Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Minimize Trace for non-symmetric matrix Let $A$ be a given matrix and $X$ a symmetric positive definite matrix which
shall be optimized:
min $tr (AX)$ subject to $X>0$
This optimization problem is convex if $A$ is symmetric as it is the standard form of Semidefinite Programming. But is the problem convex for non-symmetric $A$ as well?
| Yes, the program is convex for any $A$.
Let $X \in S^n$ (the space of $n\times n$ dimensional symmetric matrices). Then,
\begin{align*}
\textrm{trace}(AX) &= \textrm{trace}\left(A\left(\sum_{i=1}^n\nu_i x_i x_i^T\right)\right)\\
&= \textrm{trace}\left(\sum_{i=1}^n\nu_i A x_i x_i^T\right)\\
&= \sum_{i=1}^n\nu_i \textrm{trace}\left( A x_i x_i^T\right)\\
&= \sum_{i=1}^n\nu_i \textrm{trace}\left(x_i^T A x_i \right)\\
&= \sum_{i=1}^n\nu_i \left(x_i^T A x_i \right)\\
\end{align*}
We will use the following identity: For any square matrix $A$
$$ x^TAx = \frac{1}{2}x^T(A+ A^T)x$$
Proof:
\begin{align*}
\frac{1}{2}x^T(A+ A^T)x &= \frac{1}{2}x^TAx + \frac{1}{2}x^TA^Tx\\
&= \frac{1}{2}x^T(Ax) + \frac{1}{2}(Ax)^Tx\\
&= \frac{1}{2}x^T(Ax) + \frac{1}{2}x^T(Ax)\\
&= x^TAx
\end{align*}
Thus, we get
\begin{align*}
\textrm{trace}(AX) &= \sum_{i=1}^n\nu_i \left(x_i^T \frac{1}{2}(A+ A^T)x_i \right)\\
&= \frac{1}{2}\textrm{trace}\left(\sum_{i=1}^n\nu_i x_i^T(A+ A^T)x_i \right)\\
& = \frac{1}{2}\textrm{trace}\left((A+ A^T)\sum_{i=1}^n\nu_i x_ix_i^T \right)\\
& = \frac{1}{2}\textrm{trace}\left((A+ A^T)X\right)
\end{align*}
The term $\textrm{trace}\left((A+ A^T)X\right)$ is in standard form as desired.
| {
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Inversions of a Permutation Show that the largest number of inversions of a permutation of $\{1,2,3,4,5,6,7,8\}$ equals $n(n-1)/2$. Determine the unique permutation with $n(n-1)/2$ inversions. Also determine all those permutations with one fewer inversion.
| $\{8,7,6,5,4,3,2,1\}$ has $8(8-1)/2$ inversions.
The number of inversions is one less in the following permutations: $\{7,8,6,5,4,3,2,1\}$,$\{8,6,7,5,4,3,2,1\}$,$\{8,7,5,6,4,3,2,1\}$,$\{8,7,6,4,5,3,2,1\}$,$\{8,7,6,5,3,4,2,1\}$,$\{8,7,6,5,4,2,3,1\}$,$\{8,7,6,5,4,3,1,2\}$. There are no others because we need only one pair of numbers to be in correct order. If $i$ and $j$ are in correct order, any number located between the two has to be less than $i$ and more than $j$, which is not possible so $i$ and $j$ have to be adjacent. Moreover, they should be consecutive numbers because any number more than $i$ and less than $j$ cannot be located to right of $i$ or left of $j$.
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Simplify $(A+C)(AD+AD) + AC + C$ using Boolean algebra I have solved the equation like this:
(A + C)(AD + AD) + AC + C
=(A + C)(AD) + (A + C)(AD) + AC + C
=AAD + ACD + AAD + ACD + AC + C
=AAD + AAD + ACD + ACD + AC + C
=AAD + ACD + AC + C
=AAD + ACD + C(A+1)
=AAD + ACD + C
=AD + ACD + C
=AD(1+C) + C
=AD + C
Am I right?
| It looks just fine! Well done!
A bit more simply, $$\begin{align}(A+C)(AD+AD)+AC+C &= (A+C)AD+AC+C\\ &= AAD+CAD+AC+C\\ &= AD+(AD+A+1)C\\ &= AD+C\end{align}$$
| {
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How find this inequality$\sqrt{\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)}+1$ let $x,y,z>0$,show that
$$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\dfrac{(x^2+yz)(y^2+xz)(z^2+xy)}{x^2y^2z^2}}$$
My try:
$$\Longleftrightarrow \left(\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\right)^3\ge \dfrac{8(x^2+yz)(y^2+xz)(z^2+xy)}{x^2y^2z^2}$$
let
$$a=\dfrac{x}{y},b=\dfrac{y}{z},c=\dfrac{z}{x}$$
$$\Longleftrightarrow \left(\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)(a+b+c)}+1\right)^3\ge 8\left(1+\dfrac{c}{a}\right)\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)$$
then I can't ,so I think this inequality maybe have other nice methods,Thank you
| As the inequality is homogeneous, we can normalise by $xyz=1$. Then we have:
$$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\left(x^2+\frac{1}{x}\right)\left(y^2+\frac{1}{y}\right)\left(z^2+\frac{1}{z}\right)}$$
$$\sqrt{3+\sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)}+1\ge 2\sqrt[3]{2 + \sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)}$$
Let $\displaystyle a = \sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)\ge 6$. Then the inequality is reduced to
$$f(a) = \sqrt{3+a} +1 - 2 \sqrt[3]{2+a} \ge 0$$
which is easy to do as $f(6)=0$ and $f'(a) > 0 $ for $a > 6$.
Addendum: alternate way to show $\sqrt{3+a} +1 \ge 2 \sqrt[3]{2+a}$ would be to cube, group terms and then square, to get the equivalent $(a+2)(a-6)^2 \ge 0$, which is obvious for $a \ge 6$.
| {
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Is there an easy way to factor polynomials with two variables? On a recent precal test, I saw a question involving the following expression:
$$(x+1)^2-y^2$$
Which factored out into:
$$(x+y+1)(x-y+1)$$
This wasn't very hard, considering that it was already written as the difference of two squares. I then thought "this is easy, but how would I have done it if it weren't easy?"
Let's say that I had to factor something unholy, such as
$$2 a^2 + a^3 - 2 a^4 - a^5 - 2 b - a b + 4 a^2 b - a^4 b - 2 b^2 +
a b^2 + 3 a^2 b^2 + 3 a^3 b^2 + a^4 b^2 - 2 b^3 - 3 a b^3 + a^2 b^3 +
a^3 b^3 - 2 b^4 - a b^4$$
Of course, this was reverse-engineered, so I already know that the answer is:
$$(a + 2) (a + b + 1) (b - a^2) (a - b^2 - 1)$$
This, in turn, means that the zeros of the polynomial are located at:
$$\begin{align}
\ a & = -2 \\
a & = -b-1 \\
b & = a^2 \\
a & = b^2+1 \\
\end{align}$$
One approach that I was thinking about was possibly formatting the polynomial into a table, like so:
$$
\begin{array}{c|lcr}
& a^0 & a^1 & a^2 & a^3 & a^4 & a^5 \\
\hline
b^0 & 0 & 0 & 2 & 1 & -2 & -1 \\
b^1 & -2 & -1 & 4 & 0 & -1 & 0 \\
b^2 & -2 & 1 & 3 & 3 & 1 & 0 \\
b^3 & -2 & -3 & 1 & 1 & 0 & 0 \\
b^4 & -2 & -1 & 0 & 0 & 0 & 0 \\
\end{array}
$$
Does this table tell me anything about the underlying factorization?
Given a large polynomial with two variables, is there a reasonably efficient way of factoring it?
| I know this is nothing algorithmical but for simple polynomial $F(x,y)$ e.g. with integer coefficients it is an approach. It basically consists of plotting the zeros $Z(F) = \{(x,y) \in \mathbb R^2 | F(x,y)=0\}$.
If you consider two polynomials $F(x,y)$ and $G(x,y)$ then $Z(F\cdot G) = Z(F) \cup Z(G)$. That means if you look at the plot of the zeros, you will probably be able to guess different "components" that were just multiplied together. A plot of your example:
As the algebraic curves are usually quite well behaved (usually without many singularities / self intersections $^*$) it can be easy to find low degree factors. In this case e.g. the lines we see that $a = -2$ is one line here, so $(a+2)=0$ must be a factor. We can factor this factor out
$F(a,b)=(-a^4+a^3 b^2-a^3 b+a^2 b^3+a^2 b^2+2 a^2 b+a^2-a b^3+a b^2-b^4-b^3-b^2-b)(a+2)$
and repeat the process. Another line is $b = -a-1$ or $(b+a+1)=0$
$F(a,b)=(-a^3+a^2 b^2+a^2+a b-b^3-b)(a+2)(a+b+1)$
Then we have two parabolas: $b = a^2$ and $a = b^2+1$, which can again be factored out.
What you have to consider here is that powers of those factors are not visible in the graph, but if you see the same line again (e.g. $a=-2$) after factoring it out, you know that you have to add this factor one more time, e.g. $(a+2)^2$ e.t.c. But you might also have factors that do not show up at all, e.g. $(x^2+1)$ which is never zero on $\mathbb R^2$.
This is obviously a simple example, and it can still get quite complicated, but it works quite well with lower degree integer polynomials.
Other than that the field that investigates questions like those is called algebraic geometry but that is usually even more abstract, that means using fields other than $\mathbb R$.
$^*$ There are examples e.g. $y^2 = x^3-3x+2$:
PS: Common zeros of two polynomials over a ring (e.g. another polynomial ring) can be studied using the resultant.
| {
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Superposition of Sine and Cosine functions I was wondering this:
Let a and b be real numbers. Is it always possible to find real numbers c, d, and e such that $$a\sin(x)+b\cos(x)=c\cos\left(\frac{x+d}e\right)$$
Why is this the case?
Thanks.
| $$a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\left(\dfrac{a}{\sqrt{a^2+b^2}} \sin{x} + \dfrac{b}{\sqrt{a^2+b^2}} \cos{x} \right)\\
=\sqrt{a^2+b^2}\left(\sin{\varphi} \sin{x} + \cos{\varphi} \cos{x} \right)=\sqrt{a^2+b^2}\cos{(x-\varphi)},$$ where
$$\sin{\varphi} =\dfrac{a}{\sqrt{a^2+b^2}}, \\
\cos{\varphi} =\dfrac{b}{\sqrt{a^2+b^2}}.
$$
| {
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How many ways are there to distribute 8 teachers to 4 schools where each school must get at least 1 teacher? Additional details: the teachers are considered distinct from one another.
So here is what I thought:
1) Choose four teachers to go to each one of the schools: $\binom{8}{4}\cdot4!$
2) For each of those situations, distribute the other 4 teachers to the 4 schools: $4^4$
So total: $\binom{8}{4}\cdot4!\cdot4^4$
However, I am almost 100% sure that I am over-counting but can't quite put my finger on it. Any help would be appreciated.
| Another way is using exponential generating functions.
The EGF of this problem is:
$\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+
\frac{x^7}{7!}+\frac{x^8}{8!}\right)^4$
to solve the problem we can use
$(e^x-1)^4$
for that expression. This expands out to
$ 1-4 e^x+6 e^{2 x}-4 e^{3 x}+e^{4 x}$
this can expanded by hand or taken over to Wolfram Alpha. We get:
$x^4+2 x^5+\frac{13 x^6}{6}+\frac{5 x^7}{3}+
\frac{81 x^8}{80}+...+$
The coefficient of x^8 is the one we want and we multiply that by 8!
$\dfrac{81}{80}\cdot 8! = 40824$
That is the number of ways.
| {
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Simplify $\frac{x^{2}-4x}{x^{3}+2x^{2}-5x-6}$ Simplify $$\frac{x^{2}-4x}{x^{3}+2x^{2}-5x-6}$$
Attempt.
I get $$\frac{-\frac{5}{6}}{x+1}+\frac{\frac{-4}{15}}{x+3}+\frac{\frac{21}{10}}{x-2}$$ but it's wrong.
| $\displaystyle \frac{x^3-4x}{x^3+2x^2-5x-6} = \frac{x\cdot(x-2)\cdot(x+2)}{(x+1)\cdot(x-2)\cdot(x+3)} = \frac{x.(x+2)}{(x+1)\cdot(x+3)}$
Now Using Partial fraction Method::
$\displaystyle \frac{x^2+2x}{(x+1)\cdot(x+3)} = \frac{A}{(x+1)}+\frac{B}{(x+3)}$
$\displaystyle x^2+2x = A(x+3)+B(x+1)$
Now Put $x+1 = 0\Leftrightarrow x = -1$
We Get $\displaystyle A = -\frac{1}{2}$
and again Put $x+3 = 0\Leftrightarrow x = -3$
We Get $\displaystyle B = -\frac{3}{2}$
So $\displaystyle \frac{x^2+2x}{(x+1)\cdot(x+3)} = -\frac{1}{2(x+1)}-\frac{3}{2(x+3)}$
| {
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Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?
| Yes indeed, that's the way to go about it. Now, we have $$\begin{align} \lim_{x \to 2}\frac{x-2}{\sqrt{x^2+5}-3} & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{x^2 - 4} \\ \\ & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{(x - 2)(x+2)} \\ \\ & \overset{x\neq 2}{=} \lim_{x \to 2}\dfrac {\sqrt{x^2 + 5} + 3}{x+2} \\ \\
& = \dfrac 64 = \frac 32\end{align}$$
| {
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Prove, that $\lim_{x \to \infty} 2^x \sin(\pi/(2^k))$ converges to $\pi$. Prove, that $$\lim_{x \to \infty} 2^x \sin\left(\frac \pi {2^x}\right)=\pi.$$
I've tried calculus to no avail.
I found out that every element of this sequence is the area of a regular rectangle with $2^{k+2}$ sides that's inside a circle whose radius is 1, therefor the function is monotonically rising and every time smaller than $\pi\cdot 1^2=\pi$.
Is there a simpler way?
| Using $\sin$'s MacLaurin series,
\begin{align}
\lim_{x\rightarrow\infty} \ 2^x\sin\frac{\pi}{2^x} &= \lim_{x\rightarrow\infty} \ 2^x\left(\frac{\pi}{2^x} - \frac{\pi^3}{3! \cdot 2^{3x}} + \frac{\pi^5}{5! \cdot 2^{5x} } + \ \ldots\right) \\
& = \lim_{x\rightarrow\infty} \ \pi - \frac{\pi^3}{3! \cdot 2^{2x}} + \frac{\pi^5}{5! \cdot 2^{4x} } + \ \ldots \\
& = \pi
\end{align}
| {
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If $f(3x-1)=9x^2+6x-7$, determine $f(x)$ if $f(3x-1)=9x^2+6x-7$ determine all the $f(x)$ functions.
I tried in this way :
$t=3x-1 \Rightarrow x=(t+1)/3$
$f(t)=9(t+1)^2/9-6((t+1)/3)-7((t+1)/3)\ldots$
but unfortunately I get the original function.
Thanks in advance.
| Using your method, substituting $t = 3x - 1 \iff x = \dfrac {t+1}{3} $, we have $$\begin{align} f(t) & = \dfrac{9(t+1)^2}{9} + \dfrac{6(t+1)}{3} - 7 \\ \\ & = (t+1)^2 +2(t+1) - 7 \\ \\ &= t^2 + 4t - 4 \end{align}$$
Hence, $$f(x) = x^2 + 4x - 4$$
| {
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Is my answer correct? (And what's the name of the rule?)$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!}$ Want to know if I solved this problem correctly:
$$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!} =\lim_{n \to \infty} \frac{1 \cdot 2 \ldots(n-1)n(n+1)(n+2)(n+3) - 1 \cdot 2 \ldots(n-1)n}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n}=\lim_{n \to \infty}\frac{(1 \cdot 2 \ldots(n-1) n)\cdot((n+1)(n+2)(n+3)-1)}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n} =\lim_{n \to \infty} \frac{(n+1)(n+2)(n+3)-1}{(n+1)(n+2)\cdot n} = \frac{1}{1}=1$$
The answer is 1 because we've got $n^3$ in numerator and $n^3 $ in denominator, and the constants are both equals 1 so the limit can be calculated as $\frac{1}{1}$. (Sorry I don't remember that rule. Could you remain me please?)
| You can more easily write the expression (cancelling as much as possible) as $$\frac {n+3}n-\frac 1{n(n+1)(n+2)}=1+\frac 3n-\frac 1{n(n+1)(n+2)}$$
Then the limit is very easy to see.
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Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7 Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7.
I have that such an integer n has n=3 or 7(mod 10) but don't know where to go from there.
Then show that there are infinitely many prime numbers n with n=3 or 7 (mod 10)
| Every prime number is congruent to $1,3,7$, or $9$ modulo $10$. Here is a a multiplication table modulo $10$
\begin{array}{c|cccc}
\times & 1 & 3 & 7 & 9 \\
\hline
1 & 1 & 3 & 7 & 9 \\
3 & 3 & 9 & 1 & 7 \\
7 & 7 & 1 & 9 & 3 \\
9 & 9 & 7 & 3 & 1 \\
\end{array}
Note that every integer ending in $3$ or $7$ had at least one divisor that ends in $3$ or $7$. Since a number can only have a finite number of divisors, it follows that at least one prime divisor must end in $3$ or $7$.
Since $\gcd(10,3) = \gcd(10,7) = 1$, Dirichet's theorem says that every arithmetic progression of the form $10n+3$ and $10n+7$ will contain infinitely many prime numbers.
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Extended Euclidean Algorithm in $GF(2^8)$? I'm trying to understand how the S-boxes are produced in the AES algorithm. I know it starts by calculating the multiplicative inverse of each polynomial entry in $GF(2^8)$ using the extended euclidean algorithm. However I am having some trouble understanding how to perform the euclidean algorithm with polynomials in a field. Could someone please explain how to do this with a step by step example?
| Using Rijndael's finite field, the reducing polynomial is $x^8+x^4+x^3+x+1$.
Suppose we want to compute the inverse of $x^5+1$ in this field. We want to solve the equation
$$
a(x^5+1)+b(x^8+x^4+x^3+x+1)=1
$$
I like to use the Euclid-Wallis Algorithm. Since we are dealing with polynomials, I will write things rotated by $90^\circ$.
$$
\begin{array}{c|c}
x^8+x^4+x^3+x+1&0&1&\\\hline
x^5+1&1&0\\\hline
x^4+x+1&x^3&1&x^3\\\hline
x^2+x+1&x^4+1&x&x\\\hline
\color{#C00000}{1}&\color{#C00000}{x^6+x^5+x^3+x^2+x}&\color{#C00000}{x^3+x^2+1}&x^2+x\\\hline
0&x^8+x^4+x^3+x+1&x^5+1&x^2+x+1
\end{array}
$$
The fifth row tells us that in $\mathbb{Z}_2[x]$
$$
(x^5+1)(\color{#C00000}{x^6+x^5+x^3+x^2+x})+(x^8+x^4+x^3+x+1)(\color{#C00000}{x^3+x^2+1})=\color{#C00000}{1}
$$
Thus, $x^6+x^5+x^3+x^2+x$ is the inverse of $x^5+1$ in $\left.\mathbb{Z}_2[x]\middle/(x^8+x^4+x^3+x+1)\mathbb{Z}_2[x]\right.$.
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Showing "$30$ divides $n^5-n$ for all $n\in\Bbb N$" using induction
Prove that $(n^5 - n)$ divides by $30$ for every $ n\in N$, using induction only.
How on earth do I do that? Thing is $(n^5 - n)$ can't be opened using any known formula...
| Let $f(m)=m^5-m$
$\displaystyle\implies f(1)=1^5-1=0$
Let $f(m)$ is divisible by $30$ for $m=n$
Now $\displaystyle f(n+1)-f(n)=(n+1)^5-(n+1)-(n^5-n)=5n^4+10n^3+10n^2+5n$
$\displaystyle=5n(n^3-n)+10(n^3-n)+15n^2+15n$
$\displaystyle=5(n+2)n(n+1)(n-1)+15n(n+1)$
Observe that $n(n+1)(n-1),$ being product of $3$ consecutive integers is divisible by $3$
and $n(n+1)$ is divisible by $2$
$\displaystyle\implies n(n+1)(n-1)$ is divisible by lcm$(2,3)=6$
$\displaystyle\implies f(n+1)-f(n)$ is divisible by $30$
But, $f(n)$ is divisible by $30,\implies f(n+1)$ will be divisible by $30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Given $A^2$ where A is matrix, how find A? Problem is simple.
Given $$A^2=\begin{bmatrix}13 & 9 &-9 \\ 0 & 4 & 0 \\ 12 & 12 & -8 \end{bmatrix}$$
How find $A$?
I think a method using eigenvalues and I find them.
But I can't find an actual $A$.
Is it right to use eigenvalues?
| Here is an answer obtained by simple inspection. It is not a generally applicable method, but every step below can be done very easily by hand. I have absolutely no intention to recommend this answer. I just want to see how far simple inspection can reach.
*
*Observe that the second row of $A$ has two off-diagonal zeroes. Therefore, if
$$
\pmatrix{4&0^\top\\ \color{blue}{v}&\color{red}{B}}
:=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}A^2\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}
=\left(\begin{array}{c|cc}4&0&0\\ \hline\color{blue}{9}&\color{red}{13}&\color{red}{-9}\\ \color{blue}{12}&\color{red}{12}&\color{red}{-8}\end{array}\right)
=\pmatrix{2&0^\top\\ x&C}^2
\tag{1}
$$
for some $2$-vector $x$ and some $2\times2$ matrix $C$, one can set $A=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}\pmatrix{2&0^\top\\ x&C}\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}$.
