Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| Notice, $$LHS=\cos x+\sin x\tan \frac{x}{2}$$
$$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\tan \frac{x}{2}$$
$$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan^2 \frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$=\frac{1-\tan^2\frac{x}{2}+2\tan^2\frac{x}{2}}{1+\tan^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 5
} |
Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.
Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.
My Approach:
single digit no formed = 2,4,6,8
sum= 2+4+6+8= 20
two digit... | Consider an example by taking digits as $3,4,6,8$.
total combinations possible$=4\times 3\times 2=24$ numbers possible
no of digits$=4$
hence $\dfrac{24}{4}$ each digit comes $6$ times in ones, tens and hundred's place
at ones place=$6\times (3+4+6+8)=126\equiv6$.
at ten's place$=6\times (3+4+6+8)+12(\text{carry})=138\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
How many pairs of natural numbers $(x,y)$, satisfy the equation $\space xy=x+y+\gcd(x,y)$. How many pairs of natural numbers $(x,y)$, satisfy the equation
$\space xy=x+y+\gcd(x,y)$. You may assume that $x≤y$.
| If $x=1$ then $\gcd(x,y)=1$ and $x+y+\gcd(x,y)=y+2\ne 1\cdot y$. Hence we may assume $x\ge 2$.
As $\gcd(x,y)\le x$, we have $xy\le y+2x$, so
$$(x-1)(y-2)=xy-y-2x+2\le 2 $$
which leads to $y\le 4$. As wlog. $x\le y$, we need only check the few candidates $(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)$.
Alternatively (again assu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Can this congruence be simplified? $$p(p+1) \equiv -q(q+1) \bmod pq$$
Can this be reduced to an easier format?
| Mod $pq$,
$p(p+1) \equiv -q(q+1)
\iff
p^2+p = -q^2-q
\iff
p^2+q^2 +p+q =0
$
$\begin{array}\\
(p+q+1)^2
&\equiv
p^2+q^2+1+2pq+2p+2q\\
&\equiv
p^2+q^2+1+2p+2q\\
&\equiv
(p^2+q^2+p+q)+p+q+1\\
&\equiv
p+q+1\\
\end{array}
$
Therefore,
if $n = p+q+1$,
$n^2 \equiv n$
or
$n(n-1) \equiv 0
$.
If $p$ and $q$ are distinct primes
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
My approach:
For $x=1$, $z$ is not divisible by $6$.
For $x=2$, $z$ is divisible by $6$.
For $x=3$, $z$ is not divisible by $6$.
For $x=4$, $z$ is divisible by $6$.
For $x=5$, $z$... | I think a good idea for the problem is using modulos. So, the problem can be written like this
$$
z=4-x^2 \text{ }mod(6),
$$
We know that $100=96+4=4\quad mod(6)$. So $100^2=(100)*(100)=4*4=16=12+4=4\quad mod(6)$. If $z$ is divisible for $6$, then $z=0\quad mod(6)$. For all of this we have to find $x$ such that
$$
x^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\cos \arctan 1/2 = 2/\sqrt{5}$ How can we prove the following? $$\cos \left( \arctan \left( \frac{1}{2}\right) \right) =\frac{2}{\sqrt{5}}$$
| We know that $\tan (θ)=\frac{y}{x}$.
Thus, for $\tan^{-1}(1/2)$, we have a right triangle whose $x$ and $y$ values are $2$ and $1$, respectively.
Pythagorean theorem in terms of $x$, $y$, and $r$ is written as $x^2+y^2=r^2$. Simply plug in the values $x=1$ and $y=2$ from earlier to get the following:
$$1^2+2^2=r^2$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
5-Card Poker Two-Pair Probability Calculation Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?
Solution: $$\frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}}{\binom{52}{5}} = 0.047539$$
Why doesn't... | Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Can 720! be written as the difference of two positive integer powers of 3? Does the equation:
$$3^x-3^y=720!$$
have any positive integer solution?
| As alex.jordan writes, $3^x-3^y$ factors as $3^y(3^{x-y}-1)$, so $y$ must be the number of factors of $3$ in $720!$.
I don't actually need to count the number of $3$s in $720!$, so let's just define the notation $720!_3$ for "$720!$ with all of the powers of $3$ divided out". This must yield the other factor $3^{x-y}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression:
$$
\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}
$$
The result should be a number.
I try this:
$$
\frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sq... | Would it help you to know that
${2\pm\sqrt5}=\left(\frac{1\pm\sqrt5}2\right)^3$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Balancing chemical equations using linear algebraic methods I know there are already plenty of questions on this site regarding this topic but I am having difficulty with a particular chemical equation.
I am trying to balance the following:
$$
{ C }_{ 2 }{ H }_{ 2 }{ Cl }_{ 4 }\quad +\quad { C }a{ { (OH }) }_{ 2 }\quad... | The 4th row of your matrix is missing a minus sign, i.e.
$$
\begin{pmatrix} 2 & 0 & -2 & 0 & 0 & 0 \\ 2 & 2 & -1 & 0 & -2 & 0 \\ 4 & 0 & -3 & -2 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 2 & 0 & 0 & -1 & 0 \end{pmatrix}
$$
$$ $$
But instead of forming an augmented matrix, you should drop that last column of zeroes, use ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$? How to prove:
$$\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$$
Is it possible to convert it into a finite integral?
| Given how popular these seem to be becoming, it might be worth learning a nice fact to eliminate the need for a lot of cleverness:
$$ \sum_{k=1}^n \frac{1}{k} = \log n + \gamma + O(1/n) $$
where $\gamma$ is Euler's constant.
Partial fractions tells us
$$\frac{k-1}{2k(k+1)(2k+1)} = \frac{3}{2k+1} - \frac{1}{2k} - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Proof by induction for "sum-of" Prove that for all $n \ge 1$:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$
What I have done currently:
Proved that theorem holds for the base case where n=1.
Then:
Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n... | For n=1:
$$
\sum_{k=1}^{1} 1/(k(k+1))=1/2=1/(1+1)
$$
Let's prove it for $n+1$:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}= \sum_{k=1}^{n}+\frac{1}{(n+1)(n+2)}\\
$$
But using our assumption for $P(n)$ :
$$
\sum_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1}
$$
So the sum will be:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}=\frac{n}{(n+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $z$ when $z^4=-i$? Consider $z^4=-i$, find $z$.
I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$
$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$
$|z^4|=\sqrt{(-1)^2}=1$
$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$
$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\... | Remember, that $-i = e^{i\frac{3\pi}{2}}$, then
$$ z^4 = -i = e^{i\frac{3\pi}{2}}$$
The n$^{\text{th}}$ root of a complex number is
$$z^n = re^{i\varphi} \implies z_k = r^{1/n}e^{i\left(\frac{\varphi}{n} + \frac{2\pi k}{n} \right)} \qquad \text{where} \quad k = 0, \dots, n-1$$
In your case:
$$z_0 = e^{i\left(\frac{3\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluation of $5\times 5$ determinant The following $5\times 5$ det. comes from a Russian book. I don't want to expand the det. rather than do some operations on it and extract the result.
Prove:
$$\begin{vmatrix}
-1 &1 &1 &1 &x \\
1& -1 &1 &1 &y \\
1& 1 & -1 & 1 &z \\
1& 1 & 1 & -1 & u\\
x& y & z & u &0
... | Let $Q$ be a real orthogonal matrix whose first column is $(\frac12,\frac12,\frac12,\frac12)^T$. Let also $v=(x,y,z,u)^T$ and $Q^Tv=(a,b,c,d)$. Then the matrix in question is equal to $\pmatrix{Q&0\\ 0&1}A\pmatrix{Q^T&0\\ 0&1}$, where
$$
A=\left[\begin{array}{cccc|c}
2&&&&a\\ &-2&&&b\\ &&-2&&c\\ &&&-2&d\\
\hline
a&b&c&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\... | Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int\frac{\sqrt{(x+2)(1-x)}}{x^2}dx$$
Now Let $\displaystyle (x+2) = (1-x)t^2\;,$ Then $\displaystyle x = \frac{t^2-2}{t^2+1} = 1-\frac{3}{t^2+1}$
So $$\displaystyle dx = \frac{6t}{(t^2+1)^2}dt$$ and $$\displaystyle (1-x) = \frac{3}{t^2+1}$$
So Integral $$\dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Value of $\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$
Given
$$
\tag1\frac{1}{\omega+a}+\frac{1}{\omega+b}+\frac{1}{\omega+c} = 2\omega^2
$$
and
$$
\tag2\frac{1}{\omega^2+a}+\frac{1}{\omega^2+b}+\frac{1}{\omega^2+c} = 2\omega
$$
what is the value of $\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$... | This turned out to be a cubic. There are three roots. Two of them are $x=\omega$ and $x=\omega^2$.
