to $k-1$; therefore the sum is just $x^k/k!$ multiplied by the number of terms, which is $k$:
$$
\sum_{k=1}^\infty k\cdot\frac{x^k}{k!} = x\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!} = xe^x.
$$
|
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|
Analysis - Find the Limit Let $P(x) = ax^3 + bx^2 + cx + d$, $a, b, c, d \in \mathbb{R}$, determine whether the sequence
$$
\left(
\frac{P(n)}
{P(n + 1)}
\right)
$$
it is convergent or not. If so, calculate the limit.
I managed to show that the sequence
$$\frac{an^3 + bn^2 + cn + d}{a(n+1)^3 + b(n+1)^2 + c(n+1) + d}$$
is limited $(0<x_n<1)$, but I can't prove that your limit is 1 by the definition.
|
We can use that, for $a\neq 0$
$$\frac{an^3 + bn^2 + cn + d}{a(n+1)^3 + b(n+1)^2 + c(n+1) + d}=\frac{a + b\frac1{n} + c\frac1{n^2} + d\frac1{n^3}}{a\left(1+\frac1n\right)^3 + \frac b n\left(1+\frac1n\right)^2 + \frac c {n^2}\left(1+\frac1n\right) + \frac d{n^3}} \to 1$$
For $a=0$ we can divide by $n^2$ and so on.
The assuming $a\neq 0$ by the definition we need to show that
$$\left|\frac{an^3 + bn^2 + cn + d}{a(n+1)^3 + b(n+1)^2 + c(n+1) + d}-1\right| \le \epsilon$$
$$\left|\frac{a(3n^2+3n+1) + b(2n+1) + c }{a(n+1)^3 + b(n+1)^2 + c(n+1) + d}\right| \le \epsilon$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that the solutions to an ODE are non-periodic oscillating. I have the ODE $$\ddot{x}+ \dot{x}+k^2x=0$$
where the dot means derivative with respect to $t$ ($x$ is a function of $t$) and $k$ is a constant such that $k>\frac{1}{2}$
prove that the solutions to the ODE are non-periodic and oscillating.
What I tried is the following.
I solved the characteristic polynomial equation $$\lambda^2+\lambda+k^2=0$$
which gave me $$ \lambda_{1,2} = \frac{-1 \pm 2 \sqrt{k^2-\frac{1}{4}}}{2} $$
Now since $k>\frac{1}{2}$
the solutions are real so the ODE has no periodic solutions, but I can't understand how to prove that the solutions oscillate, what does it even mean for the solutions to oscillate?
|
Given the equation
$\ddot x + \dot x + k^2x = 0, \tag 1$
the characteristic equation is indeed
$\lambda^2 + \lambda + k^2 = 0, \tag 2$
but the "quadratic formula" used to find $\lambda_{1,2}$ is misstated; it should read
$\lambda_{1,2} = \dfrac{-1 \pm \sqrt{1 - 4k^2}}{2}; \tag 3$
from this we see that when $k > 1/2$ the values of $\lambda_{1,2}$ are complex conjugate numbers with real part $-1/2$, hence the solutions exponentially decay to $0$; however the presence of a non-vanishing imaginary component in the $\lambda_{1,2}$ forces the solutions to venture
back and forth across $0$ whilst they decay; thus the solutions are non-periodic and oscillating.
We can in fact write down explicit solutions to (1); they serve to reify and illustrate the words of the preceding paragraph. Writing (3) in the form
$\lambda_{1, 2} = -\dfrac{1}{2} \pm i \dfrac{1}{2}\sqrt{4k^2 - 1}, \tag 4$
we may express two relatively simple solution as
$x_\pm(t) = \exp \left (\lambda_{1,2} t\right ) = \exp \left (\left (-\dfrac{1}{2} \pm i \dfrac{1}{2}\sqrt{4k^2 - 1} \right) t\right )$ =
$\exp \left (-\dfrac{1}{2} t\right )\exp \left (\pm i \dfrac{1}{2}\sqrt{4k^2 - 1} t\right )$
$= \exp \left (-\dfrac{1}{2} t\right )\left (\cos \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right ) \pm i \sin \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right )\right). \tag 5$
Since (1) is a real differential equation, the real and imaginary parts of $x_\pm(t)$ are individually solutions, thus for example
$x(t) = \exp \left (-\dfrac{1}{2} t\right )\left (\cos \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right ) \right) \tag 6$
is a solution which exhibits both exponential decay and oscillation.
|
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|
How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$? So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)
But after this I don't have any clue on how to proceed.
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta}
{\sin \theta+\cos \theta}
$$
Original image
|
$$\frac{\sin3x-\cos3x}{\sin x+\cos x}+1=\frac{\sin3x+\sin x+\cos x-\cos3x}{\sin x+\cos x}=\frac{\sin(2x+x)+\sin (2x-x)+\cos (2x-x)-\cos(2x+x)}{\sin x+\cos x}=\frac{\sin2x\cos x+\sin x\cos2x+\sin2x\cos x-\sin x\cos 2x+\cos 2x\cos x+ \sin 2x\sin x-(\cos 2x\cos x- \sin 2x\sin x)}{\sin x+\cos x}=\frac{2 \sin2x\cos x +2 \sin 2x\sin x}{\sin x+\cos x}=\frac{2 \sin2x(\cos x +\sin x)}{\sin x+\cos x}=2\sin2x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$. I am about to solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.
What I did is I use the partial fraction decomposition to rewrite the $f(z)$ into $f(z) = \frac{3}{z+3}+\frac{2}{z-2}$. Now based on the region to where the function $f$ should be valid on is that $1< |z-1|< 4$, I'm not sure if I did it right when I assumed that $\left|\frac{2}{z}\right|<1$ and $\left|\frac{3}{z}\right|<1$. (I just did it with brute force to arrive on the two inequalities). Can you help me with that?
Assuming my two inequalities are correct, I have now
$\frac{3}{z+3} = \frac{3/z}{1+3/z} = 3/z \sum_{n=0}^{\infty} \left(- \frac{3}{z}\right)^n = \sum_{n=0}^{\infty} (-1)^{n-1}3^nz^{-n} = \sum_{n=1}^{\infty} (-1)^n3^{n+1}z^{-(n+1)}$,
and
$\frac{2}{z-2} = \frac{2/z}{1-2/z} = 2/z \sum_{n=0}^{\infty} \left(\frac{2}{z}\right)^n = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}}.$
Then after that, I combined which I have the result as
$f(z) = \frac{z-12}{z^2+z-6} = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}} + \sum_{n=1}^{\infty} (-1)^n(3)^{n+1}z^{-(n+1)}$.
Now, did I do it correctly? or I did wrong on the inequalities of the region to where the laurent should be valud to? Thanks for those who can help.
|
Your answer cannot possibly be an answer to the given question, since the answer should be a series of the form $\sum_{n=-\infty}^\infty a_n(z-1)^n$.
Note that\begin{align}\frac{z-12}{z^2+z-6}&=\frac3{z+3}-\frac2{z-2}\\&=\frac3{z-1+4}-\frac2{z-1-1}\\&=\frac34\frac1{1+\frac{z-1}4}+2\frac1{1-(z-1)}.\end{align}Now, since $1<|z-1|$ we have$$\frac1{1-(z-1)}=-\sum_{n=-\infty}^{-1}(z-1)^n$$and, since $|z-1|<4$ we have$$\frac1{1+\frac{z-1}4}=\sum_{n=0}^\infty\frac{(-1)^n}{4^n}(z-1)^n.$$Can you take it from here?
|
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|
Probability of first to draw red ball from urn without replacement There are R red and G green balls in a box. Alice and Bob pick one ball from the box in turns without replacement. First one to draw a red ball wins. What is the probability that Alice wins if she draws first?
Here is my attempt - Could someone please advise where I went wrong?
For any draw, the probability of picking a red ball is equal to picking $k$ green balls before that and a red ball on $k+1$ draw.
$\displaystyle P(i=k)=\frac{1}{(k+1)}\frac{\dbinom{G}{k}\dbinom{R}{1}}{\dbinom{N}{k+1}}=\displaystyle \frac{R}{(k+1)} \frac{G!}{k!(G-k)!}\frac{(N-k-1)!(k+1)!}{N!}$
$\displaystyle=\frac{G!}{N!} \frac{(R+G-k-1)!}{(G-k)!}.R\frac{(R-1)!}{(R-1)!}=\displaystyle\frac{1}{\binom{R+G}{G}}\dbinom{R+G-k-1}{G-k}$
Let $P(A)$=Alice wins, $P(B)$=Bob wins. Let $P(x)=P(A) - P(B)$. Therefore $P(A) =\displaystyle\frac{1}{2}(1+P(x)) $
$\displaystyle P(x)=\frac{1}{\dbinom{R+G}{G}} \sum_{k=0}^{G} (-1)^{k} \dbinom{R+G-k-1}{G-k}$
If $G$ is even, $-1^k=-1^{G-k}$; using binomial identity
$\dbinom{n}{k} = -1^k \dbinom{k-n-1}{k}$, we get,
$\displaystyle P(x)=\frac{1}{\dbinom{R+G}{G}} \sum_{k=0}^{G} \dbinom{-R}{G-k}$
$\displaystyle P(x)=\frac{1}{2^N\dbinom{R+G}{G}} \sum_{k=0}^N \dbinom{N}{k} \sum_{k=0}^G \dbinom{-R}{G-k}$
$\displaystyle P(x)=\frac{1}{2^N\dbinom{R+G}{G}} \sum_{k=0}^N \sum_{k=0}^G \dbinom{G+R}{k} \dbinom{-R}{G-k}$
Using Chu-Vandermonde identity, we get,
$\displaystyle P(x)=\frac{1}{2^N\dbinom{R+G}{G}} \sum_{k=0}^N \dbinom{G}{G}$
$\displaystyle P(x)=\frac{N+1}{2^{N}\dbinom{R+G}{G}}$ - which seems wrong e.g. set $G=0$
|
Let random variable $R_1$ be the number of draws to get the first red ball. For $k\in\{1,\dots,G+1\}$, we have
\begin{align}
P(R_1=k)
&= \left(\prod_{d=0}^{k-2} \frac{G-d}{R+G-d}\right)\frac{R}{R+G-(k-1)} \\
&= \frac{\binom{G}{k-1}}{\binom{R+G}{k-1}}\frac{R}{R+G-(k-1)}\\
&= \frac{\binom{R+G-(k-1)}{G-(k-1)}}{\binom{R+G}{G}}\frac{R}{R+G-(k-1)}\\
&= \frac{\binom{R+G-(k-1)}{R}}{\binom{R+G}{G}}\frac{R}{R+G-(k-1)}\\
&= \frac{\binom{R+G-k}{R-1}}{\binom{R+G}{G}}.
\end{align}
So the probability that Alice wins is
\begin{align}
\sum_{k=0}^{\lfloor G/2 \rfloor} P(R_1=2k+1)
&=\sum_{k=0}^{\lfloor G/2 \rfloor} \frac{\binom{R+G-(2k+1)}{R-1}}{\binom{R+G}{G}} \\
&=\frac{1}{\binom{R+G}{G}}\sum_{k=0}^{\lfloor G/2 \rfloor}\binom{R-1+G-2k}{R-1} \\
&=\frac{1}{\binom{R+G}{G}} \sum_{k=0}^G\frac{1+(-1)^k}{2}\binom{R-1+G-k}{R-1} \\
&=\frac{1}{2\binom{R+G}{G}}\left(
\sum_{k=0}^G\binom{R-1+G-k}{R-1}
+\sum_{k=0}^G(-1)^k\binom{R-1+G-k}{R-1}
\right)\\
&=\frac{1}{2\binom{R+G}{G}}\left(
\binom{R+G}{R}
+(-1)^G \sum_{k=0}^G\binom{-(G-k)-1}{R-1}
\right)\\
&=\frac{1}{2}+\frac{(-1)^G}{2\binom{R+G}{R}}
\sum_{k=0}^G\binom{-(G-k)-1}{R-1}
\\
\end{align}
|
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|
Proving that if $(x_n) \to x$ then $p(x_n) \to p(x)$ in Real Analysis I am self-learning Real Analysis from Stephen Abott's Understanding Analysis. Exercise 2.3.8 asks to prove that when a polynomial function $p$ is applied to values of a convergent sequence $(x_n)$, then if the input sequence converges to $x$, the resulting sequence converges to $p(x)$.
Questions.
(a) Do you guys think, my proof of part (a) is technically correct and rigorous?
(b) For part (b), should I just construct any function, such that the function value is different from the limiting value?
Let $(x_n) \to x$ and let $p(x)$ be a polynomial.
(a) Show $p(x_n) \to p(x)$.
(b) Find an example of a function $f(x)$ and a convergent sequence $(x_n) \to x$ where the sequence $f(x_n)$ converges, but not to $f(x)$.
Proof.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
(a) We are given that $(x_n) \to x$, so we can make the distance $\absval{x_n - x}$ as small as we like. Suppose $p(x) = x^{k}$ where $k$ is a non-negative integer.
Observe that,
\begin{align*}
\absval{p(x_n) - p(x)} &= \absval{x_n^k - x^k}\\
&= \absval{(x_n - x)(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}\\
\end{align*}
Also,
\begin{align*}
\tiny
\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})} &\le \tiny \absval{x_n^{k-1}} + \absval{x_n^{k-2}\cdot x} + \ldots + \absval{x_n \cdot x^{k-2}} + \absval{x^{k-1}}\\
\tiny\frac{1}{\absval{x_n^{k-1}} + \absval{x_n^{k-2}\cdot x} + \ldots + \absval{x_n \cdot x^{k-2}} + \absval{x^{k-1}}} &\le \tiny\frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}}
\end{align*}
Now, there exists $N_1$ such that $\absval{x_n - x} < \frac{\absval{x}}{2}$ for $n \ge N_1$. If the distance between $x_N$ and $x$ is smaller than $\frac{\absval{x}}{2}$, then clearly $\absval{x_n} > \frac{\absval{x}}{2}$.
Thus,
\begin{align*}
\frac{1}{k\cdot \frac{\absval{x^k}}{2^{k-1}}} &< \frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}}\\
\implies \frac{2^{k-1}}{k \cdot \absval{x^k}}&< \frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}}
\end{align*}
There exists $N_2$ such that $\absval{x_n - x} < \epsilon \cdot \frac{2^{k-1}}{k \cdot \absval{x^k}} $ for all $n \ge N_2$.
Let $N = \max \{N_1,N_2 \}$. To show that this $N$ indeed works, we prove that:
\begin{align*}
\absval{p(x_n) - p(x)} &= \absval{(x_n - x)(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}\\
&< \epsilon \cdot \frac{2^{k-1}}{k \cdot \absval{x^k}} \cdot \absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})} \\
&< \epsilon \cdot \frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}} \cdot \absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})} \\
&= \epsilon
\end{align*}
for all $n \ge N$. Thus, $p(x_n) \to p(x)$.
By algebra of limits, any linear combination of the powers of the sequence $(x_n)$ must be convergent and $\lim p(x_n) = p(x)$.
|
If you are using the algebra of limits in the end of your proof, you might as well use it in the beginning. Considering $p(x)=a_0 + a_1 x + \cdots + a_k x^k$, you have that
$$
\lim p(x_n) = \lim(a_0 + a_1 x_n + \cdots + a_k x_n^k)= a_0 + a_1 \lim x_n + \cdots + a_k (\lim x_n)^k = p(x).
$$
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.
I had the following idea, we write:
$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$
Let's pretend the identity we want to prove is true, then:
$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$
We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that
$$x= \frac{2n (n+1) (2 n+1)}{3} $$
We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?
Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.
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Let $S=1^2+3^2+5^2+\cdots+(2n+1)^2$$.
Next, consider the sums
$$S_1=2^2+4^2+6^2+\cdots+(2n)^2$$
$$S_2=1^2+2^2+3^2+\cdots+(2n+1)^2$$
Thus, we can calculate $S$ as $S=S_2-S_1$.
Let's evaluate $S_1$ and $S_2$ using the identity we are given.
Clearly,
$$S_2=\frac{1}{6}(2n+1)(2n+2)(4n+3)$$
and
$$S_1=2^2(1^2+2^2+\cdots+n^2)$$
$$S_1=2^2\cdot\frac{1}{6}n(n+1)(2n+1)$$
Therefore,
$$S=S_2-S_1$$
$$S=\frac{1}{6}\left[2(n+1)(2n+1)(4n+3)-4n(n+1)(2n+1)\right]$$
$$S=\frac{(n+1)(2n+1)}{6}\left[8n+6-4n\right]$$
$$S=\frac{(n+1)(2n+1)(2n+3)}{3}$$
as desired.
Hope this helps :)
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|
Simultaneous algebra question Let $x$, $y$ be real numbers such that
$$x+3y=4$$$$x^2+3y^2=8$$
Find $$x^3+3y^3$$
Also, let $$ax+by=v_1$$$$ax^2+by^2=v_2$$
Find the value of $ax^3+by^3$ in terms of $v_1$, $v_2$
I've tried $(x+3y)(x^2+9y^2−3xy)−6y^3$ and was left with $48+20y^2−6y^3$
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$$x+3y=4 \implies x=4-3y \implies (4-3y)^2+3y^2=8 \implies 3y^2-6y+2=0 \implies y=1\pm \frac{1}{\sqrt{3}}$$
We get $x=1-\sqrt{3}$ for $y=1+\frac{1}{\sqrt{3}}$, then $F=x^3+3y^3=(1-\sqrt{3})^3+3(1+\frac{1}{\sqrt{3}})^3=16-\frac{8}{\sqrt{3}}.$
or $x=1+\sqrt{3}$ for $y=1-\frac{1}{\sqrt{3}}$
Finally, $F=x^3+3y^3=(1+\sqrt{3})^3+3(1-\frac{1}{\sqrt{3}})^3=16+\frac{8}{\sqrt{3}}$
|
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Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried to subtract one equation from the other, but got nothing. No idea what to do.
|
There are some basic division tricks. $7*13*11=1001$ so to have $xyz138\equiv 0 \pmod 7$ implies that $xyz138 \equiv xyz138 - 1001*xyz = 138-xyz \equiv 0$ so $xyz \equiv 138\pmod 5\pmod 7$.