*So, we try to solve
$$
\pmatrix{4&0^\top\\ v&B}
=\pmatrix{2&0^\top\\ x&C}^2
=\pmatrix{4&0^\top\\ (C+2I)x&C^2}.\tag{2}
$$
We need to find some $C$ such that $B=C^2$ and some $x$ such that $(C+2I)x=v$.
*Since $\det(B)=4$, we have $\det(C)=\pm2$. Let us assume that $\det(C)=2$ and see if $(2)$ is really solvable. By Cayley-Hamilton theorem, $C^2-\operatorname{tr}(C)C+\det(C)I=0$ and hence $\operatorname{tr}(C)C=B+2I$. Taking traces of both sides, we get $\operatorname{tr}(C)^2=9$. Set $\operatorname{tr}(C)=3$, we get $C=\operatorname{tr}(C)^{-1}(B+2I)=\pmatrix{5&-3\\ 4&-2}$.
*Solving $(C+2I)x=v$, we get $v=(3,4)^\top$ and hence we obtain the following solution:
$$
A=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}
\pmatrix{2&0^\top\\ x&C}
\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}
=\pmatrix{5&3&-3\\ 0&2&0\\ 4&4&-2}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the orthogonal trajectories of the following family of curves: $x^3y - xy^3 = \alpha$ The answer is $-6x^2y^2 + x^4 + y^4 = \beta$, but I get $-3x^2y^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta$ and I can't seem to find my mistake. My work:
Differentiaing the given equation, we get
$$3x^2y + x^3\frac{dy}{dx} - y^3 - 3y^2x\frac{dy}{dx} = 0. $$
So $$3x^2y - y^3 = (3y^2x -x^3)\frac{dy}{dx}, $$ and hence
$$ \frac{dy}{dx} = \frac{3x^2y - y^3}{3y^2x - x^3}. $$
Now, the slope of the other family of curves will be the negative inverse, so for the orthogonal trajectory, we have $$\frac{dy}{dx} = -\frac{3y^2x-x^3}{3x^2y-y^3}, $$ which we can write as
$$-(3y^2x-x^3)dx = (3x^2y-y^3)dy. $$ Integration of the last differential equation yields
$$-\frac{3}{2}y^2x^2 + \frac{x^4}{4} = \frac{3}{2}x^2y^2 - \frac{y^4}{4} + \beta, $$ $\beta$ being an arbitrary constant.
This simplifies to $$-3y^2x^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta. $$ which is not correct.
Where am I going wrong?
| Since this is tagged complex analysis,
let $z = x+iy$ and let $f(x+iy) = x^3y - xy^3$.
Expressed in terms of $z$, this is $f(z) = \frac 1 {16i}\left((z+\bar{z})^3(z - \bar z) + (z + \bar z)(z - \bar z)^3\right) = \frac 1{8i}(z^4 - \bar z^4) = \frac 1 4 Im(z^4) = Im(z^4/4)$.
So the curves $x^3y-xy^3 = A$ are the pullback of the horizontal lines by the holomorphic function $z \mapsto z^4/4$. Since holomorphic functions preserve angles (except when the derivative vanish, here it is at $z=0$), the curves that are always orthogonal to those, are the pullback of the vertical lines by $z \mapsto z^4/4$, i.e. the curves $Re(z^4/4) = B$.
And finally, $Re(z^4/4) = (x^4-6x^2y^2+y^4)/4$
At $z=0$, $f$ multiplies angles by $4$ and is $4$-to $1$ around $z=0$, so the two curves $Re(f(z)) = 0$ and $Im(f(z)) = 0$ both have a star-like singular point (an $8$-branched star), and the angle between the two stars is $\pi/8$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Find the solution to the following non-homogenous recurrence relation: Find the solution to the following non-homogenous recurrence relation:
$a_{n+2} + a_{n+1} - 2a_n = n$ for $a_0 = 1$, $a_1 = -2$
I have found the homogenous part with the characteristic polynomial is
$a_n = C_1(-2)^n + C_2$ but don't know how to find the non-homogenous part?
Thanks in advance.
| One way is to find a difference equation that is solved by the function $n \mapsto n$. If $b_n = n$, we see that $b_{n+1}-b_n = 1$ and so $b_{n+2}-2 b_{n+1} + b_n = 0$, with $b_0 = 0, b_1 = 1$.
Now let $x_n= (a_{n+1}, a_n, b_{n+1}, b_n)^T$, then we are looking for a solution to $x_{n+1} = A x_n$ where $A = \begin{bmatrix} -1 & 2 & 1 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 2 & -1 \\
0 & 0 & 1 & 0\end{bmatrix}$, and $x_0=(-2,1,1,0)^T$. We note that $A$ has an eigenvalue of multiplicity $3$ at $\lambda=1$ and an eigenvalue of multiplicity $1$ at $\lambda = -2$.
The solution is $x_n = A^n x_0$.
Rather than evaluating $A^n$ explicitly, since we are only interested in $a_n = e_2^T A^n x_0$, we use the eigenstructure of $A$ and assume that $a_n = C_1 (-2)^n + C_2 + C_3 n + C_4 n^2$. Since we know $a_0,a_1,a_2,a_3$, this gives four equations in $C_1,...,C_4$. Solving the equations gives $C_1 = \frac{32}{27}, C_2 = - \frac{5}{27}, C_3 = \frac{7}{18}, C_4 = \frac{1}{6}$, and so
$a_n = -\frac{{\left( -2\right) }^{n+5}}{27}+\frac{{n}^{2}}{6}+\frac{7\,n}{18}-\frac{5}{27}$.
| {
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"answer_count": 4,
"answer_id": 3
} |
Decomposition of an ideal as a product of two ideals How to show $$5\mathbb{Z}[\sqrt[3]{2}] = (5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1).$$
Firstly, I think that I can say that $$(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1)= (25,5(\sqrt[3]{2}+2),5((\sqrt[3]{2})^2+3\sqrt[3]{2}-1),5((\sqrt[3]{2})^2+\sqrt[3]{2})).$$ Since 5 divides each term, I have that $(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1) \subseteq (5)$. But now how do I show the other direction? Any help would be appreciated. Thanks in advance.
| anon gave you the elementary answer, but if we want to use more machinery, we can use the Dedekind-Kummer theorem.
Namely, to factor $5\mathcal{O}_K$ (where $K=\mathbb{Q}(\sqrt[3]{2})$) it suffices to factor $x^3-2$ in $\mathbb{F}_5[x]$. But, this factors as $(x+2)(x^2+3x+4)$. Thus, we know that
$$5\mathcal{O}_K=(5,\sqrt[3]{2}+2)(5,\sqrt[3]{2}^2+3\sqrt[3]{2}+4)$$
But, by subtracting $5$ from the second entry in the second ideal gives your factorization.
| {
"language": "en",
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"source": "stackexchange",
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} |
Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.
Here is a simple technique used to approximate square roots by Persian author Hassan be al-Hossein:
For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.
if $n=2^k$ we can use the method above.
For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 } } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 } } =\sqrt { 11\frac { 15 }{ 23 } } \\ \sqrt { 11\frac { 15 }{ 23 } } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know.
| $$\sqrt[3]{x}=1+\frac{x-1}{3}-\frac{2(x-1)^2}{9(2!)}+\sum_{n=1}^\infty (-1)^{n-1} \frac{(2+3n)(2+3(n-1))(x-1) ^n)}{3n(n!)}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "24",
"answer_count": 13,
"answer_id": 11
} |
Inversion applied on circles I'm studying for my exam and one of the questions I am stuck on is:
Show that under inversion in the unit circle a circle with centre C and radius $S$ inverts into a circle with centre $\frac{C}{C\overline{C} - s^2}$ and radius $\frac{s}{C\overline{C} - s^2}$.
And I know:
$$Z\overline{Z}-Z\overline{C}-\overline{Z}C+C\overline{C}-s^2=0$$
Can someone please explain how to do this question?
| Inversion is
$$z\mapsto \frac1{\bar z}$$
Your equation for the circle is basically the squared form of
$$ \lvert z - C\rvert = s $$
Now combine them:
\begin{align*}
\left\lvert\frac1{\bar z}-C\right\rvert&=s \\
\left(\frac1{\bar z}-C\right)\left(\frac1{z}-\bar C\right)&=s^2 \\
\frac1{z\bar z}-\frac{C}{z}-\frac{\bar C}{\bar z}+C\bar C-s^2&=0 \\
1-C\bar z -\bar Cz+z\bar z\left(C\bar C-s^2\right) &= 0 \\
z\bar z - \frac{C}{C\bar C-s^2}\bar z - \overline{\left(\frac{C}{C\bar C-s^2}\right)}z + \frac{1}{C\bar C-s^2} &= 0
\\
\left(z-\frac{C}{C\bar C-s^2}\right)
\left(\bar z-\frac{\bar C}{C\bar C-s^2}\right)
-\frac{C\bar C}{\left(C\bar C-s^2\right)^2}
+\frac{C\bar C-s^2}{\left(C\bar C-s^2\right)^2}
&= 0
\\
\left(z-\frac{C}{C\bar C-s^2}\right)
\left(\bar z-\frac{\bar C}{C\bar C-s^2}\right)
-\left(\frac{s}{C\bar C-s^2}\right)^2
&= 0
\\
\left(z-\frac{C}{C\bar C-s^2}\right)
\left(\bar z-\frac{\bar C}{C\bar C-s^2}\right)
&= \left(\frac{s}{C\bar C-s^2}\right)^2
\\
\left\lvert z-\frac{C}{C\bar C-s^2}\right\rvert
&= \frac{s}{C\bar C-s^2}
\end{align*}
From this you can read
\begin{align*}
C' &= \frac{C}{C\bar C-s^2} &
s' &= \frac{s}{C\bar C-s^2}
\end{align*}
as expected.
| {
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} |
Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$
Let $x,y$ be real numbers such that
$$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
Show that $x+y=0$.
My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\
x^2=b^2+(2b-1)y+y^2\cdots\cdots
\end{cases}$$
$(1)+(2)$
then
$$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$
so
$$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$
we take $(3)$ in $(2)$,we have
$$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$
so
$$(2ab-a+b)=0$$
or
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
if
$$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$
and if
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
I don't prove
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
| A more generalised approach over my earlier post. This is not intended to be an exhaustive proof but an experimental one. Constructive comments are most welcome.
Let
$$\sqrt{y^2-x}-x=Ay^n\qquad \cdots (1)\\
\sqrt{x^2+y}-y=\frac {y^{1-n}}A \qquad \cdots (2)\\$$
such that the original equation $$\left(\sqrt{y^2-x}-x\right)\left(\sqrt{x^2+y}-y\right)=y$$
is satisfied as required.
From $(1)$,
$$\begin{align}
\sqrt{y^2-x}&=x+Ay^n\\
y^2-x&=x^2++2Axy^n+A^2y^{2n}\\
y^{2n}A^2+2xy^nA+(x^2-y^2+x)&=0\\
A^2+\frac {2x}{y^n}A+\frac{(x^2-y^2+x)}{y^{2n}}&=0\qquad \qquad \qquad \qquad \cdots (3)
\end{align}$$
From $(2)$,
$$\begin{align}
\sqrt{x^2+y}&=y+\frac {y^{1-n}}A\\
A\sqrt{x^2+y}&=Ay+y^{1-n}\\
A^2(x^2+y)&=A^2y^2+2Ay^{2-n}+y^{2(1-n)}\\
(x^2-y^2+y)A^2-2y^{2-n}A-y^{2(1-n)}&=0\\
A^2-\frac{2y^{2-n}}{x^2-y^2+y}A-\frac{y^{2(1-n)}}{x^2-y^2+y}&=0\qquad \qquad \qquad \cdots (4)
\end{align}$$
Equating coefficients of $A^1$:
$$\begin{align}\frac{2x}{y^n}&=-\frac{2y^{2-n}}{x^2-y^2+y}\\
y^2&=-x(x^2-y^2+y)\qquad \qquad \qquad \qquad \qquad \qquad \cdots (5)\end{align}$$
Equating coefficients of $A^0$:
$$\begin{align}\frac{x^2-y^2+x}{y^{2n}}&=-\frac{y^{2(1-n)}}{x^2-y^2+y}\\
y^2&=-(x^2-y^2+x)(x^2-y^2+y)\qquad \cdots (6)\end{align}$$
(5)=(6):
$$\begin{align}x(x^2-y^2+y)&=(x^2-y^2+x)(x^2-y^2+y)\\
(x^2-y^2)(x^2+y^2-y)&=0\\
(x-y)(x+y)(x^2-y^2+y)&=0\\
\Rightarrow x-y=&0, x+y=0, x^2-y^2+y=0\end{align}$$
Checking by substitution into the original equation shows that only
$$x+y=0$$
is valid.
This graph created on desmos.com might help illustrate the approach:
https://www.desmos.com/calculator/qrlbgbalix
| {
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"source": "stackexchange",
"question_score": "25",
"answer_count": 7,
"answer_id": 5
} |
How prove this inequality $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{3^n}\right)\ge\frac{14}{25}$
show that
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{14}{25}\tag{1}$$
My try:
I only prove following not strong inequality:
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{1}{2}$$
proof: use Bernoulli inequality
$$(1+x_{1})(1+x_{2})\cdots (1+x_{n})\ge 1+x_{1}+x_{2}+\cdots +x_{n},$$where $x_{i}\ge -1$
so
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge1-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{3^n}\right)$$
so
$$LHS\ge1-\sum_{n=1}^{\infty}\dfrac{1}{3^n}=1-\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{1}{2}$$
But for $(1)$,I can't prove it,Thank you
| The product corresponds to QPochhammer[1/3,1/3,m] which is a decreasing function. For m going to infinity, its limit is QPochhammer[1/3,1/3] which is approximately 0.5601260779279490 as mentioned by MJD (14/25 = 0.56000).
| {
"language": "en",
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"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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How prove this $ab|a^8+b^4+1$
show that: there exsit infinite $(a,b)$ such
$$ab|a^8+b^4+1$$
my try: let $a^8+b^4+1=kab,k\in N^{*}$
and I can't work,Thank you
| We show that for any solution $(a, b), a, b \in \mathbb{Z}^+$, we can get another solution $(a', b'), a', b' \in \mathbb{Z}^+$ such that $a'+b'>a+b$. This will imply that starting from the solution $(1, 1)$, we can get an infinite sequence of distinct solutions $(a_n, b_n)$, with $2=a_0+b_0<a_1+b_1<a_2+b_2< \ldots$.
Consider $2$ cases.
Case $1$: $a \leq b^2$. Then we have $a \mid b^4+1$ so $b^4+1=ka$ for some $k \in \mathbb{Z}^+$ so $k \mid b^4+1$. Also $b \mid a^8+1 \Rightarrow b \mid k^8(a^8+1)=(ka)^8+k^8=(b^4+1)^8+k^8 \Rightarrow b \mid k^8+1$. Thus $(k, b)$ is another solution, and $k+b=\frac{b^4+1}{a}+b>a+b$ since $a \leq b^2$.
Case $2$: $a>b^2$. Then clearly $a^4>b^8 \geq b$. We have $b \mid a^8+1$ so $a^8+1=lb$ for some $l \in \mathbb{Z}^+$ so $l \mid a^8+1$. Also $a \mid b^4+1 \Rightarrow a \mid l^4(b^4+1)=(bl)^4+l^4=(a^8+1)^4+l^4 \Rightarrow a \mid l^4+1$. Thus $(a, l)$ is another solution, and $a+l=a+\frac{a^8+1}{b}>a+b$ since $b<a^4$.
As such, we get an infinite sequence of distinct solutions $(a_n, b_n)$, with $2=a_0+b_0<a_1+b_1<a_2+b_2< \ldots$, so indeed there are infinitely many positive integer solutions.
| {
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"source": "stackexchange",
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Inequality in triangle Let $ABC$ be a triangle and $M$ a point on side $BC$. Denote $\alpha=\angle BAM$, $\beta=\angle CAM$. Is the following inequality true?
$$\sin \alpha \cdot (AM-AC)+\sin \beta \cdot (AM-AB) \leq 0.$$
| As I already stated in a comment, your question only makes sense if you assume $0\le\alpha,\beta\le\pi$. As a consequence you get $\sin\alpha\ge0,\sin\beta\ge0$ which will be very useful.
I use $a$ to denote the length $AB$ and $b$ to denote $AC$. Yes, the letters are a bit counter-intuitive with respect to the points $B$ and $C$, but they fit in well with $\alpha$ and $\beta$. Using $a,b,\alpha$ and $\beta$ as parameters, you can deduce
$$AM=\frac{ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta}
{a\sin\alpha + b\sin\beta}$$
I computed this with explicit coordinates and computations learned from projective geometry:
$$
\left(
\begin{pmatrix}a\cos\alpha\\a\sin\alpha\\1\end{pmatrix} \times
\begin{pmatrix}b\cos\beta\\-b\sin\beta\\1\end{pmatrix}
\right) \times
\begin{pmatrix}0\\1\\0\end{pmatrix} =
\begin{pmatrix}
ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta \\
0 \\ a\sin\alpha + b\sin\beta
\end{pmatrix}
$$
But there might be other ways to obtain that distance using more basic considerations. Now plug that expression into your inequality and you obtain
$$
\sin\alpha\left(
\tfrac{ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta}{a\sin\alpha + b\sin\beta}
-b\right)
+
\sin\beta\left(
\tfrac{ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta}{a\sin\alpha + b\sin\beta}
-a\right)
\le 0
$$
Since $a\sin\alpha + b\sin\beta>0$ you can multiply everything by that denominator. You can also divide by $ab>0$. You obtain the following formula:
\begin{multline*}
\sin^2\alpha\cos\beta
+ \sin\alpha\cos\alpha\sin\beta
- \sin^2\alpha
- \tfrac ba\sin\alpha\sin\beta
\\
+ \sin\alpha\sin\beta\cos\beta
+ \cos\alpha\sin^2\beta
- \tfrac ab\sin\alpha\sin\beta
- \sin^2\beta \le 0
\end{multline*}
You can regroup terms like this:
$$
\sin^2\alpha\left(\cos\beta-1\right)
+ \sin^2\beta\left(\cos\alpha-1\right)
+ \sin\alpha\sin\beta\left(\cos\alpha+\cos\beta-\tfrac ab-\tfrac ba\right)
\le 0
$$
With $\cos\varphi\le1$ for all $\varphi$, you can show that each of the three summands is non-positive. The first two parentheses are easy, and for the last parenthesis you can use
$$\cos\alpha+\cos\beta-\tfrac ab-\tfrac ba\le
2-\tfrac ab-\tfrac ba = -\tfrac ab\left(1-\tfrac ba\right)^2\le 0$$
Thus the sum as a whole is non-positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/547374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prime factorization, distinct primes Let $n=p^eq^f$ where $p$ and $q$ are distinct primes and $e$ and $f$ are positive integers.
Show that $n$ has $(e + 1)(f + 1)$ distinct factors in $N$, and that the sum of all these factors is
$(p^{e+1} −1)(q^{f+1} −1) / (p − 1)(q − 1)$
(You may assume the uniqueness of prime factorization).
What if $n$ has more than two distinct prime factors?
(Hint: you may wish to warm up by doing n = pe first.)
Sorry I have no idea where to start :(
Would doing $n=p^e$ first be easiest?
| I'll do a warm up for you. Warm-ups help us understand the nature of a problem and how all of the relevant parts interact. Therefore when an exercise comes with warm-ups and you want to understand how to do the exercise, it's a good idea to try your hand at the warm-ups. Just for future reference. If you followed the directions and did work on the warm-up, then note you should talk about what you've tried and anywhere you got stuck whenever you post a question here.
Here are all factors of $648=2^33^4$
$$\begin{array}{|c|c|c|}\hline \color{Red}{2^0} \color{Blue}{3^0} & \color{Red}{2^1} 3^0 & \color{Red}{2^2}3^0 & \color{Red}{2^3} 3^0 \\ \hline 2^0 \color{Blue}{3^1} & 2^1 3^1 & 2^23^1 & 2^3 3^1 \\ \hline 2^0 \color{Blue}{3^2} & 2^13^2 & 2^23^2 & 2^3 3^2 \\ \hline 2^0 \color{Blue}{3^3} & 2^1 3^3 & 2^2 3^3 & 2^3 3^3 \\ \hline 2^0 \color{Blue}{3^4} & 2^1 3^5 & 2^23^4 & 2^3 3^4 \\ \hline \end{array}$$
Make sure you understand why the divisors can all be listed in this way (via prime factorization).
Homework.
*
*How many columns are there (red)? How many rows are there (blue)? Entries total?
*How are the number of rows and columns related to the powers of $2$ and $3$ in $648$?
*Can you now argue why the number of divisors of $p^eq^f$ is $(e+1)(f+1)$?
*Generalize to $p_1^{e_1}\cdots p_g^{e_g}$: how many divisors do you think it will have and why?
*In the blanks below, fill each entry of the bottom row (except the last) with the sum of the divisors in that column, fill each entry of the right column (except the last) with the divisors of that row, and fill the corner in with the sum of all of the divisors of the whole array.
*How would you sum $p^rq^0+p^rq^1+\cdots p^rq^f$ in general, where $r$ remains fixed? Or what about $p^0q^s+p^1q^s+\cdots+p^eq^s$ where $s$ remains fixed?
