Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since ... | Note that no two of $(-a+b+c)$, $(a-b+c)$, and $(a+b-c)$ can be negative. If so, then one of
$$
\begin{align}
(a-b+c)+(a+b-c)&=2a\\
(a+b-c)+(-a+b+c)&=2b\\
(-a+b+c)+(a-b+c)&=2c
\end{align}
$$
would be negative, but each of $a$, $b$, and $c$ is positive. Thus, at most one can be negative. If only one were negative, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 1
} |
Proving inequality on functions $x-\frac{x^2}{2}<\ln(1+x)To prove: $$x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)},\quad\forall x>0$$
I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0.
Is there a cleaner /... | Another way to do the lower bound, for example, would be to consider $f(x) = x - \dfrac{x^2}{2} - \ln(1+x)$. $f(0) = 0$.
But also $f'(x) = 1 - x - \dfrac{1}{1+x} = \dfrac{(1-x)(1+x) - 1}{1+x} = \dfrac{1 - x^2 - 1}{1+x} = \dfrac{-x^2}{1+x} < 0$.
So since $f(0) = 0$ and $f' < 0$, $f$ is monotonically and strictly decreas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$3\sin^2x=\cos^2x;$ $ 0\leq x\leq 2\pi$ Solve for $x$
$3\sin^2x=\cos^2x;$ $0\leq x\leq 2\pi$ Solve for $x$:
I honestly have no idea how to start this. Considering I'm going to get a number, I am clueless. I have learned about $\sin$ and $\cos$ but I do not know how to approach this problem. If anyone can go step-by-s... | Hint: $\cos^2 x=1-\sin^2 x$.${}{}{}{}{}{}{}$
Substitute. We get after some simplification $4\sin^2 x=1$. Can you finish from here?
Added: You just need to find $\sin^2 x$, then $\sin x$. You should get $\sin x=\pm\frac{1}{2}$. Then identify the angles from your knowledge about "special angles." One of the angles will... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now ... | Let's first find the general form of the terms of the sum
$1 + 2+ 4 + 7 + 11 +\ldots.$
The terms obey the recurrence
$a_n = a_{n-1} + n$, where $a_0 = 1.$
Using standard techniques we find
$$a_n = \frac{1}{2}n(n+1) + 1.$$
These are basically the triangular numbers, as indicated by @anon in the comments.
The sum of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
Please help me integrate the following: $\int \frac{y^2 - x^2}{(x^2 + y^2)^2}dy$ I'm self-studying a Cramster solution and I came across this integral and I don't know what they've done with it. Help would be appreciated.
$$\int \frac{y^2 - x^2}{(x^2 + y^2)^2} ~dy.$$
| In an integral $dy$, $x$ is a constant. Rewirite this as:
$$\int\frac{y^2-x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{y^2+x^2-2x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{dy}{x^2+y^2}-2x^2\int\frac{dy}{\left(x^2+y^2\right)^2}$$
Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating :$\int \frac{1}{x^{10} + x}dx$ $$\int \frac{1}{x^{10} + x}dx$$
My solution :
$$\begin{align*}
\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\
&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\
&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C
\end{align*}$$
Is there complete... | Let we generalise the problem with a slightly different way. Consider
$$\int\frac{\mathrm dx}{x^n+x}$$
Making substitution $x=\frac{1}{y}$ and $z=y^{n-1}$ then reverse it back we get
\begin{align}
\int\frac{\mathrm dx}{x^n+x}&=-\int\frac{y^{n-2}}{1+y^{n-1}}\mathrm dy\\[9pt]
&=-\frac{1}{n-1}\int\frac{\mathrm dz}{1+z}\\[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?
I know how to prove
$ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $
in this way:
$ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$
... | Using Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,
$$\tan20^\circ\tan(60^\circ-20^\circ)\tan(60^\circ+20^\circ)=\tan(3\cdot20^\circ)$$
and $\tan80^\circ=\cot(90^\circ-80^\circ)=\dfrac1{\tan?}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$
Prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$.
Do I use the terms $x= r^2 - s^2$, $y = 2rs$, and $z = r^2 + s^2$ to prove this problem?
Thanks for any help.
| Actually, a very strong statement is true, and quite easy: given any positive integer $n,$ such that all prime factors $p$ of $n$ satisfy $p \equiv 1 \pmod 4,$ then there is a solution to $$ x^2 + y^2 = n $$ with $$ \gcd(x,y) = 1.$$
For $n$ a prime (so itself $1 \pmod 4$) the existence of a solution was finally pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving $\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$ for $a,b>1$ I'm trying to prove rigorously the following:
$$\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$$ for integ... | This theorem is not true if $b$ is not an integer. Take $x=b=1.5$ and take $a=1$.
If $b$ is an integer, this follows from the rule $$\left\lfloor \frac{y}{b}\right\rfloor = \left\lfloor \frac{\lfloor y \rfloor}b\right\rfloor$$
Setting $y=\frac x a$.
Showing this rule, then, suffices.
Let $y = \lfloor y \rfloor + \{y\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find function to make maximum value Let ${f : [0, 1] \rightarrow [-1, 1] }$ is a continuous function such that ${ \int_{0}^{1} x f \left(x\right) dx =0}$
Find $f(x)$ such that ${ \int_{0}^{1} \left(x ^{2 } + \frac{1}{4} \right) f \left(x\right) dx}$ has the maximum value.
| Disregard continuity. Take $f = -1 + g$ so $0 \le g \le 2$ and you want to maximize $F = \int_0^1 (x^2+1/4) (g(x)-1)\ dx$ subject to $C = \int_0^1 x g(x) \ dx = \int_0^1 x\ dx = 1/2$. Since $x^2+1/4$ is strictly convex while $x$ is linear, if you have a bit of "mass" somewhere in $(0,1)$, you can increase $F$ while k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute integral $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ I want to solve $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ but I get the wrong results:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x =
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \math... | $$ I:=\int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}}\ dx$$
Let $u=e^x, du = e^x \ dx$, leaving us with:
$$\int_{e^{-6}}^{e^{6}} \frac{\left( 4u^2 + 2 \right)^2}{u^3} \ du$$
Expand the numerator to get
$$\int_{e^{-6}}^{e^{6}} \frac{16u^4 + 16u^2 + 4}{u^3} \ du$$
Since the highest power in the numerator is greater than the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Determine the sum $T=a_0+a_1+a_2+...+a_{2012}$ Let ${a_n}$, $n \ge 0$ be a sequence of positive real numbers, given by
$a_0=1$ and
$a_m<a_n$ for all $m,n \in \mathbb{N}, m<n$ with
$a_n=\sqrt{a_{n+1}a_{n-1}}+1$ and $4\sqrt{a_n}=a_{n+1}-a_{n-1}$ for all $n \in \mathbb{N}, n\neq 0$.
Help me, determining the sum $T=a_0+a_... | From the conditions given, we have
$$
(a_n-1)^2=a_{n+1}a_{n-1}\tag{1}
$$
and
$$
16a_n=(a_{n+1}-a_{n-1})^2\tag{2}
$$
Adding $4$ times $(1)$ to $(2)$ yields
$$
4(a_n+1)^2=(a_{n+1}+a_{n-2})^2\tag{3}
$$
Taking the square root of each side and subtracting $2a_n$ from both sides, we get
$$
a_{n+1}-2a_n+a_{n-1}=2\tag{4}
$$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For which angles we know the $\sin$ value algebraically (exact)? For example:
*
*$\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
*$\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
*$\sin(30^\circ) = \frac{1}{2}$
*$\sin(45^\circ) = \frac{1}{\sqrt{2}}$
*$\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt... | Starting with $\tan(\pi/3)=\sqrt{3}$ and $\tan(\pi/4)=1$ and using
$$
\tan(x/2)=\frac{\sqrt{1+\tan^2(x)}-1}{\tan(x)}\tag{1}
$$
and
$$
\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}\tag{2}
$$
and
$$
(\cos(x),\sin(x))=\frac{(1,\tan(x))}{\sqrt{1+\tan^2(x)}}\tag{3}
$$
we can construct the sine and cosine of all rationa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ Show that
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$
I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox.
If the series continues infin... | consider,$$f(n)=\frac{1}{(2n-1) \cdot (2n-1+2)}$$
where $n$ is a natural number
$$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty\frac{1}{(2n-1) \cdot (2n+1)}$$
Let, $$\sum_{n=1}^{\infty} f(n) = S$$
i.e. $$S=\sum_{n=1}^\infty\frac{1}{(2n-1)\cdot(2n+1)}$$
i.e. $$S=\sum_{n=1}^\infty(\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
What is the units digit of $13^4\cdot17^2\cdot29^3$?
