Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solution to $\frac{d}{d\frac{1}{x}} x$ If I want to solve
$$\frac{d}{d\frac{1}{x}} x$$
is my approach correct?
As
$$\begin{align*} \frac{d}{d\frac{1}{x}}x&=\\
\text{with }\frac{1}{x}&=y\\
\frac{d}{dy}\frac{1}{y}&=-\frac{1}{y^2}\\
&=-\frac{1}{\left(\frac{1}{x}\right)^2}\\
&=-x^2
\end{align*}$$
Is this approach correct, or did I miss something?
| $\frac{\mathrm{d} x}{\mathrm{d} \frac{1}{x}}=\left (\frac{\mathrm{d} \frac{1}{x}}{\mathrm{d} x} \right )^{-1}=\left ( \frac{-1}{x^{2}} \right )^{-1}=-x^{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Divergence of squared sum of Chebyshev Polynomials $\equiv L+R$ has empty point spectrum The Chebyshev Polynomials of the second kind $U_n$ are the solutions of the differential equation
$$(1-x^2)U_n''(x)-3xU_n'(x)+n(n+2)U_n(x)=0$$
Alternatively they are defined inductively:
$$U_0(x)=1 \qquad U_1(x)=2x \qquad U_{n+1}(x)=2xU_n(x) - U_{n-1}(x)$$
My question is whether or not
$$\sum_n |U_n(x/2)|^2\tag{1}$$
diverges for all $x \in \mathbb C$. The convergence of (1) is equivalent to the operator $L+R$ on $\mathscr l^2 (\mathbb N)$ having an eigenvalue $x$, where $L$ is the left shift and $R$ the right shift operator. The correspondence is explained in this answer. (This is the reason behind the functional analysis tag.)
Since $L+R$ is hermitian, bounded on a Hilbert space and has norm $≤2$, the spectrum is a subset of $[-2,2]$, and as such the sum automatically diverges if $x \notin [-2,2]$.
Experimentally it looks like as if the sum diverges on $[-2,2]$, but I would be interested in an angle allowing a proof or evidence for points at which it does converge.
| if $a_n = 0$ for $n < 0$, $a_0 = 1$ and $a_{n+1} = 2 x a_n - a_{n-1} $ for $n > 1$ then by multiplying by $z^{-n}$ and summing we get the Z-transform :
$$\sum_{n=0}^\infty (a_{n+1} - 2 x a_n + a_{n-1}) z^{-n} = \sum_{n=0}^\infty \delta(n) z^{-n} = 1$$
and we have also $\sum_{n=0}^\infty (a_{n+1} - 2 x a_n + a_{n-1}) z^{-n} = (2x + z^{-1} + z) \sum_{n=0}^\infty a_n z^{-n} $
so with $A(z) =\sum_{n=0}^\infty a_n z^{-n}$ :
$$A(z) = \frac{1}{2x + z^{-1} + z} = \frac{z}{1 -2x z + z^2} = \frac{4z}{(z-2x+2\sqrt{x^2-1})(z-2x+2\sqrt{x^2-1})} = \frac{bz}{z-2x+2\sqrt{x^2-1}} + \frac{cz}{z-2x-2\sqrt{x^2-1}}$$
(for some constants $b,c$ obtained in general by partial fraction decomposition)
so for $n \ge 0$ :
$$a_n = B (2x-2\sqrt{x^2-1})^{n} + C (2x+2\sqrt{x^2-1})^{n} $$
the solutions for other initial conditions are easily obtained by linearity and shifting,
so the convergence of $\sum_n |a_n|^2$ depends on if those two roots $2x\pm 2\sqrt{x^2-1}$ are of modulus strictly $< |1|$
(don't worry if I made some mistakes the idea is there)
| {
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Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.
Solution
I calculated the two lines represented by $ax^2+2hxy+by^2=0$ as follows;
here.
$$ax^2+2hxy+by^2=0$$
Multiplying by $a$ on both sides and adding $h^2y^2$ to both sides :
$$ax+hy=\pm y\sqrt{h^2-ab}.$$
What should I do next?
| If one of the line of
$$ax^2 +2hxy+by^2=0 \tag{1}$$
bisects the coordinate axes then the eqn. of that line might be $y=x$ OR, $y =-x$.
Combining these eqns we get $y=\pm x$ and substituting this value of $y$ in eqn. (1) we get
\begin{eqnarray}
ax^2+2hx(\pm x)+by^2=0 \\
ax^2±2hx^2+by^2=0 \\
x^2(a±2h+b)=0 \\
a±2h+b=0 \\
a+b=±2h
\end{eqnarray}
Also, $(a+b)^2=4h^2$
| {
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Prove that $a+\frac{1}{b}>2$ or $b+\frac{1}{a}>2$ for two strict positive numbers Another Olympiad Problem, let $x$ and $a$ and $b$ be strictly real positive numbers.
*
*Prove that $x$+$\frac{1}{x}$$>$$2$ (proven)
*Than conclude that $a$+$\frac{1}{b}$$>$$2$ or
$b$+$\frac{1}{a}$$>$$2$
For the second question, I know I am supposed to replace $x$ with a number in function of $a$ and $b$ , but I can't find it .
if we replace $x$ by $ab$ we still have a problem.
$ab$+$\frac{1}{ab}$$+2>$$4$ , we can't just conclude that $a$+$\frac{1}{b}$$>$$2$ or $b$+$\frac{1}{a}$$>$$2$
| Slightly different.
If $a = b$ then $a + 1/b = b + 1/a =a + 1/a \ge 2$.
If $a > b$ then $a + 1/b > a + 1/a \ge 2$.
If $a < b$ then $b + 1/a > b + 1/b \ge 2$.
(If $a = b = 1$ then $a + 1/b = 2$. Otherwise $a + 1/b > 2$ or $b + 1/a > 2$ or both.)
| {
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Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integer it has to be $k=101$ cause $gcd(k,k-101)=1\lor101$ and both $k,k-101$ can't be both squares of an integer,so $k=101t$ and $t(t-1)$ is never an square except for $t=1,0$ and $t=0$ is not possible hence $k=101$ is only possible integer solution
EDIT: So if $\gcd(k,k-101)=1$ then $k=h^2,k-101=(h-s)^2$ then $h(2h-s)=101$ which can be $s=1,h=51$ or $s=101,h=51$.$y=51^2+51\cdot50=51\cdot 101,x=51$
And since $x=2k-y$ is integer then $k=\frac{h}{2}$,if $h=2q$ then $k$ is integer otherwise if $h=2q+1$ then $$\frac{2q+1}{2}+\frac{1}{2}\sqrt{4q^2-400q-201}=\frac{2q+1}{2}+\frac{1}{2}\sqrt{(2q-10)^2-301}\\(2q-10)^2-301=r^2\\2q-10=z\\z^2-301=(z-c)^2\\c(2z-c)=301,c=1,z=151,c=7,z=25,c=43,z=25$$
The $z=151$ is impossible cause $2q-10$ is even,and $z=25$ is also impossible because $2q-10$ is even.Hence the only solutions are $(x,y)=(101,101),(51,5151),(5151,51)$ The last one clearly from symmetry
| Sneaky hint: Think about $\frac{1}{a}+\frac{1}{a(2a-1)}$
| {
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A variant of Viète's formula (the 2's replaced by 3's) I am wondering whether there exists an easy way to evaluate the following infinite product :
$\sqrt{\frac{1}{3}}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}}}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}}}}...$
or even better (assuming convergence) of :
$\sqrt{\frac{1}{a}}.\sqrt{\frac{1}{a}+\frac{1}{a}.\sqrt{\frac{1}{a}}}.\sqrt{\frac{1}{a}+\frac{1}{a}.\sqrt{\frac{1}{a}+\frac{1}{a}.\sqrt{\frac{1}{a}}}}...$
with $a>0.$
This is a variant of Viète's formula :
$V_{\infty}=\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\frac{1}{2}.\sqrt{\frac{1}{2}}}.\sqrt{\frac{1}{2}+\frac{1}{2}.\sqrt{\frac{1}{2}+\frac{1}{2}.\sqrt{\frac{1}{2}}}}...=\frac{2}{\pi}$
which is much easier to prove since we can define :
$u_{n+1}=\sqrt{\frac{1+u_n}{2}}$ and realise that for $|u_0|<1$ we can also define $cos(\theta)=u_0$.
Viète's formula is just $\prod_{n=1}^{\infty}u_n$, which can be easily evaluated using trigonometric identities, especially $cos(2\theta)=2cos²(\theta)-1$ which yields : $cos(\frac{\theta}{2})=\sqrt{\frac{1+cos(\theta)}{2}}$
So $\prod_{n=1}^{\infty}u_n=\prod_{n=1}^{\infty}cos(\frac{\theta}{2^n})$
Then using $sin(2\theta)=2sin(\theta)cos(\theta)$ mass cancellation occurs and we find $\prod_{n=1}^{\infty}u_n=\frac{sin(\theta)}{\theta}.$ Viète's formula corresponds to $u_0=0$, so $\theta=\frac{\pi}{2}$. By replacing with this value, we find Viète's value.
Unfortunately, this strategy doesn't seem to work when the 2's are replaced by something else (like 3 for example) (or at least, I can't find any tractable trigonometric formulas to help) Any idea ? Thank you !
(P.S. : I am very aware that Viète did not use this strategy to find this formula. He used geometry arguments I do not know, so perhaps there is a way to generalize using geometry, although I'd rather have a proof using only analysis...)
| Where does this problem come from? Anyway, for $a>0$, let $u_0^{(a)} = 0$ and $u_{n+1}^{(a)} = \sqrt{\frac{1+u_n^{(a)}}{a}} = \sqrt{\frac{2}{a}}\sqrt{\frac{1+u_n^{(a)}}{2}}$. We wish to determine $\prod_{n=1}^{\infty}u_n^{(a)}$. Repeating the trigonometric substitutions in your post, we find
$$\prod_{n=1}^{\infty}u_n^{(a)} = \frac{2}{\pi}\cdot \lim_{n \to \infty} \left(\sqrt{\frac{2}{a}} \right)^n $$
For $0<a<2$ this diverges, for $a=2$ it equals $2/\pi$ and for $a>2$ it equals $0$.
| {
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Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
I understand that when factored out, you have $(x+3)^3 = x^3+9x^2+27x+27$ as opposed to the other side which is $x^3+27$, so even if you plug in $0$ the left side will always be greater. Is this the right thinking process for determining the answer? Or am I missing something?
| $(x+3)^3 = x^3+3^3$
$\Leftrightarrow x^3+9x^2+27x+27 = x^3+27$
$\Leftrightarrow 9x^2+27x = 0$
$\Leftrightarrow x^2+3x = 0$
$\Leftrightarrow x(x+3)=0$
$\Leftrightarrow x=0$ or $x=-3$
| {
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If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$,
If $$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0,$$ prove that
$$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C).$$
My solution:
From the given,
$$\cos^3A+\cos^3B+\cos^3C=3\cos A\cos B\cos C$$
Now,
L.H.S$$=\cos3A+\cos3B+\cos3C$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$
$$=12\cos A \cos B \cos C.$$
My solution ends here. How should I complete?
NOTE: I am not allowed to use complex numbers at my level. So, please help me solve this problem without using complex numbers.
| we have $$sinA+sinB=-sinC$$ and $$cosA+cosB=-cosC$$ squaring and adding both we get
$$2+2cos(A-B)=1$$ that is
$$cos(A-B)=\frac{-1}{2}$$ $\implies$
$$A-B=\frac{2\pi}{3}$$ similarly
$$B-C=\frac{-4\pi}{3}$$ and
$$C-A=\frac{2\pi}{3}$$
Now $$cos3A=cos\left(3\left(B+\frac{2\pi}{3}\right)\right)=cos(3B+2\pi)=cos3B$$
similarly $$cos3C=cos3B$$
Hence
$$cos3A+cos3B+cos3C=3cos3A=3cos(A+A+A)=3cos\left(A+B+\frac{2\pi}{3}+C-\frac{2\pi}{3}\right)=3cos(A+B+C)$$
| {
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What is wrong with this infinite sum We know that:
https://www.youtube.com/watch?v=w-I6XTVZXww
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$
So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers
Furthermore, we can add it to itself but shifting the terms one place:
$$
\begin{align}
1+2+3+4+\cdots & \\ 1+2+3+\cdots & \\
=1+3+5+7+\cdots & =2S
\end{align}
$$
This is the sum of the odd numbers
If we were to now sum the odd numbers and the even numbers like below:
$$ 2+4+6+8+\cdots \\[6pt] 1+3+5+7+\cdots \\[6pt] \text{if we add the terms in a certain order we can get } 1+2+3+4+5+6+7+\cdots$$
This supposedly tells us that:
$$4S = S\\[6pt] 4 \left(\frac{-1}{12}\right)=\frac{-1}{12} \\[6pt] \frac{-1}{3} = \frac{-1}{12} $$
What is faulty with this proof.
| As pointed out in section 3, page 1191 of this article, the rules for manipulating divergent series are more restrictive than those for convergent series. As pointed out in the article, to avoid problems you should work with the power series obtained by multiplying the nth term by $x^n$, instead.
| {
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Total number of $4$ digit numbers whose product of digits is $72$
Total number of $4$ digit numbers whose product of digits is $72$
$\bf{My\; Try::}$ Here the possible factor of $72 = 2^{3}\cdot 3^2$
Now here we divide $2^3\cdot 3^3$ into product of $4$ numbers
Like $1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot 9$
and $1\cdot 2\cdot 6\cdot 6$
and $1\cdot 3\cdot 4\cdot 6$
and $1\cdot 3\cdot 3\cdot 8$
and $2\cdot 2\cdot 3\cdot 6$
and $2\cdot 2 \cdot 2\cdot 9$
So arrangement of these numbers is $$\frac{4!}{2!}+4!+\frac{4!}{2!}+4!+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{3!} = 12+12+12+24+24+12+4=100$$
But answer given is $=112$
So Where i have make mistake, Is there is any other better method,
Then plz explain here
Thanks
| $1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot 9$
and $1\cdot 2\cdot 6\cdot 6$
and $1\cdot 3\cdot 4\cdot 6$
and $1\cdot 3\cdot 3\cdot 8$
and $2\cdot 2\cdot 3\cdot 6$
and $2\cdot 2 \cdot 2\cdot 9$
and $2\cdot 3 \cdot 3\cdot 4$
So arrangement of these numbers is $$\frac{4!}{2!}+4!+\frac{4!}{2!}+4!+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{3!}+\frac{4!}{2!} = 12+12+12+24+24+12+4+12=112$$
| {
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Third degree polynomial for $ \sin(x^2+y^2)?$ Let R be an open region containing the point $(x_0,y_0).$ Let f, g, and h be functions defined on R, except possibly at$ (x_0,y_0).$ Suppose that for every $(x,y) \in R$ not equal to $(x_0,y_0)$, we have:
$g(x,y) \leq f(x,y) \leq h(x,y)$.
If
$\lim_{(x,y)\to(x_0,y_0)} g(x,y) = L = \lim_{(x,y)\to(x_0,y_0)} h(x,y)$
then:
$\lim_{(x,y)\to(x_0,y_0)} f(x,y) = L$
Use the squeeze theorem evaluate $\displaystyle \lim_{(x,y)\to(0,0)} \dfrac{\sin (x^2+y^2) }{x^2+y^2}.$
To apply the Squeeze Thoerem we must select functions g and h. Use the Taylor Polynomials of degree 1 and 3 as natural bounds of $\, \sin u $to fill in the blanks.
$? \leq \sin (x^2+y^2) \leq ?$
Okay, so what is 3rd degree Taylor Polynomial? I know the Taylor formula for sin(x) is:
$$\sum_{k=0}^n{{(-1)}^k \frac{x^{2k +1}}{(2k+1)!}}.$$
So for the 1st degree Taylor polynomial, my value is:
$(-1)^0 \cdot \frac{ (x^2+y^2)^{2 \cdot 0 +1} } {(2 \cdot 0 + 1)!} $ = $x^2+y^2$
???
Also, do I have to plug in 0 for x and y, since I'm supposed to evaluate at 0?
| Treat $x^2+y^2$ as a single variable. Recall that
$$z-\frac16z^3\le \sin(z)\le z$$
for $0<z<1$. Then, let $z=x^2+y^2$ reveals
$$(x^2+y^2)-\frac16(x^2+y^2)^3\le \sin(x^2+y^2)\le (x^2+y^2)$$
Dividing by $x^2+y^2$ yields
$$1-\frac16(x^2+y^2)^2\le \frac{\sin(x^2+y^2)}{x^2+y^2}\le 1$$
whereupon applying the squeeze theorem, we have
$$\lim_{(x,y)\to (0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1$$
| {
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Integration with double absolute value How to prove that?
$$
\int\left|\sqrt{|x|}\right|\ \mathrm{d}x
= \frac{1}{3} x
\left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x)
+ \sqrt{x} + \sqrt{-x}\right)
$$
I cannot understand this result.
Where does the $1/3$ factor come out from?
| I find the outer $| \,|$ incomprehensible since $\sqrt{|x|}\ge0$ according to the usual definition of square root. I will answer ignoring that.
For $x\ge 0$ we have the usual result $\int x^{1/2}\,dx={2\over 3}x^{3/2}$ plus constant, which you can verify by taking the derivative of ${2\over 3}x^{3/2}$.
Suppose $x\ge 0$; then $\mathrm{ sgn} (x)=1$, and
$$\frac{1}{3} x
\left((\sqrt{x} - \sqrt{-x})\mathrm{sgn}(x)
+ \sqrt{x} + \sqrt{-x}\right)={1\over 3}x(2\sqrt{x})={2\over 3}x^{3/2}$$
as desired (but that raises the issue of what $\sqrt{-x}$ is, perhaps $i\sqrt{x}$).
Now suppose $x\le 0$. We then have $\mathrm{ sgn} (x)=-1$, and we get
$$\frac{1}{3} x
\left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x)
+ \sqrt{x} + \sqrt{-x}\right)={1\over 3}x(2\sqrt{-x})=-{2\over 3}(-x)^{3/2}.$$
Differentiated this also gives the right answer, $\sqrt{-x}$, and thus our expression equals $\int |x|^{1/2}\,dx=\int (-x)^{1/2}\,dx$, for $x\le 0$.
| {
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taylor expansion and limit of a series?? $f\left(x\right)=∫_0^x\tan ^{-1}tdt$
what is the taylor expansion about the origin of this function?
and how do i use this to get the limit of the series
$1-\frac{1}{2}-\frac {1}{3}+\frac {1}{4}+\frac {1}{5}-\frac{1}{6}-\frac {1}{7}.......$
i could get the limit by using concepts like rearranging the terms and got a different limit since it is a conditionally convergent series and can be made to converge to any real number.but how do i get the limit using this taylor expansion.please somebody help?
Answer to the second part is $\frac {π}{4}-\frac {\log 2}{2}$
| The Taylor expansion can be obtained by observing
$$ f(x) = \int_0^x\arctan(t)dt=x\arctan x - \frac{1}{2}\ln(1+x^2)$$
Using the Mercator series we have
$$\ln(1+x^2) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{2k}$$
and for the first summand there is the well-know expansion
$$x\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+1}=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+2}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}x^{2k}$$
Therefore the sum of the two functions is
$$f(x)=\sum_{k=1}^\infty (-1)^{k+1}x^{2k}\left( \frac{1}{2k-1}- \frac{1}{2k}\right)= \sum_{k=1}^\infty\frac{(-1)^{k+1}}{2k(2k-1)}x^{2k} $$
$$=\frac{1}{2}x^2-\frac{1}{12}x^4+\frac{1}{30}x^6-\frac{1}{56}x^8+\frac{1}{90}x^{10} \cdots$$
Of course this result can also be obtained by termwise integrating the $\arctan$ series
$$f(x)=\int \left (\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}x^{2k-1}\right) dx
= \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}\int x^{2k-1}dx
=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{2k(2k-1)}x^{2k}$$
Unfortunately I have no clue how to relate the Taylor series to your other part of the question.
| {
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} |
Find the value of $\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90}$ Find the value of $p+q$, where $p$ and $q$ are two positive integers such that $p$ and $q$ have no common factor larger than $1$ and
$$\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} = \frac{p}{q}.$$
By using Wolfram Alpha, I obtain the sum is $\frac{3}{20}$. But I have no idea on how to obtain the sum.
Any hint would be appreaciated.
[The question is taken from SMO 2015 Junior section]
| $1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 = $
$1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
Note: $\frac{1}{n(n+1)} + \frac{1}{(n +1)(n +2)} = \frac{n+2 + n}{n(n+1)(n+2} = \frac {2n + 1}{n(n+1)(n+2)} = \frac{2}{n(n+2)}$
And $\frac{k}{n(n+k)} + \frac{1}{(n+k)(n + k + 1)} = \frac{k(n + k + 1) + n}{n(n+k)(n + k + 1)} = \frac{(k + 1)n + (k+1)k}{n(n+k)(n+k + 1)} = \frac{k+1}{n(n+ k + 1)}$.
So...
$1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
$2/4*6 + 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
$3/4*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
$4/4*8 + 1/8*9 + 1/9*10 = $
$5/4*9 + 1/9*10 = $
$6/4*10 = 3/20 = p/q$
$p + q = 3 + 20 = 23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known. Is there a way to to find out what $a^3+b^3+c^3$ evaluates to, when the values of $abc$, $ab+bc+ca$ and $a+b+c$ are given?
Alternatively, is there a way to express $a^3+b^3+c^3$ in terms of the aforementioned expressions?
| Use Newton's identities:
\begin{align}
p_1 &= e_1\\
p_2 &= e_1p_1-2e_2\\
p_3 &= e_1p_2 - e_2p_1 + 3e_3 \\
\end{align}
where
\begin{align}
e_1 &= a+b+c &\qquad p_1 &= a^1+b^1+c^1\\
e_2 &= ab+bc+ca &\qquad p_2 &= a^2+b^2+c^2\\
e_3 &=abc &\qquad p_3 &= a^3+b^3+c^3\\
\end{align}
Newton's identities give
$$
p_3 = e_1^3-3e_1e_2+3e_3
$$
that is
$$
a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
What's $\alpha+\beta$ if we have: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ ($\alpha$ and $\beta$ are Real) What's $\alpha+\beta$ if we have $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$? Here $\alpha$ and $\beta$ are real.
Firstly, I subtracted the two equations and got the following:
$$\alpha^3-\beta^3-6(\alpha^2-\beta^2)+13(\alpha-\beta)=-18$$ Then I tried to factorize the left hand side as: $$(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2-6\alpha-6\beta+13)=-18$$
At this point it seems we can't go on!
Then I tried adding the two equations as: $$(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2-6\alpha-6\beta+13)+12\alpha\beta=20$$
unfortunately, again I can't continue!
Is there any special creativity needed for solving this question?
| Let $f(x)=x^3-6x^2+13x,$ Then $f'(x)=3x^2-12x+13 = 3[(x-2)^2]+1>0$
So $f(x)$ is strictly increasing function.
