Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 = (x^2 + 3x + 2)^3 $have?
How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 =
(x^2 + 3x + 2)^3 $have?
If I didn't make any mistake that equation could be reduced to the form:$$(2+6x^2)(x+2)^3=(8–2x^2 )^3$$
However,I don't know how to ... | We are trying to solve the equation
$$(x^2 + x – 2)^3 + (8–2x^2 )^3 -(x^2 + 3x + 2)^3=0.$$
Let $P(x)$ be the polynomial on the left-hand side. Note that $x^2+x-2=(x+2)(x-1)$, and $8-2x^2=-2(x-2)(x+2)$, and $x^2+3x+2=(x+2)(x+1).$
Thus
$$P(x)=(x+2)^3[(x-1)^3-8(x-2)^3-(x+1)^3]=(x-2)^3Q(x).$$
Expand the remaining cubes. W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Trig substitution integration of $\int1/(x^2\sqrt{x^2 - 9}) dx$ - stuck on a problem I am getting stuck on this trig substitution problem.
$$\int\frac1{x^2\sqrt{x^2 - 9}}~\mathrm dx.$$
$$x = 3 \sec\theta,\qquad\theta = \sec^{-1} \sqrt{\frac{x^2}{9}},\qquad\mathrm dx = \sec\theta\tan\theta\ \mathrm d\theta$$
I can get t... | Using your substitution $x=3\sec \theta $ and cancelling $\sec \theta$ in the numerator and denominator, I got
$$I=\int \frac{1}{x^{2}\sqrt{x^{2}-9}}\ \textrm{d}x=\int \frac{ \tan \theta }{3\ \sec
\theta \ \sqrt{9\sec ^{2}\theta -9}}\ \textrm{d}\theta=\int \frac{\tan \theta }{9\ \sec
\theta \ \sqrt{\sec ^{2}\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$?
If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?
The answer says $1$, but I am not sure how to derive it.
| From the fact that Expressions $1$ and $2$ are equal, we obtain
$$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}.$$
From the fact that Expressions $2$ and $3$ are equal, we obtain
$$b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ca}.$$
From the fact that Expressions $3$ and $1$ are equal, we obtain
$$c-a=\frac{1}{b}-\frac{1}{a}=\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How could we find the sum of reciprocals of the roots of this equation?
A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How
could we find the sum of reciprocals of the roots of this equation?
I need some hints for solving this problem.
| This shall answer your question.
A polynomial is of the form,
$$P(x) = (x-r_1)(x-r_2)........(x-r_n)$$
where $r_1,r_2,.....r_n $ are the roots of the polynomial.
Then the first derivative of the polynomial will be,
$$P^{'}(x) = \frac{(x-r_1)^{'}}{(x-r_1)}P(x) + \frac{(x-r_2)^{'}}{(x-r_2)}P(x) + \frac{(x-r_1)^{'}}{(x-r_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/61690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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How to create 2x2 matrix to rotate vector to right side? I have vector u=(x,y) and i need to create matrix M: M*u=(1,0).
But that matrix has to rotate vector, instead of keep and scale the x unit. So when i apply it on different vectors, the angle between them won't change.
Btw, this isn't homework! We haven't learned... | Your problem is equivalent to find the transformation between the $x,y$
coordinates of a point and the $x^{\prime },y^{\prime }$ coordinates of the
same point in a rotated system of coordinates, followed by a multiplication
by the factor $k=1/\sqrt{x^{2}+y^{2}}$, so that $x^{\prime \prime
}=kx^{\prime }=1$ and $y^{\pri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/63503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluate $\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}\mathrm dx$ I need help to evaluate the integral: $$ \int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}\mathrm dx.$$
The procedure on Wolfram Alpha is very long and complicated. Is there any easier way ?
Thanks.
| As suggested in the comments, you can substitute to get rid of the $-1$ in the third root. Let $y = x - \frac{1}{4}$. To write $p(x) = x^2 - 2x + 7$ as a polynomial in $y$, note that $p(x) = p(y + \frac{1}{4})$ so substituting $x = y + \frac{1}{4}$ in $p(x)$ gives $x^2 - 2x + 7 = y^2 - \frac{3}{2} y + \frac{105}{16}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/63762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
| Here's one way to go about deriving a recursion relation for integrals of the form
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx$$
Split the integral like so:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx$$
and integrate by parts:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\sqrt{1+x^2} x^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/64450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
| This inequality can be solved by simple algebra
we have the equation
$$a^2+b^2+c^2 >= ab+ac+cb$$
multiply and divide 2 on both sides
$$\frac{2}{2}(a^2+b^2+c^2) >= \frac{2}{2}(ab+ac+cb)$$
putting the left side equation on the right side
$$\frac{2a^2+2b^2+2c^2 - 2(ab+ac+cb)}{2} >= 0$$
then factorize and multiply by 2 on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/64868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 8
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How can I compute $\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} $? This sum is difficult. How can I compute it, without using calculus?
$$\sum_{k = 1}^n \frac1{k + 1}\binom{n}{k}$$
If someone can explain some technique to do it, I'd appreciate it.
Or advice using a telescopic sum, I think with a telescopic co... | \begin{align*}\sum_{k=1}^n \binom{n}{k} \frac{1}{k+1}
= \frac{1}{n+1} \sum_{k=1}^n \binom{n+1}{k+1}
= \frac{2^{n+1} - 1 - (n+1)}{n+1} = \frac{2^{n+1} - n-2}{n+1}.
\end{align*}
The first step follows from the identity $\binom{n}{k} \frac{n+1}{k+1} = \frac{n!}{k! (n-k)!} \frac{n+1}{k+1} = \frac{(n+1)!}{(k+1)! (n-k)!} =... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4... | On the right side, when you say to take the positive square root of $(2-5/2)(2-5/2)$, you're taking a $-.5 [(2-5/2)]$ instead of $.5$
It's easy to see if you multiply out all the numbers in each step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating $\sum_{k=0}^n \binom{n}{k} 2^{k^2}$ Can someone please help me simplifying this sum
$$\sum_{k=0}^n \binom{n}{k} 2^{k^2}$$
Wolframalpha fails (see here).
Thanks in advance.
The sum counts the number of (labelled) digraphs (with loops) with vertex set $V \subseteq [n]$.
| I agree with J.M.'s comment above that this is not going to resolve into something simple. Asymptotically, though, the last term will dominate, and maybe that will be useful. We have
$$2^{n^2} \leq \sum_{k=0}^n \binom{n}{k} 2^{k^2} \leq 2^{n^2}\left(1 + \frac{2n^2}{4^n}\right).$$
The relative remainder term $R(n) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/75721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
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Problem with generating functions and binary recurrences I am considering the following recurrence:
$a_0 = 1$;
$a_1 = 2$
$a_{n} = 2 (a_{n - 1} + a_{n - 2})$
Then I proceeded with the generating function:
$F(x) = \displaystyle\sum_{n = 0}^\infty a_n x^n = 1 + 2x + \displaystyle\sum_{n = 2}^\infty a_{n} x^{n} = 1 + 2x +... | OK, using the recurrence equation $a_0 = 1$, $a_1=2$, $a_2 = 6$, $a_3 = 16$, $a_4 = 44$ and $a_5=120$.
Verifying this with the generating function directly:
$$
\frac{1}{1-2x - 2x^2} \sim \sum_{k=0}^5 (2 x+2 x^2)^k \sim 1+ 2x + 6 x^2 + 16 x^3 + 44 x^4 + 120 x^5 + o(x^5)
$$
Now using roots of the denominator $1-2x-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/76363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Formula for completing the square? My math teacher said that this was the formula for completing the square.
Original function: $$ax^2 + bx + c$$
Completed square: $$a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$
However, using this formula I'm not getting the same answers that I would get just by determining ... | Let me derive it for you,
$$ax^2+bx+c= a \left( x^2+\frac{b}{a} x +\frac ca \right) = a\left(x^2+2\frac{b}{2a} x + \left( \frac b{2a} \right) ^2 - \left( \frac b{2a} \right) ^2+\frac ca \right)$$
$$ = a \left\{ \left(x+\frac{b}{a}\right)^2 - \frac{(b^2-4ac)}{4a^2} \right\} = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/76772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Summation Identity: $\sum_{i=1}^ni^3 = \left( \frac{n(n+1)}{2} \right)^2$ I have to prove:
$$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2$$
Using the following:
$$n^3 = 6 {n \choose 3} + 6 {n \choose 2} + n \quad \forall n \in \mathbb{N}$$
My work is that first I substitute $n^3$ for $6 {n \choose 3} + 6... | Don't stop, just do! =)
$$
\begin{eqnarray*}
&& 6 \binom{n+1}{4} + 6\binom{n+1}{3} + \binom{n+1}{2}
\\ &=& 6 \frac{(n+1) n (n-1)(n-2)}{24} + 6 \frac{(n+1) n (n-1)}{6} + \frac{(n+1) n}{2}
\\ &=& \frac{n(n+1)}{4} \left( (n-1)(n-2) + 4 (n-1) + 2 \right)
\\ &=& \cdots
\end{eqnarray*}
$$
I presume you can simplify t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need help finding smallest value of $x^2 + y^2$ I need to find the smallest value of $x^2 + y^2$ with the restriction $2x + 3y = 6$. This chapter focuses on the vertex formula.
| Let's solve for $y$ in the equation $2x + 3y = 6$. This gives $y = 2 - \frac{2x}{3}$.
