Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove by induction that $21 | 4^{n+1} + 5^{2n-1}$ The problem that I have is: Prove by induction that 21 divides 4n+1 + 52n-1
So far I have:
Base Case:
$n = 1$
$4^{1+1} + 5^{2-1} = 4^2 + 5^1 = 16 + 5 = 21$
Inductive Step:
Assume: $4^{k+1} + 5^{2k-1} = 3m$
$4^{(k+1)+1} + 5^{2(k+1)-1} = 3m$
$4^{k+2} + 5^{2k+1} = 3m$
I'm ... | Consider this:
$$4^{n+2} + 5^{2n+1} = 4\cdot 4^{n+1} + (21+4)\cdot 5^{2n-1} = 21\cdot5^{2n-1} + 4\cdot (4^{n+1}+5^{2n-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Why is this sum equal to $0$? While solving a differential equation problem involving power series, I stumbled upon a sum (below) that seemed to be always equal to $0$, for any positive integer $s$.
$$
\sum_{k=0}^s \left( \frac{ \prod_{r=1}^k (-4r^2+10r-3) \prod_{r=1}^{s-k} (-4r^2+6r+1)}{2^s (2k)! (2s-2k+1)!} \times (2... | Let
$$
\alpha=\frac{5+\sqrt{13}}{4},\beta=\frac{5-\sqrt{13}}{4}
$$ be the roots of equation $4r^2-10r+3=0$, and
$$
\gamma=\frac{3+\sqrt{13}}{4},\delta=\frac{3-\sqrt{13}}{4}
$$
be the roots of equation $4r^2-6r-1=0$. Note that $\alpha-\gamma=\beta-\delta=\frac{1}{2}$,$\ \alpha+\beta=\frac{5}{2}$ (this will be used bel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Evaluate the Integral: $\int\frac{e^{2x}}{e^{2x}+3e^x+2}\mathrm{d}x$ $$\int\frac{e^{2x}}{e^{2x}+3e^x+2}\mathrm{d}x$$
Step one- Factor the denominator
${e^{2x}+3e^x+2}$
$e^x\quad\quad\quad +1$ $=1e^x$
$e^x\quad\quad\quad +2$ $=2e^x$
$\quad\quad\quad\quad\quad\quad\quad3e^x$
$(e^x+1)(e^x+2)$
Step two: Further Decompositi... | Set $e^x=t$, we then have that $e^xdx = dt$. This gives us that
$$I=\int \dfrac{e^{2x}}{e^{2x}+3e^x+2} dx = \int \dfrac{t}{t^2+3t+2}dt = \int\dfrac{2dt}{t+2} - \int\dfrac{dt}{t+1}=2\ln(t+2)-\ln(t+1) + c$$
Hence,
$$I = 2\ln(2+e^x) - \ln(1+e^x) + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $ \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x} $. I need to solve the following integral:
$$
I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.
$$
Wolfram Alpha gives the answer as $ \dfrac{\pi}{2 \sqrt{2}} $.
I think it’s achievable by complex analysis, but I really ... | I think I have a different answer.
Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then
$$\begin{align}
\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx
&= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\
&= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du
\end{align}$$
Now let ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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The series of reciprocals of the least common multiples of integers $X_1
Let $X_1, X_2,\dots$ be a sequence of strictly increasing positive integers.
For each $n\ge1$, let $W_n$ be the least common multiple of the first $n$ terms $X_1, X_2,\dots, X_n$ of the sequence.
I need to prove the following statement:
The seri... | $\newcommand{\lcm}{\operatorname{lcm}}$
Note that for $n>1$ the least common multiple $W_n$ of natural numbers $1\leq X_1<X_2<\ldots<X_n$ fulfills
\begin{align*}
\lcm(X_{n-1},X_n) \leq W_n
\end{align*}
The following theorem was originally conjectured by P. Erdös:
Theorem: Let $X_0,\ldots,X_n$ be integers satifying... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
Find the green function for the following BVP by using dirac delta function and solve the BVP using Green function Here's the question:
$y'' + y = x^2 + 1, y(0) = 5, y(1) = 0$
I manage to get a solution to be this:
$y = -6\cot(1)\sin(x) + 6\cos(x) + x^2 - 1 - 2\cos(1)\sin(x) + 2\cot(1)\sin(1)\sin(x)$
Can somebody help... | First of all, the last two terms of your solution cancel out since
$\cot(1)\sin(1) = \cos(1)$.
Then, $$y_p = -6\cot(1)\sin(x) + 6\cos(x) + x^2 - 1$$
$$y'_p = -6\cot(1)\cos(x) + 6\sin(x) + 2x$$
$$y''_p = 6\cot(1)\sin(x) - 6\cos(x) + 2$$
Thus, $y''_p + y_p = x^2 +1 $, and the boundary conditions are also satisfied,
$$y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equation of the plane which passes through the point of intersection of two space lines and at greatest distance from the point $(0,0,0)$ The equation of the plane which passes through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and ... | An alternative to your approach is to this geometrically. To be the furthest distance then the plane must be perpendicular to the vector $(4,3,5)$. So to find the points of the plane $(x,y,z)$ we have:
$$(4,3,5)\cdot((x,y,z)-(4,3,5))=0$$
$$4x+3y+5x-16-9-25=0$$
$$4x+3y+5x=50$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating $\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}$ Find the following limit
$$
\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}
$$
I have used natural logarithm to get
$$
\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)
$$
After this, I have tried l'opital's rule but I was... | Solution Procedure Using Taylor Series and L'Hoptial
Try to understand or prove each of the following steps:
1) $\ln \left( {\frac{{1 + x{2^x}}}{{1 + x{3^x}}}} \right) = \ln \left( {1 + x\frac{{{2^x} - {3^x}}}{{1 + x{3^x}}}} \right)$
2) $\ln (1 + x) = x + o({x^2})$
3) $o\left( {{x^2}{{\left( {{{{2^x} - {3^x}} \over {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solve modular arithmetic equation $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$ From another problem, I have reduced it to: This is the last step in solving:
Solve $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$
How should I begin, a major problem is the $1/24$
| You want to find $n$ such that
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{240}\tag1$$
By the way, $$\begin{align}(1)&\Rightarrow n(n+1)(n+2)(n+3)\equiv 4\pmod{10}\\\\&\Rightarrow n\equiv 1,6\pmod{10}\end{align}$$
So, you only need to check if each of the following 48 numbers satisfies $(1)$ :
$$1,11,21,\cdots, 231$$
$$6,16,26,\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove $\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$ using the half angle identities. Given is the sequence $x_1=0,\; x_{n+1}=\sqrt{2+x_n}$. Prove: $$\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$$
Hint:
Use the following formulas:
$$\cos\left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$$
$$\sin\left(\frac{x}{2}\... | By mathematical induction it can be proved that $x_n=2\cos(\frac{\pi}{2^n})$.
for base of induction $x_0=0=2\cos(\frac{\pi}{2})$. Also we have $x_{n+1}=\sqrt{2+x_n}=\sqrt{2+2\cos(\frac{\pi}{2^n})}=2\sqrt{\frac{1+\cos(\frac{\pi}{2^n})}{2}}=2\cos(\frac{\pi}{2^{n+1}})$. Therefore $\sqrt{2-x_n}=2\sqrt{\frac{1-\cos(\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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If $f(x)=x^3+3x+2$ and $g(x)$ is the inverse of it, find the area bounded by $g(x)$ and $x$-axis and ordinates $x=-2$ and $x=6$. Is there any way to solve it without finding inverse of the function because finding I can't find inverse?
| Here are the graphs of $f(x) = x^3 + 3 x + 2$ (red) and $g(x)$ (green) and the lines $x = -2$ and $x = 6$.
(Large version)
We see that the area $A$ between $g$ and the $x$-axis for $x \in [-2,6]$ is the same as the area between $f$ and the $y$-axis for $y \in [-2, 6]$.
The complication is the change of sign of $f$ at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum up a function series $f(1/9)+f(2/9)+\dots+f(26/9)$ for $f(x)=\frac{9^{x}}{9^{x}+27}$ Given $f(x)=\dfrac{9^{x}}{9^{x}+27}$.
Find:
$$S=f\left(\frac{1}{9}\right)+f\left(\frac{2}{9}\right)+\dotsb+ f\left(\frac{26}{9}\right)$$
Teacher did not allow us to use calculator...Use sigma notation???
| Observe that $f(x)=\frac{1}{1+\frac{27}{9^x}}$ and $f(3-x)=\frac{1}{1+\frac{9^x}{27}}$. Adding these two expressions gives $f(x)+f(3-x)=1$. (It may be easier to see this by making a substitute such as $a=\frac{27}{9^x}$ and $a^{-1}=\frac{9^x}{27}$). Therefore, $$S=f\left(\frac{1}{9}\right)+f\left(\frac{2}{9}\right)+\do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $ \sum_{k=0}^n k4^k = \frac 49((3n-1)4^n + 1) $ by induction Prove that for every position integer $n$ that
$$ \sum_{k=0}^n k4^k = \frac 49((3n-1)4^n + 1) $$
Proof: Let $P(n)$ denote the above statement.
