Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$? Let $(a_n)_{n\ge 1}$ be the sequence defined as the following :
$$a_1=1 ,\ a_{n+1}=\dfrac{a_n}{n} + \dfrac{n}{a_n} ,\ n\ge1$$
Show that for every $n\ge4,\ \lfloor a_n^2 \rfloor = n$.
My approach to this problem was tryi... | We will prove the following result instead.
$$n+{2\over n}<{a_n}^2<n+1$$
First we check it's true for $n=4$.
Now the induction step, first notice that $x>y>1$ implies $x+{1\over x}>y+{1\over y} \equiv 1>{1\over xy}$
Therefore for RHS $${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+{2\over n}\over n^2}+{n^2 \over n+{2\over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
If $x$ is a nonnegative real number, find the minimum value of $\sqrt{x^2 +4} + \sqrt{x^2 -24x+153}$ If $x$ is a nonnegative real number, then find the minimum value of
$$\sqrt{x^2 +4} + \sqrt{x^2 -24x+153}$$
How can I approach this? Thanks
| Process 1: Derivate it to zero
Process 2: Using triangle inequality,
Given,$$\sqrt{x^2 +4} + \sqrt{x^2 -24x+153}$$
$$\rightarrow\sqrt{x^2+2^2}+\sqrt{\left(-x+12\right)^2+3^2}\geq \sqrt{\left(x-x+12\right)^2+(2+3)^2} = \sqrt{{12}^2+{5}^2}$$
$$=\sqrt{169}=13$$
I.e, So,
The minimum value is $$\bbox[5px,border:2px solid re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
What is the meaning of "due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x=\frac1r$ is also a zero" I was studying this answer about factoring $x^4+x^3+x^2+x+1$:
https://socratic.org/questions/how-do-you-factor-x-4-x-3-x-2-x-1
The author says: "A cleaner algebraic approach is to no... | To answer the original question, the thinking process comes as follows:
(1) If $r$ is a solution to $x^4-x^3+x^2-x+1=0$, then $r^4-r^3+r^2-r+1=0$.
(2) Divide both sides by $r^4$ you get $({1\over r})^4-({1\over r})^3+({1\over r})^2-({1\over r})+1=0$. Therefore $1\over r$ is also a solution.
(3) Hence if $(x-r)$ is a fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end... | My steps to calculate your determinant. With $R$ I indicate the row and with $C$ the column.
$$C_2=C_2-\left(\frac{b^{2}}{a^{2}}\right)C_1 $$
$$\left| \begin{array}{ccc} a^{2} & 0 & c^{2} \\\\ a b & b c - \frac{b^{3}}{a} & a c \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} & a^{2} \end{array} \right| \tag{first step}$$
Subtr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Factorize: $x^3 + x + 2$. How do I factorize the term $x^3 + x +2$?
What I have previously tried is the middle term factor method but it didn't work...
$x^3 + x + 2$
$\Rightarrow x^3 + 2x - x + 2$
$\Rightarrow x(x^2 + x) - 2(x - 1)$
This doesn't work.
What should I do?
| Note
\begin{align}
x^3+x+2& = (x^3+x^2)-(x^2-x-2)\\
&= x^2(x+1)-(x-2)(x+1)\\
&=(x+1)(x^2-x+2)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{... | By AM-GM
$$(x+y)^2=xy(x-y)^2=\frac{1}{4}\cdot4xy(x-y)^2\leq\frac{1}{4}\left(\frac{4xy+(x-y)^2}{2}\right)^2=\frac{(x+y)^4}{16},$$
which gives $$x+y\geq4.$$
The equality occurs for $(x-y)\sqrt{xy}=x+y$ and $4xy=(x-y)^2,$ which gives $$(x,y)=(2+\sqrt2,2-\sqrt2),$$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find the smallest possible value of an equation, where $a+b+c=3$ We have the positive real numbers $a, b, c$ such that $a+b+c=3$. Find the minimum value of the equation:
$$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$
I solved it in the following fashion:
$$
\begin{align}
A&=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1... | We have
$$A = 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left[(a+b+c)^2-2(ab+bc+ca)\right]$$
$$=2(ab+bc+ca)\left(\frac{1}{abc}+1\right)-9.$$
Using know inequality $(ab+bc+ca)^2 \geqslant 3abc(a+b+c)$ and the AM-GM inequality, we get
$$A \geqslant 2\sqrt{3abc(a+b+c)}\cdot \frac{2}{\sqrt{abc}}-9=3.$$
Equality occ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$ Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$
Can somebody check my solution? Thanks!!
Let $u=11xe^{2x}$
then $\frac{du}{dx} = \frac{d}{dx}(11xe^{2x})$
$=11(e^{2x}+2xe^{2x})$
and so $du=11e^{2x}(2x+1)dx$
Now, let $dv=\frac{1}{(2x+1)^2}dx$
Then $\int dv = \int \frac{1}{(2x+1)^... | Your answer is correct, but it can be further simplified:
$$\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{4}e^{2x}$$
$$=\frac{11}{4}e^{2x} \left(\frac{-2x}{2x+1} + 1 \right) =\frac{11}{4}e^{2x} \left(\frac{-2x-1}{2x+1} + \frac{1}{2x+1} + 1 \right)$$
$$=\frac{11}{4}e^{2x} \cdot \frac{1}{2x+1} = \frac{11e^{2x}}{4(2x+1)}$$
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the second derivative of the absolute function $\left|\frac{x+1}{x+2}\right|$? I calculated the derivative of $\left|\frac{x+1}{x+2}\right|$ in the same way that I would do with $ \frac{x+1}{x+2}$ in order to study the function.
But when I verified on wolfram, I noticed it is all wrong. Wolfram uses the chain r... | As an alternative, using sign function we have that for $x\neq -1,-2$
$$\left|\frac{x+1}{x+2}\right|=\frac{x+1}{x+2}\cdot \frac{\left|\frac{x+1}{x+2}\right|}{\frac{x+1}{x+2}}=\frac{x+1}{x+2} \operatorname{sign}\left(\frac{x+1}{x+2}\right)$$
therefore by chain rule, since $(\operatorname{sign}(x))'=0 $ for $x\neq 0$, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$\cos\left(\frac{\pi}{5}\right)$ using De Moivre's Theorem This is an Exercise 3.2.5 from Beardon's Algebra and Geometry:
Show that $\cos\left(\frac{\pi}{5}\right)=\frac{\lambda}{2}$, where $\lambda$ = $\frac{1+\sqrt{5}}{2}$ (the Golden Ratio).
[Hint: As $\cos(5\theta) = 1$, where $\theta = \frac{2\pi}{5}$, we see fro... | Here is another way to prove it, we'll be using De moivre's theorem but without having to get to the $ 5^{\mathrm{th}} $ degree.
We have : $$ \cos{\left(\pi-\theta\right)}=-\cos{\theta} $$
With $ \theta $ being equal to $ \frac{2\pi}{5} $, we get : $$ \cos{\left(\frac{3\pi}{5}\right)}=-\cos{\left(\frac{2\pi}{5}\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a pr... | You properly wrote $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$ Tak into account the homegeneity and let $b=k a$ to get
$$k+\frac 1k-1=\frac{7}{6}\implies k=\frac 23 \qquad \text{and} \qquad k=\frac 32$$ and then the slutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is my calculation of the integral $\int \tan^{-1} x \, dx$ correct? Compute $\int \tan^{-1}x \,dx$.
First, set $u = \arctan(x)$ and $dv = dx$. We want to find $du$ and we already have $v = x$.
We start by taking the tangent of both sides, leaving us with
$$\tan(u) = x.$$
Next, using implicit differentiation, we get $\f... | Yes, but you can simply do like below
$$\int \arctan x~\mathrm{d}x= x\cdot \arctan x - \int \frac{x}{x^2+1}\mathrm{d}x = x\cdot \arctan x - \frac{1}{2}\log (x^2+1) +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$ for $x\in(0,\pi/2)$
Solve this equation for $x\in (0 , \frac{\pi}{2})$
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$$
I gave a try using the $\cos(a-b)$ and $\sin(a-b)$ formul... | As an alternative, since $x=k\frac \pi 4$ and $x = \frac{\pi}{6}+k\frac \pi 2$ are not solutions, we have that
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right) \iff \frac{\sin(2x)}{\cos(2x)}=\frac{\cos\left(\frac{\pi}{6} - x\right)}{\sin\left(\frac{\pi}{6} - x\right)}$$
$$\iff ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equ... | $123^{456}\equiv 2^6=64\equiv9\bmod 11$.
$123^{456}\equiv 3^0=1\equiv9\bmod 8$.