Likewise $138xyz \equiv 138xyz - 1001*138\equiv xyz -138\equiv 6 \pmod {13}$ so $xyz \equiv 144\equiv 1 \pmod {13}$.
Both together means $xyz \equiv 5+7k \equiv 1+13j\pmod{7\times 13}$ for some $k,j$. So $xyz \equiv 40\pmod{91}$
We can do the same trick for $11$ to get $x1y-3z8 \equiv 5\pmod {11}$. Or we can do the other trick that $x1y3z8\equiv 1+3+8 - (x+y+z)\equiv 5\pmod {11}$.
So we have $x+y+x \equiv 7\pmod 11$.
So $xyz = 40 + 91k$ for some $k$ where $x+y+z\equiv 7\pmod {11}$.
$40$ yield $4+0\equiv 4\pmod {11}$.
$131\implies 1+3+1\equiv 5\pmod {11}$.
$222\implies 2+2+2\equiv 6\pmod{11}$.
$313\implies 3+1+3\equiv 7\pmod {11}$.
So that'll do
$313138\equiv 0\pmod 7$
$138313 \equiv 6\pmod 13$
And $311338 \equiv 5\pmod{11}$
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How do you prove that $\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert + C}$? I would like to prove that
$$\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert} + C$$
I tried with applying derivative with respect to $x$ to $\ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert}$ and show it is equal to $\frac{1}{\sqrt{x^2+r^2}}$, but I ended up with $$ \frac{d}{dx} \ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert} = \dots = \frac{1}{x + \sqrt{x^2+r^2}}\left(1+\frac{1}{\sqrt{x^2+r^2}}\right)$$ and I do not see how it can be simplified to $\frac{1}{\sqrt{x^2+r^2}}$.
Any help would be appreciated (even involving complex calculus, if needed).
|
Your calculation of the derivative is incorrect.
Note $$\begin{align}
\frac{d}{dx}\left[ x + \sqrt{x^2 + r^2} \right]
&= 1 + \frac{1}{2}\left(x^2 + r^2 \right)^{-1/2} \cdot \frac{d}{dx}\left[x^2 + r^2\right] \\
&= 1 + \frac{1}{2 \sqrt{x^2 + r^2}} \cdot 2x \\
&= 1 + \frac{x}{\sqrt{x^2 + r^2}}. \end{align}$$ You are missing the $x$ in the numerator of the second term.
Then we obtain $$\begin{align}
\frac{d}{dx}\left[\log \left| x + \sqrt{x^2 + r^2} \right| \right]
&= \frac{1}{x + \sqrt{x^2 + r^2}} \left(1 + \frac{x}{\sqrt{x^2 + r^2}}\right) \\
&= \frac{1}{x + \sqrt{x^2 + r^2}} \left( \frac{\sqrt{x^2 + r^2} + x}{\sqrt{x^2 + r^2}} \right) \\
&= \frac{1}{\sqrt{x^2 + r^2}},
\end{align}$$ as claimed.
In order to perform the integration in the first place, there are a variety of substitutions that can be performed. One is via the tangent identity $$\tan^2 \theta + 1 = \sec^2 \theta,$$ hence the choice $$x = r \tan \theta, \quad dx = r \sec^2 \theta \, d\theta$$ yields
$$\int \frac{1}{\sqrt{x^2 + r^2}} \, dx = \int \frac{1}{\sqrt{r^2 (\tan^2 \theta + 1)}} \cdot r \sec^2 \theta \, d\theta = \int \frac{\sec^2 \theta}{\sec \theta} \, d\theta = \int \sec \theta \, d\theta.$$
Now we perform the algebraic manipulation
$$\sec \theta = \frac{\sec \theta (\sec \theta + \tan \theta)}{\sec \theta + \tan \theta} = \frac{\sec^2 \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta}.$$ Then the derivative of the denominator equals the numerator; i.e. $$\frac{d}{d\theta}[ \sec \theta + \tan \theta] = \sec \theta \tan \theta + \sec^2 \theta,$$ from which we may choose the substitution $u = \sec \theta + \tan \theta$ and obtain $$\begin{align}
\int \sec \theta \, d\theta &= \int \frac{1}{u} \, du \\ &= \log |u| + C \\
&= \log |\sec \theta + \tan \theta| + C \\
&= \log \left| \sqrt{(x/r)^2 + 1} + x/r \right| + C \\
&= \log \left| \frac{x + \sqrt{x^2 + r^2}}{r} \right| + C \\
&= \log \left| x + \sqrt{x^2 + r^2} \right| - \log |r| + C \\
&= \log \left| x + \sqrt{x^2 + r^2} \right| + C,
\end{align}$$
since $\log |r|$ is a constant with respect to $x$ and may be absorbed into $C$.
|
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Is $n=26$ the only integer satisfying $n-1$ being a square and $n+1$ being a cube? I guess it is possible to solve it by ring theory, but I would prefer an elementary method instead. I can only figure out that $n$ is an even integer, otherwise there will be contradiction w.r.t. modulo $4$. I have also tried to modulo several other integers(3, 5, 8, 9, 10) but haven't got any contradiction yet. Any discussion is welcomed!
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Lemma.
Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthermore, for any such choice of $r$ and $s$, there are coprime integers $t$ and $u$ such that $a=rt-2su$, $b=ru+st$, and $n=t^2+2u^2$ divides $bt-au$.
Proof.
Assume the theorem is false, and let $m$ be a minimal counterexample. Evidently $m > 1$ since the theorem is trivially true for $m=1$.
Note that $b$ is coprime to $m$. Let $A$ be an integer such that $Ab \equiv a\!\pmod{m}$, chosen so that $\tfrac{-m}{2} < A \le \tfrac{m}{2}$. Then $A^2+2 = lm$ for some positive integer $l < m$. Clearly $l$ cannot be a smaller counterexample than $m$, and so there exist coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$.
Let $t = \tfrac{ar+2bs}{m}$ and $u=\tfrac{br-as}{m}$. Direct calculation confirms the equations for $a$, $b$, and $n$. From $n=t^2+2u^2$, we deduce that $t$ is an integer because $u$ is an integer, and $t$ and $u$ are coprime because $\gcd(t,u)$ divides both $a$ and $b$. Finally, note that $n$ divides $bt-au=sn$.
Hence $m$ is not a counterexample, contradicting the original assumption. $\blacksquare$
Corollary.
Let $a$ and $b$ be coprime integers with $m$ an integer such that $m^3=a^2+2b^2$. Then there are coprime integers $r$ and $s$ such that $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$.
Proof.
Evidently $m$ is odd since $a^2+2b^2$ is at most singly even. And $a$ and $m$ must be coprime. Using the theorem, we have $m=r^2+2s^2$ and $m^2=t^2+2u^2$. Then $m$ divides $a(ur-ts)=t(br-as)-r(bt-au)$, and therefore $m \mid (ur-ts)$. The lemma can then be reapplied with $a$ and $b$ replaced by $t$ and $u$. Repeating the process, we eventually obtain integers $p$ and $q$ such that $p^2+2q^2=1$. The only solution is $q=0$ and $p=\pm1$. Ascending the path back to $a$ and $b$ (reversing signs along the way, if necessary) yields $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$, as claimed. $\blacksquare$
Theorem.
The Diophantine equation $X^3 = Y^2+2$ has only one integer solution, namely $(x,y) = (3, \pm 5)$.
Proof.
Evidently $y$ and $2$ are coprime. By the corollary, we must have $b=1=s(3r^2-2s^2)$ for integers $r$ and $s$. The only solutions are $(r,s)=(\pm 1,1)$. Hence $a=y=r(r^2-6s^2)=\pm 5$, so $(x,y)=(3,\pm 5)$. $\blacksquare$
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Let $x,y \in R$ such that $|x+y| + |x-y| = 2$
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.
How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.
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Hint:
If $f(x,y)=|x+y|+|x-y|-2,$
observe that $f(x,y)=f(-x,y)=f(x,-y)=f(-x,-y)$
WLOG $x+y, x-y\ge0;$
$$x=1, -x\le y\le x, y^2\le x^2=1$$
Now,
$$x^2+y^2+10y=1+(y+5)^2-25$$
$-1\le y\le1\iff5-1\le5+y\le5+1\implies4^2\le(y+5)^2\le6^2$
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Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$ Let f be the differentiable function satisfying the relation $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {1 + h} \right) + f\left( h \right) - {e^{ - 1}}}}{h}$ exist. The value of $\int\limits_0^1 {f\left( x \right)dx} = \_\_\_\_\_\_\_$.
My approach is as follow
$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
y=h
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - h}} - 1} \right) + 1}}{h}$
How will I proceed from here
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Let $g(x)=f(x)+e^{-x}$. Then given equation reduces to Cauchy's equation $g(x+y)=g(x)+g(y)$. If $f$ is assumed to be continuous (or at least measurable) then $g(x)=cx$ for some constant $c$ and $f(x)=-e^{-x}+cx$.
The given limit exists if and only if $c=1$ and this gives $\int_0^{1} f(x)dx=\frac 1 e-\frac 1 2$.
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Prove that $\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+2}\right)\cdot\ldots\cdot\left(1-\frac{1}{2n}\right)=\frac{1}{2}$ Prove the following equality:$$\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+2}\right)\cdot\ldots\cdot\left(1-\frac{1}{2n}\right)=\frac{1}{2}$$
$$\text{for all }\;n\in\mathbb{N}\;.$$
Could you tell me if my proof is correct?
Is there another way to prove it?
Does it exist a simpler proof?
We are going to prove by induction the following equality:
$\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n}\right)=\dfrac12\;.\quad\color{blue}{(*)}$
For $\;n=1\;$ we get that
$\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n}\right)=$
$=\left(1-\dfrac12\right)=\dfrac12$.
Now we suppose that $(*)$ is true and prove that
$\left(1-\dfrac{1}{n+2}\right)\left(1-\dfrac{1}{n+3} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)=\dfrac12\;.$
It results that
$\left(1-\dfrac{1}{n+2}\right)\left(1-\dfrac{1}{n+3} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)=$
$=\dfrac{\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)}{\left(1-\dfrac{1}{n+1}\right)}=$
$=\dfrac{\dfrac12\left(1-\dfrac{1}{2n+1}\right)\left(1-\dfrac{1}{2n+2}\right)}{1-\dfrac{1}{n+1}}=$
$=\dfrac{\dfrac12\left(\dfrac{2n}{2n+1}\right)\left(\dfrac{2n+1}{2n+2}\right)}{\dfrac{n}{n+1}}=$
$=\dfrac12\cdot\dfrac{\color{red}{2}\color{blue}{n}}{\color{green}{2n+1}}\cdot\dfrac{\color{green}{2n+1}}{\color{red}{2}(\color{brown}{n+1})}\cdot\dfrac{\color{brown}{n+1}}{\color{blue}{n}}=$
$=\dfrac12\;.$
Hence, by induction, the equality $(*)$ is true for all $n\in\mathbb{N}\;.$
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HINT
Notice that each denominator cancels the next numerator, whence we get
$\begin{align}
&\left(1 - \frac{1}{n+1}\right)\left(1 - \frac{1}{n+2}\right)\cdot\ldots\cdot\left(1 - \frac{1}{2n}\right) =\\
&=\left(\frac{n}{n+1}\right)\left(\frac{n+1}{n+2}\right)\cdot\ldots\cdot\left(\frac{2n-1}{2n}\right) = \frac{n}{2n}=\frac{1}{2}\;.
\end{align}$
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|
Pythagoras Theorem Proof Is my logic correct below to prove the Pythagoras Theorem? Thanks.
Area Rectangle R
\begin{align*}
R &= WL\\
&=(2a+b)(2b+a)\\
&=4ab+2a^2+2b^2+ab\\
&=5ab+2a^2+2b^2
\end{align*}
Total Area Yellow Triangles T
\begin{align*}
T &= 10(\frac{ab}{2})\\
&=5ab
\end{align*}
Calculate Area $c^2$
\begin{align*}
c^2 &= R-T-a^2-b^2 \\
&=5ab+2a^2+2b^2-5ab-a^2-b^2\\
&=a^2+b^2
\end{align*}
∎
Proof by rearrangement
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Yes. This is correct provided you have shown that all yellow triangles are congruent.
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Matrix exponential of a skew-symmetric Toeplitz matrix From today's exam:
Given the following matrix,
$$B = \begin{pmatrix} 0 & -2/3 & 1/3\\ 2/3 & 0 & -2/3\\ -1/3 & 2/3 & 0\end{pmatrix}$$ Prove that $$\exp(aB) = I + \sin(a) B + (1 -\cos(a)) B^2$$
I have tried expanding the exponential
$$\exp(aB)=I+aB+\frac{a^2}{2}B^2+\frac{a^3}{3!}B^3 + \cdots$$
But could not manage to find the $B^k$ expression.
Can anyone help with this?
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The characteristic polynomial of the given matrix is $x^3 +x$. Hence by Cayley-Hamilton theorem,
\begin{align*}
B^3+B&=0\\
B^3 &= -B\\
B^4 &= -B^2\\
B^5 &= B\\
\vdots\\
\exp(aB)&= I + aB+\frac{a^2B^2}{2!}+\frac{a^3B^3}{3!}+\cdots\\
\exp(aB) &= I + \left(a-\frac{a^3}{3!}+\cdots \right) B + \left( \frac{a^2}{2!}-\frac{a^4}{4!}+\cdots \right)B^2 &&\square\\
\end{align*}
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Showing that if $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$ I need to show that if $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$.
Here is what i have done:
$$P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0 \\ (X+i)^{2n+1}-(X-i)^{2n+1} = 0*2i = 0 \\ (X+i)^{2n+1} = (X-i)^{2n+1} \\ \frac{(X+i)^{2n+1}}{(X-i)^{2n+1}} = 1 \\ \left(\frac{X+i}{X-i}\right)^{2n+1} = 1$$
Is my reasoning correct ?
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Please see the discussion here where it is mentioned that $z=x+i$, $z^\ast=x-i$ and $z^{2n+1}={z^{\ast}}^{2n+1}$. Another way to write this is $$\left(\frac{x+i}{x-i}\right)^{2n+1} = 1.$$
|
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If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$ If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$
Is this true? This simple algebra should hold in order to finish my problem on Complex Analysis. Computing few numbers suggests that this is true, but I don't seem to be able to prove it.
|
Hint: If $x$, $y\le1$ then $$0\le(1-x)(1-y)=1-x-y+xy$$
|
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|
Find $a_0$ and $a_1$ when given a generating function. The recursive sequence $a_n$ has the following generating function
$$\begin{align}
&f(x)={\frac{x}{1-2x}+{\frac{4}{1+3x}}}
\end{align}$$
I am to find $a_0$ and $a_1$:
Rule: ${\frac{1}{1-a}}=1+a+a^2+a^3+...+a^n$
So we have
${\frac{1}{1-2x}}=1+2x+4x^2+8x^3+...+2^{n-1}x^{n-1}+2^nx^n$
${\frac{1}{1+3x}}=1-3x+9x^2-27x^3+...+(-3)^{n-1}x^{n-1}+(-3)^nx^n$
Thus,
$$\begin{align}
&f(x)={\frac{x}{1-2x}+{\frac{4}{1+3x}}}\\
&f(x)= x(1+2x+4x^2+8x^3+...) + 4(1-3x+9x^2-27x^3+...)\\
&f(x)= x+2x^2+4x^3+8x^4+... + 4-12x+36x^2-108x^3+...
\end{align}$$
For $a_0$ we have: $x+4=0+4=4$ , but when I try to calculate $a_1$ I get the wrong answer. The right answer should be $a_1=-11$, but I get
$$\begin{align}
&a_1=x+2x^2+4-12x=1+2+4-12=-5
\end{align}$$
What am I doing wrong? What is the correct way to find $a_1$?
|
\begin{align}f(x) &= x+2x^2 + 4x^3+ \dots + 4 - 12 x + 36 x^2 - 108x^3 + \dots \\&= 4-11x+38x^2-104x^3+\dots\\&=a_0+a_1x+a_2x^2+a_3x^3+\dots\end{align}
where the last line is precisely the definition of a generating function. There is no need to substitute any values of $x$.
Alternatively we have $a_n = \dfrac {f^{(n)}(0)}{n!}$, where $f^{(n)}(x)$ is the $n$th derivative. This might be useful for some other $f$.
|
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|
$a + b + c, \ ab + bc + ca, \ abc$ Prove a, b and c are real. I have found this problem in a book. If a, b and c are complex numbers, such that $|a| \neq |b| \neq |c| \neq |a|$ and $a + b + c, \ ab + bc + ca, \ abc$ are all real numbers, we must prove $a, \ b, \ c$ are also real numbers. We can get $(a + b + c) ^ 2 \in \ \mathbb {R}$, so $a ^ 2 + b ^ 2 + c ^ 2 \in \mathbb{R}$. Then, using $a ^ 3 + b ^ 3 + c ^ 3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab -bc-ac) = 3(a + b)(b + c)(a+c)$ I can get $a ^ 3 + b ^ 3 + c ^ 3 \in \mathbb{R}$ and $(a+b)(b+c)(a+c) \in \mathbb{R}$. But I cannot fit these to prove that a, b or is real.
|
By Viète's relations, $a,b,c$ are the roots of
$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
All the coefficients of this polynomial are real. If the polynomial has complex roots, there must be one such pair of them related by conjugation, but conjugate roots have the same magnitude, which is disallowed by the question. Thus the polynomial has only real roots, i.e. $a,b,c$ are real.
|
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|
Powers of complex numbers : Which is the general formula? Let $z=\frac{1}{2}\left (\sqrt{3}+i\right )\in \mathbb{C}$.
I want to calcualte the powers $z^n$ for $n\in \mathbb{Z}$.