Base case: $n=1$ : Note that $$ \sum_{k=1}^1 k4^k = \frac 49((3(1)-1)4^{(1)} + 1) $$
$\frac 49((3(1)-1)4^{(... | All you need to do is expand the expression you got after using the inductive hypothesis (in general, if you don't know where to go, at least do something you know you can do!). You get:
$$
\begin{split} \frac{4}{9}((3s-1)4^s+1)+(s+1)4^{s+1}&= \frac{1}{9}(3s-1)4^{s+1}+\frac{4}{9}+s\cdot 4^{s+1}+4^{s+1} \\ &=\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Simplify $2 \sqrt[3]{50x^2 z^5} × 3 \sqrt[3]{15 y^3 z}$ Image of problem
As you can see the answer is $30 y z^2 \sqrt[3]{6x^2}$ . I understand mostly everything in the problem, but one thing that I am having confusion on is where the "30" came from in the answer. I tried many ways to find out how the "30" got there.
| You have:
$$6 \sqrt[3]{5^2\cdot 2 \cdot x^2 \cdot z^5 \cdot 3 \cdot 5 \cdot y^3 \cdot z} = 6 \sqrt[3]{5^3\cdot 6 \cdot x^2 \cdot z^6 \cdot y^3} = 30 \cdot y \cdot z^2 \sqrt[3]{ 6 \cdot x^2} $$
I hope its clear now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Double integral over the region enclosed by 4 curves. I'm a bit stuck on a problem that involves trying to integrate the area between 4 curves.
Problem:
$\int \int_D x^2+y^2 dxdy$
Where D is the region enclosed by the curves $xy =2, xy=7, y= 2x^2$ and $y=5x^2$.
For this, I set $u=y/x^2$, $v=xy$ and then work through th... | With $u=y/x^2$, $v=xy$, we first need to write $x^2+y^2$. Solving for $y$ on both equalities we obtain $x^2u=v/x$, so $x^3=v/u$. Then $x^2=(v/u)^{2/3}$. Squaring the second equality, $v^2=x^2y^2$, so $y/u=v^2/y^2$, or $y^3=v^2/u$. Then
$$
x^2+y^2=\left(\frac{v}{u}\right)^{2/3}+\left(\frac{v^2}{u}\right)^{2/3}.
$$
The J... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Real polynomials with $P(x^3-2)=P(x)^3-2$ Find all real polynomials $P(x)$ such that $P(x^3-2)=P(x)^3-2$.
Clearly $P(x)=x$ works. If $P(x)=ax+b$ is linear, then $P(x^3-2)=ax^3-2a+b$ and $P(x)^3-2=a^3x^3+3a^2bx^2+3ab^2x+b^3-2$, so $ab=0$ and $a^3=a$ and $-2a+b=b^3-2$.
If $b=0$, then the last equation implies that $a=1$... | This is not a complete answer as well but I hope the following result will help. This result guarantees there exist exactly one solution for polynomials of degree which is a power of $3$. If one can prove there exist no solution for polynomials with degree not a power of $3$ then the whole proof can be completed.
Lemma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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How to find matrix from general solution?
Find a $2 \times 3$ system (two equations with three unknowns) such that its general solution has form $\begin{pmatrix}1\\1\\0 \end{pmatrix}+s\begin {pmatrix}1\\2\\1\end{pmatrix},\ s \in \Bbb R.$
I tried thinking that $s\begin {pmatrix}1\\2\\1\end{pmatrix}$ is solution to ke... | One can translate the first two components of $u= (x,y,z)^T= (s+1, 2s+1, s)^T$ into the following system $Au=b$, the third component stays free and acts as $z=s$:
$$
x = s + 1 = z + 1 \iff x - z = 1 \\
y = 2s + 1 = 2 z + 1 \iff y - 2 z = 1 \\
$$
Or written as augmented matrix
$$
(A\mid b) =
\left(
\begin{array}{rrr|r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| In what follows, I will show how to start from the l.h.s. written as
\begin{equation*}
\sqrt{A+\sqrt{B}}
\end{equation*}
and arrive to the r.h.s written as
\begin{equation*}
\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ \ with\ \ \ C=\sqrt{A^{2}-B}.
\end{equation*}
To this end, make use of the following standard ide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Proofs of basilar powers identities We all know that a simple and intuitive way to show what $2^n$ is (for $n$ an integer number) is to write it as
$$2^n = \underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}$$
My question is: what is the most intuitive method to show the meaning of these two identiti... | By the first relation
$$2^{n}=\underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}=\underbrace{2\times 2\times 2\times \cdots \times 2}_{n-1\ \text{times}}\times2=2^{n-1}\times2.$$
Then $$2^{n-1}=\frac{2^n}2,\\
2^{n-2}=\frac{2^n}{2^2},\\
\cdots\\
2^{n-m}=\frac{2^n}{2^m}.$$
By the first relation
$$2^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the expression $( \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 ) / ( \cos 5x + 5 \cos 3x + 10 \cos x ) $ equal to? $$\frac{ \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 }{ \cos 5x + 5 \cos 3x + 10 \cos x }$$
My approach so far : Tried to represent the denominator as a factor of numerator by manipulating numerator's $\cos 6x =... | In a concept similar to the Chebyshev Polynomial,
For a fixed angle $x$, We let $T(n)=\cos nx$. Note the relation of $T(n+1)=2T(1)T(n)-T(n-1)$.
Now we have the numerator as $$T(6)+6T(4)+15T(2)+10=2T(1)T(5)-T(4)+6T(4)+15T(2)+10 = 2T(1)T(5)+5T(4)+15T(2)+10 = 2T(1)T(5)+5(2T(1)T(3)-T(2))+15T(2)+10=2T(1)T(5)+10T(1)T(3)+10T(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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What is the strategy to finding $\sum_{k=2}^{300} k^k \pmod 7$? I am stuck on this modular arithmetic problem for homework practice:
$$\sum_{k=1}^{300} k^k \pmod 7$$
I am not quite sure how to approach this problem. I've tried finding a pattern between the sums but i do not think there is one. I know how to find mods ... | It does repeat, but only after $7*6 = 42$ terms (the bases repeat in cycles of 7 while the exponents repeat in cycles of 6). Try breaking it up like this
$$\sum_{k=1}^{43} 1^{1 + 7(k-1)} + \sum_{k=1}^{43} 2^{2+7(k-1)} + \sum_{k=1}^{43} 3^{3+7(k-1)} + \sum_{k=1}^{43} 4^{4+7(k-1)} + \sum_{k=1}^{43} 5^{5+7(k-1)} + \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
if $2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$ then what is $f(x)$? assume that:
$$2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$$
Then what is $f(x)$?
| Assume that $u=\frac{x}{x^2+x+1}=\frac{1}{1+x+\frac{1}{x}}$. We have $\frac{1}{u}-1=x+\frac{1}{x}$, and by squaring:
$$ \frac{1}{u^2}-\frac{2}{u} = x^2+\frac{1}{x^2}+1,$$
hence:
$$ \frac{u^2}{1-2u} = \frac{1}{1+x^2+\frac{1}{x^2}} = \frac{x^2}{1+x^2+x^4} $$
and:
$$ f(u) = \color{red}{\frac{u^2}{2-4u}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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solution to a complex equation I'm asked to solve this equation :
$$z^4+7(1+i)z^2+25i=0$$
by taking $z^2 = u $ :
$$u^2+7(1+i)u+25i=0$$
this turns out to be a quadratic equation. solving it gives:
$$u_1=-6-8i$$
$$u_2=-8-6i$$
so how are we supposed to solve $z^2=-6-8i$ and $z^2=-8-6i$ ?
answeres are :
$$z=\pm \sqrt5(sin\... | $$z^4+7(1+i)z^2+25i=0\Longleftrightarrow$$
$$z^4+(7+7i)z^2+25i=0\Longleftrightarrow$$
Substitute $x=z^2$:
$$x^2+(7+7i)x+25i=0\Longleftrightarrow$$
$$(x+(3+4i))(x+(4+3i))=0\Longleftrightarrow$$
$$x+(3+4i)=0\Longleftrightarrow\space\space\vee\space\space x+(4+3i)=0\Longleftrightarrow$$
$$x=-3-4i\Longleftrightarrow\spac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$ To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.
| L'Hopital's rule may be used repeatedly until you find an answer:
$$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\lim_{x \to 0}\frac{2x-2\sin x}{\sin^2 x(\sin x+3x\cos x)}$$
$$= \lim_{x \to 0}\frac{2-2\cos x}{ 6\sin ^2(x)\cos (x) - 3x(\sin^3(x)-2\sin(x)\cos^2(x))}$$
$$= \lim_{x \to 0}\frac{2\sin (x)}{ 3(-3\sin^3 (x) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find two solutions of $299x+247y=13$ Applying Euclidean algorithm I got
$299=247+52$
$247=4\cdot 52+39$
$52=39+13$
$39=3\cdot 13+0$
Applying this in reverse order I got
$52-39=13$
$52-(247-4\cdot 52)=13$
$5(299-247)-247=13$
$5\cdot 299-6\cdot 247=13$
$x=5, y=-6$, this is one solution, what is the other solution?
