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What's wrong in my calculation of $\int \frac{\sin x}{1 + \sin x} dx$? I have the following function:
$$f: \bigg ( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg ) \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{\sin x}{1 + \sin x}$$
And I have to find $\displaystyle\int f(x) dx $. This is what I did:
$$\int \dfrac{\sin x}{1 + \sin x}dx=
\int \dfrac{1+ \sin x - 1}{1 + \sin x}dx =
\int dx - \int \dfrac{1}{1 + \sin x}dx =
$$
$$ = x - \int \dfrac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx$$
$$= x - \int \dfrac{1 - \sin x}{1 - \sin ^2 x} dx$$
$$= x - \int \dfrac{1 - \sin x}{\cos^2 x} dx$$
$$= x - \int \dfrac{1}{\cos^2x}dx + \int \dfrac{\sin x}{\cos^2 x}dx$$
$$= x - \tan x + \int \dfrac{\sin x}{\cos^2 x}dx$$
Let $u = \cos x$
$du = - \sin x dx$
$$=x - \tan x - \int \dfrac{1}{u^2}du$$
$$= x - \tan x + \dfrac{1}{u} + C$$
$$= x - \tan x + \dfrac{1}{\cos x} + C$$
The problem is that the options given in my textbook are the following:
A. $x + \tan {\dfrac{x}{2}} + C$
B. $\dfrac{1}{1 + \tan{\frac{x}{2}}} + C$
C. $x + 2\tan{\dfrac{x}{2}} + C$
D. $\dfrac{2}{1 + \tan{\frac{x}{2}}} + C$
E. $x + \dfrac{2}{1 + \tan{\frac{x}{2}}} + C$
None of them are the answer I got solving this integral. What is the mistake that I made and how can I find the right answer? By what I've been reading online, you can get different answers by solving an integral in different ways and all of them are considered correct. They differ by the constant $C$. I understand that, but I don't see how to solve this integral in such a way to get an answer among the given $5$. And, even more importantly, how can I recognize the right answer in exam conditions if the answer provided by my solution is not present among the given options? Is solving in a different manner my only hope?
|
Your answer is correct and it matches with choice (E). You have to use half angle formulas:
$$\sin A = \frac{2 \tan \frac{A}{2}}{1+\tan^2 \frac{A}{2}} \quad \cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}}$$
Observe that
\begin{align*}
\frac{1}{\cos x}-\tan x&=\frac{1-\sin x}{\cos x}\\
& = \frac{\left(1-\tan \frac{x}{2}\right)^2}{1-\tan^2 \frac{x}{2}}\\
& = \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\\
& = 1+\frac{2}{1+\tan \frac{x}{2}}\\
\end{align*}
|
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|
Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$. Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.
So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod 1000$. I don't know how to proceed since 1000 seems to be a bit high.
|
$$n(n-1)\equiv0\pmod{2^35^3}$$
As $(n,n-1)=1,$ we can have following four cases
$$n\equiv0\pmod{2^35^3}$$
$$n-1\equiv0\pmod{2^35^3}$$
$$n-1\equiv0\pmod{2^3}\text{ and } n\equiv0\pmod{5^3}$$
$$n-1\equiv0\pmod{5^3}\text{ and } n\equiv0\pmod{2^3}$$
For the last two cases use Chinese Remainder Theorem
|
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|
Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2a}
$$
I think I am getting stuck here, is there a better substitution that I can chose so that the integral becomes more simple to evaluate ?
Solution as per my reference: $\dfrac{2(\sqrt{x}-1)}{\sqrt{1-x}}$
Note: I'd prefer to choose a substitution which does not make use of partial fractions, as there seems to be 4 terms for the substitution $\sqrt{x}=y\implies \frac{dx}{2\sqrt{x}}=dy$.
$$
I=\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}\\
=\int\frac{2dy}{y(1+y)(1-y^2)}=\int\frac{2dy}{y(1+y)^2(1-y)}\\
\frac{2}{y(1+y)^2(1-y)}=\frac{A}{y}+\frac{B}{1-y}+\frac{C}{1+y}+\frac{D}{(1+y)^2}
$$
|
What's wrong with partial fractions? They are easy if you use the Heaviside method. If $\sqrt x=y$ then
$$\int\frac{dx}{\left(1+\sqrt x\right)\left(x-x^2\right)}=\int\frac{2dy}{y(1+y)^2(1-y)}$$
And if
$$\frac2{y(1+y)^2(1-y)}=\frac Ay+\frac B{1-y}+\frac C{1+y}+\frac D{(1+y)^2}$$
Then
$$\begin{align}A&=\left.\frac2{(1+y)^2(1-y)}\right|_{y=0}=2\\
B&=\left.\frac2{y(1+y)^2}\right|_{y=1}=\frac12\\
C&=\left.\frac d{dy}\frac2{y(1-y)}\right|_{y=-1}=\left.\frac{-2}{y^2(1-y)^2}(1-2y)\right|_{y=-1}=-\frac32\\
D&=\left.\frac2{y(1-y)}\right|_{y=-1}=-1\end{align}$$
So
$$\begin{align}\int\frac{dx}{\left(1+\sqrt x\right)\left(x-x^2\right)}&=\int\left(\frac 2y+\frac{1/2}{1-y}-\frac{3/2}{1+y}+\frac1{(1+y)^2}\right)dy\\
&=2\ln|y|-\frac12\ln|1-y|-\frac32\ln|1+y|-\frac1{1+y}+C_1\\
&=\ln x-\frac12\ln\left|1-\sqrt x\right|-\frac32\ln\left(1+\sqrt x\right)+\frac1{1+\sqrt x}+C_1\end{align}$$
So at this point maybe you can see your way through to a clever $u$-substitution to achieve this result.
|
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|
Calculating real integrals using complex functions and residue theory. $\int_0^{\pi}\frac{\cos^2x}{1-a\sin^2x}dx$, where $0<a<1$.
My professor thought us some ways of calculating the following type of real integrals using residue theorem in complex analysis:
$\int_0^{2\pi}P(\sin x,\cos x)dx=\int_Cf(z)dz$, $C$ is the unit circle, and we are using $z=e^{ix}$, $x\in[0,2\pi]$, and so on.
First, in order to match the bounds of the integral I change the variable $x=2t$, so $t\in[0,2\pi]$. Then, I used the following formulas:
$\cos x=\frac{z^2+1}{2z}$ and $\sin x=\frac{z^2-1}{2iz}$
But the integrand was so ugly, I could not solve the integral. Is there any other way of calculating this integral? Can you help me? Thanks for any help.
|
Use the half-angle formulas to simplify the integrand:
$$
\int_0^\pi \frac{\cos^2 x}{1-a\sin^2 x}dx = \int_0^\pi \frac{1+\cos(2x)}{2-a+a\cos(2x)}dx = \frac{1}{2a}\int_0^{2\pi} \frac{1+\cos(t)}{\frac{2}{a}-1+\cos(t)}dt.
$$
Then apply $z = e^{it}$:
$$
\frac{1}{2a}\int_0^{2\pi} \frac{1+\cos(t)}{b+\cos(t)}dt = \frac{1}{2a}\int_C \frac{1+(z^2+1)/(2z)}{b+(z^2+1)/(2z)}\frac{dz}{iz} = \frac{1}{2ia}\int_C\frac{(z+1)^2}{z(z^2 + 2bz + 1)}dz,
$$
where I've written $b = 2/a -1$ to save space. This has residues at $0$, $-b+\sqrt{b^2-1}$, and $-b-\sqrt{b^2-1}$. Only the first two are in the unit circle and contribute to the integral, so expressing in terms of residues we have
$$
\int_0^\pi \frac{\cos^2 x}{1-a\sin^2 x}dx =\frac{\pi}{a}\left(\mathrm{Res}\left[\frac{(1+z)^2}{z(z^2+2bz+1)},0\right] + \mathrm{Res}\left[\frac{(1+z)^2}{z(z^2+2bz+1)},-b+\sqrt{b^2-1}\right]\right).
$$
I'll let you do the fancy residue stuff, but the end result is
$$
\int_0^\pi \frac{\cos^2 x}{1-a\sin^2 x}dx = \frac{\pi}{1+\sqrt{1-a}}
$$
|
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|
let $f(x)=x+\frac{1}{x} \ \ x \geq 1$ and $g(x)=x^2+4x-6$ the find Min of $g(f(x))=?$
Let $f(x)=x+\frac{1}{x}$ for all $x \geq 1$ and $g(x)=x^2+4x-6$. Find minimum of $g\circ f$.
My try:
The domain of $g\circ f$ is $[1,+\infty)$ and we have
$$g(f(x))=\left(x+\frac{1}{x}\right)^2+4\left(x+\frac{1}{x}\right)-6.$$
At $x=-2$, the minimum of $g(f(x))=-10$. Is that right?
|
Since for positive numbers $a,b$ we have $a+b\geq 2\sqrt{ab}$ we have also $$f(x) \geq 2$$ (put $a=x$ and $b=1/x$) with equality iff $x=1$.
Now $g(x)= (x+2)^2-10$, so for $x\geq -2$ it is increasing and thus the minimum of $g$ is at $x=2$.
So the minimum of $g(f(x))$ is at $f(x)=2$ i.e. $x=1$.
|
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|
Find $\lim\limits_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \,dx$. I have the following limit to find:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx$$
And I have to choose between the following options:
A. $0$
B. $1$
C. $\dfrac{3}{4}$
D. $\dfrac{1}{2}$
E. $\dfrac{1}{4}$
This is what I did:
$$\int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx =
\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx +
\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx$$
*
*If we consider the first term of this sum:
$$\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\int_0^\frac{1}{2} \sqrt[n]{x^n+x^n} \, dx =
\int_0^\frac{1}{2}2^{\frac{1}{n}}x \, dx$$
And that means:
$$\lim_{n \to \infty} \int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\lim_{n \to \infty} \int_0^\frac{1}{2}2^{\frac{1}{n}}x \, dx =
\int_0^\frac{1}{2}x \, dx =
\frac{1}{8} \tag 1$$
*
*If we consider the second term of that sum we have:
$$\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\int_\frac{1}{2}^1 \sqrt[n]{(1-x)^n+(1-x)^n} \, dx =
\int_\frac{1}{2}^1 2^{\frac{1}{n}}(1-x) \, dx$$
And that means:
$$\lim_{n \to \infty} \int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge \lim_{n \to \infty} \int_\frac{1}{2}^1 2^{\frac{1}{n}}(1-x) \, dx = \lim_{n \to \infty} \int_\frac{1}{2}^1 (1-x) \, dx = \frac{1}{8} \tag 2$$
Now we can sum $(1)$ and $(2)$ and we have:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge \frac{1}{4}$$
So now we have a lower bound. We can do something similar for the upper bound.
*
*Again, let's consider the first term of that sum:
$$\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \le
\int_0^{\frac{1}{2}} \sqrt[n]{(1-x)^n+(1-x)^n} \, dx =
\int_0^{\frac{1}{2}} 2^\frac{1}{n}(1-x) \, dx$$
That means:
$$\lim_{n \to \infty} \int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \le
\lim_{n \to \infty} \int_0^{\frac{1}{2}} 2^\frac{1}{n}(1-x) \, dx =
\int_0^{\frac{1}{2}} (1-x) \, dx =
\frac{3}{8} \tag 3$$
*
*And if we consider the second part of that sum:
$$\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+x^n} \, dx =
\int_{\frac{1}{2}}^1 2^\frac{1}{n} x \, dx$$
And that means:
$$\lim_{n \to \infty} \int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\lim\_{n \to \infty} \int_{\frac{1}{2}}^1 2^\frac{1}{n} x \, dx =
\lim_{n \to \infty} \int_{\frac{1}{2}}^1 x \, dx =
\frac{3}{8} \tag 4$$
And now if we sum $(3)$ and $(4)$ we get:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\frac{3}{4}$$
So after all of that, we have:
$$\frac{1}{4} \le
\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le \frac{3}{4}$$
But this didn't help me all that much. Choices C, D and E are still consistent with this inequality that I got. So what should I do to find the exact answer? Or did I do something wrong in my calculations?
|
The present problem is proposed by Prof. Ovidiu Furdui and appeared in Teme de analiza matematica. The solution presented is straightforward (using Squeeze theorem)
$$\small\frac{3}{4}=\int_0^{1/2}(1-x)\textrm{d}x+\int_{1/2}^1 x \textrm{d}x\le I_n \le \int_0^{1/2} \sqrt[n]{(1-x)^n+(1-x)^n}\textrm{d}x +\int_{1/2}^1\sqrt[n]{x^n+x^n}\textrm{d}x=\frac{3}{4}\sqrt[n]{2},$$
where the integral under the limit is denoted by $I_n$.
Letting $n\to\infty$ the desired limit follows, that is $3/4$.
More information: If interested in limits with such a structure, you may consult the book Limits, Series, and Fractional Part Integrals
Problems in Mathematical Analysis by Ovidiu Furdui.
For example, on page $7$ you may find the trigonometric version
$$\lim_{n\to\infty} \int_0^{\pi/2}\sqrt[n]{\sin^n(x)+\cos^n(x)}\textrm{d}x,$$
which leads to $\sqrt{2}$ and that can be finished by a similar strategy.
|
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|
How to derive this series
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.
I know the first series and it is easy to obtain it. But the second series is strange. It is not a power series, not a Taylor series. How does one obtain this series?
|
We can show the identity
\begin{align*}
\sum_{n=0}^\infty \frac{n!x^n}{\prod_{j=1}^n(1+jx)}=\frac{1}{1-x}\qquad\qquad0<x<1\tag{1}
\end{align*}
with the help of Gauss' summation formula.
We obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty \frac{n!x^n}{\prod_{j=1}^n(1+jx)}}
&=\sum_{n=0}^{\infty}\frac{n!}{\left(1+\frac{1}{x}\right)^{\overline{n}}}\tag{2}\\
&=\sum_{n=0}^{\infty}\frac{1^{\overline{n}}1^{\overline{n}}}{\left(1+\frac{1}{x}\right)^{\overline{n}}}\,\frac{1}{n!}\tag{3}\\
&={}_2F_1\left(1,1;1+\frac{1}{x};1\right)\tag{4}\\
&=\frac{\Gamma\left(\frac{1}{x}+1\right)\Gamma\left(\frac{1}{x}-1\right)}{\Gamma\left(\frac{1}{x}\right)\Gamma\left(\frac{1}{x}\right)}\tag{5}\\
&=\frac{\frac{1}{x}\Gamma\left(\frac{1}{x}\right)\Gamma\left(\frac{1}{x}-1\right)}
{\Gamma\left(\frac{1}{x}\right)\,\left(\frac{1}{x}-1\right)\Gamma\left(\frac{1}{x}-1\right)}\tag{6}\\
&=\frac{\frac{1}{x}}{\frac{1}{x}-1}\tag{7}\\
&\,\,\color{blue}{=\frac{1}{1-x}}
\end{align*}
and the claim (1) follows.
Comment:
*
*In (2) we expand with $\frac{1}{x^n}$ and use the rising factorial notation $q^{\overline{n}}=q(q+1)\cdots (q+n-1)$.
*In (3) we write $1^{\overline{n}}=n!$ and prepare the representation for use of hypergeometric series.
*In (4) we use the hypergeometric series notation
\begin{align*}
{}_2F_1\left(a,b;c;z\right)=\sum_{n=0}^{\infty}\frac{a^{\overline{n}}b^{\overline{n}}}{c^{\overline{n}}}\,\frac{z^n}{n!}
\end{align*}
with $a=b=z=1$ and $c=1+\frac{1}{x}$.
*In (5) we use Gauss' summation formula
\begin{align*}
{}_2F_1\left(a,b;c;1\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}
\end{align*}
with $a=b=1$ and $c=1+\frac{1}{x}$ valid for $\Re\left(\frac{1}{x}\right)>1$.
*In (6) we use the identity $\Gamma(x+1)=x\Gamma(x)$ for all $x\in\mathbb{C}\setminus\{0,-1,-2,\ldots\}$.
*In (7) we finally cancel terms.
|
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|
Prove that there no $k$ such that : $3n^{2}+3n+7=k^{3}$ Problem :
Let $n,k$ be natural numbers
Prove that there no $k$ such that :
$$3n^{2}+3n+7=k^{3}$$
I see this short solution from a book :
We use $\pmod{9}$
$k^{3}\equiv 0,1,8\pmod{9}$
And $3n^{2}+3n+7\equiv 4,7\pmod{9}$
$\implies $ no solution ?
But I don't understand why he chose mod $9$ and how we know that $k^{3}\equiv 0,1,8\pmod{9}$ ?
And how he know $3n^{2}+3n+7\equiv 4,7\pmod{9}$ ?
Can someone explain to me or give me another solution !
|
The reason for choosing 9 is that n is not divisible by 3 , let $n=3m+r$ , we can write:
$n^2=9m^2+6mr+r^2$
⇒ $$3n^2+3n+7 ≡(R=3r^2+3r+7) \ mod (9)$$
$r= 0, 1, 2$
$r=0$ ⇒ $R=7$ ⇒ $3n^2+3n+7 ≡7 \mod(9)$
$r=1$ ⇒ $R=13$ ⇒ $$3n^2+3n+7 ≡13 \mod(9)≡4 \mod(9)$$
$r=2$ ⇒ $R=7 \mod (9)$
⇒ $$3n^2+3n+7 ≡(4, 7) \mod(9)$$
$k≡(0, 1, 2, . . .8)\ mod (9)$
⇒ $K^3≡(0, 1, 8, 0, 1, 8, 0, 1, 8)\ mod (9) ≡(0, 1, 8)\ mod (9)$
|
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|
Evaluating $\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$ without l'Hopital's rule or Taylor series Can anyone please help me find this limit without l'Hopital's rule, I already used it to evaluate the limit, but I didn't know how to calculate it without l'Hopital's rule.
$$\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$$
Any tips will be helpful.
Sorry, but I don't want to use the Taylor series as well.
|
Hint:
Use Taylor expansion at order $4$: as
$$\cos x=1-\frac{x^2}2+\frac{x^4}{24}+o(x^4),$$
setting $u=-\dfrac{x^2}2+\dfrac{x^4}{24}+o(x^4)$, we have to expand $\ln (1+u)$ at order $2$ in $u$ and truncate the result at order $4$ (in $x$):
\begin{align}
\ln(\cos x)&=\ln(1+u)=u-\frac{u^2}2+o(u^2)=-\dfrac{x^2}2+\dfrac{x^4}{24}-\frac12\biggl(-\dfrac{x^2}2+\dfrac{x^4}{24}\biggr)^2+o(x^4)\\
&=-\dfrac{x^2}2+\dfrac{x^4}{24}-\dfrac{x^4}{8}+o(x^4)=-\dfrac{x^2}2-\dfrac{x^4}{12}+o(x^4)
\end{align}
so that the numerator is
$$x^2+2\ln(\cos x) =-\dfrac{x^4}{6}+o(x^4)\sim_0 -\dfrac{x^4}{6}.$$
|
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|
How to evaluate $\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin{x})^2}\,\textrm{d}x\,$? Problem:
Let $f$ be a bounded continuous function on the interval [0,1].
(a) show that $\int_{0}^{\pi} xf\,(\sin{x})\,\textrm{d}x = \frac{\pi}{2}\int_{0}^{\pi}f\,(\sin{x})\,\textrm{d}x\,$.
(b) Hence evaluate $\int_{0}^{\pi} \dfrac{x}{1+\sin{x}}\,\textrm{d}x\,$.
(c) Hence deduce that $\int_{0}^{\pi} \dfrac{2x^3-3\pi x^2}{(1+\sin{x})^2}\,\textrm{d}x = -\dfrac{2\pi ^3}{3}\,$.
I have completed Part (a) and Part (b). The answer of Part (b) is $\pi\,$. I have tried to do Part (c) in the following way:
$\textrm{Let }I=\int_{0}^{\pi} \dfrac{2x^3-3\pi x^2}{(1+\sin{x})^2}\,\textrm{d}x\\
\quad\;\;\;\,=\int_{0}^{\pi} \dfrac{2(\pi-x)^3-3\pi (\pi-x)^2}{(1+\sin{(\pi-x)})^2}\,\textrm{d}x\quad(\textrm{by letting }u=\pi-x\textrm{)}\\
\quad\;\;\;\,=\int_{0}^{\pi} \dfrac{2(\pi^3-3\pi^{2}x+3\pi x^2-x^3)-3\pi (\pi-x)^2}{(1+\sin{(\pi-x)})^2}\,\textrm{d}x\\
\quad\;\;\;\,=\int_{0}^{\pi} \dfrac{-\pi^3+3\pi^{2}x-2x^3}{(1+\sin{x})^2}\,\textrm{d}x\\
\quad\;\;\;\,=-\pi^3\int_{0}^{\pi}\dfrac{1}{(1+\sin{x})^2}\,\textrm{d}x-I\\
\quad\;\;\;\,=-\dfrac{\pi^3}{2}\int_{0}^{\pi}\dfrac{1}{(1+\sin{x})^2}\,\textrm{d}x\\
\quad\;\;\;\,=-\dfrac{\pi^3}{2}\cdot\dfrac{4}{3} \quad\textrm{(by letting }t=\tan{(\frac{x}{2})}\textrm{)}\\
\quad\;\;\;\,=-\dfrac{2\pi^3}{3}\\ \textrm{However, I did not make use of any results from Part (a) and Part (b).}\\
\textrm{I hope there is another solution that follows the hints of the problem.}$
|
I guess it is really just one of approach, with your approach being perfectly fine.
Alternatively, note that
$$(x - \pi)^3 = x^3 - 3\pi x^2 + 3 \pi^2 x - \pi^3.$$
Thus
\begin{align}
I &= \int_0^\pi \frac{2x^3 - 3\pi x^2}{(1 + \sin x)^2} \, dx\\
&= 2\int_0^\pi \frac{(x - \pi)^3}{(1 + \sin x)^2} \, dx + 3\pi \int_0^\pi \frac{x^2}{(1 + \sin x)^2} \, dx\\
& \qquad -6 \pi^2 \int_0^\pi \frac{x}{(1+ \sin x)^2} \, dx + 2\pi^3 \int_0^\pi \frac{dx}{(1 + \sin x)^2}\, dx.
\end{align}
Using the result in part (a) on the third integral appearing in the second line of equality, we have
$$I = 2 \int_0^\pi \frac{(x - \pi)^3}{(1 + \sin x)^2} \, dx + 3\pi \int_0^\pi \frac{x^2}{(1 + \sin x)^2} \, dx - \pi^3 \int_0^\pi \frac{dx}{(1 + \sin x)^2}.$$
If a substitution of $x \mapsto \pi - x$ is now enforced in the first of the integrals appearing above, one finds
$$I = - I - \pi^3 \int_0^\pi \frac{dx}{(1 + \sin x)^2},$$
or
$$I = -\frac{\pi^3}{2} \int_0^\pi \frac{dx}{(1 + \sin x)^2},$$
agreeing with what you found by enforcing a substitution of $x \mapsto \pi - x$ to begin with.
|
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|
sum of coefficient of all even power of $x$
The sum of all Coefficient of even power of $x$ in
$(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n}),n\in \mathbb{N}$
what i try
for $n=1,$ we have
$(1-x+x^2)(1+x+x^2)=(1+1)-(1)+(1+1)=3$
for $n=2,$ we have
$(1-x+x^2-x^3+x^4)(1+x+x^2+x^3+x^4)$
$=(1+1+1)-(1+1)+(1+1+1)-(1+1)+(1+1+1)=5$
so in this way , get sum of coefficient in original expression is $2n+1$
but How do i solve it without substituting value of $n$, Help me
|
$$f(x)=(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n})$$
$$f(x)=\Bigr[(1+x^2+\cdots +x^{2n})-(x+x^3+\cdots +x^{2n-1})\Bigr]\times \Bigr[(1+x^2+\cdots +x^{2n})+(x+x^3+\cdots +x^{2n-1})\Bigr]$$
$$f(x)=(1+x^2+\cdots +x^{2n})^2-(x+x^3+\cdots +x^{2n-1})^2$$
$$f(x)=\biggr(\frac{1-x^{2n}}{1-x^2}\biggr)^2-\biggr(\frac{x-x^{2n-1}}{1-x^2}\biggr)^2$$
$$f(x)=\frac{1+x^{4n}-2x^{2n}-x^2-x^{4n-2}+2x^{2n}}{(1-x^2)^2}$$
$$f(x)=\frac{1-x^2+x^{4n}-x^{4n-2}}{(1-x^2)^2}$$
$$f(x)=\frac{(1-x^2)(1-x^{4n-2})}{(1-x^2)^2}$$
$$f(x)=\frac{1-x^{4n-2}}{1-x^2}$$
|
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|
$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
x^3+y^3=7(x+y) \end{cases}$$
by reducing the system to a system of second degree.
We can factor:
$$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
(x+y)(x^2-xy+y^2)=7(x+y) \end{cases}$$
I really don't want to divide the equations by $x-y$ and $x+y$, respectively. I am taught to divide by expressions containing variables only in special cases. Is there any other way here?
|
The thing that catches my eye is that $7,19$ are primes congruent to $1 \pmod 3,$ therefore integrally represented by both $x^2 + xy+y^2, \; x^2 -xy + y^2,$ meaning there are integer points on the two ellipses. It is worth drawing both, by hand (a valuable skill), to see whether that gives a simplified answer when $y \neq \pm x.$
|
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|
Solutions to $\prod_{n=1}^\infty \left ( 1+ \frac1{f(n)} \right ) = \varphi$ The constant $\varphi = \frac{1 + \sqrt5}{2}$ seems to show up everywhere. I am wondering what non-trivial function(s) $f$ satisfy
$$\prod_{n=1}^\infty \left ( 1+ \frac1{f(n)} \right ) = \varphi$$
I was suprised when I could not find any solutions. The 'closest' solution I have found is $f(n)=n^2+3n+2$, for which the infinite product is approximately $1.61847$.
|
This is a little too long for a comment and may not be a complete answer, but here is one such $f(n)$.
