Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Help to solve and understand Simultaneous equations. Could anyone help me solve this simultaneous equation:
$$6x + 7y = 12.5\tag1$$
$$7x + 5y = 14\tag2$$
I have watched loads of videos on YouTube, and still don't understand how to do it. I am trying to learn it. And practice with different equations. But don't understa... | We have \begin{align}6x + 7y &= 12.5\tag{1}\\
7x + 5y &= 14\tag{2}\end{align}
One way to solve this, would be to rearrange equation $(1)$ for $x$:
\begin{align}6x + 7y &= 12.5\\
6x&=12.5-7y\\
x&=\frac{12.5-7y}{6}\end{align}
We can then put this into equation $(2)$ and solve for $y$:
\begin{align}7x + 5y &= 14\\
7\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ wil... | If I were you, I would try induction. First establish that it is true for $n=1$:
$$1*2*3*4=24=\frac{1(2)(3)(4)(5)}{5}$$
Then use induction by first assuming that, for some $k$,
$$1*2*3*4+...+k(k+1)(k+2)(k+3)=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$
Then add $(k+1)(k+2)(k+3)(k+4)$ to both sides:
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Find short and simple methods to solve $24x^4+1=y^2$
Find all this diophantine equation $$24x^4+1=y^2\tag{1} $$ postive integers solution
it is clear $(x,y)=(1,5)$
I know $y^{2}=Dx^{4}+1$, where $D>0$ and is not a perfect square, has at most two solutions in positive integers (cf. L. J. Mordell, Diophantine equat... | $\frac{y-1}{2} \cdot \frac{y+1}{2} = 6x^4$
So $\frac{y+1}{2} = p a^4, \frac{y-1}{2} = q b^4$ where $pq = 6$ and we need to solve the equation $p a^4 - q b^4 = 1
Case 1: $p = 6, q = 1$
This is impossible modulo 3.
Case 2: $p = 2, q = 3$
This is impossible modulo 3.
Case 3: $p = 1, q = 6$
We will show that there is no so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Characteristic curves : Why do I get different results? With a given $h(x)$ we want to solve $$xu_y-yu_x=u \\ u(x,0)=h(x)$$
I have solved it using two ways.
$$$$
First way:
For $x\neq 0$ we get $u_y-\frac{y}{x}u_x=\frac{u}{x}$.
We have that $$\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{... | They are a few typos in your answer.
And :
Also, the arctan and arccos must be considered as multivalued in the final equations. With convenient choice of signs $\quad \cos^{-1}\left(\frac{y}{\pm\sqrt{x^2+y^2}}\right)\pm\frac{\pi}{2}= \tan^{-1}\left(\frac{y}{x}\right)$
So, the result is the same for both ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Existence/Limit of Improper Integral I got the improper integral:
$$ \int_0^\infty \frac{x^2}{x^4+1} \, dx $$
On one hand one could of course just evaluate the integral and then apply limits.
However, it's not always practical to find an antiderivative.
I got a hint that one can trace back improper integrals to sequen... | $$I=\int_{0}^{1}\frac{x^2}{x^4+1}\,dx + \int_{1}^{+\infty}\frac{x^2}{x^4+1}\,dx = \int_{0}^{1}\left(\frac{x^2}{x^4+1}+\frac{1}{x^4+1}\right)\,dx \tag{1}$$
leads to:
$$ I = \int_{0}^{1}\frac{1+x^2-x^4-x^6}{1-x^8}\,dx = \sum_{k\geq 0}\left(\frac{1}{8k+1}+\frac{1}{8k+3}-\frac{1}{8k+5}-\frac{1}{8k+7}\right)\tag{2}$$
but du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmat... | Eliminate $bc$ from the first equation (the uppercase letters are the RHS),
$$a^2+\frac{BC}{(a+d)^2}=A,$$ and from the first and the last, $$d^2-a^2=D-A.$$
Let $p:=(d+a)/2,m:=(d-a)/2$, giving
$$(p-m)^2+\frac{BC}{4p^2}=A,\\mp=4(D-A)$$
and multiplying by $4p^2$,
$$4(p^2-mp)^2+BC=4(p^2-4(D-A))^2+BC=4Ap^2.$$
This biquadrat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Evaluate the closed form of $\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)$ Last question of this form
I am very curious to what is the closed form of:
Assume where $a,b,c > 0$
$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)\tag1$$
$$b\sin^2x+c\cos^2x$$... | We will use Fourier transformation and Parseval's identity.
Say that
\begin{equation*}
\hat{g}(\xi) = \int_{-\infty}^{\infty}e^{-i\xi x}g(x)\, dx .
\end{equation*}
Then the Fourier transform of $g(x) = \arctan\dfrac{2}{x^2}$ will be
\begin{equation*}
\hat{g}(\xi) = 2\pi\dfrac{\sin \xi}{\xi}e^{-|\xi|}\tag{1}
\end{equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Circular track problem with LCM I am not able to understand this question based on circular tracks:
$A$ and $B$ start together from the same point on a circular track, and walk in the same direction until they both again arrive together at the starting point. $A$ completes one round in $224$ seconds and $B$ in $364$ s... | Lets make a table, of laps completed, and the floor of laps ahead (times when they must have passed):
$\begin{array}{|c|c|c|}
\hline
A & B& \lfloor{\text {laps ahead}}\rfloor \\ \hline
1 & \frac{8}{13}& 0\\ \hline
2 & 1\frac{3}{13}&0\\ \hline
3 & 1\frac{11}{13}&1\\ \hline
4 & 2\frac{6}{13}&1\\ \hline
5 & 3\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A right triangle has leg $n$, hypotenuse $2n+6$, and perimeter $60$. Find the lengths of all sides. This problem comes from a Grade 11 math textbook in a section on solving quadratic functions by factoring. What follows is my attempt at a solution.
A right triangle has one leg, $n$ and a hypotenuse, $2n+6$. The challe... | Let's correct your solution. Following your notation, the other leg will have length $x$. Since the perimeter is $60$,
\begin{align*}
x + n + 2n + 6 & = 60\\
x + 3n + 6 & = 60\\
x & = 54 - 3n
\end{align*}
By the Pythagorean Theorem,
\begin{align*}
x^2 + n^2 & = (2n + 6)^2\\
x^2 + n^2 & = 4(n + 3)^2\\
x^2 + n^2 & = 4(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
geometry question - Prove that $KB = KD$ Let $\Delta ABC$ have $AB = AC$. On the line perpendicular to $AC$ at $C$ take $D$ so that points $B, D$ is different side $AC$. Let $K$ is the intersection of the straight line through $B$ perpendicular to $AB$ and the line through the center of $M$ of the $CD$ perpendicular to... | Let $AB=AC=a$, $CD=2b$ and $ML\perp AD$, where $L\in AD$.
Hence, $MD=b$ and since $\Delta MLD\sim \Delta ACD$, we obtain:
$$\frac{LD}{CD}=\frac{MD}{AD}$$ or
$$\frac{LD}{2b}=\frac{b}{\sqrt{a^2+4b^2}},$$ which gives
$$LD=\frac{2b^2}{\sqrt{a^2+4b^2}}.$$
Hence, $$AL=AD-LD=\sqrt{a^2+4b^2}-\frac{2b^2}{\sqrt{a^2+4b^2}}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots } = \frac 1 {z\big(1 - (\frac{z^2}{3!} - \frac{z^4}{... | The radius of convergence of the Laurent series of $1/\sin z$ about $z=0$ is $\pi$. The reason is that its poles are at $n\pi$, $n\in\Bbb Z$
(because that's where the sine has its zeroes). The nearest poles to zero
are at $\pm \pi$, as distance $\pi$ from the origin.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve $\sin(5\theta)=1$, $0<\theta<2\pi$. Show that the roots of $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, $r=0,2,3,4$. This is a very interesting problem that I came across. I know it's got something to do with trigonometry identities, polynomials and complex numbers, but other than that, I'm not t... | Use$$\sin5\theta=\sin{x}\cos4\theta+\cos{x}\sin4\theta=\sin{\theta}(2\cos^22\theta-1)+4\cos^2\theta\cos2x\sin{\theta}=$$
$$=\sin{\theta}(2\cos^22\theta-1+2(1+\cos2\theta)\cos2\theta))=\sin{\theta}(4\cos^22\theta+2\cos2\theta-1)$$
and $16x^4+16x^3-4x^2-4x+1=(4x^2+2x-1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$? $$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$
in terms of summation i can write it as
$$S_{n}=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$$
How to continue from this point?
used partial frac... | A route. One may recall the standard series representation of the digamma function,
$$
\psi(z+1)+\gamma=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+z}\right),\quad \text{Re}\: z>-1,
$$
giving
$$
2\sum _{n=1}^{\infty } \frac{(-1)^n}{n+z}=\psi\left(\frac{z}{2}+\frac{1}{2}\right)-\psi\left(\frac{z}{2}+1\right),\quad \text{Re}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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Base $5$ representation is the last $k$ digits of its base $2$ representation for some $k$
Compute the largest integer between $1$ and $1000$ whose base $5$ representation consists of the last $k$ digits of its base $2$ representation, for some $k$.
