Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Diophantine Equation: $xy+ax+by+c=0$ How to find integer solutions $x,y$ of $xy+ax+by+c=0$ for given $a,b,c \in \mathbb{Z}$?
Is there somewhere a treatise on this kind of equations?
| (I'm guessing that what was intended was $xy+ax+by+c=0$.)
\begin{align}
u & = x+y \\
v & = x-y \\[8pt]
\text{so that }x & = \frac{u+v}{2} \\[8pt]
y & = \frac{u-v}{2}
\end{align}
Then $xy = \dfrac{u^2-v^2}{4}$. So
$$
xy + ax + by + c =0
$$
becomes
$$
\frac{u^2-v^2}{4} + a\frac{u+v}{2} + b \frac{u-v}{2} + c =0,
$$
and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
The evaluation of the infinite product $\prod_{k=2}^{\infty} \frac{k^{2}-1}{k^{2}+1}$ How does one show that$$ \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} =\frac{\pi}{\sinh \pi} ?$$
My attempt: $$ \begin{align} \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} &= \lim_{n \to \infty} \prod_{k=2}^{n}\frac{(k-1)(k+1)}{(k-i)(k... | If you know the product representation
$$\sin (\pi z) = \pi z \prod_{k = 1}^\infty \left(1 - \frac{z^2}{k^2}\right),\tag{1}$$
it is rather easy.
Setting $z = i$, we obtain
$$\frac{\sin (\pi i)}{\pi i} = \frac{\sinh \pi}{\pi} = \prod_{k=1}^\infty \left(1 - \frac{i^2}{k^2}\right) = \prod_{k=1}^\infty \left(1 + \frac{1}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Sequence problem, find root the equation $x^3-5x+1=0$ has a root in $(0,1)$. Using a proper sequence for which $$|a(n+1)-a(n)|\le c|(a(n)-a(n-1)|$$ with $0<c<1$ , find the root with an approximation of $10^{-4}$.
| You can attempt this: define a sequence $x_n$ by
$$
x_{n+1} = \frac 15\left(1 + x_n^3\right).
$$
We know that if $x_1 \in (0, 1)$, then $x_n \in (0, 1)$ for all $n$. Next, we show that it satisfies the said condition:
\begin{align*}
x_{n+1} - x_n & = \frac 15(1 + x_n^3 - 5x_n)\\
& = \frac 15\left(1 + x_n^3 - (1 + x_{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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The map $p(x)\mapsto p(x+1)$ in vector space of polynomials I encountered this question in a previous year paper of an exam. I wish somebody could help me how to go by this question.
Choose the correct option
Let $N$ be the vector space of all real polynomials of degree at most $3$. Define
$$S:N \to N \ \text{by} \ (Sp... | So, you can begin computing the images of $1, x, x^2$ and $x^3$. For instance:
$$
S(1) = 1 \ ,
$$
because it doesn't matter wherever you evaluate the constant polynomial $1$, its value will always be $1$. Next, you compute also
$$
S(x) = x + 1 \ , \qquad S(x^2) = (x + 1)^2 = x^2 + 2x + 1
$$
and
$$
\qquad S(x^3) = (x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$n$ and $n^5$ have the same units digit? Studying GCD, I got a question that begs to show that $n$ and $n^5$ has the same units digit ...
What would be an idea to be able to initiate such a statement?
testing
$0$ and $0^5=0$
$1$ and $1^5=1$
$2$ and $2^5=32$
In my studies, I have not got "mod", please use other means, i... | Without using any modular arithmetic:
$$n^5-n=n(n-1)(n+1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)=n(n-1)(n+1)(n^2-4)+5n(n-1)(n+1)=$$
$$=(n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1)$$
$(n-2)(n-1)n(n+1)(n+2)$ is the product of 5 consecutive integers thus divisible by 2 and 5.
$5n(n-1)(n+1)$ is multiple of $5$ and even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding a Liapunov function. Is my analysis correct? Consider the following system of differential equations $x'=f(x)$:
\begin{equation}
x_1' = -x_1 -x_2^3 \\
x_2' = x_1-x_2
\end{equation}
Here is the Liapunov function I wish to use: $V(x_1, x_2) = \alpha x_1^2 + \beta x_2^4$. So I need to find $\alpha$ and $\beta$ tha... | We are given
$$x_1' = -x_1 -x_2^3 \\
x_2' = x_1-x_2$$
with
$$V(x_1, x_2) = \alpha x_1^2 + \beta x_2^4$$
We have:
\begin{align}
V' &= 2 \alpha x_1 x'_1 + 4 \beta x_2^3 x'_2 \\
&=2 \alpha x_1(-x_1 -x_2^3) + 4 \beta x^3_2(x_1-x_2)\\
&=-2 \alpha x_1^2 -2 \alpha x_1 x^3_2 + 4 \beta x_1 x^3_2 - 4 \beta x^4_2 \\
&=-2 \alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
the strategy about a $0-1$ matrix game Given a $4\times4$ binary matrix as following,
$\left( \begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{matrix} \right)$
if you choose $a_{ij}$ in this matrix, then all the number in i th row and j th column will be changed( f... | This is yet another variant of the Lights Out puzzle. For each button (i,j) we define a toggle matrix $T_{ij}$ where the entry is 1 if the button in that location changes state, or if it doesn't.
For example, $T_{11}=\left(\begin{matrix}1&1&1&1&\\1&0&0&0&\\1&0&0&0\\1&0&0&0\end{matrix}\right)$
The initial matrix
$A=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$ $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, $\p... | You're quite right about $\phi_2$. For the others, you should see whether you can compute what one of the three values ought to be just using the two given values. So for instance $\begin{pmatrix}2\\1\end{pmatrix}=4\begin{pmatrix}1\\-1\end{pmatrix}+\begin{pmatrix}-2\\5\end{pmatrix}$. Can you conclude anything from this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
closed form of $\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$ closed form of $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$$
for $a,b>0$
n=1 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}$$
n=2 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{... | Here is a different solution using differentiation under integral sign.
We firstly look at the the case $n=1$, then
\begin{align*}
I_1(\alpha,\beta) & = \int_{0}^{2\pi}\frac{dx}{\alpha \cos^2x+\beta \sin^2x} \\
& = \int_{0}^{2\pi} \frac{\sec^2x}{\alpha + \beta \tan^2x} \,\mathrm{d}x\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
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Summation and proof by induction question: $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$ I can't figure this out based on examples in textbooks, etc.
Show via induction that $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
So far, I have:
(a) base case
$P(1)= 1(1+1)(1+2) = \frac{1(1+1)(1+2)(1+3)}{4} = 6 ... | If $$\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$$
then $$\sum_{j=1}^{n+1}j(j+1)(j+2)=\sum_{j=1}^{n}j(j+1)(j+2)+(n+1)(n+2)(n+3)$$
$$=\frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)=$$
$$=\frac{n(n+1)(n+2)(n+3)}{4}+\frac{4(n+1)(n+2)(n+3)}{4}=$$
$$=\frac{(n+1)(n+2)(n+3)(n+4)}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/455197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Sequence and Series - If $a_n =\int^{\frac{\pi}{2}}_0 \frac{\sin^2nx}{\sin^2x}dx,$..... If $\displaystyle a_n =\int^{\frac{\pi}{2}}_0 \frac{\sin^2nx}{\sin^2x}dx, $ then find the value of
$$\begin{vmatrix}
a_1 & a_{51} & a_{101} \\ a_2 &a_{52} & a_{102}\\ a_3 & a_{53}&a_{103}\\
\end{vmatrix}.$$
My approach :
We know t... | HINT:
As $\displaystyle a_n =\int^{\frac\pi2}_0 \frac{\sin^2nx}{\sin^2x}dx, $
If $n=1, \displaystyle a_1 =\int^{\frac\pi2}_0 dx=\frac\pi2$
Else we are interested in $n>1$
$\displaystyle a_{n+1}-a_n=\int^{\frac\pi2}_0 \frac{\sin^2nx-\sin^2(n-1)x}{\sin^2x}dx$
Using $\sin^2A-\sin^2B=\sin(A-B)\sin(A+B),$
$\displaystyle a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\fra... | $\sqrt{\left(4 - \frac 9 2 \right)^2} = 4 - \frac 9 2 = -0.5.$
It's not true.
