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How to prove this trignometrical Identities? The following two identities comes from my trigonometry module without any sort of proof,
If $A + B + C = \pi $ then,
$$\tan A + \tan B + \tan C = tan A \cdot tan B \cdot tan C$$
and,
$$ \tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1 $$
PS:I am not much sure about whether the first one is fully correct or not, so if not please suggest the correct one and also I will be grateful if somebody suggest a suitable method (may be using mathematica) to verify an identity like this prior to proving.
|
If $A+B+C= \pi \Longrightarrow \tan(A+B) = \tan(\pi -C) =-\tan(C)$. So we have $$\tan(A+B)= \frac{\tan(A) + \tan(B)}{1 - \tan(A)\cdot \tan(B)} = -\tan(C) $$ $$\Longrightarrow \tan(A)+\tan(B) = -\tan(C) \cdot \Bigl[1 - \tan(A)\tan(B)\Bigr]$$ from which the first one follows.
And for the second one, we have $\displaystyle\frac{A}{2} + \frac{B}{2} =\frac{\pi}{2}- \frac{C}{2} \Longrightarrow \tan\Bigl(\frac{A+B}{2}\Bigr)= \cot\Bigl(\frac{C}{2}\Bigr)$ Now expanding we have $$\tan\Bigl(\frac{A+B}{2}\Bigr)= \frac{\tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr)}{1- \tan\Bigl(\frac{A}{2}\Bigr)\cdot \tan\Bigl(\frac{B}{2}\Bigr)} = \cot\Bigl(\frac{C}{2}\Bigr)$$ Multiplying both sides by $\tan\frac{C}{2}$ we have $$\tan\Bigl(\frac{C}{2}\Bigr) \cdot \Bigl[ \tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr) \Bigr] = 1 \cdot \Bigl[ 1 - \tan\Bigl(\frac{A}{2}\Bigr) \cdot \tan\Bigl(\frac{B}{2}\Bigr)\Bigr]$$
|
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|
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
|
$\sin(x)\cos(x) = \frac{1}{2} \sin(2x)$.
$I = 2\int \csc(2x)$ let $u = 2x$ then:
$I = \int \csc(u) du = - \log(\cot(2x) + \csc(2x)) + C$
|
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|
$n^2 + 3n +5$ is not divisible by $121$ Question:
Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.
|
As $121=11^2,$ we need $11|(n^2+3n+5)$
Let us find $x,y$ such that $x-y=3,x+y=11\implies x=7,y=4$
$$n^2+3n+5=(n+7)(n-4)+33$$
As $33$ is divisible by $11,$ so must be $(n+7)(n-4)$ to make $11|(n^2+3n+5)$
Now $11|(n-4)\iff 11|(n+7)$ as $(n+7)-(n-4)=11$
So in that case, $11^2|(n+7)(n-4),$ but $11^2\not|33$
|
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|
$x^y = y^x$ for integers $x$ and $y$ We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
|
Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y \log x = x \log y$; rearranging, $(\log x)/x = (\log y)/y$. Let $f(x) = (\log x)/x$; then this is $f(x) = f(y)$.
Now, $f^\prime(x) = (1-\log x)/x^2$, so $f$ is increasing for $x<e$ and decreasing for $x>e$. So if $x^y = y^x$ has a solution, then $x > e > y$. So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.
(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)
If $x > 0 > y$, then $0 < x^y < 1$ and $y^x$ is an integer, so there are no such solutions.
If $0 > x > y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.
|
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|
How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ How can we prove the following trigonometric identity?
$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$
|
You can find the solution in this page:
*
*http://natto.2ch.net/math/kako/1002/10029/1002903143.html
Translation of the page into English.
$I = \tan (3π/11) +4 \sin (2π/11)$
and $t = 3π/11 $
$11t = 3π$
⇔ $6t = 3π-5t$
⇒ $\sin (6t) = \sin (3π-5t)$ taking sin of both sides
⇔ $2\sin (3t) \cos (3t) = \sin(5t)$ double angle formula
⇔ $[3\sin(t)-4 \sin^3 (t)] [4 \cos^3 (t)-3\cos(t)] = 16 \sin^5(t) -20 \sin^3(t) +5 \sin(t)$
⇔ $[3-4 \sin^2 t ] [4 \cos^3 t -3\cos t] = 16 \sin^4 t - 20 \sin^2 t +5$ dividing by $\sin t ≠ 0$
⇔ $32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = 0$, where $\sin^2 t = 1 - \cos^2 t$, $x = \cos t$
Thus $x = \cos (3π/11)$ is a solution of
$ 32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = 0 $
Since $(2π/11) = [1 - (9 / 11)] π = (π-3t)$, so
$I = \tan (3π/11) +4 \sin (2π/11)$
$ = \tan t +4 sin (π-3t)$
$ = \tan t +4 \sin (3t)$
$ = (\sin t / \cos t) +4 [3\sin t-4 \sin^3 t ]$
$ = (\sin t / \cos t) [16 \cos^3 t- 4 \cos t +1]$
$I ^ 2 = (\sin t / \cos t) ^ 2 [16 \cos^3 t -4 \cos t +1]^2$
$ = [(1 - \cos^2 t) / \cos^2 t] [16 \cos^3 t -4 \cos t +1] ^ 2$
$ = [(1-x^2) (16x^3-4x +1)^2]/x^2$, where $x = \cos t$
Molecule {(1-x ^ 2) (16x ^ 3-4x +1) ^ 2} a
{32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1} is divided by
← 2 11x ^ quotient remainder omitted
|
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classification up to similarity of complex n-by-n matrices Classify up to similarity all 3 x 3 complex matrices $A$ such that $A^n$ = $I$.
|
It is an Hoffman Kunze exercise problem. It will be for $3\times 3$ matrices $A$, $A^3=I$.
My answer is, the minimal polynomial of $A$ will divide $X^3-1=0$. Now $x^3-1=(x-1)(x-\omega)(x-\omega^2)$ where $\omega^3=1$. So the minimal polynomial can be of the forms
*
*$m=x-a$
*$m=(x-a)(x-b)$
*$m=(x-a)(x-b)(x-c)$
Now if $m=(x-a)$,then characteristic polynomial of $A$ will be equal to $(x-a)^3$. Hence $A$ is similar to a diagonal matrix with all entries equal.
If $m=(x-a)(x-b)$, then characteristic polynomial of $A$ will be equal to either $(x-a)^2(x-b)$ or $(x-a)(x-b)^2$. Hence $A$ is similar to a diagonal matrix with two entries equal.
If $m=(x-a)(x-b)(x-c)$, then characteristic polynomial of $A$ will be equal to either $(x-a)(x-b)(x-c)$. Hence $A$ is similar to a diagonal matrix with all entries are unequal.
|
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Exercise 9.2 from Apostol's Mathematical Analysis book. Uniform convergence of product This is a problem (Exercise 9.2) from Apostol's Mathematical Analysis (second edition) which I couldn't solve.
$\bullet$ Define two sequences $\{f_{n}\}$ and $\{g_{n}\}$ as follows:
$f_{n}(x) = x \Bigl(1 + \frac{1}{n}\Bigr)$ for $x \in \mathbb{R}$, $n \in \mathbb{N}$
$g_{n}(x)= \begin{cases}
\frac{1}{n} & x=0 \ \text{or}\ x \in \mathbb{R} - \mathbb{Q} \\
b+\frac{1}{n} & x \in \mathbb{Q},\ \text{say}\ x=\frac{a}{b}, b>0
\end{cases}$
$\mathbb{R}$ is the set of reals, $\mathbb{Q}$ is the set of rational and $\mathbb{R} - \mathbb{Q}$ is the set of irrationals.
I have to show that if $h_{n}(x)=f_{n}(x)g_{n}(x)$, then $h_{n}(x)$ does not converge uniformly on any bounded interval.
How can I show this?
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Assume that $h_n \to h$ uniformly on the domain $D$ of $\{h_n\}$, where
$$h_n(x) =
\begin{cases}
\frac{x}{n} \biggl(1 + \frac{1}{n} \biggr), & \text{if } x \text{ is irrational} \\
a + \frac{a}{n} + \frac{a}{b} \biggl(1 + \frac{1}{n} \biggr) \biggl(\frac{1}{n} \biggr), & \text{if } x \text{ is rational, say } x = a/b, b > 0
\end{cases}$$
and $$h(x) =
\begin{cases}
0, & \text{if }x\text{ is irrational} \\
a, & \text{if }x\text { is rational, say }x = a/b, b > 0.
\end{cases}$$
If there is a positive rational in $D$ we can find an interval $A = [s,t]$ contained in $D$ with $s$ rational because $D$ is a bounded interval. Then given $\epsilon = 1$ there exists a positive integer $M$ such that $n \ge M$ implies that
\begin{align}
\biggl|h_n\biggl(\frac{a}{b}\biggr) - h\biggl(\frac{a}{b}\biggr) \biggr| &= \
\biggl| \frac{a}{n} + \frac{a}{b} \biggl( \frac{1}{n} + \frac{1}{n^2} \biggr) \biggr| &< 1
\end{align}
whenever $n \ge$ M and $a/b \in A$ where $a$ and $b$ are positive integers.
But then
\begin{align}
\frac{a}{M} &< \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \
&= \biggl| \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \biggr| \
&< 1.
\end{align}
Writing $s = c/d$ where $s$ is the left endpoint of $A$ and $c$ and $d$ are positive integers, we can choose a positive integer $k$ large enough so that both $(ck + 1)/d \in A$ and $ck + 1 > M$.
This contradicts $(ck + 1)/M < 1$. Therefore the assumption that $h_n \to h$ uniformly is false. A similar argument holds if no positive rationals are in $D$
|
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|
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$ Hello :)
I hit a problem.
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x + \cot x$?
|
HINT.
*
*$\tan{x}+\cot{x} = \frac{1}{\sin{x} \cdot \cos{x}}$
*$(\sin{x}+\cos{x})^{2} -1 = 2 \sin{x} \cdot \cos{x}$
|
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How can I find $\int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$? My question is ; How can I solve the following integral question?
$\displaystyle \int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$
Thanks in advance,
|
You've had some time to study this, so let's look closer at the two evident approaches: (a) conversion to polar coordinates, (b) integrate directly with a standard hyperbolic substitution.
(a) Conversion to polar coordinates: Let
$$I = \int_{1/\sqrt{2}}^1 \, \int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y \, \text{d}x.$$
If you draw a diagram of the situation you will see that
$$I = \int_0^{\pi/4} \, \int_{r=1}^{r=\sec \theta} \frac{1}{r} r \text{d}r \, \text{d} \theta = \int_0^{\pi/4} \sec \theta \, - 1 \textrm{ d} \theta $$
$$= \left[ \frac{1}{2} \log \left| \frac{1+ \sin \theta}{1- \sin \theta} \right| - \theta \right]_0^{\pi/4} = \log(1+\sqrt{2}) - \frac{\pi}{4}.$$
(b) Standard hyperbolic substitution: To evaluate the inner integral set $y= x \sinh u$ and noting that $\sinh^{-1} u = \log(u+ \sqrt{1+u^2})$ we have
$$\int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y =
\left[ \log \left( \frac{y}{x} +
\sqrt{ 1+ \frac{y^2}{x^2} }\right) \right]_{\sqrt{1-x^2}}^x $$
$$= \log(1+ \sqrt{2}) + \log x - \log( 1+ \sqrt{1-x^2}). \quad (1)$$
Both the logs can be integrated by parts. The first is standard, $\int \log x \textrm{ d}x = x \log x - x + C$ and the second
$$ \int \log( 1+ \sqrt{1-x^2} ) \textrm{ d}x = x \log( 1+ \sqrt{1-x^2}) -x + \sin^{-1} x + C.$$
You will need (do the integration) to note that
$$\frac{1}{\sqrt{1-x^2}} - 1 = \frac{x^2}{(1-x^2) + \sqrt{1-x^2}}.$$
And so upon integrating $(1)$ between $1/\sqrt{2}$ and $1$ you obtain
$$I = \left(1 - \frac{1}{\sqrt{2}} \right) \log(1+ \sqrt{2}) +
\left[ x \log x - x\log( 1+ \sqrt{1-x^2}) - \sin^{-1} x \right]_{1/\sqrt{2}}^1$$
$$= \log(1+\sqrt{2}) - \frac{\pi}{4}.$$
|
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|
Approximation For Difference Of Two Sides Of A Triangle I have been trying to derive this approximation but have been unsuccessful in doing so. Any help would be greatly appreciated.
|
The vectors ${\mathbf{a}}$ and ${\mathbf{c}}$, in terms of their magnitudes and the angles $\theta$ and $\gamma$, are
$$
{\mathbf{a}} = (a \sin \theta, a \cos \theta)
$$
and
$$
{\mathbf{c}} = (c \cos \gamma, c \sin \gamma).
$$
Assuming $c \ll a$, we can expand $b-a$ in powers of the small parameter $\varepsilon = c/a$ as follows:
$$
\begin{eqnarray}
b-a &=& -a + \lVert{\mathbf{b}}\rVert \\
&=& -a + \lVert{\mathbf{a}} - {\mathbf{c}}\rVert \\
&=& -a + \sqrt{(\mathbf{a} - \mathbf{c})\cdot(\mathbf{a} - \mathbf{c})} \\
&=& -a + \sqrt{a^2 - 2{\mathbf{a}}\cdot{\mathbf{c}} + c^2} \\
&=& -a + a \sqrt{1 - 2 \varepsilon\left(\sin\theta \cos\gamma + \cos\theta \sin\gamma\right) + \varepsilon^2} \\
&=& -a + a \sqrt{1 - 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2} \\
&=& -a + a \left(1 + \frac{1}{2}\left(- 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2\right) - \frac{1}{8}\left(- 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2\right)^2 + O(\varepsilon^3)\right) \\
&=& a \left(- \varepsilon\sin(\theta+\gamma) + \frac{1}{2}\varepsilon^2 - \frac{1}{2}\varepsilon^2\sin^2(\theta+\gamma) + O(\varepsilon^3)\right) \\
&=& - c\sin(\theta+\gamma) + \frac{c^2}{2a}\cos^2(\theta+\gamma) + O\left(\frac{c^3}{a^2}\right),
\end{eqnarray}
$$
which is the desired result. The series expansion $\sqrt{1+x} = 1 + (1/2)x - (1/8)x^2 + O(x^3)$ was used to approximate the square root.
|
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A geometric look at $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$? Is there a geometric way of looking at the relationship between the positive real numbers $a$, $b$ and $c$ if $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?
|
If you have two poles of length $a$ and $b$, and string wire from the top of one pole to the bottom of the other, and vice versa, the height $c$ at which the wires intersect will be given by half the harmonic mean of the height of the two poles, $\displaystyle \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$, no matter how far apart they are. This can be easily shown using similar triangles.
I can also think of one more place where the harmonic mean is commonly found in a geometric context: Given a right triangle with legs $a$, $b$, the squared height of the altitude to the hypotenuse is given by half the harmonic mean of the squared lengths. Wikipedia provides a handy diagram so I'll just quickly write it out:
(Here $A$, $B$, $C$ are the angles opposite the lowercase sides, as is usual, and $P$ is the point where the altitude intersects the hypotenuse.) Then, since $\bigtriangleup ABC \; \sim \; \; \bigtriangleup CBP \;$,
$$\dfrac{a}{f} = \dfrac{c}{b} \Rightarrow \dfrac{a^2}{f^2} = \dfrac{c^2}{b^2} = \dfrac{a^2 + b^2}{b^2}$$
So, by cross-multiplying we get,
$$\dfrac{1}{f^2} = \dfrac{a^2 + b^2}{a^2b^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$$
half the harmonic mean of the squared lengths.
|
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Limit of algebraic function $\ \lim_{x\to\infty} \sqrt[5]{x^5 - 3x^4 + 17} - x$ How to solve this limit?
$$\lim_{x \to \infty}{\sqrt[5]{x^5 - 3x^4 + 17} - x}$$
|
If you want to solve ab initio, then the way to go about is the way Arturo suggested.
Note that $a^5 - b^5 = (a-b)(a^4 + ba^3+ b^2a^2 + b^3a + b^4)$ and hence $$(a-b)= \frac{a^5 - b^5}{a^4 + ba^3+ b^2a^2 + b^3a + b^4}$$
Now take $a=\sqrt[5]{x^5-3x^4+17}$ and $b=x$
$\sqrt[5]{x^5-3x^4+17}-x = \frac{(x^5-3x^4+17)-x^5}{\left(\sqrt[5] {x^5-3x^4+17} \right)^4 + x \left(\sqrt[5] {x^5-3x^4+17} \right)^3 + x^2 \left(\sqrt[5] {x^5-3x^4+17}\right)^2 + x^3 \left(\sqrt[5] {x^5-3x^4+17}\right)+ x^4}$
Simplifying, we get,
$\sqrt[5]{x^5-3x^4+17}-x = \frac{-3x^4+17}{x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + x^4}$
$\sqrt[5]{x^5-3x^4+17}-x = \frac{-3+\frac{17}{x^4}}{\left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + 1}$
Hence, we have
$$\sqrt[5]{x^5-3x^4+17}-x = \frac{Nr(x)}{Dr(x)}$$ where $Nr(x) = -3+\frac{17}{x^4}$ and $Dr(x) = \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + 1$
$\displaystyle \lim_{x \rightarrow \infty} Nr(x) = -3$ and $\displaystyle \lim_{x \rightarrow \infty} Dr(x) = 5$.
Hence, we have $\displaystyle \lim_{x \rightarrow \infty} Nr(x)$ and $\displaystyle \lim_{x \rightarrow \infty} Dr(x)$ exists as a real number and hence
$$\displaystyle \lim_{x \rightarrow \infty} \sqrt[5]{x^5-3x^4+17}-x = \frac{\displaystyle \lim_{x \rightarrow \infty} Nr(x)}{\displaystyle \lim_{x \rightarrow \infty} Dr(x)} = \frac{-3}{5} = - \frac{3}{5}$$
|
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|
Help solving another integral $\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$ $$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$$
Complete the square?
$$\int \frac{1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$$
Not sure what do do next, or if I should try something else?
Big help if you can show as "step-by-step" possible as you can.
Thanks in advance!
@Chandru1:
I'm confused at what happens to the 2 that gets factored out when completing the square (the 2 in the denominator) because it looks like Arturo left it out and instead substituted:
$\int \frac {1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$
$u = x+1$
$du =dx$
which changes the integral to $\int \frac {1}{(2(u^2-4))^\frac{3}{2}}$
Would I be best factoring it out as a 1/2 in front of the integral or leaving it in and then using trig. substitution like this?
$u= \sqrt{2}sec(t)$
$du = \sqrt{2}sec(t)*tan(t) dt$
Which changes it to:
$\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(2sec^2(t)-4))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(tan^2(t)))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2tan^3(t))} \ dt$
My $\sqrt{2}$ factors out like yours, but I have no 1/8 so figure I factor it out as a 1/2 earlier like I thought? But then the constant in front of my integral would be $\frac{\sqrt{2}}{2}$ and you have $\frac{\sqrt{2}}{8}$
I know what's wrong, just not how I went wrong.
|
First, factor out a $2$:
$$(2x^2 + 4x - 2) = 2(x^2+2x-1)\quad\text{so}\quad (2x^2+4x-2)^{-3/2} = 2^{-3/2}(x^2+2x-1)^{-3/2}.$$
Do this because you can pull it out of the integral and it will make things easier.
So, up to a constant, this is the same as doing the integral
$$\frac{1}{2^{3/2}}\int\frac{dx}{(x^2+2x-1)^{3/2}}.$$
Completing the square is a good first step. $x^2+2x-1 = (x^2+2x+1) - 2 = (x+1)^2 - 2$. So we get the integral
$$\frac{1}{2^{3/2}}\int\frac{dx}{\bigl( (x+1)^2 - 2\bigr)^{3/2}}.$$
Doing a change of variable $u=x+1$ changes it to
$$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}}.$$
Now try a trigonometric substitution to get rid of that pesky square root in the exponent. Use $\tan^2\theta + 1 = \sec^2\theta$, or $\sec^2\theta - 1 =\tan^2\theta$. So set $u = \sqrt{2}\sec\theta$ to get
$$u^2-2 = 2\sec^2\theta - 2 = 2(\sec^2\theta - 1) = 2\tan^2\theta$$
and $du = \sqrt{2}\sec\theta\tan\theta\,d\theta$, so
$$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{(2\tan^2\theta)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{2^{3/2}\tan^3\theta}.$$
Now work the trigonometric integral.
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|
Using generating functions to find a formula for the Tower of Hanoi numbers So the Tower of Hanoi numbers are given by the recurrence $h_n=2h_{n-1}+1$ and $h_1=1$. I let my generating function be
$$
g(x)=\sum h_nx^n
$$
Then
$$
g(x)=\sum h_n x^n=\sum (2h_{n-1}+1)x^n=\sum 2h_{n-1}x^n+\sum x^n=2xg(x)+\frac{1}{1-x}.
$$
Solving for $g(x)$ I find
$$
g(x)=\frac{1}{(1-2x)(1-x)}=\frac{2}{1-2x}-\frac{1}{1-x}=2\sum (2x)^n-\sum x^n.
$$
It seems then that the coefficient $h_n$ of $x^n$ is $2^{n+1}-1$, but wolfram mathworld says it should be $h_n=2^n-1$. What did I do wrong here? Thanks.
|
Setting $$g(x) = \sum_{n=0}^{\infty} h_{n+1} x^n,$$ we obtain using $h_n = 2 h_{n-1} +1$ and $h_1 =1$
$$g(x) = \sum_{n=0}^{\infty} h_{n+1} x^n = h_1 + \sum_{n=1}^{\infty} (2 h_{n} +1) x^n = 1 + \sum_{n=0}^{\infty} (2 h_{n+1} +1) x^{n+1} =1 +2 x g(x) + \frac{x}{1-x}.$$
Solving for $g(x)$, we obtain $$g(x) = \frac{1}{(1-x) (1-2x)} = \frac{2}{1-2x} - \frac{1}{1-x} = \sum_{n=0}^{\infty} (2^{n+1} -1) x^n$$ such that $h_n = 2^n -1$.
|
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|
Finding the distance between two triangles, one inside the other I have two right triangles One is a $6$-$8$-$10$ and inside is a $3$-$4$-$5$ and the space between the two triangles is a uniform amount.
I made a really awkward and weird pic of the diagram and I need to solve for $X$
How would I go about solving this? I couldn't figure out any good approaches.
|
Draw lines connecting the "respective vertices". Add up the areas to solve for $x$.
Area of a right triangle with the non-hypotenuse sides being $a$ and $b$ is $\frac{1}{2}(a \times b)$ while the area of a trapezium with parallel sides being $a$ and $b$ and the height being $h$ is $\frac{1}{2}h(a+b)$
$$\frac{1}{2}x(5+10) + \frac{1}{2}x(3+6) + \frac{1}{2}x(4+8) + \frac{1}{2}(3 \times 4) = \frac{1}{2} (6 \times 8)$$
$$36x + 12 = 48 \Rightarrow x = 1$$
EDIT:
Let us try to look at a slightly general case. Take a triangle and scale it to another similar triangle with the scale factor being $t$.
Let the sides of the inner triangle be $a$, $b$ and $c$.
