Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find number of real roots of a transcendental equation? The number of real roots of the equation
$$2\cos\left(\frac{x^2+x}6\right)=2^x+2^{-x}$$
Another question is...
can we use descartes rule of sign in here or in any transcendental equation ?
| Notice that
$$2\cos\left(\frac{x^2+x}6\right) \le 2$$
$$2^x+2^{-x} \ge 2$$
First is true since $\cos\left(\frac{x^2+x}6\right) \le 1$ and the second is equivalent to $\left(2^{\frac{x}{2}} - \frac{1}{2^{\frac{x}{2}}} \right)^2 \ge 0$ which is always true. So in order to have an equality the second inequality also has t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Use partial fractions to find the integral. Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{ali... | Hint: Write out the fraction given as $\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$ and equate the numerator to $2x^2+5x+3$. And the ensuing integral should be easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/564845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do I simplify this exponential expression?: $ 2(-3x^{-2}y^3)^{-1} \cdot (-3x^{-3}\cdot y)^2 $ How do I simplify this expression?
Simplify:
$$ 2(-3x^{-2}y^3)^{-1} \cdot (-3x^{-3}\cdot y)^2 $$
I tried and didn't get the answer.
| Hint: Note that $$ \left( ax^p \cdot bx^q \right)^e = \left( ab \cdot x^{p+q} \right)^e = \left( ab \right)^e \cdot x^{e\cdot(p+q)}. $$These are just usages of exponent properties.
Solution, here: $$ \begin {align*} 2 \cdot \left( - 3 x^{-2} y^3 \right)^{-1} \cdot \left( -3 x^{-3} y \right)^2 &= 2 \cdot \left( - \dfra... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 1,
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Proving a trigonometric identity: $\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x$ I really need some help with this question. I need to prove this identity:
$$\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x.$$
| $\displaystyle\frac{2\sin^3x}{1-\cos x}=2\sin^2x\cdot\frac{\sin x}{1-\cos x}=2\sin^2x\cdot\frac{1+\cos x}{\sin x}$(using this)
$\displaystyle=2\sin x(1+\cos x)=2\sin x+2\sin x\cos x=\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/569408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help With Series (Apostol, Calculus, Volume I, Section 10.9 #9) I am looking for help finding the sum of a particular series from Apostol's Calculus (Volume I, Section 10.9, Problem 9). The trouble is that I can find the correct answer, but only using methods that aren't available at this point in the text, or with too... | Take a look at the sum of
the first $m$ terms.
$\begin{align}
\sum_{n = 1}^{m} \frac{(-1)^{n - 1}(2n + 1)}{n(n + 1)}
&=\sum_{n = 1}^{m} (-1)^{n - 1}\left(\frac{1}{n} + \frac{1}{n + 1}\right)\\
&=\sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n} + \sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n + 1}\\
&=\sum_{n = 1}^{m} (-1)^{n - 1}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplification of $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}}$ I'm having trouble understanding how this expression:
$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}} \cdot \left(\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}\right)=$$
got to this one:
$$\frac2{\sqrt2... | Now that I am reading this correctly I think I can make some sense of the claim.
Let $x=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$ and $ y=\sqrt{2-\sqrt{2+\sqrt{2+\cdots}}}$ The "equation" you ask about is $$y=y\frac{x}{x}=\frac{2}{\sqrt{2}\cdot\sqrt{2+\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\ \cdots} $$ Since the righth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving an inequality by induction and figuring out intermediate inductive steps? I'm working on proving the following statement using induction:
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
Fair enough. I'll start with the basis step:
Basis Step: (n=1)
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
$$ \fra... | For the inductive step you have to show that $ \sum_{r=1}^{n} \frac{1}{r^2} \le \frac{2n}{n+1} $ implies $ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n+2}{n+2} $.
Start with $ \sum_{r=1}^{n+1} \frac{1}{r^2}$ and show that is less or equal than $\frac{2n}{n+1}+\frac{1}{(n+1)^2}$ (use the inductive hypothesis). Then, show... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_... | $\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}~dy~dx$
$=\int_0^b\left[\dfrac{y\sqrt{a^2-x^2-y^2}}{2}+\dfrac{a^2-x^2}{2}\sin^{-1}\dfrac{y}{\sqrt{a^2-x^2}}\right]_0^x~dx$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions)
$=\int_0^b\dfrac{x\sqrt{a^2-2x^2}}{2}dx+\int_0^b\dfrac{a^2-x^2}{2}\sin^{-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
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Boolean Algebra simplify minterms I have this equation
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
and need to simplify it. I have got as far as I can and spent a good 2 hours at it. I've realized I now need to use De Morgan's law to continue however I am baffl... | If you use a karnaugh map:
$$
\begin{array}{c|c|c|c|c}
C, AB & 00 & 01 & 11 & 10 \\ \hline
0 & 1 & & 1 & \\ \hline
1 & & & 1 & 1\\ \hline
\end{array}
\equiv
\bar{A}\bar{B}\bar{C} + A\bar{B} C + AB\bar{C} + ABC
$$
Which suggests Xor of the 2 groups:
$$
\begin{array}{c|c|c|c|c}
C, AB & 00 & 01 & 11 & 10 \\ \hl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$ For any real numbers $a,b,c$ show that:
$$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \... | Note that LHS does not change if you replace $(a,b,c)$ by $(a-t,b-t,c-t)$. Thus we can first minimize RHS with respect to $t$,after we replace $(a,b,c)$ by $(a-t,b-t,c-t)$. It turns out that RHS is minimized when $\displaystyle t = \frac{a+b+c}{3}$, with minimum value being
$$\frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.
| So, consider the following:
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \int_0^{1} dx \frac{\log{x}}{(1+x^2)^2} + \int_1^{\infty} dx \frac{\log{x}}{(1+x^2)^2}$$
Sub $x \mapsto 1/x$ in the latter integral on the RHS and get that
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \int_0^{1} dx \frac{1-x^2}{(1+x^2)^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 0
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How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
... | Let the best of $n$ series be decided after $k$ games. This will happen if in the preceding $k-1$ games $A$ also wins $\lfloor n/2 \rfloor$ and wins the $k^{th}$ game. Hence,
$$C(k) = \dbinom{k-1}{\lfloor n/2 \rfloor}$$
Hence, the total number of ways is
$$\sum_{k=\lfloor n/2 \rfloor+1}^n C(k)$$
In your case, set $n=7$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Evaluating $\iint_D \sqrt{4x^2-y^2}\;\ \mathrm dx \ \mathrm dy$ I have to evaluate $\displaystyle\iint_Df(x,y)\;dxdy$ for $f(x,y) = \sqrt{4x^2-y^2}$ with $D = \{(x,y)\in\mathbb{R}^2: 0\leq x \leq 1, 0\leq y \leq x\}$.
It seems that i can't solve for $\displaystyle\int_0^1 \displaystyle\int_0 ^x\sqrt{4x^2-y^2} dydx$ but... | I am posting this about finding the proper limits not to solve the definite integrals. We have $$D = \{(x,y)\in\mathbb{R}^2: 0\leq x \leq 1, 0\leq y \leq x\}$$ and it seems you want to change the order of the integrals. OK! As you might already do, the following plot makes the region of $D$ clear.
Changing the order o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$? Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this:
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.
| Your idea is good, write
$$\frac{4x+4}{x^4+x^3+2x^2}=4\frac{x+1}{x^2 (x^2+x+2)}$$
Then, partial fraction will be of the form
$$\frac{x+1}{x^2 (x^2+x+2)}=\frac{a}{x} + \frac{b}{x^2} + \frac{cx+d}{x^2+x+2}$$
There are several methods to identify coefficients from this. The simplest IMHO is to rewrite the right hand side ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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limit problem (with roots) Is it possible to evaluate this limit without graphing or guessing
(ie to replace it by a simpler function)
$$\lim_{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I tried normalizing by multiplying by the conjugate (both denominator and numerator) didn't work.
| $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{(\sqrt{6-x}-2)\times (\sqrt{6-x}+2)\times(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)\times(\sqrt{3-x}+1)\times (\sqrt{6-x}+2)}$$
$$=\frac{\left((\sqrt{6-x})^2-4\right)\times(\sqrt{3-x}+1)}{\left((\sqrt{3-x})^2-1\right)\times (\sqrt{6-x}+2)}=\frac{\overbrace{\left((6-x)-4\right)}^{(2-x)}\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I al... | $$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$$
$$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^2\tan^2x}\right)$$
$$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)\times \lim_{x\to 0}\left(\frac{x}{\tan x}\right)^2=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)$$
Now, using L' Hospital Rule successively... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3\cdot4^{n+2} +5\right)$ using congruences Using congruence theory, show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3 \cdot 4^{n+2} +5\right)$. The proof is quite simple with induction, but how can it be proved with congruences?
| HINT:
As $10\equiv1\pmod 9, 10^n \equiv1^n\equiv1$
$\displaystyle 4^{n+2}=(1+3)^{n+2}$
$\displaystyle=1+\binom{n+2}13+\binom{n+2}23^2+\cdots+3^{n+2}\equiv1+(n+2)3\pmod 9\equiv3n+7$
Can you take it from here?