After rewriting $x^2 + y^2$ in terms of $x$, we have
$x^2 + y^2 = x^2 + (2 - \frac{2x}{3})^2 = x^2 + (4 - 2 \cdot \frac{4x}{3} +\frac{4x^2}{9})$
This is a quadratic "in $x$" that we would like to minimize. You can write it in standar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Difficult integral: $\sin^3\theta / (\sin^3\theta - \cos^3\theta)$ The question is, find the integral to the function:
$\sin^3\theta / (\sin^3\theta - \cos^3\theta)$
The only thing I could think of was to factor the denominator. But then I couldn't make any further progress.
| We take advantage of the symmetry, indeed expand on it. Let
$$I=\int \frac{\sin^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta} \qquad\text{and}\qquad J=\int \frac{\cos^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta}.$$
Note that
$$\frac{\sin^3\theta}{\sin^3\theta-\cos^3\theta}=1+ \frac{\cos^3\theta}{\sin^3\theta-\cos^3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/81525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving sequence equality using the binomial theorem The problem: Prove that for $n \in \mathbb N$:
$$ \left(1 + \frac{1}{n} \right)^n = 1 + \sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right). $$
The hint is to use the binomial theorem. So the ... | Take your second sum
$$
\sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m
$$
and write it as
$$
1+\sum_{m=1}^n \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m
$$
to get the indices to match.
In your first sum
$$
\sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/81887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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alternative proof to induction, binomial sum identity
$\displaystyle \sum_{q=0}^{k} \begin{pmatrix}n-1+q\\ n-1 \end{pmatrix} = \begin{pmatrix}n+k\\n \end{pmatrix} $
induction
$\begin{pmatrix}n \\ k \end{pmatrix} := \frac{n!}{(n-k)!k!}$
Beginning of induction : $k=0 \rightarrow \begin{pmatrix} n-1 \\ n-1 \end{pma... | Another way is to use snake oil:
$\begin{align}
\sum_{n \ge 1} z^n \sum_{0 \le q \le k} \binom{n + q - 1}{n - 1}
&= \sum_{0 \le q \le k}
z^{1 - q} \sum_{n \ge 1} \binom{n + q - 1}{q} z^{n + q - 1} \\
&= \sum_{0 \le q \le k}
z^{1 - q} \sum_{s \le 0} \binom{s}{q} z^s \\
&= \sum_{0 \le q \le k}
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $8^{x+1} = 32 \cdot\sqrt2$ without $\log$ I need help solving the equation $8^{x+1} = 32 \cdot\sqrt2$. The obvious answer is to use log, but that is reserved for the next section. The example given for this section of questions is:
$4^x = 8$
$(2^2)^x = 2^3$
$2x = 3$
$x = \dfrac{3}{2}$
The example looks obvious an... | $$
8^{x+1} = 32 \sqrt{2}
$$
$$
8\times 8^x = 32 \sqrt{2}
$$
$$
8^x =2^{3x}= 4\sqrt{2}=2^2\times 2^{1/2}= 2^{2,5}
$$
$$
3x=2,5 \to x=\frac{5}{6}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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The last digit of $n^5-n$ What will be the last digit of
$$n^5 - n \bmod 1000,$$
where $n$ is a natural number and $m \bmod 1000$ is the remainder when $m$ is divided by $1000$.
| The unit digit of $n^{4k+1}$ is same as the unit digit of $n$
This is because the unit digit is periodic with period $4$,or $2$, or $1$, For instance
$2^1=2 _{U.D=2}$
$2^2=4_{U.D=4}$
$2^3=8_{U.D=8}$
$2^4=16_{U.D=6}$
$2^5=32_{U.D=2}$
$2^6=64_{U.D=4}$
Hence we observe that the unit digit is $2,4,8,6$ and it oscillates wi... | {
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"timestamp": "2023-03-29T00:00:00",
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Is this proof about Mersenne numbers acceptable? I want to prove following property of Mersenne numbers :
If $p > 3$ then $M_p\equiv 1 \pmod {6\cdot p}$
So, according to Fermat's Little Theorem we may write :
$2^p\equiv 2 \pmod p \Rightarrow 2^p-2=a\cdot p \Rightarrow 2^p-1=a\cdot p +1 \Rightarrow$
$\Rightarrow 2^p-1... | There is nothing incorrect, but there are a few things that could be changed. We only need $p>2$.
From $2^p \equiv 2 \pmod {p}$ one should conclude $M_p=2^p -1\equiv 1 \pmod{p}$ immediately, without the detour through $a\cdot p +1$.
A similar needless detour is made when from $M_p\equiv 7\pmod{24}$ it is argued that $... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the limit of $\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$ $$\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$$
Can't find a means to resolve. The answer is $10$ by graphing.
| Divide through by $x, \lim \limits_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\lim \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}}$ then use the expansion around $x=\infty \lim \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}} \approx \frac{1-[1+... | {
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"timestamp": "2023-03-29T00:00:00",
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Congruence and diagonalizations How does one find matrix $M\in M_3(\mathbb R)$ such that $M^TAM=I$ where $$A=\left(\begin{array}{rrr}2&0&0\\ 0&3&-1\\ 0&-1&3\end{array}\right)$$
and simultaneously, $M^TBM$ is some diagonal matrix where $$B=\left(\begin{array}{rrr}1&3&-3\\ 3&3&1\\ -3&1&3\end{array}\right)$$?
Thanks.
| As noted in my answer to your related question, there is a theorem regarding simultaneous diagonalization by congruence: if $A,B$ are real symmetric and $A$ is nonsingular, then they are simultaneously congruent to diagonal matrices if and only if $C=A^{−1}B$ is diagonalizable by similarity transform (Horn and Johnson,... | {
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Compositeness of number $k\cdot 2^n+1$? Every odd prime number can be expressed in the form $k \cdot 2^n+1$ ,where $k$ is an odd number .
For $n>2$ number $k \cdot 2^n+1$ is composite if :
$1.$ $k\equiv 1 \pmod {30} \land (n\equiv 2 \pmod 4 \lor n \equiv 1 \pmod 2 ) $
$2.$ $k\equiv 3 \pmod {30} \land n\equiv 3 \pmod 4... | Your findings have been generalized in this paper. Here, a covering system of integers is constructed and that allows for infinitely many numbers of the form $k\times2^n+1$ to be composite in the sequence of Lucas nubers.
So, for you, I guess the important part is that there are infinitely many Sierpinski numbers i.e.... | {
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} |
recurrence relation, linear, second order, homogeneous, constant coefficients, generating functions How to solve this by using the generating functions? What is the possible solution for this?
recurrence relation $$ a_n = 5a_{n-1} – 6a_{n-2}, n \ge 2,\text{ given }a_0 = 1, a_1 = 4.$$
Thanks.
| $$
\begin{align}
a_n & = 5 a_{n-1} - 6 a_{n-2}\\
a_n - 3a_{n-1} & = 2 (a_{n-1} - 3 a_{n-2})
\end{align}
$$
Letting $T_n = a_n - 3a_{n-1}$ we get that $T_n = 2T_{n-1}$ and $T_1 = a_1 - 3a_0 = 1$. Hence, $$T_n = 2T_{n-1} = 2^2 T_{n-2} = \cdots = 2^{n-1} T_1 = 2^{n-1}$$
Hence,
$$
\begin{align}
a_n - 3a_{n-1} & = 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/91418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to find $\sqrt[3]{8i}$ How do I find the following cube root?
$$\sqrt[3]{8i} = ?$$
I tried by adding $\sqrt[3]{i^3 + 8i + i}$ but that is where my imagination quits.
| If you want a general method for finding roots of complex numbers, you may try my answer here, for instance.
