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Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $ Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $
Try:: Let $x = 2009$, Then expression convert into $(x+4)^4-4(x+2)^4+6x^4-4(x-2)^4+(x-4)^4$
$\left\{(x+4)^4+(x-4)^4\right\}-4\left\{(x+2)^4+(x-2)^4\right\}+6x^4$
But This is very Complicated for calculation.
can anyone explain me better idea for that expression
If yes plz explain me
Thanks
| Let's calculate the coefficients at powers of $x$.
First, it's clear the coefficient at $x^4$ is zero ($1+1-4-4+6=0$).
Next, from your grouping of terms it's obvious that coefficients at $x^3$ and $x$ are zero, too. Or from another point of view, our function is even.
Then again, the coefficient at $x^2$ is $2\cdot 6\cdot 4^2 - 2\cdot 4\cdot 6\cdot 2^2=0$.
Finally, we have the coefficient at $x^0$, which is $2\cdot 4^4-4\cdot 2\cdot2^4 = 3\cdot 128 = 384$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving a formula about binomial coefficients I found the following formula in a book without any proof:
$$\sum_{j=k}^{\lfloor\frac n2\rfloor}{\binom{n}{2j}}{\binom{j}{k}}=\frac{n}{n-k}\cdot2^{n-2k-1}{\binom{n-k}{k}}$$
where $n$ is a natural number and $k$ is an integer which satisfies $0\le k \le\frac n2$.
I've tried to prove this, but I'm facing difficulty. Could you show me how to prove this?
| We will use induction to prove the identity (equivalent to the one given) that
1) $\displaystyle\sum_{j\ge0}\binom{n}{2j}\binom{j}{k}=2^{n-2k-1}\bigg[\binom{n-k}{k}+\binom{n-k-1}{k-1}\bigg]$ $\;\;$for $0\le k\le\frac{n}{2}$ and the identity
2) $\displaystyle\sum_{j\ge0}\binom{n}{2j+1}\binom{j}{k}=2^{n-2k-1}\binom{n-k-1}{k}$ $\;\;$for $0\le k\le\frac{n-1}{2}$.
If $n=1$, $k=0$ and both sides of both identities are 1;
so assume that both identities are valid for some $n\in\mathbb{N}$.
1) $\displaystyle\sum_{j\ge0}\binom{n+1}{2j}\binom{j}{k}=\sum_{j\ge0}\bigg[\binom{n}{2j}+\binom{n}{2j-1}\bigg]\binom{j}{k}=\sum_{j\ge0}\binom{n}{2j}\binom{j}{k}+\sum_{j\ge0}\binom{n}{2j-1}\binom{j}{k}$
$\;\;\;\displaystyle=\sum_{j\ge0}\binom{n}{2j}\binom{j}{k}+\sum_{l\ge0}\binom{n}{2l+1}\binom{l+1}{k}$
$\;\;\;\displaystyle=\sum_{j\ge0}\binom{n}{2j}\binom{j}{k}+\sum_{l\ge0}\binom{n}{2l+1}\bigg[\binom{l}{k}+\binom{l}{k-1}\bigg]$
$\;\;\;\displaystyle=\sum_{j\ge0}\binom{n}{2j}\binom{j}{k}+\sum_{l\ge0}\binom{n}{2l+1}\binom{l}{k}+\sum_{l\ge0}\binom{n}{2l+1}\binom{l}{k-1}$
$\;\;\;\displaystyle=2^{n-2k-1}\bigg[\binom{n-k}{k}+\binom{n-k-1}{k-1}\bigg]+2^{n-2k-1}\binom{n-k-1}{k}+2^{n-2k+1}\binom{n-k}{k-1}$
$\;\;\;\displaystyle=2^{n-2k-1}\bigg[\binom{n-k}{k}+\binom{n-k-1}{k-1}+\binom{n-k-1}{k}+4\binom{n-k}{k-1}\bigg]$
$\;\;\;\displaystyle=2^{n-2k-1}\bigg[\binom{n-k+1}{k}+\binom{n-k}{k}+\binom{n-k}{k-1}+2\binom{n-k}{k-1}\bigg]$
$\;\;\;\displaystyle=2^{n-2k-1}\bigg[2\binom{n-k+1}{k}+2\binom{n-k}{k-1}\bigg]=2^{n-2k}\bigg[\binom{n-k+1}{k}+\binom{n-k}{k-1}\bigg]$,
$\;\;\;\;$ so identity 1) holds for $n+1$.
2) $\displaystyle\sum_{j\ge0}\binom{n+1}{2j+1}\binom{j}{k}=\sum_{j\ge0}\bigg[\binom{n}{2j+1}+\binom{n}{2j}\bigg]\binom{j}{k}=\sum_{j\ge0}\binom{n}{2j+1}\binom{j}{k}+\sum_{j\ge0}\binom{n}{2j}\binom{j}{k}$
$\;\;\;\displaystyle=2^{n-2k-1}\binom{n-k-1}{k}+2^{n-2k-1}\bigg[\binom{n-k}{k}+\binom{n-k-1}{k-1}\bigg]$
$\;\;\;\displaystyle=2^{n-2k-1}\bigg[\binom{n-k}{k}+\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\bigg]=2^{n-2k-1}\bigg[\binom{n-k}{k}+\binom{n-k}{k}\bigg]$
$\;\;\;\displaystyle=2^{n-2k}\binom{n-k}{k}$,
$\;\;\;$so identity 2) holds for $n+1$.
| {
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Prove $2^b-1$ does not divide $2^a + 1$ for $a,b>2$ I'm trying to prove $2^b-1$ does not divide $2^a + 1$ for $a,b>2$. Can someone give a hint in the right direction for this?
| Assume otherwise and let $a$ be the nonnegative integer such that $2^b-1\mid 2^a+1$ for some $b>2$.
If $2^a+1=k(2^b-1)$ with $k\ge 1$, then $$2^{a-b}+1 = 2^a+1-2^{a-b}(2^b-1)=(k-2^{a-b})(2^b-1).$$
If $a\ge b$, this shows that $2^b-1\mid 2^{a-b}+1$, contradicting minimality of $a$.
Therefore $a\le b-1$. But then we have $$2^b-1=2^{b-1}+2^{b-1}-1\ge 2^{b-1}+3>2^a+1$$ contradicting $2^b-1\mid 2^a+1$.
Note that $b=2$ allows us to find $2^2-1=3\mid 9=2^3+1$ or $33=2^5+1$ and so on.
| {
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Solve the simultaneous equations $x + \frac{3x-y}{x^2+y^2} = 3 $, $ y – \frac{x+3y}{x^2+y^2} = 0$ Find all solution in $\mathbb{R}$ for the following system of equations:
\begin{cases} x + \frac{3x-y}{x^2+y^2} = 3 \\ y – \frac{x+3y}{x^2+y^2} = 0 \end{cases}
I've tried few method, but none bring a success.
First I tried to substitute, and tried to express $y$ in terms of $x$, but I wasn't able to do that. I tried to use $y = \frac{x+3y}{x^2+y^2}$, but as you guess that substitution made the things even more complicated.
Next I tried to multiply the second equation with $\frac{3x-y}{x+3y}$, assuming it's not zero. And I end up with:
\begin{cases} x + \frac{3x-y}{x^2+y^2} = 3 \\ y\left(\frac{3x-y}{x+3y}\right) – \frac{x+3y}{x^2+y^2} = 0 \end{cases}
Now I add the together and I end up with:
$$x + y\left(\frac{3x-y}{x+3y}\right) = 3$$
But solution to this equation aren't solution to the system, because there are infinite amount of them.
Next I tried just to add them and I end up with:
$$x+y \frac{2x-3y}{x^2+y^2} = 3$$
But again I faced the same problem, this equation has an infinite amount of solution, unlike the system of equations.
So how can I solve this system of equations?
| Complexify. Let $z = x+iy$. Then for $x^2+y^2 \neq 0$:
$$\begin{align}
&&\left(x + \frac{3x-y}{x^2+y^2}\right) + i\left(y - \frac{x+3y}{x^2+y^2}\right) &= (3+i0)\\
&\iff& (x+iy) + \frac{(3-i)(x-iy)}{x^2+y^2} &= 3\\
&\iff& z + \frac{3-i}{z} &= 3\\
&\iff& z^2 - 3z + (3-i) &= 0\\
&\iff& \left(z-\frac32\right)^2 &= i - \frac34\\
&\iff& z &= \frac{3 \pm \sqrt{4i-3}}{2}.
\end{align}$$
Now, $(1+2i)^2 = -3+4i$, so the solutions are
$$z = \frac{3 \pm (1+2i)}{2},$$
either $z = 2+i$ or $z = 1-i$, which, in real form, yields $(x,y) = (2,1)$ or $(x,y) = (1,-1)$.
| {
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Finding the equation of a circle given two points and the radius Can't seem to figure this out - the question is:
There are exactly two circles of radius $r = \sqrt{5}$ through the points $(6,3)$ and $(7,2)$. Find the equations of both circles.
I was thinking that I would find the equation of the line passing through these two points which would give me a chord on the circle. I could then find a line perpendicular to this by taking the negative reciprocal. This perpendicular bisector of the chord would pass through the center of the circle (which I'm assuming I need to find).
Am I making this more complicated than it actually is...? I don't know where to go from here
| Let A be (6,3) and B be (7,2).
Gradient of AB = $\frac { 2-3 }{ 7-6 }$ = -1
So gradient of perpendicular bisector of AB = $-\frac { 1 }{ 1 }$ = 1
Midpoint of AB is ($\frac { 6+7 }{ 2 }$, $\frac { 3+2 }{ 2 }$), which is ($\frac { 13 }{ 2 }$, $\frac { 5 }{ 2 }$).
Hence, the equation of the perpendicular bisector is $y-\frac { 5 }{ 2 }= 1(x-\frac { 13 }{ 2 })$
The equation can be written as $y=x-4$.
Let the centre of the circles be (x,y).
Substitute $y=x-4$ as the centre lies on the perpendicular bisector.
Then the centre becomes $(x,x-4)$.
Take the distance between any point on the circle and the centre, which is the radius.
${ (x-6) }^{ 2 }+{ (x-4-3) }^{ 2 }={ \sqrt { 5 } }^{ 2 }\\ { (x-6) }^{ 2 }+{ (x-7) }^{ 2 }=5\\ { x }^{ 2 }-12x+36+{ x }^{ 2 }-14x+49=5\\ 2{ x }^{ 2 }-26x+80=0\\ (x-8)(x-5)=0\\ So\quad x=5\quad or\quad x=8\\ As\quad y=x-4,\\ y=1\quad or\quad y=4\\ \\ $
Hence, the possible coordinates of centres are (5,1) and (8,4).
The equations are
${ (x-5) }^{ 2 }+{ (y-1) }^{ 2 }=5$
and ${ (x-8) }^{ 2 }+{ (y-4) }^{ 2 }=5\\$
| {
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Prove that $a_n=na/(1+n^2 b)$ is strictly decreasing given a, b >0
I have tried proving this directly, by setting
$\frac{na}{1+n^2b} > \frac{(n+1)a}{1+(n+1)^2b}$
and end up with
$\frac{na+n^3ab+2n^2ab+nab}{(1+n^b)(1+n^2b+2bn+b)} > \frac{na+a+n^3ab+n^2ab}{(1+n^b)(1+n^2b+2bn+b)}$
I simplify that down to...
$\frac{n^2b+nb}{(1+n^2b)(1+n^2b+2bn+b)} > \frac{1}{(1+n^2b)(1+n^2b+2bn+b)} $
But I have no particular reason to believe that the inequality is true. Am I making an algebraic error? Is there a better way to do this? Is there a way to do this by induction?
Thanks!
| Just rewrite it as
$$
a_n = \frac{a}{\frac{1}{n} + nb}.
$$
Since $\frac{1}{n} + nb$ is increasing for sufficiently large $n$ and for $b > 0$, it shows $a_n$ must be decreasing for sufficiently large $n$ and for $a > 0$.
To be rigorous we need to say precisely what sufficiently large means and prove it. By sufficiently large, we mean that there is a $B$ such that $n > B$ implies $\frac{1}{n + 1} + (n + 1)b > \frac{1}{n} + nb$. But this is equivalent to $(n + 1)b - nb > \frac{1}{n} - \frac{1}{n + 1} \iff b > \frac{1}{n(n + 1)}$. But this condition is fulfilled if we have the stronger condition $b > \frac{1}{n^2} \implies n > \frac{1}{\sqrt{b}}$. Hence a $B$ that works is $B = \frac{1}{\sqrt{b}}$.
| {
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Findinf volume enclosed using triple integrals I have to find the volume cut off from the paraboloid $4z=x^2+y^2$ by the plane $z=4$
I understand that the paraboloid will have its vertex at the orgin and its axis as the $z-axis$. now, if it is cut off by the plane $z=4$, i get the following limits for the triple intgral
$z$ from $0$ to $4$
$x$ from $-\sqrt{16-y^2}$ to $\sqrt{16-y^2}$
$y$ from $-16$ to $16$
However, this is double the correct answer. Where am i going wrong ?
| We can interpret this volume as $$\int_0^4 \text{area of disk at height } z\ dz.$$ More formally, we convert to a cylindrical coordinate system, $$\int_0^4 \underbrace{\int_0^{2\pi} \int_0^{\sqrt{4z}} r\ dr\ d\theta}_{\text{area of disk}}\ dz.$$
The area of a disk is $\pi r^2$ where $r$ is the radius, in this case $r=\sqrt{4z}$, so the area is $4\pi z$, and the integral becomes $$\int_0^4 4\pi z\ dz=\left[2\pi z\right]_0^4=32\pi.$$
Alternatively, we can find the volume of solid of revolution via $$2\pi \int_0^{4} x|f(x)-g(x)|\ dx$$ where $f(x)=4$ and $g(x)=\tfrac{1}{4}x^2$. This gives the volume
\begin{align*}
2\pi \int_0^{4} 4x-\tfrac{1}{4}x^3\ dx &= 2\pi \left[2x^2-\tfrac{1}{16} x^4\right]_0^{4} \\
&= 2\pi(2 \times 4^2-\tfrac{4^4}{16}) \\
&= 32\pi.
\end{align*}
| {
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Which of the following are complete residue systems modulo $11$? Which of the following are complete residue systems modulo 11?
$(a)\quad 0,1,2,4,8,16,32,64,128,256,512$
$(b)\quad 1,3,5,7,9,11,13,15,17,19,21$
$(c)\quad 2,4,6,8,10,12,14,16,17,20,22$
$(d)\quad -5,-4,-3,-2,-1,0,1,2,3,4,5$
I have the answer which is that they are all complete residue systems modulo 11. However, I do not fully understand the definition of residue system or how to determine what sets make up a complete residue system?
| The set of integers {0, 1, 2, ..., n - 1} is called the least residue system modulo n. Any set of n integers, $\bf no\ two\ of\ which\ are\ congruent\ modulo\ n$, is called a $\bf complete\ residue\ system\ modulo\ n$.
http://en.wikipedia.org/wiki/Complete_residue_system_modulo_m#Residue_systems
{0,1,2,3,4,5,6,7,8,9,10} is a complete residue system modulo 11.
Since $1 \equiv 12 \pmod{11}$,$3 \equiv 14 \pmod{11}$, ... , $10 \equiv 21 \pmod{11}$.
After dividing 11 for each of them, the residues are:
(a)0,1,2,4,8,5,10,9,7,3,6;
(b)1,3,5,7,9,0,2,4,6,8,10;
(c)2,4,6,8,10,1,3,5,6,9,0; (c isn't)
(d)6,7,8,9,10,0,1,2,3,4,5.
| {
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Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:
$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$
(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor
| This is a sequel to my previous work. The aim is to get the relation $(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$ directly rather than just verifying it. In order to do so, I have to re-arrange the original question as
“If $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor $x^2+px+1$, then $(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$.”
[I can do that because the second part of the original question has been proved independently.]
For some polynomials P(x) and Q(x) of degree 3, we let
$f(x) = ax^5 + bx^2 + c = (x^2 + px + 1)P(x)…………..(1)$
$g(x) = cx^5 + bx^3 + a = (x^2 + px + 1)Q(x)…………..(2)$
Let $m$ and $n$ (where $m.n = 1$) be the roots of $x^2+ px +1= 0$, then m is also a root of $f(x) = 0$ and of $g(x) = 0$. Thus,
$ am^5 + bm^2 + c = 0…..........(1.1)$
$ cm^5 + bm^3 + a = 0…......….(2.1)$
Do [a*(2.1) – c*(1.1)] and get $abm^3 – cbm^2 + (a^2 – c^2) = 0$
This means m is also a root of $F(x) = abx^3 – cbx^2 + (a^2 – c^2) = 0$ if we define
$F(x) = a*g(x) – c*f(x)$
On one hand $F(x) = abx^3 – cbx^2 + (a^2 – c^2)$
On the other hand $F(x) = (x^2+px+1).[aQ(x)] – (x^2+px+1)[cP(x)] = (x^2+ px+1).[R(x)$, a polynomial of degree 1 in x]
i.e. $F(x) = abx^3 – cbx^2 + (a^2 – c^2) = (x^2 + px + 1).[R(x)]$
Using the similar reasoning as before, $(a^2 – c^2)x^3 – cbx^1 + ab$ also has the same quadratic factor $(x^2+ px+1)$
For some polynomial, S(x) of degree 1 in x, define $G(x) = (a^2 – c^2)x^3 – cbx^1 + ab = (x^2 + px + 1).[S(x)]$
Since m is a root of $x^2 + px + 1 = 0$, m is also a root of $G(x) = (a^2 – c^2)x^3 – cbx^1 + ab = 0$
From $(a^2 – c^2)*F(x) – (ab)*G(x) = 0$, we have
$$x^2 – \frac {ab^2c} {bc(a^2 – c^2)}x + \frac {a^2b^2 – (a^2 – c^2)} {bc(a^2 – c^2)} = 0$$
The LHS of the above is actually that quadratic factor.
Hence, $[a^2b^2 – (a^2 – c^2)] / [bc(a^2 – c^2)] = 1$
The required relation in a, b and c then follows.
| {
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integrating $\int \frac{dt}{(t+2)^2(t+1)}$ I'm practicing to solve a whole, and I am not able to solve this one, could you help me? $$\int \frac{dt}{(t+2)^2(t+1)}$$I tried $$\frac{1}{(t+2)^2(t+1)}=\frac{A}{(t+2)^2}+\frac{B}{(t+2)}+\frac{C}{(t+1)}\\1=A(t+1)+B(t+2)(t+1)+C(t+2)^2\\t=-2\Longrightarrow 1=-A\Longrightarrow \fbox{$A=-1$}\\t=-1\Longrightarrow1=C(-1)^2\Longrightarrow\fbox{$C=1$}$$Making $t = 0$ and substituting $A$ and $C$ we have $$1=A+2B+4C=-1+2B+4=2B+3\\\fbox{$B=-1$}$$THEN$$\int \frac{dt}{(t+2)^2(t+1)}=\int -\frac{1}{(t+2)^2}-\frac{1}{(t+2)}+\frac{1}{(t+1)}\;dt\\=-\int \frac{1}{(t+2)^2}-\int\frac{1}{(t+2)}+\int\frac{1}{(t+1)}\;dt=-\int u^{-2\;}du-\ln|t+2|+\ln|t+1|\\=-\frac{u^{-1}}{-1}-\ln|t+2|+\ln|t+1|=\\\fbox{$\frac{1}{t+2}-\ln|t+2|+\ln|t+1|+c$}$$Only I could not do the derivative to "take the test", can you help me? Or contains an error in my resolution?
| $$D_t\;\;\frac{1}{t+1}-\ln|t+2|+\ln|t+1|+c=\\=\frac{(t+2)\cdot0-1\cdot1}{(t+1)^2}-\frac{1}{t+2}+\frac{1}{t+1}+0=\\=\frac{-(t+1)-(t+1)(t+2)+(t+2)^2}{(t+2)^2(t+1)}=\\=\frac{-t-1-t^2-3t-2+t^2+4t+4}{(t+2)^2(t+1)}=\\=\frac{1}{(t+2)^2(t+1)}$$
Correct .. -
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$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0$ mod n iff $n \equiv \pm 1$ mod 6 The problem is as follows: Prove that $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0 \, \text{mod n} \ \text{if and only if } n \equiv \pm 1 \, \text{mod} 6.$$
My idea is to of course rewrite the summation. We have $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6}.$$
Now, in order for this to be an integer, must we have in particular that $6 \mid n-1 \Rightarrow n \equiv 1 \, \text{mod} 6$?
We also know that the sum $$1^2 + 2^2 + 3^2 + \ldots + (n-1)^2 + n^2 = \frac{n(n+1)(2n+1)}{6}.$$ Must we have in particular that then $6 \mid n+1 \Rightarrow n \equiv -1 \, \text{mod} 6$?
I think the statement holds if and only if the whole way so we are done with both ways?
| For the proof, we use the following observations:
(i) Suppose that $n$ divides $\frac{(n-1)(n)(2n-1)}{6}$. Then
$6n$ divides $(n-1)(n)(2n-1)$, and therefore $6$ divides $(n-1)(2n-1)$.
(ii) Conversely, if $6$ divides $(n-1)(2n-1)$, then $6n$ divides $(n-1)(n)(2n-1)$, and therefore $n$ divides $\frac{(n-1)(n)(2n-1)}{6}$.
We first show that if $\gcd(n,6)\gt 1$, then $6$ cannot divide $(n-1)(2n-1)$.