The perimeter and area of the inner triangle is $P$ and $A$ respectively.
The sides of the outer triangle are $ta$, $tb$ and $tc$ while the perimeter and area are $tP$ and $t^2A$.
As before join the "respective" vertices and summing the areas give us,
$$A + \frac{1}{2}x(ta + a) + \frac{1}{2}x(tb + b) + \frac{1}{2}x(tc + c) = t^2A$$
$$A + \frac{1}{2}x(t+1)P = t^2A \Rightarrow \frac{1}{2}x(t+1)P = (t^2-1)A$$
$$x = \frac{2A}{P}(t-1)$$
In the problem asked, $t=2$ with $P = 2A = 12$ and hence we get $x=1$.
Also, $t=1$ gives $x=0$ as expected.
This also gives a nice proof that the radius of the incircle of a triangle is $$r_{in} = \frac{2A}{P}$$
This is got by plugging in $t=0$ and realizing that $\left| x \right|$ is nothing but the radius of the incircle.
Hence, $$x = (t-1)r_{in}$$
|
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|
Trivial: Rationalize fraction with a third-degree root This is a pretty trivial question. How do you rationalize a function with a denominator that contains a third degree root?
Edit: My expression is $\displaystyle{\frac{1}{\sqrt[3]{2}-1}}$.
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To rationalize when you have a denominator of the form $a^{1/3}-b^{1/3}$ (as you do here, with $a=2$ and $b=1$), use the identity
$x^3-y^3 = (x-y)(x^2+xy+y^2)$.
So
$$2-1 = \left(\sqrt[3]{2}-\sqrt[3]{1}\right)\left(\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}\right) = \left(\sqrt[3]{2} - 1\right)\left(\sqrt[3]{4}+\sqrt[3]{2} + 1\right).$$
So, multiply both numerator and denominator by $\sqrt[3]{4}+\sqrt[3]{2}+1$. You get:
$$\begin{align*}
\frac{1}{\sqrt[3]{2}-1} &= \frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}\\
&= \frac{\sqrt[3]{4} + \sqrt[3]{2}+1}{(\sqrt[3]{2})^3 - 1^3)}\\
&= \frac{\sqrt[3]{4} + \sqrt[3]{2}+1}{2-1}\\
&= \sqrt[3]{4} + \sqrt[3]{2} + 1.
\end{align*}$$
If you had a denominator of the form $a^{1/3}+b^{1/3}$, you can instead use the identity $(x^3+y^3) = (x+y)(x^2-xy+y^2)$.
|
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|
How to prove $\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor$? everybody, how can I prove that, for natural $m$ and $n$,
$$
\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor \; ?
$$
Thanks a lot.
|
The claim is false for real $n$ and $m$. Take $n = \pi$ and $m = -3$ for a counter-example. It isn't even true for integral $n$ and $m$. Take $n = 3$ and $m = -3$ for a counter-example.
Suppose $n$ and $m$ are positive integers. Let $\chi_{\pm,X}(x)$ denote the characteristic function of $X_{\pm}$ defined to be $1$ if $x \in X_{\pm}$ and $0$ otherwise. Write
\begin{align}
\lceil \tfrac{n}{m} \rceil = \tfrac{n}{m} - \lbrace \tfrac{n}{m} \rbrace + \chi_{+, \mathbb{R} \setminus \mathbb{Z}}(\tfrac{n}{m})
\end{align}
and, similarly,
\begin{align}
\lfloor \tfrac{n + m - 1}{m} \rfloor = \tfrac{n}{m} + 1 - \tfrac{1}{m} - \lbrace \tfrac{n-1}{m} \rbrace
\end{align}
by the $1$-periodicity of the fractional part function. If the claim is to be true, then it must be that
\begin{align}
\lbrace \tfrac{n}{m} \rbrace - \lbrace \tfrac{n - 1}{m} \rbrace = \tfrac{1}{m} + \chi_{+, \mathbb{R} \setminus \mathbb{Z}}(\tfrac{n}{m}) - 1.
\end{align}
which is indeed an identity. The extra factors handle the case when $m$ divides $n$, in which case
\begin{align}
\lbrace \tfrac{n}{m} \rbrace - \lbrace \tfrac{n - 1}{m} \rbrace = \tfrac{1}{m} + 0 - 1.
\end{align}
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|
Solving indetermination in limit I'm trying to solve this limit, but I can't get it out of a $\frac{0}{0}$ indetermination:
$$\displaystyle \lim_{x \to 4} \; \frac{x-4}{5-\sqrt{x^2+9}}$$
Maybe there is something I'm missing. Thanks a lot in advanced.
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Assuming a = numerator conjugated and b = denominator conjugated. You can simplify this by doing the following:
$$
\frac{numerator}{denominator} \times \frac{a}{b} \times \frac{b}{a}
$$
Applying this in your case
$$
\lim_ {x \to 4} \frac{x - 4}{5 - \sqrt{x^2 + 9}} =
$$
$$
\lim_ {x \to 4} \frac{x - 4}{5 - \sqrt{x^2 + 9}} \times \frac{x + 4}{5 + \sqrt{x^2 + 9}} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =
$$
$$
\lim_ {x \to 4} \frac{x^2 - 16}{25 - (x^2 + 9)} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} = $$
$$
\lim_ {x \to 4} \frac{x^2 - 16}{-(x^2 - 16)} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =
$$
$$
\lim_ {x \to 4} \frac{1}{-1} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =
$$
$$
\lim_ {x \to 4} -1 \times \frac{5 + \sqrt{x^2 + 9}}{x + 4}
$$
Now you can apply the Direct Substitution Property
$$
\lim_ {x \to 4} -1 \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} = -1 \times \frac{5 + \sqrt{4^2 + 9}}{4 + 4} = -1 \times \frac{10}{8} = - \frac{5}{4}
$$
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|
How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$? Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$
Apart from induction, I tried with Wolfram Alpha to check the validity, but I can't think of an easy (manual) alternative.
Please suggest an intuitive/easy method.
|
Note that
$$\begin{align*}\frac{(n+1)\binom{n}{r}}{r+1} &= \frac{(n+1)n!}{(r+1)r!(n-r)!} = \frac{(n+1)!}{(r+1)!(n-r)!}\\ &= \frac{(n+1)!}{(r+1)!((n+1)-(r+1))!} = \binom{n+1}{r+1}.\end{align*}$$
Therefore,
$$\begin{align*}
(n+1)\sum_{r=0}^{n}(-1)^r\frac{\binom{n}{r}}{r+1} &= \sum_{r=0}^{n}(-1)^r\frac{(n+1)\binom{n}{r}}{r+1}\\
&= \sum_{r=0}^{n}(-1)^r\binom{n+1}{r+1}\\
&= -\sum_{r=0}^n(-1)^{r+1}\binom{n+1}{r+1}\\
&= -\sum_{s=1}^{n+1}(-1)^s\binom{n+1}{s}\qquad(\text{setting }s=r+1)\\
&= \left(-\sum_{s=0}^{n+1}(-1)^s\binom{n+1}{s}\right) + (-1)^0\binom{n+1}{0}\\
&= -(1-1)^{n+1} + 1\\
&= 1.
\end{align*}$$
Dividing through by $n+1$ gives the desired result.
Once you realize that
$$\frac{(n+1)\binom{n}{r}}{r+1} = \binom{n+1}{r+1},$$ it should be obvious that you are dealing with a binomial expansion of some $(n+1)$st power. Then it's just a matter of figuring out which power, and whether any terms are missing.
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|
Determinant of Abstract Matrix Given an $n \times n$ matrix $A$, where $x$ is any real number:
$A = \left[
\begin{array}{ c c c c c c c c }
1 & 1 & 1 & 1 & 1 & 1 & \cdots & 1 \\
1 & x & x & x & x & x & \cdots & x \\
1 & x & 2x & 2x & 2x & 2x & \cdots & 2x \\
1 & x & 2x & 3x & 3x & 3x & \cdots & 3x \\
1 & x & 2x & 3x & 4x & 4x & \cdots & 4x \\
1 & x & 2x & 3x & 4x & 5x & \cdots & 5x \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & x & 2x & 3x & 4x & 5x & \cdots & (n-1)x
\end{array} \right]$
Find the determinant.
By using $n=2,3,4,5,...$ and random $x=1,2,3,4...$ I have found that $det(A) = (x-1)(x)^{n-2}$ through observing a pattern.
However, I would like to be able to prove this through a proof, yet I have no idea where to start.
When I try to solve for the determinant using the abstract matrix A and using the property that the determinant of a square matrix is $(-1)^r * (\text{products of pivots})$, where r is the number of row interchanges, my answer is of the form $(x-1)(x)(x)(x)(x)...(n-?)x$ where "?" depends on how many rows I include in the abstract form of A. How do I show that $(x)(x)(x)...(n-?)x$ equals $(x)^{n-2}$?
Here is my work: http://i.imgur.com/tinDw.jpg
Any hints? Thanks for the help!
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I think what you did is almost perfectly correct. In any case, here is another way that uses induction. Let $A_n$ refer to the $n\times n$ matrix. The base case of the $2\times 2$ matrix is $\det A_2 =(x-1)$.
Now, for the inductive step lets look at $A_{n+1}$ and use the linearity of the determinant on the last entry of the last row. We have that $$\det\left[\begin{array}{ccccc}
1 & 1 & \cdots & 1 & 1\\
1 & x & \cdots & x & x\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
1 & x & \cdots & (n-1)x & (n-1)x\\
1 & x & \cdots & (n-1)x & nx\end{array}\right]=$$ $$\det\left[\begin{array}{ccccc}
1 & 1 & \cdots & 1 & 1\\
1 & x & \cdots & x & x\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
1 & x & \cdots & (n-1)x & (n-1)x\\
1 & x & \cdots & (n-1)x & (n-1)x\end{array}\right]+\det\left[\begin{array}{ccccc}
1 & 1 & \cdots & 1 & 1\\
1 & x & \cdots & x & x\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
1 & x & \cdots & (n-1)x & (n-1)x\\
0 & 0 & \cdots & 0 & x\end{array}\right]$$ The first determinant will be zero since two rows are the same. The second is $x\det A_n$.
Hope that helps,
|
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|
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
|
I will refer to Qiaochu's excellent answer here as proof that if we define
$$f(N):=\sum\limits_{n=0}^N n^2$$
then $f$ is a polynomial of degree $3$.
It is easy to calculate the first few values of this sum. Namely,
$\begin{align}
f(0) &= 0 \\
f(1) &= 1 \\
f(2) &= 5 \\
f(3) &= 14
\end{align}$
I claim that these four points are sufficient to uniquely determine $f$.
Since $$\deg f = 3$$, we have in general that
$$f(x)=\sum\limits_{k=0}^3 c_kx^k$$
which when be combined with the four computed values above results in the following system of equations:
$$\begin{pmatrix}
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 \\
1 & 2 & 4 & 8 \\
1 & 3 & 9 & 27
\end{pmatrix}
\begin{pmatrix}
c_0 \\ c_1 \\ c_2 \\ c_3
\end{pmatrix}
=
\begin{pmatrix}
0 \\ 1 \\ 5 \\ 14
\end{pmatrix}$$
Note that $c_0 = 0$ trivially, and so we can help ourselves by writing the reduced matrix equation for the other three coefficients as
$$\begin{pmatrix}
1 & 1 & 1 \\
2 & 4 & 8 \\
3 & 9 & 27
\end{pmatrix}
\begin{pmatrix}
c_1 \\ c_2 \\ c_3
\end{pmatrix}
=
\begin{pmatrix}
1 \\ 5 \\ 14
\end{pmatrix}$$
Call the above matrix $V$. This matrix is a Vandermonde matrix which has a well-known determinant
$$\begin{align}
\det(V) &= 1\cdot 2\cdot 3\cdot(2-1)(3-1)(3-2) \\
&\neq 0
\end{align}$$
Because its determinant is nonzero, the matrix is invertible, and so we have
$$\begin{pmatrix}
c_1 \\ c_2 \\ c_3
\end{pmatrix}
=
V^{-1}\cdot\begin{pmatrix}
1 \\ 5 \\ 14
\end{pmatrix}$$
from which $f(x)$ can be determined directly.
However, if you're like most people, inverting matrices doesn't exactly tickle your fancy!
Luckily, now that we see that the interpolating cubic is unique, we could find it through the described matrix multiplication, but we would get to the same result if we proceeded a different route as well. This is where Lagrange polynomials come to the rescue.
Using the general formula, we have immediately that
$$\begin{align}
f(x) &= 0\cdot(\dots)+1\cdot\frac{x(x-2)(x-3)}{1(1-2)(1-3)}+5\cdot\frac{x(x-1)(x-3)}{2(2-1)(2-3)}+14\cdot\frac{x(x-1)(x-2)}{3(3-1)(3-2)} \\
&= \frac{1}{2}\left(x^3-5x^2+6x\right)-\frac{5}{2}\left(x^3-4x^2+3x\right)+\frac{14}{6}\left(x^3-3x^2+2x\right) \\
&= \frac{1}{6}\left(2x^3+3x^2+x\right) \\
&= \frac{1}{6}x\left(x+1\right)\left(2x+1\right)
\end{align}$$
You can generalize this approach to find expressions for $\sum n^p\quad\forall p\in\mathbb{N}$.
Or, you know, there's always Faulhaber's formula.
|
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|
Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$? Given a series, how does one calculate that limit below? I noticed the numerator is an arithmetic progression and the denominator is a geometric progression — if that's of any relevance —, but I still don't know how to solve it.
$$\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$$
I did it "by hand" and the result should be $\frac{9}{4}.$
|
For summing $\sum\frac{k}{3^k}$ use the following formula: $$(1-x)^{-2} = 1 + 2x + 3x^{2} + 4x^{3} + \cdots \qquad \Bigl[\because \small (1-x)^{-n} = 1+nx +\frac{n\cdot (n-1)}{2!}\cdot x^{2} + \cdots \Bigr]$$ Multiplying the above equation by $x$ and then putting $x=\frac{1}{3}$ we have $$\frac{1}{3} + \frac{2}{9}+\frac{3}{27} + \frac{4}{3^{4}} + \cdots = \frac{1}{3}\Bigl(1-\frac{1}{3}\Bigr)^{-2} = \frac{3}{4} \qquad\quad \cdots (1)$$
Also you know that $$\sum\limits_{k=1}^{\infty} \frac{1}{3^k}= \frac{\frac{1}{3}}{1-\frac{1}{3}} =\frac{1}{2} \qquad\qquad \cdots (2)$$
Add equations $(1)$ and $(2)$ to get your answer.
|
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$x^4+4$ is composite for $x>1$
$x^4+4$ is composite for $x>1$
I know the Sophie Germain indentity and the get the factorization $$x^4+4 = (x^2+2-2x)(x^2+2+2x)$$
But I am stuck here. I cannot see any general factor here.
|
HINT $\ \ x\:(x+2)+2\ \ge\ x\:(x-2)+2\ \ge\ 2\ $ for $\ x\ge 2$
You may find of interest that this is a special case of a class of cyclotomic factorizations due to Aurifeuille, Le Lasseur and Lucas, the so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:
$$\begin{array}{rl}
x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\frac{x^6 + 3^3}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{array}$$
|
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|
The integral $\int_0^8 \sqrt{x^4+4x^2}\,dx$
$\displaystyle \int\nolimits_0^8 \sqrt{x^4+4x^2}\,dx$.
Alright, so I thought I had this figured out. Here's what I did:
*
*I factor out an $x^2$ to get $\sqrt{x^2(x^2+4)}$.
*I let $x = 2\tan(\theta)$, therefore the integrand is $\sqrt{4\tan^2(\theta) (4\tan^2(\theta) + 4)}$.
*Factor out a 4 and it becomes $\sqrt{(16\tan^2(\theta) (\tan^2(\theta) + 1))}$
*Which equals $\sqrt{16\tan^2(\theta) \sec^2(\theta)}$
*This is easy to take the sqrt of. The integrand becomes $4\tan(\theta)\sec(\theta)$.
*Now, the integral of this is $4\sec(\theta)$
*And it's evaluated from $0$ to $\arctan(4)$ right? Because as $x$ goes to $0$, so does $\theta$, and as $x$ goes to $8$, $\theta$ goes to $\arctan(4)$...
*But the end result $(4 (\sec(\arctan(4)) - 1) )$ isn't the correct answer
I put it into WolframAlpha and I get $(8/3) (17\sqrt{17} - 1)$, which is the right answer. How did they get that? (there's no "show steps" option)
Any help is greatly appreciated!
PS, what's the syntax for doing sqrts and exponentials?
|
There are other ways of doing this integral, but let me try to fix your attempt, which is certainly a fine idea as far as it goes.
The main problem I spot with your development is that you forgot to change the $dx$ when you did the change of variable. (And you should be able to evaluate $\sec(\arctan a)$ as well; we'll get to that shortly).
So: you start with
$$\int_0^8 \sqrt{x^4+4x^2}\,dx = \int_0^8 \sqrt{x^2(x^2+4)}\,dx.$$
Then you do the change of variable $x=2\tan(\theta)$. If you do this, then
$$dx = 2\sec^2\theta\,d\theta;$$
when $x=0$, you want $\theta=0$, and when $x=8$ you want $\theta=\arctan(4)$ (you are correct there). So the integral actually becomes, after changing integrand, limits, and the $dx$:
$$\begin{align*}
\int_0^8\sqrt{x^2(x^2+4)}\,dx &= \int_0^{\arctan(4)}\sqrt{4\tan^2\theta(4\tan^2\theta+4)}2\sec^2\theta\,d\theta\\
&= \int_0^{\arctan(4)} \sqrt{16\tan^2\theta(\tan^2\theta+1)}2\sec^2\theta\,d\theta\\
&= \int_0^{\arctan(4)}8\sec^2\theta\sqrt{\tan^2\theta\sec^2\theta}\,d\theta\\
&= 8\int_0^{\arctan(4)}\sec^2\theta|\tan\theta\sec\theta|\,d\theta.
\end{align*}$$
Now, on $[0,\arctan(4)]$, both tangent and secant are positive, so we can drop the absolute value signs (something else you were not careful with), and the integral becomes
$$8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta.$$
Set $u=\sec\theta$. Then $du=\sec\theta\tan\theta$, so we have
$$\begin{align*}
8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta &= 8\int_{\sec(0)}^{\sec(\arctan(4))}u^2\,du\\
&= \frac{8}{3}u^3\Biggm|_{\sec(0)}^{\sec(\arctan(4))}\\
&=\frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right).
\end{align*}$$
Now, $\sec(0) = 1$. What about $\sec(\arctan(4))$?
Say $\psi$ is an angle with $\tan(\psi)=4$. Take a right triangle with this angle; by scaling, we may assume the opposite side has length $4$ and the adjacent side has length $1$. Then the hypotenuse has length $\sqrt{17}$, so the cosine of $\psi$ is $\frac{1}{\sqrt{17}}$, hence the secant has value $\sqrt{17}$. So $\sec(\arctan(4)) = \sec(\psi) = \sqrt{17}$. Thus, the integral is:
$$\begin{align*}
\int_0^8\sqrt{x^4+4x^2}\,dx &= \frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right)\\
&=\frac{8}{3}\left( \sqrt{17}^3 - 1^3\right)\\
&= \frac{8}{3}\left(17\sqrt{17} - 1\right).
\end{align*}$$
In summary: your mistake was that when you did the change of variable, you forgot to change the differential as well; and at the end you could have simplified $\sec(\arctan(4))$.
Of course, the better way of doing this is to factor out $x$ from the square root, and then recognize that you can do
$$\int_0^8\sqrt{x^4+4x^2}\,dx = \int_0^8x\sqrt{x^2+4}\,dx$$
with the change of variable $u=x^2+4$, like DJC suggested. But I thought you might like to know where exactly your approach went wrong (the substitution), and whether it could be brought to a correct conclusion (it could).
|
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|
Finding the smallest positive integer a Can we find the smallest positive integer $a$ such that $1971|50^n+a.23^n$ where n is odd?
Source:Problem Solving Strategies by Arthur Engel
|
Since $50^2\equiv23^2\equiv529\pmod{1971}$ and $(529,1971)=1$, we have
$$
\begin{align}
50^{2n+1}+a\cdot23^{2n+1}&\equiv0\pmod{1971}\\
50\cdot529^n+a\cdot23\cdot529^n&\equiv0\pmod{1971}\\
50+a\cdot23&\equiv0\pmod{1971}
\end{align}
$$
Using the Euclid-Wallis Algorithm
$$
\begin{array}{r}
&&85&1&2&3&2\\\hline
1&0&1&-1&3&-10&23\\
0&1&-85&86&-257&857&-1971\\
1971&23&16&7&2&1&0\\
\end{array}
$$
we get that $857\cdot23\equiv1\pmod{1971}$. Therefore,
$$
a\equiv-50\cdot857\equiv512\pmod{1971}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
trigonometric system In order to show that $ e^{ix}+e^{iy}+e^{iz}=0 \Longrightarrow e^{2ix}+e^{2iy}+e^{2iz}=0 $, I want to prove that $ \cos x+\cos y+\cos z=0 $ and $ \sin x+\sin y+\sin z=0 \Longrightarrow \cos 2x+\cos 2y+\cos 2z=0$ and $ \sin 2x+\sin 2y+\sin 2z=0 $
$ \cos 2x=2\cos^2 x-1=2(\cos y+\cos z)^2-1 $
$ \cos 2x+\cos 2y+\cos 2z=2(\cos^2x+\cos^2y+\cos^2z+$
$(\cos^2x+cos^2y+\cos^2z+2(\cos x\cos y+\cos y\cos z+\cos x\cos z)))-3 $
$ \cos 2x+\cos 2y+\cos 2z=2(\cos^2x+\cos^2y+\cos^2z)-3 =3-2(\sin^2x+\sin^2y+\sin^2z)$ ...
Any idea?
|
You can prove it with pure complex number manipulations itself (and if needed a "pure trigonometric" proof can be read off from that).
Hint:
If $z_1 + z_2 + z_3 = 0$ then $\overline{z_1} + \overline{z_2} + \overline{z_3} = 0$ and if $|z| = 1$, then $\overline{z} = \frac{1}{z}$.
Now try squaring something and use the above two facts.
Since OP seems to ignore this, will forget about giving further hints. For the sake of completeness:
Spoiler
If $z_1 = e^{ix}$, $z_2 = e^{iy}$ and $z_3 = e^{iz}$, then we have $z_1 + z_2 + z_3 = 0$. We also have $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} = 0$ which gives $z_1 z_2 + z_2 z_3 + z_3 z_1 = 0$, multiplying by $z_1z_2z_3$. Squaring the first gives $z_1^2 + z_2^2 + z_3^2 + 2 (z_1 z_2 + z_2 z_3 + z_3 z_1) = 0$, which implies $z_1^2 + z_2^2 + z_3^2 = 0$ which is what you wanted to prove.
|
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|
Optimizing $a+b+c$ subject to $a^2 + b^2 + c^2 = 27$
If $a,b,c \gt 0$ and $a^2+b^2+c^2=27$, find the maximum and minimum values of $a+b+c$.
How to solve this one?
