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find the minimum of $A=\frac{x^3}{3y+1}+\frac{y^3}{3z+1}+\frac{z^3}{3x+1}$ with $x^3+y^3+z^3=3$ With $x,y,z \ge 0, x^3+y^3+z^3=3$: find the minimum of $A=\dfrac{x^3}{3y+1}+\dfrac{y^3}{3z+1}+\dfrac{z^3}{3x+1}$
My attempts: $A=\dfrac{x^6}{3yx^3+x^3}+\dfrac{y^6}{3zy^3+y^3}+\dfrac{z^6}{3xz^3+z^3} \ge \dfrac{(x^3+y^3+z^3)^2}{3(yx^3+zy^3+xz^3)+x^3+y^3+z^3}=\dfrac{3}{yx^3+zy^3+xz^3+1}$ and I don't know what to do next (I think my approach is not right)
|
The key observation is the term $3x,3y,3z$ at the denominators. This suggests the application of the AM-GM inequality somehow. Indeed, by AM-GM inequality: for $a \ge 0 \implies a^3 + 2 = a^3 + 1 +1 \ge 3a\implies a^3+ 3 \ge 3a+1$. Using this fact for $a = x, y, z$ in the denominators of $A$. Thus: $A \ge \dfrac{x^3}{y^3+3} + \dfrac{y^3}{z^3+3}+\dfrac{z^3}{x^3+3}$. Now let $m = x^3, n = y^3, p = z^3 \implies A \ge \dfrac{m}{n+3}+\dfrac{n}{p+3}+\dfrac{p}{m+3}$, with $m+n+p=3$ and $m,n,p \ge 0$. Using Cauchy-Schwarz inequality twice: $\displaystyle \sum_{\text{cyclic}}\dfrac{m}{n+3} = \displaystyle \sum_{\text{cyclic}}\dfrac{m^2}{mn+3m}\ge \dfrac{3(m+n+p)^2}{3(mn+np+pm)+9(m+n+p)}\ge \dfrac{3\cdot3^2}{(m+n+p)^2+9\cdot 3}=\dfrac{27}{9+27}=\dfrac{27}{36}= \dfrac{3}{4}$. Equality occurs in the original inequality when $m = n = p = 1 \implies x = y = z = 1$.
|
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Show that there is no pair of coprime positive integers $(x,y)$ such that $(x+y)^3 \mid (x^n+y^n)$
Let $n$ be a square-free integer. Show that there is no pair of coprime positive integers $(x,y)$ such that $$(x+y)^3 \Big| (x^n+y^n)$$
The problem can be apparently solved by LTE, but I don't know what are the cases to break it up to.
If I first take $n$ to be odd and $p >2$, then I need to assume that $p \mid x-y$ and now by LTE we get that $$v_p((x+y)^3)=3v_p(x+y) \le v_p(x^n+y^n)=v_p(x+y)+v_p(n) = v(x+y)+1$$ as $v_p(n)=1$ since $n$ was square-free. This implies that $2v_p(x+y) \le 1 \iff v_p(x+y) \le \frac12$ which is a contradiction.
I think I need to consider some other cases now such as $n$ being even and $p=2$ at least? How can I approach this one?
|
$(x+y)^3 | (x^n+y^n), (x, y)=1.$
\begin{align}
\text{Case 1. } \; & n=1: \\
&(x+y)^2|1, x+y=\pm1. \Rightarrow \text{cannot be positive.} \\
\ \\
\text{Case 2. } \; & n=2: \\
&(x+y)^3 | x^2+y^2. \\
&\text{We can easily find that $(x+y)^3$ will be absolutely bigger than $x^2+y^2$ when $x, y$ increases.} \\
& \therefore (x, y)=(1, 1). (\because (1+2)^3>1^2+2^2.) \\
& \text{But it doesn't satisfies.} \\
\ \\
\text{Case 3. } \; & n\geq 3: \\
\ \\
\text{(i) } \; & n=\text{odd}, \exists p \text{ s.t. }p|(x+y), p \neq 2(p: \text{prime}): \\
& \text{Using LTE: } \\
& \text{Contradiction, as you did.} \\
\ \\
\text{(ii) } \; & n=\text{odd}, \exists! p \text{ s.t. } p|(x+y), p:\text{prime}(\text{which is } 2): \\
& 3v_2(x+y) \leq v_2(x^n+y^n)=v_2(x+y). \\
& \therefore v_2(x+y)\leq0, \text{Which is contradiction.} \\
\ \\
\text{(iii) } \; & n=\text{even}: \\
&\text{let } p: \forall \text{prime number s.t. } p|x+y. \\
\ \\
& \text{if }p=2: \\
& x^n=(x^2)^k \equiv 0, 1, 5 (\mod 8)(\because k \geq 2.) \\
& \therefore \text{if } 8|x^n+y^n, x^n \equiv y^n \equiv 0 (\mod 8.) \Rightarrow \text{Contradiction.} \\
\ \\
& \text{if } p\neq2: \\
& x \equiv -y(\mod p) \Rightarrow x^n-y^n \equiv x^n-(-x)^n \equiv 0 (\mod p) (\because n: \text{even.}) \\
& \text{if } p|y, x \equiv 0 (\mod p), \text{Contradiction.} \\
&\therefore p \not|y. \\
& (p, x^n+y^n)=(p, x^n-y^n+2y^n) = (p, 2y^n) = 1, \text{Contradiction.} \\
\ \\
\ \\
\ \\
&\text{All of them are contradiction.} \\
&\therefore \not\exists x, y \text{ s.t. } (x, y)=1, (x+y)^3 | (x^n+y^n).
\end{align}
|
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Find inverse of rational function $f(x) = \frac{x^3+2}{x-3}$ when x not equal to 3 Question:
I'm trying to find inverse of $f(x) = \frac{x^3+2}{x-3}$ for an assignment.
$y = \frac{x^3+2}{x-3}, x\neq3$
Replace $x$ and $y$.
$x = \frac{y^3+2}{y-3}$
$x(y-3) = y^3 + 2$
$xy - 3x = y^3 + 2$
I don't know how to proceed after this. Ideally I want to get $y = \text{(in terms of x)}$. But I can't seem to do that due to term $xy$ and lack of other common factors. Please help!
|
If $x\neq 3$ $$y= \frac{x^3+2}{x-3}\quad \implies \quad x^3-y \,x+(3y+2)=0$$ For $y>0$, following the steps given here, we have
$$\Delta=4 y^3-27 (3 y+2)^2$$ which is negative; so, only one real root.
Using the hyperbolic solution for one real root, we end with
$$x=-\frac{2 \sqrt{y} }{\sqrt{3}}\cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{ 3\sqrt{3}}{2}\frac{ (3
y+2)}{y\sqrt y}\right)\right)$$
If you dont know the solution of cubic equations, there is more likely a typo in the textbook. It is not impossible the the problem was in fact $y= \frac{x^\color{red}{\large 2}+2}{x-3}$ (as you wrote in the first version of the post).
|
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|
Why am I getting extraneous solutions solving $\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$? I came across a beautiful problem:
Solve $$\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$$
My approach:
Obviously $x=0$ is not a solution.
Now we have:
$$1+\sqrt{1-x}=2x^2-2x\sqrt{1-x^2} \tag1$$
Squaring the above equation both sides we get
$$2(1+\sqrt{1-x})-x=4x^4+4x^2(1-x^2)-8x^3\sqrt{1-x^2} \tag2$$
From $(1)$ we get
$$2(2x^2-2x\sqrt{1-x^2})-x=4x^2-8x^3\sqrt{1-x^2} \tag3$$
$\implies$
$$(8x^3-4x)\sqrt{1-x^2}=x \tag4$$
So we have:
$$(8x^2-4)\sqrt{1-x^2}=1 \tag5$$
Letting $\sqrt{1-x^2}=p, p>0$ We get
$$8p^3-4p+1=0 \tag6$$
which gives $p = 0.5, \frac{\sqrt{5}-1}{4}$
So using $x=\pm \sqrt{1-p^2}$ we get four values of $x$ as:
$$x=\frac{\pm \sqrt{3}}{2}, \frac{\pm \sqrt{10+2 \sqrt{5}}}{4} \tag7$$
Out of which only $$x=\frac{-\sqrt{3}}{2}, \frac{-\sqrt{10+2 \sqrt{5}}}{4}$$ Will satisfy.
But at which step the positive extraneous roots boiled down and how to get rid them in this approach?
|
we get four values of $x:$
$$x=\frac{\pm \sqrt{3}}{2}, \frac{\pm \sqrt{10+2 \sqrt{5}}}{4} \tag7$$
out of which only $$x=\frac{-\sqrt{3}}{2}, \frac{-\sqrt{10+2 \sqrt{5}}}{4}$$ will satisfy.
Actually, among these four solutions, only $x=\frac{\sqrt{3}}{2}$ is extraneous; it was introduced, together with the other extraneous solution $x=0,$ in step $(3)$. Every other step—including the two squaring steps—turns out to be “reversible” (preserve the solution set), even as you have not justified how (for example, by specifying a relevant implicit constraint).
$$\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$$
$$1+\sqrt{1-x}=2x^2-2x\sqrt{1-x^2} \tag1$$
Squaring the above equation, we get
$$2(1+\sqrt{1-x})-x=4x^4+4x^2(1-x^2)-8x^3\sqrt{1-x^2} \tag2$$
From $(1),$ we get $$2(2x^2-2x\sqrt{1-x^2})-x=4x^2-8x^3\sqrt{1-x^2}\tag3$$
Substituting equation $(1)$ into equation $(2)$ to create equation $(3)$ has expanded the candidate solution set, as $(1)$ is a conditional equation rather than an identity (like $(x+3)^2=x^2+6x+9,$ which is universally true).
Since equation $(1)$ holds at our required solution points, they will not be discarded. However, at every other point, equation $(1)$ is false, so deductive explosion will occur, so a spurious conclusion (extraneous solution) may ensue even as our subsequent argument remains valid.
Here's a silly clearer example to illustrate. We want to solve the equation $$x^2+1=5.\tag i$$ Thus, $$x^2=4.\tag {ii}$$ Substituting $(\mathrm{ii})$ into $(\mathrm i):$ $$5=5\\x+5=x+5\\x=x.$$ Therefore, every number is a potential/candidate solution of equation $(\mathrm i).$
To be clear, neither of the above substitution steps is invalid: they do not discard actual solutions, and merely turn out to have created extraneous solutions which then need to be sifted out.
|
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Find the maximum of $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$,where$a,b,c>0$ $a,b,c>0$, find the maximum of :
$$\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$$
I try to find the minimum of $\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}=\frac{4a+1}{\sqrt{a}}\cdot\frac{9a+b}{\sqrt{ab}}\cdot\frac{4b+c}{\sqrt{bc}}\cdot\frac{9c+1}{\sqrt{c}}
=\left(4\sqrt{a}+\frac{1}{\sqrt{a}}\right)\left(9\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(4\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}}\right)\left(9\sqrt{c}+\frac{1}{\sqrt{c}}\right) \geq 4 \times 6\times 4\times 6$
when I try to use AM-GM inequality, I find I can't take equality in the above $4$ parentheses.so I only find a upper bound of the expression.: $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)} < \frac{1}{576}$
|
Hint
Try instead to minimize
$$\Phi=\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}$$ Compute the partial derivatives
All of them being equal to $0$, the solution is immediate.
|
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|
Find minimum of $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$ Let $a, b,c$ be the lengths of the sides of a triangle such that $a+b+c=2$, find the minimum value of $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$$
I don't have many ideas for this problem, my attemps:
The minimum value is $2\sqrt{2}+\dfrac{4}{3}$, the equality occurs for $a=b=c=\dfrac{2}{3}$
By Cauchy–Schwarz inequality, we have: $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge \dfrac{1}{\sqrt{2}}(a+b+b+c+c+a)=2\sqrt{2}$$
Thus, it suffices to prove that $$ab+bc+ca\ge\dfrac{4}{3}$$
This cannot be true because $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}=\dfrac{4}{3}$
Does anyone have any ideas, please give me a hint
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Denote the expression by $f$.
Using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&(\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2})^2\\
=\,& 2a^2 + 2b^2 + 2c^2
+ 2\sqrt{(a^2 + b^2)(c^2 + b^2)}
+ 2\sqrt{(c^2 + b^2)(c^2 + a^2)}\\
&\quad
+ 2\sqrt{(c^2 + a^2)(b^2 + a^2)}\\
\ge\,& 2a^2 + 2b^2 + 2c^2 + 2(ac + b^2) + 2(c^2 + ab) + 2(bc + a^2)\\
=\,& 4a^2 + 4b^2 + 4c^2 + 2ab + 2bc + 2ca.
\end{align*}
Thus, we have
$$f \ge \sqrt{4a^2 + 4b^2 + 4c^2 + 2ab + 2bc + 2ca} + ab + bc + ca. \tag{1}$$
Let $a = x + y, b = y + z, c = z + x$
for $x, y, z > 0$ (the so-called Ravi's substitution).
Then $x + y + z = 1$.
From (1), we have
\begin{align*}
f &\ge \sqrt{10x^2 + 10y^2 + 10z^2 + 14xy + 14yz + 14zx}\\
&\quad + x^2 + y^2 + z^2 + 3xy + 3yz + 3zx\\
&= \sqrt{10(x + y + z)^2 - 6(xy + yz + zx)} + (x + y + z)^2 + xy + yz + zx\\
&= \sqrt{10 - 6q} + 1 + q \tag{2}
\end{align*}
where $q = xy + yz + zx$.
Since $x + y + z = 1$ and $x, y, z> 0$, we have $0 < xy + yz + zx \le 1/3$. Thus, $0 < q \le 1/3$.
It is easy to prove that $\sqrt{10 - 6q} + 1 + q \ge \sqrt{8} + \frac43$.
From (2), we have $f \ge \sqrt{8} + \frac43$.
Also, when $a = b = c = 2/3$, we have $f = \sqrt{8} + 4/3$.
Thus, the minimum of $f$ is $2\sqrt 2 + \frac43$.
|
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Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
Solutions in the answers.
$\ \\ \ \\ \ \\ \ \\$
Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment says.
This identity has a special feature, that the form of the expression maintains from LHS to RHS.
Also, for addition, we can expanse this identity to:
$$ (a^2+nb^2)(c^2+nd^2)=(ac\pm nbd)^2+n(ad\mp bc)^2. $$
or:
$$ X=xz-Cyw, Y=axw+a'yz+BYw. \\ (ax^2+Bxy+a'Cy^2)(a'z^2+Bzw+aCw^2)=aa'X^2+BXY+CY^2 $$
, from the answer of @Will Jagy.
This can be proved by various solutions, for example, just multiplying out this identity, or with trigonometric functions, or with the imaginary number "$i$".
I want you to prove this identity with more solutions.
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\begin{align}& (a^2+b^2)(c^2+d^2) \\= \; & (ac)^2+(bc)^2+(ad)^2+(bd)^2 \\= \; & \Big((ac)^2+(bd)^2\Big)+\Big((ad)^2+(bc)^2\Big) \\= \; & \Big((ac)^2+2(ac)(bd)+(bd)^2\Big)+\Big((ad)^2-2(ad)(bc)+(bc)^2\Big) \\= \; & (ac+bd)^2+(ad-bc)^2\end{align}
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Calculating complex integral with Residue - where is my fault? \begin{align}\int_0^{2\pi} \frac{\cos(x)}{13+12\cos(x)} dx & = \displaystyle\int_0^{2\pi} \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)\frac{1}{2}}\frac{1}{iz} dz \\
& = \cdots \\
&= -i\displaystyle\int_0^{2\pi} \frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)z}dz \\
&= 2\pi i (-i) \operatorname{Res}\left(\frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)z};0\right) \\
&=2\pi \lim_{z\to0}\frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)}\\
&=2\pi \frac{0^2+1}{\left(0+\frac{2}{3}\right)\left(0+\frac{3}{2}\right)}\\
&=2\pi
\end{align}
But Wolfram Alpha says it's $\frac{4\pi}{15}$.
What am I doing wrong ?
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On the right-hand side of the first line, the integral should no longer be over the real interval $[0,2\pi]$; rather, it should be over the unit circle in the complex plane. In addition to correcting that calculational step and subsequent ones, that also explains why the residue at $z=-\frac23$ is relevant: it's inside the unit circle.
|
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limits of 2 lebesgue integrals
Compute the following limits, with justification, where the integrals denote Lebesgue integrals:
*
*$\lim\limits_{m\to\infty} \int_0^\infty \dfrac{m\sin (y/m)}{y(1+y^2)} dy$.
*$\lim\limits_{m\to\infty} \int_0^1 \dfrac{1+my^2}{(1+y^2)^m}dy.$
For 1), I think it's useful to use the Taylor expansion of $\sin(x) = x - \frac{x^3}{3!} + \dfrac{\sin c}{4!} x^4,$ where $c\in (0,x)$. Fix $m\ge 1$. Applying the above expansion to $\sin(y/m),$ we get $\frac{y}m - \frac{(y/m)^3}{3!} + \frac{(y/m)^4}{4!}\ge \sin(y/m) \ge \frac{y}m - \frac{y^3}{3! m^3} - \frac{(y/m)^4}{4!}$. Hence letting $g_m(y) = \frac{y}m - \frac{y^3}{3! m^3}, f_m(y) = \dfrac{m\sin (y/m)}{y(1+y^2)}$, we have that for all $y\ge 0,$ $h_m^-(y) := \dfrac{1-y^2/(6m^2) - y^3/(24m^3)}{1+y^2}= \dfrac{m(g_m(y) - \frac{1}{24}(y/m)^4)}{y(1+y^2)} \leq f_m(y)\leq \dfrac{m(\frac{y}m - \frac{y^3}{3! m^3} + \frac{1}{24}(y/m)^4)}{y(1+y^2)} =: h_m^{-}(y).$
Note that both $h_m^-(y)$ and $h_m^+(y)$ converge pointwise to $\frac{1}{1+y^2}$ as $m\to \infty$, which implies by the Squeeze theorem that $\dfrac{m \sin(y/m)}{y(1+y^2)}$ converges pointwise to $\frac{1}{1+y^2}$ as $m\to\infty$. Also, $|\dfrac{m\sin(y/m)}{y(1+y^2)}|\leq |\dfrac{m(\frac{y}m - \frac{y^3}{3! m^3} + \frac{1}{24}(y/m)^4)}{y(1+y^2)}|$. But I need to upper bound $f_m(y)$ by an integrable function (a function $g$ so that $\int_0^\infty |g| dy < \infty$) to use the Dominated convergence theorem.
For the second limit, for any $y > 0, $ one has L'Hopital's rule that $\lim\limits_{m\to\infty} \dfrac{1+my^2}{(1+y^2)^m} = \lim\limits_{m\to\infty} \dfrac{my^2}{(1+y^2)^m} = 0.$ The issue is that to apply the Dominated convergence theorem, I need to find an upper bound for the integrand that's integrable.
|
For 1) use the fact that $|\sin t| \leq t$ for all $t >0$ so a dominating function is $\frac 1 {1+y^{2}}$.
For 2) use the fact that $(1+y^{2})^{m} \geq 1+my^{2}$ so a dominating function is the constant function $1$.
[The function $(1+x)^{m}-1-mx$ vanishes when $x=0$ and its dervative is $m[(1+x)^{m-1}-1]$ which is non-negative for $m >1, x \geq 0$. Hence, $(1+x)^{m}-1-mx\geq 0$].
|
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Proof for a prime-generating sequence Let $k$ be a positive integer.
Let $n$ be an integer such that $n=6k-1$
Let $r$ be the remainder of the division of $(n-1)!-n$ by $(n+2)$
Conjecture: if $6k+1$ is prime $r=3k+2$
For example the first 25 values of $r$ are:
${5,8,11,2,17,20,23,2,2,32,35,38,41,2,2,50,53,56,2,2,65,2,71,2,77}$
And we have:
$8+5=13=6(2)+1$
$11+8=19=6(3)+1$
$20+17=37=6(6)+1$
$23+20=43=6(7)+1$
$35+32=67=6(11)+1$
$38+35=73=6(12)+1$
$41+38=79=6(13)+1$
$53+50=103=6(17)+1$
$56+53=109=6(18)+1$
I have already proved that if $6k+1$ is a composite number $r=2$ but I failed to prove this conjecture.
Is it hard to prove it?
Thanks.
|
$$n+2=6k+1$$ by Wilson's theorem $$(n+1)!\equiv -1\pmod {n+2}$$ it follows that $$n!\equiv 1\pmod{n+2}$$ and that $$(n-1)!\equiv -(2^{-1})\equiv 3k\pmod {n+2}$$ and $$-n\equiv 2\pmod {n+2}$$ so together we have $$3k+2$$ as the remainder $\blacksquare$
As to the addition chains $$(3k+2)+(3(k+1)+2)=6(k+1)+1$$ so what.
|
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Evaluating $\lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$
Problem statement:
If
$$ A = \lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}, $$
Find $A^2$.
Solution:1)
\begin{align*}
\prod_{k=0}^{n} \binom{n}{k}
&= \prod_{k=1}^{n} \frac{n!}{k!(n-k)!}
= \frac{(n!)^{n+1}}{(1! \cdot 2!\cdots n!)^2} \\
&= \prod_{k=1}^{n} (n+1-k)^{n+1-2k} \\
&= \prod_{k=1}^{n} \left(\frac{n+1-k}{n+1}\right)^{n+1-2k},
\end{align*}
since $\sum_{k=1}^{n} (n+1-2k) = 0$. Taking log and limit, we get
\begin{align*}
& \frac{1}{n}\sum_{k=1}^{n}\left( 1 - \frac{2k}{n+1} \right) \ln\left( 1 - \frac{k}{n+1} \right) \\
&\to \int_{0}^{1} (1 - 2x)\log(1-x) \, \mathrm{d}x
= \frac{1}{2}.
\end{align*}
Source: FITJEE AITS 2020 FT-8 Paper 1 of JEE Advanced question 53
Could anybody explain the solution?
I tried taking log both sides to bring the power down but got stuck on evaluating $$\sum_{r=0}^n \ln\binom{n}{r}$$ further I think the answer should be $e$ instead of $0.5$ which is given.
|
Let $A_n = \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$ denote the expression before the limit is taken. Then
\begin{align*}
\log A_n
&= \frac{1}{n(n+1)} \log \left[\prod_{r=0}^n \binom{n}{r}\right] \tag{1} \\
&= \frac{1}{n(n+1)} \sum_{r=0}^{n} \log \binom{n}{r} \tag{2} \\
&= \frac{1}{n(n+1)} \sum_{r=0}^{n} \left[ \log n! - \log r! - \log(n-r)! \right] \tag{3} \\
&= \frac{1}{n(n+1)} \left[ (n+1) \log n! - 2 \sum_{r=1}^{n} \log r! \right] \tag{4} \\
&= \frac{1}{n(n+1)} \left[ (n+1) \left( \sum_{k=1}^{n} \log k \right) - 2 \sum_{r=1}^{n} \sum_{k=1}^{r} \log k \right] \tag{5} \\
&= \frac{1}{n(n+1)} \left[ (n+1) \left( \sum_{k=1}^{n} \log k \right) - 2 \sum_{k=1}^{n} \sum_{r=k}^{n} \log k \right] \tag{6} \\
&= \frac{1}{n(n+1)} \left[ (n+1) \left( \sum_{k=1}^{n} \log k \right) - \sum_{k=1}^{n} 2(n+1-k) \log k \right] \tag{7} \\
&= \frac{1}{n(n+1)} \sum_{k=1}^{n} (2k-n-1) \log k \\
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \log k \tag{8}
\end{align*}
Note:
*
*$\text{(1)}$ : $\log(a^b) = b \log a$ for $a, b > 0$.
*$\text{(2)}$ : $\log(ab) = \log a + \log b$.
*$\text{(3)}$ : $\binom{n}{r} = \frac{n!}{r!(n-r)!}$
*$\text{(4)}$ : $\sum_{r=0}^{n} \log (n-r)! = \log n! + \log(n-1)! + \cdots + \log 1! + \log 0! = \sum_{r=0}^{n} \log r!$. (Or, substitute $r' = n - r$.)
*$\text{(5)}$ : $k! = 1 \cdot 2 \cdots (n-1) \cdot n $ and then use the property of $\log$.
*$\text{(6)}$ : Swapping the order of summation; the double sum runs over $(r, k)$ where $1 \leq k \leq r \leq n$. This is the same as $1 \leq k \leq n$ and $ k \leq r \leq n$.
*$\text{(7)}$ : The inner sum is $\sum_{r=k}^{n} \log k = (\log k) \sum_{r=k}^{n} 1 = (\log k)(n+1-r) $.
Now, $\text{(8)}$ almost looks like a Riemann sum, and it will be indeed a Riemann sum if $\log k$ were $\log \frac{k}{n+1}$ instead. So, it is natural to replace $\log k$ by $\log\frac{k}{n+1}$ and then examine how the difference behaves. Proceeding,
\begin{align*}
\log A_n
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \left[ \log\left(\frac{k}{n+1}\right) + \log (n+1) \right] \\
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \log\left(\frac{k}{n+1}\right) + \frac{\log (n+1)}{n(n+1)} \sum_{k=1}^{n} (2k - n - 1). \tag{9}
\end{align*}
However, we have
$$ \sum_{k=1}^{n} (2k - n - 1)
= n(n+1) - n(n+1)
= 0. $$
So the second sum in $\text{(9)}$ is zero and
\begin{align*}
\log A_n
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \log\left(\frac{k}{n+1}\right) \\
&\to \int _{0}^{1} (2x - 1)\log x \, \mathrm{d}x \\
&= \left[ x - \frac{x^2}{2} + (x^2 - x) \log x \right]_{0}^{1} \\
&= \frac{1}{2}.
\end{align*}
Therefore $ \log A = \frac{1}{2} $ and hence $A^2 = e^{2\log A} = e$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4483861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Probability of rolling exactly 1 number exactly 3 times in 6 rolls of a fair die I'm trying to calculate the probability based on the size of the event space divided by the size of the sample space $P=\frac{|E|}{|S|}$
I know that $|S|=6^6$, but am not sure what exactly the event space consists. Currently my thoughts are that we have 6 choices for our favorable event(the triples) and for the remaining 3 numbers we have $5\times5\times4=100$, $4$ because we do not want to include the possibility of having 3 same numbers two times, and to consider all possible arrangements, there are then $\frac{6!}{3!1!1!1!}$ possibilities.
This leads to our final equation of : $P=\frac{|E|}{|S|}=\frac{6!}{3!1!1!1!} \times \frac{6\times100}{6^6}$
But the problem is that this exceeds 1, which is clearly wrong but I couldn't really figure out what is the fix for my equation.
Thanks:)
|
Since there are six possible values for each of the six rolls, there are indeed $6^6$ elements in the sample space.
For the favorable cases, since we want to calculate the number of cases in which exactly one number appears three times, there are two possibilities:
*
*One number appears three times and three other numbers each appear once.
*One number appears three times, a second number appears twice, and a third number appears once.
One number appears three times and three other numbers each appear once: There are six ways to select the number which appears three times, $\binom{6}{3}$ ways to select which three of the six positions in the sequence of rolls that number occupies, $\binom{5}{3}$ ways to select which three of the other five numbers each appear once, and $3!$ ways to arrange those three distinct numbers in the remaining positions of the sequence. Hence, there are $$\binom{6}{1}\binom{6}{3}\binom{5}{3}3!$$
such cases.
One number appears three times, a second number appears twice, and a third number appears once: There are six ways to select the number which appears three times, $\binom{6}{3}$ ways to select which three of the six positions in the sequence of rolls that number occupies, five ways to select which of the remaining numbers appears twice, $\binom{3}{2}$ ways to select which two of the three positions in the sequence of rolls that number occupies, and four ways to select which of the remaining numbers fills the remaining position in the sequence. Hence, there are
$$\binom{6}{1}\binom{6}{3}\binom{5}{1}\binom{3}{2}\binom{4}{1}$$
such cases.
Since the above cases are mutually exclusive and exhaustive, the number of favorable cases is
$$\binom{6}{1}\binom{6}{3}\binom{5}{3}3! + \binom{6}{1}\binom{6}{3}\binom{5}{1}\binom{3}{2}\binom{4}{1}$$
Hence, the probability that one number appears exactly three times in six rolls of a fair six-sided die is
$$\frac{\dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{3}3! + \dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{1}\dbinom{3}{2}\dbinom{4}{1}}{6^6}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4485857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Finding the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ with other approaches It is a problem from a timed exam,
What is the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ ?
$1)1\qquad\qquad2)\sqrt[4]2\qquad\qquad3)2\qquad\qquad4)2\sqrt[4]2$
I solved it with two approaches.
First approach,
$$\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt{1+\sqrt7}=\sqrt[4]{\frac{2}{(1+\sqrt7)^2}}\times\sqrt{1+\sqrt7}=\sqrt[4]2$$
Second approach,
$$\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt{1+\sqrt7}=\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt[4]{8+2\sqrt7}=\sqrt[4]{\frac{2(4+\sqrt7)}{4+\sqrt7}}=\sqrt[4]2$$
I'm wondering is it possible to solve this problem with other efficient approaches?
|
If we call $ \ u \ = \ 4 + \sqrt7 \ \ , $ then we have $ \ 1 + \sqrt7 \ = \ u - 3 \ $ and $ \ \frac{1}{u} \ = \ \frac{4 - \sqrt7}{9} \ \ . $ Consequently,
$$\sqrt[4]{(4+\sqrt7)^{-1}} \ · \ \sqrt{1+\sqrt7} \ \ = \ \ \sqrt[4]{\frac{(u \ - \ 3)^2}{u}} \ \ = \ \ \sqrt[4]{ \ u \ - \ 6 \ + \ \frac{9}{u} \ }$$
$$ = \ \ \sqrt[4]{ \ (4 + \sqrt7) \ - \ 6 \ + \ 9· \left(\frac{4 - \sqrt7}{9} \right) \ } \ \ = \ \ \sqrt[4]{ \ 4 \ - \ 6 \ + \ 4 \ } \ \ = \ \ \sqrt[4]{ 2 } \ \ . $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4487244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Find $\lim_{n \to \infty}\frac{x_n}{\sqrt{n}}$ where $x_{n+1}=x_n+\frac{n}{x_1+x_2+\cdots+x_n}$ Assume a positive sequence $\{x_n\} $ satisfies $$x_{n+1}=x_n+\frac{n}{x_1+x_2+\cdots+x_n}.$$ Find $\lim\limits_{n \to \infty}\dfrac{x_n}{\sqrt{n}}$.