| As $13|(299,247)=13$
from this, we shall have $13$ in-congruent solutions $\pmod{13}$
$299x+247y=13\iff23x+19y=1=23\cdot5-19\cdot6$
$\iff23(x-5)=-19(y+6)$
As $\dfrac{19(y+6)}{23}=5-x$ which is an integer,
$\implies23\mid(y+6)$ as $(19,23)=1$
$\iff y\equiv-6\pmod{23}$
and similarly, $x\equiv5\pmod{19}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form of $\sum\limits_{p = k}^{n - k + 1} C^{k - 1}_{p - 1} C^{k - 1}_{n - p}$ Does anyone has an idea how to prove that $$\sum\limits_{p = k}^{n - k + 1} \binom{p - 1}{k - 1} \binom{n - p}{k - 1} = \frac{n!}{(2 k - 1)! (n - 2 k + 1)!}?$$
| Suppose we seek to evaluate
$$\sum_{p=k}^{n-k+1} {p-1\choose k-1} {n-p\choose k-1}.$$
Introduce
$${n-p\choose k-1} = {n-p\choose n-k+1-p} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+2-p}} (1+z)^{n-p} \; dz.$$
Observe that this controls the range being zero when $p\gt n-k+1$ so
we may extend $p$ to infinity... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Locus of vertice of equilateral triangle. Here is a question:
On two mutually perpendicular lines, two points $A$ and $B$ are taken such that $A$ is a variable point and $B$ is a fixed point. An equilateral triangle $ABC$ is completed. Find the locus of point $C$.
I'm getting a quadratic equation but the back answer... | The midpoint between A and B is $$M=(\frac{a}{2},\frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$\vec{n} = (\frac{b}{\sqrt{a^2+b^2}},\frac{a}{\sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $\ell = \sqrt{a^2+b^2}$ is $$h = \frac{\sqrt{3}}{2} \sqrt{a^2+b^2}$$
Point C i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluate the following integral $\int_{-1/2}^{1/2}\big(\frac{\sin(n\pi f)}{\sin(\pi f)}\big)^4 df$ There are similar questions out there, but I was hoping someone could show how to would evaluate the following integral
$$\int_{-1/2}^{1/2}\bigg(\frac{\sin(n\pi f)}{\sin(\pi f)}\bigg)^4 df$$
I've seen the approach of usin... | Assuming $n \gt 0$. I would rewrite as
$$\frac1{2 \pi} \int_{-\pi}^{\pi} dt \frac{\sin^4{n t}}{\sin^4{t}} $$
We can express this as an integral over the unit circle in the complex plane and use the residue theorem. Thus, the integral is equal to, upon subbing $z=e^{i t}$:
$$\frac1{i 2 \pi} \oint_{|z|=1} \frac{dz}{z} ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I simplify $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ )$? $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ ) =$ ?
The answer is $3$.
My progress so far:
\begin{align}
\sqrt{3}(\cot 70^\circ+4\cos 70^\circ)&= \sqrt{3}(\tan 20^\circ+4\sin 20^\circ) \\
&= \sqrt{3}(\sin 20^\circ)\left(\frac{1}{\cos 20^\circ}+4\right)\\
&=... | To continue from your method:
$$\sqrt{3}\frac{(sin20+2sin40)}{cos20}=2\left(\frac{\frac{\sqrt{3}}{2}sin20+2\frac{\sqrt{3}}{2}sin40}{cos20}\right)$$ Using $\frac{\sqrt{3}}{2}=cos30$ we get
$$2\left(\frac{\frac{\sqrt{3}}{2}sin20+2\frac{\sqrt{3}}{2}sin40}{cos20}\right)=\left(\frac{2cos30sin20+2 \times 2cos30sin40}{cos20}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find out the number of solutions to the equation- $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ under certain constraints How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ where each $x_i$ is a non-negative integer, $3 \leq x_1 \leq 10, \ 2 \leq x_2 \leq 7$ and $x_3 \geq 5$
I have been ab... | The generating function would be
$$(x^3+x^4+...+x^{10})(x^2+x^3+...+x^7)(x^5+x^6+...)(1+x+x^2+...)^3$$
Simplify,
$$x^{10}(1-x^8)(1-x^6)({1\over 1-x})^6$$
Now we are finding the coefficient of $x^{25}$ of the above, which is same as $x^{15}$ in
$$(1-x^8)(1-x^6)({1\over 1-x})^6=(1-x^8-x^6+x^{14})({1\over 1-x})^6$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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$n^{th}$ determinant Find determinant $D_n$ of matrix $$
\begin{bmatrix}
1 & 1 & \cdots & 1 & -n \\
1 & 1 & \cdots & -n & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
1 & -n & \cdots & 1 & 1 \\
-n & 1 & \cdots & 1 & 1
\end{bmatrix}
$$
After multiplying first... | Let $D_m$ be the determinant of the $m\times m$ matrix like this with $-n$ in the antidiagonal and $1$ everywhere else. You are seeking $D_n$.
Note for $k\leq n$, we can find a relation on $D_k$ as follows. Multiply the top row by $n-k+2$, thus scaling the determinant by $n-k+2$.
Then add rows $2$ through $k$ to the fi... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$... Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$.
So far I have the base case completed, and believe I am close to completing the proof itself.
Base case:$(n=1)$
$3^1 + 7^1 - 2 = 8/8 = 1 $
I... | We have that $3^n+7^n-2=8k$ for some $k$. Now we substitute this expression into $3(3^n)+7(7^n)-2$ to get:
$3(8k-7^n+2)+7(7^n)-2=24k+4(7^n)+4=24k+4(7^n+1)$ where $7^n+1$ is even so we may rewrite $24k+4(7^n+1)=24k+4(2m)=24k+8m$ which is divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the conditi... | A partial counterexample $a=-3,b=-1,c=-4$ thus $b^2=1$ while ca+1=$-11$ thus i think the condition should be $a=<b=<c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Limit problem with roots I'm struggling solving the following limit problem:
$$\lim_{x \to \infty} \left(\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\right)$$
At first I thought I could
Multiply by:
$$\frac{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}$$
But that doesn'... | Alternatively:
\begin{align}
&\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\\
=&\frac{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\left(\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\right)\\
=&\frac{\left({x^9+x^8}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove this sum $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$ Show that
$$\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$$
| We can compute the generating function:
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k2^{2n-2k}\binom{2n-k+1}{k}x^{2n+1}
&=\frac12\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k2^{2n-2k+1}\binom{2n-k+1}{2n-2k+1}x^{2n+1}\tag{1}\\
&=\frac12\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{k+1}2^{2n-2k+1}\binom{-k-1}{2n-2k+1}x^{2n+1}\tag{2}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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What is the difference between the two real numbers that satisfy this equation?
What is the absolute difference between the two real numbers $x$ for which $(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$? Express your answer in simplest radical form
I tried guessing solutions but seeing how there are no common zeroes to both the ... | We have
$$(x^2-1)(x-2) = (x^2-9)(x+2) \implies x^3-2x^2-x+2 = x^3+2x^2-9x-18$$
which simplifies to
$$4x^2-8x-20 = 0 \implies x^2-2x-5 = 0 \implies (x-1)^2 = 6$$
Hence, the roots are $1\pm\sqrt6$, which means that the difference between the roots is $2\sqrt6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Euclidean Algorithm for Polynomials in $\mathbb F_7$
I have two polynomials $x^4+5x+1$ and $x^2-1$ and I'm trying to find their GCD over $\mathbb F_7$.
My attempt is:
$x^4+5x+1$ = $(x^2+1)(x^2-1)+5x+2$
$x^2-1 =\ ?$
I get lost on the second step...
| $x^4+5x+1=(x^2+1)(x^2-1)+5x-5=(x-1)(x^3+x^2+x+6)$
$x^2-1=(x-1)(x+1)$
$x^3+x^2+x+6$ has no $-1$ as a root
$\gcd(x^4+5x+1,x^2-1)=x-1$
Using the Euclidean algorithm: $x^2-1=(5x+2)(3x+1)+3(x-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate this definite double integral? $$\int\int_D 6x\sqrt{y^2-x^2}dA, D=\{(x,y)|0\leq y\leq 2, 0 \leq x \leq y\}$$
I tried: $$\int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy$$
But that is incorrect.
| Notice, $$\int_{0}^2\int_0^y 6x\sqrt{y^2-x^2} \ dxdy=\int_{0}^2\left(\int_0^y 6x\sqrt{y^2-x^2} \ dx\right)dy $$
$$=\int_{0}^2\left(-3\int_0^y (y^2-x^2)^{1/2} \ d(y^2-x^2)\right)dy $$
$$=\int_{0}^2dy\left(-3\frac{2(y^2-x^2)^{3/2}}{3}\right)_{0}^{y}$$
$$=\int_{0}^2\left(2y^3\right)\ dy$$
$$=\left(2\frac{y^4}{4}\right)_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The set of real solutions to an equation $$4^x - 7(2^{\frac{x-3}{2}}) = 2^{-x}$$
Set of real solutions is in which interval:
*
*$(-9, -2)$
*$(0, 3]$
*$(-2, 0]$
*$(7, 12]$
*$(3, 7]$
I tried the following. Dividing by $2^{-x}$ I get $2^{3x} - \frac{7\sqrt(2)}{4}2^{\frac{3x}{2}} - 1 = 0$. Substituing
$$
2^{3x} =... | Notice, $$4^x-7\cdot 2^{\frac{x-3}{2}}=2^{-x}$$
$$(2\sqrt2)2^{3x}-7\cdot 2^{\frac{3x}{2}}-2\sqrt2=0$$
let $2^{\frac{3x}{2}}=t$,
$$(2\sqrt 2)t^2-7t-2\sqrt 2=0$$
$$t=\frac{-(-7)\pm\sqrt{(-7)^2-4(2\sqrt 2)(-2\sqrt 2)}}{2(2\sqrt2)}=\frac{7\pm9}{4\sqrt 2}$$ but $t(=2^{\frac{3x}{2}})>0$ hence, one should get
$$2^{\frac{3x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Given a polynomial find the minimum value of the variable. If $x^5 - x^3 + x = a. $
Then we have to find the minimum value of $x^6$ in terms of a.
