Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Simplify $(\sqrt{x}) + x + 2 = (\sqrt{y}) + y + 2$ $\sqrt{x} + x + 2 = \sqrt{y} + y + 2$
I've simplified as follows:
$\sqrt{x} + x = \sqrt{y} + y$, square both sides
$x + x^2 = y + y^2$
It seems obvious that $x = y$ but I can’t get to that solution by algebraic means.
| both $\sqrt{x}$ and $\sqrt{y}$ are roots of a quadratic equation:
$$
t^2+t + c=0
$$
hence either $\sqrt{x}=\sqrt{y}$ or $\sqrt{x}+\sqrt{y}=-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$168... | I noted that $9,40$ and $41$ are pythagorean triple, i.e $9^2+40^2=41^2$. This fact can be used to solve the question easily. I'll show the solution for the general case:
$a\cos\theta+b\sin\theta=c,$ given that $a^2+b^2=c^2$
$$a\cos\theta+b\sin\theta=c$$
$$b\sin\theta=c-a\cos\theta$$
Squaring both the sides:
$$b^2... | {
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"timestamp": "2023-03-29T00:00:00",
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Why is $2^{n/2}(5/3)^n(\cos(n\pi/4)+\sin(n\pi/4))$ an alternate form of the complex number $(5/3)^n(1+i)^n$? How do I get to that point?
I am aware of the formula
$$z = r (\cos \alpha + i \sin \alpha)$$
and that
$$z^n = r^n(\cos n\alpha+ i\sin n\alpha)$$
But I don't know how to get to what is in the title, particula... | Let
$$z = 1+i = \sqrt{2}\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right) = 2^{1/2}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)$$
Now use de Moivre:
$$z^n = 2^{n/2}\left(\cos\left(\frac{n\pi}{4}\right)+i\sin\left(\frac{n\pi}{4}\right)\right)$$
And finally:
$$\left(\frac{5}{3}\right)^n... | {
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Let a,b $\in$ $\mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$
Let $a,b \in \mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$
The question is from the inequalities section of An Excursion in mathematics by Bhaskaraycharya Pratisthanan. My heuristics include using the AM-GM inequality. I am unable to design the problem to proceed... | You're on right path just a small trick using AM-GM
Write $$a^4+b^4+8=a^4+b^4+4+4$$
And then apply AM-GM
$$\frac{a^4+b^4+4+4}{4}\ge \sqrt[4]{16a^4b^4}$$
$$\frac{a^4+b^4+4+4}{4}\ge 2ab$$
$${a^4+b^4+8}\ge 8ab$$
Or You can just SOS it
$$(a^2-b^2)^2\ge0\implies a^4+b^4-2a^2b^2\ge0$$
$$2(ab-2)^2\ge0\implies2a^2b^2+8-8ab\ge0... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\forall n\in\mathbb{N}, \exists m\in\mathbb{N}; n=\pm1^2\pm2^2\pm\cdots\pm m^2.$ And we choose the positive and negative signs in a way that the equation becomes true.
I think it can be proved with mathematical induction. So here's how I begin:
For $n=1$, $1=+1^2$ which is true. It's also true for 2 and 3, e.g. ... |
Lemma A. We may represent $0,1,2,3$.
$$\begin{eqnarray*}0 &=& 1^2 + 2^2 - 3^2 + 4^2 - 5^2 - 6^2 + 7^2\\ 1 &=& 1^2\\2&=&-1^2-2^2-3^2+4^2\\3&=&-1^2+2^2.\end{eqnarray*}$$
$\phantom{}$
Lemma B. If we are able to represent some $n$, we may represent $n+4$ as well.
Since for any $m$ we have $(m+1)^2-(m+2)^2-(m+3)^2+(m+4)^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic How do I show, step by step, that $\sqrt [3]{2}-\sqrt [3]{4}$ is a root of $x^3+6x+2$?
Start with $x=\sqrt [3]{2}-\sqrt [3]{4}$ do not use the cubic, the cubic is given for convenience.
( This is example 4.1.3 from Introductory ANT by Alaca/Williams )
| Let $a = 2^{1/3}$ and $b = 4^{1/3} = (2^2)^{1/3} = (2^{1/3})^2 = a^2$. Then $$(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3a^3(a-b),$$ where upon substituting values we obtain $$(a-b)^3 = 2-4 - 3(2)(a-b) = -2 - 6(a-b).$$ Thus if $x = a-b = 2^{1/3} - 4^{1/3}$, then $$x^3 + 6x + 2 = 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)} $ Let $x,y,z$ be real numbers in the range of $(0,1)$.
Prove that
$$\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)}.$$
| \begin{eqnarray}
&&[\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)}][xyz+(1-x)(1-y)(1-z)]\\
&=&[\frac{yz}{1-y}+\frac{(1-x)(1-z)}{x}]+[\frac{xz}{1-z}+\frac{(1-x)(1-y)}{y}]+[\frac{xy}{1-x}+\frac{(1-y)(1-z)}{z}]\\
&\ge&[\frac{z}{1-y}+\frac{1-z}{x}-1]+[\frac{x}{1-z}+\frac{1-x}{y}-1]+[\frac{y}{1-x}+\frac{1-y}{z}-1]\\
&... | {
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"timestamp": "2023-03-29T00:00:00",
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convergence series geometric test Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get
$$\frac{2^n ... | You teacher increased the numerator (from $2^n+3$ to $2^n+ 2\cdot 2^n$, since $3\leq 2\cdot 2^n$), and decreased the denominator (from $3^n-1$ to $3^n-3^n/2$, since $1<3^n/2$), thus obtains a fraction (i.e. $\frac{2^n+2\cdot 2^n}{3^n-3^n/2}$) which is bigger than the original (i.e. $\frac{2^n+3}{3^n-1}$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove or disprove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ Is it true that $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ for any positive integer $n$?
We cannot prove the statement using induction as it is, because the left-hand side is increasing while the right-hand side stays constant. So we need to mo... | We want to show that
$$\sum_{k=1}^{\infty} \dfrac{\ln(k+1)}{2^{k}} < \ln(3)$$
We have $\ln(1+k) \leq k$ for all $k \in \mathbb{N}$. Hence, we have
$$\sum_{k=1}^{\infty} \dfrac{\ln(k+1)}{2^k} = \sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} + \sum_{k=11}^{\infty} \dfrac{\ln(k+1)}{2^k} \leq \sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified? I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?
| Note that $f\left(\pm \frac{1}{2}\right) = 16\left(\pm \frac{1}{2}\right)^6 - 1 = -\frac{3}{4}$, so $\pm \frac{1}{2}$ are not roots of $f$.
One way to proceed with the original problem is to write $v^6 := 16 x^6$, that is, $v = \sqrt[3]{4} x$, so that the original polynomial is $v^6 - 1$. This clearly has roots $v = \p... | {
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"timestamp": "2023-03-29T00:00:00",
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The inequality $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+3/4 >0$ holds for all $x\in\mathbb R$
Show
$\forall \ x \in \mathbb{R}:\quad x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\dfrac{3}{4}>0$
My attemps:
Case $x=-1$
That is true for this case
Then for $x \neq - 1$:
$$\dfrac 3 4 - x + x^2 - x^3 + x^4 - x^5 + x^6 = \dfrac{1 + x^7}{1 +... | Another way would be to prove $(1 + x^7) - {1 \over 4}(1 + x) > 0$ for $x > -1$, and $(1 + x^7) - {1 \over 4}(1 + x) < 0$ for $x < -1$, using calculus on $f(x) = (1 + x^7) - {1 \over 4}(1 + x) = x^7 - {1 \over 4}x + {3 \over 4}$. Since $\lim_{x \rightarrow \infty} f(x) = \infty$ and $\lim_{x \rightarrow -\infty} f(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution ... | Note that the polynomial can be written as
$$x^4+2x^2+1 - 3x(x^2+1) = (x^2+1)^2 - 3x(x^2+1) = (x^2+1)(x^2-3x+1)$$
I trust you can finish the rest.
| {
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$8x +9y = 5$ where $x,y \in \mathbb{Z}$
Solve the following Diophantine equation algebaically: $$8x+9y=5$$
Give 3 possible solutions for the equation
I have the following:
The Diophantine equation has solutions $x,y \iff 8x=5\mod{9}$ has a solution $x \equiv\mod{9}$
Since $\gcd(8,9)=1$, by Bezout's Lemma, for $r,t ... | $8x+9y=5\iff8~(x+y)+y=5=8\cdot0+5\iff x+y=0$ and $y=5\iff x=-5$. Then all numbers of the form $x=-5-9k$ and $y=5+8k$ are solutions to the above equation.
| {
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"url": "https://math.stackexchange.com/questions/1025155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I hav... | It's easy. Differentiate
$$\frac{1}{4 (-1 + 2 x^2)} - \frac{\sqrt{2 + x^2 - x^4}}{6 (-1 + 2 x^2)} -
{1\over 4} \log[3 + 2 \sqrt{2 + x^2 - x^4}]$$
To get your integral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$ How can I prove $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$?