Assume the limit we want is $L$. Then by Stolz theorem, one can obtain
\begin{align*}
L&=\lim_{n \to \infty}\frac{x_n}{\sqrt{n}}=\lim_{n \to \infty}\frac{x_{n+1}-x_n}{\sqrt{n+1}-\sqrt{n}}=\lim_{n \to \infty}\frac{2n\sqrt{n}}{x_1+x_2+\cdots+x_n}\\
&=\lim_{n \to \infty}\frac{2n\sqrt{n}-2(n-1)\sqrt{n-1}}{x_n}=\lim_{n \to \infty}\frac{3\sqrt{n}}{x_n}=\frac{3}{L},
\end{align*}
which implies $L=\sqrt{3}$. But how to prove the limit exists?
|
This is a community-wiki answer illustrating Iosif Pinelis's solution (with some simplifications).
*
*If you like this, please give kudos to his original solution as well.
*Also, feel free to improve this answer as you please!
Setting. We will write $s_n = \sum_{k=1}^{n} x_k$, so that the recurrence relation takes the form
$$ x_{n+1} = x_n + \frac{n}{s_n}. \tag{RE} $$
Also, let $\alpha$ and $\beta$ by
$$ \alpha = \liminf_{n\to\infty} \frac{x_n}{\sqrt{n}}
\qquad\text{and}\qquad
\beta = \limsup_{n\to\infty} \frac{x_n}{\sqrt{n}}. $$
Our goal is to prove that $\alpha = \beta = \sqrt{3}$. To make use of these quantities, we will frequently utilize the following inequalities:
Lemma. Let $(a_n)$ be any sequence of real numbers, and let $p > 0$. Then
$$ \color{navy}{\liminf_{n\to\infty} \frac{a_n}{n^p}}
\geq \liminf_{n\to\infty} \frac{a_n - a_{n-1}}{pn^{p-1}} \tag{SC1} $$
and
$$ \color{navy}{\limsup_{n\to\infty} \frac{a_n}{n^p}}
\leq \limsup_{n\to\infty} \frac{a_n - a_{n-1}}{pn^{p-1}}. \tag{SC2} $$
Proof. This is an immediate consequence of the Stolz–Cesàro theorem together with the asymptotic formula $n^p - (n-1)^p \sim pn^{p-1}$.
Step 1. Since $(x_n)$ is increasing, we know that $s_n \leq n x_n$. Then by the lemma,
\begin{align*}
\alpha^2
= \liminf_{n\to\infty} \frac{x_n^2}{n}
&\geq \liminf_{n\to\infty} \left( x_{n+1}^2 - x_n^2 \right)
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\
&\geq \liminf_{n\to\infty} \frac{2nx_n}{s_n}
\tag*{by $\color{#2E8B57}{\text{(RE)}}$} \\
&\geq 2.
\tag*{$\because \ s_n \leq n x_n$}
\end{align*}
Now, by the Stolz–Cesàro theorem again,
\begin{align*}
\beta
= \limsup_{n\to\infty} \frac{x_n}{\sqrt{n}}
&\leq \limsup_{n\to\infty} \frac{x_{n+1} - x_n}{\frac{1}{2}n^{-1/2}}
\tag*{by $\color{#2E8B57}{\text{(SC2)}}$} \\
&= \limsup_{n\to\infty} \frac{2n^{3/2}}{s_n}
\tag*{by $\color{#2E8B57}{\text{(RE)}}$} \\
&= \left[ \liminf_{n\to\infty} \frac{s_n}{2n^{3/2}} \right]^{-1} \tag{1} \\
&\leq \left[ \liminf_{n\to\infty} \frac{x_n}{3\sqrt{n}} \right]^{-1}
= \frac{3}{\alpha}.
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$}
\end{align*}
These altogether show that $0 < \alpha \leq \beta < \infty$.
(Remark. Starting from $\alpha$ and applying a similar argument as above, we can show that $\alpha \beta = 3$. However, this does not determine the value of $\alpha$ and $\beta$. So, we will not bother to prove this.)
Step 2. Using the recurrence relation $\color{#2E8B57}{\text{(RE)}}$, we find that
\begin{align*}
s_{n+1}^2 - 2s_n^2 + s_{n-1}^2
&= (s_n + x_{n+1})^2 - 2s_n^2 + (s_n - x_n)^2 \\
&= 2 s_n(x_{n+1} - x_n) + x_{n+1}^2 + x_n^2 \\
&= 2n + x_{n+1}^2 + x_n^2. \tag{2}
\end{align*}
Using this, we get
\begin{align*}
\liminf_{n\to\infty} \frac{s_n^2}{n^3}
&\geq \liminf_{n\to\infty} \frac{s_{n+1}^2 - s_n^2}{3n^2}
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\
&\geq \liminf_{n\to\infty} \frac{s_{n+1}^2 - 2s_n^2 + s_{n+1}^2}{6n}
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\
&\geq \frac{1}{3} + \frac{1}{6} \biggl( \liminf_{n\to\infty} \frac{x_{n+1}^2}{n} \biggr) + \frac{1}{6} \biggl( \liminf_{n\to\infty} \frac{x_n^2}{n} \biggr)
\tag*{by $\color{#2E8B57}{\text{(2)}}$} \\
&= \frac{1 + \alpha^2}{3}.
\end{align*}
Plugging this into $\color{#2E8B57}{\text{(1)}}$,
\begin{align*}
\beta
\leq \left[ \liminf_{n\to\infty} \frac{s_n}{2n^{3/2}} \right]^{-1}
\leq 2\sqrt{\frac{3}{1+\alpha^2}}. \tag{3}
\end{align*}
A similar calculation also shows that
$$ \alpha \geq 2\sqrt{\frac{3}{1+\beta^2}}. \tag{4} $$
Step 3. Define $f(x) = 2\sqrt{\frac{3}{1+x^2}}$, and note that $f$ is decreasing on $[0, \infty)$. So, using $\color{#2E8B57}{\text{(3)}}$ and $\color{#2E8B57}{\text{(4)}}$, we get
$$ \beta \leq f(\alpha) \implies f(f(\alpha)) \leq f(\beta)
\qquad\text{and}\qquad
\alpha \geq f(\beta) \implies f(\alpha) \leq f(f(\beta)) $$
and hence
$$ \beta \leq f(f(\beta)) \qquad\text{and}\qquad f(f(\alpha)) \leq \alpha. $$
It is not hard to check that these inequalities yield $\beta \leq \sqrt{3} \leq \alpha$:
So, using the obvious relation $\alpha \leq \beta$, we conclude that
$$ \sqrt{3} \leq \alpha \leq \beta \leq \sqrt{3} $$
and therefore $\alpha = \beta = \sqrt{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is independent of $n$.
Taking
$$f(n)=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
I could prove that
$$f(n+25)=f(n)$$
But, I have no idea how to show
$$f(n+k)=f(n)\\\forall k\in \{1,2,\dots 24\}$$
Of course, we can check these finite number of cases by force. But is there any other elegant method to handle this?
|
This is an amendment of my solution inspired by fleablood's comment:
To simplify the expression $\left \lfloor \frac{n-17}{25} \right \rfloor$, let $n-17=25k+r$ i.e. $n=25k+r+17$ where $0 \le r \le 24$.
Then
$$\left\lfloor \frac{8n+13}{25}\right \rfloor=\left\lfloor \frac{8(25k+r+17)+13}{25} \right\rfloor = 8k+5+\left\lfloor \frac{8r+24}{25} \right\rfloor$$
and $$\left\lfloor \frac{n-12-\left\lfloor \frac{n-17}{25} \right\rfloor}{3} \right\rfloor =\left\lfloor \frac{25k+r+17-12-\left\lfloor \frac{25k+r+17-17}{25}\right \rfloor}{3} \right\rfloor=8k+1+\left\lfloor \frac{r+2}{3}\right \rfloor $$
The job is done if we can prove that
$$\left\lfloor \frac{8r+24}{25}\right \rfloor - \left\lfloor \frac{r+2}{3} \right\rfloor $$ is independent of $r$ when $0 \le r \le 24$
For this, put $r=3m+h$ where $0 \le h \le 2$ and $0 \le r \le 24$.
It is not difficult to check that
$$\left\lfloor \frac{8r+24}{25}\right \rfloor = \left\lfloor \frac{r+2}{3} \right\rfloor $$
under the said conditions.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Is $\frac{x^2-3x-4}{-3x-15}\Big/\frac{x^2-16}{x^2-x-30}$ defined at $x=6$ or not? If a number is in the denominator of a rational expression that itself is the denominator of another rational expression, such that the number causes the latter denominator to be undefined, then isn't the entire expression undefined?
For example, isn't $$\frac{\frac{x^2-3x-4}{-3x-15}}{\frac{x^2-16}{x^2-x-30}}=\frac{\frac{(x-4)(x+1)}{-3(x+5)}}{\frac{(x+4)(x-4)}{(x-6)(x+5)}}$$ undefined at $x=6\;?$ $$\frac{\frac{x^2-3x-4}{-3x-15}}{\frac{x^2-16}{x^2-x-30}} \implies\quad\frac{(x-4)(x+1)(x-6)(x+5)}{-3(x+5)(x+4)(x-4)},x\neq{-5},{-4},4,6\quad?$$
If so, why do my scientific calculator and Desmos compute the expression as zero at $x=6\;?$
|
$$\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}} \implies\quad\frac{(x-4)(x+1)(x-6)(x+5)}{-3(x+5)(x+4)(x-4)},x\neq{-5},{-4},4,6\quad?$$
Rather: $$f(x)=\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}}\iff\bigg(\color\red{x\neq-5,4,6}\quad\text{and}\quad f(x)=\frac{(x+1)(x-6)}{-3(x+4)} \bigg).$$
isn't $$\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}}$$ undefined at $x=6\;?$
This expression is indeed undefined at $x=6;$ Wolfram also agrees.
why do my scientific calculator and Desmos compute the expression as zero at $x=6\;?$
Most likely, Desmos determines a graphing function's domain only after getting rid of any compound (multi-storey) fraction: $$\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}}$$ contains four implicit conditions $(\color\red{x\ne{-}5,{-}4,4,6}),$ while $$\frac{(x-4)(x+1)(x-6)(x+5)}{-3(x+5)(x+4)(x-4)}$$ contains only three implicit conditions $(\color\red{x\ne{-}5,{-}4,4}).$
Because of this, Desmos is selectively noticing the compound fraction's implicit conditions, mistaking $f(6)$ as $0$ even while it doesn't similarly think that $f(-5)= \frac{44}3$ or that $f(4)=\frac5{12}.$
|
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"url": "https://math.stackexchange.com/questions/4491310",
"timestamp": "2023-03-29T00:00:00",
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|
Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$
I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sure how to find the root for negative values of $x$.
By multiplying both sides by $(x+1)^2$ we get,
$$x^2(x^2+2x+1)+x^2-\frac54(x^2+2x+1)=0$$
$$4x^4+8x^3+3x^2-10x-5=0 \implies (x-1)(4x^3+12x^2+15x+5)=0$$
But I can't factor the third degree polynomial.
|
Let $t=x+1$
$\displaystyle (t-1)^2 + \left( \frac{t-1}{t} \right)^2 = \frac{5}{4}$
$(t^2 + 2 + \frac{1}{t^2}) - 2\left(t + \frac{1}{t} \right) = \frac{5}{4}$
Let $s = \left(t + \frac{1}{t} \right)$
$s^2 - 2s - \frac{5}{4} = (s+\frac{1}{2})(s-\frac{5}{2}) = 0$
We reduced solving quartic for x to quadratic for s, then quadratic for t
$\displaystyle s=-\frac{1}{2}\qquad ⇒ x = \frac{-5 ± i \sqrt{15}}{4}$
$\displaystyle s=+\frac{5}{2}\qquad ⇒ x = \frac{1 ± 3}{4}$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $P(P(x))=Q(x)^2$, then $P(x)=R(x)^2$? Let $P$ and $Q$ are polynomials from $\mathbb{R}[x]$. Suppose that $P(P(x))=Q(x)^2$. Does it mean that for any such $P(x)$ exists some $R(x)\in \mathbb{R}[x]$ such that $P(x)=R(x)^2$?
|
Yes, $P(x)$ must be a square. Suppose we have a counterexample $(P(x), Q(x))$, and let $d$ be the degree of $P(x)$. Then $P(P(x))$ has degree $d^2$, so in order for it to be a square $d$ must be even. Write $d = 2n$ and $P(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_1 x + a_0$. The leading coefficient $a_{2n}$ is nonzero, and if it were negative then so would be the leading coefficient of $P(P(x))$; thus $a_{2n} > 0$.
We will now construct an approximate square root $R(x) = b_n x^n + \cdots + b_0$ of $P(x)$. Set $b_n = \sqrt{a_{2n}}$. Then by repeatedly completing the square, we can find real numbers
\begin{align*}
b_{n-1} & = \frac{a_{2n}}{2b_n}, \\
b_{n-2} & = \frac{a_{2n-2} - b_{n-1}^2}{2b_n}, \\
& \dots \\
b_0 & = \frac{a_n - 2b_1 b_{n-1} - 2b_2 b_{n-2} - \cdots}{2b_n}
\end{align*}
such that $R(x)^2$ agrees with $P(x)$ in degrees $n$ and above; that is, the polynomial $S(x) := P(x) - R(x)^2$ has degree less than $n$. But then we have
$$
Q(x)^2 = P(P(x)) = R(P(x))^2 + S(P(x)),
$$
where $S(P(x))$ has degree $< 2n^2$ and $Q(x)^2$ and $R(P(x))^2$ both have degree $d^2 = 4n^2$. These are too close together to be distinct squares, so we must have $Q(x) = R(P(x))$. Then $S(P(x)) = 0$, so $S(x) = 0$, and thus $P(x)$ is indeed equal to $R(x)^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\sum_{\ell=0}^k \binom{2k+1}{2\ell} = 2^{2k}$ In acquainting myself with the Cauchy product, I am trying to derive the power series of $\sin(x)\cos(x)$. Of course, $\sin(x)\cos(x)=\frac12\sin(2x)$ is easy to work with so I know what to expect.
We have
$$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1} \implies \frac12\sin(2x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} 2^{2n} x^{2n+1}$$
Then the Cauchy product is
$$\begin{align*}
\sin(x)\cos(x) &= \sum_{k=0}^\infty \sum_{\ell=0}^k \frac{(-1)^\ell}{(2\ell)!} \frac{(-1)^{k-\ell}}{(2(k-\ell)+1)!} x^{2k+1} \\
&= \sum_{k=0}^\infty (-1)^k x^{2k+1} \sum_{\ell=0}^k \frac1{(2\ell)! (2k-2\ell+1)!} \\
&= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k+1} \sum_{\ell=0}^k \binom{2k+1}{2\ell}
\end{align*}$$
and the titular identity follows. How can one prove it?
Induction is a bit tricky. We have
$$\begin{align*}
\sum_{\ell=0}^{k+1} \binom{2k+3}{2\ell} &= \binom{2k+3}0 + \sum_{\ell=1}^k \binom{2k+3}{2\ell} + \binom{2k+3}{2k+2} \\[1ex]
&= 2k+4 + \sum_{\ell=1}^k \left[\binom{2k+1}{2\ell-2} + 2 \binom{2k+1}{2\ell-1} + \binom{2k+1}{2\ell}\right] \\[1ex]
&= 2^{2k} + 2k+3 + \sum_{\ell=1}^k \left[\binom{2k+1}{2\ell-2} + 2 \binom{2k+1}{2\ell-1}\right]
\end{align*}$$
but I'm not sure where to go from here. Any other suggestions/methods are appreciated.
|
Hint: Since $\dbinom{2k+1}{2\ell} = \dbinom{2k+1}{2k+1-2\ell}$, the sum (which we'll denote by $S$) satisfies:
\begin{align*}
S &= \dbinom{2k+1}{0}+\dbinom{2k+1}{2}+\cdots+\dbinom{2k+1}{2k-2}+\dbinom{2k+1}{2k}
\\
S &= \dbinom{2k+1}{2k+1}+\dbinom{2k+1}{2k-1}+\cdots+\dbinom{2k+1}{3}+\dbinom{2k+1}{1}
\end{align*}
Now, add these two equations together and see what you get.
|
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|
How does one show this complex expression equals a natural number? We have:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3}+ \frac{7}{3 \left(\frac{10}{3^{3/2}}i-3\right)^{1/3}}=2$$
This comes from solving the cubic equation of $x^3-7x+6=0$ which factors as $(x-2)(x-1)(x+3)=0$
We can simplify the problem into finding just one part of this, namely:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3} = 1 + \frac{2}{\sqrt{3}} i$$
Now, is there any general method (which doesn't involve factorising a cubic!) in which we get from the LHS of the above equation to the RHS? Or do we simply say that the cubic formula fails in this case and we have to resort to trial-and-error factorisation to find the result?
In which case, if there is a general method, could we not express this method as another cubic formula circumvents the intermediate complex steps?
BTW, I am only interested in cases in which the solutions of the cubic are rational or real algebraic solutions (without complex sub-parts). So we might be able to use this fact in a general method.
(Some might say that even finding $27^{1/3}$ is trial-and-error in a way, since we could try numbers 1,2,3... to see which one works. But we shall ignore and just say that finding the cube root of an integer is "allowed"!)
|
Let $\,u = \left(-3 + \dfrac{10}{3^{3/2}}i\right)^{1/3}\,$ and $\,v = \left(-3 -\dfrac{10}{3^{3/2}}i\right)^{1/3}\,$ be the principal values of the cube roots. Being the principal value, $\,u\,$ is the root with the greatest real part of $\,
z^3 = -3 + \dfrac{10}{3^{3/2}} i\,$.
Taking complex conjugates, $\,\overline{u}\,$ is a root of $\,z^3 = -3 - \dfrac{10}{3^{3/2}} i\,$. Given that the real part is invariant under complex conjugation, it follows that $\,\overline{u}\,$ is the root with the greatest real part of the conjugate equation, and therefore $\,v = \overline{u}\,$ (see also Sum of cube roots of complex conjugates).
We'll further note that $\,|u|^2 = \dfrac{7}{3}\,$ since $\,\displaystyle
\left|- 3 \pm \dfrac{10}{3^{3/2}}i\right|^2 = 9 + \dfrac{100}{27} = \dfrac{343}{27} = \left(\dfrac{7}{3}\right)^3\,$.
Then the LHS of the top equality can be written as $\,a = u + \dfrac{|u|^2}{u} = u + \overline u\,$, and:
$$
\require{cancel}
\begin{align}
a^3 = (u+\overline{u})^3 &= u^3 + \overline{u}^3 + 3u\overline{u}(u+\overline{u})
\\ &= -3 + \cancel{\dfrac{10}{3^{3/2}}i} -3 - \cancel{\dfrac{10}{3^{3/2}}i} + 3|u|^2\,a
\\ &= -6 + 7a
\end{align}
$$
Piecing it all together, $\,a = 2\,\text{Re}(u)\,$ is a root of $\,a^3-7a+6 = (a-2)(a-1)(a+3)\,$, and by the choice of the principal value of the cube root it is the largest real root, so in the end $\,a=2\,$.
|
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|
Pythagorean Quadruples (Integer Question) We have $x^2+y^2+z^2=m^2$ with $x,y,z,m$ integers, $\gcd(x,y,z)=1$, $z$ is odd, $x$ and $y$ are even.
Set $x_1:=\frac{x}{2}$ und $y_1:=\frac{y}{2}$. We get
\begin{gather*}
x_1^2+y_1^2=\frac{1}{4}(x^2+y^2)=\frac{1}{4}(m^2-z^2)=\left(\frac{m+z}{2}\right)\left(\frac{m-z}{2}\right).
\end{gather*}
Set $f:=\gcd(x_1,y_1)$, $f_1:=\gcd(f,\frac{m+z}{2})$, $f_2:=\gcd(f,\frac{m-z}{2})$.
We can proof, that $\gcd(f_1,f_2)=1$ is true.
If $d=\gcd(f_1,f_2)>1$, then we have $d\mid f$, $d\mid \frac{m+z}{2}$, $d\mid \frac{m-z}{2}$ and consequently also $d\mid \frac{m+z}{2}-\frac{m-z}{2}=z$. Moreover we get $d\mid \frac{x}{2}$, especially $d\mid x$. Analog $d\mid y$, which creates a contradiction to $\gcd(x,y,z)=1$.
Set $x_2:=\frac{x_1}{f}$, $y_2:=\frac{y_1}{f}$, $z_1:=\frac{m+z}{2f_1^2}$ and $z_2:=\frac{m-z}{2f_2^2}$.
Can someone explain to me why $z_1$ and $z_2$ are integers?
I unterstand that for example $\frac{m+z}{2f_1}$ is an integer, but I don't understand why $z_1$ is an integer.
Thanks for your help.
|
Since $f_1|x_1,y_1$, we have
$$f_1^2\,|\,x_1^2+y_1^2\ =\ \left(\frac{m+z}2\right)\left(\frac{m-z}2\right)$$
and $f_1$ is coprime to $f_2$, hence also to $\frac{m-z}2$, so $f_1^2\,|\,\frac{m+z}2$.
|
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|
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $()^4$.
I tried to rewrite $8x^4=(2x)^4\times\frac{1}{2}$, but It wasn't helpful :
$$\left(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\right)((2x)^4+(3y)^4+(4z)^4)\ge (x+y+z)^4$$
So, is there any nice transform to apply C-S inequality?
or should I apply Lagrange Multiplier?
|
Using Lagrange multiplier method:
$$\frac{\partial}{\partial x}\left(8x^4+27y^4+64z^4-\lambda(x+y+z-\frac{13}4)\right)=0\\4\cdot8x^3=\lambda$$
Similarly for $y$ and $z$ you get:
$$4\cdot 27y^3=\lambda\\4\cdot 64z^3=\lambda$$
If we use $$\frac\lambda4=\mu^3$$
one can take the cubic root and get $$x=\frac\mu2\\y=\frac\mu3\\z=\frac\mu 4$$
Plugging it into the constraint:
$$\frac\mu2+\frac\mu3+\frac\mu4=\frac{13}{4}\\\mu\frac{6+
4+3}{12}=\mu\frac{13}{12}=\frac{13}4$$
So $\mu=3$. Then $$x=\frac32\\y=1\\z=\frac34$$
The minimum value of the function is then $$8\frac{3^4}{2^4}+27+64\frac{3^4}{4^4}=\frac{351}4$$
|
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|
Proving that $\frac {(n+1)n(n-1)...(n-i)}{i+1}$ can be used as the summ of $k^h$ I have previously proved that $$\sum_{k=i}^{n} k(k-1)(k-2)\cdots(k-i+1) = \frac{(n+1)\cdot n\cdot (n-1)\cdots(n-i+1)}{i+1}$$
From here I am supposed to show that the formula above can be used to compute the sum of $k^h$, for example, $\sum_{k=1}^{n}k^2$ or $\sum_{k=1}^{n}k^3$, starting with $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$
I cannot figure out the connection between these and do not know where to start...
|
Yes, it can be used.
Example with $\sum_{k=1}^{n}k^2$
$\sum_{k=1}^{n}k^2$
$=\sum_{k=1}^{n}k(k-1+1)$
$=(\sum_{k=1}^{n}k(k-1)) + \sum_{k=1}^{n}k$
$=(\sum_{k=2}^{n}k(k-1)) + \sum_{k=1}^{n}k$ ( means that $i = 2$)
$=\frac{(n+1)n(n-1)}{3} + \frac{n(n+1)}{2}$
$=\frac{(n+1)n(2(n-1) + 3)}{6}$
$=\frac{(n+1)n(2n+1)}{6}$
Example with $\sum_{k=1}^{n}k^3$
$\sum_{k=1}^{n}k^3$
$=\sum_{k=1}^{n}k(k-1+1)(k-2+2)$
$=\sum_{k=1}^{n}k((k-1)(k-2) + 2(k-1)+(k-2)+2)$
$=\sum_{k=1}^{n}k((k-1)(k-2) + 3k-2)$
$=\sum_{k=1}^{n}k(k-1)(k-2) + 3\sum_{k=1}^{n}k^2 - 2\sum_{k=1}^{n}k$
$=\sum_{k=3}^{n}k(k-1)(k-2) + 3\sum_{k=1}^{n}k^2 - 2\sum_{k=1}^{n}k$ ( means that $i=3$)
$=\frac{(n+1)n(n-1)(n-2)}{4} + 3\frac{n(n+1)(2n+1)}{6} - 2\frac{n(n+1)}{2}$
$=\frac{n^2 (n + 1)^2}{4}$
Same way in higher power
|
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|
Find $f:\mathbb{R} \to [-1, 1]$ which satisfies these two conditions:
Find $f:\mathbb{R} \to [-1, 1]$ which satisfies these two conditions:
$f(x+y)=f(x)f(a-y)+f(y)f(a-x)$, $f(a-x-y)=f(a-x)f(a-y)-f(x)f(y).$ ($a$: constant.)
My expectation of $f$ is $f(x)=\sin(\frac {\pi x} {2a} +2\pi n).$ But, how can we prove this? Or, are there other functions that satisfy those conditions?
If we substitute $f(x)=\sin(\frac{\pi x}{2a}+2\pi n)$, we get:
$\newcommand{\x}{\dfrac{\pi x}{2a}} \newcommand{\y}{\dfrac{\pi y}{2a}} \newcommand{\pin}{\pi n} \sin\left(\x+\y+2\pin\right)=\sin\left(\x+\y\right)\\
=\sin\left( \x+2\pin \right)\sin\left( \dfrac {\pi}{2}-\y+2\pin \right)+\sin\left(\y+2\pin\right)\sin\left(\dfrac{\pi}{2}-\x+2\pin\right) \\
=\sin\left(\x\right)\cos\left(\y\right)+\sin\left(\y\right)\cos\left(\x\right). \\
\ \\
\ \\
\sin\left(\dfrac{\pi}{2}-\x-\y+2\pin\right)= \cos\left(\x+\y\right)\\
=\sin\left(\dfrac{\pi}{2}-\x+2\pin\right)\sin\left(\dfrac{\pi}{2}-\y+2\pin\right)-\sin\left(\x+2\pin\right)\sin\left(\y+2\pin\right) \\
=\cos\left(\x\right)\cos\left(\y\right)-\sin\left(\x\right)\sin\left(\y\right).$
Uh... Can we show that $f'(x)=f(a-x)?$
|
If we let $g(x) = f(a-x) + i.f(x)$, the two equations are equivalent to $g(x+y) = g(x).g(y)$. This resembles Cauchy exponential functional equation from $\mathbb{R}$ to $\mathbb{C}$, and there are infinitely many solutions via Hamel's basis.
If $f$ is continuous on the other hand, plugging $x=0$ yields $g(y) = g(0)g(y)$. Excluding trivial solution $g = 0$, pick some $g(y) \ne 0$, so $g(0) = 1$. Plugging $x \rightarrow a-x$ to $g$'s definition yields $g(a-x) = i.g(x)$, so $g(a) = i.g(0) = i$. For integers $m, n$ with $n \ne 0$, we have $g(\frac{m}{n}a)^{n} = g(ma) = g(a)^m = i^m = e^{\frac{\pi i m}{2}}$, so $g(\frac{m}{n}a) = e^{\frac{\pi i m}{2n} + \frac{2 \pi i k}{n}}$ for some $k \in \mathbb{Z}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and $g$ is continuous, we "plug $x = \frac{m}{n} a$" to get $g(x) = e^{(\frac{1}{2} + 2k(x)) \frac{\pi i x}{a}}$ for some integer function $k$. Notice that in order for $g$ to be continuous, $k$ must be continuous except when $x \in a \mathbb{Z}$, where $k$ is free to be any integer. In another words, $k(x) = \sum k_n 1_{x \in (na, (n+1)a)} + \sum l_n 1_{x = na}$ for constant integers $\{k, l\}_n$. Now, $g(na)$ stays the same when $l_n$ change, so WLOG $l = 0$, and $k(x) = \sum k_n 1_{x \in (na, (n+1)a)}$.
Thus $f(x) = Im(g(x)) = \sin(\frac{(4k(x)+1) \pi x}{2a})$, which along with $f(x) = 0$ satisfies the original equations.
|
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|
How to show $\sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}} = 0$? I came across this sum while doing the homework my teacher gave us on series. It was originally $$a_k = \frac{k}{81}, S = \sum_{k=1}^{80} {a_k^2 \over 1+2a_k^2-2a_k}$$
I then decomposed it into partial fractions and ended up with
$40+ \sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}}$. I know the answer is 40 from WolframAlpha, however I don't know how to go about showing that the sum with the complex numbers is 0.
How should I proceed?
|
There is no need to use complex numbers. Here the keyword is symmetry: note that $a_{81-k}=1-a_k$ and therefore
\begin{align}
2S&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{a_{81-k}^2}{1-2a_{81-k}(1-a_{81-k})}\\
&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{(1-a_k)^2 }{1-2(1-a_k)a_k}\\
&=\sum_{k=1}^{80} \underbrace{\frac{a_k^2+(1-a_k)^2}{1-2a_k(1-a_k)}}_{=1}=80
\end{align}
which implies that $S=40$.
|
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|
Probability that a $5$ cards hand from a standard deck of $52$ contains at least $1$ ace $\frac{C(4, 1)C(51, 4)}{C(52,5)}$? I understand the direct and indirect methods for solving this problem, but I do not understand why $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ is wrong. The answer that $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ results is about 4% bigger than the answer I got with $1 - \frac{C(4, 0)C(48, 5)}{C(52,5)}$ or the direct approach. What are the extra scenarios that $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ incorrectly cover that causes the $4\%$ difference?
|
Since there are
$$\binom{4}{k}\binom{48}{5 - k}$$
ways to select exactly $k$ aces and $5 - k$ non-aces, the correct count of favorable cases should be
$$\binom{4}{1}\binom{48}{4} + \binom{4}{2}\binom{48}{3} + \binom{4}{3}\binom{48}{2} + \binom{4}{4}\binom{48}{1}$$
Your failed attempt counts each hand with more than one ace as many times as an ace appears in the hand, once for each way you could have designated one of the aces in the hand as the ace in that hand.