We know that $2\cos\left(\frac{\pi}{5}\right)=\phi$, hence $\frac{\sin\left(2\pi/5\right)}{\sin\left(\pi/5\right)}=\phi.$ Then by Euler's product for the sine function
$$\begin{align*}
\frac{\sin\left(\frac{2\pi}{5} \right )}{\sin\left(\frac{\pi}{5} \right )} &=\frac{\frac{2\pi}{5}\prod_{n=1}^\infty\left(1-\frac{(2\pi/5)^2}{n^2\pi^2} \right )}{\frac{\pi}{5}\prod_{n=1}^\infty\left(1-\frac{(\pi/5)^2}{n^2\pi^2} \right )} \\
&=2\prod_{n=1}^\infty\frac{(5n)^2-4}{(5n)^2-1} \\
&= 2\prod_{n=1}^\infty\frac{(5n+2)(5n-2)}{(5n+1)(5n-1)}\\
&= \prod_{n=0}^\infty\frac{(5n+2)(5n+3)}{(5n+1)(5n+4)}.\\
\end{align*}$$
Solving for the $f(n)$ in your product gives
$$\phi=\prod_{n=0}^\infty\left(1+\frac{1}{f(n)}\right)$$
where $f(n)=\frac{25}{2}n^2+\frac{25}{2}n+2.$ The only difference is the index shift, which is no big deal.
|
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|
Not sure what to do after trying to simplify the inequality and getting nowhere?
Here is the question:
If $x, y, z \in {\displaystyle \mathbb {R} }$, then prove
$$x^2+y^2+z^2+4\ge2(x+y+z)$$
Here is what I tried doing:
I tried simplifying the inequality and getting all the terms on one side, like so:
$$x^2+y^2+z^2+4\ge2x+2y+2z$$
$$x^2+y^2+z^2+4-2x+2y+2z\ge0$$
$$(x^2-2x)+(y^2-2y)+(z^2-2z)+4\ge0$$
$$(x^2-2x)+(y^2-2y)+(z^2-2z)\ge-4$$
Is it possible to use the fact that it is always true that $x^2-2x\ge-4$, $y^2-2y\ge-4$, and $z^2-2z\ge-4$, to show that inequality holds? Other than that, I'm not sure what to do next. I can't think of any manipulation that might work in this case.
Any help would be greatly appreciated!
|
\begin{align}
x^2 - 2x + y^2 - 2y + z^2 - 2z + 4
= (x-1)^2 + (y-1)^2 + (z-1)^2 + 1
\ge 1 > 0.
\end{align}
|
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|
Verify that the equation of a plane is this determinant. Verify that the equation of a plane that passes through three points $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{13})$ of the space is:
$$\begin{vmatrix}
1 & x_{1} & x_{2} & x_{3} \\
1 & a_{1} & a_{2} & a_{3} \\
1 & b_{1} & b_{2} & b_{3} \\
1 & c_{1} & c_{2} & c_{3} \\
\end{vmatrix}= 0$$
I have tried to reduce the determinant by making zeros, but it continues being so enormous to solve by Sarrus:
$$\begin{vmatrix}
1 & x_{1} & x_{2} & x_{3} \\
0 & a_{1}-x_{1} & a_{2}-x_{2} & a_{3}-x_{3} \\
0 & b_{1}-x_{1} & b_{2}-x_{2} & b_{3}-x_{3} \\
0 & c_{1}-x_{1} & c_{2}-x_{2} & c_{3}-x_{3} \\
\end{vmatrix}= 0$$
Is there a simple way to prove that?
Thanks in advance.
|
One approach is as follows. Let $a$ denote the vector $(a_1,a_2,a_3)$ and so-forth. It suffices to show that $x$ lies in the plane through $a,b,c$ if and only if the system of equations
$$
\begin{cases}
k_1 a + k_2 b + k_3 c = x\\
k_1 + k_2 + k_3 = 1
\end{cases}
$$
has a solution for $k_1,k_2,k_3 \in \Bbb R$. Indeed, the $x$ of the form $x = k_1 a + k_2 b + k_3 c$ for some $k_i$ with $k_1+k_2+k_3 = 1$ form the affine space generated by $a,b,c$.
Another approach: $x$ lies in the plane through $a,b,c$ if and only if the vectors $x-c,a-c,b-c$ are linearly dependent.
|
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|
A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation. A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation.
My attempt is as follows:-
Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$
As center lies on the line $x+y=2$
$$-g-f=2$$
$$g+f=-2\tag{1}$$
As circle passes through the point $(0,1)$
$$1+2f+c=0$$
$$2f+c=-1\tag{2}$$
As $4x – 3y + 4 = 0$ is tangent to the circle
$$\dfrac{\left|-4g+3f+4\right|}{5}=\sqrt{g^2+f^2-c}$$
Squaring both sides
$$16g^2+9f^2-24gf+16+8(-4g+3f)=25g^2+25f^2-25c$$
$$9g^2+16f^2+24gf+32g-24f-25c-16=0$$
Eliminating $g$ with the help of equation $(1)$
$$9(-2-f)^2+16f^2+24(-2-f)f+32(-2-f)-24f-25c-16=0$$
$$9(4+f^2+4f)+16f^2-48f-24f^2-64-32f-24f-25c-16=0$$
$$f^2-68f-44-25c=0$$
Eliminating $c$ with the help of equation $(2)$
$$f^2-68f-44-25(-1-2f)=0$$
$$f^2-18f-19=0$$
$$f^2-19f+f-19=0$$
$$f=19,-1$$
$$(g,f,c)\equiv (-1,-1,1),(-21,19,-39)$$
So equations are $x^2+y^2-2x-2y+1=0$, $x^2+y^2-42x+38y-39=0$
But this got too long, any shorter method?
|
Let $(a,2-a)$ be a center of the circle.
Thus, $$\frac{|4a-3(2-a)+4|}{\sqrt{4^2+(-3)^2}}=\sqrt{(a-0)^2+(2-a-1)^2}.$$
It's easier than your attempt.
After squaring of the both sides we obtain:
$$a^2-22a+21=0$$ and the rest is smooth.
|
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|
How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(4(1-(\cos x)^2 + 2\cos x -1) & = \cos 9x + \cos 8x + 1
\end{align*}
I don't see any way to get rid of $8x$ and $9x$ as arguments to have same angles from there
|
As $5-4=2-1$
$$2(\sin x\sin2x-\sin5x\sin4x)=2(\cos^24x-\cos^2x)$$
Using http://mathworld.wolfram.com/WernerFormulas.html
$$\cos x-\cos3x-(\cos x-\cos9x)=-2(\sin^4x-\sin^2x)$$
As $\dfrac{9-3}2=4-1,$
using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html and
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$$2\sin\dfrac{9x-3x}2\cdot\sin\dfrac{9x+3x}2=2\sin(4x+x)\sin(4x-x)$$
I should leave it here for you!
|
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|
Prove that $\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$ I need to prove that: $$\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$$
I am new to inequalities so I only tried C-S and AM-GM but none of those work.
Any hints on how to proceed here?.
|
For $xy=0$ it's obvious.
But for $xy\neq0$ by AM-GM we obtain:
$$\left|\frac{x^3y^3}{9x^4+y^4}\right|\leq\frac{|x^3y^3|}{2\sqrt{9x^4y^4}}=\frac{|xy|}{6}\leq\frac{\frac{x^2+y^2}{2}}{6}\leq\frac{x^2+y^2}{6}.$$
|
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|
Logarithm Subtraction and Division with Same Bases I'm rusty on logarithms. What is the approach to a problem like this? Any hints would be appreciated.
I'm thinking the subtraction on the numerator and denominator can become division since the bases are the same?
|
Often better to factor the arguments of the logarithms to get to much simpler logarithms more quickly... \begin{align*}
&\frac{\log_2 24 - \frac{1}{2} \log_2 72}{\log_3 18 - \frac{1}{3} \log_3 72} \\
&\quad{}= \frac{\log_2 (2^3 \cdot 3) - \frac{1}{2} \log_2 (2^3 \cdot 3^2)}{\log_3(2\cdot 3^2) - \frac{1}{3} \log_3 (2^3 \cdot 3^2)} \\
&\quad{}= \frac{\log_2 2^3 + \log_2 3 - \frac{1}{2} \left( \log_2 2^3 + \log_2 3^2 \right)}{\log_3 2 + \log_3 3^2 - \frac{1}{3} \left( \log_3 2^3 + \log_3 3^2 \right)} \\
&\quad{}= \frac{3\log_2 2 + \log_2 3 - \frac{1}{2} \left( 3\log_2 2 + 2\log_2 3 \right)}{\log_3 2 + 2\log_3 3 - \frac{1}{3} \left( 3\log_3 2 + 2\log_3 3 \right)} \\
&\quad{}= \frac{3 + \log_2 3 - \frac{1}{2} \left( 3 + 2\log_2 3 \right)}{\log_3 2 + 2 - \frac{1}{3} \left( 3\log_3 2 + 2 \right)} \\
&\quad{}= \frac{3 + \log_2 3 - \frac{3}{2} - \log_2 3 }{\log_3 2 + 2 - \log_3 2 - \frac{2}{3}} \\
&\quad{}= \frac{3 - \frac{3}{2}}{2 - \frac{2}{3}} \cdot \frac{6}{6} \\
&\quad{}= \frac{18-9}{12-4} \\
&\quad{}= \frac{9}{8} \text{.}
\end{align*}
|
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|
Angles in triangle with inscribed circle that touches one side in a ratio A triangle $\mathrm{ABC}$ has $\angle B=3\times\angle C$. The side $BC$ is devided by the touch point of the inscribed circle in the ratio 1:5. Can the angles be calculated?
With the side $a=\mathrm{BC}=6y$ I tried the "touch point" formula
$$d(B,T_a)=y=\frac{1}{2}(a+b-c)$$
combined with the cosinus theorem
$$
(6y)^2
=b^2+c^2-2bc\cos(\pi-4x)
=b^2+c^2+2bc\cos(4x)
$$
but didn't get anywhere with that. Any hints? TIA.
|
Let ∠C = $x$, $r$ the radius of the incircle, and D the touch point. Then
$$\tan\frac{\angle B}2 = \tan \frac{3x}2 = \frac r{BD},\>\>\>\>\>
\tan\frac{\angle C}2=\tan \frac{x}2 = \frac r{CD}$$
Take the ratio of above equations,
$$\frac{\tan\frac{3x}2 }{\tan\frac{x}2} = \frac{CD}{BD}=5$$
Apply the identity $\tan 3a = \frac{\tan a(3-\tan^2 a)}{1-3\tan^2 a}$ to get
$$\frac{3-\tan^2 \frac x2}{1-3\tan^2 \frac x2} = 5$$
Solve to obtain $\tan^2 \frac x2 = \frac17$, which yields $x= 2\tan^{-1}\frac1{\sqrt7}$. Thus, the angles are
$$\angle C = x = 2\tan^{-1}\frac1{\sqrt7},\>\>\>\>\>\>\>\angle B = 3x = 6\tan^{-1}\frac1{\sqrt7}$$
|
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|
Conditional expectation of a geometric variable
$X, Y$ are two independent random variables which are both
Geometry($p$). We also define random variables $Z = |X-Y|, W = \min
\{X,Y\}$. Calculate $E [W | Z = 1]$.
I tried to use $\min\{X,Y\}=(|X+Y|-|X-Y|)/2$. At first I thought $Z,W$ are independent, because $Z$ is the absolute value of the difference, so $Z$ doesn't tell me which one of $X$ or $Y$ is the minimum. Am I wrong?
the answer is (1-p)/(2-p)^2 and I don't understand how to get to this
|
For $Z$, first note that $\{|X-Y|\geqslant 0\}$ has probability one, and $|X-Y|= 0$ if and only if $X=Y$. So first we compute
\begin{align}
\mathbb P(Z=0) &= \mathbb P(X=Y)\\
&= \sum_{n=1}^\infty \mathbb P(X=Y\mid X=n)\mathbb P(X=n)\\
&= \sum_{n=1}^\infty \mathbb P(Y=n)\mathbb P(X=n)\\
&= \sum_{n=1}^\infty \mathbb P(X=n)^2\\
&= \sum_{n=1}^\infty (1-p)^{2(n-1)}p^2\\
&= \frac p{2-p}.
\end{align}
For $n\geqslant 1$, $$\{|X-Y|=n\} = \{X-Y=n\}\cup\{X-Y=-n\}.$$ Hence
\begin{align}
\mathbb P(Z = n) &= \mathbb P(X-Y=n) + \mathbb P(X-Y=-n)\\
&= \sum_{k=1}^\infty\mathbb P(X-Y=n\mid Y=k)\mathbb P(Y=k) + \sum_{k=n+1}^\infty \mathbb P(X-Y=-n)\mathbb P(Y=k)\\
&= \sum_{k=1}^\infty \mathbb P(X=n+k)\mathbb P(Y=k) + \sum_{k=n+1}^\infty \mathbb P(X=k-n)\mathbb P(Y=k)\\
&= \sum_{k=1}^\infty (1-p)^{n+k-1}p(1-p)^{k-1}p + \sum_{k=n+1}^\infty \mathbb (1-p)^{k-n-1}p(1-p)^{k-1}p\\
&= \frac{p (1-p)^n}{2-p} + \frac{p (1-p)^n}{2-p}\\
&= \frac{2p (1-p)^n}{2-p}.
\end{align}
For $W$, note that
$$
\{X\wedge Y=n\} = \{X=n,Y=n\}\cup \{X=n,Y>n\}\cup \{X>n,Y=n\}.
$$
By symmetry, $\mathbb P(X=n,Y>n) = \mathbb P(X>n,Y=n)$. So we have
\begin{align}
\mathbb P(W=n) &= \mathbb P(X=n,Y=n) + 2\mathbb P(X=n,Y>n)\\
&= \mathbb P(X=n)^2 + 2\sum_{k=n+1}^\infty \mathbb P(X=n,Y=k)\\
&= (1-p)^{2(n-1)}p^2 + 2\sum_{k=n+1}^\infty (1-p)^{n-1}p(1-p)^{k-1}p\\
&= (1-p)^{2(n-1)}p^2 + 2p (1-p)^{2 n-1}\\
&= p(2-p) (1-p)^{2 (n-1)}.
\end{align}
That should help you in computing the desired conditional expectation.
|
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|
A pastry shop sells 4 kind of pastries. How many distinct sets of 7 pastries can one buy? I HAVE SEEN stars and bars reference from this sum.
But still I dont get it .
I dont get it whether it is permutation or combination.
I have done the sum in this way.
All arrangements
C1 C2 C3 C4
7 0 0 0. = 4p1 (from 4 cakes choosing one)
5 2 0 0. = 4p2 (from 4 cakes choosing two)
6 1 0 0 = 4p2 (from 4 cakes choosing 2)
5 1 1 0= 4p3
4 2 1 0 = 4p3
4 1 1 1= 4p4
3 3 1 0= 4p3
3 2 2 0 = 4p3
3 2 1 1 = 4p4
2 2 2 1 = 4p4
Summing all of these will give me the correct answer???
|
What matters here is how many of each type of pastry are selected. Selecting three pecan pies, two apple strudel, one croissant, and one baklava is different from selecting four baklava, two apple strudels, and one pecan pie.
Let's label the types of pastries 1, 2, 3, and 4. Let $x_i$, $1 \leq i \leq 4$, be the number of pastries of type $i$ that are selected. Since a total of seven pastries are selected from the four types,
$$x_1 + x_2 + x_3 + x_4 = 7 \tag{1}$$
The number of ways the pastries can be selected is the number of solutions of equation 1 in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of three addition signs in a row of seven ones. For instance,
$$1 1 + 1 1 1 + 1 + 1$$
corresponds to the solution $x_1 = 2$, $x_2 = 3$, $x_3 = 1$, $x_4 = 1$, while
$$1 + + 1 1 1 1 + 1 1$$
corresponds to the solution $x_1 = 1$, $x_2 = 0$, $x_3 = 4$, $x_4 = 2$.
The number of solutions of equation 1 in the nonnegative integers is the number of ways $4 - 1 = 3$ addition signs can be placed in a row of $7$ ones, which is
$$\binom{7 + 4 - 1}{4 - 1} = \binom{10}{3}$$
since we must select which three of the ten positions required for seven ones and three addition signs will be filled with addition signs.
What was wrong with your attempt?
The partitions of $7$ into at most four parts are
\begin{align*}
7 & = 7\\
& = 6 + 1\\
& = 5 + 2\\
& = 5 + 1 + 1\\
& = 4 + 3\\
& = 4 + 2 + 1\\
& = 4 + 1 + 1 + 1\\
& = 3 + 3 + 1\\
& = 3 + 2 + 2\\
& = 3 + 2 + 1 + 1\\
& = 2 + 2 + 2 + 1
\end{align*}
Your counts are correct for the cases in which a different number of each type of pastry is selected. However, they are incorrect when the same number of pastries are selected from two or more types.
$5 + 1 + 1$: There are four ways to select the type of pastry from which five pieces of pastry will be selected. There are $\binom{3}{2}$ ways to select the two types of pastry from which one piece of pastry each will be selected. Hence, there are
$$\binom{4}{1}\binom{3}{2}$$
such selections.
$3 + 3 + 1$: There are $\binom{4}{2}$ ways to select the two types of pastries from which three pieces of pastry will be drawn and two ways to select the type of pastry from which one piece of pastry will be drawn. Hence, there are
$$\binom{4}{2}\binom{2}{1}$$
such selections.
Notice that
$$\binom{4}{1}\binom{3}{2} = \binom{4}{2}\binom{2}{1}$$
This is because in both the $5 + 1 + 1$ case and the $3 + 3 + 1$ case, there are three types of pastry drawn, with equal amounts of exactly two of them. We could have done the $3 + 3 + 1$ case by first selecting the type of pastry from which one piece of pastry will be drawn, then selecting from which two of the three types of pastry three pieces of pastry each would be drawn, which would have yielded the count
$$\binom{4}{3}\binom{3}{2}$$
$3 + 2 + 2$: The argument above shows that there are also
$$\binom{4}{1}\binom{3}{2}$$
such cases.
$4 + 1 + 1 + 1$: There are four ways to select the type of pastry from which four pieces will be selected. We must select one of each the other types. Hence, there are
$$\binom{4}{1}$$
such selections.
$2 + 2 + 2 + 1$: There are four ways to select the type of pastry from which one piece will be selected. We must select two pieces each from each of the remaining types of pastry. Hence, there are
$$\binom{4}{1}$$
such selections.
$3 + 2 + 1 + 1$: There are four ways to select the type of pastry from which three pieces will be drawn and three ways to select the type of pastry from which two pieces will be drawn. We must select one piece each from each of the remaining types of pastry. Hence, there are
$$\binom{4}{1}\binom{3}{1}$$
such selections.
With these corrections, we obtain a total of
$$\binom{4}{1} + \binom{4}{1}\binom{3}{1} + \binom{4}{1}\binom{3}{1} + \binom{4}{1}\binom{3}{2} + \binom{4}{1}\binom{3}{1} + \binom{4}{1}\binom{3}{1}\binom{2}{1} + \binom{4}{1} + \binom{4}{2}\binom{2}{1} + \binom{4}{1}\binom{3}{2} + \binom{4}{1}\binom{3}{1} + \binom{4}{1} = \binom{10}{3}$$
ways to select seven pastries of four types.
What type of problem is this?
This is a combination with repetition problem since we are selecting $k$ objects from $n$ types of objects, where we may take the same type of object repeatedly.
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|
Minimum value of $p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$ where $x,y,z\in \mathbb{R}^+$.
Find the minimum value of $$p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$$ where $x,y,z\in \mathbb{R}^+.$
Applying the AM-GM inequality,
$$
\begin{aligned}\frac{p}{6} & \geqslant\left(3x\cdot\frac{1}{15x}\cdot 5y\cdot\frac{25}{y}\cdot z\cdot \frac{1}{36z}\right)^{1/6} \\
\frac{p}{6} & \geqslant \left(\frac{5}{6}\right)^{1/3}\\
p & \geqslant 6\left(\frac{5}{6}\right)^{1/3}
\end{aligned}$$
$$\implies \text{The minimum value of the expression is } 6\left(\frac{5}{6}\right)^{1/3}
$$
Now, consider $f(x)=3x+\dfrac{1}{15x},\ g(y) = 5y+\dfrac{25}{y}$ and $h(z)=z+\dfrac{1}{36z}.$
$$\begin{aligned}f'(x) &= \frac{\mathrm d}{\mathrm dx}\left(3x+\frac{1}{15x}\right) = 3 - \frac{1}{15x^2}\\f''(x)&=\frac{2}{15x^3}\end{aligned}$$
At the critical points, $f'(x) = 0 \implies x = \dfrac{\pm1}{3\sqrt{5}}.$
$f''\left(\dfrac{1}{3\sqrt5}\right) > 0\implies f(x)$ has a local minima at $x = \dfrac{1}{3\sqrt5}.$
Similarly, the local minima of $g(y)$ is at $y = \sqrt5$ and the local minima of $h(z)$ is at $z=1/6.$
Substituting these values into the original expression, the minimum value of the expression comes out to be
$$
\begin{aligned}
p & = f\left(\dfrac{1}{3\sqrt5}\right)+g\left(\sqrt5\right)+h\left(\frac{1}{6}\right) \\
& = \frac{1}{3} + \frac{52}{\sqrt5}.
\end{aligned}
$$
The answer according to the AM-GM inequality is $\approx5.646$ and according to calculus is $\approx23.588$, which are way off. Also, the functions have only two points of inflection, one is the maxima (for values less than $0$) and the other is the minima (for values greater than $0$). As the question clearly states "for $x,y,z \in \mathbb{R}^+$", the "correct" answer should be $\approx23.588$, shouldn't it?
Why are the answers different? Also, is it possible to figure out the individual $x, y$ and $z$ values for which the expression has the minimum value (in case of AM-GM inequality)?
Note: This question is from a Test and the correct answer according to the "test creators" is $$6\left(\dfrac{5}{6}\right)^{1/3}.$$
$\text{Graph made using Desmos.}$
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The calculus solution is correct.
Equality holds for AM-GM when all the values are the same, but this cannot be the case. (If 3x=5y, then it is not true that 1/(15x)=25/y.)
|
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|
If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ or $y$ is divisible by $3$, but not both, then:
$\begin{align}x \equiv \pm 1 \pmod{3} \land y \equiv 0 \pmod{3}
&\iff x^2 \equiv 1 \pmod{3} \land y^2 \equiv 0 \pmod{3}
\\
&\iff x^2 - y^2 \equiv 1 \pmod{3}\end{align}$
So if $x$ or $y$ is divisible by $3$ but not both then no values for $x$ and $y$ will ever make $x^2 - y^2$ divisible by $3$.
*If $x$ and $y$ are divisible by $3$:
Since both $x$ and $y$ are divisible by $3$ they can be expressed $x = 3m$ and $y = 3n$ for some integers $m$, $n$.
$(3m)^2 - (3n)^2 = 1995 \iff 9m^2 - 9n^2 = 1995 \iff 3(m^2-n^2) = 665$
And it's easy to see that there are no whole number solutions for m and n, and thus there are no whole number solutions for x and y such that $x^2 - y^2 = 1995$ where x or y (or both) is divisible by 3.
Is this a reasonable proof? Are there more elegant ways to do it?
|
The question is a bit ambiguous. Anyway, you can indeed answer both interpretations.
If both $x$ and $y$ are divisible by $3$, then the left-hand side is divisible by $9$, but the right-hand side isn't (which is your proof).
Suppose $x$ is divisible by $3$. Then $-y^2\equiv0\pmod{3}$, because $1995\equiv0\pmod{3}$. Thus implies $y\equiv0\pmod3$ and we already have excluded the case.
Similarly if $y$ is divisible by $3$.
|
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|
Largest integer such that $n+5$ divides $n^3+89$ I received a calendar with daily math problems as a Christmas gift, and so far most of the problems have been fun and challenging but not too difficult. However, I am completely stumped on the one for today, January 31. The problem reads:
What is the largest positive integer $n$ such that $n^3+89$ is divisible by $n+5$?
I struggled with it for a bit and found a few $n$'s which worked before throwing it at a computer. I tested up to $n=200,000$ and the only ones that satisfy the above condition are $\{ 1,4,7,13,31\}$. I then tried to prove that it doesn't work for $n\geq 31$ with induction for a bit before coming here. The calendar usually has the day of the month as the answer, so the answer should be 31, but I wasn't able to prove this.
Also, for all $n\geq 31$, at least up to $200,000$, the remainder on division of $n^3+89$ by $n+5$ is always $n-31$, if that is useful. Thanks for any help or tips!
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Synthetic division.
$n^3 + 89 = n^3 + 5n^2 - 5n^2 + 89=$
$(n+5)n^2 - 5n^2 + 89=$
$(n+5)n^2 - 5n^2 - 25n + 25n +89=$
$(n+5)n^2 - (n+5)5n + 25n + 89 =$
$(n+5)n^2 - (n+5)5n + 25n + 125 -125+89=$
$(n+5)n^2 - (n+5)5n + (n+5)25 - 36=$
$(n+5)(n^2-5n + 25) -36$.
