Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find a $2 \times 2$ matrix $A$ that satisfies $A^3 = \begin{pmatrix} -1 & -1 \\ 1 & -1 \\ \end{pmatrix}$
Find a $2 \times 2$ matrix $A$ such that
$$A^3 =
\begin{pmatrix}
-1 & -1 \\
1 & -1 \\
\end{pmatrix}$$
I'm trying to think about this geometrically. Could this have something to do with a rotation... | There are several ways of tackling this problem, a nice one of which you have already thought of. Since $\ \begin{pmatrix}-1&-1\\1&-1\end{pmatrix}\ $ is indeed a dilated rotation matrix, namely $\ \sqrt{2}\begin{pmatrix}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{pmatrix}\ $, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Understanding the constraints to find a $2\times 2$ non-zero matrix $A$ such that $A^2=0$ Using the rules for matrix multiplication, I have found four algebraic equations as "constraints" for getting every element in the resulting matrix to be zero. Assuming that the elements in the resulting matrix are $a, b, c$ and $... | With
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \tag 1$
we have
$A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & (a + d)b \\ (a +d)c & d^2 + bc \end{bmatrix} = 0, \tag 2$
whence
$a^2 + bc = 0, \tag 3$
$d^2 + bc = 0, \tag 4$
$(a + d)b = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Use ratio test to determine convergence or divergence. Find general term first. I have this series:
$$\frac{2}{3} + \frac{2*5}{3*5} + \frac{2*5*8}{3*5*7} ...$$
I'm confused how to find the general term here to then apply the ratio test.
So it starts with a $$\frac{2}{3}$$ and the next term is $\frac{2*5}{3*5}$
So each ... | Find formulae for the product terms in the numerator and the denominator. The questions seems to suggest that they are arithmetic progressions. In particular, the $n$th term (not the $n$th partial product) in the product $2 \cdot 5 \cdot 8 \cdot \ldots$ is supposedly $2 + 3(n - 1)$. Similarly, the $n$th term in the pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generating function of $\frac{h(x)}{(1-x)^2}$
If $h(x)$ is the generating function for $a_r$, what is the generating function of $$\frac{h(x)}{(1-x)^2}$$
Let $h(x)$ be written as
$$h(x) = \sum_{r} a_r x^r $$
Consider more simply
$$\frac{h(x)}{1-x} = \frac{1}{1-x} h(x) =\sum_{r} x^r \sum_{r} a_r x^r$$
I tried to expan... | Another way.
Since
$(1-x)^2 = 1-2x+x^2$,
if
$\dfrac{h(x)}{(1-x)^2}
=\sum_{n=0}^{\infty} a_nx^n$
then
$\begin{array}\\
h(x)
&=(1-x)^2\sum_{n=0}^{\infty} a_nx^n\\
&=(1-2x+x^2)\sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} a_nx^n-2x\sum_{n=0}^{\infty} a_nx^n+x^2\sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} a_nx^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\fr... | Hint:
Let $\arctan x=y,-\pi/2<y<\pi/2,\cos y>0$
$\arccos\dfrac{\tan y}{\sec y}=\arccos(\sin y)=\dfrac\pi2-\arcsin(\sin y)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\sum^{20}_{r=0} \binom{20}{r}\cos r \theta$
Evaluate $$\sum^{20}_{r=0} \binom{20}{r}\cos r \theta$$
To do this my initial thought was to set up an alternative series, $S$, $$S=\sum^{20}_{r=0} \binom{20}{r}\sin r \theta$$ If yoy let the original series be $C$, then $$C+iS = \sum^{20}_{r=0} \binom{20}{r}\cos ... | You did very well. Now
$$
\eqalign{
& C + i\,S = \sum\limits_r {\left( \matrix{
20 \cr
r \cr} \right)e^{\,ir\,\theta } } = \sum\limits_r {\left( \matrix{
20 \cr
r \cr} \right)\left( {e^{\,i\,\theta } } \right)^{\,r} } = \cr
& = \left( {1 + e^{\,i\,\theta } } \right)^{\,20} \cr
& \left| {1 + e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Occurance of Prime Factor 3 Let $P^{3}(a, b)$ return the sum of the frequency of Prime Factor 3 of all Integers in $[a, b]$, e.g: $P^{3}(5, 10) = 3, P^{3}(12, 20) = 4, ...$
Then
$\forall x>10 \in \mathbb N: P^{3}(2^{x}, 2^{x+1}) > P^{3}(1, 2^{x}) + 1$
?
| Partial
Notice that:
$$P^3(a,b)=P^3(b,1)-P^3(a,1)$$
So your question becomes:
$$P^3(2^{k+1},1)>2P^3(2^k,1)+1$$
There is a standard formula for $P^3(N,1)$:
$$ P^3(N,1)=\sum_{i=1}^{\lfloor \log_3(N) \rfloor} \lfloor \frac{N}{3^i} \rfloor $$
So:
$$\sum_{i=1}^{\lfloor \log_3(2^{k+1}) \rfloor} \lfloor \frac{2^{k+1}}{3^i} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $8 | x^2 + y^2$ iff x,y are both even is false Per the question above I am trying to prove this statement false. As such only one of two conditions have to be met both x and y being odd, or them both being odd. I've seen a lot of examples on this site regarding x^2 - y^2 for a similar example but none where ... | You want to disprove the statement
$$8\mid x^2+y^2\iff 2\mid x, y $$
As @J.W. Tanner pointed out in the comments, a simple counterexample - such as $2^2+4^2$ or its generalization $(2n)^2+(2n+2)^2=8n^2+8n+4\not\equiv 0\pmod 8$ - is enough. This proves that $$2\mid x, y\color{red}{\not\Rightarrow}8\mid x^2+y^2$$
Howev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
In the task which I do I need record $\sin (\sin x-x)$ in a way to have $ax^3$. So: $$\sin (\sin x -x)=\sin x -x +r(\sin x -x)=x-\frac{x^5}{6}+\frac{x^5}{120}+r(x)-x +r(\sin x -x)$$ I know that $r(x)=o(x^5)$ and it i... | You can note that $$\sin x - x=-\frac{x^3}{6}+o(x^3)\tag{1}$$ and replacing $x$ with $\sin x - x$ we get $$\sin(\sin x - x) - (\sin x - x) =-\frac{(\sin x - x)^3}{6}+o((\sin x - x) ^3)$$ Using $(1)$ we can see that both the terms in right hand side of above equation are $o(x^5) $ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$, at $a = 0$ Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$ , at $a = 0$
So I start of with:
$f(0) = \frac{1}{(1-0)}=1$
$f'(0) = \frac{1}{(1-0)^2 }= 1$
$f''(0) = \frac{2}{(1-0)^3 }= 2$
$f'''(0) = \frac{6}{(1-0)^4 }= 6$
$f^{(4)}(0) = \frac{24}{(1-0)^... | Its easy to do, take the inverse of (1-x) and then expand
$(1-x)^{-1}$ = 1 + $x + x^{2} + x^{3}$ + .....
So coefficient of each term is 1.
From your method, you are skipping the factorial term in the denominator. Calculator is right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many cricket teams can the manager choose if one particular batsman refuses to play when one particular bowler does?
A certain country has a cricket squad of 16 people, consisting of 7 batsmen, 5 bowlers, 2 all- rounders and 2 wicket-keepers. The manager chooses a team of 11 players consisting of 5 batsmen, 4 bowl... | There are $5$ from $7 = 21$ ways to choose batsmen. For everyone of those there are $4$ from $5 = 5$ bowlers and for every one of those there are $2$ choices of wicket keeper and $2$ choices of all-rounders. Subtracting combinations that contain a certain bowler and batter involves counting the number of ways a batter ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now,... | We get from $$3x^2-4x-2=0$$ to $$3x^2-4x=2$$ by adding two on both sides. Then dividing by $3$ we obtain
$$x^2-\frac{4}{3}x=\frac{2}{3}$$ then we can write $$x^2-2\cdot \frac{2}{3}x+\frac{4}{9}=\frac{2}{3}+\frac{4}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Integral $\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx$
Prove that$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$
I've found this integral in my notebook and perhaps I encountered it before since it looks quite familiar.