*Write the sum of all divisors of $p^eq^f$ as a double summation. Evaluation the double summation using the geometric sum formula. Note $\sum_{i,j}a_ib_j=(\sum_ia_i)(\sum_jb_j)$.
$$\begin{array}{|c|c|c|c|}\hline \color{Red}{2^0} \color{Blue}{3^0} & \color{Red}{2^1} 3^0 & \color{Red}{2^2}3^0 & \color{Red}{2^3} 3^0 & \phantom{--} \\ \hline 2^0 \color{Blue}{3^1} & 2^1 3^1 & 2^23^1 & 2^3 3^1 \\ \hline 2^0 \color{Blue}{3^2} & 2^13^2 & 2^23^2 & 2^3 3^2 \\ \hline 2^0 \color{Blue}{3^3} & 2^1 3^3 & 2^2 3^3 & 2^3 3^3 \\ \hline 2^0 \color{Blue}{3^4} & 2^1 3^5 & 2^23^4 & 2^3 3^4 \\ \hline \phantom{r} \\ \hline \end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/547995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
A problem on finding dy/dx If $a+b+c=0$ and $$y=\frac{1}{x^b+x^{-c}+1}+\frac{1}{x^c+x^{-a}+1}+\frac{1}{x^a+x^{-b}+1}$$then $\frac{dy}{dx}$=?
The only way which I can think of solving this is by differentiating each term. However, is there a simpler way?
| $$y=\frac{1}{x^b+x^{a+b}+1}+\frac{1}{x^{-a-b}+x^{-a}+1}+\frac{1}{x^a+x^{-b}+1}$$
$$y=\frac{1}{x^b+x^{a+b}+1}+\frac{x^{a+b}}{1+x^{b}+x^{a+b}}+\frac{x^b}{x^{a+b}+1+x^b}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove this inequality?$\frac{1}{n+1} + \frac{1}{n+2} + \cdots+\frac{1}{n+n} + \frac{1}{4n} > \ln 2$ $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} > \ln 2$$
$n$ is positive integer.
Thank you !
| Because $\frac 1t$ is concave up, its integral is overestimated by the trapezoidal approximation. In this light, we have
$$
\begin{align}
\ln 2 = \int_n^{2n}\frac 1t dt &<
\frac 12 \left( \frac 1{n} + \frac 1{n+1} \right) +
\frac 12 \left( \frac 1{n+1} + \frac 1{n+2} \right) + \cdots+
\frac 12 \left( \frac 1{2n-1} + \frac 1{2n} \right)\\
&= \frac 1{2n} + \frac 12 \left( \frac 1{n+1} + \frac 1{n+1} \right) +
\frac 12 \left( \frac 1{n+2} + \frac 1{n+2} \right) + \cdots \\
&\qquad+
\frac 12 \left( \frac 1{2n-1} + \frac 1{2n-1} \right) + \frac 12 \frac 1{2n}
\\
&= \frac{1}{n+1} + \frac{1}{n+2} + ... +\frac{1}{2n} + \frac{1}{4n} &=
\end{align}
$$
I hope that's clear.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ Is it true that for $x\in[0,2\pi]$ we have
$$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$$
How can I prove it?
For other intervals what is the value of above series if is convergent?
| Here there is a way using Fourier series. Consider the orthonormal basis of $L^{2}([0,2\pi])$, given by the functions
$$ e_{n}(x) = \frac{1}{\sqrt{2\pi}} e^{-i n x}, \quad n \in \mathbb{Z}. $$
Since for $x \in [0,2\pi]$ the indicator function $\mathbf{1}_{[0,x]}$ is in $L^{2}([0,2\pi])$, it can be expressed in its development in the basis $(e_{n})_{n}$. If we denote $(f,g)$ the dot-product of two functions $f,g$ in $L^{2}([0,2\pi])$, that is $$(f,g) = \int_{[0,2\pi]}f(x)\overline{g(x)}dx,$$ then we have
$$ x = (\mathbf{1}_{[0,x]} , \mathbf{1}_{[0,x]} ) = \sum_{n \in \mathbb{Z}} ( \mathbf{1}_{[0,x]} , e_{n} ) ( e_{n} , \mathbf{1}_{[0,x]}) = \sum_{n \in \mathbb{Z}} |( \mathbf{1}_{[0,x]} , e_{n} )|^{2}. $$
The rest is just computing integrals and separating absolutely convergence series. We have
$$(\mathbf{1}_{[0,x]} , e_{n} ) = \frac{1}{\sqrt{2\pi}}\int_{0}^{x}e^{iny}dy = \begin{cases} \frac{e^{inx} - 1}{\sqrt{2\pi} ni} & n \neq 0, \\
\frac{x}{\sqrt{2\pi}} & n = 0.
\end{cases} $$
Hence,
$$ x = \frac{x^{2}}{2\pi} + \frac{1}{2\pi}\sum_{\substack{n \in \mathbb{Z} \\ n \neq 0}} \frac{2-2\cos(nx)}{n^2}. $$
Using symetric arguments ($ \frac{2-2\cos(nx)}{n^2} = \frac{2-2\cos(-nx)}{(-n)^2}$), we obtain
$$ x = \frac{x^{2}}{2\pi} + \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{1}{n^{2}} - \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{2}} = \frac{x^{2}}{2\pi} + \frac{\pi}{3} - \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{2}}, $$
from where we finally obtain
$$ \sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{2}} = \frac{x^{2}}{4} - \frac{\pi x}{2} + \frac{\pi^{2}}{6}. $$
Outside the interval, just use the periodicity of $\cos$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 4
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What is the maximum value of $4(\sin x)^2 + 3(\cos x)^2$ The question is: What is the maximum value of: $4\sin^2\theta + 3\cos^2\theta$
This is the way I did it:
$4\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3$
The max value of $\sin^2\theta$ is $1$, so the answer must be $4$. However my book says that the answer is $5$. Where did I go wrong?
| Here's a geometric "solution". With the curve $x(t)= \sqrt{3} \cos t$ and $y(t)=\sqrt{4} \sin t$, $0\leq t \leq 2 \pi$, we get an ellipse with major semiaxis of 2 along the y-axis and minor semiaxis of $\sqrt{3}$ along the x-axis. Then the distance squared to the origin $x(t)^2+y(t)^2=4\sin^2 t + 3 \cos^2 t$ must be maximized to the value of the major semiaxis squared. This is because we can imagine an ellipse inscribed inside a circle to see that it must be 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/552838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Volume of a cuboid whose diagonal and surface area is known The sum of length, breadth and depth of cuboid is $19$cm and its diagonal is $5\sqrt{5}$cm.
Its volume is:
a) 125
b) 236
c) 361
d) 486
Solution:
$$\ell^2 + b^2 +h^2 = 125\quad\text{ and }\quad \ell+b+h=19,$$
How can i find volume from this information?
| Define
$$\begin{cases}
p_1 = & \ell + b + h\\
p_2 = & \ell^2 + b^2 + h^2\\
p_3 = & \ell^3 + b^3 + h^3
\end{cases}
\quad\text{ and }\quad
\begin{cases}
s_1 = & \ell + b + h\\
s_2 = & \ell b + \ell h + b h\\
s_3 = & \ell b h
\end{cases}
$$
We know $p_1 = s_1 = 19$ and $p_2 = 125$.
Apply $AM \ge GM$ to the three numbers $\ell, b, h$, we get an upper bound for $s_3$:
$$s_3 = \ell b h \le \left(\frac{\ell + b + h}{3}\right)^3 = \frac{p_1^3}{3} = \frac{19}{3}^3 \sim 254.037$$
This rules out choices (c) and (d).
Apply Cauchy Schwarz inequality to $( \sqrt{\ell}, \sqrt{b}, \sqrt{h} )$ and $( \sqrt{\ell}^3, \sqrt{b}^3, \sqrt{h}^3 )$, we get
$$p_2^2 \le p_1 p_3$$
Together with the Newton's identities
*
*$p_1 = s_1$,
*$p_2 = s_1 p_1 - 2 s_2$,
*$p_3 = s_1 p_2 - s_2 p_1 + 3s_3$
We obtain a lower bound for $s_3$:
$$s_3 = \frac13 \left( p_3 - s_1 p_2 + s_2 p_1 \right)
\ge \frac13 \left( \frac{p_2^2}{p_1} - p_1 p_2 + \frac{p_1^2 - p_2}{2} p_1 \right)
= \frac13 \left( \frac{p_2^2}{p_1} + \frac{p_1^2 - 3 p_2}{2} p_1 \right)
= \frac13 \left( \frac{125^2}{19} + \frac{19^2 - 3\cdot 125}{2} \cdot 19\right)
= \frac{4366}{19} \sim 229.790
$$
This rules out choice (a) and leaves us choice (b) $s_3 = 236$ as the only possible answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integral $\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$I=\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$$
It also can be represented as
$$I=\int_0^{\pi/4}\frac{\phi^2}{\cos \phi\,\sqrt{\cos 2\phi}}\mathrm d\phi$$
| Okay, finally I was able to prove it.
Step 0. Observations. In view of the following identity
$$ \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \, d\theta = 2 \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \right), $$
Vladimir's result suggests that there may exists a general formula connecting
$$ I(r, s) = \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \arctan (s \sin\theta) \, d\theta $$
and the Legendre chi function $\chi_{2}$. Indeed, inspired by Vladimir's result, I conjectured that
$$ I(r, s) = \pi \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \cdot \frac{\sqrt{1+s^{2}} - 1}{s} \right). \tag{1} $$
I succeeded in proving this identity, so I post a solution here.
Step 1. Proof of the identity $\text{(1)}$. It is easy to check that the following identity holds:
$$ \arctan(ab) = \int_{1/b}^{\infty} \frac{a \, dx}{a^{2} + x^{2}}. $$
So it follows that
\begin{align*}
I(r, s)
&= \int_{1/r}^{\infty} \int_{1/s}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2}\theta}{(x^{2} + \sin^{2}\theta)(y^{2} + \sin^{2}\theta)} \, d\theta dy dx \\
&= \int_{1/r}^{\infty} \int_{1/s}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{x^{2} - y^{2}} \left( \frac{x^{2}}{x^{2} + \sin^{2}\theta} - \frac{y^{2}}{y^{2} + \sin^{2}\theta} \right) \, d\theta dy dx \\
&= \frac{\pi}{2} \int_{1/r}^{\infty} \int_{1/s}^{\infty} \frac{1}{x^{2} - y^{2}} \left( \frac{x}{\sqrt{x^{2} + 1}} - \frac{y}{\sqrt{y^{2} + 1}} \right) \, dy dx.
\end{align*}
For the convenience of notation, we put
$$ \alpha = \frac{\sqrt{r^{2} + 1} - 1}{r} \quad \text{and} \quad \beta = \frac{\sqrt{s^{2} + 1} - 1}{s}. $$
Then it is easy to check that $\mathrm{arsinh}(1/r) = - \log \alpha$ and likewise for $s$ and $\beta$. Thus with the substitution $x \mapsto \sinh x$ and $y \mapsto \sinh y$, we have
\begin{align*}
I(r, s)
&= \frac{\pi}{2} \int_{-\log\alpha}^{\infty} \int_{-\log\beta}^{\infty} \frac{\sinh x \cosh y - \sinh y \cosh x}{\sinh^{2}x - \sinh^{2}y} \, dy dx.
\end{align*}
Applying the substitution $e^{-x} \mapsto x$ and $e^{-y} \mapsto y$, it follows that
\begin{align*}
I(r, s)
&= \pi \int_{0}^{\alpha} \int_{0}^{\beta} \frac{dydx}{1 - x^{2}y^{2}} \\
&= \pi \sum_{n=0}^{\infty} \left( \int_{0}^{\alpha} x^{2n} \, dx \right) \left( \int_{0}^{\beta} y^{2n} \, dx \right)
= \pi \sum_{0}^{\infty} \frac{(\alpha \beta)^{2n+1}}{(2n+1)^{2}} \\
&= \pi \chi_{2}(\alpha \beta)
\end{align*}
as desired, proving the identity $\text{(1)}$.
EDIT. I found a much simpler and intuitive proof of $\text{(1)}$. We first observe that $\text{(1)}$ is equivalent to the following identity
$$ \int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \arctan\left( \frac{2s\sin\theta}{1-s^{2}} \right) \, d\theta = \pi \chi_{2}(rs). $$
Now we first observe that from the addition formula for the hyperbolic tangent, we obtain the following formula
$$
\operatorname{artanh}x - \operatorname{artanh} y = \operatorname{artanh} \left( \frac{x - y}{1 - xy} \right) $$
which holds for sufficiently small $x, y$. Thus
\begin{align*}
\arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right)
&= \frac{1}{i} \operatorname{artanh}\left( \frac{2ir\sin\theta}{1-r^{2}} \right)
= \frac{\operatorname{artanh}(re^{i\theta}) - \operatorname{artanh}(re^{-i\theta})}{i} \\
&= 2 \Im \operatorname{artanh}(re^{i\theta})
= 2 \sum_{n=0}^{\infty} \frac{\sin(2n+1)\theta}{2n+1} r^{2n+1}.
\end{align*}
We readily check this holds for any $|r| < 1$. Therefore
\begin{align*}
&\int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \arctan\left( \frac{2s\sin\theta}{1-s^{2}} \right) \, d\theta \\
&\quad = 4 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \int_{0}^{\frac{\pi}{2}} \sin(2m+1)\theta \sin(2n+1)\theta \, d\theta\\
&\quad = 2 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \int_{0}^{\frac{\pi}{2}} \{ \cos(2m-2n)\theta - \cos(2m+2n+2)\theta \} \, d\theta\\
&\quad = \pi \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \delta_{m,n} \\
&\quad = \pi \chi_{2}(rs).
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
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Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
| Alternatively, take
$$ \frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}, $$
and integrate three times with lower limit $0$, giving
$$ \begin{align*}
-\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} \\
x + (1-x)\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} \\
\frac{3}{4}x^2 - \frac{1}{2}x - \frac{1}{2} (1-x)^2 \log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)(n+2)},
\end{align*}$$
and (as @Clement C reminds me) we then apply Abel's theorem to take the limit as $x \to 1$, which gives $1/4$ as the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/560816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 10,
"answer_id": 3
} |
Calculating squared reciprocals of arithmetic series Let $n>0$ be an integer. Is it possible to calculate the value of the sum
$$1+\frac1{(1+n)^2}+\frac1{(1+2n)^2}+\ldots$$?
| One can prove Marko Riedel's claim using a simple Taylor series argument. We have
$$
\sum_{k=1}^\infty \frac{1}{(kn+1)^2} = \sum_{k=1}^\infty \frac{1}{(kn)^2} \left(\frac{1}{(1+(kn)^{-1})^2}\right),
$$
as well as the Taylor series
$$
\frac{1}{(1+x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots,
$$
absolutely convergent for $|x| < 1$. The right hand side of the penultimate display is
$$
\sum_{k=1}^\infty \frac{1}{(kn)^2} \left(1 - \frac{2}{kn} + \frac{3}{(kn)^2} - \frac{4}{(kn)^3} + \cdots \right) = \sum_{k=1}^\infty \sum_{q=2}^\infty \frac{(q-1)(-1)^q}{(kn)^q}.
$$
Since the Taylor series is absolutely convergent, we may interchange the order of summation. The sum over $k$ gives $\zeta(q)$, which yields Riedel's claim.
In the special case, $n = 2$, we can exploit a bit of symmetry to obtain
$$
\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{4n^2} = \frac{3}{4}\zeta(2) = \frac{\pi^2}{8}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Prove that if $ac + bc + c^2 < 0$ then equation (usual notation) has two roots We have $a, b, c$ real parameters, $a ≠ 0$.
Prove that $ax^2 + bx + c = 0$ has two different roots ($b^2-4ac > 0$), if $ac + bc + c^2 < 0$
| $ac+bc+c^2<0\Rightarrow -ac>bc+c^2$. Now $\Delta=b^2-4ac>b^2+4bc+4c^2=(b+2c)^2>0\Rightarrow$ there are $2$ distinct real roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$ for $n\geq 2$ Prove, for all $n\geq 2$ that
$$\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$$
Let us prove the inductive base for $n = 2$
$$\rm{LHS} = 1^2 + 2^2= 1 + 4 = 5$$
$$\rm{RHS} = \binom{3}{2} + 2\binom{3}{3} = 3 + 2\cdot 1 = 5$$
$$\rm{LHS} = \rm{RHS}$$
as desired.
Now, assume for some $k$,
$$1^2 + 2^2 + \dots + k^2 = \binom{k+1}{2} + 2\binom{k+1}{3}$$
To prove the inductive step, it is sufficient to prove that,
$$\binom{k+2}{2} + 2\binom{k+2}{3} - \binom{k+1}{2} - 2\binom{k+1}{3} = (k+1)^2$$
The LHS can be simplified as:
$$\frac{(k+2)!}{2!k!} + 2\left[\frac{(k+2)!}{3!(k-1)!}\right] - \frac{(k+1)!}{2!(k-1)!} - 2\left[\frac{(k+1)!}{3!(k-2)!}\right]$$
$$=\frac{3(k+2)! + 2k(k+2)! -3k(k-1)! - 2k(k-1)(k+1)!}{6k!}$$
$$=\frac{(k+1)!(3k + 6 - 2k^2 - 2k) + (k-1)!(2k^3 + 2k^2 - 3k)}{6k!}$$
$$=\frac{(k+1)!(-2k^2+k+6) + k!(2k^2+2k - 3)}{6k!}$$
$$=\frac{(k+1)(-2k^2 + k + 6) + 2k^2 +2k - 3}{6}$$
$$=\frac{-2k^3 + k^2 + 9k + 3}{6}$$
I'm pretty sure I'm making a mistake somewhere, but I can't figure it out. If someone could complete this inductive proof for me, I will be grateful.
Also, I don't feel satisfied with this ugly proof. Please add a combinatorial proof for this in your answer if possible.
| If you want a short proof, there are few shorter than using difference calculus:
$$\sum_{k=1}^nk^2=\sum_1^{n+1}(k^{\underline{2}}+k^{\underline 1})\delta k=\frac{1}{3}k^{\underline{3}}+\frac{1}{2}k^{\underline{2}}\big|_1^{n+1}=\left(\frac{1}{3}(n+1)^{\underline{3}}+\frac{1}{2}(n+1)^{\underline{2}}\right)-(0+0)={n+1\choose 3}+{n+1\choose 2}$$
For an introduction, see the excellent book Concrete Mathematics.
However, if your homework is to prove this by induction, then you should proceed as you have laid out, which is a good approach.
| {
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How do you prove the inequality $x^2+y^2+z^2-y(x+z)>0$? This actually comes out of a matrix multiplication of $x^TAx$, but when you multiply it out, you get the following.
$x^2+y^2+z^2-y(x+z)>0$
I just can't figure out how to actually prove that that inequality holds for all $x,y,z\neq0$. Also, note that this is for a linear algebra class, so I don't think you're allowed to use derivatives. Also, in case it is important, here are the original matrices used to get that inequality.
$A=$ $
\left( \begin{array}{ccc}
2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2 \end{array} \right)$
and $x = \left(\begin{array}{c}
x\\
y\\
z\ \end{array}\right)$
| Since $(x-y)^2 \geq 0$, it follows easily that $xy \leq \frac{1}{2}(x^2+y^2)$. Similarly, $yz \leq \frac{1}{2}(y^2+z^2)$. That gives you
$$x^2+y^2+z^2-y(x+z) \geq \frac{1}{2}(x^2+z^2). $$
The right-hand-side above is positive unless $x=z=0$. But if $x=z=0$, then
$$ x^2+y^2+z^2-y(x+z)=y^2,$$
and that will be positive unless $x=y=z=0$.
| {
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$a_1=1,a_{n+1}=\frac{n}{a_n}+\frac{a_n}{n}$. Prove that for $n\ge4$, $\lfloor{a_n^2}\rfloor=n$ Define a sequence $\left\lbrace a_{n}\right\rbrace$ by
$\displaystyle{a_{1} = 1\,,\ a_{n + 1} = {n \over a_n} + {a_n \over n}.\quad}$ Prove that for $n \geq 4,\,\,\left\lfloor a_{n}^{2}\right\rfloor=n$
The substitution $b_{n} = a_{n}^{2}$ might be helpful, but I still haven't proved the assertion yet.
| First,we use Mathematical induction have following inequality
$$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$
we easy prove this function $$f(x)=\dfrac{x}{n}+\dfrac{n}{x} $$ is decreasing on $(0,n)$
becasue $$f'(x)=-\dfrac{n}{x^2}+\dfrac{1}{n}\le 0$$
since $a_{1}=1,a_{2}=2,a_{3}=2$,so
$$\sqrt{3}\le a_{3}\le\dfrac{3}{\sqrt{2}}$$
Assmue that
$$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$
then
$$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}}=\dfrac{n}{\sqrt{n-1}}>\sqrt{n+1}$$
$$a_{n+1}=f(a_{n})\le f(\sqrt{n})=\dfrac{n+1}{\sqrt{n}}$$
In fact,we can prove
$$a_{n}\le \sqrt{n+1},n\ge 4$$
since
$$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}})=\dfrac{n}{\sqrt{n-1}},n\ge 3$$
so
$$a_{n}\ge\dfrac{n-1}{\sqrt{n-2}},n\ge4$$
and note
$$a_{n+1}=f(a_{n})\le f\left(\dfrac{n-1}{\sqrt{n-2}}\right)=\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}$$
it suffces prove that
$$\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}\le\sqrt{n+2},n\ge 4$$
$$\Longleftrightarrow n^3-n^2-2n+1<(n^2-n)\sqrt{n^2-4}$$
$$\Longleftrightarrow 2n^3-6n^2+4n-1=2n^2(n-3)+4n-1>0$$
| {
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Simplyfying the equation
It is given that, $a+b+c=0$ we will have to prove that,
$$\frac{1}{x^b + x^{-c}+1} + \frac{1}{x^a + x^{-b}+1} + \frac{1}{x^c +
x^{-a}+1} =1$$
What I have done is, $a=-(b+c)$ into the equation and also I have tried with multiplying $x^a, x^b x^c$ but it didn't look simple. What should be done here?
| Keeping the first term as is, I am trying to set the numerator of the rest two terms as parts of the denominator of the first term.
$$\frac1{x^a+x^{-c}+1}=\frac{x^b}{x^{b+a}+1+x^b}=\frac{x^b}{x^{-c}+1+x^b}\text{ as }b+a=-c$$
$$\frac1{x^c + x^{-a}+1}=\frac{x^{-c}}{1+x^{-(c+a)}+x^{-c}}=\frac{x^{-c}}{1+x^b+x^{-c}}\text{ as }c+a=-b$$
| {
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Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1+1/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{16/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1}\cdot\frac{15}{16}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underbrace{a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{15}{16}}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{15}\\
\end{align}$$
$$\therefore \boxed{16^m - 1=\frac{b}{15}}$$
Or is this the wrong way and I have to do it with mathematical induction?
| HINT:Use induction and
$$16^{n+1}-1=16(16^n-1)+15$$
| {
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$\lim_{x \to 0} \frac{x^n}{\cos\sin x -\cos x}=l$, find $n$ such that $l$ is non zero finite real Problem: Consider $\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$. The value of $n$ such that $l$ is non zero finite real is:
(a) $3$, (b) $4$, (c) $5$, (d) $6$
Solution: The given limit can be written as $$-\frac{1}{2}\lim_{x \to 0} \frac{x^n}{\sin\bigl(\frac{\sin x+x}{2}\bigr) \sin\bigl(\frac{sinx-x}{2}\bigr)}.$$
Please suggest how to proceed further ... getting no clue on this. Thanks.
| Clearly we have
$\displaystyle \begin{aligned}l &= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{{\displaystyle \sin\left(\dfrac{\sin x + x}{2}\right)\sin\left(\dfrac{\sin x - x}{2}\right)}}\\
&= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{\left(\dfrac{\sin x + x}{2}\right)\left(\dfrac{\sin x - x}{2}\right)}\cdot\dfrac{\dfrac{\sin x + x}{2}}{\sin\left(\dfrac{\sin x + x}{2}\right)}\cdot\dfrac{\dfrac{\sin x - x}{2}}{\sin\left(\dfrac{\sin x - x}{2}\right)}\\
&= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{\left(\dfrac{\sin x + x}{2}\right)\left(\dfrac{\sin x - x}{2}\right)}\cdot 1\cdot 1\\
&= -2\lim_{x \to 0}\dfrac{x^{n}}{(\sin x + x)(\sin x - x)}\\
&= -2\lim_{x \to 0}\dfrac{x^{n}}{x\left(\dfrac{\sin x}{x} + \dfrac{x}{x}\right)x^{3}\left(\dfrac{\sin x - x}{x^{3}}\right)}\\
&= -2\lim_{x \to 0}\dfrac{x^{n - 4}}{(1 + 1)(-1/6)}\\
&= 6\lim_{x \to 0}x^{n - 4}\end{aligned}$
Clearly we can see that if $n > 4$ then $l = 0$ and if $n < 4$ then the above limit does not exist. Hence $n = 4$ and $l = 6$.