So if $n+5$ divides that, then we must have $n+5| 36$
If $n=31$ then $31+5 = 36$ and division is possible, but if $n > 31$ then $n+5>36$ and division is impossible.
....
Alternatively $n^3 + 89 \equiv 0 \pmod {n+5}$ so
$(-5)^3 + 89= -36 \equiv 0\pmod {n+5}$
So $n+5|36$
|
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|
Using Ratios for proof making If
$$a(y+z)=x,b(z+x)=y,c(x+y)=z$$
prove that
$$\frac{x^2}{a(1-bc)}=\frac{y^2}{b(1-ac)}=\frac{z^2}{c(1-ba)}$$
I haven't been able to get to the form in the proof, any suggestions on how to do it are welcome.
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If $y+z=0,$ we have $x=0,$ which leads to $y=z=0.$ Assuming that the fractions in the final relation are well-defined, we have the desired equality.
We arrive to the same conclusion if $z+x=0$ or $x+y=0.$
Assume that $(x+y)(y+z)(z+x)\neq 0.$
Rewrite $$a(y+z)=x,\;b(z+x)=y,\;c(x+y)=z$$ as $$a=\frac{x}{y+z},\;b=\frac{y}{z+x},\;c=\frac{z}{x+y}.$$
Then $$\begin{aligned}a(1-bc)=&\frac{x}{y+z}\left(1-\frac{yz}{(z+x)(x+y)}\right)\\=&\frac{x^2(x+y+z)}{(y+z)(z+x)(x+y)}\end{aligned}$$
We have analogous expressions for $b(1-ca)$ and $c(1-ab).$
Finally, we obtain $$\frac{x^2}{a(1-bc)}=\frac{(y+z)(z+x)(x+y)}{x+y+z}=\frac{y^2}{b(1-ac)}=\frac{z^2}{c(1-ba)}.$$
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$y'(1)$ and $z'(1)$ of $x^2+y^2-2z^2 = 0, x^2+2y^2+z^2 = 4$ Given this equation system
$$x^2+y^2-2z^2 = 0$$
$$x^2+2y^2+z^2 = 4$$
locally at $(1,1,1)$
It can be solved by:
$$ x^2+y^2-2z^2 = 0 \tag 1$$
$$x^2+2y^2+z^2 = 4 \tag 2$$
$2 \times (1) - (2)$ gives
$$x^2-5z^2 = -4 $$
$2 \times (2) + (1)$ gives
$$3x^2+5y^2 = 8 $$
In the neighbourhood of $(1,1,1)$ the intersection can be parametrized as
$$I(x) = \bigg (x, \sqrt{ \frac {8-3x^2}5} , \sqrt{ \frac {4+x^2}5 } \bigg ) $$
But how can one find out $y'(1)$ and $z'(1)$?
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From $x^2-5z(x)^2 = -4$ we derive $2x-10z(x)z'(x)=0.$ With $x=1$ and $z(1)=1$ we get
$$2-10z'(1)=0.$$
Hence $z'(1)=1/5.$
It is now your turn to compute $y'(1)$ with the aid of $3x^2+5y^2 = 8.$
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|
Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induction:
(1) Base: $\tau(1)$.
$$
\begin{align}
a_2 &=\frac{(2+1)a_1-2+1}{2+1-(2-1)a_1} \\
&=\frac{3a_1-1}{3-a_1}.
\end{align}
$$
(2) Assumption: Let
$$ a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1} $$
hold for some $n\in\mathbb N$.
(3) Step $\tau(n+1)$:
$$
\begin{align}
a_n &= \frac{3\cdot\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}-1}{3-\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}} \\
&= \frac{3\cdot2^{n-1}a_1+3a_1-3\cdot2^{n-1}+3-2^{n-1}-1+2^{n-1}a_1-a_1}{3\cdot 2^{n-1}+3-3\cdot 2^{n-1}a_1+3a_1-2^{n-1}a_1-a_1+2^{n-1}-1} \\
&= \frac{(4\cdot 2^{n-1}+2)a_1-4\cdot 2^{n-1}+2}{4\cdot 2^{n-1}+2-(4\cdot 2^{n-1}+2)a_1} \\
&= \frac{(2^n+1)a_1-2^n+1}{2^n+1-(2^n+1)a_1}
\end{align}
$$
Is this correct and is there a more efficient method than induction?
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Embed $a_n$ in a projective coordinate $[a_n,1]$ for each $n$. Then the given recursive relation may be written as
$$\left[\begin{array}{c}a_{n+1}\\ 1\end{array}\right]=\left[\begin{array}{cc}3&-1\\ -1&3\end{array}\right]\left[\begin{array}{c}a_n\\ 1\end{array}\right].$$
It follows by recursion that $$\left[\begin{array}{c}a_n\\ 1\end{array}\right]=A^{n-1}\left[\begin{array}{c}a_1\\ 1\end{array}\right],n\geq 1~\quad (1)$$ where $$A=\left[\begin{array}{cc}3&-1\\ -1&3\end{array}\right].$$
Diagonalize $A$ to get $$ \frac 12\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]A\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]=\left[\begin{array}{cc}2&0\\ 0&4\end{array}\right]$$ $$\Rightarrow A^{n-1}=\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]\left[\begin{array}{cc}2^{n-1}&0\\ 0&2^{2n-2}\end{array}\right]\frac 12\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]$$
$$=\left[\begin{array}{cc}2^{n-2}+2^{2n-3}&2^{n-2}-2^{2n-3}\\ 2^{n-2}-2^{2n-3}&2^{n-2}+2^{2n-3}\end{array}\right].$$
Substituting this back to (1), one gets that
$$\left[\begin{array}{c}a_n\\ 1\end{array}\right]=\left[\begin{array}{cc}2^{n-2}+2^{2n-3}&2^{n-2}-2^{2n-3}\\ 2^{n-2}-2^{2n-3}&2^{n-2}+2^{2n-3}\end{array}\right]\left[\begin{array}{c}a_1\\ 1\end{array}\right]$$
$$\Leftrightarrow a_n=\frac{(2^{n-2}+2^{2n-3})a_1+(2^{n-2}-2^{2n-3})}{(2^{n-2}-2^{2n-3})a_1+(2^{n-2}+2^{2n-3})}=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1},$$ as required. QED
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limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
Now it is $e^{(0\cdot \infty)}$ which we can not conclude about the limit
|
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
From your work we have:
$$\lim_{x\to 7}\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)=\lim_{x\to 7}\left(\frac{(x-8)(x-10)}{1}\right)\frac{\ln\left(1+\color{red}{\left(\frac{x}{7}-1\right)}\right)}{\color{red}{\left(\frac{x}{7}-1\right)}}\cdot\frac{\left(x-7\right)}{7\left(x-7\right)}={\frac{3}{7}}$$
So the limit would be $\color{red}{\exp\left(\frac{3}{7}\right)}$
|
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Convergence of $u_{k+1} = \frac{1}{2-u_k u_{k-1}}$ Let $u_0, u_1 \in \mathbb{R}$ be arbitrary. I am interested in the sequence defined by:
$$u_{k+1} = \frac{1}{2-u_k u_{k-1}}.$$
In particular I would like to find $\lim_{k\rightarrow \infty} u_k$ for this sequence.
For a value $\bar u$ to be a potential limiting point, this value should at least satisfy:
$$
\bar u = \frac{1}{2-\bar u^2},
$$
it is not hard to see that this equation has $3$ solutions, namely:
$$
\bar u = 1, -\frac{1}{2}(1+\sqrt{5}), \frac{1}{2}(-1+\sqrt{5}).
$$
I can numerically verify that the first two candidates do not attract the sequence $(u_k)_k$, thus we only have $u_k \rightarrow 1$ resp. $u_k\rightarrow -\frac{1}{2}(1+\sqrt{5})$ when $u_0,u_1 = 1$ resp. $u_0,u_1=-\frac{1}{2}(1+\sqrt{5})$ whilst the third seems to be an attractor for the sequence.
Furthermore there is the special case where $u_0 \cdot u_1 = 2$, which we exclude as $u_2$ is not defined in this case.
However, I am unable to prove that for any $u_0, u_1 \notin \{-1, -\frac{1}{2}(1+\sqrt{5})\}$ and $u_0 \cdot u_1 \neq 2$, we have $u_k \rightarrow \frac{1}{2} (-1+\sqrt{5})$.
A related problem was studied here, and I understand the proof given there, but I do not see how to extend this new problem.
|
We are lucky : one can find a closed-form expression for $u_n$ (see (1) below). This will allow us to show that the sequence always converges and to determine the limit.
Let us compute a first few terms. Putting $x=u_1,y=u_0$ we have $u_2=\frac{1}{2-xy}$, $u_3=\frac{2-xy}{4-x-2xy}$, $u_4=\frac{4-x-2xy}{7-2x-4xy}$, $u_5=\frac{7-2x-4xy}{12-4x-7xy}$, $u_6=\frac{12-4x-7xy}{20-7x-12xy}$.
We start to see a pattern here. First, the denominator of $u_n$ seems to coincide with the numerator of $u_{n+1}$, and the coefficients $1,2,4,7,12,20$
seem to be the Fibonacci numbers minus 1. To summarize, the following holds for
$n\leq 6$ :
$$
u_n=\frac{v_n}{v_{n+1}}, \ \textrm{where} \ v_n=(F_{n+1}-1)-(F_{n-1}-1)x-(F_{n}-1)xy \tag{1}
$$
Now that we have guessed (1), we can prove it rigorously by induction : if (1) holds for $n$ and $n+1$, we have
$$u_{n+2}=\frac{1}{2-u_nu_{n+1}}=\frac{1}{2-\frac{v_n}{v_{n+2}}}=\frac{v_{n+2}}{2v_{n+2}-v_n} \tag{2}$$
Notice that the Fibonacci sequence satisfies
$$2F_{n+2}-F_n=F_{n+2}+(F_{n+2}-F_n)=F_{n+2}+F_{n+1}=F_{n+3} \tag{3}$$
Since $(v_n)$ is a linear combination of $F_{n-1},F_n$ and $F_{n+1}$, it will satisfy this linear recurrence as well : $2v_{n+2}-v_n=v_{n+3}$, so that (2) gives $u_{n+2}=\frac{v_{n+2}}{v_{n+3}}$ which finishes the proof of (1) by induction.
Let $\phi$ be the golden ratio, $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=1-\phi=\frac{1-\sqrt{5}}{2}$ is the conjugate of $\phi$. Using the well-known formula $F_n=\frac{\phi^n-\psi^n}{\sqrt{5}}$, we deduce
$$
v_n=\frac{\phi^2-x-\phi xy}{\sqrt{5}}\phi^{n-1}-\frac{\psi^2-x-\psi xy}{\sqrt{5}}\psi^{n-1}+(xy+x-1)\tag{4}
$$
There are now several cases to consider. If $c_1=\frac{\phi^2-x-\phi xy}{\sqrt{5}}$ is nonzero, then $v_n \sim c_1 \phi^{n-1}$ as $n\to\infty$, so that $u_n$ converges to $\phi$.
Next, assume that $c_1=0$ but $c_2=xy+x-1$ is nonzero, then $v_{n} \to c_2 $ as $n\to\infty$, so that $(u_n)$ converges to $1$.
Finally, if both $c_1$ and $c_2$ are zero, then $x=y=1$ and $(u_n)$ is constant equal to $1$.
|
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|
Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
|
Hint:
Simplify
$$\int{\frac{1}{(x^3+1)^2}}dx=\frac x{3(x^3+1)}+\frac13\int\frac2{x^3+1}dx$$
Then, decompose
$$\frac2{x^3+1}=\frac1{x+1}+\frac{1}{x^2-x+1}- \frac{x^2}{x^3+1}$$
|
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|
Asymptotic behaviour of an integral whose integrand cannot be expanded How does one evaluate how a limit is approached in the following case?
Consider the function
$$
f(z) = \int_{0}^1 d x \int_{0}^1 d y \left( \sqrt{1 - \frac{x y}{\sqrt{x^2 + z^2}\sqrt{y^2 + z^2}}} + \sqrt{1 + \frac{x y}{\sqrt{x^2 + z^2}\sqrt{y^2 + z^2}}} \right).
$$
This has limiting values
$$
\lim_{z \to 0}f(z) = \sqrt{2} \quad \text{and} \quad \lim_{z \to \infty}f(z) = 2.
$$
and interpolates monotonically between them for intermediate values. I am interested in establishing the expansions of $f(z)$ about these limits. Specifically I want to know the leading order behaviour of
$$
f(z) - \sqrt{2} \quad \text{as} \quad z \to 0, \qquad \text{and} \qquad f(z) - 2 \quad \text{as} \quad z \to \infty.
$$
In the case of $z \to \infty$ the expansion about $z = \infty$ is easily established by Taylor expansion:
$$
f(z) = \int_{0}^1 d x \int_{0}^1 d y \left( 2 - \frac{x^2y^2}{4z^4} + \mathrm{O}(z^{-6}) \right) = 2 - \frac{1}{36 z^4} + \mathrm{O}(z^{-6}).
$$
However the same trick fails in the limit $z \to 0$
$$
f(z) = \int_{0}^1 d x \int_{0}^1 d y \left( \sqrt{2} + \frac{z\sqrt{x^2+y^2}}{\sqrt{2} x y} + \mathrm{O}(z^2) \right)
$$
as the second term in the integral does not converge. This makes sense as the integrand is not Taylor expandable on the lines $x=0,y=0$ when $z=0$.
How does one determine how the limit is approached in this case?
Numerically it appears that $(f(z) - f(0)) \sim z \log z$ as $z \to 0$ but I have been unable to show this formally
|
Let $g(z) = f(z) - \sqrt{2}$, and consider the substitution
$$1-s = \frac{x}{\sqrt{x^2+z^2}}, \qquad 1-t = \frac{y}{\sqrt{y^2+z^2}}, \qquad w = 1-\frac{1}{\sqrt{z^2+1}}.$$
Then from the computation
$$\mathrm{d}x=-\frac{z}{s^{3/2}(2-s)^{3/2}} \mathrm{d}s, \qquad \mathrm{d}y=-\frac{z}{t^{3/2}(2-t)^{3/2}} \mathrm{d}t, $$
we obtain the following integral representation:
\begin{align*}
g = g(z)
&= z^2 \int_{w}^{1} \int_{w}^{1} \frac{\sqrt{s+t-st} + \sqrt{2-(s+t-st)} - \sqrt{2}}{(st)^{3/2} (2-s)^{3/2}(2-t)^{3/2}} \, \mathrm{d}s\mathrm{d}t \\
&= 2 z^2 \int_{w}^{1} \int_{t}^{1} \frac{\sqrt{s+t-st} + \sqrt{2-(s+t-st)} - \sqrt{2}}{(st)^{3/2} (2-s)^{3/2}(2-t)^{3/2}} \, \mathrm{d}s\mathrm{d}t.
\end{align*}
Now by noting that $w \sim \frac{z^2}{2}$ as $z \to 0$, we show that $g \sim c\sqrt{w}\log w$ as $w \to 0^+$ for some constant $c \neq 0$. Indeed,
\begin{align*}
&\lim_{w \to 0^+} \frac{g}{\sqrt{w}\log w} \\
&= \lim_{w \to 0^+} \frac{4}{w^{-1/2}\log w} \int_{w}^{1} \int_{t}^{1} \frac{\sqrt{s+t-st} + \sqrt{2-(s+t-st)} - \sqrt{2}}{(st)^{3/2} (2-s)^{3/2}(2-t)^{3/2}} \, \mathrm{d}s\mathrm{d}t \\
&= \lim_{w \to 0^+} \frac{8}{w^{-3/2}\log w} \int_{w}^{1} \frac{\sqrt{s+w-sw} + \sqrt{2-(s+w-sw)} - \sqrt{2}}{(sw)^{3/2} (2-s)^{3/2}(2-w)^{3/2}} \, \mathrm{d}s \\
&= \lim_{w \to 0^+} \frac{8}{\log w} \int_{1}^{1/w} \frac{\sqrt{w(r+1-wr)} + \sqrt{2-w(r+1-wr)} - \sqrt{2}}{\sqrt{w} r^{3/2} (2-wr)^{3/2}} \, \mathrm{d}r,
\end{align*}
where the L'Hospital's Rule is applied in the second step and the substitution $s=wr$ is utilized in the last step. Now it is not hard to show that the last limit is $-1$, and therefore,
$$ g(z) \sim -\sqrt{w}\log w \sim -\sqrt{2}z\log z \qquad \text{as} \qquad z \to 0^+. $$
|
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|
Evaluate $\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
An approach I can think about is to expand $\cos$ using taylor series, is there another approach?
|
This is another way to get the limit. The idea is the same as Quanto's. In fact.
\begin{eqnarray}
&&\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}\\
&=&\lim _{x\to 0}\frac{2\sin^2(\frac{x}{2(1-x^2)})-\frac{x^2}{2}}{x^4}\\
&=&\frac12\lim _{x\to 0}\frac{4\sin^2(\frac{x}{2(1-x^2)})-x^2}{x^4}\\
&=&\frac12\lim _{x\to 0}\frac{(2\sin(\frac{x}{2(1-x^2)})-x)(2\sin(\frac{x}{2(1-x^2)})+x)}{x^4}\\
&=&\frac12\lim _{x\to 0}\frac{2\sin(\frac{x}{2(1-x^2)})-x}{x^3}\cdot\frac{2\sin(\frac{x}{2(1-x^2)})+x}{x}\\
&=&\lim _{x\to 0}\frac{2(\frac{x^3}{2(1-x^2)}-\frac1{3!}(\frac{x}{2(1-x^2)})^3+O(x^5))}{x^3}\\
&=&\frac{23}{24}.
\end{eqnarray}
|
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|
Question: Using the Cauchy-Schwarz Inequality to compare between 2 expressions Use the Cauchy-Schwarz Inequality to determine whether $a^2+b^2+c^2$ is bigger than/smaller than/equal to $ab+bc+ac$, where $a,b,c$ are integers and $a<b<c$.
Cauchy-Schwarz Inequality:
$$(\sum_{i=1}^{n}a_ib_i)^2 \leq {\left(\sum_{i=1}^{n}a_i^2\right ) \left ( \sum_{i=1}^{n}b_i^2 \right ) }$$
My attempt:
$n=3$
$a_1=\sqrt{ab}$, $a_2=\sqrt{bc}$, $a_3=\sqrt{ac}$
$b_1=\frac{\sqrt{a}}{\sqrt{b}}$, $b_2=\frac{\sqrt{b}}{\sqrt{c}}$, $b_3=\frac{\sqrt{c}}{\sqrt{a}}$
Plugging it in,
$$ab+bc+ac+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq a^2 + b^2 + c^2$$
There are $3$ unwanted fractions. Is there any way to remove them?
|
Scalar product:
$|(u,v)|\le ||u||$ $||v||$.
$|(a,b,c)\cdot (c,a,b)|\le$
$ ||(a,b,c)||$ $||(c,a,b)||$;
$ac+ba +bc \le |ac+ba+bc|\le$
$ a^2+b^2+c^2.$
|
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|
Integral of modified Bessel function of second kind first order multiply by incomplete gamma function? Is there any possible solution or approximation for that given integral? $$\int_0^\infty {\big(v^{\frac{m}{2}-\frac{1}{4}}\big)}K_1\Bigg[\frac{2\sqrt[4]{v}}{l}\Bigg]\Gamma\left[m,-\frac{a+b v}{c}\right]\text{d}v.$$
|
I suppose that the approach could be the same as for your previous post.
If $a=0$, we have
$$\frac{1}{4 \pi }\left(-\frac{b}{c}\right)^{-\frac{2m+3}{4}}\,\,G_{2,5}^{4,2}\left(-\frac{c}{16\, b\, l^4}|
\begin{array}{c}
\frac{1-6 m}{4} ,\frac{1-2m}{4} \\
-\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{3}{4},-\frac{2m+3}{4} )
\end{array}
\right)$$
Now, write
$$\Gamma\left[m,-\frac{a+b v}{c}\right]=\sum_{p=0}^\infty d_n\, a^n$$ with
$$d_0=\Gamma \left(m,-\frac{b v}{c}\right)\qquad d_1=\frac 1c\left(-\frac{b}{c}\right)^{m-1} e^{\frac{b }{c}v}$$
$$d_{n+2}=\frac{(n+1) (b v+c (m-n-1))\,d_{n+1}+n\, d_n}{b c (n+1) (n+2) v}$$
The problem is that I am unable to compute even the first integral.
|
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|
On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$
What I have managed to do so far is to convert $S$ to two rather difficult integrals as follows.
Starting with the result
$$\frac{H_{2n}}{2n} = -\int_0^1 x^{2n - 1} \ln (1 - x) \, dx \tag1$$
Multiplying (1) by $(-1)^n H_n/n$ then summing the result from $n = 1$ to $\infty$ gives
$$S = -2 \int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n}. \tag2$$
From the following generating function for the harmonic numbers
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \frac{1}{2} \ln^2 (1 - x) + \operatorname{Li}_2 (x),$$
replacing $x$ with $-x^2$ leads to
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n} = \frac{1}{2} \ln^2 (1 + x^2) + \operatorname{Li}_2 (-x^2).$$
Substituting this result into (2) yields
$$S = -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx,$$
or, after integrating the first of the integrals by parts twice
$$S = -\frac{5}{2} \zeta (4) + 4 \zeta (3) \ln 2 - 8 \int_0^1 \frac{x \operatorname{Li}_3 (x)}{1 + x^2} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx. \tag3$$
I have a slim hope the first of these integrals can be found (I cannot find it). As for the second of the integrals, it is proving to be a little difficult.
Can someone find each of the integrals appearing in (3)? Or perhaps an alternative approach to the sum will deliver the closed-form I seek, I am fine either way.
Update
Thanks to Ali Shather, the first of the integrals can be found. Here
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \ dx &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{2n-1}\ln(1-x)\ dx\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n^3}\\
&=-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}\\
&=-4 \operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3}.
\end{align}
And using the result I calculated here, namely
$$\operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3} = \frac{5}{8} \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{195}{256} \zeta (4) + \frac{5}{192} \ln^4 2 - \frac{5}{32} \zeta (2) \ln^2 2 + \frac{35}{64} \zeta (3) \ln 2,$$
gives
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx &= -\frac{5}{2} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{195}{64} \zeta (4) - \frac{5}{48} \ln^4 2\\
& \qquad + \frac{5}{8} \zeta (2) \ln^2 2 - \frac{35}{16} \zeta (3) \ln 2.
\end{align}
|
Remark: I have noticed too late that this integral has already been solved (in the update of omegadot).
Nevertheless I don't delete the contribution because, together with this information, it shows that the hypergeometric functions appearing here can be simplified appreciably which gives hope for other cases.
Original post
A closed expression of the integral
$$i = \int_0^1 \frac{x \operatorname{Li}_3(x)}{x^2+1}\tag{1}$$
can be found in terms of (sorry Ali) hypergeometric function as follows.
Partial integration gives
$$i=s_{0}-\int_0^1 \frac{\text{Li}_2(x) \log \left(x^2+1\right)}{2 x} \, dx\tag{2a}$$
where
$$s_0 = \frac{1}{2} \zeta (3) \log (2)\tag{2b}$$
Expanding the denominator of the integrand we find that $i=s_{0}+\sum a_{k}$ where
$$a_{k} =-\frac{1}{2} \int_0^1 \frac{(-1)^{k+1} x^{2 k-1} \text{Li}_2(x)}{k} \, dx=-\frac{(-1)^{k+1} \left(\pi ^2 k-3 H_{2 k}\right)}{24 k^3}\tag{3}$$
The two sums are
$$s_{1}=\frac{1}{24} \left(-\pi ^2\right) \sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{k^2}=-\frac{\pi ^4}{288}\tag{4}$$
$$s_{2} = +\frac{1}{8} \sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_{2 k}}{k^3}=\frac{1}{32} \left(-2 \,_P\tilde{F}_Q^{(\{0,0,0,0\},\{0,0,1\},0)}(\{1,1,1,1\},\{2,2,2\},-1)\\-\sqrt{\pi } \,_P\tilde{F}_Q^{(\{0,0,0,0,0\},\{0,0,0,1\},0)}\left(\left\{1,1,1,1,\frac{3}{2}\right\},\left\{2,2,2,\frac{3}{2}\right\},-1\right)\\+3 \zeta (3) (\gamma +\log (2))\right)\tag{5}$$
Where $\,_P\tilde{F}_Q$ is the regularized hypergeometric function. For more details see https://math.stackexchange.com/a/3544006/198592.
Two terms appear in $s_{2}$ due to the relation
$$H_{2 k}=\frac{1}{2} \left( H_{k-\frac{1}{2}}+ H_k \right)+\log (2)$$
The complete integral is then given by
$$i = s_{0}+s_{1}+s_{2}$$
The numeric check shows good agreement.
Discussion
I'm almost sure that the sum
$$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_k}{k^3}$$
has a simpler expression, and so might
$$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_{k-\frac{1}{2}}}{k^3}$$
and I'd be happy to replace the hypergeometric constructs.
No need to conjecture: omegadot has done it, see https://math.stackexchange.com/a/3290607/198592
|
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I can't understand a how the podium is built in my problem So we are going to build a podium with blocks with the measurements $1 \times 1 × 1$ inch the question is as follows.