Now above we have given $f(\alpha)=1$ and $f(\beta) = 19\;,$ Where $\alpha<\beta$
Then $f(\alpha)<f(\beta)$, because the function is increasing.
Now \begin{align*}f(4-x)&=(4-x)^3-6(4-x)^2+13(4-x)\\&=64-x^3-48x+12x^2-96-6x^2+48x+52-13x\\&=-x^3+6x^2-13x+20=-f(x)+20\end{align*}
So if $x=\alpha$, Then $f(4-\alpha)=-f(\alpha)+20=19=f(\beta)$
The function is continuous and monotonically increasing, thus $f(x)=f(y)\implies x=y$, hence $$4-\alpha=\beta\Rightarrow \boxed{\alpha+\beta=4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to rigorously prove the following continued fraction identity? The following identity seems to be satisfied for any $z \in C$
$$\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cdots}}}}=\begin{cases}1 & |z| \geq 1\\z^2 & |z| \leq 1\end{cases}$$
I checked it numerically for real and for complex numbers.
If we assume that the limit exists, we have:
$$\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cdots}}}}=y$$
$$\cfrac{z^2}{1+z^2-y}=y$$
$$y_1=1,~~~~~~~y_2=z^2$$
This solution doesn't tell us when to use $y_1$ and when to use $y_2$.
I think I can prove the second solution for $|z| < 1$ by transforming the continued fraction into geometric series by Euler's formula:
$$\frac{1}{1-y}=\cfrac{1}{1-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cdots}}}}=1+z^2+z^4+z^8+\dots=\frac{1}{1-z^2}$$
But I'm not sure how to prove the case of $|z| \geq 1$.
Edit
Of course, I can use the same way for $|z| > 1$:
$$\frac{1}{1-\frac{y}{z^2}}=\cfrac{1}{1-\cfrac{\frac{1}{z^2}}{1+\frac{1}{z^2}-\cfrac{\frac{1}{z^2}}{1+\frac{1}{z^2}-\cdots}}}=1+\frac{1}{z^2}+\frac{1}{z^4}+\frac{1}{z^8}+\dots=\frac{1}{1-\frac{1}{z^2}}$$
$$y=1$$
And the case for $|z| = 1$ seems to be self-evident. Still, I'm not sure if those proofs are rigorous.
| The convergents of this continued fraction are easy to compute explicitly:$$\underbrace{\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{\ddots}{\ddots-\cfrac{z^2}{1+z^2}}}}}_{n\text{ numerators/denominators}}=\frac{z^2(1-z^{2n})}{1-z^{2n+2}}$$ (understood in the limit sense if $z^{2n+2}=1$, and proven by induction). Now taking $n\to\infty$ is as rigorous as one gets. This confirms your result for $|z|\neq 1$ and $z=\pm 1$; for other values of $z$, the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $\lim _{x\to 0}\left\lfloor\frac{\tan x \sin x}{x^2}\right\rfloor$ How do I evaluate the limit
$$\lim _{x\to 0}\left\lfloor\frac{\tan x \sin x}{x^2}\right\rfloor$$
where $\lfloor\cdot\rfloor$ denotes greatest integer function.
I know that $x>\sin x$ and $x < \tan x$ but how do I use these results here?
$\tan x \over x$ tends to $1+$ whereas $\sin x \over x$ tends to $1-$
| Using Taylor–Young expansions: we know that
$$\tan(x)=x+\frac{x^3}3+o(x^4),\qquad\sin(x)=x-\frac{x^3}6+o(x^4),$$
hence
$$\tan(x)\sin(x)=x^2+\frac{x^4}6+o(x^5),$$
and hence
$$\frac{\tan(x)\sin(x)}{x^2}=1+\frac{x^2}6+o(x^3).$$
From here, we conclude that there exists a punctured neighborhood $V$ of $0$ such that
$$\forall x\in V,\ \frac{\tan(x)\sin(x)}{x^2}>1.$$
Since
$$\lim_{x\to0}\frac{\tan(x)\sin(x)}{x^2}=1$$
we can assume that $V$ has been chosen such that
$$\forall x\in V,\ 1<\frac{\tan(x)\sin(x)}{x^2}<2.$$
Hence
$$\forall x\in V,\ \left\lfloor\frac{\tan(x)\sin(x)}{x^2}\right\rfloor=1,$$
from which we conclude that
$$\lim_{x\to0}\left\lfloor\frac{\tan(x)\sin(x)}{x^2}\right\rfloor=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$
Using partial fractions
$$-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}=\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{2v}{v^2+1}-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}$$
$$=\frac{1}{5}ln|v^2+1|-\frac{4}{5}\arctan(v)+\frac{1}{5}ln|-v^2+v+1|$$ from $\frac{\pi}{8}$ to $0$
$0.02-0+0.299-0+0.04-0=0.359$
But it should come out 0.32
| You have made a mistake in simplifying the integrand. You should have
$$\int \frac{\frac{2dv}{1+v^2}}{2+\frac{2v}{1-v^2}} = \int \frac{1-v^2}{(1+v^2)(1+v-v^2)}dv.$$
Now do partial practions. You get
$$ \frac{1-v^2}{(1+v^2)(1+v-v^2)} = \frac{2v-1}{5(v^2-v-1)} -\frac{2v-2}{5(v^2+1)} .$$
The first integral is a U-substitution. For the second, split the numerator up, then U-substitution and arctan respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
| Let I = 3(x² + xy + y²)(y² + yz + z²)(z² + zx + x²)
Then, By Holder's Inequality, it follows that
I >= 81x²y²z²
Thus, we need to prove that
81x²y²z² >= (x + y + z)²(xy + yz + zx)²
Or 9xyz >= (x + y + z)(xy + yz + zx)
Expanding R.H.S. gives
9xyz >= xy(x + y) + yz(y + z) + zx(z + x) + 3xyz
Or xy(x + y) + yz(y + z) + zx(z + x) >= 6xyz
Which is evidently true by AM>=GM. QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Algebraic proof by contradiction I have the following problem:
For the following predicate from domain of all positive real numbers
$$ \frac{}{x+1} < \frac{+1}{x+2}$$
a. Write down the predicate logic statement for the theory above
b. Negate your predicate logic statement in (a)
c. Show that negated predicate statement in (b) cannot be correct and
therefore predicate in part (a) must be correct (prove by contradiction)
My solution is as follows:
a)
$ \Bbb R _>0$ = {x=$\Bbb R $ | $ x > 0 $}
P(y) = $ \frac{y}{y+1} < \frac{y+1}{y+2}$
$\forall x \in \Bbb R_>0 (P(x)) $
b)
$\exists x \in \Bbb R_>0 (\lnot P(x)) $
c)
This is where I'm stuck. How should I go about doing a proof by contradiction here?
| Assume that $x$ is positive and:
$$
\frac{x}{x+1} \geq \frac{x + 1}{x + 2}
$$
This leads to:
\begin{align}
\frac{x}{x+1} \geq&\ \frac{x + 1}{x + 2}\\
x(x + 2) \geq&\ (x + 1)^2 \\
x^2 + 2x \geq&\ x^2 + 2x + 1 \\
0 \geq 1
\end{align}
This is universally false meaning that the original statement, $\frac{x}{x+1} < \frac{x + 1}{x + 2}$ is universally true...but for all real numbers--not just positives.
p.s. My above reasoning is wrong...
You need to do the following:
\begin{align}
\frac{x}{x+1} \geq&\ \frac{x + 1}{x + 2}\\
\frac{x}{x +1} - \frac{x + 1}{x + 2} \geq&\ 0 \\
\frac{x^2 + 2x - x^2 - 2x - 1}{(x + 1)(x + 2)} \geq&\ 0 \\
\frac{-1}{(x + 1)(x + 2)} \geq&\ 0
\end{align}
You then need to make a sign chart:
$$
\frac{-1}{(x + 1)(x + 2)} \begin{cases}
< 0 & -\infty < x < -2 \\
> 0 & -2 < x < -1 \\
< 0 & -1 < x < \infty
\end{cases}
$$
Meaning that assuming that $\frac{x}{x+1} \geq \frac{x + 1}{x + 2}$ leads to the conclusion that $-2 < x < -1$ which contradicts the original assumption that $x$ is a positive real...although I don't see how this result is amenable to proof by contradiction since, at least the way I did it, required directly proving the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$x^4 -10x^2 +1 $ is irreducible over $\mathbb Q$ I have seen the thread Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ but this didn't really have a full solution.
Is it true that if it is reducible then it can be factored into a linear factor or quadratic factor in the form $x^2 - a$. Is $a$ in the rationals? And what exactly is a linear factor.
What I did was the following:
Let $y=x^2$. Then $y^2 - 10y +1=0 \iff y^2 -10y = -1$, add $25$ to both sides:
$$y^2 - 10y + 25 = -1 + 25 \iff (y - 5)^2 = 24 \iff y = 5 \pm 2\sqrt6$$
So $x^2=5 \pm 2 \sqrt6= (\sqrt2 \pm \sqrt3 )^2$
So $x^4 -10x^2 +1 =(x^2 -(\sqrt2 + \sqrt3 )^2)(x^2 - (\sqrt2 - \sqrt3)^2)$
Then difference of two squares:
$ (x-(\sqrt2 + \sqrt3))(x+(\sqrt2 + \sqrt3))(x-(\sqrt2 - \sqrt3))(x+(\sqrt2 - \sqrt3))=0$
Expanding this out gives $x^4 -10x^2 +1 =0$ so we have found all the roots and we can confirm that they are the roots right? None of these roots are rational so it must be irreducible?
Please can you advise me on this method that I am using and not direct me to a theorem or any type of shortcut.
Thanks in advance.
| First note that it suffices to show the polynomial is irreducible over $\mathbb{Z}$. We then note that $p(x)=x^4-10x^2-1$ splits either as
$$(x-a)(x^3+bx^2+cx+d)$$
or as
$$(x^2+bx+c)(x^2+dx+e)$$
In the first case $a$ would be a root of $p$, but by the rational root theorem the polynomial has no roots in $\mathbb{Q}$ (and hence not in $\mathbb{Z}$). In the second case
$$(x^2+bx+c)(x^2+dx+e)=x^4+(b+d)x^3+(e+c+bd)x^2+(be+dc)x+ec$$
so $e=c=\pm 1$, $b=-d$, filling this in we find
$$x^4+(b+d)x^3+(e+c+bd)x^2+(be+dc)x+ec=x^4+(2-b^2)+1$$
and $2-b^2=-10$ has no solution in $\mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| Since the OP does not know L'Hospital's rule (yet), this approach is probably not the best way. When you multiply top and bottom by $x^2\sqrt{x+2}+8$ you get the fraction: $\frac{x^5+2x^4-64}{(4-x^2)(x^2\sqrt{x+2}+8)}$ Both NUM and DENOM have a factor $x-2$ In case of the NUM, you could perform long division to arrive at $x^4+4x^3+8x^2+16x+32$ and in case of the DENOM, it is of course straight forward. Can you finish it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Solving $\int {\frac{2 \sin(x) \cos(x)}{\sin^4(x) + \cos^4(x)}}\, dx$ See this link here. I am having trouble with this one as well. Any hints or help appreciated.
$$\int{\frac{2\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}}\,dx$$
| Hint
For the denominator
$$\sin^4(x)+\cos^4(x)=(\sin^2(x)+\cos^2(x))^2-2\sin^2(x)\cos^2(x)=1-\frac 12\sin^2(2x)=\frac 12(1+\cos^2(2x))$$
For the numerator $$2\sin(x)\cos(x)=\sin(2x)$$
So $$\int {\frac{2 \sin(x) \cos(x)}{\sin^4(x) + \cos^4(x)}}\, dx=\int \frac{2\sin(2x)}{1+\cos^2(2x)}\,dx$$
Changing variable $t=\cos(2x)$ looks quite promising.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\angle{BEC}=\angle{DAC}$. A convex quadrilateral $\quad{ABCD}$ has $AD=CD$ and$\angle{DAB}=\angle{ABC}<90$. The line through $D$ and the midpoint of $BC$ meets line $AB$ in $E$. Prove that $\angle{BEC}=\angle{DAC}$.
I have to approaches: Either we can prove that $EB.EA=EC^2$ or we can prove that line $AD$ is the tangent to the circumcircle of $\triangle{ACE}$ as shown, but I am not getting how to prove them. Pure geometrical method is preferred.
|
Let $F$ be the intersection of line $AD$ and line $BC$. Since $\angle BDA = \angle ABC$, $AF=FB$.
Let $O$ be the center of the circumcircle of $\triangle ABC$ and let $G$, $I$ be the projections of $O$ onto $AF$ and $CD$. Then
*
*$OG = OM$ because $O$ and $F$ lie on the perpendicular bisector of $AB$.
*$OG = OI$ because $O$ and $D$ lie on the perpendicular bisector of $AC$.
It follows that $OM=OI$ and that $CO$ is the angle bisector of $\angle DCB$.
Next, consider $\triangle ABF$ with $D$, $M$, and $E$ colinear. Menelaus' theorem implies that
$$\frac{EA}{EB}\cdot\frac{MB}{MF}\cdot\frac{DF}{DA}=1.$$
Similarly, $\triangle ABF$ with $H$, $C$, and $E$ colinear. Menelaus' theorem gives us
$$\frac{EA}{EB}\cdot\frac{CB}{CF}\cdot\frac{HF}{HA} = 1.$$
Thus we have
$$\frac{MB}{MF}\cdot\frac{DF}{DA} = \frac{CB}{CF}\cdot\frac{HF}{HA},$$
or
$$\begin{aligned}\frac{HF}{HA} &= \frac{\color{blue}{MB}}{MF}\cdot\frac{DF}{DA}\cdot\frac{CF}{\color{blue}{CB}}\\
&=\frac{DF}{DA}\cdot\frac{CF}{2MF}
\end{aligned}$$
Since $AD = CD$ and $2MF = DF+CD+CF$, we have
$$\frac{HF}{HA}=\frac{DF\cdot CF}{CD\cdot(DF+CD+CF)}.$$
So
$$\frac{HF}{HF+HA}=\frac{DF\cdot CF}{CD(DF+CD+CF)+DF\cdot CF},$$
or
$$\frac{HF}{HA} = \frac{DF\cdot CF}{(CD+DF)(CD+CF)}.$$
Since $HA = DA+DF = CD+DF$, we have
$$HF = \frac{DF\cdot CF}{CD+DF}.$$
So $CH$ is the angle bisector of $\angle CDF$. Thus $CO\perp CH$, equivalently $EC$ is tangent to the circumcircle of $\triangle ABC$ at $C$.
It follows that $\angle BCE = \angle BAC$, and so $\angle DAC=\angle BEC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove convergence of trignometric sum from $-\infty$ to $\infty$
Prove convergence of $$\pi^2\sum_{n=-\infty}^{\infty}\left ( \frac{1}{\cos^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}-\frac{1}{\sin^2\left (\pi\left (n-\frac{1}{2}\right )\tau\right )}\right )$$ and $$\pi^2\sum_{n=-\infty}^{\infty}\left ( \frac{1}{\cos^2\left (\pi n\tau \right )}-\frac{1}{\sin^2\left (\pi\left (n-\frac{1}{2}\right )\tau\right )}\right )$$
where $\tau$ is a constant.
The proof I have which I am trying to follow goes like this (Ahlfors text):
The series are strongly convergent for both $n\rightarrow +\infty$ and $n\rightarrow -\infty$ for $|\cos n\pi\tau|$ and $|\sin n\pi\tau|$ are comparable to $e^{|n|\pi\text{Im}(\tau)}$ and hence the convergence is uniform for $\text{Im}(\tau)\geq \delta>0$. We can now take the limits termwise and we find that the first sum converges to zero whilst the second conerges to $\pi^2$ (from the $n=0$ term).
I don't understand most of the proof; I can see how changing to $e^{|n|\pi\text{Im}(\tau)}$ satisfies convergence for $\text{Im}(\tau)>0$. However I don't see how that relates to $\cos$ and $\sin$ and why taking limits termiwise gives the result it does (I understand we are allowed to as we have uniform convergence).
| (1) How $e^{|n|\pi\text{Im}(\tau)}$ relates to $\cos$ and $\sin$ :
For simplicity we first argue about $$\sum_{n=1}^\infty \frac{1}{\cos^2 \left (\pi\left (n-\frac{1}{2}\right )\tau \right )}.$$
Since $$\left|e^{i\pi\left(n-\frac{1}{2}\right)\tau}\right|=
e^{-\pi\left(n-\frac{1}{2}\right)\sigma}
\quad\text{and}\quad
\left|e^{-i\pi\left(n-\frac{1}{2}\right)\tau}\right|=
e^{\pi\left(n-\frac{1}{2}\right)\sigma}
,$$ where $\sigma=\text{Im}(\tau)>0$, we have
\begin{align}
2\left|\cos \left (\pi\left (n-\frac{1}{2}\right )\tau \right )\right|&=\left|e^{i\pi\left(n-\frac{1}{2}\right)\tau}+ e^{-i\pi\left(n-\frac{1}{2}\right)\tau} \right|\\
&\ge e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}.
\end{align}
Furthermore we have
\begin{align}
2\left|\cos \left (\pi\left (n-\frac{1}{2}\right )\tau \right )\right|&\ge e^{\pi\left(n-\frac{1}{2}\right)\sigma} \left(1- e^{-\pi(2n-1)\sigma} \right)\\
&\ge \frac{1}{2} e^{\pi\left(n-\frac{1}{2}\right)\sigma}\ge \frac{1}{2} e^{\pi(n-1)\sigma}
\end{align}
for $n\ge n_0,$ where $n_0$ is an integer such that $e^{-(2n_0-1)\pi\sigma}\le \frac{1}{2}$.
Therefore $$
\sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|
\le \sum_{n=1}^{n_0} \frac{4}{ \left(e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}\right)^2 } +\sum_{n=n_0+1}^\infty 4^2\cdot \,e^{-2(n-1)\pi\sigma}<\infty,
$$
which ensures the absolute convergence of $\sum_{n=1}^\infty 1/\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )$.
(2) Taking limits termwise or Uniform convergence:
If $\sigma\ge \delta >0$, we can take $n_0$ (depending only on $\delta $) for arbitrary $\varepsilon >0$ so that $$
\sum_{n=n_0+1}^\infty 4^2\cdot \,e^{-2(n-1)\pi\sigma}\le \sum_{n=n_0+1}^\infty 4^2\cdot \,e^{-2(n-1)\pi\delta }<\varepsilon.
$$
Then we have $$
\sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|
\le \sum_{n=1}^{n_0} \frac{4}{ \left(e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}\right)^2 } +\varepsilon .
$$
Taking limits as $\sigma \to \infty$, we have
\begin{align}
\lim_{\sigma\to\infty} \sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|
&\le \lim_{\sigma\to\infty}\sum_{n=1}^{n_0} \frac{4}{ \left(e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}\right)^2 } +\varepsilon \\
&\le \varepsilon .
\end{align}
Since $\varepsilon $ is arbitrary, we see $$
\lim_{\sigma\to\infty} \sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|=0,$$
which implies $$
\lim_{\sigma\to\infty} \sum_{n=1}^\infty \frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}
=0.$$
Similarly (we omit details, but just similarly) we have $$
\lim_{\sigma\to\infty} \sum_{n=-\infty}^0 \frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}
=0$$
and $$
\lim_{\sigma\to\infty} \sum_{n=-\infty}^{\infty} \frac{1}{\sin ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}
=0,\quad
\lim_{\sigma\to\infty} \sum_{n=-\infty,\, n\ne 0}^\infty \frac{1}{\cos ^2\left (\pi n\tau \right )}
=0.$$
Thus we have $e_3-e_2\to 0,$ $e_1-e_2 \to \pi^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make sense:
Let $n = 1.$ Then $1 + a + a^2 + \ldots +1 < \frac {1 + a + a^2 + \ldots +1 + a}{2} = \frac{2 + 2a + a^2+ \ldots}{2} = 1 + a + \frac{a^2}{2} + \ldots$
Then I did this below, but it's unclear if the difference is positive:
$\frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1} - \frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} = \frac {n + na +na^2+ \ldots + na^{n - 1} + na^n - n - 1 - an - a- na^2 - a^2 - \ldots -na^{n - 1} - a^{n - 1}}{n(n + 1)} = \frac{n(1 + a + a^2+ \ldots + a^{n - 1} + a^n - 1 -a - a^2 -a^{n - 1}) - 1 - a - a^2 - \ldots - a^{n - 1}}{n(n+1)} $
$ = \frac{- 1 + a(n(a^{n - 1}) - 1 - a - \ldots - a^{n - 2})}{n(n+1)}$
What can I try now?
| You have a more general result:
If $\{a_n \}$ is strictly increasing then also the sequence of the avarages over the first $N$ terms is strictly increasing.
In fact we have
$$
S_N=\frac 1 N \sum_{k=1}^N a_k<\frac 1 N \sum_{k=1}^N a_N=a_N<a_{N+1}
$$
and therefore
$$
S_{N+1}=\frac {\sum_{k=1}^N a_k+ {a_{N+1}}} {N+1}=\frac{N S_N+a_{N+1}}{N+1}>
\frac{N S_N+S_N}{N+1}=S_N
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Calculating $\int_0^{\pi/2} \sqrt{\cot x} + \sqrt{\cos x} dx$ How should I solve the following integral:
$$\int_0^{\pi/2} (\sqrt{\cot x} + \sqrt{\cos x} )\,\mathrm dx$$
| I think can use
\begin{align}I&= \int_0^{\frac{\pi}{2}}\left(\sqrt{\cot (x)}+\sqrt{\cos (x)}\right)\,\mathrm dx\\&=\int _0^1\frac{1}{\cos(x)}\left(\sqrt{\cot (x)}+\sqrt{\cos (x)}\right)\, \mathrm d(\sin(x))\\&=\int _0^1\frac{1}{\sqrt{1-a^2}}\left(\sqrt{\frac{\sqrt{1-a^2}}{a}}+\sqrt{\sqrt{1-a^2}}\right)\,\mathrm da\\ &=\int _0^1 a^{-\frac{1}{2}}(1-a^2)^{-\frac{1}{4}}\,\mathrm da+\int _0^1 (1-a^2)^{-\frac{1}{4}}\,\mathrm da\\& =\frac{1}{2}\int _0^1 (a^2)^{-\frac{3}{4}}(1-a^2)^{-\frac{1}{4}}\,\mathrm d(a^2)+\frac{1}{2}\int _0^1 (a^2)^{-\frac{1}{2}}(1-a^2)^{-\frac{1}{4}}\,\mathrm d(a^2)\\ &=\frac{1}{2}B \left(\frac{1}{4}, \frac{3}{4}\right)+\frac{1}{2}B\left(\frac{1}{2},\frac{3}{4}\right)\end{align} with $B$ is beta function. Hope it can help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to evaluate this integral $\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$ How to evaluate this integral
$$\mathcal{I}=\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$$
Mathematica gave me the answer below
$$\mathcal{I}=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$
where $\mathbf{G}$ is Catalan's constant.
| Lemma 1::$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2\pi}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$
Use this well known
$$\int_{0}^{+\infty}\dfrac{\ln{(x^2+2\sin{a}\cdot x+1)}}{1+x^2}dx=\pi\ln{2\cos{\dfrac{a}{2}}}+a\ln{|\tan{\dfrac{a}{2}}|}+2\sum_{k=0}^{+\infty}\dfrac{\sin{(2k+1)a}}{(2k+1)^2}$$
this indentity proof is very easy consider $\ln{(x^2+2\sin{a}\cdot x+1)}$ Fourier expansions(possion fourier).
then you can take
$a=-\dfrac{\pi}{6}$
then we have
$$\pi\ln{2\cos{\dfrac{\pi}{12}}}=\dfrac{\pi}{2}\ln{(2+\sqrt{3})}$$
$$-\dfrac{\pi}{6}\ln{\tan{\dfrac{\pi}{12}}}=\dfrac{\pi}{6}\ln{(2+\sqrt{3})}$$
and
$$2\sum_{k=0}^{3N}\dfrac{\sin{(2k+1)-\pi/6}}{(2k+1)^2}=-\sum_{k=0}^{3N}\dfrac{(-1)^k}{(2k+1)^2}-3\sum_{k=0}^{N-1}\dfrac{(-1)^k}{(6k+3)^2}\to -G-\dfrac{G}{3}=-\dfrac{4}{3}G$$
so
$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2\pi}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$
By done!