HINT: Do you know how to find the sum of the roots of a polynomial, without solving the polynomial itself?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$ I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac... | First of all we know that:
$$
\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac{1}{2}\ln^2(1-z), \quad z \notin (1,\infty).\tag{$\diamondsuit$}
$$
Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:
$$\operatorname{Li}_2\left(e^{i\theta}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
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Evaluating $\int \frac{dx}{x^3+x+1}$ $$\int \frac{dx}{x^3+x+1}$$
I have no idea how to solve this. How do I evaluate it?
Any advice, hint or well-thought solution will be appreciated.
| Let $\zeta_1,\zeta_2,\zeta_3$ be the roots of $x^3+x+1$. We have:
$$ \text{Res}\left(\frac{1}{z^3+z+1},z=\zeta_i\right) = \frac{1}{3\zeta_i^2+1},\tag{1}$$
hence:
$$ \frac{1}{z^3+z+1}=\sum_{i=1}^{3}\frac{1}{3\zeta_i^2+1}\cdot\frac{1}{z-\zeta_i} \tag{2} $$
and:
$$ \int \frac{dz}{z^3+z+1} = C+\sum_{i=1}^{3}\frac{\log(z-\z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the matrix A
Find the matrix A that has two rows and two columns and has
$$A\pmatrix{1\\ 1}=\pmatrix{2\\ 1}\text{ and }A\pmatrix{-1\\1}=\pmatrix{1\\-1}.$$
Question:
How do I write out the corresponding system of linear equations with $4$ equations and $4$ unknowns so I can determine $A$?
| If you want to write a system of linear equations, then you can write $$A=\begin{pmatrix}a & b\\ c&d\end{pmatrix}$$
and then see that $$A\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}a & b\\ c&d\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}a+b \\ c+d\end{pmatrix}$$
so that gives you the first two... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequation: quadratic difference equations Given:
$$\frac{(x - 3)}{(x-4)} > \frac{(x + 4)}{(x + 3)}$$
Step 1:
$$(x + 3)(x - 3) > (x + 4)(x - 4)$$
Step2 : Solving step 1:
$$x^2 - 3^2 > x^2 - 4^2$$
*Step 3:
$ 0 > -16 + 9$ ???
As you see, I can delete the $x^2$, but there is no point in doing that.
What should be the nex... | One way to do this more carefully is to take $$\frac {x-3}{x-4}-\frac {x+4}{x+3}=\frac 7{(x-4)(x+3)}\gt 0$$ and this is clearly true iff $(x-4)(x+3)\gt 0$
To do this more formally, note that from $a\gt 0$ we have $a^2\gt 0$ and we can deduce $\frac {a}{a^2}=\frac 1a\gt 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Where I'm wrong? Why my answer is different from the book? The question is to integrate
$$\int x\cos^{-1}x dx$$
Answer in my book
$$(2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+C$$
I'm learning single variable calculus right now and at current about integration with part. I'm confused in a problem from sometim... | Your answer is not false, but you didn't thoroughly simplify.
This is why both answers are the same:
*
*$\arcsin x+\arccos x=\dfrac\pi2$
*$\sin(2\arcsin x)=2\sin(\arcsin x)\cos(\arcsin x)=2x\sqrt{1-x^2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Computing $E(XY)$ for finding $Cov(X,Y)$ Consider tossing a cubic die once and let $n$ be the smallest number of dots that appear on top. Define two random variables $X$ and $Y$ such that:
*
*$X=1$ if $n \in \left \{1,2 \right \}$, $X=2$ if $n \in \left \{3,4 \right \}$ and $X=3$ if $n \in \left \{5,6 \right \}$... | I think I've figured it out. The support of $E(XY) = \left \{0,1,2,3,4,6 \right \}$.
So $E(XY) = \frac{1}{3} (0) + \frac{1}{3} (1) +\frac{1}{3} (2) + \frac{1}{3} (3) + 0 + 0 = 2$
So $Cov(X,Y) = 2-2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Simple limit of a sequence
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - ver... | You can use the binomial theorem to expand this.
$$\lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)$$
Here 1 and 2 are the smallest terms and they can be ignored.
\begin{align}
&\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)
\\=& \lim _{x\to \infty \:}\left(\sqrt[3]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\int \tan^6x\sec^3x \ \mathrm{d}x$ Integrate $$\int \tan^6x\sec^3x \ \mathrm{d}x$$
I tried to split integral to $$\tan^6x\sec^2x\sec x$$ but no luck for me. Help thanks
| You can use the substitution $u = \sec + \tan$.
Note that $\dfrac 1u = \sec - \tan$, because $(\sec + \tan)(\sec - \tan) = \sec^2 - \tan^2 = 1$.
We have:
$$u^2 + 1 = [(\sec + \tan)^2] + [1] \\
= [\sec^2 + 2 \sec \tan + \tan^2] + [\sec^2 - \tan^2]\\
= 2\sec^2 + 2 \sec \tan \\
= \sec[2(\sec + \tan)] \\
= \sec(2u) \\
\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Use the inclusion-exclusion principle to determine Use the inclusion-exclusion principle to determine :
(a) the number of ways there are to choose nineteen balls (identical apart from their colour) from a pile of red, blue, yellow and green balls if there have to be at most seven balls of each colour?
(b) the number of... | $a)$
How many ways are there to select them so that two colors have more than $7$ balls? There are $\binom{4}{2}=6$ ways to choose the two colors. After that we can go ahead and take out $8$ balls of each of those colors. We must now select $19-16=3$ balls out of $4$ colors. By stars and bars there are $\binom{6}{3}=20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
When is $f(y)$ prime? Find all prime numbers that can be expressed as :
$f(y)=y^{2015}+y+1$, where $y$ is a natural number. $y=1$ gives us 3, but how do we find others or prove that there can be no other??
| Hint
Use the fact that
$$x^{2015}+x+1 = x^{2015}-x^2+(x^2+x+1)=x^2(x^{3\times 671}-1)+(x^2+x+1) = x^2(x^{3}-1)(x^{3\times 670}+x^{3\times 669}+\dots+x^3+1)+(x^2+x+1) = (x^2+x+1)(x^2(x-1)A+1)$$
where $A = x^{3\times 670}+x^{3\times 669}+\dots+x^3+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Rearange an expression I am learning induction. At one step I have to show that:
$$ 1 - \frac{1}{(1+n)} + \frac{1}{(n+1)(n+2)} $$
can be transformed to
$$ 1 - \frac{1}{n+2} $$
Theese are the steps for the transformation, but I cant understand them:
$$ 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)... | $$\begin{align}1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}&=1-\frac{n+2}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\\&=1+\frac{-(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\\&=1+\frac{-(n+2)+1}{(n+1)(n+2)}\\&=1+\frac{-(n+1)}{(n+1)(n+2)}\\&=1+\frac{-1}{n+2}\\&=1-\frac{1}{n+2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Verify Mean Value Theorem for $\,f\left(x\right)=x^3\,$ I am trying to verify the Mean-Value Theorem for $\,f\left(x\right)=x^3.\,$ So far I have:
$$f'\left(c\right)=\frac{b^3-a^3}{b-a}=b^2+ab+a^2=3c^2$$
Question: since we know that $b>a$, how can we show that at least one solutions of $c$ lies between $a$ and $b$?
| I could not find a simpler solution than the following one.
Mean Value Theorem: There exists $c\in (a,b)$ such that $3c^2=a^2+ab+b^2$.
Denote $c_1 = \sqrt{\frac{a^2+ab+b^2}{3}} > 0$ and $c_2 = -\sqrt{\frac{a^2+ab+b^2}{3}} < 0$. We need to prove that: at least one of the following two inequalities holds:
$$(1) \quad a<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Condition for quartic polynomial coefficients given at least one real root
Find the minimum possible value of $a^2+b^2$ where $a$, $b$ are two real numbers such that the polynomial
$$x^4+ax^3+bx^2+ax+1,$$
has at least one real root.