The top layer should measure $2 \times 2$ inches and then each layer should project $1/2$ inch on all four sides relative to the layer above (so that the podium will be staircase-shaped)
and the sum for n layer is $\dfrac13n^3+\dfrac32n^2+\dfrac{13}6n$
before I used the equation I thought it was 4 blocks in the top layer and 12 blocks the next layer, but when I tried to confirm that with the equation, it showed for layer 1 4 blocks and for layer 2 13 blocks. So did I misunderstand how to build the podium or did i miss calculate the equation?
|
Since the podium projects $\frac{1}{2}$ inch on each of the four sides, going from the top layer of $2 \times 2$ inches to the next layer would increase the size by $\frac{1}{2} + \frac{1}{2} = 1$ inch in its width, and the same in its height, to give that each dimension is now $2 + 1 = 3$ inches, i.e., the second layer dimensions are $3$ inches by $3$ inches. This results in $3 \times 3 = 9$ blocks for the second layer, for a total of $4 + 9 = 13$ blocks for the $2$ layers combined, just as the equation for the sum gives.
Note also when you wrote "for layer 2 13 blocks" that the equation is for the sum of the blocks up to layer $n$, not just the number of blocks in the $n$'th layer.
To confirm, have $f(n)$ be the total # of blocks up to layer $n$ to get
$$\begin{equation}\begin{aligned}
f(n) & = \frac{1}{3}n^3 + \frac{3}{2}n^2 + \frac{13}{6}n \\
& = \frac{2n^3 + 9n^2 + 13n}{6}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, $f(1) = \frac{2 + 9 + 13}{6} = \frac{24}{6} = 4$, and $f(2) = \frac{16 + 36 + 26}{6} = \frac{78}{6} = 13$.
|
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prove a sequence of functions that are partial sums of a series converges let
$f_n:[-\frac{\pi}{2},\frac{\pi}{2}]\to \mathbb{R}$
be defined by
$\displaystyle f_n(x):=\sum_{k=1}^n \left(\frac{x}{2}\right)^k\!\sin(kx)$.
Prove that the sequence $\{f_n\}_{n=1}^\infty$ is uniformly convergent.
My attempt:
$\sum_{k=1}^n|(\frac{x}{2})^k\sin(kx)| \leq \sum_{k=1}^n|(\frac{x}{2})^k| \;$ since $ |\sin(x)| \leq 1. \\ $
This is geometric series and $|\frac{x}{2}| \leq 1$ on this interval, so $\sum_{k=1}^n|(\frac{x}{2})^k|$ is convergent for all $x \in [-\frac{\pi}{2},\frac{\pi}{2}]$, and it follows that
$\sum_{k=1}^n|(\frac{x}{2})^k\sin(kx)|$ is convergent and therefore $\sum_{k=1}^n(\frac{x}{2})^k\sin(kx)$ is convergent.
writing out some of the functions of the sequence:
$f_1: (\frac{x}{2})\sin(x) \\f_2: (\frac{x}{2})\sin(x) + (\frac{x^2}{2^2})\sin(2x) \\f_3: (\frac{x}{2})\sin(x) + (\frac{x^2}{2^2})\sin(2x) + (\frac{x^3}{2^3})\sin(3x)\\f_n: (\frac{x}{2})\sin(x) + (\frac{x^2}{2^2})\sin(2x) + (\frac{x^3}{2^3})\sin(3x) + \cdots + (\frac{x^n}{2^n})\sin(nx) \\$
as $n$ appraoches infinity, the functions are converging to the infinite sum of $\sum_{k=1}^\infty \left(\frac{x}{2}\right)^k\!\sin(kx)$, and since $\sum_{k=1}^\infty \left(\frac{x}{2}\right)^k\!\sin(kx)$ is convergent, the sequence $\{f_n\}_{n=1}^\infty$ is a Cauchy sequence.
Let $S(x) = \lim_{n \to \infty} f_n(x)$
$|S(x) - f_n(x)| = \lim_{k \to \infty}|\frac{x^{n+1}}{2^{n+1}}\sin((n+1)x) + \frac{x^{n+2}}{2^{n+2}}\sin((n+2)x) + \cdots + \frac{x^{n+k}}{2^{n+k}}\sin((n+k)x)|< \epsilon \text{ because it is a convergent series}$
I was also trying to apply Weierstrass M-test for this, but the m-test seems to apply for an infinite series of functions, but for this question it is a sequence of functions that are all partial sums of a series, so I was not sure how to apply it.
|
$|(\frac x 2)^{k} \sin (kx)| \leq (\frac {\pi} 4)^{k}$ and $\frac {\pi} 4 <1$ so M-test applies.
|
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Decomposition of $2+2^2+2^{2^2}+2^{2^{2^2}}+2^{2^{2^{2^2}}}$ It's a little question about number theory :
Remark that we have :
$$2+2^2=2*3$$
$$2+2^2+2^{2^2}=2*11$$
$$2+2^2+2^{2^2}+2^{2^{2^2}}=2*32779$$
Do we have :
$$2+2^2+2^{2^2}+2^{2^{2^2}}+2^{2^{2^{2^2}}}=2*p\quad?$$
Where $p$ is a prime number .
Thanks a lot for your comment or answer .
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No.$$2+2^2+2^{2^2}+2^{2^{2^2}}+2^{2^{2^{2^2}}} \equiv 2 + 1 + 1 +1 + 1 \equiv 0\pmod{3}$$ Because $2^{2n}\equiv (-1)^{2n}\equiv 1\pmod{3}$
|
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Solving $\frac{5}{8} \cot36^\circ = \cos^3x$ without substituting the trig values for $36^\circ$
Find the value of $x$ such that
$$\frac{5}{8} \cot36^\circ = \cos^3x$$
The answer is $x=18^\circ$.
It's really messy to plug in the standard values of $\cos36^\circ$, $\sin36^\circ$ and miraculously guess a suitable value of $x$ and prove that our guess is correct.
Is there any nice way to find the value of $x$?
|
$$\dfrac58\cdot\cot36^\circ=\dfrac{5\cos36^\circ}{8\sin36^\circ}=\dfrac{5\cos^236^\circ}{4\cos18^\circ}$$
Using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
$$\cos36^\circ-(2\cos^236^\circ-1)=\frac12\iff5\cos^2 36^\circ=(1+\cos36^\circ)^2=(2\cos^218^\circ)^2$$
Can you take it from here?
|
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Which expression about power series is correct? I found something weird about power series.
In my textbook,
$$ \frac{1}{1-x} = 1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n \quad |x|<1 $$
and $ \frac{1}{2+x} $ could be expressed using above equation.
$$ \frac{1}{2+x} = \frac{1}{2 \left(1 + \frac{x}{2} \right)} = \frac{1}{2 \left[ 1- \left( - \frac{x}{2} \right) \right]} \\ = \frac{1}{2} \sum_{n=0}^\infty \left( - \frac{x}{2} \right)^n = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n $$
which has interval of convergence $ (-2,2) $.
And below is my solution.
$$ \frac{1}{2+x} = \frac{1}{1-(-1-x)} = 1+(-1-x)+(-1-x)^2+(-1-x)^3+\cdots \\ =\sum_{n=0}^\infty (-1-x)^n \quad |-1-x|<1 $$
which has interval of convergence $ (-2,0) $
Here are my questions.
*
*Why above two power series have different interval of convergence from one equation $ \frac{1}{2+x} $? Why does it appear?
*How do I find out right answer?
|
Notice that $\frac{1}{x+2}$ is undefined at $x = -2$. Any series you produce will not converge at this point.
We can write the series $\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n$ as $\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}(x-0)^n$. We say this series is centered at $0$. We are obstructed at $x = -2$ and $|(-2)-0| = 2$ is the radius of convergence.
We can write the series $\sum_{n=0}^\infty (-1-x)^n$ as $\sum_{n=0}^\infty (-1)^n(x--1)^n$. We say this series is centered at $-1$. We are obstructed at $x = -2$ and $|(-2)- -1| = 1$ is the radius of convergence.
Note that, in general, the obstructions can be anywhere in the complex plane, not just on the real line, so the bound on the radius of convergence may be set by features of the function that are not as easy to see as in this example.
|
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|
For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$ For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$
I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}{4n}$.
|
For $n=2$, we have:
$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2} < \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{3^2}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}<\frac{5}{8}$$
Now assume $n\geq 3$. Since:
$$\frac{1}{x^2} < \frac{1}{x(x-1)}=\frac{1}{x-1}-\frac{1}{x}$$
we have
$$\sum_{n}^{3n-1}\frac{1}{x^2} < \sum_{n}^{3n-1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{n-1}-\frac{1}{3n-1}=\frac{2n}{(n-1)(3n-1)}$$
It remains to prove:
$$\frac{2n}{(n-1)(3n-1)}< \frac{5}{4n}\Leftrightarrow 7n^2-20n+5 > 0$$
which holds when $n \geq 3$.
|
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|
how many five-digit integers can one make from the digits 2,2,6,0,9,4? I want to ask two questions:
(i) How many five-digit integers can one make from the digits $2, 2, 6, 0, 9, 4$?
(ii) How many digits in (i) satisfy that there is no $6$ followed by $2$?
My Approach
(i) I divided it into cases
*
*if we have $2$ in the first place so we have $1 \times 5 \times 4 \times 3 \times 2$ ways to choose the other 4 digits.
*if we have $2$ in the second place so we have $3 \times 1 \times 4 \times 3 \times 2$ ways to choose the 5 digits.
*if we have only one 2 so we have $3 \times 4 \times 4 \times 3 \times 2$ way to choose
and the result is the sum of the 3 cases.
(ii) I considered 62 as one digit, but couldn't solve it.
|
There seem to be some issues in your reasoning. The first question can be solved by considering three scenarios:
*
*The integer contains one $2$. There are four possibilities for the first digit (it cannot be $0$), after which the remaining four digits need to be assigned. The number of possibilities equals: $$4 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 96$$
*The integer contains two $2$s and no $0$. We choose the positions of the $2$s and then assign the positions of the remaining digits: $${5 \choose 2} \cdot 3 \cdot 2 \cdot 1 = 60$$
*The integer contains two $2$s and a $0$. We choose the position of the $0$, then choose the positions of the two $2$s, then choose the two remaining digits and finally assign their positions: $${4 \choose 1} \cdot {4 \choose 2} \cdot {3 \choose 2} \cdot 2 = 144$$
The total number of valid integers dus equals:
$$96 + 60 + 144 = 300$$
To answer the second question, we can consider all cases where the combination $62$ occurs. Using the same distinction as before:
*
*$62$ in front with $3 \cdot 2 \cdot 1 = 6$ possibilities or $62$ in second, third or fourth position, so assign $62$, then $0$, then the remaining digits for a total of $3 \cdot 2 \cdot 2 \cdot 1 = 12$ possibilities
*Consider $62$ as one of four digits, for a total of $4 \cdot 3 \cdot 2 \cdot 1 = 24$ possibilities
*$62$ in front with $3 \cdot {2 \choose 1} \cdot 2 \cdot 1 = 12$ possibilities or $62$ in second, third or fourth position, so assign $62$, then $0$, then the remaining digits for a total of $3 \cdot 2 \cdot {2 \choose 1} \cdot 2 \cdot 1 = 24$ possibilities
The number of possible integers thus equals:
$$300 - 6 - 12 - 24 - 12 - 24 = 222$$
|
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|
Why is the "solving for cubic equation roots general rule" sometimes not applicable while the equation obviously has roots? the general rule:
we have $ax^3+bx^2+cx+d=0$
$\Delta_0=b^2-3ac$
$\Delta_1=2b^3-9abc+27a^2d$
$C=\sqrt{\Delta_1^2-4\Delta_0^3}$
$D=(\frac{\Delta_1+C}{2})^\frac{1}{3}$
$x=-\frac{1}{3a}(b+D+\frac{\Delta_0}{D})$
imagine $x^3-6x^2+11x-6=0$
we know its roots are $x=1$, $x=2$ and $x=3$.
but when you use the general rule, you will find a negative $\Delta_1^2-4\Delta_0^3$ and thus; you can't continue the process!
Also, when you use this rule, you will just find ONE real root (remember the last process to find final $x$); while that equation has 3 real roots and no imaginary roots. so how to find the other real roots using the general rule?!
|
You use the Cardano formula to solve a cubic equation. Your formula for $x$ contains real numbers $C$ and $D$ if $\Delta_1^2-4\Delta_0^3 \ge 0$ and thus obviously gives you a real root. If $\Delta_1^2-4\Delta_0^3 > 0$, there are two additional non-real complex roots which are complex conjugate. If $\Delta_1^2-4\Delta_0^3 = 0$, there is one additional real root of multiplicity two.
The case $\Delta_1^2-4\Delta_0^3 < 0$ is known as the casus irreducibilis. In that case there exist three distinct real roots, but your formula represents them via non-real complex numbers $C$ and $D$. See my answer to Is there really analytic solution to cubic equation? where the complete story is told.
Edited:
In your example $x^3- 6x^2 + 11x - 6 = 0$ we have the three real roots $1,2,3$. Transformation as in the above link with $x = y - \frac{1}{3}(-6) = y + 2$ yields $y^3 - y = 0$. This shows that $y = 0$ is a solution and reduces the problem to $y^2 - 1 = 0$ which gives $y = \pm 1$.
Nevertheless we are in the casus irreducibilis: We have $a = -1, b = 0$, thus $R = -1/27 < 0$. Thus $\sqrt{R} = i\sqrt{1/27}$ and $w = \eta /\sqrt{3}$, where $\eta$ is one of the three complex third roots of $i$. We may take $\eta = \sqrt{3}/2 + i/2$ which gives $w = 1/2 + i/2\sqrt{3}$. Thus $w' = 1/(3w) = 1/2 - i/2\sqrt{3}$ and $y = w + w'= 1$ is a solution found by Cardano's formula.
Your formula yields $\Delta_0 = 3, \Delta_1 = 0$. Similar computations as above give a solution of your equation: $C = i\sqrt{27}$, $D = \eta \sqrt{3}/\sqrt[3]{2}$.
|
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|
A closed form for $\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$ Mathematica does give an analytic form for $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}.$$
The question here is: How to find a simpler closed form for this alternating summation by hand. The summation of the absolute seies has been discussed earlier in MSE: Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$
|
Use Binomial identity:
$$
(1+t)^n=\sum_{k=0}^{n} {n \choose k}t^n.
\tag{1}
$$
Integration of $(1)$ from $t=0$ to $t=x$ gives
$$
\frac{(1+x)^{n+1}-1}{n+1}= \sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}.\tag{2}
$$
We can change $x$ to $-1/x$ in $(2)$ to get
$$
\frac{(1-1/x)^{n+1}-1}{n+1}= \sum_{k=0}^n (-1)^k {n \choose k}\frac{x^{-k-1}}{k+1}
.
\tag{3}
$$
Multiplying $(2)$ and $(3)$ and collecting terms free of $x$ on RHS, we get
$$
\frac{x^{-n-1}\big[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}\big]}{(n+1)^2}=\sum_{k=0}^{n}\frac{ (-1)^k {n \choose k}^2}{(k+1)^2} x^0+\dots
$$
Equating the coefficients yields
$$S_n = \sum_{j=0}^{n} \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$$
$$S_N=[x^{n+1}]~\frac{[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}]}{(n+1)^2} $$
$$S_n=\frac{-(-1)^{(n+1)/2}{n+1 \choose (n+1)/2}+1}{(n+1)^2}, \text{when $n$ is odd}$$
$$S_n= \frac{1}{(n+1)^2}, \text{when $n$ is even} $$
|
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|
Distance from $P(4,3)$ to tengent point on curve $x^2 + y^2 -2x-4=0$
Tangent line to curve $$x^2 + y^2 -2x-4=0$$ at A passes $P(4,3)$.
Find distance A and P.
Tangent of the line is $\frac{2-2x}{\sqrt{2x+4-x^2}}$
A(k,l) means
$l = \sqrt{2k+4-k^2}$
And
$l = \frac{2-2k}{\sqrt{2k+4-k^2}}$
So, I get $k=\frac{3\pm\sqrt{21}}{2}$
$AP = \sqrt{(k-4)^2 + (l-3)^2}$
Its so complicated, am i wrong somewhere?
|
Let $O(1,0)$ be the center of the given circle $(x-1)^2 + y^2 =5$. Then,
$$PO^2 = (4-1)^2+(3-0)^2 = 18, \>\>\>\>\>OA^2 = 5$$
Use the fact that PAO is a right triangle to calculate,
$$PA = \sqrt{PO^2-OA^2} = \sqrt{18-5}=\sqrt{13}$$
|
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|
For non-negative reals such that $a+b+c\geq x+y+z$, $ab+bc+ca\geq xy+yz+zx$, and $abc\geq xyz$, show $a^k+b^k+c^k\geq x^k+y^k+z^k$ for $0
Let $a$, $b$, $c$, $x$, $y$, $z$ be non-negative real numbers such that
$$a+b+c \geq x+y+z,$$
$$ab+bc+ca \geq xy+yz+zx,$$
$$ abc \geq xyz$$
Show that
$$a^k+b^k+c^k \geq x^k+y^k+z^k, \quad 0 < k < 1$$
This is case $n=3$ of a problem posted by Ji Chen in the Art of Problem Solving forums, 2008.
I have posted a partial result in an answer below.
I have a proof when $r = \frac12,$ for weaker conditons
$$a+b+c = x+y+z,$$
$$\min(x, y, z) \leqslant \min(a, b, c),$$
$$\max(a, b, c) \leqslant \max(x, y, z).$$
Indeed, if $u, v > 0$ it's easy check
$$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$
Assume $x \geqslant y \geqslant z$ and $a \geqslant b \geqslant c$ then $x \geqslant a, \; c \geqslant z.$ Therefore
$$\begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}$$
|
Thank @Michael Rozenberg, I have a proof when $k = \frac12,$ for weaker conditons
$$a+b+c = x+y+z,$$
$$\min(x, y, z) \leqslant \min(a, b, c),$$
$$\max(a, b, c) \leqslant \max(x, y, z).$$
Indeed, if $u, v > 0$ it's easy check
$$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$
Assume $x \geqslant y \geqslant z$ and $a \geqslant b \geqslant c$ then $x \geqslant a, \; c \geqslant z.$ Therefore
$$\begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}$$
|
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|
Roots of quadratic with $a$ and $8a+3b+c$ of the same sign cannot lie in (2,4) I have the second degree equation $ax^2+bx+c=0$ with $a$ non-zero such that $a$ and $8a+3b+c$ have the same sign and I need to prove that both the roots cannot lie in $(2,4)$.
My try:
I assume two roots are in $(2,4)$
$2 < \frac{-b+\sqrt{b^2-4ac}}{2a} < 4$
$2 < \frac{-b-\sqrt{b^2-4ac}}{2a}< 4$
$2+\frac{b}{2a}<\frac{\sqrt{b^2-4ac}}{2a} < 4+\frac{b}{2a}$
and
$2+\frac{b}{2a}<-\frac{\sqrt{b^2-4ac}}{2a} < 4+\frac{b}{2a}$
$\implies 2+\frac{b}{2a}<|\frac{\sqrt{b^2-4ac}}{2a}|<4+\frac{b}{2a}$
$\implies (2+\frac{b}{2a})^2<\frac{b^2-4ac}{4a^2}<(4+\frac{b}{2a})^2$
$\implies 4+4\frac{b}{a}+\frac{b^2}{4a}<\frac{b^2-4ac}{4a^2}<16+4\frac{b}{a}+\frac{b^2}{2a^2}$
$\implies 4+4\frac{b}{a}<\frac{-c}{a}<16+4\frac{b}{a}$
What to do next? How to apply condition?
|
Hint: Let's say the roots are $x_1$ and $x_2$. If the roots lie in $(2,4)$ simultaneously, then:
$$(x_1-2)(x_2-4)+(x_1-4)(x_2-2)<0$$
and then Vieta's. Can you end it from here?
|
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|
Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$
Evaluate $$\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$$
Let $$y=x + \frac{1 \cdot 3 \cdot}{2!} x^2 + \frac{1 \cdot 3 \cdot 5}{3!} x^3+\ldots$$ be the given expression.(replacing $2/5$ with $x$)
After some manipulations,
$$y+1=(1-2x)\frac{dy}{dx}$$
Integrating and substituting $x=\dfrac{2}{5}$, we get $y=\sqrt{5}-1$.
Is there any other way to solve this question?
|
To carry out the integration of
$y+1=(1-2x)\frac{dy}{dx}
$
we have
$\dfrac1{1-2x}
=\dfrac{y+1}{(y+1)'}
=(\ln(y+1))'
$
so
$\ln(y+1)
=\int \dfrac{dx}{1-2x}
=-\frac12\ln(1-2x)+c
$
so
$y+1
=\dfrac{C}{\sqrt{1-2x}}
$
so
$y
=\dfrac{C}{\sqrt{1-2x}}-1
$.
At $x=0, y=0$ so
$C=1$.
Then,
for $x = \frac25$,
$y
= \dfrac{1}{\sqrt{1/5}}-1
=\sqrt{5}-1
$.
|
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|
simple Application of Gram-Schmidt Orthogonalization I wanted to apply the Gram-Schmidt orthogonalization to the following simple example of polynomials $1,x,x^2,x^3$ in $L^2[-1,1]$, is it correct?
$e_1 = 1$
$e'_2 = x - \int_{-1}^{1}xdx = x \implies e_2 = \frac{x}{\|x\|}$
$e'_3 = x^2 - \int_{-1}^{1}x^2dx -\int_{-1}^{1}x^2xdx = x^2 - \frac{1}{2} \implies e_4 = \frac{x^2 - \frac{1}{2}}{\|x^2 - \frac{1}{2}\|}$
$e'_4 = x^3 - \int_{-1}^{1}x^3dx -\int_{-1}^{1}x^3xdx -\int_{-1}^{1}x^3x^2dx = x^3 - \frac{2}{5} \implies e_3 = \frac{x^3 - \frac{2}{5}}{\|x^3 - \frac{2}{5}\|}$
|
For $e'_3$ the first integral is wrong. $\int_{-1}^1x^2dx=\frac23$. But you forgot to apply the norm at every step. For example
$$e'_3=x^2-\frac{\int_{-1}^1x^2\cdot 1dx}{\int_{-1}^11\cdot1dx}-x\frac{\int_{-1}^1x^2\cdot xdx}{\int_{-1}^1x\cdot xdx}=x^2-\frac 13$$
And so on. These are the Legendre polynomials.
$$e'_4=x^3-\frac{\int_{-1}^1x^3\cdot 1dx}{\int_{-1}^11\cdot1dx}-x\frac{\int_{-1}^1x^3\cdot xdx}{\int_{-1}^1x\cdot xdx}-x^2\frac{\int_{-1}^1x^3\cdot x^2dx}{\int_{-1}^1x^2\cdot x^2dx}=x^3-\frac {\frac25}{\frac23}x=x^3-\frac35x$$
|
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|
Find the value of $\int_\frac{-\pi}{2}^\frac{\pi}{2} 2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx$
Find the value of $$\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx.$$
I tried to solve it using the concept: $$\int_{a}^b f(x)\text dx=\int_{a}^b f(a+b-x)\text dx.$$
My step is as follow:
$$I=\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x}dx=\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{-\sin x}dx.$$
$$I=\int_\frac{-\pi}{2}^\frac{\pi}{2} f(x)dx=\int_\frac{-\pi}{2}^\frac{\pi}{2}f(-x)dx$$
|
It seems this problem is designed to use the formula
$$\int_a^bf(x)\;dx + \int_{f(a)}^{f(b)}f^{-1}(x)\;dx = bf(b)-af(a)$$
(See here)
You only need to rewrite the second integral using $x\mapsto x-2$ as
$$\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \;dx= \int_{\frac 12}^{2}\sin^{-1}(\log_2(x))\;dx$$
So, you get
$$\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx$$
$$=\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x} \text dx + \int_{\frac 12}^{2}\sin^{-1}(\log_2(x))\;dx= \frac{\pi}{2}\cdot 2+ \frac{\pi}{2}\cdot\frac 12 = \frac 54\pi$$
|
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|
Verify the integral $\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$ I'm stuck solving the integral
$$\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$$
This is what I got so far
\begin{align}
\int_{1}^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}} - u} \,du &= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{u - u\cdot(1 + \sqrt{u^2 + 1})} \,du \\
&= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{-u\cdot \sqrt{u^2 + 1}} \,du = -\bigg( \underbrace{\int_1^z \frac{1}{u\sqrt{u^2 + 1}} \,du}_{(*)} + \int_1^z \frac{1}{u}\,du\bigg) \\
&= - (*) - \ln(z)
\end{align}
Calculation of $(*)$ yields
\begin{align*}
(*) &= \int_1^z \frac{1}{u \cdot \sqrt{u^2 + 1}} \,du = \int_1^z \frac{u}{u^2 \cdot \sqrt{u^2 + 1}} \,du = \int_1^z \frac{\varphi'(u)}{\varphi(u)^2 - 1} \,du = \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u^2 -1} \,du \\
&= \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u-1} \,du - \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u+1} \,du = \frac{1}{2} \bigg[\ln(u-1) - \ln(u+1)\bigg]_{\sqrt{2}}^{\sqrt{z^2 + 1}} \\
&= \frac{1}{2} \left(\ln(\frac{\sqrt{z^2 + 1} - 1}{\sqrt{z^2 + 1} + 1}) - \ln(\frac{\sqrt{2} - 1}{\sqrt{2} + 1})\right)
\end{align*}
Thus we have the result
\begin{align*}
&-\frac{1}{2} \left(\ln(\frac{\sqrt{z^2 + 1} - 1}{\sqrt{z^2 + 1} + 1}) - \ln(\frac{\sqrt{2} - 1}{\sqrt{2} + 1})\right) - \ln(z) \\
&= -\frac{1}{2} \left(\ln(\frac{(\sqrt{z^2 + 1} - 1)^2}{(\sqrt{z^2 + 1} + 1) \cdot (\sqrt{z^2 + 1} - 1)}) - \ln(\frac{(\sqrt{2} - 1)^2}{(\sqrt{2} + 1)\cdot(\sqrt{2} - 1)})\right) - \ln(z) \\
&= -\frac{1}{2} \left(\ln(\frac{z^2 + 1 - 2\sqrt{z^2 + 1} + 1}{z^2}) - \ln(\frac{2 - 2\sqrt{2} + 1}{2 - 1})\right) - \frac{1}{2}\ln(z^2) \\
&= -\frac{1}{2} \left(\ln(z^2 - 2\sqrt{z^2 + 1} + 2) - \ln(3 + 2\sqrt{2})\right) \\
&= \frac{1}{2} \ln(3 + 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)
\end{align*}
The correct solution is given by
$$\ln(\frac{\sqrt{z^2 + 1} + 1}{z^2(1 + \sqrt{2})})$$
I'm not able to simplify my solution to the given solution. Did I integrate wrong or is there a way to verify the solution?
