Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Urn with balls - Finding the joint distribution
Consider an urn containing $4$ red and $6$ white balls. We draw two times with replacement. Let $X$ be the number of red balls we draw, and $Y$ the number of white balls we draw. Find the joint probability distribution of $X$ and $Y$.
Since we draw exactly two times:
$$\begin{align}P(X=0, Y= 0) &= P(X = 0, Y= 1) \\[1ex]&= P(X = 1, Y= 0) \\[1ex]&= P(X = 1, Y= 2) \\[1ex]&= P(X = 2, Y= 1) \\[1ex]&= P(X = 2, Y= 2) \\[1ex]&= 0\end{align}$$
Now we have $3$ cases left:
$P(X = 0, Y = 2) = (\dfrac{6}{10})^2 = 0.36.$
$P(X = 1, Y = 1) = \dfrac{4}{10} \cdot \dfrac{6}{10} = 0.24.$
$P(X = 2, Y = 0) = (\dfrac{4}{10})^2 = 0.16.$
These don't add up to $1$, so there must be something wrong with my solution. Where's my mistake?
| Hint: $~~0.36+{0.24}+0.16=1-{0.24}$
It's because $\mathsf P(X{=}1,Y{=}1)= 2\cdot\tfrac{4}{10}\cdot\tfrac{6}{10}=0.48$ by reason that we are selecting 1 red and 1 white ball in any order of extraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\cos x-\sin 3x=\cos 2x$? My attempt:
$$\begin{align} (\cos x- \cos 2x) - \sin 3x &= 0 \\
2\sin \frac{3x}{2}\sin \frac{x}{2} - 2 \sin \frac{3x}{2}\cos \frac{3x}{2}&=0\\
\sin\frac{3x} {2} \left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)&=0 \end{align}$$
Now either $\sin\frac{3x} {2} = 0$ or $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$. Solving for the former,
$$\frac{3x}{2}=n\pi\rightarrow x = \frac{2n\pi}{3}$$
How can I solve $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$? I know the elementary identities involving trigonometric ratios, but not complex numbers or calculus.
| We have $\sin\frac12x=\cos\frac32x=\sin(\frac12\pi-\frac32x)$. The general solution to this is$$\tfrac12x=n\pi+(-1)^n(\tfrac12\pi-\tfrac32x)\quad (n\in\Bbb Z).$$The cases of even or odd $n$ then give rise to two sets of solutions:$$x=(k+\tfrac14)\pi\quad\text{or}\quad x=(2k-\tfrac12)\pi\quad(k\in\Bbb Z).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the Limit.? What is the limit of
$Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2} $
for $n > 2$ and $Y_{1}<Y_{2}$.
I have tried solving, but the limit tends to exist between $y_{1}$ and $y_{2}$ and hence I am not able to evaluate what would be the ratio between which they would lie?
| $Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2}$ is homogenous linear diference equation which has characteristic equation $x^2=\frac{1}{3}x+\frac{2}{3}$ or $3x^2-x-2=0$, $(3x+2)(x-1)=0$. Solution is $$Y_{n}=C_1(-\frac{2}{3})^n+C_2(1)^n=C_1(-\frac{2}{3})^n+C_2.$$
Constants $C_1$, $C_2$ can be found from $Y_1, Y_2$:
$$-\frac{2}{3}C_1+C_2=Y_1$$
$$\frac{4}{9}C_1+C_2=Y_2$$
from where $C_1=\frac{9}{10}(Y_2-Y_1)$ and $C_2=\frac{2}{5}Y_1+\frac{3}{5}Y_2$ and
$$Y_n=\frac{9}{10}(-\frac{2}{3})^n(Y_2-Y_1)+\frac{2}{5}Y_1+\frac{3}{5}Y_2.$$
Therefore, $$\lim_{n\to\infty} Y_n=\frac{2}{5}Y_1+\frac{3}{5}Y_2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question about convergence of series using comparison test I am wondering how to determine whether the following series converges using the comparison test:
$$\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$$
In class, my professor used the idea that $\frac{\sin x}{x}\to 1$ as $x$ gets large.
So, my professor said that since $$\frac{\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}}{\frac{1}{n}}\to\frac{2}{3}$$by the comparison test, $\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$ converges. Can someone explain why this is true? I though that the comparison tests says that if $\lim_{n\to\infty} \frac{a_n}{b_n}\to L$ where $0<L<\infty$, then $a_n$ and $b_n$ follow the same convergence. So wouldn't this mean that $\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$ diverges since $\frac{1}{n}$ diverges?
| What we are using is limit comparison test and since
$$\frac{\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}}{\frac{1}{n}}=\frac{\sin{\frac{2}{\sqrt{n}}}}{\frac{2}{\sqrt{n}}}\frac{2}{3+2\frac{\sqrt n}n} \to \frac23$$
as you noticed the series diverges.
Note that limit comparison test also works for the two limiting cases
*
*$b_n$ converges and $\frac{a_n}{b_n} \to 0 \implies a_n$ converges
*$b_n$ diverges and $\frac{a_n}{b_n} \to \infty \implies a_n$ diverges
As an alternative by direct comparison test, using that $\sin x\ge x-\frac16 x^3$ we have
$$\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2} \ge \frac{\frac{2}{\sqrt{n}}-\frac{4}{3n\sqrt{n}}}{3\sqrt{n}+2}=\frac{2-\frac{4}{3n}}{3n+2\sqrt{n}}\ge \frac{\frac23}{3n+3n}=\frac19\frac1n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882523",
"timestamp": "2023-03-29T00:00:00",
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Volume with spherical polar coordinates Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.
I convert to spherical polar coordinates.
$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$
$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$
I can't figure out how to take $\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$, which makes me think I made a mistake somewhere.
EDIT: Thanks for all the answers.
Now I understand how the limits of $\theta ,r,z$ works.
I don't fully understand where the function "disappear".
$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$
Why isn't it then:
$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$
| Using spherical coordinates, you would have to split up $K$ into two regions,
$$K_1=\left\{(r,\theta,\phi)\mid 0\le r\le2,0\le\theta\le\frac{3\pi}4,0\le\phi\le\frac\pi6\right\}$$
$$K_2=\left\{(r,\theta,\phi)\mid0\le r\le\sqrt{\csc\phi},0\le\theta\le\frac{3\pi}4,\frac\pi6\le\phi\le\frac\pi2\right\}$$
(where $x=r\cos\theta\sin\phi$, $y=r\sin\theta\sin\phi$, and $z=r\cos\phi$). The upper limit on $\phi$ for $K_1$ and lower limit for $K_2$ come from the intersection of the cylinder $x^2+y^2=1$ and the sphere $z=\sqrt{4-x^2-y^2}$. On the sphere, $r=2$, so we have
$$2\cos\phi=\sqrt3\implies\phi=\cos^{-1}\left(\frac{\sqrt3}2\right)=\frac\pi6$$
The upper limit for $r$ in $K_2$ is obtained by converting the equation of the cylinder $x^2+y^2=1$ into spherical coordinates:
$$(r\cos\theta\sin\phi)^2+(r\sin\theta\sin\phi)^2=r^2\sin^2\phi=1\implies r=|\csc\phi|=\csc\phi$$
Then the volume is
$$\int_0^{\frac\pi6}\int_0^{\frac{3\pi}4}\int_0^2r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi+\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi$$
The first integral is trivial. For the second, integrating with respect to $r$ yields
$$\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi=\frac13\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\csc^2\phi\,\mathrm d\theta\,\mathrm d\phi$$
and observing that $\csc^2\phi=\frac{\mathrm d}{\mathrm d\phi}(-\cot\phi)$, it turns out the second integral is, too.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving Vector Identities Let r=$(x,y,z)$ and $r=$||r||.
(A) Prove that $\nabla^2r^3=12r$.
(B) Is there a value of $p$ for which the vector field f(r) = r/$r^p$ is solenoidal?
What I have tried:
For part (a) I think that $r^3=\sqrt{27} $ but I am unsure of what to do next in terms of the del operator.
| $$
\nabla^2r^3=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})(x^2+y^2+z^2)^{3/2}\\
= \frac{\partial }{\partial x }(3x(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial y }(3y(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial z }(3z(x^2+y^2+z^2)^{1/2}) \\
=\frac{3(2x^2+y^2+z^2)}{(x^2+y^2+z^2)^{1/2}}+\frac{3(x^2+2y^2+z^2)}{(x^2+y^2+z^2)^{1/2}} +\frac{3( x^2+y^2+2z^2)}{(x^2+y^2+z^2)^{1/2}}\\=3\frac{ (4 x^2+4y^2+4z^2)} {(x^2+y^2+z^2)^{1/2}} =12r
$$
$$
\nabla\cdot\big[\frac{(x, y, z)}{(x^2 + y^2 + z^2)^{p/2}}\big] =\frac{3-p}{(x^2+y^2+z^2)^{p/2}}=\frac{3-p}{r^{p }}
$$
So $\nabla\cdot\frac{{\bf r}}{r^p}=0$ when $p=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884835",
"timestamp": "2023-03-29T00:00:00",
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How many 4 digit number can be formed by using 1,2,3,4,5,6,7 if at least one digit is repeated My approach for this was selecting 4 numbers from which one is being repeated
$$\boxed{A}\boxed{B}\boxed{C}\boxed{A} $$
$$\boxed{A}\boxed{B}\boxed{A}\boxed{A} $$
$$\boxed{A}\boxed{A}\boxed{A}\boxed{A} $$
and permitting them which gives 7$C_4$×4!/2! +7$C_4$×4!/3!+7$C_4$×4!/4! Since 2,3,4 no. Are being repeated numbers are being repeated
What they have done is
$7^4$-7$P_4$ subtracting cases where all 4 no. Are different from total cases. I agree but what is the flaw in my method
| You’ve miscounted all three of your types with repetitions. This is most obvious in the case of the $AAAA$ type: you counted $\binom74\cdot\frac{4!}{4!}=35$ of them, but there are clearly only $7$: $1111$, $2222$, $3333$, $4444$, $5555$, $6666$, and $7777$. For your $ABAA$ type you give a figure of $\binom74\cdot\frac{4!}{3!}=140$, but this is an underrestimate: there are $7$ ways to choose $A$, then $6$ ways to choose $B$, and finally $4$ possible places to put the $B$ digit, for a total of $7\cdot6\cdot4=168$ possibilities. For the $ABCA$ type you calculate $\binom74\cdot{4!}{2!}=420$; however, there are $7$ ways to choose the $A$ digit, then $\binom42=6$ ways to choose $2$ places for them, $6$ ways to choose the first of the $2$ remaining places, and $5$ ways to choose a digit for the one remaining place, for an actual total of $7\cdot6\cdot6\cdot5=1260$ numbers of this type.
The most obvious problem with your calculation is the binomial coefficient $\binom74$: that’s the number of ways to choose $4$ different digits from the set $\{0,1,2,3,4,5,6,7\}$, and for these types you’re choosing fewer than $4$ different digits. For the first type you’re actually choosing only $3$ different digits, and there are only $\binom73$ ways to do this. Then you have to choose which one of them is to appear twice; there are $3$ ways to do this. Then you can multiply by $\frac{4!}{2!}$ to get the correct figure of $1260$. Similarly, in your second type you’re choosing only $2$ different digits, something that can be done in $\binom72$ ways. Then you have $2$ ways to choose which one of those digits is used only once and $4$ ways to choose where to use it, for a total of $168$.
And as noted in the comments, you missed the $AABB$ type with $2$ distinct digits, each appearing twice. There are $7$ choices for the first digit, $3$ choices for the position of its mate, and $6$ choices for the other digit, for a total of $7\cdot3\cdot6=126$ possible numbers of this type. And sure enough, $7+168+1260+126=1561$.
| {
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Finding $1998x + 5y + 3z$ given $\frac{1}{x} + \frac{1}{y} = \frac{3}{z}$ Suppose $x,y,z$ are positive integers so that $y$ and $z$ are prime and $x = yz.$ If they satisfy $$\frac{1}{x} + \frac{1}{y} = \frac{3}{z},$$ find $1998x + 5y + 3z.$
I first moved $\frac{1}{y}$ over to the other side and expanded out, but from there I got stuck. Can someone help me?
| You are on the right track. After that you should have
$$\frac{1}x=\frac{3}z-\frac{1}y\Longrightarrow \frac{1}{yz}=\frac{3y-z}{yz}\Longrightarrow 1=3y-z\Longrightarrow 3y=z+1$$
Now note that if $y$ is odd, then $z$ has to be even. And since $z$ is a prime, $z$ has to be $2$, but in this case we will have $y=1$ which is not a prime.
Therefore $y$ has to be an even prime. That is, $\color{red}{y=2}$.
Can you end it now?
| {
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prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.
The case $a\ge c\ge b$ is analogous.
Thus Proved!
Is it correct?...And any other alternative ways possible?
| I have an alternative proof.
We need to prove that
$$
\frac{a}{b} +
\frac{b}{c} +
\frac{c}{a} \geqslant
\frac{a + 2}{b + 2} +
\frac{b + 2}{c + 2} +
\frac{c + 2}{a + 2}
$$
Here we can write this inequality in $2$ forms:
$$
\frac{c}{a} -
\frac{c + 2}{a + 2} \geqslant
\frac{a + 2}{b + 2} -
\frac{a}{b} +
\frac{b + 2}{c + 2} -
\frac{b}{c}
$$
$$
\frac{c - a}{a^2 + 2a} =
\frac{b - a}{a^2 + 2a} + \frac{c - b}{a^2 + 2a}
\geqslant
\frac{b - a}{b^2 + 2b} +
\frac{c - b}{c^2 + 2c}
$$
And
$$
\frac{b}{c} - \frac{b + 2}{c + 2} +
\frac{c}{a} - \frac{c + 2}{a + 2} \geqslant
\frac{a + 2}{b + 2} - \frac{a}{b}
$$
$$
\frac{b - c}{c^2 + 2c} +
\frac{c - a}{a^2 + 2a} \geqslant
\frac{b - a}{b^2 - 2b}
$$
Let $\min{(a,b,c)} = a$.
Case I: $c\geqslant b\geqslant a$: Write inequality in the first form.
Case II: $b\geqslant c\geqslant a$: Write inequality in the second form.
| {
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Inequality with integers I am trying to find an analytical way to prove that for integers $n\ge 6$,
$$\sqrt[n]{n+1}\le \frac{1}{n}+\frac{1}{\sqrt[n]{n+1}}+\frac12$$
| Denote $x = \sqrt[n]{n+1}$. The desired inequality is written as
$\frac{-2nx^2 + (n+2)x + 2n}{2nx} \ge 0$.
It suffices to prove that $-2nx^2 + (n+2)x + 2n \ge 0$ which is written as
$$2n(x - x_1)(x - x_2)\le 0$$
where
$$x_1 = \frac{n+2 + \sqrt{17n^2 + 4n + 4}}{4n},
\quad x_2 = \frac{n+2 - \sqrt{17n^2 + 4n + 4}}{4n}.$$
Since $x > 0$ and $x_2 < 0$, it suffices to prove that $x \le x_1$, namely,
$$\sqrt[n]{n+1} \le \frac{n+2 + \sqrt{17n^2 + 4n + 4}}{4n}. \tag{1}$$
For $n = 6, 7, \cdots, 11$, (1) is verified directly.
For $n \ge 12$, since $\sqrt{17n^2 + 4n + 4} \ge 4n + 1$, it suffices to prove that
$$\sqrt[n]{n+1} \le \frac{n+2 + 4n + 1}{4n}$$
or
$$n+1 \le (1 + \tfrac{1}{4})^n (1 + \tfrac{3}{5n})^n.$$
By the binomial theorem, we have $(1 + \tfrac{1}{4})^n \ge 1 + \frac{1}{4}n + (\frac{1}{4})^2\frac{n(n-1)}{2}$
and
$(1 + \tfrac{3}{5n})^n \ge 1 + \frac{3}{5n}\cdot n = 1 + \tfrac{3}{5}$.
It suffices to prove that
$$n + 1 \le [1 + \tfrac{1}{4}n + (\tfrac{1}{4})^2\tfrac{n(n-1)}{2}] (1 + \tfrac{3}{5})$$
or
$$\frac{1}{20}(n-1)(n-12)\ge 0$$
which is true.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that a matrix raised to a power is equal to itself Let $F$ be a field of cardinality $q=2^k$ for some $k \in \mathbb{N}$ and let
$$E= \begin{bmatrix}
1 & 0 & & & & & &\\
1 & 0 & & & & & &\\
& & \ddots & & & & & &\\
& & & 1 & 0 & & &\\
& & & 1 & 0 & & \\
& & & & & & 1 & 0 & a_{n-1}a_{n-2}\\
& & & & & & 1& 0 & a_{n-2} \\
& & & & & & & & 1+a_{n-1}
\end{bmatrix} \in M_n(F).$$
I want to show that $E^q=E.$
I was able to show for $q=2,4,8$ that
$$E^q= \begin{bmatrix}
1 & 0 & & & & & &\\
1 & 0 & & & & & &\\
& & \ddots & & & & & &\\
& & & 1 & 0 & & &\\
& & & 1 & 0 & & \\
& & & & & & 1 & 0 & a_{n-1}^qa_{n-2}\\
& & & & & & 1& 0 & a_{n-2} \\
& & & & & & & & 1+a_{n-1}^q
\end{bmatrix}$$
through brute force computation. This gives the desired result as $x^q=x$ in $F$. Note that this is not the case for $E^3, E^4, E^5, E^6, E^7.$
I want to generalize this for any $q=2^k$ and I have no idea how. Any help would be appreciated.
| Write $x=a_{n-1}$ and $y=a_{n-2}$ for notational ease. Let $M$ represent the matrix formed by the first $n-1$ columns of $E$ with zeroes in column $n$, and let $N=E-M$ be the $n$th column of $E$ with zeroes in the first $n-1$ columns. It is not hard to check that
$$M^2=M,\ N^2=(x+1)N,\ NM=0.$$
In particular, when we expand the sum $(M+N)^q$, all terms with an $NM$ vanish, so
\begin{align*}
(M+N)^q
&=\sum_{i=0}^q M^{q-i}N^i\\
&=M^q+N^q+\sum_{i=1}^{q-1} M^{q-i}N^i\\
&=M+(x+1)^{q-1}N+MN\sum_{i=1}^{q-1}(x+1)^{i-1}\\
&=M+(x+1)^{q-1}N+MN\left(\frac{(x+1)^{q-1}-1}x\right),
\end{align*}
where the last fraction is as a polynomial in $x$. We need to show that
$$(x+1)^{q-1}N+MN\left(\frac{(x+1)^{q-1}-1}x\right)=N,$$
which is equivalent to
$$MN\left(\frac{(x+1)^{q-1}-1}x\right)=N\left((x+1)^{q-1}-1\right).$$
If $x\neq 0,1$, this is true, since both sides are $0$. When $x=1$, we require that $MN=N$, which can be checked directly (we can find that $MN$ is $0$ everywhere except for the bottom of the last column, where it is $\dots 0,xy,xy,0$). When $x=0$, the right side is still $0$, so it suffices for $MN$ to be $0$, which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functions which satisfy $f(m^2+n^2)=f(m)^2+f(n)^2$ Find all $f$: $f(m^2+n^2)$=$f(m)^2+f(n)^2$ for all $m,n\in \mathbb{N_0}$ where
$f(1)>0$,
$f\in\mathbb{N_0}$ for all inputs
I did $f(0)=2f(0)^2\implies f(0)=0$
$f(1)=f(1)^2\implies f(1)=1$
From this we can get $f(2)=2, f(4)=4$ etc. easily and also use some tricks like
$f(2^2+1^2)=f(5)=5, f(4)=4$
so $f(3)=\sqrt{5^2-4^2}=3$
I tried using the parametrization of Pythagorean triples but am unable to prove that $f(7)=7$
Please help
| We see that $f(2n^2) = 2f^2(n)$. So $f(50)=2f^2(5)=2(25)=50$
So $f(50)=50$. But $50 = 1+49$. So $f(50) = 1 + f^2(7)$.
So $f(7)=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3900226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Fitting circle in $f(x)=x^3-3x^2$ what is the radius?
The whole problem is the above.
I tried to solve it analytically but I couldnot find enough equation to solve it.
Assume the center is $(a,-r)$ where $r$ is the radius of the circle.
The points on the sphere $(x-a)^2+(y+r)^2=r^2$
and $(x,f(x))$ must satisfy it so if we denote the intersection points $(x_1,y_1),(x_2,y_2)$:
$$(x_1-a)^2+(x_1^3-3x_1^2+r)^2=r^2$$
$$(x_2-a)^2+(x_2^3-3x_2^2+r)^2=r^2$$
Are two equation with 4 unknown.
Can you give me elegant relations between these two unknowns?
|
In figure suppose AC is straight line.We first find the angle BAC the curve makes with x axis:
$y=x^3-3x^2$ ⇒ $y'=3x^2-6x$
⇒ $tan (\widehat{BAC})=y'$ at point A(3, 0):
$tan (\widehat{BAC})=3^3-6\times 3=9$
⇒ $ \widehat{BAC} ≈83^o$
⇒$ \widehat{BAC}=\widehat BAD ≈41.5^o$
So the center of circle is on line $y=(m=tan 41.5)(x-3)$
Now we use plain geometry. we have:
$AB^2+(BD=r)^2=AD^2$
$BD^2+r^2=OD^2$
$AB+BO=3$
$(AB+BO)^2=AB^2+OB^2+2\cdot AB\cdot BO$
⇒ $AB^2+OB^2-2r^2=9-(DA^2+OD^2)$
$tan(41.5)=\frac r{AB}$
$OD^2=OB^2+r^2=(3-AB)^2+r^2$
These two relations give $OD^2$ in terms of r.
$sin(41.5)=\frac r{AD}$
This relation gives AD in term of r.
Now in relation:
$AB^2+OB^2-2r^2=9-(DA^2+OD^2)$
$OB=3-AB$
So OB can also be found in term of r. Plugging, DA, OA, AB and OB in above relation give an equation in terms of r
Note, r ≈1.1 and coordinates of its center is D( ≈1.8, ≈-1.1)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3900605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
A bag contains 7 blue balls and 3 green balls. What is the probability of the following events? A bag contains 7 blue balls and 3 green balls.
I am not sure about my answer but please allow me to share.
*
*If 5 balls are drawn in succession at random without replacement, what is the probability of drawing 1 green, 1 blue, 1 green, and 1 blue in order?
There are 10 balls in the bag initially, 7 of which are blue and 3 balls are green. All balls are equally likely to be drawn from the bag.
Since the order is important, we need to compute the probability of getting 1 green, 1 blue, 1 green, and 1 blue in order.
Hence,
$GBGB = (\frac{3}{10})(\frac{7}{9})(\frac{2}{8})(\frac{6}{7}) = \frac{1}{20}$
The probability of getting 1 green, 1 blue, 1 green, and 1 blue in order is $\frac{1}{20}$.
*If 3 balls are drawn at the same time, what is the probability that all the 3 balls drawn are blue?
$BBB = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8}) = (\frac{7}{24})$
*If 5 balls are drawn at the same time, what is the probability that 3 balls are blue and 2 balls are green?
Work it out first where the order they are drawn is important.
$BBBGG = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8})(\frac{3}{7})(\frac{2}{6}) = (\frac{1}{24})$
Then work out possible different orders they could be drawn, there are 20 permutations $P(5,2)$. so the answer is
$20 × (\frac{1}{24}) = (\frac{5}{6})$
Any comments or suggestions will be much appreciated. Thank you in advance.
|
If five balls are drawn in succession at random without replacement, what is the probability of drawing one green, one blue, one green, one blue in order?
You have correctly calculated that the probability that the sequence GBGB occurs in the first four draws is
$$\Pr(GBGB) = \frac{3}{10} \cdot \frac{7}{9} \cdot \frac{2}{8} \cdot \frac{6}{7}$$
However, it could also happen in the second four draws after being preceded by either a green or a blue ball.
$$\Pr(GGBGB) = \frac{3}{10} \cdot \frac{2}{9} \cdot \frac{7}{8} \cdot \frac{1}{7} \cdot \frac{6}{6}$$
and
$$\Pr(BGBGB) = \frac{7}{10} \cdot \frac{3}{9} \cdot \frac{6}{8} \cdot \frac{2}{7} \cdot \frac{5}{6}$$
Since these three cases are mutually exclusive and exhaustive, the probability that the sequence GBGB appears in the first five draws is found by adding the above probabilities.
If three balls are drawn at the same time, what is the probability that all three balls drawn are blue?
Your answer is correct.
An alternate method is to observe that we can select three of the ten balls in
$$\binom{10}{3}$$
ways. Of these, we can select three blue balls in
$$\binom{7}{3}$$
ways. Hence, the probability that a random selection of three balls from a bag containing seven blue and three green balls results in the selection of three blue balls is
$$\Pr(\text{three blue}) = \frac{\dbinom{7}{3}}{\dbinom{10}{3}}$$
If five balls are drawn at the same time, what is the probability that three balls are blue and two balls are green.