Another way
$$10^n+3\cdot4^{n+2}+5=(10^n-1^n)+3\left(4^{n+2}-1^{n+2}\right)+1+3+5$$
Using congruence,
$\displa... | {
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"timestamp": "2023-03-29T00:00:00",
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Fibonacci sequence proof Prove the following:
$$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$
For $n \ge 2$
Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$
So, here are my first steps:
$$ \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = ... | One way to derive/prove these identities is to start from
$$\begin{pmatrix} f_{n-1} & f_n \\ f_n &f_{n+1} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^n$$
Note that
\begin{align}
& \begin{pmatrix} 0 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} f_2+f_5+ \ldots +f_{3n-1} & f_3+f_6+ \ldots +f_{3n} \\ f_3+f_6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove that $x_1^n+x_2^n$ is an integer and is not divisible by $5$ If $x_1$ and $x_2$ are the roots of the polynomial $x^2-6x+1$ then , for every non-negative integer, prove that $x_1^n+x_2^n$ is an integer and is not divisible by $5$ .
My trying:
$ x_1 = 3+2\sqrt{2}$ and $ x_1 = 3-2\sqrt{2}$
So $ x_1^n +x_2^n = (3+2... | We have $x_1 + x_2 = \text{ Sum of roots }=6$. From the equation, we have
$$x_1^2 -6x_1 + 1 = 0 \text{ and }x_2^2 -6x_2 + 1 = 0$$
Adding both we get
$$x_1^2 + x_2^2 = 6\underbrace{(x_1+x_2)}_{\text{Integer}} - 2 = \text{Integer}$$
Now use strong induction and make use of the fact that
$$x_1^{n+2} -6x_1^{n+1} + x_1^n = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| $$x = \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
$$x = \sqrt{7x}$$
$$x^2 - 7x = 0$$
$$x(x - 7) =0 \implies x = 7$$
Because $\sqrt{7} > 0$ we reject the $x=0$ solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "69",
"answer_count": 7,
"answer_id": 3
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How prove this inequality $\sin{\sin{\sin{\sin{x}}}}\le\frac{4}{5}\cos{\cos{\cos{\cos{x}}}}$ Nice Question:
let $x\in [0,2\pi]$, show that:
$$\sin{\sin{\sin{\sin{x}}}}\le\dfrac{4}{5}\cos{\cos{\cos{\cos{x}}}}?$$
I know this follow famous problem(1995 Russia Mathematical olympiad)
$$\sin{\sin{\sin{\sin{x}}}}<\cos{\co... | We still start from the original Russian Olympiad Problem:
$\cos \cos \cos \cos x> \sin \sin \sin \sin x$. It could have another numerical proof simply by doing in a calculator:
We have $-1\leq \cos x \leq 1, \text{that is }\cos 1\leq \cos \cos x\leq 1,
\text{that is }\cos 1 \leq \cos \cos \cos x \leq \cos \cos 1$.
Fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "59",
"answer_count": 2,
"answer_id": 1
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Why is $\sum_{k=0}^{n} f(n,k) = F_{n+2}$? If $f(n,k)$ is the number of $k$ size subsets of $[ n ] = { 1 , \ldots , n }$ which do not contain a pair of consecutive numbers, how can I show that $\sum_{k=0}^{n} f(n,k) = F_{n+2}$?
($F_{n}$ is the nth Fibonacci number: $F_{0} = 0, F_{1} = 1, F_{n} = F_{n-1} + F_{n-2}$ for $... | Here is a solution that uses generating functions. Suppose the subsets
are ordered with the smallest element first. Choosing this element
corresponds to the generating function
$$\frac{z}{1-z}.$$
The remaining elements of the subset are chosen by adding a series of
$k-1$ gap values $\ge 2$ consecutively starting w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $... | The way I would approach it is to first observe that the zeroes of $y^2-3 y+2$ are at $y=1$ and $y=2$. Because these zeroes are integers, then $x=1$ and $x=2$ are part of the solution set.
To get the rest of the solution set, we must find all other $x$ near these zeroes that do not change the values of either $\lfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
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Integer $a$ , If $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots If the equation $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots. Then the integer value of $a$ is
$\bf{My\; Try}::$ Let $\alpha,\beta\in \mathbb{Z}$ be the roots of the equation . Then $\alpha+\beta = (6-a)$ and $\alpha\cdot \beta = a$
Now... | Another way:
We can set $a^2-16a+36$ to $(a-n)^2$ where $n$ is some integer
$\displaystyle \implies n^2=36-16a+2an$ which is even $\implies 2|n^2\iff 2|n$ as $2$ is prime.
So, we can set $n=2m$ where $m$ is some integer
$\displaystyle \implies 4m^2=36-16a+4am\implies a=\frac{m^2-9}{m-4}=m+4+\frac7{m-4}$
$\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
| First way:
$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})$
set a=2, b=1
Edit: Second way.
Set $X=2^{(n-1)} + 2^{(n-2)} + … + 2 + 1$
$2X=2^n+2^{(n-1)} + 2^{(n-2)} + … + 2$
$2X=2^n+2^{(n-1)} + 2^{(n-2)} + … + 2 +1 -1$
$2X=2^n+X-1$
$X=2^n-1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility induction proof: $8\mid 7^n+3^n-2$ I'm stuck on the following proof by induction: $$8\mid3^n +7^n -2$$
And this is how far I've gotten:
$$\begin{aligned}3&\cdot3^n+7\cdot7^n-2\\3&(3^n+7^n-2)+7^n(7-3)-2\end{aligned}$$
Any help on where to go after this would be great!
| Note that by hypothesis $$8|3^n+7^n-2$$ then $8|3^{n+1}+7^{n+1}-2$ if and only if also $$8|(3^{n+1}+7^{n+1}-2)-(3^{n}+7^{n}-2)=3^n(3-1)+7^n(7-1)=\underbrace{2\cdot 3^n+2\cdot 7^n-4}_{\mathrm{divisible\ by\ 8}}+\underbrace{4+4\cdot 7^n}_{\mathrm{divisible\ by\ 8}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I find the Jordan normal form of a matrix with complex eigenvalues? I'm trying to obtain the Jordan normal form and the transformation matrix for the following matrix:
$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$
I've calculated its characteristic and mi... | We are given:
$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$$
We find that characteristic polynomial by solving $|A - \lambda I| = 0$, yielding:
$$(\lambda -1)^2 (\lambda^2 + \lambda +1) = 0$$
This yields a double and a complex conjugate pair of eigenvalues:
$$\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Using the method of induction to show How can I use the method of induction to show for any real number $r$ does not equal $1$ and any positive integer $n$
show that
$$1+r+r^2+\cdots+r^n=\frac{1r^{n+1}-1}{r-1}$$
for $n=1$ it seems to work
$$1+r+\cdots+r^n=(1+r)$$
then $\dfrac{r^2-1}{r-1}$ for the right side
$$\frac{(r-... | Assuming $r\ne1$
Let $\displaystyle F(n): 1+r+r^2+....+r^n=\frac{r^{n+1}-1}{r-1}$ holds true for $n=m $
$\displaystyle\implies 1+r+r^2+....+r^m=\frac{r^{m+1}-1}{r-1}$
$\displaystyle \implies 1+r+r^2+....+r^m+r^{m+1}=\frac{r^{m+1}-1}{r-1}+r^{m+1}=\frac{r^{m+1}-1+r^{m+2}-r^{m+1}}{r-1}$
$\displaystyle \implies 1+r+r^2+...... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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combination problem. A small company employs $3$ men and $5$ women. If a team of $4$ employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly $2$ women?
