Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
The sum of powers of $2$ between $2^0$ and $2^n$ Lately, I was wondering if there exists a closed expression for $2^0+2^1+\cdots+2^n$ for any $n$?
| Clue:
If you are searching closed formulas to a series like:
$a+ab+ab^2+ab^3+\ldots+ab^n = S_n$
You can use the follow trick:
$a+ab+ab^2+ab^3+\ldots+ab^n = S_n$
so,
$b(a+ab+ab^2+ab^3+\ldots+ab^n) = bS_n = ab+ab^2+ab^3+ab^4+\ldots+ab^n+ab^{n+1}=$
$a+ab+ab^2+ab^3+\ldots+ab^n -a +ab^{n+1}= bS_n \iff$
$S_n-a+ab^{n+1}=bS_n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/86838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove that if $a\equiv b \pmod{kn}$ then $a^k\equiv b^k \pmod{k^2n}$ What I have done is this:
$a\equiv b \pmod{2n}$,
$a=b+c\times2n$, for some $c$,
$a^2=b^2+2b\times c\times2n+c^2\times2^2n^2$,
$a^2-b^2=(b\times c+c^2n)\times4n$, then
$a^2\equiv b^2\pmod{2^2n}$.
I think that this is right: what I DON’T understa... | Since
$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\ldots b^{k-1})$
we need only prove that
$a^{k-1}+a^{k-2}b+\ldots + b^{k-1}$ is divisible by $k$. But since $a \equiv b \ (\text{mod} \ k)$ we see that
$ a^{k-1}+a^{k-2}b+\ldots + b^{k-1} \equiv a^{k-1}+a^{k-1}+\ldots +a^{k-1} \equiv ka^{k-1} \equiv 0 \ (\text{mod} \ k). $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Compositeness of number $k\cdot 2^n+1$? Every odd prime number can be expressed in the form $k \cdot 2^n+1$ ,where $k$ is an odd number .
For $n>2$ number $k \cdot 2^n+1$ is composite if :
$1.$ $k\equiv 1 \pmod {30} \land (n\equiv 2 \pmod 4 \lor n \equiv 1 \pmod 2 ) $
$2.$ $k\equiv 3 \pmod {30} \land n\equiv 3 \pmod 4... | There are zillions of such relations. For example: $2^6+1$ is a multiple of $13$, so $k2^n+1$ is composite if $n\equiv6\pmod{12}$ and $k\equiv1\pmod{13}$. You can make as many of these as you want.
EDIT: In general, take any numbers $q$ and $s$, and any prime $p$ dividing $q2^s+1$, and any $r$ such that $2^r\equiv1\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89871",
"timestamp": "2023-03-29T00:00:00",
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$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$
Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?
| $$
\begin{align*}
\sum_{n=1}^\infty n(2/3)^n &=
\sum_{m=1}^\infty \sum_{n=m}^\infty (2/3)^n \\ &=
\sum_{m=1}^\infty \frac{(2/3)^m}{1-2/3} \\ &=
\frac{2/3}{(1-2/3)^2} = 6.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/90637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
Sum of squares of roots of a polynomial $P(x)$ Well, I recently proved a formula (at least, I think) to the sum of the inverse of the roots $x_{1}, x_{2}, x_{3},\ldots, x_{n} \in \mathbb{C}$, and $\neq 0$. It starts:
Let a polynomial $P(x) = a_nx^n+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_1x + a_0$ of roots $x_{1}, x_{2}, x_{3... | Consider this:
Equate your functionto zero: $\large a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x^1+a_0x^0=0$
With Vieta's formula, we get the value
*
*$\large x_1+x_2+x_3+...+x_n=\frac{a_{n-1}}{a_n}$
and
*$\large\left(x_1x_2+x_1x_3+x_1x_4+...+x_1x_n\right)+\left(x_2x_3+x_2x_4+x_2x_5+...+x_2x_n\right)+...+\left(x_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/90914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Are we unable to find the extrema of the function $f(x,y) = \frac{y}{x^2+y^2}$ using the second partial test?
$$\begin{align*}
&f(x,y) = \frac{y}{x^2+y^2}\\
&f_{xx} = \frac{∂}{∂x}\left(-\frac{2xy}{(x^2+y^2)^2}\right)=-\frac{2y(y^2-3x^2)}{(x^2+y^2)^3}\\
&f_{yy} = \frac{∂}{∂x}\left(\frac{x^2-y^2}{(x^2+y^2)^2}\right)=\fr... | First, you’ve some errors in your third line of calculations. The correct partials are:
$$\begin{align*}
&f_x(x,y)=-\frac{2xy}{(x^2+y^2)^2}\\
&f_y(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}\\
&f_{xx}(x,y)=\frac{2y(3x^2-y^2)}{(x^2+y^2)^3}\\
&f_{xy}(x,y)=\frac{2x(3y^2-x^2)}{(x^2+y^2)^3}\\
&f_{yy}(x,y)=\frac{3y(y^2-2x^2)}{(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/91654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Systems of equations finding right triangles I need help setting up the equation for the question, "Find all right triangles for which the perimeter is $24$ units and the area is $24$ square units."
I know that the area is $A = \frac12 b h$ and perimeter is $P = a + b + h$.
Using this, would the system look something ... | Let A=24. I am assuming c as hypotenuse and a, b the other two sides of the triangle $\implies c^2=a^2+b^2$.
We now have $1/2*a*b= A$ and $a+b+c=A$
$(a+b)^2=(A-c)^2$
$\implies a^2+b^2+2ab= A^2-2Ac+c^2$
$\implies c=\dfrac{A^2-2ab}{2A}=10$
There we have ab=48 and c=10. So the possible right triangle is 6,8,10.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$?
Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$ ?
$$\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbb{Q}\}$$
$$\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \math... | $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.
Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence
$\sqrt{2} + \sqrt{3} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "84",
"answer_count": 6,
"answer_id": 5
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Proving by mathematical induction: $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2(\sqrt{n+1}-1)$
Possible Duplicate:
Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with inductio... | There are ways other than induction (such as comparison with an integral) that have much clearer intuitive content. But let's stick with induction. We need to deal with the induction step. Suppose that we know for a given $k$ that
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{k}}>2(\sqrt{k+1}-1).\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/96444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Isolate yaw-pitch-roll from rotation I have a transformation that acts as such:
$$RX=Y$$
Where $R_{3\times3}$ is a yaw-pitch-roll rotation matrix, and $X, Y$ are 3D vectors.
Yaw ($\alpha$)-pitch ($\beta$)-roll ($\gamma$) rotations are the product of the 3 rotation matrices:
$$R_x(\gamma)R_y(\beta)R_z(\alpha)$$
Given $X... | The first answer to any question on how to find Euler angles is: don't; use quaternions instead.
If external factors force you to use Euler angles, you could proceed like this: If $X$ and $Y$ are identical, the rotation can be a rotation around $X$ through any angle. Otherwise, the rotation can be a rotation through an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/96522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$
Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:... | No idea how Wolfram does it. But I guess you just start with power series and play around.
\begin{equation}
X=\sum_{n=1}^\infty(-1)^n\frac{n}{(2n+2)!}=-\frac{1}{4!}+\frac{2}{6!}-\frac{3}{8!}+\dots
\end{equation}
The annoying part is the numerators. Try doubling it so that the numerators "keep up" with the denominators.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/96571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Selecting multiple balls of same color from different urns Urn I contains 10 balls: 4 red and 6 blue. Urn II contains 20 balls: 16 red and 4 blue. A single ball is drawn from urn I, and two balls are drawn from urn II. What is the probability that all three balls are the same color?
| Numer of all possible ways of selecting 1 ball from urn 1 and 2 balls from urn 2 is ${\left( \begin{gathered}
10 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
20 \\
2 \\
\end{gathered} \right)}$.
