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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-2=-7 \Leftrightarrow n = -5 $
*$n-2=-1 \Leftrightarrow n = 1 $
*$n-2=7 \Leftrightarrow n = 3 $
*$n-2=-1 \Leftrightarrow n = 9 $
So $S = \{ 1, 3, 9 \}$
I decompose $n+1$ the exact same way:
$$n+1 = 3n+11 - 2(n+5)$$
But I'd get stuck as $2(n+5)$ since only $-2$, $-1$, $1$ and $2$ are divisors, which don't satisify the equation as I'd hoped:
*
*$3n+11 = -2 \Leftrightarrow n =\frac{-13}{3}$
*$3n+11 = -1 \Leftrightarrow n = -4$
*$3n+11 = 1 \Leftrightarrow n =\frac{-10}{3}$
*$3n+11 = 1 \Leftrightarrow n = -3$
Any clues?
|
$n+1$ certainly divides $3n+3$. If it divides $3n+11$, it must also
divide $(3n+11)-(3n+3)$.
|
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|
How is this Taylor expansion computed? I am reading a paper, where we consider the following function,
$$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)$$
where $b>0$ is a constant such that $r>2/b.$ After stating this definition the author writes that the taylor expression of $F$ is,
$$F(r) = b\frac{r^2}{4}-\frac{\ln(r)}{b} -\frac{1}{2b} -\frac{\ln(2b)}{b} + \mathcal{O}(b^{-3}r^{-2}).$$
I know how to compute taylor series, but I am not sure what the taylor series is centered around. I have computed the derivatives upto order $2$.
$$F'(r) = \frac{1}{2}\sqrt{b^2r^2-4}, \quad F''(r) = \frac{b^2r}{2\sqrt{b^2r^2-4}}.$$
Any explanations regarding the expansion would be much appreciated.
|
Here you need the following expansions centered at $x=0$:
$$\sqrt{1+x}=1+\frac{x}{2}+O(x^2)\quad,\quad\ln(1+x)=x+O(x^2).$$
Hence, for $r$ large enough and $x=-\frac{4}{b^2r^2}$,
$$\begin{align*}F(r)&=\frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)\\
&=\frac{br^2}{4}\sqrt{1-\frac{4}{b^2r^2}}-\frac{1}{b}\ln(br(\sqrt{1-\frac{4}{b^2r^2}} + 1))\\
&=\frac{br^2}{4}\sqrt{1-\frac{4}{b^2r^2}}-\frac{\ln(br)}{b}-\frac{1}{b}\ln(\sqrt{1-\frac{4}{b^2r^2}}+1)\\
&=\frac{br^2}{4}\left(1-\frac{4}{2b^2r^2}\right)+O(b^{-3}r^{-2})-\frac{\ln(br)}{b}-\frac{1}{b}\ln(2-\frac{4}{2b^2r^2}+O(b^{-4}r^{-4}))\\
&=\frac{br^2}{4}-\frac{1}{2b}+O(b^{-3}r^{-2})-\frac{\ln(br)}{b}-\frac{\ln(2)}{b}-\frac{1}{b}\ln(1-\frac{1}{b^2r^2}+O(b^{-4}r^{-4}))\\
&=\frac{br^2}{4}-\frac{1}{2b}-\frac{\ln(2b)}{b}+\frac{\ln(r)}{b}+O(b^{-3}r^{-2})
\end{align*}$$
|
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Calculate $\frac{1}{2\pi i} \int_{\gamma} \frac{e^{iz}}{z^6(z^2+2)} dz$, where $\gamma$ is the circle with center 1/4 and radius 1/2 I need to calculate the following complex integral:
$$
\frac{1}{2\pi i} \int_{\gamma} \frac{e^{iz}}{z^6(z^2+2)} dz
$$
where $\gamma$ is the circle with center $1/4$ and radius $1/2$. It seems natural to use Cauchy's Integral Formula, since we can split the integrand into the two terms,
$$
\frac{e^{iz}}{z^6(z^2+2)}
=\frac{e^{iz}}{2\sqrt{2}iz^6(z-\sqrt{2}i)}
-\frac{e^{iz}}{2\sqrt{2}iz^6(z+\sqrt{2}i)}
$$
but before using the formula, somehow the $z^6$ in both denominators has to be dealt with. I've tried splitting up the terms once again, but without any succes so far. I've also tried to insert $e^{iz}=\sum \frac{(iz)^n}{n!}$ and interchange integral and sum, this removes $z^6$ from both denominators, but the resulting series, I think, cannot be calculated exactly.
Help would be appreciated.
|
Without residues you may proceed as follows:
The function $f(z) = \frac{e^{iz}}{z^2+2}$ is holomorph on your disc and contains $z= 0$.
So, Cauchy integral formula gives
$$f^{(5)}(0) = \frac{5!}{2\pi i }\oint_{\gamma}\frac{\frac{e^{iz}}{z^2+2}}{z^6}dz$$
Now, you know that $\frac{f^{(5)}(0)}{5!}$ is the coefficient of $z^5$ in the McLaurin-expansion of $f$
\begin{eqnarray*} [z^5]\frac{e^{iz}}{z^2+2}
& = & [z^5]\frac{1}{2}\frac{e^{iz}}{1 - \left(-\frac{z^2}{2}\right)} \\
& = & [z^5]\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{(iz)^n}{n!}\right)\left(\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{2^n}\right) \\
& = & \frac{1}{2}\left( \frac{i^5}{5!} + \frac{i^3}{3!}\left(-\frac{1}{2}\right)+\frac{i}{4}\right) \\
& = & \boxed{\frac{41}{240}i}
\end{eqnarray*}
|
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|
Real Analysis Inf and Sup question I am hung up on this question for real analysis ( intro to anaylsis ).
Find $\inf D$ and $\sup D$
$$\mathrm{D}=\left\{\frac{m+n\sqrt{2}}{m+n\sqrt{3}} :m,n\in\Bbb{N}\right\}$$
I have spent enough time staring at this thing that I know the $\sup D=1$ and $\inf D=\frac{\sqrt{2}}{\sqrt{3}}$.
for $\sup D$:
$$m+n\sqrt{2}<m+n\sqrt{3}\implies\frac{m+n\sqrt{2}}{m+n\sqrt{3}}<1$$
so $1$ is an upper bound for $D$, and then for the confirmation that 1 is the least upper bound I can prove by contradiction that $\sup D$ cannot be less than $1$, because I could always find a $d \in D$ such that $$\sup D<d<1$$, which is the contradiction since no $d \in D$ can be greater than $\sup D$.(proof omited)
So my problem is with $\inf D$. I am having trouble establishing that $\frac{\sqrt{2}}{\sqrt{3}}$ is a lower bound. I am just not seeing it. The intuition is that if $m$ is small and $n$ is large than the fraction $\frac{\sqrt{2}}{\sqrt{3}}$ dominates the expression, however it will always be slightly greater than $\frac{\sqrt{2}}{\sqrt{3}}$. Analytically I am just not able to show it.
Any help would be greatly appreciated
|
Well, $\frac {\sqrt 2}{\sqrt 3} < \frac{m+n\sqrt 2}{m+n\sqrt3} \iff$
$m\sqrt 2 + n\sqrt 6 < m\sqrt 3 + n\sqrt 6 \iff$
$m\sqrt 2 < m\sqrt 3$ which is always the case if $m > 0$.
.... so $\frac {\sqrt 2}{\sqrt 3}$ is a lower bound of D....
And $\frac{m+n\sqrt 2}{m+n\sqrt3} < \frac {\sqrt 2}{\sqrt 3} + \epsilon\iff$
$m\sqrt 3 + n\sqrt 6 < m\sqrt 2 + n\sqrt 6 + \sqrt 3\epsilon(m+n\sqrt3)\iff$
$m(\sqrt 3-\sqrt 2)< \sqrt3 \epsilon(m+n\sqrt 3)\iff$
$m\frac {\sqrt 3-\sqrt 2}{\sqrt 6\epsilon}-m < n$
If we set $m=1$ and $n>\frac {\sqrt 3-\sqrt 2}{\sqrt 6\epsilon}-1$ we can find this for any $\frac {\sqrt 3-\sqrt 2}3 > \epsilon > 0$.
So now, $\frac {\sqrt 2}{\sqrt 3} + \epsilon > \frac {\sqrt 2}{\sqrt 3}$ is not a lower bound.
So... that was a mess but... $\inf D =\frac {\sqrt 2}{\sqrt 3}$
=====
In hindsight: I should have just taken my advice and just done it.
$\frac {m+n\sqrt 2}{m+n\sqrt 3} -\frac {\sqrt 2}{\sqrt 3}=$
$\frac {\sqrt 3(m + n\sqrt 2) -\sqrt 2(m+n\sqrt 3)}{\sqrt 3(m+n\sqrt 2)} =$
$\frac {m(\sqrt 3-\sqrt 2)}{\sqrt 3(m+n\sqrt 2)}:= \Delta(m,n)$
So as $\sqrt 3 > \sqrt 2$ and all other terms are positive, $\Delta(m,n) > 0$. And so $\frac {\sqrt 2}{\sqrt 3}$ is a lower bound.
For any $\epsilon > 0$ we can ensure
$\Delta(m,n) =\frac {m(\sqrt 3-\sqrt 2)}{\sqrt 3(m+n\sqrt 2)}< \epsilon$
by fixing $m$ and letting $n> \frac m{\sqrt 2}(\frac {(\sqrt 3-\sqrt 2)}{\sqrt 3\epsilon}-1)$.
And so $\inf D = \frac {\sqrt 2}{\sqrt 3}$.
The does require that we select an $\epsilon$ so that $\frac {\sqrt 3-\sqrt 2}{\sqrt 3} > \epsilon > 0$ but we can, wolog, assume that.
|
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|
Existence of integers s.t. $Y^2 = X^3 + n^3 - 4m^2$ Given $m, n \in \mathbb Z$ with $n = -1 (mod 4)$. Let's say that $m$ has no prime divisors that are congruent to $3 (mod 4)$.
Why are there no integers $X, Y$ such that $Y^2 = X^3 + n^3 -4m^2$ ?
|
Assume for the sake of contradiction that a solution $(X,Y)$ exists. Write $$Y^2+(2m)^2=X^3+n^3=(X+n)(X^2-nX+n^2).$$ We first claim that any prime factor $p$ of $Y^2+(2m)^2$ satisfies $p=2$ or $p\equiv 1\pmod{4}$.
To show this, let $p\equiv 3\pmod{4}$ be a prime factor of $Y^2+(2m)^2$. Because $\left(\frac{-1}{p}\right)=-1$, we conclude that $p$ divides both $Y$ and $m$. But this contradicts the assumption that $m$ has no prime divisor congruent to $3$ modulo $4$.
If $Y^2+(2m)^2$ is even, then $Y$ is even, and therefore $X^3+n^3$ is even and divisible by $4$. Because $X^2-nX+n^2$ is always odd regardless of the parity of $X$, $4$ divides $X+n$. Because $n\equiv-1\pmod 4$, we get $X\equiv 1\pmod 4$ so $$X^2-nX+n^2\equiv 3\pmod{4}.$$ Since $X^2-nX+n^2>0$, $X^2-nX+n^2$ has a prime divisor congruent to $3$ mod $4$. This is a contradiction.
Assume now that $Y^2+(2m)^2$ is odd. Then $X$ must be even. Since $X+n>0$ is odd, we get $X+n\equiv 1\pmod{4}$. Thus, $X\equiv 2\pmod{4}$, but then
$$X^2-nX+n^2\equiv 3\pmod{4}.$$
This is another contradiction.
|
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|
Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$$
I'm eagerly interested in learning one method which assumes $c\not\equiv {\rm mid}(\!a, b, c\!)$. But if $c\equiv {\rm mid}(\!a, b, c\!)$:
$$2\sqrt{(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)(\!a+ b+ c\!)}\leqq c(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)+ \frac{a+ b+ c}{c}= \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}$$
We need to prove
$$\begin{align} \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}\leqq \frac{a+ b}{c}+ \frac{b+ c}{a} & + \frac{c+ a}{b}\Leftrightarrow 1+ \frac{c^{2}}{ab}\leqq \frac{c}{a}+ \frac{c}{b}\Leftrightarrow \\ & \Leftrightarrow (\frac{c}{a}- 1)(\frac{c}{b}- 1)\leqq 0\Leftrightarrow \frac{(c- a)(c- b)}{ab}\leqq 0 \end{align}$$
Who can teach me what would we do if $c\not\equiv {\rm mid}(\!a, b, c\!)$ ? I am goin' to set a bounty, thank u so much
|
As I have written, the idea of proof taking $c=mid(a,b,c)$ is just a matter of notation. Without loss of generality (WLOG), You can take the other mid one if $c$ is not the one. Note that there is always something in the middle. Even if $a=b=c$ you can take any of them.
|
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|
Solving $3\sin(2x+45^\circ)=2\cos(x+135^\circ)$ for $x$ between $0^\circ$ and $360^\circ$ Please find the value of $x$ in degree from this equation, with explanation
$$3\sin(2x+45^\circ)=2\cos(x+135^\circ)$$
For $x$ between $0^\circ$ and $360^\circ$.
|
Starting from S. Dolan's answer
$$6\sin x\cos x+6\cos^2 x-3=-2\cos x-2\sin x$$ let $c=\cos x$ ans $s=\sin x$ to make
$$6sc+6c^2-3=-2c-2s$$
Solving for $s$ gives $$s=\frac{-6 c^2-2 c+3}{2 (3 c+1)}$$, Square both sides and remember that $c^2+s^2=1$ to get
$$1-c^2=\left(\frac{-6 c^2-2 c+3}{2 (3 c+1)} \right)^2$$ Assuming $3c+1 \neq 0$, cross multiply, expand and simplify to get
$$72 c^4+48 c^3-64 c^2-36 c+5=0$$ which could be solved with (nasty) radicals (have a look here).
I wonder if there could be a typo in the problem since the solutions are quite ugly.
|
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|
Equilateral triangle ABC has side points M, D and E. Given AM = MB and $\angle$ DME $=60^o$, prove AD + BE = DE + $\frac{1}{2}$AB I would like to ask if someone could help me with solving the following probelm.
The triangle $ABC$ is equilateral. $M$ is the midpoint of $AB$. The points $D$ and $E$ are on the sides $CA$ and $CB$, respectively, such that $\angle DME=60^o$
Prove that $AD+BE = DE + \frac{1}{2}AB$.
Thank you in advance.
|
To simplify, assume all lengths relative to $|AM|$. In other words, $AM=1$
Let $x = CD, \quad y = CE,\quad z=DE$
Assume the statement to be proved is true, this follows:
$AD + BE = (2-x)+(2-y) = z + 1 \quad → x+y+z=3$
$ΔCDE:\; z^2 = x^2 + y^2 - x y$
$$(3-x-y)^2 = x^2+y^2-xy$$
$$x^2+y^2+2xy-6(x+y)+9 = x^2+y^2-xy$$
$$3xy - 6(x+y) + 9 = 0$$
$$xy - 2(x+y) + 3 = 0$$
Let $T = xy-2(x+y)+3$. If $T=0$, we proved $AD+BE=DE+{1\over2}AB$
Laws of Cosines:
$ΔADM:\; DM^2 = 1 + (2-x)^2 - (2-x) = x^2-3x+3$
$ΔBEM:\; EM^2 = 1 + (2-y)^2 - (2-y) = y^2-3y+3$
$ΔDEM:\; z^2 = x^2+y^2-xy = DM^2 + EM^2 - DM·EM$
$$DM·EM = xy-3(x+y)+6 = T+(3-(x+y))$$
$$(DM·EM)^2 = (T+(3-(x+y)))^2$$
$$T^2 + (x+y)T + z^2 = T^2+2(3-(x+y))T+(3T+z^2) $$
$$3(x+y-3)\;T = 0$$
All is left is to show $(x+y-3)$ is non-zero.
If $x+y=3$, we have $DM·EM = xy - 3(x+y) + 6 = xy-3$
Using AGM inequality $\large \sqrt{xy} ≤ {x+y \over 2} = 1.5$
This implied $DM·EM ≤ 1.5^2 - 3 < 0$, thus $x+y≠3$
|
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|
Differentiation using l´Hopital I need to use L´Hopital's rule with this functions:
$$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$$
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$$
I take the exponent down using the properties of logarithms and then make the denominator like: $\lim_{x\rightarrow\frac{\pi}{2}} \frac{\cos(x)}{\frac{1}{\ln(1-\sin(x))}}$ but I still get stuck.
|
Here is my take on the second and a remark on the first limit:
$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$:
So, consider
\begin{eqnarray*} \tan(2x)\ln(\tan x)
& \stackrel{\tan(2x)= \frac{2\tan x}{1-\tan^2x}}{=} & 2\frac{\tan x\cdot \ln(\tan x)}{1-\tan^2x}\\
& \stackrel{t = \tan x, t\to1}{=} & 2\frac{t \ln t}{1-t^2} \\
& \stackrel{L'Hosp.}{\sim} & 2\frac{\ln t + 1}{-2t}\\
& \stackrel{t\to 1}{\longrightarrow} & -1
\end{eqnarray*}
Hence, $\boxed{\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)} = e^{-1}}$.
$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$:
I recommend just to consider the logarithmic limit and not to write all time $e^{\lim ...}$. This is tedious and not necessary:
\begin{eqnarray*} \cos x \ln(1-\sin x)
& = & \frac{\ln(1-\sin x)}{\frac{1}{\cos x}}\\
& \stackrel{L'Hosp.}{\sim} & -\frac{\frac{\cos}{1- \sin x}}{-\frac{\sin x}{\cos^2 x}}\\
& = & \frac{\cos^3 x}{\sin x (1-\sin x)} \\
& = & \frac{\cos x (1-\sin^2 x)}{\sin x (1-\sin x)} \\
& \stackrel{\frac{1-\sin^2 x}{1-\sin x} = 1+\sin x}{=} & \frac{\cos x (1+\sin x)}{\sin x}\\
& \stackrel{x\to \frac{\pi}{2}}{\longrightarrow} & 0
\end{eqnarray*}
Hence, $\boxed{\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)} = e^0 = 1}$.
|
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|
Prove hard inequality Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$
I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can do. I think this inequality is too tight to use AM-GM or Cauchy Schwarz alone.
|
After full expanding we need to prove that:
$$\sum_{cyc}(4a^{5}b+4a^{5}c-12a^{4}b^{2}+12a^{4}c^{2}+5a^{3}b^{3}+8a^{4}bc-19a^{3}b^{2}c+5a^{3}c^{2}b-7a^{2}b^{2}c^{2})\geq0$$ or
$$6\sum_{cyc}ab(a^{2}-b^{2}-2ab+2ac)^{2}+
\sum_{sym}(2a^{5}b-a^{3}b^{3}-4a^{4}bc+10a^{3}b^{2}c-7a^{2}b^{2}c^{2})\geq0.$$
Thus, it's enough to prove that:
$$\sum_{cyc}(a^5b+a^5c-a^3b^3-4a^4bc+5a^3b^2c+5a^3c^2b-7a^2b^2c^2)\geq0.$$
Can you end it now?
|
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Prove that $(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$ is an integer and find its value I had proceed this question by taking
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Then
$$x + (5^{\frac 12} -2)^{\frac 13} =(5^{\frac 12} + 2)^{\frac 13}$$
And then cubing both sides and then solving for $x$ by using Cardano's method but I got the same equation at last that is
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Now I don't know how to solve this
|
Some naive bounding gives
$(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{\frac 13} < (\sqrt 5 + 2)^{\frac 13} < (3+2)^{\frac 13} < 8^\frac 13 = 2$.
It is also easy to show that this quantity is positive, so if it is an integer then it must be $1$.
Some Galois theory shows that if $(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{\frac 13}$ really is an integer then $(\sqrt 5 + 2)$ is a cube in the ring of integers of $\Bbb Q(\sqrt 5)$, so $(\sqrt 5 + 2) = ((a + b\sqrt 5)/2)^3$ where $a$ and $b$ are some integers with $a \equiv b \pmod 2$. Conversely, if this is the case then one can quickly show $(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{\frac 13} = (a + b\sqrt 5)/2 - (-a + b\sqrt 5)/2 = a$.
Since we know this can only be $1$, we must have $a=1$. Then writing $(1 + b\sqrt 5)^3 = 16 + 8\sqrt 5$, we get the equations $1+15b^2 = 16$ and $3b+5b^3 = 8$. The first one gives $b^2=1$ and then the second one gives $8b = 8$ so $b=1$, which is compatible with $b^2=1$, so everything works out well.
If you don't know Galois theory, assuming that the cube root of $\sqrt 5 +2$ has such a nice form sounds like a gamble, but it is a gamble that is secretly rigged in our favour.
|
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Solve Euler Project #9 only mathematically - Pythagorean triplet The "Euler Project" problem 9 (https://projecteuler.net/problem=9) asks to solve:
$a^2$ + $b^2$ = $c^2$
a + b + c = 1000
I find answers solving it with brute-force and programmatically, but is there a way to solve the problem ONLY mathematically? Can someone help, please?
Problem as explained in Project Euler website:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
|
Hint Euclid's parameterization of the Pythagorean triples (Elements, Book X, Proposition XXIX) is:
$$a = k (m^2 - n^2), \qquad b = 2 k m n, \qquad c = k (m^2 + n^2),$$
where $m > n > 0$ and $m, n$ coprime and not both odd.
Substituting in our condition gives
$$1000 = a + b + c = 2 k m (m + n),$$ and clearing the constant leaves $$\phantom{(\ast)} \qquad 500 = k m (m + n) . \qquad (\ast)$$
Now, notice that (1) $500 = 2^2 5^3$ has only two distinct prime factors, and (2) since $m$ and $n$ are coprime, so are $m$ and $m + n$.
So, one of $m, m + n$ must be one of $1, 2, 4$ (in fact one of $2, 4$, since $m > n > 0$ implies $m + n > m > 1$) and the other must be one of $1, 5, 25, 125$. Because $m + n > m$, we must have $m \in \{2, 4\}$, and so $m + n < 2 m \leq 8$. Thus, $m + n = 5$, and $2 m > m + n = 5$ implies $m \geq 3$, leaving $m = 4$ as the only possibility. So, $n = 1, k = 25$, and $$\color{#df0000}{\boxed{(a, b, c) = (375, 200, 425)}} .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How how do I find the measure of the sides of an equilateral triangle inscribed in a 30-60-90 triangle?
I am completely lost, I have no idea where to even start. I am sure someone here would be able to do this easily which is why I'm posting this here. Basically the point of the problem is to find the measure of line X. All that is given is that the two segments on the bottom are congruent and equal to 1 unit each which totals to 2 units for the base of the 30-60-90 triangle. If someone could tell me the answer and an explanation on how to get there it would be very much appreciated.
Just to clarify, the 30s and 60s are degrees, but I made this illustration and I was too lazy to add in the degree symbol. Also please let me know if there is not enough information to solve the problem, but I doubt that is the case because this was a challenge problem that my teacher gave me.
|
Let the side length of the equilateral triangle be $x$, and let the tilt angle $\theta$ be as shown. Then
$ x \cos \theta = 1 $
And the top right vertex of the equilateral triangle has coordinates
(Assuming the right angle vertex is the origin)
$P = (1 - x \cos(\theta + 60^\circ) , x \sin(\theta + 60^\circ) ) $
And this point lies on the line whose equation is $y = 2 \sqrt{3} - \sqrt{3} x $
Hence,
$ x \sin(\theta + 60^\circ) = 2 \sqrt{3} - \sqrt{3} (1 - x \cos(\theta+60^\circ)) $
Expanding, and multiplying through by $2$,
$ x ( \sin \theta + \sqrt{3} \cos \theta ) = 2 \sqrt{3} + \sqrt{3} x ( \cos \theta - \sqrt{3} \sin \theta ) $
Simplifying,
$ x \sin \theta = \dfrac{\sqrt{3}}{2}$
And since $x \cos \theta = 1 $, then $\tan \theta = \dfrac{\sqrt{3}}{2} $
And therefore, $\cos \theta = \dfrac{1}{\sqrt{ 1 + \dfrac{3}{4}}} = \dfrac{2}{\sqrt{7}} $
So that, $ x = \dfrac{\sqrt{7}}{2} $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int_0^1 \frac{2x-1}{1+x-x^2}\left(4\ln x\ln(1+x)-\ln^2(1+x)\right)dx$ The following problem was posted earlier this year by Cornel Ioan Valean:
Prove that
$$I=\int_0^1 \frac{2x-1}{1+x-x^2}\left(4\ln x\ln(1+x)-\ln^2(1+x)\right)dx=\frac{127}{20}\zeta(3)-\frac{8\pi^2}{5}\ln(\varphi)$$
My idea was to consider the following integral:
$$\mathcal J(a)=\int_0^1 \frac{2x-1}{1+x-x^2}\ln(a+x)\ln(1+x)dx$$
So $I=4\mathcal J(0)-\mathcal J(1)$. In order to evaluate $\mathcal J(a)$ I tried to apply Feynman's trick:
$$\mathcal J'(a)=\int_0^1 \frac{\ln(1+x)}{a+x}\frac{dx}{\varphi-x}-\int_0^1 \frac{\ln(1+x)}{a+x}\frac{dx}{\frac{1}{\varphi}+x},\quad \varphi=\frac{1+\sqrt 5}{2}$$
$$\small =\frac{1}{a+\varphi}\int_0^1 \frac{\ln(1+x)}{a+x}dx+\frac{1}{a+\varphi}\int_0^1 \frac{\ln(1+x)}{\varphi-x}dx+\frac{1}{a-\frac{1}{\varphi}}\int_0^1 \frac{\ln(1+x)}{a+x}dx-\frac{1}{a-\frac{1}{\varphi}}\int_0^1 \frac{\ln(1+x)}{\frac{1}{\varphi}+x}dx$$
But I gave up on this idea after I realised that:
$$\int_0^1 \frac{\ln(1+x)}{t+x}dx=\ln 2\ln \left(\frac{t+1}{t-1}\right)+\operatorname{Li}_2\left(\frac{2}{1-t}\right)-\operatorname{Li}_2\left(\frac{1}{1-t}\right)$$
Where $\operatorname{Li}_2(x)$ is the Dilogarithm.