So, first, how many possible combinations are there. Well there are $8$ people and a group of $4$ ... | There are $8$ employees. $4$ are chosen from the $8$ to organize the retreat: there are ${8 \choose 4} = 70 $ ways to do this. How many of these ways have just $2$ women? Well, remember that we have $5$ women. There are ${5 \choose 2} = 10$ ways to pick them. Then we have $3$ men. If in a team of $4$ we have $2$ women,... | {
"language": "en",
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"source": "stackexchange",
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Induction: $n^{n+1} > (n+1)^n$ and $(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$ How do I prove this by induction:
$$\displaystyle n^{n+1} > (n+1)^n,\; \mbox{ for } n\geq 3$$
Thanks.
What I'm doing is bunch of these induction problems for my first year math studies.
I tried using Bernoulli's inequality at som... | Supposed for $n\geq 3$ you have that $n^{n+1}>(n+1)^n$. WTS $(n+1)^{n+2}>(n+2)^{n+1}$. Since $n^{n+1}>(n+1)^n$ you get that $n^{n+1} \cdot \frac{(n+1)^{n+2}}{n^{n+1}}>(n+1)^n\cdot \frac{(n+1)^{n+2}}{n^{n+1}}=\frac{(n+1)^{2n+2}}{n^{n+1}}>(n+2)^{n+1}$, where the last inequality follows since $(n+1)^2>n(n+2)$. Hence $(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Calculate how many ways to get change of 78 I been asked to calculate how many ways there are to get change of 78 cents with the coins of 25,10,5,1.
I been able to write this down:
$25a + 10b + 5c + d = 78$
But I do not know how to continue. Can you help me please?
| Consider the product in alex.jordan's answer. The product of the last two factors is
$$1 + x^{10} + x^{20} + x^{25} + x^{30} + x^{35} + x^{40} + x^{45} + 2 x^{50} + x^{55}
+ 2 x^{60} + x^{65} + 2 x^{70} + 2 x^{75} + \ldots)$$
The product of the first two is
$$ 1 + \ldots + x^4 + 2 (x^5 + \ldots + x^9) + 3 (x^{10}+\ldo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Ellipse problem : Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r. Problem :
Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r.
Few concepts about Ellipse :
Equation of Ta... | You have done sufficient hard work.
As $m$ is the slope of the common tangent,
we have $\displaystyle \pm \sqrt{a^2m^2+b^2} = \pm r\sqrt{1+m^2}$
Squaring we get $\displaystyle a^2m^2+b^2=r^2(1+m^2)$
What is $m$, compare with my other answer?
| {
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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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What is $\lim_{x\to0} \frac{(\cos x + \cos 2x + \dots+ \cos nx - n)}{\sin x^2}$? What is the limit of $$\lim_{x\to0} \frac{\cos x + \cos 2x + \dots+ \cos nx - n}{\sin x^2}$$
| $$
\begin{align}
& \lim_{x\to0}\frac{\cos x+\cos 2x+\cdots+\cos nx-n}{\sin x^2} \\[10pt]
& =\lim_{x\to0} \frac{\cos x -1+\cos 2x -1+\cdots+\cos nx-1}{x^2} \cdot\frac{x^2}{\sin x^2} \\[10pt]
& = -\left(\frac{1}{2} + \frac{4}{2}+\frac{9}{2}+\cdots+\frac{n^2}{2}\right)\cdot1=-\frac{n(n+1)(2n+1)}{12}.
\end{align}
$$
We app... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Probability that committee chosen from 8 men and 7 women has more men
A board of trustees of a university consists of 8 men and 7 women. A committee of 3 must be selected at random and without replacement. The role
of the committee is to select a new president for the university. Calculate the
probability that the... | In the line "Probability that the first two selected are men with the third a woman," the probability of choosing a woman third, after choosing two men, is 7/13. You need to multiply by that as well. Similarly, the probability of choosing the woman first is missing from the following line. Finally, you also need to add... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/611138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $f(x)$ where $ f(x)+f\left(\frac{1-x}x\right)=x$ What function satisfies $ f(x)+f\left(\frac{1-x}x\right)=x$ ?
| $f(x)+f\left(\dfrac{1-x}{x}\right)=x$
$f(x)+f\left(\dfrac{1}{x}-1\right)=x$
$\because$ The general solution of $T(x+1)=\dfrac{1}{T(x)}-1$ is $T(x)=\dfrac{(\sqrt5-1)^{x+1}+\Theta(x)(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2\Theta(x)(-\sqrt5-1)^x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
$\therefor... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
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System of quadratic Diophantine equations Is there a method for determining if a system of quadratic diophantine equations has any solutions?
My specific example (which comes from this question) is:
$$\frac{4}{3}x^2 + \frac{4}{3}x + 1 = y^2$$
$$\frac{8}{3}x^2 + \frac{8}{3}x + 1 = z^2$$
I want to know if there are any p... | Your specific example can be formalized in the following way:
$$
\begin{eqnarray}
4x^2+4x+3=3y^2\\
8x^2+8x+3=3z^2\\
4x^2+4x=3(z^2-y^2)\\
(z^2-y^2)=4x(x+1)/3
\end{eqnarray}
$$
let $x+1=3n$
$$
\begin{eqnarray}
(z^2-y^2)=4(3n-1)n
\end{eqnarray}
$$
Let $z=4n-1$ and $y=2n-1$ (more generally if $3n^2-n=uv$, then $z=u+v$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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calculation of $\int_{0}^{1}\tan^{-1}(1-x+x^2)dx$
Compute the definite integral
$$
\int_{0}^{1}\tan^{-1}(1-x+x^2)\,dx
$$
Failed Attempt:
Let $1-x+x^2=t$. Then
$$
\begin{align}
(2x-1)\,dx &= dt\\
dx &= \frac{1}{(2x-1)}dt
\end{align}
$$
Changing the limits of integration, we get
$$\int_{1}^{1}\tan^{-1}(t)\cdot \frac{... | HINT:
As $\displaystyle 1-x+x^2=\frac{(2x-1)^2+3}4>0$ for real $x,$
using this, $\displaystyle \tan^{-1}(1-x+x^2)=\cot^{-1}\left(\frac1{1-x+x^2}\right)$
Now, we know $\displaystyle\cot^{-1}\left(\frac1{1-x+x^2}\right)=\frac\pi2-\tan^{-1}\left(\frac1{1-x+x^2}\right)$
Again, $\displaystyle\tan^{-1}\left(\frac1{1-x+x^2}\r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}... | Use Riemann sums to show that $S\in\Big[\tfrac{\ln2}2,\tfrac12\Big]$ :
$$\sum_{k=1}^n\frac k{k^2+n^2}<\sum_{k=1}^n\frac k{k+n^2}<\sum_{k=1}^n\frac k{n^2}\quad\iff\quad\int_0^1\frac x{1+x^2}dx<S<\int_0^1xdx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/617407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
} |
Need help simplifiying a rational expression There's a math question on an online test which asks the following
Multiply the following expression, and simplify:
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I ca... | As mentioned in the comments, by factoring out $x$ in the second fraction, $x^2+4xy$ becomes $x(x+4y)$. Which leaves you with:
$$\frac{x^2+16y^2}{x} \cdot \frac{x(x+4y)}{x-4y}=\frac{(x^2+16y^2)(x+4y)}{x-4y}$$
Alternately if you had not seen the cancellations and had ended up with $x^3+16xy^2+4x^2y+64y^3$ as your numera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617594",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$? How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$, where $\lfloor \cdot \rfloor$ denotes the greatest integer function?
