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Find the range of $A$ if $A=\sin^{20}x+\cos^{48}x$
Find the range of $A$ if $$A=\sin^{20}x+\cos^{48}x$$
$$
A'=20\sin^{19}x\cos x-48\cos^{47}x\sin x=0\implies5\sin^{19}x\cos x=12\cos^{47}x\sin x\\
\implies5\sin^{18}x=12\cos^{46}x
$$
How do I proceed further and prove that $A\in(0,1]$ ?
Is it possible to find the range of $A$ without using differentiation ?
Note: $\sin^2 x,\cos^2 x\in[0,1]\implies A\in[0,2]$ but $2$ is not "the" maximum value of $A$
|
Apart from the trivial upper bound $A\le 2$, we have the stronger (and sharp - try $x=0$) bound
$$ \tag1A\le 1.$$
Consider $f(x):=(1-x)^{10}+x^{24}$ for $0\le x\le 1$. Then $f'(x)=24x^{23}-10(1-x)^9$ is strictly increasing (as each summand is) on $[0,1]$, hence has at most one root there. As $f'(0)=-10$ and $f'(1)=24$, we conclude that there is exactly one such root $\alpha$. As $f'$ goes from negative to positive, $f$ must have a local minimum there. We conclude that $f$ has its only minimum at $\alpha$ and its maximum at the boundary - in fact, at both ends of the boundary since $f(0)=f(1)$. As $A=f(\cos^2 x)$ and $\cos^2 x$ ranges from $0$ to $1$, inclusive, we conclude that the maximal value of $A$ is also $1$ (thus proving $(1)$), and the minimum value of $A$ is $f(\alpha)$.
Using $(1-\alpha)^9=\frac{12}5\alpha^{23}$, we have
$$ f(\alpha)=(1-\alpha)^{10}+\alpha^{24}=(1-\alpha)\cdot\frac{12}5\alpha^{23}+\alpha^{24}=\alpha^{23}\cdot \frac{12-7\alpha}5=(1-\alpha)^9\cdot \frac{12-7\alpha}{12},$$
so certainly $$\min A=\min f>0,$$ but not by much.
From $f'(\frac 35)=24\frac{3^{23}}{5^{23}}-10\frac{2^{9}}{5^{9}}=\frac{24\cdot 3^{23}-10\cdot 2^95^{14}}{5^{23}}<0$(!), we conclude that $\alpha>\frac35$ and hence
$$ f(\alpha)=(1-\alpha)^9\cdot \frac{12-7\alpha}{12}<(1-\tfrac35)^9\cdot \frac{12-7\cdot\frac35}{12}=\frac{1664}{9765625}\approx 0.00017$$
(whereas the true minimal value is $\approx 0.000058575$)
|
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"timestamp": "2023-03-29T00:00:00",
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|
find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
|
Let $f(x)=x^4+x^3−1$ and $F(x)=x^6+x^4+x^3−x^2−1$. One has
$$F(x)=(x^2-x+2)f(x)-x^3-x+1\\F(x)=0\iff f(x)=\frac{-x^3-x+1}{x^2-x+2}$$
We look at the values for which $$\frac{-x^3-x+1}{x^2-x+2}=x^4+x^3-1$$
The problem suggests that these values are a simple function of $a$ and $b$. Proving with $a + b$ and with $ab$, this second value is good. In fact
$$\frac{a^3b^3+ab-1}{a^2b^2-ab+2}=a^4b^4+a^3b^3-1$$ it is equivalent to
$$(ab)^6+(ab)^4+(ab^3-(ab)^2-1=0$$
Then $F(ab)=0$ which (because of $ab(\dfrac{-1}{ab})=-1)$ suggests that $\dfrac{-1}{ab}$ could be another root. It is actually as we can verify.
|
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|
Calculate side lenght of triangle from two rectangle on top of each other I want to calculate the side lenght of $b$. I have two rectangles with one at 0° (screen) and I have one rectangle at 20° (turned image). With respect to the middle point. Both rectangles have a height of 6 and a width of 8.
Beceause the image is rotated, there will be black triangles in the corners of the screen. Now I want to calculate the lenght of $B$, as showed on the drawining. How can I do that?
|
Using the center where you've drawn the dot as the origin, and letting $c \approx .940$ and $s\approx .342$ denote the cosine and sine of $20$ degrees, respectively, the equation of the top horizontal line is
$$
y = 3,
$$
and the equation of the tilted top line is
$$
\pmatrix{-s \\ c} \cdot \pmatrix{x\\y} = 3,
$$
which is
$$
-sx + cy - 3 = 0.
$$
To compute the intersection, we plug in $y = 3$ into the second equation, getting
$$
-sx + 3c - 3 = 0 \\
3(c-1) = sx \\
\frac{3(c-1)}{s} = x
$$
so that
$$
(x, y) \approx (-0.530, 3)
$$
is the intersection point on the top edge.
Doing the same for the left edge, whose equation is
$$
x = -4,
$$
we get
$$
4s + cy - 3 = 0 \\
cy = 3 - 4s \\
y = \frac{3 - 4s}{c} \approx \frac{3 - 4\cdot .342}{.940} \approx 1.737
$$
so that the left-hand intersection point is at
$$
(x, y) \approx (-4, 1.737).
$$
In short form:
Assuming inches, the top intersection is about .53 inches to the left of the middle of the top of the card; the left-hand intersection is about $1.74$ inches above the middle of the left edge of the card.
Now..about that length that you asked for: it's the distance between the two points. So we need to compute
\begin{align}
d
&= \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \\
&= \sqrt{ (-0.53 - (-4))^2 + (3 - 1.737)^2 } \\
&\approx 3.693
\end{align}
If you want an exact formula, that'd be
\begin{align}
d
&= \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \\
&= \sqrt{ (\frac{3(c-1)}{s} - (-4))^2 + (3 - \frac{3 - 4s}{c})^2 } \\
&= \sqrt{ (\frac{3(c-1)}{s} + 4)^2 + (\frac{3c}{c} - \frac{3 - 4s}{c})^2 }\\
&= \sqrt{ (\frac{3(c-1)}{s} + \frac{4s}{s})^2 + (\frac{3c}{c} - \frac{3 - 4s}{c})^2 } \\
&= \sqrt{ (\frac{3(c-1) + 4s}{s})^2 + (\frac{3c - 3 + 4s}{c} )^2 } \\
&= \sqrt{ (\frac{3c- 3 + 4s}{s})^2 + (\frac{3c - 3 + 4s}{c} )^2 } \\
&= \sqrt{ \frac{(3c- 3 + 4s)^2}{s^2} + \frac{(3c - 3 + 4s)^2}{c^2} } \\
&= \sqrt{ \frac{(3c- 3 + 4s)^2 c^2}{s^2 c^2} + \frac{(3c - 3 + 4s)^2s^2}{s^2c^2} } \\
&= \sqrt{ \frac{(3c- 3 + 4s)^2 (c^2+s^2)}{s^2 c^2}} \\
&= \sqrt{ \frac{(3c- 3 + 4s)^2}{s^2 c^2}} \\
&= \frac{|3c- 3 + 4s|}{s c}
\end{align}
Plugging that into an online calculator, I get $d \approx 3.69378132923$.
|
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|
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n+1} - a_n = \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}$.
Since $\lim (a_{n+1} - a_n) = \lim S_n - a_1$, suffice to compute $\lim_{n\to\infty}(a_{n+1} - a_n)$.
$$
\begin{aligned}
\lim_{n \to \infty} (\frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2}) &=\lim_{n \to \infty} (\frac{1}{4n+1}) + \lim_{n \to \infty} (\frac{1}{4n+3}) - \lim_{n \to \infty} (\frac{1}{2n+2}) \\
&= \frac{5}{6}
\end{aligned}
$$
And $a_1 = S_3 = 5/6$, thus $\lim S_n = 5/3$. Am I right?
|
It seems to me the pattern is you are adding the following triplets: $\frac 1 {4k+1} +\frac 1 {4k+3}-\frac 1 {2^{k+1}} $.
If this converged we could rearrange the terms. The infinite sum of $\sum\frac {-1} {2^{k+1}}$ is $-1$ which is finite so that would mean the sum $\sum (\frac 1 {4k+1}+\frac 1 {4k+3})= \sum \frac 1 {2k+1} $ is finite.
But it's not because it is harmonic.
So the sum does not converge.
Unless the pattern is something else.
|
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|
Finding dependencies such that $0 > \frac{2b^2r^2}{z}-\left(2r ^2-2br\sqrt{1-\frac{b^2}{z^2}}\right)z$ I'm trying to solve an inequality with 3 variables.
$$0 > \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Basically, I want to know under which dependencies the formula is less than zero.
I tried to transform it in many ways, but it seems I cannot get a nice result. Especially the root seems to make problems:
$$2r^2z - 2brz \sqrt{1 - \frac{b^2}{z^2}} > \frac{2b^2r^2}{z} \tag{1}$$
$$rz - bz \sqrt{1 - \frac{b^2}{z^2}} > \frac{b^2r}{z} \tag{2}$$
$$r - b \sqrt{1 - \frac{b^2}{z^2}} > \frac{b^2r}{z^2} \tag{3}$$
I know that all variables are > 0.
So: $$r > 0 \qquad b > 0 \qquad z > 0$$ I also know that $$b \leq z$$
Do you have a hint, what I can try?
Do you think it is possible to calculate a nice solution, in which relation the 3 variables have to be, such that the formula is negative?
Thank you very much.
|
Note that $b=z$ violates the strict inequality (the right-hand side becomes zero). Consequently, we have $0<b < z$, which allows us to write
$$b=z \sin\theta \tag{1}$$
for some $0^\circ < \theta < 90^\circ$. Then the square root reduces immediately to $\cos\theta$, and your inequality simplifies to
$$0 > 2 r^2 z \sin^2\theta - z \left( 2 r^2 - 2 r z \sin\theta\cos\theta \right)
\quad\to\quad 0 > 2 r z \cos\theta\;(z\sin\theta - r \cos\theta) \tag{2}$$
Now, since $z$, $r$, $\cos\theta$ are strictly-positive, $(2)$ implies $0 > z \sin\theta - r\cos\theta$, so that
$$r > z \tan\theta \tag{3}$$
|
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|
How to find the number of real roots for a polynomial? How can you find of real roots for $P(x) = x^4 - 4x^3 + 4x^2 - 10$?
Using the Descartes' rule of signs:
The polynomial $P(x) = x^4 - 4x^3 + 4x^2 - 10$ has three sign changes between the first, second, third and fourth terms (the sequence of pairs of successive signs is +, −, +, -), therefore it has exactly three positive roots.
The polynomial $P(-x) = x^4 + 4x^3 + 4x^2 - 10$ has one sign change between the third and fourth terms (the sequence of pairs of successive signs is +++, -), therefore it has exactly one negative root.
Finally, summing up, the polynomial $P(x) = x^4 + 4x^3 + 4x^2 - 10$ has four real roots.
However, the actual answer is two real roots. How can you find this answer and what is wrong with my implementation of the Descartes' rule of signs?
|
Let $\>x=t+1\>$ then
$$
\\P(x)=P(t+1)=(t+1)^4-4(t+1)^3+4(t+1)^2-10=
\\=(t^4+4t^3+6t^2+4t+1)-(4t^3+12t^2+12t+4)+(4t^2+8t+4)-10=
\\=t^4-2t^2-9
\\P(t+1)=0=>t^4-2t^2-9=0=>
$$
if $\>x\>$ is real then $\>t\>$ is real =>
$$
\\t^2=\dfrac{2+\sqrt{40}}{2}=1+\sqrt{10}
$$
Answer $\>x=1+\sqrt{1+\sqrt{10}}\>$ or $\>x=1-\sqrt{1+\sqrt{10}}\>$
|
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|
How to prove Fibonacci recurrence holds mod p? Let $$J_n \equiv c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n \Big) \ \text{(mod p)},$$ with $c$ and $c^{-1}$ integers such that $c^2 \equiv 5 \ \text{(mod p)}$ and $cc^{-1} \equiv 1 \ \text{(mod p)}$. And $c$ is an odd integer.
It is easy to check that $J_1 \equiv J_2 \equiv 1 \ \text{(mod p)}$. However, I can't prove that $J_{n} \equiv J_{n-1} + J_{n-2} \ \text{(mod p)}$.
My attempt: $$J_{n-1} + J_{n-2} = c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^{n-2} - \Big( \dfrac{1-c}{2} \Big)^{n-2}\Big) \Big( \dfrac{6+2c}{4} \Big) + k_3 p = c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n\Big) + c^{-1} \Big( \dfrac{1+c}{2} \Big)^{n-2} \dfrac{k_1p}{4} - c^{-1} \Big( \dfrac{1-c}{2} \Big)^{n-2} \Big(1+ \dfrac{k_2p}{4} \Big) + k_3 p$$ which after a long try still I can't reduce it to the desired result because of factor 4 in the denominator.
PS I always keep $k_ip$'s to avoid mistakes working in mod p.
Please help!
Edit. If $\dfrac{c^{-1}}{2^2} ((1+c)^{n-2} - (1-c)^{n-2})$ were an integer for any $n$ then we are done! But that also I couldn't prove.
|
$$J_n \equiv c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n \Big) \ \text{(mod p)}$$
In order to show that $$J_n = J_{n-1}+J_{n-2}$$ it suffices to show that both $ (\frac {1+c}{2})^n$ and $ (\frac{1-c}{2})^n$ satisfy the same relation.
Note that for $ (\frac {1+c}{2})^n =(\frac{1+c}{2})^{n-1} +(\frac{1+c}{2})^{n-2} $ it suffices to show that $\frac{1+c}{2}$ satisfies $$x^2-x-1=0$$
Solving for $c$ we get $c=\sqrt 5$ which solves the problem.
|
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|
Finding $B^{225}$ without many computations I have that $B$ is a 4x4 matrix. $B-5I=\begin{pmatrix} -2 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 3 & 0 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The question asks to find $B^{225}$ without performing many computations, leaving the answer as a product of 3 matrices (univerted matrices are acceptable).
My thinking is that this problem may have to do with diagonalizing $B$ based on eigenvectors. $\lambda=5$ is clearly an eigenvalue because $det(B-5I)=0$, and I also reasoned that possible eigenvectors could be $\begin{pmatrix} 3 \\ 0 \\ -2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ 0 \\ 2 \\ 0 \end{pmatrix}$, just based on the way the matrix multiplication would work to yield linear combinations of these eigenvectors. So for diagonalization we need to bases $\alpha=...$ and $\beta=\left\{\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right\}$. I believe these eigenvectors should belong in $\alpha$ but I don't know which other eigenvectors belong or how to find them.
Even beyond that point, I don't quite understand the meaning of this question and how we are supposed to make the logical leap to $B^{225}$. What am I missing?
|
Let $C = B - 5I$. Notice it can be written as a outer product of two column vectors (or matrix product between a column and a row vector):
$$C = B - 5 I = u \otimes v = u v^T\quad\text{ where }\quad u = \begin{bmatrix}-2\\ 0 \\ 3 \\ 0\end{bmatrix}\quad\text{ and }\quad v = \begin{bmatrix}1 \\ 0 \\ -1 \\0\end{bmatrix}$$
This leads to
$$C^2 = (u v^T)(u v^T) = u(v^Tu) v^T = u((-2)(1) + (3)(-1))v^T = -5uv^T = -5C$$
As a result,
$$B^2 = (5I+C)^2 = 25I +10C + C^2 = 25I + 5C = 5B$$
Start from this, we can use induction to prove $B^n = 5^{n-1}B$ for all $n > 1$.
In particular, we have
$$B^{225} = 5^{224}B = 5^{224}\begin{bmatrix}
3 & 0 & 2 & 0\\
0 & 5 & 0 & 0\\
3 & 0 & 2 & 0\\
0 & 0 & 0 & 5\end{bmatrix}$$
|
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|
How can one calculate $342342^{1001}$ mod $5$? How can one calculate $342343^2$ mod $3$? I know that the answer is $1$.
And $342342^{1001}$ mod $5$.
I know that
$
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
$
So 1001 = 250 + 250 + 250 + 250 + 1, which is why the answer is also 1?
|
This is an observation but you can take it as an answer.
Let $n\in\mathbb{N}$ then $3|n\iff$ $3$ divides the sum of all the digits of $n$. Now consider $n=342343$ and sum of all of its digits $=19$. So clearly $n\equiv-2 (\mod 3)\implies n^2\equiv1(\mod 3)$ since $4\equiv 1(\mod 3)$.
For the second part observe this, $2^4\equiv 1(\mod 5)\implies 2^{250\times4}\equiv 1(\mod 5)\implies 2^{250\times 4+1}\equiv 2(\mod 5)\implies 2^{1001}\equiv 2(\mod 5)$
Now $342342\equiv 2(\mod 5)\implies 342342^{1001}\equiv 2^{1001}(\mod 5)\equiv 2(\mod 5)$
|
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|
Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have no clue on how to approach the problem. Most surely not by direct substitution, but I'm not seeing how. Some hint will be appreciated.
|
Let $\alpha = \sqrt[3]{\frac 19}$ and $\beta = 1-\sqrt[3]2+\sqrt[3]{2^2}$, so that $c = \alpha\beta$. Define $$Q(x) = P(\alpha x) = \frac{x^3}9+\frac{\sqrt[3]2}3 x^2-1$$ and note that $P(c) = 0$ iff $Q(\beta) = 0$.
Now, notice that $(1+\sqrt[3]2)(1-\sqrt[3]2+\sqrt[3]{2^2}) = 1^3+\sqrt[3]2^3 = 3$, so $\beta = \frac{3}{1+\sqrt[3]2}$.
Finally, substitute that into $Q(x)$:
\begin{align}
Q(\beta) &= \frac 19 \cdot\frac {27}{(1+{\sqrt[3]2})^3} + \frac{\sqrt[3]2}3\cdot\frac 9{(1+{\sqrt[3]2})^2}-1 \\
&=\frac 1{(1+{\sqrt[3]2})^2}\left(\frac{3}{1+\sqrt[3]2} +3\sqrt[3]2 - (1+{\sqrt[3]2})^2 \right)\\
&= \frac 1{(1+{\sqrt[3]2})^2}\left(1-\sqrt[3]2+\sqrt[3]{2^2} +3\sqrt[3]2 - 1-2{\sqrt[3]2} -\sqrt[3]{2^2} \right) = 0.
\end{align}
|
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|
Algebraic manipulation with indices The question is:
For a>0 and $\sqrt{a}+\frac{1}{\sqrt{a}}=3$, find the value of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$
So I first squared the given equation and got:
$$a+\frac{1}{a}+2=9$$
$$a+\frac{1}{a}=7$$
Then to get the form of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$:
$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$$
And I just got stuck right here because I didn't really know what to do with the $\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$. So I looked into the solutions and apparently it's
$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\sqrt{a}+\frac{1}{\sqrt{a}}$$
I'm not sure how those two are equal...
|
$\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}=\sqrt{a}+\frac{1}{\sqrt{a}}=3$ !
|
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|
Evaluate $\lim\limits_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$ Problem
Evaluate
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$$
Attempt
First, we may obtain
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right]=\lim_{x \to 0+}\frac{6e^{\sin x\ln x}-6e^{x\ln\sin x}+x^3\ln x}{6x^3}.$$
Here, you can apply L'Hôpital's rule, but it's too complicated. Moreover, you can also apply Taylor's formula, for example
$$e^{\sin x\ln x}=1+\sin x\ln x+\frac{1}{2}(\sin x\ln x)^2+\cdots,\\e^{x\ln\sin x}=1+x\ln\sin x+\frac{1}{2}(x\ln\sin x)^2+\cdots,$$
but you cannot cancel the terms, thus you cannot avoid differentiating, either. Is there any elegant solution?
P.S. Please don't suspect the existence of the limit. The result equals $\dfrac{1}{6}.$
|
The key point is that $x\log \sin x \to 0$ and $\sin x \log x \to 0$ then by Taylor's series we have
*
*$x^{\sin x}=e^{\sin x \log x}=1+x\log x+\frac12x^2\log^2 x+\frac16x^3\log x(\log^2 x -1)+O(x^4\log^2 x)$
*$(\sin x)^{x}=e^{x \log (\sin x)}=1+x\log x+\frac12x^2\log^2 x+\frac16x^3(\log^3 x -1)+O(x^4\log x)$
then
$$\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}=\frac{\frac16x^3\log^3 x-\frac16x^3\log x-\frac16x^3\log^3 x +\frac16x^3+O(x^4\log x)}{x^3}+\frac{\ln x}{6}=$$
$$=\frac16+O(x\log x) \to \frac16$$
To see how obtain the Taylor's expansion, let consider the first one, then since
*
*$\sin x =x-\frac16 x^3+O(x^5) \implies \sin x \log x=x\log x-\frac16 x^3\log x+O(x^5\log x)$
*$e^t = 1+t+\frac12 t^2+\frac16t^3+O(t^4)$
we obtain that
$$x^{\sin x}=e^{\sin x \log x}
=1+\left(x\log x-\frac16 x^3\log x+O(x^5\log x)\right)+\frac12\left(x\log x-\frac16 x^3\log x+O(x^5\log x)\right)^2+\frac16\left(x\log x-\frac16 x^3\log x+O(x^5\log x)\right)^3+O(x^5\log^4 x)=$$
$$=1+x\log x-\frac16 x^3\log x+\frac12x^2\log^2x-\frac16x^4\log^2x+\frac16x^3\log^3x+O(x^4\log^2x)=$$
$$=1+x\log x+\frac12x^2\log^2x+\frac16x^3\log x(\log^2x-1)+O(x^4\log^2x)$$
and for the second one since
*
*$\log (1+t)= t-\frac12t^2+\frac13t^3+O(t^4)$
*$\sin x =x-\frac16 x^3+O(x^5)\implies \frac{\sin x}x=1-\frac16 x^2+O(x^4)$
*$\log \sin x=\log x+\log \frac{\sin x}x=\log x+\log \left(1-\frac16 x^2+O(x^4)\right)=\log x-\frac16 x^2+O(x^4)$
*$x\log \sin x=x\log x-\frac16 x^3+O(x^5)$
we obtain that
$$(\sin x)^x=e^{x\log \sin x}=1+\left(x\log x-\frac16 x^3+O(x^5)\right)+\frac12\left(x\log x-\frac16 x^3+O(x^5)\right)^2+\frac16\left(x\log x-\frac16 x^3+O(x^5)\right)^3+O(x^4\log^4x)$$
$$=1+x\log x-\frac16 x^3+\frac12x^2\log^2x-\frac16x^4\log x+\frac16x^3\log^3 x+O(x^4\log x)=$$
$$=1+x\log x+\frac12x^2\log^2x+\frac16x^3(\log^3 x-1)+O(x^4\log x)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$
Evaluate $$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$$
My Solution
Denote
$$f(t):=\arctan t.$$
By Lagrange's Mean Value Theorem，we have
$$f\left(\frac{2x^2+5}{x^2+1}\right)-f\left(\frac{2x^2+7}{x^2+2}\right)=f'(\xi)\left(\frac{2x^2+5}{x^2+1}-\frac{2x^2+7}{x^2+2}\right)=\frac{3}{(1+\xi^2)(x^2+1)(x^2+2)}$$
where $$\frac{2x^2+5}{x^2+1}\lessgtr \xi \lessgtr \frac{2x^2+7}{x^2+2}.$$
Here, applying the Squeeze Theorem, it's easy to see$$\lim_{x \to \infty}\xi=2.$$
It follows that
$$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)=\lim_{x \to \infty}\frac{3x^4}{(1+\xi^2)(x^2+1)(x^2+2)}=\frac{3}{5}.$$
Hope to see other solutions.THX.
|
When $x\to \infty$,
$$\frac{2x^2+5}{x^2+1}=2+\underbrace{\frac{3}{x^2+1}}_{\to 0},\qquad \frac{2x^2+7}{x^2+2}=2+\underbrace{\frac{3}{x^2+2}}_{\to0},$$
and the Taylor approximation of $\arctan(2+t)$ around $0$ is given by
\begin{align*}
\arctan(2+t)=\arctan 2+(\arctan(t+2))'|_{t=0}t+o(t)=\arctan 2+\frac t5+o(t).
