Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Two inequalities Let$n$ be an integer, $n\geq 2$ and $a_1,a_2,...,a_n$ positive real numbers.
For each$\{2,3,...,n\}$ we define the numbers as follows:
$s_k=a_1a_2...a_k)^{1/k}$.
a.) Pro thafor all$k$ in the set $\{2,3,...,n-1\}$ we have $s_{k+1}\geq s_k$
b.) Prove, that $$eq i<j\leq n} (\sqrt{a_i}-\sqrt{a_j})^2$$ (CAL... | a) $s_{k+1} - s_k = a_{k+1} - (k+1)(a_1a_2...a_{k+1})^{\frac{1}{k+1}} + k(a_1a_2...a_k)^{\frac{1}{k}}$.
Apply AM-GM inequality we have: $a_{k+1} + k(a_1a_2...a_k)^{\frac{1}{k}} \geq (k+1)(((a_1a_2...a_k)^{\frac{1}{k}})^ka_{k+1})^{\frac{1}{k+1}} = (k+1)(a_1a_2...a_{k+1})^{\frac{1}{k+1}}$, this means: $s_{k+1} \geq s_k$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A square number and a positive cubic number which differ by six Is there a square number and a positive cubic number (both positive integers) which differ by six? If not, how do we prove this?
| Here is a proof that $y^2 = x^3 - 6$ has no solutions, taken from here.
First note modulo $8$ that $y^2$ is one of $0,1, 4$ and $x^3$ is one of
$0, 1, 3, 5, 7$.
Thus, we must have $y^2 \equiv 1$ and $x^3 \equiv 7$.
I.e., $y$ is odd and $x \equiv 7 \pmod 8$.
Now write
$$
y^2 - 2 = (x - 2)(x^2 + 2x + 4)
$$
Since $x - 2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Ground plan of Backward direction (<=) - Let $p$ be an odd prime. Prove $x^{2} \equiv -1 \; (mod \, p)$ has a solution $\iff p\equiv 1 \; (mod 4)$ Apply the identity $p-i \equiv -i \mod p$ for $i=1, \ldots$ to the pink factors
$ \begin{align}
\color{seagreen}{ (p-1)! } = 1\times 2\times\cdots\times \dfrac{p-1}{2} & \... | Here is problem I liked on math.SE: look at Primes congruent to 1 mod 6
a prime $p$ can be written in the form $p = a^2 + ab + b^2 $ for some $a,b \in \mathbb{Z}$ iff $p \equiv 1 \mod 6$
One part of the solution says it is elementary to show there exists an integer $d$ such that $d^2 \equiv -3 \mod p$. This was not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/787917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solutions to a particular equation of perfect squares Can we prove (or is it otherwise known) that for positive integers $a>b>c>1$, the only solutions to the equation:
$$a^2+1=b^2(c^2+1)$$
is given by the following family for any fixed $c$:
$$ a_1 = 4 c^3 + 3 c, \; \; \; b_1 = 4 c^2 + 1,$$
$$ a_2 = 16 c^5 + 20 c^3... | Yes. Given $c$, this is a "Pell equation" $a^2 - (c^2+1) b^2 = -1$
whose general solution is
$$
a_n + b_n \sqrt{c^2+1} = (c + \sqrt{c^2+1})^{2n-1}
$$
because $c + \sqrt{c^2+1}$ is clearly the fundamental unit in
${\bf Z}[\sqrt{c^2+1}]$.
I just answered this on stackexchange.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how can we get from $x \equiv 4 \pmod 9$ to $x \equiv 1 \pmod 3$ I want to solve the system:
$$ x \equiv 1 \pmod 3 , x \equiv 2 \pmod 5, x \equiv 3 \pmod 7, x \equiv 4 \pmod 9, x \equiv 5 \pmod {11}$$
The numbers $3,5,7,9,11$ are not pairwise coprime,especially the numbers $3,9$ are not coprime.
According to my notes, ... | $x\equiv4(\text{mod }9)$ is equivalent to:
$x-4=9k (\text{ for some integer k})\implies x-3-1=3(3k)\implies x-1=3(3k+1)$
or, $x\equiv1(\text{mod }3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Recurrence with multiplication Let $\{a_{n}\}$ be a sequence of nonnegative numbers such that $a_{n} = 2^{n}a_{n - 1}^{3/2}$. If $a_{1}$ is sufficiently small, why must $a_{n} \rightarrow 0$ as $n \rightarrow \infty$?
| If $a_{n_0}=0$ for some positive integer $n_0$ then $a_n=0$ for every $n\geq n_0$ and we are done.
So let us suppose that $a_n>0$ for every $n$. Let $b_n=\ln(a_n)$. We have
$$
b_n=n\ln2+\frac{3}{2}b_{n-1}
$$
So
$$
\left(\frac{2}{3}\right)^nb_n=\left(\frac{2}{3}\right)^{n-1}b_{n-1}+ (\ln2)n\left(\frac{2}{3}\right)^n
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all non-singular $3 \times 3$ matrices, such as $A$ and $A^{-1}$ elements are non-negative Task is to describe all non-singular $3 \times 3$ matrices $A$ for which holds: all elements of $A$ and $A^{-1}$ is non-negative.
As I discovered linear algebra is the most problematic part of math for me. Expect to get b... | We know that $A A^{-1}= I$.
Then suppose $X=A^{-1}$ => Elements of X could be derived from following set of linear systems:
$\begin{pmatrix}
a & 0 & 0 & | & 1\\
a_{21} &a_{22} &a_{23} & | & 0\\
a_{31} &a_{32} &a_{33} & | & 0\\
\end{pmatrix} => x_{1,1}= \frac 1 a$
$ \begin{pmatrix}
a & 0 & 0 & | ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/791394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Solving implicit equation for x or y I want to solve the following equation for $x$ or $y$ (does not matter wich one) analytically.
$$\sqrt[3]{x+y} + \sqrt[3]{x-y} = 1$$
Wolframalpha returns following solution, but I could not think of a way how to get there:
$$ x+1 \neq 0 \qquad y = \frac{(x+1) \sqrt{8 x-1}}{3 \sqr... | Suppose $a+b+c=0$ and that $a,b,c$ are the roots of the equation $$f(x)=x^3-px^2+qx-r=0$$
Then $p=a+b+c=0, r=abc$ and adding $f(a)+f(b)+f(c)=0$ we obtain:$$a^3+b^3+c^3+q(a+b+c)-3r=0$$ whence $$a^3+b^3+c^3=3abc$$
Now let $a=\sqrt[3]{x+y}, b=\sqrt[3]{x-y}, c=-1$ to obtain:$$2x-1=-3\sqrt[3]{x^2-y^2}$$ You can then cube th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Binomial Sum Related to Fibonacci: $\sum\binom{n-i}j\binom{n-j}i=F_{2n+1}$ How would I prove
$$
\sum\limits_{\vphantom{\large A}i\,,\,j\ \geq\ 0}{n-i \choose j} {n-j \choose i}
=F_{2n+1}
$$
where $n$ is a nonnegative integer and $\{F_n\}_{n\ge 0}$ is a sequence of Fibonacci numbers?
Thank you very much! :)
| Here is a solution using two complex variables.
Suppose we seek to evaluate in terms of Fibonacci numbers
$$\sum_{p,q\ge 0} {n-p\choose q} {n-q\choose p}.$$
We use the integrals
$${n-p\choose q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1-z)^{q+1} z^{n-p-q+1}} \; dz$$
and
$${n-q\choose p} =
\frac{1}{2\pi i}
\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/801730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 5
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How to show if $m$ and $n$ are coprime and $m-n$ is odd, then $(m^2-n^2)^2$, $(2mn)^2$, $(m^2+n^2)^2$ have a common factor? We know that the Pythagorean triples can be generated by the Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$ for any positive integers $m,n$ and $m>n$.
I am trying to prove the statement:
The t... | Informal outline Suppose for example that $(m^2-n^2)^2$ and $(2mn)^2$ are not relatively prime. Then there is a prime $p$ that divides $m^2-n^2$ and $2mn$. Since $m$ and $n$ have opposite parity, $p\ne 2$.
Since $p$ divides $2mn$, and $p\ne 2$, it follows that $p$ divides one of $m$ or $n$, say $m$. Since $p$ divides... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$
Let $a,b,c>0$ such that
$$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$
Show that
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$
My try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$... | Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\frac{3}{\sqrt7}$, $\frac{1}{a}=\frac{kx}{\sqrt7}$, $\frac{1}{b}=\frac{ky}{\sqrt7}$ and $\frac{1}{c}=\frac{kz}{\sqrt7}$, where $k>0$ and $x+y+z=3$.
Hence, $k>1$ and $\frac{1}{3}=\sum\limits_{cyc}\frac{1}{a^2+2}=\sum\limits_{cyc}\frac{1}{\frac{7}{k^2x^2}+2}>\sum\limits_{cyc}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/806224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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"answer_id": 0
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Strictly decreasing sequence Let $$a_n=\sum_{k=1}^n\sqrt{k^2+1}$$
Define the sequence $b_n=\dfrac{2a_n}{n} - n$.
