Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Compute $A^5-4A^4-7A^3+11A^2-A-10I$ for $A= \begin{pmatrix}1&4\\2&3\end{pmatrix}$ by Cayley-Hamilton Question: Use Cayley-Hamilton theorem to express $A^5-4A^4-7A^3+11A^2-A-10I$, where $$A= \begin{pmatrix}1&4\\2&3\end{pmatrix}$$
Attempt: I only know how to get the characteristic polynomial equation which is $\lambda^2-4\lambda -5$ for that matrix. By definition $A^2-4A-5=0$. But where does that 5 degree polynomial come from?
| Continuing your analysis,
$$A^2 - 4A -5I=0\implies A^2 = 4A+5I$$
Thus,
$$\begin{array}{cc}
A = A\\
A^2 = 4A + 5I\\
A^3 = 4A^2 + 5A &= 4(4A+5I) + 5A = 21A +20I\\
\vdots
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $A: l_2 \to l_2$ is a bounded operator Let us consider the following linear operator acting on $l_2$:
$$ A(x_1,x_2,x_3,\ldots) ~\colon=~ \left(x_1,\frac{x_1+x_2}{2},\frac{x_1+x_2+x_3}{3},\ldots\right) $$
I need to show that $A$ is a bounded operator, that is $||Ax|| \leq C~||x||$ for some constant $C$ and all $x \in l_2$. In other words, I need to prove the inequality
$$ \sum_{n=1}^{\infty}{\left( \frac{x_1+\ldots+x_n}{n} \right)^2} \leq C
\sum_{n=1}^{\infty}{x_n^2} $$
I tried to use the fact that
$$ \frac{x_1+\ldots+x_n}{n} \leq \sqrt{\frac{x_1^2+\ldots+x_n^2}{n}} $$
but it doesn't work because in that case we get
$$ \sum_{n=1}^{\infty}{\left( \frac{x_1+\ldots+x_n}{n} \right)^2} \leq \sum_{n=1}^{\infty}~{\frac{x_1^2+\ldots+x_n^2}{n}} = \sum_{n=1}^{\infty}{\left( \frac{1}{n} + \frac{1}{n+1} + \cdots \right)x_n^2} $$
and coefficients of $x_n^2$ diverge.
--
Thank you.
| *
*We show that $\left|\sum_{j=1}^nx_j\right|\leq \left(\sum_{j=1}^nx_j^2\sqrt j\right)^{1/2}\left(\sum_{j=1}^n\frac 1{\sqrt j}\right)^{1/2}$, applying Cauchy-Schwarz inequality $\sum_j a_jb_j\leq \sqrt{\sum_ja_j}\sqrt{\sum_jb_j}$ to $a_j=x_j\sqrt j$ and $b_j=\frac 1{\sqrt j}$.
*We have $$\sum_{j=1}^n\frac 1{\sqrt j}\leq \sum_{j=1}^n\int_{j-1}^jx^{-1/2}dx=\sum_{j=1}^n2(\sqrt j-\sqrt{j-1})=2\sqrt n.$$
*Using the last inequality
\begin{aligned}
\frac 1{n^2}\left(\sum_{j=1}^nx_j\right)^2&\leq\frac 1{n^2}\sum_{j=1}^nx_j^2\sqrt j2\sqrt n\\
&=2n^{-3/2}\sum_{j=1}^nx_j^2\sqrt j,
\end{aligned}
so
\begin{aligned}
\sum_{n=1}^{+\infty}\frac 1{n^2}\left(\sum_{j=1}^nx_j\right)^2&\leq 2\sum_{1\leq j\leq n\leq +\infty}n^{-3/2}x_j^2\sqrt j\\
&=2\sum_{j=1}^{+\infty}\sum_{n=j}^{+\infty}n^{-3/2}x_j^2\sqrt j\\
&\leq 2\sum_{j=2}^{+\infty}\sum_{n=j}^{+\infty}\int_{n-1}^nt^{-3/2}dtx_j^2\sqrt j+2\sum_{n=1}^{+\infty}n^{-3/2}x_1^2\\
&=2\sum_{j=2}^{+\infty}\sum_{n=j}^{+\infty}[-2t^{-1/2}]_{n-1}^nx_j^2\sqrt j+2\sum_{n=1}^{+\infty}n^{-3/2}x_1^2\\
&=4\sum_{j=2}^{+\infty}\sum_{n=j}^{+\infty}((n-1)^{-1/2}-n^{-1/2})x_j^2\sqrt j+2\sum_{n=1}^{+\infty}n^{-3/2}x_1^2\\
&=4\sum_{j=2}^N(j-1)^{-1/2}x_j^2\sqrt j+2x_1^2\sum_{n=1}^{+\infty}
n^{-3/2}\\
&\leq 2\max(2\sqrt 2,\sum_{n=1}^{+\infty}
n^{-3/2})||x||_2^2.
\end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding root of a polynomial Let's say I've got $ax^3 + bx^2 + cx + d = 0$
I want to find x for a general-case.
Mathematica gave me a very long solution, and even longer for one which has $kx^4$.
Can you please help me reproduce the algorithm used by Mathematica to generate formula of:
$$P(x) = a_0x^0 + a_1x^1 + a_2x^2 + a_3x^3 + \cdots+ a_nx^n$$
i.e., general case of a polynomial.
| Here's a (standard) method to treat the cubic case, though it hides some of what is going on, which is basically Galois Theory. You might as well suppose that $a \neq 0$ (otherwise you would be dealing with a quadratic at worse). Hence you can divide through and assume that $a = 1.$ So let's work instead with the cubic $x^3 + ax^2 + bx + c.$ Now set $y = x + \frac{a}{3}.$ Then $x^3 + ax^2 + bx + c$ has the form $y^3 + ey + f,$ and if we can solve for $y,$ we can easily recover $x$ (the constants $e$ and $f$ can be calculated routinely in terms of $a,b$ and $c,$ but we omit the rather tedious details. Hence if we can solve equations of the form $x^3 + ax + b =0,$ we can deal with the general cubic, and we may as well suppose that $b \neq 0,$ for the case $b =0$ is easy. To solve the equation $x^3 + ax + b = 0$ when $b \neq 0$, set $x = u+v.$ The equation may then be written as
$(u^3 + v^3) + b + (u+v)(3uv+a) = 0.$ We can try to solve separately, but simultaneously, $(u+v)(3uv+a) = 0$ and $u^3 +v^3 = -b.$ Note that $u +v \neq 0$ since $b \neq 0,$ so we need $uv = -\frac{a}{3}$ and $u^3 + v^3 = -b.$ Hence $(t -u^3)(t-v^3) = t^2 +bt -\frac{a^3}{27}$, and two solutions do have a simultaneous solution, with $u^3$ and $v^3$ being the roots of a quadratic.
| {
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"url": "https://math.stackexchange.com/questions/112514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Integrating $\int \sin^n{x} \ dx$ I am working on trying to solve this problem:
Prove: $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$
Here are the steps that I follow in the example that I am reading:
$u = \sin^{n - 1}{x}$
$du = (n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx$
$v = -\cos{x}$
$dv = \sin{x} \ dx$
$\int \sin^n{x} \ dx = \sin^{n - 1}{x} \cdot \sin{x} \ dx$
$\int \sin^n{x} \ dx = \underbrace{\sin^{n - 1}{x}}_{u} \cdot \underbrace{-\cos{x}}_{v} - \int \underbrace{-\cos{x}}_{v} \cdot \underbrace{(n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx}_{du}$
$\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \cos^{2}{x} \ dx$
$\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$
Here is where I get lost. How did we go from $\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$ to $\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$? Even more specifically, where did $\sin^{n}{x}$ come from?
$\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$
I get this part.
$n\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx$
$\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot x \ \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$
Could someone please explain what I am missing?
Thank you for your time.
| First let us denote
$$I_n = \int \sin^n{x} \ dx $$
$$ \int u(x) v'(x) dx = u(x) v(x) - \int v(x) u' (x) dx $$
Here $u(x) = \sin^{n-1}{x} \hspace{3pt}$ and $\hspace{3pt} v'(x) = \sin x $
$ \Rightarrow v(x) = -\cos x$
Therefore
$$
\begin{align*}
I_n &= -\cos x \hspace{3pt} \sin^{n-1}x + \int \cos^2 x \hspace{4pt} (n-1) \sin^{n-2} x \hspace{4pt} dx \\
&= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx - (n-1) \int \sin^{n} dx\\
&= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx - (n-1) I_n
\end{align*}
$$
$$ \Rightarrow (1+n-1)I_n = -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x dx $$
$$ \Rightarrow I_n = \frac{-\cos x \hspace{3pt} \sin^{n-1}x}{n} + \frac{(n-1)}{n} \int \sin^{n-2} x dx $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Prove that $a^2 + b^2 + c^2 $ is not a prime number I am having difficulty solving this problem:
Let $a, b, c \in\mathbb{Z}$, $abc \neq 0$ and $a\neq c$ be such that
$$\frac{a}{c} = \frac{a^2+b^2}{c^2+b^2}.$$
Prove that $a^2 + b^2 + c^2$ is not a prime number.
Thanks in advance!
| W.l.o.g we may assume that $a,c>0$. The equation
$$
\frac{a}{c}=\frac{a^2+b^2}{c^2+b^2}
$$
together with the assumption $a\neq c$ quickly gives us $ac=b^2$ as a corollary. Therefore we have
$$
a^2+b^2+c^2=a^2+ac+c^2
$$
and the extra condition that $ac=b^2$ must be a perfect square.
There are two main cases. If $gcd(a,c)>1$, then that common divisor is also a divisor of $a^2+ac+c^2$, so this latter number won't be a prime. If the numbers $a$ and $c$ are coprime, then the equation $ac=b^2$ and unique factorization force both $a$ and $c$ to be squares. So we can assume that $a=p^2, c=q^2$ for some integers $p,q$. But then we see that
$$
a^2+b^2+c^2=p^4+p^2q^2+q^4=(p^2+q^2)^2-p^2q^2=(p^2+pq+q^2)(p^2-pq+q^2).
$$
Here $p\neq q$, so both these factors are $>1$, and the claim follows in this case, too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/113962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer.
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.
| A trick I learnt from Bill Dubuque.
Write it as
$$\frac{n^5 - n}{5} + \frac{n^4 - n}{2} + \frac{n^3 - n}{3} + n$$
(See Bill's answer here: Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
| Since powers of x and y is always greater than or equal to zero, You can prove it by mathematical induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Chinese Remainder theorem with non-pairwise coprime moduli Let $n_1,...,n_k \in \mathbb{N}$ and let $a_1,...,a_k \in \mathbb{Z}$.
How to prove the following version of the Chinese remainder theorem (see here):
There exists a $x \in \mathbb{Z}$ satisfying system of equations:
$$x=a_1 \pmod {n_1}$$
$$x=a_2 \pmod {n_2}$$
$$\ldots$$
$$x=a_k \pmod{n_k}$$
if and only if $a_i=a_j \pmod{\gcd(n_i,n_j)}$ for all $i,j=1,...,k$?
If numbers $n_i$, for $i=1,...,k$, are pairwise coprime, it is a classical version of Chinese remainder theorem.
Thanks.
| If we factor $n_k$ into primes, $n_k = p_{1}^{b_{1}}\cdots p_r^{b_{r}}$, then the Chinese Remainder Theorem tells us that $x\equiv a_k\pmod{n_k}$ is equivalent to the system of congruences
$$\begin{align*}
x&\equiv a_k\pmod{p_1^{b_{1}}}\\
x&\equiv a_k\pmod{p_2^{b_{2}}}\\
&\vdots\\
x&\equiv a_k\pmod{p_r^{b_{r}}}
\end{align*}$$
Thus, we can replace the given system of congruences with one in which every modulus is a prime power, $n_i = p_i^{b_i}$.
Note that the assumption that $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ "goes through" this replacement (if they were congruend modulo $\gcd(n_i,n_j)$, then they are congruent modulo the gcds of the prime powers as well).
So, we may assume without loss of generality that every modulus is a prime power.
I claim that we can deal with each prime separately, again by the Chinese Remainder Theorem. If we can solve all congruences involving the prime $p_1$ to obtain a solution $x_1$ (which will be determined modulo the highest power of $p_1$ that occurs); and all congruences involving the prime $p_2$ to obtain a solution $x_2$ (which will be determined modulo the highest power of $p_2$ that occurs); and so on until we obtain a solution $x_n$ for all congruences involving the prime $p_n$ (determined modulo the highest power of $p_n$ that occurs), then we can obtain a simultaneous solution by solving the usual Chinese Remainder Theorem system
$$\begin{align*}
x &\equiv x_1 \pmod{p_1^{m_1}}\\
&\vdots\\
x &\equiv x_n\pmod{p_n^{m_n}}
\end{align*}$$
(where $m_i$ is the highest power of $p_i$ that occurs as a modulus).
So we are reduced to solving figuring out whether we can solve the system
$$\begin{align*}
x &\equiv a_1\pmod{p^{b_1}}\\
x &\equiv a_2\pmod{p^{b_2}}\\
&\vdots\\
x & \equiv a_n\pmod{p^{b_n}}
\end{align*}$$
with, without loss of generality, $b_1\leq b_2\leq\cdots\leq b_n$.
When can this be solved? Clearly, this can be solved if and only if $a_i\equiv a_j\pmod{p^{b_{\min(i,j)}}}$: any solution must satisfy this condition, and if this condition is satisfied, then $a_n$ is a solution.
For example: say the original moduli had been $n_1 = 2^3\times 3\times 7^2$, $n_2= 2^2\times 5\times 7$, $n_3=3^2\times 5^3$. First we replace the system with the system of congruences
$$\begin{align*}
x&\equiv a_1 \pmod{2^3}\\
x&\equiv a_2\pmod{2^2}\\
x&\equiv a_1\pmod{3}\\
x&\equiv a_3\pmod{3^2}\\
x&\equiv a_2\pmod{5}\\
x&\equiv a_3\pmod{5^3}\\
x&\equiv a_1\pmod{7^2}\\
x&\equiv a_2\pmod{7}.
\end{align*}$$
Then we separately solve the systems:
$$\begin{align*}
x_1&\equiv a_1 \pmod{2^3} &x_2&\equiv a_1\pmod{3}\\
x_1&\equiv a_2\pmod{2^2}&x_2&\equiv a_3\pmod{3^2}\\
\strut\\
x_3&\equiv a_2\pmod{5}&x_4&\equiv a_1\pmod{7^2}\\
x_3&\equiv a_3\pmod{5^3}&x_4&\equiv a_2\pmod{7}.
\end{align*}$$
Assuming we can solve these, $x_1$ is determined modulo $2^3$, $x_2$ modulo $3^2$, $x_3$ modulo $5^3$, and $x_4$ modulo $7^2$, so we then solve the system
$$\begin{align*}
x &\equiv x_1\pmod{2^3}\\
x &\equiv x_2\pmod{3^2}\\
x&\equiv x_3 \pmod{5^3}\\
x&\equiv x_4\pmod{7^2}
\end{align*}$$
and obtain a solution to the original system.
Hence, if the condition $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ holds in the original system, then we obtain a solution for each prime, and from the solution for each prime we obtain a solution to the original system by applying the usual Chinese Remainder Theorem twice.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exercise on irreducible polynomials I know that this is not the right place for questions like that, but I need someone that explain me step-by-step how can I resolve this exercise (I've exam in the next days):
Write as products of irreducible factors the polynomial $f=x^3-3x^2+x-3 \in\mathbb{R}[x]$ and $f=x^3-\overline{3}x^2+x-\overline{3} \in\mathbb{Z}_5[x]$.
Thank you for helping.
| This polynomial factors, and that can be seen using the grouping method. If your polynomial has four terms like this one, the grouping method might work. You group terms together two at a time and factor out a Greatest Common Factor:
$$
\begin{align*}
x^3-3x^2+x -3 & = x^2(x-3)+1(x -3)\\
& = (x^2+1)(x-3)
\end{align*}
$$
And voila, the polynomial is at least partially factored. Over $\mathbb{R}$, we cannot factor further, since $-1$ has no square root.
Next, now that we have exhausted factoring over $\mathbb{R}$, maybe factoring can continue over $\mathbb{F}_5$. Indeed, here we have $$
\begin{align*}
x^3-3x^2+x -3 & = x^2(x-3)+1(x -3)\\
& = (x^2+1)(x-3)\\
& = (x^2-4)(x-3)\\
& = (x-2)(x+2)(x-3)
\end{align*}
$$
Lastly, it's nice to choose residues from the same neighborhood, so $$
\begin{align*}
x^3-3x^2+x -3 & = x^2(x-3)+1(x -3)\\
& = (x^2+1)(x-3)\\
& = (x^2-4)(x-3)\\
& = (x-2)(x+2)(x-3)\\
& = (x-2)(x-3)(x-3)\\
& = (x-2)(x-3)^2
\end{align*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the integral $\int_{-\infty}^{\infty}\frac{dx}{1+ae^{bx^2}}$ Could somebody tell me how to find the integral
$$\int_{-\infty}^{\infty}\frac{dx}{1+ae^{bx^2}}$$
for constants $a$ and $b$?
Thanks!
| I will assume that $a, b$ are positive constants, in order to circumvent singularity issues. By the substitution $z = \sqrt{b} \, x$, the integral in question is equal to
$$ \frac{1}{\sqrt{b}} \int_{-\infty}^{\infty} \frac{e^{-z^2}}{a + e^{-z^2}} \; dz.$$
From the identity
$$ 1 + x^{2n+1} = (1 + x)(1 - x + \cdots - x^{2n-1} + x^{2n}), $$
we obtain
$$ \frac{1}{1 + x} = 1 - x + \cdots - x^{2n-1} + x^{2n} - \frac{x^{2n+1}}{1 + x}.$$
Now we temporary assume further that $a > 1$, so that $\alpha = a^{-1} \in (0, 1)$. Then
$$\begin{align*}
\int_{-\infty}^{\infty} \frac{e^{-z^2}}{1 + \alpha e^{-z^2}} \; dz
&= \int_{-\infty}^{\infty} \left( \sum_{k=1}^{2n+1} \alpha^{k-1} e^{-kz^2} - \frac{\alpha^{2n+1}e^{-(2n+2)z^2}}{1 + e^{-z^2}} \right) \; dz \\
&= \sum_{k=1}^{2n+1} (-1)^{k-1} \alpha^{k-1} \int_{-\infty}^{\infty} e^{-kz^2} \, dz - \alpha^{2n+1} \int_{-\infty}^{\infty} \frac{e^{-(2n+2)z^2}}{1 + e^{-z^2}} \; dz \\
&= \sum_{k=1}^{2n+1} (-1)^{k-1} \alpha^{k-1} \sqrt{\frac{\pi}{k}} - \alpha^{2n+1} \int_{-\infty}^{\infty} \frac{e^{-(2n+2)z^2}}{1 + e^{-z^2}} \; dz
\end{align*}$$
Now taking $n\to\infty$, the remainder term vanishes. Hence we have
$$ \int_{-\infty}^{\infty} \frac{e^{-z^2}}{1 + \alpha e^{-z^2}} \; dz
= \sum_{k=1}^{\infty} (-1)^{k-1} \alpha^{k-1} \sqrt{\frac{\pi}{k}}
= -a \sqrt{\pi} \, \mathrm{Li}_{1/2} \left( -\tfrac{1}{a}\right),$$
where
$$ \mathrm{Li}_{s}(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^s}$$
is the polylogarithm of order $s$, primarily defined on $|z| < 1$. Although we have proved this identity only for $a > 1$, the equality above can be used to define an analytic continuation of the right hand side, thus (by tautology) it holds for all $a > 0$.
It has special value at $\alpha = 1$, given by
$$ \int_{-\infty}^{\infty} \frac{1}{1 + e^{z^2}} \; dz
= -\sqrt{\pi} \, \mathrm{Li}_{1/2}(-1)
= \sqrt{\pi} (1 - \sqrt{2}) \zeta \left( \tfrac{1}{2} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/122411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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outer automorphisms of $S_6$ $$
\begin{array}{|l|c|c|}
\hline
\text{cycle structure} & \text{number of permutations} & \text{order} \\
\hline
6 & 120 & 6 \\
5+1 & 144 & 5 \\
4+2 & 90 & 4 \\
4+1+1 & 90 & 4 \\
3+3 & 40 & 3 \\
3 + 2 + 1 & 120 & 6 \\
3 + 1 + 1 + 1 & 40 & 3 \\
2 + 2 + 2 & 15 & 2 \\
2 + 2 + 1 + 1 & 45 & 2 \\
2 + 1 + 1 + 1 + 1 & 15 & 2 \\
1 + 1 + 1 + 1 + 1 + 1 & 1 & 1 \\ \hline
\end{array}
$$
In trying to understand outer automorphisms of $S_6$ at the most concrete level of arithmetic, it appears to me that the $15$ generators whose cycle structure is $2+1+1+1+1$ get mapped to the $15$ permutations whose structure is $2+2+2$ (which also generate the group). And the $40$ permutations whose structure is $3+1+1+1$ similarly get interchanged with the $40$ whose structure is $3+3$; likewise the $120$ whose structure is $3+2+1$ with the $120$ whose structure is $6$.