Lemma 2:$$\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx=\dfrac{\pi}{4}\ln{2}+G$$
\begin{align*} \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{1}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx \\
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{0}^{1} \frac{\log (x^{-1} + 1)}{x^2 + 1} \, dx \quad (x \mapsto x^{-1}) \\
&= 2 \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx - \int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx\\
&=\dfrac{\pi}{4}\ln{2}+G
\end{align*}
so
$$\int_{0}^{+\infty}\dfrac{\ln{(1+x^3)}}{1+x^2}dx=\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx+\int_{0}^{+\infty}\dfrac{\ln{(x^2-x+1)}}{1+x^2}dx=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
} |
Can we prove $\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} \le \frac{1}{(n-1)e}$ and arrive at an upper bound for $k$? Does this inequality hold true?
Given $\frac{n}{2} \ge k$,
$$\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} \le \frac{1}{(n-1)e}$$
where $e = 2.71828\dots$
I have reduced this to
$$\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \sum_{i=1}^{k}\frac{(n-i)*(n-i-1)}{n*(n-1)}$$
$$ \le \sum_{i=1}^{n}\frac{(n-i)^{2}}{n*(n-1)} $$
$$ \le \frac{n}{n-1}\sum_{i=1}^{k}(1-\frac{i}{n})^2 $$
Using $1 + x \le e^x $, we get
$$ \le \frac{n}{n-1}\sum_{i=1}^{k}e^{\frac{-2*i}{n}} $$
Then I can apply sum of a GP and arrive at
$$ LHS = \frac{n}{n-1} e^{\frac{-2}{n}}(\frac{1 - e^{\frac{2k}{n}}}{1 - e^{\frac{-2}{n}}}) $$
But now I am unable to relate this with the RHS and get stuck here.
Basically once this inequality is proven, I want to manipulate the terms and arrive at an upper bound for the value of k.
| Without restrictions on $n$ (and/or $k$), this will not be true. In particular, for $k=n-1$ you get
$$
\sum_{i=1}^{n-1} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \frac{n-2}{3}.
$$
and for $k=\frac{n}{2}$ something that behaves asymptotically as
$$
\sum_{i=1}^{n-1} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \frac{7}{6} n + o(n).
$$
You have
$$
\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \sum_{i=1}^{k}\frac{(n-i)(n-i-1)}{n(n-1)}
= \frac{1}{n(n-1)}\sum_{i=1}^{k}(n-i)(n-i-1)
$$
Now,
$$\begin{align}
\sum_{i=1}^{k}(n-i)(n-i-1) &= \sum_{i=1}^{k} \left(i^2 -i(2n-1) + (n-1)(n)\right) \\
&= \sum_{i=1}^{k} i^2 - (2n-1) \sum_{i=1}^{k}i + k(n-1)(n) \\
&= \frac{k(k+1)(2k+1)}{6} - (2n-1) \frac{k(k+1)}{2} + k(n-1)(n) \\
\end{align}$$
so
$$\begin{align}
\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}}
&= \frac{k(k+1)(2k+1)}{6n(n-1)} - \frac{2n-1}{n(n-1)}\frac{k(k+1)}{2} + k \\
&= \frac{1}{6n(n-1)}\left(k(k+1)(2k+1) - 3(2n-1)k(k+1) + 6n(n-1)k\right)
\end{align}$$
As mentioned above, for $k=n-1$ this simplifies to
$$
\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \frac{n-2}{3}.
$$
and for $k=\frac{n}{2}$ we get
$$
\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \frac{7n^2-18n+8}{6(n-1)} \operatorname*{\sim}_{n\to\infty} \frac{7n}{6} .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$ using generating function I'm preparing to an exam and trying to solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$, where $a_0=0$ and $a_1=1$.
This is my approach:
Let $A(z)=\sum_{n \geq 0} a_{n+2} z^{n+2}$, then:
$$\sum a_{n+2} z^{n+2} = 2 \sum a_{n+1} z^{n+2} - \sum a_n z^{n+2} + \sum (2z)^{n+2}$$
$$A(z)-a_0-a_1 z = 2z(A(z)-a_0)-z^2 A(z) + z^2 \frac{1}{1-2z}$$
$$A(z)(1-2z+z^2)=z+\frac{z^2}{1-2z}$$
$$A(z)=\frac{z(1-z)}{(1-2z)(1-2z+z^2)}=\frac{z}{(1-2z)(1-z)}=\frac{1}{1-2z}-\frac{1}{1-z}=\sum (\underbrace{2^n - 1}_{=a_n}) z^n$$
According to Wolfram, the result is different. Can you tell me where did I make the mistake?
| While the generating function approach works systematically, you can solve this example by hand directly without too much trouble. First I'll define $a_{-1}=1$ (so that the recurrence holds down to $n=1$) and define $b_n=a_n-a_{n-1}$. Then your recurrence becomes $b_n=b_{n-1}+2^n$ for all positive $n$, and from the first few terms ( e.g. -1,1,5,13,...) it's easy to see that the solution is given by $b_n=2^{n+1}-3$. From this we obtain $a_{n\geq 0}$:
\begin{align}
a_n
&=b_n+a_{n-1}\\
&=b_n+b_{n-1}+a_{n-2}\\
&\cdots \\&=b_n+b_{n-1}+\cdots+b_1+a_0\\
&=(2^{n+1}-3)+(2^{n}-3)+\cdots +(2^{2}-3)+0\\
&=\fbox{$4(2^{n}-1)-3n$}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
How to find the limit of this hard sequence? How to find this limits
$$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{n^2-(i-1)^2}{n\left\{[2n(a+b)]^2+[a(2i-1)]^2\right\}}$$
Where $a$ and $b$ are real positive constants.
My friend asked this question,but I have no idea how to do it.
| I'm sorry I post this as an answer when it is only a lower bound, but I do not have enough reputation to comment.
First note that, after the swap $i \to n-i+1$ the sum is
$$
\frac{1}{n} \sum_{i=1}^n \frac{i(2n-i)}{4n^2(a+b)^2+a^2(2n+1-2i)^2}
$$
By AM-GM, the $n$-th term is greater than or equal to
\begin{align*}
\left (\prod_{i=1}^n \frac{i(2n-i)}{4n^2(a+b)^2+a^2(2n+1-2i)^2} \right )^{1/n} &= \left [\frac{(2n)!}{2} \right ]^{1/n} \left (\prod_{i=1}^n \frac{1}{4n^2(a+b)^2+a^2(2n+1-2i)^2} \right )^{1/n} \\ &\geq \left [\frac{(2n)!}{2} \right ]^{1/n} \left (\prod_{i=1}^n \frac{1}{4n^2[(a+b)^2+a^2]} \right )^{1/n} \\ &= \left [\frac{(2n)!}{2} \right ]^{1/n} \frac{1}{4n^2[(a+b)^2+a^2]}
\end{align*}
which tends to
$$
\frac{1}{e^2[(a+b)^2+a^2]}
$$
unless I was wrong using Stirling.
So, taking in account Daniel Fischer's useful comment, one should have that your limit (if it exists) is between
$$
\frac{1}{e^2[(a+b)^2+a^2]} \text{ and } \frac{1}{4(a+b)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
| $$\begin{align} & \sin^7x \\ &
=\sin^6x\cdot \sin x \\ &
=\frac{1}{16}\cdot 16\sin^6x\cdot \sin x \\ &
=\frac{1}{16}\cdot (4\sin^3x)^2 \cdot \sin x \\ &
=\frac{1}{16}\cdot (3\sin x-\sin 3x)^2 \cdot \sin x \\ &
=\frac{1}{16}\cdot (9\sin^2 x-6\sin x\sin 3x+\sin^2 3x) \cdot \sin x \\ &
=\frac{1}{16}\cdot [\frac{9}{2}(1-\cos 2x)-3(\cos 2x-\cos 4x)+\frac{1}{2}(1-\cos 6x)] \cdot \sin x \\ &
=\frac{9}{32}(\sin x-\sin x\cdot \cos 2x)-\frac{3}{16}(\sin x\cdot \cos 2x-\sin x\cdot \cos 4x)+\frac{1}{32}(\sin x-\sin x\cdot \cos 6x) \\ &
=\frac{9}{32}[\sin x-\frac{1}{2}(\sin 3x- \sin x)]-\frac{3}{32}[(\sin 3x- \sin x)-(\sin 5x - \sin 3x)]+\frac{1}{32}[\sin x-\frac{1}{2}(\sin 7x-\sin 5x)] \end{align}$$
The coefficient of $\sin x$ in the above expression is $$\color{red}{\frac{9}{32}+\frac{9}{64}+\frac{3}{32}+\frac{1}{32}}=\color{blue}{\frac{35}{64}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Solving system of three quadratic equations $$\begin{cases}
x^2 = yz + 1 \\
y^2 = xz + 2 \\
z^2 = xy + 4
\end{cases}
$$
How to solve above system of equations in real numbers? I have multiplied all the equations by 2 and added them, then got $(x - y)^2 + (y - z)^2 + (x - z)^2 = 14$, but it leads to nowhere.
| Multiplying both sides of the first equation by $z$, of the second by $x$, and of the third by $y$, we get
$$x^2z=z^2y+z, \quad y^2x=x^2z+2x,\quad z^2y=y^2x+4y.$$
Adding up and cancelling, we get
$$2x+4y+z=0.$$
Similarly,
$$x^2y=y^2z+y, \quad y^2z=z^2x+2z, \quad z^2x=x^2y+4x,$$
giving
$$4x+y+2z=0.$$
To finish, use the linear equations to express $y$ and $z$ in terms of $x$, and substitute in $x^2=yz+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Comparing Coefficients for Partial Fractions See this simple example :
$$\frac{x+1}{(x-1)(x-2)}\equiv \frac{A}{(x-1)}+\frac{B}{(x-2)}$$
Then we can get
$x+1 \equiv A(x-2)+B(x-1)$ for $x \neq 1 ,2$
My Question :
Is it correct to put $x=1$ to find $A$ ,
Is it correct to put $x=2$ to find $B$, other than comparing coefficients ?
Both methods give the same answer. But we get $x+1 \equiv A(x-2)+B(x-1)$ for $x \neq 1 ,2$ . So is it okay if I use $x=1$ to find $A$ ?
| Yes, it is perfectly valid to do that. Notice that while the fractions are not defined at x= 1 or x= 2, once you have multiplied by (x- 1)(x- 2) those fractions are no longer a problem. If you set x= 1, 1+ 1= 2= A(1- 2)+ B(0) or 2= -A so A= -2. If you set x= 2, you get 2+ 1= 3= A(0)+ B(2- 1) or 3= B.
Another thing you can do is actually get "common denominators" on the right and add the fractions: $\frac{A}{x- 1}+ \frac{B}{x- 2}= \frac{A(x- 2)}{(x- 1)(x- 2)}+ \frac{B(x-1)}{(x- 1)(x- 2)}= \frac{(A+ B)x+ (-2A- B)}{(x- 1)(x- 2)}= \frac{x+ 1}{(x-1)(x- 2)}$. Since those fractions have the same denominator, they must have the same numerator: (A+ B)x+ (-2A- B)= x+ 1 for all x. Since they are equal for all x, the corresponding coefficients must be equal. That means we must have A+ B= 1 and -2A- B= 1. Adding the two equations, -A= 2 so A= -2 as before. And then, -2+ B= 1 so B= 3 as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Maximizing $\sqrt6xy+4yz$ Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$.
Let $A$ be maximum value of $\sqrt6xy+4yz$. Find $2A^2-4$.
An initial approach was trying out the inequality $\text{RMS-AM-GM}$. Then I tried a parametric substitution: $x=\cos\alpha, y=\sin\alpha cos\beta, z=\sin\beta \sin\alpha$.
How would you go about this?
| HINT
$$1=x^2+y^2+z^2=x^2+ay^2+(1-a)y^2+z^2 \ge 2\sqrt{a}xy+2\sqrt{(1-a)}yz \quad (\because \text{AM-GM})$$
$$a=\frac{3}{11}$$
Note that $2\sqrt{a}:2\sqrt{(1-a)}=\sqrt{6}:4$.
Let $2\sqrt{a}= \sqrt{6}k$, and multiply $k$ on each side.
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed?
| Suppose $n$ is a multiple of 6. Then we are done. Suppose $n$ is neither a multiple of 2 or 3. Then $n+1$ must be even, and if $n+1$ is still not a multiple of 3 then $n+5$ is. Note $n+5\equiv n+2$ (mod 3) and since neither $n$ nor $n+1$ is a multiple of 3, then $n+2$ is. If $n$ is even but not a multiple of 3, we proceed the same way to show either $n+1$ or $n+2$ is. The only case left is when $n$ is an odd multiple of 3 then $n+1$ is even and we are done. In any case, we get a factor of 2 and a factor of 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
Linear Factorization of Complex Polynomials I am trying to find a linear factorization of the polynomial $$p(z) = 1 +z+z^2 +z^3 +z^4 +z^5 + z^6 +z^7 +z^8$$
I know what it means by linear factorization in the sense of non-complex polynomials, but i'm not sure where to begin for a complex polynomial of degree 8. I tried some trial and error by factoring out $(z-1)$ and $i$ but didn't seem to have much luck! Could someone help me get on the right track please!
| The polynomial factorizes, over the reals as ($z\neq 1$)
$$p(z) = 1 +z+z^2 +z^3 +z^4 +z^5 + z^6 +z^7 +z^8=\frac{1-z^9}{1-z}=(1+z+z^2)(1+z^3+z^6)$$
and its roots are the (complex) $9$-th roots of unity (except $1$ of course):
$$-(-1)^{\frac{1}{9}},(-1)^{\frac{2}{9}},-(-1)^{\frac{3}{9}}, (-1)^{\frac{4}{9}},-(-1)^{\frac{5}{9}}, (-1)^{\frac{6}{9}},
-(-1)^{\frac{7}{9}},(-1)^{\frac{8}{9}}$$
which should make the $8$ linear factors clear. So:
$$p(z)=\big(z+(-1)^{\frac{1}{9}}\big)\big(z-(-1)^{\frac{2}{9}}\big)\big(z+(-1)^{\frac{3}{9}}\big)\big(z-(-1)^{\frac{4}{9}}\big)\big(z+(-1)^{\frac{5}{9}}\big)\big(z-(-1)^{\frac{6}{9}}\big)\big(z+(-1)^{\frac{7}{9}}\big)\big(z-(-1)^{\frac{8}{9}}\big)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Multivariable partial differentiation
$f(r,\theta)=r^5\cos \theta$,$x=r\cos \theta$, $y=r\sin \theta$
Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial x}$ in therms of $r$ and $\theta$
I know that $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial r}\cdot \frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\cdot \frac{\partial \theta}{\partial x}$$
so I have $\frac{\partial f}{\partial r}=5r^4\cos \theta$
$x=r\cos \theta \Rightarrow r=\frac{x}{cos\theta}$
so $\frac{\partial r}{\partial x}=\frac{1}{\cos \theta}$
Or because $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}(\frac{y}{x})$
It is $r=\frac{x}{\cos \theta}=\frac{x}{\frac{x}{\sqrt{x^2+y^2}}}=\sqrt{x^2+y^2}$ so $\frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=cos\theta$
?
| \begin{align}
f &= r^5 \cos \theta = r^4 \cdot r \cos \theta = x \cdot \left ( x^2 + y^2\right)^2 \implies \\
f_x &= \left ( x^2 + y^2\right)^2 + x \cdot 2(x^2 + y^2) \cdot 2x = \left ( x^2 + y^2\right) \left( x^2 + y^2 + 4x^2\right) = \\
&= \left ( x^2 + y^2\right) \left ( 5x^2 + y^2\right) = r^4 \left ( 5 \cos^2 \theta + \sin^2 \theta\right) \\
f_y &= 4x y \left ( x^2 + y^2\right) = 4r^2 \cos \theta \sin \theta \cdot r^2 = 4r^4 \cos \theta \sin \theta
\end{align}
You can simplify even more, if you want, using trig identities.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Taylor series of $\ln{\sqrt[4]{\frac{x-2}{5-x}}}$ to $o((x-x_0)^n)$ when $x_0 = 3$ Well I have tried to get it as
$$f(x) = f(x_0) + \frac{f'(x_0)(x-x_0)}{1!} + \frac{f''(x_0)(x-x_0)^2}{2!} + ... + o((x-x_0)^n)$$
and got wrong results:
First: $$f'(x) = \frac{3}{4(x-2)(5-x)}$$
Second: $$f''(x) = \frac{3(2x-7)}{4(x-2)^2(5-x)^2} , $$
Third: $$f'''(x) = \frac{3}{4}*\frac{2(x-2)(5-x) - 2(5-x)(2x-7) + 2(x-2)(2x-7)}{(x-2)^3(5-x)^3}$$
Then I have tried changing it to $$\ln{\sqrt[4]{\frac{x-2}{5-x}}} = \frac{ln{(x-2)}}{4} - \frac{ln{(5-x)}}{4}$$ and got nothing useful. So I need an advice how to start. Or at least a hint.
| Answer is $$-\frac{ln2}{4} + \sum^n_{k=1}{\frac{(-1)^{k-1} + 2^{-k}(x-3)^k}{4k}} + o((x-3)^n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stuck. The general idea I know is to get each of the 3 terms above into the form $D\cdot \frac{1}{1-z}$, where $D$ is a constant. But for the first term, it would be $\frac{1}{4}\cdot\frac{1}{(0-z)}$. I can't get rid of the zero.
| Let $\zeta = z - 2$. We are trying to find the Laurent series for $1/(\zeta+2)\zeta^2$ centred at $\zeta = 0$, so
$$ f = \frac{1}{(\zeta+2)\zeta^2} = \zeta^{-2} \frac{1}{\zeta + 2}
$$
The latter is holomorphic in a neighbourhood of $0$, so let $(a_j)_{j=0}^\infty$ such that
$$ \frac{1}{\zeta + 2} = \sum_{j=0}^\infty a_j \zeta^j, 1 = 2\sum_{j=0}^\infty a_j \zeta^j + \sum_{j=0}^\infty a_j \zeta^{j+1}
$$
Therefore,
$$ 1 = 2a_0 + \sum_{j=1}^\infty 2a_j \zeta^j + \sum_{j=1}^\infty a_{j-1} \zeta^j = 2a_0 + \sum_{j=1}^\infty (2a_j + a_{j-1}) \zeta^j
$$
so $a_0 = \frac{1}{2}$, and we establish the recurrence relation $a_{j+1} = -\frac{1}{2}a_j$. Hence $a_j = (-1/2)^j$.
The original series becomes
$$ \zeta^{-2} \frac{1}{\zeta + 2} = \zeta^{-2} \sum_{j=0}^\infty (-\frac{1}{2})^j \zeta^j
$$
Then substitute $\zeta = z-2$.
To find the region of convergence, notice that $f$ is holomorphic except at the poles $z = 0, z = 2$. The result follows easily from Laurent's theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to find the following limit Problem: Find the following limit $${\displaystyle \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)\sin\frac{\pi}{n^{2}}+\left(1+\frac{2}{n}\right)\sin\frac{2\pi}{n^{2}}+\ldots+\left(1+\frac{n-1}{n}\right)\sin\frac{\left(n-1\right)}{n^{2}}\pi\right]}$$.
Attempt:
Observe that
\begin{align*}
& \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)\sin\frac{\pi}{n^{2}}+\left(1+\frac{2}{n}\right)\sin\frac{2\pi}{n^{2}}+\ldots+\left(1+\frac{n-1}{n}\right)\sin\frac{\left(n-1\right)}{n^{2}}\pi\right]\\
= & \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)\left(\frac{\pi}{n^{2}}+O\left(\frac{\pi^{3}}{n^{6}}\right)\right)+\ldots+\left(1+\frac{n-1}{n}\right)\left(\left(\frac{n-1}{n^{2}}\right)\pi+O\left(\frac{\left(n-1\right)^{3}\pi^{3}}{n^{6}}\right)\right)\right]\\
= & \lim_{n\rightarrow\infty}\left[\left(\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{k\pi}{n^{2}}\right)+\left(\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)O\left(\frac{k^{3}\pi^{3}}{n^{6}}\right)\right)\right]\\
= & \lim_{n\rightarrow\infty}\left[\left(\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{k\pi}{n^{2}}\right)+\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)O\left(\frac{1}{n^{3}}\right)\right]\tag{1}
\end{align*}
Note that
\begin{align*}
\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{1}{n^{3}} & =\lim_{n\rightarrow\infty}\left(\frac{1}{n^{2}}+\sum_{k=1}^{n}\frac{k}{n^{4}}\right)\leq\lim_{k\rightarrow\infty}\left(\frac{1}{n^{2}}+\frac{1}{n^{3}}\right)=0.
\end{align*}
Hence
\begin{align*}
\left(1\right) & =\lim_{n\rightarrow\infty}\left[\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{k\pi}{n^{2}}\right]=\pi\lim_{n\rightarrow\infty}\left[\sum_{k=1}^{n}\left(\frac{k}{n}+\left(\frac{k}{n}\right)^{2}\right)\frac{1}{n}\right]\\
& =\pi\int_{0}^{1}x+x^{2}dx=\frac{5\pi}{6}.
