Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Taylor series expansion for $f(x)=\sqrt{x}$ for $a=1$ I seem to be stuck defining an alternating sequence of terms in this series because $f^{(0)}(x)=f(x)$ is positive, as well as $f'(x)$, but then every other term starting with $f''(x)$ is negative. How can I define $f^{(n)}(x)$ given this?
\begin{array}{ll}
f(x)=... | This is a formula which won't display so well in a comment.
$$1\cdot3\cdot5\cdot7=\frac {1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}{2\cdot4\cdot6}=\frac {7!}{2^33!}$$ You should be able to work out the general term from there.
Note also that there is no reason that every term of the sum has to fit the same neat formula. Yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_2^\infty{\frac{3x-2}{x^2(x-1)}}$ To be shown that $\int_2^\infty{\dfrac{3x-2}{x^2(x-1)}}=1-\ln2$
My thought: $\dfrac{3x-2}{x^2(x-1)}=\dfrac{3x}{x^2(x-1)}-\dfrac{2}{x^2(x-1)}$
• $\dfrac{3x}{x^2(x-1)}=\dfrac{3}{x(x-1)}=\ldots=-\dfrac{3}{x}+\dfrac{3}{x-1}$
• $\dfrac{2}{x^2(x-1)}=\ldots=-\dfrac{2}{x^2}+\dfra... | HINT:
Using Partial Fraction Decomposition,
$$\dfrac{3x-2}{x^2(x-1)}=\frac A{x^2}+\frac Bx+\frac C{x-1}$$
$$\implies 3x-2=A(x-1)+Bx(x-1)+Cx^2=(B+C)x^2+x(A-B)-A$$
Comparing the constants in the above identity $A=2$
Comparing the coefficients of $x,A-B=3\implies B=A-3=-1$
Comparing the coefficients of $x^2,B+C=0\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding indefinite integral $$\int\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{x}}\, dx$$
Can someone give a hint as to how to break this apart into a do-able integral?
Just spent ages trying to substitute $\sqrt{a-x}$ and $\sqrt{x}$ as $u$, without success.
Advice/help much appreciated.
| Note that $$\frac{{\sqrt {a - x} }}{{\sqrt a - \sqrt x }} = \frac{{\sqrt {a - x} }}{{a - x}}\left( {\sqrt a + \sqrt x } \right) = \frac{{\sqrt a + \sqrt x }}{\sqrt{a - x}}$$
This allows to get an easy integral in the first summand. For the second one, we make a series of elementary yet (maybe) non-obvious manipulati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the determinant of the following general matrix Let $A_r$ and $B_r$ be the $r\times r$ matrix blocks
$A_r=\left(
\begin{array}{A}
1-t & t^2 & 0 & 0 & 0 & \cdots & 0 \\
t^2 & 0 & 0 & 0 & 0 & \cdots & 0 \\
0 & 0 & t^2 & 0 & 0 & \cdots & 0 \\
0 & 0 & 0 & t^2 & 0 & \cdots & 0 \\
0 & 0 & 0 & 0 & t^2 & \cdots & 0... | Here is a partial answer. For convenience, let us drop the subscript $r$. Write $C=\pmatrix{A&R\\ S&T}$, where $T$ is $r(s-1)\times r(s-1)$. Note that
$$
T^{-1} =
\left[\begin{array}{rrrrr}
A^{-1}&-A^{-1}BA^{-1}&A^{-1}BA^{-1}BA^{-1}&\cdots&(-1)^{s-2}(A^{-1}B)^{s-2}A^{-1}\\
&A^{-1}&-A^{-1}BA^{-1}&\ddots&\ddots\\
&&\ddot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/444318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.
I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $v^2$ and we need to find a maximum of
$$abc-ab-ac-bc+a+b+c-1$$ or
$$6u-3v^2-1,$$
which is linear function of $v^2$,
which says that it's enough to find a maximum for an extremal value of $v^2$,
which happens for equality case of tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/445293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Is it possible to calculate $3^{-1}\equiv ?\pmod{10}$? If I wanted to calculate $3^{-1}\equiv ?\pmod{10}$ would I first calcuate $3^1$ which is just
$3\equiv 3\pmod{10}$ and then divide both sides by $3^2$ which would get $3^{-1}\equiv 3^{-1} mod{10}$ Then im not sure what to do next. My book states that $3^{-1}\equi... | A number $x^{-1}$ is one such that $x \cdot x^{-1} = 1$ (here $\mod 10$). In this case it is easy, because there is only 10 possibilities: $0,1,2,3,4,5,6,7,8,9$, with product being $0,3,6,9,2,5,8,1,4,7$ respectively. That means there exists, and there exists only one number such that $$x \cdot x^{-1} = 1 \mod 10.$$ The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/446364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Simplifying the expression $(1+\sqrt[4]3)/(1-\sqrt[4]3)+1/(1+\sqrt[4]3)+2/(1+\sqrt{3})$ Can anyone give provide me some help to simplify this expression?
The three denominators are pretty much different, and I can't find a common denominator.
$$\frac{1+\sqrt[4]3}{1-\sqrt[4]3}+\frac1{1+\sqrt[4]3}+\frac2{1+\sqrt{3}}$$
Th... | Putting $\sqrt[4]3=x$
$$\frac{1+\sqrt[4]3}{1-\sqrt[4]3}+\frac1{1+\sqrt[4]3}+\frac2{1+\sqrt{3}}$$
$$=\frac{1+x}{1-x}+\frac1{1+x}+\frac2{1+x^2}$$
$$=\frac x{1-x}+\frac1{1-x}+\frac1{1+x}+\frac2{1+x^2}$$
$$=\frac x{1-x}+\frac{1+x+1-x}{(1-x)(1+x)}+\frac2{1+x^2}$$
$$=\frac x{1-x}+2\left(\frac1{1-x^2}+\frac1{1+x^2}\right)$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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For what natural numbers is $n^3 < 2^n$? Prove by induction Problem
For what natural numbers is $n^3 < 2^n$?
Attempt @ Solution
*
*For $n=1$, $1 < 2$
*Suppose $n^3 < 2^n$ for some $n = k \ge 1$
*It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$
*But, how can I prove this through induction?
|
Claim: If $n \in \{0,1\} \cup \{n \in \Bbb{N} \mid n \ge 10\}$, then $n^3<2^n$.
You have verified the claim for $n=0$ and $n=1$. The next time the inequality holds is at $n=10$, since $10^3=1000<1024=2^{10}$. This is our base case.
Induction Hypothesis: Assume that $n^3<2^n$ holds true for $n=k\ge10$.
It remains to p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$
*
*Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$.
*
*Represent $I(x)$ by a power series $\sum^{\infty}a_{n}x^{n}$.(Find $a_{n}$)
*What is the radius of convergence of $I(x)$ ?
*Two curves are generated by polar equations $r=1+\sin\theta$ and $r=-\sin\theta$.
*
*Find the area o... | HINT:
For the first problem, $$\frac x{1-x^8}=\frac x2\left(\frac1{1-x^4}+\frac1{1+x^4}\right)$$
$$=\frac x4\left(\frac1{1-x^2}+\frac1{1+x^2}\right)+\frac12\frac x{1+x^4}$$
$$=\frac x4\left(\frac1{1-x^2}\right)+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$
$$=\frac 18\left(\frac{1+x-(1-x)}{1-x^2}\right)+\frac14\frac x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $ \frac{2x}{2+x}$ provide a particularly tight lower bound for $\ln(1+x)$ for small positive values of $x$? EDIT:
My question was poorly worded.
I wasn't asking about showing $\ln(1+x) > \frac{2x}{2+x}$ for $x>0$.
What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for sma... | As I mentioned in a comment, the power series near $x=0$ are
$$
\begin{align}
\log(1+x)&=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\\
\frac{2x}{2+x}&=x-\frac{x^2}{2}+\frac{x^3}{4}+O(x^4)\\
\log(1+x)-\frac{2x}{2+x}&=\frac{x^3}{12}+O(x^4)
\end{align}
$$
So near $x=0$, the value, and the first and second derivatives match. That... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The indu... | For the second part note that $$2^{2n}=(1+1)^{2n}=1+\binom {2n}1 +\dots + \binom {2n}n+\dots\gt \binom{2n}n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 0
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Show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing. How do you show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing?