(Here's the source of inspiration for the problem.)
|
Here's a geometric way of looking at it... the points where $x,y,z > 0$ where $x^2 + y^2 + z^2 = 27$ is the "upper right" 1/8 of the outside of the sphere centered at the origin, of radius $3\sqrt{3}$. Call this surface $S$. If you connect the corners $(3\sqrt{3},0,0)$, $(0,3\sqrt{3},0)$, and $(0,0,3\sqrt{3})$ of $S$, you get a triangle in the plane with equation $x + y + z = 3\sqrt{3}$. If you replace $3\sqrt{3}$ by any $r < 3\sqrt{3}$, then the plane $x + y + z = r$ will no longer intersect $S$. So the minimum is $3\sqrt{3}$.
Similarly, the plane $x + y + z = 9$ intersects $S$ at $(3,3,3)$ only, and replacing $9$ by any $r > 9$ will result in a plane $x + y + z = r$ that doesn't intersect $S$ at all. So the maximum is $9$.
|
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|
Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
|
Let P(n) be the given statement. You'll see why in the following step.
$$P(n):1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$
Step 1. Let $n = 1$.
Then $\mathrm{LHS} = 1^3 = 1$, $\mathrm{RHS} = \frac{1^2(1+1)^2}{4} = \frac{4}{4} = 1
$.
So LHS = RHS, and this means P(1) is true!
Step 2. Let $P(n)$ be true for $n = k$; that is,
$$1^3 + 2^3 + 3^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4}$$
We shall show that $P(k+1)$ is true too!
Add $(k+1)^3$, which is $(k+1)^{\mathrm th}$ term of the LHS to both sides of (1); then we get:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
&= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\
&= \frac{(k+1)^2(k+2)^2}{4}.\\
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{(k+1)^2(k+2)^2}{4}
\end{align*}$$
I think this statement is the same as $P(n)$ with $n = k+1$.
|
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|
Simplify with no calculator $\dfrac{(8^3)(-16)^5}{4(-2)^8}$
$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{4\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$
$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$
$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$
is it equal to... ?
$$
\dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2}
$$
I am bit confused, How can I handle this problem ?
|
Hint: Just take everything to powers of 2 using the laws of exponents
$$\frac{8^3(-16)^5}{4(-2)^8}=\frac{-(2^3)^3(2^4)^5}{2^2\cdot2^8}=?$$
|
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|
Solving the equation $x + \sqrt{2x+1} = 7$ I can't solve
$x + \sqrt{2x+1} = 7$.
Well, I know the answer is 4, but that is from just reasoning it out. I can't algebraically solve it.
Thus, a step by step is what I really need.
Thanks in advance!!
|
$7-x= \sqrt{2x+1}$
$ (7-x)^2 = 2x+1$
$ x^2-14x+49 = 2x+1 $
$x^2-16x+48=0$
$(x-8)^2-64+48=0$
$(x-8)^2=16$
$x-8=\pm 4$
$x=12$ or $x=4$
But $x=12$ does not work in the original equation. So the answer is $x=4$.
(Or the original equation requires $7-x\ge 0$ and so $x\le 7$.)
|
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|
If $X, Y \sim N(0,1)$, find the CDF of $\alpha X + \beta Y$
Possible Duplicate:
Proof that the sum of two Gaussian variables is another Gaussian
Let $X,Y$ be independent normally distributed $N(0,1)$ random variable, and $\alpha,\beta\in \mathbb{R}$. What is the cumulative distribution function of $\alpha X+\beta Y$?
Thank you very much for your help.
|
It looks from your comment as if the meaning of your question is different from what I thought at first. My first answer assumed you knew that the sum of independent normals is itself normal.
You have
$$
\exp\left(-\frac12 \left(\frac{x}{\alpha}\right)^2 \right) \exp\left(-\frac12 \left(\frac{z-x}{\beta}\right)^2 \right)
= \exp\left(-\frac12 \left( \frac{\beta^2x^2 + \alpha^2(z-x)^2}{\alpha^2\beta^2} \right) \right).
$$
Then the numerator is
$$
\begin{align}
& (\alpha^2+\beta^2)x^2 - 2\alpha^2 xz + \alpha^2 z^2 \\ \\
& = (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz\right) + \alpha^2 z^2 \\ \\
& = (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz + \frac{\alpha^4}{(\alpha^2+\beta^2)^2}z^2\right) + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2 \\ \\
& = (\alpha^2+\beta^2)\left(x - \frac{\alpha^2}{\alpha^2+\beta^2}z\right)^2 + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2,
\end{align}
$$
and then remember that you still have the $-1/2$ and the $\alpha^2\beta^2$ in the denominator, all inside the "exp" function.
(What was done above is completing the square.)
The factor of $\exp\left(\text{a function of }z\right)$ does not depend on $x$ and so is a "constant" that can be pulled out of the integral.
The remaining integral does not depend on "$z$" for a reason we will see below, and thus becomes part of the normalizing constant.
If $f$ is any probability density function, then
$$
\int_{-\infty}^\infty f(x - \text{something}) \; dx
$$
does not depend on "something", because one may write $u=x-\text{something}$ and then $du=dx$, and the bounds of integration are still $-\infty$ and $+\infty$, so the integral is equal to $1$.
Now look at
$$
\alpha^2z^2 - \frac{\alpha^4}{\alpha^2+\beta^2} z^2 = \frac{z^2}{\frac{1}{\beta^2} + \frac{1}{\alpha^2}}.
$$
This was to be divided by $\alpha^2\beta^2$, yielding
$$
\frac{z^2}{\alpha^2+\beta^2}=\left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2.
$$
So the density is
$$
(\text{constant})\cdot \exp\left( -\frac12 \left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2 \right) .
$$
Where the standard deviation belongs we now have $\sqrt{\alpha^2+\beta^2}$.
|
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|
The product rule with square roots I am suppose to find the derivative of $H(u) = (u - \sqrt{u})(u - \sqrt{u})$
I know the formula is the derivative of the second function times the first function plus the derivative of the first function times the second function.
I know that it will be $(1-(1/2) u^{-1/2})(1-(1/2) u ^{-1/2})$ I am pretty certain my problem comes from trying to multiply the $(1/2)u^{-1/2}$ term with itself. What exactly happens? I am getting$(1/2)u^{-1/2}$ I am pretty bad at math so I probably made some simple error. All together I get that the derivative is $2u - u^{-1/2} +1$
|
If you are going to use the Product Rule, you have
$$\begin{align*}
H'(u) &= \left(u-\sqrt{u}\right)'\left(u-\sqrt{u}\right) + \left(u-\sqrt{u}\right)\left(u-\sqrt{u}\right)'\\
&= \left(u - u^{1/2}\right)'\left(u-u^{1/2}\right) + \left(u-u^{1/2}\right)\left(u-u^{1/2}\right)'\\
&= \left( 1 - \frac{1}{2}u^{-1/2}\right)\left(u - u^{1/2}\right) + \left(u-u^{1/2}\right)\left(1 - \frac{1}{2}u^{-1/2}\right)\\
&= 2\left(u - u^{1/2}\right)\left(1 - \frac{1}{2}u^{-1/2}\right).
\end{align*}$$
The first step is the Product Rule. The second step is just the fact that $\sqrt{u}=u^{1/2}$. The third step uses the Sum Rule and the Power Rule. The fourth and final step is just the fact that the two summands are equal.
If you want to further multiply out the product, we have:
$$\begin{align*}
H'(u) &= 2\left(u - u^{1/2}\right)\left(1 - \frac{1}{2}u^{-1/2}\right)\\
&= \left(u - u^{1/2}\right)\left(2 - u^{-1/2}\right)\\
&= 2u - uu^{-1/2} - 2u^{1/2} + u^{1/2}u^{-1/2}\\
&= 2u - u^{1/2} -2u^{1/2}+1\\
&= 2u-3u^{1/2} + 1.
\end{align*}$$
For the function you identify in the comments as the "correct one":
$$H(u) = (u-\sqrt{u})(u+\sqrt{u})$$
assuming you want to exercise the Product Rule, we have:
$$\begin{align*}
H'(u_) &= \left( u - \sqrt{u}\right)'\left(u+\sqrt{u}\right) + \left(u-\sqrt{u}\right)\left(u+\sqrt{u}\right)'\\
&= \left( u - u^{1/2}\right)'\left(u+u^{1/2}\right) + \left(u-u^{1/2}\right)\left(u+u^{1/2}\right)'\\
&=\left(1 - \frac{1}{2}u^{-1/2}\right)\left(u+u^{1/2}\right) + \left(u - u^{1/2}\right)\left(1 + \frac{1}{2}u^{-1/2}\right)\\
&= u+u^{1/2}-\frac{1}{2}u^{-1/2}u - \frac{1}{2}u^{-1/2}u^{1/2} + u + \frac{1}{2}uu^{-1/2} - u^{1/2} - \frac{1}{2}u^{1/2}u^{-1/2}\\
&= u + u^{1/2} - \frac{1}{2}u^{1/2} - \frac{1}{2} + u + \frac{1}{2}u^{1/2} - u^{1/2}-\frac{1}{2}\\
&= 2u-1.
\end{align*}$$
You can verify this is correct, since
$$(u-\sqrt{u})(u+\sqrt{u}) = u^2 - u,$$
and
$$(u^2-u)' = 2u-1.$$
|
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|
Inverse Laplace Transform -s domain How can I find the inverse Laplace transforms of the following function?
$$ G\left(s\right)=\frac{2(s+1)}{s(s^2+s+2)} $$
I solved so far. After that, how do I do?
$$ \frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}=G\left( s \right)$$
|
Your function $$ G(s) = \frac{2(s+1)}{s(s^2 + s + 2)} $$ has the partial fraction decomposition $$ G(s) = \frac{A}{s} + \frac{Bs + C}{(s^2 + s + 2)} $$The way how I will solve this is to use complex analysis. Your original function can be broken down into three distinct linear factors by solving for the zeros of $(s^2 + s + 2)$ using quadratic formula.
$$\frac{b\pm\sqrt{b^2 -4ac}}{2a}$$ with a = b = 1 and c = 2. The zeroes are $s = -\frac{1}{2}+j\frac{\sqrt7}{2}$ and $s = -\frac{1}{2}-j\frac{\sqrt7}{2}$. Your original function can now be written as $$ G(s) = \frac{2(s+1)}{s(s-\frac{1}{2}+j\frac{\sqrt7}{2})(s-\frac{1}{2}-j\frac{\sqrt7}{2})} $$ fixing the brackets, we have $$ G(s) = \frac{2(s+1)}{s(s-[\frac{1}{2}-j\frac{\sqrt7}{2}])(s-[\frac{1}{2}+j\frac{\sqrt7}{2}])} $$
The new partial fraction decomposition now is
$$G(s) = \frac{A}{s}+\frac{B}{[s-(\frac{1}{2}-j\frac{\sqrt7}{2})]}+\frac{C}{[s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$ Using cover-up method, if we set $s = 0$ we will be able to solve for $A$. Thus
$$A = \frac{2(s+1)}{[s-(\frac{1}{2}-j\frac{\sqrt7}{2})][s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$
$$A = \frac{2}{2} = 1$$ Utilize FOIL method to get this. To solve for B and C, we set $s = (\frac{1}{2}-j\frac{\sqrt7}{2}) $ and $s = (\frac{1}{2}+j\frac{\sqrt7}{2}) $ respectively. Since B and C are complex conjugates, C's imaginary part will be the negative of B. Solving for B,
$$B = \frac{2(s+1)}{s[s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$
$$B = \frac{(1 + j\sqrt7)}{{-j\sqrt7}}$$
$$B = 1 -j\frac{\sqrt7}{7}$$
Because C is the complex conjugate of B, C is therefore
$$C = 1 +j\frac{\sqrt7}{7}$$
Thus, your original function is decomposed into
$$G(s) = \frac{1}{s} + \frac{1 - j\frac{\sqrt7}{7}}{[s-(\frac{1}{2} - j\frac{\sqrt7}{2})]} + \frac{1 + j\frac{\sqrt7}{7}}{[s-(\frac{1}{2} +j \frac{\sqrt7}{2})]}$$
The inverse Laplace Transform of this is
$$ g(t) = 1 + (1 - j\frac{\sqrt7}{7})e^{(\frac{1}{2} - j\frac{\sqrt7}{2})t} + (1 + j\frac{\sqrt7}{7})e^{(\frac{1}{2}+j\frac{\sqrt7}{2})t} $$
Do note that this is not yet the final answer. What I will be doing is I will make use of sine and cosine's equivalent complex functions, namely:
$$ sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})$$
$$ cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})$$
We will then factor out $e^{\frac{1}{2}t}$ and distribute $e^{j\frac{\sqrt7}{2}t}$ to each terms. Upon doing so, we get
$$g(t) = 1 + e^{\frac{1}{2}t}[(1 - j\frac{\sqrt7}{7})e^{-j\frac{\sqrt7}{2}t}] + [(1 + j\frac{\sqrt7}{7})e^{j\frac{\sqrt7}{2}t}] $$
$$g(t) = 1 + e^{\frac{1}{2}t}[(e^{-j\frac{\sqrt7}{2}t} - j\frac{\sqrt7}{7}e^{-j\frac{\sqrt7}{2}t}) + (e^{j\frac{\sqrt7}{2}t} + j\frac{\sqrt7}{7}e^{j\frac{\sqrt7}{2}t}) ] $$
We will then collect the real and imaginary parts of $e^{j\frac{\sqrt7}{2}t}$ and $e^{-j\frac{\sqrt7}{2}t}$ because we will be able to get the inverse laplace transform from those. By doing so, we have
$$g(t) = 1 + e^{\frac{1}{2}t}[(e^{j\frac{\sqrt7}{2}t} + e^{-j\frac{\sqrt7}{2}t}) - j\frac{\sqrt7}{7}(e^{j\frac{\sqrt7}{2}t} - e^{-j\frac{\sqrt7}{2}t})] $$
$$g(t) = 1 + 2e^{\frac{1}{2}t}cos(\frac{\sqrt7}{2}t) - j\frac{\sqrt7}{7}(j2)e^{\frac{1}{2}t}sin(\frac{\sqrt7}{2}t) $$
$$g(t) = 1 + 2e^{\frac{1}{2}t}cos(\frac{\sqrt7}{2}t) + \frac{2\sqrt7}{7}e^{\frac{1}{2}t}sin(\frac{\sqrt7}{2}t) $$
|
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|
Strong Induction: Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$ Can you please help me and tell, how should I move on?
Can this be proved by induction?
Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$.
Thank you in advance
|
We can avoid an explicit appeal to induction by using the fact that every natural number $n$ has remainder $0$, $1$, or $2$ on division by $3$. Let $n \ge 8$.
If $n$ has remainder $2$ on division by $3$, then $n-8$ is divisible by $3$, say $n-8=3m$. Represent $8$ using $8=3\cdot 1+5\cdot 1$. Then add $m$ $3$'s.
If $n$ has remainder $0$ on division by $3$, then $n-9$ is divisible by $3$, say $n-9=3m$. Represent $9$ using $9=3\cdot 3 +5\cdot 0$. Then add $m$ $3$'s.
If $n$ has remainder $1$ on division by $3$, then $n-10$ is divisible by $3$, say $n-10=3m$. Represent $10$ using $10=3\cdot 0 +5\cdot 2$. Then add $m$ $3$'s.
The argument for remainder $0$ was a little silly, since if $n$ has remainder $0$ on division by $3$, we can clearly use a bunch of $3$'s to represent $n$. But we wanted the solutions for the three cases to use a single template.
|
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|
Proof of $\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1}$? How do I prove that
$$\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1} \>?$$
I saw this in a book discussing generating functions.
|
Suppose we seek to verify that
$$\sum_{0\le k\le t} {t-k\choose r} {k\choose s}
= {t+1\choose r+s+1}.$$
Introduce
$${t-k\choose r} = {t-k\choose t-k-r} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{t-k-r+1}} (1+z)^{t-k} \; dz.$$
This controls the range so we may extend $k$ to infinity,
getting for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{t-r+1}} (1+z)^{t}
\sum_{k\ge 0} {k\choose s} \frac{z^k}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{t-r+1}} (1+z)^{t}
\sum_{k\ge s} {k\choose s} \frac{z^k}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{t-r+1}} (1+z)^{t} \frac{z^s}{(1+z)^s}
\sum_{k\ge 0} {k+s\choose s} \frac{z^k}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{t-r+1}} (1+z)^{t} \frac{z^s}{(1+z)^s}
\frac{1}{(1-z/(1+z))^{s+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{t-r-s+1}} (1+z)^{t+1}
\frac{1}{(1+z-z)^{s+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{t-r-s+1}} (1+z)^{t+1}
\; dz.$$
This evaluates to
$${t+1\choose t-r-s} = {t+1\choose r+s+1}$$
by inspection.
|
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|
Can $x^{n}-1$ be prime if $x$ is not a power of $2$ and $n$ is odd? Are there any solutions to $x^{n}-1=p$ with p prime, integers $x,n>1$ and $x$ not a power of $2$?
$x$ must be even. $n$ is odd since if $n=2m$ then $p=x^{n}-1=(x^{m}+1)(x^{m}-1)$ hence $p=x^{m}+1$ and $1=x^{m}-1$, which has solution $p=3$ given by $x=2^{2}$.
This generalises; for any solution to $x^{n}-c^{2}=p$, $n$ must be odd. And for any solution to $x^{n}-c=p$, $x$ is odd iff $c$ is odd unless $p=2$ and, if $a$ divides $c$ and $p$ does not divide $c$ then $x$ divides $c$.
|
The following result can be found in most introductions to Number Theory.
Theorem: Let $x$ and $n$ be integers greater than $1$. If $x^n-1$ is prime, then $x=2$ and $n$ is prime.
Proof: Note that $x-1$ divides $x^n-1$, for
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+ \cdots + 1).$$
If $x>2$ and $n>1$, each of the above factors of $x^n-1$ is $>1$. It follows that if $x>2$ and $n>1$, then $x^n-1$ cannot be prime. So if $x^n-1$ is prime, then $x=2$.
Next we show that if $2^n-1$ is prime, then $n$ itself must be prime. For suppose to the contrary that $n=ab$ where $a$ and $b$ are greater than $1$. Then
$$2^n-1=2^{ab}-1=(2^a)^b-1.$$
Let $y=2^a$. Then $2^n-1=y^b-1$. But $y-1$ divides $y^b-1$. Thus
$2^a-1$ divides $2^n-1$. It is easy to see that in fact $2^a-1$ is a proper divisor of $2^n-1$, so $2^n-1$ cannot be prime.
This concludes the proof. By calculating, we can verify that $2^n-1$ is prime for $n=2$, $3$, $5$, and $7$. But one should not jump to conclusions. When $n=11$, $2^n-1$ is not prime, for $23$ divides $2^{11}$.
Primes of the form $2^n-1$ (where $n$ is necessarily prime) are called Mersenne primes.
There has been interest in what would later be called Mersenne primes ever since the time of Euclid, because of their connection with the even perfect numbers. Despite a search spanning millenia, only $47$ Mersenne primes are currently known. The current record holder is $2^{43,112,609}-1$. From the computational evidence, it appears that for "most" primes $p$, the number $2^p-1$ is not prime. It is not known whether there are infinitely many Mersenne primes.
|
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|
A pencil approach to find $\sum \limits_{i=1}^{69} \sqrt{\left( 1+\frac{1}{i^2}+\frac{1}{(i+1)^2}\right)}$ What is the fastest, paper-pencil method of finding $$\sum \limits_{i=1}^{69} \sqrt{\left( 1+\frac{1}{i^2}+\frac{1}{(i+1)^2}\right)}?$$
This is actually a quantitative aptitude problem, and hence the solutions should be fast enough and probably under a minute.
Using wolframalpha, the sum seems to be $\frac{4899}{70}$; however, I am not sure how to find this quickly in the paper-pencil way.
|
Edit: Now in beautiful high-definition technicolor!
$$\sum \limits_{k=1}^n \sqrt{\color{red}1+\color{Green}{\frac{1}{k^2}}+\color{Blue}{\frac{1}{(k+1)^2}}}$$
$$=\sum_{k=1}^n\sqrt{\frac{\color{red}{k^2(k+1)^2}+\color{Green}{(k+1)^2}+\color{Blue}{k^2}}{k^2(k+1)^2}}$$
$$=\sum_{l=1}^n \sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}}$$
$$=\sum_{k=1}^n\sqrt{\frac{\color{Red}1k^4+(\color{Red}1+\color{Green}1)k^3+(\color{Red}1+\color{Green}1+\color{Blue}1)k^2+(\color{Green}1+\color{Blue}1)k+\color{Blue}1}{k^2(k+1)^2}} $$
$$=\sum_{k=1}^n\sqrt{\frac{\color{Red}{k^2(k^2+k+1)}+\color{Green}{k(k^2+k+1)}+\color{Blue}{(k^2+k+1)}}{k^2(k+1)^2}} $$
$$=\sum_{k=1}^n\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}$$
$$=\sum_{k=1}^n\frac{\color{Purple}{k^2+k}+\color{Blue}1}{k^2+k}$$
$$=\sum_{k=1}^n \left(\color{Purple}1+\color{Blue}{\frac{1}{k(k+1)}}\right) $$
$$=\sum_{k=1}^n \left(1+\frac{1}{k}-\frac{1}{k+1}\right)$$
$$=n+1-\frac{1}{n+1}.$$
Specializing $n=69$ gives $69+1-1/70=4899/70$. Note that it's highly useful to memorize this:
$$\frac{1}{k(k+a)}=\frac{1}{a}\frac{(k+a)-k}{k(k+a)}=\frac{1}{a}\left(\frac{1}{k}-\frac{1}{k+a}\right).$$
|
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|
General formula to obtain triangular-square numbers I am trying to find a general formula for triangular square numbers. I have calculated some terms of the triangular-square sequence ($TS_n$):
$TS_n=$1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056
1882672131025, 63955431761796, 2172602007770041, 73804512832419600
I have managed to find that:
$m^2 = \frac{n(n+1)}{2}$
where $m$ is the $m^{th}$ term of the square number sequence, and $n$ is the $n^{th}$ term of the triangular numbers sequence.
Would anyone be able to point out anything that could help me derive a formula for the square-triangular sequence?
|
Since all triangular numbers are of the form $\frac{n(n+1)}{2}$, we must have the condition cited
$$
m^2=\frac{n(n+1)}{2}\tag{1}
$$
Equation $(1)$ is equivalent to
$$
2 = \frac{(2n+1)^2-1}{(2m)^2}\tag{2}
$$
According to standard continued fraction theory, a rational approximation to $\sqrt{2}$ as good as $(2)$ must also be an approximant for the continued fraction of $\sqrt{2}$. So we must find an overestimate for $\sqrt{2}$ with an even denominator (continued fraction approximants alternate between over- and under-estimates). The continued fraction for $\sqrt{2}$ is $\{1;2,2,2,\dots\}$, so the sequences of numerators and denominators for the approximants satisfy
$$
\begin{array}{}a_k=2a_{k-1}+a_{k-2}&\text{ and }&b_k=2b_{k-1}+b_{k-2}\end{array}\tag{3}
$$
where $a_0=a_1=b_1=1$ and $b_0=0$. Computing the first few approximants, we get
$$
\frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\frac{99}{70},\dots\tag{4}
$$
It follows from $(3)$ that every other denominator is even, and that they correspond to the overestimates. Therefore, every other approximant yields a square triangular number.
