Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find integer a,b > 1 such that $2^a + 3^b = 2^{a+b} +1$ I would like to know if it is possible to find an integer solution to
$2^a + 3^b = 2^{a+b} +1$
with $a,b > 1$
| We have
$$2^a(2^b - 1) = (3-1)(1 + 3 + 3^2 + \ldots + 3^{b-1}),$$ so
$$2^{a-1}(2^b - 1) = (1 + 3 + 3^2 + \ldots + 3^{b-1}).$$
Considering the second equation mod $3$ we thus see that $2^{a-1}$ and $2^b - 1$ aren't zero. As $2^b$ isn't zero mod $3$ either, it must be $2$ mod $3$, i.e. $b$ must be odd.
But this implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Improper fraction to proper fraction in algebra This question is in regards to the very first part of a question titled 'Express the following in partial fractions'
From my lecture notes I have got the following written down:
$$\frac{x(x+3)}{x^2+x-12}=1+\frac{2x+12}{x^2+x-12}$$
I cant understand how the numerator goes ... | Working backwards:
$$1+\frac{2x+12}{x^2+x-12}= \frac{x^2+x-12}{x^2+x-12}+\frac{2x+12}{x^2+x-12}=\frac{x^2+x-12+2x+12}{x^2+x-12}=\frac{x^2+3x}{x^2+x-12}=\frac{x(x+3)}{x^2+x-12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve inequalities with more than one absolute value expression There are two parts to this question.
1. I'm seeing the correct method to solve these types of inequalities as something to do with "transition points". I don't quite understand this method. How do we find the specific inequalities, and how do we wo... | We need to rewrite the inequation without using the absolute value. Let's recall the definition of the absolute value:
$$
\lvert x \rvert =
\begin{cases}
-x & \text{if $x < 0$,} \\
x & \text{if $x \geq 0$.}
\end{cases}
$$
The transition points are the solutions of the equations $x-3 = 0$ and $2x+5 = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving that $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} =\tan \left ( \frac{\alpha+\beta}{2} \right )$ Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.
$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \f... | Another approach:
Put: $\tan (\alpha/2)=a$ and $ \tan (\beta/2)=b$.
Than:
$$
\sin \alpha= \dfrac{2a}{1+a^2} \qquad \cos \alpha=\dfrac{1-a^2}{1+a^2}
$$
$$
\sin \beta= \dfrac{2b}{1+b^2} \qquad \cos \beta=\dfrac{1-b^2}{1+b^2}
$$
and:
$$
\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \dfrac{2a(1+b^2)+2b(1+a^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluating the indefinite integral $\int\frac{\sqrt{1-x}}{x\sqrt{1+x}}\,dx$ $$\int\frac{\sqrt{1-x}}{x\sqrt{1+x}}\,dx$$
Looking at the term under the square root, I tried out the subsitution $x = \cos\theta$. This reduced the expression to $\int(1 - \sec\theta)d\theta$. This doesn't tally with the answer provided though... | Let $$\displaystyle I = \int\frac{\sqrt{1-x}}{\sqrt{1+x}}\cdot \frac{1}{x}dx = \int\frac{\sqrt{1-x}}{\sqrt{1+x}}\cdot \frac{\sqrt{1-x}}{\sqrt{1-x}}\cdot \frac{1}{x}dx$$
So $$\displaystyle I = \int\frac{1-x}{x\sqrt{1-x^2}}dx = \int\frac{1}{x\sqrt{1-x^2}}dx-\int\frac{1}{\sqrt{1-x^2}}dx$$
Now Put $1-x^2=t^2$ Then $xdx = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the the number of functions $f:A\to B$ whose domain is $A$ such that if $x_1\geq x_2$ then $f(x_1)\geq f(x_2)\forall x_1,x_2\in A$ Let set $A=\left\{1,2,3,4,5\right\}$ and $B=\left\{-2,-1,0,1,2,3,4,5\right\}$
Find the the number of functions $f:A\to B$ whose domain is $A$ such that if $x_1\geq x_2$ then $f(x_1)\ge... | Here we have to find number of non-decreasing function from $A = \left\{1,2,3,4,5\right\}$ to
$B=\left\{-2,-1,0,1,2,3,4,5\right\}.$
Means $f(1)\leq f(2)\leq f(3)\leq f(4)\leq f(5).$
Where $f(1),f(2),f(3),f(4),f(5)\in \left\{-2,-1,0,1,2,3,4,5\right\}$
Now we will break into Different cases.
$\bullet\; $ If $f(1)<f(2)<f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the sum of the moduli of the roots of $x^4-5x^3+6x^2-5x+1=0$ is $4$ Prove that the sum of the moduli of the roots of $x^4-5x^3+6x^2-5x+1=0$ is $4$.
I tried various methods to find its roots,but no luck.Is there any method to solve it without actually finding the roots of the equation.Please help me.
Thanks.... | \begin{align}
x^4-5x^3+6x^2-5x+1&=(\color{red}1\times x^4-\color{red}4\times x^3+\color{red}1\times x^2)\\
&-(\color{red}1\times x^3-\color{red}4\times x^2+\color{red}1\times x)\\
&+(\color{red}1\times x^2-\color{red}4\times x+\color{red}1)\\
&=x^2(x^2-4x+1)-x(x^2-4x+1)+(x^2-4x+1)\\
&=(x^2-4x+1)(x^2-x+1)
\end{align}
In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Compute $\int\frac{1}{1+x^4}dx$ I was given the following hint and I can solve the problem by using the following equation. But I'm curious about how one can get the equation. Can someone give me some wikipedia links about it or hints to manipulate the equation?$$\frac{1}{1+x^4}=\frac{x-\sqrt{2}}{2\sqrt{2}(-x^2+\sqrt{2... | Notice, $$\int\frac{dx}{1+x^4}=\int\frac{dx}{x^2\left(\frac{1}{x^2}+x^2\right)}$$
$$=\int\frac{\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx=\frac{1}{2}\int\frac{\frac{2}{x^2}}{x^2+\frac{1}{x^2}}dx$$
$$=\frac{1}{2}\int\frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}dx$$
$$=\frac{1}{2}\int\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number Theory: Mod Simultaneous Congruences I have this problem assigned for homework. I completed it but I wanted to see if there was a simpler way of solving it. Here it is:
Problem: Solve the set of simultaneous congruences:
$2x\equiv 1\pmod{5}$,
$3x\equiv 9\pmod{6}$,
$4x\equiv 1\pmod{7}$,
$5x\equiv 9\pmod{11}$.
Sol... | There is a classic formula to "write down" the solution after solving some auxiliary congruences. In your case, solve for the following inverses using your favorite algorithm for modular inversion:
$b_1 \equiv (2 \cdot 2 \cdot 7 \cdot 11)^{-1} \mod 5$
$b_2 \equiv (5 \cdot 7 \cdot 11)^{-1} \mod 2$
$b_3 \equiv (4 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\ln(a+b) =\ ln(a) + \ln(b)$ when $a = \frac{b}{b-1}$ Show that $\ln(a+b) = \ln(a) + \ln(b)$ when $a = \frac{b}{b-1}$
My attempt at a solution was
$$
\ln(a+b) = \ln \left(a(1+\frac{b}{a})\right)
$$
and by setting $a = \frac{b}{b-1}$ we get
$$
\ln \left(\frac{b}{b-1}(1+(b-1))\right) = \ln \left(\frac{b}{b-1}+b... | $$\begin{align}\ln (a + b) &= \ln \left(\frac{b}{b - 1} + b\right)\\&=\ln\left(\frac{b + b^2 - b}{b - 1}\right)\\&=\ln\left(\frac{b^2}{b - 1}\right)\\&= \ln \left(\frac{b}{b - 1} \cdot b\right)\\&= \ln\left(\frac{b}{b - 1}\right) + \ln b\\&= \ln a + \ln b\end{align}$$
The trick is that the $b$ terms in the numerator g... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Rationalise the denominator and simplify $\frac {3\sqrt 2-4}{3\sqrt2+4}$ Does someone have an idea how to work $\dfrac {3 \sqrt 2 - 4} {3 \sqrt 2 + 4}$ by rationalising the denominator method and simplifying?
| $\dfrac{3\sqrt 2 - 4}{3\sqrt 2 + 4} \cdot \dfrac{3\sqrt 2 - 4}{3\sqrt 2 - 4}=$
$=\dfrac{18-24\sqrt 2 + 16}{(3\sqrt 2)^2-4^2}=\dfrac{34-24\sqrt 2}{18-16}=\dfrac{2(17-12\sqrt 2)}{2}=$
$=17-12\sqrt 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Infinite inequality with logarithm Solve inequality
$3 - \log_{0.5}x - \log^2_{0.5}x - \log^3_{0.5}x - \cdots \ge 4\log_{0.5}x$
Any suggestions how to start?
| We have
$$\begin{align}
3-\log_{0.5}x-\log^2_{0.5}x-\log^3_{0.5}x-\cdots&=4-\sum_{n=0}^{\infty}\log^n_{0.5}x\\\\
&=4-\frac{1}{1-\log_{0.5}x}\\\\
&=\frac{3-4\log_{0.5}x}{1-\log_{0.5}x}
\end{align}$$
for $|\log_{0.5}x|<1\implies 0.5<x<2$. Now note that the inequality
$$\frac{3-4\log_{0.5}x}{1-\log_{0.5}x}\ge 4\log_{0.5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose $a,b$ are integers. If $4\mid(a^2 +b^2)$, then $a$ and $b$ are not both odd. I've tried using direct proof and contrapositive proof to prove this.