For instance, by designating a particular ace in the hand as the ace that hand contains, you count the hand $\color{red}{A\heartsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit, 2\clubsuit$ twice.
\begin{array}{l l}
\text{designated ace} & \text{additional cards}\\ \hline
\color{red}{A\heartsuit} & A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit, 2\clubsuit\\
A\spadesuit & \color{red}{A\heartsuit}, \color{red}{7\diamondsuit}, 4\clubsuit, 2\clubsuit
\end{array}
Similarly, you count the hand $\color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit$ three times,
\begin{array}{l l}
\text{designated ace} & \text{additional cards}\\ \hline
\color{red}{A\heartsuit} & \color{red}{A\diamondsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit\\
\color{red}{A\diamondsuit} & \color{red}{A\heartsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit\\
A\spadesuit & \color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, \color{red}{7\diamondsuit}, 4\clubsuit
\end{array}
For the same reason, you count the hand $\color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\spadesuit, A\clubsuit, \color{red}{7\diamondsuit}$ four times.
\begin{array}{l l}
\text{designated ace} & \text{additional cards}\\ \hline
\color{red}{A\heartsuit} & \color{red}{A\diamondsuit}, A\spadesuit, A\clubsuit, \color{red}{7\diamondsuit}\\
\color{red}{A\diamondsuit} & \color{red}{A\heartsuit}, A\spadesuit, A\clubsuit, \color{red}{7\diamondsuit}\\
A\clubsuit & \color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\spadesuit, \color{red}{7\diamondsuit}\\
A\spadesuit & \color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\clubsuit, \color{red}{7\diamondsuit}
\end{array}
Notice that
$$\binom{4}{1}\binom{48}{4} + \color{red}{2}\binom{4}{2}\binom{48}{3} + \color{red}{3}\binom{4}{3}\binom{48}{2} + \color{red}{4}\binom{4}{4}\binom{48}{1} = \color{red}{\binom{4}{1}\binom{51}{4}}$$
|
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|
Solving a functional equation $ g(x) = \frac{1}{1-x} g\left(\frac{ax}{1-x}\right) $ Consider a following functional equation
$$ g(x) = \frac{1}{1-x} g\left(\frac{ax}{1-x}\right) $$
for a given fixed real parameter $a$.
I don't know whether the solution for $g$ is unique. However, assuming it has a Taylor series around $x=0$, we get, expanding the both sides and comparing each power of $x$,
$$ \frac{1}{n!} g^{(n)}(0) = \sum_{k=0}^n \frac{a^k}{k!} \binom{n}{k} g^{(k)}(0)$$
Upto a constant factor $g(0)$, we see clearly that all of the derivatives of $g$ at zero are uniquely determined.
On the other hand, note that a simple guess $$g(x) = \frac{1-a}{1-a-x}g(0)$$ satisfies the functional equation for $g$. Together with the uniqueness of the Taylor coefficients, we then deduce it is the solution of the initial problem.
However, is there a generic approach how to arrive to this solution? Based on manipulation of the functional equation for example?
|
I will assume that $g$ is continuous at $0$ and $a\in (0,1)$.
Let's prove by induction the following: for $n\in \Bbb N$,
\begin{align}
g(x) &= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}g\left(\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}\right)
\end{align}
For $n=0$ the statement is equivalement to the functional equation. Assume it is true for $n\in \Bbb N$,
\begin{align}
g(x) &= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}g\left(\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}\right)\\
&= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x} \frac{1}{1-\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}}g\left(\frac{a\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}}{1-\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}}\right)\\
&= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x} \frac{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}{1 - \left(\sum\limits_{k=0}^{n+1}a^k\right)x}g\left(\frac{a^{n+2}x}{1-\left(\sum\limits_{k=0}^{n+1}a^k\right)x}\right)\\
&= \frac{1}{1 - \left(\sum\limits_{k=0}^{n+1}a^k\right)x}g\left(\frac{a^{n+2}x}{1 - \left(\sum\limits_{k=0}^{n+1}a^k\right)x}\right)
\end{align}
So for fixed $x$, $$g(x) = \lim\limits_{n\to \infty} \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}g\left(\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}\right) = \frac{1}{1 - \left(\sum\limits_{k=0}^{\infty}a^k\right)x}g\left(0\right) = \frac{1}{1-\frac{x}{1-a}} g(0)$$
|
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|
How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta}$$
Adding two versions together yields
$$
\begin{aligned}
2 I &=2 a \int_{0}^{\pi} \frac{d \theta}{a^{2}-b^{2} \cos ^{2} \theta} \\
&=4 a\int_0^{\frac{\pi}{2}} \frac{\sec ^{2} \theta}{a^{2} \sec ^{2} \theta-b^{2}} d \theta \quad \textrm{( By symmetry )}\\
&=4 a{\int_{0}^{\infty}} \frac{d t}{\left(a^{2}-b^{2}\right)+a^{2} t^{2}} \\
&=\frac{4}{\sqrt{a^{2}-b^{2}}}\left[\tan^{-1} \left(\frac{at}{\sqrt{a^{2}-b^{2}}}\right)\right]_{0}^{\infty} \\
&=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
We can now conclude that $$
\boxed{I=\frac{\pi}{\sqrt{a^{2}-b^{2}}}}
$$
Are there any other methods to deal with the integral?
|
By Contour Integration
Considering the integral
$$
J=\int_{0}^{2 \pi} \frac{d \theta}{a+b \cos \theta}.
$$
Let $z=e^{\theta i}$, then $\cos \theta=\frac{1}{2}\left(y+\frac{1}{z}\right)$ and $d z =i e^{\theta i} d \theta=i z d \theta .$
$$
\begin{aligned}
\\
J &=\int_{K(0,1)} \frac{1}{a+\frac{b}{2}\left(z+\frac{1}{z}\right)} \cdot \frac{d z}{i z} \\
&=-2 i \int_{K(0,1)} \frac{1}{b z^{2}+2 a z+b} d z \\
&=-2 i \int_{K(0,1)} \frac{1}{b(z-\alpha)(z-\beta)} d z
\end{aligned}
$$
where K(0,1) is the unit circle with centre O, $\alpha=\frac{-a-\sqrt{a^{2}-b^{2}}}{b}$ and $\beta=\frac{-a+\sqrt{a^{2}-b^{2}}}{b}$.
Noting that $|\alpha|>1$ and $|\beta |<1$, therefore $\beta$ is only one residue in $K(0,1).$
$$
\begin{aligned}
J &=-2 i\left[2 \pi i \operatorname{Res}\left(\frac{1}{b(z-\alpha)(z-\beta)}, \alpha\right)\right] \\
&=4 \pi \lim _{z \rightarrow \beta}(z-\beta) \frac{1}{b(z-\alpha)(z-\beta)} \\
&=4 \pi \frac{1}{b(\beta-\alpha)} \\
&=4 \pi \frac{1}{2 \sqrt{a^{2}-b^{2}}} \\
&=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
By symmetry, \begin{equation}
I=\frac{1}{2} J=\frac{\pi}{\sqrt{a^{2}-b^{2}}}
\end{equation}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$
Show that for any non negative real numbers $x_1,x_2,\cdots x_n,$
$$\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
My work:
Let$$S(n)=\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
By theorem of triviality, if any of $x_i$'s are $0$ the inequality is certainly true. So assume all numbers are $\gt0$
$S(1)$ says ${x_1}^3\le {x_1}^3$ which is certainly true.
$S(2)$ says $({x_1}^2+{x_2}^2)x_1x_2\le\frac18(x_1+x_2)^4$ which reduces to $0\le(x_1-x_2)^4$ which is certainly true.
Assume $S(k)$ is true. Now we just needs to prove that $S(k+1)$ is true. But I'm having a hard time in doing that.
Any help is greatly appreciated. Or is there any better method than induction$?$
|
Another simple solution by induction, motivated by OP's questions and by the currently most voted answer (which is also by induction but a bit complicated).
\begin{equation}
\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}
\end{equation}
First, the inequality is clearly true for $n=2$, as it can be reduced to $(x_1-x_2)^4\ge 0$. Denote $p = x_1\dots x_n, q = (x_1^2+\dots + x_n^2)/n, s = (x_1+\dots+x_n)/n$, and $x=x_{n+1}$. Assume that the inequality is true for $n$ numbers, that is, $s^{n+2} \ge pq$. We will show that it is also true for $n+1$ numbers, that is, $f(x) \ge 0$, where
\begin{equation}
f(x) = \left(\frac{x+ns}{n+1}\right)^{n+3} - \frac{px(x^2 + nq)}{n+1}.
\end{equation}
WLOG, assume that $x_{n+1}$ is the maximum component. Then, it suffices to show that $f(x) \ge 0$ whenever $x\ge s$. Taking derivatives:
\begin{align}
f'(x) &= \frac{n+3}{n+1}\left(\frac{x+ns}{n+1}\right)^{n+2} - \frac{p(3x^2 + nq)}{n+1},\\
f''(x) &= \frac{(n+3)(n+2)}{(n+1)^2}\left(\frac{x+ns}{n+1}\right)^{n+1} - \frac{6px}{n+1}.
\end{align}
We can observe that $f(s)\ge 0$ and $f'(s) \ge 0$. Indeed, as $s^{n+2} \ge pq$ and $s^n\ge p$ (AM-GM), we have
\begin{align}
f(s) &= s^{n+3} - \frac{ps(s^2 + nq)}{n+1} = \frac{ns(s^{n+2}-pq) + s^3(s^n-p)}{n+1} \ge 0,\\
f'(s) &= \frac{(n+3)s^{n+2}}{n+1} - \frac{p(3s^2 + nq)}{n+1} = \frac{n(s^{n+2}-pq)}{n+1} + \frac{3s^2(s^n-p)}{n+1} \ge 0.
\end{align}
Next, notice that $\left(\frac{x+ns}{n+1}\right)^{n+1} \ge xs^n \ge xp$ (AM-GM), we have
\begin{equation}
f''(x) \ge \frac{(n+3)(n+2)px}{(n+1)^2} - \frac{6px}{n+1} = \frac{n(n-1)px}{(n+1)^2} \ge 0.
\end{equation}
Therefore, $f'$ is increasing (on $[s,+\infty)$) and thus $f'(x) \ge f'(s) \ge 0$, hence $f$ is increasing and thus $f(x)\ge f(s) \ge 0$. QED.
|
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|
Can anyone explain this process of solving? (Differentiation) I'm at differentiation of algebraic functions. There's an example in the module that I couldn't quite get how it led to that.
$y=\frac {(x+1)^3}{x^2}$
It's solved by using a combination of quotient and power rules. I'll enumerate how it's solved.
$(1) y′= \frac{(x^2(3(x+1)^2))–(x+1)^32x}{x^4}$
$(2) y′= \frac{x(x+1)^2 (3x–2(x+1))}{x^4}$
$(3) y′= \frac {(x+1)^2 (3x-2x-2)}{x^3}$
$(4) y′= \frac {(x+1)^2 (x-2)}{x^3}$
How did $(2)$ came to be? Why was it done like that? I just can't wrap my head around it. It just looked like it skipped a couple steps (at least to me). Can anyone help me with this?
|
Here is one way to solve it. Use the fact that the derivative of $(ax+b)^n=n(ax+b)^{n-1}(ax+b)'$. This just follows from the chain rule.
Rewrite the equation as $y=(x+1)^3x^{-2}$. Then apply the product rule. The product rule says $(f(x)g(x))'=f(x)g'(x)+g(x)f'(x)$.
Thus, we get $((x+1)^3x^{-2})'=((x+1)^3\cdot-2x^{-3})+3(x+1)^2x^{-2}$.
Writing this out in a clean way we get: $((x+1)^3x^{-2})'=\frac {-2(x+1)^3} {x^3}+\frac {3(x+1)^2} {x^2}=\frac {-2(x+1)^3+3x(x+1)^2} {x^3}$.
Now factor out $(x+1)^2$ to get $\frac {(x+1)^2(-2(x+1)+3x)} {x^3}=\frac {(x+1)^2(-2x-2+3x)} {x^3}=\frac {(x+1)^2(x-2)} {x^3}$
|
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|
Squeeze theorem to prove that the sequence $a_n= \frac{3^n}{n!}$ is convergent? I want to know if the following proof of the convergence of a sequence is correct.
Proof. Let $a_n= \frac{3^n}{n!}$. Firstly, it is trivial to see that $a_n \geq 0$ for all $n \in \mathbb{N}$. Secondly, see that
$$a_n = \frac{3^n}{n!}= \frac{3 \times 3\times3 \times \cdot\cdot\cdot \times 3 \times 3 \times 3}{n \times (n-1) \times (n-2) \times \cdot \cdot \cdot 3 \times 2 \times 1} = \frac{3\times3\times3\times3}{n\times3\times2\times1} \times \frac{3 \times 3 \times...\times 3}{(n-1) (n-2)\cdot\cdot\cdot \times 4}$$
$$= \frac{3^4}{6n} \times \Big(\frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4}\Big)$$
Because $0 < \frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4} <1$ we know $a_n=\frac{3^4}{6n} \times \Big(\frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4}\Big) \leq \frac{3^4}{6n}$. Using the fact that $a_n \geq 0$ we conclude that, since
$$0 \leq a_n \leq \frac{3^4}{6n}$$
and $0 \to 0, \frac{3^4}{6n} \to 0$ when $n \to 0$,
$$\lim_{n\to\infty}a_n=0$$
I skipped some trivial steps (e.g., showing that $\frac{3^4}{n}$ tends to $0$ if $n \to \infty$) because I assume theorems that prove such properties. I am mostly concerned about whether my manipulation of the factorial expression is correct and if the squeeze theorem is properly applied.
Thanks in advance!
|
Alternatively replacing all $n \geq 4$ to 4: $$0\leq a_n\leq \frac{9}{2}(\frac{3}{4})^{n-3}, $$ you will get the result. This is stronger that you can see the sum of this series also converges.
|
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|
Show that $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$. The problem asks us to show that the following equation holds true. $$4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$$
This equation has been verified on my calculator.
Perhaps some basic trigonometric formulas will be enough to solve the problem. I've tried the following: $$\begin{align} 4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ})&=16\sin^2(12^{\circ})\cos^2(12^{\circ})+8\sin^2(12^{\circ})
\cos(12^{\circ})\\
\\
&=8\sin^2(12^{\circ})\cos(12^{\circ})\Big(2\cos(12^{\circ} ) + 1\Big)\end{align}$$
As you can see, I was trying to simplify the expression so that it'll contain only $\sin(12^{\circ})$ and $\cos(12^{\circ})$, since I thought by unifying the angles I would have a bigger chance of solving it. However, I couldn't find a way to make any further progress. Can someone show me the way?
|
We can avoid values of $\cos36^\circ,\sin18^\circ$ to derive at the identity.
Let $x=6^\circ$
using Prosthaphaeresis Formulas
$$4\sin^24x+4\sin4x\sin2x=4\sin4x(\sin4x+\sin2x)=8\sin4x\sin3xcos x$$
$$\implies\sin24^\circ\sin18^\circ\cos6^\circ=\sin24^\circ\cos(90-18)^\circ\sin84^\circ$$
Using Prove that: $\sin\beta\sin\left(\dfrac\pi3-\beta\right)\sin\left(\dfrac\pi3+\beta\right)=\frac {1}{4}\sin 3\beta$,
$$4\sin24^\circ\sin(60+24)^\circ\sin(60-24)^\circ=\sin(3\cdot24)^\circ $$
$$\implies4\sin24^\circ\sin(60+24)^\circ=\dfrac{\sin72^\circ}{\sin36^\circ}=\dfrac{\cos(90-72)^\circ}{2\sin18^\circ\cos18^\circ}=?$$
|
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|
$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $ Find all x's
It is given that $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $$ Find all possible values of $x$.
What I did: $$\frac{1}{\log_2(x-2)}+\frac{1}{\log_2(x+2)}=\frac{5}{6} $$ $$\log_{(x-2)}2 +\log_{(x+2)}2 =\frac{5}{6}$$
What is the next ? I stuck in here..
I am looking for algebraic approach. The answer is $+6,-6$
|
Let $a = \log_2(x-2)$ and $b = \log_2(x+2)$. Then your equation becomes:
$$\frac{1}{a} + \frac{1}{b} = \frac{5}{6}$$
$$\frac{a + b}{ab} = \frac{5}{6}$$
$$6(a + b) = 5ab$$
$$b = \frac{6a}{5a - 6}$$
$$\log_2 (2^a + 4) = \frac{6a}{5a - 6}$$
This has the integer solution $a = 2$. Unfortunately, while this solution is easy to verify, it's hard to see how to explicitly solve for $a$.
Anyhow, with $\log_2(x-2) = 2$, we get $x - 2 = 4$, or $x = 6$.
|
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|
Equation has two distinct solutions We have $\displaystyle{ a,b,c }$ real numbers such that $\displaystyle{ a^2+b^2+c^2>0 }$. Which condition must hold so that the equation $\displaystyle{ ax^2+bx+c=0 }$ has two different solutions?
$$$$
I have done the following :
\begin{align*}&ax^2+bx+c=0 \Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0 \text{ if } a\neq 0 \\ &a=0 \Rightarrow ax^2+bx+c=0 \Rightarrow bx+c=0 \Rightarrow x=-\frac{c}{b}\ \text{ In this case we have only one solution, so it must be } a\neq 0\end{align*}
We have that \begin{equation*}x^2+\frac{b}{a}x+\frac{c}{a}=0 \Rightarrow x=-\frac{b}{2a}\pm \sqrt{\left (\frac{b}{2a}\right )^2-\frac{c}{a}}\end{equation*}
We have two different solution if $\frac{b^2}{4a^2}\neq \frac{c}{a}$, i.e. if $b^2\neq 4ac$.
Is the condition correct?
|
for any degree two polynomial in order for it to have real roots we must have :
$$\Delta = b^2-4ac > 0$$
and if $\Delta \neq 0$ it always has two distinct solutions real or complex so that I believe suffices.
|
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|
Infinite summation with multiple terms in denominator and variables in exponents I need to find $\sum\limits_{n=1}^{\infty} \frac{6^n}{2^{1 + 2 n} + 3^{1 + 2 n} - 5\cdot 6^n}$, but I don't know how to do infinite summation with multiple terms in the denomiator and variables in exponents. Can anybody give me a hint? Thanks :)
|
Hint
Let $x=2^n$ and $y=3^n$ to make
$$\frac{6^n}{2^{1 + 2 n} + 3^{1 + 2 n} - 5\times 6^n}=\frac{x y}{2 x^2-5 x y+3 y^2}=\frac{x y}{(2 x-3 y) (x-y)}$$ Now, as @person commented, using partial fractions
$$\frac{x y}{(2 x-3 y) (x-y)}=\frac{x}{y-x}-\frac{2 x}{3 y-2 x}=\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$ which seems to telescope.
|
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|
Maximize $f(x,y)=x+y$ subject to $\sqrt{4-2 x}+\sqrt{4-2 y}=\sqrt{x y}$ Given that $x,y \in [0,2]$ find the maximum value of $x+y$ if $$\sqrt{4-2 x}+\sqrt{4-2 y}=\sqrt{x y}$$
Looking for an elementary approach. No Lagrange multipliers please.
My try:
I tried using CS inequality:
$$\sqrt{4-2x}+\sqrt{4-2y}\leq \sqrt{2}\times \sqrt{8-2(x+y)}$$
$\implies$
$$\sqrt{xy} \le \sqrt{2} \times \sqrt{8-2(x+y)}$$
But not able to proceed?
|
Remark: Alternatively, we can use the substitutions $\sqrt{4 - 2x} = u, \sqrt{4-2y} = v$.
From $\sqrt{4-2x} + \sqrt{4-2y} = \sqrt{xy}$, we have
$$(8 - 2x - 2y) + 2\sqrt{(4-2x)(4-2y)} \ge xy$$
or
$$4(8 - 2x - 2y) + 8\sqrt{(4-2x)(4-2y)} \ge 4xy$$
or
$$4(8 - 2x - 2y) + 8\sqrt{(4-2x)(4-2y)} \ge (4-2x)(4-2y) - 16 + 8x + 8y$$
or
$$ 64 - 16x - 16y \ge \Big(4 - \sqrt{(4-2x)(4-2y)}\Big)^2$$
or
$$\sqrt{64 - 16x - 16y} \ge 4 - \sqrt{(4-2x)(4-2y)}$$
or
$$\sqrt{64 - 16x - 16y} + \sqrt{(4-2x)(4-2y)} \ge 4.$$
Using $\sqrt{(4-2x)(4-2y)} \le \frac{4-2x + 4 - 2y}{2}$, we have
$$\sqrt{64 - 16x - 16y} + \frac{4-2x + 4 - 2y}{2} \ge 4$$
or
$$\sqrt{64 - 16(x + y)} \ge x + y$$
or
$$64 - 16(x + y) \ge (x + y)^2$$
which results in
$$x + y \le 8\sqrt 2 - 8.$$
Also, if $x = y = 4\sqrt 2 - 4$, we have $x, y\in [0, 2]$
and $\sqrt{4-2x} + \sqrt{4-2y} = \sqrt{xy}$
and $x + y = 8\sqrt 2 - 8$.
Thus, the maximum of $x + y$
is $8\sqrt 2 - 8$.
|
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|
Darboux sum $\int_{2}^{5}1-x+3x^2dx$ calculate the Darboux sum of
$$\int_{2}^{5}1-x+3x^2dx$$
I think i did the calculation good but I need help in getting my partition more formal please tell me if you see any errors and mis
$P=\{2,2+\frac{1}{n}....5\}$,
$\Delta x_i =\frac{1}{n} $
\begin{align}
&=\int_{2}^{5}1dx-\int_{2}^{5}xdx+\int_{2}^{5}3x^2dx\\
&=\sum_{0}^{n-1}\left(2+\frac{i}{n}\right)\cdot \Delta x_i-\sum_{0}^{n-1}\left(2+\frac{i}{n}\right)\cdot \Delta x_i+\sum_{0}^{n-1}3\left(2+\frac{i}{n}\right)^2\cdot \Delta x_i\\
&=5-2 -\frac{1}{n^2}\sum_{0}^{n-1}\lim_{n\rightarrow\infty}\left(\frac{2}{n}+i\right)+3\cdot\frac{1}{n^3}\cdot\sum_{0}^{n-1} \lim_{n\rightarrow\infty}\left( \frac{2}{n} + i^2\right)\\
&=3 -\frac{1}{n^2}\sum_{0}^{n-1}i+3\cdot\frac{1}{n^3}\cdot\sum_{0}^{n-1}i^2\\
&=\lim_{n\rightarrow\infty} \left[ \ \ \ \ 3 -\frac{1}{n^2}\cdot\frac{n(n-1)}{2}+3\cdot\frac{1}{n^3}\cdot\frac{n(n-1)(2n-3)}{6} \ \ \ \ \right]\\
&=3 -\frac{1}{2}+1\\
&=3.5
\end{align}
what do you think? about my solution?
|
There are plenty of errors.
*
*Take $x_i=2+\frac3n$, not $x_i=2+\frac1n$. Hence $\Delta x_i=\frac3n$, not $\frac1n$.
*Always keep "lim" in front of your "Darboux" (Riemann) sums because as such, they are not "equal" to your integrals.
*Don't put anything before $\Delta x_i$ inside your first sum (the one which corresponds to $\int_2^51dx$).
*The square inside your third sum (the one which corresponds to $\int_2^53x^2dx$) does not give what you write in the following line.
*Etc.
Here is a correct (and more compact) version:
$$\begin{align}\int_2^5(1-x+3x^2)dx&=\lim_{n\to\infty}\frac3n\sum_0^{n-1}\left(1-(2+3i/n)+3(2+3i/n)^2\right)\\
&=\lim_{n\to\infty}\frac3n\sum_0^{n-1}\left(11+\frac{33}ni+\frac{27}{n^2}i^2\right)\\
&=\lim_{n\to\infty}\frac3n\left(11n+\frac{33(n-1)}2+\frac{9(n-1)(2n-1)}{2n}\right)\\
&=3\left(11+\frac{33}2+9\right)\\
&=\frac{219}2,
\end{align}$$
as you can check by direct integration:
$$\int_2^5(1-x+3x^2)dx=\left[x-\frac{x^2}2+x^3\right]_2^5=3-\frac{21}2+117=\frac{219}2.$$
|
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|
Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.
|
$$
\begin{aligned}
\tan^2 \left(\frac{3 \pi}{4}+\alpha\right) &=\tan ^2\left(\pi-\frac{\pi}{4}+\alpha\right) \\
&=\tan ^2\left(\frac{\pi}{4}-\alpha\right) \\
&=\left[\frac{\sin \left(\frac{\pi}{4}-\alpha\right)}{\cos \left(\frac{\pi}{4}-\alpha\right)}\right]^2 \\
&=\left(\frac{\sin \alpha-\cos \alpha}{\cos \alpha+\sin \alpha}\right)^2 \\
&=\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}
\end{aligned}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
finding integral of $ \int \frac{1}{\sin x + \sqrt{3} \cos x}\ dx $ In $ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx $, If I multiply and divide by $1/2$ I get $$ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx ,$$
then I can write $ 1/2=\sin (\frac{\pi}{6}) $ and $ \sqrt{3}/2=\cos (\frac{\pi}{6}) $ and apply $\cos(a-b)=\cos a \cos b + \sin a \sin b$,
I'll get $ \int\frac{1}{\cos(\frac{\pi}{6} - x)}\ dx $, I get $ \int \sec(\frac{\pi}{6} - x)\ dx $, now we know $ \int \sec(x)\ dx = \log|\tan(\frac{\pi}{4}+\frac{x}{2})| + c $ ,
so I finally get $$ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx = \frac{1}{2}\log|\tan(\frac{\pi}{4}+\frac{\pi}{6}-\frac{x}{2})| + c $$
However if at this step $ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx $
I take $ 1/2=\cos (\frac{\pi}{3}) $ and $ \sqrt{3}/2=\sin (\frac{\pi}{3})$, then what I get is $ \int\frac{1}{\sin(\frac{\pi}{3} + x)}\ dx $, which is equal to $ \int \csc(\frac{\pi}{3} + x)\ dx $, now we know $ \int \csc(x)\ dx = \log|\tan(\frac{x}{2})| + c $
so I finally get $$ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx = \frac{1}{2}\log|\tan(\frac{\pi}{6}+\frac{x}{2})| + c $$
why answer is different, I have rechecked and could not find any mistake
|
You should not get different answers (up to a constant of integration.) In fact, this is simply a matter of a calculation error.
$$\int \sec x\, \mathrm dx = \log\left|\tan\left(\frac\pi 4 + \frac x2 \right) \right| + C$$
gives
$$\frac12 \int \sec \left(x - \frac\pi 6 \right)\mathrm dx = \frac12 \log\left|\tan\left(\frac\pi 6 + \frac x2 \right) \right| + C$$
as $\frac\pi 4 + \frac x 2 - \frac {\pi}{12} = \frac x 2 + \frac \pi 6$. On the other hand,
$$\int \operatorname{cosec} x\, \mathrm dx = \log\left|\tan\left(\frac x2 \right) \right| + C$$
gives
$$\frac12 \int \operatorname{cosec}\left(x + \frac\pi 3 \right) \, \mathrm dx = \frac12 \log\left|\tan\left(\frac\pi 6 + \frac x2 \right) \right| + C$$
as $\frac12 \left( x + \frac\pi 3\right) = \frac x2 + \frac \pi 6$.
|
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|
Derivation of the behavior of solutions to $\tan x = x$ This question is related to Chapter IV, Note IV.36 of Flajolet & Sedgewick's Analytic Combinatorics, and The question: Sum of the squares of the reciprocals of the fixed points of the tangent function as well:
Let $x_k$ be the $k^{th}$ positive root of the equation $\tan z = z$. Then, the sum $S_r = \sum_k x_k^{-2r}$ are rational numbers for $r \ge 1$. For instances, $S_1 = 1/10$, $S_2 = 1/350$, $S_3 = 1/7875$ (from Note IV.36, pp. 269.)
I follow the approach in the book for Bernoulli numbers (Chapter IV, IV.6.1, pp. 268.) Consider the function:
\begin{align}
f(z) = \frac{1}{\tan z - z} \tag{1}
\end{align}
It is obvious that $\{x_k\}$ are poles of $f(z)$, with $k \in \mathbb{Z} \backslash \{0\}$. Further,
\begin{align}
Res[f(z);z=x_k] &= \frac{1}{\frac{d}{dz}(\tan z - z)|_{z=x_k}} \\
&= \frac{1}{\tan^2 x_k} \\
&= \frac{1}{x_k^2} \tag{2}
\end{align}
Hence,
\begin{align}
\frac{1}{\tan z - z} \sim \frac{1}{x_k^2}\cdot \frac{1}{z-x_k}, \text{ for } z \to x_k \tag{3}
\end{align}
By the Cauchy's coefficient formula, we have
\begin{align}
f_n = [z^n]f(z) &= \int_C \frac{f(z)}{z^{n+1}} dz \tag{4} \\
&= -\sum_{k \in \mathbb{Z} \backslash \{0\}} \frac{1}{x_k^2}\cdot \frac{1}{x_k^{n+1}}
\end{align}
where $C$ is a contour encircling all poles. Since $f(z)$ is an odd function, the poles are in pairs like $\pm x_k$. Then, $f_n = 0$, when $n$ is even. As a result,
\begin{align}
f_{2n-1} &= -2 \sum_{k=1}^\infty x_k^{-2(n+1)}, \text{ for } n \ge 1 \tag{5}
\end{align}
Therefore, the following relation is established:
\begin{align}
S_r = \sum_{k=1}^\infty x_k^{-2r} = -\frac{1}{2}f_{2r-3}, \text{ for } r > 1 \tag{6}
\end{align}
From the expansion of (1), that is,
\begin{align}
\frac{1}{\tan z - z} = \frac{3}{z^3} - \frac{6}{5z} - \frac{1}{175}z - \frac{2}{7875}z^3-\frac{37}{3031875}z^5+O(z^6) \tag{7}
\end{align}
It shows that $S_2$ and $S_3$ calculated from (6) are correct. But $S_1$ cannot be calculated from (6) since $f_{-1}$ is related to the pole at $z=0$, which is excluded from (6).
Then, for $S_1$, I follow the approach in an answer to The question to consider the integral in the region including $z=0$. That is,
\begin{align}
\int_C f(z)dz = 2i\pi \sum_{k \in \mathbb{Z}} Res[f(z); z=x_k] \tag{8}
\end{align}
From the expansion in (7), we have $Res[f(z);z=0]=-\frac{6}{5}$. Together with (2), (8) becomes
\begin{align}
\int_C f(z)dz = 2i\pi \left(-\frac{6}{5} + 2\sum_{k=1}^\infty x_k^{-2}\right) \tag{9}
\end{align}
Since it is found that (details in an answer to The question)
\begin{align}
\int_C f(z)dz = -2i\pi \tag{10}
\end{align}
(9) becomes
\begin{align}
-\frac{6}{5} + 2\sum_{k=1}^\infty x_k^{-2} = -1
\end{align}
Or
\begin{align}
S_1 = \sum_{k=1}^\infty x_k^{-2} = \frac{1}{10} \tag{11}
\end{align}
This result is well known as proved in The question.