\end{align*}
Question I don't know if I did it correctly.
| Yes, you did this perfectly except for one (inconsequential) mistake. In "note that", you should have
$$ \begin{align*}
\sum_{k=1}^n \left( 1 + \dfrac{k}{n} \right) \dfrac{1}{n^3} & = \dfrac{1}{n^2} + \sum_{k=1}^n \dfrac{k}{n^4} \\
& < \dfrac{1}{n^2} + \mathbf{ \sum_{k=1}^n \dfrac{n}{n^4} } \\
& = \dfrac{1}{n^2} + \dfrac{1}{n^2}
\end{align*} $$
which still tends to 0.
If you'd like some empirical confidence boosting,
(source: pages.iu.edu)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Rationalize a surd $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ How can I rationalize the following surd
$$\frac{1}{1+\sqrt{2}-\sqrt{3}}$$
What would be the conjugate of the denominator
| Rationalise twice, because after rationalising once , there would still remain a surd in the denominator.
First multiply and divide by $1+\sqrt{2}+ \sqrt{3}$
$$\frac{1}{1+\sqrt{2}-\sqrt{3}}\times\frac{1+\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}$$
$$= \frac{1+\sqrt{2}+\sqrt{3}}{1+2\sqrt{2}+2-3}=\frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$$
Now multiply and divide by $\sqrt{2}$
$$=\frac{\sqrt{2}+2+\sqrt{6}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Real part of $(1+2i)^n$ Is it true that for all $n\in \mathbb{N}$, $n\ge 2$ we have
$$|\textrm{Re}((1+2i)^n)|>1?$$
I do know de Moivre's Theorem.
I do not know how to show that $|\sqrt{5}^n\cos(n\arccos\left ( \frac{1}{5} \right ))|>1$ because the value $\cos(n\arccos\left ( \frac{1}{5} \right ))$ can become (theoretically) arbitrarily small.
| Recursion for $\boldsymbol{r_n=\mathrm{Re}\!\left((1+2i)^n\right)}$
Both $1+2i$ and $1-2i$ satisfy the equation $z^2-2z+5=0$. Therefore, the real part of $(1+2i)^n$
$$
r_n=\frac{(1+2i)^n+(1-2i)^n}2
$$
satisfies
$$
r_n=2r_{n-1}-5r_{n-2}
$$
mod $\boldsymbol{4}$
Since $r_0=r_1=1$, we have that $r_n\equiv1\pmod{4}$ for all $n$.
Therefore, if $\left|r_n\right|\le1$, we must have $r_n=1$.
mod $\boldsymbol{25}$
Compute the first $24$ terms:
$$
\small r_n=1,1,\underbrace{22,14,18,16,17,4,23,\overset{\normalsize r_9}{1},12,19,3,11,7,9,8,21,2,24,13,6}_{\text{period of $20$}},22,14,\dots\pmod{25}
$$
This means that if $n\ge2$ and $r_n=1$, we must have $n\equiv9\pmod{20}$.
mod $\boldsymbol{11}$
Since $(1+2i)^3=-11-2i\equiv9i\pmod{11}$, Little Fermat says
$$
(1+2i)^{60}\equiv(9i)^{20}=81^{10}\equiv1\pmod{11}
$$
Therefore, $r_n\pmod{11}$ repeats every $60$ terms. However,
$$
\begin{align}
r_9&\equiv0\pmod{11}\\
r_{29}&\equiv2\pmod{11}\\
r_{49}&\equiv9\pmod{11}
\end{align}
$$
Thus, if $n\equiv9\pmod{20}$, $r_n\not\equiv1\pmod{11}$.
Conclusion
$\left|r_n\right|\gt1$ for $n\ge2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$ If the points $\big(\frac{a^3}{a-1}, \frac{a^2-3}{a-1}),(\frac{b^3}{b-1}, \frac{b^2-3}{b-1}) ,\big(\frac{c^3}{c-1}, \frac{c^2-3}{c-1}\big)$ are collinear for three distinct values of $a,b,c$ and $a,b,c\neq1$, then prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$.
What should be the smart approach to this question. If I try equating area of triangle formed by three points equal to $0$, then it gets too complicated. Please provide some insight.
| HINT.- I do not know if my answer could be “the smart approach to this question” as you want. Anyway here I give you the following.
Equalizing slopes
$$\frac{\frac{a^3-3}{a-1}-\frac{b^3-3}{b-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{\frac{3}{b-1}-\frac{3}{a-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{3}{ab(a+b)-(a^2+ab+b^2)}$$
So, by symmetry, $$a^2b+ab^2-(a^2+ab+b^2)=a^2c+ac^2-(a^2+ac+c^2)=b^2c+bc^2-(b^2+bc+c^2)$$
It follows $$\begin{cases}\color{green}{(a^2b+ab^2)}+(a^2+ac+c^2)=a^2c+ac^2+\color{green}{(a^2+ab+b^2)}\\\color{green}{(a^2b+ab^2)}+(b^2+bc+c^2)=b^2c+bc^2+\color{green}{(a^2+ab+b^2)}\end {cases}\qquad(*)$$
Hence, by subtracting the equalities in (*),
$$(a^2+ac+c^2)- (b^2+bc+c^2)= (a^2c+ac^2)-(bc^2+bc^2)\Rightarrow(c-1)(a+b+c)=0$$
Thus, because of $c\ne 1$, $$\color {red}{a+b+c=0}$$
Consequently you have to prove the easier relation
$$abc-(ab+ac+bc)=0$$
There are several ways to solve now. Try to finish the question. If you cannot (I believe you can), I will return.
| {
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Stuck on definite integral problem due to inappropriate $\log$ I have this definite integral problem which I have solved correctly but I'm stuck in one of the steps. I have manipulated it but I think it's not feasible to solve it that way.
$$\int_0^a(a^2 + x^2)^\frac{5}{2} dx$$
I have substituted $$x = a\cot\theta$$
and then the integral will look something like this $$\int_0^a (cosec\theta)^{5} dx$$
I don't know how do I proceed further to solve this problem. Kindly help me solving this further.
| When you want do integrate a function of $a^2+x^2$ you could let $x=$ any of $a\tan\theta, a\cot\theta, a\sinh\theta, \text{or} \,\,a\,\text{csch}\,\theta$. The substitution $x=a\sinh\theta$ has many advantages for this kind of problem. Then $a^2+x^2=a^2\cosh\theta$ and $dx=a\cosh\theta d\theta$ and so
$$\int_0^a\left(a^2+x^2\right)^{\frac52}dx=a^6\int_0^{\sinh^{-1}1}\cosh^6\theta d\theta$$
Since $\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right)$, $\sinh^{-1}1=\ln\left(1+\sqrt2\right)$ and since $\cosh(n+1)x=2\cosh nx\cosh x-\cosh(n-1)x$ we can run up Chebyshev polynomials pretty fast:
$$\begin{align}\cosh2x & =2\cosh^2x-1 \\ \cosh3x & =4\cosh^3x-3\cosh x \\ \cosh4x & =8\cosh^4x-8\cosh^2+1 \\ \cosh5x & =16\cosh^5x-20\cosh^3x+5\cosh x \\ \cosh6x & = 32\cosh^6x-48\cosh^4x+18\cosh^2-1\end{align}$$
We can solve this for $$\cosh^6x=\frac1{32}\left(\cosh6x+6\cosh4x+15\cosh2x+10\right)$$ Then
$$\begin{align}\int_0^a\left(a^2+x^2\right)^{\frac52}dx & =\frac{a^6}{32}\int_0^{\ln\left(1+\sqrt2\right)}\left(\cosh6\theta+6\cosh4\theta+15\cosh2\theta+10\right)d\theta \\ & =\frac{a^6}{32}\left.\left(\frac16\sinh6\theta+\frac32\sinh4\theta+\frac{15}2\sinh2\theta+10\right)\right|_0^{\ln\left(1+\sqrt2\right)}\end{align}$$
$\sinh0=0$ and $$\sinh6\ln\left(1+\sqrt2\right)=\frac12\left[\left(\sqrt2+1\right)^6-\left(\sqrt2-1\right)^6\right]=70\sqrt2$$
$$\sinh4\ln\left(1+\sqrt2\right)=\frac12\left[\left(\sqrt2+1\right)^4-\left(\sqrt2-1\right)^4\right]=12\sqrt2$$
$$\sinh2\ln\left(1+\sqrt2\right)=\frac12\left[\left(\sqrt2+1\right)^2-\left(\sqrt2-1\right)^2\right]=2\sqrt2$$ So
$$\int_0^a\left(a^2+x^2\right)^{\frac52}dx=\frac{a^6}{32}\left(\frac{70}6\sqrt2+18\sqrt2+15\sqrt2+10\ln\left(1+\sqrt2\right)\right)=\frac{a^6}{48}\left[67\sqrt2+15\ln\left(1+\sqrt2\right)\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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There are 3 red balls and 7 blue balls in a bag. We take 4 balls randomly from the bag without replacement. What is the probability of getting at least $2$ blue balls?
When it asks for specific quantities of balls, i.e. $2$ red balls and $2$ blue balls, I understand to do
$\dfrac{\binom{3}{2}\cdot\binom{7}{2}}{\binom{10}{4}}$
But how do I solve this when it says at least? I have tried computing $\binom{7}{2} + \binom{7}{3} + \binom{7}{4} \,etc. / \binom{10}{4}$, but this is incorrect.
The correct answer is $\frac{29}{30}$. Could someone please provide the steps on how to get to this solution?
| Let $X$ be the random variable denoting how many blue balls are selected.
The question asks us to find $Pr(X\geq 2)$
Breaking apart via cases:
$Pr(X\geq 2)=Pr(X=2)+Pr(X=3)+Pr(X=4)$
In other words, to have at least two, this is the same as asking having exactly two or exactly three or exactly four.
You say you know how to calculate each of these specific cases.
It continues then as:
$$Pr(X\geq 2)=\frac{\binom{3}{2}\binom{7}{2}+\binom{3}{1}\binom{7}{3}+\binom{3}{0}\binom{7}{4}}{\binom{10}{4}}$$
Each term of which comes from multiplication principle: Pick the red balls followed by pick the blue balls.
Alternatively, one can see that $Pr(X\geq 2) = 1-Pr(X<2) = 1-Pr(X=1)$ in this specific case since $Pr(X=0)=0$ (there are only three red balls to pick, we can't have picked four red)
We have then $1-\frac{\binom{3}{3}\binom{7}{1}}{\binom{10}{4}} = 1-\frac{1}{30}=\frac{29}{30}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality involving circumradii Let $ABC$ be a triangle and $M$ a point on the side $BC$. Let $R_1$,$R_2$, and $R$ be the circumradii of the triangles $ABM, ACM$, and $ABC$. Show that $\max\{R_1,R_2\} \geq R\cos\frac A 2$.
| In $\Delta ABC$ we have:
$$\frac{a}{\sin A}=2R$$
$\Rightarrow 2R\cdot\sin A=a\Rightarrow2R(2\sin\frac{A}{2}\cdot\cos\frac{A}{2})=a
\Rightarrow R\cdot\cos\frac{A}{2}=\dfrac{1}{4}\cdot\dfrac{a}{\sin\frac{A}{2}}$
Now assume $AM$ divides $\angle A$ into $\angle A_1$ and $\angle A_2$, also consider
$|\alpha|\lt\dfrac{A}{2}\lt 90^\circ$
$\max\{R_1,R_2\}\ge\dfrac{1}{2}(R_1+R_2)=\dfrac{1}{4}(\dfrac{BM}{\sin A_1}+\dfrac{CM}{\sin A_2})\ge\dfrac12(\dfrac{BM+CM}{\sin A_1+\sin A_2})$
$=\dfrac12\cdot\dfrac{a}{\sin(\frac{A}{2}-\alpha)+\sin(\frac{A}{2}+\alpha)}
\ge \dfrac12\cdot\dfrac{a}{2\sin\frac{A}{2}\cos\alpha}\ge\dfrac14\cdot\dfrac{a}{\sin\frac{A}{2}}=R\cdot\cos\frac{A}{2}$
P.S: Proof of the used inequality: Suppose $a=\sin A_1$, $b=BM$, $c=\sin A_2$, $d=CM$ $(b,d,b+d\neq0)$
By using Chebyshev's sum inequality (since $\sin A_1\ge\sin A_2 \equiv BM\ge CM$) on the third step we have:
$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}\ge\dfrac{(a+c)(b+d)}{2bd}
=\dfrac{(a+c)(b+d)^2}{(b+d)\cdot2bd}=\dfrac{(a+c)(b^2+d^2+2bd)}{(b+d)\cdot2bd}
\ge\dfrac{(a+c)\cdot4bd}{(b+d)\cdot2bd}=2\cdot\dfrac{a+c}{b+d}$
| {
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I'm a new to modular maths and I need to find 'a' given $(x+y)^p \equiv a^b+c^d\pmod p$ I'm new to modular maths and I've been asked to do the following:
given p is prime and $(x+y)^p \equiv a^b+c^d\pmod p$ , find a=?,b=?,c=?,d=?
Can anyone help me with the same?
Or atleast point me in the right direction?
Or give me a solution for the same?
Thanks.
[UPDATE 1]
Using congruence property and then fermat's little theorem,
i was able to get to:
$(x+y)^p \pmod p \equiv a^b+c^d$
$(x+y) \pmod p \equiv a^b+c^d$
which means that :
$(x+y) = a^b+c^d$ (am I on the right path?)
[UPDATE 2]
I've got to:
$(x+y) \pmod p \equiv a^b+c^d$, which can be written as
$x \pmod p \equiv a^b$
$y \pmod p \equiv c^d$
| Binomial coefficients
You may know these formulas:
\begin{align*}
(a+b)^2 &= a^2 + 2ab + b^2 \\
(a+b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 \\
(a+b)^4 &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \\
\text{etc.}
\end{align*}
Each term in the expansion of $(a+b)^n$ clearly has the form $c \cdot a^k b^{n-k}$, where the coefficient $c$ arises from the number of different ways you can choose $k$ copies of $a$ and $(n-k)$ copies of $b$ in the product:
$$\underbrace{(\color{brown}a+\color{teal}b)(\color{brown}a+\color{teal}b)\dots(\color{brown}a+\color{teal}b)}_{\text{$n$ times}}.$$
For example, for $n=4$ and $k=2$, there are six choices:
$$\color{brown}{aa}\color{teal}{bb}+\color{brown}a\color{teal}b\color{brown}a\color{teal}b + \color{brown}a\color{teal}{bb}\color{brown}a+\color{teal}b\color{brown}{aa}\color{teal}b + \color{teal}b\color{brown}a\color{teal}b\color{brown}a + \color{teal}{bb}\color{brown}{aa} = 6\color{brown}a^2\color{teal}b^2.$$
We call $6$ “the binomial coefficient $\binom 42$”.
Computing $\binom nk$
We can count the number of anagrams of $a^k b^{n-k}$, by saying:
*
*We count all $n!$ permutations of the string $a^k b^{n-k}.$
*But now we will have counted each possible string $k! (n-k)!$ times: once for each possible choice of both
*
*a permutation of the $k$ occurences of $a$, and
*a permutation of the $(n-k)$ occurences of $b$.
(That is, shuffling the $a$s among themselves, and shuffling the $b$s among themselves, does not actually give us distinct strings.)
If we count $n!$ strings when counting each string $k! (n-k)!$ times, then there must be $$\binom nk=\frac{n!}{k!(n-k)!} \textbf{ distinct}\textrm{ strings}.$$
The binomial formula
We can now write, in general:
$$(a+b)^n = \sum_{k=0}^n \binom nk a^k b^{n-k}.$$
For which $k$ is $\binom pk$ a multiple of $p$?
The fraction looks like this:
$$\binom pk = \frac{p!}{k!(p-k)!} = \frac{1 \cdot 2 \cdot (\ldots) \cdot k \cdot (k+1) \cdot (\ldots) \cdot p}{1 \cdot 2 \cdot (\ldots) \cdot k \cdot 1 \cdot (\ldots) \cdot (p-k)}.$$
We know that $0 \leq k \leq p$. (That is the range we take the sum over.)
We also know that $\binom pk$ must be an integer.
There is a term $p$ in the numerator: if we can’t cancel it out somehow, the fraction will be a multiple of $p$.
But since $p$ is a prime, dividing by any number in $\{1, \dots, p-1\}$ can’t harm the $p$ on top. Thus, we need a $p$ in the denominator to cancel out the $p$.
The terms in the denominator range from $1$ to $k$; and then, from $1$ to $p-k$. Thus, we can cancel out the $p$ precisely when $k=p$ or $p-k=p$ (i.e. $k=0$). In all other cases ($1 \leq k \leq p-1$), the binomial coefficient will be a multiple of $p$!
That means, working modulo $p$, all of the terms $1 \leq k \leq p-1$ will vanish, and we get:
$$(a+b)^p = \sum_{k=0}^p \binom nk a^k b^{p-k} \equiv a^0 b^{p-0} + a^p b^{p-p} = a^p + b^p \pmod p.$$
| {
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How to prove $P(x)=1-\frac{x^2}{2}$ is a good approximation of order $3$ for $f(x)=\cos x$ near $x=0$? Let $f$ be a function and we want to approximate $f$ using a different function $P$ near $x=0$. The error of approximation is $E(x)=f(x)-P(x)$. If the approximation is going to be any good, we want $\lim\limits_{x\rightarrow 0}E(x)=0$. In fact, we want $E(x)$ approaches $0$ fast as $x\rightarrow 0$
Definition for $good \ approximation$: we say that $P$ is a good approximation of order $n$ for $f$ near $x=0$ when $E(x)$ approaches $0$ faster than $h_n(x)=x^n$.
Definition for $approache\ 0\ faster$: Assume $\lim\limits_{x\rightarrow 0}f(x)=\lim\limits_{x\rightarrow 0}g(x)=0$. We say $\lim\limits_{x\rightarrow 0}f(x)$ approaches $0$ faster than $g(x)$, as $x\rightarrow 0$, when $\lim\limits_{x\rightarrow 0}\frac{f(x)}{g(x)}=0$.
How to prove $P(x)=1-\frac{x^2}{2}$ is a good approximation of order $3$ for $f(x)=\cos x$ near $x=0$?
Here is my try:
We have $E(x)=f(x)-P(x)=\cos x -1+\frac{x^2}{2}$. Since $E(x)$ and $x^3$ are continuous and $E(0)=0^3=0$, then $\frac{E(x)}{x^3}=\frac{\cos x -1+\frac{x^2}{2}}{x^3}=\frac{\frac{\cos x}{x^2}-\frac{1}{x^2}+\frac{1}{2}}{x}$. Clearly $\lim\limits_{x\rightarrow 0}\frac{E(x)}{x^3}=0$. I think we can conclude that $P$ is a good approximation as $E(x)$ approaches $0$ faster than $x^3$.
However I am sure the steps shouldn't be this simple. Could someone help?
| Your mostly done already. The only jump in logic is the statement that
"Clearly $\lim\limits_{x\rightarrow 0}\frac{E(x)}{x^3}=0$."
This is not clear based on the previous line as the prior line if you were to plug in 0 would get you 0 over 0. You can deal with this by using L'Hospital's rule.
$\lim\limits_{x\rightarrow 0} \frac{E(x)}{x^3} = \lim\limits_{x\rightarrow 0} \dfrac{\frac{\cos x}{x^2} - \frac{1}{x^2} + \frac{1}{2}}{x}$
$= \lim\limits_{x\rightarrow 0} -\frac{x^2\sin x + 2x\cos x}{x^4} +\frac{2}{x^3} = \lim\limits_{x\rightarrow 0} \frac{2}{x^3} - \frac{x\sin x + 2\cos x}{x^3} =
\lim\limits_{x\rightarrow 0} \frac{2-x\sin x -2\cos x}{x^3}$
Applying L'Hospital's rule a few more times and,
$\lim\limits_{x\rightarrow 0} -\frac{x\cos x + \sin x - 2\sin x}{3x^2} = \lim\limits_{x\rightarrow 0} -\frac{x\cos x - \sin x}{3x^2}
= \lim\limits_{x\rightarrow 0} -\frac{\cos x - x\sin x - \cos x}{6x}
=\lim\limits_{x\rightarrow 0} -\frac{\sin x}{6}
= 0$
which completes the problem. In practice, you would normally just apply taylor's theorem assuming you know your function has enough derivatives (and $\cos x$ has as many derivatives as you can want).
| {
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"source": "stackexchange",
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Divisibility of $n^4 -n^2$ by 4 (induction proof) We have to show that $$ n^4 -n^2 $$ is divisible by 3 and 4 by mathematical induction
Proving the first case is easy however I do not know how what to do in the inductive step. Thank you.
| Doing it by the book, though you were already told there are simpler/faster ways:
For $\;n=1:\;\;1^4-1^2=0\;\color{green}\checkmark\;$
Suppose it is true for $\;n\;$ and prove for $\;n+1\;$
$$(n+1)^4-(n+1)^2=(n+1)^2\left((n+1)^2-1\right)=(n+1)^2(n+1-1)(n+1+1)=$$
$$(n^2+2n+1)(n^2+2n)=n^4+4n^3+5n^2+2n=n^4-n^2+2n(2n^2+3n+1)=$$
$$=n^4-n^2+2n(n+1)(2n+1)$$
Now observe that the factor $\;2n(n+1)(2n+1)\;$ is always divisible by $\;4\;$ (as either $\;n\;$ or $\;n+1\;$ is even), and also by $\;3\;$ (since either $\;n\;$ is, or $\;n=1\pmod 3\;$ and then $\;2n+1=0\pmod 3\;$ , or $\;n=2\pmod3\;$ and then $\;n+1=0\pmod3\;$)
| {
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The graph of the equation $x+y=x^3+y^3$ is the union of The graph of the equation $x+y=x^3+y^3$ is the union of
$(A)$line and an ellipse$(B)$line and a parabola$(C)$line and hyperbola$(D)$line and a point
I tried to factorize the given equation.
$x^3-x+y^3-y=0$
$x(x^2-1)+y(y^2-1)=0$
The answer given is a line and an ellipse but i do not understand how this equation is split into a line and an ellipse equation.
| Note that
$$x+y=x^3+y^3=(x+y)(x^2-xy+y^2)$$
Setting $$x=\cos\left(\frac{\pi}{4}\right)X+\sin\left(\frac{\pi}{4}\right)Y=\frac{X+Y}{\sqrt 2}$$$$y=-\sin\left(\frac{\pi}{4}\right)X+\cos\left(\frac{\pi}{4}\right)Y=\frac{-X+Y}{\sqrt 2}$$
gives
$$x^2-xy+y^2=1\Rightarrow 3X^2+Y^2=2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Induction for divisibility: $3\mid 12^n -7^n -4^n -1$ I must use mathematical induction to show that
$a_{n} = 12^n −7^n −4^n −1$ is divisible by 3 for all positive integers n.