Anyway I thought it's quite a trivial integral so I'm gonna s... | Solution 3. Consider the following integral:
$$ I(a)=\int_0^\frac{\pi}{2}\frac{\arctan(a\tan x)}{\sin x}dx\Rightarrow I'(a)=\int_0^\frac{\pi}{2}\frac{\sec x}{1+a^2\tan^2 x}dx$$
$$ =\int_0^\frac{\pi}{2}\frac{\cos x}{\cos^2 x+a^2\sin^2 x}dx\overset{\sin x=y}=\int_0^1 \frac{dy}{1+(a^2-1)y^2}=\frac{\arctan\sqrt{a^2-1}}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when... | Let
$f = \sqrt{x^2 + 25} + \sqrt{y^2 + 16}$
$g = x+y -12 =0 $
By Lagrange's Undetermined multipliers method,
$F = f + g\lambda = \sqrt{x^2 + 25} + \sqrt{y^2 + 16} + \lambda(x+y -12)$
Differentiating partially w.r.t x and y and using $F_x = 0$ and $F_y = 0 $ at extremum,
$F_x = \frac{x}{\sqrt{x^2 + 25}} +\lambda = 0$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Efficiently compute $f'(a)f'(b)f'(c)$ for the roots of a cubic polynomial Let $a,b, c$ be the zeroes of $f(x) = x^3+3x^2-7x+1$. Find $f'(a)f'(b)f'(c)$.
My idea involved substituting in $a,b,c$ into $f'(x)$ then using Vieta's but that would take far too long of a time, especially since I'm training for competitions. Wha... | If $f(x)=(x-r)g(x)$, then $f'(x)=g(x)+(x-r)g'(x)$ and hence $f'(r)=g(r)$.
As $f(x)=(x-a)(x-b)(x-c)$, $f'(a)f'(b)f'(c)=(a-b)(a-c)(b-a)(b-c)(c-a)(c-b)$.
Let $g(x)=f(x-1)$ and $\alpha$, $\beta$ and $\gamma$ be the roots of $g(x)=0$. Then $\alpha=a+1$, $\beta=b+1$ and $\gamma=c+1$. Therefore, $g'(\alpha)g'(\beta)g'(\gamma)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Problem with $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}$ How to simplify $$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$
Rationalise the denominator
$$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$
This is still not simplify.
| We have $$6+4\sqrt{2}=4+2+4\sqrt{2}$$ so $$2\sqrt{6+4\sqrt{2}}=4+2\sqrt{2}$$ and we get
$$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Arc length of $f(t)=(cos^4(t),sin^4(t))$ from $t=0$, to $t=2\pi$. I'm trying to compute the arc length $L$ of $f(t)=(cos^4(t),sin^4(t))$ from $t=0$, to $t=2\pi$. Using the regular formula we have that:
$$L=\int_{0}^{2\pi}\sqrt{16\cos^2(t)\sin^2(t)(\cos^4(t)+\sin^4(t))}dt$$
Then after some (long) calculi it reduced to ... | It is better to write $$\sin 2t \sqrt{3 + \cos 4t} = \sin 2t \sqrt{2 + 2 \cos^2 2t},$$ hence with the substitution $$u = \cos 2t, \quad du = -2 \sin 2t \, dt,$$ we obtain $$\int_{t=0}^{\pi/2} \!\!\sin 2t \sqrt{2 + 2 \cos^2 2t} \, dt = -\frac{1}{2} \int_{u=1}^{-1} \!\!\!\!\sqrt{2 + 2u^2} \, du = \frac{1}{\sqrt{2}} \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing volume inside a ball and outside a cylinder I try to calculate the volume inside the ball $\ x^2 + y^2 + z ^2 = 4 $ the outside the cylinder $\ x^2+y^2=2x $ using double integral.
Since the shape is symmetric I chose $\ z = 0 $ as bottom limit and the ball is the upper limit. I try to convert to polar form a... | You've correctly set up the integral representing the volume of the inside cylinder. The scalar $4$ on the left represents
*
*the reflectional symmetry about the $xy$-plane; and
*the reflectional symmetry about the $yz$-plane.
\begin{align}
& 4 \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to
$$100!\times\left(1+\sum_{n=1}^{10... | Want
$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)
$
Going up to $m$,
$\begin{array}\\
\sum_{n=1}^{m} \dfrac{(-1)^n (n^2+n+1)}{n!}
&=\sum_{n=1}^{m} \dfrac{(-1)^n (n^2-n+2n+1)}{n!}\\
&=\sum_{n=1}^{m} \dfrac{(-1)^n (n^2-n+2n+1)}{n!}\\
&=\sum_{n=1}^{m} (-1)^n(\dfrac{ n^2-n}{n!}+\dfrac{2n}{n!}+\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^4}$
Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$
The answer stated is $\displaystyle {1 \over 6}$.
What I've tried:
$$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x... | By using trigonometry identity and Taylor series,\begin{align} &\lim_{x\to0} \frac{\cos (\sin x)-\cos(x)}{x^4}\\
&=-2\lim_{x\to 0}\frac{\sin \left( \frac{\sin x - x}2\right) \sin\left(\frac{\sin x + x}2 \right)}{x^4}\\
&= -2\lim_{x \to 0 }\frac{\sin \left( \frac{-x^3}{2(6)}\right)\sin \left( \frac{x+x}{2}\right)}{x^4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $
Find antiderivative $$\int (2x^3+x)(\arctan x)^2dx $$
My try:
$$\int (2x^3+x)(\arctan x)^2dx =(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int \frac{2\arctan x}{1+x^2}(\frac{1}{2}x^4+\frac{1}{2}x^2)dx=(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int (\arctan x) (x... | Let $t=x^2$. Your integral is $\int \frac{1}{6}\frac{tdt}{t+1}=\int (1-\frac{1}{t+1})dt=\frac{t}{6}-\int\frac{1}{t+1}dt$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x =... |
In the figure, let $AB=AC=x$. Note that $AD=CD=2$. As $\triangle ABC\sim\triangle CDB$, $\dfrac x2=\dfrac 2{x-2}$
So, $x^2-2x-4=0$ and hence $x=1+\sqrt{5}$.
$\sin18^\circ=\dfrac 1x=\dfrac{\sqrt{5}-1}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Why is this integration method not valid? Let $$I=\int \frac{\sin x}{\cos x + \sin x}\ dx \tag{1}$$
Now let $$u=\frac{\pi}{2} - x \tag{2}$$ so $$I=\int \frac{\sin (\frac{\pi}{2} - u)}{\cos (\frac{\pi}{2} - u)+\sin (\frac{\pi}{2} - u)}\ du \tag{3}$$
$$=\int\frac{-\cos u}{\sin u + \cos u} \ du \tag{4}$$
$$= \int\frac{-\c... | $$\int_{a}^{b}\frac{\sin \ x}{\cos \ x + \sin \ x}dx\neq \int_{a}^{b}\frac{-\cos \ x}{\cos \ x + \sin \ x}dx$$
but
$$\int_{a}^{b}\frac{\sin \ x}{\cos \ x + \sin \ x}dx= \int_{\pi/2-a}^{\pi/2-b}\frac{-\cos \ x}{\cos \ x + \sin \ x}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3232250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 6,
"answer_id": 2
} |
integration of a gaussian with $x^2$ I need to integrate
$$\int_{-\infty}^{\infty} x^2 e^{-ax^2} \qquad \text{where } a\in R$$
The book does the following:
I don't understand what's happening. I tried solving the integral using integration by parts and this is what I got
$$
\begin{align}
\int_{-\infty}^{\infty} x^2 e^... | note that the derivative of $e^{-ax^2}$ is $-2ax e^{-ax^2}$, so
$$ x^2 e^{-ax^2} = \frac{-1}{2a}x \,(-2ax e^{-ax^2}) = \frac{-1}{2a}x\, \frac{d}{dx} e^{-ax^2}$$
Insert this into the integral and integrate by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Next Term Of Strange Sequence I tutored a 10th grader and I was asked this puzzle and I had spent nearly an hour with it and got “no where”. Any one can crack it? Please let me know. Thank you.
Question: Find the $14$ th term of the sequence: $$ \frac{1}{2}, \frac{3}{7}, \frac{1}{3}, \frac{5}{19}, \frac{3}{14}, .... $... | $\displaystyle\frac24\ ,\ \frac37\ ,\ \frac4{12}\ ,\ \frac5{19}\ ,\ \frac6{28}\ ,...$
differences between denominators are $3,5,7,9,...$
(and numerators are consecutive integers $2,3,4,5,6,...$ )
$\displaystyle\frac24,\frac37,\frac4{12},\frac5{19},\frac6{28}, \\
\displaystyle\frac7{39},\frac8{52},\frac9{67},\frac{10}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me pleas... | Upon completing the square the equation $$(x^2+6x+7)^2+6(x^2+6x+7)+7=x$$
$$(x^2+6x+7)^2+6(x^2+6x+7)+9=x+2$$
$$[(x^2+6x+7)+3]^2=x+2$$
Which simplifies to $$[(x+3)^2+1]^2=x+2$$ which has no real solutions.