It looks the flow of logic leading to the results (6) and (11) are both correct. However, the point I don't understand is: both (4) and (9) include an integral around the same contour and both give the same result ($-2i\pi$). Nevertheless, by Cauchy's residue theorem, they equal to the sum of different sets of residues - one with the pole at $0$ (9) while the other without (4). The difference in residue set is due to the difference in the regions in concern ($\mathbb{C}$ vs. $\mathbb{C} \backslash \{0\}$.) But why the contour integrals of a function at these different regions give the same result?
Could you please indicate what is wrong in the above logic? Thank you.
|
You don't say precisely what contour $C$ you are using. For definiteness, I'm going to assume it's the square $C_m$ with side length $2m\pi$ ($m$ a positive integer), center 0, and sides parallel to the axes given in robjohn's answer to your linked question. Then by the residue theorem, for all nonnegative integers $s$,
$$\int_{C_m} \frac{f(z)}{z^s} \, dz =2\pi i\left(f_{s-1}+\sum_{\hbox{$j\ne 0$, $x_j$ inside $C_m$}} \frac{1}{x_j^{s+2}}\right),$$
where $x_{-1}$, $x_{-2}$, $\dots$ are the negative real roots of $\tan x=x$, and $f_k=[z^k]f(z)$.
So, since the contour is the same, regardless of $s$, the pole at 0 is always included; it corresponds to $f_{s-1}$. The difference between the $s=0$ and $s>0$ cases is in the evaluation of the integral. As robjohn showed in his answer, for large $m$, $f(z)$ is approximately $-z^{-1}$ on $C_m$, so $f(z)/z^s$ is approximately $-z^{-(s+1)}$ and
$$\lim_{m\to\infty} \int_{C_m} \frac{f(z)}{z^s} \, dz =\left\{\begin{array}{cl}
-2\pi i, & \hbox{if $s=0$,}\\
0, & \hbox{if $s>0$,}\end{array}\right.$$
so, letting $m\to\infty$,
$$\sum_{j\ne 0} \frac{1}{x_j^{s+2}} = -f_{s-1}-\left\{\begin{array}{cl}
1, & \hbox{if $s=0$,}\\
0, & \hbox{if $s>0$.}\end{array}\right.
$$
|
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|
Evaluating Surface Integral with Divergence Theorem
Evaluate $\displaystyle\int_S \mathbf{F\cdot n}\ dS$ over the entire surface of the region above the $xy$ plane bounded by the cone $z^2=x^2+y^2$ and the plane $z=4$ if $\mathbf F=x\hat i+y\hat j+z^2\hat k$.
Solution: By the divergence theorem, since $\nabla \cdot\mathbf F=2+2z$, we have that
$$\begin{align}
\int_S \mathbf{F\cdot n}\ dS&=\int_A \int_{z=\sqrt{x^2+y^2}=r} ^4 (2+2z)\ dz\ dx\ dy\\
&=\iint_A [2z+z^2]_r ^4 \ dx\ dy\\
&=\int_\theta^{2\pi} \int_{r=0} ^{4} (24-2r-r^2)r\ dr\ d\theta\\
&=\int_{\theta=0}^{2\pi}\left[24\frac{r^2}{2}-2\frac{r^2}{3}-\frac{r^4}{4}\right]_0^4\\
&=\int^{2\pi}_0 \frac {256}{3}\ d\theta\\
&=\frac{512}{3} \pi
\end{align}$$
The given answer is $\dfrac{ 128\pi}{3}$. Where did I mistake, exactly?
|
Your computation is correct. Indeed, the direct computation gives
$$
\begin{align}\int_S \mathbf{F\cdot n}\ dS&=\iint_{x^2+y^2\leq 4^2}(x,y,4^2)\cdot(0,0,1) dx dy
\\&\;+\iint_{x^2+y^2\leq 4^2}(x,y,x^2+y^2)\cdot(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},-1) dx dy
\\&=16\cdot 16\pi+\int_{\theta=0}^{2\pi}\int_{r=0}^4(r-r^2)r dr d\theta\\
&=256\pi +2\pi\left[\frac{r^3}{3}-\frac{r^4}{4}\right]_0^4
=256\pi-\frac{256\pi}{3}=\frac {512\pi}{3}.
\end{align}$$
P.S. In the book's answer (see the comment below), the volume of the cone is $\frac{\pi}{3}\cdot 4^2\cdot 4$ which is equal to $\frac{64\pi}{3}$, it is not $\frac{16\pi}{3}$. Therefore a factor $4$ is missing, and $4\cdot \frac{ 128\pi}{3}$ gives you the correct answer $\frac{512\pi}{3} $.
|
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|
A question on Brahmagupta Identity The Brahmagupta Identity states that
$$
\begin{align}
N & = (a^2+nb^2 )(c^2+nd^2 ) \\
& = (ac-nbd)^2+n(ad+bc)^2 \\
& = (ac+nbd)^2+n(ad-bc)^2 \\
\end{align}
$$
Knowing only $N$, is there a way to find $n$ such that the two cofactors of $N$ (not necessarily prime) can each be represented in the form $x^2 + ny^2$.
Any $n$ that is a solution must satisfy the condition
$$
\begin{align}
N & \equiv (ac + nbd)^2 & \mod {n} \\
& \equiv k^2 & \mod {n}
\end{align}
$$
i.e., $N$ is a quadratic residue modulo $n$.
Also,
$$
\begin{align}
N & \equiv (ac + bd)^2 + (ad - bc)^2 & \mod {(n-1)} \\
& \equiv u^2 + v^2 & \mod {(n-1)}
\end{align}
$$
i.e., $N$ must be the sum of two quadratic residues modulo $n-1$.
Assuming, the factorization of $N$ is unknown,
Q1: Are there any other conditions that can help with finding such candidate $n$ efficiently (and eliminating ones that are not fit)?
Q2: Does an $n$ always exist for all $N$ or are there any obstructions?
|
$N=35$ is represented as $x^2 + n y^2 $ for
$$ n = 10, 19, 26, 31, 34, 35 \; . $$
In particular, it is not represented by $x^2 + y^2, x^2 + 2 y^2, x^2 + 3 y^2, x^2 + 4 y^2, x^2 + 5 y^2$
For any of the $n$ above, $5,7$ are not represented by $x^2 + n y^2 $
Same idea for $$N = 55583= 11 \cdot 31 \cdot 163 $$
which is represented as $x^2 + n y^2 $ for
$$ n = 358, \; 827, \; 1294, \; 1759, \; 2222, \; 2683, \ldots $$
but for no smaller $n.$
|
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|
Prove $\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1$ for $x,y,z > 0 $
Let $x, y, z > 0$. Prove that
$$\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1.$$
I encountered this problem today and thought that it was straightforward, and to be honest it still might be I just can't figure it out for some reason.
I approached this problem using the famous Cauchy-Schwarz inequality. I applied the inequality to every denominator. For example for the first fraction:
$$(1+xy+xz)(1 + \frac{1}{x}+ \frac{1}{x}) \ge (1+y+z)^2$$
After doing this for every fraction I switched the denominator for every one of them with the left side of the Cauchy-Schwarz inequality. Reducing each fraction we are left with
$$\frac{1}{1+2/x} + \frac{1}{1+2/y} + \frac{1}{1+2/z} \ge 1 $$
I am not sure how to continue from here. I tried to use Titu's lemma but it didn't work. I will be very happy if someone can help me out here.
Edit: My Cauchy was wrong.Thanks to all who contributed :)
|
Hint :
Can you show ? For $a,b,c>0$
$$\frac{1+ba+bc}{(1+a+c)^{2}}-\frac{b^{2}+ba+bc}{(b+a+c)^{2}}=\frac{\left(b-1\right)^{2}\left(a+c\right)}{\left(a+c+1\right)^{2}\left(a+b+c\right)}\geq 0$$
|
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|
How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations and substituting into the first. None of these methods worked. How do I do this?
|
From the first two equations we get $b,d$ in terms of $c$:
$$b=\frac12\left(c^2-6-\frac{1}{c}\right)$$
$$d=\frac12\left(c^2-6+\frac{1}{c}\right)$$
Now substitute into the third:
$$\frac14\left((c^2-6)^2-\frac{1}{c^2}\right)=6$$
Let $a=c^2$:
$$(a-6)^2-\frac{1}{a}=24$$
$$a^3-12a^2+12a-1=(a-1)(a^2-11a+1)=0$$
Since $c$ is integral, so is $a$ and hence $a=1$ and $c=\pm1$; setting $c=1$ leads to $b=-3$ and $d=-2$ while $c=-1$ gives $b=-2$ and $d=-3$. Thus the integer solutions are
$$(b,c,d)=(-3,1,-2)\lor(b,c,d)=(-2,-1,-3)$$
|
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|
Evaluating $\lim\limits_{x \to 0} \frac{\cos\left(\sin^2 x\right)-1}{\sin x}$ without l'Hôpital I'm stuck trying to evaluate this limit without l'Hôpital's rule:
$$
\lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x}
$$
Could anyone give me a hint on what I need to do? I tried many trig identities but I can't seem to find one that doesn't give me an indeterminate form.
|
My attempt, using asymptotic estimates. As $u \to 0$,
$$
\cos u = 1 -\frac12 u^2 + O(u^4) .
$$
As $x \to 0$, also $\sin x \to 0$, so
\begin{align}
\sin x &= x + O(x^3)
\\
\frac{1}{\sin x} &= \frac{1}{x}+O(x)
\\
\sin^2 x &= x^2 + O(x^4)
\\
\cos(\sin^2 x) &= 1 -\frac12 \sin^2 x + O(\sin^4 x)
= 1 -\frac12 (x^2 + O(x^4)) + O(x^4) = 1 -\frac12 x^2 + O(x^4)
\\
\cos(\sin^2 x) - 1 &= -\frac12 x^2 + O(x^4)
\\
\frac{\cos(\sin^2 x) - 1}{\sin x} &=
\big(\frac{1}{x}+O(x)\big)\;\big(-\frac12 x^2 + O(x^4)\big)
= -\frac12 x + O(x^3)
\end{align}
so
$$
\lim_{x\to 0} \frac{\cos(\sin^2 x) - 1}{\sin x} = 0 .
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate $\mathbb{E}(X^2)$ when $X \sim Geo(\frac{1}{2})$ I am reading Mark Joshi's "Quant Job Interview Questions And Answers" question 3.7, and I am having trouble understanding how he proved that $\mathbb{E}(X^2)$ of $X \sim Geo(\frac{1}{2})$ is 6.
Here is his original question:
QUESTION 3.6. Suppose we toss a fair coin, and let N denote the number of tosses until we get a head (including the final toss). What is $\mathbb{E}(N)$ and $Var(N)$?
Here is his proof:
In particular, he says "applying the same technique (twice) gives $\mathbb{E}(X^2) = \frac{2 - 0.5}{0.5^2} = 6$". How did that work? How does applying the same technique twice gives this answer?
|
Applying the trick once gives:
\begin{align*}
E(X^2) &= 0.5^1+2^2\times0.5^2 + 3^2\times0.5^3+\dots\\
\implies 0.5E(X^2) &= 0.5^2+2^2\times0.5^3 + 3^2\times0.5^4+\dots\\
\implies 0.5E(X^2) &= 0.5^1 +(2^2-1)\times0.5^2 + (3^2-2^2)\times0.5^3+\dots
\end{align*}
But $k^2-(k-1)^2$ simplifies to $2k-1$, so we can rewrite this as
$$0.5E(X^2) = 0.5^1 +(2\times2 - 1)\times0.5^2 + (2\times3 - 1)\times0.5^3+\dots$$
Doing the trick again:
\begin{align*}
0.5E(X^2) &= 0.5^1 +(2\times2 - 1)\times0.5^2 + (2\times3 - 1)\times0.5^3+\dots\\
\implies 0.25E(X^2) &= 0.5^2 +(2\times2 - 1)\times0.5^3 + (2\times3 - 1)\times0.5^4+\dots\\
\implies 0.25E(X^2) &= 0.5 +2\times0.5^2 + 2\times0.5^3+ 2\times 0.5^4\dots\\
\end{align*}
i.e.
\begin{align*}
0.25E(X^2) + 0.5 &= 2\times 0.5 + 2\times0.5^2 + 2\times0.5^3+ 2\times 0.5^4\dots\\
&= 2.
\end{align*}
|
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|
How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$? How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$?
I'm trying to solve $(2x - 4y + 6)dx + (x+y-3)dy = 0$. The two lines intersect at $(1,2)$. After substituting to $(\tau = x-1, \mu = y-2)$ and then to $(\tau, u = \mu/\tau)$ I get $-\dfrac{d\tau}{\tau} = \dfrac{(u+1)du}{u^2 - 3u +2}$ and then I integrate. On the left I have $- \ln |x-1| + C$. On the right, as the denominator is the same as $(u-3/2)^2 - 1/4$ I substitute with $z = (u-3/2)$ and finally get $-2\ln |z + 1/2| + 3 \ln |z - 1/2| + C \equiv -2 \ln |\frac{y}{x} - 1| + 3\ln |\frac{y}{x} -2| + C $. After raising to powerof $e$ I get $|x-1| = \dfrac{(\frac{y}{x} - 1)^2}{|\frac{y}{x} - 2|^3}*e^C \equiv \dfrac {|x|(y-x)^2}{|y-2x|^3}*e^C$ but I'm not able to obtain the form in the answer key.
|
Don't complete the square; instead from
$$-\frac1\tau\,d\tau=\frac{u+1}{u^2-3u +2}\,du$$
decompose $\frac{u+1}{u^2-3u+2}$ through partial fractions, integrate and temporarily leave the equation in terms of $\tau,u$ to get
$$K-\ln\tau=3\ln(u-2)-2\ln(u-1)$$
Then simplify:
$$K=\ln\frac{\tau(\mu/\tau-2)(\mu/\tau-2)^2}{(\mu/\tau-1)^2}=\ln\frac{(\mu-2\tau)^3}{(\mu-\tau)^2}$$
$$C=\frac{(\mu-2\tau)^3}{(\mu-\tau)^2}=\frac{(y-2x)^3}{(y-x-1)^2}$$
$$(y-2x)^3=C(y-x-1)^2$$
|
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|
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Firstly, $\sin x - 1=0 \implies x= \frac{\pi}{2}$
Next, $(\sqrt{2} \cos x +1)=0 \implies x= \frac{3\pi}{4}$
Why is this solution wrong? Is there more than one $x$ value for $\sin x - 1=0$ or $(\sqrt{2} \cos x +1)=0$ in the given interval? How do I know that?
|
Can you find out where did you miss? Look at the blue graph. It's the graph of $y=\sqrt{2} \cos x + 1$. So, the root of the blue graph should be $x=\pi \pm \dfrac{\pi}{4} \Rightarrow x = \dfrac{3\pi}{4}, \dfrac{5\pi}{4}$.
|
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|
You roll 5 dice. Find the probability of 2 pairs. I have seen this answered on here before. I have a slightly different form of answer, also my answer differs from the solution given.
My answer is
$${6 \choose 3} {5 \choose 2,2,1}6^{-5}.$$
${6 \choose 3}$ comes from choosing 3 faces out of 6 i.e $(2,3,1)$. We then arrange 5 symbols two pairs of which are identical. i.e $(2,2,3,3,1), (1,1,3,2,2), (3,3,2,1,1)$. This gives us the term ${5 \choose 2,2,1}$.
The answer provided in the book gives me
$$4{6 \choose 2} {5 \choose 2,2,1}6^{-5}.$$
Since $ {5 \choose 2,2,1} = {5 \choose 4}{4 \choose 2}$ this can be written
$$4{6 \choose 2} {5 \choose 4}{4 \choose 2}6^{-5}.$$
I just want to understand where I am going wrong. Also, I would like to know how to get
$$4{6 \choose 2} {5 \choose 4}{4 \choose 2}6^{-5}.$$
from first principles.
EDIT
The following
$${6 \choose 3} {5 \choose 2,2,1}6^{-5},$$
is incorrect. Because ${6 \choose 3} $ would "include" only $\{11223\}$ instead of both $\{22331\}$ and $\{11223\}$
|
Expanding on @user51547's comment, after choosing the 3 faces and grouping of the 5 dice, you also need to decide the assignment of the faces to the dice. In this case, we just need to decide which face is the unique one (not paired), and the remaining two must be the pair faces. So there is an extra $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ choice, meaning your answer becomes
$$\begin{pmatrix} 6 \\ 3 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2,2,1 \end{pmatrix} 6^{-5}$$
If you expand the first two factors and rearrange, you get the form that book provided you.
For your second question on getting from first principles, I'll first make a slight modification
$$4 \begin{pmatrix} 6 \\ 2 \end{pmatrix} \begin{pmatrix} 5 \\ 4 \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} 6^{-5} = \begin{pmatrix} 6 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 4 \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} 6^{-5}$$
Now it can be explained as
*
*Among the 6 possible faces, choose the 2 pairing faces
*Among the remaining 4, choose the 1 unique face
*Among the 5 dice, choose the 4 to have the pairing faces
*Among the 4 pairing dice, choose which 2 have the first pairing face
Note that here I am assigning which faces are pairing/non-pairing while I am choosing them.
|
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|
How to prove $(1 + x)^{n} \leq 1 + 2nx$ I am currently working on a math problem, and it boils down to proving $(1 + x)^{n} \leq 1 + 2nx$, for a small $x$
By the Binomial expansion, it is clear that $$(1 + x)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$
However, how can we prove that $$nx \geq \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$
which would prove my claim.
Any ideas?
|
One can continue from the binomial expansion, for $x>0$ this gives estimates against a geometric series
\begin{align}
(1+x)^n&= 1+nx\left(1+\frac{(n-1)x}{2}+\frac{(n-1)(n-2)x^2}{2·3}+...\right)
\\
&\le 1+nx\left(1+\frac{(n-1)x}{2}+\frac{(n-1)^2x^2}{2^2}+...\right)
\\
&\le 1+\frac{nx}{1-\frac{(n-1)x}{2}}
\end{align}
For the convergence of the series one needs $\frac{(n-1)x}{2}<1$, then for the claim $\frac{(n-1)x}{2}\le\frac12$, thus for $n\ge 2$ and $0\le x\le\frac1{n-1}$ the claimed inequality is true, for $n=0,1$ it is true for all $x>0$.
For $x<0$ the claim is wrong, as by the Bernoulli inequality
$$
(1-|x|)^n\ge 1-n|x|>1-2n|x|=1+2nx.
$$
|
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|
Derive that $r=a\cos θ$ from $(x-h)^2 + (y-k)^2 = r^2$ I am a beginner here so please kindly tell me how I can do better when asking my questions so I can improve next time and if there are any complaints to this question.
I want to ask, how do I obtain the formula $r=a\cos θ$ from the circle formula $(x-h)^2 + (y-k)^2 = r^2$.
More preferably how do I obtain it algebraically?
This problem I have comes from I believe Polar Curves/Equations; where the circle is tangent to the $\frac{1}{2}π$ axis with diameter $a$.
Please explain to me how you would have taught it to yourself if you were new to this kind of problem.
|
$$x=r \cos\theta\space; y=r \sin\theta\space; r=\frac{a}{2}\space;
C(\frac{a}{2}, 0)$$
$$ (x-h)^2 + (y-k)^2 = r^2$$
$$ (r \cos\theta-\frac{a}{2})^2 + (r \sin\theta-0)^2=(\frac{a}{2})^2$$
Using the identity $ (a-b)^2 = a^2-2ab+b^2$
$$
r^2\cos^2\theta - a r \cos\theta + \frac{a^2}{4} + r^2\sin^2\theta
=\frac{a^2}{4}
$$
Simplifying and rearranging the terms we get...
$$r^2\cos^2\theta + r^2\sin^2\theta - a r \cos\theta = 0$$
$$r^2(\cos^2\theta+\sin^2\theta) - a r \cos\theta = 0$$
$ \cos^2\theta + \sin^2\theta = 1$
$$r^2 - a r \cos\theta = 0$$
$$ r^2 = a r \cos\theta$$
$$ r = a \cos\theta$$
Tried to explain it the way I understood from all the responses, thanks guys!
|
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|
Solve $y'-\frac{y}{x}=\frac{(x+y)^2}{x^2}$ $$
\begin{aligned}
y'-\frac{y}{x} &=\frac{(x+y)^2}{x^2}\\\\
x^2y'-xy &=x^2+2xy+y^2\\
\end{aligned}$$
I have no idea what is the next step.
Updated:
Let $y=vx$, which $y′=v+v′x$
$$
\begin{aligned}
y'-\frac{y}{x} &=\frac{(x+y)^2}{x^2}\\
xv'&=(v+1)^2\\
\frac{v'}{(v+1)^2}&=\frac{1}{x}\\
\int\frac{1}{(v+1)^2}\,dv &=\int\frac{1}{x}\, dx\\
-\frac{1}{v+1}&=\ln x+C\\
v&=-\frac{\ln x+1+C}{\ln x+C}\\
y&=-\frac{x(\ln x+1+C)}{\ln x+C}
\end{aligned}$$
|
$$ax^2+bx+c=0\implies x=(-b\pm \sqrt{b^2-4ac})/(2a)$$
substitute: $ y^{(1)}=\frac{-(3/x)+/- \sqrt{9/x^2-4(1/x^2)(1)}}{2(1/x^2)}$
solution: $y=\int \frac{-(3/x)+/- \sqrt{9/x^2-4(1/x^2)(1)}}{2(1/x^2)} dx$
|
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|
Finding the derivative of $y=\sin(\cos^{-1}(2x))$ Question
Find the derivative of $y=\sin(\cos^{-1}(2x))$.
My Working
So I let $u=\cos^{-1}(2x)$, so
\begin{align}
\frac{du}{dx}&=-\frac{1}{\sqrt{1-(2x)^2}}\times2\\
&=-\frac{2}{\sqrt{1-4x^2}}
\end{align}
Also,
$$\frac{dy}{du}=\cos u$$
I feel like I should use the chain rule somewhere, but I am not sure where. Could anyone please help? Thanks!
|
This differentiation can also be accomplished in a couple of other ways by constructing the right triangle implied by the argument of the sine function. If we call $ \ \theta \ = \ \cos^{-1}(2x) \ \ , \ $ then $ \ \cos \theta \ = \ 2x \ = \ \frac{2x}{1} \ \ . \ $ With the hypotenuse of the right triangle taken as $ \ 1 \ \ , \ $ the "leg" opposite to $ \ \theta \ $ has length $ \ \sqrt{1 - (2x)^2} \ = \ \sqrt{1 - 4x^2} \ \ . $
We can now write our function as $ \ y \ = \ \sin \theta \ = \ \sin(\cos^{-1}(2x)) \ = \ \frac{\sqrt{1 - 4x^2}}{1} \ = \ (1 - 4x^2)^{1/2} \ \ . \ $ The derivative is then $$ y' \ \ = \ \ \frac{1}{2} · (1 - 4x^2)^{-1/2} · (1 - 4x^2)' \ \ = \ \ \frac{1}{2} · (- \ 8x) · (1 - 4x^2)^{-1/2} \ \ = \ \ \frac{- \ 4x}{\sqrt{1 - 4x^2}} \ \ . $$
If we instead begin by re-arranging the expression as $ \ \sin^{-1} y \ = \ \cos^{-1}(2x) \ \ $ (an equation of two angles), implicit differentiation of the left side produces
$$ \frac{1}{\sqrt{1 - y^2}} \ · \ y' \ \ = \ \ -\frac{1}{\sqrt{1 - (2x)^2}} · (2x)' \ \ \Rightarrow \ \ y' \ \ = \ \ -\frac{2 · \sqrt{1 - y^2}}{\sqrt{1 - 4x^2}} \ \ . $$
We then "clear" $ \ y \ $ from the expression for the derivative by applying our earlier result from the constructed right triangle:
$$ y' \ \ = \ \ -\frac{2 · \sqrt{1 - [ \ (1 - 4x^2)^{1/2} \ ]^2}}{\sqrt{1 - 4x^2}} \ \ = \ \ -\frac{2 · \sqrt{1 - (1 - 4x^2)}}{\sqrt{1 - 4x^2}} $$ $$ = \ \ -\frac{2 · \sqrt{ 4x^2 }}{\sqrt{1 - 4x^2}} \ \ = \ \ -\frac{4x}{\sqrt{1 - 4x^2}} \ \ . $$
|
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|
Evaluate $\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!}$ I want to evaluate $\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!}$ but while doing my research, I noticed that
\begin{align*}
&\sin(z) = \sum^{\infty}_{n=1} (-1)^n \frac{z^{2n+1}}{(2n+1)!} \\
&\sinh(z) = \sum^{\infty}_{n=1} \frac{z^{2n+1}}{(2n+1)!}
\end{align*}
So
\begin{align*}
\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!} = \frac{\sin(z)+\sinh(z)}{2}
\end{align*}
Is it that easy?
|
Consider the following.
\begin{align}
\frac{1}{\sqrt{2}} \, \cosh\left(\frac{x}{\sqrt{2}}\right) \, \sin\left(\frac{x}{\sqrt{2}}\right) &= \sum_{r} \frac{x^{2 r}}{2^{r} \, (2 r)!} \times \sum_{s} \frac{(-1)^s \, x^{2 s + 1}}{2^{s+1} \, (2 s + 1)!} \\
&= \sum_{n} \sum_{s=0}^{n} \frac{(-1)^s \, x^{2 n +1}}{2^{n+1} \, (2n - 2s)! \, (2 s + 1)!} \\
&= \sum_{n} \left( \sum_{s=0}^{n} (-1)^s \, \binom{2 n +1}{2 s+1} \right) \, \frac{x^{2 n +1}}{2^{n+1} \, (2 n+1)!} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} (-1)^{\lfloor{n/2\rfloor}} \, \frac{x^{2 n+1}}{(2 n+1)!}.
\end{align}
In a similar manor it can be found that
$$ \frac{1}{\sqrt{2}} \, \cos\left(\frac{x}{\sqrt{2}}\right) \, \sinh\left(\frac{x}{\sqrt{2}}\right) = \frac{1}{2} \, \sum_{n=0}^{\infty} (-1)^{\lfloor{n/2\rfloor}} \, \frac{(-1)^n \, x^{2 n+1}}{(2 n+1)!}. $$
Now
\begin{align}
S &= \frac{1}{\sqrt{2}} \, \cosh\left(\frac{x}{\sqrt{2}}\right) \, \sin\left(\frac{x}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \, \cos\left(\frac{x}{\sqrt{2}}\right) \, \sinh\left(\frac{x}{\sqrt{2}}\right) \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} (-1)^{\lfloor{n/2\rfloor}} \, \frac{(1 + (-1)^n) \, x^{2 n+1}}{(2 n+1)!} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n \, x^{4 n+1}}{(4 n+1)!}.
\end{align}
A different series can be obtained from:
\begin{align}
\frac{1}{2} \left( \sin(x) + \sinh(x)\right) &= \sum_{n=0}^{\infty} \frac{1 + (-1)^n}{2} \, \frac{x^{2n+1}}{(2n+1)!} \\
&= \sum_{n=0}^{\infty} \frac{x^{4 n+1}}{(4 n +1)!}
\end{align}
|
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|
Probability of drawing two colors after four draws You have a bin with 5 balls of different colors: Red, Orange, Yellow, Green, and Blue, all of which are equally likely to be drawn. After a ball is drawn, it is put back afterwards. What is the probability of drawing both the yellow and red balls at least once after drawing 4 times from the bin?
I attempted this problem by finding the probability that a yellow and red ball is not drawn as 3/5 for a single draw. Then, I did $1-(3/5)^4$ to find the probability that a yellow and red ball is drawn at least once in the 4 draws. Is this correct?
|
As Flip Tack pointed out in the comments, you have calculated the probability that at least one yellow ball or at least one red ball is selected. We wish to find the probability that ball of both colors are selected.
Method 1: We use the Inclusion-Exclusion Principle.
Let $R$ be the event that a red ball is chosen; let $Y$ be the event that a yellow ball is chosen. We wish to find
$$\Pr(R \cap Y) = 1 - \Pr(R' \cup Y')$$
By the Inclusion-Exclusion Principle,
$$\Pr(R' \cup Y') = \Pr(R') + \Pr(Y') - \Pr(R' \cap Y')$$
Hence,
$$\Pr(R \cap Y) = 1 - \Pr(R') - \Pr(Y') + \Pr(R' \cap Y')$$
$\Pr(R')$: Since four of the five balls are not red, the probability that a red ball is not selected in four draws is
$$\Pr(R') = \left(\frac{4}{5}\right)^4$$
$\Pr(Y')$: Since four of the five balls are not yellow, the probability that a yellow ball is not selected in four draws is
$$\Pr(Y') = \left(\frac{4}{5}\right)^4$$
$\Pr(R' \cap Y')$: Since three of the five balls are neither red nor yellow, the probability that neither a red nor yellow is selected in four draws is
$$\Pr(R' \cap Y') = \left(\frac{3}{5}\right)^4$$
Therefore, the probability that at least one red ball and at least one yellow ball are selected after drawing four times from the bin with replacement is
$$\Pr(R \cap Y) = 1 - \left(\frac{4}{5}\right)^4 - \left(\frac{4}{5}\right)^4 + \left(\frac{3}{5}\right)^4$$
Method 2: In the comments, Vadim Chernetsov asked how to perform a direct count. We count directly.
One red, one yellow, and two from the other three colors: Choose one of the four positions for the red ball and one of the remaining three positions for the yellow ball. There are then three choices for each of the remaining two positions. Hence, there are
$$\binom{4}{1}\binom{3}{1} \cdot 3^2$$
such cases.
Two red, one yellow, and one from the other three colors: Choose two of the four positions for the red balls and one of the remaining two positions for the yellow ball. There are three choices for the remaining position. Hence, there are
$$\binom{4}{2}\binom{2}{1} \cdot 3$$
such cases.
One red, two yellow, and one from the other three colors: By symmetry, there are
$$\binom{4}{2}\binom{2}{1} \cdot 3$$
such cases.
Two red and two yellow: Choose two of the four positions for the red balls. The other positions must be filled by yellow balls. Hence, there are
$$\binom{4}{2}$$
such cases.
Three red and one yellow: Choose three of the four positions for the red balls. The other position must be filled by a yellow ball. Hence, there are
$$\binom{4}{3}$$
such cases.
One red and three yellow: By symmetry, there are
$$\binom{4}{3}$$
such cases.
Since there are five choices for each of the four positions in the sequence, there are $5^4$ possible outcomes in the sample space.