I understand that the binomial theorem should be used here but I don't see how we can use it to prove that $a_{n+1} < a_n$.
I will rewrite the sequence as,
\begin{align*}
(1 - \frac{1}{n})^{-n}... | Let
$$a_n=(1 - \frac{1}{n})^{-n}=\frac{n^n}{(n-1)^n}$$
Then
$$\frac{a_n}{a_{n+1}}=\frac{n^n}{(n-1)^n}\frac{n^{n+1}}{(n+1)^{n+1}}=\frac{n^{2n+2}}{(n-1)^{n+1}(n+1)^{n+1}}\frac{n-1}{n}$$
$$=\left( \frac{n^{2}}{n^2-1} \right)^{n+1} \frac{n-1}{n}=\left( 1+\frac{1}{n^2-1} \right)^{n+1} \frac{n-1}{n}$$
By Bernoully or Binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Finding $\frac{a+b}{a-b}$ such that $a^2+b^2=6ab$ For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$
| We have $a^2+b^2=6ab$.
To both sides, add $2ab$ to obtain, $(a+b)^2 = 8ab$.
Similarly, subtract $2ab$ to obtain, $(a-b)^2=4ab$.
Thus, $\left(\dfrac{a+b}{a-b}\right)^2 = 2$.
So ultimately $\frac{a+b}{a-b}= \pm\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"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... | Very fundamental theorem says that linerar mapping from a vector space is the same thing as any function from some basis. Any such function can be uniquelly extended to whole vector space. In your case for example vectors $(2, 0)$, $(1, 1)$ are lineary independent and so form a basis of $\mathbb{R}^2$. So there exists ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :
... | Setting
$$
u_n(x)=\cos\frac{x}{2}\cdot\cos\frac{x}{4}\cdot\cdots\cos\frac{x}{2^n},
$$
we have
$$
v_n(x):=u_n(x)\cdot\sin\frac{x}{2^n}=u_{n-1}(x)\cdot\cos\frac{x}{2^n}\cdot\sin\frac{x}{2^n}=\frac12u_{n-1}(x)\cdot\sin\frac{x}{2^{n-1}}=\frac12v_{n-1}(x).
$$
It follows that
$$
v_n(x)=\frac{1}{2^{n-1}}v_1(x)=\frac{1}{2^{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
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$2+2 = 5$? error in proof $$\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... | Here's what your "proof" would look like correcting all the errors. As you can see, it's not nearly as impressive as a proof that 2+2=5.
$$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= -\sqrt{(4-\frac92)^2} +\frac92\\
&= -\sqrt{16 -2\times4\times\frac92 +(\frac92)^2} + \frac92\\
&= \left(-\sqrt{16 -36 + (\frac92)^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Question involving approximation, taylor series and proving Question:
Consider the approximation $$\ln(2)\approx 2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}} \right )$$
Prove that the error in this approximation is less than $$\frac{1}{7\times 2^{2} \times 3^{5}}$$
Attempt:
It looks like the exp... | Since $9\gt 7$, and $11\gt 7$, and $13\gt 7$, and so on, the tail
$$\frac{2}{7\cdot 3^7}+\frac{2}{9\cdot 3^9}+\frac{2}{11\cdot 3^{11}}+\frac{2}{13\cdot 3^{13}}+\cdots$$
is less than the sum of the geometric series
$$\frac{2}{7\cdot 3^7}\left(1+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots\right).$$
But
$$1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/459146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to check whether this sum converge or diverge? Consider the following sum:
$$\sum_{k=1}^\infty\prod_{j=1}^k\frac{1}{\sqrt{j+1}-\sqrt{j}+1}$$
How could I check whether this sum converge or diverge? Root and ratio tests are inconclusive...
| A different solution, which is not mine, is the following:
Let $\alpha_{k}=\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\,,\; k\in\mathbb{N}$, then
\begin{align*}
1-\frac{\alpha_{k+1}}{\alpha_k}&=1-\frac{\prod_{j=1}^{k+1}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}{\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}\\
&=1-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$n\text{ odd}\implies n^2=8k+1$ for some $k\in \mathbb{Z}$ So this girl tells me "Did you know that if $n$ is odd, then $n^2=8k+1$ for some $k\in \mathbb{Z}$?" And so I was like, "Really?" She said, "Yeah!" So I wrote this down:
If $n$ is odd, then $n=2m+1$, and so by squaring $n$ we get
$$n^2=(2m+1)^2=8(\frac{1}{2}m^... | The problem is that your proof doesn't handle the case when $m$ is not a multiple of $4$ - but the claim is still true then (e.g. $n = 5$, so that $m = 2$).
A much easier way to proceed is as follows: If $n$ is odd, then $n$ leaves a remainder of either $1$ or $3$ upon division by $4$. If the remainder is $1$, we find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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can someone explain this limit i have,
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand.
i tried this
$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {... | You can also write your expression as
$$\frac{7}{3}\frac{\sqrt{1+x/49}-1}{1-\sqrt{1+x/9}}$$
If $x$ is small, then $\sqrt{1+x/49}=1+x/98, \sqrt{1+x/9}=1+x/18$
and finally we have $-3/7.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$ How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
I don't know the solution for this.
Help me!
Thank all!
| The use of the "multiple-angle" formulas is probably the surest way to establish the solutions to this equation; however, it requires a goodly amount of algebraic manipulations. We can also say something about the solutions by investigating the properties of the terms in the equation
$$( \ \sin x \cdot \sin 2x \cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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How can I calculate this integral I solve integral considerable number, but the following integral I could not solve. It is about this integral:
$$
\int\frac{\ln^2(a+bx)dx}{x^n}
$$
I want to make the solution through partial method: $\int udv=uv-\int vdu$
I've solved the integral $$\int\frac{\ln(a+bx)dx}{x^n}$$.
Pleas... | This is just an (failed) attempt to find a reduction formula, and as @mtiano noted, the close form requires some hypergeometric functions. Let
$$\begin{align}
u =& \frac{\ln(a+bx)}{x^n} \\
du =& \frac{\frac{bx^n}{a+bx}-n x^{n-1}\ln(a+bx)}{x^{2n}}dx
= \left[ \frac{b}{x^n(a+bx)}-\frac{n\ln(a+bx)}{x^{n+1}}\right]dx\\
dv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{... | HINT
Go in the opposite direction. Let $u = x^2$, $dv = \cos \frac{x}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$. Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$.
I'm not sure, but I suppose that $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < \sum_{n=0}^{\inft... | Notice that
$$\sum_{n=0}^\infty \frac{1}{2^n}=\sum_{n=1}^\infty \frac{1}{2^{n-1}}=2$$
and the result follows since
$$\frac{1}{2^{n-1}}\geq \frac{1}{2^{n}-1}\iff 2^{n-1}\geq1,\; \forall n\geq1\quad\text{which's true}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$ let $a,b,c\ge 0$, such that $a+b+c=1$, prove that
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This problem is simple as 2005, china west competition problem
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge 1$$
see:(http://www.artofproblemsolving.com/Forum/viewtopi... | Set $c=1-a-b$ and find the three lines on which $\frac{\partial f(a,b)}{\partial a\partial b}=0$. Now derive $f(a,b(a))$ according to $a$ to find the extrema. This is a bit tedious, but you fill get the desired $9/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/470296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to ... | If $7x=\pi,4x=\pi-3x$
$\implies \sin4x=\sin(\pi-3x)=\sin3x$
$\implies 2\sin2x\cos2x=3\sin x-4\sin^3x$
$\implies 4\sin x\cos x\cos2x=3\sin x-4\sin^3x$
If $\sin x\ne0,$ we have $4\cos x\cos2x=3-4\sin^2x\implies 4\cos x(1-2\sin^2x)=3-4\sin^2x$
On squaring & rearrangement, $64(\sin^2x)^3-112(\sin^2x)^2+56\sin^2x-7=0$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Prove by induction $\frac{n^n}{3^n}$\dfrac{n^n}{3^n}<n!<\dfrac{n^n}{2^n}$
The case $n!<\dfrac{n^n}{2^n}$ is easier.
| 1) Show that $n!<\frac{n^n}{2^n}$ for $n\ge6$
a) This is true for $n=6$, since $6!=720<729=3^6$.
b) Assume that $n!<\frac{n^n}{2^n}$ for some integer $n\ge6$.