What is the units digit of $13^4\cdot17^2\cdot29^3$?
I saw this on a GMAT practice test and was wondering how to approach it without using a calculator. Thanks.
| $13^1$ will give $3$ at the units place
$13^2$ will give $9$ at the units place, we get $9$ by taking the units digit of the result $3^2=9$
$13^3$ will give $7$ at the units place, we get $7$ by taking the units digit of the result $3^3=27$
$13^4$ gives $1$ at the units place, we get $1$ by taking the units digit of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ I'm having trouble computing the integral:
$$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$
I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.
An... | We can write the integrand as
$$\begin{equation*}
\frac{1}{1+\cot x}
\end{equation*}$$
and use the substitution $u=\cot x$. Since $du=-\left( 1+u^{2}\right) dx$ we reduce it to a rational function
$$\begin{equation*}
I:=\int \frac{\sin x}{\sin x+\cos x}dx=-\int \frac{1}{\left( 1+u\right)
\left( u^{2}+1\right) }\,du.
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "88",
"answer_count": 8,
"answer_id": 2
} |
$f(n)=7^{6n} - 6^{6n}$, where $n$ is a positive integer. $f(n)=7^{6n} - 6^{6n}$, where $n$ is a positive integer. find the divisors of $f(n)$ for odd and even values of $n$. Is there a general solution for the divisors.
$$f(1)=7^6-6^6=(7^3)^{2}-(6^3)^{2}$$
$$f(1)=(7-6)(7^2+(7)(6)+6^2)(7^3+6^3)$$
$$f(1)=(1)(127)(7^3+6^3... | As you have said, you always have $7^{6n}-6^{6n}=(7^{3n})^2-(6^{3n})^2=(7^{3n}-6^{3n})(7^{3n}+6^{3n})$
As Mark Bennet has said $7^{3n}+6^{3n}=(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$
Also, $7^{3n}-6^{3n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})$
So we have $7^{6n}-6^{6n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/182809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)$? I would like to evaluate the sum
$$
\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)
$$
Where $\operatorname{Si}$ is the sine integral, defined as:
$$\operatorname{Si}(x) := \int_0^x \frac{\sin t}{t}\, dt$$... | A bit late, but here's a more general calculation, which explains where the mysterious $\frac{\pi}{4}$ in Raymond Manzoni's answer comes from:
We will use the Fourier series
$$ \sum \limits_{n=1}^\infty \frac{\cos (ny)}{n} = - \log \left[2 \left \lvert \sin \left(\frac{y}{2}\right) \right \rvert \right] \, , \, y \in \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 4
} |
Integrating $\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}$ I am bugged by this problem: how do I evaluate this?
$$\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}.$$
A closed form will be convenient and fine. Thanks (it does not seem particularly inpiring).
| If you sub $u=x+\sqrt{x^2+1}$, that alone will make the integral a simple rational integral. We can solve for $x$ in terms of $u$:
$$\begin{align}u - x & =\sqrt{x^2+1}\\
u^2 - 2ux + x^2& =x^2+1\\
- 2ux& =1-u^2\\
x & = \frac{u^2-1}{2u}\\
dx & = \frac{2u(2u)-2(u^2-1)}{4u^2} du\\
dx & = \frac{u^2+1}{2u^2} du
\end{align}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Sum of Stirling numbers of both kinds Let $a_k$ be the number of ways to partition a set of $n$ elements $orderly$,which means that order of subsets matters, but order of elements in each subset does not.
My task:
Prove, that$$\sum_{k=1}^n \left[n\atop k\right] a_k = n!2^{n-1}$$
The only thing I got so far is:
$$a_k ... | Start by observing that the species of orderly partitions has the
specification
$$\mathfrak{S}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z})).$$
This gives the bivariate generating function
$$G(z, u) = \frac{1}{1-u(\exp(z)-1)}.$$
Hence the generating function of the $a_n$ is
$$G(z) = G(z, 1) =
\frac{1}{2-\exp(z)}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluating a simple definite integral I'm currently teaching myself calculus and am onto the Mean Value Theorem for Integration.
I am finding the value of $f(c)$ on the function $f(x)=x^3-4x^2+3x+4$ on the interval $[1,4]$.
So, with the equation $(b-a)\cdot f(c)=\int_1^4f(x)dx $, you get
$(4-1)\cdot f(c)=\int_1^4(x^3-4... | $\int_1^4(x^3-4x^2+3x+4)dx=\int_1^4x^3dx -4\int_1^4x^2dx +3\int_1^4xdx +4\int_1^4dx=:I$
$$\begin{align*}I &= \left.\left((1/4)x^4 -(4/3)x^3+(3/2)x^2 +4x\right)
\right|_1^4\\ &=
(1/4)(4^4-1^4) -(4/3)(4^3-1^3)+(3/2)(4^2-1^2) +4(4-1)
\\ &= (1/4)(255)-(4/3)(63)+(3/2)(15)+4(3)\\ &= 14.25\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
all integer $(n,m)$, such that $n^3+2^m\cdot n$ is a perfect square What are all integer $(n,m)$, such that $n^3+2^m\cdot n$ is a perfect square
| I’ve left a bit for you to finish, but here’s a pretty detailed guide to a solution.
Notice that $n^3+2^mn=n(n^2+2^m)$, which is a perfect square iff either (1) $n$ and $n^2+2^m$ are both perfect squares, or (2) $n^2+2^m=na^2$ for some positive integer $a$.
In case (1) say $n=a^2$ and $n^2+2^m=b^2$. Then $2^m=b^2-a^4=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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On the number of possible solutions for a quadratic equation. Solving a quadratic equation will yield two roots:
$$\frac{-\sqrt{b^2-4 a c}-b} {2 a}$$
and:
$$\frac{\sqrt{b^2-4a c}-b}{2 a}$$
And I've been taught to answer it like:
$$\frac{\pm\sqrt{b^2-4a c}- b}{2 a}$$
Why does it yields only two solutions? Aren't there i... | One can see that there are only two solutions from the way it is solved. If $ax^2+bx+c=0$ with $a\neq 0$, it follows that $ax^2+bx=-c$, so that $x^2+\frac{b}{a}x=-\frac{c}{a}$. Trying to complete the square, one obtains $(x+\frac{b}{2a})^2=x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}=\frac{b^2-4ac}{4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Solving the cubic polynomial equation $x^3+3x^2-5x-4=0$
How can I solve the cubic polynomial equation $$x^3+3x^2-5x-4=0$$
I simplified it to:
$$x(x^2+3x-5)=4$$
But I don't know where to go from here.
| The manipulation that you performed does not help. Use the rational root test: since the leading coefficient is $1$ and the constant term is $4$, the only possible rational roots of the cubic are fractions of the form $a/b$, where $a$ is a divisor of $4$ and $b$ is a divisor of $1$. In other words, if the cubic has any... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates? How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ other than changing into polar coordinates? It is possible to prove it using infinite series?
| Method 1. Let $I$ denote the integral. Then for $s > 0$, the substitution $t = x\sqrt{s}$ gives
$$ \int_{0}^{\infty} e^{-sx^2} \; dx = \frac{1}{\sqrt{s}} \int_{0}^{\infty} e^{-t^2} \; dt = \frac{I}{\sqrt{s}}. $$
Thus for $u = x^2$,
$$\begin{align*}I^2
&= \int_{0}^{\infty} I e^{-x^2} \; dx
= \int_{0}^{\infty} \frac{I}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 3
} |
Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ ... | Consider the polynomial $f(x)=x^n-b^n.$ Then $f(b)=b^n-b^n=0.$ So $b$ is a root of $f$ and this implies $(x-b)$ divides $f(x)$. Put $x=a$, then $a-b$ divides $f(a)=a^n-b^n.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 8,
"answer_id": 4
} |
solving for a coefficent term of factored polynomial.
Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$
I expanded it and got
$64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{
6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4... | The coefficient of $x^2$ is obtained by adding three contributions :
*
*$a\, 2^6$ : the coefficient of $x^2$ at the left and $2\,x^0$ at the right ($2$ at power $6$)
*$2(6\cdot 2^5(-1))$: the $x$ coefficients on both sides ((constant term $2$)$^5$ $\times\ x$ coef. at the right)
*$1\binom{6}{2}2^4$ : the constant ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How may I prove this inequality? Let $a, b, c$ be positive real, $abc = 1$. Prove that:
$$\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a} \le \frac{1}{2+a} + \frac{1}{2+b}+\frac{1}{2+c}$$
I thought of Cauchy and AM-GM, but I don't see how to successfully use them to prove the inequality. Any hint, suggestion will b... | $p=a+b+c$ and $q=ab+bc+ca$. Using $AM-GM$ we obtain that : $p,q \geq 3.$
The inequality is equivalent with:
$$\dfrac{3+4p+q+p^2}{2p+q+p^2+pq} \leq \dfrac{12+4p+q}{9+4p+2q} \Leftrightarrow$$
$$3p^2q+pq^2+6pq-5p^2-q^2-24p-3q-27 \geq 0 \Leftrightarrow \\\left(3p^2q-5p^2-12p\right)+\left(pq^2-q^2-3p-3q\right)+\left(6pq-9p-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| Let $x=a+bi$, where $a$ and $b$ are real. Then $(a+bi)^3 = -1$. Expanding the left-hand side gives $$a^3 +3a^2bi -3ab^2 -b^3i = -1.$$
We can separate the real and imaginary parts of this equation:
$$\begin{align}
a^3 -3ab^2 & = -1 \\
3a^2b - b^3 & = 0
\end{align}$$
Taking the obvious $b=0$ solution gives us $a=-1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 4
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| Beferore we prove this inequality first prove that the following inequality:
For $a>0$, $b>0$ is true this inequality: $\frac{a}{b}+\frac{b}{a}\geq 2$
$\frac{a}{b}+\frac{b}{a}-2$=$\frac{a^2+b^2-2ab}{ab}$=$\frac{(a-b)^2}{ab}$.
Since $a>0$, $b>0$ $\Rightarrow$ $a-b>0$ $\Rightarrow$ $(a-b)^2>0$, $ab>0$.
Means, $(a-b)^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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More and more limits for sequences So here goes a bit of homework:
$$\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}$$
Well, this would trivially lead to:
$$\lim_{n\to\infty}{\left(\frac{3+\frac{2}{n}+\frac{1}{n^2}}{3+\frac{5}{n^2}}\right)^{\frac{n\left(1+\frac{2}{n^2}\right)}{2+\frac{1}{n... | Reduce to the limit for the exponential:
$$
\lim_{n \to \infty} \left( 1 + \frac{2}{3 n}\frac{1 - \frac{2}{n}}{1 + \frac{5}{3n}} \right)^{\frac{n}{2} - \frac{n-4}{4n+2}} = \underbrace{\lim_{n \to \infty} \left( 1+\frac{2}{3n} \right)^{n/2}}_{\exp\left(\frac{1}{3}\right)} \cdot \underbrace{\lim_{n \to \infty} \left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer solutions to $ x^2-y^2=33$ I'm currently trying to solve a programming question that requires me to calculate all the integer solutions of the following equation:
$x^2-y^2 = 33$
I've been looking for a solution on the internet already but I couldn't find anything for this kind of equation. Is there any way to c... | Suppose that $x=y+n$; then $x^2-y^2=y^2+2ny+n^2-y^2=2ny+n^2=n(2y+n)$. Thus, $n$ and $2y+n$ must be complementary factors of $33$: $1$ and $33$, or $3$ and $11$. The first pair gives you $2y+1=33$, so $y=16$ and $x=y+1=17$. The second gives you $2y+3=11$, so $y=4$ and $x=y+3=7$. As a check, $17^2-16^2=289-256=33=49-16=7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
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how to calculate the exact value of $\tan \frac{\pi}{10}$ I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$.
From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.
Thank you in advance,
Greg
| $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Prove $\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$ using AM-GM I need to proof this inequality by AM-GM method.
Any ideas how to do it?
$$\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ ... | By AM-GM we obtain:
$$(1+abc)\sum_{cyc}\frac{1}{a(1+b)}+3=\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}=$$
$$=\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{a(1+b)}\geq\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc},$$
which says that
$$\sum_{cyc}\frac{1}{a(1+b)}\geq\frac{\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}-3}{1+abc}=\frac{3\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/196176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Generating integral triangles with two equal sides How can I generate all triangles which have integral sides and area, and exactly two of its three sides are equal?
For example, a triangle with sides ${5,5,6}$ satisfies these terms.
| Area of isosceles triangle with sides $a,a,b$ is $\frac{b\sqrt{4a^2-b^2}}{4}$
Now $\sqrt{4a^2-b^2}$ must be integer$=c$(say),
$\implies 4a^2-b^2=c^2\implies b^2+c^2=4a^2$
Clearly, $b,c$ are of same parity.
If $b,c$ are odd $=2C+1,2D+1$ respectively, then $b^2+c^2=(2C+1)^2+(2D+1)^2$
$=4(C^2+D^2+C+D)+2\equiv 2\pmod 4$,no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/198034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing vectors span a vector space by definition I need to show that the vectors $v_1 = \langle 2, 1\rangle$ and $v_2 = \langle 4, 3\rangle$ span $\mathbb R^2$ by definition. By definition if I can write any vector in $\mathbb R^2$ as a linear combination of $v_1$ and $v_2$ then the vectors span $\mathbb R^2$. How do ... | We want to show that any $\mathbf{v} = (x,y) \in \mathbb{R}^2$ can be written as $v = a\mathbf{v_1}+b\mathbf{v_2}$. This equation can be written more explicitly like this: $\begin{pmatrix}x \\ y\end{pmatrix} = a\begin{pmatrix}2 \\ 1\end{pmatrix} + b\begin{pmatrix}4 \\ 3\end{pmatrix}$, because $\mathbf{v_1} = \begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Completing the square with negative x coefficients I know how to complete the square with positive $x$ coefficients but how do you complete the square with negative $x$ coefficients?
For example:
\begin{align*}
f(x) & = x^2 + 6x + 11 \\
& = (x^2 + 6x) + 11 \\
& = (x^2 + 6x + \mathbf{9}) + 11 - \mathbf{9} \\
& = (x+3... | $
\begin{split}
f(x) &= -3(x^2-5x/3 - 1/3)\\
&= -3( x^2 - 5x/3 + (5/6)^2 - (5/6)^2 - 1/3)\\
&= -3( (x - 5/6)^2 - 25/36 - 12/36) \\
&= -3(x - 5/6)^2 -3(- 25/36 - 12/36) \\
&= -3(x - 5/6)^2 + 37/12.
\end{split}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/199970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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What is the remainder when $4^{100}$ is divided by 6? I am trying to find the remainder when $4^{96}$ is divided by 6.
SO using the cyclicity method,
Dividing $4^1$ by 6 gives remainder 4.
Dividing $4^2$ by 6 gives remainder 4.
Dividing $4^3$ by 6 gives remainder 4.
Dividing $4^4$ by 6 gives remainder 4.
Dividing $4^5... | The first method is correct. Since $4\times 4 \equiv 4 \mod 6$, you can conclude $4^n \equiv 4 \mod 6$ for $n \ge 1$.
The second method is wrong. A simple example is that it suggests the remainder when $8$ is divided by $6$ is the same as the remainder when $4$ is divided by $3$.
If you know that $$2^{2n-1} \equiv ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
"answer_count": 1,
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Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| Changing into polar form we have $ 1+ i \sqrt{3} = 2 e^{i\pi/3}$ and $1 - i \sqrt{3} = 2e^{-i\pi/3}$ so the left hand side is $$ \sqrt{2} \left( e^{i\pi/6} + e^{-i\pi/6} \right)= 2 \sqrt{2} \cos(\pi/6)= 2\sqrt{2} \cdot \frac{\sqrt{3}}{2}= \sqrt{6}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 3
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How to solve this recurrence relation? $f_n = 3f_{n-1} + 12(-1)^n$ How to solve this particular recurrence relation ?
$$f_n = 3f_{n-1} + 12(-1)^n,\quad f_1 = 0$$
such that $f_2 = 12, f_3 = 24$ and so on.
I tried out a lot but due to $(-1)^n$ I am not able to solve this recurrence?
Any help will be highly appreciated.