In your case, it would work as follows: write your complex number in exponential form,
$$
8i = 8 e^{i\pi/2} \ .
$$
Then,
$$
\sqrt[3]{8e^{i\pi/2}} = \sqrt[3]{8} e^{i(\pi/2 + 2k\pi)/3} \qquad \text{for} \quad k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Can a regular heptagon be constructed using a compass, straightedge, and angle trisector? Euclid has a magical compass with which he can trisect any angle. Together with a regular compass and a straightedge, can he construct a regular heptagon?
| Gleason's article "Angle Trisection, the Heptagon, and the Triskaidecagon" (also available here) mentions a construction due to Plemelj:
Draw the circle with center $O$ passing through $A$ and on it find $M$ so that $AM=OA$. Bisect $OM$ at $N$, and trisect at $P$, and find $T$ on $NP$ so that $\angle NAT=\frac13\angl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 0
} |
Dice Question Probability A fair die is rolled until a $6$ appears. What is the probability that is must be cast more than 5 times?
So this is $1- P(\text{dice has to be cast less than or equal to}\ 5 \ \text{times})$. So this probability is equal to $$ P =\frac{1}{6}+ \frac{5}{6} \frac{1}{6}+\left(\frac{5}{6} \right)... | The probability of getting a non-$6$ the first five times is $(5/6)^5$.
$$
\begin{align}
& \Pr(\text{a non-}6\text{ on the 1st }5\text{ trials}) \\ \\
& = \Pr(\text{a non-}6\text{ on the first trial and a non-}6\text{ on the 2nd trial and a non-}6\text{ on the 3rd trial and }\ldots) \\ \\
& = \Pr(\text{a non-}6\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Why do we assume $a+b=b+a$ in a Ring with 1. Also, is it true with Rings without 1? I was wondering why do some people use redundant axioms in definitions?
If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutati... | Without a $1$, you cannot deduce the condition. Take $(R,\cdot,\times)$ where $(R,\cdot)$ is any nonabelian group with identity element $e$, and let $a\times b = e$ for all $a$ and $b$ (the "zero multiplication ring"). This satisfies all the axioms of a ring without $1$, except for commutativity of the first operation.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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solve complex equation $x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$
What's the way to solve this kind of equation? I think there must be 8 solutions.
I tried to solve the following two equations
$a^6 = 1+... | Any equation of the form $z^n=\alpha$, where $\alpha$ is a fixed complex number can be solved by switching $\alpha$ to polar form
$$\alpha= r (\cos(\theta)+i\sin(\theta) ) \,.$$
Now, can you find the polar forms of $z$?
For second equation, since the equation only has powers of $a^6$, a substitution $t=a^6$ should work... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/95226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Geometric inequality: $2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$ Suppose $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $R$ and $r$ are its circumradius and inradius respectively. How can one prove the following inequality? $$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$
| We need to prove that
$$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$ or
$$2\cdot\frac{4S^2}{(a+b+c)^2}+8\cdot\frac{abc}{4S}\cdot\frac{2S}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$\frac{\prod\limits_{cyc}(a+b-c)}{2(a+b+c)}+\frac{4abc}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$(a^2+b^2+c^2)(a+b+c)-8abc+\sum_{cyc}\left(a^3-a^2b-a^2c+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/95388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
} |
Showing that $(2^a - 1)\bmod (2^b - 1) = 2^{a \; \bmod \; b} - 1 $ I've been thinking on this proof for two days. I'm stuck.
Show that,
$$ (2^a - 1)\bmod (2^b - 1) = 2^{a \! \! \mod b} - 1 $$
where $a,b \in \mathbb{Z}^+$.
I would be happy if someone can help me.
Thanks.
| I have another approach which I think it should be easier.
We must prove that $2^a-1 \equiv 2^{a mod b}-1 \pmod{2^b-1}$
Thus, We should prove
$$2^a \equiv 2^{a mod b} \pmod{2^b-1}$$
Proof: Using dividing algorithm, we have:
$a = bq + r$ then $r = a - bq$
Thus we have: $2^a \equiv 2^r \pmod{2^b-1}$ then $2^a \equiv 2^{a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$
Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:... | let $m=n+1$, so your series reduces to $$\sum_{m=2}^\infty(-1)^{m-1}\frac{m-1}{(2m)!}=-\sum_{m=2}^\infty(-1)^{m}\frac{m}{(2m)!}+\sum_{m=2}^\infty(-1)^{m} \frac{1}{(2m)!}$$
$$=-\sum_{m=2}^\infty(-1)^{m}\frac{1}{2(2m-1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$
Note that $\frac{m}{(2m)!}=\frac{m}{(2m)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/96571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Let $f(x)=x^2+2x$ and $g(x)=x^3$, finding the roots of $f\circ g(x)=g\circ f(x)$ Let $f(x)=x^2+2x$ and $g(x)=x^3$, how many different roots does $f\circ g(x)=g\circ f(x)$ have?
I solved it and found $x=-1$ but it says it has two different roots, could someone clarify?
| Other than the nice answer from Alex, you can also do it concretely:
$$f\circ g(x)=f(x^3)=x^6+2x^3=x^3(x^3+2)$$ and
$$g\circ f(x)=g(x^2+2x)=(x^2+2x)^3=x^3(x+2)^3.$$
Therefore, $f\circ g(x)=g\circ f(x)$ can be written as
$x^3(x^3+2)-x^3(x+2)^3=0$, or equivalently,
$$x^3\Big[(x+2)^3-(x^3+2)\Big]=x^3\Big[(x^3+6x^2+12x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/101586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Perfect Squares ending in 576 I want to find out perfect squares ending in 576, after the number 576.
Here is my derivation to arrive at such a number. Let the perfect square ending in $576$ be $1000k+576$. Every perfect square can be expressed as a the sum of a certain number of consecutive odd numbers. For eg: $2^2 =... | If $x^2=24^2\pmod{1000}$, then $(x-24)(x+24)=0\pmod{1000}$. Thus, exactly one of $x-24$ or $x+24$ must be divisible by $125$ because their product is and if one is divisible by 5, the other is not. Furthermore, both $x-24=x+24=0\pmod{4}$ because they are equal $\!\!\pmod{4}$ and none of $1^2$, $2^2$, or $3^2$ are $0\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Three maximal ideals lying over $3\mathbb{Z}$? A few weeks ago I asked a question about finding the number of maximal ideals lying above $3\mathbb{Z}$ in $B$, where $B$ is the integral closure of $\mathbb{Z}$ in a splitting extension $E\supset\mathbb{Q}$ for the polynomial $f(x)=x^3+x+1$.
I'm reading over the following... | Yes. The integral closure of $\mathbb{Z}$ in a finite extension of $\mathbb{Q}$ is a Dedekind domain, where each (nonzero) ideal factors uniquely as a product of (maximal) prime ideals. For proofs of these facts (and much more), see Robert Ash's notes here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Using the complex logarithm to find the sum of angles in a triangle. Suppose you have a triangle with vertices $a$, $b$, and $c$. I asked earlier how you can define the angles in a triangle based on the $\log$ function. I received the answer that, for instance, the angle at $a$ is found as $\left|\Im\log\left(\frac{c-a... | Flipping the $\frac{a-c}{b-c}$ (or any of the fractions in your angle expressions) actually isn't cheating. First, perhaps more intuitively, when you said that the measure of angle $a$ is $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|$, swapping $b$ and $c$ should not change the measure of the angle, so you should ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
How to find$\int x^2 \arctan(x) \;dx $ I was solving $\int x^2 \arctan(x) \;dx $
I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.
From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$
I got
$$\begin{align*}&\int x^2 \arctan(x) \; dx =\\
&x^3\arctan(x) -\frac12\ln(1+x^{2})-\in... | $$\int{ x^2 \cdot \tan^{-1} x} dx = $$
$$ \tan^{-1} x dx = du $$
$$ x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) = u $$
$$ x^2 = v $$
$$ 2xdx = dv $$
$$\int{ x^2 \cdot \tan^{-1} x} dx = x^2\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)-\int 2x\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)dx$$
$$I = {x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/106827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Finding the value of $\sum k/3^k$
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
please help me with this
$$\sum\limits_{k=0}^{\infty}\frac{k}{3^k}$$
I need just a hint, not a full answer. Thanks!
| There are several ways to do this. One involves the fact that
$$
\frac{k}{3^k} = kx^k = x\cdot kx^{k-1} = x\cdot \frac{d}{dx} x^k
$$
where $x=1/3$, and $\displaystyle\sum_{k=0}^\infty x\cdot \frac{d}{dx} x^k$ can be found.