Then $(n+1)!=(n+1)n!<(n+1)\cdot\frac{n^n}{2^n}$, and
$\frac{(n+1)^n}{n^n}=\big(\frac{n+1}{n}\big)^n=\big(1+\frac{1}{n}\big)^n\ge1+n(\frac{1}{n})=2$ by Bernoulli'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find $f(x)$ such that $f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$
Given a polynomial $f(x)$ of n degree such that
$$f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$$
Find the polynomial
I've tried considering $f(x)=\sum \limits_{i=1}^{n} a_{i}x^{i}$ and after a number of st... | $$f(y) = 1+y$$
$$f(x)+f\left(\frac{1}{x}\right) = 1+x+1+\frac{1}{x} = 2 + x + \frac{1}{x}$$
$$f(x) f\left(\frac{1}{x}\right) = (1+x) \left ( 1+\frac{1}{x}\right) = 1 + x + \frac{1}{x} + x \frac{1}{x} = 2 + x + \frac{1}{x}$$
In fact, $f(y) = 1+y^k$ for $k \in \mathbb{N}$ seems to work for the same reason.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Find the value of the the term The sequence $a_1,a_2,a_3,\ldots$ satisfies $a_1=1$, $a_2=2$, and $$a_{n+2}=\frac2{a_{n+1}}+a_n\;;$$ find the value of
$$\frac{a_{2012}2^{2009}}{2011}$$
| HINT:
$$a_{n+2}=\frac2{a_{n+1}}+a_n\iff a_{n+2}a_{n+1}=a_{n+1}a_n+2$$
$$\implies a_{n+2}a_{n+1}=2n+a_2a_1=2n+2\ \ \ \ (1)$$
$$\text{Replacing }n\text{ with } n-1,\implies a_{n+1}a_n=2n$$
$$\implies a_{n+2}a_{n+1}\cdot2n=a_{n+1}a_n(2n+2)$$
$$\implies a_{n+2} =a_n\frac{(n+1)}n$$
$$n=2m-2\implies a_{2m} =a_{2m-2}\frac{(2m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots$ is positive? Let $0<x<\pi$. $n$ be a natural number.
How to prove
$$\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots+ \frac{\sin nx}{n}>0$$
| Let $f_n(x)=\sum\limits_{k=1}^n\frac{\sin{kx}}{k}$.
For $n=1$ it's obvious.
Let $f_n(x)>0$ for any $0<x<\pi.$
It's enough to prove that $f_{n+1}(x)>0.$
Indeed, let $f_{n+1}(x)\leq0$ for some value of $x\in(0,\pi)$.
Since $f_{n+1}$ is a continuous function on $[0,\pi],$ $f_{n+1}(0)=f_{n+1}(\pi)=0,$
there is $x_0\in(0,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
The integer $c_n$ in $(1+4\sqrt[3]2-4\sqrt[3]4)^n=a_n+b_n\sqrt[3]2+c_n\sqrt[3]4$ For non-negative integer $n$, write
$$(1+4\sqrt[3]2-4\sqrt[3]4)^n=a_n+b_n\sqrt[3]2+c_n\sqrt[3]4$$
where $a_n,b_n,c_n$ are integers. For any non-negative integer $m$, prove or disprove
$$2^{m+2}|c_n\iff2^m|n$$
So far I have
$$\left[\begin{... | $$
x \equiv \sqrt[3]{2}
\quad\Longrightarrow\quad
x^{2} = \sqrt[3]{4},\ x^{3} = 2,\ x^{4} = 2x,\ x^{5} = 2x^{2}, x^{6} = 4,\ x^{7} = 4x\ldots
$$
$$\left(1 + 4\sqrt[3]{2} - 4\sqrt[3]{4}\right)^{n}
=
\left(1 + 4x - 4x^{2}\right)^{n}
=
a_{n} + b_{n}x + c_{n}x^{2}
$$
$$
{1
\over
1- z\left(1 + 4x - 4x^{2}\right)}
=
\over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
A real solution to a cubic equation What is the easiest way to find the real solution of the equation $x^3-6x^2+6x-2=0$?
I know the solution to be $x=2+2^{2/3}+2^{1/3}$ (Mathematica) but I would like to find it analytically. If possible, not by plugging the coefficients in Cardano's or similar formula.
| $$\begin{equation*}
x^{3}-6x^{2}+6x-2=0 \tag{1}
\end{equation*}$$
Set $x=t+2.$ Then
$$\begin{equation*}
t^{3}-6t-6=0\tag{2}
\end{equation*}$$
Set $t=u+v.$ Then
\begin{equation*}
( u+v) ^{3}-6\left( u+v\right) -6=0,
\end{equation*}
\begin{eqnarray*}
\left( u+v\right) ^{3}-6\left( u+v\right) -6 &=&( u^{3}+v^{3}-6)
+(3u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\ta... | Maybe a little bit obvious but what I would do is the following:
$\textrm{Let:}$
$$\omega + \phi + \psi = \pi$$
Then take the tangent function to both sides.
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \pi \right)$$
Since $\tan \pi = 0$, then:
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \... | {
"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": 9
} |
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!
| Using the hint that OP stated,
$$ \begin{array} { l l}
(a^2 + 3b^2 )^3 & = a^6 + 9 a^4 b^2 + 27a^2 b^4 + 27b^6 \\
& = (a^6 + 6 a^4 b^2 + 9 a^2 b^4) + 3( a^4 b^2 + 6a^2 b^4 + 9b^6 ) \\
& = (a^3 + 3ab^2)^2 + 3 (a^2b + 3b^3)^2 \end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Implicit differentiation question Given that $x^n + y^n = 1$, show that $$\frac{d^2y}{dx^2} = -\frac{(n-1)x^{n-2}}{y^{2n-1}}.$$
I found that $\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx} = 0$ so that $\displaystyle y'=\frac{-x^{n-1}}{y^{n-1}}$.
Then $$n(n-1)x^{n-2}+n(n-1)y^{n-2}\left(\frac{dy}{dx}\right)^2 + \frac{d^2y... | Expanding your final expression, we have
\begin{align*}
y'' &= \frac{-n(n-1)x^{n-2} -n(n-1)y^{n-2}\dfrac{x^{2n-2}}{y^{2n-2}}}{ny^{n-1}}\\
&= \frac{-n(n-1)x^{n-2} -n(n-1)y^{-n}x^{2n-2}}{ny^{n-1}}\\
&= \frac{-n(n-1)x^{n-2}(1 + x^{n}y^{-n})}{ny^{n-1}}\\
&= \frac{-(n-1)x^{n-2}(1 + x^{n}y^{-n})}{y^{n-1}}.
\end{align*}
Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??
| for $n=3$, they are solutions ($x_1=1,y_1=1$).
If you have odd solutions $x_n$, $y_n$ for $2^n$, then
$$\left(\frac{7y_n-x_n}{2}\right)^2+7\left(\frac{x_n+y_n}{2}\right)^2=2x_n^ 2+14y_n^2=2(x_n^2+7y_n^2)=2^{n+1}$$
$$\left(\frac{7y_n+x_n}{2}\right)^2+7\left(\frac{x_n-y_n}{2}\right)^2=2x_n^ 2+14y_n^2=2(x_n^2+7y_n^2)=2^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Evaluation Of Maximum Value Without Calculus $x$,$y$,$z$ are non-negative numbers.
