Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
determine $\left(\sum\limits_{g=1}^{p-1} g \times ((ag \mod p))\right) \bmod p^2$ Let $p$ be a prime, and let $((x))$ be the least positive residue of $x$ modulo $p$. For $a \in \{1, 2, ..., p-1\}$, consider $f(a) = \sum\limits_{g=1}^{p-1} g \times ((a g))$.
Does the following hold: $f(a)=f(b)$ if and only if $a=b$ or ... | First,
$f(a) = \frac{1}{2} \sum_{g=1}^{p-1} [ (g + a(g))^2 - g^2 - a(g)^2 ] = \frac{1}{2} \sum_g (g + a(g))^2 - \sum_g g^2$ .
Next, note that $g + a(g) = c(g)$ or $c(g) + p$ for $a \neq p-1$, where $c = a+1 ($mod $p)$, and $g + a(g) = p$ for $a=p-1$.
So for $a=p-1$,
$f(a) = \frac{p^2(p-1)}{2} - \frac{(p-1)p(2p-1)}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/72328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Need help computing $(f \circ g)(x)$ $$f(x) = \frac{1}{x^2} + 1$$
$$g(x) = \frac{1}{x - 1}$$
Do I just insert the $g(x)$ function into the $f(x)$ function giving me this terrible looking thing?:
$$\frac{1}{\frac{1}{x^2}+1-1}$$
Thanks for any help, I really just need confirmation I'm going in the right direction.
| Actually, it would be the other way around. $(f \circ g)(x) = f(g(x))$, and $f(g(x)) = \frac{1}{g(x)^2}+1 = \frac{1}{\frac{1}{(x-1)^2}}+1$. This last expression is almost equal to $(x-1)^2+1$. The difference is that in the former, $ f \circ g$ is undefined at $x = 1$, while in the latter, it is. The correct function wo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/73680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the Laurent series of $f(z)=1/((z-1)(z-2))$ Let
$$f(z)=\frac{1}{(z-1)(z-2)}$$
and let
$$R_1=\Bigl\{z\Bigm| 1<|z|<2\Bigr\}\quad\text{ and }\quad R_2=\Bigl\{z\Bigm| |z|>2\Bigr\}.$$
How do you find the Laurent series convergent on $R_1$? Also how do you do it for $R_2$?
I'm having serious trouble with this as I ... | The function $f(z)$ can be expanded into two partial fractions
$$
f(z):=\frac{1}{\left( z-1\right) \left( z-2\right) }=\frac{1}{z-2}-\frac{1}{z-1}.
$$
We now expand each partial fraction into a geometric series. On $R_{2}$
these series are
$$
\begin{eqnarray*}
\frac{1}{z-2} &=&\frac{1}{z\left( 1-2/z\right) }=\frac{1}{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/79012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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Proving $f(z)=(e^{3iz}-3e^{iz}+2)/z^3$ has a simple pole at $0$ Let $f(z)=(e^{3iz}-3e^{iz}+2)/z^3$, this clearly has a singularity at $z=0$, how do you show that this is a simple pole? i.e. a pole of order one, every way it definity expands to something with a minimum power of $-3$, so I come to the conclusion there is... | For $|z|\ll 1$, we have
$$
\begin{split}
e^{3iz} &= 1 + 3 i z - \frac{9}{2} z^2 + O(z^3) \\
e^{iz} &= 1 + i z - \frac{1}{2} z^2 + O(z^3)
\end{split}
$$
so
$$
\begin{split}
\frac{1}{z^3} (e^{3iz} - 3 e^{iz} +2 )
&= \frac{1}{z^3} \left( 1 - 3 + 2 + (3 - 3) i z + (-\frac{9}{2} + \frac{3}{2}) z^2
+ O(z^3) \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/79069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Showing whether two numbers are equal or not $\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0<x,y\le\dfrac{\pi}{4}$ .
Can I show that $x=y $ or find two numbers $x,y$ such that $x\not=y$?
| Start with
$$
\frac{\sin (2x+y)}{\sin (2x)} =\frac{\sin (x+2y)}{\sin (2y)}\tag{1}
$$
Regrouping $(1)$, we get
$$
\frac{\sin\left(\frac{3}{2}(x+y)+\frac{1}{2}(x-y)\right)}{\sin\left((x+y)+(x-y)\right)}=\frac{\sin\left(\frac{3}{2}(x+y)-\frac{1}{2}(x-y)\right)}{\sin\left((x+y)-(x-y)\right)}\tag{2}
$$
Expanding $(2)$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Rolling infinitely many dice For every integer $n>1$, I have a die $D_n$ with $a_n$ sides labeled $1$ to $a_n$.
If $a_n=n^k$, for integer $k>0$, and I roll all the dice at once, what is the probability none of them lands on the side labeled $1$? What is the probability exactly one lands on the side labeled $1$?
| First, suppose there are total of $q$ dice.
Given that, for fixed $m$, $D_m = 1$, the probability that no other die rolled at one is:
$$
\mathbb{P}(\forall_{n \not= m} \, {D_n \not= 1} ; D_m = 1 ) = \frac{1}{1-\frac{1}{a_m}} \prod_{n=2}^q \left( 1- \frac{1}{a_n} \right)
$$
The probability that exactly one die land... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Modular exponentiation by hand ($a^b\bmod c$) How do I efficiently compute $a^b\bmod c$:
*
*When $b$ is huge, for instance $5^{844325}\bmod 21$?
*When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$?
*When $(a,c)\ne1$, for instance $6^{103... | Let's try $5^{844325} \bmod 21$:
$$
\begin{align}
5^0 & & & \equiv 1 \\
5^1 & & &\equiv 5 \\
5^2 & \equiv 25 & & \equiv 4 \\
5^3 & \equiv 4\cdot 5 & & \equiv 20 \\
5^4 & \equiv 20\cdot 5 & & \equiv 16 \\
5^5 & \equiv 16\cdot 5 & & \equiv 17 \\
5^6 & \equiv 17\cdot 5 & & \equiv 1
\end{align}
$$
So multiplying by $5$ six... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/81228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "128",
"answer_count": 11,
"answer_id": 2
} |
Find x: $2^{2x^2}+2^{x^2+2x+2}=2^{4x+5}$ $$2^{2x^2}+2^{x^2+2x+2}=2^{4x+5}$$
I have to find x in this exponential equation.
I tried to write it in another way, like this:
$2^{2x^2}+2\cdot2^{(x+1)^2}=2\cdot2^{4(x+1)}$
But I don't think this would help.
| Divide by $2^{2x^2}$ to get $1 + 2^{-(x-1)^2} \cdot 2^3 = 2^{-2 (x-1)^2} \cdot 2^7$. Substitute $y = 2^{-(x-1)^2}$, and solve the quadratics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/84617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proof about an infinite sum: $\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$ Hello I have a pretty elementary question but I am a bit confused.
I am trying to prove that $$\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$$
thanks,
Thrasyvoulos
| $$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\sum_{k=1}^\infty\frac{1}{k^2+3k+2}=\sum_{k=1}^\infty\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=$$
$$\lim_{n\to\infty}\;\;\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$$
Note that
$$\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/84899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Find the minimum of $x^{2}+5y^{2}+8z^{2}$ if $xy+yz+zx=-1$
If $xy+yz+zx=-1$，find the minimum of $x^{2}+5y^{2}+8z^{2}$.
How to solve it use Elementary mathematics methods？
| Taking the first variation of $xy+yz+zx=-1$, we get
$$
(y+z)\delta x+(z+x)\delta y+(x+y)\delta z=0\tag{1}
$$
Setting the first variation of $x^2+5y^2+8z^2$ to $0$, we get
$$
2x\delta x+10y\delta y+16z\delta z=0\tag{2}
$$
At an extreme $\{x,y,z\}$, any $\{\delta x,\delta y,\delta z\}$ that satisfies $(1)$, must als... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/85434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one? I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?
| You can do long division with polynomials almost the same way you would for integers. For example, $4x^2+5$ can be multiplied by $-\frac{1}{2}x$ to get a leading term of $-2x^3$, so we might say that $4x^2+5$ goes into $-2x^3 + 4x^2 - 3x + 5$ about $-\frac{1}{2}x$ times. As with long division, we then multiply $4x^2+5$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/86190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Help with integrating $\int \frac{t^3}{1+t^2} ~dt$ What am I doing wrong on this integration problem?