Number of ways to select 1 red ball from urn 1 and 2 red balls from urn 2 is ${\left( \begin{gat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/98281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\sum\limits_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}$ Just as the topic ask how to evaluate $$\sum_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}.$$
| Notice that
$$
\frac{(k-1)!^2}{(2k)!} = \frac{\Gamma(k) \Gamma(k)}{ \Gamma(2k+1) } = \frac{1}{2k} B(k,k) = \frac{1}{2k} \int_0^1 x^{k-1} (1-x)^{k-1} \mathrm{d} x
$$
Thus
$$ \begin{eqnarray}
\sum_{k=1}^\infty \frac{(k-1)!^2}{(2k)!} &=& \int_0^1 \left( \sum_{k=1}^\infty \frac{1}{2k} x^{k-1} (1-x)^{k-1} \right) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/98507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{4}{x+1}+\frac{2}{x-2} \leq 3$ Solve
$$\frac{4}{x+1}+\frac{2}{x-2} \leq 3$$
I must be making a very stupid mistake somewhere ... been stuck on this for 1hr+ or even more ...
| $$
\begin{eqnarray}
\frac{2(x-1)}{(x+1)(x-2)} &\leq& 1 \implies \\
\frac{2(x-1)}{(x+1)(x-2)} - 1 &\leq& 0 \\
\frac{(2x-2)-(x+1)(x-2)}{(x+1)(x-2)} &\leq& 0 \\
\frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)} &\leq& 0 \\
\frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)} &\leq& 0 \\
\frac{-(x^2-3x)}{(x+1)(x-2)} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/100092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Perfect Squares ending in 576 I want to find out perfect squares ending in 576, after the number 576.
Here is my derivation to arrive at such a number. Let the perfect square ending in $576$ be $1000k+576$. Every perfect square can be expressed as a the sum of a certain number of consecutive odd numbers. For eg: $2^2 =... | Your question definitely deserves a more precise answer than this, see the answer by Peter Taylor.
But I'd like to point out that what you are looking for are squares that are $576 \pmod{1000}$, i.e. solutions to $x^2 = 24^2 \pmod{1000}$.
This reduces to the equations
$$ \begin{align*} x^2 &= 24^2 \pmod{125} \\ x^2 &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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An expression of $1\cdot2 + 2\cdot3 + \cdots + n\cdot(n+1)$ I got a question in my homework, which is:
Find the following sum and prove your claim:
$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n\cdot(n+1).$$
I want to prove this by mathematical induction, but I couldn't find an expression of the sum. If anyone has ... |
Find the following sum and prove your claim:
$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n\cdot(n+1).$$
$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + i(i+1) + \cdots + n(n+1) = \sum_{i=1}^n i(i+1)=\sum_{i=1}^n i^2+\sum_{i=1}^n i \tag1$$
It is easy to prove
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
But how you derive $\sum_{i=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/106042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
How to find$\int x^2 \arctan(x) \;dx $ I was solving $\int x^2 \arctan(x) \;dx $
I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.
From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$
I got
$$\begin{align*}&\int x^2 \arctan(x) \; dx =\\
&x^3\arctan(x) -\frac12\ln(1+x^{2})-\in... | Let $u=\arctan x$, $dv = x^2\,dx$. Then $du = \frac{dx}{1+x^2}$, $v = \frac{1}{3}x^3$, so
$$\begin{align*}
\int x^2\arctan x\,dx &= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\frac{x^3}{1+x^2}\,dx\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\left(\frac{x^3+x}{1+x^2}-\frac{x}{1+x^2}\,dx\right)\\
&= \frac{1}{3}x^3\arc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/106827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the minimum possible value of $(a+b+c)$? $a, b$ and $c$ are real positive numbers satisfying $ \frac 13 \le ab+bc+ca \le 1 $ and $abc \ge \frac 1{27}$ then what is the minimum possible value of $(a+b+c)$?
Applying AM $\ge$ GM gives $(a+b+c) \ge 1$ and if we apply AM $\ge$ HM gives $(a+b+c) \ge\frac 13$ but appa... | Answer is 1
just expand $(a+b+c)^2$
and use the AM-GM inequality for $a^2+b^2+c^2$
so $a^2+b^2+c^2$ has a minimum value of $2(abc)^{3/2}$ which is $1/3$
and $2(ab+bc+ca)$ has a minimum value of $2*1/3=2/3$
adding both you will get 1, which is minimum value of $(a+b+c)^2$ and hence $(a+b+c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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For a polynomial $P$ for which $P(x+5) - P(x) = 2 ,\forall x$. What is the least possible value of $P(4) - P(2)$? There are infinite number of polynomials $P$ for which $P(x+5) - P(x) = 2,\forall x \in \mathbb{R}$.
How could we determine the least possible value of $P(4) - P(2)$?
| The condition $P(x+5)-P(x)=2$ for all $x$ can be rewritten as "$(x,y)$ is on the graph of $y=P(x)$ if and only if $(x+5,y+2)$ is." This implies that if $(x,y)$ is on the graph of $y=P(x)$, so are $(x+5k,y+2k)$ for all $k\in\mathbb{Z}$. All of those points lie on a line of the form $y=\frac{2}{5}x+b$ for some $b\in\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/108058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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Constructing Isomorphism between finite field Consider $\mathbb{F}_3(\alpha)$ where $\alpha^3 - \alpha +1 = 0$ and $\mathbb{F}_3(\beta)$ where $\beta^3 - \beta^2 +1 =0$.
I know these two fields are isomorphic but I have difficulty buliding an isomorphism between them.
I know I have to determine where $\alpha$ is mapped... | One way to do it is just to slug through it:
An isomorphism $\mathbb{F}_3(\alpha)\to\mathbb{F}_3(\beta)$ is completely determined by the image of $\alpha$; and, as you note in comments, the image of $\alpha$ must be of the form $a\beta^2 + b\beta + c$, with $a,b,c\in\mathbb{F}_3$. Moreover, the image of $\alpha$ must s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/108717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Steps to solve this system of equations: $\sqrt{x}+y=7$, $\sqrt{y}+x=11$ I want to solve this system of equations, I have been out of Maths for a long!!
$$\sqrt{x}+y=7$$
$$\sqrt{y}+x=11$$
Just wondering easiest step to find values for $x$ and $y$ from the above equations?
| Let
$x=\tan^2\theta$
and $y=\sec^2\theta$
*
*$\tan\theta+\sec^2\theta=7$
*$\tan^2\theta+\sec\theta=11$
Take 1) $\tan\theta+1/\cos^2\theta=7$
Solve it and you will get
$$7\sin^2\theta+\sin\theta-6=0$\tag{A}$$
Now similarly take equation (2) solve it
you will get
$$12\cos^2\theta-\cos\theta-1=0\tag{B}$$
Take (A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/108962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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How can I make the following 2 fractions integers? Let $m,n$ be integers. I want to find the possible values of $m,n$ such that $4(m+n)\over (2m+n)^2+3n^2$ and $4n\over (2m+n)^2+3n^2$ are both integers too. Would someone please help? Of course letting $(2m+n)^2+3n^2=4$ gives some good values, but is this all the $m,n$ ... | Since $(2m+n)^2 + 3n^2 = 4m^2 + 4mn + n^2 + 3n^2 = 4m^2 + 4mn + 4n^2$, then
$$\frac{4(m+n)}{(2m+n)^2+3n^2} = \frac{4(m+n)}{4m^2+4mn+4n^2} = \frac{m+n}{m^2+mn+n^2}.$$
Similarly,
$$\frac{4n}{(2m+n)^2+3n^2} = \frac{4n}{4m^2+4mn+4n^2} = \frac{n}{m^2+mn+n^2}.$$
For both to be integers, you need $m^2+mn+n^2$ to divide both $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/109003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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unit circle, derive number for any degree, cosinus and sinus $\sin(90°)= \sin(\frac{1}{2}\pi)= 0$
$\cos(90°)= \cos(\frac{1}{2}\pi)= 1$
$\sin(60°)= \sin(\frac{1}{3}\pi)=\frac{\sqrt{3}}{2}$
$\cos(60°)= \cos(\frac{1}{3}\pi)=\frac{1}{2} $
$\sin(45°)= \sin(\frac{1}{4}\pi)=\frac{\sqrt{2}}{2}$
$\cos(45°)= \cos(\frac{1}{4}\p... | Using the usual formulas for $\sin(x+y),\cos(x+y)$, one obtains $\cos(3x) = 4\cos^3x-3\cos x$. Thus $\cos(10°)$ is a root of the polynomial $4x^3-3x-\sqrt{3}/2$. Cardano's formula now yields
$$\cos(10°) = \frac{1}{2}\big (\;\sqrt[3]{\frac{\sqrt{3} + i}{2}} + \sqrt[3]{\frac{\sqrt{3} - i}{2}}\;\big)$$
where the 3rd root... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/110856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n=... | One quick way is to recognize that $$\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}$$ is nothing but $1^4 + 2^4 +\cdots + n^4$.