... | $f(1)=0$; $f(n)=3f(n-1)+12(-1)^n=3\Big(3f(n-2)+12(-1)^{n-1}\Big)+12(-1)^n$
For $n\ge 3$ you have
$$\begin{align*}
f(n)&=3f(n-1)+12(-1)^n\\
&=3\Big(3f(n-2)+12(-1)^{n-1}\Big)+12(-1)^n\\
&=9f(n-2)+36(-1)^{n-1}+12(-1)^n\\
&=9f(n-2)+12(-1)^{n-1}\Big(3+(-1)\Big)\\
&=9f(n-2)+24(-1)^{n-1}\\
&=9f(n-2)-24(-1)^n\\
&=\begin{cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/205372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 1
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Real jordan form to complex jordan form then compute P matrix. I have the matrix
$$A =
\begin{bmatrix}
5 & 0 & 1 & 0 & 0 & -6 \\
3 & -1 & 3 & 1 & 0 & -6 \\
6 & -6 & 5 & 0 & 1 & -6 \\
7 & -7 & 4 & -2 & 4 & -7 \\
6 & -6 & 6 & -6 & 5 & -6 \\
2 & 1 & 0 & 0 & 0 & 0
\end{bmatrix}$$
This can be brought in the followi... | Notice that, with every complex pair of eigenvalues $\lambda = a \pm ib$, there exists a complex pair of eigenvectors $u \pm i v$. If you look at the columns of your matrix $T$, you can observe that you can pair up your eigenvectors according to complex conjugates in this precise way.
In real canonical form, each of yo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Unsure about applying series comparison test Does this converge or diverge?
$$
\sum\limits_{n=1}^\infty (a_{n} = \frac{1}{2\sqrt{n} + \sqrt[3]{n}})
$$
The answer is: diverges by limit comparison to $\sum (b_{n} = \frac{1}{\sqrt{n}})$
If I look at $\lim_{n \to \infty}\frac{a_{n}}{b_{n}}$ I get
$$
\frac{\sqrt{n}}{2\sqrt{... | $$\frac{\sqrt{n}}{2\sqrt{n}+\sqrt[3]{n}}=\frac{\sqrt{n}}{\sqrt{n}\left(2+\frac{\sqrt[3]{n}}{\sqrt{n}}\right)}=\frac{1}{2+\frac{\sqrt[3]{n}}{\sqrt{n}}}\to\frac12,\text{ as $n\to\infty$}$$
because
$$\frac{\sqrt[3]{n}}{\sqrt{n}} = n^{1/3}/n^{1/2} = n^{1/3-1/2}=n^{-1/6}$$
An other argument goes like this $$\sqrt[3]{n}\le... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$ It's known that $\lim\limits_{x \rightarrow x_0}f(x) = A$, how to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$?
Here's what I've got now:
When $A = 0$, to prove $\lim\limits_{x \rightarrow ... | Hint: For $A\neq0$,
$$\left|\sqrt[3]{f(x)}-\sqrt[3]{A}\right|
=\frac{|f(x)-A|}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}}+A^{\frac{2}{3}}|}
=\frac{|f(x)-A|}{\bigg(\sqrt[3]{f(x)}+\frac12\sqrt[3]{A}\bigg)^2+A^{\frac{2}{3}}}\leq\frac{|f(x)-A|}{A^{\frac{2}{3}}}.$$
Or just use the inequality:
$$\left|\sqrt[3]{f(x)}-\sqrt[3]{... | {
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"url": "https://math.stackexchange.com/questions/213386",
"timestamp": "2023-03-29T00:00:00",
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Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation wi... | I'm assuming $ab\neq 0$, since otherwise trivial. $a\sin x+b\cos x-c=0$
$$\iff a\left(2\sin\frac{x}{2}\cos\frac{x}{2}\right)+b\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right)-c\left(\sin^2\frac{x}{2} +\cos^2\frac{x}{2}\right)=0$$
*
*Assume $\cos\frac{x}{2}\neq 0$. Then
$$\stackrel{:\cos^2 \frac{x}{2}\neq 0}\... | {
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"url": "https://math.stackexchange.com/questions/213545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ I would like to try and prove
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$
using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.
I found a quick way by ... | $\frac{1+\sin x}{\cos x}$
$=\frac{(1+\sin x)(1-\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1-\sin^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{\cos^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{(\cos x)(1+\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1+\sin x+\cos x}{1-\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Prove that the sequence is convergent How can we show that the sequence $$a_n=\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3-n^2}$$ is convergent?
| $a_n =\sqrt[3]{n^3+n^2} - \sqrt[3]{n^3-n^2}$, so when $n$ is large enough,
$$
\begin{align}
a_n &= n\cdot \sqrt[3]{1+\frac{1}{n}} - n\cdot \sqrt[3]{1 - \frac{1}{n}} \\
&= n\cdot \left(1+\frac{1}{3n} - \frac{1}{9n^2} + \mathcal{O}(n^{-3})\right) - n\cdot \left(1 - \frac{1}{3n} - \frac{1}{9n^2} + \mathcal{O}(n^{-3})\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/220415",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Try to solve root in inequality got wrong result I am very confused. So I have to solve this inequality. The result is $13/24$.
But if I try to solve it myself, I get $17/24$. Because:
$$\sqrt{\left(\frac{-5}{24}\right)^2 + \frac{1}{4}} = \frac{5}{24} + \frac{1}{2} = \frac{5}{24} + \frac{12}{24} = \frac{17}{24}.$$
The ... | $\sqrt{(-5/24)^2+1/4} = \sqrt{25/24^2+1/4} = \sqrt{(25+6\cdot 24)/24^2} = \sqrt{169}/\sqrt{24^2} = 13/24$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Iterated exponentials in modular arithmetic: $a^{b^{c^d}}-a^{b^c}\equiv 0 \mod 2^{2^{2^2}}-2^{2^2}$? How to prove $a^{b^{c^d}}-a^{b^c}\equiv 0 \mod 2^{2^{2^2}}-2^{2^2}$ for all positive integers $a,b,c,d\in \mathbb{N}$? I tried to reduce $d$ to $2$ by induction. But I kind of stuck at the rest of the parameters. Any t... | Clearly, this is true if at least one of $b,c,d$ is $1$, so we can safely focus
on $b,c,d>1$.
$2^{2^{2^2}}-2^{2^2}=2^{16}-2^4=2^4(2^{12}-1)=2^4(2^6-1)(2^6+1)=2^4\cdot7\cdot 9\cdot 13\cdot 5$
We know,for any prime $p$ either $p\mid a$ or $(p,a)=1$
So, if $p\mid a,p\mid(a^e-a)$ for $e\ge 1$ else $(p,a)=1,$
(i)For $p=1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many strings of length $n$... Let $n$ be a positive integer. If each $a_i$ is chosen from $\left\{ 0,1,2, \dots, 9 \right\},\;$ $1\leq i\leq n$, then how many strings of length $n$ ($a_1a_2\dots a_n$) are there such that the number of occurrences of $0$ is even? Thanks!
| Let us fix the alphabet as $\{0, 1, 2, \dots, M\}$ (we'll later take $M = n$). Let $a_n$ be the number of strings of length $n$ which contain an even number of $0$s, and $b_n$ be the number of strings of length $n$ which contain an odd number of $0$s. Note that there are $(M+1)^n$ strings of length $n$ over that alphab... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$ Could you explain the operation in the third step?
$$\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$$
It comes from the sumation $$\sum_{n=1}^\infty\frac1{4n^... | The motivation is to write $\dfrac1{(2n+1)(2n-1)}$ as $\dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$. Hence, we need $A$ and $B$ such that
$$\dfrac1{(2n+1)(2n-1)} = \dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$$
Simplifying the right hand side, we get that
$$\dfrac1{(2n+1)(2n-1)} = \dfrac{A(2n+1) + B(2n-1)}{(2n+1)(2n-1)} = \dfrac{2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simple linear recursion $x_n=\frac{x_{n-1}}{a}+\frac{b}{a}$ with $a>1, b>0$ and $x_0>0$
I tried to solve it using the generating function but it does not work because of $\frac{b}{a}$, so may you have an idea.
| Using generating functions here is easy. Write your recurrence as:
$\begin{equation*}
a x_{n + 1}
= x_n + b
\end{equation*}$
Define the generating function $g(z) = \sum_{n \ge 0} x_n z^n$, multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} x_{n + 1} z^n
&= \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limits with trig, log functions and variable exponents Would someone mind verifying this?
$$ \lim_{x\to \infty} \frac{2 \cdot 3^{5x} + 5}{3^{5x} + 2^{5x}}
= \lim_{x\to \infty} \frac{3^{5x}(2 + \frac{5}{3^{5x}})}{3^{5x}(1 + (\frac{2}{3})^{5x})}
= \lim_{x\to \infty} \frac{2 + \frac{5}{3^{5x}}}{1 + (\frac{2}{3})^{5x}}
= \... | Yes those are both right.
$$\lim_{x\to\infty}\frac{2\cdot 3^{5x}+5}{3^{5x}+2^{5x}}=\lim_{x\to\infty}\frac{2+5\cdot3^{-5x}}{1+\left(3/2\right)^{-5x}}=2$$
and
$$\lim_{x\to0}\frac{e^{2x}-e^{x\ln\pi}}{\sin 3x}=\lim_{x\to0}\frac{(2-\ln\pi)x+O(x^2)}{3x}=\frac{1}{3}(2-\ln\pi).$$
In the second one, you can use that $e^u=1+u+O(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the solution to this system of equation? $$
xy^3z^3 = yx^3z^3 = zx^3y^3
$$
Is there a way to solve this system?