Hence, the probability that at least one red ball and at least one yellow ball are selected in four draws with replacement from the bin with one ball each from the five colors red, orange, yellow, green, and blue is
$$\Pr(R \cap Y) = \frac{\dbinom{4}{1}\dbinom{3}{1}\cdot 3^2 + 2\dbinom{4}{2}\dbinom{2}{1} \cdot 3 + \dbinom{4}{2} + 2\dbinom{4}{3}}{5^4}$$
|
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|
if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$ Let $x \in \mathbb{R}^*$ and $y \geq \frac{1}{4}$.
Show that if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$
I tried the following idea:
$$
2\sqrt{y-\frac{1}{4}}\leq \sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y
$$
which leads to:
$$
y^2 - 4 y +1 \geq 0
$$
then $y \leq 2-\sqrt{3} $ or $ y \geq 2+\sqrt{3} $
|
Let $y=z^2+\frac14$ and
$$y=\sqrt{\left(x+\tfrac12\right)^2+z^2} + \sqrt{\left(\tfrac12-x\right)^2+z^2}$$
Now by triangle/Minkowski inequality,
$$\implies y \geqslant \sqrt{(1)^2+(2z)^2} = \sqrt{4z^2+1}=\sqrt{4y}$$
$\implies y(y-4)\geqslant 0 \implies y\geqslant 4$
Hence among the options provided, $y\leqslant1$ clearly fails. As $(x,y)=(0,4)$ satisfies the original equation, this is indeed the minimum possible.
|
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|
Evaluate $\int x^2\log(1-x^2)dx$, and hence prove that $\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+...=\frac23\log2-\frac89$
Evaluate $\int x^2\log(1-x^2)dx$, and hence prove that $\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+...=\frac23\log2-\frac89$
My Attempt:
Integrating by parts,
$$\log(1-x^2)\cdot\frac{x^3}3-\int\frac{-2x}{1-x^2}\cdot\frac{x^3}3dx\\=\frac{x^3}3\log(1-x^2)-\frac23\int\frac{1-x^4-1}{1-x^2}dx\\=\frac{x^3}3\log(1-x^2)-\frac23\int1+x^2-\frac1{1-x^2}dx\\=\frac{x^3}3(\log(1-x)+\log(1+x))-\frac23(x+\frac{x^3}3)+\frac13\log|\frac{1-x}{1+x}|+c\\=\frac{x^3-1}3\log(1-x)+\frac{x^3+1}3\log(1+x)-\frac23(x+\frac{x^3}3)+c$$
If we apply limits from $0$ to $1$, we get
$$\frac23\log2-\frac89,$$
which is the required RHS.
How to get the required LHS?
|
Using the Taylor series of $\log(1-t)$:
$$\log(1-t)=-\sum_{n=1}^\infty\frac{t^n}n,\qquad t\in[0,1),$$
we have
$$\int_0^1 x^2\log(1-x^2)dx=-\int_0^1x^2\sum_{n=1}^\infty\frac{x^{2n}}n\,dx=-\sum_{n=1}^\infty\frac1n\int_0^1x^{2+2n}\,dx=-\sum_{n=1}^\infty\frac1{n(3+2n)},$$
which is exactly the required LHS (except for the minus sign). Here, the reason why we can interchange the summation and integration is the Tonelli's theorem, or that we can always integration a power series term by term, see here.
So, we have proven that
$$-\sum_{n=1}^\infty\frac1{n(3+2n)}=\frac23\log2-\frac89,$$
which differs from the original problem by a minus sign. Indeed, the original problem is not correct. Clearly we have $\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+\cdots>0$, but $\frac23\log2-\frac89< \frac23-\frac89<0.$ Therefore, the right version is
$$\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+\cdots=\frac89-\frac23\log2.$$
Wlofram alpha agrees with our computation.
|
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|
Finding $\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$ Consider the sequences $a_n$ and $b_n$ such that $a_n=\sum_{k=1}^{n}\frac{1}{k^2}$ and $b_n=\sum_{k=1}^{n}\frac{1}{(2k-1)^2}$. Compute $$\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$$
When I saw this question, my first thought is that to use $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$ and $\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}$
So $a_n=\frac{\pi^2}{6}-\sum_{k=n+1}^{\infty}\frac{1}{k^2}$ and $b_n=\frac{\pi^2}{8}-\sum_{k=n+1}^{\infty}\frac{1}{(2k-1)^2}$
But after substituting $a_n$ and $b_n$ into the $\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$, I could not proceed.
Is my approach a correct way to start or any other method to solve?
|
Hint: Notice that
\begin{align}
(ab-a_nb_n)n = n(a-a_n)b+a_n(b-b_n)n.
\end{align}
Next, observe that
\begin{align}
n(a-a_n) = n\left(\frac{\pi^2}{6}-a_n\right) = n\sum^\infty_{k=n+1}\frac{1}{k^2}
\end{align}
and that
\begin{align}
\frac{n}{n+1}=n\int^\infty_{n+1}\frac{1}{x^2} dx\le n\sum^\infty_{k=n+1}\frac{1}{k^2}\le n\int^\infty_{n}\frac{1}{x^2} dx =1.
\end{align}
You can do the rest.
|
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|
Solving differential equation with step function without using Laplace Transforms. Suppose we have the differential equation:
$$ \ddot{y} + y = H(x - \pi) - H(x - 2\pi) $$
where $ H(x)$ is the Heaviside step function with initial conditions $ y(0) = \dot{y}(0) = 0 $ as initial conditions, and $ y(x) $ and $ \dot{y}(x) $ are continuous everywhere. I know how to solve this using Laplace Transforms, but I was curious if there are any alternative approaches for solving this and differential equations involving step functions in general.
|
Without the Laplace transform the solution is as follows.
Using
$$ H(t - \pi) - H(t - 2 \pi) = \begin{cases} 0 & t < \pi \\ 1 & \pi < t < 2 \pi \\ 0 & t > 2 \pi \end{cases} $$
then
$$\begin{cases} y_{1}^{''} + a^2 \, y_{1} \\ y_{2}^{''} + a^2 \, y_{2} \\ y_{3}^{''} + a^2 \, y_{3} \end{cases} = \begin{cases} 0 & t < \pi \\ 1 & \pi < t < 2 \pi \\ 0 & t > 2 \pi \end{cases}. $$
This gives the general solutions
$$ y(t) = \begin{cases} y_{1} = A_{0} \, \cos(a t) + A_{1} \, \sin(a t) & t < \pi \\ y_{2} = B_{0} \, \cos(a t) + B_{1} \, \sin(a t) + \frac{1}{a^2} & \pi < t < 2 \pi \\ y_{3} = C_{0} \, \cos(a t) + C_{1} \, \sin(a t) & t > 2 \pi \end{cases}. $$
The given boundary condition, $y(0) = y^{'}(0) = 0$ apply to $y_{1}$ and leads to
$$ y(t) = \begin{cases} y_{1} = 0 & t < \pi \\ y_{2} = B_{0} \, \cos(a t) + B_{1} \, \sin(a t) + \frac{1}{a^2} & \pi < t < 2 \pi \\ y_{3} = C_{0} \, \cos(a t) + C_{1} \, \sin(a t) & t > 2 \pi \end{cases}. $$
Now using $y_{1}(\pi) = y_{2}(\pi)$ and $y^{'}_{1}(\pi) = y_{2}^{'}(\pi)$ then $0 = B_{0} \, \cos(a \pi) + B_{1} \, \sin(a \pi) + \frac{1}{a^2}$ and $0 = -a \, (B_{0} \sin(a \pi) - B_{1} \, \cos(a \pi) )$ which gives
$$ y(t) = \begin{cases} y_{1} = 0 & t < \pi \\ y_{2} = \frac{2}{a^2} \, \cos^{2}\left(\frac{a t - a \pi}{2}\right) & \pi < t < 2 \pi \\ y_{3} = C_{0} \, \cos(a t) + C_{1} \, \sin(a t) & t > 2 \pi \end{cases}. $$
The remaining conditions $y_{2}(2 \pi) = y_{3}(2 \pi)$ and $y_{2}^{'}(2 \pi) = y_{3}^{'}(2 \pi)$ lead to the final solution of
$$ y(t) = \begin{cases} 0 & t < \pi \\ \frac{2}{a^2} \, \cos^{2}\left(\frac{a t - a \pi}{2}\right) & \pi < t < 2 \pi \\ \frac{1}{a^2} \, \left( \cos(a t - a \pi) + \cos(a t - 2 a \pi) \right) & t > 2 \pi \end{cases}. $$
This can also be seen in the form
$$ y(t) = \frac{2}{a^2} \, \cos^{2}\left(\frac{a t - a \pi}{2}\right) \, H(t - \pi) - \frac{2}{a^2} \, \sin^{2}\left(\frac{a t - 2 a \pi}{2}\right) \, H(t - 2 \pi). $$
By using the Laplace transform the solution is seen as follows.
Fist note that
$$ L\left\{ H(t - a) \right\} = \int_{0}^{\infty} e^{-s t} \, H(t -a) \, dt = \int_{a}^{\infty} e^{-s t} \, dt = \frac{e^{- a s}}{s}. $$
Now,
\begin{align}
L\{ y^{''} + a^2 \, y \} &= L\left\{ H(t - \pi) - H(t - 2 \pi) \right\} \\
(s^2 + a^2) \, \overline{y} - s \, y(0) - y^{'}(0) &= \frac{e^{-\pi s}}{s} - \frac{e^{-2 \pi s}}{s} \\
\overline{y} &= \frac{e^{- pi s}}{s \, (s^2 + a^2)} - \frac{e^{- pi s}}{s \, (s^2 + a^2)} + \frac{s}{s^2 + a^2} \, y(0) + \frac{y^{'}(0)}{s^2 + a^2}.
\end{align}
Since
$$ \frac{1}{s \, (s^2 + a^2)} = \frac{1}{a^2 \, s} - \frac{1}{a^2 \, (s^2 + a^2)} $$
then
$$ \overline{y} = \frac{1}{a^2} \left( \frac{e^{- \pi s}}{s} - \frac{s \, e^{- \pi s}}{s^2 + a^2} - \frac{e^{- 2 \pi s}}{s} + \frac{s \, e^{- 2 \pi s}}{s^2 + a^2} \right) + \frac{s}{s^2 + a^2} \, y(0) + \frac{y^{'}(0)}{s^2 + a^2}. $$
The Laplace inversion gives
\begin{align}
y(t) &= \frac{1}{a^2} \, \left( (1+\cos(a t - a \pi)) \, H(t - \pi) - (1 - \cos(a t - 2 a \pi)) \, H(t - 2 \pi) \right) \\
& \hspace{10mm} + y(0) \, \cos ( a t) + \frac{y^{'}(0)}{a} \, \sin(a t)
\end{align}
or
\begin{align}
y(t) &= \frac{2}{a^2} \, \cos^{2}\left(\frac{a t - a \pi}{2}\right) \, H(t - \pi) - \frac{2}{a^2} \, \sin^{2}\left(\frac{a t - 2 a \pi}{2}\right) \, H(t - 2 \pi) \\
& \hspace{15mm} + y(0) \, \cos ( a t) + \frac{y^{'}(0)}{a} \, \sin(a t).
\end{align}
With the conditions $y(0) = y^{'}(0) = 0$ and making use of the Heaviside step function definition then the solution can be seen in regions as:
$$ y(t) = \begin{cases} 0 & t<\pi \\ \frac{1}{a^2} & t = \pi \\ \frac{2}{a^2} \, \cos^{2}\left(\frac{a t - a \pi}{2}\right) & \pi < t < 2 \pi \\ \frac{2}{a^2} \, \cos^{2}\left(\frac{a \pi}{2}\right) & t = 2 \pi \\ \frac{1}{a^2} \, \left( \cos(a t - a \pi) + \cos(a t - 2 a \pi) \right) & t > 2 \pi \end{cases}. $$
|
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|
Show ${\pm 3^i}, 1\leq i \leq 2^{n-2}$ is reduced residue system mod $2^n$ Show set ${\pm 3^i}$, $1\leq i \leq 2^{n-2}$ is reduced residue system mod $2^n$.
For this, I believe I need to prove that for any integer $a$ such that $gcd(a,2^n)=1$, there exists an element $b$ in the set ${\pm 3^i}$, $1\leq i \leq 2^{n-2}$ so that $a \equiv b$ mod $2^n$. As part of a problem related to this question, I've managed to show that the order of $3$ modulo $2^n$ is $2^{n-2}$. By the definition of order, this should fulfill the requirement of distinct elements in the given set.
How can I proceed further?
|
You're correct regarding what you need to prove, and have made a good start to determining a solution. Note that since there are $2^{n-1}$ odd integers between $1$ and $2^n$, inclusive, i.e., $\varphi(2^n)=2^{n-1}$, and the set $S = \{\pm 3^{i}, 1 \le i \le 2^{n-2}\}$ has $2(2^{n-2}) = 2^{n-1}$ elements, each of which are odd (i.e., coprime to $2^n$), we only need to prove that each element of $S$ is distinct modulo $2^n$.
To do that, first with $n = 2$, we have $\pm 3$ which is a reduced residue system for $2^2 = 4$. Thus, consider only $n \gt 2$. Using the $2$-adic order function, for even integers $m$, we have $3^m \equiv 1 \pmod{8}$, so $3^m + 1 \equiv 2 \pmod{8}$ giving $\nu_2(3^m + 1) = 1$. With $m$ odd, we have $\nu_2(3^m + 1) = 2$, with this determined by using the Lifting-the-exponent lemma (LTE lemma) with $3^m - (-1)^m$, or noting that $3^m + 1 = (3 + 1)(\color{blue}{3^{m-1} - 3^{m-2} + \ldots - 3 + 1})$, with the blue part having $m$ (i.e., an odd number) terms, each of which are odd, so their sum is odd. Altogether, this means $\nu_2(3^m + 1) \le 2$.
In addition, as you've noted, plus this can be shown using the LTE lemma and is also explained in if $3^k \equiv \ 1\pmod{2^n}$ then $2^{n-2} | k$, the multiplicative order of $3$ modulo $2^n$ is $2^{n-2}$.
Consider that $2$ distinct elements of $S$ have the same congruence modulo $2^n$. First, with them having the same sign and $1 \le j \lt k \le 2^{n-2}$, we have
$$\pm 3^{j} \equiv \pm 3^{k} \pmod{2^n} \; \to \; 1 \equiv 3^{k-j} \pmod{2^n} \tag{1}\label{eq1A}$$
However, $0 \lt k - j \lt 2^{n-2}$, but the multiplicative order being $2^{n-2}$ means this is impossible.
For the second case, with them having opposite signs and $1 \le j, k \le 2^{n-2}$, we get
$$\pm 3^{j} \equiv \mp 3^{k} \pmod{2^n} \; \to \; 3^{j} + 3^{k} \equiv 0 \pmod{2^n} \tag{2}\label{eq2A}$$
WLOG, have $j \le k$, so \eqref{eq2A} becomes $3^{j}(1 + 3^{k-j}) \equiv 0 \pmod{2^n} \; \to \; 1 + 3^{k-j} \equiv 0 \pmod{2^n}$. This requires $\nu_2(1 + 3^{k-j}) \ge n$, but $n \gt 2$ and, as shown previously, $\nu_2(1 + 3^{k-j}) \le 2$, so this is not possible.
The results of these $2$ cases show that all of the elements of $S$ have distinct congruences modulo $2^n$ so, as explained originally, $S$ forms a reduced residue system mod $2^n$.
|
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|
Can one deduce $\lim_{n\rightarrow\infty}P(X_n>1)=0$ from $P(X_n>1+\frac{x}{2\log n})\leq e^{-x}\quad\forall x$ I want to control the tail probability of some random variable $(X_n)_n$. (In particular, $X_n = \max_{1\leq i\leq n} N_i$, where $N_i$ are independent standard normal)
I come up with the following two upper bounds:
(1)$$P(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n})\leq e^{-x}\quad\forall x>0.$$
Can I deduce $\lim_{n\rightarrow\infty}P(\frac{X_n}{\sqrt{2\log n}}>1)=0$ from it?
I only know that for every $\epsilon_1,\epsilon_2>0$, there exists $N_1$ such that for every $n>N_1$, we have
$$P(\frac{X_n}{\sqrt{2\log n}}>1+\epsilon_1)\leq \epsilon_2.$$ I have no idea how to do next.
(2)$$P(\frac{X_n}{\sqrt{2\log n}}\leq \alpha)\leq (1-n^{-\alpha+o(1)})^n\quad 0<\alpha< 1 .$$
Can I deduce $\lim_{n\rightarrow\infty}P(\frac{X_n}{\sqrt{2\log n}}< 1)=0$ from it?
I only know $\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)})^n=0$. Don't know how to deal with $\alpha$ inside the probability.
(3) Suppose we have $$P(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n})\leq e^{-x}\quad\forall x>0.$$ and $$P(\frac{X_n}{\sqrt{2\log n}}\leq \alpha)\leq (1-n^{-\alpha+o(1)})^n\quad 0<\alpha< 1 .$$
Can we deduce $\frac{X_n}{\sqrt{2\log n}}\rightarrow 1$ in distribution?
|
First, we determine the probability function of $X_n$, we have:
$$\begin{align}
P(X_n<a)&=P(\max_{i=1,..,n}N_i<a)\\
&=P^n(N_i<a)\\
&=\Phi^n(a)
\end{align}$$
With $\Phi(\cdot)$ the cumulative distribution function of $\mathcal{N}(0,1)$.
Return back to the main problem, it suffices to prove that for all $\epsilon>0$, there exists $N$ such that for all $n>N$, we have $P(\frac{X_n}{\sqrt{2\log n}}>1)<\epsilon$.
We have
$$\begin{align}
P(\frac{X_n}{\sqrt{2\log n}}>1)&=P\left(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n}\right)+P\left(1<\frac{X_n}{\sqrt{2\log n}}<1+\frac{x}{2\log n}\right) \\
&=P\left(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n}\right)+P\left(\sqrt{2\log n}<X_n<\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}}\right) \tag{1}\\
\end{align}$$
We know already how to bound the first term of $(1)$, so let's study the second term, we have
$$\begin{align}
L&:=P\left(\sqrt{2\log n}<X_n<\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}}\right) \\
&=\Phi^n\left(\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}} \right) -\Phi^n\left(\sqrt{2\log n} \right) \\
\end{align}$$
Applying the mean value theorem with $(a,b) = \left(\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}},\sqrt{2\log n} \right)$ and $f=\Phi^n(\cdot)$, there exists $c$ such that $\sqrt{2\log n} <c<\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}}$ and
$$L =\frac{x}{\sqrt{2\log n}}\left( \Phi^n(c) \right)' = \frac{x}{\sqrt{2\log n}}\cdot n \cdot \Phi^{n-1}(c)\cdot\phi(c)< \frac{x}{\sqrt{2\log n}}\cdot n \cdot \phi(c)\tag{2}$$
with the normal density function $\phi(x) =\frac{\exp{-\frac{x^2}{2}}}{\sqrt{2\pi}}$.
As $\phi(\cdot)$ is decreasing for $x>0$, then
$$\phi(c)<\phi(\sqrt{2\log n}) = \frac{\exp{\left(-\frac{2\log n}{2} \right) }}{\sqrt{2\pi}} = \frac{1}{n\sqrt{2\pi}}\tag{3}$$
From $(2),(3)$, we deduce that
$$L<\frac{x}{2\sqrt{\pi \log n}}\tag{4}$$
From $(1),(4)$, we have
$$P(\frac{X_n}{\sqrt{2\log n}}>1)< e^{-x}+\frac{x}{2\sqrt{\pi \log n}} \tag{5}$$
For all $\epsilon >0$, set
$$x= - \ln \left(\frac{\epsilon}{2} \right)$$ and $$N>\exp\left(-\frac{1}{\pi\epsilon^2} \ln \left( \frac{\epsilon}{2} \right) \right)$$
Then, for all $n>N$, the two terms on RHS of $(5)$ are both smaller than $\frac{\epsilon}{2}$. Then, for all $n>N$,
$$P\left(\frac{X_n}{\sqrt{2\log n}}>1\right)< \epsilon$$
We can conclude that
$$\lim_{n\to +\infty}P\left(\frac{X_n}{\sqrt{2\log n}}>1\right) = 0$$
PS: $\frac{X_n}{\sqrt{2\log n}} \xrightarrow{n\infty}{1}$ in distribution if and only if $\forall x$
$$\lim_{n\to +\infty}P\left(\frac{X_n}{\sqrt{2\log n}}<x\right) = \mathbb{I}_{\{ x \ge 1 \}}$$
So, your last result is not sufficient to prove that $\frac{X_n}{\sqrt{2\log n}}$ converge to $1$ in distribution.
|
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|
Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$x^3+y^3 = 4(x+y)$
$x^3+y^3 = 8$
Then:
$8 = 4(x+y)$
$x+y = 2$
However if we replace that value in $(x+y)^2$ we have:
$(x+y)^2 = 2^2 =4$
$(x+y)^2 = x^2+y^2+2xy = 5 + 2 = 7$
As you can see $4 \neq 7$, what is happening?
|
Alternative approach:
Let $z = (x + y)$.
The following equations over-determine the value of $z$.
*
*E-1: $~x^2 + y^2 = 5.$
*E-2: $xy = 1.$
*E-3: $x^3+y^3 = 8.$
Since $~z^2 = x^2 + y^2 + 2xy,~$
E-1,E-2 collectively imply that
$\displaystyle z^2 = 7 ~\implies~ z ~\in ~\left\{+\sqrt{7}, -\sqrt{7}\right\}.$
So, $\displaystyle ~~\left\{+\sqrt{7}, -\sqrt{7}\right\}~$ are the only candidate values for $z$.
Since $z^3 = x^3 + y^3 + 3xy(z)$,
E-2,E-3 collectively imply that $z^3 = 8 + 3z \implies z^3 - 3z - 8.$
Trying each candidate value in turn gives:
*
*$\displaystyle \left[+\sqrt{7}\right]^3 - 3\left[+\sqrt{7}\right] = 4\sqrt{7} \neq 8.$
*$\displaystyle \left[-\sqrt{7}\right]^3 - 3\left[-\sqrt{7}\right] = -4\sqrt{7} \neq 8.$
So, neither of the two candidate values collectively implied by E-1 and E_2, satisfy the constraint collectively implied by E-2 and E-3.
|
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|
Evaluate $\sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n n! (2n+1)^2}$.
Evaluate $\sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n n! (2n+1)^2}$.
I am attempting to evaluate this sum as an alternate form of an integral I was trying to calculate. Can anyone give any insight? WolframAlpha says it's $\frac{1}{2}\pi\log(2)$ but I'm not sure how to get there.
|
Consider using $ (2 n -1)!! = 2^n \, \left(\frac{1}{2}\right)_{n}$, where $(a)_{n}$ is the Pochhammer symbol, to obtain
$$ f(x) = \sum_{n=0}^{\infty} \frac{(2 n -1)!! \, x^{2 n +1}}{2^n \, n! \, (2n + 1)^2} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, x^{2 n +1}}{ n! \, (2n + 1)^2}. $$
With some manipulations the series becomes:
\begin{align}
x \, \frac{d f}{d x} &= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, x^{2 n + 1}}{n! \, (2 n + 1)} \\
\frac{d}{dx} \, \left(x \, \frac{d f}{dx} \right) &= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, x^{2 n}}{n!} = \frac{1}{\sqrt{1-x^2}} \\
x \, \frac{d f}{dx} &= \int^{x} \frac{d u}{\sqrt{1- u^2}} = \sin^{-1}(x) + c_{0} \\
f(x) &= \int^{x} \frac{\sin^{-1}(u)}{u} \, du + c_{0} \, \ln x + c_{1} \\
&= \sin^{-1}(x) \, \ln\left( 1 - e^{2 \, i \, \sin^{-1}(x)} \right) - \frac{i}{8} \, \left( \left(\sin^{-1}(x)\right)^2 + \text{Li}_{2}\left(e^{2 \, i \, \sin^{-1}(x)} \right) \right) + c_{0} \, \ln(x) + c_{1}.
\end{align}
Since $f(0) = 0$ then $c_{0} = 0$ and $c_{1} = \frac{i \, \zeta(2)}{2}$ which gives
$$ f(x) = \sin^{-1}(x) \, \ln\left( 1 - e^{2 \, i \, \sin^{-1}(x)} \right) - \frac{i}{8} \, \left( \left(\sin^{-1}(x)\right)^2 + \text{Li}_{2}\left(e^{2 \, i \, \sin^{-1}(x)} \right) \right) + \frac{i}{2} \, \zeta(2). $$
When $x = 1$ this becomes
$$ f(1) = \frac{\pi}{2} \, \ln 2. $$
Hence,
$$ \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{ n! \, (2n + 1)^2} = \frac{\pi}{2} \, \ln 2. $$
|
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|
Showing that $\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$ is the square of a Fibonacci number I was experimenting with products of the form
$$\prod_{k=1}^{n} \left( a + b\cos(ck) \right)$$
when I found that the expression
$$\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$$
seems to always return a perfect square. (verified numerically).
$$\left(3 + 2\cos\left(\frac{2\pi}{2}\right)\right) = 1 = F_2^2$$
$$\left(3 + 2\cos\left(\frac{2\pi}{3}\right)\right) \left(3 + 2\cos\left(\frac{4\pi}{3}\right)\right) = 4 = F_3^2$$
$$\left(3 + 2\cos\left(\frac{2\pi}{4}\right)\right)\left(3 + 2\cos\left(\frac{4\pi}{4}\right)\right) \left(3 + 2\cos\left(\frac{6\pi}{4}\right)\right) = 9 = F_4^2$$
$$\dots$$
Furthermore, it appears these squares are the squares of Fibonacci numbers. ($F_1 = F_2 = 1$, $F_n = F_{n-1} + F_{n-2})$
I've tried to prove that this expression always gives a perfect square by considering the identity (Cassini's identity)
$$F_{n}^2 = F_{n+1}F_{n-1} + (-1)^{n-1}$$
and using induction to show that the above product satisfies this relation, and since the base cases ($F_1, F_2...$) are that of the Fibonacci sequence then this would prove that the expression is the square of Fibonacci numbers. However, this approach led nowhere.
Is there a clever way to show that the above expression is related to Fibonacci numbers?
|
This was a blast solving. Never in a million years would I have guessed these to be true. But the $n+1$ in the denominator really hints at what to use. I will be skipping the more well known proofs.
We start with the Chebyshev polynomials of the second kind $U_n(x)$ which satisfy the following recursion relations
$$U_0(x)=1,\ U_1(x)=2x,\ U_n(x)=2xU_{n-1}(x)-U_{n-2}(x)$$
Observe that $U_n(x)$ is a $n$-degree polynomial. Easy proof. It is true for $n=0,1$. Then use induction.
Claim 1 : $$U_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin\theta}$$
Then the roots of the polynomial $U_n(x)$ are exactly where the RHS of the above vanishes, namely $\frac{k\pi}{n+1}$, and you can produce $n$ of them.
Proposition 2 : $$U_n(x)=2^n\prod_{k=1}^n\left(x-\cos\left(\frac{k\pi}{n+1}\right)\right)$$
Proof : All that's left to prove is the fact that the leading coefficient of $U_n$ is $2^n$. It is true for $n=0,1$. By induction, $U_{n-1},\ U_{n-2}$ has leading coefficients $2^{n-1},\ 2^{n-2}$ respectively and thus $U_n$ has leading coefficient $2\cdot 2^{n-1}=2^n$. The result follows. $\ \square$
Let $F_n$ denote the $n$-th Fibonacci number, $n\ge 0$.
Theorem 3 : $$F_n=i^{-n}U_n\left(\frac{i}{2}\right)$$
Proof : Induction yet again. $$1=F_0=i^{-0}U_0\left(\frac{i}{2}\right)=1$$ $$1=F_1=i^{-1}U_1\left(\frac{i}{2}\right)=i^{-1}2\left(\frac{i}{2}\right)=1$$
Let this be true for all $0\le k\le n-1$. Then for $k=n$
\begin{align*}
F_n&=F_{n-1}+F_{n-2}\\&=i^{-n+1}U_{n-1}\left(\frac{i}{2}\right)+i^{-n+2}U_{n-2}\left(\frac{i}{2}\right)\\&=i^{-n+1}2\left(\frac{i}{2}\right)U_{n-1}\left(\frac{i}{2}\right)+i^{-n+2}U_{n-2}\left(\frac{i}{2}\right)\\&=i^{-n}\left\{2\left(\frac{i}{2}\right)U_{n-1}\left(\frac{i}{2}\right)+i^2U_{n-2}\left(\frac{i}{2}\right)\right\}\\&=i^{-n}\left\{2\left(\frac{i}{2}\right)U_{n-1}\left(\frac{i}{2}\right)-U_{n-2}\left(\frac{i}{2}\right)\right\}\\&=i^{-n}U_n\left(\frac{i}{2}\right)
\end{align*}
and we are done by induction. $\ \square$
Main Result : $$F_n^2=\prod_{k=1}^n\left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$$
Proof : We first look at $U_{n}^2\left(\frac{i}{2}\right)$.
We have
\begin{align*}
U_n^2\left(\frac{i}{2}\right)&=2^{2n}\left[\prod_{k=1}^n\left(\left(\frac{i}{2}\right)-\cos\left(\frac{k\pi}{n+1}\right)\right)\right]^2\\&=\left[2^n\prod_{k=1}^n\left(\left(\frac{i}{2}\right)-\cos\left(\frac{k\pi}{n+1}\right)\right)\right]\left[2^n\prod_{k=1}^n\left(\left(\frac{i}{2}\right)-\cos\left(\frac{k\pi}{n+1}\right)\right)\right]\\&=\left[\prod_{k=1}^n\left(i-2\cos\left(\frac{k\pi}{n+1}\right)\right)\right]\left[\prod_{k=1}^n\left(i-2\cos\left(\frac{k\pi}{n+1}\right)\right)\right]
\end{align*}
Now to calculate this, pair up the $j$-th term of the first product with the $n+1-j$-th term of the second product. This produces terms like
$$\left(i-2\cos\left(\frac{j\pi}{n+1}\right)\right)\left(i-2\cos\left(\frac{(n+1-j)\pi}{n+1}\right)\right)=\left(i-2\cos\left(\frac{j\pi}{n+1}\right)\right)\left(i+2\cos\left(\frac{j\pi}{n+1}\right)\right)=-\left(1+4\cos^2\left(\frac{j\pi}{n+1}\right)\right)$$
Plugging this back, we have
$$U_n^2\left(\frac{i}{2}\right)=(-1)^n\prod_{k=1}^n\left(1+4\cos^2\left(\frac{k\pi}{n+1}\right)\right)$$
Using Theorem 3, we get
$$F_n^2=(-1)^{-n}\prod_{k=1}^n\left(1+4\cos^2\left(\frac{k\pi}{n+1}\right)\right)$$
Now using the standard cosine double angle identity, we get $$F_n^2=\prod_{k=1}^n\left(3+2\cos\left(\frac{2k\pi}{n+1}\right)\right)$$ and we are done. $\ \square$
Some random musings:
This seems generalizable in quite a few directions. What if we use $\sin$ instead of $\cos$? Do we still get something that has some recurrence? What if we used Chebyshev polynomials of the first kind? What kind of identities will that lead to?