Edit: I think, that I made an mistake somewhere, since the solution should equal zero for $z = 1$ but $1$ is not root of my solution..
Edit 2: I found the mistake. I switched "+" and "-" in the line of the second last equality. My question remains how to simplify my solution to the given solution.. My solution then looks like this
$$\frac{1}{2} \ln(3 - 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)$$
|
HINT: (comment)
Substitution $ u= \sinh x $ makes it easy as it gives three log terms.
$$ \int \dfrac{-(1+\cosh x)dx}{\sinh x} $$
|
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|
$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$ Prove the following inequality
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$
I tried by multying both sides by the denominator $(a+b+c)^2$ and then applying Holder for the left side but I couldn’t work it out. I would prefer a proof without never ending computations (a.k.a. Brute Force / opening up the brackets) because I know how to do it this way already. A proof using C-B-S, Holder, Titu’s Lemma or their generalizations or other well-known inequalities would be ideal. Thank you!
|
This seems to be deceptively cumbersome, needing a tight bound on the LHS while symmetrisation. The following is from my old notes, unfortunately no way to attribute it correctly....
First we need a well known inequality $4(x+y+z)^3 \geqslant 27(x^2y+y^2z+z^2x+xyz)$, which follows from AM-GM, as WLOG we may assume $y$ is in between $x, z$:
$$\frac{\frac{x+z}2+y+\frac{x+z}2}3\geqslant \sqrt[3]{\frac{x+z}2\cdot y \cdot \frac{x+z}2}$$
$$\implies \frac4{27}(x+y+z)^3 = x^2y+y^2z+z^2x+xyz - z(y-x)(y-z) \\\geqslant (x^2y+y^2z+z^2x+xyz)$$
Using the above with $x = \frac{a}b, y=\frac{b}c, z = \frac{c}a$, we get
$$\left(\frac{a}b+\frac{b}c+\frac{c}a \right)^3 \geqslant \frac{27}4\left( \frac{a^3+b^3+c^3}{abc}+1\right)$$
Hence to prove the original inequality, it is enough to show instead the symmetric
$$\frac{a^3+b^3+c^3}{abc} +1\geqslant 108 \frac{(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{1}$$
$$\iff \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{abc}+4\geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{2}$$
Using the obvious $(ab+bc+ca)^2 \geqslant 3abc(a+b+c)$,
$$\frac{a+b+c}{abc} \geqslant \frac{3(a+b+c)^2}{(ab+bc+ac)^2}$$
and hence, it is enough to show
$$\frac{3(a+b+c)^2(a^2+b^2+c^2-ab-bc-ac)}{(ab+bc+ac)^2} + 4 \geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{3}$$
Setting
$t=\dfrac{3(a^2+b^2+c^2)}{(a+b+c)^2}\, \in [1, 3)$
we need to equivalently show
$$\frac{54(t-1)}{(3-t)^2}+4\geqslant 4t^3 \tag{4}$$
which is $(t-1)(9-6t-8t^2+10t^3-2t^4)\geqslant 0$
and hence holds true as
$$9-6t-8t^2+10t^3-2t^4=2(3+3t-t^2)(t-1)^2+3>0$$
|
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|
Show $\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta \leq \frac{C}{(1+\vert v \vert)^2}$
$\textbf{Problem}$
\begin{equation*}
\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta
\leq \frac{C}{(1+\vert v \vert)^2}
\end{equation*}
For obtaining the above upper bound, I tried to change the variable $\eta \rightarrow v+ \sigma \rho $ for $\sigma \in \mathbb{S}^2, \rho \in \mathbb{R}^{+}$.
However, I stuck to handle the part $\displaystyle{\frac{1}{(1+\vert{\eta\vert})^4}}$.
$\textbf{Attempt}$
\begin{align*}
\frac{1}{(1+\vert \eta \vert)^4} &= \frac{1}{(1+\vert v+\sigma \rho\vert)^4}\\
\int_{\mathbb{R}^3} \frac{1}{\vert \eta - v \vert^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta &= \int_0^{\infty} \int_{\mathbb{S}^2} \frac{1}{\rho^2} \frac{1}{(1+\vert v + \sigma \rho\vert)^4} \rho^2 d\sigma d\rho
\end{align*}
I don't know how to derive $(1+\vert v \vert)^2$ from the last integral term in my attempt.
Any help is appreciated.
Thank you!
|
We can compute the integral by rotating the $\sigma$-coordinate system in such a way that $v$ points towards the north pole of the sphere (i.e. in $e_3$-direction) and introducing polar coordinates $(\theta,\phi)$. Then the norm becomes $\lvert v + \rho \sigma \rvert = \sqrt{\lvert v \rvert^2 + \rho^2 + 2 \lvert v \rvert \rho \cos(\theta)}$ and the integral turns into
$$ \int \limits_0^\infty \int \limits_0^\pi \int \limits_0^{2\pi} \frac{\sin(\theta) \, \mathrm{d} \phi \, \mathrm{d} \theta \, \mathrm{d} \rho}{\left(1 + \sqrt{\lvert v \rvert^2 + \rho^2 + 2 \lvert v \rvert \rho \cos(\theta)}\right)^4} = 2 \pi \int \limits_0^\infty \int \limits_{-1}^1 \frac{\mathrm{d} t \, \mathrm{d} \rho }{\left(1 + \sqrt{\lvert v \rvert^2 + \rho^2 + 2 \lvert v \rvert \rho t}\right)^4} \equiv f(\lvert v \rvert) \, .$$
Clearly, $f(0) = \frac{4 \pi}{3}$. For $r > 0$ we can let $s = \sqrt{r^2 + \rho^2 + 2 r \rho t}$ to obtain
$$ f(r) = 2 \pi \int \limits_0^\infty \int \limits_{\lvert r - \rho \rvert}^{r + \rho} \frac{s}{r \rho (1+s)^4} \, \mathrm{d} s \, \mathrm{d} \rho \, .$$
The remaining integrals are not particularly hard, albeit slightly tedious, and the final result looks rather nice:
$$ f(r) = \frac{4\pi}{3} \begin{cases} 1, & r = 0 \\ \frac{1}{3}, & r = 1 \\ \frac{1-r^4 + 4 r^2 \log(r)}{(1-r^2)^3} , & r \in (0,\infty) \setminus \{1\}\end{cases} \, .$$
We want to find some $C > 0$ such that $f(r) \leq \frac{C}{(1+r)^2}$ holds for every $r \geq 0$.
*
*If any constant is fine, we can simply use the inequality $\frac{-\log(r)}{1-r} \geq \frac{2}{1+r}\, ,r > 0,$ which follows from $\frac{\tanh(t)}{t} \leq 1\, , t \in \mathbb{R}$, with $t = \frac{1}{2} \log(r)$.
It leads to
$$ \frac{3}{4\pi} f(r) = \frac{1 + r + r^2 + r^3 - 4 r^2 \frac{-\log(r)}{1-r}}{(1+r)^3 (1-r)^2} \leq \frac{1 + r + r^2 + r^3 - 4 r^2 \frac{2}{1+r}}{(1+r)^3 (1-r)^2} = \frac{1 + \frac{2 r}{(1+r)^2}}{(1+r)^2} \leq \frac{3}{2 (1+r)^2} $$
and therefore $C = 2 \pi$.
*In order to find the optimal constant we need to use the more complicated inequality $$\frac{-\log(r)}{1-r} \geq \frac{-1 + 7r + 7r^2-r^3}{12 r^2} \, , \, r > 0,$$
which is a consequence of
$$ \frac{\tanh(s)}{s} \leq \frac{1}{4} \left[\frac{\sinh(s)}{s} + \frac{3}{\cosh(s)}\right]\, , \, s \in \mathbb{R},$$
with $s = \log(r)$ (the latter can be proven by rewriting it as $1 - \frac{\sinh(s) [4- \cosh(s)]}{3 s} \geq 0$ and noting that the Maclaurin series of the left-hand side contains only non-negative terms). It follows that
$$ \frac{3}{4\pi} f(r) \leq \frac{1 + r + r^2 + r^3 - \frac{-1+7r+7r^2-r^3}{3}}{(1+r)^3 (1-r)^2} = \frac{4}{3 (1+r)^2} $$
and the corresponding constant $C = \frac{16 \pi}{9}$ is the smallest possible, since $f(1) = \frac{4\pi}{9} = \frac{16 \pi}{9 (1+1)^2}$.
It would be nice to obtain the estimate without evaluating the integral explicitly, but I have not found a way to do that yet.
|
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|
Use Taylor series to compute $\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $ Use Taylor series to solve $$\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $$
This equals $$\lim\limits_{x \to 0} \left( \sum_{n=0}^{\infty}\frac{(2n+1)!}{x^{2n+1}(-1)^{n+1}} - \sum_{n=0}^{\infty}\frac{1}{(-x+1)^n} \right)$$
I am unsure how to proceed. I tried writing out the first few terms, but nothing seemed to cancel.
|
Let $ x $ be a real from $ \mathbb{R}^{*} \cdot $
Since $ \sin{x}=\sum\limits_{n=0}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n+1}}{\left(2n+1\right)!}} $, we get that $ \frac{x-\sin{x}}{x^{3}}=\sum\limits_{n=0}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n}}{\left(2n+3\right)!}}=\frac{1}{6}+x^{2}\sum\limits_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n-2}}{\left(2n+3\right)!}}\underset{x\to 0}{\longrightarrow}\frac{1}{6} \cdot $
Which means \begin{aligned} \lim_{x\to 0}{\left(\frac{1}{\sin{x}}-\frac{1}{x}\right)}&=\lim_{x\to 0}{x\left(\frac{x-\sin{x}}{x^{3}}\right)\left(\frac{x}{\sin{x}}\right)}\\ &=0\times\frac{1}{6}\times 1\\ \lim_{x\to 0}{\left(\frac{1}{\sin{x}}-\frac{1}{x}\right)}&=0 \end{aligned}
|
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|
For which value of x $\sum_{n=1}^\infty \frac{n^{nx}}{n!}$ converges I want to know for which values of $x$ this series
$$\sum_{n=1}^\infty \frac{n^{nx}}{n!}$$ converges.
This series is defined $ \forall x \in R$.
$a_n=\frac{n^{nx}}{n!}= \frac{(n^x)^{n}}{n!}= \frac{(n^x)}{n} \frac{(n^x)}{n-1}
\frac{(n^x)}{n-2}...\frac{(n^x)}{3} \frac{(n^x)}{2} \frac{(n^x)}{1}= n^{x-1} \frac{n^{x-1}}{1-\frac{1}{n}}...\frac{(n^x)}{3} \frac{(n^x)}{2} \frac{(n^x)}{1}
\sim 0 \Leftrightarrow x<1$.
Applying the ratio test:
$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{(n+1)x}}{(n+1)!} \frac{n!}{(n)^{nx}}=\frac{(n+1)^{(n+1)x}}{n+1} \frac{1}{(n)^{nx}}=\frac{(n+1)^{nx+x-1}}{n^{nx}} \sim \frac{(n)^{nx+x-1}}{n^{nx}}= (n)^{x-1}<1 \Leftrightarrow x-1<0 \Leftrightarrow x<1 $
Is it right?
|
For $ n\in\mathbb{N}^{*} $, denoting $ f_{n}:x\mapsto\frac{n^{nx}}{n!} \cdot $
If $ x\geq 1 $, observe that $ f_{n}\left(x\right)\geq\frac{n^{n}}{n!}=\prod\limits_{k=1}^{n-1}{\frac{n}{k}}\geq\prod\limits_{k=1}^{n-1}{\frac{k+1}{k}}=n $, meaning $ \lim\limits_{n\to +\infty}{f_{n}\left(x\right)}=+\infty\neq 0 $, which means $ \sum\limits_{n\geq 1}{f_{n}\left(x\right)} $ diverges.
If $ x<1 $, using d'Alembert's ratio test, $ \lim\limits_{n\to +\infty}{\frac{f_{n+1}\left(x\right)}{f_{n}\left(x\right)}}=\lim\limits_{n\to +\infty}{\frac{\left(n+1\right)^{\left(n+1\right)x}}{n^{nx}\left(n+1\right)}}=\lim\limits_{n\to +\infty}{\left(n+1\right)^{x-1}\left(1+\frac{1}{n}\right)^{nx}}=0<1 $, meaning $ \sum\limits_{n\geq 1}{f_{n}\left(x\right)} $ converges.
|
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|
$y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes? $y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes?
My attempt is as follows:-
Horizontal asymptotes are the horizontal lines which signify the values of $y$ which graph cannot ever attain.
There are two ways to find horizontal asymptotes for the rational function:
$1)$ If degree of numerator is equal to degree of denominator, then $y=\dfrac{\text {leading coefficient of numerator }}{\text{leading coefficient of denominator}}$
So here we have degree of denominator and numerator as equal, so $y=-\dfrac{1}{2}$ will be the horizontal asymptote.
It means graph cannot ever touch $y=-\dfrac{1}{2}$
$2)$ Another way is by finding range, elements which are not in the range will correspond to horizontal asymptotes.
Let's find out the range,
$$-4x^2y+3xy+2y=2x^2+3x-4$$
$$x^2(2+4y)+3x(1-y)-4-2y=0$$
$$D\ge0$$
$$9(1+y^2-2y)+4(4+2y)(2+4y)\ge0$$
$$9+9y^2-18y+4(8+20y+8y^2)\ge0$$
$$41y^2+62y+41\ge0$$
This is always greater than equal to zero because $D=62^2-4\cdot1681<0$
So this indicates that the range is $\left(-\infty,\infty\right)$. So it means there should be no horizontal asymptotes, but by the first way we got $y=-\dfrac{1}{2}$ as horizontal aysmptote.
What am I missing here?
|
Another way of finding the horizontal asymptotes is by multiplying by $\frac{\frac1{x^2}}{\frac1{x^2}}$
$$
y = \dfrac{2x^2+3x-4}{-4x^2+3x+2} \sim \dfrac{2x^2+3x-4}{-4x^2+3x+2}\frac{\frac1{x^2}}{\frac1{x^2}} = \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}}
$$
and find the limit as $x\to\infty$
$$
\lim_{x\to\infty} y \sim \lim_{x\to\infty} \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}} = \dfrac{2 + 0 + 0}{-4 + 0 + 0} = -\frac24 = \color{green}{\boxed{-\frac12}}
$$
|
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|
Elementary proof of tangent half angle formula
Hi all, I am interested to find elementary proof of tangent half angle formula.
My solutions are the following:
Triangle $AOB$ is such that $|AB|=1$ and $\angle AOB=\theta$. We then extend $OB$ to $P$ and $Q$ such that $|OP|=|OQ|=1$. Thus we will have two isosceles triangles: $AOP$ and $AOQ$.
From the picture, $\tan{\left(\frac{\theta}{2}\right)}=\frac{AB}{BP}=\frac{\sin{\left(\theta\right)}}{1+\cos{\left(\theta\right)}}\ \ \ =\frac{BQ}{AB}=\frac{1-\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}$
Could You guys please check my solution. I am also wondering if there are other elementary solutions, please share, thanks!
|
Here's another proof:
$\tan(\frac{A}{2}) = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \cdot \left(\frac{2\sin\frac{A}{2}}{2\sin\frac{A}{2}}\right) = \frac{2\sin^2(\frac{A}{2})}{2\sin(\frac{A}{2})\cos(\frac{A}{2})} = \frac{2 \cdot \frac{1-\cos A}{2}}{\sin \left(2 \cdot \frac{A}{2}\right)} = \frac{1-\cos A}{\sin A}$
And yes, to me, your proof is correct.
|
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|
Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6t+1=0$$
Or,$$3t^2+6t-1=0$$
On solving I got the roots
$$\dfrac{-3+2\sqrt6}3$$ and $$\dfrac{-3-2\sqrt6}3$$
Then I plugged in the values and found only 2 roots are real use discriminant
|
You have committed a mistake. The correct solution after substitution is
$$3(t^2-2)+6t+1=0\\
\implies3t^2+6t-5=0$$
Note that it takes value $3\times4+12-5>0$ at $2$ and $-5<0$ at $-2$. Hence, there is precisely one root between $2$ and $-2$. But we have $2$ solutions of the original quartic equation for each value of $|t|\ge2$. Hence, the original quartic equation has $2$ real solutions.
|
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|
Finding integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square I am tring to find "integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square". Make this statement into equation I have
\begin{align}
x + 1 = a^2, \qquad \frac{x}{2} + 1 = b^2
\end{align}
where $x,a,b \in \mathbb{N}$.
From knowledge of odd perfect square is of the form of $8k+1$. I noticed $x$ should be multiple of 16.
From trial and errors I found the smallest integer $x$ is $48$. i.e.,
\begin{align}
48 +1 = 7^2, \qquad 24+1 = 5^2
\end{align}
and the second one is 1680. What is the general form of $x$ and how one can find that?
|
We have:
$$2b^2-1=(x+2)-1=x+1=a^2 \implies a^2-2b^2=-1$$
This is a Pell Equation. We can observe that:
$$(a_1^2-2b_1^2)(a_2^2-2b_2^2)=(a_1a_2+2b_1b_2)^2-2(a_1b_2+a_2b_1)^2=A^2-2B^2$$
This means that multiplying values of the form $x^2-2y^2$ will result in values of the same form. It is easy to see that the first solution is $(a,b)=(1,1)$. Now, we can see that since $a^2-2b^2=1$, the expression $$(a^2-2b^2)^{2k-1}=-1$$
will generate solutions when $k \in \mathbb{N}$. For example,
$$k=2 \implies (1^2-2\cdot1^2)^3=(3^2-2\cdot2^2)(1-2\cdot1^2)=7^2-2\cdot5^2=-1$$
This gives the solution $(a,b)=(7,5)$ that you had gotten.
$$k=3 \implies (1^2-2\cdot1^2)^5=(7^2-2\cdot5^2)(3^2-2\cdot2^2)=41^2-2\cdot29^2=-1$$
This gives $(a,b)=(41,29)$ and so on...
|
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|
Function and Floor Properties Call a function $f : x \mapsto \lfloor x / 2\rfloor$. Let $x_0 = n$, then $x_1 = \lfloor n / 2 \rfloor$ and $x_2 = \lfloor \lfloor n / 2 \rfloor \rfloor / 2 \rfloor$.... $x_k = f(x_{k-1}).$
How do we prove that $x_k = \left\lfloor \frac{n}{2^k} \right\rfloor$?
My best guess is induction but I'm not sure where to begin. I started off with using the floor definition but it got messy.
|
Induction is a good idea.
Note if $x_k = [\frac n{2^k}]$ then
$x_k \le \frac n{2^k} < x_k + 1$ so
$\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12$
Now if $x_k$ is even than $\frac {x_k}2$ is an integer and
$\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12<\frac {x_k}2+1$
And $x_{k+1} = [\frac {x_k}2] = \frac {x_k}2$ and $[\frac n{2^{k+1}}] = [\frac {x_k}2]=x_{k+1}$
And if $x_k$ is odd then $\frac {x_k}2 \pm \frac 12$ are integesr.
$\frac {x_k}2-\frac 12 <\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12$
And both $x_{k+1} = [\frac {x_k}2] = \frac {x_k}2-\frac 12 $ and $[\frac n{2^{k+1}}] =\frac {x_k}2-\frac 12 = [\frac {x_k}2] =x_{k+1}$
|
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"url": "https://math.stackexchange.com/questions/3600887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
equation with quadratic power I was thinking how to solve:
*
*If $x^{(x-1)^2}=2x+1$, find $x-\frac{1}{x}$
*Solve $x^{x-x^2+13} = x^2-12$
I noticed that in both problems, the linear part can be constructed in the quadratic exponent, I tried a few change of variable and nothing.
|
Here's a possible approach to the second problem:
Observe that $x-x^2+13=-(x^2-12)+x+1$, so let $a=x^2-12$.
Now, we wish to solve the equation $x^{-a+x+1}=a$. Note that $x=0$ is not a solution; this is by inspection.
It is easy to see that $x=a$ actually satisfies the preceding equation. Now, we argue that all other values of $x$ does not work:
$x>a \Rightarrow x-a >0 \Rightarrow x-a+1>1 \Rightarrow x^{x-a+1}>a^{x-a+1} >a \ \forall \ x$, if $x \in \mathbb{R^+}$. So, equality does not hold.
Likewise, $x<a \Rightarrow x-a <0 \Rightarrow x-a+1<1 \Rightarrow x^{x-a+1}<a^{x-a+1} <a \ \forall \ x$, if $x \in \mathbb{R^+}$, and $x-a+1 \geq 0$. Otherwise, if $x-a+1<0$, $ x^{x-a+1}>a^{x-a+1} >a$. So equality never happens.
Thus, given that $x=a$, we have $x^2-12=x \Rightarrow (x+4)(x+3)=0 \Rightarrow x=4$ or $ x=-3$, and these are the two solutions to our equation, so we are done.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $\big|\frac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\frac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\big|$ If $$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}=1,$$ where $\theta_1$ and $\theta_0$ do not differ by an odd multiple of $\pi$, then find the value of $$\left|\dfrac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|.$$
My attempt is as follows:
Attempt $1$:
$$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}$$
$$\dfrac{\sin\left(\theta_1+\theta_2\right)}{\cos\theta_2\sin\theta_2}=\dfrac{\sin\left(\theta_0+\theta_2\right)}{\cos\theta_2\sin\theta_2}$$
$$\sin\left(\theta_1+\theta_2\right)-\sin\left(\theta_0+\theta_2\right)=0$$
$$2\sin\dfrac{\left(\theta_1-\theta_0\right)}{2}\cos\dfrac{2\theta_2+\theta_1+\theta_0}{2}=0$$
either $\theta_1-\theta_0=2n\pi$ or $2\theta_2+\theta_1+\theta_0=2n\pi$
Unfortunately it is given that $\theta_1-\theta_0$ is not equal to odd multiple of $\pi$ but we are getting $\theta_1-\theta_0$ as even multiple of $\pi$ so we cannot rule out one of the factor. Due to this reason I didn't find any way ahead.