I think the answer is 1, but i can't verify my intuition.
| Another way of looking at it, clearly $x=0$, or $y=0$, or $z=0$ is a solution. So, starting with
\begin{alignat*}{3}
xy^3z^3 &= yx^3z^3 &&= zx^3y^3
\end{alignat*}
And considering the case where $x \ne 0$, $y \ne 0$, $z \ne 0$
\begin{alignat*}{3}
y^2z^2 &= x^2z^2 &&= x^2y^2
\end{alignat*}
Consideri... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integration by parts from Apostol I working through Apostol's calculus, and I need to prove integrating by parts that :
$\int (a^2 - x^2)^n \,dx = \frac{x (a^2 - x^2)^n}{2n + 1} + \frac{2 a^2 n}{2n+1} \int (a^2 - x^2)^{n-1} \,dx + C $
Now, using the integration by parts formula after first division the integral to par... | NOTE: You can only pull the constant part out of an integral, but not your variable:
*
*This is good: $\int 3x^3 dx = 3 \int x^3 dx$
*This is BAD: $\int 3x^3 dx = \color{red}x \int 3x^2 dx$
So far, so good:
$\int (a^2 - x^2)^n \,dx = x(a^2 - x^2)^n + 2n \int \color{red}{x^2} (a^2 - x^2)^{n-1} \,dx$
The red part $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
| Let's use the method of annihilators to turn this into a third-order, homogeneous recurrence, and then solve with a characteristic equation. First, we write the recurrence so $n$ is the least index:
$$T(n) - 2T(n-1)= n \implies T(n+1)-2T(n) = n+1$$
Then, we rewrite the recurrence in terms of the shift operator $E$:
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Random variable $Z = |X-Y|$ Let's take two independent, random variables $X$ and $Y$ from set $\{1,\ldots,n\}$
probability, that $X =i$ AND $Y = j$ is $\frac{1}{n^2}$.
I want to determine the expected value, so I started:
$$E[Z] = E[|X-Y|] = \sum_i (a_i) \cdot P(|X-Y|=a_i),$$ (sum is to $n$) but I don't now how to cou... | Draw the $n\times n$ grid of points with coordinates $(i,j)$, where $i$ and $j$ range from $1$ to $n$. The points $(x,y)$
such that $x-y=k$ are the points on the line $x-y=k$. For $k=0$ there are $n$ such points, the points on the diagonal. For $x-y=1$, there are $n-1$ points, for $x-y=2$ there are $n-2$ points, and ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find X given Y in a cubic function. Having asked this question on the math overflow boards one of the contributors suggested this may be a more appropriate forum.
I have a cubic function in the form:
$$y = ax^3 + bx^2 + cx + d$$
...(where a, b, c and d are all known constants e.g -0.3, 3.5, 3.83 and 0 respectively) tha... | As your hint says, it is easy to solve your particular equation when $y=0$ as you have $$-0.3 x^3+0.5 x^2 + 3.83x =0$$ so either $x=0$ or $-0.3 x^2+0.5 x + 3.83 =0$ and you can solve the latter as a quadratic.
It is usually not so easy when $y \not = 0$ or in general for a cubic. Wikipedia gives the general (sometime... | {
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"timestamp": "2023-03-29T00:00:00",
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Find three polynomials whose squares sum up to $x^4 + y^4 + x^2 + y^2$ Prove that $$p(x,y) = x^4 + y^4 + x^2 + y^2$$ can be written as a sum of squares of three polynomials over $x,y$ for real numbers.
| $(\sqrt{2\sqrt{2}-2}y^2+x)^2 + (-\sqrt{2\sqrt{2}-2}xy + y)^2 + (x^2 + (1-\sqrt{2})y^2)^2 = x^4+y^4+x^2+y^2$.
In fact, there are many other solutions by using brute force, as suggested by Will Jagy in the comments.
Step 1
Let's suppose that $x^4+y^4+x^2+y^2 = \sum_{i=1}^3 (a_i x^2 + b_i xy + c_i y^2 + d_i x + e_i y)^2$... | {
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How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? The numbers in the AP must be distinct.
For example if $n=6$ then number of AP's possible are $6$
*
*$1,2,3$
*$2,3,4$
*$3,4,5$
*$4,5,6$
... | One can select any tow distinct elements of the same parity. Together with their mean, they make such a progression.
There are $\left\lfloor \frac n2\right\rfloor$ even numbers $\le n$ and $\left\lfloor \frac{n+1}2\right\rfloor$ odd numbers.
Since we need to select unordered pairs (or ordered? But you seem not to disti... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Generating Pythagorean triples for $a^2+b^2=5c^2$? Just trying to figure out a way to generate triples for $a^2+b^2=5c^2$. The wiki article shows how it is done for $a^2+b^2=c^2$ but I am not sure how to extrapolate.
| Here is a cute way to construct a family of solutions for the given diophantine:
Key Theorem: Let $a, b, c, d \in \mathbb Z$. Then, $$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$$
Proof: Expand.
Consider a pythagorean triple $(a, b, c)$, which you can easily generate. A simple application of the key theorem yiel... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Is there an easy way of finding the taylor series for $1/(1+x^2)$? I was trying to calculate the fourth derivative and then I just gave up.
| Recall that a geometric series can be represented as a sum by
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty}x^n \quad \quad|x| <1$$
Then we can simply manipulate our equation into that familiar format to get
$$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty}(-x^2)^{n} = \sum_{n=0}^{\infty}... | {
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"timestamp": "2023-03-29T00:00:00",
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Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} ... | $\frac{a}{b}\times\frac{1}{x}=\frac{a}{b}\div x=\frac{\frac{a}{b}}{x}\neq\frac{a}{\frac{b}{x}}=a\div\frac{b}{x}=a\times \frac{x}{b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/251317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a quadratic equation via a tangent half-angle formula (Maybe I'll post my own answer here, but maybe others will make that redundant.)
This is a fun (?) trivia item that fell out of a bit of geometry I was thinking about.
One of the tangent half-angle formulas says
$$
\tan\frac\alpha2 = \frac{\sin\alpha}{1+\cos... | I've up-voted marty cohen's answer, but I would add this:
Maybe the first omission could be said to be the use of $\arctan b$ instead of using both of the points on the circle corresponding to $\tan=b$. For the conventional value of $\arctan b$, the positive square root of $1+b^2$ is the only one that's right.
When yo... | {
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"source": "stackexchange",
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If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that... If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that $\gcd(a,b,c)=1$ implies that $\gcd(a,b)=1$
What I know:
A Pythagorean Triple is a triple of positive integers $a$,$b$,$c$ such that $a^2+b^2=c^2$
A Primitive... | If $gcd(a,b)>1$, then there is a prime number $p$ dividing $gcd(a,b)$, i.e., $p|a$ and $p|b$.
Since $a^4-b^4=c^4$, we have $p|c$. Thus $p|gcd(a,b,c)$. But $gcd(a,b,c)=1$, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solving a trig equation $1+ \sin (x)=2 \cos(x)$? How would I solve the following equation?
$$
1+ \sin (x)=2 \cos(x)
$$
I am having difficult with it.
| We have $1 + \sin(x) = 2 \cos(x)$. Recall that $1 - \sin^2(x) = \cos^2(x)$. Hence, we get that $$(1 + \sin(x))(1- \sin(x)) = \cos^2(x)$$ i.e.
$$2 \cos(x) (1 - \sin(x)) = \cos^2(x)$$
If $\cos(x) \neq 0$, then we get that $$1 - \sin(x) = \dfrac{\cos(x)}2$$
Hence, have
\begin{align}
1 + \sin(x) & = 2\cos(x)\\
1 - \sin(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ Evaluate
$$
\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)
$$
The answer is $\frac{1}{2}$, have no idea how to arrive at that.
| Method I
By simply applying l'Hôpital's rule, we have
$$\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x \right)=\lim_{x \to \infty}\frac{\displaystyle\sqrt{\frac{x}{x-1}}-1}{\left (\displaystyle \frac{1}{x}\right)}=\lim_{x \to \infty}\frac{x^2}{2(x-1)^2\displaystyle\sqrt{\frac{x}{x-1}}}=\frac{1}{2}.$$
Done.
