Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$? How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?
| $$\begin{align}\frac{d}{dy}(f(y)) &= \frac{d}{dy} \Bigg(\int_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx\Bigg) \\&= \int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} . 1\end{align}$$
Now $$\begin{align}&\int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} = 0 \\&\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Sum of a series of a number raised to incrementing powers How would I estimate the sum of a series of numbers like this: $2^0+2^1+2^2+2^3+\cdots+2^n$. What math course deals with this sort of calculation? Thanks much!
| Interestingly, this sort of summation is not too difficult to prove either.
If we swap $2$ for a generic number $a$:
$$
\text{(1) } \sum\limits_{i = 1}^n {a^i} = a^0+a^1+a^2+a^3+...+a^n
$$
Multiplying $(1)$ by $a$ gives:
$$
\text{(2) } a\sum\limits_{i = 1}^n {a^i} = a^1+a^2+a^3+a^4+...+a^{n+1}
$$
Subtracting $(1)$ from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Coordinates in a rectangular triangle Numbers from $1$ to $\frac{n^2+n}{2}$ are formed into a triangle: numbers from $1$ to $n$ form first column, numbers from $n+1$ to $2n - 1$ form second and so on untill last column has just one number.
For example with $n = 5$ we get
$\begin{matrix}
1&&&&\\
2&6&&&\\
3&7&10&&\\
4&8... | If we add up entire columns we can form a trapezoid with the exact same area:
$$
\begin{align}
5&=1\cdot 5\\
9&=2\cdot\frac{5+4}2\\
12&=3\cdot\frac{5+3}2
\end{align}
$$
This can be continuously interpolated by $x=y\cdot\frac{n+(n+1-y)}{2}$ which then leads to
$$
y^2-(2n+1)y+2x=0
$$
which is a quadratic equation in $y$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How prove this diophantine equation $(x^2-y)(y^2-x)=(x+y)^2$ have only three integer solution? HAPPY NEW YEAR To Everyone! (Now Beijing time 00:00 (2015))
Let $x,y$ are integer numbers,and such $xy\neq 0$,
Find this diophantine equation all solution
$$(x^2-y)(y^2-x)=(x+y)^2$$
I use Wolf found this equation only have... | $$
\begin{array}{l}
\left( {x^2 - y} \right)\left( {y^2 - x} \right) = \left( {x + y} \right)^2 \\
\Leftrightarrow xy\left( {xy - 1} \right) = x^3 + y^3 + x^2 + y^2 \\
\Leftrightarrow xy\left( {xy - 1} \right) = x^2 \left( {1 + x} \right) + y^2 \left( {1 + y} \right) \cdots \left( * \right) \\
\left\{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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Does this polynomial exist? I'm looking for a polynomial $P(x)$ with the following properties:
*
*$P(0) = 0$.
*$P\left(\frac13\right) = 1$
*$P\left(\frac23\right) = 0$
*$P'\left(\frac13\right) = 0$
*$P'\left(\frac23\right) = 0$
From 1 and 3 we know that $P(x) = x\left(x - \frac23\right)Q(x)$. From 4 and 5 we k... | This is the same solution as Ross's, but from a more linear algebra focused perspective.
You have five linearly independent conditions, so a polynomial with five parameters should work.
$P(x) = ax^4 + bx^3 + cx^2 + dx + e$
Now your five conditions can be written as such:
*
*$e = 0$
*$a(\frac{1}{3})^4 + b(\frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How many sets of two factors of 360 are coprime to each other? My attempt:
$360=2^3\cdot3^2\cdot5^1$
Number of sets of two factor coprime sets for $2^3$ and $3^2$ only $=12+6=18$
With that if we add the effect of $ 5^1$, number of sets $=18+2\cdot 18-1=53$.
Is this ok?
The answer given is $56$.
| I'll assume you mean sets with exactly two elements, so the two factors $1$ and $1$ (represented by the singleton $\{1\}$) won't count.
Case 1: One of the factors is $1$.
$\gcd(1,n)=1$ for all $n$, so this is just the number of factors of $360$, minus $1$. The number of factors of $360=2^3\cdot3^2\cdot5$ is simply t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What is meant by positive root of $x^3-x^3-1$? I am a bit confused. I think there must be a mistake.
In a text I read:
The entropy is $2\ln p$, where $$p=\frac{1}{3}\left(\sqrt[3]{\frac{29+9\sqrt{31/3}}{2}}+\sqrt[3]{\frac{29-9\sqrt{31/3}}{2}}+1\right)$$ is the positive root of $x^3-x^3-1$.
But $x^3-x^3-1=-1$... I ... | There is no positive root, or any root at all, of $x^3-x^3-1=0$, since, as you noted, $x^3-x^3-1=-1$ for all $x$, and there is no $x$ that satisfies $-1=0$. This is definitely a typo.
The positive root of $$x^3-x^2-1=0$$
is exactly
$$\frac{1}{3}\left(\sqrt[3]{\frac{29+9\sqrt{31/3}}{2}}+\sqrt[3]{\frac{29-9\sqrt{31/3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is sy... | Another way to prove is to use series:
\begin{eqnarray}
\int_{-\infty}^{\infty} \frac{1}{x^4 + x^2 + 1} dx&=&2\int_{0}^{\infty} \frac{1}{x^4 + x^2 + 1} dx\\
&=&2\int_{0}^{1} \frac{1}{x^4 + x^2 + 1} dx+2\int_{1}^{\infty} \frac{1}{x^4 + x^2 + 1} dx\\
&=&2\int_{0}^{1} \frac{1}{x^4 + x^2 + 1} dx+2\int_{0}^{1} \frac{x^2}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
How to solve $ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $ I need some help to solve the next equation:
$$ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $$
Where $ \left \lfloor \cdot \right \rfloor $ is the floor function.
What I've tried:
$$ x^2 - x - 2 - x ... | $$\left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor \tag{1.}$$
Let's start with the observation that
$\left \lfloor y \right \rfloor = \left \lfloor x \right \rfloor
\iff 0 \le y-x < 1$ or $-1 \lt y-x \le 0$
The proof follows from the observation that, for any integer n,
$$ n \le x \le y < n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Coordinate of the excentre of a triangle I am just wondering that how the coordinate of the excentre comes out if we know the coordinates of vertices of the triangle.
| Consider the diagram:
$d=\overline{CE}=\overline{CF}$. Note that $c=\overline{AB}=(d-a)+(d-b)$. Therefore,
$$
d=\frac{a+b+c}2\tag{1}
$$
Furthermore, $d=\overline{CD}\cos(\theta/2)$ and $\overline{CH}=d\cos(\theta/2)$; therefore, $\overline{CH}=\overline{CD}\cos^2(\theta/2)$.
The Law of Cosines gives
$$
\cos(\theta)=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Calculate the limit of $(1+x2^x)/(1+x3^x)$ to the power $1/x^2$ when $x\to 0$ I have a problem with this: $\displaystyle \lim_{x \rightarrow 0}{\left(\frac{1+x2^x}{1+x3^x}\right)^\frac{1}{x^2}}$.
I have tried to modify it like this: $\displaystyle\lim_{x\rightarrow 0}{e^{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}}$ and ... | let $f(x)=\ln\left(\frac{1+x2^x}{1+x3^x}\right)$ and $g(x)=x^2$ then we have $0/0$ for $x$ tends to $0$ you must calculate $$\frac{f'(x)}{g'(x)}$$ and look if the limit exists.
you will get this here $$1/2\,{\frac {{2}^{x}\ln \left( 2 \right) {3}^{x}{x}^{2}-{2}^{x}{3}^{x
}\ln \left( 3 \right) {x}^{2}+x{2}^{x}\ln \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
My attempt:
I have used a hit and trial approach. I found out that only for x=2 and x=3 this is happening. But how can I be sure of it? ... | you can use the following expission
$$\displaylines{
x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = x\left( {x + 3} \right)\left( {\left( {x + 1} \right)\left( {x + 2} \right)} \right) \cr
= x\left( {x + 3} \right)\left( {x^2 + 3x + 2} \right) \cr}$$
$$\left( {x\left( {x + 3} \right) + 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplifying fractions with exponents I'm revising for an exam, which I have the solutions to. One of the questions asks me to prove that a sequence is a Cauchy sequence, sequence is written as:
$ a_n = \frac{2^{n+2}+1}{2^n},$
and then in the solutions, it has:
$\left | a_n - a_m \right | = \left | \frac{2^{n+2}+1}{2^n}... | The important equality is:
$$\frac{2^{n+2} + 1}{2^n} = \frac{2^{n+2}}{2^n} + \frac{1}{2^n} = 2^2 + \frac{1}{2^n}$$
The same process is performed on both fractions and the $2^2$ drops out. (One fraction contributes a $+4$ while the other fraction contributes a $-4$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Trouble with definite integral calculating probabilities I cannot solve this:
$$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \tan ^{-1}(a+\tan (x)) \, dx$$
it apeared when trying to find out the probability:
$$P\{\tan a - \tan b \leq 2x\},\ \ 0 < x < 1\sqrt{3}$$
Knowing that the joint distribution $f(a,b)$ is
$f(a,b) = \f... | $$\tan{u} = a + \tan{x} \implies x=\arctan{(\tan{u}-a)} \implies dx = \frac{\sec^2{u}}{1+(\tan{u}-a)^2} du $$
Then the integral is equal to
$$\begin{align}\int_{-\pi/2}^{\pi/2} du \frac{u \sec^2{u}}{1+(\tan{u}-a)^2} &= \int_{-\pi/2}^{\pi/2} du \frac{u}{\cos^2{u} + (\sin{u}-a \cos{u})^2} \\ &= \frac12 \int_{-\pi}^{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Factor the polynomial $z^5 + 32$ in real factors The question that I have trouble solving is the following:
Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sq... | HINT:
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$
$$a^4-a^3b+a^2b^2-ab^3+b^4=(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2$$
If $ab\ne0,$
$$(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2=a^2b^2\left[\left(\frac{a^2+b^2}{ab}\right)^2-\frac{a^2+b^2}{ab}-1\right]$$
So, we can express $a^2+b^2$ in terms of $ab$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Pythagorean triple problem I am doing research on perfect cuboids, and I'm looking for values $a,b,c$ such that the following is integer, and I'm not sure how to continue this. Any suggestions are appreciated!