Upon comparison of derivatives for $x>-2$, we see that function on the LHS is always above the one on the RHS so t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Rotation Matrix and Triple Angle Formulas? Define $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ as the rotation matrix by angle $\theta$, where
$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
Observe that
$$
(R_\theta)^2 = \begin{pmatrix} \cos^2\theta-\sin^2\the... | You may use these identities
$\cos x \cos y - \sin x \sin y = \cos(x+y)$
$\sin x \cos y + \cos x \sin y = \sin(x+y)$
and use $\theta $ and $2\theta$ in place of $x$ and $y$. For $R_{n\theta}$, try proving it by induction, assuming $(R_{\theta})^n = R_{n\theta}$ to be true and find $(R_{\theta})^{n+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
Evaluate the following integral:
$ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$.
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\f... | Another way:
Integrate by parts
$$\int\dfrac{x^4}{(2-x^2)^{3/2}}dx=\int x^3\cdot\dfrac x{(2-x^2)^{3/2}}dx$$
$$=x^3\int\dfrac x{(2-x^2)^{3/2}}dx-\int\left[\dfrac{d(x^3)}{dx}\int\dfrac x{(2-x^2)^{3/2}}dx\right]dx$$
Finally, $$\dfrac{x^2}{\sqrt{2-x^2}}=\dfrac{x^2-2+2}{\sqrt{2-x^2}}=\dfrac2{\sqrt{2-x^2}}-\sqrt{2-x^2}$$
Use... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Identity of tan(x) I came across the following formulas for analytical expressions of fundamental modes of asymmetric dielectric waveguide. $$ \tan(x) = x\frac{\pi^2-x^2}{\pi^2-4x^2} $$
This approximation is not present in Abramowitz's handbook. It has the same poles and zeros as $\tan(x)$.
Can someone guide me on its ... | There are several ways to think about approximating $\tan x$ as a rational function.
Taking the log-derivative of $\sin x=x\prod_{k\ge 1}\left(1-\frac{x^2}{k^2\pi^2}\right)$ gives $$\cot x=\frac{1}{x}-2\sum_{k\ge 1}\frac{x}{k^2\pi^2-x^2}.$$Keeping only the $k=1$ term,$$\tan x\approx\frac{x(\pi^2-x^2)}{\pi^2-3x^2}.$$Thi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\,
Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$
Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\i... | If we take the contour $\sf C$ to be the upper half-plane (with $\sf{\Im z>0}$), then we can write the integrand as $$\sf{\frac{z^2}{(z^2+1)^2}=\frac{\left(\frac{z}{z+i}\right)^2}{(z-i)^2}}$$ since the singularity in $\sf C$ is at $\sf{z=i}$ (of order two). Thus, using Cauchy's Integral Formula, we obtain $$\sf{\int_{-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$.
$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$
We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfra... | $\large \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b}$
Taking $a+b = c$ for some c; $0<c\leq 1$, and expressing $b=c-a$; the above expression becomes (after modification)
$-2 + \{2(c+1) +(c^2+c)\}(\dfrac{1}{-a^2+ac+(c+1)} ) +1/c$
here only variable is a ; c is assumed constant.
Differentiating it and equati... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$. I need to show that $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$.
This happens if $(x+1)\ln\Big(1 + \frac 2 {2x+1}\Big) > 1$ so let's study the function $f(x) = (x+1)\ln\Big(1 + \frac 2 {2x+1}\Big)$
Edit
We have abandoned the previous metho... | Consider the function $f(y):=(1+\frac{2}{y})^{y+1}$. Clearly, $\lim_{y\rightarrow\infty}f(y)=e^2$. Moreover, $f$ is monotone decreasing since
\begin{align*}
f'(y)<0 &\iff \left({1+\frac{2}{y}}\right)\log\left({1+\frac{2}{y}}\right) < \left({\frac{2}{y}}\right)+\frac{1}{2}\left({\frac{2}{y}}\right)^2 \\
& \iff (1+t)\log... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrating factor for $(x^2-y^2-y)dx-(x^2-y^2-x)dy=0$ I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.
Solve the differential ... | The given differential equation is not exact and I think you can't find the integrating factor by known way and an easier way, rather you can solve it as follows
${}$
$(x^2-y^2-y)dx - (x^2-y^2-x)dy=0$
$\implies (x^2-y^2)(dx-dy)+xdy-ydx=0$
$\implies (1-\frac{y^2}{x^2})(dx-dy)+\frac{xdy-ydx}{x^2}=0$
$\implies (1-\frac{y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17 $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equi... | $$x^2+5x+6\equiv187\equiv0 \pmod {187=11\times17}$$
$$(x+2)(x+3)\equiv 0 \pmod {11 , 17}$$
$$x\equiv-2 \text { or } -3 \pmod {11, 17}$$
Now use the Chinese Remainder Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find the limit of a multiplying term function when n tends to infinity. How do I evaluate the limit,
$$\lim_{n \to \infty} \ \ \ \left(1-\frac1{2^2}\right)\left(1-\frac1{3^2}\right)\left(1-\frac1{4^2}\right)...\left(1-\frac1{n^2}\right)$$
I tried to break the $n^{\text{th}}$ term into $$\frac{(n+1)(n-1)}{n.n}$$ and th... | The following will not be of help for such questions "in general", but it is too long for a comment.
Following Euler, we have
$$
\begin{align}
\frac{\sin x}{x} &=
\left( 1 - \frac{x}{\pi} \right) \left( 1 + \frac{x}{\pi} \right)
\left( 1 - \frac{x}{2 \pi} \right) \left( 1 + \frac{x}{2 \pi} \right)
\left( 1 - \frac{x}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Calculate the maximum value of $\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$ where $3a + 4b + 5c = 12$
$a$, $b$ and $c$ are positives such that $3a + 4b + 5c = 12$. Calculate the maximum value of $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$$
I want to know if ... | Also, we can use C-S and AM-GM:
$$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}\leq$$
$$\leq\frac{ab}{(1+2)^2}\left(\frac{1^2}{ab}+\frac{2^2}{a+b}\right)+\frac{2ca}{9}\left(\frac{1}{ca}+\frac{4}{c+a}\right)+\frac{3bc}{9}\left(\frac{1}{bc}+\frac{4}{b+c}\right)=$$
$$=\frac{2}{3}+\frac{4}{9}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If the HCF of the polynomials $x^3+px+q $ and $x^3+rx^2+lx+x$ is $x^2+ax+b$, then their LCM is? (provided that $r≠0$) I tried multiplying the two polynomials together and then dividing them by the HCF (as product=HCF*LCM), but reached nowhere.
Then, I used the factor theorem but also got stuck.
Can somebody help?
| If $b \neq 0$,
$$x^2+ax+b \mid x(x^2+rx+l+1) $$
This would imply that $x^2+ax+b = x^2+rx+l+1 $.
And the LCM would be,
$$\frac{x(x^2+rx+l+1)(x^3+px+q)}{x^2+ax+b}=x^4+px^2+q$$
If $b=0$,
$$x+a|x^2+rx+l+1$$
if you divide you'll find the quotient to be $x+(r-a)$ and remainder to be $l+1-a(r-a)=0$
LCM is,
$$(x^3+px+q)(x+r-a)... | {
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"timestamp": "2023-03-29T00:00:00",
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Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2... | By the power mean inequality, with $x\geq 0, n \geq k$, we have
$$ \left( \frac{ x^n + 1 } {2} \right) ^\frac{1}{n} \geq \left( \frac{x^k + 1 } { 2} \right) ^ \frac{1}{k}. $$
Hence, all that we need to show is:
$$ \frac{ x^{m+1} + 1 } { x^m + 1} \geq \left( \frac{ x^{2m+1 } + 1 } {x^0 + 1 } \right) ^ \frac{1}{2m+1} \q... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
This has become the norm now... This problem is adapted from a recent competition.
We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \g... | If we write $x+y+z=1$ (we can do that) then we have to prove $$1+9xyz\geq 4(xy+yz+zx)$$ Clearly one of $x,y,z$ is smaller than $4/9$, we can assume $y<4/9$.
If we put $z=1-x-y$ we get $$0\geq (9y-4)x^2+x(9y^2-13y+4)-(2y-1)^2=:f(x)$$
Since the discirminant for $f(x)$ is: $$\Delta= (9y-4)y(3y-1)^2$$
is also nonegative f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $\iiint_\omega (x+y+z)^2\,dxdydz$. Problem: Calculate $$\displaystyle\iiint_\Omega (x+y+z)^2\,dxdydz,$$ where $\Omega$ is the domain bounded by the sphere $x^2+y^2+z^2\le 3a^2$ and the paraboloid $x^2+y^2\le 2az$, where $z\ge 0$.
