Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$, prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
My attempts:
Attempt 1:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\sin^2 x+\cos^2 x-2\sin x \cos x}{\sin^2 x+\cos^2 x+2\sin x \cos x}}$
$\implies y=\sqrt{\frac{(\sin x -\cos x)^2}{(\sin x+\cos x)^2}}$
$\implies y=\frac{\sin x -\cos x}{\sin x +\cos x}$
$\therefore \frac{dy}{dx}=\frac{(\sin x +\cos x)(\cos x+\sin x)-(\sin x -\cos x)(\cos x -\sin x)}{(\sin x +\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{(\sin x+\cos x)^2+(\sin x -\cos x)^2}{(\sin x+\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{2(\sin^2 x+\cos^2 x)}{(\cos(\frac{\pi}{2}-x)+\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{2}{(2\cos(\frac{\pi}{4})\cos(\frac{\pi}{4}-x))^2}$
$\implies \frac{dy}{dx}=\frac{2}{4\cos^2(\frac{\pi}{4})\cos^2(\frac{\pi}{4}-x)}$
$\implies \frac{dy}{dx}=\frac{2}{4\times\frac{1}{2}\cos^2(\frac{\pi}{4}-x)}$
$\implies \frac{dy}{dx}=\sec^2(\frac{\pi}{4}-x)$
$\implies \frac{dy}{dx}-\sec^2(\frac{\pi}{4}-x)=0$
But i have to prove $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
Attempt 2:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\cos^2 x+\sin^2 x-2\cos x \sin x}{\cos^2 x+\sin^2 x+2\cos x \sin x}}$
$\implies y=\sqrt{\frac{(\cos x -\sin x)^2}{(\cos x+\sin x)^2}}$
$\implies y=\frac{\cos x -\sin x}{\cos x +\sin x}$
$\therefore \frac{dy}{dx}=\frac{(\cos x +\sin x)(-\sin x-\cos x)-(\cos x -\sin x)(-\sin x +\cos x)}{(\cos x +\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{-(\cos x+\sin x)^2+(\cos x -\sin x)^2}{(\cos x+\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{(\cos x -\sin x)^2-(\cos x+\sin x)^2}{(\cos x+\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{(\cos x+\cos(\frac{\pi}{2}-x))^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{(2\cos(\frac{\pi}{4})\cos(x-\frac{\pi}{4}))^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{4\cos^2(\frac{\pi}{4})\cos^2(x-\frac{\pi}{4})}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{4\times \frac{1}{2}\cos^2(x-\frac{\pi}{4})}$
$\implies \frac{dy}{dx}=\frac{-2\sin x\cos x}{\cos^2(x-\frac{\pi}{4})}$
Attempt 3:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{(1-\sin 2x)(1-\sin 2x)}{(1+\sin 2x)(1-\sin 2x)}}$
$\implies y=\frac{1-\sin 2x}{\cos 2x}$
$\therefore \frac{dy}{dx}=\frac{\cos 2x(-2\cos 2x)-(1-\sin 2x)(-2\sin 2x)}{\cos^2 2x}$
$\implies \frac{dy}{dx}=\frac{-2\cos^2 2x+2\sin 2x(1-\sin 2x)}{\cos^2 2x}$
My questions:
(i) Why does Attempt 2 give a result different from that given by Attempt 1.
(ii) How do i prove the result?
|
"Why does Attempt 2 give a result different from that given by Attempt 1."
Your mistake is that You conclude $\sqrt{a^2}=a$. Actually, one time You conclude that $\sqrt{a^2}=a$ and another time that $\sqrt{(-a)^2}=-a$ although $\sqrt{a^2}$ and $\sqrt{(-a)^2}$ are the same. Correct would be $\sqrt{a^2}=\pm a$, so two values.
"(ii) How do i prove the result?"
You don't need to do any transformation of the kind, You can directy differentiate the given expression.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2813208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Find out if polynomial has inverse in quotient ring Find out if polynomial $x^3-x^2+x-1$ has inverse element in quotient ring ${\displaystyle \mathbb {Z} }_{11}/(x^4+3x^3-3x^2-4x-1)$, if yes find this inverse.
I know that $x^3-x^2+x-1$ has inverse element when $gcd(x^4+3x^3-3x^2-4x-1;x^3-x^2+x-1)$~$1$. I also know that I can use Euclidean algorithm. But I'm not sure about my result, I have got result that the g$cd$~$5$ it means that $x^3-x^2+x-1$ does not have inverse. Am I correct?
Thanks for any help!
|
Since Jose already covered the fact that your approach is valid, I'll present a cool alternative approach. First, construct the companion matrix of $x^4+3x^3-3x^2-4x-1$, call it $A$.
Here $$A = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -3
\end{pmatrix}_.$$
Now consider the matrix, $B = A^3 - A^2 + A - I$. Then $B$ is invertible (recall that the coefficients of the matrix lie in $\mathbb{Z}_{11}$) iff $x^3 - x^2 + x - 1$ is invertible. Even stronger, if you express $B^{-1}$ as a linear combination of $I, A, A^2, A^3$, this tells you the inverse of $(x^3-x^2 + x -1)$!
More explicitly, if $B^{-1} = c_3A^3 + c_2A^2 + c_1A + c_0I$, then $$(x^3-x^2+x-1)^{-1} = c_3x^3+c_2x^2+c_1x+c_0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2815784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Simplify: $\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$ Simplify:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$$
So what I've tried was:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{1+\cos^2x-\sin^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$=\frac{2\cos^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{2\cos^2x\,\tan^2x-4\cos^2(\frac{x}{2})(1-\cos x)}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{2\sin^2x-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=\frac{2\cdot2\sin^2(\frac{x}{2})\cos^2(\frac{x}{2})-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos x)}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(-1+\cos^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{-2\sin^2(x)}{(1-\cos x)(\tan^2x)}$$
This is farthest I got but the answer is supposed be: $0$,
according to the solutions, I might've missed something on the process even though I checked it a couple of time..
I'm preparing for the Tel-Aviv univeristy math entry test and I took this exercise from math-exercises.com to practise (that is where the solution is from).
|
We have that
$$\cos2x=\cos^2x-\sin^2x=2\cos^2x-1$$
$$\cos^2\left(\frac{x}{2}\right)=\frac12(\cos x +1)$$
$$\tan^2x=\frac{1-\cos^2x}{\cos^2 x}$$
thus
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$\frac{2\cos^2x}{1-\cos x}-\frac{2(\cos x +1)}{\frac{1-\cos^2x}{\cos^2 x}}=$$
$$\frac{2\cos^2x}{1-\cos x}-\frac{2\cos^2x}{1-\cos x}=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2818753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Bitstrings and set of solutions There is something about counting bitstrings and the format of the solutions I didn't really understand yet.
Given a bitstring problem that asks to elementarily count how many bitstrings of length 36 there are that contain exactly 5 $\{1\}$ in the first 10 posititions and the $\{10100101\}$ substring in the last 20 positions, finding it out is pretty straightforward.
Since the first 10 positions need to contain exactly 5 $\{1\}$, you find out the number of all the possible combinations containing maximum five ones:
$${10 \choose 5}$$
We have no information about the next 16 positions, so that is $2^{16}$ more combinations, and thus far we have:
$${10 \choose 5} \cdot 2^{16}$$
many combinations.
Now the substring:
*
*there are $\{10100101\} \cdot 2^{12}$ bitstrings with no overlapping; $20 - 8 + 1 = 13 \rightarrow {13 \choose 1} \cdot 2^{12}$
*there are $\{1010010100101\} \cdot 2^{7}$ bitstrings with single overlapping, same goes as you overlap the starting $1$ and the one at the end, so ${8 \choose 1} \cdot 2^{7}, {6 \choose 1} \cdot 2^{5}$
*there are $2\cdot \{10100101\} \cdot 2^{4}$ bitstrings as the substring repeats itself once, so ${6 \choose 2} \cdot 2^{4}$
*there are $\{101001010010100101\} \cdot 2^{2}$ bitstrings with double overlapping, and considering also all the other possibilities you have ${3 \choose 1} \cdot 2^{2}, {1 \choose 1} \cdot 2^{0}, {1 \choose 1} \cdot 2^{0}$
Since $\{10100101\} \cdot 2^{12}$ "spans" all the other combinations below and are thus included within it, shouldn't we remove all the overlapping and repeating substrings from the universe $\{10100101\} \cdot 2^{12}$? Given the number of substrings above, how should I then build the solution?
|
In order to calculate the number of binary strings of length $20$ containing the substring $10100101$ pretty much of all the hard work is already done. We just have to use the inclusion-exclusion principle and put all the intermediate results together.
The number of wanted strings is
\begin{align*}
&\binom{13}{1}2^{12}-\left(\binom{8}{1}2^7+\binom{6}{1}2^5+\binom{6}{2}2^4\right)
+\left(\binom{3}{1}2^2+\binom{1}{1}2^0+\binom{1}{1}2^0\right)\\
&\qquad=53\,248-(1024+192+240)+(12+1+1)\\
&\qquad\,\,\color{blue}{=51\,806}
\end{align*}
which counts the number of binary strings of length $20$
*
*containing the substring $10100101$ at least once
*minus the number of strings where the substring occurs at least twice, including overlaps.
*Since we have subtraced strings where the substring occurs at least three times (here in form of overlaps only) too often we have to add them for compensation.
We can check the result by applying another approach called the Goulden-Jackson Cluster Method.
We consider the set words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{0,1\}$$ and the set $B=\{010100101\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $f(s)$ with the coefficient of $s^n$ being the number of searched words of length $n$.
According to the paper (p.7) the generating function $f(s)$ is
\begin{align*}
f(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\tag{1}
\end{align*}
with $d=|\mathcal{V}|=2$, the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[10100101])
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C}[10100101])&=-s^8-s^5\text{weight}(\mathcal{C}[10100101])\\
&\qquad\quad-s^7\text{weight}(\mathcal{C}[10100101])\\
\end{align*}
and get
\begin{align*}
\text{weight}(\mathcal{C})=-\frac{s^8}{1+s^5+s^7}
\end{align*}
It follows
\begin{align*}
f(s)&=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-2s+\frac{s^8}{1+s^5+s^7}}\\
&=\frac{1+s^5+s^7}{1-2s+s^5-2s^6+s^7-s^8}\\
\end{align*}
The coefficient $[s^n]$ in $f(s)$ gives the number of binary words of length $n$ which does not contain $10100101$. Since we want to count the number of words of length $20$ which do contain the word we take the generating function
\begin{align*}
\frac{1}{1-2s}=1+2s+4s^2+8s^3+\cdots
\end{align*}
which counts all binary words and subtract $f(s)$ from it.
We obtain with some help of Wolfram Alpha
\begin{align*}
\frac{1}{1-2s}-\frac{1+s^5+s^7}{1-2s+s^5-2s^6+s^7-s^8}&=s^8+4s^9+12s^{10}+32s^{11}+\cdots\\
&\qquad+\color{blue}{51\,806s^{20}}+\cdots
\end{align*}
in accordance with the calculation above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2820787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.
What is the probability that he wears all three shirts after 6 days?
My try:
$$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$
First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.
Thanks in advance.
|
Time to use inclusion-exclusion: the probability that he'll use all three is the probability that the sequence cannot come from a corpus of just a given two.
So we have: $$P(\text {all of }RGW)=P(RGW)-P(RG)-P(RW)-P(GW)+P(R)+P(G)+P(W)-P(\emptyset)$$. Which is relatively simple: each $P$ is the summed probability of getting a shirt that's in the list, to the sixth power (because six days).
Replacing and doing the obvious simplification of using a common denominator: $$\begin {align}&\frac{6^6-5^6-4^6-3^6+3^6+2^6+1^6-0^6}{6^6}\\=&\frac{46656-15625-4096-729+729+64+1-0}{46656}\\=&\frac{27000}{46656}\\=&\frac{125}{216}
\end {align}$$
... Which I'll be honest is a much nicer number then I expected
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2821454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Finding $C_n=4C_{n-1}+6n-1$ 2.$$\begin{cases}
C_n=4C_{n-1}+6n-1\\
C_0=2\\
\end{cases} $$
$\begin{cases}
C_n=D_n+dn+e\\
D_n=\lambda D_{n-1}\end{cases}$
$D_n+dn+e=4C_{n-1}+6n-1$
But $C_{n-1}=D_{n-1}+d(n-1)+e$
So $D_n+dn+e=4(D_{n-1}+d(n-1)+e)+6n-1$
$D_n+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$
But for $\lambda=4$ we get $4D_{n-1}+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$ and
$0=-dn-e+4dn-4d+4e+6n-1$ or $0=(3d+6)n+3e-4d-1$
So $d=-2,e=1.5$
$C_n=D_n+dn+e$ So $C_n=D_n-2n+1.5$ And $C_0=D_0+1.5$ So $2=D_0+1.5$ And $D_0=-0.5$
$D_n=4D_{n-1}$
So
$C_n=D_n+dn+e=4^n-2n+1.5$ which is $O(4^n)$
Is it correct?
|
$$C_1=4C_0+6\cdot1-1,$$
$$\frac{1}{4}C_2=C_1+\frac{1}{4}(6\cdot2-1),$$
$$\frac{1}{4^2}C_3=\frac{1}{4}C_2+\frac{1}{4^2}(6\cdot3-1),$$
$$.$$
$$.$$
$$.$$
$$\frac{1}{4^{n-1}}C_n=\frac{1}{4^{n-2}}C_2+\frac{1}{4^{n-1}}(6n-1),$$
which after summing gives
$$\frac{1}{4^{n-1}}C_n=4\cdot2+6\left(1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{n}{4^{n-1}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{n-1}}\right)=$$
$$=8+6\left(1+2x+3x^2+...+4x^{n-1}\right)_{x=\frac{1}{4}}-\frac{\left(\frac{1}{4}\right)^n-1}{\frac{1}{4}-1}=$$
$$=\frac{20}{3}+\frac{4}{3}\left(\frac{1}{4}\right)^n+6(x+x^2+...+x^n)'_{x=\frac{1}{4}}=$$
$$=\frac{20}{3}+\frac{4}{3}\left(\frac{1}{4}\right)^n+6\left(\frac{x(x^n-1)}{x-1}\right)'_{x=\frac{1}{4}}=$$
$$=\frac{20}{3}+\frac{4}{3}\left(\frac{1}{4}\right)^n+\frac{6\left(nx^{n+1}-(n+1)x^n+1\right)}{(x-1)^2}_{x=\frac{1}{4}}=$$
$$=\frac{20}{3}+\frac{4}{3}\left(\frac{1}{4}\right)^n+\frac{6\left(n\left(\frac{1}{4}\right)^{n+1}-(n+1)\left(\frac{1}{4}\right)^n+1\right)}{\left(\frac{1}{4}-1\right)^2}=$$
$$=\frac{52}{3}+\left(\frac{1}{4}\right)^n\left(-8n-\frac{28}{3}\right).$$
Thus, $$C_n=\frac{52}{3}\cdot4^{n-1}-2n-\frac{7}{3}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2823765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
How do I solve this system of non-linear differential equations? How do I solve this system of differential equations?
\begin{equation}
\frac{d^2x}{dt^2}+2\frac{1}{y}\frac{dx}{dt}\frac{dy}{dt}=0 \quad\quad
\frac{d^2y}{dt^2}+\frac{1}{y}
\left[\left(\frac{dx}{dt}\right)^2-\left(\frac{dy}{dt}\right)^2\right]=0
\end{equation}
I would also like to know if there is some resource that is just a catalogue of different types of differential equations and their solutions (without any proofs and such) just for practical use. I don't usually work with D.E.s and forgot everything.
|
Rearranging the first equation and integrate
\begin{align} \frac{\ddot x}{\dot x} + \frac{2\dot y}{y} &= 0 \\
\ln(\dot x) + 2\ln(y) &= \ln(a)
\end{align}
$$ \implies \dot x = \frac{a}{y^2} $$
Plugging this into the second equation
$$ \ddot y + \frac{1}{y}\left[\frac{a^2}{y^4} - (\dot y)^2\right] = 0 $$
This equation is autonomous, so we substitute $v = \dot y$ and obtain a first-order ODE
$$ v\frac{dv}{dy} - \frac{v^2}{y} + \frac{a^2}{y^5} = 0 $$
which is a Bernoulli equation, so another substitution $z = v^2$ gives
$$ \frac{dz}{dy} - \frac{2}{y}z + \frac{2a^2}{y^5} = 0 $$
which has the solution
$$ z = b^2y^2 - \frac{a^2}{3y^4} $$
Going back to $\dot y$ and rearrange
$$ \dot y = \frac{\sqrt{3b^2y^6 - a^2}}{\sqrt{3}y^2} $$
$$ \frac{y^2\dot y}{\sqrt{y^6 - \dfrac{a^2}{3b^2}}} = b $$
which can be integrated to give
$$ y = \left[\frac{a}{\sqrt{3}b}\cosh(3bt + c)\right]^{1/3} $$
and
$$ \dot{x} = \left[\frac{3b^2a}{\cosh^2(3bt+c)}\right]^{1/3} $$
unfortunately $x$ does not have a nice form
Note: A special case occurs when $a=0$, then we have a different solution set
$$ x = c_1, \quad y = c_2e^{bt} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2827823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Integral Related to Exponential Function and Modified Bessel Function 1 I'm looking for a closed form solution for the following integral:
$$S(x) = \displaystyle \int\limits_0^x r \,I_0(r) \,e^{-b^2r^2} \,dr $$
where $I_0()$ is the $0$-th order modified Bessel function of first kind, $b>0$, and $x>0$.