$PED$ is a very large constant=$899231100768000$
$$
\begin{align}
&\exp\left(\sigma_1+\sigma_2+\frac{\ln(a^2+b... | assuming $a^2+b^2$, $a^2+c^2$, $b^2+c^2$, and $a^2+b^2+c^2$ are perfect squares, then it's impossible.
$\sqrt{(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)}$
must be an integer, therefor
$(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)$
must be a square, which allows us to assume that all terms are squares, and that makes the fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Closed form for ${\large\int}_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$ I'm interested in a closed form for this simple looking integral:
$$I=\int_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$$
Numerically,
$$I\approx0.235708612100161734103782517656481953570915076546754616988...$$
Note that if we try to... | This is another solution just for reference.
The laplace transform of $\displaystyle\frac{x-\sin{x}}{x^2}$ is given by
\begin{align}
\mathcal{L}_s\left(\frac{x-\sin{x}}{x^2}\right)
&=\int^s_\infty\int^t_\infty\frac{1}{u^2}-\frac{1}{1+u^2}\ du\ dt\tag1\\
&=\int^s_\infty-\frac{1}{t}-\arctan{t}+\frac{\pi}{2}\ dt\\
&=-\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
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Taylor expansion about a point I need help with the following calculus problem:
Use completing the square and the geometric series to get the Taylor expansion about ${x=2}$ of ${\frac{1}{x^{2}+4x+3}}$
So far I have the following:
By completing the square. It can be shown that
${\frac{1}{x^{2}+4x+3} = \frac{1}{(x+2)^{... | $$\frac{1}{x^2+4x+3}=\frac{1}{(x+1)(x+3)}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)=\frac{1}{2}\left(\frac{1}{(x-2)+3}-\frac{1}{(x-2)+5}\right)$$
Now just use:
$$\frac{1}{z+3}=\frac{\frac{1}{3}}{1+\frac{z}{3}}=\sum_{n\geq 0}\frac{(-1)^n}{3^{n+1}}z^n $$
to get:
$$\frac{1}{x^2+4x+3}=\frac{1}{2}\sum_{n\geq 0}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the following inequality without using differentiation Let $a,b,c$ be real numbers that satisfy $0\le a,b,c\le 1$. Show that
$$\frac a{b+c+1} + \frac b{a+c+1} + \frac c{a+b+1} + (1-a)(1-b)(1-c) \le 1.$$
I don't know where to start. Multiplying everything by the denominators creates extreme mess.
| WLOG: $a\le b\le c$, then
\begin{align*}&\dfrac{a}{b+c+1}+\dfrac{b}{c+a+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&\le\dfrac{a}{a+b+1}+\dfrac{b}{a+b+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(a+b+1)(1-a)(1-b)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{(1+a)(1+b)(1-a)(1-b)(1-c)}{a+b+1}\\
&=\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Combination questions Q: A team of $11$ is to be chosen out of $15$ cricketers of whom $5$ are bowlers and $2$ others are wicket keepers. In how many ways can this be done so that the team contains at least $4$ bowlers and at least $1$ wicket keeper?
| $$\binom{5}{4}\binom{2}{1}\binom{8}{6}+\binom{5}{5}\binom{2}{1}\binom{8}{5}+\binom{5}{4}\binom{2}{2}\binom{8}{5}+\binom{5}{5}\binom{2}{2}\binom{8}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
My proof of: $|x - y| < \varepsilon \Leftrightarrow y - \varepsilon < x < y + \varepsilon$ Is it reasonable to prove the following (trivial) theorem?
If yes, is there a better way to do it?
Let $x, y \in \mathbb{R}$.
Let $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$.
$\textbf{Theorem.}$
We have
\begin{equation*}... | This is correct. A neater way of putting it would be to note that
$$
\begin{align}
\mid x-y \mid < \epsilon &\iff x-y \lt \epsilon \text{ and } y-x \lt \epsilon\\
&\iff x-y \lt \epsilon \text{ and } x-y \gt -\epsilon\\
&\iff-\epsilon \lt x-y \lt \epsilon \\
&\iff y-\epsilon \lt x \lt y+\epsilon
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one... | $$\begin{gathered}
\frac{{x + 2}}
{{{x^3} - x}} = \frac{{x + 2}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}
{x} + \frac{B}
{{x - 1}} + \frac{C}
{{x + 1}} \hfill \\
= \frac{{A\left( {{x^2} - 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right)}}
{{x\left( {x - 1} \right)\left( {x + 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
| The response of @TonyK is perfect and exemplary. Since some seem not to follow it, let me reinterpret it. By using Euclidean Division on $X^2+X+1$ and $X^3$, you get the identity $X^3=(X^2+X+1)(X-1) + 1$. This identity is still valid when any number is substituted for $X$. In particular, when you substitute that $x$ fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Proof that a Limit equals a Function Let f(x) = $\frac{x}{|x|}$ if x $\neq$ 0, and define f(0)=0. Show that
f(x) = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \tan^{-1}(nx)$
My work:
$\frac{x}{|x|}$ = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \tan^{-1}(nx)$
$\frac{x}{|x|}$ = $\lim_{n \rightarrow \infty} \frac{2}{\... | If $x = 0$, $\frac{2}{\pi} \tan^{-1}(nx) = 0$ for all $n$, and thus $\lim_{n\to \infty} \frac{2}{\pi}\tan^{-1}(nx) = 0$ when $x = 0$. If $x > 0$, then $$\lim_{n\to \infty} \frac{2}{\pi} \tan^{-1}(nx) = \frac{2}{\pi}\lim_{u\to +\infty} \tan^{-1}(u) = \frac{2}{\pi}\cdot\frac{\pi}{2} = 1 = \frac{x}{|x|}.$$ If $x < 0$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{x\to\infty}(x(\log(1+\sqrt{1+x^2}-\log(x)))$ can someone give me a hint/solution for:
$$\lim_{x\to\infty}\left(x(\log(1+\sqrt{1+x^2}-\log(x))\right)$$
Shall I do a derivative ?
But there's no L'Hospital to use here..
Shall I change its form ?
$x(\log(1+(1+x^2)^{1/2})-\log(x))$ (minimal change)
Or what shal... | for $x$ large
$\begin{align}
\ln(1+\sqrt{x^2 + 1}) - \ln x &= \ln \left( 1+ x(1+1/x^2)^{1/2} \right) - \ln x\\
&=\ln[ 1 + x(1 + \frac{1}{2x^2} + \cdots )] - \ln x\\
&= \ln[x(1 + \frac{1}{x} + \frac{1}{2x^2}+\cdots)] -\ln x\\
&= \ln(1 + \frac{1}{x} + \frac{1}{2x^2}+\cdots) \\
&= \frac{1}{x} +\cdots
\end{align}$
t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show m... | L'Hospital's rule:
$$\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{\tan^3(x)}=\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{x^3}\cdot\frac{\tan^3 x}{x^3}=$$
$$=\lim_{x\to0} \frac{(\ln(x+\sqrt{1+x^2})-x)'}{(x^3)'}=\lim_{x\to0}\frac{\frac{1}{\sqrt{1+x^2}}-1}{3x^2}=\lim_{x\to0}\frac{-x^2}{3x^2\sqrt{1+x^2}(1+\sqrt{1+x^2})}=-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Determining $\gcd(94, 27)$ I want to determine $\gcd(94, 27)$. Using the Euclidean algorithm, I got
\begin{align}
94 &= 27 (3) + 13 \\
\implies 27 &= 13 (2) + 1 \\
\implies \;\;2 &= 2 (1)
\end{align}
Does this mean the GCD is $2$? Clearly $2$ doesn't divide $27$, so what am I doing wrong?
| As mentioned in the comments, the last non-zero remainder is the GCD, not the quotient corresponding to the expression with zero remainder. To highlight, that is the red/boxed number below:
\begin{align}
94 &= 27 (3) + 13 \\
27 &= 13 (2) + \color{red}{\boxed{1}} \\
13 &= 13 (1) + 0
\end{align}
For another example, co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove relations between the roots of 3 quadratic equations
Let $x_1, x_2$ be the roots of the equation $x^2 + ax + bc = 0$, and $x_2, x_3$ the roots of
the equation $x^2 + bx + ac = 0$ with $ac \neq bc$. Show that $x_1, x_3$ are the roots of the
equation $x^2 + cx + ab=0$.