I have done the following: \begin{align*}&z^1=\frac{1}{2}\left (\sqrt{3}+i\right ) \\ &z^2=\left (\frac{1}{2}\left (\sqrt{3}+i\right )\right )^2=\frac{1}{4}\left (3+2\sqrt{3}i-1\right )=\frac{1}{4}\left (2+2\sqrt{3}i\right )=\frac{1}{2}\left (1+\sqrt{3}i\right ) \\ &z^3=z^2\cdot z=\frac{1}{2}\left (1+\sqrt{3}i\right )\cdot \frac{1}{2}\left (\sqrt{3}+i\right )=\frac{1}{4}\left (\sqrt{3}+i+3i-\sqrt{3}\right )=\frac{1}{4}\cdot 4i=i \\ & z^4=z^3\cdot z=i\cdot \frac{1}{2}\left (1+\sqrt{3}i\right )=\frac{1}{2}\left (-\sqrt{3}+i\right ) \end{align*} Is therea specific formula that we get for an arbitrary $n$ ?
|
It is easiest to express the general formula using the polar representation of $z = r e^{i\theta}$
$$
z^n = r^n e^{in\theta} = r^n\left(\cos(n\theta) + i \sin(n\theta)\right).
$$
See also Euler formula and De Moivre's formula.
|
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|
A question of Number Theory+Binomial Theorem Let the sum of all divisors of the form $2^{p} \cdot 3^{q}$ (p,q are positive integers) of the number $19^{88} - 1$ be $\lambda$ . Find unit digit of $\lambda$
I noticed one fact that $2^{p} \cdot 3^{q}$ means any number which is not of form $6k\pm 1$.
Tried to expand it $(19^{44}+1)(19^{22}+1)(19^{11}+1)(19^{11}-1)$ then found that $19^{11}+1$ has one factor as $20$ but this approach didn't work.
Then I tried to expand it using Binomial Theorem but it turned out be very lengthy process.
|
We just need to find the largest factor of $19^{88}-1$ of the form $2^p3^q$. The rest shall follow from considering its factors.
Using the factorization $19^{88}-1 = (19^{44}+1)(19^{22}+1)(19^{11}+1)(19^{11}-1)$, we first check the value of $p$. Taking mod $4$:
$$19^{88}+1 \equiv (-1)^{88}+1 \equiv 2 \equiv 19^{44}-1 \pmod 4$$
$$19^{11}+1 \equiv (-1)^{11}+1 \equiv 0 \pmod 4$$
$$19^{11}-1 \equiv (-1)^{11}-1 \equiv 2 \pmod 4$$
Now we check mod $8$ for $19^{11}+1$: $19^{11}+1 \equiv 361^5 \times 19 + 1\equiv 1^5 \times 3 + 1 \equiv 4 \pmod 8$.
This shows that the greatest power of $2$ that divides $19^{88}$ is $2 \times 2 \times 2 \times 4 = 2^5$.
Now for $q$, notice that:
$$19^{k}+1 \equiv 1^k+1\equiv 2 \pmod 3$$
the factor of $3^q$ comes from the factor $19^{11}-1$. Now we use the factorization:
$$19^{11}-1 = (19-1)(19^{10}+19^9+19^8+\dots +19 + 1)$$
The latter factor $\equiv 11 \pmod 3$. Hence the greatest power of $3$ that divides $19^{11}-1$ is $9$ (from $19-1=18$).
Hence the largest factor of $19^{88}-1$ of the form $2^p3^q$ is precisely $2^53^2$.
|
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|
prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from there to apply the Euler's theorem
|
By Fermat's little theorem, we have
$$2^{18} \equiv 1 \pmod{19} \implies 2^{36} = (2^{4})^{9} \equiv 1 \pmod{19} \tag{1}\label{eq1A}$$
Since $2^6 = 64 \equiv 1 \pmod{9}$, we thus get $2^{6k} = (2^6)^k \equiv 1 \pmod{9}$, so there's an integer $j$ where $2^{6k} = 9j + 1$. Thus,
$$(2^4)^{2^{6k}} = (2^4)^{9j + 1} = ((2^4)^{9})^{j}(2^4) \equiv 2^4 \pmod{19} \tag{2}\label{eq2A}$$
|
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|
Limit of a cube root Given $\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1}-\alpha x-\beta)=0$ for some $\alpha, \beta \in \mathbb{R}$. I have tried equating it as $\lim_{x \to +\infty}\alpha x+\beta=\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1})$. Cubing both sides would give$$(\alpha x)^3+\beta^3+3\alpha^2x^2\beta+3\beta^2\alpha x=x^3+x^2+x+1$$
Hence,
$$\alpha^3=1; \beta^3=1;3\beta^2\alpha=1;3\alpha^2\beta=1$$
However, I am unsure if I am doing it right.
|
For large $x$, it is guaranteed that $\left|\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right| < 1$, and so we may use a Binomial expansion :
$$ \left(1 + \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right)\right)^\frac13 = 1 + \frac13 \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right) + O\left( \frac{1}{x^2} \right) = 1 + \frac{1}{3x} + O\left( \frac{1}{x^2} \right).$$
Therefore,
$$\sqrt[3]{x^3+x^2+x+1} = x \ \sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\alpha x-\beta$$
$$ = x\left(1 + \frac{1}{3x} + O\left( \frac{1}{x^2} \right)\right) -\alpha x -\beta = x(1 - \alpha) + (\frac{1}{3} - \beta) + O\left( \frac{1}{x} \right) ,$$
which tends to $0$ as $x \to \infty \iff \alpha = 1$ and $\beta = \frac13.$
$$$$
Addendum:
To answer your question for general $f(x), g(x)$ (assuming they both $\to \infty$ as $x \to \infty$): if $f(x)$ and $g(x)$ are very close for large values of $x$, such that $ f(x)-g(x) \to 0$, then it is true that, for all $ \varepsilon > 0,$ no matter how small, $\exists N$ such that $|f(x) - g(x)| < \varepsilon$ for all $x > N$.
However, this does not imply that $\left|f^3(x) - g^3(x)\right| < \varepsilon.$ In fact,
$$\left|f^3(x) - g^3(x)\right| = \left|f(x) - g(x)\right|\left(f^2(x) + f(x)g(x) + g^2(x)\right)$$
$$ = \left|f(x) - g(x))\right|\left((f(x) - g(x))^2 + 3f(x)g(x)\right) < \varepsilon\left(\varepsilon^2 + 3f(x)g(x)\right),$$
but $3f(x)g(x)$ is not necessarily bound by $\varepsilon.$ If, for example, $3f(x)g(x) > \frac{1}{\varepsilon},$ then there may be a significant/noticable difference between $f^3(x)$ and $g^3(x)$.
Using numbers as an example, consider $10^6$ and $10^6 + \frac{1}{10^3}.$ You might think that their cubes are very close, but in fact they differ by a whooping $ \approx 3 \times 10^9$ (!)
|
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|
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$
*$\left(1- \frac {x+1}{4x+3}\right)^x$
*$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$
*$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$
*$e^{\infty} = \infty$
answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!
|
Note that $$\lim_{u\to \infty}\left( 1-\frac{1}{u}\right)^{u}=\frac{1}{e}$$
|
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|
Solving a 2-D SDE Need to solve the system of SDE
\begin{align*}
dX^1&=-X^2dt+dW^1\\
dX^2&=X^1dt+dW^2
\end{align*}
I wrote the first eq. in integral form as $$X_t^1=X_0^1-\int_0^tX^2(s)ds+W_t^1$$
Then I plugged it into the second eq. and took the integral form of that which turned out to be
$$X_t^2=X_0^2+X_0^1t-\int_0^t\left(\int_0^tX^2(s)ds\right)ds+\int_0^tW^1(s)ds+W_t^2$$
Is this going in the right direction? If so, how much more can it be simplified?
|
Good question!
It may first be helpful to rewrite as
\begin{align*}
\begin{pmatrix} dX_{t}^{(1)} \\ dX_{t}^{(2)} \end{pmatrix}
=
\begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} X_{t}^{(1)} \\ X_{t}^{(2)} \end{pmatrix}\, dt
+ \begin{pmatrix} dW_{t}^{(1)} \\ dW_{t}^{(2)} \end{pmatrix}.
\end{align*}
This is similar to vector ODE: $\vec{X}'(t) = A\vec{X}(t) + \vec{b}(t)$.
*
*First diagonalize
\begin{align*}
A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} :&= PDP^{-1} = \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix}.
\end{align*}
*Next compute
\begin{align*}
e^{At} &= Pe^{Dt}P^{-1} \\
&= \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} \begin{pmatrix} e^{it} & 0 \\ 0 & e^{-it} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} \\
&= \begin{pmatrix} e^{it} & e^{-it} \\ ie^{it} & -ie^{-it} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} \\
&= \begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix}
\end{align*}
*Solve using d'Alembert's principle.
\begin{align*}
X_{t} &= e^{-At}\vec{X}_{0} + \int_{0}^{t} e^{-A(t-s)} \begin{pmatrix} dW_{s}^{(1)} \\ dW_{s}^{(2)} \end{pmatrix}
\end{align*}
This yields
\begin{align*}
&X_{t}^{(1)} = \cos(t)X_{0}^{(1)} - \sin(t)X_{0}^{(2)} + \int_{0}^{t} \cos(t-s)\, dW_{t}^{(1)} - \int_{0}^{t} \sin(t-s)\, dW_{t}^{(2)} \\
&X_{t}^{(2)} = \sin(t)X_{0}^{(1)} + \cos(t)X_{0}^{(2)} + \int_{0}^{t} \sin(t-s)\, dW_{t}^{(1)} + \int_{0}^{t} \cos(t-s)\, dW_{t}^{(2)}.
\end{align*}
You may wish to check, for example, that
\begin{align*}
dX_{t}^{(1)} &= -\sin(t)X_{0}^{(1)}\, dt - \cos(t)X_{0}^{(2)}\, dt + \left (\int_{0}^{t} \partial_{t}(\cos(t-s)) \, dW_{s}^{(1)}\right )\, dt \\
&\quad+ \cos(t-t)\, dW_{t}^{(1)} - \left (\int_{0}^{t} \partial_{t}(\sin(t-s)) \, dW_{s}^{(2)} \right )\, dt - \sin(t-t)\, dW_{t}^{(2)} \\
&= -X_{t}^{(2)}\, dt + dW_{t}^{(1)}
\end{align*}
as desired.
|
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How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$ The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$
My attempt
I managed to figure out that the denominator is given out as a perfect square of $$\{1+2\sin(x + \frac{\pi}{3})\}$$ and broke the $\sin(x+\frac{\pi}{3})$ so it looks like
$$\int \frac{\cos(x) + \sqrt 3}{(1+ \sin(x) +\sqrt 3 \cos(x))^2}{\rm d}x$$
I can't figure out how to approach further. Please guide me through this question.
|
Let $t=x-\frac\pi6$ to get
\begin{align}
& I= \int \frac{\cos x+ \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{d}x
=\int \frac{\frac{\sqrt3}2(2+\cos t)-\frac12\sin t }{(1+ 2 \cos t)^2}{d}t
\end{align}
Note that $\left( \frac{\sin t}{1+2\cos t}\right)’ = \frac{2+\cos t}{(1+2\cos t)^2}$ and $\left( \frac1{1+2\cos t}\right)’ = \frac{2\sin t}{(1+2\cos t)^2}$. Then
$$I = \frac{\sqrt3}2 \frac{\sin t}{1+2\cos t}- \frac14\frac{1}{1+2\cos t} = \frac{2\sqrt3\sin t-1}{4(2\cos t+1)}+C
$$
|
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|
Find the minimum value of the ratio $\frac{S_1}{S_2}$ using given data
3 points $O(0, 0) , P(a, a2 ) , Q(b, b2 )$ are on the parabola $y=x^2$. Let S1 be the area bounded by the line PQ and the parabola and let S2 be the area of the triangle OPQ, then find min of $\frac{S1}{S2}$
$$S_1 = \frac 12 (a+b)(a^2+b^2) -\int_{-b}^a x^2.dx$$
$$=\frac 16 (a^3+b^3) +\frac 12 (ab^2+a^2b)$$
And $$S_2=\frac 12 (a^2b+ab^2)$$
How do I solve from here?
|
In general, if you have a fraction involving $a,b$ where each term in the numerator and denominator has the same degree $d$, you can divide through by say $b^d$ to get an expression in $\frac{a}{b}$. Here, this would look like $$\frac{S_1}{S_2} =\frac{ \frac 16 (a^3+b^3) +\frac 12 (ab^2+a^2b)}{\frac 12 (a^2b+ab^2)} =\frac{ \frac 16 ((a/b)^3 +1) +\frac 12 (a/b+a^2/b^2)}{\frac 12 (a^2/b^2+a/b)} $$ Let $a/b =t$, then $$\frac{S_1}{S_2} = \frac{ t^3/6 +1/6 + t/2 +t^2/2}{t^2/2 +t/2} $$ And then you can follow the usual optimization procedure.
|
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Calculate $X_A(x) $ and $m_A(x) $ of a matrix $A\in \mathbb{C}^{n\times n}:a_{ij}=i\cdot j$
*
*Caclulate the characteristic & the minimal polynomial of the matrix:
$$A\in\mathbb C^{n\times n}:a_{ij}=i\cdot j ,\forall i,j=1,..,n$$
$$\text{i.e for $n=3$, } $$
$$A=\left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\\3 & 6 & 9\end{matrix}\right]$$
$$\text{I've calculate $X_A(x),m_A(x)$ for the following cases:}$$
$$ \left[\begin{matrix}1 & 2\\2 & 4\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\\3 & 6 & 9\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4\\2 & 4 & 6 & 8\\3 & 6 & 9 & 12\\4 & 8 & 12 & 16\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4 & 5\\2 & 4 & 6 & 8 & 10\\3 & 6 & 9 & 12 & 15\\4 & 8 & 12 & 16 & 20\\5 & 10 & 15 & 20 & 25\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4 & 5 & 6\\2 & 4 & 6 & 8 & 10 & 12\\3 & 6 & 9 & 12 & 15 & 18\\4 & 8 & 12 & 16 & 20 & 24\\5 & 10 & 15 & 20 & 25 & 30\\6 & 12 & 18 & 24 & 30 & 36\end{matrix}\right]$$
$$\xrightarrow{X_A(x)} \left [ x \left(x - 5\right), \quad - x^{2} \left(x - 14\right), \quad x^{3} \left(x - 30\right), \quad - x^{4} \left(x - 55\right), \quad x^{5} \left(x - 91\right)\right ]$$
$$\xrightarrow{m_A(x)} \left [ x \left(x - 5\right), \quad x \left(x - 14\right), \quad x \left(x - 30\right), \quad x \left(x - 55\right), \quad x \left(x - 91\right)\right ]$$
$$\text{Hence, i assumed that if }A\in\mathbb{C}^{n\times n}:$$
$$\fbox{$X_A(x)=(-1)^nx^{n-1}\big( x-\operatorname{tr}(A)\big)$},$$
$$\fbox{$m_A(x)=x\big( x-\operatorname{tr}(A)\big)$} .$$
How can i prove this in general case?
any suggestions?
|
Let $\mathbf{v}:=(1,2,\ldots,n)$ and $\mathbf{w}_i:=(i,0,\ldots,0,-1,0,\ldots)$ for $i=2,\ldots,n$. These vectors are non-zero orthogonal and hence form a basis of $\mathbb{R}^n$. They are also eigenvectors of $A$ since $$A\mathbf{v}=|\mathbf{v}|^2\mathbf{v},\qquad A\mathbf{w}_i=\mathbf{0}$$ Note that $|\mathbf{v}|^2=\mathrm{trace}(A)$.
Hence $A$ has eigenvalues $|\mathbf{v}|^2$ and $0$ repeated $(n-1)$ times. The characteristic polynomial is thus $$X_A(x)=(-1)^n(x-\lambda_1)\cdots(x-\lambda_n)=(-1)^nx^{n-1}(x-|\mathbf{v}|^2)$$
The minimal polynomial can be found by multiplying $A$ with itself: $$A^2=\begin{bmatrix}\mathbf{v}\\ 2\mathbf{v}\\ \vdots\\n\mathbf{v}\end{bmatrix}\begin{bmatrix}\mathbf{v}&2\mathbf{v}&\cdots&n\mathbf{v}\end{bmatrix}=|\mathbf{v}|^2A$$ $$\therefore\ m_A(x)=x^2-|\mathbf{v}|^2x=x(x-|\mathbf{v}|^2)$$ $m_A$ is the minimal polynomial since it is the least degree polynomial containing both eigenvalues.
These results for $X_A$ and $m_A$ generalize to any matrix of the type $A={\mathbf{v}\mathbf{v}}^\top$.
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Difference of equal derivative functions value $f_1(x) = \text{tan}^{-1}(x)$
$f_2(x) = \text{tan}^{-1}\big(\frac{x-1}{x+1}\big)$
Where
$f_1^\prime(x) = f_2^\prime(x)=\frac{1}{x^2+1}$
Therefore
$f_1(x) - f_2(x) = \theta$
How to find values of $\theta$ mathematically. I have solved it graphically and the answers are $\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ in my opinion. Thanks in advance.
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Hint:
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
$$\arctan x+\arctan(-1)=\begin{cases} \arctan\frac{x-1}{1-(-1)x} &\mbox{if } x\cdot(-1)<1\\ \pi+\arctan\frac{x-1}{1-(-1)x} & \mbox{if } x\cdot(-1)>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } x\cdot(-1)=1\end{cases} $$
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Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC As title,
I could only figure out that
$$a^2 + r^2 = 9$$
$$b^2 + r^2 = 16$$
$$c^2 + r^2 = 25$$
$$r^2 = \frac{abc}{a+b+c}$$
But couldn't get how to derive $2(a+b+c)$.