But apparently the set of $90$ whose structure is $4+2$ is invariant, as is the set of $90$ whose structure is $4+1+1$.
*
*Is this in accord with the experience of those who've actually thought about this?
*Does some web page somewhere walk through this? I would think the routine arithmetic would have been figured out.
| I think the best source for understanding the outer automorphisms of $S_6$ is the following paper:
Combinatorial Structure of the automorphism group of $\mathbf{S_6}$ by T.Y. Lam and David B. Leep, Expositiones Mathematicae 11 (1993), no. 4, pp. 289-903.
It describes exactly how to think about the exchange map between $2$-cycles and "pseudo-transpositions" $(ab)(cd)(ef)$, and it gives nice ways to interpret them combinatorially. I highly recomment it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under Trigonometry, I assumed the following.
Since $a,b$ are positive real numbers with $0 \leq a,b \leq 1$, I can assume that for some $x,y, a=\tan(x), b=\tan(y)$ and therefore it is to be shown that
$$\frac{1}{\sec x} + \frac{1}{\sec y} = \cos x+ \cos y \geq \frac{2\cos x \cos y}{\sqrt{cos(x-y)}}$$
(Originally posted without that $2$ on the right - Sorry!)
I do know that
$$\cos x + \cos y \geq 2 \sqrt{\cos x \cos y}$$
Now how to proceed? Just give me hints !
| Edit: Note that when the answer below was posted, and for a long time thereafter, the expression on the right was $\dfrac{1}{\sqrt{1+ab}}$.
I am puzzled. Suppose without loss of generality that $a\le b$. Then $1+a^2\le 1+ab$, and therefore $\dots$. So the desired inequality is true for any positive $a$, $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Expressing an algebraic function as a single expression I have the following question in my text book:
Express as a single fraction
$$ \frac{3}{x-4} - \frac{2}{(x-4)^2} $$
The answer the book gives is this:
$$ \frac{3x-14}{(x-4)^2} $$
I understand how they get to this through these steps:
$$ \frac{3(x-4)-2}{(x-4)^2} = \frac{3x-12-2}{(x-4)^2} $$
My question is why can you not cancel the $$ (x-4) $$ instead like this:
$$ \frac{3(x-4)-2}{(x-4)^2} = \frac{3-2}{(x-4)} = \frac{1}{(x-4)} $$
| It is never allowed to cancel across a plus or minus in a fractional expression. If it were possible, strange things would happen. Consider your expression when $x = 6$:
$$
\frac{3(x-4) - 2}{(x-4)^2} = \frac{3(6-4)-2}{(6-4)^2} = \frac{3(2) - 2}{2^2}
= \frac{4}{4} = 1.
$$
But if you tried canceling the $x-4$ first:
$$\frac{3(x-4) - 2}{(x-4)^2} \stackrel{?}{=} \frac{3-2}{(x-4)} = \frac{1}{x-4} = \frac{1}{6-4} =\frac{1}{2}.
$$
Not the same, right?
Hope this helps!
| {
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Induction Proof: Proof of Strict Inequality involving Exponents of 3 in the Denominator. I have been trying to do this problem by using induction but I became stuck halfway through:
Use induction to show that
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{n^3}\right) \lt 3 - \frac{1}{n^2}$$
for $n\geq 2$.
Does the series $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$ converge? Justify your conclusions.
So far I have this:
Base Case of Induction, $n=2$
$$\begin{align*}
\frac{1}{n^3} &\lt 3- \frac{1}{n^2}\\
\frac{1}{8} &\lt 3 - \frac{1}{4}\\
\frac{1}{8} &\lt \frac{11}{4}
\end{align*}$$
Induction Step: Assume true for some $k \geq 2 :$
$$
2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}$$
Show true with $n= k+1$
$$ 2 + 4 + \frac{2}{27} + \cdots + \frac{2}{k^3} + \frac{2}{(k+1)^3} \lt 3-\frac{1}{(k+1)^2}.$$
From there, I have no idea what to do.
I was planning to have
$$3 - \frac{1}{k^2} + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}$$
but I feel like it won't work since
$$2\left(1 + \frac{1}{8} + \frac{1}{27}+\cdots+\frac{1}{k^3}\right)$$
does not equal
$$3 - \frac{1}{k^2}.$$
| Your sum converges to
$$
\sum_{n=1}^\infty \frac{1}{n^3} = \zeta(3),
$$
because the series $ \sum_{n=1}^\infty \frac1{n^{1+\varepsilon}} $ (cf. Riemann zeta function) converges for every ε > 0, because
$$
\int_1^M\frac1{x^{1+\varepsilon}}\,dx =-\frac1{\varepsilon x^\varepsilon}\biggr|_1^M= \frac1\varepsilon\Bigl(1-\frac1{M^\varepsilon}\Bigr) \le\frac1\varepsilon \quad\text{for all }M\ge1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving the maximum of $(x^2-y^2-1)^2+4x^2y^2$ occurs at $x=0$, $y=\pm 1$. I'm trying to get some practice using the Maximum Modulus theorem, and want to use it to conclude that the maximum of $(x^2-y^2-1)^2+4x^2y^2$ occurs at $x=0$, $y=\pm 1$, supposing $x^2+y^2\leq 1$.
My thinking is I want to find some suitalbe complex function $f(z)$ such that $|f(z)|^2=(x^2-y^2-1)^2+4x^2y^2$, and then apply the Maximum Modulus principle, since I know the maximum will occur somewhere on the boundary $x^2+y^2=1$. However, I can't find such a function, so maybe I"m approaching it incorrectly.
I did manage to find that
$$
z^2-1=(x+iy)^2-1=(x^2-y^2-1)+2xyi
$$
which looked somewhat close, but no cigar. How can this be done better? Thanks.
| $x^2+y^2=1$
$(x^2-y^2-1)^2+4x^2y^2=(x^2+y^2)^2-2(x^2-y^2)+1=4-4x^2\leq 4$
$x=0,y=±1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$
Evaluate $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$
I was just wondering what I am doing wrong, as I don't seem to be arriving at the correct answer for $\sin(\frac{\pi}{8})$
What I did:
Let $\theta = \frac{\pi}{8}$
$\cos(2\theta) = 2\cos^2(\theta) - 1$
$\therefore \cos(\theta) = \sqrt{\frac{\cos(2\theta) + 1}{2}} = \frac{\sqrt{\sqrt{2} + 2}}{2}$
Now, $\sin(2\theta) = 2\cos(\theta)\sin(\theta)$
$\therefore \sin(\theta) = \frac{\sin(2\theta)}{2\cos(\theta)}$
Solving for $\sin(2\theta)$ and substituting in my answer for $\cos(\theta)$, I get:
$\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}$ but I have an answer saying that $\sin(\frac{\pi}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ and I couldn't seem to arrive at that.
| The trick is to rationalize the denominator by multiplying by the conjugate, along the following lines:
$$\begin{align}\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}&=\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}\cdot\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\\&=\dfrac{\sqrt{2}\sqrt{2-\sqrt{2}}}{2 \sqrt {4-2}}=\dfrac{\sqrt{\not 2}\sqrt{2-\sqrt 2}}{2 \sqrt {\not 2} }\\\ &=\frac{\sqrt{2-\sqrt{2}}}{2}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$. I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome
Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble by $2^{n+1}$.
here n belongs to natural numbers (positive integers).
| This is similar to Henry's answer, but things are simpler if we use $4+2\sqrt{3}=(1+\sqrt{3})^2$ and $4-2\sqrt{3}=(1-\sqrt{3})^2$ instead. These satisfy $x^2-8x+4=0$.
The sequence defined by
$$
a_0=2\text{, }a_1=8\text{, and }a_n=8a_{n-1}-4a_{n-2}\tag{1}
$$
has the solution
$$
\begin{align}
a_n
&=(4+2\sqrt{3})^n+(4-2\sqrt{3})^n\\
&=(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}\\
&=\left\lceil(\sqrt{3}+1)^{2n}\right\rceil\text{ for }n>0\tag{2}
\end{align}
$$
Let $b_n=a_n2^{-n-1}$. Then, $(1)$ becomes
$$
b_0=1\text{, }b_1=2\text{, and }b_n=4b_{n-1}-b_{n-2}\tag{3}
$$
Recursion $(3)$ insures that $b_n\in\mathbb{Z}$ for all $n\ge0$. Thus, for $n>0$, $(2)$ yields
$$
\begin{align}
\left\lceil(\sqrt{3}+1)^{2n}\right\rceil
&=a_n\\
&=2^{n+1}b_n\\
&\in2^{n+1}\mathbb{Z}\tag{4}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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Finding $\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx$ The question I am working on: Evaluate
$$\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx$$
So using integration by parts to solve:
(letting $u=(1-x)^4$ and $dv=x^4$)
$$\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x-1) - \frac{4}{5}\left(\frac{x^7}{7} - \frac{x^6}{6}\right)+c$$
Is it correct so far? If so ...
$$\int^1_0{ x^4 (1-x)^4 } dx = \frac{4}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac2{105}$$
$$\frac{1}{2} \int^1_0{ x^4 (1-x)^4 } dx = \frac{2}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac1{105}$$
But
$$\int\frac{{x^4(1-x)^4}}{1+x^2} dx = ??$$
Since I found $\int{{x^4(1-x)^4}} dx$. I thought of integration by parts, letting $u=\frac{1}{1+x^2}$, $dv = x^4(1-x)^4$. But I will get a very complicated $v$ to integrate later? Same if I did it the other way around?
| Hint :
Rewrite integral into form :
$$I=\int \frac{x^8+x^6}{1+x^2} \,dx + \int \frac{x^6+x^4}{1+x^2} \,dx +4\cdot \int \frac{x^6}{1+x^2} \,dx -4\cdot \int \frac{x^7+x^5}{1+x^2} \,dx$$
for third integral do long division .
| {
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"timestamp": "2023-03-29T00:00:00",
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Probabilistic method: what is the probability of the event that randomly selected composition of $n$ Can anyone help me out with the following question?
Q. What is the probability of the event that randomly selected composition of $n$ has a second part and that second part is $1$?
I know that the expected number, $E(X)$ that the first part of a randomly selected composition of $n$ is $2- 1/2^{n-1}$.
This problem is from 'A walk through Combinatorics' by Bona.
Thanks,
| Obviously for $n=1$ the probability is $0$. I suggest that you do some experimenting with small values of $n$, say $n=2,3,4$, and $5$. The compositions that meet the requirement are as follows.
$n=2$: $$1+1$$
$n=3$: $$\begin{align*}&1+1+1\\&2+1\end{align*}$$
$n=4$: $$\begin{align*}&1+1+1+1\\&1+1+2\\&2+1+1\\&3+1\end{align*}$$
$n=5$:
$$\begin{align*}
&1+1+1+1+1\\
&1+1+1+2\\
&1+1+2+1\\
&2+1+1+1\\
&1+1+3\\
&3+1+1\\
&2+1+2\\
&4+1
\end{align*}$$
Presumably you know already that there are $2^{n-1}$ compositions of $n$, so with these experimental data you ought to be able to make a good conjecture. And once you’ve made that conjecture, ask yourself why the number of ‘good’ compositions of $n$ is the same as ...
| {
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"timestamp": "2023-03-29T00:00:00",
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Basketball Team Combinatorial Question: A basketball team has 5 players, 3 in forward position (which includes a center) and 2 in guard position. In how many ways can we make a team if there are 6 forwards, 4 guards and 2 people who can play forward or guard?
The way I am approaching this question is as follows:
We need to pick 3 people out of 6 for the forward position: $\binom{6}{3}$.
We also need to pick 2 guards out of possible 4: $\binom{4}{2}$.
So far, the answer is: $\binom{6}{3} \times \binom{4}{2}$. (assuming that the 2 people who can play forward or guard statement is disregarded).
If two people can play forward or guard, I am thinking of assuming both play forward, add that to assuming one is forward one is guard, plus assuming two are guards. So:
$\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$
Would this yield the right answer? If not, why?
| When you write $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$, you consider the possibility of not using the free positioned player at all, thrice. So, I think you should subtract $2\times \binom{6}{3}\times \binom{4}{2}$ from $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$ How to prove: $$\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$$ Help please. Don't know where to start.
| $$\sin2\theta$$
$$2\cdot\sin\theta\cdot \cos\theta$$
multiply and divide by $\cos\theta$
$$ 2\cdot \dfrac {\sin\theta}{\cos\theta}\cdot \cos^2\theta$$
$$2\cdot\tan\theta\cdot\cos^2\theta$$
$$\dfrac{2\cdot\tan\theta}{\sec^2\theta}$$
$$\dfrac{2\cdot\tan\theta}{1+\tan^2\theta}$$
| {
"language": "en",
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"source": "stackexchange",
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How to $\int \sqrt{10x-x^2}dx$ Theres a hint to use $x=5+5\sin{t}$. Ok, but how do I know what substitution to use if a hint wasn't given? Is it "trivial" or perhaps, its very unlikely that that will appear?
Anyways, I did:
$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\
= \int \sqrt{50+50\sin{t} - (25+50\sin{t} + 25\sin^2{t})} dx\\
= \int \sqrt{ 25-25\sin^2{t} } dx \\
= 5 \int \sqrt{1-\sin^2{t}} dx \\
= 5\sin^{-1}{\sin{t}} \\
= 5t \\
= 5 \sin^{-1}{\frac{x-5}{5}}$
But the answer was:
$$\frac{25}{2}\sin^{-1}{\frac{x-5}{5}}+\frac{x-5}{2}\sqrt{10x-x^2}+c$$
What did I do wrong? Or is the answer wrong perhaps?
| For these types of problems, try to substitute the variable $x$ so that the integral reduces into a form that is easier to integrate.
For this integral, I would start by completing the square in the integrand:
$\sqrt{10x-x^2} = \sqrt{-(x^2-10x)}= \sqrt{-((x-5)^2-25)}=\sqrt{25-(x-5)^2}$.
Note that, so far, I have made no substitution... I have just fiddled around with the integrand.
Now I am ready to substitute. Let $x-5 = t$.
Then the integrand turns into $\sqrt{25-t^2} = 5\sqrt{1-\frac{t^2}{5^2}}$ (which you ought to know how to integrate... either off hand or by further substitution)
and also
$\frac{dx}{dt} = 1 $ which implies $dx = dt$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify a trigonometric expression I've been simplying a slew of trigonometric expressions and most of them fall out pretty clearly, but this one has been giving me fits:
$$\frac{(\sec x - \tan x)^{2} + 1}{\sec x\csc x - \tan x\csc x}$$
Here's my best attempt so far:
$$\begin{align*}
&=\frac{\left(\frac{1}{\cos x}-\frac{\sin x}{\cos x} \right)^{2} + 1}{\frac{1}{\cos x\sin x}-\frac{1}{\cos x}}\\\\
&=\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}}
\end{align*}$$
But I can't help but think I've already gone wrong since I feel like I should start factoring things out again. Am I on the right track here?
Thanks for any suggestions.
| You have
$$\begin{align*}
&=\frac{(\frac{1}{\cos x}-\frac{\sin x}{\cos x})^{2} + 1}{\frac{1}{\cos x\sin x}-\frac{1}{\cos x}}\\
&=\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}}
\end{align*}$$
Your next steps might be to note that on top, $\sin x/\cos x = \tan x$ and $1 + \tan^2 x= \sec^2 x$, leaving you with $2\dfrac{(1 - \sin x)}{\cos ^2 x}$
On the bottom, you have $\dfrac{\cos x}{\sin x}\dfrac{(1 - \sin x)}{\cos ^2 x}$
So cancel all that you can, and I think you'll be able to finish from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Show that $\sum\limits_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\frac{N+1}{2}$ I have recently found this excercise and was not able to solve it so far.
Show that
$$\sum_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\frac{N+1}{2}\;,$$ where $m \in \lbrace1,2,...,N\rbrace$.
This was one of my attempts:
$$\sum_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\sum_{i=1}^N\frac{1}{2}\left[\cos(0)-\cos\frac{2m\pi i}{N+1}\right]=\frac{1}{2}-\frac{1}{2}\sum_{i=1}^N\cos\frac{2m\pi i}{N+1}$$
It seems the last sum should be equal to $-N$. Can anyone give me a clue?
| $$
\sum_{k=1}^{n} \cos\left( \frac{2 m \pi k}{n+1}\right) = \Re \sum_{k=1}^{n} \exp\left( i \frac{2 m \pi k}{n+1}\right) = \Re\left( \sum_{k=1}^n z^k \right) = \Re\left( \frac{1-z^n}{1-z} z \right)
$$
where $z= \exp\left( i \frac{2 m \pi}{n+1}\right)$. Since $z^{n+1} = 1$:
$$
\Re\left(\frac{z-1}{1-z}\right) = -1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Root equation - What am I missing? There's a problem of which I know the solution but not the solving process:
$(\sqrt{x} + 7)(\sqrt{x} - 1) = \frac{105}{4}$
I'm convinced that up to:
$x + 6\sqrt{x} - 7 = \frac{105}{4}$
everything is correct.
But afterwards, I never seem to be able to get to the solution of $x = \frac{49}{4}$
How can this equation be solved?
| Hint $\ $ Put $\:z = \sqrt{x},\:$ clear denominators and rearrange to obtain the quadratic equation$\:0 = 4 z^2 + 24 z - 133 = (2z-7)(2z+19),\:$ solvable by the quadratic formula, or the Rational Root Test or AC method, etc.
Alternatively, purely arithmetically, if it has rational root $\:\sqrt{x} = a/b\:$ in lowest terms then
$$ (\sqrt{x} +7)(\sqrt{x}-1)\ =\ \frac{a+7b}b\ \frac{a-b}{b}\ =\ \frac{3\cdot 5\cdot 7}4 $$
Since $\:b\:$ is coprime to $\:a,\:$ also $\:b\:$ is coprime to $\:a+7b,\ a-b,\:$ so we deduce $\:b = 2.\:$ Hence
$$ (a + 14)\:(a-2)\ =\ 3\cdot 5\cdot 7 $$
So we seek a factorization of $\:3\cdot 5\cdot 7\:$ whose factors differ by $\:16 = a+14-(a-2).\:$ Checking the few possible factorizations we find $\:21\cdot 5\:\Rightarrow\:a=7,\:$ and $\:-5\:(-21)\:\Rightarrow\:a = -19.\:$ Notice that this solution depends on uniqueness of factorization (as does RRT and the AC method).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/145498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Square root of negative numbers If:
$$a = \sqrt{ b^2 - b }$$
The problem I have is that for values of:
$0 < b < 1$
the result of:
$b^2 - b$
Is a negative number which gives rise to an error on Excel and my calculator.
I understand that negative numbers don't have square roots (I read it on Wikipedia at least), so how do I solve this for values of $b$ less than 1?
Thanks! :)
| If $b^2-b<0$, then $b-b^2>0$ and $$a = \pm i\sqrt{ b - b^2 },$$ where $i$ is the imaginary unit, which by definition is the unique complex number that satisfies $$i^2=-1\Leftrightarrow i=\pm\sqrt{-1}.$$
The complex numbers are numbers of the form $a+bi$, where $a$ and $b$ are real numbers. They appear e.g. in the solution of a quadratic equation with negative discriminant, such as this one $$x^2+x+1=0,$$ whose solutions are $$x=\dfrac{-1\pm\sqrt{1-4}}{2}=\dfrac{-1\pm\sqrt{-3}}{2}=\dfrac{-1\pm\sqrt{3}\ i}{2}.$$
Example: For $b=1/2$, we have $b^2-b=1/4-1/2=-1/4$ and
$$a = \pm i\sqrt{ \frac{1}{2} - \frac{1}{4 }}=\pm i\sqrt{ \frac{1}{4} }=\pm \frac{1}{2}i .$$
We could have computed as follows
$$a = \sqrt{ \frac{1}{4 }-\frac{1}{2} }=\sqrt{ -\frac{1}{4} }=\sqrt{ -1}\sqrt{ \frac{1}{4} }=\sqrt{ -1}\frac{1}{2}=\pm i \frac{1}{2}=\pm \frac{1}{2}i .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Is there a quicker way of doing this integral? I have an integral $\int_0^\infty {x^2\over 1+x^4} dx$. I gave it a go and it turned out quite messy, so I consulted Wolfram Alpha but the steps given there seem rather long winded too. Is there is a faster way of doing the integral?
| $$\int_0^\infty\frac{x^2}{1+x^4}dx$$
Let, $t=\frac{1}{x}$, and, $dt=\frac{-dx}{x^2}$
$$\int_\infty^0 \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$$
$$=\int_\infty^0\frac{-dt}{1+t^4} =\int_0^\infty \frac{dt}{1+t^4}$$
Follow Norberts solution after that.
