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How to show that $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}]=9$?
Fraleigh, Sec31, Ex9. Show that $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}]=9$.
Here is my trial: It is obvious that $\sqrt[3]2$ is algebraic of degree 3 over $\mathbb{Q}$, since $x^3-2$ is irreducible over $\mathbb{Q}$ by Eisenstein crieterion with $p=2$. Then we need to show that $\sqrt[3]3$ is algebraic of degree 3 over $\mathbb{Q}(\sqrt[3]2)$. Since $\sqrt[3]3$ is a zero of $x^3-3$, its degree is at most 3.
To show that $\sqrt[3]3$ is not of degree 1, i.e. $\sqrt[3]3 \not\in \mathbb{Q}(\sqrt[3]2)$, suppose that $\sqrt[3]3=a+b\sqrt[3]2+c\sqrt[3]4$, where $a, b, c \in \mathbb{Q}$. (The usual degree argument is not available since $\deg(\sqrt[3]3,\mathbb{Q})=3$ divides $\deg(\sqrt[3]2,\mathbb{Q})=3$.) Cubing both sides, $3=p+q\sqrt[3]2+r\sqrt[3]4$ with some $p, q, r$ in $\mathbb{Q}$, so $\sqrt[3]2$ is a zero of $rx^2+qx+p-3$, which is a contradiction to $\deg(\sqrt[3]2,\mathbb{Q})=3$.
Now to prove $\sqrt[3]3$ is not of degree 2, suppose that $\sqrt[3]3$ is a zero of quadratic polynomial. This means that $\sqrt[3]9=p\sqrt[3]3+q$ for some $p,q \in \mathbb{Q}(\sqrt[3]2)$. Cubing both sides, $9=3p^3+q^3+3p\sqrt[3]3q(\sqrt[3]3p+q)=3p^3+q^3+9pq:=a+b\sqrt[3]2+c\sqrt[3]4$ for some $a,b,c\in\mathbb{Q}$, which leads to the same contradiction.
Actually I didn't know how to solve it but while writing out this question, it seems that I solved the problem. But is the above solution right? And is there any other way to solve it? At first, I tried to show that $x=\sqrt[3]{2}+\sqrt[3]{3}$ is algebraic of degree 9 over $\mathbb{Q}$. Cubing yields that $x^3=5+3 \sqrt[3]{6}x$, so $(x^3-5)^3=162x^3$, so $x^9-15x^6-87x^3-125=0$ has $\sqrt[3]{2}+\sqrt[3]{3}$ as a zero. But I couldn't show that it is irreducible (Eisenstein criterion with $p=5$ fails.)
Edit: As Alex pointed out, it is sufficient to show that $x^3-3$ has no roots in $\mathbb{Q}(\sqrt[3]{2})$. And as Gerry pointed out, this process some more work than the above(check the nonzero condition). Suppose $(a+b\sqrt[3]2+c\sqrt[3]4)^3=3$. I did the heavy computation, $a^3+2b^3+4c^3+12abc+3\sqrt[3]2(a^2b+2b^2c+2c^2a)+3\sqrt[3]4(a^2c+b^2a+2c^2b)=3$, and stuck on here. How can I proceed here?
| Since this question has been bugging me for a few years, I decided to go ahead and chug out the calculations necessary to show "$q$ and $r$ can't both be zero," as Gerry Myerson said.
Since from your cubing, we get
$$p=a^3+2b^3+4c^3+12abc$$
$$q=3a^2b+6b^2c+6c^2a$$
$$r=3b^2a+6c^2b+3a^2c$$
and in Gerry Myerson's case we would have the second and third of these equations equal to zero and thus the first would have to be 3, we get
$$a^3+2b^3+4c^3+12abc=3$$
$$3a^2b+6b^2c+6c^2a=0$$
$$3b^2a+6c^2b+3a^2c=0$$
and the goal is to show that this can't happen with rational $a,b,c.$
Dividing the second and third equations by 3 and completing the square, we find that
$$(ba+c^2)^2=c^4-2b^3c$$
$$(ab+c^2)^2=c^4-ca^3$$
so that $a^3=2b^3.$ Then solving the first of the three equations for $abc,$ we get
$$abc=\frac{3-4b^3-4c^3}{12}.$$
Multiplying the second of the three equations through by $b,$ and substituting for $abc,$ we find that
$$6a^2b^2+8b^3c+3c-4c^4=0.$$
This is ready to be multiplied through by $c^2,$ and then substituting for $abc$ once again, we get the equation of the sixth degree in $b$ and in $c$
$$16b^6+(224c^3-24)b^3+12c^6+48c^3+9=0.$$
We can let $B=b^3, C=c^3,$ and then this becomes a quadratic, where you can view one of the variables (say $B$) as what we're solving for and the other as a parameter. Remembering that the quadratic would need to have a rational square for a discriminant in order to have a rational root, we find that
$$(224C-24)^2-4\cdot16\cdot(12C^2+48C+9)=r^2$$
for some rational number $r.$ Now we can view this as a quadratic in $C$ and apply the same discriminant trick, and find that
$$4\cdot240^2+4\cdot193r^2=s^2$$
for a rational number $s.$ Letting $r=m/n,$ where $m,n$ are integers,
$$240^2 n^2 + 193 m^2 = (ns/2)^2.$$
Mod 5, this says $3m^2$ is a square, which is impossible.
I hope someone who knows Galois theory well can provide a more easily generalizable answer.
| {
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Sierpinski's Conjecture The Sierpinski's conjecture states that for all integer $n>1$, we have $\frac{5}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ where $(a,b,c) \in \mathbb{N}_*^3$.
But is it easier to prove that $\frac{5}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ ?
Thanks,
B.L.
| Well, first of all, this conjecture is implied by the Erdos-Straus Conjecture $(\forall n>1, \frac{4}{n} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ for some $(a,b,c) \in \mathbb{N}_*^3)$. So any counterexamples must also be counterexamples of Erdos-Straus -- and hence very unlikely.
But specifically, any counterexample must have $n \equiv 1 \mod 24$.
Furthermore, $n \neq 0 \mod 5$ for obvious reasons.
Next, if $n = 5k + 4$, take $a = k+1, b = n*a$. (We can ignore c and d, because we can always replace b by b+1 and set c = b*(b+1) and similarly for d.)
If $n = 5k + 3$, take $a = b = c = 3k+2$ and $d = n*a$.
If $n = 5k + 2$, take $a = b = 2k+1$ and $c = n*a$.
So we're left with $n = 1 \mod 120$ as potential counterexamples.
| {
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Relation between determinant of two matrices $A$ is a matrix with $1$ along diagonal and arbitrary unequal numbers less than $1$ on its non-diagonal. Let $B$ is a matrix with same $1$ along diagonal but having maximum non-diagonal element of $A$ as its non-diagonal element. What would be the relation between $\det (A)$ and $\det(B)$? Suppose $A$ and $B$ are correlation matrices. Does increasing the off-diagonal entries i.e. correlation coefficient decreases determinant?
| Enumerate the matrices $n \times n$ matrices $A_m$ with diagonal elements 1 and off-diagonal elements either $b$ or $0$, with at least one $0$ (there are $2^{n^2-n} - 1$ such matrices, but we can use symmetry to make the calculations somewhat more tractable). For each $m$, $\det(A_m) = 1 - k b^2 + O(b^3)$ where $k$ is the number of pairs $(i,j)$ such that
$(A_m)_{ij} = (A_m)_{ji} = b$, while $\det(B) = 1 - \frac{n^2-n}{2} b^2 + O(b^3)$. Thus $\det(A_m) \ge \det(B)$ for $0 \le b \le b_m$ where $ b_m > 0$. Then any $A$ with
diagonal elements 1 and off-diagonal elements in $[0,b]$ will have $\det(A) \ge \det(B)$
as long as $0 \le b \le \min_m b_m$. I don't know if there's a closed-form formula for $\epsilon(n) = \min_m b_m$. The first few values, obtained by exhaustive enumeration, are $\epsilon(2) = \infty$, $\epsilon(3) = 1/2$, $\epsilon(4) = (4 - \sqrt{10})/3$, $\epsilon(5) = -3/4\,\cos \left( 1/3\,\arctan \left( 4/7\,\sqrt {2} \right) \right)
+5/4-3/4\,\sqrt {3}\sin \left( 1/3\,\arctan \left( 4/7\,\sqrt {2}
\right) \right)$. This last is the least positive root of $-2+12 b-15 b^2+4 b^3$, corresponding to $A_m = \left[ \begin {array}{ccccc} 1&b&b&b&b\\ b&1&b&b&0\\ b&b&1&0&b\\ b&b&0&1&b
\\ b&0&b&b&1\end {array} \right]$
| {
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Help with finding the surface area of a revolution Here's the problem I'm stuck on:
Find the surface are of this revolution about the y-axis
$x = \sqrt{9-y^2}; -2\leq y\leq2$
What I've done so far:
$A= 2\pi \int_{-2}^2 \sqrt{9-y^2} \sqrt{1 + (\frac{1}{2}(9-y^2)^\frac{-1}{2}(-2y))^2} dy$
$ = 4\pi \int_{0}^2 \sqrt{9-y^2} \sqrt{1 + (\frac{1}{2}(9-y^2)^\frac{-1}{2}(-2y))^2} dy$
$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (9-y^2)(\frac{1}{2}(-2y)(9-y^2)^\frac{-1}{2})^2} dy$
$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (9-y^2)((-y)^2(9-y^2)^{-1})} dy$
$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (9-y^2)(\frac{(-y)^2}{(9-y^2)})} dy$
$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (-y)^2} dy$
$ = 4\pi \int_{0}^2 \sqrt{9-y^2 -y^2} dy$*
$ = 4\pi \int_{0}^2 \sqrt{9-2y^2} dy$
The answer in the book says its:
$24\pi$
Which means that I needed to get the integral to be:
$ = 4\pi \int_{0}^2 \sqrt{9} dy$
But I just don't see how I can manipulate the problem with algebra to get there... Any guidance?
EDIT:
Added in some steps to show where my algrbra went wrong
*This is where I made the mistake.
| $$
\begin{align}
& \sqrt{9-y^2} \sqrt{1 + \left(\frac{1}{2}(9-y^2)^\frac{-1}{2}(-2y)\right)^2}
= \sqrt{9-y^2} \sqrt{1 + \frac{4y^2}{4(9-y^2)}} \\
& = \sqrt{9-y^2} \sqrt{1 + \frac{y^2}{9-y^2}}
= \sqrt{9-y^2} \sqrt{\frac{(9-y^2) + y^2}{9-y^2}} \\
& = \sqrt{9-y^2} \sqrt{\frac{9}{9-y^2}} = \sqrt{9}.
\end{align}
$$
| {
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partial fraction question
$ \frac{125x^{2}+x+3}{x^{2}(x-5)} =
> \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-5)} | * x^{2}(x-5)$
$125x^{2}+x+3 = Ax(x-5) + B(x-5) + C (x^{2})$
$125x^{2}+x+3 = A x^{2} - 5Ax + Bx -5B +Cx^{2}$
$125x^{2}+x+3 = x^{2}(A+C) -x(A+B)-5B$
$3 = -5B \Rightarrow B = \frac{-3}{5}$
$-1 = A+B \Rightarrow A = -1 - B \Rightarrow A = \frac{-5}{5} - \frac{-3}{5} \Rightarrow A=\frac{-8}{5}$
$125 = A+C$
Where I did wrong in calculating of variable $A$, because correct answer is $A = \frac{-8}{25}$, but I get $A = \frac{-8}{5}$.
| From
$$\frac{125x^{2}+x+3}{x^{2}(x-5)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-5}=
\frac{x(x-5)A+(x-5)B+x^{2}C}{x^{2}(x-5)}$$
it should be
$$125x^{2}+x+3=\left( A+C\right) x^{2}+\left( B-5A\right) x-5B$$
instead of
$$125x^{2}+x+3=\left( A+C\right) x^{2}+\left( A+B\right) x-5B.$$
Hence
$$\left\{
\begin{array}{c}
3=-5B \\
1=B-5A \\
125=A+C
\end{array}
\Leftrightarrow \right. \left\{
\begin{array}{c}
B=-\frac{3}{5} \\
1=-\frac{3}{5}-5A \\
125=A+C
\end{array}
\Leftrightarrow \right. \left\{
\begin{array}{c}
B=-\frac{3}{5} \\
A=-\frac{8}{25} \\
C=\frac{3133}{25}
\end{array}
\right. $$
and the expansion into partial fractions is $$\frac{125x^{2}+x+3}{x^{2}(x-5)}=-\frac{8}{25x}-\frac{3}{5x^{2}}+\frac{3133}{25\left( x-5\right) }.$$
Second method. Multiply
$$\frac{125x^{2}+x+3}{x^{2}(x-5)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-5}
\qquad (\ast )$$
by $x-5$
$$\frac{125x^{2}+x+3}{x^{2}}=\frac{A(x-5)}{x}+\frac{B(x-5)}{x^{2}}+C$$
and let $x\rightarrow 5$
$$\lim_{x\rightarrow 5}\frac{125x^{2}+5+3}{x^{2}}=\frac{3133}{25}=C.$$
Multiply $(\ast )$ by $x^{2}$
$$\frac{125x^{2}+x+3}{x-5}=Ax+B+\frac{x^{2}}{x-5}C$$
and let $x\rightarrow 0$
$$\lim_{x\rightarrow 0}\frac{125x^{2}+x+3}{x-5}=-\frac{3}{5}=B.$$
Substitute $C$ and $B$ in $(\ast )$
$$\frac{125x^{2}+x+3}{x^{2}(x-5)}=\frac{A}{x}-\frac{3}{5}\frac{1}{x^{2}}+
\frac{3133}{25}\frac{1}{x-5}$$
and set, say, $x=1$
$$-\frac{125+1+3}{4}=A-\frac{3}{5}-\frac{3133}{25}\frac{1}{4},$$
to find $A=-\dfrac{8}{25}$.
| {
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Puzzled by how to determine when a function takes on its maximum (or minimum) I apologize for the specificity of the my question, but I'm concerned that I'm having trouble grasping an important concept.
I'm puzzled by the answer provided for exercise 1.(v) in chapter 7 of Spivak's Calculus (4E, p.129):
For $a>-1$ and
$$f(x) =\begin{cases}
x^{2}, & x ≤ a \\
a+2, & x>a
\end{cases},\qquad x\in(-a-1,a+1),$$
where where does $f(x)$ take on its maximum and minimum?
I get
$$\begin{array}{cc}
Range & Max & Min\\
-1<a<-\frac{1}{2} & a+2 & a+2 \\
-\frac{1}{2}≤a<0 & a+2 & a^{2} \\
0≤a≤\frac{\sqrt{5}-1}{2} & a+2 & 0 \\
\frac{\sqrt{5}-1}{2}<a & - & 0 \\
\end{array}$$
but the answer key has $a^{2}$ as a minimum only for $-\frac{1}{2}<a\lt 0$.
What am I missing?
| I have the 2nd Spanish edition (Editorial Reverté, S.A.), translated from the second English edition. The problem is the same, but did not include the condition $a\gt -1$ until the answer key. But the answer key there reads (translated):
It is bounded above and below. It is understood that $a\gt -1$ (so that $-a-1\lt a+1$). If $-1\lt a\leq -\frac{1}{2}$, then $a\lt -a-1$, so $f(x)=a+2$ for all $x\in (-a-1,a+1)$, so $a+2$ is the maximum and the minimum. If $-\frac{1}{2}\lt a\leq 0$, then $f$ has minimum $a^2$, and if $a\geq 0$, then it has minimum $0$. Since $a+2\gt (a+1)^2$ only for $\frac{-1-\sqrt{5}}{2}\lt a \lt \frac {1+\sqrt{5}}{2}$, when $a\geq -\frac{1}{2}$ the function $f$ has a maximum only for $a\leq \frac{1+\sqrt{5}}{2}$ (when this maximum is $a+2$).
So it looks like your answer matches exactly with this one.
Added. Oh, I see; the problem is what happens when $a=-\frac{1}{2}$.
If $a=-\frac{1}{2}$, then the function is
$$f(x) = \left\{\begin{array}{ll}x^2 & \text{if }x\leq -\frac{1}{2}\\
\frac{3}{2} &\text{if }x\gt -\frac{1}{2}
\end{array}\right.\qquad x\in\left(-\left(-\frac{1}{2}\right)-1,-\frac{1}{2}+1\right).$$
What you seem to be missing is that since the domain is $(-\frac{1}{2},\frac{1}{2})$, the first case never occurs, so you are always in the second case.
| {
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A "fast" approach to solve $2^{133} \equiv x \mod 133 $ I have to solve this equation $2^{133} \equiv x \mod 133 $.Using Euler's theorem I reduced it to $2^{25} \equiv x \mod 133$ but I couldn't think off any fast way to proceed after this.
Any ideas?
| Use $133 = 7 \times 19$. Use $2^p = 2 \mod p$ for $p \in \mathbb{P}$.
Now $2^{133} = (2^7)^{19} \mod 7 = 2^{19} \mod 7 = (2^7)^2 \times 2^5 = 2^2 \times 2^5 = 2^7 = 2 \mod 7$. Similarly $2^{133} = (2^{19})^7 = 2^7 = 14 \mod 19$. Thus
Let $x = 2^{133} \mod 133$, then from $x = 14 \mod 19$, it follows that $x = 14 + 19 \times k$. From $x = 2 \mod 7$, it follows $(14 + 19 \times k) = 5 \times k = 2 \mod 7$, and $k = 6 \mod 7$. From this I get $2^{133} = 128 \mod 133$.
| {
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Inductive proof for the Binomial Theorem for rising factorials I want to proove the following equality containing rising factorials
$$(x+y)^\overline{n}\overset{(*)}{=}\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k}.$$
For $n=1$ this equality is obviously correct. Now I tried to create the transition $n-1\rightarrow n$ this way
$$(x+y)^\overline{n}=(x+y)^\overline{n-1}+\underbrace{\binom{n}{n}x^\overline{n}y^\overline{n-n}}_{x^\overline{n}}$$
$$(x+y)^\overline{n}=\sum_{k=0}^{n-1}\binom{n-1}{k}x^\overline{k}y^\overline{n-k-1}+x^\overline{n}.$$
As I am not that familiar with binomials and such sums, can anyone explain me how transform this equality to achieve a representation like in $(*)$?
| The notation is a little neater if we do the induction step from $n$ to $n+1$ instead of from $n-1$ to $n$. My induction hypothesis is that for all $x$ and $y$, $$(x+y)^{\overline n} = \sum\limits_{k=0}^n \binom{n}{k}x^{\overline k}y^{\overline{n-k}}.$$
I want to show that for all $x$ and $y$, $$(x+y)^{\overline {n+1}} = \sum\limits_{k=0}^{n+1}\binom{n+1}{k}x^{\overline k}y^{\overline{n+1-k}}.$$
I’ll be using the fact that $u^{\overline {m+1}} = u(u+1)^{\overline m}$.
$$\begin{align*}
\sum\limits_{k=0}^{n+1}\binom{n+1}{k}x^{\overline k}y^{\overline{n+1-k}} &= \sum\limits_{k=0}^n\binom{n+1}{k}x^{\overline k}y^{\overline{n+1-k}}+x^{\overline{n+1}}\tag{1}\\
&= \sum\limits_{k=0}^n \left(\binom{n}{k-1}+\binom{n}{k}\right)x^{\overline k}y^{\overline{n+1-k}}+x^{\overline{n+1}}\tag{2}\\
&= \sum\limits_{k=0}^n \binom{n}{k-1}x^{\overline k}y^{\overline{n+1-k}} + \sum\limits_{k=0}^n \binom{n}{k}x^{\overline k}y^{\overline{n+1-k}} + x^{\overline{n+1}}\\
&= \sum\limits_{k=0}^{n-1} \binom{n}{k}x^{\overline {k+1}}y^{\overline{n-k}} + \sum\limits_{k=0}^n \binom{n}{k}x^{\overline k}y^{\overline{n+1-k}} + x^{\overline{n+1}}\tag{3}\\
&= \sum\limits_{k=0}^n \binom{n}{k}x^{\overline {k+1}}y^{\overline{n-k}} + \sum\limits_{k=0}^n \binom{n}{k}x^{\overline k}y^{\overline{n+1-k}}\tag{4}\\
&= x \sum\limits_{k=0}^n \binom{n}{k} (x+1)^{\overline k}y^{\overline{n-k}} + y \sum\limits_{k=0}^n \binom{n}{k} x^{\overline k} (y+1)^{\overline {n-k}}\tag{5}\\
&= x(x+1+y)^{\overline n} + y(x+y+1)^{\overline n}\tag{6}\\
&= (x+y)(x+y+1)^{\overline n}\\
&= (x+y)^{\overline{n+1}}
\end{align*}$$
At step $(1)$ I split off the last term of the sum. At step $(2)$ I used the basic binomial recursion. At step $(3)$ I did an index shift on the first sum: the $k=0$ term is $0$, so $k$ might as well run from $1$ to $n$ and can then be replaced by $k-1$, which runs from $0$ to $n-1$. At step $(4)$ I combined the separated term from step $(1)$ with the first summation. At step $(5)$ I used the fact mentioned just before the computation, and at step $(6)$ I applied the induction hypothesis. The rest is just algebra and again the fact mentioned before the computation.
| {
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Find the limit of a function $$\lim_{x\to 3}\frac{\sqrt{3x} - 3}{\sqrt{2x-4} - \sqrt{2}}.$$
Letting
$$F(x) = \frac{\sqrt{3x} - 3}{\sqrt{2x-4}-\sqrt{2}},$$
we have
$$F(x) = \frac{\sqrt{3}(\sqrt{x} - \sqrt{3})}{\sqrt{2}(\sqrt{x-2}-1)}.$$
Multiplying numerator and denominator by $\sqrt{x-2} + 1$,
$$F(x) = \frac{
(3)^{1/2} ((x(x-2)^{1/2})+(x)^{1/2}-(3(x-2)^{1/2})-(3)^{1/2})}
{\sqrt{2}(x-3)}.$$
Dividing numerator and denominator by $x$ and substituting $3$ for $x$, I get $\frac{0}{\sqrt{2}} = 0$.
Is it correct? My textbook does not have answer, one of the site gives the answer as $\frac{1}{\sqrt{2}}.$.
| Your error is that after rationalizing the denominator, you cannot just plug in $3$ for $x$, because the denominator still evaluates to $0$ (not $\sqrt{2}$, as you seem to think).
$$F(x) = \frac{\sqrt{3}}{\sqrt{2}}\cdot\left(\frac{\quad\frac{\sqrt{x}-\sqrt{3}}{x-3}\quad}{\quad\frac{\sqrt{x-2} - 1}{x-3}\quad}\right).$$
So
$$\lim_{x\to 3}F(x) = \frac{\sqrt{3}}{\sqrt{2}}\frac{\lim\limits_{x\to 3}\frac{\sqrt{x}-\sqrt{3}}{x-3}}{\lim\limits_{x\to 3}\frac{\sqrt{x-2}-1}{x-3}} = \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{g'(3)}{h'(3)},$$
where $g(x) = \sqrt{x}$ and $h(x) = \sqrt{x-2}$.
Since $g'(x) = \frac{1}{2\sqrt{x}}$, $g'(3) = \frac{1}{2\sqrt{3}}$. Since $h'(x)=\frac{1}{2\sqrt{x-2}}$, then $h'(3) = \frac{1}{2}$. So
$$\lim_{x\to 3}F(x) = \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{\frac{1}{2\sqrt{3}}}{\frac{1}{2}} = \frac{\quad\frac{1}{2}\quad}{\frac{\sqrt{2}}{2}}=\frac{1}{\sqrt{2}}.$$
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How to prove the identity $\frac{1}{\sin(z)} = \cot(z) + \tan(\frac{z}{2})$? $$\frac{1}{\sin(z)} = \cot (z) + \tan (\tfrac{z}{2})$$
I did this:
First attempt: $$\displaystyle{\frac{1}{\sin (z)} = \frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} = \frac{\cos (z) }{\sin (z)} + \frac{2\sin(\frac{z}{4})\cos(\frac{z}{4})}{\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4})}} = $$
$$\frac{\cos (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))+2\sin z \sin(\frac{z}{4})\cos(\frac{z}{4})}{\sin (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))}$$
Stuck.