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In a triangle $ABC$ with sides $a,b,c$ opposite to $A,B,C$, respectively, and with incenter $I$, one has $$AI^2 = bc\cdot \frac{b+c-a}{a+b+c}, \quad BI^2 = ca\cdot \frac{c+a-b}{a+b+c}, \quad CI^2 = ab\cdot \frac{a+b-c}{a+b+c}.$$
The conditions $AI=3$, $BI=4$ and $CI=5$ can be rewritten as
$$\lambda := \frac{2abc}{a+b+c} = bc-9 = ca-16 = ab - 25.$$
One gets $$a^2=\frac{ab\cdot ac}{bc} = \frac{(\lambda+16)(\lambda+25)}{\lambda+9}$$ and similarly $$b^2=\frac{(\lambda+9)(\lambda+25)}{\lambda+16}, \quad c^2=\frac{(\lambda+9)(\lambda+16)}{\lambda+25}.$$
Substituting this into $\lambda = \dfrac{2abc}{a+b+c}$ one obtains
$$\lambda = \frac{2(\lambda+9)(\lambda+16)(\lambda+25)}{(\lambda+9)(\lambda+16)+(\lambda+9)(\lambda+25)+(\lambda+16)(\lambda+25)}$$
which, according to wolframalpha, is equivalent to $$\lambda^3-769\lambda-7200=0,$$
hence the answer is
$$\sqrt{\frac{(\lambda+16)(\lambda+25)}{\lambda+9}} + \sqrt{\frac{(\lambda+9)(\lambda+25)}{\lambda+16}} + \sqrt{\frac{(\lambda+9)(\lambda+16)}{\lambda+25}}$$ where $$\lambda = \frac{\sqrt[3]{32400 + i \sqrt{314509827}}}{3^{2/3}} + \frac{769}{\sqrt[3]{3(32400 + i \sqrt{314509827})}} \approx 31.5756803606316$$
is the positive root of $\lambda^3-769\lambda-7200=0$.
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Is $x=2,y=13$ the unique solution? Problem:
Find all positive integers $x$ and $y$ satisfying:
$$12x^4-6x^2+1=y^2.$$
If $x=1, 12x^4-6x^2+1=12-6+1=7,$ which is not a perfect square.
If $x=2, 12x^4-6x^2+1=192-24+1=169=13^2$, which is a perfect square. Thus, $x=2,y=13$ is a solution to the given Diophantine equation.
However, after testing a few more small cases, it seems as if $12x^4-6x^2+1$ can never be a perfect square if $x>2$. I have tried to prove this, but to no avail. Here is the gist of what I considered:
$12x^4-6x^2+1=y^2 \iff 12x^4-6x^2 = (y+1)(y-1)$. Since the L.H.S. is a multiple of $2$, it follows that $2 \mid (y+1)(y-1) \Rightarrow y$ is odd $\Rightarrow y+1$ and $y-1$ are both even.
Hence, $4 \mid (y+1)(y-1) \Rightarrow 4 \mid 12x^4-6x^2 \Rightarrow 4 \mid 6x^2 \Rightarrow x $ is even. Let $x=2m$ and $y+1=2k$, so $12x^4-6x^2=192m^4-24m^2=4k(k-1) \Rightarrow 48m^4-6m^2=k(k-1)$. At this point, I am at a loss of how to continue. We could continue with divisibility arguments, but it seems to be a never-ending process?
Another method I tried was to let $y=x+k$, thus $12x^4-6x^2+1 = x^2 + 2xk+k^2 \iff k^2+2xk-(12x^4-7x^2+1)=0$, which is a quadratic in terms of $k$. However, stuff like the discriminant or sum and product of roots did not seem to yield any important information.
Any hints provided to point me in the right direction will be much appreciated.
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We try to get a perfect square term on the LHS.
Note that multiplying both sides by 4 keeps the right side as a square:
$$48x^4 - 24x^2 + 4 = (2y)^2$$
Now, completing the square yields: $$3(4x^2 -1)^2 +1 = (2y)^2$$
Try to now find the solutions to $3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$
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Equation using $f(x)=\frac{\sin\pi x}{x^2} $ If $f(x)= \frac{\sin \pi x}{x^2}$, $x>0$
Let $x_1<x_2<x_3<\cdots<x_n<\cdots$ be all the points of local maximum of $f$.
Let $y_1<y_2<y_3<\cdots<y_n<\cdots$ be all the points of local minimum of $f$.
Then which of the options is(are) correct.
(A) $x_1<y_1$
(B) $x_{n+1}-x_n>2$ for all $n$
(C) $x_n \in (2n,2n+\frac{1}{2})$ for every $n$
(D) $|x_n-y_n|>1$ for every $n$
The official answer is $B,C,D$
My approach is as follow
$$y' = \frac{x^2 \times \pi \times \cos \pi x - 2x\sin \pi x}{x^4}$$
$$y' = \frac{x^2 \times \pi \times \cos \pi x - 2x\sin \pi x}{x^4} = \frac{x \times \pi \times \cos \pi x - 2\sin \pi x}{x^3} = \frac{2\cos \pi x \left( \frac{\pi x}{2} - \tan \pi x \right)}{x^3}$$
We should consider $x\in\frac{1}{2},\frac{3}{2},\ldots.$ as inflection point even though $\tan\pi x$ is not defined because we need to consider the numerator part $\pi x\cos \pi x - 2\sin \pi x$
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Only the option $A)$ is correct.
$$x_1=\frac 12\;\;,\; y_1=\frac 32$$
Let $$f(x)=\frac{\sin(\pi x)}{x^2}$$
it is clear that
$$(\forall x>0)\;\; -\frac{1}{x^2}\le f(x)\le \frac{1}{x^2}$$
The local maximums are such that
$$f(x_i)=\frac{1}{x_i^2}$$
or
$$\sin(\pi x_i)=1$$
thus
$$x_i=\frac 12+2n$$
with $n\ge 0$.
So, $$x_i\in\left\{\frac 12,\frac 52,\frac 92,\ldots\right\}$$
by the same,
$$y_i\in\left\{\frac 32,\frac 72,\frac{11}{2},\ldots\right\}$$
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Evaluate and explain $\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{\left(1+2\cos x\right)}\right)dx$ $$\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{1+2\cos x}\right)dx$$
I found this integral on a math discord server. I was unable to apply any standard integration technique to solve it (I am just an advanced high school student so my experience is limited). What methods can I apply to solve such an integral?
The expression does not seem to have an analytical solution.
graph of the function
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As @K.defaoite notes, this is Coxeter's integral, equal to $5\pi^2/24$. The usual way (see Inside Interesting Integrals Secs. 6.2 and 6.3, where this takes Nahin just over 11 pages) to evaluate it (whose details I've half-provided as they're worth working through to appreciate how hard this is!) is to prove it's four times Ahmed's integral $A(1)$ where $A(u):=\int_0^1\tfrac{\arctan\left(u\sqrt{2+x^2}\right)}{(1+x^2)\sqrt{2+x^2}}dx$, so$$A(\infty)=\left[\tfrac{\pi}{2}\arctan\tfrac{x}{\sqrt{2+x^2}}\right]_0^1=\tfrac{\pi^2}{12}$$and$$A^\prime=\tfrac{1}{1+u^2}\left[\arctan x-\tfrac{u}{\sqrt{1+2u^2}}\arctan\tfrac{xu}{\sqrt{1+2u^2}}\right]_0^1$$so$$A(\infty)-A(1)=\tfrac{\pi^2}{16}-\int_1^\infty\tfrac{u}{(1+u^2)\sqrt{1+2u^2}}\arctan\tfrac{u}{\sqrt{1+2u^2}}du.$$With $u\mapsto1/u$, this simplifies to$$A(\infty)-A(1)=\tfrac{\pi^2}{16}-A(\infty)+A(1)\implies A(1)=\tfrac{5\pi^2}{96}.$$Now call your integral $C$ so, with some trigonometric identities,$$C=\int_0^{\pi/2}2\arctan\sqrt{\tfrac{1+\cos x}{1+3\cos x}}dx\stackrel{y=\tfrac{x}{2}}{=}4\int_0^{\pi/4}\arctan\tfrac{\cos y}{\sqrt{2-3\sin^2y}}dy.$$Since $\arctan b=\int_0^1\tfrac{bdt}{1+b^2t^2}$,$$C=\int_0^{\pi/4}\int_0^1\tfrac{4\cos y\sqrt{2-3\sin^2y}}{t^2+2-(t^2+3)\sin^2y}dtdy.$$With $\sin y=\sqrt{\tfrac23}\sin w$ followed by $s=\tan w$ (who'd guess that?),$$C=\int_0^{\sqrt{3}}\int_0^1\tfrac{8\sqrt{3}dtds}{(1+s^2)(t^2s^2+3t^2+6)},$$which simplifies by partial fractions to$$C=\tfrac{2\pi^2}{9}-4\int_0^1\tfrac{t\arctan\tfrac{t}{\sqrt{t^2+2}}}{(t^2+3)\sqrt{t^2+2}}dt.$$With $u=\arctan\tfrac{t}{\sqrt{t^2+2}}$, this becomes $C=4A(1)$, as desired.
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$a,b,c > 0$ and $a+b+c=1$. Prove that $\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)} $ $a,b,c > 0$ and $a+b+c=1$. Prove that
$$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)} $$
This inequality can be solved by computer ( essentially, remove the constrain through homogenizing, clear denominator, expanding and proving positive terms). Since this mechanical strategy is very simple to do but impractical during math contest, I am looking forward to solution using classical inequalities approach.
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I would not call this an answer, but perhaps could be helpful. Observe that:
\begin{align*}
(1-a)(1-b)(1-c)&=\color{red}{1-a-b-c}+ab+ac+bc-abc\\
&=ab+ac+bc-abc \\
&=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}-abc \\
&=\frac{1-(a^2+b^2+c^2)}{2}-abc \\
&=1-abc-\frac{1+a^2+b^2+c^2}{2} \\
&=1-abc-\frac{a+b+c+a^2+b^2+c^2}{2}.
\end{align*}
Thus, $$a+b+c+a^2+b^2+c^2=2[(1-abc)-(1-a)(1-b)(1-c)].$$
By Cauchy Schwartz inequality, $$(a+b+c+a^2+b^2+c^2)\left(\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\right)\geq3^2=9.$$
As a result, $$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geq\frac{9}{2[(1-abc)-(1-a)(1-b)(1-c)]}.$$
If we can show $$\frac{9}{2[(1-abc)-(1-a)(1-b)(1-c)]}\geq\frac{13}{2(1-abc)},$$ then we are done. Suppose this is true. It gives $$13[(1-abc)-(1-a)(1-b)(1-c)]\leq 9(1-abc),$$ or equivalently, $$(1-abc)\leq\frac{13}{4}(1-a)(1-b)(1-c).$$
If this inequality is true under the constraint $a+b+c=1$ and $a,b,c>0$, then we are done. I checked that the equality precisely holds when $a=b=c=1/3$. Hope anyone can finish it.
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Indefinite integral $\displaystyle \int \frac 1 {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\mathrm{d}x$ Working on $\displaystyle \int \frac 1 {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\mathrm{d}x$.
According to Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968) item $14.422$ it should work out as:
$$\frac {-1} {a \sqrt {p^2 + q^2} } \tan \left({\frac \pi 4 - \frac {ax + \arctan (q/p)} 2}\right)$$
First thing I do is a Weierstrass substitution: $u = \tan \dfrac {a x} 2$ which leads (after algebra) to the expression:
$$\dfrac 2 {a (\sqrt {p^2 + q^2} - q) } \int \frac {\mathrm d u} {\left({u + \dfrac p {\sqrt {p^2 + q^2} - q} }\right)^2}$$
This is a standard integral, giving (again after algebra):
$$\frac {-2} {(\sqrt {p^2 + q^2} - q) u + p}$$
Replacing $u = \tan \dfrac {a x} 2$ gives us:
$$\frac {-2} {(\sqrt {p^2 + q^2} - q) \tan \dfrac {a x} 2 + p}$$
which is worlds away from what the book gives.
I can't reconcile the two expressions. I'm fairly sure of the exactness of the square on the bottom, because it's the result of a quadratic in $u$ with a discriminant of zero. Fiddly and messy, but ultimately tractable.
I have tried deriving the expression in the book w.r.t. $x$ but all I get is a complicated and daunting squared-secant expression which may work out in the end to what I started with, but eugh.
How on earth does one arrive at Spiegel's result?
(Sidenote: It's a special case of this integral:
$$\displaystyle \int \frac {\mathrm d x} {p \sin a x + q \cos a x + r} = \begin{cases}
\dfrac 2 {a \sqrt {r^2 - p^2 - q^2} } \arctan \left({\dfrac {p + (r - q) \tan \dfrac {a x} 2} {\sqrt {r^2 - p^2 - q^2} } }\right) + C & : p^2 + q^2 < r^2 \\
\dfrac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \left| {\dfrac {p - \sqrt {p^2 + q^2 - r^2} + (r - q) \tan \dfrac {a x} 2} {p + \sqrt {p^2 + q^2 - r^2} + (r - q) \tan \dfrac {a x} 2} }\right| + C & : p^2 + q^2 > r^2
\end{cases}$$
but where the discriminant equals zero and so cannot be used.)
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Integrate as follows
\begin{align}
& \int \frac {dx} {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\\
=&\frac1{\sqrt {p^2 + q^2} } \int \frac {dx} {\cos(\tan^{-1}\frac pq -ax)+1 }\\
=&\frac1{\sqrt {p^2 + q^2} } \int \frac {dx} {\cos(\frac\pi2-ax -\tan^{-1}\frac qp )+1 }\\
=&\frac1{2\sqrt {p^2 + q^2} } \int \sec^2\left(\frac\pi4-\frac {ax+\tan^{-1}\frac qp}2\right)dx\\
=&-\frac1{a\sqrt {p^2 + q^2} } \tan\left(\frac\pi4- \frac {ax+\tan^{-1}\frac qp}2\right)+C\\
\end{align}
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How to calculate the second derivative of this example? I am trying to calculate the second derivative of the following equation:
$$f(x)=\frac{6}{(x^2+3)}$$
$$f'(x)=\frac{-12x}{(x^2+3)^2}$$
The correct answer given was $$f''(x) = \frac{36(x+1)(x-1)}{(x^2+3)^3}$$ but the second derivative that I have gotten was different from the answer.
I have shown the calculations below and would like to understand where is the mistake. Would anyone be able to point out my mistake?
$$f'(x) =\frac{-12x}{(x^2+3)^2} = (-12x)(x^2+3)^{-2}$$
$$= (-12x)\frac{d}{dx} [(x^2+3)^{-2}]+(x^2+3)^{-2}\frac{d}{dx}[-12x]$$
$$= 48x^2(x^2+3)^{-3} -12(x^2+3)^{-2}$$
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Keep going...
$$48x^2(x^2+3)^{-3}-12(x^2+3)^{-2}$$
$$=\frac{48x^2}{(x^2+3)^3}-\frac{12}{(x^2+3)^2}$$
$$=\frac{48x^2-12(x^2+3)}{(x^2+3)^3}$$
$$=\frac{36x^2-36}{(x^2+3)^3}$$
Factorise the numerator to match the given answer.
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$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$ Let $a,b,c>0$ and $a^2+b^2+c^2=3$, prove
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$$
This inequality looks simple but I do not know how to solve it. The straightforward method is to bash the inequality with brutal computation method and assume $x = \min \{ x,y,z \}$, $y = x+a$, $z = x + b$, etc. I wonder if we can prove it using classical inequality.
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If you are unable to solve this equation by other means, this may help
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\ge 3
\implies(a,b,c)\\= (-3,3,3), (0,1,1), (1,0,1),(1,1,0), (1,1,1), (3,-3,3), (3,3,-3) $$
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}= 4\implies
(a,b,c)\\
=(-5,3,7), (-4,4,4), (-3,9,5),(-2,2,0), (-2,4,2), (0,-2,2), (2,-2,4), (2,0,-1) $$
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}= 5\implies
(a,b,c)\\
= (-5,5,5), (5,-5,5), (5,5,5) $$
You can continue this for $6,7,8,\cdots$ and get different solutions for each. This may not be the "proper" way to solve an inequality but it may help in revealing "patterns".
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|
$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?
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The Carmichael function value for $100$, $\lambda(100)=\text{lcm}(\lambda(25),\lambda(4)) = \text{lcm}(20,2) = 20$ gives us $1234 \equiv 14 \bmod \lambda(100) \implies 3^{1234} \equiv 3^{14} \bmod 100$.
Probably the easiest way to calculate $3^{14} \bmod 100$ is as $81^3 \cdot 9$ discarding higher digits, so $81^2 \equiv 61$ and $81\cdot 61 \equiv 41$ leading to $3^{1234} \equiv 3^{14} \equiv 69 \bmod100$.
Starting alternatively from your observation that $3^{15} \equiv 7 \bmod 100$, we could find $3^{14} \equiv 7\cdot 3^{-1} \equiv 7\cdot 67 \equiv 469 \equiv 69 \bmod 100$.
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difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 & 4 \\ \hline
1 & 6 & 6 & 0 \\ \hline
2 & 26 & 30 & -4 \\ \hline
3 & 94 & 102 & -8 \\ \hline
4 & 306 & 318 & -12 \\ \hline
\end{array}
We can deduce that $f(n) - g(n) = 4 - 4n$
But now we have to prove it.
Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$
We can simply check this by induction but I will skip it, so the question won't be so long.
Now lets put it together:
$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$
As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.
How to prove that:
$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$
All I could do is this:
\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}
And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$
What to do next? Did I go the wrong direction anywhere?
Thank you for your fast responses.
|
I think you can just use an induction proof, since you already have an intuition of the result, it is much easier to check :
$$\text{Let : } H_n :"w_n=f(n)-g(n)=4-4n" $$
First, for $n=0$ :
$$w_0=4=4-4\times 0 $$
Hence, $H_0$ is true.
Let $n\in\mathbb{N} $, such that $H_n$ is true, let us show $H_{n+1}$
$$w_{n+1}=3w_n+8n-12=3(4-4n)+8n-12\\
= 12-12n+8n-12=-4n=4-4(n+1) $$
Hence, $H_{n+1} $ is true, so we can conclude that :
$$\forall n \in \mathbb{N}, f(n)-g(n)=4-4n $$
|
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|
Finding the derivative of $ \cos(\arcsin x)$ I study maths as a hobby. I am trying to find the derivative of $ \cos(\arcsin x)$
This is how I have been proceeding:
Let u = $\arcsin x$
Then $\sin u = x$
Differentiating:
\begin{align}
\cos u \frac{du}{dx} &= 1 \implies \frac{du}{dx} = \frac{1}{\cos u} \\[4pt]
\cos^2 u + \sin^2 u &= 1 \\[4pt]
\cos u &= \sqrt {1 - \sin^2 u} = \sqrt {1 - x^2}
\end{align}
But that is as far a I get.