If $a \geq 0$, then $\sqrt{a} \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 3
} |
Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even:
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cd... | For $n=4$ you need to show that $1^2-3^3+5^2-7^2 = -2\cdot 16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Convert $\sin\theta/2$ using trig identities $$\sin\dfrac \theta2 = \sin^2θ+\cos^2θ-1$$
$$\sinθ = 2\sin^2θ+\cos^2θ-1$$
Am I on the right track?
| In the event that you are being asked to solve for $\theta$ given the equation $$\sin\left(\frac \theta2\right) = \sin^2 \theta + \cos^2 \theta - 1,$$
...note that by the Pythagorean Theorem, we know that $\sin^2 \theta + \cos^2 \theta = 1.\;$
$$\begin{align}\sin\left(\frac \theta2\right) & = \underbrace{(\sin^2 \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the function equation $g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0$ let $g(x)\in \Bbb R$ and for any $x\in \Bbb R$ such that
$$g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0, g(0)=0$$
find $g(x)$
my idea
let $x\longrightarrow x+1$, then we have
$$g^2(x+1)-g(x+2)-\dfrac{(x+1)^2+2(x+1)-6}{4}=0$$
$$g^2(x+2)-g(x+3)-\dfrac{(x+2)^2+2(... | Take $\,x=-1\;$:
$$g(-1)^2-g(0)-\frac{(-1)^2+2(-1)-6}4=0\implies g(-1)^2=-\frac74$$
which of course is impossible unless there are further conditions or we're working with complex numbers and not with the reals, as written, say.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a_n$ given, then evaluate $\lim_{n\to \infty}a_n$. If $$a_n=\sqrt{4+\sqrt{4+3\sqrt{4+5\sqrt{4+7\sqrt{\cdots\sqrt{4+(2n-1)\sqrt{4+(2n+1)}}}}}}}$$for any natural number $n$,
then evaluate $\lim_{n\to \infty}a_n$.
Note that $a_1=\sqrt{4+\sqrt{4+3}}$ and $a_2=\sqrt{4+\sqrt{4+3\sqrt{4+5}}}$.
I don't have any good idea. I ... | Define
$$
f(x)=\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+(x+4)\sqrt{4+\dots}}}}
$$
then $f(x)^2=4+xf(x+2)$. This indicates we should look at $f(x)=x+2$.
Considering $f(x)=x+2$, we are lead to show inductively that
$$
x+2=\small\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+\dots+\sqrt{4+(x+2k-4)\sqrt{4+(x+2k-2)(x+2k+2)}}}}}
$$
Define
$$
f_{k,x}(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Using partial fractions to find an antiderivative of $(x^2+2x)/(x+1)^2$
Evaluate $$
\int\frac{x^2+2x}{(x+1)^2}dx
$$
My solution
Let $u =x+1$,
$
du=dx
$.
Then
$
du(x^2+2x)=(x^2+2x)dx
$ and
$
x=u-1
$.
We get
$$
\int\frac{(u-1)^2+2(u-1)}{u^2}du
=
\int\frac{u^2-2u+1+2u-2}{u^2}du
=
\int\frac{u^2-1}{u^2}du
$$
which sim... | That's perfectly fine. We can absorb the $1$ into the constant (for example, define $D=C+1$). Recall that general antiderivatives may differ by any arbitrary constant. To see this, try checking your work by taking the derivative of your answer (as well as the "official" answer). In either case, you'll end up with:
$$
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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what is the remainder when $(17^{3}+19^{3} + 21^{3}+23^{3})$ is divided by 83? what is the remainder when $(17^{3}+19^{3} + 21^{3}+23^{3})$ is divided by 83?
NOTE:$a^{3}+b^{3}=(a+b)(a^2-ab+b^2)$
| For every $a$, $(a+1)^3+(a-1)^3=2(a^3+3a)$ and $(a+3)^3+(a-3)^3=2(a^3+3\cdot9a)$ hence $s(a)=(a-3)^3+(a-1)^3+(a+1)^3+(a+3)^3$ is
$$
s(a)=4a^3+2\cdot3\cdot(1+9)a=4a(a^2+15).
$$
Since $83$ is prime, $83$ divides $s(a)$ if and only if $83$ divides $a$ or $83$ divides $a^2+15$. In your post, $a=20$, then $83$ does not divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the sum of this : $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....$ How to find the sum of the following :
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\fr... | Hint: Try writing it as a series, then do a little bit of "algebra work".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
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Solve : $\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$ Solve : $$\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$$
Can we take like this :
Let us take (k+1)th term = $$\frac{(k+1)^2(k+1)!}{2^{k+1}} ; \frac{(k+1)(k... | The "2010" in
$\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$
is a MacGuffin.
If it is replaced by $n$,
this becomes
$
\frac{(n+2)!}{2^n}-\sum^{n}_{k=1} \frac{k^2k!}{2^k}-\sum^{n}_{k=1} \frac{k\cdot k!}{2^k}
$.
Adding the last two terms
(as Kunnysan did)
$\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Finding unit's digit in exponentiation Could someone please explain to me how to find the unit's digit in the following expression:
$$7^{95} - 3^{58}$$
| Note that $7^2$ "ends" in a $9$. So $7^4=7^2\cdot 7^2$ ends in a $1$, So $7^8=7^4\cdot 7^4$ ends in a $1$, and therefore $7^{12}=7^8\cdot 7^4$ ends in a $1$, and therefore $7^{16}$ ends in a $1$, and so on. In general, if $n$ is divisible by $4$, then $7^n$ ends in a $1$.
It follows that $7^{92}$ ends in a $1$. We are ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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can someone explain this limit i have,
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand.
i tried this
$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {... | You did the double multiplication together. That invites mistakes:
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}=\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}\frac{\sqrt{x+49}+7}{\sqrt{x+49}+7}=$$
$$=\lim_{x\to 0}\frac x{\left(\sqrt{x+49}+7\right)\left(3-\sqrt{x+9}\right)}\frac{3+\sqrt{x+9}}{3+\sqrt{x+9}}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Trace zero matrix that can't be written as $AB - BA$? According to this paper, every trace zero matrix over a field can be written in the form $AB - BA$. However, here's a basic counterexample:
Let $A = diag(a, -a)$ for some nonzero number a. Then $A = BC - CB$ if and only if a system of equations involving entries ... | Consider
$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ a & 0 \end{bmatrix} -\begin{bmatrix} 0 & 0 \\ a & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
What made you conclude that the first and last of those equations can't be true?
| {
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"url": "https://math.stackexchange.com/questions/466018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
Compute $ax^3+by^3$ given $ax^2+by^2$ and $ax+by$. Given are positive real numbers $A$ and $B$ and positive integers $a$ and $b$ such that
$$ \begin{aligned} ax+by &= A\\ ax^2+by^2 &= B.\end{aligned} \tag{*}$$
What are the possible values of
$$ ax^3 + by^3 = C?$$
Think of it as $a+b$ numbers, $a$ of them are $x$ and $... | Well, you have 3 equations,
$$ax+by = A\tag{1}$$
$$ax^2+by^2 = B\tag{2}$$
$$ax^3+by^3 = C\tag{3}$$
If you want to express C in terms of $a,b,A,B$, you can eliminate $x,y$ between $(1), (2), (3)$ to have an equation purely in $a,b,A,B,C$ (easily done in Mathematica using the Resultant command).
However, we get a quadra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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what is the taylor expansion of $\ln(\sin x/x)$ what is the Taylor expansion of the function
$$f(x) = \ln \frac{\sin x}x$$
around the point $x=0$? Ignore powers of $x$ which are greater than $6$.
Here is my method:
$$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$
so we should get the function
$$g(x)=\frac{\sin x... | Your instincts are correct. Write
$$\frac{\sin{x}}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$$
We wish to maintain the series to $O(x^6)$. Then
$$\begin{align}\ln{\left ( \frac{\sin{x}}{x}\right)} &= \left (\frac{\sin{x}}{x} - 1\right ) - \frac12 \left (\frac{\sin{x}}{x} - 1\right )^2 + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{... | $$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
Let $u = x^2$ and $\mathrm{d}u = 2x\,dx$ and let $dv = \cos\frac x2\,dx \implies v = 2 \sin \frac x2 $.
$$ 2x^2 \sin \frac x2 - \int 2x\cdot 2\sin \frac x2\,dx $$
You'll only need to do integration by parts one additional time. Let me know if you get stuck after that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Finesse vs. brute force in solving quadratic equations In Higher Algebra by Hall and Knight, the following "artifice" for solving a certain type of equations is given:
Solve: $\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x-11} = 9$
They make use of the fact that $(3x^2-4x+34) - (3x^2-4x-11) = 45$, and utilizing the formula $a^2 ... | Oops! Realized that I had written $-11$ as $11$. Sorry for the mistake! I'll try again and re-post if needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Conjugacy classes and orders of matrices. The following are prime decompositions in $\Bbb{Z}_7[x]$:
$x^8+1= (x^2-x-1)(x^2+x-1)(x^2+3x-1)(x^2+4x-1)$
$x^4+1= (x^2+3x+1)(x^2+4x+1)$
(a) Give representatives for the conjugacy classes of elements of order dividing 16 in $Gl_2(\Bbb{Z}_7)$ and give the order of each.