Second attempt:
$$\displaystyle{\frac{1}{\sin z} = \left(\frac{1}{2i}(e^{iz}-e^{-iz})\right)^{-1} = 2i\left(\frac{1}{e^{iz}-e^{-iz}}\right)}$$
Stuck.
Does anybody see a way to continue?
| Let $w = \frac{z}{2}$. Then
$$
\cot(2w) + \tan(w) = \frac{\cos^2(w)-\sin^2(w)}{2 \sin(w) \cos(w)} + \frac{\sin(w)}{\cos(w)} = \frac{1}{\cos(w)} \left( \frac{\cos^2(w)-\sin^2(w) + 2 \sin^2(w)}{2 \sin(w)} \right)
$$
The numerator becomes 1, and we arrive at the result $\frac{1}{2 \sin(w) \cos(w)} = \frac{1}{\cos(2w)} = \frac{1}{\cos(z)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Evaluating an expression using snake oil and convolutions gives different answers I have to evaluate this expression $\sum \limits_{k=0}^n(-1)^k\binom{n}{k}\binom{m+n-k}{n-k}$ using snake oil and convolutions. The problem is that I got two different results, could you help me to find the mistake?
(Notation: $[x^n]$ is the operator "take the coefficient of $x^n$.")
Convolution:
$$
\begin{eqnarray*}
\sum_{k=0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k}
&=& [x^n] \left( \sum_k(-1)^k \binom{n}{k}x^k \right) \left( \sum\binom{m+k}{k}x^k \right)
\\ &=& [x^n] (1-x)^n \frac{1}{(1-x)^{m+1}}
\\ &=& [x^n] (1-x)^{n-m-1}
\\ &=& (-1)^n\binom{n-m-1}{n}
\end{eqnarray*}
$$
Snake oil:
$$
\begin{eqnarray*}
\sum_m \sum_{k=0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k} x^m
&=&
\sum_{k=0}^n (-1)^k \binom{n}{k} \sum_m \binom{m+n-k}{n-k} x^m
\\ &=&
\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{(1-x)^{n-k}}
\\ &=&
\frac{1}{(1-x)^n} \sum_{k=0}^n (-1)^k \binom{n}{k}(1-x)^k
\\ &=&
\frac{x^n}{(1-x)^n}
\\ &=&
\sum_m \binom{m-1}{n-1} x^m ,
\end{eqnarray*}
$$
and so the value should be $\binom{m-1}{n-1}$.
Could you help me to find the mistake, please?
| I think there's an error in the "snake oil" part. You should have
$$\sum_{m = 0}^{\infty} \binom{m+n-k}{n-k} x^m = \frac{1}{(1-x)^{n+1-k}}$$
Which leads you to
$$\sum_{m = 0}^{\infty} \sum_{k = 0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k} x^m = \frac{x^n}{(1-x)^{n+1}} = \sum_{m = 0}^{\infty} \binom{m}{n} x^m$$
And finally, it's easy to check
$$\binom{m}{n} = (-1)^n \binom{n-m-1}{n}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Definite integral calculation with poles at $0$ and $\pm i\sqrt{3}$ $$\int_0^\infty \frac{\sin(2\pi x)}{x(x^2+3)} \, dx$$
I looked at $\frac{e^{2\pi i z}}{z^{3}+3z}$, also calculated the residues, but they don't get me the right answer. I used that $\int_{-\infty}^\infty f(z)\,dz = 2\pi i (\sum \operatorname{Res} z_r) + \pi i \operatorname{Res}_0$, but my answer turns out wrong when I check with wolframalpha.
Residue for $0$ is $1$, for $z=\sqrt{3}i$ it's $-\frac{e^{-2\pi}}{2}$ . . .
In a worse attempt I forgot $2\pi$ and used $z$ only (i.e. $\frac{e^{iz}}{z^{3}+3z}$) and the result was a little closer, but missing a factor of 2 and and $i$.
Can anyone see the right way? Please do tell.
| Resolving $\frac{1}{x(x^2+1)}$ into partial fractions yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\sin (2 \pi x)}{x\left(x^{2}+3\right)} d x &=\frac{1}{3}\left( \int_{0}^{\infty} \frac{\sin (2 \pi x)}{x} dx-\int_{0}^{\infty} \frac{x^{2} \sin (2 \pi x)}{x^{2}+3} dx \right)\\
&=\frac{1}{6}\left(\pi-\int_{-\infty}^{\infty} \frac{x \sin (2 \pi x)}{x^{2}+3} d x\right)
\end{aligned}
$$
By Jordan’s Lemma,
$$
\begin{aligned}
& \int_{-\infty}^{\infty} \frac{x \cos (2 \pi x)}{x^{2}+3} d x+i \int_{-\infty}^{\infty} \frac{x \sin (2 \pi x)}{x^{2}+3} d x \\
=& 2 \pi i \operatorname{Res}\left(\frac{z e^{2 \pi zi}}{z^{2}+3}, z=\sqrt 3 i\right)\\
=& 2 \pi i \frac{\sqrt{3} i e^{-2 \sqrt{3} \pi}}{2 \sqrt{3} i} \\
=& \pi ie^{-2 \sqrt{3} \pi},
\end{aligned}
$$
using contour integration along anti-clockwise direction of the path $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) $$
Comparing the imaginary parts of both sides gives
$$\int_{-\infty}^{\infty} \frac{x \sin (2 \pi x)}{x^{2}+3} d x = \pi e^{-2 \sqrt{3} \pi} $$
Hence $$\int_{0}^{\infty} \frac{\sin (2 \pi x)}{x\left(x^{2}+3\right)} d x = \frac{\pi}{6}\left(1-e^{-2 \sqrt{3} \pi}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/83828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 4
} |
Hard elementary combinatorics problem How does one compute (without brute force) the smallest integer $n$ such that
$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?
| Here is a hint. Prove the identity
\begin{align}
\sum_{i = 1}^{n} \binom{2n}{2i-1} x^{i-1} = \frac{(1 + 2 \sqrt{x} + x)^{n} - (1 - 2 \sqrt{x} + x)^{n} }{2 \sqrt{x}}.
\end{align}
Setting $x = -3$, we have
\begin{align}
\frac{(1 + 2 \sqrt{-3} -3)^{n} - (1 - 2 \sqrt{-3} -3)^{n} }{2 \sqrt{-3}} = \frac{4^{n}}{\sqrt{3}} \sin (\tfrac{2 \pi n}{3})
\end{align}
Using an argument involving periodicity, you should conclude that this sum vanishes infinitely often and you can therefore determine the smallest $n$ accordingly, which is $n = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Prove a linear combination of tensor product can not be written as a single tensor product Prove that in $\mathbb{C}^{3}\otimes\mathbb{C}^{3}$, the state vector $$\mathbf{h}=\frac{1}{\sqrt{8}}=e_{1}\otimes e_{1}+e_{2}\otimes e_{2}+e_{1}\otimes e_{2}+e_{2}\otimes e_{1}+e_{1}\otimes e_{3}+e_{3}\otimes e_{1}+e_{2}\otimes e_{3}+e_{3}\otimes e_{2}$$
cannot be written in the form $h_{1}\otimes h_{2}.$
=========
My proof goes as follows:
Without loss of generality, let $e_{1}=\left(\begin{array}{c}
1\\
0\\
0\end{array}\right), e_{2}=\left(\begin{array}{c}
0\\
1\\
0\end{array}\right), e_{3}=\left(\begin{array}{c}
0\\
0\\
1\end{array}\right),$
then $e_{1}\otimes e_{1}=\left(\begin{array}{c}
1\\
0\\
0\\
0\\
0\\
0\\
0\\
0\\
0\end{array}\right), e_{2}\otimes e_{2}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
1\\
0\\
0\\
0\\
0\end{array}\right), e_{1}\otimes e_{2}=\left(\begin{array}{c}
0\\
1\\
0\\
0\\
0\\
0\\
0\\
0\\
0\end{array}\right),$
$e_{2}\otimes e_{1}=\left(\begin{array}{c}
0\\
0\\
0\\
1\\
0\\
0\\
0\\
0\\
0\end{array}\right), e_{1}\otimes e_{3}=\left(\begin{array}{c}
0\\
0\\
1\\
0\\
0\\
0\\
0\\
0\\
0\end{array}\right), e_{3}\otimes e_{1}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
0\\
0\\
1\\
0\\
0\end{array}\right), e_{2}\otimes e_{3}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
0\\
1\\
0\\
0\\
0\end{array}\right), $
$e_{3}\otimes e_{2}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
0\\
0\\
0\\
1\\
0\end{array}\right), $ and $\mathbf{h=\frac{1}{\sqrt{8}}}\left(\begin{array}{c}
1\\
1\\
1\\
1\\
1\\
1\\
1\\
1\\
0\end{array}\right)$
And eventually we will get some contradiction. But can this method be really applied "without loss of generality"? What is the way to do this without writing all these big vectors?
| Consider ${\mathbb C}^3 \otimes {\mathbb C}^3$ as corresponding to $3 \times 3$ matrices with complex entries. $u \otimes v$ corresponds to the matrix $u v^T$,
which has rank 1 (if $u$ and $v$ are nonzero column vectors). So what is the rank of the matrix corresponding to $\bf h$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all real numbers $x$ such that $||x+1| - |3x - 1|| < 1$ This is on my final exam review. I have the solution, but I do not understand it. When looking at $|x+1| - |3x - 1|$, I see four cases:
a) $x + 1 > 0$ and $3x - 1 > 0$
$x > -1$ and $x > \frac{1}{3}$
$x > -1$
b) $x + 1 < 0$ and $3x - 1 > 0$
$x < -1$ and $x > \frac{1}{3}$ (cannot happen)
c) $x + 1 > 0$ and $3x - 1 < 0$
$x > -1$ and $x < \frac{1}{3}$
$-1 < x < \frac{1}{3}$
d) $x + 1 < 0$ and $3x - 1 < 0$
$x < -1$ and $x < \frac{1}{3}$
$x < \frac{1}{3}$
So when solving the equation, I would look at cases a, c, and d. However, the solution says to use $x < -1$, $-1 \leq x < \frac{1}{3}$, and $\frac{1}{3} \leq x$
What am I missing here?
| Case (a) is $x\gt\frac13$, case (b) cannot happen, case (c) is $-1\lt x\lt\frac13$ and case (d) is $x\lt-1$.
In case (d), [$x\lt-1$ and $x\lt\frac13$] is equivalent to [$x\lt-1$], not to [$x\lt\frac13$]. In case (a), [$x\gt-1$ and $x\gt\frac13$] is equivalent to [$x\gt\frac13$], not to [$x\gt-1$].
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/90802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Odd numbers expressed as : $x^2-y^2$ How to prove following statement :
Conjecture:
An odd number $n$ , $(n>1)$ can be uniquely expressed as : $n= x^2-y^2$ ; $x,y \in \mathbb{Z}^{*}$
if and only if $n$ is a prime number .
If $x-y=m$ , where $m>1$ then $m \mid n$
Proof :
$n=x^2-y^2=(y+m)^2-y^2=y^2+2\cdot y\cdot m +m^2-y^2 \Rightarrow$
$\Rightarrow n=m\cdot (2\cdot y+m) \Rightarrow m \mid n$
Therefore , if $m \neq 1$ it follows that $n$ is a composite number , but how to prove that every odd
composite number ,other than $1$ , has representation : $x^2-y^2$ , where $x-y>1$ ?
| Write $n=ab$ with $b>1$ and $x^2-y^2=(x+y)(x-y)$. Solve $x+y=a$, $x-y=b$. This system has integer solutions $x=(a+b)/2$ and $y=(a-b)/2$ because $a$ and $b$ are odd and hence $a\pm b$ is even. Finally, $x-y=b>1$, as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/97893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Use induction to prove that $n^3 + (n+1)^3 + (n+2)^3 $ is divisible by $9$ Prove that for all integers $n\geq 0, n^3 + (n+1)^3 + (n+2)^3 $ is divisible by 9.
*
*If $ n=1, 1+8+27 = 36 = 9 * x $
*Suppose $ n = k, k^3 + (k+1)^3 + (k+2)^3 $ is divisible by 9.
*Find out $ n = k + 1, $ is divisible by 9.
Because $2$ is divisible by $9$. if $3$ is divisible by $9$, $3$ - $2$ will be divisible by $9$
and the result of $3$ - $2$ = $ 9(k^2 + 3k + 3) $ which is divisible by $9$. So, it is proved.
What I found out the answer so far is them. Are they right way?
| The write up is rather confused; it is particularly bad to use "2" and "3-2" as you do, since it seems you are saying that the number $2$ is divisible by $9$, that $3-2$ (that is, that $1$) is divisible by $9$, etc.
Spend the time writing out complete, coherent, self-contained sentences! Confused writing usually indicates confused thinking.
More specific comments:
*
*Bad use of $x$ in your first step. Better to write out explicitly that $36$ is equal to $9\times 4$, hence divisible by $9$.
*Write out explicitly what your assumption is: namely, your assumption is that there exists an integer $q$ such that
$$ k^3 + (k+1)^3 + (k+2)^3 = 9q.$$
*Write out explicitly what you need to prove. Namely, you need to prove that
$$(k+1)^3 + (k+2)^3 + (k+3)^3\text{ is divisible by }9.$$
Then you make a very bad mistake: you affirm the consequent. You are trying to prove that the expression in point 3 is divisible by $9$. You assume that it is divisible by 9, and then point out that if it is divisible by 9, then so is
$$\Bigl( (k+1)^3 + (k+2)^3 + (k+3)^3\Bigr) - \Bigl( k^3 + (k+1)^2 + (k+2)^3\Bigr).$$
Then you note that this is indeed divisible by $9$, and conclude that the assumption must be true.
This is a logical fallacy. You are saying: "If $P$, then $Q$; since $Q$ is true, then $P$ must be true." If I fall into a pool, I'll get wet. I'm wet. Therefore, I fell into a pool. (Well, no, there are other reasons why I may be wet).
Don't assume what you are trying to prove. Deduce it.
Instead, what you want to do is to write out the expression you want, and try to use the fact that $k^3 + (k+1)^3 + (k+2)^3$ is divisible by $9$ to conclude the expression you want is divisible by $3$. For instance, you can write:
$$\begin{align*}
(k+1)^3 + (k+2)^3 + (k+3)^3 &= (k+1)^3 + (k+2)^3 + \Bigl( k^3 + 9k^2 + 27k + 81\Bigr)\\
&= k^3 + (k+1)^3 + (k+2)^3 + \Bigl( 9k^2 + 27k + 81\Bigr)
\end{align*}$$
and go from there to conclude that if $k^3 + (k+1)^3 + (k+2)^3$ is divisible by $9$, then $(k+1)^3 + (k+2)^3 + (k+3)^3$ is also divisible by $9$.
This completes the Inductive Step. Together with the Base (step 1), this establishes the desired result for all integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove by arithmetical means that $\sum\limits_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum\limits_{k=1}^{\infty}\frac{1}{k^{2}}$ I've been trying to prove, by arithmetical means, that
$$\sum_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum_{k=1}^{\infty}\frac{1}{k^{2}}$$
without success.
When I say "by arithmetical means" I mean to say, go from the left to the right expression just by symbolic manipulation.
Can anyone devise a way of doing this?
| First, let's compute a slightly simpler sum
$$
\begin{align}
\sum_{k=1}^\infty\frac{\Gamma(k)^2}{\Gamma(2k)}(2x)^{2k-1}
&=\sum_{k=1}^\infty\mathrm{B}(k,k)(2x)^{2k-1}\\
&=\sum_{k=1}^\infty\int_0^1t^{k-1}(1-t)^{k-1}(2x)^{2k-1}\mathrm{d}t\\
&=\int_0^1\frac{2x}{1-4x^2t(1-t)}\mathrm{d}t\\
&=\frac{1}{2x}\int_0^1\frac{1}{t^2-t+\frac{1}{4x^2}}\mathrm{d}t\\
&=\frac{1}{2x}\int_0^1\frac{1}{(t-\alpha)(t-1+\alpha)}\mathrm{d}t\text{ where }2\alpha-1=\sqrt{1-\frac{1}{x^2}}\\
&=\frac{1}{2x}\frac{1}{2\alpha-1}\int_0^1\left(\frac{1}{t-\alpha}-\frac{1}{t-1+\alpha}\right)\mathrm{d}t\\
&=\frac{1}{2\sqrt{x^2-1}}\left[\log\left(\frac{\alpha-1}{\alpha}\right)-\log\left(\frac{\alpha}{\alpha-1}\right)\right]\\
&=\frac{1}{\sqrt{x^2-1}}\log\left(\frac{\alpha-1}{\alpha}\right)\text{ where }\frac{\alpha-1}{\alpha}=\left(\sqrt{1-x^2}+ix\right)^2\\
&=\frac{-2i}{\sqrt{1-x^2}}i\,\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\\
&=\frac{2}{\sqrt{1-x^2}}\sin^{-1}(x)\tag{1}
\end{align}
$$
Integrating both sides of $(1)$ yields
$$
\frac12\sum_{k=1}^\infty\frac{\Gamma(k)^2(2x)^{2k}}{\Gamma(2k+1)}
=\left[\sin^{-1}(x)\right]^2\tag{2}
$$
Plugging $x=\frac12$ into $(2)$ gives
$$
\sum_{k=1}^\infty\frac{(k-1)!^2}{(2k)!}=\frac{\pi^2}{18}\tag{3}
$$
In this answer, it is shown that
$$
\sum_{k=1}^\infty\frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}\tag{4}
$$
Combining $(3)$ and $(4)$ yields
$$
\sum_{k=1}^\infty\frac{(k-1)!^2}{(2k)!}=\frac13\sum_{k=1}^\infty\frac{1}{k^2}\tag{5}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
Prove $\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$ Q. Prove
$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$$
My try: expand and got:
$$\frac{5^{2x-2}+2(5^{x^2-1})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}$$
Now what? I find my pre-calculus skills esp with Indices, Logarithms & Trigo lacking ... need to know how to apply the formulas more
| To find $x$ from
$\displaystyle\frac{5^{2x-2}+2(5^{2x})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}=\frac{338}{313}$
i.e.
$\displaystyle 1+\frac{2\times 5^{2x}}{5^{2x-2}+5^{2x+2}}=1+\frac{25}{313}$
i.e.
$\displaystyle\frac{2\times 5^{2x}}{5^{2x-2}+5^{2x+2}}=\frac{25}{313}$
i.e. $\displaystyle\frac{2}{5^{-2}+5^2}=\frac{25}{313}$, an identity. So the above equation is valid for all $x\epsilon \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/102260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Two theorems about an inscribed quadrilateral and the radius of the circle containing its vertices I think those two theorem are two of the most complicated formulas I have ever seen; please prove it because I am not able to find proofs on the internet:
It is known that if the sides of an inscribed quadrilateral $ABCD$ (that is in the order $AB,BC,CD,DA$) have lengths $a,b,c,d$ respectively and $p$ is the semi perimeter of the quadrilatral, then:
Theorem 1: The length of diagonal $AC$ of the quadrilatral is equal to $$\sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}\;.$$
Theorem 2: The radius of the circle that contains all the vertices of the quadrilateral is equal to $$\frac14\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(p-a)(p-b)(p-c)(p-d)}}\;.$$
By the way, has anyone seen those theorems in a geometry textbook with solution?
| Using cosine rule on triangles $ABC$ and $ACD$ gives $$\frac{a^2+b^2-AC^2}{2ab}+\frac{c^2+d^2-AC^2}{2cd}=\cos(\angle{ABC})+\cos(\angle{CDA})=0$$
Thus $(ab+cd)AC^2=(a^2+b^2)cd+(c^2+d^2)ab=(ac+bd)(ad+bc)$ so $AC=\sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}$.
The area of a triangle with sides $a, b, c$ and circumradius $R$ is $\frac{abc}{4R}$, so we have by Brahmagupta's formula (where $A$ is the area of the cyclic quadrilateral)
$$\sqrt{(p-a)(p-b)(p-c)(p-d)}=A=\frac{ab(AC)}{4R}+\frac{cd(AC)}{4R}=\frac{(ab+cd)AC}{4}\frac{1}{R}$$
Thus $$R=\frac{(ab+cd)AC}{4\sqrt{(p-a)(p-b)(p-c)(p-d)}}=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(p-a)(p-b)(p-c)(p-d)}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A question about proving that there is no greatest common divisor I have to answer this question:
Prove that in the ring $\mathbb Z[\sqrt{-5}]$ there's no gcd to $6$ and $2\cdot (1+\sqrt{-5})$.
I have no clue how to do this but however I've tried to prove that their sum $$6+2\cdot (1+\sqrt{-5}) = 8+2\cdot\sqrt{-5}$$ is not in $\mathbb Z[\sqrt{-5}]$ but I was not able to do this.
How should I proceed?
| I'm assuming a greatest common divisor of $a$ and $b$ means something that divides both $a$ and $b$, such that every divisor of $a$ and $b$ divides the gcd.
Use the norm function $N(r + s\sqrt{5}i) = r^2 + 5s^2$, and show that $N(ab) = N(a)N(b)$.
This leads to the important fact that if $c$ divides $d$ in the ring, then $N(c)$ divides $N(d)$ as integers. This places restrictions on the possible norms of a gcd $r + s\sqrt{5}i$: you need $r^2 + 5s^2$ to divide $N(2(1 + \sqrt{5}i)) = 24$ and $N(6) = 36$. Thus $r^2 + 5s^2$ divides $12$. Going through the possibilities, this can only happen if $(r,s) = (\pm2,0)$, $(1,1)$, $(-1,-1)$, and $(\pm 1,0)$. So the only possibilities for the greatest common divisor are $\pm 2$, $\pm(1 + \sqrt{5}i)$, and $\pm 1$.
To see $\pm 2$ can't be a gcd: Note that $6 = (1 + \sqrt{5}i)(1 - \sqrt{5}i)$. So $1 + \sqrt{5}i$ divides both $6$ and $2(1 + \sqrt{5}i)$. Thus it would have to divide the gcd. Since $1 + \sqrt{5}i$ has norm $6$ and $\pm 2$ has norm $4$, $\pm 2$ can't be gcd: $6$ does not divide $4$.
To see $\pm(1 + \sqrt{5}i)$ can't be gcd: Since $2$ divides $6$ and $2(1 + \sqrt{5}i)$, $2$ divides the gcd. If $\pm(1 + \sqrt{5}i)$ were the gcd, then $N(\pm2) = 4$ would have to divide $N(1 + \sqrt{5}i) = 6$, again a contradiction.
Lastly, we consider the cases $\pm 1$. In this case, since $2$ divides the gcd, one would have to have that $N(2)$ divides $N(\pm 1) = 1$, or that $4$ divides $1$. Once again we have a contradiction.
Thus we see all potential gcd's don't work: we conclude that $6$ and $2(1 + \sqrt{5}i)$ have no gcd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/103593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Computing determinant of a matrix with non-zero values on three diagonals let $A$ be an $n\times n$ matrix with entries $a_{ij}$ such that
$a_{ij}=2$ if $i=j$.