I'm stuck. Is anyone able to help solve or give me some hints?
| If $4|a^2+b^2$ then $a^2+b^2=4m$ for some integer $m$.
Now suppose that $a$ and $b$ are both odd. Then $a=2i+1$ and $b=2j+1$ for some integers $i,j$.
Then $a^2+b^2=4i^2+4i+1+4j^2+4j+1=4(i^2+i+j^2+j)+2$, which is a contradiction.
If you are comfortable with modular arithmetic, we would say that we can't have both $a^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve recurrence $T(n) = nT(n - 1) + 1$ Assume $T(n) = \Theta(1)$ for $n \leq 1$. Using iterative substitution.
So far I have:
\begin{align*}
&T(n) = nT(n - 1) + 1\\
&= n((n - 1)T(n - 2) + 1) + 1\\
&= n(n - 1)T(n - 2) + n + 1
\end{align*}
I'm stuck on how I'm supposed to get the asymptotic value from this. Or ho... | Method 1: Iteration
What you have so far is good. Just keep going!
Here are the next two iterations, to help you:
\begin{align*}
T(n) &= n(n-1)(n-2)T(n-3) + n(n-1) + n + 1 \\
T(n) &= n(n-1)(n-2)(n-3)T(n-4) + n(n-1)(n-2) + n(n-1) + n + 1
\end{align*}
Do you see the pattern? (It's not so obvious!)
Method 2: Generating Fu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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How to show that $\sum_{n=0}^\infty\frac{(2n)!}{(n+1)!n!}(\frac{a}{2})^{2n}\sum_{m=0}^n b^{2m}=\frac{1}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}$? How to show that $$\sum_{n=0}^\infty\frac{(2n)!}{(n+1)!n!}(\frac{a}{2})^{2n}\sum_{m=0}^n b^{2m}=\frac{1}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}$$
I substitute $\sum_{m=0}^n b^{2m}=\frac{1-b^{2n... | The right-hand side of the equation should actually be
$$\frac{\color{red}{2}}{\sqrt{1 - a^2} + \sqrt{1 - a^2b^2}}.\tag{*}$$
I'll assume $a$ and $b$ are real numbers such that $|a| < 1$ and $|ab| < 1$. Since
$$(2n)! = 2^{2n}n!\left(n - \frac{1}{2}\right)\cdots \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)$$
and
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Problem related to integration I tried using partial fractions but I am not sure whether it is the right approach or not. I need help with this problem
Evaluation of Integral $$\int\frac{3x^4-1}{(x^4+x+1)^2}dx $$
| Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2}dx $$
So $$I= \underbrace{\int\frac{3x^4+4x^3}{(x^4+x+1)^2}dx}_{J}-\int\frac{4x^3+1}{(x^4+x+1)^2}dx$$
for second Integral Put $x^4+x+1 = t\;,$ Then $(4x^3+1)dx = dt$
Now for calculation of $\displaystyle J = \int\frac{3x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Division of polynomials to find unknown coefficients The polynomial $-x^4 -x^3 + ax^2 -17x +b$, where $a$ and $b$ are constants, is denoted by $p(x)$. It is given that $x^2 + 4x + 1$ is a factor of $p(x)$.
(i) Find the values of $a$ and $b$.
(ii) With theses values of $a$ and $b$, show that the equation $p(x)=0$ has ex... | by using the long division of $\frac{-x^4-x^3+ax^2-17x+b}{x^2+4x+1}$
$$\frac{-x^4-x^3+ax^2-17x+b}{x^2+4x+1}=(-x^2+3x+a-11)+\frac{(24-4a)x+b-a+11}{x^2+4x+1}$$
the reminder should be zero
$$(24-4a)x+b-a+11=0$$
$$24-4a=0$$
$$a=6$$
$$b-a+11=0$$
$$b-6+11=0$$
$$b=-5$$
$$-x^2+3x+a-11=-x^2+3x+6-11=-x^2+3x-5$$
or
$$x^2-3x+5$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding right inverse matrix Given a $3\times 4$ matrix $A$ such as
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
\end{pmatrix}
,$$
find a matrix $B_{4\times 3}$ such that
$$AB =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
... | Consider the following case:
$$\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}\times \begin{bmatrix}0&0&0\\?&?&?\\?&?&?\\?&?&?\end{bmatrix}=I$$
Removing the first row, the remaining matrix is square:
$$\begin{bmatrix}?&?&?\\?&?&?\\?&?&?\end{bmatrix}=\begin{bmatrix}1&1&1\\1&1&0\\0&1&1\end{bmatrix}^{-1}=\begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Divisibility by 41 and 5-digit number How to prove that if a $5$-digit number is divisible by $41$,then all the numbers generated from it by cyclic shift are also divisible by $41$
| The reason is that $10^5 \equiv 1 \bmod 41$:
$$
0 \equiv 10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0 \bmod 41
\implies\\
0 \equiv 10(10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0) \equiv 10^4 a_3 + 10^3 a_2 + 10^2 a_1 + 10 a_0 + a_4 \bmod 41
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Help with a Differential Equation / Variation of Parameters - Wrong Answer Problem:
Solve the following differential equation.
\begin{eqnarray*}
y'' + y &=& \cot x \\
\end{eqnarray*}
Answer:
The solution I seek is $y = y_c + y_p$ where $y_c$ is the solution to
the corresponding homogeneous differential equation. To fin... | You have the correct answer. (You merely forgot to account for the absolute values when integrating to get the logarithm function.) Combine the two logarithms in your answer, and use the trigonometric identity:
$$\tan(x/2)=\csc(x)-\cot(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Verify $\int\sec x\ dx=\frac12 \ln \left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert + C$ Question says it all, how can I verify the following?
$$\int\sec x\ dx=\frac12 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C$$
| You can verify this by integrating $\sec (x)$ and applying some trig identities:
\begin{align}
\displaystyle
LHS &=
\int \sec(x) \; \mathrm{d}x \\
&= \int \sec(x) \left ( \frac{\sec (x) + \tan (x)}{\sec (x) + \tan (x)} \right ) \; \mathrm{d}x \\
&= \int \frac{\sec^2 (x) + \sec (x) \tan (x)}{\sec (x) + \tan (x)} \; \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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What is the maximum value of $ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}I would appreciate if somebody could help me with the following problem
Q: What is the maximum value of $$ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}<x<\frac{2\pi}{3})$$
I have done my work... | HINT: Notice, $$\frac{2\sin(3x)}{3\sin(x)+3\sqrt{3}\cos (x)}=\frac{1}{3}\left(\frac{\sin(3x)}{\frac{1}{2}\sin(x)+\frac{\sqrt{3}}{2}\cos (x)}\right)=\frac{1}{3}\left(\frac{\sin(3x)}{\sin(x)\cos\frac{\pi}{3}+\cos (x)\sin \frac{\pi}{3}}\right)=\frac{1}{3}\frac{\sin(3x)}{\sin \left(x+\frac{\pi}{3}\right)}$$
Now, $$f'(x)=\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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how to find this limit : lim x to infinity how can I find this limit which is become infinite
$\lim _{x\to \infty }\left(x(\sqrt{x^2+1}-x)\right)$
can I use conjugate method ?
that what I'm doing until now
$= x\left(\frac{\sqrt{x^2+1}-x}{1}\cdot \:\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)\:$
$= x\left(\frac{1}{\sqrt... | For the edited and corrected question, i.e. the limit at $\infty$ of
$$
x\left(\sqrt{x^2+1} - x\right)
$$
indeed you can deal with this multiplying by the conjugate (cf. jeantheron's answer). However, I would advocate for a more systematic approach, simple enough and highly generalizable. Assuming you have heard or wi... | {
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What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
| $$\small\begin{align}&\frac{x+1}{(x-1)(x+2)}\gt\frac{x}{(x-1)(x-3)}\\\\&\iff \frac{x+1}{(x-1)(x+2)}\cdot (x-1)^2(x+2)^2(x-3)^2\gt\frac{x}{(x-1)(x-3)}\cdot (x-1)^2(x+2)^2(x-3)^2\\\\&\iff (x+1)(x-1)(x+2)(x-3)^2\gt x(x-1)(x+2)^2(x-3)\\\\&\iff (x-1)(x+2)(x-3)((x+1)(x-3)-x(x+2))\gt 0\\\\&\iff (x-1)(x+2)(x-3)(-4x-3)\gt 0\\\\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 5,
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Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods wou... | Let $\displaystyle \frac{x}{y} =a \;\;,\frac{y}{z} = b\;\;,\frac{z}{x} = c\;,$ Then $a+b+c=7$ and $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9$
and $abc=1$. Then we have to calculate $a^3+b^3+c^3 = $
Now Using $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9\Rightarrow \frac{ab+bc+ca}{abc} = 9\Rightarr... | {
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"source": "stackexchange",
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What is the smallest possible value of $\lfloor (a+b+c)/d\rfloor+\lfloor (a+b+d)/c\rfloor+\lfloor (a+d+c)/b\rfloor+\lfloor (d+b+c)/a\rfloor$? What is the smallest possible value of
$$\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\f... | Since $\lfloor x\rfloor \gt x-1$, we have
$$\begin{align}&\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\frac{d+b+c}{a}\right\rfloor\\\\&\gt \frac{a+b+c}{d}-1+\frac{a+b+d}{c}-1+\frac{a+d+c}{b}-1+\frac{d+b+c}{a}-1\\\\&=\left(\frac a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Powers of a root in terms of basis of field extension I'm wondering if my solution correct for these types of problems
Problem: given irreducible $$f(x)=x^3-2x-2 \in \mathbb{Q}[x]$$
Let $\theta$ be a root of $f$, and $K=\mathbb{Q}(\theta)$ extension over $\mathbb{Q}$. $K$ has $\{1,\theta,\theta^2\}$ as a $\mathbb{Q}$ b... | Divide the polynomial $P=X^5$ by $T=X^3-2X-2$.