There are
$$\binom{10}{5}$$
ways to select five of the ten balls. Of these, we can select three of the seven blue balls and two of the three green balls in
$$\binom{7}{3}\binom{3}{2}$$
ways. Hence, the probability that a random selection of five balls from a bag with seven blue and three green balls results in the selection of three blue and two green balls is
$$\Pr(\text{three blue and two green balls}) = \frac{\dbinom{7}{3}\dbinom{3}{2}}{\dbinom{10}{5}}$$
Your answer resulted in too large a probability since the number of ordered sequences of three blue and two green balls is
$$\binom{5}{2}$$
since we must choose two of the five positions in the sequence for the green balls. Notice that
$$\binom{5}{2}\left(\frac{7}{10}\right)\left(\frac{6}{9}\right)\left(\frac{5}{8}\right)\left(\frac{3}{7}\right)\left(\frac{2}{6}\right) = \frac{\dbinom{7}{3}\dbinom{3}{2}}{\dbinom{10}{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the obvious mistake of this double integration? There is a definite double integral in Cartesian coordinates as follows:
$$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 2\sqrt{4-x^2}\ dydx\Rightarrow \int_{-2}^{2} 2y\sqrt{4-x^2}\rvert_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \ dx \Rightarrow 4\int_{-2}^{2} (4 - x^2)dx \Rightarrow 4(4x - \frac{x^3}{3})|_{-2}^{2} = \frac{128}{3} $$
and there is another double integral in polar coordinates which is supposed to be the exact same integral as the above, but the final answer is not 128/3. it took me several hours, but I cant figure out which step of this integration is wrong.
$$ \int_{0}^{2\pi}\int_{0}^{2} 2r\sqrt{4-r^2{\cos^2\theta}} drd\theta
\xrightarrow[]{u = 4-r^2{\cos^2\theta}} \int_{0}^{2\pi} - \frac{2}{3} \frac{(4-r^2{\cos^2\theta})^\frac{3}{2}}{\cos^2\theta}\Big|_{0}^{2} \ d\theta \Rightarrow \int_{0}^{2\pi} (-\frac{2}{3}\frac{(4-4\cos^2\theta)^\frac{3}{2}}{\cos^2\theta} + \frac{2}{3}\frac{4^\frac{3}{2}}{\cos^2\theta} ) d\theta \Rightarrow \frac{16}{3}\int_{0}^{2\pi} \frac{1-(1-\cos^2\theta)^\frac{3}{2}}{\cos^2\theta} d\theta \Rightarrow \frac{16}{3} \int_{0}^{2\pi} \frac{1-{\sin^3\theta}}{\cos^2\theta} d\theta \Rightarrow \frac{16}{3}\int_{0}^{2\pi} \frac{1}{\cos^2\theta} - \frac{\sin^3\theta}{\cos^2\theta} d\theta \Rightarrow \frac{16}{3}(\int_{0}^{2\pi} \frac{1}{\cos^2\theta} d\theta - \int_{0}^{2\pi} \frac{\sin^3\theta}{\cos^2\theta} d\theta) \Rightarrow \frac{16}{3}(\tan\theta - \int_{0}^{2\pi} \frac{\sin^3\theta}{\cos^2\theta} d\theta) \xrightarrow[]{u=\cos\theta} \frac{16}{3}(\tan\theta - \int \frac{1-u^2}{u^2} du) \Rightarrow \frac{16}{3}(\tan\theta - (-\sec\theta - \cos\theta))\Big|_{0}^{2\pi}\Rightarrow \frac{16}{3} ((0+1+1) - (0+1+1)) = 0 $$
I know there is a huge problem with the second one and my apologies if it's an elementary problem. But I would really appreciate if you could help me find the problem.
| The problem is that $(1-\cos^2)^\frac{3}{2}=|\sin^3\theta|$ and not $\sin^3\theta$. Then you have to consider intervals for $\theta$ to discard absolute value.
The second problem is at the second to the last line, the change of variable $u=\cos\theta \Rightarrow d\theta = -\frac{du}{\sin\theta}$ and you forgot a negative sign.
It is very nice that you try to verify your answer by considering other possible ways of solving and in this problem it is possible to solve it in both coordinates but sometimes it is not possible to find answer in other coordinates and the coordinate system that you choose is completely dependent on the geometry of problem, like finding the electrostatic potential function in spherical or cylindrical capacitors, just try to solve it with simplest way possible and be done with it dude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A confusion in the $\epsilon-\delta$ proof of $\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0$ I want to prove
$$
\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0
$$
The proof is: Let $\epsilon > 0$ be given. Then
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
provided $|x - 2| < \delta = \epsilon$.
My question is we have
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| < \epsilon \text{ and }\frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2|.$$
How do we conclude
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
from these two inequalities? What if $\epsilon < |x - 2|$?
| You have made an incorrect statement which I think needs to be corrected:
it is not true that you "have"
$$ \left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon \tag1 $$
at the point in the proof you are questioning.
Instead, this inequality is the thing you have to prove at that point.
The real point of difficulty, however, seems to be how to recognize the implications that are used in the proof, and to understand which one follows from which.
Note that when we write, "Blah blah blah provided $|x - 2| < \delta = \epsilon$,"
it means
If $\delta = \epsilon$ and $|x - 2| < \delta,$ then blah blah blah.
So the steps of the proof are:
*
*For whatever $\epsilon$ you are given, choose $\delta = \epsilon.$
*Now you know that if $|x - 2| < \delta,$ then $|x - 2| < \epsilon.$
*As separate facts, we see that $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2}$ and that $\frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert.$
*Putting these facts together, if $|x - 2| < \delta,$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert < \epsilon.$
*Notice that if $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert < \epsilon$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon.$
*From the two implications in the previous two steps, you can conclude that if $|x - 2| < \delta,$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon,$ which is what you need to show for the critical step of your delta-epsilon proof.
In the proof as you saw it, many of the steps above are rephrased or condensed, which might cause confusion.
You may well ask, "What if $\left\lvert x - 2\right\rvert > \epsilon$?"
In that case, since you have already said that $\delta = \epsilon,$
if $\left\lvert x - 2\right\rvert > \epsilon$ then
$\left\lvert x - 2\right\rvert > \delta$
and the "if $\left\lvert x - 2\right\rvert < \delta$" condition is false.
For a delta-epsilon proof where $x\to 2$ you only need to show what happens if
$\left\lvert x - 2\right\rvert < \delta$ is true;
it does not matter what happens when $\left\lvert x - 2\right\rvert < \delta$ is false.
So if $\left\lvert x - 2\right\rvert > \epsilon$ then we are exploring cases that are completely irrelevant to whether the limit exists and what it is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3907306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help solving the limit of the sequence: $\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$ Given the following sequence,
$$\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$$
I am asked to determine to which $l \in \mathbb{R}$ the sequence converges.
This is what I have tried so far:
$$
\lim_{n} \left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( \frac{2n^2 + 1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( \frac{2n^2}{2n^2 - 1} + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( 1 + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies ? $$
My objective was to algebraically modify the sequence such that I could be able to reduce it to the form: $$\lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e$$
but seem to be stuck at the ? point.
Any suggestion? Thank you.
| Use$$\ln l=\lim_{n\to\infty}2n^2\underbrace{\frac{1-1/(2n^2)}{1+3/n}}_{\sim1}\underbrace{\ln\underbrace{\frac{1-1/(2n^2)}{1+1/(2n^2)}}_{1-1/n^2+o(1/n^2)}}_{\sim-1/n^2}=-2\implies l=e^{-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the minimal area under the curve of parabola $y=ax-bx^2$ between $x = 0$ and $x=\frac{a}{b}$ such that it passes through (1,1)?
Let $a > b >0$. The area under a parabola $y=ax−bx^2$ between its two roots, $x=0$ and $x=\frac{a}{b}$, is given by $\frac{a^3}{6b^2}$. Determine the $a$ and $b$ such that the parabola passes through the point $(1,1)$ and this area is minimised.
So far I have calculated the maximum value of this parabola to be at $x = \frac{a}{2b}$ giving $y = \frac{a^2}{4b^2}$. The restriction of $(1,1)$ gives that $1 = a-b$.
I don't know where to go from here so that the area under $y$ is minimised. Any help?
| Hint. Since $b=a-1>0$, you have to minimize the function
$$f(a)=\frac{a^3}{6b^2}=\frac{a^3}{6(a-1)^2}$$
over $(1,+\infty)$.
You may also let $a=b+1$ and minimize the function
$$g(b)=\frac{a^3}{6b^2}=\frac{(b+1)^3}{6b^2}$$
over $(0,+\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the condition for system of linear equations to be solvable It is clear that if the system of linear equations
$$
\left\{
\begin{array}{c}
x_1-x_3=c_1 \\
x_2-x_1=c_2 \\
x_3-x_2=c_3
\end{array}
\right.
$$
is solvable, then we have $c_1+c_2+c_3=0$.
How could we prove for the backward direction? That is, how could we prove that if $c_1+c_2+c_3=0$, then the above system of linear equations must be solvable without any corner cases?
| The augmented form of our system is
$$
\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
-1 & 1 & 0 & c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right]
$$
Row-reducing gives
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
-1 & 1 & 0 & c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right]
\xrightarrow{R_2+R_1\to R_2}\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
0 & 1 & -1 & c_{1} + c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right] \\
\xrightarrow{R_3+R_2\to R_3}\left[\begin{array}{rrr|r}
\fbox{1} & 0 & -1 & c_{1} \\
0 & \fbox{1} & -1 & c_{1} + c_{2} \\
0 & 0 & 0 & \fbox{$c_{1} + c_{2} + c_{3}$}
\end{array}\right]
\end{align*}
The first two boxed positions are guaranteed to be pivots in the reduced form of our system. The third boxed position is a pivot position if and only if $c_1+c_2+c_3\neq 0$. This third position is in the augmented column, so our system is solvable if and only if $c_1+c_2+c_3=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$ Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
I couldn't do this question and hence I looked at the solution which goes as follows:
if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$
I state that $y=lx$. Hence the system becomes:
$x^2(1+2l)=5$, $x^2(l^2-3l)=-2$ which becomes $5l^2-11l+2=0$, hence $l=2$ or $l=\frac{1}{5}$.
Hence the possible solutions are:
$(1, 2), (-1,-2), (\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}), (-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}})$.
My question is why was I supposed to think of substituting $y=lx$? Why was this supposed to be intuitive and is there a more intuitive approach?
| $\begin{align}
x^2+2xy & = 5 \\
y^2-3xy & = -2
\end{align}$
$\begin{align}
y = & \frac{5-x^2}{2x} \\
x = & \frac{2+y^2}{3y}
\end{align}$
$\begin{align}
y = & \frac{5-\left( \frac{2+y^2}{3y} \right)^2}{2 \left(\frac{2+y^2}{3y}\right)} \\
\end{align}$
$\begin{align}
y = & \frac{5 \cdot(9y^2)-(2+y^2)^2}{2 \cdot 3y (2+y^2)} \\
\end{align}$
$\begin{align}
{2 \cdot 3y^2 (2+y^2)} = & {5 \cdot(9y^2)-(2+y^2)^2} \\
\end{align}$
$\begin{align}
7y^4-29y^2+4=0
\end{align}$
$\begin{align}
y \in \{2,-2,\frac{1}{\sqrt{7}},\frac{-1}{\sqrt{7}}\}
\end{align}$
Plugging it into $x = \frac{2+y^2}{3y} $ we finish the problem:
$(x,y) \in \left\{(1, 2), (-1, -2),
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right),
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$n$-th derivative of $f(x)=x\cos(x)$ Can someone help me prove that the $n$-th derivative of $f(x)=x\cos(x)$ is:
$$f^n(x)=x\cos\left(x+\frac{n\pi}{2}\right)+n\cos\left(x+(n-1)\frac{\pi}{2}\right)?$$
Thanks, and stay safe.
| Just using induction using the product rule:
$f(x) = x\cos x$ so
$f'(x) = \cos x + x(-\sin x) = \cos x - x\sin x$
$f^2(x) = -\sin x - \sin x - x\cos x= -2\sin x - x\cos x$
$f^3(x) = -2\cos x-\cos x - x(-\sin x) = -3\sin x +x\sin x$.
By pattern recognition we sort of see that $f^n = \pm n \operatorname{trig} x \pm x\operatorname{other\ trig} x$. Bearing in mind that $\sin x = \cos (\frac \pi 2 - x)$ and that the pattern of the $\pm$ has a periodicity of $4$ and so on we can see that
$f^n(x) =\pm n \operatorname{trig} x \pm x\operatorname{other\ trig} x=$
$x\cos(x+\frac{n\pi}{2})+n\cos(x+(n-1)\frac{\pi}{2})$
and we can feel grateful that someone else hammered the details and fine polished it for us.
But we have to verify it.
For $n= 0$ this fits
$f(x) = x\cos x = x\cos(x+\frac{0\pi}{2})+0\cos(x+(0-1)\frac{\pi}{2})$
No question of that.
And for $n=1$ we have
$f'(x) = \cos x - x\sin x= x\cos(x+\frac{1\pi}{2})+1\cos(x+(1-1)\frac{\pi}{2})$
$=1\cos (x + 0) + x\cos(x+\frac {\pi}2) = \cos x- x \sin x$.
It works. Now to verify in general we must use induction.
If $f^k(x) =x\cos(x+\frac{k\pi}{2})+k\cos(x+(k-1)\frac{\pi}{2})$
then
$f^{k+1}(x) = \cos(x+\frac{k\pi}{2}) + x(-\sin(x+\frac{k\pi}2) -k\sin(x+(k-1)\frac \pi 2)=$
$=\cos(x+\frac{k\pi}{2}) +k\cos(x+k\frac {\pi}2) + x(\cos(x+\frac {(k+1)\pi}2))=$
$x\cos (x+\frac {(k+1)\pi}2)+ (k+1)\cos(x+k\frac \pi 2)$
So the result follows by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for all $n \in \mathbb{N}$, $\sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$ $$ \sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$$
To be honest, I can't really get started,
I would like to ask everyone to give me an idea of how to solve it, give me a starting push, thank you.
| we rewrite the sum like this ;$$\sum_{k=0}^{k=n-1}\binom{n+k-1}{n-1}\frac{1}{2^{n+k}}$$
this is just the coefficient of $x^{n-1}$ in expansion $$\frac{{(1+x)}^{n-1}}{2^n}+\frac{{(1+x)}^n}{2^{n+1}}...+\frac{{(1+x)}^{2n-2}}{2^{2n-1}}$$
Recognise this as a GP: hence we seek the coefficient of $x^{n-1}$ in $$\frac{1}{2^{n-1}}{(1+x)}^{n-1}\frac{1-{(\frac{x+1}{2})}^n}{1-x}$$
or $$\frac{{(1+x)}^{n-1}(1+x+x^2..)-{(1+x)}^{2n-1}(1+x+x^2+x^3....)}{2^{2n-1}}$$ The coefficent is $$\frac{\left(\binom{n-1}{0}+\binom{n-1}{1}..+\binom{n-1}{n-1}\right)-\left(\binom{2n-1}{0}+\binom{2n-1}{1}+...\binom{2n-1}{n-1}\right)}{2^{2n-1}}$$
$$=\frac{2^{2n-1}-\frac{1}{2}2^{2n-1}}{2^{2n-1}}=\frac{1}{2}$$
| {
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There are $5$ different apples, $4$ different oranges, and $3$ different bananas. Number of ways to select at least one of each kind?
There are $5$ different apples, $4$ different oranges, and $3$ different bananas. Number of ways to select at least one of each kind?
My first approach: I selected one of each kind first and then I have $9$ fruits remaining, of which I can choose in $2$ ways [either include or exclude].
$${5\choose1} \cdot {4\choose1} \cdot {3\choose1} \cdot (2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2)$$
My second approach:
$$\bigg({5\choose1}+{5\choose2}+{5\choose3}+{5\choose4}+{5\choose5}\biggl)\bigg({4\choose1}+{4\choose2}+{4\choose3}+{4\choose4}\bigg)\bigg({3\choose1}+{3\choose2}+{3\choose3}\bigg)$$
but the two methods give different answers.
I know my second approach is correct, but what is going wrong in my first approach?
You can help me by giving small example.
| There are a total of $5 + 4 + 3 = 12$ fruits. You can either select or not select each fruit. Thus, if we did not have the restriction that at least one fruit of each type must be selected, there would be $2^{12} = 4096$ possible ways to select the fruits (including the choice of not selecting any of them). Your first approach gives
$$\binom{5}{1}\binom{4}{1}\binom{3}{1} \cdot 2^9 = 30,720$$
an indication you are over counting.
Let's focus on the apples. According to the count in your first approach, there are
$$\binom{5}{1} \cdot 2^4 = 80$$
ways to select at least one apple rather than
$$\binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 31$$
ways to select at least one apple. This is because you count each selection with exactly $k$ apples $k$ times, once for each way you could have designated one of those $k$ apples as the apple you selected. To illustrate this, consider the following two examples.
Suppose the five apples are Ambrosia, Fuji, Gala, Granny Smith, and Honeycrisp. You count the selection of an Ambrosia apple and a Fuji apple twice, depending on which of those apples you designate as the apple you have selected.
$$
\begin{array}{l l}
\text{apple} & \text{additional apples}\\ \hline
\text{Ambrosia} & \text{Fuji}\\
\text{Fuji} & \text{Ambrosia}
\end{array}
$$
You count the selection of an Ambrosia apple, a Fuji apple, and a Gala apple three times, depending on which of those apples you designate as the apple you have selected.
$$
\begin{array}{l l}
\text{apple} & \text{additional apples}\\ \hline
\text{Ambrosia} & \text{Fuji, Gala}\\
\text{Fuji} & \text{Ambrosia, Gala}\\
\text{Gala} & \text{Ambrosia, Fuji}
\end{array}
$$
Since your first approach counts each selection of exactly $k$ apples $k$ times, your first approach yields
$$1 \cdot \binom{5}{1} + \color{red}{2} \cdot \binom{5}{2} + \color{red}{3} \cdot \binom{5}{3} + \color{red}{4} \cdot \binom{5}{4} + \color{red}{5} \cdot \binom{5}{5} = \color{red}{5 \cdot 2^4} = \color{red}{80}$$
ways to select at least one apple,
$$1 \cdot \binom{4}{1} + \color{red}{2} \cdot \binom{4}{2} + \color{red}{3} \cdot \binom{4}{3} + \color{red}{4} \cdot \binom{4}{4} = \color{red}{4 \cdot 2^3} = \color{red}{32}$$
ways to pick at least one orange, and
$$1 \cdot \binom{3}{1} + \color{red}{2} \cdot \binom{3}{2} + \color{red}{3} \cdot \binom{3}{3} = \color{red}{3 \cdot 2^2} = \color{red}{12}$$
ways to pick at least one banana, giving your total of
$$\color{red}{80 \cdot 32 \cdot 12} = \color{red}{30,720}$$
ways to select at least one fruit of each type.
A final note: There are $2^n$ subsets of a set with $n$ elements since we can either choose to include or not include each of the $n$ elements in a subset. Of these $2^n$ subsets, $2^n - 1$ are nonempty. Hence, there are $2^5 - 1$ ways to select at least one of the five apples, $2^4 - 1$ ways to select at least one of the four oranges, and $2^3 - 1$ ways to select at least one of the three bananas. Therefore, there are
$$(2^5 - 1)(2^4 - 1)(2^3 - 1) = 31 \cdot 15 \cdot 7 = 3255$$
ways to select at least one fruit of each type. Notice that
$$\left[\binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5}\right]\left[\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}\right]\left[\binom{3}{1} + \binom{3}{2} + \binom{3}{3}\right] = 31 \cdot 15 \cdot 7 = 3255$$
| {
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Determine the multiplicity of a root in determinant Consider the determinant
$$\det(A(x,z))=\begin{vmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{vmatrix}.$$
It is easy to see that if we write the determinant as a bivariate polynomial $p(x,z)$, then $x=0$ and $z=0$ are its roots, i.e. $p(0,z)=p(x,0)=0$. But how can we show that these roots have multiplicity 2?
Specifically, I'm interested if it is possible to determine the multiplicity of a root in a determinant without computing the determinant explicitely. One possibility would be to use the Jacobi formula to compute, e.g., $\frac{\partial}{\partial x}\det(A(x,z))$, but this task seems to be even more complicated. Probably there are some tricks that allow us to make a conclusion about the multiplicity of a root with less effort?
EDIT: The solution is supposed to follow this scheme:
*
*Show that $x=0$ ($z=0$) is a root of the polynomial.
*Show that $x=0$ ($z=0$) is a double root of the polynomial.
*Deduce that $p(x,z)=x^2z^2$.
It is the second item, where I have a problem.
| Adding or subtracting one row to another does not change the value of the determinant. So $$\det\begin{bmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{bmatrix}=\det\begin{bmatrix}1&1&1&1+z\\0&x&0&-z\\0&0&z&z\\0&0&x&x+xz\end{bmatrix}=x^2z^2$$
NB It is $x=0$ and $z=0$ that are the roots of the polynomial.
EDIT: Let's expand $\det(A(x))$ as a series in $x$:
$\det(A(0))=0$ since it contains a repeated column of $1$s. This shows that $x=0$ is a root of $p(x,z)$.
To get the first order terms in $x$, recall that determinants are linear in each column separately, hence expanding the first and second columns gives \begin{align}\det(A(x))&=\det\begin{bmatrix}1&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&1&1&1\\0&1+x&1&1\\0&1&1-z&1\\0&1&1&1+z\end{bmatrix}\\
&=\det\begin{bmatrix}1&0&1&1\\1&x&1&1\\1&0&1-z&1\\1&0&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&1&1&1\\0&1&1&1\\0&1&1-z&1\\0&1&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&0&1&1\\0&x&1&1\\0&0&1-z&1\\0&0&1&1+z\end{bmatrix}\\
&=-x^2\det\begin{bmatrix}1-z&1\\1&1+z\end{bmatrix}\end{align} In the second equation, the first two determinants cancel out by noticing that the second can be obtained from the first by swapping the first two columns and rows. It is this cancellation that annihilates the $x$ term and produces the double root $x^2$.
| {
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Finding $abc$ for distinct reals satisfying $\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$ The problem is
Let $a$, $b$, $c$ be three distinct real numbers satisfying
$$\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$$ Find $abc$.
I can solve this in the case of $a, b, c \ge 0$ by assuming $a =\max\{a, b, c\}$. In that case, $abc =0$. But I haven't solved it yet in the general case.
Please help me. Thanks.
| As @fleablood observed, none of $a,b,c$ can be zero. Hence, notice that the condition is equivalent to $$\frac{1+ab}{a}=\frac{1+bc}{b}=\frac{1+ca}{c}\iff \frac1a+b=\frac1b+c=\frac1c+a$$ Yet this implies that $$\frac1a+b=\frac1b+c\iff b-c=\frac{a-b}{ab}$$ Similarly $$\frac1b+c=\frac1c+a\iff c-a=\frac{b-c}{bc}\quad\text{and}\quad \frac1c+a=\frac1a+b\iff a-b=\frac{c-a}{ca}$$ Multiply the three resulting equations to obtain $$(b-c)(c-a)(a-b)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2}$$ Since $a,b,c$ are distinct real numbers, we have that $a-b\not= 0, b-c\not=0, c-a\not=0$ and thus $$a^2b^2c^2=1\iff abc=\pm1$$ Can you take it from here? Can both $1$ and $-1$ be attained?
| {
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$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $ Did I do it right? How to do the last step? I need to find:
$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $$
So I calculate:
$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| = \lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} | $$
Denominator grows to $\infty$, so I can use stolz theorem:
$$\lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} | = \lim_{n \to +\infty} |\frac{\sqrt{2n}}{\sqrt{n} - \sqrt{n-1}} | \implies[\frac{\sqrt{2}}{0} ] $$
So I have to transform that a bit:
$$\lim_{n \to +\infty} |\frac{\sqrt{2n}(\sqrt{n} + \sqrt{n-1})}{n - n + 1} | = \lim_{n \to +\infty} |\sqrt{2}n + \sqrt{2n^2-2n}| \implies \infty - \infty$$
Another transformation:
$$\lim_{n \to +\infty} |\frac{2n^2 - 2n^2 + 2n}{\sqrt{2}n - \sqrt{2n^2-2n}}| = \lim_{n \to +\infty} |\frac{2n}{n(\sqrt{2} - \sqrt{2-\frac{2}{n}})}| = \lim_{n \to +\infty} |\frac{2}{\sqrt{2} - \sqrt{2-\frac{2}{n}}}| \implies[\frac{2}{0} ]$$
... and I feel like it doesn't lead nowhere. Can somebody tell what do I do wrong?
| $$\begin{split}
\sum_{k=1}^{2n} (-1)^k \sqrt{k} &= \sum_{k=1}^{n} \left((-1)^{2k-1} \sqrt{2k-1} + (-1)^{2k} \sqrt{2k}\right)\\
&= \sum_{k=1}^{n} \left( \sqrt{2k}-\sqrt{2k-1}\right)\\
&= \sum_{k=1}^{n} \frac{1}{\sqrt{2k}+\sqrt{2k-1}}
\end{split}$$
Thus $$\sum_{k=1}^n\frac{1}{2\sqrt{2k}}\leq \sum_{k=1}^{2n} (-1)^k \sqrt{k}\leq \sum_{k=1}^n\frac{1}{2\sqrt{2k-1}}$$
We can use integrals to estimate the bounds
$$\frac 1 {2\sqrt 2}\int_1^{n+1}\frac{dx}{\sqrt x}\leq \sum_{k=1}^{2n} (-1)^k \sqrt{k}\leq \frac 1 2+ \int_2^{n+1}\frac{dx}{2\sqrt{2x-1}}$$
to see that $$\sum_{k=1}^{2n} (-1)^k \sqrt{k}\sim \sqrt{\frac n {2}}$$
Thus
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}|=\frac 1 {\sqrt 2}$$
| {
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Prove using mathematical induction Sequence $b_1, b_2,...$ where $b_1=b_2=1$ and for every natural number $k$ bigger than $2$, $b_k = b_{k-1}+b_{k-2}$.
Prove using mathematical induction that, for every $k \in \mathbb N, b_k \le (\frac{7}{4})^{k-1} $
I started with if $k=3$, $k_3=1+1=2$ and $(\frac{7}{4})^{k-1} =(\frac{7}{4})^2=3 \frac{1}{16}$, $2<3 \frac{1}{16}$. This is the induction base?