I thought about taking cases based on the number of leading $1$s suc... | If a number $n$ has $k$ digits (quits?) in base $5$, then let
$$n = \sum_{i=0}^{k-1} a_i 5^i = a_{k-1} 5^{k-1} + \cdots + a_15 + a_0,$$ where $a_{k-1} > 0$. Since the same digits $a_i$ are used as bits, $a_i \in \{0, 1\}$.
Consider the last $k$ bits in base $2$:
$$\begin{align*}
n &\equiv \sum_{i=0}^{k-1} a_i 2^i \pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x $ $$
\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x
$$
I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and
$(x)^2-1/(x)^2=z $ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x) $
Please help me out
| Consider the ${}_{2}F_{1}$ hypergeometric integral form given by
$${}_{2}F_{1}(a, b; c; x) = \frac{\Gamma(c)}{\Gamma(b) \, \Gamma(c-b)} \, \int_{0}^{1} t^{b-1} \, (1-t)^{c-b-1} \, (1-x \, t)^{-a} \, dt$$
leads to, with $a=-1/2$, $b=1/4$, $c=5/4$, $x=-1$,
$${}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
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Prove that the product is never a perfect square
Prove that for nonnegative integers $x_1,\ldots,x_{2011}$ and $y_1,\ldots,y_{2011}$ the product $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_{2011}^2+3y_{2011}^2)$$ is never a positive perfect square.
I thought about generalizing this question to any odd subscript $n$ i... | Modulo $3$, the equation for $n=1$ reduces to $2x^2=z^2 \pmod3$.
If $x$ were not divisible by $3$ then we would find $2=u^2 \pmod3$ for some $u$. But $2$ is not a quadratic residue $\pmod 3$ so this is impossible.
But if $x$ were divisible by $3$ this would make $z$ divisible by $3$, and thus $z^2$ divisible by $3^2 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
simplify :$\frac{1}{f_5}+\frac{1}{f_6}+\frac{1}{f_{12}}+\frac{1}{f_{20}}$
let :
$$f_n=\sqrt[4]{2}+\sqrt[n]{4} \ \ \ n\geq2 \in \mathbb{N}$$
then simplify :
$$\dfrac{1}{f_5}+\dfrac{1}{f_6}+\dfrac{1}{f_{12}}+\dfrac{1}{f_{20}}=?$$
MyTry :
$$f_5=\sqrt[4]{2}+\sqrt[5]{4} \ \ \ , \ \ f_6=\sqrt[4]{2}+\sqrt[6]{4} \ \ , \ \ f... | You get that $$\frac1{\sqrt[20]{32}+\sqrt[20]{256}}+\frac1{\sqrt[12]{8}+\sqrt[12]{16}}+\frac{1}{\sqrt[12]{8}+\sqrt[12]{4}}+\frac{1}{\sqrt[20]{32}+\sqrt[20]{4}}$$
If we group first and fourth and second and third term we get
$$\frac1{\sqrt[20]{32}(1+\sqrt[20]{8})}+\frac{1}{\sqrt[20]{4}(1+\sqrt[20]{8})}+\frac1{\sqrt[12]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For $a,b,c>0$ . Prove $ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9+8\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}$ For $a,b,c>0$ . Prove
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9+8\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}$$
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\frac{(a+b+c)(ab+ac+bc)}{abc}\geq9+\frac{16(a^2+b^2+c^2-ab-ac-bc)}{(a+b+c)^2}$$ or
$$\frac{9uv^2}{w^3}\geq9+\frac{16(u^2-v^2)}{u^2},$$
which is $f(w^3)\geq0$, where $f$ is an decreasing function.
Thus, it's enough to prove our inequality for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find complex number Z in $\lvert Z\rvert= Z+3-2i$ $$\lvert Z\rvert = Z+ 3-2i$$
what I did so far is
let $Z = a +bi$
so $$\sqrt{a^2 + b^2} = a+bi+3-2i$$
$$\sqrt{a^2 + b^2} = a+3 + i (b-2)$$
now what I'm thinking is squaring both sides but that doesn't work, any tips?
| Solve
$$
\lvert z \rvert = z + 3 - 2i
\tag{1}
$$
1
Observation: $\lvert z \rvert$ is real, while the right hand side has the imaginary term $-2i$.
For the equality to hold, the imaginary part of $z$ must be $2i$ to cancel the imaginary component, that is
$$z=x+2i\tag{2}$$
2
Use $(2)$ in $(1)$ to obtain
$$
\lvert ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find Angle Between Two Curves at Point of Intersection Find angle between these two curves at point of intersection :
$$K_1: x^2y^2 + y^4 = 1$$ and
$$K_2 : x^2 + y^2 = 4 $$
Thanks!
| Problem
Given the equations
$$
\color{blue}{x^{2} + y^{2} = 4}
\tag{1}
$$
and
$$
x^{2}y^{2} + y^{4} = 1
\tag{2}
$$
find the angles between the curves at the points of intersection.
Symmetry analysis
Each term in both equations has a definite parity; this invites investigation.
The $x^{2}$ terms are invariant under ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding the number of ways of choosing 4 distinct integers from first n natural numbers so that no two are consecutive. Question:
Show that the number of ways of choosing 4 distinct integers from the first $n$ numbers so that no two are consecutive is $\binom{n-3}{4}$.
I tried to find it by the principle of inclusion a... | You must have done something wrong, since for $n=7$ your expression is $35-60+20-4=-9.$
To use inclusion-exclusion, your events should be something like "includes $i$ and $i+1$". Now the number of choices which include $i$ and $i+1$ is $\binom{n-2}{2}$, and there are $n-1$ possible $i$, so you should start $$\binom{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
A polynomial is divisible with another one For what $m$ and $n$ is the polynomial $2X^{19}+X^{13}+mX^{11}+X^8+2X^6+nX^2+2$ divisible by $X^4+X^3+X^2+X+1$.
I tried to find the real solutions for g but couldn't
| Note that $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
\begin{align}
&\;2x^{19}+x^{13}+mx^{11}+x^8+2x^6+nx^2+2\\
=&\;2x^4(x^5-1+1)^3+(x^3+mx)(x^5-1+1)^2+(x^3+2x)(x^5-1+1)+nx^2+2\\
=&\;2x^4[(x^5-1)^3+3(x^5-1)^2+3(x^5-1)+1]\\
&\quad+(x^3+mx)[(x^5-1)^2+(x^5-1)+1]+(x^3+2x)[(x^5-1)+1]+nx^2+2\\
=&\;P(x)(x^5-1)+2x^4+x^3+mx+x^3+2x+nx^2+2
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
| If $f(x)$ has degree $n $ then $ f'(x) $ has $n-1$ . Here highest degree is $3$ thus $ f (x) $ is a polynomial of degree $3$. Thus let $f (x)=ax^3+bx^2+cx+d $ thus we have $f (x)+f'(x)=ax^3+bx^2+cx+d+3ax^2+2bx+c $ Now comparing we have $ a=1,3a+b=5,c+2b=1,d+c=2$ thus $ f(x)=x^3+2x^2-3x+5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&... | The characteristic polynomial of $A$ is $x^2-6x-1$, whose roots are $3\pm\sqrt{10}$. An example of an eigenvector with norm $1$ whose eigenvalue is $3+\sqrt{10}$ is $\frac1{\sqrt{20-2\sqrt{10}}}\bigl(\sqrt{10}-1,3\bigr)$ and an example of an eigenvector with norm $1$ whose eigenvalue is $3-\sqrt{10}$ is $\frac1{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Generalised Stirling number that each partition has more than one component We know that the Stirling number of the second kind is the number of ways to partition a set of $n$ objects into $k$ non-empty subsets and is denoted by $s(n,k)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\begin{pmatrix}
k\\
j
\end{pmatrix}j^n$
This f... | The first question you should be asking is where does the formula
$$ s(n,k) = \frac1{k!} \sum_{j = 0}^k \binom{k}{j} j^n (-1)^{k - j} $$
come from? Then once you understand that, perhaps you will be able to generalize.