$x+y+z=3$
Find the maximum value of $~$ $x^{2}y+y^{2}z+z^{2}x$ $~$ without calculus.
| in general
if $x,y,z\ge 0$,and such $x+y+z=3$, then we have
$$x^ky+y^kz+z^kx\le\max{\{3,\dfrac{3^{k+1}k^k}{(k+1)^{k+1}}\}}$$
For $k=2$ I have nice methods
with out loss of let $x=\max{(x,y,z)}$
then we use Benoulli inequality,we have
$$(1+\dfrac{z}{x})^2\ge 1+\dfrac{2z}{x}$$
so
$$(x+z)^2y=x^2y(1+\frac zx)^2\ge x^2y(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/489292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$ Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}... | $$\begin{aligned}\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| As was already mentioned, we have
$$0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$
$$|x|=|e^{2\pi i/5}|=1\implies x^2=x^{-3}\;,\;x=x^{-4}\;,\;x^{-k}=\overline{x^k}\implies$$
$$(x^2-x^3)^2(x^4-x)^2=(x^2-x^{-2})^2(x^{-1}-x)^2=(x-x^{-1})^4(x+x^{-1})^2=$$
$$=\left(2i\sin\frac{2\pi}5\right)^4\left(2\cos\frac{2\pi}5\right)^2=64\sin^4\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Solving a system of equation modul0 5 Consider the system of linear equations $$\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3\\ 1 \end{pmatrix} $$
a) Solve the system in $\mathbb{F}_5$
I just want to make sure my solution is correct:
We have: $$A=\begin{pmatrix} ... | Note that modulo $5$ one has
$$
A =
\begin{bmatrix} 6 & -3\\ 2 & 6 \end{bmatrix}
=
\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}.
$$
The latter matrix has determinant $-3 \equiv 2$, whose inverse is $3$, so
$$
A^{-1} =
\begin{bmatrix} 3 & -1\\ -1 & 3 \end{bmatrix}.
$$
Now
$$
\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}
=
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$ How can I evaluate
$$I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx$$
for $a,b>0$?
My methods:
Let $a,b > 0$ and let
$$I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.$$
Then
$$I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,d... | The integral can be evaluated as follows
$$
\begin{align}
\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx&=2\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 1
} |
Smallest number with specific number of divisors Is there a general method for finding smallest number of specific number of divisors? I am doing "Higher Algebra by Barnard JM Child" and came across a question that "find the smallest number with 24 divisors", that's how I tried to solve it, alert me if I am wrong:
Sinc... | Take any number $N=p_1^{m_1}...p_k^{m_k}$, where $p_i$ are its prime divisors, and compute how many divisors it has. Each divisor would be a product of the same primes in varying powers, $D=p_1^{d_1}...p_k^{d_k}$, where $0\leq d_i \leq m_i$. Different divisors have different collections of powers, so the number of divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
How to solve this matrix using gauss-elimination by hand I feel like i am having a brain fart. I have been given this $Ax=b$ system:
$A= \begin{pmatrix} 0.913 & 0.659 \\ 0.780 & 0.563 \end{pmatrix}$
$b= \begin{pmatrix} 0.254 \\ 0.217 \end{pmatrix}$
I know the answer is $x_1 = 1$ and $x_2 = -1$ but for some reason when ... | Alright I worked it out step by step, so bear with me (I'll use A followed by b notation):
\begin{pmatrix} 0.913 & 0.659 \\ 0.780 & 0.563 \end{pmatrix} \begin{pmatrix} 0.254\\ 0.217 \end{pmatrix}
We multiply the top row by the bottom and vice versa:
\begin{pmatrix} 0.71214 & 0.51402 \\ 0.71214 & 0.514019 \end{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fermat's Combinatorial Identity: How to prove combinatorially? $$\binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} + \dotsb + \binom{n}{r} = \binom{n+1}{r+1}$$
I don't have much experience with combinatorial proofs, so I'm grateful for all the hints.
(Presumptive) Source: Theoretical Exercise 1.11, P18, A First Course in ... | This is analogous to https://math.stackexchange.com/a/357087/53259. The RHS here imports the number of ways of picking $r+1$ numbers out of $\{1,2,...,\color{magenta}{r}, \color{green}{r +1},...,\underbrace{n}_{= \color{magenta}{r} + (n - r)},\color{green}{n +1}\} \qquad (*)$
Now for any given choice of $r+1$ numbers,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to solve the following equation $\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$ I am trying to solve this equation:
$$\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$$
I would like to get some advice, how to solve it.
Thanks.
| First note that $x=0$ is a solution. Now we consider $x\neq 0$ and we divide by $\sqrt[3]{x}$. We get $\sqrt[3]{1+\frac{3}{x}}+1=\sqrt[3]{8+\frac{3}{x}}$, so lets define $y=\frac{3}{x}$ and write
$$1+\sqrt[3]{1+y} = \sqrt[3]{8+y}.$$
Now we cube both sides to obtain
$$1+3\sqrt[3]{1+y}+3\sqrt[3]{1+y}^2+1+y = 8+y$$
or
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
How can I solve this equation if I know the solution for a similar equation For equation
\begin{equation}
-y''-2a^2\operatorname{sech}^2(ax)~y=k^2y
\end{equation}
I know there is the solution
\begin{equation}
y(x)=C_1\frac{-a \tanh (a x)+i k}{a+i k}e^{i k x} +C_2\frac{-a \tanh (a x)-i k}{a-i k}e^{-i k x}
\end{equation}... | Your reference book is lying!
Let $u=-\tanh ax$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-a(\text{sech}^2ax)\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-a(\text{sech}^2ax)\dfrac{dy}{du}\right)=-a(\text{sech}^2ax)\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the area between the curves $r=2\sin\theta$ and $r=\sin\theta+\cos\theta$
So the problem asked me to find the area of the region that lies inside both of the circles
$$r=2\sin\theta, \quad r=\sin\theta +\cos\theta $$
I know that $r=2\sin\theta$ is $x^2+(y-1)^2=1,$but the second one is a little bit harder ... | To clarify the common area, you need to plot two graphs. Both are circles.
$\cos{\theta} + \sin{\theta} = \sqrt{2}\sin(\theta + \pi/4)$. So draw a circle $r = \sqrt{2} \sin {\theta}$, and rotate it $\pi/4$ clockwise.
Then, it is the circle whose cernter is (1/2,1/2) and the radius is $\frac{\sqrt{2}}{2}$.
Purely alge... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Express sequence in closed type Given the sequence $a_n = \sqrt{2+a_{n-1}}$. Is there anyway to find a closed form for this sequence?
Thank you for your time.
| We are given
$a_n = \sqrt{2+a_{n-1}}$.
I will assime that
$0 \le a_0 < 2$.
If $a_{n-1} < 2$,
$a_n < \sqrt{2+2} = 2$,
so all subsequent $a_n < 2$.
Since $a_1 = \sqrt{2+a_0} < 2$,
all $a_n < 2$.
Let $d_n = 2-a_n$,
so $a_n = 2-d_n$.
$2-d_n = \sqrt{2+(2-d_{n-1})}
=\sqrt{4-d_{n-1}}
$
or,
since $1-x < \sqrt{1-x} < 1-x/2$
i... | {
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Quantum Fourier Transform and roots of unity. I need to find $QFT_{6}$ for the state quantum state $\frac{1}{\sqrt2}(|0\rangle + |3\rangle)$. I received a very sufficient answer recently on simplifying nth roots of unity, but I am having a lot of trouble applying it to this problem. I have obtained the $QFT_6$ matrix b... | Position a regular hexagon on the unit circle with vertices at $\pm1$. Let $\zeta$ be the first complex root of unity counterclockwise from $1$. Then $\zeta^2$ is a third full rotation away from $1$, in other words it is a third root of unity. And $\zeta^3=-1$ since it sweeps out exactly half a rotation. Algebraically ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what $n$ does $x^n \equiv 2\pmod{13}$ have a solution? I want to ask for what values of $n$ the congruence $$x^n \equiv 2 \pmod{13}$$ has a solution for $x$.
| Hints: Note that $x^{12k+m}\equiv x^m\pmod{13}$. So you will "only" have to examine the cases $m=0$ to $11$.
If $m$ and $12$ are relatively prime, then the congruence $x^m\equiv a\pmod{13}$ always has a solution.
So we are down to $m=0$, $2$, $3$, $4$, $6$, $8$, $9$, $10$.