$a_{ij}=1$ if $|i-j|=2$
and $a_{ij}=0$ otherwise.
compute the determinant of $A$.
using the famous formula $\det A=\sum_{i=1}^{n}(-1)^{i+j}a_{ij}\det A^{(ij)}$ where $A{(ij)}$ is the submatrix obtaining from $A$ by omiting it's $i$th row and $j$th colomn, I reached to the formula $\det A=\frac{1}{4}n^2+n+\frac{7}{8}+\frac{1}{8}(-1)^n$. is it correct?
| If $A$ is of this form
$$A_{7\times 7}=
\begin{pmatrix}
2 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 2 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 2 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 2 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 2 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 2 \\
\end{pmatrix},
$$
I get
$$
\det(A) =
\begin{cases}
\frac{n^2}{4}+n+1,\quad n \text{ even}\\
\quad\\
\frac{n^2}{4}+n+\frac{3}{4},\quad n \text{ odd}
\end{cases}
$$
This is just an alternative way of stating the questioner's own succint answer:
$$\det(A) = \frac{n^2}{4}+n+\frac{7}{8} +(-1)^n\frac{1}{8}.$$
This is a nice exercise in symbolic $LU=A$ factorization and would be a good exam question to separate the good from the mediocre students.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/106208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Rewriting a function so it is in a form that can be easily integrated. I am trying to find an integral of the function $f(x)=\frac{2}{3}x+3$. I am a bit stumped on how to find the integral of $\frac{2}{3}x$, specifically.
I looked at the answer thinking I could work backwards but got stuck doing this as well. I suspect I am missing something pretty obvious.
The answer provided is: $\frac{1}{3}x^2+3x+c$
So by using the rule for indefinite integrals I have been given for $x^n$:
$$\frac{1}{n+1}x^{n+1}+c$$
I thought I could just take the answer and work out $n$ by reversing this formula. But if my answer is $\frac{1}{3}x^2+c$ then by the rule above $n=1$ and $n=2$ for the answer provided.
This can't be right, which means I am using the rule incorrectly. I think I have to rewrite $\frac{2}{3}x$ as $x$ to the power of something say $y$ so they are equivalent i.e. $\frac{2}{3}x=x^y$ but I do not understand how to do this.
I am basing this on the answer provided being in the form of the rule stated above which applies to functions in the form of $x^n$.
| Writing $I(f(x))$ for the indefinite integral of $f(x)$, we need the following rules:
$$\begin{align}I(f(x) + g(x)) = I(f(x)) + I(g(x)) \tag{1}\end{align}$$
$$I(a \cdot f(x)) = a \cdot I(f(x)), \quad \text{for constants $a$} \tag{2}$$
$$I(x^n) = \frac{1}{n+1}{x^{n+1}} + c \tag{3}$$
Applying these to your function $f(x)$, we get:
$$\begin{eqnarray}I(f(x)) &=& I\left(\frac{2}{3} x + 3\right) \\ &\stackrel{(1)}{=}& I\left(\frac{2}{3} x\right) + I(3) \\ &\stackrel{(2)}{=}& \frac{2}{3} \cdot I(x) + 3 \cdot I(1) \\ &\stackrel{(3)}{=}& \frac{2}{3} \cdot \left(\frac{1}{2} x^2 + c_1\right) + 3 \cdot (x + c_2) \\ &=& \frac{1}{3} x^2 + 3x + \left(\frac{2}{3} c_1 + 3 c_2\right) \\ &=& \frac{1}{3} x^2 + 3x + c, \end{eqnarray}$$
where we wrote $c = \frac{2}{3} c_1 + 3 c_2$ for a new constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$ How can I evaluate the integral
$$\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$$
I tried manipulating the known integral
$$\int_0^1 \frac{\ln(1 - x)}{x}dx = -\frac{\pi^2}{6}$$
but couldn't do anything with it.
| Following is an elementary proof.
I assume only that $\displaystyle \int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}$
\begin{align}J&=\int_0^1 \frac{\ln(1-x)}{1+x}dx\\
&\overset{y=\frac{1-x}{1+x}}=\int_0^1 \frac{\ln\left(\frac{2y}{1+y}\right)}{1+y}dy\\
&=\int_0^1 \frac{\ln\left(\frac{2}{1+y}\right)}{1+y}dy+\int_0^1 \frac{\ln t}{1+t}dt\\
&\overset{u=\frac{1-y}{1+y}}=\int_0^1 \frac{\ln\left(1+u\right)}{1+u}du+\int_0^1 \frac{\ln t}{1+t}dt\\
&=\frac{1}{2}\ln^2 2+\int_0^1 \frac{\ln t}{1+t}dt\\
\int_0^1 \frac{\ln t}{1+t}dt&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2t\ln t}{1-t^2}dt\\
&\overset{w=t^2}=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln w}{1-w}dw\\
&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\
&=-\frac{1}{12}\pi^2
\end{align}
Therefore,
$\boxed{\displaystyle J=\frac{1}{2}\ln^2 2-\frac{1}{12}\pi^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 3
} |
Gradient of linear scalar field $X \mapsto \operatorname{tr}(AXB)$ Could someone explain the following?
$$ \nabla_X \operatorname{tr}(AXB) = BA $$
I understand that
$$ {\rm d} \operatorname{tr}(AXB) = \operatorname{tr}(BA \; {\rm d} X) $$
but I don't quite understand how to move ${\rm d} X$ out of the trace.
| These are the main equations to remember:
*
*Let $\mathbf{A} \in \mathbb{R}^{n\times m}$, $\mathbf{X} \in \mathbb{R}^{m\times n}$. Then
\begin{equation}
\frac{d}{d\mathbf{X}}\text{Tr}(\mathbf{AX}) = \frac{d}{d\mathbf{X}}\text{Tr}(\mathbf{XA}) = \mathbf{A}^T
\end{equation}
*Let $\mathbf{A} \in \mathbb{R}^{n\times m}$, $\mathbf{X} \in \mathbb{R}^{n\times m}$. Then
\begin{equation}
\frac{d}{d\mathbf{X}}\text{Tr}(\mathbf{AX^T}) = \frac{d}{d\mathbf{X}}\text{Tr}(\mathbf{X^TA}) = \mathbf{A}
\end{equation}
Proof 1.
\begin{equation}
\left[ \frac{d}{d\mathbf{X}} \text{Tr}(\mathbf{AX}) \right]_{i,j} = \frac{d}{dx_{i,j}} \text{Tr}(\mathbf{AX}) = \frac{d}{dx_{i,j}} \sum_{k,l} a_{k,l} x_{l,k} = a_{j,i} = \left[\mathbf{A}^T\right]_{i,j}
\end{equation}
Proof 2
\begin{equation}
\left[ \frac{d}{d\mathbf{X}} \text{Tr}(\mathbf{AX^T}) \right]_{i,j} = \frac{d}{dx_{i,j}} \text{Tr}(\mathbf{AX^T}) = \frac{d}{dx_{i,j}} \sum_{k,l} a_{k,l} x_{k,l} = a_{i,j} = \left[\mathbf{A}\right]_{i,j}
\end{equation}
Once you have these, you can derivate crazy things like the following:
Example 1. Let $\mathbf{A} \in \mathbb{R}^{m\times m}$, $\mathbf{X} \in \mathbb{R}^{m\times n}$. Then
\begin{equation}
\frac{d}{d\mathbf{X}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{X}) = (\mathbf{A} + \mathbf{A}^T) \mathbf{X}
\end{equation}
We can derive it as follows:
\begin{split}
\frac{d}{d\mathbf{X}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{X}) =& \frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{Y}^T \mathbf{A} \mathbf{X}) + \frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{Y})\\
=& \mathbf{A} \mathbf{X} + (\mathbf{X}^T \mathbf{A})^T\\
=& \mathbf{A} \mathbf{X} + \mathbf{A}^T \mathbf{X} = (\mathbf{A} + \mathbf{A}^T) \mathbf{X}
\end{split}
Example 2. Consider now this example.
\begin{equation}
f(\mathbf{X}) = \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{X}^T \mathbf{C})
\end{equation}
where $\mathbf{X} \in \mathbb{R}^{n\times m}$, $\mathbf{A} \in \mathbb{R}^{n\times n}$, $\mathbf{B} \in \mathbb{R}^{m\times m}$, $\mathbf{C} \in \mathbb{R}^{n\times m}$.
\begin{equation}
\begin{split}
\frac{d}{d\mathbf{X}} f(\mathbf{X}) =& \frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{Y}^T \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{X}^T \mathbf{C})\\
+& \frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{Y} \mathbf{B} \mathbf{X}^T \mathbf{C})\\
+& \frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{Y}^T \mathbf{C})
\end{split}
\end{equation}
Calculating these:
\begin{split}
\frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{Y}^T \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{X}^T \mathbf{C}) = \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{X}^T \mathbf{C}
\end{split}
\begin{split}
\frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{Y} \mathbf{B} \mathbf{X}^T \mathbf{C}) =& \frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{Y} \mathbf{B} \mathbf{X}^T \mathbf{C} \mathbf{X}^T \mathbf{A})\\
=& (\mathbf{B} \mathbf{X}^T \mathbf{C} \mathbf{X}^T \mathbf{A})^T\\
=& \mathbf{A}^T \mathbf{X} \mathbf{C}^T \mathbf{X} \mathbf{B}^T
\end{split}
\begin{split}
\frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{Y}^T \mathbf{C}) = \frac{d}{d\mathbf{Y}} \text{Tr}(\mathbf{Y}^T \mathbf{C} \mathbf{X}^T \mathbf{A} \mathbf{X} \mathbf{B}) = \mathbf{C} \mathbf{X}^T \mathbf{A} \mathbf{X} \mathbf{B}
\end{split}
So the result is:
\begin{equation}
\frac{d}{d\mathbf{X}} \text{Tr}(\mathbf{X}^T \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{X}^T \mathbf{C}) = \mathbf{A} \mathbf{X} \mathbf{B} \mathbf{X}^T \mathbf{C} + \mathbf{A}^T \mathbf{X} \mathbf{C}^T \mathbf{X} \mathbf{B}^T + \mathbf{C} \mathbf{X}^T \mathbf{A} \mathbf{X} \mathbf{B}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 4,
"answer_id": 1
} |
Complex Numbers $x,y,z$ Find $x^{2007}+y^{2007}+z^{2007}$ Let $x,y,z$ be complex numbers such that
$$x+y+z = x^{5}+y^{5}+z^{5} = 0, \hspace{10pt}
x^3+y^3+z^3=2$$
Find all possible values of
$$x^{2007}+y^{2007}+z^{2007}$$
| It $x+y+z$ = 0, then $x,y,z$ are roots of $t^3 + at-b = 0$.
Then, we can show that, $x^5 + y^5 + z^5 = -5ab$ and $x^3 + y^3 + z^3 = 3b$, either using Newton's Identities or as in my answer here: https://math.stackexchange.com/a/115534/1102
Since $b \ne 0$, we must have that $a = 0$.
Thus $x,y,z$ are roots of $t^3 = b$. We know that $b = \frac{2}{3}$ and thus can compute the expression you need easily as $3b^{669} = \dfrac{2^{669}}{3^{668}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Help evaluating $\int \frac{dx}{(x^2 + a^2)^2}$ I have following integral and it should be simple, however whatever substitution I use and no matter how many times I integrate it by parts (or combine both) I never get the correct solution (or any alternative solution):
$$\int \frac{dx}{(x^2 + a^2)^2}$$
I'm looking for what is on the Wolfram|Alpha in the alternative solutions section:
$$\frac{\arctan(\frac{x}{a})}{2a^3} + \frac{x}{2a^2(a^2 + x^2)} $$
| Substitute
$$x=a \hspace{3pt} \tan \theta$$
$$ dx = a\hspace{3pt} \sec^2 \theta \hspace{3pt} d\theta$$
The integral
$$
\begin{align*}
\int \frac{1}{(x^2+a^2)^2} \hspace{3pt}dx &= \int \frac{\hspace{3pt}a \hspace{3pt}\sec^2 \theta }{a^4 \hspace{3pt} \sec^4 \theta}\hspace{3pt} d\theta\\
&= \frac{1}{a^3} \int \frac{1}{\sec^2 \theta}\hspace{3pt} d\theta\\
&= \frac{1}{a^3} \int \cos^2 \theta\hspace{3pt} d\theta\\
&= \frac{1}{2a^3} \int 2\hspace{3pt}\cos^2 \theta\hspace{3pt} d\theta\\
&= \frac{1}{2a^3} \int (1+\cos2 \theta)\hspace{3pt} d\theta\\
&= \frac{1}{2a^3} \theta + \frac{1}{2a^3} \frac{\sin 2\theta}{2} + \text{constant} \\
&= \frac{1}{2a^3} \tan^{-1}\frac{x}{a} + \frac{1}{2a^3} \frac{\sin 2\theta}{2} + \text{constant} \\
\tag{A}\\
\end{align*}
$$
But since we substituted $x=a \hspace{3pt} \tan \theta$, which is equivalent to
$$\sin \theta = \frac{x}{\sqrt{x^2+a^2}}$$ and $$\cos \theta = \frac{a}{\sqrt{x^2+a^2}}$$
$$ \sin2\theta = 2 \sin\theta \cos\theta = \frac{2xa}{x^2+a^2}$$
The integral therefore simplifies to
$$\frac{1}{2a^3} \tan^{-1}\frac{x}{a} + \frac{1}{2a^3} \frac{ax}{x^2+a^2} + \text{constant}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Proof of uniform continuity I seem to have hit a dead-end in the following proof.
Define $f:\mathbb{R}\to\mathbb{R}$ by:
$f(x)=\frac{1}{1+x^2}$
Show that $f$ is uniformly continuous.
My proof:
Let $x_{0}\in \mathbb{R}$.
Also let $\epsilon >0$
Choose $\delta = ?$
Then, for $x\in \mathbb{R}$, such that $|x-x_{0}|<\delta$, we have:
|$f(x)-f(x_{0})|=|\frac{1}{1+x^2}-\frac{1}{1+x_{0}^2}|=|\frac{x_{0}^2-x^2}{(1+x^2)(1+x_{0}^2)}|=\frac{|x_{0}-x||x_{0}+x|}{(1+x^2)(1+x_{0}^2)}\le \delta|x+x_{0}|$
The last line uses the fact that $x^2, x_{0}^2\ge 0$. How can I finish the proof?
| Write
$$|f(x)-f(y)|=\left|\frac 1{1+x^2}-\frac 1{1+y^2}\right|=\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|\leq |x-y|\frac{|x|+|y|}{(1+x^2)(1+y^2)}\\\
\leq \frac{|x-y|}2\frac{x^2+1+y^2+1}{(1+x^2)(1+y^2)}\leq \frac{|x-y|}2\left(\frac 1{1+y^2}+\frac 1{1+x^2}\right)\leq |x-y|$$
and now the $\delta$ you have to choose is clear. In fact, $f$ is lipschitz continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2$? I have tried something to solve the series
$$\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2.$$
My approach is :
$$(1+x)^n=\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n.$$
Differentiating the above equation
$$n(1+x)^{n-1} = \binom{n}{1} + \binom{n}{2}x + \cdots + n\binom{n}{n}x^{n-1}$$
Also,
$$
\left(1+\frac{1}{x}\right)^n =\binom{n}{0} + \binom{n}{1}\frac{1}{x} + \binom{n}{2}\left(\frac{1}{x}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{x}\right)^n$$
Multiplying above two equation I get,
$$\begin{align*}
&{n(1+x)^{n-1}\left(1 + \frac{1}{x}\right)^n}\\
&\quad= \left(
\binom{n}{1}^2 + 2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2 + \cdots + n\binom{n}{n}^2\right)\left(\frac{1}{x}\right) + \text{other terms}
\end{align*}$$
So I can say that coefficient of $\frac{1}{x}$ in expansion of $n(1+x)^{n-1}(1+\frac{1}{x})^n$ will give me the required answer.
Am I doing it correct,please correct me if I'm wrong ?
If I'm right,please tell me how to calculate the coefficient of $\frac{1}{x}$ ?
Based on the answers,I tried to implement the things in a C++ code.
I tried implementing the code using extended euclidean algorithm so that the problem of truncated division can be eliminated but still not abled to figure out why am I getting wrong answer for n>=3. This is my updated code : http://pastebin.com/imS6rdWs I'll be thankful if anyone can help me to figure out what's wrong with this code.
Thanks.
Solution:
Finally abled to solve the problem.Thanks to all those people who spent their precious time for my problem.Thanks a lot.This is my updated code :
http://pastebin.com/WQ9LRy6F
| First recall that the coefficient of $x^n$ in $(1+x)^n(1+x)^n=(1+x)^{2n}$ implies
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}^2
&=\sum_{k=0}^n\binom{n}{k}\binom{n}{n-k}\\
&=\binom{2n}{n}\tag{1}
\end{align}
$$
and then note that
$$
\begin{align}
\sum_{k=0}^nk\binom{n}{k}^2
&=\sum_{k=0}^nk\binom{n}{n-k}^2\\
&=\sum_{k=0}^n(n-k)\binom{n}{k}^2\tag{2}
\end{align}
$$
Adding the first and last parts of $(2)$ yields
$$
\begin{align}
2\sum_{k=0}^nk\binom{n}{k}^2
&=n\sum_{k=0}^n\binom{n}{k}^2\\
&=n\binom{2n}{n}\tag{3}
\end{align}
$$
Therefore,
$$
\sum_{k=0}^nk\binom{n}{k}^2=\frac{n}{2}\binom{2n}{n}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/122147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Distance between orthocenters Given a triangle $\Delta ABC$ with angle $\angle ACB=90^\circ$. The bisector of $\angle ACB$ intersects $AB$ at $L$. If $AC=4 \operatorname {cm}$. and $BC=5 \operatorname {cm}$, find the distance between the orthocenter $H_1$ of $ALC$ and the orthocenter $H_2$ of $BLC$.
| A fun approach, using a bit of geometry:
Let $a = AC$ and $b = BC$. Assume that $a < b$, as in your example. Let $D$ be the intersection of the altitude from $L$ to $AC$. Let $E$ be the intersection of the altitude from $L$ to $BC$. (Sorry for the lack of a diagram, but I don't have a scanner available, and I don't want to teX it out.) Since $\angle DCL = 45^{\circ}$, it's easy to see that $CD = DL = EL = CE$. Let $x$ be the common length. By similar triangles ($\triangle ADL \sim \triangle LEB$), we have
$$ \frac{a-x}{x} = \frac{x}{b-x} \;\Rightarrow\; x = \frac{ab}{a+b}.$$
Furthermore, $\angle ADH_1 = \angle BEH_2 = 45^{\circ}$ (which is easy enough to verify by labeling angles). Therefore $AD = DH_1$ and $BE = EH_2$. Consider $\triangle H_1LH_2$. Note, $H_1$ is in the interior of $\triangle ABC$, $H_2$ exterior, and $\angle H_1LH_2 = 90^{\circ}$. Thus, the required length is found by the Pythagorean Theorem:
$$\begin{align*}
(H_1H_2)^2 &= (LH_1)^2 + (LH_2)^2 \\
&= (LD - DH_1)^2 + (EH_2 - LE)^2 \\
&= \left( \frac{ab}{a+b} - \left[a - \frac{ab}{a+b}\right]\right)^2
+ \left( \left[b - \frac{ab}{a+b}\right] - \frac{ab}{a+b}\right)^2\\
&= \textrm{a few steps of algebra...}\\
&= \left(\frac{b-a}{a+b}\right)\sqrt{a^2 + b^2}.
\end{align*}
$$
I like the formula, as it involves a relationship between the sum and difference of the legs, and the hypotenuse of the triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/123990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing the identity: $\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$ My knowledge of trigonometry is still insufficient to resolve this problem. Any help would be greatly appreciated.
$$\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$$
| This follows easily from the identity:
$$\cot x - 2 \cot 2x = \tan x$$
Which can easily be seen by using $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$ as follows:
$$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin ^2 x}{\sin x \cos x} = 2 \cot 2x$$
We now have
$$\cot \alpha - 2 \cot 2\alpha = \tan \alpha$$
$$2\cot 2\alpha - 4 \cot 4\alpha = 2\tan 2\alpha$$
$$4\cot 2\alpha - 8 \cot 4\alpha = 4\tan 4\alpha$$
Adding gives us the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving a linear system with an initial condition So there is a linear system
$
\frac{du}{dt} = - \left( \begin{array}{cc}
4 & -3 \\
6 & -5 \end{array} \right) \frac{du}{dx}
$
with an initial condition
$
u(x, 0)= \left( \begin{array}{c}
-tanh(\lambda x) \\
0 \end{array} \right)
$
Now to start this off I have manipulated the first equation to get this
$
\left( \begin{array}{ccc}
1 & \frac{-3}{4} & -\frac{1}{4} \frac{du_{1}}{dt}\\
0 & \frac{1}{12} & \frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt}\end{array} \right) \frac{du}{dx}
$
which can be rewritten as
$
\left( \begin{array}{ccc}
1 & \frac{-3}{4} & -\frac{1}{4} \frac{du_{1}}{dt}\\
0 & \frac{1}{12} & \frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt}\end{array} \right) \frac{du}{dx} =>\left( \begin{array}{cc}
1 & \frac{-3}{4}\\
0 & \frac{1}{12}\end{array} \right) \frac{du}{dx} =
\left( \begin{array}{c}
-\frac{1}{4} \frac{du_{1}}{dt} \\
\frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt} \end{array} \right)
$
but I'm not sure how to deal with the derivatives. Would I use what I already have but just plug in the initial condition to $u_{1}$ and $u_{2}$ or am I using the wrong method?
| If one writes $u=(u_1,u_2)$ and one diagonalizes the matrix $A$, one gets the simpler differential system $\frac{\partial v_1}{\partial t}=-\frac{\partial v_1}{\partial x}$ and $\frac{\partial v_2}{\partial t}=2\frac{\partial v_2}{\partial x}$, with $v_1=u_1+u_2$ and $v_2=u_1+2u_2$.
For any $a$, the differential equation $\frac{\partial v}{\partial t}=a\frac{\partial v}{\partial x}$ means that $v$ is a function of $x+at$, hence that $v(x,t)=v(x+at,0)$. Here, $v_1(x,t)=v_1(x-t,0)$ and $v_2(x,t)=v_2(x+2t,0)$.
Finally, $u_1=2v_1-v_2$ and $u_2=v_2-v_1$ hence
$$
u_1(x,t)=2u_1(x-t,0)+2u_2(x-t,0)-u_1(x+2t,0)-2u_2(x+2t,0),
$$
and
$$
u_2(x,t)=-u_1(x-t,0)-u_2(x-t,0)+u_1(x+2t,0)+2u_2(x+2t,0).
$$
In the special case when $u_1(x,0)=u_0(x)$ and $u_2(x,0)=0$, one gets
$$
u_1(x,t)=2u_0(x-t)-u_0(x+2t),\quad
u_2(x,t)=u_0(x+2t)-u_0(x-t).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$ Can you please help me with this integral
$$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$$
Thanks!
| $$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$$
Substitution:$x=\sin\theta$
$$I=\int_{-\pi/2}^{\pi/2}e^{\sin\theta}d\theta$$
$$e^{\sin\theta}=1+\frac{\sin\theta}{1!}+\frac{\sin^2\theta}{2!}+\frac{\sin^3\theta}{3!}.......$$
The odd powers of sine when used in the integral will produce zero.
Again for even values of n we have
$$\int_{-\pi/2}^{\pi/2}\sin^n\theta d\theta=2 \times \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}....\frac{3}{4}\frac{1}{2}\frac{\pi}{2}$$
Therefore,
$$I=\int_{-\pi/2}^{\pi/2}e^{\sin\theta}d\theta=\pi+2\times\Sigma\frac{1}{n!}[\frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}....\frac{3}{4}\frac{1}{2}\frac{\pi}{2}]$$
$$=\pi+\pi\Sigma [\frac{1}{n(n-2)(n-4)......2}]^2$$
["n" running from two through even values ]
Writing n=2m we have,
$$\pi<I=\pi\Sigma[\frac{1}{2^m}\frac{1}{m!}]^2<4$$
[m running from 0 to infinity]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/129743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Please check my answer to $\sum\limits_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$ $$\sum_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$$
I have this answer, please let me know if there is a more beautiful proof.