Note that the generating function for nonempty sets is $e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Proof that $x^3+y^4=z^{31}$ has infinitely many solutions This is a question from RMO 2015.
Show that there are infinitely many triples (x,y,z) of integers such that $x^3+y^4=z^{31}.$
This is how I did my proof:
Suppose $z=0,$ which is possible because $0$ is an integer. Then $x^3+y^4=0 \Rightarrow y^4=-x^3.$ Now, supp... | This is the solution they gave...
Choose $x = 2^{4r}$ and $y = 2^{3r}$. Then the left side is $2^{12r+1}$. If we take$ z = 2^k$,
then we get$ 2^{12r+1} = 2^{31k}$. Thus it is sufficient to prove that the equation 12r + 1 = 31k
has infinitely many solutions in integers. Observe that$ (12.18) + 1 = 31.7$. If we choose
$r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Series expansion of Elliptic integral $F(\sin^{-1}(1-x)|-1)$ where $0I understand the leading term $F(\pi/2\mid-1)$ would be $K(-1)$. I also want the next term contributing from $x$. Actually I could not expand $\sin^{-1}(1-x)$ (when $0<x\ll1$) around 1. Any help would be highly appreciated.
Thanks a lot.
| We may put $1-y=\sin{\theta}$ in the integral, so $dy = -\cos{\theta} d\theta $, $\cos{\theta}=\sqrt{y(2-y)}$ on the interval in question, to find
$$ F(\arcsin{(1-x)}\mid -1) = \int_x^1 \frac{dy}{\sqrt{y(2-y)(1+(1-y)^2)}} = \int_x^1 \frac{dy}{\sqrt{y(2-y)(2-2y+y^2)}}. $$
Of course,
$$F(\arcsin(1) \mid -1) = F(\pi/2 \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Area of a circle inscribed in a polygon If a circle is inscribed in a polygon, show that,
$$\dfrac{\text{(Area of inscribed circle)}}{\text{(Perimeter of inscribed circle)}} = \dfrac{\text{(Area of Polygon)}}{\text{(Perimeter of Polygon)}}$$
| For a regular polygon with $n$ sides with side length $l$. The ends of each side when connected to the centre of the polygon forms a triangle with an angle of $\frac{2\pi}{n}$ at the centre. There will be $n$ such triangles. The altitude of each triangle starting from the centre of the polygon has a length of $\frac{a}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$... | Let $u=x^4+y^2$. Then $|x|\le u^{1/4}$ and $|y|\le u^{1/2}$. So
$|x^3y^4|\le u^{11/4}$ and, for $(x,y)\ne(0,0)$, $|f(x,y)|\le u^{3/4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Combinatorial proof of Recurrence for the number of permutations of odd order Let $a_n$ denote the number of permutations of odd order in $S_n$. This is sequence A000246 in OEIS. Then there is the recurrence relation on OEIS that $a_{2n} = (2n-1)a_{2n-1}$ and $a_{2n+1} = (2n+1)a_{2n}$, or equivalently $a_n = a_{n-1} + ... | These permutations are precisely the ones that do not contain even
cycles, giving the species
$$\mathfrak{P}(\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathfrak{C}_{=3}(\mathcal{Z})
+ \mathfrak{C}_{=5}(\mathcal{Z})
+ \mathfrak{C}_{=7}(\mathcal{Z})
+ \cdots)$$
This yields the EGF
$$G(z) = \exp\left(\sum_{q\ge 1} \frac{z^q}{q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding angle in triangle I have a triangle $\triangle ABC$ with vertices $A$, $B$ and $C$ with $\angle BAC = 45^\circ$ and $\angle ABC = 30^\circ$. Point $M$ is the center of the $BC$. Find $\angle AMC$.
This is not my homework, it's a little girl's, so she can't use trigonometric equations or anything after special ... |
Let $X$ on $AB$ be the foot of the altitude through $C$.
*
*Since $\angle BAC = 45^\circ$, $\angle ACX = 45^\circ$, $\triangle AXC$ is isoceles and $AX = CX$.
*Since $\angle CBA = 30^\circ$, $\angle XCB = 60^\circ$, $\triangle XBC$ is half of an equilateral triangle and $CX : BC = 1 : 2$.
*Since $M$ is midpoin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that : if $a, b \in \mathbb{R}^+$ : then : $a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$ prove that : if $a, b \in \mathbb{R}^+$ : then :
$$a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2b+b^2a-a^2$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2+b^3a-b^2$$
now what ?
| it is equivalent to $$(a-b)^2((ab)^2-ab+1)+2ab(ab-1)^2\geq 0$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to ... | $$3x^2+3(y-1)^2+3(x-y)^2-1=(x+y-1)^2+(2x-y)^2+(x-2y+1)^2\geq0.$$
The equality occurs for
$$x+y-1=2x-y=x-2y+1=0,$$ which gives $x=\frac{1}{3}$ and $y=\frac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Coefficient in expansion - Discrete Math
Let $\alpha$ be the coefficient of $x^2y^2z^2$ in $ (x^2+3y-2z)^5$. Show that $800\leq|\alpha|<1100$.
By my calculation, $\alpha=\binom {5}{2} \times (3^{2})\binom {3}{2} \times(-2)^2 \binom {1}{2}$
However $\binom {1}{2}=0$, so $\alpha$ should be $0$.
What am I doing wrong?
| $$
\begin{align}
1^1\cdot3^2\cdot(-2)^2\cdot\overbrace{\binom{5}{2,2}}^{\substack{\text{trinomial}\\\text{coefficient}}}
&=1^1\cdot3^2\cdot(-2)^2\cdot\frac{5!}{2!2!1!}\\
&=1080
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Range in an Interval I want to find the range of the function, $$f(x) = {{x + 1} \over {{x^2} + 1}}\,\,\;{\rm{when }}\;x \in \left[ { - 1,1} \right].\;\;\;$$
I have isolated $x$ as - $$x = {{1 \pm \sqrt {1 - 4y(y - 1)} } \over {2y}}$$
But I am unable to solve this inequality, $$ - 1 \le {{1 \pm \sqrt {1 - 4y(y - 1)} } ... | $$\frac{x+1}{x^2+1}\geq0.$$
The equality occurs for $x=-1$, which says that $0$ is a minimum.
Now, by AM-GM for $x\neq-1$ we obtain:
$$\frac{x+1}{x^2+1}=\frac{x+1}{(x+1)^2-2(x+1)+2}=\frac{1}{x+1+\frac{2}{x+1}-2}\leq\frac{1}{2\sqrt2-2}=\frac{1}{2}(\sqrt2+1).$$
The equality occurs for $x+1=\frac{2}{x+1}$, which says that... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Condition on a matrix so that its determinant is $2$
Is the following statement true or false? There exists $A \in M_{3,3}(\mathbb{Z})$ with determinant $2$ such that $$A\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-8\\16\end{pmatrix}$$
I first thought of eigenvalues but it doesn't look like the second vec... | False. Let us say $A=(a_{i,j})$ with $1\le i,j\le 3$. Then for all $i$
$$
4\,\,\text{ divides }\,\, 2a_{i,1}+a_{i,2}+4a_{i,3},
$$
which happens if and only if $a_{i,2}$ is even and $a_{i,1}$ has the same parity of $a_{i,2}/2$.
Then
\begin{align}
\mathrm{det}(A)&=
\begin{bmatrix}a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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$\sin(x) = \sin(x − \frac\pi3)$, solve for $x$ on interval $[-2\pi, 2\pi]$ According to the answer sheet:
$\sin(x) = \sin(x -\frac\pi3)$ gives:
$x = x-\frac\pi3 + k \cdot 2\pi$ or $x = \pi-(x-\frac\pi3) + k \cdot 2\pi$
^ How did they go from $\sin(x) = \sin(x-\frac\pi3)$ to the equations above?
Thanks in advance!
| The first comes from the fact that $\sin(x) = \sin(x \pm 2\pi n) \,\, n\in \mathbb{Z}$, because $\sin$ has period $2\pi$ The second comes from $\sin(x) = \sin(\pi - x)$.