From Vieta's I have:
$\begin{cases} x_1+... | Subtract the first two equations to get $x_2$
$$\begin{align}x_2^2+ax_2+bc&=0\\x_2^2+bx_2+ac&=0\end{align}$$ to get $$(a-b)x_2-(a-b)c=0$$ from where $x_2=c$.
So, $$\begin{align}x_1&=b=-a-c\\x_3&=a=-b-c\end{align}$$ from where $$\begin{align}x_1+x_3&=a+b=c\\x_1x_3&=ab\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^a{f(x)}\, dx$ SMT 2013 Calculus #8:
The function $f(x)$ is defined for all $x\ge 0$ and is always nonnegative. It has the additional property that if any line is drawn from the origin with any positive slope $m$, it intersects the graph $y=f(x)$ at precisely one point, which is $\frac{1}{\sqrt{m}}$ units ... | First Approach
If $y=mx$ and $x^2+y^2=\frac1m$, then $x^2+y^2=\frac xy$, which implies
$$
yx^2-x+y^3=0\tag{1}
$$
Taking the implicit derivative of $(1)$:
$$
y'=\frac{1-2xy}{x^2+3y^2}\tag{2}
$$
$y'=0$ means $1=2xy$; then applying $y=mx$, we get
$$
1=2xy=2mx^2\tag{3}
$$
applying $x^2+y^2=\frac1m$ gives
$$
1+2my^2=2mx^2+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$? I searched a prime of the form $4^n+n^4$ with $n\ge 2$ and did not find one with
$n\le 12\ 000$.
*
*If $n$ is even, then $4^n+n^4$ is even, so it cannot be prime.
*If $n$ is odd and not divisible by $5$ , then $4^n+n^4\equiv (-1)+1\equiv 0 \pm... | An alternative to @labbhattacharjee’s exemplary response (for the case of odd $n$ only):
(1) $X^4 + 4=(X^2-2X+2)(X^2+2X+2)$; (2) $X^4+4a^4=(X^2-2aX+2a^2)(X^2+2aX+2a^2)$; (3) $X^4+4^{2k+1}=X^4+4\cdot2^{4k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Probability of Seven (Distinct) car accidents occurred on the same day Seven (Distinct) car accidents occurred in a week. What is probability that they all occurred on the same day?
My Solution:
All 7 accident occurs in 1 day in $\binom{7}{1}$ ways
All 7 accident occurs in 2 days in $\binom{7}{2}$ ways
All 7 accident ... | The probability that $1$ accident occurs on a given day is $\frac17$
The probability that $7$ accidents occur on a given day is $(\frac17)^7$
The probability that $7$ accidents occur on the same day is $(\frac17)^7\cdot7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this infinitely nested radical? $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}$
Apparently, the answer is 3.
| Hint: Observe $$(x+1)^2 - 1 = x(x+2),$$ or $$x+1 = \sqrt{1 + x(x+2)}.$$ Now recursively substitute: $$\begin{align*} x+1 &= \sqrt{1 + x\sqrt{1 + (x+1)(x+3)}} , \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)(x+4)}}}, \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)\sqrt{1 + (x+3)(x+5)}}}}, \ldots \end{align*}$$ Of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| The homogeneous form of the inequality is
$$(a+b+c)^2(ca^2+ab^2+bc^2)^2\geq 27(a^2+b^2+c^2)\,a^2 b^2 c^2 $$
that is equivalent to:
$$ (a^2+b^2+c^2+2ab+2bc+2ac)(c^2 a^4+a^2 b^4 + b^2 c^4 + 2 a^3 b^2 c+2 b^3 c^2 a+2 c^3 a^2 b) \geq 27(a^4 b^2 c^2+a^2 b^4 c^2 + a^2 b^2 c^4) $$
that can be proved by combining Schur's inequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Given that a,b,c are distinct positive real numbers, prove that (a + b +c)( 1/a + 1/b + 1/c)>9
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9$
This is how I tried doing it:
Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.
Using A... | Consider
$$(a-b)^2 \ge 0$$
$$a^2-2ab+b^2 \ge 0$$
$$a^2+b^2 \ge 2ab$$
$$\frac{a^2+b^2}{ab} \ge 2$$
$$\frac{a}{b}+\frac{b}{a}\ge 2$$
without a loss of generality, we can also say that
$$\frac{a}{c}+\frac{c}{a}\ge 2$$
and
$$\frac{b}{c}+\frac{c}{b}\ge 2$$
of course
$$3=1+1+1=\frac{a}{a}+\frac{b}{b}+\frac{c}{c}$$
So, puttin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Where am I going wrong in my linear Diophantine solution? Let $-2x + -7y = 9$. We find integer solutions $x, y$.
These solutions exist iff $\gcd(x, y) \mid 9$. So,
$-7 = -2(4) + 1$
then
$-2 = 1(-2)$
so the gcd is 1, and $1\mid9$. OK.
In other words,
$(-7)(1) + (-2)(-4) = 1$
Multiply by 9 on each side to get
$(-7)(9) + ... | 1125588
$-2x-7y=9$
$2x+7y=-9$
$2x=-9-7y$
$x={-\frac{9+7y}2}=-5-4y+{\frac{1+y}2}$
New variable: $a={\frac{1+y}2}$
$y={2a-1}$
$x=-\frac{9+7(2a-1)}2=\frac{-9-14a+7}2=-1-7a$
Check it out: $-2(-1-7a)-7(2a-1)=2+14a-14a+7=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $f_{1}, f_{2}$ in $\mathbb{Z_{6}}[x]$ such that deg$(f_{1})$ = deg$(f_{2}) = 2$ and deg$(f_{1}+f_{2})=1$
*
*Find $f_{1}, f_{2}$ in $\mathbb{Z_{6}}[x]$ such that deg$(f_{1})$ = deg$(f_{2}) = 2$ and deg$(f_{1}+f_{2})=1$
*Find $g_{1}, g_{2}$ in $\mathbb{Z_{6}}[x]$ such that deg$(g_{1})$ = deg$(g_{2}) = 1$ and deg$... | I have came up with this solution.
(1) Let $f_{1}(x) = \bar{3}x^2+\bar{2}x+\bar{1}$ and $f_{2}(x) = \bar{3}x^2+x+\bar{2}$, which are of degree $2$.
Then, $(f_{1}+f_{2}) = \bar{6}x^2 + \bar{3}x + \bar{3} = \bar{0}x^2 + \bar{3}x + 3$ is of degree $1$.
(2) Let $g_{1}(x) = 3+2x$ and $g_{2}(x) = 2+3x$, which are of degree... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Elementary number theory , when is $12n^2 + 1$ a square Prove that if
$$k = 2 + 2\sqrt{12n^2 + 1}$$
is an integer then it is a square.
Can anyone help me with this? All I know is that k is an integer if and only if ${12n^2 + 1}$ is a square. What do I do next?
| $m^2 = 12 n^2 + 1$ implies $m^2 - 12 n^2 = 1$ so
$$(m + n\sqrt{12} )= (7 + 2\sqrt{12} )^N$$
for some $N\ge 0$ (see also http://en.wikipedia.org/wiki/Pell%27s_equation, just mind you, their $n$ is our $12$).
We also have the conjugate equality:
$$(m - n\sqrt{12} )= (7 - 2\sqrt{12} )^N$$
and , therefore, for $\sqrt{12n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating Exponents
Answer choices:
22
15
10
358
None of the above
This is my solution:
| You are correct, the answer is 10.
\begin{align*} 2^2 + 9^ \frac{1}{2} + 2^1 + 225^0 &= 4 + 9^ \frac{1}{2} + 2^1 + 225^0 \\
&= 4 + {\sqrt{9}} + 2^1 + 225^0 \\
&= 4 + 3 + 2^1 + 225^0 \\
&= 4 + 3 + 2 + 225^0 \\
&= 4 + 3 + 2 + 1 \\
&= 10
\end{align*}
Did some extended working out for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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find the length of rectangle based on area of frame A rectangular picture is $3$ cm longer than its width, $(x+3)$. A frame $1$ cm wide is placed around the picture. The area of the frame alone is $42 \text{cm}^2$. Find the length of the picture.