Assume true for $n=k$
$a_{k} = 12^k -7^k -4^k -1$
Prove true for $n=k+1$
$a_{k} = 12^{k+1} -7^{k+1} -4^{k+1} -1$
$ = (12^k)(12) - (7^k)(7) - (4^k)(4) -1$
$ = (12^k)(12) - (7^k)(3+4) - (4^k)(3+1) -1$
I'm not really sure about the last step, as someone just told me to do it. Am I supposed to find the right addends to use and then distribute the exponent terms until I get a multiple of the original $a_{k}$? Because I can't get it to work out evenly, and the -1 at the end gives me trouble. Also, I know that $12^n$ is a multiple of three already, but I don't know how to implement that fact to my advantage. Can I prove that $7^{n}-4^{n}-1$ is also a multiple of three and go from there?
| First, show that this is true for $n=1$:
$12^1−7^1−4^1−1=0$
Second, assume that this is true for $n$:
$12^n−7^n−4^n−1=3k$
Third, prove that this is true for $n+1$:
$12^{n+1}−7^{n+1}−4^{n+1}−1=$
$\color\red{12^n−7^n−4^n−1}+11\cdot12^n-6\cdot7^n-3\cdot4^n=$
$\color\red{3k}+11\cdot12^n-6\cdot7^n-3\cdot4^n=$
$3k+11\cdot12\cdot12^{n-1}-6\cdot7^n-3\cdot4^n=$
$3k+132\cdot12^{n-1}-6\cdot7^n-3\cdot4^n=$
$3(k+44\cdot12^{n-1}-2\cdot7^n-1\cdot4^n)$
Please note that the assumption is used only in the part marked red.
| {
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"source": "stackexchange",
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Find the volume of the areas bounded by following:
*
*$(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$ with $a = const$
*$z=x^2+y^2, z^2=2(x^2+y^2), xy=a^2 , xy=2a^2,x=2y, 2x=y$ and $x > 0, y> 0$
First of all, I have never heard of this first geometric shape defined in (1)
This was given as a homework assignment but we haven't done assignment like this one, just simpler equations, inequalities. Does anyone know how this is done, or know a set of similar problems online that are done that I can read through.. Help very much needed.
| 1.
Start by making the replacement $x' = x/a, y' = y/a, z' = z/a$, then the equation becomes $(x'^2+y'^2+z'^2)^2=(x'^2+y'^2-z'^2)$, and $dx\,dy\,dz = a^3\,dx'\,dy'\,dz'$, so the volume is related by $V = a^3V'$. We can assume $a=1$, and get the more general case by multiplying the volume by $a^3$.
Next, switch to cylindrical coordinates: $$x = r\cos \theta\\y= r\sin\theta$$ Then $dV = r\,dr\,d\theta\,dz$ and the volume is
$$V = \iiint dV = \iiint r\,dr\,d\theta\,dz$$
The equation defining the volume becomes $$(r^2 + z^2)^2 = r^2 - z^2$$. Note that does not involve $\theta$ at all, so $\theta$ is independent of $r$ and $z$. Hence its limits are $[0, 2\pi]$ and we can integrate it out:
$$V = 2\pi\int_z\int_r rdr\,dz$$
If we do the integration on $r$ first, holding $z$ constant, then integrate $z$, we can make this substitution: $u = r^2 + z^2$, which gives $u^2 = u - 2z^2$. So $du = 2rdr$ and the limits on $u$ are $$u_l=\frac12 - \frac{\sqrt{1 - 8z^2}}2 \le u \le \frac12 + \frac{\sqrt{1 - 8z^2}}2 = u_u$$
And this also provides the limits on $z$, which must have $8z^2 \le 1$. So $$\begin{align}V &= 2\pi\int_z\int_{u_l}^{u_u} \frac12du\,dz\\&=\pi\int_zu_u - u_l\,dz\\&=\pi\int_{-\sqrt{2}/4}^{\sqrt{2}/4}\sqrt{1 - 8z^2}\,dz\end{align}$$
I'll leave the rest to you.
*The first graph is of $r= \sqrt{x^2 + y^2}$ vs $z$. The second is the $x-y$ plane (where $a = 1$). If you take the solid of revolution of the first about the $z$-axis, and the intersect it with the prism based on the second extended orthogonally up and down for all $z$, the resultant volume is what you are calculating.
Now making the cylindrical substitutions into the various restrictions gives:
*
*$z = r^2$
*$z^2 = 2r^2$ (this is a problem - maybe your prof mean $z = 2r^2$, not $z^2 = 2r^2$?)
*$r^2\sin\theta\cos\theta = a^2 \implies r^2\sin 2\theta = 2a^2$
*$r^2\sin\theta\cos\theta = 2a^2 \implies r^2\sin 2\theta = 4a^2$
*$r\cos\theta = 2r\sin\theta \implies \tan\theta = 1/2$
*$2r\cos\theta = r\sin\theta \implies \tan\theta = 2$
Which gives us the following limits:
*
*$r^2 \le z \le \sqrt2r$
*$\frac{\sqrt2a}{\sqrt{\sin 2\theta}} \le r \le \frac{2a}{\sqrt{\sin 2\theta}}$
*${1\over2} \le \tan \theta \le 2$
The volume is given by $$V = \iiint dV =\iiint dz\,rdr\,d\theta$$
Now $$\int_{r^2}^{\sqrt2r}dz = r^2 - \sqrt2r$$
$$\int r^3 - \sqrt2r^2\,dr = \left.\frac{r^4}4 - \frac{\sqrt2}{3}r^3\right|_{\sqrt2a/\sqrt{\sin 2\theta}}^{2a/\sqrt{\sin 2\theta}} = \frac{3a^4}{\sin^22\theta} -\frac{(8\sqrt2-4)a^3}{3\sin^{3/2}2\theta}$$
(note that the power of $r$ in the integral increases because of the extra $r$ already there from the Jacobian).
The $\sin^22\theta$ term can be integrated easily, but the term in $\sin^{3/2}2\theta$ does not have a nice antiderivative. In fact Wolfram Alpha expresses in terms of the elliptic integral of the 2nd kind.
I strongly suspect there are errors in the statement of the problem. If you didn't make them, your professor did. There is no nice closed form answer.
| {
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Prove that $\lim_{x \to 3} \sqrt{x+1} = 2$ Prove that $\displaystyle\lim_{x \to 3} \sqrt{x+1} = 2$
Attempt:
$0 < |x - 3| < \delta \Rightarrow |\sqrt{x+1} - 2| < \epsilon$
Well $|\sqrt{x+1} - 2| = |(\sqrt{x+1} - 2) \cdot \displaystyle\frac{\sqrt{x+1} + 2}{\sqrt{x+1}+2}| = |\frac{x-3}{\sqrt{x+1} + 2}| = |x-3| \cdot \frac{1}{|\sqrt{x+1}+2|}$
Here, the second term is the nuisance so maybe do something like the following:
Suppose we want $|x-3| < 1$. Then, $2 < x < 4$.
So $|x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3|$
Looks like we want $\delta = \min((\sqrt{3}+2)\epsilon, 1)$
Then $|\sqrt{x+1} - 2| = |x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3| < \frac{1}{\sqrt{3}+2} \cdot (\sqrt{3} + 2)\epsilon = \epsilon$
| Let $x = y+3$,
so that
$x \to 3$
is the same as
$y \to 0$.
$\begin{array}\\
\sqrt{x+1}-2
&=\sqrt{y+4}-2\\
&=(\sqrt{y+4}-2)\dfrac{\sqrt{y+4}+2}{\sqrt{y+4}+2}\\
&=\dfrac{(\sqrt{y+4}-2)(\sqrt{y+4}+2)}{\sqrt{y+4}+2}\\
&=\dfrac{(y+4)-4}{\sqrt{y+4}+2}\\
&=\dfrac{y}{\sqrt{y+4}+2}\\
&\to 0 \text{ as } y \to 0\\
\end{array}
$
Move on - nothing original here.
| {
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Steps to solve $\lim_{n \to \infty} (\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} ) = 4 $ $$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$
I think $\sqrt{n^4+1} - \sqrt{n^4-1}$ is approaching to zero, but it is not correct. What steps can evaluate above limit to 4?
| Set $1/n=h\implies h\to0^+, h>0$
$$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \dfrac1{(2n+1)^2}} \right) =\lim_{ h\to0^+}\dfrac{(\sqrt{1+h^4}-\sqrt{1-h^4})(2+h)^2}{h^4} $$
$$=\lim_{ h\to0^+}(2+h)^2\cdot\dfrac1{\lim_{ h\to0^+}(\sqrt{1+h^4}+\sqrt{1-h^4})}\cdot\lim_{ h\to0^+}\dfrac{1+h^4-(1-h^4)}{h^4}=?$$
Cancel out $h$ as $h\ne0$ as $h\to0^+$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the gcd of $f(x)$ and $g(x)$ Use the division algorithm to the find the gcd of $f(x)= 2x^4 + 5x^3 -5x -2$ and $g(x)= 2x^3 -3x^2 -2x$ in $\mathbb Q[x]$.
I used long division and found the following
$$f(x) = (x+4)g(x) + \left( 14x^2 + 3x -2 \right)$$
So I have found the polynomials $q(x)$ and $r(x)$ such that $$f(x) = q(x)g(x) + r(x)$$ where $\deg [r(x)] < deg[g(x)]$.
This is where I am stuck. How do I go about finding the GCD of $f$ and $g$? I know that it must be some monic ploynomial $d(x)$
| They key property is that if $f=gq+r$, then $$\tag{1}\gcd(f,g)=\gcd(g,r).$$
So, you already found that
$$
f(x) = (x+4)g(x) + \left( 14x^2 + 3x -2 \right).
$$
Now
$$
g(x)=\left(\frac17\,x-\frac{12}{49}\right)(14x^2+3x-2)+\left(-\frac{48}{49}x-\frac{24}{49}\right).
$$
And
$$
14x^2+3x-2=\left(-\frac{343}{24}x+\frac{49}{12}\right)\left(-\frac{48}{49}x-\frac{24}{49}\right).
$$
Then, using property $(1)$,
$$
\gcd(f,g)=\gcd(g,14x^2+3x-2)=\gcd\left(14x^2+3x-2,-\frac{48}{49}x-\frac{24}{49}\right)=-\frac{48}{49}x-\frac{24}{49}
$$
(the last equality because $-\frac{48}{49}x-\frac{24}{49}$ is a factor of $14x^2+3x-2$). To get the monic version, we multiply by $-49/48$, to get
$$
\gcd(f,g)=x+\frac12.
$$
| {
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If $A+B+C = \pi$, show that $\sum_{\text{cyclic}} \tan A \tan B \neq 0$
If $A+B+C = \pi$, and $A,B,C > 0$ show that $$\sum_{\text{cyclic}} \tan A \tan B \neq 0$$
without using the fact that if $A+B+C = \pi$, then $\tan A + \tan B + \tan C = \tan A \tan B \tan C$
My Work
I tried to show it as follows:
$$\tan(A+B+C) = \dfrac{\tan A + \tan B + \tan C - \tan A \tan B \ tan C}{1-\sum_{\text{cyclic}} \tan A \tan B}$$
For $$A+B+C = \pi$$
$$0 = \dfrac{\tan A + \tan B + \tan C - \tan A \tan B \ tan C}{1-\sum_{\text{cyclic}} \tan A \tan B}$$
I can't proceed from here.
| First
$$
\tan C=\tan(\pi-A-B)=-\tan(A+B).
$$
Hence
$$
0=\tan A\tan B+\tan A\tan C+\tan C\tan B=\tan A\tan B+\tan C(\tan A+\tan B) \\=\tan A\tan B-\frac{(\tan A+\tan B)^2}{1-\tan A\tan B}
$$
Thus
$$
(\tan A+\tan B)^2=\tan A\tan B(1-\tan A\tan B)
$$
or
$$
(\tan A)^2+(\tan B)^2+\tan A\tan B=-(\tan A)^2(\tan B)^2.
$$
Contradiction, since the right-hand side is negative, while the left hand side is positive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use an expression for $\frac{\sin(5\theta)}{\sin(\theta)}$ to find the roots of the equation $x^4-3x^2+1=0$ in trigonometric form
Question: Use an expression for $\frac{\sin(5\theta)}{\sin(\theta)}$ , ($\theta \neq k \pi)$ , k an integer to find the roots of the equation $x^4-3x^2+1=0$ in trigonometric form?
What I have done
By using demovires theorem and expanding
$$ cis(5\theta) = (\cos(\theta) + i\sin(\theta))^5$$
$$ \cos(5\theta) + i \sin(5\theta) = \cos^5(\theta) - 10\cos^3(\theta)\sin^2(\theta) + 5\cos(\theta)\sin^4(\theta) +i(5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta) + \sin^5(\theta)$$
Considering only $Im(z) = \sin(5\theta)$
$$ \sin(5\theta) = 5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta) + \sin^5(\theta) $$
$$ \therefore \frac{\sin(5\theta)}{\sin(\theta)} = \frac{5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta) + \sin^5(\theta)}{\sin(\theta)}$$
$$ \frac{\sin(5\theta)}{\sin(\theta)} = 5\cos^4(\theta) -10\cos^2(\theta)\sin^2(\theta) + \sin^4(\theta) $$
How should I proceed , I'm stuck trying to incorporate what I got into the equation..
| HINT:
Using Prosthaphaeresis Formula,
$$\sin5x-\sin x=2\sin2x\cos3x=4\sin x\cos x\cos3x$$
If $\sin x\ne0,$
$$\dfrac{\sin5x}{\sin x}-1=4\cos x\cos3x=4\cos x(4\cos^3x-3\cos x)=(4\cos^2x)^2-3(4\cos^2x)$$
OR replace $\sin^2x$ with $1-\cos^2x$ in your $$ 5\cos^4x-10\cos^2x\sin^2x + \sin^4x$$
Now if $\sin5x=0,5x=n\pi$ where $n$ is any integer
$x=\dfrac{n\pi}5$ where $n\equiv0,\pm1,\pm2\pmod5$
So, the roots of
$\dfrac{\sin5x}{\sin x}=0\implies x=\dfrac{n\pi}5$ where $n\equiv\pm1,\pm2\pmod5$
But $$\dfrac{\sin5x}{\sin x}=(4\cos^2x)^2-3(4\cos^2x)+1$$
| {
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Do there exist other tricks for trig with Cauchy's Theorem? I have noticed that $$\int_{\theta=0}^{2\pi} \frac{\mathrm{d}\theta}{a+b\cos\theta}=\frac{2\pi}{\sqrt{a^2-b^2}}$$ with $|b|<|a|$
Are there any other tricks like this for say $\int_{\theta=0}^{2\pi} \frac{\mathrm{d}\theta}{a+b\sin\theta}$?
| If one wishes to use contour integration to evaluate the integral of interest, then one simply enforces the substitution $z=e^{i\theta}$. Thus, $\cos (\theta)=\frac{z+z^{-1}}{2}$, $d\theta =\frac{1}{iz}\,dz$ and we have
$$\begin{align}
\int_0^{2\pi}\frac{1}{a+b\cos(\theta)}\,d\theta&=\oint_{|z|=1}\frac{1}{a+b\left(\frac{z+z^{-1}}{2}\right)}\,\left(\frac{1}{iz}\right)\,dz\\\\
&=\frac{2}{ib}\oint_{|z|=1}\frac{1}{\left(z+(a/b)+\sqrt{(a/b)^2-1}\right)\left(z+(a/b)-\sqrt{(a/b)^2-1}\right)}\,dz \tag 1\\\\
&=2\pi i \left(\frac{2}{ib}\frac{1}{2\sqrt{(a/b)^2-1}}\right) \tag 2\\\\
&=\frac{2\pi}{\sqrt{a^2-b^2}}
\end{align}$$
as expected!
NOTE:
In going from $(1)$ to $(2)$ we applied the Residue Theorem. Note that in proceeding, we recognizing that for $|a|>|b|$, the only pole inside $|z|=1$ is at $z=-(a/b)+\sqrt{(a/b)^2-1}$.
| {
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What's the sum of the series $\sum _{n=0}^{\infty }\left(n^2a^n\right)$ $\sum _{n=0}^{\infty }\left(n^2a^n\right)$, I have seen a similar sum with just $n$ instead of the $n^2$, yet I'm not sure how to solve this one.
| I assume $0<|a|<1$. Let $S = \sum_n a^n$, $T = \sum_n n^2 a^n$. Observe that $S(n) = \frac{1}{1-a}$ (geometric series) which implies $\frac{d}{da} S = \frac{1}{(1-a)^2}$ and $\frac{d}{da} S = \frac{-2}{(1-a)^3}$. Then
$$\frac{d^2}{da^2} S = \sum_n n(n-1)a^{n-2} = \frac{1}{a^2} \underbrace{\sum_n n^2 a^n}_{=T}- \frac{1}{a}\underbrace{\sum_n n a^{n-1}}_{=\frac{d}{da} S} = \frac{1}{a^2} T-\frac{1}{a} \frac{d}{da}S.$$ Now let us replace $\frac{d}{da} S$ and $\frac{d^2}{da^2} S$ with the terms we calculated above and we get:
$$\frac{-2}{(1-a)^3} = \frac{1}{a^2} T - \frac{1}{a} \frac{1}{(1-a)^2}$$
This can now easily be solved for $T$.
| {
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Find solution set $\sqrt{x-5} + 5 =0$ Find solution set :
$\sqrt{x-5} + 5 =0$
Is there a solution to a set of complex numbers ?
| No. There is no solution because $\sqrt{}$ is by definition non-negative.
However $\pm\sqrt{x-5} + 5 = 0$ does have a solution.
$\pm\sqrt{x-5} + 5 = 0$
$\pm\sqrt{x-5} = -5$
$(\pm\sqrt{x-5})^2 = (-5)^2$
$x-5 = 25$
$x = 30$.
But seriously. You are NOT allowed to say $\sqrt{x-5} = -5$. You just can't. It's against the rules.
| {
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How can I solve this using prime factors? I'm stuck with this problem: $2^x \cdot 3^3 \cdot 26^y = 39^z$ for $x, y, z \in \mathbb{N}$.
I know that there isn't a natural solution for the equation, but I need to "prove" it using prime factors. I'm stuck here:
$$
2^x \cdot 3^3 \cdot (2\cdot 13)^y = (3\cdot 13)^z \\
\implies 2^x \cdot 3^3 \cdot 2^y \cdot 13^y = 3^z \cdot 13^z \\
\implies 2^{x+y} \cdot 3^3 \cdot 13^y = 3^z \cdot 13^z
$$
Could anyone give me a hint on this problem? (Sorry for the formatting)
| $2^x \cdot 3^3 \cdot 26^y = 39^z\iff 2^{x+y}\cdot3^3\cdot13^y=3^z\cdot13^z\Rightarrow2^{x+y}=3^{z-3}\cdot13^{z-y}$
This is possible just when you have $1=1\cdot1$ i. e. $x+y=z-3=z-y=0$. Thus $$(x,y,z)=(-3,3,3)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of series $\sum\limits_{n=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}$. Using the power series $\sum\limits_{n=2}^{+\infty}\frac{1}{n^2+n-2}x^{n}$, the first derivative gives $\sum\limits_{n=2}^{+\infty}\frac{n}{n^2+n-2}x^{n-1}$ where the sum doesn't exist.
Integration of function $\frac{1}{n^2+n-2}x^n$ given $\frac{1}{(n+1)(n^2+n-2)}x^{n+1}$, and after forming the series, $\sum\limits_{n=2}^{+\infty}\frac{1}{(n+1)(n^2+n-2)}x^{n+1}$. After multiple integrations, I didn't found any simplifications.
| $$\begin{eqnarray*}\sum_{n\geq 2}\frac{(-1)^n}{(n+2)(n-1)}&=&\int_{0}^{1}\int_{0}^{1}\sum_{n\geq 2}(-1)^n x^{n+1} y^{n-2}\,dx\,dy\\&=&\int_{0}^{1}\int_{0}^{1}\frac{x^3}{1+xy}\,dy\,dx\\&=&\int_{0}^{1}x^2\log(1+x)\,dx\\&=&\frac{\log 2}{3}-\int_{0}^{1}\frac{x^3}{3(1+x)}\,dx\end{eqnarray*}$$
The last integral is straightforward to compute by writing $x^3$ as $(x^3+1)-1$.
The final outcome is:
$$\sum_{n\geq 2}\frac{(-1)^n}{(n+2)(n-1)} = \color{red}{-\frac{5}{18}+\frac{\log(4)}{3}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c>0\;,$ Then value of $\displaystyle \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\rfloor $,
If $a,b,c>0\;,$ Then value of $\displaystyle \bigg \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg\rfloor $, Where $\lfloor x \rfloor $ Rep. floor function of $x$.
$\bf{My\; Try::}$ Using $b+c>a$ and $c+a>b$ and $a+b>c$
So $\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<3$
And Using $\bf{A.M\geq H.M}\;,$ we get
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{9}{\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}}$$
I did not Understand How can I calculate its Lower bond, Help Required
Thanks
| $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<\frac{a+a}{b+c+a}+\frac{b+b}{c+a+b}+\frac{c+c}{a+b+c}=$$
$$=\frac{2(a+b+c)}{a+b+c}=2$$
And
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{c+a+b}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$$
So
$$ \bigg\lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg\rfloor =1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For $\triangle ABC,$ $r_1+r_3+r=r_2$, find $\sec^2A+\csc^2B-\cot^2C.$ In a triangle $ABC,$ if $r_1+r_3+r=r_2$,then find the value of $\sec^2A+\csc^2B-\cot^2C.$,where symbols have their usual meanings.
Here $r_1=\dfrac{\Delta}{s-a},r_2=\dfrac{\Delta}{s-b},r_3=\dfrac{\Delta}{s-c},r=\dfrac{\Delta}{s}$
I put these values and simplified $r_1+r_3+r=r_2$ to get $\dfrac{c^2}{\sin^2\dfrac{C}{2}}=\dfrac{b^2}{\cos^2\dfrac{B}{2}}$
I am stuck here. Please help.
| $$r_1+r_3+r-r_2=0$$
We can write $$\frac{\Delta}{s-a}+\frac{\Delta}{s-c}+\frac{\Delta}{s}-\frac{\Delta}{s-b}=0$$
to have
$$\frac{1}{b+c-a}+\frac{1}{a+b-c}+\frac{1}{a+b+c}-\frac{1}{c+a-b}=0$$
and so
$$\frac{4b(a^2+c^2-b^2)}{(b+c-a)(a+b-c)(a+b+c)(c+a-b)}=0$$
Since we have $b^2=a^2+c^2$, we know that $\triangle{ABC}$ is a right triangle with $\angle B=90^\circ$.
Therefore,
$$\sec^2A+\csc^2B-\cot^2C=\left(\frac bc\right)^2+1^2-\left(\frac ac\right)^2=\frac{(c^2+a^2)+c^2-a^2}{c^2}=\color{red}{2}.$$
| {
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Compute $\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$ I have to compute $$\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$$ I know that $\sinh t$ can be represented as a series1, but for that I require only odd powers $t^{2n-1}$, but I have no idea how to get them or what to do with the even part. Can anybody give me a hint how to start?