Please help. I have no idea about this.
| First, notice that the domain of $x$ is
$$2-x^2\ge0\iff-\sqrt2\le x\le \sqrt 2.$$
And the floor function $\lfloor x\rfloor$ is defined as
$$\lfloor x\rfloor=t\iff t\le x\lt t+1.$$
So, in your question, we have
$$\lfloor \sqrt{2-x^2}\rfloor =t\iff t\le\sqrt{2-x^2}\lt t+1$$
Now note that $t\ge 0 \in\mathbb Z$ (This is b... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I ... | Let $x=\frac{a-b}{a+b}$ and so on. We then have: $(1-x)(1-y)(1-z)=(1+x)(1+y)(1+z)$
The terms with an even absolute degree will cancel, therefore: $x+y+z=-xyz$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 1
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find general solution to the Differential equation Find the general solution to the differential equation
\begin{equation}
\frac{dy}{dx}= 3x^2 y^2 - y^2
\end{equation}
I get
\begin{equation}
y=6xy^2 + 6x^2 y\frac{dy}{dx} - 2y\frac{dy}{dx}
\end{equation}
rearrange the equation
\begin{equation}
\frac{dy}{dx} = \frac{y-... | HINT :
$$\frac{1}{y^2}dy=(3x^2-1)dx$$
$$-\frac 1y=x^3-x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $a,b,c$ are positive integers and $a^2+b^2=c^2$ and $a$ is a prime, what can we conclude about primeness of b and c? Let $a,b,c$ be positive integers and they satisfy $a^2+b^2=c^2$, and if $a$ is prime, can we conclude whether $b$ and $c$, are both prime, composite or neither? If yes, why, if not why not?
I can conc... | $5^2+12^2=13^2$ and $7^2+24^2=25^2$, so I don't think there is anything interesting about primality of $b$ and $c$...
What you may say is that, except for trivial cases, $c+b\neq c-b$, hence we must have $c-b=1$ and $c+b=2b+1=a^2$, so $b=\frac{a^2-1}2$ and $c=\frac{a^2+1}2$. Hence, you find that exists exactly one such... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it i... | Let $\zeta_n = e^{2\pi i/n} = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$ be a primitive $n^{\rm th}$ root of unity, so in particular, the roots of $z^5 - 1$ are $\zeta_5^k$ for $k = 0, 1, 2, 3, 4$. Since the sum of the roots of a polynomial of degree $n$ is equal to the negative of the coefficient of the degree $n-1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Calculation of $ \int_{0}^{\sqrt{n}}\lfloor t^2 \rfloor dt\;$, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$. Calculation of $\displaystyle \int_{0}^{\sqrt{n}}\lfloor t^2 \rfloor dt\;$, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$.
$\bf{My\; Try}::$ Let $t^2 = u$ a... | You can obtain $$\frac{1}{2}\int_0^n\frac{\lfloor u\rfloor}{u^{1/2}}\text{d}u =\frac{1}{2}\int_0^n\frac{u-\{u\}}{u^{1/2}}\text{d}u = \frac{1}{2}\int_0^n u^{1/2}\text{d}u - \frac{1}{2}\int_0^n\frac{\{u\}}{u^{1/2}}\text{d}u,$$ which gives
$$\frac{1}{2}\int_0^n\frac{\lfloor u\rfloor}{u^{1/2}}\text{d}u =\frac{1}{2}\left[\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$ Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$
Solution :
$4(2y-x-3)^2 = 4x^2-16xy+24x+16y^2-48y+36$
and $9(2x+y-1)^2 = 36x^2+36xy-36x+9y^2-18y+9$
$\therefore 4(2y-x-3)^2 -9(2x+y-1)^2 = 7y^2+60x -52xy-32x^2-30y+27 =80$
Can we have othe... | With a translation of the origin in the center of the conic (which is the point of intersection of the two lines, you get the equation in the form
$$
4(-X+2Y)^2-9(2X+Y)^2=80
$$
Now the lines are indeed orthogonal: since $1^2+2^2=5$, you can write the equation as
$$
4\cdot5\left(-\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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compute limit (no l'Hospital rule) I need to compute
$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}.$$
I can not use the l'Hospital rule.
| $$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{\cos x} -1+1- \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{1+\cos x-1} -1- (\sqrt[3]{1+\cos x-1}-1)}{\cos x-1}\cdot\frac{\cos x-1}{\sin^2x}=\lim_{x\to 0}[\frac{(1+\cos x-1)^\frac{1}{2}-1}{\cos x-1}-\frac{(1+\cos x-1)^\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove that $\sum_{i=0}^{a}\frac{i\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=\frac{ac(a+b)}{b(b+1)}$ let $$b\ge c,a,b,c\in N^{+}$$
Show that
$$\sum_{i=0}^{a}\dfrac{i\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=\dfrac{ac(a+b)}{b(b+1)}$$
This sum is similar to Hypergeometric distribution,
bu... | Your result follows from some standard results about binomial coefficients
(as found in [1], for example).
I'm not sure of the minimal conditions needed to make the argument work,
but I will assume that $a\geq 0$ and $b>c\geq 1$. I first rewrite your sum, without
the factor $i$ as
$$\sum_{i\geq 0}{a+b-c-i\choose b-c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares. If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.
My work:
$3n+1=x^2$
$3n+3=x^2+2$
$3(n+1)=x^2... | If $\displaystyle 3n+1=a^2, (a,3)=1\implies a$ can be written as $\displaystyle3b\pm1$ where $b$ is an integer
So we have $\displaystyle 3n+1=(3b\pm1)^2\implies n=3b^2\pm2b$
$\displaystyle n+1=3b^2\pm2b+1=b^2+b^2+b^2\pm2b+1=b^2+b^2+(b\pm1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How find this postive integer $a$ such $a(x^2+y^2)=x^2y^2$ always have roots Find all postive integer numbers of $a$,such this equation
$$a(x^2+y^2)=x^2y^2,xy\neq0$$
always have integer roots $(x,y)$
my try: since
$$\dfrac{x^2y^2}{x^2+y^2}\in N$$
and I can't
Thank you
| If $(x,y)=1$, then $(x^2+y^2,x^2y^2)=1$. Suppose $(x,y)=d$, then $\left(\frac xd,\frac yd\right)=1$ and therefore
$$
\left(\frac{x^2+y^2}{d^2},\frac{x^2y^2}{d^4}\right)=1\tag{1}
$$
which implies that both
$$
\left(\frac{x^2+y^2}{d^2},\frac{x^2}{d^2}\right)=1\quad\text{and}\quad\left(\frac{x^2+y^2}{d^2},\frac{y^2}{d^2}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that $n^7\equiv n^3\mod40,\forall n\in\mathbb{Z}$ I have a problem when I try to induction on $\mathbb{Z}$. I don't know how to solve when I try $n-1$ or $n+1$.
| $$n^7-n^3=n^3(n^4-1)=n^2(n^5-n)$$
Using Fermat's Little Theorem, $5|(n^5-n)$
If $n$ is even $8|n^3$
Else $\displaystyle n$ is odd, $=2m+1,$(say),
Both $n+1,n-1$ are even, one is divisible by $4$ and the other is by $2$
Algebraically, $\displaystyle(2m+1)^2=4m^2+4m+1=8\frac{m(m+1)}2+1\implies 8|(n^2-1)$ if $n$ is odd
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Calculate GCD$(x^4+x+1,x^3+x^2)$ and a Bezout Identity in $\mathbb{F_2}$ A really short task:
Calculate GCD$(x^4+x+1,x^3+x^2)$ and a Bezout Identity in $\mathbb{F_2}.$
I've tried it but my GCD is $1$ and I cannot see where my mistake is.
$x^4+x+1= x \cdot (x^3+x^2) + x^3 +x + 1$
$x^3+x^2 = 1 \cdot (x^3 + x + 1) + x^2... | This is easy when using the augmented-matrix form of the extended Euclidean algorithm.
$\begin{eqnarray}
(1)&& &&x^4\!+x+1 \,&=&\, \left<\,\color{#c00}1,\color{#0a0}0\,\right>\ \ \ {\rm i.e.}\,\ \ x^4\!+x+1 = \color{#c00}1\cdot (x^4\!+x+1) + \color{#0a0}0\cdot(x^3\!+x^2)\\
(2)&& && x^3\!+x^2 \,&=&\, \left<\,\color{#c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs
and I perform my steps I get what Wolfram Alpha shows as an alternate solution.
Any help is greatly appreciated
The problem is the following:
S... | $$
\sum_{i=1}^n\frac1{i^2}=1+\sum_{i=2}^n\frac1{i^2}\leqslant1+\sum_{i=2}^n\frac1{i(i-1)}=1+\sum_{i=2}^n\left(\frac1{i-1}-\frac1i\right)=\ldots
$$
Edit: (About the Edit to the question 2014-01-14 21:25:21)
I'd just like to know the mistake I made or next step I should take.