Progress: So, by sketching the domain, one can see that the domain includes a pa... | Note that $(x+y+z)^2 = x^2+y^2+z^2 +2(xy+yz+zx)$ and, due to the symmetry of the integration region, the integral simplifies to
$$I=\iiint_\Omega (x+y+z)^2\,dxdydz
= \iiint_\Omega (x^2+y^2+z^2)\,dxdydz
$$
Then, in cylindrical coordinates, the integration region is the circular area of radius $r=\sqrt 2 a$ and the integ... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding solutions to a system of equations using elementary symmetric polynomials Find the value of a, b, and c given:
$a^2 + b^2 + c^2 = 129$
$ab + ac + bc = -4$
$(a^2)(b^2) + (a^2)(c^2) + (b^2)(c^2)$ = 984
I attempted this problem using elementary symmetrical polynomials and found that the second equation can be writ... | Let $P(t) = (t-a)(t-b)(t-c) = t^3-e_1t^2+e_2t-e_3$.
Trivially, $e_1 = a+b+c$, $e_2 = ab+bc+ca = -4$, and $e_3 = abc$.
As noted in the comments, $129 = a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) = e_1^2-2e_2 = e_1^2+8$, so $e_1^2 = 121$, and thus, $e_1 = \pm 11$.
Next, note that $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
Find limit of $ \underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
How can I do that? It is interesting due to mathematica says that
$$\underset{\{x,y\}\to \{0,0\}}... | Using the Maclaurin formula
$$
\sqrt {A + 1} - 1 - \frac {A}2 \sim -\frac 18 A^2 [A \to 0],
$$
we get
$$
\frac {\sqrt { xy+ 1} - 1 - \frac {xy}2}{ y \sqrt {x^2 + y^2}} \sim -\frac 18\cdot \frac {x^2 y^2} {y \sqrt {x^2 + y^2}}.
$$
Now let
$$
x = r \cos t, y =r \sin t,
$$
then
$$
\frac {x^2 y^2}{y \sqrt {x^2 + y^2}} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, ... | For the equalty,
By telescoping: We have,
$$(a-b)\sum_{i=1}^na^{n-i}b^{i-1}=a\sum_{i=1}^na^{n-i}b^{i-1}-b\sum_{i=1}^na^{n-i}b^{i-1},$$
which can be rewritten
$$\sum_{i=1}^na^{n+1-i}b^{i-1}-\sum_{i=1}^na^{n-i}b^i=a^n+\sum_{i=2}^na^{n+1-i}b^{i-1}-b^n-\sum_{i=1}^{n-1}a^{n-i}b^i,$$
which on simplifying gives
$$a^n-b^n+\sum... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Find the degree of a ODE $(y''')^{\frac{4}{3}}+(y')^{\frac{1}{5}}+ y = 0$ Find the degree of the differential equation.
$$(y''')^{\frac{4}{3}}+(y')^{\frac{1}{5}}+ y = 0$$
The answer is available (order $= 3$; degree $= 60$). I need help with the steps. I'm stuck in eliminating the radicals of the differential coefficie... | Went through @Lutz's answer Find the degree of the differential equation $\left( \frac{d^3y}{dx^3} \right)^{\frac{4}{3}} + \left( \frac{dy}{dx} \right)^{\frac{1}{5}} + y = 0.$. But despite his elaborate explanations, I found it really hard to understand.
Then, I tried it the hard way. It's only a 2-step solution. Noth... | {
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I need to define the cubic equation $f(x)= ax^3 + bx^2 + cx + d$ It is given that if cubic function $f(x)$ is divided by $(x^2+4)$ and $(x+3)$, the remainders will be $7x-20$ and $-2$ respectively. Also it is said show that $f(x)$ is $x^3 + 6x^2 + 11x +4$. I tried but can i find the $a,b,c$, and $d$ without knowing the... | $f(x) = (x^2 + 4)(ax +b) +7x - 20$
$f(-3)=-2 = 13(-3a+b)-41 $
therefore $b= 3a+3$
so, $f(x)= (x^2+4)(ax +3a+3)+ 7x-20 $ (non-zero $a$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear quadratic cubic system of equation I have been trying to make something out of this system of equations for quite sometimes now
$x+y+z=4\\x^2+y^2+z^2=38\\x^3+y^3+z^3=106$
I have tried direct substitution but the equation keeps expanding, and keeps getting complicated. I have also tried multipling by $x,$ $y,$ $z... | $$16=38+2(xy+xz+yz),$$ which gives
$$xy+xz+yz=-11.$$
Thus, since
$$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz,$$ we obtain:
$$106=64-3\cdot4\cdot(-11)+3xyz,$$ which gives $$xyz=-30$$ and $x$, $y$ and $z$ are roots of the equation:
$$t^3-4t^2-11t+30=0.$$
Can you end it now?
I got $(x,y,z)=(2,-3,5)$ and all symmetric ... | {
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Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:
Find the polynomial $P(x)$ of minimal degree such that
$$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$
and the answer in the book is
$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}... | What they are asking you is to find a polynomial $p(x)$ such that
$$
\lim_{x\rightarrow0}\frac{|\sin(x-x^2)-p(x)|}{x^6}=0
$$
No to your statements:
*
*$o((x-x^2)^2) =o(x^5)$ is in general false. For example $h(x)=x(x-x^2)^2=o((x-x^2)^2$ as $x\rightarrow0$ but $x^{-5}h(x)\nrightarrow0$ as $x\rightarrow0$.
*$o(x^5)... | {
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Rotated Ellipse It is well known that the equation
$$\frac {(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac {(x\sin\alpha-y\cos\alpha)^2}{b^2}=1\tag{1}$$
(where $\beta\neq\alpha$) represents an ellipse centred at the origin with semimajor/minor axes $a,b$, and rotated by $\alpha$.
Question
The equation
$$\frac {(x\cos\alp... | If you rewrite your equation as
$$
ux^2 + vy^2 +2wxy = 1
$$
then the squared reciprocal lengths of the semimajor/semiminor axes of the ellipse are the eigenvalues of the matrix
$$
\begin{pmatrix} u & w \\ w & v \end{pmatrix}
$$
You get
$$
\begin{eqnarray*}
u & = & \frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \beta}{b^2} \\... | {
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proof that $\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer I would appreciate if somebody could help me with the following problem:
Q: How to proof?
If $\{a_n\}$ satisfy $a_{1}=a$, $a_2=b$, $a_{n+2}=a_{n+1}+a_{n}$($a,b$: positive integers)
then proof that
$\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer
I try start by mathematical ... | $R=\frac{a_{4n}-a_2}{a_{2n+1}}$
$a_1=a$
$a_2=b$
$a_3=a+b$
$a_4=a+2b$
$a_5=2a+3b$
$a_6=3a+5b$
$a_7=5a+8b$
$a_8=8a+13b$
For n=2 we have:
$\frac{8a+13b-b}{2a+3b}=4$
By experimental induction if R is true for n=2 it must also be true for n+1=3, we check this:
$\frac{a_{12}-b}{a_7}=\frac{55a+89b-b}{5a+8b}=11$
This true for... | {
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"answer_id": 2
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Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+... | After clearing the denominators we have $$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)(*)$$
as you obtained.
Indeed, $A=\frac{1}{3}$ if we let $x=1$.
Differentiate $(*)$ to get that $$4x^3-3x^2+4x-1=\frac{4}{3}x(x^2+2)+B(4x^3-3x^2+4x-2)+C(3x^2-2x+2)+D(2x-1)+E(**)$$
Differentiate the above relation a... | {
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Does any (right) triangle exist such that $a^3+b^3=c^3$?
Does any right triangle exist such that $a^3+b^3=c^3$? Does any triangle exist such that $a^3+b^3=c^3$?
I'm stuck on this problem; I tried applying the Pythagorean theorem in three dimensions, but in vain. Any tips?
| We need $a\ne0$ and $b\ne0$. If it holds, then we have
$(a^3+b^3)^2=c^6=(a^2+b^2)^3$
$a^6+2a^3b^3+b^6=a^6+3a^4b^2+3a^2b^4+b^6$
$2ab=3a^2+3b^2$
$(a-b)^2+2a^2+2b^2=0$
$a=b=0$
| {
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If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0$ then c is equal to? If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0,$ then $c$ is equal to?
I have tried solving it by putting $x = \frac{-4y}{3}$ inside the given equation, as $3x + 4y$ is a solution of $6x^2 – xy + 4cy^2 = 0$, but ... | Your idea is good; plugging in $x=-\tfrac43y$ yields
$$0=\frac{96}{9}y^2+\frac43y^2+4cy^2=\frac49\left(36+12c\right)y^2,$$
which must hold for all $y$. Then $36+12c=0$ so $c=-3$.