We have made use of $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$ and by L'Hospital's Rule $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = \lim_{x \to 0}\frac{\cos x - 1}{3x^{2}} = -\frac{1}{3}\cdot\frac{1}{2} = -\frac{1}{6}$$
| {
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Integral $\int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}dx$ Could you suggest any ideas how to evaluate this integral? Is there a closed-form result?
$$\int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}dx$$
| This integral can be solved by brutal force. I will use Vladimir's suggestion as a starting point:
$$I = \int_{1}^{\infty} \frac{\log(x + \sqrt{x^{2} - 1}) \log(1 + \sqrt{x^{2} + 1})}{x^{2}} \, dx. $$
By the knowledge in trigonometry, we obtain
\begin{align*}
I
&= \int_{1}^{\infty} \frac{\operatorname{arcosh} x \cdot (\operatorname{arsinh}(1/x) + \log x)}{x^{2}} \, dx \\
&= \int_{0}^{1} \operatorname{arcosh} (1/x) \cdot (\operatorname{arsinh} x - \log x) \, dx.
\end{align*}
Note that
$$ \int (\operatorname{arsinh} x - \log x) \, dx = x\operatorname{arsinh} x - x \log x + x + 1 - \sqrt{x^{2} + 1}. $$
Here, the constant of integration is chosen so that the integrand becomes $O(x\log x)$ as $x \searrow 0$. Since $\operatorname{arcosh}(1/x) = O(\log x)$ as $x \searrow 0$, we can perform integration by parts to obtain
$$ I = \int_{0}^{1} \left( \operatorname{arsinh} x - \log x + 1 + \frac{1 - \sqrt{x^{2} + 1}}{x} \right) \, \frac{dx}{\sqrt{1-x^{2}}}. $$
Plug $x = \sin\theta$. Then
$$ I = \int_{0}^{\frac{\pi}{2}} \left( \operatorname{arsinh} (\sin\theta) - \log \sin\theta + 1 + \frac{1 - \sqrt{1 + \sin^{2}\theta}}{\sin\theta} \right) \, d\theta. $$
We divide the integral into 4 parts and consider them separately.
Part 1. By the Taylor expansion of $\operatorname{arsinh} x$,
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \operatorname{arsinh} (\sin\theta) \, d\theta
&= \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{1}{2n+1} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \, d\theta \\
&= \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2n+1} \frac{\Gamma(\frac{1}{2})}{\Gamma(n+1)\Gamma(\frac{1}{2}-n)} \frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{2}+n)} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}}
= G.
\end{align*}
Part 2. We know so many ways to prove that
$$ -\int_{0}^{\frac{\pi}{2}} \log \sin\theta \, d\theta = \frac{\pi}{2}\log 2. $$
Part 3. Do you need an explanation for this?
$$ \int_{0}^{\frac{\pi}{2}} d\theta = \frac{\pi}{2}. $$
Part 4. Again, by the Taylor expansion,
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{1 - \sqrt{1 + \sin^{2}\theta}}{\sin\theta} \, d\theta
&= - \sum_{n=1}^{\infty} \binom{1/2}{n} \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}\theta \, d\theta \\
&= - \frac{1}{2} \sum_{n=1}^{\infty} \frac{\Gamma(\frac{3}{2})}{\Gamma(n+1)\Gamma(\frac{3}{2}-n)} \frac{\Gamma(n)\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2}+n)} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)(2n-1)}
= \frac{\log 2}{2} - \frac{\pi}{4}.
\end{align*}
Putting all these together, we obtain the answer as Cleo claimed.
| {
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Prove $\lim_{x\to 3}\frac{1}{x+1}=\frac{1}{4}$ using $\epsilon$-$\delta$ definition of a limit First of all, I'd just like to say I'm fairly new to proving limits using $\epsilon-\delta$ definition, so I apologize in advance if I ask a very obvious question, or make an elementary error.
Given $\displaystyle\lim_{x\to 3}\frac{1}{x+1}=\frac{1}{4}$, I want to prove it using the $\epsilon-\delta$ definition of a limit:
The definition: $\forall\epsilon>0, \exists\delta>0\ni0<|x-c|<\delta\implies|f(x)-L|<\epsilon$
The definition applied to my example: $\forall\epsilon >0, \exists\delta >0\ni0<|x-3|<\delta\implies\left|\dfrac{1}{x+1}-\dfrac{1}{4}\right|<\epsilon$.
$\text{Scratch Work For Determining}\space \delta$:
$$\left|\dfrac{1}{x+1}-\dfrac{1}{4}\right|=\left|\dfrac{4-(x+1)}{(x+1)\cdot4}\right|=\left|\dfrac{4-x-1}{(x+1)\cdot 4}\right |=\left|\dfrac{3-x}{(x+1)\cdot 4}\right|=\dfrac{|x-3|}{|4(x+1)|}<\epsilon$$
I can let the latter inequality be $|x-3|<\epsilon\cdot|4(x+1)|=4\epsilon\cdot|(x+1)|$, and then try to bound it:
Let $\delta=1$:
$$|x-3|<\delta=1\implies -1<x-3<1\implies 3<x+1<5$$. Plugging the value $5$ instead of $|(x+1)|$, gives the following:
$$|x-3|<4\epsilon\cdot 5=20\epsilon.$$
At this stage, I'd then say $\delta=\min\{1,20\epsilon\}$.
I was told this is not the correct answer though. Can anyone explain why it's not right?
| To maximize a fraction, we want to minimize its denominator. With this in mind, given any $\epsilon>0$, let $\delta = \min\{1,12\epsilon\}$. Then if $0 < | x-3| <\delta$, notice that:
\begin{align*}
0<|x-3|<\delta \leq 1
&\implies -1 < x-3 < 1 \\
&\implies 3 < x+1 < 5 \\
&\implies 12 < 4(x+1) < 20 \\
&\implies \frac{1}{12} > \frac{1}{4(x+1)} > \frac{1}{20} \\
&\implies \frac{1}{20} < \frac{1}{4(x+1)} < \frac{1}{12} \\
&\implies \frac{-1}{12} < \frac{1}{20} < \frac{1}{4(x+1)} < \frac{1}{12} \\
&\implies \left| \frac{1}{4(x+1)} \right| < \frac{1}{12} \\
\end{align*}
Hence, it follows that:
\begin{align*}
\left| \frac{1}{x+1} - \frac{1}{4} \right|
&= \left|\frac{1}{4(x+1)}\right||x-3| &\text{using your work}\\
&< \frac{1}{12}|x-3| & \text{from above}\\
&< \frac{1}{12}(12\epsilon) & \text{since $|x-3|<\delta \leq 12\epsilon$}\\
&= \epsilon
\end{align*}
as desired.
| {
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Solve a Linear Equation Please help to solve this problem.what is the answer of this equation. Is it any famous problem?
If $a,b,c\in\mathbb{C}$ such that
$$a+b\neq0,\quad b+c\neq0,\quad a+c\neq0\\a+b+c=-23,\quad ab+ac+bc=-324$$
Then what is the value of $x$ in the following equation?
$$\frac{x^2+2b^2+c^2}{a+b}+\frac{x^2+2c^2+a^2}{b+c}+\frac{x^2+2a^2+b^2}{a+c}=0$$
| HINT:
$$\frac{x+2b^2+c^2}{a+b}=\frac{x+a^2+b^2+c^2+b^2-a^2}{a+b}=\frac{x+a^2+b^2+c^2}{a+b}+b-a$$
and $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$$
| {
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What is the $n$ times iterate of $f(x)=\frac{x}{\sqrt{1+x^2}}$? We were asked to determine the composition $f \circ f \circ f \circ \cdots \circ f $, $n$ times, where $f(x)=\dfrac{x}{\sqrt{1+x^2}}$.
Has anyone an idea?
| Well, try to find some pattern:
$$\begin{align*}f^2(x)=f(f(x))&=f\left(\frac x{\sqrt{1+x^2}}\right)=\frac{\frac x{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac x{\sqrt{1+2x^2}}\\
f^3(x)=f(f^2(x))&=f\left(\frac x{\sqrt{1+2x^2}}\right)=\frac{\frac x{\sqrt{1+2x^2}}}{\sqrt{1+\frac{x^2}{1+2x^2}}}=\frac x{\sqrt{1+3x^2}}\ldots\ldots\end{align*}$$
Now a little induction could help here...
| {
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Using euclidean algo to find d (RSA encryption) The questions says "let p = 5, q =11, n = 55 tocient(n) = 40. e=7.
Use the Euclidean algo to find the value of d.
This is driving me crazy. Here's what I did:
40=5(7) +5
7= 1(5) + 2
5= 2(2) +1
2 = 2(1) + 0
So the GCD is 1.
Now to get 1 = de +f*tocient(n) to find d.
1 = 5 - 2(2) => 1 = 5 - 2(7-5) => 1= -2(7) + 3(5) => 1 = -2(7) +3(40 - 5(7))
=> 1 = -17(7) +3(40)
Therefore d = -17. But how can that be when in order to decode a message I have to use c^d mod n.
I don't think d can be negative but I don't know what I'm doing wrong. Any help would be great!
Regards
OSFTW
| Mathematically, it's fine for $d$ to be negative, but because positive powers don't require computing the inverse, you may wish to make it positive. To do so, add and subtract $7 \cdot 40$ from the rightmost side:
$$
\begin{align}
1 &= -17 \cdot 7 + 3 \cdot 40 \\
&= -17 \cdot 7 + \color{red}{40 \cdot 7} + 3 \cdot 40 - \color{red}{7 \cdot 40 }\\
1 &= \ \qquad 23 \cdot 7 \qquad+ \quad -4 \cdot 40
\end{align}
$$
| {
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Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}$$
I have $$\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}$$
Now, I am not able to see what $$\sum_{n=1}^{\infty} \frac{n}{(n+2)!}$$ would be.
I would be thankful if some one can help me to clear this.
Thank you.
| $$\text{From }\frac{n+1}{(n+2)!},$$
$$=\frac{n+2-1}{(n+2)!}=\frac1{(n+1)!}-\frac1{(n+2)!}$$
Can you identify the Telescoping series?
The survivor will be $$\frac1{2!}$$
| {
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$a^2+b^2+c^2 \geq 3 abc$ I have to solve this Prove that
$a^2+b^2+c^2 \geq 3 abc$
given that
$a+b+c \geq 3 abc$
I am stuck!
|
If $a,b,c\in \Bbb R$, and if $a+b+c\ge 3abc$, then: $$a^2+b^2+c^2\ge 3abc$$
Proof:
*
*Case 1:
If $abc\le 1$, then we use AM-GM inequality, and have:
$$a^2+b^2+c^2=|a|^2+|b|^2+|c|^2\ge 3\sqrt[3]{|abc|^2}\ge 3abc$$
this is true because
$$\Longleftrightarrow |abc|^2\ge (abc)^3$$
*Case 2:
If $abc\ge 1$, since: $$a+b+c\ge 3abc>0$$
we can use Cauchy-Schwarz inequality in the following way:
$$a^2+b^2+c^2\ge\dfrac{1}{3}(a+b+c)^2\ge 3a^2b^2c^2=3(|abc|)^2\ge 3abc$$
| {
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How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$ I'm having a bit of trouble solving this integral: $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$
Here is my attempt at a solution:
I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^2}}{x}dx.$$
Further simplification resulted in $$\int\frac{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}{x}dx.$$
Using trigonometric substitution, I set $$x-\frac{1}{2}=\frac{1}{2}\sin\theta$$ and solving for the differential $dx$ got $$dx=\frac{1}{2}\cos\theta.$$ Substituting this all back into $\int\frac{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}{x}dx$ (and some simplification later) yielded $$\frac{1}{2}\int{\frac{\cos^2\theta}{\sin\theta+1}}d\theta.$$ By substituting $1-\sin^2\theta$ for $\cos^2\theta$ I obtained $$\frac{1}{2}\int{\frac{1}{\sin\theta+1}-\frac{\sin^2\theta}{\sin\theta+1}d\theta}.$$ The issue I'm having is trying to solve this resultant integral. If there is an easier method to solve the problem, that would be graciously accepted.
| Integrate $\int \frac{\sqrt{1-x}}{\sqrt x} dx$
let
$$
\begin{equation} \tag 1
x=t^2, \,\,\,\,\, dx=2t\, dt
\end{equation}
$$
now; $\int \frac{\sqrt{1-t^2}}{t} dt$
$$
\begin{equation} \tag{as $dt=\frac{dx}{2t}$}
\int 2 \sqrt{1 - t^2} dt
\end{equation}
$$
$$
\begin{equation} \tag 2
2 \int \sqrt{1 - t^2} dt
\end{equation}
$$
$$
\begin{equation} \tag 3
2 \left[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}\sin^{-1} (t) \right]
\end{equation}
$$
put $t=\sqrt{x}$, to get answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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} |
Cofactor expansion method for finding the determinant of a matrix Use the determinant properties to simplify the given matrix and show that $\det(A) = (x -
y)(x - z)(x - w)(y - z)(y - w)(z - w)$ for
$$A =
\begin{pmatrix}
1 & x & x^2 & x^3 \\
1 & y & y^2 & y^3 \\
1 & z & z^2 & z^3\\
1 & w & w^2 & w^3 \\
\end{pmatrix}
$$
For this problem, I noticed that the area of interest is the second column of matrix $A$. Will using the cofactor expansion method of finding the determinant yield the desired result if I choose the second column? The second column has $x$,$y$,$z$,$w$ and what I am looking for is $\det(A) = (x-y)(x-z)(x-w)(y-z)(y-w)(z-w)$. Thanks for any help!
| Here is a non-clever approach that is still better than brute force. Subtract the bottom row from each of the preceding rows. This does not change the determinant. The result is
$$\begin{pmatrix}
0 & x-w & x^2-w^2 & x^3-w^3 \\
0 & y-w & y^2-w^2 & y^3-w^3 \\
0 & z-w & z^2-w^2 & z^3-w^3 \\
1 & w & w^2 & w^3
\end{pmatrix}$$
Expanding along the first column, find that the determinant is
$$-\det \begin{pmatrix}
x-w & x^2-w^2 & x^3-w^3 \\
y-w & y^2-w^2 & y^3-w^3 \\
z-w & z^2-w^2 & z^3-w^3
\end{pmatrix}$$
Factor out $(x-w)(y-w)(z-w)$:
$$-(x-w)(y-w)(z-w) \det \begin{pmatrix}
1 & x+w & x^2+xw+w^2 \\
1 & y+w & y^2+yw+w^2 \\
1 & z+w & z^2+zw+w^2
\end{pmatrix}$$
Again subtract the bottom row from the first two.
$$-(x-w)(y-w)(z-w) \det \begin{pmatrix}
0 & x-z & x^2-z^2+(x-z)w \\
0 & y-z & y^2-z^2+(y-z)w \\
1 & z+w & z^2+zw+w^2
\end{pmatrix}$$
This becomes a $2\times 2$ determinant, from which you can factor
$(x-z)(y-z)$:
$$-(x-w)(y-w)(z-w)(x-z)(y-z) \det \begin{pmatrix}
1 & x+z+1 \\
1 & y+z+1
\end{pmatrix}$$
and conclude.
| {
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"source": "stackexchange",
"question_score": "1",
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} |
$\left(\frac1\alpha-\frac1\beta\right)^2$ for $p(x)=x^2+x-2$ If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x)=x^2+x-2$,
then $\left(\frac1\alpha-\frac1\beta\right)^2 is:$
A) $\frac94$
B) $\frac{-9}4$
C) $\frac25$
D) $\frac{-2}5$
This is a homework question from our school's home assignment for class 10. I tried this:
We can write $p(x)=x^2+x-2$ as $$p(x)=\underbrace{(1)x^2}_{ax^2}
+\underbrace{(1)x}_{bx}+\underbrace{(2)}_c$$ Therefore, $$a=1\quad
b=1\quad c=2$$ We know that $$\alpha\beta=\frac ca=2$$ and
$$\alpha+\beta=\frac{-b}{a}=-1$$ Now $$\left(\frac1\alpha-\frac1
\beta\right)^2=\left(\frac{\alpha-\beta}{\alpha\beta}\right)^2=
\frac{(\alpha-\beta)^2}{(\alpha\beta)^2}=\frac{\alpha^2+\beta^2-2
\alpha\beta}4=\frac{\alpha^2+\beta^2-4}4$$
I got stuck here. What to do now?
| You can solve this problem in a single swoop. Since $\alpha$ and $\beta$ are the roots of the polynomial $x^2+x-2$ therefore we must have $\alpha$=-2 and $\beta$=1(or vice versa). Hence
$$\left(\frac1\alpha-\frac1\beta\right)^2=(-\frac12-1)^2$$
| {
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"source": "stackexchange",
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Proving by induction that $\sum_{i=2}^n(i^2-i) = \frac{n(n^2-1)}{3}$ for all $n \ge 2$ Doing proof by induction exercises, everything was fine until I tried to do one with $\sum$
:
Prove that
$$\sum_{i=2}^n(i^2-i) = \frac{n(n^2-1)}{3}$$
holds for all $n \ge 2$.
Now, my real problem is converting an algebraic expression to another (see bottom of question), but if you're curious here's the whole process.
Test it for $n = 2$:
$$\sum_{i=2}^2(i^2-i) = \frac{2(2^2-1)}{3}$$
$$(2^2-2) = \frac{2\cdot3}{3}$$
$$2 = \frac{6}{3}$$
We have to prove that it holds for $n + 1$, that is:
$$\sum_{i=2}^{n+1}(i^2-i) = \frac{(n+1)((n+1)^2-1)}{3}$$
Assume
$$\sum_{i=2}^n(i^2-i) = \frac{n(n^2-1)}{3}$$
Prove it. Have:
$$\sum_{i=2}^{n+1}(i^2-i)$$
Okay so, I normally attempt to use the hypothesis to make some replacement, but to do so I need to expand a bit this sum. I will try separating the last element in the sum, which I think would be like this:
$$\sum_{i=2}^{n}(i^2-i) + \color{blue}{((n+1)^2-(n+1))}$$
Because, it seems to me, that in the last element of the sum, the $i$ would be $(n+1)$.
Now I can make the replacement with the hypothesis that I wanted:
$$\color{blue}{\frac{n(n^2-1)}{3}} + ((n+1)^2-(n+1))$$
$$\frac{n(n^2-1)+3((n+1)^2-(n+1))}{3}$$
Not sure. I'll just try expanding that quadratic.
$$\frac{n(n^2-1)+3(n^2+2n+1-n-1)}{3}$$
$$\frac{n(n^2-1)+3(n^2+n)}{3}$$
$$\frac{n(n^2-1)+(3n^2+3n)}{3}$$
$$\frac{n^3-n+3n^2+3n}{3}$$
$$\frac{n^3+3n^2+2n}{3}$$
$$\frac{n^3+3n^2 + 2n}{3}$$
$$\frac{n^2(n+3)+2n}{3}$$
It does seem to me like this last expression is indeed equivalent to what I want to prove $\frac{(n+1)((n+1)^2-1)}{3}$, but I can't seem to figure how to transform it into that. How can I proceed?
| MY approach:
first show that $\sum^n i = \frac{n(n+1)}{2} $ and $\sum^n i^2 = \frac{n(n+1)(2n +1 )}{6} $.