First step : $P=X^5=\color{red}{X^2}\,T+2X^3+2X^2$.
Second step : $P=\color{red}{X^2}T+\color{blue}{2}T+2X^2+4X+4$.
Conclusion
$$P=\underbrace{(X^2+2)}_QT+\underbrace{2X^2+4X+4}_{R}.$$
This is the Euclidian division of polynomials, the remainder $R$ is what you want beca... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Find the angle of complex number Let
$$
z = \frac{a-jw}{a+jw}.
$$
Then the angle of $z$ is
$$
-\tan^{-1}z\left(\frac{w}{a}\right) -\tan^{-1}\left(\frac{w}{a}\right) = -2\tan^{-1}\left(\frac{w}{a}\right).
$$
How is that so?
| I suppose that it could be easier to see writing $$z = \frac{a-jw}{a+jw}= \frac{a-jw}{a+jw} \times \frac{a-jw}{a-jw}=\frac{(a-jw)^2}{a^2+w^2}=\frac{a^2-w^2}{a^2+w^2}-\frac{2 a w}{a^2+w^2}j$$ which makes the angle to be such that $$\tan(\theta)=-\frac{2aw}{a^2-w^2}=-\frac{2\frac aw}{1-(\frac aw)^2}$$ and to remember th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Linear equations with parameters
For which $a$ the following had one solution, no solution, infinite solutions:
$$\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a & 4 \\
4a-8 & 16 & 4a-2 & a+14 \\
10a-20 & -2a+36 & 8a-4 & 2a+30 \\
\end{array}
\right]$$
I manage to come to this:
$$\
\left[
\begin{array}{ccc|c}
2a-4 & ... | What about that: divide the second row by $2$ and the third row by $3$
$$
\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a & 4 \\
0 & 2a+4 & 2a-2 & a+6 \\
0 & 3a+6 & 3a-4 & 2a+10 \\
\end{array}
\right]=\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a & 4 \\
0 & a+2 & a-1 & \frac12 a+3 \\
0 & a+2 & a-\frac43 & \frac23a+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $P(1)/P(-1)$
The polynomial $f(x)=x^{2007}+17x^{2006}+1$ has distinct zeroes $r_1,\ldots,r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\left(r_j+\dfrac{1}{r_j}\right)=0$ for $j=1,\ldots,2007$. Determine the value of $P(1)/P(-1)$.
Let $P(x) = a_{2007}x^{2007} + ... + a_0$.
$r_j + \frac... | Continuing after step $(2)$ in @Jack D'Aurizio's answer.
$$\begin{align}\frac{P(1)}{P(-1)}&=\prod_{i=1}^{2007}\frac{r_i^2-r_i+1}{r_i^2+r_i+1}=\prod_{i=1}^{2007}\frac{(r_i+\omega)(r_i+\omega^2)}{(r_i-\omega)(r_i-\omega^2)}=\prod_{i=1}^{2007}\frac{(-\omega-r_i)(-\omega^2-r_i)}{(\omega-r_i)(\omega^2-r_i)}\\ &=\prod_{i=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding more critical points If we have the function $f$ defined:
$$f(x,y)=2\sin(x)+2\sin(y)+\sin(x+y)$$ for $-\pi \le x\le \pi$ and $-\pi \le y \le \pi$
Find the critical points and determine the nature of each.
I'm a bit stuck on this.
I've found:
$\frac {\partial f}{\partial x}=2\cos(x)+\cos(x+y)$
$\frac {\partial... | Based on your comment the function is $f(x,y)=2\sin(x)+2\sin(y)+\sin(x+y)$, to solve critical points,
$f_x=2cos(x)+cos(x+y)=0$ and
$f_y=2cos(y)+cos(x+y)=0$
Subtract equations we have $\cos(x)=\cos(y)$, when $x$ and $y$ is constrained on given range, that means $x=y$ or $x=-y$.
When $x=y$, $2\cos(x)+\cos(2x)=0\Rightarro... | {
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"source": "stackexchange",
"question_score": "2",
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$ x+y+z = 3, \; \sum\limits_{cyc} \frac{x}{2x^2+x+1} \leq \frac{3}{4} $
For positive variables $ x+y+z=3 $, show that $ \displaystyle \sum_{cyc} \dfrac{x}{2x^2+x+1} \leq \dfrac{3}{4} $.
Apart from $ (n-1)$ EV, I could not prove this inequality. I've tried transforming it into a more generic problem - it looks fairly ... | $\left(\frac{x}{2x^2+x+1}\right)''=\frac{2(4x^3-6x-1)}{(2x^2+x+1)^3}<0$ for all $0<x<1$.
Hence, by Vasc's LCF Theorem it remains to prove our inequality for $y=x$ and $z=3-2x$,
which gives $(x-1)^2(8x^2-14x+9)\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the remainder of $31^{2008}$ divided by $36$? What is the remainder of $31^{2008}$ divided by $36$?
Using Euler's theorem, we have:
$$
\begin{align*}
\gcd(31,36) = 1 &\implies 31^{35} \equiv 1 \pmod{36} \\
&\implies 31^{2008} \equiv 31^{35(57) + 13} \equiv (31^{35})^{57+13} \equiv 1^{57}*31^{13} \\
&\implies 31... | Euler's Theorem states $a^{\phi (n)} \equiv 1 \left(\mod \, \, n\right)$ if $a \perp n$, or $\gcd(a,n)=1$. Since $\gcd(31,36)=1$,
\begin{equation}
31^{\phi (36)} = 31^{12} \equiv 1 (\mod\,\,36)
\end{equation}
Then,
\begin{align}
(31^{12})^2 &= 31^{144} \equiv 1 (\mod \, \, 36) \\
31^{144} \cdot (31^{12})^{155} &= 31^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a binomial identity Prove that ($k\le m$)
$$\sum_{j=k}^m(_{2m+1}^{2j+1})(_k^j)=\frac{2^{2(m-k)}(2m-k)!}{(2m-2k)!k!} , (k\le m)$$
Please help me with this identity, I've spent a lot of time on it but didn't solve the problem.
| According to OPs expression we show:
The following is valid
\begin{align*}
\sum_{j=k}^{m}\binom{2m+1}{2j+1}\binom{j}{k}=4^{m-k}\binom{2m-k}{k}\qquad\qquad 0\leq k \leq m
\end{align*}
In the following we use the coefficient of operator $[x^k]$ to denote the coefficient of $x^{k}$ in a series $\sum_{j=0}^{\infty}a_jx^... | {
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"source": "stackexchange",
"question_score": "1",
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Can a pre-calculus student prove this?
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square
So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite... | Given any $(a,b)$ such that $a^3 + 4a^2b = 4a^2 + b^4$.
(1) If $a = 0$, then $b$ is clearly 0, $\sqrt{a} - 1 = -1$ is not a rational square.
(2) If $b = 0$, then $a^3 = 4a^2 \implies a = 0, 4$.
*
*In the first sub case $a = 0 \implies \sqrt{a}-1 = -1$ is not a square.
*In the second sub case, $a = 4 \implies \sq... | {
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"source": "stackexchange",
"question_score": "33",
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The Diagonal Elements Of A Special Symmetric Matrix
A $n \times n$ matrix $M$ is a symmetric matrix,where $n$ is odd($i.e.n=2k+1,k\in \mathbb{Z}^{+}\cup{\{0\}}$). Every row of $M$ is a permutation of $\{1,2,\cdots,n\}$.