Attempt $2$:
$$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=1$$
$$\sin\left(\theta_1+\theta_2\right)=\cos\theta_2\sin\theta_2$$
$$2\sin\left(\theta_1+\theta_2\right)=\sin2\theta\tag{1}$$
In the similary way
$$2\sin\left(\theta_0+\theta_1\right)=\sin2\theta\tag{2}$$
But by doing this we are tending towards result obtained in Attempt $1$:
Attempt $3$:
$$\left|\left(1-\dfrac{\sin\theta_0}{\sin\theta_2}\right)\left(1-\dfrac{\sin\theta_1}{\sin\theta_2}\right)+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|$$
$$\left|1-\dfrac{\sin\theta_1}{\sin\theta_2}-\dfrac{\sin\theta_0}{\sin\theta_2}+\dfrac{2\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|$$
Also not getting anything from here, what to do?
|
We have $$\sin\dfrac{\theta_0-\theta_1}2\cos\dfrac{\theta_0+\theta_1-2\theta_2}2=0$$
If $\sin\dfrac{\theta_0-\theta_1}2=0,$ this will make both equations identical
$$\implies\cos\dfrac{\theta_0+\theta_1-2\theta_2}2=0\implies\dfrac{\theta_0+\theta_1-2\theta_2}2=\dfrac{(2n+1)\pi}2$$ for some integer $n$
$\implies\theta_0+\theta_1=2n\pi+\pi-2\theta_2$
$$1=\left(\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}\right)\left(\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}\right) =\dfrac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}+\dfrac{\sin(\theta_0+\theta_1)}{\sin\theta_2\cos\theta_2}$$
Replace the value of $\theta_0+\theta_1$ in the last term
|
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|
In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to
My attempt is as follows:-
Squaring both equations and adding
$$9+16+24\sin(P+Q)=37$$
$$\sin(P+Q)=\dfrac{1}{2}$$
either $P+Q=\dfrac{\pi}{6}$ or $P+Q=\dfrac{5\pi}{6}$
If $P+Q=\dfrac{\pi}{6}$, then $R=\dfrac{5\pi}{6}$ otherwise $R=\dfrac{\pi}{6}$
Let's see case $1$: $P+Q=\dfrac{\pi}{6}$
$$3\sin P+4\cos\left(\dfrac{\pi}{6}-P\right)=6$$
$$3\sin P+4\left(\dfrac{\sqrt{3}}{2}\cos P+\dfrac{1}{2}\cdot\sin P\right)=6$$
$$3\sin P+2\sqrt{3}\cos P+2\sin P=6$$
$$5\sin P+2\sqrt{3}\cos P=6\tag{1}$$
$$4\left(\dfrac{1}{2}\cdot\cos P-\sin P\cdot\dfrac{\sqrt{3}}{2}\right)+3\cos P=1$$
$$-2\sqrt{3}\sin P+5\cos P=1\tag{2}$$
$$\cos P=\dfrac{12\sqrt{3}+5}{37}$$
$$\sin P=\dfrac{30-2\sqrt{3}}{37}$$
Using calculator I found $\cos P=0.69$, this means $P>\dfrac{\pi}{6}$ because $\cos \dfrac{\pi}{6}=0.866$, this mean $Q$ will be negative because $Q=\dfrac{\pi}{6}-P$. So this cannot be the case hence $P+Q$ would be $\dfrac{5\pi}{6}$ and $R$ will be $\dfrac{\pi}{6}$
This is the correct answer also , but I want to know does there exist any better way to decide on the value of $P+Q$. I am asking this because I had to use the calculator for finding the value of $\cos P$.
|
If $R=\frac{5\pi}6$, we have
$$3\sin P + 4\cos(\frac\pi6-P)=6$$
Note that the RHS is an increasing function of $P$ for $P\in(0,\frac\pi6]$, whose maximal value is at $P=\frac\pi6$, i.e.
$$RHS_{max}=3\cdot \frac12+4\cdot 1 = 5.5 <6$$
Thus, $R\ne \frac{5\pi}6$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Studying convergenve of an integral I need to study the convergence of an integral
$$
\int_{2}^{\infty} \tan{(\frac{\arctan(x^2)}{x^2})}\frac{\sqrt{x^4+1}}{1+\sin^2(x)}\,dx
$$
We are only interested in the case where $x\rightarrow\infty$.
$$
\frac{\arctan(x^2)}{x^2} \rightarrow 0 \Rightarrow \tan{(\frac{\arctan(x^2)}{x^2})} \sim \frac{\arctan(x^2)}{x^2}\\
\frac{\sqrt{x^4+1}}{1+\sin^2(x)} \sim \frac{x^2}{1+\sin^2(x)}\\
\tan{(\frac{\arctan(x^2)}{x^2})}\frac{\sqrt{x^4+1}}{1+\sin^2(x)} \sim \frac{\arctan(x^2)}{1+\sin^2(x)}
$$
Im not exactly sure what do I do next. Can I say that
$$
\frac{1}{x}\leq\frac{\arctan(x^2)}{1+\sin^2(x)}
$$
and since $\int_{2}^{\infty}\frac{1}{x}$ diverges, so does my initial integral? In case I cannot, what do I do next?
|
You can even estimate the integrand from below by only a constant, because for $x\geq 2$ you have
$$\frac{\arctan x^2}{1+\sin^2x}\geq \frac{\arctan 4}{2}$$
Starting from the beginning you can use $\tan x\geq x$ for $0<x<\frac{\pi}2$. So, for $x\geq 2$ you surely have
$$\tan \left(\frac{\arctan x^2}{x^2}\right) \geq \frac{\arctan x^2}{x^2}$$
Hence,
$$\int_{2}^{\infty} \tan\left(\frac{\arctan(x^2)}{x^2}\right)\frac{\sqrt{x^4+1}}{1+\sin^2(x)}\,dx \geq \int_{2}^{\infty} \frac{\arctan(x^2)}{x^2}\frac{x^2}{2}\,dx$$
$$\geq \int_{2}^{\infty} \frac{\arctan(4)}{2}\,dx =+\infty$$
So, the integral is divergent.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quadratic Diophantine equation $x^2+6y^2-xy=47$ has no solutions. I am trying to show that $x^2 + 6y^2 - xy = 47$ has no integer solutions. I know that the an efficient way is to look at this equation modulo $n$; other equations can be easily be solved this way. I tried this for $n = 2,3,4,5,6$ so far and I still cannot conclude that no solutions exist. Is there an efficient way of knowing what $n$ to try? Can you give some ideas for $n$ not large? Thanks.
|
Let $z=x/y$. Since the roots of $z^2-z+6$ are $\frac{1\pm i\sqrt{23}}2$, we get
$$
z^2-z+6=\left(z-\frac{1+i\sqrt{23}}2\right)\left(z-\frac{1-i\sqrt{23}}2\right)
$$
Multiplying by $y^2$ gives
$$
\begin{align}
47
&=x^2-xy+6y^2\\[9pt]
&=\left(x-\frac{1+i\sqrt{23}}2\,y\right)\left(x-\frac{1-i\sqrt{23}}2\,y\right)\\
&=\left(x-\frac12\,y\right)^2+\left(\frac{\sqrt{23}}2\,y\right)^2
\end{align}
$$
Multiplying by $4$ yields
$$
x^2+6y^2-xy=47\iff(2x-y)^2+23y^2=188
$$
Since $23\cdot3^2=207\gt188$, the only choices for $y$ are $\{0,\pm1,\pm2\}$.
$$
\begin{array}{r|c}
y&188-23y^2\\\hline
0&188\\
\pm1&165\\
\pm2&96
\end{array}
$$
None of the numbers in the right column are perfect squares, so none can be $(2x-y)^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Source: ISI BMath UGB 2010
My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1-c<1.$$
Thus by AM-GM inequality we have $$\frac{(1-a)+(1-b)+(1-c)}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{(1-a)(1-b)(1-c)}\ge 27.$$
Also by AM-GM inequality we have $$\frac{a+b+c}{3}\ge (abc)^\frac{1}{3}\\ \implies \frac{2}{3}\ge (abc)^\frac{1}{3}\\\implies abc\le \frac{8}{27}\\\implies \frac{1}{abc}\ge \frac{27}{8}.$$
This implies that, we have $$\frac{1}{abc(1-a)(1-b)(1-c)}\ge \frac{27^2}{8}\\ \implies \frac{(abc)^2}{abc(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}\\\implies \frac{abc}{(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}.$$
How to proceed after this?
|
You may set
$$u=1-a,\; v= 1-b,\; w = 1-c$$
$$\Rightarrow u+v+w= 1 \text{ and } u,v,w\in (0,1)$$
To show is now
$$(1-u)(1-v)(1-w)\geq 8uvw$$
or, after expanding and using $u+v+w=1$
$$uv+vw+wu \geq 9uvw$$
or, because auf AM-HM
$$\frac 1 3 = \frac{u+v+w}{3}\geq \frac 3{\frac 1w + \frac 1u + \frac 1v}$$
which is true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
System of 2 equations Solve the system of equations:
$$
\left\{\begin{matrix}2\sqrt{x+y} = y^2+y-x & \quad(1) \\ x(y^2+y)=(y^4-y^2)^2-2 & \quad(2) \end{matrix}\right.
$$
My attempt:
From $(1)$ I get:
$$(1) \implies (\sqrt{x+y}+y+2)(\sqrt{x+y}-y) = 0$$
So either $\sqrt{x+y}+y+2 = 0 \implies x = y^2+3y+4$ or $\sqrt{x+y}-y = 0 \implies x = y^2-y$. If I replace either of them into $(2)$, It would be very messy since this will become a $8$th-degree polynomial.
How can I solve this problem?
|
If $x=y^2-y$ so
$$y^4-y^2=(y^4-y^2)^2-2,$$
which gives $$y^4-y^2=-1,$$ which is impossible, or
$$y^4-y^2=2.$$
Can you end it now?
I got that the second case: $$\sqrt{x+y}+y+2=0$$ is impossible.
Indeed, $y\leq-2$ and $$x+y=(y+2)^2$$ or
$$x=y^2+3y+4,$$ which gives
$$(y^2+3y+4)(y^2+y)=(y^4-y^2)^2-2$$ or
$$y^8-2y^6-4y^3-7y^2-4y-2=0$$ or
$$y^8-4y^6+2y^6-8y^4+8y^4-32y^2-4y^3+25y^2-100-4y+98=0$$ or
$$(y^2-4)(y^6+2y^4+8y^2+25)-4y^3-4y+98=0,$$ which is impossible for $y\leq-2$.
|
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|
For all real $x, y$ that satisfy $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ I started with $x+y=a$ and $xy=b$ and I rewrote the equations with a and b.
I got
$$b=a^2+a-4$$
$$x^3+y^3=2a^3+3a^2-12a^2=7$$
$$f(a)=-2a^3+3a^2-12a^2-7=0$$
I factorised it to get
$$f(a)=-(a-1)(a-1)(2a+7)=0$$
So $a=1,b=-2$ or $a=\frac{-7}{2},b=\frac{19}{4}$
Now I know that values of a and b but how do I get x and y?
Note:
The textbook solution says that if we consider a=1 and b=-2 then x and y are the roots of
$$t^2+t-2=0$$
But shouldn't it be
$$t^2-t-2=0$$
from Vieta's formulas?
I would like a clarification for this or any other solutions would be fine too.
|
Let $t^2+u t+v$ be a quadratic polynomial with roots $a$ and $b$. Then
$$t^2+u t+v=(t-a)(t-b)=t^2-(a+b)t+ab,$$
and hence $v=ab$ and $a=-(a+b)$. Note the $-$-sign. In you particular case with $a=1$ and $b=-2$
$$u=-(a+b)=1\qquad\text{ and }\qquad v=ab=-2,$$
corresponding to the polynomial $t^2+t-2$.
|
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|
$\alpha,\beta,\gamma$ are the roots of the equation $x^3 − 9x + 9 = 0$. Find the value of $ \alpha^{-5}+\beta^{-5}+\gamma^{-5}$ I've simplified the expression to get
$$\frac{(\alpha\beta)^5+(\beta\gamma)^5+(\gamma\alpha)^5}{(\alpha\beta\gamma)^5}.$$
Now all I need to find is $\sum (\alpha\beta)^5$ given that $\sum \alpha\beta=-9$ (by Vieta's relations).
If it helps, I found that:
$$\sum (\alpha\beta)^2=81,$$ and
$$\sum (\alpha\beta)^3=-486.$$
|
Let $a=\alpha^{-1}, b=\beta^{-1},c=\gamma^{-1}.$ Then, we have $a+b+c=\alpha^{-1}+\beta^{-1}+\gamma^{-1}=(\alpha\beta+\beta\gamma+\gamma\alpha)/(\alpha\beta\gamma)=1,$ $abc=(\alpha\beta\gamma)^{-1}=-1/9,$ and $ab+bc+ca=(\alpha+\beta+\gamma)/(\alpha\beta\gamma)=0.$ Then, we have
\begin{equation}
\begin{split}
\alpha^{-5}+\beta^{-5}+\gamma^{-5}&=a^5+b^5+c^5\\
&=(a+b+c)^5-5(ab+bc+ca)(a+b+c)^3\\
&\ \ \ \ \ +5abc(a+b+c)^2+5(ab+bc+ca)^2(a+b+c)\\
&\ \ \ \ \ -5abc(ab+bc+ca),
\end{split}
\end{equation}
and all that is left to do is substitute.
P.S. The equation above can be obtained by repeated use of Newton's identity, which can be used to express $a^i+b^i+c^i$ using the basic symmetric polynomials $a+b+c,ab+bc+ca,abc.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$ Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$
Listing out a few terms and simplifying the denominators gets me $$-1+\sqrt{3}+2-\sqrt{2}-\sqrt{3}+\sqrt{5}\cdots-\sqrt{97}+3\sqrt{11}-7\sqrt{2}+10-3\sqrt{11}+\sqrt{101}.$$
A lot of these terms cancel, and I think that the first and last terms will be left.
I feel like I'm close, but missing something.
|
Let $ m \in\mathbb{N}^{*} : $
\begin{aligned}\sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n}\right)}\\ &=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n+1}\right)}+\sum_{n=1}^{m}{\left(\sqrt{n+1}-\sqrt{n}\right)}\\ \sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sqrt{m+2}-\sqrt{2}+\sqrt{m+1}-1\end{aligned}
|
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|
Given $4$ variables and $5$ pairwise products, find the $6$th pairwise product?
Consider four positive numbers (not necessarily integers). The pairwise products are $2$, $3$, $4$, $5$, $6$, plus one more number.
What is the 6th product? What are the numbers?
I found this from Quora and I would be interested in a nice solution!
If we name the four numbers $x_1, x_2, x_3, x_4$ and the missing product $p_6$, then all of the possible products are:
$$x_1 x_2,\quad
x_1 x_3,\quad
x_1 x_4,\quad
x_2 x_3,\quad
x_2 x_4,\quad\text{and}\quad
x_3 x_4$$
There are six equations and five unknowns, but I don't know how to assign the six different numbers to each of them.
I understand that the partial products which do not share a common factor (for example, $x_1 x_2$ and $x_3 x_4$) should not be assigned to numbers which do have a common factor, for example $2$ and $4$, or $2$ and $6$, or $3$ and $6$.
|
You can separate the six products into three pairs with each pair having different factors
$$x_1\cdot x_2\quad \ x_3\cdot x_4\\
x_1\cdot x_3\quad \ x_2 \cdot x_4\\
x_1 \cdot x_4\quad \ x_2 \cdot x_3$$
When we multiply the partial products on each line, we should get the same result. The only two pairs that have the same product are $2 \cdot 6$ and $3 \cdot 4$, so the product of the last line must also be $12$. The sixth partial product is $$\frac {12}5$$
Now we can check that the solution works. By symmetry we can assign the first line $2 \cdot 6,$ the second $3 \cdot 4$ and the last $5 \cdot \frac {12}5$ but we cannot be sure of the order of the last.Then $\frac {x_3}{x_2}=\frac 32.$ If the last is $5 \cdot \frac {12}5$ then $\frac {x_4}{x_2}=\frac 52, \frac {x_1}{x_2}=\frac 54$. The product of them all is $12$, so we have
$$\frac 32\cdot \frac 52 \cdot \frac 54 x_2^4=12\\x_2=\sqrt{\frac 85}\\
x_1=\frac 54\sqrt {\frac 85}\\x_3=\frac 32\sqrt{\frac 85}\\x_4=\frac 52 \sqrt{\frac 85}$$
If we switch the products in the last line, we get another solution
$$x_2=\sqrt{\frac {10}3}\\
x_1=\sqrt {\frac 65}\\x_3=\sqrt{\frac{15}2}\\x_4=\sqrt{\frac{24}5}$$
We can permute the assignment of the variables at will.
|
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|
Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
|
We seek a quartic whose roots are $a^4, b^4, c^4 , d^4$.
Naively, this is $ x + px^{3/4} + qx^{1/2} + rx^{1/4} + 1 = 0 $, but the powers need to be integers.
Shifting terms, $x + qx^{1/2} + 1 = - px^{3/4} - rx^{1/4}$.
Squaring both sides, $x^2 + q^2x + 1 + 2x + 2qx^{3/2} + 2qx^{1/2} = p^2 x^{3/2} + 2pr x + r^2 x^{1/2}$.
Shifting terms, $x^2 + (q^2 - 2pr +2)x + 1 = x^{1/2} ((p^2-2q)x + (r^2 -2q))$.
Squaring again, $x^4 + (q^2 - 2pr +2)^2x^2 + 1 + 2(q^2 - 2pr +2)x^3 + 2(q^2 - 2pr +2)x + 2x^2 = x( (p^2-2q)^2x^2 + (r^2 -2q)^2 + 2(p^2-2q)(r^2 -2q)x)$
Let
$$ g(x) = x^4 + (q^2 - 2pr +2)^2x^2 + 1 + 2(q^2 - 2pr +2)x^3 + 2(q^2 - 2pr +2)x + 2x^2 - x( (p^2-2q)^2x^2 + (r^2 -2q)^2 + 2(p^2-2q)(r^2 -2q)x).$$
Then $g(x) = ( x - a^4)(x-b^4)(x-c^4)(x-^4)$,
so $g(-1) = \prod ( -1 - a^4) = \prod (a^4 + 1) $.
On the other hand, by substituting in $ x = -1$,
$ g(-1) = 1 + (q^2 - 2pr +2)^2 + 1 - 2(q^2 - 2pr +2) - 2(q^2 - 2pr +2) + 2 + ( (p^2-2q)^2 + (r^2 -2q)^2 - 2(p^2-2q)(r^2 -2q))$
$ = p^4 + 2p^2r^2 + r^4 + q^4 - 4pq^2 r $ (E.g. expand by Wolfram)
which is the form that you want.
|
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|
Prove $\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$
Prove
$$\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$$
Not sure how to go about this proof, was thinking of putting numerator and denominator in mod arg form straight away and then using binomial expansion but doesn't seem to work out... Then I tried realising the denominator first and ended up with
$$\frac{(1+cis2θ)^{2n} }{ 2^n(1+cos2θ)^n}$$ and not sure where to go from there?? Any help much appreciated, thanks :)
|
$$\begin{align}
e^{i2\theta n} \cdot (1+e^{-i2\theta})^n
&=e^{i2\theta n}\cdot \left(\binom{n}{0} e^{-i2\theta \cdot 0} + \binom{n}{1} e^{-i2\theta \cdot 1}+\binom{n}{2} e^{-i2\theta \cdot 2} + \dots + \binom{n}{n-1} e^{-i2\theta \cdot (n-1)}+ \binom{n}{n} e^{-i2\theta \cdot n}\right) \\
&= \binom{n}{0} e^{i2\theta \cdot(n- 0)} + \binom{n}{1} e^{i2\theta \cdot (n- 1)}+\binom{n}{2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{n-1} e^{i2\theta \cdot (n-(n-1))}+ \binom{n}{n} e^{i2\theta \cdot(n- n)} \\
&= \binom{n}{0} e^{i2\theta \cdot n} + \binom{n}{1} e^{i2\theta \cdot (n- 1)}+\binom{n}{2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{n-1} e^{i2\theta \cdot 1}+ \binom{n}{n} e^{i2\theta \cdot 0}\\
&=\binom{n}{n} e^{i2\theta \cdot n} + \binom{n}{n-1} e^{i2\theta \cdot (n- 1)}+\binom{n}{n-2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{1} e^{i2\theta \cdot 1}+ \binom{n}{0} e^{i2\theta \cdot 0}\\
&=(1+e^{i2\theta})^n
\end{align}$$
Thus,
$$
e^{i2\theta n}= \frac{
(1+e^{i2\theta})^n}{(1+e^{-i2\theta})^n} = \left(\frac{
1+e^{i2\theta}}{1+e^{-i2\theta}} \right)^n=\left(\frac{
1+\cos2\theta +i \sin2\theta}{1+\cos2\theta -i \sin2\theta} \right)^n \ .$$
|
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Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.
By trial I found $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?
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Without loss of generality $a\le b\le c$, so $a!|2^d\implies a!\in\{1,\,2\}$.
If $a!=1$, $b!+c!$ is odd so $b!=1$ and $c!=2^d-2$, so $c!\nmid4$ and $c\le3$. This gives the solutions $c=2$ and $c=3$.
If $a!=2$, $b!+c!$ isn't a multiple of $4$, so $b\le3$. In particular, if $a=b-2$ then $c!=2^d-4$ is a multiple of $4$ but not $8$ so $4\le c\le7$, and similarly if $a=2,\,b=3$ then $8\le c\le15$. I'll leave you to check these cases.
|
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Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove this afterwards, but failed. I would love to see another proof to this Problem.
-A proof-: we know that $\displaystyle\sum_{i=1}^{n}\cos{a_i}= \sum_{k=1}^{n}\cos{(a+(k-1)x)}=\frac{\cos{\frac{a+a_n}{2}}\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}=T(x)\quad(1)$
(Here's the source for $(1)$).
Also, $\displaystyle\sum_{i=1}^{n}a^2_i=\sum_{k=1}^{n}(a+(k-1)x)^2\\=\displaystyle\sum_{k=1}^{n}(a^2+2ax(k-1)+x^2(k-1)^2)\\=\displaystyle\sum_{k=1}^{n}a^2+2ax\sum_{k=1}^{n}(k-1)+x^2\sum_{k=1}^{n}(k-1)^2\\=na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\quad(2)$
We consider the function $f(x)=\frac{x^2}{2}+\cos{x}\implies f''(x)=1-\cos{x}\geq{0}$ therefore $f(x)$ is a concave function. From Jensens inequality for concave functions:$ \displaystyle\sum_{i=1}^{n}f(a_i)\geq{nf\left(\frac{\sum_{i=1}^{n}a_i}{n}\right)}\iff\displaystyle\sum_{i=1}^{n}\left(\frac{a^2_i}{2}+\cos{a_i}\right)\geq n\Big(\frac{1}{2}\left(\frac{\frac{n}{2}(a+a_n)}{n}\right)^2+\cos{\frac{\frac{n}{2}(a+a_n)}{n}}\Big)\iff\frac{1}{2}\sum_{i=1}^{n}a^2_i+\sum_{i=1}^{n}\cos{a_i}\geq \frac{n}{2}\left(\frac{a+a_n}{2}\right)^2+n\cos{\frac{a+a_n}{2}}\overset{(1),(2)}{\iff}\frac{1}{2}\left(na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\right)+T(x)\geqslant \frac{n\left(2a+(n-1)x\right)^2}{8}+n\cos{\frac{a+a_n}{2}}\overset{\ldots}{\iff} \frac{x^2n(n^2-1)}{3}\geqslant n\cos{\frac{a+a_n}{2}}-T(x)\overset{(1)}{\iff} \frac{x^2n(n^2-1)}{3}\geq \cos{\frac{a+a_n}{2}}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\iff \frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\quad(3) $
$Lemma.$ For every dinstict $x,x\in\mathbb{R}_{\neq kπ}$ there exists at least one value of $a$ such that $\cos{(a+\frac{(n-1)x}{2}})=1\quad(5)$
Proof of the lemma: $(5)\iff (n-1)x=2-2a+4kπ\iff a=2kπ+1-\frac{(n-1)x}{2}$.
Hence, for every $x,x\in\mathbb{R}_{\neq k\pi}$
we have $\frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\overset{(4,x\to 2x)}{\iff}\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}} \square$
*
*To me it looks like $n-\frac{x^2n(n^2-1)}{6}$ is a good approximation function of $\frac{\sin{nx}}{\sin{x}}$ for small values of $x$ (or for small values of $x+ z\pi$, $z\in\mathbb{Z}$, it depends on $n$, because the second one is periodic with period multiple of $\pi$).
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Alternative solution:
Remark: $n - \frac{n(n^2-1)x^2}{6}$ is the second order Taylor approximation of $f(x) = \frac{\sin nx}{\sin x}$ around $x = 0$.
First, we give the following result. The proof is given later.
Fact 1: Let $n\ge 3$ be a positive integer and $x\in (0, \frac{\pi}{2})$. Then
$\frac{\sin nx}{\sin x} \ge n - \frac{n(n^2-1)x^2}{6}$.
The remaining cases are easy to prove:
For $n=1$, clearly the inequality is true.
For $n=2$, the inequality is equivalent to $2\cos x - 2 + x^2\ge 0$ which is true.
For $n\ge 3$ and $x\in [\frac{\pi}{2}, \infty)$,
we have $-n \ge n - \frac{n(n^2-1)x^2}{6}$.
It is easy to prove that $-n \le \frac{\sin nx}{\sin x} \le n$ for $x\in \mathbb{R}$ (by math induction).
The inequality is true.
$\phantom{2}$
Proof of Fact 1: The inequality is written as
$$\frac{\sin n x}{nx} \ge \frac{\sin x}{x} - \frac{n^2-1}{6}x^2\cdot \frac{\sin x}{x}.$$
To proceed, we need the following results (Facts 2 through 3). Their proof is not hard and hence omitted.
Fact 2: $\frac{\sin y}{y} \ge \frac{-7y^2+60}{3y^2 + 60}$ for $y\in \mathbb{R}$.
(Pade $(2,2)$ approximation)
Fact 3: $\frac{\sin y}{y} \le \frac{6}{6+y^2}$ for $y \in (0, \frac{\pi}{2})$.
(Pade $(0,2)$ approximation)
By using Facts 2 and 3, it suffices to prove that
$$\frac{-7n^2x^2+60}{3n^2x^2 + 60}
\ge \frac{6}{6+x^2} - \frac{n^2-1}{6}x^2\cdot \frac{-7x^2+60}{3x^2 + 60}$$
or (after clearing the denominators)
$$(-7x^4+18x^2+360)n^4+(7x^4-200x^2-1200)n^2+140x^2 \ge 0.$$
Since $-7x^4+18x^2+360 > 0$, we have
\begin{align}
&(-7x^4+18x^2+360)n^4+(7x^4-200x^2-1200)n^2+140x^2\\
\ge \ & (-7x^4+18x^2+360)n^2\cdot 3^2+(7x^4-200x^2-1200)n^2+140x^2\\
= \ & (-56x^4-38x^2+2040)n^2+140x^2\\
\ge \ & 0
\end{align}
where we have used the fact that $-56x^4-38x^2+2040 > 0$. We are done.
|
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|
Evaluation of $\int^{1}_{0}\frac{x^4}{1+x^8}dx$
How can I Integrate $\displaystyle \int^{1}_{0}\frac{x^4}{1+x^8}dx$?