Every other term of a sequence that satisfies $(3)$, also satisfies
$$
\begin{array}{}a_{2k}=6a_{2k-2}-a_{2k-4}\text{ and }b_{2k}=6b_{2k-2}-b_{2k-4}\end{array}\tag{5}
$$
where $a_0=1$, $a_2=3$, $b_0=0$, and $b_2=2$. Let $2n_k+1=a_{2k}$ and $2m_k=b_{2k}$. Applying $(5)$ yields
$$
\begin{array}{}n_k=6n_{k-1}-n_{k-2}+2\text{ and }m_k=6m_{k-1}-m_{k-2}\end{array}\tag{6}
$$
where $n_0=m_0=0$ and $n_1=m_1=1$.
Using standard recurrence methods, we can solve $(6)$ for $m_k$ to get
$$
\begin{align}
m_k^2
&=\left(\frac{(3+2\sqrt{2})^k-(3-2\sqrt{2})^k}{4\sqrt{2}}\right)^2\\
&=\frac{(17+12\sqrt{2})^k+(17-12\sqrt{2})^k-2}{32}\tag{7}
\end{align}
$$
Thus, the sequence $TS_k=m_k^2$.
|
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|
Characteristic polynomial of companion matrix I have a matrix in companion form,
$$A=\begin{pmatrix} 0 & \cdots & 0& -a_{0} \\ 1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}$$
where $A \in M_{n}$. I want to prove by induction that the characteristic polynomial is
$$p_{A}=t^{n}+a_{n-1}t^{n-1}+\cdots +a_{0}$$
The part that is confusing me is if we assume this hold for $A_{1}\in M_{n-1}$ how do we transform, or "add" to $A_{1}$ to get the n by n matrix $A\in M_{n}$? I hope this makes sense.
|
As suggested in the comment above, expand along the first row:
$$\mathrm{det}(tI_n-A) = \mathrm{det} \begin{pmatrix} t & 0 & \cdots & 0 & a_0 \\
-1 & t & \cdots & 0 & a_1 \\
\vdots & \ddots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -1 & t+a_{n-1} \end{pmatrix} = $$
$$ t \cdot \mathrm{det} \; \begin{pmatrix} t & 0 & \cdots & 0 & a_1 \\
-1 & t & \cdots & 0 & a_2 \\
\vdots & \ddots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -1 & t+a_{n-1} \end{pmatrix} +
(-1)^{1+n} a_0 \cdot \mathrm{det} \begin{pmatrix} -1 & t & 0 & \cdots & 0 \\
0 & -1 & t & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & -1 & \end{pmatrix} $$
By induction we can replace the determinant on the left by $a_1+a_2t+\cdots+a_{n-1}t^{n-2}+t^{n-1}$ and the second matrix's determinant is the product of its diagonals (since it's upper-triangular). The product of the diagonals is $(-1)^{n-1}$. Therefore, the determinant is $t(a_1+a_2t+\cdots+a_{n-1}t^{n-2}+t^{n-1})+(-1)^{n+1}(-1)^{n-1}a_0$. Which simplifies to $a_0+a_1t+\cdots+a_{n-1}t^{n-1}+t^n$.
[When writing this up, don't forget the base case for the induction.]
|
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|
Help deriving an function from (Lagrangian?) properties I'm trying to derive a function f(x) that has the following properties:
$$f_x-\frac{f}{x}=g_x$$
(You might call this the Lagrangian?)
where
$$\begin{align*}
f&=f(x)\\
f_x&=\frac{df}{dx}\\
g_x&=\frac{d}{dx}\left(\frac{f}{x}\right)
\end{align*}$$
By rearranging I've gotten to the point where:
$$\frac{f}{f_x}=\frac{x^2-x}{x+1}.$$
But can't figure out what function satisfies this.
Thanks
|
Your derivation is incorrect (you've got a wrong sign). I'll use $f'$ instead of $f_x$ (too many $x$'s around...)
We have:
$$f' - \frac{f}{x} = \frac{d}{dx}\frac{f}{x}.$$
If $f=0$, we get a solution. So assume that $f$ is not always $0$. Then the above equation
is equivalent to
$$\begin{align*}
f' - \frac{f}{x} &= \frac{xf' - f}{x^2}\\
x^2f' - xf &= xf' - f\\
x^2f' - xf' &= xf - f\\
(x^2-x)f' &= (x-1)f\\
\frac{f'}{f} &= \frac{x-1}{x^2-x} = \frac{x-1}{x(x-1)} = \frac{1}{x}.
\end{align*}$$
(you can see that you had $x+1$ instead of $x-1$).
Now, integrating we have:
$$\begin{align*}
\ln|f| &= \ln|x|+C\\
|f| &= Ax &&A\gt 0\\
f(x) &= Bx &&B\neq 0.
\end{align*}$$
Adding in $B=0$ we get that the solutions are $f(x)=Cx$, with $C$ a constant. Indeed, notice that if $f(x) = Cx$, then
$$f'-\frac{f}{x} = C - C = 0,$$
and
$$\frac{d}{dx}\frac{f}{x} = \frac{d}{dx}C = 0,$$
so they all satisfy your desired equation.
(If you actually had $\frac{f'}{f} = \frac{x+1}{x^2-x}$, then this can be solved by integration as well:
$$\begin{align*}
\ln|f| &= \int\frac{x+1}{x^2-x}\,dx\\
&= \int\left(\frac{-1}{x} + \frac{2}{x-1}\right)\,dx\\
&= -\ln|x| + 2\ln|x-1| + C\\
&= \ln\left|\frac{(x-1)^2}{x}\right| + C,
\end{align*}$$
and from this you can likewise obtain a formula for $f$.)
|
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|
Integrate $\int\frac{1}{x^6} \sqrt{(1-x^2)^3} ~ dx$ How to integrate the following?
$$\int\frac{\sqrt{(1-x^2)^3}}{x^6} \;dx .$$
|
Let $x = \sin(\theta)$. We then get $dx = \cos(\theta) d \theta$. Hence, $$I= \int \frac{\cos^3(\theta)}{\sin^6(\theta)} \cos(\theta) d \theta = \int \cot^4(\theta) cosec^2(\theta) d \theta$$
Let $\cot(\theta) = t$, then $-cosec^2(\theta) d \theta = dt$. Hence, $$I = -\int t^4 dt = -\frac{t^5}{5} + c = -\frac{\cot^5(\theta)}{5} + c = -\frac15 \left( \frac{\sqrt{1-x^2}}{x} \right)^5 + c$$
|
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|
How many factors of $2010^{2010}$ have last digit $2$?
How many factors of $2010^{2010}$ have last digit $2$?
It is not difficult to solve this using python/mathematica, but I want to know how to (smartly) solve this one with paper and pencil?
|
Since $2010=2\cdot3\cdot5\cdot67$, the factors of $2010^{2010}$ are all of the form $F=2^a\cdot3^b\cdot5^c\cdot67^d$ with $0\le a,b,c,d\le2010$. Since we want the last digit of $F$ to be $2$, we must have $a\ge1$ and $c=0$.
*
*The last digits of $2^a$ ($a\ge1$) are $2,4,8,6$, and they repeat
periodically.
*The last digits of $3^b$ ($b\ge0$) are $1,3,9,7$, and they repeat periodically.
*The last digits of $67^d$ ($d\ge0$) are $1,7,9,3$, and they repeat periodically.
Let $\alpha$ be the last digit of $2^a$, $\beta$ be the last digit of $3^b$ and $\delta$ be the last digit of $67^d$. For the last digit of $F=2^a\cdot3^b\cdot67^d$ to be $2$, the last digit of $\alpha\cdot\beta\cdot\delta$ must be also $2$. When does this happen? There are $16$ possibilities for the triple $\{\alpha,\beta,\delta\}$: $\{2,1,1\}$, $\{2,3,7\}$, $\{2,9,9\}$, $\{2,7,3\}$, $\{4,1,3\}$, $\{4,3,1\}$, $\{4,9,7\}$, $\{4,7,9\}$, $\{8,1,9\}$, $\{8,3,3\}$, $\{8,9,1\}$, $\{8,7,7\}$, $\{6,1,7\}$, $\{6,3,9\}$, $\{6,9,3\}$ and $\{6,7,1\}$. All we have to do is count how many factors there are with each of the $16$ possibilities. For instance, for $\{2,1,1\}$ we must have
$$
a\equiv 1 \mod4,\quad b\equiv0\mod4\quad\text{and}\quad d\equiv0\mod 4.
$$
This gives $503$ possible values for each of $a$, $b$ and $c$, for a total of $503^3$ factors.
I'll leave it to you to do the computations for the other $15$ cases.
|
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|
How to determine $x$ and $y$ intercepts for $y = 4(x - 2)^2(x + 2)^3$ I need help to determine $x$ and $y$ intercepts for
$$
y = 4(x - 2)^2(x + 2)^3
$$
I guess my first question is, do I need to get the equation into
$$
ax^3 + bx^2 + cx + d
$$
form before starting?
|
The x-intercepts can be calculated like this:
$$ y = 4(x - 2)^2(x + 2)^3 $$
$$ y = 0$$
$$ a \cdot b \cdot c = 0 \rightarrow a = 0 \vee b=0 \vee c=0 $$
$$ 4 = 0 \vee (x-2)^2 = 0 \vee (x+2)^3 = 0$$
$$ None \vee x = 2 \vee x \ -2$$
The y-intercept can be found by substituting $0$ for $x$ in $4(x - 2)^2(x + 2)^3$.
|
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|
Which is the "fastest" way to compute $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $?
I am looking for the "fastest" paper-pencil approach to compute $$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $$
This is a quantitative aptitude problem and the correct/required answer is $3.75$
In addition, I am also interested to know how to derive a closed form for an arbitrary $n$ using mathematica I got $$\sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} = \frac{5 \times \left(3 \times 2^n-2 n-3\right)}{2^{n+2}}$$
Thanks,
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I would do it like this. Using $x \frac{\mathrm{d}}{\mathrm{d} x}\left( x^k \right) = k x^k$, and $\sum_{k=1}^n x^k = x \frac{x^n-1}{x-1}$. Then
$$\begin{eqnarray}
\sum_{k=1}^n \left( a k +b\right) x^k &=& \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + b\right) \circ \sum_{k=1}^n x^k = \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + b\right) \circ \left( x \frac{x^n-1}{x-1} \right) \\
&=& x \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + a + b\right) \circ \left(\frac{x^n-1}{x-1} \right) \\
&=& x \left( (a+b) \frac{x^n-1}{x-1} + a x \frac{ n x^{n-1}(x-1) - (x^n-1) }{(x-1)^2} \right) \\
&=& x \left( (a+b) \frac{x^n-1}{x-1} + a x \frac{ (n-1) x^{n} - n x^{n-1} + 1 }{(x-1)^2} \right)
\end{eqnarray}
$$
Now applying this:
$$
\begin{eqnarray}
\sum_{k=1}^n \frac{10 k -5}{2^{k+2}} &=& \frac{5}{4} \sum_{k=1}^n (2k-1)\left(\frac{1}{2}\right)^k \\
&=& \frac{5}{4} \frac{1}{2} \left( -(2^{1-n} - 2) + (n-1) 2^{2-n}- n 2^{1-n} + 4 \right) \\
&=& \frac{5}{4} \left( 3 + (n-3) 2^{-n} \right)
\end{eqnarray}
$$
|
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|
Compacting a Matrix Suppose you have a matrix that is zero-valued everywhere except the diagonal.
As an example, take the identity matrix, $I$. For this example, let's say you are using the $4\times 4$ version.
Is there an operation that can produce a $2\times 2$ matrix consisting of only the non-zero values? If not, is there a notation to express this intent?
Example:
$$ I = \pmatrix{
i_{11} &0 &0 &0\\
0 &i_{22} &0 &0\\
0 &0 &i_{33} &0\\
0 &0 &0 &i_{44}}
$$
$$ I_2 =\pmatrix{
i_{11} &i_{22}\\
i_{33} &i_{44}}$$
|
Here is a solution that you can find by solving a linear system. First we will obtain a $4\times 4$ matrix that looks like your target matrix ($I_2$) but padded. We let
$$A = \begin{bmatrix}a_1 & 0 & 0 & 0\\0 & a_2 & 0 & 0\\0 & 0 & a_3 & 0\\0 & 0 & 0 & a_4\end{bmatrix} \quad{}\text{and}\quad X = \begin{bmatrix}
x_{1,1} & x_{1,2} & x_{1,3} & x_{1,4} \\
x_{2,1} & x_{2,2} & x_{2,3} & x_{2,4} \\
x_{3,1} & x_{3,2} & x_{3,3} & x_{3,4} \\
x_{4,1} & x_{4,2} & x_{4,3} & x_{4,4}
\end{bmatrix}$$
When you multiply these two two matrices you obtain
$$AX =
\begin{bmatrix}
a_1 x_{1,1} & a_1 x_{1,2} & a_1 x_{1,3} & a_1 x_{1,4} \\
a_2 x_{2,1} & a_2 x_{2,2} & a_2 x_{2,3} & a_2 x_{2,4} \\
a_3 x_{3,1} & a_3 x_{3,2} & a_3 x_{3,3} & a_3 x_{3,4} \\
a_4 x_{4,1} & a_4 x_{4,2} & a_4 x_{4,3} & a_4 x_{4,4}
\end{bmatrix}
$$
You want all the entries of this matrix to be $0$ except the four entries in the upper left corner (which we want to be $a_1,a_2,a_3$ and $a_4$). Solving the equations formed leads to
$$ X =
\begin{bmatrix}
1 & a_2/a_1 & 0 & 0 \\
a_3/a_2 & a_4/a_2 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
and when you multiply $A$ and $X$ together you obtain
$$AX = \begin{bmatrix}
a_1 & a_2 & 0 & 0 \\
a_3 & a_4 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}$$
Note that the solution assumes that $a_1 \neq 0$ and $a_2 \neq 0$. If one or more of these is zero, we cannot solve the system. In your case it is okay because you say that the entries on the diagonal are non-zero. As a final step we need to remove two of the columns and two of the rows. Let $B$ be given by
$$B =
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 & 0 \\
0 & 0
\end{bmatrix}.$$
Then $B^T A X B = I_2$. Note that the matrix $X$ depends on the entries of matrix $A$. I am not sure if there is a matrix that will work on all such diagonal matrices.
|
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|
I have a problem understanding the proof of Rencontres numbers (Derangements) I understand the whole concept of Rencontres numbers but I can't understand how to prove this equation
$$D_{n,0}=\left[\frac{n!}{e}\right]$$
where $[\cdot]$ denotes the rounding function (i.e., $[x]$ is the integer nearest to $x$). This equation that I wrote comes from solving the following recursion, but I don't understand how exactly the author calculated this recursion.
$$\begin {align*}
D_{n+2,0} & =(n+1)(D_{n+1,0}+D_{n,0}) \\
D_{0,0} & = 1 \\
D_{1,0} & = 0
\end {align*}
$$
|
(This argument is adapted from page 195 of Concrete Mathematics, Second Edition)
We start with the more conventional representation for the Rencontres number (subfactorial):
$$D_{n,0}=!n=n!\sum_{k=0}^n \frac{(-1)^k}{k!}$$
We also know that
$$\frac{n!}{e}=n!\sum_{k=0}^\infty \frac{(-1)^k}{k!}$$
The difference is
$$\begin{align*}\frac{n!}{e}-!n&=n!\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}\\&=\frac{(-1)^{n+1}}{n+1}\left(1-\frac1{n+2}+\frac1{(n+2)(n+3)}-\cdots\right)\end{align*}$$
and since
$$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right| \leq \frac1{n+1}$$
along with knowing that $!n$ is an integer, rounding $n!/e$ to the nearest integer gives the subfactorial.
We have
$\small \begin{align}(n+2)!\sum_{k=0}^{n+2} \frac{(-1)^k}{k!}&=(n+1)\left[(n+1)!\sum_{k=0}^{n+1} \frac{(-1)^k}{k!}+n!\sum_{k=0}^n \frac{(-1)^k}{k!}\right]\\(n+2)(n+1)\sum_{k=0}^{n+2} \frac{(-1)^k}{k!}&=(n+1)\left[(n+1)\sum_{k=0}^{n+1} \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\right]\\(n+2)\sum_{k=0}^{n+2} \frac{(-1)^k}{k!}&=(n+1)\sum_{k=0}^{n+1} \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\\(n+2)\left(\frac{(-1)^{n+2}}{(n+2)!}+\frac{(-1)^{n+1}}{(n+1)!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\right)&=\frac{(-1)^{n+1}}{n!}+(n+1)\sum_{k=0}^n \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\\(n+2)\left(\frac{(-1)^n}{(n+2)!}+\frac{(-1)^{n+1}}{(n+1)!}\right)+(n+2)\sum_{k=0}^n \frac{(-1)^k}{k!}&=\frac{(-1)^{n+1}}{n!}+(n+1)\sum_{k=0}^n \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\\(n+2)\left(\frac{(-1)^n}{(n+2)!}+\frac{(-1)^{n+1}}{(n+1)!}\right)&=\frac{(-1)^{n+1}}{n!}\\(-1)^n+(-1)^{n+1}(n+2)&=(-1)^{n+1}(n+1)\\1-(n+2)&=-(n+1)\end{align}$
and the last bit is easily established, thus proving the recursion relation for the Rencontres numbers.
|
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|
$5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ Work out the values of $\frac{1}{x+y}$ $5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$
Work out the values of $\frac{1}{x+y}$
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Remark : This answer is based on WolframAlpha calculations .
As Sp300 pointed out in his comment by inspection we can see that $x=-1$ is a solution. I will try to show that this is the only solution.
So :
$$2^y=\frac{7}{10}-5^x \Rightarrow y= \log_2{\left(\frac{7}{10}-5^x\right)}$$
$$5^y=\frac{7}{10}-2^x \Rightarrow y= \log_5{\left(\frac{7}{10}-2^x\right)}$$
Therefore we may conclude that :
$$\log_2{\left(\frac{7}{10}-5^x\right)}=\log_5{\left(\frac{7}{10}-2^x\right)}$$
Let's define $f(x)$ as :
$$f(x)=\log_5{\left(\frac{7}{10}-2^x\right)}-\log_2{\left(\frac{7}{10}-5^x\right)}$$
Note that $f(x)$ is defined for $x\in \left(-\infty,\log_2{\left(\frac{7}{10}\right)}\right)$
WolframAlpha gives following results :
*
*$\frac{d}{dx}f(x)=0 \Rightarrow x\in \varnothing$ (no real
solutions)
*$\displaystyle\lim_{x \to -\infty} \frac{d}{dx}f(x)=0 $
*$\displaystyle\lim_{x \to
\left(\log_2{\left(\frac{7}{10}\right)}\right)^{-}}
\frac{d}{dx}f(x)=-\infty$
From this above we may conclude that $f'(x) < 0$ for all $x\in\left(-\infty,\log_2{\left(\frac{7}{10}\right)}\right)$ , so $f(x)$ is decreasing on interval $x\in \left(-\infty,\log_2{\left(\frac{7}{10}\right)}\right)$ therefore , $x=-1$ is the only solution .
From the equation $y= \log_2{\left(\frac{7}{10}-5^x\right)}$ we can obtain value of $y$ , so $y=-1$
Finally : $\frac{1}{x+y}=-\frac{1}{2}$
|
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|
Problem based on Range Find $a$ and $b$ such that the inequality $a \le 3 \cos{x} + 5\cos\left(x - \frac{\pi}{6}\right) \le b$ holds good for all x.
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We can take for example $a=-100$ and $b=100$. However, it might be interesting to find sharp bounds $a$ and $b$. That is a standard max/min problem. We will solve the problem without using the calculus.
Note that
$$\cos(x-\pi/6)=\cos x\cos(\pi/6)+\sin x\sin(\pi/6)=(\cos x)(\sqrt{3}/2)+(\sin x)(1/2).$$
Thus
$$3\cos x+5\cos(x-\pi/6)=\frac{6+5\sqrt{3}}{2}\cos x +\frac{5}{2}\sin x.$$
To make the structure clearer (and make typing easier), let $p=\frac{6+5\sqrt{3}}{2}$ and $q=\frac{5}{2}$.
Note that
$$(p\cos x+q\sin x)^2+(p\sin x-q\cos x)^2=p^2+q^2.$$
Thus $(p\cos x+q\sin x)^2$ can never be bigger than $p^2+q^2$, and is equal to $p^2+q^2$ precisely if $p\sin x -q\cos x=0$, that is, if $\tan x=q/p$.
In the first quadrant, $\tan x=q/p$ at roughly $0.33$ radians. At that value of $x$, the numbers $\cos x$ and $\sin x$ are positive, so our function is maximized.
There is also a solution of $\tan x=q/p$ in the third quadrant, where $\cos x$ and $\sin x$ are both negative, so our function is minimized. Thus
$$-\sqrt{p^2+q^2} \le 3\cos x +5\cos(x-\pi/6) \le \sqrt{p^2+q^2},$$
and these bounds are best possible.
Comment: One can obtain the same result more mechanically by using the calculus. The derivative of $p\cos x+q\sin x$ is $-p\sin x+q\cos x$. This derivative is $0$ when $\tan x=q/p$.
Another way of looking at things is to rewrite our expression as
$$\sqrt{p^2+q^2} \left(\frac{p}{\sqrt{p^2+q^2}}\cos x+\frac{q}{\sqrt{p^2+q^2}}\sin x\right).$$
Let $\theta$ be the angle whose sine is $p/\sqrt{p^2+q^2}$ and whose cosine is $q/\sqrt{p^2+q^2}$.
Then our expression is equal to
$$\sqrt{p^2+q^2} \sin(x+\theta),$$
and sharp bounds follow easily.
|
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|
Determinant of a matrix with diagonal entries $a$ and off-diagonal entries $b$ I have the following $n\times n$ matrix:
$$A=\begin{bmatrix} a & b & \ldots & b\\ b & a & \ldots & b\\ \vdots & \vdots & \ddots & \vdots\\ b & b & \ldots & a\end{bmatrix}$$
where $0 < b < a$.
I am interested in the expression for the determinant $\det[A]$ in
terms of $a$, $b$ and $n$. This seems like a trivial problem, as the
matrix $A$ has such a nice structure, but my linear algebra skills are
pretty rusty and I can't figure it out. Any help would be
appreciated.
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Add row 2 to row 1, add row 3 to row 1,..., add row $n$ to row 1, we get
$$\det(A)=\begin{vmatrix}
a+(n-1)b & a+(n-1)b & a+(n-1)b & \cdots & a+(n-1)b \\
b & a & b &\cdots & b \\
b & b & a &\cdots & b \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & \ldots & a \\
\end{vmatrix}$$
$$=(a+(n-1)b)\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 \\
b & a & b &\cdots & b \\
b & b & a &\cdots & b \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & \ldots & a \\
\end{vmatrix}.$$
Now add $(-b)$ of row 1 to row 2, add $(-b)$ of row 1 to row 3,..., add $(-b)$ of row 1 to row $n$, we get
$$\det(A)=(a+(n-1)b)\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 \\
0 & a-b & 0 &\cdots & 0 \\
0 & 0 & a-b &\cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & a-b \\
\end{vmatrix}=(a+(n-1)b)(a-b)^{n-1}.$$
|
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|
How to get closed form from generating function? I have this generating function:
$$\frac{1}{2}\, \left( {\frac {1}{\sqrt {1-4\,z}}}-1 \right) \left( \,{
\frac {1-\sqrt {1-4\,z}}{2z}}-1 \right)$$
and I know that $\frac {1}{\sqrt {1-4\,z}}$ is the generating function for the sequence $\binom {2n} {n}$, and $\frac {1-\sqrt {1-4\,z}}{2z}$ is the generating function for the sequence $\frac {1}{n+1}\binom{2n} {n}$.