I need to show that the sequence $\langle b_n\rangle_{n=1}^\infty$ is strictly decreasing.
| Try to show $b_{n+1} - b_n < 0$:
$b_{n+1} - b_n = \left[\dfrac{2a_{n+1}}{n+1} - (n+1)\right] - \left[\dfrac{2a_n}n - n\right]$
$ = \dfrac1{n(n+1)}\left[2n \sum_{k=1}^{n+1}\sqrt{k^2+1} - 2(n+1) \sum_{k=1}^n\sqrt{k^2+1} - n(n+1)\right]$
$ = \dfrac1{n(n+1)} \left[2n \sqrt{(n+1)^2 + 1} - 2 \sum_{k=1}^n\sqrt{k^2+1} - n(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/807786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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odd/even binomial coefficient identity For all n\geq1
:
$$\left(\begin{matrix}2n\\
0
\end{matrix}\right)
+\left(\begin{matrix}2n\\
2
\end{matrix}\right)
+\left(\begin{matrix}2n\\
4
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2k
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2n
\end{matrix}\... | Hint:
From binomial formula $$(1+x)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^k$$ for $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find a formula for $0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + \dots +n(n + 1)(n + 2)$, for $n \in \mathbb N$ $$\sum\limits_{i=1}^n i(i + 1)(i + 2)$$
$$\sum\limits_{i=1}^n i^3 + 3i^2 + 2i$$
$$\sum\limits_{i=1}^n i^3 + 3\sum\limits_{i=1}^ni^2 + 2\sum\limits_{i=1}^ni$$
$$= (\frac14)n^4 + (\frac12)n^3 + (\frac14)n^2 + n^3 + 3(\f... | $$
0 \cdot 1 \cdot 2 + 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots +n(n + 1)(n + 2)
=
3!\ \sum_{k=1}^{n+2} \binom{k}{3}
=
6 \binom{n+3}{4}
$$
The key sum is the sum of a column of Pascal's triangle, which is found by induction using Pascal's rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a Pellian connection in the divisibility condition $(a^2+b^2+1) \mid 2(2ab+1)$ I'm trying to prove that all integer solutions $a > b \ge 0$ to the divisibility condition in the title, namely
$$(a^2+b^2+1) \mid 2(2ab+1),$$
are given by
$$(a,b)=(1,0),(4,1),(15,4),(56,15),(209,56),\dots$$
where $a_n$ is in http://... | Your link does not like me. Solutions for hard to find such equations. So easy.
the equation: $\frac{m+1}{n}+\frac{n+1}{m}=a$
Can be solved using the equation Pell:
$p^2-(a^2-4)s^2=1$
Solutions have the form:
$n=2(p-(a+2)s)s$
$m=-2(p+(a+2)s)s$
Solutions have the form:
$n=\frac{2p(p+(a-2)s)}{a-2}$
$m=\frac{2p(p-(a-2)s)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Show that $7\mid(3^{2n+1}+2^{n+2})$ for all $n\in\mathbb{N}$
Prove that the following is true for every $n∈ℕ$:
$$7\mid(3^{2n+1}+2^{n+2}).$$
I've noticed
$$3^{2n+1}+2^{n+2} =3^{2n} \cdot 3+2^{n} \cdot 4.$$
Any suggestions how to continue from there to get something like $7k$ for $k\in\mathbb{N}$.
Thank you in advance!... | \begin{align}3^{2n+1}+2^{n+2}=3\cdot 9^n + 4\cdot 2^n&=7\cdot 9^n -4(9^n-2^n)\\
&=7\cdot 9^n-4(9-2)(9^{n-1}+9^{n-2}\cdot2+\cdots+2^{n-1})\\
&=7[ 9^n-4(9^{n-1}+9^{n-2}\cdot 2\cdots+2^{n-1})]\end{align}
Which is divisible by $7$
Here is a solution based on congruences,
$$3\cdot 9^n +4\cdot 2^n \equiv 3\cdot 2^n +4\cdot 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to evaluate $\lim\limits_{n\to\infty}(n^3)\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2})$?
How to evaluate
$\lim\limits_{n\to\infty}(n^3\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}))$?
My though is, that since $\lim\limits_{n\to\infty}n^3=\infty$, the limit must be $\infty$, if the right hand side is greater than or equal... | Hint: \begin{align}\\&\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}
\\\\=&\dfrac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}
\\\\=&\dfrac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}
\\\\=&\dfrac{n^4+1-n^4}{\left(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}\right)\cdot\left(\sqrt{n^4+1}+n^2\right)}\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(4/5)^{\frac{4}{5}}$ is irrational Prove that $(4/5)^{\frac{4}{5}}$ is irrational.
My proof so far:
Suppose for contradiction that $(4/5)^{\frac{4}{5}}$ is rational.
Then $(4/5)^{\frac{4}{5}}$=$\dfrac{p}{q}$, where $p$,$q$ are integers.
Then $\dfrac{4^4}{5^4}=\dfrac{p^5}{q^5}$
$\therefore$ $4^4q^5=5^4p^5$
I... | A slightly modified approach.
$(\frac{4}{5})^\frac{4}{5} = (\frac{4}{5})^{1-\frac{1}{5}} = (\frac{4}{5})^1 \times(\frac{5}{4})^\frac{1}{5}$
Since the first factor is rational we need only show that the latter term $(\frac{5}{4})^\frac{1}{5}$ is irrational.
Let us assume that the latter is rational and write (where $p$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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if A diagonalizable then show that $a=0$ Let $A= \begin{pmatrix}
2 & 0 & 0\\
a & 2 & 0 \\
b & c & -1\\
\end{pmatrix}$
if A diagonalizable then show $a=0$.
$P_A(x)=|xI-A|=(x-2)^2(x+1)=x^3-3x^2+4$
since A diagonalizable there is a P matrix such as $A=P^{-1}DP$
$D= \begin{pmatrix}
2 & 0 & 0\\
0... | A matrix is diagonalisable if and only if its minimal polynomial is the product of distinct factors. Since the characteristic polynomial is $(x-2)^2(x+1)$, we see that $A$ is diagonalisable if and only if the polynomial $(x-2)(x+1)$ is the minimal polynomial.
Calculating this we get
$$\begin{pmatrix}
0&0&0 \\
a&0&0\\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Values of $n$ for which $\lfloor 2 x\rfloor +\lfloor 4 x\rfloor +\lfloor 8 x\rfloor +\lfloor 20 x\rfloor =n$ has a solution $$\lfloor 2 x\rfloor +\lfloor 4 x\rfloor +\lfloor 8 x\rfloor +\lfloor 20 x\rfloor =n$$
How would you find the values of $n$ for which the equation has a solution under the condition that $n \leq 5... | Let's make a change of variables, that $x \mapsto \tfrac{x}{40}$, so our new function reads:
$$ f(x) = \lfloor \tfrac{x}{20} \rfloor +
\lfloor \tfrac{x}{10} \rfloor +
\lfloor \tfrac{x}{5}\rfloor +
\lfloor \tfrac{x}{2} \rfloor $$
This is the sum of 4 step functions and we notice this function behaves kind of like a lin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ordered pair satisfying the equation $x^2 + 6x + y^2 = 4$ How many ordered pairs of integers $(x,y)$ satisfy the equation $x^2+6x + y^2 = 4$?
I can't approach this problem. Please help me.
| By completing the square we get $(x+3)^2+y^2=13$.
If $x$ and $y$ are integers, so are $x+3$ and $y$ so we require two square numbers that add to $13$. The only option is $9$ and $4$. We can choose which is $9$ and which is $4$ as well as whether we use the positive or negative square roots. Hence $8$ possibilities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx$ I am having trouble showing this equality is true$$
\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\... | Replace $\alpha$, $\beta$ and $\gamma$ with $a$, $b$ and $c$ respectively.
With the substitution $x=a\sinh t$, the integral can be written as:
$$\begin{aligned}
I & = \sqrt{2a}\int_0^{\infty} e^{-ab\cosh t}\sin(ac\sinh t)\sinh \left(\frac{t}{2}\right)\,dt \\
&=-\sqrt{2a}\Im\left(\int_0^{\infty} e^{-ab\cosh t}\cos\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 0
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Find the row echelon form of a 4x4 matrix I want to row reduce the following matrix into an echelon form:
\begin{pmatrix}
-(-\beta+\alpha)^{1/2} & 0 & 1 & 0\\
0& -(-\beta+\alpha)^{1/2}& 0 & 1\\
-\beta & \alpha & -(-\beta+\alpha)^{1/2}&0 \\
\alpha & -\beta & 0 & -(-\beta+\alpha)^{1/2}
\end{pmatrix}
where $\beta = ... | Set $\gamma=-(-\beta+\alpha)^{1/2}$; then your matrix becomes
$$
\begin{pmatrix}
\gamma & 0 & 1 & 0\\
0& \gamma & 0 & 1\\
-\beta & \alpha & \gamma &0 \\
\alpha & -\beta & 0 & \gamma
\end{pmatrix}
$$
Now, multiply the first row by $\gamma^{-1}$:
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0& \gamma & 0 & 1\\
... | {
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"url": "https://math.stackexchange.com/questions/819994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
Is it possible to find the limit without l'Hopital's Rule or series Is it possible to find $$\lim_{x\to0}\frac{\sin(1-\cos(x))}{x^2e^x}$$
without using L'Hopital's Rule or Series or anything complex but just basic knowledge (definition of a limit, limit laws, and algebraic expansion / cancelling?)
| We can proceed as follows: $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{x^{2}e^{x}}\\
&= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}e^{x}}\\
&= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}}\cdot \frac{1}{e^{x}}\\
&= \lim_{t \to 0}\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Solving a separable differential equation Solve the differential equation:
$$y'=\frac{1-y^2}{1-x^2}$$
My book says the solution is: $$y=\frac{x+c}{cx+1},$$ where $c$ is a constant.