This is obvious if $n$ is even, since then $n-1$ and $2n-1$ are odd. So suppose that $3$ divides $n$. Then $3$ cannot divide $n-1$ or $2n-1$.
We have shown that if $1^2+2^2+\cdots+(n-1)^2$ is divisible by $n$, then $n$ and $6$ must be relatively prime, and therefore $n\equiv \pm 1\pmod{6}$.
Next we show that if $n\equiv \pm 1\pmod{6}$, then $6$ divides $(n-1)(2n-1)$. This is obvious if $n\equiv 1\pmod{6}$.
If $n\equiv -1\pmod{6}$, then $2n\equiv -2\pmod{6}$, and therefore $2n-1\equiv -3\pmod{6}$, and therefore $3$ divides $2n-1$. Moreover, $n$ is odd, so $n-1$ is even, and therefore $6$ divides $(n-1)(2n-1)$.
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Is this operation legal? Is this operation allowed?
Going from this:
$\left ( \frac{x^{2}+6}{x^{2}-4} \right )^{2}= \left ( \frac{5x}{4-x^{2}} \right )^{2}$
To this: $\left ( \frac{\left (x^{2}+6 \right )\left ( 4-x^{2} \right )}{\left (x^{2}-4 \right )5x} \right )^{2}= 1$
| Notice that the denominators are equal, save for a factor of $-1$: which has no impact since the fraction is squared. $$\; (4 - x^2)^2 = (-(x^2 - 4))^2 = (x^2 - 4)^2$$
So multiply both sides by $(x^2 - 4)^2$, and you'll cancel both denominators. Then there's no need to divide by $5x$.
After multiplying both sides by $(x^2 - 4)^2$, we get
$$\left (x^{2}+6\right )^{2}= \left ( 5x \right )^{2}\iff (x^4 + 12x^2 + 36) - 25x^2 = 0 \iff x^4 - 13x^2 + 36 = (x^2 - 9)(x^2 - 4) = (x+3)(x-3)(x+2)(x-2) = 0 $$
But we need to throw out the solutions $x = 2, -2$ because the equation is not defined there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\det(A-xI)$
in case of $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ we have $\det(A-xI)=\det\left( \begin{array}{cccc}
-x & 0 & c \\
1 & -x & b \\
0 & 1 & a-x \end{array} \right)=-(x^3-ax^2-bx-c)$
So, i have got the characteristic polynomial as $x^3-ax^2-bx-c$.
Now, the problem is how do i find minimal polynomial.
As $a,b,c$ are arbitrary, I can not factorize $x^3-ax^2-bx-c$ so as to see which factor gives me minimal polynomial.
I am confused.
please suggest me some hint.
EDIT : This is just after Mr.Will Jagyy's hint :
I have $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ then, $A^2= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)$
Now,
$A^2+rA+sI=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)+r\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+s\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
s & c & * \\
r & b+s & * \\
1& * & * \end{array} \right)$
As element of $3^{rd}$ row $1^{st}$ column is $1$ in above matrix, this can never be $0$
i.e., $A^2+rA+sI$ can never be $0$.
Now, $A+rI=\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+r\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
r & *& * \\
\color{magenta}{1} & * & * \\
* & \color{magenta}{1} & * \end{array} \right)\neq 0$ if $r\neq 0$
Thus, $A^2+rA+sI\neq 0$ and $A+rI\neq 0$ for any $r,s$.
Thus, minimal polynomial for $A$ can not be of order less than $3$.
Thus, minimal polynomial for $A$ has to be $x^3-ax^2-bx-c$.
I have written this just to make sure i have tried in correct way as i can not write this in a comment.
I would be thankful if there is any other way to proceed further.. Thank you :)
| The matrix of that form is called the "companion matrix"
Let $\{e_1,e_2,e_3\}$ be the standard basis for $\mathbb{F}^3$.
Then we see that $Ae_1=e_2, Ae_2=e_3, Ae_3=ce_1+be_2+ae_3$
Give a $\mathbb{F}[x]$-module structure on $\mathbb{F}^3$ by $x$-action defined as $A$-multiplication.
Denote that $\mathbb{F}[x]$-module structure as $M^A$.
Then $M^A\simeq \mathbb{F}[x]/(x^3-ax^2-bx-c)$ as $\mathbb{F}[x]$-module.
The RHS clearly shows that $A$ has minimal polynomial $x^3-ax^2-bx-c$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
The Fejer kernel has this $\sin$ closed form. Let $D_N$ be the $N$th Dirichlet kernel, $D_N = \sum_{k = -N}^N w^k$, where $w = e^{ix}$. Define the Fejer kernel to be $F_N = \frac{1}{N}\sum_{k = 0}^{N-1}D_k$. Then $$F_N = \frac{1}{N}\frac{\sin^2(N x/2)}{\sin^2(x/2)}$$.
So far I have $D_k = \frac{w^{k+1} - w^{-k}}{w-1}$, and so
$$
\begin{align*}
F_N &= \frac{1}{N}\sum_{k=0}^{N-1} D_k \\
&= \frac{1}{N(w-1)}\sum_{k=0}^{N-1} (w^{k+1} - w^{-k}) \\
&= \frac{1}{N(w-1)}\left ( w\sum_{k=0}^{N-1} w^k - \sum_{k=-N+1}^0 w^k \right ) \\
&= \frac{1}{N(w-1)}\left ( \frac{w(w^N - 1)}{w-1} - \frac{1-w^{-N + 1}}{w-1} \right ) \\
&= \frac{1}{N(w-1)^2}\left ( w^{N+1} +w^{-N + 1} - (w + 1) \right )
\end{align*}
$$
| Remember that $NF_N(x)=D_0(x)+\dots+D_{N-1}(x)$ where $D_n(x)$ is the Dirichlet kernel.
Therefore , if $\omega = e^{ix}$ we have
$$NF_N(x)=\sum_{n=0}^{N-1}\frac{\omega^{-n}-\omega^{n+1}}{1-\omega}$$
Then we have
$$\sum_{n=0}^{N-1}\omega^{-n}=1+1/\omega+\dots+1/\omega^{N-1}=\frac{1-\omega^{-N}}{1-\omega^{-1}}=\omega\frac{\omega^{-N}-1}{1-\omega}$$
and
$$\sum_{n=0}^{N-1}\omega^{n+1}=\omega(1+\omega+\dots+\omega^{N-1})=\omega\frac{1-\omega^N}{1-\omega}$$
Thus $$NF_N(x)=\frac{\sum_{n=0}^{N-1}\omega^{-n}-\sum_{n=0}^{N-1}\omega^{n+1}}{1-\omega}
=\omega\frac{\omega^N-2+\omega^{-N}}{(1-\omega)^2}=\frac{(\omega^{N/2}-\omega^{-N/2})^2}{(\omega^{1/2}-\omega^{-1/2})^2}=\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$$
So have get
$$F_N(x)=\frac{1}{N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$. Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.
| Just making sure what has to be done here, as mentioned above: we will multiply in the following way:
$$(11 \cdot 5 \cdot 6)\left(\frac{1}{5}+\frac{1}{6}\right)=\left((11 \cdot 5 \cdot 6)\frac{2}{11}\right)$$, in $\mathbb{Z_p}$
then we get $11\cdot6 + 11\cdot 5= 121 = 5 \cdot 6\cdot 2 = 60$, but this is only possible inside the $\mathbb{Z_{61}}$, observe that in this ring (because $p$ is prime), we have that $61 = 0 \in \mathbb{Z_{61}} $ and that $122 = 0$, so $121 = 60$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I prove by induction? For example if i wanted to prove:
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start.
Thanks.
| Prove for the base case, n=1:
$$1^{2} = \frac{1(1+1)(2+1)}{6} = \frac{2\cdot3}{6}=1$$
The "sum" of just $1^2$ is indeed 1. Base case proven.
Now for the induction step, proving that it holds for n+1:
Observe that for $1^{2} + 2^{2} +... + n^{2}$, the sum including $n+1$ equals the original summation plus the $n+1$ term: $1^{2} + 2^{2} + ... + n^{2} + (n+1)^{2}$, i.e. equals $ \frac{n(n+1)(2n+1)}{6} + (n+1)^{2}$.
So all we need to do is show that $\frac{n(n+1)(2n+1)}{6} + (n+1)^{2} = \frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$.
(Edit: To be clear: $\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$ is $\frac{n(n+1)(2n+1)}{6}$ with $n+1$ substituted in for $n$.)
Let's do that now:
$$\frac{n(n+1)(2n+1)}{6} + (n+1)^{2} = \frac{n(n+1)(2n+1)}{6} + \frac{6(n+1)^{2}}{6}$$
$$ = \frac{n(n+1)(2n+1)+6(n+1)^2}{6}$$
$$ = \frac{n(n+1)(2n+1)+6(n^{2}+2n+1)}{6}$$
$$ = \frac{n(2n^{2}+3n+1)+6n^{2}+12n+6)}{6}$$
$$ = \frac{2n^{3}+3n^{2}+n+6n^{2}+12n+6}{6}$$
$$ = \frac{2n^{3}+9n^{2}+13n+6}{6}$$
Having simplified this, we'll now show that it is equal to $\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$:
$$\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} = \frac{(n+1)(n+2)(2n+3)}{6}$$
$$ = \frac{(n^{2}+3n+2)(2n+3)}{6}$$
$$ = \frac{2n^{3}+6n^{2}+4n+3n^{2}+9n+6}{6}$$
$$ = \frac{2n^{3}+9n^{2}+13n+6}{6}$$
And we're done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the sequence ${a_n}$ converges where $a_n = \sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{n}}}}$ for $n\geq 1$. The original question was to determine whether the sequence converges, but I have checked for extremely high values of $n$ and it seems as though it does converge. This lead me to wonder if there was an "easy" method of showing that the sequence is bounded (since it is monotone, it then follows that it converges). Thanks.
| Term by term, is less than
$$\sqrt{1+\sqrt{2^1+\sqrt{2^2+\sqrt{2^3...}}}}=\sqrt{1+\sqrt{2}\sqrt{1+2^{-1}\sqrt{2^2+\sqrt{2^3...}}}}$$
$$<\sqrt{1+\sqrt{2}\sqrt{1+2^{-0.5}\sqrt{2^2+\sqrt{2^3+...}}}}$$
$$=\sqrt{1+\sqrt{2}\sqrt{1+\sqrt{2+2^{-1}\sqrt{2^3...}}}}$$
$$\cdots$$
$$<\sqrt{1+\sqrt{2}\sqrt{1+\sqrt{2+\sqrt{2^2...}}}}$$
So we now have to show that $a_0=1$,$a_{n+1}=\sqrt{1+\sqrt{2}a_{n}}$ converges(this series is strictly greater than the last equation).We proceed by induction.
Assume $a_n\le\frac{1+\sqrt3}{\sqrt2}$, then
$$a_{n+1}\le\sqrt{2+\sqrt{3}}=\frac{1+\sqrt3}{\sqrt2}$$
And since our sequence is monotone increasing and $1\le\frac{1+\sqrt3}{\sqrt2}$ the result follows.
Note: The number $1.757933$ of ncmathsadist is fairly close to this limit, $\approx 1.931851$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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The identity $ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin(anx) \, \mathrm dx$ Let $p(x)$ be a polynomial, and assume that $ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx $ converges.
How do you prove that $$ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin(anx) \, \mathrm dx? $$
I can verify that this identity is true in particular cases, but I'm not sure how to prove it.
EDIT:
The lower limit of the integral and the integrand parameter don't need to be the same.
So the identity could be written as $$ \int_{a}^{b} p(x) \cot \left(\frac{ r x}{2} \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin( r nx) \, \mathrm dx .$$
And as was mentioned below, $p(x)$ doesn't need to be a polynomial.
There are three other similar identities:
$$ \begin{align*} &\int_{a}^{b} p(x) \tan \left(\frac{rx}{2} \right) \, \mathrm dx = -2 \sum_{n=0}^{\infty} (-1)^{k} \int_{a}^{b} p(x) \sin(rnx) \, \mathrm dx \\ &\int_{a}^{b} p(x) \csc \left(rx \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin[(2n+1)rx] \, \mathrm dx \\ &\int_{a}^{b} p(x) \sec \left(rx \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} (-1)^{k} \int_{a}^{b} p(x) \cos[(2n+1)rx] \, \mathrm dx \end{align*} $$
They can all be derived in the way Daniel Fischer derived the original one by using the following finite sums:
$$ \begin{align} &\sum_{n=0}^{N} (-1)^{n} \sin(rnx) = - \frac{1}{2} \tan \left(\frac{rx}{2}\right) + \frac{(-1)^{N} \sin [(N+\frac{1}{2})rx]}{2\cos (\frac{rx}{2})} \\ &\sum_{n=0}^{N} \sin [(2n+1)rx] = \frac{1}{2} \csc (rx) - \frac{\cos [2(N+1)rx]}{2 \sin (rx)} \\ &\sum_{n=0}^{N} (-1)^{n} \cos [(2n+1)rx] = \frac{1}{2} \sec(rx) + \frac{(-1)^{N}\cos [2(N+1)rx]}{2 \cos (rx)} \end{align}$$
| Basically, because of the Riemann-Lebesgue lemma. By summing a geometric sum, or by induction using trigonometric identities, one finds
$$\sum_{n=0}^N 2\sin (anx) = \cot \frac{ax}{2} - \frac{\cos \left(a(N+\frac12)x\right)}{\sin \frac{ax}{2}}.$$
So that yields
$$\int_a^b p(x) \cot \frac{ax}{2}\,dx = 2\sum_{n=0}^N \int_a^b p(x)\sin (anx)\,dx + \int_a^b p(x)\frac{\cos \left(a(N+\frac12)x\right)}{\sin \frac{ax}{2}}\,dx.$$
Now if $\int_a^b p(x)\cot \frac{ax}{2}\,dx$ converges, the same is true for
$$\begin{align}
\int_a^b p(x)\frac{\cos \left(a(N+\frac12)x\right)}{\sin \frac{ax}{2}}
&= \int_a^b p(x) \frac{\cos (aNx)\cos \frac{ax}{2} - \sin (aNx)\sin \frac{ax}{2}}{\sin \frac{ax}{2}}\,dx\\
&= \int_a^b p(x) \cot \frac{ax}{2}\cos (aNx)\,dx - \int_a^b p(x)\sin (aNx)\,dx,
\end{align}$$
and by the Riemann-Lebesgue lemma, both of these integrals converge to $0$ for $N \to \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the equation of the plane that passes through the line of intersection of the planes... Find the equation of the plane that passes through the line of intersection of the planes $4x - 2y + z - 3 = 0$ and $2x - y + 3z + 1 = 0$, and that is perpendicular to the plane $3x + y - z + 7 = 0$.
This is what I got: $3x + 4y - z + 15 = 0$.
Can you please tell me if this is right?
Thanks in advance!
Here is my work:
EDIT:
Changed my answer to: 2x + 3y + 9z - 9 = 0
| Hint: The line of intersection of the 2 planes is parallel to
$$ \begin{pmatrix} 4 \\ -2 \\ 1 \\ \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 3 \\ \end{pmatrix} = \begin{pmatrix} -5 \\ -10 \\ 0 \end{pmatrix}.$$
Hint: The plane that you are interested in is parallel to $\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$ and parallel to $\begin{pmatrix}3 \\ 1 \\ -1 \end{pmatrix}$. Hence, it also is perpendicular to
$$ \begin{pmatrix} 1 \\ 2 \\ 0 \\ \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \\ -1 \\ \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\-5 \end{pmatrix}.$$
Hint: You found that the point $(3,4,-1)$ lies on both planes, hence lies on the line of intersection, hence lies on the plane that you are interested in.
Thus, the equation of the plane is $$-2x+y-5z = 3.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.
I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x\equiv2\pmod3$$What $x$ value?
| Given
$x=6a+5=6(a+1)-1$
$x=5b+4=5(b+1)-1$
$x=4c+3=4(c+1)-1$
$x=3d+2=3(d+1)-1$
therefore x will be of the form $(\text{L.C.M(3,4,5,6)}k-1)$ or,
$x=60k-1$ for some $k$.
Can you guess that $k$?
ANSWER:$k=1$, or $x=59$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
| we can prove this follow
$$3^{14}x^4y^4z^4(x^3+y^3+z^3)\le (x+y+z)^{15}$$
then assuming $x+y+z=1$,and denoting
$t=3(xy+yz+xz),q=xyz$
$$\Longleftrightarrow 3^{14}q^4(1-t+3q)\le 1\Longleftrightarrow 1-3^{14}q^4(1-t)-3^{15}q^5\ge 0$$
since
$$3xyz(x+y+z)\le (xy+yz+xz)^2\Longleftrightarrow q\le\dfrac{t^2}{3^3}$$
so
$$1-3^{14}q^4(1-3p)-3^{15}q^5\ge 1-3^{14}\left(\dfrac{t^2}{3^3}\right)^4(1-t)-3^{15}\left(\dfrac{p^2}{3^3}\right)^5=1-9t^8(1-t)-t^{10}$$
and
$$1-9t^8+9t^9-t^{10}=(1-t)(t^9+(1+t+t^2+\cdots+t^7-8t^8))=t^9(1-t)+(1-t)^2(1+2t+3t^2+4t^3+\cdots+8t^7)\ge 0$$
since
$$t=3(xy+yz+xz)\le(xy+yz+xz)^2=1$$
By done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that x is rational Let $x$ be a real number with the properties that $x^3+x$ and $x^5+x$ are rational.
Prove that $x$ is rational. Denote $a=x^3+x$; $b=x^5+x$. We can multiply and add them together until we get the desired result. I also know some non-elementary proofs of this, but have you some nice elementary proofs?
Thank you.
| $$x^5+x^3=ax^2$$
$$x^5+x=b$$
$$x^3-ax^2-x-b=0$$
If $a=0$ then the only option is $b=0$ and $x=0$. Assume $a\neq0$.
$$x^3+x=a$$
$$a-x-ax^2-x-b=0$$
$$ax^2+2x+b-a=0$$
So we get that $$x=\frac{-2\pm2\sqrt{1-a(b-a)}}{2a}$$
If $\sqrt{1-a(b-a)}$ is rational we are done. Assume it is not. Plug in $x^3+x=a$.
We get $$\frac{-4}{a^3}-\frac{4}{a}-a+\frac{3 b}{a^2}+\frac{4\sqrt{1-a (-a+b)}}{a^3}+\frac{2\sqrt{1-a (-a+b)}}{a}-\frac{b\sqrt{1-a (-a+b)}}{a^2}$$
So, $0=\frac{4}{a^3}+\frac{2}{a}-\frac{b}{a^2}=\frac{4+2a^2-ba}{a^3}$, i.e. $4+2a^2-ba=0$.
We should also have $0=\frac{-4}{a^3}-\frac{4}{a}-a+\frac{3 b}{a^2}=\frac{-4-4a^2-a^3+3ab}{a^3}$, i.e. $-4-4a^2-a^3+3ab=0$
I am lazy. Solve for $a$ and $b$ from these two equations (cubic in $a$ so it is easy). Then solve for $x$ (the cubic so it is easy). Check if the irrational solutions for $x$ satisfy the quintic with the value of $b$. Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Multiplication by $x^2$ linear maps $T:P(\mathbb{R}) \mapsto P(\mathbb{R})$ defined by
$(Tp)(x) = x^2p(x)$
Verify that multiplication by $x^2$ is a linear map.
Additivity: $x^2(p+q) = x^2p+x^2q$
Homogeneity: $x^2(ap) = a(x^2p)$
Is this a correct verification?
| It seems that you've just written down the definition of linearity, but I expect that the problem is looking for a more thorough proof using the definition of $P(\mathbb{R})$ - so it's really asking why the two equalities you wrote are true.
Let's start with polynomials $f = a_0 + a_1 x + a_2 x^2 + \cdots+ a_n x^n$, and $g = b_0 + b_1 x + \cdots + b_n x^n$ (we don't require $b_n \ne 0$ or $a_n \ne 0$). Then
\begin{align*}
x^2 (f + g) &= x^2 \Big((a_0 + b_0) + (a_1 + b_1) x + \cdots + (a_n + b_n)x^2\Big) \\
&= (a_0 + b_0) x^2 + (a_1 + b_1) x^3 + \cdots + (a_n + b_n) x^{n + 2} \\
&= \Big(a_0 x^2 + a_1x^3 + \cdots + a_n x^{n + 2}\Big) + \Big(b_0 x^2 + b_1 x^3 + \cdots+ b_n x^{n + 2}\Big) \\
&= x^2 (a_0 + \cdots + a_n x^n) + x^2 (b_0 + \cdots + b_n x^n) \\
&= x^2 f + x^2 g
\end{align*}
as desired. Now do something similar for multiplication by a scalar.
| {
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Finding the limit of $\frac{3^{n+1}}{4^{n-1}}$ as $n\rightarrow\infty$ How would one find the limit of the following.
as $n\rightarrow\infty$
$$\frac{3^{n+1}}{4^{n-1}}$$
I did the following
$$\frac{3\cdot 3^m}{1/4 \cdot 4^m}$$
$$\frac{3}{\frac{1}{4}}\ \frac{3^m}{4^m}$$
The limit if zero.