I have tries:
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0... | Let $H$,$W$,$FH$,$FW$ be the height, width, frame height and frame width.
We're given the following equations:
$H = W+3$
$W + 2 = FW$
$H + 2 = FH$
$FW \cdot FH - W\cdot H = 42$
It follows:
$(W+2)\cdot (H+2) - W\cdot H = 42$
$\Leftrightarrow$ $((H-3)+2)\cdot (H+2) - ((H-3)\cdot H)= 42$
$\Leftrightarrow$ $H^2+2H-H-2-H^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Inverse of diagonally dominant matrix with equal off-diagonal entries Is there an explicit expression for the inverse of strictly diagonally dominant matrix with identical off-diagonal elements?
For example:
$$ \begin{pmatrix} a & -b & -b \\
-b & c & -b \\
-b & -b & d \end{pmatrix... | The Sherman-Morrison formula gives the inverse of any rank one update of a matrix for which the inverse is known.
Here we can write your matrix as follows:
$$ \begin{pmatrix} a & -b & -b \\ -b & c & -b \\ -b & -b & d \end{pmatrix}
= \begin{pmatrix} a+b & 0 & 0 \\ 0 & c+b & 0 \\ 0 & 0 & d+b \end{pmatrix}
- b \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $x^3 - 9x^2 + 18x - 12$
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $$x^3 - 9x^2 + 18x - 12.$$
This is not a homework problem. I'm not even a student. I'm going through an old textbook. I know this is a simple problem. Can't seem to crack it though.
| It makes sense to let $y = x - 3 = 3^{1 \over 3} + 3^{2 \over 3}$. You are to compute
$$(y + 3)^3 -9(y + 3)^2 + 18(y + 3) - 12$$
This works out to
$$y^3 - 9y - 12$$
Note that $y^3 = (3^{1 \over 3})^3(1 + 3^{1 \over 3})^3 = 3(1 + 3*3^{1 \over 3} + 3*3^{2 \over 3} + 3) = 3(4 + 3y) = 12 + 9y$. Thus $y^3 - 9y - 12 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
An inequality with fractional parts $$
\frac{n^k-n}{2}
\leq
\left\{\sqrt[k]{1}\right\} + \left\{\sqrt[k]{2}\right\} + \dots + \left\{\sqrt[k]{n^k}\right\}
\leq
\frac{n^k-1}{2}
$$
how it can be proven?
| Checking at wolfram it actually seems to be the opposite that $\frac{n^k-1}{2}$ is a lower bound and I will prove it below.
We can separate the sum into intervals like this by using that $\{n\}=n-\lfloor n\rfloor$ and the only integers are the $k$'th powers so they are equal to $0$. $$\sum_{d=2}^{2^k-1}(\sqrt[k]{d}-1)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Let $F$ be a field in which we have elements satisfying $a^2+b^2+c^2 = −1$. Show that there exist elements satisfying $d^2+e^2 = −1$. Let $F$ be a field in which we have elements $a, b$, and $c$ satisfying $a^2+b^2+c^2 = −1$. Show that there exist elements $d$ and $e$ of $F$, satisfying $d^2+e^2 = −1$.
Any hint?
This i... | Hint $\ $ Let $\ d=1\ $ in $\,(a^2+b^2)(\overbrace{a^2+b^2+c^2+d^2}^{\large =\, 0})\, =\, (\overbrace{a^2+b^2}^{\large x})^2 + (\overbrace{ac-bd}^{\large y})^2 + (\overbrace{ad+bc}^{\large z})^2$
That yields $\ x^2+y^2+z^2 = 0\ $ which, divided by $\,x^2,\,$ yields the result.
Remark $\ $ The latter two summands aris... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Sum of $\lfloor k^{1/3} \rfloor$ I am faced with the following sum:
$$\sum_{k=0}^m \lfloor k^{1/3} \rfloor$$
Where $m$ is a positive integer. I have determined a formula for the last couple of terms such that $\lfloor n^{1/3} \rfloor^3 = \lfloor m^{1/3} \rfloor^3$. For example if the sum is from 0 to 11 I can find the ... | Note: This answer provides a solution for the general case $\lfloor \sqrt[p]{k}\rfloor$ with $p\geq 1$.
The following is valid for $m\geq 1, p\geq 1$
\begin{align*}
\sum_{k=1}^{m}\lfloor \sqrt[p]{k}\rfloor
=m\lfloor \sqrt[p]{m}\rfloor^{p+1}
-\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\lfloor \sqrt[p]{m}\rfloor^{p+1-j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Equivalent of adding to a denominator? Given the inequality $\frac{n}{m} \ge \frac{1}{2}$, I want to add $1$ to both $n$ and $m$:
$$\frac{n+1}{m+1}.$$
What would be the equivalent operation on the RHS of the equation?
Adding $1$ to $n$ is equivalent to $+\frac{1}{m}$ on both sides, but what about adding $1$ to $m$?
| Hint 1
$$\frac{n}{m}=\frac{n+1}{m+1}+\left(\frac{n}{m}-\frac{n+1}{m+1}\right)$$
and
$$ a \leq b+c \iff a-b \leq c $$
Hint 2
$$\frac{n}{m}=\frac{n+1}{m+1}\cdot\left(\frac{n}{m}\cdot\frac{m+1}{n+1}\right)$$
and,
$$ a \leq b\cdot c \iff \frac{a}{b} \leq c \qquad \text{ if }\ b>0 \\ a \leq b\cdot c \iff \frac{a}{b} \geq c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Pell Equations: $a^2+4=5b^2$ This is a challenge problem in the Pell Equations chapter of my number theory book, but I'm not seeing the connection to Pell Equations. The Pell Equation with the coefficient $5$ is $5b^2+1=a^2$, but it doesn't look like the one I have.
Thanks if you can help me.
| I know full well the problem was asked and answered 3 years ago. Yet I wish to contribute. Please critique and comment so I learn something.
$$\begin{align}
a^2+4&=5b^2 \\
a^2-5b^2&=-4 \quad \text{initial solution} \quad (a_1, b_1) =(1,1)\\
(a-b \sqrt5)(a+b \sqrt 5)&=-4\\
\text{since} \quad &(1-\sqrt 5)(1+\sqrt 5)=-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Solve the following systems of equations ( Matrices) Solve the following systems of equations:
\begin{Bmatrix}
x_1 & -x_2 & -x_3 & +0x_4 & = 2 \\
-x_1 & +2x_2 & +0x_3 & +3x_4 &= 1 \\
x_1 & +0x_2 & +x_3 & +0x_4 & =5\\
\end{Bmatrix}
I try to make a zero triangle but always reach a step whe... | You should now do backwards elimination; first divide the third equation by $3$ and add the third equation to the second and the first, getting
$$
\begin{bmatrix}
1 & -1 & 0 & 0 & 7/3\\
0 & 1 & 0 & 3 & 10/3\\
0 & 0 & 1 & 0 & 1/3
\end{bmatrix}
$$
Now add the second equation to the first one:
$$
\begin{bmatrix}
1 & 0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Help me to solve this trigonometrical equation I want to solve the equation
$$\sin(3x+a)+\sin(3x-a)+\sin(a-x)-\sin(a+x)=2\cos a$$
I solved up to this part
$$\begin{align}
2 \sin 3x \cos a - 2 \cos a\sin x &= 2 \cos a \\
2 \cos a \left(\;\sin 3x -\sin x \;\right) &= 2 \cos a \\
\sin 3x - \sin x &= 1
\end{align}$$
| HINT:
$\sin3x=4\sin x-4\sin^3x$
Alternatively,
using Prosthaphaeresis Formula,
$$\sin3x-\sin x=2\sin x\cos2x=2\sin x(1-2\sin^2x)=2\sin x-4\sin^3x$$
Then you can use this
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$ I dont have an idea to prove it because of exist $\sin(4n+2)^4$
$$\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$$
| One first calculates$$\sin\left(\theta\right)^{4}=\frac{1}{8}\cos\left(4\theta\right)-\frac{1}{2}\cos\left(2\theta\right)+\frac{3}{8}.$$
Then$$\sum_{n=0}^{+\infty}\frac{\sin\left(2\left(2n+1\right)\right)^{4}}{\left(2n+1\right)^{2}}=\frac{1}{8}\sum_{n=0}^{+\infty}\frac{\cos\left(8m\right)}{m^{2}}-\frac{1}{2}\sum_{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Why is $\int$ $1\over(a^2+b^2)^{3/2}$ $da$ $ = $ $ a\over b^2\sqrt {(a^2+b^2)}$ $$\int\frac{da}{(a^{2}+b^{2})^{3/2}} =\frac{a}{b^{2}\sqrt{(a^{2}+b^{2})}}.$$
Found this in the solution to a problem in my physics textbook, and left clueless.