Another way to do this is
$$
I=\int_0^\infty\frac{x^2}{1+x^4}dx
$$
Let, $x= \sqrt{\tan\theta}$, then $dx=\frac{1}{2\sqrt{\tan\theta}}\sec^2\theta d\theta$
$$
I=\int_0^{\frac{\pi}{2}} \frac{\tan\theta}{1+\tan^2\theta}\times\frac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$$
$$
I=\frac{1}{2}\times\int_0^{\frac{\pi}{2}}\sqrt{\tan\theta}
$$
also,
$
I=\frac{1}{2}\times\int_0^{\frac{\pi}{2}}\sqrt{\cot\theta}
$
hence,
$$
4I=\int_0^{\frac{\pi}{2}}\sqrt{\tan\theta}+\sqrt{\cot\theta}
$$
$$
4I=\int_0^{\frac{\pi}{2}} \frac{\sin\theta + \cos\theta}{\sqrt{\sin\theta\cos\theta}}
$$
$$=\sqrt2 \int_0^{\frac{\pi}{2}} \frac{(\sin\theta + \cos\theta)}{\sqrt{1-(\sin\theta - \cos\theta)^2}}$$
Let $t=\sin\theta - \cos\theta$, then
$$4I=\sqrt2 \int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}}$$
$$I=\frac{1}{2\sqrt2}\left(\sin^{-1}(1)-\sin^{-1}(-1)\right) =\frac{\pi}{2\sqrt2}$$
:)
$$ \int_0^1 \left(\sqrt[3]{1-x^7} \right)$$
$$\frac{a}{d}$$
| {
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Showing this sequence converges. Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?
I did the following.
Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$
\begin{align}
\left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \frac{2n^3+n}{n^3}-2 \right \rvert+ \lvert \frac{3n}{n+1}-3 \rvert\\
& = \left \lvert \frac{1}{n^2} \right \rvert+ \left \lvert -\frac{3}{n+1} \right \rvert\\
& =\frac{1}{n^2}+\frac{3}{n+1} \\
& = \frac{n+1+3n^2}{n^2+n^3} <?<\epsilon
\end{align}
Hence $x_n \rightarrow 2+3i$.
| You reached the point where you want to prove $\,\displaystyle{\frac{3n^2+n+1}{n^3+n^2}<\epsilon}\,$ , but $$\frac{3n^2+n+1}{n^3+n^2}\leq \frac{5n^2}{n^3}\leq \frac{5}{n}$$ so $\,\displaystyle{\frac{5}{n}<\epsilon \Longrightarrow n>\frac{5}{\epsilon}}\,$ , and thus it is enough to choose $\displaystyle{\,N_\epsilon:=\left[\frac{5}{\epsilon}\right] + 1}\,$ to have that
for any $\,n>N_\epsilon\,$ we have the wanted inequality (with [x] = the integer part of x)
| {
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} |
Determinant of matrix composition Considering a matrix given by composition of square matrices like:
$$
M = \begin{pmatrix}
A & B\\
C & D
\end{pmatrix}
$$
I want to calculate its determinant $|M|$. Consider that all components are square matrices of course.
Can this be related somehow to the components $A$, $B$, $C$ and $D$?
| Note that $$\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \begin{pmatrix} A & 0 \\ C & I \end{pmatrix} \begin{pmatrix} I & A^{-1}B \\ 0 & D - CA^{-1}B \end{pmatrix} = \begin{pmatrix} A - BD^{-1}C & BD^{-1} \\ 0 & I \end{pmatrix} \begin{pmatrix} I & 0 \\ C & D\end{pmatrix}$$
The middle factorization is valid assuming $A$ is invertible and the last factorization is valid assuming $D$ is invertible. Further, $$\det \left( \begin{pmatrix} X_{11} & X_{12}\\ 0 & X_{22} \end{pmatrix} \right) = \det(X_{11}) \det(X_{22})$$ and $$\det \left( \begin{pmatrix} X_{11} & 0\\ X_{21} & X_{22} \end{pmatrix} \right) = \det(X_{11}) \det(X_{22})$$
Hence, $$\det \left( \begin{pmatrix} A & B \\ C & D \end{pmatrix} \right) = \det(A) \det(D-CA^{-1}B) = \det(A-BD^{-1}C) \det(D)$$
The middle one assumes $A$ is invertible, and the last one assumes $D$ is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/147227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_{4,4} & a_{4,5} & a_{4,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{5,4} & a_{5,5} & a_{5,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{6,4} & a_{6,5} & a_{6,6} & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & a_{7,7} & a_{7,8} & a_{7,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{8,7} & a_{8,8} & a_{8,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
Question.
What about its determinant $|A|$?.
Another question
I was wondering that maybe matrix $A$ can be expressed as a product of particular matrices to have such a structure... maybe using these matrices:
$$
A_1 =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{4,4} & a_{4,5} & a_{4,6}\\
a_{5,4} & a_{5,5} & a_{5,6}\\
a_{6,4} & a_{6,5} & a_{6,6}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{7,7} & a_{7,8} & a_{7,9}\\
a_{8,7} & a_{8,8} & a_{8,9}\\
a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
I can arrange $A$ as a compination of those: $A = f(A_1,A_2,A_3)$
Kronecker product
One possibility can be the Kronecker product:
$$
A=
\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_1 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_2 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1\\
\end{pmatrix} \cdot A_3
$$
But what about the determinant??? There are sums in this case which is not good...
| *
*The determinant of a block diagonal matrix is equal to the product of the determinants of the diagonal blocks. In your case, you have a block diagonal matrix of the form
$$A=\left(\begin{array}{ccc}
A_1 & 0 & 0\\
0 & A_2 & 0\\
0 & 0 & A_3
\end{array}\right)$$
so $\det(A) = \det(A_1)\det(A_2)\det(A_3)$.
*You can, though it is a bit ad-hoc. For example, note that if we let
$$T_1 = \left(\begin{array}{ccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0
\end{array}\right)$$
then $T_1^tA_1T_1$ is the block diagonal matrix
$$\left(\begin{array}{ccc}
A_1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right).$$
Likewise, if we let
$$\begin{align*}
T_2 &=\left(\begin{array}{ccccccccc}
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0
\end{array}\right)\\
T_3 &= \left(\begin{array}{ccccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{array}\right)
\end{align*}$$
then the resulting matrix is
$$f(A_1,A_2,A_3) = T_1^tA_1T_1 + T_2^tA_2T_2 + T_3^t A_3 T_3.$$
| {
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"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
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Analyze the convergence or divergence of $\{1/n^2\}$
Analyze the convergence or divergence of $\{1/n^2\}$
$$\left|\frac{1}{m^2} - \frac{1}{n^2}\right| < \left|\frac{1}{m^2}\right| < \frac{1}{N^2} < \varepsilon$$
whenever $N > \dfrac{1}{\sqrt{\varepsilon}}$ (first step because $n^2 > 0$ and hence $\dfrac{1}{n^2} > 0$ and $\dfrac{1}{m^2} - \dfrac{1}{n^2} < \dfrac{1}{m^2}$)
So the sequence is Cauchy and it certainly converges to zero.
Is my approach correct?
| The reasoning is not quite correct, but it is in fact a Cauchy sequence that converges to $0$.
$\frac{1}{n^2}>0$ is true when $n$ is a positive integer, and this does imply that $\frac{1}{m^2}-\frac{1}{n^2}<\frac{1}{m^2}$. However, that does not mean that $\left|\frac{1}{m^2}-\frac{1}{n^2}\right|<\frac{1}{m^2}$. For example, consider $n=1$, $m=4$.
Instead, you could note that $\left|\frac{1}{m^2}-\frac{1}{n^2}\right|$ is equal to either $\frac{1}{m^2}-\frac{1}{n^2}$ or $\frac{1}{n^2}-\frac{1}{m^2}$, and find a number that is an upper bound for both of these.
The reasoning could also be made more complete by specifying how $m$ and $n$ are related to $N$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to compute this integral involving a cdf? $\int_0^\infty\Phi(\frac{-x}{\sqrt{2}})d\Phi(x)=?$ where $\Phi(x)$ is the cumulative distribution function of a standard normal random variable.
| Consider $I(a) = \int_0^\infty \Phi(a x) \mathrm{d} \Phi(x)$. Differentiate with respect to $a$, and denote $\phi(x) = \Phi^\prime(x)$:
$$
I^\prime(a) = \int_0^\infty x \phi(a x) \phi(x) \mathrm{d} x = \frac{1}{2 \pi} \int_0^\infty x \mathrm{e}^{-\frac{(1+a^2) x^2}{2}} \mathrm{d} x = \frac{1}{2 \pi} \frac{1}{1+ a^2}
$$
Now, noting that $I(0) = \int_0^\infty \frac{1}{2} \mathrm{d} \Phi(x) = \frac{1}{4}$:
$$
I\left(a\right) = \frac{1}{4} + \frac{1}{2 \pi} \int_{0}^{a} \frac{\mathrm{d} a}{1+a^2} = \frac{1}{4} + \frac{1}{2 \pi} \arctan(a)
$$
Now $I\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{4} - \frac{1}{2 \pi} \arctan\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2 \pi} \arctan\left(\sqrt{2}\right) \approx 0.152043 $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate sums of inverses of binomial coefficients How to calculate the sum of sequence $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=？$$ How about its limit?
| As an alternative, starting from this result by robjohn's answer [tag (7)]
$$\sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}=\frac{\sum_{k=1}^{n+1}\frac{2^k}{k}}{\frac{2^{n+1}}{n+1}}$$
by Stolz-Cesaro we obtain
$$\frac{\frac{2^{n+2}}{n+2}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\frac{2}{2-\frac{n+2}{n+1}}\to 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/151441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 5,
"answer_id": 4
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Simultaneous equations My question is: Solve simultaneously:(anwers are in integers)
$$\begin{align} y^3 - 9x^2 + 27x - 27 &= 0 \\
z^3-9y^2+27y-27 &= 0 \\
x^3-9z^2+27z-27 &= 0 \end{align}$$
Any hints to solve this question would be greatly appreciated.
| Edit: It turns out that it is only integer solutions that are being asked about. Then there is a simpler approach that involves play with inequalities. Below is the original answer, followed by a solution of the integer version of the problem.
A start: Note that the first equation can be rewritten as
$$y^3+(x-3)^3=x^3.$$
Similarly, the other two can be rewritten as
$$z^3+(y-3)^3=y^3$$
and
$$x^3+(z-3)^3=z^3.$$
Add up, cancel. We get
$$(x-3)^3+(y-3)^3+(z-3)^3=0.$$
If you are interested in integer or rational solutions, the problem was long ago settled by Euler, and (perhaps) earlier by Fermat. If you want to examine the system over the reals, there is more work to do.
Integer solutions, another approach:
First we note that by the case $n=3$ of Fermat's Last Theorem, any solution of $a^3+b^3=c^3$ in integers (with some possibly negative) has at least one of $a$, $b$, or $c$ equal to $0$. The case $n=3$ of FLT has a relatively elementary, though by no means simple proof that goes back to Euler. A proof may even have been given by Fermat, though there is documentary evidence only for the case $n=4$. Using FLT, we can easily classify all the solutions of our system.
Now we show how to solve the problem without using the case $n=3$ of FLT. First we do a simplification that makes the coefficients smaller.
From the first equation, it is clear that $y^3$ is divisible by $3$, so $y$ must be. Let $y=3v$. Similarly, $z$ is divisible by $3$, say $z=3w$, and $x$ is divisible by $3$, say $x=3u$.
After the substitutions have been done, we arrive at the equations
$$\begin{align} v^3 &=3u^2 -3u+1 \\
w^3 &=3v^2 -3v+1\\
u^3 &=3w^2 -3w+1
\end{align}$$
Note that for any real number $t$, we have $3t^2-3t+1\ge \frac{1}{4}$. So since we are looking for integer solutions, all of $u$, $v$, and $w$ are positive integers.
If $t$ is a positive integer, then $3t^2-3t+1 \lt 3t^2$. So from our system of equations we conclude that
$$v^3 \lt 3u^2, \qquad w^3 \lt 3v^2,\qquad u^3 \lt 3w^2.$$
From the first inequality we get $v^{27}\lt 3^9 u^{18}$. But by the third inequality, $u^{18}\lt 3^6 w^{12}$. But by the second inequality $w^{12} \lt 3^4 v^8$. Putting things together, we get
$$v^{27}\lt 3^{19}v^8,$$
from which it follows that $v^{19} \lt 3^{19}$, or more simply $v\lt 3$. By symmetry we also have $u \lt 3$ and $w \lt 3$. Thus each of our variables $u$, $v$, and $w$ must be $1$ or $2$, and now we are essentially finished.
| {
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"timestamp": "2023-03-29T00:00:00",
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Matrices with $A^3+B^3=C^3$ Problem: Find infinitely many triples of nonzero $3\times 3$ matrices $(A,B,C)$ over the nonnegative integers with
$$A^3+B^3=C^3.$$
My proposed solution is in the answers.
| OR OR OR, given
$$ x, y > 0, $$
let
$$ R \; = \;
\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
x & 0 & 0
\end{array}
\right) , \; \;
S \; = \;
\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
y & 0 & 0
\end{array}
\right) , \; \;
T \; = \;
\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
x + y & 0 & 0
\end{array}
\right) ,
$$
then
$$ R^3 = x I, \; \; S^3 = y I, \; \; T^3 = (x+y) I $$
and
$$ R^3 + S^3 = T^3. $$
OR
$$ S \; = \;
\left( \begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
2n^2 & 0 & 0
\end{array}
\right) , \; \;
T \; = \;
\left( \begin{array}{rrr}
0 & 0 & 1 \\
2 n & 0 & 0 \\
0 & 2 n & 0
\end{array}
\right) .
$$
Then
$$ S^3 = 2 n^2 I, \; \; T^3 = 4 n^2 I, $$
and
$$ S^3 + S^3 = T^3. $$
OR OR, given a Pythagorean triple
$$ a^2 + b^2 = c^2, $$
let
$$ R \; = \;
\left( \begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
a^2 & 0 & 0
\end{array}
\right) , \; \;
S \; = \;
\left( \begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
b^2 & 0 & 0
\end{array}
\right) , \; \;
T \; = \;
\left( \begin{array}{rrr}
0 & 0 & 1 \\
c & 0 & 0 \\
0 & c & 0
\end{array}
\right) ,
$$
then
$$ R^3 = a^2 I, \; \; S^3 = b^2 I, \; \; T^3 = c^2 I $$
and
$$ R^3 + S^3 = T^3. $$
| {
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Mod operator and divisibility by $3$? I wanted to prove that $n(n+1)(2n+1)$ is always divisible by three, for this I used the following approach:
$n = 0, 1\pmod2$
for $n = 0$;
$$n(n+1)(2n+1) \equiv 0\pmod3$$
for $n = 1$;
$$n(n+1)(2n+1) \equiv 0\pmod3$$
But now I am unsure if this the correct way of doing because of the following eg:
Let $k = \text{an odd number}\equiv 1\mod3$
for $k = 1$;
$2k + 1 \equiv 0\pmod3$, therefore for every odd number $2k + 1$ is divisible by $3$, but this is not true.
So my question is the approach I am using to prove $n(n+1)(2n+1)$ divisible by $3$ is wrong or only the second example is wrong?, and why?
| Every number $n$ is congruent to $0$, $1$, or $2$ mod $3$.
If $n \equiv 0 \pmod 3$, then $3 \mid n$.
If $n \equiv 1 \pmod 3$, then $2n \equiv 2 \pmod 3$ which implies $2n + 1 \equiv 3 \pmod 3$ which implies $2n + 1 \equiv 0 \pmod 3$.
If $n \equiv 2 \pmod 3$, then $n + 1 \equiv 3 \pmod 3$ which implies $n + 1 \equiv 0 \pmod 3$.
Hence for all for all $n$, $n(n + 1)(2n + 1)$ is divisible by $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Integral of $\int \tan^4 x \, \sec^6 x dx$ I am trying to find the integral of $$\int \tan^4 x \, \sec^6 x dx$$
I tried to rewrite as trig identities using $\sec^2 - \tan^2 = 1$ but that got me nowhere so I wrote it like this.
$$\int \frac{\sin^4x}{\cos^4 x} \frac{1}{\cos^6 x} dx$$
$$\int \frac{\sin^4x}{\cos^{10} x} dx$$
Then I use the idea that making u substitutions for cos will get rid of a power of sin so I just say that the power will go $4 3 2 1 0$ and I will get a $- + - + -$ sign change. I am not sure if this is correct or really how this works exactly but I did a few steps of it and it seemed to work correctly.
$$-1\int \frac{1}{u^{10}} dx$$
$$-1\int u^{-10} dx$$
$$-1 \times \frac{u^{-9}}{-9}$$
$$ \frac{u^{-9}}{9}$$
$$ \frac{\cos^{-9}}{9}$$
This is wrong and I am not sure why.
| Here is a relatively easier way. Remember that $\sec^2(x) = 1 + \tan^2(x)$. Hence the integral $$I = \int \tan^4(x) \sec^6(x) dx = \int \tan^4(x) \sec^4(x) \sec^2(x) dx\\ = \int \tan^4(x) \left(1 + \tan^2(x) \right)^2 \sec^2(x) dx $$
Let $\tan(x) = t$. Hence, $\sec^2(x) dx = dt$. Hence, we get that
$$I = \int t^4 \left( 1+t^2 \right)^2 dt = \int t^4 \left( 1+2t^2 +t^4 \right) dt = \int \left(t^4 + 2t^6 + t^8 \right) dt\\
=\dfrac{t^5}{5} + \dfrac{2t^7}{7} + \dfrac{t^9}{9} + C$$
Hence, we get that $$I = \dfrac{\tan^5(x)}{5} + \dfrac{2\tan^7(x)}{7} + \dfrac{\tan^9(x)}{9} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$ My question is: Solve $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$
I deduced that:$LHS= x+1-(x-2)$
I am unable to solve this equation. I would like to get some hints to solve it.
| $\sqrt {x^2 +2x + 1}-\sqrt { x^2-4x+4}= \sqrt{(x+1)^2} - \sqrt{(x-2)^2}=|x+1|-|x-2|=3$
$|x+1|-|x-2|=3$
1) $x\in(-\infty, -1)$$\Rightarrow$$|x+1|=-(x+1)=-x-1$, $|x-2|=-(x-2)=2-x$.
$|x+1|-|x-2|=3$$\Rightarrow$ $-x-1-2+x=3$$\Rightarrow$$-3=3$, this is a contradiction.
In this interval equation has no solution.
2) $x\in[-1, 2)$$\Rightarrow$ $|x+1|=x+1$, $|x-2|=-(x-2)=2-x$.
$|x+1|-|x-2|=3$$\Rightarrow$ $ x+1-2+x=3$ $\Rightarrow$$2x=4$$\Rightarrow$$x=2$.
$2\notin [-1, 2)$. Also in this interval equation has no solution.
3) $x\in(2, \infty)$$\Rightarrow$ $|x+1|=x+1$, $|x-2|=x-2$.
$|x+1|-|x-2|=3$$\Rightarrow$ $ x+1-x+2=3$ $\Rightarrow$$3=3$.
On this interval equation has infinity solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Strategies to find the set of functions $f:\mathbb R\to\mathbb R$ satisfing the functional equation: $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))$ My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case where this applies is the following:
Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equation:
$$f\big(x^3\big)+f\big(y^3\big)=(x+y)\Big(f\big(x^2\big)+f\big(y^2\big)-f(xy)\Big):\forall x, y\in\mathbb R$$
I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis.