Thanks.
|
$\int_0^xrI_0(r)e^{-b^2r^2}~dr$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{r^{2n+1}e^{-b^2r^2}}{4^n(n!)^2}~dr$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{r^{2n}e^{-b^2r^2}}{2^{2n+1}(n!)^2}~d(r^2)$
$=\int_0^{x^2}\sum\limits_{n=0}^\infty\dfrac{t^ne^{-b^2t}}{2^{2n+1}(n!)^2}~dt$
$=-\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{t^ke^{-b^2t}}{2^{2n+1}b^{2n-2k+2}n!k!}\right]_0^{x^2}$ (according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_of_polynomials)
$=\sum\limits_{n=0}^\infty\dfrac{1}{2^{2n+1}b^{2n+2}n!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{x^{2k}e^{-b^2x^2}}{2^{2n+1}b^{2n-2k+2}n!k!}$
$=\dfrac{e^\frac{1}{4b^2}}{2b^2}-\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{x^{2k}e^{-b^2x^2}}{2^{2n+1}b^{2n-2k+2}n!k!}$
$=\dfrac{e^\frac{1}{4b^2}}{2b^2}-\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{x^{2k}e^{-b^2x^2}}{2^{2n+2k+1}b^{2n+2}(n+k)!k!}$
$=\dfrac{e^\frac{1}{4b^2}}{2b^2}-\dfrac{e^{-b^2x^2}}{2b^2}\Phi_3\left(1,1;\dfrac{1}{4b^2},\dfrac{x^2}{4}\right)$ (according to https://en.wikipedia.org/wiki/Humbert_series)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2828319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
integration by parts $\int \frac{x^{2}+4x}{x+2}\,dx$ $$\int \frac{x^{2}+4x}{x+2}\,dx$$
I have written it in this form:
$$\int \frac{(x+2)^{2}-4}{x+2}\,dx$$: on this stage I try to do integration by parts, which gets me to :
$$\frac{x^{2}}{2}+2x-4\ln\left | {x+2} \right |$$
but it's wrong for some reason. The right answer is : $$\frac{x^{2}+4x+4}{2}-4\ln\left | x+2 \right |$$
Where am I wrong?
|
Also, it should be $$\frac{x^{2}+4x+4}{2}-4\ln\left | x+2 \right |+C_1=\frac{x^{2}+4x+4}{2}-4\ln(x+2)+C_1$$ for $x>-2$ and
$$\frac{x^{2}+4x+4}{2}-4\ln\left | x+2 \right |+C_2=\frac{x^{2}+4x+4}{2}-4\ln(-x-2)+C_2$$ for $x<-2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2828775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
|
Write the product of the four consecutive integers starting at some $n-1$, so that
$$m=(n-1)n(n+1)(n+2)+1,$$
and expand:
\begin{align}
m&=(n^2-1)(n^2+2n)+1=n^2(n^2-1)+2n(n^2-1)+1 \\
&=(n^2-1)^2+2n(n^2-1)+\not 1+ n^2{-}\!\not1 \\
&= \bigl((n^2-1)+n\bigr)^2.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 8
}
|
Showing that $\sin(\sqrt{4 \pi^{2}n^{2} + x})$ converges uniformly on $[0,1]$
Suppose we are considering the sequence of functions $f_{n}(x)=\sin(\sqrt{4 \pi^{2}n^{2} + x})$ and I am having trouble showing that that $f_{n}$ converges uniformly on the interval $[0,1]$.
An idea, I've tried is to consider the Taylor series:
$$\sin(\sqrt{4 \pi^{2}n^{2} + x}) = (\sqrt{4 \pi^{2}n^{2} + x})- \frac{(\sqrt{4 \pi^{2}n^{2} + x})^{3}}{6} + O((\sqrt{4 \pi^{2}n^{2} + x})^{5})$$
but I haven't gotten anything useful as of yet.
|
Fix $x\in[0,1]$. The Mean Value Theorem shows that $$|\sqrt{4\pi^2n^2+x}-2\pi n|<\frac x{4\pi n}\le\frac1{4\pi n}.$$ So $2\pi n \le \sqrt{4\pi^2n^2+x} < 2\pi n +1/(4\pi n)$. Then apply MVT again to show$$|\sin(\sqrt{4\pi^2n^2+x})|
=|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|<\frac1{4\pi n}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2833814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Evaluate: $\lim_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$ without L'Hopitals I have to find the limit of $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$
here is my try $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}=\lim\limits_{x \to \pi/4}\frac{4(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}$
now observe that for $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$,
$2.5-(1-\cos^5(\frac{\pi}{4}- x))\leq\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5-(\sin^2 (\frac{\pi}{4}- x))$
then
$\lim\limits_{x \to \pi/4}(2.5-(1-\cos^5(\frac{\pi}{4}- x)))\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq \lim\limits_{x \to \pi/4} (2.5-(\sin^2 (\frac{\pi}{4}- x)))$
$2.5\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5$
this implies $4\lim\limits_{x \to \pi/4}\frac{(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}=4\cdot 2.5=10$
is my answer correct? wolfram gave the same answer. is there any other simpler method?
|
Observe that
$$
\cos x+\sin x=\sqrt{2}\cos(x-\pi/4)
$$
so with the substitution $x=t+\pi/4$ and noting that $\sin2x=\sin(2t+\pi/2)=\cos2t$, the limit becomes
$$
\lim_{t\to0}\frac{8-8\cos^5 t}{1-\cos2t}=
8\lim_{t\to0}\frac{1-\cos t}{1-\cos 2t}(1+\cos t+\cos^2t+\cos^3t+\cos^4t)
$$
that's easy to manage.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2835175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Final all $x$ such that $|\sqrt{x} - x| < 1$ (simple but can’t resolve) I’m a beginner so be kind with me, I tried to resolve this for 2 hours now but always give me the same wrong result. I think I’m missing something big...
Can someone help me with this?
I began by assuming $x\geq 0$ (root) and then I started to make 2 systems, one with the argument $(\sqrt{x} - x) \geq 0$ and the other with the argument $<0$.
Once the systems are resolved I find a different result...
The result given by the textbook is $0\leq x<\frac{3 + \sqrt5}{2}$
|
For $x\geq 0$ you have that
$|x-\sqrt{x}|=|(\sqrt{x}-\frac{1}{2})^2-\frac{1}{4}|<1$ if and only if
$-\frac{3}{4}<(\sqrt{x}-\frac{1}{2})^2<1+\frac{1}{4}=\frac{5}{4}$
So
$-\frac{\sqrt{5}}{2} <\sqrt{x}-\frac{1}{2}<\frac{\sqrt{5}}{2}$
and
$\frac{1}{2}-\frac{\sqrt{5}}{2} <\sqrt{x}<\frac{1}{2}+\frac{\sqrt{5}}{2}$
$0\leq x<(\frac{1}{2}+\frac{\sqrt{5}}{2})^2=\frac{3+\sqrt{5}}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2835637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?
|
The spectral mapping theorem gives $$\{\lambda^2 +\lambda : \lambda \in \sigma(X)\} = \sigma(X^2 + X) = \sigma\begin{pmatrix}1&1\\1&1\end{pmatrix} = \{0,2\}$$
It follows that we have one of the four options:
*
*$\sigma(X) = \{0,1\}$ so $\operatorname{Tr} X = 1$.
A calculation with $X = \pmatrix{a & b \\ c & 1-a}$ yields
$$\pmatrix{a+a^2+bc & 2b \\ 2c & a^2-3a+bc+2} = X^2 + X = \begin{pmatrix}1&1\\1&1\end{pmatrix}$$
so the only solution is $a = b = c = \frac12$.
*
*$\sigma(X) = \{0,-2\}$ so $\operatorname{Tr} X = -2$.
A calculation with $X = \pmatrix{a & b \\ c & -2-a}$ yields
$$\pmatrix{a+a^2+bc & -b \\ -c & a^2+3a+bc + 2} = X^2 + X = \begin{pmatrix}1&1\\1&1\end{pmatrix}$$
so the only solution is $a = b = c = -1$.
*
*$\sigma(X) = \{-1,-2\}$ so $\operatorname{Tr} X = -3$.
A calculation with $X = \pmatrix{a & b \\ c & -3-a}$ yields
$$\pmatrix{a+a^2+bc & -2b \\ -2c & a^2+5a + bc + 6} = X^2 + X = \begin{pmatrix}1&1\\1&1\end{pmatrix}$$
so $a = -\frac32$, $b = c = -\frac12$ is the only solution.
*
*$\sigma(X) = \{-1,1\}$ so $\operatorname{Tr} X = 0$.
A calculation with $X = \pmatrix{a & b \\ c & -a}$ yields
$$\pmatrix{a+a^2+bc & b \\ c & a^2-a+bc} = X^2 + X = \begin{pmatrix}1&1\\1&1\end{pmatrix}$$
so $a = 0$, $b = c = 1$ is the only solution.
So the solutions for $X$ are
$$\pmatrix{\frac12 & \frac12 \\ \frac12 & \frac12},\quad \pmatrix{-1 & -1 \\ -1 & -1},\quad \pmatrix{-\frac32 & -\frac12 \\ -\frac12 & -\frac32},\quad \pmatrix{0 & 1 \\ 1 & 0}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 1
}
|
Finding the recursive formula for calculating the solution of the linear equation system $A$x = b. Let $K$ be a field and n $\in$ $\mathbb{N}$. It's given that $A$ = $
\begin{bmatrix}
0 & p_{1} & 0 & 0 & \dots & 0 \\
q_{2} & 0 & p_{2} & 0 & \dots & 0 \\
0 & q_{3} & 0 & p_{3} & \dots & 0 \\
\vdots & & \ddots & \ddots & \ddots & \vdots \\
0 & \dots & 0 & q_{n-1} & 0 & p_{n-1} \\
0 & \dots & 0 & 0 & q_{n} & p_{n}
\end{bmatrix}
$ $\in$ $K^{n,n}$,
$b$ = $
\begin{bmatrix}
b_{1} \\
\vdots \\
b_{n} \\
\end{bmatrix}
$ $\in$ $K^{n,1}$ with $p_{i}$,$q_{i}$ $\neq$ 0 for all i = 1, $\dots$ ,n.
Find a recursive formula for calculating the solution of the linear equation system $A$x = b.
My ideas and thoughts:
I thought about writing the matrix as the following:$$
\left[\begin{array}{cccccc|c}
0 & p_{1} & 0 & 0 & \dots & 0 & b_{1} \\
q_{2}&0&p_{2}&0&\dots&0&\vdots\\
0&q_{3}&0&p_{3}&\dots&0&\vdots\\
\vdots&&\ddots&\ddots&\ddots&\vdots&\vdots \\
0 & \dots & 0 & q_{n-1}&0&p_{n-1}&\vdots \\
0&\dots&0&0&q_{n}&p_{n}&b_{n} \end{array}\right]
$$ so that \begin{equation}
=
\begin{cases}
0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1} \\
q_{2} \cdot x_{1} + 0 \cdot x_{2} + p_{2} \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{2} \\
0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3} \\
\vdots \\
0 \cdot x_{1} \dots 0 \cdot x_{n} + q_{n-1} \cdot x_{n+1} + 0 \cdot x_{n+2} + p_{n-1} \cdot x_{n+3} = b_{n-1} \\
0 \cdot x_{1} \dots 0 \cdot x_{n} + 0 \cdot x_{n+1} + q_{n} \cdot x_{n+2} + p_{n} \cdot x_{n+3} = b_{n}
\end{cases}
\end{equation} Now I could calculate the solution for each row:
(1) $0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1}$
$\Leftrightarrow$ $ p_{1} \cdot x_{2} = b_{1}$ $\quad$ $\mid$ $\div$ $p_{1}$
$\Leftrightarrow$ $x_{2}$ = $\frac{b_{1}}{p_{1}}$
(3) $ 0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3}$
$\Leftrightarrow$ $q_{3} \cdot x_{2} + p_{3} \cdot x_{4} = b_{3}$
$\Leftrightarrow$ $q_{3} \cdot \frac{b_{1}}{p_{1}} + p_{3} \cdot x_{4} = b_{3}$ $\quad$ $\mid$ $- q_{3} \cdot \frac{b_{1}}{p_{1}}$
$\Leftrightarrow$ $p_{3} \cdot x_{4} = b_{3} - q_{3} \cdot \frac{b_{1}}{p_{1}}$$\quad$ $\mid$ $\div p_{3}$
$\Leftrightarrow$ $x_{4}$ = $\frac{b_{3}-q_{3} \cdot \frac{b_{1}}{p_{1}}}{p_{3}}$
$\Leftrightarrow$ $\frac{p_{1} \cdot b_{3}-q_{3} \cdot b_{1}}{p_{1} \cdot p_{3}}$
And so on $\dots$
I don't know how to continue at this point to find a recursive formula for calculating the solution of the linear equation system. My ideas did not really lead me anywhere (I don't even think that they are correct) and im not sure if I even understand the matrix correctly. So how should I start instead?
Any hints guiding me to the right direction I much appreciate.
|
Here are $A^{-1}$ for $n$ from $1$ to $4$:
$$(1/p_1),\ \pmatrix{ -p_2/(p_1 q_1) & 1/q_1\cr 1/p_1 & 0\cr},\
\pmatrix{p_2 q_2/(p_1 q_1 p_3) & 1/q_1 & -p_2/(q_1 p_3) \cr 1/p_1 & 0 & 0\cr -q_2/(p_1 p_3) & 0 & 1/p_3},\
\pmatrix{-p_2 q_2 p_4/(p_1 q_1 p_3 q_3) & 1/q_1 & p_2 p_4/(q_1 p_3 q_3) & -p_2/(q_1 q_3)\cr
1/p_1 & 0 & 0 & 0\cr q_2 p_4/(p_1 p_3 q_3) & 0 & -p_4/(p_3 q_3) & 1/q_3\cr
-q_2/(p_1 p_3) & 0 & 1/p_3 & 0} $$
See a pattern?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2837975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
$(\dim_K\ker f)^2 + (\dim_K\operatorname{im} f)^2 = (\dim_K X )^2$ So I have $X$, $Y$ vector spaces over a field $K$ and $f : X\to Y$ linear operator.
The problem asks me to define $\ker f$. Next, it says that given
$(\dim_K\ker f)^2 + (\dim_K\operatorname{im} f)^2 = (\dim_K X)^2$ and $\dim_K\ker f \neq 0$
I must show that $\dim_K\operatorname{im} f = 0$
Any ideas?
|
By the Rank-Nullility Theorem:
$$\dim_K \ker f + \dim_K \operatorname{im} f = \dim_K X\tag{1}$$
Now
$$(\dim_K\ker f)^2 + (\dim_K\operatorname{im} f)^2 = (\dim_K X)^2\tag{2}$$
and $\dim_K\ker f \neq 0$. Leting $\dim_K \ker f=a$, $\dim_K \operatorname{im} f =b$, and $\dim_K X=c$, for some $a$, $b$, $c\in\Bbb{Z}_{\ge0}$then by $(1)$,
$$a+b=c\tag{3}$$
But by $(2)$,
$$a^2+b^2=c^2\tag{4}$$
The question then is: is $(4)$ possible in light of $(3)$ with $b\neq0$.
By $(3)$ you get
$$a^2=c^2-2bc+b^2\tag{5}$$
By $(4)$ you get
$$a^2=c^2-b^2\tag{6}$$
So by $(5)$ and $(6)$
$$c^2-2bc+b^2=c^2-b^2\implies b^2=bc$$
or $b=c$, which is impossible by $(3)$ as $\dim_K\ker f =a>0$, and so we must have $\dim_K \operatorname{im} f=b=0$ as required.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2839882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$5^x+7^x+11^x=(5+1)^x+(7+1)^x+(11-2)^x$$
and tried to think of some AM-GM mean inequality or to divide everything by $11^x$, but those don't seem like the way to go. Any hints?
|
Another elemenatary solution, using that $x$ is supposed to be a natural number. It even uses a technique the OP considered:
The equation is equivalent to
$$\left(\frac{5}{11}\right)^x + \left(\frac{7}{11}\right)^x +\left(\frac{11}{11}\right)^x = \left(\frac{6}{11}\right)^x +\left(\frac{8}{11}\right)^x +\left(\frac{9}{11}\right)^x$$
All terms are positive and the left hand side contains a summand 1 in the form of $\left(\frac{11}{11}\right)^x$.
From the terms on the right hand side, $\left(\frac{9}{11}\right)^x$ is the biggest, but it will of course still decrease for increasing $x$. Using a calculator will show you that $\left(\frac{9}{11}\right)^6 < \frac13$.
That means for $x \ge 6$, the right hand side consists of the sum of 3 values, the highest of which is less than $\frac13$. That means the right hand side is less than 1, while the left hand side is bigger than 1, leading to a contradiction.
Now 'only' the cases $x=2,3,4,5$ need to be checked by hand, and they don't lead to the equation being fullfilled.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2840394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
}
|
When is $n^2-1$ a sum of two squares? I am trying to work out when $n^2-1$ is a sum of two squares. Is there a formula for such $n$? I have found $n=1$, $n=3$ and $n=9$ so far but am struggling to find a pattern that will generalise. If there is no such pattern then how might I prove that there are infinitely many such $n$ that have this property?
E.g. For $n=3$ we have $n^2-1=9-1=8=2^2+2^2$ and for $n=9$ we have $n^2-1=9^2-1=80=8^2+4^2$.
|
Here is a proof that there are infinitely many such $n$. Note that for any $m\in\mathbb{N}$, both $m^2=m^2+0^2$ and $m^2+1=m^2+1^2$ are a sum of two squares. It follows that both $2m^2$ and $2m^2+2$ are a sum of two squares (since a product of two sums of two squares is a sum of two squares). Letting $n=2m^2+1$, then, $n^2-1=(n-1)(n+1)=2m^2(2m^2+2)$ is a sum of two squares. So, $n^2-1$ is a sum of two squares whenever $n$ has the form $2m^2+1$.