Another looks like this:
$$
\sum_{k=0}^\infty \frac{k}{3^k} = \sum_{k=1}^\infty \frac{k}{3^k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/107468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I make the following 2 fractions integers? Let $m,n$ be integers. I want to find the possible values of $m,n$ such that $4(m+n)\over (2m+n)^2+3n^2$ and $4n\over (2m+n)^2+3n^2$ are both integers too. Would someone please help? Of course letting $(2m+n)^2+3n^2=4$ gives some good values, but is this all the $m,n$ ... | We can't have $m=n=0$. Let's suppose for a moment that $n=0$ then (using Graphth's answer) $n/(m^2+mn+n^2)=0$ and $m/(m^2+mn+n^2)=1/m$ and $1/m$ is integer only if $m=-1$ or $m=1$. Since the problem is symmetric in $m$ and $n$ we found the solutions $(m,n)=(0,-1),(0,1),(1,0),(-1,0)$.
We know that if $a|b$ and $a,b\neq0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/109003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$ Calculate below limit
$$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$
| The following is an elementary consideration, which shows how to compute the limit in terms of an infinite series. It's evaluation requires usage of Euler's summation formula, already covered by Dane.
Consider the following transformation
$$
\sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \left(\frac{1}{\sqrt{k}} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/109660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\frac{d}{dx} \frac{x^2}{y}$ $$\frac{d}{dx} \frac{x^2}{y}$$
According to Wolframalpha
I "factor out constants"
$$\frac{\frac{d}{dx} x^2}{y}$$
Then I will get $\frac{2x}{y}$. Is that right? But $y$ is not a constant? What I did actually (quotient rule got me stuck)
The actual question is "Find $\frac{d^2y}{dx^2}... | $$
2x^3-3y^2 = 8.
$$
Differentiate both sides with respect to $x$:
$$
6x^2 - 6yy' = 0.
$$
Solve for $y'$:
$$
y' = \frac{x^2}{y}.
$$
Differentiate again with respect to $x$:
$$
y'' = \frac{y(2x)- x^2y'}{y^2}.
$$
Put $x^2/y$ in place of $y'$ and simplify:
$$
y'' = \frac{2xy - x^2\frac{x^2}{y}}{y^2} = \frac{2xy^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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How to prove this property of congruences? Prove that for each $n \in \mathbb{N}, s \in \mathbb{N}$ the following is true
(i) $n \equiv Q_s(n) \left(\bmod\ 10^s - 1\right)$
(ii) $n \equiv Q'_s(n)\left(\bmod\ 10^s + 1\right)$
where
$$Q_s(n) = \sum_{i=0}^{\infty}(a_{is+s-1}\dots a_{is+1}a_{is})$$
for example
$$Q_3 (6154... | Hint: mod $10^s -1$ we have $10^s\equiv 1$, and mod $10^s + 1$ we have $10^s\equiv -1$. Use these congruences as rewrite rules to replace powers of $10^s$ by powers of $\pm 1$ in radix $10^s$ representation.
This leads to simple universal divisibility tests: evaluate a radix polynomial in nested Horner form, using modu... | {
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If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$ $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\... | Change variable, $g(u) = f(\sqrt{u})$. You need to decide what you want for negative $u$. Then this functional equation becomes $g(u+v)=g(u)g(v)$.
| {
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"url": "https://math.stackexchange.com/questions/115784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, ... | Sorry but I don't speak English well but I will try to write as correctly as possible.
Your demonstration is correct, but your idea of cancellation is incorrect.
$x^n−y^n=(x−y)(x^\left(n−1\right)+x^\left(n−2\right)y+⋯+xy^\left(n−2\right)+y^\left(n−1\right))$
When we multiply first for 'x' to our terms of the second par... | {
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"url": "https://math.stackexchange.com/questions/117660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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"answer_id": 4
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Limit of a Sequence involving $\frac {n^2} {\sqrt{n^{6}+k}}$ I am stuck on this problem:
Compute the limit of the sequence $(a_{n})_{n=1}^{\infty}$ defined by
$$a_{n}:=\frac {n^2} {\sqrt{n^{6}+1}}+\frac {n^2} {\sqrt{n^{6}+2}}+\cdot \cdot \cdot + \frac {n^2} {\sqrt{n^{6}+n}}=\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}... | I think this should work, $$\frac{n\cdot n^2}{\sqrt{n^6+k}}\leq \sum_{k=1}^n\frac{n^2}{\sqrt{n^6+k}}\leq \frac{n\cdot n^2}{\sqrt{n^6+1}}$$Taking limits at both sides says the limit is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Divide with remainder $\frac{x^2}{x^2 + x + 2}$ I am having a hard time long dividing:
$$\frac{x^2}{x^2 + x + 2}.$$
Could someone please show a step by step way to divide this, as I can only get it down to : $1 + \frac{x^2}{x + 2}$.
Thank you for your time!
| Since $x^2 - 1 \cdot (x^2 + x + 2) = -x - 2$ (one step of polynomial division), we get $x^2 : (x^2 + x + 2) = 1$, remainder $-x-2$, or equivalently, $$\frac{x^2}{x^2 + x + 2} = 1 + \frac{-x-2}{x^2 + x + 2}.$$ This is, in fact, the final result, since the degree of $-x-2$ is less than that of $x^2 + x + 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Exponent Law of Addition? Is there a rule for adding exponential terms of like bases just like there are rules for multiplying and dividing such terms?
For example we know that:
$x^1 \cdot x^2 = x^{1+2} = x^3$
But what about for addition (or subtract for that matter)?
$x^1 + x^2 = x^?$
If no such pattern exists, why is... | Observe that
$ \log(A+B) = \log(A) + \log \left(\frac{B}{A} + 1 \right) $. Using this fact, one can show that
$
\begin{align}
\log (x^a + x^b) &= \log(x^a) + \log ( x^{b-a} + 1 ) \\
&= a\log x + \log ( x^{b-a} + 1 ) \\
&= \log x \left( a + \frac{\log ( x^{b-a} + 1 )}{\log x} \right) \\
\Rightarrow x^a + x^b &= x^{\lef... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$ Given that $\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}$, where $a,b,c$ are integers. Show $a=b=c=0$
Do I use modular arithmetic?
| Let $\alpha=\sqrt[3]{2}$
Then $m_{\alpha}({x})=x^3-2$ is the minimal polynomial of $\alpha$ over $\Bbb{Q}$.
$f(x)=a+bx+cx^2$ is an anhilating polynomial of $\sqrt[3]{2}$ as $f(\alpha) =0$
Hence $m_{\alpha}(x) \mid f(x) $
$x^3-2\mid a+bx+cx^2$ implies $a=b=c=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework:
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
| Since the integrand is a function of $x$ and $\sqrt{ax^{2}+bx+c}$ another
option is to use the Euler substitution $\sqrt{ax^{2}+bx+c}=\pm
\sqrt{a}x\pm t$, with $a>0$. Choosing $\sqrt{1+x^{2}}=t-x$, squaring both
sides and solving for $x$, we obtain $x=\frac{t^{2}-1}{2t}$ and $dx=\frac{
t^{2}+1}{2t^{2}}dt$. The integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
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$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$ is a $C^1$-function I would like to show that the function:
$$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$$
is a $C^1$-function.
$$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$$
$$ \frac{\partial f... | A partial answer to at least show that $f$ is differentiable at $(0,0)$. Then it remains to show that its derivative is continuous. I didn't check if that also follows from this bound directly.
Combine $\tan(x)\geq x$ and $\sin(x) \leq x$ for $x \in [0,\pi/2)$ to get
$$
\cos(x) \leq \frac{\sin(x)}{x} \leq 1
$$
for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding inverse of polynomial in a field I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take:
In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$.
My understanding is that one needs to use the (Extended?) Euclidean Algorithm a... | Write $f := x^3+2x+1$ and $g := x^2+1$. We want to find the inverse of $g$ in the field $\mathbb F_3[x]/(f)$ (I prefer to write $\mathbb F_3$ instead of $\mathbb Z_3$ to avoid confusion with the $3$-adic integers), i.e. we are looking for a polynomial $h$ such that $gh \equiv 1 \pmod f$, or equivalently $gh+kf=1$ for s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 4
} |
Finding $x^2+y^2+z^2$ given that $x+y+z=0$, $x^3+y^3+z^3=3$ and $x^4+y^4+z^4=15$ I just ran into this:
$$\begin{align}
x^1+y^1+z^1&=0\\
x^3+y^3+z^3&=3\\
x^4+y^4+z^4&=15\\
x^2+y^2+z^2&=\text{?}
\end{align}$$
I know the answer, but can this be solved without guessing?
| Note that $0 = (x+y+z)^2 = (x^2+y^2+z^2) + 2(xy+xz+yz)$, so $x^2+y^2+z^2 = -2(xy+xz+yz)$.