We have
$(a+b-c)^2+=a^2+b^2+b^2+2ab-2bc-2ca$,
$(b+c-a)^2+=b^2+c^2+a^2+2bc-2ca-2ab$
$(c+a-b)^2+=c^2+a^2+b^2+2ca-2ab-2bc$
But I don't know how to show the required result?
Please give hint. Thank you
| this inequality is equivalent to $a^2+b^2+c^2\geq ab+bc+ca$ which is
$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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dividing polynomials using long division I'm not following logic of using long division on polynomials.
If you are using regular long division, we would do the following:
+----------
2 | 86
How many times does 2 go into 8? 4. Multiply 2 by 4 and drop down the next tenth power. So we end up with 6.
... | Let's take more easily comparable examples such as $\dfrac{156}{13}$ and $\dfrac{x^2+5x+6}{x+3}$
So for the numerical example, you first subtract $10\times 13$ and then $2 \times 13$ to get $156 = (10+2)\times 13$ so $\dfrac{156}{13}=12$, or laying it out
10+2
+----------
13 | 156
130
... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$
Evaluate $$\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$$
This is a geometric series and since $a = \dfrac{1}{2}$ Then the infinite sum is jsut $S = \dfrac{1}{1-\frac{1}{2}} = 2$ Then I multiply by $2$ to get $4$ right? But the actual answer should just be $2$. Am I missing... | $$\sum\limits_{n = 1}^\infty {\frac{2}{{{2^n}}}} = 2\sum\limits_{n = 1}^\infty {{{\left( {\frac{1}
{2}} \right)}^n}} = 2.\frac{{1/2}}
{{1 - 1/2}} = 2.1 = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Pair of die and probability A pair of dice is loaded. The probability that a 2 appears on the first
die is 3/13 and the probability that a 4 appears on the second die is
3/13. Other outcomes for each die appear with probability 2/13. What
is the probability of 6 appearing as the sum of the numbers when the
two dice are... | Add up the following:
*
*Probability of $1$ on 1st dice and $5$ on 2nd dice: $\frac{2}{13}\cdot\frac{2}{13}$
*Probability of $2$ on 1st dice and $4$ on 2nd dice: $\frac{3}{13}\cdot\frac{3}{13}$
*Probability of $3$ on 1st dice and $3$ on 2nd dice: $\frac{2}{13}\cdot\frac{2}{13}$
*Probability of $4$ on 1st dice and... | {
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"timestamp": "2023-03-29T00:00:00",
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$ \mathop {\lim }\limits_{n \to + \infty } \frac{{v_{n + 1} }}{{v_n }} = 2$ help me please
true or fulse
(1)$$\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1?
$$
\begin{array}{l}
u_n = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\qu... | $$
\begin{array}{l}
n! = \sqrt {n\pi e} \left( {\frac{n}{e}} \right)^n \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\sqrt {\left( {2n} \right)\pi e... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding the generating function of a series with a binomial coefficient and a exponential coefficient So I am given this series
$$2^8, 2^7 \binom{8}{1}, 2^6 \binom{8}{2}, 2^5 \binom{8}{3}, 2^4 \binom{8}{4}, 2^3 \binom{8}{5}, 2^2 \binom{8}{6}, 2^1 \binom{8}{7}, \binom{8}{8}, 0, 0, 0, 0, ...$$
which I converted to the su... | $$\sum_{n=0}^\infty \frac{2^8\binom 8n x^n}{2^n}=\sum_{n=0}^8\binom 8n 2^{8-n}x^n
=(2+x)^8\qquad \blacksquare$$
OR
$$(a+x)^N=a^N\left(1+\frac xa\right)^N\\
=a^N\sum_{n=0}^N \binom Nn \left(\frac xa\right)^n\\
=\sum_{n=0}^\infty a^N\binom Nn \left(\frac xa\right)^n$$
Put $a=2, N=8$:
$$(2+x)^8=\sum_{n=0}^\infty 2^8\bin... | {
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"timestamp": "2023-03-29T00:00:00",
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Does $x,y,z>0$ and $x+y+z=1$ imply $\left(1+\frac 1x\right)\left(1+\frac 1y \right)\left(1+\frac 1z \right)\ge 64$? If $x,y,z$ are positive real numbers such that $x+y+z=1$ then is it true that
$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$ ?
| By inequality between harmonic and geometric mean we have:
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\ge\left(\frac{3}{\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}}\right)^3
$$
Now if we prove that
$$
\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\le\frac{3}{4}
$$
we are done. But it is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding inverse of a function $h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$ I have a function:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$
With just pen and paper, how can I determine if there exists an inverse function? Am I supposed to sketch it on paper to see if it can have an invers? Or is there another/simplier way to do i... | Nice work!
But this is how you should have done it!
$$\begin{align}1-y&=y\sqrt x+\sqrt x\\
1-y&=(y+1) \sqrt x\\
\frac{1-y}{y+1}&=\sqrt x\\
\sqrt x&=\frac{1-y}{y+1}\\
x&=\frac{(1-y)^2}{(y+1)^2}\\
\end{align} $$
Hence $f^{-1}(x)$
$$f^{-1}(x) = \frac{(1 - x)^2}{(x+ 1)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $f(x)= \frac{1}{4}(x+4)^2-2$ Find vertex, $ y$ intercept etc. Given $f(x)= \frac{1}{4}(x+4)^2-2$
Find: vertex, $y$-intercept, $x$-intercepts (if any), axis of symmetry
What I have so far:
Vertex: $(-4,-2)$
$y$-intercept: $(0,2)$
$x$-intercept: $2$
Axis of symmetry $x=-4$
If you could please tell me if these are ... | Because the function is in parabola vertex-form the vertex is at $\left( -4,-2\right)$
The axis of symmetry is about the $x-$value of the vertex, because the function is a vertical parabola.
$$ \therefore x_{axis}=-4 $$
for the $y-$intercept you need $f\left( 0\right)$
$$ f\left( 0\right) = \frac{1}{4}\left( 0+4\right)... | {
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"timestamp": "2023-03-29T00:00:00",
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Number of real solutions of the equation $1+8^x+27^x = 2^x+12^x+9^x$
Find the number of real solutions $x\in\mathbb{R}$ of the equation
$$
1+8^x+27^x = 2^x+12^x+9^x
$$
My Attempt:
Let $2^x=a>0$ and $3^x=b>0$ where $x\in \mathbb{R}$. This allows us to change the equation to
$$
1+a^3+b^3 = a+a^2b+b^2
$$
This can be r... | Hint: by rearrangement inequality
$$a^3+b^3+c^3 = a^2b+b^2c+c^2a$$
happens iff $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Why is $(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000\cdots$? Apparently,
$$(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000000000000\cdots$$
Since it is a $2.0000000000\cdots$ instead of $2$, it isn't exactly $2$.
Why is that?
| $$\begin{align}
1-\cot{\left(x\right)}
&=1-\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}}\\
&=\frac{\sin{\left(x\right)}-\cos{\left(x\right)}}{\sin{\left(x\right)}}\\
&=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}.\\
\end{align}$$
Therefore,
$$\begin{align}
1-\cot{\left(\frac{\pi}{4}-x\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find all solutions in positive integers of the diophantine equation $w^2+x^2+y^2=z^2$ It's an exercises of the text book Elementary Number Theory and It's Applications 6th Edition by Kenneth H.Rosen. I wanted to solve it using the method in solving the diophantine equation $x^2+y^2=z^2$. But some difficult gaps showed ... | Here I have a solution just using Pythagorean triples. we know that any solution of $$x^2+y^2=z^2$$ can be written as $$x=2ab,\,\,\,\,\, y=a^2-b^2,\,\,\,\,\ z=a^2+b^2.$$
Therefore for your equation, we can choose
$$w=4abc,\,\,\,\,\, x=2(a^2-b^2)c,\,\,\,\,\ y=(a^2+b^2)^2-c^2,\,\,\,\,\,z=(a^2+b^2)^2+c^2.$$
Isn't it in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is $s_3$ and $s_4$ for $x$ $\sum_{i=0}^n i^k = s_k(n)$,
$s_k$ polynomial from degree $k+1$
I have already shown for $s_2(x) = \frac{x(x+1)(2x+1)}6$
How from the sum and $s_2(x)$ can be shown for $s_3(x)$ and $s_4(x)$ respectively.
| if $f(x)$ be the generator function of ${(a_n)}_{n \in \mathbb N}$ then $\frac{f(x)}{1-x}$ is generator function of $\sum_{i=1}^{k}a_i $.