More explicitly, we have $$(2m^2+1)^2-1=4m^4+4m^2=(2m^2)^2+(2m)^2.$$
In some sense, all examples must come from an idea similar to this one (and indeed, that is how I came up with this one). If $n^2-1$ is a sum of two squares, then mod $4$ considerations require that $n$ is odd. We then have $n^2-1=4a(a+1)$ where $a=\frac{n-1}{2}$ is an integer. By the classification of sums of two squares, $n^2-1$ is a sum of two squares iff $a$ and $a+1$ are both sums of two squares (since $a$ and $a+1$ share no prime factors, and in particular their prime factors that are $3$ mod $4$ are distinct). So every example comes from a pair of consecutive integers that are both sums of two squares; my method above just came from the observation that $m^2$ and $m^2+1$ always gives you such a pair.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2841530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction:
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, thanks!
|
Hint: by polynomial euclidean division $\;\dfrac{4x^2+8x+13}{6(x+1)} = \dfrac{1}{6}\left(4(x+1) + \dfrac{9}{x+1}\right) \ge \ldots\,$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2842418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
If $a_n\sim b_n$ decrease to $0$ and $a_{n+1}b_n-a_nb_{n+1}$ changes sign i.o. , does it imply the same for $(1-a_{n+1})(1-b_n)-(1-a_n)(1-b_{n+1})$?
Suppose $a_n<b_n$ for all $n$, both are strictly decreasing to $0$ and $a_n\sim b_n$. If $A_n=a_{n+1}b_n-a_nb_{n+1}$ changes sign infinitely often, does it follow that $B_n=(1-a_{n+1})(1-b_n)-(1-a_n)(1-b_{n+1})$ also does?
In some sense, the implication should not be there, because while $A_n<0$ does not seem to be crucial for $B_n$, I have found that whenever $A_n>0$ we have $B_n<0$: the former is equivalent to $$a_{n+1}>a_n\frac{b_{n+1}}{b_n}$$ and since for large enough $n$, $0<a_n<1, a_n<b_n<1, 0<b_{n+1}<b_n$ this is stronger than $$a_{n+1}>1-\frac{1-a_n}{1-b_n}(1-b_{n+1}),$$ equivalent to $B_n<0$. $A_n<0$ does not seem to be crucial because it should be possible to have this happen without making $a_{n+1}$ so small that $B_n$ becomes positive.
On the other hand, maybe the fact that $A_n$ changes sign infinitely often gives more weight to the instances of $A_n<0$. Not for too long, but I have tried to construct some simple $A_n,B_n$ contradicting the claim and I have not found them yet. Are there any, with $B_n<0$? Besides, how does the situation change if $a_n-b_n$ is also assumed to change sign infinitely often?
|
$\def\peq{\mathrel{\phantom{=}}{}}$Take an arbitrary $c > 1$ and define$$
a_n = \frac{1}{c^n + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2}}}, \quad b_n = \frac{1}{c^n}. \quad \forall n \geqslant 1
$$
It is easy to verify that $a_n < b_n$, $a_n \sim b_n\ (n → ∞)$, and both $\{a_n\}$ and $\{b_n\}$ strictly decrease to $0$. Now define$$
x_n = \frac{1}{a_n} = c^n + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2}}, \quad y_n = \frac{1}{b_n} = c^n, \quad z_n = x_n - y_n,\quad \forall n \geqslant 1
$$
then\begin{align*}
A_n > 0 &\Longleftrightarrow \frac{x_{n + 1}}{x_n} < \frac{y_{n + 1}}{y_n} \Longleftrightarrow \frac{c^{n + 1} + c^{\frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2}}}{c^n + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2}}} < c\\
&\Longleftrightarrow c^{n + 1} + c^{\frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2}} < c^{n + 1} + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2} + 1}\\
&\Longleftrightarrow \frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2} < \frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2} + 1\\
&\Longleftrightarrow n \text{ is even}.
\end{align*}
and\begin{align*}
B_n < 0 &\Longleftrightarrow (a_n - a_{n + 1}) - (b_n - b_{n + 1}) + (a_{n + 1} b_n - a_n b_{n + 1}) < 0\\
&\Longleftrightarrow y_n y_{n + 1}(x_{n + 1} - x_n) - x_n x_{n + 1}(y_{n + 1} - y_n) + (x_n y_{n + 1} - x_{n + 1} y_n) < 0.
\end{align*}
Since\begin{align*}
&\peq y_n y_{n + 1}(x_{n + 1} - x_n) - x_n x_{n + 1}(y_{n + 1} - y_n) + (x_n y_{n + 1} - x_{n + 1} y_n)\\
&= y_n y_{n + 1}\bigl( (y_{n + 1} - y_n) - (z_{n + 1} - z_n) \bigr) - (y_n + z_n)(y_{n + 1} + z_{n + 1})(y_{n + 1} - y_n)\\
&\peq + \bigl( (y_n + z_n) y_{n + 1} - (y_{n + 1} + z_{n + 1}) y_n \bigr)\\
&= \color{blue}{y_n y_{n + 1}}\bigl( \color{red}{(y_{n + 1} - y_n)} - (z_{n + 1} - z_n) \bigr) - (\color{blue}{y_n y_{n + 1}} + y_n z_{n + 1} + y_{n + 1} z_n + z_n z_{n + 1})\color{red}{(y_{n + 1} - y_n)}\\
&\peq + \bigl( (y_n + z_n) y_{n + 1} - (y_{n + 1} + z_{n + 1}) y_n \bigr)\\
&= y_n y_{n + 1} (z_{n + 1} - z_n) - (y_n z_{n + 1} + y_{n + 1} z_n + z_n z_{n + 1})(y_{n + 1} - y_n) + (y_{n + 1} z_n - y_n z_{n + 1})\\
&= y_n y_{n + 1} (z_{n + 1} - z_n) - (y_n z_{n + 1} + y_{n + 1} z_n) y_{n + 1} + (y_n z_{n + 1} + y_{n + 1} z_n) y_n\\
&\peq - z_n z_{n + 1}(y_{n + 1} - y_n) + (y_{n + 1} z_n - y_n z_{n + 1})\\
&= y_n^2 z_{n + 1} - y_{n + 1}^2 z_n - z_n z_{n + 1}(y_{n + 1} - y_n) + (y_{n + 1} z_n - y_n z_{n + 1})\\
&= (y_n^2 - y_n) z_{n + 1} - (y_{n + 1}^2 - y_{n + 1}) z_n - z_n z_{n + 1}(y_{n + 1} - y_n),
\end{align*}
and\begin{gather*}
\frac{y_{n + 1}^2 - y_{n + 1}}{y_n^2 - y_n} = \frac{y_{n + 1}}{y_n} \cdot \frac{y_{n + 1} - 1}{y_n - 1} > c^2,\\
\frac{z_{n + 1}}{z_n} = c^{\left( \frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2} \right) - \left( \frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2} \right)} = c^{\frac{3}{4} \left( \left[ \frac{n + 1}{2} \right] - \left[ \frac{n}{2} \right] \right) + \frac{1}{2}} < c^2\\
\Longrightarrow \frac{y_{n + 1}^2 - y_{n + 1}}{y_n^2 - y_n} > \frac{z_{n + 1}}{z_n} \Longrightarrow (y_n^2 - y_n) z_{n + 1} - (y_{n + 1}^2 - y_{n + 1}) z_n < 0,
\end{gather*}
then $B_n < 0$ for all $n \geqslant 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2845464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
Geometric or matrix intuition on $A(A + B)^{-1}B = B (A + B)^{-1} A$ I am curious about a seemingly simple identity in matrix algebra. Though matrix multiplication is not commutative (the classic example of noncommutativity, it does allow a commutativity of sorts around a very specific third matrix:
$$
\color{blue}{A (A + B)^{-1} B} = \color{blue}{B (A + B)^{-1} A}
$$
There is a very simple algebraic proof:
$$
\begin{eqnarray}
\color{blue}{A (A + B)^{-1} B} + \color{red}{B (A + B)^{-1} B} &=& (A + B)(A + B)^{-1} B\\
&=& B\\
&=& B (A + B)^{-1} (A + B) \\
&=& \color{blue}{B (A + B)^{-1} A} + \color{red}{B (A + B)^{-1} B}\\
\end{eqnarray}
$$
because matrix addition is commutative. The identity follows by cancelling $B (A + B)^{-1} B$. (If there is a simpler linear proof, say, without needing cancellation, please say so. That extra weird matrix, while obviously looking right and obviously doing the job, just pops out of nowhere. Or does it?)
Algebra is blind manipulation of symbols. The identity holds in any abstract ring with multiplicative inverses. But for a given model, the identity says something about that model. I've only seen the identity in the context of matrices, but I don't see what's so special about it there.
What does this identity do for matrices?
Is there something, in matrix theory, for which this is special? Does it ever really come up in proofs? $(A+B)^{-1}$ can't be the only matrices that allow such quasi-commutativity, can they? Is there a visualization or geometric interpretation or a meaningful anything interpretation?
|
Just put
$$
A = \left( {A + B} \right) - B
$$
to get
$$
\eqalign{
& \left( {\left( {A + B} \right) - B} \right)\left( {A + B} \right)^{\, - \,1} B = B\left( {A + B} \right)^{\, - \,1} \left( {\left( {A + B} \right) - B} \right) \cr
& \quad \Downarrow \cr
& B - B\left( {A + B} \right)^{\, - \,1} B = B - B\left( {A + B} \right)^{\, - \,1} B \cr}
$$
Maybe a better insight on what is going on can be obtained by putting
$$
\left\{ \matrix{
A = {1 \over 2}\left( {A + B} \right) - {1 \over 2}\left( {B - A} \right) \hfill \cr
B = {1 \over 2}\left( {A + B} \right) + {1 \over 2}\left( {B - A} \right) \hfill \cr} \right.
$$
which then gives
$$
\eqalign{
& {1 \over 4}\left( {\left( {A + B} \right) - \left( {B - A} \right)} \right)\left( {A + B} \right)^{\, - \,1} \left( {\left( {A + B} \right) + \left( {B - A} \right)} \right) = \cr
& = {1 \over 4}\left( {\left( {A + B} \right) + \left( {B - A} \right)} \right)\left( {A + B} \right)^{\, - \,1} \left( {\left( {A + B} \right) - \left( {B - A} \right)} \right) = \cr
& \quad \Downarrow \cr
& \left( {I - \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} } \right)\left( {\left( {A + B} \right) + \left( {B - A} \right)} \right) = \cr
& = \left( {I + \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} } \right)\left( {\left( {A + B} \right) - \left( {B - A} \right)} \right) \cr
& \quad \Downarrow \cr
& \left( {A + B} \right) - \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} \left( {B - A} \right) = \cr
& = \left( {A + B} \right) - \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} \left( {B - A} \right) \cr}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2852002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Is there a way to know if a row reduction of a matrix has been done correctly? I'm an undergrad taking the class of "Linear algebra 1". I came across a problem:
sometimes we need to apply Gaussian elimination for matrices. Very quickly this skill is not much necessary as it's not a thinking skill but purely Technic.
Yet, often in exams there's a question that requires you to apply row reduction
to a matrix.
I am looking for a way to know if the Gaussian elimination has been done properly, meaning - with no Calculation mistakes, other than going over all the steps and check that the Calculation has been done correctly. as this processes will double the time I will spend on a given question, and due to the lack of time in a big course exam - which is also very stressful - such method could be much helpful for me.
note: we're allowed to use a simple scientific calculator (not a graph calculator)
|
The actual algorithm for Gaussian Elimination looks like this.
I answered a similar answer showing how to perform the LU decomposition which is Gaussian Elimination without pivoting The purpose is to zero out the row beneath is with $ \ell_{jk}$. That is why you have the operation below $ u_{j,k:m} = u_{j,k:m} - \ell_{jk} u_{k,k:m}$ It is subtracting off the ratio you just computed.
Which yielded this.
Suppose that
$$ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ A = LU $$
$$ U =A, L=I$$
$$ k=1,m=3,j=2$$
$$\ell_{21} = \frac{u_{21}}{u_{11}} = \frac{a_{21}}{a_{11}} = 3 $$
$$ u_{2,1:3} = u_{2,1:3} - 3 \cdot u_{1,1:3} $$
Then we're going to subtract 3 times the 1st row from the 2nd row
$$ \begin{bmatrix} 3 & 5 & 6 \end{bmatrix} - 3 \cdot \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3\end{bmatrix} $$
Updating each of them
$$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
$$k=1,j=3,m=3 $$
$$\ell_{31} = \frac{u_{31}}{u_{11}} = \frac{-2}{1} = -2 $$
$$ u_{3,1:3} = u_{3,1:3} +2 \cdot u_{1,1:3} $$
Then we add two times the first row to the third row
$$ \begin{bmatrix} -2 & 2 & 7 \end{bmatrix} + 2 \cdot \begin{bmatrix} 1 & 1& 1 \end{bmatrix} = \begin{bmatrix}0 & 4 & 9 \end{bmatrix} $$
Updating
$$ U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} $$
$$ k=2, j=3,m=3 $$
$$ \ell_{32} = \frac{u_{32}}{u_{22}} = \frac{4}{2} = 2$$
We're subtracting out little blocks
$$ u_{3,2:3} = u_{3,2:3} - 2 \cdot u_{2,2:3} $$
$$ \begin{bmatrix} 4 & 9 \end{bmatrix} - 2 \cdot\begin{bmatrix} 2& 3 \end{bmatrix} = \begin{bmatrix} 0 & 3 \end{bmatrix} $$
Updating
$$ U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix} $$
It now terminates
$$ A = LU $$
$$ \underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}}_{A} = \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}}_{L} \underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}}_{U} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2852287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
}
|
Is there a way to simplify $\prod_{i=1}^n\cos(a^i\theta)$, where $a<1$? I recently came upon the following expression in an attempt at getting a closed-form solution for a recursive relation:
$$\prod_{i=1}^n \cos(a^i\theta)$$
where $a<1$. Is there a way to make this product into a sum or otherwise make it simpler, or approximate it? In particular for the problem I was looking at, $a$ was $\frac{3}{4}$ and $\theta$ was $\frac{\pi}{4}$. Thanks!
|
Note this is @saulspatz answer with the computations added (I couldn't resist).
With
$$
f(\theta)=\prod_{k=1}^n\cos(a^k\theta)
$$
Then noting
$$
-\int_0^{a^k\theta}\tan x\, \mathrm{dx}=\log\cos(a^k\theta)
$$
we have on taking the logarithm
\begin{align*}
\log f(\theta) &= \sum_{k=1}^n\log\cos(a^k\theta)=\log\cos(a\theta)+\dotsb+\log\cos(a^n\theta)\\
&=-\sum_{k=1}^n\int_0^{a^k\theta}\tan x \,\mathrm{dx}\\
&=-\sum_{k=1}^n\int_0^{a^k\theta}x + \frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\dots\, \mathrm{dx}\quad\text{(by the Maclaurlin series for $\tan{x}$)}\\
&=-\left(\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^4}{12}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^6}{45}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{17x^8}{2520}\Big{]}_0^{a^k\theta}
+\dots\right)\\
\end{align*}
Note each of these sums are all finite geometric sums: For the first
\begin{align*}
\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta} &=\frac{\theta^2}{2}(a^2+a^4+\dotsb+a^{2n})\\
&=\frac{a^2\theta^2}{2}(1+a^2+\dotsb+a^{2(n-1)})\\
&= \frac{a^2\theta^2}{2}\cdot\frac{1-a^{2n}}{1-a^2}=\frac{a^2\theta^2}{2}\cdot\frac{(1-a^{n})(1+a^{n})}{(1-a)(1+a)}\\
\end{align*}
Hence
\begin{align*}
\log f(\theta) &=-\left(\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^4}{12}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^6}{45}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{17x^8}{2520}\Big{]}_0^{a^k\theta}
+\dots\right)\\
&=\frac{a^2\theta^2}{2}\cdot\frac{a^{2n}-1}{1-a^2}+\frac{a^4\theta^4}{12}\cdot\frac{a^{4n}-1}{1-a^4}+\frac{a^6\theta^6}{45}\cdot\frac{a^{6n}-1}{1-a^6}+\frac{17a^8\theta^8}{2520}\cdot\frac{a^{8n}-1}{1-a^8}+\dotsb
\end{align*}
Now take the exponential of the sum to compute $f(\theta)$:
\begin{align*}
f(\theta) &=
\exp\left(\frac{a^2\theta^2}{2}\cdot\frac{a^{2n}-1}{1-a^2}\right)\cdot
\exp\left(\frac{a^4\theta^4}{12}\cdot\frac{a^{4n}-1}{1-a^4}\right)\cdot
\exp\left(\frac{a^6\theta^6}{45}\cdot\frac{a^{6n}-1}{1-a^6}\right)\\
&\qquad\qquad\qquad\qquad\cdot\exp\left(\frac{17a^8\theta^8}{2520}
\cdot\frac{a^{8n}-1}{1-a^8}\right)\dotsb\\
&=\prod_{k=1}^{\infty}\exp\left(\frac{\tan^{(2k-1)}(0)\,a^{2k}\theta^{2k}}{2k(2k-1)!}\cdot\frac{a^{2kn}-1}{1-a^{2k}}\right)
\end{align*}
where $\tan^{(2k-1)}(0)$ is the $(2k-1)$th derivative of $\tan x$ evaluated at $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2852547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$
I solved it as follows but I seek better or quicker way:
$\text{ Assume }x, y \in \mathbb Z^+\\ 1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x}=1+\frac{y+1}{y}+4 \\
\Rightarrow \left(1+\frac{x}{y}\right)\left(1+\frac{y+1}{x}\right)=6+\frac{1}{y}\\
\Rightarrow (x+y)(x+y+1)=x(6y+1)\\
\Rightarrow x\mid (x+y) \;\text{ or }\; x\mid (x+y+1)\\
\Rightarrow x\mid y \;\text{ or }\; x\mid (y+1)\\
\text{Put}\; y=nx ,n \in \mathbb Z^+\;\Rightarrow\; \frac{x}{nx} +\frac{nx+1}{x}=4 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x}=4-n \;\rightarrow\;(1)\\
\text{But}\; \frac{1}{n} +\frac{1}{x} \gt 0 \;\Rightarrow\; 4-n \gt 0 \;\Rightarrow\; n \lt 4\\
\text{Also}\; \frac{1}{n},\frac{1}{x}\le 1 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x} \le 2 \;\Rightarrow\; 4-n \le 2 \;\Rightarrow\; n \ge 2\\
\;\Rightarrow\; n=2 \;\text{ or }\; 3, \;\text{substituting in eq. (1), we find no integral values for } x.\\ \text{The same for the other case.}\\
$
So is there any other better or intelligent way to get this result?