To answer the question in your title, however, about the intersections on the interval $[-2\pi, 2\pi]$, we'll reduce this to a formula for $x$.
\beg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of
$(2x-1)(x+3)^{\frac{1}{2}}$
My try -
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $
My numerator is wrong and should be
$6x+... | The derivative of $2x-1$ is $2$.
The derivative of $(x+3)^{1/2}$ is
$$
\frac{1}{2}(x+3)^{-1/2}\cdot \color{red}{1}=\frac{1}{2(x+3)^{1/2}}
$$
Not $(x+0)$ as you wrote, but this disappeared in the second step.
The product rule gives
$$
2(x+3)^{1/2}+(2x-1)\frac{1}{2(x+3)^{1/2}}
=\frac{4x+12+2x-1}{2(x+3)^{1/2}}
=\frac{6x+1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to properly simplify and trig substitute this integral? $\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then substitute this $x$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, d\theta $
Take the root, simplified to $\int\frac { {5-(5 \sin\theta... | Following your steps, with corrections:
$\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then $dx=5\cos\theta d\theta$
Then substitute this $x$ and $dx$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, 5\cos\theta d\theta $
Using the trig identity, si... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Find volume bounded by 3 equations using integration The prompt is to find the volume of the solid bounded by the equations $x^2 + y^2 -2y = 0$, $z = x^2 + y^2$ and $z \ge 0$
Plotting the equations, I get something like this
We get a paraboloid and a cylinder.
How to know if I have to use double or triple integration... | First, complete the square to see that
$$
x^2+y^2-2y=x^2 +(y-1)^2-1=0.
$$
So $x^2+(y-1)^2=1$. The cylindrical change-of-coordinate we will be using is:
$$
x=r\cos \theta, \hspace{4mm} y=1+r\sin \theta,\hspace{4mm} z=z, \hspace{4mm} \mbox{ where } 0\leq r\leq 1,\hspace{4mm} 0\leq \theta\leq 2\pi.
$$
Note that the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate the sum of two complex conjugates to the power of $n$ Please help me out with the following problem for I'm stuck and don't even have any idea on how to proceed.
Find all integers $n$ such that
$$\left(\frac {-1 +i{\sqrt3} } {2}\right)^n+\left(\frac {-1 -i{\sqrt3} } {2}\right)^n=2$$
| Hint: let $\displaystyle z_{1,2}=\frac {-1 \pm i{\sqrt3} } {2}\,$, then $z_1+z_2=-1$ and $z_1z_2=1\,$, so $z_{1,2}$ are the roots of $z^2+z+1\,$ which implies that they are roots of $z^3-1=(z-1)(z^2+z+1)$ i.e $z_1$ and $z_2 = \overline{z_1}$ are the complex cube roots of unity, therefore $z_1^n = z_1^{n \bmod 3}\,$.
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find $x$ of this equation I am a beginner to logarithms.
By using logarithms, in this equation, I want to find the value of $x$:
$$ \ 3^x = 4^{x-1} \ $$
I am using 10 as base of logarithm.
And I want $x$ in terms of $\log2$ and $\log3$, as far as possible.
| As mentioned logarithm use is desired. Here, for a solution in terms of natural logarithms,
\begin{align}
\ln(3^{x}) &= \ln(4^{x-1}) \\
x \, \ln(3) &= (x-1) \, \ln(4) \\
x \, (\ln(3) - \ln(4)) &= - \ln(2^{2}) = - 2 \, \ln(2) \\
x &= - \frac{2 \, \ln(2)}{\ln\left(\frac{3}{4}\right)}
\end{align}
Converting from natural l... | {
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"url": "https://math.stackexchange.com/questions/2340968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$ I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have... | LHS:
$$\frac{(1 + \sin\theta)(1-\sin\theta)}{\cos\theta(1-\sin\theta)} + \frac{\cos\theta}{1 - \sin\theta} = \frac{2\cos\theta}{1 - \sin\theta}.$$
RHS:
$$2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)=2\frac{\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}+1}{1-\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}}=2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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If $x$ is real, evaluate $k$ in absolute inequality
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$
My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to pr... | $$\frac{x^2+1}{x^2+x+1}-\frac 43=1-\frac x{x^2+x+1}-\frac 43=-\frac 13-\frac x{x^2+x+1}$$
So you need to find $m$ such that $$\frac{x}{x^2+x+1}\leq m$$
Since both $m,x^2+x+1>0$ we can multiply it out we get $$ mx^2+(m-1)x+m\geq 0$$
Now we must have that $m>0$ and that $\Delta\leq 0$(discriminant), now just find the min... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
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Baby Rudin: Theorem 3.20 (c) and (d) In theorem 3.20, Rudin offers a proof to the limits of the following sequences:
$$u_n = n^{\frac{1}{n}}$$
$$v_n = \frac{n^{\alpha }}{(1+p)^n}$$ with $\alpha$ a real number and $p > 0$.
In both the proofs, Rudin uses inequalities which I am not familiar with. For $u_n$, the following... | The first inequality can be derived, for $x\ge0$, from the binomial theorem:
$$
(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k
$$
If we remove all the (non negative) terms except the one for $k=2$, we get
$$
(1+x)^n\ge \binom{n}{2}x^2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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BMO2 1997 - Combinatorics Find the number of polynomials of degree 5 with distinct coefficients from the set {1, 2, 3, 4, 5, 6, 7, 8} that are divisible by $x^2-x+1$.
I tried multiplying $x^2-x+1$ by $ax^3+bx^2+cx+d$ to get $ax^5+(b-a)x^4+(a-b+c)x^3+(b-c+d)x^2+(c-d)x+d$, but I am struggling to count from here.
| It's divisible by $x^2-x+1$ if and only if it vanishes at $\zeta=e^{\pi i/3}$. Now $$a\zeta^5+b\zeta^4+c\zeta^3+d\zeta^2+e\zeta+f=(d-a)\zeta^2+(e-b)\zeta+f-c$$ so we must have $d-a=b-e=f-c$. So you just have to count solutions of that system with the variables taking on distinct values from $1,\dots,8$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Solution of differential equation using Laplace transform Can someone please help me identify my mistake as I am not able to find it.
$y''+y'=3x^2$ where $y(0)=0,y'(0)=1$
$L[y'']+L[y']=3L[x^2]$
$L[y'']=p^2L[y]-py(0)-y'(0)=p^2L[y]-1$
$L[y']=pL[y]-y(0)=pL[y]$
$p^2L[y]-1+pL[y]=\frac{6}{p^3}$
$p^2L[y]+pL[y]=\frac{6}{p^3}+1... | Using
\begin{align}
\mathcal{L}\{y'\} &= s \, \overline{y} - y(0) \\
\mathcal{L}\{y''\} &= s^2 \, \overline{y} - s \, y(0) - y'(0)
\end{align}
then
$$y'' + y' = a \, t^2$$
transforms to
\begin{align}
s \, (s+1) \, \overline{y} &= s \, y(0) + y'(0) + y(0) + \frac{2 a}{s^3} \\
\overline{y} &= \frac{y(0) + y'(0)}{s \, (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality using AM-GM only. By considering "Arithmetic mean $\geq $ Geometric mean" prove the trigonometric inequality:
$$\sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}. $$
where $A+B+C=180°$.
My try: By using transformation formulae, I proved that
$$\sin A + \sin B + \sin C = 4\cos\left(\frac{A}{2}\rig... | You can use Jensen Inequality on concave function.
Let $f(x)=\sin x$, then $f''(x)=-\sin x<0$, as $x\in (0,\pi)$.
Then $$\frac{f(A)+f(B)+f(C)}{3}\leq f\Big(\frac{A+B+C}{3}\Big)\\\text{i.e.}\space\frac{\sin A+\sin B+\sin C}{3}\leq \sin \Big(\frac{A+B+C}{3}\Big)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\\\text{i.e.}\space \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Solving Telescoping Series $$\ \sum_{n=5}^\infty \frac{8}{n^2-1} $$
I tried the following:
$$\ \sum_{n=5}^\infty \frac{8}{n^2-1} = \sum_{n=5}^\infty \frac{8}{n-1} - \frac{8}{n} =$$ $$\left(2-\frac{8}{5}\right) + \left(\frac{8}{5} - \frac{8}{6}\right) + \left(\frac{8}{6} - \frac{8}{7}\right) + \cdots + \left(-\frac{8}{... | Your partial fraction is wrong:
$$\frac{8}{n^2-1}=\frac{-4}{n+1}+\frac{4}{n-1}$$
Using similar trick, you will see that most terms cancel out and you will left with $1+\frac45.$
Edit:
to obtain the partial fractions,
Since $n^2-1=(n-1)(n+1)$,
$$\frac{8}{(n-1)(n+1)}= \frac{A}{n+1}+\frac{B}{n-1}.$$
We can for instance eq... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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what is smallest possible value of range of $7$ values given that their mean is $12$ and their median is $9$
what is smallest possible value of range of $7$ values given that their mean is $12$ and their median is $9$ :-
a) $3$
B) $6$
C) $7$
D) $8$
What is the proper approach to solve this problem , is there any r... | List the numbers in order: $a_1\le a_2\le a_3\le a_4\le a_5\le a_6\le a_7$.