Alternatively, you could note that the point $(x,y)=(4,-3)$ is on the line, and hence on the curve, so
$$6\cdot4^2-4\cdot(-3)+4\cdot c\cdot(-... | {
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Find the base system, $x$, such that $\frac{1}{5}$ and $\overline{.17}$ are numerals for the same number. So far I know that $\frac{1}{5} = 0.2$ in base 10. However, I'm not really sure how to convert it into another base. Any suggestions would be greatly appreciated.
| $$\begin{align} (0.1717...)_b =& \left(\frac{1}{b}+\frac{7}{b^2}\right) +
\left(\frac{1}{b^3}+\frac{7}{b^4}\right) + \cdots \\
=& \frac{b+7}{b^2}+\frac{b+7}{b^4} +\cdots \\
=& \sum_{n=1}^{\infty} \frac{b+7}{b^{2n}} \\
=& (b+7)\sum_{n=1}^{\infty}(b^{-2})^n \\
=& \frac{b+7}{b^2-1}
\end{align}$$
and this is equal to $1/5$... | {
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Solve $\int_{-\pi}^{\pi}\frac{1}{1+\sin^{^{2}}t}dt$ I need to solve $\int_{-\pi}^{\pi}\frac{1}{1+\sin^{^{2}}t}dt$. This is what I did, but I think the answer should be $\sqrt{2}\pi$.
${\sin^{^{2}}t}=\frac{1-\cos(2t)}{2}$
$z=e^{2ti}=\cos(2t)+i\sin(2t)$
$-2\pi \leq 2t\leq 2\pi$
$\cos(2t)=\Re(z)=\frac{z+\bar{z}}{2}$
$z^{-... | Here is an alternative approach without using complex analysis.
The change of variables $\sin(t)=\dfrac{2u}{1+u^2}$ with $u=\tan\left(\dfrac{t}{2}\right)$ gives $\mbox{d}u=\dfrac{1}{2}\left(1+\tan\left(\dfrac{t}{2}\right)^2\right)\mbox{d}t$.
And therefore
\begin{align*}I&=\int_{-\pi}^{\pi}\dfrac{1}{1+\sin(t)^2}\mbox{d... | {
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What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$? $\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}=
2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi... | What you want is the set of solutions to the equation
$$
x^7 = (-\sqrt3 -i)^5.
$$
Now, taking the polar form of $-\sqrt3 -i$ you get
$$
(-\sqrt3 -i)^5 = \left(2\, e^{-5\pi/6}\right)^5
= 2^5 e^{-25\pi/6}
= 2^5 e^{-\pi/6}
$$
therefore the solutions are
$$
\{
2^{5/7} e^{-\pi/6 + 2 k \pi/7}, k=0,1,...,6
\}.
$$
| {
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Three cevians in a triangle create four sub-triangles of area $1$. Find the area of a non-triangular region.
I'm having trouble proving that all the white and green areas have the same area, from there on we can obtain the answer $1+\sqrt5$ by proving that the inner red triangle points are midpoints.
| Let $AD$, $BE$ and $CF$ be our chevians of $\Delta ABC$.
Also, let $AD\cap BE=\{P\}$, $AD\cap CF=\{R\}$ and $BE\cap CF=\{Q\}.$
Thus, $$S_{\Delta APE}=S_{\Delta BQF}=S_{\Delta CRD}=S_{\Delta PQR}=1.$$
Let $S_{AFQP}=x,$ $S_{BDRQ}=y$ and $S_{CEPR}=z$.
Thus, $$S_{\Delta BDR}=\frac{S_{\Delta BDR}}{S_{\Delta CDR}}=\frac{BD}{... | {
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Show that $r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$ In triangle $\Delta~ ABC$,
$~r~$ is the in-radius and
$~[ABC]~$ is the area.
Please explain $$ r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$$ thanks.
|
\begin{align}
[ABC]&=[IAB]+[IBC]+[ICA]
.
\end{align}
\begin{align}
[IAB]&=\tfrac12\cdot|AB|\cdot|IC_t|
=\tfrac12(|AC_t|+|BC_t|)\cdot|IC_t|
=\tfrac12(r\cot\tfrac\alpha2+r\cot\tfrac\beta2)\cdot r
=r^2(\tfrac12\,\cot\tfrac\alpha2+\tfrac12\,\cot\tfrac\beta2)
.
\end{align}
Similarly,
\begin{align}
[IBC]&=
r^2(\tfrac12\,... | {
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Why $ \frac{n+2}{n-2}<(n+2)^{2/n}$ for $n\geq 7$. In some paper, the authors mentioned the following statement:
One can easily check that for $n\geq 7$,
$$ \frac{n+2}{n-2}<(n+2)^{2/n}.$$
This statement is correct, and their objective was to find an upper bound of $\frac{n+2}{n-2}$, eventually starting from some inte... | (1) The inequality is true for $n=7$.
(2) Assume $(k+2)^{\frac{k-2}{k}} < k - 2$
(3) We must show $(k+3)^{\frac{k-1}{k+1}} < k - 1$.
Consider $f(x) = (x+3)^{\frac{x-1}{x+1}} - x + 1$. (We want to show $f(x) < 0$ for $x \geq 7$.)
Well, $(x+3)^{\frac{x-1}{x+1}} - x + 1 < 0$ iff $(x+3)^{\frac{x-1}{x+1}} < x - 1$ iff $\f... | {
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"answer_count": 5,
"answer_id": 3
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Change of variable in integrals I am trying to solve a definite integral of a positive function, but I keep getting 0.
| On using $ \int^a_{-a} f(x) dx = 2 \int^a_0 f(x) dx$ and $ \int^{2a}_0 f(x) dx = \int^a_0 f(x) dx + \int^a_0 f(2a - x) dx$
$$I = \int^{\pi}_{-\pi} \dfrac{1}{1 + \sin^2 (x)} dx = 4 \int^{\pi /2}_0 \dfrac{1}{1 + \sin^2 (x) } dx $$
Multiply, both denominator and numerator by $\sec^2 (x)$ , Use $\sec^2 (x) = \tan^2 (x) + 1... | {
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If I roll a dice 4 times then what is the probability of getting 20 total dots? So, here the total evets = $6^4$
If I get $a,b,c,d$ doing each rolling then, $$a+b+c+d=20$$
$$where\ a,b,c,d\in\mathbb{N} \ and\ 1\le a,b,c,d\le6$$
Now, how can I solve this by stars and bars method? Or there are other easier ways to solve ... | Using generating functions: The number of ways to roll a total of $20$ on four dice is $$\begin{align}[x^{20}](x+x^2+x^3+x^4+x^5+x^6)^4 &= [x^{16}](1+x+x^2+x^3+x^4+x^5)^4 \\ &= [x^{16}]\left({1-x^6 \over 1-x}\right)^4 \\
&= [x^{16}]{\binom40-\binom41x^6+\binom42x^{12}-\cdots \over (1-x)^4} \\
&= [x^{16}]{\binom40 \ove... | {
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Rearranging the formula Transpose this formula to make $y$ the subject.
$$x=\sqrt{x^2y^2+1-y}$$
My try:
$$x^2=x^2y^2+1-y$$
$$x^2-x^2y^2=1-y$$
$$x^2(1-y^2)=1-y$$
Here I got 2 $y$ terms, but I am not sure what to do next.
| You can write $$0=y^2-\frac{1}{x^2}y+\frac{1-x^2}{x^2}$$ using the quadratic formula we obtain
$$y_{1,2}=\frac{1}{2x^2}\pm\sqrt{\left(\frac{1}{2x^2}\right)^2-\frac{1-x^2}{x^2}}$$
| {
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Let $a,b,c$ be the sides of a triangle. The find maximum value $ \frac{A}{S^{2}}$ Let $a,b,c$ be the sides of a triangle, $A$ is the area and $S$ is the semi-perimeter $(a+b+c)/2$.
Find the maximum value $\frac{A}{S^{2}}$.
My Approach:
Method 1:
Applying AM-GM inequality on $S,S-a,S-b,S-c$
$$\frac{4S-2S}{4} \ge \sqrt... | Your Method 2 is fine and it gives the exact maximum value that you are looking for.
Since $A=\sqrt{S(S-a)(S-b)(S-c)}$, the AM-GM inequality in Method 2 yields
$$\frac{(S-a)+(S-b)+(S-c)}{3} \ge \sqrt[3]{(S-a)(S-b)(S-c)}$$
that is
$$\frac{S^3}{3^3}\geq(S-a)(S-b)(S-c)=\frac{A^2}{S},$$
and therefore
$$\frac{1}{3\sqrt{3}}\... | {
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How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}... | $\log_3 (x)=\log_{1/3} (x)+8 $
$\iff \log_3 (x)=-\log_3 (x)+8$
$\iff 2\log_3 (x)=8$
$\iff \log_3 (x)=4$
$\implies x=3^4=81$
| {
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"answer_count": 5,
"answer_id": 4
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Haar functions are basis in $L^2[0,1]$ Define the Haar functions as $e_0^0=1$ and for $n\ge 1$, $k=1,\ldots,2^n$
$$
e_n^k(t)=\left\{ \begin{array}{ll}
2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K-1}{2^n},\frac{K}{2^n}\big)$};\\
-2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K}{2^n},\frac{K+1}{2^n}... | We can use $f_n^j := \frac1{2^{\frac{n-1}2}} e_n^j$ since they have the same linear span, denote it by $S$.