$$
\begin{align*}
\int\frac{t^3}{1+t^2} &=
\frac14 t^4 (\ln(1+t^2) (t+\frac13 t^3))
\\ &= \frac14 t^4(t \ln(1+t^2)+\frac13 t^3 \ln(1+t^2)
\\ &= \frac14 t^5 \ln(1+t^2)+\frac{1}{3}t^7 \ln(1+t^2)
\end{align*}$$
Ans... | $$
\int \frac{t^3}{1+t^2}\, dt = \int \frac{t^2}{1+t^2} \Big( \underbrace{{}\quad t\, dt\quad{}}_{\text{HINT}}\Big) = \int \frac{u}{1+u} \Big(\frac 1 2 \, du \Big) = \int \left( 1 - \frac{1}{1+u} \right) \Big(\frac 1 2 \, du \Big)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to prove :If $p$ is prime greater than $3$ and $\gcd(a,24\cdot p)=1$ then $a^{p-1} \equiv 1 \pmod {24\cdot p}$? I want to prove following statement :
If $p$ is a prime number greater than $3$ and $\gcd(a,24\cdot p)=1$ then :
$a^{p-1} \equiv 1 \pmod {24\cdot p}$
Here is my attempt :
The Euler's totient function ca... | HINT $\ $ By Carmichael's simple generalization of Euler $\phi$, since prime $\rm\:p\:$ is coprime to $2,3$
$\rm\ \ \lambda(8\cdot3\cdot p) = lcm(\lambda(8),\lambda(3),\lambda(p)) = lcm(\phi(8)/2,\phi(3),\phi(p)) = lcm(2,2,p-1) = p-1\ $
therefore $\rm\quad gcd(n,24\:p) = 1\ \ \Rightarrow\ \ 1\ \equiv\ n^{\:\lambda(24\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Simplifying Quadratic Equation Quick question, say I'm simplying a solution I got using the quadratic equation and I run into this:
Original version (as posted by OP):
x = -7 +- 3 sqrt(5) over 3
Edited version:
$$
x = \frac{-7\pm 3 \sqrt{5} }{3}
$$
Would the two $3$s cross each out leaving the answer to be $x = -7... | I'm a little late on the scene, but from your most recent comments you still haven't seen the light. Maybe this will help.
$$\frac{-7\pm 3 \sqrt{5} }{3} \;\; =\;\; \left(\frac{1}{3}\right) \left(\frac{-7\pm 3 \sqrt{5}}{1}\right) \;\;= \;\; \left(\frac{1}{3}\right)\left(-7 \; \pm \; 3 \sqrt{5}\right) $$
$$= \;\; \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/88859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prime Partition A prime partition of a number is a set of primes that sum to the number. For instance, {2 3 7} is a prime partition of $12$ because $2 + 3 + 7 = 12$. In fact, there are seven prime partitions of $12$: {2 2 2 2 2 2}, {2 2 2 3 3}, {3 3 3 3}, {2 2 3 5}, {2 5 5}, {2 3 7}, and {5 7}. The number of prime part... | You need to learn a bit about generating functions. The text associated with A000607 means the following. For each prime $p$ expand the function $1/(1-x^p)$ as a power series:
$$
\frac1{1-x^p}=1+x^p+x^{2p}+x^{3p}\cdots=\sum_{k=0}^\infty x^{kp}.
$$
Call that series $f_p(x)$.
Then you multiply these power series togeth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 1
} |
Coefficients of $(1+x+\dots+x^n)^3$? Consider the following polynomial:
$$ (1+x+\dots+x^n)^3 $$
The coefficients of the expansion for few values of $n$ ($n=1$ to $5$) are:
$$ 1, 3, 3, 1 $$
$$ 1, 3, 6, 7, 6, 3, 1 $$
$$ 1, 3, 6, 10, 12, 12, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10,... | Denote by
$$\sum_{k=0}^{3n}{\binom{3}{k}}_{n+1}x^{k}=(1 + x + ... + x^{n})^3\,$$
the expansion of multinomial of degree $n$
From that relation follow that
$$ \sum_{k=0}^{3n)}{\binom{3}{k}}_{n+1}x^{k}=\left(\frac{1-x^{n+1}}{1-x}\right)^{3}=(1-x^{n+1})^{3}(1-x)^{-3}\,$$
from binomial formula we have
$$(1-x^{n+1})^{3}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/91516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
General method for factorizing matrix determinants I'm learning how to factorize determinants of a square matrix in school, but we haven't learnt a general method to do that, besides 'creating zeros'. So I thought maybe I'll ask here if someone does know a method that generalizes the factorization of a square matrix.
A... | There is generally no easy way. Concise (closed) formula for the determinants of a general matrix is rare. For your example, it is too special, the determiant is equal to $\left|
\begin{array}{cccc}1 & 1 & 1 & 1\\
x & y & 1 & 0\\
x^2 & y^2 & 1 &0\\
x^3 & y^3 & 1&0
\end{array}
\right|$, where the matrix is a Van... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/92219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does iterating $n \to 2n+1$ always eventually produce a prime number? Is it the case that for every non-negative integer $n$, iterating $n \to 2n+1$ eventually produces a prime number? (This is the same as asking whether for every positive integer $n$, there is a non-negative integer $k$ such that $2^k n - 1$ is prime... | (Following up on the connection to Riesel numbers mentioned in other answers ...)
For an infinite set of counterexamples, here is a proof-sketch (adapted from Baczkowski, et al.) showing that
$$n \equiv 33737173 \pmod{3\cdot 5\cdot 7\cdot 13\cdot 19\cdot 37\cdot 73} \implies \forall k\in \mathbb{N}\ \big(f^k(n-1) \te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 2
} |
What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$? Find the limit:
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$
I am not able to find it because I don't know how to prove or disprove $0$ is the answer.
| Simplify to have $$\frac{\sin x-x }{x\sin x}$$ and consider Maclaurin's series for $$\sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}-...$$
So you have $$\frac{(x-\frac {x^3}{3!}+\frac {x^5}{5!}-...)-x}{x(x-\frac {x^3}{3!}+\frac {x^5}{5!}+...)}=\frac{(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 6
} |
prove$\frac{ \sin a\vphantom{(}}{\sin b} +\frac{\cos a\vphantom{(}}{\cos b} = \frac{2\sin (a+b)}{\sin 2b}$ I've got this far but don't understand where the $2$ on the numerator comes from:
$$\dfrac{\sin a \cos b + \cos a \sin b}{\sin b \cos b}\overset{?}{=}\dfrac{\sin(a+b)}{\sin 2b}$$
| $$\dfrac {\sin a}{\sin b}+\dfrac{\cos a}{\cos b}=\dfrac{\sin a\cdot\cos b+\cos a\cdot\sin b}{\sin b\cdot \cos b}$$
After getting this far, you need to observe that, $\boxed{\sin 2b=2 \cdot\sin b \cdot \sin b}$, you'll have to multiply the numerator and denominator by $2$, you'll have the following,
$$\dfrac{2(\sin a\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/99408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$.
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
| $n^3+6n^2+11n+6=n^3+n^2+5n^2+5n+6n+6=n^2(n+1)+5n(n+1)+6(n+1)=$
$=(n+1)(n^2+5n+6)=(n+1)(n^2+3n+2n+6)=(n+1)(n+2)(n+3)$
Now , since this last expression represents a product of three consecutive integers it has to be divisible by $3$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 7
} |
how to work out a closed form of a sequence
Consider the following linear recurrence sequence.
$x_1 = 11$, $x_{n+1} = -0.8x_n + 9,\quad n = 1,2,3, \ldots.$
Find a closed form for this sequence.
| Use generating functions. Define $X(z) = \sum_{n \ge 0} x_{n + 1} z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\frac{X(z) - x_1}{z} = - \frac{4 X(z)}{5} + \frac{9}{1 - z}
$$
Thus:
$$
X(z) = \frac{55 - 10 z}{5 - z - 4 z^2}
= 5 \cdot \frac{1}{1 - z} + 6 \cdot \frac{1}{1 + 4 z / 5}
$$
Two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\sin^3{\left(\cot^{-1}{\left(x\right)}\right)}$ How can the following function such that no trigonometric functions are present:
$\sin^3{\left(\cot^{-1}{\left(x\right)}\right)}$
Wolfram|Alpha shows the result as $\frac{1}{{\sqrt{x^2+1}}^3}$.
Thank you for your time.
| You can show that for $x > 0$
$${\cot ^{ - 1}}x = {\sin ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}$$
Then
$$\sin {\cot ^{ - 1}}x = \frac{1}{{\sqrt {1 + {x^2}} }}$$
and thus
$${\left( {\sin {{\cot }^{ - 1}}x} \right)^3} = \frac{1}{{1 + {x^2}}}\frac{1}{{\sqrt {1 + {x^2}} }}$$
The proof:
$$x = \cot y$$
$$1+x^2 = \csc^2 y $$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/107702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find remainder when dividing $9^{{10}^{{11}^{12}}}-5^{9^{10^{11}}}$ by $13$.