\end{align*}
Set $t=\dfrac{3}{x^2+1}$ and $t=\dfrac{3}{x^2+2}$, we have $o(t)=o(x^{-2})$ and
\begin{align*}
\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}&=\frac 35\underbrace{\left(\frac{1}{x^2+1}-\frac{1}{x^2+2}\right)}_{:=g(x)}+o(x^{-2}).\\
\end{align*}
Therefore,
\begin{align*}
L=\frac35\underbrace{\lim_{x\to\infty}x^4g(x)}_{=1}+\underbrace{\lim_{x\to\infty}\frac{o(x^{-2})}{x^{-4}}}_{=\lim\limits_{t\to0}\frac{o(t)}{t^2}=0}=\frac35.
\end{align*}
|
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|
Prove that $C_{3 \over 2}^n$ is bounded given $C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}$
Let:
$$
\begin{cases}
C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}\\
C_{a}^0 = 1
\end{cases}
$$
Prove $C_{3 \over 2}^n$ is bounded.
I've started with finding a reduced formula:
$$
C_{3\over 2}^n = \frac{{3\over 2}\left({3\over 2} - 1\right)\left({3\over 2} - 2\right)\left({3\over 2} - 3\right)\cdots\left({3\over 2} - n+1\right)}{n!} =\\
= \frac{3\cdot 1 \cdot(-1)\cdot(-3)\cdots(3-2n)}{2^n\cdot n!} = \\
\frac{1}{2^n \cdot n!}\prod_{k=1}^n\left(5-2k\right)
$$
This seems to converge to $0$ and therefore the sequence should be bounded, but how do i formally show that using inequalities? Moreover the elements change their sign depending on $n$ is even/odd (should i consider absolute values?).
Please note the precalculus tag. Thank you!
|
After some pondering i'm going to try to answer this myself.
I've been thinking about considering two cases for $n$ is odd and $n$ is even. Start with even $n$. We know that for any $n \ge 2 \implies C_{3/2}^n > 0$ in case $n$ is even. Lets inspect a subsequence for even indices:
$$
C_{3/2}^n = \frac{1}{2^nn!}\prod_{k=1}^n(5-2k) \\
C_{3/2}^{n+2} = \frac{1}{2^{n+2}(n+2)!}\prod_{k=1}^{n+2}(5-2k)
$$
Consider the fraction:
$$
\frac{C_{3/2}^{n+2}}{C_{3/2}^n} = \frac{2^nn!}{2^{n+2}(n+2)!}(5-2(n+1))(5-2(n+2)) = \\
=\frac{1}{2^2(n+1)(n+2)}\cdot(3-2n)(1-2n) = \frac{(2n-3)(2n-1)}{(2n+2)(2n+4)}
$$
Clearly this is less than $1$ and hence the sequence is decreasing towards $0^+$. With that being said the maximum is obtained for $n=1$ which is $3 \over 2$. Also note that for odd indices the value for $C_{3/2}^n < 0$.
Now consider the case for odd $n$:
$$
C_{3/2}^n = -\left|\frac{1}{2^nn!}\prod_{k=1}^n(5-2k)\right| \\
C_{3/2}^{n+2} = -\left|\frac{1}{2^{n+2}(n+2)!}\prod_{k=1}^{n+2}(5-2k)\right|
$$
Now using the results from the case for even $n$ we obtain that:
$$
\frac{|C_{3/2}^{n+2}|}{|C_{3/2}^{n}|} < 1
$$
At the same time $C_{3/2}^n$ for odd $n$ is less than $0$ hence for odd $n$:
$$
0 < C_{3/2}^{n} < C_{3/2}^{n+2}
$$
This means $C_{3/2}^n$ is increasing and tending to $0^-$. Thus the minimum value of $C_{3/2}^n$ is obtained at $n=3$
Compiling both cases gives us that:
$$
C_{3/2}^3< C_{3/2}^n < C_{3/2}^1
$$
or:
$$
−0.0625 < C_{3/2}^n < {3 \over 2}
$$
I would appreciate any comments on the given approach.
|
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|
Irrational equation $\sqrt{9-4x}=p-2x$
The equation
$$\sqrt{9-4x}=p-2x$$
has exactly 2 real and different solutions only if parameter $p$ belongs to which set?
So what I see here, to have the solutions be real in the first place,
$9-4x\ge0$
So $-4x\ge-9 \Rightarrow x\le{9\over4}$
$x\in\left(-\infty,{9\over4}\right]$
From here I quadrate both sides so I get 2 different possibilities:
1. For $p\ge2x$:
$9-4x=p^2-4x+4x^2\\9-4x-p^2+4x-4x^2=0\\-4x^2-p^2+9=0$
And for this to have 2 real and different solutions, $D>0$
$-4(-4)(-p^2+9)>0\\16(-p^2+9)>0\\-16p^2+144>0\\-16p^2>-144\\p^2<9\\|p|<3 \Rightarrow p\in(-3,3)$
2. For $p<2x:$
$9-4x=-p^2+4x-4x^2\\9-4x+p^2+4x^2-4x=0\\4x^2-8x+p^2+9=0$
Again $D>0$
$(-8)^2-4\cdot4(p^2+9)>0\\64-16p^2-144>0\\p^2<-5\\p\notin\Bbb R$
The problem here is I don't know how to use $x\in\left(-\infty,{9\over4}\right]$ in both cases, so I don't exactly know the solution to this problem. From the textbook I got this question from the result is $p\in\left[{9\over2},5\right)$
|
As already pointed out in a comment, $(p-2x)^2 = p^2-4px+4x^2$, which results in
$$4x^2 - 4(p-1)x+(p^2-9) = 0,$$
or
$$x = \frac{(p-1)\pm\sqrt{10-2p}}{2}.$$
For two real distinct roots, $10-2p > 0 \implies \color{red}{p < 5}$.
Also, $p \ge 2x$ implies $$(p-1)\pm\sqrt{10-2p} \le p \implies \sqrt{10-2p}\le 1 \implies \color{red}{p \ge \frac{9}{2}}.$$
Note that
|
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|
Probably of winning with 2 dice (maximum of them) against another one could some of you help me to find out what is the probability of A) obtain with two dice a greather number than another die? B) and if the dice are 3 how can I do? Not the sum of the 2 dice, but the greatest value of those 2 against another die
|
When rolling $1$ die, the probability of throwing "N or less" is $P(X \le N) =\frac{N}{6}$. When rolling $n$ dice, the probability of throwing "N or less" is $P_n(X\le N) = (\frac{N}{6})^n$.
So with $n$ dice, the probability of throwing "at least one N and nothing higher than N" is $P_n(N) = (\frac{N}{6})^n-(\frac{N-1}{6})^n$. This means the probability of someone with $n$ dice rolling a higher number than someone with $1$ die would be: $$P = P_n(6)\cdot P(X \le 5) + P_n(5)\cdot P(X \le 4)+ \ldots + P_n(2)\cdot P(X \le 1)$$
or
$$P =\left((\frac{6}{6})^n-(\frac{5}{6})^n\right)\frac{5}{6}+\left((\frac{5}{6})^n-(\frac{4}{6})^n\right)\frac{4}{6}+ \ldots$$
or $$P =\frac{1}{6^{n+1}} \left(6^n\cdot 5 - 5^n-4^n-3^n-2^n-1^n \right)$$ or
$$P= \frac{5}{6} - \frac{1}{6^{n+1}} \sum_{i=1}^5 i^n$$
For $n=1$, i.e two people with one die each, we get $P= \frac{15}{36}$, a known result. For $n=2$ we get $P= \frac{125}{216}$ and for $n=3$ we get $P= \frac{855}{1296}$.
EDIT
It makes sense that the probability tops out at $P=\frac{5}{6}$ as there is always $\frac{1}{6}$ probability that the person with $1$ die throws a $6$, in which case it doesn't matter how many dice the other person has as they will never beat the $6$.
|
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|
Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$
Prove that
$$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$
So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum_{k=0}^{n-1} \frac{(n-1)!n!}{(n-k-1)!k!(n-k-1)!(k+1)!}
= n^2 \sum_{k=0}^{n-1} \frac{(n-1)!^2}{(n-k-1)!^2k!^2(k+1)}
=n^2 \sum_{k=0}^{n-1} \binom{n-1}{k}^2\frac{1}{k+1}.
$$
I do not know what to do with $\frac{1}{k+1}$, how to get rid of that.
|
We present a slight variation using formal power series and the
coefficient-of operator. Starting from
$$\sum_{k=1}^n k {n\choose k}^2
= \sum_{k=1}^n k {n\choose k} [z^{n-k}] (1+z)^n
\\ = [z^n] (1+z)^n \sum_{k=1}^n k {n\choose k} z^k
= n [z^n] (1+z)^n \sum_{k=1}^n {n-1\choose k-1} z^k
\\ = n [z^n] z (1+z)^n \sum_{k=0}^{n-1} {n-1\choose k} z^k
= n [z^{n-1}] (1+z)^n (1+z)^{n-1}
\\ = n [z^{n-1}] (1+z)^{2n-1} = n \times {2n-1\choose n-1}$$
which is the claim.
|
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|
Inequality equation of a Triangle I have divided the Problem into two parts, a) and b):
a) Let $a$, $b$ and $c$ be the side lengths of a triangle with a perimeter of $4$.
I need to prove that
$a^2 + b^2 + c^2 + abc < 8$.
b) And is there a real number $d<8$ such that for each triangle with the perimeter $4$, the inequality equation
$a^2 + b^2 + c^2 + abc <d$
is valid?
I tried to use Heron's Formula somehow for the first inequality equation.
$s \, = \, \frac{a+b+c}{2}$
and
$F_{\triangle} = \sqrt{s(s-a)(s-b)(s-c)}$
Now we have:
$2s=4$ so that $s=2$
Thus:
$F_{\triangle} = \sqrt{2(2-a)(2-b)(2-c)}$
Maybe we can exchange $F_{\triangle}$ somehow?
Has anyone another approach or an idea how to continue with Heron's formula?
Thx
|
a) Because of $a+b+c=4$, you have
$${a^2}+{b^2}+{c^2}={(a+b+c)^2}-2(ab+bc+ac)=4(a+b+c)-2(ab+bc+ac)$$
Therefore
$${a^2}+{b^2}+{c^2}+abc=4(a+b+c)-2(ab+bc+ac)+abc=8-(2-a)(2-b)(2-c)$$
By triangle-inequality $b+c>a$ $$\Rightarrow4=a+b+c>a+a=2a\Rightarrow2>a$$
Analugously you can prove that $2>b$ and $2>c$.
Thus
$${a^2}+{b^2}+{c^2}+abc=8-(2-a)(2-b)(2-c)<8$$
b) This inequality for $d<8$ isn't necessarily valid. Let for instance $k=1-\frac{d}{8}>0$.
Let furthermore $$a=b=2-k>0 $$
Hence $${a^2}+{b^2}+{c^2}+abc>{a^2}+{b^2}=2{(2-k)^2}=8-8k+{k^2}>8-8k=d$$ which contradicts the condition $${a^2}+{b^2}+{c^2}+abc<d$$
|
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|
$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
|
Let $a=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}. . . .\frac{99}{100}$
and:
$b=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}. . . \frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$\frac{1}{2}<\frac{2}{3}, \frac{3}{4}<\frac{4}{5}. . . \frac{99}{100}<\frac{100}{101}$
⇒ $a^2 < ab=(\frac{1}{2}.\frac{2}{3}).(\frac{3}{4}.\frac{4}{5}). . . .(\frac{99}{100}.\frac{100}{101})=\frac{1}{101}$
$a^2<\frac{1}{101}$⇒$a<\frac{1}{\sqrt{101}}<\frac{1}{10}$
Also:
$2a=\frac{3}{4}.\frac{5}{6}. . . .\frac{99}{100}$
$\frac{3}{2}.b=\frac{4}{5}.\frac{6}{7}. . . \frac{100}{101}$
$2a<\frac{3}{2}b$⇒$2a^2<\frac{3}{2}ab=\frac{3}{2}.\frac{3}{101}$
Or $a^2<\frac{3}{4}ab=\frac{3}{4}.\frac{3}{101}$
Since $9>4$ then $a^2 >\frac{4}{9\times 101}$ and therefore:
$a>\frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=\frac{1}{101}$⇒$a^2>\frac{1}{202}$⇒$a>\frac{1}{\sqrt{202}}>\frac{1}{\sqrt{225}}=\frac{1}{15}$
|
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|
${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$ In a proof, the author states that it is clear that:
Given $x\geq 1$ and $ n-x \geq 1$ and finally also $n\geq 2$
$${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$$
This is not immediately clear to me. Of course, If I have $n$ objects and I split them up in $x$ and $n-x$ objects and then I choose to form pairs, amongst these
two subsets, I end up with fewer pairs than if I would consider the bigger set, so it is certainly smaller than $ n \choose 2$.
I just don't see how it is smaller than $ {n-1\choose 2}$.
|
\begin{align}
{x\choose2} +{n-x\choose2}
&=\frac{(x-1)x}2 +\frac{(n-x)(n-x-1)}2 \\
&=\tfrac12 (x^2-x +n^2-nx-n -nx+x^2+x) \\
&=\tfrac12 (2x^2 -2nx+n^2-n) \\
&=\tfrac12 (2x(x-n) +n(n-1))
\end{align}
Given $1 \leq x \leq n-1$ (and $x-n \leq -1$), then
\begin{align}
&\leq \tfrac12 (2(n-1)(-1) +n(n-1)) \\
&=\frac{(n-1)(n-2)}2 \color{blue}{\cdot \frac{(n-3)!}{(n-3)!}} \\
&=\frac{(n-1)!}{2!(n-3)!} \\
&={n-1\choose2}.
\end{align}
|
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|
Prove that number $4p^2+1$ can show as sum of squares of three different numbers Let $p>3$ is prime number. Prove that number $4p^2+1$ can show as sum of squares of three different numbers.
Only what I know that every prime number $p>3$ can show as $p=6k+1$ or $p=6k-1$, such that $k \in \mathbb Z$.
If I put $p=6k+1$, then $4(6k+1)^2+1=(12k)^2+(24k+3)^2-(24k+2)^2$, here I did not show what they want in task.
For $p=6k-1$ things do not change, do you have some idea?
|
Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$\therefore 144=a^2+b^2+c^2 \qquad$ and $\qquad 2b+4c=48$
By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$
Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2+(ck-2)^2=(a^2+b^2+c^2)k^2-(2b+4c)k+5$
$\therefore 144=a^2+b^2+c^2 \qquad$ and $\qquad -48=-(2b+4c)$
So, $4(6k-1)^2+1=(4k)^2+(8k-1)^2+(8k-2)^2$
|
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|
$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\
&= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}
But the answer is $\dfrac{1}{2}$ by L'Hopital's Rule.
|
Your problem arises from the fact that you used $\color{red}{\lim_\limits{x \to 0} \frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $\color{red}{\infty-\infty}$...
Only split an initial limit into a product if the individual limits are defined.
|
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|
What am I doing wrong finding the derivative of $\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$? $$y=\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$$
For convenience, let
$$A=\frac{3-x}{2}\sqrt{1-2x-x^2},$$
$$B=2\arcsin{\frac{1+x}{\sqrt{2}}}.$$
$$y'=A'+B'$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2})+(-2x)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2})+(-x)(\frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$
$$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$
$$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$
$$A'=\frac{2x^2-x-1}{{2\sqrt{1-2x-x^2}}}$$
$$B'=2\bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$
$$B'=2 \bigg( \frac{1+x}{\sqrt{2}} \bigg)' \bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$
$$B'=\sqrt{2} \bigg( \frac{1}{\sqrt{1- \frac{1+2x+x^2}{2}} } \bigg)$$
$$B'=\sqrt{\frac{4}{1-2x-x^2}}$$
$$B'=\frac{4}{2\sqrt{1-2x-x^2}}$$
$$y'=A'+B'=\frac{2x^2-x+3}{2\sqrt{1-2x-x^2}}$$
The answer in the book is
$$y'=\frac{x^2}{\sqrt{1-2x-x^2}}$$
|
You got A' wrong.
$$ A = \frac{3-x}{2}\cdot(\sqrt{1-2x-x^2}) = f\cdot g$$
where $ f =\frac{3-x}{2} $ and $ g = \sqrt{1-2x-x^2} $.
Looking at $$A'= f'g + g'f$$the first part
$$ f'g = (-\frac{1}{2})(\sqrt{1-2x-x^2})$$ is correct, but the second part is $$ g'f = (-2x -2)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$ rather than
$(-2x)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$ as you wrote.
|
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|
Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$
Evaluate : $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$
I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$ but still couldnt't get any answer as when I substitute numbers $(n)$ into the $Sn \left(-\frac{1}{6}+\frac{1}{2}-\frac{1}{8}+\frac{1}{4}-\frac{1}{10}\right)$ , $Sn$ just keeps going on and on. So how do I solve this problem and what does this series converge to?
|
$$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$
Taking $\frac{1}{2}$ common ,
$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$
Write the first few terms of the series as :
$\frac{1}{2}( 1 - \frac{1}{3} +\frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \cdot \cdot \cdot$
You can see that except $1$ and $\frac{1}{2}$ every terms get cancelled out and the $n$-th term tends to zero. So the result inside the bracket is $\frac{3}{2}$ . But there is a $\frac{1}{2}$ outside the bracket which needs to be multiplied. So the answer is $\frac{3}{4}$.
|
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|
How to find a matrix with given eigen values Find a $3\times 3$ matrix $B$ which contains $A$ as a sub-matrix and has eigen values $0$, $1$ only
where $$A=
\begin{bmatrix} -2& -12 \\1&5 \end{bmatrix}$$
I cant find a way how to construct $B$
I took the eigen values to be $0$, $0$, $1$ and then took $$B =
\begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&e\end{bmatrix}$$
$\implies 3+e=1\implies e=-2$
But how to take $a$, $b$, $c$, $d$?
Please help.
|
Let's start with your $$B = \begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&e\end{bmatrix}$$
Let's take the eigenvalues $0,1,1$. Then $\operatorname{Tr}(B)=3+e=2 \Rightarrow e=-1$. And the characteristic polynomial must be $\lambda(\lambda -1)^2=\lambda^3-2\lambda^2+\lambda$.
$$B = \begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&-1\end{bmatrix}$$
It's characteristical polynomial is:
$$\lambda^3-2\lambda^2-(ac+bd+1)\lambda - (-5ac+ad-12bc+2bd-2)$$
So:
$$\begin{cases}ac+bd+1=-1 \\ -5ac+ad-12bc+2bd-2=0\end{cases} \Rightarrow
\begin{cases}ac+bd=-2 \\ -5ac+ad-12bc+2bd=2\end{cases}$$
Let's try $c=0$ to get rid of the largest coefficients.
$$\begin{cases}bd=-2 \\ ad+2bd=2\end{cases}\Rightarrow
\begin{cases}bd=-2 \\ ad=6\end{cases}$$
Now pick for instance $a=3,b=-1, d=2$ to find:
$$B = \begin{bmatrix} -2& -12& 3\\1&5 &-1\\0&2&-1\end{bmatrix}$$
|
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|
How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}}$, and then observe $t$ is a root of $f(x) = x^3 + 3x - 36 = 0$. Also observe that $f(3) = 0$. Now $f(x)$ has only one real root, since $f'(x) = x^2+3$ has no real solution. So $t =3$, as desired.
But I couldn't find any direct/algebraic way of fiddling it for a while. I guess morever this is true: $\left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}+\left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}} = t$, which can possibly be proved in the same indirect way but I don't know how to prove this by direct manipulation.
One way maybe is to write $z_+ = \left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, $z_- = \left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, and writing $t = z_+ + z_-$, observe $z_+z_- = -1$, so $(x-z_+)(x-z_-) = x^2 - tx -1$, and then with Fundamental Theorem of Symmetric Polynomials, maybe calculating $(z_+^3 - z_-^3)^2$ and $(z_+^3 + z_-^3)$ in terms of $t$ would help ?
|
Let first root be $A$ and second be $B$, then we need to find the value of $A + B = x$
*
*$A^3 + B^3 = -36$
*$AB = -1$
Expand the cube: $A^3 + B^3 = (A+B)*(A^2 - AB + B^2) = (A+B)*((A+B)^2-3AB) = x*(x^2+3) = x^3 + 3x = -36$.
Easy guess $x = -3$
Another approach: assume that you have a cube under the root, then, say, you take the right equation where you factor the minus out:
$18 + 5\sqrt{13} = (a+b\sqrt{13})^3$
Then you gotta solve the system of equations made from matching the summands with the $\sqrt{13}$ and without:
*
*$a^3 + 39ab^2 = 18$
*$3a^2b + 13b^3 = 5$
Wolfram yields $a = 1.5$ and $b = 0.5$, by hand calculations led me nowhere
|
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|
Prove the sequence $a_n =\frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}$ has a limit I have several questions to ask:
1) Show increasing, find the upper bound if you can of
$\sqrt{(n^2-1)}/n$.
$\sqrt{(n^2-1)}/n= |n|\sqrt{1-1/n^2}/n$ if $n$ is positive than $\sqrt1$ else $-\sqrt1$;
bound: $\sqrt{(n^2-1)}/n \le \sqrt {n^2}/n = |n|/n=1$
To show if it is increasing should I do $\frac{\sqrt{(n+1)^2-1}}{n+1} \ge \frac{\sqrt{(n)^2-1}}{n}$?
2)Prove the sequence $a_n = \frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}$ has a limit
$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
|
Cross-multiplication yields, for $k\ge1$,
$$
\left(\frac{2k-1}{2k}\right)^2\le\frac{2k-1}{2k+1}
$$
Therefore,
$$
\begin{align}
\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2
&\le\prod_{k=1}^n\frac{2k-1}{2k+1}\\
&=\frac1{2n+1}
\end{align}
$$
Thus,
$$
\prod_{k=1}^\infty\frac{2k-1}{2k}=0
$$
For $n\ge1$,
$$
\begin{align}
\frac{\frac{\sqrt{n^2-1}}n}{\frac{\sqrt{(n+1)^2-1}}{n+1}}
&=\sqrt{\frac{(n+1)^3(n-1)}{n^3(n+2)}}\\
&=\sqrt{\frac{n^4+2n^3-2n-1}{n^4+2n^3}}\\[18pt]
&\lt1
\end{align}
$$
Thus, $\frac{\sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
\begin{align}
\lim_{n\to\infty}\frac{\sqrt{n^2-1}}n
&=\lim_{n\to\infty}\sqrt{1-\frac1{n^2}}\\[6pt]
&=1
\end{align}
$$
|
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|
Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation Prove that $11^{11}+12^{12}+13^{13}$ is divisible by $10$.
Obviously you could just put that in to a calculator and see the results, but I was wondering about some of the other approaches to this? I have not studied modulus', so if you could explain it without them, it would be better for me. Thanks!
|
You can easily prove it without any modular arithmetic. Just look at the last digits.
$$3^1 = \color{blue}{3} \quad 3^2 = \color{blue}{9} \quad 3^3 = 2\color{blue}{7} \quad 3^4 = 8\color{blue}{1} \implies 3, 9, 7, 1, 3, 9, 7, 1, ...$$
$$2^1 = \color{green}{2} \quad 2^2 = \color{green}{4} \quad 2^3 = \color{green}{8} \quad 2^4 = 1\color{green}{6} \implies 2, 4, 8, 6, 2, 4, 8, 6, ...$$
The pattern loops every $4^{th}$ exponent, as you can see. Notice that $13^{13}$ must end with whatever $3^{13}$ ends with, and $2^{12}$ must end with whatever $2^{12}$ ends with.
$$13 = \color{purple}{3}(4)+\color{blue}{1} \quad\text{Three loops done; first exponent}$$
$$12 = \color{purple}{2}(4)+\color{green}{4} \quad\text{Two loops done; fourth exponent}$$
Thus, $13^{13}$ ends with $\color{blue}{3}$ and $12^{12}$ ends with $\color{green}{6}$. $11^{11}$ obviously ends with $1$, so what does the unit digit become?
|
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|
Compute $\int_{-1}^{1} 5x^{6}(1 - |x|^{5}) \mathop{dx}$
Compute $$\int_{-1}^{1} 5x^{6}(1 - |x|^{5}) \mathop{dx}$$
$$\int_{-1}^{1} 5x^{6}(1 - |x|^{5}) \mathop{dx}$$
$$=\int_{-1}^{1} 5x^{6}- 5x^{6}|x|^{5} \mathop{dx} $$
$$= \int_{-1}^{1} 5x^{6} - 5\int_{-1}^{1}x^{6}|x|^{5}\mathop{dx}$$
$$= \frac{10}{7} - 5\left(\int_{-1}^{0}x^{6}|x|^{5} \mathop{dx} + \int_{0}^{1} x^{6}|x|^{5} \mathop{dx}\right) $$
$$= \frac{10}{7} -5\left(\int_{-1}^{0} x^{6}(-x)^{5} \mathop{dx} + \int_{0}^{1} x^{6} \cdot (x)^{5} \mathop{dx}\right) $$
$$= \frac{10}{7} - 5\left(\frac{1}{12} + \frac{1}{12}\right) $$
$$= \frac{25}{42}.$$
Is it ok ?
|
Another way:
As $x^6(1-|x|^5)$ is an even function,
$$I=\int_{-1}^15x^6(1-|x|^5)\ dx=2\int_0^15x^6(1-|x|^5)\ dx$$
$$\dfrac I{2\cdot5}=\int_0^1(x^6-x^{11})\ dx=?$$
|
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|
Co-ordinate Geometry : Circle Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, \frac{1}{2})$. Find the slope of the line.