More algebraically, does this suggest some deeper connection between cyclotomic extensions and quadratic extensions (Since Fibonacci numbers are related to $\mathbb Z[\sqrt 5]$ and we have a bunch of products of cosines)? [Kronecker-Weber is a step in this direction, but I am not sure how it related honestly]. Or more intriguing, is there a relation between cyclotomic extensions and Chebyshev polynomials?
I am not learned enough to even try and make a guess. Maybe people in the comment can guide me?
I just now saw the comment by Thomas Andrews mentioning Chebyshev polynomials. Interesting how we came up with the same idea. The $n+1$ denominator suggested this approach. I would be interested in learning how they came up with it though.
|
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|
About the equivalence of $\cos(x) = \sqrt{1-\sin^2(x)}$ and $\cos(x) = \cos(2x-x)$ By the title, my questions may sound trivial or silly, but I have problems with those outputs.
I need the sine of $15$ degrees. Now I thought of: $15 = 45-30$ hence:
$$\sin(15) = \sin(45-30) = \sin(45)\cos(30) - \cos(45)\sin(30) = \dfrac{\sqrt{6}-\sqrt{2}}{4}$$
So far so good.
Yet, when calculating the cosine, I have
Mode 1
$$\cos(15) = \sqrt{1 - \sin^2(15)} = \sqrt{1 - \dfrac{8 - 4\sqrt{3}}{16}} = \dfrac{\sqrt{2+\sqrt{3}}}{2}$$
Mode 2
$$\cos(15) = \cos(45-30) = \cos(45)\cos(30) + \sin(45)\sin(30) = \dfrac{\sqrt{6}+\sqrt{2}}{4}$$
The second method returns something more clean than the previous one. I did not really see the numerical equivalence though, so I computed the difference and I found that
$$\dfrac{\sqrt{2+\sqrt{3}}}{2} - \dfrac{\sqrt{6}+\sqrt{2}}{4} = 1.11\cdot 10^{-16}$$
(On W. Mathematica).
Can someone please explain me this infinitesimal difference between them, and how to pass from Mode 1 to Mode 2 in writing?
|
$$
\frac{\sqrt{2+\sqrt{3}}}{2} = \frac{\sqrt{8+4\sqrt{3}}}{4}
$$
Solving
$$
8+4\sqrt{3} = (a+b\sqrt{3})^2
$$
results in
$$
\begin{cases}
a^2+3b^2=8 \\
ab = 2
\end{cases}
$$
and solving that results in
$$
(a, b)= (\sqrt2, \sqrt2), (\sqrt6, \sqrt\frac{2}{3})
$$
so
$$
8+4\sqrt3 = (\sqrt6+\sqrt2)^2
$$
and thus
$$
\frac{\sqrt{8+4\sqrt3}}{4} = \frac{\sqrt6+\sqrt2}{4}
$$
So it's probably a rounding error in the software.
|
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|
Expected value of highest die roll Daniel will roll a fair, six-sided die until he gets a 4. What is the expected value of the highest number he
rolls through this process?
It seems the expected number of rolls is 6. Can we reword this question to what is the expected value of the maximum of 6 die rolls?
|
Let $P(6)$ denote the probability of getting a $(6)$ before the first $(4)$.
Let $P(5)$ denote the probability of getting a $(5)$ before the first $(4)$ and simultaneously, not getting a $(6)$ before the first $(4)$.
Then, the computation for the expected value of the highest roll is
$$4 +~ \left[ ~2 \times P(6) ~\right] ~+~ \left[ ~1 \times ~P(5) ~\right]. \tag1 $$
$$P(6) = \frac{1}{6} + \left[\frac{4}{6}P(6)\right] \implies \frac{2}{6}P(6) = \frac{1}{6} \implies P(6) = \frac{1}{2}.$$
$$P(5) = \frac{1}{6} + \left[\frac{3}{6}P(5)\right] \implies \frac{3}{6}P(5) = \frac{1}{6} \implies P(5) = \frac{1}{3}.$$
Thanks to lulu, for her comment, immediately following this posting, which indicated a flaw in the above analysis, pertaining specifically to the computation of $P(5).$
$\color{red}{\text{The above analysis is wrong because:}}$
The above equation assumes that if the first roll is a $(5)$, that it is game over, and that the high roll will be a $(5)$. This overlooks the fact that the die rolls do not stop, if the first roll is a $(5)$.
Instead, the die rolls continue, and by the already computed $P(6)$, (1/2) the (subsequent) time, there will be a $(6)$ before the $(4)$.
Therefore, the correct equation for $P(5)$ is
$$P(5) = \left[ ~\frac{1}{6} \times \left( ~1 - P(6) \right) ~\right] + \left[\frac{3}{6}P(5)\right] \implies $$
$$\frac{3}{6}P(5) = \left[\frac{1}{6} \times \left( ~1 - \frac{1}{2} ~\right) ~\right] \implies P(5) = \frac{1}{6}.$$
Note:
As indicated in the comment of lulu, an alternative (less convoluted) approach is to:
*
*Let $P(4)$ denote the probability that the high roll will be $(4)$.
*Recognize that $P(4) + P(5) + P(6) = 1$.
*Recognize that
$\displaystyle P(4) = \frac{1}{6} + \left[\frac{3}{6}P(4)\right] \implies \frac{3}{6}P(4) = \frac{1}{6} \implies P(4) = \frac{1}{3}.$
Plugging these $\color{red}{\text{now corrected}}$ results back into (1) above, the computation of expected value becomes
$$4 +~ \left[ ~2 \times \frac{1}{2} ~\right] ~+~ \left[ ~1 \times \frac{1}{6} ~\right] = \frac{31}{6}. $$
|
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|
A Nonrectifiable Curve Do Carmo's book Page 11.
Let $\alpha : [0,1] \to \mathbb{R}^2$ be given as $$\alpha(t)=
\begin{cases}
(t,t \sin (\pi/t)) & \text{ for } t \ne 0\\
(0,0) & \text{ for } t=0 \,.
\end{cases}$$ Show that the arc length of the portion of the curve corresponding to $\dfrac{1}{n+1} \leq t \leq \dfrac{1}{n}$ is at least $2/(n+1/2)$. Use this to show that the length of the curve in the interval $1/N \leq t \leq 1$ is greater than $2 \sum_{n=1}^N 1/ (n+1)$.
For the first part, I considered the distance between $\bigl(\alpha(1/n), \alpha(2/(2n+1))\bigr)$ and $\bigl(\alpha(2/(2n+1)), \alpha(1/(n+1))\bigr)$ and proved that it should be at least $2/(n+1/2)$.
In the second part, we can see that
\begin{align}
[1/N, 1] & = [1/N, 1/(N-1)]\cup [1/(N-1), 1/(N-2)]\cup \dots \cup [1/2,1] \,.
\end{align}
Therefore,
\begin{align}
|\alpha(1)-\alpha(1/2)|+ \dots + |\alpha(1/(N-1))-\alpha(1/N)| & \geq \sum_{n=1}^{N-1}\dfrac{2}{n+1/2} \,,
\end{align} and that won't give the desired solution. I don't know where I have made a mistake.
|
Everything you have is correct. All that's remaining is to show that your lower bound is at least as large as the desired lower bound. Equivalently, we need to show that
$$\sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^N \frac{1}{n+1} > 0 \,.$$
We have the following.
\begin{align*}
\sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^N \frac{1}{n+1}
&= \sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^{N-1} \frac{1}{n+1} - \frac{1}{N+1} \\
&= \sum_1^{N-1} \biggl(\frac{1}{n+0.5} - \frac{1}{n+1}\biggr) - \frac{1}{N+1} \\
&= \sum_1^{N-1} \biggl(\frac{n+1}{(n+0.5)(n+1)} - \frac{n+0.5}{(n+0.5)(n+1)}\biggr) - \frac{1}{N+1} \\
&= \sum_1^{N-1} \frac{n+1-n-0.5}{(n+0.5)(n+1)} - \frac{1}{N+1} \\
&= \sum_1^{N-1} \frac{0.5}{(n+0.5)(n+1)} - \frac{1}{N+1} \\
&= \sum_1^{N-1} \frac{1}{(2n+1)(n+1)} - \frac{1}{N+1}
\end{align*}
The final term is approximately equal to $0.07$ for $N=4$ and is an increasing function in $N$, so the claim holds for all $N \geq 4$.
This doesn't necessarily show the statement is false for $N \leq 3$ but doesn't prove it true either.
|
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|
$\int \frac{\cos 4x}{4 \sin 2x} dx$ $\int \frac{\cos 4x}{4 \sin 2x} dx$
Let $u=2x$, $dx = 1/2 du$
$\int \frac{\cos 2u}{4 \sin u} \frac{1}{2} du = \frac{1}{8} \int \frac{1-2\sin^2 u}{\sin u}du \frac{1}{8} \int \frac{1}{\sin u} du - \frac{1}{8} \int 2 \sin u$
How do I integrate $\int \frac{1}{\sin u} du$ to get $\ln (\tan x)$ ?
The online calculator told me to use Weierstrass Substitution which I have not learnt before. Is there any other way to solve this ?
|
$$\int \frac{dx}{\sin x}=\int \frac{d(\cos x)}{\cos^2x-1}
= \frac12\ln \frac{1-\cos x}{1+\cos x}=\frac12\ln\frac{\sin^2\frac x2}{\cos^2\frac x2}=\ln\tan\frac x2
$$
|
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|
Show that the diagonal elements of $V$ coincide Let $V$ be an $n\times n$ diagonal positive definite matrix, and suppose that
$$[I_n-XX']VX=0$$
for all $X\in\mathbb R^{n \times k}$ satisfying $X'X=I_k$, where $k<n$ is fixed.
Can I show that the diagonal elements of $V$ are all equal in this case?
|
As you said you're interested in the case $k <n$, here it is. Denote $V=\mathrm{Diag}(a_1,a_2,\cdots,a_n)$ and $(e_1,\cdots,e_n)$ be the canonical basis of $\mathbb{R}^n$. Take $X \in \mathcal{M}_{n,k}(\mathbb{R})$ the matrix whose columns are $(e_1+e_2,e_3,e_4,\cdots,e_{k+1})$ It is possible as $k+1\leq n$. This is
$$X = \begin{pmatrix}1&0&0&\cdots & 0\\
1&0&0&\cdots & 0\\
0&1&0&\cdots & 0\\
0&0&1&\cdots & 0\\
\vdots & \vdots &\vdots &\ddots &\vdots \\
0&0&0&\cdots & 1\\
0&0&0&\cdots & 0\\
\vdots & \vdots &\vdots &\ddots &\vdots\\
0&0&0&\cdots & 0\\
\end{pmatrix}$$
Take $X' \in \mathcal{M}_{k,n}(\mathbb{R})$ the matrix whose rows are $(\dfrac{1}{2}(e_1+e_2),e_3,e_4,\cdots,e_{k+1})$. AS the coefficient of $X'X$ are the inner products of the rows of $X'$ with the columns of $X$, we get $XX' = I_k$.
We then have
$$XX'= \begin{pmatrix}\frac{1}{2}&\frac{1}{2}&0&\cdots & 0&0&\cdots & 0\\
\frac{1}{2}&\frac{1}{2}&0&\cdots & 0&0&\cdots & 0\\
0&0&1&\cdots & 0&0&\cdots & 0\\
\vdots & \vdots &\vdots &\ddots &\vdots&0&\cdots & 0 \\
0&0&0&\cdots & 1&0&\cdots & 0\\
0&0&0&\cdots & 0&0&\cdots & 0\\
\vdots & \vdots &\vdots &\ddots &\vdots&0&\cdots & 0\\
0&0&0&\cdots & 0&0&\cdots & 0\\
\end{pmatrix}.$$
$$I_n -XX' = \begin{pmatrix}\frac{1}{2}&-\frac{1}{2}&0&\cdots & 0&0&\cdots & 0\\
-\frac{1}{2}&\frac{1}{2}&0&\cdots & 0&0&\cdots & 0\\
0&0&0&\cdots & 0&0&\cdots & 0\\
\vdots & \vdots &\vdots &\ddots &\vdots&0&\cdots & 0 \\
0&0&0&\cdots & 0&0&\cdots & 0\\
0&0&0&\cdots & 0&1&\cdots & 0\\
\vdots & \vdots &\vdots &\ddots &\vdots&0&\vdots & 0\\
0&0&0&\cdots & 0&0&\cdots & 1\\
\end{pmatrix}.$$
Let's look at the first column of $(I_n -XX')VX$, which is the column given by $(I_n -XX')V(e_1+e_2)$.We have
$$\begin{array}{rll}
(I_n -XX')V(e_1+e_2) &=& (I_n -XX')(a_1.e_1+a_2.e_2) \\
&=&a_1.(I_n -XX')e_1+a_2.(I_n -XX')e_2)
&=& a_1.\begin{pmatrix}{\frac{1}{2} \\ -\frac{1}{2} \\0\\ \vdots \\ 0}\end{pmatrix} + b_1.\begin{pmatrix}{-\frac{1}{2} \\ \frac{1}{2} \\0\\ \vdots \\ 0}\end{pmatrix} \\
&=& \begin{pmatrix}{\frac{1}{2}a_1-\frac{1}{2}a_2 \\ -\frac{1}{2}a_1+\frac{1}{2}a_2 \\0 \\ \vdots \\ 0}\end{pmatrix}.
\end{array}$$
By assumption, this must be $0$, so $a_1=a_2$.
$\rhd$ We can now do the exact same proof to prove $a_p=a_q$, by taking $X \in \mathcal{M}_{n,k}(\mathbb{R})$ the matrix whose columns are $(e_p+e_q,e_1,e_2,\cdots,e_{\text{something}})$ where $e_p$ and $e_q$ do not appear among the vectors $e_1,e_2,\cdots,e_{\text{something}}$ ; and similarly for $X'$.
|
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|
If $\frac{1+\sin{\theta}}{\cos{\theta}} = n$, find $\tan{\frac{1}{2}\theta}$. The question is:
If $\cfrac{1+\sin{\theta}}{\cos{\theta}} = n$, find $\tan{\frac{1}{2}\theta}$.
I'm quite confused on how to solve this problem. What trigonometric identities do I use? I tried squaring them and ended with $n^2=\dfrac{1+\sin{\theta}}{1-\sin{\theta}}$ but I suppose that doesn't really help the problem. Any suggestions would be appreciated.
|
If we also have the "tangent half-angle" identity $ \ \tan \frac{\theta}{2} \ = \ \frac{1 \ - \ \cos \theta}{\sin \theta} \ \ , \ $ we can approach the problem in this way. From $ \ n \ = \ \frac{1 \ + \ \sin \theta}{\cos \theta} \ = \ \sec \theta + \tan \theta \ \Rightarrow \ \tan \theta \ = \ n - \sec \theta \ \ , \ $ construct a right triangle with this tangent "value". (We will produce an expression for the hypothenuse $ \ H \ $ shortly.)
It is clear that $ \ \sec \theta \ = \ H \ \ . \ $ We then have
$$ \ \tan \frac{\theta}{2} \ \ = \ \ \csc \theta \ - \ \cot \theta \ \ = \ \ \frac{H \ - \ 1}{n \ - \ H} \ \ . $$
From the "Pythagorean theorem", we find
$$ H^2 \ - \ 1 \ \ = \ \ (n \ - \ H)^2 \ \ = \ \ n^2 \ - \ 2nH \ + \ H^2 \ \ \Rightarrow \ \ nH \ \ = \ \ \frac{n^2 \ + \ 1}{2} \ \ . $$
Consequently,
$$ \ \tan \frac{\theta}{2} \ \ = \ \ \frac{nH \ - \ n}{n^2 \ - \ nH} \ \ = \ \ \frac{\left(\frac{n^2 \ + \ 1}{2} \right) \ - \ n}{n^2 \ - \ \left(\frac{n^2 \ + \ 1}{2} \right)} \ \ = \ \ \frac{ n^2 \ + \ 1 \ - \ 2n}{2n^2 \ - \ ( n^2 \ + \ 1 )} $$
$$ = \ \ \frac{ n^2 \ - \ 2n \ + \ 1 }{ n^2 \ - \ 1 } \ \ = \ \ \frac{ (n \ - \ 1)^2 }{ (n \ + \ 1) · (n \ - \ 1) } \ \ = \ \ \frac{ n \ - \ 1 }{ n \ + \ 1 } \ \ . $$
[This contains elements appearing in the other answers, arrived at in a different way. This is perhaps inevitable since all of them hinge on the "Pythagorean identity" $ \ \tan^2 \theta \ + \ 1 \ = \ \sec^2 \theta \ $ in some manner.]
|
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|
Improper integrals with bounds 0 to infinity $$\int_{0}^{\infty }\frac{1}{x^{1/2}+x^{3/2}}dx$$
I managed to integrate correctly and get $2\arctan(\sqrt{x})+C$, but I was wondering why my teacher split it up into two pieces, one from $0$ to $1$ with a limit at $0$ evaluating $2\arctan(\sqrt{x})+C$ and the other $1$ to $\infty$ with a limit at infinity evaluating $2\arctan(\sqrt{x})+C$. Can I just take $2\arctan(\sqrt{x})$ and evaluate from $0$ to $\infty$ with a limit at infinity?
|
If not wrong: You could solve the problem but you want to know why did your teacher split the above integral as
$$\int_0^\infty = \int_0^1 + \int_1^\infty$$
or as
$$\arctan(x)|_0^\infty = \arctan(x)|_0^1 + \arctan(x)|_1^\infty$$
then Comment by @Accelerator
however, for this particular case
$$\begin{align*}\int_0^\infty \frac {dx}{x^{\frac 12} + x^{\frac 32}}
&=\left(\int_0^1 + \int_1^\infty\right)\frac {1}{x^{\frac 12} + x^{\frac 32}}dx
\\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \color{blue}{\int_1^\infty\frac {dx}{x^{\frac 12} + x^{\frac 32}}}
; \text{ Let } \color{green}{x = \frac 1y }
\\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \int_1^0\frac {\frac{-dy}{y^2}}{y^{\frac {-1}2} + y^{\frac {-3}2}}
\\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \color{blue}{\int_0^1 \frac {dy}{y^{\frac 12} + y^{\frac {3}2}}}
\\& = 2\times\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}}
\\& = 2\times\int_0^1 \frac {x^{-\frac 12}dx}{1 + x}
\\& = 2\frac {\beta\left(\frac 12, \frac 12\right)}{2} = \pi
\end{align*}$$
Why did I use $\beta$ function?: Master Theorem
|
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|
Minimize function $\sqrt{x^2-4x+5}+\sqrt{4+x^2}$ Suppose I want to find
$$
\min f(x) = \min(\sqrt{x^2-4x+5}+\sqrt{4+x^2})
$$
I'd start with computing derivative and set it to $0$
$$
f'(x) = \frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}}
$$
Then
$$
\frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}} = 0
$$
$$
(x-2)\sqrt{4+x^2} + x\sqrt{x^2-4x+5}=0
$$
My first instinct is to rewrite it as
$$
(2-x)\sqrt{4+x^2} = x\sqrt{x^2-4x+5}
$$
and square both sides. The thing is, I get
$$
3x^2 -16x +16=0
$$
with $x_1 = 4/3$ and $x_2 = 4$. So $f'(x_1^-) <0$ and $f'(x_1^+) > 0$ thus in $x = 4/3$ there is a minimum of our function.
Question: is it valid to square both sides of an equation as I did above? Is there a possibility that because of that I lose or introduce a solution that shouldn't exist and might mess everything up?
|
You can square, but only after ensuring that you're equating numbers that are either both positive or both negative.
In your case you have
$$\textstyle
(2-x)\sqrt{4+x^2}=x\sqrt{x^2-4x+5}
$$
The square roots give no problem: they are both defined anywhere and positive. The sign is determined by the other factors.
If $2-x\ge0$, we also need $x\ge0$, that is, $0\le x\le2$.
If $2-x\le 0$, we also need $x\le0$, but this cannot happen.
Thus we can square, but we must discard roots that are not in $[0,2]$.
The equation you get is indeed $3x^2-16x-16=0$, with roots $4/3$ and $4$. The latter must be discarded.
By the way, the problem can be solved in a different way. Your function is
\begin{align}
f(x)
&=\textstyle\sqrt{x^2+4}+\sqrt{(x-2)^2+1^2} \\[6px]
&=\textstyle\sqrt{(x-0)^2+(0-2)^2}+\sqrt{(x-2)^2+(0-(-1))^2}
\end{align}
which is the sum of the distances of a point on the line $x=0$ from the points $(0,2)$ and $(2,-1)$.
You don't even need to compute the point of minimum, because the minimum is the distance between the two points, hence
$$\textstyle
\sqrt{(0-2)^2+(2+1)^2}=\sqrt{4+9}=\sqrt{13}
$$
If you want to find the point of minimum, the line joining the two points has equation
$$
\frac{y+1}{2+1}=\frac{x-2}{0-2}
$$
The only point $(x,0)$ on the $x$-axis satisfies
$$
\dfrac{1}{3}=\dfrac{2-x}{2}
$$
hence $x=4/3$.
|
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|
Usual method of partial fractions decomposition over the reals seems to fail. I assumed that it would be straightforward to find the partial fraction decomposition over the reals of the rational function $$f(x) = \frac{1}{(x^2 +1)^2}.$$ However, when I try what I thought would be the usual method of writing it as $$\frac{1}{(x^2+1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2+1)^2},$$ I find that the only choice of constants is $A = B = C = 0$, and $D = 1$, simply reproducing what I started with. Typically, one might think that this would be a logical approach to finding the indefinite integral of $f(x)$, but it seems to fail here. Could someone explain why this happens, or where I have made a mistake?
For the record, the integral is elementary using the substitution $x = \tan \theta$ and a trig identity in the result.
|
The method didn't "fail". Partial Fractions had already reached the point that you want it to reach, namely one in which you have to do integrals all of which are of one of the following forms:
*
*$\displaystyle \int \frac{A}{(ax+b)^k}\,dx$, $k\geq 1$. These yield to a substitution $u=ax+b$.
*$\displaystyle \int\frac{Ax}{q(x)^k}\,dx$, $k\gt 1$, $q(x)$ an irreducible quadratic; which can be solved by completing the square and doing some substitutions.
*$\displaystyle \int\frac{B}{q(x)}\,dx$ where $q(x)$ is irreducible quadratic, which can be solved by completing the square and doing some substitutions.
*$\displaystyle\int \frac{B}{(q(x))^k}\,dx$ with $k\gt 1$. These can be solved using the Reduction formulas.
The corresponding reduction formula,
$$\int\frac{dw}{(w^2+b^2)^n} = \frac{1}{2b^2(n-1)}\left(\frac{w}{(w^2+b^2)^{n-1}} + (2n-3)\int\frac{dw}{(w^2+b^2)^{n-1}}\right)$$
gives
$$F(x) = \int \frac{dx}{(x^2+1)^2} = \frac{x}{2(x^2+1)} + \frac{1}{2}\int\frac{dx}{x^2+1}= \frac{x}{2(x^2+1)} + \frac{1}{2}\arctan(x) + C.$$
(See the long discussion here; the case of an irreducible quadratic denominator raised to some power with constant numerator is item 5 towards the end).
You can verify this holds by differentiation:
$$
F'(x) = \frac{1}{2}\left( \frac{(x^2+1) - 2x^2}{(x^2+1)^2} + \frac{1}{x^2+1}\right) = \frac{1}{2}\left(\frac{1-x^2+x^2+1}{(x^2+1)^2}\right) = \frac{1}{(x^2+1)^2}.$$
|
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|
Solving $\sqrt[2k+1]{x+1}+\sqrt[2k+1]{x+2}+\sqrt[2k+1]{x+3}+...+\sqrt[2k+1]{x+n}=0$ I noticed that when graphing the function $f(x)=\sqrt[2k+1]{x+1}+\sqrt[2k+1]{x+2}+\sqrt[2k+1]{x+3}+...+\sqrt[2k+1]{x+n}$, its roots seemed to behave predictably, that is, $x=-\frac{n+1}{2}$ seems to always be the only real zero of $f(x)$. However, I'm not quite sure how to prove this. There is no way for me to expand this into a polynomial with integer roots so instead I tried to give it a bound using Am-Gm.
This gives, $f(x)≥n\big((x+1)(x+2)...(x+n)\big)^{\frac{1}{2k+1}}$. Setting $f(x)=0$ gives $\big((x+1)(x+2)...(x+n)\big)^{\frac{1}{2k+1}}≤0$ which implies that $(x+1)(x+2)...(x+n)≤0$ since $2k+1$ is odd and preserves the signs. However, I can't deduce anything from this result.
I also found $f'(x)=\frac{1}{2k+1}\Big((x+1)^{\frac{-2k}{2k+1}}+(x+2)^{\frac{-2k}{2k+1}}+...+(x+n)^{\frac{-2k}{2k+1}}\Big)$ and
$f''(x)=\frac{-2k}{(2k+1)^2}\Big((x+1)^{-\frac{4k+1}{2k+1}}+(x+2)^{-\frac{4k+1}{2k+1}}+...+(x+n)^{-\frac{4k+1}{2k+1}}\Big)$. However, nothing significant jumps out about these two results that I could use.
So could someone help identify a way to "naturally" find the root, namely, not an educated guess that seems to work, and prove that it is the only real root which is what desmos' graph suggests?
|
Since $g(x)=\sqrt[2k+1]{x}$ is a strictly increasing function, therefore $f(x)$ is also a strictly increasing function. This means that, if there exist a real solution, then there is only one real solution.
We know that, the function $g(x)$ is odd. Using this property and setting $x+1=-(x+n)$ leads to:
$$\begin{align}&x+1=-(x+n)\\
&x+2=-(x+(n-1))\\
&x+3=-(x+(n-2))\\
&\cdots\cdots\cdots\\\
&x+n=-(x+1)\end{align}$$
This implies,
$$\begin{align}&\sqrt [2k+1]{x+1}=-\sqrt [2k+1]{(x+n)}\\
&\sqrt [2k+1]{x+2}=-\sqrt [2k+1]{(x+(n-1))}\\
&\sqrt [2k+1]{x+3}=-\sqrt [2k+1]{(x+(n-2))}\\
&~~~\cdots\cdots\cdots\\\
&\sqrt [2k+1]{x+n}=-\sqrt [2k+1]{(x+1)}\end{align}$$
Summing up the equations side-by-side, we have:
$$f(x)=-f(x)\implies f(x)=0.$$
Therefore, the only real solution is $$\begin{align}&x+1=-(x+n)\\
\implies &x=-\frac {n+1}{2}.\end{align}$$
|
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|
Calculus limit question If $f:(0,+\infty)\rightarrow\mathbb{R}$ continuous
$$f(x)=\lim_{n\to\infty}\frac{3^n(x^3+ax^2+3x+1)+x^n(2x^3+6x^2+6x+2)}{2x^n+3^n} \hspace{0.5em} \forall x\in(0,3)\cup(3,+\infty) ,a\in \mathbb{R}$$Prove $f(x)=(x+1)^3 \hspace{0.2em},x>0 $
Ok so $$\Leftrightarrow f(x)=\lim_{n\to\infty}\frac{3^n(x^3+ax^2+3x+1)+2x^n(x+1)^3}{3^n+2x^n}$$
I've tried picking separate cases for $x>3$ while factoring $x^n$ and I got $f(x)=(x+1)^3 $ but for $x\in(0,3)$ factoring $3^n$ i get $f(x)=x^3+ax^2+3x+1$. What am I missing?
|
Something is wrong here. For $a=0$ and $x=1$ we have
$$f(1)=\lim_{n\to\infty}\frac{3^n(1+3+1)+1^n(2+6+6+2)}{2\cdot 1^n+3^n}=\lim_{n\to\infty}\frac{5\cdot 3^n+16}{3^n+2}=5\neq 8=(1+1)^3$$
The formula only works if $a=3$.
EDIT: As @enzotib pointed out, the function $f(x)$ is assumed to be continuous. Really, this exercise should have been phrased: Given that $f(x)$ is continuous and defined by
$$f(x)=\lim_{n\to\infty}\frac{3^n(x^3+ax^2+3x+1)+2x^n(x+1)^3}{3^n+2x^n}$$
for some $a\in\mathbb{R}$, find $a$ and show that with this choice of $a$ $f(x)=(x+1)^3$.
Then what you are missing is showing that
$$\lim_{x\to 3^{-}}f(x)=37+9a$$
$$\lim_{x\to 3^{+}}f(x)=64$$
This lets you find $a$ and complete the problem.
|
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|
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead anywhere. My actual approach, which I will post as an answer below, uses the law of Cosines. Please share your own approaches especially if they use a different method!
|
We can use Law of Tangents
Let $t_1 = \tan\frac{A-B}{2}$
Let $t_2 = \tan\frac{A+B}{2}$
$\displaystyle \cos(A-B) = \frac{7}{8} = \frac{1-t_1^2}{1+t_1^2}
\quad → t_1^2 = \frac{8-7}{8+7} = \frac{1}{15}$
$\displaystyle \frac{\tan\frac{A-B}{2}}{\tan \frac{A+B}{2}}
=\frac{a-b}{a+b}$
$\displaystyle \frac{t_1}{t_2} = \frac{5-4}{5+4} = \frac{1}{9}
\qquad\qquad\quad → t_2^2 = 81×t_1^2 = \frac{27}{5}$
$\displaystyle \cos(C) = -\cos(A+B) = -\frac{1-t_2^2}{1+t_2^2}
= \frac{27-5}{27+5} = \frac{11}{16}$
|
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|
Are there any other fraction-to-decimal conversions like $\frac{5}{2} = 2.5$? I noticed that to convert 2.5 to a fraction, it's the number after the decimal divided by the number before the decimal. I wondered if there are any others like these, and I cannot find any nor prove that they don't exist.
So the problem, explicitly, is: are there any pairs $(a,b)$ of positive integers, besides $(2,5)$ such that
$\frac{b}{a} = a + \frac{b}{10^{\lfloor\text{log}_{10}b\rfloor+1}}$
|
I don't have a complete answer to the problem, but I have shown that $(2, 5)$ is the only coprime pair.