Find remainder when dividing
$$9^{{10}^{{11}^{12}}}-5^{9^{10^{11}}} \hspace{1cm} \text{by} \hspace{1.2cm} 13.$$
I tried transforming these who numbers separately to form $13k+n$ but failed.
| Write this number as $9^N-5^M$.
Since $3^3=1\pmod{13}$, $9^3=1\pmod{13}$. Since $10=1\pmod{3}$ and $N$ is a power of $10$, $N=1\pmod{3}$. Hence $9^N=9\pmod{13}$.
Since $5^2=-1\pmod{13}$, $5^4=1\pmod{13}$. Since $9=1\pmod{4}$ and $M$ is a power of $9$, $M=1\pmod{4}$. Hence $5^M=5\pmod{13}$.
Finally, $9^N-5^M=9-5=4\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/109963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
kronecker product property I'm confused about this concept. Is this right? If not what is the correct version
$$
\begin{bmatrix}\mathbf{I}\otimes\mathbf{x}_{1}^{\prime}\\
\vdots\\
\mathbf{I}\otimes\mathbf{x}_{i}^{\prime}\\
\vdots\\
\mathbf{I}\otimes\mathbf{x}_{n}^{\prime}
\end{bmatrix}=\ma... | We have
$$\begin{bmatrix} \mathbf I \otimes \mathbf x_1' \\ \vdots \\ \mathbf I \otimes \mathbf x_n' \end{bmatrix} =
\begin{bmatrix}
\mathbf y_1 \\ \vdots \\ \mathbf y_n
\end{bmatrix}$$
where
$$\mathbf y_i = \mathbf I \otimes \mathbf x_i' =
\begin{bmatrix}
\mathbf x_i' & 0 & 0 & \cdots & 0 \\
0 & \mathbf x_i' & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Simultaneous Differential Equations How can we solve the simultaneous equations:
$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{x\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$
$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot y}{\sqrt{\dot x^2+\dot y^2}}\... | Take the difference between the two equations and get
$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot x-\dot y}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{(x-y)\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$
and the result follows straightforwardly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of $ u_{n}=\sin(\frac{1}{n+1})+\cdots+\sin(\frac{1}{2n})$ How can I find the limit of
$$ u_n =\sin\left(\frac{1}{n+1}\right)+\cdots+\sin\left(\frac{1}{2n}\right)$$
when $n\rightarrow\infty$?
We have:
$$ \sum_{n=1}^\infty u_{n+1}-u_n =u_\infty -\sin\left(\frac{1}{2}\right)$$
So how can I find $$ \sum_{n=1}^\infty... | We can make use of the fact that $$\sin x = \sum\limits_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
and note that when $x\leq 1$, the absolute value of each term is more than the sum of the absolute values of the later terms. Thus $x-\frac{x^3}{6}<\sin x<x$ for $x\leq 1$, so we have $$\sum\limits_{k=n+1}^{2n}\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\varlimsup _{n\rightarrow \infty}(u_n)^{\frac{1}{n}}=1$ where $u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}$ and $u_0=1$. Prove that
$$\varlimsup_{n\rightarrow \infty} (u_n)^{\frac{1}{n}}=1,$$
where $u_0=1$ and $$u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\;.$$
| We can do even better: $\lim\limits_{n\to\infty} (u_n)^{1/n}=1$. Note that for $x>0$ we have
$$\frac{d}{dx}\frac{2x^3+2x^2+x}{2x^2+3x+1}=\frac{4x^4+12x^3+10x^2+4x+1}{(2x^2+3x+1)^2}>0$$
thus since clearly each $u_n>0$, $\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}$ is an increasing function of $u_n$. This allows us to show t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
| First $(x^3 + y^3 +z^3)(x+y+z) = x^4 + y^4 + z^4 + xy^3 + yx^3 + xz^3 +zx^3 +yz^3+y^3z = 0 $ which means
$$ x^4 + y^4 + z^4 = -xy(x^2+y^2) - yz(y^2+z^2) -zx (z^2 + x^2) \ \ \text{(1)}$$
$x+y+z=0$ also implies $x^2+y^2=z^2-2xy$, substitute all the sum of square we have
$ x^4 + y^4 + z^4 = -xy(z^2-2xy) - yz(x^2-2xz)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series. Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$
Computing the Fourier series of $f$ and using Parseval's identity, I have computed $\z... | One method is to consider the generating function of $\zeta(2k)$:
$$
\begin{align}
f(x)
&=\sum_{k=1}^\infty\zeta(2k)\,x^{2k}\\
&=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{x^{2k}}{n^{2k}}\\
&=\sum_{n=1}^\infty\frac{x^2/n^2}{1-x^2/n^2}\\
&=\sum_{n=1}^\infty\frac{x^2}{n^2-x^2}\\
&=-\frac{x}{2}\sum_{n=1}^\infty\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 6,
"answer_id": 0
} |
Partial Fraction Decomposition of $\frac{x^4+2}{x^5+6x^3}$ I tried answering the following question but I'm getting it wrong for some reason. I would appreciate any help.
$$\frac{x^4+2}{x^5+6x^3}$$
My answer:
$$\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{Dx+E}{x^3}+\frac{Fx+G}{x^2+6}$$
What am I doing wrong?
| There are two ways to deal with a repeated factor in the denominator, such as your $x^3$:
*
*You can put a single fraction, with denominator the full repeated factor, and undetermined denominator of degree one less. In your example, since the denominator factors as $x^3(x^2+6)$, you would set up the partial fraction... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/116199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How does one find the inequality $\tau(n) < (\frac{2}{\log 2}\log n)^{2^x}n^{\frac{1}{x}}$? Context (Though I don't see how this might help deriving the final inequality):
Example : For $p$ prime, $x$ greater than $0$ a real number, $n$ greater or equal to $2$ an integer: for $p< 2^{x}$ it holds that $$1+\text{ord}_... | I think this might work. Given a real $x > 0$ partition the prime divisors of $n$ into two classes, those which are bounded from above by $2^{x}$ and those bounded from below by $2^x$. We have
\begin{align}
\tau(n) = \prod_{2^{x} \geqslant p \mid n}( \text{ord}_{p}(n) + 1) \prod_{2^{x} < p \mid n}( \text{ord}_{p}(n) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $a+b+c+d=16$, then $(a+\frac{1}{c})^2+(c+\frac{1}{a})^2 + (b+\frac{1}{d})^2 + (d+\frac{1}{b})^2 \geq \frac{289}{4}$
If $a,b,c,d$ are positive integers and $a+b+c+d=16$, prove that
$$\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2 \geq \frac{289... | First apply the Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality (from http://tinyurl.com/84o57u4)
$$\sqrt{\frac{\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2}{4}} \geq \frac{\left(a+\frac{1}{c}\right)+\left(c+\frac{1}{a}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes?
How do we show that For $x,y \ge 0$ real numbers, there exists a constant C suchthat: $$\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$$ Where $\pi(.)$ denotes thes prime counting function, is true?
the hint is to sieve n with $y<... | Hint: Which numbers modulo $6$ can be prime? (Certainly not those divisible by $2$ or $3\ldots$)
So on an interval of width $x$ from $y$ to $y + x$, how many primes ($\pi(y+x) - \pi(y)$) do we expect at most on this interval?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Prove the equation without solving for X My niece asked me this - If $x=1/(5-x)$ prove $x^3 + \dfrac{1}{x^3}=110$ without solving for x. I said its not possible since solving for x itself gives me two roots for x (one being $\approx4.79$) and substituting for it in the second equation approx gives me 110. So proving al... | From your equation you get $5x-x^2=1$, so $5=\frac{x^2+1}{x}$. Now remind the following cubic fromula $(a+b)^3=a^3+b^3+3ab(a+b)$, then
$$
5^3=\left(\frac{x^2+1}{x}\right)^3=\left(x+x^{-1}\right)^3=x^3+x^{-3}+3xx^{-1}(x+x^{-1})=
$$
$$
x^3+x^{-3}+3\frac{x^2+1}{x}=x^3+x^{-3}+15.
$$
So you get
$$
x^3+x^{-3}=125-15=110... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what... | $10^{0} +3⋅4^{0} +5=\color{red}{9}$ and $10^{n+1} +3⋅4^{n+1} +5=10^n +3⋅4^n +5+\color{red}{9}\cdot(10^n+4^n)$ for every $n\geqslant0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 7
} |
First order differential equation homework help Hi I'm really stuck with some homework:
Find the general solution of the differential equation,
$$\left(x+\dfrac{1}x\right)\dfrac{dy}{dx} + 2y = 2\left(x^2+1\right)^2$$
So far, I've divided both sides by $x+\dfrac{1}x$ and integrated $\dfrac{2y}{x + \frac{1}{x}}$ to get $... | $$
\begin{align*}
(x+(1/x))\frac{dy}{dx} +2y &= 2(x^2+1)^2\\
\left(\frac{x^2+1}{x}\right)\frac{dy}{dx} +2y &= 2(x^2+1)^2\\
\frac{dy}{dx} +\frac{2xy}{x^2+1} &= 2x(x^2+1) \tag{A}\\
\frac{dy}{dx} +P(x)y &= Q(x)\\
\end{align*}
$$
where $\displaystyle{P(x) = \frac{2x}{x^2+1}}$ and $\displaystyle{Q(x) = 2x(x^2+1)}$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Gaussian proof for the sum of squares? There is a famous proof of the Sum of integers, supposedly put forward by Gauss.
$$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$$
$$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$$
$$S=\frac{n(1+n)}{2}$$
I was looking for a similar proof for when $S=\sum\limits_{i=1}^{n}i^2$
I've tried th... | You can use something similar, though it requires work at the end.