The text book says the answer is $\frac{x}{\sqrt {1 - x^2}}$ but I cannot see how this is arrived at.
|
With your notation, $x = \sin u$, we have $$\frac{d}{dx}\left[\cos (\sin^{-1} x)\right] = -\sin (\sin^{-1} x) \cdot \frac{d}{dx} \left[ \sin^{-1} x \right] = -x \frac{d}{dx} \left[ \sin^{-1} x \right].$$ Since $$\frac{dx}{du} = \cos u,$$ we have $$\frac{du}{dx} = \frac{1}{\cos u} = \frac{1}{\sqrt{1 - \sin^2 u}} = \frac{1}{\sqrt{1-x^2}}.$$ Therefore, $$\frac{d}{dx}\left[\cos(\sin^{-1} x)\right] = - \frac{x}{\sqrt{1-x^2}}.$$
|
{
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"url": "https://math.stackexchange.com/questions/3973505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the volume of the set in $\mathbb{R}^3$ defined by the equation $x^2 + y^2 \leq z \leq 1 + x + y$. Let $D=\left\{\left(x,y,z\right) \in \mathbb{R}^3: x^2 + y^2 \leq z \leq 1 + x + y \right\}$. I tried to integrate with cartesian coordinates, so that
$$\iiint_Ddxdydz = \int_{x=\frac{1-\sqrt{5}}{2}}^{{\frac{1+\sqrt{5}}{2}}} \int_{z=x^2}^{1+x} \left( \int_{y=z-(1+x)}^{\sqrt{z-x^2}}dy\right)dxdz$$
but the calculation seems to be quite laborious. Is there a way to solve it easily, maybe with another coordinate system?
Edit: After your suggestions I finally solved the problem. If someone is interested in the calculation itself I'll report the main steps below:
$$\int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{\frac{3}{2}}}\left(\int_{z=r^2+r(\sin\theta+\cos\theta)+\frac{1}{2}}^{r(\sin\theta+\cos\theta)+2}dz\right)rdrd\theta = \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{\frac{3}{2}}} \left( \frac{3}{2}r-r^3\right)drd\theta = \pi\int_{0}^{\frac{3}{2}}\left(\frac{3}{2}-t\right)dt = \frac{\pi}{2} \left[3t-t^2\right]_{0}^{\frac{3}{2}}=\frac{9\pi}{8}$$
|
Define $$x=r\cos(\theta)+\frac{1}{2}$$ $$y=r\sin(\theta)+\frac{1}{2}$$ $$z=z$$ The volume of your solid is the triple integral $$\int_0^{2\pi} \int_0^{\sqrt{3/2}} \int_{r^2+r\big(\sin(\theta)+\cos(\theta)\big)+\frac{1}{2}}^{r\big(\sin(\theta)+\cos(\theta)\big)+2}rdzdrd\theta$$
|
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|
Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root? this might sound like a stupid question,
bear with me it probably is.
I know the solutions for $x=x^6$ are 1 and 0.
Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot x$ to get $ 1=x^5 $, how can I find $x=0$?
Thank you so much in advance.
|
Don't divide. Factorise:
$x=x^6$
$x-x^6=0$
$x(1-x^5)=0$
Either $x=0$ or $1-x^5=0 \Rightarrow x=1$
|
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|
Want to verify my approach
If repetition of digit is not allowed, then how many five digit numbers which are divisible by $3$ can be formed
using the digits?
My approach is that i made 3 sets of digits according to the remainder they give on dividing by $3$ -
$ A:(0,3,6,9)$ , $B:(1,4,7)$ , $C:(2,5,8)$, Now the following cases are possible:
*
*Any $1$ digit from $A$ $+$ Any $2$ digit from $B$ $+$ Any $2$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 2} \cdot {3 \choose 2}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 1} \cdot {3 \choose 2} \cdot {3 \choose 2}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$
*Any $2$ digit from $A$ $+$ Any $3$ digit from $B$ $+$ Any $0$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1\cdot {3 \choose 1} \cdot {3 \choose 3} \cdot {3 \choose 0}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 2} \cdot {3 \choose 3}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$
*Any $2$ digit from $A$ $+$ Any $0$ digit from $B$ $+$ Any $3$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 1}\cdot {3 \choose 0} \cdot {3 \choose 3}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 2} \cdot {3 \choose 3}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$
*Any $3$ digit from $A$ $+$ Any $1$ digit from $B$ $+$ Any $1$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 2}\cdot {3 \choose 1} \cdot {3 \choose 1}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 3} \cdot {3 \choose 1}\cdot {3 \choose 1}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$.
Total $= 9072$
pls verify my approach and answer, other methods will also be appreciated ,Thanks.
|
The count can be simplified using probability as under:
Without zeroes, there will be $\binom9 5 = 126$ digit strings, and the same number with zeroes, at$\,\binom9 4 = 126$
$\frac 1 3$ of each, i.e. $42$ will be divisible by $3$, but since zeroes can only be allowed at $4$ of the $5$ digits, the multiplication factor while combining them won't be $2$ but $1.8$
Thus answer $= 42\times1.8\times5! = 9072\,$, which confirms your answer
|
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|
Transformation to complex spherical basis. According to this Wikipedia page about the complex spherical basis, it is stated that the spherical basis vectors are written in terms of the Cartesian basis vectors the following way:
$$
\left(
\begin{array}{c}
e_{+} \\
e_{-} \\
e_{0}
\end{array}
\right)
=
\left(
\begin{array}{ccc}
-\frac{1}{\sqrt{2}} & -\frac{i}{\sqrt{2}} & 0 \\
+\frac{1}{\sqrt{2}} & -\frac{i}{\sqrt{2}} & 0 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{c}
e_{x} \\
e_{y} \\
e_{z}
\end{array}
\right)
=
\textbf{U}
\left(
\begin{array}{c}
e_{x} \\
e_{y} \\
e_{z}
\end{array}
\right)
$$
However, the "components" of a vector transform by:
$$
\left(
\begin{array}{c}
A_{+} \\
A_{-} \\
A_{0}
\end{array}
\right)
=
\left(
\begin{array}{ccc}
-\frac{1}{\sqrt{2}} & +\frac{i}{\sqrt{2}} & 0 \\
+\frac{1}{\sqrt{2}} & +\frac{i}{\sqrt{2}} & 0 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{c}
A_{x} \\
A_{y} \\
A_{z}
\end{array}
\right)
=
\textbf{U*}
\left(
\begin{array}{c}
A_{x} \\
A_{y} \\
A_{z}
\end{array}
\right)
$$
I understand that this is what you get by substituting the components of the vectors in the basis definition but, is there a better way to explain why the transformation matrix is conjugated when I go from transforming the basis vectors to transforming the "components" of a vector?
|
Let $B_c$ denote the Cartesian basis as a column vector. Let $B_s$ be the spherical basis. Then we have $B_s=U B_c.$ If $v$ is a vector, let $v_c$ and $v_s$ be the coordinate representation in each basis. Then
$$
{B_c}^T v_c={B_s}^T v_s=(UB_c)^T v_s={B_c}^T U^T v_s.
$$
But because $B_c$ is a basis this gives $v_c=U^T v_s$ or $v_s=(U^T)^{-1} v_c.$ But because $U$ is unitary,
$$
(U^T)^{-1}=(U^{-1})^T=((U^*)^T)^T=U^*.
$$
|
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|
Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with the following values for the constants
$$
\left(x-\frac{1}{2}\right)^2
\left(x+\frac{5}{2}\right)^2+\frac{5}{2} \left(x+\frac{1}{5}\right)^2+\frac{107}{80}
$$
I suppose there are simpler ways to prove that the polynomial is positive, perhaps by using some inequalities.
Please, advice.
|
Here's a slightly more motivated way to find a sum-of-squares representation.
We should want the first term to be the square of a quadratic with leading coefficient $1$, and to cancel out the $x^3$ term we need it to be $(x^2+2x+a)^2$ for some real $a$. This will leave us to show that
$$-2ax^2-(4a+4)x+(3-a^2)\geq 0;$$
the simplest case of this is to choose $a=-1$ (to remove the $x$ term entirely), which gives us that
$$x^4+4x^3+4x^2-4x+3=(x^2+2x-1)^2+2x^2+2>0.$$
|
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|
Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$. Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$.
attempt:
Let $a$ and $n$ be integers with $n>0$.
Let $m=\frac{n}{\gcd(a,n)}$.
Then,
\begin{equation*}
ma = \left(\frac{n}{\gcd(a,n)}\right)a =
n\left(\frac{a}{\gcd(a,n)}\right)
\equiv 0 \pmod{n}.
\end{equation*}
Since $d=\gcd(a,n)$ is the greatest positive integer such that $d\mid a$ and $d \mid n$, then $m$ is the smallest positive integer such that $n \mid ma$.
Hence, $m$ is the additive order of $a$ modulo $n$.
Am I true?
|
The additive order of $a$ modulo $n$ is the smallest positive integer $m$ such that $ma \equiv 0 \pmod{n}$.
Now, $ma = kn$ for some integer $k$.
Then, $m = \frac{kn}{a}$.
Let $\gcd(a,n) = d$. Then,
$m = k\frac{\frac{n}{d}}{\frac{a}{d}}$.
Since $\frac{a}{d}$ and $\frac{n}{d}$ are relative prime, then $\frac{a}{d} \mid k$. Hence, the order of $a$ is a multiple of
$\frac{n}{d}$. Since the order must be minimum, then the order of $a$ is
$\frac{n}{d}$.
Is the approach above true?
|
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|
If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
I found this in a Facebook group.
I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.
Then we set $z=2-x-y$ and replace:
I get LHS = $x^2y+xy^2+2(x+y)-x^2-y^2-3xy-1$ but don't know how to proceed.
Thank you for your help.
|
We let $a=1-x,b=1-y,c=1-z$ so that we have $a+b+c=1$ and $0 < a,b,c < 1$.
Our new inequality now becomes $$8abc \leq (1-a)(1-b)(1-c)\\ \iff 8 \leq \left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right) \\ \iff 8 \leq \left(\dfrac{b+c}{a}\right)\left(\dfrac{c+a}{b}\right)\left(\dfrac{a+b}{c}\right) $$ which means $(a+b)(b+c)(c+a)\geq 8abc$ which is true by A.M-G.M inequality.
Equality occurs at $x=y=z=\dfrac{2}{3}$.
As for your approach, it is also correct. If you are substituting the value of $z$, making it a two variable inequality, you can form a quadratic equation taking everything to one side and set it's discriminant less than or equal to $0$.
|
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|
Analytic Proof of Wilson's Theorem I'm trying to find a reference for the following (analytic) proof of Wilson's Theorem. I would appreciate it immensely if some of you knew and told me the person who first came up with it. Here is the proof:
We start by considering the Maclaurin series of the function $f(x)=\ln \left( \frac{1}{1-x} \right)$, that is,
$\displaystyle \ln \left( \frac{1}{1-x} \right)= \sum_{n=1}^{\infty} \frac{x^n}{n} \tag*{}$
for $x \in [-1, 1).$
By simply using the definition of $\ln$ on the equation above, we get
$\displaystyle e^{x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots}=\frac{1}{1-x}=1+x+x^{2}+ \cdots + x^{p} + \cdots \tag{1}$
We can then rewrite the leftmost side of $(1)$ as
$\displaystyle \begin{align} e^{x+\frac{x^2}{2}+\frac{x^3}{3}+…} & = e^{x}e^{\frac{x^2}{2}}e^{\frac{x^3}{3}} \cdots \\ & = \prod_{n=1}^{\infty} e^{\frac{x^n}{n}} \\ & = \prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\left( \frac{x^n}{n} \right)^k}{k!} \\ & = \left( 1+\frac{x}{1!}+\frac{x^2}{2!}+ \cdots \right) \cdot \left( 1+\frac{x^{2}/2}{1!}+\frac{(x^{2}/2)^2}{2!} + \cdots \right) \cdots \left(1+\frac{x^{p}/p}{1!}+\frac{(x^{p}/p)^2}{2!}+ \cdots \right) \cdots \\ & = 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots \end{align} \tag{2}$
Combining the rightmost sides of $(1)$ and $(2)$ together gives us
$\displaystyle 1+x+x^{2}+ x^{3}+ \cdots + x^{p} + \cdots= 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots , \tag{3}$
from which follows that the coefficient of $x^{p}$ on the RHS of $(3)$ is $1$. That is,
$\displaystyle \frac{1}{p!}+\frac{1}{(p-2)!} \frac{1/2}{1!}+ \cdots + \frac{1/p}{1!} = 1. \tag{4}$
Notice that such a coefficient is of the form $\frac{1}{p!}+\frac{r}{s}+\frac{1}{p}$, where $\frac{r}{s}$ is the sum of a finite number of rationals, none of which has the factor $p$ in its denominator. Therefore, if $\frac{r}{s}$ is irreducible, then $p$ doesn’t divide $s$. We then proceed by rewriting $(4)$ as
$\displaystyle \frac{1}{p!}+\frac{r}{s}+\frac{1}{p}=1. \tag*{}$
Equivalently,
$\displaystyle 1-\frac{r}{s}=\frac{1}{p!}+\frac{1}{p}=\frac{1+(p-1)!}{p!}, \tag*{}$
and
$\displaystyle (s-r) \cdot (p-1)! = \frac{s(1+(p-1)!)}{p}. \tag*{}$
Hence, $\frac{s(1+(p-1)!)}{p}$ is an integer. Moreover, since $p$ doesn’t divide $s$, $\frac{1+(p-1)!}{p}$ is also an integer. Equivalently, $(p-1)! \equiv -1 \text{ (mod } p)$, as desired.
|
An alternative viewpoint.
*
*Analytic part
In $\Bbb{Q}[[x]]$ we have
$$\frac1{1-x}=\prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{x^{nk}}{n^k k!}$$
*
*Algebraic part
Let $R=\Bbb{Z}_{(p)}[x,x^p/p]+x^{p+1}\Bbb{Q}[[x]]$ the subring of formal series such that $p$ doesn't divide the denominator of the coefficients of $x^m,m<p$ and $p^2$ doesn't divide the denominator of the coefficient of $x^p$.
In the quotient ring $$R/(p,x,x^{p+1}\Bbb{Q}[[x]])\cong \Bbb{F}_p+\frac{x^p}p\Bbb{F}_p$$ we get
$$1=\frac1{1-x}=\prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{x^{nk}}{n^k k!}=(1+\frac{x^p}p)(1+\frac{x^p}{p!})=1+(1+\frac{1}{(p-1)!}) \frac{x^p}p$$
And hence $$1+\frac{1}{(p-1)!}=0\bmod p$$
|
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|
Finding integers $s$ and $t$ such that $s+t=\frac{\alpha}{b}+b$ and $st=\frac{\beta}{b}+b$ for integers $\alpha$, $\beta$, $b$ In my research I need to find integers $s$ and $t$ with the following properties (respect to $\alpha, \beta, b$) :
\begin{align}
s+t=\frac{\alpha}{b}+b\\
st=\frac{\beta}{b}+b,
\end{align}
where $\alpha, \beta, b\in\mathbb{Z}$ and $b\neq0$. Do there exist any integers $s$ and $t$ with the above properties?
Anyone can help me? Thanks in advance.
|
$$s+t=\frac{\alpha}{b}+b \qquad st=\frac{\beta}{b}+b\\
\implies {α + b^2 = b (s + t) \qquad b^2 + β = b s t}\\
\implies α = -b^2 + b s + b t \qquad β = b s t - b^2\qquad b\ne0
$$
We can see infinite solutions in terms of $\quad b, s, t\quad$ but it appears $\quad \alpha,\beta\subset\mathbb{Z}.$
Example of $(b,s,t,\alpha, \beta)$
$$(1,1,1,3,0)\quad
(2,1,1,8,-2)\quad
(3,1,1,15,-6)\quad
(4,1,1,24,-12)\\
(1,1,2,4,1)\quad
(2,1,2,10,0)\quad
(3,1,2,18,-3)\quad
(4,1,2,28,-8)\\
(1,2,2,5,3)\quad
(2,2,2,12,4)\quad
(3,2,2,21,3)\quad
(4,2,2,32,0)\\ $$
For $\quad s,t \quad\text{in terms of}\quad b, \alpha, \beta\quad$ as in Hagen von Eitzen's answer, we an immediately see that few of the lower number combinations yield integers. Some $\quad (b, \alpha, \beta )\quad$ values are
$$(1,3,2,3,1)\quad (1,3,3,2,2)\quad (1,4,3,4,1)\quad (1,4,5,3,2)\quad (1,5,4,5,1)\\$$
$$(2,4,2,3,1)\quad (2,4,4,2,2)\quad (2,6,4,4,1)\quad (2,6,8,3,2)\quad (2,8,6,5,1)\\ $$
$$(3,3,3,2,2)\quad (3,6,3,4,1)\quad (3,6,9,3,2)\quad (3,9,6,5,1)\quad (3,9,15,4,2)\\$$
|
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|
Find the Maximum Trigonometric polynomial coefficient $A_{k}$
Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.
I don't know if this question has been studied
If $n=2$ it is easy to solve it.
|
I'll just do a subcase of the $n=3$ case.
$A_{k}=1$ is the maximum for k=2, 3.
First notice that letting $A_{k}=1$ and taking the other values of $A_{j}=0$ we have
$$cos(kx)\leq 1$$
which holds for all x. So each $A_{k}$ is at least 1. Now we will show that each $A_{k}$ is at most 1 for $k=2,3$.
Plugging in $x=0$ gives us $A_{1}+A_{2}+A_{3} \leq 1$.