1 The expansion is $$\sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} +\cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}.$$
| Let's carry this computation out for instructional purposes.
Let
$$f(t) = \sum_{n=1}^{\infty} \frac{t^n}{(2 n)!} = \cosh{\left (\sqrt{t}\right )}-1$$
Then the sum we seek is
$$2 \sum_{n=1}^{\infty} \frac{n^3 t^n}{(2 n)!} = \left (t \frac{d}{dt} \right )^3 f(t) = 2 t f'(t) + 6 t^2 f''(t) + 2 t^3 f'''(t)$$
$$f'(t) = \frac{\sinh{\left (\sqrt{t}\right )}}{2 \sqrt{t}} $$
$$f''(t) = -\frac14 t^{-3/2} \sinh{\left (\sqrt{t}\right )} + \frac14 t^{-1} \cosh{\left (\sqrt{t}\right )}$$
$$f'''(t) = \frac18 t^{-5/2} (3+t) \sinh{\left (\sqrt{t}\right )} - \frac38 t^{-2} \cosh{\left (\sqrt{t}\right )} $$
Putting this all together, I get that the sum equals
$$\sum_{n=1}^{\infty} \frac{n^2 t^n}{(2 n-1)!} = \frac14 t (1+t) \frac{\sinh{\left (\sqrt{t}\right )}}{\sqrt{t}} + \frac34 t \cosh{\left (\sqrt{t}\right )} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{\frac{2}{\cos\theta} +\frac{3\sin\theta}{\cos\theta}+5\sin\theta-7\cos\theta+5}{\frac{2\sin\theta}{\cos\theta} +\frac{3}{\cos\theta}+5\cos\theta+7\sin\theta+8}$$
$$=\frac{2+3\sin\theta+5\sin\theta\cos\theta-7\cos^2\theta+5\cos\theta}{2\sin\theta+3+5\cos^2\theta+7\sin\theta\cos\theta+8\cos\theta}$$
I am stuck here.I tried to factorize numerator and denominator but does not succeed.
| hint $\frac{1-\cos(\theta)}{sin(\theta)}=\tan(\theta/2)$ so now convert everything to tan by using $\sin(2x)=\frac{2tan(x)}{1+\tan^2(x)},\cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}$ then just simplify it. as $1+\tan^2(x)$ cancels off from the expression by taking lcm then use $tan(x)=t$ for algebraic simplifications. I dont think there's a nice short way to do it.
| {
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Bézout's Identity of polynomials?
Let $P=X^3−7X+6$, $Q = 2X^2+ 5X − 3$ and $R = X^2 − 9 ∈\mathbb Q[X]$. What are $S$ and $T ∈\mathbb Q[X]$ such that $PS + QT = R$?
I have calculate the greatest common divisor of $P,Q,R$ are $(x+3)$,
But this can only prove that the existence of $S,T$.
So what should I do next?
| .We know that the gcd is $x+3$, as you have found. Divide both polynomials by $(x+3)$ to get: $(x^2-3x+2)$ and $2x-1$ respectively. Now the gcd of these two polynomials is $1$. We do the following "reverse of the Euclidean algorithm":
$x^2-3x+2 = 0.5x(2x-1) + (-2.5x + 2)$ and,
$2x-1 = -0.8(-2.5x+2)+0.6$. Therefore,now letting $G=x^2-3x+2$ and $H=2x-1$, $$0.6=2x-1 +0.8(-2.5x+2) = 2x-1 + 0.8((x^2-3x+2) - 0.5x(2x-1)) = 0.8H + (1-0.4x)G.$$
Multiplying by $\frac{10}{6}(x+3)$, we get that $$1=\frac{25}{12}(P) + \frac{10}{6}(1-0.4x)(Q).$$ Now multiply both sides by $x^2-9$ to get the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| You can use double integral to calculate. Let $f(x)=\sum_{n=1}^\infty\frac{1}{n(n+3)}x^n$. Then $f(1)=\sum_{n=1}^\infty\frac{1}{n(n+3)}$. Clearly
$$ f'(x)=\sum_{n=1}^\infty\frac{1}{n+3}x^{n-1}$$
and
$$ (x^4f'(x))'=\sum_{n=1}^\infty x^{n+2}=\frac{x^3}{1-x}.$$
Thus
\begin{eqnarray}
f(1)&=&\int_0^1\left(\int_0^x\frac{1}{x^4}\frac{t^3}{1-t}dt\right)dx\\
&=&\int_0^1\left(\int_t^1\frac{1}{x^4}\frac{t^3}{1-t}dx\right)dt\\
&=&-\frac13\int_0^1\left(1-\frac{1}{t^3}\right)\frac{t^3}{1-t}dt\\
&=&\frac13\int_0^1(t^2+t+1)dt\\
&=&\frac{11}{18}.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$ Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
| Let $x=\cos2y$
WLOG $0\le2y\le180^\circ\implies\sin y\ge0$
$$\sqrt2\sin y=\cos4y+\sin4y\iff\sin\left(4y+45^\circ\right)=\sin y$$
$\implies$
either $4y+45^\circ=360^\circ n+y\iff y=120^\circ n-15^\circ\implies y=(120-15)^\circ\implies x=\cos2y=?$
or $4y+45^\circ=(2n+1)180^\circ-y\iff y=72^\circ n+27^\circ\implies y=27^\circ\implies x=\cos2y=?$
| {
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"url": "https://math.stackexchange.com/questions/1741924",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what series to compare. I've tried to compare with $\sum \frac{1}{4n+1} - \frac{1}{4n + 3}$ because this serie converges to $\frac{\pi}{4}$, but we have exactly the opposite inaquality $\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} > \frac{1}{4n + 1} - \frac{1}{4n + 3}$.
Can you help me?
| Another possible solution:
$$
0 < a_n := \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}
= \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) - 1}} \\
< \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) + 1}} =: b_n
$$
and $\sum b_n$ is convergent as a telescoping series.
It follows that $\sum a_n$ is convergent and
$$
\sum_{n=0}^\infty a_n < \sum_{n=0}^\infty b_n = 1 \quad .
$$
Or: The terms are positive, and the partial sums
$$
\sum_{n=0}^N \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} =
1 + \sum_{n=1}^N \left(-\frac{1}{\sqrt{4n - 1}} + \frac{1}{\sqrt{4n + 1}} \right) - \frac{1}{\sqrt{4N + 3}} < 1
$$
are bounded. (This is essentially what achille hui suggested
in the above comment.)
| {
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Lucas numbers proof I'm running through some example problems and encountered this one:
Define a sequence of integers $L_n$ by $L_1=1, L_2=3, L_{n+1}=L_n+L_{n-1}.$ Show that $L_n = a\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n + b\cdot \left(\frac{1-\sqrt{5}}{2}\right)^n$ and find the values of $a$ and $b$.
I wasn't able to prove the formula using induction first since the LHS has $a$ and $b$ in it so I cannot show the base case. How can I show the equation holds, and then find $a$ and $b$?
| The recurrence is $L_{n+1}=L_{n}+L_{n-1}$.Now I am using the characteristic
equation of this above recurrence to solve the recurrence. So the characteristic
equation of this above recurrence is as follows $\implies$
Put $L_{n}={r^n}$ and solve the equation $\implies ({r^2}-r-1)=0$
By solving this equation we get $r=\frac{1+_-\sqrt{5}}{2}$(+ and -).
Hence $r=\frac{1+\sqrt{5}}{2}\;\;and\;\;\frac{1-\sqrt{5}}{2}$.Now denote these
two roots by $r_1\;and\;r_2$. It is given that $L_1=1\;and\;L_2=3$.
Hence $L_n=A{r_1^n}+B{r_2^n}$ where A and B are constants.Hence we have shown
that $L_n=a{(\frac{1+\sqrt5}{2})^n}+b{(\frac{1-\sqrt5}{2})^n}$. Now put $n=1$
and$n=3$, then you get two equations with two unknowns which is solvable.
Two equations are $\implies$ $1=A{r_1}+B{r_2}\;\;and\;\;3=A{r_1^2}+B{r_2^21}$.
Solving this two equations we get $A=1\;\;and\;\;B=1$. Hence $a=1\;and\;b=1$.
Finally we get $L_n={(\frac{1+\sqrt5}{2})^n}+{(\frac{1-\sqrt5}{2})^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$ I came across this problem that I'm supposed to be able to solve in under 5 minutes (for a competition).
$$\lim_{n\to \infty} \sum_{k=0}^n \frac {n}{n^2+k}$$
I tried solving this for small sums, $\sum_{k=0}^2 \frac {n}{n^2+k}$, $\sum_{k=0}^3 \frac {n}{n^2+k} $ and made this:
$$\sum_{k=0}^2 \frac {n}{n^2+k} = \frac {n}{n^2+1} + \frac{n}{n^2+2} = \frac {n(n^2+2)+n(n^2+1)}{(n^2+1)(n^2+2)} = \frac {2n^3+...}{n^4+...}$$
And it's limit is $0$, for $k$ up to $3$ is $0$ too, so I assumed that the original limit must be $0$. Wrong, my book says that is $1$, I graphed the function and indeed that is correct. How should I solve it?
| Slightly more involved solution:
$$
\frac{1}{n} \sum_{k=0}^{n} \frac{1}{1+\frac{k}{n^2}} = \frac{1}{n} \sum_{k=0}^{n} e^{- \log (1+\frac{k}{n^2}) } \sim \frac{1}{n} \sum_{k=0}^{n} e^{-\frac{k}{n^2}} = \frac{1}{n} \times \bigg(\frac{1-e^{-\frac{1}{n^2}(n+1)}}{1-e^{-\frac{1}{n^2}}} \bigg) \sim \frac{1}{n} \times \bigg( \frac{\frac{1}{n} + \frac{1}{n^2}}{\frac{1}{n^2}} \bigg) = 1 +\frac{1}{n} \to_n 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to calculate the series $-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$? $-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$
After rearrangement the series looks like $\sum^{\infty}_{n=2}\frac{(-1)^{n+1}}{n}$.
My way of doing this is using Taylor series of $\ln{(1+x)}=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}x^n}{n}$.
Therefore let $x=1$, $\sum^{\infty}_{n=2}\frac{(-1)^{n+1}}{n}=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}x^n}{n}-\sum^{1}_{n=1}\frac{(-1)^{n+1}x^n}{n}=\ln{2}-\sum^{1}_{n=1}\frac{(-1)^{n+1}x^n}{n}=\ln{2}-1$
Is my solution correct?
| You cannot rearrange an infinite series which is not absolutely convergent. We can evaluate it as follows. Your infinite series is
\begin{align}
\sum_{n=1}^{\infty}\left(-\dfrac1{4n-2} - \dfrac1{4n}+\dfrac1{2n+1}\right) & = \sum_{n=1}^{\infty}\left(-\dfrac1{4n-2} - \dfrac1{4n}+ 2\cdot \dfrac1{4n+2}\right)\\
& = \sum_{n=1}^{\infty}\left(-\int_0^1x^{4n-3}dx- \int_0^1x^{4n-1}dx + 2\int_0^1x^{4n+1}dx\right)\\
& = \sum_{n=1}^{\infty}\int_0^1 x^{4n}\left(2x-\dfrac1x - \dfrac1{x^3}\right)dx\\
& = \int_0^1\left(2x-\dfrac1x - \dfrac1{x^3}\right)\left(\sum_{n=1}^{\infty}x^{4n}\right)dx\\
& = \int_0^1\left(2x-\dfrac1x - \dfrac1{x^3}\right) \dfrac{x^4}{1-x^4}dx\\
& = \int_0^1 \left(\dfrac{x}{x^2+1}-2x\right)dx = \dfrac{\ln(2)}2-1
\end{align}
To justify the swapping of the limit and integral, we have
\begin{align}
I_m & = \sum_{n=1}^m \int_0^1 x^{4n}\left(2x-\dfrac1x-\dfrac1{x^3}\right)dx = \int_0^1 \dfrac{x^4\left(1-x^{4m}\right)}{1-x^4}\left(2x-\dfrac1x-\dfrac1{x^3}\right)dx\\
& = \int_0^1 \left(\dfrac{x}{x^2+1}-2x\right)dx + \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}\cdot x^{4m}dx
\end{align}
Now $$\lim_{m\to\infty} \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}\cdot x^{4m}dx = 0$$
There are multiple ways to prove this. One is to just invoke dominated convergence theorem, since $x^{4m} \leq 1$ and $\displaystyle \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}dx < \infty$. Other is to simply note that $\dfrac{x(2x^2+1)}{x^2+1} \leq \dfrac32$ on $[0,1]$. Hence,
$$\lim_{m\to\infty} \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}\cdot x^{4m}dx \leq \lim_{m\to\infty} \int_0^1 \dfrac32\cdot x^{4m}dx = \lim_{m\to\infty} \dfrac3{2(4m+1)} = 0$$
Hence,
$$\lim_{m \to \infty} I_m = \dfrac{\log(2)}2-1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\sin\frac{3\pi}{14}-\sin\frac{\pi}{14}-\sin\frac{5\pi}{14}$ I have interesting trigonometric expression for professionals in mathematical science. So, here it is:
$$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$
Okay! I attempt calculate it:
\begin{gather}
\sin\dfrac{3\pi}{14}-\left(\sin\dfrac{\pi}{14}+\sin\dfrac{5\pi}{14}\right)=\\
=\sin\dfrac{3\pi}{14}-\left(2\sin\dfrac{3\pi}{14}\cdot\cos\dfrac{\pi}{7}\right)=\\
=\sin\dfrac{3\pi}{14}\left[1-2\cdot\cos\dfrac{\pi}{7}\right]=...
\end{gather}
Tried everything... Here deadlock. I really do not know what to do next. Help somebody, please.
| For simplicity, let $x = \frac{\pi}{14}$, then we want to simplify:
$$\sin 3x-\sin x -\sin 5x$$
Multiply by $\cos x$ to get:
$$\color{blue}{\sin 3x\cos x}-\color{green}{\sin x\cos x} -\color{red}{\sin 5x\cos x} \quad (*)$$
With $\sin\alpha\cos\beta = \tfrac{1}{2}\left( \sin(\alpha+\beta)+\sin(\alpha-\beta) \right)$, you have:
$$\color{blue}{\sin 3x\cos x = \tfrac{1}{2}\left( \sin 4x +\sin 2x \right)} \quad \mbox{and} \quad \color{red}{\sin 5x\cos x = \tfrac{1}{2}\left( \sin 6x +\sin 4x \right)} $$
and $\color{green}{\sin x\cos x = \tfrac{1}{2}\sin 2x}$; so:
$$\require{cancel} (*) \quad
\tfrac{1}{2}\left( \cancel{\sin 4x} +\bcancel{\sin 2x} \right) - \bcancel{\tfrac{1}{2}\sin 2x} - \tfrac{1}{2}\left( \sin 6x +\cancel{\sin 4x} \right) = - \tfrac{1}{2}\sin 6x
$$
Divide again by $\cos x$:
$$- \tfrac{1}{2}\frac{\sin 6x}{\cos x} = - \tfrac{1}{2}\frac{\sin \frac{6\pi}{14}}{\cos \frac{\pi}{14}}= - \tfrac{1}{2}\frac{\cos\left( \frac{\pi}{2}-\frac{6\pi}{14}\right)}{\cos \frac{\pi}{14}} =- \tfrac{1}{2}\frac{\cos\frac{\pi}{14}}{\cos \frac{\pi}{14}} = -\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Hint solving an integral using trig substitution I want a hint for the following integral:
$$\int \frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} dx$$
where $a$ is a real constant.
My attempt:
$$\frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} = \frac{1}{x^{2}\sqrt{1- a^{2}+x^{2}}}(1-\frac{1}{a^2-x^2})$$
I know from the post
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}} = -\frac{\sqrt{1- a^{2}+x^{2}}}{x(1- a^{2})}.$$
But how shoud i solve the following integral???
$$\int \frac{1}{x^{2}\sqrt{1- a^{2}+x^{2}}}\frac{dx}{a^2-x^2} $$
Thanks for your help.
$$$$
| Let be
$$
I=\int
\frac{\sqrt{1-a^2+x^2}}{x^2(a^2-x^2)}\mathrm d x
$$
with the substitution
$
\frac{\sqrt{1-a^2+x^2}}{x}=u
$ and $x^2=\frac{1-a^2}{u^2-1}$ and $\mathrm d u=\frac{a^2-1}{x^2\sqrt{1-a^2+x^2}}\mathrm d x$ and then $\frac{x^2\sqrt{1-a^2+x^2}}{a^2-1}\mathrm d u=\mathrm d x$ the integral $I$ becomes
$$
J=\int \frac{u^2}{1-u^2a^2}\mathrm d u
$$
and with the substitution $au=t$ the integral $J$ becomes
$$
K=\frac{1}{a^3}\int \frac{t^2}{1-t^2}\mathrm d t=\begin{cases}\frac{1}{a^3}\left[\tanh^{-1}(t)-t\right]+c & \text{for }|t|<1\\
\frac{1}{a^3}\left[\coth^{-1}(t)-t\right]+c & \text{for }|t|>1
\end{cases}
$$
observing that $\left(\tanh^{-1}(t)\right)'=\frac{1}{1-t^2}$ for $|t|<1$ and $\left(\coth^{-1}(t)\right)'=\frac{1}{1-t^2}$ for $|t|>1$.
Thus
$J=K(au)$ and $I=K\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)$, that is
$$
I=\begin{cases}\frac{1}{a^3}\left[\tanh^{-1}\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|<1\\
\frac{1}{a^3}\left[\coth^{-1}\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|>1
\end{cases}
$$
or
$$
I=\begin{cases}\frac{1}{a^3}\left[\tanh^{-1}\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|<1\\
\frac{1}{a^3}\left[\tanh^{-1}\left(\frac{x}{a\sqrt{1-a^2+x^2}}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|>1
\end{cases}
$$
observing that $\tanh^{-1}\left(\frac{1}{z}\right)=\coth^{-1}(z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose $P (1) + P (2) = 3P (3)$ and $P (2) + P (3) = P (1)$ for a sample space $(S,P)$ . Find the probability function. So we have the following equations:
$P(1) + P(2) = 3P(3)$
$P(2) + P(3) = P(1)$
I know that by definition, $P(1) + P(2) + P(3) = 1$
I'm not entirely sure what I'm looking for as my answer though. Are there any other equations hiding that could be useful here? Hints are fine.
I've worked out that I can end up with $4P(3) = 1$ by using those three equations.
| Here's one way to solve this problem.
Since we know that $P(1) + P(2) + P(3) = 1$ by definition, we have the following three equations:
$P(1) + P(2) = 3P(3)$
$P(2) + P(3) = P(1)$
$P(1) + P(2) + P(3) = 1$
Plugging the first equation into the third, we get $4P(3) = 1$. So $P(3) = \frac{1}{4}$.
From the last equation, we get $P(2) = 1 - P(3) - P(1) \Rightarrow P(2) = \frac{3}{4} - P(1)$.
Now substitute that into the second equation to find that $P(1) = \frac{1}{2}$.
Lastly, we can return to the first equation to find that $P(2) = \frac{1}{4}$.
So we have:
$P(1) = \frac{1}{2}$
$P(2) = \frac{1}{4}$
$P(3) = \frac{1}{4}$
More formally, for all $x \in S = \{1, 2, 3\}$,
$$P(x) =
\begin{cases}
\frac{1}{2}, & \text{$x$ = 1} \\[2ex]
\frac{1}{4}, & \text{$x \in \{2, 3\}$}
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate this infinite product: $\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)$
$$\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)\;=\;\text{???}$$
I took the LCM and split the numerator as $(n+2)(n-2)$ and then took the product of the numerator and the denominator separately but I was not able to get the answer from that so can you please help me in what to do next.
| $$ \prod_{n=3}^{\infty} \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{(n-2)(n+2)}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{n-2}{n} \right)\prod_{n=3}^{\infty} \left( \frac{n+2}{n} \right) = \lim_{n\rightarrow \infty} \frac{2^2}{4!} \frac{(n-2)!}{n!} \frac{(n+2)!}{n!} $$
So now we pull out Stirling's Approximation
$$ \frac{1}{3!} \lim_{n \rightarrow \infty} \frac{2 \pi \sqrt{n^2-4}\left( \frac{n-2}{e} \right)^{n-2}\left( \frac{n+2}{e} \right)^{n+2} }{2 \pi n \left( \frac{n}{e} \right)^{2n}} $$
Dividing out common terms we have:
$$ \frac{1}{3!} \lim_{n \rightarrow \infty} \frac{\sqrt{n^2-4}\left( \frac{n-2}{e} \right)^{n-2}\left( \frac{n+2}{e} \right)^{n+2} }{n\left( \frac{n}{e} \right)^{2n}} $$
We can take a natural logarithm and then exponentiate our answer to recover the original. So now consider:
$$ \lim_{n \rightarrow \infty} \frac{1}{2} \ln \left(\frac{n^2 - 4}{n} \right) + n \ln \left( \frac{(n+2)^2}{n^{n}(n-2)^2} \right)+n - \ln(3!) $$
Our problem then simplifies to:
$$ \lim_{n \rightarrow \infty} n \ln \left( \frac{(n+2)^2}{(n-2)^2} \right)+n - \ln(3!) - n \ln(n)$$
Which at limit yields:
$$ \lim_{n \rightarrow \infty} O(n) - O(n \ln n)$$
And so obviously tends to $-\infty$. This means that the original tends to 0.
So we conclude that the product tends to 0 (using our approximation).
An easier method.
Note that
$$ \frac{1 \times 2 ... \times n-2}{3 \times 4 ... \times n} = \frac{2}{(n-1)n} $$
On the other hand we have:
$$ \frac{5 \times 6 ... \times n+2}{3 \times 4 ... \times n} = \frac{(n+1)(n+2)}{3 \times 4} $$
So their combined product is
$$ \frac{2(n+1)(n+2)}{12(n-1)n} = \frac{1}{6} \frac{(n+1)(n+2)}{n(n-1)}$$
Now it's easy to prove $\lim_{n\rightarrow \infty}
\frac{(n+1)(n+2)}{n(n-1)} = 1$ and you get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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How to evaluate this limit using Taylor expansions? I am trying to evaluate this limit:
$\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$
I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their first term removed so the series starts at $n=1$ rather than at 0.