Another one I did is
$$\frac{2^{3n-1}}{7^{n+2}}$$
I did the following
$$\frac{\frac{1}{2}}{49}\ \frac{8^n}{7^n}$$
which means n approach infinity
But I am not sure if I did this correctly the numbers I got on the let side.
| $\lim_{n\to\infty}\frac{3^{n+1}}{4^{n-1}}=\lim_{n\to\infty}\frac{3^{n}\cdot 3}{4^{n}\cdot 4^{-1}}=\lim_{n\to\infty}\frac{3^{n}\cdot 3\cdot 4}{4^{n}}=12\lim_{n\to\infty}(\frac{3}{4})^{n}$=$12\cdot 0$=$0$, because $\lim_{n\to\infty}(\frac{3}{4})^{n}=0$.
Your answer is correct.
and the another
$\lim_{n\to\infty}\frac{2^{3n-1}}{7^{n+2}}=\lim_{n\to\infty}\frac{2^{3n}\cdot 2^{-1}}{7^{n}\cdot 7^{2}}=\lim_{n\to\infty}\frac{2^{3n}}{7^{n}\cdot 7^2\cdot 2}=\frac{1}{98}\lim_{n\to\infty}(\frac{8}{7})^{n}$=$\infty$, because $\lim_{n\to\infty}(\frac{8}{7})^{n}=\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of a sequence $\frac{1}{1+n^3}$ How can I prove by integral test that the sequence $1+ \dfrac{1}{1+2^3} + \dfrac{1}{1+3^3} + \dots + \dfrac{1}{1+n^3}$ is convergent?
Thank you. Is there a way that I can integrate $\dfrac{1}{1+n^3}$ ?
| Just in case somebody doesn't feel like doing calculus today:
$$\sum_{n=2}^\infty \frac{1}{1+n^3} < \frac{1}{2} \sum_{n=2}^\infty \frac{2}{n^3-n} = \frac{1}{2} \sum_{n=2}^\infty \left(\frac{1}{(n)(n-1)} - \frac{1}{(n+1)(n)}\right) = \frac{1}{2} \cdot \frac{1}{2(2-1)} = \frac{1}{4}$$
| {
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$50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$. The $50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$ after the decimal point.
$\underline{\bf{My\; Try}}::$ Let $\left(\sqrt{50}+7\right)^{50} = I+f$, where $I = $Integer part and $f = $ fractional part. and $0\leq f<1$
Now Let $\left(\sqrt{50}-7\right)^{50} = g$ and here $0\leq g<1$
Now $\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = 2\left(\binom{50}{0}\cdot (\sqrt{50})^{50}+\binom{50}{2}\cdot (\sqrt{50})^{48}\cdot 7^2+..............+\binom{50}{50}\cdot 7^{50}\right)$
$\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = $Integer quantity.
$\displaystyle I+f+g = $ Integer Quantity. and $0\leq f+g<2$ So $(f+g) = 0$ or $(f+g) = 1$
So $I=$Integer quantity or $I = $ Integer quantity$\;\; -1$.
Now I did not understand how can i get it.
Help required.
Thanks
edited.
| Consider the number $N=(\sqrt{50}+7)^{50}+(\sqrt{50}-7)^{50}$.
Using the Binomial Theorem, or otherwise, we can show that $N$ is an integer.
Our number $(\sqrt{50}+7)^{50}$ is a little below $N$. How much below? Courtesy of the calculator, $(\sqrt{50}-7)^{50}\approx 4\times 10^{-58}$. So the $50$-th digit after the decimal point of $(\sqrt{50}+7)^{50}$ is $9$.
| {
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Another Basic Integration Question, Possibly By Substitution What's the integral of $f(x)=(1-x^2)^{1/2}$? I tried making $x=\sin(t)$ and doing integration by substitution but I don't think I arrived to the correct answer. All responses are appreciated...
| Recall the three most common trigonometric substitutions and how they proceed.
Let $x=\sin\theta$. Then $\,dx = \cos\theta \,d\theta$. And $\theta = \arcsin x$.
$$\begin{align} \int\sqrt{1-x^2}\,dx
& =\int\sqrt{1 - \sin^2\theta}\,\cos\theta \,d\theta \\ \\
& =\int\sqrt{\cos^2\theta}\,\cos\theta \,d\theta \\ \\
& = \int\cos^2\theta \,d\theta \\ \\
& = \int\frac12(1+\cos2\theta) \,d\theta\\ \\
& =\frac12\Big(\theta+\frac{\sin2\theta}2\Big)+C\\ \\
& =\frac12\Big(\theta+\sin\theta\cos\theta\Big)+C \\ \\
& = \frac 12 \Big(\arcsin x + x\sqrt{1-x^2}\Big) + C\tag{$\dagger$}\end{align}$$
$(\dagger)$. Since $x = \sin \theta$, $\theta = \arcsin x$. And $\cos x = \sqrt{1 - x^2}$.
How do we know $\cos x = \sqrt{1-x^2}\;?\;$ Note: $\sin^2 \theta + \cos ^2 \theta = 1 = x^2 + (\sqrt{1 - x^2})^2 = x^2 + 1 - x^2$.
| {
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If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy... Problem :
If $2005 +x =y^2 ; 2005+y =x^2$ then find xy...
My approach :
Let $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii) $
Now from (i) we get :
$ y = \sqrt{x + 2005}$
Now putting this value of y in (ii) we get :
$ \Rightarrow 2005 +\sqrt{x + 2005} =x^2$
$ \Rightarrow \sqrt{x + 2005} =x^2 -2005 $
Now squaring both sides we get :
$\Rightarrow (\sqrt{x + 2005})^2 =(x^2 - 2005)^2$
Is there any other way I can solve this problem please suggest... thanks.
| Hint: $$2005 + x- (2005+y)=y^2 - x^2 \quad \Longrightarrow \quad (x-y)(x+y+1) = 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\log^25 + \log^27 > \log12$. Prove that $\log^25 + \log^27 > \log12$.
What I tried so far:
$\log^25 + \log^27 > \log3 + \log4$
$(\log5 + \log7)^2 - 2 \cdot \log5 \cdot\log7 > \log3 + \log4$
But it seems that I'm not even near the result.
Every suggestion / hint would be appreciated :)
Thanks in advance.
EDIT: $\log$ means $\log_{10}$
| From $5^3=125$ and $7^6=117649$, we deduce that ${\sf log}(5) \geq \frac{2}{3}$
and ${\sf log}(7) \geq \frac{5}{6}$.
From $3(6^7)=839808$ and $5^9=1953125$, we deduce that $3(6^7) \leq 5^9$ and hence
$12^8 \leq 10^9$. So ${\sf log}(12) \leq \frac{9}{8}$.
Finally, we have
$$
{\sf log}(5)^2+{\sf log}(7)^2 \geq \big(\frac{2}{3}\big)^2+
\big(\frac{5}{6}\big)^2=\frac{41}{36}=\frac{82}{72}\geq\frac{81}{72} \geq \frac{9}{8} \geq {\sf log}(12)
$$
| {
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How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended up with
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1-a & a-1 & 0 & 0 \\
0 & 1-a & a-1 & 1 \\
0 & 0 & 1-a & a-1 \\
\end{bmatrix}
$
I then factored out (1-a) to get this
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}
$
which will then make my determinant a multipile of $(1-a)^3$:
$\det(A) = (1-a)^3x$
But now here I don't know what to do. I have a feeling that my approach is wrong. Any help or guide please?
| The LDU-decomposition gives
$$ L=\small \begin{bmatrix}
1 & . & . & . \\
1/a & 1 & . & . \\
1/a & 1/(a+1) & 1 & . \\
1/a & 1/(a+1) & 1/(a+2) & 1 \\
\end{bmatrix} \\ D= \small \begin{bmatrix}
a & . & . & . \\
. & (a^2-1)/a & . & . \\
. & . & (a^2+a-2)/(a+1) & . \\
. & . & . & (a^2+2a-3)/(a+2) \\
\end{bmatrix} \\ U=\small \begin{bmatrix}
1 & 1/a & 1/a & 1/a \\
. & 1 & 1/(a+1) & 1/(a+1) \\
. & . & 1 & 1/(a+2) \\
. & . & . & 1
\end{bmatrix} $$
Here L and U are simply transposed (which reflects the symmetry of the original matrix), and all of the determinant-compuation happens in the diagonal of D.
The determinant is $$ a {a^2-1\over a} {a^2+a-2\over a+1}{a^2+2a-3\over a+2} $$
Expanding gives
$$ a\cdot {(a-1)(a+1)\over a} \cdot {(a-1)(a+2)\over a+ 1}\cdot {(a-1)(a+3)\over a+2}$$ and cancelling over the diagonal gives $$
1{a-1\over 1}\cdot {a-1\over 1}\cdot {(a-1)(a+3)\over 1}$$
which is the same as that in the answer of triplesec above
This representation has also the advantage that it is easily extensible to larger matrices
| {
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How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.
The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form.
I need help.
Update :
Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?
Thank you for your attention
| I first would eliminate the negative exponents of $x$ by multiplying $f(x)$ by $x^3$ to get
$$g(x)=x^3f(x)=x^6-12\,x^4+48\,x^2-64$$
If you are right in assuming that $f(x)=h(x)^3$ then $g(x)=p(x)^3$ and $g'(x)=3p'(x)p(x)^2$
Now we assume that $p$ is a polynomial then
$$p(x)^2 \mid \text{gcd}(g(x),g'(x))$$
and
$$\frac{g(x)}{\text{gcd}(p(x),p'(x))} \mid p(x)$$
We get
$$p'(x)=6\,x^5-48\,x^3+96\,x$$
and
$$\text{gcd}(p(x),p'(x))=x^4-8\,x^2+16$$
$$\frac{g(x)}{\text{gcd}(p(x),p'(x))}=x^2-4=(x-2)(x+2)$$
Because the degree of the last polynomial is $2$ we must have
$$p(x)=x^2-4$$
(ignoring constant factors)
We can check that $$x^6-12\,x^4+48\,x^2-64=(x^2-4)^3=(x-2)^3(x+2)^3$$
and so
$$f(x)=(x-\frac{4}{x})^3=(\sqrt{x}-\frac{2}{\sqrt{x}})^3 (\sqrt{x}+\frac{2}{\sqrt{x}})^3$$
Now let's solve
$$(x-2/x)=y$$
for $x$. By multiplying the equation by $x<$ we finally get the quadratic equation
$$x^2-yx-2=0$$
and get
$$x=-{{\sqrt{y^2+8}-y}\over{2}} $$
$$x=+{{\sqrt{y^2+8}+y}\over{2}}$$
So we have two inverse images for each value $y$.
| {
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Partial Fractions and power of a factor with $x^2$ I just started working with partial fractions and hit a wall with splitting this one:
$$ \frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} $$
I get here:
$$ \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}$$
Then on to:
$$ (Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2 $$
I find that $E = 1$ by using $x = -2$. I am unsure how to proceed from here.
| You already have
\begin{align*}
&\frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2}\\
&= \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}\\
&= \frac{(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2}{(x+2)(x^2+x+1)^2}.
\end{align*}
Multiplying both sides by $(x+2)(x^2+x+1)^2$ we have $$3x^2 + 2x + 1 = (Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2.$$ The right hand side is a quartic equation. If you expand all the brackets, you can collect all the like terms to get an equation as follows: $$3x^2 + 2x + 1 = k_4x^4 + k_3x^3+k_2x^2+k_1x+k_0$$ where $k_i$ depends on $A, B, C, D,$ and $E$. Then you can compare the coefficients of $1, x, x^2, x^3,$ and $x^4$ to get five equations in five unknowns which will be enough to determine $A, B, C, D,$ and $E$.
| {
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Show that $\frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )}$ is decreasing in $y > 1 $. I am interested in the function
$f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )},$
for values of $c \in (0,1)$, and $y > 1$, and have been trying to show that the function is decreasing.
I have tried differentiating the function, but this does not yield a particularly amenable formula to decide whether $f'(y) < 0$. Similarly I have considered the ratios $f(y)/f(\tilde y)$, but again have been unable to ascertain an inequality for this.
I'd appreciate any other approaches to the problem.
| We want to show that $$ f(y) = \dfrac {\left( 1 - y^{-1} \right)^2}{\left( 1 - y^{-1-c} \right) \cdot \left( 1 - y^{1+c} \right)} $$ is decreasing in $ y > 1 $. Indeed, we can differentiate. It suffices to show that $ f'(y) < 0 $, for $ y > 1 $, if $ c \in (0, 1) $.
So, we differentiate using the Quotient Rule: $$ f'(y) = \dfrac {(1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) - (1-y^{-1})^2\left((1-y^{-1-c})(1-y^{1+c})\right)'}{\left( 1 - y^{-1-c} \right)^2 \cdot \left( 1 - y^{1+c} \right)^2}. $$Now, let $h(x)$ be the numerator of this function. That is: $$ h(y) = (1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) - (1-y^{-1})^2\left((1-y^{-1-c})(1-y^{1+c})\right)'. $$Since the denominator of $f'(y)$ is always positive, $ f'(y) < 0 \implies h(y) < 0 $.
Let $ g(y) = \left( 1 - y^{-1-c} \right) \cdot \left( 1 - y^{1+c} \right) $.
Then, $$ g(y) = 1 - y^{-1-c} - y^{1+c} + 1 = 2 - y^{1+c} - y^{-1-c}, $$ so $$ g'(y) = - \left( 1+c \right) y^c - \left( -1 - c \right) y^{-2-c} = - \left( 1 + c \right)y^c + (1+c) y^{-2} - c, $$ $$ \implies g'(y) = (1+c) \cdot \left( y^{-2-c} - y^c \right). $$Now, going back to $h(y)$, we have: $$ (1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) - (1-y^{-1})^2(1+c)(y^{-2-c}-y^c) < 0. $$
$ \implies (1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) < (1-y^{-1})^2 (1+c)(y^{-2-c} - y^c) $
$ \stackrel{y \ne 1}{\implies} (2 - y^{-1-c} - y^{1+c})(2)(y^{-2}) < (1-y^{-1})(1+c)(y^{-2-c}-y^c) $
$ \implies 2 \cdot \left( 2y^{-2} - y^{-3-c} - y^{-1+c} \right) < (1+c) \cdot (y^{-2-c} - y^c - y^{-3-c} + y^{-1+c}) $
$ \implies 4y^{-2} - 2y^{-3-c} - 2y^{-1+c} < y^{-2-c} - y^c - y^{-3-c} + y^{-1+c} + cy^{-2-c} - cy^c - cy^{-3-c} + y^{-1+c}) $
From here, group like terms and we get that, if $ c \in (0, 1) $, the result is true for $ y > 1 $.
| {
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A Non-recursive Fibonacci Sequence How can I determine the general term of a Fibonacci Sequence? Is it possible for any one to calculate F2013 and large numbers like this?
Is there a general formula for the nth term of the Fibonacci Sequence?
If there is a combinatorial formula, that is more preferable; some formula like
Fn = xCy or xPy or something like that for some x and y.
| When
(1) $F_{n+2} = F_{n+1} + F_{n}$
one can try $F_n = \phi^n$, then we obtain
(2) $\phi^{n+2} - \phi^{n+1} - \phi^n = 0$,
thus
(3) $\phi^n \Big( \phi^2 - \phi - 1 \Big) = 0$,
ignoring the trivial case $F_n = 0$ we obtain
(4) $\phi = \frac{1}{2} \pm \frac{1}{2} \sqrt{5}$
So we get
(5) $F_n = a \left( \frac{1}{2} + \frac{1}{2} \sqrt{5} \right)^n + b \left( \frac{1}{2} - \frac{1}{2} \sqrt{5} \right)^n $
As $F_0 = 0$ and $F_1 = 1$, we get
(6a) $ a + b = 0 $
(6b) $ \frac{1}{2} \Big( a + b \Big) + \frac{1}{2} \Big( a - b \Big) \sqrt{5} = 1$
So
(7a) $\displaystyle a = \frac{1}{\sqrt{5}} $
(7b) $\displaystyle a = -\frac{1}{\sqrt{5}} $
Then
(8) $\displaystyle F_n = \frac{ \left( \frac{1}{2} + \frac{1}{2} \sqrt{5} \right)^n - \left( \frac{1}{2} - \frac{1}{2} \sqrt{5} \right)^n }{\sqrt{5}} $
| {
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If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$? All is in the title:
If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?
I can't conclude from the fact that $k^2 - 1$ is divisible by $8$, that then $k^4-1$ is divisible by $16$.
| Hint: $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = (k^2 - 1)\Big((k^2 - 1) + 2\Big)$$
ADDED per comment: So yes, we have that if $(k^2 - 1)$ is divisible by $8$, then $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = 8b(k^2 + 1)$$ for some integer $b$.
And now, if $k^2 - 1$ is divisible by 8, it is even, then so is $k^2 + 1$.
That is, $k^2 + 1 = (k^2 - 1) + 2 = 8b + 2$. So $$(k^2 - 1)(k^2 + 1) = 8b(8b + 2) = 16b (4b + 1)$$
| {
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Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
| let the nos be
n-1, n, n+1, n+2
we need to prove that (n-1)(n)(n+1)(n+2)+1= square no.
rearranging it : (n-1)(n+2)(n)(n+1)+1= square no.
LHS= (n-1)(n+2)(n)(n+1)+1
=(n^2+n-2)(n^2 +n)+1
= (k-2)(k) + 1 (by replacing (n^2 +n) by k)
= k^2 -2k +1
= (k-1)^2
hence proved , LHS is a square no.
on an interesting note,
(k-1)^2 = (n^2 +n-1)^2= {(n)(n+1)-1} ^2
therefore, the no. formed is always the square of, product of the two middle terms -1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
How find the value of $\cos{x}+\cos{y}$ let
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}+\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=1$$
Find the value $$cos{x}+\cos{y}=?$$
this following My ugly solution: let
$$\tan{\dfrac{x}{2}}=a,\tan{\dfrac{y}{2}}=b$$
then
$$\cos{x}=\dfrac{1-a^2}{1+b^2},\cos{y}=\dfrac{1-b^2}{1+b^2}$$
then
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}=\dfrac{1}{1+\tan{x}\tan{\dfrac{y}{2}}}=\dfrac{1-a^2}{1-a^2+2ab}$$
and
$$\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=\dfrac{1-b^2}{1-b^2+2ab}$$
so
$$\dfrac{1-a^2}{1-a^2+2ab}+\dfrac{1-b^2}{1-b^2+2ab}=1$$
then
$$\Longrightarrow (1-a^2)(1-b^2+2ab)+(1-b^2)(1-a^2+2ab)=(1-a^2+2ab)(1-b^2+2ab)$$
then
$$(1-b^2)(1-a^2+2ab)=2ab(1-b^2+2ab)$$
$$a^2+b^2=1-3a^2b^2$$
so
$$\cos{x}+\cos{y}=\dfrac{2-2a^2b^2}{1+a^2+b^2+a^2b^2}=1$$
My question: this problem have nice methods? Thank you,because My methods is very ugly.Thank you
| Here is a slight simplification of your method : one has
$$
\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}=\frac{1}{1-\alpha},
\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=\frac{1}{1-\beta} \tag{1}
$$
with $\alpha=\tan(x)\tan(\frac{y}{2})$ and $\beta=\tan(y)\tan(\frac{x}{2})$.
Your equation then becomes $\frac{1}{1-\alpha}+\frac{1}{1-\beta}=1$, or equivalently
$\alpha\beta=1$, i.e.
$$
\tan(x)\tan(\frac{x}{2})\tan(y)\tan(\frac{y}{2})=1
$$
or (if we put $a=\tan(\frac{x}{2})$ and $b=\tan(\frac{y}{2})$),
$$
\frac{2a^2}{1-a^2}\frac{2b^2}{1-b^2}=1 \tag{2}
$$
so $4a^2b^2=(1-a^2)(1-b^2)$ and hence
$$
3a^2b^2+a^2+b^2=1 \tag{3}
$$
We deduce
$$
(1+a^2)(1+b^2)=1+(a^2)+(b^2)+(a^2b^2)=1+(1-3a^2b^2)+(a^2b^2)=
2(1-a^2b^2)
$$
Then
$$
\cos(x)+\cos(y)=\frac{1-a^2}{1+a^2}+\frac{1-b^2}{1+b^2}=
\frac{2(1-a^2b^2)}{(1+a^2)(1+b^2)}=\frac{2(1-a^2b^2)}{2(1-a^2b^2)}=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve the following non-homogeneous recurrence relation: Find the solution to the following non-homogenous recurrence relation:
$a_{n+2} - 4a_{n+1} + 4a_{n} = 2^n$ for $a_0=1, a_1 = 2$.
I have found from the characteristic polynomial the general homogenous solution is:
$a_{n} = c_{1}2^n + c_{2}n2^n$ where $c_1, c_2$ are constants.