I have searched in that forum and get the result for $\displaystyle \int^{\infty}_{-\infty}\frac{x^4}{1+x^8}dx$ or for $\displaystyle \int^{\infty}_{0}\frac{x^4}{1+x^8}dx$
Although i know the formula $\displaystyle \int^{\infty}_{0}\frac{x^{m-1}}{1+x^n}dx=\frac{\pi}{n}\cdot \csc\frac{\pi m}{n}$
But did not understand How can i use in my original Integration
Help me please Thanks
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Knowing that the roots of the denominator are the primitives $x_k=e^{i\pi/8 + i2\pi k/8}$ with $k=0,1,2,...,7$ the integral is quickly calculated by partial fractions decomposition i.e.
$$\int_0^1 \frac{x^4}{1+x^8} \, {\rm d}x = \int_0^1 {\rm d}x \sum_{k=0}^7 \frac{A_k}{x-x_k} = \sum_{k=0}^7 A_k \log\left(1-\frac{1}{x_k}\right)$$
where the $A_k=\frac{1}{8x_k^3}$ are the residues and the result is $$\sum_{k=0}^7 \frac{e^{-i3\pi/8 - i6\pi k/8}}{8} \, \log\left(1-e^{-i\pi/8 - i2\pi k/8}\right) \, .$$
Complex conjugates are $k=[0,7],[1,6],[2,5],[3,4]$, so grouping these terms gives
$$\frac{1}{4}\sum_{k=0}^3 \left\{ {\cos\left(\frac{3\pi}{8}+\frac{3\pi k}{4}\right)\log\left(2\sin\left(\frac{\pi}{16}+\frac{\pi k}{8}\right)\right)} + \sin\left(\frac{3\pi}{8}+\frac{3\pi k}{4}\right) \left(\frac{7\pi}{16} - \frac{\pi k}{8}\right) \right\} \\
=\frac{\pi}{8} \left\{ \cos\left(\frac{\pi}{8}\right) - \sin\left(\frac{\pi}{8}\right) \right\} + \frac{1}{4} \left\{ \sin\left(\frac{\pi}{8}\right) \log\left(\tan\left(\frac{\pi}{16}\right)\right) - \cos\left(\frac{\pi}{8}\right) \log\left(\tan\left(\frac{3\pi}{16}\right)\right) \right\}$$
which is relatively symmetric. The $\sqrt{2}$ expressions arise by using $$\cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2+\sqrt{2}}}{2} \\
\sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2-\sqrt{2}}}{2} \\
\tan\left(\frac{\pi}{16}\right) = \sqrt{\frac{2-\sqrt{2+\sqrt{2}}}{2+\sqrt{2+\sqrt{2}}}} \\
\tan\left(\frac{3\pi}{16}\right) = \sqrt{\frac{2-\sqrt{2-\sqrt{2}}}{2+\sqrt{2-\sqrt{2}}}} \, .$$
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|
Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem, but $\frac{x}{(x^2+a^2)}$ and $\frac{x^2}{(x^2+a^2)}$ aren't in $L^1(\mathbb{R})$.
Any hints hold be appreciated.
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HINT: Unless you’re dead-set on using the Fourier transform, I would try using that
$$\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{1}{a^2-b^2}\bigg(\frac{a^2}{x^2+a^2}-\frac{b^2}{x^2+b^2}\bigg)$$
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Is there a nice way to reconcile the antiderivatives of $\frac{1}{x^2+a^2}$, $\frac{1}{x^2-a^2}$, and $\frac{1}{x^2}$ as $a\to0$? If $a$ is a (WLOG) positive real number, ignoring all constants of integration we have
$$
\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)
$$
$$
\int \frac{1}{x^2-a^2}\,dx = \frac{1}{2a}\log \left(\frac{a-x}{a+x}\right)
$$
$$
\int \frac{1}{x^2}\,dx = \frac{-1}{x}
$$Is it the case that the first two functions should approach $-x^{-1}$ in the limit as $a\to0$, and if so why? I can't imagine why or how we'd be able to pass the limit inside the integral first.
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Remember that antiderivatives are only defined up to constant. One can select a sequence of constants $C_a$ so that $F_a(x)+C_a\to-1/x$ when $a\to0$ and $x\neq0$ for both those functions. It is easier to see with definite integrals like $\int_1^x \frac{1}{t^2+a^2}\,dt$, where it is justified to move the limit inside the integral because the convergence on $[1,x]$ is uniform. This gives us for $a\to0$:
$$
\int_1^x \frac{1}{t^2+a^2}\,dt=\frac{1}{a}\arctan\left(\frac{x}{a}\right)-\frac{1}{a}\arctan\left(\frac{1}{a}\right)\\
\to\int_1^x \frac{1}{t^2}\,dt=-\frac{1}{x}+1.
$$
Therefore,
$$
\frac{1}{a}\arctan\left(\frac{x}{a}\right)-\left(\frac{1}{a}\arctan\left(\frac{1}{a}\right)+1\right)\to -\frac{1}{x}.
$$
The other case is analogous. There is a deeper sense to reconciling those two cases that comes from passing to the complex domain, namely:
$$
\arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+C.
$$
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How can I get the $\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=3$?
Let the triangle $ABC$ be inscribed in a circle, let $P$ denote the centroid of the triangle and let $O$ denote the circumcenter. Suppose that $A,B,C$ have coordinates $(0,0),(a,0)$ and $(b,c)$ respectively.
a) Express the coordinates of $P$ and $O$ in terms of $a,b,c$.
b) Extend line segments $AP,BP,CP$ to meet the circle in points $D,E$,and $F$ respectively. Show that
$$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=3\,.$$
I have done in part a, but for part b, since $$CP\cdot PF=BP\cdot PE=AP\cdot PD=R^2-x^2\,,$$ so $$PD=\frac{CP\cdot PE}{AP}\,,$$ $$PE=\frac{CP\cdot PE}{BP}\,,$$
and $$PF=\frac{BP\cdot PE}{CP}=\frac{CP\cdot PE}{CP}\,.$$
Hence,
$$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=\frac{AP^2}{CP\cdot PE}+\frac{BP^2}{CP\cdot PE}+\frac{CP^2}{CP\cdot PE}=\frac{AP^2+BP^2+CP^2}{CP\cdot PE}$$
How to get $$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=\frac{AP^2+BP^2+CP^2}{R^2-x^2}\,?$$
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I am certain that the theorem below has been proven on this site. Unfortunately, I could not find it.
Theorem. Let $ABC$ be a triangle with circumcenter $O$ and centroid $G$. If $R$ is the circumradius and $P$ is an arbitrary point on the plane (or in the space), then
$$\begin{align}PA^2+PB^2+PC^2&=GA^2+GB^2+GC^2+3\,PG^2\\&=3\,\left(R^2-OG^2+PG^2\right)\,.\end{align}$$ Furthermore, the value $PA^2+PB^2+PC^2$ is minimized if and only if $P=G$, and this minimum value is $$GA^2+GB^2+GC^2=3\,\left(R^2-OG^2\right)\,.$$
Without loss of generality, suppose $O$ is at the origin. Identify each point $T$ on the plane as the vector $\overrightarrow{OT}$. We have
$$G=\frac{A+B+C}{3}\text{ and }\|A\|=\|B\|=\|C\|=R\,.$$
That is,
$$\begin{align}PA^2+PB^2+PC^2&=\left\|\overrightarrow{PA}\right\|^2+\left\|\overrightarrow{PB}\right\|^2+\left\|\overrightarrow{PC}\right\|^2
\\&=\|A-P\|^3+\|B-P\|^2+\|C-P\|^2
\\&=\|A\|^2+\|B\|^2+\|C\|^2-2\,(A+B+C)\cdot P+3\,\|P\|^2
\\&=\|A\|^2+\|B\|^2+\|C\|^2-2\,(3\,G)\cdot P+3\,\|P\|^2
\\&=\|A\|^2+\|B\|^2+\|C\|^2-3\,\|G\|^2+3\,\|G-P\|^2
\\&=3\,R^2-3\,\left\|\overrightarrow{OG}\right\|^2+3\,\left\|\overrightarrow{PG}\right\|^2=3\,\left(R^2-OG^2+PG^2\right)\,.\end{align}$$
When $P=G$, we recover the identity
$$GA^2+GB^2+GC^2=3\,(R^2-OG^2)\,,$$
whence we may also write
$$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3\,PG^2\,.$$
Remark. The theorem above also holds in general. There is also an integral version (which physicists call it the Parallel Axis Theorem).
Theorem. For a positive integer $n$, let $A_1$, $A_2$, $\ldots$, $A_n$ be $n$ points in a Euclidean space, whose barycenter (centroid) is $G$. Then,
$$\begin{align}\sum_{i=1}^n\,PA_i^2&=\sum_{i=1}^n\,GA_i^2+n\,PG^2\,.\end{align}$$ Furthermore, the value $\sum\limits_{i=1}^n\,PA_i^2$ is minimized if and only if $P=G$.
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How are the factors obtained at *this point* in the solution of proving $2^{2^n}+ 1 \mid 2^{2^m} -1$ I understood everything up until "BUT". My question is how are the factors on the right side for the bolded line of equation under the word "BUT" obtained? I tried to get these factors using property (5), but I was unsuccessful, can someone please show me how to get those factors? Thanks in advance.
This is example 2 from "Problems of Number Theory in Mathematical Competitions" by You Hong-Bing on page 2.
Some properties the example will be using are:
(1) If $b\mid c$, and $c\mid a$, then $b\mid a$, that is, divisibility is transitive.
(5) If $n$ is a positive odd number, then
$x^n-y^n= (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$
The problem, example 2 is as follows:
Let $\mathbf{m > n}$ and $\mathbf{n\ge 0}$, show that $\mathbf{(2^{2^n}+1)\mid(2^{2^m}-1)}$
The following is the proof from the book:
Take $x=2^{2^{n+1}}$,$\,\,y=1$ in factorization (5), and substitute $n$ by $2^{m-n-1}$
WE GET:
$2^{2^m}-1 = (2^{2^{n+1}}-1) \left[(2^{2^{n+1}})^{2^{m-n-1}-1} +\ldots + 2^{2^{n+1}}+1 \right]$
THUS,
$(2^{2^{n+1}}-1) \mid (2^{2^m}-1)$
BUT,
$\mathbf{(2^{2^{n+1}}-1) = (2^{2^n} - 1)(2^{2^n} + 1)}$
This is the equation I'm having trouble with so everything else after this didn't make much sense to me
HENCE,
$(2^{2^n} + 1)\mid(2^{2^{n+1}} - 1)$
FURTHER, BY PROPERTY (1) WE HAVE:
$(2^{2^n} +1)\mid(2^{2^m} - 1)$
REMARK: sometimes it is difficult to prove $b\mid a$ directly when dealing with divisibility problems. Therefore, we can attempt to choose an "intermediate number" $c$ and prove $b\mid c$ and $c\mid a$ first, then use property (1) of divisibility to deduce the conclusion.
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More generally,
$x^{2^m}-1
=(x-1)\prod_{n=0}^{m-1}(x^{2^n}+1)
$.
For this,
put $x = 2$.
Proof.
For $m=1$ this is
$x^{2^1}-1
=(x-1)\prod_{n=0}^{0}(x^{2^n}+1)
$
or
$x^2-1
=(x-1)(x+1)
$.
If it is true for $m$ then
$\begin{array}\\
(x-1)\prod_{n=0}^{(m+1)-1}(x^{2^n}+1)
&=(x-1)\prod_{n=0}^{m}(x^{2^n}+1)\\
&=(x-1)(x^{2^m}+1)\prod_{n=0}^{m-1}(x^{2^n}+1)\\
&=(x^{2^m}+1)(x-1)\prod_{n=0}^{m-1}(x^{2^n}+1)\\
&=(x^{2^m}+1)(x^{2^m}-1)
\qquad\text{(induction hypothesis)}\\
&=x^{2^{m+1}}-1
\qquad (x^a+1)(x^a-1)=x^{2a}-1
\text{ and } 2(2^m) = 2^{m+1}\\
\end{array}
$
|
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|
If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle? If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle?
I was doing strange math yesterday, and I can across the realization that two different rectangles can’t have the same area and same diagonal. But I haven’t been able to solve a equation that shows this. I also haven’t be able to prove myself wrong? So maybe higher math gods will help?
$$\text{Diagonal}=\sqrt{x^2+y^2}$$
$$\text{Area}=xy$$
|
Sure:
You have $xy = A$ a known constant.
And $\sqrt{x^2 +y^2} = d$ a known constant.
So just substitute.
$x = \frac Ay$ (assuming $y\ne 0$ which if the area is positive must be so... or we could do $y = \frac Ax$.... it doesn't matter.
And $\sqrt{(\frac Ay)^2 + y^2} = d$.
So $(\frac Ay)^2 + y^2 = d^2$
So $A^2 + y^4 =d^2 y^2$
$y^4 - d^2y^2 +A^2 =0$. Use the quadratic formula:
$y^2 = \frac {d^2 \pm \sqrt{d^4-4A^2}}2$ We know $y$ is positive
So $y =\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}$ and $x =\frac {A}{\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}}=\frac {A}{\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}}\frac {\sqrt{\frac {d^2 \mp \sqrt{d^4-4^2A}}2}}{\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}}=\frac {A}{\sqrt{\frac {d^4-(d^4-4A^2)}4}}\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}=\frac AA\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}=\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}$
....
So if we take the rectangle with sides $3$ and $4$ and area $A=12$ and diagonal $d=5$ and pretended we didn't know the sides we'd have.
$y = \sqrt{\frac {5^2+\sqrt{5^4-4*12^2}}2}=\sqrt{\frac {25+\sqrt{(25+2*12)(25-2*12)}}2}=\sqrt{\frac{25+\sqrt{49*1}}2}=\sqrt{\frac {25+7}2}=\sqrt{16}=4$
And $x = \sqrt{\frac {5^2-\sqrt{5^4-4*12^2}}2}=\sqrt{\frac {25-\sqrt{(25+2*12)(25-2*12)}}2}=\sqrt{\frac{25-\sqrt{49*1}}2}=\sqrt{\frac {25-7}2}=\sqrt{9}=3$
|
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|
Restrictions on a Function If $f(x) =$
\begin{cases}
x^2-4 &\quad \text{if } x \ge -4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
then for how many values of $x$ is $f(f(x)) = 5$?
I'm not sure how to really start from this question other than bashing values. I need some help on a start, thanks!
|
First find the values such that $f(x) = 5$:
$x^2-4=5\iff x^2=9 \iff x=\pm 3$ (both are greater than or equal to $-4$)
$x+3=5\iff x = 2$ (but this value is not less than $-4$).
Hence we must have $f(x) = \pm 3$
So now we find the values such that $f(x) = 3$
$x^2-4=3\iff x^2=7 \iff x=\pm \sqrt{7}$ (both are greater than or equal to $-4$)
$x+3=3\iff x = 0$ (but this value is not less than $-4$).
We now find the values such that $f(x) =-3$
$x^2-4=-3\iff x^2=1 \iff x=\pm 1$ (both are greater than or equal to $-4$)
$x+3=-3\iff x = -6$ (this value is less than $-4$).
Hence the values are $\pm\sqrt{7},\pm1$ and $-6$
|
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|
Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$ Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$
My Attempt :
$$(y-px)^2 (1=p^2)=a^2p^2$$
$$y-px=\frac {ap}{\sqrt {1+p^2}}$$
$$y=px+\frac {ap}{\sqrt {1+p^2}} .....(1)$$
This is Clairaut's Equation so we differentiate both sides with respect to $x$
$$\frac {dy}{dx} = p + x\cdot \frac {dp}{dx} + \frac {\sqrt {1+p^2} \cdot a \cdot \frac {dp}{dx} + ap\cdot \frac {1}{2\sqrt {1+p^2}} \cdot 2p \cdot \frac {dp}{dx}}{(1+p^2)}$$
$$p=p+x+\frac {a\sqrt {1+p^2} + \frac {ap^2}{\sqrt {1+p^2}}}{(1+p^2)} \cdot \frac {dp}{dx} $$
$$(x+\frac { a+2ap^2 }{(1+p^2)^{\frac {3}{2}}}) \cdot \frac {dp}{dx}=0$$
Either,
$$\frac {dp}{dx}=0$$
$$\textrm {so p}=c$$
Using $p=c$ in $(1)$ we get:
$$y=cx+\frac {ac}{\sqrt {1+c^2}}$$ which is the required general solution.
How do I evaluate the singular solution ?
|
The clairaut Eq.:
$$y=xy'+f(y')$$
D.w.r.t. $x$ to get
$$y''(x+f'(y')=0$$
Here $f'(z)$ is d.w.r. t to $z$ (the argument)
the general sdolution is gicen by $y''=0 \implies y/=C$
For the present case
$$y=xy'+\frac{ay'}{\sqrt{1+y'^2}}~~~~(1)$$
$$f'(y')=\frac{a}{(1+y'^2)^{3/2}}$$
$$\frac{x}{a}+\frac{1}{(1+y'^2)^{3/2}}=0, x<0,\implies y(x)= \pm \int \sqrt{(-a/x)^{2/3}-1} dx~~~~(2)$$
There will be no constant of integration. So the singular solution of (1) is given by (2).
|
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|
y=x^3sinx,what's the 6th derivative of y at x = pi/6? question 1： $y=x^3\sin x$, what's $y^{\left(6\right)}\left(0\right)$ ?
I have solved it use Taylor's Formula in the follow way.
step1:
$y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n$
and x_0 = 0 for the question1, So $y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(0\right)}{n!}\left(x\right)^n$
step2:
$y=x^3\sin x=x^3\left(x-\frac{1}{3!}x^3+\ldots \right)=x^4-\frac{1}{6}x^6+\ldots$
[PS: $\sin x=\displaystyle \sum _{n=0}^{\infty }\:\frac{\left(-1\right)^n}{\left(2n+1\right)!}x^{2n+1}$]
step3:
$\dfrac{y^{\left(6\right)}\left(0\right)}{6!}\:=\:-\dfrac{1}{6}. \quad $
So $y^{\left(6\right)}\left(0\right)=-120$
The above method is so useful. But for question2 I find that I can't use the method above.Is there a way to use Taylor's formula to solve this problem, such as substitution? I know leibniz formula can solve it but I want to know if Taylor's formula can be made better in this case?
Question2: $y=x^3\sin x$, what's $y^{\left(6\right)}\left(\dfrac{\pi }{6}\right)$?
|
Equivalently, we want $6^3(\sqrt{3}\sin y+\cos y)/2$, where $[y^k]f(y)$ is the $y^k$ coefficient in $f(y)$. So the result is$$\begin{align}&360[y^6]\left(y^3+\frac{\pi}{2}y^2+\frac{\pi^2}{12}y+\frac{\pi^3}{216}\right)\left(1+y\sqrt{3}-\frac12y^2-\frac{\sqrt{3}}{6}y^3+\frac{1}{24}y^4+\frac{\sqrt{3}}{120}y^5-\frac{1}{720}y^6\right)\\&=360\left(-\frac{\sqrt{3}}{6}+\frac{\pi}{48}+\frac{\pi^2\sqrt{3}}{1440}-\frac{\pi^3}{216\times720}\right)\\&=-60\sqrt{3}+\frac{15\pi}{2}+\frac{\pi^2\sqrt{3}}{4}-\frac{\pi^3}{432}.\end{align}$$
|
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|
Vieta's formulas for quadratic equation problem I'm using one hack, which I never though of why it works. But now I'm curious why it's works and how I can prove it.
Here's the deal: we have quadratic equation $ax^2 + bx + c = 0$, to find roots I just multiply $c$ by $a$ and solving $y^2 + by + ca = 0$, and then I divide roots by $a$.
For example:
$$-6x^2+7x+5=0$$
I solve
$$
y^2 + 7y -30 = 0\\
y_1=-10\\
y_2=3
$$
And then divide the roots:
$$
x_1=\frac{-10}{-6}=\frac{5}{3}\\
x_2=\frac{3}{-6}=-\frac{1}{2}
$$
Which gives me a correct answer. But I want to know why it's so. For now what I figured out is only that:
$$
\text{for } a\neq 0 \text{ :}\\
ax^2 + bx + c = 0 \Leftrightarrow x^2 + \frac{b}{a}x + \frac{c}{a} = 0\\
x_1 + x_2 = -\frac{b}{a} \Leftrightarrow a(x_1 + x_2) = -b\\
x_1 \cdot x_2 = \frac{c}{a} \Leftrightarrow a(x_1 \cdot x_2) = c\\
$$
$$
y^2 + by + ca = 0\\
y_1 + y_1 = -b\\
y_1 \cdot y_2 = ca \Leftrightarrow c = \frac{y_1 \cdot y_2}{a}\\
$$
$$
a(x_1 + x_2) = y_1 + y_1 \Leftrightarrow x_1 + x_2 = \frac{y_1}{a} + \frac{y_2}{a}\\
a(x_1 \cdot x_2) = \frac{y_1 \cdot y_2}{a} \Leftrightarrow x_1 \cdot x_2 = \frac{y_1}{a} \cdot \frac{y_2}{a}\\
$$
Any ideas?
|
Take the original equation $ax^2+bx+c=0$ then multiply by $a$ to obtain $$a^2x^2+abx+ac=0$$Now set $y=ax$ so that $$y^2+by+ac=0$$ and you are done.
|
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|
Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$ Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$
My Attempt:
The given equation is
$$(D^2+4)y=x\sin^2 x$$
It's auxiliary equation is
$$m^2+4=0$$
$$m^2=-4$$
$$m=\pm 2i$$
$$\textrm {Complementary Function (C.F)}=c_1 \cos (2x) + c_2 \sin (2x)$$
Now, the particular integral is given by
$$\textrm {P.I.}=\frac {x\sin^2 x}{D^2+4}$$
$$=\frac {x}{D^2+4}\cdot \frac {1-\cos (2x)}{2}$$
$$=\frac {x}{2(D^2+4)} - \frac {x\cdot \cos (2x)}{2(D^2+4)}$$
$$=\frac {1}{2} \cdot \frac {1}{4} \cdot (1+\frac {D^2}{4})^{-1} \cdot x - \frac {1}{2}(x\cdot \frac {\cos (2x)}{D^2+4} - \frac {2D \cos (2x)}{(D^2+4)^2})$$
$$=\frac {x}{8} - \frac {1}{2} (x\cdot \frac {x \cos (2x)}{2D} - \frac {2D \cos (2x)}{(D^2+4)^2} )$$
How do I solve further?
|
Starting from the 3rd step
$$P.I =\frac {x}{2(D^2+4)} - \frac {x\cdot \cos (2x)}{2(D^2+4)}
$$
$$=\frac{x}{4} -\frac {x\cdot \cos (2x)}{2(D^2+4)}...(1)$$
First we evaluate
$$\frac{cos {2x}}{D^2+4} $$
$$=\frac{1}{D^2+4}\frac{e^{2ix}+e^{-2ix}}{2}$$
$\text{ You can do the easy calculation. it will give. }$
$$=\frac{x\sin {2x}}{4}$$
Similarly
$$\frac{\sin {2x}}{D^2+4} =\frac{x\cos {2x}}{-4}$$
Now let $$V=\frac {x\cdot \cos (2x)}{(D^2+4)}$$
$$=[x\cdot \frac {\cos (2x)}{D^2+4} - \frac {2D \cos (2x)}{(D^2+4)^2}]$$
$$=\frac{1}{4}x^2\sin {2x} -\frac{1}{2}\frac{1}{D^2+4}D(x\sin {2x})$$
$$=\frac{1}{4}x^2\sin {2x} -\frac{1}{2}\frac{1}{D^2+4}(\sin {2x}+2x\cos {2x}$$
$$=\frac{1}{4}x^2\sin {2x} -\frac{1}{2}\frac{1}{D^2+4}\sin {2x}--\frac{1}{D^2+4}x\cos {2x}$$
$$V=\frac{1}{4}x^2\sin {2x} -\frac{1}{2}(\frac{x \cos {2x}}{-4})-V$$
This gives
$$ V=\frac {x^2 sin 2x}{4}+\frac{ x cos 2x}{16}$$
Thus from $(1)$
$$P.I=\frac{x}{8} -\frac{1}{2}(\frac {x^2 sin 2x}{8}+\frac{ x cos 2x}{16})$$
|
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|
Show that $\int_{0}^{\infty} \cos(x^4+x+1) dx$ converges I need to show that $\int_{0}^{\infty} \cos(x^4+x+1) dx$ converges. I showed that $\int_{0}^{\infty} \cos(x^4) dx$ converges but I don't know how to continue. I can't say that $\cos(x^4) \sim \cos(x^4+x+1)$ to conclude as $\cos(x^4)$ keeps changing sign.
If I write $\cos(x^4+x+1) = \cos(x^4)\cos(x+1)-\sin(x^4)\sin(x+1)$ it's the same problem, as the integral of $|\cos(x^4)|$ diverges.
Any help would be greatly appreciated
|
I've often used the technique of introducing a factor that allows me to do integration by parts. Setting $dv=\cos(x^4+x+1)\,dx$ is intractable, but if we introduce $4x^3+1$ then $dv=(4x^3+1)\cos(x^4+x+1)\,dx$ works great:
$$
\int _0^{\infty} \underbrace{\frac{1}{4x^3+1}}_{u}\cdot\underbrace{(4x^3+1)\cos(x^4+x+1)\,dx}_{dv}
$$
$$=\left.\frac{\sin(x^4+x+1)}{4x^3+1}\right|_0^{\infty}+
\int _0^{\infty} \frac{12x^2}{(4x^3+1)^2}\cdot\sin(x^4+x+1)\,dx
$$The boundary term evaluates to $-\sin(1)$ and the new integral is easily seen to be absolutely convergent by considering $\int_{0}^{\infty} \frac{12x^2}{(4x^3+1)^2}\,dx=\int_{1}^{\infty} \frac{1}{z^2}\,dz=1$.
|
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|
$ABCD$ is a convex quadrilateral with $\angle CAB=60^\circ$, $\angle CAD=20^\circ$, $\angle ABD=50^\circ$, $\angle DBC=10^\circ$. Find $\angle ACD$. Can you help me solve this Olympiad angles problem?