The answer given is $2a - 1$ if that gives any idea.
I have no idea how to approach this problem.
A hint would do fine.
| I reinterprete the problem a bit differently (based upon the proposed answer) to show that
$$\tag1x^5-x^3+x=a\implies x^6\ge 2a-1. $$
Note that $(1)$ is trivially true for $a\le \frac12$. Hence we may assume that $a$ is positive.
We have $$(x^2+1)a=(x^2+1)(x^5-x^3+x)=x^7+x $$
so that $x$ must be positive.
Divide by $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Evaluating limit by sandwich theorem: $\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$ $$\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$$
For using Sandwich theorem I need two functions such that $g(x)<f(x)<h(x)$
$$\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{... | Hint:
$$(1+2+3+\cdots+n)\frac{1}{n^2+n} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}\leq (1+2+3+\cdots+n)\frac{1}{n^2+1} $$
Limit is $\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Suppose $n|a^2-1$ Show that $n=$gcd$(a-1,n)$gcd$(a+1,n)$ Suppose $n|a^2-1$ where $a>1$ and n is odd. Show that $n=$gcd$(a-1,n)$gcd$(a+1,n)$.
Part 2
Show that if $a<n-1$ then this gives a nontrivial factorization of n
What I did:
I found the gcd$(a-1,a+1)$ which is $2$ if $a$ is odd and $1$ if $a$ is even. And that's it... | Since $n|(a^2-1)$, we have $\gcd(a^2-1, n)=n$.
First assume that $a$ is even. Then $\gcd(a-1, a+1)=1$. Therefore
$$\gcd(a^2-1, n)=\gcd(a-1, n)\gcd(a+1, n)$$
and we are done.
Now suppose $a$ is odd.
Then $\gcd(a^2-1, n)=\gcd((a^2-1)/4, n)$. This is because $n$ is odd.
Since $\gcd((a-1)/2, (a+1)/2)=1$, we get that
$$\gcd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Generating Functions To Deal With I've been working on producing a closed-form generating function for the
coefficients $a_n = \binom{n}{2}.$ I was wondering what might be a good
procedure to start on this. I get that
$$\sum_{n=0}^{\infty} x^n = x^0 + x^1 + x^2 + ... = \frac{1}{1-x}$$
by the convergence of geometric se... | Differentiate term-by-term twice:
$$
\begin{aligned}
\frac{d}{dx} \sum_{n=0}^\infty x^n
&= \sum_{n=1}^\infty n x^{n-1} \\
\frac{d^2}{dx^2} \sum_{n=0}^\infty x^n
&= \sum_{n=2}^\infty n(n-1) x^{n-2}.
\end{aligned}
$$
Next, since $a_n = \binom{n}{2} = \frac{n(n-1)}{2}$ and $a_0 = a_1 = 0$, we have
$$
\begin{aligned}
\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Roots of complex numbers - true/false Here's another one of the problems I'm having trouble with. Below are two statements and the book is asking me to tell for each whether it is true or false:
*
*All the solutions to the equation $\ \ z^3=i\ \ $ are:
$z_1=-i\ ,\ z_2=\frac{\sqrt3}{2}+\frac{i}{2}\ ,\ z_3=-\frac{\s... | $$z^2=i-\sqrt{3}\Longleftrightarrow$$
$$z^2=|i-\sqrt{3}|e^{\arg(i-\sqrt{3})i}\Longleftrightarrow$$
$$z^2=2e^{\frac{5\pi i}{6}}\Longleftrightarrow$$
$$z=\left(2e^{\left(2\pi k+\frac{5\pi}{6}\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$
$$z=\sqrt{2}e^{\frac{1}{2}\left(2\pi k+\frac{5\pi}{6}\right)i}$$
With $k\in\mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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I want to calculate the limit of: $\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $ I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$
or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be great... | Let $f(x) = \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $. For $x \to 0$, We have
$$\begin{align*}
\log f(x) &= \frac{1}{x} \log \left[ \frac{1}{2}\left(e^{x \log 2} + e^{x \log 8}\right) \right] \\
&= \frac{1}{x}\log\left(1 + x \log 4 + o(x)\right) \\
&= \frac{1}{x}\left(x \log 4 + o(x)\right) \\
&= \log 4 + o(1). ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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Find the Range of $f(a,b)=2a+b-3ab$ Let $a,b>0$,and such $$a^2+b^2-ab=4$$ Find the range
$$f(a,b)=2a+b-3ab$$
I try let $a=x+y,b=x-y$,then
$$a^2+b^2-ab=4\Longrightarrow x^2+3y^2=16,x>y,x>-y$$
so we Let
$$\begin{align}
x &=4\cos{t}, \qquad y=\dfrac{4}{\sqrt{3}}\sin{t} \\
f(a,b) &= 2a+b-3ab \\
&= \dfrac{3x}{2}+\dfrac{y}{2... | I think you have made a mistake in your computations as I did the arithmetic with my CAS. However, if we make the substitutions
$$\begin{align}
a &= 2 \cos t + \frac{2}{\sqrt{3}} \sin t = \frac{4}{\sqrt{3}} \sin(t+\frac{\pi}{3}) \\
b &= 2 \cos t - \frac{2}{\sqrt{3}} \sin t = \frac{4}{\sqrt{3}} \sin(t+\frac{2\pi}{3})
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}$
Find $\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}$
Using Taylor series:
$$\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}=\lim\limits_{x\to 0}\frac{(1+x^2+O(x^{2n}))-(1-\frac{x^{2\sqrt{2}}}{2}+O(x^{\sqrt{2}(2n+1)})}{x^2}=\lim\lim... | Starting from where you left off, we divide the numerator and denominator by $x^2$ we obtain $\frac{2+x^{2\sqrt{2}-2}}{2}$. Now $\lim_{x\to 0}{\frac{2+x^{2\sqrt{2}-2}}{2}}=\frac{2+0}{2}=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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$S1 = 1 + {x^3 \over 3!} + {x^6 \over 6!} + ...$ In one of my lecturer's problem sheets we were asked to evaluate the following sums:
$$S1 = 1 + {x^3 \over 3!} + {x^6 \over 6!} + \dots $$
$$S2 = {x^1 \over 1!} +{x^4 \over 4!} +{x^7 \over 7!} + \dots$$
$$S3 = {x^2 \over 2!} +{x^5 \over 5!} +{x^8 \over 8!} + \dots$$
In c... | Hint: We have
$$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$$ and
$$e^{\omega x}=1+\omega x+\frac{1}{2!}\omega^2 x^2+\frac{1}{3!}x^3+\cdots$$ and $$e^{\omega^2 x}=1+\omega^2 x+\frac{1}{2!}\omega x^2+\frac{1}{3!}x^3+\cdots.$$
Add. After finding $S_1(x)$ we can get the others by differentiating.
| {
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Calculate $\int_{-\infty}^{\infty}\;\left( \frac{x^2}{1+4x+3x^2-4x^3-2x^4+2x^5+x^6}\right) \;dx$
Calculate $$\displaystyle \int_{-\infty}^{\infty}\;\left(
\frac{x^{2}}{1+4x+3x^{2}-4x^{3}-2x^{4}+2x^{5}+x^{6}}\right) \;dx$$
The answer given is $\pi$. How does one calculate this?
| Let $F(x) = \frac{x^2}{P(x)}$ where
$$P(x) = x^6+2x^5-2x^4-4x^3+3x^2+4x+1 = (x^3+x^2-2x-1)^2 + (x^2+x)^2$$
Change variable to $u = \frac{1}{x+1} \iff x = \frac{1}{u}-1$. The integral at hand becomes
$$\int_{-\infty}^\infty F(x) dx
= \left(\int_{-\infty}^{-1^{-}} + \int_{-1^{+}}^\infty\right) F(x) dx
= \left(\int_{0^{-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimum integer solution of $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$ If $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$
find minimum integer N such that true for this inquality.
or maybe in the other cases for g... | The RMS-AM-GM-HM inequality is very unlikely to help with this problem as the terms are not close to each other. Here is an alternative approach:
The solution to the Basel problem tells us that $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.