If $A_{3}>1$ then this forces $A_{1}+A_{2} \leq 1-A_{3} < 0$. However, plugging in $x=\frac{2\pi}{3}$ we have $A_{3} + \frac{-A_{1}-A_{2}}{2} \leq 1$
So $A_{3}\leq 1+\frac{A_{1}+A_{2}}{2} < 1$
Therefore $A_{3}\leq 1$ thus $A_{3}=1$.
If $A_{2}>1$ then $A_{1}+A_{3} \leq 1-A_{2} < 0$.
Plugging in $x=\pi$ we have $A_{2} + -A_{1}-A_{3} \leq 1$.
So $A_{2}\leq 1+ A_{1} + A_{3} < 1$.
Therefore $A_{2}\leq 1$ thus $A_{2}=1$.
|
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|
Prove that $x/(1 + \frac{2}{\pi} \cdot x) < \arctan(x)$ $\forall x > 0$ The first thing that I tried to do is to differentiate both functions and try to see if there establishes inequality that we want (considering that they are equal when $x = 0$). This attempt failed because firstly it really is true but after that we get opposing inequality. Also, I have noticed that $\lim \frac{x}{1 + x \cdot \frac{2}{\pi}} = \lim \arctan(x) = \frac{\pi}{2}$ as $x \rightarrow + \infty$ but it didn't lead me to solution. I also tried to apply Taylor's formula but it didn't help much either.
So, what are available ways to solve this problem?
|
With the idea of @trancelocation: the map $x\mapsto x/(1 + a x)= \frac{1}{a} ( 1 - \frac{1}{1+a x})$ is increasing on $(0, \infty)$ ( for $a>0$), so the inequality is equivalent to
$$x < \frac{ \arctan x}{ 1 - \frac{\arctan x}{\pi/2}} \textrm{ or } \tan t < \frac{t}{1- \frac{t}{\pi/2}}$$
We have the product expansions
$$\cos t = \prod_{n=1}^{\infty} \left(1- \frac{t^2}{(n-1/2)^2\pi^2} \right), \ \ \sin t = \prod_{n=1}^{\infty} \left(1- \frac{t^2}{n^2 \pi^2}\right)$$
so
$$\frac{\tan t }{t} = \frac{1}{1- (\frac{t}{\pi/2})^2} \cdot\prod_{n=1}^{\infty} \frac{\left(1- \frac{t^2}{n^2 \pi^2}\right)}{ \left(1- \frac{t^2}{(n+1/2)^2 \pi^2}\right)}< \frac{1}{1- (\frac{t}{\pi/2})^2}$$
for $0< t <\pi/2$. So we have in fact
\begin{equation}
\boxed{\color{indigo}{
\frac{1-(\frac{t}{\pi})^2 }{1- (\frac{t}{\pi/2})^2}<\frac{\tan t}{t} < \frac{1}{1- (\frac{t}{\pi/2})^2} < \sec t
}}
\end{equation}
for real $t$, $0<|t|< \frac{\pi}{2}$.
In the above inequalities, we have also the corresponding inequalities for the corresponding coefficients of the Taylor series, clear if we examine the product formulas.
From the second inequality we get
\begin{eqnarray}
\boxed{\color{indigo}{
\frac{\frac{\pi}{2} x}{ \sqrt{x^2 + (\frac{\pi}{4 })^2} + \frac{\pi}{4 }} < \arctan(x)
}}
\end{eqnarray}
for $x > 0$, an improvement over the inequality $\frac{\frac{ \pi }{2} x}{ x+\frac{\pi}{4}+\frac{\pi}{4}} < \arctan x$.
|
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|
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)
|
Don't rush to square equations to get rid of square roots. It's almost always more work and you end up missing opportunities for simplification. Instead eliminate square roots by substitution. Here, let $a\ge0$ and $b\ge0$ be the values of the two square roots:
$$
\begin{cases}
x+y=23\\
a+b=33\\
a^2 = x^2 + 12y\\
b^2 = y^2 + 12x
\end{cases}
$$
While this turns your two equations into four equations, it makes simplifications easier to spot. Subtracting the last two equations will let you factor $a^2 - b^2$ and $x^2 - y^2$ and make use of the fact that $x+y$ and $a+b$ are known:
$$33(a-b) = 23(x-y) - 12(x-y)$$
$$3(a-b) = x-y$$
Now this becomes very simple. Let $z = a-b$ and solve for $z$. For example, express $x$, $y$, $a$ in terms of $z$ and substitute into $(a-x)(a+x) = 12y$. Everything will neatly cancel out and you will get $z^2 = 1$ or $a=b\pm1$ and $x=y\pm3$. The two solutions are $x=10, y=13$ and $x=13,y=10$
|
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|
$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you
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Although $\sum\frac{k^2}{n^3+kn}$ and $\sum\frac{k^2}{n^3}$ have the same limit, they differ to $O(1/n)$. So when they are multiplied by $n$, they give different limits
|
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|
How do we integrate the function $\frac{b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}}{4(\theta-1)}$? I came across the problem of determining the following integral:
\begin{align*}
I(\theta) = \int_{0}^{1}\dfrac{b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}}{4(\theta-1)}\mathrm{d}b
\end{align*}
where $b\in[0,1]$, $\theta > 0$ and $\theta\neq 1$. Here, we are going to study the case where $0 < \theta < 1$.
I managed to rearrange the square root argument as $1 + 4b(1-b)(\theta-1)$.
One possible approach consists in trying the substitution $u = 2b(\theta -1)$, from whence we get
\begin{align*}
I(\theta) = \frac{1}{16(\theta - 1)^{3}}\int_{0}^{2(\theta-1)}u\sqrt{1 + 2u\left(1 - \dfrac{u}{2(\theta-1)}\right)}\mathrm{d}u
\end{align*}
Can someone help me to take it from here? Any help is appreciated.
|
Rearranging
\begin{equation}
[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^2=(1-\theta)(2b-1)^2+\theta
\end{equation}
the integral can be written as
\begin{align}
I(\theta) &= \frac{1}{4(\theta-1)}\int_{0}^{1}b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}\,db\\
&=\frac{1}{4(\theta-1)}\int_{0}^{1}b\sqrt{(1-\theta)(2b-1)^2+\theta}\,db\\
&=\frac{\sqrt \theta}{16(\theta-1)}\int_{-1}^{1}(x+1)\sqrt{1+\frac{(1-\theta)}{\theta}x^2}\,dx
\end{align}
(substitution $x=2b-1$ is used to obtain the latter expression). Due to parity reasons it can be simplified as
\begin{equation}
I(\theta)=\frac{\sqrt \theta}{8(\theta-1)}\int_{0}^{1}\sqrt{1+\frac{(1-\theta)}{\theta}x^2}\,dx
\end{equation}
Now, with $x=\sqrt{\theta/(1-\theta)}\sinh t$,
\begin{equation}
I(\theta)=\frac{ -\theta}{8(1-\theta)^{3/2}}\int_{0}^{\operatorname{arcsinh} \sqrt{\frac{1-\theta}{\theta}}}\cosh^2t\,dt
\end{equation}
After some simple manipulations, we find
\begin{equation}
I(\theta)=\frac{-1}{16(1-\theta)^{3/2}}\left[\sqrt{1-\theta}+\theta\operatorname{arcsinh}\sqrt{\frac{1-\theta}{\theta}}\right]
\end{equation}
|
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|
Does the following lower bound improve on $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the abundancy index
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical sum of divisors of $x$.
Since $q$ is prime, we have the bounds
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$
which implies, since $N$ is perfect, that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
By considering the negative product
$$\bigg(I(q^k) - \frac{2(q-1)}{q}\bigg)\bigg(I(n^2) - \frac{2(q-1)}{q}\bigg) < 0,$$
since we obviously have
$$\frac{q}{q-1} < \frac{2(q-1)}{q},$$
then after some routine algebraic manipulations, we arrive at the lower bound
$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$
Now, a recent MO post improves on the lower bound for $I(n^2)$, as follows:
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$
Added February 3, 2021 - 8:10 PM (Manila time) Here is a quick way to show the improved lower bound for $I(n^2)$. We have
$$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} - 1}\bigg)$$
$$=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{(q^{k+1} - 1) + 1}{q^{k+1} - 1}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - 1}\bigg)$$
$$>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)$$
since $q^{k+1} > q^{k+1} - 1$. But, of course, we obtain
$$I(n^2)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg).$$
QED.
Repeating the same procedure as above, we have the negative product
$$\Bigg(I(q^k) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg)\Bigg(I(n^2) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg) < 0.$$
This implies, after some algebraic manipulations, that
$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}.$$
But WolframAlpha says that the partial fraction decomposition of the new lower bound is given by
$$\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}} = \frac{3q^2 - 4q + 2}{q(q - 1)} + \frac{2(q - 1)}{q^{k+2}} - \frac{q}{(q - 1)(q^{k+1} + 1)}.$$
So essentially, my question boils down to:
QUESTION: Is it possible to produce an unconditional proof (that is, for all $k \geq 1$ and for all special primes $q \geq 5$) for the following inequality?
$$\frac{2(q - 1)}{q^{k+2}} > \frac{q}{(q - 1)(q^{k+1} + 1)}$$
MY ATTEMPT
I tried to ask WolframAlpha for a plot of the above inequality, it gave me the following GIF image:
So it does appear that the inequality is unconditionally true, which would mean that the new lower bound for $I(q^k) + I(n^2)$ improves on the old. Is it possible to prove this analytically?
And lastly: Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?
|
Suppose to the contrary that there exists an integer $k \geq 1$ and a (special) prime $q \geq 5$ such that
$$\frac{q}{(q-1)(q^{k+1}+1)} \geq \frac{2(q-1)}{q^{k+2}}.$$
This inequality is equivalent to
$$q^{k+3} \geq 2(q-1)^2 (q^{k+1}+1) = 2q^{k+1} - 4q^{k+2} + 2q^{k+3} + 2q^2 - 4q + 2,$$
which in turn is equivalent to
$$0 \geq q^{k+3} - 4q^{k+2} + 2q^{k+1} + 2q^2 - 4q + 2 = q^{k+2} (q - 4) + 2q^{k+1} + 2q(q - 2) + 2 > 0,$$
a contradiction.
We therefore conclude that
$$\frac{2(q-1)}{q^{k+2}} > \frac{q}{(q-1)(q^{k+1}+1)}$$
for all integers $k \geq 1$ and all (special) primes $q \geq 5$.
Hence, the new lower bound
$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}$$
does indeed improve on the old lower bound
$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q - 1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$
It remains to be seen whether this implies that there does exist an integer $a$ such that $k \leq a$, if $q^k n^2$ is an odd perfect number with special prime $q$.
|
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|
Prove that $0<\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}$ for $x>0$ So far, I've tried using induction:
Let P(n): $0<\sum_{i=2}^n\ (-1)^i \frac{x^{2i}}{2i!}$
Let $n=2$. Thus, P(2): $\frac{x^4}{4!}>0$ for $x>0$. Then, let P(k) be true (Induction Hypothesis). Now, prove the validity of P for n=k+1:
$P(k+1) : \sum_{i=2}^{k+1}\ (-1)^i\frac{x^{2i}}{2i!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{k+1}\frac{x^{2(k+1)}}{2(k+1)!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}$
Let $k=2m-1$:
$\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}=\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!}$ which is $>0$ since $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}>0$ and $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!} > \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}$ for $x>0$ according to the Induction Hypothesis.
I'm lost for the case where $k=2m$, I'd like some help there, please.
|
$$S_n=\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}=\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}-\sum_{i=0}^2\ (-1)^n \frac{x^{2n}}{2n!}$$
$$S_n=\frac{x^2-1}2+\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}$$ If you are aware of the complete and incomplete gamma function
$$\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}=\frac{e^{-x^2}\,\, \Gamma \left(n+1,-x^2\right)}{2 \Gamma (n+1)}$$
$$S_n=\frac{1}{2} \left(e^{-x^2}\frac{ \Gamma \left(n+1,-x^2\right)}{\Gamma
(n+1)}+x^2-1\right)$$ which, for a given $n$ or a given $x$ can cancel.
As @bjorn93 did, let us use $x=10$ and $n=5$. $S_5$ will become negative as soon as $x>1.96259$. Checking for $x=2$, we have $-\frac{4}{15}$
|
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|
(Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola I searched but couldn't find the proof.
Isosceles hyperbola equation:
$${H:x^{2}-y^{2} = a^{2}}$$
And let's take any point $P(x, y)$ on this hyperbola. Now, the product of the distances of this point $P(x, y)$ to the foci of the isosceles hyperbola is equal to the square of the distance from point $P$ to the center of the hyperbola.
Proof?
I took this question from my analytical geometry project assignment. I tried various ways (I found the foci $F(x,y)$ and $F^{'}(x,y)$ terms of $x$, $y$ and chose any point on the hyperbola...) but I couldn't prove.
I request your help.
|
$$PS_1.PS_2=\sqrt{[(x-ae)^2+y^2][(x+ae)^2+y^2]}$$
$$=\sqrt{(x^2-a^2e^2)^2+y^2(x^2+a^2e^2+2aex)+y^2(x^2+a^2e^2-2aex)+y^4} $$
$$=\sqrt{x^4+a^4e^4-2a^2e^2x^2+y^2x^2+y^2a^2e^2+2aexy^2+y^2x^2+y^2a^2e^2-2aexy^2}$$
$$=\sqrt{x^2+2x^2y^2+y^4+a^4e^4-2a^2e^2(x^2-y^2)}$$
$$=\sqrt{(x^2+y^2)^2+a^4e^4-2a^4e^2}$$
Since $e=\sqrt{2}$ for this hyperbola, so finally we prove that
$$PS_1.PS_2= x^2+y^2=OP^2.$$
|
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|
Proving $\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq\frac{3\sqrt{3}}2$ for $a$, $b$, $c$, $s$ the sides and semi-perimeter of a triangle
If $a, b, c$ are the lengths of the sides of a triangle and s is the semiperimeter, prove that:
$$\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq \frac{3\sqrt{3}}{2}$$
My attempt: $$\sum_{cyc} cos \,\frac{A}{2}=\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \frac{3 \sqrt{3}}{2}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. As the first inequality is obvious, it is enough to show that $$\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. We prove that $$\sqrt{\frac{s(s-a)}{bc}} \leq\frac{\sqrt{s(s-a)}}{a}$$ or equivalently $a^{2}\leq bc$. Similary $b^{2}\leq ac$ and $c^{2}\leq ab$ and adding the inequalities we get $(a-b)^{2}+(b-c)^{2}+ (c-a)^{2}\leq 0$.
|
After using Ravi's substituition $a=x+y,b=y+z,c=x+z$ as nguyenhuyen_ag did we have to prove $$\sum \frac{\sqrt{z(x+y+z)}}{y+x} \geq \frac{3\sqrt 3}{2}.$$ Now we will use the method of Isolated fudging. We guess that we might be able to prove the follwing inequality $$\frac{\sqrt{z(x+y+z)}}{y+x}\ge \frac{3\sqrt{3}}{2}\frac{z}{x+y+z}$$ Indeed after squaring this inequality is equivalent to $$\frac{{(x+y+z)}^3}{27}\ge z(\frac{x+y}{2})(\frac{x+y}{2})$$ which is evident by AM_GM. Hence $$\sum \frac{\sqrt{x(x+y+z)}}{y+z} \ge\frac{3\sqrt{3}}{2} \sum \frac{z}{x+y+z}=\frac{3\sqrt{3}}{2}$$ done
|
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}
|
Three unit vectors whose sum is zero Let $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ be unit vector such that $\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$. Which of the following is correct ?
(A) $\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a = \overrightarrow 0 $
(B) $\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a \ne \overrightarrow 0 $
(C) $\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow a \times \overrightarrow c \ne \overrightarrow 0 $
(D) $\overrightarrow a \times \overrightarrow b ,\overrightarrow b \times \overrightarrow c ,\overrightarrow c \times \overrightarrow a $ are mutually perpendicular
The official answer is (B).
My approach is s follow
$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$
$\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow b + \overrightarrow c \times \overrightarrow b = 0$
$\overrightarrow a \times \overrightarrow b = - \overrightarrow c \times \overrightarrow b = \overrightarrow b \times \overrightarrow c $
$\overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow c = 0 \Rightarrow \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a $.
Now I am confused between option (A) and (B) if $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are parallel to each other then option (A) is correct or option (B)
Vector being a unit vector plays any vital role in deciding between (A) and (B)
|
(A) would imply the vectors are pairwise parallel or antiparallel, so their sum is of length $1$ or $3$.
|
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|
Calculating the improper integral $\int _{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$ I want to calculate the value of $$\int_{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$$
I was able to prove using Dirichlet's test that it does converge, but how can I calculate its value?
|
We have that, integrating by parts
$$
\eqalign{
& I = \int_1^\infty {e^{\,\sin x} {{\cos x} \over x}dx}
= \int_1^\infty {{1 \over x}d\left( {e^{\,\sin x} } \right)} = \cr
& = \left. {{{e^{\,\sin x} } \over x}} \right|_1^\infty
- \int_1^\infty {e^{\,\sin x} d\left( {{1 \over x}} \right)}
= - e^{\,\sin 1} + \int_1^\infty {{{e^{\,\sin x} } \over {x^{\,2} }}dx} = \cr
& = - e^{\,\sin 1} + J \cr}
$$
with the advantage of having replaced $1/x$ with $1/x^2$.