Show that the diagonal elements of $M$ is also a permutation of $\{1,2,\cdots,n\}.$
$e.g.$ wh... | A simple parity argument will do. By assumption, every element of $S=\{1,2,\ldots,n\}$ must occur an odd number of times (or more precisely, $n$ times). Yet, if some element of $S$ does not appear on the diagonal, it must appear in $M$ an even number of times, because the matrix is symmetric and off-diagonal elements o... | {
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"source": "stackexchange",
"question_score": "5",
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Matrices such that $(A+B)^n = A^n + B^n$ How can I prove that $(A+B)^n = A^n + B^n$ for all integers $n \geq 1$ ?
I have been thinking about induction? Start for example with basecase 2 and then assume it's true for n = k, which would imply it's true for n = k + 1. But exactly how would this look, could someone help me... | Once you establish $AB = BA = 0$ the induction is easy:
We will take as our base case $n = 2$ (so we may assume later $n$ is at least $2$ -the case $n = 1$ is trivial).
$(A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2 = A^2 + 0 + 0 + B^2 = A^2 + B^2$.
Suppose $(A + B)^{n-1} = A^{n-1} + B^{n-1}$.
Then $(A + B)^n = (A +... | {
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"source": "stackexchange",
"question_score": "2",
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Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$ Prove that:
$$
\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}
$$
if $a,b$ are positive real numbers such that $a+b=1$.
I have tried expanding the squares and rewriting them such that $a+b$ is a term/part of a term but what I get is completely... | For $E=(a+1/a)^2+(b+1/b)^2=a^2+b^2+1/a^2+1/b^2+4$ you have $1=(a+b)^2=a^2+b^2+2ab\leq 2(a^2+b^2)$, so $a^2+b^2\geq 1/2$. Moreover, $\frac{a+b}{2}\geq 2\sqrt{ab}$ so $\frac{1}{(ab)^2}\geq 16$. This implies $$E=a^2+b^2+\frac{a^2+b^2}{a^2b^2}+4\geq 9/2+8=\frac{25}{2},$$
because $\frac{a^+b^2}{a^2b^2}\geq \frac{1}{2}\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solve complex equation for $x$ and $y$ How to solve this?
I have tried to put all $z$s on one side,
but I don't have an idea to continue.
$$z^3-i(z-2i)^3=0$$
| $$
z^3 = i\cdot(z-2i)^3.
$$
Therefore $z$ is one of the three cube roots of $i(z-2i)^3$.
Observe that $(-i)^3 = i$ so $z$ is one of the three cube roots of $(-i)^3 (z - 2i)^3 = (-2-iz)^3$.
One of those three cube roots is $-2-iz$.
So either $z=-2-iz$ or $z$ is one of the other two cube roots.
If $z=-2-iz$ then $z(1+i) ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find solutions to $x^2 \equiv 4 \pmod{91}$? As the title says, I'm looking to find all solutions to $$x^2 \equiv 4 \pmod{91}$$ and I am not exactly sure how to proceed.
The hint was that since 91 is not prime, the Chinese Remainder Theorem might be useful.
So I've started by separating into two separate congruen... | We begin with your system of equations: $$\begin{cases} x^2 \equiv 4 \pmod{7} \\ x^2 \equiv 4 \pmod{13} \end{cases}$$
Then, solving each congruence, we obtain the system: $$\begin{cases} x \equiv \pm 2 \pmod{7} \\ x \equiv \pm 2 \pmod{13} \end{cases}$$
We therefore have four systems of linear congruences, each with a u... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that if $\gcd(a,b)=1$, then $\gcd(a+b,a^2+b^2-ab)=1\text{ or } 3$. I come across things like the following but I don't quite know how to use them.
$$\gcd(a+b,a^2+b^2-ab)|3ab$$
$$\gcd(a,b)=1\Rightarrow \gcd(a+b,ab)=1$$
| $a^2 + b^2 - ab = (a + b)^2 - 3ab$
so
$$\gcd(a+b,a^2 + b^2 - ab) = \gcd(a + b， 3ab)$$
If $3$ does not divide $ a + b$ then
$$\gcd(a+b,a^2 + b^2 - ab) = \gcd(a + b， ab) = \gcd(a,b) = 1$$
otherwise let $a + b = 3k,\,\, k\in \mathbb{Z}$, then $b = 3k - a$ and
$$\gcd(a+b,a^2 + b^2 - ab) = 3\gcd(k， a(3k - a)) = 3\gcd(k,a^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A good way to solve this trigonometric equation $$\sin x+\cos x=\frac{1}{2}$$
What is the value of $\tan x$? I tried using
$$\sin2 x=\frac{2\tan x}{1+\tan^2x}$$ and $$\cos2 x=\frac{1-\tan^2x}{1+\tan^2x}$$ but we get a quadratic for $\tan\left(\frac{x}{2}\right)$ . So any better approach would be much appreciated. Than... | HINT:
$$\cos(x)+\sin(x)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\left(\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\left(\sin\left(\frac{\pi}{4}\right)\cos(x)+\cos\left(\frac{\pi}{4}\right)\sin(x)\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\sin\left(\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral $\sqrt{1+\frac{1}{4x}}$ $$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$
This integral came up while doing an arc length problem and out of curiosity I typed it into my TI 89 and got this output
$$\frac{-\ln(|x|)+2\ln(\sqrt\frac{4x+1}{x}-2)-2x\sqrt\frac{4x+1}{x}}{8}\ $$
How would one get this answer by integrating m... | $$I=\int\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$$ Use $$\sqrt{x}=t$$ $\implies$ $$\frac{dx}{2\sqrt{x}}=dt$$
we get $$I=\int \sqrt{4t^2+1}dt=2\int\sqrt{t^2+0.5^2}dt$$
now use standard integral $$\int\sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln|x+\sqrt{x^2+a^2}|$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent Let $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\cdot A=A$ and $B\cdot B=B$?
Steps I took:
*
*Let $A= \left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right]$ and let $B= \left[\begin{array}{rr}
w ... | $$A \cdot B = A \\
\implies (A \cdot B) \cdot A
= A \cdot A \\
\implies A \cdot (B \cdot A) = A^2 \\
\implies A \cdot B = A^2 = A$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\fr... | Your proof is correct, but I feel that this could be proved by contradiction.
Assume for contradiction $\exists x>0$ such that the equation $\sqrt{x+2}-\sqrt{x+1}=\sqrt{x+1}-\sqrt{x}$ is true. Then,
\begin{align}
\sqrt{x+2}-\sqrt{x+1}&=\sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{x+1-x}{\sqrt{x +1... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof (cases & induction): Find the set of positive integers such that $n! \geq n^3$ I need to find the set of positive integers such that $n! \geq n^3$, and then prove my answer is true using cases and induction on $n$.
There is a lemma that I will need to prove and use for this proof.
The lemma is :
$n^2+2n+1\leq n... | Let $f(x) = x^3 - (x + 1)^2$
$f'(x) = 3x^2 - 2x + 2$. $f'(x) = 0$ when $x = \frac{2 \pm \sqrt(4 - 24}{6}$ so $f'(x)$ never equals $0$ for any real x. So $f(x)$ has no extrema and thus only one real root.
$ f(2) = -1; f(3) = 18$ so $f(x)$ has only one real root is between 2 and 3. So $f(n) > 0$ iff $n^3 > (n+1)^2$ ... | {
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"question_score": "2",
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"answer_id": 1
} |
Evaluate thus limit using series: $\lim_{x\to0} (\sin x-\tan x)/x^3$ Evaluate thus limit using series:
$$\lim_{x\to0} \frac{\sin x-\tan x}{x^3}$$
I know the value of this limit is -1/2, and I also know the series expansion for $\sin x$ is $$x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$$
I am having trouble being able t... | Expand both $\sin$ and $\tan$: around $0$,
$$\begin{align}
\sin x &= x - \frac{x^3}{6} + o(x^3) \\
\tan x &= x + \frac{x^3}{3} + o(x^3)
\end{align}$$
See below for more.
$$\begin{align}\frac{\sin x - \tan x}{x^3} &= \frac{x - \frac{x^3}{6} - \left(x + \frac{x^3}{3}\right) + o(x^3)}{x^3} = \frac{-\frac{x^3}{3} - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving the limit of a radical using epsilon delta Define $f:(0,1)->R$ by $f(x)=\frac{\sqrt{9-x}-3}{x} $
I know that $f(x)=\frac{\sqrt{9-x}-3}{x}*\frac{\sqrt{9-x}+3}{\sqrt{9-x}+3}=\frac{9-x-9}{\sqrt{9-x}+3}=-\frac{1}{\sqrt{9-x}+3} $
Since $f$ is only defined in $(0,1)$, $-\frac{1}{5}<-\frac{1}{\sqrt{9-x}+3}<-\frac{1}{6... | From what you wrote I do not see by what you make at the choice. Instead:
If $0 < x < 1$, then
$$
\bigg| \frac{-1}{\sqrt{9-x}+3} + \frac{1}{6} \bigg|
=
\frac{3 - \sqrt{9-x}}{6\sqrt{9-x}+18} < 3 - \sqrt{9-x} =: s_{x} \leq 3;
$$
given any $0 < \varepsilon \leq 3$, we have $s_{x} < \varepsilon$ if $x < 9 - (3-\varepsilon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1? I can show for any given value of n that the equation
$$\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$$
is true and I can see that geometrically it is true. However, I can not seem to prove it out analytically. I have spent most of my time trying induction... | One approach is to write
$$\cos\left(\frac{2\pi k}{n}\right)=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(k+1)}{n}\right)-\sin\left(\frac{2\pi(k-1)}{n}\right)\right)$$
Now, we have converted the sum into a telescoping sum, which we can evaluate directly as
$$\begin{align}
\sum_{k=1}^n\cos\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Random Variables, Die toss A fair die is tossed twice. Let $X$ = the ssm of the faces, $Y$= the maximum of the two faces, and $Z$=|face 1 - face 2|.
write down the value of $X,Y,$ and $W=XZ$ for each outcome $w\in\ S$
I already found the value and range of $X,Y$ but I'm not sure how to find $W=XZ$.