None, neither mistake nor next step. Actual... | {
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Fourier Series of $f(x) = 0$ from $(-\pi, 0)$, $x$ from $(0,\pi)$ I need to determine the fourier series of the following function, (using trig method, not complex)
$$ f(x) = \begin{cases} 0 & \text{if } -\pi < x < 0, \\
x & \text{if } 0 < x < \pi \end{cases} $$
and then use it to show that $$\sum_{n=1}^{\infty} \fra... | The Fourier coefficients are
$$
\begin{align}
a_{0} & = \frac{1}{\pi}\int_{0}^{\pi}xdx = \frac{1}{\pi}\frac{\pi^{2}}{2}=\frac{\pi}{2},\\
a_{n} & = \frac{1}{\pi}\int_{0}^{\pi}x\cos(nx)dx = \frac{1}{\pi}\left[\left.\frac{1}{n}\sin(nx)x\right|_{x=0}^{\pi}-\frac{1}{n}\int_{0}^{\pi}\sin nx\,dx\right] \\
& = \frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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conditions on $\{a_n\}$ that imply convergence of $\sum_{n=1}^{\infty} a_n$ (NBHM 2011) Question is :
For a sequence $\{a_n\}$ of positive terms, Pick out the cases which imply convergence of $\sum_{n=1}^{\infty} a_n$.
*
*$\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$
*$\sum_{n=1}^{\infty} n^2a_n^2<\... | Comment on your first case:
$n^{1.5}a_n$ can converge to 1.5 while always staying above 1.5. But it must eventually go below (say) 3 so that eventually $a_n \leq 3/n^{1.5}$.
Comment on your third case:
You wrote since $a_n \geq 1/n^2$ (which isn't true anyway), comparison test applies which is also false. What you real... | {
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"source": "stackexchange",
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Prove that $f(x)=x$ if the following holds true Let $f\colon\mathbb R \to \mathbb R$ be a continuous odd function such that
1) $f(1+x)=1+f(x)$
2) $x^2f(1/x)=f(x)$ for $x\ne0$.
Prove that $f(x)=x$.
| If we do it in a more "functional equation"-ish approach, we notice that
$$\left(\frac{x+1}{x}\right)^2f\left(\frac{x}{x+1}\right) = f\left(\frac{x+1}{x}\right)$$
On the other hand,
$$\begin{align}f\left(\frac{x}{x+1}\right) &= f\left(1 - \frac{1}{x + 1}\right)\\
&= 1 + \left(-\frac{1}{x+1}\right)\\
&= 1 - \left(\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Coefficient of $x^4$ in multinomial expansion What is the coefficient of $x^4$ in $(1 + x - 2x^2)^7$? What is a quick way to solve this problem using the binomial theorem (I have not learned multinomial theorem)?
| An alternative approach, if you absolutely, positively have to use the binomial theorem, would be to let $a=1+x$ and $b=-2x^2$. Note that in the expansion of $(a+b)^7$ only the terms $a^7$, $a^6b$ and $a^5b^2$ will, when expanded again, eventually contain $x^4$. But how many? $a^7 = (x+1)^7$ should be easy, $a^6b=-2x^2... | {
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Proving $~\sum_\text{cyclic}\left(\frac{1}{y^{2}+z^{2}}+\frac{1}{1-yz}\right)\geq 9$ $a$,$b$,$c$ are non-negative real numbers such that $~x^{2}+y^{2}+z^{2}=1$
show that $~\displaystyle\sum_\text{cyclic}\left(\dfrac{1}{y^{2}+z^{2}}+\dfrac{1}{1-yz}\right)\geq 9$
| Letting $a = x^2, b = y^2, c = z^2$,
it suffices to prove that, for all $a, b, c > 0$ with $a + b + c = 1$,
$$\sum_{\mathrm{cyc}} \left(\frac{1}{b + c} + \frac{1}{1 - \sqrt{bc}}\right) \ge 9.$$
We have
$$\frac{1}{1 - \sqrt{bc}} = \frac{1 + \sqrt{bc}}{1 - bc}
\ge \frac{1 + \frac{2bc}{b + c}}{1 - bc} = \frac{b + c + 2bc}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $x^2 + y^2 + z^2 = 3$ prove that $x/\sqrt{x^2+y+z} + y/\sqrt{y^2+x+z} + z/\sqrt{z^2+x+z} \le \sqrt3$ Given $x^2 + y^2 + z^2 = 3$
Then prove that $${x\over\sqrt{x^2+y+z}} + {y\over\sqrt{y^2+x+z}} + {z\over\sqrt{z^2+x+y}} \le \sqrt 3$$
I tried using the Cauchy-Schwarz inequality but the inequality is coming in oppo... | Use $(a+b+c)^2 \leq 3(a^2+b^2+c^2)$. Then multiply out the denominators and solve.
Alternatively, maximize the left hand side subject to the constraint.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Relative error machine numbers Suppose we are working with a machine that does arithmetical calculations with a relative accuracy of $\xi, |\xi| \leq \xi '$. We want to calculate $a^2 - b^2$ in the following two ways; $(A);$ $a^2 - b^2 = (a - b)(a+b)$ and $(B)$; $a^2 - b^2 = a \times a - b \times b$. Suppose $a$ and $b... | You have to use different $\xi$ with absolute value bounded by the same $\xi'$.
I use the symbols $\oplus$ and $\odot$ for the machine operations. We get
$$(a \odot a )\ominus (b \odot b) = \\
(a^2(1+\xi_1) - b^2(1+\xi_2)) (1+\xi_3) = \\
a^2-b^2+a^2(\xi_1+\xi_3+\xi_1\xi_3)-b^2(\xi_1+\xi_2+\xi_1\xi_2)
$$
so the relativ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve for matrix $A$ in $AB = I$ Given $B$ = $\begin{bmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
1 & 1 & 1
\end{bmatrix}$
I know that $B$ is equal to inverse of $A$, how can I go backwards to solve for $A$ in $AB = I$?
| Using Gauss-Jordan to invert the given matrix $B$ you get
$$\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
1&1&0&0&1&0\\
1&1&1&0&0&1
\end{array}\right)
\leadsto
\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&-1&1&0\\
0&1&1&-1&0&1
\end{array}\right)
\leadsto
\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&-1&1&0\\
0&0&1&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that it is a solution for all values of $t$ Show that $(1-t, 2+3t, 3-2t)$ is a solution for all values of $t$ to following linear system:
$a + b + c = 6$
$a - b - 2c = -7$
$5a + b - c = 4$
I have found that these rows are linearly dependent but I do not know how to continue. Can you help me? Here is after Gaussian... | Hint: One approach is to use Gaussian Elimination, you end up with:
$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & -\dfrac{1}{2} \\0 & 1 & \dfrac{3}{2} & \dfrac{13}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
Can you take it from here?
Update
We have:
*
*$a = -\dfrac{1}{2} +\dfrac{1}{2}c$
*$b = \dfrac{13}{2} -\dfrac{3}{2}c$
... | {
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How prove this $x^2+y^2+z^2+3\ge 2(xy+yz+xz)$ let $x,y,z$ be real numbers, and such $$x+y+z+xyz=4$$ show that
$$x^2+y^2+z^2+3\ge 2(xy+yz+xz)$$
My try: let
$$x+y+z=p,xy+yz+xz=q,xyz=r$$
then
$$p+r=4$$
then
$$\Longleftrightarrow p^2+3\ge 4q$$
But I can't.Thank you
| Note that it is sufficient to consider $x, y, z \ge 0$. Then the condition yields $4 = x+y+z+xyz \ge 3\sqrt[3]{xyz}+xyz \implies xyz \le 1$.
Also note that among $x-1, y-1, z-1$, at least two have the same sign. WLOG let $(y-1)(z-1) \ge 0$. Then we have:
$$(x-1)^2+(y-z)^2+2x(y-1)(z-1) \ge 0$$
$$\implies x^2+y^2+z^2+... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$,
$$4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$$
I've tried expressing it as a sum of squares, but haven't got anywhere.