My answer:
at first, we prove two inequalities:
*
*If $x\in (x,\pi)$ then $\sin x \leq x$
*When $x\in(0,\frac{\pi}{2})$, $\sin x \geq \frac{2x}{\pi}$
1) first, let $y = \sin x -x $
$y^{\prime} = \cos x -1 \leq 0$
so $\sin x - x \leq \sin 0 -0 = 0$
which can be rewritten as
$\sin x \leq x$
2) Let $y=\sin x - \frac{2x}{\pi}$
thus $y^{\prime} = \cos x - \frac{2}{\pi}$ because $x\in (0, \frac{\pi}{2})$
so y at first decreases and then increases on the boundary of $x \in (0,\frac{\pi}{2})$
so $ \sin x - \frac{2}{\pi}\leq \max \{{\sin 0 - \frac{2}{\pi}0, \sin (\frac{\pi}{2} - \frac{2}{\pi}\frac{\pi}{2}) \}}$
so $\sin x \leq \frac{2x}{\pi}$
Then select $M\in N$
$\frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m+1} + \ldots + \frac{\sin ((m+n)x)}{m+n} \leq \frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m} + \ldots + \frac{\sin ((m+n)x)}{m} $
=> $\frac{1}{2M} \times \frac{\sin ((m-\frac{1}{2})x) - \sin ((n+\frac{1}{2})x)}{\sin \frac{x}{2}} < \frac{1}{m \times \sin \frac{x}{2}} \times \sin x + \frac{\sin 2x}{2} + \ldots + \frac{\sin ((m-1)x)}{m-1} < x + \frac{2x}{2} + \ldots + \frac{(m-1)x}{m-1} $
so just need to prove that
$(m-1)x + \frac{1}{m \times \sin \frac{x}{2}} \leq 2\sqrt{\pi}$
select M which satisfies
$ \frac{\sqrt{\pi}}{x} \leq m < \frac{\sqrt{\pi}}{x} + 1 $
so $ (m-1)x < [ \frac{\sqrt{\pi}}{x} \times x = \sqrt{\pi} ] $
thus
$\frac {1}{m \times \sin(\frac{x}{2})}\leq[ \frac{1}{\sqrt{\pi}}\times \frac{2}{\frac{ \sin (0.5x)}{0.5x}} = \frac{1}{\sqrt{\pi}} \times \frac{2}{\frac{\sin 0.5x}{0.5x}} ]$
because $x\in (0, \pi)$ thus $\frac{x}{2} \in (0, \frac{\pi}{2})$
$ (m-1)x + \frac{1}{m \times \sin(0.5x)} \leq 2\sqrt{\pi} $
thanks, for viewing and commenting.
ps. I'm still learning latex and mathematics, so my answer isn't pretty to read, nor is the latex I wrote.
| I was correct.
Read the comments for better ideas.
Answering this out of a need for my question to have an answer, and I wrote a correct answer in my post.
Yea, for me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| If $a=\cos\frac{\pi}5$ and $b=\cos\frac{2\pi}5=2a^2-1$ (by the double-angle identity for cosine), we want to show that
$$
0=(x-a)(x+b)=x^2-(a-b)x-ab=x^2-\tfrac12x-\tfrac14,
$$
i.e., that $a$ and $-b$ are roots of $4x^2-2x-1$.
Note also that $4x^2+2x-1=4(x+a)(x-b)$ has roots $-a$ and $b$.
What is special about the numbers $\{a,b,-b,-a\}$?
They are the $x$-coordinates (real parts) of the nonreal $10$th roots of unity,
$$
x+iy=e^{\pm\pi i\cdot\frac{k}5}
\qquad
\text{for}
\qquad
k\not\equiv0\pmod5.
$$
But these satisfy the equation $(x+iy)^5=(-1)^k=\pm1$.
Taking the imaginary part, we have
$$
\eqalign{
0 &= \Im\left[(x+iy)^5\right] \\
&= \Im\left[x^5+5x^4(iy)+10x^3(iy)^2+10x^2(iy)^3+5x(iy)^4+(iy)^5\right] \\
&= \Im\left[iy\left(5x^4+10x^2(iy)^2+(iy)^4\right)\right] \\
&= y\left[5x^4-10x^2y^2+y^4\right] \\
\implies
0 &= \left[5x^4-10x^2(1-x^2)+(1-x^2)^2\right] \\
&= 16x^4-12x^2+1 \\
&= 16x^4-8x^2+1 ~-~4x^2 \\
&= \left( 4x^2-1 \right) - \left( 2x \right)^2 \\
&= \left( 4x^2+2x-1\right)\left( 4x^2-2x-1\right) \,.
}
$$
So far, we have shown this has roots $\{\pm a,\pm b\}$.
It only remains to show that
$\{-a,b\}$ are the roots of the first factor
and that the desired pair $\{a,-b\}$
splits the second quadratic factor into linear terms
(as we wrote at the outset).
Perhaps there is a clever way to infer the correct linear order
of this set of roots by noticing that $a>\cos\frac\pi4>b$ so that $a^2>\frac12>b^2$ and combining this with our identity $b=2a^2-1$ above.
I propose in stead to use that $a>b$ but then to notice
that our quadratic factors are in fact parabolas,
with roots that are very easy to order on the $x$-axis.
If we let $t=2x$ (which preserves order), then we have
$$
t^2\pm t-1 = t\,(t\pm1)-1
$$
which both pass through $(0,-1)$
and alternately pass through $(\mp1,-1)$,
shown respectively in red and blue below;
the result follows.
It's only a small extra step to note that the roots of our factor,
$t^2-t-1$, are $t=\frac{1\pm\sqrt5}{2}$, or $x=\frac{1\pm\sqrt5}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 18,
"answer_id": 6
} |
Factorize polynomial over $GF(3)$ I want to factorize $x^{11}-1$ over $GF(3)$ but I'm stuck at
$(x-1)(x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1).$
I have tried to do it trial and error but failed. Is
$$
x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1
$$
already irreducible over $GF(3)$?
| I don't want at all to depreciate the cleverness and the mastery of the previous answer, however, you should be aware that this kind of questions – it is hard to define what is this kind, it grows every day – can be answered by computer algebra.
For example, using sage :
sage: R.<x> = PolynomialRing(GF(3))
sage: factor(x^11 - 1)
(x + 2) * (x^5 + 2*x^3 + x^2 + 2*x + 2) * (x^5 + x^4 + 2*x^3 + x^2 + 2)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If a group is $3$-abelian and $5$-abelian, then it is abelian In a group $(Z,*)$, $(a*b)^{5}=a^{5}*b^{5},\forall a,b\in Z$ and $(a*b)^{3}=a^{3}*b^{3}$ then prove that $Z$ is abelian.
I know that for three consecutive integer if $(a*b)^{i}=a^{i}*b^{i},\forall a,b\in Z$ holds then $Z$ is abelian.
I know i have to use this property and i have to use three consecutive integer $3,4$ and $5$. But i am stuck.
| Here, consider $(G,\circ) \equiv (Z,*)$
Let, $a, b \in G$.
Now, using $a^3 \circ b^3 =(a \circ b)^3$ and $a^5 \circ b^5 =(a \circ b)^5$ :
$(a^3 \circ b^3) \circ (a\circ b)^2 =a^5 \circ b^5$
$ \implies a^3 \circ b^3 \circ (a \circ b)^2 = a^2 \circ (a \circ b)^3 \circ b^2 $
$\implies a \circ b^3 \circ (a \circ b)^2 = (a \circ b)^3 \circ b^2 \implies (a\circ b) \circ b^2 \circ (a \circ b)^2 = (a \circ b)^3 \circ b^2 \implies b^2 \circ (a \circ b)^2 = (a \circ b)^2 \circ b^2$
As $a \circ b= c \in G \forall a, b \in G$, therefore:
$b^2 \circ c^2 = c^2 \circ b^2 \implies c \circ b^2 \circ c^2 \circ b= c^3 \circ b^3 \implies (c\circ b)\circ ( b\circ c) \circ (c\circ b) = (c\circ b)^3 \implies b\circ c = c\circ b$, i.e the Group is abelian
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Computation of a Particular Convolution Let $\xi_{1}, \xi_{2}, \xi_{3}$ be i.i.d. $N(0,1)$. I'm attempting to compute the density of $\max \{\xi_{1}, \xi_{2}\} + \xi_{3}$. I know the density of $\max \{\xi_{1}, \xi_{2}\} $ is $2\Phi(y) \phi(y)$ and the density of $\xi_{3}$ is $\phi(s)$ so that the density of interest is the convolution $\int_{\mathbb{R}} 2\Phi(y) \phi(y) \phi(z-y)dy$. Is there any way to get at this expression, say, express it in terms of $\Phi$?
| The density of $\max\{\xi_1,\xi_2\}$ is given by
$$g(t):=\frac 1{2\pi}e^{-t^2/2}\int_{-\infty}^te^{-s^2/2}ds$$
so we want to compute
$$f(x):=\frac 1{2\pi}\frac 1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-t^2/2}\int_{-\infty}^te^{-s^2/2}ds\cdot e^{-(x-t)^2/2}dt.$$
We can write
\begin{align*}
e^{-t^2/2}e^{-(x-t)^2/2}&=\exp\left(-\frac 12(t^2+x^2-2xt+t^2)\right)\\\
&=\exp\left(-(t^2-xt)-\frac{x^2}2\right)\\\
&=\exp\left(-\frac{x^2}2\right)\exp\left(-\left(t-\frac x2\right)^2+\frac{x^2}4\right)\\\
&=\exp\left(-\frac{x^2}4\right)\exp\left(-\left(t-\frac x2\right)^2\right)
\end{align*}
hence
\begin{align*}f(x)&=(2\pi)^{-3/2}\exp\left(-\frac{x^2}4\right)\int_{-\infty}^{+\infty}\exp\left(-\left(t-\frac x2\right)^2\right)\int_{-\infty}^te^{-s^2/2}dsdt\\\
&=(2\pi)^{-3/2}\exp\left(-\frac{x^2}4\right)\int_{-\infty}^{+\infty}\exp\left(-y^2\right)\int_{-\infty}^{y+x/2}e^{-s^2/2}dsdy.
\end{align*}
Put $h(x):=e^{x^2/4}(2\pi)^{3/2}f(x)$. We have
\begin{align*}
h'(x)&=\frac 12\int_{-\infty}^{+\infty}\exp\left(-y^2\right)e^{-(y+x/2)^2/2}dy
\end{align*}
and
\begin{align*}
\exp\left(-y^2\right)e^{-(y+x/2)^2/2}&=\exp\left(-\frac 32 y^2-xy-x^2/4\right)\\\
&=\exp\left(-\frac 32\left(y^2+\frac 23xy+\frac{x^2}6\right)\right)\\\
&=\exp\left(-\frac 32\left(\left(y+\frac 13x\right)^2-\frac{x^2}9+\frac{x^2}6\right)\right)\\\
&=\exp\left(-\frac{x^2}{12}\right)\exp\left(-\frac 32\left(y+\frac 13x\right)^2\right)
\end{align*}
hence
\begin{align*}
h'(x)&=\frac 12\exp\left(-\frac{x^2}{12}\right)\int_{-\infty}^{+\infty}e^{-3t^2/2}dt\\\
&=\frac 12\frac 1{\sqrt 3}\exp\left(-\frac{x^2}{12}\right)\int_{-\infty}^{+\infty}e^{-s^2/2}ds\\\
&=\frac 12\frac 1{\sqrt 3\sqrt{2\pi}}\exp\left(-\frac{x^2}{12}\right).
\end{align*}
We deduce that
$$f(x)=\frac 12(2\pi)^{-3/2}\frac 1{\sqrt 3\sqrt{2\pi}}e^{-x^2/4}\int_{-\infty}^x\exp\left(-\frac{t^2}{12}\right)dt.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Systematically solving $ax^2+bx+c>0$ and the like We know that if $ax^2+bx+c=0$ with $a\ne0$, then the solution(s) can be given by the quadratic formula $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
But what if we want to solve a quadratic inequality such as $ax^2+bx+c>0$?
First, if $ax^2+bx+c$ can be factored into $(x+p)(x+q)$, then we have $(x+p)(x+q)>0$, so we know:
*
*Both $(x+p)$ and $(x+q)$ are positive; or,
*Both $(x+p)$ and $(x+q)$ are negative.
For the first case, we have $(x+p)>0$ and $(x+q)>0$, which simplify to $x>-p$ and $x>-q$, which is just $x>\max\{-p,-q\}$.
For the second case, we have $(x+p)<0$ and $(x+q)<0$, which simplify to $x<-p$ and $x<-q$, which is just $x<\min\{-p,-q\}$.
So the final answer is just $x\in\{x:x>\max\{-p,-q\}\text{ or }x<\min\{-p,-q\}\}$.
Next suppose $\sqrt{b^2-4ac}$ is imaginary. Then we have two cases:
*
*$a>0$ so that the graph of $f(x)=ax^2+bx+c$ is completely above the $x$-axis. Then all real numbers are solutions to $ax^2+bx+c>0$.
*$a<0$ so that the graph of $f(x)=ax^2+bx+c$ is completely below the $x$-axis. Then no real number is a solution.
Now suppose $ax^2+bx+c$ does not factor. Then "force-factor" using the quadratic equation into $(x-\frac{-b + \sqrt{b^2-4ac}}{2a})$ and $(x-\frac{-b - \sqrt{b^2-4ac}}{2a})$ and repeat the steps above. Another way to think about this: since we are guaranteed that $\sqrt{b^2-4ac}$ is real (we took care of the imaginary cases already), we can divide the real number line into three intervals: $(-\infty,\frac{-b - \sqrt{b^2-4ac}}{2a})$, $(\frac{-b - \sqrt{b^2-4ac}}{2a},\frac{-b + \sqrt{b^2-4ac}}{2a})$, and $(\frac{-b + \sqrt{b^2-4ac}}{2a},+\infty)$ (or the other way round if $a<0$). Then check some number in each interval, and if that number is greater than $0$, then include that entire interval in the final answer. Otherwise don't include it. We can also do this by taking the outer two intervals when $a>0$ and taking the middle interval when $a<0$ (I think).
Questions: (1) Is this approach correct? (2) If so, is there a more efficient way to do it, i.e., how might one go about programming an algorithm for this type of problem?
| $$\text{If }~ f>0 ~\text{then}~ \begin{cases}
x \in(-\infty,x_1) \cup (x_2,+\infty), & \text{if } a>0 ~\text{and}~ D>0 \\
x \in (x_1,x_2), & \text{if } a<0 ~\text{and}~ D>0 \\
x \in \mathbb{R}, & \text{if } a>0 ~\text{and}~ D < 0 \\
x \in \mathbb{R} \backslash \{x_{1,2}\} ,& \text{if } a>0 ~\text{and}~ D = 0 \\
x \in \emptyset & \text{if } a<0 ~\text{and}~ D \leq 0
\end{cases}$$
where $x_{1,2}$ are solutions of $ax^2+bx+c=0$ , and $D=b^2-4ac$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Sufficient condition for a matrix to be positive definite Is a sufficient condition for a $2\times 2$ matrix
$$\left(\begin{array}{cc}a&b\\b&d\end{array}\right)$$ to be positive definite that $a >0$ and $ad > b^2$ ?
| Slightly high-brow route: if a symmetric matrix $\mathbf A$ is positive definite, then the matrix $\mathbf X\mathbf A\mathbf X^\top$ is positive definite as well (Sylvester). Now, consider the decomposition
$$\begin{pmatrix}a&b\\b&d\end{pmatrix}=\begin{pmatrix}1&0\\\frac{b}{a}&1\end{pmatrix}\cdot\begin{pmatrix}a&0\\0&d-\frac{b^2}{a}\end{pmatrix}\cdot\begin{pmatrix}1&0\\\frac{b}{a}&1\end{pmatrix}^\top$$
How does one check if a diagonal matrix is positive definite?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Irreducible Polynomial over $\mathbb{Q}$ I'm trying to see if the following polynomials are irreducible over $\mathbb Q$:
$f(x) = x^4 - x^2 + 2x -1$
$g(x) = x^3 + 7x^2 -8x +1$
$h(x) = x^4 + x^3 + x^2 + x + 1.$
Now, for $h(x)$, I can write $(x+1)$ for $x$ and get :
$x^4 + 5x^3 + 10x^2 + 10x +5$, for which I can set $P=5$ and by Eisenstein, this is irreducible over $\mathbb Z$, therefore irreducible over $\mathbb Q$.
For $g(x)$, since it is cubic and primitive, there is either a root or it is irreducible. I get to the stage $g(x)=(x-a)(x^2 + bx +c)$ but then cannot equate the coefficients.
Similar situation for $f(x)$, I get to the stage $f(x)=(x^2 + ax +b)(x^2 +cx + d)$ but get stuck when trying to figure out the coefficients. Am I doing something wrong here?
I would really appreciate it if someone could explain this to me please, since I am wondering if I am overlooking a silly mistake. Thanks.
| If $f(x)=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$, then we have
$a+c=0$, $ac+b+d=-1$, $ad+bc=2$ and $bd=-1$. By Gauss's theorem of primitive polynomial, we have $a,b,c,d$ must all be integers, hence $c=-a$, and $d=1$, $b=-1$ or $d=-1$, $b=1$. We may assume $b=1$, $d=-1$ by symmetry of these two factors, then $-a^2=-1$, $-2a=2$ hence $a=-1$, $c=1$, so we have $f(x)=(x^2-x+1)(x^2+x-1)=x^4-(x-1)^2=x^4-x^2+2x-1$. Hence $f$ is reducible in $\mathbb{Q}[t]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that the fields $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4 \rangle$ are isomorphic I have been stuck in this problem for some time now.
Prove that $x^2+2$ and $x^2+x+4$ are irreducible over $\mathbb{Z}_{11}$. Also, prove further $\mathbb Z_{11}[x]/\langle x^2+1\rangle$
and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic, each
having $121$ elements.
The first part is easy to prove since there is no element of $\mathbb Z_{11}$ that satisfies either of the polynomials given in the question. However, proving that $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic has been a challenge for me. How do I proceed? Moreover, how do I show that the the fields $\mathbb {Z}_{11}[x]/\langle x^2+1\rangle$ and $\mathbb {Z}_{11}[x]/\langle x^2+x+4\rangle$ each have $121$ elements?
| Here's an explicit isomorphism $f:\mathbb{Z}_{11}[x]/\langle x^2 + 1\rangle\rightarrow \mathbb{Z}_{11}[x]/\langle x^2+x+4\rangle$ between the 2 fields. Abusing notation, I'm going to refer to an element in the domain by it's natural preimage in $\mathbb{Z}_{11}[x]$, and likewise in the range.
First note that $1$ is uniquely determined in a field, so we must send $1$ to $1$. Using additivitiy, $f(n) = n$ for any $n\in\mathbb{Z}/11\mathbb{Z}$. So, the only remaining question is what $f(x)$ will be. Notice that $x^2 = -1$, so $f(x)$ must be a squareroot of $-1$ in the other field.
Writing $f(x) = ax+b$, we get $$-1 = f(x)^2 = a^2 x^2 +2abx + b^2 = a^2(-x-4)+2abx + b^2 = (2ab-a^2)x + b^2-4.$$
Now, $a\neq 0$ (else $f$ is not injective), so we learn that $2b=a$ and $b^2-4 = -1$.
The second equation, $b^2-4 = -1$ has 2 solutions (mod 11), $b= 5$ and $b = 6 (=-5)$. We'll pick $b=5$ (the choice doesn't matter). Then since $2b=a$, we get $a = 10 = -1$.
Thus, we have $f(x) = -x + 5$.
Putting this altogether, we have $f(ax+b) = -ax + 5a + b$ defining our isomorphism.
To see $f$ is injective, assume $f(ax+b) = 0$. Then, since $-ax + 5a+b = 0$, we must have $-a = 0$, so $a=0$. Once we know $a=0$, $b=0$ follows. So $f$ is injective. To see it's surjective, notice $f(-ax + 5a+b) = ax + b$, so $f$ is surjective.
Finally, we check it's a homomorphism. We have \begin{align*}f((ax+b) + (cx+d)) &= -(a+c)x + 5(a+c) + (b+d)\\\ &= -ax + 5a +b + -cx + 5c + d\\\ &= f(ax+b) + f(cx+d).\end{align*}
We also have \begin{align*} f((ax+b)(cx+d)) &= f(acx^2 + (ad+bc)x + bd)\\\ &= f((ad+bc)x + bd-ac)\\\ &=-(ad+cb)x + 5(ad+bc) + bd-ac\end{align*}
while \begin{align*} f(ax+b)f(cx+d) &= (-ax + 5a +b)(-cx+5c+d)\\\ &=acx^2 -5acx -adx -5acx +25ac+5ad-bcx+5bc+bd \\\ &= ac(-x-4) +x(-10ac-ad-bc) + 25ac+5ad+5bc+bd \\\ &= x(-11ac-ad-bc) + (21 ac+5ad+5bc+bd) \\\ &= x(-ad-bc) + 5ad+5bc +bd-ac\end{align*}
so $f((ax+b)(cx+d)) = f(ax+b)f(cx+d)$. Thus, $f$ is the desired isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Summing a binomial series Consider the following sum:
$$S(n)=\sum_{k=0}^{\infty}\frac{\binom{2k+n}{k}}{2k+n}\frac{1}{2^{2k}};n=0,1,2,3,...$$ Is there a closed form for $S(n)$?
| The sum at hand is a hypergeometric series. Let
$$
c_k = \frac{1}{n+2k} \binom{n+2k}{k} \frac{1}{2^{2k}} = \frac{(n-1+2k)!}{k! (n+k)!} \frac{1}{4^k}
$$
Indeed, the hypergeometric certificate is:
$$
\frac{c_{k+1}}{c_k} = \frac{1}{4} \frac{(n+2k)(n+2k+1)}{(n+1+k) (k+1)}
$$
Meaning that
$$
\sum_{k=0}^\infty c_k = c_0 \sum_{k=0}^\infty \frac{\left(n/2\right)_k \left(n/2+1/2\right)_k}{(n+1)_k} \frac{1}{k!} = \frac{1}{n} \cdot {}_2 F_1 \left(\frac{n}{2}, \frac{n+1}{2} ; n+1 ; 1 \right)
$$
where $(a)_k$ denotes Pochhammer symbol.
Using Gauss's theorem, applicable for $\Re(c-a-b)>0$
$$
{}_2 F_{1} \left(a,b; c; 1\right) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}
$$
we have
$$
\sum_{k=0}^\infty c_k = \frac{1}{n} \frac{ \Gamma(n+1) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2} + 1 \right) \Gamma\left(\frac{n+1}{2} \right) } \stackrel{\text{duplication}}{=} \frac{1}{n} \frac{ 2^{n} \pi^{-1/2} \Gamma\left(\frac{n+1}{2}\right) \Gamma\left(\frac{n}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2} \right) \Gamma\left(\frac{n+1}{2} \right) } = \frac{2^n}{n}
$$
Since $\Gamma(1/2) = \sqrt{\pi}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\int \frac{1}{\sqrt{2-x^2}}\, dx$ Somewhere in the provided answer:
$$\int \frac{1}{\sqrt{2-x^2}} dx = \sin^{-1}{\frac{x}{\sqrt{2}}}$$
How did they get that? What I have:
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2(1-\frac{x^2}{2})}} = \frac{1}{\sqrt{2} \sqrt{1-\frac{x^2}{2}}}$$
$$\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}} = \frac{1}{\sqrt{2}} \sin^{-1}{\frac{x}{\sqrt{2}}}$$
So I have an extra $\frac{1}{\sqrt{2}}$ ... I probably had some stupid mistakes?
| $$\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}$$
$$\int\frac{dx}{\sqrt{2-x^2}}=\int\frac{dx}{\sqrt{{(\sqrt2})^2-x^2}} = \sin^{-1}\frac{x}{\sqrt2}$$
or, doing it other way
$$\int\frac{1}{\sqrt{2-x^2}}dx=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}}dx$$
now put , t=$\frac{x}{\sqrt2}$
$$\frac{1}{\sqrt{2}}\int \frac{\sqrt2 \, dt}{\sqrt{1-t^2}}= \int\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t+c = \sin^{-1}\frac{x}{\sqrt2}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For what values of x does the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ converge? For what values of x does the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ converge?