We shall prove by induction on $n$ that $$\chi_{\left(\frac{k-1}{2^n},\frac{k}{2^n}\right)} \in S, \quad \text{ for all } n \ge 0, 1 \le k \le 2^n.$$
Indeed, if $n=1$ then
$$\chi_{(0,1)} \stackrel{\text{a.e.}}{=}... | {
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Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
My solution:
Set $f(z)=3$.
For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^... | Note that on $|z| < 1$ there are no roots as $|z^5+2z^3| < 3$ and so the roots of $z^5+2z^3+3$ on $|z| \leq 1$ lie on the boundary $|z| = 1$. Let $z = cos(\theta) + isin(\theta)$ satisfy the polynomial relation $P(z)$ stated above.
We must have $|(e^{i\theta})^5 + 2(e^{i\theta})^3|^2 = 9$ and hence $|e^{i2\theta}+2|^... | {
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Prove That $3^n + 8^n$ is Not Divisible by $5$ (Using Induction) Prove that $3^n+8^n$ is not divisible by 5.
I know that this can be proved by using congruence and I am providing the proof by congruence below. But is there any way to Prove It By Induction.
The proof by congruence goes like this:
$3\equiv 3\pmod 5 \\ 3^... | You can go like this. For the base case, let $n=1$. Then
$$3^n+8^n = 3+8 = 11,$$
which is not divisible by $5$.
For the induction step, let $n\geq 1$ be arbitrary and assume that $3^n+8^n$ is not divisible by $5$. Now
\begin{align*}
3^{n+1} + 8^{n+1} &= 3\cdot 3^n + 8\cdot 8^n\\
&= 3\cdot 3^n + (3+5)\cdot 8^n\\
&= 3\cd... | {
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Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\... | Because for $n\geq1$ we have: $$10^{2n+1}+1=(10+1)(10^{2n}-10^{2n-1}+...-10+1).$$
For $n=0$ it's obvious.
| {
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"answer_id": 1
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shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{2... | Hint
$3x=12-4y=4(3-y)$
WLOG any point on the line $P(4m,3-3m)$
$\dfrac43=\dfrac{3-3m-0}{4m-0}$
$m=?$
Can you find the distance of $P$ from the origin
Subtract the length of the radius from the distance
| {
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When going from $(x+2)^2=5$ to $x+2=\pm \sqrt{5}$, why isn't there also a $\pm(x+2)$? Say I am solving the following equation:
$$(x+2)^2 = 5$$
$$x + 2 = \pm \sqrt{5}$$
$$x = -2 \pm \sqrt{5}$$
However, when I took the positive and negative square root of $5$ in the second line, I did not take the positive and negative s... | When solving $(x+2)^2=5$, recall in general that for $x\in\mathbb{R}$ we have $\sqrt{x^2}=|x|$. And since clearly $x+2\in\mathbb{R}$, we have by the last identity that by taking the square root of both sides of $(x+2)^2=5$ $$\sqrt{(x+2)^2}=|x+2|=\sqrt{5}\tag1$$
Then we have reduced the problem to solving $$|x+2|=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left... | There is no function satisfying $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $ x \ne 0$.
For any $a \ne 0$ the equation $x - \frac 1x = a$ has two solutions $x_1, x_2$, which are related by $x_1 x_2 = -1$. Then
$$
x_1^2 - \frac{1}{x_1^2} = h(a) = x_2^2 - \frac{1}{x_2^2} = \frac{1}{x_1^2} - x_1^2 \\
\implies x_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find partial fractions of $\frac{3x^2-2x+1}{x^2(1-x^2)}$ and then deduce partial fractions of $\frac{3x^2-8x+6}{x(x-1)^2(2-x)}$ Find partial fractions of, $$\frac{3x^2-2x+1}{x^2(1-x^2)}$$
Hence deduce partial fractions of $$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}$$
My Try
I was able to do the first part,$$\frac{3x^2-2x+1}{x^2... | Let $X=x-1$ then
$$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}=-\frac{3(X+1)^2-8(X+1)+6}{(X+1)X^2(X-1)}=
-\frac{3X^2-2X+1}{X^2(X^2-1)}$$
Now you can use the first part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Knowing that $u_1 = 1, u_2 = 3$ and $u_{n + 2} = 2u_{n + 1} - u_n + 1, \forall n \in \mathbb Z^+$, prove that $4u_{n + 2}u_n + 1$ is a square number.
Knowing that $$\large \left\{ \begin{align*} u_1 = 1&, u_2 = 3\\ u_{n + 2} = 2u_{n + 1} - u_n + 1&, \forall n \in \mathbb Z^+ \end{align*} \right.$$, prove that $\large ... | Use characteristic polynomials, like here, to get to the conclusion that
$$(x-1)(x^2-2x+1)=0$$
is the characteristic polynomial and $x=1$ is a root of multiplicity 3. Then
$$u_n=(an^2+bn+c)\cdot 1^n$$
and
$$u_1=1=a+b+c$$
$$u_2=3=4a+2b+c$$
$$u_3=6=9a+3b+c$$
which resolves into $a=\frac{1}{2}$, $b=\frac{1}{2}$ and $c=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Cauchy Schwarz: Proving this inequality For positive reals a,b,x and y, prove that
$(5a^2 + 2ab + 3b^2)(5x^2 + 2xy + 3y^2) \ge (5ax + ay + bx + 3by)^2$
My attempt: By Cauchy we can show that the LHS $\ge(5ax + 2(abxy)^{1/2} + 3by)^2$
I can then show that $ay + bx \ge 2(abxy)^{1/2}$ using AM-GM, but this doesn't help.
... | By C-S we obtain:
$$(5a^2+2ab+3b^2)(5x^2+2xy+3y^2)=\left(\frac{1}{5}(5a+b)^2+\frac{14}{5}b^2\right)\left(\frac{1}{5}(5x+y)^2+\frac{14}{5}y^2\right)\geq$$
$$\geq\left(\frac{1}{5}(5a+b)(5x+y)+\frac{14}{5}by\right)^2=(5ax+ay+bx+3by)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$ Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$
I thought it could be good to use the function $f: ]-\infty, 81] \rightarrow \mathbb{R}$ given by $f(x) = \sqrt{81-x}$
Because this function is of class $C^{\infty}$, we can compute its Taylor... | Barry Cipra's answer does everything that is required with no fuss at all, so it gets my vote, but I was still curious to know how much work it would be to get the result using the binomial series. Not a lot, I think:
$$
1 - \sqrt{1 - x} = \tfrac{1}{2}\sum_{k=0}^\infty a_kx^{k+1} \quad (|x| < 1),
\qquad a_k = \frac{1\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 8
} |
Find the inverse in $F = \frac{\Bbb{Z}_{7}[x]}{g(x)}$ of a polynomial $p(x)$ $g(x) = x^{3}+2x^{2}+1$
$p(x) = x^{2}+x+2$
I don't know what I did wrong, hope you figured it out:
$g(x) = p(x)(x+1) + 4x + 6$
$p(x) = (4x+6)(2x+6)+1$
$1=p(x)-(2x+6)(4x+6)$
$1=p(x)-(2x+6)[g(x) - p(x)(x+1)]$
$1=p(x)+(5x+1)[g(x) + p(x)(6x+6)]$
... | $F$ can be seen as the algebra generated by the matrix $M = \begin{pmatrix}
0 & 0 & -1 \\
1 & 0 & 0 \\
0 & 1 & -2 \\
\end{pmatrix}$( with entries in $GF(7)\; $), called the companion matrix of the polynomial $g(x) = x^3 + 2x^2 + 1$. This matrix describes the action of the multiplication with $x$ namely $1 \mapsto x$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can I solve this Entrance Exam problem? This question is from NEST 2018.