To show then, it is much easier the induction. So, therefore
$$ \sum^n (i^2 - i) = \sum^n i^2 - \sum^n i = \frac{n(n+1)(2n +1 )}{6} - \frac{n(n+1)}{2} = $$
$$ = \frac{n(n+1)(2n-1) - 3n(n+1)}{6} = \frac{(n+1)(2n^2 + n - 3n )}{6} = \frac{(n+1)(2n^2 - 2n)}{6} =$$
$$ = \frac{(n+1)(n^2 - n)}{3} = \frac{n(n+1)(n-1)}{3} = \frac{n(n^2-1)}{3}$$
| {
"language": "en",
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"answer_count": 2,
"answer_id": 1
} |
Writing $\frac{(\sqrt{2}+1)^{2n+1}+(\sqrt{2}-1)^{2n+1}}{2\sqrt{2}}, n\geq2$ as sum of two perfect squares I tried to show that
$$
{\left(\sqrt{2\,} + 1\right)^{2n+1} + \left(\sqrt{2\,} - 1\right)^{2n+1}
\over
2\,\sqrt{2\,}}\,,\qquad n\geq2
$$
is written as the sum of two perfect squares. We used Newton's binomial formula and we did. Is there another way ?. Thanks you !.
| Let us look at a quadratic whose roots are
$$(\sqrt2 + 1)^2 \text{ and }(\sqrt2 - 1)^2$$
i.e.,
$$3+2\sqrt2 \text{ and }3 - 2\sqrt2$$
The quadratic is $x^2-6x+1$. Now let us look at the recurrence
$$a_{n+1} = 6a_n - a_{n-1}$$
where $a_0 = 1$ and $a_1 = 5$. This gives the sequence you are after. Now note that
$$a_0 = 0^2 + 1^2$$
$$a_1 = 1^2 + 2^2$$
$$a_2 = 2^2 + 5^2$$
$$a_3 = 5^2 + 12^2$$
$$a_4 = 12^2 + 29^2$$
Hence, it looks like if $a_{n-1} = x^2 + y^2$, then $a_n = y^2 + (2y+x)^2$. Prove this is the case using the recurrence and induction. The identity
$$(2(2y+x) + y)^2 + (2y+x)^2 = 6(y^2 + (2y+x)^2) - (x^2+y^2)$$ will be helpful in the process.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating the integral $\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$ As suggested in the title, I am given the following integral:
$$\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$$
I have tried to solve it several times, but I wind up with the wrong answer, although its close. I assume there is a blunder somewhere.
Applying long division, we have that
$$\frac{2x^3+x^2+1}{x^2+x-2} = 2x - 1 + \frac{-3x-1}{x^2+x-2}$$
I note now that I may write this as
$$\frac{2x^3+x^2+1}{x^2+x-2} = 2x - 1 + \frac{(-1)3x+1}{x^2+x-2} = 2x - 1 - \frac{3x+1}{x^2+x-2}$$
I did not notice this while doing the task originally, but I am having a hard time seeing how this will make answer different. (After calculating it on paper: it didn't.)
Thus, we have the integral
$$\int 2x - 1 \space dx - \int \frac{3x+1}{x^2+x-2}\space dx$$
The first integral is trivial. For the second one, we have that
$$\frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} = \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}$$
We have the set $A + B = 3, \space 2A - B = 1$.This gives us the solutions $A = \frac{4}{3}$ and $B = \frac{5}{3}$
Thus, we are left with
$x^2 - x - \int \frac{3x+1}{x^2+x-2}\space dx = x^2 - x - \int \frac{4/3}{x-1} + \frac{5/3}{x+2} \space dx = x^2 - x - \frac{4}{3}\ln{|x-1|}-\frac{5}{3}\ln{|x+2|} + C$
I'm having a hard time finding my mistake here.
EDIT: Seems like I made my mistake in the long division part, confirming now.
| You're off at the start, with your long division: in particular, the numerator of the remainder needs to be $5x - 1$.
$$\frac{2x^3+x^2+0\cdot x +1}{x^2+x-2} = 2x - 1 + \frac{5x - 1}{x^2+x-2} = 2x - 1 + \frac{5x-1}{(x+2)(x-1)}$$
Procedurally, your approach is just fine: once you correct for this mistake, you should obtain the correct result.
Note also that in the logarithm components of your answer, we can "consolidate":
$$a\ln |f(x)| + b\ln |g(x)| = \ln|(f(x))^a| + \ln|(g(x))^b| = \ln|(f(x))^a\cdot (g(x))^b|$$
and$$a\ln |f(x)| - b\ln |g(x)| = \ln|(f(x))^a| - \ln|(g(x))^b| = \ln\left|\dfrac{(f(x))^a}{(g(x))^b}\right|$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Generating function for division of $n$ into smaller subsets. I need to find the generating function for the number of ways of dividing $n$ into parts out of even numbers. The ways are only different if their parts are different, meaning that $2+1+2$ and $2+2+1$ are the same, because only their arrangmenent is different.
If we would like all of them to be distinct, then we would have $F(x)=(1+x^2)(1+x^4)(1+x^6)...$, but what if some can be repeated? Is it $\prod\limits_{n=0}^{\infty}(1+x^2+x^4+x^6...)$?
| If the parts can be repeated, this generating function can be written as
\begin{align}
F(x) & = (1 + x^{2} + x^{2+2} + \ldots)\cdot (1 + x^{4}+x^{4+4}+\ldots)\cdot (1 + x^{6}+x^{6+6}+\ldots)\cdot \ldots\\[0.7em]
& = \dfrac{1}{1-x^{2}}\cdot \dfrac{1}{1-x^{4}} \cdot \dfrac{1}{1-x^{6}} \cdot \ldots\\[0.9em]
& = \prod_{n=1}^{\infty}\dfrac{1}{1-x^{2n}}
\end{align}
| {
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"source": "stackexchange",
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Can Master Theorem be applied on any of these? 1) $T(n) = 6T(n/2) + 2^{3 \log(n)}$
2) $T(n) = 8T(n/2) + \frac{n^3}{(\log(n))^4}$
3) $T(n) = 9T(n/3) + n(\log(n))^3$
Can the complexity for these be calculated with the Master Theorem?
I am not sure how to decide upon which case they fit in.
| For problem three, lets solve the recurrence
$$T(n) = 9 T(\lfloor n/3\rfloor) + n \times (1+\lfloor\log_3 n\rfloor)^3$$
where we set $T(0)=0.$
Let the base three representation of $n$ be
$$n = \sum_{k=0}^{\lfloor\log_3 n\rfloor} d_k 3^k.$$
Then we get the exact formula
$$T(n) = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 9^j \times
(1+\lfloor\log_3 n\rfloor- j)^3 \times
\sum_{k=j}^{\lfloor\log_3 n\rfloor} d_k 3^{k-j}
\\ = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times
(1+\lfloor\log_3 n\rfloor - j)^3 \times
\sum_{k=j}^{\lfloor\log_3 n\rfloor} d_k 3^k .$$
Now for an upper bound consider a string of two digits, which gives
$$T(n)\le \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times
(1+\lfloor\log_3 n\rfloor - j)^3 \times 2 \times
\sum_{k=j}^{\lfloor\log_3 n\rfloor} 3^k
\\ = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times
(1+\lfloor\log_3 n\rfloor - j)^3 \left(3^{\lfloor\log_3 n\rfloor + 1}-3^j\right)
\\ = \sum_{j=1}^{\lfloor\log_3 n\rfloor+1}
3^{\lfloor\log_3 n\rfloor+1-j} \times j^3 \times
\left(3^{\lfloor\log_3 n\rfloor + 1}-3^{\lfloor\log_3 n\rfloor + 1- j}\right)
\\ = 3^{2(\lfloor\log_3 n\rfloor + 1)}
\sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 3^{-j} j^3 (1- 3^{-j}).$$
Now the sum term converges to a constant and we get the upper bound
$$\frac{7917}{2048} \times 3^{2(\lfloor\log_3 n\rfloor + 1)}.$$
For the lower bound take a one followed by a string of zeros, which gives
$$T(n)\ge \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times
(1+\lfloor\log_3 n\rfloor - j)^3 \times 3^{\lfloor\log_3 n\rfloor}
\\ = 3^{\lfloor\log_3 n\rfloor}
\sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 3^{\lfloor\log_3 n\rfloor +1 -j} \times
j^3
= 3^{2\lfloor\log_3 n\rfloor + 1} \sum_{j=1}^{\lfloor\log_3 n\rfloor+1} j^3 3^{-j}.$$
The sum term again converges to a constant and we get for the lower bound asymptotics the formula
$$\frac{33}{8} 3^{2\lfloor\log_3 n\rfloor + 1} .$$
Note that for the upper bound we have $\lfloor\log_3 n\rfloor+1\to\log_3 n,$ so that it is in fact $$\frac{7917}{2048} n^2 \approx 3.86572265625\times n^2$$ but for the lower bound $\lfloor\log_3 n\rfloor = \log_3 n$, so that it is $$\frac{33}{8} \times 3 \times n^2 = \frac{99}{8} n^2 \approx 12.375 \times n^2.$$
Joining the two bounds we may conclude that $$T(n)\in
\Theta\left(3^{2\lfloor\log_3 n\rfloor}\right) = \Theta(n^2).$$
A similar computation which is a bit simpler can be used to analyse the cost of Strassen matrix multiplication.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$
my solution:
let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$
then
$$t(\sqrt{x+1}-\sqrt{x})=1$$
$$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$
$(1)-(2)$ we have
$$2\sqrt{x}=t-\dfrac{1}{t}\Longrightarrow x=\dfrac{1}{4}(t-\dfrac{1}{t})^2$$
so
$$dx=\dfrac{1}{2}(t-\dfrac{1}{t})(1+\dfrac{1}{t^2})dt=\dfrac{t^4-1}{2t^3}dt$$
$$I=\int\dfrac{1}{1+t}\cdot\dfrac{t^4-1}{2t^3}dt=\dfrac{1}{2}\int\left(1+\dfrac{1}{t}+\dfrac{1}{t^2}+\dfrac{1}{t^3}\right)dt=\dfrac{1}{2}\left(t+\ln{t}-\dfrac{1}{t}-\dfrac{1}{2t^2}+C\right)$$
so
$$I=\dfrac{1}{2}\left(\sqrt{x}+\sqrt{x+1}+\ln{(\sqrt{x}+\sqrt{x+1})}-\dfrac{1}{\sqrt{x}+\sqrt{x+1}}-\dfrac{1}{2(\sqrt{x}+\sqrt{x+1})^2}+C\right)$$
My question: have other methods? Thank you very much
| How about: substitute $x=u^2.$ Then (dropping absolute values for now), you get
$$\int \frac{2 u}{1+u+ \sqrt{u^2+1}} du$$
Now, substitute $u=\tan \theta,$ to get
$$2 \int \frac{\tan \theta \sec^2 \theta}{1+\tan \theta + \sec \theta} d\theta =
\int \frac{\sin \theta}{\cos^2\theta (\sin \theta + \cos \theta + 1)} d\theta.$$
Now, make the substitution $t = \tan \frac{\theta}2,$ and you have a rational function.
| {
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"url": "https://math.stackexchange.com/questions/596467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Is $\lim_{n \to \infty}{\left(\frac{3}{1+\frac{1}{n}}\right)^n} = \lim_{n \to \infty}{3^n}$? I started out with the following series:
$\sum_{n=1}^{\infty}{\left(\frac{3}{n}\right)^nn!}$
The goal is to find out if the sequence converges or diverges. It went quite well, hopefully, until I got stuck here:
$\lim_{n \to \infty}{\left(\frac{3}{1+\frac{1}{n}}\right)^n} = \lim_{n \to \infty}{3^n}$
The problem is that I cannot figure out if this is a valid equality. Based on the property of multiplication of limits, i would assume it is, but the fact that it is raised to $n$ makes me wonder.
| Using the Ratio Test (expounding):
$$\frac{a_{n+1}}{a_n}= \frac{3^{n+1} \cdot (n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n \cdot n!} = \frac{3^n \cdot 3 \cdot(n+1)n! \cdot n^n}{(n+1)^n \cdot (n+1)^1 \cdot 3^n \cdot n!} = \frac{3 \cdot n^n}{(n+1)^n} = 3 \cdot \left(\frac{n}{n+1} \right)^n$$ $$= 3 \cdot \left(\frac{1}{1+1/n} \right)^n = \frac{3}{(1+1/n)^n}$$
$$\lim_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$$
Since the $L > 1 \implies$ the series diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many integers between $100$ and $500$ are divisibly by both $6$ and $12$, but not by $9$? The numbers divisible by both $6$ and $12$ and just the numbers divisible by $12$, so:
$12 \times 9 = 108$
$12 \times 41 = 492$
$\frac {492-108}{12} = 32 + 1 = 33$ numbers divisible by $12$, and in turn div. by $6$
$\frac {495-108}{9}+1 = 44$ numbers divisible by $9$
So there are $500-44 = 466$ numbers not divisible by $9$
Here is where I am stuck. Any help?
| The number of integers divisible by $12$ between $[100,500]$ is $$N_1=\left\lfloor\frac{500}{12}\right\rfloor-\left\lfloor\frac{100-1}{12}\right\rfloor$$
As lcm$(12,9)=18,$ among these $N_1$ numbers, the number of integers that are also divisible by $9$ between $[100,500]$ will be $$N_2=\left\lfloor\frac{500}{18}\right\rfloor-\left\lfloor\frac{100-1}{18}\right\rfloor$$
So, the required number of integers will be $N_1-N_2$
| {
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Find value of integral: $I=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$ Find value of integral: $$I_1=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$$ and $$I_2=\int_0^{2\pi}\frac{dx}{(2+\sin x)^2}$$
I don't know how, i need a solution, please
| We can also use contour integration.
Let $z=e^{ix}$, then $\mathrm{d}x=-i\,\mathrm{d}z/z$ and $\cos(x)=\frac{z+1/z}{2}$
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}x}{(2+\cos(x))^2}
&=\oint\frac{-i\,\mathrm{d}{z}/z}{\left(2+\frac{z+1/z}{2}\right)^2}\\
&=\oint\frac{-4iz\,\mathrm{d}{z}}{(z^2+4z+1)^2}\tag{1}
\end{align}
$$
The zeros of $z^2+4z+1$ are $-2\pm\sqrt3$. $-2-\sqrt3$ is outside the unit circle, so the only pole that matters is at $-2+\sqrt3$.
Partial fractions gives
$$
\frac{z}{(z-a)^2(z-b)^2}=\frac{\frac{a+b}{(b-a)^3}}{z-a}+\frac{\frac{b+a}{(a-b)^3}}{z-b}+\frac{\frac{a}{(a-b)^2}}{(z-a)^2}+\frac{\frac{b}{(b-a)^2}}{(z-b)^2}\tag{2}
$$
With $a=-2+\sqrt3$ and $b=-2-\sqrt3$, we get the residue of $(2)$ at $z=a$ to be $\frac1{6\sqrt3}$. Plugging this into $(1)$ gives
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}x}{(2+\cos(x))^2}
&=(2\pi i)(-4i)\frac1{6\sqrt3}\\
&=\frac{4\pi}{3\sqrt3}\tag{3}
\end{align}
$$
My original inclination was to use Igor's substitution, just to verify the previous answer, I will compute using the substitution $z=\tan(x/2)$, where $\mathrm{d}x=\frac{2\,\mathrm{d}z}{1+z^2}$ and $\cos(x)=\frac{1-z^2}{1+z^2}$ :
$$
\begin{align}
&\int_0^{2\pi}\frac{\mathrm{d}x}{(2+\cos(x))^2}\\
&=\int_{-\infty}^\infty\frac{\frac{2\,\mathrm{d}z}{1+z^2}}{\left(2+\frac{1-z^2}{1+z^2}\right)^2}\\
&=\int_{-\infty}^\infty\frac{2(1+z^2)\,\mathrm{d}z}{\left(3+z^2\right)^2}\\
&=\int_{-\infty}^\infty\frac{2\,\mathrm{d}z}{3+z^2}-\int_{-\infty}^\infty\frac{4\,\mathrm{d}z}{\left(3+z^2\right)^2}\\
&=\frac2{\sqrt3}\left[\tan^{-1}\left(\frac{z}{\sqrt3}\right)\right]_{-\infty}^\infty-\frac2{3\sqrt3}\left[\tan^{-1}\left(\frac{z}{\sqrt3}\right)+\frac{\sqrt3z}{3+z^2}\right]_{-\infty}^\infty\\
&=\frac{4\pi}{3\sqrt3}\tag{4}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$2^5 \cdot a^b=2,5ab$ I came across this problem in an elementary number theory book, and I think I solved it. Well, the question is posed as
$2^5 \cdot 9^2 = 2,592$. Are there any other pairs $a,b \in \mathbb{Z}$ such that $2^5 \cdot a^b = 2,5ab$ is satisfied ($a,b$ are the tens and one representative, not a product)?
So at first I just set up a $9\times9$ grid and sieved the answer (which is "No") mainly by eliminating such obvious answers as $a\ne1,2$, and $a^b\le82$. Then I quickly realized that $2^5|25ab$, and there are only a select few integers where $32\cdot x=25ab$, for some integers $a,b,x$. For example $32 \cdot 81 = 2,592 = 2^5 \cdot 9^2$, but this does not work for $32 \cdot 80=2,560 \neq 2^5 \cdot 6^0$ and $32 \cdot 79 = 2,528 \neq 2^5 \cdot 2^8$
So i think that, no, this is not possible except where $a=9, b=2$.
Now that I think i solved it, is there an algorithm for doing this quickly or does it require this sort of analysis for all $a^b \cdot c^d=a,bcd$?
| There is a 'simple' way of doing this. Notice that $2^5\cdot a^b=25ab$ are both integers. Therefore, since $a^b$ is an integer, so too must be
$$
a^b=\frac{25ab}{2^5}=\frac{25ab}{32}
$$
Now this can be done by hand very simply. We know that $2500<a^b<2600$. How many multiples of $32$ fit this inequality? This is simple to check by hand, the only possibilities are $32\cdot 79=2528$, $32\cdot 80=2560$, and $32\cdot 81=2592$. Now just factor the numbers:
$$
2528=2^5\cdot79^1\;\;,\;\;2560=2^9\cdot5^1\;\;,\;\;2592=2^5\cdot 3^4
$$
factoring out a $2^5$ yields the choices $79^1$, $5^1$, and $3^4$ for the value of $a^b$. Since $0\leq a \leq 9$, $79^1$ is eliminated. It is easy to check that $5^1$ cannot work for then $a=5$ and $b=1$ and the $2^5\cdot 5^1=160 \neq 2551$.
This leaves $a^b=3^4$. We can also write this $a^b=(3^2)^2=9^2$ or $a^b=81^1$. Since $a,b$ are integers with $0\leq a \leq 9$, either $a=3$ and $b=4$ or $a=9$ and $b=2$. We know the last to be true and the case of $a=3$ and $b=4$ is a simple matter to show doesn't work. This method does not require a computer or calculator and took only about $5$ minutes using simple pen and paper.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/606584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate: $\int \frac{1}{(x+a)(x+b)}$ Evaluate:
$$\int \frac{1}{(x+a)(x+b)}$$
My attempt:
$$\int \frac{1}{(x+a)(x+b)} = \frac{A}{x+a} + \frac{B}{x+b}$$
$$1 = A(x+b) + B(x+a)$$
$$x = -b$$
$$1 = A(-b + b) + B(-b + a)$$
$$1 = B(-b + a)$$
$$x = -a$$
$$1 = A(-a + b) + B(-a + a)$$
$$1 = A(-a + b)$$
At this point I have no idea how to proceed. Can someone help me with this? Please.
| $$\displaystyle {1\over (x+a)(x+b)}={1\over (b-a)}\left[{1\over x+a}-{1\over x+b}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Conic matrix and diagonalization If I have the conic $C$:
$$
5x^2 - 4xy + 8y^2 = 36
$$
How would I express it as a matrix of the form:
$$
\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}a&b/2\\b/2&c\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = k
$$
Also, how to find a unitary matrix $P$ such that
$$
P^* \begin{bmatrix}a&b/2\\b/2&c\end{bmatrix}P
$$
is diagonal.
Thanks for your help.
| Let
$$
A\equiv\left[\begin{array}{cc}
a & \frac{1}{2}b\\
\frac{1}{2}b & c
\end{array}\right]
$$
so that
\begin{align*}
\left[\begin{array}{cc}
x & y\end{array}\right]A\left[\begin{array}{c}
x\\
y
\end{array}\right] & =\left[\begin{array}{cc}
x & y\end{array}\right]\left[\begin{array}{cc}
a & \frac{1}{2}b\\
\frac{1}{2}b & c
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]\\
& =\left[\begin{array}{cc}
x & y\end{array}\right]\left[\begin{array}{c}
ax+\frac{1}{2}by\\
\frac{1}{2}bx+cy
\end{array}\right]\\
& =ax^{2}+\frac{1}{2}byx+\frac{1}{2}bxy+cy^{2}\\
& =ax^{2}+bxy+cy^{2}.