Step:
Let us assume that S(m) is true, $b_m \le (\frac{7}{4})^{m-1}$ for every m $\in$ {3, ..., k}.
We have to show that $b_{k+1} \le (\frac{7}{4})^k$.
Using assumption and $b_{k+1} = b_{k}+b_{k-1}$ I got
$$b_k+b_{k-1}\le (\frac{7}{4})^{k+1}+(\frac{7}{4})^{k}$$
I don't know what to do next.
|
I don't know what to do next.
$ \left(\dfrac{7}{4}\right)^{k+1}+\left(\dfrac{7}{4}\right)^{k}=\left(\dfrac74\right)^k\left(\dfrac74+1\right)=\left(\dfrac74\right)^k\left(\dfrac{11}4\right)=\left(\dfrac74\right)^k\left(\dfrac{44}{16}\right)$
$<\left(\dfrac74\right)^k\left(\dfrac{49}{16}\right)=\left(\dfrac74\right)^k\left(\dfrac{7}{4}\right)^2=\left(\dfrac74\right)^{k+2}$
| {
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Range of quadratic function using discriminant
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.
Let $2x^2-2xy+y^2=a$.
Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.
We divide both side by $y^2$ and let $t=\frac{x}{y}$.
Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.
Since its discriminant is not negative, $\frac{\Delta}{4}\ge 0\implies a^2-7a-4\ge 0$. It gives us $a$ can have negative values like $-1$. But if clearly contracts $a-4=x^2+4y^2\ge 0\implies a\ge 4$. Where did I mistake?
| Method$\#1:$
For real $t,$ $$a^2-7a-4\ge0$$
If $x=ty$
$$4=x^2-2xy-3y^2=y^2(t^2-2t-3)$$
$$t^2-2t-3=\dfrac4{y^2}>0$$
$$\implies(t-3)(t+1)>0$$
Either $t<-1$ or $t>3$
So, the values of $a$ must satisfy this condition as well
Method$\#2:$
$$4=(x-y)^2-(2y)^2$$
WLOG $y=\tan t, x-y=2\sec t\implies x=2\sec t+\tan t$
$$a=2(2\sec t+\tan t)^2-2(2\sec t+\tan t)\tan t+\tan^2t$$
Multiplying both sides by $\cos^2t=1-\sin^2t,$
$$(1+a)\sin^2t+4\sin t+8-a=0$$
What if $a+1=0?$
Else
$$\sin t=\dfrac{-2\pm\sqrt{a^2-7a-4}}{a+1}$$
As $\sin t$ is real, the discriminant must be $\ge0$
But that is not sufficient, we need $$-1\le\dfrac{-2\pm\sqrt{a^2-7a-4}}{a+1}\le1$$ as well for real $t$
Also, $\dfrac xy=2\csc t+1$ $\implies\dfrac xy\ge2+1$ or $\dfrac xy\le-2+1$
| {
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$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan x-x^2)}{x^4}= \frac{ \arctan x-x}{x^3} \sim \frac{x-x}{x^3} \sim 0$$
But the final result should be $-\frac{4}{3}$.
Any help would be appreciated.
| $$L=\lim_{x \to 0}\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
You need to use Taylor series:
$$L=\lim_{x \to 0}\frac{\ln (1+x (x-\dfrac {x^3}3+...)-(1+x^2+\dfrac {x^4}{2}+o(x^4))+1}{x^4}$$
$$L=\lim_{x \to 0}\frac{ (x^2-\dfrac {x^4}3-\dfrac {x^4}2+o(x^4))-x^2-\dfrac {x^4}{2}+o(x^4)}{x^4}$$
$$L=\lim_{x \to 0}\frac{ -\dfrac {x^4}3-\dfrac {x^4}2-\dfrac {x^4}{2}+o(x^4)}{x^4}$$
$$L=\lim_{x \to 0}\frac{ -\dfrac {x^4}3- {x^4}+o(x^4)}{x^4}$$
$$L= -\dfrac {1}3-1=-\dfrac 43$$
Where :
$$\sqrt{1+2x^4}=1+x^4+o(x^4)$$
$$e^{x^2}=1+x^2+\dfrac {x^4}2+o(x^4)$$
$$\ln (1+x)=x-\dfrac {x^2}2 +...$$
$$\arctan x = x-\dfrac {x^3}3+o(x^4)$$
for the arctan and log term
$$S=\ln (1+x (x-\dfrac {x^3}3+...)$$
Remember that we have the Taylor serie for log:
$$\ln (1+z)=z-\dfrac {z^2}2$$
So that we have:
$$S=\ln (1+x (x-\dfrac {x^3}3+...)$$
$$S=\ln (1 +(x^2-\dfrac {x^4}3+...)$$
$$S=(x^2-\dfrac {x^4}3+...)-\dfrac 12(x^2-\dfrac {x^4}3+...)^2$$
We only keep terms with exponent less or equal to four...
$$S=(x^2-\dfrac {x^4}3+...)-\dfrac 12(x^4 \ \color {red}{ \text { + more terms with exponent> 4}})$$
We don't need the terms that have exponent greater than four. Because we take the limit at zero and all these terms will be zeros. They play no role. So that:
$$S=x^2-\dfrac {x^4}3-\dfrac {x^4}2+o(x^4)$$
| {
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$\lim_{x \to + \frac{1}{2}} [ \tan(\pi x^2)+(2x-1)4^x+6x-4 ] \cot(x-\frac{1}{2})$ I've tried to solve this limit:
$\lim_{x \to + \frac{1}{2}} \left [ \tan(\pi x^2)+(2x-1)4^x+6x-4 \right ] \cot(x-\frac{1}{2})$
the first parentesis should tend to $\frac{\sqrt{2}}{2}-1$ but I don't know how to procede.
The final result should be $2 \pi + 10$
| Let $~f(x) = \left[\tan(\pi x^2)+(2x-1)4^x+6x-4 \right]~$ and
let $~g(x) = \tan[x - (1/2)].$
Note that $g(x)$ and each term of $f(x)$ are continuous functions.
You want $\lim_{x\to (1/2)} \frac{f(x)}{g(x)}.$
Before applying L'Hopital's rule, first check $f(1/2)$ and $g(1/2)$.
$$f(1/2) = \tan(\pi/4) + (0)4^{(1/2)} + 3 - 4 = 1 + 3 - 4 = 0.$$
$$g(1/2) = \tan(0) = 0.$$
Applying L'Hopital's rule:
$$f'(x)|_{x=(1/2)}
~=~\frac{(2\pi)x}{\cos^2(\pi x^2)} + (2)(4^x) + (2x - 1)\log(4)[4^x] + 6 ~|_{x = (1/2)}
$$
$$= \frac{\pi}{(1/\sqrt{2})^2} + (2)(2) + (0)\log(4)(2) + 6
~=~ 2\pi + 4 + 0 + 6 ~=~ 2\pi + 10.$$
and
$$g'(x) ~|_{x = (1/2)}
~=~ \frac{1}{\cos^2(x - 1/2)} ~|_{x = (1/2)} = 1.$$
Thus, $$\frac{f'(1/2)}{g'(1/2)} = \frac{2\pi + 10}{1}.$$
| {
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Determinant and inverse matrix calculation of a special matrix Is there any smart way of calculating the determinant of this kind matrix?
\begin{pmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}
I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.
I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.
I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.
| If it’s the determinant you are looking for, then what if you subtract column $j+1$ from column $j$ to get the matrix
$$\begin{pmatrix}
-1 & -1 & -1& \cdots & n \\
1& -1 & -1& \cdots & n-1 \\
1& 1& -1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1& 1& 1& \cdots & 1 \end{pmatrix}_{n\times n}$$
Then you could add the bottom row to the top row to get $[0,0,...,n+1]$ then the determinant will be
$$(-1)^{n+1}(n+1)\begin{vmatrix}
1& -1 & -1& \cdots & -1\\
1& 1& -1 & \cdots & -1\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1& 1& 1& \cdots & 1 \end{vmatrix}_{(n-1)\times(n-1)}$$
From which you can always add the bottom row to the top to get $[2,0,...,0]$, so you end up getting
$(-1)^{n+1}(n+1)2^{n-2}$?
It looks like for the $n \times n$ matrix $A$ given, the inverse has the general form
$$B=
\begin{pmatrix}
\frac{-n}{2(n+1)}& \frac{1}{2} & 0 & 0 &\cdots & 0 & \frac{1}{2(n+1)} \\
\frac{1}{2} & -1 & \frac{1}{2} & 0 & \cdots & 0& 0 \\
0 & \frac{1}{2} & -1 & \frac{1}{2} & \cdots & 0&0 \\
0 & 0 & \frac{1}{2} & -1 & \cdots & 0&0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots &\vdots\\
0 & 0 & 0 & 0 & \cdots &-1&\frac{1}{2}\\
\frac{1}{2(n+1)} & 0 & 0 & 0& \cdots &\frac{1}{2}& \frac{-n}{2(n+1)} \end{pmatrix}
$$
I don't really have proof of how I got this except plugging in the matrix to an inverse matrix calculator for various values of $n$ and noticing a pattern. But one can certainly check that it is indeed the inverse, multiplying on the left and getting $B\cdot A=I$ ensures it is the inverse.
To check this, when multiplying this matrix on the left, it’s not hard to check for the rows of the form $[0,...,\frac{1}{2},-1,\frac{1}{2},...,0]$ because either we have three consecutive integers $n,n+1,n+2$ or $n,n-1,n-2$ $$\frac{n}{2}-(n+1)+\frac{n+2}{2}=0$$ $$\frac{n}{2}-(n-1)+\frac{n-2}{2}=0$$ or $$2\cdot\frac{1}{2}+(-1)\cdot 1+ 2\cdot \frac{1}{2}=1$$ which always is a diagonal entry. Then considering the top row:
$$\frac{-n}{2(n+1)} + 2\cdot \frac{1}{2} + \frac{n}{2(n+1)}=1$$
$$(i+1)\cdot\frac{-n}{2(n+1)}+ i\cdot \frac{1}{2} + (n-i)\cdot \frac{1}{2(n+1)}=0$$ for $i=1,2,...,n-1$.
The bottom row acts similarly. Thus we indeed get $I$. This probably yields to a nice method to find the inverse through $[A|I] \rightsquigarrow [I|B]$.
Hope this helps!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve integral $\int^1_0 \frac{1-x^2}{{(1+x^2)}\sqrt{1+x^4}}dx$ using subsituition $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ Hi this question has been posed in my integration book where it has been asked to solve it using the given substitution or any other substitution, I've found identical question posted here {1} with different substitutions
The problem is when using the given substitution I end up with integral
$$\int^{\pi/4}_0 \frac{\sin{\theta}}{2x}d\theta $$
Solving for $x$ : $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$
According to WolframAlpha: $$x = \frac{|\csc{\theta}||1 \pm \sqrt{\cos{2\theta}}|}{\sqrt{2}} $$
I do know the sign for x is $+ve$ so I'll take $+ve$ solutions after opening the modulus, I do not know which sign to prefer in $ \pm \sqrt{\cos{2\theta}} $ term
Eitherway, I solved for both cases :
$$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 + \sqrt{\cos{2\theta}}} d\theta -(I) $$
$$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 - \sqrt{\cos{2\theta}}} d\theta -(II) $$
Which leads to :
$$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0 1 \pm \sqrt{\cos{2\theta}} d\theta = \frac{\pi}{4\sqrt{2}}
\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}} $$
Here I cheated and used WolframAlpha as I was exhausted.
Both the cases lead to a similar answer with difference of a function unknown to me: $\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}}$
The correct answer is :
$$\frac{\pi}{4\sqrt{2}}$$
My main questions are:
*
*What am I doing wrong?
*How did author arrive at this ingenious substitution $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ , this doesn't appear to me trial n' error kind of substitution, I have spent hours doing algebraic transformation on this and no matter how you procced and plug the variable your integrand will only have one of these terms only $2x$, $1-x^2$, $1+x^2$ and $ \sqrt{1+x^4}$ and $2x$ being the simplest.
Related question: How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$
| Let's begin with the substitution suggested in the question. We have $$\cos t=\frac{\sqrt{1+x^4}} {1+x^2}$$ ($\theta$ is replaced by $t$ to reduce burden of typing mathjax). Differentiating we get
\begin{align}
-\sin t \, dt&=\dfrac {(1+x^2)\cdot \dfrac{4x^3} {2\sqrt{1+x^4}} - 2x\sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\
&=\dfrac {(1+x^2)\cdot \dfrac{2x^3} {\sqrt{1+x^4}} - 2x\sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\
&= 2x\cdot\dfrac {(1+x^2)\cdot \dfrac{x^2} {\sqrt{1+x^4}} - \sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\
&= 2x\dfrac { \dfrac{x^2(1+x^2)-(1+x^4)} {\sqrt{1+x^4}}} {(1+x^2)^2} \, dx\notag \\
&= 2x\dfrac { \dfrac{x^2-1} {\sqrt{1+x^4}}} {(1+x^2)^2} \, dx\notag \\
\implies \sin t \, dt&=\frac{2x(1-x^2)}{(1+x^2)^2\sqrt{1+x^4}}\, dx\notag\\
\end{align}
You need to double check your calculations as there is no cancellation in above to reduce the $(1+x^2)^2$ in denominator to $1+x^2$.
Next we note that $$\sin t =\frac{\sqrt {2}x}{1+x^2}$$ and therefore $$dt=\sqrt{2}\cdot\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx$$ or $$\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx=\frac{dt}{\sqrt{2}}$$ Integrating the above (noting that interval $[0,1]$ for $x$ maps to $[0,\pi/4]$ for $t$) we get $$I=\int_0^1\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx=\frac{1}{\sqrt {2}}\int_0^{\pi/4}\,dt=\frac{\pi}{4\sqrt{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
A question related to $f(n)=n^4+n^2+1$ If $$f(n)=n^4+n^2+1$$ then we have to evaluate $$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}$$ which, when run in Desmos, returns 61. But obviously we can't evaluate this with brute force. The farthest I have reached is that $$n^4+n^2+1=n^2(n+1)+\frac{n+1}{n+1}=(n^2+\frac{1}{n+1})(n+1)$$ Any hints?
| By using Sophie German Identity
$$f(n)=n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$$
$$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}=\frac{7\times 13\times 21 \times 31\times 43\times 57\times 73\times 91\times 111\times 133\times 157\times 183}{3\times 7\times 13 \times 21\times 31\times 43\times 57\times 73\times 91\times 111\times 133\times 157}=61$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove by definition that $\frac{1-2n^3}{4n^3-2n^2-1} \to -\frac12$ $a_n = \dfrac{1-2n^3}{4n^3-2n^2-1}$
My attempt (of the scratch):
$\left| \dfrac{1-2n^3}{4n^3-2n^2-1} + \dfrac{1}{2}\right|=\left| \dfrac{2-4n^3+4n^3-2n^2-1}{8n^3-4n^2-2}\right|=\left|\dfrac{-2n^2+1}{8n^3-4n^2-2} \right|=\dfrac{2n^2-1}{8n^3-4n^2-2}<\dfrac{2n^2}{8n^3-4n^2-2}=\dfrac{n^2}{4n^3-2n^2-1}<\epsilon$
But I don't know what to do next to choose my $N \in \Bbb{N}$
| For large $n$, we have $4n^3 - 2n^2 - 1 > n^3$, in which case $\frac{n^2}{4n^3-2n^2-1} < \frac{n^2}{n^3} = \frac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given rank-$1$ matrix $A$, how to compute $A^{100}$?
$$A = \begin{bmatrix} 6 & 4\\ -6 & -4\end{bmatrix}$$ Find $A^{100}$.
I tried to find it using diagonalization, but as it is a singular matrix so one of eigenvectors came out zero. How $A^{100}$ can be calculated of same matrix?
| $$|\lambda I - A| = \lambda^2 - 2\lambda$$
A matrix satisfies its characteristic equation:
$$A^2 -2A =0$$
$$A^2 = 2A$$
$$A^4 = 4A^2 = 8A$$
$$A^8 = 16A^4= 2^7 A$$
$$A^{16} = (A^4)^4 = 2^{12}A^4=2^{15}A$$
$$A^{32}= (A^16)^2 = 2^{30}A^2 = 2^{31}A$$
$$A^{64} = (A^16)^4 = 2^{60} A^4 = 2^{63}A$$
$$A^{100}= A^{64+32+4} = 2^{63+31+3}A^3 = 2^{97}A^3=2^{99}A$$
$$A^{100} \approx \pmatrix{3.8&2.5\\-3.8&-2.5}\cdot10^{30}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Convergence of the recursive sequence $x_{n+1} = \frac{1}{4-x_n}$ I'm doing exercise 2.4.1 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, if my proof is rigorous and technically correct.
(a) Prove that the sequence defined by $x_1 = 3$ and
\begin{align*}
x_{n+1} = \frac{1}{4 - x_n}
\end{align*}
converges.
(b) Now that we know $\lim x_n$ exists, explain why $\lim x_{n+1}$ must also exist and equal the same value.
(c) Take the limit of each side of the recursive equation in part (a) to explicitly compute the $\lim x_n$.
Proof.
(a) By direct computation, we find that $x_1 = 3$, $x_2 = 1$. Let us prove that $(x_n)$ is a decreasing sequence, that is $x_{n+1} < x_{n}$ for all $n \in \mathbf{N}$. This is true for $n=1$. By induction, let's assume that $x_{k+1} < x_k$. Therefore,
\begin{align*}
x_{k+1} &< x_{k} \implies \frac{1}{4 - x_{k+1}} < \frac{1}{4 - x_k} \implies x_{k+2} < x_{k+1}
\end{align*}
So, $(x_{n})$ is a monotonically decreasing sequence.
Moreover, we can show that $(x_n)$ is bounded. We are interested to show that $x_n > 0$ for all $n \in \mathbf{N}$. This holds for $n=1$. Assume that $x_k > 0$, then
\begin{align*}
x_{k+1} = \frac{1}{4 - x_k} > \frac{1}{4} > 0
\end{align*}
Thus, the sequence $(x_n)$ has a lower bound $0$. By the Monotone Convergence Theorem, the sequence $(x_n)$ converges.
(b) The sequence $(x_{n+1})=x_2,x_3,x_4,\ldots$ also converges and has the same limiting value because the $1$-tail of $(x_n)$ is a (i) monotonically decreasing sequence (ii) has the same lower bound $0$. It is the infinite tail of the sequence, that ultimately determines the convergence of a sequence.
(c) We have:
\begin{align*}
\lim (x_n) &= \frac{1}{4 - \lim x_{n+1}}\\
L &= \frac{1}{4 - L} \\
4L - L^2 &= 1\\
L^2 - 4L + 1 &= 0\\
(L - 2)^2 - 3 &= 0\\
L&= 2 \pm \sqrt{3}
\end{align*}
As $0 < L < 1$, we have $L = 2 - \sqrt{3}$.
| An alternative solution.
It is clear from the form of the recursion that there is a sequence $\{y_n\}_{n=1}^\infty$ such that $x_n=\frac{y_n}{y_{n+1}}$.
$$\frac{y_{n+1}}{y_{n+2}}=\frac{1}{4-\frac{y_n}{y_{n+1}}}=\frac{y_{n+1}}{4y_{n+1}-y_{n+2}}\implies y_{n+2}=4y_{n+1}-y_n$$.
The base case is $y_1=3$ and $y_2=1$.
$$\lambda^2=4\lambda-1\implies \lambda=2\pm\sqrt{3}\implies y_n=C_+(2+\sqrt{3})^n+C_-(2-\sqrt{3})^n$$
We don't need to solve for the two constants, but their values are $C_{\pm}=\frac{33\mp 19\sqrt{3}}{6}$.
Nevertheless, $2-\sqrt{3}<1\implies\lim_{n\to\infty}(2-\sqrt{3})^n=0$ and so:
$\lim_{n\to\infty}x_n=\lim_{n\to\infty}\frac{C_+(2+\sqrt{3})^n+C_-(2-\sqrt{3})^n}{C_+(2+\sqrt{3})^{n+1}+C_-(2-\sqrt{3})^{n+1}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$.
Notice that this essentially solves all three parts together.
(a) The computation shows that the limit exists.
(b) We are just shifting the index.
(c) The limit's existence was established by actually finding it to be $2-\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$ I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.
One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)
| Making the problem more general
$$I_n=\int_0^t \frac {dx}{\sum_{k=0}^n x^k}=\int_0^t \frac {1-x}{1-x^{n+1}}\,dx$$
$$x=y^{\frac{1}{n+1}} \quad \implies \quad I_n=\frac 1{n+1}\int_0^{t^{n+1}} \frac{y^{-\frac{n}{n+1}}-y^{-\frac{n-1}{n+1}}} {1-y} \,dy$$ Using
$$\int \frac {y^a}{1-y}\,dy=\frac{y^{a+1} }{a+1}\,\, _2F_1(1,a+1;a+2;y)$$
$$\int_0^s \frac {y^a}{1-y}\,dy=B_s(a+1,0) \qquad \text{if} \qquad s<1\land a>-1$$
$$I_n=\frac 1{n+1}\Bigg[B_{t^{n+1}}\left(\frac{1}{n+1},0\right)-B_{t^{n+1}}\left(\frac{2}{n+1},0\right) \Bigg]$$
For the specific case where $t=1$,
$$I_n=\frac{\psi \left(\frac{2}{n+1}\right)-\psi \left(\frac{1}{n+1}\right)}{n+1}=\frac{H_{-\frac{n-1}{n+1}}-H_{-\frac{n}{n+1}}}{n+1}$$ If $n$ is large
$$I_n=\frac{1}{2}+\frac{\pi ^2}{6 n^2}-\frac{9 \zeta (3)+\pi^2
}{3n^3}+\frac{810 \zeta (3)+45 \pi ^2+7 \pi ^4}{90 n^4}+O\left(\frac{1}{n^5}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$ How to evaluate:
$$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$$
Personally, i always have trouble to evaluate sum of trignometric series, if you know a paper to recommend, or part of a book, about this, please let me know.
Anyway
$$\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right) = \frac{1}{n^2}\left(\sum_{k=1}^{n} \frac{1}{\sin^2\left(\frac{\pi k}{2n+1}\right)} - \sum_{k=1}^{n} 1\right)$$
What now?
| Let $P_n(X)=\frac{(X+i)^{2n+1}-(X-i)^{2n+1}}{2i}$, then
$$ \begin{aligned} P_n(z)=0&\iff \left(\frac{z+i}{z-i}\right)^{2n+1}=1 \\
&\iff \exists k\in[\![1,2n+1]\!],\frac{z+i}{z-i}=e^{\frac{2ik\pi}{2n+1}} \\
&\iff\exists k\in[\![1,2n+1]\!],z=i\frac{e^{\frac{2ik\pi}{2n+1}}+1}{e^{\frac{2ik\pi}{2n+1}}-1} \\
&\iff\exists k\in[\![1,2n+1]\!],z=\cot\left(\frac{k\pi}{2n+1}\right)
\end{aligned}$$
Therefore the roots of $P_n$ are $\alpha_k:=\cot\left(\frac{k\pi}{2n+1}\right)$. Moreover, using the binomial theorem, we have
$$ P_n(X)=\frac{1}{2i}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\left(i^k-(-i)^k\right)X^{2n+1-k}=\sum_{k=0}^n\binom{2n+1}{2k+1}(-1)^k X^{2n-2k}$$
because the even terms cancel each other. Let $\displaystyle Q_n(X)=\sum_{k=0}^n\binom{2n+1}{2k+1}(-1)^k X^{n-k}$, then $Q_n(X^2)=P_n(X)$ and the roots of $Q_n$ are the $\alpha_k^2=\cot^2\left(\frac{k\pi}{2n+1}\right)$. But since $\alpha_{2n+1-k}=-\alpha_k$ it remains only $n=\deg Q_n$ roots which are the $\alpha_k^2$ for $k\in[\![1,n]\!]$. Therefore, the lead coefficient of $Q_n$ being $2n+1$, we have
$$ Q_n(X)=(2n+1)\prod_{k=1}^n(X-\alpha_k^2)$$
Thus, the sum $\sum_{k=1}^n\alpha_k^2$ is the coefficient of degree $n-1$ in $Q_n$ divided by $-\frac{1}{2n+1}$, that is
$$\sum_{k=1}^n\alpha_k^2=\frac{1}{2n+1}\binom{2n+1}{3}=\frac{n(2n-1)}{3}$$
Finally,
$$ \lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^n\cot^2\left(\frac{k\pi}{2n+1}\right)=\frac{2}{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the absolute maximum and minimum of $f(x) = \left|\cos^2(x) - \frac{3}{4}\right|$ on $[0,\pi]$
Find the absolute maximum and minimum of $f(x) = \left|\cos^2(x) - \frac{3}{4}\right|$ on $[0,\pi]$
I have learnt that in order to find the absolute maximum, I need to check the value of the function at the:
*
*sides of $[0,\pi ]$
*in points that the function isn't differentiable.