For the particular solution I think I should substitute $a_{n} = c_3n^22^n$ where $c_3$ is also a constant. However when I make that substitution I can't seem to solve the equation for $c_3$, can someone help please? Thanks
| Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and add over $n \ge 0$. Recognize e.g.
\begin{align}
\sum_{n \ge 0} a_{n + k} z^k
&= \frac{A(z) - a_0 - a_1 z - \ldots - a_{k - 1} z^{k - 1}}{z^k} \\
\sum_{n \ge 0} 2^n z^n
&= \frac{1}{1 - 2 z}
\end{align}
to get
$$
\frac{A(z) - 1 - 2 z}{z^2} - 4 \frac{A(z) - 1}{z} + 4 A(z)
= \frac{1}{1 - 2 z}
$$
As partial fractions:
$$
A(z) = \frac{1}{4} (1 - 2 z)^{-3}
- \frac{1}{2} (1 - 2 z)^{-2}
+ \frac{5}{4} (1 - 2 z)^{-1}
$$
Using the generalized binomial theorem you can read off the coefficients:
\begin{align}
a_n &= \frac{1}{4} \binom{-3}{n} (-2)^n
- \frac{1}{2} \binom{-2}{n} (-2)^n
+ \frac{5}{4} \binom{-1}{n} (-2)^n \\
&= \frac{1}{4} \binom{n + 3 - 1}{3 - 1} 2^n
- \frac{1}{2} \binom{n + 2 - 1}{2 - 1} 2^n
+ \frac{5}{4} \binom{n + 1 - 1}{1 - 1} 2^n \\
&= \frac{1}{4} \left(
\frac{(n + 2) (n + 1)}{2}
- 2 \frac{n + 1}{1}
+ 5
\right) \cdot 2^n \\
&= (n^2 - n + 8) \cdot 2^{n - 3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/535946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$123^{561}$ find last $2$ digits Modular Exponentiation mod $100$? $123^{561}$
Find the last $2$ digits
Can't I work this out using modular exponentiation working mod $100$?
$123^2 = 29\pmod{100}$
$123^4 = (123^2)^2 = 41\pmod{100}$
$123^8 = (123^4)^2 = 81\pmod{100}$
$123^{16} = (123^8)^2 = 61\pmod{100}$
$123^{32} = (123^16)^2 = 21\pmod{100}$
$123^{64} = (123^32)^2 = 41\pmod{100}$
$123^{128} = (123^64)^2 = 81\pmod{100}$
$123^{256} = (123^128)^2 = 61\pmod{100}$
$123^{512} = (123^256)^2 = 21\pmod{100}$
Then $561 = 512 \cdot 32 \cdot 16 \cdot 1$
so $21 \cdot 21 \cdot 61 \cdot 29 = 29\pmod{100}$
The answer is meant to be $21$!
What am I doing wrong?
Do I have to use Eulers?
| $123^1$ is $23$, not $29$ mod $100$. Your solution is otherwise correct, although Euler's theorem would be more efficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/536497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.
Here is a simple technique used to approximate square roots by Persian author Hassan be al-Hossein:
For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.
if $n=2^k$ we can use the method above.
For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 } } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 } } =\sqrt { 11\frac { 15 }{ 23 } } \\ \sqrt { 11\frac { 15 }{ 23 } } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know.
| The continued fraction method works like this: Suppose $x = a^2 + b$, where $a = \lfloor \sqrt x \rfloor$. Then
$$
\begin{align}
x &= \sqrt{a^2 + b}\\
x-a &= \sqrt{a^2 + b} - a\\
\frac{1}{x-a} &= \frac{1}{\sqrt{a^2 + b} - a}\\
&= \frac{1}{\sqrt{a^2 + b} - a}\frac{\sqrt{a^2 + b} + a}{\sqrt{a^2 + b} + a}\\
&= \frac{\sqrt{a^2 + b} + a}{b}\\
&= \frac{x + a}{b}
\end{align}
$$
Substitute, and get:
$$
\begin{align}
x &= a + (x-a)\\
&= a + \frac{b}{a+x}\\
%= a + \frac{b}{2a+\frac{b}{a+x}}\\
x &= a+\cfrac{b}{2a+\cfrac{b}{2a+\cfrac{b}{2a + \dots}}}
\end{align}
$$
Now, this is not a simple continued fraction. However, if one divides the numerator and denominator of $\frac{b}{2a+x}$ by $b$, then one can eventually get a periodic simple continued fraction, and one approximates by the convergents. The above expression turns out to be faster.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/538051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 13,
"answer_id": 1
} |
When to stop doubling down? My question is similar to this one but very specificly different
When to stop in this coin toss game?
Imagine a game where you would start with $100. Every time you can roll a die (d6), if it is 1-5 you double the winnings, but if it is a 6 you lose everything.
How would you calculate the ideal number of rolls to make? Lets define ideal as "if preformed 1000 times, would have the highest average winning"
The question above is similar but the reward is linear. With a linear reward it seems very clear, play until the winning odds become worse than the reward. In this case though the reward always keeps up with the risk. To me it seems like at any one moment the logical thing is to keep playing as the odds are in your favor. It is obvious though that following that you are guaranteed a result of $0.
| After first roll:
$P_1(\$200) = \frac{5}{6}$, $P_1(0) = \frac{1}{6}$.
Expected value (average winning) $E_1[X] = P_1(\$200)\cdot 200 + P_1(0) \cdot 0 = \frac{5}{6}\cdot 200 = \frac{5}{3}\cdot 100$.
After $2$ rolls:
$P_2(\$400) = \frac{5^2}{6^2}$, $P_2(0) = 1 - P_2(\$400) = \frac{6^2-5^2}{6^2}$.
$E_2[X] = P_2(\$400)\cdot 400 + P_2(0) \cdot 0 = \frac{5^2}{6^2}\cdot 400= \frac{5^2}{3^2}\cdot 100$.
...
After $n$ rolls:
$P_n(\$100 \cdot 2^n) = \frac{5^n}{6^n}$, $P_n(0) = 1-\frac{5^n}{6^n} = \frac{6^n-5^n}{6^n}$.
$E_n[X] = P_n(\$100\cdot 2^n)\cdot 2^n \cdot 100 + P_n(0) \cdot 0 = \frac{5^n}{6^n}\cdot 2^n \cdot 100 = \frac{5^n}{3^n} \cdot 100$.
Even if I see, that $E_n[X]$ increases with $n$,
I'd stop on the last $n$, when $P_n(\$100\cdot 2^n)>\frac{1}{2}$.
Because if $E_n(\$100\cdot 2^n)$ is very large, but probability to win is very small (when $n$ is large).
Let calculate it:
$P_1(\$200) = \frac{5}{6}>\frac{1}{2}$.
$P_2(\$400) = \frac{5^2}{6^2}=\frac{25}{36}>\frac{1}{2}$.
$P_3(\$800) = \frac{5^3}{6^3}=\frac{125}{216}>\frac{1}{2}$.
$P_3(\$1600) = \frac{5^4}{6^4}=\frac{625}{1296}<\frac{1}{2}$.
So, I'd stop on $3$rd step (if have fortune to get to $3$rd step) with $\$800$ or $0$.
Well, well, maybe on the $4$th step, because $P_4(\$3200) \lesssim \frac{1}{2}$.
But (!!!)
if you can repeat this game again, again (starting of $\$100$) $-$ then there are no doubts to play it endless!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/541081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Denoting the set of initial segments of a binary sequence The index is an infinite, innumerable binary sequence in $\{0,1\}$. $ I= \{f \mid f: \Bbb {N} \to \{0,1\}\} $
Is there a way to get a set $X_i$ from the infinite index number $10110\ldots$
$ X_i = \{1,10,101,1011,10110,\ldots\}$ where $i \in I$ and $i=10110\ldots$
| $${ X }_{ a }={ { \left\{ \left\lfloor \frac { a }{ { 10 }^{ n } } \right\rfloor ,\left\lfloor \frac { a }{ { 10 }^{ n-1} } \right\rfloor ,...,\left\lfloor \frac { a }{ { 10 }^{ 0 } } \right\rfloor \right\} } }\\n=\left\lfloor \log _{ 10 }{ a } \right\rfloor $$ This works for only finite sequences but you can divide the sequence and join. For example divide the total sequence such as $n=5$. The first five term would be $${ X }_{ a_{1-5} }={ { \left\{ \left\lfloor \frac { a_{1-5} }{ { 10 }^{ 5 } } \right\rfloor ,\left\lfloor \frac { a_{1-5} }{ { 10 }^{ 4} } \right\rfloor ,...,\left\lfloor \frac { a_{1-5} }{ { 10 }^{ 0 } } \right\rfloor \right\} } }$$. The next $5$ term:
$${ X }_{ a_{6-10} }={ { \left\{ { X }_{ a_{1-5} }\left\lfloor \frac { a_{6-10} }{ { 10 }^{ 5 } } \right\rfloor ,{ X }_{ a_{1-5} }\left\lfloor \frac { a_{6-10} }{ { 10 }^{ 4} } \right\rfloor ,...,{ X }_{ a_{1-5} }\left\lfloor \frac { a_{6-10} }{ { 10 }^{ 0 } } \right\rfloor \right\} } }$$.
This may not be very practical.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution verification: $\lim\limits_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$ I am trying to find the following limit
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2}$$
and I did the following steps:
\begin{align}
\require{cancel}
&\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} \\
&\lim_{x\rightarrow \infty} \frac{x^2\left(\frac{\sqrt{x}}{x^2}+1\right)}{x^2\left(\frac{2x}{x^2}-1\right)} \\
& \lim_{x\rightarrow \infty} \frac{\cancel{x^2}\left(\frac{\sqrt{x}}{x^2}+1\right)}{\cancel{x^2}\left(\frac{2\cancel{x}}{\cancel{x^2}}-1\right)}\\
& \lim_{x\rightarrow \infty} \frac{\left(\frac{\sqrt{x}}{x^2}+1\right)}{\left(\frac{2}{x}-1\right)} \\
\end{align}
Now here, the top portion goes to $0$ because the there is a larger power of $x$ in the denominator leaving only a $+1$ on top. On the bottom, the same thing happens, $\frac{2}{x}$ goes to $0$ and we left with $-1$ in the denominator. Therefore
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$$
Is my solution correct and did I take the right steps with the correct logic?
Thanks!
| Excellent work. You've made a correct and tight case for the limit being $-1$.
Just replace
Now here, the top portion goes to $0$ because the there is a larger power of $x$ in the denominator leaving only $(+1)$ on top.
With
Now here, the top portion goes to $1$ because in the first term of the numerator there is a larger power of $x$ in the denominator leaving only the term of $(+)1$ on top. On the bottom...
You could also argue that in the numerator, $\dfrac {\sqrt x}{x^2} = \dfrac 1{x^{3/2}} \to 0$ as $x \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/546922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Find the Area of the ellipse Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
where $a>0$, $b>0$
I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx$$
I can't get the answer $\pi ab$
| from the equation of the ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
$$ y = \sqrt{\frac{a^2b^2 - x^2b^2}{a^2}} = \frac{b}{a}\sqrt{a^2-x^2}$$
since proceding with change of variable in integration is, using x = a*cos(Θ)::
$$
\\
\sqrt{a^2 - x^2} = \sqrt{a^2-a^2cos^2\theta} =asin\theta
\\
\int_{}^{}\sqrt{a^2 - x^2} = a\int_{}^{}sin\theta \
dx \
= a^2\int_{}^{}sin^2\theta d\theta = a^2\int_{}^{}\frac{1 - cos 2\theta}{2} d\theta = \frac{a^2\theta}{2}-\frac{a^2sin^22\theta}{2}+C = \frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x\sqrt{1-\frac{x^2}a^{}}}{a})+C
$$
so the area in the first quadrant is:
$$ \frac{b}{a} \int_{0}^{a} \sqrt{a^2-x^2} dx$$
that is:
$$ \frac{b}{a}((\frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x \sqrt{1-\frac{x^2}{a^2}}}{a}))|^a_0) = \frac{ab \pi}{4}$$
so for the area of the entire ellipse:
$$ \frac{ab\pi}{4} 4 = ab\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/548876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
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How do I find the total area of total area of the red circles C_n,n=1,2,3 (Calculus II)
I am very confused with this problem. I have been working on it for about an hour and still nothing. Help would be greatly appreciated!
| One possible solution for this uses Decartes' Circle Theorem, which states that for four mutually externally tangent circles with radii $a,b,c,d$, then the following equality holds:
$$
2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2
$$
Let $r_n$ be the radius of circle $n$. Use the distance formula between the center of $C_1$ and the center of one of the big circles to obtain
$$
(1-r_1)^2+1^2=(1+r_1)^2\Rightarrow r_1=\frac{1}{4}
$$
We claim by induction that $r_n=\frac{1}{2n(n+1)}$. The base case is trivial, and given $r_n$ is decreasing, it suffices to show that
$$
2(4(n-1)^2n^2+4n^2(n+1)^2+2)=(2(n-1)n+2n(n+1)+2)^2
$$
Or equivalently that
$$
(4n^2(n^2+1)+1)=(2n^2+1)^2
$$
which is true.
Now we want to find
$$
\frac{\pi}{4}\left(\displaystyle\sum_{n\ge 1} \frac{1}{n^2(n+1)^2}\right)
$$
But since
$$
\frac{1}{n^2(n+1)^2} = \frac{1}{n^2}+\frac{1}{(n+1)^2}+2\left(\frac{1}{n+1}-\frac{1}{n}\right)
$$
The answer is
$$
\frac{\pi}{4}\left(\frac{\pi^2}{3}-3\right)=\frac{\pi^3}{12}-\frac{3\pi}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/549354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $x_{n+1} = \frac12(x_n + \frac2{x_n}).$
Let $x_1=1$ and $$x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).$$ Prove or disprove $(x_n)$ is convergent and show the limit.
When I tried working on it I found the sequence was bounded by
square root of 2 and it is was monotone. But apparently the sequence is not bounded by square root of two and is not monotone. But I have no idea why. Any help would be greatly appreciated! Thanks!
| Let $y_n = x_{n}^2$, then
\begin{align*}
\\y_{n+1} = x_{n+1}^2 &= \frac{1}{4}\left(x_n + \frac2{x_n}\right)^2
\\ &=\left(\frac{x_n}{2}\right)^2 +\left(\frac{1}{x_n}\right)^2 + 1 \ge 2\sqrt{\frac{x_n}{2}\cdot\frac{1}{x_n}} +1 = \sqrt{2}+1 \gt 2 \tag{1}
\end{align*}
Thus by (1), we have $x_n \gt \sqrt{2}~~(n\ge 2)$. Also, by the given condition, we have:
$$x_{n+1}-x_n=\frac{1}{2}x_n+\frac{1}{x_n}-x_n=\frac{1}{x_n}-\frac{x_n}{2} \tag{2}$$
(2) implies that the sequence is decreasing when $x_n \gt \sqrt{2}$ and we have already known that for $n \ge 2$, $x_n \gt \sqrt{2}$. So $x_n$ is decrease and bound below by $\sqrt{2}$. Hence, $x_n$ is convergent.
Note: Let $f(x) = \frac{1}{x} +\frac{x}{2}$, then
$$f(x)\ge 0 \space\text{when } x \in (-\infty,-\sqrt2] \cup (0,\sqrt2]$$
$$f(x)\lt 0 \space\text{when } x \in (-\sqrt2,0) \cup (\sqrt2, +\infty)$$
| {
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"source": "stackexchange",
"question_score": "3",
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Show that prime $p=4n+1$ is a divisor of $n^{n}-1$
Show that the prime number $p=4n+1$ is a divisor of $n^{n}-1$
Ok, the question itself is simple as hell, but I couldn't think of a simple way to solve this question. I tried to solve the question by using $p\equiv 1 \pmod n$ but only to fail miserably... I couldn't use quadratic residue concept neither since n could be odd.
It would be really glad if some could figure it out using methods of elementary number theory such as primitive root, Jacobian symbols and so on (since I'm just a beginner for number theory stuff).
Thanks!
| We'll use this lemma which I'll leave for you to prove:
Ic $p=4n+1$ is prime, then $a^n\equiv 1\pmod {p}$ if and only if $a\equiv x^4\pmod p$ for some $x$ not divisible by $p.$
We'll show that $-4$ is always a fourth power, $\pmod p$.
If $n$ is even, then $p\equiv 1\pmod 8,$ so $2$ is a square, and hence $4$ is a fourth power. Also $(-1)^n\equiv 1\pmod p$, $-1$ is a fourth power modulo $p$ by our lemma, and so $-4$ is a fourth power.
If $n$ is odd, then $-1\equiv x^2$ where $x$ is not a square. This is because if $x$ were a square, then $x^4=1$ and $x^{(p-1)/2}=1$, and hence $x^2=1,$ since $\gcd\left(4,\frac{p-1}{2}\right)=2.$
Similarly, $2$ is not a square modulo $p,$ So $2x$ is a square, $\pmod p$. So $4x^2\equiv -4\pmod p$ is a fourth power.
Knowing that $-4$ is a fourth power is all you need, since:
$$-4n\equiv 1\pmod p$$
So $n$ is a fourth power, too, so, again by our lemma, $n^n\equiv 1\pmod p$.
An alternative approach to showing that $x^4+4\equiv 0\pmod{p}$ for some $x$ is to use that:
$$x^4+4 = (x^2+2)^2-(2x)^2=(x^2-2x+2)(x^2+2x+2)=((x-1)^2+1)((x+1)^2+1)$$
Thus we have that $x^4+4\equiv 0\pmod p$ when $(x\pm 1)^2\equiv -1\pmod p.$ But since $p\equiv 1\pmod{4},$ $-1$ is a square modulo $p$, we know such $x$ exists.
Using that last approach as a jumping off point, we can start with $y$ such that $y^2\equiv -1\pmod{p}.$
Then $$\begin{align}(y+1)^4 &= y^4+4y^3+6y^2+4y+1 \\&\equiv 1+4y(-1)+6(-1)+4y+1\pmod{p}\\&=1+(-6)+1\\&=-4\pmod{p}\end{align}$$
| {
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"answer_id": 1
} |
A problem on indefinite integration $$\int\frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx$$
I tried some substitutions, but none succeeded in simplifying the expression. Please help.
| Disclaimer I'm cheating, I get the final answer from WA and reverse engineering out the steps. People have any intuition how to get the steps without cheating, please update this answer.
$$\begin{align}
\int \frac{x^4 - 2}{x^2 \sqrt{x^4+x^2+2}} dx
= & \int \frac{2x^4 + x^2 - (x^4 + x^2 + 2)}{x^2 \sqrt{x^4+x^2+2}} dx\\
= & \int \frac{\frac{x}{2}(x^4+x^2+2)' - (x^4+x^2+2)}{x^2 \sqrt{x^4+x^2+2}} dx\\
= & \int \left[
\frac{1}{x} \frac{d}{dx}\left(\sqrt{x^4+x^2+2}\right)
+ \frac{d}{dx}\left(\frac{1}{x}\right) \sqrt{x^4+x^2+2} \right]
dx\\
= & \frac{\sqrt{x^4+x^2+2}}{x} + \text{constant.}
\end{align}$$
| {
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How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy} \right)(x^2+y^2+z^2+3xyz)\ge (x+y+z)^2$$
then we only prove
$$\dfrac{9}{x^2+y^2+z^2+3xyz}\ge\dfrac{3}{2}$$
$$\Longleftrightarrow x^2+y^2+z^2+3xyz\le 6$$
Then I can't,and for $n$ how prove it?
| For the proof in case $n\geq 2$, I found Muirhead's inequality very useful.
*
*Expand the inequality. You have $$3xyz+\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^3yz}-3x^2y^2z^2 \geq 0$$, or stated another way, $$xyz(3-x^2+y^2+z^2)+(\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2})\geq0.$$
Therefore it is enough to prove that $$x^2+y^2+z^2\leq3$$ and $$\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2}\geq0.$$
*Showing $x^2+y^2+z^2\leq3$ is easy. We have $x^n+y^n+z^n=3$ for some $n\geq2$. Think about a function $f(x)=x^{n/2}$. It is convex because $n/2\geq1$, so we can use Jensen's inequality. Therefore, $$1=\frac{x^n+y^n+z^n}{3}=\frac{f(x^2)+f(y^2)+f(z^2)}{3}\geq f(\frac{x^2+y^2+z^2}{3})$$, and this implies $$1\geq \frac{x^2+y^2+z^2}{3}.$$
*Now we have to prove $$\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2}\geq0.$$ Notice that $1=\frac{x^n+y^n+z^n}{3}\geq x^{n/3}y^{n/3}z^{n/3}$, so $1 \geq xyz$. Because $xyz\leq1$, we have $$\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2}\geq \sum_{cyc}{x^{8/3}y^{8/3}z^{2/3}}-\sum_{cyc}{x^2y^2z^2}.$$ Here, we're done because Muirhead's inequality says $$\sum_{cyc}{x^{8/3}y^{8/3}z^{2/3}}\geq\sum_{cyc}{x^2y^2z^2}.$$
| {
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$\sum\sin(\frac{k\pi}{4})$ absolute convergent, conditional convergent,divergent? Would the following be abs convergent, conditional convergent or divergent.
$\sum\sin(\frac{k\pi}{4})$
I know sin(x) is between $-1<x<1$
y=sin(x) is oscillating would it be $(-1)^n$
| Sorry about the old answer, that was just garbage.
The real answer is that it's not convergent. You can try grouping them in terms of $8$ like I had done, but the problem is that, because it ends at $\infty$, you don't have a point to end at. That means the answer could be the same as any of the following eight:
$\displaystyle \sum_{k=1}^1 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^2 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2} + 1$.
$\displaystyle \sum_{k=1}^3 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^4 \sin \left(\frac{k \pi}{4}\right) = 0$.
$\displaystyle \sum_{k=1}^5 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^6 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2} - 1$.
$\displaystyle \sum_{k=1}^7 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^8 \sin \left(\frac{k \pi}{4}\right) = 0$.
Because they're different, and because you have no sure way of determining which is the right answer, the answer is that the series does not converge.
This is actually a classic problem of trying to find the sum of an alternating series. Let's take a similar problem, where you're trying to find the answer to $$1 - 1 + 1 - 1 + \cdots$$ (This is called Grandi's Series, btw.)