| Using the trig substitution $a = b\tan \theta$, we get $a^2 + b^2 = b^2\sec^2\theta$ (since $\tan^2\theta + 1 = \sec^2\theta$), so $(a^2 + b^2)^{3/2} = b^3\sec^3\theta$. Since $da = b\sec^2\theta\, d\theta$, we obtain
$$\int \frac{1}{(a^2 + b^2)^{3/2}}\, da = \int \frac{1}{b^3\sec^3\theta} b\sec^2\theta\, d\theta = \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}... | Find the fourier series of $f(x)=x^2$ on $(-1,1)$. It can give you the accurate solution which is $\dfrac {\pi ^2}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$ Unless I am making a mistake, I am calculating that:
$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$
Here's my reasoning:
*
*$\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y ... | $$\frac{x^m+y^m}{x+y}+(xy)\frac{x^{m-2}+y^{m-2}}{x+y}=\frac{x^m+y^m+(xy)(x^{m-2}+y^{m-2})}{x+y}= \\
=\frac{x^m+y^m+x^{m-1}y+xy^{m-1}}{x+y}=\frac{x^m+x^{m-1}y+xy^{m-1}+y^m}{x+y}\\
=\frac{x^{m-1}(x+y)+y^{m-1}(x+y)}{x+y}=x^{m-1}+y^{m-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\g... | Since $(a,b,c)$ and $\left(\frac{1}{b+c-a},\frac{1}{a+c-b},\frac{1}{a+b-c}\right)$ are the same ordered, by Chebyshov and C-S we obtain:
$\sum\limits_{cyc}\frac{a}{b+c-a}\geq\frac{1}{3}(a+b+c)\sum\limits_{cyc}\frac{1}{b+c-a}=\frac{1}{3}\sum\limits_{cyc}(b+c-a)\sum\limits_{cyc}\frac{1}{b+c-a}\geq\frac{1}{3}\cdot9=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} =
\frac {1-2\cos... | The derivative of the function that is constantly $1$ should be zero, not $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Equation with complex numbers Solve the following equation in $ \mathbb{C} $:
$ |z - |z + 1|| = |z + |z - 1|| $
I started it but I don't know how to finish it. Here is what I did so far:
$ |z - |z + 1||^2 = (z - |z + 1|)(\bar{z} - \overline{|z + 1|}) = |z|^2 - z \cdot \overline{|z + 1|} - \bar{z} \cdot |z + 1| + |z + 1... | $z+\overline{z}=|z+1|-|z-1|$
You might square twice
$$(z+\overline{z})^2=2z\overline{z}+2-2|z+1||z-1|\\
4(z+1)(\overline{z}+1)(z-1)(\overline{z}-1)=(2-z^2-\overline{z}^2)^2\\
4(z^2-1)(\overline{z}^2-1)=4-4z^2-4\overline{z}^2+z^4+2z^2\overline{z}^2+\overline{z}^4\\
0=(z^2-\overline{z}^2)^2$$
so $z^2$ is real, so $z=x$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
A Polygon is inscribed in a circle $\Gamma$ A regular polygon P is inscribed in a circle $\Gamma$. Let A, B, and C, be three consecutive vertices on the polygon P, and let M be a point on the arc AC of $\Gamma$ that does not contain B. Prove that
$MA\cdot MC=MB^2-AB^2$
I inscribed the polygon P in the unit circle and l... | We can place the diagram in the complex plane so that $\Gamma$ is the unit circle, point $A$ goes to the complex number $a$, point $B$ goes to the complex number 1, and point $C$ goes to the complex number $\overline{a} = 1/a$. Let $m$ be the complex number corresponding to point $M$.
Then
\begin{align*}
MA \cdot MC &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Inequality involving exponential of square roots How can I show that:
$$ 2e^\sqrt{3} \leq 3e^\sqrt{2} $$
? (that's all I have)
Thank you so much!
| Note that
$$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1$$
and
$$(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}>5+2\cdot 2=9$$
Thus $\sqrt{3}+\sqrt{2}>3$, which gives $\sqrt{3}-\sqrt{2}<\frac{1}{3}$.
Now we claim that $e^{\frac{1}{3}}<\frac{3}{2}$, this is because
$$e<3<\frac{27}{8}=\left(\frac{3}{2}\right)^3$$
It follows that
$$e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Integration of $\int \tan^2 x \ dx $ without the use of trig identities I am trying to integrate $\int \tan^2 x \ dx $ without the use of trig identities, although I know that using $ 1 + \tan^2 x \equiv \sec^2 x $ makes the question trivial.
Using the exponential form of tan(x) I arrived at
$$ -\int \dfrac{(e^{2ix}-1... | $$\int\tan^2xdx=\int\frac{\sin^2{x}}{\cos^2{x}}dx=\int\frac{1-\cos{2x}}{2\cos^2{x}}dx=*$$ here we used trigonometric formula $\sin^2{x}=1-\cos{2x},$ $$*= \int\frac{1-\cos{2x}}{2\cos^2{x}}dx=\frac{1}{2}\int\frac{1}{\cos{2x}}dx - \frac{1}{2} \int \frac{\cos{2x}}{\cos^2{x}}dx=*$$ now we will use this one $\cos{2x}=2\cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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What am i doing wrong when solving this differential equation $$
f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}
$$
Given that $f(x) = cx$
$$
\frac{c}{2}x^2 + k_1 = \ln(e^y y')
$$
$$
k_2\int e^{\frac{c}{2}x^2} dx = e^y
$$
$$
y = \ln(k_2\int e^{\frac{c}{2}x^2} dx)
$$
Therefore
$$
y'^2 - cxy' + y'' = 0
$$
$$
y' = ... | Why does this not reduce to: y'' + (1-cx)y'=0?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
| You can do it using L'Hopital's rule this way:
$$\begin{align}
\lim\limits_{x \to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2} = \lim\limits_{x \to \infty} \exp\ln\Big(\frac{x^2+1}{x^2-1}\Big)^{x^2} &= \lim\limits_{x \to \infty} \exp x^2\ln\Big(\frac{x^2+1}{x^2-1}\Big) \\ &= \exp \lim\limits_{x \to \infty} \frac{\ln(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
CAL I - Quotient Rule to find Derivative I'm stuck on this problem $h(x) = \frac{\sqrt{x}}{x^3+1}$.
I get to here: $\frac{x^3+1-x^{1/2}(3x^2)}{2x^{1/2}(x^3+1)^2}$.
I started going this route: $\frac{x^3+1-3x^{5/2}}{2x^{1/2}(x^3+1)^2}$.
The book's solution is getting this in their step process: $\frac{x^3+1-6x^3}{2x^{1... | Using
$$
\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}
$$ we have
$$
\left(\frac{\sqrt{x}}{x^3+1}\right)'=\frac{\frac{1}{2\sqrt{x}}(x^3+1)-\sqrt{x}(3x^2)}{(x^3+1)^2}=\frac{1}{2\sqrt{x}}\frac{(x^3+1)-2x\:(3x^2)}{(x^3+1)^2}=\frac{1}{2\sqrt{x}}\frac{-5x^3+1}{(x^3+1)^2}
$$ as desired.
Hoping it's clear for you now.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $ \cos x - \cos y = -2 \sin ( \frac{x-y}{2} ) \sin ( \frac{x+y}{2} ) $ Prove that $ \cos x - \cos y = -2 \sin \left( \frac{x-y}{2} \right) \sin \left( \frac{x+y}{2} \right) $ without knowing cos identity
We don't know that $ \cos0 = 1 $
We don't know that $ \cos^2 x + \sin^2 x = 1 $
I have managed to prove i... | You may use the complex definition of the sin and cos functions derived from the Euler's formula.
$$\begin{array}{l}\cos x = \frac{{{e^{ix}} + {e^{ - ix}}}}{2},\sin x = \frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}};\\ - 2\sin (\frac{{x - y}}{2})\sin (\frac{{x + y}}{2})\\ = - 2\left( {\frac{{{e^{i(\frac{{x - y}}{2})}} - {e^{ -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1170630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Number of distinct terms in the expansion of $\big(x+\frac{1}{x}+x^2+\frac{1}{x^2}\big)^{15}$
Number of distinct terms in the expansion of $\bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15}$ is equal to ?
We can write the above as,
$$ \bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15} = \dfrac{1}{x^{30}}(1+x+x^3+x^4... | Your expression is equal to
$$
\frac1{x^{30}}(x^3+1)^{15}(x+1)^{15}
$$
$(x^3+1)^{15}$ contains $16$ terms, each containing a power of $x^3$ with a positive coefficient. Since $(x+1)^{15}$ contains $16$ terms of consecutive powers of $x$ with positive coefficients, the product will fill in the gaps in $(x^3+1)^{15}$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Volume of Solid in Calculus Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^5, y=1$ and the $y$-axis and whose cross sections perpeddicular to the $x$ axis are semicircles
I am tutoring someone, and we keep getting the wrong answer, any help would be appreciated
| The diameter of the semicircular slice is $1 - x^5$, and its area then is $\pi(1-x^5)^2/8.$
The curve $y=x^5$ intersects the line $y=1$ at $x=1$ in the first quadrant, so your limits of integration are $[0,1]$.