You can bound $\dfrac{1}{n(n+1)} \le \dfrac{1}{n^2} \l... | {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{4\sqrt{7}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{1}{\sqrt{28}}}\right)}+\frac{1}{3}$ If $\dfrac{2\sqrt{19}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{7}{\sqrt{76}}}\right)}-\dfrac{1}{3}$ can be simpified to $2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right)$.
How to simpl... | Call $\theta=\frac{1}{3}\arccos \frac{1}{\sqrt {28}}$. We can use the formula
$\cos (3\theta)=4\cos^3\theta -3\cos \theta$
to deduce
$$\frac{1}{\sqrt {28}}=4\cos^3\theta -3\cos \theta. \tag{1}$$
The question is asking us to evaluate the quantity
$$ y=\frac{4 \sqrt 7}{3}\cos \theta+\frac{1}{3}$$
with $\cos \theta$ satis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Derive the generating function: $\frac1{(1-x)^3}= \sum_{n=0}^\infty \binom{n+2}{2}x^n$
$$\frac1{(1-x)^3} = \sum_{n=0}^\infty \binom{n+2}2x^n$$
I am not sure how to begin.
$$\begin{aligned}\frac1{1-x}&= \sum_{n=0}^{\infty} x^n\\\frac1{(1-x)^2}&=\sum_{n=0}^{\infty} nx^{n-1}\end{aligned}$$
Then you would get:
$$\frac{1}... | Take the derivative of both sides
$$\frac{d}{dx} \left (\frac{1}{(1-x)^2} \right) = \frac{d}{dx} \left ( \sum_{n=0}^{\infty} n x^{n-1} \right)$$
$$-(-2)\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} n (n-1) x^{n-2}$$
$$\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} \frac{n (n-1)}{2} x^{n-2} = \sum_{n=0}^{\infty} \frac{(n +2) (n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$
Attempt
Expanding the LHS we obtain $\left... | Use the AM GM inequality : $$a+a+b+c\geq 4(a^2bc)^\frac{1}{4}$$ $$a+b+b+c\geq 4(ab^2c)^\frac{1}{4}$$ $$a+b+c+c\geq 4(abc^2)^\frac{1}{4}$$ Multiply the three inequalities and then divide $abc$ on both sides to get the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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simple proof that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ It is well known that for $x>0$ that $\left(1+\frac{1}{x}\right)^x\le e\le\left(1+\frac{1}{x}\right)^{x+1}$ (see wikipedia). However, one can obtain the stronger inequality
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}... | I found a proof that seems well-motivated enough to satisfy me. The key lemma is the Hermite-Hadamard inequality (see wikipedia, or Old and new on the Hermite-Hadamard inequality).
Lemma (Hermite-Hadamard): Let $f:[a,b]\to\mathbb{R}$ be convex. Then $$
f\left(\frac{a+b}{2}\right)
\le\frac{1}{b-a}\int_a^b f(x)dx
\le \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
| $$LHS = \frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta}$$
$$\Longleftrightarrow\frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta} = \frac{1+\sin\theta}{\cos\theta}$$
$$\Longleftrightarrow(\sin\theta+1-\cos\theta)\cos\theta = (\sin\theta - 1+ \cos\theta)(1+\sin\theta)$$
$$\Longleftrightarrow\sin\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
| $$\lim _{ x\to -\infty } \left( \frac { x\left( \sqrt { x^{ 2 }-x } -\sqrt { x^{ 2 }-1 } \right) \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } } \right) =\\ =\lim _{ x\to -\infty } \left( \frac { x\left( x^{ 2 }-x-x^{ 2 }+1 \right) }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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For all nonnegative real numbers $x,y$ and $z$, prove that $\dfrac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$
For all nonnegative real numbers $x,y$ and $z$, prove that $$\frac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$$
It seems like AM-GM works here. We have $\dfrac{(x+y+z)^2}{3} \geq \dfrac{(... | Using AM-GM inequality we have:
$\displaystyle \sum_{\text{cyclic}} x\sqrt{yz}\leq \displaystyle \sum_{\text{cyclic}}\dfrac{x(y+z)}{2}=\displaystyle \sum_{\text{cyclic}} xy\leq \dfrac{2\displaystyle \sum_{\text{cyclic}} xy + \displaystyle \sum_{\text{cyclic}} x^2}{3}=\dfrac{\displaystyle \left(\sum_{\text{cyclic}} x\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\alpha= \frac{\pi}{7}$ and also if $\frac{3- \tan^2 (\alpha)}{1 - \tan^2 (\alpha)} = \lambda \cos(\alpha)$. Then find $\lambda$. Given that
$$\frac{3- \tan^2 \left(\dfrac{\pi}{7}\right)}{1 - \tan^2 \left(\dfrac{\pi}{7}\right)} = \lambda \cos\left(\dfrac{\pi}{7}\right)$$
Then find the value of $\lambda$.
| Notice, setting the value of $\alpha=\pi/7$, one can simplify LHS as follows $$LHS=\frac{3-\tan^2(\alpha)}{1-\tan^2(\alpha)}$$
$$=\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}$$
$$=\frac{1-\tan^2\frac{\pi}{7}+2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2\tan\frac{\pi}{7}}{\left... | {
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Solve $ \int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx $ I don’t know how to solve this integral:
$$\int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx$$
I know this is a rational integral but I don’t know how to write it in a different way
|
$$ \int{\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}}dx$$
Partial fractions:
You need to solve the equation: $$ \color{green}7x^2 +\color{red}0\cdot x +\color{blue}1 = A(x-1)(x+3) + B(x+1)(x+3) + C(x+1)(x-1) $$
$$=A(x^2+2x-3)+B(x^2+4x+3)+C(x^2-1)\\=\color{green}{(A+B+C)}x^2+\color{red}{(2A+4B)}x+\color{blue}{(-3A+3B-C)} $$
To ... | {
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Proof of Multiple angle trigonometri ratios If $\tan^2{A}=1+2\tan^2B$ then prove that $\cos2B=1+2\cos2A$
I could not relate from $\tan$ to $\cos$
| Notice, given that $$\tan^2A=1+2\tan^2B$$
$$\tan^2A+1=2+2\tan^2B$$
$$1+\tan^2A=2(1+\tan^2B)$$
$$\frac{1}{1+\tan^2B}=\frac{2}{1+\tan^2A}$$
$$\frac{1-\tan^2 B}{1+\tan^2B}=\frac{2(1-\tan^2 B)}{1+\tan^2A}$$
$$\cos 2B=\frac{2-2\tan^2B}{1+\tan^2A}$$
setting $2\tan^2B=\tan^2A-1, $
$$\cos 2B=\frac{2-\tan^2A+1}{1+\tan^2A}$$
$$\... | {
"language": "en",
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Positive definite binary quadratic forms Please help me to solve this question or introduce references that help me:
Let $f(x,y) = ax^2 + bxy + cy^2$ be a reduced positive definite form.
Suppose that $\gcd(x, y) = 1$ and that $f(x, y) ≤ a + |b| + c$. Show
that $f(x,y)$ must be one of the numbers $a, c, a - |b| + c$ ... | This answer breaks the problem into three cases: $y = 0$, $y = \pm 1$, and $|y| \ge 2$. Recall that because $f$ is reduced and positive definite, $|b| \le a \le c$.
Assume first that $y = 0$. Then because $\gcd(x, y) = \gcd(x, 0) = 1$, $x$ must equal $\pm 1$. In this case, $f(x, y) = f(\pm 1, 0) = a$, which satisfie... | {
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How to factor $x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1$ by hand? I know that
$x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1 = (x^4-x^3+x+1)(x^2+2x-1)$
but I would not know how to do that factoring without a software.
Some idea? Thank you!
| The equation is palindromic (well, almost), so:
We can write it as $$x^3\left[x^3+x^2-3x+2+\frac 3x+\frac{1}{x^2}-\frac{1}{x^3}\right]$$
$$=x^3\left[(x^3-3x+\frac 3x-\frac{1}{x^3})+\left((x-\frac 1x)^2+2\right)+2\right]$$
$$=x^3\left[u^3+u^2+4\right],$$ where $u=x-\frac 1x$
And hence the factorization is $$x^3(u+2)(u^2... | {
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How to solve $\lim\limits_{x \to +\infty} \left(\frac{(\ln(x+1))^x}{(2x)^{x/3}}\right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution with asymptotic approximations, or with variable change(so without Hopital)? Thanks
$$\lim\limits_{x \to +\infty} \left(\fr... | Using Taylor expansions:
$$
\ln(x+1) = \ln x + \ln\left(1+\frac{1}{x}\right) = \ln x + \frac{1}{x} + o\left(\frac{1}{x}\right)
$$
and therefore
$$
\ln(x+1)^x = e^{x\ln\left(\ln x + \frac{1}{x} + o\left(\frac{1}{x}\right)\right) }
= e^{x\ln\ln x + x\ln\left(1 + \frac{1}{x\ln x} + o\left(\frac{1}{x\ln x}\right)\right) }
... | {
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How to compute $\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution ?
$$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}... | Let $\frac{1}{x}=t\implies t\to 0$ as $x\to \infty$, $$\lim_{x\to +\infty}\frac{x^4+x^5\sin\left(\frac 1x\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x}=\lim_{x\to +\infty}\frac{\frac 1x+\sin\left(\frac 1x\right)}{\frac 1x\ln\left(\frac{1}{2+\frac1x}\right)-\frac 1{x^4}}$$
$$=\lim_{t\to 0}\frac{t+\sin t}{-t\ln(2+t)-t^4}$... | {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this inequality given the equation Given $ a>0 $ and $ a^5-a^3+a=3$, how can I prove the inequality: $ a^6 \ge 5 $ ?