It's been ten minutes I tried to verify if it was correct but I'm pretty sure the book is wrong. Can someone confirm it?
| $$y'=\frac{cx+1-c(x+c)}{(cx+1)^2}=\frac{1-c^2}{(cx+1)^2},$$
$$\frac{1-y^2}{1-x^2}=\frac{\frac{(cx+1)^2-(x+c)^2}{(cx+1)^2}}{(1-x^2)}=\frac{(c^2-1)x^2+(1-c^2)}{(cx+1)^2(1-x^2)}=\frac{1-c^2}{(cx+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Integrating $(3x + 1) / x^\frac{1}{2}$ I am trying to integrate the following equation, but my answer is different from the textbook and I cannot see where I am going wrong:
\begin{align} \int_1^2\frac{ 3x + 1}{x^{1/2}}dx
&= \int_1^2 \frac{3x}{x^{1/2}}dx + \int_1^2 \frac{1}{x^{1/2}}dx \\
&= \int_1^2 3x^{1/2}dx + \i... | Continuing from where you left off, which is correct so far...
\begin{align} 2x^{3/2}\vert_1^2 + 2x^{1/2}\vert_1^2&=[2(2\sqrt{2})-2(1)]+[2(\sqrt{2})-2(1)] \\
&=4\sqrt{2}-2+2\sqrt{2}-2 \\
&=6\sqrt{2}-4\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to arrange $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$ in the increasing order? For these six numbers, $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$, how to arrange them in the increasing order?
This problem is taken from the today test: National Higher Education Entrance Examination.
I think we can consider
$$f(x)=\dfrac{\ln{x}... | We start by assuming that somehow we know that
$$ \color{green}{e < 3 < \pi}$$.
This immediately gives the relations
\begin{align*}
\color{blue}{e^3 }&\color{blue}{< e^\pi < 3^\pi}\\
\color{blue}{3^e }&\color{blue}{< \pi ^ e < \pi^3}
\end{align*}
Next, we can consider $$f(x) = \frac {\ln x}x$$ as you stated.
Different... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
$ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ $\implies$ the triangle is equilateral? If in a triangle $ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ , then is the triangle equilateral ?
| This is a 1990 Russian maths contest problem
$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, so
$$a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $$
$$\Longleftrightarrow$$
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)=0\tag{1}$$
There are several ways to show (1)
In fact
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)$$
$$=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Help needed in solving a system of DE The system of DE is:
$$\frac{dI}{db}=-\frac{b}{c}\frac{dJ}{db}-\frac{2ab+1}{2c}J$$
$$\frac{dJ}{db}=\frac{b}{c}\frac{dI}{db}-\frac{2ab-1}{2c}I$$
Assume that $a$ and $c$ are constants and both $I$ and $J$ tends to zero as $b$ tends to $\infty$.
To be honest, I have never solved a sy... | Removing mixed derivative terms is possible by inserting the second equation's $dJ/db$ into the first equation and vice versa:
$$
\frac{dI}{db} = -\frac{b^2}{c^2} \frac{dI}{db} +
\frac{b}{c}\frac{2ab-1}{2c} I -
\frac{2ab+1}{2c} J \\
\frac{dJ}{db} = -\frac{b^2}{c^2} \frac{dJ}{db} -
\frac{2ab-1}{2c} I -
\frac{b}{c}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/824304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ Please help me to prove this trigonometric equation.
$\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$
Thank you.
| As $\displaystyle\cos(180^\circ-y)=-\cos y$
$$S=\cos36^\circ-\cos72^\circ=\cos36^\circ+\cos(180^\circ-108^\circ)=\cos36^\circ+\cos108^\circ$$
Now multiplying the numerator & the denominator by $\displaystyle2\sin\frac{(108-36)}2^\circ,$
$$S=\frac{2\sin36^\circ\cos36^\circ+2\sin36^\circ\cos108^\circ}{2\sin36^\circ}$$
Us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ This is a simple question. Since $\cos(\theta)^2 + \sin(\theta)^2 = 1$
Can I take the inverse of this $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \frac{1}{1}$?
Finally getting $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \cos(\theta)^2 ... | The (multiplicative) inverse of $\sin^2\theta+\cos^2\theta$ is
$$\frac{1}{\sin^2\theta+\cos^2\theta},$$
and in general
$$\frac{1}{a+b}\neq \frac{1}{a}+\frac{1}{b},$$
for 'numbers' $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
formula for number triangles Hi, I have a triangle starting from $0$ and going up by one on the bottom row until there are $r$ items on the bottom row and there are $r$ rows a number is formed by adding the two numbers towards the fat end of the triangle together.
Example:
$$x \downarrow\ $$
$$
\begin{matrix}12\en... | Consider the first two 'columns'. Notice any patterns?
$$0 \space 1\\1 \space 3 \\4 \space 8\\ 12\space 20$$
What is the rule?
$$a_{i+1} = a_i + b_i$$
$$b_{i+1} = a_{i+1} + 2 (b_i - a_i) = 3b_i - a_i$$
Try find a formula for $a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using Cylindrical Coordinates to find volume of a solid How do I use Cylindrical Coordinates to find volume of a solid in the first Octant that is bounded by the cylinder $x^2 + y^2 = 2y$, the half cone $z = \sqrt{x^2 + y^2}$, and the $xy$-plane.
I have drawn the region of integration and obtained this:
$\int_0^2 \int_... | The question makes only sense if the solid is the volume outside the half cone.
Otherwise the volume is infinite, because (almost) half of the cylinder which extends
to $+\infty$ in $z$ is in it. So we have to integrate over $x^2+y^2\ge z^2$.
For constant $z$ the section through the solid
is the area inside the circle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/830530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integrals $\int \sin{x} \cot^2{x} \,dx$ and $\int \cos{x} \cot^2{x} \,dx$. Can you please show how to evaluate the integrals $$\int \sin{x} \cot^2{x} \,dx$$ and $$\int \cos{x} \cot^2{x} \,dx.$$
I know that $\cot x=\dfrac{\cos x}{\sin x}$, which can simplify the integrands a bit.
$$\sin x\cot^2x = \frac{\c... | Hint :
\begin{align}
\int\sin x\cot^2x\ dx&=\int\sin x\cdot\frac{\cos^2x}{\sin^2x}\ dx\\
&=\int\frac{\cos^2x}{\sin x}\ dx\\
&=\int\frac{1-\sin^2x}{\sin x}\ dx\\
&=\int \frac1{\sin x}\ dx-\int\sin x\ dx\\
&=\color{red}{\int \frac{\sin x}{1-\cos^2 x}\ dx}-\int\sin x\ dx\tag1
\end{align}
and
\begin{align}
\int\cos x\cot^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/830609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How would I factor $a^3+b^3+c^3-6abc$ How would I factor the polynomial $a^3+b^3+c^3-6abc$? The values are homogenous, so so must be the factors. I don't know where to go from there.
| Try solving the cubic for $c=x+y$ so that $c^3=(x^3+y^3)+3xy(x+y)=(x^3+y^3)+3xyc$
Whence $x^3+y^3=-a^3-b^3$ and $xy=2ab$ so that $x^3y^3=8a^3b^3$ and $x^3, y^3$ are roots of the quadratic $$z^2+(a^3+b^3)z+8a^3b^3$$ so that $$x^3,y^3=\frac {-a^3-b^3\pm \sqrt {a^6+b^6-30a^3b^3}}2$$
And use this to build your factors $(c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is there a simpler/better proof of this simple trigonometric property? The sine function has the following nice property : for any
$x,y$, we have $\sin(x)+\sin(y)=\sin(x+y)$ iff at least one of
$x,y,x+y$ is $0$ modulo $2\pi$.
I sketch below my current proof of it, which I find somewhat
unsatisfying. Does anyone know a... | Well here is (yet another) solution, this using only real variables.
By the sum formula we have
$$\sin x(1-\cos y)+\sin y(1-\cos x)=0$$
multiplying by $(1+\cos x)(1+\cos y)$ we get
$$\sin x \sin y [\sin y(1+\cos x)+\sin x(1+\cos y)]=0$$ so
either $\sin x=0$ or $\sin y=0$ or
$$\sin y(1+\cos x)+\sin x(1+\cos y))=0$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Prove summation related to cycles Let $b_r(n,k)$ be the number of n-permutations with $k$ cycles, in which numbers $1,2,\dots,r$ are in one cycle.