The solution states:
The general term is of the form $u_n(x)=\frac{x^{n-1}}{(2n-1)}$, and hence
$$\frac{|u_{n+1}|}{|u_n|}=\frac{|x^n|}{(2n+1)}\cdot\frac{(2n-1)}{|x^{n-1}|}$$
------edit start-------
$$=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$$
------ edit end -------
$$=\frac{(2n-1)}{(2n+1)}|x|$$
clearly
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=|x|$$
My question is:
*
*How do you get from $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ to $u_n(x)=\frac{x^{n-1}}{(2n-1)}$?
*Why do the absolute restriction only apply to the $|x^n|$ and $|x^{n-1}|$ in the next line and not to the rest of the equation?
*and lastly, We go from $=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$ to $=\frac{(2n-1)}{(2n+1)}|x|$ I.E. how does $\frac{|x^n|}{|x^{n-1}|}=|x|$ in the next line? (I understand that once you apply the limits, that $\frac{(2n-1)}{(2n+1)} = 1$)
Edits
I have added a line to the equation that was not there earlier and clarified my last question.
| *
*They are defining the terms of your sum. $u_1(x)=\frac{x^{1-1}}{2-1}=1$ and $u_2(x)=\frac{x^{2-1}}{4-1}=\frac{x}{3}$ and $u_3(x)=\frac{x^{3-1}}{6-1}=\frac{x^2}{5}$ and so on and so forth.
2.$|u_n|=\left|\frac{x^n}{2n+1}\right|$ but since $2n+1>0$ because we are dealing in positive integers then we can remove the absolute value around them resulting in $|u_n|=\frac{|x^n|}{2n+1}$ similar reasoning for $u_{n+1}$
3.basic exponent rules for division and a little bit of absolute value rules thrown in $\frac{|x^n|}{|x^{n-1}|}=\frac{|x|^n}{|x|^{n-1}}=|x|$
| {
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Find the inverse of a $4\times4$ matrix My matrix looks like this:
$$\left(\begin{array}{rrrr}
1& 1 & 1 & 1\\
1& -1 & 1 & 0\\
1& 1 & 0 & 0\\
1& 0 & 0 & 0
\end{array}\right)$$
The right lower half are all zeros.
Is there a quick way to find an inverse of this matrix?
I have the solution, but I'm unable to find the algorithm to get the inverse.
| Another way. Let your matrix be $A$. Then $A = J B$ where $$J = \pmatrix{0&0&0&1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0\cr}, \ B = JA = \pmatrix{1 & 0 & 0 & 0\cr
1 & 1 & 0 & 0\cr 1 & -1 & 1 & 0\cr 1 & 1 & 1 & 1\cr}$$ since multiplying by $J$ on the left flips the matrix vertically (interchanging first and fourth rows and second and third rows).
Now $B$ is a lower triangular matrix. We can write it as $I + N$ where the nonzero entries of $N$ are all below the main diagonal. Thus $N^4 = 0$ and
$$ B^{-1} = I - N + N^2 - N^3 =\pmatrix{ 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr} - \pmatrix{ 0 & 0 & 0 & 0 \cr
1 & 0 & 0 & 0 \cr 1 & -1 & 0 & 0 \cr 1 & 1 & 1 & 0 \cr}
+ \pmatrix{ 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & 0 \cr -1 & 0 & 0 & 0 \cr 2 & -1 & 0 & 0 \cr}
- \pmatrix{ 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr -1 & 0 & 0 & 0 \cr}
= \pmatrix{ 1 & 0 & 0 & 0 \cr
-1 & 1 & 0 & 0 \cr -2 & 1 & 1 & 0 \cr 2 & -2 & -1 & 1 \cr}
$$
We'll then have $$A^{-1} = B^{-1} J^{-1} = B^{-1} J = \pmatrix{ 0 & 0 & 0 & 1 \cr
0 & 0 & 1 & -1 \cr 0 & 1 & 1 & -2 \cr 1 & -1 & -2 & 2 \cr}
$$ where multiplying by $J$ on the right flips the matrix horizontally (interchanging first and fourth columns and second and third columns).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/141936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculus question about integration. Consider $ f$ from $[0,1]$to $[0,1] $ be continuous and non constant .
Then, is there $c\in[0,1]$ such that $f(c) =\int^1 _0 f^2(t) dt $ ?
| Counterexample
Consider $f(x) = \frac{1}{2} (1-x) + \frac{1}{4} x = \frac{1}{4} \left(2-x\right)$. Then
$$
\int_0^1 f^2(x) \mathrm{d} x = \frac{1}{16} \int_0^1 \left(2-x\right)^2 \mathrm{d} x = \frac{1}{16} \int_1^2 y^2 \mathrm{d} y = \frac{1}{16} \cdot \frac{2^3-1^3}{3} = \frac{7}{48}
$$
But the equation
$$
\frac{1}{4} \left(2-x\right) = \frac{7}{48} \quad x = 2 - \frac{7}{12} = \frac{24-7}{12} = \frac{17}{12}
$$
but the solution $ x= \frac{17}{12}$ lies outside $[0,1]$ interval.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing the trace and determinant of $A+B$, given eigenvalues of $A$ and an expression for $B$ Let $A$ be $4\times 4$ matrix with real entries such that $-1$, $1$, $2$, and $-2$ are its eigenvalues.
If $B = A^4 - 5A^2+5I$, where $I$ denotes $4\times 4$ identity matrix, then what would be determinant and trace of matrix $A+B$?
| Since $A$ is $4\times 4$ and its eigenvalues are $2$, $-2$, $1$, and $-1$, the minimal and characteristic polynomials of $A$ agree and are both equal to
$$(t-1)(t+1)(t-2)(t+2) = (t^2-1)(t^2-4) = t^4 - 5t^2 + 4.$$
In particular, by the Cayley-Hamilton Theorem,
$$A^4 - 5A^2 + 4I = 0,$$
and therefore
$$B+A = A^4 - 5A^2 + 5I + A = (A^4-5A^2+4I) + (A+I) = A+I.$$
Now notice that $\lambda$ is an eigenvalue of $A$ if and only if $\alpha\lambda+\beta$ is an eigenvalue of $\alpha A+\beta I$, to conclude that the eigenvalues of $B+A=A+I$ are $0$, $-1$, $2$, and $3$. Therefore, the trace is $0-1+2+3 = 4$, and the determinant is $0$ (since $A+I$ is not invertible, or since the determinant is the product of the eigenvalues).
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $f'(x)$ when $f(x)=\int^1_0 e^{xy+y^2}dy$
If $f(x) = \int^1_0 e^{xy+y^2}dy$, find $f'(0)$.
I understand that this is function defined by an integral, and $e^{y^{2}}$ does not integrate into an elementary function. So, I will need to take $f'(x)$ which yields:
$$\int^1_0 ye^{xy+y^2}dy$$
I am trying to integrate this, but I am failing. I take it I should use integration by parts, but I can't because I still have $e^{y^2}$ term. Any help?
| You are given
$$f(x) =\int_0^1 \exp(xy+y^2)dy $$
Differentiating gives
$$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy $$
Since we need $f'(0)$ we might as well plug in $x=0$. This gives
$$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy $$
But this integral is quite striaghtforward,
$$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy =\left. \frac 1 2 e^{y^2} \right|_0^1 = \frac 1 2 (e-1) $$
By the OP's request:
$$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy $$
We complete the square
$$f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_0^1 {y\exp \left[ {{{\left( {y + \frac{x}{2}} \right)}^2}} \right]dy} $$
We change variables
$$\eqalign{
& y + \frac{x}{2} = u \cr
& dy = du \cr} $$
$$\eqalign{
& f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\left( {u - \frac{x}{2}} \right)} \exp \left( {{u^2}} \right)du \cr
& f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du} \cr} $$
Let's focus on the first integral:
$${I_1} = \int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du = \left. {\frac{1}{2}\exp {u^2}} \right|_{x/2}^{1 + x/2} = \frac{1}{2}\exp \frac{{{x^2}}}{4}\left[ {\exp \left( {x + 1} \right) - 1} \right]$$
So we have
$$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du} $$
The other integral evaluates in terms of the error function
$$\operatorname{erf} (x) = \frac{2}{{\sqrt \pi }}\int_0^x {{e^{ - {t^2}}}} dt.$$
with a change or variables $u=-v$,
$$\int_{ - \left( {1 + \frac{x}{2}} \right)}^{ - \frac{x}{2}} {\exp \left( { - {v^2}} \right)dv} = \frac{{\sqrt \pi }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$$
So the function is
$$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\frac{{\sqrt \pi }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$$
For large values of $x$, you can neglect the last result (since the value will be smaller and smaller), and you can estimate $f'$ with
$$f'(x) \approx \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right]$$
Cheers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq (a_1 + a_2 + \ldots + a_k)^2$ I need to prove that
$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq (a_1 + a_2 + \ldots + a_k)^2\;,$$ where $a_1, a_2, \dots, a_k$ is some set of reals.
Firstly:
Can I presume without the loss of generality that $a_1 \leq a_2 \leq \ldots \leq a_n$ ?
This is how far I got:
I used the formula $\left \langle a,b \right \rangle \leq |a||b|$:
$$\begin{align*}\left \langle a,1 \right \rangle &\leq |a||1|\\
(a_1 + a_2 + \ldots + a_k) &\leq \sqrt{(a_1^2 + a_2^2 + \ldots + a_k^2)}\sqrt{k}
\end{align*}$$
Square it:
$$(a_1 + a_2 + \ldots + a_k)^2 \leq k(a_1^2 + a_2^2 + \ldots + a_k^2)$$
Now I have to prove that:
$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq k(a_1^2 + a_2^2 + ... + a_k^2)$$
But I'm not sure how. Any pointers?
| You can simply use the inequality of quadratic and arithmetic mean for $k$ elements $\frac{a_1}k$, $k-1$ elements $\frac{a_2}{k-1}$ etc. For the inequality between quadratic and arithmetic mean
see e.g. Jensen inequality
and Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality at AoPS.
Arithmetic mean is $$a=\frac{a_1+\dots+a_k}{\frac{k(k+1)}2}.$$
Quadratic mean is $$q=\sqrt{\frac{\frac{a_1^2}k+\frac{a_2^2}{k-1}+\dots+a_k^2}{\frac{k(k+1)}2}}.$$
So from $q^2\ge a^2$ you get
$$\frac{\frac{a_1^2}k+\frac{a_2^2}{k-1}+\dots+a_k^2}{\frac{k(k+1)}2} \ge \left(\frac{a_1+\dots+a_k}{\frac{k(k+1)}2}\right)^2$$
and
$$\frac{k(k+1)}2 \left(\frac{a_1^2}k+\frac{a_2^2}{k+1}+\dots+a_k^2\right) \ge (a_1+\dots+a_k)^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How did they get this result? Please, explain these computations:
1) $-\left(\frac{1}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{3}}{2} = \cos x$
How did we get $\frac{\sqrt{3}}{2}$ from $-\left(\frac{1}{2}\right)^2 +1$?
2) $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{2}}{2} = \cos x$
How did we get $ \frac{\sqrt{2}}{2}$ from $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 $?
| $$\sqrt{-\left(\frac12\right)^2 +1} = \sqrt{-\frac14 +1} = \sqrt{\frac34} = \frac{\sqrt{3}}{2}$$
$$\sqrt{-\left(\frac{\sqrt{2}}{2}\right)^2 +1} = \sqrt{-\frac24 +1} = \sqrt{\frac24} = \frac{\sqrt{2}}{2}$$
though you should also consider the negative square roots.
| {
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"url": "https://math.stackexchange.com/questions/146785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Constant functions
Let $f$ , $g$ , $h$ be three functions from the set of positive real numbers to itself satisfying $$f(x)g(y) = h\left((x^2+y^2)^{\frac{1}{2}}\right)$$ for all positive real numbers $x$ , $y$ . Show that $\dfrac{f(x)}{g(x)}$ , $\dfrac{g(x)}{h(x)}$ and $\dfrac{h(x)}{f(x)}$ are all constant functions .
I have proved that $\dfrac{f(x)}{g(x)}$ is constant and can see that proving either of the last two will prove the final one , but I am not able to prove any of the last two .
Thanks for any help .
| forall $x,y,z>0$, $h(\sqrt{x^2+y^2})f(z) = f(x) g(y) f(z) = f(x) h(\sqrt{y^2+z^2})$,
thus $\frac {h(\sqrt{x^2+y^2})}{h(\sqrt{z^2+y^2})} = \frac{f(x)}{f(z)} $.
Therefore, forall $x,y,z,t > 0$ : $\frac {h(\sqrt{x^2+z^2})}{h(\sqrt{y^2+z^2})} = \frac {f(x)}{f(y)} = \frac {h(\sqrt{x^2+z^2+t^2})}{h(\sqrt{y^2+z^2+t^2})} = \frac {f(\sqrt{x^2+z^2})}{f(\sqrt{y^2+z^2})}$, thus $\frac {h(\sqrt{x^2+z^2})}{f(\sqrt{x^2+z^2})} = \frac{h(\sqrt{y^2+z^2})}{f(\sqrt{y^2+z^2})} $, which proves that $h/f$ is a constant function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Determining whether this congruence is solvable: $3x^2+6x+5 \equiv 0\pmod{89}$ I'm trying to figure out whether the following quadratic congruence is solvable: $3x^2+6x+5 \equiv 0\pmod{89}$.
It's impossible to divide $3x^2+6x+5$ to a form of $f(x) \cdot g(x)=3x^2+6x+5$ and then to check whether $f(x)\equiv 0
\pmod{89}$ or $g(x)\equiv 0(89)$, but $3x^2+6x+5 \equiv 0\pmod{89}$ is equal to $3(x+1)^2+2 \equiv 0\pmod{89}$ or $3(x+1)^2 \equiv -2\pmod{89}$ or $3(x+1)^2 \equiv 87\pmod{89}$ or
$(x+1)^2 \equiv 29\pmod{89}$. for $y=x+1$, I need to determine whether $y^2 \equiv 29\pmod{89}$ is solvable, and it is not. Am I able to conclude something about the original equation in this way? what is the correct way to solve this problem?
Thanks a lot!
| If you know about the law of quadratic reciprocity, it gives you a way to tell whether $y^2\equiv29\pmod{89}$ is solvable.
Here's how it goes: Since 29 and 89 are both prime and congruent 1 mod 4, $y^2\equiv29\pmod{89}$ is solvable if and only if $y^2\equiv89\pmod{29}$ is solvable. That reduces to $y^2\equiv2\pmod{29}$. Then there's a little result that says if $p$ is an odd prime then $y^2\equiv2\pmod p$ is solvable if and only if $p\equiv\pm1\pmod8$.
There's more to quadratic reciprocity than what I've done here. It's in every intro number theory text, and all over the web.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/147350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to evaluate $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$ I need to calculate the length of a curve $y=2\sqrt{x}$ from $x=0$ to $x=1$.
So I started by taking $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$, and then doing substitution: $\left[u = 1+\frac{1}{x}, \text{d}u = \frac{-1}{x^2}\text{d}x \Rightarrow -\text{d}u = \frac{1}{x^2}\text{d}x \right]^1_0 = -\int\limits^1_0 \sqrt{u} \,\text{d}u$ but this obviously will not lead to the correct answer, since $\frac{1}{x^2}$ isn't in the original formula.
Wolfram Alpha is doing a lot of steps for this integration, but I don't think that many steps are needed.
How would I start with this integration?
| Note that the integrand $\sqrt{1 + {1 \over x}}$ decreases from infinity to $\sqrt{2}$ as $x$ goes from $0$ to $1$. The area under the graph is therefore equal to the area of the box $[0,1] \times [0,\sqrt{2}]$ plus the area under the graph of the inverse function $g(y)$ to $\sqrt{1 + {1 \over x}}$ from $y = \sqrt{2}$ to $y = \infty$. Note that $g(y) = {1 \over y^2 - 1}$. So the answer is
$$\sqrt{2} + \int_{\sqrt{2}}^{\infty} {1 \over y^2 - 1}\,dy$$
This integral is easily computed, using partial fractions for example. The result is
$$\sqrt{2} + {1 \over 2}\ln\bigg({y - 1 \over y + 1}\bigg)\bigg|_{\sqrt{2}}^{\infty}$$
$$=\sqrt{2} - {1 \over 2}\ln\bigg({\sqrt{2} - 1 \over \sqrt{2} + 1}\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 6
} |
Equivalent statement of Fermat's last theorem I'd like to know how to show that, if there are no integer solutions to $a^n + b^n = c^n$ for $a, b, c, n \in Z$ and $n > 2$ then this is equivalent to either $a$ or $b = 0$ are the only rational solutions to $a^n + b^n = 1$ for $n > 2$. Not sure if this is a simple proof or not, could someone provide some pointers?
| You got the statements slightly wrong. I believe what you meant to write as the two statements is:
*
*Statement 1. If $n\gt 2$, and $a,b,c\in\mathbb{Z}$ are integers such that $a^n+b^n=c^n$, then one of $a$, $b$, and $c$ are equal to $0$. (There are no nonzero integer solutions to $a^n+b^n = c^n$ if $n\gt 2$).
*Statement 2. If $n\gt 2$, then the only rational solutions to $a^n+b^n = 1$ have $a=0$ or $b=0$.
Statement 1*$\implies$*Statement 2
Suppose $a$ and $b$ are rational numbers such that $a^n+b^n = 1$. Finding a common denominator, we can write $a=\frac{c}{m}$ and $b=\frac{d}{m}$, with $c,d$ integers, $m\gt 0$. Therefore, we have
$$\frac{c^n}{m^n} + \frac{d^n}{m^n} = 1.$$
Multiplying through by $m^n$ we obtain
$$c^n + d^n = m^n$$
with $c,d,m$ integers, $n\gt 2$. By Statement 1, at least one of $c$, $d$, and $m$ must be equal to $0$. We know that $m$ is not equal to $0$, which means either $c=0$ or $d=0$. If $c=0$, then $a=\frac{c}{m} = 0$; if $d=0$, then $b = \frac{d}{m} = 0$. So we conclude that if Statement 1 holds, and $a^n + b^n = 1$ with $a$ and $b$ rationals, $n\gt 2$, then either $a=0$ or $b=0$, as claimed.
Statement 2*$\implies$*Statement 1
Let $n\gt 2$ and let $a,b,c$ be integers such that $a^n+b^n=c^n$. We want to show that at least one of $a$, $b$, and $c$ are equal to $0$. If $c=0$, then we are done. If $c\neq 0$, then dividing through by $c^n$ we have
$$\frac{a^n}{c^n} + \frac{b^n}{c^n} = 1$$
or
$$\left(\frac {a}{c}\right)^n + \left(\frac{b}{c}\right)^n = 1.$$
By Statement 2, we must have $\frac{a}{c}=0$ or $\frac{b}{c}=0$. In the first case, we have $a=0$; in the second, we have $b=0$. So, either $a=0$, $b=0$, or $c=0$, as claimed. $\Box$
Note that there are certainly integer solutions to $a^n+b^n=c^n$: take $a=c=1$, $b=0$. The point is that there are no nonzero solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/151606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Lagrange multiplier - find minima of a function satisfying a condition I am supposed to find the points (x,y,z) satisfying the condition $x^2+2y^2-z^2-1=0$ that are the closest to origin (0,0,0).
So basically, the idea was to find the minima of
$$\Lambda(x,y,z,\lambda) = \sqrt{x^2+y^2+z^2}+\lambda(x^2+2y^2-2z^2-1)$$
For that I determined
$$\Lambda_x(x,y,z,\lambda) = \frac{x}{\sqrt{x^2+y^2+z^2}} + 2\lambda x$$
$$\Lambda_y(x,y,z,\lambda) = \frac{y}{\sqrt{x^2+y^2+z^2}} + 4\lambda y$$
$$\Lambda_z(x,y,z,\lambda) = \frac{z}{\sqrt{x^2+y^2+z^2}} - 2\lambda z$$
But that would mean for $\lambda \neq 0$
$$\lambda_1 = -\frac{1}{2\sqrt{x^2+y^2+z^2}}$$
$$\lambda_2 = -\frac{1}{4\sqrt{x^2+y^2+z^2}}$$
$$\lambda_3 = \frac{1}{2\sqrt{x^2+y^2+z^2}}$$
Since $\lambda_1 \neq \lambda_2 \neq \lambda_3$ doesn't that mean that this function doesn't have any minima? I'm probably having a huge misunderstanding somewhere, but I just can't figure out where I went wrong, so please help me with this.
| The problem with your reasoning is that it assumes that neither $x$, $y$ or $z$ is $0$, but this is false!
It's easier to solve this problem using the square of distance instead. The distance is always positive, so if you minimize the square of distance, you'll minimize the distance too.
We have:
$$\Lambda(x,y,z,\lambda) = x^2+y^2+z^2 + \lambda(x^2+2y^2-z^2-1)$$
Therefore:
\begin{align*}
\Lambda_x(x,y,z,\lambda) &= 2x + 2\lambda x &= 0 \\
\Lambda_y(x,y,z,\lambda) &= 2y + 4\lambda y &= 0 \\
\Lambda_z(x,y,z,\lambda) &= 2z - 2\lambda z &= 0 \\
\Lambda_\lambda(x,y,z,\lambda) &= x^2+2y^2-z^2-1 &= 0
\end{align*}
Or:
\begin{align*}
(1+\lambda)x &= 0 \\
(1+2\lambda)y &= 0 \\
(1-\lambda)z &= 0 \\
x^2+2y^2-z^2-1 &= 0
\end{align*}
For each of the first three equations use: If $a \cdot b = 0$, then $a = 0$ or $b = 0$. It'll enable you to find $(x, y, z)$ that minimizes the distance.
Also, notice that $x^2+2y^2-z^2-1 = 0$ is a hyperboloid of one sheet. Look at its plot and try to guess the points that minimize the distance. Compare with your answer above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/151738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.
I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem:
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok.
If you're answering with an example then use one pair per answer.
I just want to see that clock with another pair of numbers :)
Notes for the current clock:
1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.