You can find shortcuts to deal with some of... | {
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Independence of $\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$ from $k$. How to one can show that the value of the following expression
$$\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$$
doesn't depend to values of $k$?
| We have $\frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)}$ and what bothers me at first sight is that $k$ multiplies $\sin$ in numerator but $\cos$ in denominator. So, let's turn $\cos$ into $\sin$:
$$
\begin{align}\frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)} &= \frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$ How can I solve this trigonometric equation?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
| Using this solution,
$$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$
Putting $x=12^\circ,$ $$\tan12^\circ\tan48^\circ\tan72^\circ=\tan36^\circ$$
$$\implies \tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ =\tan 54^\circ \tan36^\circ=\tan(90^\circ-36^\circ)\tan36^\circ=\cot36^\circ\tan36^\circ=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the Limit $\lim_{x\to 0} \frac{\sin x(1 - \cos x)}{x^2}$ Here's the problem.
$$\lim_{x\to 0} \frac{\sin x(1 - \cos x)}{x^2}$$
I really don't know where to start with this. Please help.
| I assume you know that $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$
Now, $$\lim_{x \to 0} \frac{1-\cos x}{x^2} = \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x)}{x^2 (1+\cos x)}=\lim_{x \to 0} \frac{\sin^2 x}{x^2(1+\cos x)},$$
and therefore $$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \tim... | {
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Finding the curve whose asymptotes are given?
Find the equation of the cubic curve whose asymptotes are $x+a=0$,
$y-a=0$, $x+y+a=0$ and which touches the axis of X at origin and
passes through the point (-2a,-2a)?
My solution: The general equation of the curve must be $(x+a)(y-a)(x+y+a)+g(x,y)=0$
Its given that ... | Consider the most general form of a cubic planar curve $$Ax^3+By^3+Cx^2y+Dxy^2+Ex^2+Fxy+Gy^2+Hx+Ky+L=0.$$
Since we know that the curve must pass through the origin we can conclude that $L=0$. Now using the implicit differentiation we obtain $$y'_x=-\frac{3Ax^2+2Cxy+Dy^2+2Ex+Fy+H}{3By^2+Cx^2+2Dxy+Fx+2Gy+K}=\tag{1}$$ $$-... | {
"language": "en",
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Homework - ERO with Unknown in Matrix I was having a problem with how to properly perform elementary row operation (ERO) on a matrix.
In the question, we were given an augmented matrix $\begin{bmatrix}1&1&-1&1\\2&3&a&3\\1&a&3&2\end{bmatrix}$ and require us to determine all possible values of $a$ such that the system is... | Hint: consider what you have after your step 3 and the simplification, that is,
$\qquad\begin{bmatrix}1 & 1 & -1 & 1 \\ 0 & 1 & a+2 & 1 \\ 0 & 0& (3+a)(2-a) & 2-a \end{bmatrix}$
*
*If $a=2$, then …
*If $a=-3$, then …
*If $a\ne2$ and $a\ne 3$ then …
Explanation of the error
You can't do $\frac{R_{3}}{2-a} \rightar... | {
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Valid values of constant for probability mass function (PMF)
For what values of constant $c$ do the following functions define a valid PMF for random variable $X$ on supports $X=\{1,2,...\}$
(1) $f(x) = c/2^x$
(2) $f(x) = c2^x/x!$
I was thinking $\sum_{i=1}^\infty f(x) = 1$ so,
$$\begin{aligned}
\sum_{i=1}^\infty ... | For the first, your start was correct, we want
$$c\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots\right)=1.$$
The inner sum above is an infinite geometric series, with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$.
It has sum is $\dfrac{1/2}{1-1/2}$, which is $1$.
So we want $(c)(1)=1$, and therefore $... | {
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How to find sum of 3 perpendiculars of a triangle? Q. ABC is an equilateral triangle with side 10cm and P is a point inside the triangle, at a distance of 2cm from AB. If PD, PE and PF are perpendiculars to the three sides, find sum PD+ PF+PE.
What I've done:
In triangles ADP and AFP,
$AP^2=x^2+PF^2$
$AP^2=y^2+4$
In ... | From equilateral triangle and from $DB=FC$ we see $x=y$.
By using Carnot's theorem:
$$(10-z)^2+(10-y)^2+y^2=z^2+x^2+(10-y)^2$$
$$100-20z=x^2-y^2=0$$
$$5=z$$
From here we can say that the points $A,P,E$ are on a line and $AE=BE=EC=5$.
So $AP$ is bisector of $DAF$ and from there we can conclude that $FP=2$ and $AP=4$.
H... | {
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"timestamp": "2023-03-29T00:00:00",
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Need to prove that $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $99$ for all $n \in \mathbb{N} $, using induction. First, obviously, I figured out the base case. So I have $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5 = 99k$ for some $k \in \mathbb{N} $. As for the inductive step, I was thinking about splitting it... | A number is a multiple of $9$ iff the sum of its decimal digits is a multiple of $9$. Your numbers satisfy this: $4+9+5=18$.
A number is a multiple of $11$ iff the alternating sum of its decimal digits is a multiple of $11$. Your numbers satify this: $4-9+5=0$.
Explicitly, we have $4900\cdots05=9\cdot 544\cdots45$ and ... | {
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Prove by induction that $1^3 + \dots + n^3 = (1 + \dots + n)^2$ I'm suppose to prove by induction:
$1^3 + \dots + n^3 = (1 + \dots + n)^2$
This is my attempt; I'm stuck on the problem of factoring dots.
| Your attempt looks OK as far as it goes (except for a missing superscript $2$ at one point, but that's not causing any further errors).
To prove that something is equal to $1^3+2^3+3^3+4^3+\cdots+n^3$, one must show that at each step the amount that gets added is the next cube.
So how much has to be added to $(1+2+\cdo... | {
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If $17! = 355687\underline{ab}8096000$. Then value of $(a,b)$ is
*
*If $17! = 355687\underline{ab}8096000$, what is the value of $(a,b)$?
*If $34! = 295232799\underline{cd}9604140847618609643\underline{ab}0000000$, what is the value of $(a,b,c,d)$?
My Attempt:
We know that
$$
17! = 1\times 2 \times 3 \times ... | To avoid unanswered questions, I will compile the tips from the comments here, along with a bit more.
Instead of using divisibility by $3$, use divisibility by $9,$ so that we can actually conclude that $\frac{57+a+b}{9}=6+\frac{3+a+b}9$ is an integer. Since $0\le a+b\le 18,$ then it follows that either $a+b=6$ or $a+b... | {
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Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$? It is an exercise on the lecture that i am unable to prove.
Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$?
| HINT:
Let prime $p$ divides $a+b, a^2-ab+b^2$
$\implies p$ divides $\{(a+b)^2-(a^2-ab+b^2)\}=3ab$
If $p|a,$ as $p|(a+b),p$ must divide $(a+b)-a=b\implies p|(a,b)$
But as $(a,b)=1,p$ can not divide $a$
Similarly, $p$ can not divide $b$
$\implies p|3$
$\implies (a+b,a^2-ab+b^2)|3$
and $(a+b,a^2-ab+b^2)=3$ if $3|(a+b)\i... | {
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Evaluating $\int_{-2}^{2} 4-x^2 dx$ with a Riemann sum I'm having problems with a Riemann sum ... I need to find the integral:$$\int_{-2}^2 (4-x^2)\;dx$$Clearly we have $$\int_{-2}^{2}(4-x^2)\;dx=4x-\frac{x^3}{3}\mid_{-2}^{2}=(4\cdot2-\frac{2^3}{3})-(4\cdot(-2)-\frac{(-2)^3}{3})=\frac{32}{3}$$OK.
On the other hand, we... | The problem is an accidental algebra mistake, as I pointed out in comments: $n(n+1)(2n+1)=2n^3+3n^2+n\neq n^3+3n^2+n$, and this should do it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Identity proofs I'm strugling with these two combinatorial identities:
$$ \binom{n+2}{3}=\sum_{i=1}^{n} i(n+1-i) $$
and
$$\binom{n+1}{2}^2=\sum_{i=1}^{n} i^3$$
Please give me some footholds and hints
| For the first problem split the RHS:
$$\sum_{i=1}^n i(n+1-i) = \sum_{i=1}^n in +i - i^2 = n\sum_{i=1}^n i + \sum_{i=1}^n i - \sum_{i=1}^n i^2$$
Now we know that the some for consecutive integer from $1$ to $n$ is $\frac{n(n+1)}{2}$ and for the squares of consecutive integers from $1$ to $n$ is $\frac{n(n+1)(2n+1)}{6}$.... | {
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Is there a way to calculate $\int \limits_0^1\frac{x^3}{\sqrt{x^2-1}}\frac{1}{1-a^2x^2}\frac{1}{1-b^2x^2}\frac{1}{c-x}\mathrm dx$ I want to calculate
$\displaystyle \int \limits_0^1\dfrac{x^3}{\sqrt{x^2-1}}\dfrac{1}{1-a^2x^2}\dfrac{1}{1-b^2x^2}\dfrac{1}{c-x}\mathrm dx$
$a$ and $b$ are real parameters, c could be comple... | Let us denote$A:=\arccos(a)$, $B:=\arccos(b)$ and $C:=\arccos(1/c)$.