Thanks in advance.
| Using the equation, we can see that $f(0) = 0$ is necessary. Let $y = -x$, then
\begin{gather}
f(x^3) + f(-x^3) = 0
\end{gather}
This implies that $f$ must be an odd function. Also letting $y = x$ we see that
\begin{equation}
f(x^3) = x f(x^2)
\end{equation}
Plugging this relationship into the LHS gives
\begin{align}
xf(x^2) + yf(y^2) & = (x + y)(f(x^2) + f(y^2) - f(xy)) \\
0 & = x f(y^2) + y f(x^2) - (x+y)f(xy) \tag{*}
\end{align}
If we let $y \to -y$ then by the symmetry of the function we have
\begin{align}
0 = xf(y^2) - yf(x^2) + (x-y)f(xy) \tag{**}
\end{align}
Adding $(^*)$ and $(^{**})$ we see that
\begin{equation}
x f(y^2) = y f(xy)
\end{equation}
If we let $y = 1$ we obtain
\begin{equation}
f(x) = x f(1)
\end{equation}
EDIT:
At this point we should check whether this function indeed satisfies the above relation. Letting $f(1) = c$, and plugging in we have
\begin{equation}
c(x^3 + y^3) = c(x+y)(x^2 + y^2 - xy) = c(x^3 + xy^2 -x^2 y + x^2 y + y^3 - xy^2) = c(x^3 + y^3)
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find the vertical and horizontal asymptotes of the function. I am asked to find the vertical and horizontal asymptotes of the equation:
$$f(x)=(a^{-1}+x^{-1})^{-1}$$
I simplify this to
$$f(x)=\frac{1}{a^{-1}+x^{-1}}$$
$$f(x)=a^1+x^1$$$$f(x)=a+x$$Which is some constant, graphed as horizontal line - that will not have a vertical or horizontal asymptote. Is my algebra terribly off?
| \begin{align}
f(x) & = (a^{-1} + x^{-1})^{-1} = \dfrac1{a^{-1} + x^{-1}}\\
& = \dfrac1{\dfrac1a + \dfrac1x} = \dfrac1{\dfrac{x+a}{ax}}\\
& = \dfrac{ax}{x+a} = \dfrac{ax + a^2 - a^2}{x+a} \text{(Adding and subtracting $a^2$ to the numerator)}\\
& = \dfrac{ax+a^2}{x+a} - \dfrac{a^2}{x+a} = \dfrac{a(x + a)}{x+a} - \dfrac{a^2}{x+a}\\
& = a - \dfrac{a^2}{x+a}
\end{align}
Now note that as $x \to -a^{-}$, we have $f(x) \to + \infty$ and as $x \to -a^+$, we have $f(x) \to -\infty$.
Hence, $x=-a$ is a vertical asymptote.
Now letting $x \to +\infty$, we get that $f(x) \to a^{-}$ and $x \to -\infty$, we have that $f(x) \to a^+$.
Hence, $f(x) = a$ is a horizontal asymptote.
Below is a plot of this curve in blue with $a=2$. The red line indicates the horizontal asymptote i.e. $f(x) = a$. The pink line indicates the vertical asymptote i.e. $x=-a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| $$\int\frac 1{1+x^4}dx=\frac12\int\frac{1+x^2+1-x^2}{1+x^4}dx$$
$$\int\frac{1+x^2}{1+x^4}dx=\int\frac{\frac1{x^2}+1}{\left(x-\frac1x\right)^2+2}dx$$
Set $x-\frac1x=\sqrt2\tan\phi$
$$\int\frac{1-x^2}{1+x^4}dx=-\int\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}dx$$
Set $x+\frac1x=\sqrt2\sec\psi$
Reference: Trigonometric substitution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
When does $q^b = (p^a - 1)/(p - 1)$? $\DeclareMathOperator{\Ord}{Ord}$When does $q^b = (p^a - 1)/(p - 1)$ for $p, q$ odd primes and $a, b$ odd integers $> 1$ ? If no examples are possible, please give a simple proof.
A proof of this for $q=3, p=5$ might be
Assume $3^b = (5^a - 1)/4$; then
\begin{equation}
5^a - 4 \cdot 3^b = 1. \tag{1}
\end{equation}
We show that the exponents $a,b$ in (1) are even.
We see that $3^b \mid 5^a - 1$ $\Rightarrow$ $5^a \equiv 1 \bmod 3$. Also $5 \mid 4\cdot3^b+1$ or $4 \cdot 3^b \equiv -1 \bmod 5$ so that $3^b \equiv 1 \bmod 5$.
In short:
\[
3^b\equiv1 \bmod 5, \quad 5^a\equiv1 \bmod 3.
\]
By a well known theorem: if $X^c \equiv 1 \bmod p$, then $\Ord(c,p) \mid c$.
By inspection, $\Ord(3,5) = 4$ so that $4 \mid b$.
Similarly, $\Ord(5,3) = 2$ so that $2 \mid a$.
This violates our assumption that $a,b$ are odd. Proof complete.
However, moving further:
As $a,b$ are both even we can write (1) as $1 = 5^{2A} - 4\cdot 3^{2B} = (5 - 2\cdot3^B)(5 + 2\cdot3^B)$. Thus, $5 + 2\cdot3^B = 1$ but there is no positive value of $B$ satisfying this. Therefore (1) is not true.
Note: I want to be clear that the above proof, if it is correct, is not mine but was given to me by someone I am not free to name. If it is incorrect, the fault is entirely mine.
| q^b = (p^a - 1)/(p - 1)
Now for q^b to be defined p is ne 1
p is an odd prime number therefore p = 3 or p = (6k - 1) or p = (6k + 1) where k is a positive integer.
when p = 3
q^b = (3^a - 1)/(3 - 1)
when p = 6k - 1
q^b = ((6k - 1)^a - 1)/(6k - 2)
= [(6k - 1 - 1)(6n + 1)]/(6k - 2) in atleast one of the instances.
= [(6k - 2)(6n + 1)]/(6k - 2)
= 6n + 1 where n is a positive integer
when p = 6k + 1
q^b = ((6k + 1)^a - 1)/(6k)
= [(6k + 1 - 1)(6N - 1)]/(6k) in atleast one of the instances.
= [(6k)(6N - 1)]/(6k)
= 6N - 1 where N is a positive integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
a representation of $f''$ only using terms of $f$ and constants. Using Taylor formulare How can I prove the following result?
$
f''\left( x \right) \approx Af\left( x \right) + Bf\left( {x + h} \right) + Cf\left( {x + 2h} \right) + Df\left( {x + 3h} \right)\,\,h > 0
$
Obviously I have to consider the taylor expansion and "playing" with the terms, but that's my problem . Sorry for asking this stupid things
| As for $1 \le i \le 3$
\[ f(x+ih) = f(x) + ihf'(x) + \frac 12 {i^2}{h^2}f''(x) + \frac 16{i^3}{h^3}f'''(x) + O(h^4)
\]
So we have for your right hand side
\begin{align*}
f(x)\bigl(A + B + C + D)\\
{} + f'(x)\bigl(B + 2C + 3D\bigr)h\\
{} + f''(x)\frac 12\bigl(B + 4C + 9D\bigr)h^2\\
{} + f'''(x)\frac 16 \bigl(B + 8C + 27D)h^3\\
{} + O(h^4)
\end{align*}
So we want to have
\begin{align*}
A + B + C + D &= 0\\
B + 2C + 3D &= 0\\
B + 4C + 9D &= 0
\end{align*}
Wlog we can let $A = 1$. Then our System gives
\begin{align*}
B + C + D &= -1\\
C + 2D &= 1\\
3C + 8D &= 1
\end{align*}
hence
\begin{align*}
B + C + D &= -1\\
C + 2D &= 1\\
2D &= -2
\end{align*}
so $D = -1$, $C = 3$, $B = -3$. Then $B + 8C + 27D = -3+24-27 = -6$. That gives
\[
f'''(x) = \frac{-f(x) + 3f(x+h) - 3f(x+2h) + f(x+3h)}{h^3} + O(h).
\]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$ How could we prove that
$$ \frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$
I have reduced it the form
$$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$
I am not sure what to do next to get snappy solution. Any ideas?
| This refers to the original question, which had the left hand side equal to $\frac{1}{2}$ instead of $\frac{1}{12}$.
They are not equal.
Your simplification is correct. Then we can rewrite the left hand side as
$$\left(\frac{4^{\ln(4)}}{3^{\ln(3)}}\right)^{1/\ln(3/4)}$$
so raising both sides of the equation to the $\ln(3/4)$ power, we get that the equation would be equivalent to
$$\frac{4^{\ln(4)}}{3^{\ln(3)}} \stackrel{?}{=} \frac{1}{2}^{\ln(3/4)}.$$
Rewriting $4^{\ln(4)}$ as $e^{(\ln 4)^2}$, $3^{\ln(3)}$ as $e^{(\ln(3))^2}$, and $\left(\frac{1}{2}\right)^{\ln(3/4)}$ as $e^{-\ln(2)\ln(3/4)}$, the equality would be equivalent to
$$\left(\ln 4\right)^2 - \left(\ln 3\right)^2 \stackrel{?}{=} -\ln(2)\ln\frac{3}{4}.$$
Now, $\ln(4) = 2\ln(2)$, and $\ln\frac{3}{4} = \ln 3 - 2\ln 2$. So the left hand side is equal to
$$4(\ln 2)^2 - (\ln 3)^2$$
while the right hand side is equal to
$$-\ln(2)(\ln 3 - 2\ln 2) = 2(\ln 2)^2 - (\ln 2)(\ln 3).$$
But $$4(\ln 2)^2 - (\ln 3)^2 \approx 0.714863$$
and
$$2(\ln 2)^2 - (\ln 2)(\ln 3) \approx 0.199406$$
As corrected, the right hand side now be, after the simplification
$$
\left(\frac{1}{12}\right)^{\ln(3/4)} = \exp\left(-\ln(12)\ln(3/4)\right).$$
The exponent can be simplified:
$$\begin{align*}
-\ln(12)\ln(3/4) &= -\left(\ln(3)+2\ln(2)\right)\left(\ln(3)-2\ln(2)\right)\\
&= \left(\ln(3)+2\ln(2)\right)\left(2\ln(2)-\ln(3)\right)\\
&= \left(2\ln(2)\right)^2 - \left(\ln 3\right)^2\\
&= 4(\ln 2)^2 - (\ln 3)^2.
\end{align*}$$
Since this is the same as the exponent of $e$ on the left hand side, we do indeed have
$$\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}.$$
There's nothing special about $3$ and $4$. Replacing them with arbitrary positive numbers $a$ and $b$ will lead to
$$\exp\left((\ln(a))^2 - (\ln(b))^2\right) \stackrel{?}{=} \exp\left(-\ln(ab)(\ln(b/a)\right)$$
which of course holds, since
$$-\ln(ab)\ln(b/a) = (\ln a + \ln b)(\ln a - \ln b)$$
giving the equality you have in the comment:
$$\frac{a^{1/\log_a(b/a)}}{b^{1/\log_b(b/a)}} = \frac{1}{ab}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all integer solutions for $x^2 - 2y^2 =2 $ I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2 =1$ $y=3, x=4$, so form Pell I get that $\alpha= (4+3\sqrt{2})^n$ for every integer $n$, and a private solution for $-\frac{1}{2}x^2+ y^2 =-1$ is $y=1, x=2$, so the total solution is $\alpha= (1+\sqrt{2}) \cdot (+/- (4+3\sqrt{2})^n)$. Are all these steps correct? and if not- how should I solve this one?
Thank you!
| The standard way is to find one solution to $x^2-2y^2=2$, e.g., $x=2$, $y=1$, and find the fundamental solution to $x^2-2y^2=1$, which is $x=3$, $y=2$, and then go $(2+\sqrt2)(3+2\sqrt2)^n$, etc., etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For prime $p>2: 1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$
Possible Duplicate:
Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}\pmod p$?
If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$
I'd love your help with proving the following claim:
For prime $p>2$:
$$1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p.$$
I instantly thought of Wilson Theorem which says that $1\cdot2\cdot3\cdot\cdot\cdot\cdot(p-1) \equiv (-1) \pmod p$, but I can't see how to use it.
I also tried to divide it to two cases, for $p \equiv 1 \pmod4$, and $p \equiv 3 \pmod4$, but again I didn't reach the conclusion.
Thanks a lot!
| $$1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} \pmod p$$
is the same as
$$1^2 2^2 3^2 \cdots
\left(\frac{p-1}{2}\right)^2
\equiv (-1)^{\frac{p-1}{2}} \pmod p,$$
since we are working modulo $p$.
Now we can split each square itself as $ i^2 = i(p-i)(-1) $
$$(-1)^{\frac{p-1}{2}} 123\cdot\cdot\cdot(p-1)\equiv (-1)^{\frac{p+1}{2}} \pmod p$$
from Wilson's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing the value of $(\frac{p-1}{2})!$ modulo $p$. I want to prove that for $p \geq 3$, and for $a=(\frac{p-1}{2})!$, if $p \equiv1\pmod 4$, then $a^2\equiv -1 \pmod p$, and if $p \equiv 3\pmod4$, then $a \equiv +/-1 \pmod p$.
For the first part, I used Wilson's theorem which says that for prime $p$, $(p-1)!=-1 \pmod p$.
so $a^2=((\frac{p-1}{2})!)^2=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))^2$ and since $p-k \equiv -k \pmod p$ we get that $a^2=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))((p-1)(p-2)(p-3)\cdot\cdot\cdot(-1)^{(p-1)/2}).$
so since $p \equiv 1\pmod 4$ and using Wilson I get the what I need. Is this correct? How should I prove the second part?
(what is the latex for +/- symbol?)
Thank you.
| You got the hard part. To summarize, we have $a^2\equiv (-1)^{(p+1)/2}\pmod{p}$.
If $p$ is of the form $4k+1$, then $(p+1)/2$ is odd, and therefore $a^2\equiv -1\pmod{p}$.
If $p$ is of the form $4k+3$, then $(p+1)/2$ is even, and therefore $a^2\equiv 1\pmod{p}$. It follows that $a\equiv 1\pmod{p}$ or $a\equiv -1\pmod{p}$.
This is because from $p$ divides $a^2-1$, we see that $p$ divides $(a-1)(a+1)$. Therefore either $p$ divides $a-1$, in which case $a\equiv 1\pmod{p}$ or $p$ divides $a+1$, in which case $a\equiv -1\pmod{p}$.
Remark: Note that for example when $p=3$, then $a=1$, so $a\equiv 1\pmod{p}$. It is also the case that when $p=23$, we have $a\equiv 1\pmod{p}$. But, for example, when $p=7$, and when $p=11$, we have $a=\equiv -1\pmod{p}$. So for $p$ of the form $4k+3$, both $a\equiv 1\pmod{p}$ and $a\equiv -1\pmod{p}$ can occur. Whether $a$ in this case is congruent to $1$ or $-1$ turns out to be connected with the solvability of the congruence $x^2\equiv 2\pmod{p}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/166295",
"timestamp": "2023-03-29T00:00:00",
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If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
This seems clear, but I don't know how to prove this..
I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
| Lemma: $\gcd(a,b)=1\implies\gcd\left(a^2,b^2\right)=1$
Proof: By Bezout, $\gcd(a,b)=1$ implies
$$
\begin{align}
ax+by&=1\\
a^2x^2&=1-2by+b^2y^2\\
\left(2y-by^2\right)b&=1-a^2x^2\\
\left(2y-by^2\right)^2b^2&=1-2a^2x^2+a^4x^4\\
\left(2x^2-a^2x^4\right)a^2+\left(2y-by^2\right)^2b^2&=1
\end{align}
$$
Therefore, $\gcd\left(a^2,b^2\right)=1$.$\qquad\square$
Corollary: $\gcd(a,b)=1\implies\gcd\left(a^{2^n},b^{2^n}\right)=1$
Proof: Induction using the Lemma.$\qquad\square$
Thus, for any $k$, we can find an $n$ so that $k\le2^n$. The Corollary says that there are $x_n$ and $y_n$ so that
$$
a^{2^n}x_n+b^{2^n}y_n=1
$$
and therefore,
$$
a^k\left(a^{2^n-k}x_n\right)+b^k\left(b^{2^n-k}y_n\right)=1
$$
Thus, $\gcd(a,b)=1\implies\gcd\left(a^k,b^k\right)=1$.
| {
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"answer_id": 8
} |
Infinite product How do I solve the infinite product of $$\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}?$$
I know that I have to factorise to $$\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)},$$
but how do I do the partial product?
Thanks a lot in advance.
If I'm not mistaken the Answer is 2/3
| After factorization, the product looks like $(\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}\frac{5}{7}\cdots)(\frac{7}{3}\frac{13}{7}\frac{21}{13}\frac{31}{21}\frac{43}{31}\cdots)=(2)(1/3)=2/3$.$$$$ Here terms in first () are from expression $\frac{n-1}{n+1}$ and terms in second () from expression $\frac{n^2+n+1}{n^2-n+1}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Write each expression in the form $ca^pb^q$ Write each expression in the form $ca^pb^q$
c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$
\begin{align*}
&= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b}*\frac{b}{1}=\frac{2a^2b}{3}=\frac{2}{3}a^2b^1
\end{align*}
e) $\dfrac{a^{-1}}{(b^{-1})\sqrt{a}}$
\begin{align*}
&= \frac{1}{(b^{-1})a\sqrt{a}}=\frac{1b}{1a^1a^{\frac{1}{2}}}=\frac{1b^1}{1a^{\frac{2}{3}}}=1a^{\frac{-2}{3}}b^1
\end{align*}
These are my steps. Any corrections help.
| On what grounds did you move $b$ to the top on (c)? It's incorrect. You cannot just multiply by $\frac{b}{1}$ because it pleases you to do so. And, after you suddenly create a factor of $\frac{b}{1}$ ex nihilo, it would have cancelled with the denominator. So both the penultimate and antepenultimate equality signs are incorrect.
$$\begin{align*}
\frac{a(\frac{2}{b})}{\frac{3}{a}} &= \frac{2a}{b}\frac{a}{3}\\
&= \frac{2}{3}\frac{a^2}{b}\\
&= \frac{2}{3}a^2b^{-1}.
\end{align*}$$
(d) is almost correct, except that $1+\frac{1}{2}=\frac{3}{2}$, not $\frac{2}{3}$. So the exponent of $a$ is incorrect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Verify trigonometry equation $\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A$ Sorry for asking so many of these type of questions.
How would I verify the following trigonometry identity:
$$\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A.$$
My work is
$$\frac{\sin A + \frac{\sin A}{\cos A}}{\frac{\cos A}{\sin A} + \frac{1}{\sin A}}.$$
Do I have to use a common denominator between the sin and tan to solve the identity?
| $\begin{align}
\dfrac{\sin A + \tan A}{\cot A + \csc A} &= \dfrac{\sin A + \dfrac{\sin A}{\cos A}}{\dfrac{\cos A}{\sin A} + \dfrac{1}{\sin A}}\\
&= \dfrac{\dfrac{\sin A}{\cos A}\color{green}{\left(\cos A + 1\right)}}{\dfrac{1}{\sin A}\color{green}{(\cos A + 1)}}\\
&=\dfrac{\sin A}{\cos A} \div \dfrac{1}{\sin A}\\
&= \sin A \tan A
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
} |
Proving $\frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{z^2+zx+x^2}}+\frac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$
Let $x,y,z >0$. Prove
that:$$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}}
\ge \sqrt{3}$$
My solution:
By Hölder,
$$\left(\sum\frac{x}{\sqrt{4y^2+yz+4z^2}}\right)^2(\sum x(4y^2+yz+4z^2)) \ge (x+y+z)^3$$
Let us denote $\sum_{sym} x^2y = X$, so we have to prove
$$(x+y+z)^3 \ge 4X+3xyz \iff \sum x^3 +3xyz \ge X$$
which is Schur.
How to prove it in a different way ?
| The function $f(x) = \frac 1 {\sqrt x}$ is convex, by Jensen's inequality
$$
\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}} \geq \\ \frac {(x + y + z)^{3/2}} { \sqrt{x(y^2 + yz + z^2) + y(z^2 + zx + x^2) + z(x^2 + xy + y^2)} }
$$
Our inequality becomes
$$
x(y^2 + yz + z^2) + y(z^2 + zx + x^2) + z(x^2 + xy + y^2) \leq \frac 1 3 (x + y + z)^3
$$
Expanding both sides, it is reduced to
$$
xyz\leq \frac {x^3 + y^3 + z^3} {3}
$$
that is true by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
evaluate the integral: $\int_0 ^\sqrt5 \frac{4x}{\sqrt{x^2+4}}dx$ $$\int_0 ^\sqrt5 \frac{4x}{\sqrt{x^2+4}}dx$$
I got $4(5^\frac{1}{4})$ but I'm not sure if thats right
| Your computation is incorrect.