\end{align*}
Therefore,$a=5$, $b=-4$ and $c=8$. Naturally, $k=36$. This gives us
$$
A\equiv\left[\begin{array}{cc}
a & \frac{1}{2}b\\
\frac{1}{2}b & c
\end{array}\right]=\left[\begin{array}{cc}
5 & -2\\
-2 & 8
\end{array}\right].
$$
We want to diagonalize $A$. Note that the eigenvalues of $A$ are $\lambda_{1}=4$ and $\lambda_{2}=9$ (you can verify this). Constructing $P$ by placing the corresponding eigenvectors in the columns,
$$
P\approx\left[\begin{array}{cc}
-0.89443 & -0.44721\\
-0.44721 & 0.89443
\end{array}\right].
$$
You can verify that
$$
P^{\star}AP\approx\left[\begin{array}{cc}
\lambda_{1}\\
& \lambda_{2}
\end{array}\right].
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a weird recurrence relation using exponential generating functions and differential equations So the recurrence relation is:
$a(n) = (n-1)\cdot a(n-2) + (n-1)(n-2)\cdot a(n-3)$
I've tried several things, but the instructions are to use a differential equation and nothing I'm doing seems to work. Here's what I have so far:
$a(n) = (n-1)\cdot a(n-2) + (n-1)(n-2)\cdot a(n-3)$
$a(n)\frac{x^n}{n!} = (n-1)\cdot a(n-2)\frac{x^n}{n!} + (n-1)(n-2)\cdot a(n-3)\frac{x^n}{n!}$
$\sum \limits_{n\ge 0}a(n)\frac{x^n}{n!} = \sum\limits_{n\ge 0}(n-1)\cdot a(n-2)\frac{x^n}{n!} + \sum\limits_{n\ge 0}(n-1)(n-2)\cdot a(n-3)\frac{x^n}{n!}$
$A(x)= \sum\limits_{n\ge 0}(n-1)\cdot a(n-2)\frac{x^n}{n!} + \sum\limits_{n\ge 0}(n-1)(n-2)\cdot a(n-3)\frac{x^n}{n!}$
but now I'm stuck. Can anyone offer any advice?
| Let $A(x)$ be the exponential generating function. For convenience I’ll write $a_n$ instead of $a(n)$.
$$\begin{align*}
A(x)&=\sum_{n\ge 0}a_n\frac{x^n}{n!}\\\\
&=\sum_{n\ge 0}(n-1)a_{n-2}\frac{x^n}{n!}+\sum_{n\ge 0}(n-1)(n-2)a_{n-3}\frac{x^n}{n!}\\\\
&=x^2\sum_{n\ge 0}\frac{(n-1)a_{n-2}}{n(n-1)}\cdot\frac{x^{n-2}}{(n-2)!}+x^3\sum_{n\ge 0}\frac{(n-1)(n-2)a_{n-3}}{n(n-1)(n-2)}\cdot\frac{x^{n-3}}{(n-3)!}\\\\
&=x^2\sum_{n\ge 0}\frac{a_n}{n+2}\cdot\frac{x^n}{n!}+x^3\sum_{n\ge 0}\frac{a_n}{n+3}\cdot\frac{x^n}{n!}\\\\
&=\sum_{n\ge 0}\frac{a_n}{n+2}\cdot\frac{x^{n+2}}{n!}+\sum_{n\ge 0}\frac{a_n}{n+3}\cdot\frac{x^{n+3}}{n!}
\end{align*}$$
Differentiating the righthand side with respect to $x$ yields
$$\sum_{n\ge 0}a_n\frac{x^{n+1}}{n!}+\sum_{n\ge 0}a_n\frac{x^{n+2}}{n!}=(x+x^2)A(x)\;,$$
so $A'(x)=(x+x^2)A(x)$. There’s your differential equation, and I’ll leave the rest to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/610037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Infinite Series $\sum\limits_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}$ How can we prove that?
$$\sum_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}=\frac{1}{3}e^{\frac{-x}{2}}x\left(e^{\frac{3x}{2}}-2\sin\left(\frac{\pi+3\sqrt{3}x}{6}\right)\right).$$
I think if we write the taylor expansion of $\sin(u)$ and $e^u$, we can arrive from RHS to LHS, but I am looking for a way to prove it from LHS.
| This is actually calling for Laplace transform.
Set $f(x)=\sum_{n=1}^\infty \frac{x^{3n}}{(3n-1)!}$ then
\begin{align}
F(s) &= \int_0^\infty \frac{f(x)}{x} \, e^{-sx} \, {\rm d}x \\
&= \sum_{n=1}^\infty \frac{1}{s^{3n}} \int_0^\infty \frac{u^{3n-1} \, e^{-u}}{(3n-1)!} \, {\rm d}u \\
&= \frac{1}{s^3-1}
\end{align}
Now
\begin{align}
\frac{f(x)}{x} = \frac{1}{2\pi i} \int_{\gamma -i\infty}^{\gamma + i\infty} F(s) \, e^{xs} \, {\rm d}s
\end{align}
where $\gamma > 1$. The contour can be closed along the half-circle to the left and using the residue-theorem you will get
\begin{align}
\frac{f(x)}{x} &= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}} + e^{\frac{i\sqrt{3}x}{2}+\frac{2\pi i}{3}} + e^{-\frac{i\sqrt{3}x}{2}-\frac{2\pi i}{3}} \right) \\
&= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}} + 2\cos\left(\frac{\sqrt{3}x}{2}+\frac{2\pi}{3}\right) \right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/610526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Taylor series of $(1+x)\ln(1+x)$ in $x=0$ How to determine the Taylor series of $(1+x)\ln(1+x)$ in $x=0$?
My idea is finding the second derivative of the expression, which is $\frac{1}{1+x}$.
The Taylor series of this expression is $1-x+x^2-x^3$ and so on. If I integrate then this series I get $\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}$ and so on. But the solution is $x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}$. My question is, how does the $x$ at the beginning of the series appear?
| You might want to use the fact that
$$
\frac{\mathrm{d}^2}{\mathrm{d}x^2} \left( (1+x) \ln(1+x) \right) =
\frac{\mathrm{d}}{\mathrm{d}x} \left( \ln(1+x) + 1 \right) = \frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n
$$
Integrating term-wise twice, and using initial conditions of 0 for the function, and 1 for the first derivative:
$$
(1+x) \ln(1+x) = x + \sum_{n=0}^\infty \frac{(-1)^n x^{n+2}}{(n+1)(n+2)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/612166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n>0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$
This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$
I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$
It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$
or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$
So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here).
| Writing ${\displaystyle {x_i \over 2 - x_i} = -1 + {2 \over 2 - x_i}}$, you have
$$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = -n + \frac{2}{2-x_1} + \frac{2}{2-x_2} +\ldots + \frac{2}{2-x_n}$$
So it suffices to show
$$ \frac{2}{2-x_1} + \frac{2}{2-x_2} +\ldots + \frac{2}{2-x_n} \geq n + \frac{n}{2n-1}$$
This is the same as
$$\frac{1}{2-x_1} + \frac{1}{2-x_2} +\ldots + \frac{1}{2-x_n} \geq \frac{n^2}{2n - 1}$$
Letting $y_i = 2 - x_i$, this is the same as showing
$$\frac{1}{y_1} + \frac{1}{y_2} +\ldots + \frac{1}{y_n} \geq \frac{n^2}{2n - 1}$$
The condition $\sum_i x_i = 1$ translates into $\sum_i y_i = 2n - 1$.
By the arithmetic-harmonic mean inequality,
$$n\bigg(\frac{1}{y_1} + \frac{1}{y_2} +\ldots + \frac{1}{y_n}\bigg)^{-1} \leq {1 \over n}\sum_i y_i $$
$$= {2n - 1 \over n}$$
This is equivalent to what you want to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Evaluating an Infinite Product Does anyone know how to evaluate the infinite product
$$
\left(1 - \frac{4}{1}\right) \prod_{k = 3}^{\infty} \left( 1 - \frac{4}{k^2} \right)
$$
| For $k \ge 3$:
$$\prod_{k=3}^{\infty} \left ( 1-\frac{4}{k^2}\right )$$
This is equal to
$$\frac{1\cdot 5}{3 \cdot 3} \frac{2\cdot 6}{4 \cdot 4} \frac{3\cdot 7}{5 \cdot 5} \cdots$$
With cancellations: note that only $1\cdot 2$ survives in the numerator, and a single $3 \cdot 4$ survives in the denominator. Thus, the product is $1/6$. The front factor produces a $-3$, so the stated product is $-1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Asymptotics of probability of Newton-Pepys problem In Newton-Pepys problem one is interested in probability $p_n$ of getting at least $n$ sixes in $6 n$ independent throws of regular 6-sided die.
The number of sixes $S_m$ obtained in $m$ throws follows a Binomial distribution $\operatorname{Bin}(m,p)$, where $p=\frac{1}{6}$ is the probability of getting a six in a single throw. Thus
$$
p_n = \Pr\left(S_{6n} \geqslant n\right)
$$
Notice that $\mathsf{E}(S_{6n}) = 6 n p = n$, and variance $\mathbb{Var}(S_{6n}) = \sqrt{5 n/6}$, hence in the large $n$ limit
$$
\lim_{n \to \infty} p_n = \lim_{n \to \infty} \Pr\left(\frac{S_{6n}-n}{\sqrt{5 n/6} }\geqslant 0\right) \stackrel{\text{CLT}}{=} \Pr(Z \geqslant 0) = \frac{1}{2}
$$
In fact $p_n$ is monotonically decreasing sequence:
Q.: How can one find the large $n$ asymptotics of $p_n$?
| The probability $p_n$ has a representation in terms of regularized beta-function, which incidentally is the cumulative distribution function of beta distribution.
$$
p_n = \Pr(S_{6n} \geqslant n) = I_{p}\left(n, 5n+1\right) = \Pr(X < p ) = \Pr\left(X < \frac{1}{6} \right))
$$
where $X$ follows $\mathcal{Be}\left(n,5n+1\right)$. It is well-known that $X \stackrel{d} = \frac{Y}{Y+Z}$ where $Y$ and $Z$ are independent random variables following gamma-distirbutions $\Gamma(n,1)$ and $\Gamma(5n+1,1)$ respectively.
$$ \begin{eqnarray}
p_n &=& \Pr\left(\frac{Y}{Y+Z} < \frac{1}{6}\right) = \Pr(5 Y < Z) = \Pr\left(5 (Y-n) < Z -5 n\right) \\ &=& \Pr\left(5 \frac{Y-n}{\sqrt{n}} < \frac{Z -5 n}{\sqrt{n}}\right) = \Pr\left(5 \frac{Y-n}{\sqrt{n}} < \frac{Z -5 n-1}{\sqrt{n}} + \frac{1}{\sqrt{n}}\right) \\
&=& \Pr\left(5 \frac{Y-n}{\sqrt{n}} < \frac{Z -5 n-1}{\sqrt{5n+1}} \sqrt{5 + \frac{1}{n}} + \frac{1}{\sqrt{n}}\right)
\end{eqnarray}
$$
Since mean and standard deviation of $Y$ equal $n$ and $\sqrt{n}$ and mean and standard deviation of $Z$ equal $5n+1$ and $\sqrt{5n+1}$.
Approximating $\frac{Y-n}{\sqrt{n}}$ and $\frac{Z -5 n-1}{\sqrt{5n+1}}$ with iid standard normal random variables:
$$
p_n \approx \Pr\left(5 Z_1 - Z_2 \sqrt{5+\frac{1}{n}} < \frac{1}{\sqrt{n}}\right) = \Phi\left(\frac{1}{\sqrt{30 n + 1}}\right)
$$
Unfortunately, the above approximation is not very accurate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry equation $\sin(x)+\cos(x)-\tan(x)=0.4$ There's some way to find $x$ here ?
$$\sin(x)+\cos(x)-\tan(x)=0.4$$
| This is a difficult problem to solve. If you write $\tan x$ as the quotient of $\sin x$ and $\cos x$ you get
$$\sin x + \cos x + \frac{\sin x}{\cos x} = \frac{2}{5}$$
It is possible to rearrange this formula, to get $\sin x$ as the subject:
$$\sin x = \frac{(2-5\cos x)\cos x}{5+5\cos x}$$
Squaring both sides, and using the fact that $\cos^2x+\sin^2x \equiv 1$, gives:
$$1-\cos^2x = \frac{(2-5\cos x)^2\cos^2 x}{25+25\cos^2 x}$$
We can cross multiply, and give ourselves a quartic equation to solve:
$$50\cos^4x + 30\cos^3x + 4\cos^2x - 50\cos x - 25 = 0 $$
We can perform as similar procedure and get ourselves the quartic equation:
$$50\sin^4x-70\sin^3x-46\sin^2x+70\sin x+21=0$$
Each of these quartics, when counted with multiplicity, will have exactly four solutions. Each of the four solutions will give four equations, e.g. $\cos x = u_1$, $\cos x = u_2$, $\cos x = u_3$ and $\cos x = u_4$. Similarly, $\sin x = v_1$, $\sin x= v_2$, $\sin x = v_3$ and $\sin x = v_4$. In each case, the $x$ values that solve your original problem will solve some of the $\cos x = u_i$ and some of the $\sin x = v_j$. The method I used has given us "If $x$ solves your equation then $\cos x$ and $\sin x$ will solve the two resulting quartics." The converse need not be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/617464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
integer $a$ for which $x^3-3x+a = 0$ has $3$ distinct real roots in $x\in (0,1)$ If the equation $x^3-3x+a=0$ has $3$ distinct roots between $0$ and $1$ then the integer value of $a$ is
Can we solve it without Calculus.
$\bf{My\; Try}::$ Let $x=2\cos \phi$, where $-1\leq \cos \phi\leq 1\Rightarrow 0\leq \phi \leq \pi$(Principle value, Inverse exists)
$\displaystyle 8\cos^3 \phi-6\cos \phi+a = 0\Rightarrow 4\cos^3 \phi-3\cos \phi = -\frac{a}{2}\Rightarrow \cos 3\phi = -\frac{a}{2}$
Now $-1\leq \cos 3\phi\leq 1$, So $\displaystyle -1 \leq -\frac{a}{2}\leq 1\Rightarrow -2\leq a\leq 2$
So we get $a = \left\{-2,-1,0,1,2\right\}$
But answer is no integer value of $a$
Help me
Thanks
| HINT:
If two roots are same we can write $$x^3-3x+a=(x-c)(x-b)^2$$
Compare the coefficients of the different powers of $x$ to determine $a,b,c,d$
If three roots are same we can write $$x^3-3x+a=(x-d)^3$$
Comparing the coefficients of $x^2, d=0\implies a=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Which is wrong in the proof? Consider the following steps:
\begin{align}
a &= x \\
a+a &= a+x && \text{[add }a\text{ to both sides]}\\
2a &= a+x && \text{[}a+a = 2a\text{]}\\
2a-2x &= a+x-2x && \text{[subtract }2x\text{ from both sides]}\\
2(a-x) &= a+x-2x && \text{[}2a-2x = 2(a-x)\text{]}\\
2(a-x) &= a-x && \text{[}x-2x = -x\text{]}\\
2 &= 1 && \text{[divide both sides by }a-x\text{]}
\end{align}
Which step is wrong in this proof?
| $$\begin{align}
a & = x \\
a + a & = a + x \\
2a & = a+x \\
2a - 2x & = a + x - 2x\\
2 (a -x) &= a-x
\end{align}$$
All is OK so far but in the next step you are trying to divide $0$ by $0$ which is not allowed $\frac{0}{0}$ is undefined.
Note: $a = x$ so $a - x = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the minimum of a function of two variables
Find the smallest value of $\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}\quad \quad$ for $a,b$ real and positive.
What I've done so far:
Let $F(a,b)=\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}$
Then
$$\begin{align}
\displaystyle F_a=\frac{2a-7{\sqrt{2}}\ a}{2\sqrt{49+a^2-7{\sqrt{2}}\ a}}+\frac{2a-\sqrt{2}\ b}{2\sqrt{a^2+b^2-{\sqrt{2}}\ ab}}=0\\
\displaystyle F_b=\frac{2b-{\sqrt{2}}\ a}{2\sqrt{a^2+b^2-{\sqrt{2}}\ ab}}+\frac{2b-10}{2\sqrt{50+b^2-10b}}=0
\end{align}$$
I then tried to simplify the two equations as follows:
$$\begin{align}
\left({2a-7{\sqrt{2}}\ a}\right)\left(\sqrt{a^2+b^2-{\sqrt{2}}\ ab}\right)+\left({2a-\sqrt{2}\ b}\right)\left(\sqrt{49+a^2-7{\sqrt{2}}\ a}\right)=0\\
\left({2b-{\sqrt{2}}\ a}\right)\left(\sqrt{50+b^2-10b}\right)+\left(2b-10\right)\left(\sqrt{a^2+b^2-{\sqrt{2}}\ ab}\right)=0\\
\end{align}$$
But then I'm stuck.
Do I need to include a constraint?(i.e $a$ and $b$ are positive and real); if so what will the constraint equation be?
Is there an altogether different way to do this other than using Lagrange??
| Hint: Consider vectors $X, A, B, Y$ where $|X| = 7$, $|A| = a$, $|B| = b$, $|Y| = \sqrt{50}$ and $\angle XOA = 45^\circ$, $\angle AOB = 45^\circ$, $\angle BOY = 45^\circ $.
How does your expression relate to these vectors? Think of distances. Apply Cosine rule.
Hence, the minimum is $ \sqrt{7^2 + (\sqrt{50})^2 - 2 \times 7 \times \sqrt{50} \times \cos 135^\circ} = \sqrt{49 + 50 + 70} = 13 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find five consecutive odd integers such that their sum is $55$. So my professor asked us to do an Olympiad exercise which says that the sum of five consecutive odd integers is $55$, find those integers. But I've never seen such an exercise so it is quite new and if I learn how then I will be able to solve similar exercises.
I tried to think of it in terms of geometry but that didn't help me.
| Hint: Choose one of those equations and solve it:
$$\begin{align}
& \text{The first integer:} & \qquad x+\color{blue}{x+2+x+4+x+6+x+8}=55 \\\,\\
& \text{The second integer:} & \qquad \color{red}{x-2}+x+\color{blue}{x+2+x+4+x+6}=55 \\\,\\
& \text{The third integer:} & \qquad \color{red}{x-4+x-2}+x+\color{blue}{x+2+x+4}=55 \\\,\\
& \text{The fourth integer:} & \qquad \color{red}{x-6+x-4+x-2}+x+\color{blue}{x+2}=55 \\\,\\
& \text{The fifth integer:} & \qquad \color{red}{x-8+x-6+x-4+x-2}+x=55
\end{align}$$
Obviously the simplest choice is to solve for the third integer, since the $2$s and $4$s cancel out, and you're left with the simple equation, $5x=55$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 10,
"answer_id": 8
} |
Finding integer solutions for $6x+15y+20z=1$ Problem: Find integers $x$, $y$, and $z$ that satisfy the equation $6x+15y+20z=1$.
I noticed that $\gcd(6,15)=3$, $\gcd(15,20)=5$, and that $\gcd(6,20)=2$. And of course $\gcd(6,15,2)=1$.
Of course I know to set one of the variables (x, y, or z) to be zero. But I want to know how to get the more trivial answers. But where do I go from here to find the solutions?
| Hint: Using Euclidean Algorithm you can find the gcd of the 3 numbers. Try using Euclidean Algorithm in the inverse direction some way.
gcd(a, b, c) = gcd(gcd(a, b), c)
$15=6 \times 2+3$ so $3= 1 \times 15-6 \times 2$ (+)
$20=6 \times 3+2$
$3=2 \times 1+1$
$1=3-2=3-20+6 \times 3=-20+7 \times 3=-20+7 \times 15-14 \times 6$ using (+)
So $z=-1$,$y=7$,$x=-14$ is a solution
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Real integral by keyhole contour Evaluate
$$\int_0^\infty \frac{\log x \; dx}{x^{2} + 2x + 2}$$
by integrating a branch of $(\log z)^{2}/(z^{2} + 2z +2)$ along a keyhole contour.
The thing I have trouble with is why I should be examining the square of the log - I guess it has something to do with ln x in fact NOT being the real part of log z, since log z is (or can be) defined on the entire negative real axis and ln x can't. But looking at the square would definitely not have been my first plan of attack :(
| Here are some details of the calculation. Let the keyhole contour be oriented counterclockwise and have the slot on the positive real axis. Call the segment above the real axis $\Gamma_1$, the large circle of radius $R$ $\Gamma_2$, the segement below the real axis $\Gamma_3$ and the circle of radius $\epsilon$ around the origin $\Gamma_4.$ We use the branch of the logarithm with the cut along the positive real axis and returning an argument from zero to $2\pi.$ Note that the poles of $$\frac{1}{z^2+2z+2}$$ are at
$$(z+1)^2 + 1 = 0$$ or
$$\rho_{0,1} = -1 \pm i.$$
Calling the desired integral $I$, we have
$$\left(\int_{\Gamma_1} + \int_{\Gamma_2} + \int_{\Gamma_3} + \int_{\Gamma_4}\right)
\frac{\log^2 z}{z^2+2z+2} dz
\\= 2\pi i
\left(\mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 + i\right)
+ \mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 - i\right)\right).$$
Now along the large circle we have
$$\left|\int_{\Gamma_2} \frac{\log^2 z}{z^2+2z+2} dz\right|
\sim 2\pi R \times \frac{\log^2 R}{R^2} \rightarrow 0$$
as $R\rightarrow\infty.$
Along the small circle we get
$$\left|\int_{\Gamma_4} \frac{\log^2 z}{z^2+2z+2} dz\right|
\sim 2\pi \epsilon \frac{\log^2\epsilon}{2} \rightarrow 0$$
as $\epsilon\rightarrow 0$ by repeated application of L'Hôpital's rule.