Therefore, by the constant case of the Chinese Remainder Theorem, $123^{456}\equiv9\bmod88$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?
I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the ... | Yes, we need to find $ab+ac+bc=7$ before.
Now, $a$, $b$ and $c$ are roots of the equation:
$$(x-a)(x-b)(x-c)=0$$ or
$$abc=x^3-5x^2+7x.$$
Now, $$(x^3-5x^2+7x)'=(x-1)(3x-7),$$ which gives that a maximal value of $abc$ for which the equation $$abc=x^3-5x^2+7x$$ has three real roots holds for $x=1$, which gives:
$$\max_{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$ .
What I Tried :- I tried ... | $$\dfrac{1}{\log_{ab}a}-\dfrac{1}{\log_{ab}b}=\log_ab\,-\log_ba \tag{1}$$
Let $\log_b a=x$, we get the first expression to be $$x +\frac{1}{x}=\sqrt{1229} \tag{2}$$,
Now, squaring (1) and (2), we see (1) becomes:
$$x^2+\frac{1}{x^2} + 2 = 1229\tag{3}$$
and expression (2) becomes:
$$x^2-2+\frac{1}{x^2} = A^2\tag{4}$$
Su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Primality test for specific class of $N=k \cdot 2^n+1$ Can you prove or disprove the following claim:
Let $N=k \cdot 2^n+1$ be a natural number that is not a perfect square such that $ 2 \nmid k$ , $n>2$ . Let $c$ be the smallest odd prime number such that $\left(\frac{c}{N}\right)=-1$ , where $\left(\frac{}{}\right)$... | This fails for $N=22577=1411\cdot 2^4+1=107\cdot 211$. The value of $c$ that is taken is $3$, but
$$(1-\sqrt3)^{11288}\equiv 11502+11502\sqrt 3\bmod 22577.$$
It is true, however, if $N$ is prime. Setting $x=\sqrt c$, we have that $x^N$ is the Galois conjugate of $x$, so $x^N=-x$, whence
$$(1+x)^{N+1}+(1+x)^N(1+x)=(1+x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluating $\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$ without a calculator? Is there a way to get this value without calculator?
$$\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$$
I'm currently studying AIME.
| Consider
$$S_n=\sum_{a=1}^n\sum_{b=1}^n\sum_{c=1}^n\frac{ab(3a+c)}{2^a\,2^b\,2^c\,(a+b)(b+c)(c+a)}$$ and compute $S_n$ for the very first values of $n$. This gives the sequence
$$\left\{\frac{1}{16},\frac{27}{128},\frac{343}{1024},\frac{3375}{8192}\right\}$$ and you can notice that the numerators are cubes and that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Computing a finite sum involving binomial coefficients I would like to prove the following: given the sequence $$a_m = \begin{cases}1 & \text{if $m = 0, 1$} \\\frac{(\alpha+2)(\alpha+4)\cdots(\alpha+2(m-1))}{(\alpha+1)(\alpha+2)\cdots(\alpha+m-1)} & \text{otherwise}\end{cases}$$ then, $$\sum_{m=0}^{n}(-1)^m\binom{n}{m}... | From the definition, of the Pochhammer symbol, $$(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$$we can express
\begin{equation}
a_m=2^m\frac{ \left(\frac{\alpha }{2}\right)_{m}}{(\alpha )_{m}}
\end{equation}
with $a_0=a_1=1$. (The expression given by @ClaudeLeibovici can be retrieved by remarking that
\begin{equation}
(s+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Can we find $ \sum_{n=1}^{\infty}\frac{1+2+\cdots +n}{n!} $? Consider the sequence $$ a_{n} = \sum_{r=1}^{n}\frac{1+2+\cdots +r}{r!} $$
Then we have, $$ a_{n} = \sum_{r=1}^{n}\frac{1}{r!} \ + 2\sum_{r=2}^{n}\frac{1}{r!} \ + 3\sum_{r=3}^{n}\frac{1}{r!} \ + \cdots + n\sum_{r=n}^{n}\frac{1}{r!} \ \geq \ 1 + \sum_{r=1}^{n... | Note that
$$ \sum_{k = 1}^\infty \frac{1 + 2 + \cdots + k}{k!} = \frac{1}{2} \sum_{k = 1}^{\infty} \frac{k + 1}{(k - 1)!} = \frac{1}{2} \left( \sum_{k = 1}^\infty \frac{(k - 1) + 1}{(k - 1)!} + e \right) = \frac{1}{2} \left( \sum_{k =2 }^\infty \frac{1}{(k - 2)!} + \sum_{k = 1}^{\infty} \frac{1}{(k-1)!} \right) + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
How do i form the inequality $ -5\le 3\cos x - 4\sin x\le5$? We know that
$-1\le \cos x\le1$ so we if we multiply 3 so we get $-3\le 3\cos x\le 3$ and we also know that $-1\le \sin x\le 1$ and and again we multiply 4 we get $-4\le 4\sin x\le 4$ and we can add these two inequality we get $-7\le 3\cos x - 4\sin x\le 7$ b... | Hint :
\begin{eqnarray*}
3 \cos \theta -4 \sin \theta = R \cos ( \theta +\phi) .
\end{eqnarray*}
$R=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Probability of taking white balls until a black shows up. (No replacement) I have an urn with $n$ white balls, and 1 black ball.
What's the probability of taking $x-1 \in \{0,n\}$ white balls, and the last extraction being a black ball? (No replacement)
My reasoning:
$$P(ww...b)=\frac{n}{n+1}\frac{n-1}{n}\cdots\frac{n-... | In order to select the black ball on the $k$th draw, we must first select $k - 1$ of the $n$ white balls while selecting $k - 1$ of the $n + 1$ balls, then select the only black ball from the remaining $n + 1 - (k - 1) = n + 2 - k$ balls. Hence, the probability of selecting the black ball on the $k$th draw is
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$ For $a,b,c>0.$ Prove$:$ $$4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}$$
My proof is using SOS$:$
$${c}^{2}{a}^{2} {b}^{2}\Big( \sum a\B... | After using the nguyenhuyen_ag's reasoning it's enough to prove our inequality for positive variables,
and we can end the proof by $uvw$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\frac{4(9v^4-6uw^3)}{w^6}+\frac{81}{9u^2}\geq\frac{21u}{w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Rational Roots (with Lots of Square Roots!) Find the smallest positive integer $a,$ greater than 1000, such that the equation $$\sqrt{a - \sqrt{a + x}} = x$$ has a rational root.
Squaring both sides we have $a-\sqrt{a+x}=x^2.$ We shall not square again as that gives a quartic in $x.$ Rearranging, we have $$-x^2+a-\sqrt... | Since the quartic equation is monic, i.e. it has the form
$$x^4+ \dots =0$$
by the rational root theorem, $x$ must be an integer. Moreoever
$$x=\sqrt{a-\sqrt{a+x}}>0$$
so that $x$ must be a positive integer.
Now, consider the equation
$$\sqrt{a+x}=a-x^2$$
and call $$y=\sqrt{a+x}=a-x^2$$
Note that $y=\sqrt{a+x}>0$ is a ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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proof that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $
Proof that
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $$
by induction.
Proof
Base case: Statement clearly holds for $n = 1$. Now assume that st... | As $\frac 1{n(n-1)}$ is not defined for $n =1$ and also because the first term is $\frac 1{2\cdot 1} = \frac 1{2(2-1)}$ and so the first term is for $n = 2 > 1$, then it clearly does NOT work for $n= 1$.
If the statement were true for any $n$ it would be true for any subsequent natural numbers but it is not true for an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate below expression of modular arithmetic? How to evaluate (10^18)%(10^9 + 7) using modular arithmetic? How to proceed and what will be the steps?
| Building on dave's answer above but in a way that I can understand -
Given
10^18 = (10^9 + 7) * (10^9 - 7) + 49
10^18 % (10^9 + 7) = 49
....
Generalising ...