*in points that satisfy $f'(x)=0$
How do I approach this question? I have no idea how to differentiate it with the absolute value, and how could I find the points that aren't differentiable in $[0,\pi ]$. any help is appreciated.
| Well, I will give you some (big) hints:
*
*Notice that when $x\in\left[0,\frac{\pi}{6}\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\cos^2\left(x\right)-\frac{3}{4}$;
*Notice that when $x\in\left[\frac{\pi}{6},\frac{5\pi}{6}\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\frac{3}{4}-\cos^2\left(x\right)$;
*Notice that when $x\in\left[\frac{5\pi}{6},\pi\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\cos^2\left(x\right)-\frac{3}{4}$;
*Notice that:
$$\cos^2\left(x\right)=\frac{1+\cos\left(2x\right)}{2}\tag1$$
*Notice that:
$$\frac{\text{d}}{\text{d}x}\left(\cos^2\left(x\right)-\frac{3}{4}\right)=\frac{1}{2}\cdot\left\{\frac{\text{d}}{\text{d}x}\left(1\right)+\frac{\text{d}}{\text{d}x}\left(\cos\left(2x\right)\right)\right\}-\frac{3}{4}\cdot\frac{\text{d}}{\text{d}x}\left(1\right)=$$
$$\frac{1}{2}\cdot\left\{0+\frac{\text{d}}{\text{d}x}\left(\cos\left(2x\right)\right)\right\}-\frac{3}{4}\cdot0=-\frac{1}{2}\cdot\sin\left(2x\right)\cdot\frac{\text{d}}{\text{d}x}\left(2x\right)=$$
$$-\frac{1}{2}\cdot\sin\left(2x\right)\cdot2\cdot\frac{\text{d}}{\text{d}x}\left(x\right)=-\frac{1}{2}\cdot\sin\left(2x\right)\cdot2\cdot1=-\sin\left(2x\right)\tag2$$
*Notice that, likewise $(2)$ only with a flipped sign:
$$\frac{\text{d}}{\text{d}x}\left(\frac{3}{4}-\cos^2\left(x\right)\right)=\sin\left(2x\right)\tag3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric problem (problem from a Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two angles are $135^\circ$ and the sum of their tangents are $5$. Calculate the angles.
Is there a shorter/simpler solution than the one presented below that I made some months ago? It seems rather ‘lengthy’.
Solution
Let the angles be $\alpha$ and $\beta$.
We have
$$
\left\{
\begin{aligned}
\alpha+\beta&=135°,\\
\tan(\alpha)+\tan(\beta)&=5.
\end{aligned}
\right.
$$
Since
$$
\tan(x)+\tan(y)
=\frac{\sin(x)}{\cos(x)}+\frac{\sin(y)}{\cos(y)}
=\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos(x)\cos(y)}
=\frac{\sin(x+y)}{\cos(x)\cos(y)}
$$
we have
$$
5
=\tan(\alpha)+\tan(\beta)
=\frac{\sin(\alpha+\beta)}{\cos(\alpha)\cos(\beta)}
=\frac{\sin(135°)}{\cos(\alpha)\cos(\beta)}
=\frac{\frac{1}{\sqrt{2}}}{\cos(\alpha)\cos(\beta)}
$$
which gives
$$
\cos(\alpha)\cos(\beta)=\tfrac{1}{5\sqrt{2}}.
$$
Further
$$
-\tfrac{1}{\sqrt{2}}
=\cos(135°)
=\cos(\alpha+\beta)
=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}-\sin(\alpha)\sin(\beta)
$$
which gives
$$
\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}+\tfrac{1}{\sqrt{2}}
=\tfrac{6}{5\sqrt{2}}.
$$
Since $\alpha+\beta=135°$ we have
\begin{align*}
\sin(\alpha)\sin(\beta)
&
=\sin(\alpha)\sin(135°-\alpha)
\\&=\sin(\alpha)\bigl(\sin(135°)\cos(\alpha)-\cos(135°)\sin(\alpha)\bigr)
\\&=\sin(\alpha)\bigl(\tfrac{1}{\sqrt{2}}\cos(\alpha)+\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigl(\cos(\alpha)+\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\bigl(\sin(\alpha)\cos(\alpha)+\sin^2(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\Bigl(\tfrac{1}{2}\sin(2\alpha)+\tfrac{1}{2}\bigl(1-\cos(2\alpha)\bigr)\Bigr)
\\&=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\end{align*}
why
\begin{gather*}
\tfrac{6}{5\sqrt{2}}=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\\\quad\Leftrightarrow\quad
\tfrac{12}{5}=\sin(2\alpha)+1-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sin(2\alpha)-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sqrt{2}\sin(2\alpha-\tfrac{\pi}{4})
\\\quad\Leftrightarrow\quad
\tfrac{7}{5\sqrt{2}}=\sin(2\alpha-\tfrac{\pi}{4})
\end{gather*}
which gives
\begin{gather*}
2\alpha-\tfrac{\pi}{4}=
\begin{cases}
\arcsin(\tfrac{7}{5\sqrt{2}})+n_12\pi\\
\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+n_22\pi
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\bigl(\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_12\pi\bigr)\\
\tfrac{1}{2}\bigl(\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_22\pi\bigr)
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{8}+n_1\pi\\
-\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{5\pi}{8}+n_2\pi
\end{cases}
\end{gather*}
where $\beta=135°-\alpha=\frac{3\pi}{4}-\alpha$ and $n_1,n_2\in\mathbb{Z}$, and vice versa since the problem is ‘symmetrical’.
The original exam
| Take $\tan$ of both sides of first equation, $$\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$
$$\Rightarrow \tan 135 = -1 = \dfrac{5}{1-\tan \alpha \tan \beta} $$
$$\Rightarrow \tan \alpha \tan \beta=6$$
Setting $\tan \alpha = x$, $\tan \beta = y$ we get two equations : $$x+y=5 \quad xy=6$$
I believe you can easily solve from here.
| {
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"url": "https://math.stackexchange.com/questions/3962548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number? Is there a way to show that
$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$
is a rational number?
I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.
| Note $\sqrt[3]{20-14 \sqrt{2}}=\sqrt2 \cdot \sqrt[3]{5 \sqrt{2}-7}$ and assume $\sqrt[3]{5\sqrt{2}\pm7}=p\sqrt2\pm q$. Then
$$\alpha
=(p\sqrt2+q)+\sqrt2(p\sqrt2-q)=(q+2p)+(p-q)\sqrt2
$$
For $a$ to be rational requires $p=q=\frac{\sqrt[3]{5 \sqrt{2}+7}}{\sqrt2+1}
= \sqrt[3]{\frac{ 5 \sqrt{2}+7}{(\sqrt2+1)^3 }}=1$ to be rational, which is indeed.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality for a two variables function I have tried to solve this exercise but I am having difficulties.
I have to find a function greater than $$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}}$$ as the parameter $\alpha$ changes.
That's what I tried:
Since $$sin(xy)\leq1$$
$$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{(x^2+y^2)^{3/2}}$$
Then:
$$(x^2+y^2)^{3/2} \geq y^3 $$ so that:
$$\frac{|y|^\alpha}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{y^{3}}$$
Now I have problems and I don't know what I did wrong. Following my reasoning I should have:
$$\frac{|y|^\alpha}{y^{3}}=|y|^{\alpha - 3}$$
But I Know it's wrong.
| $x^2, y^2 \geq 0$ thus $x^2 + y^2 \geq 0$ and $x^2+y^2\geq y^2$ because $x^2 \geq 0$.
Then put both sides to the power $\frac{3}{2}$
$(x^2+y^2)^\frac{3}{2}\geq (y^2)^\frac{3}{2}=|y|^3$ instead of $y^2$
and we can do this because $y^2 \geq 0$ and $x^2 + y^2 \geq 0$ and $\frac{d}{dz}[z^\frac{3}{2}]=\frac{3}{2}z^\frac{1}{2}\geq 0$ for $z\geq 0$, i.e. $z^\frac{3}{2}$ is an increasing function, so if $z_2\geq z_1$ then $z_2^\frac{3}{2} \geq z_1^\frac{3}{2}$.
Other than that, it doesn't look like the typo really affected your answer.
| {
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| (This has a fairly obvious improvement
of the inequality.)
Here is a simple algebraic proof of the stronger inequalities
$x^2+y^2=1
\implies x^3+y^3
\ge \dfrac{x+y}{2} \ge \sqrt{xy}.$
More generally,
without the restriction on
$x^2+y^2$,
this gives
$x^3+y^3
\ge \dfrac{(x+y)(x^2+y^2)}{2}
$.
Proof that this inequality is stronger:
$xy \le (x^2+y^2)/2 = 1/2$
so
$1/\sqrt{xy} \ge \sqrt{2}$.
Therefore
$\sqrt{xy} = xy/\sqrt{xy}
\ge xy \sqrt{2}$.
Proof of the inequality.
$\begin{array}\\
x^3+y^3
&=(x+y)(x^2-xy+y^2)\\
&= (x+y) \dfrac{x^2+y^2+x^2-2xy+y^2}{2}\\
&= (x+y) \dfrac{1+(x-y)^2}{2}\\
&\ge \dfrac{x+y}{2}\\
&\ge \sqrt{xy}\\
\end{array}
$
In all these inequalities, there is equality when x=y and strict inequality otherwise.
Example.
If
$x=3/5, y=4/5$
then
$x^3+y^3 = (27+64)/125 =91/125=0.728,\\
\dfrac{x+y}{2} = 7/10 =0.7\\
\sqrt{xy} = \sqrt{12/25} = 2 \sqrt{3}/5 = 0.6928...,\\
xy \sqrt{2} = 0.6788...
$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Plotting graph given parametric form satisfying $x=f(t)$ and $y=-f(-t)$ I had to plot the curve given by the parametric definition as follows. In attempting to simplify doing so, here is what I see holds between $x$ and $y$.
Plot the curve given by $$x=\frac{(t+2)^2}{t+1} ,y=\frac{(t-2)^2}{t-1}$$
$$\boxed{x=f(t)\\ y=-f(-t)}$$
Is there somehow we can use that to plot the graph easily. Any hints are appreciated. Thanks.
| write $x$ as $\dfrac{(1+t+1)^2}{t+1} = t+1+2+\dfrac{1}{t+1}$ and $y = t-1 -2 + \dfrac{1}{t-1} $
$$\begin{align} x+y &= 2t \Big(\dfrac{t^2}{t^2-1} \Big) \\
y - x &= 6 - \dfrac{2}{t^2-1}\\
x+y &= t^3 \cdot (6+x-y) \end{align}$$
and now you can just put value of $t$ in terms of $x,y$
to get $$(8+x-y)^3 = (x+y)^2 (6+x-y) $$
here is what geogebra gave
| {
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Maclaurin expansion for sech$(x)$ I am a bit unsure where I have gone wrong in working this out.
Sech$(x)=2/(e^x+e^{-x}).$
Maclaurin expansions:
$e^x = 1+ x + x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 - x^4/24;$
so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$ (I think this is where I have gone wrong.)
The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).
How would I work this out?
| The simplest way t get it is to write $\;\operatorname{sech} x=\frac1{\cosh x}$ and to use the MacLaurin expansion of $\cosh x$ at order $4$:
$$\cosh x=1+\frac{x^2}2+\frac{x^4}{24}+o(x^4)$$
You obtain the expansion of its reciprocal dividing $1$ by the MacLaurin expansion of $\cosh x$ along increasing powers, up to order 4, truncating every term at order $4$ in this process:
$$\begin{array}[t]{r|}
1\phantom{{}-\frac{x^2}2-\frac{x^4}{24}} \\
-1-\frac{x^2}2-\frac{x^4}{24}\\\hline
-\frac{x^2}2-\frac{x^4}{24} \\
\frac{x^2}2+\frac{x^4}{4} \\ \hline
\frac{5x^4}{24}\\ -\frac{5x^4}{24}\\\hline 0
\end{array}
\begin{array}[t]{l}
1+\frac{x^2}2+\frac{x^4}{24} \\\hline
\color{red}{1 -\frac{x^2}2+\frac{5x^4}{24}}
\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\lim_{(x,y)\rightarrow (0,0)}\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2}$ if it exists I'm trying to calculate the above limit. I ran a few paths and found out that the limit is $\pi$ (which I also confirmed through WolframAlpha), but to prove it I use polar coordinates to get an expression of the form $F(r) \cdot G(\theta)$ where $F(r)\rightarrow$ when $r\rightarrow 0$ and $G$ is blocked. So far I have:
$$\lim_{r\rightarrow 0} \left|f(r\cos \theta,r \sin \theta)\right| = \lim \left|\frac{1-\cos(\pi r^2 \cos\theta \sin \theta ) + \sin (\pi r^2 (\sin^2\theta + \cos ^2 \theta ))}{r^2\cos ^2 \theta + r^2 \sin^2 \theta}\right|
\\ \underset{\sin ^2 \theta + \cos ^2 \theta = 1}{=} \lim \left|\frac{1 - \cos(\pi r^2 \cos\theta \sin\theta ) + \sin (\pi r^2 )}{r^2}\right|$$
However, I don't know how to separate $r$ from the rest of the expression at this stage. But it occurred to me that this only works if the limit of the function is zero. So, how do I confirm this is the correct limit in a case such as this?
| $$\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2} = \frac{1 - \cos(\pi x y ) }{x^2 + y^2}+\pi\frac{ \sin (\pi(x^2 + y^2))}{\pi(x^2 + y^2)}$$
For first
$$\left|\frac{1 - \cos(\pi x y ) }{x^2 + y^2}\right|\leqslant\left|\frac{2\sin^2(\pi \frac{x^2 + y^2}{2} ) }{x^2 + y^2}\right| \to 0$$
While second is $\pi$
| {
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A clarification in a limit While reading this answer I do not understand that how can we claim that there exist a polynomial with the properties, even if it exist then how is it same for the function and the inverse.
| Start with $ f ( x ) = x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 \cdots $ and $ f ^ { − 1 } ( x ) = x + A _ 2 x ^ 2 + A _ 3 x ^ 3 + A _ 4 x ^ 4 \cdots $ and substitute into $ f ^ { - 1 } ( f ( x ) ) = x $ (or into $ f ( f ^ { - 1 } ( x ) ) = x $ for the same final result):
$$ \eqalign { x & = ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) + A _ 2 ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) ^ 2 \\ & \quad { } + A _ 3 ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) ^ 3 + A _ 4 ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) ^ 4 + \cdots \\
& = x + ( a _ 2 + A _ 2 ) x ^ 2 + ( a _ 3 + 2 a _ 2 A _ 2 + A _ 3 ) x ^ 3 + ( a _ 4 + 2 a _ 3 A _ 2 + a _ 2 ^ 2 A _ 2 + 3 a _ 2 A _ 3 + A _ 4 ) x ^ 4 + \cdots \text . } $$
This holds for all $ x $ (at least sufficiently close to $ 0 $), so
$$ \displaylines { a _ 2 + A _ 2 = 0 \text , \\ a _ 3 + 2 a _ 2 A _ 2 + A _ 3 = 0 \text , \\ a _ 4 + 2 a _ 3 A _ 2 + a _ 2 ^ 2 A _ 2 + 3 a _ 2 A _ 3 + A _ 4 = 0 \text , \\ \cdots \text . } $$
We can now start solving these:
$$ \displaylines { A _ 2 = - a _ 2 \text , \\ A _ 3 = - a _ 3 - 2 a _ 2 A _ 2 = - a _ 3 + 2 a _ 2 ^ 2 \text , \\ A _ 4 = - a _ 4 - 2 a _ 3 A _ 2 - a _ 2 ^ 2 A _ 2 - 3 a _ 2 A _ 3 = - a _ 4 + 5 a _ 2 a _ 3 - 5 a _ 2 ^ 3 \text , \\ \cdots \text . } $$
So we have $ A _ n = - a _ n + P _ n ( a _ 2 , \ldots , a _ { n - 1 } ) $, where $ P _ 2 = 0 $, $ P _ 3 ( a _ 2 ) = 2 a _ 2 ^ 2 $, $ P _ 4 ( a _ 2 , a _ 3 ) = 5 a _ 2 a _ 3 - 5 a _ 2 ^ 3 $, etc.
So these polynomials exist; and it doesn't matter whether you call the coefficient sequences $ a $ and $ A $, or $ b $ and $ B $; it's the same polynomial sequence $ P $ either way. As long as a function $ f $ is analytic at $ 0 $ with $ f ( 0 ) = 0 $ and $ f ' ( 0 ) = 1 $, then the coefficients of the function and its inverse will be related by these same polynomials.
| {
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Find the integer solutions to $4^x - 9^y = 55$ I want to find the integer solutions of:
$$ 4^x - 9^y = 55$$
For now, I see that $x = 3, y = 1$ is an integer solution to the equation. How can I rigorously prove there are no other solutions for $x, y$ integers?
I tried to solve for $y$, but to no avail. WolframAlpha tells me that it is the only solution, but it doesn't provide an explanation.
| This factors as $(2^x + 3^y)(2^x-3^y) = 55$, so $2^x + 3^y \in \{1, 5, 11, 55\}$. This leaves us only finitely many cases of $x$ and $y$ to check: $x \le 5$ (because $2^6$ is bigger than $55$) and $y \le 3$ (because $3^4$ is bigger than $55$).
A check of all $(x,y) \in \{0,1,2,3,4,5\} \times \{0,1,2,3\}$ gives only one solution: $(x,y) = (3,1)$.
We can be more clever and eliminate some cases, reducing the amount of casework we have to do, but that's already enough.
To minimize casework: first, because $2^x + 3^y > 2^x - 3^y$, we know that $2^x + 3^y$ is either $11$ or $55$, with $2^x - 3^y$ being either $5$ or $1$ respectively. But adding them together gives a value of either $11+5=16$ or $55+1 = 56$ for $2 \cdot 2^x$. Only the first of these is a power of $2$, so we conclude that $2^x + 3^y = 11$.
To get this, $3^y$ can be either $1$, $3$, or $9$. This leaves $10$, $8$, or $2$ for $2^x$. We reject $2^x = 10$ because $x$ must be an integer. We reject $2^x = 2$ with $3^y=9$ because the original equation $4^x-9^y=55$ is not satisfied when $x=1$ and $y=2$. Finally, the case $2^x=8$ with $3^y=3$ gives us the solution $(x,y) = (3,1)$.
It's harder, but still possible, to solve the more general equation $2^x - 3^y = 55$, to which we now know that $(x,y) = (6,2)$ is a solution.
With the aim of rejecting solutions with $y>2$, consider the equation modulo $27$: if $y$ is $3$ more, then $3^y \equiv 0 \pmod{27}$, so $2^x \equiv 55 \equiv 1 \pmod{27}$. The order of $2$ modulo $27$ is $18$, so $2^x \equiv 1 \pmod{27}$ exactly when $x$ is a multiple of $18$.
In that case, take the equation modulo $73$, which is one of the prime factors of $2^{18}-1$, so we know that $2^x \equiv 1 \pmod{73}$. This tells us that $3^y \equiv 1 - 55 \equiv 19 \pmod{73}$. This has no solutions: the powers of $3$ can take on the values $3, 9, 27, 8, 24, 72, 70, 64, 46, 65, 49, 1$, after which they repeat.
So it's impossible to have $y \ge 3$. We can try $y=0$ (in which case $2^x=56$ has no solution), $y=1$ (in which case $2^x = 58$ has no solution), or $y=2$ (in which case $2^x = 64$ gives us $(x,y) = (6,2)$).
| {
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Manipulating an ODE into a full differential Find the general solution of:
$$(x-2)\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + \frac{4y}{x^2} = 0$$
If I first multiply through by $x^2$ then I can write the above as
$$x^2(x-2)\frac{d^2y}{dx^2} + 3x^2\frac{dy}{dx} + 4y $$
$$ =\left[ x^3\frac{d^2y}{dx^2} + 3x^2\frac{dy}{dx}\right]-2x^2 \frac{d^2y}{dx^2} +4y $$
$$ =\frac{d}{dx}\left( x^3\frac{dy}{dx}\right) -2x^2 \frac{d^2y}{dx^2} +4y = 0$$
and I'm not sure how to express the rest of the equation as the derivative of some function (I tried $\frac{d}{dx} 2x^2y$ to no avail) . Any help would be appreciated.
| Hint:
$$\frac{d}{dx}\left( x^3\frac{dy}{dx}\right) -2x^2 \frac{d^2y}{dx^2} +4y = 0$$
$$\frac{d}{dx}\left( x^3\frac{dy}{dx}\right) -2x^2 \frac{d^2y}{dx^2} \color {red}{-4xy'+4xy'} +4y = 0$$
$$\left( x^3y'\right)' -(2x^2 y')' +4(xy)' = 0$$
| {
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Find the missing angle in triangle In the below triangle, we are looking for the value of angle $φ$.
We are given $α=30, β=18, γ=24$ and also that $CD=BD$.
I have solved it with trigonometry (sine law) and found the required angle to be 78 but I need to solve it with Geometry only.
What I have tried so far:
First of all, the angle is constructible, which means to me that there must be a geometrical solution. I first drew triangle ABC; easy, since we know 2 of its angles. We are not interested in the lengths of the sides. Then, with side AC as a base, and angle of 24 degrees, we can draw a ray from the point A.
Then, since $CD=BD$, triangle DCB is isosceles, therefore D must lie on the perpendicular bisector of CB, which we can draw. The point of intersection of the ray from A and the perpendicular bisector, is point D.
From triangle FEB we have that
angle AFD = 108.
From triangle AFD,
$ADC+CDE+54+108=180$ so $ADC+CDE=18$
We also have $24+ACD+ADC=180$
$ACB=132$
$132+φ+ACD=180$
$18+φ+54+ADC+2CDE=180$
I am always one equation short.
Any ideas?
Many thanks in anticipation!
EDIT:
Sine law in triangle ABD:
$\frac {sin (φ+18)}{AD} = \frac {sin (54)}{BD}$
Sine law in triangle ACD:
$\frac {sin (360-132-φ)}{AD} = \frac {sin (24)}{CD} = \frac {sin (24)}{BD}$
so
$\frac {sin (φ+18)}{sin (228-φ)} = \frac {sin (54)}{sin (24)}$
hence $φ=78$.
| Consider a regular $30$-gon $X_1X_2X_3X_4X_5X_6X_7X_8X_9X_{10}X_{11}X_{12}X_{13}X_{14}X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}X_{22}X_{23}X_{24}X_{25}X_{26}X_{27}X_{28}X_{29}X_{30}$ and place it on the plane so that $X_1 \equiv A$, $X_6\equiv B$, and that $X_2$ and $C$ lie on different halfplanes determined by the line $AB$. Denote $K=X_2$, $L=X_3$, $M=X_4$, $N=X_5$, and $X_{15}=R$.
Build regular pentagon $KLOPQ$ as in the picture. We shall prove that $P\equiv C$.
Note that $\angle QKA = \angle LKA - \angle LKQ = 168^\circ - 108^\circ = 60^\circ$. Since $QK=KL=AK$, it follows that the triangle $AKQ$ is equilateral. In particular, $AQ=KQ=QP$, so $Q$ is the circumcenter of $AKP$. Angle chasing yields $\angle AQP = 360^\circ - 2\angle PKA = 360^\circ - 2(60^\circ + 36^\circ) = 168^\circ$, so by SAS triangle $AQP$ is congruent to $KLM$, $MNB$, and by symmetry it is congruent to $MOP$. Continuing angle chasing, $\angle PAQ = 6^\circ$, and finally $\angle BAP = \angle KAQ - \angle PAQ - \angle KAB = 60^\circ - 6^\circ - 24^\circ = 30^\circ$.
On the other hand, by congruency of $KLM$, $MNB$ and $MOP$, we have $MK=MP=MB$, so $M$ is the circumcenter of $KPB$ and therefore $\angle BMP = 2\angle BKP = 2(\angle LKP - \angle LKB) = 2(72^\circ - 18^\circ) = 108^\circ$, hence $\angle PBM = 36^\circ$ and $\angle PBA = \angle PBM - \angle ABM = 36^\circ - 18^\circ = 18^\circ$.
Since $\angle BAP = 30^\circ$ and $\angle PBA = 18^\circ$, we have that $P\equiv C$.
We shall prove now that $R\equiv D$. First of all, we have $\angle CAR = \angle BAR - \angle BAC = 54^\circ - 30^\circ = 24^\circ$. Secondly, since $\angle LKC = 72^\circ = \angle LKR$, we have that $K$, $C$, $R$ are collinear. Since $M$ is the circumcenter of $CKB$, we have $\angle BCR = \frac 12 \angle BMK = \frac 12 \cdot 156^\circ = 78^\circ$. We also have $\angle RBC = \angle RBA - \angle CBA = 96^\circ - 18^\circ = 78^\circ$. Since $\angle BCR = \angle RBC$, it follows that $R$ lies on the perpendicular bisector of $CB$, which along with $\angle CAR = 24^\circ$ means that $R\equiv D$. The answer follows:
$$\varphi = \angle BCD = \angle BCR = 78^\circ.$$
| {
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"url": "https://math.stackexchange.com/questions/3976315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Jacobian Transformations of Two Random Variables ( probability density function) Let ${Y}_{1}$ and ${Y}_{2}$ be independent,each with density $\frac{1}{{y}^{2}}$ , $y>1$. Consider the transformation ${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$. Find probability density function of random variables $\left({U}_{1},{U}_{2}\right)$.
Here is my solution:
Since ${Y}_{1}$ and ${Y}_{2}$ are independent then $f({y}_{1},{y}_{2})=\frac{1}{{\left({y}_{1}{y}_{2}\right)}^{2}}$ , ${y}_{1}>1$ , ${y}_{2}>1$.
${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$ then ${Y}_{2}={U}_{1}{U}_{2}$ , ${Y}_{1}={U}_{2}-{U}_{1}{U}_{2}$
From here Jacobian is $J=\left|\begin{array}{cc}{u}_{2}& {u}_{1}\\ -{u}_{2}& 1-{u}_{1}\end{array}\right|={u}_{2}\Rightarrow \left|J\right|={u}_{2}$
Then I stuck here because I found the density function $f({u}_{1},{u}_{2})=\frac{{u}_{2}}{{\left({u}_{2}-{u}_{1}{u}_{2}\right)}^{2}{\left({u}_{1}{u}_{2}\right)}^{2}}$ but when I want to check it this function with integral definition I can not define the interval of integral. Any help will be appreciated.
| When defining joint support like this, you need to consider to specify the marginal support of which variable first. If you specify $u_1$ first, since
$$ u_1 = \frac {y_2} {y_1 + y_2} $$
then we see that
when $y_1$ fixed, $y_2 \to +\infty$, $u_1 \to 1$;
when $y_1 \to +\infty$, $y_2$ fixed, $u_1 \to 0$.