However, when I actually try to evaluate I get stuck at:
$\lim_{x \to 0} \dfrac{(-\frac{1}{3!}+\frac{x^2}{5!}...)(-\frac{9x^2}{2!}+\frac{81x^4}{4!}...)}{(\frac{1}{x}+\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}...)^4}$
I don't know how to proceed from here.
| Just in case you wanted something to check your answer off of:
\begin{align}
\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4} &= \lim_{x\to 0} \frac{\left(\frac{-x^3}{3!}+\frac{x^5}{5!} - \cdots\right)\left(\frac{-x^2}{2!} + \frac{x^4}{4!}- \cdots\right)}{x\left(x+\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right)^4}\\
&= \lim_{x\to 0} \frac{x^3\left(\frac{-1}{3!}+\frac{x^2}{5!} - \cdots\right)x^2\left(\frac{-1}{2!} + \frac{x^2}{4!}- \cdots\right)}{x^5\left(1+\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)^4}\\
&= \lim_{x\to 0} \frac{\left(\frac{-1}{3!}+\frac{x^2}{5!} - \cdots\right)\left(\frac{-1}{2!} + \frac{x^2}{4!}- \cdots\right)}{\left(1+\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)^4}\\
&= \left(\frac{-1}{3!}\right)\left(\frac{-1}{2!}\right)\\
&= \frac{1}{12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Showing H is a normal subgroup by calculating left and right coset If $G = \begin{pmatrix} a & b \\
0 & 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0$)
and assume G is a group under matrix multpication
Prove that H = ($\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}: t\in\mathbb{R}$) is a normal subgroup of G
I'm meant to show that H is a normal subgroup by showing that the left coset is equal to the right coset.
So far I have shown the following:
take $g=\begin{pmatrix} a & b \\
0 & 1 \end{pmatrix}$
then
$gH=\begin {pmatrix}a &at+b \\ 0 &1 \end{pmatrix}$
and take $g'=\begin{pmatrix} a' & b' \\
0 & 1 \end{pmatrix}$
then $Hg'=\begin{pmatrix} a' & b'+t \\
0 & 1 \end{pmatrix}$
I don't quite understand how $gH=Hg'$
| First let me solve the problem the way you asked: proving that every left coset is a right coset.
Given $g=\begin{pmatrix}a & b \\ 0 & 1\end{pmatrix}$ one must prove that $gH=Hg$ (introducing an additional symbol $g'$ is wrong). Your question correctly describes the general format of matrices in the sets $gH$ and $Hg$, so I'll work with that.
Consider $M$ in $gH$, and so there exist $t$ such that
$$M = \begin{pmatrix}a & at+b \\ 0 & 1 \end{pmatrix} \in gH
$$
For $M$ to be in $H g$, one must find $t'$ so that
$$M = \begin{pmatrix}a & b+t' \\ 0 & 1 \end{pmatrix} \in Hg
$$
Three of the matrix entries are already equal, so we must simply find $t'$ so that
$$at+b = b+t'
$$
That equation is easily solved: $t'=at$, which proves that $M \in Hg$.
The reverse inclusion $Hg \subset gH$ is proved similarly.
However, despite the definition of a normal subgroup $H < G$ being
*
*"every left coset of $H$ is a right coset of $H$"
and despite the utility of that definition in applying normality, for purposes of proving normality it is almost always easier to work from the equivalent statement
*
*"$gHg^{-1}=H$ for every $g \in G$"
or even easier
*
*"$g H g^{-1} \subset H$ for every $g \in G$"
So, given $g = \begin{pmatrix}a & b \\ 0 & 1 \end{pmatrix}$ and $h = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$, I simply do a calculation
$$g h g^{-1} = \begin{pmatrix}a & b \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\begin{pmatrix}a^{-1} & -b/a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & at+b \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a^{-1} & -b/a \\ 0 & 1\end{pmatrix} = \begin{pmatrix} 1 & at \\ 0 & 1\end{pmatrix}
$$
And this is obviously in $H$, by setting $t'=at$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show convexity of a function via inequalities I am stuck with deriving the convexity of the function
$$
f(x) = \sqrt{1 + x^2}
$$
from first principles, that is I would like to show that for any $x,y \in \mathbb R$ and $\lambda \in (0,1)$ we have
$$
f(\lambda x + (1 - \lambda)y) \le \lambda f(x) + (1 - \lambda) f(y)
$$
The fact that it is a convex function is easy to see from the second derivative test so I am ok with the statement, its just that I cannot derive it using basic inequalities!
Here is what I tried so far:
Attempt 1
Use the convexity of the squaring function to write
\begin{align*}
\sqrt{1 + (\lambda x + (1 - \lambda)y)^2} &\le \sqrt{1 + \lambda x^2 + (1 - \lambda)y^2} \\
&= \sqrt{\lambda (1 + x^2) + (1 - \lambda)(1 + y^2)}
\end{align*}
Now use the fact that
$$
\sqrt{A + B} \le \sqrt{A} + \sqrt{B}
$$
to bound the the last expression by
\begin{align*}
\sqrt{\lambda (1 + x^2) + (1 - \lambda)(1 + y^2)} &\le \sqrt{\lambda (1 + x^2)} + \sqrt{(1 - \lambda)(1 + y^2)}
\end{align*}
but now I'm stuck because I cannot take out the factors and bound it above as both $\lambda$ and $1 - \lambda$ are less than $1$ .. :(
Attempt 2:
Don't use the convexity of the squaring function, instead expand. Then
\begin{align*}
\sqrt{1 + (\lambda x + (1 - \lambda)y)^2} &= \sqrt{1 + \lambda^2x^2 + 2\lambda(1 - \lambda)xy + (1 - \lambda)^2y^2} \\
&\le \sqrt{1 + 2\lambda^2x^2 + 2(1 - \lambda)^2y^2}
\end{align*}
so now I need to find a way to show that
$$
1 + \lambda^2x^2 + (1 - \lambda)^2y^2 \le \lambda^2 + (1 - \lambda)^2
$$
but somehow I am too blind / tired to find the right argument .. :(
Thanks for hints!
| To make the two sides of the inequalities easier to compare, put $a:=\lambda$ and $b:=(1-\lambda)$ and then square both sides (a valid operation since both sides are nonnegative and therefore $x\mapsto x^{2}$ is invertible and monotonic, thus inequality preserving).
After doing this, we have on the one hand
$$\begin{align*}
[f(ax+by)]^{2}
&=1+(ax+by)^{2}\\
&=1+a^{2}x^{2}+2abxy+b^{2}y^{2}\\
&=a^2+2ab+b^2+a^2x^2+2abxy+b^2y^2&\text{($a^{2}+2ab+b^{2}=(a+b)^{2}=1$})\\
&=a^2(1+x^2)+b^2(1+y^2)+2ab(1+xy),
\end{align*}$$
and on the other
$$\begin{align*}
[af(x)+bf(y)]^{2}
&=a^{2}f^{2}(x)+b^{2}f^{2}(y)+2abf(x)f(y)\\
&=a^{2}(1+x^{2})+b^{2}(1+y)^{2}+2ab\sqrt{1+x^{2}}\sqrt{1+y^{2}}\\
&=a^{2}(1+x^{2})+b^{2}(1+y)^{2}+2ab\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}\\
\end{align*}$$
We now see that to bridge this chain, we only need to verify that
$$2ab(1+xy)\leq2ab\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}},$$
or (after dividing both sides by $2ab>0$ [note that we are done if $2ab=0$])
$$1+xy\leq\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}},$$
or (after squaring both sides assuming $1+xy\geq0$ [note that we are done if $1+xy<0$])
$$1+2xy+x^{2}y^{2}\leq 1+x^{2}+y^{2}+x^{2}y^{2},$$
or (after subtracting $1+x^{2}y^{2}$ from both sides)
$$2xy\leq x^{2}+y^{2},$$
which is true since $0\leq(\alpha+\beta)^{2}=\alpha^{2}+2\alpha\beta+\beta^{2}$ is of course valid for any $\alpha,\beta\in\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How do I simplify $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ using long division? The infinite series $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ is supposed to simplify to $1-\frac{x^2}{2}+\frac{7x^4}{48}+\frac{19x^6}{576}\cdots$ but I don't know how this was calculated.
Did I make an error or did Zill make an error on the $x^6$ term (I got $\frac{11}{576x^6}$.)
| Actually, I got
$$1-\frac{x^2}{2}+\frac{7x^4}{48}-\frac{19x^6}{576}\cdots$$
Let
$$D = 1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots,$$
your divisor.
First remainder is
$$R_1 = 1 - 1D = -\frac{x^2}{2} - \frac{5x^4}{48} - \frac{7x^6}{576} + O(x^8).$$
Second remainder (for the $-x^2/2$ term) is
$$R_2 = R_1 - \left(-\frac{x^2}{2}\right)D = \frac{7x^4}{48} + \frac{23x^6}{576} + O(x^8).$$
Third remainder (for the $5x^4/48$ term) is
$$R_3 = R_2 - \left(\frac{7x^4}{48}\right)D = -\frac{19x^6}{576} + O(x^8),$$
which gives the last term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$ Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$
Let $d=\gcd(4n^2+1,24)$ then we have:
$$d|24n^2+6,24n^2\ \Rightarrow\ d|6\ \Rightarrow\ d|6n^2,4n^2+1\ \Rightarrow\ d|12n^2,12n^2+3\ \Rightarrow\ d|3\ \Rightarrow\ d=1\ or\ 3$$
Using modular equivalence it's very easy to show that $d$ can't be 3,but how can I show it WITHOUT using modular equivalence???
| If
$n=0,1,2,3,4,5$,
$4n^2+1
=1, 5, 17, 37, 65, 101
$,
and all of these
are relatively prime to 24.
If
$n = 6m+k$
where
$0 \le k \le 5 $,
then
$\begin{array}\\
4n^2+1
&=4(6m+k)^2+1\\
&=4(36m^2+12mk+k^2)+1\\
&=4(36m^2+12mk)+4k^2+1\\
&=48(3m^2+mk)+4k^2+1\\
\end{array}
$
so
if $d$ divides both
$4n^2+1$ and $24$,
it also divides
$4k^2+1$.
Since
$24$ and
$2k^2+1$are relatively prime
for
$0 \le k \le 5$,
the only integer
that divides both
$24$ and $4k^2+1$
is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I show that $\sum_{i = 1}^n \frac 1{\sqrt{a_n}} \lt \frac {\sqrt 3}6$ for $a_n = 4n(4n + 1)(4n + 2)$? Let $a_n = 4n(4n + 1)(4n + 2)$, show that $$\sum_{i = 1}^n \frac 1{\sqrt{a_i}} \lt \frac {\sqrt 3}6 \quad \forall n \in \mathbb{N}^+.$$
I know I need to find an upper bound for $1/\sqrt{a_n}$ but I can't see how, especially with the square root. Any hints will be appreciated!
| $$
\begin{align}
\frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right)
&=\sum_{k=1}^\infty\frac2{\sqrt{4k-1}\sqrt{4k+3}\left(\sqrt{4k-1}+\sqrt{4k+3}\right)}\tag{1}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k-1}\sqrt{4k+3}\sqrt{4k+1}}\tag{2}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k}\sqrt{4k+2}\sqrt{4k+1}}\tag{3}
\end{align}
$$
Explanation:
$(1)$: arithmetic
$(2)$: concavity of $\sqrt{x}$ says that $\frac{\sqrt{x}+\sqrt{y}}2\le\sqrt{\frac{x+y}2}$
$(3)$: $(4k-1)(4k+3)\le4k(4k+2)$ by expanding
This says that
$$
\begin{align}
\sum_{k=1}^\infty\frac1{\sqrt{4k(4k+1)(4k+2)}}
&\le\frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right)\\
&=\frac12\frac1{\sqrt3}\\
&=\frac{\sqrt3}6\tag{4}
\end{align}
$$
Motivation
Since the terms of the sum are $\sim\frac18k^{-3/2}$, it is often useful to consider a telescoping series where the terms are differences of something $\sim\frac14k^{-1/2}$ because such a difference is $\sim\frac18k^{-3/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finite sum $\sum_{n=2}^N\frac{1}{n^2}\sin^2(\pi x)\csc^2(\frac{\pi x}{n})$ I was looking for a closed form but it seemed too difficult. Now I'm seeking help to simplify this sum. The 50 bounty points or more will be awarded for any meaningful simplification of this sum.
I found this function that has very interesting property to show if $n \nmid x$. And the sum of this function from $2$ to $N$ can show all the primes within $N$ on 1, and primes beyond $n$ and within $N^2$ as zeros, and prime sieve beyond that
$$
f(n,x)=\frac{1}{n}F_n\left(\frac{2\pi}{n}x\right)=\frac{1}{n^2}\left(\frac{1-\cos(2\pi x)}{1-\cos\left(\frac{2\pi}{n}x\right)}\right)=\frac{1}{n^2}\left(\frac{\sin(\pi x)}{\sin\frac{\pi x}{n}}\right)^2
$$
This is a prime test function and I try to find a closed form formula to simplify the calculation of $$\sum_{n=2}^Nf(n,x)=\sum_{n=2}^N\frac{1}{n^2}\left(\frac{\sin(\pi x)}{\sin\left(\frac{\pi x}{n}\right)}\right)^2=\sum_{n=2}^N\frac{1}{n^2}\sin^2(\pi x)\csc^2\left(\frac{\pi x}{n}\right)$$
Is there a close form formula (i.e. no or few terms of summation) for this sum? It's known that $$\int_{2}^{N} f(t,x)dt=\sin^2(\pi x)\int_{2}^{N}\frac{1}{t^2}\csc^2\left(\frac{\pi x}{t}\right)dt=\left. \left(\frac{1}{\pi x}\sin^2(\pi x)\cot\left(\frac{\pi x}{t}\right) \right)\right|_{t=2}^{t=N}$$
Does the Euler–Maclaurin Summation Formula help here?
Here's the example graphic of $\sum_{n=2}^{50}f(n,x)$ shows all primes within $2500$.
| An idea:
Note the identity
$$\sin n \theta=U_{n-1}\left( \cos \theta \right) \sin \theta ,$$
where $U_i$ is the $i^{th}$ Chebyshev orthogonal polynomial of the second kind. This gives
$\sin^2 \pi x=\sin^2 \left(n \frac{\pi x}{n}\right)=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right) \sin^2 \frac{\pi x}{n},$
and then
$\sin^2 \pi x \csc^2 \frac{\pi x}{n}=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right) \sin^2 \frac{\pi x}{n} \csc^2 \frac{\pi x}{n}=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right),$
so your sum becomes
$\sum \limits_{n=2}^N \frac{1}{n^2} U_{n-1}^2\left( \cos \frac{\pi x}{n} \right).$
There may be an identity on the Chebyshev polynomials to further simplify this sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
"answer_id": 0
} |
Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1})$$
Here is what I tried:
$$\lim_{x->0}{\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{4!}+o(x^{2*5}))}{x^{10}+o(x^{10*2})}}=\lim_{x->0}{\frac{-\frac{x^{10}}{4!}-o(x^{12})}{x^{10}+o(x^{20})}}$$
This is where I am stuck. I figured that the problem occurs when expending $o()$. What am I missing here?
| $$\lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2\cos(x^2)}{\sin(x^{10})} = \lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{24}-\frac{x^{12}}{720}+\cdots)}{x^{10}-\frac{x^{30}}{6}+\cdots}$$
$$=\lim_{x\to0}\frac{x^2-\frac{x^6}{2}+(-x^2+\frac{x^6}{2}-\frac{x^{10}}{24}+\frac{x^{12}}{720}-\cdots)}{x^{10}-\frac{x^{30}}{6}+\cdots}$$
Canceling terms in the numerator simplifies the equation to:
$$\lim_{x\to0}\frac{-\frac{x^{10}}{24}+\cdots}{x^{10}-\cdots}$$
The other terms in the series are unnecessary, so
$$\lim_{x\to0}(-\frac{x^{10}}{24}*\frac{1}{x^{10}}) = -\frac{1}{24}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach Evaluate $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach.
Using the formula $$\int_{a}^{b} f(x)\:dx=(b-a) \times \lim_{n \to \infty} \frac{1}{n} \times \sum_{k=1}^{n}f\left(a+\frac{(b-a)k}{n}\right)$$ we have
$$I=\int_{0}^{3} f(x)\:dx=(3-0) \times \lim_{n \to \infty} \frac{1}{n} \times \sum_{k=1}^{n}f\left(0+\frac{(3-0)k}{n}\right)=3 \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{3k}{n}\right)$$ So
$$I=3\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} \sqrt{1+\frac{3k}{n}}$$ Now if we expand the summation we get limits of the form
$$\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{3}{n}}+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{6}{n}}+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{9}{n}}+\cdots+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{3n}{n}}$$
But each limit is clearly zero and hence $I=0$. I know the answer is wrong but what is my mistake?
| Consider
$$\int_0^3 \sqrt{1+x} \, dx = \lim_{n \to \infty} \frac{3}{n}\sum_{k=1}^n\sqrt{1 + \frac{3k}{n}} = \lim_{n \to \infty} \frac{3}{n^{3/2}}\sum_{k=1}^n\sqrt{n + 3k}.$$
Using the binomial expansion, we have
$$(n + 3k - 3)^{3/2} = (n + 3k)^{3/2} \left(1 - \frac{3}{n+3k} \right)^{3/2} \\ = (n + 3k)^{3/2}\left(1 - \frac{3}{2}\frac{3}{n+3k} + O(1/n^2)\right),$$
and
$$\sqrt{n+3k} = \frac{2}{9}\left[(n+3k)^{3/2} - (n+3k-3)^{3/2}\right] +O(1/\sqrt{n}).$$
Summing we get
$$\lim_{n \to \infty}\frac{3}{n^{3/2}}\sum_{k=1}^n\sqrt{n + 3k} = \lim_{n \to \infty}\frac{3}{n^{3/2}}\sum_{k=1}^n\frac{2}{9}\left[(n+3k)^{3/2} - (n+3k-3)^{3/2}\right] \\ = \lim_{n \to \infty}\frac{3}{n^{3/2}}\frac{2}{9}\left[(4n)^{3/2} - (n)^{3/2}\right] \\ = \frac{2}{3}\left[4^{3/2} - 1\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$
If $x,y\in \mathbb{R}$ and $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ and $f'(0)=0\;,$ Then $f(x)$ is
$\bf{My\; Try::}$ Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)}{h}$$
Now Put $x=y=0$ in $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}\;,$$ We get $f(0)=0$
So we get $f(0)=0$
So $$f'(x) = \sqrt{f(x)}+\lim_{h\rightarrow 0}\frac{f(h)}{h}=\sqrt{f(x)}$$
So $$\int\frac{f'(x)}{\sqrt{f(x)}}dx = 1\int dx\Rightarrow 2\sqrt{f(x)}=x+c$$
Now Put $x=\;,$ We get $c=0$
So we get $2\sqrt{f(x)}=x\Rightarrow 4f(x)=x^2\Rightarrow \displaystyle f(x)=\frac{x^2}{4}$
Can we solve it some short way, If yes then please explain here, Thanks
| $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}=f(y+x)=f(y)+f(x)+x \sqrt{f(y)}$$
Subtracting $f(x)+f(y)$ from each side and squaring , we have that $$y^2f(x)=x^2f(y) \Leftrightarrow \frac{f(x)}{x^2}=\frac{f(y)}{y^2}$$
So we have $\frac{f(x)}{x^2}$ is a constant function.
Now put $f(x)=cx^2$ in the original equation, where $c$ is a constant. Simplifying gives us that $$2cx=\sqrt{c} |x|$$
Note that if $c \neq 0$, $c$ will take different values depending on the value of $x$, This is a contradiction, as $c$ is a constant. So we have $c=0$. Thus, $f(x)=0$ is the only solution. In order for $\frac{x^2}{4}=f(x)$, to be a solution we must have a constraint that $x \ge 0$, or the functional equation should be altered so: $$f(x+y)=f(x)+f(y)+y\operatorname{sgn}(x)\sqrt{f(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
$\sum_{k=1}^{n}|1+z_{k}|+\frac{1}{n-1}\sum\limits_{1\le iLet $n\ge 2$is a integer,$z_{1},z_{2},\cdots,z_{n}$ are $n$ complex numbers
Prove that
$$\color{crimson}{\sum_{k=1}^{n}|1+z_{k}|+\dfrac{1}{n-1}\sum_{1\le i<j\le n}|1+z_{i}z_{j}|\ge\sum_{k=1}^{n}|z_{k}|}$$
Proof for $n=2$:
$$
\DeclareMathOperator{\Re}{Re}
| 1+x |+ | 1+y |+ | 1+xy |- | x |- | y |=
\frac{(| 1+x |+ | 1+y|+| 1+xy |)^{2}-(| x |+|y |)^{2}}{| 1+x|+ | 1+y|+|1+xy|+|x|+|y|}=\frac{A}{| 1+x|+|1+y |+ |1+xy |+ |x |+ |y|}$$where
$$A=( | 1+x |+ | 1+y |+ | 1+xy |)^{2}- ( | x |+ | y | )^{2}
\\=
| 1+x |^{2}+ | 1+y |^{2}+ | 1+xy |^{2}+2 | 1+x | | 1+y|+2 | 1+y | | 1+xy|+2 | 1+xy | | 1+x|- | x |^{2}- | y |^{2}-2 | x | | y |
\\=1+ | x |^{2}+2\Re(x)+1+ | y |^{2}+2\Re(y)+ 1+ |xy |^{2}+2\Re(xy)+2 | 1+x | | 1+y|+2 | 1+y | | 1+xy|+2 | 1+xy | | 1+x|- | x |^{2}- | y |^{2}-2 | x | | y |
\\=2\Re ( 1+x ) ( 1+y )+2 | 1+x | | 1+y|+ ( 1- | xy | )^{2}+ 2 | 1+y | | 1+xy|+2 | 1+xy | | 1+x|
\\ \geq2 | 1+y | | 1+xy|+2 | 1+xy | | 1+x|
\\ \Longrightarrow | 1+x |+ | 1+y |+ | 1+xy |\geq | x |+ | y |
$$
Is it true for a general $n$?
| For $z, w \in \Bbb C$ we have (and this is what you already did)
$$
\DeclareMathOperator{\Re}{Re}
\bigl( \lvert 1+z \rvert + \lvert 1+w \rvert + \lvert 1+zw \rvert \bigr)^2 \\
= 1 + \lvert z \rvert ^2 + 2 \Re z + 1 + \lvert w \rvert ^2 + 2 \Re w + 1 + \lvert zw \rvert ^2 + 2 \Re (zw) \\
+ 2 \lvert 1+z \rvert \lvert 1+w \rvert + 2 \bigl(\lvert 1+z \rvert + \lvert 1+w \rvert \bigr) \lvert 1+zw \rvert \\
= 2 \Re \bigl((1+z)(1+w) \bigr) + 2 \lvert 1+z \rvert \lvert 1+w \rvert \\
+ \bigl(\lvert z \rvert + \lvert w \rvert \bigr)^2
+ 2 \bigl(\lvert 1+z \rvert + \lvert 1+w \rvert \bigr) \lvert 1+zw \rvert \\
\ge \bigl(\lvert z \rvert + \lvert w \rvert \bigr)^2
+ 2 \bigl(\lvert 1+z \rvert + \lvert 1+w \rvert \bigr) \lvert 1+zw \rvert \\
\ge \bigl(\lvert z \rvert + \lvert w \rvert \bigr)^2
$$
and therefore
$$
\lvert 1+z \rvert + \lvert 1+w \rvert + \lvert 1+zw \rvert \ge \lvert z \rvert + \lvert w \rvert \, ,
$$
which proves the assertion for $n = 2$.