Prove that for $n \geq r $ there is:
$$
\sum_{k=1}^{n} {b_r(n,k)x^k=(r-1)!\frac{x^\overline{n}}{(x+1)^\overline{r-1}}}
$$
| As Brian Scott points out I have the wrong interpretation of the
question. We want permutations where $1,2,\ldots r$ are on one cycle
plus possibly some other elements, say $q.$
We obtain
$$b_r(n, k) = \sum_{q=0}^{n-r} {n-r\choose q}
\frac{(r+q)!}{r+q} \left[n-r-q\atop k-1\right].$$
Recall the bivariate gener... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the orthogonal trajectories of the family of curves given by $x^2 + y^2 + 2Cy =1$. Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$
The ordinary differential equation for the family of curves is given by $y'=\frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for th... | For the given family of curves
$$
x^2+y^2+2cy=1
$$
which can be written as
$$
\frac{1-x^2-y^2}{2y} = c
$$
differentiating both sides
$$
\frac{dy}{dx}=\frac{2xy}{x^2-y^2-1}
$$
and the family of curves orthogonal to this will be
$$
-\frac{dx}{dy}=\frac{2xy}{x^2-y^2-1}
$$
or,
$$(x^2-y^2-1)dx+(2xy)dy=0
$$
this is not an e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculation of $ \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots(1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $ Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $
My Trial After... | By L'Hospital's rule,
\begin{align*}
\lim_{x \to 1} \frac{1 - x^{n + k}}{1 - x^k} &= \lim_{x \to 1} \frac{-(n + k) x^{n + k}}{-k x^k} = \frac{n + k}{k}
\end{align*}
Hence the desired limit is
$$\frac{n + 1}{1} \cdot \frac{n + 2}{2} \cdot \frac{n + 3}{3} \cdots \frac{n + n}{n} = \frac{(2n)!}{(n!)^2}$$
Alternatively, fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/836841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\alpha$ and $\beta$ are solution of $a \cdot \tan\theta + b \cdot \sec\theta = c$ show $ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$ If $\alpha$ and $\beta$ are the solution of $$a \cdot \tan\theta + b \cdot \sec\theta = c$$, then show that $$ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$$
I did the following:
$$a ... | Using Double-Angle Formula,
$$\sec\theta=\dfrac{1+t^2}{1-t^2},\tan\theta=\dfrac{2t}{1-t^2}$$ where $t=\tan\dfrac\theta2$ to form a Quadratic Equation in $t$ whose roots will be $$\tan\dfrac\alpha2,\tan\dfrac\beta2$$
Can you find $$\tan\left(\dfrac\alpha2+\dfrac\beta2\right)$$ and subsequently, $$\tan(\alpha+\beta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/838082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculus Area Cubic curve Why area bounded between the "line $AB$" and the "cubic curve" and area bounded between the "line $BC$" and the "cubic curve" is $16$ times?
| Given is
$$
f(x) = x^3
$$
For the point $A$, $x=x_o$, we have a tangent
$$
3 x_o^2
$$
So that line is given by
$$
y = \Big(3 x - 2 x_o \Big) x_o^2
$$
The intersection is given by the equation
$$
x^3 - 3 x x_o^2 + 2 x_o^3 = 0
$$
which can be written as
$$
\Big( x - x_o \Big)^2 \Big( x + 2 x_o \Big) = 0
$$
As for point ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/838480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$ I wish to show
$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$$
I've tried substitution and integration by parts to get a recursive formula for the integra... | I will present you the main steps to a final solution and the details on each step can be your job.
First notice that a simple substiution $n=2k$ gives
$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \lim_{k\rightarrow \infty} \int_0^{\pi/2} 2^{2k} \sqrt{2k} \sin^{2k}(x) \cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/839343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integration by parts or substitution for $\int x \cos (x^2) dx$ With $$\int x \cos (x^2) dx$$ I can use substitution to solve this integral.
$$u=x^2, dx =\frac{du}{2x}$$
$$\int x \cos (x^2) dx = \int x \cos (x^2) \frac{du}{2x} = \frac{1}{2} \int \cos(u) = \frac{1}{2}\sin(u)=\frac{1}{2}sin(x^2)+C$$
But can't I also solv... | $$\int \sin(x^2)\,dx \neq -\frac{1}{2}\cos(x^2)+C$$
$$\int \cos(x^2)\,dx \neq \frac{1}{2}\sin(x^2)+C$$
You can confirm that by differentiating the RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/839823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$... | By AM-GM, $4x + \dfrac{9\pi^2}{x} \ge 2\sqrt{4x \cdot \dfrac{9\pi^2}{x}} = 12\pi$, with equality iff $4x = \dfrac{9\pi^2}{x}$, i.e. $x = \dfrac{3\pi}{2}$.
Also, $\sin x \ge -1$, with equality iff $x = \dfrac{3\pi}{2} + 2\pi k$ for some integer $k$.
Therefore, $f(x) = 4x + \dfrac{9\pi^2}{x} + \sin x \ge 12\pi - 1$ wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/843107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$
Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$ from the definition (using $\epsilon-\delta$)
Why can't I do something like this?
We want:
$|\frac{x^4-2x+1}{x-1} + \sqrt{x}-3| = |\frac{x^4-2x+1}{x-1} + \frac{x-9}{\sqrt{x}+3}| ... | You went off the track when you introduced the term
$$\left|\frac{x-9}{\sqrt x+3}\right|\ .$$
As $x\to1$, this term will tend to $2$ and therefore it can't be made arbitrarily small. Something like
$$\eqalign{\left|\frac{x^4-2x+1}{x-1}+\sqrt{x}-3\right|
&\le\left|\frac{x^4-2x+1}{x-1}-2\right|+\left|\sqrt{x}-1\right|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Algebraic manipulation of floors and ceilings I am trying to solve the summation
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \sum_{i=3}^{n} \left\lceil\frac{i-2}{2}\right\rceil
$$
I will list some of the simplifications that I've found so far, and then get around to asking my question. Please feel free to point out any logica... | Start with the generating function for $\displaystyle a_i=\left\lceil \frac{i-2}{2}\right\rceil\; , i\ge 3$
which turns out to be
\begin{align*}
G(x) &= \frac{x^3}{1-x-x^2+x^3}
\end{align*}
and for the sum of the coefficients, it's
\begin{align*}
H(x) &= \frac{1}{1-x}\, G(x)
\end{align*}
and the required answer i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Existence of some type matrix Is there square matrix $A$ of size $3$ with real entries such that
$$
\operatorname{tr}(A)=0\text{ and }A^2+A^T=I.
$$
I have proved that there is not with size $2$ using definition of "trace", but for size $3$ it becomes complicated.
Here is the sketch of the proof for $2$.
$$
a_{11}+a_... | The answer is no.
If we substitute $A^T=I-A^2$ into the transpose of $A^2+A^T=I$, namely $(A^T)^2+A=I$, we get $$(I-A^2)^2+A=I$$
Hence $I-2A^2+A^4+A=I$, or $$A^4-2A^2+A=0 ~~~~~~(\star)$$
must be satisfied by our matrix $A$. This has just four roots: $0,1,\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$.
Now, consider $B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Show that $\sum_{k=0}^{\infty}k^2x^k= \frac{x(1+x)}{(1-x)^3}$ for $0 < |x| < 1$ Show that $$\sum_{k=0}^{\infty}k^2x^k= \frac{x(1+x)}{(1-x)^3}$$ for $0 < |x| < 1$.
(This is appendix question A.1-3 from Introduction to Algorithms by Cormen)
| To find the required sum you could just find the sum of the geometric progression $1,x,x^2,x^3,... \space where \space |x|\lt 1,$ and then double differentiate it .$$\sum_{k=0}^\infty x^k=\frac 1{1-x}\\=>\frac d{dx}\left(\sum_{k=0}^\infty x^k\right)=\frac d{dx}\left(\frac 1{1-x}\right)\\=>\sum_{k=0}^\infty kx^{k-1}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of two odd squares What is the characterization of numbers expressible as a sum of two odd squares? I showed that it must be congruent to $2$ mod $4$, but obviously this is insufficient since not all numbers $2$ mod $4$ can be written this way.
| Analyzing this question, I came to the startling discovery:
If we define $S(X)$ as "X is the sum of two squares", then, for any integer $X$,
$$
S(X) \iff S(2X)
$$
We'll get back to that later. For this question, we have $X = a^2 + b^2$, where $a$ and $b$ are odd integers.
Let
$$
c = {a + b\over 2}\\
d = {a - b\over 2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Finding Integral of $y = e^{ax}\sin (bx)$ I was solving questions like $ \int e^{2x} \sin x dx. $ I decided to find the general term for $$\int e^{ax} \sin (bx) $$ which can be directly used in questions like (above).
