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Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to use the rearrangement inequality and conclude that $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$ is more than or equal to $\dfrac{\sin^3x}{\sin x} +\dfrac{\cos^3x}{\cos x}$ which is equal to $1$? Thanks.
|
Your proof is right.
Also, by C-S we obtain:
$$\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}=\frac{\sin^4x}{\sin{x}\cos x} +\frac{\cos^4x}{\sin x\cos{x}}\geq\frac{(\sin^2x+\cos^2x)^2}{2\sin{x}\cos{x}}=\frac{1}{\sin2x}\geq1.$$
The equality occurs for $x=\frac{\pi}{4},$ which says that $1$ is a minimal value.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$. Problem
Evaluate $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}.$$
Solution
Notice that $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right].$$
According to Taylor's Formula $(1+x)^{\alpha}=1+\dfrac{\alpha}{1!}x+\mathcal{O}(x)$, we have $$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1+\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)$$and $$\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1-\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$Therefore,$$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$As a result,
\begin{align*}
\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}&=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right]\\
&=\lim\limits_{x \to \infty}x \cdot \left[
\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)\right]\\
&=4.
\end{align*}
Hope to see another solution. Thanks!
|
$$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{t \to 0}\frac{(1+\sqrt{2}t)^{\sqrt{2}}-(1-\sqrt{2}t)^{\sqrt{2}}}{t}\\
=2\sqrt2\lim\limits_{s \to 0}\frac{(1+s)^{\sqrt{2}}-(1-s)^{\sqrt{2}}}{2s}$$
is obviously the derivative of $2\sqrt2r^{\sqrt 2}$ taken at $r=1$, i.e. $\color{green}4$.
It can also be justified by the generalized binomial theorem,
$$(1+s)^{\sqrt2}=1+\sqrt2s+\sqrt2(\sqrt2-1)\frac{s^2}2+\cdots$$
|
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|
Solving recurrence relation of this form by iteration for closed form The recurrence relation is of the form:
$$ G(n) = 5G(n-1) +\frac{4^n}{4^2},\quad G(1)=3$$
My answer always differs from that on wolfram Alpha where is my mistake??
My steps are
$ G(n) = 5^2G(n-2) +\frac{5\cdot 4^n}{4^3}+\frac{4^n}{4^2} $
$G(n)= 5^3G(n-3) +\frac{5^2\cdot 4^n}{4^4}+\frac{5\cdot 4^n}{4^3}+\frac{4^n}{4^2} $
and so on after $n-1$ steps of the process we reach $G(1)$ and
$G(n)= \frac{5^n}{5}G(1) +4^n(\frac{5^n}{5^2\cdot 4^n\cdot 4}+\dots+\frac{5}{4^3}+\frac{1}{4^2}) $
then I use the sum of geometric series and sub 3 for $G(1)$. I think the mistake is some where in the above.
|
Following your approach we have:
$$
\begin{align*}
G(n) &= 5G(n-1) +4^{n-2}=5\left(5G(n-2) +4^{n-3}\right)+4^{n-2}\\&=
5^2G(n-2) +5\cdot 4^{n-3}+4^{n-2}\\
&=5^{n-1}G(1) +5^{n-2}\cdot 4^{0}+\dots+5^1\cdot 4^{n-3}+5^0\cdot4^{n-2}\\
&=5^{n-1}G(1)+5^{n-1}-4^{n-1}=4\cdot 5^{n-1}-4^{n-1}\end{align*},$$
where at the end we used the fact that
$$\frac{x^{n-1}-y^{n-1}}{x-y}=x^{n-2}\cdot y^{0}+\dots+x^1\cdot y^{n-3}+x^0\cdot y^{n-2}.$$
|
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|
Trouble proving the trigonometric identity $\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$ I have become stuck while solving a trig identity. It is:
$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$
I have simplified the left side as far as I can:
\begin{align}
\frac{1-2\sin(x)}{\sec(x)}
&=\frac{1-2\sin(x)}{1/\cos(x)}=(1-2\sin(x))\cos(x)\\
&=\cos(x)-2\sin(x)\cos(x)=\cos(x)-\sin(2x)
\end{align}
However, I'm not sure what to do on the right side. I know I can use a compound angle formula to break $\cos(3x)$ into $\cos(2x)\cos(x)-\sin(2x)\sin(x)$; however, I do not know where to go after that. My main problem is with the denominator of the right side, I can't figure out how to get rid of it, either by multiplying, or by using a trig identity. Any help in solving this identity would be greatly appreciated!
|
First, in case you don't know how, I'll expand $\cos{3x}$.
$$\require{cancel}\begin{aligned}\cos{3x}&=\cos\left(x+\left(x+x\right)\right)\\&=\cos x\cos\left(x+x\right)-\sin x\sin\left(x+x\right)\\&=\cos{x}\left(\cos^{2}x-\sin^{2}x\right)-\sin{x}\left(2\sin{x}\cos{x}\right)\\&=\cos^3 x-\sin^{2}{x}\cos{x}-2\sin^{2}{x}\cos{x}\\&=\cos^3 x-3\sin^{2}{x}\cos{x}\\&=\cos^2 x\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}\left(1-\sin^2{x}\right)-3\sin^{2}{x}\cos{x}\\&=\cos{x}-\sin^2{x}\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}-4\sin^2{x}\cos{x}\end{aligned}$$
Let's rock!
$$\begin{aligned}\frac{1-2\sin{x}}{\sec{x}}&=\frac{1-2\sin{x}}{\frac{1}{\cos x}}\\&=\cos{x}\left(1-2\sin x\right)\\&=\cos x-2\sin{x}\cos{x}\\&=\cos x-2\sin{x}\cos{x}\cdot\frac{1+2\sin x}{1+2\sin x}\\&=\frac{\cos x\cancel{+2\sin{x}\cos{x}}\cancel{-2\sin{x}\cos{x}}-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos x-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos{3x}}{1+2\sin{x}}\end{aligned}$$
|
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|
The value of $\sum_{1\leq l< m How to solve this summation ? Also, I'm not sure what does $1\leq l< m <n$ supposed to imply in the development of summation form.
$$\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}$$
This one is from the Galois-Noether Contest in 2018:
Galois-Contest
|
Outline:
You have
$$
\sum_{1\leq \ell<m<n} \frac{1}{5^\ell3^m2^n}
= \sum_{n=1}^\infty\sum_{m=1}^{n-1}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell3^m2^n}
= \sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{3^m}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell} \tag{1}
$$
Since $$
\sum_{\ell=1}^{m-1} \frac{1}{5^\ell} = \frac{1}{5}\cdot\frac{1-1/5^{m-1}}{1-1/5} = \frac{1-1/5^{m-1}}{4} = \frac{1-5/5^{m}}{4}
$$
you can rewrite it as
$$
\sum_{1\leq \ell<m<n} \frac{1}{5^\ell3^m2^n}
= \frac{1}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1-5/5^{m}}{3^m}
= \frac{1}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{3^m}
- \frac{5}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{15^m}\,.
$$
Since $\sum_{m=1}^{n-1}\frac{1}{3^m}$ and $\sum_{m=1}^{n-1}\frac{1}{15^m}$ can be computed as [...], you can rewrite [...]
Can you continue?
|
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|
Prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ using the roots of a polynomial I have to
prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ by using the roots of the polynomial $x^3-21x^2+35x-7=0$
I tried to factor the polynomial but it didn't work and later found it cannot factorize with rationals. and I saw some similar problems in StackExchange. But the answers are very complex to me. I cannot use Euler's complex number formula since it's not in the syllabus. I do not want the exact answer but guidance to the answer.
|
let $t=\tan(\theta)$, we have
\begin{eqnarray*}
\tan(7 \theta) =\frac{ 7t-35t^3+21t^5-t^7}{1-21t^2+35t^4-7t^6}.
\end{eqnarray*}
Set $\tan(7 \theta) =0$ then the polynomial
\begin{eqnarray*}
7t-35t^3+21t^5-t^7=0
\end{eqnarray*}
has roots $t=0, \tan( \pi/7), \cdots ,\tan( 6 \pi/7)$. So
\begin{eqnarray*}
x^3-21x^2+35x-7=0
\end{eqnarray*}
has roots $x= \tan^2(\pi/7),\tan^2(2\pi/7),\tan^2(3\pi/7)$. Now let $y=x+1$.
|
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|
How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FOR LUNCH.
PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $
I LOVE YOU HONEY.
Anyone knows? Are we gonna get an integer number?
My attempt:
Does this help?
$$\frac{3x^3-x^2+2x-4}{x-1}=3x^2+2x+4$$
(long division)
\begin{align*}
I&=\int\frac{3x^3-x^2+2x-4}{[(x-1)(x-2)]^{1/2}} dx = \\
&=\int\frac{(3x^2+2x+4)(x-1)^{1/2}}{(x-2)^{1/2}} dx = \\
&=\int 3(u^4-4u^2-4)(u^2+1)^{1/2}du \times 2
\end{align*}
after the substitution
\begin{gather*}
(x-2)^{1/2}=u\\
du=\frac1{2(x-2)^{1/2}}dx\\
u^2=x-2\\
(x-1)^{1/2}=(u^2+1)^{1/2}
\end{gather*}
Update: This may help us proceed.
I tried to proceed:
$$6\int (u^4-4u^2-4)(u^2+1)^{1/2} du = 6\int ((t-3)^2-8)t \frac{dt}{2u}$$
after $u^2+1=t$ and $dt=2udu$
\begin{align*}
u^4-4u^2-4
&= (u^2+1)^2-(6u^2+5) \\
&= (u^2+1)^2-6(u^2+1)+1 \\
&= ((u^2+1)-3)^2-8
\end{align*}
I wonder whether this question can be solved from here?
Update:
This has been getting a lot of views, and I think most people came for the sort of problem mentioned in the title (where I got stuck) rather than the original problem itself.
Keepin this in mind, I'm reopening the question and here's the kind of answers I expect — Solutions to the original problem are good, but I'd prefer solutions that continue from the part where I got stuck — the polynomial in $u$ — that's the sort of problem mentioned in the title.
|
Noticing that
$ 3 x^3-x^2+2 x-4=\frac{3}{2}(2 x-3)\left(x^2-3 x+2\right)+\frac{5}{2}\left(5 x^2-7 x+2\right),$
we have $$
I=\frac{3}{2} \underbrace{\int_0^1(2 x-3) \sqrt{x^2-3 x+2} d x}_J+\frac{5}{2} \underbrace{\int_0^1 \frac{5 x^2-7 x+2}{\sqrt{x^2-3 x+2}} d x}_K
$$$$
\begin{aligned}
J =\frac{3}{2} \int_0^1 \sqrt{x^2-3 x+2} d\left(x^2-3 x+2\right) =\left[\left(x^2-3 x+2\right)^{\frac{3}{2}}\right]_0^1=-2 \sqrt{2}
\end{aligned}
$$
Again, decompose the numerator of the integrand of $K$ into 3 parts as
$$
5 x^2-7 x+2=5\left(x^2-3 x+2\right)+4(2 x-3)+4,
$$
we get
$$
\begin{aligned}
K&=5 \int_0^1 \sqrt{x^2-3 x+2} d x+4 \int_0^1 \frac{d\left(x^2-3 x+2\right)}{\sqrt{x^2-3 x+2}} +4 \int \frac{d x}{\sqrt{x^2-3 x+2}}\\&= 5 \cdot \frac{1}{8}(6 \sqrt{2}-\ln (3+2 \sqrt{2}))-8 \sqrt{2}+4 \ln (3+2 \sqrt{2}) \cdots(*)\\&= -\frac{17}{4} \sqrt{2}+\frac{27}{8} \ln (3+2 \sqrt{2})
\end{aligned}
$$
where $(*)$ comes from my post.
Plugging back yields
$$
\boxed{I=\frac{135}{16} \ln (3+2 \sqrt{2})-\frac{101}{8} \sqrt{2}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding The Zeros Of $\frac{z^2\sin z}{\cos z -1}$ $$f(z)=\frac{z^2\sin z}{\cos z -1}$$
$$\frac{z^2\sin z}{\cos z -1}=0\iff z^2\sin z=0$$ so $z=\pi k$ and $z=0$ are zeros, to find the order we must derive $\frac{z^2\sin z}{\cos z -1}$?
|
Since
$\sin(2x)=2\sin(x)\cos(x)$
and
$\cos(2x)=\cos^2(x)-\sin^2(x)
=1-2\sin^2(x)$
so
$\cos(2x)-1
=-2\sin^2(x)
$,
$f(2z)
=\frac{4z^2\sin 2z}{\cos 2z -1}
=\frac{4z^22\sin(z)\cos(z)}{-2\sin^2(z)}
=\frac{-2z^2\cos(z)}{\sin(z)}
=-2\frac{z}{\sin(z)}z\cos(z)
$.
The zeros are
$z=0$
(taking the limit
since it is undefined there)
and the zeros of
$\cos(z)$
which are
$(k+\frac12)\pi$.
|
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|
Find minimal polynomial of element Find minimal polynomial of element $3-2\sqrt[3]{2}-\sqrt[3]{4}$ $\in {\displaystyle \mathbb {Q}(\sqrt[3]{2}) } $ over the field $ {\displaystyle \mathbb {Q} } $.
Thank you for any help.
|
Here's a systematic way to find the minimal polynomial, which works in general setting. We know that $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is a basis of $\mathbb{Q}(\sqrt[3]{2})$ as a vector space over $\mathbb{Q}$. Now consider the following relations:
$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot1 = 3-2\sqrt[3]{2}-\sqrt[3]{4}$$
$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot \sqrt[3]{2} = -2 + 3\sqrt[3]{2} - 2\sqrt[3]{4}$$
$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot \sqrt[3]{4} = -4 - 2\sqrt[3]{2} + 3 \sqrt[3]{4}$$
These three equations can be written in matrix notation in the following way:
$$ (3-2\sqrt[3]{2}-\sqrt[3]{4})\left[
\begin{array}{c}
1\\
\sqrt[3]{2}\\
\sqrt[3]{4}
\end{array}
\right] = \left[
\begin{array}{ccc}
3&-2&-1\\
-2&3&-2\\
-4&-2&3
\end{array}
\right]\left[
\begin{array}{c}
1\\
\sqrt[3]{2}\\
\sqrt[3]{4}
\end{array}
\right] $$
This gives us that $(3-2\sqrt[3]{2}-\sqrt[3]{4})$ is an eigenvalue of the matrix on the right. Hence it's a zero of the polynomial $\det(xI-M)$. Obviously as the degree of it is $3$ and $(3-2\sqrt[3]{2}-\sqrt[3]{4}) \not \in \mathbb{Q}$ we get that it must be the minimal polynomial of $(3-2\sqrt[3]{2}-\sqrt[3]{4})$. Evaluating the determinant isn't a hard thing to do and you will get that the minimal polynomial is:
$$f(x) = x^3 - 9x^2 + 15x + 29$$
|
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|
solving a linear diophantine equation in two variables I'm a bit confused in applying the Euclidean algorithm to solve the following equation:
$$11x + 34 = 13y + 35$$
$$x,y\in\Bbb Z; \qquad0 < x,y$$
How can I go about solving it, if there are solutions?
|
$11x - 13 y = 1$
$$ \gcd( 13, 11 ) = ??? $$
$$ \frac{ 13 }{ 11 } = 1 + \frac{ 2 }{ 11 } $$
$$ \frac{ 11 }{ 2 } = 5 + \frac{ 1 }{ 2 } $$
$$ \frac{ 2 }{ 1 } = 2 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccc}
& & 1 & & 5 & & 2 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 6 }{ 5 } & & \frac{ 13 }{ 11 }
\end{array}
$$
$$ $$
$$ 13 \cdot 5 - 11 \cdot 6 = -1 $$
But you want
$11 \cdot 6 - 13 \cdot 5 = 1$ which is true, $66-65$
|
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|
How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1+\sqrt{x}}{1+\sqrt{x}}$$
With $(a + b)(a - b) = a^2 - b^2$, I can remove the irrational denominator.
$$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}$$
I the multiply the two parenthesis
$$\lim_{x\to1}\dfrac{\sqrt{x} \cdot x^2+\sqrt{x}\cdot x- \sqrt{x} \cdot 2 + x^2 + x - 2}{1-x}$$
I'm not sure where to continue to solve this limit.
|
$$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}=\\
\lim_{x\to1}(1+\sqrt{x})\lim_{x\to1}\dfrac{(x^2+x-2)}{1-x}\\
=2\lim_{x\to1}\frac{(x+2)(x-1)}{1-x}=-6
$$
|
{
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"url": "https://math.stackexchange.com/questions/2818925",
"timestamp": "2023-03-29T00:00:00",
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|
Finding reason for removal of the logarithm in ODE I'm given an ODE
$$y' = \frac{x + 2y + 1}{2x + 3},~y(5/2) = 1/4.$$
To solve this I'm doing the following:
*
*Solve system of equations
$$2x_0 + 2y_0 = 1,$$
$$2x_0 = 3,$$
this gives me solutions $x_0 = \frac{3}{2}$, $y_0 = -\frac{1}{4}$. Then, I'm transforming variables $s = x + \frac{3}{2}$, $t = y - \frac{1}{4}$.
*Transform the ODE using those new variables to the form
$$t' = \frac{s + 2t}{2s}$$
and using substitution of the form $t = s\cdot u(s)$ the equation is reduced to the
$$u' = \frac{1}{2s} \implies u' = \frac{1}{2}\ln|C\cdot s|.$$
Using initial conditions constant C can be found using $y(5/2) = 1/4$, $s(5/2) = 4$, $t(4) = 0$, $u(4) = 0$, hence $0 = \ln|4C| \implies C = \pm \frac{1}{4}$.
*Putting it all together yields me
$$y = \frac{1}{2}\left(x + \frac{3}{2}\right)\ln\left|\frac{x + 3/2}{4}\right| + 1/4$$
with two domains $I_1 = (-3/2, +\infty)$ and $I_2 = (-\infty, -3/2)$.
But my textbook states that the answer is
$$y = \frac{1}{2}\left(x + \frac{3}{2}\right)\ln\frac{x + 3/2}{4} + 1/4,~I = (-3/2, +\infty)$$
Hence the following two questions:
*
*What was the reason for removal of absolute value in the logarithm?
*Why is mine domain of validity wrong?
|
Since the initial condition is given for $x=\frac52$ we are assuming the solution for the domain $x>-\frac32\implies \frac{x + 3/2}{4}>0$.
|
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|
Problem related to probability from Sheldon Ross' book The stumbled upon following problem from Sheldon Ross's book:
Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue and 18 green balls. What is probability that either exactly 3 red balls or exactly 3 blue balls are withdrawn?
In this pdf, the solution is given as $p=\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.
But I guess it should be $\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\color{red}{2\times}\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.
Am I right? (The book has not given answer.)
|
I'd interpret the question as that exactly one of the two events (3 red or 3 blue) occurs but not both ("either or" strongly hints at that). This amounts to $P(3 \text{ red}) + P(3 \text{blue}) - 2P(3 \text{ red and } 3 \text{ blue })$ which is
$${{\binom{12}{3}\binom{16+18}{4}}\over{\binom{12+16+18}{7}}} + {{\binom{16}{3}\binom{12+18}{4}}\over {\binom{12+16+18}{7}}} - 2{{\binom{12}{3}\binom{16}{3}\binom{18}{1}} \over {\binom{12+16+18}{7}}}$$
In short, I concur.
|
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|
Area of Shaded Region Inside Quadrilateral I recently saw a similar problem online but found the area in a completely different way.
Problem:
There is a unit square, i.e. a square with side length equal to $\text 1$. Two lines are placed inside the square. Lets call them $\color{red}{\text {Line 1}}$ and $\color{blue}{\text {Line 2}}$.
$\color{red}{\text {Line 1}}$ is straight - It starts at a vertex and intersects the square exactly half way on the opposite edge.
$\color{blue}{\text {Line 2}}$ is curved - It is a quarter circle with radius, $r$, the same length as the square, i.e. $r=1$.
The diagram below shows the representation of the description provided above (I am not a geometric mathematician so forgive my elementary descriptions - please):
The goal is to find the area $A$.
My Attempt:
My intuition was to straight away subtract areas. So find the area of the quarter circle and subtract off the area under the intersection of $\color{red}{\text {Line 1}}$ and $\color{blue}{\text {Line 2}}$. So in the following diagram I find area $A+B+C$ and subtract area $B$ and area $C$:
Now clearly the area $A+B+C$ is simply the area of a quarter circle. This is: $$Area(A+B+C)=\frac {\pi r^2}4=\frac \pi4$$
Next, I find the point of intersection in the Cartesian plane of the $\text 2$ lines (where the red and blue lines intersect). I assume the bottom left point on the square represents the coordinates $(0,0)$.
Therefore the $\color{red}{\text {Line 1}}$ has equation:
$$y=-2x+1$$
And $\color{blue}{\text {Line 2}}$ lines has equation:
$$y=\sqrt{1-(x-1)^2}$$
The intersection is at:
$$-2x+1=\sqrt{1-(x-1)^2} \implies (-2x+1)^2=1-(x-1)^2$$
$$\implies 4x^2-4x+1=-x^2+2x$$
$$\implies 5x^2-6x+1=0 \implies (5x-1)(x-1)=0 \implies x=-\frac 15$$
Then we have $y=-2x+1=-2\frac 15 +1 = \frac 35$ so the coordinates of intersection (inside the square) are $\left(\frac 15, \frac 35 \right)$.