Comments: First I write the family of lines passing through the given point i.e., $y - \frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 \sqrt{1 + m^2}$ but the solution of this equation is lengthy.
I am seeking a shorter solution.
|
*
*Line $$2mx-2y+4m+1=0 \tag{1}$$
*First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=\left| \frac{4m+1}{2\sqrt{m^2+1}} \right|$$
*Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=\sqrt{4^2-11}=\sqrt{5}$
distance of $P$ from $(1)$: $$d_2=\left| \frac{12m+1}{2\sqrt{m^2+1}} \right|$$
*Equating semi-chord length: $r^2-d^2$
\begin{align}
r_1^2-d_1^2 &= r_2^2-d_2^2 \\
1-\frac{(4m+1)^2}{4(m^2+1)} &=
5-\frac{(12m+1)^2}{4(m^2+1)} \\
m &= \frac{-1\pm \sqrt{29}}{14}
\end{align}
Only the negative root gives intersections:
|
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|
Proof using AM-GM inequality The questions has two parts:
Prove
(i) $ xy^{3} \leq \frac{1}{4}x^{4} + \frac{3}{4}y^{4} $
and
(ii) $ xy^{3} + x^{3}y \leq x^{4} + y^{4}$.
Now then, I went about putting both sides of $\sqrt{xy} \leq \frac{1}{2}(x+y)$
to the power of 4 and it left me with
$$-x^{3}y \leq \frac{1}{4}x^{4} + \frac{1}{4}y^{4} + xy^{3} + \frac{5}{2}x^{2}y^{2}. $$
Curiously squaring and multiplying $\sqrt{xy} \leq \frac{1}{2}(x+y)$ I've tried merging the results with my other inequality a few times to no avail - I just can't seem to get the signs right and nor can I seem to make the coefficients of the $x^{4}$ and $y^{4}$ different, as they are in (i). I feel like there's something I'm missing. Does anyone see a nice way about this problem?
|
Because by AM-GM $$3y^4+x^4\geq4\sqrt[4]{\left(y^4\right)^3x^4}=4|xy^3|\geq4xy^3$$ and
$$x^4+y^4-x^3y-xy^3=(x-y)(x^3-y^3)=(x-y)^2(x^2+xy+y^2)\geq$$
$$\geq(x-y)^2\left(2\sqrt{x^2y^2}+xy\right)=(x-y)^2\left(2|xy|+xy\right)\geq0.$$
|
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|
If $\frac{x}{y^\frac{n-1}{n}}$ is constant, how do I prove $\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$? From:
$\frac{x}{y^\frac{n-1}{n}}=constant$
To:
$\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$
It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.
|
Let $x=c\cdot y^\frac{n-1}{n}$. Then
\begin{align}
\frac{dx}{dy}&=\frac{n-1}{n}c\cdot y^\frac{-1}{n}\\
&=\frac{n-1}{n}c\cdot y^\frac{-1}{n} \cdot \frac{y}{y}\\
&=\frac{n-1}{n}c\cdot y^\frac{n-1}{n} \cdot \frac{1}{y}\\
&=\frac{n-1}{n} \frac{x}{y}\\
\end{align}
|
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|
Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.
$$
= \int_0^\frac{\pi}{4} \frac{\sec^4(x)}{3\tan^4(x)+3-\sec^4(x)} \text{d}x
= \frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t
$$
Which looks a little tough but reminiscent of the well-known integral:
$$ \int_0^\infty \frac{1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac12\int_0^\infty \frac{x^2+1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac{\pi}{4\cos(a)} $$
Where we choose $a=\frac\pi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?
|
Note that the integrand function $f(x)=\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}$ is even with period $\pi/2$, so
$$\int_0^\frac{n\pi}{4} f(x) dx=
\frac{1}{2}\int_{-\frac{n\pi}{4}}^\frac{n\pi}{4} f(x) dx=
\frac{1}{2}\int_{0}^\frac{n\pi}{2} f(x) dx
=\frac{n}{2}\int_{0}^\frac{\pi}{2} f(x) dx.$$
Now, following your approach, we have that
$$\frac{t^2+1}{t^4-t^2+1} =
\frac{2}{1+(2t+\sqrt{3})^2} +\frac{2}{1+(2t-\sqrt{3})^2}.$$
Therefore
$$\int\frac{t^2+1}{t^4-t^2+1}\,dt=\arctan(2t+\sqrt{3})+\arctan(2t-\sqrt{3})+c$$
Hence, for $t=\tan(x)$,
$$\int_0^\frac{\pi}{2} f(x) dx
= \frac12 \int_0^{\infty} \frac{t^2+1}{t^4-t^2+1} dt=\frac{1}{2}\left[\arctan(2t+\sqrt{3})+\arctan(2t-\sqrt{3})\right]_0^{+\infty}=\frac{\pi}{2}.$$
|
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|
Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$
Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$
If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and
$$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq {3\over 8}$$
Since $f(x)=\Big({1\over 1+x}\Big)^3$ is convex we get, by Jensen,: $$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq 3f({x+y+z\over 3})$$
Unfortunately, since $f$ is decreasing we don't have $f({x+y+z\over 3}) \geq f(1) = {1\over 8}$.
Some idea how to solve this?
|
This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.
Added:
We can rewrite the condition as $xyz=1$ and $\frac{(1+x)^4}{x} = \frac{(1+y)^4}{y} = \frac{(1+z)^4}{z}$
The function $g(x)=\frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $\infty$. This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$. It remains to solve $\frac{(1+x)^4}{x} = \frac{(1+1/x^2)^4}{1/x^2}$. This time, it's not hard to check $x=1$ is the only solution and we are done.
|
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|
How to evaluate $\int\frac{1}{3+4x+4x^2}$?
How to evaluate
$$\int\frac{1}{3+4x+4x^2}\quad ?
$$
This is what I've done so far:
$$
\frac{1}{4} \int\frac{1}{x^2+x+\frac{3}{4}}
=\frac{1}{4} \int\frac{1}{(x+\frac{1}{2})^2 + \frac{1}{2}}
$$
$$
y = \frac{1}{a}\arctan\frac{u}{a},\quad \frac{dy}{du} = \frac{1}{a^2 + u^2}
$$
$$
a = \sqrt{\frac{1}{2}},\quad
u = (x+\frac{1}{2}),\quad \frac{du}{dx} = 1
$$
so
$$ \frac{dy}{dx} = \frac{dy}{du} \cdot 1
$$
I don't know how to proceed.
Could I also have some help with
$$
\int\frac{1}{\sqrt{-4x^2-4x+3}}\quad ?
$$
|
What can be done is a double U-Substitution
We have $$\int\frac{1}{3+4x+4x^2}dx$$
By completing the square we see $$\int\frac{1}{3+4x+4x^2}dx = \int\frac{1}{(2x+1)^2+2}dx$$
Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$\frac{1}{2}\int\frac{1}{u^2+2}du = \frac{1}{2}\int\frac{1}{2(\frac{u^2}{2}+1)}du = \frac{1}{4}\int\frac{1}{(\frac{u^2}{2}+1)}du$$
Now we substitute $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}du$
in the integrand. $$\frac{1}{2\sqrt{2}}\int\frac{1}{s^2+1}ds$$
$$\int\frac{1}{s^2+1}ds = \tan^{-1}(s)$$ So we are left with $$\frac{\tan^{-1}(s)}{2\sqrt{2}}$$
Back substituting $s = \frac{u}{\sqrt{2}}$, $$\frac{\tan^{-1}(s)}{2\sqrt{2}} = \frac{\tan^{-1}(\frac{u}{\sqrt{2}})}{2\sqrt{2}}$$
Now back substitute $u = 2x+1$ and $$\frac{\tan^{-1}(\frac{u}{\sqrt{2}})}{2\sqrt{2}} = \frac{\tan^{-1}(\frac{2x+1}{\sqrt{2}})}{2\sqrt{2}}$$,
so our final answer is $$\boxed{\frac{\tan^{-1}(\frac{2x+1}{\sqrt{2}})}{2\sqrt{2}}+ C}$$
|
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|
Find $A$ given $A^2$ and $B$ and $(A^2)^{-1} = A^{-1}B$ This excersice took place in class I had today. The exercise was the following:
Let the regular matrix $A$,$B$:
$$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right] \hspace{2cm} B = \left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1&3&-1 \end{matrix} \right]$$
Knowing that $(A^2)^{-1} = A^{-1}B$, find A.
My Attempt
During the exam I tried the following:
$(A^2)^{-1} = A^{-1}B \Leftrightarrow A^2(A^2)^{-1} = A^2A^{-1}B \Leftrightarrow I_n = AB \Leftrightarrow A = B^{-1}$
And procedeed to find the inverse of B, which is:
$A = B^{-1} = \left[ \begin{matrix} \frac{1}{2} &\frac{-5}{4} & 1\\ 0 & \frac{1}{2} & 0 \\ \frac{1}{2}& \frac{1}{4}&0 \end{matrix} \right]$
But, as I tend to make silly mistakes, I figured out that it would be a better way to do find $A$ just by using matrix products. We had $I_n = AB \Leftrightarrow A = A^2B$. Using this method I got that
$A = A^2B = \left[ \begin{matrix} 2 & 0 & 2\\ 2 & 12 & -6 \\ -4& -8&12 \end{matrix} \right]$
But as you can see the $2$ matrix are different. Where is the mistake in my logic?
Another thing I noticed is that $I_n = AB \Leftrightarrow A*I_n*B = A*(AB)*B \Leftrightarrow I_n = AB = A^2B^2$, but $A^2B^2$ isn't the identity matrix, so maybe the exercise is wrong, but I rather not rush into that thinking.
|
The problem is broken, here is a precise statement showing how it is broken.
Let $B=\left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1&3&-1 \end{matrix} \right].$
For every invertible matrix $A$, at least one of the following are false:
*
*$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right]$
*$(A^2)^{-1}=A^{-1}B$
Proof.
If 2. holds, then
$$
\det \bigl((A^2)^{-1}\bigl)=\det (A^{-1}B).
$$
By direct computation, $\det B=-4$. On the other hand,
$$
\det\ \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right]=-64.
$$
Consequently if 1. and 2. both hold, then $(-64)^{-1}=(\det A)^{-1}\cdot (-4),\text{ or equivalently, } \det A=256.$ But this contradicts 1. Thus, it there does not exist any invertible matrix satisfying both 1. and 2. simultaneously.
|
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|
Simplified form of $\cos^{-1}\big[\frac{3}{5}\cdot\cos x+\frac{4}{5}\cdot\sin x\big]$, where $x\in\big[\frac{-3\pi}{4},\frac{3\pi}{4}\big]$
Find the simplified form of $\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]$, where $x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]$
My reference gives the solution $\tan^{-1}\frac43-x$, but is it a complete solution ?
My Attempt
Let $\alpha=\cos^{-1}\dfrac{3}{5}\implies \dfrac{3}{5}=\cos\alpha,\;\dfrac{4}{5}=\sin\alpha$
$$
\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]=\cos^{-1}\bigg[\cos\alpha\cdot\cos x+\sin\alpha\cdot\sin x\bigg]\\
=\cos^{-1}\bigg[\cos\Big(\alpha-x\Big)\bigg]=2n\pi\pm(\alpha-x)=2n\pi\pm\Big(\tan^{-1}\frac{4}{3}-x\Big)\\
=\tan^{-1}\frac{4}{3}-x\quad\text{iff }\tan^{-1}\frac{4}{3}-x\in[0,\pi]
$$
$$
-x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]\quad\&\quad\alpha=\tan^{-1}\frac{4}{3}\in\Big(0,\frac{\pi}{2}\Big)\\
\implies\alpha-x\in\big[\frac{-3\pi}{4},\frac{5\pi}{4}\big]\not\subset[0,\pi]
$$
|
$$-\dfrac{3\pi}4\le x\le\dfrac{3\pi}4$$
$$\iff-\dfrac{3\pi}4-\cos^{-1}\dfrac35\le x-\cos^{-1}\dfrac35\le\dfrac{3\pi}4-\cos^{-1}\dfrac35$$
Now $\dfrac{3\pi}4-\cos^{-1}\dfrac35\le\pi$ as $\cos^{-1}\dfrac35>0>\dfrac{3\pi}4-\pi$
So, $\cos^{-1}\bigg[\cos\Big(x-\cos^{-1}\dfrac35\Big)\bigg]=x-\cos^{-1}\dfrac35$ if $x-\cos^{-1}\dfrac35\ge0\iff x\ge\cos^{-1}\dfrac35$
Again we can prove $-2\pi<-\dfrac{3\pi}4-\cos^{-1}\dfrac35<-\pi$
For $-\pi<x-\cos^{-1}\dfrac35<0,$ $\cos^{-1}\bigg[\cos\Big(x-\cos^{-1}\dfrac35\Big)\bigg]=-\left(x-\cos^{-1}\dfrac35\right)$
For $-2\pi<x-\cos^{-1}\dfrac35<-\pi,$ $\cos^{-1}\bigg[\cos\Big(x-\cos^{-1}\dfrac35\Big)\bigg]=2\pi+x-\cos^{-1}\dfrac35$
|
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|
Quadratic form as a homogenous polynomial Let $Q(v)=v'Av$, where$v=\begin{pmatrix}x&y&z&w\end{pmatrix}\ \ A=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$, then does there exist an invertible matrix $P$ such that $Q(Pv)=x^2+y^2-zw$? Note that all terms are in $\mathbb{R}$. I think we should use orthogonal diagonalization of $A$. But, the term $zw$ is intriguing
|
Note that $Q(Pv)=v^{\top}P^{\top}APv$, so one approach is to find $P$ such that
$$P^{\top}AP=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&-\frac{1}{2}\\0&0&-\frac{1}{2}&0\end{pmatrix}.$$
As the first two rows and columns of $P^{\top}AP$ are the same as those of $A$, it makes sense to have $P$ leave those alone, i.e. to take $P$ of the form
$$P=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&a&b\\0&0&c&d\end{pmatrix}.$$
This reduces the problem to a problem on $2\times2$-matrices, which is very manageable without any theory. We get the system of equations
\begin{eqnarray*}
ac+ba&=&0,\\
c^2+ad&=&-\frac{1}{2},\\
b^2+ad&=&-\frac{1}{2},\\
cd+bd&=&0
\end{eqnarray*}
which immediately implies $b^2=c^2$, and $b+c=0$ or $a=d=0$. This immediately shows the easy solution $a=d=0$ and $b=c=\frac{1}{\sqrt{2}}i$. The remaining (family of) solutions is not hard to determine from the system above.
|
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|
If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$
Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of
$$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$
Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)
My attempt:
Developing expression $P(1)$, replacing the 1 by $x$, follows
$$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$
As $x_1,x_2,\ldots,x_n$ are the roots, it must be true that
$$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$
and
$$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$
Therefore the denominator of $P(1)$ is
$$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.
Another fact that is probably useful is that
$$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$
with roots that are 1 in addition of the given roots $x_1,x_2,\ldots,x_n$ for the original equation, that is
$$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$
This is as far as I could go...
Hints and full answers are welcomed.
|
We notice that the equation:
$$x^{n} + x^{n-1} + ... +x +1 = 0 $$
is a geometric sequence and can be rewritten as:
$$x^{n} + x^{n-1} + ... +x +1 = \sum_{i=0}^{n}x^n = \frac{1-x^{n+1}}{1-x} = 0 $$
From the denominator $x-1 $ we conclude that $x\ne1$, but we can rewrite $ 1 = e^{2\pi k} $. So from:
$$ 1-x^{n+1} = 0 \rightarrow x^{n+1} = e^{2\pi k} $$
$$ x_k = e^{2\pi k/(n+1)} $$
Where $k \ne 0$ and goes from $1$ to $n$. Because $k\ne 0$ then $x\ne1$ and there is no problem with the denominator anymore. The sum becomes:
$$ \sum_{k=1}^{n} \frac{1}{1-x_k} = \sum_{k=1}^{n} \frac{1}{1-e^{2\pi k/(n+1)}} = \sum_{k=1}^{n} \frac{e^{-\pi k/(n+1)}}{e^{-\pi k/(n+1)}-e^{\pi k/(n+1)}} = \sum_{k=1}^{n} \frac{\cos(\pi k/(n+1))-i\sin(\pi k/(n+1))}{-2i\sin(\pi k/(n+1))} = $$
$$ \sum_{k=1}^{n} \frac{1}{2} + i\frac{\cot(\pi k/(n+1))}{2} = \frac{n}{2} + \frac{i}{2}\sum_{k=1}^{n}\cot(\pi k/(n+1)) = \frac{n}{2} $$
From Wolfram alpha $ \sum_{k=1}^{n}\cot(\pi k/(n+1)) = 0 $ but you can prove it easily. An example:
$$ \cot(\frac{n\pi}{n+1}) = \cot(\frac{(n+1)\pi-\pi}{n+1}) = \cot(\pi- \frac{\pi}{n+1}) = -\cot(\frac{\pi}{n+1}) $$
which will cancel out with the first term and so on with other terms.
NOTE: I conducted the calculations for the sum $ \sum_{k=1}^{n} \frac{1}{1-x_k} $.
|
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|
Finding a distribution with a given correlation Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
answer to the problem.
Thanks,
Bob
Problem:
Let $X$ and $Y$ be uniformly distributed independent variables on the interval $(-1,1)$. Let $K$ be
a real number. Let $Z = Y + KX$ such that the correlation of $X$ and $Z$ is $\frac{1}{2}$. Find $K$.
Answer:
If we select $K = 0$ then we get a correlation of $0$. If we select $K$ to be a very large number
then the correlation will be close to $1$.
\begin{align*}
\rho &= \frac{1}{2} \\
u_x &= 0 \\
u_y &= 0 \\
u_z &= u_y + K(u_y) = 0 + K(0) = 0 \\
\sigma_x^2 &= \frac{(1 - -1)^2}{12} = \frac{4}{12} \\
\sigma_x^2 &= \frac{1}{3} \\
\sigma_x &= \frac{1}{\sqrt{3}} \\
\sigma_y &= \frac{1}{\sqrt{3}} \\
\sigma^2_z &= \sigma^2_y + K^2 \sigma_x^2 + K(0) \\
\sigma^2_z &= \frac{1}{3} + K^2 \left( \frac{1}{3} \right) \\
\rho &= \frac{\sigma_x \sigma_z}{\sigma_{xz}} \\
\sigma_{xz} &= \frac{\sigma_x \sigma_z}{\rho} =
\frac{ \left( \frac{1}{\sqrt{3}} \right) \left( \frac{1}{3} + K^2 \left( \frac{1}{3} \right) \right) }{\frac{1}{2} }\\
\sigma_{xz} &= \left( \frac{2}{\sqrt{3}} \right) \left( \frac{1}{3} + K^2 \left( \frac{1}{3} \right) \right) \\
\sigma_{xz} &= \left( \frac{2}{3 \sqrt{3}} \right) \left( K^2 + 1 \right) \\
\end{align*}
|
As you show, Pearson's $r$ should be a function of $k$. We will find that function and solve for
$r(k) = \frac{1}{2}$.
$r(A,B)$ may be calculated from the covariance $cov(A,B)$ and the standard deviations $\sigma_A$ and $\sigma_B$ as $\frac{cov(A,B)} { \sigma_A \sigma_B}$.
The covariance itself may be calculated as $E(AB) - E(A)E(B)$.
With $A$ and $B$ independent and each symmetric around $0$, the second term is $0$.
So with $Z = KX + Y$, we have $cov(X,Z) = E[X(kX+Y)] = E(KX^2) + E(XY)$.
Again, with $X$ and $Y$ independent and each symmetric around $0$, the second term is $0$.
$$ cov (X,Z) = E(KX^2) = KE(X^2) $$
The second moment $E(X^2)$ of a continuous uniform distribution $(a^2 + ab + b^2)/3$ where $a$ and $b$
are the bounds of the distribution. In this case, $(-1)^2 + (-1)*1 + 1^2 = 1 -1 + 1 = 1$.
$$ cov(X,Z) = K \frac{1}{3} = \frac{K}{3}$$
The variance $V(X)$ of a uniform distribution is $(b-a)^2/12$.
\begin{align*}
V(X) &= \frac{[1-(-1)]^2}{ 12 } = \frac{2^2}{12} = \frac{1}{3} \\
V(Z) &= V(KX+Y) = K^2*V(X)*V(Y) = K^2V(X) + V(X) = (K^2+1)V(X) \\
V(Z) &= \frac{(K^2+1)}{3} \\
V(X)V(Z) &= \frac{K^2+1}{9} \\
\sigma(X)\sigma(Z) &= \sqrt{ \frac{K^2+1}{9} } = \frac{ \sqrt{K^2+1} }{ 3 }\\
r(X,Z) &= \frac{ \frac{K}{3}}{ \frac{\sqrt{K^2+1}}{3}} = \frac{K}{ \sqrt{K^2+1)} } \\
\end{align*}
So for $r = \frac{1}{2}$, $\frac{K}{\sqrt{K^2+1}} = \frac{1}{2}$.
$$ 2K = \sqrt{K^2+1} $$
We will square both sides, which will give two solutions for $K$, only one of which will be relevant.
\begin{align*}
4K^2 &= K^2 + 1 \\
3K^2 - 1 &= 0 \\
\end{align*}
using $a^2 - b^2 = (a+b)(a-b)$, we see $(\sqrt{3}K +1)(\sqrt{3}K - 1) = 0$.
$$ \sqrt{3}K = 1 \text{ OR } -1$$
Since we can see in the original problem K must be positive to yield a positive correlation,
$\sqrt{3}K = 1$.
$$ k = \frac{1} {\sqrt{3}} $$
|
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|
Find all Pythagorean triples $x^2+y^2=z^2$ where $x=21$ Consider the following theorem:
If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite
parity.
According to the previous theorem,My try is the following:
since $x = r^2-s^2$, $x$ is difference of two squares implying that $x \equiv 0 \pmod 4$. But $x=21 \not \equiv 0 \pmod 4$. Hence, there are no triangles having such $x$.
Is that right?
Added:
My argument is false here. Please refer to the appropriate answer.
|
We have $21=x=k(m^2-n^2),\, y=2kmn,\, z=k(m^2+n^2)$ where $m,n, k \in \Bbb N$ with $\gcd (m,n)=1$ and $m,n$ not both odd.
So $(m^2-n^2,k)\in \{(1,21),(3,7),(7,3),(21,1)\}.$ Now $m^2-n^2=1$ is impossible, so $(m,n,k)\in \{(2,1,7), (4,3,3),(11,10,1),(5,2,1)\},$ giving $$(x,y,z)\in \{ (21,28,35), (21,72, 75),(21,220, 221),(21, 20, 29)\}.$$ We have $m\leq 11$ because if $m\geq 12$ then $x\geq m^2-n^2\geq m^2-(m-1)^2=2m-1\geq 23>21...$ There are 2 solutions $(11,10)$ and $(5,2)$ to $m^2-n^2=21.$
|
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|
Find the Arc length of the parametric curve $$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = \int_{0}^{2\pi} \sqrt{(6-6cost)^2+(6sint)^2}dt\\
=\int_{0}^{2\pi} \sqrt{36-72cost+36cos^2t+36sin^2t}dt \\
=\int_{0}^{2\pi} 6 \sqrt{1-2cost+cos^2t+sin^2t}dt\\
=\int_{0}^{2\pi} 6\sqrt{2-2cost}dt\\
=\int_{0}^{2\pi} 6\sqrt{2}\sqrt{1-cost}dt\\
= 6\sqrt{2}\int_{0}^{2\pi}\sqrt{1-cost}dt\\
=6\sqrt{2}\int_{0}^{2\pi}\sqrt{1-cost}\frac{\sqrt{(1+cost)}}{\sqrt{(1+cost)}} dt\\
=6\sqrt{2}\int_{0}^{2\pi}\frac{\sqrt{(1-cos^2t)}}{\sqrt{(1+cost)}} dt\\
=6\sqrt{2}\int_{0}^{2\pi}\frac{sint}{\sqrt{(1+cost)}} dt\\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6\sqrt{2}\int_{0}^{2\pi}\frac{1}{\sqrt{u}}\\
=-12\sqrt{2}[u^\frac{1}{2}]\\
=0$$
But the answer is 48.
|
Leaving the coefficient $6$ on the side, we have
$$s=\int_0^{2\pi}\sqrt{(1-\cos t)^2+(\sin t)^2}\,dt=\int_0^{2\pi}\sqrt{2-2\cos t}\,dt=2\int_0^{2\pi}\left|\sin\frac t2\right|\,dt.$$
|
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|
How can we solve the diophantine equation? How can we find all the primitive solutions of the diophantine equation $x^2+3y^2=z^2$ ?