Now, I thought that I could substitute those in there, and where they multiply I'll use a summation like this:
$$\frac{1}{2}\left( 1-\frac{1}{n+1}\binom{2n} {n}-\binom{2n} {n} +
\sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k}\binom{2(n-k)}{n-k}
\right)$$
Could this be right? It doesn't seem to work when I try in Maple. What else could I do?
I already know that the end sequence will be $\binom{2n-1}{n-2}$ if this can help...
|
With this type of problem Lagrange inversion is the preferred
approach. Suppose we seek to extract coefficients from
$$Q(z) = \frac{1}{2}
\left(\frac{1}{\sqrt{1-4z}}-1\right)
\left(\frac{1-\sqrt{1-4z}}{2z}-1\right).$$
The closed form for the coefficients is
$$[z^n] Q(z) = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{2}
\left(\frac{1}{\sqrt{1-4z}}-1\right)
\left(\frac{1-\sqrt{1-4z}}{2z}-1\right) \; dz.$$
Now put $1-4z=w^2$ so that $1/4-z=1/4 \times w^2$ or $z=1/4\times(1-w^2)$
and $dz = -1/2 \times w\; dw.$
This gives for the integral
$$-\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{w}{4}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(\frac{1}{w}-1\right)
\left(\frac{1-w}{2\times 1/4\times(1-w^2)}-1\right) \; dw
\\ = -\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^n}{(1-w^2)^{n+1}}
(1-w)
\left(\frac{2}{1+w}-1\right) \; dw
\\ = -\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^n}{(1-w)^n (1+w)^{n+1}}
\left(\frac{2}{1+w}-1\right) \; dw
\\ = -\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^n}{(1-w)^n (1+w)^{n+1}}
\frac{1-w}{1+w} \; dw
\\ = -\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^n}{(1-w)^{n-1} (1+w)^{n+2}} \; dw.$$
Prepare for coefficient extraction.
$$-\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^n}{(1-w)^{n-1} (2+(w-1))^{n+2}} \; dw
\\= -\frac{1}{2\pi i}
\int_{|w-1|=\epsilon} \frac{4^n}{2^{n+2}}
\frac{1}{(1-w)^{n-1} (1+(w-1)/2)^{n+2}} \; dw
\\= \frac{1}{2\pi i}
\int_{|w-1|=\epsilon} 2^{n-2}
\frac{(-1)^n}{(w-1)^{n-1}}
\sum_{q\ge 0} {q+n+1\choose n+1} (-1)^q
\frac{(w-1)^q}{2^q} \; dw.$$
We need the coefficient $[(w-1)^{n-2}]$ which is
$$2^{n-2} (-1)^n {n-2+n+1\choose n+1} (-1)^{n-2}
\frac{1}{2^{n-2}} = {2n-1\choose n+1}$$
or alternatively
$${2n-1\choose n-2}.$$
|
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|
How to prove :If $p$ is prime greater than $3$ and $\gcd(a,24\cdot p)=1$ then $a^{p-1} \equiv 1 \pmod {24\cdot p}$? I want to prove following statement :
If $p$ is a prime number greater than $3$ and $\gcd(a,24\cdot p)=1$ then :
$a^{p-1} \equiv 1 \pmod {24\cdot p}$
Here is my attempt :
The Euler's totient function can be written in the form :
$n=p_1^{k_1}\cdot p_2^{k_2} \ldots \cdot p_r^{k_r} \Rightarrow \phi(n)=p_1^{k_1}\cdot\left(1-\frac{1}{p_1}\right)\cdot p_2^{k_2}\cdot\left(1-\frac{1}{p_2}\right)\ldots p_r^{k_r}\cdot \left(1-\frac{1}{p_r}\right)$
So,
$\phi(24 \cdot p)=2^3\cdot \left(1-\frac{1}{2}\right)\cdot3^1\cdot\left(1-\frac{1}{3}\right)\cdot p\cdot\left(1-\frac{1}{p}\right)=8\cdot(p-1)$
Euler's totient theorem states that :
if $\gcd(a,n)=1$ then $a^{\phi(n)} \equiv 1 \pmod n$
Therefore we may write :
$a^{\phi(n)}-1 \equiv 0 \pmod n \Rightarrow a^{\phi(24\cdot p)}-1=a^{8\cdot(p-1)}-1 \equiv 0 \pmod{24\cdot p} \Rightarrow$
$\Rightarrow \left(a^{p-1}\right)^8-1=(a^{p-1}-1)\cdot \displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
So we may conclude :
$(a^{p-1}-1) \equiv 0 \pmod {24\cdot p}$ , or $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
How can I prove that $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \not\equiv 0\pmod{24\cdot p}$ ?
|
From $p$ is prime, $p>3$ and $\gcd(a,24p)=1$, we infer $\gcd(a,p)=1$, $\gcd(a,3)=1$, $\gcd(a,8)=1$, and $\gcd(p,24)=1$.
From $\gcd(a,p)=1$ we have $a^{p-1} \equiv 1\pmod{p}$.
From $\gcd(a,3)=1$ we have $a^2 \equiv 1\pmod{3}$, and so $a^{p-1} \equiv 1\pmod{3}$ as $p-1$ is even.
From $\gcd(a,8)=1$ we have $a^2 \equiv 1\pmod{8}$ (since $a$ is odd), and so $a^{p-1} \equiv 1\pmod{8}$.
Therefore, $a^{p-1}-1$ is a common multiple of $3,8,p$, and so also a multiple of $24p$. $\blacksquare$
|
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|
Parallelogram trigonometry (Sorry for the ambiguous title, couldn't think of a better one)
While leafing through a highschool textbook, I found what looked like an interesting question in trigonometry. My trigonometry skills are borderline 0, but I didn't expect it to be too much of a challenge. Well, I was wrong:
The sides of a parallelogram are $a$ and $b$ and its sharp angle is $\alpha$. The diagnols are $n$ and $m$, and the sharp angle between them is $\beta$.
A. Prove: $\frac{mn}{2ab} = \frac{\sin\alpha}{\sin\beta}$
B. Let: $\alpha = \beta$, $a < b$, $m < n$
Prove: $6a^2 + 2b^2 = 3m^2+n^2$
And in (rough) drawing:
Following the law of cosines (and that $\cos(180-\theta) = -\cos(\theta)$):
$n^2 = a^2 + b^2 - 2ab \cos\alpha$ (in $\Delta ABC$)
$m^2 = a^2 + b^2 - 2ab \cos(180-\alpha) = a^2 + b^2 + 2ab \cos(\alpha)$ (in $\Delta DAC$)
$a^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(\beta)$ (in $\Delta AEB$)
$b^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(180 - \beta)$ (in $\Delta BEC$)
Expanding the last two equations:
$$a^2 = \frac{m^2}{4} + \frac{n^2}{4} - \frac{mn \cos(\beta)}{2}$$
$$b^2 = \frac{m^2}{4} + \frac{n^2}{4} + \frac{mn \cos(\beta)}{2}$$
$$\Rightarrow$$
$$a^2 + b^2 = \frac{m^2}{2} + \frac{n^2}{2}$$
And that's where I hit a wall. I have six variables, and can't find a way to express them in a fashion which resembles the end result. A major setback is that I couldn't find a way to express both alpha and beta in the same triangle - if I could, then the law of sines will probably be a rescuer.
If possible, I'd like that instead of solving it, maybe you can show me a guideline - where I went wrong, or what I'm missing. Thank you in advance.
|
For part A, try counting the area of the parallelogram in two different ways, as suggested by Jim Belk.
For part B, notice that your diagram has $n$ and $m$ reversed, since $m<n$. In particular, $\alpha$ should be opposite $m$, not $n$. The modified version of your formula for $m$ is then
$$m^2 = a^2+b^2 -2ab\cos\alpha.$$
Try combining this with your formula
$$4a^2 = m^2 + n^2 -2mn\cos\beta$$
and use the result from part A.
|
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|
Coefficients of $(1+x+\dots+x^n)^3$? Consider the following polynomial:
$$ (1+x+\dots+x^n)^3 $$
The coefficients of the expansion for few values of $n$ ($n=1$ to $5$) are:
$$ 1, 3, 3, 1 $$
$$ 1, 3, 6, 7, 6, 3, 1 $$
$$ 1, 3, 6, 10, 12, 12, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1 $$
Is there a closed-form formula for the $i$th element of this sequence (for different values of $n$)?
Edit This looks similar to the sequence A109439 on OEIS corresponding to the coefficients of the expansion of: $ \left( \frac{1 - x^n}{1 - x} \right)^3 .$
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If the polynomial were replaced by an infinite sum, the coefficient of $x^i$ would be equal to the number of ways to choose $(a,b,c) \in \{0,1,2,...\}^3$ such that $a+b+c=i$. This is equal to the number of ways to choose two numbers $\le i$, or $\frac{1}{2}(i+1)(i+2)$. The only difference here is that you want to limit the count to cases where $a,b,c \le n$. Using the inclusion-exclusion principle, you can subtract the cases where one of the numbers is greater than $n$, add back the cases where two of the numbers are greater than $n$, and subtract the cases where all three are greater than $n$. For each of these counts, subtract $n+1$ (or $2n+2$ or $3n+3$) from the target sum first, returning you to the case where all three numbers are unrestricted. The result is
$$
\begin{eqnarray}
a_{i,n}&=&\frac{1}{2}(i+1)(i+2) - \frac{3}{2}(i-n)(i-n+1)\Theta(i-n-1) \\
&=& + \frac{3}{2}(i-2n-1)(i-2n)\Theta(i-2n-2) - \frac{1}{2}(i-3n-2)(i-3n-1)\Theta(i-3n-3),
\end{eqnarray}
$$
where $\Theta(x)=1$ for $x\ge 0$ and $0$ for $x<0$.
|
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|
Two players alternately flip a coin; what is the probability of winning by getting a head?
Two players, $A$ and $B$, alternately and independently flip a coin and the first player to get a head wins. Assume player $A$ flips first. If the coin is fair, what is the probability that $A$ wins?
So $A$ only flips on odd tosses. So the probability of winning would be $$ P =\frac{1}{2}+\left(\frac{1}{2} \right)^{2} \frac{1}{2} + \cdots+ \left(\frac{1}{2} \right)^{2n} \frac{1}{2}$$
Is that right? It seems that if $A$ only flips on odd tosses, this shouldn't matter. Either $A$ can win on his first toss, his second toss, ...., or his $n^{th}$ toss. So the third flip of the coin is actually $A$'s second toss. So shouldn't it be $$P = \frac{1}{2} + \left(\frac{1}{2} \right)^{2} + \left(\frac{1}{2} \right)^{3} + \cdots$$
|
\mathsrc{A_k}={A losing in his first k tosses and B losing in his first k tosses and A winning in his k+1 toss}
$$\begin{align}
P(\text{A winning}) &= P(\text{A winning in his first toss or } \mathrm{A_1} \text{ or } \mathrm{A_2} \text{ or } \ldots)\\
&= P(\text{A winning in his first toss}) + P(\mathrm{A_1}) + P(\mathrm{A_2}) + \dots\\
&= 0.5 + (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5)(0.5)(0.5) + \dots \\
&=0.5 + (0.5)^3 + (0.5)^5 + .........\\
&= \frac{0.5}{1 - 0.25} = \frac{0.5}{0.75} = \frac{50}{75} = \frac{2}{3}\\
\end{align}$$
|
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|
Continuity of a function I was trying to do an exercise: proving that $\frac{x^2}{1-x^2}$ is continuous on $(0,1)$. I did it but I want to be sure that it's right, could you tell me if my argument is wrong?
$\frac{x^2}{1-x^2}-\frac{a^2}{1-a^2}=\frac{(x+a)(x-a)}{(1-x^2)(1-a^2)}$, now $x+a\leq 1+a$. $1-x^2=1-x^2+a^2-a^2=1-a^2-(x^2-a^2)=1-a^2-(x-a)(x+a)\geq 1-a^2-(x-a)a\geq$ $1-a^2+\delta a$. So $\frac{(x+a)(x-a)}{(1-x^2)(1-a^2)}\leq \frac{(1+a)\delta}{(1-a^2+\delta a)(1-a^2)}\leq\varepsilon$ and so we can just take $\delta\leq\frac{(1-a^2)^2}{1+a-a\varepsilon}$. Is that right?
|
Here is the definition of continuity in terms of the epsilon-delta definition: $f$ is continuous at $a$ if and only if for any $\epsilon>0$, there exists $\delta>0$ such that if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.
Now we have $f(x)=\displaystyle\frac{x^2}{1-x^2}$. Then for any $a\in(0,1)$, we have (as you have calculated)
$$\tag{1}\left|\frac{x^2}{1-x^2}-\frac{a^2}{1-a^2}\right|=\left|\frac{(x+a)(x-a)}{(1-x^2)(1-a^2)}\right|=\frac{|x+a|\cdot|x-a|}{|(1-x^2)(1-a^2)|}\leq \frac{2|x-a|}{[1-(\frac{1+a}{2})^2](1-a^2)}$$
if $x\in(\displaystyle\frac{a}{2},\frac{1+a}{2})$.
Therefore, for any $\epsilon>0$, there exists $\delta=\min\{\displaystyle\frac{\epsilon}{2}[1-(\frac{1+a}{2})^2](1-a^2),\frac{a}{2},\frac{1-a}{2}\}>0$ such that if $|x-a|<\delta$, then
$$-\delta<x-a,\mbox{ or equivalently }, x>a-\delta>a-\frac{a}{2}=\frac{a}{2}$$
and
$$x-a<\delta,\mbox{ or equivalently }, x<a+\delta<a+\frac{1-a}{2}=\frac{1+a}{2}.$$
That is
$$\tag{2} x\in(\frac{a}{2},\frac{1+a}{2}).$$
Hence, using $(1)$ and $(2)$, we have
$$|f(x)-f(a)|=\left|\frac{x^2}{1-x^2}-\frac{a^2}{1-a^2}\right|<\frac{2\delta}{[1-(\frac{1+a}{2})^2](1-a^2)}\leq\epsilon.$$
|
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Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^{x+1} = 2 \cdot 3^{2x+2}$
and tried to write the polynomial as a binomial square, without success.
I know I should solve it using logarithm, but I don't see how to continue.
EDIT: WolframAlpha factors it as: $(3 \cdot 2^x - 2 \cdot 3^x)(2^{x+2} - 3^{x+2}) = 0$ and then the solution is straightforward. Any hint about how to reach that?
|
The following substitution may have to work:
$$2^x=t; ~~3^x=s$$
Note that the equation simplifies to, $$12 t^2-35st+ 18s^2=0$$
This factorises to $$(3t-2s)(4t-9s)=0$$
Therefore,
$$3\cdot2^x=2 \cdot 3^x ~~\text{or}~~2^{x+2}=3^{x+2}$$ Since, $(2,3)=1$, we have that $\boxed{x=1~~ \text{or}~~-2}$
Edited to add:
You can view that as a quadratic equation in $t$:
So, the solution will have to be, $$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}$$
This is a bit numerically taxing and honestly, I did not do it this way. Rather, I resorted to something that is equivalent to this. You need to write $18 \times 12$ as product of two numbers whose sum is $35$. Later, prefix a minus sign to these numbers and use them here.
With a little bit of playing around, with factorisations, you'll see they should be $27 \times 8$.
Added by dindoun
$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}
=\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24}
=\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s$$
|
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|
Infinite Series: Fibonacci/ $2^n$ I presented the following problem to some of my students recently (from Senior Mathematical Challenge- edited by Gardiner)
In the Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, 21, 34, 55,\ldots$ each term after the first two is the sum of the two previous terms. What is the sum to infinity of the series:
$$\frac{1}{2} + \frac{1}{4}+ \frac{2}{8} + \frac{3}{16} + \frac{5}{32} +\frac{8}{64} + \frac{13}{128} +\frac{21}{256} +\frac{34}{512}+ \frac{55}{1024} + \cdots$$
Now, I solved this using an infinite geometric matrix series (incorporating the matrix version of the relation $a_n= \frac{a_{n-1}}{2}+ \frac{a_{n-2}}{4}$), and my students, after much hinting on my part, googled the necessary string to stumble across Binet's formula (which allows one to split the series into two simple, if rather messy, geometrics).
Both of these are good methods, but neither really seems plausible for a challenge set for 15-18 year olds under exam conditions. So how is one supposed to do it?
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Let $\displaystyle S = \sum_{n=1}^{\infty} \frac{F_n}{2^n}.$ Then
$$ S = \sum_{n=1}^{\infty} \frac{F_n}{2^n} = \frac{1}{2} + \frac{1}{4} + \sum_{n=3}^{\infty} \frac{F_n}{2^n} = \frac{3}{4} + \sum_{n=3}^{\infty} \frac{F_{n-1}+F_{n-2} }{2^n}$$
$$ = \frac{3}{4} + \frac{1}{2} \sum_{n=3}^{\infty} \frac{ F_{n-1} }{2^{n-1} } + \frac{1}{4} \sum_{n=3}^{\infty} \frac{F_{n-2} }{2^{n-2} } $$
$$ = \frac{3}{4} + \frac{1}{2} \left( S - \frac{1}{2} \right) + \frac{1}{4} S.$$
Thus we have $ S = 2.$
To prove the series converges, we prove by induction that $ F_n \leq \phi^n$ where $ \phi = \frac{1+ \sqrt{5} }{2} \approx 1.618.$
The base cases are simple to check. Now assume there exists some integers $n-2$ and $n-1$ such that $ F_{n-2}\leq \phi^{n-2} $ and $ F_{n-1} \leq \phi^{n-1}.$ Then $$ F_n = F_{n-1}+ F_{n-2} \leq \phi^{n-1} + \phi ^{n-2} $$
$$= \phi^{n-2} ( \phi + 1) = \phi^{n-2} \phi^2 = \phi^n$$
which proves the claim. Note we used the fact that $\phi + 1 = \phi^2 $, the defining property of the golden ratio.
|
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Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series. Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$
Computing the Fourier series of $f$ and using Parseval's identity, I have computed $\zeta(2)$ and $\zeta(4)$.
How can I compute $ \zeta(6) $ now?
Fourier series of $ f $:
$$ S(f)= \frac{\pi^2}{6}-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$
$$ x=0, \zeta(2)=\pi^2/6$$
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Let $f(t):=t^3\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. The Fourier series of this function is
$$t^3=\sum_{k=1}^\infty {2(-1)^{k-1}(k^2\pi^2-6)\over k^3}\sin(kt)\qquad(-\pi\leq t\leq \pi).$$
We now use Parseval's formula
$$\|f\|^2=\sum_{k=0}^\infty |c_k|^2.$$
But
$$\|f\|^2={1\over\pi}\int_{-\pi}^\pi x^6\>dt={2\pi^6\over7}$$
and
$$c_k^2={4(k^2\pi^2-6)^2\over k^6}=4\left(\frac{36}{k^6}-\frac{12\pi^2}{k^4}+\frac{\pi^4}{k^2}\right),k\geq1.$$
Noting that
$$ \sum_{k=1}^\infty \frac{12\pi^2}{k^4}=12\pi^2\frac{\pi^4}{90}=\frac{2\pi^6}{15}, \sum_{k=1}^\infty\frac{\pi^4}{k^2}=\pi^4\frac{\pi^2}{6}=\frac{\pi^6}{6}$$
we have
$$ \sum_{k=1}^\infty \frac{1}{k^6}=\left({\pi^6\over14}+\frac{2\pi^2}{15}-\frac{\pi^4}{6}\right)/36=\frac{\pi^6}{945}. $$
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If $a+b+c+d=16$, then $(a+\frac{1}{c})^2+(c+\frac{1}{a})^2 + (b+\frac{1}{d})^2 + (d+\frac{1}{b})^2 \geq \frac{289}{4}$
If $a,b,c,d$ are positive integers and $a+b+c+d=16$, prove that
$$\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2 \geq \frac{289}{4}.$$
I know I have to use some inequality, not sure AM GM will work here or Minkowski inequality. But I only need hints, not a complete solution. I want to work on it myself.
Please provide only Hints.
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Since $f(x)=x^2+\frac{1}{x^2}$ is a convex function, by Jensen and AM-GM we obtain:
$$\sum_{cyc}\left(a+\frac{1}{c}\right)^2=\sum_{cyc}f(a)+2\sum_{cyc}\frac{a}{c}\geq 4f\left(\frac{a+b+c+d}{4}\right)+8=\frac{289}{4}$$
|
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How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes?
How do we show that For $x,y \ge 0$ real numbers, there exists a constant C suchthat: $$\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$$ Where $\pi(.)$ denotes thes prime counting function, is true?
the hint is to sieve n with $y< n \le x+y $:
$$\pi (x+y) \le 1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + \sum_{n\le x+y} 1 = $$
$$1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + [x+y]$$
because: $1\le n = dm \le x+y \Leftrightarrow \frac{1}{d}\le m \le \frac{x+y}{y} $
so: $$\sum_{n\le x+y , d|n}1 = [\frac{x+y}{d}]$$
then that gives: $$\pi (x+y) < 1+ [x+y] - [\frac{x+y}{2}] - [\frac{x+y}{3}] + [\frac{x+y}{6}]$$
so that will give: $\pi (x+y) < \frac{x+y}{3} + 3$ but also we get : $\pi(y) < \frac{y}{3} + 3$ so for any constant $C\ge 0$ it will surely hold that:
$$\pi(x+y) - \pi(y) < \frac{x}{3} \le \frac{x}{3} + C$$
Is this correct?
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You are asking us to show that your inequality holds for any $x, y \ge 0$ and for any constant $C$, which is obviously false. Perhaps this is what you mean:
Show that there exists a constant $C$ such that for any real numbers $x, y \ge 0$, $\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$.
|
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How I can find the value of $abc$ using the given equations? If I have been given the value of
$$\begin{align*}
a+b+c&= 1\\
a^2+b^2+c^2&=9\\
a^3+b^3+c^3 &= 1
\end{align*}$$
Using this I can get the value of
$$ab+bc+ca$$
How i can find the value of $abc$ using the given equations?
I just need a hint.
I have tried by squaring the equations.
But could not get it.
Thanks in advance.
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You can get a term involving $abc$ by cubing $a+b+c$:
$$\begin{align*}
(a+b+c)^3 &= (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3\\
&= a^3+3a^2b+3ab^2 + b^3 + 3a^2c+\color{blue}{6abc} + 3b^2c + 3ac^2 + 3bc^2 + c^3.
\end{align*}$$
Now use the other information you have to try to find the value of $abc$.
For example, you know this whole thing equals $(a+b+c)^3 = 1$. You also know the value of $a^3+b^3+c^3$...
|
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|
Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use.