The substitution $u=x^2+4$ gives $du = 2x\,dx$. When $x=0$, we have $u=4$; when $x=\sqrt{5}$, we get $u=9$. So
$$\begin{align*}
\int_0^{\sqrt{5}}\frac{4x}{\sqrt{x^2+4}}\,dx &= 2\int_0^{\sqrt{5}}\frac{2x\,dx}{\sqrt{x^2+4}}\\
&= 2\int_4^9\frac{du}{\sqrt{u}}\\
&= 2 \int_4^9 u^{-1/2}\,du\\
&= 4u^{1/2}\Bigm|_4^9\\
&= 4(9^{1/2} - 4^{1/2})\\
&= 4(3-2)\\
&= 4.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to conceptualize conditional expectation inductively? Attempting the solve the following problem
A fair die is successively rolled. Let $X$ and $Y$ denote,
respectively, the number of rolls necessary to obtain a 6 and a 5.
Find
$E[ X|Y = 5]$
I am confused about how to conceptualize such a problem?
I know $E[X|Y= 1] = 1 + \frac{ 1}{ p} = 1 + \frac{ 1}{ 1/6} = 7$ building off the that we have a geometric random variable. But, I am unsure of how to extend this to a higher given expectation for $Y$.
I would like to solve this in an inductive mannor but
*
*I am unable to conceptualize an inductive solution?
*I am not sure that an inductive solution would even valid?
| I am not quite sure what kind of induction you are looking for, so I will just try giving a more general solution to this problem, i.e. $E[X|Y=y]$.
The point is to notice that random variable X distributed differently before and after the y-th roll, for tosses before the first one which yield 5 cannot have 5 as result, hence there are 1 out of 5 chances to get a 6.
The conditional probability is
$$P_{X|Y}(x,y)=\left\{ \begin{array}{1}(4/5)^{x-1}\times 1/5 ~ ~(X<Y)\\0 ~ ~(X=Y)\\(4/5)^{y-1}\times(5/6)^{x-y-1}\times1/6~~(X>Y)\end{array} \right.$$
Then follow the definition of conditional expectation we have:
$E[X|Y=y]\\=\sum\limits_{x}xP\{X=x|Y=y\}\\=\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}\frac{1}{5}+0+\sum\limits_{x=y+1}^\infty x(\frac{4}{5})^{y-1}(\frac{5}{6})^{(x-y-1)}\frac{1}{6}\\=\frac{1}{5}\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}+\frac{1}{6}(\frac{4}{5})^{y-1}\sum\limits_{x=y+1}^\infty x(\frac{5}{6})^{(x-y-1)}\\=\frac{1}{5}[5^{1-y} (-5(4^y)+5^{1+y}-4^yy)-y(\frac{4}{5})^{y-1}]+\frac{1}{6}(\frac{4}{5})^{y-1}6(y+6)\\=\frac{5}{4}(4-(\frac{4}{5})^y(y+4))+(\frac{4}{5})^{y-1}(y+6)$
To quickly verify this, plug in y=1 we will get 7 and for y=5 we have 5.8192.
Note:
The part of the equation to obtain $\sum_{x=y+1}^\infty x(\frac{5}{6})^{x-1}$ may not be obvious but the trick here is to use the series of $\sum_{k=0}^\infty(k+1)x^k=\frac{1}{(1-x)^2}$, which can be obtained by geometric series in the following way:
$$
x\sum (k+1)x^k-\sum (k+1)x^k=(x-1)\sum (k+1)x^k =1+x+x^2+...=\frac{1}{1-x}\Rightarrow\sum (k+1)x^k=\frac{1}{(1-x)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/176692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
My problem is with the extra $Y$ in this problem. What can I do about I think I know a solution which is to do $\tan(A+B)$ then $\tan(B+Y)$ but I am not sure how to apply it.
| We have
$$
\tan(A+B+C)=\tan(A+(B+C))=\frac{\tan A+\tan(B+C)}{1-\tan A \tan(B+C)}=
$$
$$
\frac{\tan A+\frac{\tan B+\tan C}{1-\tan B \tan C}}{1-\tan A\frac{\tan B+\tan C}{1-\tan B\tan C}}=
\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}
$$
In the last step we multiplied the numerator and the denominator by $1-\tan B\tan C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/177640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| For a more geometric proof:
Let $C$ be a cuboid whose edges are $a$, $b$ and $c$. If $\text{diam}(C)=\sqrt{a^2+b^2+c^2}$ is fixed, then $a+b+c$ is at most $\sqrt{3}\text{diam}(C)$. So $a+b+c \leq \sqrt{3(a^2+b^2+c^2)}$.
Then, if $\text{diam}(C)$ is fixed, $\text{Vol}(C)=abc$ is at most $\displaystyle \left( \frac{a^2+b^2+c^2}{3} \right)^{3/2}$ (you maximize the volume of a cuboid inscribed into a sphere of radius $\sqrt{a^2+b^2+c^2}/2$). So $a^2+b^2+c^2 \geq 3$.
Finally, $a+b+c \leq a^2+b^2+c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 3
} |
Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS=
$$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$
$$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$
$$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$
Now, considering the fact that I must have a common denominator to subtract, would this be correct:
$$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$
I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.
| $$\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}=\frac1{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
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Is this 3D curve a circle? The following is a curve in $3$ dimensions:
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/3) \\ z & = & \cos(\theta - 2\pi/3) \end{eqnarray}$$
Is the curve a circle?
If it is, what about this curve in $4$ dimensions?
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/4) \\
z & = & \cos(\theta - 2\pi/4) \\ w & = & \cos(\theta - 3\pi/4) \end{eqnarray}$$
I don't know if there is something like a circle in $4$-D. If there is, is this curve the $4$-D version of a circle?
P.S.: Is two-dimensional subspace the generalized plane? I want to learn more about this. What should I read?
| $\cos3\theta=cos3(\theta)$
$\cos3(\theta-\frac{\pi}{3})=cos(3\theta-\pi)=-\cos3\theta$
As $-y=-\cos(\theta-\frac{\pi}{3})=cos(\theta+\frac{2\pi}{3})$,
$\cos3(\theta+\frac{\pi}{3})=\cos(2\pi+3\theta)=\cos3\theta$
$\cos3(\theta-\frac{2\pi}{3})=cos(3\theta-2\pi)=\cos3\theta$
Now, $\cos3\theta=4cos^3\theta-3\cos\theta$
If $\cos3\theta=a$ and $\cos\theta=t$
So, $x,-y,z$ are the roots of $4t^3-3t-a=0$
$=>x+(-y)+z=0$
$=>x(-y)+(-y)z+zx=\frac{3}{4}$
$=>x^2+y^2+z^2=(x+(-y)+z)^2-2(x(-y)+(-y)z+zx)=0+2\frac{3}{4}$
$=>x^2+y^2+z^2=\frac{3}{2}$
Observe that $(x,y,z)$ satisfy a general plane equation $Ax+By+CZ+D=0$ where $A,B,C,D $ constants, not all zeros.
Also, satisfies the equation of the general circle in 3-D, $(x-a)^2+(y-b)^2+(z-c)^2=d^2 $. $a=b=c=0, d^2=\frac{3}{2}$
In case of $x,y,z,w$,
$z=\cos(\theta-\frac{2\pi}{4})=\sin\theta$,
$\sqrt2 y=\cos\theta+\sin\theta$,
$\sqrt2 w=-\cos\theta+\sin\theta$
So,$x^2+z^2=1$ and $w^2+y^2=1$
$x^2+y^2+z^2+w^2=2$
Now $\sqrt2 (y+w)=2\sin\theta=2z=>\sqrt 2z=y+w$
Similarly, $y-w=\sqrt 2x$
Observe that $(x,y,z,w)$ satisfies two general plane equations $Ax+By+CZ+Dw=E$ where $A,B,C,D,E $ constants, not all zeros.
Also, satisfies the equation of the general circle in 4-D, $(x-a)^2+(y-b)^2+(z-c)^2+(w-d)^2=e^2 $ . $a=b=c=d=0, e^2=2$
Again, we know $\cos nx=$Real part of $(\cos x+i\sin x)^n=(\cos x)^n+^nC_2(\cos x)^{n-2}(\sin x)^2+^nC_4(\cos x)^{n-4}(\sin x)^4+...$
Observe there is no term containing $=(\cos x)^{n-1}$
As $\cos n(2x-\frac{2r_i\pi}{n})=\cos(2nx-2r_i\pi)=\cos 2nx=C(say)$
So, $\cos (2x-\frac{2r_i\pi}{n})=R_i$(say), where all $r_i$s are distinct integers with $0 ≤r_i< n$ are the roots of the equation
$2^{n-1}y^n+C_1y^{n-2}+...-C=0$
So, $\sum R_i=0$ as the coefficient of $y^{n-1}$ is 0.
If $x_i=\cos (x-\frac{r_i\pi}{n})=>R_i=2(x_i)^2-1$
So, $\sum (2(x_i)^2-1)=0 =>\sum (x_i)^2=\frac{n}{2}$
This is another way of generalization("The curve lies on a sphere") already achieved by Rahul Narain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 3
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The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$ How to find the all continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
| Let $f: \mathbb R \to \mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $\tilde f := f-g$ is also a solution and we have $\tilde f(-1) = \tilde f(1) = 0$. We will show $\tilde f \equiv 0$, so that $f = g$.
By replacing $f$ with $\tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f \equiv 0$. Let $Z = \{x \in \mathbb R: f(x) = 0 \}$. A priori we have $\{-1,+1\} \subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields
$$f(x) = \frac{f(x+1) + f(x-1)}{2},$$
which implies $\mathbb Z \subseteq Z$. If we can show
$$2x,x \in Z \Rightarrow \frac{x}{2} \in Z$$
we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $\mathbb R$ (here the continuity of $f$ is needed). So let $2x \in Z$ and $x \in Z$. Plug $(\frac{x}{2},\frac{x}{2},x)$ into the functional equation to get
$$2f\left(\frac{x}{2}\right) + f(x) + f(2x) = f(x) + 2f\left(\frac{3}{2}x\right),$$
hence $f\left(\frac{x}{2}\right) = f\left(\frac{3}{2}x\right)$. Then plugging in $(\frac{x}{2},\frac{x}{2},\frac{x}{2})$ yields
$$3f\left(\frac{x}{2}\right) + f\left(\frac{3}{2}x\right) = 3f(x),$$
hence $f(\frac{x}{2}) = 0$ and $\frac{x}{2} \in Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 3
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Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots. Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots.
for real roots, $a \gt 0$, $b^2-4{a}{c} \ge 0$
Here we have $c=1$, and $a \ge 0$
Now we need to have $b^2-4a \ge 0$
i.e. $(-2\sqrt{a} \ge b) \cup (b \ge 2\sqrt{a})$
| You have basically answered the question yourself. You just need $a>0$ which is automatic and $b^2\geq 4ac=4a$. So for $a=1$, $b$ can be $2,3,4$, for $a=2$, $b$ can be $3,4$ and for $a=3$ or $4$, $b$ can be only $4$.
Thus there are $7$ possible polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can a partial differential equation have two different solutions? Consider:
$$x^2p+y^2q=(x+y)z$$ where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$. Thus by Lagrange's Method
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$$
$$\Rightarrow \frac{dx}{x^2}=\frac{dy}{y^2}
\Rightarrow \frac{1}{x}-\frac{1}{y}=k$$
$$\Rightarrow \frac{dx}{x^2}=\frac{dz}{(x+y)z}
\Rightarrow -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$
$$\Rightarrow \phi (\frac{1}{x}-\frac{1}{y},x+y+x\ln(z))=c$$
Or if we explore further
$$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$
So can we even write $f(\frac{1}{x}-\frac{1}{y},xyz)=k$. So which one is right?
Soham
| Both are not right.
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$
$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$
$\dfrac{dz}{dt}=(x+y)z=\left(-\dfrac{1}{t}-\dfrac{1}{t+y_0}\right)z$ , we have $z(x,y)=\dfrac{f(y_0)}{t(t+y_0)}=xy~f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $a_n$ for very large $n$ where $a_n = a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3} $ I have a recurrence relation,
$$ a_n = a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3} $$
for $n>3$ and $a_1 = 0, a_2 = 0, a_3 = 1$
I have to find the value of $a_n$ for very large values of n. I tried an approach similar to that which we use for finding the fibonacci numbers using matrix method which gives us $f_n$ in $log(n)$ time complexity. But I am not able to use it here because of the $2^{n-3}$ term.
Can we do better than $O(n)$ time complexity.
| For $n \in \mathbb N$, we let $b_n = a_n - 2^n$. We get by using the equation for $(a_n)$:
\begin{align*}
b_n &= a_n - 2^n\\
&= a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3} - 2^n\\
&= b_{n-1} + 2^{n-1} + b_{n-2} + 2^{n-2} + b_{n-3} + 2^{n-3} + 2^{n-3} - 2^n\\
&= b_{n-1} + b_{n-2} + b_{n-3} + 2^{n-3}\cdot(4 + 2 + 1 + 1 - 8)\\
&= b_{n-1} + b_{n-2} + b_{n-3},
\end{align*}
with the initial values $b_1 = -2$, $b_2 = -4$, $b_3 = -7$.
So for $(b_n)$ we got rid of the $2^{n-3}$-term and can use the same trick as for the Fibonacci numbers.
| {
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Inequality. $(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $
Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality)
$$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$
Thanks :)
| The inequality is obviously true if $a = b$, because we have $2(n+1)a^{n+1}\geq 2na^{n+1}$, which is true for $a\geq 0$.
Now, if $a\neq b$, we have $a^n+a^{n-1}b+\dots+b^n=\displaystyle\frac{a^{n+1}-b^{n+1}}{a-b}$. Let's assume $a>b$ (otherwise we can simply inverse the roles of $a$ and $b$). The inequality to prove is equivalent to $(n+1)(a^{n+1}+b^{n+1})(a-b)\geq (a+b)(a^{n+1}-b^{n+1})$, or $(n+1)(a^{n+2}-b^{n+2} +ab(b^n - a^n))\geq a^{n+2}-b^{n+2}+ab(a^n-b^n)$, or again $n(a^{n+2}-b^{n+2})\geq nab(a^n-b^n)$, or $(a^{n+2}-b^{n+2})\geq ab(a^n-b^n)$, or $1-x^{n+2}\geq x(1-x^n)$ with $x = a/b$ (if $a=0$, then necessarily $b=0=a$ and this case has been covered previously).
Finally we need to prove that $\forall x\in [0,1), 1-x^{n+2}-x(1-x^n)=\varphi(x)\geq 0$.
Now $\varphi'(x) = -(n+2)x^{n+1}-1+(n+1)x^n = -1 -((n+2)x - (n+1))x^n$
We can call $g(x) = ((n+2)x - (n+1))x^n$. We have $g'(x)=(n+2)x^n + n((n+2)x-(n+1))x^{n-1}=((n+1)(n+2)x - n(n+1))x^{n-1}$. $g'(x) = 0$ when $x = x_n = \frac{n}{n+2}$, so $g$ is maximized in $x_n$ and $g(x_n) = -(n/(n+2))^n< 1$, which proves that $\varphi'<0$, and thus $\varphi$ is decreasing, which finally achieves to prove the initial inequality.
| {
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Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.
$$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$
First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?
Integrate by parts with
$$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$
which leads to
$$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$
Since $(n-1)$ is a constant, we can throw it out front of the integral:
$$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$
I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$
$$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$
According to Wikipedia as noted here, this simplifies to:
$$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$
Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?
Assuming knowledge of equation 5, I see how to rewrite it as
$$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} \tag{6}$$
and solve for $I_n$. I had tried exploiting the fact that
$$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$
and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$
which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:
$$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$
but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of
$$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
| I guess your problem is step four:
$$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$
Note that
$$\begin{align}\int(1-\cos^2 x) \ \cos^{n-2} x \ dx & = \int \cos^{n-2} x \ dx-\int \cos^2 x\; \cos^{n-2} x \ dx \\ & =\int \cos^{n-2} x \ dx-\int \cos^{n} x \ dx \end{align}$$
so we get
$$\begin{align}I_n & = \sin x \ \cos^{n-1} x + (n-1)\int \cos^{n-2} x \ dx-(n-1)\int \cos^{n} x \ dx \\ I_n & = \sin x \ \cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n \end{align}$$
or
$$nI_n = \sin x \ \cos^{n-1} x + (n-1)I_{n-2}$$
$$I_n =\frac{ \sin x \ \cos^{n-1} x}n + \frac{n-1}nI_{n-2}$$
| {
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Finding $\sum\limits_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $ How we can find
$$\sum_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $$
| Hint: Compute $(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n+1}+\sqrt[4]{n})(\sqrt[4]{n+1}-\sqrt[4]{n})$.
*
*Warm up: Simplify the expression $(a^2+b^2)(a+b)(a-b)$.
*Pre-warm up: Simplify the expression $(a+b)(a-b)$ and deduce that $\sum\limits_{n=1}^{99}\frac1{\sqrt{n+1}+\sqrt{n}}=9$.
| {
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Rational number solution for an equation Does there exist $v=(a,b,c)\in\mathbb{Q^3}$ with none of $v$'s terms being zero s.t. $ a+b\sqrt[3]2+c\sqrt[3]4=0$ ?
And I was doing undergraduate algebra 2 homework when I encountered it in my head. At first It seemed like it can be proved there can be no such $v$ like how $\sqrt{2}$, or $\sqrt{2}+\sqrt{3}$ are proved to be irrational, but this case wasn't easy like those. Or maybe I was too hasty.
| I'll try to show some more elementary approach, not using field theory. Suppose there is such a $v$. We have $c \ne 0$ (as $\sqrt[3]2$ is irrational). Rewriting the equation, we find $\alpha, \beta \in \mathbb Q$ with
\[ \sqrt[3]4 = \alpha + \beta \sqrt[3]2 \]
and $\alpha, \beta \ne 0$ (as $\sqrt[3]2, \sqrt[3]4 \not\in \mathbb Q$).
Taking the third power, we get
\[ 4 = \alpha^3 + 3\alpha^2\beta \sqrt[3]2 + 3\alpha\beta^2\sqrt[3]4 + 2\beta^3 \]
so, as $\alpha\beta^2 \ne 0$,
\[ \sqrt[3]4 = \frac{4 - \alpha^3 - 3\alpha^2\beta\sqrt[3]2 - 2\beta^3}{3\alpha\beta^2} \]
Which gives
\[ \alpha + \beta \sqrt[3]2 = \frac{4-\alpha^3 - 2\beta^3}{3\alpha\beta^2} - \frac\alpha\beta \sqrt[3]2 \]
As $\sqrt[3]2$ is irrational, we must have
\[ \alpha = \frac{4-\alpha^3 - 2\beta^3}{3\alpha\beta^2}, \beta = -\frac\alpha\beta \]
So $\beta^2 = -\alpha$, giving
\[ -3\alpha^3 = 3\alpha^2\beta^2 = 4-\alpha^3 - 2\beta^3 \iff 2(\beta^3 - \alpha^3) = 4 \iff \beta^3 - \alpha^3 = 2 \iff \beta^3 + \beta^6 = 2
\]
But $x^6 + x^3 - 2$ has no rational zeros, as $\pm 1, \pm 2$ are the only possibilities. Contradiction.
So, there is no such $v$.
| {
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Trigonometry Table Problem Evaluate the given trigonometric expression:
\[
\frac{5\sin^2 30° + \cos^245° - 4\tan^2 30°}{2\sin30°\cdot\cos 30° + \tan 45°}
\]
| HINT: Posted below are some identities/axioms. $$\begin{aligned}\sin 30^{\circ} &= 0.5 \\ \\ \sin 45^{\circ} &= {1 \over \sqrt 2} \\ \\ \cos 30^{\circ} & = {\sqrt3 \over 2} \\ \\ \cos 45^{\circ}& = {1 \over \sqrt 2} \\ \\ \tan \theta &= {\sin \theta \over \cos\theta } \\ \\ \sin^2\theta &= (\sin\theta)^2 \\ \\ \tan^2\theta &= (\tan\theta)^2 \\ \\ \cos^2\theta &= (\cos\theta)^2 \\ \\ n\sin\theta &= n \times \sin \theta \end{aligned} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Exponential modular equation I am having some trouble in proving that the only solutions to
$$ -2^{m-1} \equiv m \pmod{7} $$
are $m \equiv 3,5, 13 \pmod{42}$.