5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$
| solution for n = 1, k = 12:
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1 = 2
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1 = 3
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1 = 4
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1 = 5
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1 = 6
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1 = 7
$$
$$
1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1 = 8
$$
$$
1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1+1 = 9
$$
$$
1 \times 1 \times 1+1+1+1+1+1+1+1+1+1 = 10
$$
$$
1 \times 1+1+1+1+1+1+1+1+1+1+1 = 11
$$
$$
1+1+1+1+1+1+1+1+1+1+1+1 = 12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 10,
"answer_id": 7
} |
Integral of $\int \frac {\sqrt {x^2 - 4}}{x} dx$ I am trying to find $$\int \frac {\sqrt {x^2 - 4}}{x} dx$$
I make $x = 2 \sec\theta$
$$\int \frac {\sqrt {4(\sec^2 \theta - 1)}}{x} dx$$
$$\int \frac {\sqrt {4\tan^2 \theta}}{x} dx$$
$$\int \frac {2\tan \theta}{x} dx$$
From here I am not too sure what to do but I know I shouldn't have x.
$$\int \frac {2\tan \theta}{2 \sec\theta} dx$$
I also know I shouldn't have dx anymore.
$$dx = 2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int \frac {2\tan \theta}{2 \sec\theta} 2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int {2\tan^2 \theta} \; \mathrm d\theta$$
$$2\int {\tan^2 \theta} \; \mathrm d\theta$$
I have no idea how to find the integral of $\tan^2 \theta$
So I use Wolfram Alpha:
$$\tan \theta - \theta + c$$
Now I need to replace theta with x.
$$x = 2 \sec\theta$$
With same mathmagics I produce
$$ \frac {x}{2} = \sec \theta$$
$$ \theta = \operatorname {arcsec} \left(\frac{x}{2}\right)$$
$$\tan \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) - \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) + c$$
This is wrong but I am not sure why.
| Multiply both top and bottom by $\sqrt{x^2 - 4}$ to obtain
$$
\int\dfrac{x^2 - 4}{x\sqrt{x^2 - 4}}\,\mathrm{d}x=\int\dfrac{x}{\sqrt{x^2 - 4}} + 4\int-\dfrac{1}{x}\dfrac{1}{\sqrt{x^2\left(1 - (2/x)^2\right)}}\,\mathrm{d}x
$$
In the fisrt integral, set $u=x^2-4$ and $x\,\mathrm{d}x = \dfrac{1}{2}\,\mathrm{d}u$ to get
$$
\int\dfrac{x}{\sqrt{x^2-4}}\,\mathrm{d}x = \sqrt{x^2 - 4}+C_0.
$$
In the second, consider $x>2$:
$$
2\int-\dfrac{2}{x^2}\dfrac{\mathrm{d}x}{\sqrt{1 - (2/x)^2}}
$$
Now, set $u=\dfrac{2}{x}$ and $\mathrm{d}u = -\dfrac{2}{x^2}\,\mathrm{d}x$. Then,
$$
\begin{aligned}
2\int-\dfrac{2}{x^2}\dfrac{\mathrm{d}x}{\sqrt{1 - (2/x)^2}} &=2\int\dfrac{\mathrm{d}u}{\sqrt{1 - u^2}} \\
&=2\arcsin u + C_1
\end{aligned}
$$
Adding them up:
$$ \int\dfrac{\sqrt{x^2 - 4}}{x}\,\mathrm{d}x = \sqrt{x^2 - 4} + 2\arcsin\left(\frac{2}{x}\right) + C $$
Note that, if $x<-2$, we would obtain the integral
$$\int\dfrac{\sqrt{x^2 -4}}{x}\,\mathrm{d}x = \sqrt{x^2 - 4}-2\arcsin\left(\dfrac{2}{x}\right)+C, $$
but, since $x<-2$, $-2\arcsin\left(\dfrac{2}{x}\right)>0$, and the sign would become positive.
Therefore, we can write
$$\begin{aligned}
\int\dfrac{\sqrt{x^2 - 4}}{x}\,\mathrm{d}x &= \sqrt{x^2-4}+2\arcsin\left(\dfrac{2}{|x|}\right) + C \\&= \sqrt{x^2-4}+2\operatorname{arccsc}\left(\dfrac{|x|}{2}\right) + C
\end{aligned}
$$
Note that the arbitrary constants are not necessarily the same.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Arc length of $y^3 = x^2$ I am trying to find the arc length of $y^3 = x^2$ and I am suppose to use two formulas, one for in terms of x and one for in terms of y.
At first I need to find (0,0) (1,1) and I start with in terms of x
$$y\prime = \frac{2}{3}x^\frac{-1}{3}$$
$$\left(\frac{2}{3}x^\frac{-1}{3}\right)^2 = \frac{4}{9x^\frac{2}{3}}$$
$$\int_0^1 \sqrt{1 + \left( \frac{2}{3}x^\frac{-1}{3}\right)^2}\, dx$$
$$\int_0^1 \sqrt{1 + \frac{4}{9x^\frac{2}{3}}}\, dx$$
This is undefined at 0 so it is an improper function. I do not think I can continue.
Now in terms of y.
$$x = y^\frac{3}{2}$$
$$x \prime = \frac{3\sqrt{y}}{2}$$
$$ \left(\frac{3\sqrt{y}}{2}\right)^2 = \frac{9y}{4}$$
$$\int_0^1 \sqrt{1+ \frac{9y}{4}}\, dy$$
I have no idea how to factor that out, I have tried many ways but I can not get it.
| $$\int_0^1\sqrt{1 + {9y\over 4}}
= {4\over 9}\int_1^{13/4} \sqrt{y}\,dy
= {8\over 27}\left({13\sqrt{13}\over 8} - 1\right)
= {13\sqrt{13} - 8\over 27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit finding of an indeterminate form: $\lim\limits_{x\to0} \frac{x^3}{\tan^3(2x)}$ Here is the limit I'm trying to find out:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$$
Since it is an indeterminate form, I simply applied l'Hopital's Rule and I ended up with:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)} = \lim_{x\rightarrow 0}\frac{6\cos^3(2x)}{48\cos^3(2x)} = \frac{6}{48} = 0.125$$
Unfortuntely, as far as I've tried, I haven't been able to solve this limit without using l'Hopital's Rule. Is it possibile to algebrically manipulate the equation so to have a determinate form?
| $\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$=$\lim_{x\rightarrow 0} \frac{x^3}{\frac{\sin^3(2x)}{\cos^3(2x)}}$=$\lim_{x\rightarrow 0} \frac{x^3\cos^3(2x)}{\sin^3(2x)}$=$\lim_{x\rightarrow 0}\frac{x\cdot x\cdot x \cdot\cos^3(2x)}{\sin(2x)\cdot
\sin(2x)\cdot\sin(2x)}$=$\lim_{x\rightarrow 0}\frac{2x\cdot 2x\cdot 2x \cdot \frac{1}{8} \cos^3(2x)}{\sin(2x)\cdot\sin(2x)\cdot\sin(2x)}$=$|\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$\Rightarrow$ $\lim_{x\rightarrow 0}\frac{x}{\sin x}=1$ $\Rightarrow$ $\lim_{x\rightarrow 0}\frac{2x}{\sin 2x}=1$|=$\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$1\cdot 1\cdot 1\cdot \frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}{\cos[\lim_{x\rightarrow 0}(2x)]^3}$=$\frac{1}{8}\cos0$=$\frac{1}{8}\cdot 1$=$\frac{1}{8}$
| {
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"url": "https://math.stackexchange.com/questions/157301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots$ How to prove the following product?
$$\frac{\sin(x)}{x}=
\left(1+\frac{x}{\pi}\right)
\left(1-\frac{x}{\pi}\right)
\left(1+\frac{x}{2\pi}\right)
\left(1-\frac{x}{2\pi}\right)
\left(1+\frac{x}{3\pi}\right)
\left(1-\frac{x}{3\pi}\right)\cdots$$
| We will use Hardamard Factoriztation theorem to prove it (see https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).
Observe $|\sin(\pi z)|\le e^{\pi |z|}$, hence it has order of growth less than equal to one. Further it has simple zero at every integer $n \in Z$. Hence by Hardamard Factorization theorem we get
$$ \sin (\pi z)= z e^{az+b}\displaystyle \prod_{0 \ne n \in Z}\left(1-\frac zn\right)e^{\frac z n}, \text { for some } a, b \in C.$$ Pairing $n$ and $-n$ together in the product we get
\begin{equation}
\sin (\pi z)= z e^{az+b}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).
\label{on}
\end{equation}
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= \frac{e^{az+b}}{\pi}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Taking $z \to 0$, in both side we get $\frac{e^{b}}{\pi}=1$.
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= {e^{az}}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Again using the fact that
$\frac {\sin (\pi z)}{\pi z}$ is an even function, it can be easily seen that $a=0$.
Hence we get $$\frac {\sin (\pi z)}{\pi z}= \displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Thus $${\sin (\pi z)}= \pi z\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
| Let $a = 1+x$ and $b = 1+y$. Then we need to prove that $$\dfrac{(x+1)^2}{y} + \dfrac{(y+1)^2}{x} \geq 8$$ i.e. $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$
for $x,y \geq 0$.
Now apply AM-GM as shown below.
$$\dfrac{\dfrac{x^2}{y} + \dfrac{x}{y} + \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + \dfrac{y}{x} + \dfrac{y}{x} + \dfrac1x}{8} \geq \sqrt[8]{\dfrac{x^2}{y} \times \dfrac{x}{y} \times \dfrac{x}{y} \times \dfrac1y \times \dfrac{y^2}{x} \times \dfrac{y}{x} \times \dfrac{y}{x} \times \dfrac1x} = 1$$
Hence, we get that $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$ which is what we wanted to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show a sequence is decreasing I'm stuck trying to show that the following sequence is decreasing
$$a_{n} = \left(\frac{n+x}{n+2x}\right)^{n}$$ where $x>0$. I've tried treating $n$ as a real number and took derivatives but it didn't lead to anything promising. Any hints would be appreciated.
| $$\frac{a_{n}}{a_{n+1}}=\frac{\left(\frac{n+x}{n+2x}\right)^{n}}{ \left(\frac{n+1+x}{n+1+2x}\right)^{n+1}}= \frac{n+1+2x}{n+1+x}\left( \frac{n+1+2x}{n+1+x} \frac{n+x}{n+2x}\right)^n$$
Lets observe that
$$ \frac{n+1+2x}{n+1+x} \frac{n+x}{n+2x}=\frac{n^2+n+3nx+x+2x^2}{n^2+n+3nx+2x+2x^2}=1-\frac{x}{n^2+n+3nx+2x+2x^2}$$
Then, by Bernoulli
$$\frac{a_{n}}{a_{n+1}} \geq (1+\frac{x}{n+1+x})(1-n\frac{x}{n^2+n+3nx+2x+2x^2}) $$
An easy computation shows that
$$(1+\frac{x}{n+1+x})(1-n\frac{x}{n^2+n+3nx+2x+2x^2}) \geq 1 \Leftrightarrow$$
$$\frac{x}{n+1+x} \geq \frac{nx}{(n+1+x)(n+2x)} +\frac{x}{n+1+x}\frac{nx}{(n+1+x)(n+2x)} \Leftrightarrow $$
$$x(n+1+x)(n+2x) \geq nx(n+1+x)+nx^2 \Leftrightarrow $$
$$n^2x+nx+3nx^2+2x^2+2x^3 \geq n^2x+nx+nx^2+nx^2 \Leftrightarrow $$
$$ nx^2+2x^2+2x^3 \geq 0 $$
Thus
$$\frac{a_{n}}{a_{n+1}} \geq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Calculate: $\sum_{k=0}^{n-2} 2^{k} \tan \left(\frac{\pi}{2^{n-k}}\right)$ Calculate the following sum for integers $n\ge2$:
$$\sum_{k=0}^{n-2} 2^{k} \tan \left(\frac{\pi}{2^{n-k}}\right)$$
I'm trying to obtain a closed form if that is possible.
| Consider
$$\prod_{k = 0}^{n - 2}\cos(2^k \theta)$$
Multiplying numerator and denominator by $2\sin(\theta)$ we get,
$$\frac{2\sin(\theta)\cos(\theta)}{2\sin(\theta)}\prod_{k = 1}^{n - 2} \cos(2^k\theta) = \frac{\sin(2\theta)}{2\sin(\theta)}\prod_{k = 1}^{n - 2} \cos(2^k\theta)$$
Now, repeatedly multiplying and dividing by 2, we can reduce the above to,
$$\prod_{k = 0}^{n - 2}\cos(2^k \theta) = \frac{\sin(2^{n - 1} \theta)}{2^{n - 1} \sin(\theta)}$$
Take logs on both sides,
$$\sum_{k = 0}^{n - 2}\ln(\cos(2^k \theta)) = \ln(\sin(2^{n - 1} \theta)) - \ln(2^{n - 1}) - \ln(\sin(\theta))$$
Differentiating both sides w.r.t $\theta$ we get,
$$-\sum_{k = 0}^{n - 2}2^k\tan(2^k \theta) = 2^{n - 1}\cot(2^{n - 1} \theta) - \cot(\theta)$$
Substitute $\theta = \frac{\pi}{2^n}$ above to get,
$$\sum_{k = 0}^{n - 2}2^k\tan\left(\frac{\pi}{2^{n - k}}\right) = \cot\left(\frac{\pi}{2^n}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
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} |
How to integrate $\int{\frac{dx}{\sqrt{16-9x^2}}}$ Me again, probably someone is going to blame me. I have this:
$$\int{\frac{dx}{\sqrt{16-9x^2}}}$$
I have asked an old teacher of mine an he said me I should let $x=\frac{4}{3}\sin{\big(\frac{3}{4}x\big)}$ but I don't how he realise that. I have tried $u-substitution$ and Integration by Parts but the square root is making the whole thing much harder.
I tried multiplying both terms by $\sqrt{16-9x^2}$ but nothing. Also tried to factor to eliminate square root but nothing.
I don't want that anyone solve my integral, just give me a hint about how do it, and how I can realise what to do in this cases :)
| The given integral is:
$\int{\frac{dx}{\sqrt{16-9x^2}}}$.
This integral can remark as: $\int{\frac{dx}{\sqrt{4^2-(3x)^2}}}$.
By received substition 3x with $4\sin t$ it is $3x=4\sin t$ have:
$3x=4\sin t$ $\Rightarrow$ $x=\frac{4}{3}\sin t$.
Hence we: $dx=\frac{4}{3}\cos t$
If this data replacement the intial integral we have:
$\int{\frac{dx}{\sqrt{16-9x^2}}}$=$\int{\frac{dx}{\sqrt{4^2-(3x)^2}}}$=$\int\frac{4\cos t dt}{3\sqrt{16-(4\sin t)^2}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16-16\sin^2 t}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16(1-\sin^2 t)}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16}\cdot\sqrt{1-\sin^2 t}}$=$\frac{4}{3}\int\frac{cos t dt}{4\cdot\cos t}$=$\frac{4}{3}\cdot\frac{1}{4}\int{dt}$=$\frac{1}{3}t$
By substition of the above have:
$3x=4\sin t$ $\Rightarrow$ $\sin t=\frac{3}{4}\cdot x$ $\Rightarrow$ $ t=\arcsin \frac{3}{4}\cdot x $.
Theres definitly the given integral is:
$\int{\frac{dx}{\sqrt{16-9x^2}}}$=$\frac{1}{3}t$=$\frac{1}{3}\arcsin \frac {3}{4}\cdot x+C $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 4
} |
Factoring by grouping: $x^4 - y^4 -4x^2 + 4$ Please help me factor $x^4 - y^4 -4x^2 + 4$ by grouping terms.
Thank you.
| Group all the $x$ terms together and all the $y$ terms together:
$$(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$
using $a^2-b^2=(a-b)(a+b)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the indefinite integral of $1/(16x^2+20x+35)$ Here is my steps of finding the integral, the result is wrong but I don't know where I made a mistake or I may used wrong method.
$$
\begin{align*}
\int \frac{dx}{16x^2+20x+35}
&=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{35}{16}} \\
&=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16}} \\
&=\frac{1}{16}\int \frac{dx}{(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2}\\
&=\frac{1}{16}\frac{4}{5}\textstyle\arctan ((x+\frac{\sqrt{10}}{4})\cdot \frac{4}{5}) \\
&=\frac{1}{20}\textstyle\arctan(\frac{4x+\sqrt{10}}{5})
\end{align*}
$$
| You want to complete a square. So, remember that
$$
(x+\alpha)^2 = x^2 + 2 \alpha x + \alpha^2.
$$
You have
$$
2\alpha = \frac{20}{16},
$$
i.e. $\alpha = 5/8$. Hence
$$
x^2 +\frac{20}{16}x = \left( x + \frac{5}{8} \right)^2 - \frac{25}{64}.
$$
Can you go further, now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
| To prove
$$
\frac1{1+2b^2c}+\frac1{1+2c^2a}+\frac1{1+2a^2b}\ge1\tag{1}
$$
subtract $\frac13$ from each term on the left and multiply by $\frac32$:
$$
\frac{1-b^2c}{1+2b^2c}+\frac{1-c^2a}{1+2c^2a}+\frac{1-a^2b}{1+2a^2b}\ge0\tag{2}
$$
Multiplying $(2)$ by $\frac13\left(1+2b^2c\right)\left(1+2c^2a\right)\left(1+2a^2b\right)$ shows that $(1)$ is equivalent to
$$
1+a^2b+b^2c+c^2a-4a^3b^3c^3\ge0\tag{3}
$$
The AM-GM gives
$$
1=\frac{a+b+c}3\ge abc\tag{4}
$$
The AM-GM and $(4)$ yield
$$
\frac{1+a^2b+b^2c+c^2a}4\ge\left(a^3b^3c^3\right)^{1/4}\ge a^3b^3c^3\tag{5}
$$
which is $(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/168704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$? Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$?
| Try to find a sequence such that $\frac{n+1}{2n}=\prod_{j=2}^nx_j$ (it will do the job). We have $x_2=3/4$ and
$$x_{n+1}=\frac{\prod_{j=2}^{n+1}x_j}{\prod_{j=2}^nx_j}=\frac{n+2}{2(n+1)}\frac{2n}{n+1}=\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{(n+1)^2}<\frac{n^2+2n\color{red}{+1}}{(n+1)^2}=1.$$
So $x_n=\frac{n^2-1}{n^2}$ does the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
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$(\sin\theta+\cos\theta)^2=1+\sin2\theta$ 49) $(\sin\theta+\cos\theta)^2=1+\sin2\theta$
Left Side:
\begin{align*}
(\sin\theta+\cos\theta)^2=\sin^2\theta+2c\cos\theta\sin\theta+cos^2\theta=1+2\cos\theta\sin\theta
\end{align*}
This can either be $1$ or I can power reduce it. I don't know.
Right Side:
\begin{align*}
1+\sin2\theta=1+2\sin\theta\cos\theta
\end{align*}
Thank you!
| Remember the binomial formula. Importantly, $(a+b)^2\neq a^2+b^2$ ! Rather,
$\begin{align*}
(\sin\theta+\cos\theta)^2=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta = 1 + 2\sin\theta\cos\theta
\end{align*}$
Looking at your previous questions, I think you should be a little more careful with the 'simpler' steps in your calculations and double-check those, otherwise you get lost further down.
| {
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"url": "https://math.stackexchange.com/questions/170958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for $+r$ ; $A=2\pi r^2+2\pi rh$
Solve for $+r$
$$A=2\pi r^2+2\pi rh$$ Since $2\pi$ is common on both sides of the $+$ so I will take it out
$$A=2\pi (r^2+rh)$$
Now, divide both sides by $2\pi$
$$\dfrac{A}{2\pi}=r^2+rh$$
Then, we can divide by the $h$
$$\dfrac{Ah}{2\pi}=r^2+r$$
Then;
$$0=r^2+r-\dfrac{Ah}{2\pi}$$
Quadratic formula $=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $a=1,b=1,c=?$ What are the values of $a,b,c$
| There is a series of mistakes in your solution. I'll mark them one by one:
You correctly arrive to
$$\frac A {2\pi}=r^2+rh$$
But "dividing by $h$" produces
$$\frac 1 h \frac A {2\pi}=\frac 1 h\left(r^2+rh\right)$$
$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+\frac{rh}h$$
$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+r$$
So that step is wrong.
Similarily, if you have
$$\dfrac{Ah}{2\pi}=r^2+r$$
then "taking square roots" produces
$$\sqrt{\dfrac{Ah}{2\pi}}=\sqrt{r^2+r}$$
You then seem to assert
$$\sqrt{r^2+r}=2r$$
Let's check if it is indeed true for, say $r=1$, which gives
$$\sqrt{2}=2$$
which is manifestly false. So there is something awry there, too.
The best thing you can do is check wether each step is correct. To solve for $r$, since
$$\frac A {2\pi}=r^2+rh$$
is a quadratic we need to make a very old trick, which is called completing the square:
$$\eqalign{
& \frac{A}{{2\pi }} = {r^2} + rh \cr
& \frac{A}{{2\pi }} = {r^2} + 2r\frac{h}{2} \cr
& \frac{A}{{2\pi }} = \underbrace {{r^2} + 2r\frac{h}{2} + {{\left( {\frac{h}{2}} \right)}^2}}_{{\text{This is a perfect square!}}} - {\left( {\frac{h}{2}} \right)^2} \cr
& \frac{A}{{2\pi }} = {\left( {r + \frac{h}{2}} \right)^2} - {\left( {\frac{h}{2}} \right)^2} \cr
& \frac{A}{{2\pi }} + {\left( {\frac{h}{2}} \right)^2} = {\left( {r + \frac{h}{2}} \right)^2} \cr
& \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} = {{{\left( {r + \frac{h}{2}} \right)}^2}} \cr
& \pm \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} = {r + \frac{h}{2}} \cr} $$
Note in the last steps we take the square root. We then have to think about both the positive and negative root. So you final solution is
$$r = - \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} - \frac{h}{2}{\text{ or }}r = \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} - \frac{h}{2}$$
COMPLETING THE SQUARE:
Say we have a quadratic $$0=ax^2+bx+c$$
"Completing the square" consist of writing it in the form
$$0 = A{\left( {x + h} \right)^2} + C$$
We can accomplish this with some "trickery"
$$\eqalign{
& 0 = a{x^2} + bx + c \cr
& 0 = 4{a^2}{x^2} + 4abx + 4ac{\text{ ; multiply by }}4a \cr
& 0 = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + 4ac{\text{ ; cleverly rearrange the eqn}}{\text{.}} \cr
& {b^2} = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2} + 4ac{\text{ ; add }}{b^2} \cr
& {b^2} = \underbrace {{{\left( {2ax} \right)}^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2}}_{{\text{This is a perfect square!}}} + 4ac \cr
& {b^2} = {\left( {2ax + b} \right)^2} + 4ac \cr
& {b^2} - 4ac = {\left( {2ax + b} \right)^2} \cr
& \sqrt {{b^2} - 4ac} = 2ax + b \cr
& - b + \sqrt {{b^2} - 4ac} = 2ax \cr
& \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = x \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Verify $\frac{\sin^3A + \cos^3A}{\sin A + \cos A} = 1 - \sin A\cos A$ How can I verify the following trigonometric identity?
$$\frac{\sin^3 A + \cos^3 A}{\sin A+\cos A} = 1-\sin A\cos A.$$
My work so far is
$$\begin{align*}
&\frac{\sin\cos(\sin^2+\cos^2)}{\sin+\cos}\\
&\frac{\sin\cos(1)}{\sin+\cos}
\end{align*}$$
| First: don't use $\sin$ and $\cos$ without arguments. For instance, your last formula, it's easy to get confused and think you are computing, inter alia, $\cos(1)$, which you are not.
Second: it seems that you think that
$$\sin^3 A + \cos^3A \text{ is equal to } \sin A\cos A(\sin^2A + \cos^2A).$$
They are not equal; you can verify that by actually multiplying out the right hand side; you will see that you get
$$\sin^3A\cos A + \cos^3A\sin A\neq \sin^3A + \cos^3A.$$
There are a few algebraic formulas that show up a lot and it is good to know them. When I was in middle school, they were known as "notable products", because they were, well, notable and showed up a lot. One was the square of a binomial, $(a+b)^2 = a^2+2ab+b^2$; one was the "difference of squares" or "conjugate product": $(a+b)(a-b)=a^2-b^2$.
Two others are the difference-of-cubes and the sum-of-cubes:
$$\begin{align*}
a^3-b^3 &= (a-b)(a^2+ab+b^2)\\
a^3+b^3 &= (a+b)(a^2-ab+b^2).