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No induction necessary.
$$10 + 3\cdot 4^n + 5 \equiv 1 + 3\cdot 4^n + 5 \equiv 6 + 3\cdot 4^n \pmod9$$
Since everything including the base is divisible by three, this reduces to
$$6 + 3\cdot 4^n \pmod9 \iff 2 + 4^n \equiv 2 + 1 \equiv 0 \pmod3$$
|
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The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real numbers
The correct answer is $C$
my solution:
$\ 2x-3 = -(2x-3)$ when $2x-3$ $\geqslant$ $0$ $\Rightarrow$ $x$ = $\frac{3}{2}$
$-(2x-3) = -(2x-3)$ when $2x-3$ $<$ $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is presented in interval notation (-$\infty$ , $\frac{3}{2}$].
Any help is appreciated.
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In your second case you write $2x - 3 < 0$. I don't understand how you get $\implies 0 = 0$.
From $2x - 3 < 0$ you get $2x < 3$ and hence $x < \frac{3}{2}$.
Now you take the union of your two sets of solutions to get $x \leq \frac{3}{2}$, or in other words, $x \in (-\infty , \frac{3}{2}]$
|
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How to compute $\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}$? I got stuck at the computation of the sum
$$
\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}.
$$
I think there is no purely combinatorial proof here since the sum can achieve negative values. Could you give me solution, it seems to involve generating functions.
|
Indeed, generating function method works. Let $c_n$ denoe the given sum. Then we have
$$\begin{align*}
\sum_{n=0}^{\infty} c_n y^{2n}
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^k y^{2n}}{(2n-2k)!k!} \int_{0}^{\infty} x^{2n-k} e^{-x} \; dx \\
&= \int_{0}^{\infty} \sum_{k=0}^{n} \sum_{n=0}^{\infty} \frac{(yx)^{2n-2k}}{(2n-2k)!} \frac{(-y^2 x)^k}{k!} e^{-x} \; dx \\
&= \int_{0}^{\infty} \cosh(yx) e^{-(y^2+1)x} \; dx \\
&= \int_{0}^{\infty} \frac{1}{2} \left( e^{-(y^2-y+1)x} + e^{-(y^2+y+1)x} \right) \; dx \\
&= \frac{1}{2} \left( \frac{1}{y^2 - y + 1} + \frac{1}{y^2 + y + 1} \right) \\
&= \frac{1 - y^4}{1 - y^6} \\
&= 1 - y^4 + y^6 - y^{10} + \cdots.
\end{align*}$$
Thus by comparing the coefficients, we have
$$ c_n = \begin{cases}
1 & n \equiv 0 \ (\mathrm{mod} \ 3) \\
0 & n \equiv 1 \ (\mathrm{mod} \ 3) \\
-1 & n \equiv 2 \ (\mathrm{mod} \ 3)
\end{cases}.$$
Similar calculation also shows that
$$ \sum_{k=0}^{n} \binom{2n-k}{k} = F_{2n+1},$$
where $F_0 = 0, F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$ is the Fibonacci sequence.
Okay, here is a direct approach, motivated by Brain M. Scott's illuminating answer. By Pascal's triangle,
$$\begin{align*}
c_{n+1}
&= \sum_{k=0}^{n+1}(-1)^k \binom{2n+2-k}{k} \\
&= \sum_{k=0}^{n+1}(-1)^k \binom{2n+1-k}{k-1} + \sum_{k=0}^{n+1}(-1)^k \binom{2n+1-k}{k} \\
&= - \sum_{k=0}^{n+1}(-1)^k \binom{2n-k}{k} + \sum_{k=0}^{n+1}(-1)^k \binom{2n+1-k}{k} \\
&= - c_n + \sum_{k=0}^{n+1}(-1)^k \binom{2n+1-k}{k}.
\end{align*}$$
But applying exactly the same technique to the last sum, we have
$$\begin{align*}
\sum_{k=0}^{n+1}(-1)^k \binom{2n+1-k}{k}
&= \sum_{k=0}^{n+1}(-1)^k \binom{2n-k}{k-1} + \sum_{k=0}^{n+1}(-1)^k \binom{2n-k}{k} \\
&= - \sum_{k=0}^{n}(-1)^k \binom{2n-1-k}{k} + \sum_{k=0}^{n}(-1)^k \binom{2n-k}{k} \\
&= - \sum_{k=0}^{n}(-1)^k \binom{2n-1-k}{k} + c_n.
\end{align*}$$
Plugging back, we obtain
$$ c_{n+1} = - \sum_{k=0}^{n}(-1)^k \binom{2n-1-k}{k}.$$
Thus we have
$$c_{n+1}
= - c_n + \sum_{k=0}^{n+1}(-1)^k \binom{2n+1-k}{k}
= - c_n - c_{n+2},$$
or equivalently
$$c_{n+2} + c_{n+1} + c_n = 0.$$
So we have $c_{n+3} = c_n$ and the proof is complete.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find one side of a parallelogram given its width and two opposite corners Given a parallelogram with two vertical sides defined by a known width (distance between two parallel sides) and two opposite points, what is the equation for the length of the either side?
Real-world application:
I need to determine the cut angle on a board of a given thickness spanning between two points. Given the length of one side, I can easily determine the length of all other sides/angles.
|
Let $\Delta y = C_y - A_y$ and $\Delta x = C_x - A_x$. Then using the right triangle that you've drawn, the length of the non-vertical sides of the parallelogram is
$$\sqrt{\Delta x^2 + (\Delta y - F)^2}.$$
If you slide your blue "W" all the way up so that it touches the "F" line you've drawn, then you have a new little right triangle that's similar to the big right triangle just mentioned. We therefore have the following proportion (big hypotenuse / small hypotenuse = leg in big triangle / leg in small triangle):
$$\frac{\sqrt{\Delta x^2 + (\Delta y - F)^2}}{F} = \frac{\Delta x}{W}.$$
Now solve for $F$:
$$\begin{align}
W\sqrt{\Delta x^2 + (\Delta y - F)^2} &= F \Delta x \\
W^2 (\Delta x^2 + (\Delta y - F)^2) &= F^2 \Delta x^2 \\
W^2 (\Delta x^2 + \Delta y^2 + F^2 - 2F\Delta y) &= F^2\Delta x^2 \\
(W^2 - \Delta x ^2)\cdot F^2 -2W^2\Delta y \cdot F + W^2(\Delta x^2 + \Delta y^2) &= 0 \end{align}$$
The quadratic formula gives
$$\begin{align}F &= \frac{2W^2\Delta y \pm \sqrt{4W^4\Delta y^2 - 4\cdot(W^2-\Delta x^2) \cdot W^2 (\Delta x^2 + \Delta y^2)}}{2(W^2-\Delta x^2)}\\
&= \frac{W^2\Delta y \pm \sqrt{W^4\Delta y^2 - W^4\Delta x^2 - W^4 \Delta y ^2 + W^2 \Delta x^4 + W^2 \Delta x^2 \Delta y^2}}{W^2-\Delta x^2} \\
&= \boxed{\displaystyle\frac{W^2\Delta y \pm W \Delta{x} \sqrt{-W^2 + \Delta x^2 + \Delta y^2}}{W^2-\Delta x^2} }\end{align}.$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to Derive a Double Angle Identity. How does one derive the following two identities:
$$\begin{align*}
\cos 2\theta &= 1-2\sin^2\theta\\
\sin 2\theta &= 2\sin\theta\cos\theta
\end{align*}$$
|
The first one follows from:
$$\cos 2\theta = \cos(\theta +\theta) = \cos \theta \cos \theta - \sin \theta \sin \theta = \cos^2 \theta - \sin^2 \theta .$$ Now use the fact $\cos^2 \theta + \sin^2 \theta =1.$
The second one follows from
$$ \sin 2\theta = \sin(\theta +\theta) = \sin \theta \cos \theta + \cos \theta \sin \theta. $$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $u_n$ is arithmetic sequence if $\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$ Let $(u_n)$ be a sequence $u_i\neq0$ and
$$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$ for all $n\geq3$
Prove that the sequence $(u_n)$ is arithmetic sequence
|
You need two key features of an arithmetic progression:
$$a_n = a_1 +d(n-1)$$
and
$$a_n-a_{n-1}=d$$ (which is a consequence of the previous one).
Thus
$$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$
$$d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-
1}u_n}\right)=\frac{d(n-1)}{u_1u_n}$$
Now sum $\dfrac{u_1}{u_1 u_n}$ to get
$$\frac{u_1}{u_1 u_n} + d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}
{u_3u_4}+....+\frac{1}{u_{n-1}u_n}\right)=\frac{u_1+d(n-1)}{u_1u_n}$$
$$\frac{1}{u_n} + \frac{d}{u_1u_2}+\frac{d}{u_2u_3}+\frac{d}{u_3u_4}+....+\frac{d}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
Now replace $d$ by the differences, conveniently:
$$\frac{1}{u_n} + \frac{u_2-u_1}{u_1u_2}+\frac{u_3-u_2}{u_2u_3}+\frac{u_4-u_3}{u_3u_4}+....+\frac{u_n-u_{n-1}}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
$$\frac{1}{u_n} + \frac{1}{u_1}-\frac{1}{u_2}+\frac{1}{u_2}-\frac{1}{u_3}+-....+\frac{1}{u_{n-1}}-\frac{1}{u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
This telescopes, giving
$$\frac{1}{u_1}=\frac{u_1+d(n-1)}{u_1u_n}$$
or
$$u_n=u_1+d(n-1)$$
|
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|
Writing $1$ in form of $\frac{1}{t_1}+\cdots+\frac{1}{t_n}$
Possible Duplicate:
Prove that any rational can be expressed in the form $\sum\limits_{k=1}^n{\frac{1}{a_k}}$, $a_k\in\mathbb N^*$
Can anyone help me with this problem? It's a little strange:
Let $M$ be a natural number. Prove that we can write $1=\frac{1}{t_1}+\cdots+\frac{1}{t_n}$ such that all $t_i$'s are distinct natural numbers greater than $M$.
|
Hint. $$\frac{1}{n} +\frac{1}{n} = \frac{1}{n} + \frac{1}{n+1}+\frac{1}{n(n+1)}.$$
For example, say $N=3$. Then we can write:
$$\begin{align*}
1 &= \frac{1}{4}+\frac{1}{4} + \frac{1}{4}+\frac{1}{4}\\
&= \frac{1}{4} + \frac{1}{5}+\frac{1}{20} + \frac{1}{5}+\frac{1}{20}+\frac{1}{5}+\frac{1}{20}\\
&= \frac{1}{4}+\frac{1}{5}+\frac{1}{20} + \frac{1}{6}+\frac{1}{30} + \frac{1}{21}+\frac{1}{420} + \frac{1}{6}+\frac{1}{30} + \frac{1}{21}+\frac{1}{420}\\
&= \frac{1}{4}+\frac{1}{5}+\frac{1}{20}+\frac{1}{6}+\frac{1}{30}+\frac{1}{21}+\frac{1}{420} + \frac{1}{7}+\frac{1}{42} + \frac{1}{31}\\
&\qquad\mathop{+}\frac{1}{930} + \frac{1}{22}+\frac{1}{462} + \frac{1}{421}+\frac{1}{176820}\\
&= \frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{30}+\frac{1}{31}+\frac{1}{42}\\
&\qquad\mathop{+}\frac{1}{420}+\frac{1}{421}+\frac{1}{462}+\frac{1}{930}+\frac{1}{176820}.
\end{align*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
ordinary differential equations test the exactness of the O.D.E $(4xy+2x^2 y)dx+(2x^3+3y^2)dy=0$ and hence find the potential function which is the general solution.I tried to solve it and I reached ending up failing to get the integrating factor.please help me
|
Hint:
$(4xy+2x^2y)dx+(2x^3+3y^2)dy=0$
$(2x^3+3y^2)dy=-((2x^2+4x)y)dx$
$(2x^3+3y^2)\dfrac{dy}{dx}=-(2x^2+4x)y$
Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{(2x^3+3y^2)}{2y}\dfrac{du}{dx}=-(2x^2+4x)y$
$(2x^3+3y^2)\dfrac{du}{dx}=-(4x^2+8x)y^2$
$(2x^3+3u)\dfrac{du}{dx}=-(4x^2+8x)u$
This belongs to an Abel equation of the second kind.
Let $v=u+\dfrac{2x^3}{3}$ ,
Then $u=v-\dfrac{2x^3}{3}$
$\dfrac{du}{dx}=\dfrac{dv}{dx}-2x^2$
$\therefore3v\left(\dfrac{dv}{dx}-2x^2\right)=-(4x^2+8x)\left(v-\dfrac{2x^3}{3}\right)$
$3v\dfrac{dv}{dx}-6x^2v=-(4x^2+8x)v+\dfrac{8x^4(x+2)}{3}$
$3v\dfrac{dv}{dx}=(2x^2-8x)v+\dfrac{8x^4(x+2)}{3}$
$v\dfrac{dv}{dx}=\dfrac{(2(x-2)^2-8)v}{3}+\dfrac{8x^4(x+2)}{9}$
Let $s=x-2$ ,
Then $\dfrac{dv}{dx}=\dfrac{dv}{ds}\dfrac{ds}{dx}=\dfrac{dv}{ds}$
$\therefore v\dfrac{dv}{ds}=\dfrac{(2s^2-8)v}{3}+\dfrac{8(s+2)^4(s+4)}{9}$
|
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|
Integrating $\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$ Can someone please help me integrate $$\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$$ the question says, as a hint, use $\cos\theta = \frac{z + z^{-1}}{2}$ with $|z|=1$. I'm not really sure where to start.
|
Let $p(x) = (x-x_1) (x-x_2) \cdots (x-x_n).$ An application of the product rule shows this useful identity (for all $x$ not a root of $p$): $$\frac{p'(x) }{p(x) }= \sum_{k=1}^n \frac{1}{x-x_k}.$$
Now since $\displaystyle \cos t = \frac{e^{it} + e^{-it}}{2} $ we have $$ 1+ 8\cos^2 \left( \frac{k\pi}{n} \right) = 5 + 2a_k + 2a^{-1}_k $$ where $a_k= \exp(2k\pi i/n),$ which are precisely the roots of $x^n-1.$ Thus, we have (after some partial fractions) that
$$\sum_{k=0}^{n-1} \frac{1}{1+8\cos^2(k\pi/n)} = \frac{1}{6} \sum_{k=0}^{n-1} \frac{1}{\frac{-1}{2} - a_k} - \frac{2}{3} \sum_{k=0}^{n-1} \frac{1}{-2-a_k} .$$
Applying the identity we developed lets us evaluate these sums in closed form, and we have
$$ \sum_{k=0}^{n-1} \frac{1}{1+8\cos^2(k\pi/n)} = \frac{2n}{3} \frac{2^{n-1} }{2^n + (-1)^{n+1} } + \frac{1}{6} \frac{n (-1/2)^{n-1} }{ (-1/2)^n -1} .$$
Hence $$ \int^1_0 \frac{d\theta}{1+8\cos^2(\pi \theta)} = \lim_{n\to\infty} \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{1+8\cos^2(k\pi/n)} = \frac{1}{3}$$
which immediately implies that your integral has value $2\pi/3.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum the series: $ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \cdots \ \text{ad inf}$ How do I sum the following series?
$$ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \sin{3\alpha} + \cdots \ \text{ad inf}$$
|
Let $$ S = \frac{1}{2}\cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4}\cdot \sin{2\alpha} + \frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\cdot \sin{3\alpha} + \cdots \ \text{ad inf}$$ and $$ C = 1 + \frac{1}{2}\cdot \cos\alpha + \frac{1\cdot 3}{2 \cdot 4}\cdot \cos{2 \alpha} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \cos{3\alpha} + \cdots \ \text{ad inf}$$
From this you have $C+iS = (1-e^{\alpha \cdot i})^{-1/2}$, if $\alpha \neq 2n\pi$. Again you have
\begin{align*}
C+iS &= \{1-\cos\alpha-i\: \sin\alpha\}^{-1/2} \\\ &= \biggl\{2 \sin\frac{\alpha}{2} \biggl(\sin\frac{\alpha}{2} - i\: \cos\frac{\alpha}{2}\biggr)\biggr\}^{-1/2} \\\ &= \biggl\{2 \sin\frac{\alpha}{2}\biggr\}^{-1/2} \ \biggl\{\cos\biggl(\frac{\alpha}{2}-\frac{\pi}{2}\biggr) + i \: \sin\biggl(\frac{\alpha}{2}-\frac{\pi}{2}\biggr)\biggr\}^{-1/2} \\\ &= \biggl\{2 \sin\frac{\alpha}{2}\biggr\}^{-1/2} \: \biggl\{\cos\biggl(\frac{\pi-\alpha}{4}\biggr) + i \: \sin\biggl(\frac{\pi-\alpha}{4}\biggr)\biggr\}
\end{align*}
Now equate real and imaginary parts to get the answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Definition of derivative $f(x) = \sqrt{3-5x}$ I am not sure how to factor this out
$$f(x) = \sqrt{3-5x}$$
I then make it $f(x) = \frac {\sqrt{3-5(x+h)} - \sqrt{3-5x}}{h}$
I tried to multiply by the first time + the second term from the numerator which I called x and y
$$\frac {x - y}{h} \cdot \frac {x + y}{x+y}$$
which gives me
$$\frac {x^2 - y^2}{h(x+y)}$$
From here it gets very difficult
$$\frac {5}{ h \sqrt{3-5(x+h)} - \sqrt{3-5x}}$$
|
$$\frac{\sqrt{3-5(x+h)}-\sqrt{3-5x}}{h}\times\frac{\sqrt{3-5(x+h)}+\sqrt{3-5x}}{\sqrt{3-5(x+h)}+\sqrt{3-5x}} $$
$$=\frac{\big(3\color{Red}-5(x+\color{Red}h)\big)-\big(3-5x\big)}{h\big(\sqrt{3-5(x+h)}\color{Red}+\sqrt{3-5x}\big)}=\frac{\color{Red}-5\color{Red}h}{h\big(\sqrt{3-5(x+h)}\color{Red}+\sqrt{3-5x}\big)}.$$
After you cancel the $h$'s you can plug in $h=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Bivariate recurrence relation Consider the following recurrence relation:
$$A(h,0)=1\\
A(h,h)=c^h\\
A(h,r)=A(h-1,r)+(c-1)\cdot A(h-1,r-1).$$
Obviously, this is just a generalization of A008949, where $c=2$. Since I'm pretty sure we're not the first ones dealing with it -- is there some source where it was already solved? I'm really only looking for a source to cite, we already have the solution.
Thanks in advance!
Sacha
|
Interesting double recurrence. Define the generating function $g(x, y) = \sum_{r, s} A(r, s) x^r y^s$, write without subtraction in indices:
$$
A(r + 1, s + 1) = A(r, s + 1) + (c - 1) A(r, s)
$$
Multiply by $x^r y^s$ and recognize the sums:
\begin{align}
\sum_{r, s} A(r + 1, s) x^r y^s
&= \frac{1}{x} \left( g(x, y) - \sum_s A(0, s) y^s \right) \\
&= \frac{g(x, y) - g(0, y)}{x} \\
\sum_{r, s} A(r + 1, s + 1) x^r y^s
&= \frac{g(x, y) - g(0, y) - g(x, 0) + g(0, 0)}{x y}
\end{align}
The $g(0, 0)$ term was subtracted twice in the last expression, and has to be restored.
Too bad that your boundary conditions take the form they do. You have:
$$
g(x, 0) = \frac{1}{1 - x}
$$
while $g(0, y)$ remains unknown. In any case:
$$
\frac{g(x, y) - g(0, y) - 1 / (1 - x) + 1}{x y}
= \frac{g(x, y) - 1 / (1 - x)}{y} + (c - 1) g(x, y)
$$
Solving for $g(x, y)$ gives:
\begin{align}
g(x, y)
&= \frac{g(0, y)}{1 - x - (c - 1) x y} \\
&= g(0, y) \frac{1}{1 - ((c - 1) y + 1)x} \\
&= g(0, y) \sum_r ((c - 1) y + 1)^r x^r
\end{align}
Thus:
$$
[x^r y^r] g(x, y) = [y^r] g(0, y) ((c - 1) y + 1)^r = c^r
$$
This is:
$$
\sum_{0 \le k \le r} \binom{r}{k} (c - 1)^{r - k} A(0, k) = c^r
$$
If we now define the exponential generating function:
$$
a(z) = \sum_{s \le 0} A(0, s) \frac{z^s}{s!}
$$
multiply the last equation by $\frac{z^r}{r!}$, and sum over $r \ge 0$,
the resulting left hand side is the product of $a(z)$ and:
$$
\sum_{k \ge 0} (c - 1)^k \frac{z^k}{k!} = \exp((c - 1) z)
$$
while the right hand side is:
$$
\sum_{k \ge 0} c^k \frac{z^k}{k!} = \exp(c z)
$$
So:
$$
a(z) = \exp(z)
$$
Sneaky... it is just $A(0, s) = 1$, and thus $g(0, y) = 1 / (1 - y)$. We get:
\begin{align}
A(r, s)
&= [x^r y^s] \frac{1}{(1 - y) (1 - x - (c - 1) x y} \\
&= [x^r y^s] \frac{1}{1 - y} \sum_k (1 + (c - 1) y)^k x^k \\
&= [y^s] \frac{(1 + (c - 1) y)^r}{1 - y} \\
&= \sum_{0 \le k \le s} \binom{r}{k} (c - 1)^k
\end{align}
|
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|
Approximation for $\pi$ I just stumbled upon
$$ \pi \approx \sqrt{ \frac{9}{5} } + \frac{9}{5} = 3.141640786 $$
which is $\delta = 0.0000481330$ different from $\pi$. Although this is a rather crude approximation I wonder if it has been every used in past times (historically). Note that the above might also be related to the golden ratio $\Phi = \frac{\sqrt 5 + 1}{2} $ somehow (the $\sqrt5$ is common in both).
$$ \Phi = \frac{5}{6} \left( \sqrt{ \frac{9}{5} } + \frac{9}{5} \right) - 1 $$
or
$$ \Phi \approx \frac{5}{6} \pi - 1 $$
I would like to know if someone (known) has used this, or something similar, in their work. Is it at all familiar to any of you?
Related Question (link).
|
I have not seen it before. Note that $\pi = \sqrt{a} + a$ where $a = (1+2\,\pi -\sqrt {1+4\,\pi })/2$, and what you're saying is that a rational approximation of $a$ is
$9/5$. In fact, we have a continued fraction
$$ a = 1 + \dfrac{1}{1 + \dfrac{1}{3+ \dfrac{1}{1+\dfrac{1}{1139 + \ldots}}}}$$
and $1+1/(1+1/(3+1/1)) = 9/5$. The fact that the first omitted element, $1139$, is so large makes this a very good approximation: the error in approximating $a$ by $9/5$ is only about $3.5 \times 10^{-5}$. Four elements later comes $7574$, so an even better approximation is $1+1/(1+1/(3+1/(1+1/(1139+1/(1+1/(15+1/1)))))) = 174530/96963$ with error about $1.4 \times 10^{-14}$.