Another way is to look at $6n^5 + 15n^4 + 10n^3 - n$ and prove that $30$ divides it i.e. prove that $2$, $3$ and $5$ divide $6n^5 + 15n^4 + 10n^3 - n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 4
} |
Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, ... | You have everything right except the last line.
Maybe it is easier to do in this order:
$$(x−y)\left(x^{n−1}+x^{n−2}y+\cdots+xy^{n−2}+y^{n−1}\right)=\\
=x\cdot x^{n-1}-y\cdot x^{n-1} +x\cdot x^{n−2}y-
y\cdot x^{n−2}y+x\cdot x^{n−3}y^2-\cdots\\ \cdots -y\cdot x^2y^{n-3} +x\cdot xy^{n-2}-y \cdot y^{n-1}$$
The second t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 3
} |
Limit of a Sequence involving $\frac {n^2} {\sqrt{n^{6}+k}}$ I am stuck on this problem:
Compute the limit of the sequence $(a_{n})_{n=1}^{\infty}$ defined by
$$a_{n}:=\frac {n^2} {\sqrt{n^{6}+1}}+\frac {n^2} {\sqrt{n^{6}+2}}+\cdot \cdot \cdot + \frac {n^2} {\sqrt{n^{6}+n}}=\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}... | That looks great!
I'll show you another, slightly cleaner (in my opinion) method:
We seek $\displaystyle \lim_n \sum_{k = 1}^n \frac{n^2}{\sqrt{n^6 + k}}$
Each term in the sum shares bounds, namely $\dfrac{n^2}{\sqrt{n^6 + n}} \leq \dfrac{n^2}{\sqrt{n^6 + k}} < \dfrac{n^2}{\sqrt{n^6}}$
So then $\displaystyle 1 =\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding a CDF given a PDF The PDF for $Y$ is
$$f_Y(y) = \begin{cases}
0 & |y|> 1 \\
1-|y| & |y|\leq 1
\end{cases}$$
How do I find the corresponding CDF $F_Y(y)$? I integrated the above piecewise function to get
$$F_Y(y)=\begin{cases}
1/2 -y/2-y^2/2 & [-1,0] \\
1/2-y/2+y^2/2 & [0,1]
\end{ca... | We have that $F(y) = \displaystyle \int_{-\infty}^y f(x) dx$. In your case, we are given that $$f(x) = \begin{cases} 0 & x <-1\\ 1 + x & x \in[-1,0]\\ 1-x & x \in [0,1]\\ 0 & x > 1\end{cases}$$
*
*If $y < -1$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^y 0 dx =0 $. We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Matrix-vector multiplication before evaluating the matrix inverse Suppose I need to evaluate the expression
$$\begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}
\begin{bmatrix} s+2 & -1 & 0 & 0\\
0 & s+3 & 0 & 0\\
-1 & -2 & s & -1\\
2 & -1 & 1 & 4 \end{bmatrix}^{-1}
\begin{bmatrix} 0\\
1\\
1\\
1 \end{bmatrix}
$$
If I... | The value of $$
\begin{bmatrix} s+2 & -1 & 0 & 0\\
0 & s+3 & 0 & 0\\
-1 & -2 & s & -1\\
2 & -1 & 1 & 4 \end{bmatrix}^{-1}$$
$$\text{is}$$
$$\begin{bmatrix}
\frac{1}{s + 2} & \frac{1}{s^2 + 5 s + 6} & 0 & 0 \\
0 & \frac{1}{s + 3} & 0 & 0\\ \frac{2}{4 s^2 + 9 s + 2} & \frac{9 s + 20}{4 s^3 + 21 s^2 + 29 s + 6} & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of ways in which $38808$ can be expressed as a product of 2 coprime factors? Number of ways in which $38808$ can be expressed as a product of $2$ coprime factors ?
the answer given is $8$ ways,
what I did was,
$$38808 = 2^3 \times 3^2 \times 7^2 \times 11$$
so the number of ways of expressing $38808$ as product... | No, 8 is correct. Look at the prime factorisation. Two factors would be coprime if the four exponent-expressions in the prime factorisation were partitioned between the two factors. There are 8 ways of doing this, as follows.
$$\{2^3 , 3^2\cdot 7^2\cdot 11\}$$
$$\{2^3\cdot 3^2, 7^2\cdot 11\}$$
$$\{2^3\cdot 7^2, 3^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
How does one prove that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}?$ I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:
$\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+... | You want to show that
$$
\frac1{\sqrt{3n+1}}\;\frac{2n+1}{2n+2}\leq\;\frac1{\sqrt{3n+4}}.
$$
Since everything is positive, this inequality is the same as
$$
(3n+4)(2n+1)^2\leq (3n+1)(2n+2)^2.
$$
After expanding and cancelling the $n^3$ terms we get
$$
12n^2+19n\leq 24n^2+20n.
$$
This inequality holds trivially fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Divide with remainder $\frac{x^2}{x^2 + x + 2}$ I am having a hard time long dividing:
$$\frac{x^2}{x^2 + x + 2}.$$
Could someone please show a step by step way to divide this, as I can only get it down to : $1 + \frac{x^2}{x + 2}$.
Thank you for your time!
| You can do long division of polynomials exactly like one does long division of integers on paper.
Set it:
___________________________
x^2+x+2 | x^2
Now: look at the leading term of the divisor: $x^2$; the leading term of the dividend: $x^2$; how many times does the leading term ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$ Given that $\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}$, where $a,b,c$ are integers. Show $a=b=c=0$
Do I use modular arithmetic?
| A decade late but here's a nice elementary proof:
Take $x = a$, $y = \sqrt[3]{2}b$ and $z = \sqrt[3]{4}c$
Note the identity
$$x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)((x - y)^2 + (y - z)^2 + (x - z)^2)$$
If $x = y = z$, we have $a = b = c = 0$.
Otherwise $x + y + z = 0 \iff x^3 + y^3 + z^3 - 3xyz = 0$
Substitute... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework:
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
| If we choose to take a (purely) trigonometric route, we might start like this:
$$
\eqalign{
x&=\tan\theta\\
dx&=\sec^2\theta\,d\theta\\
I&=\int_0^1\sqrt{1+x^2}\,dx\\
&=\int_0^\frac{\pi}{4}\sec^3\theta\,d\theta\\
}
$$
However at this stage, we actually need the substitution
$t=\sec\theta+\tan\theta$, believe it or not:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Minimal polynomial of $\sqrt[3]{7-\sqrt{2}}$ I want to find the minimal polynomial (over $\mathbb{Q}$) of: $k:=\sqrt[3]{7-\sqrt{2}}$.
With simple 'tricks' I got that: $P=(x^3-7)^2+2$ is a polynomial such that $P(k)=0$.
But I don't know if, or how to prove that $P$ is the minimal polynomial. How can I prove this/find t... | Let $\alpha = \sqrt[3]{7 - \sqrt{2}}$ and $P(x) = x^6 - 14x^3 + 47$. Since $P(\alpha) = 0$, in order to show $P$ is irreducible it suffices to show $[Q(\alpha):Q] = 6$. Since $7 - \alpha^3 = \sqrt{2}$, you have that $\sqrt{2} \in Q(\alpha)$ and thus
$$[Q(\alpha):Q] =[Q(\alpha):Q(\sqrt{2})][Q(\sqrt{2}):Q] = 2[Q(\alpha)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under ... | The statement seems wrong: Put $a:=0.1$, $b:=1$. Then the left side is
$${1\over\sqrt{1.01}}+{1\over\sqrt{2}}\doteq1.702\ ,$$
and the right side is
$${2\over\sqrt{1.1}}\doteq 1.907\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Finding $x^2+y^2+z^2$ given that $x+y+z=0$, $x^3+y^3+z^3=3$ and $x^4+y^4+z^4=15$ I just ran into this:
$$\begin{align}
x^1+y^1+z^1&=0\\
x^3+y^3+z^3&=3\\
x^4+y^4+z^4&=15\\
x^2+y^2+z^2&=\text{?}
\end{align}$$
I know the answer, but can this be solved without guessing?
| Yes, this can be solved without guessing, using Newton's identities.
Since $x + y + z = 0$, they are the roots of $t^3 + at -b = 0$.