Some solutions are $(\pm n , 0 , \pm n)$ for $n\in \mathbb{N}$. How can we find also the other ones?
|
We can follow the stereographic projection method used to find Pythagorean triples. We start by dividing through by $z^2$, giving us the equation
$$\left(\frac{x}{z}\right)^2 + 3\left(\frac{y}{z}\right)^2 = 1.$$
Or, in other words, we search for rational points on the ellipse
$$x^2 + 3y^2 = 1.$$
We'll take the point $(1, 0)$ and stereographically project onto the $y$-axis. Consider an arbitrary point $(0, y)$ on the $y$-axis, and consider the line
$$r = (1, 0) + t(-1, y)$$
between $(1, 0)$ and $(0, y)$. This line passes through the ellipse at $(1, 0)$ and a second point, which will be the point that stereographically projects onto $(0, y)$. The $t$ value for this point must satisfy,
$$(1 - t)^2 + 3(ty)^2 = 1 \iff t((3y^2 + 1)t - 2) = 0.$$
The $t = 0$ solution produces $(1, 0)$, so we discard it. The other solution is
$$t = \frac{2}{3y^2 + 1},$$
which yields the point on the ellipse,
$$\left(\frac{3y^2 - 1}{3y^2 + 1}, \frac{2y}{3y^2 + 1}\right).$$
Now, when we have a rational point on the ellipse, the line will be of rational slope, and hence will stereographically project to a point $(y, 0)$ where $y$ is rational. That is, the rational point on the ellipse must take the above form where $y \in \mathbb{Q}$. If we take $y = \frac{m}{n}$, where $n \neq 0$, and $m, n \in \mathbb{Z}$, then this becomes
$$\left(\frac{3\left(\frac{m}{n}\right)^2 - 1}{3\left(\frac{m}{n}\right)^2 + 1}, \frac{2\frac{m}{n}}{3\left(\frac{m}{n}\right)^2 + 1}\right) = \left(\frac{3m^2 - n^2}{3m^2 + n^2}, \frac{2mn}{3m^2 + n^2}\right).$$
To wrap this up, suppose we have an integer solution $(x, y, z)$ to the Diophantine equation $x^2 + 3y^2 = z^2$. If $z = 0$, then $x = y = 0$, which is one solution. Otherwise, $z \neq 0$ and $\left(\frac{x}{z}, \frac{y}{z}\right)$ must be a (rational) point on the ellipse. This point may be $(1, 0)$, in which case $(x, y, z) = k(1, 0, 1)$ (the points you've mentioned). The other solutions must take the form
$$\left(\frac{x}{z}, \frac{y}{z}\right) = \left(\frac{3m^2 - n^2}{3m^2 + n^2}, \frac{2mn}{3m^2 + n^2}\right).$$
Equating the fractions, we must have, for some integers $k, l$,
\begin{align*}
x &= k(3m^2 - n^2) \\
z &= k(3m^2 + n^2) \\
y &= l(2mn) \\
z &= l(3m^2 + n^2).
\end{align*}
As such, $k = l$, so our solution is of the form
$$(x, y, z) = k(3m^2 - n^2, 2mn, 3m^2 + n^2)$$
for integers $m, n, k$. Importantly, each of these are a solution, as
$$(k(3m^2 - n^2))^2 + 3(k(2mn))^2 = k^2(9m^4 + n^4 - 6m^2 n^2 + 12m^2 n^2) = (k(3m^2 + n^2))^2.$$
Note that the $(0, 0, 0)$ solution can be obtained from $k = 0$, but there is no way to obtain the $k(1, 0, 1)$ solutions. Hence, our general solution is
$$(x, y, z) = k(1, 0, 1) \text{ or } k(3m^2 - n^2, 2mn, 3m^2 + n^2) \text{ for } k, m, n \in \mathbb{Z}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $\sum_{iIn a example about U-statistics, $h(x_1,x_2)=\frac 12(x_1-x_2)^2$, then
$$U_n=\frac{2}{n(n-1)}\sum_{i<j}\frac{(X_i-X_j)^2}{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2$$
I don't know how to prove it completely.
|
We know that (I found it here)
\begin{equation}
\left( \sum_{n=1}^N a_n \right)^2 = \sum_{n=1}^N a_n^2 + 2 \sum_{j=1}^{N}\sum_{i=1}^{j-1} a_i a_j
\end{equation}
So using the above identity
\begin{align}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\sum_{i=1}^{n}(X_i-\frac{1}{n}\sum_{j=1}^nX_j)^2\\
&=
\sum_{i=1}^{n}(X_i^2-\frac{2}{n}X_i\sum_{j=1}^nX_j + \frac{1}{n^2}(\sum_{j=1}^nX_j)^2 )\\
&=
\sum_{i=1}^{n}(X_i^2-\frac{2}{n}X_i\sum_{j=1}^nX_j + \frac{1}{n^2}(\sum_{j=1}^nX_j^2 + 2\sum_{j=1}^n\sum_{k=1}^{j-1}X_jX_k) )
\end{align}
The last term above is independent of $i$ so it sums up $n$ times as
\begin{align}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\sum_{i=1}^{n}(X_i^2-\frac{2}{n}X_i\sum_{j=1}^nX_j) + \frac{n}{n^2}(\sum_{j=1}^nX_j^2 + 2\sum_{j=1}^n\sum_{k=1}^{j-1}X_jX_k)
\end{align}
which is also
\begin{align}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\sum_{i=1}^{n}(X_i^2-\frac{2}{n}X_i\sum_{j=1}^nX_j) + \frac{1}{n}(\sum_{j=1}^nX_j^2 + 2\sum_{j=1}^n\sum_{k=1}^{j-1}X_jX_k)
\end{align}
which could also be written as
\begin{align}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
(1 + \frac{1}{n})
\sum_{i=1}^{n}X_i^2-\frac{2}{n}\sum_{i=1}^{n}X_i\sum_{j=1}^nX_j) + \frac{1}{n}( 2\sum_{j=1}^n\sum_{k=1}^{j-1}X_jX_k)
\end{align}
Rewriting differently we have
\begin{align}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
(1 + \frac{1}{n})
\sum_{i=1}^{n}X_i^2-\frac{2}{n}\sum_{i,j}X_iX_j + \frac{2}{n}\sum_{i<j}X_iX_j
\end{align}
The last two terms above are the same terms with missing terms. Notice that $\sum_{i,j}X_iX_j$ spans all $i = 1 \ldots n$ and $j = 1 \ldots n$ but the other one spans an upper triangular version of it. This means that their difference will span the lower triangular version of it as
\begin{align}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
(1 + \frac{1}{n})
\sum_{i=1}^{n}X_i^2 - \frac{2}{n}\sum_{i\geq j}X_iX_j
\end{align}
Factor $n$ on the right hand side, then divide by $n-1$ on both sides, then Multiply/divide by $2$ on the right hand side
\begin{align}
\frac{1}{n-1}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\frac{2}{n(n-1)}
\Big(
\frac{(n + 1)
\sum_{i=1}^{n}X_i^2 - 2\sum_{i\geq j}X_iX_j}{2}
\Big)
\end{align}
Notice that $i \geq j$ could be split to two summations
\begin{align}
\frac{1}{n-1}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\frac{2}{n(n-1)}
\Big(
\frac{(n + 1)
\sum_{i=1}^{n}X_i^2 - 2\sum_{i = j}X_iX_j - 2\sum_{i > j}X_iX_j}{2}
\Big)
\end{align}
but when $i = j$, it is the same as a single summation, hence
\begin{align}
\frac{1}{n-1}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\frac{2}{n(n-1)}
\Big(
\frac{(n + 1)
\sum_{i=1}^{n}X_i^2 - 2\sum_{i=1}^n X_i^2 - 2\sum_{i > j}X_iX_j}{2}
\Big)
\end{align}
which gives
\begin{align}
\frac{1}{n-1}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\frac{2}{n(n-1)}
\Big(
\frac{(n -1)
\sum_{i=1}^{n}X_i^2- 2\sum_{i > j}X_iX_j}{2}
\Big)
\end{align}
The numerator above is nothing other than $\sum_{i<j} (X_i - X_j)^2 = \sum_{i<j} X_i^2 - 2 \sum_{i<j} X_iX_j + \sum_{i<j} X_j^2$. It is easy to see the cross terms, however it is not as straightforward to see that we have $n-1$ terms of the form $X_i^2$. This should conclude
\begin{align}
\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2
=
\frac{2}{n(n-1)}\sum_{i<j}\frac{(X_i-X_j)^2}{2}
\end{align}
|
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|
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
My Attempt
\begin{align}
\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{4-3}{1+4.3}+\tan^{-1}\frac{5-4}{1+5.4}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}5-\tan^{-1}3\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{8}=\pi
\end{align}
My reference gives the solution $0$, so what's going wrong with my attempt ?
|
You started by noting $\cot^{-1}21=\tan^{-1}\frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $\cot^{-1}(-8)=-\cot^{-1}8=-\tan^{-1}\frac{1}{8}$, no $\pi$ involved. (Another way to prove $\tan^{-1}\frac{1}{21}+\tan^{-1}\frac{1}{13}=\tan^{-1}\frac{1}{8}$ is to use the identity $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}$, which for $a=1/m,\,b=1/n$ simplifies to $\tan^{-1}\frac{m+n}{mn-1}$.)
|
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|
Can't solve a quartic equation I'm trying to solve an algebraic question.The question wants me to solve $n^4+2n^3+6n^2+12n+25=m^2$.The question also states that n is a positive integer and the answer for $n^4+2n^3+6n^2+12n+25$ is a square number. Here's how I tried to solve it:
$$n^4+2n^3+6n^2+12n+25=\\n^4+6n^2+2n^3+12n+25=\\
n^2(n^2+6)+2n(n^2+6)+5^2=\\
(n\sqrt {n^2+6})^2+2n(n^2+6)+5^2.\\$$
Because $a^2+2ab+b^2=(a+b)^2$,so
$\sqrt {(n\sqrt {n^2+6})^2}\cdot\sqrt5^2=n(n^2+6)$
Then:
$$n\sqrt {n^2+6}\cdot 5=n(n^2+6)\\
\sqrt {n^2+6}\cdot 5=n^2+6\\
25(n^2+6)=n^4+12n^2+36\\
n^4+12n^2+36=25n^2+150\\
n^4-13n^2-114=0\\
(n^2+6)(n^2-19)=0\\
n^2=19\\
n=\sqrt 19$$
But $n$ is a positive integer.
Can anyone help?
|
I believe (correct me if I'm wrong) that $a^2-b^2$ only has factors $1, a^2-b^2, (a-b), (a+b)$ if $a$ and $b$ are coprime. Using this, we have that:
$$n^4+2n^3+6n^2+12n+25=m^2 \to n(n^3+2n^2+6n+12)=m^2-25$$
If we assume that $m\neq 5k,k\in\Bbb Z$, by what I stated earlier we have that either $n=1$ and $(n^3+...)=m^2-25$, which leads to $m=\pm\sqrt{46}$, or $(n^3+...)=1$ and $n=m^2-25$, which leads to $n\approx -1.896\to m\approx\sqrt{23.104}$, or that $(n^3+...)=n\pm 10\because m^2-25=(m+5)(m-5)$ and these factors are $10$ apart. Neither of these yields integer solutions, so we can safely discard them too.
In short, this is impossible unless $m$ is a multiple of $5$. Now see if you can find if it works when $m$ is.
|
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|
AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and x=((a+b)(b+c)(c+a))/abc and x<0, then x=? I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$=$\frac{-a+b+c}{a}$ and x=$\frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?
What I have done so far: Consider, if $\frac{a}{b}$=$\frac{c}{d}$, and (b+d) is non-zero, then $\frac{a+c}{b+d}$=$\frac{a}{b}$=$\frac{c}{d}$. So with that fact, I did... $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$ $\implies$
$\frac{2a}{c+b}$ and $\frac{2a}{c+b}$=$\frac{-a+b+c}{a}$$\implies$ $\frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $\frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$\frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=\frac{(2a)(2b)(2c)}{abc}$, which is $x=\frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?
|
Hint: $$a=-\frac{b}{2},b=b,c=-\frac{b}{2}$$ solves your problem and $$x=-1$$
|
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|
How to find minimal polynomial in $GF(2^3)$ I have $GF(2^3)$ field defined by $\Pi(x)=x^3+x+1$. From literature they say those are the minimal polynomial, but I can't understand the operative method to find them. Any explanation for a general method?
$$\begin{array}{lll}
\textbf{Elem.} & \textbf{Polyn.} & \color{red}{\textbf{Minimal Polyn.}} \\
0 & 0 & \color{red}{x} \\
\alpha^0 & 1 & \color{red}{x+1} \\
\alpha^1 & \alpha & \color{red}{x^3+x+1} \\
\alpha^2 & \alpha^2 & \color{red}{x^3+x+1} \\
\alpha^3 & \alpha+1 & \color{red}{x^3+x^2+1} \\
\alpha^4 & \alpha^2+\alpha & \color{red}{x^3+x+1} \\
\alpha^5 & \alpha^2+\alpha+1 & \color{red}{x^3 + x^2 + 1} \\
\alpha^6 & \alpha^2+1 & \color{red}{x^3 + x^2 + 1} \\
\end{array}$$
Note: I saw another post about minimal polynomial but there was no such method explained
|
The elements of $GF(8)$ are exactly the roots of the polynomial $X^8-X\in GF(2)[X]$. This polynomial decomposes into irreducible polynomials as follows,
$$X^8-X = X(X+1)(X^3+X+1)(X^3+X^2+1).$$
If the field $GF(8)$ is given as $GF(2)[X]/\langle X^3+X+1\rangle$ and $\alpha$ is a zero of $X^3+X+1$, then (as stated above), $\alpha,\alpha^2,\alpha^4$ are the zeros of $X^3+X+1$ and $\alpha^3,(\alpha^3)^2=\alpha^6,((\alpha^3)^2)^2=\alpha^5$ are the zeros of $X^3+X^2+1$.
|
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|
How to prove that $\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$ Through some calculation, it can be shown that
$$e = 1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{\vdots}{4}}{3}}{2}}{1}\tag{1}$$
$$2e = 1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{\vdots}{4}}{3}}{2}}{1}\tag{2}$$
$$5e = 1+\cfrac{2^2+\cfrac{3^2+\cfrac{4^2+\cfrac{\vdots}{4}}{3}}{2}}{1}\tag{3}$$
In general, how can I show that
$$\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$$
, where $B_n$ is the $n^{th}$ Bell number.
I saw a similar question on Brilliant.org, but I did not pay close attention to the proof, and I ended up forgetting how to prove this kind of problem.
I remember that the proof involves in simplifying the denominator and moving from top to bottom so that the final denominator is in the form of $n!$, which is the criteria for Maclaurin series.
Here is the background information Infinite Series $\sum\limits_{k=1}^{\infty}\frac{k^n}{k!}$
|
After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.
We will first prove the first equation. Note that
$$\begin{align}
1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\vdots}{5}}{4}}{3}}{2}}{1}&=1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1}{5}+\cfrac{\vdots}{5}}{4}}{3}}{2}}{1}\\
&=1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1}{4}+\cfrac{1}{4\cdot5}+\cfrac{\vdots}{4\cdot5}}{3}}{2}}{1}\\
&=1+\cfrac{1+\cfrac{1+\cfrac{1}{3}+\cfrac{1}{3\cdot4}+\cfrac{1}{3\cdot4\cdot5}+\cfrac{\vdots}{3\cdot4\cdot5}}{2}}{1}\\
&=1+\cfrac{1+\cfrac{1}{2}+\cfrac{1}{2\cdot3}+\cfrac{1}{2\cdot3\cdot4}+\cfrac{1}{2\cdot3\cdot4\cdot5}+\cfrac{\vdots}{2\cdot3\cdot4\cdot5}}{1}\\
&=1+\cfrac{1}{1!}+\cfrac{1}{2!}+\cfrac{1}{3!}+\cfrac{1}{4!}+\cfrac{1}{5!}+\cdots\\
&=\color{red}e\tag{1}
\end{align}$$
We will then proceed to the second equation.
$$\begin{align}
1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{5+\cfrac{6+\vdots}{5}}{4}}{3}}{2}}{1}&=1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{5+\cfrac{6}{5}+\cfrac{\vdots}{5}}{4}}{3}}{2}}{1}\\
&=1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{5}{4}+\cfrac{6}{4\cdot5}+\cfrac{\vdots}{4\cdot5}}{3}}{2}}{1}\\
&=1+\cfrac{2+\cfrac{3+\cfrac{4}{3}+\cfrac{5}{3\cdot4}+\cfrac{6}{3\cdot4\cdot5}+\cfrac{\vdots}{3\cdot4\cdot5}}{2}}{1}\\
&=1+\cfrac{2+\cfrac{3}{2}+\cfrac{4}{2\cdot3}+\cfrac{5}{2\cdot3\cdot4}+\cfrac{6}{2\cdot3\cdot4\cdot5}+\cfrac{\vdots}{2\cdot3\cdot4\cdot5}}{1}\\
&=1+\cfrac{2}{1!}+\cfrac{3}{2!}+\cfrac{4}{3!}+\cfrac{5}{4!}+\cfrac{6}{5!}+\cdots\\
&=\sum_{n=1}^{\infty}\cfrac{n^2}{n!}\\
&=\color{red}{2e}\tag{2}
\end{align}$$
Then, using the same logic, we have
$$\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$$
|
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|
Find the limit of $\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ Find the limit of the sequence $$\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$\frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}\longrightarrow\frac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.
|
$$
\begin{align}
\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^{1/3}-\left(1+\frac1{n^3}\right)^{1/3}\right)\\
&=\lim_{n\to\infty}n\left(\left[1+\frac1{3n}+O\!\left(\frac1{n^2}\right)\right]-\left[1+O\!\left(\frac1{n^3}\right)\right]\right)\\
&=\lim_{n\to\infty}\left(\frac13+O\!\left(\frac1n\right)\right)\\
&=\frac13
\end{align}
$$
|
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|
How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be multiplied together, (the $(2)(-3)$, rather than $6(-1)$?
When following the factors $-1$ and $6$. I have
$(2x^2-1x)(6x-3)$
$x(2x-1)+3(2x-1)$
Is this correct, if not; what is the best way to solve a leading coefficient when factoring?
|
$$2x^2 + 5x − 3 = 0$$
$$ac = 2(-3) = -6$$
$$\text{$^-1\times 6 =\phantom .^-6 \ $ and $ \ ^-1+6 = 5$}$$
$-1$ and $6$ are correct. What you did after that is wrong.
Here are two methods that I know of for proceeding from $-1$ and $6$.
Method 1. Replace $5x$ with $-1x+6x$ and factor-by-pairing-off.
\begin{array}{c}
2x^2 + 5x − 3 \\
2x^2 -1x + 6x - 3 \\
(2x^2 -1x) + (6x - 3) \\
x(2x-1) + 3(2x-1) \\
(x+3)(2x-1)
\end{array}
Method 2. Write out $(ax-1)(ax+6)$ and then divide out the greatest common divisors.
\begin{array}{c}
(2x-1)(2x+6) &\{\gcd(2,-1)=1 \ \text{and} \ \gcd(2,6)=2\}\\
\dfrac{(2x-1)}{1} \cdot \dfrac{(2x+6)}{2} \\
(2x-1)(x+3)
\end{array}
|
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|
Evaluate $\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$
$\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$
What are the good/ clever methods to evaluate this limit?
I tried taking $\tan^{-1} (x+5) = \theta$ to avoid inverse functions but its not helpful and makes it even more complicated.
I also tried $\tan^{-1}a - \tan^{-1}b$ formula for the terms attached to x but that does not help to get rid of other terms multiplied by $1$ and $5$.
Edit: (Please address this in your answer)
Can't we directly do this:
$\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$
$= (x+5)\dfrac{\pi}{2} - (x+1)\dfrac{\pi}{2}$
$ = \dfrac {5\pi - \pi}{2} = 2\pi$
I don't see anything wrong with it and it gives the right answer.
Is this method correct? Can it be used in other questions too?
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Use Taylor expansion:
$$\tan^{-1}(x+1)=\frac{\pi}{2} - \frac 1x + \frac 1{x^2} - \frac{2}{3 x^3} + O\left(\frac{1}{x^5}\right);\\
\tan^{-1}(x+5)=\frac{\pi}2 - \frac 1x + \frac5{x^2} - \frac{74}{3 x^3} + \frac{120}{x^4} + O\left(\frac 1{x^5}\right);\\
\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)=\\
\lim_{x\to \infty} (x+5)\left[\frac{\pi}{2}-\frac1x+\frac{5}{x^2}+O\left(\frac 1{x^3}\right)\right]- (x+1)\left[\frac{\pi}{2}-\frac1x+\frac{1}{x^2}+O\left(\frac 1{x^3}\right)\right]=\\
\lim_{x\to \infty} \left[2\pi-\frac 4x+O\left(\frac1{x^2}\right)\right]=2\pi.$$
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|
Inequality with $(x+y)(y+z)(z+w)(w+x)=1$ Let $x,y,z,w>0$ and such that
$$(x+y)(y+z)(z+w)(w+x)=1.$$
Show that
$$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le 2.$$
I'm trying to use Holder's inequality
$$(\sqrt[3]{xyz}+\sqrt[3]{yzw})^3
\le (x+y)(y+z)(z+w)$$
$$(\sqrt[3]{zwx}+\sqrt[3]{wxy})^3\le (z+w)(w+x)(x+y)$$
so
$$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le \sqrt[3]{(x+y)(y+z)(z+w)}+\sqrt[3]{(z+w)(w+x)(x+y)}.$$
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This does not seem the best, but using only Holder's inequality, we can get
$$\begin{eqnarray}
S^{12}&\le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}\left(\prod_{\text{cyc}}(x+y)\right)^3 =2^{12},
\end{eqnarray}$$for $S= \sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}.$
|
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if $x-y =\sqrt{x}-\sqrt{y}$ with $x\neq y$ then $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$? let $x\neq y$ be positive real numbers such that :$x-y= \sqrt{x}-\sqrt{y}$ , I have tried to prove this inequality $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$ that i have created but i didn't got it.
Attempt I have showed that:$(\frac{1}{x}+\frac{1}{y})\geq \frac{2}{\sqrt{xy}}$ using this identity: $(\sqrt{x}-\sqrt{y})^2\geq0$ , I also showed that :$\frac{1}{xy}\geq \frac{1}{16}$ , Now I have used both result I have got the following inequality :$(1+\frac{y}{x})(1+\frac{x}{y})\geq 25$ but not what i have claimed , any way ?
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By AM-GM
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\left(1+\frac{4}{4x}\right)\left(1+\frac{4}{4y}\right)\geq$$
$$\geq\frac{5}{\sqrt[5]{(4x)^4}}\cdot\frac{5}{\sqrt[5]{(4y)^4}}=\frac{25}{\sqrt[5]{4^{8}\left(\sqrt{xy}\right)^8}}\geq\frac{5}{\sqrt[5]{4^8\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^{16}}}=25.$$
|
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|
Factor polynomials I am having trouble with these expressions:
$$x^4 - 23x^2 + 1$$
$$2y^2 - 5xy + 2x^2 - ay - ax - a^2$$
$$2x^3 - 4x^2y - x^2z + 2xy^2 - y^2z +2xyz$$
I tried to consult a chapter on factoring in my textbook. It seems to suggest the following:
Check for expressions like $a^n - b^n$ or $(a + b)^n$.