$\frac{1}{1-x}= 1+x+ x^2+x^3 +x^4 +... $ $\rightarrow $ $\frac{x}{(1-x)^2}=x+ 2x^2+ 3x^3+4x^4+... $ $\rightarrow $ $A+\frac{x(1-x)^2-2x(1-x)}{(1-x)^4}=1+2^2x^2+3^2x^3+... $
then co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
| Since $3^3=27 > 4=2^2$, we have
$$
\left(\frac{1}{2}\right)^2
>
\left(\frac{1}{3}\right)^3
$$
Take square roots and get
$$
\left(\frac{1}{2}\right)
>
\left(\frac{1}{3}\right)^{3/2}
$$
Take cube roots and get
$$
\left(\frac{1}{2}\right)^{1/3}
>
\left(\frac{1}{3}\right)^{1/2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 9,
"answer_id": 5
} |
Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$ While doing a combinatorial problem, with $n$ being even, I came up with the expression
$$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$
for which I used wolfram to get a closed form expression of $\dfrac{1}{3}\left(2^{n+1}+(-1)^n\right)$.
Is there ... | Permit me to contribute a proof complex variables, for variety's sake,
which is an instructive exercise.
Suppose we seek to verify that
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
{n-q\choose q} 2^q
= \frac{1}{3} \left(2^{n+1} + (-1)^n\right)$$
with $n$ a positive integer.
Introduce the integral representation
$${n-q\choose q}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Continuous polar equation of Sqrt spiral Please help find polar coordinate equation $ r =f(\theta) $ or $ \theta =g(r) $
where $r$ and $\theta$ are continuous functions of $n$ as given:
SqrtSpiral
EDIT1:
n is a discrete (discontinuous) variable ( n = 1,2,3..) with
$$ r_n = \sqrt n ;$$
$$ r_{n+1}^2 -r_n^2 = 1 ; \\ ... | $r_{n+1}^2 -r_n^2 = 1 ; \\ r_n = \cot\Delta\theta =\cot (\theta_{n+1} - \theta_n ) ;$
I'll write $t$ for $\theta$ since I'm lazy.
From your equations,
$\cot^2 (t_{n+2} - t_{n+1} )
=\cot^2 (t_{n+1} - t_n ) +1
$,
so
$t_{n+2}$
is a function of
$t_{n+1}$
and
$t_{n}
$.
Since
$\cot(a-b)
=\dfrac{1+\tan a \tan b}{\tan a -tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &... | We already have that
$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$$
If $x$ is an acute angle strictly between $0$ and $\pi/2$, then $\sin x \cos x > 0$. Hence $\tan x + \cot x \geq 2$ if and only if $$1 \geq 2\sin x \cos x \ \ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Find closed form formula for $c(n,n-4)$. Find closed form formula for $c(n,n-4)$.
Where $c(n,k)$ are signless stirling numbers of first kind.
I need help.This is my last question all others problems of my exercise I have solved this is last one please help me.
Thanks.
| Here is a solution by generating functions for verification purpose
until something simpler appears.
Observe that there cannot be a cycle of length at least six because
that leaves $n-6$ items which can form at most $n-6$ cycles for a
total of $n-5$ cycles. Similarly for a cycle of length seven and so
on.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Lagrange Multipliers: Find $\min$ of $f(x,y)=3(x+1) +2(y-1)$ subject to the constraint $x^2+y^2=4$
Find the minumum value of the function $f(x,y)=3(x+1) +2(y-1)$, subject to the constraint that $x^2+y^2=4$.
The problem states to use Lagrange Multipliers. In doing so I obtained the point $(\frac 6{\sqrt{13}},\frac 4{\... | Write: $f(x,y) = 3x+2y + 1$. For a quick answer, using Cauchy-Schwarz inequality:
$|3x+2y| \leq \sqrt{3^2+2^2}\sqrt{x^2+y^2} = 2\sqrt{13}$. Thus:
$-2\sqrt{13}\leq 3x+2y \leq 2\sqrt{13} \to 1-2\sqrt{13} \leq f(x,y) \leq 1+2\sqrt{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Intersection multiplicity of the curves I want to find the intersection multiplicity of the curves $f(x,y)=x^5+x^4+y^2$ and $g(x,y)=x^6-x^5+y^2$ at the point $P=(0,0)$.
That`s what I have tried:
$f$ and $g$ have a common tangent, the $y=0$.
So $I(P, f\cap g) > m_P(f) \cdot m_P(g)=4$
$$f(x, 0)=x^5+x^4 \Rightarrow s=\deg... | $ f(x,y) =0 $ (red curve) has no real common tangent with $ g(x,y) =0 $ (blue curve) at $ P(0,0) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determine whether or not the following sequence converges. If so, find the limit. Im not sure if I am evaluating the convergence of the following sequence correctly, and I am unsure of how to determine the limit. All help is greatly appreciated.
$a_n = {1 \over (\sqrt{n^2-1}-\sqrt{n^2 +n})}$
My attempt at the converge... | We have: $\sqrt{n^2-1} - \sqrt{n^2+n} = (\sqrt{n^2-1} - n) - (-n +\sqrt{n^2+n}) = -\dfrac{1}{\sqrt{n^2-1} + n} - \dfrac{1}{\sqrt{1+\dfrac{1}{n}} + 1} \to -\dfrac{1}{2}$. Thus $a_n \to -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { \sqrt { 2 } }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt... | Both of them are positive, so let's square both of them:
first one will give:
\begin{gather}
\frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{4-2\sqrt{3}}\right) = \frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{1-2\sqrt{3}+3}\right) =\frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{(1-\sqrt{3})^2}\right) = \frac{1}{4} \left( 4 - \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
compute an integral $\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}$ How we can compute this integral:
$$\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}$$
i know that the solution is:
$$I=\frac{2 (x-2)}{\sqrt{x^2-3x+2}}$$
| Substitute $\displaystyle t=\frac{\sqrt{x^2-3x+2}}{x-1}$ to get $x=\dfrac{t^2-2}{t^2-1}$ and $dx=\dfrac{2t}{(1-t^2)^2}dt$; then
$\displaystyle\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}=\int\frac{1}{\frac{1}{(1-t^2)^2}\cdot t}\cdot\frac{2t}{(1-t^2)^2} dt=2\int dt=2t+C$
$\displaystyle=\frac{2\sqrt{x^2-3x+2}}{x-1}+C=\frac{2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using ... | Let $\theta$ be one-half of the vertex angle (less than a right angle) of the isosceles triangle.
Exercise: Show that the area of the inscribed triangle is
$A(\theta) = \dfrac{h b}{2} = \dfrac{(r + r \cos\theta)}{2} (2 r \sin\theta) = r^2 \,(1+\cos\theta) \, \sin\theta $
Differentiating $A$ and setting it to $0$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solve for $x$ in the following trigonometric equation $$\sqrt 2\cos^2 x-\cos x=0$$
Solve for $x$ algebraically, where $x$ is greater than or equal to zero, and less than $2\pi$. Answer must be an exact solution.
To be honest, I don't know where to start with this one. I know I need to isolate $\cos x$, but I have litt... | We have: $\sqrt{2}\cos^{2}(x)-\cos(x)=0$
Let's factor out a $\cos(x)$ to get:
$\Rightarrow \cos(x)\big(\sqrt{2}\cos(x)-1\big)=0$
As we have a product that equals to zero, either one of the multiples must equal to zero:
$\Rightarrow \cos(x)=0 \Rightarrow x=\arccos(0)=\dfrac{\pi}{2}$, $\hspace{1 mm}\dfrac{3\pi}{2}$
$\Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| The expressions on the right are
$$
\sqrt{4\left( 3\cdot 10^{k-1} + \frac{8}{9}\cdot 10^{k-1} - \frac{80}{9} + 9\right) }$$
for $k \geq 2$. I'm going to compare that to the expressions on the left, which have $k$ 5's.
$$
\sqrt{4\left( 3\cdot 10^{k-1} + \frac{8}{9}\cdot 10^{k-1} - \frac{80}{9} + 9\right) }
= \sqrt{4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ Proving the roots of
$$x^4-2(a+b)x^2+(a-b)^2=0$$
are......
$$x=\sqrt{a}+\sqrt{b}$$
$$x=\sqrt{a}-\sqrt{b}$$
$$x=-\sqrt{a}+\sqrt{b}$$
$$x=-\sqrt{a}-\sqrt{b}$$
When $a$ and $b$ are real numbers(negative or positive)
I prove... | One way to verify this with minimal computations is to expand the product $P(x) = \prod (x\pm\sqrt{a}\pm\sqrt{b})$ (where the product runs over all combinations of signs).
Taking as a variable $t_1 = x+\sqrt{a}$ and $t_2 = x-\sqrt{a}$, this becomes
$$P(x) = (t_1 + \sqrt{b})(t_1-\sqrt{b})(t_2 + \sqrt{b})(t_2-\sqrt{b}) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$ We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an ... | In fact there's no reason why $\varepsilon' - \frac{3}{2}$ should be positive.