And $$\begin{align*}
15 &= x^4+y^4+z^4\\
& = (x^2+y^2+z^2)^2 - 2(x^2y^2 + x^2z^2 + y^2z^2).\end{align*}$$
And
$$\begin{align*}
(xy+xz+yz)^2 &= (x^2y^2 + x^2z^2 + y^2z^2) + 2(x^2yz+xy^2z+xyz^2)\\
&= (x^2y^2+x^2z^2 + y^2z^2) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellip... | Just keep going:
$$2x^2+2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$$
$$\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 127 - (x^2+y^2)$$
$$(x^2 + (y - 1)^2)(x^2 + (y + 1)^2) = (127 - (x^2+y^2))^2$$
$$x^4 + x^2[(y - 1)^2 + (y + 1)^2] + [(y+1)(y-1)]^2 = 127^2 -2\cdot127\cdot(x^2+y^2)+ (x^2+y^2)^2$$
$$x^4 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Recurrence relation for number of ternary strings that contain two consecutive zeros The question is: Find a recurrence relation for number of ternary strings of length n that contain two consecutive zeros.
I know for ternary strings with length one, there are 0. For a length of 2, there is just 1 (00), and for a len... | Let us count the number of strings $a_n$ that do not contain $00$ as a substring. The required number of strings that contain $00$ is $3^n - a_n$.
Let us call a string that does not contain $00$ as good. Let $x_n, y_n, z_n$ be the number of good strings of length $n$ that end with $0, 1, 2$ respectively. Clearly,
\begi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
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Proving a simple inequality. I am trying to prove that
$\frac{n}{4n^2+1} > \frac{n+1}{4(n+1)^2+1}, \forall n\in\mathbb{N}$.
What I did so far was
$n < n+1\\
\Rightarrow \frac{n^2}{n} < \frac{(n+1)^2}{n+1}\\
\Rightarrow \frac{n}{n^2} > \frac{n+1}{(n+1)^2}\\
\Rightarrow \frac{n}{4n^2} > \frac{n+1}{4(n+1)^2}\\
\Righta... | Inequality is equivalent to the :
$4n^2+4n-1>0$
which is equivalent to the :
$(2n+1)^2>2$
You can use induction to prove this inequality :
*
*for $n=1$ it follows $9>2$
*suppose $(2n+1)^2>2$
*we want to to prove that $(2n+3)^2>2$
Since $(2n+1)^2>2$ it follows :
$4n^2+4n+1 > 2$
$4n^2+4n+1+8n+8>2+8n+8>2$
Hence :
$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/129290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$ How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
| Rewrite it as
$$
I =\int_0^\infty \frac{\log\left(x+\frac{1}{x}\right)}{x+\frac{1}{x}} \frac{\mathrm{d} x}{x} = \left.\frac{\mathrm{d}}{\mathrm{d}s} \mathcal{I}(s)\right|_{s=-1}
$$
where
$$
\mathcal{I}(s) = \int_0^\infty \left( x+\frac{1}{x}\right)^s \frac{\mathrm{d} x}{x} \stackrel{x=\sqrt{\frac{u}{1-u}}}{=} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How do you integrate $\int \frac{1}{a + \cos x} dx$? How do you integrate $\int \frac{1}{a + \cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.
Thanks!
| Also a generalised solution, borrowing from and expanding upon user1357113's answer,
I. For the case $|a| > |b|$, note that the substitution $t=\tan \left( \frac{x}{2} \right)$ is not injective. So to retrieve a continuous antiderivative, we have as a complete answer
$$
\int \frac{1}{a+b\cos(x)} {\rm d}x
= \frac{2}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
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Using linear algebra, how is the Binet formula (for finding the nth Fibonacci number) derived? If possible, please refrain from any type of proof besides linear algebra. So, using the recursion formula $F_{n+1} = F_{n-1} + F_n$, for $n\gt 1$, and where $F_0 = 0$ and $F_1 = 1$, and the Fibonacci matrix, derive the golde... | The Fibonacci numbers are defined by a second-order linear recurrence equation:
$$F_{n+2} = F_{n+1} + F_n$$
This means we can treat the solution of $F_n$ in terms of $n$ as a problem in linear algebra involving only $2$-dimensional vectors. In some sense, what we are doing is modelling this as a dynamical process on a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A 'should be simple' derivative Help! I can't seem to prove that $$\Bigl[\log(x+\sqrt{x^2 + 1})\Bigr]' = \frac{1}{\sqrt{x^2 + 1}}$$ I keep getting some horrible answer namely $$\frac{x + \sqrt{x^2 + 1}}{x\sqrt{x^2 + 1} + 1 + x^2}$$ does this cancel down at all?
| we have in your denominator
$$ x\sqrt{x^2 + 1} + (1 + x^2) = \bigl(x + \sqrt{x^2 + 1}\bigr)\cdot \sqrt{x^2 + 1}$$
so by canceling $x + \sqrt{x^2 + 1}$ we get $\frac 1{\sqrt{x^2 + 1}}$ as wished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/135920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to $\int \sqrt{10x-x^2}dx$ Theres a hint to use $x=5+5\sin{t}$. Ok, but how do I know what substitution to use if a hint wasn't given? Is it "trivial" or perhaps, its very unlikely that that will appear?
Anyways, I did:
$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\
= \int \sqrt{50+50\sin{t} - (25+50\sin{t} +... | put dx = 5cost dt
from your fourth step,
5∫cost * 5cost dt = 25∫(cos^2)(t) dt
= 25∫(1 + cos2t)/2 dt
= (25/2)∫(1 + cos2t)dt
= (25/2)(t + (sin2t)/2) + c
=(25/2)(t + (2 sint cost)/2) + c
and substitute the value of t and sint and cost in above. Then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Find tangent line of curve that intersects point. How do I find the tangent line of the curve $y=x^2$ that intersects the point $(8,2)$?
| Starting with Andre's equation of
$y-a^2=2a(x-a)$,
$y=2ax-2a^2+a^2 = 2ax-a^2$,
or $a^2-2ax = -y$
or $a^2-2ax + x^2 = x^2-y$
or $(a-x)^2 = x^2-y$
or $a = x \pm \sqrt{x^2-y}$.
Note that this requires $x^2 \ge y$,
which means that the point that the tangent goes through
has to be below (or on) the parabola
and that, if th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/137065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Evaluating $\int |x|^3 \; dx $ $$\int |x|^3 \; dx $$
In my module it is suggest to use integration by parts,
$$ \text{ Set }I = \int (|x|^3 \cdot 1) \; dx = |x|^3 \cdot x - \int \color{red}{\frac {x^3}{|x|^3}3x^2}\cdot x \; dx$$
$$ \implies I = |x|^3 \cdot x - \color{red}{3\int |x|^3\;} dx$$
$$ \implies I = \frac 1 4... | Let $\displaystyle I= \int|x|^3 dx$
\begin{align*}I =& \int x^2|x| dx \\
=& |x|\int x^2 dx - \int\frac{d}{dx}(|x|)\left(\int x^2 dx \right)dx \\
=&\frac{|x|x^3}{3} - \int\left(\frac{|x|}{x}\cdot\frac{x^3}{3}\right)dx \text{ since } \frac{d}{dx}(|x|) = \frac{|x|}{x} \\
=&\frac{|x|x^3}{3} - \frac{1}{3}\int|x|x^2dx \\
=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/137943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$
Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$
So, taking $\tan$ of both sides I get:
LHS $=\frac{2\tan(\tan^{-1}(2))}{1 - \tan^2(\tan^{-1}(2))} = -\frac{4}{3}$
and
RHS $= \tan(\pi - \cos^{-1}(\frac{3}{5})) = ~...$
I wasn't sure how to treat the RHS g... | From this or Ex$\#5$ of Page $\#276$ of this,
$$\tan^{-1}x+\tan^{-1}y=\begin{cases} \tan^{-1}\frac{x+y}{1-xy} &\mbox{if } xy<1 \\\pi+ \tan^{-1}\frac{x+y}{1-xy} & \mbox{if } xy> 1. \end{cases} $$
$$\implies2\tan^{-1}x=\begin{cases} \tan^{-1}\frac{2x}{1-x^2} &\mbox{if } x^2<1\iff -1\le x\le1 \\\pi+ \tan^{-1}\frac{2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$?
Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.