You know the median is 9, so you now have $a_1,a_2,a_3,9,a_5,a_6,a_7$
The mean is $12$, so you know that $7\times 12=84=a_1+a_2+a_3+9+a_5+a_6+a_7$.
So $a_1+a_2+a_3+a_5+a_6+a_7=75$
You want $a_7-a_1$ to be as small as possible. This means you wa... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explanation of hyperbola drawing Here is an image of a hyperbola:
I'm wondering why $c^2=a^2+b^2$ in this drawing given that $c$ is the distance from the origin to both of the foci and $a$ is half the length of the absolute distance between any given point and the two foci.
Sorry if this seems like a stupid question.
| Take the foci of the hyperbola $F_1=(c,0)$ and $F_2=(-c,0)$ and the two vertices : $V_1=(a,0)$ and $V_2=(-a,0)$.
From the definition we have that a point $P$ of the hyperbola is such that $\overline{PF_1}-\overline{PF_2}=k$, and using as $P$ a vertex we find that $k=2a$.
So, for $P=(x,y)$ the equation is
$$
\sqrt{(x-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Greatest term in the expansion $(2+3x)^9$ I have got a question which is stated as
Find numerically the greatest term in the expansion of $(2+3x)^9$ where $x=\frac{3}{2}$
I haven't any idea how to proceed but I just plugged in value of $x$ and got $$(2+3x)^9=\bigg(\frac{13}{2}\bigg)^9$$
How can I determine greatest t... | You have
$$(3 x+2)^9=19683 x^9+118098 x^8+314928 x^7+489888 x^6+489888 x^5+326592 x^4+145152 x^3+41472 x^2+6912 x+512$$
Plugging $x=1.5$ the terms, from degree $0$ up are
$$512,10368,93312,489888,1.65337\times 10^6,3.72009\times 10^6,5.58013\times 10^6,5.38084\times 10^6,3.02672\times 10^6,756681$$
the largest terms i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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If $b>a>0$ , how to prove that $\left(\frac {\sqrt a +\sqrt b}{2}\right )^2 < \frac 1e \left(\frac {b^b} {a^a}\right)^{\frac 1{b-a}}$? If $b>a>0$ , then how to prove that $\left(\dfrac {\sqrt a +\sqrt b}{2}\right )^2 < \dfrac 1e \left(\dfrac {b^b} {a^a}\right)^{\dfrac 1{b-a}}$ ?
I tried applying the known standard ine... | Let $b=ax$, where $x>1$.
Hence, we need to prove that
$$ea\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(\frac{(ax)^{ax}}{a^a}\right)^{\frac{1}{a(x-1)}}$$ or
$$ea\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(\frac{(ax)^{x}}{a}\right)^{\frac{1}{x-1}}$$ or
$$e\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(x^{x}\right)^{\frac{1}{x-1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Interval of convergence of $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$ I want to find the interval of convergence of $$\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$$
By the root test, $\sqrt[n]{|(\frac{n}{n+1})^{n^2}(2x)^n|}=|(\frac{n}{n+1})^{n}(2x)|$
And $(\frac{n}{n+1})^n=(1-\frac{1}{n+1})^n\to\frac1e$ as $n\to\... | One can use the Taylor expansion of $\log$ to get an equivalent of the sequence :
$$
\begin{align}
\left(\frac{n}{n+1}\right)^{n^2}e^n &= \exp{\left[n+n^2\log\frac{n}{n+1}\right]} \\
&= \exp{\left[n+n^2\left(-\frac{1}{n+1}-\frac{1}{2(n+1)^2}+o\left(\frac{1}{n^2}\right)\right)\right]} \\
&= \exp{\left[\frac{n^2+2n}{2(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verifying $|F(r)| \geq \frac{1}{1-r}\log(\frac{1}{1-r}) $ and $|F(re^{i \theta})| \geq c_{q/r}\frac{1}{1-r}\log({\log(\frac{1}{1-r})})$ I'm attempting to take a Tauberian route in verifying the proposition in $(1)$ below, which is from Complex Analysis, by Elias M Stein and Rami M. Shakarchi.
Let $F(z)$ be the followin... | For the case of (1.2), note that
$$1-r^n = (1-r)(1+r+r^2+\cdots + r^{n-1}) \le (1-r)\cdot n.$$
Therefore
$$ \sum_{n=1}^{\infty}\frac{r^n}{1-r^n}\ge \sum_{n=1}^{\infty}\frac{r^n}{(1-r)n} = \frac{1}{1-r}\sum_{n=1}^{\infty}\frac{r^n}{n}.$$
But the sum on the right is precisely $\ln\left (\dfrac{1}{1-r}\right ).$ This give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{... | $$a=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)}=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{4},$$ which gives $\sqrt{x+2}+\sqrt{x-2}=2\sqrt{a}$ and from here $\sqrt{x+2}-\sqrt{x-2}=\frac{2}{\sqrt{a}}.$
After summing of last two equalities we obtain ... | {
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"source": "stackexchange",
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What can be said if $A^2+B^2+2AB=0$ for some real $2\times2$ matrices $A$ and $B$? Let $A,B\in M_2(\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\det A= \det B$. Our goal is to compute $\det(A^2 - B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true:
*
*$\det(A^2+B... | Of course, the OP did not understand much about the proposed exercise. In particular, in $M_2(\mathbb{C})$, $A^2 +B^2+2AB=0$ implies $\det(A)=\det(B)$.
In fact, the interesting result is
Proposition. Let $A,B\in M_2(\mathbb{C})$ be such that $(1)$ $A^2+B^2+2AB=0$; then $AB=BA$. Note that the result works only in dimens... | {
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"source": "stackexchange",
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How to find $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots$? Find the sum of series $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots,$$ where the terms are the reciprocals of positive integers whos... | $$\sum_{k=0}^{\infty}\frac{1}{2^k}\sum_{k=0}^{\infty}\frac{1}{3^k}=\frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{1}{3}}=3$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the maximum number of distinct positive integer's square that sums up to $2002$?
What is the maximum number of distinct positive integer's square that sums up to $2002$ ?
My tries:
$$\frac{n(n+1)(2n+1)}{6} = 2002$$
$$\implies n\approx 17 $$ but am clueless as to how to proceed any further.
| Your estimate gives an upper bound: Even the smallest sum obtainable from $18$ distinct positive squares is $1^2+2^2+\ldots+18^2=2109>2002$, hence we can have at most $17$ squares.
Let's try to find a solution with $17$ squares.
If $k\le 17$ is the first number the sequence $x_1<x_2<\ldots< x_{17}$ omits, then $x_1^2+... | {
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"source": "stackexchange",
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Second order nonhomgeneous equation euler stuck at nonhomogeneous part there is second order nonhomogeneous differential equation, euler
Im stuck at find nonhomogeneous equation,
$x^2y"-3xy'+4y=log x$ and i have to find the solution
my attempt :
first i found solution for the homogeneous equation, it is a repeated r... | I believe your error was in finding $u_1^{\prime}$ and $u_2^{\prime}$;
you should have
$\displaystyle\frac{\ln x}{x^2}$ instead of logx in the expressions
for $W1$ and $W2$, so that $u_1^{\prime}=-\frac{(\ln x)^2}{x^3}$ and $u_2^{\prime}=\frac{\ln x}{x^3}$.
This will give $u_1=\left(\frac{1}{2}(\ln x)^2+\frac{1}{2}\ln ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequal... | your inequality is equivalent to
$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2(a+b+c)\geq 15$$
By $AM-HM$ we get
$$a+b+c\geq \frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus $$2(a+b+c)\geq \frac{18}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus we have
$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Why two characterizations of the arcsine distribution are equiv? According to Wikipedia the CDF of arcsine dist. is:
$$F(x)=\frac2{\pi}\arcsin(\sqrt{x})=\frac{\arcsin(2x-1)}{\pi}+\frac12$$
So, why are these two equivalent?