Instead notice that for all natural numbers $n$ such that $x_n\neq 0$, it holds that $$\left|\dfrac{1}{x_n}-\dfrac1 2\right|=\left|\dfrac{x_n-2}{2x_n}\right|.$$
Relating the RHS to the convergence of $(x_n)_{n\in \mathbb N}$ and to $(x_n)_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Computing the trigonometric sum $ \sum_{j=1}^{n} \cos(j) $ I have a task to compute such a sum:
$$ \sum_{j=1}^{n} \cos(j) $$
Of course I know that the answer is $$ \frac{1}{2} (\cos(n)+\cot(\frac{1}{2}) \sin(n)-1) = \frac{\cos(n)}{2}+\frac{1}{2} \cot(\frac{1}{2}) \sin(n)-\frac{1}{2} $$
but I don't have any idea how to ... | Using the formula for a geometric series:
$$\sum_{k=1}^Ne^{ik}=\frac{e^{i(N+1)}-1}{e^{i}-1}-1=\frac{e^{i(N+1)}-1}{e^{i}-1}\frac{e^{-i}-1}{e^{-i}-1}-1$$
Taking real parts :
\begin{align}
\sum_{k=1}^N\cos(k)&=\Re\left[\frac{\left(e^{i(N+1)}-1\right)\left(e^{-i}-1\right)}{2-2\cos(1)}\right]-1\\
&=\Re\left[\frac{e^{iN}-e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$ If $2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$.
This is a question from a regional math olympiad and thus there must exist solutions without application of calculus.
I have no idea how to begin.
| WLOG, we can set $\sqrt2x=\cos A,\sqrt3y=\sin A\cos B,2z=\sin A\sin B$
$\implies4x+3y+2z=2\sqrt2\cos A+\sqrt3\sin A\cos B+\sin A\sin B$
$=2\sqrt2\cos A+\sin A(\sqrt3\cos B+\sin B)$
$=2\sqrt2\cos A+2\sin A\cos\left(B-\dfrac\pi6\right)$
If $\sin A\ge0,$
$4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| Do the operations to result to $$2x^3 + 3x^2y- 3xy^2 - 2y^3 - 20 = 0 \iff \\ 2(x^3 - y^3) +3xy(x-y) - 20 = 0 \iff \\ 2(x-y)(x^2 + xy + y^2) + 3xy(x-y) - 20 = 0 \iff \\ 14(x-y)+3xy(x-y) - 20 = 0$$
That simplifies the solution so keep substituting (2) into (1) to simplify the system
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
How to compute $\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)$ I find this problem on facebook group.
$$\mbox{Is it possible to find exact value of}\quad
\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)\ {\large ?}.
$$
I think this is not telescope sum. And Wolfram A... | @Jack DAurizio answer is nice,and I have two solution for this
\begin{align*}\arctan{\dfrac{1}{2n^2}}&=\arctan{\dfrac{2}{4n^2-1+1}}=\arctan{\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}}\\
&=\arctan{(2n+1)}-\arctan{(2n-1)}
\end{align*}
solution 2:
\begin{align*}\arctan{\dfrac{1}{2n^2}}&=\arctan{\dfrac{n^2-(n^2-1)}{(n^2-n)+(n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Probability of winning a rigged coin-flipping game Betsy and Katie are playing a game with an unfair coin. The coin is rigged to come up heads with probability $\frac35$ and tails with probability $\frac25$.
Betsy goes first. The two take turns. The first player to flip a tail wins. What is Betsy's probability of winn... | Let $p$ be the probability that Betsy wins the game. We check some cases. If Betsy flips tails on her first move, then she wins and the game ends; this happens with probability $2/5$. Otherwise, Katie gets a chance to play; this happens with probability $3/5$. If Katie gets a chance to play, she wins with probability $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| An additional solution is using Groebner bases. For this problem, consider the ring $R=\mathbb{Q}[x,y,z]$ and the ideal
$$
I=\langle x+y+z-1,x^2+y^2+z^2-2,x^3+y^3+z^3-3\rangle.
$$
Using a computer algebra system we can find a Groebner basis (with the grRevLex order) as
$$
\{x+y+z-1,2y^2+2yz+2z^2-2y-2z-1,6z^3-6z^2-3z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$ Show this integral
$$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$
where $0<a,b<1$
my idea:
Let
\begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+... | \begin{eqnarray}
& &\int_{-1}^{1}\dfrac{1}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}dx\\
&=&-\frac{1}{\sqrt{ab}}\tanh^{-1}\sqrt{\frac{{b(a^2+1)-2abx}}{{a(b^2+1)-2abx}}}\bigg|_{-1}^1\\
&= &\frac{1}{\sqrt{ab}}\tanh^{-1}\dfrac{2\sqrt{ab}}{1+ab}
=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Writing $1+3x^2+8x^4+21x^6+\cdots$ as a power series representation How would I write the power series $$1+3x^2+8x^4+21x^6+\cdots$$ as a power series representation (something neat similar to $\frac{1}{1-x}$)?
This reminds me of the power series $1+x^2+x^4+x^6+\cdots$ where the power series representation for that is... | Let $f(x) = 1+2x+3x^2+5x^3+8x^4+13x^5+21x^6+\cdots$.
You probably already know a closed form for $f(x)$.
Then, $f(-x) = 1-2x+3x^2-5x^3+8x^4-13x^5+21x^6-\cdots$.
Do you see how to get the series you want from $f(x)$ and $f(-x)$?
To get a closed form for $f(x)$ try combining the following equations in a way that leave... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Help identifying the singularities of $\csc(\cos z) = \frac{1}{\sin(\cos z)}$ I am really stuck with this one:
$\frac{1}{\sin(\cos z)}$ has a singularity when $\cos z = k\pi $ since $\sin(k\pi) = 0$ but how do I solve for the value of $z$, how can i evaluate $\cos z = k\pi $?
| Solve $\cos z = w$, let $y=e^{iz}$. Then $\cos z = \frac 12\left(y+y^{-1}\right)$. So $y+\frac{1}{y} = 2w$, or $y^2-2wy+1=0$ or $$y=\frac{2w\pm\sqrt{4w^2-4}}{2}= w \pm \sqrt{w^2-1}$$
So $iz = \log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right)$.
So:
$$z = i\log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right)$$
Since $(k\pi+ \sqrt{k^2\pi^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the basis and dimension of a vector space Find the basis and dimension of vector space $ L_{1}$ spanned by
vectors $ a_{1} ,a_{2},a_{3} $, the basis and dimension of vector space $ L_{2}$ spanned by
vectors $ b_{1} ,b_{2},b_{3} $
and also $ L_{1} + L_{2} , L_{1} \cap
L_{2}$
$ a_{1} = (-10,10,-6,-2) ,a_{... | For the first, form a matrix $A$ whose columns are the vectors $a_i$ and row reduce it:
$$\begin{bmatrix}
-10 & -6 & -2 \\
10 & 4 & -12 \\
-6 & -4 & -4 \\
-2 & -4 & -20
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 0 & -4 \\
0 & 1 & 7 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
There are leading ones in the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving an algebraic inequality For any $a$, $b$, and $c$ prove
$$3a^2+3b^2-2b+2a+1>0$$
I tried the following
$$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\
(a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\
(a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\
(a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\
$$
$a^2$ and $(b-1)^2$ and $2(a^2+b^2)$ are always positive and greater... | $$3a^2+3b^2−2b+2a+1 = (3 a^2 + 2a +\frac{1}{3}) + (3 b^2 - 2b + \frac{1}{3}) + \frac{1}{3} = $$
$$ = (\sqrt{3}a + \frac{1}{\sqrt{3}})^2 + (\sqrt{3}b - \frac{1}{\sqrt{3}})^2 + \frac{1}{3} > 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
prove conjecture; the limit of iterating is $\sqrt{z^2 - 2}$ $$\lim_{n \to \infty} f_n(x)=x-\frac{1}{nx}\;\;\; g(x)=f_n^{on}(x)$$
The conjecture is for values of $|x|>\sqrt{2}$: $g(x) = \sqrt{z^2 - 2}$
This question comes from another matstack question/answer.
If $n=2^m$, then convergence is much quicker by starting wi... | Pick an $x > \sqrt{2}$ and $n > \dfrac{1}{x^2-2}$ and define a sequence by $x_0 = x$, $x_{k+1} = f_n(x_{k})$ for $k \ge 0$.
Rearrange $x_{k+1} = f_n(x_k) = x_k - \dfrac{1}{nx_k}$, to get $x_k(x_k-x_{k+1}) = \dfrac{1}{n}$.