I tried hard but nothing seems to work. :(
| Note that
$$\left(a+\frac{1}{2}\right)^n + \left(b+\frac{1}{2}\right)^n=\frac{1}{2^n}\left(c^n+d^n\right),$$
where $c=2a+1$ and $d=2b+1$.
Let $2^e$ be the highest power of $2$ that divides $c+d$.
If $n$ is even, the highest power of $2$ that divides $c^n+d^n$ is $2$.
For odd $n$, note that $u^n+v^n=(u+v)(u^{n-1}+u^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
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A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere.
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of t... | Hint: $$(\tan\theta+\cot \theta)^2=\left(\frac{\sin\theta}{\cos \theta} +\frac{\cos \theta}{\sin \theta}\right)^2$$ $$= \left(\frac{\cos^2 \theta+\sin^2\theta}{\cos \theta \sin \theta}\right)^2.$$ Now try using the fact that $\cos^2\theta+\sin^2\theta=1$, twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Inverse of $f^{-1}(x)=x^5+2x^3+3x+1$ question Let $f$ be a one-to-one function whose inverse function is $f^{-1}(x)=x^5+2x^3+3x+1$.
Compute the value of $x_0$ such that $f(x_0)=1$.
I am confused as to what this question is asking me, particularly since I don't understand the subscript under the $x$ variable.
| $f^{-1}(x)=x^5+2x^3+3x+1$.
$f(f^{-1}(x))=f(x^5+2x^3+3x+1)$.
$f(f^{-1}(x))=x$ as function property
$x=f(x^5+2x^3+3x+1)$.
$f(x^5+2x^3+3x+1)=x$
$f(x_0)=1$
You can see that $x=1$ so $x_0=x^5+2x^3+3x+1$
$x_0=1^5+2.1^3+3.1+1=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Derivative of $\ln\sqrt{\frac{e^{x^2}}{e^x+2}}$?
Being: $\ln f(x)=\log_ef(x)$
I started derivating $$\ln\sqrt{\dfrac{e^{x^2}}{e^x+2}}$$ but I get to a point that I don't know how to follow.
I try to get it by derivating the logarithm directly and by using the logarithmic properties such us: $$\ln f(x)^n=n\ln f(x)$$ a... | A check, if you trusting your integration skills more than your derivating ones:
Integrate your $f'(x)$, and if you compute your derivative correctly, you will then get your original function, $f(x)$ with some constant. (For more reading: Fundamental theorem of calculus)
$$\int\frac{x(x^2-5)}{x^2-4}\ dx=\int (x-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$ I met an inequality, I ask, do not mathematical induction to prove that:
Prove \[ \left(\frac n2\right)^n > n! > \left(\frac n3\right)^n \] without using induction
| Let $a_n =\displaystyle \frac{2^n n!}{n^n}.$ Note that $a_6 = 80/81 < 1.$ We also have $$ \frac{a_{n+1}}{a_n} = \frac{2^{n+1} (n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n}{2^n n!} = 2 \left(\frac{n}{n+1}\right)^n < 1.$$ The sequence $$x_n = \left(1- \frac{1}{n+1} \right)^n$$ is monotonically decreasing to $1/e.$ Since $e>2$,... | {
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"url": "https://math.stackexchange.com/questions/144176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
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Showing this sequence converges. Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?
I did the following.
Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$
\begin{align}
\left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \... | For sequences of complex numbers it occurs that a given sequence of complex numbers $a_{n}=x_{n}+y_{n}i$ converges to some complex number $x+yi$ iff sequence $x_{n}$ converges to $x$ and as one may expect $y_{n}$ converges to $y$. Sequence diverges if at least one of the mentioned conditions does not hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/147109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_... | First write
$$\left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] = \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
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Find multiplicative inverse existense. I cannot find whether multiplicative inverse of $x^3+x^2+x+1 \pmod{x^5+x^4+x^3-x^2-x+1}$ over $\mathrm{GF}(3)$ exists. This problem must be solved with Extended Euclidean algorithm.
I tried to divide $x^5+x^4+x^3-x^2-x+1$ by $x^3+x^2+x+1$.
I think I divided it wrong. I got $x^2-... | The problem you are trying to solve is not clearly stated, but what is clear is that you are having some difficulty with polynomial division, or (more likely) just think that you are. What probably happened is something that happens to everybody, a slip with signs.
We will do ordinary polynomial division. To start... | {
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"answer_id": 0
} |
Find an equation for the plane that contains the following line and passes through point P How do you determine the plane which contains the line
\begin{align}
x & = -1 + 3t \\
y & = 5 + 2t \\
z & = 2 + t
\end{align}
and passes through the point $P = (2,4,-1)$?
| the line is
\begin{equation}
(-1,5,2) + t(3,2,1)
\end{equation}
Then the plane contains the segment
$\overline{(-1,5,2)(2,4,-1)}$ $\quad((-3,1,3) = (-1,5,2) -((2,4,-1))$ and the plane $\Pi$ has normal $(3,2,1) \times (-3,1,3) = (5,-12,9)$. Hence the plane is $5x -12y + 9z + d = 0$. As $(2,4,-1) \in \Pi$, we have $d = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/151292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sin x + \sin x \cdot \cot^2 x = \csc x $ The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process:
$$\begin{align*}
\tan \theta &= \frac{\sin \theta}{\cos \theta}\\
(\sin \theta)^2 + (\cos \theta)^2 &... | Start with $\sin x + \sin x\cot^2 x$. Factoring out $\sin x$ we get
$$\sin x + \sin x\cot^2 x = \sin x (1 + \cot^2 x).$$
Now use the definition of $\cot x$ and rewrite the expression in the parenthesis as a single fraction:
$$\begin{align*}
\sin x + \sin x\cot^2x &= \sin x\Bigl( 1 + \cot^2 x\Bigr)\\
&= \sin x \left( 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/152583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Summing the series $ \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}$ How could I find the sum of the series
$$ \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}$$
With: $$ \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}=\frac{2\pi\Gamma(n-1/3)}{\Gamma(2/3)3^{3/2}n!}$$
(Previous post)
?... | Since integrals are taken over positive measurable functions we can interchange integration and summation
$$
\sum_{n=1}^{\infty} \int_0^{\infty} \frac{dx}{n(1+x^3)^n}=
\int_0^{\infty} \sum_{n=1}^{\infty}\left(\frac{1}{n(1+x^3)^n}\right)dx
$$
Consider the following Taylor expansion
$$
\log(1-q)=-\sum\limits_{n=1}^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral of $\int \frac {dx}{x^2 \sqrt{4-x^2}}$ I am trying to find $$\int \frac {dx}{x^2 \sqrt{4-x^2}}$$
I make $t=2\sin x$
$$\int \frac {dx}{x^2 \sqrt{4-4\sin^2 t}}$$
$$\int \frac {dx}{x^2 \sqrt{4(1-\sin^2 t)}}$$
$$\int \frac {dx}{x^2 \sqrt{4(\cos^2 t)}}$$
$$\int \frac {dx}{x^2 \cdot 2\cos t}$$
I do not really know w... | You must replace every occurrence of $x$. Let $x=2\sin t$. This is so that the square root of $4-x^2$ will be nice. Then $dx=2\cos t\,dt$, and $\sqrt{4-x^2}=2\cos t$. So we end up needing to find
$$\int \frac{2\cos t}{(4\sin^2 t)(2\cos t)}dt.$$
There is cancellation, and we end up needing to find $\int\frac{1}{4}\csc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Parametric representation of rectangular form in terms of parameters $\rho$ & $\theta$
I need to represent the cone $z=\sqrt{3x^2+3y^2}$ parametrically in terms of $\rho$ and $\theta$ where $(\rho,\theta,\phi)$ are spherical coordinates.
Attempt. I tried using: $$x=\rho\sin\phi\cos\theta \\y=\rho\sin\phi\sin\theta\\z... | If you have $z=\sqrt{3x^2+3y^2}$ and $z^2=3x^2+3y^2$, making the appropriate replacements in $\cos\phi=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$ gives
$$\cos\phi=\dfrac{z}{\sqrt{\frac{z^2}{3}+z^2}}$$
Can you see what to do next? (Hint: your Cartesian equation guarantees that $z$ is always positive, so $\sqrt{z^2}=?$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Plotting a quadratic equation in the $\,xy\,$- plane My question is:
Represent the following set of points in the $\,xy\,$- plane:
$$\left\{ (x,y)\,\, |\,\, x^2 + y^2 - 2x - 2y + 1 = 0 \right\}$$
What i got: $\,\,(x-2)^2 + (y-2)^2 = 1\,\,$
I am not getting what to do next.