|
For $x,y\in \Bbb Z^+$ we have $$\frac {x}{y}+\frac {y+1}{x}=4\implies x^2-4xy+y^2+y=0\implies x=2y\pm \sqrt {3y^2-y}\implies$$ $$\implies \exists z\in \Bbb Z^+\;( z^2=3y^2-y=y(3y-1))\implies$$ $$ \implies\exists a,b \in \Bbb Z^+\;( y=a^2 \land 3y-1=b^2)\implies$$ $$\implies\exists a,b\in \Bbb Z^+\;(3a^2-1=b^2)\implies$$ $$ \implies \exists b\in \Bbb Z^+\;(b^2\equiv -1 \mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $z\ne 0$ because $y\in \Bbb Z^+\implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist because $y$ and $3y-1$ are co-prime members of $\Bbb Z^+$ and their product is the square of the positive integer $z.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2853669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Angles on a point inside a triangle Let $ABC$ be an isosceles triangle with $AB=AC$ and $∠BAC = 100$. A point $P$
inside the triangle $ABC$ satisfies that $∠CBP=35$ and $∠PCB= 30$. Find the
measure, in degrees, of angle $∠BAP$. Attached is the figure of the triangle
I tried to Angle Chase but it seemed true for all values of $BAP$. I then tried using the sine law. In the triangle $PBC$, We have
$$\frac{PB \sin(35)}{\sin(30)}= PC$$
Trying it with triangles $APB$ ($x= BAP$) and the fact that they are isosceles
$$\frac{PC\sin(x+70)}{\sin(100-x)}=\frac{PB\sin(175-x)}{\sin (x)}$$
Which becomes,
$$\frac{\sin(35)\sin(x+70)}{\sin(30)\sin(100-x)} = \frac{\sin(175-x)}{\sin(x)}$$
Where in I don't know how to solve it. Other methods are welcome.
|
Let $x=\angle APB$, $a=AB=AB$, $b=AP$. We have
$$\frac{\sin x}{a}=\frac{\sin 5^\circ }{b}$$
and
$$\frac{\sin \left(245^{\circ} - x\right)}{a}=\frac{\sin 10^\circ }{b}$$
We have then:
$$\frac{\sin \left(245^{\circ} - x\right)}{\sin x}=\frac{\sin 10^\circ }{\sin 5^\circ }$$
Using the formula
$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$$
we obtain:
$$\cot x = \frac{\sin 10^\circ }{\sin 245^\circ \sin 5^\circ } + \cot 245^\circ$$
Thus
$$x = \text{arccot}\left(\frac{\sin 10^\circ }{\sin 245^\circ \sin 5^\circ } + \cot 245^\circ\right)$$
After wolframAlpha we have
$$x=150^{\circ}$$
Now it's easy to show, that $$\angle PAB = 25^\circ$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2854384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably I am mistaken in solving.
|
Hi I don't think this is true. For example if $a=b=c=d=1$ then we have $4/3 >= 4$ which is false.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2854810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
If $ab \mid c(c^2-c+1)$ and $c^2+1 \mid a+b$ then prove that $\{a, b\}=\{c, c^2-c+1 \}$ If $ab \mid c(c^2-c+1)$ and $c^2+1 \mid a+b$ then prove that $\{a, b\}=\{c, c^2-c+1 \}$ (equal sets), where $a$, $b$, and $c$ are positive integers.
This is math contest problem (I don't know the source). I was struggling to solve this, but I cannot find a way to make it easier. Can you help me?
All I did was like this:
$$a+b=d(c^2+1)=d(c^2-c+1+c) \\ c(a+b)=dc(c^2-c+1)+dc^2 \\ c(a+b)=deab+dc^2=d(eab+c^2)$$
I suppose this is a quadratic in $c$ and determine its discriminant. But it doesn't make the problem any easier!
|
HINT.-For some positive integers $m,n$ we have
$$\begin{cases}a+b=m(c^2+1)\\abn=c(c^2-c+1)\end{cases}\Rightarrow c^2-\frac{a+b}{m}c+abn=0$$
If $m$ and $n$ are equal to $1$ then $c$ is root of the equation $X^2-(a+b)X+ab=0$ from which clearly $c\in\{a,b\}$. It easily follows that $a$ or $b$ is equal to $c^2-c+1$.
If both $m$ and $n$ is greater than $1$ then $c$ is root of $X^2+\dfrac{a+b}{m}X+abn=0$ where the sum of the two roots shrinks while the product is enlarged. This is not possible for positive integers.
I leave for the O.P. the case just one of the $m,n$ is equal to $1$
Finally the proof is clear (it has been done already above) because necessarily the positive integers $m$ and $n$ are equal to $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2860229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
}
|
Adding always the half of the previous number to a number beginning with 1... This question blows my mind.
Let's say we start at 1 and keep adding always 0.5 * the previous number.
1 + 0.5 + 0.25 + 0.125 + .....
The question is: will it reach infinity?
I really don't know, kept thinking about this alot.
|
Consider calculating the sum in steps, at each step adding the next term. At the $n^{th}$ step, you're adding $1/2^n$ (adding $1$ is the $0^{th}$ step). Then at the $n^{th}$ step, the total sum is the following:
$$
s_n = 1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^n}
$$
Now remark that you can perform the following calculation:
\begin{align*}
1 + \frac{1}{2} s_n &= 1 + \frac{1}{2}
\left(
1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^n}
\right)
\\
&=
1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^n} + \frac{1}{2^{n+1}}
\\
&= s_n + \frac{1}{2^{n+1}}
\end{align*}
Then if you solve for $s_n$ using elementary algebra (which you can do since $s_n$ is finite), you end up with:
$$
s_n = 2 - \frac{1}{2^n} < 2
$$
What we have just shown is that $s_n \leq 2$ for any $n$ that we choose. In other words, no matter how many terms we add, the resulting sum will always be less than $2$. So, the sum will not reach infinity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2862709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Finding the minimum value of $\log _d a + \log _bd + \log _ac + \log _c b$ The question says to find the minimum value of $$\log _da + \log _bd + \log _ac + \log _cb$$ Given, $a,b,c,d \; \in R^+ -\Bigl(1\Bigl)$
My approach:
I used the AM-GM inequality and so we can see that $$\frac{\log _da + \log _bd + \log _ac + \log _cb}{4} \geq \Biggl(\log _da \cdot \log _bd \cdot \log _ac \cdot \log_cb\Biggl)^\frac{1}{4}$$ So the minimum value is, $$4\Biggl(\log _da \cdot \log _bd \cdot \log _ac \cdot \log _cb\Biggl)^\frac{1}{4}$$
Now, I wrote it as, $$\Biggl(\frac{\log_da}{\log_db} \cdot \frac{\log_ca}{\log_cb}\Biggl)^\frac{1}{4}$$
I am not sure what do to after this point. I also feel like we can use HM inequality but that will only give us,
$$\log _da + \log _bd + \log _ac + \log _cb = \log _ad + \log _db + \log _ca + \log _bc $$
|
Hint.
Assuming positive values, You can use also
$$
\log_y x = \frac{\log x}{\log y}
$$
then
$$
\frac{\log a}{\log d}+\frac{\log d}{\log b}+\frac{\log c}{\log a}+\frac{\log b}{\log c}\ge 4
$$
NOTE
Another approach.
Assuming $a\ne 0,b\ne 0,c\ne 0,d\ne 0\;\;$ and calling
$$
f(a,b,c,d) = \frac ad+\frac db+\frac ca+\frac bc
$$
we have
$$
\nabla f = \left\{
\begin{array}{c}
\frac{1}{d}-\frac{c}{a^2}=0 \\
\frac{1}{c}-\frac{d}{b^2}=0 \\
\frac{1}{a}-\frac{b}{c^2}=0 \\
\frac{1}{b}-\frac{a}{d^2}=0 \\
\end{array}
\right.
$$
and solving
$$
a = b = c = d \;\;\; \mbox{which gives}\;\;\;\; f(a,b,c,d) \ge 4
$$
because
$$
H = \left(
\begin{array}{cccc}
\frac{2 c}{a^3} & 0 & -\frac{1}{a^2} & -\frac{1}{d^2} \\
0 & \frac{2 d}{b^3} & -\frac{1}{c^2} & -\frac{1}{b^2} \\
-\frac{1}{a^2} & -\frac{1}{c^2} & \frac{2 b}{c^3} & 0 \\
-\frac{1}{d^2} & -\frac{1}{b^2} & 0 & \frac{2 a}{d^3} \\
\end{array}
\right)
$$
and for $a=b=c=d$
$$
H = \left(
\begin{array}{cccc}
2 & 0 & -1 & -1 \\
0 & 2 & -1 & -1 \\
-1 & -1 & 2 & 0 \\
-1 & -1 & 0 & 2 \\
\end{array}
\right)
$$
with eigenvalues $4,2,2,0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2863392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Between any two powers of $5$ there are either two or three powers of $2$ Is this statement true?
Between any two consecutive powers of $5$, there are either two or
three powers of $2$.
I can see that this statement is true for cases like
$$5^1 < 2^3 < 2^4 < 5^2$$
or
$$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$
But I am having a trouble figuring out the proof through generalization.
Could somebody help me?
|
Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.
Let $5^a$ and $5^{a+1}$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^{a+1}$.
Then we have $2^{b-1} \leq 5^a < 2^b$ and $2^c < 5^{a+1} \leq 2^{c+1}$.
From this we get
$$\frac{2^{c}}{2^{b}} < \frac{5^{a+1}}{5^a} \leq \frac{2^{c+1}}{2^{b-1}} $$
or, equivalently,
$$2^{c-b} < 5 \leq 4 \cdot 2^{c-b}.$$
This inequality can be rewritten as $ \frac{5}{4} \leq 2^{c-b} < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^{a+1}$ are either $2^b, 2^{b+1} = 2^c$, or $2^b, 2^{b+1}, 2^{b+2} = 2^c.$ This is what you wanted.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2863510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
}
|
$|a^{2x}+a^{x+2}-1|\ge 1$ equation for positive a If $|a^{2x}+a^{x+2}-1|\ge 1$ for all values of a(a>0), $\ne 1$. Find the domain of x.
I tried to substitute $a^x=t$ and used the following $|
t^{2}+a^2t-1|\ge 1$ but it is getting complicated.
|
Assuming that $a, t > 0$,
If $t^2 + a^2 t - 1 < 0$, then
\begin{align}
|t^2 + a^2 t - 1| &\ge 1 \\
1 - a^2 t - t^2 &\ge 1 \\
t^2 + a^2 t &< 0
\end{align}
Which has no positive real solution.
If $t^2 + a^2 t - 1 \ge 0$, then
\begin{align}
|t^2 + a^2 t - 1| &\ge 1 \\
t^2 + a^2 t - 1 &\ge 1 \\
(t + \frac 12 a^2)^2 &\ge 2 + \frac 14 a^4 \\
t + \frac 12 a^2 &\ge \frac 12\sqrt{8 + a^4} \\
t &\ge \frac 12\left(\sqrt{8 + a^4} - a^2 \right) \\
t &\ge \dfrac{4}{\sqrt{8 + a^4} + a^2} \\
a^x &\ge \dfrac{4}{\sqrt{8 + a^4} + a^2} \\
x \ln a &\ge \ln 4 -\ln(\sqrt{8 + a^4} + a^2) \\
x &\ge \dfrac{\ln 4 - \ln(\sqrt{8 + a^4} + a^2)}{\ln a} \\
&\text{OR} \\
a^x &\ge \frac 12\left(\sqrt{8 + a^4} - a^2 \right) \\
x \ln a &\ge \ln{\left(\sqrt{8 + a^4} - a^2 \right)} - \ln 2 \\
x &\ge \dfrac{\ln{\left(\sqrt{8 + a^4} - a^2 \right)}}{\ln a}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2864124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Integration of $\frac{1}{x^2-a^2}$ by trigonometric substitution?
$$\int \frac{1}{x^2-a^2}dx$$
Now, I know this can be done by splitting the function into two integrable functions,
$\displaystyle\dfrac{1}{2a}\int \bigg(\dfrac{1}{x-a} - \dfrac{1}{x+a}\bigg)dx$
And then doing the usual stuff.
My question is, how can we do this by using trigonometric substitution?
The only thing that gets in my mind is $x=a\sec\theta$, but then got stuck on proceeding further.
Any help would be appreciated.
|
We have
$$I=\int \frac{1}{x^2-a^2}dx$$
Let $x=a\cos\theta$, then $dx=-a\sin\theta d\theta$.
\begin{align}
I
&= \int \frac{1}{x^2-a^2}dx \\
&= \int \frac{-a\sin\theta}{a^2(\cos^2\theta-1)}d\theta \\
&=\int \frac{-a\sin\theta}{-a^2\sin^2\theta}d\theta \\
&= \frac1a\int \csc\theta d\theta \\
&= -\frac1a\ln\left\lvert\csc\theta+\cot\theta\right\rvert +C\\
&= -\frac1a\ln\left\lvert\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}\right\rvert +C\\
&= -\frac1a\ln\left\lvert\frac{1+x/a}{\pm\sqrt{1-x^2/a^2}}\right\rvert +C\\
&= -\frac1a\ln\left\lvert\frac{\sqrt{1+x/a}\sqrt{1+x/a}}{\sqrt{1-x/a}\sqrt{1+x/a}}\right\rvert +C\\
&= -\frac1{2a}\ln\left\lvert\frac{1+x/a}{1-x/a}\right\rvert +C\\
&= -\frac1{2a}\ln\left\lvert\frac{a+x}{a-x}\right\rvert +C\\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2866731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Find the least value for $\sin x - \cos^2 x -1$
Find all the values of $x$ for which the function $y = \sin x - \cos^2 x -1$ assumes the least value. What is that value?
At first I found the first derivative to be $y' = \cos x + 2 \sin x \cos x$.
Critical point $0$, $-π/6$ (principal)
$y'' = - \sin x + 2(\cos 2x)$
Then substitution of $x$ by critical points I found minima. But my answer is incorrect.
Correct minimum value is $-9/4$
|
The critical points in $[-\pi,\pi]$ are those for which
$$
\cos x(1+2\sin x)=0
$$
which means $x=\pi/2$, $x=-\pi/2$, $x=-\pi/6$ and $x=-5\pi/6$.
Note that
$$
y''=-\sin x+2\cos^2x-2\sin^2x
$$
Since
$$
y''(\pi/2)=-1-2=-3,
\quad
y''(-\pi/2)=1-2=-1,
\\
y''(-\pi/6)=\frac{1}{2}+2\frac{3}{4}-2\frac{1}{4}=\frac{3}{2}=y''(-5\pi/6)
$$
the points of local minimum are $-\pi/6$ and $-5\pi/6$.
Just compute
$$
y(-\pi/6)=-\frac{1}{2}-\frac{3}{4}-1=-\frac{9}{4}
$$
The value at $-5\pi/6$ is the same.
Simpler: write
$$
y=\sin x-1+\sin^2x-1=\sin^2x+\sin x-2=
\left(\sin x+\frac{1}{2}\right)^{\!2}-\frac{9}{4}
$$
Obviously, the minimum value is where $\sin x=-\frac{1}{2}$ and is $-9/4$.
The maximum is where $\sin x=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2870490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Showing that $\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4}$
I am trying to show that if $|z|=r<1$, then
$$\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
I have shown the inequality $$\left|\frac{1}{z^3+1}\right|\leq\frac{1}{1-r^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ holds under the same conditions, but I am having showing the result of $(1)$.
I considered \begin{align}
|z^4+1|&\geq\left||z^4|-|-1|\right| \\
&=\left||z|^4-1\right| \\
&=\left|r^4-1\right|
\end{align}
Now, $r^4-1$ is positive $\forall r<1$\ $\{0\}$ $(r\neq -1)$. Ideally, like in $(2)$, if $r^4-1$ was strictly negative then the result would immediately follow.
A hint would be very helpful.
|
By the reverse triangle inequality it holds that $|z^4+1| = |1-(-z^4)| \geq 1-|z|^4$ and therefore $\frac{1}{|z^4+1|} \leq \frac{1}{1-|z|^4} = \frac{1}{1-r^4}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2870848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
[Integral][Please identify problem] $\displaystyle\int \cfrac{1}{1+x^4}\>\mathrm{d} x$ Here is my attempt. The result is not right. Please help identify the issue(s).