===> a^2 = (a + b) * (a - b) + b^2
===> a^2 % (a + b) = { (a + b) * (a - b) + b^2 } % (a + b)
Consider : { (a + b) * (a - b) + b^2 } % (a + b)
===> [ { (a + b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a simpler expression for this piecewise-defined function? As a math-for-fun exercise, I challenged myself to find a globally-defined, everywhere-differentiable antiderivative of $\sqrt{1-\sin(x)}$. Because of the fundamental theorem of calculus, the problem boils down to evaluating the integral
$$f(x)=\int_{-\... | You can verify that
$$ \sqrt{1 - \sin t} =
\sqrt2 \left\lvert \sin \left(\frac12t - \frac\pi4\right)\right\rvert . $$
This suggests the substitution $u = \frac12t - \frac\pi4,$
$$ \int \sqrt{1 - \sin t} \,\mathrm dt
= 2\sqrt2 \int \lvert\sin u\rvert \,\mathrm du
\DeclareMathOperator{\sgn}{sgn}
= 2\sqrt2 \int \sgn(\... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the limit $\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$ Find the limit
$\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$
I tried substituting "h" and also multiplying $\frac{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}$
| Your idea is fine, indeed we obtain
$$\frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}\cdot \frac{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}=$$
$$=\frac{{1 - (x+h) ^2} - ({1-x^2})}{h(\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2})}=$$
$$=\frac{-2xh-h^2}{h(\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2})}$$
... | {
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What to do with exponent with different base in denominator? I‘m struggling right now with a equation. The equation is
$$\frac{2^{(5x)}}{7^{(x+2)}} = 10.$$
The solution should be $4.076$ but I don‘t know how to solve this equation.
I came to the conclusion that I can simplify the equation so it‘s $\frac{32^x}{7^{(x+2)}... | You simplified $2^{5x}$ to $\left(2^5\right)^x=32^x$.
You can also simplify $7^{x+2}$ to $7^x\cdot7^2=49\cdot7^x$, using the exponent rule $a^{b+c}=a^b\cdot a^c$.
Thus, your equation simplifies to
$$
\begin{align*}
\frac{32^x}{49\cdot7^x}&=10\\
\frac{1}{49}\cdot\frac{32^x}{7^x}&=10\\
\frac{32^x}{7^x}&=490\\
\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\lim\limits_{(x,y)\to (0,0)} \dfrac{x^4}{(x^2+y^4)\sqrt{x^2+y^2}}$ Calculate, $$\lim\limits_{(x,y)\to (0,0)} \dfrac{x^4}{(x^2+y^4)\sqrt{x^2+y^2}},$$ if there exist.
My attempt:
I have tried several paths, for instance: $x=0$, $y=0$, $y=x^m$. In all the cases I got that the limit is $0$. But I couldn't figure... | Assuming $y<1$ and using polar coordinates we have
$$\dfrac{x^4}{(x^2+y^4)\sqrt{x^2+y^2}} \le \dfrac{x^4}{(x^2+y^2)\sqrt{x^2+y^2}} =r \sin^4\theta \to 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}.$
Question: Prove that $$\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}.$$
Solution: Let $$I_n:=n^2\int_0^{\frac{1}{n}}x^{x+1}dx, \forall \in\mathbb{N}.$$ Substituting $nx=t$ in $I_n$, we have $$I_n=n\int_0^1\left(\frac{t}{n}\rig... | $$ \lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}\mathrm{d}x = \lim_{n \to \infty} n^2 I_n =\frac{1}{2}. $$
My approach is quite similar to @Sanket. Sandwiching is the basic idea.
We have that $$0 \leqslant x \leqslant \frac{1}{n} \implies 1 \leqslant x + 1 \leqslant 1 + \frac{1}{n} \implies x \geqslant x^{x + 1} \geq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to factor $ 1 - 3x + x^2 + x^3$? By inspection, I can see that one of the roots is $1$.
So we can write
$$
1 - 3x + x^2 + x^3 = (x - 1)f_2(x)
$$
where $f_n(x)$ is an n-th order polynomial. I haven't used long division for polynomials in ages, but I feel like that might be overcomplicating things and there might be ... | One method to factor it is to check whether it can be separated into several parts that have a common factor. In this case, we can separate $x^3+x^2-3x+1$ into $x^3-x$ and $x^2-2x+1$. Since $x^3-x=x(x^2-1)=x(x-1)(x+1)$ and $x^2-2x+1=(x-1)^2$, the 2 parts have a common factor of $(x-1)$ and can be factored out. Thus, $x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many $ax^2+bx+c=0$ with distinct pairs of rational roots can be made from integers $N=|ac|=|\alpha\beta|$ and $b=\alpha+\beta$? Only for illustration, let's choose an integer $N=6$. I want to create a collection of all possible quadratic equations in the form of $ax^2+bx+c=0$ where
*
*$|ac|=|\alpha\beta|=N=6$, an... | This is a partial answer.
This answer deals with $N$ such that $N\not\equiv 0\pmod 3$.
This answer proves that the number of such quadratic equations is
$$\begin{cases}2\sigma_0(N)^2&\text{if $N$ is not a square number with $N\not\equiv 0\pmod 3$}
\\\\\sigma_0(N)(2\sigma_0(N)-1)&\text{if $N$ is a square number with $N\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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finding a relation in $p:p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$
if $$p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$$ and $$p^2+ap+c=0.$$ Find $a,c$ also $|c|=2$
My progress:The general term $$T_{m+1}=\frac{(1)(3)\cdots(2m+1)}{(3)(6)\cdots(3m+3)}$... | It turns out that $p = \sqrt{3} - 1$. To see this, you need to recognize $p$ as the power series of the function $1/\sqrt{1-x} - 1$, evaluated at $2/3$. More specifically, the $n$-th derivative of $(1-x)^{-1/2}$ is
$$
\begin{align}
\frac{1}{2}\cdot \frac{3}{2}\cdots \frac{2n-1}{2} (1-x)^{-\frac{1}{2} - n} & = (1-x)^{-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the number of digits in repunit Find the number of digits in the smallest repunit divisible by $19$.
I believe a repunit number, with $N$ digits, is of the form $ \sum_{k=0}^{N-1} 10^k = \frac{10^N -1}{10-1} = \frac{10^N - 1}{9}, $ and is divisible by $n$ if $10^N - 1 \equiv 0 \pmod{9n}.$ The solution should be... | $10\equiv1\pmod9$, so $10^N\equiv 1 \pmod9$ for all $N\in\mathbb N$,
so your question becomes what is the smallest number $N$ satisfying $10^N\equiv1\pmod{19}$.
By Fermat's little theorem, we know $10^{18}\equiv1\pmod{19}$;
we just have to show that $10^6\not\equiv1\pmod{19}$ and $10^{9}\not\equiv1\pmod{19}$.
Method 1
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(ab + bc + ca) \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9}{(p + 1)(q + 1)}$ One of my friends showed me this inequality.
$$(ab + bc + ca) \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9... | The AM-GM:
$\dfrac{(q+1)(a+pb)+(p+1)(a+qb)}2 \geqslant \sqrt{(p+1)(q+1)(a+pb)(a+qb)}$
$$\implies \frac1{(a+pb)(a+qb)}\geqslant \frac{4(p+1)(q+1)}{[(q+1)(a+pb)+(p+1)(a+qb)]^2}=\frac{4(p+1)(q+1)}{[(p+q+2)a+(2pq+p+q)b]^2}$$
Defining $k$ using $2pq+p+q=k(p+q+2)$, the above can be rewritten as
$$\frac1{(a+pb)(a+qb)}\geqslan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Generating function coefficient I have a generating function $\frac{1}{(1-x-x^{3}+x^{4})^{2}}$.
I have to find the $x^{n}$ coefficient. So far it seems that the next step is to convert this function back to the sequence, but I am not sure how to proceed.
| First note that $$1-x-x^3+x^4=(1-x)(1-x^3)=(1-x)^2(1+x+x^2)\,,$$
so if nothing else, you can decompose
$$\frac1{(1-x)^4(1+x+x^2)^2}$$
into partial fractions and go from there. Alternatively, you can work with
$$\frac1{(1-x)^2}\cdot\frac1{(1-x^3)^2}\,.$$
It’s standard that
$$\frac1{(1-x)^2}=\sum_{n\ge 0}(n+1)x^n\,,$$
so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(... | We have that
$$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}
\iff \frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2}-\frac{{x}^{2}+3x+18}{({x+1})^{2}}<0 $$
$$\iff \frac{({x}^{2}(x+1)^2-({x}^{2}+3x+18)(({x+1-\sqrt{x+1}})^2) }{({x+1-\sqrt{x+1}})^2(x+1)^2} < 0$$
$$\iff ({x}^{2}(x+1)^2-({x}^{2}+3x+18)(({x+1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to Evaluate $ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} $ How can I evaluate
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} \approx - 0.198909 $$
The Sum can be given also as
$$ \frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)\sqrt[4]{(-x)^{3}}}\,\left(\,\tan^{-1}\left(\sqrt[4]{... | The result is not so bad. If
$$S=\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1}$$ For more legibility, I shall write $S$ as
$$S=\frac {A}{96}-i\frac B 4$$ where $A$ and $B$ contain real and complex parts.
$$A=24 C-5 \pi ^2+9 \log ^2\left(3-2 \sqrt{2}\right)+(24-6 i) \pi \log \left(3-2
\sqrt{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Limit related to $ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $? Let $n$ be a positive integer.