Then the area $B$ is simply the area of a triangle, with height $h$ and base $b$:
$$h=\frac 35 \ , \ b=\frac 12 - \frac 15= \frac 3{10}$$
$$\therefore Area(B)=\frac 12 bh= \frac 12 \frac 3{10} \frac 35= \frac 9{100}$$
Now for area $C$, this is simply the area under the curve of a (semi-) circle between the $x$-coordinates $x=0$ and $x=\frac 15$. Clearly this is not the neatest integral, so we just shift the circle coordinates and the area becomes:
$$Area(C)=\int_0^{1/5} \sqrt{1-(x-1)^2}dx$$
$$=\int_{-1}^{-4/5} \sqrt{1-x^2}dx=\left[ \frac 12 \left(x\sqrt{1-x^2}+\sin^{-1}(x) \right) \right]_{-1}^{-4/5}$$
$$=-\frac 6{25}+\frac \pi 4 - \frac 12 \sin^{-1}\left(\frac 45 \right)$$
Putting it all together we get:
$$Area(A)=Area(A+B+C)-Area(B)-Area(C)$$
$$=\frac \pi4- \frac 9{100} - \left[-\frac 6{25}+\frac \pi 4 - \frac 12 \sin^{-1}\left(\frac 45 \right)\right]$$
$$=\frac 3{20}+\frac 12 \sin^{-1}\left(\frac 45 \right)$$
$$\approx 0.61365$$
My Question:
Is this the best approach? Have I done this right? Is there a simpler or neater method? And what would change if we were to calculate any of the other 3 areas?
Thank you in advance.
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There is a simpler approach. Consider an extension of the diagram you drew:
Note that the desired area "A" is the sum of the areas of the circular sector and the obtuse triangle. As shown in the diagram, the angle subtended by an arc from the vertex on the circumference is half of the angle measured from the center of the circle. Since $\theta = \tan^{-1} \frac{1}{2}$, it follows that $2\theta = 2\tan^{-1} \frac{1}{2}$ and the area of the circular sector is simply $$A_1 = \tan^{-1} \frac{1}{2}.$$ Then the area of the obtuse triangle is $$A_2 = \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot \sin \left(\frac{\pi}{2} - 2\theta\right) = \frac{1}{4} \cos \left(2 \tan^{-1} \frac{1}{2} \right) = \frac{3}{20}.$$ Thus the total area is $$A = \tan^{-1} \frac{1}{2} + \frac{3}{20} \approx 0.613648.$$
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|
How many triangles have sides $a$, $a+1$, $a+2$ and angles $x$ and $2x$?
How many triangles can I draw if its arms are $a$, $a+1$, and $a+2$, and two angles are $x$ and $2x$?
I think there are only two solutions to this problem. A triangle with arms $1$, $2$, $3$ and $2$, $3$, $4$. But I am not sure.
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This can be done with $a = 1, \frac{1}{2}(1 + \sqrt{13})$ and $4$ only. Not for $a = 2$. It's a simple matter of putting a set of equations into a symbolic calculator and letting it chew through them.
If the angles are $x, y, z$. Then the Cosine Law says, for instance,
$$ a^2 = (a+1)^2 + (a+2)^2 - 2(a+1)(a+2) \cos x $$
and we also have,
\begin{align*}
0 \le x &\le \pi \\
0 \le y &\le \pi \\
0 \le z &\le \pi \\
x + y + z &= \pi
\end{align*}
Then put in the various conditions like $x = 2z$ or $y = 2x$.
For example, in Mathematica,
Reduce[{a^2 == (a + 1)^2 + (a + 2)^2 - 2 ((a + 1) (a + 2)) Cos[x],
(a + 1)^2 == a^2 + (a + 2)^2 - 2 (a (a + 2)) Cos[y],
(a + 2)^2 == a^2 + (a + 1)^2 - 2 (a (a + 1)) Cos[2 x],
3 x + y == Pi, 0 <= x, 0 <= y, 1 <= a}]
Tells you that $a = 4$ is a solution when $z = 2x$.
|
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|
Doubt in change of basis matrix Let $A=\{(1,0,5) (4, 5,5)(1,1,4)\}; B= \{(1,3,2)(-2,-1,1)(1,2,3) \}$. To find change of basis matrix.
$\boxed{ M_{A\to B} = M_{A\to e}M_{e\to B} } ---(1)\\= \begin{pmatrix}
1 & 4 & 1 \\
0 & 5 & 1 \\
5 & 5 & 4
\end{pmatrix}^{-1} \begin {pmatrix}
1 & -2 & 1 \\
3 & -1 & 2 \\
2 & 1 & 3
\end {pmatrix} = \boxed {\begin {pmatrix}
-2 & -2 & -1 \\
0 & -1 & 0 \\
3 & 4 & 2
\end {pmatrix} } -----(2)$
Is equation (1) correct??? or it is $M_{A \to B} = M_{e \to B} M_{A \to e}$ as given in solution part of LINK:- How to construct change of basis matrix (I am confused here pls correct me if i am wrong...)
To reconfirm i did it in following way:-
to find coordinates of B interms of A
$(1,3,2)= a(1,0,5)+b(4,5,5)+c(1,1,4)=(a+4b+c,5b+c,5a+5b+4c)=[-2,0,3]_B$
Similarly $(-2,-1,4) \sim [-2,-1,4]_B \\ (1,2,3) \sim [-1,0,2]_B$
THis boils down to same above matrix by (2)
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We need
$$M_{A\to B} = M_{e\to B}M_{A\to e}$$
that is
$$\begin {pmatrix}
1 & -2 & 1 \\
3 & -1 & 2 \\
2 & 1 & 3
\end {pmatrix}^{-1}\begin{pmatrix}
1 & 4 & 1 \\
0 & 5 & 1 \\
5 & 5 & 4
\end{pmatrix} $$
indeed the first matrix takes a vector in basis $A$ to the standard basis and the second takes a vector in the standard basis to the basis $B$, therefore the product matrix takes a vector from basis $A$ to basis $B$.
|
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|
Show that all eigenvalues of this pentadiagonal matrix are double degenerate I am trying to show in general that the following pentadiagonal matrix $\mathbf{M}$ has double degenerate eigenvalues,
\begin{equation}
\mathbf{M} =
\left[ \begin{array}{cccccccc}
4 & 0 & a & 0 & 0 & 0 & 0 & 0 \\
0 & 3 & 0 & a & 0 & 0 & 0 & 0\\
a & 0 & 2 & 0 & a & 0 & 0 & 0\\
0 & a & 0 & 1 & 0 & a & 0 & 0\\
0 & 0 & a & 0 & 1 & 0 & a & 0\\
0 & 0 & 0 & a & 0 & 2 & 0 & a\\
0 & 0 & 0 & 0 & a & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 0 & a & 0 & 4
\end{array} \right]\!,
\end{equation}
where $a$ is just some real number. It doesn't particularly matter what the matrix elements are on the diagonal, so long as they are symmetric about the anti-diagonal the eigenvalues will still be double degenerate. Numerically, I find that a pentadiagonal matrix of this form and of arbitrary size always yields double degenerate eigenvalues.
In block form we can express $\mathbf{M}$ as
\begin{equation}
\mathbf{M} =
\left[ \begin{array}{cc}
\mathbf{A} & \mathbf{B} \\
\mathbf{J}^{-1}\mathbf{B}\mathbf{J} & \mathbf{J}^{-1}\mathbf{A}\mathbf{J} \\
\end{array} \right]\!,
\end{equation}
where
\begin{equation}
\mathbf{A}=\left[\begin{array}{cccc}
4 & 0 & a & 0 \\
0 & 3 & 0 & a \\
a & 0 & 2 & 0 \\
0 & a & 0 & 1
\end{array} \right],
%
\; \; \mathbf{B}=\left[ \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
a & 0 & 0 & 0 \\
0 & a & 0 & 0
\end{array} \right]
\end{equation}
and $\mathbf{J}$ is the exchange matrix, hence the top-left and bottom-right blocks are permutation-similar (as are the top-right and bottom-left blocks). However, this has not helped me show that all eigenvalues are double degenerate.
So far I have that the truncated 4x4 matrix
\begin{equation}
\mathbf{M}_{\text{trunc}}=\left[\begin{array}{cccc}
2 & 0 & a & 0 \\
0 & 1 & 0 & a \\
a & 0 & 1 & 0 \\
0 & a & 0 & 2
\end{array} \right]
\end{equation}
has a characteristic polynomial
\begin{equation}
p(\lambda)=(a^2+(1-\lambda)\lambda + 2(\lambda-1))^2
\end{equation}
which has roots which are evidently double degenerate.
Any help at all would be much appreciated!
|
Shortly speaking, such a matrix can be essentially split into two submatrices formed by odd and even rows and columns, and those two matrices are obtained from one another by conjugating by $\mathbf{J}$. To be more detailed: consider the matrix
\begin{equation}
\mathbf{S} =
\left[ \begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0
\end{array} \right]\!.
\end{equation}
Then $\mathbf{S^{-1}MS}$ is of the form
\begin{equation}
\left[ \begin{array}{cc}
\mathbf{C} & \mathbf{0} \\
\mathbf{0} & \mathbf{C} \\
\end{array} \right]\!,
\end{equation}
where\begin{equation}
\mathbf{C} =
\left[ \begin{array}{cccccccc}
4 & a & 0 & 0 \\
a & 2 & a & 0 \\
0 & a & 1 & a \\
0 & 0 & a & 3
\end{array} \right]\!,
\end{equation}
so the eigenvalues of $\mathbf{S^{-1}MS}$ (which are the same as the eigenvalues of $\mathbf{M}$) are evenly degenerate. I don't know how to prove that the multiplicities of eigenvalues of $\mathbf{M}$ are exactly 2 (i.e., that the eigenvalues of $\mathbf{C}$ are non-degenerate), though, and whether it's actually true for all values of $a$.
|
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|
Solving $\cos(3x) = \cos(2x)$ I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and
$$\cos(3x) = 4\cos^3(x) -3\cos(x)$$
Plugging into the equation and we have that
$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1$$
$$4\cos^3(x) -3\cos(x) - 2\cos^2(x)+1= 0$$
Recalling $t = \cos (x)$,
$$4t^3-2t^2-3t +1 = 0$$
Which is a cubic equation. Your sincerely helps will be appreciated.
Regards!
|
Hint
You can use sum-product equivalence. Which is:
$$\cos(A)-\cos(B)=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
so,
$$\cos(3x)-\cos(2x)=0\to-2\sin\left(\frac{3x+2x}{2}\right)\sin\left(\frac{3x-2x}{2}\right)=0$$
$$\sin\left(\frac{5x}{2}\right)\sin\left(\frac{x}{2}\right)=0$$
so,
$$\sin\left(\frac{5x}{2}\right)=0 \text{ or } \sin\left(\frac{x}{2}\right)=0$$
|
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|
Finding limit using Euler number $$\lim_{x\rightarrow \infty }\left ( 1+\frac{3}{x+2} \right )^{3x-6}$$
I've tried to factor and simplfy the expression. I got:
$${\left ( 1+\frac{3}{x+2} \right )^{\frac{1}{x+2}}}^{3({x^2-4})}$$
I set $x$ to $1/t$ I get:
$${\left ( 1+\frac{3}{\frac{1}{t}+2} \right )^{\frac{1}{\frac{1}{t}+2}}}^{3 \left({\frac{1}{t}^2-4} \right)}$$
then I am left with:
$$\left ( e^{3} \right )^{3\left(\frac{1}{t^2}-4\right)}$$ which I get by using Euler number.
The answer is $e^9$, but clearly the answer I get is $(e^9)^{\text{expression}}$ which is not equal to the answer.
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Let $y = \left(1+\dfrac{3}{x+2}\right)^{3x-6} $. Then
$$\ln y=(3x-6)\ln\left(1+\dfrac{3}{x+2}\right)\text{.}$$
On the left-hand side, by continuity of $\ln$,
$$\lim_{x \to \infty}\ln y = \ln \left(\lim_{x \to \infty}y\right)\text{.}$$
On the right-hand side,
$$\lim_{x \to \infty}(3x-6)\ln[1+3/(x+2)]=\lim_{x \to \infty}\dfrac{\ln[1+3/(x+2)]}{1/(3x-6)}\text{.}$$
Apply L-Hospital's rule. The numerator changes to
$$\dfrac{1}{1+3/(x+2)} \cdot \dfrac{-3}{x^2}$$
and the denominator changes to
$$\dfrac{-1}{3x^2}$$
and furthermore,
we have
$$\lim_{x \to \infty}\dfrac{1/[1+3/(x+2)] \cdot -3/x^2}{-1/(3x^2)} = \lim_{x \to \infty}\dfrac{9}{1+3/(x+2)} = 9\text{.}$$
By L-Hospital,
$$\lim_{x \to \infty}\dfrac{\ln[1+3/(x+2)]}{1/(3x-6)} = 9\text{.}$$
Hence, taking into account the left-hand side, we have
$$\ln\left(\lim_{x \to \infty}y\right)=9 \implies \lim_{x \to \infty}y = e^9\text{.}$$
|
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|
Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y = 2b-a$, we see that $x+y = 4b$. Substituting,
$$xy+\dfrac {x+y}{2}+y=0$$
$$2xy+x+3y=0$$
From this we see that $y|x$, so we can substitute $x = ky$ for some integer $k$
$2ky^2+ky+3y = 0$
$$k = \dfrac {3}{2y+1}$$
From here we get that $y \in \{-2, -1 , 1 \}$, and each of the cases can be checked individually.
|
Hmmm...
Show that $4b^2 + 4b = a^2 + a$ has no integer solutions where none of $x, y$ are zero.
Where do $x$ and $y$ come from? Is the following what was meant?
Show that $4b^2 + 4b = a^2 + a$ has no integer solutions where none of $a, b$ are zero.
I'll proceed along the lines of the latter. Clearly $a$ must be even, because if it's odd then $a^2 + a$ is singly even but $4b^2 + 4b$ is clearly doubly even. Therefore, with $a$ even, it follows that $$\frac{a^2 + a}{4}$$ is an integer, and so the equation can be rephrased as $$\frac{a^2 + a}{4} = b^2 + b.$$ Therefore $\sqrt{a^2 + a} = 2 \sqrt{b^2 + b}$. Trouble is, this requires both $\sqrt{a^2 + a}$ and $\sqrt{b^2 + b}$ to be integers.
Okay, I'm sorry, this is looking to be much more complicated than what you have...
|
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|
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$ If the circle $(x-a)^2+(y-b)^2=r^2$ and the line $y=mx+c$ do not meet:
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$
These are the steps I have thus taken (although they may be wrong/useless):
*
*Rearranged the circle equation for y:
$y=\sqrt{r^2-x^2+2ax-a^2}+b$
*Set the circle and line equations equal to each other as in simultaneous equations.
I was then planning on finding some discriminant and as we know the line and circle do not meet I could set the discriminant < 0 and show it is the same as the long inequality.
Any solutions/pointers appreciated as I cannot seem to work this problem out.
|
Substitute the equation for the line into the equation for the circle
\begin{eqnarray*}
(x-a)^2+(mx+c-b)^2= r^2 \\
(m^2+1)x^2 -2x(a+m(b-c))+a^2+(c-b)^2-r^2=0.
\end{eqnarray*}
This is a quadratic in $x$ and for no solutions to exist we require the disciminant to be negative so
\begin{eqnarray*}
(a+m(b-c))^2-(m^2+1)(a^2+(c-b)^2-r^2) < 0
\end{eqnarray*}
Now rearrange a bit to obtain your inequality.
|
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|
For the differential equation given by $(x^2-y^2)dx+3xydy=0$ the general solution is $(x^2+2y^2)^3 =cx^2$
For the differential equation given by $$(x^2-y^2)dx+3xy \ dy=0$$ the general soultion is $$(x^2+2y^2)^3 =cx^2$$ Check this solution by differentiating and rewriting to get the original function.
I solved for $\frac{dy}{dx}$ and got $$\frac{dy}{dx}=\frac{cx-3x(x^2+2y^2)^2} {6y(x^2+2y^2)^2}$$ I was told to solve for $c$ and multiply by $dx$. but I cannot figure out how to get this to the original equation. My teacher said to use implicit differentiation so I solved for $\frac{dy}{dx}$.
|
Calling
$$
f(x,y) = (x^2 + 2 y^2)^3 -c x^2 = 0
$$
we have
$$
df=f_x dx+f_y dy = 0\to y' = -\frac{f_x}{f_y} = \frac{x \left(c-3 \left(x^2+2 y^2\right)^2\right)}{6 y \left(x^2+2 y^2\right)^2}
$$
now substituting $c = \frac{\left(x^2+2 y^2\right)^3}{x^2}$ and simplifying we obtain
$$
y' = \frac{y}{3 x}-\frac{x}{3 y}
$$
as expected.
NOTE
now making $z = \frac yx \to y' = x z' + z$ we have the transformed DE
$$
x z' = \frac 13\left(z-\frac 1z\right)-z
$$
which is separable and can be easily solved.
|
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|
Maximizing by setting Derivative equal to 0. Stuck (struggling) I'm having immense difficulty doing this question.
Want to maximize (respect to x):
$f(x)$=$(x-x_0)(x-x_1)(x-x_2)(x-x_3)$ where
*
*$x_1-x_0=h$
*$x_2-x_1=h$
*$x_3-x_2=h$.
Want to get the value of x in terms of $x_1$ and $h$.
First I take the derivative and get:
$(x-x_0)(x-x_1)(x-x_2)$ +$(x-x_0)(x-x_1)(x-x_3)+(x-x_0)(x-x_2)(x-x_3)+ (x-x_1)(x-x_2)(x-x_3)=0$
Now using the facts 1,2, and 3, I substitute for $x_0, x_2$, and $x_3 $:(Need to write everything in terms of x1 and h)
$x_0$=$x_1-h$,
$x_1=x_1$,
$x_2=x_1+h$,
$x3=h+x2=h+(x_1)+h=2h+x_1$
Then we have for the derivative:
$(x-x_1+h)(x-x_1)(x-(x_1+h))$ + $(x-x_1+h)(x-x1)(x-(2h+x_1))$+
$(x-x_1+h)(x-x_1-h)(x-(2h+x_1))$+$(x-x_1)(x-x_1-h)(x-(2h+x_1))$=0.
Then we have:
$[(x-x_1)^2+h(x-x_1)]((x-(x_1+h))$+$[(x-x_1)^2+h(x-x_1)]((x-(2h+x_1))$+
$[(x-x_1)^2-h^2](x-(2h+x_1))$+$[(x-x_1)^2-h(x-x_1)]((x-(2h+x_1))$=0
This is all I have but I can't solve for x still. Wolfram Alpha won't give me answer. Any help would be much appreciated thank you. I believe the max value of f(x) should be (like max y value): $f(x)=9/16$. or $(f(x)=72/128)$
Thank you. I have been working on this problem for over three hours so any help would really be appreciated.
Here is the actual question I'm working on for reference:
5 Part c. Where Theorem 1 is given to be:
|
let $a = \frac {x_1+x_2}{2}$ then
$x_1 = a - \frac {h}{2}\\
x_0 = a - 3\frac {h}{2}\\
x_2 = a + \frac {h}{2}\\
x_3 = a + 3\frac {h}{2}$
This will take advantage of some of the symmetry of our fuction
$f(x) = (x-a - \frac {3h}{2})(x-a + \frac {3h}{2})(x-a - \frac {h}{2})(x-a + \frac {h}{2})\\
((x-a)^2 - \frac {9h^2}{4})((x-a)^2 - \frac {1h^2}{4})\\
(x-a)^4 - \frac {10h^2}{4}(x-a)^2 + \frac {9h^4}{16}$
$f'(x) = 4(x-a)^3 - \frac {10h^2}{2} (x-a) = 0\\
(2(x-a) - \frac {h\sqrt {10}}{2})(x-a)(2(x-a) + \frac {h\sqrt {10}}{2})=0$
We have minima at $x = \frac{h\sqrt{10}}{4} + x_1+\frac {h}{2}, -\frac{h\sqrt{10}}{4} + x_1 +\frac {h}{2}$
And a maximum at $x = a = x_1 + \frac {h}{2}$
$f(a) = \frac {9h^4}{16}$
|
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|
Doubt with an inequality involving integrals I have to prove this inequality about integrals. I did it but I'm not sure if my arguments were correct. Take a look:
Prove:
$$0\leq\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$$
My try is this:
Let $f(x)=\frac{1-x^2}{x^4+1}$ with $f:[-1,1]\to\mathbb{R}$ and $g(x)=\frac{2+x^4}{x^4+1}$ with $g:[-1,1]\to\mathbb{R}$. So we have to prove: $f(x)\leq g(x)$ for all $x\in[-1,1]$.
Clearly we only need to prove that $1-x^2\leq2+x^4$ because the denominator is always a positive number in $[-1,1]$. So we get:
$$0\leq x^2(x^2+1)+1=(x^2+1)(x^2+\frac{1}{x^2+1})$$
Therefore we need:
$$x^2+1\geq0 \land x^2+\frac{1}{x^2+1}\geq0$$
$$\lor x^2+1\leq0 \land x^2+\frac{1}{x^2+1}\leq0$$
The first pair of inequalities are always true for any $x\in\mathbb{R}$, because they involve positive numbers. And the second pair is the empty set by the same reason.