Then, your infinitesimal slice has volume $(\pi(1-x^5)^2/8) dx$, and the volume then is
$$V = \frac{\pi}{8} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral of square root. We have
$$\int \sqrt{x} \text{ }dx=\int x^{1/2} \text{ }dx=\frac{2}{3}x^{3/2}+C.$$
But, let $u=\sqrt{x}$. Then $x=u^2$, and $dx=2udu$. Substituting, we have
$$\int \sqrt{x} \text{ }dx=\int 2u^2 \text{ }du=\frac{2}{3}x^3+C.$$
Which one is correct, and why?
| $$\int \sqrt{x} \text{ }dx=\int 2u^2 \text{ }du=\frac{2}{3}u^3+C.$$
Remember, after substituting, we need to calculate the integral with respect to $u$.
Once that's done, then "back"-substitute $\sqrt x = u$ to get $$\frac 23(x^{1/2})^3 + C = \frac 23 x^{3/2} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show $n^5 + 29 n$ is divisible by $30$ Show that $n^5 + 29 n$ is divisible by $30$.
Attempt:
$n^4 ≡ 1 \pmod 5$ By Fermat Little Theorem
| You want to show that $n^5+29n$ is divisible by $2$, $3$ and $5$.
*
*$n^5+29n\equiv n^5+n\equiv n^2n^2n+n\equiv nnn+n\equiv nn+n\equiv n+n\equiv0\pmod{2}$
*$n^5+29n\equiv n^5-n\equiv n^3n^2-n\equiv nn^2-n\equiv n-n\equiv0\pmod{3}$
*$n^5+29n\equiv n^5-n\equiv n-n\equiv0\pmod{5}$
All use Fermat's little theorem: $n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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For real numbers a and b, when is the equation |a + b| = |a – b| true? I put that the statement was true only when a = 0 and b = 0 but the correct answer was that it only held true for a = 0 OR b = 0. With 'and' I figured |0 + 0| = 0 and |0 - 0| = 0. Could someone explain why my answer was incorrect? Thank you.
| If we square both sides of the equation $|a + b| = |a - b|$, we obtain
\begin{align*}
|a + b|^2 & = |a - b|^2\\
a^2 + 2ab + b^2 & = a^2 -2ab + b^2\\
4ab & = 0\\
ab & = 0
\end{align*}
which holds if $a = 0$ or $b = 0$ since a product is equal to zero if and only if one of the factors is equal to zero.
The statement $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Taylor's Theorem applies at the point $a=0$ and with $n=4$.
Got no idea how to proceed. My lecture notes have one example that I barely understand. I'd really appreciate a... | $$
\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \cdots
$$
So
$$
\sin (x^2) = x^2 - \frac{x^6}6+\frac{x^{10}}{120} - \cdots
$$
and
$$
(\sin x)^2 = x^2 - \frac{x^4}3 + \frac{2x^6}{45} -\cdots.
$$
To get the last series just look at
$$
\left(x - \frac{x^3}6 + \frac{x^5}{120} - \cdots\right)\left(x - \frac{x^3}6 + \frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Proving $(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2)$. Given $a,b,c,d \in \Bbb R$, I want to prove that: $$(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2).$$
But the bound I'm getting is just too weak. I know that: $$|a| = \sqrt{a^2} \leq \sqrt{a^2+b^2} \implies |a|+|b| \leq 2\sqrt{a^2+b^2}\implies (|a|+|b|)^2\leq ... | Here's another way of proving it. From the Cauchy-Schwarz Inequality we have:
$$(1 + 1)(a^2 + b^2) \ge \left(\sqrt{1\cdot a^2} + \sqrt{1\cdot b^2}\right)^2 = \left(\mid a \mid + \mid b \mid\right)^2$$
$$(1 + 1)(c^2 + d^2) \ge \left(\sqrt{1\cdot c^2} + \sqrt{1\cdot d^2}\right)^2 = \left(\mid c \mid + \mid d \mid\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$F_{2n} = F_{2n-2}+2F_{2n-4}+\dots+n$ rigorous proof Let $F_{n}$ be n-th fibonacci number($F_{0}$ = 0) and $g_{n} = F_{2n}$ if $n > 0$ $g_{0} = 1$.
I want to prove that $g_{n} = g_{n-1}+2g_{n-2}+\dots +ng_{0}$. It's obviously seen from direct evaluation but it's not so rigorous and my attempts to use induction end in f... | There are many ways to prove this. One way is to notice that
$$
F_{2n} = F_{2n-1} + F_{2n-2} = 2F_{2n-2} + F_{2n-3} = 2F_{2n-2} + F_{2n-3} + F_{2n-4} - F_{2n-4} = 3F_{2n-2} - F_{2n-4}.
$$
Therefore
$$ g_n = 3g_{n-1} - g_{n-2}. $$
Given this, it is easy to prove the claim by induction. Let's rephrase it slightly:
$$
g_n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality: $\left|\sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right| \le \frac{1}{n + 1 + x}$ This (with $x > 0$ and $n \in \Bbb N$):
$$ \left|\sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right| \le \frac{1}{n + 1 + x}$$
Was used to prove that the series of general term the summand on the left hand side of the in... | Hint 1: The terms $$ \frac{(-1)^i}{i + x} $$
are of alternating signs and their absolute value decreases.
Hint 2: Group the terms in the sum into pairs of two.
More detailed answer:
$$
\left| \sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right|
= | (-1)^{n+1} | \left| \Bigl( \frac{1}{n + 1 + x} - \frac{1}{n + 2 + x}... | {
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"source": "stackexchange",
"question_score": "1",
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how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are
$$
\sin x \cos x= \frac{1}{2} \sin(2 x)\\
\sin^2 x + \... | You have to prove that
$$
\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}.\tag{1}
$$
Note that $(1)$ may be expressed as follows by cross-multiplying:
$$
\underbrace{(\cos x)(1+\cos x+\sin x)}_{\text{LHS}}=\underbrace{(1+\sin x)(1+\cos x-\sin x)}_{\text{RHS}}.\tag{2}
$$
Now all you have to do is expan... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Tricky 3d geometry problem We have a cube with edge length $L$, now rotate it around its major diagonal (a complete turn, that is to say, the angle is 360 degrees), which object are we gonna get?
Astoundingly the answer is D. And here is a demonstration:
Well now I'm required to calculate the volume of this monster. ... | Let's take the unit centered cube, with vertexes at $\pm 1$. To rotate it so that its main diagonal gets aligned with the $x$ axis (vertical axis in the figure) we can use two rotations along two axes, the first by 45 degrees, the second by $\tan^{-1}(\sqrt{1/2})=\sin^{-1}(\sqrt{1/3})$. We get then the rotation matrix:... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
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Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$.
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one ... | Suppose there were real solutions to $a^2 + b^2 = ab = 4$. Then $$0 \leq (a - b)^2 = a^2 - 2ab + b^2 = (a^2 + b^2) - 2ab = -4$$ which is a a contradiction. Thus there cannot be real $a, b$ satisfying the requirements.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Write coupled wave equations as a decoupled system Given the coupled wave equations:
$$
\frac{\partial }{\partial t} [u~~v] +
\begin{pmatrix}
a & c \\
c & a
\end{pmatrix} \frac{\partial }{\partial x} [u~~v] = [0~~0].
$$
I want to transform this problem into a decoupled system with variables [r,s]. Can someone give me a... | The system
$$
\frac{\partial }{\partial t} \begin{pmatrix} u \\ v \end{pmatrix} +
\begin{pmatrix}
a & c \\
c & a
\end{pmatrix} \frac{\partial }{\partial x} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.
$$
leads to the system
\begin{align}
\begin{pmatrix} u_{t} \\ v_{t} \end{pmatrix} = \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the derivative using the chain rule and the quotient rule
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.
Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the fin... | it is easier to use logarithmic differentiation on this problem. here is how you do this. $$y = \left(\frac x{x+1}\right)^4 \to \ln y = 4 \ln x - 4 \ln x + 1\to \frac{dy}{y} =4\left(\frac{dx}{x} - \frac{dx}{x+1} \right) = \frac{4dx}{x(x+1)} $$ multiplying through by $y$ gives you $$\frac{dy}{dx} = \frac{4x^3}{(x+1)^5} ... | {
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"source": "stackexchange",
"question_score": "4",
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How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried lett... | Consider the integral
\begin{align}
I = \int \sqrt{1 + x \sqrt{x^{2} + 2}} \, dx
\end{align}
Make the substitution $x = \sqrt{2} \, csch(t)$ to obtain the integral
\begin{align}
I = - \sqrt{2} \, \int \sqrt{1 + 2 \, csch(t) \, coth(t)} \cdot csch(t) \, coth(t) \, dt.