Other methods weren't promising either, such as the substitution $x=\frac{1-t}{1+t}$, to combine the integral with it's sister one that has the denominator $1-x+x^2$, or to integrate by parts which gives:
$$\small 2I=2\int_0^1 \frac{\ln(1+x-x^2)\ln x}{1+x}dx+2\int_0^1 \frac{\ln(1+x-x^2)\ln(1+x)}{x}dx-\int_0^1 \frac{\ln(1+x-x^2)\ln (1+x)}{1+x}dx$$
I believe that the factor of $4$ plays a big role into obtain nicely this result and one shouldn't split the integral into two parts, but I had no success and I would appreciate some help.
|
We start by writing
\begin{align}
I &= \int \limits_0^1 \frac{1-2x}{1+x -x^2} \log(1+x) [\log(1+x)-4\log(x)] \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \left[\log^2\left(\frac{x^2}{1+x}\right) - 4 \log^2(x)\right] \mathrm{d} x \equiv J -4K \, .
\end{align}
Then
\begin{align}
J &= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \log^2\left(\frac{x^2}{1+x}\right) \, \mathrm{d} x = \int \limits_0^1 \left[\frac{1}{1+x} - \frac{x (2+x)}{(1+x)(1+x-x^2)}\right] \log^2\left(\frac{x^2}{1+x}\right) \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{\log^2\left[\frac{\left(\frac{x}{1+x}\right)^2}{1-\frac{x}{1+x}}\right]}{1 - \frac{x}{1+x}} \, \mathrm{d} \frac{x}{1+x} - \int \limits_0^1 \frac{\log^2\left(\frac{x^2}{1+x}\right)}{1 - \frac{x^2}{1+x}} \, \mathrm{d} \frac{x^2}{1+x} = \int \limits_0^{1/2} \frac{\log^2\left(\frac{t^2}{1-t}\right) - \log^2(t)}{1-t} \, \mathrm{d} t \\
&= \int \limits_0^{1/2} \frac{3 \log^2(t) + \log^2(1-t) - 4 \log(t) \log(1-t)}{1-t} \, \mathrm{d} t \\
&= \left[6 \operatorname{Li}_3(t) - 6 \log(t) \operatorname{Li}_2(t)-3\log(1-t)\log^2(t) \vphantom{\frac{1}{3}}\right. \\
&\phantom{= \left[\vphantom{\frac{1}{3}}\right.}\left.-\frac{1}{3}\log^3(1-t) + 4 \operatorname{Li}_3(1-t) - 4 \log(1-t) \operatorname{Li}_2(1-t) \right]_{t=0}^{t=1/2}\\
&= \frac{19}{4} \zeta(3)
\end{align}
follows from the known polylogarithm values $\operatorname{Li}_2(1),\operatorname{Li}_3(1),\operatorname{Li}_2(1/2),\operatorname{Li}_3(1/2)$ and we find
\begin{align}
K &= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \log^2(x) \, \mathrm{d} x = \int \limits_0^1 \left[\frac{\log^2(x)}{\varphi^{-1} + x} - \frac{\log^2(x)}{\varphi - x}\right] \mathrm{d} x \\
&= 2 \left[- \operatorname{Li}_3(-\varphi) -\operatorname{Li}_3 (\varphi^{-1})\right] = 2\left[- \operatorname{Li}_3(-\varphi) + \operatorname{Li}_3 (-\varphi^{-1}) - \frac{1}{4} \operatorname{Li}_3 (\varphi^{-2})\right] \\
&= 2 \left[\frac{1}{6} \log^3(\varphi) + \frac{\pi^2}{6} \log(\varphi) - \frac{1}{4} \left(\frac{4}{5} \zeta(3) +\frac{2}{3} \log^3(\varphi) - \frac{2\pi^2}{15} \log(\varphi)\right)\right] \\
&= \frac{2}{5} \left[\pi^2 \log(\varphi) - \zeta(3)\right] \, .
\end{align}
using trilogarithm functional equations and the known value of $\operatorname{Li}_3(\varphi^{-2})$ (from here).
Therefore,
$$ I = J - 4 K = \frac{19}{4} \zeta(3) - \frac{8}{5} \left[\pi^2 \log(\varphi) - \zeta(3)\right] = \frac{127}{20} \zeta(3) - \frac{8}{5} \pi^2 \log(\varphi) \, .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much.
|
We have
$$
3^0 \equiv +1 \bmod 4
\\
3^1 \equiv -1 \bmod 4
\\
3^2 \equiv +1 \bmod 4
\\
3^3 \equiv -1 \bmod 4
\\
$$
Therefore, $3^{3n}+3^{2n}+3^{n} + 1 \equiv 1^n +(-1)^n + 1^n + (-1)^n \equiv 0 \bmod 4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Logarithmic equation, some variables in bases and in arguments This is the exercise, there are no clues in the book about it.
$$
40\log_{4x}x^\frac{1}{2}-14\log_{16x}x^3=-\log_{\frac{1}{2}x}x^2
$$
Solutions given by the book: $x=1; x=4; x=\frac{\sqrt{2}}{2}.$
And this is what I did so far:
*
*conditions for existence:
$$
\Biggl\{
\begin{eqnarray}
x &\gt& 0.\\
x &\ne& \frac 14; x \ne \frac{1}{16}; x \ne 2.
\end{eqnarray}
$$
*I simplified some exponent in the arguments:
$$
20\log_{4x}x-42\log_{16x}x+2\log_{\frac{1}{2}x}x=0
$$
*Change of bases:
$$
\frac{20}{\log_x4x}-\frac{42}{\log_x16x}+\frac{2}{\log_x\frac{1}{2}x}=0
$$
*Observing that: $\log_xnx=1+\log_xn$
*least common multiple:
$$
\frac{20(1+\log_x16)(1+\log_x\frac12)-42(1+\log_x4)(1+\log_x\frac12)+2(1+\log_x4)(1+\log_x16)}{(1+\log_x16)(1+\log_x4)(1+\log_x\frac12)}=0
$$
*denominator can be toggled, as for conditions for existence
*then I lost confidence in what I was doing...
Any clue is welcome, thanks :]
|
We can use that
$$\log_a^b=\frac{\log a}{\log b}$$
therefore
$$40\log_{4x}x^\frac{1}{2}-14\log_{16x}x^3=-\log_{\frac{1}{2}x}x^2$$
$$40\frac{\log x^\frac{1}{2}}{\log {4x}}-14\frac{\log x^3}{\log {16x}}=-\frac{\log x^2}{\log {\frac{1}{2}x}}$$
$$20\frac{\log x}{\log {x}+\log 4}-42\frac{\log x}{\log {x}+\log 16}=-2\frac{\log x}{\log {x}+\log \frac12}$$
and eliminating $\log x \neq 0$ (which is a solution)
$$\frac{10}{\log {x}+2\log 2}-\frac{21}{\log {x}+4\log 2}=-\frac{1}{\log {x}-\log 2}$$
then let $y=\log x$ and $a=\log 2$ to obtain
$$\frac{10}{y+2a}-\frac{21}{y+4a}+\frac{1}{y-a}=0$$
$$\frac{10(y+4a)(y-a)-21(y+2a)(y-a)+(y+2a)(y+4a)}{(y+2a)(y+4a)(y-a)}=0$$
$$\frac{10(y^2+3ay-4a^2)-21(y^2+ay-2a^2)+(y^2+6ay+8a^2)}{(y+2a)(y+4a)(y-a)}=0$$
$$\frac{-10 y^2+15 ay+10a^2}{(y+2a)(y+4a)(y-a)}=0$$
$$\frac{-5(y-2a)(2y+a)}{(y+2a)(y+4a)(y-a)}=0$$
that is
*
*$\log x = 2\log 2$
*$2\log x = -\log 2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of $a_n := \frac{5^n}{2^{n^2}}$ Consider the sequence $(a_n)$ defined by $a_n := \frac{5^n}{2^{n^2}}$.
1. Prove that the sequence $(a_n)$ is bounded below by $0$.
We note that $a_n > 0$ for $n\geq 0$. Thus, the sequence is bounded from below.
2. Prove that the sequence $(a_n)$ is strictly decreasing by showing that $a_{n+1}-a_n < 0$ for all $n\in \mathbb{N}$.
We look to $a_n = \frac{5^n}{2^{n^2}}$ and $a_{n+1} = \frac{5^{n+1}}{2^{(n+1)^2}}$. For $n\geq 1$ we see that $a_n > a_{n+1}$. Therefore, we have a strictly decreasing sequence.
3. Deduce that the sequence $(a_n)$ converges and calculate its limit.
Since we have a (monotonically) decreasing sequence which is bounded below, by the monotone convergence theorem this sequence converges. How do we find the limit? Is it the squeeze theorem? Thank you for the help!!!
|
For the first question: The are many ways to show this but we will use induction. Well $a_1 = \frac{5}{2} > 0$. Now suppose that $a_n > 0$ for some $n$. Then
\begin{equation*}
a_{n+1} = \frac{5^{n+1}}{2^{(n+1)^2}} = \frac{5^{n+1}}{2^{n^2+2n+1}} = \frac{5}{2^{2n+1}}\cdot \frac{5^n}{2^{n^2}} = \frac{5}{2^{2n+1}}\cdot a_n.
\end{equation*}
Since $a_n > 0$ and $\frac{5}{2^{2n+1}} > 0$, we see that $a_{n+1} > 0$. Hence, $a_n > 0$ as required.
For the second question: We have
\begin{equation*}
a_{n+1}-a_n = \frac{5}{2^{2n+1}}\cdot a_n-a_n = \left(\frac{5}{2^{2n+1}}-1\right)a_n.
\end{equation*}
Now $a_n > 0$ so
\begin{equation*}
a_{n+1}-a_n < 0 \Longleftrightarrow \frac{5}{2^{2n+1}}-1 < 0 \Longleftrightarrow 5 < 2^{2n+1},
\end{equation*}
which is easy to prove by induction: For $n = 1$, $2^{2\times 1+1} = 2^3 = 8 > 5$. Now if $2^{2n+1} > 5$, then
\begin{equation*}
2^{2(n+1)+1} = 2^{2n+3} = 2^2\cdot 2^{2n+1} = 4\cdot 2^{2n+1} > 4\times 5 = 20 > 5.
\end{equation*}
Thus, the sequence $(a_n)$ is strictly decreasing as required.
For the final question: Since we have a (monotonically) decreasing sequence which is bounded below, by the monotone convergence theorem $(a_n)$ converges, say to $\ell$. Then because $a_{n+1} = \frac{5}{2^{2n+1}}\cdot a_n$ we get
\begin{equation*}
\begin{split}
\ell = \left(\lim_{n\to\infty} \frac{5}{2^{2n+1}}\right)\cdot \ell = 0\cdot \ell = 0
\end{split}
\end{equation*}
because $\lim_{n\to\infty} 2^{2n+1} = \infty$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Markov Chain: Calculating Expectation Reach a Certain Set of States Suppose I have a Markov chain $Z_k$ with $6$ states, as depicted below:
The probability of moving from one node to a neighboring node is $1/2$. For example, the probability of moving from node $1$ to node $2$ is $1/2$ and the probability of moving from node $1$ to node $6$ is $1/2$ etc.
Suppose $P(Z_0=1)=1$. That is we start state $1$. We need to compute two things:
*
*Compute the expected time we first reach the bottom of our pyramid (states $3$, $4$, or $5$). That is compute $E[T_B]$ where $T_B=min\{j: Z_j \in \{3,4,5\}\}$
My attempt: I try listing out all the possibilities
I can go from:
*
*I can go from State $1-2-3$. Time is $2$ when base is reached and probability $\frac{1}{2} \cdot \frac{1}{2}$
*$1-2-1-2-3$. Time is $4$ when base is reached. Probability of occurring is $\frac{1}{16}$.
*$1-2-1-2-1-2-3$. Time is $6$ when base is reached. Probability of occurring is $\frac{1}{64}$
*$1-2-1-2-1-2-1-2-3$. Time is $8$ when base is reached. Probability of occurring is $\frac{1}{64}$.
etc...
Thus my conclusion for these types of sequences expected value is:
$2 \cdot \frac{1}{4}+ 4*\frac{1}{16}+6*\frac{1}{64}+8*\frac{1}{256} ...$
$\sum_{k=1}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{8}{9}$
*Now, I can also go from State $1-6-5$. Time is $2$ when base is reached and probability $\frac{1}{2} \cdot \frac{1}{2}$
*$1-6-1-6-5$. Time is $4$ when base is reached. Probability of occurring is $\frac{1}{16}$.
*$1-6-1-6-1-6-5$. Time is $6$ when base is reached. Probability of occurring is $\frac{1}{64}$
*$1-6-1-6-1-6-1-6-5$. Time is $8$ when base is reached. Probability of occurring is $\frac{1}{64}$.
Thus my conclusion for these types of sequences expected value is:
$2 \cdot \frac{1}{4}+ 4*\frac{1}{16}+6*\frac{1}{64}+8*\frac{1}{256} ...$
$\sum_{k=1}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{8}{9}$
But there are more possibilities:
*$1-2-1-6-5.$ The time is $4$ probability $1/16$
*$1-2-1-6-1-6-5$. The time is $6$ probability $1/64$
10.$1-2-1-6-1-6-1-6-5 etc..$. The time is $8$ with probability $1/256$.
$\sum_{k=2}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{7}{18}$
Still more possibilities....:
Too many possibilities (unfortunately gave up) as its like the Markov Chain restarts when we go back to $1$. Couldn't figure it out. Please let me know what I should do and thank you for the help
|
Taking two steps in the Markov chain can lead to one of two things, with equal probability:
*
*$1 \to 2 \to 3$ or $1 \to 6 \to 5$ and we're done.
*$1 \to 2 \to 1$ or $1 \to 6 \to 1$ and we're back where we started.
We took $2$ steps. In the first case, we have $0$ steps left, and in the second case, we have $\mathbb E[T_B]$ more steps left in expectation. Therefore
$$
\mathbb E[T_B] = 2 + \frac12 (0 + \mathbb E[T_B])
$$
and, solving, $\mathbb E[T_B] = 4$.
|
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|
Prove this formula for the $\sin\left(\frac{x}{2^n}\right), x \in [0,\frac{\pi}{2}[, n \in \Bbb{N}$ The formula in question:
$$\sin\left(\frac{x}{2^n}\right) = \sqrt{a_1-\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\dots+\sqrt{a_{n-1}+\sqrt{\frac{a_{n-1}}{2}\left(1-\sin^2(x)\right)}}}}}}$$
where
$$a_k = \frac{1}{2^{2^k-1}} \quad \forall k \in \{1,2,\dots,n-1\}, \ n \in \Bbb{N}, \ x \in \left[0,\frac{\pi}{2}\right[$$
and only the first sign (after $a_1$) is $-$, the rest is $+$.
If this holds, this is a great way to calculate the $\sin$ of small angles, specifically ones that are a power of $\frac{1}{2}$ radians.
My attempt:
I have derived at this by continously using the formula
$$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{1-\sqrt{1-\sin^2(x)}}{2}} = \sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{\sin^2(x)}{4}}}$$
Which holds true, since
$$
\begin{align}
\sin(2x) &= 2\sin(x)\cos(x) \\
\sin(x) &= 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) \\
\sin(x) &= 2\sin\left(\frac{x}{2}\right)\sqrt{1-\sin^2\left(\frac{x}{2}\right)} \\
\sin^2(x) &= 4\sin^2\left(\frac{x}{2}\right)\left(1-\sin^2\left(\frac{x}{2}\right)\right) \ \left(\text{if } x \in \left[0,\frac{\pi}{2}\right[\right) \\
\sin^2(x) &= 4\sin^2\left(\frac{x}{2}\right)-4\sin^4\left(\frac{x}{2}\right) \\
0 &= 4\sin^4\left(\frac{x}{2}\right)-4\sin^2\left(\frac{x}{2}\right)+\sin^2(x) \\
\sin^2\left(\frac{x}{2}\right)_{1,2} &= \frac{4 \pm \sqrt{16-16\sin^2(x)}}{8} = \frac{1 \pm \sqrt{1-\sin^2(x)}}{2}
\end{align}
$$
And this holds true with $-$, since:
$$
\begin{align}
\sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} \frac{1-\sqrt{1-\sin^2(x)}}{2} \\
2\sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} 1-\cos(x) \\
2\sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} 1-\cos^2\left(\frac{x}{2}\right)+\sin^2\left(\frac{x}{2}\right) \\
\sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} 1-\cos^2\left(\frac{x}{2}\right)
\end{align}
$$
Yes, since
$$
\sin^2\left(\frac{x}{2}\right) + \cos^2\left(\frac{x}{2}\right) = 1
$$
So we found that if $x \in \left[0,\frac{\pi}{2}\right[$, then
$$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{1-\sqrt{1-\sin^2(x)}}{2}}$$
Now I plugged the formula into itself a couple times, and guessed what it would look like if I had plugged it in $n$ times. However, I have only assumed the above values of $a_k$ are true by looking at the results, so I'd like a rigorous proof of the formula.
I tried induction by $n$, but I couldn't figure out the $n \rightarrow n+1$ step.
Question:
Provide a proof of the first formula or correct it if it's wrong.
|
It's a bit tedious, but nonetheless doable:
Since we have
\begin{align}\sin\left(\frac{x}{{2^{n+1}}}\right)&= \sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{\sin^2\left(\frac{x}{{2^{n}}}\right)}{4}}}
\end{align}
We can use the induction hypothesis to calculate
\begin{align*}\frac{1}{4}\cdot\sin^2\left(\frac{x}{{2^{n}}}\right)&=\frac{1}{4}\cdot \left(a_1-\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\dots+\sqrt{a_{n-1}+\sqrt{\frac{a_{n-1}}{2}\left(1-\sin^2(x)\right)}}}}}\right)\\
&=\frac{a_1}{2^2}-\sqrt{\frac{a_2}{(2^{2})^2}+\sqrt{\frac{a_3}{((2^{2})^2)^2}+\sqrt{\frac{a_4}{2^{2^4}}+\dots+\sqrt{\frac{a_{n-1}}{2^{2^{n-1}}}+\sqrt{\frac{a_{n}}{2^{2^n}\cdot 2}\left(1-\sin^2(x)\right)}}}}}\\
&=a_2-\sqrt{a_3+\sqrt{a_4+\sqrt{a_5+\dots+\sqrt{a_{n}+\sqrt{\frac{a_{n+1}}{2}\left(1-\sin^2(x)\right)}}}}}
\end{align*}
where we used that
$$\frac{a_n}{2^{2^n}}=\frac{1}{2^{2^n-1}\cdot 2^{2^n}}=\frac{1}{2^{2\cdot 2^{n}-1}}=\frac{1}{2^{2^{n+1}-1}}=a_{n+1}$$
And since
$$\color{red}{a_1=\frac{1}{2^{2^1-1}}=\frac{1}{2}}$$
and $a_2=\frac{1}{2^{2^2-1}}=\frac{1}{8}$ so that
$$\color{blue}{\frac{1}{4}-a_2=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}=a_2}$$
you get
\begin{align}\sin\left(\frac{x}{{2^{n+1}}}\right)&= \sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{\sin^2\left(\frac{x}{{2^{n}}}\right)}{4}}}\\
&= \sqrt{\color{red}{\frac{1}{2}}-\sqrt{\color{blue}{\frac{1}{4}-a_2}+\sqrt{a_3+\sqrt{a_4+\sqrt{a_5+\dots+\sqrt{a_{n}+\sqrt{\frac{a_{n+1}}{2}\left(1-\sin^2(x)\right)}}}}}}}\\
&= \sqrt{\color{red}{a_1}-\sqrt{\color{blue}{a_2}+\sqrt{a_3+\sqrt{a_4+\sqrt{a_5+\dots+\sqrt{a_{n}+\sqrt{\frac{a_{n+1}}{2}\left(1-\sin^2(x)\right)}}}}}}}
\end{align}
|
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|
Evaluate $\sum _{k=0}^{\infty } \frac{L_{2 k+1}}{(2 k+1)^2 \binom{2 k}{k}}$ How to prove
$$\sum _{k=0}^{\infty } \frac{L_{2 k+1}}{(2 k+1)^2 \binom{2 k}{k}}=\frac{8}{5} \left(C-\frac{1}{8} \pi \log \left(\frac{\sqrt{50-22 \sqrt{5}}+10}{10-\sqrt{50-22 \sqrt{5}}}\right)\right)$$
Where $L_k$ denotes Lucas number and $C$ Catalan? Any help will be appreciated.
|
This is more of an extended comment at this point until I link to the final result, but numerically the above relation is plausible.
There are two observations to make to take the main step forward:
$$L_n = \phi^n + (1-\phi)^n $$
where $\phi = (\sqrt{5}+1)/2$ is the golden ratio, and
$$\sum_{k=0}^{\infty} \frac{x^{2 k+1}}{(2 k+1) \binom{2 k}{k}} = \frac{2 \arcsin{(x/2)}}{\sqrt{1-(x/2)^2}} $$
Then after some manipulation, one may show that the above sum is equal to the following definite integral:
$$2 \int_{\arcsin{(\phi-1)/2}}^{\arcsin{(\phi/2)}} d\theta \, \theta \, \csc{\theta} $$
Note that the lower limit is equal to $\pi/10$ and the upper limit is $3 \pi/10$. This may be integrated by parts and manipulated a little further to produce
$$\frac{\pi}{5} \log{\left (\frac{\tan^3{\left ( \frac{3 \pi}{20} \right )}}{\tan{\left ( \frac{\pi}{20} \right )}} \right )} - 4 \int_{\tan{(\pi/20)}}^{\tan{(3 \pi/20)}} du \frac{\log{u}}{1+u^2}$$
What remains for us to prove is that
$$\int_{\tan{(\pi/20)}}^{\tan{(3 \pi/20)}} du \frac{\log{u}}{1+u^2} = -\frac25 C$$
which amazingly checks out numerically, and that the ratio
$$\frac{\tan^3{\left ( \frac{3 \pi}{20} \right )}}{\tan{\left ( \frac{\pi}{20} \right )}} = \frac{10-\sqrt{50-22 \sqrt{5}}}{10+\sqrt{50-22 \sqrt{5}}}$$
which also checks out numerically (and should be simple algebra).
|
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|
Isocline in dynamical system I solved differential equation and now should draw graph. So, we have:$$y'=\sqrt{3+y^2}$$ $$x'=x^2+x$$
I get isocline $x=-1$ (from $x'=0$). Is it correct or maybe here is more?
|
Solving both equations we get the following:
$$y = \sqrt{3}\sinh(t+C)$$
$$ t + C = \int \frac{1}{x^2+x}dx = \int \frac{\frac{1}{x^2}}{1+\frac{1}{x}}dx = -\log\left(1+\frac{1}{x}\right)$$
Then plugging in, we get these nice graphs which we can plot:
$$y = \frac{\sqrt{3}}{2}\left(\frac{C}{1+\frac{1}{x}} - C^{-1}\left(1+\frac{1}{x}\right)\right)$$
where this $C$ must be positive. Or equivalently,
$$ y = \frac{\sqrt{3}}{2}\left(\frac{Cx^2-C^{-1}(1+x)^2}{x^2+x}\right)$$
Of course, this is not valid when $x=0$ or $x=-1$. Those are asymptotes of this family of functions.
|
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|
Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$
This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.
Any way, I tried hard on this one but no success yet. here is what I did:
Using the identity
$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$
$$H_{n/2}=H_n-2n\int_0^1 x^{2n-1}\ln(1+x^2)\ dx$$
We can write
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$
where
\begin{align}
\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\
&=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\
&=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\
&=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)
\end{align}
Therefore
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$
The sum can be done using the following identity
$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$
Differentiate both sides with respect to $a$ then set $a=1/2$ we get
$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$
and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks
|
I asked Cornel for a solution to the nice key result from Ahmad Bow's solution. Here is a solution in large steps.