Therefore $f(x)\leq g(x)$ for any $x\in\mathbb{R}$ in particular for $x\in[-1,1]$ so we have: $\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$
The part of the zero is easy because we only need to prove: $0\leq1-x^2$, that is $x\leq|1|$, $-1\leq x\leq1$.
Is the first part correct or it is "forced"? Thanks!
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Alt. hint: $\displaystyle\;\;\int_{-1}^1\frac{-x^2}{x^4+1}dx \lt 0 \lt \int_{-1}^1\frac{x^4}{x^4+1}dx\,$, and $\displaystyle\;\int_{-1}^1\frac{1}{x^4+1}dx \lt \int_{-1}^1\frac{2}{x^4+1}dx\,$.
|
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|
Using algebraic identities to prove a number is not prime $$3^{3^n}(3^{3^n}+1)+3^{3^n +1}-1$$
I want the prove that the number is not prime. I used the identity $a^3+b^3+c^3-3abc $. I couldn't simplify to the state where the factors could be observed.
|
The other approach does show a quite large factor of the number; good to know how to approach this.
The expression $$ 3^{3^n} 3^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$
first becomes
$$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$
The first two terms really are cubes, since the exponents are divisible by 3. If we look for $a^3 = 3^{3^n},$ it is reasonable to try
$$ a = 3^x $$
because we want a power of three. Then
$$ a^3 = \left(3^x \right)^3 = 3^{3x}. $$
Then $a^3 = 3^{3^n}$ tells us
$$ 3^{3x} = 3^{3^n} $$
so
$$ 3x = 3^n $$ and
$$ x = 3^n 3^{-1} = 3^{n-1}, $$
$$ a = 3^x = 3^{3^{n-1}} $$
$$ $$
If we look for $b^3 = 9^{3^n},$ it is reasonable to try
$$ b = 9^y $$
because we want a power of nine. Then
$$ b^3 = \left(9^y \right)^3 = 9^{3y}. $$
Then $b^3 = 9^{3^n}$ tells us
$$ 9^{3y} = 9^{3^n} $$
so
$$ 3y = 3^n $$ and
$$ y = 3^n 3^{-1} = 3^{n-1}, $$
$$ b = 9^x = 9^{3^{n-1}} $$
$$ $$
We are not yet sure what we want to use for $c.$ Let us examine what we already have for $3ab,$ anticipating the eventual $-3abc.$
$$ a b = 3^{3^{n-1}} 9^{3^{n-1}} $$
MORE TO COME 11:08 am
We do need to understand $b = a^2.$ Well,
$$ a^2 = \left( 3^{3^{n-1}}\right)^2 = 3^{ 2 \cdot 3^{n-1}} $$ while
$$ b = \left( 3^2 \right)^{3^{n-1}} = 3^{ 2 \cdot 3^{n-1}} \; . $$
So $b = a^2,$ also $ab = a a^2 = a^3$ so
$$ ab = 3^{3^n} \; . $$ Next
$$ 3ab = 3 \cdot 3^{3^n} = 3^1 \cdot 3^{3^n} = \cdot 3^{3^n + 1} \; .$$
This is progress. We now have the overall expression
$$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 = b^3 + a^3 + 3ab - 1. $$
We can introduce a value of $c$ by demanding $-3abc = 3ab.$ The requires $c=-1.$ We have
$$ b^3 + a^3 - 3abc - 1. $$
Finally $c^3 = (-1)^3 = -1,$ and we get
$$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 = b^3 + a^3 - 3abc - c^3 = a^3 + b^3 + c^3 - 3abc. $$
Perhaps I should say why the new factors are both large. We have
$$a+b+c > b = 9^{3^{n-1}}$$
It is also worth remembering that
$$ a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( (b-c)^2 + (c-a)^2 + (a-b)^2 \right) $$
which is bigger than $ (b+1)^2 / 2 \; > \; b^2 / 2 \; . $
|
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Determine Minimum Value.
Find the minimum value of
$$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
for $x>0$.
When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks
I think differentiation would be really complicated
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$$\dfrac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=\dfrac{(x+1/x)^6-(x^3+1/x^3)^2}{(x+1/x)^3+(x^3+1/x^3)}=(x+1/x)^3-(x^3+1/x^3)$$ Therefore your function reduces to $3(x+1/x)$ which has its minimum when $x=1.$
|
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expected sum of the numbers that appear on two dice, each biased so that a 3 comes up twice as often as each other number? This question is already posted here,but i want to check my approach.
Question
What is the expected sum of the numbers that appear on
two dice, each biased so that a $3$ comes up twice as often
as each other number?
My Approach
Let the probability of getting number other than $3$ is $p$
so
$$\frac{1}{p}+\frac{1}{p}+\frac{2}{p}+\frac{1}{p}+\frac{1}{p}+\frac{1}{p}=1 \Rightarrow p=\frac{1}{7}$$
Proability of getting $3=\frac{2}{7}$
and rest other $=\frac{1}{7}$
let $E(X_1)$ be the expectation of getting sum on rolling $1$ dice.
$E(X_1)=1 \times \frac{1}{7}+2 \times \frac{1}{7}+3 \times \frac{2}{7}+4 \times \frac{1}{7}+5 \times \frac{1}{7}+6 \times \frac{1}{7}=\frac{24}{7}$
Now
expected sum of the numbers that appear on two dice
$$E(X_1 +X_2)=E(X_1)+E(X_2)=\frac{24}{7}+\frac{24}{7}=\frac{48}{7}\approx 6.86$$
Is my approach correct?
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Your approach is correct. To save a bit of the arithmetic and the detour through calculating the probabilities, you could also argue that the problem statement amounts to the result being uniformly randomly drawn from $\{1,2,3,3,4,5,6\}$, so the expected result for one die is the mean of that set,
$$
\frac{1+2+3+3+4+5+6}7=\frac{24}7\;.
$$
It's essentially the same solution, just not so many fractions to write :-)
|
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Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+b+c+d=x$. Then $x+e=8\implies e=8-x$. Also, $\frac{a^2+1}{2}+\frac{b^2+1}{2}+\frac{c^2+1}{2}+\frac{d^2+1}{2}\geq (a+b+c+d)=x$ by AM-GM inequality. Hence, $a^2+b^2+c^2+d^2\geq 2x-4$.
Now we have $2x-4+e^2\leq a^2+b^2+c^2+d^2+e^2=16$. Substituting $x=8-e$, we get $e^2-2e-4\leq 0$. We can easily calculate that the lowest value that $e$ can attain is $1 -\sqrt{5}$.
However, the answer given on the internet is $\frac{16}{5}$. Where am I going wrong?
EDIT $1$ -- Is this a case of how the value $1-\sqrt{5}$ can never be attained by $e$, although the inequality is true?
EDIT $2$ -- It seems that we need to find the maximum. By my method, I’ve found the maximum to be $1+\sqrt{5}$. This is greater than $\frac{16}{5}$. Have I found a sharper inequality?
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Intuitive approach:
For a given $$(a+b+c+d=k)$$
We want
$$Min(a^2+b^2+c^2+d^2)$$
In order to maximize $e$
This is true only when $a=b=c=d$
From that we have
$$4x+e=8,\; 4x^2+e^2=16$$
Giving $x=\frac{6}{5}$ and $e=\frac{16}{5}$
Leading to the maximum value
|
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|
Probability for repeated incidents . I had problems on understanding the following problem :
A die is continuously thrown until 4 is obtained. What is the probability that the dice is thrown an even number of times?
Here , a solution states that probability is :
$$\frac{4}{6}\times \frac{2}{6}+\frac{4}{6}\times \frac{4}{6}\times \frac{4}{6}\times \frac{2}{6} + ...$$
$$=\frac{2}{5}$$
I didn't understand why such process was used and what's the the concept behind it?
Another problem :
(ii) If a biased coin is tossed two times repeatedly , then what is the probability of occurring HH?
Can I do the problem (ii) similarly like (i) as :
$$\frac{1}{4}+\frac{3}{4}\times \frac{1}{4}+\frac{3}{4}\times \frac{3}{4}\times \frac{1}{4}+ ...$$
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*
*See, that the probabilty of throwing $4$ in $k$-th throw ($k\geq 1$)is equal to:
$$p(k)=\frac{1}{6}\left(\frac{5}{6}\right)^{k-1}$$
So we have to sum up all the probabilities for even $k$:
$$P=\sum_{n=1}^{\infty}p(2n)=\sum_{n=1}^{\infty}\frac{5}{6}\left(\frac{5}{36}\right)^{2n-2}=\sum_{n=0}^{\infty}\frac{5}{36}\left(\frac{25}{36}\right)^{n}=\frac{5}{36}\frac{1}{1-\frac{25}{36}}=\frac{5}{11}$$
*Tossing a coin two times makes the probability of $HH$ equal to $p(H)^2$.
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|
Evaluate integral $\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$ Evaluate $\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$
My working let I=$\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$
$=\int_0^\pi \frac18\left(\cos (4x+4\sin 3x)-4\cos(2x+2\sin 3x)+3\right)dx$
$=\frac{3\pi}{8}+\frac18\int_0^\pi\cos (4x+4\sin 3x)dx-\frac12\int_0^\pi\cos (2x+2\sin 3x)dx$
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Before we start, let us look at a related family of integrals.
For any integer $n$ and $\lambda \in \mathbb{R}$, let $J_n(\lambda)$ be the integral
$$J_n(\lambda) \stackrel{def}{=} \int_{-\pi}^{\pi} e^{in(x+\lambda\sin(3x))} dx$$
It is easy to see $J_0(\lambda) = 2\pi$ independent of $\lambda$. Furthermore, $J_n(\lambda) = 0$ unless $3$ divides $n$.
To see this, we use the fact $\sin(3x)$ is periodic with period $\frac{2\pi}{3}$.
This allows us to rewrite $J_n(\lambda)$ as
$$\left(\int_{-\pi}^{-\frac{\pi}{3}} + \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} + \int_{\frac{\pi}{3}}^{\pi}\right)e^{in(x+\lambda\sin(3x))} dx
=
\left(\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}e^{in(x+\lambda\sin(3x))} dx\right)
\left(e^{-i\frac{2\pi n}{3}} + 1 + e^{i\frac{2\pi n}{3}}\right)
$$
When $n$ is not divisible by $3$, $J_n(\lambda)$ vanishes because of the factor $e^{-i\frac{2\pi n}{3}} + 1 + e^{i\frac{2\pi n}{3}}$.
Back to the original problem. Since both $x$ and $\sin(3x)$ is an odd function, so does the sum. Together
with $\sin^4(x)$ is an even function, we find the integrand is an even function.
As a result,
$$\begin{align}\int_0^\pi \sin^4(x + \sin(3x)) dx
&= \frac12\int_{-\pi}^\pi \sin^4(x + \sin(3x))dx\\
&= \frac12\int_{-\pi}^\pi\left(\frac{ e^{i(x+\sin(3x))} - e^{-i(x+\sin(3x))}}{2i}\right)^4 dx\\
&= \frac{1}{32}\left[ J_4(1) - 4 J_2(1) + 6J_0(1) - 4J_{-2}(1) + J_{-4}(1)\right]\\
&= \frac{1}{32}\left[ 0 - 4(0) + 6(2\pi) - 4(0) + 0\right]\\
&= \frac{3\pi}{8}
\end{align}
$$
About the family of integrals mentioned in question/comment, we have
$$\int_0^\pi \cos (2^n x + k \sin (3x)) dx
= \frac12 \int_{-\pi}^\pi \cos (2^n x + k \sin (3x)) dx
= \frac14 \left(J_{2^n}(k') + J_{-2^n}(k')\right)
$$
where $k' = \frac{k}{2^n}$. Since $2^n$ is not divisible by $3$, all of them evaluate to $0$.
|
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If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?
I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appreciated!
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Hint: suppose $\,c, x \gt 0\,$, then $\,u = x + \dfrac{c}{x} \ge 2 \sqrt{c}\,$ by AM-GM with equality iff $\,x = \sqrt{c}\,$, and:
$$
f(x)=\frac{x^2-3x+c}{x^2+3x+c} \color{red}{\cdot \frac{\;\;\cfrac{1}{x}\;\;}{\cfrac{1}{x}}}=\dfrac{x+ \cfrac{c}{x}-3}{x+ \cfrac{c}{x}+3} = \dfrac{u-3}{u+3}=1 - \cfrac{6}{u+3} \ge 1 - \frac{6}{2 \sqrt{c}+3}
$$
The RHS gives a lower bound, which is actually a minimum since it is attained for $\,x = \sqrt{c}\,$.
For it to equal $\,\dfrac{1}{7}\,$, it follows that $\displaystyle\,1 - \frac{6}{2 \sqrt{c}+3}=\frac{1}{7} \iff 2 \sqrt{c} +3 = 7 \iff c = 4\,$.
What remains to be filled in is the details to cover the rest of cases, which are fairly straightforward to work out, for example $\,f(-x) = \dfrac{1}{f(x)}\,$.
|
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Solving system of polynomial equations stemming from a constrained optimization problem Constrained optimization (maximization and minimization) problem with two equality constraints.
$$\begin{array}{ll} \text{extremize} & x y z\\ \text{subject to} & x^2 + y^2 + z^2 = 9\\ & x + y - z = 3\end{array}$$
The Lagrangian, $\mathcal{L} : \mathbb{R}^5 \to \mathbb{R}$, is
$$\mathcal{L}(x,y,z,\mu_1,\mu_2) = xyz - \mu_1(x^2 + y^2 + z^2 - 9) - \mu_2(x + y - z - 3)$$
The first-order (necessary) conditions (FOCs) are
*
*$yz - 2\mu_1x - \mu_2 = 0$;
*$xz - 2\mu_1y - \mu_2 = 0$;
*$xy - 2\mu_1z + \mu_2 = 0$;
*$x^2 + y^2 + z^2 = 9$;
*$x + y - z = 3$.
Question: How can this system of equations best be solved? In other words, how can I find all candidate constrained maximizers or minimizers?
When I tried to solve the system (by first solving (1),(2),(3) for $\mu_2$ and consequently the resulting equation for $\mu_1$, or by first solving (1),(2),(3) for $\mu_1$ and consequently the resulting equation for $\mu_2$)
I arrived at $x = y = -z$, but in that case we only reach $(1,1,1)$ (by (5)), which violates the other equality constraint (4), so these approaches apparentely do not lead to any solutions, but there definitely are (several) solutions (as I've checked on WolframAlpha).
How to find these?
|
You can reduce the problem in the beginning to:
$$\text{optimize} \ xy(x+y-3) \ \text{subject to} \\
\ x^2+y^2+(x+y-3)^2=9.$$
Hence:
$$L(x,y,\lambda)=x^2y+xy^2-3xy+\lambda(9-x^2-y^2-(x+y-3)^2).$$
FOC:
$$\begin{cases}
L_x=2xy+y^2-3y-2x\lambda-2\lambda(x+y-3)=0\\
L_y=x^2+2xy-3x-2y\lambda-2\lambda(x+y-3)=0\\
L_{\lambda}=9-x^2-y^2-(x+y-3)^2=0\\
\end{cases}.$$
$(2)-(1)$:
$$x^2-y^2-3(x-y)+2\lambda(x-y)=0 \Rightarrow (x-y)(x+y-3+2\lambda)=0.$$
Case 1: $x-y=0 \Rightarrow y=x$, which is plugged to $(3)$:
$$9-2x^2-(2x-3)^2=0 \Rightarrow x_1=0,x_2=2.$$
Can you continue?
|
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$a+b+c=1.$ Show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Assume $a,b,c>0$ and $a+b+c=1.$ Show that
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq
\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Here's what I tried:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+b}+\frac{a+b+c}{a+b} \geq
\frac{1+a+b+c}{1+a}+\frac{1+a+b+c}{1+b}+\frac{1+a+b+c}{1+c}$
$\rightarrow \frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+b} \geq
\frac{b+c}{1+a}+\frac{a+c}{1+b}+\frac{a+b}{1+c}$
Let $A=b+c, B=a+c,$ and $ C=a+b,$ so $A+B+C=2.$
$\rightarrow \frac{1-A}{A}+\frac{1-B}{B}+\frac{1-C}{C} \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
Or $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} - 3 \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
I am stuck here, thought about using AM-GM-HM to get rid of the reciprocal on the RHS but it doesn't work if applied directly.
|
Hint
Note $t\mapsto \frac1t$ is convex and $(1-a,1-b,1-c)\succ (\frac12+\frac a2, \frac12+\frac b2, \frac12+\frac c2)$, so it’s Karamata’s Inequality.
|
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Find rational numbers $a,b,c$ satisfying $(2^{1/3}-1)^{1/3} = a^{1/3}+b^{1/3}+c^{1/3}$.
$\textbf{Problem} $ Find rational numbers $a,b,c$ satisfying
\begin{align*}
(2^\frac13-1)^\frac13 = a^\frac13+b^\frac13+c^\frac13
\end{align*}
My Attempt: I try to $2^\frac13-1 = (a^\frac13+b^\frac13+c^\frac13)^3$ and compare with rational numbers and irrational numbers in LHS and RHS.
Any help is appreciated...
Thank you!
Update: I found the answer. I want to find rational numbers $a,b,c$ without assumption $a=1/9,b=-2/9,c=4/9$. Thus, I found the identity:
$$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$
1) How to get the identity?
2) I want to know about uniqueness $(a,b,c)$ satisfying $(2^\frac13-1)^\frac13=a^\frac13+b^\frac13+c^\frac13$
|
Set $t:=\sqrt[3]{2}$. Then, $t^3=2$ and $t^3-1=1$, whence
$$t-1=\frac{t^3-1}{t^2+t+1}=\frac{1}{t^2+t+1}=\frac{3}{3t^2+3t+3}=\frac{3}{t^3+3t^2+3t+1}=\frac{3}{(t+1)^3}\,.$$
Therefore,
$$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}}{t+1}\,.$$
Now, $t^3+1=3$, so
$$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}\left(t^2-t+1\right)}{t^3+1}=\frac{\sqrt[3]{3}\left(t^2-t+1\right)}{3}=\frac{t^2-t+1}{\sqrt[3]{9}}\,.$$
Plugging in $t=\sqrt[3]{2}$, we obtain
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{4}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac19}\,.$$
In fact, this is the only possible way to write $\sqrt[3]{\sqrt[3]{2}-1}=\frac{1}{3}\sqrt[3]{12}-\frac{1}{3}\sqrt[3]{6}+\frac{1}{3}\sqrt[3]{3}$ as a sum $$\sum_{j=1}^n\,x_j\,\sqrt[r_j]{d_j}\,,$$ where $x_1,x_2,\ldots,x_n\in\mathbb{Q}\setminus\{0\}$, $r_1,r_2,\ldots,r_n\in\mathbb{Z}_{>0}$, and for $j=1,2,\ldots,n$, each $d_j\neq0$ is an $r_j$-power-free integer (with $d_j=1$ iff $r_j=1$, and with $d_j>1$ if $r_j>1$ is odd). This is because radicals are linearly independent over $\mathbb{Q}$. See a proof here. (As a consequence, $a$, $b$, and $c$ are unique, up to permutation.)
|
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|
Asymptotes of an implicit curve Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of
$$
x^3 + 3x^2y - 4y^3 - x + y + 3 = 0
$$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):-
$$ 2y + x = ±1 $$ can't be deduced.
Instead another erroneous line $ y = 0 $ is outputted.
Some help will be appreciated.Alternate methods of solution would work too:)
|
I'd try to give another answer. Recall from wikipedia and the most common definition of asymptotic line, to find an asymptotic line of $f(x)$ as $x\to\infty$, we need to compute
(1) $a=\lim_{x\to\infty}\frac{f(x)}{x}$
(2) $b=\lim_{x\to\infty}(f(x) - ax)$.
Then the asymptotic line would be $y=ax+b$.
-
By implicit function theorem when $x$ is large, $y$ can be written as a function of $x$. Let's begin with
$x^3 + 3x^2 y - 4y^3 -x + y + 3 = 0. \qquad(1) $
Then
$1+ 3\frac{y}{x} - 4(\frac{y}{x})^3 - \frac{1}{x^2} + \frac{y}{x}\frac{1}{x^2} + \frac{3}{x^3}=0$
if $\lim_{x\to\infty} \frac{y}{x}=:a$ exists, then it might be a solution of $1+3a-4a^3=0$. Moreover, the little consequence $\lim_{x\to\infty} \frac{y-ax}{x}=0$ is required.
-
Then we may try $x^3 + 3x^2(y-ax+ax) -4(y-ax+ax)^3 - x + (y-ax+ax) + 3 = 0$. After simplifying, we obtain
$3(y-ax) - 4(y-ax) (\frac{y-ax}{x})^2 - 12 a(\frac{y-ax}{x})^2 - 12( y-ax)a^2 - \frac{1}{x} + \frac{y-ax}{x}\frac{1}{x} + \frac{a}{x} + \frac{3}{x^2}=0$.
Thus
$(y-ax)[3 -4(\frac{y-ax}{x})^2 -12a^2] = 12a(\frac{y-ax}{x})^2 + \frac{1}{x} -\frac{y-ax}{x}\frac{1}{x} - \frac{a}{x} -\frac{3}{x^2}$. Taking limit, we obtain $b(3-12a^2)=0$, where $b=\lim_{x\to\infty} y-ax$, existing.
|
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|
Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
|
You can do the following:
$$\frac{a^2+b^2+c^2}{b(a+c)}\geq \frac{\frac{1}{2}(a+c)^2+b^2}{b(a+c)}=\frac{1}{2}\frac{a+c}{b}+\frac{b}{a+c}$$
where at the first step we used $2(a^2+c^2)\geq (a+c)^2$
If you now set $\frac{a+c}{b}=x$ then you just have to minimize $\frac{x}{2}+\frac{1}{x}$ over the positive real numbers.