Let $b = 2 n - 1$.
Let $x$ be a positive integer.
Define $f(x)$ as :
$$ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $$
Then it appears that
$... | If you are allowed to use the complex representation of the sine and Pochhammer symbols,
$$f(b) = \prod_{i = 1}^{2n-1} \Big[\sin \left(\frac{\pi i}{n}\right)+\frac{5}{4}\Big]=-\frac{4}{5} \frac{ \left(\frac{i}{2};e^{\frac{i \pi
}{n}}\right){}_{2 n} \left(2 i;e^{\frac{i \pi }{n}}\right){}_{2 n}}{4^n\,e^{i \pi n}}$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Dependent variable substitution of a differential equation. I am attempting to answer a question from a textbook. The question is as follows:
"Use the substitution $y = x^2$ to turn the differential equation $x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$ into a second order differential equation with c... | As noticed $$\left(\frac{dx}{dt}\right)^2=\frac{dx}{dt}\cdot \frac{dx}{dt}\neq \frac{d(x^2)}{dt}$$
By the suggested substitution we have
$$y=x^2 \implies \frac{dy}{dt}=2x \frac{dx}{dt} \implies \frac{d^2y}{dt^2}=2 \left(\frac{dx}{dt}\right)^2+2x \frac{d^2x}{dt^2}$$
that is
*
*$\frac{dx}{dt}=\frac1{2x}\frac{dy}{dt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a^2x^4+b^2y^4=c^6$, then the maximum value of $xy$ Simply subsisting the value of $y$ in $xy$ will not work, but I don’t know any other method to solve it. Can I get a hint?
| Method 1
You could use the parametric equation of ellipse to solve it.
Let $p=x^2,q=y^2$ then $x^4 = p^2,y^4=y^2$, the equation turns to
$\frac{p^2}{c^6/a^2} + \frac{q^2}{c^6/b^2}=1$, which is the ellipse function. Thus
$p = \frac{c^3}{a} \cos\theta$
$q = \frac{c^3}{b} \sin\theta,\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Integrate curve given in polar coords solution verification My solution to the below problem is $4\pi$ but the answer sheet says it's $8\pi$. Please verify my calculations:
$$
\rho=4\sin2\phi \implies 4\sin2\phi \ge0 \implies \sin t\ge 0
$$
Where $t = 2\phi$
Now, I'm trying to find the integration bounds like this:
$$
... | This is the bound -
i) For $0 \le \phi \le \frac{\pi}{2}, 0 \le 2\phi \le \pi$.
ii) For $\frac{\pi}{2} \le \phi \le \pi, \pi \le 2\phi \le 2\pi$.
iii) For $\pi \le \phi \le \frac{3\pi}{2}, 2\pi \le 2\phi \le 3\pi$.
iv) For $\frac{3\pi}{2} \le \phi \le 2\pi, 3\pi \le 2\phi \le 4\pi$.
This is rose curve with $4$ petals.
... | {
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"url": "https://math.stackexchange.com/questions/3839119",
"timestamp": "2023-03-29T00:00:00",
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Limit of sum of exponential functions under root How to solve this limit?
$$\underset{x\to \infty }{\text{lim}}\left(4*6^x-3*10^x+8*15^x\right)^{1/x}$$
It is equal $15$ and it seems obvious that it is so. I just can not write it mathematically.
I tried to get rid of $1/x$ in exponent:
$$\underset{x\to \infty }{\text{li... | $$\lim_{x\to \infty}\left(4\times6^x-3\times10^x+8\times15^x\right)^{1/x}=\lim_{x\to \infty}(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}$$
and we have $$(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}<15(4\times 1-3\times 0+8)^{1/x}\to 15$$
$$(15^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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maclaurin series $\ln \left(\frac{1+x^2}{1-x^2}\right)$ I know that the Maclaurin seriess is: $\sum_{n=1}^\infty c_n(x)^n$ and $c_n = \frac{f^{n}(0)}{n!}$
$\ln \left(\frac{1+x^2}{1-x^2}\right)$
$f^0 = \ln (1+0) - \ln (1-0) = 0$
$f^{(1)} = \frac{4x}{(1+x^2)(1-x^2)} = 0$
$f^{(2)} = \frac{12x^4 + 4}{(1+x^2)^2(1-x^2)^2} = ... | Since the function is defined for $|x|<1$, you can write it as
$$
f(x)=\ln(1+x^2)-\ln(1-x^2)
$$
You surely know the Maclaurin series
$$
\ln(1+t)=\sum_{n\ge1}\frac{(-1)^{n+1}t^n}{n}
\qquad
\ln(1-t)=-\sum_{n\ge1}\frac{t^n}{n}
$$
and so
$$
\ln(1+x^2)=\sum_{n\ge1}\frac{(-1)^{n+1}x^{2n}}{n}
$$
and
$$
\ln(1-x^2)=-\sum_{n\ge1... | {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $A\sin{\theta} + B\cos{\theta} = C$? I've stumbled upon a equation in the form
$$A\sin{\theta} + B\cos{\theta} = C$$
What would be the steps necessary to solving it?
Thank you.
| $$A\sin{\theta} + B\cos{\theta} = C$$
use the identity
$$\cos\theta= \pm \sqrt{1-\sin^2\theta}$$
so
$$A\sin{\theta} \pm B\sqrt{1-\sin^2\theta} = C$$
$$ \pm B\sqrt{1-\sin^2\theta} = C-A\sin{\theta}$$
$$ B^2(1-\sin^2\theta) = (C-A\sin{\theta})^2$$
let $x=\sin{\theta}$
$$ B^2(1-x^2) = (C-Ax)^2$$
$$B^2-B^2x^2=C^2-2ACx+A^2x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Power series of $\frac{x^{3}-2}{x^{2}+1}$ I have to find the power series of the function
$$f(x)=\frac{x^3-2}{x^{2}+1}$$
centered at $a=1$. I tried to write $f$ as
$$f(x)=(x^3-2)\cdot\frac{1}{x^{2}+1}$$
and then, find the power series of $\displaystyle\frac{1}{x^{2}+1}$ centered at $a=1$. I want to know if this is the ... | If you make $x=y+1$
$$\frac{x^3-2}{x^{2}+1}=\frac{y^3+3 y^2+3 y-1}{y^2+2 y+2}=1+y-\frac{y+3}{y^2+2 y+2}$$ If you want a truncated series, continue with the long division.
Looking at the coefficient of the first terms, I suppose that their definition is not the simplest.
Edit
Working
$$\frac{y+3}{y^2+2 y+2}=\sum_{n=0}^\... | {
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"source": "stackexchange",
"question_score": "3",
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Expressing $\frac{1+x+x^2}{1-x^7}$ as a closed form for the $n^{th}$ term of the associated sequence First of all I'm a little confused about the wording, am I suppose to find the sequence associated, lets say for example $(1,1,1,1,\ldots)$, and come up with an expression like $a_n=1^n$? I've gotten this far with this ... | From expanding $$\sum_{k=0}^\infty (1+x+x^2)x^{7k} = 1 + x + x^2+x^7+x^8+x^9+x^{14}+x^{15}+x^{16}+\dots$$
the sequence associated with this G.F. is $(1,1,1,0,0,0,0,1,1,1,0,0,0,0, \dots)$, which can be written as:
$$a_n=\begin{cases}1 &\text{for } n \equiv 0,1,2 &\pmod 7\\0 &\text{for } n \equiv 3,4,5,6 &\pmod 7\end{cas... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Better proof of inequality $x - (1 + x) \log(1+x) \leq -\frac{x^2}{2(1+x)}$ for $x > 0$ The following inequality is valid for all positive real $x$,
$$
x - (1+x)\log(1+x) \leq \frac{-x^2}{2(1+x)}.
$$
It is possible to show that this is true by considering the function
$$
f(x) := x - (1+x)\log(1+x)+ \frac{x^2}{2(1+x)}.