I hope it helps others :) (Just wanted to share my work. I will appreciate feedback/suggestions or an... | $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large \textbf{Method -1} $
$$
\begin{align}
I & = \sin (bx) \left( \cfrac{1}{a} e^{ax} \right) - {\int} \left( b\cos (bx) \left(\cfrac{e^{ax}}{a} \right) \right) \\
& = \sin (bx) \left( \cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a} \left[ (\cos (bx)) \cfrac{e^{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
limits using $ \epsilon - \delta $ to prove two variable function I'm trying to use the $ \epsilon - \delta $ argument to prove $\lim_{(x,y) \rightarrow (1.1)} \frac{2xy}{x^2+y^2} =1$. I know that I need to show that $\forall \epsilon>0, \exists \delta>0$ s.t. for all (x,y) in the domain of f, $| \frac{2xy}{x^2+y^... | What about substitution $x=s+1$, $y=t+1$?
You get
$$\frac{2xy}{x^2+y^2}-1=
\frac{2(s+1)(t+1)}{(s+1)^2+(t+1)^2}-1=
\frac{2(st+s+t+1)}{(s+1)^2+(t+1)^2}-1=
\frac{2(st+s+t+1)-s^2-t^2-2s-2t-2}{(s+1)^2+(t+1)^2}=
\frac{2st-s^2-t^2}{(s+1)^2+(t+1)^2}=
\frac{-(s-t)^2}{(s+1)^2+(t+1)^2}
$$
Now you want to choose $\delta>0$ in such... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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A simple conditional probability problem Assume that two fair dice are rolled one at a time. Given that the sum of the two numbers that occured was at least $7$, compute the probability that it was equal to $7$.
I tried computing the probability as follows:
$$P \left( X+Y=7| X+Y \geq 7 \right)=\frac{P \left(X+Y=7 \righ... | The total number of possible outcomes when rolling two fair dice is $6^2=36$.
Examining the denominator of your conditional probability, it is easier to deal with the case when the sum of the two dice ($X+Y$) is less than $7$, and subtract this result from $36$.
Now, there are $5$ outcomes (pairs ($5$,$1$),($4$,$2$),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \fra... | Let's put it this way
$$\int_{-\infty}^Y f(x)=
\begin{cases}
0 & -\infty < Y < -3 \\
Y^2+3Y+{9\over 2} & -3\le Y <-2 \\
{1\over 2} & -2\le Y <2 \\
{1\over 2} +\int_2^Y(3-x)\, dx & 2\le Y < 3 \\
1 & Y\ge 3
\end{cases}$$
The first term is INDEED including the 9/2, this is because it is
$$\int_{-3}^Y f(x)\,dx\int_{-3}^Y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Solving a quadratic trigonometric equation? The equation is $6 \cos^2x+\cos x=1$,
My work:
$6x^2+x-1=0$
$(3x-1)(2x+1)$
$3x-1=0 ∨ 2x+1=0$
$x=\frac{1}{3} ∨ x= \frac{-1}{2}$
But I do not know how to progress further.
| I would not use the same letter, $x$, to refer to two different things.
$$
6\cos^2 x + \cos x - 1 = 0
$$
$$
6u^2 + u - 1 = 0
$$
$$
(3u-1)(2u+1)=0
$$
$$
u = \frac 1 3\text{ or } u = \frac{-1}{2}
$$
$$
\cos x=\frac 1 3 \text{ or }\cos x = \frac{-1}2
$$
$$
\Big(x = \left(\arccos\frac 1 3\right) + 2\pi n \text{ or } \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b Problem : If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b.
Solution :
We can write the given equation... | If we complete the square we get
$$x^2 -2x +4 = (x-1)^2 +3 \geq 3$$
Further we note that $-3\cos(ax+b) \leq 3$. So it must be that we have $x^2-2x+4 = 3 = -3\cos(ax+b)$. Further, from the completing the square formula, we see that in fact we need $x=1$.
So it reduces down to the problem of solving $-3\cos(a+b) = 3$ or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Compute integral: $\int_{0}^{\pi/2}\log(a^2\sin^2 x+b^2\cos^2 x )dx$ This is a integral for a calculus exam, and I have no idea how to solve it.
$$\int_0^{\frac{\pi}{2}} \log \big( a^2 \sin^{2}(x)+b^2 \cos^{2}(x) \big) \, \mathrm{d}x$$
| Without loss of generality, let $b>a>0$. Noting
$$ \sin^2x=\frac{1-\cos(2x)}{2}, \cos^2x=\frac{1+\cos(2x)}{2} $$
we have
\begin{eqnarray*}
I&=&\int_0^{\frac{\pi}{2}} \log \big(\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2}\cos(2x)\big) \mathrm{d}x\\
&=&\frac{1}{2}\int_0^{\pi} \log \big(\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2}\cos x\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Ratio of sides of triangle $ABC$ If in a triangle $\Delta ABC$ with $a$, $b$ and $c$ as sides
$$\begin{align}\left(Cot\frac{A}{2}\right)^2 +\left(2Cot\frac{B}{2}\right)^2+\left(3Cot\frac{C}{2}\right)^2=\left(\frac{6s}{7r}\right)^2\end{align} \tag{1}$$ where $r$ is inradius and $s$ is SemiPerimeter, then find the ratio ... | You have proved that the proposed condition is equivalent to
$$
(s-a)^2+4(s-b)^2+9(s-c)^2=\frac{36}{49}s^2\tag 1
$$
Note that, by the Cauchy-Schwarz inequality, we have
$$\eqalign{
s&=(s-a)+\frac{1}{2}(2(s-b))+\frac{1}{3}(3(s-c))\cr
&\leq\sqrt{(s-a)^2+4(s-b)^2+9(s-c)^2}\sqrt{1+\frac{1}{4}+\frac{1}{9}}\cr
&\leq\frac{7}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/861030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ Question: Prove that if
$$a_n=\left\{
\begin{array}{ll}
a_1=1\\
a_{n+1}=1+\frac{1}{1+a_n}
\end{array}
\right.$$
then $a_n$ converges, and then find $\lim_{n \to \infty}a_n$.
I found that $\lim_{n \to \infty}a_n=\sqrt{2}$, and that $1\leq a_n<2$, but the... | Most inelegant method I guess:
$a_{n+2}=1+\cfrac{1}{1+a_{n+1}}=1+\cfrac{1}{1+1+\cfrac{1}{1+a_n}}=1+\cfrac{1+a_n}{3+2a_n}
$
So $a_{n+2}-a_n=2\cfrac{2-a_n^2}{3+2a_n}$
Clearly from definition $a_n>1$
Also if $a_n>\sqrt{2}$ then $a_{n+2}>1+\cfrac{1+\sqrt{2}}{3+2}=\cfrac{6}{5}+\cfrac{\sqrt{2}}{5}>\cfrac{4\sqrt{2}}{5}+\cfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/861710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 7
} |
Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{alig... | $\textbf{1)}$ Substituting Eq. 1 into Eq. 2 gives $xy(x+1)=z+1$, and then substituting Eq. 2 into Eq. 3 gives $xyz(x+1)=x+1$; so $x=-1$ or $xyz=1$. By symmetry, we have that $y=-1$ or $xyz=1$ and $z=-1$ or $xyz=1$. Since $x=-1, y=-1, z=-1$ is a solution, all other solutions must satisfy $xyz=1$.
Since $x=1,y=1, z=1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Determining the best possible constant $k$, for an Integral Inequality If $f : [0,\infty) \to [0,\infty)$ is an integrable function, then what is the best possible constant $k$, for which the following ineqality holds:
$$\int_0^{\infty}f(x)dx \leq k\left(\int_0^{\infty}\sqrt{x}f(x)dx\right)^{1/2}\cdot\left(\int_0^{\inf... | Suppose
$$
\int_0^\infty\sqrt{x}f(x)\,\mathrm{d}x=A\tag{1}
$$
and
$$
\int_0^\infty f^2(x)\,\mathrm{d}x=B\tag{2}
$$
Then, letting $g(x)^2=A^{1/2}B^{-3/4}f(AB^{-1/2}x)$, we get
$$
\int_0^\infty\sqrt{x}g(x)^2\,\mathrm{d}x=\int_0^\infty g(x)^4\,\mathrm{d}x=1\tag{3}
$$
and
$$
\int_0^\infty f(x)\,\mathrm{d}x=A^{1/2}B^{1/4}\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Global Max and Min Hi can someone please help me figure out how to start this question?
Temperature within a circular region $R = \{(x, y)| x^2 + y^2 \le 49\}$ is given by $T(x,y) = 4x^2 -4xy + y^2$
Find the global minimum and maximum temperatures in the region and the points where these extreme temperatures occur.