There are multiple answers, depending on how you group them. You can group them as $(1-1) + (1-1) + \cdots = 0 + 0 + \cdots = 0$, as $1 + (-1 + 1) + (-1 + 1) + \cdots = 1 + 0 + 0 + \cdots = 1$, or even as $\displaystyle \frac{1}{2} + \left(\frac{1}{2} - \frac{1}{2}\right) + \left(-\frac{1}{2} + \frac{1}{2}\right) + \cdots = \frac{1}{2} + 0 + 0 + \cdots = \frac{1}{2}$. Since there's no "right" way of doing this, the answer is DNE. Hence, it does not converge.
| {
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} |
If $x_n = (\prod_{k=0}^n \binom{n}{k})^\frac{2}{n(n+1)}$ then $\lim_{n \to \infty} x_n = e$ I try to prove the following:
$$x_n = \left(\prod_{k=0}^n \binom{n}{k}\right)^\frac{2}{n(n+1)}$$
$$\lim_{n \to \infty} x_n = e$$
I want to use double sided theorem, so I've proven that
$$x_n \ge \frac{n}{\sqrt[n]{n!}}$$
As it known, $\lim\limits_{n \to \infty} \frac{n}{\sqrt[n]{n!}} = e$. But I can't find such sequence $y_n$ that $x_n \le y_n$ and $\lim\limits_{n \to \infty} y_n = e$.
There is proof for lower bound:
$$x_n = \frac{(n!)^\frac{2}{n}}{(0!\cdot 1! \dots n!)^\frac{4}{n(n+1)}} \ge \frac{(n!)^\frac{2}{n}\cdot n}{(0!\cdot 1! \dots n!)^\frac{4}{n(n+1)}\cdot(n^n(n-1)^{n-1}(n-2)^{n-2}\dots2^2)^\frac{4}{n(n+1)}} = $$
$$= \frac{(n!)^\frac 2n \cdot n}{(n!)^\frac 4n} = \frac{n}{(n!)^\frac 2n} \ge \frac{n}{\sqrt[n]{n!}}$$
| Too late, more than likely too advanced but done for the fun of deriving asymptotics.
$$\prod_{k=0}^n \binom{n}{k}=\frac{\big[\Gamma (n+1)\big]^n}{G(n+1)\, G(n+2)}$$ where $G(.)$ is the Barnes G-function.
Using Stirling-like approximations
$$\log(G(p))=\frac{1}{4} p^2 (2 \log (p)-3)+\frac{1}{2} p (-2 \log (p)+2+\log (\pi )+\log (2))+$$ $$\frac{1}{12} \left(-6 \log \left(2 \pi A^2\right)+5 \log
(p)+1\right)-\frac{1}{12 p}-\frac{1}{240 p^2}+O\left(\frac{1}{p^3}\right)$$
$$\log\Bigg[\prod_{k=0}^n \binom{n}{k} \Bigg]=\frac{n^2}{2}+\frac{1}{2} n (-\log (n)+2-\log (\pi )-\log
(2))+\frac{1}{12} (24 \log (A)-4 \log (n)-1-6 \log (2 \pi
))-\frac{1}{12 n}+\frac{1}{180 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(x_n)=1-\frac 1 n \log \left(\frac{2 \pi n}{e}\right)+\frac{24 \log (A)+2 \log (n)-7}{6 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\color{blue}{x_n=e\left(1-\frac 1 n \log \left(\frac{2 \pi n}{e}\right) \right)+O\left(\frac{1}{n^2}\right)}$$
The relative error is less than $0.1$% as soon as $n>135$.
| {
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Asymptotic expansion of an integral I came up with a simpler example which illustrates the technical difficulty I have encountered in my work.
Consider an integral depending on parameter $\epsilon$:
\begin{equation}
\int\limits^{\infty}_{1 + \frac{2 \epsilon^{2}}{R^{2} - \epsilon^{2}}}
\frac{1}{t^{2}-1} \frac{1}{\sqrt{R^{2} - \epsilon^{2}\left(t+1\right)/\left(t-1\right)\,}}\,{\rm d}t.
\end{equation}
I am interested in the behaviour of this integral $\left(~\mbox{evaluated at}\ t = \infty~\right)$ as $\epsilon \to 0$.
This integral can be performed analytically, and it is straightforward to see that the end result diverges logarithmically as $\epsilon \to 0$. My question is: how to extract this logarithmic divergence without performing full integration ?. Naive expansion of the integrand in Taylor series in $\epsilon$ never leads to a logarithmic divergence.
In my real problem the integral cannot be evaluated analytically, but I am interested only in the logarithmic divergence. Is there a systematic way to extract it?
Thank you in advance!
Yegor
| First make the change of variables $x = \epsilon^2\frac{t+1}{t-1}$ to get
$$
I(\epsilon) = \int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = \frac{1}{2} \int_{\epsilon^2}^{R^2} \frac{dx}{x\sqrt{R^2-x}}.
$$
Fix $0 < \delta < R^2$ and split the integral into the two pieces
$$
I(\epsilon) = I_1(\epsilon) + I_2 = \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2-x}} + \frac{1}{2} \int_{\delta}^{R^2} \frac{dx}{x\sqrt{R^2-x}},
$$
each of which is finite for $\epsilon > 0$. We see here that the singularity comes from $I_1(\epsilon)$, whose largest contribution comes from a neighborhood of $x = \epsilon^2 \approx 0$. Informally,
$$
I_1(\epsilon) = \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2-x}} \approx \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2}} \approx - \frac{\log(\epsilon^2)}{2\sqrt{R^2}}.
$$
To make this precise, let's write
$$
\frac{1}{\sqrt{R^2 - x}} = \frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}} + \frac{1}{\sqrt{R^2}},
$$
so that
$$
I_1(\epsilon) = \frac{1}{2\sqrt{R^2}} \int_{\epsilon^2}^{\delta} \frac{dx}{x} + \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx.
$$
The integrand in the integral on the right is bounded by a constant for all $x \in [0,\delta]$ (note that $\lim_{x\to 0}$ exists), and the integral on the left is
$$
\frac{\log(\delta/\epsilon^2)}{2\sqrt{R^2}} = -\frac{\log \epsilon}{|R|} + O(1).
$$
Putting all this together, we conclude that
$$
\int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = -\frac{\log \epsilon}{|R|} + O(1)
$$
as $\epsilon \to 0$.
In fact we know what the constant in the $O(1)$ term is; it's just the finite terms we threw out along the way. Indeed,
$$
I(\epsilon) = -\frac{\log \epsilon}{|R|} + C + o(1),
$$
where
$$
C = \frac{\log \delta}{2|R|} + \frac{1}{2} \int_{0}^{\delta} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx + \frac{1}{2} \int_{\delta}^{R^2} \frac{dx}{x\sqrt{R^2-x}}.
$$
This constant depends on $R$ but not on $\delta$, so we may take the limit as $\delta \to R^2$ to obtain
$$
\begin{align}
C &= \frac{\log |R|}{|R|} + \frac{1}{2} \int_{0}^{R^2} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx \\
&= \frac{\log(2|R|)}{|R|}.
\end{align}
$$
Thus
$$
\int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = -\frac{\log \epsilon}{|R|} + \frac{\log(2|R|)}{|R|} + o(1)
$$
as $\epsilon \to 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Functional equation: $f\left(\frac{x-1}{x}\right)+ f\left(\frac{1}{1-x}\right)= 2- 2x$ There is a function given $f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x
,f\colon \Bbb R\setminus\{0,1\}\to \Bbb R$ How many fuction exist? I have no idea how to start
| $f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x\tag{1}$
Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(1)$
$\displaystyle f\left(\frac{1}{1-x}\right)+f\left(x\right) = \frac{2}{x}\tag{2}$
Similarly Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(2)$
$\displaystyle f(x)+f\left(\frac{x-1}{x}\right) = \frac{2x}{x-1}\tag{3}$
Now $(1)-(2)-(3),$ we get
$$\displaystyle -2f(x) = 2-2x-\frac{2}{x}-\frac{2x}{x-1} = -2\left(\frac{x^3+x-1}{x^2-x}\right)$$
So $\displaystyle f(x) = \frac{x^3+x-1}{x^2-x}$ for all $x\in \mathbb{R}\setminus\{0,1\}$
| {
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"question_score": "9",
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} |
convergence of a function serie to norm 2 need to show if the following function serie converge on $||.||_2$ on [1,5]:
$$\sum _1^\infty {sin^2(nx) \over n^2} $$
I have no idea how to approach that one, would like
for some directions...
| We can in fact compute what the series converges to. Note that
$$\sum_{n\ge 1}\frac{\sin^2(nx)}{n^2}
= \sum_{n\ge 1}\frac{1}{2} \frac{1-\cos(2nx)}{n^2}
= \frac{\pi^2}{12} - \frac{1}{2} \sum_{n\ge 1} \frac{\cos(2nx)}{n^2}.$$
Re-write this as follows:
$$\frac{\pi^2}{12} - 2 x^2 \sum_{n\ge 1} \frac{\cos(n2x)}{(n2x)^2}.$$
Now the sum $$S(x) = \sum_{n\ge 1} \frac{\cos(nx)}{(xn)^2}$$ is harmonic and was evaluated by inverting its Mellin transform at this MSE link, the result being
$$S(x) = \frac{\pi^2}{6x^2} -\frac{\pi}{2x} +\frac{1}{4}.$$
This gives for our sum the value
$$\frac{\pi^2}{12} - 2 x^2
\left(\frac{\pi^2}{24x^2} -\frac{\pi}{4x} +\frac{1}{4}\right)$$
which simplifies to
$$\frac{\pi}{2} x - \frac{1}{2} x^2$$
on the interval $(0,\pi).$
| {
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How to solve this system of equations. Solve the system of equations: $$\begin{cases}\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy} \\\left(\sqrt{3}+xy\right)^{\log_2x}+\dfrac{x}{\left(\sqrt{3}-xy\right)^{\log_2y}}= 1+\dfrac{x}{y}\end{cases}$$
My try:
$\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy}\\\Leftrightarrow xy\left(x^2+y^2\right)+\left(xy\right)^2= x^2+y^2+1\\\Leftrightarrow\left(xy-1\right)\left(x^2+y^2+xy+1\right)=0\\\Leftrightarrow xy=1\,\,\,\mbox{(because:}\,\, x^2+y^2+xy+1>0)$
And here I'm stuck!
| Subsitute $xy=1 , y=1/x,and \, log y=-logx$ :-
$log(3^{1/2}+1)^{logx}+ x*log(3^{1/2}-1)^{logx}=1+x^2$
then take Log at both sides of your equation then your 2nd equation becomes
$logx*log(3^{1/2}+1)+logx+ logx*log(3^{1/2}-1)=2logx$
$log(3^{1/2}+1)+ log(3^{1/2}-1)=1$
$log((3^{1/2}+1)(3^{1/2}-1))=1$
which is always true!
| {
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Which step is wrong in this proof
Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give
$$x=−1−\frac{1}{x}$$
Substitute this back into the $x$ term in the middle of the original equation, so
$$x^2+(−1−\frac{1}{x})+1=0$$
This reduces to
$$x^2=\frac{1}{x}$$
So, $x^3=1$, so $x=1$ is the solution. Substituting back into the equation for $x$ gives
$1^2+1+1=0$
Therefore, $3=0$.
What happened?
| What you have proved is that there is no real number $x$ such that $x^2+x+1=0$.
On the other hand, the two complex solutions of $x^2+x+1=0$ do indeed satisfy $x^3=1$.
| {
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Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$
are linearly dependent vectors in $C^0[0, 1]$. Be convincing in your reasoning (argument)
I was researching and found Wronskian. Using the Wronskian for three functions. The determinant of $f$, $g$ and $h$ is $W(f, g, h) = $
$$
\begin{vmatrix}
f & g & h \\
f' & g' & h' \\
f'' & g'' & h'' \\
\end{vmatrix}
$$
If $W(f, g, h) \neq 0$ then $f(x)$, $g(x)$ and $h(x)$ are linearly independent.
If $f(x)$, $g(x)$, and $h(x)$ are linearly dependent then $W(f, g, h) = 0$
My attempt
Let
$f(x) = \sin(x + \alpha)$, $g(x) = \sin(x + \beta)$ and $h(x) = \sin(x + \gamma)$
$W(f, g, h) =$
$$
\begin{vmatrix}
\sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\
\cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\
-\sin(x + \alpha) & -\sin(x + \beta) & -\sin(x + \gamma) \\
\end{vmatrix}
$$
$= \sin(x + \alpha)[-\sin(x + \gamma)\cos(x + \beta) + \cos(x + \gamma)\sin(x + \beta)] - sin(x + \beta)[-\sin(x + \gamma)\cos(x + \alpha) + \cos(x + \gamma)\sin(x + \alpha)] + \sin(x + \gamma)[-\sin(x+ \beta)\cos(x + \alpha) + \cos(x + \beta)\sin(x + \alpha)]$
$= -\sin(x + \alpha)[\sin((x + \gamma) +(x + \beta))] + \sin(x + \beta)[\sin((x + \gamma) + (x + \alpha))] - \sin(x + \gamma)[\sin((x + \beta) + (x + \alpha))] = 0$
By Wronskian, $f(x)$, $g(x)$ and $h(x)$ are linearly dependent since $W(f, g, h) = 0$
Not sure if this argument is sound?
| A short method :
Since $$\sin(x+\alpha)=(\cos \alpha) \sin x + (\sin \alpha) \cos x$$
we have $\sin(x+\alpha) \in \text{Span}\{\sin x, \cos x\}$
This implies that the rank of the family $(\sin(x+\alpha),\sin(x+\beta),\sin(x+\gamma))$ is $2$. That means that there are linearly dependants.
| {
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which one is larger $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$? Which is larger? $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$?
Algebraic justilation does not help.
| $\sqrt[n]{x+\delta}-\sqrt[n]{x}\ \boxed{\phantom{A} }\sqrt[n]{x}-\sqrt[n]{x-\delta}$
$\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}\ \boxed{\phantom{A} }2\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]{\frac{\left(\sqrt[n]{x+\delta}\right)^n+\left(\sqrt[n]{x-\delta}\right)^n}{2}}$
But $\frac{a+b}{2}\leq\sqrt[n]{\frac{a^n+b^n}{2}}$.
$\tiny{\text{some conditions apply}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Cubic Poynomial : In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the......... Question:
In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the equation are
(a) all real and equal
(b) all real and distinct
(c) one real and two imaginary
(d) all real
What I did :
Let the cubic polynomial is $ax^3+bx^2+cx+d=0$ then let p,q,r are roots of equation then $ax^3 - a(p+q+r)x^2 + a(pq+pr+qr)x - a(pqr)$
How do I relate this with the given equation.. thanks..
| The answer is (c): The equation has one real root and two nonreal complex conjugate roots.
Proof: This Wiki article classifies the nature of the roots.
The discriminant of a cubic equation $ax^3+bx^2+cx+d=0\;$ is
$$\Delta = 18 a b c d - 4b^3d + b^2 c^2 -4ac^3 - 27 a^2 d^2.$$
*
*If $\Delta > 0,$ then the equation has three distinct real roots.
*If $\Delta = 0,$ then the equation has a multiple root and all its roots are real.
*If $\Delta < 0,$ then the equation has one real root and two nonreal complex conjugate roots.
With your coefficients we get (since $b=0$)
$$\Delta = -4ac^3 - 27a^2d^2 = -108H^3-27G^2 = -27(4H^3 + G^2) < 0.$$
And therefore, because $\Delta < 0,$ the equation has one real root and two nonreal complex conjugate roots.
Note: If the question is a kind of multiple choice, the third roots of unity (1, and two conjugate-complex) would also give a hint for case (c).
| {
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$\displaystyle f(x,y,z)=3log(x^2+y^2+z^2)-2x^2-2y^3-2z^3$ $(x,y,z)\neq(0,0,0)$ has only one extreme value $\displaystyle log(\frac{3}{e^2})$ Show that the function defined by $\displaystyle f(x,y,z)=3log(x^2+y^2+z^2)-2x^2-2y^3-2z^3$
$(x,y,z)\neq(0,0,0)$ has only one extreme value $\displaystyle log(\frac{3}{e^2})$
Attempt:
$\displaystyle df=\frac{3}{x^2+y^2+z^2}(2xdx+2ydy+2zdz)-4xdx-6y^2dy-6z^2dz$
Then i tried to find the stationary points but couldn't get any further.
| Stationary points of $f$ can be found from the system
$$\nabla{f}=\vec{0}$$
which is equivalent to
$$
\begin{cases}
\dfrac{\partial{f}}{\partial{x}}=0, \\
\dfrac{\partial{f}}{\partial{y}}=0, \\
\dfrac{\partial{f}}{\partial{z}}=0,
\end{cases}
\quad\Leftrightarrow\quad
\left \lbrace {\matrix{
x\left(\dfrac{3}{x^2 + y^2 + z^2}-2 \right)=0, &(1)\\
y\left(\dfrac{y}{x^2 + y^2 + z^2}-1 \right)=0, &(2)\\
z\left(\dfrac{z}{x^2 + y^2 + z^2}-1 \right)=0. &(3)
}}\right.
$$
Added:
If $x=0$ and $y=0$ then $(3)$ implies $z=1.$
Similarly, $x=0,\;\;z=0 \Rightarrow y=1.$
If $y=z=0$ then from $(1)$ we have $x^2=\dfrac{3}{2}\Rightarrow x=\pm{\sqrt{\dfrac{3}{2}}}.$
The case $x^2 + y^2 + z^2 =\dfrac{3}{2}\;\text{and}\;y=0 $ is impossible because $(3)\Rightarrow z=\dfrac{3}{2},$ but then $ x^2 + y^2 + z^2 \geqslant {\dfrac{9}{4}} >\dfrac{3}{2}. $ Alike the case $x^2 + y^2 + z^2 =\dfrac{3}{2}\;\text{and}\;z=0$ is impossible too.
One more case is $x=0 \;\text{and}\;\dfrac{y}{x^2 + y^2 + z^2}-1=0 \;\text{and}\;\;\dfrac{z}{x^2 + y^2 + z^2}-1=0$ which leads us to $x=0,\; y=\dfrac{1}{2},\; z=\dfrac{1}{2}.$
Finally, the case $x^2 + y^2 + z^2 =\dfrac{3}{2} \;\text{and}\;\dfrac{y}{x^2 + y^2 + z^2}-1=0 \;\text{and}\;\;\dfrac{z}{x^2 + y^2 + z^2}-1=0$ is also impossible.
| {
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Getting rid of the square roots in the expression $\sqrt{a} + \sqrt{b} + \sqrt{c} = d$ I need to find an alternate form of $\sqrt{a} + \sqrt{b} + \sqrt{c} = d$ without square roots for a problem that I'm working on, but it's rather complicated to do. What we can do is
$\sqrt{a} + \sqrt{b} + \sqrt{c} = d \iff \sqrt{a} = d - \sqrt{b} - \sqrt{c} \iff a = (d - \sqrt{b} - \sqrt{c})^2$
and then calculate the right hand side, then iterate the process by putting a square root on one side, then proceed as above until all square roots are gone. However, this is a rather complicated process and I would like to do this with a computer but it exceeds WolframAlpha's server time. Can I get this done somewhere else? I would do this by hand myself but I will also need it for $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} = e$ and perhaps even more complicated expressions, so if someone can show me an easy way to proceed with calculations - or point me to some computer program that can do this - I would appreciate it. Thanks in advance.
(note: not sure how to tag my question, feel free to change the tag if you can find something more appropriate).
Note: edited as per suggestions in the comments.
| You are asking a special case of the following problem in abstract algebra: Suppose $x$ is a solution to $p(x) = 0$ and $y$ solves $q(y)=0$, for polynomials $p,q$ (with, say, integer coefficients); find a polynomial (with, again, integer or whatever coefficients) that has $x+y$ as a solution. In the "$\sqrt{a} + \sqrt{b} = c$" version of your question, you have $p(x) = x^2-a$ and $q(y) = y^2 - b$.
There is a general way to do this. Unfortunately in the version I described, the degrees of the polynomials get large. This is because of the following. Suppose that $p$ is of degree $\deg p$, and $q$ has degree $\deg q$. Then generically $p$ has $\deg p$ many complex solutions and $q$ has $\deg q$ many complex solutions. Thus (unless these solutions happen to satisfy some coincidences) there are $(\deg p)(\deg q)$ many possible values of $x+y$ if all you know is that $x$ solves $p(x)=0$ and $y$ solves $p(y)=0$. Whatever the polynomial is that $x+y$ solves, it must be solved by all of possible of these $(\deg p)(\deg q)$ numbers, since you have no way of telling it which solution you want. Therefore, this polynomial must have degree $(\deg p)(\deg q)$.
So in the example "$\sqrt a + \sqrt b = c$", we're looking for a polynomial solved by $c$ with "integer" (really, polynomial expressions in $a,b$) coefficients, and it necessarily will have degree $4$. Now adding on another square root means that in your case, your polynomial in $d$ will have degree $(4)(2) = 8$. In the next one, you get a polynomial of degree $16$. There's really nothing you can do about this.