Let $ABCD$ be a convex quadrilateral such that: $\widehat{CAB}=60^\circ$; $\widehat{CAD}=20^\circ$; $\widehat{ABD}=50^\circ$; $\widehat{DBC}=10^\circ$.
Calculate $\widehat{ACD}$.
Thank you so much for your help
|
Let $E$ be the intersection point of lines $AC$ and $BD$.
Then:
$\angle BEC=60^\circ+50^\circ=110^\circ$
$\begin{align} \angle BCE&= 180^\circ-110^\circ - 10^\circ \\ &=60^\circ \\ &= \angle{BAC} \end{align}$
So that means $\Delta ABC$ is equilateral and hence:
$AB=AC \tag 1$
You can also deduce that:
$\begin{align} \angle BDA &= 180^\circ - 60^\circ - 20^\circ - 50^\circ \\ &= 50^\circ \\ &= \angle ABD \end{align}$
So $\Delta BAD$ is isosceles and hence:
$AB=AD \tag 2$
Combining $(1)$ and $(2)$ leads to:
$AD=AC$
Meaning that:
$\triangle DAC$ is isosceles
I believe you can finish up now.
|
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|
MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} \le a+b+c$
After trying out few values I realized it is wrong.So I decided to tackle it all at your once. I tried few well known inequalities and played around but nothing came useful out of it. Any help??
|
By Jensen $$\sum_{cyc}\sqrt{a^2-ab+b^2}\leq3\sqrt{\frac{\sum\limits_{cyc}(a^2-ab+b^2)}{3}}=\sqrt{3\sum_{cyc}(2a^2-ab)}.$$
Thus, it's enough to prove that:
$$\sqrt{3\sum_{cyc}(2a^2-ab)}+9\sqrt[3]{abc}\leq4(a+b+c).$$
Now, by $uvw$ (see here: https://artofproblemsolving.com/community/c6h278791 )
it's enough to prove the last inequality for equality case of two variables.
Can you end it now?
|
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|
Proof of 3-perfect codes I am reading a the proof of a theorem that says that $3$-perfect codes can only have length $7$ or $23$. I do not understand the following:
It follows from the definition that if $n$ is the length of a $3$-perfect then we must have that $(n+1)((n+1)^3-3(n+1)+8) = 3 \cdot 2^k$ for some $k \in \mathbb{Z}^+$. Now the proof claims that $16$ can not divide $(n+1)$.
I do not why. Any help or hints appreciated.
|
Suppose the highest power of $2$ which divides $n+1$ be $m \ge 4$, i.e., $2^4 = 16$ divides $n + 1$. Thus, you have that
$$n + 1 = 2^{m}j \tag{1}\label{eq1A}$$
where $j$ is odd. Since $n + 1$ divides evenly into $3 \cdot 2^k$, this means $j = 1$ or $j = 3$. You also have the equation
$$\begin{equation}\begin{aligned}
(n+1)((n+1)^3-3(n+1)+8) & = 3 \cdot 2^k \\
2^m(j)(2^{3m}j^3 - 3j2^{m} + 8) & = 3 \cdot 2^k \\
j(2^{m+3})(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \cdot 2^k \\
j(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \cdot 2^{k-m-3}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note that both $j$ and $2^{3m-3}j^3 - 3j2^{m-3} + 1$ are odd, so their product is odd. Thus, you must have $k = m + 3$ on the right, so it's value is just $3$.
If $j = 1$, then you have
$$\begin{equation}\begin{aligned}
j(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \\
2^{3m-3} - 3(2^{m-3}) + 1 & = 3 \\
2^{3m-3} - 3(2^{m-3}) & = 2 \\
2^{3m-4} - 3(2^{m-4}) & = 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
If $m = 4$, then $3m - 4 = 8$ so you have $2^{8} - 3 = 1$, which is not true. However, if $m \gt 4$, then the left side has at least one factor of $2$ but the right side doesn't have any factors, so that can't be true.
If $j = 3$ instead, this then means that you have
$$\begin{equation}\begin{aligned}
3(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \\
2^{3m-3}(3^3) - 3(3)2^{m-3} + 1 & = 1 \\
27(2^{3m-3}) & = 9(2^{m-3}) \\
3(2^{3m-3}) & = 2^{m-3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
However, there's at least one factor of $3$ on the left but no factor of $3$ on the right, so this can't be true either.
Since all of the possibilities have been considered and shown to not be possible, this means the original assumption, i.e., that $16$ divides $n + 1$, can't be true.
Update: A somewhat shorter & simpler way would be to note that, in \eqref{eq2A}, due to the relatively large size of the first term, you have for $j \ge 1$ and $m \ge 4$ that $2^{3m-3}j^3 - 3j2^{m-3} + 1 \gt 3$, so it's not possible as $3$ is the odd factor on the right.
|
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|
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
I substituted $3x^2 = 1-x$ in $6x^3-x^2-3x+2010$ and it simplified to
$$\frac{-8x}{3}+2010$$
I know this is a simple problem but I can't solve it. I think there's some method I'm not trying. Please help me with this.
|
$6x^3 = 2x-2x^2$
so
$6x^3-x^2 -3x+2010 = 2x-2x^2-x^2-3x+2010$
And finally $-x-3x^2+2010 = -x-(1-x)+2010=2009$
|
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|
Direct product and inverse element of Group Let $(P_2[\mathbb{R}],+)$ and $(\text{GL}(2,\mathbb R), \cdot)$ be groups. How can I write direct product G of those groups. Let
$a = \left (2x^2+4x-3,\begin{pmatrix}3 & 1\\ 2 & 1 \end{pmatrix} \right)$ and
$b= \left (-x^2+x-1, \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} \right ).$ Find id element in G, a*b and a^-1.
solution:
So far I have done following:
i) id element in G
$$
\left (0, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right )
$$
ii)
$$ a*b= \left (x^2+5x-4, \begin{pmatrix}7 & -3 \\ 5 & -2 \end{pmatrix} \right )$$
Is this the same thing as direct product or I need to do some different steps?
iii)
$$ \left (x^2-x+1, \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix} \right) $$
Am I on right way to do this, and is a*b same thing as direct product?
thanks in advance
|
You have exactly the right idea!
Given two groups $(G,\star)$ and $(H, \cdot)$ we can form their Direct Product $G \times H$ whose elements are pairs $(g,h)$ with $g \in G$ and $h \in H$.
Now this object is a group "componentwise" in the following sense:
*
*$\text{Id}_{G \times H} = (\text{Id}_G, \text{Id}_H)$
*$(g_1,h_1)(g_2,h_2) = (g_1 \star g_2, h_1 \cdot h_2)$
*$(g,h)^{-1} = (g^{-1}, h^{-1})$
In your case, you correctly identified that, for $G = P_2[\mathbb R]$ and $H = \text{GL}(2,\mathbb{R})$ this means:
*
*the identity element is $(\text{Id}_G, \text{Id}_H) =
\left ( 0 + 0x + 0x^2, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right )$
*the product $ab$ is $\left ( (2x^2 + 4x - 3) + (-x^2 + x - 1),
\begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 2 & -1\\1 & 0 \end{pmatrix} \right )$
*the inverse $a^{-1} = \left ( 2x^2 + 4x - 3, \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} \right )^{-1}$ is $\left ( - (2x^2 + 4x - 3), \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}^{-1} \right )$
Notice, in particular, that we write the "inverse" of a polynomial $p$ as $-p$, while we write the inverse of a matrix $A$ as $A^{-1}$. This is because we write the operation on the matrix group as multiplication, but we write polynomial group operation as addition.
As a final remark, you seem to have accidentally taken $b^{-1}$ for part iii. Conceptually, though, your idea is correct.
I hope this helps ^_^
|
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|
Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt forward:
$tan^2(x) + 1 = {1\over cos^2(x)} = y \text{ and } \\1-cos^2(x) = ycos^2(x) - cos^2(x) = cos^2(x)(y-1)$ combined we have:
$$5y{(5^{y-1} - 1) \over y-1}$$
as $y \rightarrow 1$ $5y{(5^{y-1} - 1) \over y-1}$ is not defined. Since both, nominator and denominator are $0$ I tried L'Hopital but ended at:
$5 5^{y-1} + 5y(y-1)5^{y-2}-5$ with $\lim_{y \rightarrow 1} = 5 + 0 - 5 = 0$ and $(y-1)' = 1$
Here I have to stop with no solution. I have also tried to use the quotient rule to differentiate the expression which did not get me anywhere.
|
Hint:
If you are using L'Hopital and differentiating with respect to $y$, then
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}y\,}5^{y-1} &= \frac{\mathrm{d}}{\mathrm{d}y\,}e^{(y-1) \ln5}\\
&=\ln 5 \cdot e^{(y-1) \ln5} \\
&=\ln 5 \cdot 5^{y-1}
\end{align}
|
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|
Can every odd integer greater than $1$ be written as a product of fractions $\frac{4m+1}{2m+1}$? Can every odd integer greater than $1$ be written as a product of fractions of the form $\frac{4m+1}{2m+1}$, $m$ a positive integer?
Here is a proof if the fractions were instead $\frac{4m-1}{2m+1}$. I tried to mimic it, but it doesn't work.
Since the fractions available have odd numerator and denominator, the same will be true of any product, so even integers aren't reachable. Odd integers can be reached, however.
Let $f(q)$ be the fraction $\frac{4q-1}{2q+1}$. Then $f(1)=1$, $f(4)f(7)=3$ and $3 f(4)=5$. So $1,3,5$ are obtainable.
Suppose now that $m\ge 2$ and we've obtained all odd numbers up to and including $4m-3$. Since $2m+1\le4m-3$ we can use $2m+1$ to obtain
\begin{gather*}
4m-1 = (2m+1) f(m),\\
4m+1=(2m+1) f(3m+1).
\end{gather*}
Hence $4m-1$ and $4m+1$ are obtainable as well, so having obtained $5$ we can then reach all odd positive integers.
Any help appreciated!
|
$\def\N{\mathbb{N}}$This solution proves by induction on $n \geqslant 0$ that all odd integers $2n + 1$ can be expressed as products of numbers of the form $f(m) = \dfrac{4m + 1}{2m + 1}$.
For $n = 0$, there is $1 = f(0)$. Assume that the proposition holds for all odd integers less than $2n + 1$. Suppose $2n + 2 = 2^k (2n_0 + 1)$ ($k \in \N_+$, $n_0 \in \N$), then $n_0 < n$ and\begin{align*}
&\mathrel{\phantom{=}} (2n_0 + 1) \prod_{j = 0}^{k - 1} f\left(2^j(2^k - 1)n_0 + 2^j(2^{k - 1} - 1) \right)\\
&= (2n_0 + 1) \prod_{j = 0}^{k - 1} \frac{ 2^{j + 2}(2^k - 1)n_0 + 2^{j + 2}(2^{k - 1} - 1) + 1 }{ 2^{j + 1}(2^k - 1)n_0 + 2^{j + 1}(2^{k - 1} - 1) + 1 }\\
&= (2n_0 + 1) · \frac{ 2^{k + 1}(2^k - 1)n_0 + 2^{k + 1}(2^{k - 1} - 1) + 1 }{ 2(2^k - 1)n_0 + 2(2^{k - 1} - 1) + 1 }\\
&= (2n_0 + 1) · \frac{ 2^{k + 1}(2^k - 1)n_0 + (2^k - 1)^2 }{ 2(2^k - 1)n_0 + (2^k - 1) }\\
&= (2n_0 + 1) · \frac{ 2^{k + 1}n_0 + (2^k - 1) }{ 2n_0 + 1 }\\
&= 2^{k + 1}n_0 + (2^k - 1) = 2n + 1.
\end{align*}
End of induction.
|
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|
If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to
Let an angle between the vectors $a$ and $b$ be $\frac{2\pi}{3}$. If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to $\frac23/\frac25/\frac13/\frac15$?
My attempt: $b^2=4a^2$. Also, $$a^2+\frac{|a||b|}{2}-\frac{x|a||b|}{2}-b^2=0$$$$\implies a^2+\frac{|a|*2|a|}{2}-\frac{x|a|*2|a|}{2}-4a^2=0$$$$\implies 1+1-x-4=0$$$$\implies x=-2$$ But it doesn't match with the options.
|
There might be a problem with your dot product.
We have
$$(a+xb)\cdot(a-b) = a^2 -a\cdot b + x \ a \cdot b -x \ b^2=0 \\ \implies a^2 +(x-1)\left(2a^2 \cdot \cos \frac{2\pi}{3}\right) -4xa^2=0$$
$$\implies 1-(x-1)-4x=0\implies x=\frac 25$$
|
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|
Prove $\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$ with complex numbers (roots of unity) I want to prove that
Using $z^9=1$ and the fact that $1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8=0$ where $w=cis(\frac{2π}{9})$, that $$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$$
I am able to do this by using $w^n+w^{-n}=2\cos \frac{2nπ}{9}$ and hence expanding the LHS but it is laborious and, so, I was wondering is there is a more efficient or elegant method of proving this this identity using the ninth root of unity.
Thanks very much
|
Note
$$\frac{1-z^9}{1-z}= \prod_{k=1}^8(w^k-z)
$$
Set $z=-1$ and use $\prod_{k=1}^8w^{k}=1$
\begin{align}
1&=\prod_{k=1}^8(w^k+1)= (\prod_{k=1}^8w^{k})^{1/2}\prod_{k=1}^8(w^{k/2}+ w^{-k/2})
= 2^8 \prod_{k=1}^8\cos\frac{k\pi}9
=\left( 2^4 \prod_{k=1}^4\cos\frac{k\pi}9 \right)^2
\end{align}
Then, substitute $\cos\frac{3\pi}9=\frac12$ to obtain
$$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac18$$
|
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|
Integral $\int\frac{4x^4}{{x^8+1}}\;dx$ I'm attempting this integral but I'm unsure how to proceed. I've begun to suspect it's actually non-elementary. Can anyone do this? I could always use Taylor series after a small bit of u-sub and integration by parts, but I wanted to know if there were a somewhat more direct way of doing it.
The integral is $$\int\frac{4x^4}{{x^8+1}}\;dx$$
Thanks in advance.
|
First of all, let me precise that, for this kind of integrals, Taylor expansions could me more thatn dangerous.
Considering the integrand, you can first write (by analogy with $(x^4+1)$
$$\frac {4x^4}{x^8+1}=\frac{\sqrt{2} x^2}{x^4-\sqrt{2} x^2+1}-\frac{\sqrt{2} x^2}{x^4+\sqrt{2} x^2+1}$$
Then
$$x^4-\sqrt{2} x^2+1=\left(x^2-\frac{1+i}{\sqrt{2}}\right) \left(x^2-\frac{1-i}{\sqrt{2}}\right)$$
$$x^4+\sqrt{2} x^2+1=\left(x^2+\frac{1-i}{\sqrt{2}}\right) \left(x^2+\frac{1+i}{\sqrt{2}}\right)$$ which finally make
$$\frac {4x^4}{x^8+1}=\frac{i \sqrt{2}}{\sqrt{2}-(1-i) x^2}-\frac{i \sqrt{2}}{\sqrt{2}-(1+i) x^2}+\frac{i \sqrt{2}}{\sqrt{2}+(1-i) x^2}-\frac{i \sqrt{2}}{\sqrt{2}+(1+i) x^2}$$ and now, we face quite trivial integrals.
$$\int\frac {4x^4}{x^8+1}\,dx=\cos \left(\frac{\pi }{8}\right) \left(\frac{1}{2} \log \left(\frac{x^2+2 x \sin
\left(\frac{\pi }{8}\right)+1}{x^2-2 x \sin \left(\frac{\pi
}{8}\right)+1}\right)+\tan ^{-1}\left(\frac{2 x \sin \left(\frac{\pi
}{8}\right)}{1-x^2}\right)\right)+$$
$$\sin \left(\frac{\pi }{8}\right) \left(\frac{1}{2} \log \left(\frac{x^2-2 x \cos
\left(\frac{\pi }{8}\right)+1}{x^2+2 x \cos \left(\frac{\pi
}{8}\right)+1}\right)-\tan ^{-1}\left(\frac{2 x \cos \left(\frac{\pi
}{8}\right)}{1-x^2}\right)\right)$$
In particular
$$\int_0^\infty\frac{4x^4}{{x^8+1}}\;dx=\pi\sqrt{1-\frac{1}{\sqrt{2}}} $$
|
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|
if the polynomial $x^4+ax^3+2x^2+bx+1=0$ has four real roots ,then $a^2+b^2\ge 32?$ if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four real roots. prove or disprove $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and when $P(x)=x^4+2x^3+2x^2+2x+1=(x+1)^2(x^2+1)$ but if the $P(x)$ have four real roots,I can't it.
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We will show that with some symmetry consideration we can either reduce to the case where all the roots have the same sign and then the result follows since if say $x_k >0$, $\sum {x_k} =|a| \ge 4(x_1..x_4)^{\frac{1}{4}} =4, a^2 \ge 16$ and then $\sum {1/x_k}=|b|$ and same applies so $b^2 \ge 16$ and we are done, or we must anyway have $|a|, |b| \ge 4$ and again the result follows.
Replacing $x \to 1/x$ changes $a$ into $b$ keeping the roots real so we can assume wlog $|a| \ge |b|$; replacing $x \to -x$ changes the signs of $a,b$ so we can assume wlog $a \ge 0$. But then if $x \ge 1, ax^3 \ge \pm bx$ since $a \ge |b|$ so $x^4+ax^3+2x^2+bx+1 >0$, while if $0 \le x \le 1, b=-c, a \ge c>0$ and we get the equation:
$x^4+ax^3+2x^2-cx+1=0, 0<c\le a$ and there is at least a root $0<x_0<1$ (sum of roots is negative so some are negative, product is $1$ so this means two are neagtive, two positive and our choices imply that the posiitive ones are in $(0,1)$)
Noting that the equation can be rewritten as:
$(x^2+2x-1)^2+(a-4)x^3+(4-c)x=0$ or $(x^2+2x-1)^2 =(c-4)x+(4-a)x^3$ we immediately get that if $c<4$ we must have $a<4$ too but then $4-a \le 4-c$ since $c \le a$ hence remembering that we have a root $0<x_0<1$ we get $0 \le (x_0^2+2x_0-1)^2 =(4-a)x_0^3-(4-c)x_0 < (4-a)x_0-(4-c)x_0 \le 0$ so a contradiction! Hence $|b|=c \ge 4, a \ge c \ge 4$ and we are done!
|
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|
Prove $\sum_{n=1}^{\infty}((n+\frac{1}{2})\ln(1+\frac{1}{n})-1)=1-\ln(\sqrt{2\pi})$ I am looking for a derivation of the following sum:
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=1-\ln(\sqrt{2\pi})$$
My current derivation(s) uses the zeta function at negative integers (and or Stirling approximation/ the derivative of $\zeta'(0)$). I want to avoid those.
How I got an answer was via regularization of
$$-\sum_{i=1}^{\infty}\frac{\zeta(-i)}{i}=\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+ \frac{1}{n}\right)-1\bigg)$$
My own other try was rewriting it via:
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=\sum_{k=2}^{\infty} \zeta(k)(-1)^k \bigg(\frac{1}{k+1}-\frac{1}{2k}\bigg)$$
If this works I am already happy. If there's another simple way I'd love to hear it as well.
|
New Answer. Let $S_N$ denote the partial sum for the first $N$ terms. Then $S_N$ is related to the Stirling's Formula by the following computation:
\begin{align*}
S_N
&= \sum_{n=1}^{N} \left(n+\frac{1}{2}\right)\log(n+1) - \sum_{n=1}^{N} \left(n+\frac{1}{2}\right)\log n - N \\
&= \left(N+\frac{1}{2}\right)\log (N+1) - \log (N!) - N.
\end{align*}
Now we consider $e^{-S_N}$ instead. Using the formula $\int_{0}^{\infty}x^{n}e^{-sx}\,\mathrm{d}x=\frac{n!}{s^{n+1}}$,
\begin{align*}
\exp(-S_N)
&= \frac{N!e^{N}}{(N+1)^{N+\frac{1}{2}}} \\
&= \frac{N^{N+1}}{(N+1)^{N+\frac{1}{2}}} \int_{0}^{\infty} x^N e^{-N(x-1)} \, \mathrm{d}x \\
&= \frac{1}{(1+\frac{1}{N})^{N+\frac{1}{2}}} \int_{-\infty}^{\infty} \left(1 + \frac{u}{\sqrt{N}}\right)_{+}^N e^{-\sqrt{N}u} \, \mathrm{d}u,
\end{align*}
where we utilized the substitution $x=1+\frac{u}{\sqrt{N}}$ in the last step and $x_{+}:=\max\{0,x\}$ denotes the positive part of $x$. Then, taking limit as $N\to\infty$ and assuming for a moment that the order of limit and integral can be swapped, we get
\begin{align*}
\lim_{N\to\infty} \exp(-S_N)
&= \biggl( \lim_{N\to\infty} \frac{1}{(1+\frac{1}{N})^{N+\frac{1}{2}}} \biggr) \int_{-\infty}^{\infty} \lim_{N\to\infty} \left(1 + \frac{u}{\sqrt{N}}\right)_{+}^N e^{-\sqrt{N}u} \, \mathrm{d}u \\
&= \frac{1}{e} \int_{-\infty}^{\infty} e^{-u^2/2} \, \mathrm{d}u
= \frac{\sqrt{2\pi}}{e}.
\end{align*}
Here, the last step follows from the gaussian integral. Therefore
$$ \sum_{n=1}^{\infty} \left[ \left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)-1 \right] = \lim_{N\to\infty} S_N = 1 - \log\sqrt{2\pi} $$
provided the interchange of limit and integral is justified. For this, we note the following inequality:
$$ \log(1+x) \leq x - \frac{x^2}{2(1+x_+)}, \qquad x > -1 $$
From this, we deduce that
$$
\left(1 + \frac{u}{\sqrt{N}}\right)_{+}^N e^{-\sqrt{N}u}
\leq \exp\left(-\frac{u^2}{2(1+u_+)}\right)
$$
holds for all $N\geq 1$ and for all $u \in \mathbb{R}$. Therefore the dominated convergence theorem is applicable and the desired step is justified, completing the proof.
Old Answer. The sum converges absolutely by the Limit Comparison Test with $\zeta(2)$. Now for each given $n \geq 1$,
\begin{align*}
\left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)-1
&= \left(n+\frac{1}{2}\right)\left(\sum_{j=1}^{\infty}\frac{(-1)^{j-1}}{jn^j} \right)-1\\
&= - \frac{1}{4n^2} + \left(n+\frac{1}{2}\right)\sum_{j=3}^{\infty}\frac{(-1)^{j-1}}{jn^j}\\
&= - \frac{1}{4n^2} + \sum_{j=3}^{\infty}\frac{(-1)^{j-1}}{j}\left(\frac{1}{n^{j-1}}+\frac{1}{2n^j}\right).
\end{align*}
Using the formula $\int_{0}^{\infty}x^{s-1}e^{-nx}\,\mathrm{d}x=\frac{\Gamma(s)}{n^s}$, this may be recast as
\begin{align*}
&= \int_{0}^{\infty}\left[ - \frac{x}{4} + \sum_{j=3}^{\infty}\frac{(-1)^{j-1}}{j}\left( \frac{x^{j-2}}{(j-2)!} + \frac{x^{j-1}}{2(j-1)!} \right)\right] e^{-nx}\, \mathrm{d}x \\
&= \int_{0}^{\infty} \left( \frac{1}{x} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)(1-e^{-x}) \right) e^{-nx} \, \mathrm{d}x.
\end{align*}
Summing this for $n = 1, 2, \dots$, we get
\begin{align*}
S
&:= \sum_{n=1}^{\infty} \left[ \left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)-1 \right] \\
&= \int_{0}^{\infty} \left( \frac{1}{x} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)(1-e^{-x}) \right) \frac{1}{e^x - 1} \, \mathrm{d}x \\
&= \int_{0}^{\infty} \left( \frac{1}{x(e^x - 1)} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)e^{-x} \right) \, \mathrm{d}x.
\end{align*}
To compute the right-hand side, we consider the following regularization:
\begin{align*}
S(s)
&:= \int_{0}^{\infty} \left( \frac{1}{x(e^x - 1)} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)e^{-x} \right) x^s \, \mathrm{d}x \\
&= \int_{0}^{\infty} \left( \frac{x^{s-1}}{e^x - 1} - \frac{1}{2}x^{s-1}e^{-x} - x^{s-2}e^{-x} \right) \, \mathrm{d}x.
\end{align*}
This function is analytic for $\operatorname{Re}(s) > -1$, and $S = S(0)$. Moreover, for $s > 2$, we easily find that
\begin{align*}
S(s)
&= \Gamma(s)\zeta(s)-\frac{1}{2}\Gamma(s)-\Gamma(s-1) \\
&= \Gamma(s+1)\biggl( \frac{\zeta(s)-\frac{1}{2}-\frac{1}{s-1}}{s} \biggr).
\end{align*}
By the principle of analytic continuation, this identity must hold on all of $\operatorname{Re}(s)>-1$. So, letting $s \to 0$ to the above formula yields
$$ S = \lim_{s\to 0}S(s) = 1 + \zeta'(0). $$
Now the desired formula follows from $\zeta'(0) = -\log\sqrt{2\pi}$.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.