Thanks in advance.
| One way to prove this trigonometric identity is to $(1)$ first show that the derivatives of the functions on both sides of the equality are the same, and $(2)$ show that the two sides are equal when $x=0.$
\begin{align}
& \frac d {dx}\,\frac 2 \pi \, \arcsin\sqrt x = \frac 2 \pi\cdot\frac 1 {\sqrt{1 - x}} \cdot \frac d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis o... | Your result is corect. This is the plot in geogebra.org
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363075",
"timestamp": "2023-03-29T00:00:00",
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Simplifying Complex Numbers in Exponential Form
$(a)$ Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
$(b)$ Let $P_1 P_2 \dotsb P_{18}$ be a regular $... | Let $x=e^{i\pi /9}.$ We have $x^9=-1.$ $$\text { We have }\quad x^3-x^6=x^{-3}(x^6-x^9)=x^{-3}(x^6+1)=x^{-3}+x^3.$$ $$\text { So }\quad x^2-x^5=x^{-1}(x^3-x^6)=x^{-1}(x^{-3}+x^3).$$ $$\text { We have } \quad x^8-1=x^{-1}(x^9-x)=x^{-1}(-1-x)=(1+x^{-1})(-1).$$ $$\text { Therefore }\quad x^8-x^6-x^5+x^3+x^2-1=$$ $$=(x^8-1... | {
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Find all possible values of $x+y+z$. Let $x, y, z$ be non-zero integers such that ${x\over y}+{y\over z}+{z\over x}$ is an integer. Find all possible values of $x+y+z$. Please provide a proof with all solutions.
| If $x = \frac{z}{4}$, and $y= \frac{z}{2}$, then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} = \frac{1}{2}+\frac{1}{2} + 4 = 5$. Thus, whenever $z = 4t, t \in \mathbb{N}$, the solution set $(x,y,z) = (t,2t,4t)$ satisfies, and thus are infinitely many forms of $x+y+z$ with the given constraints - notably of the form $7t$.
Thes... | {
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Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$ Question
Determine $n$ such that the following improper integral is convergent
$$ \int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$$
I'm not sure how to go abou... | Hint: note $\ln (b^3+1) \sim \ln b^3=3\ln b$ and $\ln (13b+1) \sim \ln 13b=\ln 13+\ln b$ for $b\to+\infty$.
Hence: $$\lim_\limits{b\to +\infty} [\frac{n}{3} \ln (b^3+1)-\frac{1}{13}\ln (13b+1)]=\lim_\limits{b\to+\infty} [\left(n-\frac{1}{13}\right) \ln b - \frac{\ln 13}{13}]=\begin{cases} -\frac{\ln 13}{13}, \ if \ n=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365228",
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"source": "stackexchange",
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Clarification for a simple argument in an inequality Suppose $a,b,c,d \in \mathbb{R}^+$ are such that $$ a \leq b \\ c \leq d \\ ac = bd $$ then is the following true? $$ a= b \\ c = d $$
Attempt:
Suppose not true, i.e. $a < b$ or $c < d$, then we have $ac < bd$ thus contradicting the fact that $ac=bd$. Thus $a= b$ and... | Okay, FINAL final answer:
If $0 \le a \le b$ and $0 \le c \le d$ then
$ac \le bc$ with equality holding only if $a = b$ or $c = 0$.
And $bc \le bd$ with equality holding only if $c = d$ or $b= 0$.
So $ac \le bd$ with equality holding only if one of the four occur:
1) $a = b$ and $c = d$
2) $a =b$ and $b = 0$ so $a = b... | {
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Proof by strong induction of $(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$ I'm not very familiar with proofs by strong induction. I have a sketch for this one but Iot quite sure about it's validaty.
Let $a$, $b$, $a_n$, $b_n$ be integers such that.
$$(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$$
where $a$ is the integer closest ... | Hint.
Observe that, if
$$
(a+b\sqrt{2})^n=a_n+b_n\sqrt{2},
$$
then (it can be shown inductively)
$$
(a-b\sqrt{2})^n=a_n-b_n\sqrt{2},
$$
and if $|a-b\sqrt{2}|<1/2$, then clearly, $(a-b\sqrt{2})^n<1/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\tan(π/16)+\tan(5π/16)+\tan(9π/16)+\tan(13π/16)$ I had to calculate $\tan(π/16)+\tan(5π/16)+\tan(9π/16)+\tan(13π/16)$
I tried writing $π/16=x$ and then writing the sum as $\tan x+\tan(9x-4x)+\tan9x+\tan(9x+4x)$ and then simplifying using $\tan(x+y)$ and $\tan(x-y)$ but it didn't simplify to anything, can anybody ... | $$a=\tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{9\pi}{16})+\tan(\frac{13\pi}{16})\\=
\tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{1\pi}{16}+\frac{\pi}{2})+\tan(\frac{5\pi}{16}+\frac{\pi}{2})\\=\\
\tan(\frac{\pi}{16})-\cot(\frac{\pi}{16})+\tan(\frac{5\pi}{16})-\cot(\frac{5\pi}{16})\\=?$$
now let me no... | {
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If $\cos A+\cos B+\cos C=\frac{3}{2}$, prove $ABC$ is an equilateral triangle If in $\Delta ABC$ $$\cos A+\cos B+\cos C=\frac{3}{2}$$prove that it is an equilateral triangle without using inequalities. I tried using the cosine rule as follows:
$$\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\... | A different kind of trick : I thought you will like it.
So multiply the above equation by $a,b,c$ and add these up. Next , use the identity that $c = a \cos B + b \cos A$, and likewise for the other sides. After pairing these, you get:
()
$$
\frac{3(a+b+c)}{2} = a \cos A + b \cos B + c \cos C + a + b +c
$$
which then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ . Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ .
Answer:
The graph is above :
Since the region is symmetrical ,
$ Area =4 \times \int_{0}^{2} \int_{0}^{\sqrt{2+\sqrt{4x^2-x^4+4}}} dxdy $
Am I right ? Is there any help ?
| According to your own sketch, the integral you wrote would not cover the area $|y| > 2$, never mind that the order of integration does not match the limits of integration. So the setup is problematic in multiple ways.
The suggestion to use polar coordinates is worth exploring. Let $(x,y) = (r \cos \theta, r \sin \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How to simplify an expression involving several square roots without a calculator? $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
This type of questions are common in the university entrance examinations in our country but the calculators are not allowed... | Assuming that your expression contains a typo :)
Actually we have to simplify
$$\frac{5 \sqrt{5}}{4\left(\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}\right)}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
Notice that both fractions can be simplified multiplying numerator and denominator by
$\left(\sqrt{3\sqrt{5}}+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Is $\displaystyle\sum_{k=1}^n \frac{k \sin^2k}{n^2+k \sin^2k}$ convergent? Let $\displaystyle x_n=\sum_{k=1}^n\frac{k \sin^2k}{n^2+k \sin^2k}$ for all $n>0$. How can we prove that $(x_n)$ is convergent?
| $$
\begin{align}
\sum_{k=1}^n\frac{k\sin^2(k)}{n^2+k\sin^2(k)}
&=n-\sum_{k=1}^n\frac{n^2}{n^2+k\sin^2(k)}\\
&=n-\sum_{k=1}^n\frac1{1+\frac{k\sin^2(k)}{n^2}}\\
&=n-\sum_{k=1}^n\left(1-\frac{k\sin^2(k)}{n^2}+O\!\left(\frac{k^2}{n^4}\right)\right)\\
&=\frac1{n^2}\sum_{k=1}^nk\sin^2(k)+O\!\left(\frac1n\right)\\
&=\frac1{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $ \sum\limits_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}=\ln\frac43$
$\sum_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}$ i must prove this sum converges to $\ln(4/3)$.