Hints are also welcome.
| WLOG, let $x$ be the smallest, $y= a+x$ and $z = b+x$ for $a, b \ge 0$. Then we have
$$4(x + y + z)^3 - 27(x^2y + y^2z + z^2x) = 27x^3+27(a+b)x^2+9(a+b)^2x + (a-2b)^2 (4a+b)>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Question about writing integrals as a sum of integrals? So the question that is given is as follows:
Write the integral as a sum of integrals without absolute values and evaluate
$$ \int_{-2}^3 |x^3| \, dx$$
Can someone help me figure this out because I keep getting $20.25$ or $\dfrac{81}{4}$ but the answer is supposed... | Note that $|x^3| = -x^3$ when $x \le 0$. We can therefore split the integral into two as follows:
$$\int_{-2}^3 |x^3| dx = \int_{-2}^0 |x^3| dx + \int_0^3 |x^3| dx = -\int_{-2}^0 x^3 dx + \int_0^3 x^3 dx = -\frac{x^4}{4}\bigg|_{-2}^0 + \frac{x^4}{4}\bigg|_0^3$$
$$= -\left(\frac{0}{4}-\frac{16}{4}\right) + \left(\frac{8... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdo... | $$6\cdot7^{n+1}-2\cdot3^{n+1}=7(6\cdot7^n-2\cdot3^n)+14 \cdot 3^n-2 \cdot 3^{n+1}=7(6\cdot7^n-2\cdot3^n)+8 \cdot 3^n$$
The first term is divisible by 4 by inductive assumption and $8= 4 \cdot 2$
| {
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"timestamp": "2023-03-29T00:00:00",
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$n$th derivative of $e^x \sin x$ Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:
Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.
Taking the... | Yes. Your derivation is correct. Here is a distillation that might make it simpler to see what is going on:
$$
\begin{align}
\frac{\mathrm{d}^n}{\mathrm{d}x^n}e^x\sin(x)
&=\frac{\mathrm{d}^n}{\mathrm{d}x^n}\frac1{2i}\left(e^{(1+i)x}-e^{(1-i)x}\right)\\
&=\frac1{2i}\left((1+i)^ne^{(1+i)x}-(1-i)^ne^{(1-i)x}\right)\\
&=\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Field of order 8, $a^2+ab+b^2=0$ implies $a=0$ and $b=0$. I was able to come up with a proof for this problem however, it seems like my argument can work for any field of even order and not just odd powers of 2 so I'm convinced there is something wrong here. Can someone verify or see where the error in reasoning is?
Pr... | The mistake you make is:
$$\frac{2^n}{2}(a^2+b^2) = \frac{2^n}{2}ab \Rightarrow a^2+b^2 =ab $$
Note that your field has characteristic $2$, which means that $\frac{2^n}{2}=0$!
You divide again by $0$ in the last line.
Hint
$$a^3-b^3=(a-b)(a^2+ab+b^2)=0$$
Thus $a^3=b^3$, and you also know what $a^7, b^7$ are....
You ask... | {
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"timestamp": "2023-03-29T00:00:00",
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Area of a quadrilateral inside right angled triangle $ABC$ is right angle triangle. $AB=24 cm$, $BC=10 cm$, $AC=26 cm$.
Point $D$ on $AC$ (hypotenuse) bisects $AC$ and connects point $E$ on side $AB$ such that $ED$ is perpendicular to $AC$. Side $AC$ is folded into half so that angle $A$ falls on angle $C$, creating li... |
Let us look at $\triangle AED$ and $\triangle CED$:
*
*$\angle EDA = \angle EDC = 90\,^{\circ}$
*$DC= DA = 13$ cm
*$ED$ is common
So, by $SAS$ congruence, those two triangles are congruent, and hence $EC = AE$. Now let $BE$ be $x$. So, by Pythagoras Theorem, $(EC)^2=x^2+100$. But, $(EC)^2= (AE)^2 = (24-x)^2$. T... | {
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Evaluate of number represented by the infinite series $\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$. Evaluate of number represented by the infinite series
$$\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$$
| HINT:
Let $a = \sqrt{1/3 + ...}$
Then $a = \sqrt{1/3 + a}$
| {
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Express the following in terms of y. If $y = 2^{2x}$, express the following in terms of y.
$2^{2x-1} - 4^{2x+1} + 16^{x-1}$
I began it in this way:
$2^{2x}\cdot 2^{-1} - (2^2)^{2x+1} + (2^4)^{x-1}$
$2^{2x}\cdot \frac{1}{2} - 2^{4x}\cdot2^2 + 2^{4x}\cdot2^{-1}$
$2^{2x}\cdot\frac{1}{2}-2^{2x}\cdot2^{2x}\cdot2^2+2^{2x}\cd... | Almost. Note that $$(2^4)^{x-1}=2^{4(x-1)}=2^{4x-4}=2^{4x}\cdot2^{-4}\ne2^{4x}\cdot 2^{-1}.$$ Also note that you can gather like terms (specifically, the $y^2$ terms).
| {
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"answer_id": 0
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Polynomial $P(x)$ such that $P(3k)=2$, $P(3k+1)=1$, $P(3k+2)=0$ for $k=0,1,2,\ldots,n-1$, $P(3n)=2$, and $P(3n+1)=730$ Let $n$ be a positive integer such that there exists a polynomial $P(x)$ over $\mathbb{Q}$ of degree $3n$ satisfying the conditions below:
$$P(0) = P(3) = \ldots = P(3n) = 2\,,$$
$$P(1) = P(4) = \ldot... | Let $\omega:=\exp\left(\frac{2\pi\text{i}}{3}\right)=\frac{-1+\sqrt{-3}}{2}$. Define $$\tau(z):=\left(\frac{1-\omega^2}{3}\right)z^2+\left(\frac{1-\omega}{3}\right)z+1\text{ for all }z\in\mathbb{C}\,.$$
Note that, for each $k\in\mathbb{Z}$, we have
$$\tau(\omega^k)=\left\{
\begin{array}{ll}
2&\text{if }k\equiv0\pmod{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s e... | The way I approached this problem is dividing the sum into two halves. Where the first half contains all the numbers starting from $1$ and reaching upto $m$ such that $\sqrt{m}$ is greatest integer $<\lfloor\sqrt{n}\rfloor$, and second half contains the leftover numbers. Intuitively $m$ can be calculated by using simpl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 5
} |
Write your answer in simplest Simplify: sqrt(2009*2011*2015*2017+36)+10. Write your answer in simplest form
| Let $\displaystyle\frac{2009+2011+2015+2017}4=2013=a$
$\displaystyle\implies 2009\cdot2011\cdot2015\cdot2017=(a-4)(a-2)(a+2)(a+4)$
$\displaystyle=(a^2-16)(a^2-4)=a^4-20a^2+64$
Now, $\displaystyle 2009\cdot2011\cdot2015\cdot2017+36$
$\displaystyle=a^4-20a^2+64+36=(a^2)^2+10^2-2\cdot a^2\cdot10=(a^2-10)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Number of factors of summation Let $a(n)$ be the number of $1$'s in the binary expansion of $n$. If $n$ is a positive integer, show that
$$\Bigg|\sum_{k=0}^{2^n-1}(-1)^{a(k)}\times 2^k\Bigg|$$
has at least $n!$ divisors.
I think this can be solved using induction, but I'm not sure. The first three values of the sum a... | If we set
$$S(n) = \sum_{k=0}^{2^n-1} (-1)^{a(k)}\cdot 2^k,$$
then we note that $a(2^n + k) = a(k) + 1$, and hence
$$S(n+1) = \sum_{k=0}^{2^{n+1}-1} (-1)^{a(k)}\cdot 2^k = \sum_{k=0}^{2^n-1} (-1)^{a(k)}\cdot \left(2^k - 2^{2^n+k}\right) = -\left(2^{2^n}-1\right)\cdot S(n).$$
Thus
$$\lvert S(n)\rvert = \prod_{k=0}^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A problem about power series and big-O The problem is:
Prove: There exist constants $a$, $b$ such that $\frac{z^3-5z^2+3z}{(z+2)^3}=1+\frac{a}{z}+\frac{b}{z^2}+O(\frac{1}{z^3})$ as $z\rightarrow \infty$ and find an explicit values for $a$ and $b$.
My thought: we know the power series of $\frac{1}{1-z}$ so by taking de... | Take the series for $(1+\frac{2}{z})^{-3}$, write out the terms up to $(1/z)^3$, multiply each of the terms by $1-5/z+3/z^3$ and collect like terms (for the terms after $(1/z)^3$, note that if you hit them with a 3rd degree polynomial in $1/z$, their contributions are at most $(1/z)^4$ (more precisely, if you hit $(1/z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do I determine the sign of $\sin\theta$ in this question.