And obviously $0 < u_1 < 1$ so the support of $U_1$ is indeed $(0, 1)$
Let say we specify $u_1$ first. Then the "conditional" support of $u_2$ will be in terms of $u_1$. Since
$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_2 &> \displaystyle \frac {1} {1 - u_1} \\ u_2 &> \displaystyle \frac {1} {u_1} \end{cases} $$
So the joint support can be specified as
$$ 0 < u_1 < 1, u_2 > \max\left\{\frac {1} {1-u_1}, \frac {1} {u_1} \right\}$$
Similarly if we want to specify $u_2$ first, obviously
$$u_2 = y_1 + y_2 > 1 + 1 = 2$$
so the marginal support of $U_2$ is $(2, +\infty)$. And
$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_1 &< \displaystyle 1 - \frac {1} {u_2} \\ u_1 &> \displaystyle \frac {1} {u_2} \end{cases} $$
So the joint support can be equivalently specified as
$$ u_2 > 2, \frac {1} {u_2} < u_1 < 1 - \frac {1} {u_2} $$
Verification:
Note
$$ \int \frac {1} {u^2(1-u)^2} du = \frac {1} {1-u} - \frac {1} {u}
- 2\ln (1-u) + 2\ln u + C$$
Therefore
$$ \begin{align}
&~ \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1 \\
=&~ \frac {1} {1-1+1/u_2} - \frac {1} {1-1/u_2}
- 2\ln \left(1-1+\frac {1} {u_2}\right)
+ 2\ln\left(1- \frac {1} {u_2} \right) \\
&~ - \frac {1} {1-1/u_2} + \frac {1} {1/u_2}
+ 2\ln \left(1-\frac {1} {u_2}\right)
- 2\ln\left(\frac {1} {u_2} \right) \\
=&~ 2u_2 - \frac {2u_2} {u_2-1} + 4\ln(u_2-1)\end{align} $$
As a result,
$$ \begin{align}
&~ \int_2^{+\infty} \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1du_2 \\
=&~ \int_2^{+\infty} \left[\frac {2} {u_2^2} - \frac {2} {u_2^2(u_2-1)} + \frac {4\ln(u_2-1)} {u_2^3} \right] du_2 \\
=&~ \left. - \frac {2\ln(u_2-1)} {u_2^2} - \frac {2} {u_2} \right|_2^{+\infty} \\
=&~ 1 \end{align} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? The question:
How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? At which points do these tangent lines touch the curve?
My attempt:
To find the slope of the tangent line to the curve, we need to get the first derivative of $y = \frac{x}{x+1}$, which is $y' = \frac{1}{(x+1)^2}$.
So, the line equation of the tangent line is $y = m(x-1) + 2$, where $m = \frac{1}{(x+1)^2}$ for the given $x$.
We also know that the point where the tangent line intersects the curve must be both on the tangent line and the curve, so using that,
$$\frac{x}{x+1} = \frac{x-1}{(x+1)^2} + 2$$
$$\frac{x-1}{(x+1)^2} + 2 - \frac{x}{x+1} = 0$$
$$\frac{x - 1 + 2x^2 + 4x + 2 - x^2}{(x+1)^2} = 0$$
$$\frac{x^2 + 5x + 1}{(x+1)^2} = 0$$
And I'm stuck. I'd appreciate if you can point out where I'm wrong and how to better approach this problem.
| hint
The equation of the tangent line to the curve at the point $ (a,\frac{a}{a+1}) $ is
$$y=\frac{1}{(a+1)^2}(x-a)+\frac{a}{a+1}$$
it passes through the point $ (1,2) $ if
$$2=\frac{1}{(a+1)^2}(1-a)+\frac{a}{a+1}$$
So,
$$2(a+1)^2=(1-a)+a(a+1)$$
or
$$a^2+4a+1=0$$
thus
$$a=-2\pm \sqrt{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$ I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:
$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{2}{(x^2-1)^2}dx+\int\frac{1}{(x^2-1)^3}dx$$
I don't know how to evaluate remaining integrals.
| If the numerator is $x^{2r},$
$$x^{2r}=(1+x^2-1)^r=?$$
Had the numerator been $x^{2r+1},$
$x^{2r+1}=x(1+x^2-1)^r$ first substitute $x^2-1=y$
Now for $S=\dfrac{d(x/(x^2-1)^n)}{dx},$
$S=1/(x^2-1)^n-2(n+1)x^2(x^2-1)^{n+1}$
Write $x^2=x^2-1+1$ and integrate both sides to find
$$\dfrac x{(x^2-1)^n}=I(n)-2(n+1)(I (n) - I(n+1))$$ where $I(m)=\displaystyle\int\dfrac{dx}{(x^2-1)^m}$
Can you identify the values of $n$ here?
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the period of this sinusoid? I'm stuck trying to find the period of this sinusoid and would really like some pointers to different ways to approach this problem.
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3})$$
I would like to turn the product into a sum and do something about that $2$ in the exponent of $\sin$
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3}) = \cos(\frac{4\pi t}{5})\left[ 1 - \cos(\frac{16 \pi t}{3}) \right]\frac{1}{2} $$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \cos(\frac{4\pi t}{5}) \cos(\frac{16 \pi t}{3})\right]$$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \frac{1}{2}\cos(\frac{4\pi t}{5} + \frac{16 \pi t}{3}) \cos(\frac{16 \pi t}{3} - \frac{4\pi t}{5})\right]$$
So the sum of cosines will have a period that is the LCM of of the periods of the individuals terms we added. The angular frequencies are $\frac{4\pi}{5}, \frac{92\pi}{15}, \frac{68\pi}{15}$.
Converting this to periods -> we get $\frac{5}{2}, \frac{15}{46}, \frac{15}{34}$, but the answer is $\frac{30}{4}$
| You get 7.5 if you simply apply the LCM to $\frac{4\pi t}{5} = 2\pi$ (period = 2.5) and $\frac{8\pi t}{3} = 2\pi$ (period = 0.75).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solution verification of $\frac{1}{(4n^2)!} >\frac{1}{(4(n+1)^2)!}$ inequality \begin{align}
\frac{1}{(4n^2)!} &>\frac{1}{(4(n+1)^2)!}\\
\frac{(4(n+1)^2)!}{(4n^2)!} &>\frac{1}{1}\\
\frac{(4(n^2+2n+1))!}{(4n^2)!} &>1\\
\frac{(4n^2+8n+4)!}{(4n^2)!} &>1\\
\iff \frac{(4n^2+8n+4)}{(4n^2)}&>1\\
4n^2+8n+4&>4n^2\\
8n &> -4\\
n&>-\frac{1}{2}
\end{align}
This implies that the inequality holds for all natural numbers, thus also the primary inequality $$\frac{1}{(4n^2)!} >\frac{1}{(4(n+1)^2)!}$$ holds.
| That looks correct, but it is much easier to see that\begin{align}4(n+1)^2>4n^2&\implies(4(n+1)^2)!>(4n^2)!\\&\implies\frac1{(4(n+1)^2)!}<\frac1{(4n^2)!}.\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.
As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.
| Here is a simple and direct way using just some rough estimates:
For $x>2$ you have surely
$$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} \stackrel{x>2}{<} \frac{5x^2 + 3x^2 + x^2}{x^3 - (x^2 +1)} = \frac{9x^2}{x^3 - (x^2 +1)}$$
Since you want to have a $10$ in the numerator, note that
$$\frac 1{10}x^3 > x^2+1 \Leftrightarrow x^2(\frac x{10}-1)>1$$
which is surely satisfied for $x>20$. Hence,
$$\frac{9x^2}{x^3 - (x^2 +1)} \stackrel{x>20}{<} \frac{9x^2}{x^3 - \frac 1{10}x^3} = \frac{10}x$$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve Surface Area Formula For Radius Ok. I have a Surface Area Formula for a cylinder, but I am wanting to solve for the radius. The only information I have is the height of the cylinder, which is 8 inches. I know I can use the Quadratic Formula to convert the formula to solve for the radius, but I get stuck when doing the math.
$$ SA=2\pi r^{2}+2\pi rh$$
Any help?
| I'd just complete the square:
$$ A = 2\pi r^2 + 2\pi rh $$
$$ \frac{A}{2\pi} = r^2 + rh $$
To complete the square on the right-hand side, we need to add $\frac{h^2}{4} $ to both sides:
$$ \frac{A}{2\pi} + \frac{h^2}{4} = r^2 + rh + \frac{h^2}{4} $$
$$ \frac{A}{2\pi} + \frac{h^2}{4} = \left(r + \frac{h}{2} \right)^2 $$
Since $r > 0$ and $h > 0$, we take the positive root:
$$ r + \frac{h}{2} = \sqrt{\frac{A}{2\pi} + \frac{h^2}{4}} $$
$$ r = -\frac{h}{2} +\sqrt{\frac{A}{2\pi} + \frac{h^2}{4}} $$
| {
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How to solve this? I am having difficulty in the very last step of the problem. The general solution of $\vert\sin x\vert = \cos x$ is -
(A) $2n\pi+{\pi\over4}$, $n\in I$
(B) $2n\pi±{\pi\over4}$, $n\in I$
(C) $n\pi+{\pi\over4}$, $n\in I$
(D) None of these
So what I did was - I made a case for when sinx is greater than 0 and equated it to $\cos x$ to get $\tan x = 1$ which implies x = $\pi\over 4$. The other case was when $\cos x = -\sin x$. Here, x = $3\pi\over 4$. I don't understand how to proceed from here.
| We will solve $\sin^{2}(x) = \cos^{2}(x)$, which is obtained upon squaring the equation (so we do not have to deal with sign issues).
$$\sin^{2}(x) = \cos^{2}(x)$$
$$\cos^{2}(x) -\sin^{2}(x) = 0$$
$$\cos(2x) = 0$$
$$x = \frac{\pi}{4} + \pi n, \frac{3\pi}{4} + \pi n$$
$$x = \frac{\pi}{4} + 2\pi n, \frac{5\pi}{4} + 2\pi n, \frac{3\pi}{4} + 2\pi n, \frac{7\pi}{4} + 2\pi n$$
Here, we split into cases of $2\pi n$ periodicity to take advantage of the properties of $\sin$ and $\cos$.
Checking these solutions in the original equation $\vert \sin(x)\vert = \cos (x)$, we find that only $\frac{\pi}{4} + 2\pi n$ and $\frac{7\pi}{4} + 2\pi n$ work. $\frac{7\pi}{4} + 2\pi n$ is equivalent to $-\frac{\pi}{4} + 2\pi n$, so the answer is $\boxed{(\text{B})\ \pm\frac{\pi}{4} + 2\pi n.}$
You can do a sanity check by drawing a quick graph and seeing if the functions intersect in the general region of the conjectured solution.
| {
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how to derive the steps of this intergral for the product of two exponential terms? For homework, I was looking for closed form solutions of exponential functions from this wiki site.
https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
But I was not sure how to derive the steps for this particular integral,
$\int_{-\infty}^{\infty}e^{-ax^{2}}e^{-\frac{b}{x^{2}}}dx={\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}\quad (a,b>0)$
I am fairly new to normal distribution but I assume this has to do with Taylor series..
Any pointers would be greatly appreciated!
| Start completing the square
$$ax^2+\frac b {x^2}=a\left(x+\sqrt{\frac{b}{a}}\frac{1}{x}\right)^2-2\sqrt{ab}$$
$$x+\sqrt{\frac{b}{a}}\frac{1}{x}=t \implies x=\frac{1}{2} \left(t\pm\sqrt{t^2+4 \sqrt{\frac{b}{a}}}\right)$$
Making the calculation for the $\color{red}{-}$ branch.
This makes now the integrand to be
$$\frac{1}{2} e^{2 a \sqrt{\frac{b}{a}}} \left(e^{-a t^2}-\frac{t e^{-a t^2}}{\sqrt{t^2-4 \sqrt{\frac{b}{a}}}}\right)$$ Now, we face two "simple" integrals
$$\int e^{-a t^2}\,dt=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{a} t\right)}{2 \sqrt{a}}$$
$$\int \frac{t e^{-a t^2}}{\sqrt{t^2-4 \sqrt{\frac{b}{a}}}}\,dt=\frac{\sqrt{\pi } e^{-4 a \sqrt{\frac{b}{a}}} \text{erf}\left(\sqrt{a} \sqrt{t^2-4
\sqrt{\frac{b}{a}}}\right)}{2 \sqrt{a}}$$
Back to $x$
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a} \sqrt{b}}
\left(\text{erfc}\left(\frac{\sqrt{b}}{x}-\sqrt{a} x\right)-e^{4 \sqrt{a}
\sqrt{b}} \text{erfc}\left(\sqrt{a} x+\frac{\sqrt{b}}{x}\right)\right)}{4
\sqrt{a}}$$ Integrateing between $-p$ and $+p$
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a} \sqrt{b}} \left(e^{4 \sqrt{a} \sqrt{b}}
\text{erf}\left(\sqrt{a}
p+\frac{\sqrt{b}}{p}\right)-\text{erf}\left(\frac{\sqrt{b}}{p}-\sqrt{a}
p\right)\right)}{2 \sqrt{a}}$$ and the limit is
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a b}}}{\sqrt{a}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longrightarrow f(x) &= x^x \\
f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\
&= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\
&= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\
&= x^x\left(\ln \left(x\right)+1\right)
\end{align}
\begin{align}
n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\
f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\
&= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\
&= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\
&= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)
\end{align}
\begin{align}
n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\
f'(x) &= ... \\
&= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right)
\end{align}
However, I am not sure if I see a pattern here that can help solve the question.
| $$x^{x^{x^\cdots}}= y \implies x^y = y \implies1 = yx^{-y} = ye^{-y\log (x)}$$
which means
$$y = \frac{-W(-\log (x))}{\log (x)}$$
where $W$ is the principal branch of the Lambert-W function. Now, the problem of the successive derivatives is "much" simpler (have a look here for the derivatives of $W(t)$).
| {
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Substituting $x = \sec(u)$ to evaluate $\int\sqrt{x^2-1}\ dx$: but why is $\tan(u) = \sqrt{x^2 - 1}$ rather than $\tan(u) = -\sqrt{x^2 - 1}\ $? I tried evaluating $\int\sqrt{x^2-1}\ dx$ using the substitution $x = \sec(u).$
I know my method gets to the same answer as wolframalpha, namely:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C,\quad (*)$
but there is one step I can't justify.
When I got to $\int \tan^2(u)\sec(u)\ du = \frac{1}{2}\left(\ \tan(u)\sec(u) - \ln\left|\sec(u)+\tan(u)\right|\ \right) + C,$
I then have to substitute stuff back in in terms of $x$.
Now $x=\sec(u) \implies \tan^2(u) = x^2 - 1$. But I don't see how this implies $\tan(u) = \sqrt{x^2 - 1}$.
Comparing the graphs of $\sec(u)$ and $\tan(u)$, I don't see why not: $\ \tan(u) = -\sqrt{x^2 - 1}$, which would give:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ -x \sqrt{x^2-1} - \ln\left|x - \sqrt{x^2-1}\right|\ \right) + C,\quad (**)$
which is a different answer than $(*)$ ?
Now I noticed that $(*) = -(**)\ $ (ignoring the $C \to -C)$.
I can see from the graph of $\sqrt{x^2-1}$ that for $x>1$, the definite integral $\int^x_1\sqrt{t^2-1}\ dt = (*),$ and for $x<-1,\ \int^{-1}_x\sqrt{t^2-1}\ dt = (**)$
So is the indefinite integral sort of poorly defined, or would you say it is:
$\int\sqrt{x^2-1}\ dx = \pm \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C\ $ ?
| Let's differentiate and see by using which sign we get the integrand.
$\begin{align}\frac{d}{dx}\left[\frac{x}{2}\sqrt{x^2-1}-\frac12\ln|x+\sqrt{x^2-1}| + c\right]& = \frac{1}{2}\sqrt{x^2-1}+\frac x2\frac{x}{\sqrt{x^2-1}}-\frac{1+\frac{x}{\sqrt{x^2-1}}}{2(x+\sqrt{x^2-1})}\\
\\& = \frac{2x^2-1}{2\sqrt{x^2-1}}-\frac{1}{2\sqrt{x^2-1}}\\
\\& = \sqrt{x^2-1}\end{align}$
But using the negative sign yields $-\sqrt{x^2-1}$. So the one we've been using is correct.
| {
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Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
If I would start from the right expression I would be able to reduce it. But I'm not used to do the reverse.
I can see that Wolframalpha expands the left expression in many ways, one of which is the one I want, but are there any trick to do this expansion by hand?
This question is linked to this one.
| Apply the identities
$$\cos4\theta = 2\cos^22\theta -1,\>\>\>\>\>\cos2\theta = 1-2\sin^2\theta $$
$$\sin3\theta =\sin\theta (2\cos2\theta+1)$$
in simplifying the RHS
\begin{align}
RHS=&1+\sin\theta -\sin\theta (2\cos2\theta+1)
-\frac12(1-2\sin^2\theta)+\frac12(2\cos^22\theta-1)\\
=&\sin^2\theta -2\sin\theta\cos2\theta+\cos^22\theta\\
=&(\sin\theta -\cos2\theta)^2=LHS
\end{align}
| {
"language": "en",
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2D system of hyperbolic equation (LeVeque) Again I need help. This time it's exercise 18.1 from LeVeque Finite Volume Methods.
Consider the system $q_t+Aq_x+Bq_y=0$ with $A=\begin{pmatrix}3&1\\1&3\end{pmatrix}$ and $B=\begin{pmatrix}0&2\\2&0\end{pmatrix}$.
Show that these matrices are simultaneously diagonalizible and determine the general solution to this system with arbitrary initial data. In particular sketch how the solution evolves in the $x-y$ plane with data
$$\begin{align*}
&q^1(x,y,0)=\left\{\begin{array}{ll}1&\text{ if } x^2+y^2\leq 1\\0&\text{ otherwise }\end{array}\right.\\
&q^2(x,y,0)\equiv 0
\end{align*}$$
So far I've got $A=R\Lambda^xR^{-1}$ and $B=R\Lambda^yR^{-1}$ with
$$\begin{align*}
&R&&=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\\
&R^{-1}&&=\frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\\
&\Lambda^x&&=\begin{pmatrix}4&0\\0&2\end{pmatrix}\\
&\Lambda^y&&=\begin{pmatrix}2&0\\0&-2\end{pmatrix}
\end{align*}$$
but now I don't know how to move on. Any tips are appreciated :)
| Setting $p = R^{-1}q$, we have
$$
p_t + \Lambda_x p_x + \Lambda_y p_y = 0 \, .
$$
Since the $\Lambda^\alpha$ matrices are diagonal, this system is decoupled.
Applying the method of characteristics componentwise yields
$$
p(x,y,t) = R^{-1} q(x,y,t) = \begin{bmatrix}
F_1(\lambda^x_1 x -t, \lambda^y_1 x - \lambda^x_1 y) \\
F_2(\lambda^x_2 x -t, \lambda^y_2 x - \lambda^x_2 y)
\end{bmatrix}
$$
where $F_1$, $F_2$ are two arbitrary functions, and $\lambda_i^\alpha$ denotes the diagonal entries of $\Lambda_i^\alpha$ (see this post, where the notations correspond to $\Phi = F_i$, $a=1$, $b=\lambda_i^x$, and $c=\lambda_i^y$). Now, it remains to apply the intial condition to determine the arbitrary functions:
$$
q(x,y,0) = R\, p(x,y,0) = \begin{bmatrix}
F_1(4 x, 2 x - 4 y) + F_2(2 x, -2 (x + y)) \\
F_1(4 x, 2 x - 4 y) - F_2(2 x, -2 (x + y))
\end{bmatrix} .
$$
From the initial condition $q_1(x,y,0) = \Bbb I_{x^2 + y^2 < 1}$ and $q_2(x,y,0) = 0$ in OP, we deduce that
\begin{aligned}
\tfrac12\Bbb I_{x^2 + y^2 < 1} &= F_2(2x,-2(x+y)) \\
& = F_2(X,Y) = \tfrac12\Bbb I_{X^2 + (Y+X)^2 < 4} \\
\tfrac12\Bbb I_{x^2 + y^2 < 1} &= F_1(4x,2 x - 4 y) \\
& = F_1(\xi,\eta) = \tfrac12\Bbb I_{\xi^2 + (\eta-\xi/2)^2 < 16}
\end{aligned}
where $\Bbb I$ is the indicator function. Finally,
$$
q(x,y,t) = \frac12 \begin{bmatrix}
\Bbb I_{(x-t/4)^2 + (y-t/8)^2 < 1} + \Bbb I_{(x-t/2)^2 + (y+t/2)^2 < 1} \\
\Bbb I_{(x-t/4)^2 + (y-t/8)^2 < 1} - \Bbb I_{(x-t/2)^2 + (y+t/2)^2 < 1}
\end{bmatrix}
$$
with the above expressions.
| {
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A weird contour integral calculation I have function $\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx$.
Here to use residue thm, I rewrite the integral as $\int_{-1}^{1}\frac{\sqrt{1-z^2}}{1+z^2}dz$ with the poles $z=i$ and $-i$. However there is a given condition $-1<x<1$, so it means $z^2<1$. This is the problem because when the poles are at $|z|=1$. It's outside the boundary. So I have no idea about how I can calculate this integral
Here's my Residues
$\operatorname{Res}_{z=i}[f(z)]=\lim_{z\to i}\frac{\sqrt{1-z^2}}{i(1-zi)}dz=\frac{\sqrt{2}}{2i}$
$\operatorname{Res}_{z=-i}[f(z)]=\lim_{z\to -i}\frac{\sqrt{1-z^2}}{i(1+zi)}dz=\frac{\sqrt{2}}{2i}$
But I guess these are wrong
Edit: since I need a closed contour i replaced $x=cos\theta$ and $dx=-sin\theta d\theta$
And my integral is now $-\frac{1}{2}\oint \frac{\sqrt{1-cos^2\theta}}{cos^20+cos^2\theta}(-sin\theta) d\theta$=
$-\frac{1}{2}\oint \frac{-sin^2\theta}{cos^20+cos^2\theta}d\theta$
| By your comments in my previous question, we can also deal in another way (but I am about to tell you that is more tedious, even if apparently at the beginning it looks easier).
With $x = \cos\theta$ we get
$$-\int_{\pi}^0 \frac{\sin^2(\theta)}{1 + \cos^2(\theta)}\ \text{d}\theta$$
We can close the contour by doubling the integration range:
$$\frac{1}{2}\int_0^{2\pi} \frac{\sin^2(\theta)}{1 + \cos^2(\theta)}\ \text{d}\theta$$
From here we are ready to Complex again:
$$\sin(\theta) = \frac{1}{2i}\left(z - \frac{1}{z}\right) ~~~~~~~ \cos(\theta) = \frac{1}{2}\left(z + \frac{1}{z}\right) ~~~~~~~ \text{d}\theta = \frac{\text{d}z}{iz}$$
Whence
$$\frac{1}{2}\oint_{|z| = 1} \frac{\left(\frac{1}{2i}\left(z - \frac{1}{z}\right)\right)^2}{\left(1 + \frac{1}{4}\left(z + \frac{1}{z}\right)^2\right)}\frac{\text{d}z}{iz} \longrightarrow \frac{1}{2}\oint_{|z| = 1} \frac{i \left(z^2-1\right)^2}{z^5+6 z^3+z}$$
Here there are five poles:
$$\left\{\{z\to 0\},\left\{z\to -i \sqrt{3-2 \sqrt{2}}\right\},\left\{z\to i \sqrt{3-2 \sqrt{2}}\right\},\left\{z\to -i \sqrt{2 \sqrt{2}+3}\right\},\left\{z\to i \sqrt{2 \sqrt{2}+3}\right\}\right\}$$
But only three of them lie inside the unit circle:
$$\{z\to 0\},\left\{z\to -i \sqrt{3-2 \sqrt{2}}\right\},\left\{z\to i \sqrt{3-2 \sqrt{2}}\right\}$$
Now it's about Residues time again.
Whereas the residue at $z = 0$ is rather easy, not easy (just tedious) are the calculations of the two others. But eventually, by doing that calculation, you will get the identical final result!
| {
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"url": "https://math.stackexchange.com/questions/4003763",
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"source": "stackexchange",
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Calculating divergence for parabolic coordinates. I'm trying to calculate divergence for parabolic coordinates with a goal to check if i understand all notions of vector analysis correctly.
Definition of parabolic coordinates:
$$x = uv \\ y = \frac{1}{2}(u^2 - v^2)$$
Firstly, I calculate $du$ and $dv$:
$$
dx = v du + udv \\
dy = udu-vdv
$$
$$
dv = \frac{u}{u^2+v^2}dx - \frac{v}{u^2+v^2}dy\\
dv = \frac{v}{u^2+v^2}dx + \frac{u}{u^2+v^2}dy
$$
Then I calculate $e_u$ and $e_v$ - basis vectors for tangent plane in new variables:
$$
x_u = v \\
y_u = u \\
e_u = \frac{(x_u,y_u)}{||(x_u,y_u)||} = (\frac{v}{\sqrt{u^2+v^2}}, \frac{u}{\sqrt{u^2+v^2}})
$$
$$
x_v = u \\
y_v = -v \\
e_v = \frac{(x_v,y_v)}{||(x_v,y_v)||} = (\frac{u}{\sqrt{u^2+v^2}}, -\frac{v}{\sqrt{u^2+v^2}})
$$
Calculating $du(e_u)$ and $dv(e_v)$:
$$
du(e_u) = \frac{u^2+v^2}{\sqrt{u^2+v^2}(u^2+v^2)} = \frac{1}{\sqrt{u^2+v^2}} (=dv(e_v))
$$
Noting that $du(e_v) = dv(e_u) = 0$, let $(F, H)$ be a vector field.