(b) Show... | Part (a) Let $A\in\operatorname{GL}_2(\mathbb Z_7)$ be a matrix whose order divides $16$.
Then $A^{16} - I = 0$, so the minimal polynomial $f$ of $A$ is a divisor of $x^{16} - 1$.
We have
$$x^{16} - 1
= (x^8 + 1)(x^8 - 1)
= (x^8 + 1)(x^4 + 1)(x^2 + 1)(x+1)(x - 1)
$$
Since $A$ is $2\times 2$, the degree of $f$ is at mos... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Quadratic equation - $\alpha$ and $\beta$ Roots If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 8x - 5 = 0$, find the quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
My working out so far:
I know that $\alpha + \beta = -8$ and $\alpha \beta = -5$ (from the roots) and t... | Since we want the zeroes of the new quadratic polynomial to be the mutual reciprocals $ \ \frac{\alpha}{\beta} \ $ and $ \ \frac{\beta}{\alpha} \ \ , $ it must be palindromic, that is, of the form $ \ Ax^2 + Bx + A \ \ $ (the sort of symmetry André Nicolas was referring to in his comments). Viete tells us that $ \ \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.
Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and ... | In general
$$x^n=y^n\iff 0=x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)$$
If $\,x\neq y \;$ then we must have that the second factor to the right above equals zero.
But if $\,x,y >0\,$ ($\;x,y<0\;$), then that second factor is always positive (negative), so we can already focus on the case with d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Ideal in the ring of upper triangular matrices I'm trying to describe an ideal of the ring $R=\left\{ \begin{pmatrix}a & b\\
0 & c
\end{pmatrix}:a,b,c \in \mathbb{R}\right\} $
It's easy to prove that $I=\left\{ \begin{pmatrix}0 & a\\
0 & 0
\end{pmatrix}:a\in\mathbb{R}\right\} $ and $J=\left\{ \begin{pmatrix}a & b\\
0 &... | $\begin{pmatrix} a & b\\ 0 & c \end{pmatrix}
\begin{pmatrix} x & y\\ 0 & z \end{pmatrix} =
\begin{pmatrix} ax & ay+bz\\ 0 & cz \end{pmatrix}$.
If $a\neq 0$, you can vary $ax$ and $ay$ over $\mathbb{R}$, by varying $x$ and $y$. Similarly, if $b\neq 0$, you can vary $bz$ over $\mathbb{R}$, and if $c\neq 0$, you can vary ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the give... | Hint: $$1^2 - 2^2 + \ldots + (2n-1)^2 - (2n)^2 = 1^2 + 2^2 + \ldots + (2n)^2 - 8 (1^2 + 2^2 + \ldots + n^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Any simplification or approximation of sqrt As you know, we could write $(a+b+c)^2$ as $a^2+b^2+c^2+2ab+2ac+2bc$.
what about $(a+b+c+\cdots)^{1/2}$? is there any expansion for $(a+b+c+\cdots)^{1/2}$?
Any simplification or approximation of $(a+b+c+\cdots)^{1/2}$ could help me.
Thanks in advance
| http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem
If $|a|>|b|$ then
\begin{align}
(a+b)^{1/2} & = a^{1/2} + \frac12 a^{1/2-1} b + \frac{(1/2)(1/2-1)}{2\cdot1} a^{1/2-2} b^2 \\[12pt]
& {}\quad{} + \frac{(1/2)(1/2-1)(1/2-2)}{3\cdot2\cdot1} a^{1/2-3} b^3 \\[12pt]
& {}\quad{} + \frac{(1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Integral of $\sqrt{a-x^2}$ I know how to integrate stuff like $\sqrt{a-x^2}$, though I haven't done this in a while. So I wanted to check the answer after calculating. My calculations go like this ($\theta=sin^{-1}(\frac{x}{\sqrt{a}})$):
$$\int_{A}^{B}{\sqrt{a-x^2}}dx=\int_{A}^{B}{\sqrt{a-x^2}}\frac{1}{\frac{1}{\sqrt{a... | So you've got the integral:
$$\int_A^B \sqrt{a - x^2} \,dx = \int_A^B \sqrt{(\sqrt a)^2 - x^2}\,dx$$
Integrals in the form $\sqrt{a^2 - x^2}$ correspond nicely to using the substitution $x = a \sin \theta \implies \,dx = a \cos \theta d\theta$.
In your case, we have $\sqrt{(\sqrt{a})^2 - x^2}$, so our nice substitutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ratio of two areas in a triangle Suppose you have the following triangle $ABC$:
with the following properties:
$$|AB|=4\cdot |AA'|,\>\>\>|AC|=4\cdot |CC'|\>\>\>|BC|=4\cdot |BB'|$$
I have to find the ratio of the total area of the triangle and the red area. I tried a lot of algebraic manipulations with the lengths of th... | EDIT: Final solution.
First off, notice that $\triangle ABB',\triangle BCC', \triangle CAA'$ all have area equal to $\frac{1}{4}$ of the area of $\triangle ABC$. To see this, pick any side as the base when measuring $\triangle ABC$, say side $BC$. Then $BB'$ is a triangle with a common height and one fourth base, so it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\ta... | \begin{eqnarray}
\tan A+\tan B+\tan C&=&\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}+\tan(180^\circ-A-B)\\
&=&\frac{\sin(A+B)}{\cos A\cos B}-\frac{\sin(A+B)}{\cos(A+B)}\\
&=&\sin(A+B)\frac{\cos(A+B)-\cos A\cos B}{\cos A\cos B\cos(A+B)}\\
&=&-\frac{\sin A\sin B\sin(A+B)}{\cos A\cos B\cos(A+B)}\\
&=&-\tan A\tan B\tan(A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 7
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Prime Numbers And Perfect Squares Find all primes $p$ and $q$ such that $p^2 + 7pq + q^2$ is a perfect square.
One obvious solution is $p = q$ and under such a situation all primes $p$ and $q$ will satisfy.
Further if $p \neq q$ then we can assume without the loss of generality that $p > q$. Assuming this and that ther... | COMMENT.- Just for fun, prove easily the impossibility for twin numbers and other couples of primes.
First, not possible for $p=2\lt q$. In fact $$q^2+14q+4=z^2\Rightarrow z^2\ge55\Rightarrow z\ge 8$$ Actually $\color{red}{z\ge 9}$ because $z$ must be odd. The discriminant of $q^2+14q+4-z^2=0$ is $ 45+z^2=t^2$ so makin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove that there are infinitely many perfect cubes of the form $p^2+3q^2$ Prove that there are infinitely many perfect cubes of the form $p^2+3q^2$ where $p$ and $q$ are integers.
Hint: one approach is to set $p^2+3q^2=(a^2+3b^2)^3$ and then find $(p,q)$ in terms of $a,b$.
Any different approach is very welcome!
| Use this from cubing Eusenstein integers, of the form $a+b\sqrt{\;-3}$.
For $a, b$ as integers, then
$$(a+b\sqrt{-3})^3 = (a^3 - 9ab^2) + (3a^2b - 3b^3)\sqrt{-3} $$
Whereapon, one sees that $(a^2+3b^2)^3 = (a^3-9ab^2)^2 + 3(3a^2b-3b^3)^2$ directly follow.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ By using the substitution $p=x+\frac{1}{x}$, show that the equation $$2x^4+x^3-6x^2+x+2=0$$
reduces to $2p^2+p-10=0$.
I can't think of anything that produces a useful result, I tried writing p as $p=\frac{x^2+1}{x}$ and findin... | $$p=x+x^{-1}$$
$$p^2=x^2+2+x^{-2}$$
$$2x^4+x^3-6x^2+x+2=0$$
$$2x^2+x-6+x^{-1}+2x^{-2}=0$$
$$2(x^2+2+x^{-2})+(x+x^{-1})-10=0$$
$$2p^2+p-10=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Middle school number theory Find at least three numbers that satisfy all three conditions:
(1) there is a remainder of $1$ when the number is divided by $2$;
(2) there is a remainder of $2$ when the number is divided by $3$;
(3) there is a remainder of $3$ when the number is divided by $4$.