I saw someone post a... | Since you said you've enumerated the outcomes for $X$, do the same for $Z$. Below I made a table for the values of both $X$ and $Z$. Can you now make the corresponding table for $W = XZ$?
$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
X& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
2 & 3 & 4 & 5 & 6 & 7 & 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trigonometry equation, odd-function. So I have the following equation:
$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$
It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?
| More unorthodoxly, we first take the midpoint of $x-\frac{\pi}{6}$ and $x + \frac{\pi}{4}$ before applying the angle sum and difference identities:
Let $A = x + \frac{\pi}{24}$ (the midpoint). Then we need
$$\sin\left(A - \frac{5\pi}{24}\right) + \cos \left(A + \frac{5\pi}{24}\right) = 0$$
$$\sin A\cos\frac{5\pi}{24} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to calculate $\lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$. Let $s = \lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$.
$$st = t\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ...}}}} = \sqrt{t^{2} + \sqrt{2t^{4} + \sqrt{4t^{8} + \sqrt{8t^{16} + ...}}}}... | If $t=\frac1{\sqrt2}$, then $t^2=\frac12$, and $2t^4=\frac12$, but $4t^8=\frac14$. Your expression for $\frac s{\sqrt2}$ isn't composed entirely of $\frac12$s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Simultaneous equations How can I find the solution $(3, 3, \frac{-3}{2})$ from the following system of equations:
$x+(y-1) z = 0$
$(x-1) z+y = 0$
$x (y-1)-y+2 z = 0$
I have done eq1 - eq2 to find the other solutions. How would i get to the solution $(3, 3, \frac{-3}{2})$
| $$x+(y-1) z = 0 \implies (y-1)=-\frac xz \tag{*1} $$
$$x (y-1)-y+2 z = 0 \implies (y-1)=\frac{y-2z}{x} $$
so $$(y-1)^2=- \frac yz +2 $$
$$ (x-1) z+y = 0 \implies \frac yz = -(x-1) \tag {*2}$$
so $$(y-1)^2= x+1 \tag{*3}$$
we can get another relation between $x$ and $y$ by dividing (*1) by (*2)
$$ -\frac xy =-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate limit $\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}$ How to evaluate $$\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}?$$
Only idea I can think of is sandwich theorem, but then I get $0^0$.
| This limit does not exist. We can pick two sequences $a_n$ and $b_n$ such that $f(a_n)$ has different limit than $f(b_n)$.
\begin{align}
a_n &= 2\pi n + \frac{\pi}{2} \\
b_n &= 2\pi n - \frac{\pi}{2}
\end{align}
Then
$$
\left(1+\sin\left(2\pi n + \frac{\pi}{2}\right)\right)^{\frac1{2\pi n + \frac{\pi}{2}}} = 2^{\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Explain the following equalities I am a little stuck on coming up with geometrical explanation for why the following equalities are true. I tried arguing the $\cos(\theta)$ is the projection to the x-axis of a vector $r$ inside a unit circle, so as it goes around by $2 \pi$, the projections on both the positive and neg... | First set of equations: Here's a geometric reason. Imagine we place $1$ kg weights at the unit circle corresponding to angles of $\theta$, $\theta+\frac{2\pi}{3}$ and $\theta+\frac{4\pi}{3}$. These points form a configuration that is a rotation of
The center of mass of this configuration is at the origin, and hence t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to prove $2^{2^{6n+2}} + 21$ is a composite number? How to prove that $$ 2^{2^{6n+2}} + 21 $$ is a composite number for each $$n=1,2,3,\ldots$$
I'm stuck on a what seems like a simple problem but I don't even know where to start.
| $ 2^{2^{6n+2}} + 21=2^{4 (2^{6n} )}-2^4+2^4+21= $
$ =37+2^4\left (2^{4(2^{6n}-1 )}-1 \right )=37+2^4\left (2^{4(2^6-1 )(2^{6(n-1)}+ \dots +1 )}-1 \right )=$
$ =37+2^4\left (2^{36 \cdot 7 \cdot (2^{6(n-1)}+ \dots +1 )}-1 \right )=$
$ =37+2^4\left (2^{36}-1\right )\left (\dots \right ) \equiv 0 \bmod 37$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the equation of a tangent line to a curve at a given point. Find the equation for the tangent (in standard form) to the curve defined by $$y = \sqrt\frac{5x}{2x-1}$$
For the procedure of this question would I simply differentiate it using the quotient rule? From which you insert $x=5$ in to, and then use $y_2-y... | The point on the curve will be $(5,\frac{5}{3})$. Then differentiate $y$ using the chain rule and quotient rule
\begin{align}
\frac{dy}{dx} = \frac{1}{2}\bigg(\frac{5x}{2x - 1}\bigg)^{-\frac{1}{2}}\bigg( \frac{5(2x - 1) - (2)(5x)}{(2x - 1)^2}\bigg)
\end{align}
Then by pluggin in $x = 5$
\begin{align}
\frac{dy}{dx}\bigg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you prove that $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x$? I have the task to prove that
$$
\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x ,\left|x\right|\le 1
$$
I do not have any ideas from where I should start.
Can anyone help m... | You may just observe that, for $x \in [0,1)$, we have
$$
\left(\arcsin\left( \frac{1}{2} \sqrt{2-\sqrt {2-2x}}\right)\right)'=\frac14\frac1{\sqrt{1-x^2}}
$$ and we have
$$
\left(\frac{\pi}{8} + \frac{1}{4} \arcsin x\right)'=\frac14\frac1{\sqrt{1-x^2}}
$$ giving
$$
\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2... | $$\frac{4-x^2}{3-\sqrt{x^2+5}}=(4-x^2)\frac{3+\sqrt{x^2+5}}{9-(x^2+5)}=3+\sqrt{x^2+5}$$
$$\lim_{x\to2}\frac{4-x^2}{3-\sqrt{x^2+5}}=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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sum of solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$ What is the sum of non-negative integer solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$?
I know that $\{(x,y,z)\mid x+y+z=k\}$ has $\binom{k+3-1}{3-1}=\binom{k+2}{2}$ non-negative integer solutions. Thus, the answer to the question above is $$\su... | Here's another elegant way to simplify the expression. See that $\binom{n}{r}$ is the coefficient of $x^r$ in $(1+x)^n$. Hence, your summation can be interpreted as the $$\sum_{k=1}^N \text{coefficient of } x^2 \text{ in } (1+x)^{k+2} = \text{coefficient of } x^2 \text{ in }\sum_{k=1}^N (1+x)^{k+2}$$ where the last se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Combinatorics Inclusion - Exclusion Principle Find the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 25$ with $ 1 \leq x_1 \leq 6, 2 \leq x_2 \leq 8,
0 \leq x_3 \leq 8, 5 \leq x_4 \leq 9.$
Firstly, I defined $y_i = x_i - lower bound$ so for example $y_1 = x_1 - 1$
This gives $y_1 + y_2 + y_3 + y_4 = 17$ with... | As noted in the comments, your solution is correct. The approach is described in Balls In Bins With Limited Capacity.
To solve the problem using generating functions, write $x+x^2+\cdots+x^6=x\frac{1-x^6}{1-x}$ for $x_1$, $x^2+x^3+\cdots+x^8=x^2\frac{1-x^7}{1-x}$ for $x_2$, $1+x+\cdots+x^8=\frac{1-x^9}{1-x}$ for $x_3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
My work:
We consider the congruences $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$, $x \equiv 5 \pmod 6$. We can reduce t... | It follows from the hypothesis that $3,4,5,6$ divides $(x+1) $ and therefore $x+1$ is a common multiple of these 4 numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculating an Exponential Integral
Calculate $\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx$ where $a,c > 0$ and $b>1$.
The best I could do about this integral is to find an upperbound for it:
$\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx \leq \int_0^{\infty} \frac{x^{2N+1}}{x^{-b}} e^{-c x^2} d... | $\int_0^\infty\dfrac{x^{2N+1}e^{-cx^2}}{a+x^{-b}}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^{2n+b+2N+1}}{n!(ax^b+1)}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^{\frac{2n+2N+1}{b}+1}}{n!(ax+1)}~d\left(x^\frac{1}{b}\right)$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Solve $\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}$ How do I solve limits such as these? The $...$ always make it seem hard to me. From what I can understand from them, they both are $0/0$ limits, and I should be looking to write the numerator in such a way that the denominator should simplify it s... | Note that:
$\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}= \frac{(x-1)+(x^2-1)+\ldots +(x^n-1)}{x-1}=\frac{(x-1)[1+(x+1)+(x^2+x+1)\ldots +(x^{n-1}+x^{n-2}+\ldots+x+1)]}{x-1}$,
after simplification we find:
$n+(n-1)x+(n-2)x^2+\ldots+x^{n-1}$.