Method I... | {
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Show that $(27!)^6 \equiv 1 \pmod{899}$ Show: $(27!)^6 \equiv 1 \pmod{899}$. Note: $899=30^2 - 1$.
Could anyone show me how to go through this please? I think I have to use Wilson's theorem which is $(p-1)! \equiv -1 \pmod p$ but not exactly sure how else to tackle this. Thanks!
| Using Wilson's Theorem, $28!\equiv-1\pmod{29}\implies 27!(28)\equiv-1$
$\implies 27!(-1)\equiv-1\implies 27!\equiv1\pmod {29}$
$\implies (27!)^6\equiv 1\pmod{29}$
Again $30!\equiv-1\pmod{31} \implies 27!(28)(29)(30)\equiv-1$
$\implies 27!(-3)(-2)(-1)\equiv-1\implies 27!(6)\equiv1 \implies 27!(30)\equiv5$ (multiplying ... | {
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"source": "stackexchange",
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Limits of three functions Could you help me calculate the following limits:
$$\lim_{x \to 0} x \left[ \frac{1}{x} \right]$$
$$\lim_{x\to 0} \frac{1-\cos x \cdot \sqrt{\cos2x} }{x^2}$$
$$\lim_{x\to 10} \frac{\log _{10}(x) - 1}{x-10}$$
As to the last one I thought I could use $\lim\frac{log _{a}(1+\alpha)}{\alpha} = \... | Here is a simpler way to calculate the second limit:
$$\lim_{x\to 0} \frac{1-\cos x \sqrt{\cos2x} }{x^2}=\lim_{x\to 0} \frac{1-\cos x \sqrt{\cos2x} }{x^2} \frac{1+\cos x \sqrt{\cos2x} }{1+\cos x \sqrt{\cos2x}}$$
$$=\lim_{x\to 0} \frac{1-\cos^2 x \cos2x }{x^2} \lim_{x \to 0} \frac{1}{1+\cos x \sqrt{\cos2x}}$$
The sec... | {
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Integers that satisfy $a^3= b^2 + 4$ Well, here's my question:
Are there any integers, $a$ and $b$ that satisfy the equation $b^2$$+4$=$a^3$, such that $a$ and $b$ are coprime?
I've already found the case where $b=11$ and $a =5$, but other than that?
And if there do exist other cases, how would I find them? And if not ... | Observe that
if $a=3k,b^2=a^3-4=(3k)^3-4\equiv-1\pmod 3$ but $b^2\equiv1,0\pmod 3$
if $a=3k+1,b^2=a^3-4=(3k+1)^3-4=9(3k^3+3k^2+k)-3$ which is divisible by $3,$ but not by $9$
So, $a$ must be of the from $3k+2$
Consequently, $b^2-4=a^3-8=(3k+2)^3-8$
$(b+2)(b-2)=9k(3k^2+6k+4)$
Also, as $(a,b)=1,$ both $a,b$ must be odd ... | {
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"timestamp": "2023-03-29T00:00:00",
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The volume of the solid from the region bounded by $x=9-y^2$, $y=x-7$, $x=0$ about $y=3$ using cylindrical shells. The volume of the solid from the region bounded by $x=9-y^2$, $y=x-7$, $x=0$ about $y=3$ using cylindrical shells.
I've tried creating two separate regions:
$V_1=2\pi(3-y)(9-y^2)dy$ from 3 to 1
and
$V_2=2... | Here again is the graph of the things mentioned in the problem:
$\color{blue}{x=9-y^2}$, $\color{red}{x=y+7}$, $\color{green}{x=0}$, $y=3$ (dashed)
You're correct that the circumference of the shell at $y$ is given by $2\pi(3-y)$, but you should be integrating over all of $-3\leq y\leq 3$, with the height of the shell... | {
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Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)... | You can write $\frac{a^2}{a+b} = a - \frac{ab}{a+b}$, and similarly with the other terms.
It remains to prove (after rearranging the inequality) that:
$$\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} \leq \frac{1}{2}\left(a+b+c\right)$$
which follows from AM-HM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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How is this a property of Pascal's triangle? For all non-negative integers $k$ and $n$,
$$
\dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n}{k} = \dbinom{n +1}{k+1}
$$
How is this a property of Pascal's triangle? I do not see at all how it relates. I'm looking for an explanation, as opposed to som... | The "usual" property of the Pascal triangle is $$\dbinom{n+1}{k+1} = \underbrace{\dbinom{n}{k+1}}_{\text{Up right term}} + \underbrace{\dbinom{n}k}_{\text{Up left term}}$$
Now proceed along the up right branch i.e. $$\dbinom{n}{k+1} = \underbrace{\dbinom{n-1}{k+1}}_{\text{Up right term}} + \underbrace{\dbinom{n-1}{k}}_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Smart demonstration to the formula $ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$ Someone could give me a smart and simple solution to show the folowing identity?
$$ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$$
| $$\sum_{n=1}^N \dfrac{n}{2^n}$$ can be written as
\begin{matrix}
\dfrac12 & + \dfrac1{2^2} & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\
& + \dfrac1{2^2} & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\
& & + \dfrac1{2^3} & + \dfrac1{2^4} & + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to find $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$?
Let $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}=f(x)$, where $f(u)=c\sin u$.Find $c$.
Trial:$\sum_{k=1}^{\infty}\frac{c \sin (x+k\pi)}{2^k}=c\sin x$.Then $c=0$ is a solution. Is there any other solution of $c$? Mainly I am interested in the sum $\sum_{k=1}^{\infty... | Using a little trigonometry and the sum of a convergent infinite geometric series:
$$\sin(x+k\pi)=\sin x\cos k\pi+\sin k\pi\cos x=\sin x\cos k\pi=(-1)^k\sin x\Longrightarrow$$
$$c\sin x=\sum_{k=1}^\infty\frac{c\sin(x+k\pi)}{2^k}=c\sin x\sum_{k=1}^\infty\left(-\frac{1}{2}\right)^k=c\sin x\frac{-\frac{1}{2}}{1+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate $\sum \limits_{x=0}^{n} \frac{n!}{(n-x)!\,n^x}\left(1-\frac{x(x-1)}{n(n-1)}\right)$ What are the asymptotics of the following sum as $n$ goes to infinity?
$$
S =\sum\limits_{x=0}^{n} \frac{n!}{(n-x)!\,n^x}\left(1-\frac{x(x-1)}{n(n-1)}\right)
$$
The sum comes from CDF related to sampling with replacemen... | I get $$S = \sqrt{\frac{\pi n}{2}} + \frac{2}{3} + O(n^{-1/2+4\epsilon}).$$
An asymptotic for the summand.
If $x \leq n^{1/2+\epsilon}$, then Stirling's approximation yields $$\log\left(\frac{n!}{(n-x)!n^x}\right) = - \frac{x^2}{2n} + \frac{x}{2n} - \frac{x^3}{6n^2} + O(n^{-1+4\epsilon}) \tag{1}.$$ To obtain this, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a plane whose intersection line with a hyperboloid is a circle Find a plane $\pi$ which involves x-axis and its intersection line with $$\frac{x^2}{4}+y^2-z^2=1$$ is a circle.
Because the plane want to be find involves x-axis,so set as $By+Cz=0$,then I must to determine its value such that $$\begin{cases}By+Cz=0\... | Assume $\pi$ has an equation like $$by+cz=0$$ for some unknown $c$ and $b$. If $b=0$ then $\pi: z=0$ and then we have an intersection like $x^2/4+y^2=1$ which is an ellipse not an circle. The same story would be for $c=0$. Let $c\neq0,b\neq0$ and so $z=-by/c$ so the intersection would be $$(1/4)x^2+y^2-b^2y^2/c^2 = 1$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An arctangent inequality As in the title, how can I prove
$$
\frac{\arctan(x)}{x}\geq\frac{1}{2}
$$
for $x\in(0,1]$?
I think I can say:
$$\frac{\arctan(x)}{x}$$
is monotonically decreasing in the interval, so its value is greater than the value in $1$, which is $\frac{\pi}{4}$, greater than $\frac{1}{2}$.