\end{align*}$$
Using the second one, you can simplify the left hand side,
$$\frac{\sin^3 A + \cos^3 A}{\sin A + \cos A}$$
by letting $a=\sin A$ and $b=\cos A$, and proceed from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$ How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$
I have not been sucessful to do this.
| You wish to find the sum $$\frac{1}{2} + \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k)(2k+1)}$$
Expanding the summand using partial fractions, we get
$$\frac{1}{(2k-1)(2k)(2k+1)}=\frac{A}{2k-1}+\frac{B}{2k}+\frac{C}{2k+1}$$$$ \implies 1=A(2k)(2k+1)+B(2k+1)(2k-1)+C(2k)(2k-1)$$
Solving this gives $A=C=\frac{1}{2},B=-1$. Thus splitting up our sum, we arrive at:
$$\frac{1}{2}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2k-1}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2k+1}-\sum_{k=1}^{\infty}\frac{1}{2k}$$
Now note that $$\sum_{k=1}^{\infty}\frac{1}{2k+1}=\sum_{k=1}^{\infty}\frac{1}{2k-1}-1$$
So our halves cancel, and grouping terms leaves us with:
$$\sum_{k=1}^{\infty}\frac{1}{2k-1}-\sum_{k=1}^{\infty}\frac{1}{2k}$$
In other words, $1-\frac{1}{2}+\frac{1}{3}-\ldots$ which is known to converge to $\ln(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the remainder Corresponding terms from the sequence $1,2,3,4,5...$ and $2^1,2^2,2^3,2^4,2^5,...$, are multiplied, creating the sequence $1\times 2^1,2\times 2^2,3\times 2^3,4\times 2^4,5\times 2^5...$. Let $A$ be the sum of the first $2011$ numbers in the new sequence. Find the remainder when $A$ is divided by $1000$
My solution:
$2S-S=2010 \cdot 2^{2012}+2$.
Then $2^{2012} \equiv 2^{12} \mod{125} \equiv 96 \mod{125}$
$2010 \cdot 2^{2012} \equiv 10 \cdot 96 \mod{1000} \equiv 960 \mod{1000}$
Add 2 and the answer is $962$
| I'm answering this so the question can be marked as answered. If you ask such "am I right?" questions in the future you might want to include some more of your thinking for the benefit of the reader.
The solution is correct. The step $2^{2012}\equiv2^{12}\bmod125$ uses Euler's theorem together with $\phi(5^3)=5^3-5^2=100$. The remainder $2^{2012}\equiv96\bmod1000$ is being computed by computing the remainders modulo the prime powers $5^3=125$ and $2^3=8$ and noting that the remainder $\bmod8$ is $0$. Alternatively, you could have used $\phi(5^3\cdot2^3)=(5^3-5^2)(2^3-2^2)=400$ directly to get $2^{2012}\equiv2^{12}\equiv96\bmod1000$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution.
| There's an algebraic way (which may be a bit messy without using a CAS). It involves exploiting the symmetry of the problem, as Peter Tamaroff suggested.
$$
\begin{eqnarray}
x^2 f(x) + f(1-x) &=& 2x - x^4 \\
(1-x)^2f(1-x) + f(x) &=& 2(1-x) - (1-x)^4 \quad \textrm{subs. $1-x$ for $x$}\\
\\
-(1-x)^2x^2f(x) - (1-x)^2f(1-x) &=& -(1-x)^2(2x - x^4) \quad \textrm{Eqn.1 times $-(1-x)^2$}\\
\textrm{Now add to Eqn.2:} \quad f(x) - (1-x)^2x^2f(x) &=& 2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)\\
f(x)\left[1 - (1-x)^2x^2\right] &=& 2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)\quad \textrm{factoring out $f(x)$}\\
f(x) &=& \frac{2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)}{1 - (1-x)^2x^2}.
\end{eqnarray}
$$
This is an unreduced final answer, which (according to sage) cancels out to leave $f(x) = 1 - x^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$ Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.
Can anyone help me with this? Thank You!
| Using the formula $ \cos 3 \theta = 4 \cos ^ { 3 } \theta - 3\cos \theta,$ we have
$\qquad\qquad 4 \cos ^ { 3 } 40 ^ { \circ } - 3 \cos 40 ^ { \circ }$
$\left. { = \cos 120 ^ { \circ } } { = - \frac { 1 } { 2 } } \right. $
$\qquad \Rightarrow \quad 8 \cos ^ { 3 } 40 ^ { \circ } - 6 \cos 40 ^ { \circ } + 1 = 0 $$ \quad \Rightarrow \quad \cos 40 ^ { \circ } $ is a root of $8 x ^ { 3 } - 6 x + 1 = 0 \cdots (1) $
$\text{Similarly, }$$\cos 80 ^ { \circ }\text{and}\cos 160^{ \circ }\text{are also the roots of (1).}$$\text{Now we are going to transform (1) by $ y=x^2 $ into another equation (2) whose roots are }
$$\cos ^2 40 ^ { \circ }, \cos ^2 80 ^ { \circ } \text{ and } \cos ^2160^ { \circ }\tag*{}. $
$\text{From (1),} -1= x(8x^2–6). \text{ Squaring both sides, we have }$$1=x^2(8x^2–6)^2
\text{and hence }64y^3–96y^2+36y-1=0 \tag*{(2)} \\$$\text{whose roots are }\cos ^2 40 ^ { \circ }, \cos ^2 80 ^ { \circ } \text{and }\cos ^2 20^ { \circ }. $
However, in order to get the sum of squares of tangents of $20^{\circ}, 40^{\circ} \text{and }80^{\circ},$ we need their corresponding secant squares.
One more transformation $z=\frac{1}{y}$ is introduced to get another equation (3):
$$\quad z ^ { 3 } - 36 z^2 + 96 z - 64 = 0, $$
$\text{whose roots are }\sec ^ { 2 } 20 ^ { \circ } , \sec ^ { 2 }40 ^ { \circ } \textrm{and } \sec ^ { 2 } 80 ^ { \circ }.$
$\therefore \tan ^ { 2 } 20 ^ { \circ } + \tan ^ { 2 } 40 ^ { \circ } + \tan ^ { 2 } 80 ^ { \circ } + 3$
$=\tan ^ { 2 } 20^ { \circ } +1 + \tan ^ { 2 } 40 ^ { \circ } +1 + \tan ^ { 2 } 80 ^ { \circ } + 1$
$= \sec ^ { 2 } 20 ^ { \circ } + \sec ^ { 2 } 40 ^ { \circ } + \sec ^ { 2 } 80 ^ { \circ }$
$=36$
Hence $\boxed{\tan ^ { 2 } 20 ^ { \circ } + \tan ^ { 2 } 4 0 ^ { \circ } + \tan ^ { 2 } 80 ^ { \circ } = 33.}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
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} |
Finding the all triples How to find the all positive integer triples such that :
$$ab+c=\gcd (a^2,b^2)+\gcd(a,bc)+\gcd(b,ac)+\gcd(c,ab)=239^2$$
| Let $x=\gcd(a,b,c)$. But $\gcd(a^2,b^2)\ge x^2$ and $x$ divides $239^2$, this is impossible due to the equality and 239 is prime. Hence $x=1$.
Let $y=\gcd(c,ab)$. if $y>1$, then once again $y=239$ ($239^2$ would be too big).
Suppose $y$ divides $a$ (the other case is symmetric), $y$ does not divide $b$ ($\gcd(a^2,b^2)$ would be at least $239^2$, too big). Hence $\gcd(c,ab)=y$, $\gcd(a,bc)=y$, and $\gcd(a,bc)=\gcd(a,b)=z$, so
$$ z^2+z+2y=y^2$$ So $y$ divides $z$ (or $z+1$), this is not possible, so $y=1$
So $c$ is prime with $a$ and $b$ and the equation is
$$ab+c=\gcd(a,b)^2+2\gcd(a,b)+1=(\gcd(a,b)+1)^2=239^2$$
$$\gcd(a,b)=238$$
The only small enough possibility is $a=b=238$ and then $c=477$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving an estimate for this integral How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$
| $$
\int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x \stackrel{x=\frac{2-u}{u}}{=} 2^{4/3} \int_0^1 u^{-1/3} \left(2-u\right)^{-2} \mathrm{d} u = 2^{-2/3} \int_0^1 u^{-1/3} \left(1-\frac{1}{2} u\right)^{-2} \mathrm{d} u
$$
This is just the Euler type integral for Gauss's hypergeometric function:
$$
\frac{\Gamma(a) \Gamma(c-a)} {\Gamma(c)} {}_2F_1\left(a,b;c;x\right) = \int_0^1 u^{a-1} (1-u)^{c-a-1} (1-x u)^{-b} \mathrm{d} u
$$
where $a=\frac{2}{3}$, $c=a+1 = \frac{5}{3}$, $b=2$ and $x=\frac{1}{2}$. Since $c=a+1$ the ratio of $\Gamma$-functions simplifies to $\frac{1}{a} = \frac{3}{2}$. Now
$$
\int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x = \frac{3}{2^{5/3}} {}_2F_1\left(\frac{2}{3}, 2; \frac{5}{3}; \frac{1}{2} \right)
$$
The said hypergeometric can be computed in (not very simple, but elementary) closed form:
$$
\int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x = \sqrt[3]{2}+\frac{1}{24} \log \left(3505753+2782518 \sqrt[3]{2}+2208486\
2^{2/3}\right)-\frac{1}{2 \sqrt{3}}\arcsin\left(\frac{\sqrt{3}}{2} \sqrt{\left(7-3 \sqrt[3]{2}\right) \left(\sqrt[3]{2}-1\right)}\right) \approx 1.6695914 \approx 4.6540453^{1/3}
$$
So $6^{1/3}$ strikes me as a pretty tight bound.
| {
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"source": "stackexchange",
"question_score": "8",
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Evaluating a complex contour I need to show the following result:
$$
\int_{-\infty}^\infty \frac{1}{(1+x^2)^{n+1}}dx\, = \frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2\cdot 4\cdot\ldots\cdot(2n)}\pi
$$
With n=1,2,3,...
This function has a pole at i and -i.
I've tried a semicircle in the upperhalf of the plain, but the residue then goes to infinity. I've also tried a rectangle in the upperhalf that stays beneath i, but all 3 sides that do not include the integral we're looking for go to zero because of the R in the denominator.
Anyone with tips?
| Integrating over a large semicircle in the upper halfplane works well.
Write $$\frac{1}{(1+z^2)^{n+1}} = \dfrac{\dfrac{1}{(z+i)^{n+1}}}{\quad(z-i)^{n+1}\quad}.$$
Hence, the residue at $z=i$ is the $1/n!$ times the $n$th derivative of $1/(z+i)^{n+1}$ evaluated at $z=i$, i.e.
$$
\begin{split}
\operatorname{Res}\limits_{z=i} \frac{1}{(1+z^2)^{n+1}} &=
\frac{1}{n!}\frac{d^n}{dz^n}\left( \frac{1}{(z+i)^{n+1}} \right)\Bigg|_{z=i} \\
&= \frac{1}{n!}(-1)^n(n+1)(n+2)\cdots(2n)\frac{1}{(2i)^{2n+1}} \\
&=\frac{(n+1)(n+2)\cdots(2n)}{n!\cdot2^{2n}\cdot 2i} \\
&= \frac{n!\cdot(n+1)\cdots(2n)}{(n!)^2\cdot 2^{2n}\cdot 2i} \\
&= \frac{(2n)!}{(2\cdot 4\cdot 6\cdots(2n))^2\cdot 2i} \\
&= \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n)\cdot 2i} \\
\end{split}
$$
I think you can do the last step yourself.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute: $\lim\limits_{n\to\infty} \frac{x_n}{\ln {n}}$ Let be the sequence $(x_n)_{n\geq0}$, $x_0$ a real number, and defined as follows: $$ x_{n+1} = x_n + e^{-x_n} $$
Compute the limit:
$$\lim_{n\to\infty} \frac{x_n}{\ln {n}}$$
Luckily, I've found another post here with a very similar question. If you know other ways to solve it and want to share them then I'd be very grateful for that.
| We examine the asymptotic behavior of $(x_n)$ inch by inch.
Step 1. We first show that $x_n \to \infty$. Note that for any $x < 0$, there exists $y > 0$ such that $x + e^{-x} = y + e^{-y}$. Thus we may assume $x_0 \geq 0$. Then $(x_n)$ is monotone increasing, thus it tends to a limit $\ell \in [0, \infty]$. Taking limit to the recursive formula, we have $\ell = \ell + e^{-\ell}$, which is valid only if $\ell = \infty$.
Step 2. Let $y_n = e^{-x_n}$. Then $(y_n)$ decreases monotonely to $0$. Also, we have
$$ y_{n+1} = \exp\left(-x_{n+1}\right) = \exp\left(-x_{n}-e^{-x_n}\right) = y_n e^{-y_n}.$$
Then we have
$$ y_{n+1}^{-1} - y_n^{-1} = \frac{e^{y_n} - 1}{y_n} \xrightarrow[]{n\to\infty} 1. $$
Thus by Cesàro-Stolz theorem, we have
$$ \lim_{n\to\infty} \frac{y_n^{-1}}{n} = 1.$$
This already shows that $\lim_{n\to\infty} \frac{x_n}{\log n} = 1$.
Step 3. Since
$$ \frac{y_{n+1}^{-1} - y_n^{-1} - 1}{y_n} = \frac{e^{y_n} - 1 - y_n}{y_n^2} \xrightarrow[]{n\to\infty} \frac{1}{2}, $$
we have
$$ \frac{\left(y_{n+1}^{-1} - (n+1)\right) - \left(y_n^{-1} - n\right)}{\log(n+1) - \log n} \sim \frac{y_{n+1}^{-1} - y_n^{-1} - 1}{y_n} \xrightarrow[]{n\to\infty} \frac{1}{2}, $$
where we used the fact here that $y_n \sim \frac{1}{n} \sim \log(n+1) - \log n$. Thus by Cesàro-Stolz theorem again, we have
$$ \lim_{n\to\infty} \frac{y_n^{-1} - n}{\log n} = \frac{1}{2},$$
which in particular implies that
$$ y_n = \frac{1}{n} + O\left(\frac{\log n}{n^2}\right). $$
Step 4. Finally, we have
$$ \frac{y_{n+1}^{-1} - y_n^{-1} - 1 - \frac{1}{2}y_n}{y_n^2} = \frac{e^{y_n} - 1 - y_n - \frac{1}{2}y_n^2}{y_n^3} \xrightarrow[]{n\to\infty} \frac{1}{6}.$$
Let $s_n = y_1 + \cdots + y_n$. Since $y_n^2 \sim \frac{1}{n^2}$,
$$ \frac{\left(y_{n+1}^{-1}-(n+1)-\frac{1}{2}s_n\right) - \left(y_n^{-1}-n-\frac{1}{2}s_{n-1}\right)}{\frac{1}{n^2}} \xrightarrow[]{n\to\infty} \frac{1}{6},$$
or simply
$$ \left(y_{n+1}^{-1}-(n+1)-\frac{1}{2}s_n\right) - \left(y_n^{-1}-n-\frac{1}{2}s_{n-1}\right) = O\left(\frac{1}{n^2}\right).$$
Thus summing,
$$\begin{align*}y_n^{-1}
&= n + \frac{1}{2}s_{n-1} + O(1) = n + \frac{1}{2}s_{n} + O(1) \\
&= n + \frac{1}{2}\sum_{k=1}^{n} \left( \frac{1}{k} + O\left(\frac{\log k}{k^2}\right)\right)+O(1) \\
&= n + \frac{1}{2}\log n + O(1),
\end{align*}$$
which finally implies the asymptotic formula
$$ y_n = \frac{1}{n} - \frac{1}{2}\frac{\log n}{n^2} + O\left(\frac{1}{n^2}\right), $$
or in terms of $x_n$,
$$ x_n = \log n + \frac{1}{2}\frac{\log n}{n} + O\left(\frac{1}{n}\right). $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $|a| < |b|$ I apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.
I am looking for a way to eliminate absolute value functions in $|a| < |b|$.
I can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.
Thank you!
An example problem:
$$|x + 2| < |x - 4|$$
| $|a| < |b| \iff -|b| < a < |b| \iff
\begin{cases}
-b < a < b & \text{If $b > 0$} \\
\text{No solution.} &\text{If $b=0$} \\
b < a < -b & \text{If $b < 0$}
\end{cases}$
E. G. $|x + 2| < |x - 4|$
\begin{align}
x > 4 &\implies-x+4 < x+2 < x-4 \\
&\implies -2x+4 < 2 < -4 \\
&\implies \text{No solution.}
\end{align}
\begin{align}
x > 4 &\implies -x+4 < x+2 < x-4 \\
&\implies 4 < 2 < 2x-4 \\
&\implies x < 1
\end{align}
\begin{align}
x < 4 &\implies x-4 < x+2 < -x+4 \\
&\implies -4 < 2 < -2x+4 \\
&\implies x < 1
\end{align}
Hence $x \in (-\infty, 1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
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Generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$ I would like to find a generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$. This is a field since $x^{2}+3x+3$ is irreducible, so every coset with $bx+a\not=0$ as a representative should be a unit...
I do not understand how to go from here though... Since $b\not=0$ we have $4$ diffrent choices for $b$ and $5$ different options for coefficient $a$ hence $24$ elements in the multiplicative group of units.
Should any $bx+a$ with order $24$ be a generator? How do I go from here?
| It suffices to rule out the maximal factors of $24$, i.e. $8=24/3$ and $12=24/2$ as possible orders. The (coset of) $x$ fits the bill, because
$$
x^4=(-x^2)^2\equiv (3x+3)^2=4(x^2+2x+1)=-x^2+3x+4\equiv6x+7=x+2,
$$
and therefore neither
$$
x^8=(x^4)^2\equiv(x+2)^2=x^2+4x+4\equiv x+1
$$
not
$$
x^{12}=x^8\cdot x^4\equiv(x+2)(x+1)=x^2+3x+2\equiv-1
$$
are equal to $1$.
The point is that other possible orders ($2,3,4,6$) that are less than $24$ are factors of either $8$or $12$.
Now that we settled that $x$ is a generator, the other generators can be found by reducing the powers $x^j$, $0<j<24, \gcd(j,24)=1$ modulo the generator $x^2+3x+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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Does this Dirichlet series converge to zero? Consider the periodic Dirichlet series that has this iterative definition:
$$\text{a1}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{6}}+...$$
$$\text{a2}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1}{\sqrt{6}}+...$$
$$\text{a3}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2}{\sqrt{6}}+...$$
$$\text{a4}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2+a3}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2+a3}{\sqrt{6}}+...$$
$$...$$
continuing this iteration, will it converge to zero?
| If I understand correctly, $a_1 = \sum_{j=0}^\infty (-1)^j \left((3j+1)^{-1/2} - (3j+2)^{-1/2} - 2 (3j+3)^{-1/2}\right)$, and
$a_{j+1} = a_j + a_j b = a_j (1+b)$ where $b = \sum_{j=1}^\infty (-1)^j (3j)^{-1/2}$.
Thus $a_n = a_1 (1+b)^{n-1}$. Now $1+b \approx .6507616$, so $|1+b| < 1$ so the sequence does converge to $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When does $n$ divide $a^d+1$? $\newcommand{\ord}{\operatorname{ord}}$
For what values of $n$ will $n$ divide $a^d+1$ where $n$ and $d$ are positive integers?
Apparently $n$ can not divide $a^d+1$ if $\ord_n a$ is odd.
If $n\mid (a^d+1)\implies a^d\equiv -1\pmod n\implies a^{2d}≡1\pmod n \implies\ord_na\mid 2d$ but $\nmid d$.
For example, let $a=10$, the factor(f)s of $(10^3-1)=999$ such that $\ord_f10=3$ are $27,37,111,333$ and $999$ itself. None of these should divide $10^d+1$ for some integer $d$.
Please rectify me if there is any mistake.
Is anybody aware of a better formula?
| There are various useful bits of information that one may deduce about factors of integers of the form $\rm\:b^n\pm 1.\:$ A good place to learn about such is Wagstaff's splendid introduction to the Cunningham Project, whose goal is to factor numbers of the form $\rm\:b^n\pm 1.\:$ There you will find mentioned not only old results such as Legendre (primitive divisors of $\rm\:b^n\pm 1\:$ are $\rm\,\equiv 1\pmod{2n},$ but also newer results, e.g. those exploiting cyclotomic factorizations. e.g. see below.
Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations (e.g. see below).
$$\begin{array}{rl}
x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to solve this limit related to series? How to solve the following limit?
$$\lim_{N\rightarrow+\infty}\frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^N$$
| Fix some large $M$ and suppose $N>M.$ Split the series into the first $N-M-1$ and the last $M$ terms like this: $$ \frac{1}{N} \sum_{k=1}^{N-1} \left( \frac{k}{N} \right)^N =\frac{1}{N} \sum_{k=1}^{N-M-1} \left( \frac{k}{N} \right)^N +\frac{1}{N} \sum_{k=N-M}^{N-1} \left( \frac{k}{N} \right)^N .$$
The summand of the first sum is maximized for $k=N-M-1$ so
$$\frac{1}{N} \sum_{k=1}^{N-M-1} \left( \frac{k}{N} \right)^N< \frac{N-M-1}{N} \left( \frac{N-M-1}{N} \right)^N< \left( \frac{N-M-1}{N} \right)^N \to \exp(-M-1)$$
as $N\to \infty.$
The summand of the second sum is maximized for $k=N-1$ so $$\frac{1}{N} \sum_{k=N-M}^{N-1} \left( \frac{k}{N} \right)^N< \frac{M}{N} \left(\frac{N-1}{N} \right)^N < \frac{M}{N} \to 0$$
as $N\to\infty.$
Thus for sufficiently large $N$, we have $$ \frac{1}{N} \sum_{k=1}^{N-1} \left( \frac{k}{N} \right)^N < \exp(-M-1).$$
For any $\epsilon>0$ we could have chosen $M$ large enough so that $\exp(-M-1) < \epsilon,$ so the limit is $0.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove inequality $2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$ Assume that $a,b,c$ are real numbers from the interval $(\frac{1}{2},1)$. What is the proof that
$$2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$$
holds?
| For the first inequality
Since the sum is symmetric, without loss of generality we can assume $a \leq b \leq c$.
Then
$$\frac{a+b}{c+1} \leq \frac{c+a}{b+1} \le \frac{b+c}{a+1} $$
$$c+1 \geq b+1 \geq a+1$$
Then, by Chebyshev's sum inequality we have
$$ 3( a+b+a+c+b+c) \leq \left( \frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \right) \left( a+1+b+1+c+1 \right)$$
Thus
$$\left( \frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \right) \geq \frac{6(a+b+c)}{a+b+c+3}$$
It is easy to see that $$\frac{6(a+b+c)}{a+b+c+3} \geq 2 \Leftrightarrow a+b+c \geq \frac{3}{2}$$
Second Inequality
This is equivalent to
$$\sum (a+b)(a+1)(b+1) \leq 3 (a+1)(b+1)(c+1)$$
Again, without loss of generality we can assume that $a \leq b \leq c$.
Then
$$(a+b)(a+1)(b+1) =(a+b)(ab+a+b+1)=a^2b+ab^2+a^2+b^2+2ab+a+b$$
Thus, the inequality becomes
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2+2ab+2ac+2bc+2a+2b+2c \leq
$$
$$\leq 3abc+3ab+3ac+3bc+3a+3b+3c+3$$
Or
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2 \leq
$$
$$\leq 3abc+ab+ac+bc+a+b+c+3 (*)$$
We start by observing that $(a+1)(1-b)(1-c) \geq 0$ thus
$$abc+a+bc +1 \geq ab+ac+b+c$$
Similarly
$$abc+b+ac +1 \geq bc+ab+a+b$$
$$abc+c+ab +1 \geq ac+bc+a+b$$
Adding them we get
$$ab+ac+bc+a+b+c \leq 3abc+3$$
Thus, to prove $(*)$ it is enough to show that
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2 \leq
$$
$$\leq 2(ab+ac+bc+a+b+c) $$
which follows immediately from $a\leq 1 \Rightarrow a^2b \leq ab$ and similar ones.