My attempt: Let p be a real root.
Therefore $p^4 + a(p^3) + b(p^2) + ap + 1 = 0$... | Replacing $a$ with $-a$ only changes the signs of the zeros, so w.l.o.g. we can assume that $a\ge0$. The quadratic with the unknown $p+1/p$ (nice trick, BTW!) that you derived gives
$$
p+\frac1p=\frac{-a\pm\sqrt{a^2-4(b-2)}}2.
$$
Given the assumption $a\ge0$ we see that of these two alternatives the solution with a min... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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If $8$ does not divide $x^2-1$, then $x$ is even; prove by contrapositive If $8$ does not divide $x^2-1$, then $x$ is even
proof by contrapositive
the contrapositive of this is :
if $x$ is odd, then $8$ divides $x^2-1$
proof by contrapositive:
Assume $x$ is odd
by definition of odd $∃k∈ℤ$ such that $x=2k+1$
Well, $x^2-... | Rewrite $$(x^2-1)=(x-1)(x+1)$$ If 8 does not divide $(x^2-1)=(x-1)(x+1)$, then 2 does not devide $x^2-1=(x-1)(x+1)$. But that is the same as saying that neither $x-1$ nor $x+1$ are even. Hence $x$ must be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2... | The condition $\frac{x^2+y^2}{x+y}=4$ is equivalent to $(x+y)^2+(x-y)^2=8(x+y)$. Let $x+y=s$ and $x-y=d$. Note that this induces a bijection from $\Bbb R^2$ to itself, meaning that for every pair $(s,d)$ there exist corresponding $x,y$. We have $0\leq d^2=8s-s^2$. The nonnegative values that $8s-s^2$ can take are $[0,1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}... | the problem is that you used
(a-b)(a+b) to simplify a cubic root instead of a simple root
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How do I solve $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule because I don't know how to use derivates yet. So I tried rationalizing the denominator and numerator but it didn't work.
$\lim_{x ... | HINT 1:
$$\begin{align}
\sqrt{1+x}-\sqrt{1-x^2}&=\sqrt{1+x}\left(1-\sqrt{1-x}\right)\\\\
&=\sqrt{1+x}\,\,\left(\frac{x}{1+\sqrt{1-x}}\right)
\end{align}$$
HINT 2:
$$\begin{align}
\frac{1}{\sqrt{1+x}-1}&=\frac{\sqrt{1+x}+1}{x}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Permutation: Distribute 10 distinct items in 3 boxes: one box contain odd number, one box even number object , and all boxes at least one item.
I wish to distribute $10$ distinct toys to my children $A, B$ and $C$. Each child must get at least one toy, but $A$ must receive an even number of toys, while $C$ must receiv... | Let's do it for $n$ toys.
Each term in $(A+B+C)^n$, expanded without making use of the commutativity of $A$, $B$, and $C$, corresponds to a way of assigning the toys. We want to eliminate terms in which the total degree of $A$ is odd or the total degree of $C$ is even. This can be done by computing
$$
\frac{1}{4}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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coefficient on $s^{14}$ in generating function I have $(s+s^2+s^3+s^4+s^5+s^6)^7$, and I'm trying to find the coefficient on $s^{14}$. I've tried using the multinomial theorem, but that leads to the problem of finding all $k_1, k_2, \ldots , k_6$ such that $\sum_{n=1}^6 k_n = 7$ and $\sum_{n=1}^6nk_n = 14$, and that do... | Binomial Series Approach
Using the Binomial Theorem and negative binomial coefficients,
$$
\begin{align}
\left(s+s^2+s^3+s^4+s^5+s^6\right)^7
&=s^7\left(\frac{1-s^6}{1-s}\right)^7\\
&=s^7\sum_{k=0}^7\binom{7}{k}\left(-s^6\right)^k\sum_{j=0}^\infty\binom{-7}{j}(-s)^j\\
&=s^7\sum_{k=0}^7(-1)^k\binom{7}{k}s^{6k}\sum_{j=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the sum of the cube of the roots of $ x^3 + x^2 - 2x + 1=0$? I know there are roots, because if we assume the equation as a function and give -3 and 1 as $x$:
$$
(-3)^3 + (-3)^2 - 2(-3) + 1 <0
$$
$$
1^3 + 1^2 - 2(1) + 1 > 0
$$
It must have a root between $[-3,1]$. However, the root is very hard and it appeared ... | You know,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)$
$a^3+b^3+c^3=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc$
Now I hope you know the relation between roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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In how many ways can we add $1$'s and $2$'s to get $11$ (when the order matters)? Examples:
$1+1+1+1+1+1+1+1+1+1+1$,
$2+2+2+2+2+1$,
$1+2+2+2+2+2$ (order matters)
I tried solving it with permutations but I realized it won't work
| Our first rather clumsy solution takes advantage of the fact that $11$ is a quite small number.
We could have $0$ $2$'s, $1$ way.
We could have $1$ $2$ and $9$ $1$'s, total $10$ numbers. The location of the $2$ can be chosen in $\binom{10}{1}$ ways.
We could have $2$ $2$'a and $7$ $1$'s, total of $9$ entries. The loca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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trigonometry equation $3\cos(x)^2 = \sin(x)^2$ I tried to solve this equation, but my solution is wrong and I don't understand why. the answer in the book is: $x = \pm60+180k$. my answer is: $x= \pm60+360k$.
please help :)
3cos(x)^2 = sin(x)^2
3cos(x)^2 = 1 - cos(x)^2
t = cos(x)^2
3t=1-t
4t=1
t=1/4
cos(x)^2 = 1/4
co... | Since $\sin^2 x=1-\cos^2 x$ the given equation is equivalent to
\begin{align}
3\cos^2 x&=1-\cos^2 x\\
\iff \quad \cos^2 x&=\frac{1}{4}\\
\iff \cos x&\in\left\{-\frac{1}{2},\frac{1}{2}\right\}
\end{align}
Hence $$x=180^{\circ}\cdot k \pm 60^{\circ}$$
where $k$ is an integer number.
Silas2033: Notice that in this proble... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Calculate $\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx $ Calculate
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx
$$
My try:
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx = \left| {x + 1 = {u^2}} \... | Try
$$\left (\sqrt{x+1}+\sqrt{x-1} \right )\left (\sqrt{x+1}-\sqrt{x-1} \right ) = 2$$
Then the integral is
$$\frac12 \int dx \, \left (\sqrt{x+1}-\sqrt{x-1} \right )^2$$
which is
$$\int dx \left (x - \sqrt{x^2-1}\right ) $$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $3^c + 7^c - 2$ by induction I'm trying to prove the for every $c \in \mathbb{N}$, $3^c + 7^c - 2$ is a multiple of $8$. $\mathbb{N} = \{1,2,3,\ldots\}$
Base case: $c = 1$
$(3^1 + 7^1 - 2) = 8$ Base case is true.
Now assume this is true for $c=k$.
Now I prove this holds for $c=k+1$ $(3^{k+1}+7^{k+1}-2)$.
$(... | By the induction hypothesis,
$$
3^k+7^k-2=8m
$$
for some integer $m$. Then $3^k=8m+2-7^k$ and so
\begin{align}
3^{k+1}+7^{k+1}-2
&=3\cdot 3^k+7^{k+1}-2\\
&=3(8m+2-7^k)+7\cdot7^{k}-2\\
&=24m+6-3\cdot7^k+7\cdot7^{k}-2\\
&=4(6m+1+7^k)
\end{align}
Can you finish up?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Estimate the value $ \int_{1}^{2}\int_{3}^{4}{f(x,y)dydx} $ Using the Gauss-Legendre form. Estimate the value of$$\displaystyle\int_{1}^{2}\displaystyle\int_{3}^{4}{f(x,y)dydx}$$
where, $f(x,y)=x^3y$.