Since the LCM($2, 3, 4$) is ... | The simple answer could just be write out the sets; i.e,
$S_1=\{...,-3,\color{red}{-1},1,3,5,7,9,\color{red}{11},13,15,17,19,21,\color{red}{23},25...\}$
$S_2=\{...,-4,\color{red}{-1},2,5,8,\color{red}{11},14,17,20,\color{red}{23},26,...\}$
$S_3=\{...,-5,\color{red}{-1},3,7,\color{red}{11},15,19,\color{red}{23},27,...\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How prove this inequality $(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$ Now my question let $a_{1},a_{2},\cdots,a_{n}$ are positive numbers,and $a_{n+i}=a_{i},i=1,2,\cdots$,show that
$$(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$$
my teacher (tian275461) have prove this
$$(a^{1.5}+b^{1.5}+c^{1.5})^2\ge (a+b... | Here is a proof in the case where the $\{a_i\}$ obey a condition for the variance about their mean.
More precisely, the formulation for the above condition is the following. Let $\bf a$ be the vector composed from the $\{a_i\}$, i.e. ${\bf a} = (a_1,a_2, \dots, a_{n-1}, a_n)$. Define the mean ${m} = \frac1n \sum a_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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$x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$
Is $x^2 +y^2 + z^2$ irreducible in $\mathbb C [x,y,z]$?
As $(x^2+y^2+z^2)= (x+y+z)^2- 2(xy+yz+zx)$,
$$(x^2+y^2+z^2)=\left(x+y+z+\sqrt{2(xy+yz+zx)}\right)\left(x+y+z-\sqrt{2(xy+yz+zx)}\right).$$
But how to show that none of these factors belong to $\mathbb C [x,y,... | We can consider $X^2+Y^2+Z^2$ as a polynomial in $R[X]$, where $R=\mathbb C[Y,Z]$. Now apply the Eisenstein criterion with the irreducible $Y+\mathrm iZ$ in $R$:
*
*It divides $Y^2+Z^2$;
*It does not divide $1$, the coefficient of $X^2$;
*It does not divide $(Y^2+Z^2)^2$.
Irreducibility follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 6
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Solutions to $\sqrt{x}+y=6,x^2+y^2=90$ $$\begin{gather}
\sqrt{x}+y=6 \tag{1} \\
x^2 + y^2 = 90 \tag{2}
\end{gather}$$
WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself. However, he himself has no idea how to solve... | Another way to show there is only one real root...
$\sqrt{x} = 6 - y \implies y \le 6$
So we are looking for roots of $f(y) = (y-6)^4 + y^2 - 90$ where $y \le 6$. We already have $y=3$ as a root from observation.
Using $f(y) = (y-3)(y^3 -21y^2 +154y - 402)$, we need to check if the cubic factor has any real roots i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Ellipse arcs. Draw a tangent line in the end point or make arc longer? I read this article:
link
It describes how to draw ellipse arcs at all from svg. Each ellipse is described with the following params (and I know them):
*
*x1, y1, x2, y2 - arc from point (x1, y1) to point (x2, y2);
*rx, ry - radiuses
*x-axis-ro... | I rewrote this answer completely, see its history for earlier versions.
Change the coordinate system in such a way that the ellipse becomes a circle of radius $1$. The following transformation achieves this:
$$ f: \begin{pmatrix}x\\y\end{pmatrix} \mapsto
\begin{pmatrix}\frac1{r_x}&0\\0&\frac1{r_y}\end{pmatrix}\cdot
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/488486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factorize : $x^6 − 10x^3 + 27$ I want to factorize $$x^6 − 10x^3 + 27$$
I tried two methods , first I let $y=x^3 $ and converted it into a quadratic but the solutions are not real .
The second method I tried was getting it to the form $a^3+b^3+c^3-3abc$ but I could only get close ( $(x^2)^3+(x)^3+3^3-3(x^2)(x)(3)$ bu... | Here is a backdoor way to get the quadratic factor.
Start with your factorization
$x^6-10x^3+27=(x^3-5+\sqrt{-2})(x^3-5-\sqrt{-2}).$
This is a factorization over $\mathbb Z[\sqrt{-2}]$, so seek a quadratic factor with the form
$x^2+2ax+(a^2+2b^2)=(x+a+b\sqrt{-2})(x+a-b\sqrt{-2}).$
The constant term in the quadratic pol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Help me solve this polynomial problem I have to find $a$,$b$ and $c$ so that this expression is a polynomial:
$$\frac{x^3+ax^2+bx+c}{x-1} + \frac{x^3+bx^2+cx+a}{x-2} + \frac{x^3+cx^2+ax+b}{x-3}$$
Please give me hints. Thanks!
| Divide with a remainder:
$$
\frac{x^3+ax^2+bx+c}{x-1}= x^2+b'x+c' +\frac{p}{x-1}
$$
and similarly for other summands. Then your question is reduced to the next one: when
$$
\frac{p}{x-1} +\frac{q}{x-2} +\frac{r}{x-3}
$$
is a polynomial? The answer: only for $p=q=r=0$, since the quadratic polynomial
$$
p(x-2)(x-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find all real matrices $A$ such that $A^2 = \mathrm{diag}(1,1,2,3,5,8,13)$ Let $A \in \mathcal{M}_{7 \times 7} (\mathbb{R})$ such that
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&2&0&0&0&0 \\ 0&0&0&3&0&0&0 \\ 0&0&0&0&5&0&0 \\ 0&0&0&0&0&8&0 \\ 0&0&0&0&0&0&13\end{pmatrix} $$
How many matrices, which satisfy... | We have $$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 & \ldots & 0 \\ \sin \varphi & -\cos \varphi & 0 & \ldots & 0 \\ 0 & 0& 0& \ldots & 0\\ \vdots & \vdots& \vdots& \ddots & \vdots\\ 0 & 0& 0& \ldots & 0 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 & 0 & \ldots & 0 \\ 0 & 1 & 0 & \ldots & 0 \\ 0 & 0& 0& \ldots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| Consider the curve $y=\frac{1}{\sqrt{x}}$. We have
$$\int_1^{n+1}\frac{1}{\sqrt{x}}\,dx\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt \int_0^n \frac{1}{\sqrt{x}}\,dx.$$
Evaluate the integrals. We get
$$2\sqrt{n+1}-2\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt 2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
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Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| Note $(x^2-x^3)(x^4-x)=-x^7+x^6+x^4-x^3$. The key is that $x$ is a root of unity, particularly one that gives $x^5=1$. This means $x^6=x$ and $x^7=x^2$ while $x^4=x^*$ and $x^3=(x^2)^*$ which means that our imaginary components fall out since $z+z^*=2a$ for $z=a+bi$. Squaring both sides to get rid of the radical reduc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Solve $z^4+1=0$ algebraically I know the result and how to solve it using trigonometry and De Moivre.
However, given that the complex number $z$ can be rewritten as $a+bi$, how can I solve it algebraically?
| $$ z^4 + 1 = 0 \Rightarrow z^4 + 2z^2 + 1 - 2z^2 = 0 $$
$$ \Rightarrow (z^2 + 1)^2 -(\sqrt{2}z)^2 = 0 $$
$$ \Rightarrow (z^2 - \sqrt{2} z + 1)( z^2 + \sqrt{2}z + 1 ) = 0 $$
use $$ ax^2 + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to solve $y' = \sqrt {x+y+1}$ How does one solve $y' = \sqrt{x+y+1}$? I try substituting $v=x+y+1$ and using substitution methods, but it turned out to be so messy.
| Another substitution is $x+y+1 = z^2$, so $1+y' = 2zz'$, from which $y' = 2zz'-1$:
$$
2zz'-1 = z \\
2zz' = z+1 \\
\frac {2zdz}{z+1} = dx \\
\int \left (1-\frac 1{z+1} \right ) dz = \frac 12\int dx \\
z - \ln \left |z+1 \right | = \frac x2 + C
$$
Starting this point, it makes sense to leave a solution as an implicit equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding Coefficients of Power Series Expansion Find the coefficients of $ a_{n}$ and $b_{n}$ for $ 0 ≤ n ≤ 4$ for the power series expansion of two linearly independent solutions of the ODE: $y'' -(e^{x}-1)y=0$.