Remember that $x^2 + (a + b)x + ab =(x + a)(x + b)$.
Complete the square and check if any of the above apply.
(For example, $x^2 - 5x + 3$ add $\pm (5/2)^2$ to get the expression equivalent to $(a^n + b^n)$.)
Grouping.
Regarding the first expression I tried the third method i.e. I added $(23/2)^2$ hoping to get the expression equivalent to $a^2 - b^2$.
In the second exercise I tried to factor monomial from different expressions to see some sort of pattern. I also tried to complete the square that is to add different expressions like $\pm(5y/2)^2$. Formula itself reminds me of $(a + b + c)^2$. I am pretty sure the result is of the form $(a + b + c) (a - b + c)$ or something like that but I am still unsure how to proceed.
Regarding the third expression I don’t even know where to start. I tried grouping and other methods but I still can’t see a pattern.
Can someone give me a hint? Am I missing some rule? How to solve problems like that? What goes on in your head while you're looking at the expressions like that?
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$$x^4-23x^2+1=x^4+2x^2+1-25x^2=(x^2-5x+1)(x^2+5x+1).$$
$$2x^2-5xy+2y^2-ax-ay-a^2=$$
$$=2x^2-5xy+2y^2+\frac{1}{4}(x+y)^2-\frac{1}{4}(x+y)^2-a(x+y)-a^2=$$
$$=\frac{9}{4}(x-y)^2-\left(\frac{1}{2}(x+y)+a\right)^2=$$
$$=\left(\frac{3}{2}(x-y)-\frac{1}{2}(x+y)-a\right)\left(\frac{3}{2}(x-y)+\frac{1}{2}(x+y)+a\right)=$$
$$=(x-2y-a)(2x-y+a).$$
The third is irreducible.
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Integral $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}$ I have stumbled upon the following integral:$$I=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}=-\frac{\pi}{24}$$
Although I could solve it, I am not quite comfortable with the way I did it.
But first I will show the way. We can substitute $\ln x \rightarrow t\ $ which gives:
$$I=\int_{-\infty}^\infty \frac{t}{\pi^2+t^2}\frac{e^{\frac{t}{2}}}{(1+e^t)^2}dt\overset{t=-x}=\int_{-\infty}^\infty \frac{-x}{\pi^2+x^2}\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}dx$$
Also adding the two integral from above and simplify some of it yields:
$$2I= \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\left(\frac{e^{\frac{x}{2}}}{(1+e^x)^2}-\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}\right)dx$$
$$\Rightarrow I=-\frac{1}{4} \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\frac{\sinh \left(\frac{x}{2}\right)}{\cosh ^2\left(\frac{x}{2}\right)}dx$$
And now a round of IBP gives:
$$I=\frac12 \int_{-\infty}^\infty \left(\frac{x^2-\pi^2}{(x^2+\pi^2)^2}\right)\left(\frac{1}{\cosh \left(\frac{x}{2}\right)}\right)dx$$
Using the Plancherel theorem the integral simplifies to: $$I=\int_0^\infty \left(\sqrt{\frac{\pi}{2}}x\left(-e^{-\pi x}\right)\right)\left(\sqrt{2\pi}\frac{1}{\cosh(\pi x)}\right)dx\overset{\pi x\rightarrow x}=-\frac{1}{\pi}\int_0^\infty \frac{x}{\cosh( x)}e^{- x}dx$$
We also have the following Laplace tranform for:$$f(t)=\frac{t}{\cosh( t)}\rightarrow F(s)=\frac18\left(\psi_1\left(\frac{s+1}{4}\right)-\psi_1\left(\frac{s+3}{4}\right)\right)$$
Where $\displaystyle{\psi_1(z)=\sum_{n=0}^\infty \frac{1}{(z+n)^2}}\,$ is the trigamma function.
$$\Rightarrow I=-\frac{1}{\pi}F(s=1)=-\frac{1}{\pi}\cdot \frac18\left(\psi_1\left(\frac{1}{2}\right)-\psi_1 (1)\right)=-\frac{1}{\pi}\cdot \frac18\left(\frac{\pi^2}{2}-\frac{\pi^2}{6}\right)=-\frac{\pi}{24}$$
Have I done anything wrong, or can it be improved?
I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.
For this question I would like to see a different proof that doesn't rely on that theorem.
Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.
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From the identity
$$\Im\int_0^\infty e^{-(\pi-it)x}\,dx=\frac t{\pi^2+t^2}$$
we see that it suffices to compute the imaginary part of the integral
$$\int_0^\infty dx\int_{-\infty}^\infty dt\;
\frac{e^{\alpha t}}{(1+e^t)^2}e^{-\pi x}$$
where $\alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get
$$\int_0^\infty \pi\left(\frac12-ix\right)\frac{e^{-\pi x}}{\cosh(\pi x)}\,dx.$$
Taking the imaginary part we see that the problem boils down to compute the integral
$$\int_0^\infty \frac{x e^{-\pi x}}{\cosh(\pi x)}\,dx
=2\int_0^\infty \frac{x}{1+e^{2\pi x}}\,dx$$
which, after the substitution $v=2\pi x$, reduces to the integral representation of the eta function $\eta(2)$. Also notice that taking the real part we obtain the evaluation
$$\int_0^\infty\frac1{(\pi^2+\log^2 x)(1+x)^2} \frac{dx}{\sqrt x}=
\frac{\log2}{2\pi}.$$
This method generalises to other integrals such as
\begin{align*}
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x}
&=\frac{3\log (2)}{8 \pi }-\frac{3 \zeta (3)}{16 \pi ^3}\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x}
&=-\frac{\pi }{24}\\
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x}
&=\frac{5 \log (2)}{16 \pi }-\frac{9 \zeta (3)}{32 \pi ^3}\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x}
&=-\frac{223 \pi }{5760}\\
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x}
&=-\frac{43 \zeta (3)}{128 \pi ^3}+\frac{15 \zeta (5)}{256 \pi ^5}+\frac{35 \log (2)}{128
\pi }\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x}
&=-\frac{103 \pi }{2880}\\
\end{align*}
Incidentally, since the integrals $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^k} \frac{dx}{\sqrt x}$ both yield the same value for $k=2,3$, we also deduce
$$\int_0^\infty \frac{\sqrt x\ln x}{(\pi^2+\ln^2 x)(1+x)^3}\;dx=0.$$
This, however, should not be surprising due to the symmetry $x\mapsto1/x$.
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|
Twin Prime Formula I have a function involving polynomials and the centre of the Binomial Triangle and I'd like to prove that the function produces a positive integer infinitely many times. I don't have any interest in what values the integers take so much, merely that they exist.
The function I have is:
$$f(n) = \dfrac{15n^5+23n^4-20n^3-56n^2-48n-16}{n^3(n+1)(n+2)^4}\begin{pmatrix}2n \\ n \end{pmatrix} + 4\dfrac{3n^2+6n+4}{n^3 (n+2)^3}$$
The only method I have thought of is that I could try to prove that $\sin\left(\pi f(n)\right)$ has an infinte number of roots. But that feels like kicking the can down the road as I don't know how to do that either!
Thank you for any and all help. Ben
Using partial fractions I get:
$$f(n) = \frac{\left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} - 2n-4}{(n+2)^4} - \frac{\begin{pmatrix} 2n \\ n \end{pmatrix} - 2}{n^3} - \frac{4\begin{pmatrix} 2n \\ n \end{pmatrix}}{n+1} $$
From here I can find conditions for each fraction separately.
$\frac{\begin{pmatrix} 2n \\ n \end{pmatrix} - 2}{n^3} \in \mathbb{N} \implies n \text{ is prime > 3 (I found this on A000984)}$
$\frac{4\begin{pmatrix} 2n \\ n \end{pmatrix}}{n+1} \in \mathbb{N} \implies n \in \mathbb{N} \text{ (Always true)}$
$$f(n) = \frac{\left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} - 2n-4}{(n+2)^4} - \color{Blue}{\frac{\begin{pmatrix} 2n \\ n \end{pmatrix} - 2}{n^3} - \frac{4\begin{pmatrix} 2n \\ n \end{pmatrix}}{n+1}} $$
This leaves the final condition:
$\frac{\left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} - 2n-4}{(n+2)^4} \in \mathbb{N} \\ \implies \left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} \equiv 2n+4\pmod{n^4+8n^3+24n^2+32n+16}$
This seems to hold true when $n+2$ is a prime.
So if this function outputs an infinite number of integers, the twin-prime conjecture should be true.
If $n$ is the lower of a pair of twin primes, then $f(n)$ is an integer.
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Here is a proof that if both $n$ and $n+2$ are prime ($n>3$), then the output is integer:
clearly it suffices to show that if $n+2$ is prime then $$ \left(4n^3+28n^2+84n+76\right)\binom{2n}{n} \equiv 2(n+2) \bmod{(n+2)^4}.$$
Let $p=n+2$ be prime, then this is equivalent to showing that:
$$\tag1 2\left(p^3+p^2+5p-3\right)\binom{2p-4}{p-2} \equiv p \bmod{p^4}.$$
Proof
Since $n>3$, $p>3$ and by Wolstenhome's theorem we have:
\begin{align}
\binom{2p-1}{p-1}&\equiv 1 \bmod p^3\\
p\binom{2p-1}{p-1}&\equiv p \bmod p^4\\
p\frac{2p-1}{p-1}\frac{2p-2}{p-2}\frac{2p-3}{p-3}\binom{2p-4}{p-4}&\equiv p \bmod p^4.
\end{align}
But $$ \binom{2p-4}{p-2}=\binom{2p-4}{p-4}\frac{p-1}{p-3}\frac{p}{p-2}.$$
Then
\begin{align}
p\frac{2p-1}{p-1}\frac{2p-2}{p-2}\frac{2p-3}{p-3}\frac{p-3}{p-1}\frac{p-2}{p}\binom{2p-4}{p-2}&\equiv p \bmod p^4
\end{align}
That is
\begin{align}
\frac{(2p-1)(2p-2)(2p-3)}{(p-1)^2}\binom{2p-4}{p-2}&\equiv p \bmod p^4
\end{align}
But by doing the power expansion of $\frac{1}{(p-1)^2}$ , we see that
$$ \frac{1}{(p-1)^2} \equiv 1+2p+3p^2+4p^3 \bmod p^4$$
and we are done since it is easy to see by expansion that $$(2p-1)(2p-2)(2p-3)(1+2p+3p^2+4p^3) \equiv 2(-3+5p+p^2+p^3) \bmod p^4.$$
I does seem that for the OP function to bring an integer, the argument must be the smallest prime of a twin prime pair. But proving this seems very difficult. What about trying to disprove it by computer search of a counterexemple ?
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Trigonometric inequality $ 3\cos ^2x \sin x -\sin^2x <{1\over 2}$ I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3\cos ^2x \sin x -\sin^2x <{1\over 2}$$
And with no succes. Of course if we write $s=\sin x$ then $\cos^2 x = 1-s^2$ and we get $$6s^3+2s^2-6s+1>0$$
But this one has no rational roots so here stops. I suspect that Cardano wasn't in a mind of a problem proposer. There must be some trigonometric trick I don't see. I also tried with $$\sin 3x = -4s^3+3s$$ but don't now what to do with this. Any idea?
Offical solution is a union of $({(12k-7)\pi \over 18},{(12k+1)\pi\over 18})$ where $k\in \mathbb{Z}$
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We consider the inequality you found:
$6s^3+2s^2-6s+1>0$, for $s=\sin x$
We compare left side with following equation:
$8s^3-4s^2-4s+1=0$
Which have solutions: $s=\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}, \cos \frac{5\pi}{7}$
We have:
$$8s^3-4s^2-4s+1>2s^3-6s^2-2s$$
That means we can write:
$2s^3-6s^2-2s<0$ for $s=\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}, \cos \frac{5\pi}{7}$
Then we have:
$\sin x=\cos \frac{\pi}{7}=\sin (\frac{\pi}{2}-\frac {\pi}{7})⇒ x=\frac{5\pi}{14}$
Similarly $x=\frac{\pi}{14}$ and $x=\frac{-3\pi}{14}$.
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On calculating the limit of the infinite product $\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$
Let $S_n=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$. What is the value of $\lim_{n \to \infty} S_n$ ?
What I attempted:-
$\log S_n=\sum_{k=3}^n \log (1-\tan^4\frac{\pi}{2^k})$.
Since $\lim_{x \to 0} \frac{\tan x}{x}=1$, $\tan^4\frac{\pi}{2^k}\approx \left(\frac{\pi}{2^k}\right)^4$
Thus, $\log S_n=\sum_{k=3}^n \log (1-\frac{\pi^4}{2^{4k}})\approx \sum_{k=3}^n\left( -\frac{\pi^4}{2^{4k}}\right) $
Taking limit as $n \to \infty$, $\lim_{n \to \infty} \log S_n=\frac{-\pi^4}{3840}$.
Finally, $\lim_{n \to \infty}S_n=e^{\frac{-\pi^4}{3840}}\approx 1-\frac{\pi^4}{3840}$
Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $\frac{\pi^3}{4},\frac{\pi^3}{16}, \frac{\pi^3}{32},\frac{\pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.
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You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-\tan^4(\alpha/2))=(1+\tan^2(\alpha/2))(1-\tan^2(\alpha/2))=
\frac{4}{\cos(\alpha)}\left(\frac{\tan(\alpha/2)}{\tan(\alpha)}\right)^2.$$
Hence, as $n$ goes to infinity,
$$\prod_{k=3}^n \left(1-\tan^4(\pi/2^k)\right)=\frac{4^{n-2}}{\prod_{k=3}^n\cos(\pi/2^{k-1})}\cdot \left(\prod_{k=3}^n\frac{\tan(\pi/2^k)}{\tan(\pi/2^{k-1})}\right)^2\\=4^{n-2}\cdot 2^{n-2}\sin(\pi/2^{n-1})\cdot \left(\frac{\tan(\pi/2^n)}{\tan(\pi/2^{2})}\right)^2\to \frac{\pi^3}{32}$$
where we used the known fact that
$$\prod\limits_{k=2}^{n}\cos\left(\frac{\pi }{2^{k}}\right)=
\frac{1}{2^{n-1}\sin(\pi/2^n)}$$
(see for example How to evaluate $\lim\limits_{n\to \infty}\prod\limits_{r=2}^{n}\cos\left(\frac{\pi}{2^{r}}\right)$).
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How do people come up with solutions like this? I was going through some problems in high school textbook and stumbled on this problem.
I could be trying to solve it whole day and I wouldn't solve it. The solution seems to be too complicated for high school student (I marked the part red I can't understand). Can you explain how do you go about solving this problem and what is going through your mind when doing so?
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Through practice, you become more familiar with various techniques which you can use to make simplification easier and faster.
The previous answer provides a clever substitution which simplifies the problem greatly, but you can solve the question rather easily even without it. In the example, you have
$$\frac{-2ab^{\frac{4}{3}}+2a^{\frac{5}{3}}b^{\frac{2}{3}}}{a^2b^{\frac{4}{3}}-b^2a^{\frac{4}{3}}}$$
In order to understand the steps given for the solution, you can break the problem down into two parts. Focusing on the numerator first, you can rewrite it as
$$-2abb^{\frac{1}{3}}+2aa^{\frac{2}{3}}b^{\frac{2}{3}}$$
because we have $a^{b+c} = a^ba^c$, so, for instance $b^{\frac{4}{3}} = b^{1+\frac{1}{3}} = bb^{\frac{1}{3}}$. Using this technique again, you can split $b$ into $b^{\frac{1}{3}}b^{\frac{2}{3}}$, so you get
$$-\color{blue}{2ab^{\frac{2}{3}}}b^{\frac{1}{3}}b^{\frac{1}{3}}+\color{blue}{2a}a^{\frac{2}{3}}\color{blue}{b^{\frac{2}{3}}}$$
Notice how both sides share common factors. You have $ab+ac = a(b+c)$, so the expression can once more be rewritten as
$$2ab^{\frac{2}{3}}\left(-b^{\frac{2}{3}}+a^{\frac{2}{3}}\right) \tag{1}$$
In fact, in the picture you’ve given, the simplification wasn’t immediately made. Rather, all the steps were shown, as in $ab+ac = a\left(\frac{ab}{a}+\frac{ac}{a}\right)$ rather than immediately $ab+ac = a(b+c)$.
Repeat the same process for the denominator (rewriting and factoring):
$$a^2b^{\frac{4}{3}}-b^2a^{\frac{4}{3}} = a^{\frac{2}{3}}\color{blue}{a^{\frac{4}{3}}b^{\frac{4}{3}}}-b^{\frac{2}{3}}\color{blue}{b^{\frac{4}{3}}a^{\frac{4}{3}}} = a^{\frac{4}{3}}b^{\frac{4}{3}}\left(a^{\frac{2}{3}}-b^{\frac{2}{3}}\right) \tag{2}$$
Putting it all together, you reach
$$\frac{2ab^{\frac{2}{3}}\color{blue}{\left(-b^{\frac{2}{3}}+a^{\frac{2}{3}}\right)}}{a^{\frac{4}{3}}b^{\frac{4}{3}}\color{blue}{\left(a^{\frac{2}{3}}-b^{\frac{2}{3}}\right)}} = \frac{2ab^{\frac{2}{3}}}{a^{\frac{4}{3}}b^{\frac{4}{3}}} = \frac{2ab^{\frac{2}{3}}}{aa^{\frac{1}{2}}b^{\frac{2}{3}}b^{\frac{2}{3}}} = \frac{2}{a^{\frac{1}{3}}b^{\frac{2}{3}}}$$
As you can see, there really was nothing apart from manipulating the exponents and factoring. At first, this might seem long, tricky, and time-consuming. You need some practice in order to see these ideas and patterns quickly, and you’ll eventually solve these problems very quickly.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.
My attempt:
$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\rangle\\ \implies&\begin{cases} -x+6=\lambda(4x-24)\\-y+\frac{13}{2}=\lambda(-4y+26)\\x^2+(12-x)^2-y^2-(13-y)^2=0\end{cases}\end{align*}$
But clearly there is no solution due to the first two equations.
Using Wolfram Alpha, however, yields a maximum at $\displaystyle \left(\frac{17}{2},\,\frac{13}{2}\right)$ being $A=36$ and shows a nice little graph.
|
Let $x(12-x)=a$ and $y(13-y)=b$.
Thus, the condition gives $b=a+12.5$.
Also, we have
$$a=x(12-x)\leq\left(\frac{x+12-x}{2}\right)=36$$ and
$$b=y(13-y)\leq\left(\frac{y+13-y}{2}\right)=42.25,$$
which gives
$$a=b-12.5\leq42.25-12.5=29.75.$$
Id est, $$A(x,y)=\frac{1}{2}(a+b)=a+6.25\leq29.75+6.25=36.$$
The equality occurs for $b=42.25$ or $y=6.5$, which says that we got a maximal value.
|
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|
Not getting the right answer with alternate completing the square method on $\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$ So I've looked up how to do this problem and when they complete the square it's using the $(\frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1\over4}$$ and then
$$x^2-x+{1\over4}=(x-\frac{1}{2})^2$$
which I understand, but is not the first option that popped into my head.
What I don't understand is why all the tutorials I've looked up are not completing the square in the traditional method. Here's my work:
$\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$
denominator:
$$3+4x-4x^2$$
$$(-4x^2+4x-1)+4$$
$$-(4x^2-4x+1)+4$$
$$-(2x-1)^2+4$$
So the integral is now:
$$\int\frac{x^2}{(4-u^2)^\frac{3}{2}}$$
So I do the rest of the work by using trig substitution now:
$u=2x-1$
$x=\frac{u+1}{2}$
Now I subbed in the trig identities:
$u=2\sin\theta$
$du=2\cos\theta d\theta$
$$\int\frac{(\frac{u+1}{2})^22\cos\theta}{\sqrt{4-4\sin^2\theta}^3}d\theta$$
$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{\sqrt{4\cos^2\theta}^3}d\theta$$
$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{(2\cos\theta)^3}d\theta$$
$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{\frac{(4\sin^2\theta+4\sin\theta+1)}{4}2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{(\frac{4\sin^2\theta}{4}+\frac{4\sin\theta}{4}+\frac{1}{4})2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})}{8\cos^3\theta}*\frac{2\cos\theta}{1}d\theta$$
$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})}{4\cos^2\theta}d\theta$$
$$\int\frac{\sin^2\theta}{4\cos^2\theta}d\theta+\int\frac{\sin\theta}{4\cos^2\theta}d\theta+\int\frac{\frac{1}{4}}{4\cos^2\theta}d\theta$$
$$\frac{1}{4}\int \tan^2\theta d\theta+\frac{1}{4}\int\frac{\sin\theta}{\cos^2\theta}d\theta+\int\frac{1}{16\cos^2\theta}d\theta$$
For the second integral, I subbed $u=\cos\theta$ and $-du=\sin\theta d\theta$
$$\frac{1}{4}\int \sec^2\theta-1d\theta-\frac{1}{4}\int\frac{du}{u^2}+\frac{1}{16}\int \sec^2\theta d\theta$$
$$[\frac{\tan\theta}{4}-\frac{\theta}{4}]+[\frac{1}{4\cos\theta}]+[\frac{\tan\theta}{16}]+C$$
On a right triangle where $\sin\theta=\frac{2x-1}{2}$ and $\cos\theta=\frac{\sqrt{4-(2x-1)^2}}{2}$, I subbed these values back in to make the integral in terms of $x$:
$$\frac{2x-1}{4\sqrt{4-(2x-1)^2}}-\frac{\arcsin(\frac{2x-1}{2})}{4}+\frac{1}{4(\frac{\sqrt{4-(2x-1)^2}}{2})}+\frac{2x-1}{16\sqrt{4-(2x-1)^2}}+C$$
This does not seem to be the right answer when I compare it to a calculator answer, so I'm guessing I either completed the square wrong or made an algebraic mistake? I've done this problem three or four times now and am getting a little frustrated over what I'm guessing is a easy fix. If anyone can help, I'd really appreciate it.
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Find $I=\int\frac{x^2}{(3+4x-4x^2)^{3/2}}dx$.
First 4 steps:
*
*Change $x=t+\frac12$
$$I=\int\frac{t^2+t+\frac14}{8(1-t^2)^{3/2}}dt\\=\frac18\int\frac{t^2}{(1-t^2)^{3/2}}dt+
\frac18\int\frac{t}{(1-t^2)^{3/2}}dt+\frac1{32}\int\frac{1}{(1-t^2)^{3/2}}dt$$
*$$\int\frac{t}{(1-t^2)^{3/2}}dt=
\frac{1}{\sqrt{1-{{t}^{2}}}}$$
*$$\int\frac{t^2}{(1-t^2)^{3/2}}dt=\frac{t}{\sqrt{1-{{t}^{2}}}}-\arcsin(t)$$
*$$\int\frac{1}{(1-t^2)^{3/2}}dt=\frac{t}{\sqrt{1-{{t}^{2}}}}$$
|
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|
Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$
$$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=\pm \sqrt{\frac{2}{3}}$
But $x \ne 0$ since $xy=-1$
$$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$
Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$
Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$
Is there any other approach?
|
Using $xy = -1$, we can put the fractions over a common denominator as
$${4\over{4 - x^2}} + {9\over{9 - y^2}} = {{72 - 9x^2 - 4y^2}\over{37 - 9x^2 - 4y^2}} = 1 + {35\over{37 - (9x^2 + 4y^2)}}$$
So we need to minimize $9x^2 + 4y^2$. Since $9x^2 \times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2 = 4/x^2$, which gives the same result as you found.
|
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|
Integrating positive function on interval from -1 to 1 but result is negative I have the following function:
$$\int_{-1}^1 \frac1 {x^4} dx$$
and my result seems to be:
$$\frac {-2}3$$
Why is my result negative? My function is always positive, does it have to do with the fact that since the interval extends from -1 to 1 but the function doesn't go there that it somehow drastically affects the total integral?
|
Since
$\dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$\int_{-1}^1 \frac1 {x^4} dx
$
is
$\lim_{c \to 0^+} (\int_{-1}^{-c} \frac1 {x^4} dx+\int_{c}^1 \frac1 {x^4} dx)
$.