So,
$\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}=n+(n-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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A weird property of $\sum_{k = 1}^{n} \sin k$ I was playing around with the sum $\sum_{k = 1}^{n} \sin k$, and using very loose rigour I arrived at the following:
Proposition. Let $n \equiv n_0 \pmod {44}$ and $n_0 \equiv n_1 \pmod {6}$. Then $$\sum_{k = 1}^{n} \sin k \sim \frac {1} {2} \left ( 3 - |3 - n_1| \right ).$... | I think it appropriate to present my rigour here now that it created sufficient curiosity.
Consider $$f (n) = \sum_{k = 1}^{n} \sin k.$$ Let $a_k = \left [ \frac {2k} {\pi} \right ]$ for $1 \leqslant k \leqslant n$ and $r_1, r_2, \cdots, r_n$ be real numbers in the interval $[0, 1]$. Then, denote $$\sin k = \begin {cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding $\binom n0+\binom n3+\binom n6+\cdots $ How to get
$$\binom n0 + \binom n3 + \binom n6 + \cdots$$
MY ATTEMPT
$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$
$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$
$$(1 + 1)^n = 2^n = \binom n0 + \binom... | Let $\omega = \dfrac{-1+ i\sqrt 3} 2 = \cos120^\circ + i\sin120^\circ$.
Then $\omega^3 = 1$, and $1+\omega = \cos60^\circ + i\sin60^\circ$, so $(1+\omega)^2 = \omega$.
A bit of arithmetic shows that $n\mapsto(1+\omega)^n + (1+\omega^2)^n$ is a periodic function with period $6$:
\begin{array}{c|c}
n & (1+\omega)^n + (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence of two recursive sequence how could I find out whether those recursive sequences are convergent?
Let $a_{n+1} = a_1(1-a_n-b_n) + a_n \\ b_{n+1} = b_1(1-a_n-b_n) + b_n$, where
$a_1 = a \\ b_1 = b$ ($a, b \in (0, 1)$)
| $$
a_{n+1}=a(1-a_n-b_n)+a_n\text{ ...(1)}\\
b_{n+1}=b(1-a_n-b_n)+b_n\text{ ...(2)}\\$$
$(1)+\alpha\times(2):$
$$a_{n+1}+\alpha b_{n+1}=(1-a-\alpha b)a_n+(-a+\alpha(1-b))b_n+a+\alpha b$$
$$\frac{-a+\alpha(1-b)}{1-a-\alpha b}=\alpha$$
$$b\alpha^2+(a-b)\alpha-a=0$$
$$(\alpha-1)(b\alpha+a)=0$$
i) $(1)+(2):$
$$a_{n+1}+b_{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2... | You have:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx
$$
First we'll do a routine substitution:
$$
u = 1+x-2x^2, \qquad du = (1-4x)\,dx, \qquad \frac{-du} 4 = \left( \frac 1 4 - x \right)\, dx
$$
\begin{align}
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx & = \int \frac{\frac 1 4-x}{\sqrt{1+x-2x^2}}\,dx + \int \frac{\frac 3 4}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to prove $\sum_p p^{-2} < \frac{1}{2}$? I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me.
Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of... | We can deduce this quickly, and without knowing the numerical value of $\pi$, from the fact that
$$\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6},$$ for which there are numerous proofs available.
Let $E$ denote the set of even numbers; the sum of the squares of all such numbers is
$$\sum_{n \in E} \frac{1}{n^2} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
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Generalized Harmonic numbers I'd like to be able to prove the following inequality:
$\frac{{{H_{n, - r}}}}{{{n^r}\left( {n + 1} \right)}} \le \frac{{{H_{n - 1, - r}}}}{{n{{\left( {n - 1} \right)}^r}}}$.
It's clear that as $n \to \infty$ we get equality, the limit on each side is $1/(r+1)$, and it also seems clear that ... | We want to show
$\frac{H_{n, - r}}{n^r\left( n + 1 \right) }
\le \frac{H_{n - 1, - r}}{n\left( n - 1 \right)^r}
$
where
$H_{n, a}
=\sum_{k=1}^n \frac1{k^a}
$,
so
$H_{n, -r}
=\sum_{k=1}^n k^r
$.
From this,
$H_{n - 1, - r}
=H_{n , - r}-n^r
$,
so we want
$\frac{H_{n, - r}}{n^r\left( n + 1 \right) }
\le \frac{H_{n , - r}-n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving a differential equation by separating variables How can one solve the following differential equation by the technique of separation of variables?
$$\frac{1}{x^2}\frac{dy}{dx}=y^5\ \ \ \text{ when }, \ y(0)=-1$$
| Notice, $$\frac{1}{x^2}\frac{dy}{dx}=y^5$$
$$\frac{1}{y^2}dy=x^2dx$$
$$\int y^{-5}\ dy=\int x^2\ dx$$
$$\frac{y^{-4}}{-4}=\frac{x^3}{3}+C$$
$$-\frac{1}{4y^{4}}=\frac{x^3}{3}+C$$
now, setting $x=0$ & $y=-1$,
$$-\frac{1}{4(-1)^{4}}=\frac{0}{3}+C\implies C=-\frac{1}{4}$$
set the value of $C$ to get the solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
2 problems related to the number 2015
*
*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$
*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, bu... | 1.
$$9p=\underbrace{99\cdots9}_\text{4030}+\underbrace{99\cdots9}_\text{2015}$$
$$=10^{4030}-1+10^{2015}-1=\left(10^{2015}\right)^2+10^{2015}-2$$
$$=\left(10^{2015}-1\right)\left(10^{2015}+2\right)=\left(10^{2015}-1\right)\left(10^{2015}-1+3\right)$$
$$\therefore p=\frac{10^{2015}-1}{3}\left(\frac{10^{2015}-1}{3}+1\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve system of $n$ equations of the form $2x_k^3+4=x_k^2(x_{k+1}+3)$ Solve the system of $n$ equations, $n\geq2$:
$$
\begin{cases}
2x_1^3+4=x_1^2(x_2+3)\\
2x_2^3+4=x_2^2(x_3+3)\\
\qquad \vdots\\
2x_{n-1}^3+4=x_{n-1}^2(x_n+3)\\
2x_n^3+4=x_n^2(x_1+3)\\
\end{cases}
$$
I think that there are only two solutions, but I don'... | $$2x_n^3+4=x_n^2(x_{n+1}+3)$$
$$2x_n+\frac{4}{x_n^2}=x_{n+1}+3$$
$$x_{n+1}=2x_n+\frac{4}{x_n^2}-3$$
Define $f(x)$:
$$f(x)=2x+\frac{4}{x^2}-3=x+\frac{x^3-3x^2+4}{x^2}=x+\frac{(x-2)^2(x+1)}{x^2}$$
\begin{align}
&f(x)=x\quad(x=-1, \space 2)\\
&f(x)>x\quad(x>-1,\space x\ne2)\\
&f(x)<x\quad(x<-1)\\
\end{align}
If $x_1\ne-1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b... |
Let's use the angle bisector theorem on $C$, we get
$$\frac{b+a}{c}=\frac{a}{EB}.$$
By the angle bisector theorem on $B$, we get
$$\frac{c+a}{b}=\frac{a}{CD}.$$
Now from the law of sines
$$EB=\frac{a\sin{C/2}}{\sin y}$$
and
$$CD=\frac{a\sin{B/2}}{\sin x}.$$
Putting all together we find
$$\frac{(b+a)\sin {C/2}}{c \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to Evaluate $\lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right)$? Here is my limit to be evaluated
$\lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right)$=?
| Here is a simple step by step approch
$$\begin{align}
&\quad \lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right) \\
&= \lim_{x \to \infty} 2 + 8 \frac{x}{4} \sin\left(\frac{4}{x}\right) \\
&= \lim_{x \to \infty} 2 + 8 \lim_{x \to \infty} \frac{\sin\left(\frac{4}{x}\right)}{\frac{4}{x}} \\
&= \lim_{x \to \infty} 2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x... | One can easily see that $2$ is a primitive root of $13$.
Hence, we can write any number less than $13$ as some power of $2$.
$x^8\equiv3\equiv2^4\pmod{13}$.
Any solution will also be of the form $x=2^m$ for some $1\leq m\leq12$.
So, we have
$(2^m)^8=2^{8m}\equiv2^4\pmod{13}$.
Now, these powers will be congruent modulo ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Limit of relation sums of exponential functions Help to solve example:
$$
\lim_{n\rightarrow\infty}\frac{\sqrt[3]{7^{3n-1}+3^n}}{3^n+4 \cdot 7^{n+1}}
$$
| Hint.