Now the residues are easy to compute because the poles are simple and we obtain
$$\mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 + i\right)
= \frac{1}{2i} \log^2(-1+i)$$
and $$\mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 - i\right)
= - \frac{1}{2i} \log^2(-1-i)$$
As we actually do the computation of these residues we need to be careful to use the same branch of the logarithm as in the integral. A computer algebra system might use a different branch!
For the first residue we get
$$\frac{1}{2i} \left(\frac{1}{2}\log 2 + \frac{3}{4} i\pi\right)^2
= \frac{1}{2i}
\left(\frac{1}{4}\log^2 2 - \frac{9}{16} \pi^2 + \frac{3}{4}\log 2 \times i\pi\right)$$
and for the second one
$$-\frac{1}{2i} \left(\frac{1}{2}\log{2} + \frac{5}{4} i\pi\right)^2
= -\frac{1}{2i}
\left(\frac{1}{4}\log^2 2 - \frac{25}{16} \pi^2 + \frac{5}{4}\log 2 \times i\pi\right)$$
Adding these contributions yields
$$\frac{1}{2i} \left(\pi^2 - \frac{1}{2}i\pi \log 2\right).$$
Finally we get
$$\left(\int_{\Gamma_1} + \int_{\Gamma_3} \right)
\frac{\log^2 z}{z^2+2z+2} dz = 2\pi i \times
\frac{1}{2i} \left(\pi^2 - \frac{1}{2}i\pi \log 2\right)
= \pi^3 - \frac{1}{2}i\pi^2 \log 2$$
Observe that along $\Gamma_3$ the logarithm term produces (actually $x$ would be a better choice of variable here rather than $z$)
$$-(\log^2 z + 4\pi i \log z -4\pi^2).$$
The first of these cancels the integral along $\Gamma_1$ and the third is real so that equating imaginary parts we find
$$I = \int_{\Gamma_1} \frac{\log z}{z^2+2z+2} dz
= - \frac{1}{4\pi} \left(- \frac{1}{2} \pi^2 \log 2\right) = \frac{1}{8} \pi\log 2.$$
We also get the following bonus integral (comparing real parts)
$$ \int_{\Gamma_1} \frac{1}{z^2+2z+2} dz = \frac{\pi^3}{4\pi^2} = \frac{\pi}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
$f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$, Then Max $\left\{f(x)-g(x)\right\}$ Let $f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$.
Find the Minimum of function $\left\{f(x)+g(x)\right\}$ and the maximum of function $\left\{f(x)-g(x)\right\}$.
$\bf{My\; Try}$:: For Minimum of $\left\{f(x)+g(x)\right\}$
Using Minkowski inequality or $\triangle$ Inequality $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\geq \sqrt{(a+b)^2+(c+d)^2}$
and equality holds when $\displaystyle \frac{a}{b} = \frac{c}{d}$
$\sqrt{(1-x)^2+(5-x^2)^2} + \sqrt{(x+2)^2+(x^2+1)^2}$
$\geq \sqrt{\left(1-x+x+2\right)^2+\left(5-x^2+x^2+1\right)^2} = \sqrt{3^2+6^2} = 3\sqrt{5}$
and equality hold, when $\displaystyle \frac{1-x}{5-x^2} = \frac{x+2}{x^2+1}$
But I did not understand How can I calculate Maximum of $\left\{f(x)-g(x)\right\}$
Help Required
Thanks
| If one observes carefully $f(x)$ is nothing but distance of the point $A(1,5)$ from an arbitrary point $P(x,x^2)$ lying on the curve $y=x^2$.
Similarly $g(x)$ is nothing but distance of the point $B(-2,-1)$ from any arbitrary point $P(x,x^2)$ on the curve $y=x^2$.
Now,$AB=3\sqrt{5}$
In other words we have been asked to find maximum value of $\mid PA-PB\mid$ and the minimum value of $ PA+PB$. By triangle inequality $\mid PA-PB\mid\leq AB$ and $PA+PB\geq AB$.
Equality occurs when the points $A,P,B$ are collinear which we see happening when $x=-1,3$.
When $x=-1$, then $P(-1,1)$ lies between $A$ and $B$. Here $PA=2\sqrt{5},PB=\sqrt{5}$
Thus $\left( PA+PB\right)_{min}=AB=3\sqrt{5}$
When $x=3$, then $P(3,9)$ does not lie between $A$ and $B$.Here $PA=2\sqrt{5},PB=5\sqrt{5}$
Thus $\mid PA-PB\mid_{max}= AB=3\sqrt{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Prove that $m^{2} + n^{2} = \csc^{2}\left(\theta\right)$ $$
\mbox{If}\quad\left\lbrace%
\begin{array}{rcl}
m^{2} + m'^{2} + 2mm'\cos\left(\theta\right) & = & 1
\\
n^{2} + n'^{2} + 2nn'\cos\left(\theta\right) & = & 1
\\
mn + m'n' + (mn' + m'n)\cos\left(\theta\right) & = & 0
\end{array}\right\rbrace
\quad\mbox{then prove that}\ m^{2} + n^{2} = \csc^{2}\left(\theta\right)
$$
Don't know how go about this. Please help.
| For convenience, let $a=m$, $b=m^{\prime}$, $x=n$, $y=n^{\prime}$; so
$a^2+b^2+2ab\cos\theta=1$, $\;\;x^2+y^2+2xy\cos\theta=1$, $\;\;ax+by+(ay+bx)\cos\theta=0$.
From the 3rd equation, $(a+b\cos\theta)x+(b+a\cos\theta)y=0$, so $
y=-\frac{a+b\cos\theta}{b+a\cos\theta}x$.
Substituting into the 2nd equation gives $x^2+\left(\frac{a+b\cos\theta}{b+a\cos\theta}\right)^2x^2-\frac{2(a+b\cos\theta)\cos\theta}{b+a\cos\theta}x^2=1$, so
$\frac{(b+a\cos\theta)^2+(a+b\cos\theta)^2-2(b+a\cos\theta)(a+b\cos\theta)\cos\theta}{(b+a\cos\theta)^2} x^2=1$.
Then $x^2=\frac{(b+a\cos\theta)^2}{b^2+2ab\cos\theta+a^2\cos^{2}\theta+a^2+2ab\cos\theta+b^2\cos^{2}\theta-2ab\cos\theta-2a^2\cos^{2}\theta-2b^2\cos^{2}\theta-2ab\cos^{3}\theta}$, so
$x^2=\frac{(b+a\cos\theta)^2}{a^2+b^2-a^2\cos^{2}\theta-b^2\cos^{2}\theta+2ab\cos\theta(1-\cos^{2}\theta)}=\frac{(b+a\cos\theta)^2}{(a^2+b^2+2ab\cos\theta)\sin^{2}\theta}=\frac{(b+a\cos\theta)^2}{\sin^2\theta}$ using Equation 1.
Then $a^2+x^2=a^2+\frac{(b+a\cos\theta)^2}{\sin^2\theta}=\frac{a^2\sin^2\theta+b^2+2ab\cos\theta+a^2\cos^2\theta}{\sin^2\theta}=\frac{a^2+b^2+2ab\cos\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}=\csc^2\theta$,
(again using Equation 1).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find max: $M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}$ For $a,b,c>0$ and $abc=1$, find the maximum of
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
| Substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$ for positive numbers $x,y,z$. We will show $M\le1$, which is equivalent to$$\sum_{cyc}\frac{x^3z^2}{x^2y^3+yz^4+x^3z^2}\le1$$and after full expansion, it is equivalent to$$\sum_{cyc}x^8 y^3 z^4 + x^7 y^2 z^6\le\sum_{cyc}x^9y^3z^3+x^7y^7z$$
By rearrangement inequality, it is obvious that $\sum_{cyc}x^5z\le\sum_{cyc}x^6$ hence $\sum_{cyc}x^8y^3z^4\le\sum_{cyc}x^9y^3z^3$. Also, for variables $p,q,r$, $\sum_{cyc}p^5r\le\sum_{cyc}p^6$ is true. Let $p=xy$, $q=xz$, $r=yz$ then it is $\sum_{cyc}x^5y^6z\le\sum_{cyc}x^6y^6$. Therefore $\sum_{cyc}x^7y^2z^6=\sum_{cyc}x^6y^7z^2\le\sum_{cyc}x^7y^7z$ and it is proved!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How find this limit $\lim_{x\to 0^{+}}\int_{x}^{1}\frac{\ln{(1+t)}}{\sqrt{t}}dt$ Find this limit
$$\lim_{x\to 0^{+}}\left(\int_{x}^{1}\dfrac{\ln{(1+t)}}{\sqrt{t}}dt+\int_{0}^{x}\dfrac{\sin{2t}}{\sqrt{4+t^2}\int_{0}^{x}(\sqrt{y+1}-1)dy}dt\right)$$
My try: since
$$\int_{0}^{x}(\sqrt{y+1}-1)dy=\dfrac{2}{3}(1+x)^{3/2}-x$$
maybe
$$\lim_{x\to 0^{+}}\int_{x}^{1}\dfrac{\ln{(1+t)}}{\sqrt{t}}dt$$ exsit?
and
$$\lim_{x\to 0^{+}}\int_{0}^{x}\dfrac{\sin{2t}}{\sqrt{4+t^2}\int_{0}^{x}(\sqrt{y+1}-1)dy}dt$$
exsit?
then I can't
| Note that you can use integration by parts on $\displaystyle\int_x^1 \frac{\ln(1+t)}{\sqrt{t}}\,dt$ with $u=\ln(1+t)$ and $\,dv=\dfrac{1}{\sqrt{t}}\,dt$ to get
$$\begin{aligned}\int_x^1\frac{\ln(1+t)}{\sqrt{t}}\,dt &= \left.\left[2\sqrt{t}\ln(1+t)\right]\right|_x^1-2\int_x^1 \frac{\sqrt{t}}{t+1}\,dt\\ &= 2\ln 2 - 2\sqrt{x}\ln(1+x) - 4\int_{\sqrt{x}}^1 \frac{s^2}{s^2+1}\,ds\quad(\text{sub: }s^2=t)\\ &= \ln 4 - 2\sqrt{x}\ln(1+x) - 4 \int_{\sqrt{x}}^1 1-\frac{1}{s^2+1}\,ds\\ &= \ln 4-2\sqrt{x}\ln(1+x)-4\left.\left[s-\arctan s\right]\right|_{\sqrt{x}}^1\\ &= \ln 4 + \pi - 4 -2\sqrt{x}\ln(1+x) +4\sqrt{x}-4\arctan\sqrt{x}\end{aligned}$$
Now we see that $\displaystyle\lim_{x\to 0^+}\int_x^1\frac{\ln(1+t)}{\sqrt{t}}\,dt = \ln 4 + \pi -4$.
The other part of the limit can be rewritten as
$$\lim_{x\to 0^+}\dfrac{\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}}\,dt}{\displaystyle\int_0^x\sqrt{y+1}-1\,dy}\rightarrow \frac{0}{0}$$
Now apply L'Hopital's rule to get that
$$\begin{aligned}\lim_{x\to 0^+}\dfrac{\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}}\,dt}{\displaystyle\int_0^x\sqrt{y+1}-1\,dy} &= \lim_{x\to 0^+}\dfrac{\dfrac{\sin(2x)}{\sqrt{4+x^2}}}{\sqrt{x+1}-1}\\ &= \lim_{x\to 0^+}\frac{\sin 2x}{\sqrt{x+1}-1}\cdot\lim_{x\to 0^+}\frac{1}{\sqrt{4+x^2}}\\ &= \ldots\end{aligned}$$
Can you take things from here?
In the end, I get $\displaystyle\lim_{x\to 0^+}\frac{\sin 2x}{\sqrt{x+1}-1}=4$; thus, $\displaystyle\lim_{x\to 0^+}\dfrac{\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}}\,dt}{\displaystyle\int_0^x\sqrt{y+1}-1\,dy}=4\cdot\frac{1}{\sqrt{4}} = 2$ and therefore $\displaystyle \lim_{x\to 0^+}\left(\int_x^1 \frac{\ln(1+t)}{\sqrt{t}}\,dt+\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}\int_0^x\sqrt{y+1}-1\,dy}\,dt\right)=\ln4+\pi-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/630118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Series of inverses of binomial coefficients Can you think of a simple way of proving that
$$
\sum_{n=k+1}^\infty \frac{1}{n \choose k}
$$
is rational for any $k \geq 2$?
Here's the background. Consider a series:
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)}
$$
Elementary algebra gives us that:
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = \lim_{k\to\infty} (1 - \frac{1}{k+1}) = 1
$$
so the sum turns out to be rational.
Next, consider
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)(n+3)(n+4)}
$$
With the same method, but much more effort we can show that:
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)(n+3)(n+4)} = \sum_{n=1}^\infty \left(\frac{1}{24 n}-\frac{1}{6(n+1)}+\frac{1}{4(n+2)}-\frac{1}{6(n+3)}+\frac{1}{24(n+4)}\right)
$$
and again we'll see that stuff cancels out, and the sum is again rational.
So the obvious conjecture is that this method will work for arbitrary (but $\geq 2$) fixed number of factors in denominator, and the sum will always be rational. Indeed that's the case. I provide a solution as an answer, but I'm not fully satisfied with it (it seems for me to be too brute force), so I'm looking for alternative solutions.
|
We show the result stated by @IgorRivin is valid.
\begin{align*}
\sum_{n=k+1}^{\infty}\frac{1}{\binom{n}{k}}=\frac{1}{k-1}\qquad\qquad k\geq 2\tag{1}
\end{align*}
We start proving a related identity from which we can easily derive (1).
The following is valid for $N\geq 1,k\geq 1$
\begin{align*}
\sum_{n=1}^{N}\frac{1}{\binom{n+k+1}{k+1}}=\frac{1}{k}-\frac{\left(1+\frac{1}{k}\right)}{\binom{N+k+1}{k}}\tag{2}
\end{align*}
We obtain
\begin{align*}
\sum_{n=1}^{N}\frac{1}{\binom{n+k+1}{k+1}}&=\sum_{n=1}^{N}\frac{n!(k+1)!}{(n+k+1)!}\\
&=(k+1)!\sum_{n=1}^N\frac{1}{(n+1)\cdots(n+k+1)}\\
&=\frac{(k+1)!}{k}\sum_{n=1}^N\frac{(n+k+1)-(k+1)}{(n+1)\cdots(n+k+1)}\\
&=\frac{(k+1)!}{k}\sum_{n=1}^N\left(\frac{1}{(n+1)\cdots(n+k)}-\frac{1}{(n+2)\cdots(n+k+1)}\right)\tag{3}\\
&=\frac{(k+1)!}{k}\left(\frac{1}{(k+1)!}-\frac{1}{(N+2)\cdots(N+k+1)}\right)\\
&=\frac{1}{k}-\frac{\left(1+\frac{1}{k}\right)k!}{(N+2)\cdots(N+k+1)}\tag{4}\\
&=\frac{1}{k}-\frac{\left(1+\frac{1}{k}\right)}{\binom{N+k+1}{k}}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
In (3) we get a telescoping sum leaving only the first and last term.
We will now apply identity (2) to answer OPs question. We use the representation from (4) with $k-1$ instead of $k$ and the claim follows for $k\geq 2$:
\begin{align*}
\sum_{n=k+1}^{\infty}\frac{1}{\binom{n}{k}}&=\sum_{n=1}^{\infty}\frac{1}{\binom{n+k}{k}}
=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{\binom{n+k}{n}}\\
&=\lim_{N\rightarrow\infty}\left(\frac{1}{k-1}-\frac{\left(1+\frac{1}{k-1}\right)(k-1)!}{(N+2)\cdots(N+k)}\right)\\
&=\frac{1}{k-1}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
Note: These formulae can be found in H.W.Goulds Combinatorial Identities Vol. 3, (3.26 - 3.29).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/630360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove by induction: $10^n+3\times4^{n+2}+5$ is divisible by $9$ I dealt with this problem but I couldn't resolve.
Prove by induction that $10^n+3\times4^{n+2}+5$ is divisible by $9$ for all non-negative integers $n$.
| Let P(n) be the statement that $10^n+3\times 4^{n+2}+5$ is divisible by $9$. When $n=0$, $10^n+3\times 4^{n+2}+5=54$ is divisible by $9$, so we have $P(0)$.
Suppose we have $P(k)$ for some nonnegative integer $k$. Then $10^k+3\times 4^{k+2}+5$ is divisible by $9$. Now
$(10^{k+1}+3\times 4^{k+3}+5)-(10^k+3\times 4^{k+2}+5)$
$=10^{k+1}-10^k+3\times (4^{k+3}-4^{k+2})$
$=9\times 10^k+9\times 4^{k+2}$
is divisible by $9$. It follows that $10^{k+1}+3\times 4^{k+3}+5$ is divisble by $9$, and we have $P(k+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/631765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the limit of $\left(\dfrac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n$ I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$
As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$
But now when I approach that top limit this is what I get:
$$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$
And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?
| By Taylor series we have:
$$\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}=\frac 1 3\left(3+\frac1 n(\log a +\log b+\log c)++o\left(\frac 1 n\right)\right)=1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)$$
so
$$\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\exp\left(n\log\left(1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)\right)\right)\sim_\infty \sqrt[3]{abc} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
An inequality with $a,b,c>0$ Let $a,b,c>0$. Prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq\left(1+\sqrt{2}+\sqrt{3}\right)\cdot\left(\frac{1}{a+b\sqrt{2}+c\sqrt{3}}+\frac{1}{b+c\sqrt{2}+a\sqrt{3}}+\frac{1}{c+a\sqrt{2}+b\sqrt{3}}\right)$$
| It will suffice to show that the following (stronger) inequality holds
for positive $a,b,c,u,v,w$ :
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq
(u+v+w)\bigg(\frac{1}{au+bv+cw}+\frac{1}{av+bw+cu}+\frac{1}{aw+bu+cv}\bigg)
$$
The fonction $f : x \mapsto \frac{1}{x}$ is convex on $(0,\infty)$, so
$$
\begin{array}{lcl}
f(\frac{u}{u+v+w}a+\frac{v}{u+v+w}b+\frac{w}{u+v+w}c) &\leq& \frac{u}{u+v+w}f(a)+\frac{v}{u+v+w}f(b)+\frac{w}{u+v+w}f(c) \\
f(\frac{v}{u+v+w}a+\frac{w}{u+v+w}b+\frac{u}{u+v+w}c) &\leq& \frac{v}{u+v+w}f(a)+\frac{w}{u+v+w}f(b)+\frac{u}{u+v+w}f(c) \\
f(\frac{w}{u+v+w}a+\frac{u}{u+v+w}b+\frac{v}{u+v+w}c) &\leq& \frac{w}{u+v+w}f(a)+\frac{u}{u+v+w}f(b)+\frac{v}{u+v+w}f(c) \\
\end{array}
$$
Adding up, we obtain the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $n,k\in\mathbb N$, solve $3^k-1=x^n$. If $n,k\in\mathbb N$, solve $$3^k-1=x^n$$
This seems like an interesting problem. I've currently tried a few things (one could try a lot of things in this case):
$1)$ $x^n\equiv -1\pmod 3\Rightarrow x\equiv -1\pmod 3\Rightarrow n=2r+1$, where $r\in\mathbb N\cup\{ 0\}$.
Which brings us to $3^k-1=x^{2r}x$.
$2)$ $2\mid 3^k-1$, so $x^{2r}x$ is divisible by $2$.
$3)$ By checking congruence modulo $4$, we see that $k=2m$, where $m\in\mathbb N$.
Thus $(3^m+1)(3^m-1)=x^{2r}x$. That's just a few things we could try.
| Since $n$ is odd, it has some odd prime divisor $p$, or is $1$, if it is $1$, we have an infinite family of trivial solutions. Since if we care of only prime $n$ we are done (why?), let's suppose $n=p$.
$$3^k=(x+1)(1-x+x^2-x^3\cdots)$$
And since $x\equiv 2 \pmod{3}$, the residues on the right parentheses $\pmod3$ look like
$$1+\underbrace{-2+1+-2+1\cdots}_\text{p-1 times}\equiv 1+\underbrace{1+1+1\cdots}_\text{p-1 times}\equiv p \pmod{3}$$
Therefore we want that $p$ is a multiple of $3$, therefore $p=3$
So finally we have $x^2-x+1=3(3q^2-3q+1)$. Since the parentheses is not divisible by $3$, and it must be a power of $3$, it must be $1$. So we have $3q(q-1)=0 \iff q=0,1$. $q=0$ does not work.
Therfore the only solution is $x=3q-1=2$, $n=3$, $k=2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Proving convergence of a sequence Let the following recursively defined sequence:
$a_{n+1}=\frac{1}{2} a_n +2,$
$a_1=\dfrac{1}{2}$.
Prove that $a_n$ converges to 4 by subtracting 4 from both sides.
When I do that, I get:
$2(\frac{1}{2} a_{n+1} -2)=(\frac{1}{2} a_n -2)$, so $y=2y$,
which is true only for $0$. But I'm not sure how to formally use this in a definition of convergence?
| Set $b_n=a_n-4$. Then $b_1=-7/2$ and
$$
b_{n+1}=a_{n+1}-4=\frac{1}{2}a_n+2-4=\frac{1}{2}(a_n-4)=\frac{1}{2}b_n.
$$
Thus
$$
b_n=\frac{b_{n-1}}{2}=\frac{b_{n-2}}{2^2}=\cdots=\frac{b_{1}}{2^{n-1}}=-\frac{7}{2^n},
$$
and finally
$$
a_n=4-\frac{7}{2^n}.
$$
Hence
$$
\lim_{n\to\infty} a_n=4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
find set of points for lots of triangulations I should find a set of $n$ points $Q$ in a plane, so that $t(Q)$ (the number of possible triangulations) the following equation holds:
$$t(Q) \ge 2^{n-2\sqrt{n}}$$
I solved the problem using the vertex points of a convex polygon. Wikipedia says, that $t(Q) = C_{n-2}$ and by induction I can show, that $C_{n-2} \ge 2^{n-2\sqrt{n}}$.