If $S_n = 1^2 +2^2 + \cdots + n^2$ then
$$S_{2n}-2S_n = ((2n)^2 - 1^2) + ((2n-1)^2-2^2) +\cdots +((n+1)^2-n^2)$$
$$=(2n+1)(2n-1 + 2n-3 + \cdots +1) = (2n+1)n^2$$ using the Gaussian trick in the middle.
Similarly $$S_{2n+1}-2S_n = (2n+1)(n+1)^2$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 1
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How to find the partial sum of a given series? On my last exam there was the question if the series $\sum_{n=2}^{\infty}\frac{1}{(n-1)n(n+1)}$ converges and which limit it has. During the exam and until now, I am not able to solve it. I tried partial fraction decomposition, telescoping sum, etc. But I am not able to fi... | So let's try partial fraction decomposition. Writing
$$
\frac 1{(n-1)n(n+1)} = \frac a{n-1} + \frac bn + \frac c{n+1}
$$
we obtain
$$
1 = a(n^2 + n) + b(n^2 - 1) + c(n^2 - n)
$$
and therefore
\begin{align*}
1 &= -b\\
0 &= a - c\\
0 &= a + b + c.
\end{align*}
This gives $b = -1$, $a = c = \frac 12$. He... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$ All the angles in a triangle $A,B,$ and $C$ are less than $120^{o}$
Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$
| Consider triangle with angles $\small{A_1=120-A, B_1=120-B, C_1=120-C}$
Applying the triangle inequality in this triangle with angles $A_1, B_1,$ and $C_1$,
$$\small{B_1 C_1+C_1 A_1 > A_1 B_1}$$
$$\small{\sin A_1 +\sin B_1 > \sin C_1}$$
$$\small{\sin (120-A) +\sin (120-B) > \sin (120-C)}$$
which by applying $\sin(x-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Find if the sequence $n!/10^{6n}$ converges Find if the sequence $a_n$ converges, $a_n=\frac{n!}{10^{6n}}=\frac{1}{\frac{10^6n}{n!}}$. Now if we take the limit of numerator and denominator as $n \to \infty$, we get $\frac{1}{0}$($\lim\limits_{n\to\infty} \frac{x^n}{n!}=0 $ for any $x$), but division by zero is not de... | Hint for an alternative route,
$$\rm \frac{a_{\large n+10^6}}{a_{\large 10^6}}\normalsize =\frac{10^6+1}{10^6}\cdot\frac{10^6+2}{10^6}\cdots\frac{10^6+n}{10^6}>\left(1+\frac{1}{10^6}\right)^n \to \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
solution to $ 7^{a}+1 =3^{b}+5^{c} $ for natural $a$,$b$ and $c$ How do I solve $ 7^{a}+1 =3^{b}+5^{c} $ for natural $a$,$b$ and $c$?All I got after some modular arithmetic is that the $a$,$b$ and $c$ are all odd.The problem was posted on Art of Problem Solving(with no responses till now) and is supposedly from India I... | It looks as if the problem can be conquered by persistent use of modular arithmetic. Throughout, it is assumed that $a,b,c$ are sufficiently large.
*
*Apply $\bmod{3}$ to find that
$$ 2 \equiv 2^c \pmod{3} $$
It follows that $c \equiv 1 \pmod{2}$. Write $c = 2c_1 + 1$.
*Apply $\bmod{8}$ to find that:
$$ (-1)^a + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/128250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Limit when circumference shrinks Let $C_1$ be a fixed circumference with equation $(x-1)^2 + y^2 = 1$ and $C_2$ a
circumference to be shrinked, with center at $(0, 0)$ and radius $r$.
Let $P$ be the point $(0, r)$, $Q$ the upper intersection between $C_1$ and $C_2$
and $R$ the intersection between the line $PQ$ wit... | The upper intersection point $T=(a,b)$ is on both $C_1$ and $C_2$ and has positive $y$-component. Thus
$$\begin{cases}a^2+b^2=r^2 \\ (a-1)^2+b^2=1. \end{cases}$$
Subtracting the first from the second, we obtain $1-2a=1-r^2$, and thus $a=r^2/2$. We can then solve for $b$ as $b=\sqrt{r^2-(r^2/2)^2}=r\sqrt{1-(r/2)^2}$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/128905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Maximize volume of box in ellipsoid I need to find the dimensions of the box with maximum volume (with faces parallel to the coordinate planes) that can be inscribed in ellipsoid
$$\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$$
A hint given was:
If vertex of box in first octants is (x,y,z) then volume is 8xyz.... | Very recently, you asked for advice on a proof of the AM/GM inequality in three variables. This inequality says that for $u$, $v$, and $w$ all $\ge 0$, we have
$$\frac{u+v+w}{3}\ge \sqrt[3]{uvw},$$
with equality iff $u=v=w$.
The result you want is a direct consequence of three variable AM/GM.
Let $u=\frac{x^2}{4}$, $v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/129249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$3^k$ not congruent to $-1 \pmod {2^e}, e > 2$. $3^k \not\equiv -1 \pmod {2^e}$ for $e > 2, k > 0$. Is this true? I have tried to prove it by expanding $(1 + 2)^k$. [Notation: $(n; m) := n! / (m! (n - m)!)$] E.g., for $e = 3$ I get: $(1+2)^k + 1 = 2 + (k; 1) 2 + (k; 2) 2^2 + (k; e) 2^e + ...$ So, here it's enough to pr... | Just a bit more explanative in case the other answers are too compressed.
We do this by induction. The remarkable property, that $\small 3^2=9=8+1$ suggests, to try this modulo 8 (but that could as well be suggested by the problem and it was asked that the factor $\small 2^e $ with $\small e=2 $ should not be exceeded)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Integration Problem: $\int \frac{x^2+1}{x^5-1}dx$
$$\int \frac{x^2+1}{x^5-1}dx$$
I am unable to integrate it, nothing works. Yes, I can use partial fraction but who remembers factorization of $x^5-1$, I need a better way of doing this.
| For these cyclotomic denominators, one can get an answer pretty fast by factoring into roots of unity. Let $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, where $\theta_k=2\pi k/5$. Then
$$\frac{x^2+1}{x^5-1}=\frac{x^2+1}{\prod_{k=0}^4(x-\omega_k)}=\sum_{k=0}^4\frac{A_k}{x-\omega_k}$$
Obeserving that $\omega_k^{-k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Integrating $\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$ Can someone please help me integrate $$\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$$ the question says, as a hint, use $\cos\theta = \frac{z + z^{-1}}{2}$ with $|z|=1$. I'm not really sure where to start.
| An easy way to evaluate:
$I=\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$
$$I=4\int_0^{\frac{\pi}{2}}\frac{1}{8+\frac{1}{\cos^2\theta}}\frac{d\theta}{\cos^2\theta }= 4\int_0^{\frac{\pi}{2}}\frac{1}{9+\tan^2\theta}d\tan\theta= 4\int_0^{\infty}\frac{1}{3^2+t^2}dt= 2\pi/3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Calculating the regression equations I have four data points $(1,2), (2,4), (3,5), (5,7)$ and Im looking for the least squares regression line that best fits them.