By firstly substituting $x=\sin(y)$ and then $u=\tan(y/2)$ we get:
\begin{eqnarray}
&&\int\limits_0^1 \frac{x^3}{\sqrt{1-x^2}} \frac{1}{1-a^2 x^2} \frac{1}{1-b^2 x^2} \frac{1}{c-x} dx= \\
&&\frac{16}{c} \int\limits_0^{\tan(\pi/4)} \frac{u^3+u^5}{\prod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help with exact value of: $\tan (\sin^{-1}(-1/2) - \tan^{-1}(3/4))$ Ok, so I used the tan formula of difference of angles, and so far I've got to:
I found that $$\tan \alpha = \frac{-1/2}{\sqrt{3}/2} = \pm \frac{\sqrt3}{3}$$
so
$$\tan \left(\sin^{-1}\left(\frac{-1}{2}\right) - \tan^{-1}\left(\frac{3}{4}\right)\right) =... | Let $a=\sin^{-1}\left(-\frac{1}{2}\right)$ and $b=\tan^{-1}\left(\frac{3}{4}\right)$. We have
$$\sin^2a+\cos^2a=1 \Rightarrow \frac{1}{4}+\cos^2a=1 \Rightarrow \cos a=\frac{\sqrt 3}{2} \Rightarrow \tan a = -\frac{\sqrt 3}{3}$$
since $\sin^{-1}x$ has $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ as image.
Also,
$$\frac{\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:
$$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$
for some integers $m, n$
$$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 ... | Statement for the general case (for any $n>1$ in place of $3$)
Let $a,b$ be two positive rational numbers such that $\sqrt[n]{a}+\sqrt[n]{b}$ is a rational number for some natural number $n>1$. Then both $\sqrt[n]{a}$, $\sqrt[n]{b}$ are rational numbers.
Proof
Let $s=\sqrt[n]{a}+\sqrt[n]{b}\in\mathbb{Q}$. Then $\sqrt[n... | {
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How prove this $x+y=0$ if $\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$ Question:
let $x,y$ are real numbers,and such
$$\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$$
show that
$$x+y=0\tag{1}$$
before I have solve following problem:
if
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1,\L... | Below some characteristics of the function $f(x,y) = \left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)-y^3$ are shown. The area where $f(x,y)$ is undefined is in gray: $y^2 - x^3 < 0$ or $x^2 + y^3 < 0 $. Therefore the domain of the function is mainly restricted to $x\le 0$ and $y\ge 0$. The line $x+y=0$ is in... | {
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Integers whose sum and product are integers Let $a$, $b$ be real numbers such that $a + b$ and $ab$ are integers.
a. Prove that $a^n + b^n$ is an integer for every natural number $n$.
b. Suppose that $a$ does not equal $b$. Prove that $\dfrac{a^n - b^n}{a - b}$ is an integer for every positive integer $n$.
Attempts
I a... | For a)
$\\ { a }^{ 2 }+{ b }^{ 2 }={ \left( a+b \right) \left( a+b \right) }-2ab,\quad integer\\ { a }^{ 3 }+{ b }^{ 3 }=\left( a+b \right) { \left( { a }^{ 2 }+{ b }^{ 2 } \right) }-ab\left( a+b \right) ,\quad integer\\ { a }^{ 4 }+{ b }^{ 4 }=\left( a+b \right) { \left( { a }^{ 3 }+{ b }^{ 3 } \right) }-ab\left( ... | {
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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?
| The answer is no for k>0.
Let $b>a$, $a^2=4k+1$ and $b^2=9k+1$. Note that,
$$
5k=b^2-a^2=(b+a)(b-a)=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)\cdot \left(\sqrt{9k+1}-\sqrt{4k+1} \right)
$$
Case 1:set $k$ a prime number. Then
*
*$k=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$ and $5=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ or
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove inequalities and one-to-one function? Can anyone please help me with these questions?
1.Given
x + 1 < 0
Prove that:
i) $2x - 1 < 0 $
ii) ${2x-1\over x+1} > 2$
2.For $g(x) = {kx + 8\over 4x - 5}$
i) Find k if gg(x) = x
Is it fine if I just let any value of x for this question?
ii) Find the value of k so t... | (1 i) $x+1<0\Rightarrow 2\left( x+1 \right)<2.0\Rightarrow 2x+2<0\Rightarrow 2x+2-3<-3\Rightarrow 2x-1<-3<0$
so $2x-1<0.$
(1 ii)$$\frac{2x-1}{x+1}=\frac{2\left( x+1 \right)-3}{x+1}=2-\frac{3}{x+1}$$
because $x+1<0$ given we can write $$-\frac{3}{x+1}>0\Rightarrow 2-\frac{3}{x+1}>2 \Rightarrow\frac{2x-1}{x+1}>2$$
(2 i) ... | {
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Writing a number as power of two Number of powers of two needed to be added to the given powers of two to write the (sum of given powers and added powers of two)number of the form $2^k-1$ where $k$ is any integer.
Okay, let me explain the question.
Suppose you are given: $2^0 , 2^1 , 2^2$
Here sum of the numbers $=1+2+... | Suppose that you’re given $2^{a_1},2^{a_2},\ldots,2^{a_n}$, where $a_1\le a_2\le\ldots\le a_n$. If $a_1=a_2$, erase $2^{a_1}$ and $2^{a_2}$ and write down $2^{a_1+1}$; then sort the powers of $2$ into non-decreasing order again. Repeat until the two smallest powers are different. Then do the same thing with the second-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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... | This is not yet a full answer, but I need to get ready for work now and I'm not sure how to proceed anyway.
This statement clearly doesn't hold if $x$ or $y$ can be $0$, and negative numbers are easily flipped about, so assume that $x$ and $y$ are positive integers.
If $x$ and $y$ are both even, then dividing each by $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? It seems like an expression from a simple induction proof problem that's missing its result.
| You could write
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} \cdot \dots \cdot \frac{2^{n - 1} - 1}{2^{n - 1}} \cdot \frac{1}{2^n}$$
and then the denominator is
$$2 \cdot 4 \cdot \dots 2^n = 2^{1 + 2 + 3 + \dots + n} = 2^{n(n + 1)/2}$$
while the numerator is
$$(2 - 1) (2^2 - 1) (2^3 - 1) \cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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I have an abstract algebra proof I am a little iffy on. Help please! I was independently studying abstract algebra and found the following problem:
The problem says to show $\mathbb{Q}(\sqrt{3},\sqrt{7}) = \mathbb{Q}(\sqrt{3} + \sqrt{7})$.
The solution given was as follows:
Everything made since except line 4. Where d... | $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab} \in \mathbb{Q}(\sqrt{a}+\sqrt{b}) \Rightarrow \sqrt{ab} \in \mathbb{Q}(\sqrt{a}+\sqrt{b})$.
Then, $\sqrt{ab}(\sqrt{a}+\sqrt{b}) = a\sqrt{b} + b\sqrt{a} \in \mathbb{Q}(\sqrt{a}+\sqrt{b})$
$\Rightarrow a\sqrt{b} + b\sqrt{a} - a(\sqrt{b}+\sqrt{a}) = b\sqrt{a}-a\sqrt{a} \in \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proper method for solving quadratic equations with exponents $(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}$ + $(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$ = $2^{(x+4)/4}$
I have found out, by trial and error method, that $x=0$ and $x=4$ satisfy this equation. But is there a proper way to solve this equation and get the solut... | Put
$$
a=(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2},\\
b=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}
$$
Then
$$
a+b=2^{(x+4)/4},\\
a \cdot b=2^{x/2}
$$
By solving it we get that $a=b,$ or
$$
(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}.
$$
It is a simple equation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of $(\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$ Any Ideas/Hints?
$$\lim\limits_{n \to \infty } (\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$$
| Hint: $a^3-b^3=(a-b)(a^2+ab+b^2)$. Take $a=\sqrt[3]{n^3+n^2}$ and $b=\sqrt[3]{n^3+1}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When is the quadratic congruence $ax^2 + bx +c \equiv 0 \pmod p$ solvable? I am learning about quadratic congruences and I don't now how to decide, for which $a, b, c$ and $p$ there is a solution of the congruence. Is it sufficient if the discrminant $b^2-4ac$ has a solution in $\Bbb Z_p^*$?
| If $a = 0$, this reduces to a linear congruence which has a unique solution (provided $b \neq 0$); I assume you know how to find the solution in this case.