But you have that by AM-GM $\frac{x}{2}+\frac{1}{x}\geq \sqrt{2}$ with equality if $x=\sqrt{2}$, i.e. $a+c=\sqrt{2}b$
Going back even further we want $a=c$ from the first step
|
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|
Inverse of a circulant matrix with a specific pattern I'm trying to invert the following circulant matrix:
$$\begin{bmatrix}1 & -1/4 & 0&0 &0&\cdots&0&-1/4\\ -1/4 & 1 & -1/4 & 0&0&\cdots&0&0\\0 & -1/4 & 1 & -1/4&0&\cdots&0&0\\\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots
\\-1/4 & 0 & 0 & 0 & 0&\cdots&-1/4 & 1 \end{bmatrix}$$
As it turns out, Fuyong (2011) "The inverse of circulant matrix" proposes the following method:
1-) Find the roots of the polynomials $g(z)=g^{\prime}(z)=1 - \frac{1}{4}z- \frac{1}{4}z^{-1}$ that are inside the unit circle:
$z_1 = z_1^{\prime}= 2-\sqrt{3}$ and $z_2 = z_2^{\prime}= 2+\sqrt{3}$
Only $z_1$ and $z_1^{\prime}$ are inside the unit circle.
2-) Compute $g_1(z_1)=-\frac{1}{4}z_1^{-1}(z_1-z_2)$ and $g_1^{\prime}(z^{\prime}_1)=-\frac{1}{4}z_1^{\prime-1}(z_1^{\prime}-z_2^{\prime})$:
$g_1(z_1)=g_1^{\prime}(z^{\prime}_1)= \frac{\sqrt{3}}{2(2-\sqrt{3})}$
3-) The elements of the inverse are given by:
$b_k= \dfrac{z_1^{k_1}}{g_1(z_1)(1-z_1^n)} +
\dfrac{z_1^{\prime k_2}}{z_1^{\prime s}g_1^{\prime}(z^{\prime}_1)(1-z_1^{\prime n})} $ , $k=0,\ldots,n-1$
where:
$k_1 = \mathrm{mod}\,e (k-1,n) $
$k_2 = \mathrm{mod}\,e (n-k-1+s,n) $
I don't know how $s$ is determined and what the operator $\mathrm{mod}\,e$ does. Can anyone help?
|
The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + \cdots + a_s x^s$. Thus, $s=-1$.
Let $B=A^{-1}$:
\begin{align*}
&B= \begin{bmatrix}b_0 & b_1 & b_2&b_3 &b_4&\cdots &b_{n-2}&b_{n-1}\\ b_{n-1} & b_0 & b_1 & b_2&b_3&\cdots & b_{n-3}&b_{n-2}\\b_{n-2} & b_{n-1} & b_0 & b_1&b_2&\cdots&b_{n-4}&b_{n-3}\\\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots
\\b_1 & b_2 & b_3 & b_4 & b_5 &\cdots&b_{n-1} & b_0 \end{bmatrix}&
\end{align*}
\begin{align*}
b_k = \dfrac{2\bigl(2-\sqrt{3}\bigr)^{k}}{\sqrt{3}\bigl[1-\bigl(2-\sqrt{3}\bigr)^n\bigr]} + \dfrac{2\bigl(2-\sqrt{3}\bigr)^{n-k}}{\sqrt{3}\bigl[1-\bigl(2-\sqrt{3}\bigr)^n\bigr]}\qquad k = 0,\ldots, n-1
\end{align*}
|
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|
Find the area of the figure that is bound by the lines: $y=|x^2-1|$ and $y=3|x|-3$?
Find the area of the figure that is bound by the lines:
$y=|x^2-1|$ and $y=3|x|-3$
I tried: $$A=\int_{-1}^{1}(|x^2-1|-3|x|+3)dx$$
Is it correct? Please, help for the right solution.
I dont know, how can I evaluate $\int|x|dx.$
|
You should draw the graphs of the two given curves carefully before solving this type of problems, and to do that you have two find the points at which they intersect each other. Let's do that
First write the equation of two given curves explicitly-
$
y=|x^2-1|=
\begin{cases}
x^2-1, & \text{if $x\in(-\infty,-1]\cup[1,\infty)$}\\
1-x^2, & \text{if $x\in(-1,1)$}
\end{cases}
$
$
y=3|x|-3=
\begin{cases}
3x-3, & \text{if $x\ge0$}\\
-3x-3, & \text{if $x<0$}
\end{cases}
$
Now, find the points at which the cut each other.
Case 1: $x\ge1$
The equation of two curves become
$y=x^2-1$ and $y=3x-3$
A simple calculation shows that these two curves intersects at the points $(1,0)$ and $(2,3)$ in $[1,\infty)$.
Case 2: $x\in[0,1)$
The equation of two curves become
$y=1-x^2$ and $y=3x-3$
These two curves don't intersect each other in $[0,1)$
Case 3: $x\in(-1,0)$
The equation of two curves become
$y=1-x^2$ and $y=-3x-3$
These two curves don't intersect each other in $(-1,0)$
Case 4: $x\le-1$
The equation of two curves become
$y=x^2-1$ and $y=-3x-3$
A simple calculation shows that these two curves intersects at the points $(-1,0)$ and $(-2,3)$ in $(-\infty,-1]$.
So, combining all the cases we get, two curves intersect each other at the points $A(-2,3)$, $B(-1,0)$, $C(1,0)$ and $D(2,3)$.
Note one thing the curves $y=|x^2-1|$, $y=3|x|-3$ intersect $Y$-axis at $E(0,1)$, $F(0,-3)$ respectively.
So, the area of the region bounded by these two curves on the plane=
Area of the region $(ABA+BCEB+\Delta BFC+CDC)=$
$
=\int_{-2}^{-1} \lbrace(-3x-3)-(x^2-1)\rbrace dx\quad+\int_{-1}^{1} (1-x^2) dx\quad+\frac{1}{2} \times 2 \times 3\quad+\int_{1}^{2} \lbrace(3x-3)-(x^2-1)\rbrace dx
$
$
=2\int_{1}^{2} \lbrace(3x-3)-(x^2-1)\rbrace dx\quad+2\int_{0}^{1} (1-x^2) dx\quad+3
$
$
=\frac{1}{3}+\frac{4}{3}+3=\frac{14}{3}
$
So, the area of the region bounded by the curves is $\frac{14}{3}$ units.
|
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|
Finding prime factors of large expression (without calculator) Show that: $5^3(5^3(253)+3)+1 = 19 \times 251 \times 829$.
I tried setting $n=5$, so that $253 = 23 \times 11 = (4n+3)(2n+1)$ and going from there, but the resulting polynomial in $n$ was $8n^6 + 10n^7 + 3n^6 + 3n^3 + 1$, which turns out to be irreducible over the integers, so this doesn't help at all.
|
My eye is drawn to the $253$ in the expression being close to the factor $251$ and the fact that $5^3$ is very close to half of $251$. Given the known result I would write
$$5^3(5^3(253)+3)+1=5^3(5^3\cdot (251+2)+3)+1\\=5^6\cdot 251+5^3\cdot253+1\\=(5^6+5^3+1)251$$
then just compute $5^6+5^3+1=15625+125+1=15751$ and divide by $19$
|
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|
Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$
Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$
Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.
My own solution
$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}~\frac{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}$$
$$\lim_{n\to\infty} \frac{5n^2+4~-~(5n^2+n)}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}~=~\lim_{n\to\infty} \frac{4-n}{n\left(\sqrt{5+\frac4{n^2}}~+~\sqrt{5+\frac1{n}}\right)}$$
$$=~\frac{-1}{2\sqrt{5}}~=~-\frac{\sqrt{5}}{10}$$
|
Considering that $\frac{4}{n^2}<< \frac{1}{n}$ for big $n$ we have
$$
\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n} \equiv \lim_{h\to 0}\frac{\sqrt{5+4h^2}-\sqrt{5+h}}{h} = \lim_{h\to 0}\frac{\sqrt{5+o(h^2)}-\sqrt{5+h}}{h}
$$
now remember the derivative definition
$$
\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}{h} = \frac 12\frac{\sqrt x}{x}
$$
so the result is
$$
-\frac 12\frac{\sqrt 5}{5}
$$
|
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|
Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1,\;a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$ Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1$
$$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$
then what will be it's $\limsup (a_n)$ and $\liminf (a_n)$ and $\sup(a_n)$, $\inf(a_n)$.
I tried to determine the nature of $a_{2n+1}$ and $a_{2n}$. I got $a_{2n+1}$ is increasing and $a_{2n+1} \geq \text{something}$ where $a_{2n}$ is decreasing and $a_{2n}\leq \text{something}$. So I could not draw any conclusion.
Can anyone please help me?
|
I think the following will help the readers to understand the solution:
We know $$\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt{|a_n| \times \frac {2}{|a_n|}}=\sqrt 2$$
$$\implies \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$
Given that $~a_1 = 1~$ and for $n\geq 1$
$$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$
When $~n~$ is even, $$a_{n+1} = \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$
Again when $~n~$ is odd, $$a_{n+1} = -\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\le-\sqrt 2$$
Which gives $~\lim\sup a_n=\sqrt 2~$ and $~\lim\inf a_n=-\sqrt 2~$.
|
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|
How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
|
Set $\;\;t^2=25-x^2,\;$ the equation becomes $$\begin{align}\sqrt{24+t^2}-\sqrt{t^2}&=3\\ \sqrt{24+t^2}&=|t|+3\\24+t^2&=t^2+6|t|+9\\|t|&=\frac{5}{2}
\end{align}$$
Then $x^2=25-\frac{25}{4}=3\cdot \frac{25}{4},$ from where $x=\pm \frac{5}{2}\sqrt 3.$
|
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|
Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17
|
Solution
Notice that $$a+b=2-c\tag1,$$and $$a^2+b^2=12-c^2.\tag2$$
Since $$ 2ab \leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab \leq 2(a^2+b^2).\tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 \leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) \leq 0.$$
As a result, $$-2\leq c \leq \frac{10}{3}.$$
|
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|
Finding 4-up numbers From Mathcounts Target 2018:
*A 4-up number is defined as a positive integer that is divisible by neither 2 nor 3 and does not have 2 or 3 as any of its digits. How many numbers from 400 - 600 inclusive are 4-up numbers?
So I tried to find the complement, which is a number that is either divisible by 2 or 3 or has 2 or 3 as a digit. This I could calculute using PIE. But then I ran into some nasty casework for the intersection of the two sets: when the number is divisble by 2 or 3 and has 2 or 3 as a digit.
Any help would be great. Thanks.
|
HINT.-You have to find the numbers of three digits $abc$ with the restrictions$$\begin{cases}a=4,5\\b=0,1,4,5,6,7,8,9\\c=1,5,7,9\end{cases}$$ Then you have in a first step $2\cdot8\cdot4=64$ possibilities from which as second and final step you have to discard those such that $a+b+c\equiv 0\pmod3$. For this you can calculate the number $n$ of solutions of the eight congruences and the answer will be $64-n$
$$1+b+1\equiv 0\pmod3\\1+b+5\equiv 0\pmod3\\1+b+7\equiv 0\pmod3\\1+b+9\equiv 0\pmod3\\2+b+1\equiv 0\pmod3\\2+b+5\equiv 0\pmod3\\2+b+7\equiv 0\pmod3\\2+b+9\equiv 0\pmod3$$
(The smaller of these last numbers is $405$ ans the largest of them is $597$).
|
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|
Computing the determinant of a $4\times4$ matrix
Compute the determinant of \begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}
My try:
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}_{R_2\rightarrow2R_2-R_1\\R_3\rightarrow R_3-R_1\\R_4\rightarrow2R_4+R_1}$$
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 0 & 5 & -1 & -1 \\ 0 & 2 & 0 & -4 \\ 0 & 1 & 5 & -1 \end{vmatrix}$$
Now I took $$\begin{vmatrix} 5 & -1 & -1 \\ 2 & 0 & -4 \\ 1 & 5 & -1 \end{vmatrix}=92$$
But the answer is $46$. Where did I go wrong?
|
You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.
|
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|
Ratio of angles in a triangle came across this question in a math contest and can't quite figure out an approach to the question.
Triangle $ABC$ has $AB=AC\neq BC$ and $∠BAC ≤ 90º$. $P$ lies on $AC$, and $Q$ lies on $AB$ such that $AP = PQ = QB = B$C. Find the ratio of $∠ACB$ to $∠APQ$.
I have tried to establish some sort of congruency between the inner triangles but to no success. So far, all I have is the obvious result that:
$180º=4 \times ∠ACB - ∠APQ$
Any tips or hints are greatly appreciated.
|
Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $\Delta APQ$ gives
$$
AQ = AP\cdot\frac{\sin(180^\circ-2x)}{\sin x}
=AP\cdot\frac {\sin(2x)}{\sin x}
=AP\cdot 2\cos x
=2\cos x
\ .
$$
The sides of $\Delta ABC$ are thus $BC=1$ and $1+2\cos x$, and the last again. The sine theorem gives again
$$
\frac {\sin x}1
=\frac{\sin(90^\circ-x/2)}{1+2\cos x}\ .
$$
This gives an equation in $x$ to be solved.
We rewrite it equivalently step for step:
$$
\begin{aligned}
\sin x &=\frac{\cos(x/2)}{1+2\cos x}\ ,\\
\sin x(1+2\cos x) &= \cos\frac x2\ ,\\
\sin x + \sin 2x &=\cos\frac x2\ ,\\
2\sin \frac{3x}2\cos \frac x2 &= \cos\frac x2\ ,\\
2\sin \frac{3x}2 &= 1\ ,\\
\sin \frac{3x}2 &= \frac 12\ ,\\
\frac{3x}2 &= 30^\circ\ ,\\
x &= 20^\circ\ .
\end{aligned}
$$
Now we have the angles in the triangle(s).
$\widehat{APQ}$ has measure $180^\circ-2x=140^\circ$.
The angles at the base are both $80^\circ$.
The problem asks now for a ration that makes no sense (geometrically). It is even misleading (in a contest, because many people would first suppose it is a beautiful ration, thus loosing 5 minutes or more).
Note: I will try to give also a synthetic proof.
|
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|
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
|
Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 \Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$\begin{cases}x_1^4 + ax_1^2 + bx_1 + c = 0\\ x_2^4 + ax_2^2 + bx_2 + c = 0\end{cases} \Rightarrow \\
(x_1-x_2)(x_1+x_2)(x_1^2+x_2^2)+a(x_1-x_2)(x_1+x_2)+b(x_1-x_2)=0 \Rightarrow \\
\require{cancel}\cancel{(x_1-x_2)}[(-3)(11)+a(-3)+b]=0 \Rightarrow \\ 3a-b=-33 \ \ \ \ (1)$$
and adding:
$$(x_1^2+x_2^2)^2-2(x_1x_2)^2+a(x_1^2+x_2^2)+b(x_1+x_2)+2c=0 \Rightarrow \\
11^2-2+11a-3b+2c=0 \Rightarrow \\
119+2a+3(\underbrace{3a-b}_{=-33})+2c=0 \Rightarrow \\
a+c=-10 \ \ \ \ (2)$$
Multiply $(2)$ by $4$ and subtract $(1)$ to get:
$$a+b+4c=-7 \Rightarrow a+b+4c+100=93.$$
|
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|
$Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$ $Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$
We know:
$E(Z) = 0$
$\operatorname{Var}(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$
Thus, we need:
$E(Z^2) = \int_{-\infty}^{\infty} z^2 \frac{1}{\sqrt{2\pi}}e^{-1/2(z^2)}dz$
Using int by parts, $u = z, du = dz$, then $dv = z \frac{1}{\sqrt{2\pi}}e^{-1/2z^2}dz$, $v = -\frac{1}{\sqrt{2\pi}}e^{-1/2z^2}$
Then we have
$$\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-1/2z^2}dz$$
I'm aware that $\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} = (0-0) = 0$
No clue how to proceed.
|
The standard method for evaluating $\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that
$$\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz=2\int_{0}^{\infty}e^{-\frac{1}{2}z^2}dz$$
as $e^{-\frac{1}{2}z^2}$ is an even function.
Let $I:=\int_{0}^{\infty}e^{-\frac{1}{2}z^2}dz$. Also note that "$z$" is just a dummy variable, so $I=\int_{0}^{\infty}e^{-\frac{1}{2}x^2}dx=\int_{0}^{\infty}e^{-\frac{1}{2}y^2}dy$.
$$
\begin{align}
I^2 &= \int_{0}^{\infty}e^{-\frac{1}{2}x^2}dx \int_{0}^{\infty}e^{-\frac{1}{2}y^2}dy
\\ &= \int_{0}^{\infty}\int_{0}^{\infty}e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}y^2}dxdy
\\ &= \int_{0}^{\infty}\int_{0}^{\infty}e^{-\frac{1}{2}(x^2+y^2)}dxdy
\\ &= \int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-\frac{1}{2}r^2}rdrd\theta
\\ &= \frac{\pi}{2}\int_{0}^{\infty}re^{-\frac{1}{2}r^2}dr
\\ &= \frac{\pi}{2} \left[ -e^{-\frac{1}{2}r^2} \right]_{0}^{\infty}
\\ &= \frac{\pi}{2}
\end{align}
$$
... and hence
$$ I = \frac{\sqrt{2\pi}}{2} $$
which, together with your working, gives:
$$
\begin{align}
\mathrm{Var}(Z) &= E(Z^2)
\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz
\\ &= \frac{2}{\sqrt{2\pi}} I
\\ &= 1
\end{align}
$$
as required!
|
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|
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
|
$$ \left( x^{6} - 7 \right) $$
$$ \left( x^{2} + 1 \right) $$
$$ \left( x^{6} - 7 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x^{4} - x^{2} + 1 \right) } + \left( -8 \right) $$
$$ \left( x^{2} + 1 \right) = \left( -8 \right) \cdot \color{magenta}{ \left( \frac{ - x^{2} - 1 }{ 8 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x^{4} - x^{2} + 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{4} - x^{2} + 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ - x^{2} - 1 }{ 8 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{6} + 7 }{ 8 } \right) }{ \left( \frac{ - x^{2} - 1 }{ 8 } \right) } $$
$$ \left( x^{6} - 7 \right) \left( \frac{ 1}{8 } \right) - \left( x^{2} + 1 \right) \left( \frac{ x^{4} - x^{2} + 1 }{ 8 } \right) = \left( -1 \right) $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the dimension of subspace $S=\{ \left[\begin{smallmatrix} a & b\\ c & d\\ \end{smallmatrix}\right] \mid c = a + b, d=a \}.$
Let the subspace $S$ from $M_{2\times2}$: $$S=\left\{ \begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix} \mid c = a + b\textrm{ and }d=a \right\}.$$ What is the dimension of $S$?
According to the answer key, $\dim (S)=2$. Why is it not $\dim(S)=2\times2=4$? What am I missing?
|
The subspace $S$ consist of matrices of the form
$
\begin{pmatrix}
a & b \\
a+b& a \\
\end{pmatrix}
$. Every such matrix can be written as a linear combination
$$
\begin{pmatrix}
a & b \\
a+b& a \\
\end{pmatrix}=
a\begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix}+
b\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}.
$$
If you also check that the matrices
$\begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix}$, $\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}$ are linearly independent, you get that this is a basis for $S$.
|
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|
Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2+b^2)}$$
After partial fraction decomposition and simplifying I get:
$u = \dfrac{\pi}{2(a+b)}$
But answer is $\frac \pi 6$.
Where have I gone wrong?
|
$$\dfrac{7 - \sqrt {45}}{2}=\left(\dfrac{3 - \sqrt {5}}{2}\right)^2$$ and ...
|
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|
How to decompose $X_p = X_r +X_n$? From an original matrix $A$, I have solved for $X_p$, $C(A^T)$, and $N(A)$. The last step is the problem below.
How should I approach this? Thank you for taking the time.
Decompose $X_p = X_r + X_n$, where $X_r \in C(A^T)$ and $X_n \in N(A)$.
$$
X_p = \begin{pmatrix}
3 \\
4 \\
-2 \\
0 \\
\end{pmatrix}, \quad
C(A^T) = \text{span}\left[\begin{pmatrix}16\\2\\3\\13\end{pmatrix},\begin{pmatrix}5\\11\\10\\8\end{pmatrix},\begin{pmatrix}9\\7\\6\\12\end{pmatrix}\right], \quad
N(A) = \begin{bmatrix}-1\\-3\\3\\1\end{bmatrix}$$
|
Since I am breaking my Xp into orthogonal projections, I would just need to solve for my scalars at this point (X below).AX=B solving for X. So my final solution would be my C(AT) and N(A) multiplied by my scalar (x).