I use the normal equation
$A^tAx=A^tb$
in this form -
$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \... | Everything looks good until
$\begin{bmatrix} 4 & 10 \\ 10 & 30 \end{bmatrix}\begin{bmatrix} c \\ m \end{bmatrix} = \begin{bmatrix} 18 \\ 53 \end{bmatrix}$, but then you have got bad solutions and wolframalpha is right. Also you should check last data point, is it $(4,7)$ or $(5,7)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Sum the series: $ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \cdots \ \text{ad inf}$ How do I sum the following series?
$$ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \sin{3\alpha} + \cdots \ \text{ad inf}$$... | The coefficients in your series can be written as,
$$\frac{1}{4^n} {2n \choose n}$$
This allows us to write your series a bit more compactly as,
$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \sin(n\alpha)$
Consider the following series instead,
$$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} e^{in\alpha} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Generalised Binomial Theorem Intuition It was not until recently (why don't they teach it in secondary school?) that I've come across the Generalised Binomial Theorem, which from what I can tell is basically the same as the regular Binomial Theorem, except that the finite sum is replace by an infinite series:
$$
(x+y)^... | Here is one straightforward approach, mentioned in Lang's Analysis book.
Let $ \alpha \in \mathbb{R} $ and $ x \in (-1,1) $.
We'll try to show $$ (1+x)^\alpha = 1 + \binom{\alpha}{1}x + \binom{\alpha}{2} x^2 + \ldots $$ holds (where for $ k \in \mathbb{Z}_{>0} $, $ \binom{\alpha}{k} := \frac{\alpha (\alpha - 1) \ldots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
What is awry with this proof? Let $x=5$, $y=7$, $z=6$
$x+y = 2z$
Rearranging, $x-2z = -y$
and $x = -y+2z$
Multiply both sides respectively. $x^2-2xz = y^2-2yz$
$$x^2-2xz+z^2 = y^2-2yz+z^2$$
$$(x-z)^2 = (y-z)^2$$
$$x-z = y-z$$
Hence $x=y$, or $5 = 7$
Well, the conclusion is clearly false, but what went wrong? I think it... | You have $x=−y+2z$ and $x−2z=−y$ then you multiply both sides.
I think you are computing $-xy$ in two ways: for the first $-xy=y^2-2zy$
for the second $-xy=-2zx+x^2$ so you have $y^2-2zy=x^2-2zx$. Adding $z^2$ to both sides
gives $(y-z)^2=(x-z)^2$. But now taking square roots gives $|y-z|=|x-z|$.
If you try to remov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/137859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Advice on using the Karamata Inequality Is it possible to use the Karamata inequality to prove the following inequalities to be true ?
$$x^{4}+y^{4}+z^{4}\geq x^{2}yz + xy^{2}z + xyz^{2}$$
$$x^{5}+y^{5}\geq x^{3}y^{2} + x^{2}y^{3}$$
$$x^{4}y + xy^{4}\geq x^{3}y^{2} + x^{2}y^{3}$$
for all $x,y,z \in \mathbb{R}^{+}... | As Byron observed all your inequalities follow immediately from AM-GM inequality.
To prove the first one, you can do the following:
$$\frac{x^4+x^4+y^4+z^4}{4} \geq x^2yz \,$$
$$\frac{x^4+y^4+y^4+z^4}{4} \geq xy^2z \,$$
$$\frac{x^4+y^4+z^4+z^4}{4} \geq xyz^2 \,$$
and add them together.
The second one follows from
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Showing $\lim _{n\rightarrow \infty } \frac {S_{n}}{n^{2}} = \frac{1}{6}$ I am trying to show that if the arithmetic mean of the products of all distinct pairs of positive integers whose sum is $n$ is denoted by $S_{n}$ then $$\lim _{n\rightarrow \infty } \dfrac {S_{n}}{n^{2}} = \dfrac{1}{6}$$
Solution attempt
If $n$ ... | If $n$ is odd, then
$$\sum_{i=1}^{(n+1)/2} i (n-i) = \frac{(n-1)(n+1)(n+3)}{12}$$
Hence, you get that $$\frac{S_n}{n^2} = \frac{2}{n^2(n+1)} \times \frac{(n-1)(n+1)(n+3)}{12} = \frac{(1-1/n)(1+3/n)}{6}$$
If $n$ is even, then
$$\sum_{i=1}^{n/2} i (n-i) = \frac{n(n+2)(2n-1)}{24}$$
Hence, you get that $$\frac{S_n}{n^2} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Determine matrix of a set in a certain base I have a set
$S = \{ x^2 + 1, x + 1, 1 - x, x^3 \}$
in a polynomial vector space.
How do I write a vector matrix of $S$ in the base $B = \{ 1, x, x^2, x^3 \}$?
I attempted this using the formula: $M(S) = B^{-1} * T * B$,
where $T$ is the matrix of the set.
I got the followin... | The purpose of this question is to write the elements of $S$ as linear combinations of elements of the basis $B$, then to give a matrix representation to the result.
Firstly let us call $(1,x,x^2,x^3) = (b_0,b_1,b_2,b_3).$
Then we have
$$
x^2+1 = 1(1) + 0(x) + 1(x^2) + 0(x^3) = b_0+b_2\\\
x+1 = 1(1) + 1(x) + 0(x^2)+0(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of identity $\ln \left|\frac{\sin x}{\cos x - 1}\right| = \ln \left|\frac{\cos x + 1}{\sin x}\right|$ How do you prove this identity:
$$\ln \left|\frac{\sin x}{\cos x - 1}\right| = \ln \left|\frac{\cos x + 1}{\sin x}\right|$$
Mathematica says it's true, but if I try to simplify both sides, I wind up with
$$ \sin^... | We have
$$\left|\frac{\sin x}{\cos x-1}\right|=\left|\frac{\sin x}{\cos x-1}\cdot \frac{\cos x+1}{\cos x+1}\right|=\left|\frac{\sin x(\cos x+1))}{(\cos x)^2-1}\right|.$$
Now, using $(\cos x)^2+(\sin x)^2=1$, we get
$$\left|\frac{\sin x}{\cos x-1}\right|=\left|\frac{\sin x(\cos x+1)}{-(\sin x)^2}\right|=\left|\frac{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Multiset Combination in Combinatorics I want to buy a $k$-combination of doughnuts, where $k$ is any amount less than or equal to the total doughnuts available. At the bakery there are $n$ different types of doughnuts but there are restricted amount left for each type of doughnut.
For example, In this case, the total d... | Let $a_i$ be the number of donuts you buy of type $i$. Then you have $$a_1+a_2+\cdots+a_7=10$$ subject to the constraints, $$0\le a_1\le4,0\le a_2\le2,\dots,0\le a_7\le 8$$ If you think about how you would multiply out the following product, you'll see you want the coefficient of $x^{10}$ in $$P(x)=(1+x+x^2+x^3+x^4)(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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System of equations: $x^2+y=7, y^2+x=11$
Possible Duplicate:
Steps to solve this system of equations
During the flight from Moscow to Yerevan my neighbor gave me the following problem:
Solve the system:
$$\left\{\begin{array}{c}x^2+y=7 \\ y^2+x=11. \end{array}\right.$$
It is easy to find 1 of the 4 solutions. Is th... | Every solution of the given system
$$\left\{
\begin{array}{c}
x^{2}+y=7 \\
y^{2}+x=11
\end{array}
\right. \tag{0}$$
is a solution of
$$\left\{
\begin{array}{c}
\left( y-3\right) \left( y^{3}+3y^{2}-13y-38\right) =0 \\
x^{2}=121-22y^{2}+y^{4}.
\end{array}\tag{1}
\right. $$
The same applies to the system
$$\left\{
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
find an $A$ so that for all $n \ge A (2^n+3^n)/(4^n+5^n)/(1/1000)$ So $1000(2^n+3^n)\le 4^n+5^n$.
I take $1000\cdot2^n\le 4^n$ and $1000\cdot3^n\le5^n$ so that adding both gives the inequality, theorem in the ordered fields.
So $1000\cdot2^n\le2^2n$. This leads me to an $A$ for $n = 10$ as $1000\cdot2^n < 2^n\cdot2^n$ ... | For the similar problem, the inequality is false. If we take the base $10$ log of each side, we get $n \log 1.01 - 147 \log n \lt 2$ or $ .00432n - 147 \log n \lt 2$ Now we just need to take $n$ large enough. $n=10^6$ is easily enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/145202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant fi... | While I was writting this answer, Artem posted his. My answer is essentially his 1st one.