For all $n \in \mathbb N$
$$
\frac{\sqrt[3]{7^{3n-1}+3^n}}{3^n+4*7^{n+1}}=\frac{7^n}{7^n}\frac{\sqrt[3]{\frac{1}{7}+\left(\frac{3}{7}\right)^n}}{28+\left(\frac{3}{7}\right)^n}=\frac{\sqrt[3]{\frac{1}{7}+\left(\frac{3}{7}\right)^n}}{28+\left(\frac{3}{7}\right)^n}
$$
and $\lim\limits_{n \to +\infty} \left(\frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\binom{n}{k}}{n^k} < \frac{\binom{n+1}{k}}{(n+1)^k}$ I have this math question that I'm kind of stuck on.
Prove that for all integers $1 < k \le n$, $$\frac{\binom{n}{k}}{n^k} <
\frac{\binom{n+1}{k}}{(n+1)^k}$$
I have to use mathematical induction on $k$ to prove this. So, I first check the base ca... | Assume your calculation is correct up to the step
$\frac{n!}{n^{z+1}(z+1)!(n-z-1)!} < \frac{(n+1)!}{(n+1)^{z+1}(z+1)!(n-z)!}$
We can cancel the factor $(z+1)!$ in both denominators. Rewriting $(n+1)! = n! (n+1)$ and $(n-z)! = (n-z-1)!(n-z)$, we can simplify further to
$\frac{1}{n^{z+1}} < \frac{(n+1)}{(n+1)^{z+1} (n-z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of the following $n \times n$ determinantes Find the value of the following $n \times n$ determinantes
*
*$$\begin{vmatrix}
a_1+x & x & x & \ldots & x \\
x & a_2+x & x & \ldots & x \\
x & x & a_3+x & \ldots & x \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}$$
*$... | For 2) Let $$A=\begin{bmatrix} a_1 & a_2 & .. &a_n \\
a_1 & a_2 & .. &a_n \\
a_1 & a_2 & .. &a_n \\
...&...&...&...\\
a_1 & a_2 & .. &a_n \\
\end{bmatrix}$$
Then $A$ has rank $1$ and hence $\lambda=0$ is an eigenvalue with geometric multiplicity $n-1$, and hence has algebraic multiplicity at least $n-1$.
As the trace i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $abc$.
The product of two $3$-digit numbers with digits $abc$, and $cba$ is $396396$, where $a > c$. Find the value of $abc$.
In order to solve this, should I just find the prime factorization of $396396$ and then find the two $3$-digit factors?
| Let $M=396396$. Because $599\times 499<M$, we infer that $a>5$. Because $11|M$, we infer that $a+c-b=11$. Finally, $3|M$ and so $3|(a+b+c)$. So let's consider the possibilities for $a+b+c$:
$$
a+b+c=2b+11\in\{12,15,18,21,24,27\}
$$
which yields $b\in\{2,5,8\}$ which corresponds to $a+c\in\{13,16,19\}$. $19$ is too high... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all functions $f$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$ Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$.
So far, I've managed to prove that if $f$ is linear, then either $f(x) = x + 1$ or $f(x) = -1$ must be true. I did this by plugging in $x=0$ to the above equa... | If the functions attains $0$ at some point, it is linear and $f(n)=n+1$.
Suppose $f(y)=0$, for some $y \in \mathbb{Z}$ then we find for $x \in \mathbb{Z}$
$$f(x)=f(x-f(y))=f(f(x))-f(y)-1 = f(f(x))-1$$
So in particular $x=y$ gives $f(0)= 1$.
Now, if $f(n)=n+1$, then $x=n$ gives $$n+1 = f(n) =f(f(n))-1= f(n+1)-1$$
So $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1<c$. I am lead to believe I should be using curves but I really don't understand what curves to choose and how to properly use them. I would appreaciate some help on the ... | We have
$$a\sin(t)+b\cos(t) = \sqrt{a^2+b^2}\sin(t+\phi) = \sin(t+\phi)$$
Hence, we have
\begin{align}
I & = \int_0^{2\pi} \dfrac{dt}{\sin(t+\phi)+c} = \int_0^{2\pi} \dfrac{dt}{\sin(t)+c} = \dfrac1c \int_0^{2\pi} \dfrac{dt}{1+\dfrac{\sin(t)}c}\\
& = \dfrac1c \sum_{k=0}^{\infty}\left(-\dfrac1c \right)^k \int_0^{2\pi} \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the set of real values of $p$ for which the equation $ | 2x + 3 | + | 2x - 3 | = px + 6 $ has more than two solutions .
Options : A ) ( 4 , 0 )
B) R-{ 4 , 0 , -4 }
C) {0}
D) None
How to find p = 0 , without having to analyse the graph ?
| let $p \in \mathbb R$, $2x-3 < 2x+3$, so if $0\leq2x-3$, $x \in [\frac{3}{2},\infty)$, the equation turns to be:
$$4x=px+6$$
this is linear equation so there is at most one solution.
if $2x+3 \leq 0 $,$x \in (-\infty,-\frac{3}{2}]$ the equation turns to be:
$$-4x=px+6$$
this is linear equation so there is at most one s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
I want to show that $\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $ converges or diverges. I want to show that $$\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $$ converges (absolutly?) or diverges.
My idea was: $n=2k+1$ and then it becomes: $$\sum_{n=1}^\inft... | You can show the terms are decreasing by finding $a_{n}-a_{n+1}$ and showing it's positive:
$\dfrac{n^2+2}{n^3+n}-\dfrac{(n+1)^2+2}{(n+1)^3+(n+1)} = \dfrac{n^4+2n^3+6n^2+5n+4}{n(n+1)(n^2+1)(n^2+2n+2)}>0$
Regarding the limit:
$\dfrac{n^2+2}{n^3+n} = \dfrac{1+\frac{2}{n^2}}{n+\frac{1}{n}}$. So as n goes to infinity the n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the solution for this differential equation Solve the differential equation;
$(xdx+ydy)=x(xdy-ydx)$
L.H.S. can be written as $\frac{d(x^2+y^2)}{2}$ but what should be done for R.H.S.?
| I think that there is actually a general solution to a general kind of ODE. So I give here its general form and its solution. After that, the solution to the ODE in the thread above is also given.
Consider the following ODE
\begin{gather*}
A(x,y)(xd x+yd y)=B(x,y)(xd y-yd x)\tag{1}
\end{gather*}
where $A$ and $B$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplify $2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$ . I am trying to simplify
$$2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$$
However if I plug this in the calculator the answer is zero. Is there a way to keep on simplifying without the calculator?
I know the identity $2\cos(\theta) = (e^{i\theta} +e^{... | $$2\cos\left(\frac{6\pi}{7}\right)+2\cos\left(\frac{2\pi}{7}\right)+2\cos\left(\frac{4\pi}{7}\right)+1=$$
$$2\left(\cos\left(\frac{6\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{2\pi}{7}\right)\right)+1=$$
$$2\left(\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Implicit Solution to a differential equation I'm looking at the ODE:
$\frac{dY}{dX} - \frac{ X^2 + 2 Y^2 - 1 }{ ( Y - 2 X )X } = 0$
I'm looking for an $implicit$ solution to the above. Meaning, I want to find a relation $F(X,Y)=0$, where $\frac{\partial}{\partial x} (F(x,y)) = ( Y - 2 X )X \frac{dY}{dX} - ( X^2 + 2 Y^... | $\dfrac{dY}{dX}-\dfrac{X^2+2Y^2-1}{(Y-2X)X}=0$
$\dfrac{dY}{dX}=\dfrac{X^2+2Y^2-1}{(Y-2X)X}$
$(Y-2X)\dfrac{dY}{dX}=\dfrac{X^2+2Y^2-1}{X}$
Let $U=Y-2X$ ,
Then $Y=U+2X$
$\dfrac{dY}{dX}=\dfrac{dU}{dX}+2$
$\therefore U\left(\dfrac{dU}{dX}+2\right)=\dfrac{X^2+2(U+2X)^2-1}{X}$
$U\dfrac{dU}{dX}+2U=\dfrac{2U^2+8XU+9X^2-1}{X}$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{x}{\sqrt{1+x^2}} \lt \arctan x$ for every $x \gt 0$
Prove $$\frac{x}{\sqrt{1+x^2}} \lt \arctan x$$ for every $x \gt 0$.
I proved half of it with lagrange rule but that I got stuck. Any ideas?
I can upload my work if you want.
| METHODOLOGY $1$:
The integral definition of the arctangent function is given by
$$\arctan(x)=\int_0^x \frac{1}{1+t^2}\,dt$$
Noting that for $t\in [0,x]$, $x\ge 0$, $1+t^2\le (1+t^2)^{3/2}$, we can assert
$$\arctan(x)\ge \int_0^x \frac{1}{(1+t^2)^{3/2}}\,dt=\frac{x}{\sqrt{1+x^2}}$$
and we are done!