A = \begin{pmatrix}
16,5,9,-1 \\
2,11,7,-3 \\
3,10,6,3 \\
13,8,12,1 \\
\end{pmatrix}
B = \begin{pmatrix}
3 \\
4 \\
-2 \\
0 \\
\end{pmatrix}
X = \begin{pmatrix}
159/85 \\
1909/1360 \\
-3853/1360 \\
-31/20 \\
\end{pmatrix}
|
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|
Intersection of 3 planes along a line I have three planes:
\begin{align*}
\pi_1: x+y+z&=2\\
\pi_2: x+ay+2z&=3\\
\pi_3: x+a^2y+4z&=3+a
\end{align*}
I want to determine a such that the three planes intersect along a line. I do this by setting up the system of equations:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
x+ay+2z&=3\\
x+a^2y+4z&=3+a
\end{align*}
\end{cases}
$$
and solve for x, y and z. I subtract the first row from the second and third
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(a^2-1)y+3z&=1+a.
\end{align*}
\end{cases}
$$
I subtract $(a+1)*\textrm{row}_1$ from the third row:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(2-a)z&=1+a.
\end{align*}
\end{cases}
$$
I think I am supposed to find an $a$ such that $0=0$ in the third row, but obviously there isn't any such $a$. Do I have the right idea about how to solve this? If so, where is my mistake?
|
You mean subtract $(a+1)$ times the second row from the third row.
After that the system become
$$x+y+z=2$$
$$(a-1)y+z=1$$
$$(2-a)z=0$$
If $a=2$, then we have $y+z=1$ and $x=1$ which is a line.
If $a \ne 2$, then $z=0$, hence we have $(a-1)y=1$ and $x+y=2$, to be consistent, clearly $a \ne 1$, and we can solve for $y$ and $x$ uniquely.
Summary, $a=2$ for the solution to be a line.
|
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|
Solve $ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $ with $a,b,c \in \mathbb{Z}$ I am trying to solve the Diophantine equation:
$$ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $$
Here's what it looks like if you expand, it's variant of the Pythagorean triples:
$$ a \times (a-1) + b \times (b-1) = c \times (c-1) $$
I was able to find solutions by computer search but this could have also been checked using the Hasse principle.
\begin{eqnarray*}
\binom{3}{2}+ \binom{3}{2}&=& \binom{4}{2} \\ \\
\binom{5}{2}+ \binom{10}{2}&=& \binom{11}{2} \\ \\
\binom{15}{2}+ \binom{15}{2}&=& \binom{21}{2}
\end{eqnarray*}
and many others. Is there a general formula for the $(a,b,c) \in \mathbb{Z}^3$ that satisfy this integer constraint.
>>> N = 25
>>> f = lambda a : a*(a-1)/2
>>> X = [(a,b,c,f(a) + f(b) - f(c)) for a in range(N) for b in range(N) for c in range(N)]
>>> [(x[0],x[1],x[2]) for x in X if x[3] == 0 and x[0] > 1 and x[1] > 1 and x[2] > 1]
[( 3, 3, 4), ( 4, 6, 7) , ( 5, 10, 11), ( 6, 4, 7), ( 6, 7, 9),
( 6, 15, 16), ( 7, 6, 9) , ( 7, 10, 12), ( 7, 21, 22), ( 9, 11, 14),
(10, 5, 11), (10, 7, 12), (10, 14, 17), (10, 22, 24), (11, 9, 14),
(12, 15, 19), (12, 21, 24), (13, 18, 22), (14, 10, 17), (15, 6, 16),
(15, 12, 19), (15, 15, 21), (15, 19, 24), (18, 13, 22), (19, 15, 24),
(21, 7, 22), (21, 12, 24), (22, 10, 24)]
|
$$ (2a-1)^2 + ( 2b-1)^2 = 1 + (2c-1)^2 $$
Taking any $c,$
if the right hand side is not twice a prime, there will be nontrivial $a,b,$ while factoring the right hand side gives a method for finding them other than raw search. So that is one approach.
Parameterizing all uses the modular group $SL_2 \mathbb Z$: take integers
$$ \alpha \delta - \beta \gamma = 1. $$ We also demand $\alpha \beta + \gamma \delta $ odd. This means that one of $\alpha, \beta, \gamma, \delta$ is even and the other three odd. Then
$$ 2a-1 = \alpha \delta + \beta \gamma $$
$$ 2b-1 = \alpha \beta - \gamma \delta $$
$$ 2c-1 = \alpha \beta + \gamma \delta $$
Yet another approach is the automorphism group of the quadratic form. However, see page 124 in Magnus, Noneuclidean Tesselations and Their Groups, where Lemma 3.7 shows that this group also depends on the modular group. Well, actually 3.33 and 3.34 on the next page, now that I look more carefully.
I ought to emphasize that $x^2 + y^2 - z^2 = 0$ possesses a parametrization in just two variables, while $x^2 + y^2 - z^2 = 1$ jumps immediately to four variables with a condition, namely the modular group. Worth adding $x^2 + y^2 + z^2 - w^2 = 0$ is naturally four variables, it is parameterized by the integral quaternions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$.
Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.
I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.
|
You're gonna need to convert $\sqrt {2-i}$ to the form $\alpha + i \beta$.
To do that, convert to polar
$$ 2-i=r(\cos \theta + i \sin \theta)$$
So that
$$\sqrt{2-i}=\sqrt r \cdot (\cos \theta + i \sin \theta)^{1/2} = \sqrt r \cdot \left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right) $$
Some of the ingredients are:
$$r=\sqrt{2^2+(-1)^2}=\sqrt 5, \quad \cos \theta= \frac{2}{\sqrt 5}, \quad \sin \theta =\frac{-1}{\sqrt 5} $$
$\theta$ will be a negative angle, so that
$$\cos{\frac{\theta}{2}}=\sqrt{\frac{1+\cos \theta}{2}} \qquad \sin{\frac{\theta}{2}}=-\sqrt{\frac{1-\cos \theta}{2}}$$
Continuing
$$\cos{\frac{\theta}{2}}=\sqrt{\frac{1+\frac{2}{\sqrt 5}}{2}}= \sqrt{\frac{\sqrt 5+2}{2\sqrt 5}}=\frac{\sqrt{2+\sqrt 5}}{\sqrt 2 \sqrt r}$$
$$\sin{\frac{\theta}{2}}=-\sqrt{\frac{1-\frac{2}{\sqrt 5}}{2}}= -\sqrt{\frac{\sqrt 5-2}{2\sqrt 5}}=-\frac{\sqrt{-2+\sqrt 5}}{\sqrt 2 \sqrt r}$$
Making
$$\sqrt{2-i}=\left(\frac{\sqrt{2+\sqrt 5}}{\sqrt 2}\right)-i \left(\frac{\sqrt{-2+\sqrt 5}}{\sqrt 2 }\right) \in \alpha + i \beta$$
|
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|
Partial fraction of $\frac{2s+12}{ (s^2 + 5s + 6)(s+1)} $ then inverse transform it
Find the inverse Laplace transform of
$$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$
I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working,
After factoring the denominator, I got a case of non repeating linear factors
$\dfrac{2s+12}{ (s+2)(s+3)(s+1)} = \dfrac{A}{s+2} + \dfrac{B}{s+3} + \dfrac{C}{s+1} $
$2s + 12 = A (s+3)(s+1) + B (s+2)(s+1) + C (s+2)(s+3) $
$2s + 12 = (As^2 + Bs^2 + Cs^2) + (4As + 3Bs + 5Cs) + (3A + 2B+6C) $
So...
$A + B + C = 0$
$ 4A + 3B + 5C = 2$
$3A + 2B + 6C = 12$
how do I solve this complicated equations ? This is where i got stuck and cannot continue.
|
$A + B + C = 0\quad[1]$
$ 4A + 3B + 5C = 2\quad[2]$
$3A + 2B + 6C = 12\quad[3]$
$ $
$[2]-3[1]$
$A+2C=2\quad[4]$
$ $
$[3]-2[1]$
$A+4C=12\quad[5]$
$ $
$[5]-[4]$
$2C=10$
$ $
$\therefore C=5$
$\therefore A=-8$
$\therefore B=3$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Is there a smarter way to differentiate the function $f(x) = \sin^{-1} \frac{2x}{1+x^2}$?
Given $f(x) = \sin^{-1} \frac{2x}{1+x^2}$,
Prove that $$f'(x) = \begin{cases}\phantom{-}\frac{2}{1+x^2},\,|x|<1 \\\\ -\frac{2}{1+x^2},\,|x|>1 \end{cases}$$
Obviously the standard approach would be to use the chain rule and simplify from there.
But I noticed that some of these expressions are familiar, specifically, from the tangent half-angle formulae:
If $x = \tan \frac \theta 2$, then $\sin \theta = \frac{2x}{1+x^2}$ and $\frac{d\theta}{dx} = \frac{2}{1+x^2}$.
So my question is: can this observation be used to construct a more elegant proof?
|
Method$\#1:$
By Chain rule,
$$\dfrac{d\arcsin\dfrac{2x}{1+x^2}}{dx}=\dfrac1{\sqrt{1-\left(\dfrac{2x}{1+x^2}\right)^2}}\left(\dfrac2{1+x^2}-\dfrac{4x^2}{(1+x^2)^2}\right)$$
$$=\dfrac{2(1+x^2)(1-x^2)}{|1-x^2|(1+x^2)^2}=?$$
Method$\#2:$
Let $\arctan x=y\implies x=\tan y$
Using principal values $$\arcsin\dfrac{2x}{1+x^2}=\begin{cases}2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2 \\
\pi-2\arctan x & \mbox{if } 2\arctan x>\dfrac\pi2\\-\pi- 2\arctan x & \mbox{if } 2\arctan x<-\dfrac\pi2\end{cases}$$
|
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|
Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 x$
Solving and re-arranging we get
$-64\sin^6 x+80\sin^4 x-20\sin^2 x+1=0$
Using the substitution $\sin^2 x=t$
$-64t^3+80t^2-20t+1=0$
I am not able to solve it from hence forth
|
Hint to solve the equation you got
If you make $t=\sin^{2}x$ then you get $-64t^3+80t^2-20t+1=0.$ It has no integer solutions. So we consider $z=1/t.$ Then we have
$$z^3-20z^2+80z-64=0.$$ It easy to see that $4$ is a root. So we have
$$z^3-20z^2+80z-64=(z-4)(z^2-16z+16).$$ Solve the quadratic equation and finally use that $\sin^2x=\dfrac1z.$
|
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|
Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x - \sqrt{x},$ then we are trying to solve $f(a)=f(b).$ Using some simple calculus I found that the turning point of $f(x)$ is $(\frac{1}{4}, -\frac{1}{4})$. Hence $0 \le b \le \frac{1}{4}$ and $\frac{1}{4} \le a \le 1$. From here, I have no idea how to proceed.
I used trial and error to find that when $a$ increases, the value of $a+b$ increases as well. Hence I hypothesise that $a+b$ is at a maximum when $a=1$ and $b=0$, which implies that $a+b=1$ is a maximum. Can anyone confirm this?
|
For $a\ne b,$
$$\sqrt a+\sqrt b=1$$
WLOG $a=\cos^4t,b=\sin^4t$
$$a+b=1-2\sin^2t\cos^2t=1-\dfrac{\sin^22t}2$$
Now $0\le\sin^22t\le1$
Or $a+b=1-\dfrac{1-\cos4t}4=\dfrac{3+\cos4t}4$
Now $-1\le\cos4t\le1$
|
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|
How can I factorize this: "$X^3 + X^2 + X - 3$" I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
\begin{align}X^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \\
&= (X - 1)^2 + X(X - 1) \\
&= (X-1)(X-1+X) \\
&= (X - 1)(2X - 1)
\end{align}
But for a few months I have not been able to find a teacher around here who can factorize this:
$$X^3 + X^2 + X - 3$$
Do we have to solve it in this way?
$$X^3 + X^2 + X - 3 = X^2(X + 1) + X - 3$$
Or something else?
I'd appreciate your help with this.
|
Surely $1$ is a root of $x^3+x^2+x-3$ therefore $x-1$ is a factor of it. We have $$x^3+x^2+x-3=(x-1)(x^2+x+1+x+1+1)=(x-1)(x^2+2x+3)$$
|
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|
Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$
Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$
First natural move would be rewriting the system as:
$$x^{2}+y^{2}+2xy=4z-1$$
$$x^{2}+z^{2}+2xz=4y-1$$
$$z^{2}+y^{2}+2zy=4x-1$$
Thus,
$$2x^{2}+2y^{2}+2z^{2}+2xy+2yz+2zx-4x-4y-4z+3=0$$
And I tried to write the polynomial above as the sum of complete squares. Is this possible or do you have other ideas?
|
Solution. $\blacktriangleleft$ Subtract every pair of the equation, we would obtain something like
\begin{align*}
y-z &= (x+y) - (x+z) \\ &= \sqrt {4z-1} - \sqrt {4y-1} \\&= \frac {4z-1 - 4y+1} {\sqrt {4z-1} + \sqrt {4y-1}} \\&= 4\frac {z-y} {\sqrt {4z-1} + \sqrt {4y-1}},
\end{align*}
then $y -z$ must be zero, otherwise a sum of two square roots would be negative, which is impossible. Thus we have $x = y =z$. Solve any one of them would obtain that
$$
x = y = z = \frac 12. \blacktriangleright
$$
|
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|
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\cos^2\left(\frac{\pi x}{2}\right)=0$$
I am stuck , I am confused now what to do now
|
Solving for $x^2$ we have
$$
x^2 = \frac 12\left(2\sin^2\left(\frac{x \pi}{2}\right)\pm\sqrt{4\sin^4\left(\frac{x \pi}{2}\right)-4}\right)
$$
now we know that $-1\le \sin \left(\frac{x \pi}{2}\right)\le 1$ so the only real possible solution is for $x = \pm 1$
|
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|
Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} $$
and therefore $\ \frac{z-1}{z+1}$ is imaginary
now let me assume $\ \frac{z-1}{z+1} $ is imaginary number, how could I conclude that $\ |z| =1 $ I really can't think of any direction..
Thanks
|
In the last step of your proof, before you use the assumption that $(a^2+b^2)=1$, you've deduced
$$\frac{z-1}{z+1}=\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}$$
Now, by assumption for the other direction, you have
$$\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}=xi$$
as $\frac{z-1}{z+1}$ is assumed to be imaginary, i.e. $\frac{z-1}{z+1}=xi$ for some $x$. Thus,
$$\frac{(a^2+b^2)-1}{(a^2+b^2)+1 +2a}+\frac{2bi}{(a^2+b^2)+1 +2a}=xi$$
and therefore
$$\frac{(a^2+b^2)-1}{(a^2+b^2)+1 +2a}=0$$
i.e. $(a^2+b^2)-1=0$, and therefore$\sqrt{a^2+b^2}=1$.
|
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|
(Should be) simple algebra exercise I have the following coupled equations
\begin{equation}
\tag{1}
a_{1}\left(T-T_{m}\right)+b_{1}M^{2}+cP^{2}=0
\end{equation}
\begin{equation}
\tag{2}
a_{2}\left(T-\theta_{2}\right)+b_{2}P^{2}+cM^{2}=0
\end{equation}
\begin{equation}
\tag{3}
\theta_{2}=\frac{ca_{1}}{a_{2}b_{1}}\left(T_{m}-T_{e}\right)+T_{e}
\end{equation}
I have to show that
$$
M^{2}=\frac{a_{1}}{b_{1}}\left(T_{m}-T_{e}\right)+\frac{a_{1}b_{2}-a_{2}c}{b_{1}b_{2}-c^{2}}\left(T_{e}-T\right)
$$
Attempt:
Doing $b_{2}\left(1\right)-c\left(2\right)$ I get to
$$
M^{2}\left(b_{1}b_{2}-c^{2}\right)=a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m}-T\right)+\frac{a_{1}c^{2}}{b_{1}}\left(T_{e}-T_{m}\right)
$$
How to finish?
|
$\displaystyle
M^{2}\left(b_{1}b_{2}-c^{2}\right)=a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m}-T\right)+\frac{a_{1}c^{2}}{b_{1}}\left(T_{e}-T_{m}\right)
$
$$
\begin{align}
M^{2}\left(b_{1}b_{2}-c^{2}\right) &= a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m} \color{red}{-T_e+T_e} -T\right)+\frac{a_{1}c^{2}}{b_{1}}\left(T_{e}-T_{m}\right) \\
&= \left(a_1b_2-\frac{a_1c^2}{b_1}\right)(T_m-T_e)+\left(-a_2c+a_1b_2\right)(T_e-T) \\
&= \frac{a_1}{b_1}(b_1b_2-c^2)(T_m-T_e) + +\left(a_1b_2-a_2c\right)(T_e-T)
\end{align}
$$
|
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|
Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number. I did this :-
$$
\sqrt{i} = \sqrt{\frac{1}{2}.2i} = \sqrt{\frac{1}{2}.(1 + 2i - 1)} = \sqrt{\frac{1}{2}.(1 + 2i + i^2)} = \sqrt{\frac{1}{2}.(1+i)^2} = \frac{1}{\sqrt{2}}(1+i) \\= \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ ---------
1} \\ \mbox{Also, }
\sqrt{-i} = \sqrt{-1.i} = \sqrt{-1}.\sqrt{i} = i\sqrt{i} = i(\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}}) \mbox{ --------from 1} \\ = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}} \\ = -\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ --------- 2} \\
\mbox{Adding 1 and 2, we get} \\ \sqrt{i} + \sqrt{-i} \\ = \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \\= i.\frac{2}{\sqrt{2}} \\= i.\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}} \\= i\sqrt{2}
$$
This is a complex number, not a real number.
What am I doing wrong here? Is there another way to prove this ?
|
Using the the polar representation for complex numbers, there holds that
$$ \sqrt{i}=e^{i\frac{\pi}{4}}~~~\mbox{and}~~~\sqrt{-i}=e^{-i\frac{\pi}{4}}$$
so that
$$ \sqrt{i}+\sqrt{-i}=e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}}=2\cos\left(\frac{\pi}{4}\right)=\sqrt{2}.$$
|
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|
find the range $(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})^2$ let $x_{i}\ge 0(i=1,2,3,\cdots,2009$,and such $$(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})=4$$
find the range $\sum_{i=1}^{2009}x^2_{i}$
I guess the range is $[1.5,2]$
because I think when $x_{1}=1,x_{2}=1,x_{3}=x_{4}=\cdots=x_{2009}=0$then is maximum of value $2$
and when $x_{1}=0.5,x_{2}=1,x_{3}=0.5,x_{4}=\cdots=x_{2009}=0$ is minmum of the value $1.5$
|
I think you are right!
We'll prove that
$$(x_1+x_2+...+x_{2009})^2\geq4(x_1x_2+x_2x_3+...+x_{2009}x_1)$$ for all non-negatives $x_i$.
Indeed, let $x_2=\max\limits_{i}\{x_i\}$.
Thus, $$x_2x_{2009}\geq x_1x_{2009}$$ and by AM-GM we obtain:
$$4(x_1x_2+x_2x_3+...+x_{2009}x_1)\leq4(x_1+x_3+...+x_{2009})(x_2+x_4+...+x_{2008})\leq\left(\sum\limits_{i=1}^{2009}x_i\right)^2,$$
where the equality occurs for
$$x_1(x_4+x_6+...+x_{2008})=0,$$
$$x_2(x_5+...+x_{2007})=0,$$ which gives
$$x_5=...=x_{2007}=0.$$
Now, if $x_1=0$ then $x_2x_3=1,$ which gives
$$2=\sum_{i=1}^{2009}x_i\geq x_2+x_{3}\geq2\sqrt{x_2x_{3}}=2,$$
which gives $x_2=x_{3}=1$, $x_1=x_4=...=x_{2009}=0$ and $\sum\limits_{i=1}^{2009}x_i^2=2$.
If $x_1>0$ we obtain:
$$x_1(x_4+...+x_{2008})=0,$$
which gives $x_4=x_5=...=x_{2008}=0.$
Now, let $x_2=a$, $x_1=b$,$x_3=c$ and $x_{2009}=d$.
Thus, $$(a+b+c+d)^2=4(ab+bd+ca)=4,$$
which gives $a+b+c+d=2$ and $ab+bd+ca=1.$
Thus, by AM-GM $$1=ab+bd+ca=ab+bd+dc+ca-dc=(a+d)(b+c)-dc\leq$$
$$\leq\left(\frac{a+b+c+d}{2}\right)^2-dc=1-dc\leq1,$$
which gives $dc=0$ and $a+d=b+c=1.$
Now, let $c=0$.
Thus, $b=1=a+d$ and since $a\geq b$, we obtain $d=0$ and $\sum\limits_{i=1}^{2009}x_i^2=2$ again.
Let $d=0$.
Id est, by C-S
$$\sum_{i=1}^{2009}x_i^2=a^2+b^2+c^2=1+\frac{1}{2}(1+1)(b^2+c^2)\geq1+\frac{1}{2}(b+c)^2=1.5.$$
The equality occurs for $a=1$ and $b=c=\frac{1}{2},$ which says that we got a minimal value.
Also, $$\sum_{i=1}^{2009}x_i^2=a^2+b^2+c^2=1+(b+c)^2-2bc\leq1+(b+c)^2=2.$$
The equality occurs for $a=b=1$ and $c=0,$ which says that we got a maximal value.
|
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|
Using $\epsilon$-$\delta$ approach, prove that $\lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3} $ How to prove that $$ \lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3}$$
By $\epsilon$-$\delta$ definition.
What i did is this:
Let $\epsilon$ be grater than zero; i need to find a $\delta(\epsilon)>0$ such that $|x/y-1/3|<\epsilon$ whenever $0<\sqrt{(x-1)^2+(y-3)^2}<\delta$.