$\displaystyle f(x)=\int\cfrac{1}{x^4+1}\>\mathrm{d}x$, let $x=\tan t$, we have $ \mathrm{d}x = \sec^2 t\>\mathrm{d}t,\> t=\tan^{-1} x\in\left(-\cfrac{\pi}{2},\cfrac{\pi}{2}\right)$
\begin{align}
\displaystyle f(\tan t)&= \int\cfrac{\sec^2 t\> \mathrm{d}t}{1+\tan^4 t}=\int\cfrac{\cos^2 t\> \mathrm{d}t}{\cos^4 t+\sin^4 t}=\int\cfrac{\cfrac{1+\cos 2t}{2}\> \mathrm{d}t}{(\cos^2 t+\sin^2 t)^2-2\sin^2 t\cos^2 t} \notag\\
&=\int\cfrac{1+\cos 2t}{2-\sin^2 2t} \>\mathrm{d}t
=\int\cfrac{\mathrm{d}t}{2-\sin^2 2t} + \cfrac 12\int\cfrac{\mathrm{d}\sin 2t}{2-\sin^2 2t} \notag\\
&=\int\cfrac{\sec^2 2t \>\mathrm{d}t}{2\sec^2 2t-\tan^2 2t} + \cfrac {\sqrt{2}}8\int\cfrac{1}{\sqrt{2}-\sin 2t} + \cfrac{1}{\sqrt{2}+\sin 2t}\>\mathrm{d}\sin 2t \notag\\
&=\cfrac 12\int\cfrac{\mathrm{d}\tan 2t}{2+\tan^2 2t} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t} \notag\\
&=\cfrac {\sqrt{2}}4 \tan^{-1} \cfrac{\tan 2t}{\sqrt{2}} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t} \notag
\end{align}
As $\tan 2t=\cfrac{2\tan t}{1-\tan^2 t}=\cfrac{2x}{1-x^2}, \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}=\cfrac{\sqrt{2}\sec^2 t+\tan t}{\sqrt{2}\sec^2 t-\tan t}=\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x},$
$f(x)=\cfrac {\sqrt{2}}4 \tan^{-1} \cfrac{\sqrt{2}x}{1+x^2} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+c$
If the above holds, $\displaystyle \int_0^{\infty} \cfrac{\mathrm{d} x}{1+x^4}$ would be $0$, which is impossible(Should be $\cfrac {\sqrt{2}\pi}{4}$).
|
First of all, you made a typo in the final answer — the correct answer must be
$$f(x)=\frac{\sqrt{2}}{4}\tan^{-1}\frac{\sqrt{2}x}{1\color{red}{-}x^2}+\frac{\sqrt{2}}{8}\ln\frac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C.$$
The next issue is the introduction of $\sec(2t)$ and $\tan(2t)$ when you switched to
$$\int\frac{\sec^2 2t\,\mathrm{d}t}{2\sec^2 2t-\tan^2 2t}$$
(as part of an expression). Both $\sec(2t)$ and $\tan(2t)$ are undefined at some points within the domain $\displaystyle t\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, viz. at $\displaystyle t=\pm\frac{\pi}{4}$. Therefore, the antiderivative you find in terms of $t$ is in fact a piecewise-defined function:
$$f(x(t))=\begin{cases}
\cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\tan2t}{\sqrt{2}}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}+C_1, \text{ if } t\in\left(-\cfrac{\pi}{2},-\cfrac{\pi}{4}\right); \\
\cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\tan2t}{\sqrt{2}}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}+C_2, \text{ if } t\in\left(-\cfrac{\pi}{4},\cfrac{\pi}{4}\right); \\
\cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\tan2t}{\sqrt{2}}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}+C_3, \text{ if } t\in\left(\cfrac{\pi}{4},\cfrac{\pi}{2}\right).
\end{cases}$$
Switching back to $x$ still creates a piecewise-defined function:
$$f(x)=\begin{cases}
\cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\sqrt{2}x}{1\color{red}{-}x^2}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C_1, \text{ if } x\in(-\infty,-1); \\
\cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\sqrt{2}x}{1\color{red}{-}x^2}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C_2, \text{ if } x\in(-1,1); \\
\cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\sqrt{2}x}{1\color{red}{-}x^2}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C_3, \text{ if } x\in(1,+\infty).
\end{cases}$$
At the points $x=\pm1$, these expressions are undefined, and so the corresponding integrals have to be treated as improper. In your case, the integral $\displaystyle \int_0^{+\infty}$ has to be split at the discontinuity at $x=1$:
$$\int_0^{+\infty}\cdots\,\mathrm{d}x=\int_0^1\cdots\,\mathrm{d}x+\int_1^{+\infty}\cdots\,\mathrm{d}x,$$
and then, when evaluating the antiderivative that you found, you'll have to take the one-sided limits from the left and from the right at $x=1$, which are NOT equal to each other! And that's probably the source of your wrong answer.
More specifically:
$$\lim_{x\to1^{-}}\frac{\sqrt{2}x}{1-x^2}=+\infty \implies \lim_{x\to1^{-}}\arctan\frac{\sqrt{2}x}{1-x^2}=\frac{\pi}{2},$$
while
$$\lim_{x\to1^{+}}\frac{\sqrt{2}x}{1-x^2}=-\infty \implies \lim_{x\to1^{-}}\arctan\frac{\sqrt{2}x}{1-x^2}=-\frac{\pi}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2873537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Solve the equation $\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$ Solve the equation:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$
Things I have done so far:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$ Deducting 2 from both sides of equation
$$\Leftrightarrow (\sqrt[3]{x^2+4}-2)=(\sqrt{x-1}-1)+2x-4$$
$$\Leftrightarrow \frac{x^2+4-8}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}=\frac{x-2}{\sqrt{x-1}+1}+2(x-2)$$
$$\Leftrightarrow (x-2)(\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2)=0$$
+)$$x-2=0\Leftrightarrow x=2$$
+)$$\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2=0(*)$$
I don't know how to solve the equation (*).
By the way, I think there must be "smart" way to solve this equation.
|
$$\sqrt[3]{x^2+4}=\sqrt{x-1}+2x-3$$
Let $x=t^2+1$ with $t \ge 0$. Then
\begin{align}
\sqrt[3]{t^4+2t^2+5}-t &= 2t^2-1 \\
\sqrt[3]{t^4+2t^2+5} &= 2t^2+t-1 \\
t^4 + 2t^2 + 5 &= 8t^6 + 12t^5 - 6t^4 - 11t^3 + 3t^2 + 3t - 1 \\
8t^6 + 12t^5 - 7t^4 - 11t^3 + t^2 + 3t - 6 &= 0 \\
(8t^5 + 20t^4 + 13t^3 + 2t^2 + 3t + 6)(t - 1) &= 0 \\
\end{align}
Note, for $t \ge 0$, that $8t^5 + 20t^4 + 13t^3 + 2t^2 + 3t + 6 > 0$. It follows that the only positive solution is $t=1$. Hence $x=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2875620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
$x+y+z=n$. Finding the number of solutions. I have found two formulas. I want to connect them!
The number of ways in which a given positive integer $n≥3$ can be expressed as a sum of three positive integers $x,y,z$ (i.e. $x+y+z=n$) , subject to $x≤y≤z$ is the integer closest to $\frac{n^2}{12}$.
If $x+y+z=n$ where $x,y,z$ are postive integers then the number of ways is ${n-1 \choose 2}$ .
The difference between these two formula is the condition $x≤y≤z$. Can we connect these two ?? Please help me!
I mainly wants to know how one implies the other.
|
Just by looking at the generating functions, we can see a link (colors applied) ...
The number of solutions for $x+y+z=n, x\geq1, y\geq1, z\geq1$ is the coefficient of $x^n$ term of the
$$(x+x^2+x^3+...+x^k+...)^3=\color{red}{\frac{x^3}{(1-x)^3}}$$
But because
$$\color{blue}{\frac{1}{(1-x)^3}=
\frac{1}{2}\left(\frac{1}{1-x}\right)^{''}=
\frac{1}{2}\left(\sum\limits_{k=0}x^k\right)^{''}=
\sum\limits_{k=2}\frac{k(k-1)}{2}x^{k-2}}$$
we have
$$\frac{x^3}{(1-x)^3}=
\sum\limits_{k=2}\frac{k(k-1)}{2}x^{k+1}=
\sum\limits_{k=3}\frac{(k-1)(k-2)}{2}x^{k}$$
and the coefficient of $x^n$ is $\color{green}{\frac{(n-1)(n-2)}{2}=\binom{n-1}{2}}$.
The number of solutions for $x+y+z=n, 1\leq x \leq y\leq z$ (like in this example) can be found from the reformulated version
$$x=x_1, y=x+y_1=x_1+y_1, z=y+z_1=x_1+y_1+z_1 \Rightarrow \\
3x_1+2y_1+z_1=n \\
x_1\geq1, y_1\geq0, z_1\geq0$$
with the generating function
$$\left(x^3+x^{2\cdot 3}+...+x^{k\cdot 3}+...\right)\left(1+x^2+x^{2\cdot 2}+...+x^{k\cdot 2}+...\right)\left(1+x^2+...+x^k+...\right)=\\
\frac{x^3}{1-x^3}\cdot \frac{1}{1-x^2}\cdot \frac{1}{1-x}=
\color{red}{\frac{x^3}{(1-x)^3}}\cdot\frac{1}{1+x}\cdot\frac{1}{1+x+x^2}=\\
\frac{1}{6(1-x)^3}-\frac{1}{4(1-x)^2}-\frac{1}{72(1-x)}-\frac{1}{8(1+x)}+\frac{2+x}{9(1+x+x^2)}=\\
\color{blue}{\sum\limits_{k=2}\frac{k(k-1)}{12}x^{k-2}}-
\sum\limits_{k=1}\frac{k}{4}x^{k-1}-
\sum\limits_{k=0}\frac{x^{k}}{72}-\sum\limits_{k=0}\frac{(-1)^kx^k}{8}+
\frac{2+x}{9}\sum\limits_{k=0}U_k\left(-\frac{1}{2}\right)x^k$$
where $U_k(x)$ are Chebyshev polynomials of the second kind. The coefficient of $x^n, n\geq2$ is
$$\frac{n(n-1)}{12}-\frac{n}{4}-\frac{1}{72}-\frac{(-1)^k}{8}+\frac{2}{9}U_n\left(-\frac{1}{2}\right)+\frac{1}{9}U_{n-1}\left(-\frac{1}{2}\right)=\\
\frac{6n^2-24n-1-9(-1)^n}{72}+\frac{2\sin{\left((n+1)\theta\right)}+\sin{\left(n\theta\right)}}{9\sin{\theta}}$$
where $\cos{\theta}=-\frac{1}{2}$. This is asymptotically close to $\color{green}{\frac{n^2}{12}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2877158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
System of non-lineair equations I can't find the solutions of this system of equations. I only need the "whole number" solutions. The solutions I'm searching for:
$x=-1 \land y = 1 \land \lambda = \frac{1}{2e}$
\begin{cases} \begin{array} {rcl} e^{x+y-y^2}+2\lambda x \ = 0 \\ e^{x+y-y^2}(1-2y)+2\lambda y\ = 0 \\ x^2+y^2-2 \ = 0 \end{array} \end{cases}
Can someone give me some hints to solve this system? Thanks!
|
Given $(x,y,z)\in \mathbb{Z}$, $x^2+y^2-2=0$ is satified by $x = \pm 1,y = \pm 1$. Then
\begin{eqnarray*}
e^{x+y-y^{2}}+2\lambda x &=& 0
\end{eqnarray*}
can have the following possibilities,$e+2\lambda=0$, $\frac{1}{e}-2\lambda=0$, $\frac{1}{e}+2\lambda=0$ or $\frac{1}{e^{3}}-2\lambda=0$. This lead to $\lambda=\left\{-\frac{1}{2}e,\pm \frac{1}{2e},\frac{1}{2e^{3}}\right\}$
To satisfy,
\begin{eqnarray*}
\left(1-2y\right)e^{x+y-y^{2}}+2\lambda y &=& 0
\end{eqnarray*}
the only possibility is $x=-1,y=1,\lambda=\frac{1}{2e}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2879400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $p_A(x) = x^4 (x+3)^2 (x-4)$ then $A$ is diagonalizable iff $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$
Given the Characteristic polynomial of a matrix $A$ is
$$
p(x) = x^4 (x+3)^2 (x-4),
$$
show that $A$ is diagonalizable if and only if
$$
\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8.
$$
Given: $A$ is diagonalizable
Prove: $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$
(Help is needed in the other way around)
The geometric multiplicity equals the algebraic multiplicity, hence the diagonal matrix $D$ will look like
$$
D
= \begin{pmatrix}
-3 & & & & & & \\
& -3 & & & & & \\
& & 4 & & & & \\
& & & 0 & & & \\
& & & & 0 & & \\
& & & & & 0 & \\
& & & & & & 0
\end{pmatrix}
$$
Since $D$ and $A$ are similar matrices, their rank must be the same:
$$
\operatorname{Rank}(A) = \operatorname{Rank}(D) = 3.
$$
Also
$$
-3I - A
= \begin{pmatrix}
0 & & & & & & \\
& 0 & & & & & \\
& & -7 & & & & \\
& & & 0 & & & \\
& & & & -3 & & \\
& & & & & -3 & \\
& & & & & & -3
\end{pmatrix}
$$
and therefore
$$
\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A)
= 3 + 5
= 8.
$$
Perfect.
(Is it okay to place $D$ instead of $A$ in $\operatorname{Rank}(-3I-A)$? The rank will be the same, is it not?)
Given: $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$
Prove: $A$ is diagonalizable
No clue really…
I was thinking since geometric multiplicity is less or equal to algebraic multiplicity, I was thinking of finding all possible $D$'s and show that the one that makes the statement $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$ hold makes $A$ diagonal.
Any hint is appreciated.
|
Look at the possible Jordan form of $A$
$$
\begin{bmatrix}
-3 & * & 0 & 0 & 0 & 0 & 0\\
0 & -3 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 4 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & * & 0 & 0\\
0 & 0 & 0 & 0 & 0 & * & 0\\
0 & 0 & 0 & 0 & 0 & 0 & *\\
0 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
(* elements are $0$ or $1$) to deduce that
*
*$\operatorname{rank}A\ge 3$ and
$\operatorname{rank}A=3$ $\iff$ the block with eigenvalue $0$ is diagonal.
*$\operatorname{rank}(-3I-A)\ge 5$ and
$\operatorname{rank}(-3I-A)=5$ $\iff$ the block with eigenvalue $-3$ is diagonal.
Therefore, $\operatorname{rank}A+\operatorname{rank}(-3I-A)\ge 8$ and $\operatorname{rank}A+\operatorname{rank}(-3I-A)=8$ $\iff$ $\operatorname{rank}A=3$, $\operatorname{rank}(-3I-A)=5$. Then...
P.S. To answer your question "is it ok to place $D$ instead of $A$ in $\operatorname{rank}(-3I-A)$": yes, it is ok, since $-3I-A$ and $-3I-D$ are similar as well
$$
\lambda I-A=\lambda I-SDS^{-1}=\lambda SS^{-1}-SDS^{-1}=S(\lambda I-D)S^{-1}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2880037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
How to proof the reason? I have this statement:
If $\frac{a}{b} = \frac{c}{d},$ prove that
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$
I tried to add 1, multiply 1 and nothing.
My development was:
$\frac{a}{b} - \frac{b}{b} = \frac{c}{d} - \frac{d}{d}$
$\frac{a-b}{b} = \frac{c-d}{d}$
$\frac{b}{a-b} = \frac{d}{c-d}$ (I raised to $^{-1}$)
So far I have arrived, without much success.
How can I prove it? Thanks in advance.
|
Actually the result is not true if $a = b$ and $\frac ab =\frac cd = 1$ and $c = d$. Then $\frac {a+b}{a-b}$ and $\frac {c+d}{c-d} $ are both undefined and not equal to each other.
.......
Three ideas:
1) Proportions:
$\frac ab = \frac cd$ means there is some ratio so that $a = cr$ and $b = dr$. (That's clear, isn't it? $\frac {cr}{dr} = \frac ab$ for all $r$ and if [assuming $a\ne 0; c\ne 0$] then if $r = \frac ac$ means $\frac ab = \frac {cr}{dr} = \frac {c\frac ac}{dr} = \frac a{dr}$. So $dr = b$.)
So $\frac {a+b}{a-b} = \frac {cr +dr}{cr - dr} = \frac {r(c+d)}{r(c-d)}=\frac {c+d}{c-d}$.
[Note: we should take $a=0$ as a possiblity. $\frac 0b = \frac cd \implies c = 0$ so $\frac {a+b}{a-b} = \frac {b}{-b} = - 1$ and $\frac {c+d}{c-d}=\frac d{-d} = -1$.]
2) Substitution:
$\frac ab = \frac cd$ so $a = \frac {bc}d$ (or any of other variable)
So $\frac {a+ b}{a-b} = \frac {\frac {bc}d + b}{\frac {bc}d - b} ==\frac {b(\frac cd +1)}{b(\frac cd - 1)}= \frac {\frac cd +1}{\frac cd - 1}==\frac {d(\frac cd +1)}{d(\frac cd - 1)} = \frac {c + d}{c-d}$.
3) Consequences:
$\frac ab =\frac cd \implies ad = bc$.
And $\frac {a+b}{a-b} = \frac{c+d}{c-d} \iff$
$(a+b)(c-d) = (c+d)(a-b) \iff$
$ac + bc -ad - bd = ac +ad - bc = bd \iff$
$2bc = 2ad \iff$
$bc = ad$
And we determined that that is true.
(NOTE: It is !!!!!!VITALLLLLLLY!!!!!! important the we note that every step in that argument is an if AND ONLY if statement. Otherwise we can NOT EVER start with what we want to prove. If you EVER try to to do a proof where you start with your conclusion and get a true result and conclude your conclusion was true and your statements were not if and only if statements, I will reach through my computer screen and pound your head to the table.)