i tried to write expresion :$\frac{3^k-2^k}{k\cdot6^k}=\frac 1 k \left(\frac 1 {2^k-3^k}\right)$ and make two sums but i tink this sums is a dezvoltat... | Work with two sums separately:
\begin{align}
\sum_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k} & = \sum_{k=1}^\infty \frac{(3/6)^k} k - \sum_{k=1}^\infty \frac{(2/6)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(1/2)^k} k - \sum_{k=1}^\infty \frac{(1/3)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(-1)^k(-1/2)^k} k - \sum_{k=1}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)=0$ is If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)$ is
A) $-d/a$
B) $d/a$
C) $a/d $
D) none of these
My try
I take $f (x) = ( x^2+x+1)(x+1) = x^3+2x^2+2x+1$
Real root $x= -d/a =-1$
And also ... | If you divide $f (x)= ax^3+bx^2+cx+d$ to $x^2+x+1$ : you will have
$$f (x)= ax^3+bx^2+cx+d=(x^2+x+1)(ax+(b-a))+x(a+c-b+a)+(a-b+d)$$ when remainder is $0 $ so $$ax+(b-a)=0 \to x=\frac{a-b}{a} \tag{1}\\
\forall x:\begin{cases}a+c-b+a=0 \\a-b+d=0\end{cases}$$ now from 3rd equation we have $d=-(a-b)$ put into (1)
$$x=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to turn the ellipse $x^2 - xy + y^2 - 3y - 1 = 0$ to the canonical form using an isometric transformation? There is an exam problem I'm having trouble with, it is as follows:
Turn the equation $x^2 - xy + y^2 - 3y -1 = 0$ into the canonical form using an isometric transformation and write down the transformation.
... | First center the equation by definting the function $f(x,y) = x^2 - xy + y^2 - 3y - 1 = 0$ and solving the following system of equations for $(x,y)$
$$\left. \begin{aligned}
\frac{\partial f(x,y)}{\partial x} & = 2x-y = 0 \\
\frac{\partial f(x,y)}{\partial y} & = -x+2y-3 = 0 \\
\end{aligned} \right\} \pmatrix{x & y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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What are the roots of this equation? I have a quadratic equation $ ax^2 +bx+c =0 $, where $ a,b,c $ all are positives and are in Arithmetic Progression.
Also, the roots $\alpha$ and $\beta$ are integers.
I need to find out $ \alpha + \beta + \alpha\beta $.
I have tried taking $ a = a' - d , b = a' , c = a' + d $ becau... | We see that the equation $$ax^2+(a+d)x+a+2d=0$$
has integer roots, which says $1+\frac{d}{a}\in\mathbb Z$.
Let $\frac{d}{a}=k$.
Hence, we have $$x^2+(1+k)x+1+2k=0$$
has integer roots, which gives
$$(1+k)^2-4(1+2k)=n^2,$$ where $n$ is non-negative integer number,
which gives
$$k^2-6k-3=n^2$$ or
$$(k-3)^2=n^2+12$$ or
$$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then diff... | By using partial fractions, chance upon the solution $A_{0}=1,$ and $A_{1}=-1$. For $n>1$, in particular for all $n=7,$ it seems that
$$A_{7}=(x+7)(\frac{(7)!}{x(x+1)...(x+7)}-\frac{6!}{x(x+1)...(x+6)})
\\=6!\frac{7-(x+7)}{x(x+1)...(x+6)}=6!\frac{-x}{x(x+1)...(x+6)}
\\=-\frac{6!}{(x+1)...(x+6)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find $\lvert m\rvert$ given $\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$ I've already found the solution, but I was wondering if there is a faster or alternative method. My solution is found below:
$$\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$$
$$(m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}=3$$
$$\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$... | Go ahead and cube:
$$m+9=27+27\sqrt[3]{m-9}+9\sqrt[3]{(m-9)^2}+m-9.$$
Denote: $\sqrt[3]{m-9}=t$ to get:
$$t^2+3t+1=0 \Rightarrow t=\frac{-3\pm \sqrt{5}}{2}$$
Now:
$$m-9=t^3$$
will produce the answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) =... | Let's get a picture:
$|1 + \sin\theta + i\cos\theta| = |1 + \sin\theta - i\cos\theta|$
$\left|\frac {1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\right| = 1$
When we divide complex numbers the argument of the ratio equals the difference between the arguments.
let $\phi = \frac {\pi}{2} - \theta$
$1 + \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 6
} |
Finding $n$th power of a $3\times 3$ matrix Find the $A^n$ if $$A=\begin{bmatrix}1 & a & b \\0 & 1 &a\\0 &0 &1\end{bmatrix}$$
I tried inductive method to show $$A^n=\begin{bmatrix}1 & na & nb+\frac{n(n-1)}{2}a^2 \\0 & 1 &na\\0 &0 &1\end{bmatrix}$$
now : My question is : Is there other method (idea ) to find $A^n$ ?
T... | $$A=I+aJ+bJ^2$$
where
$$J=\pmatrix{0&1&0\\0&0&1\\0&0&0}.$$
Then $J^3=O$.
Now
$$A^n=\sum_{k=0}^n{n\choose k}(aJ+bJ^2)^n
=I+n(aJ+bJ^2)+{n\choose 2}(aJ+bJ^2)^2$$
as $(aJ+bJ^2)^3=O$. But $(aJ+bJ^2)^2=a^2J^2$ so
$$A^n=I+naJ+nbJ^2+\frac{n(n-1)}2a^2J^2
=\pmatrix{1&na&nb+\frac{n(n-1)}{2}a^2\\0&1&na\\0&0&1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$x \equiv 1$ mod $4$ iff $x^3 \equiv 1$ mod $4$. This is my solution; I was wondering if there was a better or a neater solution than this.
Suppose $x \equiv 1$ (mod $4$).
Then $x = 4k + 1$ for some integer $k$.
Hence, $x^3 = 4(integer) + 1 \equiv 1$ (mod $4$).
Now suppose $x^3 \equiv 1$ (mod $4$) $\iff$ $x^3 - 1 \... | $$x^3-1=(x-1)(1+x+x^2)$$
If $4$ divides $x-1$, then from the factorization above we can see that it also divides $x^3-1$.
If $4$ divides $x^3-1$, then $x$ must be odd. This implies that $1+x+x^2$ must also be odd, which means that $x-1$ must be divisible by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \f... | The arithmetic error is already pinpointed. The result is a special case of the Theorem below
since we have $\ \overbrace{x\!-\!1\, +\, x\!-\!4}^{\Large a\ \ +\ \ b} \, =\, \overbrace{x\!-\!2\, +\, x\!-\!3}^{\Large c\ \ +\ \ d}\ =\ \overbrace{\color{#c00}{2x\!-\!5}}^{\LARGE\color{#c00}s}\ \ $ and $\ \ \color{#0a0}{\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
Fast/smart way to write polar curve in cartesian Is there a fast way to write the curve:
$$r=\frac{a}{1-\frac{1}{\sqrt{2}}\cos(\theta)}$$
as a cartesian curve $f(x,y)=0$?
It seems I can take
$$r(1-\frac{1}{\sqrt{2}}\cos(\theta)) = a$$
$$r-\frac{x}{\sqrt{2}}=a$$
$$\sqrt{x^2+y^2}-\frac{x}{\sqrt{2}}=a$$
$$x^2+y^2=(a+\fra... | HINT: we get
$$\sqrt{x^2+y^2}=\frac{a}{1-\frac{1}{\sqrt{2}}\frac{x}{\sqrt{x^2+y^2}}}$$
and we get by squaring
$$2(x^2+y^2)=2a^2+x^2+2\sqrt{2}ax$$
and then we have $$(x-\sqrt{2}a)^2+2y^2=2a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find all real triples $(x,y,z)$ such that: $x^2+1=2y,y^2+1=2z,z^2+1=2x$
Find all real triples $(x,y,z)$ such that: $x^2+1=2y,y^2+1=2z,z^2+1=2x$.
Can it be solved avoiding equations of orders higher than 2?
| Adding all the equations, we get -
$$x^2+1+y^2+1+z^2+1=2x+2y+2z$$
$$x^2-2x+1+y^2-2y+1+z^2-2z+1=0$$
$$(x-1)^2+(y-1)^2+(z-1)^2=0$$
$$\implies x=y=z=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding the remainder when $p(x^7)$ is divided by $p(x)$, where $p(x)=x^6+x^5+ \cdots + x + 1$? I know there are one or two similar questions to this one but I did not want to look at them since I would like just a hint.
Find the remainder when $p(x^7)$ is divided by $p(x)$, where $p(x)=x^6+x^5+ \cdots + x + 1$?
I wr... | Write $p(x^7) = q(x)p(x) +r(x)$ where $deg(r) <7$.