We need two known results, that is $\displaystyle \int_0^1\frac{x^n}{1+x}\textrm{d}x=H_{n/2}-H_n+\log(2)$ and $\displaystyle \sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1$. If replacing $p$ by $i \sqrt{p}$ in the last series, make rearrangements and reindexing the series, we obtain$\displaystyle \sum _{n=0}^{\infty } (-1)^n y^n \cos ((2 n+1) x)=\frac{(1+y) \cos (x)}{1+2y \cos (2 x)+y^2}$.
Then, we have
\begin{equation*}
\sum_{n=0}^\infty (-1)^n\left(H_{n/2}-H_n+\ln2\right)\cos((2n+1)x)=\int_0^1\frac{1}{1+y}\sum_{n=0}^\infty (-1)^n y^n\cos((2n+1)x)\textrm{d}y
\end{equation*}
\begin{equation*}
=\cos (x)\int_0^1 \frac{1}{1+2y \cos (2 x)+y^2}\textrm{d}y=\cos (x)\int_{\cos(2x)}^{1+\cos(2x)} \frac{1}{t^2+\sin^2(2x)}\textrm{d}t=\frac{x}{2\sin(x)}.
\end{equation*}
End of story.
|
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|
Eigenvalues of almost symmetric matrix after discretizing a PDE are negative
Let $h > 0$. Define, for every positive integer $n$, $a_n = a + n \cdot h,$
and $a_{n+1/2} = a + \left( n + \frac{1}{2} \right) \cdot h, $where $a \in (0,1)$. How can we prove that the eigenvalues of the matrix $$A_{n+1} = \begin{pmatrix} -\frac{a_{1+1/2}+a_{0+1/2}}{a_1} & \frac{a_{1+1/2}}{a_1} & 0 & 0 & \cdots & 0 & 0 &0 \\
\frac{a_{1+1/2}}{a_2} & -\frac{a_{2+1/2} + a_{1+1/2}}{a_2} & \frac{a_{2+1/2}}{a_2} &0 & \cdots & 0 &0 &0 \\
\vdots & \vdots & \vdots &\vdots & \cdots &\vdots & \vdots & \vdots \\
0 & 0& 0 & 0& \cdots & \frac{a_{n-1 + 1/2}}{a_n} & -\frac{a_{n+1/2} + a_{n-1 + 1/2}}{a_n} & \frac{a_{n+1/2}}{a_n} \\
0 & 0& 0& 0& \cdots & 0 & \frac{a_{n+1/2} + a_{n+1+1/2}}{a_{n+1}} & -\frac{a_{n+1/2} + a_{n+1+1/2}}{a_{n+1}} \end{pmatrix}$$ are real and non-positive?
From each row $i$ from the above matrix, we can multiply that row with $a_i$ and we get a similar matrix with $A$ (so has the same eigenvalues), so we can forget about the denominator of each term. However, the matrix is not symmetric, but almost as the last two rows on the matrix do not agree on 2 entries. How can we get past this and prove the desired result?
This comes from discretizing the $y$-part of the partial differential equation (with second-order central difference formula) $$y \cdot \frac{\partial f}{\partial x}(x,y) = \frac{\partial }{\partial y} \left(y \cdot \frac{\partial f}{\partial y}(x,y) \right), $$ with conditions: $y \in (a,1+a), x
\in \mathbb{R}, f(x,a) = 1, \forall x \in \mathbb{R}, \frac{\partial f}{\partial y}(x,1+a) = 0, \forall x \in \mathbb{R}.$
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Let $\{e_1,e_2,\ldots,e_{n+1}\}$ be the standard basis of $\mathbb R^{n+1}$. Then $A=DP$ where $D=-\operatorname{diag}\left(\frac{1}{a_1},\ldots,\frac{1}{a_{n+1}}\right)$ is a negative diagonal matrix and
$$
P=a_{0+\frac12}e_1e_1^T+\sum_{i=1}^na_{i+\frac12}(e_i-e_{i+1})(e_i-e_{i+1})^T+a_{n+1+\frac12}e_{n+1}e_{n+1}^T.
$$
$P$ is positive semidefinite because it is a positively weighted sum of rank-$1$ positive semidefinite matrices. It is also nonsingular because
\begin{aligned}
&Px=0\\
&\ \Rightarrow\ x^TPx=0\\
&\ \Rightarrow\ x^Te_1e_1^Tx=x^T(e_1-e_2)(e_1-e_2)^Tx=\cdots=x^T(e_n-e_{n+1})(e_n-e_{n+1})^Tx=x^Te_{n+1}e_{n+1}^Tx=0\\
&\ \Rightarrow\ e_1^Tx=(e_1-e_2)^Tx=\cdots=(e_n-e_{n+1})^Tx=e_{n+1}^Tx=0\\
&\ \Rightarrow\ x=0.
\end{aligned}
Therefore $P$ is positive definite. Since $A$ is the product of a negative definite matrix and a positive definite matrix, it has a negative spectrum.
|
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|
Prove that $\left(\dfrac{1}{n}\sum_{i=1}^n x_i^5\right)^{1/5}\geq \dfrac{1}{n}\sum_{i=1}^n x_i$
Show that if $x_1,x_2,...,x_n$ are positive numbers then $\left(\dfrac{1}{n}\sum_{i=1}^n x_i^5\right)^{1/5}\geq \dfrac{1}{n}\sum_{i=1}^n x_i$.
I'm a little stuck on this question. I think I might need to use convex functions. I know the left sum is equal to $\dfrac{1}{n^5}(\sum_{i=1}^n x_i^5)^{1/5}$. There are rules like $(a+b+c+d+\dots)^2 = a^3 + b^3 +c^3+d^3+\dots$ for consecutive integers $a,b,c,d,...$. This is very easy to show using induction, though I don't think induction will be very useful here.
edit: apparently Jensen's inequality could be useful.
|
One may also use Tchebecheff's inequality repeatedly. If $a \ge b \ge c $ and $p \ge q \ge r$, then $$(ap+bq+cr) \ge \frac{(a+b+c)(p+q+r)}{3}~~~~(*)$$ We have
$$(a^2+b^2+c^2) \ge \frac{(a+b+c)(a+b+c)}{3} \ge \frac{(a+b+c)^2}{3} ~~~(1)$$ Next $$(a^3+b^3+c^3) \ge \frac{(a+b+c)(a^2+b^2+c^2)}{3} \ge \frac{(a+b+c)^3}{9} ~~~(2.$$
Finally from (*), (1) and (2) we get
$$(a^2~a^3+b^2~b^3+c^2~c^3)\ge \frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{3} \ge \frac{(a+b+c)^5}{81}.$$ Or
$$\frac{a^5+b*5+c^5}{3} \ge \frac{(a+b+c)^5}{3^5}$$
|
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|
Laplace Inverse of the problem. What is the Laplace Inverse of the given question?
$$\frac{\sqrt{4+s^3}}{s^3}$$
I tried solving it by expanding $\sqrt{1+\frac{s^3}{4}}$ but the terms will not have Laplace Inverse.
If I expand it like $\sqrt{1+\frac{4}{s^3}}$i.e.$$s^{-3/2}* \sqrt{1+\frac{4}{s^3}}$$
It will still have many terms of $t^n$ .
Are there other methods?
|
With CAS help I have:
$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{4+s^3}}{s^3}\right](t)=\frac{2 \sqrt{t} \,
_1F_3\left(-\frac{1}{2};\frac{1}{2},\frac{5}{6},\frac{7}{6};-\frac{4 t^3}{27}\right)}{\sqrt{\pi }}$$
For general a:
$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{a+s^3}}{s^3}\right](t)=\\\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } a^n
\left(s^3\right)^{-\frac{1}{2}-n} \binom{\frac{1}{2}}{n}\right](t)=\\\sum _{n=0}^{\infty }
\mathcal{L}_s^{-1}\left[a^n \left(s^3\right)^{-\frac{1}{2}-n} \binom{\frac{1}{2}}{n}\right](t)=\sum
_{n=0}^{\infty } \frac{a^n \sqrt{\pi } t^{\frac{1}{2}+3 n}}{2 \Gamma \left(\frac{3}{2}-n\right) \Gamma
(1+n) \Gamma \left(\frac{3}{2}+3 n\right)}=\\\frac{2 \sqrt{t} \,
_1F_3\left(-\frac{1}{2};\frac{1}{2},\frac{5}{6},\frac{7}{6};-\frac{1}{27} \left(a
t^3\right)\right)}{\sqrt{\pi }}$$
|
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|
Show that $3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other.
$3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other. Prove it!
$3^{-1}=\overline{0\color{red}{1}}_2$
$3^{-2}=\overline{000\color{red}{111}}_2$
$3^{-3}=\overline{000010010\color{red}{111101101}}_2$
$\ldots$
These are the binary representations - add a minus sign to the left hand side and the right hand side is the 2-adic representation.
|
In general, the period of $2$ in $\Bbb Z/(3^{N+1})^\times$ is $2\cdot3^N$. And we expect that the $2$-adic expansion of $-3^{N+1}$ should be purely periodic, period $2\cdot3^N$.
Indeed, since $3^{N+1}|(2^{2\cdot3^N}-1)$, say with quotient $Q_N$, we get the results
\begin{align}
Q_N&=\frac{2^{2\cdot3^N}-1}{3^{N+1}}\\
-\frac1{3^{N+1}}&=\frac{Q_N}{1-2^{2\cdot3^n}}\,,
\end{align}
in which the second line says that the number of binary digits in the repeating block of the $2$-adic expansion of $-3^{-N-1}$ is $2\cdot3^N$, and what’s in the block is the number $q_N$.
What we know is that $2^{2\cdot3^N}-1\equiv0\pmod{3^{N+1}}$, so we can factor
$$ \left(2^{3^N}-1\right)\left(2^{3^N}+1\right)\equiv0\pmod{3^{N+1}}\,, $$
but please note that since $3^{N+1}$ is odd, we see that the left-hand factor above is $\equiv1\pmod3$, in particular relatively prime to $3$, and thus to $3^{N+1}$ as well. Thus $3^{N+1}$ divides the right-hand factor, i.e. $3^{N+1}\mid(2^{3^N}+1)$, and once again to make typing easier for
myself, I’ll call the quotient $\Omega$. Thus we have:
\begin{align}
\Omega&=\frac{2^{3^N}+1}{3^{N+1}}\\
0&<\Omega<2^{3^N}\\
Q_N&=\Omega\left(2^{3^N}-1\right)\\
&=2^{3^N}(\Omega-1)+\left(2^{3^N}-\Omega\right)\\
\text{where we note }0&<2^{3^N}-\Omega<2^{3^N}\,.
\end{align}
And that gives us our expression for $Q_N=2^{3^N}a+b$ with both $a$ and $b$ in the interval $\langle0,2^{3^N}\rangle$, namely $a=\Omega-1$ and $b=2^{3^N}-\Omega$. And surenough, $a+b=2^{3^N}-1$, as we desired.
|
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How can I solve the following inequality? I have the inequality:
$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$
And I'm not sure how should I go about solving it. I wrote it like this:
$2lg(x^3-x-1) < 2lg(x^3+x-1)$
$lg(x^3-x-1) < lg(x^3 + x - 1)$ (*)
Here I have the conditions:
$x^3-x-1 > 0$
$x^3+x-1 > 0$
At first this stumped me, but then I realised I can just add the inequalities to get:
$2x^3 -2> 0$
$x^3-1>0$
$x^3>1 \Rightarrow x \in (1, + \infty)$
So that is our condition.
Going back to (*) and raising the inequality to the power of $10$, I got:
$x^3-x-1 < x^3+ x - 1$
$2x>0 \Rightarrow x \in(0, +\infty)$
So, if we also consider the condition, we have
$x \in (1, + \infty) \cap (0, + \infty)$
So $x \in (1, + \infty)$.
The problem with this answer is that it is wrong. My textbook lists the following possible answers:
A. $\mathbb{R}$
B. $(0, + \infty)$
C. $(1, + \infty)$
D. $(0, 1)$
E. Other answer
So I got answer C, but after I checked the back of the book I found that the correct answer should be E. So, what did I do wrong and what is that other answer?
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Your condition isn't quite right. When $x = 1.1$, $x^3−x−1 = 1.331 - 1.1 - 1 < 0$.
From $\log(x^3 - x - 1) < \log(x^3 + x - 1)$, subtract one side from the other, then use a logarithm property to get one logarithm. \begin{align*}
0 &< \log(x^3 + x - 1) - \log(x^3 - x - 1) \\
0 &< \log \left( \frac{x^3 + x - 1}{x^3 - x - 1} \right)
\end{align*}
Apply $f(y) = 10^y$ to both sides to obtain
\begin{align*}
1 &< \frac{x^3 + x - 1}{x^3 - x - 1} \text{,}
\end{align*}
which you demonstrated you knew how to handle in your question. (Don't forget, when you go from $a < \frac{b}{c}$ to $ac < b$ you may have inadvertently multiplied by zero for some values of $x$ that you aren't currently thinking about (yielding $0 < 0$, which is false). You should check what happens in those cases. For instance, here, what happens when $x^3 - x - 1 = 0$ in the original equation?)
|
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|
Evaluate $\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$
Evaluate
$$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$
I did this by $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5=\left(1+\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right)^5$$ and get $0$
Does anyone have another idea?
Thanks
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Using the facts that $(e^{\pi i/5})^5=e^{\pi i}=-1$ and $i^5=i$, we have
$$\begin{align}
(1+\sin\pi/5+i\cos\pi/5)^5+i(1+\sin\pi/5-i\cos\pi/5)
&=(1+ie^{-\pi i/5})^5+i(1-ie^{\pi i/5})^5\\
&=-(e^{\pi i/5})^5(1+ie^{-\pi i/5})^5+i^5(1-ie^{\pi i/5})^5\\
&=-(e^{\pi i/5}+i)^5+(i+e^{\pi i/5})^5\\
&=0
\end{align}$$
|
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|
Combinations company A company has $4$ workers who can build a wall, $5$ workers who can paint a wall and 2 workers who can do both. A job requires $ 3$ builders and $3$ painters. In how many ways can the company choose the workers for the job?
I found this resolution, is it correct?
We will have three disjoint cases:
Case 1:
When $0$ worker is present who can do both and this can be done in ${2\choose 0}$, then we must select more $3$ builders and painters and this can be done in ${4\choose 3}$ and ${5\choose 3}$ ways respectively. Hence, total number of ways is ${2\choose 0}\cdot {4\choose 3}\cdot {5\choose 3}=40$
Case 2:
When $1$ worker is present who can do both and this can be done in ${2\choose 1}$, then we must select more $2$ builders and painters and this can be done in ${4\choose 2}$ and ${5\choose 2}$ ways respectively. Hence, total number of ways is ${2\choose 1}\cdot {4\choose 2}\cdot {5\choose 2}=120$.
Case 3:
When $2$ worker is present who can do both and this can be done in ${2\choose 2}$, then we must select more $1$ builders and painters and this can be done in ${4\choose 1}$ and ${5\choose 1}$ ways respectively. Hence, total number of ways is ${2\choose 2}\cdot {4\choose 1}\cdot {5\choose 1}=20$.
Hence, total number of ways of selecting the workers are $40+120+20=\boxed{180}.$
|
Your solution is correct. I have the same result by making another approach. I´ve made a table which shows the number of builder (B), painter (P) and mixed workers (M)
B P M
1. 3 3 0
2. 2 3 1
3. 3 2 1
4. 2 2 2
5. 1 3 2
6. 3 1 2
For 1. we calculate the number of ways to select $3$ builder out of $4$ builder and $3$ painter out of 4 painter: $\binom{4}{3}\cdot \binom{4}{3}=16$. Similar calculations for 2.-6.
$2. \ \binom{4}{2}\cdot \binom{4}{3}\cdot \binom{2}{1}=48 \qquad 3. \ \binom{4}{3}\cdot \binom{4}{2}\cdot \binom{2}{1}=48 \qquad 4. \ \binom{4}{2}\cdot \binom{4}{2}\cdot \binom{2}{2}=36$
$5. \ \binom{4}{1}\cdot \binom{4}{3}\cdot \binom{2}{2}=16 \qquad 6. \ \binom{4}{3}\cdot \binom{4}{1}\cdot \binom{2}{2}=16 $
The sum is $16+48+48+36+16+16=\boxed{180}$
|
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A question about modulus for polynomials The other day my friend was asked to find $A$ and $B$ in the equation
$$(x^3+2x+1)^{17} \equiv Ax+B \pmod {x^2+1}$$
A method was proposed by our teacher to use complex numbers and especially to let $x=i$ where $i$ is the imaginary unit. We obtain from that substitution
$$(i+1)^{17} \equiv Ai+B \pmod {0}$$
which if we have understood it correctly is valid if we define $a \equiv b \pmod n$ to be $a=b+dn$. Running through with this definition we have
$$\begin{align*}
(i+1)^{17} &=\left(\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{4}\right)\right)\right)^{17}\\
&=\sqrt{2}^{17}\left(\cos\left(\frac{17\pi}{4}\right)+\sin\left(\frac{17\pi}{4}\right)\right) \tag{De Moivre}\\
&=256\left(\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{4}\right)\right)\right)\\
&=256\left(1+i\right) \\
&=256+256i\end{align*}
$$
which gives the correct coefficient values for $A$ and $B$.
Our questions are
*
*Why is this substitution valid to begin with?
*It seems here that the special case ($x=i$) implies the general case ($x$), why is that valid?
|
*
*It comes from the exponential form of complex numbers: $1+i$ has modulus $\sqrt 2$ and argument $\frac\pi 4$, so it writes as
$$1+i=\sqrt2\,\mathrm e^{\tfrac\pi 4},\quad\text{and similarly}\quad 1-+i=\sqrt2\,\mathrm e^{-\tfrac\pi 4} $$
The substitution is valid because of the meaning of the congruence:
\begin{align}&(x^3+2x+1)^{17} \equiv Ax+B\enspace (\!\bmod {x^2+1)}\\
&\qquad\qquad\iff \exists q(x):\;(x^3+2x+1)^{17} =(x^2+1)q(x)+Ax+B ,
\end{align}
so when you set $x=\pm i$, the first term in the r.h.s. cancels.
*Setting $x=i$ yields an equation for $A$ and $B$, that's all.
|
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|
Show that sequence is convergent. if $S_1$ and $S_2$ are two positive real number and $S_{n+2}=\sqrt{S_{n+1}S_n}$
If $s_1$ and $s_2$ are two positive real number and $$s_{n+2}=\sqrt{s_{n+1}.s_n}$$ then prove that sequence is convergent and find its limit.
I know that its solve by using concept of monotonic and bounded sequence but I don't know how to proceed .
Please help! The given limit: $\sqrt[3]{s_1s_2^2}$
|
Since $s_{n+2}=\sqrt{s_{n+1}.s_n}$, $\frac{s_{n+2}}{s_{n+1}}=\sqrt{\frac{s_n}{s_{n+1}}}$. Let $T_n=\frac{s_{n+1}}{s_n}$. Then $T_{n+1}=T_n^{-1/2}$, or $log \left(T_{n+1}\right) = -\frac{1}{2}log\left(T_{n}\right)$. Hence $log\left(T_n\right)={\left(-\frac{1}{2}\right)}^{n-1}log(T_1)$, and $T_n=T_1^{{\left(-\frac{1}{2}\right)}^{n-1}}$. That is $\frac{s_{n+1}}{s_n} = T_1^{{\left(-\frac{1}{2}\right)}^{n-1}}.$ Hence $s_n=s_1\frac{s_2}{s_{1}}\frac{s_{3}}{s_{2}}\cdots\frac{s_n}{s_{n-1}}=s_1\times T_1^{\sum_{k=0}^{n-2}{\left(-\frac{1}{2}\right)}^{k}} = s_1\times T_1^{\frac{1-{\left(-\frac{1}{2}\right)}^{n-1}}{1-{\left(-\frac{1}{2}\right)}}}=s_1\times T_1^{\frac{2}{3}\left(1-{\left(-\frac{1}{2}\right)}^{n-1}\right)}$. It is now obvious that $$\lim_{n \rightarrow \infty}s_n=s_1\times T_1^{2/3}=s_1 \times {\left(\frac{s_2}{s_1}\right)}^{2/3} = \sqrt[3]{s_1s_2^2}. $$
|
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|
Matrix for reflection about the line $y = \tan (\theta) \, x$ How would I show that a reflection about the line $y = \tan (\theta) \, x$ is the following?
\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
|
The transformation\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
maps any arbitrary vector $P=\begin{pmatrix}\alpha\\\beta\end{pmatrix}\in\mathbb R^2$ to the vector
$P'=\begin{pmatrix}\alpha\cos2\theta+\beta\sin2\theta\\\alpha\sin2\theta-\beta\cos2\theta\end{pmatrix}$.
It is easy to verify that the mid-point of line joining vectors $P$ and $P'$ i.e.
$M=\begin{pmatrix}\alpha\cos^2\theta+\beta\sin\theta\cos\theta\\\alpha\sin\theta\cos\theta+\beta\sin^2\theta\end{pmatrix}$
always lie on the line $y\cos\theta=x\sin\theta.$
|
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|
Find all positive integers m,n and primes $p\geq5$ such that $m(4m^2+m+12)=3(p^n-1)$ I did it like this. We can manipulate the equation to come to
$\frac{(m^2+3)(4m+1)}{3}=p^n$
Now 3 divides either $(m^2+3) or (4m+1) $
If we assume 3 divides $(m^2+3)$ and it is equal to $(4m+1)$ (my intuition) .we get
$\frac{m^2+3}{3}=4m+1$.solving this we get $m=12$ we can substitute back in equation to get $n=4:p=7$ which is correct. But i am not able to prove that
$\frac{m^2+3}{3}=4m+1$.
Please help me to complete my solution
|
You have that
$$\frac{(m^2+3)(4m+1)}{3}=p^n \tag{1}\label{eq1A}$$
I don't see any particular way to directly prove your assumption that $\frac{m^2 + 3}{3} = 4m + 1$. Instead, here is a proof it's the only solution, so your assumption is true in that manner.
Since both $m^2 + 3$ and $4m + 1$ are $\gt 3$ for positive integers $m$, and they contain just $1$ factor of $3$, then both terms must contain at least one factor of $p$. Let
$$d = \gcd(m^2 + 3, 4m + 1) \tag{2}\label{eq2A}$$
Also, you have that
$$\begin{equation}\begin{aligned}
d \; \mid \; & 16(m^2 + 3) - (4m - 1)(4m + 1) \\
& = 16m^2 + 48 - 16m^2 + 1 \\
& = 49
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
As $p \mid d$, this means $p = 7$. There are thus $2$ basic possibilities depending on which term has a factor of $3$. First, consider for some positive integers $i,j$ where
$$m^2 + 3 = 3(7^i) \implies m^2 = 3(7^i - 1) \tag{4}\label{eq4A}$$
$$4m + 1 = 7^j \implies m = \frac{7^j - 1}{4} \tag{5}\label{eq5A}$$
From \eqref{eq3A}, you have that
$$\begin{equation}\begin{aligned}
49 & = 16(m^2 + 3) - (4m - 1)(4m + 1) \\
& = 16(3)(7^i) - (7^j - 2)(7^j) \\
& = 48(7^i) - (7^j - 2)(7^j)
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Since the left side is $49 = 7^2$, then $i$ and/or $j$ must be $\le 2$ since, otherwise, the right side would have more than $2$ factors of $7$. If $i = 1$, you have $m^2 = 18$, which gives a non-integral $m$. If $i = 2$, then $m^2 = 144 \implies m = 12$, which is the result you already have. For $i \gt 2$, then $j \le 2$, so consider \eqref{eq5A}. For $j = 1$, you get $m = \frac{6}{4}$, which is not an integer. Finally, for $j = 2$, you get $m = 12$, which has already been accounted for.
Next, consider the second factor is a multiple of $3$ instead, i.e., there are some positive integers $k,q$ where
$$m^2 + 3 = 7^k \implies m^2 = 7^k - 3 \tag{7}\label{eq7A}$$
$$4m + 1 = 3(7^q) \implies m = \frac{3(7^q) - 1}{4} \tag{8}\label{eq8A}$$
Now, from \eqref{eq3A}, you have
$$\begin{equation}\begin{aligned}
49 & = 16(m^2 + 3) - (4m - 1)(4m + 1) \\
& = 16(7^k) - (3(7^q) - 2)(3)(7^q)
\end{aligned}\end{equation}\tag{9}\label{eq9A}$$
Similar to before, $k$ and/or $q$ must be $\le 2$. Note $k = 1$ gives $m^2 = 4 \implies m = 2$. However, this means from \eqref{eq8A} that $7^q = 3$, which doesn't give an integral $q$. With $k = 2$, \eqref{eq7A} gives $m^2 = 46$, so $m$ is not an integer. For $k \gt 2$, you have $q \le 2$, so consider \eqref{eq8A}. For $q = 1$, you get $m = 5$. However, from \eqref{eq7A}, you get $7^k = 28$, so $k$ is not an integer. Finally, for $q = 2$, you get $m = \frac{146}{4}$, which is not an integer.