I'll write $d=10$ in the hopes of generalizing it to other $d\in\mathbb{Z}$ with $d\ge 2$.
Multiplying by $d^{\lfloor \log_{d}(b)\rfloor+1}a$ makes it
$$d^{\lfloor \log_{d}(b)\rfloor+1}b=a^2d^{\lfloor \log_{d}(b)\rfloor+1}+ab$$
This then means that $$d^{\lfloor\log_{d}(b)\rfloor+1}(b-a^2)=ab$$
and that we want to find solutions to $d^n(b-a^2)=ab$ where $n\ge 1$ and $d^{n-1}\le b < d^n$.
First, I'll show how to solve the more general equation $k(b-a^2)=ab$ with $k\ge 1$ and $a,b$ positive integers. The following isn't entirely original, as I use this (very useful!) tool a decent amount for this part. Note that it must be that $a<k$ since otherwise $b=\frac{ka^2}{k-a}<0$. Let $a=k-X$ and $b=Y-2k^2$ with the new restriction that $0<X<k$ and $2k^2<Y$. The equation then simplifies to $kX^2-XY=-k^3\implies X(Y-kX)=k^3$. So $X$ is a small divisor of $k^3$ and $Y=\frac{k^3}{X}+kX$ (as long as $X\not=k$, $Y$ will be more than $2k^2$).
With this question, $k=d^n$. Let $X$ be a divisor of $d^{3n}$ with $0<X<d^n$, and let $Y=\frac{d^{3n}}{X}+d^nX$. Then let $a=d^n-X$ and $b=Y-2\cdot d^{2n}$.
From here, we can show that $(2, 5)$ is the only solution with $\gcd(a, b)=1$ for $d=10$. We want $\gcd\left(10^n-X, \frac{10^{3n}}{X}+10^nX-2\cdot10^{2n}\right)=1$. If $X$ was a multiple of $5$, then $10^n-X$ would also be a multiple of $5$, and the only way $b$ wouldn't be a multiple of $5$ is if $\frac{10^{3n}}{X}$ wasn't, which means that $X=5^{3n}2^m$ with $m\ge 0$, but since $X$ needs to be smaller than $10^n$, this wouldn't work. If $X$ was just $1$, then $b\ge 10^n$, so $X$ must be a power of $2$. Moreover, it needs to be $2^{3n}$ since otherwise $b$ would be a multiple of $2$. With $X=2^{3n}$, $b=5^{3n}+10^{n}\cdot2^{3n}-2\cdot10^{2n}$ which must be smaller than $10^n$. This only happens when $n$ is less than approximately $1.17$, which in the positive integers, is only $n=1$. Then $n=1$ yields $a=2$ and $b=5$.
If we don't require $\gcd(a, b)=1$, we can continue. We want $d^{n-1}\le b < d^n$, and if we plug in $Y-2\cdot d^{2n}$ for $b$ and $\frac{d^{3n}}{X}+d^nX$ for $Y$ and simplify, we get the following inequality,
$$\left(d^{-1}+2\cdot d^{n}\right)X\le d^{2n}+X^2<\left(1+2\cdot d^{n}\right)X$$
along with the restriction of $0<X<d^n$. Letting $r=d^n$, this means that $X^2-\left(\frac{1}{d}+2r\right)X+r^2\ge0$ and $X^2-(1+2r)X+r^2<0$ and $0<X<r$. Solving these quadratic inequalities yields that $$X+\sqrt{\frac{X}{d}}\le r < X+\sqrt{X}$$ or equivalently $\frac{1+2r-\sqrt{4r+1}}{2}<X\le\frac{d^{-1}+2r-\sqrt{d^{-2}+4r\cdot d^{-1}}}{2}$. This condition and the fact that $X$ divides $d^{3n}=r^3$ determines what $X$ can be. Although $X=r$ would be a divisor of $r^3$, it wouldn't work since $\frac{d^{-1}+2r-\sqrt{d^{-2}+4r\cdot d^{-1}}}{2}<r$. I think (but am not sure!) that there'll be other solutions (and I might write a program to check if there are), but they will likely have very large values of $n$. For $d=10$, it also might be easier to generate $X$ of the form $2^m5^p$ and see if there's a valid $n$.
Edit: From $X+\sqrt{\frac{X}{d}}\le d^n < X+\sqrt{X}$, we can get that $\log_d\left(X+\sqrt{\frac{X}{d}}\right) \le n < \log_d(X+\sqrt{X})$. The difference between the bounds is less than $1$ for all $d, X>0$, so there is at most one integer in that range. Let $n=\lceil \log_d(X+\sqrt{X})\rceil-1$ so that it is the closest integer that satisfies the upper bound. We then want to see whether $\log_d\left(X+\sqrt{\frac{X}{d}}\right) \le n$ or equivalently $$X\le\frac{d^{-1}+2\cdot d^n-\sqrt{d^{-2}+4\cdot d^n\cdot d^{-1}}}{2}$$ under the restriction $X|d^{3n}$. In the case of $d=10$, this means that $X=2^m 5^p$ with $m,p\le 3n$.
|
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Finding the smallest $k$ such that $a^3+b^3+c^3+kabc\leq\frac16(k+3)(a^2(b+c)+b^2(a+c)+c^2(a+b))$, where $a,b,c$ are sides of a triangle It is known that $a, b, c$ are the sides of the triangle. Determine the smallest value of $k$, so that
$$a^3+b^3+c^3+kabc\leq\frac {k+3}{6} (a^2(b+c)+b^2(a+c)+c^2(a+b))$$
My working:
$$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (a^2(b+c)+b^2(a+c)+c^2(a+b))$$
$$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)$$
$$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (a^2b+b^2a+a^2c+c^2a+b^2c+c^2b)$$
$$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (ab(a+b)+ac(a+c)+bc(b+c))$$
Can someone help me, I only process the data on the right side? Thank you
|
Since $a,b$ and $c$ are sides of a triangle, we are allowed to set $a=x+y$, $b=y+z$ and $c=x+z$. Now, let's define:
$$A=x^3+y^3+z^3 \\ B=x^2y+xy^2+z^2x+zx^2+y^2z+yz^2 \\C=xyz.$$
Note that by the Schur's inequality we already know that: $A+3C\ge B.$ Moreover it is not hard to see (by Muirhead's inequality) that $2A\ge B.$
Then, writing the inequality in terms of $A,B$ and $C$, we should have:
$$2A+3B+k(B+2C) \le \frac{k+3}{6} (2A+5B+12C). $$
By simplifying, we should have:
$$(\frac {k+3}{6})B\le (\frac {k-3}{3})A+6C.$$
If $k\ge 9$, then:
$$(\frac {k-3}{3})A+6C=2A+6C+(\frac{k-9}{3})A\ge 2B+ (\frac{k-9}{6})B=(\frac{k+3}{6})B.$$
Therefore the inequality holds if $k\ge9$.
Now, assume $k\lt 9$; then $\frac {k+3}{6} \gt \frac {k-3}{3}.$ Let's put $x=1, y=1, z=\epsilon$.
We should have:
$$(\frac{k+3}{6})(2+2\epsilon^2+2\epsilon)\le (\frac{k-3}{3})(2+\epsilon^3)+6\epsilon,$$
which is impossible for $\epsilon$ small enough.
|
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Why are there complex numbers in the exact solution of $\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}$? Knowing that
\begin{align}
\cot(z)=\frac{1}{z}-2z\cdot\sum_{n=1}^{\infty} \dfrac{1}{\pi^2n^2-z^2}
\end{align}
we can easily calculate the value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2+1}
\end{align}
by just plugging in $z=i\pi$. Therefore:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2+1}=\frac{1}{2}\cdot\left(\pi \coth(\pi)-1\right)
\end{align}
In this case, the "$i's$" cancel eachother out nicely.
I also wanted to calculate the exact value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
with an equal approach. After partial fraction decomposition:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}=\frac{1}{2i}\left(\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\dfrac{1}{n^2+i}\right)=\frac{1}{2i}\left(\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\sum_{n=1}^{\infty}\dfrac{1}{n^2+i}\right)
\end{align}
By plugging in $z=\sqrt{i}\pi$ and $z=i\sqrt{i}\pi$, I arrive at:
\begin{align}
\frac{1}{2i}\left[\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\sum_{n=1}^{\infty}\dfrac{1}{n^2+i}\right]&=\frac{1}{2i}\left[\frac{1}{2i}-\frac{\pi}{2\sqrt{i}} \cdot \cot(\sqrt{i}\pi)-\left(-\frac{1}{2i}-\frac{1}{2i\sqrt{i}}\cdot \cot(i\sqrt{i}\pi)\right)\right]=\\&=\frac{1}{4}\left[-2+\pi\sqrt{i}\cot(\sqrt{i}\pi)-\frac{\pi}{\sqrt{i}}\cot(i\sqrt{i}\pi)\right]
\end{align}
In this case I can't get completely rid of the "$i's$" and the exact value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
is a complex number. But the answer must obviously be a real number.
Wolfram Alpha gets:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}\approx0.57848+0.\times10^{-22}\,i
\end{align}
You now can argument that the imaginary part is negligible and so the value is a real number, but that doesn't satisfy my question, because there always will be a imaginary part in the answer.
If anyone can explain to me, why it is the case, that the exact value for
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
is a complex number, I would be very glad.
|
Addendum to Somos' answer: by expanding $\cot(x+iy)$ via trig identities, one can determine the real and imaginary parts of $\cot(z)$, allowing the answer to be expressed purely in terms of real-valued terms:
$$-\frac{1}{2}-\frac{\pi \sin \left(\sqrt{2} \pi \right)}{2 \sqrt{2} \left(\cos
\left(\sqrt{2} \pi \right)-\cosh \left(\sqrt{2} \pi \right)\right)}-\frac{\pi \sinh
\left(\sqrt{2} \pi \right)}{2 \sqrt{2} \left(\cos \left(\sqrt{2} \pi \right)-\cosh
\left(\sqrt{2} \pi \right)\right)}$$
Cleaning up the result a bit, with $u = \sqrt{2}\pi$ this can be written as:
$$-\frac{1}{2} + \frac{\pi}{2\sqrt{2}}\frac{\sin(u)+\sinh(u)}{\cosh(u)-\cos(u)}$$
|
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Prove or disprove the inequality if $a,b,c>0$, $a \geq b+c$. Prove or disprove the inequality
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 7abc$$ if $$a,b,c>0, a \geq b+c.$$
I thought to use this evaluation:
$$a^2b+b^2c+c^2a \geq 3abc.$$
So we have:
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 3abc+3abc=6abc,$$ which is obvious that $$6abc<7abc.$$
Is it right? I'm embarrassed that in my solution I did not have to use the condition that $$a \geq b+c.$$
Any hint would help a lot. thanks!
|
Dividing by $abc > 0,$ the relation to be shown is $$ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \overset?\ge 7$$
Set $a = b+c + d$ for $d \ge 0.$ We then need to argue that for $b,c>0, d\ge 0,$
$$ \frac{b+c}{b+c+d} + \frac{2c+d}{b} + \frac{2b+d}{c} \overset?\ge 5$$
The LHS is invariant under scaling, so wlog set $b = 1.$ We need to show that for $c > 0, d \ge 0,$ $$ J(c,d) := \frac{1 + c}{1 + c + d} + 2c + d + \frac{2 + d}{c} \overset?\ge 5.$$
For a fixed value of $c$, the derivative with respect to $d$ is $$ \partial_d J = 1 + \frac{1}{c} -\frac{(1+c)}{(1 + c + d)^2} \ge 1 + \frac{1}{c} - \frac{1}{1+c} > 0. $$ This means that for any fixed $c$, $J$ attains its minimum at $d = 0.$ So we need ot argue that $J(c,0) \ge 5$ for any $c> 0,$ i.e. that $$ 1 + 2c + \frac2c \overset?\ge 5 \iff c + \frac1c \overset?\ge 2.$$ But this holds via the AM-GM inequality. We conclude that the original inequality is true.
|
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|
$ \text{sec}^2x + 3\text{cosec}^2x =8 $ Solving this trigonometric equation. I have found two different methods of solving this trigonometric equation :
$$ \text{sec}^2x + 3\text{cosec}^2x =8
$$
But these methods give different answers.
Solution 1
$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$
$$\implies \frac{1}{\text{cos}^2x} + \frac{3}{\text{sin}^2x} =8
$$
$$\implies 3{\text{cos}^2x} + {\text{sin}^2x} =8(\text{cos}^2x ×\text{sin}^2x)$$
$$\implies 2{\text{cos}^2x} + 1 =8(\text{cos}^2x ×(1-\text{cos}^2x))$$
$$\implies 8{\text{cos}^4x} -6\text{cos}^2x +1 =0$$
$$\implies (2\text{cos}^2x-1)(4\text{cos}^2x-1)=0
$$
$$ \implies x= 2nπ±\frac{π}{4},2nπ±(π-\frac{π}{4}),2nπ±\frac{π}{3},2nπ±(π-\frac{π}{3}),
$$
Solution 2
$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$
$$ \implies 1+\text{tan}^2x + 3 + 3\text{cot}^2x =8 $$
$$\implies {\text{tan}^2x} + \frac{3}{\text{tan}^2x} =4
$$
$$ \implies \text{tan}^4x -4\text{tan}^2x + 3 =0 $$
$$ \implies (\text{tan}^2x -1)(\text{tan}^2x-3) =0 $$
$$ \implies x=nπ±\frac{π}{3},nπ±\frac{π}{4} $$
Now my question is that which one should I reject and why ?
|
Using $\cos^{2}x + \sin^{2}x = 1$, from your first solution method and first factor, $2\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{2}$, so $\sin^{2}x = 1 - \frac{1}{2} = \frac{1}{2}$ and, thus, $\tan^{2}x = \frac{1/2}{1/2} = 1$. The second factor gives $4\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{4}$, so $\sin^{2}x = 1 - \frac{1}{4} = \frac{3}{4}$, which means $\tan^{2}x = \frac{3/4}{1/4} = 3$. Note these match the $\tan^{2}x$ values obtained from your second solution method's factorization.
Your first two sets of values of $x= 2n\pi\pm\frac{\pi}{4},2n\pi\pm(\pi-\frac{\pi}{4})$ can be combined into just $n\pi\pm\frac{\pi}{4}$ (since the $+$ part of $2n\pi\pm(\pi-\frac{\pi}{4})$ is $(2n+1)\pi-\frac{\pi}{4}$, i.e., all odd values of $n$ with $n\pi-\frac{\pi}{4}$, with the $-$ part being $(2n-1)\pi+\frac{\pi}{4}$, i.e., all odd values of $n$ with $n\pi+\frac{\pi}{4}$), with this just being the solutions to $\cos^{2}x = \frac{1}{2} \; \to \; \cos x = \pm\frac{1}{\sqrt{2}}$. For the second set of solutions, $\cos^{2}x = \frac{1}{4} \; \to \; \cos x = \pm\frac{1}{2}$, which has solutions of $n\pi\pm\frac{\pi}{3}$ (your second set of values have the same basic issue as with the first set, i.e., they can be combined into just the one set of values I give here, and which you also gave in your second solution).
Thus, the sets of $x$ values from each of your $2$ solution techniques actually match each other.
|
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|
Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integrals chapter.
I did the substitution and i obtaind $-18\int \frac{{t^{2}}}{|(t^{2}-1)^{3}|}\,dt$ . At this point i thought that since the substitution has to be an invertible function, i may consider a restrinction to cancel the absolute value.
I checked on wolfram alpha both integrals but it gives different results. Wolfram solution:
$$\int \sqrt{(x^2 + x - 2)}\,dx = \frac14 (2 x + 1) \sqrt{(x^2 + x - 2)} - \frac98 \log\left(2 \sqrt{(x^2 + x - 2)} + 2 x + 1\right) + c$$
Any help?
The purpose of the exercise is to trasform the irrational function into a rational one and use the Hermite's equation to solve it
|
First, reduce to a simpler square root
\begin{align*}
\int \sqrt{x^2+x-2} dx &= \int \sqrt{\left(x+\frac{1}{2}\right)^2-\frac{9}{4}} \,dx = \\
&= \frac{3}{2}\int \sqrt{\frac{4}{9}\left(x+\frac{1}{2}\right)^2-1} \,dx = \\
&= \frac{3}{2}\int \sqrt{\left(\frac{2}{3}x+\frac{1}{3}\right)^2-1} \,dx = \\
&= \frac{9}{4}\int \sqrt{t^2-1} \,dt
\end{align*}
Then use this know result
\begin{align*}
I = \int \sqrt{t^2-1} \,dt &= t\sqrt{t^2-1} -\int t\frac{2t}{2\sqrt{t^2-1}} \,dt = \\
&= t\sqrt{t^2-1} -\int \frac{t^2-1+1}{\sqrt{t^2-1}} \,dt = \\
&= t\sqrt{t^2-1} -\int \sqrt{t^2-1} \,dt -\int \frac{1}{\sqrt{t^2-1}} \,dt= \\
&= t\sqrt{t^2-1} -\int \sqrt{t^2-1} \,dt -\operatorname{arccosh} t = \\
&= t\sqrt{t^2-1} -I -\operatorname{arccosh} t
\end{align*}
from which
\begin{align*}
I = \int \sqrt{t^2-1} \,dt &= \frac{1}{2}t\sqrt{t^2-1} -\frac{1}{2}\operatorname{arccosh} t +C = \\
&= \frac{1}{2}t\sqrt{t^2-1} -\frac{1}{2}\log\left(t+\sqrt{t^2-1}\right) +C
\end{align*}
Then come back to the original integral
\begin{align*}
& \int \sqrt{x^2+x-2} dx = \frac{9}{4}\int \sqrt{t^2-1} \,dt = \\
& \qquad = \frac{9}{8}t\sqrt{t^2-1} -\frac{9}{8}\log\left(t+\sqrt{t^2-1}\right) +C = \\
& \qquad = \frac{9}{8}\left(\frac{2}{3}x+\frac{1}{3}\right)\sqrt{\left(\frac{2}{3}x+\frac{1}{3}\right)^2-1} -\frac{9}{8}\log\left(\left(\frac{2}{3}x+\frac{1}{3}\right)+\sqrt{\left(\frac{2}{3}x+\frac{1}{3}\right)^2-1}\right) +C = \\
& \qquad = \frac{1}{4}(2x+1)\sqrt{x^2+x-2} -\frac{9}{8}\log\left(2x+1+2\sqrt{x^2+x-2}\right) +C \\
\end{align*}
|
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|
If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, $\frac{\beta^2+\gamma^2}{\alpha^2}$,$\frac{\alpha^2+\gamma^2}{\beta^2}$,$\frac{\beta^2+\alpha^2}{\gamma^2}$.
My solution goes like this:
We consider, $a=\frac{\alpha^2+\beta^2}{\gamma^2}$,$b=\frac{\beta^2+\gamma^2}{\alpha^2}$,$c=\frac{\gamma^2+\alpha^2}{\beta^2}$. Now, $$\alpha+\beta+\gamma=0,\alpha\beta+\beta\gamma+\gamma\alpha=q,\gamma\alpha\beta=-r$$ and hence,$\alpha^2+\beta^2+\gamma^2=-2q$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{-2q-\gamma^2}{\gamma^2}=\frac{-2q}{\gamma^2}-1$ or $\gamma^2=\frac{-2q}{a+1}$. Also, $$a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-\frac{2\alpha\beta\gamma}{\gamma^2\gamma}=1+\frac{2r}{\frac{-2q}{a+1}\gamma}=1-\frac{r(a+1)}{q\gamma}$$ and hence,$\gamma=\frac{r(a+1)}{q(1-a)}$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-2\frac{\alpha\beta\gamma}{\gamma^3}=1+\frac{2r}{\gamma^3}$. Thus, $\gamma^3=\frac{2r}{a-1}$. Now, we have, $\gamma^3+q\gamma+r=0$. Thus, $\frac{2r}{a-1}+\frac{r(a+1)}{(1-a)}+r=0$, which implies $a^2-a-2=0$. Thus, the required equation is $x^2-x-2=0$.
Is the above solution correct? If not, then where is it going wrong? I dont get where is the mistake occuring?
|
HINT.-From Vieta we get $$\alpha^2+\beta^2+\gamma^2=-2q\\(\alpha\beta)^2+(\alpha\gamma)^2+(\beta\gamma)^2=q^2\\(\alpha\beta\gamma)^2=r^2$$ Now instead of $a,b,c$ we consider $a+1=\dfrac{-2q}{\alpha^2},b+1=\dfrac{-2q}{\beta^2}$ and
$c+1=\dfrac{-2q}{\gamma^2}$ so the equation in $a+1,b+1,c+1$ is
$$\left(X+\dfrac{2q}{\alpha^2}\right)\left(X+\dfrac{2q}{\beta^2}\right)\left(X+\dfrac{2q}{\gamma^2}\right)=0$$ which can be easily simplified to
$$r^2X^3+2q^3X^2-8q^3x+8q^3=0$$ or better
$$r^2X^3+2q^3(X-2)^2=0$$
What remains is to go from roots to roots minus $1$. We are done.
|
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"url": "https://math.stackexchange.com/questions/4613589",
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|
Integrating $\frac{x^3}{\sqrt{x^2 + 4x + 6}}$ Question:
Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx $.
My attempt:
$\begin{align} \int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx & = \int \frac{x^3}{\sqrt{(x+2)^2 + 2}}\ dx \\& \overset{(1)}= \int \frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)\ }{\sqrt{2\tan^2(t) + 2}}\ dt\\& = \frac{1}{\sqrt{2}}\int\frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)}{\sec(t)}\ dt\\& = \int(\sqrt{2} \tan(t) - 2 )^3 \sec(t)\ dt\\& = \int -8 \sec(t) + 2 \sqrt2 \tan^3(t) \sec(t) - 12\tan^2(t) \sec(t)\\&\qquad + 12 \sqrt2 \tan(t) \sec(t)\ dt\\\\& = -8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} \\
&\qquad + 2\sqrt{2} \int \tan^3(t) \sec(t) \ dt - 12\int\tan^2(t) \sec(t) \ dt\end{align}$
Now both of these integrals can be evaluated using some sort of substitution and integration by parts rule.
This would give us,
$$\boxed{-8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} + 2\sqrt{2}\left[\frac{\sec^3(t)}{3} - \sec(t)\right] - 12\left[\frac{-1}{2} \ln|\sec t + \tan t| + \frac12 \sec t \tan t \right] + C}$$
$(1)$ Here I've made a substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$.
Wolframalpha gives answer as $$\boxed{\frac{1}{3}(x^2 - 5x + 18) \sqrt{x^2 +4x + 6} - 2 \sinh^{-1}\left(\frac{x+2}{\sqrt{2}}\right) + C}$$
How would I simplify my answer equal to this? Undoing my substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$ is also not an easy task I think.
|
For indefinite integral of the form
$$I_n=\int \frac{x^n}{\sqrt{x^2 + bx + c}}\ dx
$$
it is advised that $I_n$ be reduced first to $I_0$ before any substitution. This is achieved by the reduction formula below with $f(x)=x^2+bx+c$
$$\int \frac{f’(x)^{n}}{\sqrt{f(x)}}dx= K_n = \frac2nf’(x)^{n-1}\sqrt{f(x)}+\frac{n-1}n (b^2-4c)K_{n-2}
$$
Thus, apply it to the integral to obtain
\begin{align}
&\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx \\
=&\ \frac18\int \frac{(2x+4)^3-12(2x+4)^2+48(2x+4)-64}{\sqrt{x^2 + 4x + 6}}\ dx \\
=&\ \frac13(x^2-5x+18)\sqrt{x^2 + 4x + 6}-2\int \frac{1}{\sqrt{x^2 + 4x + 6}}\ dx
\end{align}
|
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|
Find the intersection points of two circles Find the intersection points of the circles $$k_1:(x-4)^2+(y-1)^2=9\\k_2:(x-8)^2+(y+4)^2=100$$
The intersections point (if they exist) will satisfy the equations of both the circles, so we can find their coordinates by solving the system $$\begin{cases}(x-4)^2+(y-1)^2=9\\(x-8)^2+(y+4)^2=100\end{cases}\iff\begin{cases}x^2-8x+y^2-2y=-8\\x^2-16x+y^2+8y=20\end{cases}$$
Substracting these equations, gives $4x-5y=-14\Rightarrow x=\dfrac{5y-14}{4}$. Substituting into the first, I got (if I didn't mess up the calculations) $$41y^2-332y+766=0$$ which has no real solutions (negative discriminant). Therefore the system has no solutions as well, so the circles don't intersect. Is there something else we can use in order to conclude that they do not intersect? Maybe something which requires less calculations? The center of the first is $O_1(4;1)$, the center of the second cirlce is $O_2(8;-4)$ and their radii are $r_1=3$ and $r_2=10$, respectively, if this somehow helps. The distance $O_1O_2=\sqrt{16+25}=\sqrt{41}$ and $r_1+r_2=13=\sqrt{169}$, but I am not sure how to interpret these findings.
|
Geometric point of view: The radical axis of two circles passes through the intersection points of the circles when they intersect. If we show that the distance of the centers of the circles to that line is larger than their radii, we're done.
You found the radical axis. It is $L:4x-5y+14=0$. Now, $h_1=d(L,O_1)=\frac{|4\times4-5\times1+14|}{\sqrt{4^2+5^2}}=\frac{25}{\sqrt{41}}>\frac{25}{7}>3=R_1$ and similarly $h_2=d(L,O_2)==\frac{|4\times8-5\times(-4)+14|}{\sqrt{4^2+5^2}}=\frac{66}{\sqrt{41}}>10=R_2$, since $66^2=4356>4100.$
|
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|
Proving certain properties for Babylonian method The Babylonian method for approximating square roots is divided into three steps:
*
*Guess an initial approximation $a$ of $ {\sqrt N}$, where $a$ and $N$ are postive rational numebrs and $N$ is not the square of any rational numbers.
*Let $c=\frac{N-a^2}{2a}$.
*Now $a+c$ is a new approximation of $ {\sqrt N}$.
The question is to show $a+c>\sqrt N$.
I think that since the problem did not inform us of the relationship between $a$ and ${\sqrt N}$, so no matter what number we pick for $a$ does not affect the fact that $a+c>\sqrt N$.
Then we will have
$a+c=a+\frac{N-a^2}{2a}=\frac{a^2+N}{2a}$, if $a$ >$\sqrt N$, then we will have $\frac{a^2+N} {2a} >\frac{2N}{2a}$, $\frac{2N}{2a}<\frac{2N}{2\sqrt N}=\sqrt N$. However, how do I find the relation between $\frac{a^2+N} {2a} $ and $\frac{2N}{2\sqrt N}$? Any help would be greatly appreciated.
|
\begin{align*}
a+c &= a + \frac{N - a^2}{2a}
= \frac{N + a^2}{2a}
= \frac{(\sqrt{N} - a)^2 + 2a\sqrt{N}}{2a}\\
&= \frac{(\sqrt{N} - a)^2}{2a} + \sqrt{N} \geq \sqrt{N}
\end{align*}
|
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|
prove $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$ $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$
How to deal with this problem?
Observing that when $x=\pi/2$, the above inequality becomes equality.
Firstly, denote $f(x)=(\sin x)^{-2}-x^{-2}$ and then take derivative of $f(x)$. We have
$$-2(\sin x)^{-3}\cos x+2x^{-3}$$ Next, how to analysis the sign of $f'(x)$?
Any hints are wellcome! Thanks!
|
It's related to :
$$|\frac{1}{\sin(x)}-\frac{1}{x}|\leq 1-\frac{2}{\pi}$$
Called Prestin's inequality
Use the fact :
$$\left|\frac{1}{\sin^{2}(x)}-\frac{1}{x\sin\left(x\right)}\right|-\left(1-\frac{2}{\pi}\right)\geq (\sin x)^{-2}-x^{-2}-\left(1-\frac{4}{\pi^{2}}\right)$$
And :
$$\left|\frac{1}{\sin(x)}-\frac{1}{x}\right|-\left|\sin\left(x\right)\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or using Jordan's inequality :
$$\left|\frac{1}{\sin(x)}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or using Taylor's series for $x\in(0,1]$ :
$$\left|\frac{1}{x-\frac{1}{6}x^{3}}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or :
$$\frac{\left|x\right|\left(\frac{\pi^{2}}{\left|x^{2}-6\right|}-2\pi+4\right)}{\pi^2}\leq 0$$
Now on $x\in[1,\pi/2]$ :
$$\left|\frac{1}{1-1/2(x-\pi/2)^{2}}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or :
$$-\frac{\left((2x-\pi)(4\pi x^{3}-8x^{3}-2\pi^{2}x^{2}+4\pi x^{2}+2\pi^{2}x-8\pi x+16x-\pi^{3}+8\pi)\right)}{(\pi^{2}x(4x^{2}-4\pi x+\pi^{2}-8))}\leq 0$$
|
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|
Please explain how to solve limit. I know the answer but how to explain it? Problem:
$a@b = \frac{a+b}{ab+1}$. Solve limit:
$\lim_{n \to \infty}(2@3@...@n)$.
I've tried to solve this problem by just calculating:
$$2 @ 3 = 0.714$$
$$2 @ 3 @ 4 = 1.222$$
$$2 @ 3 @ 4 @ 5 = 0.875$$
$$2 @ 3 @ 4 @ 5 @ 6 = 1.1$$
I found the pattern. The first number is less than 1, then the next is greater than 1, the next is less than 1, and so on. So the limit must be 1. But how to explain it mathematically?
I've tried to transform this:
$$(n-1)@n = \frac{2n-1}{n^2-n+1}$$
$$n@(n+1) = \frac{2n + 1}{n^2+n+1}$$
But it didn't help me to understand the method how to solve it. I think there should be a simple idea, which I don't see. I appreciate all hints.
|
$\DeclareMathOperator{@}{\operatorname@}$
As noted in a comment, the $\@$ operator is associative.
By expanding $(a\@ b\@c)$ and $(a\@b\@c\@d)$:
$$\begin{align*}
a\@b &= \frac{a+b}{ab+1}\\
(a\@b)\@c &= \frac{\frac{a+b}{ab+1}+c}{\frac{a+b}{ab+1}c+1}= \frac{a+b+c(ab+1)}{(a+b)c + ab+1} = \frac{abc + a+b+c}{ab+ac+bc+1}\\
(a\@b\@c)\@d &= \frac{\frac{abc + a+b+c}{ab+ac+bc+1}+d}{\frac{abc + a+b+c}{ab+ac+bc+1}d+1}\\
&= \frac{abc + a+b+c + (ab+ac+bc+1)d}{(abc + a+b+c)d + ab+ac+bc+1}\\
&= \frac{abc+abd+acd+bcd + a+b+c+d}{abcd + ab+ac+ad+bc+bd+cd + 1}
\end{align*}$$
Note that the numerators and denominators seem to be sums of the elementary symmetric polynomials of $2$, $3$ or $4$ variables. Define a polynomial $f_n(x)$ for $n\ge 2$,
$$\begin{align*}
f_n(x) &= (2x+1)(3x+1)\cdots (nx+1)\\
&= (2\cdot 3\cdots n)x^{n-1} + \cdots + (2+3+\cdots + n) x + 1
\end{align*}$$
(Note that, as in the question, here $n$ is in the largest operand, not the number of operands or partial terms.)