If we say $1,a_1,...a_6$ are all zeros (which are obviously different) of $x^7=1$ then for each $i$ we get:
$$ p(1) = p(a_i^7) = q(a_i)p(a_i) +r(a_i) = r(a_i)$$
Thus $r$ takes the same value for 6 different values and thus $r(x) \equiv p(1) =7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimu... | $11$ is reasonably small. So we can search exhaustively through the options to find maximum $f=xyz+xy+yz+zx$.
Take $x<y<z$. Then smallest values for $x,y$ are $1,2$ and thus greatest $z=8$. Then we can work down values of $z$ and partition the residue accordingly, avoiding duplicates.
$\begin{array}{|c|c|} \hline
(x,y,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Solve the differential equation $\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$ This is problem 9, exercise 10 from Tannebaum and Pollard's ODE book. I have deduced that the differential equation is exact, but I can't find all the integrable combinations.... | Your solution is good, here is another approach:
$$\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$$
$$\arctan(xy)dx +\frac{xy}{1+x^{2}y^{2}}dx +\frac{x^{2}}{1+x^{2}y^{2}}dy -\frac{2xy^{2}}{1+x^{2}y^{2}}dx -\frac{2x^{2}y}{1+x^{2}y^{2}}dy=0$$
$$\arctan(xy)d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1... | It's probably easier to use $$\cos^2\alpha = \frac{1}{\sec^2\alpha} = \frac{1}{1+\tan^2\alpha} = \frac{1}{1+\frac{1}{\cot^2\alpha}}$$ to find $\cos\alpha$, this gives $\cos\alpha=\pm\frac7{25}$. The given range for $\alpha$ tells which of these applies.
I'd suggest you check your calculation of $\cos\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$
Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$
My attempt. Taking $\sin x \sin y=t$, $\iint\sin x \ dx \ dy$.
I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.
... | Perform integration by parts,
$\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\sin x \sin y) \ dx \ dy\\
&=\int_0^{\frac{\pi}{2}}\left(\Big[-\cos x\arcsin(\sin x \sin y)\Big]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx\right)\ dy\\
&=\int_0^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for ... | Define
$$f(x) = \sin x - x + \frac{1}{6}x^3$$
and
$$g(x) = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \sin x$$
We want to show $f(x) >0$ and $g(x) > 0$ for $x > 0$.
Notice that $f$ and its first four derivatives vanish at $x=0$. If $f$ has a positive zero, say $f(a_0) = 0$, then by Rolle's Theorem $f'$ has a zero $a_1$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result t... | We can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$.
Since $a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2)$, we see that
$$a^2+b^2+c^2-ab-ac-bc=0\Leftrightarrow a=b=c.$$
Thus, since $\sqrt[3]{2+\sqrt5}\neq\sqrt[3]{2-\sqrt5}$, we obtaion
$$x=\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}$$ it's
$$x-\sqrt[3]{2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
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If $x^3+y^3+(x+y)^3+33 xy=2662$, $x,y\in \Bbb R$, find $S=x+y$.
If $x^3+y^3+(x+y)^3+33 xy=2662$ and $\{x,y\}\subset \Bbb R$, find $S=x+y$.
This question from an olympiad contest. Answer stated: $S=x+y=11$
Tried to develop $(x+y)^3$ to find something useful for the situation, but without success.
| $$ (x+y)^3 + (x+y)(x^2-xy+y^2)+33xy=2662 $$
$$ S^3 + S(x^2+y^2-xy) + 33xy=2662 $$
$$ S^3 + S(S^2 - 3xy) + 33xy=2662 $$
$$ 2S^3 + 3xy(11 - S) = 2662 $$
$$ \frac{3}{2}xy(11-S)=11^3 - S^3$$
if $S \ne 11$:
$$ \frac{3}{2}xy = 121 + 11S + S^2 $$
Now try to write down roots of this equation on $S$, add condition of their exis... | {
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"url": "https://math.stackexchange.com/questions/2395460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $(\sqrt{3}-3i)^6$
Evaluate $$(\sqrt{3}-3i)^6.$$
So I assume that we should write the following in polar form
$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$
$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$
So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m}... | Yes, it is correct. In fact
$$(\sqrt{3}-3i)^6=3^3(1-\sqrt{3}i)^6=2^63^3(e^{-i\pi/3})^6=1728.$$
P.S. There is no need of the term $+2\pi k$.
| {
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Find an invertible matrix $S$ and a matrix $J$ in Jordan form, such that $S^{-1}AS = J$ $A = \begin{pmatrix}
0 & -1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
3 & -1 & -2 & -1 \\
-1 & 0 & 1 & 1 \\
\end{pmatrix}$
We need to find an invertible matrix $S$ and a matrix in Jordan form $J$, such th... | The eigenvector for $-1$ you already have, call it $v_1$. For $0$ what you want is a vector $v_4$ such that $A^3v_4 = 0$ but $A^2v_4 \ne 0$. I'll let you think of how to do this. (Incidentally all three vectors you have for a basis of $N(A^3)$ work.)
Once you have $v_4$, let $v_3 = Av_4$ and $v_2 = A^2v_4 = Av_3$. Then... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A question of trigonometry on how to find minimum value.
Find The minimum value of $$2^{\sin^2 \alpha} + 2^{\cos^2 \alpha}.$$
I can easily get the maximum value but minimum value is kinda tricky. Please help.
| $x:= \sin^2(\alpha) $; $ 1-x = \cos^2(\alpha)$ .
Then: $ f(x):= 2^x + \frac{2}{2^x} $, $0 \le x \le1$.
$z := 2^x$ ;
$ g(z): = z + \frac{2}{z} , 1 \le z \le 2$.
AM GM inequality:
$ (1/2) ( z + \frac{2}{z}) \ge (z \frac{2}{z})^{1/2} =$
$ \sqrt{2}$.
$g(z) = z + \frac{2}{z} \ge 2 \sqrt{2}$.
Equality for $z = √2$.
Back s... | {
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"url": "https://math.stackexchange.com/questions/2396947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How to differentiate $\left(x^2 + \frac{1}{x^2}\right)^5$ How can I differentiate this expression
$$\left(x^2 + \frac{1}{x^2}\right)^5$$
I will appreciate any help.
| Start with the Chain Rule:$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$
where $f(x)=x^5$ and $g(x)=x^2+\frac{1}{x^2}$. $$\\$$
For $f'(g(x))$ we get:
$$f'(x)= 5(x^2+\frac{1}{x^2})^4.$$
For $g'(x)$ we can separate the expression:
$$\frac{d}{dx} (x^2) + \frac{d}{dx} (\frac{1}{x^2})$$
$\frac{d}{dx} (x^2) = 2x$ and $\frac{d}{dx} (\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functional equation : $ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$ Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying
$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$
$\forall n \in \mathbb{N}$.
Thank you, Batominovski and Guy Fabrice..
Is my und... | Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying
$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$
This at $n+1$ gives
$$ f(1)^3 + f(2)^3 + \ldots + f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$$
i.e
$$ (f(1) + f(2) + \ldots + f(n))^2+ f(n+1)^3 = (f(1) + f(2) + \ldots +... | {
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"timestamp": "2023-03-29T00:00:00",
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How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
show that there exsit infinitely many postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$
May try it is clear $(x,y,z)=(1,1,1)$ is one solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
| For the equation.
$$(x+y+z)^2+2(x+y+z)=5(xy+xz+yz)$$
It is possible to reduce the parameterization of the solutions to some equivalent to the Pell equation.
It has the form.
$$x=3a^2-(b+c)a+b^2-3bc+c^2$$
$$y=a^2-(b+3c)a+3b^2-bc+c^2$$
$$z=a^2-(3b+c)a+b^2-bc+3c^3$$
These parameters can be recorded through the solution of... | {
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"timestamp": "2023-03-29T00:00:00",
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$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\si... | We need to prove that
$$\frac{1-\cos80^{\circ}}{2}<\frac{3}{7}$$ or
$$\sin10^{\circ}>\frac{1}{7}.$$
Let $\sin10^{\circ}=x$.
Thus, $$3x-4x^3=\frac{1}{2}$$ or $f(x)=0$, where
$$f(x)=x^3-\frac{3}{4}x+\frac{1}{8}$$ and since $$f\left(\frac{1}{7}\right)=\frac{1}{343}-\frac{3}{28}+\frac{1}{8}=\frac{57}{2744}>0,$$ we are do... | {
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"url": "https://math.stackexchange.com/questions/2401341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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