Thus, in summary, there is only the one solution you already found, i.e., $m = 12$, $n = 4$ and $p = 7$.
|
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|
Prove that $\sum \frac{x}{x^2+7}\le \frac{3}{8}$ Let $x,y,z>0$ such that $xy+yz+xz=3$. Show that $$\frac{x}{x^2+7}+\frac{y}{y^2+7}+\frac{z}{z^2+7}\le \frac{3}{8}$$
We have: $$x+y+z\ge \sqrt{3\left(xy+yz+xz\right)}=3\rightarrow \frac{3}{8\left(x+y+z\right)}\le \frac{3}{8\cdot 3}$$
Then i will prove $$\sum \frac{x}{3x^2+7\left(xy+xz+yz\right)}\le \frac{3}{8\left(x+y+z\right)}$$
But it's wrong. I tried to use uvw and done. So i need another idead without uvw
|
Hint: Use the substitution $x = \sqrt 3 \tan A$ and so on for some acute $\triangle ABC$, and then Jensen’s inequality as $t \mapsto \dfrac{\sqrt 3 \tan t}{3\tan^2t + 7}$ is concave for $t \in (0, \frac\pi2)$.
|
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Is the following series $\sum_{n=0}^\infty \frac{n^3}{n!}$ convergent and if so what is the sum? I started by converting the following $\sum_{n=0}^\infty \frac{n^3}{n!}$ into
$$\sum_{n=1}^\infty \frac{n(n-1)(n-2)}{n!} + \frac{3n(n-1)}{n!} + \frac{n}{n!}.$$ Did i get it right and where should I go from here ?
|
Note that by limit comparison test with $\sum \frac 1{n^2}$
$$\frac{\frac{n^3}{n!}}{\frac 1{n^2}}=\frac{n^5}{n!}\to 0$$
therefore the series converges.
For the sum we can use that
$$\sum_{n=0}^\infty \frac{n^3}{n!}=\sum_{n=1}^\infty \frac{n^2}{(n-1)!}=\sum_{n=0}^\infty \frac{n^2-1+1}{(n-1)!}=\sum_{n=2}^\infty \frac{n+1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=$$
$$= \sum_{n=2}^\infty \frac{n-2+3}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=\sum_{n=3}^\infty \frac{1}{(n-3)!}+3\sum_{n=2}^\infty \frac{1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=5e$$
|
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Equation of a hyperbola given its asymptotes
Find the equation of the hyperbola whose asymptotes are $3x-4y+7$ and $4x+3y+1=0$ and which pass through the origin.
The equation of the hyperbola is obtained in my reference as
$$
(3x-4y+7)(4x+3y+1)=K=7
$$
So it make use of the statement, the equation of the hyperbola = equation of pair of asymptotes + constant
I understand that the pair of straight lines is the limiting case of hyperbola.
Why does the equation to the hyperbola differ from the equation of pair of asymptotes only by a constant ?
|
$$
\frac{4x+3y+1}{5}=\pm\frac{3x-4y+7}{5}\\
\implies x+7y-6=0\;;\; 7x-y+8=0\text{ which are the axis of the hyperbola with centre }(-1,1)\\
$$
Since $m_1m_2=-1\implies$ asymptotes are perpendicular $\implies$ rectangular hyperbola
$$
\frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=\pm1\\
\text{At }(0,0): \frac{18}{25a^2}-\frac{32}{25a^2}=\pm1\implies18a^2-32a^2=\pm25a^4\\
-14a^2=\pm25a^4\implies-14a^2=25a^4\text{ not possible}\\
-14a^2=-25a^4\implies \boxed{a^2=\frac{14}{25}}\\
\frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=-1\\
\frac{(7x-y+8)^2}{50a^2}-\frac{(x+7y-6)^2}{50a^2}=1\\
(7x-y+8)^2-(x+7y-6)^2=50a^2=50.\frac{14}{25}=28\\
x^2(48)+y^2(-48)+xy(-28)+x(124)+y(68)+28=28\\
\color{blue}{12x^2-7xy-12y^2+31x+17y=0}
$$
|
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|
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try :
$n=1$ we find : $1=1+1$ $×$
$n=2$ we find : $4=1+1$ $×$
$n=3$ we find : $9=1+2$ $×$
$n=4$ we find : $16=1+6$ $×$
$n=5$ we find : $25=1+24$ $√$
Now how I prove $n=5$ only the solution ?
|
A more thorough version of the answer given by fleablood, avoiding assumptions and unexamined cases.
The stated question, find all solutions for $n$ such that $n^2=(n-1)!+1$
Since for all non-negative integers, $k!\ge 1$, we conclude $n^2\ge 2 \Rightarrow n>1 \Rightarrow (n-1)>0$
Rearranging, $(n-1)!=n^2-1=(n-1)(n+1)$
Restating, $(n-1)(n-2)!=(n-1)(n+1)$
Since $(n-1)>0$, we are allowed to divide through by $(n-1)$ and obtain $(n-2)!=(n+1)$. And since $n>1 \Rightarrow n+1 > 2$, the largest integer in the factorial must be at least as large as $2$, viz: $n-2 \ge 2$
Therefore $n-2$ must have at least one divisor $d>1$. But $d\mid (n-2) \Rightarrow d\mid (n-2)! \Rightarrow d\mid (n+1)$
Any factor which divides each of $(n-2)$ and $(n+1)$ must divide their difference, $(n+1)-(n-2)=3$
The only factor greater than $1$ which divides $3$ is $3$ itself, and since $(n-2)$ has no other common factors with $(n+1)$, it must be the case that $n-2=3 \Rightarrow n=5$ is the sole solution to the stated problem.
|
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|
Find the number of ways to distribute 10 pieces of candy using this generating function The question asks:
a) Find a generating function for the number of ways to distribute identical pieces of candy to 3 children so that no child gets more than 4 pieces. Write this generating function in closed form, as a quotient of polynomials. b) Find the number of ways to distribute 10 pieces of candy using this generating function.
I figured out for part a that $(1 + x + x^2 + x^3 + x^4)^3$ is represented as the generating function:
$$f(x) = \left( \frac{1-x^5}{1-x} \right) ^3$$
But I don't know how to find the $a_{10}$ term. Please help.
|
Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain
\begin{align*}
\color{blue}{a_{10}}&\color{blue}{=[x^{10}]\left( \frac{1-x^5}{1-x} \right)^3}\\
&=[x^{10}]\frac{1-3x^5+3x^{10}-x^{15}}{(1-x)^3}\tag{1}\\
&=[x^{10}]\left(1-3x^5+3x^{10}\right)\sum_{j=0}^\infty \binom{-3}{j}(-x)^j\tag{2}\\
&=\left([x^{10}]-3[x^5]+3[x^0]\right)\sum_{j=0}^\infty \binom{j+2}{2}x^j\tag{3}\\
&=\binom{12}{2}-3\binom{7}{2}+3\binom{2}{2}\tag{4}\\
&=66-3\cdot 21+3\\
&\,\,\color{blue}{=6}
\end{align*}
Comment:
*
*In (1) we expand the numerator.
*In (2) we skip the terms of the numerator which do not contribute to $[x^{10}]$ and apply the binomial series expansion.
*In (3) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.
*In (4) we select the coefficients accordingly.
|
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|
Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \equiv 2^6 \equiv 4\pmod{10} \\
(...)$$
I can see that this goes 2,4,8,6 and then repeats. I remember that the gist of the exercise is to find the remainder based on this repetition. How do I do that? I know that $572^{42} \equiv 2^{42}\equiv ? \pmod {10}$. How do I simplify that 42 and answer this using that repetition?
|
You want $2^{42} \mod 10$. You found that $2^4\equiv6\mod 10$. Now $6^n\equiv 6$ for all $n\in\mathbb N$ (you could prove that by induction), so $2^{42}\equiv2^{40}2^2\equiv(2^4)^{10}2^2\equiv6^{10}\times4\equiv6\times4\equiv4\mod10$.
|
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Solving the integral $\int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}$
I'm trying to solve the following indefinite integral:
$$I = \int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}$$
The integral is a general case which comes from a physics problem of potential in a capacitor with a rod inside.
I tried to figure out any plausible substitution or a transform by introducing a special function, but failed to do that.
*
*An approach I tried was by introducing a function:
$h(z) = \mathrm{ln}|z| +i\phi$ for $0 < \phi < 2\pi$.
*An approach with a special function seemed applicable to the integral, such as the Dilogarithm.
Update-1:
Applying numerical integration from $0$ to $1$, I obtained (via Matlab, for self-check):
fun = @(x) log((1+sqrt(1-x.^2))./(1-sqrt(1-x.^2))).*(1./(1-x.^2))
q = integral(fun,0,1)
>> 4.9348
Which is $\frac{1}{2} \pi^2$.
|
Start by using the substitution $x=\frac{2z}{1+z^2}$ then:$$I = \int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}=-4\int \frac{\ln z}{1-z^2}dz=2\int \frac{\ln z}{z-1}dz-2\int \frac{\ln z}{z+1}dz$$
$$=-2\operatorname{Li}_2(1-z)-2\operatorname{Li}_2(-z)-2\ln z \ln(1+z)+C,\quad z=\frac{1-\sqrt{1-x^2}}{x},x< |1|$$
The same substitution yields:
$$\int_0^1 \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}=-4\int_0^1 \frac{\ln z}{1-z^2}dz$$$$=-4\sum_{n=0}^\infty \int_0^1 z^{2n}\ln zdz=4\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{2} $$
|
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Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g(x)+5$ where $g(x)=(x+1)(x+2)(x+3)(x+4)$
Let's find the maximum value of g(x)
It can be clearly seen that from $x=-6$ to $x=6$, maximum value of $g(x)$ is at $x=6$.
$$g(6)=5040$$
Let's find the minimum value of $g(x)$
From the sign scheme one can see that negative value of $g(x)$ occurs in the interval $(-4,-3)$ and $(-2,-1)$
Hence intuitively it feels that the minimum value of g(x) would be at $x=-\dfrac{3}{2}$ or at $x=-\dfrac{7}{2}$ as in case of parabola also, minimum value is at average of both the roots.
So $g\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}=-\dfrac{15}{8}$
At $x=-\dfrac{7}{2}$, $g\left(-\dfrac{7}{2}\right)=-\dfrac{5}{2}\cdot\dfrac{3}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{15}{8}$
So range of $g(x)$ would be $[\dfrac{-15}{8},5040]$
Hence range of $f(x)$=$\left[\dfrac{25}{8},5045\right]$
But it is given that the range is $[a,b]$ where $a,b\in N$
I am stuck here. I am also not able to prove mathematically that at $x=-\dfrac{3}{2}$ or $x=-\dfrac{7}{2}$, minimum value of $g(x)$ will occur.
Please help me in this.
|
$$f(x)=(x^2+5x+4)(x^2+5x+6)+5=(x^2+5x+5)^2-1+5\geq4.$$
The equality occurs for $x^2+5x+5=0,$ which happens on $[-6,6],$ which says that $4$ is a minimal value.
Now, $f(6)=5045$ and since for any $-6\leq x\leq6$
$$5045-f(x)=(6-x)(x+11)(x^2+5x+76)\geq0,$$ we see that $5045$ is a maximal value.
|
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|
Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need to find an $ N \in \mathbb{N}$ so that : $ \forall \epsilon > 0 n > N \implies \left| a_n - \frac{1}{2} \right| < \epsilon $
Next step I simplify $a_n$ : $ a_n = \frac{n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right) \cdot \sqrt{1+ \frac{1}{n}} + 1 } {\sqrt{1+ \frac{1}{n}} + 1} = \frac{1}{\sqrt{1+ \frac{1}{n}} + 1}$
And then I got stuck,what am I supposed to do with : $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|$
|
For a more accurate bound consider the following:
$$
\left|a_{n} - \frac{1}{2} \right| = \left|\frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2} \right| =
\left|\frac{2-(\sqrt{1+\frac{1}{n}}+1)}{2\left( \sqrt{1+\frac{1}{n}}+1 \right)} \right| \leq \left| \frac{2-(\sqrt{1+\frac{1}{n}}+1)}{4} \right| \implies \left| 2-(\sqrt{1+\frac{1}{n}}+1) \right| = \left| 1-\sqrt{1+\frac{1}{n}} \right| = \left| \sqrt{1+\frac{1}{n}}-1 \right| = \sqrt{1+\frac{1}{n}}-1 < 4\epsilon
$$
Then
$$
\sqrt{1+\frac{1}{n}}-1 < 4\epsilon \implies n > \frac{1}{(4\epsilon +1)^{2}-1}
$$
|
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|
Finding a closed form expression for $\sum_{k=0}^n(k^2+3k+2)$ using generating functions First of all, i know that there is already a similar question (Closed form expression for $\sum_{k=0}^{n}(k^2 + 3k + 2)$) regarding my equation in this forum, but my question is only about verification of my thought process because i am quite new to the concept of generating functions.
I have the following equation which i can rewrite to:
\begin{equation}
\sum_{k=0}^n(k^2+3k+2) = \sum_{k=0}^n k^2 + 3\sum_{k=0}^n k + \sum_{k=0}^n 2
\end{equation}
My idea was now to apply generating functions to each of the sums which will give me the same sequences as the sums would generate:
\begin{equation}
\frac{(1+x)x}{(1-x)^3} + \frac{3x}{(1-x)^2} + \frac{2}{1-x}
\end{equation}
And now simplify it and translate it back to a simple sum as follows:
\begin{equation}
\frac{2}{(1-x)^3} \leftrightarrow \sum_{k=1}^n k(k+1)
\end{equation}
Is this a sufficient and correct way of using the concepts of generating functions to get a closed form expression from the given sum or did i miss out on a crucial concept?
|
Yes, this approach of splitting into three sums is valid, but you forgot the $\sum_{k=0}^n$. Here's a correct solution. Let $a_n=\sum_{k=0}^n(k^2+3k+2)$. Then
\begin{align}
\sum_{n=0}^\infty a_n x^n &= \sum_{n=0}^\infty \left(\sum_{k=0}^n(k^2+3k+2)\right) x^n\\
&= \sum_{k=0}^\infty (k^2+3k+2) \sum_{n=k}^\infty x^n\\
&= \sum_{k=0}^\infty (k^2+3k+2) \frac{x^k}{1-x}\\
&= \frac{1}{1-x}\left(\sum_{k=0}^\infty k^2 x^k+3\sum_{k=0}^\infty kx^k+2\sum_{k=0}^\infty x^k\right)\\
&= \frac{1}{1-x}\left(\frac{x(x+1)}{(1-x)^3}+\frac{3x}{(1-x)^2}+\frac{2}{1-x}\right)\\
&= \frac{1}{1-x}\cdot\frac{2}{(1-x)^3}\\
&= \frac{2}{(1-x)^4}\\
&= 2\sum_{n=0}^\infty \binom{n+3}{3} x^n,
\end{align}
so $$a_n = 2 \binom{n+3}{3} = \frac{(n+3)(n+2)(n+1)}{3}.$$
More generally, note that if $B(x)=\sum_{n=0}^\infty b_n x^n$, then the generating function for the partial sums $\sum_{k=0}^n b_k$ is $B(x)/(1-x)$.
|
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|
What is the interpretation of the difference? We have the linear maps \begin{equation*}f:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+x_2 \\ x_3 \\ x_1+x_2\end{pmatrix} \ \text{ and } \ h:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+ x_2 \\ x_3 \\ x_1-x_2\end{pmatrix}\end{equation*}
We also have the vectors \begin{equation*}\vec{v}:=\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix} \ \text{ und } \ \vec{w}:=\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}\end{equation*}
I want to check if $f(\vec{v}$ and $f(\vec{w})$ are linearly (in)dependent and then the same for $h(\vec{v})$ and $h(\vec{w})$ and then I want to interpret the difference.
With $f$ we get \begin{equation*}f(\vec{v})=f\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} \ \ \text{ and } \ \ f(\vec{w})=f\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}\end{equation*}
The vectors $f(\vec{v})$ and $f(\vec{w})$ are linearly dependent, since $f(\vec{v})=0 \cdot f(\vec{w})$.
With $h$ we get \begin{equation*}h(\vec{v})=h\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix} \ \ \text{ and } \ \ h(\vec{w})=h\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}\end{equation*}
The vectors $h(\vec{v})$ and $h(\vec{w})$ are linearly independent, since the one vector has zeroes and the other one not, correct?
What is the interpretation of the difference?
|
The matrices corresponding to the linear transformations $f,h$ are $$\begin{bmatrix}1&1&0\\0&0&1\\1&1&0\end{bmatrix},\begin{bmatrix}1&1&0\\0&0&1\\1&-1&0\end{bmatrix}$$
respectively. Obviously the determinant of the former is $0$ since there are two identical (and hence linearly dependent) rows, while you can show that the rows of the latter are linearly independent and hence the determinant is nonzero. What this means is that $h$ represents an injective (i.e. invertible) transformation while $f$ does not. So $f$ can and must take linearly independent vectors in its domain and "collapse" them to become dependent, since the dimension of its image is smaller than its domain, and conversely $h$ will take linearly independent vectors in its domain to linearly independent vectors in its codomain, which you can prove by considering the basis vectors.
|
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|
Find the domain of $f(x)=\sec^{-1}\dfrac{x}{\sqrt{x-[x]}}$
Domain of $f(x)=\sec^{-1}\dfrac{x}{\sqrt{x-[x]}}$ is:
$$
x-[x]\neq0\implies x\notin \mathcal{Z}\implies x\in\mathcal{R}-\mathcal{Z}\\
\sec^{-1}:\mathcal{R}-(-1,1)\to(0,\pi)\implies \dfrac{x}{\sqrt{x-[x]}}\in\mathcal{R}-(-1,1)\\
\dfrac{x}{\sqrt{x-[x]}}\notin(-1,1)\\
x-[x]\in(0,1)\implies\sqrt{x-[x]}\in(0,1)
$$
The solution given in my reference is $\mathcal{R}-\{(-1,1)\cup\mathcal{Z}\}$, but how do I prove the part $x\notin (-1,1)$ ?
|
In order to take the square root (and have a real result) we need $x - \lfloor x \rfloor \ge 0$ , but that is not a problem.
We cannot divide by $0$
$x - \lfloor x \rfloor \ne 0 \implies x\notin \mathbb Z$
For inverse secant to be defined we need
$\left|\frac {x}{\sqrt{x - \lfloor x \rfloor}}\right|\ge 1$
$|x| \ge \sqrt{x - \lfloor x \rfloor}\\
x^2 \ge x - \lfloor x \rfloor\\
x^2 - x + \lfloor x \rfloor \ge 0$
We know $x - 1 <\lfloor x \rfloor < x$
If $|x| > 1$ we are safe.
if $0<|x| < 1$
$x^2 - x + \lfloor x \rfloor = x^2 - x < 0$
if $-1<|x| < 0$
$x^2 - x - 1 = (x - \frac 12 + \frac {\sqrt 5}{2})(x - \frac 12 - \frac {\sqrt 5}{2})$
$x^2 - x - 1\le 0 \implies \frac 12 - \frac {\sqrt{5}}{2}\le x\le \frac 12 + \frac {\sqrt{5}}{2}$
What does that leave us with?
The domain is:
$x = \mathbb R - \mathbb Z - (0,1) - [\frac12 - \frac {\sqrt {5}}{2}, 0)$
|
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|
How do I prove that $\sum_{i=3}^{n} \frac{i-2}{\binom{i}{2}} < \frac n 4$ for all natural $n > 3$? How do I prove:
$$\frac{3-2}{\binom 3 2} + \frac {4 - 2}{\binom 4 2} + \dots + \frac{n-2}{\binom n 2} < \frac n 4$$
I tested this sum on a variety of $n$ from $2$ to $100$ and they all seem to be less than $\frac n 4$, however I can't find a way to prove it.
|
The inequality is equivalent to:
$$\sum_{i=3}^{n} \frac{i-2}{i(i-1)} < \frac n 8$$
Using partial fractions, $\frac{i-2}{i(i-1)} = \frac 2 n - \frac 1 {n-1} = \frac 1 n - \left(\frac 1 {n-1} - \frac 1 n\right)$.
The given sum becomes:
$$\sum_{i=3}^{n} \frac 1 i - \sum_{i=3}^{n} \left(\frac 1{i-1} - \frac 1 i\right)$$
$$= H_n - \frac 3 2 - \left( \frac 1 2 - \frac 1 3 + \frac 1 3 - \frac 1 4 + \dots + \frac 1 {n-1} - \frac 1 n \right)$$
$$= H_{n} + \frac 1 n - 2$$
Where $H_n$ is the $n$th harmonic number.
The problem then transforms to proving the following inequality:
$$H_{n} + \frac 1 n - 2 < \frac{n}{8}, \qquad n \ge 3$$
which can be done by checking cases for $n < 8$ and induction for $n \ge 8$.
|
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|
prove that $\frac{n!}{n^n}\le(\frac{1}{2})^k$ where $k=[\frac{n}{2}]$, the greatest integer $\le\frac{n}{2}$
1) For $n\ge 2$, prove that $\frac{n!}{n^n}\le(\frac{1}{2})^k$ where $k=[\frac{n}{2}]$, the greatest integer $\le\frac{n}{2}$.
2) Deduce the value of $\underset{n\rightarrow\infty}{\lim}\frac{n!}{n^n}$.
3) Show that the series $\underset{n=1}{\overset{\infty}{\sum}}\frac{n!}{n^n}$ is convergent and again deduce $\underset{n\rightarrow\infty}{\lim}\frac{n!}{n^n}$.
My attempt
*
*$\forall n\in\mathbb{N}$
$1<1+\frac{1}{n}\le 2\Rightarrow 1>\frac{1}{1+\frac{1}{n}}\ge \frac{1}{2}\Rightarrow 1>\frac{1}{(1+\frac{1}{n})^n}\ge \frac{1}{2^n}$
$\Rightarrow 1>\frac{a_{n+1}}{a_n}\ge\frac{1}{2^n}\Rightarrow a_n\ge a_{n+1}\ge\frac{a_n}{2^n}$
Using induction we eventually get, $a_1\ge a_n\ge\frac{a_1}{2^{\frac{n(n-1)}{2}}}\Rightarrow 1\ge a_n\ge\frac{1}{2^{\frac{n(n-1)}{2}}}$
*Using Sandwich theorem, we get $\underset{n\rightarrow\infty}{\lim}a_n=1$.
I couldn't get the required bound for $\frac{n!}{n^n}$ although, with or without using the result from 1 I can conclude the limit to be $1$.
I did get an answer to my doubt at https://math.stackexchange.com/a/1226238/562589 but I couldn't understand how they concluded $\frac{k}{n}\frac{k-1}{n}...\frac{1}{n}\le(\frac{1}{2})^k$.
*Let $a_n=\frac{n!}{n^n}$.
Then, $\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=(\frac{n}{n+1})^n=\frac{1}{(1+\frac{1}{n})^n}$
$\Rightarrow\underset{n\rightarrow\infty}{\lim}\frac{a_{n+1}}{a_n}=\underset{n\rightarrow\infty}{\lim}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}<1\Rightarrow\underset{n=1}{\overset{\infty}{\sum}}\frac{n!}{n^n}$ is convergent
I am not sure how to use this to deduce $\underset{n\rightarrow\infty}{\lim}\frac{n!}{n^n}$ differently(?) than done before.
|
If $n$ is even, the first $\frac n2$ factors of $$\frac {1\cdot 2 \cdot 3 \ldots n}{n^n}$$ are less than or equal to $\frac 12$ and the rest are less than or equal to $1$, so the fraction is less than or equal to $\left(\frac 12\right)^{n/2}$. If $n$ is odd we just need to account for one more factor $\frac 12$. Note that $\frac {3!}{3^3}=\frac 6{27}\lt \left(\frac 12\right)^2$ and for any $n$ higher than $3$ the first term is less than $\frac 14$ which gives one more factor $\frac 12$ and we are done.
|
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|
Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow \frac{7}{5} \rightarrow \frac{7}{12}$.
Repeating this process appears to map every fraction to
$\phi$ and $\frac{1}{\phi}$:
$$
\begin{array}{ccccccccccc}
\frac{2}{5} & \frac{7}{5} & \frac{7}{12} & \frac{19}{12} & \frac{19}{31} & \frac{50}{31} & \frac{50}{81} & \frac{131}{81} & \frac{131}{212} & \frac{343}{212} & \frac{343}{555} \\
0.4 & 1.40 & 0.583 & 1.58 & 0.613 & 1.61 & 0.617 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Another example:
$$
\begin{array}{ccccccccccc}
\frac{11}{7} & \frac{18}{7} & \frac{18}{25} & \frac{43}{25} & \frac{43}{68} & \frac{111}{68} & \frac{111}{179} & \frac{290}{179} & \frac{290}{469} & \frac{759}{469} & \frac{759}{1228} \\
1.57143 & 2.57 & 0.720 & 1.72 & 0.632 & 1.63 & 0.620 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Q. Why?
|
Instead of representing $\frac{a}{b}$ as a fraction, represent it as the vector $\left( \begin{array}{c} a \\ b \end{array} \right)$.
Then, all you are doing to generate your sequence is repeatedly multiplying by the matrix $\left( \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right)$. One of the eigenvectors of this matrix is $\left( \begin{array}{c} \frac{\sqrt{5}-1}{2} \\ 1 \end{array} \right)$, which has a slope equal to the "golden ratio".