I have tried factorizing the equation, solving for $ x, x^2, x^3 $ and then adding equations made together. Last but not least I tried, to solve for $x$ and then raise it to the power of $ x... | Given $a^5 - a^3 + a = 3$, we can factorise the left hand side to write $a(a^4-a^2+1)=3$.
The second factor here is reminiscent of cyclotomic polynomials/finite geometric series - in any case, we can write it as $\frac{a^6+1}{a^2+1}$, and clear fractions to get to $a(a^6+1)=3(a^2+1)$.
Now, we spot the $a^6$ we're look... | {
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Find the length of latus rectum of the conic $7x^2+12xy-2y^2-2x+4y-7=0$. Find the length of latus rectum of the conic $7x^2+12xy-2y^2-2x+4y-7=0$.
The given conic $7x^2+12xy-2y^2-2x+4y-7=0$ is a hyperbola because when i compare it with $ax^2+2hxy+by^2+2gx+2hy+c=0$ and found $h^2-ab>0$,
Then i tried to change $7x^2+12xy... | All of your algebraic work is correct, but you will want a different approach. The problem is in your conclusion, when you change variables back to $x,y$:
$$\frac{(5x+1)^2}{6}-\frac{(3x-y+1)^2}{3}=1$$
Now this is in the standard form, so $\ldots$
This is not in standard form because the numerator of the second frac... | {
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Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}.$
Let $a,b,$ and $c$ be positive real numbers such that $a+b+c = abc$. Prove that $$\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}.$$
This question seems tricky because how do w... | Let $a=\tan(A)$, $b=\tan(B)$ and $c=\tan(C)$. We then have $$\tan(A)+\tan(B)+\tan(C) = \tan(A)\tan(B)\tan(C) \,\,\,\, (\spadesuit)$$
Recall that $(\spadesuit)$ is true iff $A+B+C = \pi$, i.e., $A$, $B$ and $C$ are angles of a triangle. (Infact the part we need is trivial, since we have $\tan(C) = -\dfrac{\tan(A)+\tan(B... | {
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How to find the possible square roots of the two rowed unit matrix I?
How to find the possible square roots of the two rowed unit matrix I ?
I took a matrix like this $$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
and wrote,$A.A=I$
and got a set of equations
$a^2+bc=1$
$ab+bd=0$
$ac+cd=0$
$cb+d^2=1$
After that... | (Elaborating on Did's comment:) You already got that for
$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
$A^2 = I$ is equivalent to the four equations
$$
\begin{aligned}
a^2+bc &=1 \\
(a+d)b &=0 \\
(a+d)c &= 0 \\
d^2 + bc &= 1
\end{aligned}
$$
Now distinguish two cases:
Case 1: $a+d \ne 0$. Then $b=c=0$ must hold. I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that $\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=32$
If $11 \gamma = \pi$ then prove that
$
\sec(\gamma) \sec(2\gamma) \sec(3\gamma) \sec(4\gamma) \sec(5\gamma) = 32.
$
I could not use the relation $11 \gamma = \pi$.
| $$\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=\frac{1}{\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}}=$$
$$=\frac{1}{\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\left(-\cos\frac{8\pi}{11}\right)\cos\frac{4\pi}{11}\left(-\cos\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve the congruence $3x^2+x+8\equiv 0 \pmod{11}$ How to find the solutions of this congruence? $$3x^2+x+8\equiv 0 \pmod{11}$$
I need to find the inverse of $3$, and there I have a problem.
| In general, if $p$ is an odd prime and $p\nmid a$, then $$ax^2+bx+c\equiv 0\pmod{p}\iff 4a\left(ax^2+bx+c\right)\equiv 0\pmod{p}$$
$$\iff (2ax+b)^2\equiv b^2-4ac\pmod{p}$$
At this point, you're just left with finding the square roots of $b^2-4ac$ mod $p$.
In your case, $$3x^2+x+8\equiv 0\pmod{11}\iff (6x+1)^2\equiv 4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Calculating: $I=\int_{-1}^1{(1+x)^{m+n}P_n(x)}dx$ Where $P_n$ is a Legendre Polynomial. Calculating:
$$I=\int_{-1}^1{(1+x)^{m+n}P_n(x)}dx$$
Where $P_n$ is a Legendre Polynomial.
My progress:
Using Rodrigues formula:
$$I=\dfrac{1}{2^n n!} \int_{-1}^1{(1+x)^{m+n}\large(\dfrac{d}{dx}\large)^n(x^2-1)^n}dx$$
A first interat... | I can follow your derivation up to the point where
$$ I= \frac{1}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n} \frac{d^n}{dx^n} (x^2-1)^n.$$
I assume for the following that $m,n\in \mathbb{N}_0$ (as was probably intended).
Starting from there, your idea is to integrate by parts ($n$-times). Let us observe what happens after th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(x+y)(x+1)(y+1) = 3$ and $x^3 + y^3 = \frac{45}{8}$ I've came across this problem :
If $x$ and $y$ are real numbers such that
$(x+y)(x+1)(y+1) = 3$ and
$x^3 + y^3 = \frac{45}{8}$,
find $xy$.
This is what I've tried so far:
$$x^3 + y^3 = (x+y)(x^2-xy+y^2) =\frac{45}{8}$$
So $$\frac{45}{8(x^2 - xy + y^2)} = \f... | See, the whole thing is symmetric, so it would be natural to switch to the symmetric polynomials. Say, $xy=a$ and $x+y=b$. Then $(x+1)(y+1)=xy+x+y+1=a+b+1$, and $x^3+y^3=(x+y)^3-3x^2y-3xy^2=b^3-3ab$.
So $b(a+b+1)=3$ and $b^3-3ab={45\over8}$. From the first equation we have $a={3\over b}-b-1$. Plugging this into the sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Perfect square is $0$ or $1$ modulo $4$ . Prove that for every integer $n$ either
*
*$n^2 \equiv 0\pmod{4}$ or
*$n^2\equiv 1\pmod{4}$
| $n$ can be even or odd.
When $n=2k$ we get
$(2k)^2=4k^2 \equiv 0 \pmod 4$.
When $n=2k+1$ we get
$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Factorials/Binomial Coefficients (Finding Integer Solutions)
Question
There are many integer solutions to the equation $\begin{pmatrix}n\\r\\ \end{pmatrix} = \begin{pmatrix}n+1\\r-1\\ \end{pmatrix}$ including $n = r = 1$.
Find an expression for $n$ in terms of $r$, and hence find another of the integer solutions.
W... | Notice that the last equation implies that you have to find another solution to $5r^2-2r+1=z^2$, where $r,z$ are both naturals.
Multiplying by 5 gives us $25r^2-10r+5=5z^2$. $\therefore (5r-1)^2+4=5z^2$.
Since you are merely trying to find another set, say
$5r-1=2k,z=2m$. The question becomes, $k^2-5m^2=-1$.
However... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $6^{1000} \mod 23$
Find $6^{1000} \mod 23 $
Having just studied Fermat's theorem I've applied $6^{22}\equiv 1 \mod 23 $, but now I am quite clueless on the best way to proceed.
This is what I've tried:
Raising everything to the $4th$ power I have $$6^{88} \equiv 1 \mod 23 $$ $$6^{100} \equiv 6^{12} \mod 23 $$ $$... | We may exploit the fact that $1000\equiv 10\pmod{22}$ to get:
$$ 6^{1000}\equiv 6^{10} \equiv (6^5)^2 \equiv 2^2 \equiv\color{red}{4}\pmod{23}.$$
As an alternative, we may notice that $a^{11}\pmod{23}$ is the Legendre symbol $\left(\frac{a}{23}\right)$, so:
$$ 6^{11} \equiv \left(\frac{2}{23}\right)\cdot\left(\frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all complex roots of $T^4-{1/2}T^2-\sqrt{15}T+{69/16}$ I want to find all complex roots of $T^4-{1/2}T^2-\sqrt{15}T+{69/16}$.
The only way I can think to do it is to find 1 complex root, $\alpha$, by inspection, so we can rearrange the polynomial to be $(T-\alpha)(T^3+pT^2+qt+r) $then show the cubic equation is ... | It is better, I think, to try with $x^4-2x^2-8\sqrt{15}\space x+69$ where $x=2T$.