Newton's identities give us (in a straightforward mechanical manner) that
$$x^3 + y^3 + z^3 = 3b$$
$$x^4 + y^4 + z^4 = 2a^2$$
and
$$x^2 + y^2 + z^2 = -2a$$
This gives us $b=1$ and $2a^2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Difficult conversion from polar equation to rectangular equation. How do we convert this into rectangular equation?
$r=5\theta$
| I get this easily, check that if i am wrong:
$r = a\theta$, so in terms of Cartesian coordinates…
$$x = a\theta\cos\theta,\qquad y =a\theta\sin\theta.$$
We can also solve for $x$ and $y$ through simple algebraic manipulation
We know: $r^2=x ^2+ y^2$. Let's solve for $x$ first:
$$\begin{align*}
r &= a\theta\\
r^2&=a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding pairs of triangular numbers whose sum and difference is triangular The triangular numbers 15 and 21 have the property that both their sum and
difference are triangular. There are another 4 pairs less than 1000. To complete this problem,
I have done like this:
To satisfy the conditions, we have to find $u$, $v$... | There are infinitely many solutions. I'll first give infinitely many, and then talk about describing the set of all solutions.
Let $(a,u)$ be a solution to Pell's equation
$$a^2-24u^2 =1$$
with $a$ and $u$ odd. All the integer solutions to this equation are of the form $a+u\sqrt{24} = (5+\sqrt{24})^n$ (using the metho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/127514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
} |
How can I find power series of $f(x)$? $$f(x)=\dfrac{1}{1+\dfrac{x}{1+\dfrac{x^2}{1+\dfrac{x^3}{1+\dfrac{x^4}{\ddots}}}}}$$
How can a power series be found given the continued fraction $f(x)$?
I'm trying to find $f(x) =1+a_1x+\dfrac{a_2x^2}{2!}+\dfrac{a_3x^3}{3!}+\dfrac{a_4x^4}{4!}+\cdots$
I tried some ways to define $... | You are looking for the reciprocal of the classic Rogers-Ramanujan Continued fraction. A listing of some of the power series coefficients can be found here: OEIS. You might be able to find more references there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\sum\limits_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$ I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-... | The Heaviside Method gives the partial fraction decomposition
$$
\begin{align}
&\frac{1}{(n-2)(n-1)n(n+1)(n+2)}\\[6pt]
&=\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\tag{1}
\end{align}
$$
Notice that
$$
\frac{1}{24}-\frac{1}{6}+\frac{1}{4}-\frac{1}{6}+\frac{1}{24}=0\tag{2}
$$
There... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $\frac{4^{x+1}-9\cdot2^x+2}{4^x-5\cdot 2^x-24}\le 0$ Solve $$\frac{4^{x+1}-9\cdot2^x+2}{4^x-5\cdot 2^x-24}\le 0$$
So letting $\alpha = 2^x$,
$$\frac{4\alpha^2-9\alpha+2}{\alpha^2-5\alpha-24}\le 0$$
$$\frac{(4\alpha - 1)(\alpha-2)}{(\alpha+3)(\alpha-8)}\le 0$$
Multiply both sides by denominator squared:
$$(4\alph... | $$\frac{f(\alpha)}{g(\alpha)}=\frac{(4\alpha - 1)(\alpha-2)}{(\alpha+3)(\alpha-8)} \leq 0$$
$1.$ $f(\alpha) \leq 0$ and $g(\alpha)>0$
Hence :
$\left(\alpha \in \left[\frac{-1}{4},2\right]\right) \cap (\alpha \in (-\infty,-3) \cup (8,+\infty)) \Rightarrow \alpha \in \emptyset$
$2.$ $f(\alpha) \geq 0$ and $g(\alpha) <0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but ... | a) $\small n^5-n = n \cdot(n^4-1) $ and $\small (n^4-1) $ is divisible by 5 by little fermat.
b) $\small n^5-n = n \cdot(n^2-1)(n^2+1) $
$ \small \qquad \qquad =n \cdot (n-1)(n+1)(n^2+1) $
$\small \qquad \qquad = (n-1)n(n+1) \cdot (n^2+1) $
$\qquad$ and $\small (n-1)n(n+1) $ is divisble by 6 (because by 2 and by 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 18
} |
How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$ How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
| Put $x = \tan{\theta}$. Then you have
\begin{align*}
I &= \int_{0}^{\pi/2} \log(\tan\theta + \cot\theta) \ d\theta \\\ &= \int_{0}^{\pi/2} \log\frac{2}{\sin{2\theta}} \ d\theta \\\ &= - \int_{0}^{\pi/2} (\log{\sin\theta} +\log{\cos\theta}) d\theta \\\ &= -2 \int\limits_{0}^{\pi/2} \log\: {\sin\theta} \ d\theta = \pi \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding $\int e^{2x} \sin{4x} \, dx$ Finding $$\int e^{2x} \sin 4x \, dx$$
I think I should be doing integration by parts...
If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) - \color{red}{\int (-\frac{1}{4} \cos ... | I would jump into complex values,
and noting that
$\sin 4x = \Im e^{4 i x}$
(I wrote \Im there, wanting to get the imaginary part, and got the $\Im$),
$$\int e^{2x} \sin 4x \, dx
= \Im \int e^{2x+4ix} \, dx
= \Im \int e^{x(2+4i)} \, dx
= \Im \frac{e^{x(2+4i)}}{2+4i}.
$$
Since $\frac{1}{2+4i} = \frac{2-4i}{20} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Finding closed forms for $\sum n z^{n}$ and $\sum n^{2} z^{n}$
Using the identity $\frac{1}{1-z} = 1 + z + z^2 + \ldots$ for $|z| < 1$, find closed forms for the sums $\sum n z^n$ and $\sum n^2 z^n$.
My solution: Because $\displaystyle1 + z + z^{2} + \ldots = \frac{1}{1-z}$, $\displaystyle1 + 2z + 3z^2 + \ldots = \su... | I'd do the second one like this
$$
\begin{align}
\sum_{n=1}^\infty n^2 z^{n-1} & = z\sum_{n=2}^\infty n(n-1) z^{n-2} + \sum_{n=1}^\infty nz^{n-1} \\[10pt]
& = z\frac{d^2}{dz^2} \frac{1}{1-z} + \frac{d}{dz} \frac{1}{1-z}= \cdots\cdots
\end{align}
$$
(and then multiply by $z$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solving $x^3-y^3=xy+61$ in integers I'm trying to solve the following equation:
$$x^3-y^3=xy+61$$
I got:
$$(x-y)(x^2+xy+y^2)=xy+61$$
But I can't go any further. I am looking for a solution in integers.
I need some hints to proceed, please!
| If we restrict ourselves to positive solutions, we can do it by brute force. In that case clearly $x>y>0$.
If $x-y=1$, we have $x^2+y^2=61$, which has the unique solution $x=6,y=5$.
If $x-y=2$, we have $2(x^2+xy+y^2)=xy+61$, or $2(x^2+y^2)+xy=61$; clearly $x$ and $y$ must both be odd, with $x^2+y^2<30$. The only candid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Possible typo in problem involving quotients of powers Am I correct in stating that if $$\frac{x^n}{x^{n-1}}=\frac{x^n}{x^n\cdot x^{-1}}=\frac{1}{x^{-1}}=x$$ then
$$\begin{align*}
\left|\frac{(x^2-5x+2)^n}{2^{n+1}}\cdot\frac{2n}{(x^2-5x+2)^{n-1}}\right|&=\left|\frac{(x^2-5x+2)^n}{(x^2-5x+2)^{n-1}}\cdot\frac{2n}{2^{n+1}... | With the structure of the problem as you've given it, if the supposed answer is $\left|(x^2-5x+2)\right|\frac{1}{2}$, I'd agree that the $2n$ should have been $2^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a quicker way of doing this integral? I have an integral $\int_0^\infty {x^2\over 1+x^4} dx$. I gave it a go and it turned out quite messy, so I consulted Wolfram Alpha but the steps given there seem rather long winded too. Is there is a faster way of doing the integral?
| As SauravTomar pointed out
$$
\int\limits_{0}^\infty\frac{x^2}{x^4+1}dx=\int\limits_{0}^\infty\frac{1}{x^4+1}dx
$$
so
$$
\int\limits_{0}^\infty\frac{x^2}{x^4+1}dx=
\frac{1}{2}\int\limits_{0}^\infty\frac{x^2+1}{x^4+1}dx=
\frac{1}{2}\int\limits_{0}^\infty\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx
$$
$$
=\frac{1}{2}\int\... | {
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What's the sum of $\sum_{k=1}^{\infty} e^{-k(x-k)^{2}}$? Absolute convergence and uniform convergence are easy to determine for this power series. What could be a possible approach to find the sum of this series
$\sum_{k=1}^{\infty} e^{-k(x-k)^{2}}$
(the sum or an estimate of the sum)?
| The approach here is heuristic.