This is a standard example of a linear discrete dynamical system, and asymptotic convergence to an eigenvector is one of the typical things that can happen. You can also guess at the long-term behavior of the system by looking at its vector field.
https://kevinmehall.net/p/equationexplorer/#%5B-100,100,-100,100%5D&v%7C(x+y)i+(x+2y)j%7C0.1
In this case you see everything that starts in the first quadrant diverges to infinity along the path of the eigenvector I mentioned before. For your sequence, you started at $\left( \begin{array}{c} 2 \\ 5 \end{array} \right)$, which lies in the first quadrant.
Side note: There is nothing particularly special about the golden ratio, the matrix above, or the starting point of $\left( \begin{array}{c} 2 \\ 5 \end{array} \right)$ for this sequence. You can change the starting point to be in the negative quadrant if you want to diverge in the opposite direction, and you can change the matrix if you want to diverge along a differently sloped eigenvector.
|
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|
Strategy for the Limit: $\lim_{n\to\infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n} $ I do not understand how to properly solve this limit:
$$
\lim_{n\to\infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}
$$
I thought of breaking it up:
$$
\lim_{n\to\infty} \frac{2^{n+1}}{2^n+3^n} +\lim_{n\to\infty} \frac{3^{n+1}}{2^n+3^n}
$$
But I do not see how this will allow me to use any of the limit rules to reduce. I know that the series converges though.
Thanks.
|
Since the 3s dominate,
I would expect the limit
to be 3.
Check:
$\begin{array}\\
\dfrac{2^{n+1}+3^{n+1}}{2^n+3^n}-3
&=\dfrac{2^{n+1}+3^{n+1}-3(2^n+3^n)}{2^n+3^n}\\
&=\dfrac{2^{n+1}+3^{n+1}-3\cdot 2^n-3^{n+1}}{2^n+3^n}\\
&=\dfrac{2^{n+1}-3\cdot 2^n}{2^n+3^n}\\
&=\dfrac{-2^n}{2^n+3^n}\\
&\to 0
\qquad\text{since } 2^n/3^n \to 0\\
\end{array}
$
|
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|
Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1}>=0 \text { and } \dfrac{x^2-1-x^2-1}{x^2+1}<=0$$
$$\dfrac{2x^2}{x^2+1}>=0 \text { and } \dfrac{-2}{1+x^2}<=0$$
$$x\in R$$
$$x^2-1\ne0$$
$$x\ne\pm1$$
$$\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$$
$$\pi-\cos^{-1}\dfrac{1-x^2}{1+x^2}-\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac{2\pi}{3}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}$$
Substituting $x$ by $\tan\theta$
$$x=\tan\theta$$
$$\tan^{-1}x=\theta$$
$$\theta\in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)-\{-\dfrac{\pi}{4},\dfrac{\pi}{4}\}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-\tan^2\theta}{1+\tan^2\theta}+\tan^{-1}\dfrac{2\tan\theta}{1-\tan^2\theta}$$
$$\dfrac{\pi}{3}=\cos^{-1}(\cos2\theta)+\tan^{-1}(\tan2\theta)$$
$$2\theta\in(-\pi,\pi)-\{-\dfrac{\pi}{2},\dfrac{\pi}{2}\}$$
Now we break the range of $2\theta$ into various parts:-
Case $1$: $2\theta\in\left(-\pi,-\dfrac{\pi}{2}\right)$,$\theta\in\left(-\dfrac{\pi}{2},-\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\pi+2\theta+\pi+2\theta$$
$$-\dfrac{8\pi}{3}=4\theta$$
$$-\dfrac{2\pi}{3}=\theta$$
But it is not the range of $\theta$ we assumed
Case $2$: $2\theta\in\left(-\dfrac{\pi}{2},0\right]$,$\theta\in\left(-\dfrac{\pi}{4},0\right]$
$$\dfrac{\pi}{3}=-2\theta+2\theta$$
$$\dfrac{\pi}{3}=0 \text { not possible }$$
Case $3$: $2\theta\in\left(0,\dfrac{\pi}{2}\right)$,$\theta\in\left(0,\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\theta+2\theta$$
$$\dfrac{\pi}{12}=\theta$$
It is coming in the range of $\theta$, so its a valid solution.
$$\tan^{-1}x=\dfrac{\pi}{12}$$
$$x=\tan\dfrac{\pi}{12}$$
$$x=2-\sqrt{3}$$
Case $4$: $2\theta\in\left(\dfrac{\pi}{2},\pi\right)$, $\theta\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$
$$\dfrac{\pi}{3}=2\pi-2\theta+2\theta-\pi$$
$$\dfrac{\pi}{3}=\pi \text { not possible }$$
So only solution is $2-\sqrt{3}$, but actual answer is $2-\sqrt{3}, \sqrt{3}$
|
Another way:
From $\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac\pi3$
Let $\cos^{-1}\dfrac{1-x^2}{1+x^2}=2y\implies0\le2y\le\pi$
$\cos2y=?,x^2=\tan^2y$
Case$\#1:$
If $x=\tan y\ge0$
$$\tan^{-1}(\tan2y)=\dfrac\pi3-2y\implies\tan(2y)=\tan\left(\dfrac\pi3-2y\right)$$
$\implies2y=n\pi+\dfrac\pi3-2y\iff y=\dfrac{\pi(3n+1)}{12}$ such that $0\le n<4$ and $\tan y\ge0$
$0\le \dfrac{\pi(3n+1)}{12}\le\dfrac\pi2\iff0\le3n+1\le6\implies n=0,1$
Case$\#2:$
If $x=-\tan y<0$
$$\tan^{-1}(\tan(-2y))=\dfrac\pi3-2y\implies\tan(-2y)=\tan\left(\dfrac\pi3-2y\right)$$
$\implies-2y=m\pi+\dfrac\pi3-2y\iff3m+1=0$ which is untenable as $m$ is any integer
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt[3]{(n^3+2n^2+1)^2} + \sqrt[3]{(n^3+2n^2+1)(n^3-1)} + \sqrt[3]{(n^3-1)^2}}$$
$$\lim\limits_{n \to \infty} \dfrac{2n^2+2}{\sqrt[3]{n^6+4n^5+4n^4+2n^3+4n^2+1} + \sqrt[3]{n^6+2n^5-2n^2-1} + \sqrt[3]{n^6-2n^3+1}}$$
And I saw that we can factor $n^2$ in the denominator and if we do the same in the numerator, we'd get that the limit is equal to $2$. The problem is that my textbook claims that this limit is actually $\dfrac{2}{3}$. I don't see why should I have a $3$ in the denominator since the coefficient of $n^2$ would be $1$ if I would carry out the factoring to detail. So, what did I do wrong?
|
Divide all the terms by $n^{2}$ and take the limit. you will see that the limit is $\frac 2 {1+1+1}$
|
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|
Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statement where $n=2k-1$ to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume $n=k$ is odd and prove $n=k+2$ is divisible by 15?
|
Hint
Like Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$
If $f(m)=4^{2m+1}+5^{2m+1}+6^{2m+1},$
$$f(n+1)-4^2f(n)=5^{2n+1}(5^2-4^2)+6^{2n+1}(6^2-4^2)$$ will be clearly divisible by $15$ if $n\ge0$
So, if $15$ divides $f(n),15$ will divide $f(n+1)$
Now establish the base case i.e., $m=0$
|
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|
If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then Find $6xyz$. If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then
Find $6xyz$.
How can I approach this problem?
I need some hints. Thanks.
|
$x + y + xy = 3 \implies ( 1 + x ) ( 1 + y ) = 4\tag1$
$y + z + yz = 8 \implies ( 1 + y ) ( 1 + z ) = 9\tag2$
$z + x + zx = 15 \implies ( 1 + z ) ( 1 + x ) = 16\tag3$
Multiplying $(1)$ and $(3)$, we have
$$( 1 + x )^2 ( 1 + y )( 1 + z )=4\cdot 16\implies ( 1 + x )^2= \dfrac{64}{9}\qquad \text{{using $(2)$}}.$$
$$\implies x=\pm \dfrac{8}{3}-1=\dfrac{5}{3}~, -\dfrac{11}{3}$$
From $(1)$, $~y=-1+\dfrac{4}{1+x}=\dfrac{1}{2},~-\dfrac{5}{2}~$ respectively for $~x=\dfrac{5}{3}~, -\dfrac{11}{3}~$
From $(3)$, $~z=-1+\dfrac{16}{1+x}=5,~-7~$ respectively for $~x=\dfrac{5}{3}~, -\dfrac{11}{3}~$
So case I: When $~x=\dfrac{5}{3},~y=\dfrac{1}{2},~z=5~,$
$$~6xyz=6\times \dfrac{5}{3} \times\dfrac{1}{2}\times 5=25~.$$
and case II: When $~x=-\dfrac{11}{3},~y=-\dfrac{5}{2},~z=-7~,$
$$~6xyz=6\times (-)\dfrac{11}{3} \times(-)\dfrac{5}{2}\times(-) 7=-385~.$$
|
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|
Stuck on a probability law problem I'm currently trying to solve a problem, I completed the first question but I am stuck at the second one, here is the problem:
One person roll a die until the result is a $1$, a second person toss a coin until he gets $3$ tails.
*
*How many tries are they going to make on average.
$X$ = number of die rolls
$Y$ = number of time we toss a coin
$X$ follows a geometric law with parameter $1/6$, $E(X) = 6$ and $V(X) = 30$. $Y$ follows a negative binomial law with $n = 3$ and $p = 1/2,$ $E(Y) = 3/(1/2) = 6$ and $V(Y) = n(1-p)/p^2 = 6.$
*What is the probability that they both stop at the same time $(p(X = Y)).$
I found $p(X=Y) = \sum_{k=3}^\infty p(X=k, Y=k) $ so the same sum with $p(X=k)p(Y=k)$ since they are independent. Then $p(X=k)p(Y=k) = (5/6)^{k-1} \cdot\frac16 \begin{pmatrix} k-1 \\ 2 \\ \end{pmatrix}
\left(\frac12\right)^{k-3}\left(\frac12\right)^3$
But now i don't know how to compute it so i can get the value of that probability? Any help would be very appreciated.
Thanks
|
We have $$\begin{align}
\Pr(X=Y)&=\sum_{k=3}^\infty\frac16\left(\frac56\right)^{k-1}\left(\frac12\right)^k\binom{k-1}{2}\\
&=\frac15\sum_{k=3}^\infty\left(\frac{5}{12}\right)^k\binom{k-1}{2}\\
&=\frac{1}{10}\sum_{k=3}^\infty\left(\frac{5}{12}\right)^k(k-1)(k-2)\\
&=\frac{1}{24}\sum_{k=3}^\infty\left(\frac{5}{12}\right)^{k-1}(k-1)(k-2)
\end{align}$$
Now write $$f(x)=\frac{1}{24}\sum_{k=3}^\infty x^{k-1}$$
Differentiating twice,
$$f''(x)=\frac{1}{24}\sum_{k=3}^\infty (k-1)(k-2)x^{k-3}$$ and
$$x^2f''(x)=\frac{1}{24}\sum_{k=3}^\infty (k-1)(k-2)x^{k-1}$$
So, sum the geometric series for $f$, differentiate twice, multiply by $x^2$, and then set $x=\frac{5}{12}$.
|
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|
Find $\lim_{x\to 0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$
If $$f(x)= \frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$$ is continuous at $x=0$ then find $f(0)$
$$
f(0)=\lim_{x\to0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}\\
=\lim_{x\to0}\big[\frac{a^x-1}{x}\big]^3.\frac{x\log a}{\sin(x\log a)}.\frac{x^2}{\log a.\log(1+x^2\log a^2)}\\
=(\log a)^3.1.\lim_{x\to0}\frac{x^2}{\log a.\log(1+x^2\log a^2)}\\
=(\log a)^2.1.\lim_{x\to0}\frac{x^2}{\log(1+x^2\log a^2)}=?
$$
Can I use the fact that $\log(1+x)=x-x2/2+x^3/3-....$ but is it not for $-1<x\leq1$ ?
|
Using series expansions and $o$-notation you get for $x\to 0$:
$$\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)} = \frac{(e^{x\log a}-1)^3}{(x\log a +o(x))(2x^2\log a+ o(x^3))}$$
$$= \frac{(x\log a+o(x))^3}{2x^3\log^2 a+ o(x^3)} = \frac{x^3\log^3 a+o(x^3)}{2x^3\log^2 a+ o(x^3)}$$
$$=\frac{\log^3 a+o(1)}{2\log^2 a+ o(1)} \stackrel{x\to 0}{\longrightarrow}\frac{\log a}{2}$$
|
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|
Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin x=\frac{-1}{2}$
and so $x=\frac{7\pi}{6},\frac{11\pi}{6}$
Solving for the other factor,
$\sin x-1=0 \Rightarrow \sin x=1$
And so $x=\frac{\pi}{2}$
Now we have found all our base solutions, and so ALL the solutions can be written as so:
$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$
|
Other way is used identities double angle and sum-product
\begin{eqnarray*}
2\cos^2x+\sin x& = & 1 \\
2\cos^2x-1+\sin x& = & 0\\
\cos(2x)+\sin x& = & 0\\
\cos(2x)+\cos\left(\frac{\pi}{2}-x\right) & = & 0\\
2\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)\cos\left(\frac{3x}{2}-\frac{\pi}{4}\right) & = & 0\\
\end{eqnarray*}
|
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|
If $\frac{\sin^3 x -\cos^3 x}{\sin x - \cos x}-\frac{\cos x}{\sqrt{1+\cot^2x}}-2\tan x \cot x=-1$ where $x\in [0,2\pi]$ If we solve the above expression, it’s very clear that the equation doesn’t depend upon x. Then all that is left to is to get
$$\sin x -\cos x \not =0$$
$$x\not = \pi /4, 5\pi/4$$
But the answer given is $$x\in (0,\pi)-{\pi/4,\pi/2}$$
What is the reason for leaving so many values out ie. $\pi , 2\pi$?
|
Since
$$\frac{\sin^3 x -\cos^3 x}{\sin x - \cos x}-\frac{\cos x}{\sqrt{1+\cot^2x}}-2\tan x \cot x=-1$$
$$\sin^2x+\sin x\cos x+\cos^2x-\frac{\cos x}{\sqrt{\frac{\sin^2 x+\cos^2x}{\sin^2x}}}-2=-1$$
$$\sin x\cos x-|\sin x|\cos x=0$$
which is always true for $\sin x\ge 0$ otherwise it is true for $x=0,\frac \pi 2, \frac{3\pi}2,2\pi$.
Therefore, since we also need $x\neq \frac \pi 2,\frac {3\pi} 2$ for $\tan x$ and $x\neq 0,2\pi$ for $\cot x$ all the solutions are given by $x\in (0,\pi)$ with $x\neq\pi/4$ and $x\neq \pi/2$.
|
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|
Is there an integer-sided right triangle with square perimeter and square hypotenuse? Is there an integer-sided right triangle with square perimeter and square hypotenuse?
Frenicle[89] noted (pp. 71-8) that if the hypotenuse and perimeter of a right triangle both are squares, the perimeter has at least 13 digits.
"History Of The Theory Of Numbers Vol-II" by Leonard Eugene Dickson
Chapter 4, p187
https://archive.org/details/HistoryOfTheTheoryOfNumbersVolII/page/n213
Let's suppose the right triangle be $\left(x,y,q^2\right)$,
then perimeter is $\left(x+y+q^2\right)$
\begin{align*}
x\!^{\phantom{1}}\,+\,y\!^{\phantom{1}}&=p^2-q^2\\
x^2+y^2&=q^4
\end{align*}
(not sure) I am pessimistic that there is no such integer-sided right triangle.
|
An elementary proof for primitive triangles
A primitive triangle has sides $2mn,m^2-n^2,m^2+n^2$ where $m,n$ are coprime and of opposite parity. We require $m^2+n^2$ and $2m(m+n)$ to be squares.
Then $m=2u^2, n=v^2-2u^2$ and $$v^4-4v^2u^2+8u^4=z^2,$$ where $u,v,z$ are coprime in pairs and $v,z$ are odd.
Then $(v^2-2u^2)^2-z^2=-4u^4$ and so
$$\left(\frac{v^2-2u^2+z}{2}\right)\left(\frac{v^2-2u^2-z}{2}\right)=-u^4$$
The two bracketed expressions differ by $z$ and so have no common factor with $u$. Therefore there are coprime integers $r,s$ such that $u=rs$ and $$ \left\{\frac{v^2-2u^2+z}{2},\frac{v^2-2u^2-z}{2}\right\}=\{r^4,-s^4\}$$
Then $v^2-2u^2=r^4-s^4$ and $r^4+2r^2s^2-s^4=v^2.$ Complete the square again.
$$(r^2+s^2)^2-v^2=2s^4$$ $$(r^2+s^2+v)(r^2+s^2-v)=2s^4$$
Similarly to before, we obtain coprime integers $X,Y$ such that $s=2XY$, with
$$\{r^2+s^2+v,r^2+s^2-v\}=\{2X^4,16Y^4\}$$
Then $$X^4-4X^2Y^2+8Y^4=r^2.$$
Now $u=2rXY$ and so comparing this equation with $v^4-4v^2u^2+8u^4=z^2$ shows that the existence of one solution implies the existence of a smaller solution ad infinitum. So, there is no such primitive triangle.
The general case
We require $\lambda(m^2+n^2)$ and $2\lambda m(m+n)$ to be squares and then$$8m(m+n)(m^2+n^2)\text { is a square}.$$
Let $x=\frac{2n}{m}$ then $x^3+2x^2+4x+8$ is a rational square. However
$$y^2=x^3+2x^2+4x+8$$
is an elliptic curve which according to CoCalc has rank $0$ with $(-2,0)$ as the only rational point.
|
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|
Show function is continuous at $(0,0)$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function with
$$\space f(x, y) = \begin{equation}
\begin{cases}
\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}, & (x, y) \neq (0, 0)\\
0, & (x, y) = 0
\end{cases}\end{equation}$$
Show that f is continious.
I've already shown that f is continous for $(x, y) \neq (0,0)$ but I am having trouble with finding a proof for $(0, 0)$ using epsilon-delta or limits of sequences. Can anyone help?
This is how far I got with simplifying for epsilon delta
$$
\Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{y^4*(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{y^2\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{\log(1+x^2y^2)}{\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\leq \big{|}\log(1+x^2y^2)\big{|}
$$
But now I am clueless on how to connect
$$
\big{|}log(1+x^2y^2)\big{|} < \epsilon \\
$$
and
$$
\sqrt{x^2+y^2} < \delta
$$
|
To conclude your proof we can use that by AM-GM
$$x^2y^2 \le \left(\frac{x^2+y^2}2\right)^2$$
$$\big{|}\log(1+x^2y^2)\big{|} \le x^2y^2 \le \left(\frac{x^2+y^2}2\right)^2=\frac{\left(\sqrt{x^2+y^2}\right)^4}{4}<\epsilon $$
as $\sqrt{x^2+y^2} < \sqrt[4]{4\epsilon}$.
As an alternative, we have that
$$\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}=\dfrac{\log(1+x^2y^2)}{x^2y^2}\dfrac{x^2y^4}{\sqrt{x^4+y^4}}\to 0$$
indeed by standard limits $\dfrac{\log(1+x^2y^2)}{x^2y^2} \to 1$ and
$$\dfrac{x^2y^4}{\sqrt{x^4+y^4}}=y^4\dfrac{\sqrt{x^4}}{\sqrt{x^4+y^4}}\le y^4 \to 0$$
|
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|
How to prove $\int^{\infty}_0 e^{-c^2/a^2}c^4\,dc=\frac{3}{8}a^5\sqrt\pi$? How do i prove this integral can someone give me proof i've been trying this too long i tried direct integration by parts, differentiation all to no avail.Please support us :)
$$\int^{\infty}_0 e^{\frac{-c^2}{a^2}}c^4\,dc=\frac{3}{8}a^5\sqrt\pi$$
|
Two times by parts:
$\int\limits_0^\infty e^\frac{-x^2}{a^2}dx=
\left.xe^\frac{-x^2}{a^2}\right|_0^\infty+
\int\limits_0^\infty \frac{2x^2}{a^2}e^\frac{-x^2}{a^2}dx=
\frac{2}{a^2}\int\limits_0^\infty e^\frac{-x^2}{a^2}d\left(\frac{x^3}{3}\right)=$
$
\frac{2}{a^2}\left(\left.\frac{x^3}{3}e^\frac{-x^2}{a^2}\right|_0^\infty+\int\limits_0^\infty \frac{x^3}{3}\cdot\frac{2x}{a^2}e^\frac{-x^2}{a^2}dx\right)=
\frac{4}{3a^4}\int\limits_0^\infty x^4e^\frac{-x^2}{a^2}dx$
Let alone $\int\limits_0^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$ or here be a known result, we obtain
$$\int\limits_0^\infty x^4e^\frac{-x^2}{a^2}dx=\frac{3a^4}{4}\int\limits_0^\infty e^\frac{-x^2}{a^2}dx=
\frac{3a^4}{4}\cdot a\int\limits_0^\infty e^{-x^2}dx=\frac{3a^5\sqrt{\pi}}{8}\hbox{, QED}$$
|
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|
Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\displaystyle\lim_{x\to0}{\frac{\cos{(2x)}\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{(2x)}}}$
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{2x}}}=\displaystyle\lim_{x\to0}{\frac{\cosh(3x^2)\cdot e^{8x^3}-1}{x^2\sin{(2x)}}}\cdot\displaystyle\lim_{x\to0}{\frac{1}{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1}}$
All of my attempts failed.
$$\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1\;\;\text{is continuous &decreasing }$$
Source: Matematička analiza 1, 2. kolokvij
|
$${\sqrt{\cosh(3x^2)}e^{4x^3}-1\over x^2\tan(2x)}={1\over\sqrt{\cosh(3x^2)}e^{4x^3}+1}\cdot{x\over\tan(2x)}\cdot{\cosh(3x^2)e^{8x^3}-1\over x^3}$$
and
$${\cosh(3x^2)e^{8x^3}-1\over x^3}={e^{8x^3}-1\over x^3}+e^{8x^3}{\cosh(3x^2)-1\over x^3}={e^{8x^3}-1\over x^3}+{e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over x^2\right)^2x$$
Now it's easy to see that
$${1\over\sqrt{\cosh(3x^2)}e^{4x^3}+1}\to{1\over1+1}={1\over2}$$
$${x\over\tan(2x)}={\cos x\over2}\cdot{2x\over\sin(2x)}\to{1\over2}\cdot1={1\over2}$$
$${e^{8x^3}-1\over x^3}=8\cdot{e^{8x^3}-1\over8x^3}\to8\cdot1=8$$
and
$${e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over x^2\right)^2x={e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over3x^2\right)^2(9x)\to{1\over1+1}\cdot1^2\cdot0=0$$
and thus
$${\sqrt{\cosh(3x^2)}e^{4x^3}-1\over x^2\tan(2x)}\to{1\over2}\cdot{1\over2}(8+0)=2$$
|
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|
$4x≡2\mod5$ can you divide both sides by $2$ to get $2x≡1\mod5\,?$ Since gcd$(2,5)=1$ , could you treat $4x$ as $2(2x)$ and cancel the $2$ on both sides?
i.e. $$2(2x)≡2\mod5\implies 2x≡1\mod5$$
Thanks!
|
Not really.
$4x \equiv 2\pmod 6$ (so $x$ could be $2\pmod 6$ or $x$ could be $5\pmod 6$) does not mean $2x \equiv 1\pmod 6$ (which is impossible).
But you can say
$ka \equiv kb \pmod n \implies a \equiv b \pmod {\frac n{\gcd(k, n)}}$
And so
$4x \equiv 2 \pmod 6$ meams $\frac {4x}2\equiv \frac 2{2}\pmod {\frac {6}{\gcd(2,6)}}$ or $2x =1 \pmod 3$ and that's fine. $x \equiv 2\pmod 3$ so $x\equiv 2\pmod 6$ or $x \equiv 3 \pmod 6$.
ANd to get to your problem:
$4x \equiv 2 \pmod 5$ means
$\frac {4x}2 \equiv \frac 22 \pmod {\frac 5{\gcd(2,5)}}$ so
$2x \equiv 1 \pmod {\frac 51}$ and
$2x \equiv 1 \pmod 5$.
So you CAN but only because $2$ and $5$ are relatively prime. You couldn't if they weren't. (but you could if you devide the modulus but the gcd.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Checking if series converge: $\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$ and $\sum_{k=1}^\infty \frac{(1+i)^k}{k!}$ etc. I want to see if these series converge:
$$\text{1. }\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$$
$$\text{2. }\sum_{k=1}^\infty \frac{(1+i)^k}{k!}$$
$$\text{3. }\sum_{k=1}^\infty \frac{k^2+2}{k^4+1}$$
$$\text{4. }\sum_{k=1}^\infty \frac{k^2 +2}{k^3+1}$$
Regarding $1.$ I used the alternating series test and got $a_k = \frac{1}{\sqrt{k}}$. The limit is $\lim_{k \to \infty} a_k = \frac{1}{\sqrt{k}} = 0$. So I have to show that $$a_k \geq a_{k+1}$$ $$\Leftrightarrow \frac{1\sqrt{k+1}}{\sqrt{k}} \geq 1$$ $$\Leftrightarrow \frac{\sqrt{k+1}}{\sqrt{k}} \geq 1$$ which is true.