We have
$\int \dfrac{dx}{x^4}
=\int x^{-4} dx
=\dfrac{x^{-3}}{-3}
=-\dfrac{1}{3x^3}
$
so
$\begin{array}\\
\int_{-1}^{-c} \frac1 {x^4} dx
&=-\dfrac{1}{3x^3}|_{-1}^{-c}\\
&=-(\dfrac{1}{3(-c)^3}-\dfrac{1}{3(-1)^3})\\
&=-(-\dfrac{1}{3c^3}+\dfrac{1}{3})\\
&=\dfrac{1}{3c^3}-\dfrac{1}{3}\\
\end{array}
$
and
$\begin{array}\\
\int_{c}^{1} \frac1 {x^4} dx
&=-\dfrac{1}{3x^3}|_{c}^{1}\\
&=-(\dfrac{1}{3}-\dfrac{1}{3c^3})\\
&=\dfrac{1}{3c^3}-\dfrac{1}{3}\\
\end{array}
$
Adding these we get
$\dfrac{2}{3c^3}-\dfrac{2}{3}
$,
and the
$\dfrac{2}{3c^3}
$
overwhelms the
$-\dfrac{2}{3}
$.
|
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|
Solve recurrence relation: $T(n) = \frac{n}{n+1}T(n-1) + 1$ I am not able to solve this recurrence relation by substitution and variable change method.
$$T(n) = \frac{n}{n+1}T(n-1) + 1;\ \ T(1) = 1 $$
|
(n+1)T(n) = nT(n-1) + n+1
Let
S(n) = (n+1)T(n) S(0) = T(0) = c
S(n) = S(n-1) + n+1
Since, S(n-1) = S(n-2) + n
therefore, S(n) = S(n-2) + n+1 + n
for k terms...
S(n) = S(n-k) + (n+1) +n +.....+ (n-k+2)
k = n
S(n) = S(0) + (n+1) + n + ...+ (n-k+2)
= c + (n+1) + n + (n-1) + ......+ 2 + 1 - 1
= (n+1)(n+2)/2 + (c-1)
= (n^2 + 3n + 2)/2 + c - 1
= n^2/2 + 3n/2 + c
S(n) = O(n^2)
from our substitution -
T(n)(n+1) = O(n^2)
T(n)=1/(n+1) O(n^2)
We can neglect +1 in n+1
then,
T(n) = O(n)
|
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|
What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\right)\,$ independently, but that didn't go well for me.
*Multiplying and dividing by $\cos^2x$ or $\sin^2x$.
*Expressing $\cos^2x$ as $1-\sin^2x$ and splitting the integral, and I was stuck with $\int \left(\frac{\sin^3x}{\sin x - \cos x}\right)\, dx$ which I rewrote as $\int \frac{\csc^2x}{\csc^4x (1-\cot x) } dx,\,$ and tried a whole range of substitutions only to fail.
*I tried to substitute $\frac{1}{ \sin x - \cos x}$, $\frac{\sin x}{ \sin x - \cos x}$, $\frac{\cos x \sin x}{ \sin x - \cos x}$ and $\frac{\cos^2x \sin x}{\sin x - \cos x},$ independently, none of which seemed to work out.
*I expressed the denominator as $\sin\left(\frac{\pi}{4}-x\right)$ and tried multiplying and dividing by $\sin\left(\frac{\pi}{4}+x\right)$, and carried out some substitutions. Then, I repeated the same with $\cos\left(\frac{\pi}{4}+x\right)$. Neither of them worked.
|
$$\int\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}dx=$$
$$=\int\left(\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}+\frac{1}{2}\sin{x}(\sin{x}+\cos{x})\right)dx-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$
$$=\frac{1}{2}\int\frac{\sin{x}}{\sin{x}-\cos{x}}-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$
$$=\frac{1}{2}\int\left(\frac{\sin{x}}{\sin{x}-\cos{x}}-\frac{1}{2}\right)dx-\frac{1}{2}\int\left(\sin{x}(\sin{x}+\cos{x})-\frac{1}{2}\right)dx=$$
$$=\frac{1}{4}\int\frac{\sin{x}+\cos{x}}{\sin{x}-\cos{x}}dx-\frac{1}{2}\int\left(\sin{x}(\sin{x}+\cos{x})-\frac{1}{2}\right)dx=$$
$$=\frac{1}{4}\ln|\sin{x}-\cos{x}|-\frac{1}{4}\int\left(2\sin^2{x}-1+\sin2x\right)dx.$$
Can you end it now?
|
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|
How to find the equation of the conic before applying the rotation? Given the rotation matrix:
$$Q=\begin{bmatrix}\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\\-\frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{bmatrix},$$ I want to find the equation of the conic $C$ given by $$x^2+4xy+4y^2-10\sqrt{5}x=0,$$ which I know is a parabola, before applying the rotation to it.
I was thinking about solving the system of equation $$\begin{cases}x=\frac{2}{\sqrt{5}}\tilde{x}+\frac{1}{\sqrt{5}}\tilde{y}\\ y=-\frac{1}{\sqrt{5}}\tilde{x}+\frac{2}{\sqrt{5}}\tilde{y}\end{cases},$$ and then substituting $x$ and $y$ in the initial equation of the conic. The result should be $$5\tilde{y}^2-20\tilde{x}-10\tilde{y}=0,$$ but it seems that my results are different.
|
Your approach is entirely correct; plugging in your substitutions into the first equation for $C$ yields the result you claim it should be:
\begin{eqnarray*}
x^2+4xy+4y^2-10\sqrt{5}x&=&
\left(\frac{2}{\sqrt{5}}\tilde{x}+\frac{1}{\sqrt{5}}\tilde{y}\right)^2+4\left(\frac{2}{\sqrt{5}}\tilde{x}+\frac{1}{\sqrt{5}}\tilde{y}\right)\left(-\frac{1}{\sqrt{5}}\tilde{x}+\frac{2}{\sqrt{5}}\tilde{y}\right)\\
&\ &+4\left(-\frac{1}{\sqrt{5}}\tilde{x}+\frac{2}{\sqrt{5}}\tilde{y}\right)^2-10\sqrt{5}\left(\frac{2}{\sqrt{5}}\tilde{x}+\frac{1}{\sqrt{5}}\tilde{y}\right)\\
&=&\frac{1}{5}(4\tilde{x}^2+4\tilde{x}\tilde{y}+\tilde{y}^2)+\frac{4}{5}(-2\tilde{x}^2+3\tilde{x}\tilde{y}+2\tilde{y}^2)\\
&\ &+\frac{4}{5}(\tilde{x}^2-4\tilde{x}\tilde{y}+4\tilde{y}^2)-10(2\tilde{x}+\tilde{y})\\
&=&5\tilde{y}^2-20\tilde{x}-10\tilde{y}.
\end{eqnarray*}
You seem to have made a mistake in your calculation.
|
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|
Help with how to show aritmetic progression question. How can I show that if $(\chi_{n})$ is a aritmetic progression, then:
$$\frac{1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{2}} } +
\frac{1}{ \sqrt{\chi_{2}} + \sqrt{\chi_{3}} } + \cdots +
\frac{1}{ \sqrt{\chi_{n-1}} + \sqrt{\chi_{n}} } =
\frac{n-1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{n}} }
$$
|
$ \frac{1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{2}} } +
\frac{1}{ \sqrt{\chi_{2}} + \sqrt{\chi_{3}} } + \cdots +
\frac{1}{ \sqrt{\chi_{n-1}} + \sqrt{\chi_{n}} } = \frac{\sqrt{\chi_{2}} - \sqrt{\chi_{1}} }{ \chi_2-\chi_1 } + \frac{\sqrt{\chi_{3}} - \sqrt{\chi_{2}} }{ \chi_3-\chi_2 } + \cdots +
\frac{\sqrt{\chi_n}-\sqrt{\chi_{n-1}}}{ \chi_{n}-\chi_{n-1} } $
But $\chi_i-\chi_{i-1}=d$, the common difference of the arithmetic sequence.
$\frac{\sqrt{\chi_{2}} - \sqrt{\chi_{1}} }{ \chi_2-\chi_1 } + \frac{\sqrt{\chi_{3}} - \sqrt{\chi_{2}} }{ \chi_3-\chi_2 } + \cdots +
\frac{\sqrt{\chi_n}-\sqrt{\chi_{n-1}}}{ \chi_{n}-\chi_{n-1} }=\frac{1}{d}\{\sqrt{\chi_{2}} - \sqrt{\chi_{1}}+\sqrt{\chi_{3}} - \sqrt{\chi_{2}}+\cdots+\sqrt{\chi_{n}} - \sqrt{\chi_{n-1}}\}$
$=\frac{\sqrt{\chi_{n}} - \sqrt{\chi_{1}}}{d}$
But since $\chi_n=\chi_1+(n-1)\cdot d$, we have $\frac{1}{d}=\frac{n-1}{\chi_n-\chi_1}$
Hence $\frac{\sqrt{\chi_{n}} - \sqrt{\chi_{1}}}{d}=\frac{(n-1)\cdot {\left(\sqrt{\chi_{n}} - \sqrt{\chi_{1}}\right)} }{\chi_n-\chi_1}=\frac{n-1}{\sqrt{\chi_{1}} + \sqrt{\chi_{n}}}$
|
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|
Write a Limit to calculate $f'(0)$ Let $f(x) = \frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx
((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx
((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx
(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)
(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)
(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)
that's as far as I have gotten. Any input would be great.
|
it is $$\frac{f(x+h)-f(x)}{h}=\frac{\frac{2}{1+(x+h)^2}-\frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2\,{\frac {h \left( h+2\,x \right) }{ \left( {h}^{2}+2\,xh+{x}^{2}+1
\right) \left( {x}^{2}+1 \right) }}
$$
|
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|
Finding real $a$, $b$, $c$ such that $x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$ has $1+i$ as a zero, and one negative integer as a zero with multiplicity $2$
Find $a, b, c \in \mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$$
I know that one zero of $p$ is $1-i$, so I can reduce it one degree down after dividing with $x-2$, but after that I'm stuck.
I tried using Vieta's formulas, but with no result.
|
Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$
We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
$$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
$$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
$$b=3-a,$$ $$c=2a-2$$ and
$$p(x)=(x^2-2x+2)((x^3-2x-1+a(x+1)).$$
Now, let $x^3-2x-1+a(x+1)=q(x).$
Thus, $q$ and $q'$ have the same negative integer root.
$$q'(x)=3x^2-2+a,$$ which gives $$a=2-3x^2.$$
Id est, for the same root we obtain the following equation.
$$x^3-2x-1+(2-3x^2)(x+1)=0$$ or
$$2x^3+3x^2-1=0$$ or
$$2x^3+4x^2+2x-x^2-2x-1=0$$ or
$$(x+1)^2(2x-1)=0,$$
which gives $x=-1$ and $a=-1$.
|
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|
$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$ when $a,b$ are integers? Let $a$ and $b$ be positive integers.
If $b$ is even, then we have $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$$
I think the equality also hold when $b$ is odd. What could be a proof for it?
|
The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then
$$\left\lfloor\frac02\right\rfloor+\left\lceil\frac02\right\rceil=0$$
and
$$\left\lfloor\frac12\right\rfloor+\left\lceil\frac12\right\rceil=1$$
are enough as a proof.
|
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}
|
If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
|
Hint — using this changing variable $y = cos^2(x)$ —
$$y^3 + y^2-2y = 0 \Rightarrow y(y^2+y-2) = 0 \Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 \leq y \leq 1$:
$$y = 0, 1 \Rightarrow cos^2(x) = 0, 1 \Rightarrow x = 0, \frac{\pi}{2},$$
for $x \in [0, \frac{\pi}{2}]$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Idea for $\lim\limits_{x \to \frac{\pi}{2}} \left( \tan \left( \tfrac{\pi}{4} \sin x\right)\right)^{1/ ( \tan (\pi \sin x))}$ $$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital
|
Using trigonometric identities, by $y=\frac \pi 2 \sin x \to \frac \pi 2$ and $t=\frac \pi 2 -y \to 0$, we obtain:
*
*$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 +
\cos \theta} \implies \tan \left(\frac{\pi}{4}\sin x\right)=\frac{\sin y}{1 +
\cos y}=\frac{\cos t}{1 +
\sin t}$
*$\tan (2\theta) = \frac{2 \tan \theta} {1 - \tan^2 \theta} \implies \tan(\pi \sin x)=\frac{2 \tan y} {1 - \tan^2 y}=\frac{2 \cot t} {1 - \cot^2 t}=\frac{2\cos t \sin t}{\sin^2 t-\cos^2t}$
then
$$\tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=e^{{\dfrac {\log \left(\tan \left(\frac{\pi}{4}\sin x\right)\right)}{\tan(\pi \sin x)} }}
=e^{\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\log\left(\frac{\cos t}{1 +
\sin t} \right)}\to e^\frac12$$
indeed
$$\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot \log\left(\frac{\cos t}{1 +
\sin t} \right)=\\=\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot\frac{\cos t-\sin t-1}{1+\sin t}\cdot\log\left(1+\frac{\cos t-\sin t-1}{1 +
\sin t} \right)^{\frac{1+\sin t}{\cos t-\sin t -1}}\to\frac 12 \cdot \log e=\frac 12$$
since
$$\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot\frac{\cos t-\sin t-1}{1+\sin t}=\frac{\cos^2t-\sin^2 t}{\cos t(1+\sin t)}\cdot\frac{\sin t+1-\cos t}{2\sin t}=1\cdot \frac12$$
and by standard limits
$$\frac{\sin t+1-\cos t}{2\sin t}=\frac{ \frac{\sin t}{t}+t\frac{1-\cos t}{t^2} } {2\frac {\sin t} t}\to \frac{1+0\cdot 1}{2\cdot 1}=\frac12$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
|
Hint:
As the roots of the denominator are the complex fifth roots of unity, you find the factorization to be
$$1+x+x^2+x^3+x^4=\left(x^2+\frac{1+\sqrt5}2x+1\right)\left(x^2+\frac{1-\sqrt5}2x+1\right).$$
Then we can find a linear combination to obtain a quadratic polynomial with equal coefficients,
$$\left(\sqrt5+1\right)\left(x^2+\frac{1-\sqrt5}2x+1\right)+\left(\sqrt5-1\right)\left(x^2+\frac{1+\sqrt5}2x+1\right)=2\sqrt5(x^2+x+1)$$ which leads us to the decomposition in simple fractions.
Now,
$$\int_0^1\frac{dx}{x^2+2ax+1}=\int_0^1\frac{dx}{(x+a)^2+1-a^2}=\left.\frac1{\sqrt{1-a^2}}\arctan\frac{x+a}{\sqrt{1-a^2}}\right|_0^1.$$
|
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|
System of linear recurrences; finding an explicit description for both sequences involved. $$\left\{\begin{aligned} a_n &&= &&2a_{n-1} + b_{n-1} + a_{n-2} - b_{n-2} && n \ge 2 && (1)\\ b_n &&=&& b_{n-1} + b_{n-2} - a_{n-2} && n \ge 2&& (2)\end{aligned}\right.$$
with $a_0 = 5, a_1 = 3, b_0 = 0, b_1 = 3$.
From (2) I get: $a_{n-2} = b_{n-1}+b_{n-2} - b_n$. Substituting in (1): $a_n = 2a_{n-1} + 2b_{n-1} - b_n$.
Now I'm stuck. I don't see what I can do next...
|
Hint:
Adding the two equations gives
$$
a_n + b_n = 2(a_{n-1} + b_{n-1}) \quad {\rm{for}} \quad n \ge 2
$$
so
$$
a_n + b_n = 2^{n-1}(a_{1} + b_{1}) = 6 \cdot 2^{n-1}
$$
Plugging this into the second recursion gives
$$
b_n = b_{n-1} + 2 b_{n-2} - (a_{n-2} + b_{n-2}) = b_{n-1} + 2 b_{n-2} -6 \cdot 2^{n-3}
$$
Likewise,the first equation gives
$$
a_n = a_{n-1} + (a_{n-1}+ b_{n-1}) + 2 a_{n-2} - (a_{n-2} + b_{n-2}) = \\
= a_{n-1} + 6 \cdot 2^{n-2} + 2 a_{n-2} - 6 \cdot 2^{n-3} = \\
= a_{n-1} + 2 a_{n-2}+ 3 \cdot 2^{n-2}
$$
So there are two single-variable recursions which can be solved.
|
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|
Improving the bound for $\sigma(q^k)/q^k$ where $q^k n^2$ is an odd perfect number given in Eulerian form Let $x$ be a positive integer. (That is, let $x \in \mathbb{N}$.)
We denote the sum of divisors of $x$ as
$$\sigma(x) = \sum_{d \mid x}{d}.$$
We also denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.
If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, i.e. $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $k \equiv 1 \pmod 4$, then we have
$$\frac{q+1}{q} = I(q) \leq I(q^k) = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$
Since $q$ is prime and satisfies $q \equiv 1 \pmod 4$, we have $q \geq 5$, from which we obtain
$$\frac{1}{q} \leq \frac{1}{5} \implies -\frac{1}{q} \geq - \frac{1}{5} \implies \frac{q-1}{q} = 1 - \frac{1}{q} \geq 1 - \frac{1}{5} = \frac{4}{5}.$$
We get that
$$I(q^k) < \frac{q}{q - 1} \leq \frac{5}{4}$$
from which we conclude that
$$I(q^k) < \frac{5}{4}.$$
We also obtain
$$I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < \sqrt{I(n^2)} = \sqrt{\frac{2}{I(q^k)}},$$
from which we get
$$\bigg(I(q^k)\bigg)^2 < \frac{2}{I(q^k)} \implies \bigg(I(q^k)\bigg)^3 < 2 \implies I(q^k) < \sqrt[3]{2}.$$
Here is my question:
Is it possible to improve on the inequality
$$I(q^k) < \sqrt{\frac{2}{I(q^k)}},$$
to something like (say)
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}?$$
Note that $1/3 = 0.\overline{333}$.
MOTIVATION
Here is the reason why I think it might be possible to improve on the bound for $I(q^k)$.
Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.
The Descartes-Frenicle-Sorli Conjecture predicts that $k=1$. Suppose that this Conjecture holds.
Then
$$I(q^k) = I(q) = \frac{q+1}{q} = 1+\frac{1}{q} \leq 1+\frac{1}{5}=\frac{6}{5}$$
and
$$I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} \geq \frac{2}{\frac{6}{5}} = \frac{5}{3}.$$
Let
$$\frac{6}{5} = \bigg(\frac{5}{3}\bigg)^y.$$
Note that we then have that
$$I(q) \leq \frac{6}{5} = \bigg(\frac{5}{3}\bigg)^y \leq \bigg(I(n^2)\bigg)^y = \bigg(\frac{2}{I(q)}\bigg)^y$$
where
$$y = \frac{\log\bigg(\frac{6}{5}\bigg)}{\log\bigg(\frac{5}{3}\bigg)} \approx 0.356915448856724.$$
WLOG, if we assume that $k>1$ and let
$$\frac{5}{4} = \bigg(\frac{8}{5}\bigg)^z,$$
then we have that
$$I(q^k) < \frac{5}{4} = \bigg(\frac{8}{5}\bigg)^z < \bigg(I(n^2)\bigg)^z = \bigg(\frac{2}{I(q^k)}\bigg)^z$$
where
$$z = \frac{\log\bigg(\frac{5}{4}\bigg)}{\log\bigg(\frac{8}{5}\bigg)} \approx 0.474769847356948651282146696312271.$$
|
It turns out that
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}$$
implies
$$1 + \frac{1}{q} = I(q) \leq I(q^k) < \sqrt[4]{2}$$
from which we obtain
$$q > \bigg(\sqrt[4]{2} - 1\bigg)^{-1} \approx 5.2852135$$
thereby giving
$$q \geq 13$$
since $q$ is a prime satisfying $q \equiv 1 \pmod 4$. Thus, the implication
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}} \implies q \geq 13$$
holds.
If the reverse inequality
$$I(q^k) > \sqrt[3]{\frac{2}{I(q^k)}}$$
holds, then we get
$$\frac{q}{q - 1} > I(q^k) > \sqrt[4]{2}$$
which implies that
$$1 < q < \frac{\sqrt[4]{2}}{\sqrt[4]{2} - 1} \approx 6.28521$$
from which we conclude that $q = 5$.
Since $q \geq 5$ and $q$ is a prime satisfying $q \equiv 1 \pmod 4$, by the contrapositive of the last implication, we get the implication
$$q \geq 13 \implies I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}.$$
We therefore have the biconditional
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}} \iff q \geq 13.$$
(Note that we cannot have
$$I(q^k) = \sqrt[3]{\frac{2}{I(q^k)}}$$
as equality implies that $I(q^k) = \sqrt[4]{2}$, contradicting the fact that
$$I(q^k) = \frac{q^{k+1} - 1}{q^k (q - 1)}$$
is rational.)
Thus, to prove the inequality
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}$$
we need to rule out $q=5$. This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $k=1$.
|
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|
$ 0=1 $ ? Where is the mistake? I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=\sqrt{0-0+\sqrt{0-0+....}}$$
$$0=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equation I get,
$$1=\sqrt{1^{2}-1+\sqrt{1^{2}-1+.....}}$$
$$1=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$,—————->2
Which gives me that 0=1.Where is the mistake?
Edit:
Equation 1 can also be derived by the following method
$$1=\sqrt{0+\sqrt{1}}$$
$$1=\sqrt{0+\sqrt{0+\sqrt{1}}}$$
.
.
.
$$1=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$
Ok so let me tell you how I derived it.
I came across this particular expression,
$$x=\sqrt{1+\sqrt{1+\sqrt{1+.....}}}$$ which actually gives me the golden ratio,
So I chose n to be an integer and let x be the value of the following expression,
$$x=\sqrt{n+\sqrt{n+\sqrt{n+...}}}$$.
After solving for x you get it’s value to be!
$$x=\frac{1+\sqrt{1+4n}}{2}$$
(I took a positive sign since x is greater than or equal to 0)
So x can be an integer whenever n=0,2,6,10,20.....
When n=2,x=2(since x>0) and we get,
$$2=\sqrt{2+\sqrt{2+......}}$$
Similarly for n=6,x=3,
$$3=\sqrt{6+\sqrt{6+......}}$$
So x is an integer whenever $$n=k^{2}-k$$ where k is a non-negative integer.Substituting $$n=k^{2}-k$$ you get x=k.
After all this I get,
$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
|
Assuming convergence:
$$a=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
$$a=\sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=\sqrt{n^2-n+a}$ and is positive is correct.
|
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|
Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$
It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=\dfrac{1}{2}$, but I don't know how to use that.
|
the series expansion at $x=0$ of $ \ \cos(ax)=1-\frac{a^2 x^2}{2}+o(x^4) \ \ $ and
$\cos(a x) \cos(bx)=\frac{1}{2}\cos(x(a-b))+\frac{1}{2}\cos(x(a+b))$
the series of $ \ \frac{1}{2}\cos(x(a-b))+\frac{1}{2}\cos(x(a+b))=\frac{1}{2}(1-\frac{(a-b)^2 x^2}{2}+o(x^4) )+\frac{1}{2}(1-\frac{(a+b)^2 x^2}{2}+o(x^4) )=$
$\cos(a x) \cos(bx)=1-\frac{(a^2+b^2)x^2}{2}+o(x^4)$
so the series of $\cos(1 x) \cos(2x) \cos(3x)… \cos(n x) =1-\frac{(1^2+2^2+...+n^2) x^2}{2}+ o(x^4) = 1-\frac{n(n+1)(2n+1) x^2}{12}+ o(x^4)$
because $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
|
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|
Quick way of solving the contour integral $\oint \frac{1}{1+z^5} dz$ Consider the contour integral in the complex plane:
$$\oint \frac{1}{1+z^5} dz$$
Here the contour is a circle with radius $3$ with centre in the origin. If we look at the poles, they need to satisfy $z^5 = -1$. So the solutions of the poles are given by:
\begin{align*}
z_0 &= \cos(\frac{\pi}{5}) + i \sin(\frac{\pi}{5})\\
z_1 &= \cos(\frac{3\pi}{5}) + i \sin(\frac{3\pi}{5})\\
z_2 &= \cos(\pi) + i \sin(\pi) = -1\\
z_3 &= \cos(\frac{7\pi}{5}) + i \sin(\frac{7\pi}{5})\\
z_4 &= \cos(\frac{9\pi}{5}) + i \sin(\frac{9\pi}{5})
\end{align*}
So one can use Cauchy's formula or the residue theorem to calculate for every solution the integral and then adding them up to get the full integral. But I have the feeling that there needs to be a more simple way of calculating the full contour integral. Can one just calculate the integral for one solution $z_i$ (like the simple solution $-1$) and then multiply by it $5$, suggesting that the others have the same value. This would make the calculation much efficienter.