This is what I've tried so far:
$\sum_{n=2}^{\infty} n(n-1)c_{n}x^{n-2} - \sum_{n=0}^{\infty}\frac{1}{n!}x^... | Hint: $e^x = 1 + x + \dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\ldots$, $y_1(x) = x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \ldots$, and $y_2(x) = 1 + b_2 x^2 + b_3 x^3 + b_4 x^4 + \ldots$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
| I will find the $\cos\frac{2\pi}{5}$ with basic complex number theory.
Let $z = \cos(\frac{2\pi}{5}) + i \sin(\frac{2\pi}{5})$, then $z^5=1$. (de Moivre's Theorem) Let's find the solution of this quintic equation.
This equation is factored into $(z-1)(z^4 + z^3 + z^2 + z +1 ) = 0$. $z=1$ is trivial solution.
Now let ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $ \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots + \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2} $ by induction
Prove by induction that for all $n > 0$,
$$
\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots
+ \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2}
$$
I have done the ba... | (Someone please check for error ... I'm not confident if I got this correct)
Since for the base case $n=1$: $\dfrac{\sqrt2}2>\dfrac{\sqrt 1}2$, you just have to prove that $\dfrac{\sqrt{n+1}}{2n}\geq\dfrac{\sqrt{n}}2-\dfrac{\sqrt{n-1}}2$ for $n>1$ since, if $a>b$ then $a+c>b+d$ for $c\geq d$.
\begin{align*}
\frac{\sqrt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Finding the determinant of a $4\times4$ matrix How does one find the determinant of a $4\times 4$ matrix? I am using Cramer's rule to solve a system of linear equations but don't know how to find the determinant of a $4\times 4$ matrix. Our matrix is with variables and not actual values so the answer will be in terms o... | A "quick" method to compute the determinant of a $4 \times 4$ matrix by hand, is by using cofactor expansion.
Let $ D =
\begin{pmatrix}
1 & 1 & -2 & 3 \\
3 & 1 &-6 & 8 \\
0 & 7 & 3 & 0 \\
-1 & 8 & 7 & -4
\end{pmatrix}$
Here we are going to choose which row/column contains the most $0$'s Clearly, row 3 contains the mos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rate of convergence for sequences I have numerically determined that the sequence $\{f_x\} = \frac{\sin(x^2)}{x^2}$ approaches $1$ (as $x$ approaches $0$) faster than the sequence $\{g_x\} = \frac{\sin^2(x)}{x^2}$. However, I am stuck for determing the rate of convergence of each sequence.
Should I use the small angle ... | Let's start with the definition of rate of convergence of functions (There is also definition of rate of convergence of sequences):
Definition. Let $f$ be a function defined on the interval $(a,b)$ that
contains $x=0$, and suppose $\lim_{x\to 0}{f(x)=L}$. If there exists a
function $g$ for which $\lim_{x\to 0}{g(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $f'(0)$ for $f(x)=(2x+1)^3(3x+3)^2$. Find $f'(0)$ for $f(x)=(2x+1)^3(3x+3)^2$.
Do I use the chain rule for each or do I use the derivative product rule first
Please Help!!!
| $f(x) = (2x+1)^3\cdot (3x+3)^3$ and $f(0) = 27$
Taking $\log_{e}$ on both side
$\log_{e}f(x) = \log\left\{(2x+1)^3 \cdot (3x+3)^3\right\} = 3\cdot \log_{e}(2x+1)+3\cdot \log_{e}(3x+3)$
Now Diff. both side w.r.to $x$
$\displaystyle \frac{1}{f(x)}\cdot f^{'}(x) = \frac{3}{2x+1}\cdot 2+\frac{3}{3x+3}\cdot 3$
Put $x = 0$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/506519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integer $x$ for which $x^4+x^3+x^2+x+1$ is perfect square. Integer values of $x$ for which $\bf{x^4+x^3+x^2+x+1}$ is a Perfect Square.
$\underline{\bf{My\; Try}}$:: Let $\bf{x^4+x^3+x^2+x+1 = k^2}$, where $k\in \mathbb{Z}$
$4x^4+4x^3+4x^2+4x+1 = 4k^2 = (2k)^2$
Now How can I proceed after that
Help Required,
Thanks
| Hint: For all but finitely many integer values of $k$, we have
$$ (2k^2 + k)^2 < 4k^4 + 4k^3 + 4k^2 + 4k + 4 < (2k^2 + k + 2 ) ^ 2 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Damped Harmonic Oscillator and Response Function This is another one of those questions that I feel like I am almost there, but not quite, and it's the math that gets me. But here goes:
For a driven damped harmonic oscillator, show that the full width at half maximum of the response function $| R(\omega)|^2$ is $\gamm... | The problem is not so much the calculation, as that is all about quadratic functions, but to select suitable auxiliary variables so that the calculation gets a readable and thus verifiable organization.
I'll interpret the task, as did @Aurey, that the task asks for the lowest order terms of the answer assuming $0<γ=2β\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Need to prove that $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $99$ for all $n \in \mathbb{N} $, using induction. First, obviously, I figured out the base case. So I have $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5 = 99k$ for some $k \in \mathbb{N} $. As for the inductive step, I was thinking about splitting it... | Let $f_{n}\equiv4\cdot10^{2n}+9\cdot10^{2n-1}+5$. Then $f_{1}=4\cdot100+9\cdot10+5=495$,
and $99\mid495$ so $99\mid f_{1}$. Now, suppose that for some $n\in\mathbb{N}$,
$99\mid f_{n}$. Now, since $100\equiv1\text{ mod }99$,
\begin{align*}
f_{n+1} & \equiv 4\cdot10^{2\left(n+1\right)}+9\cdot10^{2\left(n+1\right)-1}+5\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Implicit differentiation of a lemniscate at a point: So here's the problem:
Find the slope of the tangent line of :
$2(x^2 +y^2)^2 = 25(x^2 - y^2)$ at the point (3,1)
Cool:
So here's what I did:
Simplification step:
$2(x^2 +y^2)^2 = 25(x^2 - y^2) \Rightarrow 2(x^2+y^2)^2 = 25x^2 - 25y^2$
Differentiate both sides:... | We start from
$$2(x^2+y^2)^2=25(x^2-y^2).$$
Don't hesitate, Differentiate. We get
$$4(x^2+y^2)(2x+2yy')=50x-50yy'.$$
Substitute $x=3$, $y=1$. We get
$$4(10)(6+2m)=150-50m,$$
and therefore $130m=-90$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Distinct Mersenne numbers are coprime How can you prove that if $p$ and $q$ are distinct primes, then the following holds?:
$$(M_p,M_q)=1$$
Note: $M_n=2^n-1$, with $n$ prime number
| Let $z < x < y$ all be integers with $y = kx + z$ for some positive integer $k$.
I claim that $2^y - 1 \equiv 2^z - 1 \pmod{2^x - 1}$.
Proof:
$(2^y - 1) - 2^{y-x}(2^x - 1) = 2^y - 1 - 2^y + 2^{y-x} = 2^{y-x} - 1$, so
$2^y - 1 \equiv 2^{y-x} - 1 \pmod{2^x-1}$. By similar reasoning, it is also congruent to $2^{y-2x} - 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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cubic equations which have exactly one real root Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c... | $(x^3+x+1)'=3x^2+1>0$ for all real $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculation of sum by using residue theory I am studying an example about the calculation of a summation by using residue theory. I understand how to calculate the sum in general -the frame of the solution way-, but I dont know how some parts of the solution to be found. I Will write the question and answer, please exp... | You have a cubic in the denominator:
$$8 z^2-(1-z)^3 = z^3+5 z^2+3 z-1$$
You should be able to see that $z=-1$ is a root of this cubic by inspection. By synthetic division, you will see that
$$z^3+5 z^2+3 z-1 = (z+1)(z^2+4 z-1)$$
So we see that the other poles are at $z_{\pm}=-2\pm\sqrt{5}$. Now use the fact that
$$\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that $\mathbb Z[\sqrt{3}]$ is a Euclidean Domain Let $R_d$ be the ring defined as $R_d=\left \{ x+y\omega : x,y\in \mathbb{Z} \right\}$, where
$$\omega =
\begin{cases}
\sqrt{d}, & \text{if } \quad d \not \equiv 1\mod 4 \\
\frac{1+\sqrt{d}}{2}, & \text{if } \quad d\equiv 1\mod 4.
\end{cases}$$
It has been proven t... | Define the norm on $\mathbb Z[\sqrt 3]$ to be $N(a + b \sqrt 3) = \vert a^2 - 3 b^2 \vert$.
Let $\alpha, \beta \in \mathbb Z[\sqrt 3]$ with $\beta \neq 0$.