Regarding $2.$ I used the ratio test which would give
$$|\frac{a_k+1}{k!}| = \frac{\frac{(1+i)^{k+1}}{(k+1)!}}{\frac{(1+i)^k}{k!}} = \frac{(1+i)^{k+1}\cdot k!}{(k+1)! \cdot (1+i)^k} = \frac{(1+i)^{k+1} \cdot k!}{k! (k+1) \cdot (1+i)^k} = \frac{(1+i)^{k+1}}{(k+1)\cdot (1+i)^k}$$, but where do I go from here?
Regarding $3.$ I used the ratio test as well and got
$$\lim_{k \to \infty} |\frac{a_{k+1}}{a_k}| = \frac{\frac{(k+1)^2+2}{(k+1)^4+1}}{\frac{(k^2+2)}{(k+1)^4+1}} = \frac{((k+1)^2+2) \cdot ((k+1)^4+1)}{((k+1)^4+1) \cdot (k^2 +2)} = \frac{(k+1)^2+2}{k^2+2} = \frac{k^2+2k+3}{k^2+2} = \frac{k^2 \cdot (\frac{2k}{k} + \frac{3}{k^2})}{k^2 \cdot \frac{2}{k}} = 2$$
Regarding $4.$ I just rewrote it as
$$\lim_{k \to \infty} \frac{k^2 \cdot \frac{2}{k^2}}{k^3 \cdot \frac{1}{k^3}} = 0$$
Can someone please help me? (I just started learning about sequences)
|
*
*What you did is fine.
*The absolute value of a complex number is always real and non-negative. So you have
$$
\left|\frac{a_{k+1}}{a_k}\right|= \left|\frac{(1+i)^{k+1}}{(k+1)(1+i)^k} \right|
=\left|\frac{(1+i)}{k+1}\right|=\frac{|1+i|}{k+1}=\frac{\sqrt2}{k+1}.
$$
*Your limit is wrong. It's $1$, which gives you no information. Here you would usually do comparison. Something like
\begin{align}
\sum_{n=1}^\infty\frac{k^2+2}{k^4+1}
&\leq\sum_{n=1}^\infty\frac{k^2+2}{k^4}
=\sum_{n=1}^\infty\frac{1}{k^2}+\sum_{n=1}^\infty\frac{2}{k^4}<\infty.
\end{align}
*Here you also want to do comparison. The terms are basically $1/k$, so you want should expect this to diverge. You can do
$$
\sum_{n=1}^\infty\frac{k^2+2}{k^3+1}\geq\sum_{n=1}^\infty\frac{k^2}{k^2+k^3}=\frac12\,\sum_{n=1}^\infty\frac1k=\infty.
$$
===========================
If you are weak with inequalities, you can do the comparison in 3 and 4 by taking limits instead. Namely, if the number $a_n$ and $b_n$ are positive and $\lim_{n\to\infty}\tfrac{a_n}{b_n}$ exists and is not zero, then $\sum_na_n$ converges if and only if $\sum_nb_n$ converges. So in 3,
$$
\frac{\frac{k^2+2}{k^4+1}}{\frac1{k^2}}=\frac{k^2(k^2+2)}{k^4+1}=\frac{1+2/k^2}{1+1/k^4}\to1
$$
tells you that the series behaves like $\sum_k\frac1{k^2}$. And in 4 you can do
$$
\frac{\frac{k^2+2}{k^3+1}}{\frac1k}=\frac{k(k^2+2)}{k^3+1}=\frac{1+2/k^2}{1+1/k^3}\to1.
$$
|
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|
Solve $(x^2+1)y''-2xy'+2y=0$
Solve $$(x^2+1)y''-2xy'+2y=0$$
Seems I can't use Euler Differential method. I tried it
\begin{align}
(x^2+1)y''-2xy'+2y&=0\\
\text{Let }y&=xv\\
(x^2+1)(xv''+2v')-2x(xv'+v)+2(xv)&=0\\
x(x^2+1)v''+2v'&=0\\
\frac{v''}{v'}&=-\frac{2}{x(x^2+1)}\\
\frac{v''}{v'}&=-\frac{2}{x}+\frac{2x}{x^2+1}
\end{align}
Can I integrate both side and treat LHS as $\int\frac{1}{v'}dv'?$ Any help will be appreciated.
Edit: Actually letting $y=xv$ made the work easy. But I suddenly think Is there any intuitive reason behind it$?$ Because I just take it as a guess without any investigate.
|
Hint: Put $x^2+1 = t$ and differentiate and put back in your ODE
Think backward, let $y=x^n$ maybe one of solution of your ODE. Then it must satisfy the ODE,
\begin{align}
y'&=nx^{n-1}\\
y''&=n(n-1)x^{n-2}
\end{align}
\begin{align}
(x^2+1)n(n-1)x^{n-2}-2xnx^{n-1}+2x^n&=0\\
(n^2-3n+2)x^n+n(n-1)x^{n-2}&=0\\
(n-1)(n-2)x^n+n(n-1)x^{n-2}&=0\\
(n-1)((n-2)x^n+nx^{n-2})&=0
\end{align}
Yes we luckily got that $y=x$ is one of the solution of your ODE. The rest of your work is simply Variation of parameters method .
|
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|
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Tried dividing by $y$ to get $$y^\prime+\frac{x}{y}=\frac{\sqrt{x^2+y^2}}{y}$$
$$(y^\prime)^2+2\frac{x}{y}+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2}$$
$$(y^\prime)^2+2\frac{x}{y}=1$$
$$y^\prime=\sqrt{1-2\frac{x}{y}}$$
Tried using $v=\frac{y}{x}$
$$y^\prime=v^\prime x+v$$
$$v^\prime x + v=\sqrt{1-2v^{-1}}$$
Not sure what to do now or if I've even done something right until now.
|
If $f = x^2 + y^2$ then we have $f' = 2\sqrt{f}$. That can be solved easily.
|
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|
how to prove the identity that $2^{ab}-1=(2^a-1)(2^{a(b-1)}+2^{a(b-2)}+.....+2^a+1)$ I am really confused.
The identity is given but I really want to know why$?$
Show that if $2^n -1$ is a prime , then $n$ is a prime . [Hint : Use the identity]
$$2^{ab} -1 = (2^a-1)\cdot(2^{a(b-1)} + 2^{a(b-2)} + \cdots + 2^a +1 )$$
|
Consider the following G.P :
$$1,2^a , 2^{2a} ,\cdots ,2^{a(b-2)} , 2^{a(b-1)} $$
We use the Formula to find it's sum :
$$S = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}$$
Or
$${\color{#d05}{1+2^a + 2^{2a}+\cdots +2^{a(b-2)} + 2^{a(b-1)} = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}}}$$
Rearranging this gives us the deisred result :
$${\color{#20f}{({2^a-1})(1+2^a + 2^{2a}+\cdots +2^{a(b-2)} + 2^{a(b-1)}) = {(2^{ab}-1)}}}$$
|
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|
Can any number of squares sum to a square? Suppose
$$a^2 = \sum_{i=1}^k b_i^2$$
where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct).
Can any positive integer be the value of $k$?
The reason I am interested in this: in a irreptile tiling where the smallest piece has area $A$, we have $a^2A = \sum_{i=1}^k b_i^2A$, where we have $k$ pieces scaled by $b_i$ to tile the big figure, which is scaled by $a$. I am wondering what constraints there are on the number of pieces.
Here is an example tiling that realizes $4^2 = 3^2 + 7 \cdot 1^2$, so $k = 8$.
|
Solutions exist for every $k>0$. The simplest forms use most of the $b_n$ values as $1$. I will list them as "$b_*$". I suspect there are infinitely many distinct answers for each $k\ge2$, but I can't prove it.
if $k = 2n$ (k is even) and large enough ($k\ge4$ or $n>1$):
*
*$ a = n $
*$ b_1 = n-1 $
*$ b_* = 1 $
\begin{align}
\sum_{i=1}^k b_i^2 & = (n-1)^2 + (2n-1)\cdot1^1 \\
& = n^2 - 2n + 1 + 2n - 1 \\
& = n^2 \\
& = a^2 \\
\end{align}
if $k = 4n + 1$ and large enough ($k\ge9$ or $n>1$)
*
*$ a = n+1 $
*$ b_1 = n-1 $
*$ b_* = 1 $
\begin{align}
\sum_{i=1}^k b_i^2 & = (n-1)^2 + ((4n+1)-1)\cdot1^1 \\
& = n^2 - 2n + 1 + 4n \\
& = n^2 +2n + 1 \\
&= (n + 1)^2 \\
& = a^2 \\
\end{align}
if $k = 4n + 3$
*
*$ a = 2n+3 $
*$ b_1 = 2n+2 $
*$ b_2 = 2 $
*$ b_* = 1 $
\begin{align}
\sum_{i=1}^k b_i^2 & = (2n+2)^2 + 2^2 + ((4n+3)-2)\cdot1^1 \\
& = 4n^2 + 8n + 4 + 4 + 4n + 1 \\
& = 4n^2 + 12n + 9 \\
&= (2n + 3)^2 \\
& = a^2 \\
\end{align}
The only values that don't work with these patterns are $k$ = 1, 2, or 5.
For $k=1$, $a=b_1$ for any values.
For $k=2$, we have we have a well known case, with minimal value $5^2=4^2+3^2$
For $k=5$, the minimal value is $4^2=3^2+2^2+1^2+1^2+1^2$
|
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|
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
|
I will be giving a $\varepsilon -\mathcal{N}$ proof
notice
$$\begin{align}\frac{n}{1+n^2} + \frac{n}{2+n^2} + \cdots + \frac{n}{n+n^2} - 1=
n\Big( \frac{1}{1+n^2} - \frac{1}{n^2} + \cdots + \frac{1}{n+n^2} - \frac{1}{n^2} \Big) = h_n\end{align}$$
Claim $\dfrac{n}{n+n^2} > \dfrac{r}{r+n^2}$
So
$|h_n| < \Big|n\cdot(\dfrac{1}{n^2}\cdot\dfrac{n^2}{n+n^2})\Big|=\dfrac{1}{n+1}<\dfrac{1}{n}$
so $\forall n > \dfrac{1}{\varepsilon}$
$|h_n| < \varepsilon$
|
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|
Limit of sequence-term not always positive Let the sequence $a_n=\frac{3^n+2(-4)^n}{(-4)^n-2^n}$.
We have $$\frac{3^n+2(-4)^n}{(-4)^n-2^n}=\frac{3^n+2(-1)^n 4^n}{(-1)^n 4^n-2^n}=\frac{(-1)^n 4^n \left( 2+ \frac{3^n}{(-1)^n 4^n}\right)}{(-1)^n 4^n \left( {1-\frac{2^n}{(-1)^n 4^n}}\right)}$$
In this case the ineqaulity $-1 \leq (-1)^n \leq 1 \Rightarrow -4^n \leq (-1)^n 4^n \leq 4^n$ does not help, since we cannot have an inequality with these terms as denominators, because we don't know if $(-1)^n 4^n$ is positive or negative. Right?
How else can we we compute the limit of the sequence?
|
You have
\begin{align}
\frac{3^n+2(-4)^n}{(-4)^n-2^n}
&=\frac{3^n}{(-4)^n-2^n}+\frac{2(-4)^n}{(-4)^n-2^n}\\ \ \\
&=\frac{3^n}{4^n((-1)^n-(1/2))^n}+\frac{2}{1-2^n/(-4^n)}\\ \ \\
&\to 0+2=2
\end{align}
|
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|
Finding the eigenvalues of a $3 \times 3$ matrix without the determinant I am trying to find the eigenvalues of
$$A = \begin{pmatrix} 0 & -2 & -3 \\ -1 & 1 & -1 \\ 2 & 2 & 5\end{pmatrix}$$
Without using the determinant. I first tried doing something like I did here Finding the eigenvalues of $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ matrix without the determinant
And from the system
$\begin{cases}
-\lambda x_1 & -2x_2 & -3x_3 &= 0 \\
-x_1 &+ \ (1-\lambda)x_2 &- x_3 &= 0 \\
2x_1 &+ \ 2x_2 &+ \ (5 - \lambda)x_3 &= 0
\end{cases}$
Tried to solve for
$\cfrac{-\lambda x_1 - 2x_2}{3} = -x_1 + (1 - \lambda)x_2 = \cfrac{2x_1 + 2x_2}{\lambda - 5}$
But nothings seems to come of that. I then proceeded to try to find $\lambda$ such that
$$\begin{pmatrix} -\lambda & -2 & -3 \\ -1 & 1-\lambda & -1 \\ 2 & 2 & 5-\lambda\end{pmatrix}$$ is non-invertible, by looking at the reduced row echelon form and see for what $\lambda$ the rank is less than 3.
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & \lambda^2 - \lambda - 2 & \lambda - 3 \\ 0 & 2(2-\lambda) & 3 - \lambda \end{pmatrix}
\rightarrow
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 2(2-\lambda) & 3 - \lambda \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 2$. Another operation yields:
$\begin{equation}
\begin{split}
\rightarrow
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 0 & 3 - \lambda \Big (1 + \cfrac{2(2-\lambda}{\lambda ^2 - \lambda - 2} \Big) \end{pmatrix} \rightarrow
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 0 & 1 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 1, 3$ (because these are roots of the polynomial).
Now it happens that the eigenvalues are indeed $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 3$. But I'm not sure how what I've done makes sense as a mathematical process for finding eigenvalues. Moreover it doesn't even seem to fully work in this other example
$B = \begin{pmatrix} 1 & 5 \\ 3 & 3 \end{pmatrix}$
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 - \lambda & 5 \\ 3 & 3 - \lambda \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 3 & 3 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 5$.
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 0 & \cfrac{-15}{1 - \lambda} + 3 - \lambda \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 0 & 1 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq -2, 6$.
Here it happens that $B$ has eigenvalues $\lambda_1 = -2, \lambda_2 = 6$, but there is no eigenvalue for $1$. So what's going on here?
|
\begin{cases}
-\lambda x_1 & -2x_2 & -3x_3 &= 0 \\
-x_1 &+ \ (1-\lambda)x_2 &- x_3 &= 0 \\
2x_1 &+ \ 2x_2 &+ \ (5 - \lambda)x_3 &= 0
\end{cases}
What you are doing above is finding the null space of the matrix $A-\lambda I$. This consists of all vectors $v$ such that $Av=\lambda v$. This gives you equations that will allow you to find the eigenvalues of the matrix.
|
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|
Approximating Integrals within an error with Maclaurin Series "Use Maclaurin's Series to approximate the integral to 3dp accuracy"
$$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}} $$
I was wondering if it is possible to solve this question by identifying out the number of terms needed to obtain a 3dp accuracy. I thought of using the Taylor Series Remainder Theorem, but it seems quite difficult with this binomial expression.
|
Using the binomial expansion, we have
$$ \frac{1}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} x^{2 n}$$ which, integrated gives
$$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \frac{2^{-(2 n+1)}}{2 n+1} \binom{-\frac{1}{4}}{n}$$
So, summing $p$ terms, you look for $p+1$ such that
$$R_p=\Big| \frac{2^{-(2 p+3)}}{2 p+3} \binom{-\frac{1}{4}}{p+1}\Big| \leq \epsilon$$
Using the representation in terms of the gamma function we have
$$R_p=\frac{2^{-(2 p+3)}}{2 p+3} \frac{\Gamma \left(\frac{3}{4}\right)}{\Gamma (p+2) \left|\Gamma \left(-p-\frac{1}{4}\right)\right|}$$ and using the reflection formula this becomes
$$R_p=\frac{2^{-(2 p+3)} \Gamma \left(\frac{3}{4}\right) \Gamma \left(p+\frac{5}{4}\right)
\left|\sin \left(\left(p+\frac{5}{4}\right) \pi \right)\right|}{\pi (2 p+3)
\Gamma (p+2)}=\frac{2^{-(2 p+\frac{7}{2})} \Gamma \left(\frac{3}{4}\right) \Gamma
\left(p+\frac{5}{4}\right)}{\pi (2 p+3) \Gamma (p+2)}$$ Now, using Stirling approximations and continuing with Taylor series to $O\left(\frac{1}{p}\right)$,we have to solve for $p$
$$-2 p \log (2)-\frac{7 \log (p)}{4}+\log \left(\frac{\Gamma
\left(\frac{3}{4}\right)}{16 \sqrt{2} \pi }\right)=\log(\epsilon)$$ and the solution is given in terms of Lambert function
$$p=\frac{7 }{8 \log (2)}W\left(k \right)\qquad \text{where} \qquad k=\frac{2^{3/7} \log (2)}{7} \left(\frac{\Gamma \left(\frac{3}{4}\right)}{16 \sqrt{2} \pi \epsilon }\right)^{4/7}$$ Numerically, this gives
$$p\sim 1.26236 \, W\left(\frac{0.0130926 } { \epsilon^{4/7}}\right)$$ Computed for $\epsilon=10^{-10}$, this would give $p=8.70$ while the exact solution would be $p=8.76$.
|
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|
Differentiation by Substitution Find $\frac{dy}{dx}$ if $$y=sin^{-1}\frac{1-x^2}{1+x^2}$$
Let us put $t=\frac{1-x^2}{1+x^2}$ then $y=sin^{-1}t$ So $$\frac{dy}{dx}=\frac{dy}{dt}×\frac{dt}{dx}$$ so we get $$\frac{dy}{dx}=\frac{d}{dt}(sin^{-1}t)×\frac{d}{dx}(\frac{1-x^2}{1+x^2})=\frac{1}{\sqrt{1-(\frac{1-x^2}{1+x^2})^2}}×\frac{-4x}{(1+x^2)^2}=\frac{1}{\sqrt{\frac{4x^2}{(1+x^2)^2}}}×\frac{-4x}{(1+x^2)^2}=\frac{1+x^2}{\sqrt{4x^2}}×\frac{-4x}{(1+x^2)^2} =\frac{1+x^2}{2x}×\frac{-4x}{(1+x^2)^2}=\frac{-2}{1+x^2}$$
$$$$ But I want to know that here we used $\sqrt{4x^2}=2x$ but if we use $\sqrt{4x^2}=-2x$ then we will get $$\frac{dy}{dx}=\frac{2}{1+x^2}$$, But if we substitute $x=tan\theta$ then we will get $$\frac{dy}{dx}=\frac{-2}{1+x^2}$$ for every $x$. Also in the method used above we have used $\frac{x}{x}=1$ but this is only true for $x$ not equal to $0$. How to explain all this?
|
Both your questions can be explained by looking at the graph of $y$.
*
*You'll notice that $y$ is not differentiable at $x = 0$ so your concern about $x/x = 1$ doesn't matter because there is no derivative there.
*You'll notice that for $x > 0$ the derivative is negative and for $x < 0$ the derivative is positive. On the other hand, $2/(1 + x^2)$ is always positive and $-2/(1+x^2)$ is always negative. So actually the derivative is
$$
\frac{dy}{dx} = \begin{cases}
\frac{2}{1+x^2} & x < 0 \\
\text{undefined} & x = 0 \\
-\frac{2}{1+x^2} & x > 0
\end{cases}
$$
What's going on here is that $\sqrt{u}$ is by definition always the non-negative square root. (See why is the range of the function $\sqrt x$ the set of numbers $\geq 0$). Therefore,
$$ \sqrt{x^2} = |x| $$
and, whereas $\lim_{x\to 0} x/x = 1$, the function $|x|/x$ does not have a limit as $x \to 0$ and hence the whole derivative is undefined at $x = 0$.
|
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|
Showing that for $p_k\ \ge 31, p_{k-1}\# > (p_{k+1})^2$ I am trying to show that for $p_k \ge 31, p_{k-1}\# > (p_{k+1})^2$
Does the following work? Is there a stronger or more straight forward result that can be shown?
Let:
*
*$p_k$ be the $k$th prime
*$p\#$ be the primorial for $p$.
Base Case: $7\# = 210 > 13^2 = 169$
Assume that $p_{k-1}\# > (p_{k+1})^2$
If $p_k \ge 31$, there are at least $4$ primes $q$ where $p_{k-1} < q < 2p_{k-1}$ [See here for details]
It follows that $p_{k-1} < p_{k+1} < p_{k+2} < p_{k+3} < 2p_{k-1}$
Let $c = p_{k+2} - p_{k+1} < 2p_{k-1} - 2 - p_{k-1} = p_{k-1}-2$
$(p_k-1)(p_{k-1}\#) > (p_{k}-1)(p_{k+1})^2> c(p_{k+1})^2 = c(2p_{k+1}) + c(p_{k+1}-2)(p_{k+1}) > 2cp_{k+1} + c^2$
$p_k\# > (p_{k+1})^2 + 2cp_{k+1}+c^2 = (p_{k+1}+c)^2 = (p_{k+2})^2$
|
I think your solution if fine. But you might find this a bit tighter ?
We have
\begin{eqnarray*}
p_{k-1} < p_{k+1} < p_{k+2} < p_{k+3} < 2p_{k-1}.
\end{eqnarray*}
So
\begin{eqnarray*}
p_{k} \#= p_k p_{k-1} \# >p_k p_{k+1}^2 > 4p_k^2 > p_{k+2}^2.
\end{eqnarray*}
|
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|
Would this be a legal move in a Matrix? Here I have a matrix:
\begin{pmatrix}
b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{pmatrix}
I was wondering can I do the following operation on this:
$C2 \to C2 \times C2$
So it would give me the following:
\begin{pmatrix}
b^2c^2 & b^2c^2 & b+c \\
c^2a^2 & c^2a^2 & c+a \\
a^2b^2 & a^2b^2 & a+b \\
\end{pmatrix}
This is needed in order to prove the determinant is equal to zero.
|
You need to think in terms of homogeneous symmetric polynomials' degrees. Subtract $ab+bc+ca$ times the second column from the first, thus changing $b^2c^2$ to $-abc(b+c)$. Now add $abc$ times the third column to the first; it should vanish then. Explicitly$$\left(\begin{array}{c}
b^{2}c^{2}\\
c^{2}a^{2}\\
a^{2}b^{2}
\end{array}\right)-\left(ab+bc+ca\right)\left(\begin{array}{c}
bc\\
ca\\
ab
\end{array}\right)+abc\left(\begin{array}{c}
b+c\\
c+a\\
a+b
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right).$$
|
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|
If $\tan2\theta=\frac{b}{a-c}$, then $\cos 2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}}$ and $\sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}}$ I am studying general equation of the second degree. While studying that chapter I came across
$$\tan2\theta=\frac{b}{a-c} \tag{1}$$
Now from (1), the author computed
$$\cos2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}} \quad\text{and}\quad \sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}} \tag{2}$$
I don't understand how did the author compute these terms?
If any member knows the correct answer may reply to this question?
|
From
$$\sin^2\alpha+\cos^2\alpha=1,$$
draw
$$\tan^2\alpha+1=\frac1{\cos^2\alpha}$$
and
$$\cos\alpha=\pm\frac1{\sqrt{\tan^2\alpha+1}}.$$
Then
$$\sin\alpha=\tan\alpha\cos\alpha=\pm\frac{\tan\alpha}{\sqrt{\tan^2\alpha+1}}.$$
|
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|
Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n+1}$
I couldn't come up with anything like that. I feel like I'm confused with something.
|
And if you want to generate
$(a_i)_{i=1}^m$ repeatedly,
the i-th term is generated by
$a_ix^i$,
and for this to repeat every $m$
this needs
$\dfrac{a_ix^i}{1-x^m}
$
so the final result is
$\sum_{i=1}^m\dfrac{a_ix^i}{1-x^m}
$.
In your case,
with $(1, 2)$
repeated,
this is
$\dfrac{x}{1-x^2}+\dfrac{2x^2}{1-x^2}
=\dfrac{x+2x^2}{1-x^2}
$.
This starts the first term at $x^1$.
To start it at the constant term,
just reduce the exponents
of the $x^i$ by one giving
$\sum_{i=1}^m\dfrac{a_ix^{i-1}}{1-x^m}
$,
so this gives,
in this case
$\dfrac{1}{1-x^2}+\dfrac{2x}{1-x^2}
=\dfrac{1+2x}{1-x^2}
$.
|
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|
There is no continuous surjective multiplicative map from $M_n(\mathbb H)$ to $\mathbb H$ Let $\mathbb H$ denote the field of quaternions. I would like to prove that there does not exist any function $f:M_n(\mathbb H)\rightarrow \mathbb H$ for $n\geq 2$ that is continous surjective and multiplicative.
I have been thinking about this problem for a while but I can't find any contradiction assuming that such a function does exist. I tried considering preimages for $1,i,j,k$ and toying with them, I also tried infering the values of some specific matrices (the $\lambda I_n$, the nilpotent matrices, etc...) but I couldn't reach any conclusion. Mostly, I fail to see how to make use of the continuity here.
Would somebody have a hint as to how to proceed with this problem?
|
I will work out the case $n = 2$ in detail. The same proof works for general $n$, I just want to save the labor of typing $n$ by $n$ matrices...