Let $f(t)=\sqrt{2+t^{2}}$. Since $f(-t)=$ $f(t)$, we have
$$\begin{equation*}
I=\int_{-20}^{20}\sqrt{2+t^{2}}dt=2\int_{0}^{20}\sqrt{2+t^{2}}dt
\end{equation*}$$
As the integrand $f(t)$ is a quadratic irrational we can also use th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/147788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Matrix transformation mapped onto itself Question:
$$A= \begin{pmatrix} k & -2 \\ 1-k & k \end{pmatrix}\text{, where k is a constant}$$
$$\text {A transformation } T : \mathbb{R}^2 \rightarrow \mathbb{R}^2 \text{ is represented by the matrix A.}$$
$$\text {Find the value of k for which the line } y = 2x \text{ is m... | On the same line (!) of thought: the line $\,l:y=2x\,$ is the same as the vector space $\,\operatorname{Span}\{(1,2)\}\leq\mathbb{R}^2$ , or if you prefer: $\,l:\{(r,2r)\,/\,r\in\mathbb{R}\}\,$ , and then what we really want to happen is $$\begin{pmatrix}k&-2\\1-k&k\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Finding square roots of $\sqrt 3 +3i$ I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.
$w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}... | Denesting the radical $\:\sqrt{(1+\sqrt{-3})/2}\:$ can be tackled by employing an easy radical denesting formula that I discovered as a teenager.
Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Factorize the polynomial $P(z) = z^4 - 2z^3-z^2+2z+10$, into linear and/or quadratic factors with real coefficients $2+i$ is given to be one of the roots of the polynomial.
I am doing this as a practice for exam prep.
Since $2+i$, is a root, then $(z-2-i)$ is a factor?
So I have:
$(z-2-i)(z^3-Az^2-Bz+C) = z^4-2z^3-z^2+... | Since the question has the "calculus" tag, letus use a slightly different method. We start with $$(z^2-4z+5)(z^2+bz+c)= z^4-2z^3-z^2+2z+10.$$ as smanoos does. When we plug zero int both sides, we get $5c=10$, so $c=2$. This gives us the equation,
$$(z^2-4z+5)(z^2+bz+2)= z^4-2z^3-z^2+2z+10.$$. Now letus take the deriva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/152290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integral of $\int \sqrt{1-4x^2}$ I know I am messing up something with the substitutions but I am not sure what.
$$\int \sqrt{1-4x^2}$$
$$u = 4x, du = 4 \,dx$$
$$\frac{1}{4}\int \sqrt{1-u^2}$$
$u = \sin \theta$
$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \the... | Some things to watch out for:
*
*As already mentioned, if you substitute $u = 4x$, then you get $u^2 = (4x)^2 = 16x^2$ instead of $u^2 = 4x^2$. If you use $u = 2x$, then $u^2 = (2x)^2 = 4x^2$.
*An annoying but important part of practicing good mathematics is being precise.
*
*In your integrals, you forgot the $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A problem about rational number for right triangles. If a right triangle has hypotenuse of length 5, and the remaining two sides $a, b$ have rational lengths, what can we say about $a$ and $b$?
| We want to generate the rational solutions of $u^2+v^2=25$. Equivalently, we
generate the solutions of the equation $x^2+y^2=25z^2$ in integers, with $z\ne 0$, and set $u=x/z$, $v=y/z$. Ultimately the procedure we get is a close analogue of the standard way of generating primitive Pythagorean triples. The case where $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 ... | A very simple way to check if the answer to an indefinite integral is correct is to differentiate the answer. If you get the original function, your answer is correct, and is equal, up to a constant, with any other solutions.
We have
$$\begin{align*}
\frac{d}{dx}\left(\ln|\sec x| + \frac{\cos 2x}{4} + C\right) &= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 3
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Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| You can use
$$x^4+1=(x^2+1)^2-2x^2=(x^2+1+\sqrt 2x)(x^2+1-\sqrt 2x).$$
Then, find $A,B,C,D$ such that
$$\frac{1}{x^4+1}=\frac{Ax+B}{x^2+1+\sqrt 2x}+\frac{Cx+D}{x^2+1-\sqrt 2x}.$$
You'll find
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt 2}\left\{ \frac{x+\sqrt 2}{x^2+\sqrt2 x+1}+\frac{-x+\sqrt 2}{x^2-\sqrt 2x+1}\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 18
} |
Discriminant for $x^n+bx+c$ The ratio of the unsigned coefficients for the discriminants of $x^n+bx+c$ for $n=2$ to $5$ follow a simple pattern:
$$\left (\frac{2^2}{1^1},\frac{3^3}{2^2},\frac{4^4}{3^3},\frac{5^5}{4^4} \right )=\left ( \frac{4}{1},\frac{27}{4},\frac{256}{27},\frac{3125}{256} \right )$$
corresponding to ... | Yes. Sketch: $b$ is a symmetric polynomial of degree $n-1$ in the roots and $c$ is a symmetric polynomial of degree $n$, whereas the entire discriminant is a symmetric polynomial of degree $n(n-1)$. It follows that the discriminant is a linear combination of $b^n$ and $c^{n-1}$, and the coefficients can be determined b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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If $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $, then either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.
| Either $\frac{1}{2a}\leq \frac{1}{2b}$, in which case
$$\frac{1}{b}=\frac{1}{2b}+\frac{1}{2b}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2a}+\frac{1}{2a}=\frac{1}{a},$$
or $\frac{1}{2b}\leq \frac{1}{2a}$, in which case
$$\frac{1}{a}=\frac{1}{2a}+\frac{1}{2a}\geq\underbrace{\frac{1}{2a}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proof an inequality I'm trying to prove that
$$ \frac{3-2\sqrt{1-15 m^2}}{1+12 m^2}\geq 1+3 m^2$$
I have obtained in a CAS software the Taylor expansion in $m=0$
One posibility to prove the inequality is showing coeficients in Taylor expansion are non-negative, by I don't find how.
Really I want only to obtain ineq... | $$ \frac{3-2\sqrt{1-15m^2}}{1+12m^2} \geq 1+3m^2 \iff $$
$$ 3-2\sqrt{1-15m^2} \geq (1+3m^2)(1+12m^2) \iff $$
$$ 2\sqrt{1-15m^2} \leq 3-(1+3m^2)(1+12m^2) \iff $$
$$ \sqrt{1-15m^2} \leq \frac{3-(1+3m^2)(1+12m^2)}{2} \iff $$
$$ \sqrt{1-15m^2} \leq \frac{2-15m^2-36m^4}{2} $$
Note that on the interval you're concerned about... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplest method to find $5^{20}$ modulo $61$ What is the simplest method to go about finding the remainder of $5^{20}$ divided by $61$?
| Note that $5^3=125 = 3\pmod{61}$, so $5^{20}=5^{18}.25=(5^3)^6.25=3^6.25\pmod {61}$.Now $3^5=243\pmod {61}=-1\pmod {61}$ $\implies$ $3^6.25\pmod {61}= -3.25\pmod{61}=-75\pmod{61}=47\pmod {61}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Factoring $x^4z-2z^2-4x^6+x^2z$ We want to factor $8x^4y^4-2y^8-4x^6+x^2y^4 = -2y^8 + (8x^4+x^2)y^4 -4x^6$. We substitute $x^4$ with $z$:
Now we want to compute this $8x^4z-2z^2-4x^6+x^2z = -(x^2-2z)(4x^4-z)$ by hand.
Therefore we transform it into $-(2z^2-(8x^4+x^4)z+4x^6z^0)$ and use the quadratic formula on the (inn... | If I understand your question correctly, the reason is because when you factor a polynomial using its roots or zeroes, you have to keep the main coefficient as a factor also. More explicitly, if you have a polynomial
$$F(x) = a_n x^n + \cdots + a_1 x + a_0$$
with zeroes $c_1, \dots , c_n$ then
$$F(x) = a_n (x - c_1)\c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 2,
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How to prove: $S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$ If $$m_a, m_b, m_c$$ are the medians of a triangle and let $$m=\frac{m_a+ m_b+ m_c}{2}$$
then Area $S$ of triangle is given by
$$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$ This looks very similar to Heron's formula. How to prove this formula?
| $$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$
$$ =\frac{4}{3}\sqrt{ \frac{(m_a+m_b+m_c)}{2}\frac{(-m_a+m_b+m_c)}{2}\frac{(m_a-m_b+m_c)}{2}\frac{(m_a+m_b-m_c)}{2}}$$
$$ =\frac{1}{3}\sqrt{(m_a+m_b+m_c)(-m_a+m_b+m_c)(m_a-m_b+m_c)(m_a+m_b-m_c)}$$
$$ =\frac{1}{3}\sqrt{[(m_a+m_b)^2-m_c^2] [m_c^2-(m_b-m_a)^2] } $$
$$ =\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/168961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Some method to solve $\int \frac{1}{\left(1+x^2\right)^{2}} dx$ and some doubts. First approach.