I'm... | We use Lagrange multipliers. Let, $F(x,y) = 4x^2 - 4xy + y^2 + \lambda(x^2 + y^2 - 49)$. Thus,
$$
\begin{cases}
\dfrac{\partial F}{\partial x} = 8x - 4y + 2\lambda x = 0 \qquad (1)\\
\dfrac{\partial F}{\partial y} = 8y - 4x + 2\lambda y = 0 \qquad(2)\\
\dfrac{\partial F}{\partial \lambda} = x^2 + y^2 -49= 0 \qquad (3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Interview riddle On the Mathematics chat we were recently talking about the following problem @Chris'ssis had to solve during an interview :
$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$
We have not managed to solve it so far, all we know is the solution (which was given after w... | Easy, just define
$$\begin{array}{rcl}a \times b &=&
\hspace{10.5pt}(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/72 + \\&& 25(a-3)(b-4)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/18 + \\&& 15(a-3)(b-4)(a-4)(b-5)(a-6)(b-7)(a-7)(b-8)/8 \hspace{5.25pt}+ \\&& 49(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-7)(b-8)/36 + \\&&\hspace{5.5pt}7(a-3)(b-4)(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 15,
"answer_id": 0
} |
How to calculate $\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx$? I want to calculate the following integral: $$I:=\displaystyle\int_{-\infty}^\infty\underbrace{\frac{x^2+2x}{x^4+x^2+1}}_{=:f(x)}dx$$ Of course, I could try to determine $\int f(x)\;dx$ in terms of integration by parts. However, I don't think that's th... | $$\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx$$
$$f(x)=\frac{x^2+2x}{x^4+x^2+1}$$
$$\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx=2\pi i\sum_{\Im x_i>0} \text{Res}(f(x);x_i)$$
$$x_i=\left\{\sqrt[3]{-1},(-1)^{2/3}\right\}$$
so we get $$\left\{\frac{1}{12} \left(3-5 i \sqrt{3}\right),\frac{1}{4} i \left(\sqrt{3}+i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to compute power series by composition Is it possible to compute the power series of every function (e.g. around $0$) just by composing of the power series of its arguments?
For example:
The power series of $\sin(x^2)$ around $0$ is the composition of power series of $\sin(x)$ and $x^2$:
$compose(x-\frac{x^3}{6}+\... | Use the series for $\sqrt{1+x}$ and $\cos x - 1$
$$\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3\pm\dots$$
$$\cos x - 1 = -\frac{1}{2}x^2+\frac{1}{24}x^4 \pm \dots$$
Now substitute the second into the first:
$$\sqrt{\cos x} =\sqrt{1 + (\cos x-1)} \\=
1+\frac{1}{2}\left(-\frac{1}{2}x^2+\frac{1}{24}x^4\pm \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Suggestion for Computing an Integral Let $$A=\left\{(x,y,z)\in \mathbb R^3:\dfrac{x^2}{2}+\dfrac{y^4}{4}+\dfrac{z^6}{6}\leq1\right\}.$$
Then I want to compute the following integral:
$$\frac{1}{\operatorname{vol}(A)}\displaystyle\int_{\partial A}^{}\!\frac{1}{\sqrt{x^2+y^6+z^{10}}}\, d(x,y,z)$$
Should I use spherical c... | Let us express the surface element in terms of $y,z$:
$$dS=\sqrt{1+\left(\frac{\partial x}{\partial y} \right)^2+\left(\frac{\partial x}{\partial z} \right)^2} \,dy dz,\tag{1}$$
where
$$\frac{x^2}{2}+\frac{y^4}{4}+\frac{z^6}{6}=1. \tag{2}$$
Now from (2) it follows that
$$\frac{\partial x}{\partial y}=-\frac{y^3}{x},\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\int_0^\pi \frac{dx}{2+2\sin x+\cos x}=\log3$? How can we prove that: $$\int_0^\pi \frac{dx}{2+2\sin x+\cos x}=\log3$$
I don't have any ideas, the $f(\pi-x)$ thing doesn't work as well. Please help :)
| $$
\begin{aligned}
&\int_0^\pi\frac{1}{2+2\sin x+\cos x}dx
\\=& \int_0^\pi \frac{dx}{2[\cos^2(x/2)+\sin^2(x/2)]+4\cos(x/2)\sin(x/2) + \cos^2(x/2)-\sin^2(x/2)}\\
=&\int_0^\pi\frac{dx}{3\cos^2(x/2)+4\cos(x/2)\sin(x/2)+\sin^2(x/2)}\\
=&\int_0^\pi \frac{\sec^2(x/2)dx}{\tan^2(x/2)+4\tan(x/2)+3}\\
=&\int_0^\infty \frac{2du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How find this integral $I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\frac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$
Find the value:
$$I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\dfrac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$$
I use computer have this reslut
$$I=-\dfrac{4\pi}{\sqrt{2}}C+\dfrac{13\pi^3}{24\sqrt{2}}+\dfrac{9}{... | You have made a mistake with the denominator, which cannot be zero. See, $2-sin(2x)^2$ is always bigger or equal than $1$.
I think that the mistake was in expanding $2-(\frac{2t}{1+t^2})^2$: you have changed that initial $2$ for a $1$ and then completed the square. Repeat that expansion slowly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/873333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Proving $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers which are not quadratic squares
Prove that $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers $0<a<\frac{p}{2}$ which are not quadratic squares $\pmod p$ ($p\equiv3\bmod4$)
I don't know really from where to start (we know from w... | I believe that theorem only holds for $p \equiv 3 \pmod{4}$.
In fact for $p = 5$ then $((p-1)/2)! \equiv 2$ and for $p = 13$ the factorial yields $5$, not a power of $-1$.
Not let $p \equiv 3 \pmod{4}$.
$$(p-1)! \equiv \left( 1·2·\ldots ·\frac{p-1}{2}\right)\left( \frac{p+1}{2}\ldots (p-1)\right)$$
But $p-a \equiv (-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Compute limit of a function Compute: $$\lim_{x \rightarrow 0^+} \frac{\arctan(e^x+\arctan x)-\arctan(e^{\sin x}+\arctan(\sin x))}{x^3}$$
WolframAlpha tells me it's 1/6. Any nice idea how to rewrite that expression? Thanks!
| We can proceed as follows $$\begin{aligned}L\, &= \lim_{x \to 0^{+}}\frac{\tan^{-1}(e^{x} + \tan^{-1}x) - \tan^{-1}(e^{\sin x} + \tan^{-1}(\sin x))}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\tan^{-1}\left(\dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving Fibonacci inequality I didn't see a question regarding this particular inequality, but I think that I have shown by induction that, for $n>1$. I am hoping someone can verify this proof.
$$\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}<F_{n+1}<\left(\frac{1+\sqrt{5}}{2}\right)^{n}$$
Here we state that $F_0=0, F_1=1$. ... | I think your proof would be clearer if you wrote things in terms of the golden ratio $\phi=\dfrac{1+\sqrt{5}}{2}$. As a demonstration of why that's helpful, let me give a non-inductive proof of the upper bound. Note that this inequality is equivalent by the Binet formula to
$$ \frac{1}{\sqrt{5}}\left(\phi^{n+1}-(-\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomial representation Why is the polynomial $P(x)$ represented as
$$ P(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots+ a_2 x^2 + a_1 x + a_0 \text{ ?}$$
A polynomial can be $5x^4 + 3x^3 + 7x^2 + 10x -2$ and it is not necessarily $a_n x^n + a_{n-1} x^{n-1}+\cdots$ I.e if $ a_n x^n$ is $ 5x^4$, doesn't $a... | $$a_{n-1}$$ is not the same as $$a_n -1 $$
The $-1$ is in the subscript in $a_{n-1}$.
Think of it as a function $a$, and we are looking at $a(1), a(2), \dots, a(n-1), a(n)$, so the polynomial is
$$a(n) x^n + a(n-1) x^{n-1} + \dots + a(1) x + a(0)$$
So for instance, if we take $a(m) = m^2$ (can also be written as $a_{m}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this inequality $\frac{3(x^2+y^2+z^2)}{(x+y+z)^2+2(yz+xz+xy)}\ge\sum\limits_{cyc}\frac{x^2}{x^2+(y+z)^2}$ Question:
let $x,y,z\ge 0$.prove or disprove
$$\dfrac{3(x^2+y^2+z^2)}{(x+y+z)^2+2(yz+xz+xy)}\ge\sum\limits_{cyc}\dfrac{x^2}{x^2+(y+z)^2}$$
My idea: let $x+y+z=1$,
then we can only
$$\sum_{cyc}\dfrac{x^2... | It's wrong! Try $x=4$ and $y=z=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
| With no quadratic steps, only using trig identities:
Multiplying through by $\frac1{\sqrt{2}}$:
$$\begin{align}
\sin(x)\frac1{\sqrt{2}}+\cos(x)\frac1{\sqrt{2}}&=\frac{1}{\sqrt{8}}\\
\sin(x)\cos(\pi/4)+\cos(x)\sin(\pi/4)&=\frac{1}{\sqrt{8}}\\
\sin(x+\pi/4)&=\frac{1}{\sqrt{8}}
\end{align}$$
Given that $x$ is in $[0,\pi]$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 8
} |
Finding common ratio from two sums I'm struggling with this very basic question on the binomial theorem:
The sum of the first and second terms of a geometric progression is 12, and the sum of the third and fourth term is 48. Find the two possible values of the common ratio and the corresponding values of the first ter... | Continuing with the notation you've used, given $a + ar = 12$, $ar^2 +ar^3 = 48$, we see that
$ar^2 + ar^3 = r^2(a + ar)$
$\implies 48 = r^2*12$
$\implies r = \pm 2$
Back-substituting this gives us $r = 2$ and $a = 4$ or $r=-2$ and $a = -12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $e^{iy} = 1 + iy + \frac{\mu_1 y^2}{2}$ for all $y \in \mathbb{R}$ with $|\mu_1| \leq 1$ How to show the expansions
\begin{gather*}
e^{iy} = 1 + iy + \frac{\mu_1 y^2}{2}\\
e^{iy} = 1 + iy - \frac{1}{2}y^2 + \frac{\mu_2 |y|^3}{3!}
\end{gather*}
where $y \in \mathbb{R}$ and $|\mu_1| \leq 1$ and $|\mu_2| \leq 1$... | You made a mistake in your second expansion. The correct one is
\begin{align*}
\cos y = 1 - \cos(\theta_1y)\frac{y^2}{2}, &&
\sin y = y - \sin(\theta_2 y)\frac{y^3}{6}
\end{align*}
then
\begin{align*}
e^{i y} &= \cos y + i \sin y \\
& = 1 + i y - \cos(\theta_1y)\frac{y^2}{2}- i\sin(\theta_2 y)\frac{y^3}{6}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$a,b,c \geq 0$,prove that $a^2+b^2+c^2+abc+5 \geq3(a+b+c) $
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^2+b^2+c^2+abc+5 \geq3(a+b+c).$$
I'm certain that this problem could be solved by using dirchlet's theory.but I do not know how to apply it exactly.