By some tedious calculation of undetermined coefficients we have
$x^4-2x^2-8\sqrt{15}\space x+69=[x^2-2\sqrt {5}\space x +(9-2\sqrt 3)]\cdot[x^2+2\sqrt {5}\space x +(9+2\sqrt 3)]$
Hence $x_{1,2}=\sqrt 5\pm\sqrt{4-2\sqrt 3}\space i$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$
Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$.
This question reminds me of rearrangement, but I can't really find two sequences that fit. Maybe there is a way usi... | Since $f(a,b,c)=LHS-RHS$ is homogeneous, we can asumme $abc=1$. Thus, inequalitie is equivalent to $a^5+b^5+c^5\ge a^2+b^2+c^2$ subject to $abc=1$.
Now, if $a\ge b\ge c$, by Tchebyshev's inequality,
$$\begin{eqnarray}\dfrac{a^5+b^5+c^5}{3}&=&\dfrac{a^3a^2+b^3b^2+c^3c^2}{3}\\&\ge&\dfrac{a^3+b^3+c^3}{3}\dfrac{a^2+b^2+c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute $\cos(\arctan(2)) = 1/\sqrt{5}$ I'm doing matrices and I rotated a line about an angle. The gradient of my line I'm rotating to the $x$-axis is $2$, from $y=2x$. So obviously the angle that the line $y=2x$ makes with the $x$-axis is $\arctan(2)$.
My question is how do I arrive at $\cos(\arctan(2)) = 1/\... | Well..
Let $\arctan 2 = x$.
$\arctan (2) = x \implies \tan x = \sin x/\cos x = 2 \implies \sin x = 2 \cos x \implies \sin^2 x + \cos^2 x = 4\cos^2 x + \cos^2 x = 1 \implies \cos^2 x = 1/5 \implies \cos x = \pm 1/\sqrt{5}$.
By convention $\arctan$ takes values in (-$\pi/2, \pi/2$) where $\cos$ is presumed to be positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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To evaluate the sum $\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots$ Right now I am working through archived papers of a math aptitude quiz. For some reason I seem to be haveing a hard time with these series problems. I have managed to write the above series in a compact f... | Lab Bhattacharjee is right. This is just an instance of the usual binomial series. More precisely,
$$
(1+x)^{-1/3}=1-\frac13x+\frac{1\cdot4}{3^2\cdot 2!}x^2-\frac{1\cdot4\cdot7}{3^3\cdot 3!}x^3+\cdots,
$$
that converges and has the prescribed sum, whenever $-1<x< 1$.
Plugging in $x=3/5$ gives
$$\frac12\root 3\of5=1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why it's $2^8-127$, not $2^8-(-127)$ when "two's complement of $a$" is $2^n-a$ in binary notation? "Table 2.5.1 Powers of 2
$$\begin{array} {|c|c|c|c||c|}
\hline
\text {Power of 2} & 2^{10}& 2^9 & 2^8 & 2^7& 2^6&2^5 &2^4 &2^3 &2^2 &2^1 &2^0 \\ \hline
2 & 1024 & 512 & 256 & 128 & 64 & 32 & 16 & 8 & 4 & 2 & 1\\ \hline
\... | Taking two's complement implies inversing a number ie binary digits ie if $0$ then $1$ and vice -versa and adding $1$ to LSB so binary $8$ -bit representation of $1$ is $00000001$ so taking inverse we get $11111110$ and adding $1$ we get $11111111$ if on subtraction the answer is negative then its in the form of 2's c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many non-negative integer solutions does the equation $3x + y + z = 24$ have? If the equation is $x + y + z = 24$ then it is solvable with stars and bars theorem. But what to do if it is $3x + y + z = 24$?
| Observe that $3x + y + z = 24$ implies that $y + z = 24 - 3x$. For a fixed value of $x$, a particular solution of the equation $y + z = 24 - 3x$ corresponds to where we place an addition sign in a row of $24 - 3x$ ones. For instance, if $x = 5$, $y + z = 9$ and
$$+ 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Evaluate $\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}dx$ Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$
What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$
| HINT:
Setting $x=\cos2y$
$$I=\int\dfrac{2-3x}{2+3x}\sqrt{\dfrac{1+x}{1-x}}dx=-2\int\dfrac{2-3\cos2y}{2+3\cos2y}\cdot\cot y\sin2y\ dy$$
$$=-2\int\dfrac{2(1+\tan^2y)-3(1-\tan^2y)}{2(1+\tan^2y)+3(1-\tan^2y)}\cos^2y\ dy$$
$$=-2\int\dfrac{5\tan^2y-1}{(5-\tan^2y)(1+\tan^2y)^2}\sec^2y\ dy$$
Set $\tan y=u$ to find $$I=-2\int\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Quartic with $4 $ equidistant roots Today I got the problem $(x^2 -1)(x^2 -4)=k$, and I have no idea how to prove this algebraically. $K$ is a real, non-zero number that makes the equation have $4$ distinct real equidistant roots. Solve for $k$ algebraically.
What I have attempted:
I have tried solving for x and then ... | As the roots are equidistant, let the roots be $a\pm d,a\pm3d$
using Vieta's formula, immediately we have $$a+d+a-d+a+3d+a-3d=0\iff a=0$$
and $5=(-d)(d)+(-d)(3d)+(-d)(-3d)+d(3d)+(d)(-3d)+(3d)(-3d)$
Now $4-k= d\cdot-d\cdot3d\cdot-3d$
Can you take it from here?
In fact $$x^4-5x^2+4-k=(x^2-d^2)(x^2-9d^2)=x^4-10d^2x^2+9d^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of value of infinite series, using definition of the integral How do I prove that $$\lim_{n \to \infty }n^2\left ( \frac{1}{(n^2+1)^2} +\frac{2}{(n^2+2^2)^2}+...+\frac{n}{(n^2+n^2)^2} \right )=\frac{1}{4}$$ using the definition of the integral?
| $$
n^2\sum_{k=1}^n\frac{k}{(n^2+k^2)^2}=n^2\sum_{k=1}^n\frac{k}{n^4(1+(k/n)^2)^2}=\frac{1}{n}\sum_{k=1}^n\frac{k/n}{(1+(k/n)^2)^2}\ ,
$$
which is a Riemann sum, yielding
$$
\int_0^1 dx \frac{x}{(1+x^2)^2}=1/4
$$
in the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding roots of cubic (trig)
The question is
By putting $x$ $=$ $\frac 23 \cos (\theta)$ Find the exact roots of the equation in terms of $\pi$
$$ 27x^3 - 9x = 1 $$
What I have attempted:
$$ 27x^3 - 9x = 1 $$
$$x=\frac 23 \cos (\theta)$$
$$ x^3=\frac 8{27} \cos^3 (\theta)$$
$$ \therefore 27\left(\frac 8{27} \cos... | Recall that
$$\cos^3 \theta = \frac 3 4 \cos \theta + \frac 1 4 \cos 3\theta$$
Hence you can rewrite your equation as
\begin{align*}
0 &= 8 \cos^3 \theta - 6 \cos \theta - 1 \\
&= 6 \cos \theta + 2 \cos 3\theta - 6 \cos \theta - 1 \\
&= 2 \cos 3\theta + 1
\end{align*}
and hence reduce to $\cos 3\theta = -\frac 1 2$. Af... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How can I calculate the sum $\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}$? I have to calculate the partial sum for an equation. How can I calculate the sum for
$$\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}\ \text{?}$$
Thanks.
| Notice, use partial fractions: $\frac{1}{16n^2-8n-3}=\frac{1}{4}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)$ as follows $$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-3}$$
$$=\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)$$
$$=\frac{1}{4}\lim_{n\to \infty}\left(\left(\frac{1}{1}-\frac{1}{5}\right)+\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Area of the surface of revolution of the ellipsoid I need to find the surface area of an ellipsoid using the equation of an ellipse. I believe my calculations are correct but the formulas I meet on the Internet are complex and have $\arcsin$ or $\arctan$ in the equation, which makes me believe that there is something ... | The graph $y = \sqrt{a^{2} b^{2} - x^{2}}$ is the upper half of the circle
$$
x^{2} + y^{2} = a^{2} b^{2}
$$
of radius $|ab|$. Revolving gives a portion of a sphere, so it's not entirely a surprise the area element came out to be independent of $x$.
The correct area integral for a spheroid comes from
$$
y = \frac{b}{a}... | {
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"timestamp": "2023-03-29T00:00:00",
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How can I compare the numbers $2^{39}$, $5^{19}$ and $52^7$? I have to compare the numbers $2^{39}$, $5^{19}$ and $52^7$. I don't know how to do that because their exponents don't have anything in common.
| $$ 52^7 = (26^7 \cdot 2^7) > ((2^{4.7})^7 \cdot 2^7) = 2^{39.9} > 2^{39} $$
As $26 = 2^{\frac{\ln 26}{ \ln 2}} \approx 2^{4.7004}$.
Also,
$$ 52^7 < 6^7 \cdot 10^7 = 6^7 5^7 2^7 < 6^7 5^7 5^4 = 5^{7 \cdot \frac{\ln 6}{\ln 5}} 5^7 5^4 = 5^{7.793 + 7 + 4} < 5^{19}$$
Using $2^7 < 5^4$.