Then being $\exp(\sin x)$ periodic
$$
\eqalign{
& J = \int_1^\infty {{{e^{\,\sin x} } \over {x^{\,2} }}dx}
= \int_1^{1 + 2\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx}
+ \int_{1 + 2\pi }^{1 + 4\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx} + \cdots = \cr
& = \sum\limits_{0\, \le \,k} {\int_{1 + k2\pi }^{1 + \left( {k + 1} \right)2\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx} }
= \sum\limits_{0\, \le \,k} {\int_0^{2\pi }
{{{e^{\,\sin \left( {x + 1} \right)} } \over {\left( {x + 1 + k2\pi } \right)^{\,2} }}dx} } = \cr
& = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi } {e^{\,\sin \left( {x + 1} \right)}
\left( {\sum\limits_{0\, \le \,k}
{{1 \over {\left( {{{\left( {x + 1} \right)} \over {2\pi }} + k} \right)^{\,2} }}} } \right)dx} = \cr
& = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi }
{e^{\,\sin \left( {x + 1} \right)} \zeta \left( {2,{{x + 1} \over {2\pi }}} \right)dx} = \cr
& = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi }
{e^{\,\sin \left( {x + 1} \right)} \psi ^{\left( 1 \right)} \left( {{{x + 1} \over {2\pi }}} \right)dx} = \cr
& = {1 \over {2\pi }}\int_0^{2\pi }
{e^{\,\sin \left( {x + 1} \right)} \psi ^{\left( 1 \right)} \left( {{{x + 1} \over {2\pi }}} \right)
d\left( {{{x + 1} \over {2\pi }}} \right)} = \cr
& = {1 \over {2\pi }}\int_{{1 \over {2\pi }}}^{{1 \over {2\pi }} + 1}
{e^{\,\sin \left( {2\pi t} \right)} \psi ^{\left( 1 \right)} \left( t \right)d\left( t \right)} \cr
}
$$
where $\psi ^{\left( 1 \right)} \left( t \right)$ denotes the
Trigamma Function.
So we have reduced the improper integral to a proper integral,
much more manageable by numeric integration.
My CAS gives:
$$
\eqalign{
& J \approx 1 .94017 \cr
& {\rm I} \approx - \,0.3796 \cr}
$$
|
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|
doubt about the calculation of the rest for a taylor's series I'm trying to calculate some numbers with the help of series.
My problem is the calculation of the rest of the series.
For example I have to calculate $log 2$.
I can consider the function $$log (\frac{1+x}{1-x})$$ in which I can put $x= \frac{1}{3}$.
The series of Taylor for this function is:
$$log (\frac{1+x}{1-x}) = 2*\sum_{k=0}^ \infty \frac{x^{2k+1}}{2k+1}$$
$$log (\frac{1+x}{1-x}) = 2*\sum_{k=0}^ n \frac{x^{2k+1}}{2k+1}+ 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} = 2*\sum_{k=0}^ n \frac{x^{2k+1}}{2k+1}+ R_n$$
The passage that I can't understand it this:
$$R_n = 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} < \frac{2 |x|^{2n+3}}{2n+3}\sum_{h=0}^ \infty x^{2k}$$ and then it becomes $\frac{2 |x|^{2n+3}}{2n+3} \frac{1}{1-x^2}$
NOTE
$$R_n = 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} = 2*\sum_{h=0}^ \infty \frac{x^{2h+3}}{2h+3} =2 x^3 \sum_{h=0}^ \infty \frac{(x^2)^h}{2h+3}$$
|
For $k \ge n+1$, $\vert x \vert^{2k+1} = \vert x \vert^{2n+3} x^{2(k-n-1)}$ and $\frac{1}{2k+1} \le \frac{1}{2n+3}$. Hence
$$2\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} \le \frac{2 |x|^{2n+3}}{2n+3}\sum_{h=0}^ \infty x^{2h}.$$
Finally $\sum_{h=0}^ \infty x^{2h}$ is the sum of the terms of a geometric sequence and therefore
$$\sum_{h=0}^ \infty x^{2k} = \frac{1}{1-x^2}$$ which provides the desired result.
|
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|
Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$ The problem goes as follows:
Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.
I first came across the problem in a school book targeted towards students just exposed to factorisation and it intrigued me how one could prove this without Sturm's Theorem. Maybe one could try to factorise into squares, but I can get no result from that. Thanks in advance.
|
Define the real polynomial
$$ p(x)\!:=\!x^8\! + \!x^7\! + \!x^6\! - \!5x^5\!-
\!5x^4\!- \!5x^3\! + \!7x^2\! + \!7x\!+\!7. \tag{1} $$
By observation, the polynomial easily factors as the product
$$ p(x) = (x^2+x+1)(x^6-5x^3+7) = p_1(x)p_2(x^3). \tag{2} $$
If we can prove for all real $\,x\,$ that $\,p(x)>1\,$ then
also $\,p(x)-1>0\,$ and hence $\,p(x)-1,\,$ the polynomial
in question, has no real roots.
Note that
$$ p_1(x) := x^2 + x + 1 = (x+1/2)^2+(3/4) \ge 3/4
\tag{3} $$
and
$$ p_2(x) := x^2-5x+7 = (x-5/2)^2+(3/4) \ge 3/4.
\tag{4} $$
Note that $\,p_1(x)\le 1\,$ iff $\,-1\le x\le 0\,$ and
$\,p_2(x^3)\le 1\,$ iff $\,2\le x^3\le3\,$ which are disjoint intervals. Now prove that when $\,p_1(x)<1\,$ that $\,p_2(x^3)>4/3\,$ and also when $\,p_2(x^3)<1\,$ that $\,p_1(x)>4/3.\,$
|
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|
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.
There are $2$ cases to consider:
Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:
\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) + abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}
Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?
|
For $a,b,c$ sides of a triangle, let $$\begin{matrix}a=y+z\\b=z+x\\c=x+y\end{matrix}$$
Then the inequality becomes $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ This is true by Cauchy's inequality: $$x+y+z\le\sqrt{\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}}\sqrt{y+z+x}$$
|
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|
Prove by induction that $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $120$.
Prove by induction that $$(n+1)(n+2)(n+3)(n+4)(n+5)$$ is divisible by $120$.
I tried to solve it, but could do so only till the inductive step.
I assumed: $$p(k)=(k+1)(k+2)(k+3)(k+4)(k+5), and $$ $$p(k+1)=(k+2)(k+3)...(k+6)$$
Then, I distributed it as: $$p(k+1)=(k+2)(k+3)...(k+5)(k+1+5)
\\=(k+1)(k+2)...(k+5)+5(k+2)(k+3)...(k+5)\\=p(k)+5(k+2)(k+3)...(k+5).$$
I got stuck over here.
Am I right till here? Can someone tell what to do next?
|
$120=2\cdot 3\cdot 4\cdot 5$. $\Rightarrow 120|1\cdot 2\cdot 3\cdot 4\cdot 5$ and $120|2\cdot 3\cdot 4\cdot 5\cdot 6$.
Now, if $120|k(k+1)(k+2)(k+3)(k+4)$ and $120|(k+1)(k+2)(k+3)(k+4)(k+5)$, then,
$120|(k+1)(k+2)(k+3)(k+4)\{(k+5)-k\}$
$\Rightarrow 24|(k+1)(k+2)(k+3)(k+4)$
Now, use normal induction and proceed.
|
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|
How to $2^n/n! \leq 4/n$ by induction I'm trying to prove that $\frac{2^n}{n!} \leq \frac{4}{n}$ for all positive integers $n$.
It's obvious this holds for $n=1$.
My inductive hypothesis is that $\frac{2^n}{n!} \leq \frac{4}{n}$.
I want to show that $\frac{2^{n+1}}{(n+1)!} \leq \frac{4}{(n+1)}$.
I can see that $\frac{2^{n+1}}{(n+1)!} = \frac{2^n \cdot 2}{n! \cdot (n+1)} \leq
\frac{4 \cdot 2}{n(n+1)}$, where the inductive hypothesis is applied in the last step.
All I need to show is that $\frac{4 \cdot 2}{n(n+1)} \leq \frac{4}{n+1}$ and I'm done. But I can't think of a good justification for this.
Any hints or suggestions?
|
$\frac{4\cdot 2}{n(n+1)} = \frac{2}{n}\frac{4}{n+1}\leq \frac{4}{n+1}$, since $\frac{2}{n}\leq 1$ for $n\geq 2$.
In the case $n=1$, $2/1 \leq 4/1$.
|
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|
Proving a positive integer Suppose that a positive integer $n$ ends with $k$ $9\text{'s}$. How can we prove that the integer $3n^4+4n^3$ ends with at least $2k$ $9\text{'s}$?
|
Firstly note that saying a number ends with $k$ $9$s is equivalent to saying that it is congruent $-1\ (\text{mod }10^k)$, as for example $99$ ends with two $9$s and $99\equiv -1\ (\text{mod }100)$.
So write $n=A\cdot 10^k -1$, for an integer $A$. Then note:
\begin{align}
n^3&\equiv 3A\cdot10^k-1\ \quad&(\text{mod }10^{2k})\\
n^4&\equiv -4A\cdot10^k+1 \quad&(\text{mod }10^{2k})
\end{align}
As all the other terms include $10$ to a $\geq2k$ integer power. So adding:
\begin{align}
3n^4+4n^3 &\equiv (-12A\cdot10^k+3) + (12A\cdot10^k-4)\quad&(\text{mod }10^{2k})\\
&\equiv -1 \quad&(\text{mod }10^{2k})
\end{align}
Which is the required answer.
|
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Positive integer solutions of equations of the form $ x^4+bx^2y^2+dy^4=z^2$ A method used to answer the post Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution? showed that the only positive integer solutions of the equations $$ x^4-3x^2y^2+3y^4=z^2$$ $$X^4+6X^2Y^2-3Y^4=Z^2,$$ are $(1,1,1)$ and $(1,1,2)$, respectively. Furthermore these solutions are related in that each can be deduced from the other.
In general, let $b,d,D$ be integers such that $b^2=4d+D$ and consider the pair of equations $$ x^4+bx^2y^2+dy^4=z^2\tag 1$$ $$X^4-2bX^2Y^2+DY^4=Z^2\tag 2$$ where we can assume that $(x,y)=(X,Y)=1$. Then for the method of the aforementioned post to produce an infinite sequence of solutions, we require a solution $(x,y,z)$ of $(1)$ to generate the solution $(\frac{z}{x},y,|\frac{x^4-dy^4}{x^2}|)$ of $(2)$ and we require a solution $(X,Y,Z)$ of $(2)$ to generate the solution $(\frac{Z}{2X},Y,|\frac{X^4-DY^4}{4X^2}|)$ of $(1)$.
I can prove that for this to occur $y=Y$ is odd. Also, since $x$ must be a factor of $z$, $x^2$ is a factor of $d$. Therefore there are only a finite number of possibilities for $x$ and so we are dealing with a finite loop of solutions. Beyond that, I only know that when solving specific equations these loops are rare and, when they occur, seem to have period $2$.
Questions
What periods are possible for such loops of solutions?
What else can be determined about them?
|
SOME PROGRESS
Let the loop of solutions have period $2n$. We can suppose that successive values taken by $x$ and $X$ are $a_1,a_2,...a_n$ and $A_1,A_2,...A_n$, respectively, where for each $i$ there are integers $c_i$ and $C_i$ such that $d=a_i^2c_i$ and $D=A_i^2C_i$.
Then the conditions given in the post simplify to:-
$$a_i^2+by^2+c_iy^4=A_i^2 \tag 1$$
$$A_i^2-2by^2+C_iy^4=4a_{i+1}^2 \tag 2$$
$$2a_{i+1}A_i=|a_i^2-c_iy^4| \tag 3$$
$$4a_{i+1}A_{i+1}=|A_i^2-C_iy^4| \tag 4$$
$y^2$ is a factor of $4^n-1$
Consider equations $(1)$ and $(2)$ modulo $y^2$.
$$a_1^2\equiv A_1^2\equiv 4a_2^2\equiv 4A_2^2\equiv 16a_3^2\equiv 16A_3^2\equiv ... $$ Hence $a_1^2\equiv 4^na_1^2$ and so, since $x$ and $y$ are coprime, $y^2$ is a factor of $4^n-1.$
This significantly limits the values of $y$. For periods of less than or equal to $18$ the only possible values for $y$ are $1$ and $3$.
A general formula for loops of period $2$
If $n=1$, then $y=1$ and the equations further simplify to
$$a^2+b+c=A^2 \tag 1$$
$$A^2-2b+C=4a^2 \tag 2$$
$$2aA=|a^2-c| \tag 3$$
$$4aA=|A^2-C| \tag 4$$
From equation $(3)$ we have $c=a^2\pm 2aA$. Then $b=A^2\mp 2aA-2a^2$ and $C=A^2\mp4aA.$
The original equations are then $$ x^4+\left(A^2\mp 2aA-2a^2\right)x^2y^2+\left(a^4\pm2a^3A\right)y^4=a^2A^2$$ $$X^4-2\left(A^2\mp 2aA-2a^2\right)X^2Y^2+\left(A^4\mp4aA^3\right)Y^4=4a^2A^2,$$
with a loop of solutions $(a,1,aA)$ and $(A,1,2aA).$
|
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|
Prove that $\angle AEF =90^\circ$ given a square $ABCD$ Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.
My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish to prove that $AE^2 +EF^2 = AF^2$. I’ve tried a lot of methods to reach this point but none of them worked.
|
Let $AB = BC = CD = DA = 4x$.
Then: $DE = EC = 2x$, $CF = x$, and $FB = 3x$.
Now,
*
*In triangle $ABF$, we have $AF^2 = AB^2 + FB^2 = (4x)^2 + (3x)^2 \Rightarrow AF = 5x$.
*In triangle $ADE$, we have $AE^2 = DA^2 + ED^2 = (4x)^2 + (2x)^2 \Rightarrow AE = 2\sqrt{5}x$.
*In triangle $ECF$, we have $EF^2 = EC^2 + CF^2 = (2x)^2 + x^2 \Rightarrow EF = \sqrt{5}x$.
Now, if you look at triangle $AEF$, we have $\underbrace{AE^2}_{(2\sqrt{5}x)^2} + \underbrace{EX^2}_{(\sqrt{5}x)^2} = \underbrace{AF^2}_{(5x)^2}$. Since Pythagoras identity holds here, we can deduce that angle $AEF = 90$.
|
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|
These functions are very similar. Is there an explanation? Blue: $x \ln x-x$
Brown: $\ln \frac{\Gamma(x+1/2)}{\sqrt{\pi}}$
|
Using Sterling’s approximation for the Gamma function, $$\Gamma(z)\sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z$$
In this case \begin{align*}
\Gamma\left(x+\frac{1}{2}\right)&\sim \sqrt{\frac{2\pi}{x+\frac{1}{2}}}\left(\frac{x+\frac{1}{2}}{e}\right)^{x+\frac{1}{2}}\\
\ln\Gamma\left(x+\frac{1}{2}\right)&\sim \frac{1}{2}\ln 2+\ln \sqrt \pi -\frac{1}{2}\ln \left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)\ln\left(x+\frac{1}{2}\right)-x-\frac{1}{2}\\
&=\frac{1}{2}\ln 2+\ln \sqrt \pi +x\ln \left(x+\frac{1}{2}\right)-x-\frac{1}{2}\\
&\sim \ln \sqrt\pi +x\ln x-x\\
\ln \left(\frac{\Gamma\left(x+\frac{1}{2}\right)}{\sqrt\pi}\right)&\sim x\ln x-x
\end{align*}
|
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|
Solving recurrence with substitution I am trying to solve this recurrence $T(n) = 4T(n − 2) + 2^{2n}$ by using substitution and knowing $T(1)=1,T(2)=2$ and here is my attempt:Expanding $T(n)=4T(n-2)+2^{2n}$ using $T(n-2)=4T(n-4)+2^{2(n-2)}$, we get $T(n)=4(4T(n-4)+2^{2(n-2)})+2^{2n}=4^2T(n-4)+2^{2n-2}+2^{2n}=4^2(4T(n-6)+2^{2(n-4)})+2^{2n-2}+2^{2n}=4^3T(n-6)+2^{2n-4}+2^{2n-2}+2^{2n}=...=4^k(T(n-2k)+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$.Substituting $T(n-2k)=2, i.e. T(2)=2$, $n-2k=2$ then $k=\frac{n-2}{2}$, so our equation looks like $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$ with $k=\frac{n-2}{2}$, $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}=2^{n-2}+2^{n+4}+...+2^{2n}=2^{n-2}+\frac{2^{n+4}(1-2^{\frac{n-4}{2}})}{1-2}=2^{n-2}+2^{\frac{3n+4}{2}}-2^{n+4}$. However I think this answer doesn't satisfy $T(2)=2$ and I really don't know where went wrong. Is there another way to do this problem?
|
This is a linear difference equation. You can get the solution as follows:
*
*Compute the general solution of the homogenous equation $T_n = 4 T_{n-2}$, which is $H_n = c_1 \cdot 2^n +c_2 \cdot (-2)^n$.
*Obtain a particular solution of the complete equation by trying something similar to the RHS, i.e. $P_n = k \cdot 4^n$, and compute $k$ substituting in the equation. One possibility is $P_n = \frac{4}{3} \cdot 4^n$.
*The general solution to our equation is given by
$$
T_n = H_n + P_n = c_1 \cdot 2^n +c_2 \cdot (-2)^n + \frac{4}{3} \cdot 4^n
$$
*From the conditions on $T_1, T_2$, compute $c_1, c_2$ to get the final answer:
$$
T_n = -\frac 72 2^{n}-\frac 43 (-2)^{n} + \frac 43 \cdot 4^n
$$
|
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|
Finding the volume of a solid through 2$\pi$ radians With reference to this question
Curve $C$ with the equation:
$$y = (2x-1)^{\frac{3}{4}}, \quad x \ge\frac{1}{2}$$
The finite region $S$, is bounded by the curve $C$, the $x$-axis. The $y$-axis and the line $y = 8$. This region is rotated through $2\pi$ radians about the $x$-axis.
Find the exact value of the volume of the solid generated.
Here's what I have tried:
$\mathbf{k}$ or $\mathbf{x}$ was found to be at $\frac{17}{2}$
$$\int\limits^{17/2}_{1/2}(2x-1)^{3/4}dx=\frac{256}{7}$$
I'm wondering if I approached this correctly?
|
Curve $C$ is given by $y = (2x-1)^{\frac{3}{4}}$ for $x \ge\frac{1}{2}$.
Note that the region S, is bounded by the curve C, the x-axis, the y-axis and the line $y = 8$. This region is rotated through $2\pi$ about the x-axis.
a) If we find the volume using shell method, it is one integral.