\end{align}
Now Wolfram can calculate the integral ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Integration of fraction with square root I have a problem with integrating of fraction
$$
\int \frac{x}{x^2 + 7 + \sqrt{x^2 + 7}}
$$
I have tried to rewrite it as $\int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{x^4 + 13x^2 + 42} = \int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{(x^2 + 6)(x^2 + 7)}$ and then find some partial fraction... | Consider the integral
\begin{align}
I = \int \frac{x}{x^{2} + 7 + \sqrt{x^2 + 7}} \, dx
\end{align}
Make the substitution $u = \sqrt{x^{2} + 7}$ for which $x = \sqrt{u^{2} - 7}$ and the integral becomes
\begin{align}
I &= \int \frac{\sqrt{u^{2}-7}}{u^{2} + u} \cdot \frac{u \, du}{\sqrt{u^{2}-7}} \\
&= \int \frac{du}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does every $9 \times 9$ Latin square contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
Q: Does every $9 \times 9$ Latin square on the symbol set $\{1,2,\ldots,9\}$ contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
This one has $1728$ such submatrices, which is... | How about this one?
$$
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & 1 & 4 & 3 & 6 & 7 & 8 & 9 & 5 \\
3 & 4 & 2 & 1 & 8 & 9 & 5 & 6 & 7 \\
4 & 3 & 1 & 2 & 7 & 8 & 9 & 5 & 6 \\
5 & 6 & 9 & 8 & 2 & 3 & 4 & 7 & 1 \\
6 & 7 & 5 & 9 & 1 & 2 & 3 & 4 & 8 \\
7 & 8 & 6 & 5 & 9 & 1 & 2 & 3 & 4 \\
8 & 9 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Find the values of m and n(Trigononetry in series) $$\sin ^6(1)+\sin ^6(2)+... ...+\sin ^6(89)=\frac{m}{n}$$
Find $\frac{m}{n}$ in its simplest form , and hence find both values. (All angles are in degree)
I've no idea how to start to solve this questions. Need some guidance for it. Thanks in advance.
| $$\begin{align} 1 &= (\sin^2 t + \cos^2 t)^3 \\
&= \sin^6 t + 3\sin^4t \cos^2 t + 3 \sin^2t+ \cos^4t + \cos^6 t \\
&= \sin^6 t + \cos^6 t + 3\sin^2t \cos^2 t \\
&=\sin^6 t + \cos^6 t + \frac34\sin^2 (2t) \\
\end{align} $$
let $$ S= \sin^6 1^\circ + \sin^6 2\circ + \cdots + \sin^6 89^\circ. $$ then we have $$\begin{al... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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Taylor series for the function $f(z) = \frac{1}{(z-5)(z-7)}$ on a disc centered at point $z_0=3$ I started by expressing the function as sum of two fractions using partial fraction decomposition to get $\frac{-1}{2(z-5)} + \frac{1}{2(z-7)}$
However I could only then end up writing that as the sum of two power series :
... | Hint: For example,
$$
\frac{1}{z-7} = \frac{1}{(z-3)-4} = -\frac{1}{4} \frac{1}{1-(z-3)/4} = -\frac{1}{4} \sum_{i=0}^\infty \left(\frac{z-3}{4}\right)^i.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is continuous and integrable function bounded? I have a function $f: \mathbb R \rightarrow \mathbb R$ continuous and integrable on $\mathbb R$. Is $f$ bounded?
| No! See this example here If $f\in L^1(\Bbb R,dx)$ then prove that for almost every $x\in\Bbb R$ $\lim\limits_{n\to \infty} f(nx) = 0.$
Where the function is a Polynomial on each interval $[n, n+\frac{1}{2^n}]$ and zero elsewhere.
Consider $f:\Bbb R\to \Bbb R$
\begin{align*}
f(x)= \begin{cases}
2^{n/2}P_n(x)& \text{if}... | {
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"source": "stackexchange",
"question_score": "3",
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Continued Fraction Algorithm for 113/50 The numbers $a_k$ can be found for $\frac{113}{50}$ by using a continued fraction algorithm. Note that $\frac{113}{50}$ is rational, and as a result it will have to terminate. Can anyone help me determine the numbers $a_k$ for $\frac{113}{50}$?
What I have Done:
$\frac{113}{50} =... | Meanwhile, here is the calculation of the "convergents." It begins with the formal fractions $0/1$ and the fake (but necessary) $1/0.$ Then, separately for numerator and denominator, for each "digit" or "partial quotient" $a_k,$ you get then next number from the evident rule, now visible: for example, $0 + 2 \cdot 1 = ... | {
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"source": "stackexchange",
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Critical points of a function $f(x) = x\sqrt{x-a}$ Find the critical points of a function $f(x) = x\sqrt{x-a}$.
A function $f(x)$ is said to have critical points at points $c$ such that $f^\prime(c)$ is $0$ or undefined.
For a function $f(x) = x\sqrt{x-a}$ :
$$\hspace{-0.5 in} f^\prime(x) = \frac{3x - 2a}{2\sqrt{x - a}... | I normally don't consider a number a critical number unless it is part of the domain of the original function. So looking at $f$ and $f'$ domain we want $x>a$. You found $x=\frac{2a}{3}$ to be a critical number. So that means we want $\frac{2a}{3}>a$ What do you get when you solve that for $a$.
| {
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Show that $\pi =4-\sum_{n=1}^{\infty }\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}$ Show that $$\pi =4-\sum_{n=1}^{\infty }\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}$$
I found the formula of $\pi$ by using the numerical calculation but I dont have the proving. Any help would be appreciated.
| Note that $$4-\pi = \frac{4}{3}-\frac{4}{5}+\frac{4}{7}-\ldots$$ and by the Euler transform
$$
4-\pi = \frac{2}{3} + \frac{2}{3\cdot 5}+\frac{ 2\cdot 2}{3\cdot 5\cdot 7} +\frac{2\cdot 3 \cdot 2}{3\cdot 5\cdot 7\cdot 9}+\ldots=\sum_{n=1}^{\infty}\frac{2^{n+1}n!(n-1)!}{(2n+1)!}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Limit of $\sqrt{25x^{2}+5x}-5x$ as $x\to\infty$ $\hspace{1cm} \displaystyle\lim_{x\to\infty} \left(\sqrt{25x^{2}+5x}-5x\right) $
The correct answer seems to be $\frac12$, whereas I get $0$.
Here's how I do this problem:
$$ \sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 2... | $$\sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \dfrac{5x}{\sqrt{25x^{2}+5x} +5x}$$
Your work is fine so far. Next factor out $x$ from denominator and cancel it with numerator
$$ \dfrac{5x}{\sqrt{25x^{2}+5x} +5x} = \dfrac{5x}{\sqrt{x^2(2... | {
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To calculate the expectation of a variable with a given pdf Let the pdf of a random variable X be given by $f(x)=ae^{-x^2-bx}, -\infty<x<\infty$. If $E(X)=-\frac{1}{2}$, then
(A)$a=\frac{1}{\sqrt{\pi}}e^{-1/4},b=1$
(B)$a=\frac{1}{\sqrt{\pi}}e^{-1/4},b=-1$
(C)$a=\sqrt{\pi}e^{-1/4},b=1$
(D)$a=\sqrt{\pi}e^{-1/4},b=-1$
My ... | Recall that a normally distributed random variable $X$ with mean $\mu$ and standard deviation $\sigma$ has the probability density function $$f_X(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad -\infty < x < \infty.$$ So if we set $\mu = -1/2$, then $\operatorname{E}[X] = \mu = -1/2$ as desired. N... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does this series $\sum_{i=0}^n \frac{4}{3^n}$ diverge or converge? I a newbie to series, and I have not done too much yet. I have an exercise where I have basically to say if some series are convergent or divergent. If convergent, determine (and prove) the sum of the series.
This is the first series:
$$\sum_{i=0}^n \fr... | Let $S_N=\sum_{i=0}^{N} r^i=1+r+r^2+\cdots r^N$. Note that
$$\begin{align}
rS_N&=\sum_{i=0}^{N} r^{i+1}\\
&=r+r^2+r^3+\cdots+r^N+r^{N+1}
\end{align}$$
So, if we form the difference $S_N-rS_N$ we find $S_N-rS_N=(1-r)S_N=1-r^{N+1}$ where upon solving for $S_N$ we find
$$\begin{align}
S_N&=\frac{1-r^{N+1}}{1-r}\\
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$. $$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\
&= \frac{2x-\sec^2{x... |
Hint to find derivative: Use FTC, http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements, $F'(x)=\frac{d}{dx}\int^x_a f(t)dt=f(x)$
Answer to original question: HINT to integrate: \begin{align}
\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt &= \frac{1}{\sqrt2} \int_{\tan{x}}^{x^2} \frac{1}{\s... | {
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"url": "https://math.stackexchange.com/questions/1213093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving $(n+1)!>2^{n+3}$ for all $n\geq 5$ by induction I am stuck writing the body a PMI I have been working on for quite some time.