Clearly, $x_0 \ge \sqrt{x^2}$. Now, suppose that $x_{m-1} \ge \sqrt{x^2-\dfrac{2(m-1)}{n}}$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
What are the intermediate fields of $\mathbb Q(\sqrt[3]2,e^{\frac{2i\pi}{3}})$ (Galois group) The elements of Galois group are
\begin{align*}
\sigma _1:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmap... | The subfields are the fields fixed by a subgroup of the Galois group. Not every subgroup is generated by a single element, whence it is no problem that there is no element of order $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simplifying sum of 2 square roots I might be a bit slow on math, but I haven't seen this explicitly described anywhere, I'm reading a paper and in the paper it has the equation:
$$\Delta d=\sqrt{y^2+(x+D/2)^2}-\sqrt{y^2+(x-D/2)^2}$$
After which it says
After some simplifications, this equation can be rewritten in a mo... | $$\Delta d=\sqrt{y^2+(x+D/2)^2}-\sqrt{y^2+(x-D/2)^2}$$$$\therefore\sqrt{y^2+(x+D/2)^2}=\Delta d+\sqrt{y^2+(x-D/2)^2}$$$$\therefore y^2+(x+D/2)^2=\left(\Delta d+\sqrt{y^2+(x-D/2)^2}\right)^2$$$$\therefore y^2+x^2+Dx+D^2/4=\Delta d^2+y^2+(x-D/2)^2+2\Delta d\sqrt{y^2+(x-D/2)^2}$$$$=\Delta d^2+y^2+x^2-Dx+D^2/4+2\Delta d\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
solving for sin of the sum of two angles of a triangle In triangle $ABC$,
$3\sin B+4\cos C=6$ and $4\sin C+3\cos B=1$.
Show that $\sin(B+C)=0.5$.
Can we assume $\angle A = 180 - ( B + C)$ and use sum formula.
| Squaring both equations yields:
$9\sin^2 B + 24 \sin B\cos C + 16 \cos^2 C = 36$
$9\cos^2 B + 24 \cos B\sin C + 16 \sin^2 C = 1$
Now what happens when you add the two together?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Summation identity involving the floor function (Kömal November B. 4666)
Prove that $\sum_{k=1}^n (2k-1) [\frac{n}{k}]=\sum_{k=1}^n [\frac{n}{k}]^2$ for every positive integer $n$, where $[n]$ is the largest integer greater than or equal to $n$.
| The trick here is to introduce a second summation and then reverse the order of summation. Introducing the second summation is easy:
$$\sum_{k=1}^n(2k-1)\left\lfloor\frac{n}k\right\rfloor=\sum_{k=1}^n(2k-1)\sum_{j=1}^{\lfloor n/k\rfloor}1=\sum_{k=1}^n\sum_{j=1}^{\lfloor n/k\rfloor}(2k-1)\;.\tag{1}$$
Reversing the order... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to solve $y=(xy'+2y)^2$? What kind of differential equation is this thing and how to solve it?
$$y=(xy'+2y)^2$$
$$y=x^2y'^2+4xyy'+4y^2$$
| Lets try a substitution $y = x^{\alpha}u$
this results in
$$
x^{\alpha}u = x^2\left(\alpha x^{\alpha-1}u + x^{\alpha}u'\right)^2 + 4x\left(\alpha x^{\alpha-1}u + x^{\alpha}u'\right)x^{\alpha}u + 4x^{2\alpha}u^2
$$
dividing by $x^{\alpha}$ and collecting like terms we find
$$
u = \left(\alpha^2+4\alpha + 4\right)u^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Synthetic division by $ax^2+bx+c$. I know that synthetic division can be used in order to find quotient $q(x)$ and remainder $r(x)$ of a polynomial $p(x)$ when it is divided by some linear polynomial like $x-c$. Now, does exist some procedure (another than long division) in order to find $q(x)$ and $r(x)$ when the divi... | Assume your divisor is $x-d$ rather than $x-c$ since $c$ is already used in the formation of the original polynomial. You can write the dividend $y$ as:
$y = a(x-d)^2 + b(x-d) + 2adx - ad^2 + c + bd = a(x-d)^2 + b(x-d) + 2ad(x-d) + 2ad^2 -ad^2 + c + bd = a(x-d)^2 + (b+2ad)(x-d) + ad^2 + bd + c = (x-d)(a(x-d) + b+2ad) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Sum $\sum_{n=1}^\infty \frac{n^2}{(n+2)!}$ Problem is to find sum
$$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$
What I knew doesn't apply on this problem
Some series are telescoping, some types are solvable using binomial , both look useless here
Binomial gives
$$n (n-1) (n-2)x^3=1$$
$$n (n-1)( n-2)(n-3)x^4=4$$
Wha... | Replace the range so that it starts at $n = 0$ and substitute $n^2 = (n+2)(n+1) - 3(n+2) + 4$ to obtain
$$
\begin{align*}
\sum_{n=0}^\infty \frac{n^2}{(n+2)!} &=
\sum_{n=0}^\infty \frac{(n+2)(n+1)}{(n+2)!} -
\sum_{n=0}^\infty \frac{3(n+2)}{(n+2)!} +
\sum_{n=0}^\infty \frac{4}{(n+2)!} \\ &=
\sum_{n=0}^\infty \frac{1}{n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality $\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2 $ I cannot prove the following inequality.
Let $a,b,c$ be positive numbers.Prove that:
$\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2. $
I tried to use Cauchy inequality. I think the function... | $$\begin{align}&\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b} - (a-b)^2-(b-c)^2-(c-a)^2 \\ =& \sum\limits_{cyc} \left(\frac{a^3}{b+c} - 2a^2 +a(b+c)\right) \\ =& \sum\limits_{cyc} \frac{a(a-b-c)^2}{b+c} \ge 0 \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Matrices, determinants, and applications to identities involving Fibonacci numbers Preamble
It is well known that since:
$$
\begin{pmatrix}
F_{n+1} \\
F_n \\
\end{pmatrix} =
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
F_n & F_{n-1} \\
\end{pmatrix}
$$
it is valid that:
$$
\begin{pma... | Following statement and proof are interesting:
Prove that
$$F_{2n-1} = {F_{n}}^2 + {F_{n-1}}^2$$
Let us start with
$$
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^n =
\begin{pmatrix}
F_{n+1} & F_n \\
F_n & F_{n-1} \\
\end{pmatrix}
$$
If we split $n$ into $p$ and $q$, this can be rewriten as:
$$
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $ \sum _{r=0} ^{2n} r ( ^{2n}C _r) ( \frac 1{r+2} ) $
Find the value of $$ \sum _{r=0} ^{2n} r ( ^{2n}C _r ) ( \frac 1{r+2} )$$
In order to solve this I am trying to make the term(s) of the series independent of $r$. However I'm unable to solve this further: $$ \sum _{r=0} ^{2n} 2n (^{2n-1}C_{r-1})... | HINT:
$$r\binom{2n}r\cdot\frac1{r+2}=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\frac{(2n+2)!}{[(2n+2)-(r+2)]!\cdot(r+2)!}$$
$$=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\binom{2n+2}{r+2}$$
Now let $r(r+1)=(r+2)(r+1)+A(r+2)+B$
$r^2+r=r^2+r(3+A)+2+2A+B$
$\implies A+3=0\iff A=-3$
and $B+2A+2=0\iff B=-2A-2=4$
$$\implies(2n+2)(2n+1)r\cdot \bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solve the ODe with five variables How to solve $$\frac{dx}{2p}=\frac{dy}{2q}=\frac{du}{2(p^2+q^2}=\frac{dp}{2up}=\frac{dq}{2uq}=dt$$
as functions $$x=x(t), y=y(t), u=u(t), p=p(t), q=q(t)$$
My method is use the last three equalities to deduce $$\frac{d^2u}{dt^2}+u\frac{du}{dt}=0$$
But this nonlinearity troubles me...
| Based on your last line:
Consider $F = u^2$
$$ \frac{df}{dt} = 2 u \frac{du}{dt} $$
Thus suppose we re-arrange
$$ \frac{d^2u}{dt^2} + u \frac{du}{dt} = 0 $$
Into
$$ \frac{d^2u}{dt^2} = - u \frac{du}{dt} $$
Then we can re-write it as
$$ \frac{d}{dt} \left[ \frac{du}{dt} \right] = \frac{d}{dt}\left[ - \frac{1}{2}u^2\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the derivative $f(x)=\sqrt{x^2 -9}$, I need to find the slope at a=5, using the definition for the function $f(x)=\sqrt{x^2 -9}$,
$$f'(x) = \lim_{\Delta x \to 0} {f(x+\Delta x)\over \Delta x}$$
The answer book says the slope is ${1\over 4}$
Here's what I did,
$$f'(x) = \lim_{\Delta x \to 0} {(\sqrt{(x+\Delta x)... | (5)to (6) step !!!you have error $$\sqrt{(x+\Delta x)^2-9}+\sqrt{(x)^2-9} \neq \sqrt{(x)^2-9+(x)^2-9} $$in fact $$\sqrt{a+b} \neq \sqrt{a} +\sqrt{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$ What is the value of $\displaystyle\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$?