EDIT:
I now see that $4$ solutions are symmetric (the solutions except $z=-1$) in the complex plane. If one approximates the solutions of the poles in decimals, one finds:
\begin{align*}
z_0 &= 0.81 + 0.58i\\
z_1 &= -0.31 + 0.95i\\
z_2 &= -1\\
z_3 &= -0.31 -0.95i\\
z_4 &= 0.81 -0.58i
\end{align*}
So there are four symmetric solutions. For instance $z_0$ is symmetric with $z_4$, they are mirrored around the x-axis. Could this mean they cancell each other out so we only need to calculate the integral for $z_2 = -1$?
|
Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.
Let $\zeta$ be a primitive fifth root of unity. If you want to be explicit, let $\zeta = e^{2\pi/5}$. But the important thing is that
*
*the roots of $z^5+1$ are $-1$, $-\zeta$, $-\zeta^2$, $-\zeta^3$, and $-\zeta^4$.
*Since $(z^5 - 1) = (z-1)(z^4+z^3+z^2+z+1)$, we know that $1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4 = 0$.
Now for something about partial fractions that I didn't know until just now.
If $f(z) = (z-\alpha_1) \cdots (z-\alpha_n)$, and all the roots are distinct, then
$$
\frac{1}{f(z)} = \frac{1}{f'(\alpha_1)(z-\alpha_1)} + \frac{1}{f'(\alpha_2)(z-\alpha_2)} + \dots + \frac{1}{f'(\alpha_n)(z-\alpha_n)}
$$
To show this, start from
$$
\frac{1}{(z-\alpha_1) \cdots (z-\alpha_n)} = \sum_{i=1}^n \frac{A_i}{z-\alpha_i}
$$
for some constants $A_1, \dots, A_n$. Clear out the denominators, and you get
$$
1 = \sum_{i=1}^n \frac{A_i f(z)}{z-\alpha_i}
= \sum_{i=1}^n A_i \frac{f(z)-f(\alpha_i)}{z-\alpha_i}
\tag{$*$}
$$
Now for any $i$ and $j$,
$$
\lim_{z\to \alpha_j} \frac{f(z)-f(\alpha_i)}{z-\alpha_i} =
\begin{cases} 0 & i \neq j \\ f'(\alpha_i) & i = j \end{cases}
$$
So if we take the limit of both sides of ($*$) as $z\to z_j$ we get $1 = A_j f'(\alpha_j)$ for each $j$.
This means that
\begin{align*}
\frac{1}{z^5+1}
&= \frac{1}{5(z+1)} + \frac{1}{5(-\zeta)^4(z+\zeta)} + \frac{1}{5(-\zeta^2)^4(z+\zeta^2)} + \frac{1}{5(-\zeta^3)^4(z+\zeta^3)} + \frac{1}{5(-\zeta^4)^4(z+\zeta^4)}
\\&= \frac{1}{5}\left(\frac{1}{z+1} + \frac{1}{\zeta^4(z + \zeta)} + \frac{1}{\zeta^3(z + \zeta^2)} + \frac{1}{\zeta^2(z + \zeta^3)} + \frac{1}{\zeta(z + \zeta^4)}\right)
\\&= \frac{1}{5}\left(\frac{1}{z+1} + \frac{\zeta}{z + \zeta} + \frac{\zeta^2}{z + \zeta^2} + \frac{\zeta^3}{z + \zeta^3} + \frac{\zeta^4}{z + \zeta^4}\right)
\end{align*}
You wanted an argument involving symmetry; what could be more symmetric than that equation?
Now $C$ encloses all those roots, so by the Cauchy Integral Formula,
$$
\oint_C \frac{dz}{z^5+1} = \frac{2\pi i}{5}\left(1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4\right)
$$
and as we showed above, the latter factor is zero.
|
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|
Find all possible values of $x$ if $\frac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real
If the expression $\dfrac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real, find the set of all possible values of $x$.
My Attempt
$$
-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}=0\\
\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos x}+2\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}=0\\
\sin\frac{x}{2}=0\text{ or }\frac{\cos\frac{x}{2}}{\cos x}+\sin\frac{x}{2}+\cos\frac{x}{2}=0\\
\frac{x}{2}=n\pi\text{ or }\cos\frac{x}{2}+\sin\frac{x}{2}.\cos x+\cos\frac{x}{2}.\cos x=0\\
\boxed{x=2n\pi}\text{ or _____________}
$$
My reference gives the solution $x=2n\pi$, but what about the other remaining expression ?
|
The second expression $\cos \frac{x}{2} +\sin \frac{x}{2}\cos x+\cos \frac{x}{2}\cos x=0$ reduces to :
$$\cos \frac{x}{2} + \cos x(\sin \frac{x}{2}+ \cos \frac{x}{2})=0$$
$$\cos \frac{x}{2} = -\cos x(\sin \frac{x}{2}+ \cos \frac{x}{2})$$
$$\cos x = -\frac{1}{1+\tan \frac{x}{2}}$$
Apply the half angle formula for $$\cos x = \frac {1- \tan^2 \frac{x}{2}}{1+ \tan^2 \frac{x}{2}}=-\frac{1}{1+\tan \frac{x}{2}} $$
Now put $\tan \frac{x}{2} = t$ to get:
$$\frac {1- t^2 }{1+ t^2 }=-\frac{1}{1+t}$$
$$1+t-t^2-t^3=-1-t^2$$
$$t^3-t-2=0$$
Solve the equation to get the value for $\tan \frac{x}{2}$
You will get the other angle. Hope this helps .....
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why does the CRT formula yield a solution of a congruence system? I understand there is a method for solving simultaneous modular equations. For example;
$$x = 2 \mod{3}$$
$$x = 3 \mod{5}$$
$$x = 2 \mod{7}$$
We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ and $3 \cdot 5$. We then find the multiplicative inverses of these numbers with modulo equal to the number missing from the product. The numbers found are then 2, 1 and 1 in this case. The value of x is then given by:
$$x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1 = 233 = 23 \mod{3\cdot5\cdot7}$$
But I do not understand how this method correctly gives the value of $x$. I understand that the Chinese remainder theorem proves that there is a unique value of $0\le x \lt 3\cdot5\cdot7 \mod{3\cdot5\cdot7}$ but can someone please explain why this method finds this value of x?
|
Taking Bill Dubuque's graphic answer and graphically expanding on it:
$x = 2 \cdot\overbrace{ (5 \cdot 7) \cdot 2}^{\equiv 1 \pmod 3\\ \equiv 0 \pmod 5\\ \equiv 0 \pmod 7} + 3 \cdot \overbrace{(3 \cdot 7) \cdot 1}^{\equiv 0 \pmod 3\\ \equiv 1 \pmod 5\\ \equiv 0 \pmod 7} + 2 \cdot \overbrace{(3 \cdot 5) \cdot 1}^{\equiv 0 \pmod 3\\ \equiv 0 \pmod 5\\ \equiv 1\pmod 7}\equiv\, \overbrace{2,\,3,\,2\pmod{3,5,7}}^{\equiv 2 + 0 +0\pmod 3\\ \equiv0+3+0 \pmod 5\\ \equiv 0+0+2\pmod 7}$
======
Think about what you, yourself just stated.
If take this sum $x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1$ and $\mod 3$ it, then $(5\cdot 7)$ and $2$ are inverses so $2\cdot[(5\cdot 7)\cdot 2]\pmod 3\equiv 2\cdot 1\pmod 3 \equiv 2 \pmod 3$. ANd the other terms are multiples of $3$ so they are $\equiv 0 \pmod 3$. So $x\equiv 2 \pmod 3$.
If you take that term $x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1$ and $\mod 5$ it, then $3\cdot 7$ and $1$ are inverses so $3\cdot[(3\cdot 7) \cdot 1] \equiv 3 \cdot 1 \equiv 3 \pmod 5$. ANd the other terms are multiples of $5$. So the sum $x \equiv 3 \pmod 5$.
And so on.
....
If you want to solve
$x \equiv a \pmod m$
$x \equiv b \pmod n$
$x \equiv c \pmod v$ then
And assuming you were able find $(nv)^{-1}\mod m; (mv)^{-1}\mod n; $and $(nm)^{-1}\mod v$
Then Let $K = a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)$
Note: $K \pmod m \equiv$
$a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)\pmod m\equiv$
$a*1 + [b(mv)^{-1}v]m + [c(nm)^{-1}n]m \pmod m\equiv$
$a*1 + 0 + 0 \equiv a\pmod m$.
And likewise:
$K \pmod n \equiv$
$a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)\pmod n\equiv$
$[a(nv)^{-1}v]n + b*1 + [c(nm)^{-1}m]n \pmod n\equiv$
$0 + b*1 + 0 \equiv b\pmod n$.
And
$a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)\pmod v\equiv$
$[a(nv)^-1n]v + [b(mv)^{-1}m]v + c*1 \pmod v\equiv$
$0 + 0 + c \equiv c\pmod m$.
So $K$ is A solution.
If $m,n,v$ are pairwise relative prime then $K$ is a unique solution upto $\mod nmv$.
|
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|
Convergence and divergence of a Complex Series I've been given the following series:
$$\frac{z}{1-z^2} + \frac{z^2}{1-z^4} + \frac{z^4}{1-z^8} + ...$$
and been told to investigate the convergence. Clearly this diverges if $z=1$ (possibly if $\vert{z}\vert = 1$?), but other than that I am at a loss as to how to proceed. Wolfram Alpha tells me that this converges to $\frac{z}{1-z}$ if $\vert{z}\vert<1$ and to $\frac{1}{1-z}$ if $\vert{z}\vert>1$ but how would one go about showing this?
|
We can show inductively that
$$\begin{align*}
\sum_{j=0}^n \frac{z^{2^j}}{1-z^{2^{j+1}}}&=\frac{\sum\limits_{k=1}^{2^{n+1}-1}z^k}{1-z^{2^{n+1}}}\\&=\frac{z-z^{2^{n+1}}}{(1-z)(1-z^{2^{n+1}})}\\&=\frac{z-1+1-z^{2^{n+1}}}{(1-z)(1-z^{2^{n+1}})}\\&=-\frac{1}{1-z^{2^{n+1}}}+\frac1{1-z}.
\end{align*}$$ The base case $n=0$ is quite obvious. We find that
$$\begin{align*}
-\frac{1}{1-z^{2^{n+1}}}+\frac1{1-z}+\frac{z^{2^{n+1}}}{1-z^{2^{n+2}}}&=\frac{-1-z^{2^{n+1}}}{1-z^{2^{n+2}}}+\frac1{1-z}+\frac{z^{2^{n+1}}}{1-z^{2^{n+2}}}\\&=-\frac{1}{1-z^{2^{n+2}}}+\frac1{1-z}.
\end{align*}$$ By induction, the claim is proved.
We can easily see that the given series converges for $|z|<1$ and $|z|>1$, to $\frac{z}{1-z}$ and $\frac1{1-z}$, respectively.
If $|z|=1$, then the given series converges if and only if
$$
\exists \lim_{n\to\infty} z^{2^n} \ne 1.
$$ If we denote such limit by $L$, then it should satisfy
$$
L^2 =\lim_{n\to\infty} z^{2^{n+1}}=\lim_{n\to\infty} z^{2^n}=L
$$ giving $L=1$ or $L=0$. Since $L\ne 0,1$, the series does not converge for all $|z|=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Asking about $\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]$ $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]=\frac{61}{184320}\pi^7\tag1$$
Step 1:
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]+\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n^2+n}{(2n+1)^7(n+1)}\right]\tag2$$
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]+\sum_{n=2}^{\infty}(-1)^n\left[\frac{1}{(2n+1)^7}\right]\tag3$$
Recall $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n+1)^7}=\frac{61}{184320}\pi^7$$
It looks like that this sum $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]$$
is a rational number
I am not able to show that sum $(1)=\frac{61}{184320}\pi^7$
Any help.Thank you!
|
It comes from telescoping series:
$$
\frac{(-1)^n}n\left(\frac{1}{(2 n+1)^7 (n+1)}+\frac{1}{(2 n-1)^7 (n-1)}\right) = (-1)^n\left(\frac{1}{n-1} + \frac{1}{n}\right) +
(-1)^n\left(\frac{1}{n}+\frac{1}{n+1}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^7}+\frac{1}{(2 n+1)^7}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^5}+\frac{1}{(2 n+1)^5}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^3}+\frac{1}{(2 n+1)^3}\right)
- 4(-1)^n\left(\frac{1}{2 n-1}+\frac{1}{2 n+1}\right)
$$
Each parenthesis has two terms. The second term of $n=n_0$ is equal to the first term of $n=n_0+1$ with the opposite sign (coming from $(-1)^n$). So only the first term of $n=2$ survives.
Edit. P.S. By the way, it's not necessary to calculate partial fractions, if you just want to show the rationality. We can see that the function is odd ($n\to -n$), so there are no fractions with even exponent and all $(2n\pm1)^{-k}$ parentheses have the same coefficient. We only need to calculate the coefficients of $n^{-1}$, $(n+1)^{-1}$ and $(n-1)^{-1}$. We can easily do it with multiplying by corresponing monomial and letting $n=-1,0,1$.
|
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|
Proving Table of Integral Integral (Trigonometric Substitution) I need help with the proof of the table of integral that:
$$\int \frac{\sqrt{a^2 + u^2}}{u} du = \sqrt{a^2 + u^2} + a \ln \left|\frac{\sqrt{a^2 + u^2} - a}{u}\right| + c$$
Must solve using trigonometric substitution.
I understand that you have to use $a^2\tan(\theta)$ as a substitution for $u$.
|
Let $u=a \tan x \tag{1}$ Then:- $u^2=a^2\tan^2 x \implies 2u \cdot du= 2a^2\tan x \cdot \sec^2x \cdot dx $
The integral reduces to :- $$ a\int \sec^2x \csc x \cdot dx $$
$$ = a\int (\tan^2x +1) \csc x \cdot dx = a\left(\int \sec x \tan x \cdot dx + \int \csc x \cdot dx \right)\\
= a\left(\sec x - \ln(\cot x + \csc x)\right) + C $$Note that from $(1)$ we have $\sec x = \frac{\sqrt{u^2+a^2}}{a}$ and $\csc x= \frac{\sqrt{u^2+a^2}}{u}$ and $\cot x = \frac{a}{u} $
Thus , by undoing substitution , we get $$\boxed{ I= \sqrt{a^2+u^2} + a\cdot \ln \left|\frac{\sqrt{a^2+u^2}-a}{u}\right| + C}$$
|
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|
Why $\frac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \frac{\pi}{4}$? Since
$$\dfrac{d}{dx} \left( \dfrac{1}{2} \arctan(x) \right) = \dfrac{d}{dx} \left( \arctan(x-\sqrt{x^2+1}) \right) $$
then the format of their graphs are the same, but separated by a constant, which is $\dfrac{\pi}{4}$. That is,
$$\dfrac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \dfrac{\pi}{4} $$
My question is: why is this constant and what procedure is done to discover it.
|
I'm going to use $y$ instead of $x$.
Let $P(\theta) = (1, y)$. Then $$\theta = \arctan(y).$$
Then
$$\sin \theta = \dfrac{y}{\sqrt{1+y^2}}
\qquad \text{and} \qquad
\cos \theta = \dfrac{1}{\sqrt{1+y^2}}$$
Then \begin{align}
\tan \bigg( \frac 12 \theta - \frac{\pi}{4} \bigg)
&= \dfrac{\tan \frac 12 \theta - \tan \frac{\pi}{4}}
{1 + \tan(\frac 12 \theta) \tan(\frac{\pi}{4})}\\
&=\dfrac{\left(\dfrac{\sin \theta}{1+\cos \theta}-1 \right)}
{\left( 1+\dfrac{\sin \theta}{1+\cos \theta} \right)} \\
&= \frac{\sin \theta - 1 - \cos \theta}
{1 + \cos \theta + \sin \theta} \\
&= \frac{\left( \dfrac{y}{\sqrt{1+y^2}} - 1 - \dfrac{1}{\sqrt{1+y^2}} \right)}
{\left( 1 + \dfrac{1}{\sqrt{1+y^2}} + \dfrac{y}{\sqrt{1+y^2}} \right)} \\
&= \frac{y - \sqrt{1+y^2} - 1}{\sqrt{1+y^2} + 1 + y} \\
&= \frac{(y - 1) - \sqrt{1+y^2}}{(y + 1) + \sqrt{1+y^2}} \cdot
\frac{(y + 1) - \sqrt{1+y^2}}{(y + 1) - \sqrt{1+y^2}} \\
&= \dfrac{2y^2 - 2y\sqrt{1+y^2}}{2y} \\
&= y - \sqrt{1+y^2}
\end{align}
Since $\tan \bigg( \dfrac 12 \theta - \dfrac{\pi}{4} \bigg) = y - \sqrt{1+y^2}$,
then $$\dfrac 12 \theta - \dfrac{\pi}{4} = \arctan\left(y - \sqrt{1+y^2}\right).$$
In other words
$$\dfrac 12 \arctan(y) = \arctan\left(y - \sqrt{1+y^2}\right) + \dfrac{\pi}{4}.$$
|
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|
Find a generating function for which $A(n)={n \choose 2}$ In the book I'm using, $A(x)$ denotes the formal power series (generating function), $A(x) = \sum a_ix^i$.
I'm really stuck on this problem. Thanks for any help.
My attempt after the given hint:
$$\begin{align}
A(x)&=\sum_{n\geq 0} \binom{n}{2}x^n\\
&=\sum_{n\geq 0} \frac{x^2}{2}\frac{d^2}{dx^2}(x^{n})\\
&=\frac{x^2}{2} \frac{d^2}{dx^2}\sum_{n\geq 0} x^{n}\\
&=\frac{x^2}{2} \frac{d^2}{dx^2}\frac{1}{1-x}\\
&=\frac{x^2}{2} \frac{2}{(1-x)^3}
\end{align}$$
Sorry, I'm new to $\rm \LaTeX$.
|
Since $\binom{n}{2}x^n=\frac12 x^2\frac{d^2}{dx^2}x^n$, $$\sum_{n\ge 0}\binom{n}{2}x^n=\frac12 x^2\frac{d^2}{dx^2}\frac{1}{1-x}=\frac{x^2}{(1-x)^3}.$$We can double-check by the binomial theorem: the $x^n$ coefficient is $$\frac{(-1)^n}{(n-2)!}\prod_{j=1}^{n-2}(-2-j)=\frac{n!}{(n-2)!2!}=\binom{n}{2}.$$
|
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|
In a triangle $ABC$, $\cos 3A + \cos 3B + \cos 3C = 1$, then find any one angle.
In a triangle $ABC$, $\cos 3A + \cos 3B + \cos 3C = 1$, then find any one angle.
HINT: The answer is $\frac{2\pi}{8}$
I have tried these steps and got stuck in the middle.
$$ A +B + C = \pi$$
$$ A + B = \pi - C$$
$$\cos 3A + \cos 3B + \cos 3C = 1 \longrightarrow$$
$$2\cos{\frac{3A + 3B}{2}}\cos{\frac{3A - 3B}{2}} + \cos 3C = 1$$
$$2\cos{\frac{3A + 3B}{2}}\cos{\frac{3A - 3B}{2}} + \cos{(\frac{3\pi}{2}-\frac{3A - 3B}{2})} = 1 \text{ [since $A - B = \pi -C$] } $$
$$2\cos{\frac{3A + 3B}{2}}\cos{\frac{3A - 3B}{2}} - \sin{\frac{3A + 3B}{2}} = 1$$
From here onwards I do not know how to continue. Please check if my method is correct and help me solve the problem. Thanks :)
|
Given $$\cos 3A+\cos3B+\cos3C=1$$
$$\implies\sum\cos3A=1$$
$$\implies\sin\dfrac{3A}{2}\cdot\sin\dfrac{3B}{2}\cdot\sin\dfrac{3C}{2}=0$$
Therefore,$$A=\dfrac{2\pi}{3}\mbox{ or }B=\dfrac{2\pi}{3}\mbox{ or }C=\dfrac{2\pi}{3}$$
|
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|
How to prove these two functions are always equal? If $a$ and $b$ are positive integers and $/$ stands for integer division, we have these two functions:
$$f(a,b) = (a + b - 1) / b$$
and
$$g(a,b) =
\begin{cases}
a/b, & \text{if $a \mod b = 0$} \\[2ex]
a / b + 1, & \text{if $a \mod b \neq 0$}
\end{cases}
$$
We can see $f(a,b)$ equals to $g(a,b)$ by filling a and b with actual numbers, but how do you prove that they are always equal? I've answered this question here but I think I was over-complicating it so not really convinced by myself.
This problem is quite common in real life. Consider we have 10 students and now we need to divide them into several groups each of which has the same number of students, say, that number is 3. Now we need to calculate how many groups there will be, and the answer is 4. If we put it into math function then $g(a,b)$ is a natural way of thinking, but $f(a,b)$ also does the job. Why?
|
Notice that we can express integer division in terms of vanilla-flavoured division using the floor function. For example, your function $f$ could be expressed
$$f(a,b)=\left\lfloor\frac{a+b-1}{b}\right\rfloor=\left\lfloor\frac{a-1}{b}+1\right\rfloor=\left\lfloor\frac{a-1}{b}\right\rfloor+1$$
and $g$ could be expressed
$$g(a,b)=\left\{\begin{array}{ll}
\displaystyle\left\lfloor\frac a b\right\rfloor,&\left(\exists k\in \mathbb Z\right)\left(a=kb\right)\\
\displaystyle\left\lfloor\frac{a}{b}\right\rfloor+1,&\text{otherwise}
\end{array}\right.$$
Before we see why $f$ and $g$ are equivalent it might help to notice the following:
*
*If $a$ can be expressed as $kb+r$ where $k\ge 0$ and $0\le r<b$ then $$\left\lfloor\frac{a}{b}\right\rfloor=\left\lfloor\frac{kb+r}{b}\right\rfloor=k+\left\lfloor\frac r b\right\rfloor=k$$
*We can "complete the remainder" to get the identity:$$\left\lfloor\frac a b\right\rfloor = \left\lfloor\frac{kb+r\color{darkorange}{+(b-r)}}{b}\right\rfloor-1=k$$
(For example, if $a=28$ and $b=5$ so that $k=5$ and $r=3$. Notice that $\left\lfloor\frac {28} 5\right\rfloor=5$ and that $\left\lfloor\frac {28\color{darkorange}{+2}} 5\right\rfloor-1=5$.)
Now, if we consider the cases in $g$ separately:
Case 1: $a$ is a multiple of $b$
$$\begin{align}
\left\lfloor\frac{a-1}{b}\right\rfloor+1
&=\left(\left\lfloor\frac{a-1\color{darkorange}{+1}}{b}\right\rfloor-1\right)+1\\\\
&=\left\lfloor\frac a b\right\rfloor
\end{align}$$
Case 2: otherwise
In this case there exist unique $k\ge 0$ and $0<r<b$ such that $a=kb+r$, so
$$\begin{align}
\left\lfloor\frac{a-1}{b}\right\rfloor+1
&=\left(\left\lfloor\frac{kb+r-1+\color{darkorange}{(b-r + 1)}}{b}\right\rfloor-1\right)+1\\\\
&=\left\lfloor \frac{b(k+1)}{b}\right\rfloor\\\\
&=\left\lfloor k+1\right\rfloor\\\\
&=\left\lfloor\frac ab\right\rfloor+1
\end{align}$$
|
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|
Reduction Formulae Integral $x(1-x^3)$ The question asks us:
"If
$$u_n=\int_0^1x(1-x^3)^ndx$$
show that
$$u_n= \frac {3n}{3n+2}u_{n-1}$$
I've tried integration by parts using a coefficient of $1, x$ and even tried reducing the $1-x^3$ term into its factors but with no progress. Any help would be greatly appreciated.
|
\begin{align}
u_n &= \int_0^1 \frac{1}{2}x^2 n(1-x^3)^{n-1} 3x^2 dx \\
&= \frac{3}{2}n\int_0^1 x x^3 (1-x^3)^{n-1} dx \\
&= \frac{3}{2} n \int_0^1 x(x^3-1+1)(1-x^3)^{n-1}dx \\
&= -\frac{3}{2}n u_n + \frac{3}{2}n u_{n-1}
\end{align}
Therefore,
$$
u_n = \frac{3n}{2+3n} u_{n-1}
$$
|
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|
Integral of $\frac{1}{\sqrt{(z-z')^2 + s^2}}$ I have a question about the signs of the antiderivative when one integrate $\frac{1}{\sqrt{(z-z')^2 + s^2}}$.