The general case $n \ge 2$ now follows from the case $n=2$:
If $z_1, \ldots z_n \in \Bbb C$ then for $1 \le j < k \le n$
$$
\lvert 1+z_j \rvert + \lvert 1+z_k \rvert + \lvert 1+z_j z_k \rvert \ge \lvert z_j \rvert + \lvert z_k \rvert \, .
$$
Adding all these inequalities gives
$$
(n-1) \sum_{j=1}^n \lvert 1+z_j \rvert + \sum_{1 \le j < k \le n} \lvert 1+z_j z_k \rvert
\ge (n-1) \sum_{j=1}^n \lvert z_j \rvert
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can we show that $ \sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4 ?$ We proposed this sum, but we are lacking in knowledge of this area of maths and we would ask if any of the authors would be willing to show us step by step how to go about proving this sum.
$$
\sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4
$$
| A quite boring approach. We can write your series as $$\left(\pi^{3}+1\right)\sum_{n\geq0}\frac{2^{n}n^{4}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}+\left(\pi^{3}+\pi^{2}+1\right)\sum_{n\geq0}\frac{2^{n}n^{3}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}
$$ $$+\left(-\pi^{3}+\pi^{2}-1\right)\sum_{n\geq0}\frac{2^{n}n^{2}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}-\pi\sum_{n\geq0}\frac{2^{n}n}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}.
$$ Now using the identity $$\sum_{n\geq0}\frac{4^{n}x^{2n+2}}{\left(2n+1\right)\dbinom{2n}{n}}=\arcsin^{2}\left(x\right)
$$ we have that, integrating both side and manipulating a bit, $$\sum_{n\geq0}\frac{2^{n}\left(\sqrt{2}x\right)^{2n}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=\frac{2\sqrt{1-x^{2}}\arcsin\left(x\right)-2x+x\arcsin^{2}\left(x\right)}{x^{3}}
$$ then if we differentiate $$-\pi\sum_{n\geq0}\frac{2^{n}n}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=-\frac{\pi}{2\sqrt{2}}\frac{d}{dx}\left.\frac{2\sqrt{1-x^{2}}\arcsin\left(x\right)-2x+x\arcsin^{2}\left(x\right)}{x^{3}}\right|_{x=1/\sqrt{2}}
$$ $$ =-\frac{\pi}{2\sqrt{2}}\left.\frac{-6\sqrt{1-x^{2}}\arcsin\left(x\right)+6x-2x\arcsin^{2}\left(x\right)}{x^{4}}\right|_{x=1/\sqrt{2}}=\frac{\pi^{3}}{8}+\frac{3\pi^{2}}{2}-6\pi.
$$ We can iterate this process and from $$\sum_{n\geq0}\frac{2^{n}n\left(\sqrt{2}x\right)^{2n}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=\frac{-3\sqrt{1-x^{2}}\arcsin\left(x\right)+3x-x\arcsin^{2}\left(x\right)}{x^{3}}
$$ we can differentiate again $$\sum_{n\geq0}\frac{2^{n}n^{2}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=\frac{1}{2\sqrt{2}}\frac{d}{dx}\left.\frac{-3\sqrt{1-x^{2}}\arcsin\left(x\right)+3x-x\arcsin^{2}\left(x\right)}{x^{3}}\right|_{x=1/\sqrt{2}}
$$ and so on. I'm too lazy to complete the calculations so I leave the details to a willing person.
| {
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"url": "https://math.stackexchange.com/questions/1765705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Evaluate the $\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$ Evaluate : $$\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$$
I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.
This is what I've done so far
\begin{align}
\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x}) &= \lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})\left(\frac{x-\sqrt{x^2 + 2x}}{x-\sqrt{x^2 + 2x}}\right)\\ \\
&= \lim_{x \to \ -\infty} \left(\frac{x^2 - (x^2 + 2x)}{x-\sqrt{x^2 + 2x}}\right)\\ \\
&= \lim_{x \to \ -\infty} \left(\frac{-2x}{x-\sqrt{x^2 + 2x}}\right)\\
\\
\end{align}
And that's as far as I've gotten. I've tried applying L'Hopitals Rule, but it still results in an indeterminate form.
Plugging it into WolframAlpha shows that the correct answer is $-1$
Any suggestions on what to do next?
| $$\lim _{ x\to -\infty } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x } } \right) =\lim _{ x\rightarrow -\infty }{ \left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x } \right) } } \right) =\lim _{ x\rightarrow -\infty }{ \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x } } } = } } \\ =\lim _{ x\rightarrow -\infty }{ \frac { -2x }{ x+x\sqrt { 1+\frac { 2 }{ x } } } = } \lim _{ x\rightarrow -\infty }{ \frac { -2x }{ x\left( 1+\sqrt { 1+\frac { 2 }{ x } } \right) } = } -1$$
| {
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Find LU decomposition of a matrix using partial pivoting I've the following matrix:
$$ A= \begin{bmatrix}
0& 7& 5& 1
\\ 4& 3& 2& 1
\\0 &0& 0& 1
\\ 0& 0& -1& -2 \end{bmatrix} $$
and I need to find the matrices $P, L$ and $U$, such that
$$PA = LU$$
what I need was to find the matrices representing the guassian elimination operations to obtain the upper triangle, and I found the following two:
$$l_2 l_2 A = U$$
specifically:
$$ l_1 =\begin{bmatrix}
0& 1& 0& 0
\\ 1& 0& 0& 0
\\ 0& 0& 1& 0
\\ 0& 0& 0& 1 \end{bmatrix} $$
$$ l_2 = \begin{bmatrix}
1& 0& 0& 0
\\ 0& 1& 0& 0
\\ 0& 0& 0& 1
\\ 0& 0& 1& 0
\end{bmatrix}
$$
such that
$$ l_2 l_1 A = PA = U $$
where
$$ U = \begin{pmatrix}
4 & 3& 2& 1
\\ 0 & 7& 5& 1
\\0 & 0& -1& -2
\\ 0 & 0& 0& 1
\end{pmatrix}
$$
So, what's the lower triangular matrix of this $LU$ decomposition? Is it the identity matrix?
| For $A= \begin{pmatrix}
0& 7& 5& 1
\\ 4& 3& 2& 1
\\0 &0& 0& 1
\\ 0& 0& -1& -2 \end{pmatrix} $, we have:
$$\text{ P = }\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right), \text{ L = }\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right), \text{ U = }\left( \begin{array}{cccc} 4 & 3 & 2 & 1 \\ 0 & 7 & 5 & 1 \\ 0 & 0 & -1 & -2 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$
You can easily verify that:
$$PA = LU$$
Next, we want to use this result to solve $Ax = b$ such that:
$$ \begin{pmatrix}
0& 7& 5& 1
\\ 4& 3& 2& 1
\\0 &0& 0& 1
\\ 0& 0& -1& -2 \end{pmatrix}x =
\begin{pmatrix}
26 \\ 9 \\1 \\ -3 \end{pmatrix}$$
The general process is:
*
*Construct $L, U, P$ as shown above
*Compute $Pb$
*Solve $Ly = Pb$ for $y$ using Forward Substitution
*Solve $Ux = y$ for $x$ using Back Substitution
So, forming $Ly = P b$ and using forward substitution gives:
$$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) y = \begin{pmatrix}
9 \\ 26 \\ -3 \\ 1 \end{pmatrix} \implies y = \begin{pmatrix}
9 \\ 26 \\ -3 \\ 1 \end{pmatrix}$$
Next, we form $Ux = y$ in order to be able to solve for $x$ using back substitution:
$$\begin{pmatrix}
4& 3& 2& 1 \\ 0& 7& 5& 1 \\ 0& 0& -1& -2 \\ 0 &0& 0& 1 \end{pmatrix} \begin{pmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix}
9 \\ 26 \\ -3 \\ 1 \end{pmatrix}$$
Using back substitution, this yields:
$$ x = \begin{pmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix}
-\dfrac{9}{14} \\ \dfrac{20}{7} \\ 1 \\ 1 \end{pmatrix}$$
| {
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Decompose $V \otimes V \otimes V$ into irreducible representations of $SL_2(\mathbb{R})$ Let $V=\mathbb{C^2}$ be the standard representation of $SL_2(\mathbb{R})$
Decompose $V \otimes V \otimes V$ into irreducible representations of $SL_2(\mathbb{R})$
I will just consider $SL_2(\mathbb{C})$ since there is a 1:1 correspondence between $SL_2(\mathbb{R})$ and $SL_2(\mathbb{C})$
Weight vectors of $V$ are $v_1$ and $v_{-1}$ with weights 1 and -1 respectively
Weight vectors of $V \otimes V \otimes V$:
Weight 3: $v_1 \otimes v_1 \otimes v_1$
Weight 1: $v_1 \otimes v_1 \otimes v_{-1}$, $v_1 \otimes v_{-1} \otimes v_1$ and $v_{-1} \otimes v_1 \otimes v_1$
Weight -1: $v_1 \otimes v_{-1} \otimes v_{-1}$, $v_{-1}\otimes v_{-1} \otimes v_1$ and $v_{-1} \otimes v_1 \otimes v_{-1}$
Weight -3: $v_{-1} \otimes v_{-1} \otimes v_{-1}$
How do I use this information to find the irreducible representations?
| One such decomposition is
$
A=\begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\cr -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\cr 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\end{pmatrix}$
$ B= \begin{pmatrix}0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\cr -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\cr 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1\cr 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\end{pmatrix}$
$C= \begin{pmatrix}0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\cr -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\cr 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\end{pmatrix}$
Then you have
$A.A=B.B=C.C= -I(8)$
| {
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Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$. Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$.
Since we know that the minimum and maximum values of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$
I applied this formula to get the minimum and maximum values of $(2\sqrt3+4)\sin x+4\cos x$ are $-\sqrt{(2\sqrt3+4)^2+(4)^2}$ and $\sqrt{(2\sqrt3+4)^2+(4)^2}$
$\sqrt{(2\sqrt3+4)^2+(4)^2}=\sqrt{12+16+16+16\sqrt3}=\sqrt{44+16\sqrt3}=2\sqrt{11+4\sqrt3}\neq 2(2+\sqrt5)$
I do not know where i am wrong or is there some other method possible?Please help.
| $$2\sqrt{11+4\sqrt{3}} \approx 8.46834180469$$
$$2(2+\sqrt(5))\approx 8.472135955$$
$$2(2+\sqrt(5)) >2\sqrt{11+4\sqrt{3}}$$
Alternatively, (ignoring the factor of 2, squaring, and then subtracting 9 from both sides):
$$4\sqrt{5} \overset{?}{>} 2+4\sqrt{3}$$
Square again
$$80 \overset{?}{>} 52 + 16\sqrt{3}$$
Subtract 52 from both sides
$$28 \overset{?}{>} 16\sqrt{3}$$
Square a third time:
$$784>768.$$
QED
| {
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$(f(1 - f(x)) = 1 - x^9$, $f(1) = 0$ and $f'(1) < 0$, then where is the real number $r$ such that $f(r) = r^{99}$? If $f(1 - f(x)) = 1 - x^9$, $f$: R $\to$ R is differentiable, $f(1) = 0$ and $f'(1) < 0$, how to show there is a real number $r$ such that $$f(r) = r^{99}?$$
Edit: Taylor Theorem makes no use. I try to take $a = 1 - {1 \over n}$, where n is also a real no. Then as $f'(1) < 0$, $f$ is decreasing at $x = 1$ and $f(a) > f(1) = 0$. By mean-value theorem, $$f(a) - f(1) = f'(a_0)(a - 1), a < a_0 < 1$$
Let $g(x) = f(x) - x^{99}$, then clearly $g(1) < 0$ and $g(r) = 0$.
How about showing $g(a) > 0$? Or to take $a$ as something else?
| After almost 4 years I finally think of these. Let me know anywhere wrong.
Firstly if $f(1 - f(x)) = 1 - x^9$, then let $a \in \mathbb{R}$ so that $f(a) = 1$, hence \begin{align} f(0) = f(1 - f(a)) = 1 - a^9. (1) \end{align}
Also we rearrange the terms in expressing $f(1 - f(x))$ to be $f(x) = 1 - f^{-1}(1 - x^9)$, then \begin{align} f(0) = 1 - f^{-1}(1 - 0) = 1 - a. (2) \end{align}
Combining $(1)$ & $(2)$, we have $a^9 - a = 0$, which $a = 1$ (rejected, or $f(1)$ has $2$ values), $0$ or $-1$ (rejected).
If $a = -1$, i.e. $f(-1) = 1$, then $f(1 - f(-1)) = f(0) = 1 - (-1)^9 = 2$. If we let $g(x)$ as I tried above, $g(0) = 2 > 0$. However, by differentiating $f(1 - f(x))$ against $x$, we have $f'(1 - f(x))\cdot -f'(x) = -9x^8$.
\begin{align} \Rightarrow f'(1 - f(-1)) \cdot -f'(-1) = f'(0)f'(-1) &= -9; \\ f'(1 - f(0)) \cdot -f'(0) = f'(-1)f'(0) &= 0 \end{align}
produces contradiction, so we have to reject this possibility.
If $a = 0$, i.e. $f(0) = 1$, then $g(0) = 1 > 0$ as well without the confusion on (the product of) the derivatives. Now the mean-value theorem becomes useful that
*
*differentiabilty of $f$ (so does of $g$) guarantees continuity on at least $[0, 1]$.
*$g(0) = 1 > 0 > -1 = g(1)$.
So $\exists \text{ } r \in [0, 1] \Rightarrow g(r) = 0$, i.e. $f(r) = r^{99}$. $\Box$
| {
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"url": "https://math.stackexchange.com/questions/1769898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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limit of the function of two variables ?? Here is the function :
\begin{equation}
f(x,y)=a^2 \left(\frac{x}{a^2-3 x^2}-\frac{y}{3 y^2-a^2}\right).
\end{equation}
We know that $y\geq x \geq 0$, $~a$ is a constant and $a^2=x^2+xy+y^2$.
I want to derive the limit of the function in the case $x\rightarrow a/\sqrt{3}=r_0$ and $y\rightarrow a/\sqrt{3}=r_0$.
One method is to take $x=r_0-\delta$ and $y=r_0+\delta$ and $a=\sqrt{3}r_0$. Then substituting them into $f(x,y)$, I can obtain
\begin{equation}
f(r_0,\delta)=-\frac{2 r_0^3}{4 r_0^2-\delta ^2}.
\end{equation}
Thus in the limit of $\delta \rightarrow 0$, we can get the result $-r_0/2$.
The another method is to substitute $a^2=x^2+xy+y^2$ into $f(x,y)$ directly, I get
\begin{equation}
-\frac{(x+y) \left(x^2+x y+y^2\right)}{(2 x+y) (x+2 y)},
\end{equation}
then set $x=y=r_0$. I get the limit is $-2r_0/3$.
Obviously, the two results are different. So which method is reliable?
| The first method makes no sense: once you decide that $x=r_0-\delta$, $y=r_0+\delta$, you have
$$
x^2+xy+y^2=(r_0+\delta)^2+(r_0-\delta)(r_0+\delta)+(r_0-\delta)^2=3r_0^2+\delta^2=a^2+\delta
$$
and your condition $x^2+xy+y^2=a^2$ is not satisfied.
In the second method you are not showing your work so I cannot comment on it, but you are getting the right result.
| {
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$a,b,c >0$, prove $\sqrt[2]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{a+b}} \geqslant \frac{7}{12} \cdot2^{\frac67} \cdot 3^{\frac47}$ $a,b,c >0$, prove
$$\sqrt[2]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{a+b}} \geqslant \frac{7}{12} \cdot2^{\frac67} \cdot 3^{\frac47}$$
What I tried:
1) It seems like a Nesbitt inequality. But I tried to imitate some Nesbitt proofs but was not successful.
2) Update: Following Frank's suggestion: the inequality is homogeneous, and can assume $a+b+c=1$ to reduce variables.
3) Test on Excel a lot of pairs $(a,b,c)$ and convinced myself the inequality is true and can be achievable.
4) Try Bernoulli's inequality but failed miserably.
Edit by Andreas, 2018-06-15: Sorry, but this is a wrong claim (counterexamples can be -and were - given), however it is a very interesting task. What should really be proved is $$\sqrt[2]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{a+b}} \geqslant 3 \cdot (\frac12)^{\frac23}$$ where the RHS is lower than the claimed $\frac{7}{12} \cdot2^{\frac67} \cdot 3^{\frac47}$, and the RHS is tight, i.e. values of $(a,b,c)$ can be given such that the RHS is indeed an infimum.
| A solution will be given by first showing that two of the arguments of the roots are always less than 1, and that for any fixed sum $s = a+b$ (or of any other two variables), the minimum of the LHS is obtained if one variable is taken to be zero. After that, the LHS can be minimized.
It shows (following the comment by fedja) that $\underline{\rm{the \; OP's \;claim \;is \;actually \;false.}}$ A correct lower bound will be given which is $3 \cdot (1/2)^{2/3} \simeq 1.8899$.
Let us give names to the three arguments of the roots:
$$f = \frac{a}{b+c}\\
g = \frac{b}{c+a}
\\
h = \frac{c}{a+b}
$$
Observe that the sum expression is homogenous in $(a,b,c)$. So we can demand $a+b+c=6$.
(The number 6 was taken to avoid fractions later on.)
This gives for the three arguments:
$$f(a) = -1 +\frac{6}{6-a}\\
g(b) = -1 +\frac{6}{6-b}\\
h(c) = -1 +\frac{6}{6-c}\\
$$
Any argument function rises with its variable, i.e. $f(a)$ rises with $a$ etc.
All three argument functions are equal ($= 1/2$) for $a=b=c=2$. The argument functions obtain the value $f=1$ for $a=3$, etc. This means that we will always have 2 or 3 argument functions which are less than 1. For if we had at least two argument functions which are greater than 1, we would have $a+b >6$ which is a contradiction.
From the discussion above, for any given $(a,b,c)$, let us choose the two argument functions (for the sake of the discussion, say, $f$ and $g$)) which have the smallest values (which are less than 1). Then we have $a+b = s \le 4$. For if $s >4$, these were not the two argument functions which have the smallest values, as $c <2$.
Now for any fixed $s\le 4$, we have that $\sqrt{f(a)}+\sqrt[3]{g(s-a)}$ takes the smallest value (infimum) at the boundary, i.e. either at $a=0$ or $a = s$, i.e. $b=0$ (likewise for a pair of other roots). It takes a lengthy discussion on values and derivatives to show this. Notably, the statement is false for some $s>4$.
The following picture illustrates the situation for $s = 3$ (lower curve) and $s = 4$ (upper curve):
This means that, for a lower limit, one variable out of $(a,b,c)$ will be zero, and the corresponding root will be zero. All the dicussions above did not specify yet which one. So we are left with the three cases of choosing that variable.
When taking $a=0$, one obtains:
$$
\sqrt[2]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{a+b}} \\
\geqslant \sqrt[3]{\frac{b}{c}}+\sqrt[4]{\frac{c}{b}} = x^{1/3} + x^{-1/4}
$$
where we introduced $x = \frac{b}{c}$. It is now easy to find the minimum w.r.t. $x$, resulting in $x = (3/4)^{12/7}$. So
$$
\sqrt[2]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{a+b}} \\
\geqslant \left(\frac{4}{3}\right)^{3/7}+\left(\frac{3}{4}\right)^{4/7}=\left(\frac{3}{4}\right)^{4/7} \left(1+\frac{4}{3} \right)=\frac{7}{12} \cdot2^{\frac67} \cdot 3^{\frac47} \simeq 1.9796
$$
which is the lower bound conjectured by the OP. However, this is not the best option. As indicated by fedja, taking $b=0$ will give a lower bound (which, again, cannot be improved by taking $c=0$). fedja may post his own solution on that, I do not want to steal it. The lower bound (infimum) is then, with the same method as described above, $3 \cdot (1/2)^{2/3} \simeq 1.8899$. This is obtained when $a = c \cdot (1/2)^{4/3}$.
As a "fun fact", compare this to the Nesbitt limit which is obtained for $a=b=c$. The LHS value is then $\sqrt[2]{\frac12}+\sqrt[3]{\frac12}+\sqrt[4]{\frac12} \simeq 2.3417$ so this is far from the true minimum.
This completes the proof. $\quad \quad \Box$
| {
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How to prove $\cos^4x - \sin^4x - \cos^2x + \sin^2x$ is always $0$? So I have a small problem here where I have to prove the following :
$$\cos^4x - \sin^4x - \cos^2x + \sin^2x = 0 $$
I know that the 2nd part is always $1$, so I need to prove that the first part also equals $1$. So how should I prove it ?
Edit : Sorry, the equation itself was wrong. I've edited it.
| The edited identity is correct.
Use $a^2 - b^2 = (a+b)(a-b)$ to rewrite $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(cos^2 x + \sin^2 x)$.
The second factor is one ($\sin^2 x + \cos^2 x = 1$), reducing the original expression to:
$\cos^2x - \sin^2x - \cos^2x + \sin^2x = 0$ as required.
| {
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Integral involving inverse of a function If $f(x)=x^3-x^2+x$, evaluate $$\lim_{n \to \infty} \int _{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)}$$ I substituted $x=f(t)$ but was unable to convert it back to $x$ and I think there would be a better approach. Some hints please. Thanks.
| For any $x>1$ we have $\left(x-\frac{1}{3}\right)^3 \leq f(x) \leq x^3 $, hence $f^{-1}(2n)-f^{-1}(n)$ is $O(\sqrt[3]{n})$ for large $n$ and $\frac{f^{-1}(2n)}{f^{-1}(n)}\to \sqrt[3]{2}$ as $n\to +\infty$
Moreover,
$$ \int \frac{3x^2-2x+1}{x^3+x}\,dx = C-2\arctan(x)+\log(x)+\log(1+x^2)\tag{1}$$
hence we may deal with
$$ \int_{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} = \int_{f^{-1}(n)}^{f^{-1}(2n)}\frac{3x^2-2x+1}{x^3+x}\,dx \tag{2}$$
through the identities:
$$ \arctan(Cx)-\arctan(x) = \left(1-\frac{1}{C}\right)\frac{1}{x}+O\left(\frac{1}{x^2}\right), $$
$$ \log(Cx)-\log(x) = \log C, $$
$$ \log(1+C^2 x^2)-\log(1+x^2) = 2\log C+O\left(\frac{1}{x^2}\right)\tag{3}$$
with $C=\sqrt[3]{2}$. It follows that:
$$ \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} = 3\log C = \color{red}{\log 2}.\tag{4}$$
A simpler approach comes from noticing that $f^{-1}(x)^3+f^{-1}(x)=x+f^{-1}(x)^2$.
Since $f^{-1}(x)^2 = o(n)$ for any $x\in[n,2n]$,
$$ \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} = \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{x+f^{-1}(x)^2} = \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{x}=\color{red}{\log 2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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Fractions in Questions and Answers
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