EDIT: Perhaps even more remarkable are
$$ \eqalign{\pi - \sqrt{1 + \dfrac{47}{35} \pi} &\approx \dfrac{6}{7}\cr
\pi - \sqrt{\dfrac{3}{5} + \dfrac{5}{2} \pi } &\approx \dfrac{216}{923}\cr}$$
corresponding to the continued fractions
$$ \eqalign{\pi - \sqrt{1 + \dfrac{47}{35} \pi} &= \dfrac{1}{1+ \dfrac{1}{6 + \dfrac{1}{126402+ \ldots}}}\cr
\pi - \sqrt{\dfrac{3}{5} + \dfrac{5}{2} \pi} &=
\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{19+\dfrac{1}{133286+\ldots}}}}}}}\cr}$$
|
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|
Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant find this one. Perhaps it's wrong? Could anyone confirm it?
I tried Sage, and it calculates it correctly, but not with steps.
|
The typical way to start dealing with an integral that has $\sqrt{a^2+t^2}$ (in your case, $a=\sqrt{2}$) is to use a trigonometric substitution or a hyperbolic substitution.
If we use hyperbolic substitution, we want to use the fact that
$$1 + \sinh^2 z = \cosh^2 z.$$
Set $t=\sqrt{2}\sinh z$. Then
$$\sqrt{2 + t^2} = \sqrt{2 + 2\sinh^2z} = \sqrt{2(1+\sinh^2 z)} = \sqrt{2\cosh^2 z} = \sqrt{2}\cosh z.$$
Also, if $t=\sqrt{2}\sinh z$, then $dt =\sqrt{2}\cosh z\,dz$. Therefore,
$$\int\sqrt{2+t^2}\,dt = \int \sqrt{2}\cosh z \sqrt{2}\cosh z\,dz = 2\int\cosh^2 z\,dz.$$
Now we need to solve the integral of $\cosh^2 z$. We can do this by parts, similarly to how we solve the integral of $\cos^2t$. Let $u=\cosh z$, $dv=\cosh z\,dz$. Then $du=\sinh z\,dz$, $v=\sinh z$, so
$$\begin{align*}
\int\cosh^2 z\,dz &= \cosh z\sinh z - \int \sinh^2z\,dz\\
&= \cosh z \sinh z - \int(cosh^2z - 1)\,dz\\
&= \cosh z\sinh z +z - \int\cosh^2z\,dz.
\end{align*}$$
Hence
$$\begin{align*}
\int\cosh^2z\,dz &= z + \cosh z\sinh z - \int\cosh^2z\,dz\\
2\int\cosh^2z\,dz &= z+ \cosh z\sinh z + C\\
\int\cosh^2 z\,dz &= \frac{1}{2}z + \frac{1}{2}\cosh z\sinh z + C.
\end{align*}$$
Then to get it back into a function of $t$ we remember that $t=\sqrt{2}\sinh z$, so $\sinh z = \frac{\sqrt{2}}{2}t$. Then $z= \mathop{\mathrm{arcsinh}}\left(\frac{\sqrt{2}}{2}t\right)$, and
$$\sinh z\cosh z = \sinh z \sqrt{1+\sinh^2z} = \frac{\sqrt{2}}{2}t\sqrt{1 + \frac{1}{2}t^2} = \frac{1}{2}\sqrt{2+t^2}.$$
Therefore,
$$\int\sqrt{2+t^2}\,dt = 2\int\cosh^2z\,dz = 2\mathop{\mathrm{arcsinh}}\left(\frac{\sqrt{2}}{2}t\right) + \sqrt{2+t^2}+C.$$
Now plugging into the definite integral gives you a solution.
If we use trigonometric substitutions, we want to use the fact that
$$1 + \tan^2 \theta = \sec^2\theta$$
Set $t=\sqrt{2}\tan\theta$, with $-\frac{\pi}{2} \lt \theta\lt\frac{\pi}{2}$. Then
$$\sqrt{2+t^2} = \sqrt{2+2\tan^2\theta} = \sqrt{2}\sqrt{1+\tan^2\theta} = \sqrt{2}\sqrt{\sec^2\theta} = \sqrt{2}|\sec\theta|.$$
Since $\sec\theta\gt 0$ on $-\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2}$, we get that $\sqrt{2+t^2} = \sqrt{2}\sec\theta$.
Also, if $t=\sqrt{2}\tan\theta$, then $dt = \sqrt{2}\sec^2\theta\,d\theta$. Therefore,
$$\int\sqrt{2+t^2}\,dt = \int \sqrt{2}\sec\theta\sqrt{2}\sec^2\theta\,d\theta = 2\int\sec^3\theta\,d\theta.$$
So we need to find $\int\sec^3\theta\,d\theta = \int\frac{1}{\cos^3\theta}\,d\theta$.
This can be done any number of ways. Using integration by parts, we get
$$\int\frac{d\theta}{\cos^3\theta} = \frac{1}{2}\tan\theta + \frac{1}{2}\int\frac{d\theta}{\cos\theta}.$$
And
$$\int\frac{d\theta}{\cos\theta} = \int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta|+C.$$
Thus, we get that
$$\int\sqrt{2+t^2}\,dt = \tan\theta + \ln|\sec\theta+\tan\theta|+C.$$
To change it into a formula using $t$, we use the fact that $t=\sqrt{2}\tan\theta$. Therefore, $\tan\theta = \frac{\sqrt{2}}{2}t$; and
$$\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + \frac{t^2}{2}} = \frac{\sqrt{2}}{2}\sqrt{2+t^2}.$$
Plugging in and evaluating gives the desired result.
|
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"url": "https://math.stackexchange.com/questions/147788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding square roots of $\sqrt 3 +3i$ I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.
$w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.
But how did they get from $\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$?
And can one just 'let $z^2=w$' as above?
Edit:
$w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$
|
I hope these will answer your direct questions:
The first step is to find the modulus-argument form: the modulus of the complex number comes from $\sqrt{x^2 + y^2}$.
They have taken this as a factor, leaving an 'obvious' way to find the argument by knowing the trigonometric exact values. It's perhaps easier to find the argument using arctan with reference to a picture of the Argand diagram.
Let $z^2 = w$ is from the definition of the square root. Really, they are saying: we want to find a $z$ so that $z^2 = w$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/148871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Equality involving Appell hypergeometric function After some algebra, Wolfram online integrator gave me the following:
$$\tag{1} \int (1-a-t)^{N-2}\ \sqrt{2t-t^2}\ \text{d} t = c\ \cdot t^{3/2}\ \operatorname{F}_1 \left( \frac{3}{2}; -\frac{1}{2}, 2-N; \frac{5}{2}; \frac{t}{2}, \frac{t}{1-a}\right)+C$$
where:
*
*$\operatorname{F}_1$ is the Appell hypergeometric function of two variables defined by the expansion:
$$\operatorname{F}_1(\alpha; \beta, \beta^\prime ; \gamma; x,y):= \sum_{m,n=0}^\infty \frac{(\alpha)_{m+n}\ (\beta)_m\ (\beta^\prime)_n}{m!\ n!\ (\gamma)_{m+n}}\ x^m\ y^n$$
($(a_k):=\frac{\Gamma (a+k)}{\Gamma(a)}$ is the Pochhammer symbol), which converges in the region $|x|,|y|<1$;
*$N\geq 3$ an integer, $a\in ]0,1[$ and $t\in [0,1-a]$ (so the RHside makes sense);
*$c=c(a,N)$ is a known constant and $C$ is an arbitrary constant (coming from indefinite integration).
My question is:
How can I get (1) without using any software?
I suppose a series expansion of the integrand and term by term integration should be used, but I could not figure out how to do explicit computations.
Neither I succeeded in simplifying the derivative of the LHside to get the integrand...
Any hint will be appreciated.
|
I found also the following term-by-term integration solution.
One has:
$$(1-a-t)^{N-2}\ \sqrt{2t-t^2} = \sqrt{2}\ (1-a)^{N-2}\ \sqrt{t}\ \left( 1-\frac{t}{1-a}\right)^{N-2}\ \sqrt{1-\frac{t}{2}}$$
and by the binomial theorem:
$$\begin{split}
\left( 1-\frac{t}{1-a}\right)^{N-2} &= \sum_{k=0}^{N-2} (-1)^k\ \binom{N-2}{k}\ \left(\frac{t}{1-a}\right)^k\\
&= \sum_{k=0}^{N-2} \frac{(2-N)_k}{k!}\ \left(\frac{t}{1-a}\right)^k\\
&= \sum_{k=0}^\infty \frac{(2-N)_k}{k!}\ \left(\frac{t}{1-a}\right)^k\\
\sqrt{1-\frac{t}{2}} &= \sum_{k=0}^\infty (-1)^k\ \binom{1/2}{k}\ \left( \frac{t}{2}\right)^k\\
&= \sum_{k=0}^\infty \frac{(-1/2)_k}{k!}\ \left( \frac{t}{2}\right)^k\; ;
\end{split}$$
now, since both series in the rightmost sides converge absolutely, Mertens theorem applies and one can take the Cauchy product:
$$\begin{split} \left( 1-\frac{t}{1-a}\right)^{N-2}\ \sqrt{1-\frac{t}{2}} &= \left( \sum_{k=0}^\infty \frac{(2-N)_k}{k!}\ \left(\frac{t}{1-a}\right)^k\right)*\left( \sum_{k=0}^\infty \frac{(-1/2)_k}{k!}\ \left( \frac{t}{2}\right)^k\right)\\
&= \sum_{k=0}^\infty \sum_{h=0}^k \frac{(2-N)_h}{h!}\ \left(\frac{t}{1-a}\right)^h\ \frac{(-1/2)_{k-h}}{(k-h)!}\ \left( \frac{t}{2}\right)^{k-h}\\
&= \sum_{k=0}^\infty \sum_{h=0}^k \frac{(2-N)_h\ (-1/2)_{k-h}}{h!\ (k-h)!}\ \frac{1}{2^h\ (1-a)^{k-h}}\ t^k
\end{split}$$
with the rigthmost side converging absolutely and uniformly on compact subsets.
Therefore one gets:
$$(1-a-t)^{N-2}\ \sqrt{2t-t^2} = \sqrt{2}\ (1-a)^{N-2}\ \sum_{k=0}^\infty \sum_{h=0}^k \frac{(2-N)_h\ (-1/2)_{k-h}}{h!\ (k-h)!}\ \frac{1}{2^h\ (1-a)^{k-h}}\ t^{k+1/2}$$
and a rearrangement of the RHside yields:
$$(1-a-t)^{N-2}\ \sqrt{2t-t^2} = \sqrt{2}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(2-N)_m\ (-1/2)_n}{m!\ n!}\ \frac{1}{2^m\ (1-a)^n}\ t^{m+n+1/2}\; ;$$
thus term-by-term integration shows that:
$$\begin{split}
\int (1-a-t)^{N-2}\ \sqrt{2t-t^2}\ \text{d} t &= \sqrt{2}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(2-N)_m\ (-1/2)_n}{m!\ n!}\ \frac{1}{2^m\ (1-a)^n}\ \int t^{m+n+1/2}\ \text{d} t\\
&= \sqrt{2}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(2-N)_m\ (-1/2)_n}{m!\ n!\ (3/2 +m+n)}\ \frac{1}{2^m\ (1-a)^n}\ t^{m+n+3/2}\ + C\\
&= \frac{2\sqrt{2}}{3}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(3/2)_{m+n}\ (2-N)_m\ (-1/2)_n}{m!\ n!\ (5/2)_{m+n}}\ \frac{1}{2^m\ (1-a)^n}\ t^{m+n+3/2}\\
&\phantom{=} +C\\
&= \frac{2\sqrt{2}}{3}\ (1-a)^{N-2}\ t^{3/2}\ F_1\left( \frac{3}{2}; -\frac{1}{2}, 2-N; \frac{5}{2}; \frac{t}{2}, \frac{t}{1-a}\right) +C
\end{split}$$
because:
$$\frac{3}{2}\ \frac{1}{m+n+3/2} = \frac{(3/2)_{m+n}}{(5/2)_{m+n}}\; .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/151548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.
I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem:
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok.
If you're answering with an example then use one pair per answer.
I just want to see that clock with another pair of numbers :)
Notes for the current clock:
1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.
5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$
|
For $n=9$ and $k=9$ here is a solution:
$1=\left(9+\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+\left(9-99\right)\right)\right)}\right)}\right)$
$2=\frac{9}{\left(9+\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{99}\right)}\right)}\right)}$
$3=\left(9-\left(9+\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)\right)\right)$
$4=\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+\left(9+\left(9+9\right)\right)\right)\right)}\right)}\right)$
$5=\left(9+\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+99\right)\right)}\right)}\right)\right)$
$6=\left(9+\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)}\right)$
$7=\left(9+\left(9+\frac{9}{\left(9 \times \frac{9}{\left(9-\left(9+99\right)\right)}\right)}\right)\right)$
$8=\left(9+\left(9 \times \frac{9}{\left(9+\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)\right)$
$9=\left(9 \times \left(9 \times \frac{9}{\left(9-\left(9+\left(9+\left(9-99\right)\right)\right)\right)}\right)\right)$
$10=\left(9-\left(9 \times \frac{9}{\left(9+\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)\right)$
$11=\frac{9}{\left(9 \times \frac{9}{\left(9-\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)}$
$12=\left(9-\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/152855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 10,
"answer_id": 0
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|
Integral of $\int \sqrt{1-4x^2}$ I know I am messing up something with the substitutions but I am not sure what.
$$\int \sqrt{1-4x^2}$$
$$u = 4x, du = 4 \,dx$$
$$\frac{1}{4}\int \sqrt{1-u^2}$$
$u = \sin \theta$
$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \theta = \frac{\sin \theta}{4}$$
Replace $\theta$ with $u$
$$\frac{\sin (\arcsin u)}{4} = \frac{u}{4} = \frac{4x}{4} = x$$
This is wrong and I have no idea why.
|
Let $2x = u$. So, $dx = \frac{1}{2}du$
$$\frac{1}{2}\int \sqrt{1-u^2} du$$
Replace $u = \sin \theta$, $du = cos \theta\ d\theta$
So, equation will be
$$\frac{1}{2}\int \sqrt{1-sin^2 \theta}\ cos\theta\ d\theta$$
$$\frac{1}{2}\int \sqrt{\cos^2\theta}\ cos\theta\ d\theta$$
$$\frac{1}{2}\int cos^2\theta\ d\theta$$
Add and subtract $\frac{1}{4}\int d\theta$
$$\frac{1}{4}\int 2cos^2\theta\ d\theta - \frac{1}{4}\int d\theta + \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int \cos2\theta\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{8} (\sin2\theta) + \frac{1}{4} \theta + C$$
$$\frac{1}{8} (2\sin\theta.\cos\theta) + \frac{1}{4} \theta + C$$
As u = $\sin\theta$
$$\frac{1}{4} (u\sqrt{1-u^2}) + \frac{1}{4} \sin^{-1}u + C$$
As $2x = u$
$$\frac{1}{4} (2x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$
$$\frac{1}{2} (x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/153838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solution of $T(n)=2T(n/2) + n\log(\log n)$ I am struggling to solve this equation:
$$T(n)=2T(n/2) + n\log(\log n).$$ I concluded that the Master Theorem does not apply in this situation so I tried to successively substitute the terms in order to solve this equation but cannot proceed. Could somebody tell me what would be the best way to solve this?
Thanks.
|
We are given that $$T(n) = 2 T(n/2) + n \log_2(\log_2(n))$$ (Note that if we take $\log$ to the base $2$ or to the base $e$, the only difference is in the coefficient of the leading order term).
Let $n = 2^k$. Call $T(n) = g(k)$. Then we have
\begin{align}
g(k) & = 2g(k-1) + 2^k \log_2(k)\\
& = 2 (2g(k-2) + 2^{k-1} \log_2(k-1)) + 2^k \log_2(k)\\
& = 4 g(k-2) + 2^k \log_2(k-1) + 2^k \log_2(k)\\
& = 4 (2g(k-3) + 2^{k-2} \log_2(k-2))+ 2^k \log_2(k-1) + 2^k \log_2(k)\\
& = 8 g(k-3) + 2^k \log_2(k-2) + 2^k \log_2(k-1) + 2^k \log_2(k)
\end{align}
Hence, in general,
\begin{align}
g(k) & = 2^k g(0) + 2^k \left( \log_2(1) + \log_2(2) + \log_2(3) + \cdots + \log_2(k)\right)\\
& = 2^k g(0) + 2^k \log_2(k!)
\end{align}
Now from Stirling's formula, we have that $$k! \sim C \sqrt{k} \left(\dfrac{k}{e} \right)^k$$
Hence, $$\log_2(k!) = k \log_2(k) - k \log_2(e) + \dfrac12 \log_2(k) + \mathcal{O}(1)$$
\begin{align}
g(k)& = 2^k g(0) + 2^k \left( k \log_2(k) -k \log_2(e) + \dfrac12 \log_2(k) + \mathcal{O}(1) \right)\\
& = 2^k k \log_2(k) - k2^k \log_2(e) + 2^{k-1} \log_2(k) + \mathcal{O} \left(2^k \right)
\end{align}
Now plug in $k = \log_2(n)$, to get
\begin{align}
T(n) & = n \log_2(n) \log_2(\log_2 n) - n \log_2(n) \log_2(e) + \dfrac{n}2 \log_2(\log_2 n) + \mathcal{O}(n)
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 x}{ \cos x}dx\\
&=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\
&=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\
&=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\
&=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\
&=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\
&=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\
&=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\
&=\ln|\sec x| + \frac{\cos 2x}{4} + C
\end{align}$$
This is the wrong answer, I have went through and back and it all seems correct to me.
|
Notice that:
$$
\frac{1}{2} \cos^2 x = \frac{1}{2} \left(\frac{1}{2} + \frac{1}{2} \cos 2x\right) = \frac{1}{4} + \frac{1}{4} \cos 2x
$$
And:
$$
-\ln|\cos x| = \ln|(\cos x)^{-1}| = \ln|\sec x|
$$
So your answer is correct. It just differs by a constant from the answer you expect.
|
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"url": "https://math.stackexchange.com/questions/155829",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ Compute
$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$
|
If $(1+x)(1+y)=2$, then
$$\begin{align}
x&=\frac{1-y}{1+y}\\
1+x^2&=2\frac{1+y^2}{(1+y)^2}\\
\frac{1+x^2}{1+x}&=\frac{1+y^2}{1+y}
\end{align}\tag{1}
$$
and since $(1+y)\,\mathrm{d}x+(1+x)\,\mathrm{d}y=0$ we get
$$
\frac{\mathrm{d}x}{1+x^2}=-\frac{\mathrm{d}y}{1+y^2}\tag{2}
$$
Therefore,
$$
\begin{align}
\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x
&=\int_0^1\frac{\log(2)-\log(1+y)}{1+y^2}\mathrm{d}y\tag{3}
\end{align}
$$
Adding the left side to both sides and dividing by $2$ yields
$$
\begin{align}
\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x
&=\frac12\int_0^1\frac{\log(2)}{1+y^2}\mathrm{d}y\\
&=\frac\pi8\log(2)\tag{4}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/155941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "102",
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|
Centroid of a region $$y = x^3, x + y = 2, y = 0$$
I am suppose to find the centroid bounded by those curves. I have no idea how to do this, it isn't really explained well in my book and the places I have looked online do not help either.
|
The region you are interested is the blue shaded region shown in the figure below.
The coordinates of the centroid denoted as $(x_c,y_c)$ is given as $$x_c = \dfrac{\displaystyle \int_R x dy dx}{\displaystyle \int_R dy dx}$$ $$y_c = \dfrac{\displaystyle \int_R y dy dx}{\displaystyle \int_R dy dx}$$
where $R$ is the blue colored region in the figure above.
Let us compute the denominator in both cases i.e. $\int_R dy dx$. Note that this is nothing but the area of the blue region. Hence, we get that
\begin{align}
\int_R dy dx & = \int_{x=0}^{x=1} \int_{y=0}^{y=x^3} dy dx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} dy dx = \int_{x=0}^{x=1} x^3 dx + \int_{x=1}^{x=2} (2-x) dx\\
& = \left. \dfrac{x^4}{4} \right \vert_{0}^{1} + \left. \left(2x - \dfrac{x^2}2 \right)\right \vert_{1}^{2} = \dfrac14 + \left( 2 \times 2 - \dfrac{2^2}{2} \right) - \left(2 - \dfrac12 \right) = \dfrac14 + 2 - \dfrac32 = \dfrac34
\end{align}
Now lets compute the numerator for both cases.
To find $x_c$, we need to evaluate $\int_R x dy dx$. We get that
\begin{align}
\int_R x dy dx & = \int_{x=0}^{x=1} \int_{y=0}^{y=x^3} x dy dx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} x dy dx = \int_{x=0}^{x=1} x^4 dx + \int_{x=1}^{x=2} x(2-x) dx\\
& = \left. \dfrac{x^5}{5} \right \vert_{0}^{1} + \left. \left( x^2 - \dfrac{x^3}{3}\right) \right \vert_1^2 = \dfrac15 + \left( 2^2 - \dfrac{2^3}3\right) - \left( 1^2 - \dfrac{1^3}3\right) = \dfrac15 + \dfrac43 - \dfrac23 = \dfrac{13}{15}
\end{align}
To find $y_c$, we need to evaluate $\int_R x dy dx$. We get that
\begin{align}
\int_R y dy dx & = \int_{x=0}^{x=1} \int_{y=0}^{y=x^3} y dy dx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} y dy dx\\
& = \int_{x=0}^{x=1} \left. \dfrac{y^2}{2} \right \vert_0^{x^3} dx + \int_{x=1}^{x=2} \left. \dfrac{y^2}{2} \right \vert_{0}^{2-x} dx\\
& = \int_{x=0}^{x=1} \dfrac{x^6}{2} dx + \int_{x=1}^{x=2} \dfrac{(2-x)^2}{2} dx = \left. \dfrac{x^7}{14} \right \vert_{0}^{1} + \left. \dfrac{(x-2)^3}{6} \right \vert_{1}^{2}\\
& = \dfrac1{14} + \left( \dfrac{(2-2)^3}{6} - \dfrac{(1-2)^3}{6} \right) = \dfrac1{14} + \dfrac16 = \dfrac5{21}
\end{align}
Hence, $$x_c = \dfrac{\displaystyle \int_R x dy dx}{\displaystyle \int_R dy dx} = \dfrac{13/15}{3/4} = \dfrac{52}{45}$$ $$y_c = \dfrac{\displaystyle \int_R y dy dx}{\displaystyle \int_R dy dx} = \dfrac{5/21}{3/4} = \dfrac{20}{63}$$
|
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|
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
|
Let $$I=\int\frac{dx}{x^4+1}$$
Enforce the substitution $x:=\frac{1}{y}\implies dx=-\frac{dy}{y^2}$ so that $$I=-\int\frac{dy}{y^2\left(\frac{1}{y^4}+1\right)}=-\int\frac{dy}{y^2+\frac{1}{y^2}}=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}+1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\tag1$$
Then observe that $$y^2+\frac{1}{y^2}=\left(y-\frac{1}{y}\right)^2+2=\left(y+\frac{1}{y}\right)^2-2\tag2$$
Now splitting $I$ at the end of $(1)$ into two integrals $$I=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}+1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\frac{1}{2}\int\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\tag3$$
Observe $$\frac{d}{dy}\left(y+\frac{1}{y}\right)=1-\frac{1}{y^2}$$ and $$\frac{d}{dy}\left(y-\frac{1}{y}\right)=1+\frac{1}{y^2}$$ The latter integrals in $(3)$ now become by combining the last identities & $(2)$:$$I=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right)^2-2} dy-\frac{1}{2}\int\frac{1+\frac{1}{y^2}}{\left(y-\frac{1}{y}\right)^2+2} dy=-\frac{1}{2}\left(\int\frac{dt}{t^2-2}+\int\frac{du}{u^2+2}\right)$$
upon the substitutions $t=y+\frac{1}{y}$ and $u=y-\frac{1}{y}$.
$$I=-\frac{1}{2}\left(\int\frac{dt}{-2\left(1-\left(\frac{t}{\sqrt{2}}\right)^2\right)}+\int\frac{du}{2\left(\left(\frac{u}{\sqrt{2}}\right)^2+1\right)}\right)=-\frac{1}{4}\left(\int\frac{du}{\left(\frac{u}{\sqrt{2}}\right)^2+1}-\int\frac{dt}{1-\left(\frac{t}{\sqrt{2}}\right)^2}\right)$$ $$=-\frac{\sqrt{2}}{4}\left(\int\frac{dw}{w^2+1}-\int\frac{dz}{1-z^2}\right)=-\frac{\sqrt{2}}{\sqrt{2^4}}\left(\arctan(w)-\text{arctanh}\right(z))+C$$
Note that $$w=\frac{u}{\sqrt{2}}=\frac{y-\frac{1}{y}}{\sqrt{2}}=\frac{\frac{1}{x}-x}{\sqrt{2}}$$ and $$z=\frac{t}{\sqrt{2}}=\frac{y+\frac{1}{y}}{\sqrt{2}}=\frac{\frac{1}{x}+x}{\sqrt{2}}$$
Thus $$I=\frac{1}{2\sqrt{2}}\left(\text{arctanh}\left(\frac{\frac{1}{x}+x}{\sqrt{2}}\right)-\text{arctan}\left(\frac{\frac{1}{x}-x}{\sqrt{2}}\right)\right)+C$$
|
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|
Discriminant for $x^n+bx+c$ The ratio of the unsigned coefficients for the discriminants of $x^n+bx+c$ for $n=2$ to $5$ follow a simple pattern:
$$\left (\frac{2^2}{1^1},\frac{3^3}{2^2},\frac{4^4}{3^3},\frac{5^5}{4^4} \right )=\left ( \frac{4}{1},\frac{27}{4},\frac{256}{27},\frac{3125}{256} \right )$$
corresponding to the discriminants
$$(b^2-4c, -4b^3-27c^2,-27b^4+256c^3,256b^5+3125c^4).$$
Does the pattern for the ratios extend to higher orders? (An online reference would be appreciated.)
|
Use the relation between the disciminant of $f$ and the resultant of $f$ and $f'$. The resultant is easy to calculate since $f'$ is so simple.
|
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|
Prove that $K$ is a field Let be $K$ the set of real numbers that can be written as $a+b\sqrt2$, with $a$ and $b$ rational numbers. Prove that $K$ is a field.