Say $\alpha = a + b \sqrt 3$ and $\beta = c + d \sqrt 3$.
Notice that
\begin{align*}
\frac\alpha\beta &= \frac{a + b \sqrt 3}{c + d \sqrt 3} \cdot \frac{c - d \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/517348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How is $\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$ I read somewhere that $$\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$$I don't know how to have it. Please let me know how this is approximated.
| Let $x>1$ and $q_n = \frac{x^n-1}{x^{n+1}-1}$.
We see that $q_n = \frac{1-\frac{1}{x^n}}{x-\frac{1}{x^n}}$, from which we have $\lim_n q_n = \frac{1}{x}$.
In this case we have $x= \frac{3}{2}$, hence $\lim_n \frac{(\frac{3}{2})^n-1}{(\frac{3}{2})^{n+1}-1} = \frac{1}{(\frac{3}{2})}$.
$n=99$ is sufficiently 'large' that ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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F ind the seven solutions to $x^{7} \equiv 1 \pmod{29}$
Use the fact that $2$ is a primitive root modulo 29 to find the seven solutions to $x^{7} \equiv 1 \pmod{29}$
As $2$ is primitive root modulo $29$ then $$2^{28} \equiv 1 \pmod{29} $$ $$2^{4*7} \equiv 1 \pmod{29} $$ $$16^{7} \equiv 1 \pmod{29} $$ where 16 is a s... | We know that $2^{28} \equiv 1 \bmod 29$, i.e. $16^7 \equiv 1 \bmod 29$. However we can also conclude from this that:
$16^{14},16^{21},16^{28},16^{35},16^{42},16^{49} \equiv 1 \bmod 29$.
So $16^2, 16^3, 16^4, 16^5, 16^6, 16^7$ are the other $6$ solutions.
Alternatively: we know every $x\in\mathbb{Z}/29\mathbb{Z}$ can b... | {
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Writing u as a linear combination of the vectors in S. Write vector
u = $$\left[\begin{array}{ccc|c}2 \\10 \\1\end{array}\right]$$
as a linear combination of the vectors in S. Use elementary row operations on an augmented matrix to find the necessary coefficients.
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\ri... | You need to solve the system
$$ c_1\left[\begin{matrix}1\\2\\2\end{matrix}\right] + c2\left[\begin{matrix}4\\2\\1\end{matrix}\right]+
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right] .$$
| {
"language": "en",
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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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Show that if a and b are positive integers, and $a^3 | b^2$, then a | b Show that if a and b are positive integers, and $a^3 | b^2$, then a | b.
If p is a prime divisor of a, and $p^r$ is the highest power of p dividing a. Then $p^{3r} | a^3$, and so $p^{3r} | b^2$. If $p^s$ is the highest power of p dividing b, then 3... | Let $a,b$ be elements of $Z^+$ (i.e., the positive integers) and suppose $a^3|b^2$.
Then, by definition $b^2=c*a^3$ for some $c$ such that $c$ is an element of $Z^+$.
Hence, $\sqrt{b^2} = b = \sqrt{c*a^3} = (\sqrt{a^2})*(\sqrt{c*a}) = a*\sqrt{c*a}$
This works for both roots of $b^2$ and $a^2$.
Consider case-by-case:... | {
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Find the value of : $\lim_{x\to\infty} \sqrt{x+\sqrt{x}}-\sqrt{x}$ I tried to multiply by the conjugate:
$\displaystyle\lim_{x\to\infty} \frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{x-x+\sqrt{x}}{\sqrt{x+\sqrt{x}}+... | $\displaystyle \lim_{x\to\infty}\left(\sqrt{x+\sqrt x}-\sqrt x\right) = \lim_{x\to\infty}\sqrt{x}\left\{\left(1+\frac{1}{\sqrt{x}}\right)^{\frac{1}{2}}-1\right\}$
Now expand Using Binomialy
$\displaystyle = \lim_{x\to\infty}\sqrt{x}.\left\{\left(1+\frac{1}{2.\sqrt{x}}+\frac{1}{2}.\left(\frac{1}{2}-1\right).\frac{1}{(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?
| In
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
let $x = y+2$.
This becomes
$\begin{align}
\lim_{y\to 0} \frac{y}{\sqrt{(y+2)^2+5}-3}
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+4+5}-3}\\
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\\
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\frac{\sqrt{y^2+4y+9}+3}{\sqrt{y^2+4y+9}+3}\\
&=\li... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$ Let $\mathbb{Z}_x$ denote the ring of integers with the operations $\odot, \oplus $ defined as $a \odot b$ = $ a+b -a b$ and $a \oplus b$ = $a+b -1$. Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$.
Not sure how to start.
| So we want to find a bijection $f \colon \def\Z{\mathbb Z}\Z \to \Z$ such that
$$ f(a+b) = f(a) \oplus f(b) , \quad f(ab) = f(a) \odot f(b) $$
holds for all $a,b \in \Z$. Let's suppose we have such an $f$ and want to find $f(0)$, for all $a$ we must have
\begin{align*}
f(a) &= f(a+0)\\
&= f(a) \oplus f(0)\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the largest rectangle
A rectangle is formed by bending a length of wire of length $L$ around four pegs. Calculate the area of the largest rectangle which can be formed this way (as a function of $L$).
How am I supposed to do this? If I'm interpreting the question correctly, a square would have an are... | You will get a rectangle of sides $a$ and $b$, whose area is $A=a\cdot b$, and perimeter $L=2a+2b$.
One approach is calculus: Let $x=a$, then $b=\frac L2-x$ and area is $A=x(\frac L2-x)=\frac L2x-x^2$, have the derivative equal $0$, and voila.
Second approach: $(a-b)^2\ge 0$ (equality only holds when $a-b=0$, so:
\begi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $n$ divides $2^{2^{n} +1}+1$ $\to$ $n$ divides $2^{n}+1$? Find a counterexample to show that the following implication is not valid.
if $n$ divides $2^{2^{n} +1}+1$ $\to$ $n$ divides $2^{n}+1$
And show how to use it.
This question appeared on the topic Does $n \mid 2^{2^n+1}+1$ imply $n \mid 2^{2^{2^n+1}+1}+1$?
UPDA... | The post you linked bounced around $n = 57$, and it happens to be a counterexample.
First, I'll show that $57\ |\ 2^{2^{57} + 1} + 1$. To do this, I'll show that it is divisble by 19 and 3.
$\phi(3) = 2$, so $2^{2^{57} + 1} \equiv 2^1 \bmod{3}$, and thus, we know that $2^{2^{57} + 1} + 1 \equiv 2^1 + 1 \equiv 0\bmod{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ in which every number is made by inserting $48$ in the middle of previous as indicated, each number is the square of an integer.
| $44...488...89$ has $n+1$ numbers "$4$", $n$ numbers "$8$", and the "$9$". So:
$$
44...488...89 = 4\cdot\frac{10^n-1}{9}\cdot 10^{n+1} + 8\cdot\frac{10^n-1}{9}\cdot 10 + 9
$$
Now, we say $10^n = y$ so
$$
\begin{align}
&\frac{4\cdot (10y-1)\cdot 10y + 8\cdot 10(y-1) + 81}{9}\\
&=\frac{400y^2 + 40y +1}{9}\\
&=\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to approximate this series? How to approximate this series, non-numerically?
$ S_n = \sum_{n=1}^{50} \sqrt{n}$
| Consider that:
$$2\left((n+1)\sqrt{n+1}-n\sqrt{n}\right)=3\sqrt{n+1}+\frac{n-1-\sqrt{n(n+1)}}{\sqrt{n}+\sqrt{n+1}}.$$
The last term is negative, but greater than $$-\frac{3}{2(\sqrt{n}+\sqrt{n+1})}$$
by the AM-GM inequality. This gives:
$$\sum_{n=1}^{50}\sqrt{n}\geq\frac{2}{3}\sum_{n=1}^{50}\left(n\sqrt{n}-(n-1)\sqrt{n... | {
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"url": "https://math.stackexchange.com/questions/530432",
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"source": "stackexchange",
"question_score": "1",
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How find this maximum $S_{\Delta ABC}$ in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.