This is a question I came up with myself. It is not homework.
I constructed this example to make the following technique work:
Integrate $\... | Let's actually do the integral using the keyhole contour. It may be time-consuming but it is not as bad as it looks.
We can begin by simplifying the integral using the substitution $u=x^2$:
$$I = \frac18 \int_0^{\infty} du \frac{\log^2{u}}{u^2+u+1} $$
Consider
$$\oint_C dz \frac{\log^3{z}}{z^2+z+1}$$
where $C$ is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Surface integral of $x^4+y^4+z^4$ over the sphere $x^2+y^2+z^2=a^2$ After doing regular methodology have reached upto integral shown in figure , but when i eliminate z from it it becomes very complicated to solve .Is there any other way to solve this .Thanks
| It seems the following.
Use spherical coordinates $$x=a\sin\varphi\cos\theta,$$ $$y=a\sin\varphi\sin\theta,$$ $$z=a\cos\varphi$$ ($0\le\varphi\le\pi$, $0\le\theta\le 2\pi$). Then $$dS=a^2\sin\varphi d\varphi d\theta$$ and
$$x^4+y^4+z^4=$$
$$a^4(\sin^4\varphi\cos^4\theta+\sin^4\varphi\sin^4\theta+\cos^4\varphi)=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simplify $\cos 1^\circ + \cos 3^\circ + \cdots+ \cos 43^\circ$? I am currently working on a problem and reduced part of the equations down to
$$\cos(1^\circ)+\cos(3^\circ)+\cdots+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ)$$
How can I calculate this easily using the product-to-sum formula for $\cos(x)+\cos(y)$?
| let $S = \cos(1^\circ)+\cos(3^\circ)+.....+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ).$ then
\begin{align} 2S\sin 1^\circ &= 2\cos 1^\circ \sin 1^\circ + 2 \cos 3^\circ \sin 1^\circ+\cdots + 2 \cos 43^\circ \sin 1^\circ \\
& = (\sin 2^\circ - \sin 0^\circ)+(\sin 4^\circ - \sin 2^\circ ) + \cdots
+(\sin 44^\circ- \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$\sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3}$ using complex analysis Evaluate:
$$S = \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3} \space \text{using complex analysis}$$
This my question: we need to consider a $f(z)$ such that,
$$\frac{1}{2\pi i} \cdot\oint_{C_N} f(z) dz = \text{something (maybe residue)} + \sum_{n=1}^{N} ... | Here is an unorthodox answer that needs some more work to make it rigorous.
Consider $$f(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}.$$
We can certainly integrate around a circular contour as ML gives $2\pi R\times \log R/R^4.$
Now we have
$$\mathrm{Res}_{z=0} f(z) = \frac{1}{8},$$
$$\mathrm{Res}_{z=-1} f(z) = -\gamma,$$
$$\mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
ODE Separable Equation Let $y = Φ(x)$ be a solution to $y' = y(5-y)(8-y)$ subject to $y(0) = 7$. Determine $\lim_{x \to ∞} Φ(x)$.
Workings:
I'm thinking I have to solve the differential equation.
$y' = y(5-y)(8-y) dy$
$y' = y(y-5)(y-8) dy$
$∫\frac{1}{(y-5)(y-8)y} dy = ∫1 dx$
$∫\frac{1}{(y-5)(y-8)y} dy$
$\frac{1}{(y-5)(... | to find the limit there is no need to solve the equation. look at the steady state solutions $y = 0, y = 5 y = 8.$
this is called the phase portrait:
(a) on a number line mark these points.
(b) on each region bounded by these points, determine the sign of $y^\prime$ by evaluating $y^\prime$ at a test point.
here is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Approximate $\coth(x)$ around $x = 0$ I'm trying to approximate $\coth(x)$ around $x = 0$, up to say, third order in $x$. Now obviously a simple taylor expansion doesn't work, as it diverges around $x = 0$. I'm not quite sure how to proceed from here. I had a look at
Series expansion of $\coth x$ using the Fourier tra... | The series representation must reflect that divergence at $x=0$. Thus,
$$\begin{align}\coth{x} &= \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+O(x^7)} \\ &= \frac1{x} \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{1+\frac{x^2}{3!}+\frac{x^4}{5!}+O(x^6)}\\&= \frac1{x} \left (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Show that if $a$ is an integer, then 3 divides $a^3 - a $ Show that if $a$ is an integer, then 3 divides $a^3 - a $
we can write, where $k$ is an integer;
$a^3 - a = 3k $
$a(a^2 - 1) = 3k $
Now if $a = k$ then
$a^2 -1 = 3$ and $a= \pm2 $ so $ a^3 - a = 24 = 3 \times 8$
If $ a $ is not equal to $k$;
then
$a(a^2 - 1) = a... | In fact, divisible by $6$. The product of three consecutive integers is divisible by $3!$. Might as well take them naturals. Note that the quotient is
$$\frac{(a+1)a(a-1)}{1\cdot 2\cdot 3} = \binom{a+1}{3}$$
Similarly you can show that the product of $n$ consecutive integers is divisible by $n!$.
Another important ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Problem with the rational root theorem Consider this polynomial: $f(x)=(2x+5)(x-3)(x+8/3)=0$. Then $f(x)=2x^3+...+(-40)$
Here is a list of all factors of $40$ and $2$:
$40$: $±1$, $±2$, $±4$, $±5$, $±8$, $±10$, $±20$
$2$: $±2$, $±1$
Now, $3$ is clearly a root, but there is no combination between factors of $40$ divided... | The problem is that for the rational root theorem to work, we need all the coefficients to be integral.
Expanding, we have
$$f(x)=(2x+5)(x-3)(x+8/3)=2x^3+\frac{13}{3}x^2-\frac{53}{3}x-40$$
Now notice that the solutions to $f(x)=0$ are the same as the solutions to $3f(x)=0$, so we can safely multiply through by $3$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help me find the following limit : $\lim_{{n}\to{\infty}} (\frac{2^x+3^x+\cdots+n^x}{n-1})^\frac{1}{x} = ?$ I have no idea where to start.$$\begin{align}\lim_{{n}\to{\infty}} \left(\dfrac{2^x+3^x+\cdots+n^x}{n-1}\right)^{1/x} = ?, n >1\\\end{align}$$
| Hint: Assume that $x > 0$, $\ln f(x) = \dfrac{1}{x}\cdot \ln\left(\dfrac{2^x+3^x+\cdots n^x}{n-1}\right)=\dfrac{1}{x}\cdot \left(\ln\left(\dfrac{2^x+3^x+\cdots n^x}{n^x}\right)+x\ln\left(\dfrac{n}{n-1}\right)\right)$.
Now we find the limit as $n \to \infty$ of the quotient: $q_n(x)=\dfrac{2^x+3^x+\cdots n^x}{n^x}$ by u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Integral of $\cos^4(2t)\,dt$ with bounds from $0$ to $\pi$ $$\int_0^\pi\cos^4(2t)\,dt=?$$ I have attempted this problem two different ways and got two different answers that are nowhere near the correct answer. Could you please show me detailed steps on how to work this problem out. The final answer is $\frac{3\pi}{8}$... | Expanding on Alex Silva's answer:
Since
$\cos^2(x) = \frac{1+\cos(2x)}{2}$,
$\begin{array}\\
\cos^4(x)
&= \frac{(1+\cos(2x))^2}{4}\\
&= \frac{1+2\cos(2x)+\frac{1+\cos(4x)}{2}}{4}\\
&= \frac{2+4\cos(2x)+1+\cos(4x)}{8}\\
&= \frac{3+4\cos(2x)+\cos(4x)}{8}\\
\end{array}
$
so
$\cos^4(2x)
=\frac{3+4\cos(4x)+\cos(8x)}{8}
$.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.
| $(x+y+z)^2 \geq 3 \iff x^2+y^2+z^2 + 2(xy+yz+zx) \geq 3 \iff x^2+y^2+z^2+ \dfrac{2}{\sqrt{x^2+y^2+z^2}}\geq 3 \iff t^2 + \dfrac{2}{t} \geq 3$. By AM-GM inequality: $t^2+\dfrac{2}{t} =t^2+ \dfrac{1}{t}+\dfrac{1}{t}\geq 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Conditional Distributions Choose a random integer $X$ from the interval
$[0, 4]$. Then choose a random integer $Y$ from the interval
$[0, x]$, where $x$ is the observed value of $X$. Make assumptions
about the marginal pmf $f_X(x)$ and the conditional
pmf $h(y | x)$ and compute $P(X + Y > 4)$.