| I am not sure if you can apply "dirchlet's theory". But another way is:
for $f(x)=px^2+qx+r, \Delta=q^2-4pr$, if $p>0 \cap\Delta \le 0 \implies f(x) \ge 0$
WLOG,let $ a \ge b \ge c$
$a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff a^2+(bc-3)a+b^2+c^2+5-3(b+c) \ge 0 \iff \Delta_a =(bc-3)^2-4(b^2+c^2+5-3(b+c)) \le 0 \iff (4-c^2)b^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Area for an ellipsoid "Calculate the area for the rotation ellipsoid you get by rotating the ellipsoid $\frac{x^2}{2}+y^2 = 1$ around the x-axis."
I solved for x:
$$ y = \pm \sqrt{1-\frac{x^2}{2}} $$
Then did $ y = 0$ to get the integration limits, $\pm \sqrt(2)$.
So I've ended up with an integral I don't know if it's ... | Yours is not correct. Since $\frac{dy}{dx}=-\frac{x}{2y}$ and $4y^2=2(2-x^2)$, we have
$$\begin{align}2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{1-\frac{x^2}{2}}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx&=2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{1-\frac{x^2}{2}}\sqrt{1+\left(-\frac{x}{2y}\right)^2}\ dx\\&=2\pi\int_{-\sqrt 2}^{\sqrt 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Does $x^2+ x^4/(3\cdot4) + x^6/(3\cdot4\cdot5\cdot6) + \cdots$ have any compact form? Is there any compact form for the following series
$$F_1(x) = x^2+ \frac{x^4}{3\cdot4} + \frac{x^6}{3\cdot4\cdot5\cdot6} + \cdots$$
$$F_2(x) = x+ \frac{x^3}{2\cdot3} + \frac{x^5}{2\cdot3\cdot4\cdot5} + \cdots$$
$$F_3(x) = F_2(x) - F_1... | Yes,
$F_1(x) = 2(\cosh x - 1)$;
$F_2(x) = \sinh x$.
(see Taylor series).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to evaluate the integral $\int e^{2x} \sin^2x\ dx$ How can I evaluate the following integral?
$$\int e^{2x} \sin^2x\ dx$$
| Here are the steps
\[
\int e^{2x}\sin^{2} x\ dx=\frac{1}{2}\int e^{2x}(1-\cos(2x))\ dx
\]
Let $u=2x$
\[
\frac{d}{dx}u=2\Rightarrow du=2\ dx \Rightarrow \frac{1}{2}du=dx
\]
Then
\[
\frac{1}{2}\int e^{2x}(1-\cos(2x))\ dx= \frac{1}{4}\int e^{u}(1-\cos u)\ du = \frac{1}{4}\int e^{u}-e^{u}\cos u\ du
\]
\[
= \frac{1}{4}\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Finding the number of solutions to $x+2y+4z=400$ My question is how to find the easiest way to find the number of non-negative integer solutions to $$x+2y+4z=400$$
I know that I can use generating functions, and think of it as partitioning $400$ with $x$ $1$'s, $y$ $2$'s, and $z$ $4$'s. The overall generating function ... | You were on the right track, but instead of truncating the factors, just consider the coefficient of $x^{400}$ in:
$$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^4+x^8+x^{12}+\ldots)=\frac{1}{(1-x)(1-x^2)(1-x^4)},\tag{1}$$
then write the RHS of $1$ as a sum of terms like $\frac{A}{(1-\xi x)^k}$, with $\xi\in\{1,-1,i,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Conditional extreme value of a function Let $x,y,z$ be the positive real numbers, if $x^2+y^2+z^2=1$, then how can we find the minimal value of this function $f(x,y,z)=\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}$.
| Hint First, note that $$\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}=xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).$$ Using $a+b+c\geq 3\sqrt[3]{abc}$, try to estimate lower bounds for $xyz$ and $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$. If you are stuck in the second, consider $$(x^2+y^2+z^2)\left(\frac{1}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $(x^2-9)+\sqrt{2-x}=0$ Solve the equation $(x^2-9)+\sqrt{2-x}=0$
*
*$(x+3)(x-3)+\sqrt{2-x}=0$
Conditions: $x\neq\pm3 \wedge x\leq2$
*$(x+3)(x-3) = -\sqrt{2-x}$
*$(x+3)^2(x-3)^2 = 2-x$
*$x^4-18x^2+81 = 2-x$
*$x^4-18x^2+x+79 = 0$
I'm positive this isn't going to the right direction.
Please... | The answer will be $x\approx -2.6175$. (see here. You can find its closed form if you want.)
Here, notice that we have $-3\le x\le 2$ because of $9-x^2=\sqrt{2-x}\ge 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Matrix Multiplication Understanding I have to multiply two 3x3 matrices. I don't understand the answer.
$$A=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
$$B=\begin{pmatrix}1&0&-1\\ 0&1&-1\\ 0&0&1 \end{pmatrix}$$
The answer given online is $$AB=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
I don't unders... | I agree with your work, and if they posted $AB=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$ as the answer for the product, it seems they're incorrect.
Another fast way to see that their answer is not feasible is to notice that they are writing $AB=A$. Since $A$ is an upper triangular matrix with no zeros on the... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Number of elements of order $2$ in $S_n$
How many elements of order $2$ are there in $S_n$?
Using combinatorics I arrived at this:
For $n$ even ($n=2k$) there are ${n\choose2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2} {n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}... | it is probably not any help, but i think your sum may be written $f(\sqrt{2})$ where
$$
f(x) = \sqrt{2}^{-n} \sum_{k=1}^{\lfloor \frac{n}2 \rfloor} \frac1{k!}\frac{d^{2k}}{dx^{2k}}x^n
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$ Let $a,b,c>0$, prove that
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$
I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.
Thanks.
| I would like to use @math 110 idea with a little difference when we prove that
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$
Since $\dfrac{a^2+b^2+c^2}{ab+bc+ac}=\dfrac{(a+b+c)^2}{ab+bc+ac}-2$, we can rewrite the inequality as
$\dfrac{(a+b+c)^2}{ab+bc+ac}\ge 4-\dfrac{4ac}{(a+c)^2}=\dfrac{4(a^2+ac+c^2)}{(a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many ways can 5 dice produce a total of 20?
How many ways can $5$ dice produce a total of $20$?
I set up the equation $x_1+x_2+x_3+x_4+x_5 = 20$. The total possible number of combinations is $\binom{19}4$. From there I subtracted the number of possibilities where $1$ of the variables is greater than $6$, which I ... | The maximum of the five dice is either $4, 5$, or $6$. Reasoning:
If the maximum of all five is $6$, then the maximum of the remaining four is $4, 5$ or $6$, because $6+3+3+3+3 < 20.$ If the maximum of all five is $5$, then the maximum of the remaining four is $4$ or $5$, because $5+3+3+3+3 < 20.$ If the maximum of ... | {
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"source": "stackexchange",
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Find the derivative: Do I use the Quotient Rule, Product Rule, or Chain Rule? I have to find the derivative of:
$$y=\frac{(5x^6-1)}{x^2}$$
I keep on getting this problem wrong. Should I use the quotient rule
$$\frac{f(x)g'(x) - g(x)f'(x)}{g(x)^2}$$
However, my answer is:
$$y'= 30x^2 +\frac{6}{x^4} - 120{x^3}$$
Am I ut... | $$\frac{dy}{dx}=\frac{(5x^6-1)'x^2-(5x^6-1) (x^2)'}{x^4}=\frac{30x^5 \cdot x^2-2x(5x^6-1)}{x^4}=\frac{30x^7-10x^7+2x}{x^4}=\frac{20x^7+2x}{x^4}=20x^3+\frac{2}{x^3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that each integer of the form $a^2+b^2$ has all the factors of this form, where $(a, b)$ are distinct integers and relatively prime
Show that each integer of the form $a^2+b^2$ has all the factors of this form, where $(a, b)$ are distinct integers and relatively prime
Progress
If $a^2+b^2$ is prime then it is al... | Assuming that you are talking about numbers of the form $n=a^2+b^2$ with $\gcd(a,b)=1$, then every odd prime $p$ that divides $n$ has the property:
$$\left(\frac{-1}{p}\right)=1,\tag{1}$$
since $(1)$ is a consequence of $a^2+b^2\equiv 0\pmod{p}$. However, $(1)$ is equivalent to
$$ p\equiv 1\pmod{4},\tag{2}$$
since the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A cycle as a product of transpositions Can someone please explain how a cycle $(1234)$ can be written as a product of transpositions: $(14)(13)(12)$? And how they can be multiplied to (1234)? Thanks in advance.