My approach: We can approximate the integral $\int_{-1}^{1}{f(x)dx}=\sum_{i=1}^{n}{A_{i}f(x_{i})}$ with the form. Gauss... | Some pretty good progress. Since the integrand is at most cubic in $x$ and $y$ the result should be exact for the Gauss $2$-point formula. In the $x$-dimension, we want $u=-1$ when $x=1$ and $u=1$ when $x=2$. So
$$\frac{x-1}{u+1}=\frac{2-1}{1+1}=\frac12$$
So $x=\frac12u+\frac32$. In the $y$-dimension we want $v=-1$ whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$
Steps I took:
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ \log _{ 2 }{ 8 } } =\log _{ 2 }{ x-1 } $$
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ 3 } =\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$
$$2\log _{ ... | Using a change of base: $$\log_b a = \frac{\log_n a}{\log_n b}$$
Change $\log_8 x^2$ base to 2: $$\log_8 x^2 = \frac{\log_2 x^2}{\log_2 8} = \frac{2\log_2 x}{3} $$
Let $\log_2 x = a$
$$ \frac{2a}{3}= a-1 \ \implies \ a=3$$
$$\log_2 x = 3\ \implies \ x=2^3 = 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Where am I going wrong while trying to solve this logarithmic equation? $$\log _{ 0.2 }{ x } +\log _{ \sqrt { 5 } }{ x } =\log _{ 25 }{ x } +1$$
Steps I took:
$$\log _{ \frac { 1 }{ 5 } }{ x } +\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ \log _{ \frac { 1 }{ 5 } }{ \sqrt { 5 } } } =\frac { \log _{ \frac { 1 }{ 5 ... | Finding a common base might be the best approach. Re-write your equation as
$$\frac{\log x}{\log 1/5} +\frac{\log x}{\log 5^{1/2}} = \frac{\log x}{\log 5^2} +1. $$
$$\log x \left(\frac{1}{\log 1/5} +\frac{1}{\log 5^{1/2}}-\frac{1}{\log 5^2}\right) = 1 $$
Then the solution is $x = e^{1/a}$ in which $a$ is the numerical... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fundamental Identities I was given a task to prove:
$$\frac{\sec\theta}{\sec\theta\tan\theta} = \sec\theta(\sec\theta + \tan\theta)$$
Then I replaced them with their Ratio and Reciprocal Identities
\begin{align*}
\sec\theta & = \frac{1}{\cos\theta}\\
\tan\theta & = \frac{\sin\theta}{\cos\theta}
\end{align*}
so I came u... | It's wrong.
Here's why:
$$\text{LHS}=\frac{\sec\theta}{\sec\theta\tan\theta}=\frac1{\tan\theta}=\frac{\text{adjacent}}{\text{opposite}}\phantom{....}$$
\begin{align}
\text{RHS}&=\sec\theta(\sec\theta+\tan\theta) \\[0.5ex]
&=\frac1{\cos\theta}\left(\frac 1{\cos\theta}+\frac{\sin\theta}{\cos\theta}\right) \\[0.6ex]
&=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to sketch the subset of a complex plane? The question asks to sketch the subset of $\{z\ \epsilon\ C : |Z-1|+|Z+1|=4\}$
Here is my working:
$z=x+yi$
$|x+yi-1| + |x+yi+1|=4$
$\sqrt{ {(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$
${ {(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$
$x^2 - 2x+1+y^2+x^2+2x+1+y^2=16$
$2x^2+2y^2+2=... | Hint:
The equations
$$\sqrt{{(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$$
$${{(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$$
are not equivalent.
Squaring the first one yields
\begin{align}
\left[\sqrt{{(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}\right]^2&=4^2\\
(x-1)^2+y^2+2\sqrt{{(x-1)}^2 + y^2}\sqrt{{(x+1)}^2+y^2}+(x+1)^2+y^2&=16
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to solve irrational inequality? Irrational inequality wolfram alpha
I want to find $x$ such that
$\sqrt{x-3}+(9-x)^{1/4}>\sqrt{3}$.
Yeah, I know the answer but I don't know how to find this answer
| The existence of roots implies that $3\le x\le 9$. Put $y=(9-x)^{1/2}$. Then $0\le y\le \sqrt{6}$ and
$$\sqrt{6-y^2}+\sqrt{y}>\sqrt{3}$$
Squaring both sides we obtain
$$6-y^2+y+2\sqrt{6y-y^3}>3$$
$$2\sqrt{6y-y^3}>y^2-y-3$$
Then $y^2-y-3\le 0$ or $4(6y-y^3)>(y^2-y-3)^2$.
*
*$y^2-y-3\le 0$ implies $\frac{1-\sqrt{13}}... | {
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"timestamp": "2023-03-29T00:00:00",
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Min of $\frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $ I got this problem I tried several time to solve it by many inequalities but I got stuk. My question is how I get the minimum value of $$ \frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^... | $2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3)$ with chebishev inequality, then $$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{a^{10}+c^{10}}{a^7+c^7} \ge a^3+b^3+c^3 \ge 3 (\frac{a+b+c}{3})^3=\frac{1}{9}$$ because with power mean we have $ (\frac{a^3+b^3+c^3}{3})^\frac{1}{3} \ge (\frac{a+b+c}{3})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find A of the matrix Find $$(2A)^T = \left[\begin{array}{cc}1&-1\\2&3\end{array}\right]^{-1}$$
My solution
$$2A^T = \left[\begin{array}{cc}3&-2\\1&1\end{array}\right] = \frac12\left[\begin{array}{cc}3&-2\\1&1\end{array}\right], A = \left[\begin{array}{cc}3&-2\\1&1\end{array}\right]$$
Apparently this is completely wron... | Determinant = ad - bc = (3 $\times$ 1) - (-2 $\times$ 1) = 5
The inverse of the matrix = $\frac{1}{determinant}$adj(matrix) = $\frac {1}{5}$adj(matrix)
Bring the 2 over, it becomes $\frac{1}{10}$.
| {
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"url": "https://math.stackexchange.com/questions/1468813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$ The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$
$(A)1\hspace{1cm}(B)2\hspace{1cm}(C)3\hspace{1cm}(D)4$
I simplified the expression to $4\sin^6x-12\sin^4x+9\sin^2x-2=0$
But i could not s... | Let $t=\sin^2 x$. Then, by your simplified expression,
$$p(t)=4t^3 - 12t^2 + 9t - 2 =0 -----(*)$$
The polynomial $p$ factors as
$$p(t)=(2t-1)^2(t-2).$$
So, the solution to $(*)$ is
$$t=\frac{1}{2} \mbox{ or } t=2.$$
Now, we try to solve
$$\sin ^2x = \frac{1}{2} \mbox{ or } \sin^2 x = 2.$$
Well, the rightmost equation d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the limit of a sequence using definitions This is a review that my professor posted and I want to make sure I'm on the right path as I study
*
*$\cdot \lim \limits_{n \to \infty} n - \sqrt{2n^2+1} = n -\sqrt{2n^2+1}*\frac{n+\sqrt{2n^2+1}}{n+\sqrt{2n^2+1}} = \frac{-n^2-1}{n+\sqrt{2n^2+1}}$
So can I say that... | For the first sequence: If $n \geq 1$, then
$$
n - \sqrt{2n^{2}+1} = \frac{n^{2}-2n^{2}-1}{n+\sqrt{2n^{2}+1}} = \frac{-n^{2}-1}{n + \sqrt{2n^{2}+1}} < \frac{-n^{2}}{n + \sqrt{2n^{2}+1}} < \frac{-n^{2}}{n+4n} = \frac{-n}{5};
$$
given any $M < 0$, we have $\frac{-n}{5} < M$ if $n > 5|M|$; so for all $n \geq \lceil 5|M| \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?
| You can either use the quadratic formula for a general quadratic $ax^2+bx+c$ which is
$$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ which I will say no more of as JMoravitz has rigorously explained it.