Any help to solve this question would be grea... | When given an equation of the form $x^2-2x+y^2-2y+1=0$ the first step is to complete the square for $x$ and for $y$.
The idea is that if we have $x^2-2x$ we can write it as $(x+C)^2+D$ instead. Since know those that the coefficient of $x$ is $2C$, we know that $C=-1$, so we have: $$(x-1)^2=x^2-2x+1\implies x^2-2x = (x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Multiplication in the field $F = \mathbb{Z}_2[x]/f(x)$ Let $f(x) = x^6 + x + 1$ and define the field $F = \mathbb{Z}_2[x]/f(x)$
Compute the following in this field:
1. $(x^5 + x + 1)(x^3 + x^2 +1)$
I start by multiplying (in $\mathbb{Z}_2[x]$):
$(x^5 + x + 1)(x^3 + x^2 +1)$ = $(x^8 + x^7 + x^5 +x^4 + x^2 + x + 1)$
T... | Your idea for (1) is correct.
As for (2), it is the same idea as for the ring of integers modulo $n$. The $gcd$ of $x+1$ and $x^6+x+1$ will be $1$ since $x+1$ and $x^6+x+1$ are coprime. Using the extended Euclidean algorithm you'll get
$$
A(x+1) + B(x^6+x+1) = 1
$$
where $A, B \in \Bbb{F}_2[x]/f(x)$. So as you can see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sum of Cosine/Exponential I try to simplify to get rid of sum
$$ \sum_{k=0}^{n-1}\cos(2 \pi fk)$$
I discover I shall use euler equation to form:
$$ \sum_{k=0}^{n-1}\frac{1}{2}(e^{2 \pi fki}+e^{-2 \pi fki})$$
but how to sum exponentials?
| You could proceed using the fact that $\cos(k \theta) = \dfrac{\exp(i k \theta) + \exp(-i k \theta)}{2}$, where $\theta = 2 \pi f$, as others have suggested.
Here is another method. Let $I_n = \displaystyle \sum_{k=0}^{n-1} \cos(k \theta)$. Now multiply by $\sin \left( \dfrac{\theta}{2}\right)$. Hence,
\begin{align}
\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Hypergeometric functions inequality Let $_2F_1(a,b;c,z)$ be the (Gauss) hypergeometric function, and $m$ and $n$ positive integers.
From a simple plot it looks like
$_2F_1(m+n,1,m+1,\frac{m}{m+n})>\frac{m}{n} \,_2F_1(m+n,1,n+1,\frac{n}{m+n})$
when $m<n$, but how do I prove this?
Hope this is not too trivial, I am not v... | Using Euler's transformation of the Gauss's hypergeometric function:
$$
{}_2 F_1\left(a,b; c; z\right) = (1-z)^{c-a-b} \cdot {}_2F_1\left(c-a,c-b; c; z\right)
$$
for $b=1$, $a=m+n$ and $c=m+1$ we have a "closed form" expression:
$$ \begin{eqnarray}
{}_2 F_1\left(m+n,1; m+1; \frac{m}{m+n}\right) &=& \left(\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| My hint:
$$\int \frac{1}{1+x^4}dx=\frac{1}{2}\left[\int \frac{1+x^2}{1+x^4}dx+\int\frac{1-x^2}{1+x^4}dx\right]=\frac{1}{2}\left[\int \frac{1}{\left(x-\frac{1}{x}\right)^2+2}d\left(x-\frac{1}{x}\right)+\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}d\left(x-\frac{1}{x}\right)\right]=\frac{1}{2}\left[\frac{1}{\sqrt{2}}\arct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 2
} |
To simplify $f_a(x)= \int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}\;dt$ Let $x\leq0$, then $$ f_a(x)= \int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}\;dt$$
$$ f'(x)= -\int_{-a}^{+a} \frac{1}{t^2-a^2} e^ {-\frac{x}{t^2-a^2}}dt$$
$$ f'(x)= -\int_{-a}^{+a} \frac{t^2-(t^2-a^2)}{a^2(t^2-a^2)} e^ {-\frac{x}{t^2-a^2}}dt$$
$$ f'(x)= ... | How about
$$
f_a(x) = \frac{\operatorname{e} ^{x/(2a^2)} x}{a} \biggl(\mathrm K_0 \biggl(\frac{-x}{2 a^{2}}\biggr) - \mathrm K_1 \biggl(\frac{-x}{2 a^{2}}\biggr)\biggr),\qquad x<0
$$
in terms of Bessel functions.
added
It should work like this: The function $f$ listed on the right-hand-side satisfies the differential ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding all integer solutions for $x^2 - 2y^2 =2 $ I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2... | Another way is to note that $x$ has to be even. Hence if $x=2x_1$, we get that
$$2x_1^2-y^2 = 1 \implies y^2 - 2x_1^2 = -1 \implies \left(y + \sqrt2 x_1\right)\left(y - \sqrt2 x_1\right) = -1$$
Now clearly, one solution is $(x_1,y) = (1,1)$.
Raise both sides to any odd power, i.e.,
$$\left(y + \sqrt2 x_1\right)^{2n-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim \limits_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$ I have difficulty with this limit. Where to start?
$$\lim_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$$
| By simple integration by parts, we have
$$ \int_{0}^{y} \frac{\log x}{\sqrt{1+x^2}} \; dx = \log y \, \sinh^{-1}y - \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx. $$
Now by the substitution
$$x = \frac{u^2-1}{2u} \quad \Longleftrightarrow \quad u = x + \sqrt{x^2+1},$$
and the easy equality $ \sinh^{-1} y = \log \left( y + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Count the number of solutions of the inequality $x + y + z \leq N$
Problem
Given $A, B, C $ and $N$. How many integer solutions are there of the following inequality:
$$x + y + z \leq N$$
where $0 \leq x \leq A, 0 \leq y \leq B, 0 \leq z \leq C$?
When $A + B + C \leq N$, the solution is obvious $(A + 1) \cdot (... | The main trick is to add a fourth variable to take up the slack between $x+y+z$ and $N$. You know that there are $\binom{N+3}3$ solutions to $w+x+y+z=N$ in non-negative integers, so there are also $\binom{N+3}3$ solutions to $x+y+z\le N$ in non-negative integers. Now we have to throw out the solutions that exceed one o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/165547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the value of a succession of additions $$1^2+2^2+3^2+...+10000$$
How do you find the exact value of that?
I'm studying induction, and I'm still not sure how to get that value.
| The sum of squares formula is
$\sum_{i= 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$
Clearly it holds for $n = 1$.
Suppose it held for $n$. That is
$\sum_{i= 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$
Then $\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2 = \frac{2n^3 + 2n + n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/166870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead... | We have that
$$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n>0 \iff \ln \left(1+\frac{1}{n+1}\right)^{(n+1)} - \ln \left(1+\frac{1}{n}\right)^n>0$$
and
$$(n+1) \ln \left(1+\frac{1}{n+1}\right) - n \ln \left(1+\frac{1}{n}\right)
\ge(n+1)\left[ \ln \left(1+\frac{1}{n+1}\right) - \ln \left(1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 14,
"answer_id": 10
} |
Pythagorian quadruples From my work on hyperelliptic equations I found how to get infinitely many solutions of the equation $a^4+b^4+c^2=d^4$. I call these solutions harmonic:
$$\begin{array}{rcccccl}
1^4 &+& 2^4 &+& 8^2 &=& 3^4\\
2^4 &+& 3^4 &+& 48^2 &=& 7^4\\
3^4 &+& 4^4 &+& 168^2 &=& 13^4\\
4^4 &+& 5^4 &+& 440^2 &=&... | EDIT
There seems to be a lot of non-harmonic solutions.
Here are couple of one parameter family of non-harmonic solutions.
$b = a(a-1)$, $d = a^2 - a + 1$ and $c = (a-1)(2a^2 - a +1)$ which relies on the identity $$a^4 + \left( a(a-1)\right)^4 + \left((a-1)(2a^2-a+1) \right)^2 = (a^2 - a + 1)^2$$ You could scale thes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the area enclosed by the curve $r=2+3\cos \theta$. the question is
Find the area enclosed by the curve:
$r=2+3\cos \theta$
Here's my steps:
since when $r=0$, $\cos \theta=0$ or $\cos\theta =\arccos(-2/3)$.
so the area of enclosed by the curve is 2*(the area bounded by $\theta=\arccos(-2/3)$ and $\theta=0$)
the a... | Area of curve, $$A=2\int_0^{\arccos(-2/3)}\frac{r^2}{2}d\phi$$ $$\implies A=2\int_0^{\arccos(-2/3)} \frac{(2+3\cos\phi)^2}{2}d\phi$$ $$=2\int_0^{\arccos(-2/3)} \frac{4+9\cos^2\phi+12\cos\phi }{2}d\phi$$ $$=\int_0^{\arccos(-2/3)} (4+9\cos^2\phi+12\cos\phi) d\phi$$ $$=3\sqrt{5}+\frac{17}{2}\cos^{-1}(\frac{-2}{3}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral of $e^{(x-x^3)/3n}$ from $0$ to $\infty$ How can you compute the following integral assuming $n>0$?
$$\int_{x=0}^{\infty}e^{\frac{x -x^3}{3n}}dx $$
Mathematica etc. fail to produce anything useful.