Now suppose $a \neq 0$.
If $\underline{p \neq 2}$, multiply the congruence by $4a$ to obtain $4a^2x^2 + 4abx + 4ac \equiv 0 \pmod p$. Completing the square gives $(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that if $x>0$, then $\ln(x)\geq 1-\frac{1}{x} $ Show that if $x>0$, then
$$ \ln(x)\geq 1-\dfrac{1}{x}. $$
I tried a few things but so far nothing has worked, I could use a hint.
| You can use mean value theorem for $\ln$.
*
*$x = 1$, then equality holds.
*$0 < x < 1$:
$$\ln x \ge \frac{x-1}{x} \\
\frac{\ln x - \ln 1}{x- 1} \le \frac{1}{x} \\
\frac{1}{c} \le \frac{1}{x} \\
c \ge x$$
Which indeed is true, because $c \in (x,1)$. Change from $\ge$ to $\le$ is due to division by $(x-1)$ which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/551858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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trigonometric equation opening Solve:
$$ \sin x + \sin 3x + \sin 5x = 0 . $$
Attempt at a solution: applying formulas for summation of sine we get after a series of operations: $ \sin x(8 \cos x \cos 2x \cos x + 1) = 0$ equaling sine to $0$ we get one solution $180k$. comparing the other factor we eventually get $... | Using $e^{xi}=\cos x+i\sin x$, we can easily derive a canonical form for $\sin(ax)$, for $a\in\mathbb{N}$.
For example, $e^{3xi}=(\cos x+i\sin x)^3=\cos 3x+i\sin 3x$. Equating the imaginary parts of each expression, we obtain $$3\cos^2x \sin x-\sin^3x=\sin 3x$$ If we reuse this technique for 5, $e^{5xi}=(\cos x+i\sin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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converting decimals to base negative-10 I have a decimal (base $10$) number, $44$, and would like to convert it to base $-10$. I know how to convert
$$
164_{-10} \mapsto 44_{10},
$$
but not the other way around.
| The simplest way is to simply carry and 'borrow' as needed, keeping track of odd vs. even digit positions. For your example, $44 = 4\times 10^1+4\times 10^0$. The ones' digit is clearly the same, so now we can concentrate on the tens' digit. Since $4=10-6$, we can express $4\times 10^1=40$ by borrowing from (or carr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you find the smallest possible value of aan equation with two unknowns? I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$:
$$ x^2 - 3x + ... | EDIT:
$$ x^2 - 3x + 2y^2 + 4y + 2$$
can be re-written as
$$ x^2-2*x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$
Obeserve that the first two terms in the expression are of the form
$$a^2+2a.b+b^2=(a+b)^2$$
Hence we get
$$ x^2 - 3x + 2y^2 + 4y + 2=x^2-2x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$
$$\Rightarr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$ Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$?
Squaring both sides will give me something but I could not go any further.
| You can also use the inequality between the arithmetic mean and the quadratic mean: $$ \frac{x+y}{2} \leq \sqrt{\frac{x^2 + y^2}{2}}. $$ Setting $x = \sqrt{5}$ and $y = \sqrt{7}$ gives you $$\frac{\sqrt{5} + \sqrt{7}}{2} \leq \sqrt{6},$$ which is equivalent to $$ \sqrt{7} - \sqrt{6} \leq \sqrt{6} - \sqrt{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 15,
"answer_id": 10
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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... | The trouble comes from saying that $n^3+3n^2+2n=n^3+3n^2+2n-(n^3-n)$. Instead we can consider $(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1$. We rewrite this as $(n^3-n)+(3n^2+3n)=(n^3-n)+3(n^2+n)$. The first term is our induction hypothesis which we can rewrite as $3k$ where $k\in\mathbb{Z}$ and the last term is divisible by three... | {
"language": "en",
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"source": "stackexchange",
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Congruence-What is remainder when $a^2+2b$ is divided 3? Suppose $a\equiv 7 \pmod 9 $, $b\equiv 1\pmod 6$. What is remainder when $a^2+2b$ is divided by 3 ?----- So, since $a=9m+7$, $b=6n+1$, $m,n\in \mathbb{Z}$, $a^2+2b=(9m+7)^2+2(6n+1)$,=$81m^2+12n+177$=$3(27m^2+4n+59)$....I am not sure what I am getting here.. Can I... | Another solution using the properties of modular arithmetic.
$a \equiv 7 \pmod 9 \Rightarrow a^2 \equiv 49 \equiv 4 \pmod 9$
$b \equiv 1 \pmod 6 \Rightarrow 2b \equiv 2 \pmod 6$
So $a^2 = 9n + 4$ and $2b = 6m + 2$, thus $a^2 + 2b = 9n + 6m + 6 = 3(3n + 2m + 2)$ which is definitely divisible by 3.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Questions about $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ We have $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ is a group with $\varphi(2^n)=2^{n-1}$ elements. Prove that $x^2=1$ has exactly four solutions in $\mathbb{Z}/2^n\mathbb{Z}$. Moreover, can we show that $q$ is a perfect square in $(\mathbb{Z}/2^n\mathbb{Z})^{\times}... | For the first question, we need $n\ge 3$. We show that there are precisely $4$ integers $x$ between $0$ and $2^n-1$ such that $x^2\equiv 1\pmod{2^n}$, that is, such that $2^n$ divides $(x-1)(x+1)$.
Note that if $a$ and $b$ are two consecutive even integers, then one of them is congruent to $2$ modulo $4$.
Suppose that ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$. I am trying to prove the following statement:
For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$.
So far I have figured out that $n^4 = 8m$ or $n^4 = 8m + 1$.
Any help or hints are appreciated... | An easier way is by using the powerful tool "$mod\ 8$" for any $ n \in \mathbb{N}: n \equiv 0,1, 2, 3 ,4 , 5 , 6 \ or\ 7 \ mod\ 8 \equiv 0, 1, 2, 3, 4, -3, -2 \ or\ -1 \ mod \ 8$
so for ${n^4}$ we have
${n^4}={n^2}^2 = n^3 n \equiv 0^4, 1^4, 2^3 \times 2 , {3^2}^2, {4^2}^2, {-3^2}^2, -2^3 \times 2 \ or\ -1^4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for: $8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2$ Solve for: $$8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2$$
My try:
$8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2\\\Leftrightarrow \log_2\left(x^2-9\r... | $$\log_2(x+3)=y$$
$$y^2+3y-10=0$$
$$y_{1,2}=\frac{-3\pm7}{2},y_1=2,y_2=-5$$
$$\log_2(x+3)=2,x+3=4,x=1$$
$$\log_2(x+3)=-5,x+3=2^{-5},x=2^{-5}-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/573004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$ Find this sum
$$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}$$
then we find th... | I undeleted my answer and fixed it up since one of the other answers refers to it.
In the comments, Lucian mentioned the generalization $$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)(n+c)} = \frac{(b-c) \, \psi(a)+ (c-a) \, \psi(b)+(a-b) \, \psi(c)}{(a-b)(b-c)(c-a)} \tag{1}$$ where $\psi(x)$ is the digamma function; $a,b,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/573119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 4,
"answer_id": 2
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How to prove(or disprove) $\begin{vmatrix} A&B\\ B&A \end{vmatrix}=|A^2-B^2|$ Let $A$ and $B$ be square matrices of the same size.
(1) If $f$ is not invertible and $AB=BA$, show that
$$\begin{vmatrix}
A&B\\
B&A
\end{vmatrix}=|A^2-B^2|.$$
(2) If $A$ is invertible and $AB\neq BA$, then do we have
$$\begin{vmatrix}
A&B\\
... | Let
$$
J=\left(\begin{array}{cc}I&0\\0&-I\end{array}\right)
$$
and
$$
M(A,B)=\left(\begin{array}{cc}A&B\\B&A\end{array}\right).
$$
Then $\det J^2=\det I=1$, so $\det(JM(A,B)J)=\det M(A,B)$. But (check this)
$$JM(A,B)J=M(A,-B),$$
so $\det M(A,B)=\det M(A,-B)$.