We find an asymptotic formula for the sum for large $x$, Eq. (1) below.
Let $f(x) = \sum_{k=1}^\infty f_k(x)$ where $f_k(x) = e^{-k(x-k)^2}$.
Each term $f_k(x)$ is an unnormalized Gaussian distribution with mean $k$ and standard deviation $\sigma_k = 1/\sqrt{2 k}$.
For $k \ge 18$ we find... | {
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"source": "stackexchange",
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Area and orthocenter I need help with the following proble.
Given a triangle ABC with orthocenter H and altitudes $CM= 2 \sqrt 2$ and $AN=3$. If H divides the altitude BP into segments with ratio 5:1, i.e.BH:HP=5:1, find the area of $\triangle ABC$.
| Let $CH=x$ and $HM=Y$. Let $AH=u$ and $HN=v$. Let $BH=5t$ and $HP=t$.
Triangles $HPC$ and $HMB$ are similar. It follows that $\frac{t}{x}=\frac{y}{5t}$ and therefore $xy=5t^2$. In a similar way we can show that $uv=5t^2$. So far we have the equations
$$x+y=2\sqrt{2},\qquad u+v=3,\qquad xy=uv.$$
We need another equati... | {
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Showing this sequence converges. Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?
I did the following.
Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$
\begin{align}
\left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \... | The sequence, as you have guessed right, converges to $2+3i$. Your argument is also almost fine. You need to clarify which norm you are using to prove convergence. Usually one uses the Euclidean norm, in which case, you get
\begin{align}
\left \lVert x_n - (2+3i) \right \rVert_2 & = \left \lVert \left( \frac{2n^3+n}{... | {
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"source": "stackexchange",
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General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_... | As stated as a comment, the result is here. It really makes block diagonal matrices wonderful, hence finding canonical forms important.
| {
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"source": "stackexchange",
"question_score": "14",
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how to calculate this infinite integral of infinite product of cosine What is the value of this nontrivial itegral:
$$\int_0^{+\infty} \left( \prod_{n = 1}^{+\infty} \cos \frac{x}{n}\right) \, \mbox d x$$
I don't know if there is nice closed answer with known constants.
| Beginning of an answer. Use these:
$$\begin{align}
\cos x &= \prod_{k=0}^\infty \left(1-\frac{4x^2}{(2k+1)^2 \pi^2}\right)
\\
\frac{\sin x}{x} &= \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2 \pi^2}\right)
\end{align}$$
so that
$$\begin{align}
f(x) &:= \prod_{n=1}^\infty \cos \frac{x}{n} = \prod_{n=1}^\infty \prod_{k=0}^\... | {
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"source": "stackexchange",
"question_score": "5",
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The calculation of a series Calculate the series
\begin{equation}
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.
\end{equation}
| Hint:
$$\frac 1 {(4n+1)(4n+3)}=\frac 1 2\left(\frac{1}{4n+1}-\frac 1{4n+3}\right) $$
Then note that
$$\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}} = \frac{1}{{2\left( {2n+1} \right) + 1}} - \frac{1}{{2\left( {2n} \right) + 1}}$$
from where
$$\sum\limits_{n = 1}^m {\left( {\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}}} \right)} =... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculating $\int \frac{dx}{1+3\sin x+\cos x}$ Can I get a detailed answer on integrating this? I am currently stumped.
$$
\int \frac{dx}{1+3\sin x+\cos x}
$$
Thanks.
| We can solve this example by taking trigonometric substitution.
Let $$\tan{\frac{x}{2}}=t$$
$$\therefore\frac{1}{2}\sec^{2}{\frac{x}{2}}dx=dt$$
$$\Rightarrow dx=\frac{2dt}{1+t^{2}}$$
Now,
$$\sin x=\frac{2\tan{\frac{x}{2}}}{1+\tan^{2}\frac{x}{2}}=\frac{2t}{1+t^{2}}$$
and
$$\cos x=\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\... | {
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Showing inequality for harmonic series. I want to show that $$\log N<\sum_{n=1}^{N}\frac{1}{n}<1+\log N.$$ But I don't know how to show this.
| It is a good excercise to show that
$$\tag 1 \frac{x}{{x + 1}} \leqslant \log \left( {1 + x} \right) \leqslant x$$
One alternative is expanding $\log$ into powers of $x$ and $\dfrac{x}{x+1}$, which gives
$$\eqalign{
& \log \left( {1 + x} \right) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} +- \cdo... | {
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Application of Cauchy's convergence criterion Use Cauchy's convergence criterion to prove convergence of $x_n$
$$x_n=1-\frac{1}{2}+\frac{1}{3}-\cdots+(-1)^{n+1}\frac{1}{n}$$
as far as I am concerned,supposing that $m>n$, it's apparently warranted that $|x_m-x_n|=\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}$
but, furthermore, ... | My post will going to aim and clarify that inequality in your textbook by using some elementary approach. First of all, notice that each term in your series that is set on an odd position is positive. Let's fix n as an even number and fix an even number of terms of your series and then we get:
$$S=\sum_{k=n+1}^m (-1)^... | {
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"source": "stackexchange",
"question_score": "2",
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Solving $\frac{dy}{dx} = xy^2$ This problem appears to be pretty simple to me but my book gets a different answer.
$$\frac{dy}{dx} = xy^2$$
For when y is not 0
$$\frac{dy}{y^2} = x \, dx$$
$$\int \frac{dy}{y^2} = \int x \, dx$$
$$\frac{-1}{y^1} = \frac{x^2}{2}$$
$$\frac{-2}{x^2} = y$$
Is there anything wrong with thi... | Continuing from what you did, we have
$$\int \frac{dy}{y^2} = \int x\,dx$$
$$-\frac{1}{y} + C_2= \frac{x^2}{2}+C_1$$
Where $C_1$ and $C_2$ are constants of integration. Letting $C_3=C_1-C_2$
$$-\frac{1}{y} + C_2= \frac{x^2}{2}+C_1=-\frac{1}{y}= \frac{x^2}{2}+C_3$$
Solving for $y$, we have:
$$-\frac{1}{y}= \frac{x^2}{2... | {
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"source": "stackexchange",
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a representation of $f''$ only using terms of $f$ and constants. Using Taylor formulare How can I prove the following result?
$
f''\left( x \right) \approx Af\left( x \right) + Bf\left( {x + h} \right) + Cf\left( {x + 2h} \right) + Df\left( {x + 3h} \right)\,\,h > 0
$
Obviously I have to consider the taylor expansion ... | Write
\begin{align}
f(x+h) & = f(x) + hf'(x) + \frac{h^2}2 f''(x) + \frac{h^3}{6}f'''(\zeta_1) \\\
f(x+2h) & = f(x) + 2hf'(x) + \frac{(2h)^2}2 f''(x) + \frac{(2h)^3}{6}f'''(\zeta_2) \\\
f(x+3h) & = f(x) + 3hf'(x) + \frac{(3h)^2}2 f''(x) + \frac{(3h)^3}{6}f'''(\zeta_3). \\\
\end{align}
Then after you work out the equati... | {
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"source": "stackexchange",
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prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that :
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
| I assume that by $x,y,z$ you mean $a,b,c$.
Without loss of generality, we can assume that $c$ is the smallest of the three i.e. $a, b \geq c >0$. Then
\begin{align}
\dfrac{b+c}{a^2} + \dfrac{c+a}{b^2} + \dfrac{a+b}{c^2} - \dfrac1a - \dfrac1b - \dfrac1c & = \dfrac{b+c-a}{a^2} + \dfrac{c+a-b}{b^2} + \dfrac{a+b-c}{c^2}\\
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$ How could we prove that
$$ \frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$
I have reduced it the form
$$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$
I am not sure what to do next to get snappy solution. Any ideas?
| This is another way, using the basics about logarithms.