$\alpha$ and $\beta$ are two different values of $\theta$ lying between $0$ and $2\pi$ which satisfy the equation $\begin{equation}6\cos\theta + 8\sin\theta = 9\end{equation}$. Find $\sin(\alpha+\beta).$
I solved and got $\cos(\alpha+\beta) = \frac{-28}{... | Your book is correct. We don't need to find either $\alpha$ or $\beta$, i.e. to solve the given linear equation in $\cos\theta$ and $\sin\theta$, which could be transformed into a quadratic equation in $\tan\frac{\theta}{2}$.
[EDIT. Since you've commented that your course was a beginner's one, I edited this answer to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Double integrals in polar coordinates Determine the domain of
$$\!\!\!\!\!\!\!{\small
D \equiv
\left\{\left(x,y\right) \in \mathbb{R}^{2}\
{\large\mid}\
x \in \left[\,{-\,\frac{1}{\,\sqrt{\,{2}\,}\,},
\frac{1}{\,\sqrt{\,{2}\,}\,}}\,\right],\
y \in \left[\,{\left\vert\,{x}\,\right\vert,
\,\sqrt{\,{1 - x^{2}}\,}\,}\,\rig... | Draw a picture. A simple plot reveals that the domain $D$ is simply the sector of the circle $r=1$ between two values of $\theta$. A little thought provides those values of $\theta$ (i.e., what purpose does the absolute value serve?).
The integrand you show is also wrong, as $1+r^2 \ne 2$.
The answer I get is $(\pi/4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Agai... | I think you can check this inequality by Lagrange multiplier method. Consider the following Lagrange function:
\begin{equation}
L(a,b,c)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)+\lambda(a+b+c-3)
\end{equation}
For finding the extreme value, we have:
\begin{equation}
\frac{\partial L}{\partial a}=2ab^2c^2+2(3-2b)(3-2c)+\lambda=0\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Taylor series of $\frac 1 {1+x^2}$ I have to construct the Taylor series of
$$\frac 1 {1+x^2}$$
around $0$ and $1$ and analyze the convergence in both cases. Also (but this is a consequence of the previous series) I have to construct the Taylor series of
$$arctan(x)$$
What I have so far:
*
*I know that the Taylor po... |
If I have the power series of a function f such as $F'=f$, I can construct the power series of $F$ with $F(x)=F(a)+\sum_{n=1}^{\infty}\frac {a_n} {n+1} (x-a)^{n+1}$
Have you tried this? Note that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$.
You have the power series for $\frac{1}{1+x^2}$ centered at $0$, for which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve this trig problem? $\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$ Basic trig problem my brother ask me, but I don't know how to do it:
$$\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$$
| $\sec\left(\sin^{-1}\left(\dfrac{-5}{13}\right)-\tan^{-1}\left(\dfrac{4}{3}\right)\right)$
Let $\alpha = \sin^{-1}\left(\dfrac{-5}{13}\right)$
Then $\alpha$ is the angle in the fourth quadrant that corresponds to the point
$(x,y) = (12,-5)$ with amplitude $r = 13.$
So $\cos(\alpha) = \dfrac xr = \dfrac{12}{13}$ and
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/678609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
| By Holder
$$\sum_{cyc}\frac{a^3}{bc}\geq\frac{(a+b+c)^3}{3(ab+ac+bc)}=\frac{(a+b+c)\cdot(a+b+c)^2}{3(ab+ac+bc)}\geq a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/679544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Finding intersection points of 2 functions. My method is incomplete. These are the 2 functions :
$y = x^{4}-2x^{2}+1$
$y = 1-x^{2} $
Here's how I solved It :
$x^{4}-2x^{2}+1 = 1-x^{2}$
$x^{4}-x^{2} = 0$
$x^2(x^2-1)=0$
$x^2-1=0$
$x=\pm \sqrt{1} $
Value of $y$ when $x=1$
$y=1-x^2\\y=1-1\\y=0$
Value of $y$ when $x=(-1)$
$... | Functions:
a.) y = x^4 - 2x^2 + 1
b.) y = 1 - x^2
Solving:
x^4 - 2x^2 + 1 = 1 - x^2
x^4 - x^2 = 0
x^2 * ( x^2 - 1) = 0
Now set each part equal to 0
x^2 = 0
x = 0
x^2 - 1 = 0
x = +/-1
Now plug in each value of x to get the corresponding value of y
X = 0
y = 1 - (0)^2
y = 1
Intersection at (0,1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z},\ n \neq 0} \frac{1}{z-n}+ \frac{1}{n}$ I'm reading the proof that $$\pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z},\ n \neq 0} \frac{1}{z-n}+ \frac{1}{n}$$
There is a function $$h(z) =\pi \cot (\pi z) -[ \frac{1}{z} + \sum_{n \in \mathbb{Z},\ n \neq 0... | Consider the function
$$d(z) = \pi \cot \pi z - \frac{1}{z-1}$$
in a neighbourhood of $1$. Writing $z = 1+h$, we have
$$\begin{align}
d(1+h) &= \pi\frac{\cos \pi(1+h)}{\sin \pi(1+h)} - \frac{1}{h}\\
&= \pi \frac{-\cos \pi h}{-\sin \pi h} - \frac{1}{h}\\
&= \frac{1 - \frac{\pi^2h^2}{2} + O(h^4)}{h\left(1 - \frac{\pi^2h^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do i solve the cubic equation?
$$x^3 - 3x^2 - 3x +2 = 0$$
The rational root test does not work; there are no rational roots.
| Here is a step by step procedure.
First shift $x$ so there is no quadratic term. To do this we shift by 1/3 the quadratic term (assuming leading coefficient is 1) This give
$$
x=y+1 \Rightarrow y^3-6\,y-3 =0
$$
We now reduce the cubic by the substitution $y = A(z+1/z)$
$$
y = A(z+1/z) \Rightarrow \frac{z^6\,A^3+3\,z^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Functional equation of function defined over non-negative reals satisfying $f \big(xf(y)\big)f(y)=f(x+y)$, $f(2)=0$ and $f(x)\ne0$ for $x\in[0, 2)$
Find all $ f:[0,\infty)\to [0,\infty) $ such that $ f (2)=0 $, $ f (x)\ne 0 $ for $ x\in [0, 2) $ and
$$ f \bigl(xf (y)\bigr) f (y)=f (x+y) $$
for all $ x, y\ge 0 $.
I tr... | Substitute in $x=y=0$, we get that $ f(0) f(0) = f(0)$. Since $f(0) \neq 0$, thus $f(0) = 1$.
Substitute in $x=x, y = 2$, $0 = f(x\cdot 0) \cdot 0 = f(x+2)$. Hence for $x \geq 2, f(x) = 0 $.
We now focus our attention to the region $x\leq 2$.
Substitute in $x=2-y, y = y < 2$. We get that $ f[ (2-y) f(y) ] f(y) = f(2) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Problem related to DE Solve and comment on the solution behaviour at $|x|$ approaches infinity (bounded or unbounded as $x$ approaches $\pm\infty$.
$$\frac{1}{1+x^2} + \sin y + y'\left(x\cos y + \frac{y^2}{2}\right)=0,\quad y(1)=\Pi.$$
Can I have some hints on how to get started on this problem? Thanks
I cannot even se... | Deal with the LHS of the ODE and we can obtain the solution
\begin{equation}
\frac{1}{1+x^2}+(\sin(y)+y'x\cos(y))+y'\frac{y^2}{2}=0\\
\frac{1}{1+x^2}+(x\sin(y)+\frac{y^3}{6})'=0\\
\int(x\sin(y)+\frac{y^3}{6})'dx+\int\frac{1}{1+x^2}=0\\
x\sin(y)+\frac{y^3}{6}+\arctan(x)+C=0
\end{equation}
which $C$ is a constant given b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/682918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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I cannot find the derivatives of these functions using the product rule? I need help finding the derivative of $y=x^3(3x+7)^2$ at $x=-2$
I tried to simplify the function to $y=x^3 (3x+7)(3x+7)$ and the simplify it into two terms and the derivatives of those terms using the product rule, but that doesn't work.