There exists ... | When $-1\le x\le 1$, we have
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots,$$
and therefore if $-1\le x\le 1$ and $x\ne 0$,
$$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots. \tag{$1$}$$
The series $(1)$ is an alternating series, and therefore if we truncate just after the term $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/270277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find an expression, in terms of n, for $\sum_{r=2}^n {1\over r-1}-{1\over r+1} $ Working:
$$\sum_{r=2}^n {1\over r-1}-\sum_{r=2}^n {1\over r+1} $$
$$={1\over1} + {1\over2} + {1\over3}+\cdots+{1\over n-1}$$
$$-\left({1\over3}+\cdots+{1\over n-1}+{1\over n+1}\right)$$
This should then make:
$$1+{1\over2}-{1\over n+1}$$
B... | We want to find a simple expression for
$$\sum_{r=2}^n \left(\frac{1}{r-1}-\frac{1}{r+1}\right).$$
To get an idea about your lost $-\dfrac{1}{n}$, let us add a few terms together, maybe up to $r=7$, to see what's going on. We get
$$\left(1-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+ \left(\frac{1}{3}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/272437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the density function of the random variable $Z=X+Y$ Let $X$ and $Y$ be independent and uniformly distributed random variable on the intervals $[0,3]$ and $[0,1]$, respectively. Find the density function of the random variable $Z=X+Y$.
I find $f(x,y)=1/3$ and the range of $Z$ to be $[0,4]$, but I cannot find the de... | In general, to find the pdf, you should find the cdf first, and then differentiate the function.
$F[z] = P(Z \leq z) = \begin{cases} \frac {1}{6} z^2 & 0\leq z \leq 1 \\
\frac {1}{6} + \frac {1}{3} (x-1) & 1 \leq z \leq 3 \\
1- \frac {1}{6} (4-z)^2 & 3 \leq z \leq 4 \\ \end{cases}$
This gives that
$f(z) = \begin{cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/272546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $|x-5|=|2x+6|-1$? $|x-5|=|2x+6|-1$.
The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph?
$|x-5|=|2x+6|-1\\
(|x-5|)^2=(|2x+6|-1)^2\\
...\\
9x^4+204x^3+1188x^2+720x=0?$
| Look for where the expressions inside the absolute values change sign: $x-5$ changes sign at $x=5$, and $2x+6$ changes sign at $x=-3$. Thus, when $x<-3$, $x-5$ and $2x+6$ are both negative, and the equation is
$$-(x-5)=-(2x+6)-1\;.$$
When $-3\le x<5$, $x-5$ is negative and $2x+3$ is non-negative, so the equation is
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Evaluating the series: $1 +(1/3)(1/4) +(1/5)(1/4^2)+(1/7)(1/4^3)+ \cdots$
Sum the series:
$$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots$$
How can I solve this? I am totally stuck on this problem.
| Let $$S = \sum_{n=0}^{\infty} \dfrac1{2n+1} \dfrac1{4^n}$$
Note that
$$\dfrac1{2n+1} = \int_0^1 x^{2n} dx$$
Hence, we can write
\begin{align}
S & = \sum_{n=0}^{\infty} \dfrac1{4^n}\int_0^1 x^{2n} dx = \int_0^1 \sum_{n=0}^{\infty} \left(\dfrac{x^2}{4} \right)^n dx\\
& = \int_0^1 \dfrac{dx}{1-\dfrac{x^2}4} = \int_0^1 \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$. How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?
| Putting $x=a\sin\theta,dx=a\cos\theta d\theta$ and $x=\pm a,\theta=\pm\frac\pi2 $
$$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx =\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\cos^2\theta}{1+a^2\sin^2\theta}d\theta$$
$$=\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\sec^2\theta}{(1+\tan^2\theta)(1+(a^2+1)\tan^2\theta)}d\theta$$ (Di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 4
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Forms $apq +b = r^{n} $ where p,q,r are primes Some small results for $2pq +3 = r^{n} $ p,q,r primes; written in the form (p,q,r,n):
$(3,1093,3,8)
(59,997,7,6)
(73,107,5,6)
(7,223,5,5)
(3,13,3,4)
(11,109,7,4)
(109,131,13,4)
(277,1667,31,4)
(5,491,17,3)
(89,137,29,3)
(11,13,17,2)$
Some small results for $2pq +1 = r^{n} ... | Primes are $1$ or $3 \mod 4$, else they would be divisible by $2$ or by $4$. Squares are always $0$ or $1 \mod 4$ since $2^2=4 = 0 \mod 4$ and $3^2=9=1 \mod 4$ and so if $a = 2\text{ or }3 \mod 4$, then $a^2= 0\text{ or }1 \mod 4$ and if $a=0\text{ or }1 \mod 4$ then $a^2= 0\text{ or }1 \mod 4$.
1) If $p$ and $q$ are b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$ I am trying to prove that
$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$
I know how to ... | Let's introduce the parameter $\alpha$, and then differentiate with respect to $\alpha$ that yields
$$I(\alpha)=\int_0^1 \frac{t^\alpha-1}{(t^2+1)\ln t}dt $$
$$I'(\alpha)=\int_0^1 \frac{t^{\alpha}}{(t^2+1)}dt=\frac{1}{4} \left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 0
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Solving lyapunov equation, Matlab has different solution, why? I need to solve the lyapunov equation i.e. $A^TP + PA = -Q$. With $A = \begin{bmatrix} -2 & 1 \\ -1 & 0 \end{bmatrix}$ and $Q = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$.
Hence...
$$
\begin{bmatrix}
-2 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
P... | X = lyap(A,B,-C) solves the continuous-time Sylvester equation
AX + XB = C
and
X = lyap(A’,Q) solves the continuous-time Lyapunov equation
ATP + PA + Q = 0
so, you can solve the lyapunov function.
A = [-2 1; -1 0];
Q = [1 0; 0 1];
P = lyap(A',Q)
P = [0.5000 -0.5000]
[-0.5000 1.5000]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple question - Proof How is $\frac{1}{2}ln(2x+2) = \frac{1}{2}ln(x+1) $ ?
As $\frac{1}{2}ln(2x+2)$ = $\frac{1}{2}ln(2(x+1))$, how does this become$ \frac{1}{2}ln(x+1)$?
Initial question was $ \int \frac{1}{2x+2} $
What I done was $ \int \frac{1}{u}du $ with $u=2x+2$ which lead to $\frac{1}{2}ln(2x+2) $ but WolframA... | We have $\ln(2x+2)=\ln(2(x+1))=\ln 2+\ln(x+1)$. Thus
$$\frac{1}{2}\ln(2x+2)=\frac{1}{2}\ln 2+\frac{1}{2}\ln(x+1).$$
The two functions thus are definitely not equal. But they differ by a constant.
So the answer to $\int \frac{dx}{2x+2}$ can be equally well put as $\frac{1}{2}\ln(|2x+2|)+C$ and $\frac{1}{2}\ln(|x+1|)+C$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quintic polynomial with Galois Group $A_5$ A recent question asks what makes degree 5 special when considering the roots of polynomials with integer coefficients etc. One answer is that the Galois Group of $S_5$ is not solvable. What I am looking for is the most straightforward example (with proof) of a polynomial with... | Here are some found by computer search
$$x^5 - 55x - 88$$
$$x^5 - 55x + 88$$
$$x^5 + 20x - 16$$
$$x^5 + 20x + 16$$
$$x^5 + 95x - 76$$
$$x^5 + 95x + 76$$
$$x^5 + 3x^3 + 5x - 10$$
$$x^5 + 3x^3 + 5x + 10$$
$$x^5 + 6x^3 - 7x - 8$$
$$x^5 + 6x^3 - 7x + 8$$
$$x^5 + 10x^3 - 10x - 4$$
$$x^5 + 10x^3 - 10x + 4$$
$$x^5 - x^4 + x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Evaluate integral with quadratic expression without root in the denominator $$\int \frac{1}{x(x^2+1)}dx = ? $$
How to solve it? Expanding to $\frac {A}{x}+ \frac{Bx +C}{x^2+1}$ would be wearisome.
| And just to make this problem unnecessarily complicated:
$$I=\int\frac{dx}{x^2\left(x+\frac{1}{x}\right)}=\int\frac{dx}{\left(x+\frac{1}{x}\right)}-\int\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)}=\frac{1}{2}\int\frac{d\left(x^2+1\right)}{x^2+1}-\int\frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF int... | \begin{align}
e^{e^x-1}&= 1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}+\cdots+\frac{(e^x-1)^n}{n!}+ \cdots \\[8pt]
&= \cdots\cdots+\frac{e^{nx}+xe^{(n-1)x}+\binom n2 e^{(n-2)x}+\cdots+ne^x + 1}{n!}+ \cdots\cdots
\end{align}
One of the terms in the expansion of $(e^x-1)^n$ is $e^x$. When that is expanded, one of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/289097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Trigonometric Functions The question is to show that $A\sin(x + B)$ can be written as $a\sin x + b\cos x$ for suitable a and b.