Suppose ${2+ \sqrt{3}}$ and $1-i$ are roots of the equation $(x^2+px+1)(x^2-2x+q)=0$ where $p,q$ are integers and $i=\sqrt{-1}$. Then what is $p+q$ ?
| If you substitute ${2+ \sqrt{3}}$ and $1-i$ in $(x^2+px+1)(x^2-2x+q)=0$, I obtain:
$$\left\{\begin{matrix}
(({2+ \sqrt{3}})^2+p({2+ \sqrt{3}})+1)(({2+ \sqrt{3}})^2-2({2+ \sqrt{3}})+q)=0
\\ ((1-i)^2+p(1-i)+1)((1-i)^2-2(1-i)+q)=0
\end{matrix}\right.$$
From the first equation, I obtain: $(2+\sqrt3)(p+4)(3+\sqrt3+q)=0$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the min value of $\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$, if $x\ne y \ne z$ and $x,y,z\in {1,2,3,4,5}$ My answer is $\frac{5}{29}$, I just use logic to substitute numbers in the expression, but I can't prove my answer If this expression be minimum then the denominator should be the greatest, so I just let x=5 and I... | It's just
$$\frac{1}{5+\frac{1}{1+\frac{1}{4}}}=\frac{5}{29}.$$
Indeed, we need to prove that
$$\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\geq\frac{5}{29}$$ or
$$\frac{1}{x+\frac{z}{yz+1}}\geq\frac{5}{29}$$ or
$$\frac{yz+1}{xyz+x+z}\geq\frac{5}{29}$$ or
$$5yz(5-x)+5(5-x)+4z(y-1)+4-z\geq0.$$
Now, if $z<4$ so $$5yz(5-x)+5(5-x)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to find least square solution to Ax=b when columns of A are not linear independent? Let $A =\begin{bmatrix} 0 & -1 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \\\end{bmatrix}$ and $b =\begin{bmatrix} 1 \\2\\ 3 \\\end{bmatrix}$
I couldn't just solve $A^TAx=A^Tb$, since the columns of A are not independent, I was told by my profe... | I don’t understand either what you are having issues with, or what your professor told you to do.
The first thing to notice here is that the system has solutions! So a “least squares solution” really just means a regular solution. If you solve the system directly, you get $x=4-t$, $y=-1$, $z=t$. So those are “least squ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Given $x, y$ that $xy-\frac{x}{y^2}-\frac{y}{x^2}=3$, work out $xy-x-y$. Today I had a competition in Xiamen, China. I know how to do the questions except this strange equation.
Given $x, y$ that $xy-\dfrac{x}{y^2}-\dfrac{y}{x^2}=3$, work out $xy-x-y$.
Such a strange question, right? I have found the integral solutio... | Making $y = \lambda x$ we have
$$
xy -\frac{x}{y^2}-\frac{y}{x^2}=3\Rightarrow \lambda x^3-3x=\lambda+\frac{1}{\lambda^2}
$$
and solving for $x$
$$
x = \left\{
\begin{array}{c}
\frac{\lambda +1}{\lambda } \\
-\frac{\lambda ^2+\lambda +\sqrt{3} \sqrt{-(\lambda -1)^2 \lambda ^2}}{2 \lambda ^2} \\
-\frac{\lambda ^2+\la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$ Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$.
My approach is to do $\lim_{n\to\infty} \frac{\left(\frac{3... | $$\dfrac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}
=\dfrac1{\left(\frac{2}{3}\right)^{3n-1}+\frac{3}{2}}$$
Since $\dfrac{2}{3} <1$ , its limit will be zero.
Thus the answer is $\dfrac{1}{0+\frac{3}{2}}=\dfrac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to isolate $x$ when $\cos(x/2)\cos(3x)/2−3\sin(x/2)\sin(3x)= 0$ Asked to find all relative and absolute extrema of $f(x) = \sin\left(\frac{1}{2}x\right)\cos(3x)$ on the interval $[0,\pi]$. I've gotten the derivative, $$f'(x)=\frac{\cos(\frac{x}{2})\cos(3x)}{2} - 3\sin(\frac{x}{2})\sin(3x),$$ just fine, but how wo... | Trying approximate solutions.
First, using trigonometric formulae, the derivative can write
$$f'(x)=\frac{1}{4} \left(7 \cos \left(\frac{7 x}{2}\right)-5 \cos \left(\frac{5
x}{2}\right)\right)$$ Forget the $\frac{1}{4}$ and plot the function. You should notice that, for the range of concern, the solution are "close"... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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initial value problem $y'=-y^2, y(1)= \sqrt{2}$ hey I cant seem to solve this at all, not sure how you do it with no x values
$$y'=-y^2, y(1)= \sqrt{2}$$
| We have $y'=-y^2 \iff \frac{y'}{-y^2}=1 \iff \frac{d}{dx}(\frac{1}{y})=\frac{dx}{dx} \iff \frac{1}{y}=x+c$
$y(1)=\sqrt{2} \iff c=\frac{\sqrt{2}-2}{2}$
So $1/y=\frac{2x+\sqrt{2}-2}{2}\iff y=\frac{2}{2x+\sqrt{2}-2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
System of three non-linear equations Can anyone help me solve this system of equations?
$$a_1 a_3 -a_2 ^2=0$$
$$a_1+a_3-2a_2-16=0$$
$$a_1 a_3 +64a_1 - a_2 ^2 -16 a_2 -64 =0$$
After couple of steps I got $4a_1-a_2-4=0, (a_2-a_3)^2=16$. Then we have two cases $a_2-a_3=4$ and $a_2-a_3=-4$, but I couldn't finish this. Than... | I'm going to use $x=a_1, \ y=a_2, \ z=a_3$
$$xz - y^2=0 \tag A$$
$$x + z - 2y - 16 = 0 \tag B$$
$$xz + 64x - y^2 - 16y - 64 = 0 \tag C$$
Substitute $xz = y^2$ from (A) into (C)
\begin{align}
xz + 64x - y^2 - 16y - 64 &= 0 \\
y^2 + 64x - y^2 - 16y - 64 &= 0 \\
64x - 16y - 64 &= 0 \\
y &= 4x - 4 \tag D
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find the sum of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$ up to $n$ terms
Find the sum of first $n$ terms of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$
*
*When $n$ is even.
*When $n$ is odd.
This sum can be written as
$$\sum_{1}^n (2k-1)^{3} +3... | Details for your comment above:
$n$ is even:
$$\sum_{1}^{n/2} (2k-1)^{3} +3 \sum_{1}^{n/2} (2k)^{2}$$
Example:
$$1^3 + 3\cdot 2^2 = \sum_{k=1}^{2/2}(2k-1)^3+3\sum_{k=1}^{2/2}(2k)^2$$
$n$ is odd:
$$\sum_{1}^{(n+1)/2} (2k-1)^{3} +3 \sum_{1}^{(n+1)/2-1} (2k)^{2}$$
Example:
$$1^3 + 3\cdot 2^2 + 3^3 = \sum_{k=1}^{(3+1)/2}(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Discrete Mathematics - Combinatorics proof $\\$ I need to prove $\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot\left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{5^n+(-3)^n}{2}$
my Attempt below
$\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot \left(\begin{array}{c}... | Observe that
$$\sum_{k = 0}^{n} \left(\frac{1^k + (-1)^k}{2}\right)\binom{n}{k}4^k = \frac{1}{2}\left[\sum_{k = 0}^{n} \binom{n}{k}4^k + \sum_{k = 0}^{n} \binom{n}{k}(-4)^k\right]$$
By the Binomial Theorem,
$$(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k}x^k$$
For the first summation, $x = 4$; for the second summation $x =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Is $\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$ the same as $\frac{\frac{2-b}{2b}}{b-2}$? Have I subtracted 2 fractions correctly?
$\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$
Starting with the numerator which is a difference of fractions, the least common denominator is 2b? So:
$\frac{2(1)}{2b}-\frac{b(1)}{2b}$ = $\frac{2}{2b}-\fra... | It is correct, since $$\frac{1}{b}-\frac{1}{2}=\frac{2}{2b}-\frac{b}{2b}=\frac{2-b}{2b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Quickly evaluating this limit: $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right).$$
If one attacks this with naive applications of L'Hospi... | $$
\frac {1}{x^2}-\frac{1}{\sin^2x} = \left(\frac {1}{x}+\frac{1}{\sin x}\right)\left(\frac {1}{x}-\frac{1}{\sin x}\right) = \frac{1}{x^2}\left(\frac {x}{x}+\frac{x}{\sin x}\right)\left(\frac {x}{x}-\frac{x}{\sin x}\right)
$$
so
$$
\lim_{x\to 0}\left(\frac {1}{x^2}-\frac{1}{\sin^2x}\right) =2\lim_{x\to 0}\frac{1}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Finding a formula for $(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$
I need a formula for this sum:
$$(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$$
I have found this formula :
$$1\cdot n +2\cdot(n-1)+3\cdot(n-2)+ \cdots +(n-1)\cdot 2 +n\cdot1= \frac16n(n+1)(n+2)$$
... | I can rewrite the sum as $S_n=\sum\limits _{k=1}^{n}k^3(n+1-k)=(n+1)\sum\limits _{k=1}^{n}k^3 - \sum\limits _{k=1}^{n}k^4$. Now, we can show by induction that $A_n=\sum\limits _{k=1}^{n}k^3=\frac{n^4+2n^3+n^2}{4}$ and $B_n=\sum\limits _{k=1}^{n}k^4=\frac{6n^5+15n^4+10n^3-n}{30}$. Substituing this formulas in: $S_n=(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary number theory, 'concert-ticket-arithmetic' Four friends, call them A,B,C and D are planning to go to a concert, but they realize that they
are a few dollars short to buy tickets.(50 $ per ticket).