(Less important but still important. We can not end with $bc = ad \iff \frac ab = \frac cd$ as it is possible that $b=0$ or $d=0$ and it will not be true that $\frac ab = \frac cd$. In particular we coulde have $a=0; d=5; b=0;c = 6$ and it certainly isnt true that $\frac 00 = \frac 65$. We could say $bc = ad \iff \frac ab = \frac cd$ or $bc = ad = 0$.)
... or less dramatic than threatening to pound heads into table:
Once we figured this, we work in the real direction:
$\frac ab = cd \implies$
$ad = bc \implies$
$2ad = 2bc \implies$
$ad -bc = bc - ad \implies$
$ac + ad - bc - bd = ac +bc -ad -bd \implies$
$a(c+d) - b(c+d) = c(a+b) - d(a+b) \implies$
$(a-b)(c+d) = (c-d)(a+b)$
Assumming $a - b \ne 0$ and $c-d \ne 0$ then
$\frac {a+b}{a-b} = \frac {c+d}{c-d}$.
However if $a - b = 0$ then $a = b$ and $\frac ab = 1$ and $\frac cd = 1$ and $c = d$ and then.... well, the result simply isn't true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2883782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
$x,y,z>0$, $x+y+z=1$. Prove $x^2+y^2+z^2+3xyz \geq \frac{4}{9}$
Let $x$, $y$, $z$ be positive real numbers such that $x+y+z=1$, then
$$ x^2+y^2+z^2+3xyz \geq \frac{4}{9}.$$
From Cauchy–Schwarz inequality, we have
$$ \left( x^2+y^2+z^2 \right)\left(1^2+1^2+1^2\right)\geq(x+y+z)^2 = 1 \implies x^2+y^2+z^2 \geq \frac{1}{3}. $$
It suffices to prove $$ xyz \geq \frac{1}{27}. $$
However by AM–GM inequality,
$$ \frac{1}{3} = \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \implies xyz \leq \frac{1}{27}. $$
Am I missing something?
As commented by @mfl, we have
$$ \begin{aligned}
&9(x+y+z)\left(x^2+y^2+z^2\right) + 27xyz \geq 4(x+y+z)^3 \\
\iff &5\left(x^3+y^3+z^3\right) + 3xyz \geq 3\left(x^2(y+z)+y^2(x+z)+z^2(x+y)\right) \\
\iff &4\left(x^3+y^3+z^3\right) \geq 2\left(x^2(y+z)+y^2(x+z)+z^2(x+y)\right)
\end{aligned} $$
by Schur’s inequality. How to go further from this?
|
Rewrite your last line as:
$$2\sum_{x,y,z}(x^3+y^3-xy^2-x^2y) = 2\sum_{x,y,z}(x-y)^2(x+y)\geq 0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2884284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Proving divisibility by $3$ For all integers $a$, there exists an integer $b$ so that $3 | a + b$ and $3 | 2a + b$.
So far I have been able to find an integer $b$ that satisfies both of them separately, but not at the same time. (For the first one I have $b = 6-a$, and for the second I've found $12-2a$)
|
Divisibilty by 3 can be easily be proved by modular arithmetic.
Suppose we have a number $A$ which is :
$A = a_n \cdot 10^n+a_{n-1} \cdot 10^{n-1}+a_{n-2} \cdot 10^{n-2}+ \cdots + a_2 \cdot 10^2 +a_1 \cdot 10+ a_0 $
We know
$1 \equiv 10 \pmod{3}$
Therefore we substitute 10 s with 1 s :
$A = a_n \cdot 10^n+a_{n-1} \cdot 10^{n-1}+a_{n-2} \cdot 10^{n-2}+ \cdots + a_2 \cdot 10^2 +a_1 \cdot 10+ a_0 \equiv a_n \cdot 1+a_{n-1} \cdot 1 + a_{n-2} \cdot 1 + \cdots + a_2 \cdot 1 +a_1 \cdot 1 + a_0$
You can see that if a number wants to be divisible by 3 sum of all of its digits should be divisible by 3
Hope that helped!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2886422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$
Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$.
I tried taking $2x$ as $a$, and then $2y$ as $b$, and then finding the possibilities. However, I am not sure about the approach I've used.
A little hint would be appreciated.
$x$, $ y$ and $z$ belong to whole numbers.
|
I assume that the equation $2x+2y+z=n$ has $28$ solutions when $x,y,z$ are positive integers.
Note that this is same as finding the powers of $t$ with coefficient $28$ in the following expansion.
$$(t^2+t^4+t^6+t^8+.......)^2(t+t^2+t^3+t^4+......)$$
$$\left(\dfrac{t^2}{1-t^2}\right)^2\times\dfrac{t}{1-t}=t^5\times\left(\dfrac{1}{1-t}\right)^3=t^5\times(1+t)^3\times\left(\dfrac{1}{1-t^2}\right)^3$$
$$=t^5(1+t)^3\left(\dbinom{2}{2}+\dbinom{3}{2}t^2+\dbinom{4}{2}t^4+\dbinom{5}{2}t^6+..........\right)$$
Note that the last part has the general term of $\dbinom{k}{2}t^{2k-4}$. Since $\dbinom{8}{2}=28$, the last part contains the term $28t^{12}$. So we have $28t^{17}+28t^{18}$
So, the value of $n$ can be either $17$ or $18$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2887290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$? Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$ ?
For $a$ and $b$ are both positive or both negative,I proved this.
But I am not able to prove for all $a,b\in\mathbb R$ ?
Please help me !
|
The first inequality is equivalent to $a^2+b^2\leq 2( a^2+b^2)+2ab$, i.e., to $0\leq (a+b)^2$, which obviously is true.
The second is equivalent to $2( a^2+b^2+ab)\leq 3( a^2+b^2)$, i.e., to $0\leq (a-b)^2$ which again is obviously true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2887404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos{\theta})(1-\tan{\theta})-(1+\tan{\theta})(\cos^2{\theta}-\sin^2{\theta})}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}=
\frac{\cancel1\cancel{-\tan{\theta}}+2\sin^2{\theta}\cancel{+2\sin{\theta}\cos{\theta}}\cancel{-\cos^2{\theta}+\sin^2{\theta}}\cancel{-\sin{\theta}\cos{\theta}}\cancel{+\tan{\theta}}\cancel{-\sin{\theta}\cos{\theta}}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}
=\frac{2\sin^2{\theta}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}$$
It should be zero, but it isn't? Here's the proof:
$$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}}
\\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}
\\{\therefore}\text{Left=Right}$$
|
Use
\begin{equation}
2 \sin \theta \cos \theta
=
\sin 2\theta
=
\frac{2 \tan \theta}{1 + \tan^2 \theta}
\end{equation}
\begin{equation}
\cos^2 \theta - \sin^2 \theta
=
\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}
\end{equation}
Replacing the above quantities, we get
\begin{equation}
\frac{1 + \frac{2 \tan \theta}{1 + \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}}
=
\frac{1 + \tan^2 \theta + 2 \tan \theta}{1 - \tan^2 \theta}
=
\frac{(1 + \tan \theta)^2}{( 1 - \tan \theta) (1 + \tan \theta)}
=
\frac{1 + \tan \theta}{1 - \tan \theta}
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2887740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
Value of a triangle inscribed in a rectangle I was doing a little trigonometry challenge given by my professor, which is not anything especially complicated. However, it is completely based on getting the area of the green triangle in relation to the side of the square and the triangle, equal to $x$.
Reference image:
I've already got a bit of data from some trig. relations. What doesn't seem to be doable is finding a relation between $x$ and the sides $L_1$ and $L_2$ or isolating the sine of $\beta$ from the area of that triangle, in relation to $x$ as well.
Anyway, the following is the data gathered so far.
*
*Sine of $\alpha$ (both $\alpha_1$ and $\alpha_2$):
$$ \begin{aligned}
h_1^2 &= x^2 + \frac{x^2}{4} = \frac{5x^2}{4} \\
h_1 &= \frac{x \sqrt{5}}{2}
\end{aligned} \\
\begin{aligned}
\sin {\alpha} &= \frac{x}{h_1} = \frac{x}{\frac{x \sqrt{5}}{2}} = \frac{2x}{x \sqrt 5} = \frac{2}{\sqrt 5} = \frac{2 \sqrt 5}{5}
\end{aligned} $$
*Relation between sides $L_1$ and $L_2$, as well as sine of $\beta$ (through the law of sines):
$$ \begin{aligned}
\frac{\sin \alpha}{L_2} &= \frac{\sin \beta}{\frac{x}{2}} = \frac{\sin 30^\circ}{L_1} \\
\frac{2 \sqrt 5}{5L_2} &= \frac{2 \sin \beta}{x} = \frac{1}{2L_1}
\end{aligned} \\
\begin{aligned}
L_1 &= \frac{L_2 \sqrt 5}{4} = \frac{x}{4 \sin \beta} \\
L_2 &= \frac{4L_1 \sqrt 5}{5} = \frac{x \sqrt 5}{5 \sin \beta}
\end{aligned} $$
*Relation between $h_2$ and sides $L_1$ and $L_2$:
$$ \begin{aligned}
\sin \alpha = \frac{2 \sqrt 5}{5} &= \frac{h_2}{L_1} \\
\sin 30^\circ = \frac{1}{2} &= \frac{h_2}{L_2}
\end{aligned} \\
\begin{aligned}
h_2 &= \frac{L_2}{2} = \frac{2L_1 \sqrt 5}{5} = L_1 \sin \alpha
\end{aligned} $$
*Calculating the area with the data so far (1) (using sine of $\beta$):
$$ \begin{aligned}
A &= \frac{\frac{x}{2} \cdot L_1 \sin \alpha}{2} \\
&= \frac{x}{4} \cdot L_1 \sin \alpha \\
&= \frac{x}{4} \cdot \frac{x}{4 \sin \beta} \sin \alpha \\
&= \frac{x^2}{16} \cdot \frac{\sin \alpha}{\sin \beta}
\end{aligned} $$
*Calculating the area with the data so far (2) (using the sides' values)
$$ \begin{aligned}
A &= \frac{\frac{x}{2} \cdot \frac{2L_1 \sqrt 5}{5}}{2} \\
&= \frac{x \sqrt 5}{5} \cdot \frac{L_1}{2} \\
&= \frac{x \sqrt 5}{5} \cdot \frac{L_2 \sqrt 5}{8} \\
&= x \cdot \frac{L_2}{8}
\end{aligned} $$
Additionally, I've tried assuming (for the sake of an approximation) that $\beta$ is a right angle, since it is roughly equal to 86 degrees, and applied a theorem I've noticed in right triangles while meddling with this challenge.
*
*Given a triangle $\Delta ABC$, where $a = BC$, $b = CA$ and $c = AB$, the height $h$ of the triangle, perpendicular to the base (assumed to be AB), is equal to the product of the product of the cathetuss divided by the hypotenuse. That is,
$$ \begin{aligned}
h = \frac{a \cdot b}{c}.
\end{aligned} $$
*Moreover, with a substitution of these values, it is possible to get the value of the sides from $x$, as well as confirm the angle of $\beta$.
$$ \begin{aligned}
h_2 &= \frac{2L_1L_2}{x} \\
&= \frac{L_2}{2} = L_1 \sin \alpha
\end{aligned} \\
L_1 = \frac{x}{4},\ L_2 = \frac{x \sqrt 5}{5} \\
x = 4L_1 = L_2 \sqrt 5 \\
\sin \beta = \sin 90^\circ = 1
$$
*And finally, calculate an estimated approximation for the area, either with the value of $\beta$ or the value of the sides in respect to $x$.
$$ \begin{align}
A &= \frac{x^2}{16} \cdot \frac{\sin \alpha}{\sin \beta} \\
&= \frac{x^2}{16} \cdot \frac{2 \sqrt 5}{5} \\
&= \frac{2x^2 \sqrt 5}{80} = \frac{x^2 \sqrt 5}{40} \\ \newline
A &= x \cdot \frac{L_2}{2} \\
&= x \cdot \frac{ \frac{x \sqrt 5}{5} }{8} \\
&= x \cdot \frac{x \sqrt 5}{40} = \frac{x^2 \sqrt 5}{40} \\
\end{align} $$
How one would come to derive $x$ into the sides' values or the sine of $\beta$, in order to get the area only through $x$? What sort of hindsight is required to do so, even?
|
You can try this simpler alternative, which is in the spirit of finding the area of $\triangle GHB$ in terms of $x$.
Since we basically have a right triangle $ADE$, we can show that:
$$x^2+(2x)^2=DE^2\implies DE=\sqrt{x^2+5x^2}=x\sqrt5 \,\,\text{and:}\\
\sin \angle ADE=\frac{2x}{x\sqrt5}\implies \angle ADE=\arcsin\left(\frac2 {\sqrt{5}} \right)$$
It follows that $\angle CDG=90-\angle ADE$ and $\angle CDG=\angle DEA$.
Now we need to find $DG$:
$$\cos \angle CDG=\frac x{DG}\\
\cos\left(90-\arcsin\left(\frac2 {\sqrt{5}} \right)\right)=\sin\left(\arcsin\left(\frac2 {\sqrt{5}} \right)\right)=\frac x{DG}\\
\implies DG=\frac{x\sqrt5}{2}$$
It follows that:
$$x^2+CG^2=DG^2=\frac{5x^2}{4}\implies CG=\sqrt{\frac{x^2}{4}}=\frac{x}{2} \,\,\text{implying}:\\
GB=CG=\frac x2$$
Now we need to find $\angle DGC$, since $\angle DGC=\angle BGH$:
$$\sin \angle DGC= \frac{2x}{x\sqrt5}\implies \angle DGC=\angle BGH=\arcsin\left(\frac2 {\sqrt{5}} \right)$$
Since $\angle CBA$ is right, and $\angle FBE=60^\circ$, then it follows that $\angle GBH=30^\circ$
Now, since we know $\angle FBE=30^\circ$ and $\angle BEH=\angle DEA=90-\arcsin\left(\frac2 {\sqrt{5}} \right)$, then $\angle BHE$ is:
$$\angle BHE=30^\circ+\arcsin\left(\frac2 {\sqrt{5}} \right) \,\, \text{thus:}\\
\angle GHB=150^\circ -\arcsin\left(\frac2 {\sqrt{5}} \right)$$
Now, we need to find $BH$. By the law of sines, we have:
$$\frac{\sin\angle GHB}{GB}=\frac{\sin BGH}{HB}\,\,\,\, \text{implying:}\\
HB=\frac{GB\sin\angle BGH}{\sin\angle GHB}=\frac{\frac x2\sin\left(\arcsin\left(\frac2 {\sqrt{5}} \right)\right)}{\sin\left(150^\circ-\arcsin\left(\frac2 {\sqrt{5}} \right)\right)}$$
Keeping in mind that: $\cos(\arcsin(x))=\sqrt{1-x^2}$, we get that:
$$HB=\frac{2x}{1+2\sqrt3}$$
Now, finally, recalling that:
$$\triangle=\frac12 ab\sin A \,\,\text{where:}\\
$$
For $a:=GB=\frac x2,
b:=HB=\frac{2x}{1+2\sqrt3},
A:=\angle GBH=30^\circ$
Thus:
$$\therefore \triangle=\frac{x^2}{4(1+2\sqrt3)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2893767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.
This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'.
It seems fairly obvious that the series expansion $e^x$ will be used. However, I am unsure where to start. Should I consider other series?
|
$$e=(1+1)+\left(\frac {1}{2!}+\frac {1}{3!}\right) +\left(\frac {1}{4!}+\frac {1}{5!}\right) +\left(\frac {1}{6!}+\frac {1}{7!}\right) +\cdots=2\left(1+\frac {2}{3!}+\frac {3}{5!}+\frac {4}{7!}+\cdots\right) $$
Q. E. D
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2894991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
}
|
find angle between two lines between $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6$ how to find to angle between these two lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$
i tried so far like this
$y-\sqrt{3}x-5=0$
$y=\sqrt{3}x+5$
in the form of $y=mx+b$
got the value for $m_1=\sqrt{3}$
and for
$\sqrt {3}y-x+6=0$
$y=\dfrac {x-6} {\sqrt {3}}$
$y=\dfrac {1} {\sqrt {3}}x-6$
$m_{2}=\dfrac {1} {\sqrt {3}}$
the formula to find angle between two lines
$\tan \theta =\dfrac {m_{2-}m_{1}} {1+m_{1}m_{2}}$
$\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3}.\frac{1}{\sqrt{3}}}$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 1 + 1 }$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 2 }$
$- \frac { 2 } { 2 \sqrt { 3 } }$
$- \frac { 1 } { \sqrt { 3 } }$
is this right till now and how to find angle after wards . should i use tan inverse of $- \frac { 1 } { \sqrt { 3 } }$ or my algebra calculation is wrong . help me thank you
|
It's a very simple problem in essence. The gradients of the two lines are $\sqrt 3$ and $\frac 1{\sqrt 3}$. Recall that gradients equal the tangents of the angles made by lines with the $x$-axis. The special angles that give those tangents can quickly be recognised to be $60^{\circ}$ and $30^{\circ}$ (both in the first quadrant), giving the difference as $30^{\circ}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2896035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Is $ \sin 2x \ge 0 $ same as $\sin x\cos x\ge 0 $? Or differently phrased do $$ \sin 2x \ge 0$$ $$\sin x\cos x\ge 0 $$ have the same solution.
$\sin 2x$ is positive from $ 0\le x \le \pi/2$ , while $\sin x\cos x$ are positive in the first and third quadrant. Why is this, when $\sin 2x=2\sin x\cos x$, am i making a mistake.
Does this mean i cannot use double angle identity under the square root?