Of course, there will be some patterns. Let me focus on your case of sums of square roots. The two solutions to $x^2-a = 0$ are, of course, $\pm \sqrt a$, and the two solutions to $y^2-b$ are $\pm \sqrt b$. Thus the four possible values of $x+y$ are $\pm \sqrt a \pm \sqrt b$. Therefore the polynomial we want is the degree-$4$ polynomial vanishing on these four points, namely:
$(z - \sqrt a - \sqrt b)(z - \sqrt a + \sqrt b)(z + \sqrt a - \sqrt b)(z + \sqrt a + \sqrt b) = ((z - \sqrt a)^2 - b)((z + \sqrt a)^2 - b) = (z^2 - a)^2 + b^2 - b((z - \sqrt a)^2 + (z + \sqrt a)^2) = z^4 - 2z^2a + a^2 + b^2 - 2z^2 b - 2ab = z^4 - 2(a+b)z^2 + (a-b)^2$
This illustrates, for example, that only even powers of the new variable (in your case, $d$) will appear — exactly because the set of solutions to this polynomial will necessarily be symmetric under $d \mapsto -d$. Such a polynomial is called "even".
In all cases, let me henceforth call the new variable $z$ — so you asked about $\sqrt a + \sqrt b + \sqrt c = z$ or $\sqrt a + \sqrt b + \sqrt c + \sqrt d = z$, and so far I've discussed $\sqrt a + \sqrt b = z$. Let's say there are $n$ terms on the left, so that we're looking for an even degree-$2^n$ polynomial in $z$; and I will set $a = a_1$, $b = a_2$, $c = a_3$, and on up to $a_n$. The above calculation also illustrates that the coefficient on $z^{2^n - 2k}$ will be a symmetric polynomial in the $a_i$, homogeneous of degree $k$. That it's symmetric is clear: the problem as posed is symmetric in the $a_i$. That it is homogeneous of degree $k$ follows from rescaling all $a_i$ to $\lambda a_i$; then the solutions $z$ uniformly rescale to $\sqrt \lambda z$.
So in the case that you originally asked about, with $a=3$, we're looking for:
$z^8 + p_1(a_1,a_2,a_3) z^6 + p_2(a_1,a_2,a_3) z^4 + p_3(a_1,a_2,a_3) z^2 + p_4(a_1,a_2,a_3)$
where each $p_j$ is a homogeneous symmetric polynomial of degree $j$. It is well-known, then, that $p_j$ is a polynomial in the polynomials $s_1 = a_1 + a_2 + a_3$, $s_2 = a_1^2 + a_2^2 + a_3^2$, $s_3 = a_1^3 + a_2^3 + a_3^3$, $\dots$, $s_j = a_1^j + a_2^j + a_3^j$. For example, $p_1 = \alpha(a_1 + a_2 + a_3)$ for some coefficient $\alpha$, and
$$ p_2 = \beta(a_1 + a_2 + a_3)^2 + \gamma(a_1^2 + a_2^2 + a_3^2) $$
There are three as-yet undetermined coefficients in $p_3 = \delta s_1^3 + \epsilon s_1 s_2 + \zeta s_3$, and five coefficients in $p_4 = \eta s_1^4 + \theta s_1^2 s_2 + \iota s_2^2 + \kappa s_1 s_3 + \lambda s_4$. All together, we have reduced the problem from computing some arbitrary degree-$8$ polynomial with polynomial coefficients to computing $11$ rational numbers.
Can we pair these down at all? When $a \neq 0$ but $b=c=0$, the two solutions need to be $z = \pm \sqrt a$. (Well, each of these is really for solutions, for "the two values of $\pm\sqrt{0}$".) So in this case our polynomial had better evaluate to $(z^2 - a)^4$. In particular, $\alpha = -4$, and $\beta + \gamma = 6$, $\delta + \epsilon + \zeta = -4$, and $\eta + \theta + \iota + \kappa + \lambda = 1$. This completely determines $\alpha$, and of the remaining $10$ unknowns, we have $3$ equations.
When $a = b$ and $c=0$, the solutions should be $\pm 2\sqrt a$ and $0$. (Actually, two each of the first two and four of the last one.) Thus the polynomial should be $(z^2 - 4a)^2z^4$. This will determine $\alpha$ again, and give you three more equations, one of which is enough to determine $\beta,\gamma$. So we've paired the space of unknowns down to being $4$-dimensional.
You can almost certainly continue in this way. For example, set $c=0$ and $b=4a$. Then $z = \pm3\sqrt a$ or $\pm \sqrt a$, each with multiplicity two, and so the polynomial in this case is $(z^2-a)^2(z^2-9a)^2$. I think this will give you two new equations when you combine it with what you already have, and I think it's enough to determine $p_3$.
One final thing to mention: the coefficient on $z^0$ is always the product (up to a sign, which is $+1$ since there are an even number of solutions) of all the solutions. Well, the product of the solutions is $(\sqrt a + \sqrt b + \sqrt c)^2(\sqrt a - \sqrt b + \sqrt c)^2(\sqrt a - \sqrt b - \sqrt c)^2(\sqrt a + \sqrt b - \sqrt c)^2$, which is a version of the $n=2$ case. Doing this gives a different way to get $p_4$.
Here's what I get, actually doing the arithmetic:
$$ z^8 - 4(a+b+c) z^6 + (2(a+b+c)^2 + 2(a^2+b^2+c^2))z^4 + (\frac43 (a+b+c)^3 - \frac83 (a^3+b^3+c^3))z^2 + (a^2 + b^2 + c^2 - 2(ab + bc + ac))^2 $$
Note that the fractions in the coefficient on $z^2$ are strong evidence that I made an arithmetic mistake there, but I think the coefficients on $z^6$, $z^4$, and $z^0$ are correct.
| {
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Finding indefinite integral $\int{ \mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$ Could anyone help me how to solve this indefinite integral?
$$\int{\mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$$
| Firstly let us find the anti-derivative in case $\alpha=0$. This is pretty straightforward:
\begin{eqnarray}
\int\frac{dx}{\sqrt{\sin(x)^3+\sin(x)}}&\underbrace{=}_{y=\sin(x)}&\int\frac{dy}{\sqrt{1-y^2} \sqrt{y} \sqrt{1+y^2}}\\
&\underbrace{=}_{z=y^4}& \frac{1}{4} \int z^{-7/8} (1-z)^{-1/2} dz\\
&=& \frac{1}{4} B_{[\sin(x)]^4}(\frac{1}{8},\frac{1}{2})
\end{eqnarray}
where $B_{.}(,)$ is the incomplete beta function.
Now let us compute the derivative at $\alpha=0$. We have:
\begin{eqnarray}
-\frac{1}{2} \int \frac{\cos(x)}{(\sin(x)^3+\sin(x))^{3/2}}dx&\underbrace{=}_{y=\sin(x)}&-\frac{1}{2} \int\frac{d y}{(y^3+y)^{3/2}}\\
&\underbrace{=}_{u=-y^2}&\frac{(-1)^{3/4}}{4} B_{-\sin(x)^2}(-\frac{1}{4},-\frac{1}{2})
\end{eqnarray}
Therefore we have:
\begin{eqnarray}
\int \frac{dx}{\sqrt{\sin(x)^3+\sin(x+\alpha)}} =\frac{1}{4} B_{[\sin(x)]^4}(\frac{1}{8},\frac{1}{2})+ \alpha \frac{(-1)^{3/4}}{4} B_{-\sin(x)^2}(-\frac{1}{4},-\frac{1}{2})+O(\alpha^2)
\end{eqnarray}
| {
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Solve for z(t) from the simultaneous equation using Laplace transform Solve for $z(t)$ from the simultaneous equation using Laplace transform
$$
y' + 2y + 6 \int\limits_0^t z \mathrm{d}t = -2 u(t) \\
y' + z' + z = 0$$
subject to $y(0) = -5$ and $z(0) = 6$.
| First note that if $u(t)$ is the Heaveside step function, then
$$\int_0^t z(s)\,dt = u(t)\ast z(t)$$
where $\ast$ denotes convolution. So we can rewrite the system as
$$\left\{\begin{aligned} y^{\prime}+2y + 6u(t)\ast z(t) &= -2u(t)\\ y^{\prime} + z^{\prime} + z &= 0\end{aligned}\right.$$
Noting that $y(0) = -5$ and $z(0) = 6$, we take Laplace transforms of both equations to see that
$$\left\{\begin{aligned} (sY(s)+5) + 2Y(s) + 6\frac{Z(s)}{s} &= - \frac{2}{s}\\ (sY(s)+5) + (sZ(s)-6) + Z(s) &= 0\end{aligned}\right.$$
which simplifies to
$$\left\{\begin{aligned}(s^2+2s)Y(s)+6Z(s) &= -5s-2\\ sY(s) + (s+1)Z(s) &= 1\end{aligned}\right.$$
We now solve for $Z(s)$ by means of elimination; multiplying the second equation by $-(s+2)$ and then adding to the first leaves us with
$$(6-(s^2+3s+2))Z(s) = -4-6s\implies Z(s) = \frac{4+6s}{s^2+3s-4}$$
Note that $\dfrac{1}{s^2+3s-4} = \dfrac{1}{(s+4)(s-1)} = \dfrac{1}{5(s-1)} - \dfrac{1}{5(s+4)}$.
Therefore,
$$Z(s) = \frac{4}{5(s-1)} -\frac{4}{5(s+4)} +\frac{6s}{5(s-1)} - \frac{6s}{5(s+4)}$$
Can you take things from here with finding the inverse Laplace Transform?
| {
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How prove this $\sqrt[5]{1782+\sqrt[3]{35+15\sqrt{6}}+\cdots}$ is positive integer numbers. Prove that
$$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}\in N$$
This problem from this
My try: let
$$x=\sqrt[3]{35+15\sqrt{6}}+\sqrt[3]{35-15\sqrt{6}}$$
then
$$x^3=70+3\sqrt[3]{35^2-(15\sqrt{6})^2}\cdot x=70-3\sqrt[3]{125}x$$
then we have
$$(x^3-70)^3=-27\cdot 125 x$$
$$\Longrightarrow x^6-15x^4-140x^3+225x^2+1050x+4900=0$$
Then I can't ,and I think this is interesting problem, Thank you
| Denotes $t=\sqrt[3]{15 \sqrt{6}+35}+\sqrt[3]{35-15 \sqrt{6}},$ then $t^3+15 t-70=0.$
$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}$
$=\sqrt[5]{1782+405t}-t$
Since $(t+2)^5-(1782+405t)=(t+5)^2 \left(t^3+15 t-70\right)=0$, we get $\sqrt[5]{1782+405t}-t=2.$
| {
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Find a power series solution centered at 0 (Differential equations Here's the problem: $$(x-1)y''+y'=0$$ This is the work that I've already done: $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n$$
$$y''=\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n$$
I then plug those into the original equation:
$$(x-1)\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n=0$$
$$\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1}-\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n=0$$
And this is where I'm stuck and don't know what to do from here. If someone could help me out that would be great.
| You have
$$
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1}-\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n = 0.
$$
Rewrite the first sum as
$$
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1} = \sum_{m=1}^{\infty}(a_{m+1})(m+1)(m)x^m.
$$
Rewrite the second sum as
$$
\begin{align}
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n &= 2a_2 + \sum_{n=1}^{\infty}(a_{n+2})(n+2)(n+1)x^n \\
&= 2a_2 + \sum_{m=1}^{\infty}(a_{m+2})(m+2)(m+1)x^m.
\end{align}
$$
Rewrite the third sum as
$$
\begin{align}
\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n &= a_1 + \sum_{n=1}^{\infty}(a_{n+1})(n+1)x^n \\
&= a_1 + \sum_{m=1}^{\infty}(a_{m+1})(m+1)x^m.
\end{align}
$$
The first expression thus becomes
$$
\sum_{m=1}^{\infty}(a_{m+1})(m+1)(m)x^m - 2a_2 - \sum_{m=1}^{\infty}(a_{m+2})(m+2)(m+1)x^m + a_1 + \sum_{m=1}^{\infty}(a_{m+1})(m+1)x^m = 0.
$$
Now combine the sums to get
$$
a_1 - 2a_2 + \sum_{m=1}^{\infty} \Bigl[ m(m+1)a_{m+1} - (m+1)(m+2)a_{m+2} + (m+1)a_{m+1} \Bigr] x^m = 0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite Integral $\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$ How can I evaluate this integral?
$$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$$
| Write $$3\sin x+2\cos x=A(2\sin x+3\cos x)+B\frac{d(2\sin x+3\cos x)}{dx}$$
$$\implies 3\sin x+2\cos x=A(2\sin x+3\cos x)+B(2\cos x-3\sin x)$$
$$\implies 3\sin x+2\cos x=(2A-3B)\sin x+(3A+2B)\cos x$$
Solve for $A,B$ equating the coefficients of $\cos x,\sin x$
So, $$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx=A+B\int\frac{d(2\sin x+3\cos x)}{2\sin x+3\cos x}$$
| {
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Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substitution $x=a\sin t$ and $x=a\sec t$ Both have looked promising but I just can't seem to finish it out.
Thanks in advance for any help.
| Calculating this integral can be done by parts:
$$I=\int x^2\sqrt{a^2-x^2} dx= -\int x(\frac{1}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}} )' dx=$$
$$=-\frac{x}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}}+\frac{1}{3}\int(a^2 -x^2)\sqrt{a^2-x^2}dx.
$$
$$\frac{4}{3}I=-\frac{x}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}}+a^2\int\sqrt{a^2-x^2} dx. (1)$$
And$$J=\int\sqrt{a^2-x^2} dx =\int(x)'\sqrt{a^2-x^2} dx =x\sqrt{a^2-x^2} -\int\frac{a^2-x^2-a^2}{\sqrt{a^2-x^2}}dx=$$
$$=x\sqrt{a^2-x^2}-J+a^2\arcsin\frac{x}{a}$$ from which is obtained$$J=\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\arcsin\frac{x}{a}.$$
Substituting formula $I$ in $(1)$ we obtain $J$.
| {
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Solving $(z+1)^5 = z^5$ The question says to solve this equation: $(z+1)^5 = z^5$
I did. Just want to find out if I did it properly and if my run-around logic makes sense.
First I begin my writing the equations as:
$$ (z+1)^5 = z^5$$
$$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$
So $$ \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i $$
$$ \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i $$
Now, because the natural logarithm is one-to-one, I write:
$$ \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|$$
So assign $ z = a +bi$
So that $|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 $
So, $z = -\frac12 + bi$ and $z+1 = \frac12 + bi$ for some $b \in \mathbb R$
Now to find $b$
$$ \mathbf{Arg}(z+1) = \mathbf{Arg}(z)$$
$$ \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}$$
I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly?
Ultimately, I get $ z = -\frac12$ which upon inspection...is wrong...
| If you divide both sides by $z^5$ (note that $z\ne0$, since for $z=0$ we get $0=1$), you get
$$
\left(1+\frac1z\right)^5=1.
$$
The expression in brackets cannot be $1$, so we are left with the four non-trivial fifth roots of unity:
$$
1+\frac1z=e^{2\pi i k/5},\ \ k=1,2,3,4.
$$
So we get four solutions, namely
$$
z=\frac1{e^{2\pi i k/5}-1},\ \ k=1,2,3,4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/577847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 1
} |
question about the limit $\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}$ Because $\sin'(x)=\cos(x)$ we can prove that $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$. but, by definition we have
$$\arcsin'(x)=\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}\tag{1}$$
therefore,
$$\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}\tag{2}$$
My question: Can we prove $(2)$ without using $(1)$? How?
| Here is one way forward. Note that $\sin (\arcsin (x))=x$ and $\cos (\arcsin(x))=\sqrt{1-x^2}$ for $|x|\le 1$. Therefore, we have for $|x|\le 1$
$$\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)}{h} \tag 1$$
Now, in THIS ANSWER, I showed using only basic inequalities that for $|x|<1$
$$\left|\frac{x}{\sqrt{1-x^2}}\right|\ge |\arcsin (x)|\ge |x| \tag 2$$
Using the right-hand side inequality of $(2)$ in $(1)$ yields
$$\begin{align}
\left|\frac{\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)}{h}\right|&\ge \left|\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h}\right|\\\\
&=\frac{(x+h)^2(1-x^2)-x^2(1-(x+h)^2)}{h\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)}\\\\
&=\frac{2x}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}} \tag 3
\end{align}
$$
And using the left-hand side inequality of $(2)$ in $(1)$ yields
$$\begin{align}
\left|\frac{\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)}{h}\right|&\le \frac{\frac{2x}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}}{\sqrt{1-\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)^2}} \tag 4
\end{align}
$$
Using the Squeeze Theorem, we obtain from $(3)$ and $(4)$
$$\lim_{h\to 0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/579170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)$ without using a calculator Evaluate without using a calculator:
$\displaystyle{\tan\left(2\sin^{-1}\left(\sqrt{5} \over 5\right)\right).}$
So I built my triangle hyp=$5$, adj=$2\sqrt{5}$, opp=$\sqrt{5}$.
$$
\tan\left(2\theta\right) = 2\sin\left(\theta\right)\cos\left(\theta\right)\,,
\qquad\qquad
\tan\left(2\theta\right) = 2\,{\sqrt{5} \over 5}\,{2\sqrt{5} \over 5} = 4.
$$
I used my calculator and it says the answer is $4/3$. I'm not sure what I'm doing wrong
|
Using the identity $\tan{2A}=\dfrac{2\tan A}{1-\tan^2 A}$, $\tan \theta=\dfrac{\sqrt{5}}{\sqrt {20}} =\dfrac{1}{2}. \text{Hence} \tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)=\dfrac{2\times \dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{4}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/579839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How can I prove this statment? I have this problem to solve:
Prove that when $$x\rightarrow 0$$ $$\sqrt{1+x}-1 \sim \frac{x}{2}$$
Can someone give me a tip? or show me the way?
Thanks in advance
| $$\sqrt { 1+x } -1=\sqrt { 1+x+\frac { { x }^{ 2 } }{ 4 } -\frac { { x }^{ 2 } }{ 4 } } -1=\sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 }-\frac { { x }^{ 2 } }{ 4 } } -1\\$$For $x$ very small$$ \frac { { x }^{ 2 } }{ 4 } <<{ \left( \frac { x }{ 2 } +1 \right) }^{ 2 },$$therefore you can ignore the term $\frac { { x }^{ 2 } }{ 4 } $ in $\sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 }-\frac { { x }^{ 2 } }{ 4 } } -1$. Then you have $$\sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 }-\frac { { x }^{ 2 } }{ 4 } } -1\sim \sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 } } -1\sim \frac { x }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/580437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$ Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuck into the abyss of incertitude.
Thus, any help to get me out of this rift is more than welcome!
| I like how Yiorgos S. Smyrlis approached to find upper limit of $L$. In similar way, you can easily observe $\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} > L$ and lets assume that it converges to some constant $c$.
Now, we can write ,
$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c$
Squaring on both sides,
$\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c^2$
Which is nothing but,
$\frac{1}{2}+c = c^2$
Solving above quadratic expression, value of $c$ will be $\dfrac{1+\sqrt{3}}{2}$ .
Thus we get slightly improved upper bound for $L$ as $ L < \dfrac{1+\sqrt{3}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/582196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 4,
"answer_id": 1
} |
Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$
Thanks in advance for any help.
| As $\displaystyle\binom n2=\frac{n(n-1)}2=\frac12\cdot n^2-\frac12\cdot n$
$$\sum_{2\le n\le m}\binom n2=\frac12 \sum_{2\le n\le m}n^2-\frac12\sum_{2\le n\le m} n$$
$$=\frac12\left( \sum_{1\le n\le m}n^2-1\right)-\frac12\left(\sum_{1\le n\le m} n-1\right)$$
$$=\frac12 \frac{m(m+1)(2m+1)}6-\frac12\frac{m(m+1)}2$$
$$=\frac{m(m+1)(2m+1)-3m(m+1)}{12}$$
$$=\frac{(m+1)m}{12}(2m+1-3)=\frac{(m+1)m(m-1)}6=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 2
} |
evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+1)}$
Now I Did not understand how can i solve after that
Help Required
Thanks
| $$\frac{x+1}{(x^2+x+1)(x^3-x^2+1)}=\frac{2x+3}{7(x^2+x+1)}-\frac{2x^2-x-4}{7(x^3-x^2+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$
let sequence $\{a_{n}\}$ such $a_{1}=1$,and
$$a_{n+1}a_{n}=n,n\ge 1$$
show that
$$2\sqrt{n}-1\le\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{2}\sqrt{n}-1$$
(2): I consider we can find this limit
$$\lim_{n\to+\infty}\dfrac{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}{\sqrt{n}}=?$$
My try:since
$$a_{n+2}a_{n+1}-a_{n+1}a_{n}=n+1-n=1$$
so
$$a_{n+2}=\dfrac{1}{a_{n+1}}+a_{n}$$
so
$$\dfrac{1}{a_{n+1}}=a_{n+2}-a_{n}$$
so
\begin{align*}\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}&=a_{1}+(a_{3}-a_{1})+(a_{4}-a_{2})+(a_{5}-a_{3})+\cdots+(a_{n+1}-a_{n-1})\\
&=a_{1}+a_{n+1}+a_{n}-a_{1}-a_{2}\\
&=a_{n+1}+a_{n}-a_{2}
\end{align*}
since $$a_{1}=1,a_{1}a_{2}=1\Longrightarrow a_{2}=1$$
so
$$a_{n+1}+a_{n}-a_{2}\ge2\sqrt{a_{n+1}a_{n}}-1\ge 2\sqrt{n}-1$$
so left hand inequality is prove it.Now consider Right hand inequality,we only prove this
$$a_{n}+a_{n+1}<\dfrac{5}{2}\sqrt{n}$$
since
$$a_{n}a_{n+1}=n\Longrightarrow a_{n}+\dfrac{n}{a_{n}}<\dfrac{5}{2}\sqrt{n}$$
so maybe we can $a_{n}<2\sqrt{n}$?
becasue $$\dfrac{5}{2}\sqrt{n}=2\sqrt{n}+\dfrac{n}{2\sqrt{n}}$$
so let $$f(x)=x+\dfrac{n}{x}$$
if we can prove $a_{n}<2\sqrt{n}$,Then
$$f(x)\le f(2\sqrt{n})=\dfrac{5}{2}\sqrt{n}$$
Thank you very much
| Hint : $a_n=\displaystyle\frac{(n-1)!!}{(n-2)!!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root
$a,b$ are real numbers. Roots may or may not be real.