Also, could somebody please show me how $f(x)=A\sin(x+B)$ satisfies $f + f ''=0$?
| If
$$
f(x) = A\sin(x+B)
$$
then
$$
f'(x) = A\cos(x+B)\cdot\frac{d}{dx}(x+B) = A\cos(x+B)\cdot1,
$$
and
$$
f''(x) = -A\sin(x+B)\cdot\frac{d}{dx}(x+B) = -A\sin(x+B).
$$
So
$$
f''(x)+f(x) = -A\sin(x+B)+A\sin(x+B) = 0.
$$
For the initial question, the standard trigonometric identity
$$
\sin(x+B) = \sin x\cos B+ \cos x\sin ... | {
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Proof of inequality involving surds Determine whether $ \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$ is greater or less than $\frac{3}{10}$
So what I did was:
Add the fraction with surds to get $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{12}}$
Then squaring both sides results in $\frac{8-4\sqrt{3}}{12} = \frac{2-\sqrt{3}}{3}$
Therefor... | $$ \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}=\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = \frac{\sqrt{2}}{\sqrt{3}(1+\sqrt{3})}=\frac{\sqrt{2}}{3+\sqrt{3}}=\frac{\sqrt{2}(3-\sqrt{3})}{6}<\frac{7\sqrt{2}}{33},$$
where we have used $\sqrt{3}>\frac{19}{11}$. Now $\frac{7\sqrt{2}}{33}<\frac{3}{10}$, since $9800... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/293485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Limits calculus very short question? Can you help me to solve this limit? $\frac{\cos x}{(1-\sin x)^{2/3}}$... as $x \rightarrow \pi/2$, how can I transform this?
| To evaluate the limit:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
we can use the fact that as $x$ approaches $\frac{\pi}{2}$, $\sin x$ approaches 1 and $\cos x$ approaches 0. Substituting these values into the expression, we get:
$$\lim_{x\to \frac{\pi }{2}... | {
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"url": "https://math.stackexchange.com/questions/293560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Factoring $3x^2 - 10x + 5$ How can $3x^2 - 10x + 5$ be factored? FOIL seemingly doesn't work (15 has no factors that sum to -10).
| You can try to Conplete the Square in this kind of problem. What you would want to do, is to manipulate $3x^2-10x+5$ into something like this $a^2 \pm 2ab + b^2 = (a \pm b)^2 \quad \mathbf{(1)}$.
First, factor the 3 out, like this $3x^2-10x+5 = 3 \left( x^2 - \dfrac{10}{3}x + \dfrac{5}{3} \right)$, now, look closely in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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How can we show the convergence of $x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$? Show that the sequence $(x_n)_{n\geq 1}$ defined by $$x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$$ converges and compute its limit.
| It smells like an easy high school question
$$\lim_{n\to\infty}\sin(2\pi (\sqrt[3]{n^3-n^2+1}-n)+2\pi n)=$$
$$\lim_{n\to\infty}\sin(2\pi (\sqrt[3]{n^3-n^2+1}-n))=-\frac{\sqrt{3}}{2}$$
because
$$\lim_{x\to0} \frac{\sqrt[3]{x^3-x+1}-1}{x}=\lim_{x\to0} \frac{3x^2-1}{3(x^3-x+1)^{2/3}}=-\frac{1}{3}$$
by the cute l'Hôpital'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/298669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$? I am finishing a proof. It seems like I can use $\cos^2 + \sin^2 = 1$ to figure this out, but I just can't see how it works. So I've got two questions.
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$?
And if it does, then how?
| Here is my favorite way to verify trigonometric identities:
First note that the equation of a circle gives us the rational parameterizations
$$\sin\theta=\frac{2t}{1+t^2}\qquad\cos\theta=\frac{1-t^2}{1+t^2}.$$
Substitute these expressions in. Now the equation we want to verify is
$$\left(\frac{2t}{1+t^2}\right)^2-\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Solve the congruence $x^3+2x-3\equiv{0}\pmod{45}$ Solve (if possible)the congruence involving polynomial
$x^3+2x-3\equiv{0}\pmod{45}$
My work:
Since $45=3^2\cdot5$, we have
$x^3+2x-3\equiv{0}\mod(3)$ and $x^3+2x-3\equiv{0}\pmod 5$
In $\mathbb{Z}_3$,
$x^3+2x-3=(x-1)(x^2+x+3)\equiv{0}\pmod 3$
We have $[0],[1],[2]$
In ... | Here is one way to find the roots $\rm\,mod\ 9.\:$ Since $\rm\,9\mid(x-1)(x^2+x+3),\:$ by unique factorization, either $\rm\,9\mid x-1,\,\ 9\mid x^2+x+3,\:$ or $\rm\:3\mid x-1,\,x^2+x+3.\:$ The last case is impossible since $\rm\:3\mid x-1\,\Rightarrow\,mod\ 3\!:\ x\equiv 1\,\Rightarrow\, x^2+x+3\equiv 2\not\equiv 0.\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/302434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sum\limits_{cyc} \frac 1 {(a-b)^2} \ge \frac{9}{4}$ $a, b, c \in [0;2]$, Prove inequality: $$\frac1{(a-b)^2}+\frac1{(b-c)^2}+\frac1{(c-a)^2}\geq \frac94$$
I tried to:
*
*Use AM-GM: $$LHS \ge \frac{(1+1+1)^2}{(a-b)^2+(b-c)^2+(c-a)^2}=\frac9{2(a^2+b^2+c^2 -ab-bc-ca)}$$ I am proving $a^2 + b^2+c^2 - ab - bc - ... | WLOG we can assume $ a < b < c$. Since multiplying $a,b,c$ all by the same constant $> 1$ decreases the left side, we may assume $c = 2$. Since decreasing $a$ and $b$ by the same positive number decreases the left side, we may assume $a=0$. Now there's only one variable left, and I'll leave it to you to minimize $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/303826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve equation $ \frac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$? $$ \dfrac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$
How can I solve this equation in the easiest way?
| Multiply by $2$ to obtain
$$\tag1\sqrt{x^2-16}+\sqrt{x^2-9}=2$$
and multiply by the conjugate $\sqrt{x^2-16}-\sqrt{x^2-9}$ to obtain
$$\tag2 -\frac72=\frac12((x^2-16)-(x^2-9))=\sqrt{x^2-16}-\sqrt{x^2-9}.$$
Add $(1)$ and $(2)$ and divide by $2$ to obtain
$$\sqrt {x^2-16}=-\frac34$$
Which has no real solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
A improper integral with Glaisher-Kinkelin constant Show that :
$$\int_0^\infty \frac{\text{e}^{-x}}{x^2} \left( \frac{1}{1-\text{e}^{-x}} - \frac{1}{x} - \frac{1}{2} \right)^2 \, \text{d}x = \frac{7}{36}-\ln A+\frac{\zeta \left( 3 \right)}{2\pi ^2}$$
Where $\displaystyle A$ is Glaisher-Kinkelin constant
I see Chris's ... | You may start with the decomposition
$$
\int_0^\infty {\frac{e^{-x}}{x^2} \left( \frac{1}{\left(e^x - 1\right)^2} - \frac{1}{x^2} + \frac{1}{x} - \frac{5}{12} + \frac{x}{12} \right)dx} - 2\int_0^\infty {\frac{e^{-x}}{x^3} \left(\frac{1}{e^x - 1} - \frac{1}{x} + \frac{1}{2}-\frac{x}{12} \right)dx} + \int_0^\infty {\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/305545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| My solution:
Now $\left( {b - c} \right)\left( {1 + {a^2}} \right) = {a^2}\left( {b - c} \right) + \left( {b - c} \right) + \left( {b - c} \right)x - \left( {b - c} \right)x = \left( {b - c} \right)\left( {x + {a^2}} \right) + \left( {b - c} \right)\left( {1 - x} \right)$.
Then $\dfrac{{\left( {b - c} \right)\left( {1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
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