Or you can complete the square, or write
$$2x^2 \color{blue}{-x}-3$$
$$=2x^2+\color{blue}{2x-3x}-3$$
$$=2x(x+1)-3(x+1)$$
$$=\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to solve this limit: $\lim_{n \to \infty} \frac{(2n+2) (2n+1) }{ (n+1)^2}$ $$
\lim_{n\to\infty}\frac{(2n+2)(2n+1)}{(n+1)^{2}}
$$
When I expand it gives:
$$
\lim_{n\to\infty} \dfrac{4n^{2} + 6n + 2}{n^{2} + 2n + 1}
$$
How can this equal $4$? Because if I replace $n$ with infinity it goes $\dfrac{\infty}{\infty}$ on... | To handle limits involving fractions you factor out the dominant term from top and bottom such that they cancel, by dominant term; I mean the term of highest degree, in this case it's $n^2$: $$\lim_{n\to\infty} \frac{4n^2 + 6n + 2}{n^2 + 2n + 1}$$
$$=\require\cancel\lim_{n\to\infty} \frac{2\cancel{n^2}\left(2 + \frac{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first ine... | an other way $$2a^4+b^4+c^4\geq 4\sqrt[4]{a^8b^4c^4}=4a^2bc$$ $$2b^4+a^4+c^4\geq 4\sqrt[4]{b^8a^4c^4}=4b^2ac$$ $$2c^4+a^4+b^4\geq 4\sqrt[4]{c^8a^4b^4}=4c^2ab$$ adding this ineqalitis we get desired one
| {
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Express the following product as a single fraction: $(1+\frac{1}{3})(1+\frac{1}{9})(1+\frac{1}{81})\cdots$
I'm having difficulty with this problem:
What i did was:
I rewrote the $1$ as $\frac{3}{3}$
here is what i rewrote the whole product as:
$$\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{3}{3}+\frac{1}{3^2}\right... | Recall the difference of squares formula $$a^2 - b^2 = (a+b)(a-b).$$ With the special case $a = 1$, $b = x^n$, we would get $$1 - x^{2n} = (1-x^n)(1+x^n).$$ What does this suggest?
| {
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"question_score": "3",
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"answer_id": 1
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How can I evaluate $ \int \frac{dx}{x(1+x^3)(1+3x^3)}$? I want to solve this problem but I have no clue on break the denominator:
$$ \int \frac{dx}{x(1+x^3)(1+3x^3)}$$
I have tried breaking the denominator into partial fractions but failed to do so.
| $$\int \frac{1}{x(1+x^3)(1+3x^3)} \text{d}x =$$
Substitue $u=x^3$ and $\text{d}u=3x^2\text{d}x$:
$$\frac{1}{3} \int \frac{1}{u(1+u)(1+3u)} \text{d}u =$$
$$\frac{1}{3} \int \left(\frac{1}{2(u+1)}-\frac{9}{2(3u+1)}+\frac{1}{u}\right) \text{d}u =$$
$$\frac{1}{3} \left(\int\frac{1}{2(u+1)}\text{d}u-\int\frac{9}{2(3u+1)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481625",
"timestamp": "2023-03-29T00:00:00",
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Find the inverse of the cubic function What is the resulting equation when $y=x^3 + 2x^2$ is reflected in the line $y=x$ ?
I have tried and tried and am unable to come up with the answer.
The furthest I was able to get without making any mistakes or getting confused was $x= y^3 + 2y^2$. What am I supposed to do after... | The general solution to the cubic equation
$$a x^3 + b x^2 + c x + d = 0$$
can be written
$$x = -\frac{1}{3 a} \left( b + \sigma C - \sigma \frac{\Delta_0}{C} \right)$$
where
$$
\Delta_0 = b^2 - 3 a c \\
\Delta_1 = 2 b^3 - 9 a b c + 27 a^2 d \\
C = \sqrt[3]{\frac{\Delta_1 \pm \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}} \\
\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Points on ellipsoid with maximum Gaussian curvature/mean curvature.
Find the points on the ellipsoid $$x^2/a^2+y^2/b^2+z^2/c^2=1$$ with maximum Gaussian curvature and mean curvature respectively.
I parametrized it as $(a\sin u\cos v,b\sin u\sin v, c\cos v)$ and managed to compute its Guassian and mean curvature, whic... | Using the implicit definition of this surface may be a better approach. Let $$F(x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1.$$ The ellipsoid is then defined by $F(x,y,z) = 0$. Then it is known that the Gaussian curvature at a point $(x,y,z)$ is given by
$$\kappa = \frac{g H^{ad}g^T}{|g|^4}.$$
and t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinite series of alternating reciprocals $\frac 1{1\cdot3}-\frac 1{3\cdot 5}+\frac 1{5\cdot 7}-\cdots $ Having misread the recent question here as
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+3)}$$
and having solved it, I thought that I would post it as a question instead. It has a rather interesting answer!
Edit
No... | Combine terms to get
$\sum \frac{(-1)^n}{(2n+1)(2n+3)} = \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} + \dots = \frac{4}{1 \cdot 3 \cdot 5} + \frac{4}{5 \cdot 7 \cdot 9} + \dots = \sum \frac{4}{(4n+1)(4n+3)(4n+5)}$
Partial fractions (and some questionable rearrangement) gives:
$\sum \frac{4}{(4n+1)(4n+3)(4n+5)} = \sum \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Laplace for a function I have problems to understand Laplace for:
$$\frac{s-1}{(s^2+4)^2}$$
I found that fraction can be wrote with Laplace like this:
$$\frac{s-1}{(s^2+4)^2}=A\cdot\frac{s}{s^2+4}+B\cdot\frac{2}{s^2+4}+C\cdot\frac{4s}{(s^2+4)^2}+D\cdot\frac{s^2-4}{(s^2+4)^2}$$
Can someone please explain me from what ar... | Sal, Alistar .
Just break your initial expression in two pieces and make use of a Laplace Transform table. The identities that encapsulate our particular case are :
$$ \bullet \mathcal{L}[t\sin(at)] = \frac{2as}{(s^2+a^2)^2} $$
$$ \bullet \mathcal{L}[\sin(at) - at\cos(at)] = \frac{2a^3}{(s^2+a^2)^2} $$
So, in or... | {
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"timestamp": "2023-03-29T00:00:00",
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Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.
| It is easy to estimate for instance:
$$
\begin{aligned}
|x^5-4x^3y^2-xy^4|
&\le |x^5|+|-4x^3y^2|+|-xy^4| \\
&= |x|\cdot( x^4+4x^2y^2+y^4) \\
&\le |x|\cdot( \ (x^2+y^2)^2 + 2(x^2+y^2)^2 + (x^2+y^2)^2\ )\\
&= 4|x|\; (x^2+y^2)^2\ .
\end{aligned}
$$
So is $f$ denotes the given function we have $|f(x,y)|\le 4|x|$, which hol... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer Let $A(-4,0)$ and $B(4,0)$.Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer,is...... | Hint: $-4 \leq y \leq 4$, so the possible $y$ values are $$\frac{1}{4}, -\frac{1}{4}, \frac{2}{4}, -\frac{2}{4}, \frac{3}{4}, - \frac{3}{4}, \dots, \frac{16}{4}, -\frac{16}{4}$$
Then, how many times is each $y$ value realized on the circle?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to use implicit differentiation to prove $\sinh^{-1}(3x)$ equal to something?
I understand that you let $y = \sinh^{-1}(3x)$ and thus $\sinh(y) = 3x$, but not sure where to go from there.
| Start from basic principles. The hyperbolic sine function can be expressed as
$$\sinh y=\frac{e^y-e^{-y}}{2}\tag 1$$
We can easily invert $(1)$ to find that
$$y=\log\left(\sinh y+\sqrt{\sinh^2 y+1}\right)$$
Letting $y=\text{arsinh}(x)$ reveals
$$\text{arsinh}(x)=\log\left(x+\sqrt{x^2+1}\right)$$
Note that
$$\begin{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A polynomial problem related to lx^2 + nx + n If the roots of $lx^2 + nx + n = 0$ are in the ratio $p:q$, find the value of $\sqrt{\frac{p}{q}}$ + $\sqrt{\frac{q}{p}}$ + $\sqrt{\frac{n}{l}}$.
How to go about this problem?
| Let $x_1$ and $x_2$ be the roots. we know that $x_1x_2=\frac n l$ or that $x_1=\frac{1}{x_2} \frac nl$. Since we have to calculate $\sqrt{\frac n l}$ then $n$ and $l$ should have the same sign. Let them both be positive.
Now if $\frac{x_1}{x_2}=\frac pq$ then we have that $\frac p q=\frac 1 {x^2} \frac n l$ therefore
... | {
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"timestamp": "2023-03-29T00:00:00",
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Chance that last digit in the product is $1,3,7 $ or $9$
If $4$ whole numbers are taken at random and multiplied
together ,what is the chance that last digit in the
product is $1,3,7 $ or $9$ ?