According to Wolfram Alpha here and here:
If one evaluates $\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz$, they get $\log (z-z' + \sqrt{(z-z')^2 + s^2}) + C$.
Evalutating $\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz'$ gives $-\log (z-z' + \sqrt{(z-z')^2 + s^2}) + C$.
However, one could rearrange the integrand before evaluating the second expression. Using the fact that $(z-z')^2 = (z'-z)^2$, one gets
\begin{equation}
\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz'= \int \frac{1}{\sqrt{(z'-z)^2 + s^2}} dz'
\end{equation}
which has the same form as the first integral, implying that the two should be equal.
What am I missing here?
|
Notice that
\begin{align}
-\log{(z-z'+\sqrt{(z-z')^2+s^2})} &= \log{\left(\frac{1}{\sqrt{(z-z')^2+s^2} -(z'-z)}\right)} \\
&= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{(\sqrt{(z-z')^2+s^2} -(z'-z))(\sqrt{(z-z')^2+s^2} +(z'-z)}\right)} \\
&= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{s^2}\right)} \\
&= \log{(\sqrt{(z-z')^2+s^2} +z'-z)} -\log{s^2} \\
\end{align}
So the two are equal except for a constant factor $\log{s^2}$
|
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|
a,b,c are three real numbers where $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Now $abc$ = ? Here will the answer be a number? I want to know whether it is possible to get a real number (not an algebraic expression) as the product of $a$, $b$ and $c$.
I tried for a long time and this is what I got.
$$3abc = a^2 b + b^2 c + c^2 a$$
But it seems like there is a way to determine the value.
|
We start with $$a + \frac 1b = b + \frac 1c = c + \frac 1a$$
This makes three equalities:
\begin{align}a + \frac 1b &= b + \frac 1c\tag{1}\\
b + \frac 1c &= c + \frac 1a\tag{2}\\
c + \frac 1a &= a + \frac 1b\tag{3}\end{align}
First consider equation $(1)$ $$a + \frac 1b = b + \frac 1c$$ which can be rearranged to give \begin{align}a + \frac 1b &= b + \frac 1c\\
a-b&=\frac 1c -\frac 1b\\
&=\frac{b-c}{bc}\end{align}
Similarly, from equation $(2)$, we can see that $$b-c=\frac{c-a}{ac}$$ and thus \begin{align}a-b&=\frac{b-c}{bc}\\
&=\frac{\frac{c-a}{ac}}{bc}\\
&=\frac{c-a}{abc^2}\end{align}
Finally, from equation $(3)$, we get $$c-a=\frac{a-b}{ab}$$ and thus \begin{align}a-b&=\frac{c-a}{abc^2}\\
&=\frac{\frac{a-b}{ab}}{abc^2}\\
&=\frac{a-b}{a^2b^2c^2}\end{align}
We can now rearrange this as follows (when $a\neq b$) \begin{align}a-b&=\frac{a-b}{a^2b^2c^2}\\
(a-b)(a^2b^2c^2)&=a-b\\
a^2b^2c^2&=\frac{a-b}{a-b}\\
a^2b^2c^2&=1\\
\sqrt{a^2b^2c^2}&=\sqrt1\\
\sqrt{a^2}\sqrt{b^2}\sqrt{c^2}&=\pm1\tag{$*$}\\
abc&=\pm1\end{align}
where $(*)$ comes from the fact that $\sqrt{a\cdot b} =\sqrt{a}\cdot\sqrt{b}$
|
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|
Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3}$
I wrote down a few cases:
$X_1 = \sin\frac{\pi}{3}-4\cos\frac{\pi}{3}= \frac{\sqrt3-4}{2} $
$X_2 = \sin\frac{2\pi}{3}-4\cos\frac{2\pi}{3}= \frac{\sqrt3+4}{2} $
$X_3 = \sin\frac{3\pi}{3}-4\cos\frac{3\pi}{3}=4 $
$X_4 = \sin\frac{4\pi}{3}-4\cos\frac{4\pi}{3}= \frac{4-\sqrt3}{2} $
$X_5 = \sin\frac{5\pi}{3}-4\cos\frac{5\pi}{3}= \frac{4-\sqrt3}{2} $
$X_6 = \sin\frac{6\pi}{3}-4\cos\frac{6\pi}{3}= -4 $
So as found above, $\min = -4$, $\max = 4$, $\inf = -4$, $\sup = 4$, $\liminf = -4$, $\limsup = 4$.
Let's check whether its convergent or not:
The sequence is bounded as stated above so lets check if its decreasing or increasing.
$X_n \geq X_{n+1}$
$\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3} \geq\sin\frac{(n+1)\pi}{3}-4\cos\frac{(n+1)\pi}{3}$
$\sin\frac{n\pi}{3} - \sin\frac{(n+1)\pi}{3} \geq -4\cos\frac{(n+1)\pi}{3} + 4\cos\frac{n\pi}{3}$
I used trigonometrical identity for $\sin\alpha+\sin\beta$ and $\cos\alpha-\cos\beta$ :
$-\cos\frac{\pi(2n+1)}{6} \geq 4\sin\frac{\pi(2n+1)}{6}$
What should I do next? I am stuck here.
Thanks, and sorry if I made mistakes.
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Computing the first $6$ terms alone does not prove that $\min X_n=-4$ nor $\max X_n=4$ but rather that $\min X_n\le-4$ and $\max X_n \ge 4$ as it does not say anything about the rest of the sequence.
Hint : $X_{n+6}=X_n$
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|
Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve.
$\sqrt{x+1} = \sqrt{x+2}$
$x+1 = x+2$
0 = 1
My question is, if there are two square roots on one side, then can I still square both sides in this way:
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$x+1 - (x+2) = x+3$
$x+1 - (x-2) = x+3$
$x = -4$
Or does squaring both sides cause something strange to happen on the left hand side?
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Hint: $(\sqrt{x+1} - \sqrt{x+2})^2 = (x+1) - 2\sqrt{x+1}\sqrt{x+2} + (x+2)$
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|
Prove that $(a+b) (a^2 + b^2) (a^4 + b^4)...(a^{32} + b^{32}) = a^{64} - b^{64}$ if $b = a-1$ Prove that if $b = a-1$, then
$(a + b) (a^2 + b^2 ) (a^4 + b^4 ) ... (a^{32} + b^{32} ) = a^{64} - b^{64}$ .
I saw this in a website and it wrote this:
hint:
Write down the equality $1 = a+b$ and use the formula $k^2 -n^2 = (k-n) (k + n) $
Decision:
We write the equality $1 = a+b$ and use the formula $k^2 -n^2 = (k-n) (k + n)$ (If you are not familiar with the formulas of reduced multiplication, prove this formula: multiply the expression on the right-hand side.). We write the expression $1*(a+b)(a^2+b^2)(a^4+b^4)…(a^{32}+b^{32})= (a-b)(a+b)(a^2+b^2)(a^4+b^4)…(a^{32}+b^{32})= (a^2-b^2)(a^2+b^2)(a^4+b^4)…(a^{32}+b^{32})= (a^4-b^4)(a^4+b^4)(a^8+b^8)(a^{16}+b^{16})(a^{32}+b^{32})= (a^8-b^8)(a^8+b^8)(a^{16}+b^{16})(a^{32}+b^{32})= (a^{16}-b^{16})(a^{16}+b^{16})(a^{32}+b^{32})= (a^{32}-b^{32})(a^{32}+b^{32})=a^{64}-b^{64}$
I don't know how this formula $k^2 - n^2 = (k-n)(k+n)$ come out?
Also, I don't understand the decision part?
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hint: $a+b = \dfrac{a^2-b^2}{a-b}, a^2+b^2= \dfrac{a^4-b^4}{a^2-b^2}, ...$, can you see the factors that cancel each other ?
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|
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \frac{\sqrt{4-0}}{\sqrt[3]{1+0}} \\
& = 2
\end{align}
Despite the steps I've taken seems plausible to me, the answer is given as $-2$.
Is dividing both the numerator and the denominator by $x$ allowed here? Where am I making a mistake?
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Dividing both the numerator and the denominator by $x$ is always allowed, but the square root creates a trap:
$$\frac{\sqrt{a}}x=\text{sgn }x\sqrt{\frac a{x^2}}$$ because the square root is always a positive number.
|
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|
Limit involving inverse functions When I am given the limit
$$\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}$$
would it be possible to evaluate it giving some substitution?
L'Hospital's rule seemed an option but I ended up going in circles.
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You may proceed as follows:
*
*Set $\tan y = \sqrt{1+x^2}$ and consider $y \to \frac{\pi}{2}^-$
\begin{eqnarray*}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}
& = & \sqrt{\tan^2y -1}\frac{y}{\tan y} \\
& = & \frac{\sqrt{\sin^2 y - \cos^2 y}}{\sin y}\cdot y\\
&\stackrel{y \to \frac{\pi}{2}^-}{\longrightarrow} & \frac{\sqrt{1 - 0}}{1}\cdot \frac{\pi}{2} = \frac{\pi}{2}
\end{eqnarray*}
|
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|
Question from the 2011 IMC (International Mathematics Competition) Key Stage III paper, about the evaluation of a quadratic equation
When $a=1, 2, 3, ..., 2010, 2011$, the roots of the equation $x^2-2x-a^2-a=0$ are $(a_1, b_1), (a_2, b_2), (a_3, b_3),\cdots, (a_{2010}, b_{2010}), (a_{2011}, b_{2011})$ respectively. Evaluate:
$$
\frac{1}{a_1} + \frac{1}{b_1} + \frac{1}{a_1} + \frac{1}{b_2} + \frac{1}{a_3} + \frac{1}{b_3} +\cdots + \frac{1}{a_{2010}} + \frac{1}{b_{2010}} + \frac{1}{a_{2011}} + \frac{1}{b_{2011}}
$$
I tried solving this question, with the use of the quadratic equation.
Using the quadratic equation, I concluded that $a_1=\frac{2+\sqrt{12}}{2}=1+\sqrt{3}$ and that $b_1=\frac{2-\sqrt{12}}{2}=1-\sqrt{3}$. The reason that I concluded to this is because the quadratic equation states:
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
So I substituted $a$ with $1$ ($1$ is multiplying $x^2$, in the original equation), $b$ with $-2$ ($x$ is getting multiplied by $-2$ in the original equation) and $c$ with $(-a^2-a)$, as they are the only ones which are not getting directly multiplied by $x$ in the original equation.
Hence, I subsequently worked out that $\frac{1}{a_1} + \frac{1}{b_1} = \frac{2}{-2}=-1$
Continuing to do the same thing I worked out $\frac{1}{a_2} + \frac{1}{b_2}= \frac{2}{-6} = -\frac{1}{3}$ and $\frac{1}{a3} + \frac{1}{b_3}= \frac{2}{-12}=-\frac{1}{6}$ and $\frac{1}{a_4} + \frac{1}{b_4} = \frac{2}{-20}=-\frac{1}{10}$ and $\frac{1}{a_5} + \frac{1}{b5} = \frac{2}{-30} =-\frac{1}{15}$.
I subsequently realised that a pattern was emerging, the denominator, each time is getting increased by the degree of $n$ at which $a$ and $b$ are (for instance $\frac{1}{a_4} + \frac{1}{b_4}= -\frac{1}{6+4}$)
Had I not been dealing with fractions, I would have solved it using arithmetic progressions, but unfortunately that is not possible.
I can think of no other way of finishing off my thoughts, nor any other way to solve this problem. Can you please help me? Can you please tell me if there is any method of finishing off my thoughts and if there isn't, can you please suggest a method of solving the problem
Thank you in advance
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My first thought is to "complete the square". The given equation, $x^2- 2x- a^2- a$ is the same as $x^2- 2x= a^2+ a$. We can "complete the square" on the left by adding 1 to both sides: $x^2- 2x+ 1= (x- 1)^2= a^2+ a+ 1$. So $x- 1= \pm\sqrt{a^2+ a}$ and $x= 1\pm\sqrt{a^2+ a}$. Now pair the reciprocals: $\frac{1}{a_i}+ \frac{1}{b_i}= \frac{1}{1+ \sqrt{a^2+ a}}+ \frac{1- \sqrt{a^2+ a}}= \frac{1- \sqrt{a^2+ a}}{1- a^2- a}+ $$ \frac{1+ \sqrt{a^2+ a}}{1- a^2- a}= \frac{2}{1- a^2- a}$. The sum of reciprocals reduces to $-2- \frac{2}{5}- \frac{2}{11}- \frac{2}{19}- ...$.
|
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|
Understanding conditional probability. Consider the problem $(b)$ (from Hwi Hsu):
This problem can be visually represented in either of the three very closely related ways:
Why is $\frac{1}{3}$ only the correct answer?
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1) Probability of rolling doubles given no other information. $P(A)$
$\begin{array}( \color{blue}{(1,1)} & (1,2)& (1,3)&(1,4)&(1,5)&(1,6)\\
(2,1) & \color{blue}{(2,2)}& (2,3)&(2,4)&(2,5)&(2,6)\\
(3,1) & (3,2)& \color{blue}{(3,3)}&(3,4)&(3,5)&(3,6)\\
(4,1) & (4,2)& (4,3)&\color{blue}{(4,4)}&(4,5)&(4,6)\\
(5,1) & (5,2)& (5,3)&(5,4)&\color{blue}{(5,5)}&(5,6)\\
(6,1) & (6,2)& (6,3)&(6,4)&(6,5)&\color{blue}{(6,6)}\\
\end{array}$
Of all the $36$ possible (black and blue), $6$ of them are doubles (blue) so probability is $\frac {6}{36} = \frac 16$.
2) Probability of rolling doubles given that the sum is at most $3$. $P(A|B)$
$\begin{array}( \color{blue}{(1,1)} & (1,2)& \color{red}{(1,3)}&\color{red}{(1,4)}&\color{red}{(1,5)}&\color{red}{(1,6)}\\
(2,1) & \color{red}{(2,2)}& \color{red}{(2,3)}&\color{red}{(2,4)}&\color{red}{(2,5)}&\color{red}{(2,6)}\\
\color{red}{(3,1)} & \color{red}{(3,2)}& \color{red}{(3,3)}&\color{red}{(3,4)}&\color{red}{(3,5)}&\color{red}{(3,6)}\\
\color{red}{ (4,1)} & \color{red}{(4,2)}& \color{red}{(4,3)}&\color{red}{(4,4)}&\color{red}{(4,5)}&\color{red}{(4,6)}\\
\color{red}{ (5,1)} & \color{red}{(5,2)}& \color{red}{(5,3)}&\color{red}{(5,4)}&\color{red}{(5,5)}&\color{red}{(5,6)}\\
\color{red}{ (6,1)} &\color{red}{ (6,2)}& \color{red}{(6,3)}&\color{red}{(6,4)}&\color{red}{(6,5)}&\color{red}{(6,6)}\\
\end{array}$
In this case all the red cases are impossible because the sum is more than $3$. So in this case of the $3$ possible cases (black or blue). $1$ of them is doubles (blue).
So the probability is $\frac 1{3}$.
=====
In your post the first image is:
What is the probability that BOTH faces are the Same and sum is at most 3. $P(A\&B)$
The second image is:
What is the probability that the Sum is at most three GIVEN that faces are the same.$P(B|A)$.
They are trying to show that $P(A|B)$ is the same thing as $\frac {P(A\&B)}{P(B|A)}$.
|
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|
Evaluating $\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}dx$
How can we prove $$\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}\mathrm{d} x=\frac{2\pi}{3\sqrt 3}?$$
Thought 1
It cannot be solved by using contour integration directly. If we replace $-1/3$ with $-2/3$ or $1/3$ or something else, we can use contour integration directly to solve it.
Thought 2
I have tried substitution $x=t^3$ and $x=1-t$. None of them worked. But I noticed that the form of $1-x+x^2$ does not change while applying $x=1-t$.
Thought 3
Recall the integral representation of $_2F_1$ function, I was able to convert it into a formula with $_2F_1\left(2/3,1;4/3; e^{\pi i/3}\right)$ involved. But I think it will only make the integral more "complex". Moreover, I prefer a elementary approach. (But I also appreciate hypergeometric approach)
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Here we piggy back off the solution posted by @pisco, organize the analysis with detail on the definitions of $\arg(z)$ and $\arg(1-z)$, and finish by evaluating the resiudes enclosed by the closed "keyhole contour."
Let $f(z)$ be the function given by
$$f(z)=\frac{z^{2/3}(1-z)^{-1/3}}{z^2-z+1}$$
where choose the branch cut from $0$ to $\infty$ along the positive real axis such that
$$\arg(z)=\begin{cases}
0&, z=x+i0^+\\\\
2\pi&,z=x+i0^-
\end{cases}$$
and we choose the branch cut from $1$ to $\infty$ along the positive real axis with $\arg(1-z)=-\pi+\arg(z-1)$ such that
$$\arg(1-z)=\begin{cases}
0&, 0<x<1\\\\
-\pi&,z=x+i0^+, 1<x\\\\
\pi&, z=x+i0^-, 1<x
\end{cases}$$
Then, the integral around the classical "key hole" contour $C$ is
$$\begin{align}
\oint_C f(z)\,dz &=(e^{i2(0)/3}e^{-i(0)/3}-e^{i2(2\pi)/3}e^{-i(0)/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\\\\
&+(e^{i2(0)/3}e^{-i(-\pi)/3}-e^{i2(2\pi)/3}e^{-i(\pi)/3})\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{x^2-x+1}\,dx\\\\
&=(1+e^{i\pi/3})\left(\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx+\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{x^2-x+1}\,dx\right)\tag1
\end{align}$$
Enforcing the substitution $x\mapsto 1/x$ in the second integral on the right-hand side of $(1)$ reveals
$$\begin{align}
\oint_C f(z)\,dz &=(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}+x^{-1/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\tag2
\end{align}$$
Using the identity $x^{2/3}(1-x)^{-1/3}+x^{-1/3}(1-x)^{2/3}=x^{-1/3}(1-x)^{-1/3}$ and observing that $x^2-x+1=(1-x)^2-(1-x)+1$ we find from $(2)$ that
$$\begin{align}
\oint_C f(z)\,dz &=3(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\\\\
&=3(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\tag3
\end{align}$$
From the residue theorem we have
$$\begin{align}
\oint_C f(z)\,dz&=2\pi i \left(\text{Res}\left(f(z), z=\frac12+i\frac{\sqrt3}2\right)+\text{Res}\left(f(z), z=\frac12-i\frac{\sqrt3}2\right)\right)\\\\
&=2\pi i \left(\frac{e^{i2\pi/9}e^{i\pi/9}}{i2\sqrt 3}+\frac{e^{i10\pi/9}e^{-i\pi/9}}{-i2\sqrt 3}\right)\\\\
&=\frac{2\pi}{\sqrt3} (1+e^{i\pi/3})\tag4
\end{align}$$
Finally, setting $(3)$ and $(4)$ equal yields the coveted result
$$\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx=\frac{2\pi }{3\sqrt 3}$$
|
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|
Integral $\int_0^1 x^n\left\{\frac{k}{x}\right\}dx$ I am trying to solve the following integral containing fractional part function (denoted by $\{.\}$)
$$\int_0^1 x^n\left\{\frac{k}{x}\right\}dx,\ 0<k\le 1,\ n\in \mathrm N^*$$
For $n=0$, it is already known that $\int_0^1 \left\{\frac{k}{x}\right\}dx=k(1-\gamma-\ln k)$.
I followed the same technique to solve for general $n$. I am stuck at the end to find the closed form of those sums if possible. Please see the picture.
EDIT: After Zacky's comment:
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I will try to redo the work from beggining and try to color in red what it's not right, let me know if I am missing something too please.$${I(n,k)=\int_0^1 x^{n}\left\{\frac{k}{x}\right\}dx }\overset{\large x=\frac{k}{t}}=k^{n+1}\int_k^\infty \frac{\left\{t\right\}}{t^{n+2}}dt=$$
$$=k^{n+1} \int_k^1 \frac{t-\lfloor t \rfloor}{t^{n+2}}dt+k^{n+1}\int_1^\infty \frac{t-\lfloor t \rfloor}{t^{n+2}}dt$$
$$=k^{n+1} \int_k^1 \frac{dt}{t^{n+1}}dt+k^{n+1}\sum_{i=1}^\infty \int_i^{i+1}\frac{\color{red}{t-i}}{t^{n+\color{red}2}}dt$$
$$=k^{n+1} \left(-\frac1n \cdot \frac{1}{t^n}\right)\bigg|_k^1+k^{n+1}\sum_{i=1}^\infty \underbrace{\int_0^1 \frac{y+\overbrace{i-i}^{=0}}{(y+i)^{n+2}}dy}_{\large t-i=y}$$
$$=\frac{k^{n+1}}{\color{red}n}\left(\frac{1}{k^n}-1\right)+k^{n+1}\sum_{i=1}^\infty \int_0^1 \left(\frac{1}{(y+i)^{n+1}}-\frac{i}{(y+i)^{n+2}}\right)dy$$
$$=\frac{k^{n+1}}{n}\left(\frac{1}{k^n}-1\right)+k^{n+1}\sum_{i=1}^\infty \left(-\frac{1}{n}\frac{1}{(y+i)^{n}}+\frac{i}{n+1}\frac{1}{(y+i)^{n+1}}\right)\bigg|_0^1$$
$$=\frac{k^{n+1}}{n}\left(\frac{1}{k^n}-1\right)+k^{n+1}\sum_{i=1}^\infty \left(\frac1n \left(\frac{1}{i^n}-\frac{1}{(i+1)^n}\right)+\frac{i}{n+1}\left(\frac{1}{(i+1)^{n+1}}-\frac{1}{i^{n+1}}\right)\right)$$
You arrived until here too. Now let's take that sum separately.
$$S=\frac1n\underbrace{\sum_{i=1}^\infty \left(\frac{1}{i^n}-\frac{1}{(i+1)^n}\right)}_{S_1}-\frac{1}{n+1}\sum_{i=1}^\infty\underbrace{\left(\frac{i}{i^{n+1}}-\frac{i}{(i+1)^{n+1}}\right)}_{S_2}$$
For $S_1$ notice that plugging in some terms yiels a nice telescoping series:
$$\require{cancel} i=1: \frac{1}{1^n}-\cancel{\frac1{2^n}}$$
$$i=2: \cancel{\frac{1}{2^n}}-\cancel{\frac1{3^n}}$$
$$i=3: \cancel{\frac{1}{3^n}}-\cancel{\frac1{4^n}}$$
$$i=4: \cancel{\frac{1}{4^n}}-\cancel{\frac1{5^n}}$$
$$\dots$$
$$\Rightarrow S_1= \frac{1}{1^n}=1$$
And for $S_2$ we will try to obtain a similar relationship.
First plugging some terms to notice a pattern for $S_2$:
$$i=1: \frac{1}{1^{n+1}}-\color{blue}{\frac{1}{2^{n+1}}}$$
$$i=2: \color{blue}{\frac{2}{2^{n+1}}}-\color{orange}{\frac{2}{3^{n+1}}}$$
$$i=3: \color{orange}{\frac{3}{3^{n+1}}}-\color{purple}{\frac{3}{4^{n+1}}}$$
$$i=4: \color{purple}{\frac{4}{4^{n+1}}}-\color{red}{\frac{4}{5^{n+1}}}$$
$$i=5: \color{red}{\frac{5}{5^{n+1}}}-\dots$$
Now guess what happens to the terms colored in the same color?
Yes, the numerator of each fraction are consecutive numbers and by reducing them we arrive at:
$$S_2=\frac{1}{1^{n+1}}+\color{blue}{\frac{1}{2^{n+1}}}+\color{orange}{\frac{1}{3^{n+1}}}+\color{purple}{\frac{1}{4^{n+1}}}+\color{red}{\frac{1}{5^{n+1}}}+\dots =\zeta(n+1)$$
$$\Rightarrow {I(n,k)=\frac{k^{n+1}}{n}\left(\frac{1}{k^n}-\cancel{1}\right)+k^{n+1}\left(\cancel{\frac1{n}}-\frac{\zeta(n+1)}{n+1}\right)}$$
$$=\boxed{I(n,k)=\frac{k}{n}-k^{n+1}\frac{\zeta(n+1)}{n+1}}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3125070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.