Now I need to calculate area form:
$$dx\wedge dy = (u^2 + v^2)du\wedge dv$$
And take interiour product with my field:
$$(u^2 + v^2)((du)(F,H) dv - dv(F,H) du) \\
= (u^2 + v^2)(\frac{F}{\sqrt{u^2+v^2}}dv - \frac{H}{\sqrt{u^2+v^2}}du) \\
= \sqrt{u^2+v^2}(Fdv-Hdu)$$
Then I apply differential on top of it to get my divergence as coefficient:
$$
d(\sqrt{u^2+v^2}(Fdv-Hdu)) \\
= (\frac{u}{\sqrt{u^2+v^2}}du + \frac{v}{\sqrt{u^2+v^2}}dv)\wedge(Fdv-Hdu) + \sqrt{u^2+v^2}(F_u du\wedge dv - H_v dv \wedge du) \\
= \frac{1}{\sqrt{u^2+v^2}}(uFdu \wedge dv - vH dv \wedge du) + \sqrt{u^2+v^2}(F_u + H_v) du \wedge dv \\
= (\frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v)) du \wedge dv
$$
So divergence of $(F, H)$ should be of form
$$
\frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v)
$$
But there I can see that another $\frac{1}{u^2 + v^2}$ was lost somewhere. But I can't find where it was lost, so I'm asking for help to find it.
| Thanks for clarifying. OK, so the vector field is dual to the $1$-form $\omega = F\omega_1 + H\omega_2$, where $\omega_1 = \sqrt{u^2+v^2}\,du$ and $\omega_2=\sqrt{u^2+v^2}\,dv$ give an orthonormal coframe. [Note that the metric $dx^2 + dy^2 = \omega_1^2 + \omega_2^2$.]
Of course we have $\star\omega_1 = \omega_2$ and $\star\omega_2=-\omega_1$. So
\begin{align*}
d(\star\omega) &= d(F\omega_2) - d(H\omega_1) = d\big(F\sqrt{u^2+v^2}dv\big) - d\big(H\sqrt{u^2+v^2}du\big)\\
&= \left(\sqrt{u^2+v^2}(F_u+H_v) + F\frac u{\sqrt{u^2+v^2}}+H\frac v{\sqrt{u^2+v^2}}\right)du\wedge dv \\
&= \left(\frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}\right)\omega_1\wedge\omega_2.
\end{align*}
Finally, using $\star(\omega_1\wedge\omega_2) = 1$, we have
$$\text{div} (F,H) = \frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}.$$
| {
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Any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$. My solution Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:
*
*If $i=0$, then
\begin{align*}
n = 5k \Rightarrow n = 5k + 4(0).
\end{align*}
*If $ i = 1 $, then
\begin{align*}
n & = 5k + 1 \\
& = 5k-5(3) +5(3) +1 \\
& = 5(k-3) + 15 + 1 \\
& = 5(k-3) +16 \Rightarrow n = 5(k-3) +4(4).
\end{align*}
*If $ i = 2 $, then
\begin{align*}
n & = 5k + 2 \\
& = 5k-5(2) +5(2) +2 \\
& = 5(k-2) + 10 + 2 \\
& = 5(k-2) +12 \Rightarrow n = 5(k-2) +4(3).
\end{align*}
*If $i=3$, then
\begin{align*}
n & = 5k + 3 \\
& = 5k-5 + 5 + 3 \\
& = 5(k-1) +8 \Rightarrow n = 5(k-1) +4(2).
\end{align*}
*If $i=4$, then
\begin{align*}
n = 5k + 4 \Rightarrow n = 5k + 4(1).
\end{align*}
Thus, every positive number can be expressed as a linear combination of $5$ and $4$. Now using that $n>11$, so we have:
\begin{align*}
n &> 11 \\
5k + i &> 5(2) +1 \\
5k-5(2) &> 1-i \\
5 (k-2) &> 1-i \\
k-2 &> \frac{1-i}{5} \\
k &> 2+\frac{1-i}{5}.
\end{align*}
So by increasing over the values that $ i $ takes, we have:
\begin{align*}
k &> 2+ \frac{1-i}{5} \geq 2+ \frac{1-0}{5}\\
k &> 2 + 0.2 = 2.2
\end{align*}
But $k\in\mathbb{Z}_0^+ \Rightarrow k \geq 3 \Rightarrow n \geq 15 $. Thus we have that every positive integer greater than or equal to $15$ is a non-negative linear combination of $5$ and $4$.
Finally, let's look at the cases that are still unverified, which are $12$, $13$ and $14$.
\begin{align*}
12 &= 5(0) +4(3) \\
13 &= 5(1) +4(2) \\
14 &= 5(2) +4(1).
\end{align*}
Therefore, any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| You also can prove it by induction on $n>11$.
Here is how the inductive step goes:
Suppose that for some $n>11$, one has $\;n=5x+4y$, $x,y \ge 0$.
*
*If $y >0$, as $1=5-4$, you can write $n+1=5(x+1)+4(y-1)$, and $y_1\ge 0 $;
*If $y=0$, $n=5x$, and since $n>11$, you have $x\ge 3$. Now you also can write $1=16-15$, so
$$n+1=5(x-3)+4\cdot 4\qquad(\text{note that } x-3\ge 0)$$
| {
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Is it possible to evaluate the integral $I=\int_0^\infty \frac{\sqrt{x}\arctan(x)}{1+x^2}dx$ with residue theorem? let's consider the following integral
$$I=\int_{0}^{+\infty}{\frac{\sqrt{x}\arctan{x}}{1+x^{2}}\,\mathrm{d}x}$$
I know exactly how to evaluate it with sneaky methods and obtained $I=\frac{\pi\sqrt{2}}{8}(\pi+2\ln{\left(2\right)})$, but would it be possible to evaluate it with contour integration (Residue theorem)? Of course I have done some simple contour integrations before for example a half circle to evaluate $J=\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+x^{2}}}$, but for some reason I cannot come up with ideas for this.
Thanks in advance.
In case it is possible, I would like to get hints.
| Notice that : \begin{aligned}I=2\int_{0}^{+\infty}{\frac{x\arctan{x}}{2\sqrt{x}\left(1+x^{2}\right)}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{x^{2}\arctan{\left(x^{2}\right)}}{1+x^{4}}\,\mathrm{d}x}\\ &=\int_{-\infty}^{+\infty}{\frac{x^{2}\arctan{\left(x^{2}\right)}}{1+x^{4}}\,\mathrm{d}x}\\ I&=\int_{0}^{1}{\int_{-\infty}^{+\infty}{\frac{x^{4}}{\left(1+x^{4}\right)\left(1+x^{4}y^{2}\right)}\,\mathrm{d}x}\,\mathrm{d}y}\end{aligned}
Now we can apply the residue theorem to $ f_{y}:z\mapsto\frac{z^{4}}{\left(1+z^{4}\right)\left(1+z^{4}y^{2}\right)} $, for $y\in\left(0,1\right] $, on the countour $ \mathscr{C}_{R}=\left[-R,R\right]\cup\Gamma_{R} $, where $ R> \frac{1}{y} $ :
We know : \begin{aligned} \int_{\mathscr{C}_{R}}{f_{y}\left(z\right)\mathrm{d}z}&=2\pi\,\mathrm{i}\left(\mathrm{Res}\left(f_{y},\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\right)+\mathrm{Res}\left(f_{y},-\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}\right)+\mathrm{Res}\left(f_{y},\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}}{\sqrt{y}}\right)+\mathrm{Res}\left(f_{y},-\frac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}{\sqrt{y}}\right)\right)\\ &=2\pi\,\mathrm{i}\left(\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}}{4\left(1-y^{2}\right)}-\frac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}{4\left(1-y^{2}\right)}-\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}}{4\sqrt{y}\left(1-y^{2}\right)}+\frac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}{4\sqrt{y}\left(1-y^{2}\right)}\right)\\ \int_{-R}^{R}{f_{y}\left(x\right)\mathrm{d}x}+\int_{\Gamma_{R}}{f\left(z\right)\mathrm{d}z}&=\frac{\pi\sqrt{2}}{2\sqrt{y}\left(1-y^{2}\right)}-\frac{\pi\sqrt{2}}{2\left(1-y^{2}\right)} \end{aligned}
We have : $$ \left|\int_{\Gamma_{R}}{f_{y}\left(z\right)\mathrm{d}z}\right|\leq\int_{\Gamma_{R}}{\left|f\left(z\right)\right|\left|\mathrm{d}z\right|}\leq\frac{R^{2}}{\left(R^{4}-1\right)\left(R^{4}y^{2}-1\right)}\int_{\Gamma_{R}}{\left|\mathrm{d}z\right|}=\frac{R^{4}\pi}{\left(R^{4}-1\right)\left(R^{4}y^{2}-1\right)}\underset{R\to +\infty}{\longrightarrow}0 $$
Thus, taking $ R $ to $ +\infty $, we get : $$ \int_{-\infty}^{+\infty}{f_{y}\left(x\right)\mathrm{d}x}=\frac{\pi\sqrt{2}}{2\sqrt{y}\left(1+\sqrt{y}\right)\left(1+y\right)} $$
That means : \begin{aligned} I=\pi\sqrt{2}\int_{0}^{1}{\frac{\mathrm{d}y}{2\sqrt{y}\left(1+\sqrt{y}\right)\left(1+y\right)}}&=\pi\sqrt{2}\int_{0}^{1}{\frac{\mathrm{d}y}{\left(1+y\right)\left(1+y^{2}\right)}}\\ &=\frac{\pi\sqrt{2}}{2}\int_{0}^{1}{\left(\frac{1}{1+y}+\frac{1-y}{1+y^{2}}\right)\mathrm{d}y}\\ &=\frac{\pi\sqrt{2}}{2}\left[\ln{\left(1+y\right)}+\arctan{y}-\frac{\ln{\left(1+y^{2}\right)}}{2}\right]_{0}^{1}\\ I&=\frac{\pi\sqrt{2}}{8}\left(\pi+\ln{4}\right) \end{aligned}
| {
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Polynomials/ remainder theorem What must be added to p(x)=x^4+2x^3-2x^2+x-1 to make it exactly divisible by g(x)=x^2+2x-3?
The book says that the remainder will be a linear polynomial and so the expression that must be added to p(x) will be of the form ax+b; the book essentially says we add the remainder. The divisor has roots 1 and -3. We can solve by putting p(1) and p(-3) equal to zero and finding the values of a and b. I know how to solve this but I am confused as to why we say that the expression to be added is of the form ax+b. How do we know this. Essentially what the book seems to be saying is add the remainder. But if we take the example7/3, the remainder is 1 but adding the remainder will not make 7 exactly divisible by 3. Here we subtract 1 from 3 and add the result to make 7 exactly divisible by 3. I believe the same principle must hold in the case of polynomials and the explanation in the book that we add the remainder to the dividend must be wrong. Please tell me the reason why we add ax+b. Please explain.
Also, can it be solved through polynomial long division? If so, how?
Are there any other easier ways to solve this? Feel free to ignore the last question and even the second last question if it will make the answer too long.
| The division algorithm says that if $p(x)$ is a polynomial of degree $n$ and $g(x)$ is a polynomial of degree $m\le 0$ then there is a unique quotient polynomial $q(x)$ of degree $n-m$ and a unique remainder polynomial of degree $< m$ so that: $p(x) = q(x)g(x) + r(x)$.
So if $p(x) = x^4+2x^3-2x^2+x-1$ is of degree $4$ and $g(x) = x^2+2x-3$ is of degree $2$ then there is a unique polynomial of degree $4-2 = 2$ called $q(x)$ and a unique polynomial $r(x)$ of degree less than $2$ so or degree of at most $1$ so that
$p(x) = q(x)g(x) + r(x)$.
Now as $r(x)$ is at most of degree $1$ it is of the form $ax + b$ (where $a$ could be $0$).
And
$x^4+2x^3-2x^2+x-1 = (x^2+2x-3)(\alpha x^2 + \beta x + \gamma) + (ax+b)$
so to make $p(x)$ exactly divisible by $g(x)$ we must add the NEGATIVE of the remainder (I don't understand why the book says to add rather than subtract the remainder) to get:
$x^4 + 2x^3 - 2x^2 + x-1 +(-ax -b) = (x^2+2x-3)(\alpha x^2 + \beta x + \gamma)$
And the most straightforwar way to do this is long division.
But typing long division on a keyboard is too hard for me so I'll talk us through:
$(x^2+2x-3)(\alpha x^2 + \beta x + \gamma)= \alpha x^4 + (2\alpha + \beta)x^3 +(\gamma+2\beta -3\alpha) x^2 + (2\gamma -3\beta)x- 3\gamma$.
We want that to be as close to $x^4 + 2x^3 -2x^2 + x-1$ as possible.
So $\alpha = 1$ and $2\alpha + \beta = 2$ and $\gamma +2\beta -3\alpha = -2$. And that's as fair as we will be able to go.
$\alpha = 1$ and $\beta = 0$ and $\gamma = 1$.
And so we have
$x^4+2x^3-2x^2+x-1 = (x^2+2x-3)(x^2 +1) + (ax+b)$
$x^4+2x^3-2x^2+x-1 = x^4 +2x^3 - 2x^2 +2x -3 +(ax+b)$.
And so the remainder is $ax + b = -x +2$
so what we must add is $-(-x+2) = x -2$.
That is to say
$(x^4+2x^3-2x^2+x-1) + (x-2) = (x^2+2x-3)(x^2 +1)$
.......
But that is the most straightforward way.
The easiest way is to note:
If $p(x) + (mx+n) = (x^4 + 2x^3 - 2x^2 + x-1) + (mx + n) = (x^2 + 2x -3)q(x)=g(x)q(x)$.
And the roots of $g(x) = x^2 + 2x -3)$ are $x = 1$ and $x=-3$ then
$p(1) + (m*1 + n) = g(1)q(1) = 0*q(1) = 0$ so
$(1^4 + 2*1^3 -2*1^2 + 1-1) + (m+n) = 0$
$1+m+n = 0$
And $p(-3) + (-3m + n) = 0$ and
$(3^4 - 2*3^3 - 2*3^2 - 3-1) -3m +n =0$
$81 - 54 -18 -4-3m+n = 0$
$5- 3m + n=0$
So $5-3m + n = 1+m+n$
$5-3m = 1+m$
$4=4m $ and $m =1$ and $1+m+n = 0$ so $1+1+n =0$ so $n=-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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General Inequality Problem. The question goes like this :
Let $a,b,c,d$ be positive reals and given that $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$
My approach goes like this:
I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$
Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$
Although I am not sure if its the correct method, I tried reducing the powers.
Then, I got:
$6(a+b)(a^2+b^2-ab) -a^2-b^2 \geqslant c^2+d^2 - 6(c+d)(c^2+d^2-cd) +\frac{1}{8} $
I tried substituting $c,d$ in terms of $a,b$ but the calculation is lengthier than I expected. Please check if my reasoning is correct, and help me solve this in a shorter method. Thanks!
| Tangent line detailed:
You can rewrite your inequality as $$f(a)+f(b)+f(c)+f(d)\geq {1\over 8}$$ where $f(x)= 6x^3-x^2$. Now we see that we have equality case if $a=b=c=d={1\over 4}$ so it is reasonable to calculate tangent line for $f$ at $x= {1\over 4}$ which is $g(x)={5x-1\over 8}$.
If you draw a graph for $f$ and $g$ you see that $f$ is all the time over $g$ for $x\in[0,1]$, so $$f(x)\geq g(x)$$ This, of course, you should verify with some calculations. So $$f(a)+f(b)+f(c)+f(d)\geq g(a)+g(b)+g(c)+g(d)= {1\over 8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove That $n(n+1)(n+2)(n+3)$ is never a square of a positive integer. $n \in \mathbb{N}$, bigger than zero.
My proof:(A Proof by Contradiction)
Alert: There is a lot of variables.
Suppose that:
$n(n+1)(n+2)(n+3)=x^{2}$, with $x \in \mathbb{N}$
$$n(n+3)(n+2)(n+1)=x^{2}$$
$$(n^{2}+3n)(n^{2}+3n+2)=x^{2}$$
let $n^{2}+3n = a$.
$$a(a+2)=x^{2} \iff a^{2} +2a=x^{2} \iff a^{2} +2a-x^{2}=0$$
And this is a quadratic equation in $a$, So:
$$a=\frac{-2 \pm \sqrt{4+4x^{2}}}{2} \implies \sqrt{4+4x^{2}} \in \mathbb{N} \iff 4+4x^{2}=b^{2}$$
$$ 4(1+x^{2})=b^{2} \iff 4 \mid b^{2} \iff 2 \mid b \iff b=2c.$$
Now let's plug this result to this equation:
$$4(1+x^{2})=(2c)^{2} \iff 4(1+x^{2})=4c^{2} \iff 1+x^{2}=c^{2} \iff (c-x)(c+x)=1$$
And $1=1 \cdot 1$, so :$(c-x)=1$ and $(c+x)=1$
$$c+x=c-x \iff x=-x \iff 2x=0 \iff x=0$$
By plugging this result into this equation : $a(a+2)=x^{2}.$
$a(a+2)=0 \iff a=0$ or $a=-2$ but $-2 \notin \mathbb{N}$ So : $x=0$
And finally:
$$n(n+1)(n+2)(n+3)=0 \iff n=0$$
And the other values $\notin \mathbb{N}$
This proof is too huge for me so perhaps I've made some mistakes, And if you have an easier proof please post it.
| Hint: Show that $ n(n+1)(n+2)(n+3) + 1$ is a perfect square.
Hence, the original expression cannot be the square of a positive integer (but could be the perfect square 0).
| {
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Prime Number Sums The prime numbers $a,b$ and $c$ are such that $a+b^2=4c^2.$ Determine the sum of all possible values of $a+b+c.$
My Attempt
$a+b^2=4c^2$.
$a=4c^2-b^2$.
$a=(2c+b)(2c-b)$.
After this, I tried testing cases but I'm not totally sure how to account for everything.
Thanks!
| Note that $c=2,3$ yield solutions: $a=7,b=3,c=2 \text{ and }a=11,b=5,c=3$
As has been noted in the comments, $2c-b=1$ is necessary for $a$ to be prime.
For prime numbers $c\ge 5$, $c$ must have the form $6k\pm 1$. Note that for $c=6k-1$ we have $2c-b=1=12k-2-b \Rightarrow b=12k-3$. Since $3\mid 12k-3$, $b$ cannot be prime and primes $c$ of the form $6k-1$ can never give rise to solutions.
Prime numbers $c=6k+1$ yield $2c-b=1=12k+2-b \Rightarrow b=12k+1$, which may or may not be prime. In cases where $b$ is in fact prime, $a=2c+b=12k+2+12k+1=24k+3$. Since $3\mid 24k+3$, $a$ cannot be prime and primes $c$ of the form $6k+1$ can never give rise to solutions.
So the only solutions are those given first, in which the asked for sums are: $a+b+c=12,19$
| {
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How do I bypass the limit for the tan function on the calculator? So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is.
Problem:
$Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$
I'll show you what we did.
First, we used the pythagorean formula to find $cos \ \theta$ (we need it later). The half-angle formula for cosine is $cos \frac {\theta}{2} = \pm \sqrt{\frac{1 + cos \ \theta}{2}}$.
These are our steps.
*
*$cos \frac {tan^{-1}\frac{3}{4}}{2} = \pm \sqrt{\frac{1 + cos \ (tan^{-1}\frac{3}{4})}{2}}$
*$ = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}}$
*$ = \pm \sqrt{\frac{\frac{9}{5}}{2}}$
*$ = \pm \sqrt{\frac{9}{10}}$
*$ = \pm \frac{3\sqrt{10}}{10}$
To decide whether we have a negative sign or a positive, we have to look at the domain.
$ \pi < \theta<\frac{3\pi}{2} \ changes \ to \ \frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$
We now know that $cos \frac {\theta}{2}$ is located in the second quadrant, which makes our final answer $cos \frac {tan^{-1}\frac{3}{4}}{2} = - \frac{3\sqrt{10}}{10}$
But then, I checked the calculator.
The values are equal except for the negative sign
My partner said that the calculator thinks tangent is in the first quadrant, and if it were, cosine would be positive. But since we are in the third quadrant, cosine is in the second quadrant. I know there are domain restrictions, but I don't entirely understand what she meant. Also, how do I get the calculator to display the same answer?
| You cannot do the inverse tangent because the range of inverse tangent is $(-\frac{\pi}{2},\frac{\pi}{2})$, and so your answer for step 2 (and also the answer from your calculator) is already incorrect.
I would suggest you do this:
Draw the triangle in the third quadrant (because it is in the range $(\pi, \frac{3\pi}{2})$, with adjacent side (on the x-axis) $x=-4$ and opposite side (on the y-axis) $y=-3$. The hypotenuse is $5$ by Pythagoras. Then, $\cos \theta = -\frac{4}{5}$. Using the cosine half-angle formula, $\cos \frac{\theta}{2}=\pm\sqrt{\frac{1+\cos \theta}{2}}=\pm\sqrt{\frac{1+(-\frac{4}{5})}{2}}=\pm\sqrt{\frac{1}{10}}=\pm \frac{\sqrt{10}}{10}$ Because the angle is in the range $(\pi, \frac{3\pi}{2})$, then half the angle is in the range $(\frac{\pi}{2}, \frac{3\pi}{4})$, so $\cos$ is negative. Therefore, your answer is $\boxed{-\frac{\sqrt{10}}{10}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$
If $a,b,c,d$ are real numbers and $a+b+c+d=0$. prove:
$$a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$$
Using Euler's identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$. in this problem substitute $x=a+b,y=c,z=d:$
$$(a+b)^3+c^3+d^3=3cd(a+b)$$
$$a^3+b^3+c^3+d^3=3(a+b)(cd-ab)=3(acd-a^2b+bcd-ab^2)$$
Although I am sure my approach is right, but I should have $3(abc+bcd+cda+dab)$ at the RHS. I don't know how to get it.
| $$3(acd-a^2b+bcd-ab^2) = 3(cd(a + b) - ab(a+b)) = 3(cd(a + b) + ab(c+d)) = 3(abc + abd + acd + bcd)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Nonlinear recurrence relation over $\mathbb{C}$ A sequence in $\mathbb{C}$ is given: $(a_n)_{n=1}^{\infty}$ s.t.
$a_1 = i$ and $a_{n+1} = \frac{3}{2 + a_n}$.
Assume $\lim(a_n)$ exists.
Using the fact that $\lim(a_n) = \lim(a_{n+1})$ and solving the resulting quadratic gives that $\lim(a_n)\in \{-3,1\}$.
I was trying to figure out which one the limit is. I tried finding a relation on the norm or the argument (using the initial condition, since clearly that is what the limit is determined by) and I also tried looking at individual terms and by this, the real part seems to stay positive and hover around one while the imaginary part we know goes to zero in any case (i.e., I am thinking the limit is 1). I wanted to ask for verification that I did this last part correctly, since $a_2 = \frac{6-3i}{5}$ if we can show that:
$Sgn[Re(a_m)] = Sgn[Re(a_{m+1})]$ $\forall m\geq 2$
we are done and the limit is 1. Suppose that $a_m = x+yi$ ($m\geq 2$) and $x > 0$. Then
$a_{m + 1}= \frac{3}{2 + x + yi}\frac{(2+x - yi)}{(2+x-yi)} = \frac{3(2+x) - 3yi}{(2+x)^2 + y^2}$
meaning that
$Sgn[Re(a_{m+1})] = Sgn\bigg[\frac{6 + 3x}{(2+x)^2 + y^2}\bigg] = 1$ since $x > 0$ and we are done. So the limit is 1. Does that look about right?
| A specific approach to convergence is to examine how $a_{n}-1$ changes in each iteration.
\begin{align}
a_{n+1} - 1
&= \frac{3}{2+a_n} - 1 \\
&= \frac{1-a_n}{3-(1-a_n)}
\end{align}
Then if $|a_n-1|$ is small relative to $3$ it is apparent that $|a_{n+1}-1|$ decreases by a factor of nearly $1/3$ on each iteration and so $a_{n+1} \to 1$. In fact, as long as $3-|1-a_n|$ is greater than some bound greater than $1$, convergence is guaranteed. For example if $|a_{n-1}-1|< 15/8$ we have,
\begin{align}
|3-(1-a_n)| &\geqslant 3 - |1-a_n| \\
&\geqslant \frac{9}{8}
\end{align}
whence $|a_{n+1} -1| \leqslant \frac{8}{9} | a_n - 1 |$.
Starting with $a_1 = i$, $|a_1-1| = \sqrt 2 < \frac{15}{8}$ so the condition is met and the sequence converges to $1$.
Another more general approach is to recognise that recurrence is a particular case of a continued fraction. Consider the expression,
\begin{align}
f = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{p_3}{\ddots}}}
\end{align}
If we define functions of $w \in \mathbb C$ by,
\begin{align}
t_0(w) = q_0 + w, \quad t_n(w) = \frac{p_n}{q_n +w}
\end{align}
then when we combine $t_0$ through $t_n$ we obtain,
\begin{align}
c_n(w) = t_0\circ t_1 \circ \cdots \circ t_n(w) = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{\ddots}{q_{n-1}+\dfrac{p_n}{q_n+w}}}}
\end{align}
and we interpret $f$ to be the limit of this as $n \to \infty$ when $w=0$ (if it exists). The number $c_n(0)$ is called the $n$th convergent.