METHODOLOGY $2$:
Rec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
I would like to calculate $\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$ I want to calculate the following limit: $$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$$
or prove that it does not exist. Now I know the result is $-3$, but I am having troub... | As this is in the form of $\frac{0}{0}$,so apply L Hospital rule
$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$
$=\lim_{x \to \frac{\pi}{6}} \frac{4 \sin x\cos x+\cos x}{4 \sin x\cos x-3 \cos x}=\frac{\sqrt3+\frac{\sqrt3}{2}}{\sqrt3-\frac{3\sqrt3}{2}}=-3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is the probability that $x^4-y^4$ is divisible by 5? Two numbers $x$ and $y$ are chosen at random without replacement from the set $\{1,2,3...,5n\}$.
What is the probability that $x^4-y^4$ is divisible by $5$?
I divided the numbers into groups of 5 $(1,2,3,4,5),(6,7,8,9,10),...$.The probability in the first group ... | $x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y)$.
If $x = 5k + j; 0 \le j < 5$ then $5|x^4 - y^4$ if $y = 5l + m; 0 \le m < 5$ where $m = j$; $m = 5 - j$; $m^2 + $j^2 = 5V$.
If $j = 0$ $5|x^4 - y^4 \iff m = 0$.
As $1 = 1 = 5 -4; 1^2 + 2^2 = 5; 1^2 + 3^2 = 10; 2 = 2 = 5 -3; 2^2 + 4^2 = 20; 3^2 + 4^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Integral $\int_0^\infty\frac{\tanh^2(x)}{x^2}dx$ It appears that
$$\int_0^\infty\frac{\tanh^2(x)}{x^2}dx\stackrel{\color{gray}?}=\frac{14\,\zeta(3)}{\pi^2}.\tag1$$
(so far I have about $1000$ decimal digits to confirm that).
After changing variable $x=-\tfrac12\ln z$, it takes an equivalent form
$$\int_0^1\frac{(1-z)^2... | Let $\psi_{1}(z)$ be the trigamma function.
Similar to the answer here, we can integrate the function $$g(z) = \psi_{1}\left(\frac{z}{\pi i } \right) \tanh^{2}(z) $$ around an rectangular contour with vertices at $\pm R, \pm R + \pi i $.
The function $\psi_{1}\left(\frac{z}{\pi i} \right) $ has a double pole at the ori... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 8,
"answer_id": 3
} |
Tangent identity given $a + b + c = \pi$ Given that $a + b + c = \pi$, that is, three angles in a triangle - then prove that $$\tan a + \tan b + \tan c = \tan a \tan b \tan c$$
Is my solution below completely rigorous? Can I justify taking the tangent of both sides of my equation (I think not, since tangent isn't an in... | We can write $a + b = \pi -c$ then taking the tangent of both sides, this yields $$\tan (a +b) = \tan(\pi -c) \iff \frac{\tan a + \tan b}{1 - \tan a \tan b} = -\tan c$$
So $$\tan a + \tan b = \tan a \tan b \tan c - \tan c$$
Hence we arrive at $$\bbox[10px, border: blue 1px solid]{\tan a + \tan b + \tan c = \tan a \tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\operatorname{spectrum}(AB) = \operatorname{spectrum}(BA)$? Suppose we have two $n \times n$ matrices $A, B$. It seems like $\delta(AB)=\delta(BA)$, but I can't generally poove it.
If $\det(A) \neq 0$ then $\det(AB - \lambda I) = 0 \Leftrightarrow $ $ \det(AB - \lambda I) \cdot \det(A) = 0 \Leftrightarrow \det(ABA - ... | I've seen the following nice proof credited to Paul Halmos. Assume that $A$ is not invertible. By performing row and column operations on $A$ and encoding them with invertible matrices, we can write
$$ A = P \begin{pmatrix} I_{r \times r} & 0 \\ 0 & 0 \end{pmatrix} Q $$
where $P, Q$ are invertible and $r = \mathrm{rank... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Is a symmetric matrix characterized by the diagonal of its resolvent? The resolvent of a square matrix $A$ is defined by $R(s) = (A-sI)^{-1}$ for $s \notin \operatorname{spect}(A)$.
Is knowing the diagonal of $R(s)$ for all $s$ sufficient to recover $A$ when $A$ is symmetric?
edit: a counter-example of two matrices $A... | No. For example, consider $$\left[ \begin {array}{cccc} 1&0&1&1\\ 0&0&1&1
\\ 1&1&0&0\\ 1&1&0&1\end {array}
\right] \ \text{and}\
\left[ \begin {array}{cccc} 1&1&0&1\\ 1&0&1&0
\\ 0&1&0&1\\ 1&0&1&1\end {array}
\right] $$
which both have diagonal of $R(s)$
$$ (s^4 - 2 s^3 - 3 s^2 + 4 s - 1) \left[ \begin {array}{c} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Comparing $\int \frac{\sqrt{1+x}}{1-x} dx$ with answer from book and sage [Edit: It turns out the answer from my book was incorrect and a later revision of the book has the correct answer:
$$
-2\sqrt{1+x} + \sqrt{2}\ln{\left| \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1+x} - \sqrt{2}} \right|} + C \tag{0}
$$
]
I'm trying to in... | setting $u= \sqrt {1+x}$ so we get
$x=u^2-1$ and $dx=2udu$ and our integral will be $$\int\frac {2u^2}{2-u^2}du$$ and is easy to solve the right result is $$-2\,\sqrt {1+x}+2\,\sqrt {2}{\rm arctanh} \left(1/2\,\sqrt {1+x}\sqrt
{2}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If two integer triples have the same sum of 6th powers, then their sums of squares agree $\bmod 9$ Given $$a^6 + b^6 + c^6 = x^6 + y^6 + z^6$$
prove that $$a^2 + b^2 + c^2 - x^2 - y^2 - z^2 \equiv 0 \bmod{9}$$
I was thinking of using $n^6 \pmod{27}$ and showing both sides have the same pattern but it's getting really ... | $(9n+a)^6=a^6\bmod27$,
so only worry about numbers between $-4$ and $4$.
Their sixth powers are
$19,0,10,1,0,1,10,0,19\pmod{27}$
and their squares are $7,0,4,1,0,1,4,7\pmod{9}$
The sixth powers are either $0$ or $9A+1$;
the squares are either $0$ or $3A+1$, the same $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
How to handle indices with fractional degree? An algebra problem ate my head!!!
$x$ and $y$ are positive real numbers such that
$$\sqrt{x^2 + \sqrt[3]{x^4 y^2}} + \sqrt{y^2 + \sqrt[3]{x^2 y^4}} = 512.$$
Find $x^{2/3} + y^{2/3}$.
It would be a great help if anybody helps me in solving this problem. I tried takin... | Let,
\begin{align}
& x^{2/3} + y^{2/3} = z \newline
\implies & \left(x^{2/3} + y^{2/3}\right)^{3/2} = z^{3/2}\newline
\implies & z^{3/2} = 512 = 8^3 \tag{1}
\end{align}
To remove the fractional index $3/2$ on the LHS of (1) to obtain the base $z$, we take root $2/3$ (i.e. two-third root) of both sides, i.e.
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Xmas Maths 2015 Simplify the expression below into a seasonal greeting using commonly-used symbols in commonly-used formulas in maths and physics. Colours are purely ornamental!
$$
\begin{align}
\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\c... | $$
\begin{align}
&\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\color{orange}{E\cdot}
\color{green}{\text{KE}}
}
{
\color{orange}{2^{\frac 23}}
\color{green}{c^2}
\color{red}{e^{i\theta}}
\color{orange}{v^2}
\color{green}{(x^2+xy+y^2)}}
\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 2,
"answer_id": 0
} |
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series... | The following proof might be longer than necessary, but it illustrates a general method that is useful for many similar problems.
Let $$A(k) = 1^2 - 2^2 + 3^2 - 4^2 + \ldots + (-1)^{k-1} k^2$$ and
$$B(k)=(-1)^{k-1} \cdot \frac{k(k+1)}2.$$ It suffices to prove that $A(0)=B(0)$ and $$A(k)-A(k-1)=B(k)-B(k-1)$$ for all $k\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Solving this indefinite integral: $\int{\frac{x\:dx} {(\sqrt{1+x} +\sqrt[3]{1+x} )}} $ I don't know how to solve this integral
I tried change of variable but had no results.
$$\int_{0}^{1}{\frac{x\:dx} {(\sqrt{1+x} +\sqrt[3]{1+x} )}} $$
| Let $1+x=t^6\implies dx=6t^5\ dt$ $$\int \frac{x}{\sqrt {1+x}+\sqrt[3]{1+x}}\ dx$$
$$=\int \frac{(t^6-1)}{t^3+t^2}(6t^5\ dt)$$
$$=6\int \frac{t^3(t^6-1)}{t+1}\ dt$$
$$=6\int \frac{t^3(t^3-1)(t^3+1)}{t+1}\ dt$$
$$=6\int \frac{(t^6-t^3)(t+1)(t^2-t+1)}{t+1}\ dt$$
$$=6\int (t^6-t^3)(t^2-t+1)\ dt$$
$$=6\int (t^8-t^7+t^6-t^5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.