Since I d... | The product rule is: if $f(x) = g(x)h(x)$, then $f'(x) = g'(x)h(x)+g(x)h'(x)$. Extending it to three terms, we get:
$$f(x) = g(x)h(x)j(x) \implies f'(x) = g'(x)\left[h(x)j(x)\right]+g(x)\left[h(x)j(x)\right]' \\ = g'(x)h(x)j(x)+g(x)\left[h'(x)j(x)+h(x)j'(x)\right] \\
= g'(x)h(x)j(x)+g(x)h'(x)j'(x)$$
as we might expect... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
I determined the coefficients of the Fourier series, which are
$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}... | From your identity
\begin{equation*}
\frac{3\pi ^{3}}{4}=\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{
\pi n^{4}}(-1)^{n}
\end{equation*}
expanding the right hand side and using the result $\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi^2}{6}$, we get
\begin{eqnarray*}
\frac{3\pi ^{4}}{4} &=&\sum_{n=1}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/687676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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Prove that $a^ab^b>\left(\frac{a+b}{2}\right)^{a+b}$ where $a\ne b$
Prove that $a^a\cdot b^b>\left(\dfrac{a+b}{2}\right)^{a+b}$ where $a\ne b$.
My work:
$$a^a\cdot b^b>\left(\frac{a+b}{2}\right)^a\cdot\left(\frac{a+b}{2}\right)^b\implies 1>\left(\frac{1+\frac{b}{a}}{2}\right)^a\cdot\left(\frac{1+\frac{a}{b}}{2}\r... | Take the function $f(x)=x\ln x,$ where $x>0$, then $f'(x)=1+\ln(x)$ and $f''(x)=\frac{1}{x}>0,$ so $f(x)$ is convex functon. Apply jensen-inequality in $f$, we have $$\dfrac{a\ln a + b\ln b}{2}\ge \dfrac{(a+b)}{2}\ln\bigg(\dfrac{a+b}{2}\bigg),$$
on simplification and taking anti-log on both sides it becomes, $a^ab^b\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/689667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$ Assume I have a function $f(n) = \frac{4n+1}{n(2n-1)}$ with $n \in \mathbb{N} \setminus \left\{ 0 \right\}$. The objective is to find all $n$ for which $f(n)$ has a proper decimal fraction. I know that any given fractiononly has a proper decimal fraction whenever the d... | Note that $\gcd(4n+1,n)=1$ and $\gcd(4n+1,2n-1)=\gcd(3,2n-1)\in\{1,3\}$.
Thus $n$ must be of the form $n=2^a5^b$ and the odd number $2n-1$ must be of the form $2n-1=5^c$ or $2n-1=3\cdot 5^c$ with $a,b,c\in\mathbb N_0$. So from $1=2\cdot n-(2n-1)$ we find
$$\tag1 1 = 2^{a+1}5^b-5^c\qquad \text{or}\qquad 1=2^{a+1}5^b-3\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/691325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Reflection across the plane Let $T: \Bbb R^3 \rightarrow \Bbb R^3$ be the linear transformation given by reflecting across the plane $S=\{x:-x_1+x_2+x_3=0\}$ (...)
Then, $S={\rm gen}[(1,1,0),(1,0,1)].$
But how can I get the matrix $R_v$ such that reflects across $S$?
Thanks!
| The vector $\vec{n}=\langle -1,1,1 \rangle$ is normal to the plane, thus $T$ maps $\vec{n}$ to $-\vec{n}=\langle 1,-1,-1 \rangle$. It also fixes the vectors $\vec{u}= \langle 1,1, 0 \rangle$ and $\vec{v} = \langle -1,1,-2 \rangle$ that are perpendicular to each other and to $\vec{n}$. More generaly, $\vec{u}$ could be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Infinite series for $ \sqrt 2 $ What is infinite series for $ \sqrt 2 $? I don't mean continued fraction. That kind of series such as like for $e, \pi, $etc.
| $$\sqrt{2}=\frac{1}{\left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{6^2}\right) \left(1-\frac{1}{10^2}\right) \left(1-\frac{1}{14^2}\right) \cdots}$$
$$\sqrt{2}=\left(1+\frac{1}{1}\right) \left(1-\frac{1}{3}\right) \left(1+\frac{1}{5}\right) \left(1-\frac{1}{7}\right) \cdots$$
$$\sqrt{2}=1+\frac{1}{2}-\frac{1}{2\cdot4}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/694699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 3
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Different methods of evaluating $\int\sqrt{a^2-x^2}dx$: Is there a simple and nice way to solve $\int\sqrt{a^2-x^2}dx$:
PS:I am not looking for a substitution like $x=a\sin p$,
| Another way to attack such kinds of integrals is using Chebyshev's substitution. Consider the integral:
$$
I_{m,p} = \int x^m (\alpha x^n + \beta)^pdx
$$
Chebyshev has shown that $I_{m, p}$ is expressed in elementary functions if and only if one of the following cases is satisfied:
$$\begin{align*}
&1)\ p \in \Bbb Z \\... | {
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"url": "https://math.stackexchange.com/questions/694853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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issues with simple algebraic equations $ab + a + b = 250$
$bc + b + c = 300$
$ac + a + c = 216$
then find $a + b + c = ?$
MY APPROACH:
(i) * c , (ii) * a , (iii) * b then we get
$abc + ac + bc = 250c$
$abc + ab + ac = 300a$
$abc + ab + bc = 216b$
(iv)+(v)+(vi)
$3abc + 2ac +2ab + 2bc = 300a + 216b +250c$
now i c... | \begin{cases}
ab+a+b=250 \\
bc+b+c=300 \\
ac+a+c=216 \\
\end{cases}
it seems of no difficulty: a system of 3 equations and 3 variables...
Let's start by adding 1 to every equations, we obtain:
\begin{cases}
ab+a+b+1=(a+1)(b+1)=251 \\
bc+b+c+1=(c+1)(b+1)=301 \\
ac+a+c+1=(a+1)(c+1)=217 \\
\end{cases}
Now let's substitute... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Simplfying $\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$ I am trying to simplify the expression:
$\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$
I tried to square the expression but I can't do that because it is not an equation so I got stuck. Can someone please give me some pointers on how to proceed?
| The general way to evaluate these kinds of expressions is to try to factor the contents of the square root into a perfect square.
One can try to find the factorization by letting $31 + 8\sqrt{15} = (a + b\sqrt{15})^2 = (a^2 + 15b^2) + 2ab\sqrt{15}$. We then test some possible values of $a$ and $b$. In this case, we see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. Show that if $m,n$ are positive integers and $m$ is odd, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
(Hint: Let $s=1^m+2^m+\cdots+(n-2)^m+(n-1)^m$. Obviously $s=(n-1)^m+(n-2)^m+\cdots+2^m+1^m$.
Consider t... | Hint: if $m$ is odd, then $(n-i)^m +i^m \equiv (-1)^mi^m+i^m\equiv -i^m+i^m\equiv 0 \bmod n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$ $$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$
I think to do :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$
but i dont get anything.
Or to divied by $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$
| Hint:
$$\sqrt3\sin x-\cos x=\sqrt2\iff \sin\frac\pi3\sin x-\cos\frac\pi3\cos x=\frac{\sqrt2}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/698964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
What is the average pathlength and probability to randomly traverse any given graph?
To get specific first off, it's about this graph:
I want to get from $A$ to $B$. Every edge has the same length (e. g. 1 m). The shortest walk from $A$ to $B$ is easily found ($A-2-5-B$ and $A-3-6-B$). But how can I calculate the av... | It's difficult to claim an average because there exist many possible infinite length paths that never reach $B$. You may want instead to calculate the probability that you have reached $B$ after $n$ steps.
@draks ... is on to something, but if you're going to determine the probably that you have "seen" node $B$ afte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/702178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove that $(1-\frac{1}{2^2}\cdots 1-\frac{1}{9\,999^2})(1-\frac{1}{10\,000^2})=0.500\,05$ Prove that $\displaystyle\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{9\,999^2}\right)\left(1-\frac{1}{10\,000^2}\right)=0.500\,05$
Here are all my attempts to solve this problem:
So the first t... | Note that you can factor $n^2 - 1 = (n - 1)(n + 1)$ to find something like
\begin{align*}
\left(1 - \frac 1 {2^2}\right)&\left(1 - \frac 1 {3^2}\right) \cdots \left(1 - \frac 1 {10000^2}\right) \\ & =\left(\frac 1 2 \cdot\frac 3 2\right) \left(\frac 2 3 \cdot\frac 4 3\right) \left(\frac 3 4 \cdot \frac 5 4\right) \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"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.