Then the denominators of the partial terms seem to be the sum of the coefficients of $x^0$ and every second term; the numerators of the partial terms seem to be the sum of the coefficients of $x^1$ and every second term.
Claim that
$$\begin{align*}
2 \@3\@\cdots \@n
&= \frac{\frac12\left[f_n(1)- f_n(-1)\right]}{\frac12\left[f_n(1)+ f_n(-1)\right]}\\
&= \frac{3\cdot4\cdots(n+1) - (-1)^{n-1}\cdot 1\cdot2\cdots(n-1)}{3\cdot4\cdots(n+1) + (-1)^{n-1}\cdot 1\cdot2\cdots(n-1)}\\
&= \frac{(n-1)!\cdot \left[\frac{n(n+1)}2+(-1)^n\right]}{(n-1)!\cdot \left[\frac{n(n+1)}2-(-1)^n\right]}\\
&= \frac{n(n+1)+2(-1)^n}{n(n+1)-2(-1)^n}
\end{align*}$$
For the base case $n=2$,
$$\begin{align*}
LHS &= 2\\
RHS &= \frac{2\cdot3+2\cdot(-1)^2}{2\cdot3-2\cdot(-1)^2} = \frac{8}4 = 2
\end{align*}$$
Assume for some integer $k\ge 2$ that the claim is true:
$$2\@3\@\cdots \@k = \frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}$$
Then for the $n=k+1$ case,
$$\begin{align*}
LHS &= 2\@3\@\cdots\@k\@(k+1)\\
&= \left[\frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}\right] \@ (k+1)\\
&= \frac{\left[\frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}\right]+(k+1)}{\left[\frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}\right](k+1)+1}\\
&= \frac{k(k+1)+2(-1)^k+\left[k(k+1)-2(-1)^k\right](k+1)}{\left[k(k+1)+2(-1)^k\right](k+1)+k(k+1)-2(-1)^k}\\
&= \frac{k(k+1)(k+2)-2(-1)^kk}{k(k+1)(k+2) + 2(-1)^kk}\\
&= \frac{(k+1)(k+2)+2(-1)^{k+1}}{(k+1)(k+2)-2(-1)^{k+1}}\\
&= RHS
\end{align*}$$
So by induction, for integers $n\ge 2$,
$$\begin{align*}
2 \@3\@\cdots \@n
&= \frac{n(n+1)+2(-1)^n}{n(n+1)-2(-1)^n}
\end{align*}$$
Taking the limit when $n\to \infty$,
$$\begin{align*}
\lim_{n\to\infty}(2 \@3\@\cdots \@n)
&= \lim_{n\to\infty}\frac{n(n+1)+2(-1)^n}{n(n+1)-2(-1)^n}\\
&= \lim_{n\to\infty}\frac{1+\frac{2(-1)^n}{n(n+1)}}{1-\frac{2(-1)^n}{n(n+1)}}\\
&= \frac{1+0}{1-0}\\
&= 1
\end{align*}$$
|
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Calculate the following integral $ \int\frac{2x+1}{x^{n+2}(x+1)^{n+2}}\ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2}+\frac7{16}\right)dx$ Hello I am trying to solve a pretty complicated integral. It is a from a set of problems, published in a monthly journal for high school students and they are exercises in preparation for a competition.
So the problem is the following: Calculate the integral $ \int \frac {2x+1}{x^{n+2}(x+1)^{n+2}} \ln\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)dx$
I first tried partial integration, integrating $\frac {2x+1}{x^{n+2}(x+1)^{n+2}}$, it is happily easy to do it and we will end up with $\int \frac {2x+1}{x^{n+2}(x+1)^{n+2}} = \frac {-1}{x^{n+1}(x+1)^{n+1}}$.
I also differentiated $\ln\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)$, ending up with $\frac {-64x^3-96x^2-96x-32}{7x^6+21x^5+71x^3+48x^2+16x}$
Now when I try to solve the integral of the product of these two, I am stuck.
I also thought about some kind of recursive relationship in terms of $n$ but I am not sure about it.
I would happily accept any help in solving this problem. Thanks in advance.
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Starting as you did, with
$$du=\frac {2x+1}{x^{n+2}(x+1)^{n+2}}\,dx\qquad \text{and}\qquad v=\log\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)$$as you wrote
$$u= -\frac {1}{(n+1)x^{n+1}(x+1)^{n+1}}$$ $$ dv=-\frac{32 (2 x+1) \left(x^2+x+1\right)}{x (x+1) \left(x^2+x+4\right) (7 x^2+7x+4)}$$
$$u\,dv=\frac {32}{n+1}\,\,\frac{ (2 x+1) \left(x^2+x+1\right)}{x (x+1) \left(x^2+x+4\right) (7 x^2+7x+4)x^{n+1}(x+1)^{n+1}}$$ Consider now that
$$A=\frac{ (2 x+1) \left(x^2+x+1\right)}{x (x+1) \left(x^2+x+4\right) (7 x^2+7x+4)}$$ can be decomposed as
$$A=-\frac{2}{x}-\frac{2}{x+1}+\frac 1{a-b}\left(\frac{2 a+1}{x-a}-\frac{2 b+1}{x-b} \right)+\frac 1{c-d}\left(\frac{2 c+1}{x-c}-\frac{2 d+1}{x-d}\right)$$ where $(a,b)$ are the complex roots of $x^2+x+4=0$ and $(c,d)$ are the complex roots of $7x^2+7x+4=0$.
So, we face by the end
$$I_1=\int \frac {dx}{ x^{n+2} (x+1)^{n+1}}=-\frac{\, _2F_1(-n-1,n+1;-n;-x)}{(n+1)x^{n+1}}$$
$$I_2=\int \frac {dx}{ x^{n+1} (x+1)^{n+2}}=-\frac{ \, _2F_1(-n,n+2;1-n;-x)}{n x^n}$$
where appears the Gaussian hypergeometric function
and four integrals of the form
$$I_k=\int \frac {dx}{x^{n+1}(x+1)^{n+1}(x-k)} \quad\quad\quad k\neq 0 \quad k \in \mathbb{C}$$
$$I_k=\frac {\frac{x F_1\left(1-n;n,1;2-n;-x,\frac{x}{k}\right)}{k^2 (n-1)}+\frac{(2 k+1) \,
_2F_1(-n,n;1-n;-x)}{k n}-\frac{(x+1)^{-n}}{n}}{(1+k)x^n }$$ where also appears the Appell hypergeometric function of two variables.
All of that seems to be complicated but as soon as you give $n$ an integer value you will have for
$$J_n=\int \frac {A}{x^{n+1}(x+1)^{n+1}}\,dx$$ things like
$$\color{blue}{J_n=\alpha_n \sqrt{7} \tan ^{-1}\left(\frac{ \sqrt{7}}{3} (2 x+1)\right)+\beta_n \sqrt{15} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{15}}\right)+}$$
$$\color{blue}{\gamma_n \log(x)+\delta_n \log(x^2+x+4)+\epsilon_n \log(7x^2+7x+4)-\frac{P_{n-1}(x)}{x^{n+1}}}$$ where $(\alpha_n,\beta_n, \gamma_n,\delta_n ,\epsilon_n)$ are rational numbers and all coefficients in the polynomials are all rational and positive.
Notice that the last integral in @user170231's answer
$$I=\int \frac{du}{\left(u-4\right)^{n+2}u}=-\frac {\, _2F_1\left(1,-n-1;-n;1-\frac{u}{4}\right) } {4 (n+1) (u-4)^{n+1} }$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determining whether $\frac{1}{2^{2}} + \frac{2}{3{^2}} + ... +\frac{n}{(n+1)^{2}}$ is a Cauchy sequence. My attempt so far: If $n \leq m$, then
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} < \frac{n+1}{(n+2)(n+1)}+\frac{n+2}{(n+3)(n+2)}+...+\frac{m}{(m+1)m} = \frac{1}{(n+2)}+\frac{1}{(n+3)}+...+\frac{1}{(m+1)} < \frac{1}{n}+\frac{1}{n}+...+\frac{1}{m} \leq \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}$.
A bit stuck here though. The hint in the book said it was not Cauchy.
Edit: If it's not Cauchy, then let me take a new approach..
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{n+1}{(n+2)^2} + \frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{1}{(m+1)^2} + \frac{1}{(m+1)^2}+...+\frac{1}{(m+1)^2}$.
The $\frac{1}{(m+1)^2}$ occurs $m-n$ times here.
|
Note that we simply have $k\geq (k+1)/2$ for any integer, so
$$
\sum_{k=1}^n \frac{k}{(k+1)^2} \geq \sum_{k=1}^n \frac{\frac{1}{2}(k+1)}{(k+1)^2} = \sum_{k=1}^n \frac{1}{2} \frac{1}{k+1} = \frac{1}{2} \sum_{k=1}^n \frac{1}{k+1}
$$
and note that the right most expression is nothing but harmonic series starting from $k=2$, and according to basic analysis results, it diverges to $\infty$, therefore the series is not Cauchy $\mathbb{R}$ (Of course you can say it is if you are consider $\mathbb{C} \cup \{\infty\}$)
|
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|
Equation of parabola passes through $4$ distinct points
Equation of axis of parabola which
passes through the point $(0,1)\ , \ (0,2)$
And $(2,0)\ ,\ (2,2)$ is
Let general equation of conic is
$ax^2+2hxy+by^2+2gx+2fy+c=0\cdots (1)$
And it represent parabola if $h^2=ab$
Parabola passes through $(0,1)$
Then put into $(1)$
$b+2f+c=0\cdots(2)$
Also parabola passes through $(0,2)$
Then $4b+4f+c=0\cdots (3)$
Also parabola passes through $(2,0)$
Then $4a+4g+c=0\cdots (4)$
Also parabola passes through $(2,2)$
Then $4a+8h+4b+4g+4f+c\cdots (5)$
$(4a+4g+c)+(4b+4f+c)+8h-c=0$
From $(2)$ and $(3)$, we get $\displaystyle h=\frac {c}{8}$
From $(2)$ and $(3)$
$\displaystyle b+2f=4b+4f\Longrightarrow 2f=-3b $
Put into $(2)$ and $(3)$
$\displaystyle b=\frac{c}{2}$ and $\displaystyle f=-\frac{3c}{4}$
And $\displaystyle h^2=ab\Longrightarrow \frac{c^2}{64}=\frac{ac}{2}\Longrightarrow a=\frac{c}{32}$
Put all into $(4)$
$\displaystyle \frac{4c}{32}+4g+c=0\Longrightarrow g=-\frac{9c}{32}$
Put all values into $(1)$
$\displaystyle \frac{c}{32}x^2+\frac{c}{4}xy+\frac{c}{2}y^2-\frac{9}{16}x-\frac{3c}{2}y+c=0$
$x^2+8xy+16y^2-18x-48y+32=0$
Buti did not know how I find axis of parabola
Please have a look
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What you've done is correct.
One way to find the axis is to write the equation as $(x+4y)^2-18x-48y+32=0,$ then you know the axis is parallel to $x+4y=0.$ The tangent at vertex is orthogonal and parallel to $4x-y.$ Now you can find where $4x-y+c$ intersects doubly to find the vertex. $$x^2+8x(4x+c)+16(4x+c)^2-18x-48(4x+c)+32=0$$ or $$289x^2+(136c-210)x+16c^2-48c+32=0$$ and this has discriminant $$(136c-210)^2-4\cdot 289\cdot (16c^2-4c+32)=(-4)(408c - 1777)$$ so the discriminant is zero for $c=\frac{1777}{408}.$ The vertex is the intersection of the tangent at vertex and the parabola. So solve $$x^2+8x(4x+\frac{1777}{408})+16(4x+\frac{1777}{408})^2-18x-48(4x+\frac{1777}{408})+32=0$$ or $$(1734x+1147)^2\frac1{10404}$$ and put back into the tangent at vertex to find $y.$
The vertex then is $(-\frac{1147}{1734},\frac{11857}{6936}).$
The axis goes through the vertex so the axis is $${ x+4y=\frac{105}{17}}.$$ As a bonus the tangent at vertex form is then $$(x+4y-105/17)^2=(96/(4\cdot 17))(4x-y+1777/408).$$ For fun I've also found the focus/directrix form $$17\cdot ((x+\frac{59}{102})^2+(y-\frac{689}{408})^2-(4x-y+\frac{113}{24})^2/17)=0$$
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|
Universal substitution or Feynman trick to solve this integral I started with an integral $ \int_{0}^{2\pi} \sqrt{2[\sin^2(t) + 16\cos^2(t) - 4\sin(t)\cos(t)]} \,dt $
And I simplified it to $ \int_{0}^{2\pi} \sqrt{17 + 15\cos(2t) - 4\sin(2t)} \, dt$
My question: I know this can be simplified with some sort of substitution that cancels the $\sin$ and $\cos$ with a $u$-sub, but I do not know how. I saw it online, with no explanation (see the first answer: find length of curve of intersection).
I think this has an exact elementary solution, if you use $\tan\left(\frac x2\right)$ substitution and possibly Feynman's trick if necessary.
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$\begin{align}
\int_0^{2\pi}\sqrt{17+15\cos2t-2\sin2t}\,dt&\overset{2t=x}{=}\frac12\int_0^{4\pi}\sqrt{17+15\cos x-2\sin x}\,dx\\
&=\int_0^{2\pi}\sqrt{17+15\cos x-2\sin x}\,dx\\
&=\int_0^{2\pi}\sqrt{17+\sqrt{229}\cos(x+\alpha)}\,dx\\
&=\int_\alpha^{2\pi+\alpha}\sqrt{17+\sqrt{229}\cos x}\,dx\\
&=\int_0^{2\pi}\sqrt{17+\sqrt{229}\cos x}\,dx\\
&=2\int_0^{\pi}\sqrt{17+\sqrt{229}\cos x}\,dx\\
&=2\int_0^{\pi}\sqrt{17+\sqrt{229}-2\sqrt{229}\sin^2(\frac x2)}\,dx\,\, (\text{By $\cos x=1-2\sin^2(x/2)$})\\
&\overset{x\rightarrow 2x}{=}4\int_0^{\pi/2}\sqrt{17+\sqrt{229}-2\sqrt{229}\sin^2x}\,dx\\
&=4\sqrt{17+\sqrt{229}}\int_0^{\pi/2}\sqrt{1-\frac{2\sqrt{229}}{17+\sqrt{229}}\sin^2x}\,dx\\
&=4\sqrt{17+\sqrt{229}}\int_0^{\pi/2}\sqrt{1-\frac{17\sqrt{229}-229}{30}\sin^2x}\,dx\\
&=4\sqrt{17+\sqrt{229}}E\left(\sqrt{\frac{17\sqrt{229}-229}{30}}\right)
\end{align}$
In agreement with Wolfram Alpha. Here, WA uses $m=\frac{17\sqrt{229}-229}{30}$ but I used $k=\sqrt m$ as the variable of the function $E$.
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|
Proof that $3 \mid 10^{n+2} - 2*10^n + 7, \forall n \in \mathbb{Z}^+$. This is what I have so far.
Proof by Induction. Let $n \in \mathbb{Z}^+$ Let $P(n)$ be the statement that $10^{n+2} - 2*10^n + 7$ is divisible by 3.
($\textit{Base Case}$): Let $n = 1$.
$$
10^{1+2} - 2*10^1 + 7 = 1000 - 20 + 7 = 987
$$
$3 \mid 987$ there for $P(1)$ is true.
($\textit{Inductive Step}$): Let $k \in \mathbb{Z}^+$. Suppose $P(k)$ is true. Now we must show that $P(k+1)$ is true.
$$
10^{(k+1) + 2} - 2*10^{k+1} + 7
$$
$$
\Rightarrow 10^{(k+2)+1} - 2*10^{k+1} + 7
$$
$$
\Rightarrow 10^{k+2}(10) - 2*10^{k}(10) - 7
$$
$$
\Rightarrow 10(10^{k+2} - 2*10^{k}) + 7
$$
I don't know how to proceed. I've tried other methods of manipulating the equation and nothing seems to work.
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Note that $10^k=99\dots(k\ times)\dots 9+1$. Now $99\dots(k\ times)\dots 9$ is always divisible by $9$ and therefore it is divisible by $3$. So, dividing $10^k$ by $9$ gives a remainder of $1$.
Therefore, $10^{n+2}\equiv 1 \mod 3$ and $10^{n}\equiv 1\mod 3$. So, $10^{n+2}-2\times 10^{n}\equiv 1-2\times 1\mod 3\equiv -1 \mod3$.
And, this implies that $10^{n+2}-2\times 10^{n}+7\equiv -1+7\mod 3\equiv 6\mod 3\equiv 0\mod 3$.
Hence, 3 divides $10^{n+2}-2\times 10^{n}+7$. Here the notation $a\equiv b\mod m$ means that $m$ divides $a-b$.
If you want to go by the method of mathematical induction then notice that you assumed that $P(k)$ is true so $3$ divides $10^{k+2}-2\times 10^{k}+7$. Let $10^{k+2}-2\times 10^{k}+7=3m$ for some integer $m$.
Now, $10(10^{k+2}-2\times 10^{k})+7=10(10^{k+2}-2\times 10^{k}+7-7)+7=10(10^{k+2}-2\times 10^{k}+7)-70+7=10(10^{k+2}-2\times 10^{k}+7)-63=30m-63$.
Here, the first term is divisible by $3$ and $63$ is also divisible by $3$. So, the whole expression is divisible by 3. This would complete your argument by induction.
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
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Here is another proof, based on the formula
$$\frac{1}{1+x}=\frac{(-1)^nx^n}{1+x}+\sum_{k=0}^{n-1}(-1)^kx^k.$$
Integrating both sides over $[0,t]$ gives
$$\ln(1+t)=\int_0^t\frac{(-1)^nx^n}{1+x}\,dx+\sum_{k=1}^n(-1)^{k+1}\frac{t^k}{k}.$$
Setting $t=1$ shows that the partial sums $s_n$ of the alternating harmonic series are given by
$$s_n=\ln2+(-1)^n\int_0^1\frac{x^n}{1+x}\,dx.$$
But on $[0,1]$, we have $0\leq x^n(1+x)^{-1}\leq x^{n}$, so
$$0\leq\int_0^1\frac{x^n}{1+x}\,dx\leq\int_0^1x^{n}\,dx=\frac{1}{n+1}.$$
Hence $$\ln2-\frac{1}{n+1}\leq s_n\leq\ln2+\frac{1}{n+1}.$$ So $s_n\to\ln 2$ as $n\to\infty$.
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Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
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Let $k$ be any number. We know that $(k+1)² - k² = 2k + 1$
Let $k = 1, 2, 3, 4, ..., n$
$$(1+1)² - (1)²= 2(1) + 1$$
$$(2+1)² - (2)²= 2(2) + 1$$
$$(3+1)² - (3)²= 2(3) + 1$$
$$.$$
$$.$$
$$.$$
$$.$$
$$(n+1)² - (n)²= 2(n) + 1$$
Adding all the equations, we get:
$$- (1)² + (n+1)² = 2(1 + 2 + 3 + ... + n) + n(1)$$
$$n² + 2n = 2(1 + 2 + 3 + ... + n) + n$$
$$2(1 + 2 + 3 + ... + n) = n² + n$$
$$\therefore (1 + 2 + 3 + ... + n) = n(n+1)/2$$
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|
What is $\sqrt{i}$? If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?
Is it used or considered often in mathematics? How is it notated?
|
More generally, if you want to compute all the $n$-th roots of a complex number $z_0$, that is, all the complex numbers $z$ such that
$$
z^n = z_0 \ , \qquad \qquad \qquad \qquad [1]
$$
you should write this equation in exponential form: $z = re^{i\theta}, \ , z_0 = r_0 e^{i\theta_0}$. Then [1] becomes
$$
\left( r e^{i \theta}\right)^n = r_0 e^{i\theta} \qquad \Longleftrightarrow \qquad r^n e^{in\theta} = r_0 e^{i\theta_0} \ .
$$
Now, if you have two complex numbers in polar coordinates which are equal, their moduluses must be equal clearly:
$$
r^n = r_0 \qquad \Longrightarrow \qquad r = +\sqrt[n]{r_0}
$$
since $r, r_0 \geq 0$.
As for the arguments, we cannot simply conclude that $n\theta = \theta_0$, but just that they differ in an integer multiple of $2\pi$:
$$
n\theta = \theta_0 + 2k\pi \qquad \Longleftrightarrow \qquad \theta = \frac{\theta_0 + 2k \pi}{n} \quad \text{for} \quad k = 0, \pm 1 , \pm 2, \dots
$$
It would seem that we have an infinite number of $n$-th roots, but we have enough with $k = 0, 1, \dots , n-1$, since for instance for $k=0$ and $k=n$ we obtain the same complex numbers. Thus, finally
$$
\sqrt[n]{r_0 e^{i\theta_0}} = +\sqrt[n]{r_0} e^{i \frac{\theta_0 + 2k\pi}{n}} \ , \quad k = 0, 1, \dots , n-1
$$
are all the complex $n$-th roots of $z_0$.
Examples
(1) For $n=2$, we obtain that every complex number has exactly two square roots:
$$
\begin{align}
\sqrt{z_0} &= +\sqrt{r_0}e^{i\frac{\theta_0 + 2k\pi}{2}} \ , k = 0,1 \\\
&= +\sqrt{r_0}e^{i\frac{\theta_0}{2}} \quad \text{and} \quad +\sqrt{r_0}e^{i\left(\frac{\theta_0}{2} + \pi \right)} \ .
\end{align}
$$
For instance, since $i = e^{i\frac{\pi}{2}}$, we obtain
$$
\sqrt{i} =
\begin{cases}
e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} +i \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \\\
e^{i(\frac{\pi}{4} + \pi)} = \cos\frac{5\pi}{4} +i \sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \ .
\end{cases}
$$
Also, if $z_0 = -1 = e^{i\pi}$,
$$
\sqrt{-1} = e^{i \frac{\pi}{2}} = i \quad \text{and} \quad e^{i\left( \frac{\pi}{2} + \pi\right)} = e^{i\frac{3\pi}{2}} = -i \ .
$$
(2) For $z_0 = 1 = e^{i \cdot 0}$ and any $n$, we obtain the $n$-th roots of unity:
$$
\sqrt[n]{1} = e^{i\frac{2k\pi}{n}} \ , \quad k= 0, 1, \dots , n-1 \ .
$$
For instance, if $n= 2$, we get
$$
\sqrt{1}= e^{i \cdot 0} = 1 \quad \text{and} \quad e^{i\pi}= -1
$$
and for $n= 4$,
$$
\sqrt[4]{1} = e^{i\frac{2k\pi}{4}} \ , \quad k = 0, 1, 2, 3 \ ,
$$
that is,
$$
\sqrt[4]{1} = 1, i, -1 , -i \ .
$$
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|
Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$
Given, such a series, how does one go about solving it. Getting an Integral Representation seems tough for me.
I thought of going along these lines, Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ but couldn't succeed.
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FWIW, Mathematica is able to sum this:
In[1]:= Sum[1/(2 n + 2 k + 1) (2 k - 1)!!/(2 k)!!, {k, 0, \[Infinity]}]
3 + 2 n
Sqrt[Pi] Gamma[-------]
2
Out[1]= -----------------------
(1 + 2 n) Gamma[1 + n]
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Algebraic puzzle: Infer $x = \frac{y + z}{2}$ from $a^{\frac{1}{x}} = b^{\frac{1}{y}} + c^{\frac{1}{z}}$ and $a = b + c$ I have a suspicion that the following expression is true, however my algebra skills aren't brilliant, so any help would be appreciated:
Is it possible to infer $x = \frac{y + z}{2}$ from $a^{\frac{1}{x}} = b^{\frac{1}{y}} + c^{\frac{1}{z}}$ and $a = b + c$
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Let $a=2$, $b=c=1$. Also let $x=1$ then $a=b+c$ and $a^{\frac{1}{x}}=2$. Regardless of the values of $y$ and $z$ then $1^{\frac{1}{y}}+1^{\frac{1}{z}}=2$.
Thus we cannot infer that $x=\frac{y+z}{2}$
Counterexample:
Let $a=2$, $b=c=1$, $x=1$ and $y=z=2$ then $b+c=2=a$ and $b^{\frac{1}{y}}+c^{\frac{1}{z}}=2=a^{\frac{1}{x}}$. But $x\neq \frac{y+z}{2}=2$.
|
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
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OK, here's my favorite. I thought of this after reading a proof from the book "Proofs from the book" by Aigner & Ziegler, but later I found more or less the same proof as mine in a paper published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9
(EDIT: ...which is actually the proof that I read in Aigner & Ziegler).
When $0 < x < \pi/2$ we have $0<\sin x < x < \tan x$ and thus
$$\frac{1}{\tan^2 x} < \frac{1}{x^2} < \frac{1}{\sin^2 x}.$$
Note that $1/\tan^2 x = 1/\sin^2 x - 1$.
Split the interval $(0,\pi/2)$ into $2^n$ equal parts, and sum
the inequality over the (inner) "gridpoints" $x_k=(\pi/2) \cdot (k/2^n)$:
$$\sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k} - \sum_{k=1}^{2^n-1} 1 < \sum_{k=1}^{2^n-1} \frac{1}{x_k^2} < \sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k}.$$
Denoting the sum on the right-hand side by $S_n$, we can write this as
$$S_n - (2^n - 1) < \sum_{k=1}^{2^n-1} \left( \frac{2 \cdot 2^n}{\pi} \right)^2 \frac{1}{k^2} < S_n.$$
Although $S_n$ looks like a complicated sum, it can actually be computed fairly easily. To begin with,
$$\frac{1}{\sin^2 x} + \frac{1}{\sin^2 (\frac{\pi}{2}-x)} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x \cdot \sin^2 x} = \frac{4}{\sin^2 2x}.$$
Therefore, if we pair up the terms in the sum $S_n$ except the midpoint $\pi/4$ (take the point $x_k$ in the left half of the interval $(0,\pi/2)$ together with the point $\pi/2-x_k$ in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into $2^{n-1}$ parts. And the midpoint $\pi/4$ contributes with $1/\sin^2(\pi/4)=2$ to the sum. In short,
$$S_n = 4 S_{n-1} + 2.$$
Since $S_1=2$, the solution of this recurrence is
$$S_n = \frac{2(4^n-1)}{3}.$$
(For example like this: the particular (constant) solution $(S_p)_n = -2/3$ plus the general solution to the homogeneous equation $(S_h)_n = A \cdot 4^n$, with the constant $A$ determined by the initial condition $S_1=(S_p)_1+(S_h)_1=2$.)
We now have
$$ \frac{2(4^n-1)}{3} - (2^n-1) \leq \frac{4^{n+1}}{\pi^2} \sum_{k=1}^{2^n-1} \frac{1}{k^2} \leq \frac{2(4^n-1)}{3}.$$
Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. This squeezes the partial sums between two sequences both tending to $\pi^2/6$. Voilà!
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"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
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|
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ Using $\text{n}^{\text{th}}$ root of unity
$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$
Prove that
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
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Here is a more "1st principles" pf. I use a hint in Marsden's book.
1st, $\cos(A-B)-\cos(A+B)=2\sin A \sin B$ (1), which follows by angle summation formulas.
Next, we use Marsden's hint to consider roots of $(1-z)^n-1$. These satisfy
$$(1-z)^n=1 \leftrightarrow (1-z) \in \left\{\cos \frac{2 \pi k}{n}+i \sin \frac{2\pi k}{n}:k=0,...,n-1 \right\}$$
(the set of nth roots of 1)
$$\leftrightarrow z \in \left\{z_k= 1-\cos \frac{2 \pi k}{n}-i \sin \frac{2\pi k}{n}:k=0,...,n-1\right\}\;\;\; (2)$$.
Since $z_0,....,z_{n-1}$ are the roots of $(1-z)^n-1$, we have by factorization that
$$(1-z)^n-1=\prod_{k=0}^{n-1}(z_k-z)=-z \prod_{k=1}^{n-1}(z_k-z) \;\;(3)$$ (since, by (2), $z_0=0$)
In (3), the LHS and RHS are polynomials in z. Equating the coeffs in front of z, we get
$$-n=-\prod_{k=1}^{n-1}z_k \leftrightarrow n=\prod_{k=1}^{n-1}z_k\,.$$
Note
$$\prod_{k=1}^{n-1} \bar{z}_k=\overline{\prod_{k=1}^{n-1}z_k}=n$$
(since $n\in \mathbb{R}$), so
$$\prod_{k=1}^{n-1}|z_k|^2=\prod_{k=1}^{n-1} z_k \bar{z}_k=\prod_{k=1}^{n-1} z_k \prod_{k=1}^{n-1} \bar{z}_k=n^2\;\; (4).$$
Next,
$$|z_k|^2=(1-\cos \frac{2 \pi k}{n})^2+ \sin^2 \frac{2\pi k}{n}=2(1-\cos \frac{2 \pi k}{n})$$
using this in (4) gives
$$2^{n-1} \prod_{k=1}^{n-1}(1-\cos \frac{2 \pi k}{n})=n^2\;\;(5)$$.
Next,
$$(\prod_{k=1}^{n-1} \sin \frac{k \pi}{n})^2=\prod_{k=1}^{n-1} \sin \frac{k \pi}{n} \prod_{k=1}^{n-1} \sin \frac{(n-k) \pi}{n}=\prod_{k=1}^{n-1} \sin \frac{k \pi}{n} \sin \frac{(n-k) \pi}{n}=$$
(where in the last 2 steps, we exploit that the order of taking a product doesn't matter)
$$=\frac{1}{2^{n-1}} \prod_{k=1}^{n-1} (\cos \frac{(n-2k) \pi}{n}-\cos \pi)=$$
(by (1))
$$=\frac{1}{2^{n-1}} \prod_{k=1}^{n-1} (1-\cos \frac{2k \pi}{n})=$$
(using $\cos (\pi -x)=-\cos x$)
$$=n^2 /2^{2(n-1)}\;.$$ Applying a sqrt to everything gives the desired result.
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"language": "en",
"url": "https://math.stackexchange.com/questions/8385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "76",
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