I tried to make a sample... | If $X=3$, then we get sum $\gt 4$ with probability $\frac{2}{4}$, and if $X=4$ we get sum $\gt 4$ with probability $\frac{4}{5}$, so the required probability is $\frac{1}{5}\cdot \frac{2}{4}+\frac{1}{5}\cdot \frac{4}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Symmetric and homogeneous three variable inequality with radicals. While trying to solve a problem, I got the following inequality which appears correct, but I cannot prove. For positive $x, y, z$,
$$\sum_{cyc} \frac{x}{y^2+z^2} \ge \sum_{cyc} \sqrt{\frac{y^2+z^2}{2(x^2+y^2)(z^2+x^2)}}$$
Equality is obviously when $x=... | By C-S $\left(\sum\limits_{cyc}\sqrt{\frac{y^2+z^2}{2(x^2+y^2)(x^2+z^2)}}\right)^2\leq\sum_(y^2+z^2)\sum\limits_{cyc}\frac{1}{2(x^2+y^2)(x^2+z^2)}=\frac{2(x^2+y^2+z^2)^2}{\prod\limits_{cyc}(x^2+y^2)}$.
Thus, it remains to prove that
$\left(\sum\limits_{cyc}(x^5+x^3y^2+x^3z^2+x^2y^2z)\right)^2\geq2(x^2+y^2+z^2)^2(x^2+y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Polynom equality modulo p I found these two equations:
(a) $$X^4 + 1 \equiv (X + 1)^4 \mod \ 2$$
(b) $$X^4 + 1 \equiv (X^2 - X - 1)(X^2 + X - 1) \mod \ 3$$
I would like to understand the concept of modulo for Polynoms.
How were they made? And how to verify these equality ?
Thanks in Advanced!
| For $(a),$
$y^{m+1}-y=y(y^m-1)$ is divisible by $y(y-1)$ which being the product of two consecutive integers is even
$\implies y^{m+1}\equiv y\pmod2$ for $m\ge1$
$\implies y^4\equiv y\pmod2$ set $y=X, X+1$
For $(b),$
$(x^2-1-x)(x^2-1-x)=(x^2-1)^2-x^2=x^4-3x^2+1\equiv x^4+1\pmod3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$ Let a and b be natural numbers so that $2a-1,2b-1$ and $a+b$ are prime numbers. Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$.
I get that $gcd(a,b)=1$.
I haven't got anything special for now but if I do I will update the question.
| Since $2a-1, 2b-1$ are primes, then $a,b\neq 1$. Since $a+b$ is a prime, then we can suppose $a$ is even and $b$ is odd.
Suppose $p=a+b|a^a+b^b$. We have: $$a^a+b^b\equiv a^a-a^b=a^b(a^{a-b}-1) \pmod {a+b}$$
Since $\gcd(a^b,a+b)=1$, then $p|a^{a-b}-1$. Let $h=ord_p(a)$, then $h|p-1,h|a-b$, then $h|p-1-(a-b)=2b-1$, thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 1
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How to evaluate $\lim_{n\to\infty}\int_0^n\frac{x^2+a^2}{x^4+b^2x^2+b^4}dx$ Evaluate this limit:
$$\lim_{n\to\infty}\int_0^n\dfrac{x^2+a^2}{x^4+b^2x^2+b^4}dx$$
I tried to simplify this fraction. I noticed that $x^4+b^2x^2+b^4$ can be written as
$$\dfrac{x^6-b^6}{x^2-b^2}$$
Then limit will be
$$\lim_{n\to\infty}\int_0^... | Maple does the indefinite integral in terms of arctangents and logarithms. So, yes, you can do it by partial fractions, where you first factor the denominator ... to factor it, complete the square, so it becomes a difference of squares
Then take limit to get the answer. Maple says:
$$
\frac{(a^2+b^2)\pi}{b^3 2\sqrt{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Question about series convergence $\sum_{n=1}^\infty \frac{1}{n}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ So I have been playing around with convergent series recently and I still have a hard time understanding why $\sum_{n=1}^\infty \frac{1}{n}$ diverges and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
I can write out t... | Think about Zeno's paradox in reverse.
$$ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots $$
You want to walk a mile.
First you walk half of it. Then take a bit of rest. Then walk half of the remaining, then take a bit of rest. And so on.
Will you ever go beyond the mile?
You have an infinite number of numbers (of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why do we put absolute brackets for ln? When writing out the final answer in $\ln$ form, why is it necessary to put absolute brackets? How does it affect the answer?
I have this answer of $-3\ln|\frac{3+\sqrt{9-x^2}}{x}|$, but why does it suddenly become
$3\ln|\frac{\sqrt{9-x^2}-3}{x}|$?
| In this case it's exactly the same:
\begin{align}
\left|\frac{3+\sqrt{9-x^2}}{x}\right|^{\!-1}
&=
\left|\frac{x}{3+\sqrt{9-x^2}}\right|\\[2ex]
&=
\left|\frac{x}{3+\sqrt{9-x^2}}\frac{3-\sqrt{9-x^2}}{3-\sqrt{9-x^2}}\right|\\[2ex]
&=\left|\frac{x(3-\sqrt{9-x^2})}{9-9+x^2}\right|\\[2ex]
&=\left|\frac{3-\sqrt{9-x^2}}{x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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What is the general term of $a_{n+1}=\frac{2a_n-1}{5a_n-1} \ , \ \ a_1=1$? I've struggled to solve this exercise
$$a_{n+1}=\frac{2a_n-1}{5a_n-1}\ , \ \ a_1=1$$
$$b_{n+1}=(5a_n-1)b_n \ , \ \ b_1=1$$
Find $b_{\ 40}$ .
$$$$
I thought 'taking inverse' will be useful, but not yet... :-(
$\color{red}{01.}$ How can I find $... | here is a partial answer:
the eigenvalues of $A = \pmatrix{2 & -1\\5&-1}$ are $\lambda_{1,2} = \dfrac{1}{2} \pm i\dfrac{\sqrt{11}}{2}=\sqrt 3(\cos t \pm i\sin t)$ and the corresponding eigenvectors are
$u_{1,2}=\pmatrix{2\\3\mp i\sqrt{11}}$
let $a, b, c$ be complex numbers such that
$z = au_1 + bu_2 = a \pmatrix{2\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Integrating: $ \;\int \frac{1}{x^2+3x+2} dx $ How can I solve the following integral:
$$ \int \frac{1}{x^2+3x+2} dx $$
Should I proceed by changing the variable (substitution)? or should I use integration by parts? Or another method altogether?
Thank you!
| This integral is "screaming": Use partial fraction decomposition!, especially after we note that the denominator factors nicely $$x^2 + 3x + 2 = (x + 2)(x+1)$$
So our integral will take the form:
$$\int \frac 1{x^2 + 3x + 2} \,dx = \int \left(\frac A{x+2} + \frac B{x+1}\right)\,dx$$
Now, we solve for $A, B$ knowing tha... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the sum $x + \frac{x^3}{3} + \frac{x^5}{5} + ... $
Evaluate the sum $$x + \frac{x^3}{3} + \frac{x^5}{5} + ... $$
I was able to notice that: $$ \sum_{n=0}^\infty \frac{x^{2n-1}}{2n-1} = \sum_{n=0}^\infty \int x^{2n-2}dx = \lim_{N\to\infty} \sum_{n=0}^N \int x^{2n-2} dx $$
Where should I take it from here? (... | You got the index wrong. The actual sum is
$$ \sum_{n = 0}^{\infty} \frac{x^{2n+1}}{2n+1} = \int \sum_{n = 0}^{\infty} x^{2n} \,dx = \int \frac{1}{1-x^2}\,dx $$
This sum converges for $|x| < 1$ and the integral evaluates to
$$ f(x) = \frac{1}{2}\, \ln \left| \frac{1+x}{1-x} \right| + C $$
The constant $C$ can be found... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to show that $\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$ is a real number at a time before the invention of complex numbers I have read this PDF from ocw.mit.edu about complex numbers. There is one interesting question: Imagine yourself at the time, when complex numbers had to be invented yet. How to show that ... | Let $$r = (-1 + \sqrt{-7})^{1/3} + (-1 - \sqrt{-7})^{1/3},$$ where by $\sqrt{-7}$ we mean some number $x$ whose square is $-7$; i.e., $x^2 = -7$. Recall the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3}).$$ Consequently, $$r^3 = -2 + 3((-1)^2 + 7)^{1/3}r = 6r - 2.$$ Now consider the trigon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Product of Divisors of some $n$ proof The function $d(n)$ gives the number of positive divisors of $n$, including n itself.
So for example, $d(25) = 3$, because $25$ has three divisors: $1$, $5$, and $25$.