| $$(14)=\begin{pmatrix}1&2&3&4 \\ 4&2&3&1\end{pmatrix}$$
$$(13)=\begin{pmatrix}1&2&3&4 \\ 3&2&1&4\end{pmatrix}$$
$$(12)=\begin{pmatrix}1&2&3&4 \\ 2&1&3&4\end{pmatrix}$$
$$(14)(13)=(14)\circ(13)=\begin{pmatrix}1&2&3&4 \\ 4&2&3&1\end{pmatrix}\circ\begin{pmatrix}1&2&3&4 \\ 3&2&1&4\end{pmatrix}=\begin{pmatrix}1&2&3&4 \\ 4&2... | {
"language": "en",
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Is there an easier way to find $\frac{\mathrm d^9}{\mathrm dx^9}(x^8\ln x)$ than using the product rule repeatedly?
Find $\dfrac{\mathrm d^9}{\mathrm dx^9}(x^8\ln x)$.
I know how to solve this problem by repeatedly using the product rule, but I was wondering if there is a short cut. Thanks.
| For any analytic function $f$,
$$\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} f(x) = f(x+z) $$
Thus $\dfrac{d^n f}{dx^n}$ is $n!$ times the coefficient of $z^n$ in the Maclaurin series of $f(x+z)$.
In this case $f(x+z) = (x+z)^8 \ln(x+z)$. Now $$
\eqalign{(x+z)^8 &= \sum_{j=0}^8 {8 \choose j} x^{8-j} z^j \cr
... | {
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"question_score": "8",
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Solving these two equations simultaneously I'm having a hard time to solve these two equations simultaneously. I'm arriving to a very long equation..
$$x_0^2+y_0^2=(7\sqrt{2})^2=98$$
$$\sqrt{25+(x_0+2)^2}+\sqrt{4+(y_0-5)^2}=7\sqrt{2}$$
| Use polar coordinates $(x,y) = (-r \sin \theta, r \cos \theta)$ to get $r=\sqrt{98}$ and
$$ \sqrt{98 \cos^2 \theta - 2 \sqrt{98} (5) \cos \theta + (-2)^2 + (5)^2} + \\
\sqrt{98 \sin^2 \theta+2 \sqrt{98} (-2) \sin\theta + (-2)^2 +(5)^2 } = \sqrt{98} $$
Numerically there are two solutions at $\theta=28.27437°$ and $\thet... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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How to find $\lim_{x\to 0}\frac{\tan 3x}{\tan 5x}$? I am asked to find the following limit:
$$ \lim_{x \to 0} \frac{\tan 3x}{\tan 5x}$$
My problem is in simplifying the function. I followed two different approaches to solve the problem. But both seems incorrect.
Apprach 1)
Since $\tan \theta = \frac{sin \theta}{cos \t... | Using L'Hopital's:
$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \lim \limits_{x \to 0} \frac{3 \sec^2 3x}{5 \sec^2 5x} = \frac{3}{5}$$
Without L'Hopital's:
$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \frac{3}{5} \lim \limits_{x \to 0} \frac{\cos 5x}{\cos 3x} \cdot \frac{\sin 3x}{3x} \cdot \frac{5x}{\sin 5x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differentiate with product rule Question: differentiate $x(x^2 +3x)^3$
I have gotten to the point where i've used the product rule and i've gotten
$$(x^2 + 3x)^3 + x\cdot(3x+9)(x^2 + 3x)^2$$
but now that it comes to the simplifying im completely at a loss, any help would greatly appreciated
| Using the product rule is not enough.
$$(x(x^2 + 3x)^3)^\prime = (x)^\prime \cdot (x^2 + 3x)^3 + x \cdot ((x^2 + 3x)^3)^\prime$$
$$= (x^2 + 3x)^3 + x \cdot ((x^2 + 3x)^3)^\prime$$
For the rigth part you need to use the chain rule.
$$= (x^2 + 3x)^3 + x \cdot 3(x^2 + 3x)^2\cdot(x^2+3x)^\prime$$
$$= (x^2 + 3x)^3 + x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905909",
"timestamp": "2023-03-29T00:00:00",
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Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins I encountered the following problem in Herber Wilf's book Generatingfunctionology:
Prove that, in country that has $1$-cent, $2$-cent, and $3$-cent coins
only, the number of ways of changing $n$ cents is exactly the integer
nearest to $ \frac{(n+3)^2}{12}... | I'll only show how to obtain a linear recurrent relation from the problem, it has been explained at various places how to solve these.
Let:
*
*$a(n)$ be solutions (ways how to change $n$) using only $1$c,
*$b(n)$ solutions using only $1,2$c and at least one $2$c,
*$c(n)$ solutions using all three coins and at leas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $a_{1}\;a_{2},a_{3}$ are the Roots of cubic eq. , Then $1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right)$
If $a_{1}\;a_{2},a_{3}$ are three real values of $a$ which satisfy the equation $$\displaystyle \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx-\frac{4a}{\pi-2}\int_{0}^{1}x\cdot \cos xdx = 2.$$ Then value of $\displ... | Continuing from JimmyK4542's answer, we find that $$C_3 = \displaystyle\int_0^1 \cos^3 x\,dx=\frac{1}{12} (9 \sin (1)+\sin (3))$$ $$C_2 = \displaystyle\int_0^1 3\sin x \cos^2 x\,dx=1-\cos ^3(1)$$ $$C_1 = \displaystyle\int_0^1 3\sin^2 x \cos x\,dx - \dfrac{4}{\pi - 2}\int_0^1 x\cos x\,dx=\sin ^3(1)-\frac{4 (-1+\sin (1)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$ if $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
My try :
$2pp' = (b^2-a^2)\sin 2\theta$
$p'^2 + pp'' = (b^2-a^2)\cos 2\theta$
Thats it ! it doesn't simplify no matter what I try. Any help ... | Since
$$p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta$$
we have
$$p=\mu (a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{1/2}, \mu=\pm 1$$
$$p + p'' =\frac{\mu a^2b^2(\cos^4\theta+\sin^4 \theta+2\cos^2\theta \sin^2\theta)}{(a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{3/2}}=\frac{a^2b^2}{p^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trigonometry and triangle proof Question:
Prove that in an acute angle triangle ABC:
$$\tan A\tan B +\tan A \tan C + \tan B \tan C \geq 9$$
I have no idea where to even begin this question. Please help me!
| Since for an acute triangle $ABC$ we have
$$\begin{align}\tan A+\tan B&=(\tan (A+B))\cdot (1-\tan A\tan B)\\&=(\tan(\pi -C))\cdot (1-\tan A\tan B)\\&=(-\tan C)\cdot (1-\tan A\tan B),\end{align}$$
we have$$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$$
Hence, we have, by AM-GM inequality,$$\tan A+\tan B+\tan C\ge 3\sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What happens to a function when it is undefined? If I have the function
$$f(x) = {x^2 - 2 \over x + \sqrt 2}$$
this is undefined for $x = -\sqrt 2$, am I correct? Since the denominator would be zero.
But the numerator is a difference of squares $x^2-2=(x - \sqrt 2)(x+\sqrt 2)$.
Now the second factor is equal to my ... | It is important to realise what you are doing when we say that a common factor "cancels" from the numerator and denominator. We are actually dividing both the numerator and denominator by that common factor. As you have already pointed out: you can't divide by zero.
It is true that
$$\frac{x^2-2}{x+\sqrt{2}} \equiv \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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Summing various rearrangements of $1-\frac12+\frac13-\frac14+\cdots$ Show that the series $1-\frac12+\frac13-\frac14+\ldots$ converges to $\log2$ but the rearranged series:
*
*$1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\frac15-\ldots$ converges to $0$.
*$1+\frac13... | Assuming that I have correctly understood the patterns occurring, I get:
(1)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{2j+1}-\frac{1}{8j+2}-\frac{1}{8j+4}-\frac{1}{8j+6}-\frac{1}{8j+8}\right)&=&\sum_{j=0}^{+\infty}\left(-\frac{1}{8j+2}+\frac{3}{8j+4}-\frac{1}{8j+6}-\frac{1}{8j+8}\right)\\&=&\int_{0}^{1}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
| For n+1 it is
$1+3+3^2+\ldots+3^{n-1}+3^n=\frac{3^{n+1}-1}{2}$
$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{3\cdot 3^{n}-1}{2}$
$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{(1+2)\cdot 3^{n}-1}{2}$
$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{1\cdot 3^{n}-1}{2}+\frac{2\cdot 3^{n}}{2}$
$\left( 1+3+3^2+\ld... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.