Theorem: $∀n∈N ≥ X$, $(n+1)!>2^{n+3}$
I will first verify that the hypothesis is true for at least one value of $n∈N$.
Consider $n=3$: (not valid)
$$(3+1)!>2^{3+3} \implies 4!>2^{6} \imp... | First, show that this is true for $n=5$:
$(5+1)!>2^{5+3}$
Second, assume that this is true for $n$:
$(n+1)!>2^{n+3}$
Third, prove that this is true for $n+1$:
$(n+2)!=\color{red}{(n+1)!}\cdot\color{blue}{(n+2)}>\color{red}{2^{n+3}}\cdot\color{blue}{(n+2)}>\color{red}{2^{n+3}}\cdot\color{blue}{2}=2^{n+4}$
Please note t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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proving recursively defined sequence by induction I would like to prove the following recursively defined sequence from $n-1$ to $n$ by induction. Im not realy sure about it. Any help or alternative ways to understand and prove it are highly appreciated :
$0,1,4,12,35,98$
$a_0=0$, $a_1=1$, $a_n=a_{n-1}+5a_{n-2}+3$ for ... | It will be easier to show $a_n \le 3^n-1$ (and hence $3^n$).
$$\begin{align}
a_n &= a_{n-1} + 5a_{n-2}+3 \\
&\le (3^{n-1}-1) + 5(3^{n-2}-1) + 3 \\
&= 3^{n-1} + 5\cdot 3^{n-2} -3\\
&= 3^{n-1} + 3\cdot 3^{n-2} + 2\cdot 3^{n-2} -3\\
&= 3^{n-1} + 3^{n-1} + 2\cdot (3^{n-2} -1)-1\\
&\le 3^{n-1} + 3^{n-1} + 2\cdot 3^{n-2} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms. A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms.
$\dfrac{a(1-r^n)}{1-r}$
$\dfrac{1(1-4^7)}{1-4} = 5461$.
The answer in the book is $21845$. How is this so?
Thank you
| The sum of the first $n$ terms of a geometric sequence $\{a_n\}$ with initial term $a_1$ and common ratio $r$ is $$s_n = a_1\frac{1 - r^{n}}{1 - r}$$
For the sequence defined by $f(n) = 4^{n - 1}$, $a_1 = 4^{1 - 1} = 4^0 = 1$, and $$r = \frac{a_{n + 1}}{a_n} = \frac{4^n}{4^{n - 1}} = 4$$ You wish to find the sum of ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots
I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
$$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{... | Let $f(x)=x^6+x^5-x^4-x^3+x^2+x-1=0$. Then $f(0)=-1$. Note for $x>0$
\begin{eqnarray}
f'(x)&=&(6x^5+2x)+(5x^4+1)-4x^3-3x^2\\
&\ge& 4\sqrt3x^3+2\sqrt5x^2-4x^3-3x^2\\
&=&(4\sqrt3-4)x^3+(2\sqrt5-3)x^2\\
&>&0
\end{eqnarray}
hance $f(x)$ is strictly increasing. Noting that $\lim_{x\to\infty}f(x)=\infty$, we have a unique ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Is there a natural number for which all the sums and differences of its factor pairs are prime? The 8 factor pairs of e.g. 462 are
$((1, 462), (2, 231), (3, 154), (6, 77), (7, 66), (11, 42), (14, 33), (21, 22))$.
Of the 16 non-negative integers which are the sums and differences of these pairs (such as $462+1=463$, $46... | It seems the following.
There is a following recursive way to search such the integer $N$.
$N=1\cdot k_1$. Since all of numbers $k_1\pm 1$ are prime, one of these numbers is odd. So we have
$N=1\cdot 2\cdot k_2$. Since all of numbers $2k_2\pm 1$, $k_2\pm 2$ are prime, if $k_2$ is not divisible by $3$, then one of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
How to factorize $(x-2)^5+x-1$? This is a difficult problem.
How to factorize this?
$$(x-2)^5+x-1$$
we can't do any thing now and we should expand it first:
$$x^5-10x^4+40x^3-80x^2+81x-33$$
but I can't factorize it.
| Using the substitution $t=x-2$, it suffices to factor $t^5 + t+1$.
But $t^2+t+1$ divides $t^5+t+1$. Why?
$t^2+t+1 = (t-\zeta)(t-\zeta')$, where $\zeta,\zeta'$ are the primitive $3rd$ roots of unity, and thus $\zeta^5 + \zeta + 1 = \zeta^2 + \zeta + 1 = 0$, similarly for $\zeta'$.
Now, the quotient $(t^5+t+1)/(t^2+t+1)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$ax^2+bx+c=0$ has roots $x_1,x_2$. what are the roots of $cx^2+bx+a=0$. Given solution: Dividing the first equation by $x^2$ we get
$c(\frac{1}{x^2})+b(\frac{1}{x})+a=0$
so $(\frac{1}{x_1}),(\frac{1}{x_2})$ are the roots of $cx^2+bx+a=0$.{How?It is not obvious to me.}
The answers so far are proving retrospectively that... | Given $ax^2 + bx + c = 0$,
$$\begin{align} ax^2 + bx + c& = 0\\ a + b\left(\frac{1}{x}\right) + c\left(\frac{1}{x^2}\right) & = 0 \tag{Divide both sides by $x^2$}\\ cy^2 + by + a & = 0\tag{Let $y = \frac{1}{x}$}\end{align}$$
We'd discover that this polynomial has roots $y_1$ and $y_2$. We know that $y = \frac{1}{x}$. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to differentiate $y=\sqrt{\frac{1+x}{1-x}}$? I'm trying to solve this problem but I think I'm missing something.
Here's what I've done so far:
$$g(x) = \frac{1+x}{1-x}$$
$$u = 1+x$$
$$u' = 1$$
$$v = 1-x$$
$$v' = -1$$
$$g'(x) = \frac{(1-x) -(-1)(1+x)}{(1-x)^2}$$
$$g'(x) = \frac{1-x+1+x}{(1-x)^2}$$
$$g'(x) = \frac{2... | Another idea:
$$\frac{1+x}{1-x}=\frac2{1-x}-1\implies \left(\frac{1+x}{1-x}\right)'=\frac2{(1-x)^2}\implies$$
$$\left(\sqrt\frac{1+x}{1-x}\right)'=\sqrt\frac{1-x}{1+x}\cdot\frac1{(1-x)^2}$$
Added on request:
$$\sqrt\frac{1-x}{1+x}\cdot\frac1{(1-x)^2}=\frac1{\sqrt{1+x}}\cdot\frac1{\sqrt{1-x}\cdot(1-x)}=\frac1{\sqrt{1-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
quadratic equation what am I doing wrong? solve
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} $$
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} \Rightarrow $$
$$ 5x+19 = (x+7) + 4\sqrt{x-5}\sqrt{x+7} + (x+5) \Rightarrow $$
$$ 3x + 17 = 4\sqrt{x-5}\sqrt{x+7} \Rightarrow $$
$$ 9x^2 + 102x + 289 = 16(x+7)(x-5) \Rightarrow $$
$... | \begin{align}
\sqrt{5x+19}&=\sqrt{x+7}+2\sqrt{x-5}\\
5x+19&=x+7+4(x-5)+4\sqrt{(x+7)(x-5)}\\
32&=4\sqrt{(x+7)(x-5)}\\
8&=\sqrt{(x+7)(x-5)}\\
64&=x^2+2x-35\\
0&=x^2+2x-99
\end{align}
which gives $x=9$ and $x=-11$ are the solutions.
But both are giving something like this..
$x=9\implies \sqrt {5(9)+19}=\sqrt{9+7}+2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
What will be the sum of the series of binomial co-efficients? What will be the sum of the following binomial co-efficent series
$$\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} = \sum\limits_{i=1}^r \binom{z+i}{z}$$
Thank you
| Hint Add $\binom{z+1}{z+1}$ to you sum and use $\binom{k}{z}+\binom{k}{z+1}=\binom{k+1}{z+1}$ repeatedly:
$$\binom{z+1}{z+1}+\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} \\
=\binom{z+2}{z+1}+\binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} \\=...$$
Here I am using the standard notat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $x + y + t \le 10$
For nonnegative integers, $x, y, t$ solve, $$x + y + t \le 10$$
This includes then: $x + y + t = 0$, ..., $x + y + t = 10$.
$x + y + t = 0$ has $1$ solution $= \binom{2}{2}$.
$x + y + t = 1$ has $3$ solutions, $= \binom{3}{2}$
$\cdots$
$x + y + t = 10$ has: $ $ solutions: $= \binom{12}{2}$
$... | You can include a slack variable $s$, and then solve $$x+y+t+s=10$$
with each variable a nonnegative integer. You can solve this with stars and bars, with solution $${10+4-1\choose 4-1}={13\choose 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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.