On the other hand using $AB=BA$ we get
$$
M(A,B)M(A,-B)=\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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stuck on a Cartesian question we have a circle $(x-1)^2+(y-2)^2=9$
Point $P=(5,2)$ lies outside the circle.
Solve the equation of the line which passes through $P$ and intersects the circle at two points whose mutual distance is $d=2$. Find the coordinates of the intersection points.
I sketched the circle and $P$ but c... | HINT:
Let the equation of the line be $$\frac{y-2}{x-2}=m\implies y=mx+2-2m\ \ \ \ (1)$$ where $m$ is the gradient
Replacing the value of $y$ in $$(x-1)^2+(y-2)^2=9$$
$$(x-1)^2+(mx-m)^2=9\iff x^2(1+m^2)-2x(1+m^2)+m^2-8=0\ \ \ \ (2)$$
If $(x_1,y_1);(x_2,y_2)$ are the points of intersection, we have $\displaystyle x_1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/577473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Boolean Algebra simplify minterms I have this equation
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
and need to simplify it. I have got as far as I can and spent a good 2 hours at it. I've realized I now need to use De Morgan's law to continue however I am baffl... | Picking up where you (initially) left off (your work thus far is correct, save for the last line were you remove $\overline B$)...
$$\begin{align} &\overline{A}\cdot\overline{B}\cdot\overline{C} + A\cdot\overline{B}\cdot C + A\cdot B\cdot \overline{C} + A \cdot B\cdot C \\ \\
&\vdots \\ \\
&=\overline A \cdot \overline... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Series: $ \sum_{n = 1}^{\infty} \left( \prod_{i = 1}^n (2i - 1) \middle/ \prod_{i = 1}^n 2i \right)^k $ In Section 10.16 of Apostol's Calculus (Problem 18), the reader is asked to prove that the series
$$ \sum_{n = 1}^{\infty} \left( \prod_{i = 1}^n (2i - 1) \middle/ \prod_{i = 1}^n 2i \right)^k $$
converges if $k > 2$... | By the generalized mean value theorem, for $0<x<1$ we have
$$
\ln(1-x) = - x - \frac1{(1-\xi)^2} \frac{x^2}2 \quad\text{and}\quad e^{-x} = 1 - x + e^{-\eta} \frac{x^2}2
$$
for some $0\le\xi,\eta\le x$. In particular, for $0<x\le\frac12$ we have
$$
-x-2x^2 \le \ln (1-x) \le -x \quad\text{and}\quad 1-x \le e^{-x} \le 1 -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Limit at infinity involving $e$ I am to find the limit of $$\lim_{x \to \infty} \left(1+\frac{x}{5x^3+x^2+8}\right)^ {\dfrac{x^3+8}{x}}$$
I could not find the proper substitution here. I would be happy if someone could shed some light. Thanks.
| $$\lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^ {x^2+8/x}=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^ {x^2+8/x}=$$
$$=\lim_{x\to\infty}\left(\left(1+\frac{1}{5x^2+x+8/x}\right)^{5x^2+x+8/x} \right)^{\frac{x^2+8/x}{5x^2+x+8/x}}=e^{1/5}$$
because
$$\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{5... | {
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"url": "https://math.stackexchange.com/questions/581138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$? Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this:
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.
| We have:
$$ \int{\frac{4x+4}{x^4+x^3+2x^2}} dx $$
We need to decompose this fraction into pieces and then integrate each one separately. We start by simplifying the denominator:
$$ \frac{4x+4}{x^2(x^2+x+2)} $$
Since we have two quadratic terms in the denominator, we can guess the form:
$$ \frac{4x+4}{x^2(x^2+x+2)} = \f... | {
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"timestamp": "2023-03-29T00:00:00",
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The indefinite integration of $\frac{1}{\sqrt{n^4-1}}$ I need the indefinite integral:
$$\int\frac{1}{\sqrt{n^4-1}}dn $$
I know it has a relation with the $\tanh^{-1}$ function, but can't find a proper substitution.
| Case $1$: $|n^4|\geq1$
Then $\int\dfrac{1}{\sqrt{n^4-1}}dn$
$=\int\dfrac{1}{n^2\sqrt{1-\dfrac{1}{n^4}}}dn$
$=\int\dfrac{1}{n^2}\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k}}{4^k(k!)^2}dn$
$=\int\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k-2}}{4^k(k!)^2}dn$
$=\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k-1}}{4^k(k!)^2(-4k-1)}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\lim\limits_{x \to a} \frac{x^m-a^m} {x-a}$ How to evaluate this:
$$\lim\limits_{x \to a} \frac{x^m-a^m} {x-a} ; m\in \mathbb{N}$$
if i take:
$ m = 1 $
$$\lim\limits_{x \to a} \frac{x^1-a^1} {x-a} =1 $$ Is this correct?
but how evaluate limit where $ m = 2 $
$$\lim\limits_{x \to a} \frac{x^2-a^2} {x-a} ... | so if i take $ m = 3 $
$$\lim\limits_{x \to a} \frac{x^3-a^3} {x-a} = \lim\limits_{x \to a} \frac{(x-a)(x^2+ax+a^2)} {x-a}= \lim\limits_{x \to a} (x^2+ax+a^2)= 3a^2 $$
so if i understood
$$\lim\limits_{x \to a} \frac{x^m-a^m} {x-a} = m \cdot a^{m-1} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I al... | $$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2 x}\right)=\lim_{x\to 0}\frac{\tan x-x}{x^3}\cdot\frac{x+\tan x}{x}\cdot\left(\frac{x}{\tan x}\right)^2=
2\cdot\lim_{x\to 0}\frac{\sec^2 x-1}{3x^2}=$$ $$=\frac{2}{3}\cdot\lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{2}{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to find algebraic simplification for recurrence relation with closed-form solution, specifically for the Lucas-Lehmer primality test I have a question based on the section Proof of correctness in the article Lucas-Lehmer primality test, see following link.
https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primalit... | The whole setup began with Lucas. For given integers $P,Q$ we get Lucas sequences $U_n, V_n,$ where $V_n = a^n + b^n.$
In your case, $P=14, \; Q=1.$
Among the relations we find $$ V_{2n} = V_n^2 - 2 Q^n, $$ the fact that $Q=1$ takes us to
$$ V_{2n} = V_n^2 - 2; $$ furthermore $V_0 = 2, V_1 = P.$
For your application... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the smallest Finite Field in which the following polynomial is factorizable to irreducible factors? What is the smallest Finite Field in which the following polynomial is factorizable to irreducible factors?
$$(x^2+x+1)(x^5+x^4+1)(x^7+x^6+x^3+1) $$
| There is no finite field in which all given three polynomials are irreducible. The last
polynomial, i.e., $x^7+x^6+x^3+1$ has always $-1$ as a root, hence is reducible over all finite fields.
On the other hand, $\mathbb{F}_3$ is the smallest finite field, in which all three polynomials are reducible: $x^2+x+1=(x+2)^2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
function question
$ f(x) = \begin{cases} -1 && \text{for}-2\le x \le0\\ x-1 && \text{for } 0<x\le2
\end{cases}\\\text{Domain of }f(x) = [-2,2]\\
\text{Let there be be another function g(x),such that}\\
g(x) = f(|x|) + |f(x)| \\
$
$\text{What is function g(x)?}$
I worked the question and got $g(x) = \begin{cases}
0 &... | There's nothing wrong with your answer.
Case 1. $-2\leq x \leq 0$
\begin{array}{rl}
g(x)&=f(-x) + |f(x)| \\
&=(-x)-1 +|-1| & \\
&=-x
\end{array}
Case 2. $0< x < 1$
\begin{array}{rl}
g(x)&=f(x) + |x-1| \\
&=x-1 + [-(x-1)] & \\
&=x-1-x+1\\
&=0
\end{array}
Case 3. $1 \leq x \leq 2$
\begin{array}{rl}
g(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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