We know that each of them has an integer amount of dollars.
If B borrowed 1$ from A,then B ... | Let $a$ be the money of $A$.
$b$ be the money of $B$. $c$ the money of $C$ and $d$ the money of $D$. I have the following system:
$$\left\{\begin{matrix}
b+1=\frac{2}{3}(a-1)
\\c+2=\frac{3}{5}(b-2)
\\d+3=\frac{5}{7}(c-3)
\end{matrix}\right.$$
Solving for $a,b,c,d$, I obtain:
$$\left\{\begin{matrix}
b=\frac{2}{3}(a-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to determine algebraically whether an equation has an infinite solutions or not? I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try... | You are trying to represent $\frac{x^2-4}{(x+5)(x-3)}$ in the form $\frac{Ax+B}{x+5}+\frac{Cx+D}{x-3}$. As explained in other responses, you have three equations in four unknowns so there are an infinite set of solutions.
The reason why there are an infinite set of solutions is that if you have any one solution $(A,B,C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 6
} |
Problem with proof by induction I am struggling with the following equation, which I need to proof by induction:
$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}= \sum_{k=n+1}^{2n}\frac{1}{k}$$
$n\in \mathbb{N}$. I tried a few times and always got stuck.
Help would be appreciated.
| The proof is simple. By induction hypothesis, assume that the statement is true for $s < n$. Then, we have
\begin{align}
\sum\limits_{k = 1}^{2n} \dfrac{\left( 1 \right)^{k + 1}}{k} &= \sum\limits_{k = 1}^{2 \left( n - 1 \right)} \dfrac{\left( 1 \right)^{k + 1}}{k} + \dfrac{1}{2n - 1} - \dfrac{1}{2n} \\
&= \sum\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Non Linear Algebraic Equation If$$ abc = 4(4+a+b+c) , find $$ $$1/(2+a) +1/(2+b) + 1/(2+c).$$
My approach; put $$a=b=c=4 $$ and we are done. Looking at the symmetry of the equation, i think it is the only value that will satisfy the given equation or if otherwise, it is too difficult to get it by hit and trail.
Any bet... | This is basically an exercise in simplification.
Rewrite the target expression as (after adding and simplifying): $$\frac{12+4(a+b+c)+ab+bc+ca}{8+4(a+b+c)+2(ab+bc+ca)+abc}=\frac{4(4+a+b+c)+ab+bc+ca-4}{4(4+a+b+c)+2(ab+bc+ca)+abc-8}.$$
Substituting using the given relationship results in $$\frac{abc+ ab+bc+ca-4}{2abc+2(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3344387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $
I need help with this limit:
$$ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $$
I've tried using the $$ \tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b} $$ with $(x+a)$ as $a$ and $(x)$ as $b$ but having difficulty getting beyond that.
| The best way to do this problem is to realize that the limit is, by definition, $\tan ' (x)$. Knowing that $\tan x = \frac{\sin x}{\cos x}$ and knowing $\sin '(x) = \cos x$ and $\cos ' (x) = -\sin x$, you can calculate the limit with the quotient rule.
As you've said in a comment, the key to doing this problem straigh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3344722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proving limits of multivariate function with epsilon-delta definition I want to solve this problem using epsilon-delta definition :
$$\lim_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^{2}+y^{2}}}}{x^{2}+y^{2}}$$
However, I have no idea where to start first - how to modify and apply some kind of inequality.
| For $r \ge 1$, you have $1 \le \frac{e^r}{r^2}$ hence $0 < e^{-r}\le \frac{1}{r^2}$.
Replace $r$ with $1/(x^2+y^2)$. You get
$$e^{-1/(x^2+y^2)} \le (x^2+y^2)^2$$ and therefore
$$0 < \frac{e^{-1/(x^2+y^2)}}{x^2+y^2}\le (x^2+y^2)$$ for
$\sqrt{x^2+y^2} \le 1$. Now, fix $\epsilon > 0$ and take $\delta = \min(1, \sqrt{\epsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For the function $f(x,y,z)= xyz$ how close to the point $(0,0,0)$ should one take the point $(x,y,z)$ in order to make $|f(x,y,z)-f(0,0,0)|<0.008$?
For the function
$$f(x,y,z)= xyz$$
how close to the point $(0,0,0)$ should one take the point $(x,y,z)$ in order to make $|f(x,y,z)-f(0,0,0)|<0.008$?
here is my solu... | To get the solution of the book you may use GM-AM:
*
*$\sqrt[3]{abc}\leq \frac{a+b+c}{3}$
$$|xyz| = \sqrt{x^2y^2z^2} \stackrel{GM-AM}{\leq}\sqrt{\frac{(x^2+y^2+z^2)^3}{3^3}} \stackrel{!}{<}0.008 = 0.2^3$$
Now, isolate $\sqrt{x^2+y^2+z^2}$:
$$\sqrt{\frac{(x^2+y^2+z^2)^3}{3^3}} <0.2^3 \Leftrightarrow \sqrt{x^2+y^2+z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Observe that all natural numbers n^3 can be written as a sum of n odd integers that are revolving around n^2 For example,
$1^3$ = $1$,
$2^3$ = $(2^2-1) + (2^2+1)$
$3^3$ = $(3^2-2) + (3^2) + (3^2 + 2)$
$4^3$ = $(4^2-3) + (4^2 - 1) + (4^2 + 1) + (4^2 + 3)$
I have been able to deduce a general formula for this behavior... | Note that:
In the expression
$n^3 = (n^2 - (n-1)) + (n^2 - (n-3)) + ... + (n^2 + (n-3)) + (n^2 + (n-1))$,
$n-1$ cancels from the first term and the last term.
$n-3$ cancels from the second term and the second to last term.
$n-5$ cancels from the thirst term and the third to last term.
And so on.
$n-a$ cancels from the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{... | Someone has already pointed out where your mistake is. This is just to point to another alternative, for the fun of it.
You can factor $x^3+y^3$ as $$(x+y)(x^2-xy+y^2).$$ To compute the sum of squares, note that $2(x^2+y^2)=(x+y)^2+(x-y)^2.$ So, you only need to compute the sum, difference and product of your numbers, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
find minimum of maximum of two functions
I need to find the angle $\theta$ so that: $$\max(\cos^2(\theta),1-\cos^2(45-\theta))$$ is minimized.
OK, so I wrote \begin{align*}f(\theta)&=\max(\cos^2(\theta),1-\cos^2(45-\theta))\\
&=\frac{\cos^2(\theta)+1-\cos^2(45-\theta)+|\cos^2(\theta)-1+\cos^2(45-\theta)|}{2}\end{ali... | Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$D=\cos^2t-(1-\cos^2(45^\circ-t))=\cos^2t-\sin^2(45^\circ-t)=\cos(45^\circ)\cos(2t-45^\circ)$$
Case $\#1:$
$\cos^2t$ will be maximum if $D\ge0$ if $\cos(2t-45^\circ)\ge0$
$\iff360^\circ n-90^\circ\le2t-45^\circ\le360^\circ n+90^\circ$
$\iff360^\circ n-45^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
| Why should the mean inequalities not work? Geometric mean vs. mean of power 4
$$
abc\le\left(\frac{a^4+b^4+c^4}3\right)^{\frac34}
$$and arithmetic mean vs. mean of power 4$$
\frac{a+b+c}3\le\left(\frac{a^4+b^4+c^4}3\right)^{\frac14}
$$
implies
$$
\frac{abc(a+b+c)}3\le \frac{a^4+b^4+c^4}3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A hard inequality indian olympiad problem If $x,y,z$ are positive real numbers, prove that:
$\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\le 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)$.
I have been stuck in it. It is an Indian Olympiad problem. Can you guys help me out, please?
| We have that:
$$x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\geq \frac{3}{4}(x+y)^2$$
Therefore
$$(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq \frac{27}{64}(x+y)^2(y+z)^2(z+x)^2$$
And it remains to prove
$$9(x+y)(y+z)(z+x)\geq 8(x+y+z)(xy+yz+zx)$$
We can prove this with AM-GM:
$$8(x+y+z)(xy+yz+zx)=8(x+y)(y+z)(z+x)+8xyz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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.