Thus assume that $f:\operatorname M_2 = \operatorname M_2(\Bbb H) \rightarrow \Bbb H$ is a surjective multiplicative map.
Lemma 1. Whenever $A\in \operatorname M_2$ is invertible, the image $f(A)$ is also invertible.
Proof: If $A$ is invertible, then multiplication by $A$ is a bijection on $\operatorname M_2$. Hence $f(A)$ cannot be zero, otherwise $f$ is constantly zero.
Lemma 2. The map $f$ restricted to $\operatorname{GL}_2 = \operatorname{GL}_2(\Bbb H)$ gives a group homomorphism from $\operatorname{GL}_2$ to $\Bbb H^\times$. In particular, we have $f\begin{pmatrix} 1 & \\ & 1\end{pmatrix} = 1$.
Proof: This is clear from Lemma 1.
We want to arrive at a contradiction, hence showing that such an $f$ does not exist.
Assumption 3. Without loss of generality, we may assume that $f\begin{pmatrix} & 1\\1 &\end{pmatrix} = 1$.
Note: for general $n$, we have the canonical embedding of the symmetric group $S_n$ into $\operatorname{GL}_n$, and this assumption becomes: $f(\sigma) = 1$ for all $\sigma \in S_n$.
Why we can make this assumption: we have $f(\sigma)^{n!} = f(\sigma^{n!}) = 1$ by Lemma 2, hence by changing $f$ to $f^{n!}$, which is still surjective multiplicative, we may make this assumption.
From now on, we always make Assumption 3.
Lemma 4. We have $f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}$ for any $\lambda \in \Bbb H$.
Proof: This comes from the identity $\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}\begin{pmatrix} & 1\\1 & \end{pmatrix} = \begin{pmatrix} & 1\\1 & \end{pmatrix}\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}$ and Assumption 3.
Lemma 5. We have $f\begin{pmatrix}1 & \\ & z\end{pmatrix} = 1$ for all $z \in \Bbb H^\times$ with $|z| = 1$.
Proof: For any $\lambda, \mu \in \Bbb H^\times$, we have: $$f\begin{pmatrix}1 & \\ & \lambda\mu\end{pmatrix} = f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}=f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}=f\begin{pmatrix}\lambda & \\ & \mu\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\lambda\end{pmatrix}.$$ Therefore we have $f\begin{pmatrix}1 & \\ & \lambda\mu\lambda^{-1}\mu^{-1}\end{pmatrix} = 1$. But any $z \in \Bbb H^\times$ with $|z| = 1$ can be written as $\lambda\mu\lambda^{-1}\mu^{-1}$ for some $\lambda, \mu \in \Bbb H^\times$.
Lemma 6. For any $a\in \Bbb R$, the value $f\begin{pmatrix}a & \\ & a\end{pmatrix}$ is real.
Proof: Since the matrix $A = \begin{pmatrix}a & \\ & a\end{pmatrix}$ is in the center of $\operatorname M_2$, we have $f(A)f(B) = f(AB) = f(BA) = f(B)f(A)$ for all $B\in \operatorname M_2$. The surjectivity of $f$ then implies that $f(A)$ lies in the center of $\Bbb H$, namely $\Bbb R$.
Assumption 7. Without loss of generality, we may assume that $f\begin{pmatrix}1 & \\ & a\end{pmatrix}$ is real for all $a \in \Bbb R$.
Why we can make this assumption: we already have $$\left(f\begin{pmatrix}1 & \\ & a\end{pmatrix}\right)^2 = f\begin{pmatrix}1 & \\ & a\end{pmatrix}f\begin{pmatrix}a & \\ & 1\end{pmatrix} = f\begin{pmatrix}a & \\ & a\end{pmatrix}\in \Bbb R.$$Therefore, by changing $f$ to $f^2$, we may make this assumption (while still keeping all required properties of $f$, including Assumption 3).
From now on, we always make Assumptions 7.
Lemma 8. We have $f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}\in \Bbb R$ for all $\lambda \in \Bbb H$.
Proof: The case $\lambda = 0$ is covered by Assumption 7. For $\lambda \neq 0$, by Lemma 5 and Assumption 7, we have: $f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}1 & \\ & |\lambda|\end{pmatrix}f\begin{pmatrix}1 & \\ & \frac \lambda {|\lambda|}\end{pmatrix}\in \Bbb R$.
Lemma 9. For any $\alpha \in \Bbb H^\times$, we have $f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = f\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}$.
Proof: This comes from the identity $\begin{pmatrix}1 & \\ & \alpha^{-1}\end{pmatrix}\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}\begin{pmatrix}1 & \\ & \alpha\end{pmatrix} = \begin{pmatrix}1 & \alpha\\ & 1\end{pmatrix}$ and the fact that $f\begin{pmatrix}1 & \\ & \alpha\end{pmatrix}$ is a real number, hence is in the center of $\Bbb H$.
Lemma 10. For any $\alpha \in \Bbb H$, we have $f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = 1$.
Proof: Let $h$ be the value of $f\begin{pmatrix}1 & 1 \\ & 1\end{pmatrix}$. By Lemma 9, we have $h = f\begin{pmatrix}1 & 2 \\ & 1\end{pmatrix} = h^2$. By Lemma 1, we get $h = 1$ and Lemma 9 tells us that $f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = 1$ for any $\alpha \in \Bbb H^\times$. The case $\alpha = 0$ is Lemma 2.
Conclusion. The map $f$ takes values in $\Bbb R$ on $\operatorname M_2$, hence is not surjective. We obtain a contradiction.
Proof: Just note that any matrix in $\operatorname M_2$ can be written as a product of matrices of the form $\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix}$, $\begin{pmatrix} & 1 \\1 & \end{pmatrix}$, $\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}$ with $\alpha, \lambda \in \Bbb H$ (by performing "row and column operations").
Final remarks.
*
*As claimed in the very beginning, the proof adapts without difficulty to general $n$.
*The continuous assumption is not used. All arguments are algebraic.
*Since it's a proof by contradiction, it doesn't show that any multiplicative map from $\operatorname M_n(\Bbb H)$ to $\Bbb H$ has image in $\Bbb R$. But it is true that any group homomorphism from $\operatorname{GL}_n(\Bbb H)$ to $\Bbb C^\times$ must factorize through $\Bbb R^\times_+$, as the abelianization of $\operatorname{GL}_n(\Bbb H)$ is isomorphic to $\Bbb R^\times_+$.
|
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|
Prove that matrix is invertible I have a symmetric real-valued matrix with the following structure:
\begin{bmatrix}
1 + (n-1)\alpha_{1}^2 & 1-\alpha_{1}\alpha_{2} & ... & 1-\alpha_{1}\alpha_{n} \\
1-\alpha_{1}\alpha_{2} & 1+(n-1)\alpha_{2}^2 & ... & 1-\alpha_{2}\alpha_{n} \\
\vdots & \vdots & \ddots & \vdots\\
1-\alpha_{1}\alpha_{n} & 1-\alpha_{2}\alpha_{n} & ... & 1+(n-1)\alpha_{n}^2
\end{bmatrix}
I am trying to show that this matrix has non-zero determinant for all $n$ and therefore is invertible. I am seeing this is the case empirically. I would appreciate any hints or ideas to prove this formally.
Thanks
Daniel
|
Let us assume that $\alpha_i\neq 0$ for $i\in\{1,\ldots,n\}.$
Let $X$ be the original matrix and
$$
A = n\cdot \begin{pmatrix}
\alpha_1^2 & & & \\
& \alpha_2^2 & & \\
& & \ddots & \\
& & & \alpha_n^2
\end{pmatrix}
$$
and
$$
U = \begin{pmatrix}
1 & \alpha_1 \\
1 & \alpha_2 \\
\vdots & \vdots \\
1 & \alpha_n
\end{pmatrix}
\;\;\; , \;\;\;
C = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\;\;\; , \;\;\;
V = \begin{pmatrix}
1 & 1 & \cdots & 1 \\
-\alpha_1 & -\alpha_2 & \cdots & -\alpha_n
\end{pmatrix}
$$
Then $X = A+UCV,$ $A$ is invertible and we can use the Woodbury matrix identity
$$
(A+UCV)^{-1} = A^{-1} - A^{-1}U(C^{-1}+VA^{-1}U)^{-1}VA^{-1}
$$
if $C^{-1}+VA^{-1}U$ is invertible.
After some calculations, we get
$$
C^{-1}+VA^{-1}U = I +
\frac{1}{n}\begin{pmatrix}
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i^2} &
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} \\
-\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} & -n
\end{pmatrix}
=
\frac{1}{n}\begin{pmatrix}
n+\sum\limits_{i=1}^{n}\frac{1}{\alpha_i^2} &
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} \\
-\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} & 0
\end{pmatrix}
$$
which is obviously invertible if and only if
$$
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} \neq 0
$$
In case that $\sum\limits\frac{1}{\alpha_i} = 0,$ it can easily be seen that
$$
X\cdot\begin{pmatrix} \alpha_1^{-1} \\ \alpha_2^{-1} \\ \vdots \\ \alpha_n^{-1}
\end{pmatrix} = 0
$$
which means that $X$ cannot be invertible in this case.
It can also easily be seen that the matrix $X$ is not invertible if two or more of the $\alpha_i$ are $0.$
So the only case left is the case in which there is exactly one $i$ such that $\alpha_i=0.$ Wolog $\alpha_1 = 0.$ We can modify the first part of the proof a little bit:
$$
A = n\cdot \begin{pmatrix}
1 & & & \\
& \alpha_2^2 & & \\
& & \ddots & \\
& & & \alpha_n^2
\end{pmatrix}
$$
and
$$
U = \begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & \alpha_2 \\
\vdots & \vdots & \vdots \\
0 & 1 & \alpha_n
\end{pmatrix}
\;\;\; , \;\;\;
C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
\;\;\; , \;\;\;
V = \begin{pmatrix}
-n & 0 & \cdots & 0 \\
1 & 1 & \cdots & 1 \\
0 & -\alpha_2 & \cdots & -\alpha_n
\end{pmatrix}
$$
Then again $X=A+UCV.$ This time we get
$$
C^{-1}+VA^{-1}U =
\frac{1}{n}\begin{pmatrix}
0 & -n & 0 \\
1 & n+1+\sum\limits_{i=2}^{n}\frac{1}{\alpha_i^2} &
\sum\limits_{i=2}^{n}\frac{1}{\alpha_i} \\
0 & -\sum\limits_{i=2}^{n}\frac{1}{\alpha_i} & 1
\end{pmatrix}
$$
As $\det(C^{-1}+VA^{-1}U)=\frac{1}{n^2}$, $X$ is always invertible if exactly one of the $\alpha_i$ is $0.$
|
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|
$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
|
Note,
$$\frac17(x^2-y^2)=\frac1{37}( x^2+xy+y^2)$$
or
$$44y^2+7xy-30x^2=0$$
which leads to $x=\frac43y$ and $x=-\frac{11}{10}y$. Plug them into $x^2-y^2=7$ to obtain the solutions
$$(4,3),\>(-4,-3), \>(-\frac{11}{\sqrt3},\frac{10}{\sqrt3}), \>(\frac{11}{\sqrt3},-\frac{10}{\sqrt3})$$
|
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|
What is the angle between the asymptotes of the hyperbola $5x^2-2\sqrt 7 xy-y^2-2x+1=0$?
What is the angle between the asymptotes of this hyperbola?
$$5x^2-2\sqrt 7 xy-y^2-2x+1=0$$
I used $S+\lambda=0$ and used straight line condition to find combined equation to asymptotes. Then how to find angle between them?
|
Rearranging lab's answer ...
Omit the lower-degree terms of $5x^2-2\sqrt 7 xy-y^2-2x+1=0$ to get $5x^2-2\sqrt 7 xy-y^2=0$. This is the equation of two lines which are parallel to your asymptotes. So find the angle between these two lines.
Factor:
$$
5x^2-2\sqrt 7 xy-y^2 =
-\frac{\left( 2\,y\sqrt {3}-\sqrt {7}y+5\,x \right) \left( 2\,y\sqrt
{3}+\sqrt {7}y-5\,x \right) }{5}
$$
Solving and rationalizing, the lines are
$$
y=(2\sqrt{3}-\sqrt{7})x,\quad y=(-2\sqrt{3}-\sqrt{7})x
$$
Use the addition formula for tangents as lab suggests.
|
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|
$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it right?
|
If you want more than the limit itself
$$y_n= \left(1+ \frac{3}{n^2+n^4}\right)^n\implies \log(y_n)=n \log\left(1+ \frac{3}{n^2+n^4}\right)$$
Using Taylor expansion
$$ \log\left(1+ \frac{3}{n^2+n^4}\right)=\frac{3}{n^4}-\frac{3}{n^6}+O\left(\frac{1}{n^8}\right)$$
$$\log(y_n)=\frac{3}{n^3}-\frac{3}{n^5}+O\left(\frac{1}{n^7}\right)$$
$$y_n=e^{\log(y_n)}=1+\frac{3}{n^3}-\frac{3}{n^5}+\frac{9}{2n^6}+O\left(\frac{1}{n^7}\right)$$ which shows the limit and how it is approached.
Moreover, this gives you a shortcut for the evaluation of $y_n$.
Consider $n=5$ and use your pocket calculator
$$y_5=\frac{118731486838493}{116029062500000}\approx 1.023291$$ while the above truncated expansion gives $\frac{31979}{31250}=1.023328$ (corresponding to a relative error equal to $3.6 \times 10^{-3}$%).
|
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|
Number of ordered pairs of $A,B$ in Probability
Let $|X|$ denote the number of elements in a set $X,$
Let $S = \{1,2,3,4,5,6\}$ be a sample space,
where each element is equally likely to occur.
If $A$ and $B$ are independent events associated with
$S,$
Then the number of ordered pairs $(A,B)$
such that $1 \leq |B| < |A|,$ equals
what i try
Let number of elements in $A,B,A \cap B$ is $x,y,z$ respectively
conditions given as $1\leq y\leq x.$
For $2$ Independent events $A,B$
is $\displaystyle P(A \cap B)=P(A)\cdot P(B)\Rightarrow \frac{z}{6}=\frac{x}{6}\cdot \frac{y}{6}\Rightarrow z=xy/6$
|
The equation
$$z=\frac{xy}{6}$$
shows that the events $A$ and $B$ can be independent only if
*
*$A=S$, or
*$|A|=3,|B|=2$
*$|A|=4,|B|=3$
In the 2nd case $A$ and $B$ have 1 common element; in the 3rd case $A$ and $B$ have 2 common elements.
In the 1st case, we have $2^6-2$ solutions ($B$ can be anything except $S$ and $\varnothing$).
In the 2nd case, we can choose $A$ by $\binom{6}{3}$ ways and then $B$ by $3\cdot 3$ ways.
In the 3rd case we can choose common elements by $\binom{6}{2}$ ways, then 2 remaining A elements by $\binom{4}{2}$ ways and the remaining B element by 2 ways.
Summing it all
$$64-2+\frac{6\cdot 5\cdot 4}{6}\cdot 3 \cdot 3 + \frac{6\cdot 5}{2}\cdot \frac{4\cdot 3}{2}\cdot 2= 422$$
PS: thanks @karthikeya kurella for noticing the 3rd case.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$. Problem: Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$.
My efforts: $(y-a)^2(x^2-a^2)=x^4+a^4\implies (y-a)^2=\dfrac{x^4+a^4}{x^2-a^2}\implies x=\pm a$ are vertical asymptotes. What are the remaining asymptotes?
|
Rewrite the equation as
$$y=a \pm \sqrt{\frac{x^4+a^4}{x^2-a^2}}$$
The vertical asymptotes $x=\pm a $ are directly identified. Since
$$\lim_{x^2\to\infty} y(x) = a\pm x$$
$y=a\pm x $ are the slant asymptotes.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Minimal polynomial with $f(\alpha)=0$ I'm learning this stuff for the first time so please bear with me.
I'm given $\alpha=\sqrt{3}+\sqrt{2}i$ and asked to find the minimal polynomial in $\mathbb{Q}[x]$ which has $\alpha$ as a root.
I'm ok with this part, quite sure I can find such a polynomial like so:
$\alpha = \sqrt{3}+\sqrt{2}i$
$\iff \alpha - \sqrt{3} = \sqrt{2}i$
squaring both sides,
$\iff \alpha^2 -2\sqrt{3}\alpha + 3 = -2$
$\iff \alpha^2+5 = 2\sqrt{3}\alpha$
again squaring both sides,
$\iff \alpha^4+10\alpha^2+25 = 12\alpha^2$
$\iff \alpha^4-2\alpha^2+25 = 0$
Therefore $f(x)=x^4-2x^2+25$ should have $f(\alpha)=0$.
My question is how can I be sure this polynomial is minimal, i.e., can I be sure there isn't some lower degree polynomial out there with $\alpha$ as a root?
|
It suffices to prove that $1,\alpha,\alpha^2,\alpha^3$ are linearly independent over $\mathbb Q$. Writing them in the basis $1,\sqrt{3},\sqrt{2}i,\sqrt{6}i$ of $\mathbb Q[\sqrt{3},\sqrt{2}i]$ we get:
$$
\begin{pmatrix}
1 \\ \alpha \\ \alpha^2 \\ \alpha^3
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 2 \\ 0 & -3 & 7 & 0
\end{pmatrix}
\begin{pmatrix}
1 \\ \sqrt{3} \\ \sqrt{2}i \\ \sqrt{6}i
\end{pmatrix}
$$
The key point is that the matrix is invertible.
You can prove this by row reduction or by computing its determinant and checking that it is not zero.
|
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|
Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos^23x+\cos x\cos3x\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\dfrac{1+\cos6x}{2}+\dfrac{1}{2}\left(\cos4x+\cos2x\right)\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{4}\left(1+\cos6x+\cos4x+\cos2x\right)}{x^2}$$
Applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-6\sin6x-4\sin4x-2\sin2x\right)}{2x}$$
Again applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-36\cos6x-16\cos4x-4\cos2x\right)}{2}=\dfrac{36+16+4}{8}=\dfrac{56}{8}=7$$
But actual answer is $3$, what am I messing up here?
|
$$P_n=\prod_{k=1}^n\cos(kx)\implies \log(P_n)=\sum_{k=1}^n\log(\cos(kx))$$ Now, by Taylor expansion
$$\log(\cos(kx))=-\frac{k^2}{2}x^2-\frac{k^4 }{12}x^4+O\left(x^6\right)$$
$$\log(P_n)=-\frac{x^2}{2}\sum_{k=1}^n k^2-\frac{x^4 }{12}\sum_{k=1}^n k^4+\cdots$$ $$\log(P_n)=-\frac{n (n+1) (2 n+1)}{12} x^2-\frac{n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)}{360} x^4+O\left(x^6\right)$$
$$P_n=e^{\log(P_n)}=1-\frac{n (n+1) (2 n+1) }{12} x^2+\frac{n (n+1) (2 n+1) (10 n^3+3 n^2-7 n+4)
}{1440}x^4+O\left(x^6\right)$$
$$1-P_n=\frac{n (n+1) (2 n+1) }{12} x^2-\frac{n (n+1) (2 n+1) (10 n^3+3 n^2-7 n+4)
}{1440}x^4+O\left(x^6\right)$$
$$\frac{1-P_n}{x^2}=\frac{n (n+1) (2 n+1) }{12} \left(1+\frac{ 10 n^3+3 n^2-7 n+4
}{120}x^2+O\left(x^4\right) \right)$$ shows the limit and how it is approached.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Normalising Beal's conjecture Beal's conjecture
Can it be shown that
If
$$ \sum_{q=0}^{u}(n+qd)^{m_{q}} =a^b $$
,where $n,u,d,a,m_q$ and $b$ are positive integers with $m_q,b> 3$, then $n,n+d,n+2d,...,n+ud$ and $a$ have a common prime factor.
Example
*
*$(n,u,d,a)=(98,2,98,98)$ and $(m_0,m_1,m_2,b)=(4,4,4,5)$
Related post
Can it be shown, $n^4+(n+d)^4+(n+2d)^4\ne z^4$?
Extending Fermat's Last Theorem
Is there any solution for $n^x+(n+d)^y+(n+2d)^z=a^b$
|
No. For example, $3^3 + 4^3 + 5^3 = 6^3$ and $\gcd(3,4,5,6) = 1$. You can find other examples from the identity
$$ 1^3 + 2^3 + \cdots + n^{3} = \left(\frac{n(n+1)}{2}\right)^{2}. $$
Since there are infinitely many triangular numbers that are also squares, there are infinitely many integers $n$ for which the right hand side of the above equation is a fourth power. (For example, $1^{3} + 2^{3} + 3^{3} + \cdots + 49^{3} = 35^{4}$.)
EDIT: The OP requested that all the exponents are at least $4$. Here's an example in that case:
$$ 1^4 + 2^4 + 3^4 + 4^4 + 5^4 + 6^4 + 7^4 + 8^4 + 9^4 + 10^5 + 11^5 + 12^5 + 13^4 + 14^4 + 15^5 + 16^4 + 17^4 = 35^4.$$
In general, one could fix an integer $N$ and consider expressions of the form $\sum_{q=1}^{N} q^{m_{q}}$ where $m_{q} \in \{k, k+1 \}$. There are $2^{N}$ such expressions, and all of these are between about $\frac{N^{k+1}}{k+1}$ and $\frac{N^{k+2}}{k+2}$. When $N$ is large enough (in terms of $k$), $2^{N}$ is much larger than $\frac{N^{k+2}}{k+2} - \frac{N^{k+1}}{k+1}$ and the chances of getting at least one $k$th power in that range are quite high.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have $x^2-y^2=5$, therefore $xy=6$. What to do next?
|
HINT
From the trigonometric point of view, one can substitute $x = r\cos(\theta)$ and $y = r\sin(\theta)$, from whence we get
\begin{align*}
\begin{cases}
r^{2}\cos(2\theta) = 5\\\\
r^{2}\sin(2\theta) = 12
\end{cases} \Longrightarrow \frac{25}{r^{4}} + \frac{144}{r^{4}} = 1 \Longrightarrow r^{4} = 169 \Longrightarrow r = \sqrt{13}
\end{align*}
since $r \geq 0$. Can you take it from here?
|
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|
How to evaluate $\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$ I am trying to calculate this integral, but I find it is very challenging
$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$$
but somehow I have managed to local its closed form to be $$(-1)^n\left(\frac{1}{2}-n\ln 2+\sum_{j=0}^{n-1}H^{*}_j\right)$$
where $n\ge 0$, $H^{*}_0=0$ and $H^{*}_k=\sum_{j=1}^{k}\frac{(-1)^{j+1}}{j}$
I have try
$$\frac{1-x}{1+x}\cdot \frac{x^n}{(x^2-1)^{1/2}}$$
$$-x^n\sqrt{\frac{x-1}{(x+1)^2}}$$
$$-\int_{0}^{1}x^n\sqrt{\frac{x-1}{(x+1)^3}}\mathrm dx$$
from this point I tried to use the binomial to expand $$\sqrt{\frac{x-1}{(x+1)^3}}$$ but it seem not possible
|
I will show that
$$\int_0^1 \frac{1 - x}{1 + x} \frac{x^n}{\sqrt{x^4 - 2x^2 + 1}} \, dx = -\frac{1}{2} + n (-1)^{n + 1} \left (\ln 2 + \sum_{k = 1}^{n - 1} \frac{(-1)^k}{k} \right ), n \geqslant 1.$$
Note here we interpret the empty sum as being equal to zero (the empty sum is the case when $n = 1$ in the finite sum) and I assume $n \in \mathbb{N}$.
For the case when $n = 0$, a direct evaluation yields: $I_0 = \frac{1}{2}$.
As already noted, since
$$\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}=\dfrac{x^n}{(1+x)^2},$$
the integral becomes
$$I_n = \int_0^1 \frac{x^n}{(1 + x)^2} \, dx.$$
For $n \in \mathbb{N}$, integrating by parts we have
\begin{align}
I_n &= \left [-\frac{x^n}{1 + x} \right ]_0^1 + n\int_0^1 \frac{x^{n - 1}}{1 + x} \, dx\\
&= -\frac{1}{2} + n \int_0^1 \frac{x^{n - 1}}{1 + x} \, dx\\
&= -\frac{1}{2} + n \sum_{k = 0}^\infty (-1)^k \int_0^1 x^{n + k - 1} \, dx\\
&= -\frac{1}{2} + n \sum_{k = 0}^\infty \frac{(-1)^k}{n + k}.
\end{align}
Note here the geometric sum for $1/(1 + x)$ of $\sum_{k = 0}^\infty (-1)^k x^k$ has been used. Reindexing the sum $k \mapsto k - n$ gives
\begin{align}
I_n &= -\frac{1}{2} + n (-1)^n \sum_{k = n}^\infty \frac{(-1)^k}{k}\\
&= -\frac{1}{2} + n(-1)^n \left (\sum_{k = 1}^\infty \frac{(-1)^k}{k} - \sum_{k = 1}^{n - 1} \frac{(-1)^k}{k} \right )\\
&= -\frac{1}{2} + n(-1)^{n + 1} \left (\ln 2 + \sum_{k = 1}^{n - 1} \frac{(-1)^k}{k} \right ),
\end{align}
where I have made use of the well-known result of $\ln 2 = -\sum_{k = 1}^\infty (-1)^k/k$.