... | Better proof:
$\log(1+y)\le y-\dfrac{1}{2}y^2\quad$ for all $\;y\in\left]-1,0\right]\;.\quad\color{blue}{(*)}$
For all $\;x\ge0\;,\;$ it results that $\;y=\dfrac{1}{1+x}-1\in\left]-1,0\right],$ hence, by applying $(*)$, we get that
$\log\left(1+\dfrac{1}{1+x}-1\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\left(\dfrac{1}{1+x}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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$\frac{\tan8°}{1-3\tan^2 8°}+\frac{3\tan24°}{1-3\tan^2 24°} + \frac{9\tan72°}{1-3\tan^2 72°} + \frac{27\tan216°}{1-3\tan^2 216°}=x\tan108°+y\tan8°$ $$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8° .$$
Find the valu... | Hint
$$f(p)=\dfrac{\tan p}{1-3\tan^2p}+z\tan p=\dfrac{(1+z)\tan p-3z\tan^3p}{1-3\tan^2p}$$
Now comparing with $\tan3p=?$ formula,
we need $$\dfrac{1+z}{3z}=\dfrac31 \iff z=\dfrac18$$
$$\implies f(p)=\dfrac{3\tan3p}8$$
Put $p=8,24,72,216^\circ$ to find the LHS to be
$$-\dfrac{\tan8^\circ}8+\dfrac{3^4\tan648^\circ}8$$
Fi... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$? From the Pre-Regional Mathematics Olympiad, 2019:
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, ... | We can rewrite $x =\sqrt{2} + \sqrt{3} + \sqrt6$ as
$$\begin{aligned}(x-\sqrt2)^2 &= (\sqrt2 + \sqrt6)^2\\
x^2 - 2\sqrt2x+2 &= 9 + 6\sqrt2\\
x^2-7&=2\sqrt2\space(x+3)\end{aligned}$$
On squaring both sides of the equation,
$$x^4-22x^2-48x-23=0$$
Therefore, $a = 0, b = -22, c = -48, d = -23$, implying $|a+b+c+d| = 93$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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What is $\tan \alpha$, if $(a+2)\sin\alpha +(2a - 1)\cos\alpha =2a + 1$? I tried the following:
$$\begin{aligned}a\sin\alpha +2\sin\alpha + 2a\cos\alpha - \cos\alpha &= 2a+1\\
a(\sin\alpha +2\cos\alpha)+(2\sin\alpha-\cos\alpha)&=2a+1\end{aligned}$$
Therefore,
$$\sin\alpha +2 \cos\alpha=2$$
$$2\sin\alpha - \cos\alpha=1$... | The OP finds one constant root that applies for all $a$. But there is a second root for most specific values of $a$, which is a function of $a$. The full answer is $\tan\alpha\in\{4/3,2a/(a^2-1)\}$.
Properly, the given equation should be combined with the identity $\sin^2\alpha+\cos^2\alpha=1$. There are two ways to... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Prove $\forall z\in\mathbb C-\{-1\},\ \left|(z-1)/(z+1)\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$ I'm trying to prove
$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$
thus showing that the solutions to $\left|(z-1)/(z+1)\right|=\sqrt2$ form the circle of center $-3$ and ... | Using thrice that $\forall c\in\mathbb C,\, \left|c\right|^2=c\,\bar c$, I got it down to
$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(z-1)\,(\bar z-1)=2\,(z+1)\,(\bar z+1)\\
&\iff0=z\,\bar z+3\,z+3\,\bar z+1\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove the convergence of the following series If $0 < d_n < 1$ with $\sum d_n$ divergent, then the two series $$\sum d_{n+1}\left[(1-d_0)\cdots(1-d_n)\right]^p$$
$$\sum\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p}$$
are convergent, for every $p>0$. I think it's enough to show the convergence of the first one, bec... | Let $S_0 = d_0$ and $S_n = \sum_{j=0}^n d_j $.
We have $(1+d_0)\cdots(1+d_n)\geqslant 1 + d_0 + \cdots + d_n \geqslant d_0 + \cdots + d_n + d_{n+1} = S_{n+1}$.
When $p > 1$ there exists a positive integer $m$ such that $\frac{1}{m} < p-1$ and
$$\sum_{n=0}^N\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p} \leqslant \... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Derivative Integral $$\frac{d}{dx} \int_{x}^{\sqrt{x}} \frac{e^{xy^2}}{y}dy$$
Where am I going wrong?
$$\begin{align*}
\frac{d}{dx} \int_{x}^{\sqrt{x}} \frac{e^{xy^2}}{y}dy = & \frac{d}{dx} ( F(\sqrt{x} ) - F(x))\\
= & \frac{F'(\sqrt{x} )}{2\sqrt{x}} - F'(x))\\
= & \frac{e^{x(\sqrt{x})^2}}{2\sqrt{x}\sqrt{x}} - \frac{e^... | We have from Leibniz's Rule for Differentiating Under the Integral
$$\begin{align}
\frac{d}{dx}\int_x^{\sqrt x}\frac{e^{xy^2}}{y}\,dy&=\color{blue}{\underbrace{\int_x^{\sqrt x}\frac{\partial}{\partial x}\left(\frac{e^{xy^2}}{y}\right)\,dy}_{\text{The Ommitted Term}}}+\color{red}{\underbrace{\left(\frac1{2\sqrt x}\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sum_{k=0}^{n} \binom{n}{k} ka^k = an(a+1)^{n-1}$ Problem:
Show that $\sum_{k=0}^{N} \binom{N}{k} ka^k = aN(a+1)^{N-1}$
Attempt:
I tried using induction but got stuck.
$N=0$ implies $\sum_{k=0}^{0} \binom{0}{k} ka^k = 0$
$N=1$ implies $\sum_{k=0}^{1} \binom{1}{k} ka^k = a = a(1)(a+1)^{1-1}$
$N=2$ implies $\s... | $$S=\sum_{k=0}^{n} k {n \choose k} a^k =\sum_{k=0}^{n} \frac{n!}{k! (n-k)!} a^k= \sum_{k=0}^{n} \frac{n (n-1)!}{(k-1)! (n-k)!}=n\sum_{k=0}^{n} {n-1 \choose k-1} a^k$$
Let $k-1=p$, then
$$S=n\sum_{p=-1}^{n-1} {n-1 \choose p} a^{p+1}=na\sum_{p=0}^{n-1} {n-1 \choose p} a^p=na(1+a)^{n-1}.$$
Note that ${m \choose -n}=0$, wh... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving limit - $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$ $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$
Since $x$ approaches $0$ and $y$ also approaches $0$ we can suspect that $0<x^2 + y^2<1$. For every $x,y\in\Bbb R$, we have that $\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$.
Now, $1\geq (x^2+y^2)^{x^2y^2}\geq (x^2+y^2)^{\frac{1}{4}(... | hint
The difference gives
$$\frac 14(x^2+y^2)^2-x^2y^2=$$
$$\frac 14\Bigl(x^4+y^4+2x^2y^2-4x^2y^2\Bigr)=$$
$$\frac 14\Bigl(x^4+y^4-2x^2y^2\Bigr)=$$
$$\frac 14(x^2-y^2)^2\ge 0$$
Other proof
Putting
$$x=r\cos(t)\;,\;y=r\sin(t)$$
We know that
$$\sin^2(2t)\le 1 \iff $$
$$4\sin^2(t)\cos^2(t)\le (\cos^2(t)+\sin^2(t))^2\iff $... | {
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"source": "stackexchange",
"question_score": "1",
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Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$ My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\t... | It's $$2\cos2\theta\cos\theta-3\cos2\theta=2\sin2\theta\cos\theta-3\sin2\theta$$ or
$$(2\cos\theta-3)(\cos2\theta-\sin2\theta)=0$$ or
$$\tan2\theta=1.$$
Can you end it now?
| {
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"source": "stackexchange",
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Solve for closed form $a_j$ given $a_j = 1 + pa_{j+1} + (1-p)a_{j-1}$ where $a_1 = 0$.
Solve for a closed form expression for $a_j$ given constant $p \in (0,1)$ and $q=1-p$ and:
\begin{align*}
a_j &= 1 + pa_{j+1} + qa_{j-1} \\
\end{align*}
Given that $a_1 = 0$.
The text says to use the hint that $a_j = c(1-j)$, but... | The intuition of the hint is that from state $j$ you need to "move to the left" a net total of $j-1$ times to reach the absorbing state $1$, and the expected number of steps to move one unit to the left is independent of $j$ because the transition probabilities are.
For an order-2 recurrence relation, you should be giv... | {
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Solving the equation $X^2 + (3 +i)X + 1 +i =0$ I have to find the roots of the equation
$$X^2 + (3 +i)X + 1 +i =0.$$
The first step is to find the discrimant which is $4 + 2i$. Then, I assume that the square root of the discriminant is in the form of $a+bi$, so we have to solve the below system
$$a^2-b^2=4\qquad\text{ ... | Hint:
Multiplying by $a^2$,
$$a^2-b^2=4\land ab=1\implies a^4-a^2b^2=a^4-1=4a^2$$ and you get a biquadratic equation in $a$.