What I tried to use:
If $-2^{m-1} \equiv m \pmod{7}$, then $2^{m-1} \equiv 6m \pmod{7}$ and we know that $2^6 \equiv 1 \pmod{7}$.
But now I can't go on. Could someone give me a hint of what should I be doing next?
| Note that modulo $7$, $2^m$ is periodic with period $3$. So $2^{m-1}+m$, modulo $7$, is periodic with period at most $21$.
Now one only needs to make a somewhat long table, starting at $m=1$. The powers of $2$, modulo $7$, are $1,2,4,1,2,4,1,2,\dots$. And $m$ is congruent to $1,2,3,4,5,6,0,1,\dots$. Add, check where we get $0$ modulo $7$.
Note that it first happens when $m=3$, since $4+3\equiv 0\pmod{7}$. Thus it happens at any $m\equiv 3\pmod{21}$. It also happens at $m=5$, since $2+5\equiv 0\pmod{7}$. Continue.
Remark: Modulo $42$, there will be more solutions, since the period is $21$.
| {
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Proof the logarithmic identity $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$ Please help me proof $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$, for $a,b,c>0$ and $a^2+b^2=c^2$. Thanks.
| Note that $\log_xy=\displaystyle\frac{\ln y}{\ln x}$, here $\ln$ is the natural log. Therefore, we have
$$\log_{b+c} a+\log_{c-b} a=\frac{\ln a}{\ln(b+c)}+\frac{\ln a}{\ln(c-b)}
=\ln a\left(\frac{\ln(b+c)+\ln(c-b)}{\ln(b+c)\ln(c-b)}\right)$$
$$=\ln a\cdot\frac{\ln[(b+c)(c-b)]}{\ln(b+c)\ln(c-b)}
=\frac{\ln a\ln(c^2-b^2)}{\ln(b+c)\ln(c-b)}=\frac{\ln a\ln(a^2)}{\ln(b+c)\ln(c-b)}$$
where we have used the assumption $a^2+b^2=c^2$ in the last equality.
Hence, we have
$$\log_{b+c} a+\log_{c-b} a=\frac{\ln a\ln(a^2)}{\ln(b+c)\ln(c-b)}
=\frac{2\ln a\ln a}{\ln(b+c)\ln(c-b)}$$
$$=2\cdot\frac{\ln a}{\ln(b+c)}\cdot\frac{\ln a}{\ln(c-b)}=2\log_{b+c}a\cdot\log_{c-b}a$$
as required.
| {
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Solving an equation with three quadratic radicals in the set of real numbers How do we solve the following equation in the set of real numbers?
$$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$
I tried putting $a=\sqrt{5x−1}$ and $b=\sqrt{5−2x}$ and then $2a^2+5b^2=23$.
| If you put $a=\sqrt{5x-1}$ and $b = \sqrt{5-2x}$, the equation says
$$ (26-x) a - (13x+14) b + 12 a b = 18 x + 32$$
The resultant of $(26-x) a - (13x+14) b + 12 a b - 18 x - 32$ and $5x-1-a^2$ with respect to $a$ is
$$ -169\,{x}^{2}{b}^{2}+5\,{x}^{3}-588\,{x}^{2}b+356\,x{b}^{2}-585\,{x}^{
2}+1808\,xb-340\,{b}^{2}+2280\,x-1520\,b-1700
$$
The resultant of this and $5 - 2 x - b^2$ with respect to $b$ is $(49 x - 10)^3 (x - 2)^3$.
So $x = 10/49$ or $2$. But now we have to check those by plugging in to the original equation. $x=2$ does work but $10/49$ doesn't work: in fact it would give you
$$ - \left( 26-x \right) \sqrt {5\,x-1}+ \left( 13\,x+14 \right) \sqrt {5
-2\,x}+12\,\sqrt { \left( 5\,x-1 \right) \left( 5-2\,x \right) }=18\,
x+32
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the trigonometric identity Please help me prove the identity:
$$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$$
| Implement the formula:
1) $1-\cos^2\alpha=\sin^2\alpha$
2) $\cos2\alpha=\cos^2\alpha-\sin\alpha$
3) $1=\sin^2\alpha+\cos^2\alpha$
Now turn the proof given identity.
$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$
$\cos^2\alpha(1-\cos^2\alpha)+\sin^4\alpha=\frac{1}{2}(1-\cos2\alpha)$
$\cos^2\alpha\sin^2\alpha+\sin^4\alpha=\frac{1}{2}(\sin^2\alpha+\cos^2\alpha-\cos^2\alpha+\sin^2\alpha)$
$\sin^2\alpha(\cos^2\alpha+\sin^2\alpha)=\frac{1}{2}\cdot 2\sin^2\alpha$
$\sin^2\alpha=\sin^2\alpha$
| {
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Find the solution of this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
Possible Duplicate:
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Help me solve this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
| Let $t_1=(\sqrt{2+\sqrt3})^x$, and $t_2=(\sqrt{2-\sqrt3})^x$. Now the given equation is:
$t_1+t_2=4$..........(1)
Above it follows that:
$t_1\cdot t_2=(\sqrt{2+\sqrt 3})^x\cdot(\sqrt{2-\sqrt3})^x=(\sqrt{({2+\sqrt3})({2-\sqrt3}})^x=(\sqrt{2^2-(\sqrt3)^2})^x=(\sqrt{4-\sqrt 3^2})^x=(\sqrt{4-3})^x=1^x=1$ $\Rightarrow$ $t_1\cdot t_2=1$...........................(2)
For (1) and (2) we have:
$t_1+t_2=4$
$t_1\cdot t_2=1$
$\Rightarrow$ $t^2-4t+1=0$
For this quadratic equation have:
$t_{1,2}=\frac{4\pm\sqrt{16-4}}{2}$
$t_{1,2}=\frac{4\pm 2\sqrt{3}}{2}$
$t_1=2+\sqrt 3$, $t_2=2-\sqrt 3$
Now return the inital substition:
$t_1=(\sqrt{2+\sqrt3})^x$
$2+\sqrt 3=(\sqrt{2+\sqrt3})^x$
$(\sqrt{2+\sqrt 3})^2=(\sqrt{2+\sqrt3})^x$ $\Rightarrow$ $x=2$, and
$(\sqrt{2+\sqrt3})^x=2-\sqrt 3$
$(\sqrt{2+\sqrt3})^x=\sqrt{(2+\sqrt 3)}^{-2}$ $\Rightarrow$ $x=-2$
Definitly $x=2$, and $x=-2$ is solve.
| {
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Multivariable limit - Two variables $ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$ How can I calculate the following limit and show that it equals $0$:
$$ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$$
Thanks in advance
| First I would change coordinates to $(x,z)$ where $z=\pi-y$. The limit becomes
$$\lim_{(x,z) \to (0, 0 ) } \frac{x^2 (\pi-z) \sin (\pi-z) } {\sin^2x + z^2 }$$
which we can evaluate by changing to polar coordinates and using the fact that near $0$, $\sin x=x+O(x^3)$. If you aren't familiar with big O notation, you can read about it on Wikipedia. This gives us
$$\begin{align}
\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {\sin^2(r\cos\theta) + r^2\sin^2\theta } &=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {(r\cos\theta+O(r^3\cos^3\theta))^2 + r^2\sin^2\theta }\\
&=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {r^2\cos^2\theta+O(r^4\cos^4\theta) + r^2\sin^2\theta }\\
&=\lim_{r \to 0 } \frac{\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {1+O(r^2\cos^4\theta)}\\
&=\frac{\cos^2\theta \cdot \pi\cdot \sin \pi } {1}=0\\
\end{align}$$
| {
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Summation manipulation. How can this be true?
\begin{align*}
\sum_{n=1}^\infty \sum_{k=n}^\infty \frac 1{k^3}
&= \sum_{k=1}^\infty \sum_{n=1}^{k} \frac 1{k^3}\\
\end{align*}
| On the left-hand side, imagine the following infinite sum of infinite sums:
$$\begin{aligned}
\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\
\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\
\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\
\frac{1}{4^3}+\cdots
\end{aligned}$$
This is equal to the infinite sum
$$\frac{1}{1^3}+\frac{2}{2^3}+\frac{3}{3^3}+\frac{4}{4^3}+\cdots$$
or
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots$$
Now look at the right-hand side. The expression $\frac{1}{k^3}$ does not depend on $n$ at all, so the inner sum is just $\frac{k}{k^3}$, and
$$\sum_{k=1}^\infty \frac{k}{k^3} = \sum_{k=1}^\infty \frac{1}{k^2}$$
which is the same as the left-hand side, as we showed above.
| {
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Is this composition of polynomial correct? Suppose $p(x)=x^2+5x+3$ and $q(x)=3x^3-3x+7$. Write the expression $(q \circ p)(x)$ as a sum of terms, each which is a constant times the power of $x$.
Here is my work for the problem:
$(q\circ p)(x)=3(x^2+5x+3)^3-3(x^2+5x+3)+7$
$(q\circ p)(x)=3(x^6+125x^3+27)-3x^2-15x-9+7$
$(q\circ p)(x)=3x^6+375x^3+81-3x^2-15x-2$
$(q\circ p)(x)=3x^6+375x^3-3x^2-15x+79$
Did I do something wrong while working with this problem? The final answer I got wasn't in the multiple choice answers. I went over this problem 3 times and cannot find what I am doing wrong.
| The mistake you made is assuming that $(a+b+c)^3=a^3+b^3+c^3$. Note that $(a+b+c)(a+b+c)=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2$. Simplify and multiply by $a+b+c$ again to get the cube. Alternatively you can rewrite $q(x)=3x(x^2-1)+7$ and work from there.
| {
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Fundamental Set of Complex Solutions I am confused on how to do this problem, it states:
Find a fundamental set of solutions and put it in general form for the given system.
$x' = \begin{pmatrix} -1/2 & 1\\ -1 & -1/2 \end{pmatrix}x$
I got the eigen values to be $\lambda_1$ = [ (-1/2) + i ] , $\lambda_2$ = [ (-1/2) - i ] so the corresponding eigen vectors are
For $\lambda_1 = v_1 = \begin{pmatrix} 1 \\ i \end{pmatrix}$
For $\lambda_2 = v_2 = \begin{pmatrix} 1\\ -i \end{pmatrix}$
But here is where I get confused on how to write the general solution using Euler's formula for $e^{it} = cost + isint$, thus we have that
$x_1(t) = e^{-t/2}(cost + isint)\begin{pmatrix} 1\\ i \end{pmatrix}$
The answer is below, but how did they get that?
$x_1(t) = \begin{pmatrix} e^{-t/2} cost\\ -e^{-t/2}sint \end{pmatrix} + i\begin{pmatrix} e^{-t/2}sint\\
e^{-t/2}cost \end{pmatrix} = u(t) + iw(t)$
$u(t) = \begin{pmatrix} e^{-t/2} cost\\ -e^{-t/2}sint \end{pmatrix} , w(t) = \begin{pmatrix} e^{-t/2}sint\\
e^{-t/2}cost \end{pmatrix}$
| Based on your derivations, the general solution has the form
$$ x(t) = c_1 \begin{pmatrix} 1 \\ i \end{pmatrix} {\rm e}^{(-\frac{1}{2}+i)t} + c_2 \begin{pmatrix} 1\\ -i \end{pmatrix} {\rm e}^{(-\frac{1}{2}-i)t}\,, $$
where $c_1,c_2$ are arbitrary constants. Simplifying further
$$ x(t) = c_1 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1 \\ i \end{pmatrix} {\rm e}^{it} + c_2 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1\\ -i \end{pmatrix} {\rm e}^{-it} $$
$$= c_1{\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1 \\ i \end{pmatrix}(\cos(t)+i\sin(t)) + c_2 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1\\ -i \end{pmatrix} (\cos(t)-i\sin(t))\,. $$
Now, make the above as a linear combination of $\cos(t){\rm e}^{-\frac{1}{2}t}$ and $\sin(t){\rm e}^{-\frac{1}{2}t}$.
$$ A\cos(t){\rm e}^{-\frac{1}{2}t} + B \sin(t){\rm e}^{-\frac{1}{2}t}\,, $$
where $A$ and $B$ are two constant vectors given by
$$ A = c_1\begin{pmatrix} 1 \\ i \end{pmatrix}+c_2 \begin{pmatrix} 1\\ -i \end{pmatrix} \,,\quad B = ic_1\begin{pmatrix} 1 \\ i \end{pmatrix}+ic_2 \begin{pmatrix} 1\\ -i \end{pmatrix} \,. $$
Or, you can write them in terms of new constants,
$$ A= \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}\,, \quad B = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/206935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Twist on log of sine and cosine integral $\int_{0}^{\frac{\pi}{2}}x\ln(\sin x)\ln(\cos x)dx$ I ran across this integral and have not been able to evaluate it.
$$\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\ln(\cos(x))dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$$
I had some ideas. Perhaps some how arrive at $\displaystyle\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{4}}=\frac{{\pi}^{4}}{192}$.
and $\displaystyle \ln(2)\int_{0}^{\frac{\pi}{2}}x\ln(2)dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}$
by using the identity $\displaystyle\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}=-x\ln(\sin(x))-x\ln(2)$
and/or $\displaystyle \ln(\cos(x))=-\ln(2)-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}$
I have used the first one to evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$, so I thought perhaps it could be used in some manner here.
I see some familiar things in the solution, but how to get there?.
Does anyone have any clever ideas?.
Thanks.
| Here is an approach that avoids using the Fourier series expansions for $\ln (\cos x)$ and $\ln (\sin x)$.
Consider
$$I = \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx.$$
Enforcing a substitution of $x \mapsto \pi/2 - x$ gives
\begin{align*}
I &= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln \left [\sin \left (\frac{\pi}{2} - x \right ) \right ] \ln \left [\cos \left (\frac{\pi}{2} - x \right ) \right ] \, dx\\
&= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln (\cos x) \ln (\sin x) \, dx\\
&= \frac{\pi^2}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx - \pi \int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx\\
& \qquad + \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx,
\end{align*}
or
$$\int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx.$$
The integral appearing on the right can be found by differentiating the beta function. As
$$\text{B}(x,y) = 2 \int_0^{\pi/2} \cos^{2x - 1} \theta \sin^{2y - 1} \theta \, d\theta,$$
we see that
$$\int_0^{\pi/2} \ln (\sin \theta) \ln (\cos \theta) \, d\theta = \frac{1}{8} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2},$$
giving
$$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{32} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2}.$$
Since
$$\partial_x \text{B}(x,y) = \text{B}(x,y) [\psi(x) - \psi (x + y)] \quad \text{and} \quad \partial_y \text{B}(x,y) = \text{B}(x,y) [\psi(y) - \psi (x + y)],$$
where $\psi (z)$ is the digamma function, we have
$$\partial_x \partial_y \text{B}(x,y) = \text{B}(x,y) \left [ \left \{\psi(x) - \psi (x + y) \right \} \left \{\psi(y) - \psi (x + y) \right \} - \psi^{(1)}(x + y) \right ].$$
Thus
\begin{align*}
\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx &= \frac{\pi}{32} \text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) \left [ \left \{ \psi \left (\frac{1}{2} \right ) - \psi (1) \right \}^2 - \psi^{(1)}(1) \right ].
\end{align*}
Since
$$\text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) = \frac{\Gamma (1/2) \Gamma (1/2)}{\Gamma (1)} = \pi,$$
and
$$\psi^{(1)}(1) = \zeta (2) = \frac{\pi^2}{6} \quad \text{and} \quad \psi \left (\frac{1}{2} \right ) - \psi (1) = - 2\ln (2),$$
we have
$$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi^2}{8} \ln^2 (2) - \frac{\pi^4}{192},$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 2
} |
Common tangent to two circles
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
and
$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.
| You can approach this with homogeneous coordinates. A circle with equation $(x-x_c)^2+(y-y_c)^2 - r^2 = 0$ is represented by a 3x3 matrix as
$$ {\rm Circle}(x_c,y_c,r) = \begin{bmatrix} 1 & 0 & -x_c \\ 0 & 1 & -y_c \\ -x_c & -y_c & x_c^2+y_x^2-r^2 \end{bmatrix} $$
This means that the equation for a circle $C_1 = \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 0 \\ -2 & 0 & -5 \end{bmatrix}$ is given by the quadratic form
$$ P^\top C_1 P = 0 $$
$$ x^2-4 x+y^2-5 = 0 $$
which is the equation for the first circle when expanded out, and $P=\begin{pmatrix} x&y&1 \end{pmatrix} ^\top $ is an arbitrary point.
The second circle is $ C_2 = \begin{bmatrix} 1 & 0 & -5 \\ 0 & 1 & -4 \\ -5 & -4 & 37 \end{bmatrix} $.
Now here is the fun stuff. A line in this notation in general is defined as $L=\begin{vmatrix}a&b&c\end{vmatrix}^\top$ such that the equation of the line is
$$ P^\top L =0 $$
$$ a x+b y+c = 0 $$
Actually $a$, $b$ above designate the direction of the line such that if the line makes an angle $\theta$ with the horizontal then the line is $L=\begin{vmatrix}-\sin\theta&\cos\theta&-d\end{vmatrix}^\top$ and $d$ is the distance of the line to the origin.
We are using the above information to find the lines that are tangent to both circles. A tangent line to the first circle has satisfies the equation
$$ L^\top C_1^{-1} L =0 $$
$$ d = \pm 3 -2 \sin \theta $$
with the two possible line equations
$$ L_A = \begin{vmatrix}-\sin\theta_A&\cos\theta_A
&2\sin\theta_A-3\end{vmatrix}^\top $$
$$ L_B = \begin{vmatrix}-\sin\theta_B&\cos\theta_B
&2\sin\theta_B+3\end{vmatrix}^\top $$
to find the orientations of these lines $\theta_A$ and $\theta_B$ we have to find the lines that are tangent to the second circle, and match the coefficients
$$ L^\top C_2^{-1} L =0 $$
$$ d = \pm 2 + 4 \cos \theta -5 \sin \theta $$
with also two possible line equations
$$ L_A = \begin{vmatrix}-\sin\theta_A&\cos\theta_A
&-4\cos\theta_A+5\sin\theta_A-2\end{vmatrix}^\top $$
$$ L_B = \begin{vmatrix}-\sin\theta_B&\cos\theta_B
&-4\cos\theta_B+5\sin\theta_B+2\end{vmatrix}^\top $$
Setting $L_A=L_A$ and solving for $\theta_A$ yields the following
$$ 2\sin\theta_A-3 = -4\cos\theta_A+5\sin\theta_A-2 $$
$$ 4\cos\theta_A-3\sin\theta_A =1 $$
$$ \sin\theta_A = \frac{8\sqrt{6}-3}{25} $$
with the solution
$$ L_A = \begin{vmatrix} \frac{3-8\sqrt{6}}{25} & \frac{4+6\sqrt{6}}{25} & \frac{16 \sqrt{6}-81}{25} \end{vmatrix} $$
$$ \left( \frac{3-8\sqrt{6}}{25}\right) x + \left(\frac{4+6\sqrt{6}}{25}\right) y + \left(\frac{16 \sqrt{6}-81}{25}\right) = 0 $$
$$ -0.664 x + 0.7479 y - 1.6723 = 0 $$
Similarly with $L_B=L_B$ yielding $4\cos\theta_B-3\sin\theta_B=-1$ or
$$ L_B = \begin{vmatrix} -\frac{3+8\sqrt{6}}{25} & \frac{6\sqrt{6}-4}{25} & \frac{16 \sqrt{6}+81}{25} \end{vmatrix} $$
$$ -\left(\frac{3+8\sqrt{6}}{25}\right) x + \left(\frac{6\sqrt{6}-4}{25}\right) y + \left(\frac{16 \sqrt{6}+81}{25}\right) = 0 $$
$$ -0.904 x + 0.4278 y + 4.808 = 0 $$
Here are the results plotted in GeoGebra for validation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/211538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Circle Least Squares Fit So my question is this:
Find the equation of the circle that gives the best least squares circle fit to the points $(-1,-2), (0,2.4), (1.1,-4),$ and $(2.4,-1.6).$
So far I have this general equation:
$2xc_1+2yc_2+(r^2-c_1^2-c_2^2)=x^2+y^2$ where $r^2-c_1^2-c_2^2 = c_3$
So then I think I create matrix:
$\begin{pmatrix} 2x_1 & 2y_1 & 1 \\ . & . & . \\ 2x_n & 2y_n & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} - \begin{pmatrix} x_1^2+y_1^2 \\ ... \\ x_n^2+y_n^2\end{pmatrix}$
and after replacing $x_1 = -1$ and $y_1 = -2$ until $x_4$ and $y_4$, I have this matrix:
$\begin{pmatrix} -2 & -4 & 1 \\ 0 & 4.8 & 1 \\ 2.2 & -8 & 1 \\ 4.8 & -3.2 & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} - \begin{pmatrix} 5 \\ 5.76 \\ 17.21 \\ 8.32\end{pmatrix}$
Which i guess I try to solve by setting equal to 0, then moving the last matrix to the other side and make it look like this:
$\begin{pmatrix} -2 & -4 & 1 \\ 0 & 4.8 & 1 \\ 2.2 & -8 & 1 \\ 4.8 & -3.2 & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} = \begin{pmatrix} 5 \\ 5.76 \\ 17.21 \\ 8.32\end{pmatrix}$
And finally solve for $c_1$, $c_2$, $c_3$. But this system has no solution as shown here. Am I missing something huge here or what am I doing wrong?