$a.)\ \dfrac{15}{653} \\
b.)\ \dfrac{12}{542} \\
c.)\ \color{green}{\dfrac{16}{625}} \\
d.)\ \dfrac{17}{625} $
I did $... | The last digits won't be evenly distributed. If any number ends in $5$ or an even number then the product ends in a $5$ or an even number. To end in a $1,3,7,$ or $9$ all numbers must end with $1,3,7,$ or $9$. The probability of that is $(4/10)^4 = (2/5)^4 = 16/625$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the solutions of diophantine eq: $x^3-2xy^2+y^3-s^2=0$ Given $x,y,s$ are natural numbers:
$$x^3-2xy^2+y^3-s^2=0$$
I found the solutions using wolfram alpha
$$(x,y,s) = (1,2,1), (6,10,4), (4,8,8)$$
But how do I prove these are the only solutions? Any tips or reference to papers that study this diophantine equat... | An infinite set of solutions is found when $y=2x$. In this case the equation $x^3 - 2xy^2 + y^3 = s^2$ reduces to:
$$x^3 = s^2$$
So we can choose $x=a^2$ for any positive integer $a$, so then $a^6 = s^2$, so $s=a^3$. So the following combination works for any positive integer $a$:
$$(x,y,s) = (a^2, 2a^2, a^3)$$
| {
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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 $$\frac{x}{y} = a,\,\,\frac{y}{z} = b,\,\,\frac{z}{x} = c$$ where $$abc = \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = 1$$
So we have $$\left\{ \begin{gathered}
a + b + c = 7 \hfill \\
ab + ac + bc = 9 \hfill \\
\end{gathered} \right.$$
Using the identity $${a^3} + {b^3} + {c^3} - 3abc = (a + b + c)({a^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Get matrix $A^{n}$ problem with eigenvalues and eigenvectors i have problem with some matrix
$A = \begin{bmatrix}
\frac{1}{2} & 0 \\
2 & \frac{1}{2}
\end{bmatrix}$
To get $A^{n} = P J^{n} P^{-1}$
$\begin{bmatrix}
\frac{1}{2} - \lambda & 0 \\
2 & \frac{1}{2} - \lambda
\end{bmatrix}$
$J^{n... | The eigenvalues are indeed $\lambda_1 = \lambda_2 = \lambda =\frac 12$. The matrix $P$ contains as its columns the right hand (generalized) eigenvectors of $A$, which are linearly independent. Solving the system $$(A- \lambda I) \mathbf{v} =\mathbf 0\iff \begin{bmatrix} 0 &0 \\ 2 & 0 \end{bmatrix}\cdot \begin{bmatrix} ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all non-negative integers $n$ satisfying $2^{n}\equiv n^{2} mod\, 5$ I'm trying to find all non-negative integers $n$ satisfying $2^{n}\equiv n^{2}\pmod{5}$.
So far, all the progress I've made is figuring out that $n^{2} mod \, 5$ for $n=1$ to $5$ has the pattern "$1,\, 4,\, 4,\, 1,\, 0$" which repeats $mod \, 5... | The powers of $2$ mod $5$ repeat with period $4$, and the squares repeat with period $5$. Thus $(2^n - n^2) \mod 5$ repeats with period $20$.
For $0 \le n \le 19$ you get
$$ \matrix{n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \cr
n^2 \mod 5 & 0 & 1 & 4 & 4 & 1 & 0 & 1 &... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expanding $(x-2)^3$ I was trying to expand $(x-2)^3$. This is what I did
*
*Expanded the term so $(x-2)(x-2)(x-2)$
*Multiplied the first term and second term systematically through each case
The answer I got did not match the one at the back of the book, can someone show me how to do this please?
| If you are not familiar with the Binomial Theorem, you can expand $(x - 2)^3$ as follows:
\begin{align*}
(x - 2)^3 & = (x - 2)[(x - 2)(x - 2)]\\
& = (x - 2)[x(x - 2) - 2(x - 2)]\\
& = (x - 2)(x^2 - 2x - 2x + 4)\\
& = (x - 2)(x^2 - 4x + 4)\\
& = x(x^2 - 4x + 4) - 2(x^2 - 4x + 4)\\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the determinant of this $5 \times 5$ matrix? How can I find the determinant of this matrix?
I know in matrix $3 \times 3$
$$A= 1(5\cdot 9-8\cdot 6)-2 (4\cdot 9-7\cdot 6)+3(4\cdot 8-7\cdot 5) $$
but how to work with a $5\times 5$ matrix?
| The Laplace expansion of the determinant can be done using any row or column of a square matrix. In each row, we multiply the $a_{ij}$ component of the matrix with the determinant of the matrix formed by deleting the $i$th row and $j$th column of our original matrix. This new matrix is called the $ij$-minor of $A$ and ... | {
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In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square ... | Set $x=z^2+1$. Then:
$$ \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+3-8\sqrt{x-1}} = \sqrt{(z-2)^2}+\sqrt{(z-3)^2} = |z-2|+|z-3| $$
equals one for every $z\in[2,3]$, hence for every $x\in[5,10]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
To show that group G is abelian if $(ab)^3 = a^3 b^3$ and the order of $G$ is not divisible by 3
Let $G$ be a finite group whose order is not divisible by $3$.
suppose $(ab)^3 = a^3 b^3$ for all $a,b \in G$. Prove that $G$ must be abelian.
Let$ $G be a finite group of order $n$. As $n$ is not divisible by $3$ ,$3$ ... | I have followed the similar method as previous answer with a few changes (without defining the $f$, thought it would be easier:) )
As $(ab)^3=a^3b^3$ for all $a,b\in G$
we have \begin{align*} ababab&=aaabbb\\
\Rightarrow baba&=aabb\\
\Rightarrow (ba)^2&=a^2b^2 \end{align*}
Consider,
\begin{align*}(ab)^4&=((ab)^2)^2\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
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... | Here's another solution that proceeds mostly by the use of divisibility properties. All steps can be justified if one assumes the fundamental theorem of arithmetic, whose statement (but not proof) will be familiar to many precalculus students. I personally, however, would have found it tough-going to follow this proo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 8,
"answer_id": 4
} |
Determine the indefinite integral for $x\tan^{-1}x^{2}$ I'm not sure how to move on after using integration by parts to arrive at
$$\frac{x^{2}\tan^{-1}x^{2}}{2} - \frac{1}{2}\int \frac{x^{2}}{x^{4} + 1}\,dx$$
| $$\int x\arctan x^2\ dx=\frac{1}{2}x^2\arctan x^2 -\int \frac{x^3}{1+x^4}\ dx$$
by integration by parts, where u=arctan(x^2) and dv=xdx.
next use a simple u-substition where $u=1+x^4$.
$$\int \arctan x^2\ dx=\frac{1}{2} x^2 \arctan x^2-\int \frac{x^3}{1+x^4}\ dx
=\frac{1}{2}x^2\arctan x^2 -\frac{1}{4}\ln(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
To determine if a polynomial has real solution I have the following polynomial : $x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$
I must determine if this polynomial has at least 1 real solution and justify why. We have a theorem which says that all polynomials with real coefficients can be decomposed in a product of polynomi... | $\begin{align*}
x^7+x^6+x^5+x^4+x^3+x^2+x+1 &= (x^4+1)(x^3+x^2+x+1) \\
&= (x^4+1)(x^2+1)(x+1)
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A quotient of trigonometric expressions in complex analysis How is $$\frac{\cos\frac{20\pi}{3}+i\sin\frac{20\pi}{3}}{\cos{\frac{15\pi}{4}}{+i\sin\frac{15 \pi}{4}}}=\cos\frac{35\pi}{12}+i\sin\frac{35 \pi }{12}\ \ \ ?$$
I think that I am having trouble understanding a fundamental concept in complex analysis, but cannot ... | $$\frac{\cos\frac{20\pi}{3}+i\sin\frac{20\pi}{3}}{\cos{\frac{15\pi}{4}}{+i\sin\frac{15 \pi}{4}}}=\frac{e^{i\frac{20\pi}{3}}}{e^{i\frac{15\pi}{4}}}$$
$$=e^{i\frac{20\pi}{3}-i\frac{15\pi}{4}}$$
$$=e^{i\frac{35}{12}}$$
$$=\cos\frac{35\pi}{12}+i\sin\frac{35 \pi }{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\epsilon-N$ for $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$. I want to prove using $\epsilon-N$ that $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Firstly we will find a sufficently large $n\in \Bbb N$:
\begin{align*}
\epsilon\quad\gt&\quad\left|\frac{3n+1}{2n+5} - \frac32 \right|\\
=&\quad \left|\frac{3n+1 -3n-7.5... | $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Let $\epsilon >0$ be arbitrary.