In this question, the aim is to find values of and b ,eg. 2,4.
| Is this what you want? If an equation $x^2+ax+b=0$ has roots $v,v^2-2$, then by Vieta's formulas, we have
$$v+(v^2-2)=-\frac{a}{1}=-a, v(v^2-2)=\frac{b}{1}=b.$$
Then, we have $(a,b)=(-v^2-v+2,v^3-2v)$.
So, you can get $(a,b)$ if you know the value of $v$.
Edit : OK. The answers for the "new" question is the followings :
Since $f(v)=0\iff v^2=-av-b,$
$$f(v^2-2)=(v^2-2)^2+a(v^2-2)+b=(-av-b-2)^2+a(-av-b-2)+b$$$$=\cdots =v(-a^3+2ab+4a-a^2)+(-a^2b+5b-ab-2a+b^2+4).$$
This has to be zero independently to $v$, so we have
$$-a^3+2ab+4a-a^2=0\ \text{and}\ -a^2b+5b-ab-2a+b^2+4=0$$
$$\iff\ a(a^2+a-2b-4)=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0$$
$$\iff\ "a=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0"\ \text{or}\ "a^2+a-2b-4=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0"$$
1) The former leads $(a,b)=(0,-1),(0,-4)$.
2) The latter leads $(a,b)=(-4,4),(-1,-2),(1,-1),(2,1)$. (you can set $b=(a^2+a-4)/2$ in the other equation. Then, you'll get an equation of $a$, so you can get $a$, then set them in $b=(a^2+a-4)/2$.)
Hence, we know that the followings are the necessary conditions :
$$(a,b)=(0,-1),(0,-4),(-4,4),(-1,-2),(1,-1),(2,1).$$
On the other hand,
1) The $(0,-1)$ case has $v=1$, so this case is sufficient.
2) The $(0,-4)$ case has $v=-2$, so this case is sufficient.
3) The $(-4,4)$ case has $v=2$, so this case is sufficient.
4) The $(-1,-2)$ case does not have any $v$, so this case is not sufficient.
5) The $(1,-1)$ case has $v=\frac{-1\pm \sqrt 5}{2}$, so this case is sufficient.
6) The $(2,1)$ case has $v=-1$, so this case is sufficient.
Hence, we now reach that the answer is either of the followings :
$(a,b,v)=(0,-1,1),(0,-4,-2),(-4,4,2),(1,-1,\frac{-1\pm\sqrt 5}{2}),(2,1,-1)$.
Note that each of the $(-4,4,2),(2,1,-1)$ cases has an relation of $v=v^2-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Find all the positive integer Find all the positive integers (x,y), such that
a) $1!+2!+3!+\cdots+ x!=y^2$
b)$1!+2!+3!+\cdots+x!=y^z$
| For part b, assuming $z$ is also a positive integer,for $x\ge 27$, first note that the last digit of summation is 3, so the last digit of $y$ must be $3$ or $7$ and $z=4k+1$ or $z=4k+3$
$1!+2!+\cdots +x!\equiv 9 \pmod{27}$
But $y^{4k+3}$ or $y^{4k+1}$ cannot be $27t+9$. So all answers, if any, must be less than $27$. This can be checked by a computer, and there is no other $y^z$ equal to the sum but $3^2$ and trivial $1^z$
Note that:
$y^{4} \bmod 27 \in \{0,1,4,7,10,13,16,19,22,25\}\\
y^{3} \bmod 27 \in \{0,1,8,10,17,19,26\}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$ Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$
To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,
$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$
But even with that I don't get that result.
| We can recognize both expressions as squares:
*
*$3+2\sqrt2=2+2\sqrt2+1=(\sqrt2)^2+2\cdot \sqrt2\cdot 1+1^2=(\sqrt2+1)^2$, and
*$4-2\sqrt3=3-2\sqrt3+1=(\sqrt3-1)^2$.
This means that $\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=2+\sqrt2-\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Establish $\int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1}}{2 \cos(\pi a /2)}$ when $-1 < a < 1$ My attempt at a solution: (this is homework, btw)
Let $f(z) = \frac{z^a}{z^2 + b^2}dz$ then the singularities of $f$ occur at $\pm ib$.
$$
Res(f; ib) = \frac{z^a}{z + ib} \biggr |_{ib} = \frac{(ib)^a}{2ib}
$$
$$
Res(f; -ib) = \frac{z^a}{z - ib} \biggr |_{-ib} = \frac{(-ib)^a}{-2ib}
$$
We sum the residues and multiply by $2 \pi i$ to give the value of a contour integral containing the two poles.
$$
2 \pi i \cdot \biggl ( \frac{(ib)^a}{2ib} - \frac{(-ib)^a}{2ib} \biggr ) = \pi b^{a-1} (i^a - (-i)^a) = \pi b^{a-1} (e^{\pi i a/2} - e^{- \pi i a/2}) = \pi b^{a-1} 2 i \sin (\pi i a/2)
$$
We use a circle contour centered at the origin with the origin cut out (since $a$ can be negative)
Also we make the branch cut for $x^a$ along the positive real axis.
$\gamma = [r, R]$
$\mu_R = Re^{i \theta}$, where $0 < \theta < 2\pi$
$\kappa = [-R, -r]$
$\mu_r = re^{i \theta}$, where $0 < \theta < 2\pi$
$$
\pi b^{a-1} 2 i \sin (\pi i a/2) = \int_r^R \frac{x^a}{x^2 + b^2}dx + \int_0^{2\pi} \frac{(Re^{i \theta})^a}{(Re^{i \theta})^2 + b^2}i R e^{i \theta} d \theta + \int_{R}^{r} -\frac{x^a}{x^2 + b^2}dx + \int_{2\pi}^0 \frac{(re^{i \theta})^a}{(re^{i \theta})^2 + b^2}i r e^{i \theta} d \theta
$$
As $r \rightarrow 0$ and $R \rightarrow \infty$ we have that the second and fourth integrals go to zero. Thus,
$$
\pi b^{a-1} 2 i \sin (\pi i a/2) = 2 \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx \implies \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1} i \sin (\pi i a/2)}{2}
$$
So where did I go wrong?
| My corrected answer (thanks to Daniel Fischer and robjohn):
We use a circle contour centered at the origin with the origin cut out (since $a$ can be negative).
(Also we make the branch cut for $x^a$ along the positive real axis.)
$\gamma = [r, R]$
$\mu_R = Re^{i \theta}$, where $0 < \theta < 2\pi$
$\kappa = [-R, -r]$
$\mu_r = re^{i \theta}$, where $0 < \theta < 2\pi$
$$
\int_{\Gamma} f = \int_r^R \frac{x^a}{x^2 + b^2}dx + \int_0^{2\pi} \frac{(Re^{i \theta})^a}{(Re^{i \theta})^2 + b^2}i R e^{i \theta} d \theta + \int_{R}^{r} \frac{e^{2 \pi i a}x^a}{x^2 + b^2}dx + \int_{2\pi}^0 \frac{(re^{i \theta})^a}{(re^{i \theta})^2 + b^2}i r e^{i \theta} d \theta
$$
As $r \rightarrow 0$ and $R \rightarrow \infty$ we have that the second and fourth integrals go to zero. Thus,
$$
\int_{\Gamma} f = (1 - e^{2 \pi i a}) \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx
$$
Now we calculate the residues of $f$ occur at $\pm ib$.
$$
Res(f; ib) = \frac{z^a}{z + ib} \biggr |_{ib} = \frac{(ib)^a}{2ib} \qquad Res(f; -ib) = \frac{z^a}{z - ib} \biggr |_{-ib} = \frac{(-ib)^a}{-2ib}
$$
We sum the residues and multiply by $2 \pi i$ to give the value of a contour integral containing the two poles.
$$
2 \pi i \cdot \biggl ( \frac{(ib)^a}{2ib} - \frac{(-ib)^a}{2ib} \biggr ) = \pi b^{a-1} (i^a - (-i)^a) = \pi b^{a-1} (e^{\pi i a/2} - e^{3 \pi i a/2})
$$
Then
\begin{align*}
\int_0^{\infty} \frac{x^a}{x^2 + b^2}dx &= \frac{ \pi b^{a-1} (e^{\pi i a/2} - e^{3 \pi i a/2}) }{1 - e^{2 \pi i a}}\\
&= \frac{ \pi b^{a-1} (e^{-\pi i a/2} - e^{ \pi i a/2}) }{e^{- \pi i a} - e^{ \pi i a}}\\
&= \frac{ \pi b^{a-1} \sin (\pi a/2)}{\sin( \pi a)}\\
&= \frac{ \pi b^{a-1} \sin (\pi a/2)}{\sin( \pi a)}\\
&= \frac{ \pi b^{a-1} \sin (\pi a/2)}{2 \sin( \pi a/2) \cos( \pi a/2)}\\
&= \frac{ \pi b^{a-1}}{2\cos( \pi a/2)}\\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\
&= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\
&= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx
\end{align}
$$
Using the identities
$$
\sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}
$$
we can transform the integral to
$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx
$$
The integral is easy to calculate from here.
My Attempt for $(2)$:
$$
\begin{align}
J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\
&= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx
\end{align}
$$
How can I solve $(2)$ from this point?
| $\displaystyle \begin{aligned}\int \frac{1}{\sin ^3 x+\cos ^3 x} d x & =\frac{1}{3} \int \left(\frac{2}{\sin x+\cos x}+\frac{\sin x+\cos x}{\sin ^2 x-\sin x \cos x+\cos ^2 x}\right) d x \\& =\frac{2}{3} J+\frac{1}{3} K \\\\J&=\int \frac{1}{\sin x+\cos x} d x \\& =\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x-\frac{\pi}{4}\right)} \\& =\frac{1}{\sqrt{2}} \ln\left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right) \right|+c_1 \\\\K&=2 \int \frac{\sin x+\cos x}{2-2 \sin x \cos x} d x \\& =2 \int \frac{d(\sin x-\cos x)}{1+(\sin x-\cos x)^2} \\& =2 \tan ^{-1}(\sin x-\cos x)+c_2 \\\\\therefore \int \frac{1}{\sin ^3 x+\cos ^3 x} d x &=\frac{\sqrt{2}}{3} \ln \left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right) \right|+2 \tan ^{-1}(\sin x-\cos x)+C \\& =\frac{\sqrt{2}}{3} \ln \left|\frac{\sqrt{2}+\sin x-\cos x}{\cos x+\sin x}\right|+\frac{2}{3} \tan ^{-1}(\sin x-\cos x)+C \\&\end{aligned}\tag*{} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Convergence of $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ I want to check, whether $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges or diverges.
I tried to use Leibniz's test :
$|a_n|= \frac{n^2}{\sqrt{n^5+1}} = \frac{n^2}{\sqrt{n^4(n+\frac{1}{n^4})}} = \frac{n^2}{n^2\sqrt{n+\frac{1}{n^4}}} = \frac{1}{\sqrt{n+\frac{1}{n^4}}}$
So $\lim\limits_{n \rightarrow \infty}{{\frac{1}{\sqrt{n+\frac{1}{n^4}}} = 0}} $
$1>|\frac{a_{n+1}}{a_n}|= \frac{(n+1)^2}{\sqrt{(n+1)^5+1}} \frac{\sqrt{n^5+1}}{n^2}= \frac{2n+1 \sqrt{n^5+1}}{\sqrt{(n+1)^5+1}}= \frac{n^2+2n+1 \sqrt{n+\frac {1}{n^4}}}{(n+1)^2\sqrt{n+1+1}} = \frac {\sqrt{n+\frac {1}{n^4}}}{\sqrt{n+2}}$
So $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges.
Could somebody please check my solution?
| I am not sure about Leibitz, but the terms are decreasing in absolute value, and the series is alternating, so you are good.
| {
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"url": "https://math.stackexchange.com/questions/596053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does the graph of $e^{1/z}$ look like a dipole? I was looking at the color wheel graph of $e^{1/z}$, and my girlfriend commented that it looked just like a dipole. Does anyone have an explanation for that, why the geometry would be so similar? I guess as we follow e.g. the red color from the left side as it goes up, and then back down to the right side, we are tracing a ray outward from the origin in $\mathbb{C}$.
| Try $e^{1/z^2}$, you will get an even better match. This is due to the following:
The potential due to the dipole you have in the figure is given by $\phi(x,y) = -\dfrac{x}{x^2+y^2}$. Hence, the electric field is given as
$$\vec{E} = \left(\dfrac{x^2-y^2}{(x^2+y^2)^2}, \dfrac{2xy}{(x^2+y^2)^2}\right) \tag{$\star$}$$
If we expand $e^{1/z}-1$, we get
$$\dfrac{\bar{z}}{r^2} + \dfrac{\bar{z}^2}{2r^4} + \mathcal{O}(1/r^3)$$
If we look at the first two terms, we get
$$\dfrac1z + \dfrac1{2z^2} = \dfrac{x-iy}{x^2+y^2} + \dfrac{(x-iy)^2}{2(x^2+y^2)^2}$$
which gives us
$$\left(\dfrac{x^2-y^2+2x(x^2+y^2)}{2(x^2+y^2)^2},- \dfrac{xy+y(x^2+y^2)}{(x^2+y^2)^2}\right)$$
If you look at $e^{1/z^2}$, and truncate with the term $\dfrac1{z^2}$, you will get the exact thing as $\star$.
We have
$$e^{1/z^2}-1 = \dfrac1{z^2} + \mathcal{O}(1/z^4)$$
And
$$\dfrac1{z^2} = \dfrac{\bar{z}^2}{r^4} = \dfrac{(x^2-y^2) -2ixy}{(x^2+y^2)^2} = \left(\dfrac{x^2-y^2}{(x^2+y^2)^2}, -\dfrac{2xy}{(x^2+y^2)^2}\right)$$
which agrees with $\star$ but for the sign.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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Find the domain of the function $f(x)=\dfrac{64+x^2+x}{9x^2+27x+9}$.
Find the domain of the function $f(x)=\dfrac{64+x^2+x}{9x^2+27x+9}$.
I don't know how to start nor how to do it. Please help me!
| The domain of a function is the set of argument values for which the function is defined.
We need to make $\displaystyle\frac{64+x^2+x}{9x^2+27x+9}$ defined
$\displaystyle\implies 9x^2+27x+9\ne0\iff x^2+3x+1\ne0$
Now, the roots of $\displaystyle x^2+3x+1=0$ are $x=\frac{-3\pm\sqrt5}2$ which will make the given function undefined
So, the domain will be $\left(-\infty, \infty\right)$ excluding those two points i.e.,
$\displaystyle\left(-\infty, \frac{-3-\sqrt5}2\right)\cup\left(\frac{-3-\sqrt5}2,\frac{-3+\sqrt5}2\right)\cup\left(\frac{-3+\sqrt5}2,\infty\right)$
Also observe that $\displaystyle\lim_{x\to\infty}\frac{64+x^2+x}{9x^2+27x+9}=\lim_{x\to\infty}\frac{\frac{64}{x^2}+1+\frac1x}{9+\frac{27}x+\frac9{x^2}}=\frac19$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Residues to solve an improper integral I'm asked to solve the following improper integral:
$$\int_0^\infty \frac{\rm {Log}^2(t)}{1+t^2}dt. $$
Do I consider the function $f(z) = \frac{\rm{Log}^2(z)}{1+z^2}$ or some variant?
Is the correct contour the so called "keyhole" that skips the $(0,+\infty)$ interval?
Any hints and tips are appreciated. I don't want this problem to be completely solved, just need the starter and some ideas that I can use in this case and similar ones.
Thanks a lot!
| To use a keyhole contour $C$, consider
$$\oint_C dz \frac{\log^3{z}}{1+z^2}$$
You can show that the integrals about the circular arcs about the origin, both large and small, vanish as theire respective radii go to $\infty$ and $0$. The contour integral is therefore equal to
$$\int_0^{\infty} dx \frac{\log^3{x} - (\log{x}+i 2 \pi)^3}{1+x^2} $$
which simplifies to
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{1+x^2} + i 8 \pi^3 \int_0^{\infty} \frac{dx}{1+x^2} $$
The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, namely $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$. (Note that it is crucial that $\arg{z} \in [0,2 \pi)$.) This sum is
$$\frac{-i \pi^3/8}{2 i} + \frac{-i 27 \pi^3/8}{-2 i} = \frac{13 \pi^3}{8} $$
Multiplying by $i 2 \pi$, we have
$$i \left [- 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + 8 \pi^3 \int_0^{\infty} \frac{dx}{1+x^2} \right ] + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{1+x^2} = i \frac{13 \pi^4}{4}$$
Equating imaginary parts and noting that the second integral in the brackets is simply $\pi/2$ (you should have permission to evaluate that without residues), we have
$$- 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} = \frac{13 \pi^4}{4} - 4 \pi^4 = -\frac{3 \pi^4}{4} $$
or
$$\int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} = \frac{\pi^3}{8}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that the polynomial is irreducible Let $a \in 3 + 4\mathbb{Z}$ and $f\left( x \right) = \left( {x + 1} \right)\left( {x - 1} \right)\left( {x + 4} \right)\left( {x - 2} \right) + a$ a polynomial. Using the homomorphism $\pi :\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$, see if $f$ is irreducible over $\mathbb{Q}$.
My attempt: It's easy to show that all rational zeroes of $f$ must be integers and that there are no integer zeroes. Also, I can show that if $\pi \left( f \right)\left( x \right) = \overline 1 {x^4} + \overline 2 {x^3} + \overline 3 {x^2} + \overline 2 x + \overline 3 $ is irreducible, then so is the $f$ (notation $\overline k = k + 4\mathbb{Z}$). However, the Eisenstein criterion fails since the only ideals in $\mathbb{Z}/4\mathbb{Z}$ are $\mathbb{Z}/4\mathbb{Z}$, $\left\{ {\overline 0 ,\overline 2 } \right\}$ and $\left\{ {\overline 0 } \right\}$, of which $\left\{ {\overline 0 ,\overline 2 } \right\}$ and $\left\{ {\overline 0 } \right\}$ are prime, and $\overline 3 $ is not contained in any of the prime ideals.
| Modulo $2$, we have
$$(x^2+x+1)^2 = x^4 + 2x^3+3x^2+2x+3$$
so any monic factor of $ x^4 + 2x^3+3x^2+2x+3$ in $\mathbf Z/4\mathbf Z$ reduces to $x^2+x+1$ modulo $2$. There are only $4$ possible factors to check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove using mathematical induction pt 2 Assumed that i asked a question like 30 min ago thinking i got the hang of this, seems not.
So $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac12n(6n^2-3n-1) \text{ for all } n\in\mathbb N$$
This time it seems way harder with the squares.
so i did the steps and got stuck on the 3rd step(Again).
Step 1: prove LHS = RHS which it does for n=1
Step 2: Assume $n=k$ is true $$1^2+4^2+7^2+\dots+(3k-2)^2=\frac12k(6k^2-3k-1)$$
Step 3: would $n = k+1$? And would $n = k+1$ work for all equations?could someone help me with the last step, would be appreciated thanks
EDIT: Cheers for the help, i know where i went wrong!
| Hypothesis:
$$
1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1)
$$
Thesis:
$$
1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2=\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)
$$
By the induction hypothesis
$$
1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2=
\frac{1}{2}n(6n^2-3n-1)+(3(n+1)-2)^2
$$
Write this as a polynomial in $n$; write
$$
\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)
$$
as a polynomial in $n$.
Verify that the expressions are the same. End.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/600608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$\textbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=\frac{1}{2}(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}})$$
$$\Rightarrow (1)-(2)=\textbf{S}_{n}-\frac{1}{2}\textbf{S}_{n}=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \textbf{S}_{n}(1-\frac{1}{2})=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \frac{1}{2^{n+1}}\rightarrow\textbf{0}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \textbf{S}_{n}\rightarrow\textbf{1}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \lim_{n \to \infty}\textbf{2}^{\textbf{S}_{n}}=2\quad\textit{when n}\rightarrow\infty$$
| It's more simple to prove that, let $\sqrt{2\sqrt{2\sqrt{2}...}}=t$. Then, $t^2=2t \rightarrow t=2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Limit of series $4\left( \frac {1}{8}+\frac {1}{12}\right) +6\left( \frac {1}{24}+\frac {1}{36}\right) +\ldots$ How to find this serie
$4\left( \dfrac {1}{8}+\dfrac {1}{12}\right) +6\left( \dfrac {1}{24}+\dfrac {1}{36}\right) +8\left( \dfrac {1}{64}+\dfrac {1}{96}\right) +\ldots $
I think it's telescopic, isn't it?
| multiple the numbers before parantheses then it's formula for the general will be obtained
$$\sum_{n=1}^\infty(\frac{1}{2\cdot 2^{n-1}}+\frac{1}{3\cdot 2^{n-1}})$$
and it's equal to following geometric serie:
$$\frac{5}{6}\sum_{n=1}^\infty\frac{1}{2^{n-1}}=\frac{10}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Trigonometry - parameter For which real parameter values $n$ equation $\frac{4n+3}{6}-\sin4x\cos4x-(n+\frac{2}{3})\sin(4x-\frac{\pi}{4})=0$ have exactly three solutions for $x \in [\frac{\pi}{16}, \frac{5\pi}{16}]$
Well I rewrite this equation as $4n+3-6\sin4x\cos4x-(3n+2)\sqrt2(\sin4x-\cos4x)=0 $ , but I don't know what's next
| First, I apologize for the length of the answer and my incapability to minimize answer as HINT only.