A useful recurrence relation exists for $c_n(w)$, $n \geqslant 0$,
\begin{align}
c_n(w) = \frac{P_n+P_{n-1}w}{Q_n +Q_{n-1} w} \tag{1}\label{eq1}
\end{align}
where we use initial conditions, $P_0 = q_0, Q_0 = 1, P_{-1} = 1, Q_{-1}=0$ and the $P_n, Q_n$ do not depend on $w$ and can be obtained from the recurrence,
\begin{align}
\begin{array}{l}
P_{n+1} = q_{n+1}P_n + p_{n+1} P_{n-1} \\
Q_{n+1} = q_{n+1}Q_n + p_{n+1} Q_{n-1} .
\end{array}\tag{2}\label{eq2}
\end{align}
This is proved by induction. We observe that the given initial conditions imply equation \eqref{eq1} holds when $n=0$. Then, if \eqref{eq1} holds for some $n \geqslant 0$, from the definition,
\begin{align}
c_{n+1}(w) &= c_n(p_{n+1}/(q_{n+1}+w)) \\
&= \frac{P_n+P_{n-1}(p_{n+1}/(q_{n+1}+w))}{Q_n+Q_{n-1}(p_{n+1}/(q_{n+1}+w))} \\
&=\frac{P_{n}q_{n+1}+P_n w + P_{n-1}p_{n+1}}{Q_{n}q_{n+1}+Q_n w + Q_{n-1}p_{n+1}} \\
&=\frac{P_{n+1}+P_n w}{Q_{n+1}+Q_n w}.
\end{align}
Thus \eqref{eq1} also holds for $n+1$ with the new coefficients $P_{n+1}, Q_{n+1}$ given by \eqref{eq2}.
Applied to the case in hand, we take $q_0 = 0, p_n = 3, q_n=2$ and $w = i$, so that $c_0(i) = i, c_1(i) = 3/(2+i)$ and in general $c_n(i) = a_{n+1}$.
We can solve the two recurrence relations for $P_n, Q_n$ using standard techniques with the initial conditions provided above, obtaining
\begin{align}
P_{n} &= \frac{3}{4} (3^{n} - (-1)^n) \\
Q_{n} &= \frac{1}{4} (3^{n+1} +(-1)^n )
\end{align}
In this form, eliminating the factor $4$, we obtain, for general $w$,
\begin{align}
c_{n-1}(w) &= \frac{3(3^{n-1}-(-1)^{n-1})+3w(3^{n-2}-(-1)^{n-2})}{3^n+(-1)^{n-1}+w(3^{n-1}+(-1)^{n-2})} \\
&=\frac{(3+w)(-3)^{n-1}-3+3w}{(3+w)(-3)^{n-1}+1-w} \\
&=1+\frac{12(1-w)}{(3+w)(-3)^{n}-3(1-w)}.
\end{align}
From which it is clear the recurrence will converge to $1$ regardless of the starting value $w$ provided not $-3$. Setting $w=i$ this becomes after multiplying numerator and denominator by $(1+i)/2$
\begin{align}
a_n &= c_{n-1}(i) \\
&=1+\frac{12}{(1+2i)(-3)^{n}-3}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$ I have recently learned about pattern that goes like this:
$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$
It is a very cool pattern, but after a bit it stops:
$$1,111,111,111^2 = 1,234,567,900,987,654,321$$
My main question:
Why does this pattern work?
A side question that's less important:
Is there an algebraic equation to describe this pattern?
| Since OP is asking for an algebraic explanation, I will try a different approach, possibly trickier than the other answer which is elegant but not formal.
We consider the polynomial $(1+x+x^2+\dots +x^n)^2$: by expanding the square we get
$$\begin{array}{|c|c|c|c|c|c|c|}
\hline
1 & x & x^2 &\cdots& x^{n-1}& x^n\\
\hline
x & x^2 & x^3 &\cdots& x^{n}& x^{n+1}\\
\hline
x^2 & x^3 & x^4 &\cdots& x^{n+1}& x^{n+2}\\
\hline
\vdots & \vdots & \vdots &\vdots& \vdots& \vdots\\
\hline
x^{n-1} & x^{n} & x^{n+1} &\cdots& x^{2n-2}& x^{2n-1}\\
\hline
x^n & x^{n+1} & x^{n+2} &\cdots& x^{2n-1}& x^{2n}\\
\hline
\end{array}$$
that is, for $n\geq 0$, and for $0\leq j\leq 2n$,
$$\begin{align}
(1+x+x^2+\dots +x^n)^2&=\sum_{j=0}^{n}\sum_{k=0}^{j} x^k\cdot x^{j-k}
+\sum_{j=n+1}^{2n}\sum_{k=j-n}^{n} x^k\cdot x^{j-k}\\
&=
\sum_{j=0}^{n}x^j\sum_{k=0}^{j} 1
+\sum_{j=n+1}^{2n}x^j\sum_{k=j-n}^{n} 1\\
&=\sum_{j=0}^{n}(j+1)x^j
+\sum_{j=n+1}^{2n}(2n-j+1)x^j.
\end{align}$$
In our case, we take $x=10$. The pattern works as soon as
$j+1\leq 9$ for $j=0,\dots,n$ AND $2n-j+1\leq 9$ for $j=n+1,\dots,2n$ where $9$ is the largest decimal digit, that is for $n\leq 8$.
For $n=8$ and $x=10$ we have that
$$(1+x+x^2+\dots +x^n)^2=111111111^2=12345678987654321$$
whereas for $n=9$ we get
$$(1+x+x^2+\dots +x^n)^2=1111111111^2=1234567900987654321.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.
I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.
| It's a variation of Candido's identities, its proof without words are as follows:
Now
\begin{align}
[x^2+y^2+(x+y)^2]^2 &= 2[x^4+y^4+(x+y)^4] \\
[a^2+(-b)^2+(a-b)^2]^2 &= 2[a^4+(-b)^4+(a-b)^4] \\
[2a^2-2ab+2b^2]^2 &= 2[a^4+b^4+(a-b)^4] \\
4(a^2-2ab+b^2)^2 &= 2[a^4+b^4+(a-b)^4] \\
a^4+b^4+(a-b)^4 &= 2(a^2-ab+b^2)^2
\end{align}
Refer to link 1 and link 2 for your further interests.
| {
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Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$ I accidentally found out that the two integrals below
$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluated explicitly. For example, the first one can be carried out via double integration, as sketched below.
\begin{align}
I_1&=\int_0^\infty \left(\int_0^1 \frac{x^2}{1+y^2x^4}dy\right)\frac{1}{1+x^2}dx\\
&= \frac\pi2\int_0^1 \left( \sqrt{\frac y2}+ \frac1{\sqrt{2y} }-1\right)\frac{1}{1+y^2}dy=\frac{\pi^2}8
\end{align}
Similarly, the second one yields $I_2=\frac{\pi^2}8$ as well.
The evaluations are a bit involved, though, and it seems an overreach to prove their equality this way, if only the following needs to be shown
$$\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx=0$$
The question, then, is whether there is a shortcut to show that the above integral vanishes.
| Since $\int_0^\infty\frac{f(x)dx}{1+x^2}=\int_0^1\frac{f(x)+f(1/x)}{1+x^2}dx$, $\int_0^\infty\tfrac{\arctan x^kdx}{1+x^2}=\tfrac{\pi}{2}\int_0^1\tfrac{dx}{1+x^2}=\tfrac{\pi^2}{8}$ for all $k\in\Bbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Impossible integral?
$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$
I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:
$$\frac {d}{dx}\int_{0}^{x^2-1} f(t) dt= \frac {d}{dx}[x^6+x^4+3x^2]$$
$$f(x^2-1)2x = 6x^5+4x^3+6x$$
$$f(x^2-1) = 3x^4+2x^2+3$$
$$f(x^2-1) = 3(x^2-1)^2+8(x^2-1)+8$$
$$f(t) = 3t^2+8t+8$$
$$\int_{0}^{x^2-1} 3t^2+8t+8 dt= x^6+x^4+3x^2-5$$
This means that there is no continuous function that satisfies the equation of the problem?
| Suppose $f(t)=at^2+bt+c$
we have
$$\int_0^{x^2-1} \left(a t^2+b t+c\right) \, dt=$$
$$=x^6+x^4+3 x^2\equiv \frac{a x^6}{3}+x^4 \left(\frac{b}{2}-a\right)+x^2 (a-b+c)-\frac{a}{3}+\frac{b}{2}-c$$
and
$$
\begin{cases}
\frac{a }{3}=1\\
\frac{b}{2}-a=1\\
a-b+c=3\\
-\frac{a}{3}+\frac{b}{2}-c=0\\
\end{cases}
$$
the system has no solution, therefore there is no polynomial $f(t)=at^2+bt+c$ which satisfies the equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6...
Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6, find distance of focus from tangent at P.
Let P and Q be $(at^2,2at)$ and $(\frac{a}{t^2}, \frac{-2a}{t})$ respectively, for parabola $y^2=4ax$
So $$0= 2(-\frac{2a}{t})+2at$$
$$\implies t^2=2$$
Now tangent at Q is
$$\frac{y}{\sqrt 2} =x+\frac a2$$
So point R is $(-a, -\frac{a}{\sqrt 2})$
Then $$QR^2 = 36$$
$$\implies a=\frac{12}{\sqrt {11}}$$
So now, tangent at P is $$\sqrt 2y=x+2a$$ whose distance from $(a,0)$ is
$$d=\left | \frac{a+2a}{\sqrt 3}\right|$$
$$d=\sqrt 3 a$$
$$d=\frac{\sqrt 3 \times 12}{\sqrt {11}}$$
But given answer is 4
Where am I going wrong?
| As you rightly obtained, $t^2 = 2 \implies t = \pm \sqrt2$
If we take $t = \sqrt2$ and rewrite,
$P(2a, 2a \sqrt2), Q(\frac{a}{2}, - a \sqrt2)$
Slope of tangent at point $Q = \frac{2a}{y} = - \sqrt2$
So equation of tangent line $y + a \sqrt2 = -\sqrt2(x-\frac{a}{2})$
or $y = - \sqrt2 x - \frac{a}{\sqrt2} $
Point $R$ where the line intersects the directrix, $R(-a, \frac{a}{\sqrt2})$.
So $QR^2 = \frac{9a^2}{4} + \frac{9a^2}{2} = 36 \implies a = \frac{4}{\sqrt3}$.
Now slope of tangent line at $P = \frac{2a}{y} = \frac{1}{\sqrt2}$.
So equation of tangent line at $P$, $y - 2a\sqrt2 = \frac{1}{\sqrt2}(x - 2a)$
or $y = \frac{x}{\sqrt2} + \sqrt2 a$
Its distance from focus $(a,0)$ is
$d = \frac{3a/ \sqrt2}{\sqrt{3/2}} = a \sqrt3 = 4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Generalization of $\sqrt{n}+\sqrt{n+2005}=m$ I was solving the math Olympiad of Belgium the year $2005$, and the last problem was :
If $n$ is an integer then find all values for $n$ for which $\sqrt{n}+\sqrt{n+2005}$ is an integer as well.
My question is just to generalize this problem, so instead of $2005$ , we can solve this over all $x\in \mathbb{N}$.
My Attempt:
If you have solved the problem you will realized that this equation :
$$\sqrt{n}+\sqrt{n+x}=m$$
Has solution if and only if $m\mid x$. But I am not sure if there are some other restrictions in this equation to has a solution.
| In general, for non-negative integers $a, b$, the sum $\sqrt a + \sqrt b$ is an integer if and only if both $\sqrt a$ and $\sqrt b$ are integers (exercise).
Solution:
Squaring, we see that $ab$ is the square of a rational number, and hence we may write $a = m^2 k$, $b = n^2 k$, with $k$ square-free. Then the sum $\sqrt a + \sqrt b$ becomes $(m + n)\sqrt k$, which is an integer if and only if $k = 1$.
Therefore, if $\sqrt n + \sqrt{n + x}$ is an integer, then we have $n = u^2$ and $n + x = v^2$ for some integers $u, v$.
This leads to $v^2 - u^2 = x$, which has a solution if and only if $x\not\equiv 2\mod 4$ (exercise).
Solution:
In one direction, we have $x = (v + u)(v - u)$ and $v + u \equiv v - u \mod 2$. Hence either both factors are even and $x$ is a multiple of $4$, or both are odd and $x$ is odd. In the other direction, if $x = 2k + 1$ is odd, then $x = (k + 1)^2 - k^2$; if $x = 4k$ is multiple of $4$, then $x = (k + 1)^2 - (k - 1)^2$.
Explicit solutions:
If $x = 2k + 1$ is odd, then taking $n = k^2$ gives $\sqrt n + \sqrt {n + x} = 2k + 1$.
If $x = 4k$, then taking $n = k^2 - 2k + 1$ gives $\sqrt n + \sqrt{n + x} = 2k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I construct a sequence of smooth functions $g_n$ s.t their integrals converge to zero? Define $g_n(x)$ to be zero on $[0,1-\frac{1}{n}]$ and $g_n(x)=ax^2+bx+c$ on $[1-\frac{1}{n},1]$
I want the $g_n$ functions to be in $C^2[0,1]$ and statifies $g_n(0)=0$ and $g_n(1)=1$, $g_n'(0)=0$ and their integrals to converge to zero as suggested in the comments.
I solved the linear system in $a,b,c$ but to give: $a=n^2$, $b=-2n^2+2n$ and $c=(n-1)^2$
\begin{align}
\int_0^1 g_n(x)= \int_0^{1-\frac{1}{n}} 0 dx + \int_{1-\frac{1}{n}}^1 n^2x^2 -2n(n+1)x+ (n-1)^2 dx = \dfrac{n^2-3n+3}{3}-\dfrac{n^3-3n^2+3n-1}{3n}=\dfrac{3}{n}
\end{align}
so Let $a_n=\frac{3}{n}$
\begin{align}
\lim_{n \to \infty} a_n=\dfrac{n^2-3n+3}{3}-\dfrac{n^3-3n^2+3n-1}{3n} =\lim_{n \to \infty} \frac{3}{n} =0
\end{align}
EDIT: Updated the body of the question
| As it was pointed out in the comments, the functions $g_n$ are not even continuous en $[0,1]$. Regarding the second part, it is true that $\int_0^1 g_n(x) dx \to 0$ and your justification is correct.
| {
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Find the whole number solutions for W+X+Y+Z = 15 where W, X, Y, Z ≤ 6 I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.
First, the number of non-negative integer solutions for $W+X+Y+Z = 15$ can be calculated using stars and bars:
$$\binom{n+k-1}{n} = \binom{15+4-1}{15}$$
Now, when value of W exceeds 6; i.e., for value of 7 violets the rule of the upper limit for W.
That means the number of violations W can have is among 15-7 = 8 identical items in 4 distinct bins.
Equation for violations becomes W+X+Y+Z = 15-7 = 8
As a result, total number of violations for W = C (n+k-1, k-1) = C (8+4-1, 4-1) = C (11, 3)
⸫ Total number of violations for all 4 bins when 1 bin cross upper limit = C (11, 3) × 4
Since, number of violations 8 is >n/2=15/2=7.5; we need to add in the subtracted repetitions.
When 2 bins cross above upper limit; W+X+Y+Z = 15-7-7 = 1
⸫ Total number of repetitions for all 4 bins when 2 bin cross upper limit = C (1+4-1, 4-1) × 4 = C (4, 3) × 4
When 3 bins cross above upper limit; W+X+Y+Z = 15-7-7-7 = -6 i.e., not possible.
Therefore, the number of integer solutions for $W+X+Y+Z = 15$ where $W, X, Y, Z \leq 6$ is $$\binom{18}{3} – \binom{11}{3} \times 4 + \binom43\times 4$$
| The “$\le6”$ constraints are annoying to work with. To fix it,we can substitute $$A=6-W, \quad B=6-X, \quad C=6-Y, \quad D = 6 - Z.$$
Now, the constraints become $A+B+C+D\le 4\times 6 - 15=9$, and $A,B,C,D\ge 0$. Now, you can solve it by standard stars and bars method.
So the answer is $$\binom{9+4-1}{4-1}=220.$$
Whereas the answer I get from your last line is $172$.
| {
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Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.
By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.
But I wonder that there is some logical ways we can find intervals that contain roots without guessing?
| We may apply Sturm's theorem https://en.wikipedia.org/wiki/Sturm%27s_theorem.
Let $\mathrm{rem}(p(x),q(x))$ denote the remainder of $p(x)$ divided by $q(x)$ for two polynomials $p(x), q(x)$.
The Sturm sequence is given by:
$p_0(x) = x^5-5x^3+4x-1$
$p_1(x) = p_0'(x) = 5x^4 - 15x^2 + 4$
$p_2(x) = -\mathrm{rem}(p_0, p_1) = 2x^3 - \frac{16}{5}x + 1$
$p_3(x) = -\mathrm{rem}(p_1, p_2) = 7x^2 + \frac{5}{2}x - 4$
$p_4(x) = -\mathrm{rem}(p2, p_3) = \frac{883}{490}x - \frac{29}{49}$
$p_5(x) = -\mathrm{rem}(p3, p4) = 1889881/779689$
As this is a constant, this finishes the computation of the Sturm sequence.
We have $V(-\infty) - V(+\infty) = 5 - 0 = 5$.
So there are five real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031227",
"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.
I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?
| Write $a = 2^x \cdot m$ and $b = 2^y \cdot n$ where $m$ and $n$ are odd. WLOG $x \geq y$. Now,
$$2^{3x} \cdot m^3 + 2^{3y} \cdot n^3 = 2^{3y} \cdot (2^{3x-3y} \cdot m^3 + n^3) = 2^c$$ If $x > y$, we have a contradiction since in that case an odd number divides LHS. So, $x = y$ and so you would arrive at your argument.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$.
Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$.
This problem seems extremely hard until you find out that $f(x)+f(1-x)=1$ and you can then evalute $f(\frac{1}{1996})+f(\frac{1995}{1996}) =1, \dots, f(\frac{998}{1996})+f(1-\frac{998}{1996})=f(\frac12)+f(\frac12)=1 \implies f(\frac{998}{1996})=\frac12$. Thus the sum is just $997+\frac12$. My question is that how can one see something like this from a question like this without having gone through some contest math training where these kinds of "tricks" are taught. Is there some kind of tell here from which one can find this "property"?
| There are two aspects which can lead us from
\begin{align*}
f(x)=\frac{9^x}{9^x+3}\tag{1}
\end{align*}
to the functional equation $f(x)+f(1-x)=1$.
Simplification: We can reduce the number of occurrences of $x$ in (1) by factoring out $9^x=3^{2x}$. We obtain
\begin{align*}
f(x)=\frac{9^x}{9^x+3}=\frac{1}{1+3^{1-2x}}\tag{2}
\end{align*}
Known pattern: The right hand side of (2) is a well known pattern which admits the following relationship
\begin{align*}
\frac{1}{1+a}+\frac{1}{1+\frac{1}{a}}=\frac{1}{1+a}+\frac{a}{1+a}=1\tag{3}
\end{align*}
The convenient relationship (3) applies to (2) with a nice twist of
\begin{align*}
\frac{1}{a}=\frac{1}{3^{1-2x}}=3^{-(1-2x)}=3^{1-2(1-x)}
\end{align*}
from which
\begin{align*}
\color{blue}{f(x)+f(1-x)=1}
\end{align*}
follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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A limit which exists in polar coordinates but not in Cartesian coordinates? Let's have look at the function
$$ f(x,y) \begin{cases}
\frac{y(x^2+y^2)}{y^2+(x^2+y^2)^2} & (x,y)\neq(0,0)
\\0 & (x,y)=(0,0)\end{cases}.$$
Switching to polar coordinates gives
$$ f(r,\theta)=\begin{cases}
\frac{r^3 \sin \theta}{r^2\sin^2\theta+r^4} & r\neq0
\\0 & r=0\end{cases}.$$
We'd like to investigate the existence of a limit for $f$ at the origin.
In Cartesian coordinates ($f(x,y)$) one can immeidately see that the limit doesn't exist because for example on the path $y=0$ we have $\lim_{x\rightarrow 0,y=0} f(x,y)=0$ and on the path $y=x^2$ we have
$\lim_{x\rightarrow 0,y=x^2} f(x,y)=\frac{1}{2}$.
However, in polar coordinates we have
$$ f(r,\theta)=\begin{cases}
\frac{r \sin \theta}{\sin^2\theta+r^2} & r\neq0
\\0 & r=0\end{cases} $$
so that $$ \lim_{r\rightarrow 0}f(r,\theta) = \begin{cases}
0 & \sin\theta = 0\\
0 & \sin\theta \neq 0
\end{cases}$$
so the limit exists and is zero regardless of $\theta$. Why does it look the limit doesn't exist in Cartesian coordinates but exists in polar coordinates?
Edit: Thanks to the insightful comments on this page and other similar questions in the site, the unboundedness of the expression with $\sin^2 \theta$ is the key to the failure of the limit existence. $\sin\theta$ can get arbitrarily small, making the whole expression arbitrarily large, effectively counteracting $r\rightarrow 0$.
| The limit seems to exist in polar coordinates since you only tested the curves with fixed $\theta$. This is equivalent to only testing the lines $y=mx$ (and the line $x=0$) in Cartesian coordinates.
Indeed for these lines we have:
$$\lim_{(x,y) \to (0,0)}\frac {y(x^2+y^2)}{y^2 + (x^2 + y^2)^2} = \lim_{x\to 0}\frac {mx^3(1+m^2)}{m^2x^2 + x^4(1+m^2)^2}=\lim_{x\to 0}\frac {mx(1+m^2)}{m^2 + x^2(1+m^2)^2} =0$$
Suppose we take another curve: $r = \theta$. Then:
$$\lim_{r\to 0}\frac {r\sin\theta}{\sin^2\theta + r^2} = \lim_{r\to 0}\frac {r\sin r}{\sin^2 r+ r^2} = \lim_{r\to0}\frac {(\sin r)/r}{1 + (\sin^2 r)/r^2} = \frac1{1+1} = \frac12$$
which is not zero, so the limit cannot exist.
| {
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Proving that $f$ verifies $f(1−x) + f(x) = 1$ I have a function which is defined for $x \in (0,1)$ and for $p>1$ with the expression
\begin{align*}
f(x) = \sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
\end{align*}
I would like to show that $f(1-x) + f(x) = 1$.
What I have done so far
With a little bit of algebra I have
\begin{align*}
f(x)
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
&& \text{using }\binom{n}{m} = \frac{n!}{m!(n-m)!}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p, j\neq-k}^{p} (x-j)\right)
&& \text{using the change of indices j=i-k}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p}^{p} (x-j)\right) \frac{1}{x+k}
&& \text{completing the product}
\\
&= \frac{1}{(2p-1)!} \left(\prod_{j=1-p}^{p} (x-j)\right) \left(\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}\right)
&& \text{factoring the product from the sum}
\end{align*}
Then I introduce the functions
\begin{align*}
F(x) &= (2p-1)! f(x) = G(x) H(x)\\
G(x) &= \prod_{j=1-p}^{p} (x-j) \\
H(x) &= \sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\end{align*}
So the initial question reduces to proving that $F(1-x) + F(x) = (2p-1)!$
Next, playing with the indices of the product I have been able to show that $G(1-x) = G(x)$.
So that now I only need to show $G(x) \left( H(1-x) + H(x)\right) = (2p-1)!$
EDIT:
I have just been able to simplify $H(1-x) + H(x)$ as follows :
\begin{align*}
H(x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\\
&=\sum_{s=p}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p+k
\end{align*}
And
\begin{align*}
H(1-x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p+k}
\\
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p-1-k}
&& \text{using } \binom{2p-1}{p+k} = \binom{2p-1}{p-1-k}
\\
&=\sum_{s=0}^{p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p-1-k
\end{align*}
So finally
\begin{align*}
H(1-x) + H(x)
&=\sum_{s=0}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
\end{align*}
EDIT2:
I have just made the same work on $G(x)$ as follows
\begin{align*}
G(x)
&=\prod_{j=1-p}^{p} (x-j) \\
&=\prod_{s=0}^{2p-1} (x-p+s)
&& \text{using a change of index } s=p-i
\end{align*}
Now I think the constant $(2p-1)!$ in the relation $F(1-x)+F(x)=(2p-1)!$ comes from the binomial factor in $H(1-x) + H(x)$ but I don't see how to prove that...
Any lead ?
| I have actually found a way to solve the question from EDIT1 and EDIT2
So
\begin{align*}
F(1-x) + F(x)
&= G(x) \Big( H(1-x) + H(x) \Big) \\
&= \Bigg(\prod_{s=0}^{2p-1}(x-p+s)\Bigg)\Bigg(\sum_{k=0}^{2p-1}\frac{(-1)^{k}}{x-p+k}\binom{2p-1}{k}\Bigg)\\
&= \sum_{k=0}^{2p-1} (-1)^{k}\binom{2p-1}{k}\Bigg(\prod_{s=0, s \neq k}^{2p-1}(x-p+s)\Bigg)
\end{align*}
This the (2p-1)-th order finite difference scheme with step $h=1$ applied to the function
\begin{align*}
g(x) = \Bigg( \prod_{s=0, s \neq k}^{2p-1}(x-p) \Bigg) = (x-p)^{2p-1}
\end{align*}
So clearly this would mean that :
\begin{align*}
F(1-x) + F(x)
&= \Delta_{h}^{2p-1} (x-p)^{2p-1} \\
&= \frac{d^{2p-1}}{dx^{2p-1}} (x-p)^{2p-1} \\
&= (2p-1)!
\end{align*}
Hot point to justify in the conclusion :
So we just need now to justify the passage from $\Delta_{h}^{2p-1}(x-p)^{2p-1}$ to $\frac{d^{2p-1}}{dx^{2p-1}} (x-p)^{2p-1}$.
Is there a property that says the nth-order finite difference scheme is exact on polynomials ?