I have already proved that $0$ and $1$ $\in K$, and that sum and product of two elements $\in K$. I have also already proved that the opposite $\in K$. I don't know how to prove that the reciprocal $\in K$. Can you help me?
|
Let $a+b\sqrt{2}\in K$ and $a+b\sqrt{2}\neq 0$. If $b=0$, then $a+b\sqrt{2}=a\neq 0$ and $1/a$ is its multiplicative inverse. Therefore, we assume $b\neq 0$, which implies that
$a^2-2b^2\neq 0$. Otherwise, if $a^2-2b^2=0$, then $\sqrt{2}=\frac{a}{b}$ which is rational since $a, b$ are rational, which contradicts to the fact that $\sqrt{2}$ is irrational.
Define
$$\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}.$$
First note that it is well-defined because $a^2-2b^2\neq 0$. Moreover, it belongs to $K$ because $\displaystyle\frac{a}{a^2-2b^2},\frac{b}{a^2-2b^2}$ are rational which follows from the fact that $a,b$ are rational numbers. Finally, we have
$$(a+b\sqrt{2})\cdot\left(\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}\right)=\frac{(a+b\sqrt{2})(a-b\sqrt{2})}{a^2-2b^2}=\frac{a^2-2b^2}{a^2-2b^2}=1.$$
That is to say, $\displaystyle\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}$ is the multiplicative inverse of $a+b\sqrt{2}$.
|
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|
Simplest method to find $5^{20}$ modulo $61$ What is the simplest method to go about finding the remainder of $5^{20}$ divided by $61$?
|
$5^3=125=2\cdot61+3$, so $5^3\equiv 3\pmod{61}$. $3^5=243=4\cdot61-1$, so $5^{15}\equiv 3^5\equiv-1\pmod{61}$. Finally, $5^{20}=5^{15}\cdot5^3\cdot5^2\equiv-1\cdot3\cdot25\equiv-75\equiv-14\equiv47\pmod{61}$.
|
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|
Factoring $x^4z-2z^2-4x^6+x^2z$ We want to factor $8x^4y^4-2y^8-4x^6+x^2y^4 = -2y^8 + (8x^4+x^2)y^4 -4x^6$. We substitute $x^4$ with $z$:
Now we want to compute this $8x^4z-2z^2-4x^6+x^2z = -(x^2-2z)(4x^4-z)$ by hand.
Therefore we transform it into $-(2z^2-(8x^4+x^4)z+4x^6z^0)$ and use the quadratic formula on the (inner) polynomial in $z$. The result is
$$z = \frac 1 2 x^2 \lor z = 4x^4.$$
So the result should be $(z-\frac 1 2 x^2)(z-4x^4)$. But the first factor appears only half? What is the reason for that? Is this a way to factor $8x^4y^4-2y^8-4x^6+x^2y^4$? The exercise is to use the recursive form $-2y^8 + (8x^4+x^2)y^4 -4x^6 \in \mathbb{Q}[y][x]$ (or alternatively $\mathbb{Q}[x][y]$).
|
$$8x^4y^4-2y^8-4x^6+x^2y^4 =(8x^4y^4-2y^8)-(4x^6-x^2y^4)=2y^4(4x^4-y^4)-x^2(4x^4-y^4)=$$
$$=(2y^4-x^2)(4x^4-y^4)=(2y^4-x^2)((2x^2)^2-(y^2)^2)=(2y^4-x^2)(2x^2-y^2)(2x^2+y^2)$$
|
{
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|
Find $\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$ $$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$$
I have tried multiplying by $\frac{1}{\sqrt{x^2+4}}$ and it's reciprocal, but I cannot seem to find the solution. L'Hospital's doesn't seem to work either, as I keep getting rational square roots.
|
Hint $$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt {{x^2} + 4} }}{{x + 4}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\dfrac{{\sqrt {{x^2} + 4} }}{x}}}{{\dfrac{{x + 4}}{x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt {\dfrac{{{x^2} + 4}}{{{x^2}}}} }}{{1 + \dfrac{4}{x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + \dfrac{4}{{{x^2}}}} }}{{1 + \dfrac{4}{x}}}$$
|
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|
Two proofs I'm having difficulty with I've been given an assignment. Almost done except the last two are tripping me up. They are as follows:
1) if $2x^2-x=2y^2-y$ then $x=y$
2) if $x^3+x=y^3+y$ then $x=y$
I imagine they use a similar tactic as they both involve powers, but I've tried factoring,completing the square, difference of squares and difference of cubes and nothing seems to help.
Any hints would be appreciated.
|
For 1) Observe that (I am assuming $x,y$ are real numbers):
\begin{align}
2x^2-x=2y^2-y\\
&\implies (2x^2-x)-2y^2+x=(2y^2-y)-2y^2+x\\
&\implies 2(x^2-y^2)=(x-y)\\
&\implies 2(x-y)(x+y)=(x-y)\\
&\implies 2(x-y)(x+y)-(x-y)=(x-y)-(x-y)=0\\
&\implies 2(x-y)(x+y)-(x-y)=0\\
&\implies (x-y)(2(x+y)-1)=0\\
&\implies (x-y)=0 \text{ or } 2(x+y)-1=0\\
\end{align} So, either $(x-y)=0$ or $2(x+y)-1=0.$ If $x-y=0$ then $x=y$ and we have our result. What if $2(x+y)-1=0$? Then there is a possiblity that $x\neq y.$ But if $x=y=1/4$ then $2(x+y)-1=2(1/4+1/4)-1=2(1/2)-1=1-1=0.$ Thus in either case $x=y$ is a solution. Like ncmathsadist pointed out $x=1/2$ and $y=0$ also works but that doesn't mean that $x=y$ doesn't work! I think for precalculus level this is enough.
For 2) do a similar analysis and Brian has done enough for you in his answer.
|
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|
Prove that there exists a natural number n for which $11\mid (2^{n} - 1)$ I'm thinking putting it into modulo form: there exists a natural number $n$ for which
$$2^{n}\equiv 1 \pmod {11}$$
but I don't know what to do next and I'm still confused how to figure out remainders when doing modulos, like $2^n\equiv \;?? \pmod{11}$. Is there some pattern to find $??$ or you would have to use specific numbers for $??$ which is divisible by $11$?
|
A natural number $n$ will have the property that $11\mid 2^n-1$ precisely when $n$ is a multiple of $10$.
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
n & \!\!\!\!\!& 1 & 2& 3& 4& 5& 6& 7&8&9&\mathbf{\Large 10}&11&12&13&14\\\hline\\
2^n\bmod 11 & \!\!\!\!\!& 2 & 4& 8 &5 &10 &9 & 7&3 &6&\mathbf{\Large 1 }&2&4&8&5
\end{array}\;\;\cdots\;\;\begin{array}{|c|c|}19 &\mathbf{\Large20}\\\hline\\ 6&\mathbf{\Large1}\end{array}\;\;\cdots$$
To be even more explicit,
Here is a proof that there exists a natural number $n$ such that $2^n\equiv 1\bmod 11$. Consider $n=10$: $$2^{10}-1=1024-1=1023=3\times \fbox{11}\times 31$$
so that $11\mid 2^{10}-1$. Thus by definition $2^{10}-1\equiv0\bmod 11$, and therefore $2^{10}\equiv 1\bmod 11$.
and
Here is a proof that there exists a natural number $n$ such that $2^n\equiv 1\bmod 11$. Consider $n=20$: $$2^{20}-1=1,048,576-1=1,048,575=3\times 5^2\times \fbox{11}\times 31\times 41$$
so that $11\mid 2^{20}-1$. Thus by definition $2^{20}-1\equiv0\bmod 11$, and therefore $2^{20}\equiv 1\bmod 11$.
|
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|
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.
|
A proof using the order inequality:
As @robjon points out, no two of $-a+b+c, a-b+c, a+b-c$ can be negative.
If exactly one of them is negative, the inequality is trivial.
Hence, assume that $-a+b+c, a-b+c, a+b-c > 0$.
Now, it's not difficult to see that $(b+c, a+c, a+b)$ and $(-a, -b, -c)$ are equally ordered.
And as $-a+b+c, a-b+c, a+b-c > 0$, it follows from the order inequality for products that
$$(-a+b+c)(a-b+c)(a+b-c)\le(b+c-b)(a+c-c)(a+b-a)=abc$$
|
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|
Complete the square and write in standard form for $3x^2+3x+2y=0$
Standard forms: $y-b=A(x-a)^2$ or $x-a=A(y-b)^2$
$3x^2+3x+2y=0$
I honestly do not know how to start this problem. I have tried a lot of things and obviously not the right one. Can someone explain to me the first step and nothing more and I will edit with my new discoveries. Thanks!
|
Since you have a factor with $x^2$, this anticipates the form you will be aiming for is
$$y-b=A(x-a)^2$$
So let's look at your eqn.:
$$3x^2+3x+2y=0$$
We need to produce a perfect square with $3x^2+3x$. So, we can do the old completing the square trick:
$$3x^2+3x=3(x^2+x)$$
$$3x^2+3x=3(x^2+2\frac 1 2 x)$$
$$3x^2+3x=3\left[x^2+2\frac 1 2 x+\left( \frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$$
$$3x^2+3x=3\left[\left( x+\frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$$
$$3x^2+3x=3\left( x+\frac1 2 \right)^2- \frac3 4$$
Can you move on?
|
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|
Prove $\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}$ How would I simplify this difficult trigonometric identity:
$$\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}.$$
I am not exactly sure what to do.
I simplified the right side to
$$\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos^2 A}{\sin^2 A}}$$
But how would I proceed.
|
Use $\sin 2A= 2\sin A \cos A$ and
$ \cos 2A= \cos^2 A- \sin^2 A$ to get
$$
\frac{\sin A \cos A}{\cos^2 A- \sin^2 A}=\frac{\sin 2A}{2\cos 2A}=\frac12\tan{2A},
$$
which is equivalent to
$$
\frac{ \tan A} {1 - \tan^2 A} .
$$
|
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|
Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now i am interested in understanding the series 1,2,4,7,11,..... In which the difference of the numbers are consecutive natural numbers.
How to find the sum of
$1+2+4+7+11+\cdots nterms$
This is my first question in MSE. If there are some guidelines i need to follow, which i am not, please let me know.
|
(Edit: Upps, I see now this is essentially solution (2) of Peter Tamaroff's answer, but because it's much shorter I just leave it here)
Your sequence can be separated into 2 sequences, where we add each pair:
$\begin{eqnarray}
&1&2&4&7&11&16&\cdots & = a_k\\
\hline
=&1&1&1&1&1&1&\cdots \\
+&0&1&3&6&10&15&\cdots \\
\hline \end{eqnarray}$
Then the partial sums are, beginning the index k at 1:
$\begin{eqnarray}
&1&3&7&14&25&41&\cdots &=&s_k\\
\hline
=&1&2&3&4&5&6&\cdots &= &&=&k\\
+&0&1&4&10&20&35&\cdots &=&\binom{1+k}{3}&=&{(k+1)!\over 3! (k-2)!}\\
\hline
=&1&3&7&14&25&41&\cdots &=&s_k&=& k+ {(k+1)!\over 3! (k-2)!}\\
\end{eqnarray}$
The last formula can be simplified to
$$ s_k = k+ {(k+1)k(k-1) \over 6} = {6k+k^3-k\over 6}
= k \cdot {k^2+5\over 6}$$
|
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|
multiple choice matrices problem If $M$ is a $3 \times 3$ matrix such that
$$
\begin{aligned}
\begin{pmatrix} 0 & 1 &2\end{pmatrix}M &= \begin{pmatrix} 1 & 0 &0\end{pmatrix} \text{ and}\\
\begin{pmatrix} 3 & 4 &5\end{pmatrix}M &= \begin{pmatrix} 0 & 1 &0\end{pmatrix} \text{ ,}
\end{aligned}
$$
then $\begin{pmatrix} 6 &7 &8\end{pmatrix}M$ is equal to
(A) $\begin{pmatrix} 2 &1 &−2\end{pmatrix}$
(B) $\begin{pmatrix} 0 &0 &1\end{pmatrix}$
(C) $\begin{pmatrix} -1 &2 &0\end{pmatrix}$
(D) $\begin{pmatrix} 9 &10 &8\end{pmatrix}$
i know that $M$ is a $3 \times 3$ matrix so we have 9 unknown and from these two equation we get six unknown so i can't solve it
|
Considering the Options,we have $\begin{pmatrix} 6 &7 &8\end{pmatrix}M=\begin{pmatrix} -1 &2 &0\end{pmatrix}$. Since If $A$ be an $m\times n$ matrix and we partition $A$ into rows then $$A = \begin{bmatrix}
a(1,:) \\ a(2,:) \\ a(3,:)
\end{bmatrix}$$ If $B$ an $n\times r$ matrix, then the $i$th row of product $AB$ is determind by multiplying the $i$th row of $A$ times $B$. Thus the $i$th row of $AB$ is $a(i,:)B$. In general the product $AB$ can be partitioned into rows as follows: $$AB = \begin{bmatrix}
a(1,:)B \\ a(2,:)B \\ a(3,:)B
\end{bmatrix}$$ Now in this question if we let $A = \begin{bmatrix}
0&1&2 \\ 3&4&5 \\ 6&7&8
\end{bmatrix}$ then have an matrix equation as follows:$$AM = \begin{bmatrix}
1&0&0 \\ 0&1&0 \\ x&y&z
\end{bmatrix}$$Now $$\det (AM)=\det(A)\det(M)=\det(\begin{bmatrix}
1&0&0 \\ 0&1&0 \\ x&y&z
\end{bmatrix})$$ and since $\det(A)=0$ we have $\det(\begin{bmatrix}
1&0&0 \\ 0&1&0 \\ x&y&z
\end{bmatrix})=0$. Therefore $z=0$. And only option is (C).
|
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$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ Evaluate Integral Here is a fun integral I am trying to evaluate:
$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$
I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series.
But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach.
$$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$
or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?.
Thanks everyone.
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Using
$$
\sin^{2n+1}(x) = \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} \sin\left((2k+1)x\right)
$$
We get
$$ \begin{eqnarray}
\int_0^\infty \frac{\sin^{2n+1}(x)}{x}\mathrm{d} x &=& \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1}\int_0^\infty \frac{\sin\left((2k+1)x\right)}{x}\mathrm{d} x\\ &=& \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1}\int_0^\infty \frac{\sin\left(x\right)}{x}\mathrm{d} x \\
&=& \frac{\pi}{2^{2n+1}}\sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \frac{\pi}{2^{2n+1}} \binom{2n}{n}
\end{eqnarray}
$$
The latter sum is evaluated using telescoping trick:
$$
\sum_k (-1)^k \binom{2n+1}{n+k+1} = \sum_k (-1)^k \frac{2n+1}{n+k+1} \binom{2n}{n+k} =
(-1)^{k+1} \binom{2n}{n+k} =: g(k)
$$
meaning that
$$
g(k+1) - g(k) = (-1)^k \binom{2n+1}{n+k+1}
$$
Hence
$$
\sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \sum_{k=0}^n \left(g(k+1)-g(k)\right) = g(n+1) - g(0) = -g(0) = \binom{2n}{n}
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/172080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 2
}
|
Prove trigonometry identity for $\cos A+\cos B+\cos C$ I humbly ask for help in the following problem.
If
\begin{equation}
A+B+C=180
\end{equation}
Then prove
\begin{equation}
\cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2)
\end{equation}
How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)$. But I am unsure what to do next.
|
What I might do is start with the right side. Since I don't remember half-angle formulas, let $a = A/2$, $b=B/2$. Note that (from the addition formulas for $\cos$) $\sin(x) \sin(y) = (\cos(x-y) - \cos(x+y))/2$, $\cos(x) \cos(y) = (\cos(x+y) + \cos(x-y))/2$, $\sin(90 - x) = \cos(x)$.
$$ \eqalign{1 &{}+ 4 \sin(a) \sin(b) \sin(90 - a - b)\cr
= & 1 + 2 (\cos(a-b) - \cos(a+b)) \cos(a+b) \cr
= & 1 + \cos(2a) + \cos(2b) - \cos(2a+2b) - \cos(0)\cr
= & \cos(2a) + \cos(2b) + \cos(180 - 2a - 2b)\cr} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/176892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
}
|
Gre Question Complex Number (plug and chug) This seems like it should be easy, but I can't seem to simplify it: If $z=e^{i\frac{2\pi}{5}}$, then what is $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$. The choices are $0, 4e^{i\frac{3\pi}{5}}, 5e^{i\frac{4\pi}{5}}, -4e^{i\frac{-2\pi}{5}}, -5e^{i\frac{3\pi}{5}},$ with the answer being $-5e^{i\frac{3\pi}{5}}.$ I can plug in the given $z$ into the equation and get $5+10e^{-i\frac{2\pi}{5}}+5e^{i\frac{2\pi}{5}}+5e^{i\frac{-4\pi}{5}}+5e^{i\frac{4\pi}{5}}$, but have been unsuccessful in simplifying it so far.
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Hint $\rm\ \ z\ne 1,\,\ z^5 = 1\:\Rightarrow\: (\color{#C00}{1\!+\!z\!+\!z^2\!+\!z^3\!+\!z^4})(1\!+\!4z^4)+5z^9 =\, \color{#C00}0+5z^9 =\, 5/z$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/179804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
what is the easiest way to represent $ \sqrt{1 + x} $ in series How to expand $ \sqrt{1 + x}$.
$$ \sum_{n = 0}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!} = 1 + \sum_{n = 1}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!}$$
How can I simplify $ \left({1 \over 2 }- n\right )! $?
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Just to motivate a more elementary solution.
Let $$f(x)=\sqrt{x+1}=(x+1)^{1/2}$$
Now, we expand around $x=0$.
$$f'(x)=\frac 1 2 (x+1)^{1/2-1}$$
$$f''(x)=\frac 1 2 \left(\frac 1 2-1 \right)(x+1)^{1/2-2}$$
$$f'''(x)=\frac 1 2 \left(\frac 1 2-1 \right)\left(\frac 1 2-2 \right)(x+1)^{1/2-3}$$
By induction, we get
$$f^{(n)}(x)=\prod_{k=0}^{n-1} \left(\frac 1 2 -k\right)(x+1)^{1/2-n}$$
Thus, we get
$$f^{(n)}(0)=\prod_{k=0}^{n-1} \left(\frac 1 2 -k\right)$$
$$f^{(n)}(0)=\prod_{k=0}^{n-1} \left(\frac {1 -2k}{2}\right)$$
$$f^{(n)}(0)=(-1)^{n}\frac{1}{2^{n}}\prod_{k=0}^{n-1} \left({2k-1}\right)$$
Now, note the product is exclusively of odd factors up to $(2n-3)$. What we do is fill in the missing even numbers, and divide to keep things the same. Note that $(2n-2)\cdots 4\cdot 2=2^{n-1} (n-1)!$
$${f^{(n)}}(0) = {( - 1)^n}\frac{1}{{{2^n}{2^{n - 1}}(n - 1)!}}\prod\limits_{k = 1}^{n - 1} {\left( {2k} \right)} \prod\limits_{k = 0}^{n - 1} {\left( {2k - 1} \right)} $$
$${f^{(n)}}(0) = \frac{{{{( - 1)}^n}}}{{{2^n}{2^{n - 1}}(n - 1)!}}\prod\limits_{k = 1}^{2n - 2} k = {( - 1)^n}\frac{1}{{{2^{2n - 1}}}}\frac{{\left( {2n - 2} \right)!}}{{\left( {n - 1} \right)!}}$$
Thus, we have that
$$\sqrt {1 + x} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{2^{2n - 1}}}}} \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}}{x^n}$$
$$\sqrt {1 + x} = 1 + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{{2^{2n - 2}}}}} \frac{{\left( {2n - 2} \right)!}}{{\left( {n - 1} \right)!\left( {n - 1} \right)!}}\frac{{{x^n}}}{n}$$
$$\sqrt {1 + x} = 1 + \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{2^{2n}}}}} \frac{{\left( {2n} \right)!}}{{n!n!}}\frac{{{x^{n + 1}}}}{{n + 1}}$$
$$\sqrt {1 + x} = 1 + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{2^{2n + 1}}}}} {2n\choose n}\frac{{{x^{n + 1}}}}{{n + 1}}$$
Note this gives that
$$\frac{1}{{\sqrt {1 + x} }} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{2^{2n}}}}} {2n\choose n}{x^n}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/180282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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.