Now by the archemedian property there exists $N \in \mathbb N$ such that $N>\frac{13}{\epsilon}$.
Observe that for each $n>N $
$|\frac{3n+1}{2n+5}-\frac{3}{2}|=\frac{13}{4n+10}< \frac{13}{n}<\frac{13}{N}<\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $a,b,c$ be non-negative real numbers , then $\frac{(a+b+c)^2}{3}\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}$? Let $a,b,c$ be non-negative real numbers , then is it true that
$\dfrac{(a+b+c)^2}{3}\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}$ ?
| Yes, note that
$$ab+bc\ge 2 b\sqrt{ac},...$$
so
$$ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}.$$
The inequality then follows from
$$\frac{(a+b+c)^2}3\ge ab+bc+ca.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to integrate this fraction: $\int\frac{1}{1-2x^2}dx$? I'm not sure how to integrate this:
$$\int\frac{1}{1-2x^2}dx$$
I think it has to be this:
$$ -2\cdot \arctan(x)$$
Or this:
$$\arctan(\sqrt{-2x^2})$$
| $$ \int \frac{1}{1-2z^2} = \frac{1}{2} \int \frac{1}{1-\sqrt{2}z} + \frac{1}{1+\sqrt{2}z} = \frac{1}{2} (-\frac{1}{\sqrt{2}} \log|1-\sqrt{2}z| + \frac{1}{\sqrt{2}} \log|1+\sqrt{2}z|) = \frac{1}{2\sqrt{2}} \log|\frac{1+\sqrt{2}z}{1-\sqrt{2}z}| $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
how to solve the following mordell equation:$ y^2 = x^3 - 3$ i just started solving mordell's equations and get a little bit stuck. For example:
$ y^2 = x^3 - 3$.
I know that $x$ must be odd, for if $x$ is even $y^2 \equiv 5 \pmod{8}$. So $x \equiv \{1,3\} \pmod{4}$. Now note the following if we add 4 at both sides:
$y... | Your solution is flawed, so here's a complete solution: As you said, $x$ is odd, because if $x$ were even, then $y^2\equiv -3\pmod{8}$, which is not a quadratic residue. Then $y=2k$ is even.
$4(k^2+1)=(x+1)\left(x^2-x+1\right)$.
$x^2-x+1$ is always odd. Let $p$ be a prime divisor of $x^2-x+1$. Then $p\equiv 1\pmod{4}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate the Integral: $\int^\pi_0\cos^6\theta\ d\theta$ $\int^\pi_0\cos^6\theta\ d\theta$
So I split the trig value into:
$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$
Then I utilized the Pythagorean theorem for $cos^5\theta$
$\int^\pi_o(1-sin^5\theta)\ cos\theta$
I utilized u-substitution:
$u=sin\ \theta$
$du=cos\ \th... | $$\int^\pi_0\cos^6\theta\ d\theta=\int^\pi_0(\frac{1+\cos2\theta}{2})^3 d\theta$$
$$\frac{1}{8}\int^\pi_0(1+\cos2 \theta)^3d\theta=\frac{1}{8}\int^\pi_0(1+3\cos2\theta+3\cos^2 2\theta+\cos^32\theta)d\theta $$
$$\frac{1}{8}\int^\pi_0(1+3\cos2\theta+\frac{3}{2}(1+cos 4\theta)+\cos2\theta(1-\sin^22\theta)))d\theta$$
see,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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... | By the MVT:
$$\sqrt {x+2} - \sqrt {x+1} = \frac{1}{2\sqrt {c_x}}\cdot 1, \ \ \ \ \sqrt {x+1} - \sqrt {x} = \frac{1}{2\sqrt {d_x}}\cdot 1.$$
Here $c_x\in (x+1,x+2), d_x\in (x,x+1).$ Because $1/\sqrt x$ strictly decreases, the left term minus the right term is negative.
Concavity: Slopes of successive chords on a strict... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
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Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the
$$\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$
series I used the criterion of integral $$\displaystyle\int_4^\infty {\frac{{x + 1}}{{(x + 5)(x + 4)(x - 3)}}dx}... | $$\dfrac{n+1}{(n+5)(n+4)(n-3)} = -\dfrac12\cdot\dfrac1{n+5} + \dfrac37\cdot\dfrac1{n+4} + \dfrac1{14} \cdot\dfrac1{n-3}$$
Hence,
\begin{align}
\sum_{n=4}^{m} \dfrac{n+1}{(n+5)(n+4)(n-3)} & = - \dfrac12 \cdot \sum_{n=4}^{m} \dfrac1{n+5} + \dfrac37 \cdot \sum_{n=4}^{m} \dfrac1{n+4} + \dfrac1{14} \cdot \sum_{n=4}^{m} \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Probability of roots of equation are real Let $A$, $B$, and $C$ be independent random variables, uniformly distributed over $[0,9], [0,3]$ and $[0,5]$ respectively. What is the probability that both roots of the equation $Ax^2+Bx+C=0$ are real?
I know I need $B^2>4AC$ but not sure how to get probability.
| This is more of a rough sketch:
With a slight abuse of notation:
$|A| = 9-0 = 9$
$|B| = 3-0 = 3$
$|C| = 5-0 = 5$
First the probability distribution of $B^2$:
$P(B^2 < x) = P(-\sqrt{x} < B < \sqrt{x}) = P(0 < B < \sqrt{x}) = \frac{\sqrt{x}}{|B|}$
$\implies f_{B^2}(x) = \frac{1}{2\sqrt{x}\ |B|}$
Now consider:
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Given Two Fibonacci numbers, predicting the median Fibonacci number Wolfram Alpha gives the $100$th fibonacci number to be $354224848179261915075$ and the $104$th fibonacci number to be $2427893228399975082453$. Just from this, can we deduce what the $102$th fibonacci will be? Is it at all possible, and if it is possib... |
Generalization:
$$F_{n+4} = F_{n+3} + F_{n+2} = (F_{n+2} + F_{n+1}) + F_{n+2} = 2\cdot F_{n+2} + F_{n+1}$$
Since
$$F_{n+2} = F_{n+1} + F_{n} \implies F_{n+1} = F_{n+2} - F_{n}$$
$$\therefore F_{n+4} = 2\cdot F_{n+2} + (F_{n+2} - F_{n}) = 3\cdot F_{n+2} - F_{n}$$
$$\implies F_{n+2} = \dfrac{F_{n+4} + F_{n}}{3}$$
Now s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so fa... | We have $$\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\frac{1}{4}\int_{0}^{a}\left(\frac{1}{\sqrt{x}\sqrt{1+4x}}+\frac{1+8x}{\sqrt{x}\sqrt{1+4x}}\right)dx
$$ for the first integral use the substitution $4x=t^{2}
$ (the second is trivial) to get $$\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\frac{1}{4}\left(\textrm{arcsinh}\left(2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Solving the recurrence $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$ using generating functions
Solve the following recurrence using generating functions: $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$.
My partial solution:
We can rewrite $a_{n+2} = 3a_{n+1} - 2a_n$, as $a_{n+2} - 3a_{n+1} + 2a_n = 0$, and we let $A(z... | Simple trick for partial fractions: Say yu have:
$$
\frac{a_0 + (a_1 - 3 a_0) z}{(1 - z) (1 - 2 z)} = \frac{A}{1 - z} + \frac{B}{1 - 2 z}
$$
This is supposed to be an identity, so for instance:
$\begin{align*}
\lim_{z \to 1/2} \left(
(1 - 2 z)
\cdot \frac{a_0 + (a_1 - 3 a_0) z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Combinatorics Identity about Catalan numbers: $\sum_{k=0}^n \frac{1}{k+1}\binom{2k}k \binom{2n-2k}{n-k}=\binom{2n+1}n$ I need to prove this identity:
$\sum_{k=0}^n \frac{1}{k+1}{2k \choose k}{2n-2k \choose n-k}={2n+1 \choose n}$
without using the identity:
$C_{n+1}=\sum_{k=0}^n C_kC_{n-k}$.