| Just multiply both sides of the equation by $A^{T}\,,$ where $A$ is your matrix, and then solve the resulting system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/214661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
A definite integral involving a parameter Does this integral seem to have a nice closed form (at least for a subset of values of $a > 0$)
$$
\int_{\sqrt{a}}^\infty \frac{y^2}{(y^2-a+1)^2} dy
$$
Using a symbolic math software and for $a$ being some small integers, I end up getting different answers in terms of $\sinh^{-1}$, $\log$, $\tanh^{-1}$ and $\coth^{-1}$ depending on $a$.
| Partial fractions does it. The partial fraction expansion of the integrand is
$$
\frac{1}{4(y-\sqrt{a-1})^2} + \frac{1}{4(y+\sqrt{a-1})^2}
+ \frac{1}{4 \sqrt{a-1} (y - \sqrt{a-1})} - \frac{1}{4 \sqrt{a-1} (y + \sqrt{a-1})}
$$
so an antiderivative is
$$ -\frac{1}{4(y-\sqrt{a-1})} - \frac{1}{4(y+\sqrt{a-1})} + \frac{1}{4 \sqrt{a-1}} \ln \left(\frac{y-\sqrt{a-1}}{y+\sqrt{a-1}}\right)$$
Thus the integral is
$$ \frac{1}{4(\sqrt{a}-\sqrt{a-1})} + \frac{1}{4(\sqrt{a}+\sqrt{a-1})} - \frac{1}{4 \sqrt{a-1}} \ln \left(\frac{\sqrt{a}-\sqrt{a-1}}{\sqrt{a}+\sqrt{a-1}}\right)$$
which simplifies to
$$ \frac{\sqrt{a}}{2} - \frac{1}{2 \sqrt{a-1}} \ln(\sqrt{a} - \sqrt{a-1})$$
This is actually valid for all $a > 0$ except $a=1$, but for $a < 1$ you may want to use an alternative form to avoid complex numbers:
$$ \frac{\sqrt{a}}{2} + \frac{1}{2 \sqrt{1-a}} \arctan\left(\sqrt{1/a-1}\right)$$
For $a=1$ you can get the answer $1$ directly, or by taking the limit of either of these as $a \to 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to integrate $\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? How to integrate $\displaystyle \int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? I Tried using partial fractions but i got lost, thanks.
| From $\frac{2x^2+ x}{(x+ 1)(x^2+ 1)}= \frac{A}{x+ 1}+ \frac{Bx+ C}{x^2+ 1}$ you can add the fractions on the right, then compare coefficients of "like terms". But we can also use the fact that this is to be true for all x. Setting x to any three values gives three equations to solve for A, B, and C.
First multiplying both sides by $x+1$ and $x^2+ 1$ simplifies to $2x^2+ x= A(x^2+ 1)+ (Bx+ C)(x+ 1).
Setting x= -1 immediately gives $2(1)+ (-1)= 1= A(2)$ so $A= \frac{1}{2}$.
Setting x= 0 gives $0= A+ C= \frac{1}{2}+ C$ so $C= -\frac{1}{2}$.
Setting x= 1 gives $2+ 1= 3= 2A+ 2(B+ C)= 1+ 2B- 1$ so $B= \frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/216742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Continued fractions with rational functions Express the following rational function in continued-fraction form:
$${4x^2+3x-7\over 2x^3+x^2-x+5}$$
The answer is :
$${4 \over 2x- \frac{1}{2}} + { \frac{23}{8} \over x-\frac{63}{92}}-{\frac{406}{529} \over x+\frac{33}{23}}\tag{inline continued fraction}$$
which means
$$
\cfrac{4}{2x- \frac{1}{2}+\cfrac{\frac{23}{8}}{x-\frac{63}{92}-\cfrac{\frac{406}{529}}{x+\frac{33}{23}}}}
$$
| Here is the solution using integer coefficients
$$
\begin{align}
\cfrac{4x^2+3x-7}{2x^3+x^2-x+5}
&=\cfrac1{\cfrac{2x^3+x^2-x+5}{4x^2+3x-7}}\\
&=\cfrac{8}{4x-1+\cfrac{23x+33}{4x^2+3x-7}}\\
&=\cfrac{8}{4x-1+\cfrac{529}{92x-63-\cfrac{1624}{23x+33}}}\\
\end{align}
$$
The book's answer can be shown to be equal by cancelling fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/218976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How did my teacher gets this result? Algebra/Limits $$\begin{align*}&\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}=\\
&\lim_{x\to0}\left(\frac{(1+x)-(1-x)}{(1+x)-(1-x)}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{(1+x)(1-x)}+\sqrt[3]{(1-x)^2}}{\sqrt{1+x}+\sqrt{1-x}}\right)
\end{align*}$$
My teacher was using this to calculate the value of the first limit as seen above.
I am not sure how he pull this huge rabbit out of the hat XD.
Also I am not sure why is it useful...
| You should use the following identities:
$$a^2-b^2=(a-b)(a+b)$$
$$c^3-d^3=(c-d)(c^2+cd+d^2)$$
(prove by opening brackets) for
$$a=\sqrt{1+x}, \hspace{5pt} b=\sqrt{1-x}, \hspace{5pt} c=\sqrt[3]{1+x}, \hspace{5pt} d=\sqrt[3]{1-x}$$
Then you multiply your function in the limit by $\frac{a+b}{c^2+cd+d^2}\cdot \frac{c^2+cd+d^2}{a+b}$, which leaves an equal expression. This is useful since instead of getting $\frac00$, you get an expression where you can substitute $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/219052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Range of a trignonmetric function I came across this in an Engineering entrance book,
What is the range of this: $a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x$
What is the method to find it? I tried the graph approach but didn't know how to proceed.
| The double angle formulas yield
$$
a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x=\frac{\color{#C00000}{b\sin(2x) + (c-a^2)\cos(2x)}+(\color{#00A000}{c+a^2})}{2}\tag{1}
$$
Since the range of $a\cos(x)+b\sin(x)$ is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$, we get that the range of $(1)$ is
$$
\left[\frac{\color{#00A000}{c+a^2}+\color{#C00000}{\sqrt{(c-a^2)^2+b^2}}}{2},\dfrac{\color{#00A000}{c+a^2}-\color{#C00000}{\sqrt{(c-a^2)^2+b^2}}}{2}\right]\tag{2}
$$
Why the range of $\mathbf{a\cos(x)+b\sin(x)}$ is $\mathbf{\left[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}\right]}$ :
The formula for rotating the point $(a,b)$ around the origin is
$$
(a,b)\stackrel{\text{rotate by }x}{\longrightarrow}(\color{#C00000}{a\cos(x)+b\sin(x)},b\cos(x)-a\sin(x))\tag{3}
$$
Thus, as $x$ varies from $0$ to $2\pi$, $(\color{#C00000}{a\cos(x)+b\sin(x)},b\cos(x)-a\sin(x))$ traces out a circle of radius $\color{#C00000}{\sqrt{a^2+b^2}}$. Thus, the first coordinate, $\color{#C00000}{a\cos(x)+b\sin(x)}$ varies between $-\color{#C00000}{\sqrt{a^2+b^2}}$ and $\color{#C00000}{\sqrt{a^2+b^2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/221927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $4w^2 + 5z^2 = 4z^2 + 5w^2$ if $|z| = |w|$ and $4z^2+5w^2=azw$ I know that:
$$|z| = |w| = p$$
$$4z^2 + 5w^2 = azw ,\qquad a \in R$$
I need to prove that $4w^2 + 5z^2 = azw$
How I solved it is:
$$|z| = p \implies |z|^2 = p^2 \implies zz^* = p^2$$
then solved as $z$ and replaced it to the original equation.
Then I wanted to prove that:
$$4z^2 + 5w^2 = 4w^2 + 5z^2$$ and find that $0 = 0$.
But this covers more than 5 pages and it contains lots of math, so what is an easier approach?
| This's strongly inspired by the other answer.
Let's denote $\cos C+i\sin C=cis C$
If $z=p(cis A),w=p(cis B)$ with $p\ne 0$
So, $\frac z w=cis(A-B),4cis 2(A-B)+acis(A-B)+5=0$
Equating real & imaginary parts(using the fact that $a$ is real ),
$4\cos2(A-B)+a\cos(A-B)+5=0--->(1)$
and $4\sin2(A-B)+a\sin(A-B)=--->(2)$
$(2)\implies \sin(A-B)\{8\cos(A-B)+a\}=0$
If $\cos(A-B)=-\frac a 8,$
from (1) we get, $4(2\{-\frac a 8\}^2-1)+a\{-\frac a 8\}+5=0$
or, $-4+5= 0$, which is impossible.
So, $\sin(A-B)=0\implies A=n\pi+B$ where $n$ is any integer.
$\cos A=\cos(n\pi+B)=(-1)^n\cos B $ and
$\sin A=\sin(n\pi+B)=(-1)^n\sin B $
So, $z=pcis A=p(-1)^n cis B=(-1)^nw\implies z^2=(-1)^{2n}w^2=w^2$
| {
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"url": "https://math.stackexchange.com/questions/222742",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluate: $\int{\frac{x^{5}-x}{x^{8}+1}}\:\mathrm dx.$
Evaluate: $$\int{\frac{x^{5}-x}{x^{8}+1}\:\mathrm dx}.$$
I am unable to see a decent starting point for this integral, there are no radicals so trigonometric substitution isn't helpful; there is no nice partial fraction decomposition to simplify the integrand, integration by parts doesn't help to simplify it much, and I cannot see any factorization or useful substitution to use.
Can anyone help shed some light on this integral?
Thanks in advance!
| Letting $y=x^2$ reduces the powers of $x$ and then dividing both the denominator and numerator of the integrand by $y^2$ yields
$$
\begin{aligned}
\int \frac{x^5-x}{x^8+1} d x&=\frac{1}{2} \int \frac{y^2-1}{y^4+1} d y \\
=& \frac{1}{2} \int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}} d y \\
=& \frac{1}{2} \int \frac{d\left(y+\frac{1}{y}\right)}{\left(y+\frac{1}{y}\right)^2-2} \\
=& \frac{1}{4 \sqrt{2}} \ln \left|\frac{y+\frac{1}{y}-\sqrt{2}}{y+\frac{1}{y}+\sqrt{2}}\right|+C \\
=& \frac{1}{4 \sqrt{2}} \ln \left|\frac{x^4-\sqrt{2} x^2+1}{x^4+\sqrt{2} x^2+1}\right|+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
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} |
How to calculate the maximum value of: $\frac{25x}{x^2+1600x+640000}$? Wolfram says it's 800, but how to calculate it?
$$
\frac{25x}{x^2+1600x+640000}
$$
| It’s straightforward as a calculus problem. To solve it without calculus, note that
$$\frac{25x}{(x^2+1600x+640000)}=\frac{25x}{(x+800)^2}\;,\tag{1}$$
so the denominator is always positive, the the function has its maximum at some positive value of $x$. That maximum will occur where
$$\frac{(x+800)^2}{25x}=\frac1{25}\left(x+1600+\frac{640000}x\right)=64+\frac1{25}\left(x+\frac{640000}x\right)\tag{1}$$
has its minimum (over the range $x>0$). This in turn occurs where $x+\dfrac{640000}x$ has its minimum.
Now $x$ and $\frac{640000}x$ are a pair of numbers whose product is $640000=800^2$; if we set $x=800$, their sum is $1600$. Suppose that we set $x=800+a$ for some $a>0$; then
$$\begin{align*}
x+\frac{640000}x&=800+a+\frac{640000}{800+a}\\
&=\frac{1280000+1600a+a^2}{800+a}\\
&=1600+\frac{a^2}{800+a}\\
&>1600\;.
\end{align*}$$
Thus, $x=800$ gives us the minimum value of of $x+\frac{640000}x$, namely, $1600$, and hence the minimum value of $(2)$ and the maximum value of $(1)$. Substituting $x=800$ into $(1)$, we find that the maximum is $$\frac{25\cdot800}{1600^2}=\frac{25}{3200}=\frac1{128}\;.$$
| {
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explain why $3|n$ with the following conditions.. If $ord_ma=3$ and if $a^n\equiv 1 \pmod{m}$ for some $n\ge1$, explain why $3|n$.
Attempt at solution:
We know that $ord_ma=3$ means $a^3\equiv 1 \pmod{m}$. Therefore, $n$ is a multiple of $3$, because $a^{3n}\equiv 1 \pmod{m}$.
| The reasoning is incomplete. First note that $\text{ord}_m(a)=3$ means that $a^3\equiv 1\pmod{m}$ and for any positive integer $k \lt 3$, we have $a^k\not\equiv 1\pmod{m}$.
Now we proceed to show that if $a^n\equiv 1\pmod{m}$, where $n\ge 1$, then $3$ divides $n$.
Suppose to the contrary that $3$ does not divide $n$. Then if we attempt to divide $n$ by $3$, we get a non-zero remainder. More precisely, $n=3q+k$ where $k=1$ or $k=2$. Then
$$a^n=a^{3q+k}=(a^3)^qa^k.$$
But since $a^3\equiv 1\pmod{m}$, it follows that
$(a^3)^q\equiv 1\pmod{m}$, and therefore $a^n\equiv a^k\pmod{m}$. Thus $a^k\equiv 1\pmod{m}$. Since $k=1$ or $k=2$, this contradicts the fact that $a$ has order $3$ modulo $m$.
| {
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} |
wave equation with initial values and boundary condititon I have a homogenious 2-dimensional wave equation:
$$ - \frac{\partial^2 u}{\partial x^2} (x, y, t) - \frac{\partial^2 u}{\partial y^2} (x, y, t) + \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} (x, y, t) = 0$$
With:
$$ 0 < x < a, 0 < y < b, t > 0$$
$$ u(x, y, 0) = 0$$
$$ \dot u(x, y, 0) = x(x-a)(y-b)$$
And boundary condition:
$$ u(0, y, t) = u(a, y, t) = u(x, 0, t) = u(x, b, t) = 0$$
It says that I should solve it using separation of variables with $u(x, y, t) = G(x)H(y)w(t)$ as a starting point. With that approach, I got this solution:
$$ u(x, y, t)
= \sin\left(\frac{n\pi}a x\right)
\sin\left(\frac{n\pi}b y\right)
\sin\left(\left(\frac{n\pi}a - \frac{n\pi}b\right) c^2 t\right)
$$
This solved everything, except the initial value for $\dot u$. A friend of mine proved that if such a seperated function would solve this initial value, the function $w(t)$ would depend on $x$ and $y$.
How do I solve this differential equation so that it satifies the $\dot u$ initial value?
| Let $u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}$ so that it automatically satisfies $u(0,y,t)=u(a,y,t)=u(x,0,t)=u(x,b,t)=0$ ,
Then $\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^2\pi^2}{a^2}C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{n^2\pi^2}{b^2}C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{1}{c^2}\dfrac{\partial^2C(m,n,t)}{\partial t^2}\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\left(\dfrac{\partial^2C(m,n,t)}{\partial t^2}+\left(\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}\right)c^2\pi^2C(m,n,t)\right)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$\therefore\dfrac{\partial^2C(m,n,t)}{\partial t^2}+\left(\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}\right)c^2\pi^2C(m,n,t)=0$
$C(m,n,t)=C_1(m,n)\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)+C_2(m,n)\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$\therefore u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_2(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u(x,y,0)=0$ :
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_2(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$C_2(m,n)=0$
$\therefore u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u_t(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u_t(x,y,0)=x(x-a)(y-b)$ :
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=x(x-a)(y-b)$
You need to handle extremely complicated double kernel inversion.
| {
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How to prove $\lim_{n \to +\infty} \sqrt{n}\int_0^\pi{\cos(\frac{t}{2})^n}dt>0$ I want to prove
$$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}dt>0.$$
First, I consider $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}\sin\left(\frac{t}{2}\right)dt,$$ which is smaller than what I want, but the second integral leads to $$\lim_{n \to +\infty}\frac{\sqrt{n}}{2(n+1)}.$$ So it does not work.
| HINT
In fact, it is not hard to compute the limit exactly. Let $I_n = \displaystyle \int_0^{\pi} \cos^n(t/2) dt$. Compute $I_n$ and use Stirling's formula to obtain asymptotics of $I_n$.
Move your cursor over the gray area for the complete solution.
Setting $t = 2x$, we get $$I_n = 2 \int_{0}^{\pi/2} \cos^n(x) dx$$ For $n = 2k+1$, we get that \begin{align} I_{2k+1} & = 2 \left( \dfrac{4^k (k!)^2}{(2k+1)!} \right)\\ & \sim 2 \cdot 4^k \times (2 \pi k) \times \left(\dfrac{k}e \right)^{2k} \times \dfrac1{\sqrt{2 \pi (2k+1)}} \left(\dfrac{e}{2k+1} \right)^{2k+1}\\ & = \dfrac{2^{2k+2} \pi k^{2k+1} e^{2k+1}}{e^{2k} \sqrt{2 \pi (2k+1)} (2k+1)^{2k+1}}\\ & = \dfrac{2 \pi (2k)^{2k+1} e}{\sqrt{2 \pi (2k+1)} (2k+1)^{2k+1}}\\ & = \dfrac{\sqrt{2\pi} e}{\sqrt{2k+1}} \left(\dfrac{2k}{2k+1} \right)^{2k+1}\\ & = \dfrac{\sqrt{2\pi} e}{\sqrt{2k+1}} \left(1-\dfrac1{2k+1} \right)^{2k+1}\\ & \sim \dfrac{\sqrt{2\pi}e}{\sqrt{2k+1}} \times e^{-1}\\ & = \sqrt{\dfrac{2 \pi}{2k+1}} \end{align} Hence, $$\sqrt{2k+1} I_{2k+1} \sim \sqrt{2 \pi}$$ You will get the same result for even $n$ as well courtesy Wallis formula.
| {
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"source": "stackexchange",
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Bijective Proof of a Fibonacci Identity Prove (Using bijections):
$F_{1}+F_{3}+\cdots+F_{2n-1}=F_{2n}$
Where $F_{i}$ is the $i$th Fibonacci number.
Apparently you use monomers and dimers to prove this, but I don't really know what to do.
| HINT: A monomer is a $1\times 1$ square, and a dimer is a $2\times 1$ rectangle, a domino. The number of ways to tile a $1\times n$ strip using monomers and dimers is $F_{2n+1}$. (For example, a $1\times 4$ strip can be tiled in $F_5=5$ ways: $1+1+1+1,1+1+2,1+2+1,2+1+1$, and $2+2$. The sum
$$F_1+F_3+F_5+\ldots+F_{2n-1}\tag{1}$$
is therefore the number of ways to tile any strip of even length less than $2n$: there are $F_1$ ways to tile a $1\times 0$ strip, $F_3$ ways to tile a $1\times 2$ strip, $F_5$ ways to tile a $1\times 4$ strip, and so on, up through $F_{2n-1}$ ways to tile a $1\times(2n-2)$ strip.