Found a justification here :
https://hal.archives-ouvertes.fr/hal-01350976/file/on_the_link_between_finite_difference_and_derivative_of_polynomials_kolosov_petro_2017.pdf
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $\tan(\alpha/2)=(1-\cos\alpha)/\sin \alpha, \,?$ Is it true that
$$\bbox[5px,border:2px solid #138D75]{\tan\left(\frac {\alpha}2\right)=\frac{1-\cos\alpha}{\sin \alpha}, \quad ?} \tag1$$
My solution:
First, we note that the expression $\tan \left ( \frac{\alpha}{2} \right )$ makes sense when $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$: henceforth, we will assume that this condition is met.
By definition of tangent function, we have: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right )}$$
When $\alpha \neq 2k \pi, k \in \mathbb{Z}$, we can multiply numerator and denominator by $\sin \left ( \frac{\alpha}{2} \right )$, because this quantity is non-zero. We therefore get:
$$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin^2 \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right )}. $$
For what we saw earlier: $$\sin^2 \left ( \frac{\alpha}{2} \right ) = \frac{1 - \cos \alpha}{2}$$and instead, applying the sine duplication formula: $$\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right ) = \frac{1}{2} \sin \alpha$$
In conclusion we get the following formula: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ 1 - \cos \alpha}{\sin \alpha}.$$
If in the expression of the tangent we had multiplied numerator and denominator by $\cos \left ( \frac{\alpha}{2} \right )$ then with steps similar to what we have done we get another formula:$$\bbox[5px,border:2px solid #118D45]{\tan \left ( \frac{\alpha}{2} \right ) = \frac{\sin \alpha}{1 + \cos \alpha}}. \tag 2 $$
We note that it always holds $\cos \left ( \frac{\alpha}{2} \right ) \neq 0$ since we have imposed $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$ (and this entitles us to multiply numerator and denominator by this term).
Is there a faster and more immediate proof of the $(1)$ or $(2)$?
| We could multiply both sides by $\sinα$
Then on the LHS use the double angle for $\sinα = 2\sin(α/2) \cos(α/2)$ and put $\tan(α/2)=\sin(α/2)/\cos(α/2)$ then the LHS becomes $2\sin^2(α/2)$.
The RHS is the same as $1-\cosα = 1-(1-2\sin^2(α/2))$ from a double angle formula for $\cosα$, so it's true.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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There's 8 black balls and 7 white balls. 3 of the balls are drawn at random. Probability of drawing 2 of one color and 1 of the other color? A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?
Here's what I tried:
Case 1: 2 black balls and 1 white ball
8/15 * 7/14 * 7/13 = 392/2730 = 28/195
Case 2: 2 white balls, 1 black ball
7/15 * 6/14 * 8/13 = 336/2730 = 24/195
28/195 + 24/195 = 52/195 = 4/15
My teacher said it was wrong and I don't get why. She didn't give me the solutions either and just told me the answer was 4/5. Can someone tell me what I'm doing wrong? Thanks!
| Case $1$: $2$ black balls and $1$ white ball. The correct probability will be
$P(2B,1W) = \displaystyle \frac{ 3 \cdot8 \cdot 7 \cdot 7}{15 \cdot 14 \cdot 13} = \frac{28}{65}$.
$\big[\text{It is basically } \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{7}{15} \cdot \frac{8}{14} \cdot \frac{7}{13}\big]$
I would suggest think as selecting $2$ black balls out of $8$ and $1$ white ball out of $7$ vs. selecting any $3$ balls out of $15$ which can be written as -
$P(2B,1W) = \displaystyle {8 \choose 2}{7 \choose 1} / {15 \choose 3}$
Case $2$: $2$ white balls, $1$ black ball
$P(2W,1B) = \displaystyle {8 \choose 1}{7 \choose 2} / {15 \choose 3} = \frac{24}{65}$.
Adding them, we get probability as $\displaystyle \frac{4}{5}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity
$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$
I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.
The answers for $c$ and $d$ are $c=29$, $d=-3$.
| $4x^3 - 24x^2+48x + d = 4(x^3 - 6x^2 + 12x - 8) + 32+ d = 4(x-2)^3 + 32+ d\implies a = 4, b = -2, c = 32+d$. Note that there are more than one right answer for this. You can take for example $c = 3, d = -29$ or $c = 0, d = -32$ as they are just two answers among many !
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Induction for binomial coefficients I would like some help to prove the following equality :
$$\sum_{i=0}^n \binom{n}i^2=\binom{2n}n$$
I wanted to do a proof by induction :
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=1}^{n+1} \binom{n+1}i^2=1 + \sum_{i=0}^{n} \binom{n+1}{i+1}^2=1+\sum_{i=0}^{n} \bigg(\binom{n+1}i+\binom{n}{i+1}\bigg)^2$$
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=0}^{n} \bigg(\binom{n}i^2+2\binom{n}i\binom{n}{i+1}+\binom{n}{i+1}^2\bigg)=1+\binom{2n}n+ \sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1} +(\sum_{i=0}^n \binom{n}i^2-1) $$
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=2\binom{2n}n+\sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1}=2\bigg(\sum_{i=0}^n \binom{n}i^2(1+\frac{n-i}{i+1})\bigg)=2(n+1)\bigg(\sum_{i=0}^n \binom{n}i^2\frac{1}{i+1}\bigg)$$
But now I'm stuck.
| One way to proceed is to prove a more general identity by induction, and then deduce the identity in the problem statement as a corollary.
The more general identity is called "Vandermonde's Identity":
$$\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} = \binom{m+n}{k} \tag{1}$$
for non-negative integers $k$, $m$ and $n$. We give a proof by induction over $n$. The case $n=0$ reduces to $\binom{m}{k} = \binom{m}{k}$. Now suppose $(1)$ holds for some $n \ge 0$. Then the $n+1$ case is
$$\begin{align}
\sum_{i=0}^k \binom{m}{i} \binom{n+1}{k-i} &= \sum_{i=0}^k \binom{m}{i} \left( \binom{n}{k-i-1} + \binom{n}{k-i} \right) \tag{2}\\
&=\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i-1} + \sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} \\
&=\binom{m+n}{k-1} + \binom{m+n}{k} \tag{3} \\
&= \binom{m+n+1}{k} \tag{4}
\end{align}$$
This completes the proof by induction. At steps $(2)$ and $(4)$ we used the identity
$$\binom{n}{m} = \binom{n-1}{m-1} + \binom{n-1}{m}$$ and at step $(3)$ we used the inductive hypothesis.
To prove $$\sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$$
from $(1)$, take the special case $k=n$, $m=n$, and apply the identity
$$\binom{n}{i} = \binom{n}{n-i}$$
| {
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Finding the integral of $\frac{1}{x(x^n+a^n)}$ I was working on the problems in Mathematical Methods for Physics and Engineering by Riley,Hobson & Bence.
In Problem 2.34 (d) I'm supposed to find this integral: $$J=\int\frac{dx}{x(x^n+a^n)}.$$
I used partial fractions and arrived at the form
$$J=\frac{1}{a^n}\left[\log x-\int \frac{dx}{x^n+a^n}\right]$$
and now I'm stuck, I don't know how to integrate $1/(x^n+a^n)$.
| We seek a decomposition of the form
$$\frac{1}{x(x^n + a^n)} = \frac{A}{x} + \frac{Bx^{n-1}}{x^n + a^n} = \frac{(A+B)x^n + Aa^n}{x(x^n+a^n)}.$$ Hence the choice $A = a^{-n}$, $B = -A = -a^{-n}$, yields
$$\frac{1}{x(x^n+a^n)} = \frac{1}{a^n} \left(\frac{1}{x} - \frac{x^{n-1}}{x^n + a^n}\right).$$
The rest is straightforward:
$$\int \frac{dx}{x(x^n+a^n)} = \frac{1}{a^n} \left(\log |x| - \frac{1}{n} \int \frac{nx^{n-1}}{x^n + a^n} \, dx \right) = \frac{1}{a^n} \left( \log |x| -\frac{1}{n} \log |x^n + a^n| \right) + C.$$
| {
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"source": "stackexchange",
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prove for bell number using induction on n hi guys I have to prove this equality $$B_n=e^{-1}\sum_{k=0}^{\infty}\frac{k^n}{k!},$$ that is called bell equality only using induction on $n$ . How can i do this? I have tried by substituting the recursive formula $\sum\limits_{k=0}^{n} \binom{n}{k} B_{k}$ in the first one. But I am completely lost with indexes.
I have already tried to search some informations on the internet but nothing using induction
| Using the Bell´s formula we have that $$B(n)=\sum\limits_{k=0}^nS(n,k).$$
This is the total number of ways to put $n$ balls in an arbitrary number of
boxes (no empty boxes remaining). To count them we look at the number of balls (at this parameter
we will call it $a$, and it can be from $0$ to $(n - 1)$) that accompany any ball (for example ball
number $1$) on your box. To do this:
*
*First we choose the $a$ companions of ball $1$ from among the $(n - 1)$ possible (which can be done in $\binom{n-1}{a}$ different ways).
*Then, for each way of selecting the companions, we count how many ways they can be entered
the remaining $(n − 1 − a)$ balls in an arbitrary number of boxes (we have called this $B(n−1−a)$)
$$\boxed{B(n)}=\sum\limits_{a=0}^{n-1}{\dbinom{n-1}{a}}\sum\limits_{k=1}^{n-1-a}{S(n-1-a)}=\boxed{\sum\limits_{a=0}^{n-1}{\dbinom{n-1}{a}B(n-1-a)}}$$
Now, if we define the $DOB$ function this way: $$DOB(n)=\sum\limits_{k=1}^{\infty}\dfrac{k^n}{n!},$$ we can see that
\begin{align*}
DOB(0)&=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots\\
DOB(1)&=\phantom{OO}\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!}+\dfrac{4}{4!}+\cdots\\
DOB(2)&=\phantom{OO}\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+\cdots\\
DOB(3)&=\phantom{OO}\dfrac{1^3}{1!}+\dfrac{2^3}{2!}+\dfrac{3^3}{3!}+\dfrac{4^3}{4!}+\cdots
\end{align*}
By combining these numbers we get that
\begin{gather*}
\color{red}{1}DOB(0)+\color{red}{2}DOB(1)+\color{red}{1}DOB(2)=1+\dfrac{4}{1!}+\dfrac{9}{2!}+\dfrac{27}{3!}\cdots=\dfrac{1^3}{1!}+\dfrac{2^3}{2!}+\dfrac{3^3}{3!}+\dfrac{4^3}{4!}+\cdots=DOB(3)\\
\color{red}{1}DOB(0)+\color{red}{3}DOB(1)+\color{red}{3}DOB(2)+\color{red}{1}DOB(3)=1+\dfrac{8}{1!}+\dfrac{27}{2!}+\dfrac{64}{3!}\cdots=\dfrac{1^4}{1!}+\dfrac{2^4}{2!}+\dfrac{3^4}{3!}+\dfrac{4^4}{4!}+\cdots=DOB(4)
\end{gather*}
What Dobinsky wanted to prove is that both sequences fulfill the same recurrence, and that have the following property:
$$DOB(n)=\sum\limits_{j=0}^{n-1}{\dbinom{n-1}{j}DOB}(j)$$
To prove this just use the binomial Theorem:
\begin{align*}
\boxed{\sum\limits_{j=0}^{n-1}{\dbinom{n-1}{j}DOB(j)}}=&\ e+\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}DOB(j)}=e+\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}\sum\limits_{k=1}^{\infty}{\dfrac{k^n}{n!}}}\\
=&\ e+\sum\limits_{k=1}^{\infty}{\dfrac{1}{k!}}\underbrace{\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}k^j}}_{=(k+1)^{n-1}-1}=\ e-\underbrace{\sum\limits_{k=1}^{\infty}{\dfrac{1}{k!}}}_{=e-1}+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^{n-1}}{k}}\\
=&1+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^{n-1}}{k!}=1+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^n}{(k+1)!}}=\sum\limits_{k=1}^{\infty}{\dfrac{j^n}{j!}}}=\boxed{DOB(j)}
\end{align*}
This tells us that:
\begin{equation}\label{Formula_Dobinsky}
eB(n)=DOB(n) \rightarrow \boxed{B(n)=\dfrac{1}{e}\sum\limits_{k=1}^{\infty}{\dfrac{k^n}{k!}}}\quad \text{for each }n=1,2,3\ldots
\end{equation}
This formula is way more friendly and computerwise more stable. To see that just check that
\begin{equation*}
B(10)=115975;\qquad\dfrac{1}{e}\sum\limits_{k=1}^{15}{\dfrac{k^{10}}{k!}}=115974.978
\end{equation*}
| {
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"question_score": "2",
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Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because
Although my answer is correct, I was apparently supposed to use some theory to answer this question.
I started by putting everything on common denominator and factoring out $n$:
$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$
then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$
$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$
$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$
$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$
$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$
$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$
to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,
and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.
Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?
| $$ \begin{align}f(n+1)-f(n)&=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac7{15}
\\&=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac7{15}\\&=\text{integer}+\frac15+\frac13+\frac 7{15}\end{align}$$
is an integer for all integers $n$ and so is $f(0)$. The claim follows by induction.
Here's a generalization for you: Suppose $p_1,p_2,\ldots,p_m$ are primes, $a_1,a_2,\ldots,a_m$ are integers and $c$ is real. Let
$$ f(n)=\frac{a_1n^{p_1}}{p_1}+\frac{a_2n^{p_2}}{p_2}+\cdots +\frac{a_mn^{p_m}}{p_m}+cn.$$
If $f(1)$ is an integer, then $f(n)$ is an integer for all integers $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$
So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?
| Hint: $\,a+b\mid a^3+b^3\Rightarrow 35\mid 16^3\!+\!19^3,\,17^3\!+\!18^3\Rightarrow 35\mid x.\,$ $x$ is even so $\,2\cdot 35\mid x$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Need help simplifying a set of equations (and understanding how to solve it) i have three algebraic expressions, each using the others. in these equations a, b, c and t are known and plugged in later:
$x = a^{-1}(t + y + z)$
$y = b^{-1}(t + x + z)$
$z = c^{-1}(t + x + y)$
i have managed to successfully solve the equations for two statements using substitution:
$x = a^{-1}(t + y)$
$y = b^{-1}(t + x)$
which substitutes and simplifies as:
$x = a^{-1}t + a^{-1}b^{-1}t + a^{-1}b^{-1}x$
$x - a^{-1}b^{-1}x = a^{-1}t + a^{-1}b^{-1}t$
$x(1 - a^{-1}b^{-1}) = a^{-1}t(b^{-1} + 1)$
therefore:
$x = a^{-1}t(1+b^{-1}) / (1 - a^{-1}b^{-1})$
$y = b^{-1}t(1+a^{-1}) / (1 - a^{-1}b^{-1})$
but introducing a third term has me stumped, as i keep getting a nonsense result when substituting leaving me with $x = x$ when i simplify the equation. (this is obviously correct, but also useless.) I'm hoping that someone would be so kind as to explain how i go about substituting more than one term and solving x, y and z in terms of t, a, b and c. does the same principle apply as with two terms?
for completeness, here is my (incorrect) work:
$x = a^{-1}t + a^{-1}y + a^{-1}z$
$a^{-1}z = x - (a^{-1}t + a^{-1}y)$
$a^{-1}z = x - a^{-1}(t + y)$
$z = ax - (t + y)$ (multiply both sides by a)
therefore, to find y i substitute z:
$x = a^{-1}t + a^{-1}y + a^{-1}(ax - (t + y))$
$x = a^{-1}t + a^{-1}y + x - a^{-1}t - a^{-1}y$
$x = (a^{-1}t - a^{-1}t) + (a^{-1}y - a^{-1}y) + x$
$x = 0 + 0 + x$
$x = x$
please forgive me if my terminology isn't correct. i don't exactly have a maths background and english isn't my first language.
| You can write your system as
$$\left\{\begin{align}
ax = (y+z) + t\\
by = (x+z) + t\\
cz = (x+y) + t
\end{align}\right.$$
and then in matrix form you'd get
$$\begin{align}
&\pmatrix{a&0&0\\0&b&0\\0&0&c}\pmatrix{x\\y\\z}
=
\pmatrix{0&1&1\\1&0&1\\1&1&0}\pmatrix{x\\y\\z}
+
t\pmatrix{1\\1\\1}\\
\implies
&\underbrace{\pmatrix{a&-1&-1\\-1&b&-1\\-1&-1&c}}_{M}\pmatrix{x\\y\\z}
=
t\pmatrix{1\\1\\1}.
\end{align}$$
Now, $M$ has determinant $\det(M) = abc - a -b -c -2$, so provided $\det(M)\neq 0$ your system admits a unique solution.
In this case, we have
$$M^{-1} = \frac1{\det(M)}\pmatrix{bc-1&c+1&b+1\\c+1&ac-1&a+1\\b+1&a+1&ab-1}$$
and
$$\pmatrix{x\\y\\z} = tM^{-1}\pmatrix{1\\1\\1}.$$
We can explicitly compute and obtain
$$\begin{align}
\pmatrix{x\\y\\z}
&= \frac t{abc-a-b-c-2}\pmatrix{b+c+bc+1\\a+c+ac+1\\a+b+ab+1}
\\
&= \frac t{abc-a-b-c-2}\pmatrix{(b+1)(c+1)\\(a+1)(c+1)\\(a+1)(b+1)}
.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the sum of all irrational roots of $4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$
Calculate the sum of all irrational roots of
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$
I'm not even sure how to begin here, I tried raising it to the power of three, tried writing $8x^{3}+ 3$ with $x^{3}+ y^{3}= \left ( x+ y \right )\left ( x^{2}- xy+ y^{2} \right ),$ but have had no meaningful results.
|
*
*If we assume "cube root" is the real-valued root, then we have
$$\begin{cases}4\sqrt[3]{8x-3}-8x^3-3=0 \\ x\in\mathbb R \end{cases} \Longleftrightarrow 512 x^9 + 576 x^6 + 216 x^3 - 512 x + 219=0$$
$$\Longleftrightarrow (2 x - 1) (4 x^2 + 2 x - 3) (64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73)=0$$
Then using standard calculus tools (but, only algebraic ways are also possible ) we get
$$64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73 > 0$$
for all $x\in\mathbb R$.
So, we have only $2$ irrational roots:
$$x_1+x_2=\frac{-2}{4}=- \frac 12$$
Then, answer is equal to $- \dfrac 12.$
*
*If we use the principal root instead of "real-valued cube root", then we have
$$4\sqrt[3]{8x-3}-8x^3-3=0 \Longleftrightarrow \begin{cases} 512 x^9 + 576 x^6 + 216 x^3 - 512 x + 219=0 \\ 8x-3≥0 \end{cases} $$
$$\Longleftrightarrow \begin{cases} (2 x - 1) (4 x^2 + 2 x - 3) (64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73)=0 \\ x≥\frac 38\end{cases}$$
The last result implies, we have only $1$ irrational root:
$$x=\frac {-1+\sqrt {13}}{4}$$
So, answer is equal to $\dfrac {\sqrt {13}-1}{4}.$
| {
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"url": "https://math.stackexchange.com/questions/4052833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability that the product of two numbers in $\{0,1,...,10^{n - 1}\}$ contains $2n - k$ digits, $k = 0,1,...$ How to find the probability that the product of two numbers in ${0,1,...,10^n - 1}$ contains $2n-k$ digits, $k = 0,1,...$? As I understand, the sum of all the combinations of two numbers is equal to $10^{2n}$. But I can't understand how to find the sum of all the positive combinations. I understand that the product of these numbers is in $10^{2n-k-1} \le x*y < 10^{2n-k}$. But it doesn't help me. Explain me please, how I can find the sum of all the positive combinations.
| This will give an approximate answer, which will be very close. The region of interest is the square $[0,10^n-1] \times [0,10^n-1]$ in the $xy$ plane. The numbers with $2n-k$ digits are the numbers from $10^{2n-k-1}$ to $10^{2n-k}-1$. Instead of counting lattice points, we can view $x$ and $y$ as continuous variables and find the area that corresponds to this range of products. This is where the approximation comes from. It is the area between the two hyperbolas $xy=10^{2n-k-1}$ and $10^{2n-k}$. Do the integral and you are done.
Added: let us do an example with $n=7, k=4$ so we are looking for factors less than or equal to $10^6$ that multiply to an eight digit number. Such numbers are in the range $[10^7,10^8)$ Our square is $10^6 \times 10^6$ for a total area of $10^{12}$. The lower boundary of the area we are looking at is $xy=10^7$ and the upper boundary is $xy=10^8$. The lower hyperbola hits the edges of the square at $(10,10^6)$ and $(10^6,10)$. The area below it and inside the square is $$10 \cdot 10^6+\int_{10}^{10^6}\frac {10^7}x\ dx=10^7+10^7 \log(10^5) \approx 1.2513\cdot 10^8$$
where the first term is the rectangle to the left of the intersection with the top of the square. Similarly the upper hyperbola hits the square at $(100,10^6)$ and $(10^6,100)$ so the area below it is
$$100 \cdot 10^6+\int_{100}^{10^6}\frac {10^8}x\ dx=10^8+10^8 \log(10^4) \approx 10.2103\cdot 10^8$$
The difference of areas is then $8.959 \cdot 10^8$ That is approximately the number of pairs that have a product with eight digits.
| {
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"url": "https://math.stackexchange.com/questions/4054465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the system of equations: $32y+32x^3=6x+17$, $16z+32y^3=6y+9$, $8x+32z^3=6z+5$ where $x,y,z\in \mathbb{R}$ Solve the system of equations: $$\begin{cases} 32y+32x^3=6x+17 \\16z+32y^3=6y+9 \\8x+32z^3=6z+5 \end{cases}$$ where $x,y,z\in \mathbb{R}$ (Bulgaria 1960)
I attempted to solve this question as follows:
$32y+32x^3=6x+17$
$y=-x^3+\frac{6x}{32}+\frac{17}{32}$
$y-\frac{1}{2}=-x^3+\frac{3x}{16}+\frac{17}{32}-\frac{1}{2}$
Here it started getting very complex, and hence I don't think it can be solved this way. After this I tried doing something similar with the other two equations, but once again it was ending up way too complex. It is obvious that the solution is $x=y=z=\frac{1}{2}$, but I can't manage to prove it. Could you please explain to me how to solve this question?
| Just for simplicity's sake, make a change of variables $$(a,b,c) = (2x,2y,2z)$$
We have the system of equations
\begin{align}
16b + 4a^3 &= 3a+17 \\
8c + 4b^3 &= 3b+9 \\
4a + 4c^3 &= 3c+5 \\
\end{align}
or
\begin{align}
16(b-1) &= - 4a^3+3a+1 \\
8(c-1) &= - 4b^3+3b+1 \\
4(a-1) &=-4c^3+ 3c+1 \\
\end{align}
or
\begin{align}
16(b-1) &= - (a-1)(4a^2+4a+1) \tag{1}\\
8(c-1) &= - (b-1)(4b^2+4b+1) \tag{2}\\
4(a-1) &=-(c-1)(4c^2+4c+1) \tag{3}\\
\end{align}
Suppose $b >1$ then from $(1)$: $a<1$ $\implies$ from $(3)$: $c>1$ $\implies$ from $(2)$: $b>1$ $\implies$ contracdiction!
Same for $b <1$, we have also a contradiction.
For $b=1$, we have $c =1$ and $a =1$.
Conclusion: the system of equations has a unique solution $(a,b,c) = (1,1,1)$ or $$(x,y,z) = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2} \right)$$
| {
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"url": "https://math.stackexchange.com/questions/4054989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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please help me finish the proof of this equation problem: prove that $ 2(\sqrt{2}+\sqrt{6})\sqrt[4]{3}\int_{0}^{1}\frac{1}{\sqrt{x}\sqrt{1-x}\sqrt{(\sqrt{2}+\sqrt{6})^{2}-x}} dx = \beta (\frac{1}{6},\frac{1}{3}) $
My attempt:
use the relationship between hypergeometric series and elliptic integral, we have
$F_{1}(\alpha,\beta,\gamma,x) = \frac{1}{\beta (\beta,\gamma-\beta)}\int_{0}^{1}t^{\beta-1}(1-t)^{\gamma -\beta -1}(1-tx)^{-\alpha} dt $
(here $ \beta() $ refers to the Beta function)
then
$ LHS = 2\sqrt[4]{3}\int_{0}^{1}x^{\frac{1}{2}-1}(1-x)^{1-\frac{1}{2}-1}(1-(\sqrt{2}+\sqrt{6})^{-2}x)^{-\frac{1}{2}} dx $
$ = 2\sqrt[4]{3}\beta (\frac{1}{2}, \frac{1}{2}) F_{1}(\frac{1}{2}, \frac{1}{2},1,(\sqrt{2}+\sqrt{6})^{-2}) $
$ = \pi \cdot 2\cdot \sqrt[4]{3} \cdot \frac{2}{\pi} \kappa ((\frac{1}{\sqrt{2}+\sqrt{6}}) $
$ = 4 \sqrt[4]{3} \kappa ((\frac{1}{\sqrt{2}+\sqrt{6}})$
(here we use $ \kappa (k) = \frac{\pi}{2}F_{1}(\frac{1}{2}, \frac{1}{2},1, k^{2}$)
Thus, we only need to prove the following equation:
$ \kappa (\frac{1}{\sqrt{2}+\sqrt{6}})=\frac{1}{4\sqrt[4]{3}}\beta (\frac{1}{6},\frac{1}{3}) $
I got stuck here, can anyone help me finish this?
note: $ \kappa () $ is the complete elliptic integral of the first kind
| $\frac1{\sqrt{2} + \sqrt{6}} = \lambda^*(3),$ where $\lambda^*$ is defined here.
Now, this seems to imply that
$$K(\frac1{\sqrt{2} + \sqrt{6}}) = \frac{3^{\frac14} \Gamma^3(1/3)}{2^{7/3}\pi},$$ as per this page.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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