More generally, the square root of $u+iv$ is obtained by solving
$$\begin{cases}a^2-b^2&=u,\\2ab&=v\end{cases}$$ and
$$a^4-ua^2-\frac{v^2}4=0.$$
So,
$$a^2=\frac{\sqrt{u^2+v^2}+u}2$$ (the negativ... | {
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"source": "stackexchange",
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Let $f(n)=\sum_{k=0}^{\left\lfloor n/2\right\rfloor} {2k \choose k}{n \choose 2k}$ . Show that $\sum_{n\geq 0}^{} f(n)x^n=\frac{1}{\sqrt{1-2x-3x^2}}$ Using the multinomial theorem, one can show that $f(n)$ is the coefficient of $x^n$ of the polynomial $(1+x+x^2)^n$. There are $3$ obvious ways to show the equation in th... | Here's a way that doesn't require knowing the result ahead of time.
\begin{align}
\sum_{n \ge 0} f(n) x^n
&= \sum_{n \ge 0}\sum_{k=0}^{\left\lfloor n/2\right\rfloor} \binom{2k}{k}\binom{n}{2k} x^n \\
&= \sum_{k \ge 0}\binom{2k}{k} \sum_{n\ge 2k} \binom{n}{2k} x^n \\
&= \sum_{k \ge 0}\binom{2k}{k} \frac{x^{2k}}{(1-x)^{... | {
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Writing a function in terms of its power series with some tricky reindexing steps Represent the following function as a power series and find it's radius of convergence:
$$\frac{x^2}{(8+x)^3}$$
By Using differentiation to find the power series of a fairly tricky function!! We know that
$\frac{1}{(8+x)^3} = \frac{1}{2}... | Yes it's correct. Since for $n=0$ and $n=1$ the terms vanish, if you want to start at $n=0$ we can write $$\frac{x^2}{(x+3)^3}= \frac{1}{2} \sum_{n=2}^\infty (-1)^{n} (n)(n-1) x^{n} (\frac{1}{8})^{n+1}=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n(n)(n-1)x^n(\frac{1}{8})^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians ... | It is indeed a theorem that can be generalized, although Mathworlddoes n't say so explicitly.
The area of a triangle formed by medians ( computed for example from Brahmagupta/Heron formula ) is three-fourths the area formed by the corresponding sides of the given triangle.
It can also be proved by projective geometry.
... | {
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"source": "stackexchange",
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Prove by induction that $\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$ What would be the right way to solve this by induction proof?
$$\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$$
This is what I've done (reference https://www.slader.com/d... | $$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2+4+6+\ldots+2n}\ge \frac{1}{2n}\tag{1} $$
can be simplified as
$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{n(n+1)}\ge \frac{1}{2n} $$
and then
$$1\cdot3\cdot5\cdot\ldots\cdot(2n-1)\ge\frac{n+1}{2}\tag{2}$$
for $n=1$ we have $1\ge 1$ true.
Now suppose $(2)$ is true and let ... | {
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Why using fewer terms of Taylor series doesn't give $0/0$ but gives a wrong answer? I was reading a Calculus book and I saw this problem which looks easy:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$
It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve t... | You are dividing by $x^3$ at the end, so you need all possible terms at least of degree $3$ in the numerator to be present, otherwise you're basically guaranteed to change the value of the limit.
Let's keep the error term in the third way, and see what happens. I will do that the following way: we have
$$
\sin(2x) = 2x... | {
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Why is $\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}$? I am trying to prove an inequality, but that itself is not the question here. Looking around for ideas how to proceed I found solutions for the same problem.
But in several occasions I couldn't follow the answers because of this step.
In both of these excerpts (taken from a... | No, the linked examples use $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, which is equivalent to the trivial $\sqrt{n+1}>\sqrt{n}$.
| {
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Area of triangle inside a $30-60-90$ triangle
In a right angle triangle ABC with hypotenuse BC and C=60 degrees, M and N are the middles of AB and AC respectively. Draw ND perpendicular to BC (D is point on the side BC). If MD=7cm calculate the area of triangle DNM.
I solved this question in the following way:
$NM//... | You have done a mistake here
$$ DN^2+DN^2*\frac{16}{\color{red}{3}}=49$$
You wrote $9$.
With correct value, I got $$ [DNM] = \dfrac{98\sqrt{3}}{19}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I found 3rd and 4th central moments of gumbel distribution with characteristic function? I tried many ways(using digamma function, partial integration etc.) to find moments using by characteristic function but I couldn't. Is there any trick or suggestion? (I couldn't calculate integral of 2nd, 3nd and 4th deriv... | The characteristic function for the Gumbel distribution with location parameter $\alpha$ and scale parameter $\beta$ is
$$ \varphi_X(t) = \mathrm{e}^{\mathrm{i} t \alpha} \Gamma(1+\mathrm{i}t \beta) \text{.} $$
The $n^\text{th}$ (noncentral) moment can be found from the characteristic function using
$$ E[X^n] = \ma... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
Showing $\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$ for $0I was trying to solve the following exercise:
Show that for $0<x<1$ the following inequalities hold: $$\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$$
I think the Mean Value Theorem might be useful, but I... | $$
\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2} \\
-\frac{1-x}{1+x^2}<\arctan (x) - \frac{\pi}{4}<-\frac{1-x}{2} \\
\frac{1-x}{1+x^2}>\frac{\pi}{4} - \arctan (x)>\frac{1-x}{2} \\
\frac{1}{1+x^2}>\frac{\frac{\pi}{4} - \arctan (x)}{1-x}>\frac{1}{2} \\
$$
where the last step assumes $x<1$.
could... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve this equation with matrices can you please give me some hints to solve the following? I really don't know how to start.
$$X^2= \begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix}.$$
I tried to express this matrix as $4\cdot I + \begin{pmatrix}
2 & 2 \\ 3 & 3
\end{pmatrix}$ And somehow solve it, but I really have ... | We have
$$
\begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix} = QDQ^{-1}$$
where
$$
Q = \begin{pmatrix} -1 & 2 \\ 1 & 3 \end{pmatrix},\ D = \begin{pmatrix} 4& 0 \\ 0 & 9 \end{pmatrix}.
$$
It follows that we may take any of the four matrices
$$
X = Q\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 3 \end{pmatrix} Q^{-1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Is my $\epsilon$-$\delta$ calculation correct? I have to show that $\lim_{x \to 1} x^4-1 =0$. Here is how i have done it:
$\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < \epsilon \qquad$ and since we are close to 1, we can assume that the $\delta$-neighborhood of $c=1$ must be havea radius of max $\delta... | You can also express $x^4-1$ in powers of $x-1$ as in
$$|x^4-1| = |((x-1)+1)^4-1| = |(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)|.$$
If we assume $|x-1| < \delta$, the triangle inequality gives
$$|x^4-1| \le |x-1|^4+4|x-1|^3+6|x-1|^2+4|x-1| < \delta^4+4\delta^3+6\delta^2+4\delta.$$
Now we want $\delta^4+4\delta^3+6\delta^2+4\delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3907681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Perfect square involving the exponential law If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?
I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?
| $$(a+b)^2=a^2+2ab+b^2$$
$$2^{10}+2^{13}+2^n=(2^5)^2+2\times2^{12}+(2^7)^2=(2^5)^2+2\times2^{5+7}+(2^7)^2=(2^5+2^7)^2$$
Hence $n=14$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that the sequence $X_n$ converges to a limit $Y$ Let $A$ be a positive definite $n\times n$ matrix. We use the iteration of the mapping $$f(X)=0.5(X^2+B), \ X\in \mathbb{R}^{n\times n}$$ where $B=I-A$.