Edit: as vadim123 said the question was wrong, so i changed it
|
Note if $M > 0$ then $M*a*b > 0$ if $a > 0; b>0$ or $a < 0; b< 0$. And $M*a*b < 0$ if $a < 0; b > 0$ or $a > 0; b < 0$. And $M*a*b = 0$ if $a = 0$ or $b =0$. It does not matter whether $M =1$ or $M = 2$ or $M = \frac {37.95}{\pi}$.
So we would expect $\cos x \sin x \ge 0$ for the exact same values of $x$ as we would expect $\sin 2x = 2\cos x \sin x\ge 0$ or for the exact same values we would expect $\frac {37.95}{\pi}\cos x \sin x \ge 0$.
And it works out to be exactly the case. $\cos x \sin x \ge 0$ if $x$ is in the first or third quadrant (as those are the two quadrants where $\sin x$ and $\cos x$ are both the same sign.) And $\sin M \ge 0$ if $M$ is in the first or second quadrant. And $M = 2x$ is in the first quadrant if $x$ is in the first half of the first or third quadrant. ANd $M$ is in the second quadrant if $x$ is in the second half of the first or third quadrant.
Or in other words:
If $0 \le x \le \frac \pi 2$ then $\sin x \ge 0$ and $\cos x \ge 0$ so $\sin x\cos x \ge 0$. Also $0 \le 2x \le \pi$ so $\sin 2x \ge 0$.
If $\frac \pi 2 \le x \le \pi$ then $\sin x \ge 0$ and $\cos x \le 0$ so $\sin x\cos x \le 0$. Also $\pi \le 2x \le 2\pi$ so $\sin 2x \le 0$.
If $\pi \le x \le \frac {3\pi} 2$ then $\sin x \le 0$ and $\cos x \le 0$ so $\sin x\cos x \ge 0$. Also $2\pi \le 2x \le 3\pi$ so $\sin 2x \ge 0$.
If $\frac {3\pi} 2 \le x \le 2\pi$ then $\sin x \le 0$ and $\cos x \ge 0$ so $\sin x\cos x \le 0$. Also $3\pi \le 2x \le 4\pi$ so $\sin 2x \le 0$.
As well $\sin 2x = 2\cos x\sin x$ and $\cos x\sin x$ and $M*\cos x \sin x; M > 0$ should all be $> 0$ if $\cos x, \sin x$ are the same sign, and should all be $< 0$ of $\cos x, \sin y$ are opposite signs, and should $= 0$ if one or the other of $\cos x, \sin x$ equals $0$. So all turns out to work exactly as it should.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2896558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Find unknown coefficient of a matrix with known rank. Let there be a matrix A = $\begin{bmatrix} 8 & 7 & 5 \\ 4 & 5 & 6 \\ 7 & 8 & λ \end{bmatrix}$.
Find the λ value, for which matrix A has a rank(A) = 2.
First I need to do the reduced row echelon form (I guess). Should I start by dividing the first row by 8 or by substracting the third row from the first? Will the result be the same? How should I proceed with the information rank(A) = 2 given?
|
Hint: You need row $3$ to be a linear combination of rows $1$ and $2$. Write $a(8,7,5)+b(4,5,6)=(7,8,\lambda) $.
Solve the system $\begin{cases}{8a+4b=7 \\7a+5b=8}\end{cases}$ for $a$ and $b$.
Then $\lambda =5a+6b$.
If we let $M=\begin{pmatrix}8&4\\7&5\end{pmatrix}$, then $M^{-1}=\frac1{12}\begin{pmatrix}5&-4\\-7&8\end{pmatrix}$.
Thus
$\begin{pmatrix} a\\b\end{pmatrix}=M^{-1}\begin{pmatrix}7\\8\end{pmatrix}=\begin{pmatrix}\frac14\\\frac54\end{pmatrix}\,\therefore \lambda =\frac{35}4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2899852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Farthest and Nearest Point in the ellipse from a focus using Derivatives Suppose I have an ellipse with major axis lying on the x- axis and center on origin $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ and two foci at $(-c,0)$ and $(c,0)$.
How can I show that the farthest and nearest points from the right focus are $(-a,0)$ and $(a,0)$ respectively using derivatives?
I tried to differentiate this with respect to x,
$$\frac{1}{a^2}2x + \frac{1}{b^2}2yy' = 0\quad (1)$$
I used the distance formula to compute the distance $d$ of every point $(x,y)$ to the focus by $$d^2 = (x-c)^2 +y^2$$. Differentiating, $$2dd' =2(x-c) +2yy'$$ Since $d'=0$ for the two points,$$0 =2(x-c) +2yy'\quad (2)$$
Combining $(1)$ and $(2)$
$$\frac{1}{a^2}x-\frac{1}{b^2}(x-c)=0$$, Then $$x=\frac{-a^2c}{b^2-a^2}$$. This is impossible because if it would, then $x\ge a$, i.e.,
$$\frac{-a^2c}{b^2-a^2}\ge a$$ $${-a^2c}\le -ac^2$$ since $(b^2-a^2)=-c^2<0$ and $a>c$. Why is this so?
|
As pointed out in the comments, $y'$ becomes infinite at the points you are looking for, causing your problems.
Instead, I would suggest you to to parameterize your ellipse as
$$\begin{cases}
x(t) = a\cos t \\
y(t) = b\sin t
\end{cases}, \quad 0 \leq t < 2\pi.$$
In this way you avoid any problems with differentiation.
Then you want to minimize
$$ d(t)^2 = (a\cos t - c)^2 + (b\sin t)^2. $$
Differentiate with respect to $t$ to get
\begin{align}
0=2d(t)d'(t) &= 2(a\cos t - c)\cdot(-a\sin t) + 2b\sin t \cdot b\cos t\\
&= -2a^2\sin t \cos t +2ac\sin t + 2b^2\sin t \cos t\\
&= 2\sin t\,(ac-(b^2-a^2)\cos t ).
\end{align}
Now there are two cases to consider, namely $\sin t = 0,$ so that $t = 0$ or $t = \pi$, or
$$ \cos t = \frac{ac}{b^2-a^2} = \frac{ac}{c^2} = \frac{a}{c} > 1, $$
which is impossible.
Finally, to decide whether the minimum is attained at $t=0$ or $t=\pi$, note that since both $a$ and $c$ are positive, we have
$$ d(0)^2 = (a\cos 0 - c)^2 + (b\sin 0)^2 = (a-c)^2 < (a+c)^2 = (a\cos \pi - c)^2 + (b\sin \pi)^2 = d(\pi)^2, $$
so that the minimum is attained at $t=0$, i.e. at $(x,y) = (a,0).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2901840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$
What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$(a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2$
Then, we have:
$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 9(a + b)^2(c + d)^2$.
Thus, we have to prove that $9(a + b)^2(c + d)^2 \geq 256$.
Then, I used the following substitution:
$c + d = t$ and $a + b = 4 - t$.
We assume wlog that $a \leq b \leq c \leq d$.
Then, $4 = a + b + c + d \leq 2(c + d) = 2t$. Thus, $t \geq 2$.
Then, what we have to prove is:
$9t^2(4 - t)^2 \geq 256$.
We can rewrite this as:
$(3t(4 - t) - 16)(3t(4 - t) + 16) \geq 0$, or
$(3t^2 + 2t + 16)(3t^2 - 12t - 16) \geq 0$,
at which point I got stuck.
|
We have
\begin{align}
(a^2+3)(b^2+3)
& =(a^2+1+2)(b^2+1+2) \\
& =(a^2+1)(b^2+1)+2(a^2+1+b^2+1)+4 \\
& = (a^2+1)(b^2+1)+2(a^2+b^2+2)+4 \\
& =(a^2+1)(b^2+1)+2(a^2+b^2)+8
\end{align}
By Cauchy Schwartz we obtain
$$(a^2+1)(b^2+1)\geq(a+b)^2\tag{$\star$}$$
$$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(a+b)^2\tag{$\star \star$}$$
Puting together ($\star$) and ($\star$) we get
$$(a^2+3)(b^2+3)=(a^2+1)(b^2+1)+2(a^2+b^2)+8\geq 2(a+b)^2+8=2[(a+b)^2+4]\tag{1}$$
In the same way we get
$$(c^2+3)(d^2+3)=(c^2+1)(d^2+1)+2(c^2+d^2)+8\geq 2(c+d)^2+8=2[(c+d)^2+4]\tag{2}$$
Multiplying (1) and (2) we get the conclusion
\begin{align}
(a^2+3)(b^2+3)(c^2+3)(d^2+3)
& \ge 4[(a+b)^2+4][(c+d)^2+4] \\
& \ge 4[2(a+b)+2(c+d)]^2 \tag{Cauchy Schwartz}\\
& =16(a+b+c+d)^2
=256
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2902750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
}
|
Where should I find the ranges for $\sum_{n=1}^k \sin n$ and other similar trigonometric series? It can be found that
$$\sum_{n=1}^k \sin n = \frac{\sin\left(\frac{k+1}{2}\right)\sin\left(\frac{k}{2} \right)}{\sin\left(\frac{1}{2}\right)},$$
$$ \sum_{n=1}^k \cos n = \frac{\cos\left(\frac{k+1}{2}\right)\sin\left(\frac{k}{2} \right)}{\sin\left(\frac{1}{2}\right)},$$
and through some manipulation of the trigonometric identities, it can be shown that
$$ -0.1277\approx\frac{\cos\left(\frac{1}{2}\right)-1}{2\sin\left(\frac{1}{2} \right)} \leq \sum_{n=1}^k \sin n \leq \frac{1+\cos\left(\frac{1}{2} \right)}{2\sin\left(\frac{1}{2}\right)} \approx 1.9582,$$
$$ -1.5430\approx\frac{-\sin\left(\frac{1}{2}\right)-1}{2\sin\left(\frac{1}{2}\right)} \leq \sum_{n=1}^k \cos n \leq \frac{1-\sin\left(\frac{1}{2} \right)}{2\sin\left(\frac{1}{2}\right)}\approx0.5430,$$
$$ -1.0597\approx\frac{\cos\left(\frac{1}{2}\right)-\sin\left(\frac{1}{2}\right)-\sqrt{2}}{2\sin\left(\frac{1}{2}\right)} \leq \sum_{n=1}^k(\cos n+\sin n) \leq \frac{\cos\left(\frac{1}{2}\right)-\sin\left(\frac{1}{2}\right)+\sqrt{2}}{2\sin\left(\frac{1}{2}\right)}\approx1.8902,$$
$$ -0.0597\approx\frac{\cos\left(\frac{1}{2}\right)+\sin\left(\frac{1}{2}\right)-\sqrt{2}}{2\sin\left(\frac{1}{2}\right)} \leq \sum_{n=1}^k(\sin n-\cos n) \leq \frac{\cos\left(\frac{1}{2}\right)+\sin\left(\frac{1}{2}\right)+\sqrt{2}}{2\sin\left(\frac{1}{2}\right)}\approx2.8902.$$
Can people please tell me where I can find and confirm these results? I check the inequalities on Desmos, and the actual maximum and minimum values are pretty close to the ideal maximum and minimum values.
|
Note that $$\sin\left(\frac{k+1}{2}\right) \sin\left(\frac{k}{2}\right) = \frac{\cos(1/2) - \cos(k+1/2)}{2} \in \left[ \frac{\cos(1/2)-1}{2}, \frac{\cos(1/2)+1}{2}\right]$$
and
$$ \cos\left(\frac{k+1}{2}\right) \sin\left(\frac{k}{2}\right) = \frac{\sin(k+1/2)-\sin(1/2)}{2} \in \left[\frac{-1-\sin(1/2)}{2}, \frac{1-\sin(1/2)}{2} \right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2904388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Partial Fraction Decomposition of quadratic factor I'm trying to break up the following equation into partial fractions:
$$\frac{1}{(x^2-1)^2}=\frac{1}{(x+1)^2(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$
I'm ending up with two equations and four unknowns though:
$$A+C=0 \\ -A + B + C+D = 0$$
(these are the $x$ and constant coefficient variables).
How do I break this up completely?
|
Alt. hint: $\;\dfrac{1}{x^2-1}=\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)\,$, and therefore:
$$
\dfrac{1}{(x^2-1)^2}=\dfrac{1}{4}\left(\dfrac{1}{(x-1)^2}+\dfrac{1}{(x+1)^2}-\dfrac{2}{x^2-1}\right)=\ldots
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2908080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Show that the series converges and find its sum Show that
$$ \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... $$
converges and find its sum.
My solution so far:
I am thinking about finding the partial sum first and show that the series converges since its finite partial sum converges.
Now
$$ S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...= \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right) + \;...+ \left( \frac{1}{N}-\frac{1}{N+1} \right)$$
but I don't know how to go on with this. Now $ \lim_{N \to\infty} \left( \frac{1}{N}-\frac{1}{N+1} \right)=0$ but the right answer should be $1$.
|
See the first two brackets. $-\frac{1}{2}$ in the first bracket and $\frac{1}{2}$ in the second bracket cancel each other out. This canceling happens all through out.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2909507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Find the Matrix X and Y $\text { Find the matrix } X \text { and } Y , \text { if } X + Y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] \text { and } X - Y = \left[ \begin{array} { c c } { 3 } & { 6 } \\ { 0 } & { - 1 } \end{array} \right]$
Adding 1 and 2
$x + y +x-y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] + \left[ \begin{array} { l l } { 3 } & { 6 } \\ { 0 } & { - 1 } \end{array} \right]$
$2 x = \frac { 1 } { 2 } \left[ \begin{array} { l l } { 8 } & { 8 } \\ { 0 } & { 8 } \end{array} \right]$
$x = \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$
to Find Y
$\left[ \begin{array} { l l } { 1 } & { 4 } \\ { 0 } & { 4 } \end{array} \right] + y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right]$
$y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] - \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$
$y = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 0 } & { 3 } \end{array} \right]$
so $x = \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$ and $y = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 0 } & { 3 } \end{array} \right]$ is this correct or not ?
|
You have several errors. The most critical one is while calculating $y$:
$$y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] - \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right] = \left[ \begin{array} { l l } { 1 } & { -2 } \\ { 0 } & { 5 } \end{array} \right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2909792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Differential equation of a vector field A vector field $X:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ is defined by the following formula:
$$ X
\begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix}
= \begin{pmatrix}
-y(x^2+y^2)\\
x(x^2+y^2)\\
(1-x^2-y^2)
\end{pmatrix}$$
Determine formulas for the solution curves $y:\mathbb{R}\rightarrow \mathbb{R}^3$ of the differential equation $ \dot{y}(t)=X(y(t))$ with the initial condition
$$y(0) = \begin{pmatrix}
0\\
0\\
0\\
\end{pmatrix}$$
Unfortunately, I don't really know how to start. Do I have to find out the eigenvalues? Or do I have to calculate the derivates? How I get the characteristic polynomial?
|
So we essentially want to solve
$$ \begin{pmatrix}
\dot x(t)\\
\dot y(t)\\
\dot z(t)\\
\end{pmatrix}
= \begin{pmatrix}
-y(t)(x(t)^2+y(t)^2)\\
x(t)(x(t)^2+y(t)^2)\\
1-x(t)^2-y(t)^2
\end{pmatrix}$$
Let $x(t) = r(t)\cos\theta(t)$ and $y(t) = r(t) \sin\theta(t)$. I will omit the argument $t$ from now on and just simply write $r$ for $r(t)$ etc.
From
$r^2 = x^2 +y^2$, we see that $\dot r r = \dot x x + \dot y y,$ and inserting $\dot x$ and $\dot y$ from our system yields
\begin{align}
\dot r r = -xy(x^2+y^2) + yx(x^2+y^2) = 0 \quad \Rightarrow \quad \dot r = 0.
\end{align}
From $\theta = \arctan y/x$, we see that $$\dot\theta = \frac{\dot y x - \dot x y }{x^2+y^2} = \frac{\dot y x - \dot x y }{r^2},$$ and hence
$$ \dot\theta = \frac{1}{r^2}(x^2(x^2+y^2)+y^2(x^2+y^2)) = x^2+y^2 = r^2. $$
Hence, in the coordinates $(r,\theta, z)$, our system becomes
$$ \begin{pmatrix}
\dot r\\
\dot \theta\\
\dot z\\
\end{pmatrix}
= \begin{pmatrix}
0\\
r^2\\
1-r^2
\end{pmatrix}.$$
The first equation implies that
$$ r(t) = A, \quad A \in \mathbb{R}. $$
Then the second equation reads $\dot\theta = A^2,$ which yields
$$ \theta(t) = A^2t+ B, \quad B\in\mathbb{R}.$$
The third one becomes $\dot z = 1-A^2,$ so
$$ z(t) = (1-A^2)t+C \quad C\in\mathbb{R}. $$
Transforming back to our original coordinates, we see that the general solution of the system is
$$ \begin{pmatrix}
x(t)\\
y(t)\\
z(t)\\
\end{pmatrix}
= \begin{pmatrix}
A\cos(A^2t+B)\\
A\sin(A^2t+B)\\
(1-A^2)t+C
\end{pmatrix}. $$
You should be able to find the values of the constants by using the initial condition.
EDIT: I might have made a mistake somewhere, but the initial condition seems to imply that the solution is $(x,y,z) = (0,0,t)$.
EDIT 2: Actually now that I look at the differential equation again, it makes sense that this is the solution given the initial condition.
In fact, solving the equation explicitly was not even necessary. We could have just noted that, at time $t=0$, $\;\dot x = \dot y = 0$, and $\dot z = 1$, so $x$ and $y$ stay in place, while $z$ increases at a constant speed.