$\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$
From this relationship, I get:
$2\int \frac{1}{\left(1+x^2\... | If we use this identity $\sin(2t)=\frac{2\tan(t)}{1+\tan^2(t)}$, the confusion should not arise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Verify trigonometry equation $\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$ How would I verify the following trigonometry identity?
$$\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$$
My work so far is
$$\frac{\sin A}{\cos A}-\frac{1}{\sin A}\frac{1}{\cos A}(1- \cos^2 A- \cos^2 A)$$
| I just want to comment on how you could take your work so far a little further.
$$\frac{\sin A}{\cos A}-\frac{1}{\sin A}\frac{1}{\cos A}(1- \cos^2 A- \cos^2 A)$$
What you need to do is to get a common denominator
$$\frac{\sin A}{\cos A}\cdot\frac{\sin A}{\sin A}-\frac{1}{\sin A\cos A}(1- \cos^2 A- \cos^2 A)$$
$$=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove $\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}$ How would I simplify this difficult trigonometric identity:
$$\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}.$$
I am not exactly sure what to do.
I simplified the right side to
$$\frac{\frac{\sin A}{\cos A}}{1-\frac{\c... | $$RHS = \frac{\tan A}{1-\tan^2 A} = \frac{\frac{\sin A}{\cos A}}{1-\frac{\sin^2 A}{\cos^2 A}}$$
$$=\frac{\frac{\sin A}{\cos A}}{\frac{\cos^2 A - \sin^2 A}{\cos^2 A}}\cdot \frac{\cos^2 A}{\cos^2 A}$$
$$=\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = LHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now ... | Let T = 1 + 2/7 + 4/7^2 + 7/7^3 + ... ----> (1)
Then
T/7 = 1/7 + 2/7^2 + 4/7^3 + ... ----> (2)
Now,
(2)-(1) => 6T/7 = 1 + 1/7 + 2/7^2 + 3/7^3 ----> (3)
Hence, (1/7)(6T/7) = 1/7 + 1/7^2 + 2/7^3 ----> (4)
So, (3)-(4)=> (1 - 1/7)(6T/7) = 1 + 0 + 1/7^2 + 1/7^3 + ...
=> (6/7)(6T/7) = 1 + (1/7^2 + 1/7^3 + ...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
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Value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$, $P(1)=10$, $P(2)=20$, $P(3)=30$
What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$
provided that $P(1)=10$, $P(2)=20$, $P(3)=30$?
I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to app... | Other way doing this:
We try to find reals $e,f,g$ such that $P(12)+P(-8)=eP(1)+fP(2)+gP(3)$. So, if we try to equal "$x^k$ evaluated", we gain a system of equations:
$$
\left\{\begin{array}{ccc}
1^ke+2^kf+3^kg&=&12^k+(-8)^k
\end{array}\right.,\quad k=0,\cdots,4
$$
In particular,
$$
\left\{\begin{array}{ccc}
e+f+g&=&1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Spivak 's Calculus (Chapter I, Problem 16c) In problem 16(c) of chapter 1 of Calculus, Spivak asks the reader to determine the conditions under which the expression $(x + y)^4$ equals $x^4 + y^4$. Clearly,
$$ (x + y)^4 = x^4 + y^4 \Leftrightarrow x = 0 \vee y = 0 \vee 4x^2 + 6xy + 4y^2 = 0 $$
From the preceding problem... | You have $(x+y)^4 - (x^4 + y^4) = 2 x y (2 y^2 + 3xy + 2 y^2)$.
To find the zeros of $2 y^2 + 3xy + 2 y^2$, use the quadratic formula to get $ y = \frac{x}{4} ( -3 \pm i \sqrt{7})$, hence the only zero (in $\mathbb{R}^2$) is $(0,0)$.
It follows that the zeros are $x=0$ or $y=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/176109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove trigonometry identity for $\cos A+\cos B+\cos C$ I humbly ask for help in the following problem.
If
\begin{equation}
A+B+C=180
\end{equation}
Then prove
\begin{equation}
\cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2)
\end{equation}
How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)... | Your observation that $C=180^\circ-(A+B)$ is a good one. Recall also the following trigonometric identities: $$\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y$$ $$\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$$ By LHS, I'll denote the expression on the left-hand side of the desired identity; by RHS, the expression on the right-han... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Why isn't this square root $+$ or $-$? I was tasked with proving the identity $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$
I used the quotient identity for tangent and the half angle identities for sine and cosine to get $ \pm \dfrac {\sqrt{\dfrac {1-\cos(x)}{2}}}{\sqrt{\dfrac {1-\cos(x)}{2}}}$
which I reduced to $\... | Avoid unnecessary squaring wherever possible.
If we use the following approach, no such confusion arises.
$$\tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}$$
Now multiply numerator & denominator by $2\cos\frac{x}{2}$
$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\frac{2\cos\frac{x}{2}\sin\frac{x}{2}}{2\cos^2\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find $x(1-t)$ consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find the value of $x(1-t)$
| Note that $(2+\sqrt{3})^6+(2-\sqrt{3})^6$ is an integer, indeed an even integer. For imagine expanding each term, using the Binomial Theorem. The terms involving odd powers of $\sqrt{3}$ cancel.
We have $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$. So $(2-\sqrt{3})^6=\frac{1}{x}$, and
$$(2+\sqrt{3})^6+(2-\sqrt{3})^6=x+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Gre Question Complex Number (plug and chug) This seems like it should be easy, but I can't seem to simplify it: If $z=e^{i\frac{2\pi}{5}}$, then what is $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$. The choices are $0, 4e^{i\frac{3\pi}{5}}, 5e^{i\frac{4\pi}{5}}, -4e^{i\frac{-2\pi}{5}}, -5e^{i\frac{3\pi}{5}},$ with the a... | Note that $z^5=1$, as $z$ is a fifth root of unity, so the expression simplifies to
$$
\begin{align}
1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4 &=5+5z+5z^2+5z^3+10z^4\\
&=5(1+z+z^2+z^3+z^4)+5z^4
\end{align}
$$
However, either by using the formula for the geometric series, or the fact that $1+X+X^2+X^3+X^4$ is the fifth cyclo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove that $\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{x^2+y^5+z^2}+\frac{z^5-z^2}{x^2+y^2+z^5}≥0 $. Given $x, y, z $ are 3 positive reals such that $xyz≥1$. Prove that
$$\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{x^2+y^5+z^2}+\frac{z^5-z^2}{x^2+y^2+z^5}≥0.$$
This question is so complicated. I failed many times to get ... | This is problem №3 from IMO 2005. Here you can find its solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg ... | As Salech pointed out in the comment above, we have $A=180^\circ-(B+C)$, so applying angle sum and difference formulas for cosine and sine, we have
$\begin{eqnarray*}
\cos\frac A2 & = & \cos\left(90^\circ-\frac{B+C}2\right)\\
& = & \cos 90^\circ\cos\left(\frac B2+\frac C2\right)+\sin 90^\circ\sin\left(\frac B2+\frac C2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
| Another way.
Let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a}{\sqrt{a+b}}=\sum_{cyc}\frac{\frac{y}{x}}{\sqrt{\frac{y}{x}+\frac{z}{y}}}=\sum_{cyc}\frac{\sqrt{y^3}}{\sqrt{x(y^2+xz)}}=\sum_{cyc}\frac{y^2}{\sqrt{xy(y^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
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Prove that $\left (\frac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$ Prove that $\left (\dfrac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$ if $a$, $b$ and $c$ are distinct natural numbers. Is it possible using induction?
| Rewrite it as:
$$
\left( a \frac{a}{a+b+c} + b \frac{b}{a+b+c} + c \frac{c}{a+b+c} \right) > a^\frac{a}{a+b+c} \cdot b^\frac{b}{a+b+c} \cdot c^\frac{c}{a+b+c}
$$
This is Jensen's inequality:
$$
\log\left(\mathsf{E}\left(X\right)\right) > \mathsf{E}\left(\log\left(X\right)\right) \quad \text{or}\quad \mathsf{E}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 0
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how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\... | put $x = \sqrt{2+x}
\implies x^2 = 2+x \implies x^2 -x-2=0$
Now just solve the quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \... | After simplifying you have $\frac{2x^2-43x-45}{x(2x+3)}=\frac{(2x-45)(x+1)}{x(2x+3)}<0$. So we consider the function on the intervals $x<-\frac{3}{2}$, $x \in (-3/2,-1)$, $x \in (-1,0)$, $x \in (0,\frac{45}{2})$, $x>\frac{45}{2}$. At $x=-2$ the function is positive and the sign alternates in each successive region.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Prove $\cos((3\pi/4)+x)-\cos((3\pi/4)-x)=(-\sqrt{2})\sin(x)$ Please help me to solve the below trigonometric function as i am trying it from the last hour.
$$\cos((3\pi/4)+x)-\cos((3\pi/4)-x)=(-\sqrt{2})\sin(x)$$
| Apply, $\cos(A+B)-\cos(A-B)=-2\sin A \sin B$
So, $\cos(3\frac{\pi}{4}+x)-\cos(3\frac{\pi}{4}-x)=-2\sin x\sin(3\frac{\pi}{4}) $
$=-2\sin x\sin(\pi-\frac{\pi}{4})=-2\sin x\sin(\frac{\pi}{4})$
as $\sin(\pi-y)=\sin y$
So, $\cos(3\frac{\pi}{4}+x)-\cos(3\frac{\pi}{4}-x)=-\sqrt2\sin x$
or apply $\cos 2C- \cos 2D=-2sin(C+D)si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $A>0$ and $B>0$ and $A^{-1}If $A>0$ and $B>0$ and $A^{-1}<B$, then can we claim that $x^T(AB)x\geq x^Tx$?