You can use just the identity
\begin{equation}
b^{\rm{log}_{\frac{b}{a}}b}=b\cdot a^{\rm{log}_{\frac{b}{a}}b}
\end{equation}
Note that $\frac{1}{\rm{log}_x{\frac{3}{4}}} =$ log$_\frac{3}{4} \ x$, you can use that to get
\begin{equation}
\frac{4^{\frac{1}{\rm{log}... | {
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"source": "stackexchange",
"question_score": "2",
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Computing the derivative from the definition Using the limit definition of the derivative which I know is:
$$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$
I am trying to solve this problem
$$f(x)= \frac{x}{x+2} $$
How do I go about properly solving this, I seemed to get
$$\frac{x}{x+2}\ $$
as my answ... | $f(x) = \frac{x+2 - 2}{x+2} = 1 - \frac{2}{x+2}$
$$\frac{f(x+h) - f(x)}{h} = \frac{\frac{2}{x+2} - \frac{2}{x+h+2}}{h} = \frac{2}{h} \cdot \frac{x+h+2 - (x+2)}{(x+2)(x+h+2)} = \frac{2}{(x+2)(x+h+2)} \xrightarrow{h\to 0} \frac{2}{(x+2)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to compute $\int\frac{x^2}{x^2+4}\,dx$ and $\int\frac{1}{\sqrt{4-x^2}}\,dx$? My calculus teacher gave us a list of integrals to solve through the substitution method. I've been trying everything for hours, and I just can't find out how to compute these:
$$∫\frac{x^2}{x^2+4}\,dx\qquad\text{and}\qquad \int\frac{1}{\s... | The first one can be solved by rewriting it first
$$\int\frac{x^2}{x^2+4}\,dx = \int\frac{x^2+4-4}{x^2+4}\,dx = \int\,dx - 4\int\frac{dx}{x^2+4}.$$
Then we can solve the second one of these integrals (first one is easy) by factoring out $4$ from the denominator and then using the substitution $u=\frac{x}{2}$:
$$\begin{... | {
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Show $|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$ for complex $a$, $b$, $c$ How to prove for any complex numbers $a$, $b$, $c$, the inequality $$|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$$ is correct?
| Both sides are non-negative, so it suffices to show that the square of the left-hand-side is at least the square of the right-hand-side. That is, we wish to show:
$$
|a|^2+|b|^2+|c|^2+|a+b+c|^2+2|ab|+2|bc|+2|ac+2(|a|+|b|+|c|)|a+b+c|
\geq\\
|a+b|^2+|b+c|^2+|a+c|^2+2(|a(a+b+c)+bc|+|b(a+b+c)+ac|+|c(a+b+c)+bc|)
$$
The squa... | {
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"source": "stackexchange",
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How to evaluate this definite integral: $\int_0^1 {(1-x^3)}/{(1-x^5)} \;\rm dx$ I need to compute:
$$\int_0^1 \frac{1-x^3}{1-x^5} dx$$
I tried integrating it by partial fractions but couldn't succeed. Is there any other way to integrate this?
| Use partial fraction:
$$
\frac{1-x^3}{1-x^5} = \frac{\sqrt{5}-1}{\sqrt{5}(2x^2+(1+\sqrt{5})x+2)}+\frac{\sqrt{5}
+1}{\sqrt{5}(2x^2+(1-\sqrt{5})x+2)}
$$
and then integrate both summands with
$$
\int 1/(ax^2+bx+c)dx= 2\frac{\tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)}{\left(\sqrt{4ac-b^2}\right)} + \text{constant},... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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An infinite product for $\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^2$? Given the Dedekind eta function,
$$\eta(\tau) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$$
where $q = \exp(2\pi i\tau)$. Consider the following "family",
$\begin{align}
\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{24} &= \frac{u^8}{(-1+16u^8)^2},\;\;... | Michael Somos just today found the identity,
$$
\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} = \frac{s}{s^2-3s-1}$$
where,
$$s=\frac{1}{q}\; \prod_{n=1}^\infty \frac{ (1-q^{13n-2})(1-q^{13n-5})(1-q^{13n-6})(1-q^{13n-7})(1-q^{13n-8})(1-q^{13n-11}) }{(1-q^{13n-1})(1-q^{13n-3})(1-q^{13n-4})(1-q^{13n-9})(1-q^{13n-10})(... | {
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"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
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proving inequality? Here is another inequality I am trying to prove:
Let $a,b,c$ be positive numbers. Prove that:
$$1) \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geqslant (a+b+c)$$
$$2) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}$$
In the book's hint, it uses t... | $\frac {bc}{a}+\frac{ca}{b}+\frac{ab}{c}=\frac{1}{2}(\frac{ca}{b}+\frac{ab}{c})+\frac{1}{2}(\frac{ab}{c}+\frac{bc}{a})+\frac{1}{2}(\frac{bc}{a}+\frac{ca}{b}\geq\sqrt\frac{ca}{b}\sqrt\frac{ab}{c}+\sqrt\frac{ab}{c}\sqrt\frac{bc}{a}+\sqrt\frac{bc}{a}\sqrt\frac{ca}{b}=a+b+c$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
This seems clear, but I don't know how to prove this..
I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
| Here's a colored, edited version of user17762's answer.
The proof is by induction. Let us denote $\gcd(a,b) = d$. It's given $\gcd(a,b) = 1
\iff$
by Bezout's Lemma, there exist $x,y \in \mathbb{Z}$ such that $ax+by = d. (\dagger)$
Assume that the statement is true for all $n \leq k$. Scilicet $\gcd(a^n,b^n) = d^... | {
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Find the value of a succession of additions $$1^2+2^2+3^2+...+10000$$
How do you find the exact value of that?
I'm studying induction, and I'm still not sure how to get that value.
| Let $$S=1+2+3+...+n$$ then, $$S=n+(n-1)+(n-2)+...+1$$ Adding corresponding terms of two equations gives , $$2S=(n+1)+(n+1)+(n+1)+...(n+1)=n(n+1)\implies S=n(n+1)/2$$ Now $$(n+1)^3-n^3=3n^2+3n+1$$ $$\implies 2^3-1^3=3(1)^2+3(1)+1$$ $$3^3-2^3=3(2)^2+3(2)+1$$ $$4^3-3^3=3(3)^2+3(3)+1$$ $$..$$$$..$$$$(n+1)^3-n^3=3(n)^2+3(n)... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x$ I have to evaluate:
$$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$
I can't get the right answer! So please help me out!
| Let $I$ denote the integral and consider the substitution $u= \frac{\pi }{2} - x.$ Then $I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u}}{\sqrt{\cos u } + \sqrt{\sin u }} du$ and $2I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u} + \sqrt{\sin u }}{\sqrt{\cos u } + \sqrt{\sin u }} du = \frac{\pi... | {
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} |
Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead... | We show that
$$
\left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}
$$
for any $n \geq 2$.
For any $n \geq 2$ we have that
$$
n \int^n_{n-1} \frac{1}{x(x+1)}dx \leq \int^n_{n-1} \frac{1}{x}dx
$$
because $\frac{n}{x+1} \leq 1$ for all $x \in [n-1,n]$.
But this inequality is equivalent to
$$
n \int^n_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 14,
"answer_id": 9
} |
Closed form solution of Fibonacci-like sequence Could someone please tell me the closed form solution of the equation below.
$$F(n) = 2F(n-1) + 2F(n-2)$$
$$F(1) = 1$$
$$F(2) = 3$$
Is there any way it can be easily deduced if the closed form solution of Fibonacci is known?
| Define $g(z) = \sum_{n \ge 0} F(n + 1) z^n$, write the recurrence as:
$$
F(n + 3) = 2 F(n + 2) + 2 F(n + 1) \qquad F(1) = 1, F(2) = 3
$$
Multiply the recurrence by $z$, sum over $n \ge 0$ and get:
$$
\frac{g(z) - F(1) - F(2)}{z^2} = 2 \frac{g(z) - F(1)}{z} + 2 g(z)
$$
Solve for $g(z)$:
$$
g(z) = \frac{1 + z}{1 - 2 z - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solving a recurrence of a sequence The sequence $a_n$ is given by recurrence $$a_0=0,a_1=1,a_n=a_{n-2}+n+4, n>2$$ How to find the general term I am interested on different approach. Can generating functions help
| The recurrence can indeed be solved using generating functions. I prefer to write the recurrence as
$$a_n=a_{n-2}+n+4-4[n=0]-4[n=1]\;,$$
using Iverson brackets and assuming that $a_n=0$ for $n<0$. Then multiply through by $x^n$ and sum over $n\ge 0$ to get the generating function, $g(x)$:
$$\begin{align*}
g(x)&=\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Why is the inradius of any triangle at most half its circumradius? Is there any geometrically simple reason why the inradius of a triangle should be at most half its circumradius? I end up wanting the fact for this answer.