$F_{2n}$, on the other hand, is the number of ways to tile a $1\times(2n-1)$ strip. We’d like somehow to set up a bijection between tilings of a $1\times(2n-1)$ strip and tilings of even strips shorter than that. Here’s a table of tilings for the case $n=3$:
$$\begin{array}{l|l}
\text{Tilings of shorter even length}&\text{Tilings of }1\times 3\\ \hline
\text{no tile}&\color{red}{1+2+2}\\ \hline
1+1&1+1+\color{red}{1+2}\\
2&2+\color{red}{1+2}\\ \hline
1+1+1+1&1+1+1+1+\color{red}{1}\\
1+1+2&1+1+2+\color{red}{1}\\
1+2+1&1+2+1+\color{red}{1}\\
2+1+1&2+1+1+\color{red}{1}\\
2+2&2+2+\color{red}{1}
\end{array}$$
I’ve arranged the table so that it represents a natural bijection between the two sets. For example, the tiling $1+1$ of a $1\times 2$ strip is paired with the tiling $1+1+1+2$ of a $1\times 5$ strip, while the tiling $1+2+1$ of a $1\times 4$ strip is paired with the tiling $1+2+1+1$ of a $1\times 5$ strip. The horizontal lines separate the different shorter even lengths; for $n=3$ those are $0,2$, and $4$, corresponding to the terms $F_1,F_3$, and $F_5$ in the sum $(1)$.
Pay close attention to the red parts of the tilings of a $1\times 5$ strip: they’re the key to how the bijection works. In case that isn’t quite enough, here’s one more bit of evidence: for $n=6$ the bijection that I have in mind would pair the tiling $$1+2+1+\color{red}{1+2+2+2}$$ of a $1\times 11$ strip with the tiling $1+2+1$ of a $1\times 4$ strip.
That should be enough information to give you a reasonable chance of discovering how the bijection works, but I can provide further help if necessary.
| {
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Encode the message $[1,1,0,1,1,0,1]$ in BCH code based on the field $\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1}$ So here's what I understand so far:
$\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1} = GF(16)$
The code is written as $[x^{14},x^{13},x^{12},x^{11},x^{10},x^{9},x^{8}$ $|$ $x^{7},x^{6},x^{5},x^{4},x^{3},x^{2},x^{1},x^{0}]$
So we write the message $[1,1,0,1,1,0,1] = x^{14}+x^{13}+x^{11}+x^{10}+x^{8}$ and we get the first part of the code:
$[1,1,0,1,1,0,1$ $|$ $x^{7},x^{6},x^{5},x^{4},x^{3},x^{2},x^{1},x^{0}]$
We divide $x^{14}+x^{13}+x^{11}+x^{10}+x^{8}$ by $(x^8+x^7+x^6+x^4+1)$ and we get $(x^6+x^4+x^2+x)$ with remainder $(x^7+x^5+x^4+x^2+x)$
Taking the coefficients of the remainder we get the encoded BCH code
$[1,1,0,1,1,0,1$ $|$ $1,0,1,1,0,1,1,0]$
My question is: Why do we divide by $(x^8+x^7+x^6+x^4+1)$? Where did that come from?:
| Let $\alpha$ be a root of $m_{\alpha}(x)=x^4+x+1$ in the field $GF(16)$. As this polynomial is primitive, $\alpha$ is of order $15$. Therefore $\alpha^3$ is of order five. Thus the minimal polynomial of $\alpha^3$ is $m_{\alpha^3}(x)=x^4+x^3+x^2+x+1$. The generator polynomial of a double-error-correcting BCH-code is
$$
m_{\alpha}(x)m_{\alpha^3}(x)=(x^4+x+1)(x^4+x^3+x^2+x+1)=x^8+x^7+x^6+x^4+1.
$$
The rest of the calculation is basic use of the generator polynomial that has undoubtedly been explained in you textbook/lecture notes.
Related topics have been discussed in another question, see Dilip Sarwate's answer and the Wikipedia article linked to in that question. It is about Reed-Solomon codes, but the algebra is more or less the same as in the case of BCH-codes.
| {
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Find k $\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$ Find k with
$$\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$$
| In other words, you want to minimize $$f(a,b,c) = (a^2 + b^2 + c^2)\left(\frac{1}{(3a-b)^2} + \frac{1}{(3b-c)^2} + \frac{1}{(3c-a)^2}\right)$$
(presumably for $a,b,c$ real numbers).
By homogeneity, we may restrict to the sphere $a^2 + b^2 + c^2 = 1$.
Using a Lagrange multiplier, we look for critical points of $F(a,b,c,\lambda) = f(a,b,c) + \lambda (a^2 + b^2 + c^2-1)$. The solution is rather unpleasant. It seems the minimum is approximately $0.808645463814078870$.
EDIT: In fact this number turns out to be the real root of the polynomial
$$2460640\,{z}^{3}-2787824\,{z}^{2}+905866\,z-210681$$
| {
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Exponential and power functions through two points I have a problem where I'm asked to determine the constants of exponential and power functions that go through both points (5, 50) and (10, 1600). I have tried to solve them below, but would appreciate it if someone could check. Would also appreciate feedback on how I could optimize my notation, if anyone has any thoughts on that.
Exponential, i.e. $f(x) = c \cdot a^x$
$50 = c \cdot a^5 \\
1600 = c \cdot a^{10}
$
$
ln(50) = ln( c ) + 5ln(a) \\
ln(1600) = ln( c ) + 10ln(a)
$
$
ln(50)-ln(1600) = 5ln(a) - 10ln(a)
\Rightarrow ln(\frac{50}{1600}) = -5ln(a)
\Rightarrow ln(\frac{1}{32}) = -5ln(a)
\Rightarrow -ln(32) = -5ln(a)
\Rightarrow \frac{-ln(32)}{-5} = ln(a)
\Rightarrow e^{ln(a)} = e^{\frac{ln(32)}{5}}
= 2
$
$
f(x) = c \cdot 2^{x}
\Rightarrow 50 = c \cdot 2^{5}, 1600 = c \cdot 2^{10}
$
$
50 = 32c
\Rightarrow c = \frac{50}{32} = \frac{25}{16}
$
$
1600 = 1024c
\Rightarrow c = \frac{1600}{1024} = \frac{25}{16}
$
$
f(x) = \frac{25}{16}2^{x}
$
Power, i.e. $f(x) = c \cdot x^r$
$
50 = c \cdot 5^r \\
1600 = c \cdot 10^r
$
$
ln(50) = ln( c ) + rln(5) \\
ln(1600) = ln( c ) + rln(10)
$
$
ln(50) - ln(1600) = r(ln(5) - ln(10))
$
$
\frac{ln(50) - ln(1600)}{ln(5) - ln(10)} = r
\Rightarrow r = 5
$
$
50 = c \cdot 5^5
\Rightarrow c = \frac{2}{125}
$
$
1600 = c \cdot 10^5
\Rightarrow c = \frac{2}{125}
$
$
f(x) = \frac{2}{125}x^{5}
$
| The steps seem to be good. In both cases, you could divide your first equation by the second one (or vice versa) and then take ln on both sides. It would save you some time.
| {
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Geometric identity, cannot show equivalence using trigonometric identities
clearly
$$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$
expanding and using the Weierstrass substitution we find that
$$\theta= 2 \arctan \frac{\left( 2ay- \sqrt{ 4a^2y^2 - ( (x-a)^2+y^2-b^2)( (x+a)^2+y^2-b^2) }\right)}{(x-a)^2+y^2-b^2} $$
if we use the law of cosines, with $c^2=x^2+y^2$
$$\theta = \arctan_2(y,x) - \arccos( (c^2+a^2-b^2) / (2ac) )$$
Is there a way to pass from one expression to the other using trigonometric identities?
| You could look at the angles at the vertex between sides $c$ and $a$. The arctan comes from the $x,y$ triangle and the arccos from the $abc$ one.
| {
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Finding sum $\sum_{i=1}^n \frac1{4i^2-1}$ I have been having problem with calculating the following summation:
$$
\sum_{i=1}^n {1\over 4i^2-1} = {1\over3} + {1\over15} + {1\over35} + \cdots + {1\over 4n^2-1}
$$
I do know the answer, but just can not find the way to get it.
Thank you in advance.
| You have,
$$
\sum_{i=1}^n {1\over 4i^2-1} = \sum_{i=1}^n\frac{1}{{(2i + 1)(2i - 1)}}
$$
$$
= \sum_{i=1}^n \left(\frac{1}{{2(2i - 1)}} - \frac{1}{{2(2i + 1)}}\right)
= \frac{1}{2}\left(\sum_{1\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n}\frac{1}{{(2i + 1)}}\right)
$$
$$
=\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right).
$$
By translation of index i=k+1, you have,
$$
=\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq k+1\leq n} \frac{1}{{(2[k+1] - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right).
$$
Remember that $2\leq k+1\leq n$ if, only if, $2-1\leq k\leq n-1$. Then
$$
=\frac{1}{2}\left( \frac{1}{1}+\sum_{1\leq k\leq n-1} \frac{1}{{(2k+1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right)
$$
$$
=\frac{1}{2}-\frac{1}{4n+2}.
$$
| {
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"question_score": "3",
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Derivatives and graphs $f(x) = \sin(x)+ \cos(x) $ for $ 0≤x≤2\pi$
I have to do the following
1) Find the intervals on which $f$ is increasing or decreasing
2) Find the local maximum and minimum values of $f$
3) Find the intervals of concavity and the inflections points
I got until this point when trying to solve problem 1: $\tan(x) = 1$. The next step in the manual says that $x = \pi/4$ or $5\pi/4$. How did they get that $x = \pi/4$ or $5\pi/4$?
| (1) $f$ is increasing or decreasing accordingly as $f'(x)>0$ or $<0$
Now, $f'(x)=\cos x-\sin x=\sqrt2\cos(x+\frac \pi 4)$
$f'(x)>0$ if $\cos(x+\frac \pi 4)>0$ if $ x+\frac \pi 4$ lies in the 1st or 4th quadrant.
As,$0\le x\le 2\pi,$ for $f'(x)>0, 0\le x<\frac \pi 4$ or $\frac{3\pi}2-\frac \pi4<x\le 2\pi$
Similarly for $f'(x)<0$
(2)For the maxima/minima, $f'(x)=0\implies \tan x=1=\tan \frac \pi 4 \implies x=m\pi+\frac \pi 4 $ where $m$ is any integer.
$f''(x)=-(\sin x+\cos x)=-\sqrt2\cos(x-\frac \pi 4)$
So, $f''(m\pi+\frac \pi 4)= -\sqrt2\cos m\pi$ which is $<0$ if $m$ is even$=2r$(say) where $r$ is any integer
So,the local maximum of $f(x)$ is at $x=2r\pi+\frac \pi 4$
As $0\le x\le 2\pi,$ for local maximum of $f(x),x=\frac \pi 4$
$f_{max}=f(\frac \pi 4)=\sqrt 2$
Similarly, for local minimum.
(3) Using this, $f(x)$ is concave up if $f''(x) > 0$
So, we need $\cos(x-\frac \pi 4)<0$ i.e., $x-\frac \pi 4$ will lie in the 2nd or in the 3rd quadrant.
$\frac \pi2<x-\frac \pi 4< \frac {3\pi}2\implies \frac \pi2+\frac \pi 4<x< \frac {3\pi}2+\frac \pi 4$ as $0\le x\le 2\pi$
Similar for the concave down if $f''(x) < 0$
For the point of inflexion, $f''(x)=0$ or $f''(x)$ does not exist.
Here clearly $f^n(x)$ exists for $n\ge 0$
So we need $f''(x)=0\implies \sin x+\cos x=0\implies \tan x=-1=\tan (-\frac{\pi}4)\implies x=s\pi-\frac{\pi}4$ where $s$ is any integer
As $0\le x\le 2\pi,x=\pi-\frac{\pi}4,2\pi-\frac{\pi}4$ for the point of inflexion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/241465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Period of linear congruential generator How can you calculate the probability distribution of the period length of a linear congruential generator? That is $X_{n+1} = (aX_n + c) \bmod m$ where $a$ is chosen uniformly at random from $\{1,\dots, m-1\}$ and $c$ is chosen uniformly at random from $\{0,\dots, m-1\}$ and $m$ is a fixed prime. Take $X_0$ to be some arbitrary value from $\{0,\dots, m-1\}$.
If it is hard to do exactly, is it possible to give good bounds for the cdf?
| This may not address the question exactly, but the results derived indicate that the final answer may depend on the factors common to $a-1$ and $m$.
A preliminary lemma and theorem
Lemma: Suppose $p$ is prime and $j\ge2$. Then, unless $p=j=2$,
$$
p^k\,|\,n\implies\left.p^{k-j+2}\,\middle|\,\binom{n}{j}\right.\tag{1}
$$
Furthermore,
$$
2^k\,|\,n\implies\left.2^{k-1}\,\middle|\,\binom{n}{2}\right.\tag{2}
$$
Proof: Unless $p=j=2$, $j\lt p^{j-1}$. Thus, $j$ has at
most $j-2$ factors of $p$. Then $(1)$ follows from the binomial identity
$$
\binom{n}{j} = \frac nj\binom{n-1}{j-1}
$$
$(2)$ follows from
$$
\binom{n}{2}=\frac n2(n-1)
$$
$\square$
Theorem: Suppose that
$$
\begin{align}
&\text{(a) for all primes $p$, }p\mid m\implies p\mid a-1\\
&\text{(b) }4\mid m\implies4\mid a-1
\end{align}
$$
Then,
$$
\left.m\,\middle|\,\frac{a^n-1}{a-1}\right.\implies m\,|\,n
$$
Proof: Assume $\left.m\,\middle|\,\dfrac{a^n-1}{a-1}\right.$. For simplicity of notation, let $r=a-1$. Then
$$
\frac{a^n-1}{a-1}=\sum_{j=1}^n\binom{n}{j}r^{j-1}\tag{3}
$$
For any odd $p\,|\,m$, assume that $p^k\,|\,n$ and $\left.p^{k+1}\,\middle|\,\dfrac{a^n-1}{a-1}\right.$. Using $(3)$, we get
$$
n\equiv-\sum_{j=2}^n\binom{n}{j}r^{j-1}\pmod{p^{k+1}}\tag{4}
$$
The Lemma and the assumption that $p\,|\,m\implies p\,|\,r$ says that $p^{k-j+2}p^{j-1}=p^{k+1}$ divides each term in $(4)$. Thus, $p^{k+1}\,|\,n$. Bootstrapping, we get that for any odd $p\,|\,m$,
$$
\left.p^k\,\middle|\,\frac{a^n-1}{a-1}\right.\implies p^k\,|\,n\tag{5}
$$
If $2\,|\,m$, then $\left.2\,\middle|\,\dfrac{a^n-1}{a-1}\right.$. Using $(3)$, we get
$$
n\equiv-\sum_{j=2}^n\binom{n}{j}r^{j-1}\pmod{2}\tag{6}
$$
The assumption that $p\,|\,m\implies p\,|\,r$ says that $2$ divides each term in $(6)$. Thus, $2\,|\,n$; that is,
$$
\left.2\,\middle|\,\frac{a^n-1}{a-1}\right.\implies2\,|\,n\tag{7}
$$
If $4\,|\,m$, then assume that $2^k\,|\,n$ and that $\left.2^{k+1}\,\middle|\,\dfrac{a^n-1}{a-1}\right.$. Using $(3)$, we get
$$
n\equiv-\sum_{j=2}^n\binom{n}{j}r^{j-1}\pmod{2^{k+1}}\tag{8}
$$
The Lemma and the assumption that $4\,|\,m\implies4\,|\,r$ says that $2^{k-j+1}4^{j-1}=2^{k+j-1}$ divides each term in $(8)$. Since $j\ge2$, we have $2^{k+1}\,|\,n$. Bootstrapping, we get that
$$
\left.2^k\,\middle|\,\frac{a^n-1}{a-1}\right.\implies 2^k\,|\,n\tag{9}
$$
$(5)$ and either $(7)$ or $(9)$ show that
$$
\left.m\,\middle|\,\frac{a^n-1}{a-1}\right.\implies m\,|\,n\tag{10}
$$
$\square$
Suppose the sequence $x_k$ is defined by the recurrence
$$
x_{k+1}=ax_k+b\tag{11}
$$
then, inductively, we have
$$
x_k=a^kx_0+\frac{a^k-1}{a-1}b\tag{12}
$$
Multiplying by $a-1$ and adding $1$ yields
$$
\frac{a^{k_1}-1}{a-1}\equiv\frac{a^{k_2}-1}{a-1}\pmod{m}\implies a^{k_1}\equiv a^{k_2}\pmod{m}\tag{13}
$$
Therefore, to investigate the periodicity of $x_k$, we look at the periodicity of $\dfrac{a^k-1}{a-1}\bmod{m}$.
To maximize the range of $x_k$ ,we will assume that $(a,m)=(b,m)=1$. This implies
$$
\begin{align}
\frac{a^{k_1}-1}{a-1}\equiv\dfrac{a^{k_2}-1}{a-1}\pmod{m}
&\implies\frac{a^{k_1-k_2}-1}{a-1}a^{k_2}\equiv0\pmod{m}\\[6pt]
&\implies\frac{a^{k_1-k_2}-1}{a-1}\equiv0\pmod{m}\tag{14}
\end{align}
$$
That is, the period of $x_k$ is the smallest positive $n$ for which
$$
\frac{a^n-1}{a-1}\equiv0\pmod{m}\tag{15}
$$
By the theorem above, $m\,|\,n$ and since there are only $m$ residue classes $\bmod{\,m}$, we must have $n=m$. Thus,
Theorem: Suppose
$$
\begin{align}
&\text{(a) for all primes $p$, }p\mid m\implies p\mid a-1\\
&\text{(b) }4\mid m\implies4\mid a-1\\
&\text{(c) }\gcd(b,m)=1
\end{align}
$$
Then the modular sequence defined by
$$
x_{n+1}\equiv ax_n+b\pmod{m}
$$
has period $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Probability problem (withdraw balls from the urn)
Problem:
An urn contains 3 red and 7 black balls. Player A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn)
I have the solution:
$$\Bbb{P}(A)=\frac{3\cdot 9!+6\cdot 3 \cdot 7!+ 7\cdot6\cdot5\cdot4\cdot3\cdot5!+7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot3\cdot3!}{10!}$$
But i have no idea how they came up with this. Can someone explain this? Thank you.
| Well, the second entry on top is wrong (there should be another factor of $7$), so that could be the source of your confusion. Let's assume it's just a typo, and see how they might have gone about it.
There are $4$ mutually exclusive possibilities to consider for ways Player A can draw the red ball:
(i) The first ball drawn is red. Probability of this is $\dfrac3{10}$
(ii) The first two balls drawn are black and the third one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac38=\dfrac{7\cdot 6\cdot 3}{10\cdot 9\cdot 8}$.
(iii) The first four balls drawn are black and the fifth one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6}$.
(iv) The first six balls drawn are black and the seventh is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36\cdot\dfrac25\cdot\dfrac34=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}$.
[Do you see why these probabilities are correct?]
Since these events are mutually exclusive, the probability that Player A gets the red ball is simply the sum of these probabilities. The least common denominator is clearly $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4=\frac{10!}{3!}$. It seems your book chose to use $10!$ as the common denominator, instead (probably to reduce the length of the answer). Rewriting the probabilities with $10!$ as denominator, we therefore have $$\begin{align}\Bbb P(A) &= \frac{3\cdot9!}{10!}+\frac{7\cdot 6\cdot 3\cdot 7!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}\\ &= \frac{3\cdot9!+7\cdot 6\cdot 3\cdot 7!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/246766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to calcualate how many unique set of 6 can i have in a given set. Hello my question is quite simple i would think but i just cant seem to find an answer. I have a set of $\{1,2,3,4,5,6,7,8,9,10\}$ and i would like to calculate how many unique given sets of $6$ can i get from this set.
In other words for the number $1$
i would end up with
$[1,2,3,4,5,6]
[1,3,4,5,6,7]
[1,4,5,6,7,8]
[1,5,6,7,8,9]
[1,6,7,8,9,10]$
I would move down the line with the number $2$ to compare to unique sets of $6$
note:
when moving to two I would no longer do this
$[2,1,3,4,5,6]$ because it repeats my first case above.
its there a formula to figure this sort of thing?
Thanks in advance.
when I work this out on paper i end up with 15 sets
here is how
for 1
[1,2,3,4,5,6]
[1,3,4,5,6,7]
[1,4,5,6,7,8]
[1,5,6,7,8,9]
[1,6,7,8,9,10]
for 2
[2,3,4,5,6,7]
[2,4,5,6,7,8]
[2,5,6,7,8,9]
[2,6,7,8,9,10]
for 3
[3,4,5,6,7,8]
[3,4,6,7,8,9]
[3,5,6,7,8,9,10]
for 4 [4,5,6,7,8,9]
[4,6,7,8,9,10]
for 5 [5,6,7,8,9,10]
after that i cant make any more groups of $6$ thus i end up with $15$ sets.
| It exactly the number of ways to choose $6$ elements out of $10$, i,e. the binomial coefficient$$\binom{10}{6}=\frac{10!}{6!4!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/249379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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