Show that the sequence $X_0:=0$ (zero matrix), $X_1=f(X_0), X_2=f(f(X_0)), \ldots $ converges to a limit $Y$ if $... | We have $$f(X) = \frac 12X^2 + \frac 12 B$$
Let's try to see what initial terms of $(X_n)$ look like (with $X_0=0$):
$$\begin{align}X_1 &= f(0) = \frac 12 0^2 + \frac 12 B = \frac 12 B\\[2mm]
X_2 &= f(f(0)) = \frac12 \bigg(\frac 12 B\bigg)^2 + \frac 12 B = \frac{1}{2^3}B^2 + \frac{1}{2}B \\[2mm]
X_3 &=f(f(f(0)))= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sum_1^n 1/i^2 \le 2 - 1/n$ for all natural $n$. I'm trying to do this by induction. It works for $n=1$ because we get $1 \le 2 - 1 = 1$. Now suppose for some natural $k \ge 1$, we have $\sum_{i=1}^k \le 2 - 1/k$. I must show $\sum_{i=1}^{k+1} \le 2 - 1/(k+1)$. I started out from the hypothesis and added $1... | $$
{1\over(k+1)^2}<{1\over k(k+1)}={k+1-k\over k(k+1)}=\frac1k-{1\over k+1}
$$
As a result, we have
$$
\sum_{i=1}^{k+1}{1\over i^2}\le2-\frac1k+\frac1k-{1\over k+1}=2-{1\over k+1}
$$
Hence, by the principle of mathematical induction, we conclude that the proposition holds for all positive integer $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given $T_1 = 0$ and for $i > 1$: $T_i = 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j$ Prove via induction that $T_i = \sum\limits_{j=1}^{i-1} \frac{1}{j}$
Given $T_1 = 0$ and for $i \in \mathbb{N}, i > 1$:
\begin{align*}
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j \\
\end{align*}
Manually computing terms $2,3,4$:
\begin... | $$\begin{align*}
T_{i+1}&=1+\frac1i\sum_{j=1}^iT_j\\
&=1+\frac1i\sum_{j=1}^i\sum_{k=1}^{j-1}\frac1k\\
&=1+\frac1i\sum_{k=1}^{i-1}\sum_{j=k+1}^i\frac1k\\
&=1+\frac1i\sum_{k=1}^{i-1}\frac1k\sum_{j=k+1}^i1\\
&=1+\sum_{k=1}^{i-1}\frac{i-k}{ik}\\
&=1+\sum_{k=1}^{i-1}\left(\frac1k-\frac1i\right)\\
&=1+\sum_{k=1}^{i-1}\frac1k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{... | Note
\begin{align}
& \tan(2\tan^{-1}2)-\tan(\pi-\cos^{-1}\frac35)\\
=&\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}+\tan(\cos^{-1}\frac35)
=\frac{2\cdot2}{1-2^2}+ \frac43=0
\end{align}
which leads to
$$2\tan^{-1}2=\pi-\cos^{-1}\frac35$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$.
After much trial and error, I was able to rearrange the above equation
as $80z^2 - 112z-4 = 0$.
$z = \frac{7\pm3\sqrt{6}}{10}$.
I substituted $\frac{7+3\sqrt{6}}{10}$ as $z$ in the... | $$\sqrt{5 z+5}=2 \sqrt{z}+\sqrt{3-3 z}\tag{1}$$
Square both sides
$$5z+5=4z+3-3z+4\sqrt{3z-3z^2}$$
$$2+4z=4\sqrt{3z-3z^2}\to 1+2z=2\sqrt{3z-3z^2}$$
Square again both sides
$$1+4z+4z^2=12z-12z^2$$
$$16 z^2-8 z+1=0\to (4z-1)^2=0\to z=\frac14$$
Which is actually a solution of $(1)$ (verify!)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3920939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the local maximum and minimum of $f(x)=x+\sin(2x)$ bounded by the interval $-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$?
Find the local maximum and minimum of $y=x+\sin(2x)$ between
$-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$.
What I have done so far:
$f'(x) = 1+2\cos(2x)$ and $f''(x)=-4\sin(2x)$.
| First, find where the derivative is zero ($y'=1+2\cos(2x)=0$).
$$1+2\cos(2x)=0$$
$$2\cos(2x)=-1$$
$$\cos(2x)=-\frac{1}{2}$$
$$2x=-\frac{2\pi}{3}, \frac{2\pi}{3}$$
(within the bounds)
$$x=-\frac{\pi}{3}, \frac{\pi}{3}$$
Checking the second derivative, you can find that $x=-\frac{\pi}{3}$ is local minimum and $x=\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Writing Fourier transform integral in terms of real-valued functions I am interested in the following integral.
$$ f(x,t) = \int \limits_{-\infty}^{\infty} \mathrm{d}k e^{-k^2}\left( e^{i(kx-pt-qk^2t)}+e^{i(kx+pt+qk^2t)} \right)$$
One way (for example, by using Mathematica) to solve it is in the following way.
$$ f(x,t... | The two parts of your answer are obviously a complex conjugate pair.
Let's just take one of the terms and double its real part:
$$\begin{align*}f(x,t) &= \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}}\\
\\
&= \sqrt{\pi}\;\Re\left[\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given $f(x) = \frac{x^3}{3} - \sqrt{x^2+1}$, proof that $f(\alpha) = \frac{\alpha^4 - 3}{3\alpha}$ I evaluated the function on $\alpha$ and removed the radical by multiplying by $\sqrt{\alpha^2+1}$ as follows:
$$ \frac{\alpha}{3} - \frac{\alpha^2 + 1}{\sqrt{\alpha^2+1}} $$
After some algebraic manipulations, I got to t... | $$\frac{x^3}{3}-\sqrt{x^2+1}=\frac{x^4-3}{3 x}$$
$$\frac{x^3}{3}-\frac{x^4-3}{3 x}=\sqrt{x^2+1}$$
$$\frac{1}{x^2}=x^2+1$$
Substitute $x^2=w$
$$\frac{1}{w}=w+1\to w^2+w-1=0\to w=\frac{1}{2}\left(\sqrt{5}-1\right)$$
the root $w_2=\frac{1}{2}\left(-\sqrt{5}-1\right)$ is discarded because is negative. The solution is:
$$x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$ Since we have $\frac00$,I Applied L'Hopital rule :
$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$
But as you can see it is getting very ugly.... | Without series, only L'Hospital
$$
\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}
$$
we get
$$
\lim_{x\to 0}\left[\left(-\frac{\ln(1+x)}{x^2}+\frac{1}{x}(1+x)^{-1}\right)(1+x)^{\frac{1}{x}}\right] \\
=\lim_{x\to 0}\left(-\frac{\ln(1+x)}{x^2}+\frac{\frac{1}{1+x}}{x}\right)\lim_{x\to 0}(1+x)^{\frac{1}{x}} \\
=\lim_{x\to 0}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
An estimate on number of solutions $e^z=a$ in $|z|\leq r.$ Let $n(r,a)$ be the number of solutions of $e^z=a$ in $|z|\leq r, $ where $a\in\mathbb{C}\setminus\left\{0\right\}.$
I have to estimate this number.
Attempt: Writing $a=|a|e^{i\theta}.$ Then $e^{z}=a$ gives $$z=\log|a| + i\left(\theta + 2k\pi\right),$$ where $k... | Your calculation is correct up to
$$
(\log|a|)^2 + (\theta + 2k\pi)^2\leq r^2 \, .
$$
That is equivalent to
$$
|\theta + 2k\pi| \le \sqrt{r^2 - (\log|a|)^2} \\
\iff - \sqrt{r^2 - (\log|a|)^2} \le \theta + 2k\pi \le \sqrt{r^2 - (\log|a|)^2} \\
\iff - \frac{\sqrt{r^2 - (\log|a|)^2}}{2 \pi} - \frac{\theta}{2 \pi}
\le k \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse function of a polynomial $x(x-1)(x+1) = 0 $ How to get the interval to get the inverse function of the polynomial $$x(x-1)(x+1) = f(x) $$, to show that the function is surjective.
The interval where the function is surjective is $x > 1$ and $x < -1$, how to get inverse on this interval ?
| Consider $$x(x-1)(x+1) =x^3-x= y$$ and then th cubic equation
$$x^3-x-y=0$$ for which the discriminant $\Delta=4-27y^2$ must be negative in order to have only one real root. This defines
$$-\frac{2}{3 \sqrt{3}} <y <\frac{2}{3 \sqrt{3}}$$
Using the hyperboic method for only one real root, we then have
$$x=\frac{2}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3936117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$ I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:
$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?
Thank you very much!
| Observe that $a_n > 0, \forall n \ge 1$, and by using the AM-GM inequality: $\forall n \ge 1 \implies (a_{n+1})^2 \ge \left(\dfrac{1}{2}\cdot 2\sqrt{a_n\cdot \dfrac{3}{a_n}}\right)^2 = \left(\sqrt{3}\right)^2 = 3 \implies a_{n+1} \ge \sqrt{3}, \forall n \ge 1$. Thus: $a_{n+1} - a_n = \dfrac{1}{2}\left(a_n+\dfrac{3}{a_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3936204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so tha... | HINT
Here it is an alternative way to solve it:
\begin{align*}
\frac{x}{(x-1)(x+4)} & = \frac{(x-1) + 1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\frac{(x+4) - (x-1)}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\left[\frac{1}{x-1} - \frac{1}{x+4}\right]\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Show that for natural $n \ge 2$ :$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$ Show that for natural $n \ge 2$ the following does hold:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
First solution:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
$$\iff$$
$$\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1... |
Show that for natural $n \ge 2$ the following does hold:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
By using the Bernoulli’s inequality
$(1+x)^{n}>1+nx$
for any natural $\;n\ge2\;$ and for any $\;x>-1\;,$
with $\;x=-\dfrac1{n^2}>-1\;,\;$
we get that
$\left(1-\dfrac{1}{n^{2}}\right)^{n}>1+n\left(-\dfrac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the best constant $c < 0$ for $c a_n = a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0$ Let's define the following recurrence relation:
$$
\begin{align}
a_0 &= 1 \\
c a_n &= a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0
\end{align}
$$
Find the best constant $c < 0, \forall n \in \mathbb{N}$ which is farest from zero... | From the recurrent relation $a_n = \frac{2c + 1}{c} a_{n-1} - a_{n-2}$ with $a_0 = 1$ and $a_1 = 1/c$, you can obtain an explicit formula for $a_n$:
$$a_n = p_1x_1^n+p_2x_2^n$$
with $x_1, x_2$ are the roots (can be non real) of the equation : $x^2 = \frac{2c + 1}{c} x - 1$ and $p_1,p_2$ are the solution of the system ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Apparently in this form the limit is "trivially" 1/2 I'll admit, I'm struggling to keep up with the material in my analysis classes, but it only demoralizes me more when the questions in my textbooks have explained solutions, but I don't even understand how you go from one step to another.