I know of two proofs of this fact.
Proof 1:
The radius of the nine-point circle is half the circu... | Compute the area of a triangle (first method):
Consider the following diagram:
$\hspace{4.5cm}$
The area of the green triangle is
$$
A=\tfrac12ab\sin(\theta)\tag{1}
$$
By the Inscribed Angle Theorem, the angle that $c$ subtends at the origin, $o$, is $2\theta$. Therefore, we get that
$$
c=2R\sin(\theta)\tag{2}
$$
Combi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Solving $\cos^2 \theta + \cos \theta = 2$ Solve the following for $\theta$:
$\cos^2 \theta + \cos \theta = 2$ [Hint: There is only one solution.]
I started this out by changing $\cos^2\theta$ to $\dfrac{1+\cos(2\theta)}{2}+\cos\theta=2$
$1+\cos(2\theta)$ turns into $1+\cos^2\theta-\sin^2\theta$ which all becomes;... | Replacing $\cos^2\theta$ with and expression involving $\cos2\theta$ is not necessarily a good idea; then you have to deal with cosines of two different angles.
A better approach is to realize that what we have is a quadratic equation: let us define $y$ to be $y=\cos\theta$. Then we can rewrite the equation as
$$y^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Write each expression in the form $ca^pb^q$ Write each expression in the form $ca^pb^q$
c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$
\begin{align*}
&= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b... | I'll do the first in two ways, $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$, and we'll see what we think. First, I'm going to separate all the 'numbers'.
*
*$\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}= \dfrac{2a\frac{1}{b}}{3\frac{1}{a}} = \frac{2}{3}\dfrac{a\frac{1}{b}}{\frac{1}{a}}$
Now I'll do the $a$ ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to solve this equation involving $()^x$? I have the equation:
$\left (\sqrt{3+2\sqrt{2}} \right )^x- \left (\sqrt{3-2\sqrt{2}} \right )^x=\frac{3}{2}$
I wrote the left side of the equation as square roots.
$(1+\sqrt{2})^x-(1-\sqrt{2})^x=\frac{3}{2}$
How do I found out the final solution? Thank you very much!
P.S. T... | First of all, $\sqrt{3-2\sqrt{2}}$ is $\sqrt{2} - 1$.
Now, Let $(\sqrt{2}+1)^x$ be y.
$y - \frac{1}{y} = \frac{3}{2}$.
So $2y^2 - 3y - 2 = 0$.
So y is 2 or $-\frac{1}{2}$. As y = $(\sqrt{2}+1)^x$, y must be positive and thus 2.
$(\sqrt{2}+1)^x = 2$
So $x = \frac{log(2)}{log(\sqrt{2}+1)}$, which is same as option c) her... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $\frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{z^2+zx+x^2}}+\frac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$
Let $x,y,z >0$. Prove
that:$$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}}
\ge \sqrt{3}$$
My solution:
By Hölder,
$$\left(\sum\frac{x}{\sqrt{4y^2+yz+4z^2}}\right)^2(\... | We need to prove that
$$\sum_{cyc}x\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq\sqrt{3\prod_{cyc}(x^2+xy+y^2)}.$$
Now, by C-S
$$(x^2+xy+y^2)(x^2+xz+z^2)=\left(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\right)\left(\left(x+\frac{z}{2}\right)^2+\frac{3}{4}z^2\right)\geq$$
$$\geq\left(\left(x+\frac{y}{2}\right)\left(x+\frac{z}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the divisors of the number $p^3q^6$ My text says that $p^3q^6$ has 28 divisors. Could anyone please explain to me how they got 28 here?
Edit:
$p$ and $q$ are distinct prime numbers Sorry for the late addition..
| Let's take $p=3$ and $q=2$ as an example. Then the number is $3^32^6 = 1728$, and the 28 divisors of 1728 are:
$$\begin{matrix}
1&2&4&8&16&32&64 \\
3&6&12&24&48&96&192 \\
9 & 18 & 36 & 72 & 144 & 288 & 576 \\
27 & 54 & 108 & 216 & 432 & 864 & 1728
\end{matrix}$$
These values are, respectively:
$$\begin{matrix}
2^03^0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
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Inequality for cosines Is the following inequality in a triangle known?
$$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$
It looks correct to me but I would appreciate if someone confirm it.
| $\cos A+\cos B+\cos C=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$
$=2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}\quad$ as $\quad A+B+C=\pi$ $\implies\cos\frac{A+B}{2}=\sin\frac{C}{2}$
$\leq 2\sin\frac{C}{2}+1-2\sin^2\frac{C}{2}\quad$ as $\quad\cos\frac{A-B}{2}\leq 1$
$=-\frac{1}{2}(2\sin\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
} |
Confusing double angle identity How would I solve the following double angle identity.
$$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$$
So far my work is
$$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$$
But how would I proceed.
| \begin{align*}
\cos^4(x) &= \left(\frac{e^{ix}+e^{-ix}}{2}\right)^4\\
&= \frac{e^{4xi} + 4e^{2xi} + 6 + 4e^{-2xi} + e^{-4xi}}{16}\\
&= \frac{3}{8} + \frac{1}{2} \frac{e^{2xi} + e^{-2xi}}{2} + \frac{1}{8} \frac{e^{4xi}+e^{-4xi}}{2}\\
&= \frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\... | Replacing $B$ by $B+Y$ in your expression for $\tan(A+B)$ gives
$$\dfrac{\tan A + \tan(B+Y)}{1 - \tan A \tan(B+Y)}$$
Expanding gives
$$\dfrac{\tan A + \frac{\tan B + \tan Y}{1 - \tan B \tan Y}}{1 - \tan A \frac{\tan B + \tan Y}{1 - \tan B \tan Y}}$$
Multiplying through by $1 - \tan B \tan Y$ gives
$$\frac{\tan A(1 - \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Polynomial coefficients in exponential-series: how can I convert this into a composite of $\exp(x)$? Assume we have the exponential-series $ \small \exp(x) = 1+ {x \over 1!} + {x^2 \over 2! } + \cdots = \sum\limits_{k=0 }^\infty {x^k \over k!} $ modified with a polynomial in the coefficients
say $ \qquad \displaystyle... | \begin{eqnarray}
\sum_{k=0}^\infty k\frac{x^k}{k!}&=&\sum_{k=1}^\infty \frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty\frac{x^k}{k!}=xe^x;\cr
\sum_{k=0}^\infty k^2\frac{x^k}{k!}&=&x\sum_{k=1}^\infty k\frac{x^{k-1}}{(k-1)!}=x\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x(x+1)e^x\cr
\sum_{k=0}^\infty k^3\frac{x^k}{k!}&=&\sum_{k=1}^\infty ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Proving that $\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2}$ How would I prove the following trig identity?
$$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$
My work thus far has been:
$$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\... | First, note that $\dfrac{\cos(A + B)}{\cos A - \cos B} \neq -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$
In particular, if we use something like $A = \pi/6, B = 2\pi/6$, then the left is $0$ as $\cos(\pi/2) = 0$ and the right side is a product of two nonzero things.
I suspect instead that you ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use the given equalities to derive trigonometric functions.
(A) $\sin(-x)=-\sin x$
(B) $\cos(-x)=\cos x$
(C) $\cos(x+y)=\cos x\cos y-\sin x\sin y$
(D) $\sin(x+y)=\sin x\cos y+\cos x\sin y$
Use these equalities to derive the following important trigonometric functions:
f) $\left|\cos\dfrac{x}{2}\right|... | Note that the identity $$\cos^2\theta=\frac{1+\cos 2\theta}2$$ is equivalent to (f), so you shouldn’t be using it: your argument will necessarily be circular. Using (C) is fine, however: just set apply it to $\cos\left(\frac{x}2+\frac{x}2\right)$. You’ll also need the Pythagorean identity $\sin^2x+\cos^2x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| Note that $$a+b+c≥3, \tag{1}$$ by AM-GM inequality
Now, by Titu's lemma $$a^2+b^2+c^2 ≥\frac{(a+b+c)^2}3 ≥ \frac{(a+b+c)^2}{a+b+c}$$ by using (1)
then $$a^2+b^2+c^2 ≥a+b+c$$ Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 4
} |
Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS=
$$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$
$$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\... | Brian's answer is short, sweet, and correct. But you might wonder if you can finish it from where you left off. You can.