So how do I prove that the product of all of the positive divisors of $n$ (including $n$ itself) is $n^{\frac{d(n)... | The solution by Pedro Tamaroff is far more compact, and better. We will give the same proof in a more long-winded way.
First we mention a fact about the private lives of the divisors of $n$. It is easy to see that $d$ is a divisor of $n$ if and only if $\frac{n}{d}$ is a divisor of $n$. It turns out that $d$ and $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that $ 17/4 \leq (a+b+c)\leq 1+ \sqrt{32}. $ If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that
$$
\frac{17}{4} \leq (a+b+c)\leq 1+ \sqrt{32}.
$$
My attempt
Tried using Vieta but it didn't work. Also I ... | WLOG, using the symmetry, we may assume $a \le b \le c$, so $a \le 1, \; c \ge 1$.
For the left inequality, note that
$$a+b+c-\frac{17}4 = \frac{5-1/c}{c}+c-\frac{17}4 = \frac{(c-2)^2(4c-1)}{4c^2} \ge 0$$
and as equality is achieved when $a=\frac14, b=c=2$ (or any permutation), this in fact gives the minimum.
Similarl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Integration of the following What is the definite integral of
$$
\int_0^1 \left(\frac{g(x)}{f(x)}\right)'\cdot\frac{1}{g(x)}\,dx,
$$
where the conditions are as follows:
$f(0) = 2 $
$f(1) = 3 $
$f'(x) $ is continuous
For all $x$ in the range $\{0,1\}, f(x)^2-g(x)^2=1$.
Thank you very much.
| By integration by parts,
\begin{align}\int_0^1 \left(\frac{g}{f}\right)' \frac{1}{g}\, dx &= \frac{g}{f}\cdot \frac{1}{g}\bigg|_{t = 0}^{t = 1} - \int_0^1 \frac{g}{f}\frac{d}{dx}\left(\frac{1}{g}\right)\, dx\\
&= \frac{1}{f(1)} - \frac{1}{f(0)} + \int_0^1 \frac{g'}{fg}\, dx\\
&= -\frac{1}{6} + \int_0^1 \frac{g'}{fg}\,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Relation between two sets of generators of SO(3) I am working with the spin 1 representation of SU(2), which is just SO(3). The ordinary generators used in quantum mechanics are:
$J_x = \left(
\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & 0 \... | As per the suggestion of Phoenix87, we can find the unitary transformation by noticing that the transformation that diagonalizes $J_z'$ will also transform $J_z'$ to into $J_z$ (since $J_z$ is diagonal). So we find the eigenvectors of $J_z'$ and then use them as the columns of our unitary transformation. We are free, h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inverse Laplace Transform of $\frac{s-3}{s[(s-3)^2+9]}$ I am having problems solving this inverse Laplace transform:
ℒ$^{-1}\Large [\frac{s-3}{s[(s-3)^2+9]}]$
I did partial fraction decomposition, but ended up with complex expressions in some denominators, and I have no idea what do do from here or if this is even how ... | Note $\dfrac{s-3}{s\left[(s-3)^2+9\right]}$ can be expressed as
\begin{align*}
\frac{s-3}{s\left[(s-3)^2+9\right]}&=\frac{s-3}{s(s^2-6s+18)}\\
&=\frac{A}{s}+\frac{Bs+C}{s^2-6s+18}
\end{align*}
For some real numbers $A$, $B$ and $C$, that means
\begin{align*}
s-3&=A(s^2-6s+18)+(Bs+C)s\\
&=(A+B)s^2+(-6A+C)s+18A\\
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Simple algebra involving trigonometry, but confusing How do I get from
$$\frac{\sqrt3}{2} + \frac12 \tan x = 2 \tan x \cdot \frac{\sqrt3}{2}$$
to
$$\frac{\sqrt3}{2 \sqrt3 - 1} = \tan x$$
and then to
$$11 \tan x = 6 + \sqrt3$$
| I will try to make it a bit more elaborate since you are going to need to be able to manipulate equations like this all the time.
Multiply both sides by 2:
$\implies \sqrt{3}+\tan(x)=2\tan(x)\sqrt{3}$
Subtract $\tan(x)$ from both sides in order isolate $\tan(x)$
$\implies \sqrt{3}=2\tan(x)\sqrt{3}-\tan(x)$
You have t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Taylor Series $(x+2)/(2-3x)$ at $x=2$ How can I find Taylor series for $$\frac{(x+2)}{(2-3x)}$$ at $x=2$?
| Hints:
$$\frac{x+2}{2-3x}=(x+2)\;\frac13\;\frac1{2-x-\frac43}=-\frac{x+2}3\;\frac34\;\frac1{1+\frac{3(x-2)}4}=$$
$$=-\frac{x+2}4\left(1-\frac{3(x-2)}4+\frac{9(x-2)^2}{16}-\ldots\right)$$
Well, now find the general expression for the above and for what values of $\;x\;$ it is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the solutions in positive integers $m^2 + 615 =2^n$ Find all positive integer solutions $m$ and $n$ of the equation
$$ m^2 + 615 =2^n. $$ ... | Note that $3 \mid 615$ and $3$ is never a factor of $2^n$, so $3 \nmid m$.
Therefore $m^2 \equiv 1 \bmod 3$ and thus also $2^n\equiv 1 \bmod 3$.
Odd powers of $2$ are $\equiv 2 \bmod 3$ and even powers of $2$ are $\equiv 1 \bmod 3$, so we know that $n$ is even. So say $n=2k$.
Now we have $m^2 +615 =2^{2k}$ so $615= 2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Under what conditions to have $ (b+1)^n \bmod 2b = (b+1)$ Under What conditions, for any integer $n \geq 1$, we will have:
$$ (b+1)^n \bmod 2b = (b+1)$$
I tested it with a couple of numbers, and it is true for $3, 5, 7$, etc.
Please prove your theory.
It is based on a comment of this question
| For $b$ odd, $b=2k+1$ and $$b^2=b(2k+1)=2b\cdot k+b \equiv b \bmod (2b)$$
Now $$(b+1)^2 = b^2+2b+1 \equiv b^2+1 \equiv (b+1) \bmod (2b)$$ as required.
Then given $(b+1)^k \equiv (b+1)\bmod(2k)$,
clearly $(b+1)^{k+1} = (b+1)^k(b+1) \equiv (b+1)(b+1) \equiv (b+1) \bmod (2b)$,
and the result holds for all $n$.
For $b$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}$ The limit is:
$$
\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}
$$
When I use calculator and substitute $h$ with $0.000001$ and $-0.000001$, the result is:
$$
\frac{1}{5x^{\frac45}}
$$
My question is:
*
*How to do it without calculator.
*Sho... | Remember that
$$
a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4).
$$
Then,
$$
a-b = \frac{a^5-b^5}{a^4 + a^3b + a^2b^2 + ab^3 + b^4}
$$
Taking $a = (x+h)^{1/5}$ and $b=x^{1/5}$ we get that
\begin{align}
\lim_{h \to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h} &= \lim_{h \to 0}\frac{(x+h) -x}{h(a^4 + a^3b + a^2b^2 + ab^3 + b^4)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Prove $\lim_{x\to \infty} \frac{4x^2 - 7}{2x^3 - 5} = 0$ using $\epsilon$-$N$ limit definition I am having difficulties manipulating the problem so that I can find a $N$ value to choose.
Suppose $x > N$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| \leq \frac{4x^2}{|2x^3 - 5|} + \frac{7}{|2x^3 - 5|} $$
This is basical... | Dividing out an $x^2$, you have to bound
$$\left | \frac{4-\frac{7}{x^2}}{2x-\frac{5}{x^2}} \right |.$$
Suppose $x>\sqrt{5}$. Then $4-\frac{7}{x^2}>0$, so $\left | 4 - \frac{7}{x^2} \right | = 4 - \frac{7}{x^2}<4.$ Additionally $\left | 2x-\frac{5}{x^2} \right | = 2x - \frac{5}{x^2} > 2x-1.$ So the whole thing is less ... | {
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"url": "https://math.stackexchange.com/questions/1138535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is $4 \times 6$ defined as $4 + 4 + 4 + 4 + 4 + 4$ or $6 + 6 + 6 + 6$? There are long debates among Indonesian netizens about this http://www.globalindonesianvoices.com/15785/is-4x6-the-same-as-6x4-this-primary-school-math-made-controversy-in-social-media/
| Are you aware, the multiplication operation is commutative over $\mathbb{N}$.
As such, $6 \times 4 = 4 \times 6$.
And we have also $4+4+4+4+4+4=24=6+6+6+6= 4 \times 6 = 6 \times 4$
EDIT: I forgot the following:
$6+6+6+6=(4+2)+(4+2)+(4+2)+(4+2)=(4+4+4+4)+(2+2+2+2)=(4+4+4+4)+(4+4)=4+4+4+4+4+4$
Thanks to other properties... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fractions in Questions and Answers
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