*P.S. * This Solution is terrible, there should be a much simpler one. The inequality $(a+1)(1-b)(1-c) \geq 0$ I used suggests that the Schur Inequality should be the Key, but I couldn't find it :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating :$((\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ))$ What is the easiest way to calculate :
$$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ)) $$
| Maybe you wanna go this way:
$$P=(\sqrt{3} + \tan (1^\circ))((\sqrt{3} +\tan(2^\circ))\cdots((\sqrt{3}+\tan(29^\circ))$$
$$\frac{P}{2^{29}}=\frac{(\sqrt{3} \cos(1^\circ) + \sin (1^\circ))}{2\cos(1^\circ)}\frac{(\sqrt{3} \cos(2^\circ) + \sin (2^\circ))}{2\cos(2^\circ)}\cdots\frac{(\sqrt{3} \cos(29^\circ) + \sin (29^\circ))}{2\cos(29^\circ)}$$
Above I used the fact that
$$\sin(60^\circ+\alpha^\circ)=\frac{\sqrt3}{2}\cos \alpha^\circ+\frac{1}{2}\sin \alpha^\circ$$
then
$$\frac{P}{2^{29}}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\cos1^\circ\cos2^\circ\cdots\cos29^\circ}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\sin89^\circ\sin88^\circ\cdots\sin61^\circ}=1$$
$$P={2^{29}}.$$
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Arc length formula for the lemniscate This question can be homework for elementary calculus.
The lemniscate of Bernoulli $C$ is a plane curve defined as follows.
Let $a > 0$ be a real number.
Let $F_1 = (a, 0)$ and $F_2 = (-a, 0)$ be two points of $\mathbb{R}^2$.
Let $C = \{P \in \mathbb{R}^2; PF_1\cdot PF_2 = a^2\}$.
Let's get the equation of $C$ in the polar coordinates.
Let $P = (r\cos\theta, r\sin\theta)$:
$$PF_1^2 = r^2 + a^2 - 2ar\cos\theta, PF_2^2 = r^2 + a^2 + 2ar\cos\theta$$
Hence:
$$(r^2 + a^2 - 2ar\cos\theta)(r^2 + a^2 + 2ar\cos\theta) = (r^2 + a^2)^2 - 4a^2r^2\cos^2\theta = a^4$$
$$r^4 + 2r^2a^2 + a^4 - 4a^2r^2\cos^2\theta = a^4$$
$$r^2 = 2a^2(2\cos^2\theta - 1) = 2a^2\cos 2\theta$$
Suppose $P \in C$ is in the first quadrant.
Let $s$ be the arc length between $O = (0, 0)$ and $P$.
Then how can we express $s$ by $r$ using an integral?
This is a related question.
| By the arc length formula in the polar coordinates,
$s = \int_{0}^{\theta}\sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta$
Since $d\theta = \frac{d\theta}{dr}dr$,
$s = \int_{0}^{r}\sqrt{r^2 + (\frac{d\theta}{dr})^{-2}} \frac{d\theta}{dr}dr$
Hence $s = \int_{0}^{r}\sqrt{1 + r^2(\frac{d\theta}{dr})^2} dr$
Since $r = a\sqrt{2\cos 2\theta}$,
$\frac{dr}{d\theta} = -\frac{2a\sin 2\theta}{\sqrt{2\cos 2\theta}}$
Hence $\frac{d\theta}{dr} = -\frac{\sqrt{2\cos 2\theta}}{2a\sin 2\theta}$
$(\frac{d\theta}{dr})^2 = \frac{\cos 2\theta}{2a^2\sin^2 2\theta}$
$\cos 2\theta = \frac{r^2}{2a^2}$
$\sin^2 2\theta = 1 - \cos^2 2\theta = \frac{4a^4 - r^4}{4a^4}$
Hence $(\frac{d\theta}{dr})^2 = \frac{r^2}{4a^4 - r^4}$
Therefore $s = \int_{0}^{r}\sqrt{\frac{4a^4}{4a^4 - r^4}}dr = \int_{0}^{r}\frac{2a^2}{\sqrt{4a^4 - r^4}}dr$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation of series of product of Fibonacci numbers What is the sum of following product of Fibonacci numbers
$$\sum_{k=1}^{n-1} Fib(k)*Fib(n+3-k)$$
can anyone suggest only approach to find general term?
| Assuming you're using the usual numbering of the Fibonacci numbers, the Binet equation
$$\text{fib}(n) = \frac{\phi^n - (-1/\phi)^n}{\sqrt{5}}$$
where $\phi = (1 + \sqrt{5})/2$ implies your sum is
$$ F(n) = \left((2 + \sqrt{5}) \frac{n}{5} - \frac{7 \sqrt{5}}{25}\right) \phi^n +
\left((2 - \sqrt{5}) \frac{n}{5} + \frac{7 \sqrt{5}}{25}\right) (-1/\phi)^n
$$
The first few terms ($n = 1$ to $10$) are $0, 3, 8, 19, 40, 80, 154, 289, 532, 965$.
EDIT: It seems your $\text{Fib}(n) = \text{fib}(n+1)$. So your sum is
$$\sum_{k=1}^{n-1} \text{fib}(k+1) \text{fib}(n+4-k) = \left(\frac{\sqrt {5} n}{2}+{\frac {11 n}{10}}-{\frac {21}{50}}\sqrt {5}-\frac{3}{2} \right) \phi^n + \left(-\frac{\sqrt {5} n}{2}+{\frac {11 n}{10}}+{\frac {21}{50}}\sqrt {5}-\frac{3}{2} \right) (-1/\phi)^n
$$
The first few terms ($n=1$ to $10$) are $0, 5, 18, 44, 96, 195, 380, 719, 1332, 2428$.
The generating function is
$$g(t) = {\frac {t \left( 3\,t+5 \right) \left( t+1 \right) }{ \left( {t}^{2}+
t-1 \right) ^{2}}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the probability that $XYZ$ is divisible by $5$? A solution of $X + Y + Z = 20$ in non-negative integers is chosen at random. What is the probability that $XYZ$ is divisible by $5$?
Edit:
This happens to be an exam question. So I can't use calculators or computers and have to get the answer in less than 20 minutes while showing systematic workings. I appreciate the answers below, but can someone instruct me on solving the question given the mentioned constraints?
| $5\mid XYZ\implies 5 \mid XY(20-X-Y) \implies 5\mid XY(X+Y)$
(1)If $5\mid X$ let $X=5a$ where $a$ is any integer, so $Y$ can have $1+20-5a$ values.
So, here the number possible values of $X,Y$ are $21+16+11+6+1=55$
(2)If $5\mid Y$ let $Y=5b$ where $b$ is any integer, leads to another $55$ values.
(3)If $5\mid Y$ and $5\mid X\implies 5(a+b)≤20\implies 0≤a+b≤4$ which has $(5+4+3+2+1)$ values= $15$ values, for example for $Y=0, X$ can be one of $0,5,10,15,20$.
So, (1)+(2)-(3) leads to $2\cdot 55 -15 =95$ values.
(4)If $5\mid (X+Y),$ but $5∤XY$
If $X=5a+r,Y=5b-r$ where $1≤r≤4$ so,$0≤a+b≤4$ and $1≤b≤4$ and $0≤a≤3$.
For each $r$, these $a,b$ can be chosen in $4+3+2+1=10$ ways,for example for $a=0, b$ can be one of $0,1,2,3$.
There can be $4$ values of $r$ ,leading to $4\cdot 10=40$ possible values of $X,Y$
So, number of the possible values of $X,Y$ such that $ 5\mid XY(X+Y)$ is $55+55-15+40=135$
As, $0 ≤ X+Y ≤ 20$, the number possible values of $X,Y$ are $(0+ 1+2+...+21)=231$, for example for $X=a, a ≤ Y ≤ 20\implies 20-a+1$ values of $Y$ .
So, the required probability $=\frac{135}{231}=\frac{45}{77}$
The answer can be validated using the following Java code:
void process() {
int all = 0;
int count = 0;
for (int x = 0; x <= 20; x++) {
boolean xMod5 = (x % 5 == 0);
for (int y = 0; y <= 20 - x; y++) {
if (xMod5 || y % 5 == 0 || (x + y) % 5 == 0) {
count++;
}
all++;
}
}
System.out.println(String.format("%d/%d", count, all));
}
| {
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"url": "https://math.stackexchange.com/questions/190869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Limit (without series expansion and l'Hôpital's rule) $$\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2}=\frac{1}{8}$$
Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.
Note:
WolframAlpha confirms that the result is $\frac{1}{8}$.
| It is easy to observe that as $x \to \infty$ the argument of both the logarithms tend to $1$ and this is exactly what we need when we are dealing with limit of logarithmic terms. Just add $1$ and subtract $1$ in the argument of log and proceed as follows:
\begin{align}
L &= \lim_{x \to \infty}\log\frac{x + \sqrt{x^{2} + 1}}{x + \sqrt{x^{2} - 1}}\left(\log\frac{x + 1}{x - 1}\right)^{-2}\notag\\
&= \lim_{t \to 0^{+}}\log\frac{1 + \sqrt{1 + t^{2}}}{1 + \sqrt{1 - t^{2}}}\left(\log \frac{1 + t}{1 - t}\right)^{-2}\text{ (putting }x = 1/t)\notag\\
&= \lim_{t \to 0^{+}}\log\left(1 + \frac{1 + \sqrt{1 + t^{2}}}{1 + \sqrt{1 - t^{2}}} - 1\right)\left\{\log\left(1 + \frac{1 + t}{1 - t} - 1\right)\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\log\left(1 + \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\right)\left\{\log\left(1 + \frac{2t}{1 - t}\right)\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\dfrac{\log\left(1 + \dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\right)}{\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}}\cdot\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\left\{\log\left(1 + \frac{2t}{1 - t}\right)\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\left\{\dfrac{\log\left(1 + \dfrac{2t}{1 - t}\right)}{\dfrac{2t}{1 - t}}\cdot\dfrac{2t}{1 - t}\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\cdot\frac{(1 - t)^{2}}{4t^{2}}\notag\\
&= \frac{1}{8}\lim_{t \to 0^{+}}\frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{t^{2}}\notag\\
&= \frac{1}{8}\lim_{t \to 0^{+}}\frac{2t^{2}}{t^{2}\left\{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}\right\}}\notag\\
&= \frac{1}{8}\notag
\end{align}
The technique used above is based on the standard limits and is well explained in the comments to this answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Compute $\lim_{x\to0} \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}$ Compute:
$$\lim_{x\to0} \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}$$
How would you tackle this problem? Thanks.
| Note that
$$
\cosh \alpha=1+\frac{\alpha^2}{2}+o(\alpha^2)
$$
for $\alpha\to0$. So for $x\to 0$ we have
$$
\prod\limits_{k=1}^n \cosh kx=
\prod\limits_{k=1}^n \left(1+\frac{k^2 x^2}{2}+o(k^2 x^2)\right)=
\prod\limits_{k=1}^n \left(1+\frac{k^2 x^2}{2}+o(x^2)\right)=
1+\sum\limits_{k=1}^n\frac{1}{2}k^2 x^2+o(x^2)
$$
Hence
$$
\lim\limits_{x\to 0}\frac{\prod\limits_{k=1}^n\cosh kx - 1}{x^2}=
\lim\limits_{x\to 0}\frac{\frac{1}{2}\sum\limits_{k=1}^n k^2 x^2+o(x^2)}{x^2}=
\frac{1}{2}\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can I prove this inequality algebraically? With $x+y\ge z$ $(x,y,z\ge0)$, prove that: $$\frac{x}{1+x}+\frac{y}{1+y}\ge\frac{z}{1+z}$$
I'm aware that using analytic view this is easy since $f(x)=\frac{x}{1+x}$ is concave in $[0,\infty)$. However I want to prove it using merely algebraic techniques. Is that possible?
| How about this? As Thomas Andrews observed, write $\frac{w}{1+w}=1−\frac{1}{1+w}$ for $w=x,y,z$ and then you need to show $\frac{1}{1+x} + \frac{1}{1+y} \le 1 + \frac{1}{z+1}$.
Now, $$1 + \frac{1}{z+1} \ge 1 + \frac{1}{x+y+1} = \frac{x+y+2}{x+y+1} \ge \frac{(x+1)+(y+1)}{1+x+y+xy} = \frac{1}{1+x} + \frac{1}{1+y}$$ and we're done. Equality holds when one of $x,y$ is $0$, and $z=x+y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove this combinatorial identity? I am wondering how to prove the following identity:
$$\sum_{i=0}^{n-r} \frac{2^i (r+i) \binom{n-r}{i}}{(i+1) \binom{2n-r}{i+1}}=1?$$
It seems this might be related to the hypergeometric distribution, but I could not convert that form back into hypergeometric distribution form.
| Note that
$$
\begin{align}
\frac{2^i(r+i)\binom{n-r}{i}}{(i+1) \binom{2n-r}{i+1}}
&=\frac{2^i(r+i)}{n}\frac{\binom{2n-r-i-1}{n-1}}{\binom{2n-r}{n}}\tag{1}\\
&=\frac{2^i}{n}\frac{2n\binom{2n-r-i-1}{n-1}-n\binom{2n-r-i}{n}}{\binom{2n-r}{n}}\tag{2}\\
&=\frac{2^{i+1}\binom{2n-r-i-1}{n-1}-2^i\binom{2n-r-i}{n}}{\binom{2n-r}{n}}\tag{3}
\end{align}
$$
$(1)\quad$ $\frac1{i+1}\frac{\color{#C00000}{(n-r)!}}{i!\color{#C00000}{(n-r-i)!}}\frac{(i+1)!\color{#00A000}{(2n-r-i-1)!}}{\color{#00A000}{(2n-r)!}}=\frac1n\frac{n!\color{#C00000}{(n-r)!}}{\color{#00A000}{(2n-r)!}}\frac{\color{#00A000}{(2n-r-i-1)!}}{(n-1)!\color{#C00000}{(n-r-i)!}}$
$(2)\quad$ $\begin{array}{l}(2n-r-i)\binom{2n-r-i-1}{n-1}=n\binom{2n-r-i}{n}\\
\Rightarrow(r+i)\binom{2n-r-i-1}{n-1}=2n\binom{2n-r-i-1}{n-1}-n\binom{2n-r-i}{n}\end{array}$
$(3)\quad$ distribute $\frac{2^i}{n}$
Next, we have
$$
\begin{align}
\sum_{k=0}^{n-m}\binom{n-k}{m}2^k
&=\sum_{k=0}^{n-m}\sum_{j=0}^k\binom{n-k}{m}\binom{k}{j}\\
&=\sum_{j=0}^{n-m}\binom{n+1}{m+j+1}\tag{4}
\end{align}
$$
Applying $(4)$ to the sum of $(3)$, we get to a nicely telescoping sum:
$$
\begin{align}
{\large\sum_{i=0}^{n-r}}\;\frac{2^i(r+i)\binom{n-r}{i}}{(i+1) \binom{2n-r}{i+1}}
&={\large\sum_{i=0}^{n-r}}\;\frac{2^{i+1}\binom{2n-r-i-1}{n-1}-2^i\binom{2n-r-i}{n}}{\binom{2n-r}{n}}\\
&=\frac1{\binom{2n-r}{n}}\sum_{i=0}^{n-r}\left(2\binom{2n-r}{n+i}-\binom{2n-r+1}{n+i+1}\right)\\
&=\frac1{\binom{2n-r}{n}}\sum_{i=0}^{n-r}\left(\binom{2n-r}{n+i}-\binom{2n-r}{n+i+1}\right)\\
&=\frac1{\binom{2n-r}{n}}\binom{2n-r}{n}\\[6pt]
&=1\tag{5}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality.$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$
Let $a,b,c \gt 0$. Prove that (Using Cauchy-Schwarz) :
$$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$$
I tried to use Cauchy-Schwarz in the following form
$$\sqrt{Ax}+\sqrt{By}+\sqrt{Cz}\leq \sqrt{(A+B+C)(x+y+z)}\tag{1}.$$
I wrote $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}=\frac{\sqrt{2a(c+a)(a+b)}+\sqrt{2b(b+c)(a+b)}+\sqrt{2c(b+c)(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}.$$
and then I applied on $(1)$:
\begin{eqnarray}
A &=& 2a(c+a) &\mbox{and}& x=a+b;\\
B &=& 2b(a+b) &\mbox{and}& y=b+c;\\
C &=& 2c(b+c) &\mbox{and}& z=c+a ,
\end{eqnarray}
but I did not obtain anything. Thanks for your help. :)
| Denote $$S = \sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}}$$
The Inequality doesn't hold. Clearly, taking $b=c=0.05, a=5$ implies $$\sqrt{2}S>\sqrt{\frac{2a}{b+c}} = 10 > 3$$ By this method, it is easily shown that no upper bound exists.
For a lower bound, note firstly that when $a = b, c \rightarrow 0$, $S \rightarrow 2$. We'll prove that $S \ge 2$ for non negative reals $a,b,c$.
We have, $$\frac{a+b+c}{2a} = \frac 12 \left(\frac{b+c}{a} + 1 \right) \ge \sqrt{\frac{b+c}{a}} \\ \implies \sqrt{\frac{a}{b+c}} \ge \frac{2a}{a+b+c}$$ Adding up the other two similar inequalities, we get the result.
| {
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Sine and Cosine equation ( diophantine ) $\cos(\frac{1}{ab} \pi) = \sin(\frac{a}{b} \pi)$
Let $a$ and $b$ be positive integers. What is the full set of solutions?
An example is $a = 2$ and $b = 5$.
I assume the best method is to take $\arccos$ on both sides and solve the resulting diophantine equation?
| Thanks to the hint given i can solve the question easily now.
$a = 1$ is trivial.
So Lets take $a <> 1$.
Using $cos(x) = sin(1/2 \pi - x)$ we get
$\frac{1}{2} + \frac{-1}{ab} = \frac{a}{b} + 2k$
$\frac{ab}{2ab} + \frac{-2}{2ab} = \frac{(2aa+4kab)}{2ab}$
$ab - 2 = 2aa + 4kab$
$2aa + (4k-1)ab = -2$
$a(2a + (4k-1)b) = -2$
$a=2$
$4+(4k-1)b = -1$
$(4k-1)b = -5$
$b=5$
Thanks for the help. I cant believe i missed the evident.
| {
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"timestamp": "2023-03-29T00:00:00",
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Three inequalities with sums of fractions over two positive integers In a proof, I arrive at three inequalities for all $p,q \geqslant 0$:
\begin{align}
\frac{p+1}{q+1} + \frac{q+1}{p+1} &\geqslant 1 +
\frac{p}{2q+1} + \frac{q}{2p+1} + \frac{1}{p+q+1};\cr
\frac{2q+3}{p+1} + \frac{p+2}{q+2} &\geqslant 2 +
\frac{2q+1}{2p+1} + \frac{2}{p+q+2};\cr
\frac{q+1}{p+1} + \frac{q+2}{p+2} + \frac{p+2}{q+2} +
\frac{p+1}{q+1} &\geqslant 2 + \frac{2p+1}{2q+2} +
\frac{2q+1}{2p+2} + \frac{2}{p+q+2}.
\end{align}
Any idea on how to attack these?
EDIT: Following the pieces of advice in comments, I expanded everything to get rid of the fractions and form bivariate polynomials which must be positive. In the first case, there is an obvious factor $pq$. Since the polynomial is zero if $p=q$ (the inequality is tight), this means that $p-q$ is another factor, yielding $pq(p-q)^2(2p+2q+3) \geqslant 0$. In the case of the second polynomial, there is a trivial factor $q$, and, again $p-q$. I didn't know how to guess the two last factors, but Wolfram|Alpha helped: $q(p-q)(p-q-1)(2p+2q+5)
\geqslant 0$.
The last one is fearsome, although the polynomial must have a factor $p-q$.
| For the last one, multiply through by $2(p+1)(q+1)(p+2)(q+2)(p+q+2)$ to get rid of the fractions; then Wolfram|Alpha factors the difference between the left-hand side and the right-hand side as
$$
(p-q)^2\left(2p^2(q+1)+p(2q^2+9q+8)+2(q+2)^2\right)\;,
$$
which is $\ge0$ for $p,q\ge0$.
| {
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solution to equation $a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$ Does the equation
$$ a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$$
have a closed-form solution for $\theta$? What about the case where $a^2 + b^2 = 1$?
| Given your use of signs I'll take it that $a,b \ge 0$. In which case, if $R = \sqrt{a^2+b^2}$ and $\alpha = \tan^{-1} \dfrac{b}{a}$ then
$$a \cos \theta - b\sin \theta = R\cos(\theta + \alpha)$$
and so if $\left| c \right| \le R$, we can solve the equation to get
$$\theta = \cos^{-1} \dfrac{c}{R} - \alpha = \boxed{\cos^{-1} \dfrac{c}{\sqrt{a^2+b^2}} - \tan^{-1} \dfrac{b}{a}}$$
Of course, this is just one solution; we can obtain the others by considering symmetries.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/194331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric Inequality. $\tan{A}+\tan{B}+\tan{C} \geq \frac{s}{r}$
For any acute-angled triangle $ABC$ show that
$$\tan{A}+\tan{B}+\tan{C} \geq \frac{s}{r},$$
where where $s$ and $r$ denote the semi-perimeter and the inradius, respectively.
Merci :)
| If $a$, $b$, and $c$ denote the sides of the corresponding triangle we have $s=s-a+s-b+s-c$. We also have $\frac{s-a}{r}=\cot\frac{A}{2}$ and similar identities for $b$ and $c$. So what we want to show is that $$ \tan A+\tan B+\tan C\geq \cot \frac{A}{2} +\cot \frac{B}{2} +\cot \frac{C}{2}. $$
This is equivalent to $$\frac{1}{\cos A} + \frac{1}{\cos B}+ \frac{1}{\cos C}\geq 6.$$ This inequality follows by application of the Jensen's inequality for the convex function $\frac{1}{\cos x}$ defined for $x\in\left(0,\frac{\pi}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/195274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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square root of $1/2 + \sqrt3/2?$ Playing with Maple, I noticed that it gives the square root of $c = 1+\frac{\sqrt3}{2}$ as equal to $a = \frac{1}{2}+\frac{\sqrt3}{2}$.
Indeed it checks out. But I got curious: how can I find that value, or more generally any square root of numbers of the form $x+y\sqrt{k}$?
I was able to do it the following way: the square of $z = x+y\sqrt{k}$ is also of the same form. Therefore, I can suppose there is a number of that form whose square is equal to $z$.
In my case, I want to find $(x,y)$ such that $(x+y\sqrt3)^2 = 1+\frac{\sqrt3}{2}$. I developed, which yields $(x^2+3y^2) + 2xy\sqrt3 = 1+\frac{\sqrt3}{2}$. Then I matched the coefficients of $1$ and of $\sqrt3$ on both sides, to get the system:
$x^2+3y^2 = 1$
$2xy = \frac{1}{2}$
Solving for $x$ and $y$, I got $x = ±\frac{1}{2}$ and $y = ±\frac{1}{2}$ (there is another pair of solutions that compute to the same number). QED.
Is my method correct? Is there any more efficient way? Is it possible to prove that a solution of the form $z = x+y\sqrt{k}$ always exist and if not, when?
Thanks.
| For rationals $x,y,a,b,k$ where $k$ is square-free and $y \ne 0$, $(a + b \sqrt{k})^2 = x + y \sqrt{k}$ if and only if $a^2 + k b^2 = x$ and $2 a b = y$, and thus $b = y/(2a)$ and
$a^2 + k y^2/(4 a^2) = x$, i.e. $(2 a^2 - x)^2 = x^2 - k y^2$. Thus $x^2 - k y^2$ must be the square of a rational, and if it is $r^2$ then $(x \pm r)/2$ must be the square of a rational.
In your example, $x=1, y=1/2, k=3$, $x^2 - k y^2 = 1 - 3/4 = 1/4 = (1/2)^2$ and
$(1 - 1/2)/2 = (1/2)^2$, so $1 + \sqrt{3}/2$ is a square.
On the other hand, for $x=3, y=1, k=5$, $x^2 - k y^2 = 9 - 5 = 2^2$ but $(3 - 2)/2 = 1/2$ and $(3+2)/2 = 5/2$ are not squares, so $3 + \sqrt{5}$ is not the square of an expression of the form $a + b \sqrt{5}$ with $a$ and $b$ rational (on the other hand, it is the square of
$\sqrt{10}/2 + \sqrt{2}/2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/195956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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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.