Can't figure out how to.
| Suppose we seek to prove that
$$\sum_{k=0}^n \frac{1}{k+1} {2k\choose k}
{2n-2k\choose n-k} = {2n+1\choose n}.$$
We get for the LHS
$$[z^n] (1+z)^{2n} \sum_{k=0}^n \frac{1}{k+1} {2k\choose k}
\frac{z^k}{(1+z)^{2k}}.$$
Here the coefficient extractor $[z^n]$ combined with the factor $z^k$
enforces the upper limit of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to show that $a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$ Let $$(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})=a_{0}+a_{1}x+\cdots+a_{200}x^{200}$$
show that
$$a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$$
I have one methods to solve this problem:
Let$$g(x)=(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})$$
... | HInt 1:
$$g(x)=(1+x^2+\cdots+x^{100})^2-x^2(1+x^2+x^4+\cdots+x^{98})^2$$
Hint 2: since
$$g(x)=\dfrac{1+x^{101}}{1+x}\cdot\dfrac{1-x^{101}}{1-x}=\dfrac{1-x^{202}}{1-x^2}=1+x^2+\cdots+x^{198}+x^{200}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book. Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book.
My method:
$\frac{9x}{(2x+1)^2(1... | You have to have
$$9x=A(2x+1)\color{red}{(1-x)}+B(1-x)+C(2x+1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subtracting square roots I'm studying trigonometry but I don't understand one example I have here:
$\cos(\frac{5}{12}\pi) = \cos(\frac{1}{6}\pi + \frac{1}{4}\pi)$
$= \cos(\frac{1}{6}\pi)\cos(\frac{1}{4}\pi) - \sin(\frac{1}{6}\pi)\sin(\frac{1}{4}\pi)$
$=(\frac{1}{2}\sqrt{3})(\frac{1}{2}\sqrt{2})-(\frac{1}{2})(\frac{1}{2... | As fast as light:
$$\frac{1}{4}\sqrt{6} - \frac{1}{4}\sqrt{2} = \frac{1}{4}(\sqrt{6} - \sqrt{2})$$
Now you will use the property $\sqrt{6} = \sqrt{3\cdot 2} = \sqrt{3}\cdot \sqrt{2}$ to get the result, i.e.
$$\frac{1}{4}\sqrt{2}(\sqrt{3} - 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 - 0.5\cos x) - 0.5\sin x}}{{{x^3}}}$ For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:
$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\rig... | Apply iteratedly l'Hôpital Rule to $$\frac{x(2-\cos x)-\sin x}{2x^3}$$ We get successively $$\frac{2-2\cos x+x\sin x}{6x^2}$$ $$\frac{3\sin x+x\cos x}{12x}$$ $$\frac{3\cos x+\cos x-x\sin x}{12}$$
Thus the searched limit is the limit of $$\frac{4\cos x-x\sin x}{12}$$ when $x$ tends to $0$ which is clearly $\frac{4}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| Applying polynomial long division, we have
$\begin{array}{rlllllllllll}
&~~1x^8-1x^7+3x^6+\dots\\
\hline
x^2+x-1&|x^{10}+0x^9+x^8+0x^7+0x^6+0x^5+0x^4+0x^3+1x^2+0x+1\\
&~x^{10}+x^9-x^8\\
\hline
&~~~~~-x^9+2x^8+0x^7\\
&~~~~~-x^9-x^8+x^7\\
\hline
&~~~~~~~~~~~~~~~~~3x^8+\dots\\
&~~~~~~~~~~~~~~~~~~\vdots
\end{array}$
event... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Solving $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$ I'm trying to solve the recurrence $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$. First I solved for the homogeneous equation $a_n=3a_{n-1}-2a_{n-2}$ and got $\alpha 1^n+\beta 2^n=a_n^h$. Solving this gives $a_n^h =1$. The particular solution, as I understand, will be $a... | $$a_n - a_{n-1}+3=2(a_{n-1}-a_{n-2}+3)$$
$$a_n - a_{n-1}+3=2^{n-1}(a_1-a_0+3)=3\cdot 2^{n-1}$$
$$\\$$
$$a_n = a_{n-1} + 3\cdot (2^{n-1}-1)$$
$$a_{n-1} = a_{n-2} + 3\cdot (2^{n-2}-1)$$
$$...$$
$$a_1 = a_0 + 3\cdot (2^0-1)$$
Adding side by side,
$$a_n=a_0+3\cdot (2^0+2^1+...+2^{n-1}-n)$$
$$=1+3\cdot (2^n-1-n)=3\cdot2^n-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$ Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$
Is there some intuitive method or formula for finding GCD of $n$ integers?
| $n=0$ gives $1+9+25 = 35$, so it's either $1, 5, 7$ or $35$.
$n=1$ gives $8+ 3^8 + 5^8 = 397194 = 2 \cdot 3 \cdot 7^3 \cdot 193$ so it could be $1$ or $7$.
Let's see what happens mod $7$ (this means we look at all numbers ignoring powers of $7$, so basically any number becomes $0,1,2,3,4,5$ or $6$ according to the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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partial fraction for complex roots While solving Laplace transform using Partial fraction expansion. I have confusion in solving partial fraction for complex roots.
I have this equation $$ \frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}$$
Please anyone help to tell me to understand the steps for solving partial fraction for comp... | It is the same story as with real roots.
Consider $$A=\frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}$$ the first partial fraction decomposition gives $$A=\frac{s+1}{s^2+2 s+10}+\frac{1}{s+2}$$ So, now, just focus on $$B=\frac{s+1}{s^2+2 s+10}=\frac{s+1}{(s+1+3i)(s+1-3i)}=\frac \alpha {(s+1+3i)}+\frac \beta {(s+1-3i)} $$ Reducin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How to Evaluate $\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$? How to find this limit without using L'Hospital rule
$$\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$
| $Step\,1.$ $(1+x)^a = 1+ax + O(x^2)$ for all a and all $x$ with $|x|<1$
$Step\,2.$ $ 4^x - 3^x= 4^x\left(1-\left(\frac34\right)^x\right)$
$Step\,3.$ $(3/4)^x = \exp(x \ln(3/4)) = 1+ x \ln(3/4)+ \frac{x^2}2 \ln^2(3/4)+ O(x^3)$.
$Step\,4.$ The limit $L= \lim_{x \to 0} \frac{(1-\frac13x + O(x^2)-1)}{4^x ( -x \ln(3/4)+O(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of the series $1+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{1}{4^2}+\frac{1}{7}\cdot\frac{1}{4^3}+\cdots$ Find the sum of the series : $$1+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{1}{4^2}+\frac{1}{7}\cdot\frac{1}{4^3}+\cdots$$
| This can be transformed to $$\sum_{n=1}^{\infty} \frac{2}{(2n-1)2^{2n-1}}$$
Let $$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)}$$
Then, we have $f'(x)=\sum_{n=1}^{\infty} x^{2n-2} = \frac{1}{1-x^2}$.
Therefore, we have $$f(x)=\int \frac{1}{1-x^2} = \frac{1}{2} \ln \frac{x+1}{1-x}+C$$
It is clear that $C=0$.
Now plug... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find generating function of sequences: $a_n=(n^2+n+1)_{n\ge0},b_n=(2^{1+[\frac{n}{3}]})_{n\ge 0}$ Find generating function of sequences: $a_n=(n^2+n+1)_{n\ge0},b_n=(2^{1+[\frac{n}{3}]})_{n\ge 0}$
For the first function, generating function is trivial:
$$f(x)=\sum\limits_{n=0}^{\infty}a_nx^n=\sum\limits_{n=0}^{\infty}(n... | $a_n$:
$$
f(x)=\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\\
\sum_{n=0}^{\infty}\left(1+n+n^2\right)x^n=f(x)+x\cdot f'(x)+x\left(x\cdot f'(x)\right)'$$
$b_n$:
$$\sum_{n=0}^{\infty}2^{n+1}\left(1+x+x^2\right)x^{3n}\\
=2\left(1+x+x^2\right)\sum_{n=0}^{\infty}\left(2x^3\right)^n\\
=2\left(1+x+x^2\right)\frac{1}{1-2x^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further... | Let $\alpha=\arctan\sqrt{\frac{x}{2}}$
$$I=\int \frac{\alpha dx}{\sqrt{x+2}}=\int\alpha d(2\sqrt{x+2})=2\alpha\sqrt{x+2}-2\int \sqrt{x+2}\space d\alpha $$
The calculation gives $$d\alpha=\frac{\sqrt {2} dx}{\sqrt x(x+2)}$$
Hence $I=2\alpha\sqrt{x+2}-2\sqrt 2\int \frac{\sqrt{x+2}\space dx}{(x+2)\sqrt x}$
$I=2\alpha\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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.