Then, rewriting $$\bigg|\frac{x}{y}-\frac{1}{3}\bigg|=\bigg|\frac{3x-y}{3y}\bigg|= \bigg|\frac{3x-y+3-3}{3y}\bigg|=
\bigg|\frac{3(x-1)+(-y+3)}{3y}\bigg| $$
Now, by triangle inequality $$\bigg|\frac{3(x-1)+(-y+3)}{3y}\bigg|\leq \frac{3|x-1|+|y-3|}{3|y|}$$
since
$$|x-1|<\sqrt{(x-1)^2+(y-3)^2}<\delta \text{ and } |y-3|<\sqrt{(x-1)^2+(y-3)^2}<\delta$$
then
$$\frac{3|x-1|+|y-3|}{3|y|}<\frac{3\delta+\delta}{3|y|}=\frac{4\delta}{3|y|}$$
But i have that $|y-3|<\sqrt{(x-1)^2+(y-3)^2}<\delta$, then $y\in (3-\delta,\delta+3)$.
So i think the next step is to take $\delta =1/3$; if i do that i get $8/3<y$ that is $1/|y|<3/8$
then
$$\frac{4\delta}{3|y|}<\frac{3}{8}\frac{4\delta}{3}=\frac{\delta}{2}.$$
So, taking $\delta=\min\{1/3, 2\epsilon\}$, we have
$$\bigg|\frac{x}{y}-\frac{1}{3} \bigg|< \frac{3}{8}\frac{4\delta}{3} = \frac{\delta}{2}= \frac{1}{2}2\epsilon = \epsilon. $$
is this reasoning right?
|
More generally: assume $\delta < 3$. Since you are working on an open circle of centre $(1,3)$ and radius $\delta$, you can say that $\lvert y \rvert > 3-\delta$. This implies $$\frac{4\delta}{3\lvert y \rvert} < \frac{4\delta}{3(3-\delta)},$$
and then you can set $\delta := \frac{9\varepsilon}{4+3\varepsilon}$
to find that
$\left\lvert \frac{x}{y}-\frac{1}{3} \right\rvert < \varepsilon.$ Explicitly:
$$\frac{4\delta}{9-3\delta} = \frac{4\frac{9\varepsilon}{4+3\varepsilon}}{9-3\frac{9\varepsilon}{4+3\varepsilon}} = \frac{36\varepsilon}{36+27\varepsilon-27\varepsilon} = \varepsilon.$$
You can now check that
$$\delta = \frac{9\varepsilon}{4+3\varepsilon} < \frac{3(4+3\varepsilon)}{4+3\varepsilon} = 3$$
which shows that our choice of $\delta$ fits with the assumption $\delta < 3$.
On the other hand, if $\delta \geq 3$ you can choose points in the circle with $\lvert y\rvert$ small enough so that $\left\lvert \frac{x}{y}-\frac{1}{3} \right\rvert$ is arbitrarily large. To see this, fix $x=1$ and observe that
$$\left\lvert \frac{x}{y}-\frac{1}{3} \right\rvert = \left\lvert \frac{3-y}{3y} \right\rvert \to +\infty, \text{ when } y \to 0.$$
|
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|
how to factor $f(x)=x^4-7x^2+1$ How do you factor this equation? I have tried to use factoring method where the expanded b term factors add to B and multiply to get C.
|
It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+\sigma)(x^2-bx+\sigma)$ with $\sigma=\pm1$. We have
$$(x^2+bx+\sigma)(x^2-bx+\sigma)=x^4+(2\sigma-b^2)x^2+1$$
so we need $2\sigma-b^2=-7$, which rewrites as $b^2=7+2\sigma$. Letting $\sigma=1$ gives $b=\pm3$, so we have the factorization
$$x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$$
Note, the other option, $\sigma=-1$, gives another valid factorization, just not over the integers:
$$x^4-7x^2+1=(x^2+\sqrt5x-1)(x^2-\sqrt5x-1)$$
|
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|
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$
Here's how I tried it
$$x+y+z=xyz $$
So, by De Moivre's Theorem,
$$(\cos a + \cos b + \cos c) + i(\sin a + \sin b + \sin c) = \cos(a+b+c) + i \sin(a+b+c)
$$
Equating real and imaginary parts,
$$\cos a + \cos b + \cos c= \cos(a+b+c)$$
and similarly for sine. Now,
$$(a-b) + (b-c) + (c-a) =0$$
What to do now? Please help. And please use De Moivre Theorem!
|
Note that $$\cos(a-b) = \cos(a)\cos(b)+\sin(a)\sin(b) = \Re(x\bar{y})$$
So $$\label{eq1}\cos(a-b)+\cos(b-c)+\cos(c-a) = \Re(x\bar{y}+y\bar{z}+z\bar{x})$$
With that in mind, note that just like $x$, $y$ and $z$, the $xyz$ complex number is of modulus 1. So if we go back to the $x+y+z=xyz$ identity, we see that $$|x+y+z|=1$$
In other words, $$\begin{split}
1 & = |x+y+z|^2 \\
& = (x+y+z)(\bar x+\bar y +\bar z) \\
& = x\bar x + x \bar y + x \bar z + y\bar x + y \bar y + y \bar z + z \bar x + z \bar y + z \bar z \\
& = 1 + x \bar y + x \bar z + 1 + y\bar x + y \bar z + 1 + z \bar x + z \bar y \\
& = 3 + (x \bar y + y \bar z + z \bar x) + (y\bar x + z \bar y + x \bar z) \\
& = 3 + 2 \Re(x \bar y + y \bar z + z \bar x)
\end{split}$$
We conclude that $$\Re(x\bar{y}+y\bar{z}+z\bar{x})+1=0$$ which yields the desired result.
|
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|
Area between the curve $y$, the $x$-axis and the lines
Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.
I have drawn the graph and concluded that:
$$\int_0^1 0 - (x^2-4x+3)\, dx + \int_1^3 0 - (x^2-4x+3)\, dx = -\frac{-26}{3}$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?
|
$\int_0^1 (x^2-4x+3)\mathbb dx-\int_1^3(x^2-4x+3)\mathbb dx=[(\frac{x^3}3-2x^2+3x)]_0^1-[\frac{x^3}3-2x^2+3x]_1^3=\frac43-[0-\frac43]=\frac83$
|
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|
Query on $\int \frac{dx}{\cos x \cos (2x+a)}$ .
$$\int \frac{dx}{\cos x \cos (2x+a)}$$
MY APPROACH:
I tried using partialization of fraction as follows :
$$ -\frac{1}{\cos a} \int \frac{dx}{\cos x} + \frac{1}{\cos \frac{\pi-2a}{4}} \int \frac{dx}{\cos(2x+a)}$$
$$=-\frac{1}{\cos a} \cdot\ln|\sec x+\tan x| +\frac{1}{\cos \frac{\pi-2a}{4}} \cdot \frac{1}{2} \cdot \ln|\sec (2x+a)+\tan(2x+a)| + c $$
I am dubious since I've never faced partialization of trigonometric fractions. Is it plausible?
Any other methods?
|
Hint: Use that $$\cos(x)\cos(2x+a)=\frac{1}{2}\left(\cos(x+a)+\cos(3x+a)\right)$$
Then use that
$$\cos(x)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2tdt}{1+t^2}$$ the so-called Weierstrass substitution.
I will post you the solution for your work!
$$\sec (a) \left(-\sqrt{2} \sqrt{\sin (a)-1} \tan
^{-1}\left(\frac{\sqrt{\sin (a)-1} \sec
\left(\frac{x}{2}\right) \left(\sin
\left(\frac{a+x}{2}\right)+\cos
\left(\frac{a+x}{2}\right)\right)}{\sqrt{2}
\left(\cos \left(\frac{a}{2}\right)-\sin
\left(\frac{a}{2}\right)\right)}\right)-\sqrt{2}
\sqrt{\sin (a)+1} \tanh ^{-1}\left(\frac{\sqrt{\sin
(a)+1} \sec \left(\frac{x}{2}\right) \left(\cos
\left(\frac{a+x}{2}\right)-\sin
\left(\frac{a+x}{2}\right)\right)}{\sqrt{2}
\left(\sin \left(\frac{a}{2}\right)+\cos
\left(\frac{a}{2}\right)\right)}\right)+\log
\left(1-\tan \left(\frac{x}{2}\right)\right)-\log
\left(\tan \left(\frac{x}{2}\right)+1\right)\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$
Calculate
$$\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$$
My Attempt:
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{x}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\frac{-2\sin^2{\frac{x}{2}}}{-2\sin^2{\frac{x}{2}}}}\cdot\frac{\frac{-2\sin^2{\frac{x}{4}}}{-2\sin^2{\frac{x}{4}}}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{\frac{x}{2}}}+\ln{2}}$$
$$\frac{1}{4}\cdot 16\cdot 1$$
$$=4$$
Am I solving this correct?
|
You are correct. This is another way to proceed: if $x\to 0^+$ then by using Taylor expansions we obatin
$$f(x):=\log_{\sin(x)}{(\cos (x))}=\frac{\log(\cos(x))}{\log(\sin(x))}=\frac{\log(1-\frac{x^2}{2}+o(x^2))}{\log (x+o(x^2))}\sim-\frac{x^2}{2\log(x)},$$
and it follows that for any $a>0$,
$$\lim_{x\to{0^+}}\frac{\log_{\sin(x)}{\cos(x)}}{\log_{\sin(ax)}\cos(ax)}=\lim_{x\to{0^+}}\frac{f(x)}{f(ax)}=\lim_{x\to{0^+}}\frac{-\frac{x^2}{2\log(x)}}{-\frac{(ax)^2}{2\log(ax)}}=\lim_{x\to{0^+}}\frac{\log(x)+\log(a)}{a^2\log(x)}=\frac{1}{a^2}.$$
In your case $a=1/2$ and the limit is $4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem This is a question that I found in a textbook:
Given that $p=q+1$, $p$ and $q$ are integers, then show that
$p^{2n} - 2nq-1$ is divisible by $q^2$ given that $n$ is a positive integer.
By taking a suitable value of $n$, $p$ and $q$, show that $3^{16}-33$ and $3^{15}+5$ are divisible by 4.
My proof:
$$p^{2n}-2nq-1=(1+q)^{2n}-2nq-1$$
$$=[1+2nq+\frac{(2n)(2n-1)}{2!} q^2+\frac{2n(2n-1)(2n-2)}{3!}q^3+...]-2nq-1$$
$$=n(2n-1)q^2+\frac{2}{3} n(2n-1)(n-1)q^3+...$$
$$=q^2[n(2n-1)+\frac{2}{3} n(2n-1)(n-1)q+...]$$
Hence the expansion has a common factor $q^2$
Taking $n=8$ and $p=3$, and given that $p=q+1, q=2$, by substitution,
$$3^{16} -33=4[120+1120+...]$$
By factoring a 3:
$$3(3^{15}-11)=4[120+1120+...]$$
Dividing both sides by 3 and adding 15 to both sides:
$$3^{15} +5=4[\frac{1}{3}(120+1120+...)+4]$$
Then it is proven that it is also divisible by 4.
The only problem I have with the proof is that how do I know that each term in the brackets $(120+1120+...)$ are divisible by 3?
|
By the Euclid's lemma: $3(3^{15}-11)=4\cdot [120+1120+...]$ implies that $3$ must divide either $4$ or the sum. It does not divide $4$, hence the conclusion.
|
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|
How can I solve $\int_0^1\frac{\arctan(x^2)}{1+x^2}\,\mathrm dx$? This integral appears very similar to $\int\frac{\arctan x}{1+x^2}\,\mathrm dx$, but this question cannot be solved through the same simple substitution of $u=\arctan x$. WolframAlpha cannot find a symbolic solution to this problem, and this Quora answer
is the only thing I can find that appears to have the exact answer of $\frac14\log^2(1+\sqrt2)$. I am not sure if there is some special trick to be used in solving this, but I have tried everything I know and nothing seems to work.
|
$\begin{align}
\displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\
&=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\
\end{align}$
Since,
$\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dt$
then,
$\begin{align}
\displaystyle K&=\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\
\displaystyle &=\int_0^1\int_0^1 \dfrac{2x^2}{(1+t^2x^2)(1+x^4)}dtdx\\
\displaystyle &=\int_0^1\int_0^1 \left(\dfrac{2t^2}{(1+t^4)(1+x^4)}+\dfrac{2x^2}{(1+x^4)(1+t^4)}-\dfrac{2t^2}{(1+t^4)(1+t^2x^2}\right)dtdx\\
&=\displaystyle 4\left(\int_0^1 \dfrac{t^2}{1+t^4}dt\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)-K
\end{align}$
Therefore,
$\displaystyle K=2\left(\int_0^1 \dfrac{x^2}{1+x^4}dx\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)$
Since,
$\begin{align}\displaystyle \int_0^1 \dfrac{x^2}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\
&=\dfrac{1}{4\sqrt{2}}\Big(\pi+\ln\left(3-2\sqrt{2}\right)\Big)
\end{align}$
and,
$\begin{align}\displaystyle \int_0^1 \dfrac{1}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\
&=\dfrac{1}{4\sqrt{2}}\Big(\pi-\ln\left(3-2\sqrt{2}\right)\Big)
\end{align}$
Therefore,
$\boxed{K=\displaystyle \dfrac{\pi^2}{16}-\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$
Therefore,
$\boxed{\displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx=\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$
NB:
$\displaystyle \left(3-2\sqrt{2}\right)=\left(1-\sqrt{2}\right)^2=\frac{1}{\left(1+\sqrt{2}\right)^2}$
therefore,
$\displaystyle \ln^2\left(3-2\sqrt{2}\right)=4\ln^2\left(1+\sqrt{2}\right)$
|
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|
Show that $\iint_S (x^2+y^2) dA = 9 \pi /4$ In exam it was asked to show that
$$\iint_S x^2+y^2 dA = 9 \pi /4$$ for $$S = {\{(x, y, z) | x>0, y>0,3>z>0, z^2 = 3(x^2 + y^2)}\}$$
I have tried many times but I don't get the $9 \pi /4$.
$$\begin{align}
\iint_S\sqrt{1+f_x^2+f_y^2}\,dA &=\int_0^{\sqrt3}\int_0^{2\pi} r^2\sqrt{1+36r^2}\,r\,d\theta\,dr\\
&=2\pi\,\int_0^\sqrt3 r^3\sqrt{1+36r^2}\,dr = 2 \pi \dfrac{\left(36r^2+1\right)^\frac{3}{2}\left(54r^2-1\right)}{9720}\Bigg|_0^{\sqrt{3}} \\&= 2 \pi\dfrac{161{\cdot}109^\frac{3}{2}+1}{9720} \ne 9 \pi/4 \end{align}$$
Where I did it wrong?
|
I am going try to reproduce your answer (at least how I think that you got there), so we can see where it went wrong.
Let
$$ x=r\cos\theta, \quad y = r\sin\theta. $$
Then
$$ z^2 = 3(x^2+y^2) = 3r^2 \quad \Rightarrow \quad z = r\sqrt{3} $$
where the final implication follows since both $z$ and $r$ are non-negative.
Hence
$$ \Psi(r, \theta) = (r\cos \theta, r\sin\theta, r\sqrt{3}), $$
parameterizes our surface, but we must also find the correct intervals for $r$ and $\theta$. First, we have
$$ 0 < z < 3 \quad \Leftrightarrow \quad 0 < r\sqrt{3} < 3 \quad \Leftrightarrow \quad 0 < r < \sqrt{3}. $$
Second, we know that $x,y > 0,$ so that they lie in the first quadrant, and we can conclude that $0 < \theta < \frac{\pi}{2}$.
Now it is only left to compute the surface element $dA$ before we can attack the integral, which can be done via
$$ dA = \left|\frac{\partial \Psi}{\partial r} \times \frac{\partial \Psi}{\partial \theta}\right| \, dr\,d\theta. $$
The formula $\sqrt{1+f_x^2+f_y^2}$ that you seem to have used works only if you have a parameterization of the form $(x,y,f(x,y)),$ which is not the case in our case here.
We have
$$ \frac{\partial \Psi}{\partial r} = (\cos\theta, \sin\theta, \sqrt{3}), $$
and
$$ \frac{\partial \Psi}{\partial \theta} = (-r\sin\theta, r\cos\theta, 0), $$
so
\begin{align}
\frac{\partial \Psi}{\partial r} \times \frac{\partial \Psi}{\partial \theta}
&= \begin{vmatrix}
e_1 & e_2 & e_3\\
\cos\theta & \sin\theta & \sqrt{3}\\
-r\sin\theta & r\cos\theta & 0
\end{vmatrix}\\[0.2cm]
&= e_1(0-\sqrt{3}r\cos\theta) + e_2(-\sqrt{3}r\sin\theta-0) + e_3(r\cos^2\theta+r\sin^2\theta)\\[0.1cm]
&= (-\sqrt{3}r\cos\theta, -\sqrt{3}r\sin\theta, r),
\end{align}
and thus
\begin{align}
\left|\frac{\partial \Psi}{\partial r} \times \frac{\partial \Psi}{\partial \theta}\right| &= \sqrt{(-\sqrt{3}r\cos\theta)^2 + (-\sqrt{3}r\sin\theta)^2+r^2}\\
&=\sqrt{3r^2\cos^2\theta+3r^2\sin^2\theta+r^2} = r\sqrt{4} = 2r.
\end{align}
We have thus found your error - your surface element was wrong!
To complete the answer, we have
\begin{align}
\iint_S x^2+y^2 \, dA &=
\int_0^{\pi/2}\int_0^{\sqrt{3}} r^2 \cdot 2r \, dr\, d\theta
= \pi \int_0^{\sqrt{3}} r^3 \, dr
= \frac{\pi}{4} [r^4]_0^{\sqrt{3}} = \frac{9\pi}{4}
\end{align}
|
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|
Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\ \left({x \over x^2+x}+{x \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1$$
Can we do this without squeeze theorem?
|
Hint :
Take, $\displaystyle a_{x,n}=\frac{x^2}{x^2+n}$. Then use Cauchy's First limit theorem.
As $a_{x,n} \to 1$ when $x\to \infty$, so $\displaystyle \frac{1}{x}\sum_{n=1}^{x}a_{x,n} \to 1$ when $x\to \infty$.
|
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|
Express polynomial as a product of real quadratic polynomials with no real roots I want to express
$$P(x) = x^4 + x^2 + 1$$
as as a product of real quadratic polynomials with no real roots.
I know that: $$(x^2 - bx + a^2)(x^2 +bx + a^2)$$
$$ =x^2 + (2a^2 - b^2)x^2 + a^4 $$
I thought that $b^2 = 2a^2 - 1$ in this case and provided a = 1, b = $\pm 1$.
However in my textbook, the answer given is $b^2 = 2a^2 + 1 \therefore b = \sqrt{3}$.
I'm not sure where I've went wrong here.
|
Note that
$$x^4+x^2+1 = (x^2-x+1)(x^2+x+1)$$
and
$$x^4-x^2+1 = (x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)$$
Considering the answer given by your textbook, it seems that your textbook wanted you to factorize $x^4-x^2+1$ into quadratic factors. If you haven't misread the problem, it's worth notifying the authors of this error/typo in their book.
|
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|
$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x=\sqrt[3]{n+x}\Rightarrow x^3=n+x\Rightarrow x^3-x-n=0$
could some help me how to solve it, Thanks
|
Using the formula for finding the roots of the general cubic equation we obtain the numerical criterion:
$$
\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+}\ldots}}}\in \mathbb{N} \Longleftrightarrow
\sqrt[3]{\dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}+\sqrt[3]{\dfrac{n}{2}-\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}
\in \mathbb{N}
$$
Let the cubic equation $a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$. In this case we have
$
a=1\qquad b=0 \qquad c=-1 \quad \mbox{ and } \quad d=-n
$
and
$
1\cdot x^3-0\cdot x^2-1\cdot x-n=0. Let
$
$Q = \dfrac {3 a c - b^2} {9 a^2}=-\dfrac{1}{3}$
$R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}=\dfrac{n}{2}$
$S = \sqrt [3] {R + \sqrt{Q^3 + R^2}}=\sqrt[3]{\dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}$
$T = \sqrt [3] {R - \sqrt{Q^3 + R^2}}=\sqrt[3]{\dfrac{n}{2}-\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}$
If $D = Q^3 + R^2\geq 0 $ ( In this case $D=\dfrac{n^2}{4}-\dfrac{1}{27} >0$ ) then the cubic equation has solutions:
$x_1 = S + T - \dfrac b {3 a}= \sqrt[3]{\dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}+\sqrt[3]{\dfrac{n}{2}-\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}$
$x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \left({S - T}\right)$
$x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \left({S - T}\right)$
If $D = Q^3 + R^2<0 $ then the cubic equation has solutions:
$x_1 = 2 \sqrt {-Q} \cos \left({\dfrac \theta 3}\right) - \dfrac b {3 a}$
$x_2 = 2 \sqrt {-Q} \cos \left({\dfrac \theta 3 + \dfrac {2 \pi} 3}\right) - \dfrac b {3 a}$
$x_3 = 2 \sqrt {-Q} \cos \left({\dfrac \theta 3 + \dfrac {4 \pi} 3}\right) - \dfrac b {3 a}$
where $\cos(\theta)=\frac{R}{\sqrt{-Q^3}}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambda & 0 & 1 \\
3 & 1-\lambda & 0 \\
-2&1&4-\lambda\end{bmatrix}$$
$$|A-\lambda I| = -(\lambda)^3 + 5(\lambda)^2 - 6(\lambda) +5$$
Using Cayley-Hamilton theorem
$$A^3-5^2+6A-5=0$$
How do I use this find $A^5 - 27A^3 + 65A^2$?
|
I don't see a way to find the answer directly but you can certainly simplify your calculations; since $A$ commutes with itself we may factor the polynomial however we want. Write
$$
A^5-27A^3+65A^2=A^2(A^3-27A+65)=A^2\left[(A^3-5A^2+6A-5)+(5A^2-33A+70)\right].
$$
Now by Cayley Hamilton you only need to compute $A^2(5A^2-33A+70)$, which cuts down on the powers of $A$ that you need to calculate directly and reduces the problem to a bunch of addition and two matrix multiplications.
|
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|
Proof Verification of a Limit in Real Analysis Is this following proof correct?