My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't
| *
*$B(0,0)$
*$P=6\,e^{i\alpha}=6\,(\cos\alpha+i\sin\alpha)\iff P_x=6\cos\alpha$ , and $P_y=6\sin\alpha$ , with $\alpha\in(0,{\pi\over3})$
*$A=A_x+iA_y$ , with $(A_x-P_x)^2+(A_y-P_y)^2=PA^2=10^2=100$ ; $\frac{A_y}{A_x}=\tan\frac\pi3=\sqrt3$
*$C(C_x,0)$ , with $(C_x-P_x)^2+P_y^2=PC^2=7^2=49$
*Heron's Formula : $S_{A... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Largest box fitting inside an ellipsoid Find the volume of the largest box with sides parallel to the $xy$, $xz$, and $yz$ planes that can fit inside the ellipsoid $(x/a)^2 + (y/b)^2 + (z/c)^2 = 1$.
My answer: We want to maximize $f(x,y,z) = xyz$ subject to $(x/a)^2 + (y/b)^2 + (z/c)^2 \leq 1$. So $\nabla f $ parallel ... | It is not difficult to verify that the largest box in the unit sphere is the cube with sides $2/\sqrt3$, whence for such a sphere you get the volume $8/3\sqrt3$. Now notice that $f:\mathbb R^3\to\mathbb R^3$, $(x,y,z)\mapsto(ax,by,cz)$ maps unit sphere to your ellipsoid, and $V(f(X))=abc V(X)$ where $V$ is the volume a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate the series: $ \frac {20} {(5-4)(5^2-4^2)} + \frac {20^2} {(5^2-4^2)(5^3-4^3)} + \frac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $ How to evaluate this infinte summation?
$$ \dfrac {20} {(5-4)(5^2-4^2)} + \dfrac {20^2} {(5^2-4^2)(5^3-4^3)} + \dfrac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $$
Telescopic cancellati... | We have:
$$
\frac{20}{(5−4)(5^2−4^2)}=\frac{4}{5-4}-\frac{4^2}{5^2-4^2}
$$
$$
\frac{20^2}{(5^2−4^2)(5^3−4^3)}=\frac{4^2}{5^2-4^2}-\frac{4^3}{5^3-4^3}
$$
$$
\frac{20^3}{(5^3−4^3)(5^4−4^4)}=\frac{4^3}{5^3-4^3}-\frac{4^4}{5^4-4^4}
$$
Means for gathering data through equations have
$$
\frac{20}{(5−4)(5^2−4^2)}+\frac{20^2}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Existence of positive integer k that are both squares Is there a positive integer k such that $4k+1$ and $9k+1$ are both squares?
| Let's find the $k$ for which $4k+1 = a^2$ is a square. $4k+1$ is odd, so $a$ is odd. Then we have
$$k = \frac{a^2-1}{4} = \frac{a-1}{2}\cdot\frac{a+1}{2} = \frac{a-1}{2}\left(\frac{a-1}{2}+1\right)$$
the product of two consecutive numbers. For convenience, let's write $k =m\cdot (m+1)$. So can $9k+1$ be a square then?
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can we calculate $\int \:\frac{1}{\sqrt{x^2-1}}dx$
Solve the following integral: $$\int \:\frac{1}{\sqrt{x^2-1}}dx$$
I attempted to solve it intergradation by parts by doing a
$$\int \:1\:\frac{1}{\sqrt{x^2-1}} \, dx$$
and set $u$ be $\frac{1}{\sqrt{x^2-1}}$ and $dv/dx$ be $1$:
but as I start doing, it gets mor... | Here's an alternative way doing it. Rewrite the integral as:
$\int \frac{1}{\sqrt{x^2-1}} dx = \int \frac{1}{\sqrt{x^2-1}} \cdot \frac{x + \sqrt{x^2-1}}{x + \sqrt{x^2-1}} dx = \int \frac{1 + \frac{x}{\sqrt{x^2-1}} dx}{x + \sqrt{x^2-1}}$.
Now use a simple substitution $u=x+\sqrt{x^2-1}$, so that the integral becomes:
$\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving two simultaneous equations Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero) such that we get the following relationships simultaneously
*
*$x + y = 1 - z$ and
*$x^3 + y^3= 1 - z^2$
Find all such $x$, $y$ and $z$.
I have tried to determine the values of $x$ and $y$ by fixing... | To find all integer solutions, we begin by substituting $z=1-x-y$, into $z^2= 1-x^3-y^3$ to get
$$(1-x-y)^2=1-x^3-y^3.$$
Since $y=-x$ is already a clear solution, we will force the factor $(y+x)$ in what follows.
\begin{align*}
& (1-x-y)^2=1-x^3-y^3 \\
\Rightarrow & 1-2x+x^2-2y+2xy+y^2=1-x^3-y^3 \\
\Rightarrow & y^3+y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$ Prove the following equation.
\begin{eqnarray}
\\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\
\end{eqnarray}
I can't prove it by many methods I use.
Please give me some hints.
Thank you for your attention
| An alternate approach
From algebra we know
if $a+b+c=0$ then $ a^3+b^3+c^3 = 3.a.b.c$
$$Derivative \begin{cases}a+b+c = 0 & \text{(1)}\\ \Rightarrow a+b=-c \\ \Rightarrow (a+b)^3=(-c)^3 & \text{cubing both sides} \\ \Rightarrow a^3+b^3+3a^2b+3ab^2=(-c)^3 \\ \Rightarrow a^3+b^3+3ab(a+b)=(-c)^3 & \text{but }a+b = -c\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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$x^2+xy+y^2$ and $x^2-xy+y^2$ are not both perfect squares
Prove that $x^2+xy+y^2$ and $x^2-xy+y^2$ cannot be both perfect squares.
Surely $x$ and $y$ are natural numbers. If $x^2+xy+y^2 =a^2$ and $x^2-xy+y^2=b^2$ simultaneously then we have to show that there are no such integers $a$ and $b$.
I have tried that:
Supp... | Adding these equations gives $2(x^2+y^2) = a^2 + b^2$; unique factoriation of Gaussian integers shows that we essentially must have $a = x+y$ and $b = x-y$. Now combine this with what you already have.
| {
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $... | If $a,b>0$, one can make the substitution $u=ab,$ where $0<u\leq \frac 1 2$ (by AM-GM). Note that $$1=a^2+b^2=(a+b)^2-2ab,$$ hence $$a+b=\sqrt{1+2u}.$$ It follows that one needs to minimize $$f(u):=a+b+\frac 1{ab}=\sqrt{1+2u}+\frac 1 u, \qquad 0<u\leq \frac 1 2.$$
By Calculus, it is easy to see that $f$ has a global mi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving the second root of a quadratic equation If $\alpha$ is a root of the equation $4x^2+2x-1=0$, then prove that $4\alpha^3-3\alpha$ is the other root. How do I proceed? The sum of the roots, the product of the roots lead me nowhere. Should I find the roots of the equation and substitute in the given two expression... | Clearly, the absolute value of both roots are $<1$
$$0=(4x^2+2x-1)(x-1)=4x^3-3x-(2x^2-1)$$
If $x=\cos y,$ we have $$\cos3y=\cos2y$$
$\implies3y=360^\circ n\pm2y$ where $n$ is any integer.
'+'$\implies y=360^\circ n\implies x=1$
'-'$\implies y=72^\circ n$ where $n\equiv0,\pm1,\pm2\pmod5$
As $x=1$ does not satisfy $4x^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Volume of a sphere with three holes drilled in it. Suppose that the sphere $x^2+y^2+z^2=9$ has three holes of radius $1$ drilled through it. One down the $z$-axis, one along the $x$-axis, and one along the $y$-axis. What is the volume of the resulting solid? I can do it for two holes but I'm stuck on three.
| The 3 holes such drilled form an intersection of 3 cylinders in the center, plus 6 cylinder/cap pairs. I will treat each problem in turn.
Here is a picture of the situation:
Intersection of 3 cylinders
The problem is to find the volume of three orthogonal, intersecting cylinders:
$$\begin{align}x^2+y^2&=1\\x^2+z^2&=1... | {
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Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity.
The way I do it is
Let ∊ > 0 be given. Notice N ∈ natural number (N) which satisfies {fill this box later}< N.
It follows that if n>=... | Try setting your last expression to $\epsilon$ and work backwards. You'll find that the following will work:
Given any $\epsilon > 0$, let $N$ be any natural number that is greater than or equal to the real number:
$$\dfrac{\dfrac{9}{5\epsilon}+2}{5} + 1$$
(by the Archimedean Property, such an $N$ must exist). Now supp... | {
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"timestamp": "2023-03-29T00:00:00",
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Volume of a cuboid whose diagonal and surface area is known The sum of length, breadth and depth of cuboid is $19$cm and its diagonal is $5\sqrt{5}$cm.