Thus,
$$ 2^{39} < 52^7 < 5^{19}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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Level Curves in multiple variabes I want to sketch some level curves to the function $f(x,y) = 3x^2 + 4xy + 3y^2$. I have tried to set the equation to C but I'm not sure how to proceed.
I've gotten hints to substitute $u = x + y$ and $v = x - y$ but it doesn't seem to make sense to me.
Thanks in advance!
| The curves are clearly elliptical.
To be more precise, notice that
$$
f(x,y) = (x,y)A\binom xy
\quad\text{where}\quad
A = \begin{bmatrix}3 & 2\\ 2& 3\end{bmatrix}
$$
Infinite many such $A$ exist, the one above is the one which is symmetric and hence diagonalizable.
Its eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 5... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$ Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$.
I begin by letting $n=1$ then $\frac{1}{2}<\frac{1}{\sqrt{3}}$. Then assume $\frac{1\cdot 3\cdots(2k-1)}{2\cdot 4 \cdots(2k)}<\frac{1}{\sqr... | First observe that
$$\frac{2n-1}{2n}<\frac{2n}{2n+1}\qquad\text{since }(2n-1)(2n+1)=4n^2-1<4n^2=(2n)(2n)$$
Then
\begin{align}
\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)^2&<\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)\left(\prod_{k=1}^{n}\frac{2k}{2k+1}\right)\\
&=\prod_{k=1}^{n}\frac{2k-1}{2k+1}\qquad\text{this product... | {
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"url": "https://math.stackexchange.com/questions/1627933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Bound on sum of combinations I came across the following inequality $\sum_{i=0}^D \binom N i \le N^D+1$.
I am not sure how to prove this. I tried to do it by induction on $D$, and started with observing the values of sum for different values of $N$, and $D$. The values are as follows:
$$
\begin{matrix}
... | We shall prove the statement
$$\sum \limits _{i=0} ^D \binom N i \le N^D + 1, \ \forall N \ge 1, \forall \ 0 \le D \le N$$
using induction on $N$.
For $N = 1$, if $D = 0$ then $\binom 1 0 = 1 \le 1 ^0 + 1$, and if $D = 1$ then $\binom 0 0 + \binom 1 0 = 1 + 1 = 2 \le 1^1 + 1$. Thus, for $N = 1$ the statement is verifie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The integral $\int\frac{{\rm d}x}{(x+3)^{\frac87}(x-2)^{\frac 67}}$ I need to solve the integral
$$\int\frac{{\rm d}x}{(x+3)^{\frac87}(x-2)^{\frac 67}}$$
I can't seem to be able to set it in a suitable form; and I know it has to do something with the fact that $\frac87+\frac67=2$ which here would be a negative integer;... | Let $$I = \int\frac{1}{(x+3)^{\frac{8}{7}}\cdot (x-2)^{\frac{6}{7}}}dx = \int\frac{1}{\left(\frac{x-2}{x+3}\right)^{\frac{6}{7}}\cdot (x+3)^2}dx$$
Now Put $$\left(\frac{x-2}{x+3}\right) = t\Rightarrow 1-\frac{5}{x+3} =t$$
So we get $$\frac{1}{(x+3)^2}dx = \frac{dt}{5}$$
So we get $$I = \frac{1}{5}\int t^{-\frac{6}{7}}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve in integers $b^{11}-1=a^{2016}+a^{2015}+\dots+1$
Find all integers $(a,b)$ satisfying $$b^{11}-1=a^{2016}+a^{2015}+\dots+1.$$
Obviously, we can get the factorisation $(b-1)(b^{10}+\dots+1)=a^{2016}+a^{2015}+\dots+1$, but I'm not sure how to proceed from there.
| Note that $a^{2016}+a^{2015}+\dots+a+1$ has only prime factors that are $0,1$ modulo $2017$.
This implies that $b^{11} \equiv 1,2 \pmod {2017}$.
Therefore, $b^{2013} \equiv 1,2^{183} \pmod {2017}$ or $b^3,2^{183}b^3 \equiv 1 \pmod {2017}$.
This implies that $b \equiv 1 \pmod {2017}$ or $b^2+b+1 \equiv 0 \pmod {2017}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof for $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$ I'm looking for a proof of $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$, optionally $\ln(n)+\frac{1}{n+1}\leq \ln(n+1)$
| If $x \ge 0$,
$$e^x = 1 + \int_0^x e^t\, dt \le 1 + \int_0^x e^x\, dt = 1 + xe^x.$$
So $(1 - x)e^x \le 1$, and thus
$$e^x \le \frac{1}{1 - x},\quad 0 \le x < 1.$$
Letting $x = \frac{1}{n+1}$, we obtain
$$e^{\frac{1}{n+1}} \le \frac{1}{1 - \frac{1}{n+1}} = \frac{n+1}{(n+1)-1} = \frac{n+1}{n} = 1 + \frac{1}{n}.$$
Equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Ways to create a quadrilateral by joining vertices of regular polygon with no common side to polygon
How many ways are there to create a quadrilateral by joining vertices of a $n$- sided regular polygon with no common side to that polygon?
It's quite easy to solve for triangles for the same question, logic remains sa... | $1)$ All $4$ consecutive $=n$ is correct
$2)$ $3$ consecutive $=n(n-5)$ is also correct
$3)$ $2\times2$ consecutive $=\frac{n(n-5)}{2}$ You were double-counting in this case $(a,b,d,e) = (d,e,a,b)$
$4)$ $2\times1\times1=\frac{n(n-6)(n-5)}{2}$ [edited - this expression is cleaner]
Collectively:
\begin{align}&=n+2n\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Complex number ( prove ) Let
$$
{x-yi\over{x+yi}}=a+bi\;\;.
$$
Prove $a^2+b^2=1$
I don't know how to start prove it, can anyone help me?
| First off, the problem posed is flawed, because it forgot to specify that $x$ and $y$ are real. If not, try $x=i$ and $y=1$ to see that the relation does not hold.
Now you may want to prove this without knowledge of deMoivre's theorem (complex exponentials). Here is how:
$$\frac{x+iy}{x-iy}=\frac{x+iy}{x-iy}\frac{x+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$11^{-1}$ modulo $91$ is $58$. Why? I am reading wiki article about Quadratic Sieve and it says
$11^{-1}$ modulo $91$ is $58$
Why? How is it been calculated?
| The reason $58 = 11^{-1} \pmod{91}$ is
$$58 \cdot 11 = 638 = 7 \cdot 91 + 1 \implies 58 \cdot 11 \equiv 638 \equiv 1 \pmod{91}$$
The reason $11$ has an inverse $\pmod{91}$ is that $\gcd(11, 91) = 1$.
To calculate the inverse, we use the extended Euclidean algorithm. First, we use the Euclidean Algorithm to solve for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
solving rectangle Diagonal of a rectangle is $13$ cm. If we extend the length of the rectangle for $4$ cm and width for $7$ cm, then diagonal will be longer for $7$ cm as well. Find sides (length and width) of the rectangle.
So $d^2=a^2+b^2 \implies b^2=d^2-a^2 \implies b^2=13^2-a^2 \implies b^2=169-a^2$ and $a^2+b^2=1... | You can substitute in $a=\sqrt{169-b^2}$ and do the algebra. I don't think the numbers will be very pretty.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$? I solved the equation $\sqrt{6-5x}=x$ as follows:
$$(\sqrt{6-5x})^2=x^2$$
$$6-5x=x^2$$
$$0=x^2+5x-6=(x+6)(x-1)$$
$$x=-6 \quad \text{or} \quad x=1$$
If I plug in $x=-6$ into the original equation, I get $\sqrt{6+30}=\sqrt{36}=\pm 6$ and if I plug in $x=1$... | You may not select all solutions from an expression $B$ derived (calculated) from expression $A$ when $B$ is not equivalent to $A$, and have to refer to the initial expression to consider the valid values.
Here (notice the / within the $\Leftrightarrow$)
$$\sqrt{6-5x}=x\quad\not\Leftrightarrow\quad6-5x=x^2$$
because re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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Find the two values of $k$ for which $2x^3-9x^2+12x-k$ has a double real root.
Find the two values of $k$ for which $2x^3-9x^2+12x-k$ has a double
real root.
I've found one method which is to equate $$2x^3-9x^2+12x-k=2(x-r)^2(x-c)$$
Expanding and equating coefficients I get the system of equations :
\begin{array}
\... | From $2x^3-9x^2+12x-k=2(x-r)^2(x-c)$, using Vieta's formulas:
$$-\frac{-9}{2} = r + r + c \implies c = \frac{9}{2} - 2r$$
$$\frac{12}{2} = r \cdot r + r \cdot c + c \cdot r \implies6=r^2 + 2(9/2 - 2r)r$$
hence $6 = r^2 + 9r - 4r^2$ or $3(r^2 - 3r + 2) = 0 \implies r = 1, 2$. Using the remainder theorem, $2(1)^3 - 9(1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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