*
*$\dfrac{\tan^2 \theta - \sin ^2 \theta}{\sin \theta \tan \theta}$
*Expand everything into $\sin$ and $\cos$ and simplify into a fraction. Along the way, we get $\dfrac{(\sin^2 \theta - \sin^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots. Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots.
for real roots, $a \gt 0$, $b^2-4{a}{c} \ge 0$
Here we h... | There are of course 16 equations of the form $ax^2+bx+1=0$ with $a,b \in X:= \{1,2,3,4\}$. Which of them have real roots?
As you note, we have to determine for which $a,b\in X$ we have $b^2-4a\ge 0$.
So if $a=1$, we have $b=2,3,4$.
If $a=2$, we have $b=3,4$, if $a=3,4$ then we must have $b=4$. Summing up there are 7 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Equilateral triangle inscribed in a triangle Consider triangle ABC. Point D is on AC, E is on BC and F is on AB. Given that triangle DEF is equilateral and that segments CD, BE and AF are equal in length, prove that triangle ABC must also be equilateral. I've played with this one for long enough using circles, sine law... | Let $a$, $b$, $c$ be the lengths of the sides opposite $A$, $B$, $C$, and let points $D$, $E$, $F$ lie on the sides opposite $A$, $B$, $C$. (The latter is slightly different than in the original question). Let $r = |DE| = |EF| = |FD|$ be the length of the sides of the equilateral triangle; let $s = |AF| = |BD| = |CE|$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Triple "Pythagorean identity" It's not hard to find multiple trigonometric functions of period $2\pi$ that added to self shifted by some constant offset result in a constant.
In classic pythagorean identity, you have
$$F(x)+F\left(x+\frac{\pi}{2} \right) = 1 $$
where $F(x)=\sin^2 x$
Or you can use symmetry of the sine... | Let $F(x)=\sin(ax+b)$ where a,b are indeterminate constants.
$F(x) + F\left(x + \frac 2 3 \pi \right) + F\left(x+\frac 4 3 \pi\right) $
$=\sin(ax+b)+\sin(a(x+\frac{2\pi}{3})+b)+\sin(a(x+\frac{4\pi}{3})+b)$
$=\sin(ax+b)+\sin(a(x+\frac{4\pi}{3})+b)+\sin(a(x+\frac{2\pi}{3})+b)$
$=(1+2\cos\frac{2\pi a}{3})\sin(a(x+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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A matrix inequality If $A=(a_{ij})$ is a real positive definite symmetric matrix of order $n$. How to show $(n-1)\prod_{i=1}^na_{ii}+\det A\ge \sum_{i=1}^na_{ii}\det A(i)$? $A(i)$ means the submatrix of $A$ by deleting the $i$th row and $i$th column.
| if $n=3$. consider the two matrix below:
$$A=\begin{pmatrix}1 & \frac{1}{4} & -\frac{1}{4}\cr \frac{1}{4} & 1 & -\frac{1}{4}\cr -\frac{1}{4} & -\frac{1}{4} & 1\end{pmatrix}$$
$$B=\begin{pmatrix}1 & -\frac{1}{4} & -\frac{1}{4}\cr -\frac{1}{4} & 1 & -\frac{1}{4}\cr -\frac{1}{4} & -\frac{1}{4} & 1\end{pmatrix}$$
they are ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
| By squaring $a+b+c=3$ we get
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=9.$$
From the AM-GM inequality
(or from the fact that $(x-y)^2=x^2+y^2-2xy\ge 0$, i.e. $2xy\le x^2+y^2$)
we have
$$ab+ac+bc \le \frac{a^2+b^2}2+\frac{a^2+c^2}2+\frac{b^2+c^2}2=a^2+b^2+c^2,$$
i.e. $\frac12(a^2+b^2+c^2) \ge \frac12(ab+ac+bc)$, which is equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outlin... | How about verifying using differentiation? We are asked to show
$$\int_0^x \cos^n t dt = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n} \int^x_0 \cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = \... | {
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"url": "https://math.stackexchange.com/questions/193435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
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Inequality.$\frac{ab}{\sqrt{ab+2c^2}}+\frac{bc}{\sqrt{bc+2a^2}}+\frac{ca}{\sqrt{ca+2b^2}} \geq \sqrt{ab+bc+ca}$ Let $a,b,c > 0$. Prove that (using Hölder's inequality):
$$\frac{ab}{\sqrt{ab+2c^2}}+\frac{bc}{\sqrt{bc+2a^2}}+\frac{ca}{\sqrt{ca+2b^2}} \geq \sqrt{ab+bc+ca}.$$
Thanks :)
I tried to apply Hölder's inequalit... | Using Hölder's inequality, we get
$$
\left(\sum_{cyc} \frac {ab}{\sqrt{ab + 2c^2}}\right) \left(\sum_{cyc} \frac {ab}{\sqrt{ab + 2c^2}}\right) \left(\sum_{cyc} ab (ab + 2c^2)\right) \geq (ab + bc + ca)^3
$$
The above inequality can be rewritten as
$$
\left(\sum_{cyc} \frac {ab}{\sqrt{ab + 2c^2}}\right)^2 (ab + bc +ca)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193820",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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} |
Finding $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$ I try to find $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$.
Left side:
$$\sin(4 \alpha + \frac{\pi}{6}) =$$
$$= \sin4\alpha \times \cos \frac{\pi}{6} + \cos 4\alpha \times \sin\frac{\pi}{6} =$$
$$= \... | Hint:
if $\sin A=\sin B$,then we have $A=k\pi +(-1)^kB$,it is easy to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\sum\limits_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $ How we can find
$$\sum_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $$
| $\frac {1}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})}=\frac {(\sqrt[4] {n+1}-\sqrt[4] {n})}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})(\sqrt[4] {n+1}-\sqrt [4]{n})}=\frac {(\sqrt[4] {n+1}-\sqrt [4]{n})}{n+1-n}=\sqrt[4] {n+1}-\sqrt[4] {n}$ so the sum is transformed to $\sum _{n=1}^{9999} \sqrt[4] {n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/196156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$? Is it possible to evaluate the following integral:$$\int \frac{\sin^3x}{(\sin^3x + \cos^3x)} \, dx$$
| Put $$I = \displaystyle\int{\dfrac{\sin^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x , \quad J = \displaystyle\int{\dfrac{\cos^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x.$$
We have $$I + J = \displaystyle\int{\mathrm{d}x} = x+ C.$$
and
\begin{equation*}
I - J = \displaystyle\int{\dfrac{\sin^3 x - \cos^3 x}{\cos^3 x +\sin^3 x ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Solving an equation with three quadratic radicals in the set of real numbers How do we solve the following equation in the set of real numbers?
$$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$
I tried putting $a=\sqrt{5x−1}$ and $b=\sqrt{5−2x}$ and then $2a^2+5b^2=23$.
| This is another solution.
Put $a=\sqrt{5x-1}$ and $b=\sqrt{5-2x}$.
We have
\begin{equation*}
26-x = \dfrac{47}{23}(5x - 1 ) + \dfrac{129}{23}(5-2x) = \dfrac{47}{23}a^2 + \dfrac{129}{23}b^2,
\end{equation*}
\begin{equation*}
13x+14= \dfrac{93}{23}(5x - 1 ) + \dfrac{83}{23}(5-2x) = \dfrac{93}{23}a^2 + \dfrac{83}{23}b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Finding the all integer solutions How to Find the all integer solutions for:
$$x+y+z=3$$
$$x^3+y^3+z^3=3$$
| We have $x^3+y^3=3-z^3$ and $x+y=3-z$. Since $x+y$ divides $x^3+y^3$, we conclude that $z-3$ divides $z^3-3$, and therefore $z-3$ divides $24$. Similar considerations apply to $x$ and $y$.
So we are down to a finite and indeed fairly short list of candidates. We can use further little tricks to winnow the list.
Rema... | {
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"url": "https://math.stackexchange.com/questions/202848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the solution of this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
Possible Duplicate:
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Help me solve this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
| HINT: Note that $2-\sqrt{3} = \dfrac1{2+\sqrt{3}}$.
Denote $\left(\sqrt{2-\sqrt{3}} \right)^x$ as $t$ and proceed to solve the quadratic in $t$ and hence solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/202997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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