I think, that this approach is way to complex. There must be a better solution, because $C_{n-2} \gg 2^{n-2\sqrt{n}}$ for big $n$.
Any ideas?
| How complex is too complex?
The central binomial coefficient $\displaystyle \binom{2n}{n}$ is bigger than $\displaystyle \frac{4^n}{2n+1}$, so
$\displaystyle t(Q) = C_{n-2} = \frac{1}{n-1} \binom{2(n-2)}{n-2} \geq \frac{1}{n-1} \frac{4^{n-2}}{2n - 3} = \frac{2^{n+2\sqrt{n} - 4}}{(n-1)(2n-3)} 2^{n-2\sqrt{n}}$
Since $2^{n+2\sqrt{n} - 4} \geq (n-1)(2n-3)$ for $n > 3$, and $C_{3-2} = 1 > 2^{3 - 2\sqrt{3}}$ we have $t(Q) \geq 2^{n-2\sqrt{n}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot y=1$, then $x^3+y^3=?$ [key says 110]
$x+y=-1$ and $x\cdot y=2$, then $x^3+y^3=?$ [key says 5]
$x-y=-1$ and $x\cdot y=2$, then $x^3 -y^3=?$ [key says -7]
$x+y=\frac{1}{3}$ and $x\cdot y=\frac{1}{9}$, then $x^3+y^3=?$ $\left[-\dfrac{2}{27}\right]$
| Hrm. How can we make $x^3 + y^3$ out of that?
What tools do we have available to us? We have $x+y$. We have $xy$. And we have arithmetic.
I don't know yet how to make $x^3 + y^3$ exactly. Let's focus on one part: let's make $x^3$, and then see what we can do with what's left over.
It's pretty easy to make $x^3$ by multiplying things together. We could take $(xy) \cdot (xy) \cdot (xy)$. We could take $(x+y) \cdot(x+y) \cdot (xy)$. There are others. Let's experiment and actually try all of them. Multiplying them out, we can make
*
*$x^3 y^3$
*$x^3 y^2 + x^2 y^3$
*$x^3 y + 2 x^2 y^2 + x y^3$
*$x^3 + 3 x^2 y + 3 x y^2 + y^3$
That last one looks intriguing: it has $x^3$ all by itself without other terms multiplied in. It even has $y^3$ in it too.
So let's go with that, and try to subtract off the rest. Now, the question is
How do we make $3x^2 y + 3 x y^2$
If we can solve that, then we can combine that answer with the above to make $x^3 + y^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
Find the value of A, B and C in the identities. $6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Find the value of A, B and C.
I started it like this:
$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Solving the right hand side:
$ (ABx^2 - Ax - Bx + 1)(x - 1) + C$
$ ABx^3 - ABx^2 - Ax^2 + Ax - Bx^2 + Bx + x - 1 + C$
$ABx^3 - (AB + A + B)x^2 + (A + B + 1)x - 1 + C$
Comparing the coefficients:
$AB = 6$
$A = \frac6 B$
$AB + A + B = 11$
Then substitute the value of A in the above equation...is this right? Is there any error?
| Hint $\ x=1\,\Rightarrow\,c = 6.\,$ Cancelling $\,x-1\,$ yields $6x^2-5x+1 = (Ax-1)(Bx-1)\ $ so $\,\ldots$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$:
$$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$
This is what I did:
$$\begin{array}{crcl}
\Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^2 x \\
\Longrightarrow &3\left(1-\sin^2 x\right)-\sin 2x &=& \sin^2 x \\
\Longrightarrow & 4\sin^2 x + \sin 2x - 3 &=& 0 \\
\Longrightarrow &2\left(1-\cos 2x\right)+\sin 2x - 3 &=& 0\\
\Longrightarrow &-2\cos 2x + \sin 2x &=& 1\end{array}$$
So, what's next?! Thank you in advance!
| Let $\displaystyle-2=r\cos\phi,1=r\sin\phi$ where $r>0$ so $\displaystyle\frac\pi2<\phi<\pi$ and $\displaystyle\tan\phi=-\frac12$
So we have $$r\cos(2x-\phi)=r\sin\phi$$
$$\implies \cos(2x-\phi)=\cos\left(\frac\pi2-\phi\right)$$
$$\implies 2x-\phi=2n\pi\pm\left(\frac\pi2-\phi\right)$$ where $n$ is any integer
Considering '+' sign, $\displaystyle2x-\phi=2n\pi+\left(\frac\pi2-\phi\right)\implies 2x=2n\pi+\frac\pi2$
Considering '-' sign, $\displaystyle2x-\phi=2n\pi-\left(\frac\pi2-\phi\right)=2n\pi-\frac\pi2+\phi\implies 2x=2n\pi+2\phi-\frac\pi2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with the algebra in for this number theory proof? For all $n\geq 1$, prove with mathematical induction
$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$
So far.. I have substituted 1 and saw that the statement is true and I have plugged in n+1 to show that the proof is true for all integers but I don't know how to go about the simplification.. right now I have
LHS: $2-\frac{1}{k}+\frac{1}{(k+1)^2} \leq 2-\frac{1}{k+1}$
Should I try to find common denominators for the left? Step by step explanation please!
| This exercise shows that the sum of the reciprocals of the squares converges to something at most $2$; in fact, the series converges to $\frac{\pi^2}{6}$.
For $n\geq 1$, denote the statement in the exercise by
$$
S(n) : 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} \leq 2 - \frac{1}{n}.
$$
Base step ($n=1$): Since $1=2-\frac{1}{1}, S(1)$ holds.
Induction step: Fix some $k\geq 1$ and suppose that $S(k)$ is true. It remains to show that
$$
S(k+1) : 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{k^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k+1}
$$
holds. Starting with the left side of $S(k+1)$,
\begin{align}
1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2} &\leq 2-\frac{1}{k}+\frac{1}{(k+1)^2}\quad(\text{by } S(k))\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k^2-k}{k(k+1)}\right)\\[1em] &\leq 2-\frac{1}{k+1},\quad(\text{since } k\geq 1, k^2-k\geq 0)
\end{align}
the right side of $S(k+1)$. Thus $S(k+1)$ is true, thereby completing the inductive step.
By mathematical induction, for any $n\geq 1$, the statement $S(n)$ is true.
| {
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"source": "stackexchange",
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Simplify the surds below. Simplify this:
$10 + 2\sqrt5 - 4\sqrt2\sqrt5 - 4\sqrt2$
Help please! $:'($
It came out like this:
$10 + 2\sqrt5 -4\sqrt10 - 4\sqrt2$
How should I do the next step? Um stuck.
| Note that $4 \sqrt{10} = 4 \sqrt 2\cdot\sqrt 5$, and $10 = 2\sqrt 5\cdot \sqrt 5$.
Hint: FACTOR!
$$\begin{align} 10 + 2\sqrt5 - \color{red}{4\sqrt2}\sqrt5 - \color{red}{4\sqrt2} & = \color{blue}{2\sqrt 5}\cdot \sqrt 5 + \color{blue}{2\sqrt 5} - \color{red}{4\sqrt 2}(\sqrt 5 + 1)\\ \\ & = \color{blue}{2\sqrt 5}\color{purple}{\bf (\sqrt 5 + 1)} - 4\sqrt 2\color{purple}{\bf (\sqrt 5 + 1)}\\ \\ & = \;\;\cdots \end{align}$$
| {
"language": "en",
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Prove that $\int\limits_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum\limits_{n=1}^\infty \frac{1}{(n+a)^2}$ Prove that $$\int_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum_{n=1}^\infty \frac{1}{(n+a)^2}$$
I know that we have a product of $x^a$, $\displaystyle\sum_{n=0}^\infty x^n$, and $\displaystyle\sum_{n=0}^\infty \frac{(1-x)^n}{n}$, but it hasn't helped me so far.
Any tips?
We are given that $a>-1$.
| $$
\begin{align}
&\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\left[\left(1-x^a\right)+\left(x-x^{a+1}\right)+\left(x^2-x^{a+2}\right)+\dots\right]\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\left[\left(1-\frac1{a+1}\right)+\left(\frac12-\frac1{a+2}\right)+\left(\frac13-\frac1{a+3}\right)+\dots\right]\\
&=-\left[\frac1{(a+1)^2}+\frac1{(a+2)^2}+\frac1{(a+3)^2}+\dots\right]
\end{align}
$$
Differentiation under the Integral
$$
\begin{align}
-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x
&=\lim_{h\to0}\int_0^1\frac{x^{a+h}-x^a}{h(1-x)}\,\mathrm{d}x\\
&=\lim_{h\to0}\int_0^1\frac{x^a}{1-x}\frac{x^h-1}{h}\,\mathrm{d}x
\end{align}
$$
Pointwise $\frac{x^h-1}{h}\to\log(x)$ and on $(0,1)$ we have $\left|\frac{x^h-1}{h}\right|\le|\log(x)|$. Dominated Convergence finishes things off.
Convergence of the Integral of the Sum
For $a\ge0$, note that on $[0,1]$, $\frac{1-x^a}{1-x}\le\max(a,1)$. Therefore,
$$
\begin{align}
\left|\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x-\sum_{k=0}^{n-1}\int_0^1(1-x^a)x^k\,\mathrm{d}x\right|
&=\left|\int_0^1\frac{1-x^a}{1-x}x^n\,\mathrm{d}x\right|\\
&\le\frac{\max(a,1)}{n+1}
\end{align}
$$
For $-1\lt a\lt0$, the term on the right becomes
$$
\begin{align}
\left|\int_0^1\frac{1-x^a}{1-x}x^n\,\mathrm{d}x\right|
&=\left|\int_0^1\frac{x^{-a}-1}{1-x}x^{n+a}\,\mathrm{d}x\right|\\
&\le\frac1{n+a+1}
\end{align}
$$
An Alternate Approach
Note that integration by parts gives us
$$
\begin{align}
\int_0^1x^a\log(x)\,\mathrm{d}x
&=\frac1{a+1}\int_0^1\log(x)\,\mathrm{d}x^{a+1}\\
&=-\frac1{a+1}\int_0^1x^a\,\mathrm{d}x\\
&=-\frac1{(a+1)^2}
\end{align}
$$
Then using the sum of a geometric series, we have
$$
\begin{align}
\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x
&=\int_0^1\sum_{k=0}^\infty x^{a+k}\log(x)\,\mathrm{d}x\\
&=-\sum_{k=0}^\infty\frac1{(a+k+1)^2}\\
&=-\sum_{k=1}^\infty\frac1{(a+k)^2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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If $|x + 1| + |x - 3| = 6$, solve for $x$ $|x + 1| + |x - 3| = 6$. Solve for X.
So I know when you have a problem like this: |x| = 6, you solve by doing x=6 and x=-6. That doesn't help us much in the above example.
You can also solve problems of this fashion by negative the variable portion of the equation. For ex:
$|x+3| = 6. $
You can solve it either by doing:
$x+3 = 6$ or
$x+3 = -6$
$x= 3,-9$
OR
$(x+3) = 6$ so $x=3$
or
$-(x+3) = 6$
$-x -3 =6$
$-x = 9$
$x = -9$
So back to the original problem of $|x + 1| + |x - 3| = 6$
if $X>3$, then $x+1 + x-3 = 6$
$2x=8$
$x=4$
or if $X<3$
$x+1 + x-3 = -6$
$2x=-4$
$x=-2$
Is that correct? Is that all I need to do? I also feel like I got the "critical points" of the equation wrong.
| First let us do it without algebra. Draw a number line. Then $|x+1|+|x-3|$ is the sum of the distances of $x$ from $-1$ and from $3$. It is clear that if $x$ is between $-1$ and $3$, the sum of the distances is $4$. The sum of the distances will be $6$ if we are $1$ unit to the right of $3$, or $1$ unit to the left of $-1$.
Now with algebra, essentially like you approached it.
If $x\ge 3$, then $|x-3|=x-3$ and $|x+1|=x+1$. So we want $(x-3)+(x+1)=6$. giving $x=4$.
If $x\gt -1$, and $x\lt 3$, then $|x+1|=x+1$ and $|x-3|=-x+3$, so we want $(x+1)+(-x+3)=6$, impossible.
If $x\lt -1$, then $|x+1|=-x-1$ and $|x-3|=-x+3$, so we want $-2x+2=6$, giving $x=-2$.
| {
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"source": "stackexchange",
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Induction: $\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}$ $$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}, \quad \text{for } x \not \in \{0,-1,-2,\dots,-n\}$$
Can somebody please help me with this?
I tried to multiply both sides with $x(x+1)\cdots(x+n)$.
$$n! = \sum_{k=0}^n (-1)^k\binom{n}{k}\frac{x(x+1)\cdots(x+n)}{x+k}$$
Then to divide both sides with $n!$.
$$1 = \sum_{k=0}^{n}\frac{(-1)^k}{k!(n-k)!}\frac{x(x+1)\cdots(x+n)}{x+k}$$
And then, assume it holds for $n$, and prove for $n+1$:
$$\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}\frac{x(x+1)\cdots(x+n+1)}{x+k} = \\ \sum_{k=0}^{n}\frac{(-1)^k}{k!(n-k)!}\frac{x(x+1)\cdots(x+n)}{x+k}\frac{x+n+1}{n+1-k} + \\
\frac{(-1)^{n+1}}{(n+1)!}x(x+1)\cdots(x+n)$$
And this $\frac{x+n+1}{n+1-k}$ term is keeping me from applying the induction hypothesis.
EDIT: @Daniel's hint was pretty useful, thanks. After doing that, it boils down to this:
$$\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}\frac{x(x+1)\cdots(x+n+1)}{x+k} = 1 + x(x+1)\cdots(x+n)\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}$$
So, proving that $\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!} = 0$ should do it.
| Hint. Use that fact that the result is an identity for all $x$ (with the stated exceptions). So, assume the result is true for a specific $n$. Then we can say
$$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}\ ,$$
and also
$$\eqalign{&\frac{n!}{(x+1)(x+2)\cdots(x+n+1)}\cr
&\qquad\qquad{}= \binom{n}{0}\frac{1}{x+1}-\binom{n}{1}\frac{1}{x+2}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n+1}\ .\cr}$$
Now write down the RHS for $n+1$, simplify and use the above two equations. You should find it just drops out.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove this is a rational number I'm not sure how to prove this is a rational number $\frac{q}{m}$, can some one show me?
$$\frac{q}{m}=\frac{(\frac{1+\sqrt5}{2})^n - (\frac{1-\sqrt5}{2})^n}{\sqrt5}$$
| Lt me consider $$S(n) = 2\frac{ \left((a+b)^n-(a-b)^n\right)}{b}$$The binomial theorem will be applied to each term and the result will be expanded and simplified. So, the results are successively $$S(1) = 4$$ $$S(2) = 8 a$$ $$S(3)= 4\left(3 a^2+b^2\right)$$ $$S(4)= 16 a \left(a^2+b^2\right)$$ $$S(5)=4 \left(5 a^4+10 a^2 b^2+b^4\right)$$ $$S(6)=8 a \left(3 a^2+b^2\right) \left(a^2+3 b^2\right)$$ $$S(7)=4 \left(7 a^6+35 a^4 b^2+21 a^2 b^4+b^6\right)$$ $$S(8)=32 a \left(a^2+b^2\right) \left(a^4+6 a^2 b^2+b^4\right)$$
So, you see, as clearly pointed out by the previous comments and answers, that if $a$ is rational and $b$ the square root of a rational, the result is rational.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Closed form integral $\int_b^c \frac{x^2}{\sqrt{(x-a)(x-b)(c-x)(d-x)}} dx$ Is there a closed form expression for the definite integral $$I=\int_b^c \frac{x^2}{\sqrt{(x-a)(x-b)(c-x)(d-x)}} dx$$ for $a<b<c<d$?
Mathematica 9.0 can do it for special cases using elliptic integrals, so I suspect that there may be a more general closed form solution. Some examples:
For $a=1$, $b=2$, $c=3$, $d=4$:
$$I=-2 E\left(\frac34\right) - K\left( \frac34 \right) + 32 K\left( -3 \right) - 20\, \Pi(-1,-3)$$
For $a=1$, $b=2$, $c=3$, $d=5$:
$$I=-\sqrt{6}\, E\left(\frac23\right) - \sqrt{6}\, K\left( \frac34 \right) + 25\sqrt{2}\, K\left( -2 \right) - \frac{33}{\sqrt{2}}\, \Pi\left(-\frac12,-2\right)$$
For $a=1$, $b=2$, $c=5$, $d=7$:
$$I=-\sqrt{20}\, E\left(\frac9{10}\right) - \sqrt{5}\, K\left( \frac9{10} \right) + 49\sqrt{2}\, K\left( -9 \right) - \frac{75}{\sqrt{2}}\, \Pi\left(-\frac32,-9\right)$$
For $a=1$, $b=2$, $c=7$, $d=11$:
$$I=-\sqrt{54}\, E\left(\frac{25}{27}\right) + \frac{103\sqrt{2}}{\sqrt{27}}\, K\left( \frac{25}{27} \right) - \frac{42\sqrt{2}}{\sqrt{27}}\, \Pi\left(-\frac59,\frac{25}{27}\right)$$
| This integral is listed in Byrd and Friedman
\begin{equation}
I^{(2)}(a,b,c,d)=\frac{d^2}{\sqrt{(d-b)(c-a)}}\big\{2K(k)+4(b^2/a^2-1)\Pi(\alpha^2,k)+
\frac{\alpha^2E(k)+(k^2-\alpha^2)K(k) + (2\alpha^2k^2+2\alpha^2-\alpha^4-3k^2)\Pi(\alpha^2,k)}{(\alpha^2-1)(k^2-\alpha^2)}\bigg\}
\end{equation}
where $\alpha^2=(c-b)/(d-b)$, $k^2=\alpha^2(d-a)/(c-a)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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If $a,b,c,d\in\mathbb N$ and $a^2+b^2\mid ac+bd$, can it be true that $\gcd(a^2+b^2,c^2+d^2)=1$? or $3$? or $74$?
If $a,b,c,d\in\mathbb N$ and $a^2+b^2\mid ac+bd$, can it be true that $\gcd(a^2+b^2,c^2+d^2)=1$? or $3$? or $74$?
That problem is complicated. I've tried some approaches, but they're useless. E.g. if $$\gcd(a^2+b^2,c^2+d^2)=1$$
Then $$\gcd((a^2+b^2)(c^2+d^2),ac+bd)=\gcd(a^2+b^2,ac+bd)\gcd(c^2+d^2,ac+bd)$$
So it would be sufficient to show that this equality can't hold. But it won't work.
Also, if $\gcd(a,b)=1$, then $\gcd(a+b,a^2+b^2)=1$ or $2$. So it'd also be sufficient to prove that it can't be true that $$\gcd(a^2+b^2+c^2+d^2,(a^2+b^2)^2+(c^2+d^2)^2)$$ is $1$ or $2$, but you can see how desperate proving this would actually be.
I'm curious to see a solution. I'll see if you could think of one.
And it's a problem from the selection of people for the IMO $2013$ phase (not sure how to say it). I can't find a solution on the Internet. I think knowing how to solve this problem could help me in the future, so I've posted it here.
Thanks.
| We have the identity
$$
b^2\bigl((a^2+b^2)-(c^2+d^2)\bigr) = (a^2+b^2)(c^2-b^2) - (ac-bd)(ac+bd).
$$
Hence $(a^2+b^2) \mid (ac+bd)$ forces $(a^2+b^2) \mid b^2(c^2+d^2)$. Since $a^2+b^2 > b^2$ for all $a,b \in \mathbb{N}$, we must have $\gcd(a^2+b^2,c^2+d^2)>1$. Furthermore, $\gcd(a^2+b^2,b^2)$ must divide $a^2\!$, so any prime $p$ involved must divide $\gcd(a,b)$, and hence $p^2 \mid (a^2+b^2)$.
Can you take it from there?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the least nonnegative residue Find the least nonnegative residue of $5^{18} \mod 11$.
To do this I took $5^2 \equiv 3 \mod 11$. Then I did $(5^2)^5 \equiv 3^5 \mod 11$. And $3^5 \equiv 1 \mod 11$.
So now I have $5^{10} \equiv 1 \mod 11$. Then I multiplied both sides by $5^8$ to get $5^{18} \equiv 5^8 \mod 11$. So I believe $5^8$ is the least nonnegative residue but I am not entirely sure. Can someone please confirm that this is correct?
| ${\rm mod}\ 11\!:\ 2^{18} 5^{18}\equiv 10^{18}\equiv (-1)^{18}\!\equiv 1,\ $ so $\ 5^{18}\!\equiv 2^{-18}\!\equiv 2^2 (2^5)^{-4}\equiv 2^2(-1)^{-4}\!\equiv 2^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$ I am trying to factor
$$x^5+4x^3+x^2+4=0$$
I've used Ruffini's rule to get
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
But I don't know what to do next.
The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give me hints?
| I would start by factoring $x^3$ out from the first two terms and noticing the pattern in the result.
$$
\begin{split}
x^5+4x^3 + x^2 + 4 &= x^3 \left(x^2+4\right) + x^2+4 \\
&= \left(x^2+4\right)\left(x^3+1\right) \\
&= \left(x^2+4\right)(x+1)\left(x^2-x+1\right), \\
\end{split}
$$
where the last step is the standard factoring of the sum of two cubes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Fractions in Questions and Answers
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