Its volume is:
a) 125
b) 236
c) 361
d) 486
Solution:
$$\ell^2 + b^2 +h^2 = 125\quad\text{ and }\quad \ell+b+h=19,$$
How can i find volume from this information?
| $l+b+h=19$
$l^2+b^2+h^2=125$
From these 2 equations, we get
$2(lb+bh+hl) = (l+b+h)^2-(l^2+b^2+h^2)
= 19^2-125 = 236 = S$
This is coincidentally the Surface area of the cuboid.
The Volume $lbh = \frac{S}{2(l^{-1}+b^{-1}+h^{-1})}$.
Assuming $l=b$, we can find $l$ and $h$ to be $\frac{19-\sqrt{7}}{3}$ and $\frac{19+2\sqrt... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\{ 1, 1-x , 1-2x + {1 \over 2} x^2\}$ a orthogonal system I have to show that the following set:
$$A = \left\{ 1, 1-x , 1-2x + {1 \over 2} x^2\right\}$$
is orthogonal system in relative to the inner product
$$\langle f, g\rangle = \int ^\infty_0 f(x)g(x)e^{-x}dx$$.
as far as I know, in order that $A$ will ... | You've forgotten the $e^{-x}$ term in $\langle f,g\rangle$.
$$\langle f,g\rangle =\int_0^\infty 1\cdot(1-2x+\frac{x^2}{2})\cdot e^{-x} \mathrm{d}x=0$$
etc.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $3\mid n^3 - n$
Prove by induction that $3\mid n^3 - n$.
I'm having an argument with my professor whether my exam was right or not. Before I sign a formal complain to get a review on my exam, I'd like to be sure it's correct. My answer:
Proof by induction:
Proposition: $\forall n \in\mathbb{ N... | Note: the following "equality" in your argument is not valid (third to fourth line):
$${\bf n^3 + 3n^2 + 2n} = {\bf n^3 + 3n^2 + 2n} \color{red}{\bf - (n^3 - n)}$$
First, your missing the inductive hypothesis:
Assume that for $n = k$, we have $p(k) = k^3 - k$ is divisible by $3$.
Then, we show based on that assumption,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562699",
"timestamp": "2023-03-29T00:00:00",
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Let $p=q+4a$. Prove that $\left( \frac{a}{p} \right) = \left( \frac{a}{q} \right)$. Here's a little number theory problem I'm wrestling with.
Let $p$ and $q$ be odd prime numbers with $p=q+4a$ for some $a \in \mathbb{Z}$. Prove that $$\left( \frac{a}{p} \right) = \left( \frac{a}{q} \right),$$
where $\left( \frac{a}{p} ... | Note that $p \equiv q \pmod{4}$, so $\frac{p-1}{2}\frac{q+1}{2} \equiv \frac{p-1}{2}\frac{p+1}{2} \equiv 0 \pmod{2}$.
\begin{align}
\left(\frac{a}{p}\right)=\left(\frac{4a}{p}\right)=\left(\frac{p-q}{p}\right)& =\left(\frac{-q}{p}\right) \\
& =\left(\frac{-1}{p}\right)\left(\frac{q}{p}\right) \\
&=(-1)^{\frac{p-1}{2}}\... | {
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"url": "https://math.stackexchange.com/questions/564839",
"timestamp": "2023-03-29T00:00:00",
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Derivative of following functions $f(x) = \cos^{-1}\left(1-2x^2\right)$ Derivative of following functions
(A) $f(x) = \cos^{-1}\left(1-2x^2\right)$
(B) $f(x) = \sin^{-1}(3x-4x^3)$
$\bf{My\; Try}::$ Given $f(x) = \cos^{-1}\left(1-2x^2\right)$ and function $f(x)$ is defined when $\left|1-2x^2\right|\leq 1\Rightarrow -1 ... | Your process seems to be fine.
Putting $x=\cos\theta,0\le \theta\le \pi$ as the principal value of inverse cosine function
I also reached at $$\cos^{-1}(1-2x^2)=\begin{cases} 2\cos^{-1}x-\pi &\mbox{if } \frac\pi2\le \theta\le \pi\implies0\ge x\ge -1 \\
\pi- 2\cos^{-1}x & \mbox{if } 0\le \theta\le \frac\pi2\implies1\ge ... | {
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How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$ let $a,b,c>0$ ,and such $a+b+c=3$.prove that
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{1+c^2}<\dfrac{11}{5}$$
if this problem find this minimum,then
$$f(x)=\dfrac{1}{x^2+1}\ge ax+b$$
where $a=f'(1),a+b=f(1)$
since $$f'(x)=-\dfrac{2x}{(1+x^2)^... | First show that $\dfrac{1}{1+x^2} < \dfrac{63 - 19x}{60}$ for $0 < x < 3$.
Then $\text{LHS} < \dfrac{3\cdot63 - 19(a + b + c)}{60} = \dfrac{189 - 57}{60} = \text{RHS}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$ for $n\geq 2$ Prove, for all $n\geq 2$ that
$$\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$$
Let us prove the inductive base for $n = 2$
$$\rm{LHS} = 1^2 + 2^2= 1 + 4 = 5$$
$$\rm{RHS} = \binom{3}{2} + 2\binom{3}{3} = 3 + 2\cdot 1 = 5$$
$$\rm{LHS}... | Perhaps this Wikipedia link might prove itself useful. The sums are computed through telescoping, each relying on the result of the previous one, which, in this case, is $\sum_1^n k=\frac{n(n+1)}2$, for instance like this.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $ S_{20} $ if $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n Define $ a_{n} $ is a sequence and all terms of $ a_{n} $ are positive. $ S_{n} $ is the summation of the first n terms. If $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n, then find $ S_{20} $
| $$\frac{a_1+2}{2}=\sqrt{2a_1} \Leftrightarrow (\sqrt{a_1}-\sqrt{2})^2=0$$
Thus
$$S_1=a_1=2$$
Next
$$\frac{ S_n-S_{n-1} +2}{2} = \sqrt{2S_n} \Leftrightarrow (\sqrt{S_n}-\sqrt{2})^2=S_{n-1}\Leftrightarrow \sqrt{S_n}=\sqrt{S_{n-1}}+\sqrt{2}$$
Let $T_n=\sqrt{S_n}$.
Then $T_1=\sqrt{2}$ and
$$T_n =T_{n-1}+\sqrt{2}$$
Can you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.
My Attempt:
We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have
$$
\begin{cases}
x^2+y^2\geq 2xy\\
y^2+z^2\geq 2yz\\
z^2+x^2\geq 2zx\\
\end{cases... | Another method:
It's clear zero is a solution and moreover $xyz\ge0$. Let's prove there are no more solutions over the integers. By the generalized mean theorem:
$$\sqrt{\frac{x^2+y^2+z^2}{3}}>\sqrt[3]{xyz}$$
Which leads to:
$$x^2+y^2+z^2 > 3(xyz)^{2/3}$$
So now we know that whenever $3(xyz)^{2/3} > 2xyz$ there are no ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,... | Just observe that $x^n-1$ factors as $(x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1)$, and make the appropriate substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/569563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How is $\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$ Can someone please explain to me how is $$\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$$
Thanks!
| Firstly$$\dfrac1{(1-x)^5}=(1-x)^{-5}$$
then from binomial theorem
$$(1-x)^{-5}=\sum_{n\geq0}(-1)^n\binom{-5}{n}x^n$$
because
$$\binom{-5}{n}=\frac{(-5)(-5-1)(-5-2)...(-5-(n-1))}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\frac{(5)(5+1)(5+2)...(5+(n-1))}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\frac{(n+4)(n+3)(n+2)...6\cdot5}{n(n-1)(n-2)..1}=$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$ Find this sum
$$I=\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}$$
My try: let
$$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{3n+3}}{(3n+1)(3n+2)(3n+3)},|x|\le 1$$
then we have
$$f^{(3)}(x)=\sum_{n=0}^{\infty}x^{3n}=\dfrac{1}{1-x^3}$$
then we find th... | OK, so let's start with
$$\frac{d^3}{dx^3} f(x) = \frac{1}{1-x^3}$$
where $f(0)=f'(0)=f''(0)=0$. Note that we may use partial fractions and a little rearranging to find that
$$\frac{1}{1-x^3} = \frac13 \left [\frac{1}{1-x} + \frac{x+\frac12}{\left (x+\frac12\right)^2+\frac{3}{4}}+\frac{3}{2} \frac{1}{\left (x+\frac12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/573119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 4,
"answer_id": 0
} |
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.