Squaring should be avoided as it immediately introduces Extraneous roots
Using $\sin2A=2\sin A\cos A$
$$4n+3-3\sin8x-(6n+4)\sin\left(4x-\frac\pi4\right)=0$$
Let $\displaystyle 4x-\frac\pi4=y$
$\displaystyle\implies(i)8x=\frac\pi2+2y,\sin8x=\sin\left(\frac\pi2+2y\right)=\cos2y=1-2\sin^2y $
and $(ii)$ as $\displaystyle\frac\pi{16}\le x\le\frac{5\pi}{16},\frac\pi4\le4x\le\frac{5\pi}4$
$\displaystyle\implies0\le4x-\frac\pi4\le\pi\implies0\le y\le\pi \ \ \ \ (2)$
So, we have $\displaystyle 4n+3-3(1-2\sin^2y)-(6n+4)\sin y=0 $
$\displaystyle\implies6\sin^2y-(6n+4)\sin y+4n=0 \ \ \ \ (1)$
$\displaystyle\implies\sin y=n,\frac23 $
If $y_1$ is a solution of $(1)$, so will be $\pi-y_1$ as $\sin(\pi-B)=\sin B$
Then we shall have either two or four solutions unless for one solution $\displaystyle y_1=\pi-y_1\iff y_1=\frac\pi2\implies \sin y_1=1$
$\displaystyle\sin y=-1$ also makes $\sin(\pi-y)=\sin y$ But from $(2),\sin y\ge0$
We already have two solutions due to $\displaystyle\sin y=\frac23$
So, $n$ must be $1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate limit of a sequence... NBHM $2013$ Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$
All I could do was to see that $$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$
Just because $\sin(2n\pi+\theta)=\sin(\theta)$..
So, we now have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin( \frac{1}{2n\pi}))$$
Now, as $\lim _{x\rightarrow \infty}x\sin(\frac{1}{x})=1$ we would have
$$\lim_{n\rightarrow \infty}2n\pi \sin( \frac{1}{2n\pi})=1$$
So, we now have $$\lim_{n\rightarrow \infty} \sin(1 + \frac{1}{2n\pi} \sin( \frac{1}{2n\pi}))$$
Now, as $\sin(x)$ is bounded and $\frac{1}{2n\pi} \rightarrow 0$ we would have
$$\lim_{n\rightarrow \infty}\frac{1}{2n\pi} \sin( \frac{1}{2n\pi})=0$$
So, we would now left with :
$$\lim_{n\rightarrow \infty} \sin(1)=\sin(1)$$
After all i would like to say that as $\sin (x)$ is continuous I can take limits inside.
So, we have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))=\sin 1$$
I would like somebody to check if I have done correctly and I would be thankful if any body can let me know if there is anything more to be specified to do so.
Thank you... :)
| Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$
As $\sin(2n\pi+\theta)=\sin(\theta)$ we would have :
$$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$
So, we now have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin( \frac{1}{2n\pi}))$$
Now, as $\lim _{x\rightarrow \infty}x\sin(\frac{1}{x})=1$ we would have
$$\lim_{n\rightarrow \infty}2n\pi \sin( \frac{1}{2n\pi})=1$$
So, we now have $$\lim_{n\rightarrow \infty} \sin(1 + \frac{1}{2n\pi} \sin( \frac{1}{2n\pi}))$$
Now, as $\sin(x)$ is bounded and $\frac{1}{2n\pi} \rightarrow 0$ we would have
$$\lim_{n\rightarrow \infty}\frac{1}{2n\pi} \sin( \frac{1}{2n\pi})=0$$
So, we would now left with :
$$\lim_{n\rightarrow \infty} \sin(1)=\sin(1)$$
After all i would like to say that as $\sin (x)$ is continuous I can take limits inside.
So, we have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))=\sin 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c \in \mathbb N$ and $\gcd(a^2-1,b^2-1,c^2-1)=1$, prove that $\gcd(ab+c,bc+a,ca+b)=\gcd(a,b,c)$. The whole problem is in the title. I'm out of ideas and can't think of a way to solve this, so I'd like to get some help. Thanks.
| *
*if $n$ divides $a$, $b$ and $c$ then it divides $(ab+c,bc+a,ca+b)$. Hence
$$gcd(a,b,c)\le gcd(ab+c,bc+a,ca+b) \quad [1]$$
*if $n$ divides $ab+c$, $bc+a$ and $ca+b$, then it divides
$$c(ca+b)-(bc+a)=a(c^2-1)$$
$$b(bc+a)-(ab+c)=c(b^2-1)$$
$$a(ab+c)-(ca+b)=b(a^2-1)$$
Hence $$gcd(ab+c,bc+a,ca+b)\le gcd(a(c^2-1),c(b^2-1),b(a^2-1))\quad [2]$$
Now, we know that $$gcd(xy,z)\le gcd(x,z).gcd(y,z)$$
and $gcd(a,a^2-1)=1$ (the same for $b$ and $c$) (So when you develop all terms you can ignore any gcd that contains $x$ and $x^2-1$)
Hence $$gcd(a(c^2-1),c(b^2-1),b(a^2-1))\le gcd(a,b,c).gcd(a^2-1,b^2-1,c^2-1)=gcd(a,b,c)$$
Hence (from $[2]$) $$gcd(ab+c,bc+a,ca+b)\le gcd(a,b,c) \quad [3]$$
Now from $[1]$ and $[3]$,
$$gcd(ab+c,bc+a,ca+b)=gcd(a,b,c)$$
| {
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} |
Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
| You can use an “unknown relation”: denote by ? either $<$ or $>$; so long as you do operations that don't change the direction of inequalities, the symbol will always mean the same; so, when you're done you'll know which one it is.
\begin{align}
1+3\sqrt{2}&\mathrel{?}3\sqrt{3}\\
(1+3\sqrt{2})^2&\mathrel{?}(3\sqrt{3})^2\\
1+6\sqrt{2}+18&\mathrel{?}27\\
6\sqrt{2}&\mathrel{?}27-1-18\\
6\sqrt{2}&\mathrel{?}8\\
3\sqrt{2}&\mathrel{?}4\\
(3\sqrt{2})^2&\mathrel{?}16\\
18&\mathrel{?}16
\end{align}
So ? was $>$.
| {
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"url": "https://math.stackexchange.com/questions/607061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to Plot $\sqrt{\frac{a^2+(b-1)^2}{a^2+(b+1)^2}}=2$ How to plot this complex division?
$$
\sqrt{\frac{a^2+(b-1)^2}{a^2+(b+1)^2}}=2
$$
| The plot of $\sqrt{\frac{a^2+(b-1)^2}{a^2+(b+1)^2}}=2$ is the following:
Work out $a^2+(b-1)^2$ and $a^2+(b+1)^2$. This will give:
$$\frac{a^2+(b-1)^2}{a^2+(b+1)^2} = \frac{1-4b}{a^2+b^2+2 b+1}$$
This would make it easier to solve.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find equation of a tangent on $y= \sin2x$ Find equation of a tangent on $y= \sin2x$ in intersections with $y=\frac{1}{2}$
What I calculated:
Intersections: $$\sin2x= \frac{1}{2}$$
...
$$0=tg^2x-4tgx-1$$
$$tgx_{1}=2+\sqrt{3}$$
$$x_{1}=75+k\pi$$
$$tgx_{2}=2-\sqrt{3}$$
$$x_{2}=15+ k\pi$$
$$T_{1}(75+k\pi,\frac{1}{2})$$ and $$T_{2}(15+k\pi,\frac{1}{2})$$
Derivative of $y= sin2x$ is: $y'= 2cos2x$
$$ y'(75)= -\sqrt {3} =k_{t_{1}} $$
$$y'(15)= \sqrt {3} =k_{t_{2}}$$
formula for calculating tangent that I used: $y-y_{1}=k_{t}(x-x_{1})$
1.)$T_{1}(75+k\pi,\frac{1}{2})$
$k_{t_{1}}=-\sqrt {3}$
$y_{1}-\frac{1}{2}=-\sqrt {3}(x-75-k\pi)$
$y_{1}=-\sqrt {3}x+75\sqrt {3}+\sqrt {3}k\pi+\frac{1}{2}$
2.)$T_{2}(15+k\pi,\frac{1}{2})$
$k_{t_{2}}=\sqrt {3}$
$y_{2}-\frac{1}{2}=\sqrt {3}(x-15 -k\pi)$
$y_{2}=\sqrt {3}x-15\sqrt {3}-\sqrt {3}k\pi+\frac{1}{2}$
Both solutions are wrong; what I should actually get:
$y_{1}=\sqrt{3}x-\frac{\sqrt{3}pi-6+12kpi\sqrt{3}}{12} $ and $y_{2}=-\sqrt{3}x-\frac{5\sqrt{3}pi+6+12kpi\sqrt{3}}{12};k=Z $
How could I possibly get that ?!
| You calculated your equations using the given $x$ (angles) in terms of degrees, not in radians: specifically, you need to multiply each angle in your equations of the lines $\,(75^\circ, 15^\circ\,)$ by the conversion factor $\frac \pi{180^\circ}$ to obtain those values in radians, which is how the solution equations are given.
Note that you've mixed degrees and radians when expressing the solutions to $x_1, x_2$:
$$x_{1}=75^\circ +k\pi$$ $$x_{2}=15^\circ+ k\pi$$
where you should express them as $$x_1 = \dfrac {5\pi}{12} + k\pi$$ $$x_2 = \dfrac{\pi}{12} + k\pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 is If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 is
$A. 1/4$
$B. 1/7$
$C. 1/8$
$D. 1/49$
I did this: Let $m>n$ (Clearly m and n can't be equal because $5$ can't divide $2*7^{m}$).
Now $7^{m}+7^{n}=7^{n}(7^{m-n}+1)$. If $5$ has to divide this, it implies that 5 has to divide $(7^{m-n}+1)$ (because it cannot divide a power of $7$). Since powers of $7$ have a cyclic order of $7,9,3,1$; $7^{m-n}$ has to therefore end in a $9$ and therefore $m-n$ can be $2,6,10,...,98$. Hence the only set of values of $n$ are $(1,2,3,...,97),(1,2,3,...,93),(1,2,3,...,89),...,(1)$. Also fixing $n$ would fix $m$.
Therefore the number of favorable cases is $97+93+89+...+2=1224$. Which means that the required probability should be $1224/100C2$ which turns out to be $68/275$, which is not matching with any of the options.. Where did I go wrong ??
Please help !!
| Shouldn't the answer be $24/100$? This is because there are 40 numbers which satisfy $7^x = 2^x = 1 \mod 5$, and $20$ each for $7^x = 2^x = 2,3,4 \mod 5$, so we can't just assign equal probabilities to each event.
( Saying equal likeliness of the events $1+4,2+3,3+2,4+1$ as the sum of the remainders out of $4\cdot 4=16$ possible combinations).
The answer should be
$$
\frac{40}{100} \cdot \frac{20}{100} + \frac{20}{100} \cdot \frac{20}{100} + \frac{20}{100} \cdot \frac{20}{100} + \frac{20}{100} \cdot \frac{40}{100} = \frac{24}{100}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove Trig Identity
For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$
This is what I've tried:
$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\right)}^y2\cos\left(\frac{B-C}2\right)\\
\sin\left(\frac{A-C}2+\frac{A-B}2\right)+\sin\left(\frac{A-C}2-\frac{A-B}2\right)\\
\left(\frac{B-C}2\right)\left[\sin\left(\frac{2A-B-C}2\right)+\sin\left(\frac{B-C}2\right)\right]\\
\cos\left(\frac{B-C}2\right)\sin\left(\frac{2A-B-C}2\right)+\cos\left(\frac{B-C}2\right)\sin\left(\frac{B-C}2\right)$$
I know it's basically turning sum into product and I previously tried to derive the LHS the same way the normal sum to product identity works but I couldn't figure out how. I tried to use the sum-product identity on RHS but couldn't quite make it work.
| You are missing a factor of $2$ in the last line of your answer. Anyway, you were on the right track. You just need to use the same type of identity (as you used in the beginning) for the first term and the double angle formula for the second.
$$
2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{2A - B - C}{2}\right) + 2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{B-C}{2}\right) \\
= \sin\left(\frac{B-C}{2} + \frac{2A - B - C}{2}\right) - \sin\left(\frac{B-C}{2} - \frac{2A - B - C}{2}\right) + \sin(B - C) \\
= \sin(A - C) - \sin(-A + B) + \sin(B - C) \\
= \sin(A - C) + \sin(A - B) + \sin(B - C)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/611324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$ After numerical analysis it seems that
$$
\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
$$
Could someone prove the validity of such identity?
| Yes, we can prove it. We can change the order of summation in
$$\begin{align}
\sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
&= \sum_{k=1}^\infty \frac{2k(2k+1)}{4^{2k+2}}\sum_{n=1}^\infty \frac{1}{n^{2k+2}}\\
&= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{2k(2k+1)}{(4n)^{2k+2}}\\
&= \sum_{n=1}^\infty r''(4n),
\end{align}$$
where, for $\lvert z\rvert > 1$, we define
$$r(z) = \sum_{k=1}^\infty \frac{1}{z^{2k}} = \frac{1}{z^2-1} = \frac12\left(\frac{1}{z-1} - \frac{1}{z+1}\right).$$
Differentiating yields $r''(z) = \frac{1}{(z-1)^3} - \frac{1}{(z+1)^3}$, so
$$1 - \sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} = \sum_{\nu = 0}^\infty \frac{(-1)^\nu}{(2\nu+1)^3},$$
and the latter sum is by an earlier answer using the partial fraction decomposition of $\dfrac{1}{\cos z}$:
$$\sum_{\nu=0}^\infty \frac{(-1)^\nu}{(2\nu+1)^3} = - \frac{\pi^3}{32} E_2 = \frac{\pi^3}{32}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value. Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.
I thought of it this way: $$f(x)=\begin{cases}2x-1-3(2x+4)+7 \,(\text{then I simplify)} & \text{if $x>0$}\\
-(2x-1-3(2x+4)+7)\,(\text{then I simplify)} & \text{if $x\le0$}\end{cases}$$
But is there some way without having to use the cases?
Edit: NEW work on this problem! I found three cases;
If $x\in ]-\infty,-2]$ then f(x)=$4x+20$
If $x\in]-2,1/2]$ then f(x)=$-8x-4$
If $x\in]1/2,+\infty[$ then f(x)=$-4x-6$
IS THIS TRUE?
Thank you very much!
| Let f(x)=|x|.We know $$f(x)=\begin{cases}
-x \text{ if x<0}\\
x\text{ if x>0}
\end{cases}$$
What if f(x)=|x-1|. We must have,
$$f(x)=\begin{cases}
-(x-1) \text{ if x<1}\\
x-1\text{ if x>1}
\end{cases}$$
Note that there are 3 terms in the given function.
$|2x-1|$, $3|2x+4|$ and 7.
The first term is postive if x$\gt \frac{1}{2}$. The second term is positive if $x\gt -2$. Hence we must have,
$$f(x)=2x-1-3(2x+4)+7 \text{ whenever x$\gt \frac{1}{2}$}$$
The first term is negative if x$\lt \frac{1}{2}$ and the second term is negative if x$\lt -2$.Hence we have a second condition,
$$f(x)=-(2x-1)+3(2x+4)+7 \text{ whenever x$\lt -2$}$$
Can you work out what will happen if x $\in (-2,\frac{1}{2})$?
! If x $\in (-2,\frac{1}{2})$ surely |2x-1|<0,hence |2x-1|=-(2x-1) and |2x+4|>0,hence |2x+4|=2x+4.
Finally we have the following for f(x)
$$f(x)=
\begin{cases}
2x-1-3(2x+4)+7~ \text{, x$\gt \frac{1}{2}$}\\
-(2x-1)+3(2x+4)+7~ \text{, x$\lt -2$}\\
-(2x-1)-3(2x+4)+7 ~\text{, x$\in(-2,\frac{1}{2})$}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$? If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must
be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$
My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using Vieta's formula
$$ \alpha+\alpha+\beta = -\frac{3b}{a}\Rightarrow 2\alpha +\beta = -\frac{3b}{a}$$
$$ \alpha \cdot \alpha +\alpha \cdot \beta +\alpha \cdot \beta = \frac{3c}{a}\Rightarrow \alpha^2+2\alpha \cdot \beta = \frac{3c}{a}$$
$$\alpha \cdot \alpha \cdot \beta = -\frac{d}{a}\Rightarrow \alpha^2 \cdot \beta = -\frac{d}{a}.$$
Now I did not understand how can I find the value of $\alpha$ in terms of $a,b$ and $c$.
Help is required.
Thanks
| First Answer:
\begin{align*}
2\alpha +\beta = -\frac{3b}{a} &\Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
&\Rightarrow {\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\\
&\Rightarrow -3\alpha^2-\frac{6b}{a}\alpha=\frac{3c}{a} \\
&\Rightarrow a{\alpha}^2+2b\alpha+c=0\tag{0}\\
&\Rightarrow 2a{\alpha}^3+4b{\alpha}^2+2c\alpha=0\tag{1}
\end{align*}
and
\begin{align*}
{\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.\tag{2}
\end{align*}
By subtracting (1) and (2) we have
$$b\alpha^2+2c\alpha+d=0\Rightarrow ab\alpha^2+2ac\alpha+ad=0.$$
Now from (0) we have
$$ab\alpha^2+2b^2\alpha+bc=0$$
Now subtract two last equations: $(2ac-2b^2)\alpha+ad-bc=0$
Thus
$$\fbox{$\alpha=\frac{bc-ad}{2(ac-b^2)}$}$$
Second Answer for alpha:
You can continue like this
$$
2\alpha +\beta = -\frac{3b}{a} \Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
$$
put it in the last one:
\begin{align*}
\left(-\frac{3b}{a}-2\alpha\right)\alpha^2=-\frac{d}{a}&\Rightarrow-2\alpha^3-\frac{3b}{a}\alpha^2=-\frac{d}{a}\\
&\Rightarrow 2a\alpha^3+3b\alpha^2-d=0\\
\end{align*}
on the other hand $\alpha$ must satisfy the original equation, thus
$$a\alpha^3+3b\alpha^2+3c\alpha+d=0$$
Then add two last equations:
$$3a\alpha^3+6b\alpha^2+3c\alpha=0$$
Since $\alpha\ne0$ we have : $a\alpha^2+2b\alpha+c=0$, thus
$$\alpha=-b+\sqrt{b^2-ac}\quad \text{ or }\quad \alpha=-b-\sqrt{b^2-ac} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove this equality $(x^2+2)(y^2+2)(z^2+2)\ge (\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2$ let $x,y,z>0$,and such
$$x+y+z=3$$
prove or disprove this
$$(x^2+2)(y^2+2)(z^2+2)\ge (\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2\tag{1}$$
I know this well know inequality 1
$$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$$
and we all know this inequality 2
$$(\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}+\sqrt{a^2+ab+b^2})^2\ge 3(a+b+c)^2$$
poof 1: since
$$(b-c)^2+2(bc-1)^2\ge 0$$
so
$$(b^2+2)(c^2+2)\ge 3\left(1+\dfrac{(b+c)^2}{2}\right)$$
Use Cauchy-Schwarz inequality ,we have
$$(a^2+2)\left(1+\dfrac{(b+c)^2}{2}\right)\ge (a+b+c)^2$$
By done!
proof 2: since
$$\sqrt{b^2+bc+c^2}=\sqrt{\dfrac{3}{4}(b+c)^2+\dfrac{1}{4}(b-c)^2}\ge\dfrac{\sqrt{3}}{2}(b+c)$$
so
$$\sum_{cyc}\sqrt{a^2+ab+b^2}\ge\sqrt{3}(a+b+c)$$
then
$$(\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}+\sqrt{a^2+ab+b^2})^2\ge 3(a+b+c)^2$$
By done!
But my inequality (1) is stronger this two inequality.Thank you
| $(\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2 \le 3(y^2+yz+z^2+z^2+zx+x^2+z^2+zx+x^2) \iff $
$(x^2+2)(y^2+2)(z^2+2)- 3(y^2+yz+z^2+z^2+zx+x^2+z^2+zx+x^2) \ge0 \iff x^2y^2z^2+2y^2z^2+2x^2z^2-2z^2-3yz-3xz+2x^2y^2-2y^2-3xy-2x^2+8\ge 0 $$ \tag 2$
let $3u=x+y+z,3v^2=xy+yz+xz,w^3=xyz,\to u=1\ge v \ge w$,
$(2)$ becomes:
$ w^6-24v^2w^3+3v^2+20 \ge 0 \iff w^6-24 w^3+3w^2+20\ge 0$ $ (3)$
$w^6+w^2+1\ge 3w^{\frac{8}{3}} \ge 3 w^3, 19+2w^2 \ge 21w^3 \implies $ $(3)$ is true.
the "=" will hold when $v=w=1=u \implies x=y=z=1$
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.