EDIT: I would be happy with an asymptotic result in $n$ if it is too hard to compute exactly. I don't know if it... | I guess Maple has one of the useless answers:
$$\int_{0}^{\infty} \operatorname{e} ^{\frac{-x (-1 + x^{2})}{3 n}} d x = \frac{i}{9} \pi \mathrm{BesselY} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) \sqrt{3} - \\
\quad{}\quad{}\frac{i}{9} \pi \mathrm{AngerJ} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Determine $a$ values allowing $x^2+ax+2$ to be divided by $x-3$ in $\mathbb Z_5$ Determine for which $a$ values $f = x^2+ax+2$ can be divided by $g= x-3$ in $\mathbb Z_5$.
I don't know if there are more effective (and certainly right) ways to solve this problem, I assume there definitely are, but as I am not aware of ... | $x-a$ is the factor of polynimial $f(x)$ iff $f(a)=0\implies (x-3)$ divides $f(x)=x^2+ax+2$ iff $f(3)=11+3a\equiv 0\pmod 5\implies 3a\equiv-11\pmod 5\implies 9\pmod 5\implies a\equiv3\pmod 5$. In your solution you are getting $(x+1)(x-3)=x^2-2x-3\equiv x^2+3x+2\pmod 5$ as $-2\equiv 3\pmod 5$ and $-3\equiv 2\pmod 5$. so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Write each expression in the form $ca^pb^q$ Write each expression in the form $ca^pb^q$
c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$
\begin{align*}
&= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b... | If you remember that dividing by a fraction is the same as multiplying by its inverse, we get at once:
$$c)\,\,\frac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}=\frac{2a}{b}\frac{a}{3}=\frac{2a^2}{3b}=\frac{2}{3}a^2b^{-1}$$
$$(e)\,\,\frac{a^{-1}}{b^{-1}a^{1/2}}=a^{-1-1/2}\,b=a^{-3/2}\,b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int^1_0 \frac{dx}{x^2+x+1}$ Here is an integral which I don't know the answer:
$$\int^1_0 \frac{dx}{x^2+x+1}$$
I tried to use complex number to solve i.e. the root of $x^2+x+1$ is $(-1/2 + \sqrt {3}i/2)$.
Let w=$(-1/2 + \sqrt {3}i/2)$ , then it becomes $ \int_{0}^{1} 1/(x-w)^2\,\mathrm{d} (x-w)$ , the answ... | \begin{eqnarray}
\int_{0}^{1} \dfrac{1}{x^2+x+1} dx &=& \int_{0}^{1} \dfrac{1}{(x+1/2)^2 +3/4} dx \\
&=& \int_{1/2}^{3/2} \dfrac{1}{y^2 +3/4}dy \\
&=& \dfrac{4}{3}\int_{1/2}^{3/2} \dfrac{1}{(\dfrac{2}{\sqrt{3}}y)^2 + 1} dy \\
&=& \dfrac{\sqrt{3}}{2} \dfrac{4}{3} \int_{\sqrt{3}/3}^{\sqrt{3}} \dfrac{1}{z^2+1} dz\\
&=& 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How to solve this equation involving $()^x$? I have the equation:
$\left (\sqrt{3+2\sqrt{2}} \right )^x- \left (\sqrt{3-2\sqrt{2}} \right )^x=\frac{3}{2}$
I wrote the left side of the equation as square roots.
$(1+\sqrt{2})^x-(1-\sqrt{2})^x=\frac{3}{2}$
How do I found out the final solution? Thank you very much!
P.S. T... | $$(\sqrt{3+2\sqrt{2}})^x-(\sqrt{3-2\sqrt{2}})^x=\frac{3}{2}$$
Multiply both side with $(\sqrt{3+2\sqrt{2}})^x$, then you get
$$(\sqrt{3+2\sqrt{2}})^{2x}-(\sqrt{(3-2\sqrt{2})(3+2\sqrt{2})})^x=\frac{3}{2}(\sqrt{3+2\sqrt{2}})^x$$
$$(\sqrt{3+2\sqrt{2}})^{2x}-(1)^x=\frac{3}{2}(\sqrt{3+2\sqrt{2}})^x$$
Let $y =(\sqrt{3+2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find center, radius and a tangent to $x^2+y^2+6x-4y+3=0$
For the circle $x^2+y^2+6x-4y+3=0$ find
a) The center and radius
b) The equation of the tangent line at the point $(-2,5)$
Now, I solved a) and got the equation
$$(x+3)^2+(y-2)^2=10$$ with center $=(-3,2)$ and radius $=\sqrt{10}$
Now, I've never learned ... | Let us complete the square. We get $x^2+6*x+9+y^2-4*y+4-9-4+3=0$ or $(x+3)^2+(y-2)^2=10$. Your circle is with center $O(-3,2)$ and radius $r=\sqrt{10}$
For the second question, let us calculate slope, which would be $\text{rise}/\text{run}$ or in our case center is $(-3,2)$ so slope=$(5-2)/(-2+3)=3$ slope of tange... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Inequality problems I have Maths test tomorrow and was just doing my revision when I came across these two questions. Would anyone please give me a nudge in the right direction?
$1)$ If $x$ is real and $$y=\frac{x^2+4x-17}{2(x-3)},$$ show that $|y-5|\geq2$
$2)$ If $a>0$, $b>0$, prove that $$\left(a+\frac1b\right)\left... | For the first problem:
Write it as
$$
\begin{eqnarray}
\left(y-5\right)^2-4&=&\left(\frac{x^2+4x-17}{2(x-3)}-5\right)^2-4\\
&=&\left(\frac{x^2+4x-17-10x+30}{2(x-3)}\right)^2-4\\
&=&\left(\frac{x^2-6x+13}{2(x-3)}\right)^2-4\\
&=&\frac{(x^2-6x+13)^2 - 16(x-3)^2}{4(x-3)^2}\\
&=&\frac{169-156 x+62 x^2-12 x^3+x^4 - 16x^2+9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding a pair of elements to satisfy an inequation Let $F$ be a field of characteristic 2 with more than 2 elements. Show that there are elements $a$ and $b$ in $F$ such that $(a+b)^3 \not= a^3 + b^3$.
$F$ couldn't possibly have less than 2 elements, and if it had exactly 2 — that is, $F = \mathbb Z_2$ —, $(a+b)^3$ w... | In a field, product is $0$ iff atleast one of the element in product is $0$. Here, if $a\neq0,b\neq 0$ and $a\neq -b$ then , $ab(a+b)\neq 0$, so choose non-zero elements of field $\Bbb F$ such that one is not the additive inverse of other one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ How would I verify the following double angle identity.
$$
\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B
$$
So far I have done this.
$$
(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)
$$But I am not sure how to proceed.
| Your question involves the basic algebra identity which says, $(a + b)(a - b) = a^2 - b^2 $. For targeting your question, it is easy to assume $ a = \sin A\cos B $ and $b = \cos A \sin B$. The process becomes easy now.
$$\begin{align}(a + b)(a - b)& =& a^2 - b^2\\
& = & (\sin A \cos B)^2 - (\cos A \sin B)^2\\ & = & \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 5
} |
Confusing double angle identity How would I solve the following double angle identity.
$$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$$
So far my work is
$$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$$
But how would I proceed.
| Notice that
\begin{eqnarray}
\cos(2x)&=& \cos^2 x - \sin^2 x \\
&=& 2 \cos^2 x - 1.\\
\end{eqnarray}
Then
\begin{equation}
\cos^2 x = \dfrac{1}{2}(1+\cos(2x)).
\end{equation}
Hence,
\begin{eqnarray}
\cos^4 x &=& (\cos^2 x)^2\\
&=& \left[\dfrac{1}{2}(1 + \cos(2x))\right]^2\\
&=& \dfrac{1}{4}(1 +2 \cos(2x)+ \cos^2(2x))\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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