This is the final from, that'... | Compute the leading terms of binomial expansions in numerator and denominator and simplify
$$
\begin{align}
& \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)= \\
& = \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$f^n(x) = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$ $f^n$(x)$ = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$
Where $f^n(x)$ is the $n^{th}$ derivative of $f(x)$. And $f(x) = (7-5x)^{\frac{1}{2}}$.
So far I have made my base case of $n=1$ and s... | The constant term in $f^n(x)$ should be $\frac{10n-15}{2}$, not $\frac{5-10n}{2}$.
It's best to rewrite the equation as
$$f^n(x) = a_n \frac{f^{n-1}(x)}{(f(x))^2}$$
This will greatly simplify our calculation.
Induction step from $n$ to $n+1$: First notice that
$$f^{n-1} (x) = \frac{1}{a_n} (f(x))^2 f^n(x)\tag1$$
and
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ ... | By using Schur's inequality and the identity
\begin{align}
&(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca)\\
=\ & a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b),
\end{align}
we have $(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca) \ge 0$ which results in
$$\frac{(a+b+c)^3 + 9abc }{4(a+b+c)} \ge ab + bc + ca. \tag{1}$$
We need to prove ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Need help to check alternating series criterion I know this is simple calculation based question but I got stuck:
I want to check convergence of the following series:
$$1-\frac{1}{2}(1+\frac{1}{3})+\frac{1}{3}(1+\frac{1}{3}+\frac{1}{5})-\cdots$$
Clearly this is alternating series. So I need satisfy alternating series c... | Doesn't answer your question, since I think you've already got the answer to it. It ads something to it though.
Let's evaluate the sum of the series $ \sum\limits_{n\geq 0}{\frac{\left(-1\right)^{n}}{n+1}\sum\limits_{k=0}^{n}{\frac{1}{2k+1}}} $.
Denoting $\left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{\left(-1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Find the minimum value of $7x-24y$ LMNAS $25^{th}$ UGM, Indonesian
Suppose that $x,y\in\mathbb{R}$, so that :
$x^2+y^2+Ax+By+C=0$
with $A,B,C>2014$. Find the minimum value of $7x-24y$
$x^2+y^2+Ax+By+C=0$
can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$
Stuck,:>
| Since
$$
x^2+Ax+\left( \frac{A}{2} \right) ^2+y^2+By+\left( \frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C\\
\left( x+\frac{A}{2} \right) ^2+\left( y+\frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C
$$
Let $\displaystyle r=\sqrt{\left( \frac{A}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$ Kazakstan 2012
Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$
prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$
My idea :
$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqr... | We have: $4(ab)^2 = (a+b)^2 \ge 4ab \implies ab \ge 1\implies LHS \ge 2\sqrt{ab} + \dfrac{1}{1+\sqrt{ab}}= f(t), t = \sqrt{ab} \ge 1\implies f'(t) = 2 - \dfrac{1}{(1+t)^2}> 0\implies f(t) \ge f(1) = 2+\dfrac{1}{2} = \dfrac{5}{2} \implies LHS \ge \dfrac{5}{2} = RHS $. Equality occurs when $a = b = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$ I'm trying to prove and compute the limit of this function.
$\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$
I've tried converting it into different functions like $\cos(\pi/2-2x)$ or multiplying by the inverse function and so on, but it keep getting back to $0/0$.
I'm s... | You can do this with regular old trig identities. Note that:
$$\sin\left(6x\right) = 6 \sin x \cos^5 x - 20 \sin^3 x \cos^3 x + 6 \sin^5 x \cos x$$
$$\sqrt{\sin\left(2x\right)} = \sqrt{2 \sin\left(x\right) \cos\left(x\right)}$$
For shorthand, let me write: $$\sin\left(x\right) = S \hspace{2.54cm} \cos\left(x\right) = C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$\frac{d}{dx}(\sin(x^{\frac{1}{3}}))$ from first principle The question contains a hint: at the appropriate point use the result $a^{3} - b^{3} =
\left(a - b\right)\left(a^{2} + b^{2} + ab\right)$.
$$
\frac{{\rm d}}{{\rm d}x}\sin\left(x^{1/3}\,\right)
$$
My attempt:
I did not use the hint as it was not immediately obv... | You have
$$\begin{aligned}
\frac{\sin(x^{\frac{1}{3}}) - \sin(c^{\frac{1}{3}}) }{x-c} &= \frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{x-c}\\
&=\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$
$y\geq-20$
==> $(99-2x)/5 \geq20$
==>... | $2x + 5y = 99$ is very rare and a strong restriction.
If $2a + 5b=99$ is a solution to find another solution $2(a\pm d) +5(b\mp e) =99=2a+5b$ we must have $2d - 5e = 0$ and that requires that $d$ be divisible by $5$ and $e$ be divisble by $2$.
so if $2a + 5b =99 $ is one answer than any $2(a + 5k) + 5(b-2k) =99$ is als... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$ For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:
*
*$\sin x \approx x$
*$\cos x \approx 1-\frac{x^2}{2}$
*$\tan x \approx x +\frac{x^3}{3}$
I am looking for a... | Proof that $\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}}=\frac16$
As shown in $(2)$ from this answer, since $\cos(x)$ is continuous, given any $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $0\lt|x|\le\delta$, we have
$$
1-\epsilon\le\cos(x)\le\frac{\sin(x)}x\le1\tag1
$$
Thus, assuming $0\lt|x|\le\delta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Ordered pairs $(m,n)$ such that ${\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m+c^m)}{m}\frac{(a^n+b^n+c^n)}{n}}$. I am unable to proceed with the following problem, : If $a+b+c=0$ then find all the ordered pairs $(m,n)$ where $m,n\in\mathbb{N}$ such that
$$\boxed{\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m... | Let us first consider the case $m\le n$.
*
*If $m=n$, then taking $(a,b,c)=(1,-1,0)$, we get $m=(1+(-1)^m)^2$ from which $m=4$ follows. However, when $m=n=4$, the equation does not hold for $(a,b,c)=(2,-1,-1)$.
*If $m=1$, then taking $(a,b,c)=(2,-1,-1)$, we get $2^{n}=(-1)^{n}$ which is impossible.
So, in the follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Integral of a rational expression involving quadratics. I want to solve this integral but I have some problems...
$$\int_2^3 \frac{(x^2-2x+1)}{(x^2+2x+1)}$$
I transformed both in $(x-1)^2$ and $(x+1)^2$ respectively but didn't find any answer. I tried as well to transform the rational expression into $1 -\frac{4x}{(x... | \begin{gather*}
I=\int ^{3}_{2}\frac{x^{2} -2x+1}{x^{2} +2x+1} dx\\
\text{Decompose $x^{2} -2x+1$ into partial fractions of $( x+1)$}\\
I=\int ^{3}_{2} 1-\frac{4x+4-4}{x^{2} +2x+1}\\
I=\int ^{3}_{2} 1-\frac{4}{x+1} +\frac{4}{x^{2} +2x+1}\\
I=\int ^{3}_{2} 1-\frac{4}{x+1} +\frac{4}{( x+1)^{2}} dx\\
\end{gather*}Can you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both ... | As the inverse trigonometric functions are multi-valued with a respective principle value
or
as sine/cosine/tangent functions are periodic, it is incorrect to apply cosines in both sides.
For example,
*
*$x$ in general $\ne -x$ but $\cos(-x)=\cos(x)$
*$x$ in general $\ne \pi-x$ but $\sin(\pi-x)=\sin(x)$
*$x\ne\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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.