The limit
\begin{align*}
\lim_{n\to\infty}\frac{t^{n+1}}{(n+1)!}
\end{align*}
tends to zero if for every $\epsilon>0$, there exists $N_\epsilon$ such that
\begin{align*}
n\geq N_\epsilon\implies \left|\frac{t^{n+1}}{(n+1)!}\right|<\epsilon
\end{align*}
For any $t\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $t<N$. Let $N$ be the least such natural number. Then
\begin{align*}
\frac{t^{N+k}}{(N+k)!}&=\frac{t\cdot t\cdots t\cdot t\cdots t\cdot t}{1\cdot 2\cdots (N-1)N\cdots(N+k-1)(N+k)}\\
&<\frac{t\cdot t\cdots t \cdot t\cdots t\cdot t}{1\cdot 1\cdots 1\cdot N\cdots N\cdot N}\\
&=t^N\left(\frac{t}{N}\right)^{k}
\end{align*}
Let $a=\frac{t}{N}$. Since $t<N$, $a=\frac{t}{N}<1$. If
\begin{align*}
\frac{t^{N+k}}{(N+k)!}<t^N\left(\frac{t}{N}\right)^{k}=t^Na^{k}<\epsilon
\end{align*}
then dividing both sides by $t^N$,
\begin{align*}
a^{k}<\frac{\epsilon}{t^N}
\end{align*}
Taking $\log_a$ of both sides reverses the inequality because it is a decreasing function. Hence
\begin{align*}
k>\log_{a}\left(\frac{\epsilon}{t^N}\right).
\end{align*}
Any value for $k$ greater than this expression will ensure the limit remains within the desired bounds. Therefore, take $N_\epsilon=N+k$. Since the choice of $t$ was arbitrary, we may conclude that the limit converges throughout the entirety of $\mathbb{R}$.
|
The proof looks good (actually very good) to me. The only improvement I can suggest is to turn it around a little, so that no cleverness is invoked.
Proof: Given $t>0$ and $\epsilon > 0$ arbitrary, put:
$$\begin{align*}
N &= \lfloor t \rfloor + 1 & (\text{$N$ is the smallest natural number strictly greater than 1})
\\
a &= \tfrac{t}{N} & (\text{so $a < 1$})
\\
k &= \left\lceil \log_{a}\left( \frac{\epsilon}{t^N} \right) \right\rceil
\\
N_{\epsilon} &= N+k
\end{align*}$$
Then if $n > N_{\epsilon}$ we have
$$\begin{align*}
\frac{t^{n+1}}{(n+1)!} &= \frac{t^{N+n-N+1}}{N!(N+1)(N+2)\dots(n+1)}
\\
&<
\frac{t^N}{1^N} \cdot \left( \frac{t}{N} \right)^{n+1-N}
\\
&=
t^N a^{n - N_{\epsilon} + N_{\epsilon} +1-N}
\\
&<
t^N a^{k+1} & (\text{via $N_{\epsilon} = N+k$, $n > N_{\epsilon}, a < 1$)}
\\
&<
t^N \frac{\epsilon}{t^N} & (\text{via construction of $k$})
\\
&=
\epsilon
\end{align*}$$
as required.
Remark: Your calculations did all the hard work of finding $N_{\epsilon}$. All this answer suggests is to summarize it into easier(?) steps from 'if $n > N_{\epsilon}$' to 'then $(\ldots) < \epsilon$'.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is where I get confused.
Base case is easy, n=1.
$$
\frac{1}{2^1}<2
$$
Induction case we assume that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} <2
$$
Then we get to fun old induction. How do I show that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} <2 ?
$$
|
You can try to do the following: write
$$ \frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} = $$
$$ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}} + $$
$$ \frac{1}{2}\left(\frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k}\right).$$
Use that the first summand is bounded by $1$, and, by induction hypothesis, the second one is strictly less than $\frac{1}{2} \cdot 2.$ This finishes the inductive step.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2) $
Let $x, y, z\in [a,b]$ and $u, v, w\in [a, b] $, where $0 <a <b $, s.t. $x^2+y^2+z^2=u^2+v^2+w^2$. Show that $$\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2) .$$
I tried to apply Cauchy- Schwartz and the fact that $ \frac {a^4+b^4}{ab (a^2+b^2)} \geq \frac {a}{b} $ but I didn't succeed.
|
The inequality is homogeneous, i.e. multiplying all variables with the same factor will not change it. This allows us to demand that the multiplied variables (for which we use the same variable letters as before) satisfy $x^2 + y^2 + z^2 = u^2 + v^2 + w^2 = 3$ and in turn $a \le 1$ and $\sqrt{3-a^2} \ge b \ge 1$. So we have to prove
$$
\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
By re-ordering, w.l.o.g., let $x \le y\le z$.
By the rearrangement inequality, the LHS gets as large as possible if we use, for $z^3$, the smallest possible $w$, which is $a$. Considering the choices of $x,y,z$ the highest value on the LHS will then be obtained for $z=b$. So we have to prove
$$
\frac {x^3}{u}+\frac {y^3}{v} + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
Again by rearrangement, the converse is true for the term $\frac {x^3}{u}$, so we need to prove
$$
\frac {a^3}{b}+\frac {y^3}{v} + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
Now the remaining values are determined: $y^2=v^2=3-a^2-b^2$. So finally we need to show
$$
\frac {a^3}{b}+(3-a^2-b^2) + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
which is
$$
(a^2+b^2)(a^4 + ab(3-a^2-b^2) + b^4) \leq 3 (a^4+b^4)
$$
Writing $3=a^2+b^2+v^2$ we have
$$
(a^2+b^2)(a^4 + ab v^2 + b^4) \leq (a^2+b^2+v^2) (a^4+b^4)\\
\rightarrow \quad 0 \leq v^2 (b^3-a^3)(b-a)
$$
which is true. This proves the claim.
Note that for more than three terms, a similar inequality holds. See here.
|
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|
Number of ways to pay (generating functions) I've just started learning generating functions.
*
*Let $a_n$ be the number of ways in which you can pay $n$ dollars using 1 and 2 dollar bills. Find the generating function for $(a_0, a_1, a_2, \ldots)$ and general term $a_n$.
*Prove that the number of ways to pay $n$ dollars using only 1, 2, and 3 dollar bills is equal to closest integer of $\frac{(n+3)^2}{12}$?
What I've got so far:
Let $f(x) = \frac{1}{1-x}$ be the generating function for $(1, 1, 1, ...)$. $1 + x^2 + x^4 + \ldots$ is the generating function for $(1, 0, 1, 0, 1, \ldots)$ which is equal to $f(x^2) = \frac{1}{1-x^2} = \frac{1}{(1-x)(1+x)}$.
$$
\begin{align}
& (1 + x + x^2 + \ldots)(1 + x^2 + x^4 + \ldots) \\
&= \frac{1}{(1+x)(1-x)^2} \\
&= \frac{1}{4} \cdot \frac{1}{1+x} + \frac{1}{4} \cdot \frac{1}{1-x} + \frac{1}{2} \cdot \frac{1}{(1-x)^2} \\
&= \frac{1}{4} \Sigma_{n \geq 0}{-1 \choose n}x^n + \frac{1}{4} \Sigma_{n \geq 0}x^n + \frac{1}{2} \Sigma_{n \geq 0}{2 + n - 1 \choose n}x^n \\
&= \Sigma_{n \geq 0}\Bigl((-1)^n \frac{1}{4} x^n + \frac{1}{4} x^n + \frac{1}{2}{n + 1 \choose n} x^n \Bigr) \\
&= \Sigma_{n \geq 0}\Bigl((-1)^n \frac{1}{4} x^n + \frac{1}{4} x^n + \frac{1 \cdot 2}{2 \cdot 2}(n + 1) x^n \Bigr) \\
&= \Sigma_{n \geq 0} \frac{1}{4} \Bigl((-1)^n x^n + x^n + 2(n + 1) x^n \Bigr)
\end{align}
$$
I don't know is this is correct nor how to proceed.
I guest that we use similar method to solve problem (2)?
|
The generating function for problem $1$ is
$$
\begin{align}
\frac1{1-x}\frac1{1-x^2}
&=\frac12\frac1{1-x}\left(\frac1{1-x}+\frac1{1+x}\right)\\
&=\frac12\left(\color{#C00}{\frac1{(1-x)^2}}+\color{#090}{\frac1{1-x^2}}\right)\\
&=\sum_{k=0}^\infty\frac{\color{#C00}{k+1}+\color{#090}{[2\mid k]}}2\,x^k\\
&=\sum_{k=0}^\infty\left\lfloor\frac{k+2}2\right\rfloor x^k\tag1
\end{align}
$$
where $[\dots]$ are Iverson brackets.
The last step of $(1)$ is justified since $1-\frac{1+[2\mid k]}2=\frac{1-[2\mid k]}2\in[0,1)$.
The generating function for problem $2$ is
$$
\begin{align}
&\frac1{1-x}\frac1{1-x^2}\frac1{1-x^3}\\
&=\color{#C00}{\frac1{6(1-x)^3}}+\color{#090}{\frac1{4(1-x)^2}}+\color{#00F}{\frac1{4\!\left(1-x^2\right)}}+\color{#740}{\frac1{3\!\left(1-x^3\right)}}\\
&=\sum_{k=0}^\infty\left(\color{#C00}{\frac{(k+2)(k+1)}{12}}+\color{#090}{\frac{k+1}4}+\color{#00F}{\frac{[2\mid k]}4}+\color{#740}{\frac{[3\mid k]}3}\right)x^k\\
&=\sum_{k=0}^\infty\frac{(k+3)^2-4+3[2\mid k]+4[3\mid k]}{12}\,x^k\\
&=\sum_{k=0}^\infty\underbrace{\left\lfloor\frac{(k+3)^2}{12}+\frac12\right\rfloor}_\text{closest integer to $\frac{(k+3)^2}{12}$}\,x^k\tag2
\end{align}
$$
The last step of $(2)$ is justified since $\frac12-\frac{-4+3[2\mid k]+4[3\mid k]}{12}=\frac{10-3[2\mid k]-4[3\mid k]}{12}\in[0,1)$.
|
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|
How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$? I learnt that $\displaystyle \sum_{n=0}^{\infty} \frac{n+1}{(2n+1)!} = \frac{e}{2}$.
I am wondering what the closed form for $\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^2}{(2n+1)!}$ is.
I tried using the fact that $ 1+3+5+\cdots+(2n-1) = n^2$, but it was not fruitful.
Could people give me some hints on how to approach this problem?
To view the general formula, please visit The value of $\sum_{n=0}^{\infty}\frac{(n+1)^k}{(2n+1)!}$, where $k\in\mathbb{W}$
|
$$\sum_{n=0, \text{odd}}^{\infty} \frac{(\frac{n+1}{2})^2}{n!}$$
$$=\frac{1}{4} \sum_{n=0, \text{odd}}^{\infty} \frac{n^2+2n+1}{n!}$$
$$=\frac{1}{4} \left( \sum_{n=0,\text{odd}}^{\infty} \frac{n}{(n-1)!}+2\sum_{n=0,\text{odd}}^{\infty} \frac{1}{(n-1)!}+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{n!} \right)$$
Now note:
$$\sum_{n=0, \text{odd}}^{\infty} \frac{1}{n!}$$
$$=\sum_{n=0}^{\infty} \frac{1}{2} \frac{1^n-(-1)^n}{n!} $$
$$=\frac{e^{1}-e^{-1}}{2}= \sinh 1$$
Also note:
$$S :=\sum_{n=0, \text{odd}}^{\infty} \frac{1}{(n-1)!}$$
$$=\sum_{n=0,\text{even}}^{\infty} \frac{1}{n!}$$
$$=\sum_{n=0}^{\infty} \frac{1}{2} \frac{1+(-1)^n}{n!}$$
$$=\cosh 1$$
Finally,
$$\sum_{n=0, \text{odd}}^{\infty} \frac{n}{(n-1)!}$$
$$=\sum_{n=0,\text{odd}}^{\infty} \frac{n-1}{(n-1)!}+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{(n-1)!}$$
$$=S+\sum_{n=0,\text{odd}}^{\infty} \frac{(n-1)}{(n-1)!}$$
$$=S+\sum_{n=3,\text{odd}}^{\infty} \frac{(n-1)}{(n-1)!}$$
$$=S+\sum_{n=3,\text{odd}}^{\infty} \frac{1}{(n-2)!}$$
$$=S+\sum_{n=1,\text{odd}}^{\infty} \frac{1}{n!}$$
$$=S+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{n!}$$
$$=S+\sinh 1$$
$$=\cosh 1+\sinh 1$$
This gives an answer of,
$$\frac{1}{4} \left(3\cosh 1+2\sinh 1 \right)$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2917988",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Getting a closed form from $\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1$ I need to get a closed form from
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1
$$
Starting from the most outer summation, I got
$$
\sum_{k=1}^{j} 1 = j
$$
But now I don't know how to proceed with:
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j
$$
Could you guys please help me?
Thanks in advance.
|
Knowing that
\begin{align}
\sum_{k=1}^{n} (1) &= n \\
\sum_{k=1}^{n} k &= \frac{n \, (n+1)}{2} \\
\sum_{k=1}^{n} k^2 &= \frac{n(n+1)(2n+1)}{6}
\end{align}
then
\begin{align}
S &= \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} (1) \\
&= \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j \\
&= \sum_{i=1}^{n-1} \left(\sum_{j=1}^{n} j - \sum_{j=1}^{i} i \right) \\
&= \sum_{i=1}^{n-1} \left(\binom{n}{2} - \frac{i(i+1)}{2}\right) \\
&= \binom{n}{2} \, \sum_{i=1}^{n-1} (1) - \frac{1}{2} \, \sum_{i=1}^{n-1} i^2 - \frac{1}{2} \, \sum_{i=1}^{n-1} i \\
&= \binom{n}{2} \, (n-1) - \frac{n(n-1)}{4} - \frac{n(n-1)(2n-1)}{12} \\
&= \frac{(n-1)(n)(n+1)}{3} = 2 \, \binom{n+1}{3}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Principal part of Laurent Series for $\frac{1}{(1-\cosh(z))^2}$ in this exercise I am asked to provide the principal part of the Laurent series of
$$\frac{1}{(1-\cosh(z))^2}$$
And i am kinda struggling with fonding a solution or even a pattern towards one
Thanks in advance to everyone keen to help
|
$$(1-\cosh z)^2=3-4\cosh z+\cosh2z$$
then
\begin{align}
\dfrac{1}{(1-\cosh z)^2}
&= \dfrac{1}{3-4\cosh z+\cosh2z} \\
&= \dfrac{1}{3-4 \left(1+\frac{1}{2!}z^2+\frac{1}{4!}z^4+\frac{1}{6!}z^6+\cdots\right) + \left(1+\frac{4}{2!}z^2+\frac{16}{4!}z^4+\frac{64}{6!}z^6+\cdots\right)} \\
&= \dfrac{1}{\frac{1}{4}z^4+\frac{1}{24}z^6+\frac{1}{320}z^8+\cdots} \\
&= \frac{4}{z^4}-\frac{2}{3z^2}+\dfrac{11}{180}+\sum_{n\geq1}a_{2n}z^{2n}\cdots
\end{align}
and you find the principal part.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\int_0^a\frac{x^4}{(a^2+x^2)^4}dx$? I have to evaluate
$$\int_0^a\frac{x^4}{(a^2+x^2)^4}\,{\rm d}x$$
I tried to substitute $x=a\tan\theta$ which then simplifies to
$$\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta\, {\rm d}\theta$$.
Now,its quite hectic to solve this. Is there any other method out?
|
This is a rather standard rational function integral, and we only have to decompose it into partial fractions.
To have lighter calculations, we'll set $T=a^2+x^2$. Then
$$x^4=T^2-2a^2x^2-a^4=T^2-2a^2T+a^4$$
which yields the decomposition
\begin{align}
\frac{x^4}{(a^2+x^2)^4}&=\frac{T^2-2a^2T+a^4}{T^4}=\frac1{T^2}-\frac{2a^2}{T^3}+\frac{a^4}{T^4}\\
&=\frac1{(a^2+x^2)^2}-\frac{2a^2}{(a^2+x^2)^3}+\frac{a^4}{(a^2+x^2)^4}.
\end{align}
Now the integrals $\;I_n=\int\frac{\mathrm d x}{(a^2+x^2)^n}$ can be computed recursively, from
\begin{cases}
I_1=\dfrac1a\arctan \dfrac xa, \\
I_{n+1}=\dfrac1{2na^2}\dfrac x{(a^2+x^2)^n}+\dfrac{2n-1}{2na^2} I_n.
\end{cases}
The recurrence relation can be obtained with an integration by parts of $I_n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
= 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
= 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
|
When you take the $90$ to the other side you need to subtract it. You'll get
$n^2 +7n +12 -90 = 0,$
$n^2 +7n -78 = 0,$
$(n - 6)(n+13) = 0,$
So $n = 6$ or $n = -13$.
Clearly $n = -13 $ doesn't make sense, so we conclude:
$n = 6$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $f:M_f \rightarrow A_f$ is bijective when $f(x)=\frac{x-1}{x+1}$ Problem
Show that $f:M_f \rightarrow A_f$ is bijective when $f(x)=\frac{x-1}{x+1}$
Edit
when $M_f \in \mathbb{R}\setminus \{-1\}$ and $A_f \subset \mathbb{R}$
Attempt to solve
$f:M_f\rightarrow A_f$ is bijective when it is injective and surjective at the same time.
Injection
$f:M_f \rightarrow Af$ is injective when:
$$ \forall (x,y) \in M_f : x \neq y \implies f(x)\neq f(y) $$
when $p= x\neq y$ and $q=f(x)\neq f(y)$. If it is possible to show $\neg p \implies \neg q$ it means: $$ \neg p \implies \neg q \iff p \implies q $$
meaning
$$ x \neq y \implies f(x) \neq f(y) \iff x = y \implies f(x)= f(y) $$
I can show:
$$ \frac{x-1}{x+1}=\frac{y-1}{y+1} $$
$$ \implies (x+1)(y+1)\frac{x-1}{x+1}=(x+1)(y+1)\frac{y-1}{y+1} $$
$$ \implies (y+1)(x-1)=(x+1)(y-1) $$
$$ \implies xy-y+x-1 = xy -x + y-1 $$
$$ \implies xy+2x = xy +2y $$
$$ \implies 2x=2y \implies x = y $$
Conclusion: $f:M_f \rightarrow A_f$ is injective since
$$ x \neq y \implies f(x) \neq f(y) \iff x = y \implies f(x)= f(y) $$
is true
Surjection
$$ f:M_f \rightarrow A_f $$ is surjective when
$$ \forall y \in A_f \exists x \in M_f : f(x)=y$$
meaning if
$$ f(x)=y $$
assuming $x \neq -1$
$$ \implies \frac{x-1}{x+1}=y $$
$$ \implies x-1=y(x+1) $$
$$ \implies x-1=yx+y $$
$$ \implies x(1-y)=1+y $$
$$ \implies x = \frac{1+y}{1-y} $$
Now we get
$$ f(x)=f(\frac{1+y}{1-y}) $$
$$\implies f(x)=\frac{(\frac{1+y}{1-y})-1}{(\frac{1+y}{1-y})+1}$$
For some reason odd reason i cannot get this into form $f(x)=y$. Should be simple elementary algebra but cannot see how this equals to $y$. In theory this expression should be equal to $y$ which would make $f:M_f\rightarrow Af$ surjective and then bijective since it was also injective.
|
Observe that
$$f(x) = \frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}$$
Thus, the range of $f$ is $\mathbb{R} - \{1\}$.
To ensure that $f: \mathbb{R} - \{-1\} \to A_f$ is surjective, we must define $A_f = \mathbb{R} - \{1\}$.
For $y \neq 1$,
\begin{align*}
f\left(\frac{1 + y}{1 - y}\right) & = \frac{\frac{1 + y}{1 - y} - 1}{\frac{1 + y}{1 - y} + 1}\\
& = \frac{\frac{1 + y}{1 - y} - 1}{\frac{1 + y}{1 - y} + 1} \cdot \frac{1 - y}{1 - y}\\
& = \frac{1 + y - (1 - y)}{1 + y + 1 - y}\\
& = \frac{2y}{2}\\
& = y
\end{align*}
Thus, the function is surjective. Since you have shown it is injective, it is bijective.