$\displaystyle y = (2x-1)^{\frac{3}{4}} \implies x = \frac{y^{4/3} + 1}{2}$
So the integral can be set up as,
$V = \displaystyle \int_0^8 \int_0^{(1 + y^{4/3})/2} 2 \pi \ y \ dx \ dy = \frac{1696 \pi}{5}$
b) If we find volume using washer method, we will have to split the integral into two, which is between $0 \leq x \leq \frac{1}{2}$ and between $\frac{1}{2} \leq x \leq \frac{17}{2}$. The volume for the first part will be a cylinder of radius $8 (0 \leq y \leq 8)$ whereas the second part will have radius as $(2x-1)^{\frac{3}{4}} \leq y \leq 8$.
$V = \displaystyle \int_{1/2}^{17/2} \int_{(2x-1)^{3/4}}^8 2\pi \ y \ dy \ dx + \int_0^{1/2} \int_0^8 2\pi \ y \ dy \ dx$
|
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|
Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.
Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.
Assuming the contrary that there exists an $|z|\geqslant2$ such that $P(z)=0$ I have that $$z^7+7z^4+4z+1 = 0 \implies 1+\frac{7}{z^3}+\frac{4}{z^6} +\frac{1}{z^7}=0 \implies 1+\frac{7}{8}+\frac{4}{64}+\frac{1}{128} = \frac{249}{128}\ne0$$
which is a contradiction so the zeros would lie inside the disk of radius $2$?
|
If there is a solution, $|z^7|=|7z^4+4Z+1|$ Suppose that $|z|\geq 2$,
$2^7=128>7 \times 2^4 +4 \times 2+1=127$
|
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|
calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$ I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$
These are my steps in hopefully deriving the correct result:
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$
Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?
EDIT:
Given that:
$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$
because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.
hence, $\frac{\sin \theta}{\theta}$ approaches 1.
So the answer is $\frac{1}{2}$
|
You are almost done!
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=$$
$$=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta} \to 1\cdot \frac{1}{1+1} = \frac{1}{2}$$
Note (and prove) that indeed
$$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$$
|
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|
A travelled inequality found by discriminant
Given three real numbers $x, y, z$ so that $1\leq x, y, z\leq 8.$ Prove that
$$\sum\limits_{cyc}\frac{x}{y}\geq\sum\limits_{cyc}\frac{2x}{y+ z}$$
I found this inequality by discriminant, we realised the homogenous, the generality of this problem. We assume $y= 1$ that leads $x^{2}+ x^{3}- zx- zx^{4}- 3z^{2}x^{2}+ z^{2}x^{4}+ z^{3}+ z^{3}x^{3}+ z^{4}- z^{4}x\geq 0.$ By jp.Wolfram|Alpha_ https://ja.wolframalpha.com/input/?i=discriminant%5Bx%5E2%2Bx%5E3-zx-zx%5E4-3z%5E2x%5E2%2Bz%5E2x%5E4%2Bz%5E3%2Bz%5E3x%5E3%2Bz%5E4-z%5E4x%2Cx%5D%3C%3D0 _of course $constant= 8$ is not the best here I think the one way that we can deal it is using substitution in $x, y, z$ but inhomogenous. I need to the help, thanks a real lot
|
Eliminating denominators, we arrive to the inequality $\;f(x,y,z)\ge 0,\;$ where
$$\begin{align}
&f(x,y,z) = x^3 y^3 + y^3 z^3 + x^3 z^3 - 3 x^2 y^2 z^2
\\[4pt]&
+ x^2 y^4 + y^2 z^4 + z^2 x^4 - x^4 y z - x y^4 z - x y z^4.
\end{align}\tag1$$
Since $(1)$ is homogenius and has the rotational symmetry, then WLOG there are two possible cases:
*
*$\mathbf{1=x\le y \le z}.\;$ Assume $\;y=1+u,\; z=y+v,\;$ then
$$\begin{align}
&f(1,1+u,1+u+v)=u^4(v^2+v)+u^3(5v^3+11v^2+7v+2)
\\[4pt]&
+u^2(9v^4+30v^3+33v^2+15v+4)
+u(7v^5+30v^4+44v^3+25v^2+4v)
\\[4pt]&
+2v^6+10v^5+18v^4+14v^3+4v^2 \ge0,
\end{align}$$
and the case is proved.
*$\mathbf{1=z\le y\le x\le 1+p}.\;$ Assume $\;y=1+pv,\; x=1+p,\;$ then
$$\begin{align}
&g(p,v)=\dfrac1{p^2}f(1+p,1+pv,1) = (p^4+p^3)v^4+(p^4+7p^3+7p^2+2p)v^3
\\[4pt]&
+(3p^3+12p^2+9p+4)v^2+(-p^3-p^2+p-4)v+2p+4.
\end{align}\tag2$$
If $\;v=\dfrac1p,\;p\to\infty,\;$ then the given inequality does not satisfy.
If $\;\mathbf{p=7},\;$ then
$$g(7,v) = 18 -389v+1684v^2+5159v^3+2744v^4
$$$$
=(18+43v)(1-12v)^2 +13(3v-14v^2)^2 + 7v^2+59v^3+196v^4 \ge 0,$$
and the case is proved.
The greatest value of $\;p\;$ can be defined from the system $\;g'_v =g=0,\;$
with the numerical solution $\;p\approx7.07031\,16307,\; v\approx 0.08248\,18190.\;$
The best constant is 8.07031 16307.
|
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|
Faster way to calculate $\frac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$ What is the value of $\cfrac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$ ?
$$1)1+\sqrt3\quad\quad\quad\quad\quad\quad2)-1+\sqrt2\quad\quad\quad\quad\quad\quad3)1-\sqrt2\quad\quad\quad\quad\quad\quad4)\sqrt2-2\sqrt3$$
It was one of the questions from timed exam (for example supposed to solve any question in average one minute). but I have difficulty to calculate this expression fast. my approach is:
$$\left(\frac{2\sqrt2+3\sqrt3}{5-\sqrt6}\times\frac{5+\sqrt6}{5+\sqrt6}\right)-\frac2{\sqrt3-1}$$
By rationalizing the second fraction it is not hard to see it is $\sqrt3+1$. but I had difficulty to accurately calculate the first one in short time. Is it possible to evaluate it quicker?
|
$$\begin{align}
\frac{\sqrt{8}+\sqrt{27}}{5-\sqrt{6}}
&= \frac{(\sqrt{8} + \sqrt{27})(5 + \sqrt{6})}{25-6} \\
&= \frac{5 \sqrt{8} + \sqrt{48} + 5\sqrt{27} + \sqrt{81(2)}}{19} \\
&= \frac{10\sqrt{2} + 4\sqrt{3} + 15\sqrt{3} + 9\sqrt{2}}{19} \\
&= \sqrt{2} + \sqrt{3}.
\end{align}$$
|
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|
Comparison test proof of convergence I want to prove that $\sum_{n=1}^\infty\frac{3^n+4^n}{3^n+5^n}$ converges. To do this, I said that $\sum_{n=1}^\infty\frac{3^n+4^n}{3^n+5^n}<3\sum_{n=1}^\infty (\frac{4}{5})^n$, which indeed converges. Hence by the comparison test, the series we are given converges. Is this valid?
|
Hint:
$$\frac{3^n+4^n}{3^n+5^n} \le \frac{3^n+4^n}{5^n} = \left(\frac 35 \right)^n + \left(\frac 45 \right)^n \le 2\left(\frac 45 \right)^n $$
|
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|
Find the probability of placing $5$ dots on an $8 \times 8$ grid. $5$ dots are placed at random on an $8 \times 8$ grid such that no cell has more than $1$ dot.
What is the probability that no row or column has more than $1$ dot?
I thought about this in the following way, the number of ways to place $5$ dots on $64$ cells is $$\begin{pmatrix} 64 \\ 5 \end{pmatrix} $$
Now if I want the $8 \times 8$ to have no row or column with more than $1$ dot then I reasoned that the dots must be placed in the diagonal which has $8$ cells, so the number of ways I can place $5$ dots into $8$ cells is $$\begin{pmatrix} 8 \\ 5 \end{pmatrix} $$
So the desired probability would be $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
However I know that this is not the answer, the correct answer is $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
but I can't see what the term $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ counts, can you explain how this term came?
|
${64 \choose 5}$ looks OK for the denominator
For distinct rows, there are ${8 \choose 5}$ ways of choosing the five rows.
*
*The dot in the first occupied row can be in any of the $8$ columns
*The dot in the second occupied row can be in any of the $7$ remaining columns
*The dot in the third occupied row can be in any of the $6$ remaining columns
*The dot in the fourth occupied row can be in any of the $5$ remaining columns
*The dot in the fifth occupied row can be in any of the $4$ remaining columns
So I would get a probability of $$\dfrac{{8 \choose 5}\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{{64 \choose 5}}$$
|
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|
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
|
We have
\begin{align}
f(x, y) &= \sqrt{4y^{2}- 12y+ 10}+ \sqrt{18x^{2}- 18x+ 5} \\
&\qquad + \sqrt{18x^{2}+ 4y^{2}- 12xy+ 6x- 4y+ 1} \\[6pt]
&= \sqrt{\frac{ ( 2y- 6 )^{2} + ( 6y- 8 )^{2}}{10}} + \sqrt{\frac{( 6x- 5 )^{2}+ ( 12x- 5 )^{2}}{10}} \\[6pt]
&\qquad + \sqrt{\frac{ ( 6x+ 2y- 1 )^{2}+ ( 12x- 6y+ 3 )^{2}}{10}}\\
&\ge \frac{6- 2y}{\sqrt{10}}+ \frac{5- 6x}{\sqrt{10}}+ \frac{6x+ 2y- 1}{\sqrt{10}} \\[10pt]
&= \sqrt{10}.
\end{align}
|
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|
Find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$ I want to find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$. It seems that would simply be to calculate the following triple integral:
$\int_0^3 \int_0^{2x} \int_0^{1-y^2} z\,dz\,dy\,dx$
This is pretty straight-forward to do without any variable substitutions etc. which makes me think it's almost too simple (for a home assignment).
Am I missing something or is the above correct?
|
I ended up doing the following (omitting some algebra steps):
$1 - y^2 \geq 0$ when $y \leq 1$ which gives an integration interval $0 \leq y \leq 1$. By dividing the integral into two parts with the intervals $0 \leq x \leq \frac{1}{2}$ and $\frac{1}{2} \leq x \leq 3$ we get the two simpler integrals:
$V =
V_1 + V_2 =
\int_0^{1/2} \int_0^{2x} \int_0^{1-y^2} dz\,dy\,dx +
\int_{1/2}^3 \int_0^1 \int_0^{1-y^2} dz\,dy\,dx$
$V_1 = \int_0^{1/2} \int_0^{2x} \int_0^{1-y^2} dz\,dy\,dx =
\int_0^{1/2} \int_0^{2x} (1-y^2)\,dy\,dx =
\int_0^{1/2} \left(2x - \frac{8}{3}x^3\right) dx =
\frac{5}{24}$
$V_2 = \int_{1/2}^3 \int_0^1 \int_0^{1-y^2} dz\,dy\,dx =
\int_{1/2}^3 \int_0^1 (1-y^2) dy\,dx =
\int_{1/2}^3 \frac{2}{3} dx =
\frac{5}{3}$
$V = V_1 + V_2 = \frac{5}{24} + \frac{5}{3} = \frac{15}{8}$
Makes sense? Thank you for the comments.
|
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|
$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$ Let $a,b,c$ be real positive number, $ab+bc+ca=3$.
Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$
My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)(b+ca)(c+a)(c+ab)}$$
$$P \ge 3\sqrt[3]{\frac{8}{9}\cdot(a+b+c)(ab+bc+ca)(a+bc)(b+ca)(c+ab)}$$
$$P \ge3\sqrt[3]{\frac{8}{3}\cdot(a+b+c)(a+bc)(b+ca)(c+ab)} $$
I can't prove $(a+bc)(b+ca)(c+ab)\ge 8$
Could you help me ?
|
Your last inequlaity is equivalent to
$$
a^2b^2c^2+abc(a^2+b^2+c^2)+a^2b^2+b^2c^2+c^2a^2+abc\ge 8.
$$
Since
$$
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2-6~\text{and}~
\\
a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=9-2abc(a+b+c),
$$
the previous inequlaity is equivalent to
$$
a^2b^2c^2+abc(a+b+c)^2-6abc+9-2abc(a+b+c)+abc\ge 8,
$$
or
$$
a^2b^2c^2+abc((a+b+c)^2-2(a+b+c))+1\ge 5abc.
$$
Now note that $(a+b+c)^2\ge 3(ab+bc+ca)=9$, so $a+b+c\ge 3$ and hence,
$$
(a+b+c)^2-2(a+b+c)=(a+b+c-1)^2-1\ge 3.
$$
Thus, it's sufficient to prove that
$$
a^2b^2c^2+3abc+1\ge 5abc\Leftrightarrow (abc-1)^2\ge 0,
$$
which is true.
|
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|
Calculating the following integral $\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$ I'm trying to find the solution of the following integral, to check for convergence or divergence. Though, I'm finding the integral itself difficult to resolve. I welcome any feedback towards my approach.
$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$$
My approach:
$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7} \implies \frac{1}{2}\int^{\infty}_{2}\frac{2x-4}{x^3+4x+7}+\frac{1}{2}\int^{\infty}_{2}\frac{2}{x^3+4x+7}$$
Then to solve for the left-side first, I'm substiting u for $x^2-4x+7$, so the denominator looks like this: $xu$
Then to find x, such that: $x = \frac{u-7}{x-4}$
To get:
$$\frac{1}{2}\int^{\infty}_{2}\frac{du}{\frac{u-7}{x-4}\cdot u}$$
Though, I'm not sure if my apporach is valid, hence I would really enjoy some support towards my approach.
|
Write $x^3+4x+7=(x-a)((x-b)^2+c^2)$ with $a\approx-1.25538,\,b\approx0.627692,\,c\approx2.2764$. Then$$\begin{align}\frac{x-1}{x^3+4x+7}&\equiv\frac{A}{x-a}+\frac{B(x-b)+C}{(x-b)^2+c^2}\\\iff x-b+b-1&\equiv A((x-b)^2+c^2)+(x-b+b-a)(B(x-b)+C)\\&=(A+B)(x-b)^2+((b-a)B+C)(x-b)+Ac^2+(b-a)C\\\iff A=-B&=\frac{a-1}{(b-a)^2},\\C&=\frac{b-1}{b-a}.\end{align}$$One antiderivative is $A\ln\tfrac{|x-a|}{\sqrt{(x-b)^2+c^2}}+\tfrac{C}{c}\arctan\tfrac{x-b}{c}$, so the integral is $A\ln\tfrac{\sqrt{(2-b)^2+c^2}}{|2-a|}+\tfrac{C}{c}\arctan\tfrac{c}{2-b}$.
|
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Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.
Third attempt:comparing $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.
Any hint or advice of how I should think the problem was very useful.
|
We can rewrite $x^2+y^2+1 \geq xy+x+y \ $ as
$2x^2+2y^2+2 - 2xy - 2x - 2y \geq 0$
or as $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$
which holds for all $x, y \in \mathbb{R}$
Or start from $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$ and expand to show that $x^2+y^2+1 \geq xy+x+y \ $.
|
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|
If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$ If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$
But the actual answer is $2.$
Am I missing something ?
|
Let $x+1=y^2$ as $x\ge0, y^2\ge1$
$$\dfrac{4x^2+8x+13}{1+x}=\dfrac{4(y^2-1)^2+8(y^2-1)+13}{y^2}=4y^2+\dfrac9{y^2}=\left(2y-\dfrac3y\right)^2+2\cdot2y\cdot\dfrac3y\ge12$$ the equality occurs if $2y-\dfrac3y=0\iff y^2=\dfrac32\iff x=?$
|
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|
Prove $2^n > n^2 + n + 1$ I did this but I'm pretty sure it's wrong
$P(5) $ is true
Assume $P(k)$ is true
for some $k \ge 5$, ie. $2^k > k^2 + k + 1$
$$2^{k+1} = (k + 1)^2 + (k + 1) + 1$$
$$2^{k+1} = k^2 + 3k + 3$$
Then:
$$2^{k+1} > (k^2 + k + 1) \cdot 2 > k^2 + 3k + 3$$
$$2^{k+1} > k^2 > k + 1$$
From the onset we know that:
$$2^{k+1} > 2^k > k^2 + k + 1$$
Therefore:
$$2^{k+1} > 2^k$$ (which is true)
Therefore, by the PMI, $P(k)$ is true for all $k \ge 5$
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You started well but the end of your proof is not right and it is unclear when you write 2 inequalities.
We want to prove $2^{k+1}>(k+1)^2+(k+1)+1$ knowing already that $2^k>k^2+k+1$.
$2^{k+1}=2\times2^k>2\times(k^2+k+1)=2k^2+2k+2$ but we wanted to prove $2^{k+1}>(k+1)^2+(k+1)+1=k^2+3k+3$ so it is enough to prove that $2k^2+2k+2>k^2+3k+3$.
Which is the same as $k^2>k+1\iff k(k-1)-1>0$ which is always true for any $k\geq5$.
Then $2^{k+1}>2\times(k^2+k+1)>(k+1)^2+(k+1)+1$ and so the proof is done.
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"language": "en",
"url": "https://math.stackexchange.com/questions/4060214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Long Division of polynomials How can we prove that $x^n$ isn't divisible by $g(x)=x^4+x+1$ without remainders?
I understand why for all $n<4$ it's working, but how with $n \ge 4$? How we can prove it formally?
Note: $x^n,\:g\left(x\right)\in \mathbb{Z}_2\left[x\right]$
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Write $x^n = q_n(x) (x^4+x+1) + r_n(x)$.
Then $$x^{n+4} = x^4 q_n(x) (x^4+x+1) + x ^4 r_n(x) = x^4 q_n(x) (x^4+x+1) + (x+1) r_n(x)$$ and so $r_{n+4} \equiv (x+1) r_n \bmod (x^4+x+1)$.
The result follows by induction because $r_n = x^n \ne 0 $ for $n=0,1,2,3$ and $x+1$ is a unit mod $(x^4+x+1)$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4065341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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