In terms of your $H^*_n$ notation for the finite sum $\sum_{k = 1}^n \frac{(-1)^{k + 1}}{k}$, this result can be re-expressed as
$$I_n = -\frac{1}{2} + n (-1)^{n + 1} (\ln 2 - H^*_{n - 1}).$$
To show my result is equivalent to the result you quote, one would need to show, after playing around with finite sums, that
$$\sum_{k = 1}^{n - 1} H^*_k = -\frac{1}{2} (1 + (-1)^n) + n H^*_{n - 1}.$$
|
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|
Find polynomials $P(x)P(x-3) = P(x^2)$
Find all polynomials $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$
I have a solution but I'm not sure about that. Please check it for me.
It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.
Consider $P(x) \neq c$ :
We have $P(x+3)P(x) = P((x+3)^2)$
If $a$ is a root of $P(x) $ then $a^2$ and $(a+3)^2$ are roots of $P(x)$.
If $|a| > 1$, then $ |a^2| = |a||a| > |a| \Rightarrow P(x)$ has infinitely many roots.
If $0 < |a| < 1$, then $|a^2| = |a||a| < |a| \Rightarrow P(x)$ has infinitely many roots.
If $a = 0 \Rightarrow (a+3)^2 = 9$ is a root of $P(x)$, so $P(x)$ has infinitely many roots.
Hence, every root $a$ of $P(x)$ has $|a| = 1$. So $1 = |(a+3)^2| = (|a+3|^2) \Rightarrow |a+3|=1$.
We have $a + 3 = \cos\alpha + i\sin\alpha + 3 \Rightarrow (\cos\alpha+3)^2 + \sin^2\alpha = 1 \Rightarrow \cos\alpha = -\frac{3}{2}\ (\text{impossible}).$
$P(x) = 0, P(x) = 1$ are all the results.
|
Your approach is sound and the arguments besides the last case $|a|=1$ are clear. Here is a slightly more stringent variant, which does not bring anything new, besides maybe the modified argument for the case $|a|=1$.
We are looking for all polynomials $P\in\mathbb{R}[x]$ such that
\begin{align*}
P(x)P(x-3)=P(x^2)\qquad \forall x\in\mathbb{R}\tag{1}
\end{align*}
Proof:
*
*Constant Polynomial $P$: At first we look at polynomials $P(x)=c$ which are constant. From (1) we obtain
\begin{align*}
c^2&=c\\
c(c-1)&=0\tag{2}
\end{align*}
Since the equation (2) is solved iff $c=0$ or $c=1$, these are the only solutions when $P$ is a constant polynomial.
In the following we consider all possible cases when $P$ has a root $a\in\mathbb{R}$. We divide the cases into
\begin{align*}
|a|>1,\qquad 0<|a|<1,\qquad a=0,\qquad a=-1\quad \text{and} \quad a=1
\end{align*}
which cover all possibilities.
*
*Root $P(a)=0$ with $|a|>1$:
Let $a\in\mathbb{R}$ with $|a|>1$. We get from (1)
\begin{align*}
P(a^2)=P(a)P(a-3)=0\cdot P(a-3)=0\tag{3}
\end{align*}
From (3) we conclude that there is an infinite sequence $(a^{2n})_{n\geq 0}$ of pairwise different roots of $P$. Thus, $P$ has infinitely many roots and it follows $P$ is the zero-polynomial: $P=0$.
In the same way we can treat the next case.
*
*Root $P(a)=0$ with $0<|a|<1$:
Let $a\in\mathbb{R}$ with $0<|a|<1$. We get from (1)
\begin{align*}
P(a^2)=P(a)P(a-3)=0\cdot P(a-3)=0\tag{4}
\end{align*}
From (4) we conclude as we did in the case before that there is an infinite sequence $(a^{2n})_{n\geq 0}$ of pairwise different roots of $P$. Thus, $P$ has infinitely many roots and it follows $P$ is the zero-polynomial: $P=0$.
Three more cases to follow:
*
*Root $a=0$:
Let $a=0$ with $P(0)=0$. We get from (1) setting $x=3$
\begin{align*}
P(9)=P(3)P(0)=P(3)\cdot 0=0
\end{align*}
We see that $9$ is also a root of $P$. We are now in the case with root $|a|>1$ and conclude $P=0$.
*
*Root $a=-1$:
Let $a=-1$ with $P(-1)=0$. We get from (1) setting $x=2$
\begin{align*}
P(4)=P(2)P(-1)=P(2)\cdot 0=0
\end{align*}
We see that $4$ is also a root of $P$. We are now in the case with root $|a|>1$ and conclude $P=0$.
*
*Root $a=1$:
Let $a=1$ with $P(1)=0$. We get from (1) setting $x=4$
\begin{align*}
P(16)=P(4)P(1)=P(4)\cdot 0=0
\end{align*}
We see that $16$ is also a root of $P$. We are now in the case with root $|a|>1$ and conclude $P=0$.
Conclusion: We have covered all types of different polynomials which fulfill the relationship (1) and conclude that $\color{blue}{P=0}$ and $\color{blue}{P=1}$ are the only solutions of (1).$\qquad\qquad\qquad\Box$
Note: When making rigorous proofs we should avoid phrases like: It is clear that ..., It is easy to see that .... It is much more preferable to prove even seemingly simple things, just to be sure that we don't overlook anything.
|
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|
How prove this $\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\frac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$ Prove or disprove:
if $f(x)\ge 0,\forall x\in [-\pi,\pi]$,show that
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\dfrac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$$
I can prove this if $2\pi$ takes the place of $\dfrac{\pi}{2}$
because use Cauchy-schwarz inequality we have
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2\le\int_{-\pi}^{\pi}\sin^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$
$$\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\int_{-\pi}^{\pi}\cos^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$
add this two inequality, we have
$$\begin{align*}\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2 &\le \int_{-\pi}^{\pi}f^2(x)dx\int_{-\pi}^{\pi}(\sin^2{x}+\cos^2{x})dx\\
&=2\pi\int_{-\pi}^{+\pi}f^2(x)dx\end{align*}$$
see this Discrete form of inequality：Prove this inequality with Cauchy-Schwarz inequality
So far, I haven't found any counterexamples，such $f(x)=1,\sin{x}+1$ it such this inequality
|
Note that
\begin{align*}
\int_{-\pi}^\pi f(x) \sin x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) + f(\pi - x) - f(-x)) \sin x \,dx \\
&= \int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx
\end{align*}
and similarly
\begin{align*}
\int_{-\pi}^\pi f(x) \cos x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) - f(\pi - x) + f(-x)) \cos x \,dx \\
&= \int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx
\end{align*}
where we define $g, h : [0, \pi/2] \to \mathbb{R}$ by $g(x) = f(x) - f(x - \pi)$ and $h(x) = f(\pi - x) - f(-x)$.
Then by Cauchy-Schwarz,
\begin{align*}
\left(\int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx\right)^2
&\leq \int_0^{\pi/2} \sin^2 x \,dx \int_0^{\pi/2} (g(x) + h(x))^2 \,dx \\
&= \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx
\end{align*}
and
\begin{align*}
\left(\int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx\right)^2
&\leq \int_0^{\pi/2} \cos^2 x \,dx \int_0^{\pi/2} (g(x) - h(x))^2 \,dx \\
&= \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx
\end{align*}
hence
\begin{align*}
\left(\int_{-\pi}^\pi f(x) \sin x \,dx\right)^2 &+ \left(\int_{-\pi}^\pi f(x) \cos x \,dx\right)^2 \\
&\leq \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx + \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx\\
&= \frac{\pi}{2} \int_0^{\pi/2} (g(x))^2 + (h(x))^2 \,dx \\
&\leq \frac{\pi}{2} \int_0^{\pi/2} (f(x))^2 + (f(x - \pi))^2 + (f(\pi - x))^2 + (f(-x))^2 \,dx \\
&= \frac{\pi}{2} \int_{-\pi}^\pi (f(x))^2 \,dx
\end{align*}
as desired, where the last inequality holds because $f$ is nonnegative.
|
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|
Proof that $n^3 \leq 2^n$ for $n \geq 10$ by induction I'm trying to prove this statement and it got a bit clumsy but at least it seems to make sense. I'd like to know if this proof is valid. Here's what I managed to do:
First the base case:
$n = 10$
$$n^3 \leq 2^n$$
$$1000 \leq 1024$$
I assumed that it's true for $k$:
$$k^3 \leq 2^k$$
Then I tried to prove that it's true for $k+1$. First I multiplied both sides by $2$:
$$2k^3 \leq 2^k \cdot 2$$
$$2k^3 \leq 2^{k+1}$$
$$2k^3 \leq 2^k + 2^k$$
$$k^3 + k^3 \leq 2^k + 2^k$$
Since $n \geq 10$:
$$(k+1)^3 \leq k^3 + k^3 \leq 2^k + 2^k$$
therefore $$(k+1)^3 \leq 2^{k+1}$$
Can someone confirm this, please?
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Your solution looks good, but you might want to add more details as to why
$$(k+1)^3\leq k^3+k^3$$
This can be seen explicitly as
$$(k+1)^3=1 + 3 k + 3 k^2 + k^3=k^3+k^3\left(\frac{3}{k}+\frac{3}{k^2}+\frac{1}{k^3}\right)$$
Since $k\geq 4$ we have
$$k^3+k^3\left(\frac{3}{k}+\frac{3}{k^2}+\frac{1}{k^3}\right)\leq k^3+k^3\left(\frac{3}{4}+\frac{3}{4^2}+\frac{1}{4^3}\right)=k^3+\frac{61}{64}k^3<k^3+k^3$$
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|
How to do partial fractions with a denominator to the power of a variable? How can I take the following sum and simplify it with partial fractions? $$\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}$$
I know the denominator can be rewritten as $(2^k)(2)$, but how do I deal with the $2^k$ when doing partial fractions?
Usually, when there is an exponent in the denominator, you put terms all the way through - if you have $x^3$ in the denominator you end up with $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}$. So, what do I do when I have $2^k$ - and I either do not know what $k$ or $k$ goes to infinity?
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TO MAKE A SENCE $$a=\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}=\frac{0}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...$$multiply by $\frac 12$
$$\frac12a=\frac{0}{4}+\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\frac{5}{128}+...$$ now notice to $a-\frac 12 a$
$$a-\frac 12 a=\frac{0}{2}+\frac{1}{4}+\frac{2}{8}-\frac 1{8}+\frac{3}{16}-\frac{2}{16}+\frac{4}{32}-\frac{3}{32}+\frac{5}{64}-\frac{4}{64}+...=\\
\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...=geometric$$
$$\frac12a=\frac{0}{4}+\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\frac{5}{128}+...=0+\sum^{\infty}_{k=2}\frac{k-1}{2^{k+1}}\\**k-1=u**\\=0+\sum^{\infty}_{u=1}\frac{u}{2^{u+2}}$$and
$$a=\frac{0}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...=0+\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{u+1}{2^{u+2}}}_{\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...}$$ now apply $a-\frac12 a$
$$a-\frac12 a=\\\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{u+1}{2^{u+2}}}_{\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...}-\sum^{\infty}_{u=1}\frac{u}{2^{u+2}}=\\
\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{\not u+1-\not u}{2^{u+2}}}_{geometric}$$
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Why do I get a different power series? My question revolves around the power series of the following function:
$$ f(x) = \frac{1}{1+x}$$
Now, it is almost immediate to have the power series of this function by substituting $-x$ into the geometric series to get:
$$\frac{1}{1 + x} = \frac{1}{1-(-x)} = \sum_{k=0}^\infty (-x)^k = \sum_{k=0}^\infty (-1)^kx^k = 1 - x + x^2 - x^3 + \ldots$$
with radius of convergence $1$. Now what got me thinking is why if I do a slight manipulation, I didn't arrive at the same result. Here was what I did:
$$\frac{1}{1+x} = \frac{1}{2 + x - 1} = \frac{1}{2}\left(\frac{1}{1 +\frac{x}{2} - \frac{1}{2}}\right) = \frac{1}{2}\left(\dfrac{1}{1-\{ -\frac{1}{2}(x-1)\}}\right)$$
And using the definition of the geometric series:
$$\frac{1}{1+x} = \frac{1}{2}\sum_{k=0}^\infty \left(-\frac{1}{2}(x-1)\right)^k = \sum_{k=0}^\infty (-1)^k\frac{1}{2^{k+1}}(x-1)^k$$
When I try to evaluate them on small values, this series doesn't look equal to the above series. Why do I get a different answer? Or are they identically the same? If so, can someone please guide/show me? Thanks.
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Note that the center of the $$\frac{1}{1+x} = \frac{1}{2}\sum_{k=0}^\infty \left(-\frac{1}{2}(x-1)\right)^k = \sum_{k=0}^\infty (-1)^k\frac{1}{2^{k+1}}(x-1)^k$$
is the point $x=1$ while the center of $$ \frac{1}{1 + x} = \frac{1}{1-(-x)} = \sum_{k=0}^\infty (-x)^k = \sum_{k=0}^\infty (-1)^kx^k = 1 - x + x^2 - x^3 + \ldots$$
is the point $x=0$
The series approximate the function about the center which in this case are two different points.
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|
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?
Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :)
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$$(-1)^3=(x^4+x^2)^3=(x^3)^4+(x^3)^2+3(x^3)^2(-1)$$
Set $x^3=y$
$$y^4-2y^2+1=0$$ whose roots are $\alpha^3$ etc.
$$(y^4-2y^2+1)^2=0$$
Set $y^2=z$
$$z^4-4z^3+\cdots+1=0$$ whose roots are $(\alpha^3)^2$ etc.
Now apply Vieta's formula
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Limit of sum $x^3+x^5+x^7+x^9+...)$ I am asked to give the limit of:
$$ x^3+x^5+x^7+x^9+... \quad x\in(-1,1)$$
So I do the following:
The sum of the first $n$ terms will be equal to:
$$x^3+x^5+x^7+...+x^{3+2(n-1)}$$
I factor out $x^3$, I get:
$$x^3(1+x^2+x^4+..+x^{2(n-1)})$$
I also factor out $x^2$, I get:
$$x^3 x^2(1/x^2+1+x^2+x^3+...+x^{n-1}=x^5(1/x^2+1+x^2+x^3+...+x^{n-1})$$
So I have:
$$x^5/x^2+x^5(1+x^2+x^3+...+x^{n-1})=x^3+\frac{x^5 (1-x^n)}{1-x}$$
So I take the limit when $n$ goes to infinity and I get:
$$x^3+\frac{x^5}{1-x}=\frac{x^3(1-x^2)}{1-x^2}+\frac{(1+x)x^5}{1-x^2}=\frac{x^3-x^5+x^5+x^6}{1-x^2}=\frac{x^3+x^6}{1-x^2}$$
But the right answer is $\frac{x^3}{1-x^2}$
Where did I go wrong?
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$x^{2(n-1)} \not= x^2 x^{n-1}$
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|
How to prove that a statement is false using principle of mathematical induction? I came across this question while solving some problems based on principle of mathematical induction ,
$P(n) : 1+2+3+....+n < \frac{(n+2)^2}{8}, n\in\mathbb{N}$, is true for
$(A) \, n\geq1\\
(B) \, n\geq2\\
(C) \text{ all } n\\
(D) \text{ none of these.}$
At first I kept some random values of $n$ in the inequality and found that it was true only for $n=1$ and not for any other natural value of $n$. Then I tried to prove the same or at least the fact that it is not true for all natural $n$ , using principle of mathematical induction. For this I first proved that $P(n)$ is true for $n=1,$ then I assumed that $P(n)$ is true for any natural number $n=k,$ i.e.
$P(k) : 1+2+3+\dots+k < \frac{(k+2)^2}{8}$ is a true statement , then I tried to prove that $P(k+1)$ is a true statement using $P(k)$ i.e. $P(k+1) : 1+2+3+...+k+k+1 < \frac{(k+1+2)^2}{8} =\frac{(k+3)^2}{8}$ should be true . So $P(k) :
1+2+3+....+k < \frac{(k+2)^2}{8},$
adding $k+1$ to both sides
$\begin{align}&\Rightarrow 1+2+...+k+k+1
< \frac{(k+2)^2}{8} + k+1\\
&\Rightarrow 1+2+...+k+k+1 < \frac{k^2+4+4k+8k+8}{8} = \frac{k^2+12k+12}{8}\end{align}$
Now $\frac{k^2+12k+12}{8} = \frac{k^2+6k+9}{8}+ \frac{6k+3}{8}= \frac{(k+3)^2}{8} + \frac{6k+3}{8} \Rightarrow
\frac{k^2+12k+12}{8}> \frac{(k+3)^2}{8}.$ So finally we have
$1+2+...+k+k+1 < \frac{k^2+12k+12}{8}$ and
$\frac{(k+3)^2}{8} < \frac{k^2+12k+12}{8},$ So from the above two inequalities we cannot prove that $P(k+1)$ is true but we also cannot prove it false , so what should we conclude from this? Also , if somehow we prove that $P(k+1)$ is false , is it not possible that the truth of $P(k)\Rightarrow$ truth of $P(k+2)$ and hence by principle of mathematical induction , $P(n)$ is true for alternate consecutive integers starting from $1$? Please help me.
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Just prove by induction that $\sum_{j=1}^nj\ge\frac18(n+2)^2$ for all $n\ge2$, using $k+1\ge(2k+5)/8$ (which is equivalent to $6k\ge3$).
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|
Cost Revenue Profit Question involving First Order Conditions A Cobb-Douglas production function is given by:
$Q(x,y) = x^{1/2}y^{1/4}$
where x and y are the input variables. The price of each product is p, and the cost of production is $C(x, y) = ax + by$.
I need to write down the profit, Π and find the stationary point for the profit Π and show that the critical values for x and y are given by
$x_* = \frac{p^4}{32a^3b}$ and $y_* = \frac{p^4}{64a^2b^2}$
So I started this question by forming an equation for the profit:
$Π = R - C = pQ(x,y) - (ax + by)$
Then the First Order Conditions are as follows:
$Π_x = \frac{1}{2}px^{-1/2}y^{1/4} - a$,
$Π_y = \frac{1}{4}px^{1/2}y^{-3/4} - b$
Anyone know where to take this from here? Cheers.
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As Henry has already commented set the derivatives equal to zero
$$\frac{1}{2}px^{-1/2}y^{1/4} - a=0\Rightarrow \frac{1}{2}px^{-1/2}y^{1/4} =a \quad (1)$$
$$\frac{1}{4}px^{1/2}y^{-3/4} - b=0\Rightarrow \frac{1}{4}px^{1/2}y^{-3/4} = b \quad (2)$$
Divide the first equation by the second equation. The laws of exponents are helpful here.
$$\frac{\frac{1}{2}px^{-1/2}y^{1/4}}{\frac{1}{4}px^{1/2}y^{-3/4} }=\frac{a}{b}$$
$$2\cdot x^{-1/2-1/2}y^{1/4-(-3/4)}=\frac{a}{b}$$
$$2\cdot x^{-1}y^{1}=\frac{a}{b}\Rightarrow 2y=\frac{ax}{b}\Rightarrow y=\frac{ax}{2b} \quad (3)$$
Next we can raise (1) to the fourth power to get an $y^1$:
$$\frac{1}{16}p^4x^{-2}y^{1} =a^4$$
Inserting the expression for $y$
$$\frac{1}{16}p^4x^{-2}\cdot \frac{ax}{2b} =a^4$$
Next you solve the equation for $x$ to obtain $x^*$. Before you start use that $x^{-2}\cdot x=\frac1x$
Finally use $(3)$ and the expression for $x^*$ to calculate $y^*$.
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Find minimum of $\frac{n}{S(n)}$ For every Natural Number like $n$ consider:$\frac{n}{S(n)}$ so that $S(n)$ is sum
of the digits of the number $n$ in base-10. find minimum of $\frac{n}{S(n)}$ when:
a)$9<n<100$
b) $99<n<1000$
c)$999<n<10000$
d)$9999<n<100000$
for $9<n<100$ I tried:
$n=10a+b$ and $Min(\frac{10a+b}{a+b})=Min(1+\frac{9a}{a+b})$ so It is obvious that $b$ should be 9. I put $a=1,2,3,...$ and realized that if $a=1$ it will be minimum so the answer of part (a) is 19 but I dont know How we can mathematically show that $a=1$
for part b,c ,d I cant find mathematically way to show when this fraction (for example for part b: $\frac{100a+10b+c}{a+b+c}$) is minimum
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For a mathematical proof of (b)
$$\frac{10a+b}{a+b}=2+\frac{8a-b}{a+b}$$
The only way this can be less than $2$ is if $a=1,b=9$. So the minimum is $2-\frac{1}{10}=1.9$.
Part (d)
$$\frac{10^4a+10^3b+10^2c+10d+e}{a+b+c+d+e}-100=\frac{9900a+900b-90d-99e}{a+b+c+d+e}$$
The numerator of the RHS is clearly positive and so the minimum will occur for $c=9$. If instead of subtracting $100$ we subtracted $10$ we would obtain $d=9$ and subtracting $1$ gives $e=9$.
However subtracting higher powers of $10$ i.e. $1000$ and $10000$ produces a fraction where the numerator can be made negative and then it is best to make $a,b$ as small as possible i.e. $a=1,b=0$
The minimum is obtained for $10999$.
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ways in which $A,B$ refuse to be the member of same team
Consider a class of $5$ Girls and $7$ boys . The number of different teams consisting of $2$ girls and $3$ boys that can be formed from this class , If there are two specific boys $A$ and $B,$ who refuse to be the member of same team, is
what i try
Method $(1)$
Ways in which $A,B$ not a member = Total -ways in which both $A,B$ included
$$=\binom{5}{2}\cdot \binom{7}{3}-\binom{5}{2}\cdot \binom{5}{1}=300$$
Method $(2)$
Ways in which $A,B$ not a member $$=\binom{5}{2}\cdot \binom{5}{3}=100$$
above we have excluded $2$ boys because they are not a member
But answer given is $300$
please explain me How i am wrong in $(2)$ Method
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As John Omielan has pointed out, your second answer is incorrect since you omitted those cases in which exactly one of the boys $A$ or $B$ is a member of the team. Let's correct your count.
$A$ is a member, but $B$ is not: Since $A$ is on the team and $B$ is not, we must choose two of the five girls and two of the five boys other than $A$ or $B$, which can be done in
$$\binom{5}{2}\binom{5}{2}$$
ways.
$B$ is a member, but $A$ is not: By symmetry, this can also be done in
$$\binom{5}{2}\binom{5}{2}$$
ways.
Neither $A$ nor $B$ is a member: This is what you actually calculated. We must choose two of the five girls and three of the other five boys, which can be done in
$$\binom{5}{2}\binom{5}{3}$$
ways.
Total: Since the three cases are mutually exclusive and exhaustive, the number of teams with two girls and three boys which can be formed given that boys $A$ and $B$ refuse to be a member of the same team is
$$\binom{5}{2}\binom{5}{2} + \binom{5}{2}\binom{5}{2} + \binom{5}{2}\binom{5}{3} = 300$$
which agrees with the answer you obtained by subtracting the number of teams that include both $A$ and $B$ from the total number of teams that could be formed without restriction.
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How do we get $(b-a)^2/12$ from factorisation? I am in the final steps of calculating the variance of the uniform distribution for $(a,b)$. I'd like to see the steps involved in getting $$\frac{1}{12}(b-a)^2$$ from $$\frac{4(b^3-a^3)-3(b-a)(a+b)^2}{12(b-a)}$$ which is the result after integrating the variable $X^2$ between $b$ and $a$ and subtracting the mean squared. I'm struggling to perform the algebra so I would appreciate any assistance you could provide in showing the steps involved. Thanks!
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$4(b^3-a^3)-3(b-a)(b+a)^2=4b^3-4a^3-3(b^2-a^2)(b+a)$
[using $(b-a)(b+a)=b^2-a^2$]
$=4b^3-4a^3-3b^3-3b^2a+3a^2b+3a^3$
$=b^3-3b^2a+3ba^2-a^3$
$=(b-a)^3\quad(*)$.
Dividing by $12(b-a)$ gives $\frac{1}{12}(b-a)^2$
So how did we do $(*)$? Well I did it because I recognize a few well-known factorisations like:
$(a+b)^2=a^2+2ab+b^2$
$(a+b)^3=a^3+3a^2b+3ab^2+b^3$
and the second of those gives you what you need if you replace $a$ by $-a$.
Suppose you don't recognize the factorisation immediately. The first thing to notice is whether every term has the same degree. It does here. So you can think of the expression as $a^3(x^3-3x^2+3x-1)$ where $x=b/a$. You may remember that checking a polynomial with integer coefficients for integer roots is easy: you just check the factors of the constant term. So the only possible integer roots of $x^3-3x^2+3x-1$ are $\pm1$. You quickly find that 1 is a root which means that $(x-1)$ is a factor. Find the other factor as $(x^2-2x+1)$. You can either factor that on sight or repeat the process to find that the polynomial is $(x-1)^3$ and hence the original is $(b-a)^3$.
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.