Alternatively, define $A_f$ as above. Define $g: \mathbb{R} - \{1\} \to \mathbb{R} - \{-1\}$ by
$$g(x) = \frac{1 + x}{1 - x}$$
To show that $f: \mathbb{R} - \{-1\} \to \mathbb{R} - \{1\}$ is bijective, show that $g$ is $f^{-1}$ by showing that
\begin{align*}
(g \circ f)(x) & = x~\text{for each $x$ in the domain of $f$}\\
(f \circ g)(y) & = y~\text{for each $y$ in the domain of $g$}
\end{align*}
Establishing that $(g \circ f)(x) = x$ for each $x$ in the domain of $f$ shows that $f$ is injective since if $(g \circ f)(x_1) = (g \circ f)(x_2)$, then
$$x_1 = (g \circ f)(x_1) = (g \circ f)(x_2) = x_2$$
Establishing that $(f \circ g)(y) = y$ for each $y$ in the domain of $g$ shows that $f$ is surjective since if $g(y) = x$, then $f(x) = y$.
|
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|
Cat / mouse probability question There exist 7 doors numbered in order from 1 to 7 (going from left to right). A mouse is initially placed at center door 4. The mouse can only move 1 door at a time to either adjacent door and does so, but is twice as likely to move to a lower numbered door than to a higher numbered door each time it moves 1 door. There are cats waiting at doors 1 and 7 that will eat the mouse immediately after the mouse moves to either of those 2 doors.
So for example, the mouse starts at door 4. He could then move to door 3, then to door 2, then back to 3, then back to 2, then to door 1 where he gets eaten. That counts as 5 moves total. Skipping doors is not allowed.
So there are 2 questions I have regarding this:
1) What is the expected average number of moves before the mouse gets eaten? (do not count the initial start at door 4 as a move but count any final move to doors 1 or 7 and any "intermediate" moves between those 2 states).
2) What is the probability that the mouse will survive for 100 or more moves?
|
I'll take a shot at this........
1) The eventual fateful outcome occurs when the total number of moves, lower versus higher, differs by 3.
So, a way to look at this as an $E(x) = n\cdot p$ type problem is to sum the $(n\cdot p)$s for all possible outcomes.
This will be:
$3(\frac{1}{3})+5(\frac{2}{9})+7(\frac{4}{27})+9(\frac{8}{81})+ .....\text{etc}$ which is an infinite arethmetico-geometric series whose infinite sum is: $$S = \frac{dg_2}{(1-r)^2} + \frac{a}{1-r} = \frac{2\cdot (\frac{1}{3}\cdot \frac{2}{3})}{\frac{1}{3}^2} + \frac{3\cdot \frac{1}{3}}{\frac{1}{3}} = 2\cdot 2 + 3 = 7$$
2) $$P(n\ge 100) = 1 - P(n<100)$$
$$P(n<100) = S_n = \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+ .......+\frac{2^{n-1}}{3^n}$$
This turns out to be a geometric series, where $a_1 = \frac{1}{3}, r = \frac{2}{3}$ and $n = 49$ (odd from $3$ to $99$) a different n from the n moves.
Example calculation for $3$rd term is: $9(\frac{2}{3})^5(\frac{1}{3})^2 + 9(\frac{1}{3})^5(\frac{2}{3})^2 = \frac{4}{27}$
$$P(n\ge 100) = 1 - \frac{\frac{1}{3}(1-(\frac{2}{3})^{49})}{1-(\frac{2}{3})}$$
$$P(n\ge 100) = 1 - .9999999976 = 2.4\cdot 10^{-9}$$
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Value of 'a' for which $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b $ is injective Let $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b \forall x \in R$.
Find the least value of a for which $f(x)$ is injective function.
My approach , for $f(x)$ to be injective either f(x) should be increasing or decreasing function.'
$Y=f'(x)=x^2+x+a$ [Increasing function]
If $Y>0$, then $x^2+x+a>0$, it is possible when $1-4a<0$ or $a>\frac{1}{4}$
If $Y<0$, then $x^2+x+a<0$, it is possible when $1-4a \ge 0$ or $a \le \frac{1}{4}$
The options are (A) $\frac{1}{4}$ which is the correct option, the other given options are (B) $1$, (C) $\frac{1}{2}$, (D) $\frac{1}{8}$
Though my optin is correct but i have one doubt when the equality=$0$, then Y=$0$ hence it is neither increasing nor decreasing function.
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If you want $f$ is injective, then $f$ should be increasing since leading part is $x^3$. So $f'(x)=x^2+x+a\geq 0$ for all $x$, so the discriminant $D\leq 0$ and that is iff $1-4a\leq 0$ so $a\geq {1\over 4}$
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Ask a about hard integral of $\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$ I want to evaluate the integral:
$$I(a,b)=\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$$
Attempt:$$\frac{\partial ^2I}{\partial a\partial b}=4ab\int_{0}^{\infty}\frac{\log x}{(a^2+x^2)(b^2+x^2)}dx=\frac{4ab}{b^2-a^2}\int_{0}^{\infty}\log x\left(\frac{1}{a^2+x^2}-\frac{1}{b^2+x^2}\right)dx$$
$$=\frac{4ab}{b^2-a^2}\frac{\pi}{2}\left(\frac{\log a}{a}-\frac{\log b}{b}\right)=\frac{2\pi(b\log a-a\log b)}{b^2-a^2}$$
Then $$I(a,b)=2\pi\int_{0}^{b}\int_{0}^{a}\frac{(y\log x-x\log y)}{y^2-x^2}dxdy$$
But this integral very hard to solve,can anyone help me,thank you!
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This derivation is slightly different from Maxim's, because I'm not fluent in Meijer G. The beginning is the same, and assume $0<b<a.$ Then
$$ I(a,b)=b\,\int_0^\infty (\log{b} + \log{x} ) \log{(1+ \big(\frac{a/b}{x}\big)^2 ) }\log{(1+1/x^2)} dx $$
Let $r=b/a \le 1$ so that some series manipulations are permissible. The integral without the $\log{x}$ is easily performed in Mathematica:
$$ \int_0^\infty \log{(1+ \big(r\,x\big)^{-2} ) }\log{(1+1/x^2)} dx = \pi\Big( (1+\frac{1}{r})(\text{arctanh(r)} - \log{(1-r^2)})-2\log{r}\Big)$$
Define
$$J(s;r)=\int_0^\infty x^s \log{(1+ \big(r\,x\big)^{-2} ) }\log{(1+1/x^2)} dx \,.$$
The objective is to find $$\frac{d}{ds}J(s;r)\Big|_{s=0} $$
Within Mathematica J(s;r) can be found in terms of elementary functions and Gauss's hypergeometric $F(a,b;c,x).$
$$\frac{J(s;r)}{\pi}=\sec{(\frac{\pi s}{2})} \Big\{\!\frac{2\,r^2}{
(s\!+\!1)(s\!\!+3)}F(1,\!\frac{s+3}{2};\! \frac{s+5}{2}, r^2) -
\frac{2\,r^{1-s}}{
(s\!+\!1)(s\!-\!1)}F(1,\!\frac{1-s}{2};\! \frac{3-s}{2}, r^2)$$
$$+ \frac{1}{s\!+\!1}\Big[ r^{-1-s}\log{(1-r^2)} + \log{(-1+1/r^2)}+\frac{2}{s+1} - \pi\,\tan{(\pi\,s/2)} \Big]
\Big\}$$
Do the derivative and take $s \to 0.$ $F(1,\frac{1}{2}; \frac{3}{2}, r^2)$ and
$F(1,\frac{3}{2}; \frac{5}{2}, r^2)$ evaluate to elementary functions. However the derivatives with respect to $s$ of the hypergeometrics do not. However, by using the series definition in terms of Pochhammer symbols, an easy calculation shows
$$ \frac{d}{ds} \frac{(3/2+s/2)_k}{(5/2+s/2)_k} \Big|_{s=0} = \frac{2k}{(2k+3)^2} \quad , \quad \frac{d}{ds} \frac{(1/2-s/2)_k}{(3/2-s/2)_k} \Big|_{s=0} = \frac{-2k}{(2k+1)^2} $$ In detail,
$$\frac{d}{ds} F(1,\frac{1-s}{2}; \frac{3-s}{2}, r^2) \Big|_{s=0}
= -\sum_{k=0}^\infty\frac{2k}{(2k+1)^2} r^{2k} = -\sum_{k=0}^\infty\frac{2k+1 -1}{(2k+1)^2} r^{2k} = $$
$$=-\frac{\text{arctanh(r)}}{r} + \sum_{k=0}^\infty \frac{r^{2k}}{(2k+1)^2}=
-\frac{\text{arctanh(r)}}{r} + \frac{1}{2r} \Big( \text{Li}_2(r) - \text{Li}_2(-r) \Big)
$$
The series with the $(2k+3)^2$ in the denominator can be brought to this form with an index shift in the summation. Collect all the results and you finally get
$$\frac{I(a,b)}{\pi\,b}= \log{b} \Big( (1+\frac{1}{r})(2\,\text{arctanh}(r) + \log{(1-r^2)} ) -2\log{r} \Big)\, +
\big(1-\frac{1}{r}\big) \big( \text{Li}_2(r) - \text{Li}_2(-r) \big) $$
$$-\Big(\frac{\pi^2}{2} + \log{(r^{-2}-1)} +2\,\big(1+\log{r}+\frac{1}{r} \big)\,\text{arctanh}(r) + \frac{1+\log{r}}{r} \, \log{(1-r^2)} \Big) $$
|
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|
Generating function problem, and distributing candy to kids. Here is the problem:
Determine how many ways I can distribute $80$ candies to $3$ kids, such that:
$\bullet$ The first kid receives an arbitrary number of candies (possibly $0$).
$\bullet$ The second kid receives an even positive number of candies.
$\bullet$ The third kid receives $0$, $2$, or $5$ candies.
$\bullet$ Every candy is distributed.
Okay, so since this is a generating function problem... or at least I think so, I do all my generating function stuff and get to where I have $\frac{2x + 2x^3 + 2x^6}{-x^3 + 3x^2 - 3x + 1}$. How can I find the $x^{80}$ coefficient from here? Hints would be appreciated!
Thanks,
Max0815
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If the option to receive $k$ candies is worth $x^k$ the options of the first child sum to $1+x+x^2+\ldots$, the options of the second child sum to $x^2+x^4+x^6+\ldots$, and the options of the third child sum to $1+x^2+x^5$. In such a situation the function
$$f(x)=(1+x+x^2+\ldots)(x^2+x^4+x^6+\ldots)(1+x^2+x^5)$$
(your $f$ looks different to me) is the generating function for this problem: Multiplying the RHS distributively produces for each legal allocation of $n_1+n_2+n_3=:n$ candies a term $x^{n_1}\cdot x^{n_2}\cdot x^{n_3}=x^n$. It follows that the number of legal allocations of $80$ candies to the three children is equal to the coefficient of $x^{80}$ in $f(x)$, when terms have been collected. A partial fraction decomposition (produced for me by Mathematica) gives
$$\eqalign{f(x)&={x^2(1+x^2+x^5)\over(1-x)(1-x^2)}\cr &=4+3x+2x^2+x^3+x^4-{23\over4}{1\over1-x}+{3\over2}{1\over(1-x)^2}+{1\over4}{1\over 1+x}\ .\cr}$$
Since ${1\over(1-x)^2}=\sum_{k=0}^\infty(k+1)x^k$ the coefficient we are after is given by
$$[x^{80}]=-{23\over4}+{3\over2}\cdot81+{1\over4}(-1)^{80}=116\ .$$
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Using power series to evaluate a limit If we want to evaluate the limit of $$\lim_{x\to 0} \frac{\ln(1-8x)+8x+32x^2}{3x^2}$$
I would imagine we need to use the power series expansion of the function $${\ln(1-8x)}$$
and then determine what cancels out.
The correct answer that I found using a calculator is $${-512/9}$$
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$$\lim_{x\to 0} \frac{\ln(1-8x)+8x+32x^2}{3x^2}=\lim_{x\to 0} \frac{\ln(1-8x)}{3x^2}+\frac{8}{3x}+\frac{32}{3}$$
$$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots-\frac{x^n}{n}$$
$$\ln(1-8x)=-8x-\frac{64x^2}{2}-\frac{512x^3}{3}-\cdots-\frac{8^nx^n}{n}$$
$$\frac{\ln(1-8x)}{3x^2}=-\frac{8}{3x}-\frac{32}{3}-\frac{512x}{9}-\cdots-\frac{8^nx^{n-2}}{3n}$$
$$\frac{\ln(1-8x)}{3x^2}+\frac{8}{3x}+\frac{32}{3}=-\frac{512x}{9}-\cdots-\frac{8^nx^{n-2}}{3n}$$
$$\lim_{x \to 0}\frac{\ln(1-8x)}{3x^2}+\frac{8}{3x}+\frac{32}{3}=\lim_{x \to 0}\left(-\frac{512x}{9}-\cdots-\frac{8^nx^{n-2}}{3n}\right)=0$$
|
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|
Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$
I am trying to find the Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$ when $0<|z-1|<1$.
I thought that if
\begin{align}
f(z)&=\frac{1}{z-1}-\frac{1}{z-2}+\frac{1}{(z-2)^2} \\
&=\frac{1}{z-1}+\frac{1}{2-z}+\frac{d}{dz}\left(\frac{1}{2-z}\right) \\
&=\frac{1}{z-1}+\frac{1}{1-(z-1)}+\frac{d}{dz}\left(\frac{1}{1-(z-1)}\right) \\
&=\frac{1}{z-1}+\sum_{n=0}^{\infty}(z-1)^n+\frac{d}{dz}\left(\sum_{n=0}^{\infty}(z-1)^n\right) \\
&=\frac{1}{z-1}+\sum_{n=0}^{\infty}\left((z-1)^n+n(z-1)^{n-1}\right) \\
&=\frac{1}{z-1}+\sum_{n=0}^{\infty}(z-1)^n\left(1+n(z-1)^{-1}\right) \\
\end{align}
But this does not agree with the series generated by wolfram. What about this method is incorrect?
alternative approach suggested by David
\begin{align}
f(z)&=\frac{1}{z-1}\left(\frac{1}{(z-2)^2}\right) \\
&=\frac{1}{z-1}\frac{d}{dz}\left(-\frac{1}{z-2}\right) \\
&=\frac{1}{z-1}\frac{d}{dz}\left(\frac{1}{1-(z-1)}\right) \\
&=\frac{1}{z-1}\frac{d}{dz}\left(\sum_{n=1}^{\infty}(z-1)^n\right) \\
&=\frac{1}{z-1}\left(\sum_{n=1}^{\infty}n(z-1)^{n-1}\right) \\
&=\sum_{n=1}^{\infty}n(z-1)^{n-2}
\end{align}
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*
*In line 2, $-\frac{d}{dz}$ should be $+\frac{d}{dz}$.
*Your final sum has many powers duplicated and these could be collected into single terms. For example when $n=5$ the first term is $(z-1)^5$ and when $n=6$ the second term is $6(z-1)^5$.
*Easier method: take $\frac1{z-1}$ as a multiplicative factor at the beginning.
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|
Classifying singular points of $\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}$
I am trying to classify the singular points of the function $$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}.$$
My attempt:
$$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}=\frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}.$$
Hence the singular points are $z=0,\frac{\pi}{4}$. For classification:
$z=\frac{\pi}{4}$ is a simple pole as $\left(z-\frac{\pi}{4}\right)$ is a simple zero.
For $z=0$, we can see that $z^2\left(z-\frac{\pi}{4}\right)$ is a zero of order $2$. Also,$\ \sin(z^2)$ is a zero of order $2$ for $z=0$. Hence, $z=0$ is a removable singularity.
I am unsure about the $z=0$ case, particularly if $\sin(z^2)$ is a zero of order $2$ for $z=0$.
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Note that $$\sin(z^2)=z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}-\cdots$$ so $$f(z)=\frac{z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}-\cdots}{z^2\left(z-\frac{\pi}{4}\right)}=\frac{1-\frac{z^4}{3!}+\frac{z^8}{5!}-\cdots}{z-\frac\pi4}$$ so $z=0$ is a removable singularity.
|
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Find the relationship of the length of triangle's sides.
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
Now determine what kind of triangle $\triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
C.Right triangle
D.Isosceles Right triangle
The only information I got is from the number in the radical need to be greater than $0$. Then $b\ge4$ and $c\ge 1$. Also $10a+2\sqrt{b-4}-22\ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
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Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|\sqrt{c-1} - 2|=(a-5)^2 + b+ |\sqrt{c-1} -2|= 2\sqrt{b-4} + 3$
And, well this seems a bit weird but that is an even number in front of the $\sqrt{b-4}$ and we have $\sqrt{b-4}$ and $b$ variables to deal with so we can complete the square again with $v = \sqrt{b-4}$ and $v^2 = b-4$.
(In the back of my mind I'm worrying about the $\sqrt{c-1}$ which is just a single term; I'm not sure at this point what will happen.)
$(a -5)^2 + (b-4) - 2\sqrt{b-4} + 1 +|\sqrt{c-1} - 2|=3-4 + 1$
$(a-5)^2 + (\sqrt{b-4} - 1)^2 +|\sqrt{c-1} - 2| = 0$.
Oh......
We have three things that can't be negative adding up to $0$. So they must each equal $0$.
So $(a-5)^2 = 0$ and $a =5$. And $(\sqrt{b-4}-1)^2=0$ and $b=5$. And $|\sqrt{c-1} -2| = 0$ so $c=5$.
Well.... okay then......
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|
Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin(u)^2}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$ , someone could help because i block here because i don't get the same resultat with this website
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$$\frac{1+\cos u}{1-\cos u}=2\cot^2\left(\frac{u}{2}\right).$$
So your integral is
$$\int\sqrt{\frac{1+x}{1-x}}\, dx=-\sqrt{2}\int \sin u \cot\left(\frac{u}{2}\right) \, du=-2\sqrt{2}\int \cos^2 \left(\frac{u}{2}\right)\, du$$
Here we are using $\sin u =2\sin \left(\frac{u}{2}\right)\cos \left(\frac{u}{2}\right)$. Now once again we can write $2\cos^2 \left(\frac{u}{2}\right)=1+\cos u$ to complete the integral.
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Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx$
Using Integration by parts
$$ I =\frac{2}{b}\bigg[\sin (x)\cdot \ln|a-b\cos x|\bigg]\bigg|^{\pi}_{0}-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln|a-b\cos x|dx$$
$$I=-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln(a-b\cos x)dx$$
Could some help me to solve it, Thanks in advance
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From $a>b$ then $\left|\dfrac{b}{a}\cos x\right|<1$ with the expansion
\begin{align}
\int_{0}^{2\pi}\frac{\sin^2(x)}{a-b\cos x}\ dx
&= \dfrac{1}{a}\int_{0}^{2\pi}\sin^2(x)\sum_{n\geq0}\left(\dfrac{b}{a}\cos x\right)^n\ dx \\
&= \dfrac{1}{a}\sum_{n\geq0}\left(\dfrac{b}{a}\right)^n\int_{0}^{2\pi}\sin^2x\cos^nx\ dx \\
&= \dfrac{1}{a}\sum_{n\geq0, \text{even}}\left(\dfrac{b}{a}\right)^n\left(\dfrac{2\pi}{2^n}{n\choose n/2} - \dfrac{2\pi}{2^{n+2}}{n+2\choose n/2+1}\right) \\
&= \color{blue}{\dfrac{2\pi}{a+\sqrt{a^2-b^2}}}
\end{align}
Using here
$$\int_0^{2\pi}\cos^n x\ dx=\dfrac{2\pi}{2^n}{n\choose n/2}$$
for even $n$ and for odd $n$ it is zero.
|
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Finding the domain and the range of $f(x) = \ln (1-2\cos x).$ Finding the domain and the range of $f(x) = \ln (1-2\cos x).$
My Attempt:
I arrived at $x < \cos^{-1}(1/2),$ then how can I complete?
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the domain of $\ln (z)$ is $z \gt 0$. So we need $1- 2\cos(x) \gt 0$.
so $$-2\cos(x) > -1$$ $$2\cos(x) \lt 1$$ $$\cos(x) < \frac{1}{2}$$
if we graph $\cos(x)$ we see that $\cos(x) \ge \frac{1}{2}$ for $x \in \big[ \frac{-\pi}{3}, \frac{\pi}{3} \big]$ but we need to exclude this to keep $\cos(x) \lt \frac{1}{2}$, so we take these $x$ values out of this domain. But this region occurs periodically so we need to exclude $x \in \big [ 2n\pi - \frac{\pi}{3},2n\pi + \frac{\pi}{3} \big]$ where $n \in \mathbb Z$
so the domain is $$x \in \mathbb R \setminus \big [2n\pi - \frac{\pi}{3},2n\pi + \frac{\pi}{3} \big]$$
As for the range: the max value of $f(x)$ occurs when $\cos(x)$ is a minimum, so the minimum value of $\cos(x)$ in the domain is $\cos(x) = -1$, so putting this into $f(x)$, we get $\ln(1-2(-1))$ = $\ln(3)$ As for the lower bound of the range, $cos(x)$ never actually reaches $\frac{1}{2}$ but it does get infintiely closer to $\frac{1}{2}$ so $1-2\cos(x)$ gets closer to 0 and from our knowledge of logarithms and exponentials, exp(z) gets closer to 0 as z gets closer to -infinity, so f(x) will tend to -infinity, therefore the range is $(-\inf,\ln(3)]$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.