($A$ and $B$ are real symmetric matrices).
I know these facts (from Matrix Mathematics book) (It seemed to me these might help but I haven't been able to use them in my advantage!),
Let $A,B\in\mathbb{F}^{n\time... | Since $B=A^{-1}+C$ with $C$ symmetric and positive, one asks whether $x^TACx\geqslant0$ for every vector $x$ and every symmetric positive matrices $A$ and $C$.
Consider $A=\begin{pmatrix}5 & 2 \\ 2 & 1\end{pmatrix}$ and $C=\begin{pmatrix}1 & -2 \\ -2 & 10\end{pmatrix}$. Then $A$ and $C$ are symmetric positive, $A^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Radius of the spherical image of a circle This is question 5 on page 20 of the book Complex Analysis by Lars Ahlfors. I have no idea how to answer that problem:
Find the radius of the spherical image of the circle in the plane whose center is $a$ and radius is $R$.
Here spherical image means: the image of a subset of... | I think I managed to write it properly.
Let's denote the stereographic projection from $\Bbb C$ to the sphere by $\varphi$. Let's denote $C(a,R)$ the circle with center $a$ and radius $R$.
Assume first that $a\in[0,\infty[$. Then since $\varphi$ maps circles into circles, the diameter of the circle image is
$$D = \sup... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/190270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\e... | Since $ab=1$, we have $b = \frac{1}{a}$, and thus your inequality is equivalent to $\displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}\leq\frac{1}{2}$.
You can simply define $f(a) = \displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}$ and maximize $f$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Solution to a second order recurrence relation with non constant coefficient I have the following equation:
$(aq^n+b+s)C_n(s)=aq^nC_{n+1}(s)+bC_{n-1}(s)$ , for $n\geq1$.
$C_0(s)=1$ , and for all $s\geq 0$ we have $0\leq C_n(s)\leq1$.
$a>0$, $b>0$, $0\leq q\leq1$.
In fact, $C_n(s)$ is a Laplace Transform of a non-neg... | As
$$
G_n(s) = \frac{1}{a q^{n-1}}\left((a q^{n-1}+b+s)G_{n-1}(s)-b G_{n-2}(s)\right)
$$
with
$$
G_0(s) = 1, \ \ \ G_1(s) = \frac{c_1}{s+b_1}
$$
the recurrence formula which follows, in MATHEMATICA, generates all the instances for $G_n(s)$
G[s, 0] = 1;
G[s, 1] = Subscript[c, 1] /(s + b_0);
G[s_, n_] := ((a q^(n - 1) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Automorphic numbers
Problem. We say that the $n$-digit number $x$ is automorphic iff $x^2\equiv x \mod(10^n)$. Prove that if $x$ is $n$-digit automorphic number then $(3x^2-2x^3)\mod(10^{2n})$ is $2n$-digit automorphic number. Hint: use Chinese reminder theorem to find the necessary and sufficient condition for number... | (3)If $x≡0\pmod{2^n}=a2^n$ for some integer $a$
$3x^2-2x^3=3(a2^n)^2-2(a2^n)^3$ is clearly divisible by $2^{2n}$
Here $x≡1\pmod{5^n}=(b5^n+1)$ for some integer $b$
$3x^2-2x^3=3(b5^n+1)^2-2(b5^n+1)^3=(b5^n+1)^2(3-2(b5^n+1))$
$≡(1+2b5^n+b^25^{2n})(1-2b5^n)≡(1+2b5^n)(1-2b5^n)=1-4b^25^{2n}≡1\pmod{5^{2n}}$
(4)If $x≡0\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Inequality with unusual constraint $a,b,c\in (0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$ Suppose as in the title that $a,b,c$ are three real positive numbers in $(0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$. Then I was asked to prove that $$a+b+c\leq \frac 32.$$ I am not very good at inequalities, so can anybo... | As $a,b,c \in (0,1)$, let $a=\cos A, b=\cos B,c=\cos C$, so that $0<A,B,C<\frac{\pi}{2}$
So, $\cos^2A+\cos A(2\cos B\cos C)+\cos^2B+\cos^2C-1=0$
$$\implies \cos A=\frac{-2\cos B\cos C±\sqrt{(2\cos B\cos C)^2-4\cdot 1\cdot (\cos^2B+\cos^2C-1)}}{2}$$
$$\implies \cos A=-\cos B\cos C±\sin B\sin C$$ as
$(2\cos B\cos C)^2-4(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Showing a sequence is convergent using the $\epsilon$-$N$ definition I need to prove that the following sequence converges:
$\lim_{n\rightarrow \infty} \frac{2n^2+3n+1}{n^2+n+1}=2$
So for the proof/solution I have the following:
Let $\epsilon >0$. Then let $N=\frac{1}{\epsilon}$. Then for all $n\geq N$,
$|\frac{2n^2+3... | There is a small error with symbols, it should be like this:
Choose $\epsilon\gt 0$
Let $N=\lceil\frac{1}{\epsilon}\rceil$. Then for all $n\gt N$,
$|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/195030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int\frac{dx}{\cos^6 x}$ Please for the substition method solving $\int\frac{dx}{\cos^6 x}$
| $\int\frac{dx}{\cos^6 x}$=$\int\frac{1}{\cos^4 x}\cdot\frac{dx}{\cos^2 x}$=$\int\frac{1}{(\frac{1}{\sqrt{1+\tan^2 x}})^4}\cdot\frac{dx}{\cos^2 x}$=$\int(1+\tan^2 x)\cdot\frac{dx}{\cos^2x}$=$|\tan x=t\Rightarrow\frac{dx}{\cos^2 x}=dt|$=$\int(1+t^2)^2dt$=$\int(1+2t^2+t^4)dt$=$t+\frac{2}{3}t^3+\frac{1}{5}t^5$=$\tan x+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality: $7a+5b+12ab\le9$ If we assume that $a,b$ are real numbers such that $9a^2+8ab+7b^2\le 6$, how to prove that :
$$7a+5b+12ab\le9$$
| We have
$$2(a-b)^2+7\left(a-\frac{1}{2}\right)^2 + 5\left(b-\frac{1}{2}\right)^2 \geq 0$$
which is equivalent to
$$7a+5b+12ab\leq 9a^2+7b^2+8ab+3 \leq 6+3=9$$
The motivation here is to search for equality case by solving the system of equation in real values $a,b$
\begin{equation*}
\begin{cases}
7a+5b+12ab=9 \\
9a^2+7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/196128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find the inverse a matrix with trigonometic entries What is the inverse of
\[
\begin{pmatrix}
1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{pmatrix}
\]
Please help me to solve the above problem.
| Implement the formula $\def\adj{\operatorname{adj}}A^{-1}=\frac{1}{\det A}\cdot \adj A$
Find $\det A$
$\det A=\begin{vmatrix} 1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{vmatrix}=-1$
Find $\adj A$
$A_{11}=(-1)^{1+1}\left\lvert \begin{array}{cc} \cos x & \sin x \\ \sin x & -\cos x \end{array} \right\rvert=-1$
$A_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $8\mid n^2-1$ if $n$ is an odd positive integer. Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.
Please help me to prove whether this statement is true or false.
| We know that any odd positive integer is of the form $4q+1$ or, $4q+3$ for some integer $q$.
So, we have the following cases:
Case 1 when $n=4q+1$: In this case, we have
$n^2 -1=(4q+1)^2 -1=16q^2 +8q+1-1=16q^2 +8q=8q(2q+1)$
$∴ n^2 -1$ is divisible by 8 [$∵ 8q(2q+1)$ is divisible by 8]
Case 2 when $n=4q+3$: In this case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 5
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.