Furthermore, since $\dot x$ and $\dot y$ are independent of $z$, they will stay both stay in $0$ at all times, and since $\dot z$ is independent of $x$ and $y$, $\;z$ will keep increasing at unit speed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2909924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the length of the curve y = $\sqrt{-x(x+1)} - \arctan \sqrt{\frac{-x}{x+1}}$ I used the formula for this example:
$\displaystyle L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$
And I start by computing the derivative:
\begin{align*}
y' &= \left(\sqrt{-x^2-x}\right)' - \left(\arctan \sqrt{\frac{-x}{x+1}}\right)' \\
&= \frac{1}{2\sqrt{-x^2-x}} \cdot (-2x-1)-\frac{1}{\left(\sqrt{\frac{-x}{x+1}}\right)^2+1} \cdot \frac {1}{2\sqrt{\frac{-x}{x+1}}} \cdot \frac{-1}{(x+1)^2} \\
&= \frac{-2x-1}{2\sqrt{-x^2-x}} - \frac{1}{{\frac{-x}{x+1}}+1} \cdot \frac {1}{2\sqrt{\frac{-x}{x+1}}} \cdot \frac{-1}{(x+1)^2} \\
&= \frac{-2x-1}{2\sqrt{-x^2-x}}+ \frac{1}{2(x+1)^2 \sqrt{\frac{-x}{x+1}}\left(1-\frac{x}{x+1}\right)}
\end{align*}
but I stopped at this moment because it does not seem to me that this derivation was so complicated, how to do it then? Maybe I made a mistake somewhere?
|
For your actual task it might be beneficial to substitute $z = \sqrt{\tfrac{-x}{x+1}}$. Then your curve is $y = \tfrac z{1+z^2}-\arctan(z)$. Note that you will also have to adapt the boundaries.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2910282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
|
Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).
What you can do instead (notice the asymmetry):
$$\lim_{x\to0}\frac{\sin^2x-x^2}{x^2\sin^2x}=\lim_{x\to0}\frac{\sin^2x-x^2}{x^4}=\lim_{x\to0}\frac{\sin x+x}{x}\lim_{x\to0}\frac{\sin x-x}{x^3}
\\=2\lim_{x\to0}\frac{\cos x-1}{3x^2}=-2\lim_{x\to0}\frac{\sin x}{6x}=-\frac13.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 3
}
|
Prove the inequality using Chebyshev's Inequality If $a,b,c \in(0,\infty)$, then prove that: $$9(a^3+b^3+c^3)\ge(a+b+c)^3$$
I was trying to prove this inequality using Chebyshev's Inequality and assuming $a\ge b \ge c$ but to no avail.
Can please someone help me in proving it using Chebyshev's Inequality?
|
Since $(a,b,c)$ and $(a^2,b^2,c^2)$ they are the same ordered, by Chebyshov twice we ontain:
$$9(a^3+b^3+c^3)=3\cdot3(a^3+b^3+c^3)\geq3(a+b+c)(a^2+b^2+c^2)\geq(a+b+c)^3.$$
Also, by Holder
$$(1+1+1)^2(a^3+b^3+c^3)\geq\left(\sqrt[3]{1^2a^3}+\sqrt[3]{1^2b^3}+\sqrt[3]{1^2c^3}\right)^3=(a+b+c)^3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2910910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Let $n \in \mathbb{N}$ such that n is a non-perfect square. Show that $\sqrt{n}$ is irrational. Can someone please check my proof?
Let $n \in \mathbb{N}$ such that n is a non-perfect square. Show that $\sqrt{n}$ is irrational.
Let's prove it by contradiction, that is, suppose there are $n,a,b \in \mathbb{N}$ such that $\frac{a}{b}=\sqrt{n}$, n is a non-perfect square and there are no common factors for a and b, that is, they are coprime. n is either even or odd. Suppose n is even, that means $n=2k$ for some $k \in \mathbb{N}$. Therefore:
$$
\begin{align*}
\frac{a^2}{b^2}&=2k\\
a^2 &= 2k\cdot b^2 = 2(k\cdot b^2)
\end{align*}
$$
Which means that $a^2$ is even and therefore a is even and that means that there is some $j\in \mathbb{N}$ such that $a=2j$. Plugging in that value for a:
$$
\begin{align*}
\frac{2j}{b^2}&=2k\\
4j^2 &= 2k\cdot b^2 \rightarrow b^2 = 2\big(\frac{2j^2}{k}\big)
\end{align*}
$$
Using the same reasoning b is also even. That contradicts the fact that a and b are coprime and therefore $\sqrt{n}$ is irrational.
Now suppose n is odd, that means that $n=2k+1$ for some $k\in \mathbb{N}$. Therefore:
$$
\begin{align*}
\frac{a^2}{b^2}&=2k+1\\
a^2 &= (2k+1)\cdot b^2 \rightarrow (2k+1)|a^2
\end{align*}
$$
Since $(2k+1)|a^2$ we can conclude that $a^2 = q(2k+1)$. Plugging in that value for a:
$$
\begin{align*}
\frac{q(2k+1)}{b^2}&=2k+1\\
q(2k+1) &= (2k+1)\cdot b^2 \rightarrow q=b^2
\end{align*}
$$
That way q is a common factor for a and b, which contradicts the fact that they are coprime, hence $\sqrt{n}$ is irrational.
Is my reasoning correct? I've seen easier and shorter proofs, by I tried to generalize the reasoning that I used to prove that $\sqrt{2}$ and $\sqrt{3}$ are irrational...
Thank you for reading it!
|
Don't bother with even or odd cases. Just do:
$\frac {a^2}{b^2} = n$ so $a^2 = nb^2$. So for any prime, $p$ that divides $b$ then $p|a^2$ so $p|a$. That's impossible because $a$ and $b$ are coprime. So there are no primes that divide $b$ (!!!!!!). So $b = 1$. And $a^2 = n$ and $n$ is a perfect square.
That's it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2912592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How to prove this inequality using AM-GM? Suppose $a,b,c$ are positive real numbers. Then prove that $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(\frac{a+b+c}{3}\Big)\Big(abc\Big)^\frac{2}{3}\tag{*}$$
My approach: From AM-GM $$\Big(\frac{a+b}{2}\Big)\ge\Big(ab\Big)^\frac{1}{2}\tag{1}$$
Similarly, $\Big(\frac{b+c}{2}\Big)\ge\Big(bc\Big)^\frac{1}{2}\tag{2}$ and $\Big(\frac{c+a}{2}\Big)\ge\Big(ca\Big)^\frac{1}{2}\tag{3}$
Multiplying $(1), (2)$ and $(3)$, we get $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(abc\Big)\tag{4}$$
Which can be written as $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(abc\Big)^\frac{2}{3}\Big(abc\Big)^\frac{1}{3}\tag{4*}$$
Now, again from AM-GM $$\Big(\frac{a+b+c}{3}\Big)\ge\Big(abc\Big)^\frac{1}{3}\tag{5}$$
But now I am stuck. What should I do to get to $(*)$ ?
|
By AM-GM: $$\prod_{cyc}(a+b)-\frac{8}{9}(a+b+c)(ab+ac+bc)=\frac{1}{9}\sum_{cyc}(a^2b+a^2c-2abc)=$$
$$=\frac{1}{9}\sum_{cyc}(a^2c+b^2c-2abc)\geq\frac{1}{9}\sum_{cyc}(2\sqrt{a^2c\cdot b^2c}-2abc)=0.$$
Id est, by AM-GM again we obtain:
$$\prod_{cyc}\left(\frac{a+b}{2}\right)^3=\frac{1}{512}\left(\prod_{cyc}(a+b)\right)^3\geq\frac{1}{512}\cdot\frac{512}{729}(a+b+c)^3(ab+ac+bc)^3\geq$$
$$\geq\frac{1}{27}\left(\frac{a+b+c}{3}\right)^3\left(3\sqrt[3]{a^2b^2a^2}\right)^3=\left(\frac{a+b+c}{3}\right)^3a^2b^2c^2$$ and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2913298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find nullspace of this matrix The question I have is this:
1) Find a basis for the solution space of the homogeneous linear system.
$$x - y + 2z = 0 \\
2x + y = 0 \\
x − 4y + 6z = 0$$
So I convert to a matrix:
$$\begin{bmatrix} 1 & -1 & 2 \\ 2 & 1 & 0 \\ 1 & -4 & 6 \end{bmatrix}$$
$$ -> \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & -4 \\ 0 & -3 & 4 \end{bmatrix}$$
$$ -> \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & -4 \\ 0 & 0 & 0 \end{bmatrix}$$
So is this right?
Let $z = t, y = \frac{4}{3}t, x = -2t + \frac{4}{3}t$
so:
$$ -> \begin{bmatrix} x & y & z \end{bmatrix}$$
$$ -> \begin{bmatrix} \frac{-2}{3}t & \frac{4}{3}t & t \end{bmatrix}$$
$$ -> t * \begin{bmatrix} \frac{-2}{3} & \frac{4}{3} & 1 \end{bmatrix}$$
So the basis is:
$$ -> \begin{bmatrix} \frac{-2}{3} & \frac{4}{3} & 1 \end{bmatrix}$$
Is this right?
More generally, the reason we do it this way is because we're trying to solve for x in the equation $Ax = 0$ so we have to reduce A, setup a system of equations that result when you multiply A by X, setup any free variables, extract the coefficients and that's the basis of the nullspace (the set of all solutions to Ax = 0). Does that sound right?
|
Looks to me like the right method to get the rational basis for the null space. Quick check also shows that
$$
\left[
\begin{matrix}
1 & -1 & 2 \\
2 & 1 & 0 \\
1 & -4 & 6 \\
\end{matrix}
\right]
\left[
\begin{matrix}
-2/3 \\
4/3 \\
1 \\
\end{matrix}
\right] =
\left[
\begin{matrix}
0 \\
0 \\
0 \\
\end{matrix}
\right]
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2914450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Simple expression for $\sum_{n=0}^{\infty} {2n+E-1 \choose 2n} x^{2n}$? The title basically says it all. If $2n$ was just $n$, I think it's relatively simple:
$\sum_{n=0}^{\infty} {n+E-1 \choose n} x^{n} = \frac{1}{(1-x)^E}$.
But what if you want every other term, i.e.
$\sum_{n=0}^{\infty} {2n+E-1 \choose 2n} x^{2n}$
Can anyone think of a simple expression for that?
|
We have
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{n+E-1}{n} x^n =\frac{1}{(1-x)^E}.
\end{eqnarray*}
Break this into its odd & even powers of $x$
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n+E-1}{2n} x^{2n} +\sum_{n=0}^{\infty} \binom{2n+E}{2n+1} x^{2n+1} =\frac{1}{(1-x)^E}=\frac{(1+x)^E}{(1-x^2)^E}.
\end{eqnarray*}
Now binomially expand the numerator and collect the even powers of $x$ and we have
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n+E-1}{2n} x^{2n} =\frac{1}{(1-x^2)^E} \sum_{i=0}^{ \lfloor \frac{E}{2} \rfloor} \binom{E}{2i} x^{2i}.
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2914746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving $\sin(3x +28^\circ)= \cos(2x-13^\circ)$
Solve the following equation :
$$\sin(3x +28^\circ)= \cos(2x-13^\circ), x \in [0^\circ, 360^\circ]$$
Solution:
$$3x+28 +2x -13 =90$$
$$x=15$$
$$3x+28 -2x +13 = 90$$
$$x= 49$$
But the value $$x = 87$$ also satisfies the equation. I do not know how it comes.
|
Alternatively:
$$\sin(3x +28^\circ)= \cos(2x-13^\circ) \Rightarrow \\
\sin(3x +28^\circ)= \sin(90^\circ -(2x-13^\circ)) \Rightarrow \\
\sin(3x +28^\circ)-\sin(103^\circ-2x)=0 \Rightarrow \\
2\cos \frac{3x+28^\circ+103^\circ-2x}{2}\sin \frac{3x+28^\circ-(103^\circ -2x)}{2}=0 \Rightarrow \\
\cos \frac{x+131^\circ}{2}\cdot \sin \frac{5x-75^\circ}{2}=0 \Rightarrow \\
1) \ \cos \frac{x+131^\circ}{2}=0 \Rightarrow \frac{x+131^\circ}{2}=\frac{\pi}{2}+\pi n,n\in Z \Rightarrow \\
x=\pi+2\pi n-131^\circ, n\in Z \Rightarrow \\
0^\circ\le \pi+2\pi n-131^\circ\le 360^\circ \Rightarrow \\
0^\circ\le 180^\circ+360^\circ n-131^\circ\le 360^\circ \Rightarrow \\
0^\circ\le 360^\circ n+49^\circ\le 360^\circ \Rightarrow \\
\color{red}{x_1}=360^\circ \cdot 0+49^\circ =\color{red}{49^\circ}.\\
2) \ \sin\frac{5x-75^\circ}{2}=0 \Rightarrow \frac{5x-75^\circ}{2}=\pi k,k\in Z \Rightarrow \\
x=\frac{2\pi k}{5}+15^\circ, k\in Z \Rightarrow \\
0^\circ\le 72^\circ k+15^\circ\le 360^\circ \Rightarrow \\
\color{red}{x_2}=72^\circ \cdot 0+15^\circ =\color{red}{15^\circ};\\
\color{red}{x_3}=72^\circ \cdot 1+15^\circ =\color{red}{87^\circ};\\
\color{red}{x_4}=72^\circ \cdot 2+15^\circ =\color{red}{159^\circ};\\
\color{red}{x_5}=72^\circ \cdot 3+15^\circ =\color{red}{231^\circ};\\
\color{red}{x_6}=72^\circ \cdot 4+15^\circ =\color{red}{303^\circ}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2915018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to find a value for a variable that makes a matrix (with said variable) equal to its own inverse I'm given $$\begin{bmatrix}3&x\\-2&-3\\\end{bmatrix}$$
and am asked to find x such that it's inverse would equal itself. To attempt this I first tried to put the question into an augmented matrix and got this:
$$\begin{bmatrix}1&x/3&1/3&0\\0&(x/3)-(3/2)&1/3&1/2\\ \end{bmatrix}$$
I found that my answer was wrong so I tried:
$$\begin{bmatrix}3&x\\-2&-3\\\end{bmatrix}$$
times
$$\begin{bmatrix}x_1&x_2\\x_3&x_4\\\end{bmatrix}$$ to try and solve for x but found similar dissatisfactory results.
The answer is listed as x = 4; how might I go about solving this? Did I just make a mistake with my methods or is this the entirely wrong way about?
|
Both methods you attempted must give correct answer.
Method 1. Finding the inverse from the augmented matrix:
$$ \left[
\begin{array}{cc|cc}
1&0&3&x\\
0&1&-2&-3
\end{array}
\right] \stackrel{R_1/3}=
\left[\begin{array}{cc|cc}
\frac13&0&1&\frac x3\\
0&1&-2&-3
\end{array}
\right] \stackrel{2R_1+R_2\to R_2}=\\
\left[\begin{array}{cc|cc}
\frac13&0&1&\frac x3\\
\frac23&1&0&\frac {2x}3-3
\end{array}
\right] \stackrel{\frac{3}{2x-9}\cdot R_2}=
\left[\begin{array}{cc|cc}
\frac13&0&1&\frac x3\\
\frac{2}{2x-9}&\frac{3}{2x-9}&0&1
\end{array}
\right] \stackrel{-\frac{x}{3}\cdot R_2+R_1\to R_1}=\\
\left[\begin{array}{cc|cc}
\frac13-\frac{2x}{3(2x-9)}&-\frac{x}{2x-9}&1&0\\
\frac{2}{2x-9}&\frac{3}{2x-9}&0&1
\end{array}
\right].$$
So, it must be:
$$\left[\begin{array}{cc}
\frac13-\frac{2x}{3(2x-9)}&-\frac{x}{2x-9}\\
\frac{2}{2x-9}&\frac{3}{2x-9}
\end{array}\right]=
\left[\begin{array}{cc}
3&x\\
-2&-3
\end{array}\right] \Rightarrow x=4.$$
Method 2. Multiply by its inverse (by condition it must be equal to the original matrix and remember the result is an identity matrix: $A\cdot A^{-1}=I$):
$$\left[\begin{array}{cc}
3&x\\
-2&-3
\end{array}\right]
\left[\begin{array}{cc}
3&x\\
-2&-3
\end{array}\right]=
\left[\begin{array}{cc}
1&0\\
0&1
\end{array}\right] \Rightarrow \\
\left[\begin{array}{cc}
9-2x&0\\
0&-2x+9
\end{array}\right]=
\left[\begin{array}{cc}
1&0\\
0&1
\end{array}\right] \Rightarrow x=4.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2915113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
Prove the inequality $\sqrt{\frac{a^2+1}{b+c}}+\sqrt{\frac{b^2+1}{a+c}}+\sqrt{\frac{c^2+1}{a+b}}\ge 3$ Let $a;b;c\in R^+$ such that $ab+bc+ca>0$. Prove that $$\sqrt{\frac{a^2+1}{b+c}}+\sqrt{\frac{b^2+1}{a+c}}+\sqrt{\frac{c^2+1}{a+b}}\ge 3$$
I have seen the similar question is $$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$ my problem is harder than it
My try: Holder: $$LHS^2\cdot \sum _{cyc}\left(a^2+1\right)^2\left(b+c\right)\ge \left(a^2+b^2+c^2+3\right)^3$$
Then we need to prove: $$\left(a^2+b^2+c^2+3\right)^3\ge 9\sum _{cyc}\left(a^2+1\right)^2\left(b+c\right)$$
After full expanding I am stuck because that inequality is not homogeneous, it hard to solve.
|
Because by AM-GM and C-S we obtain:
$$\sum_{cyc}\sqrt{\frac{a^2+1}{b+c}}\geq3\sqrt[6]{\prod_{cyc}\frac{a^2+1}{b+c}}=3\sqrt[12]{\prod_{cyc}\frac{(a^2+1)(1+b^2)}{(a+b)^2}}\geq3\sqrt[12]{\prod_{cyc}\frac{(a+b)^2}{(a+b)^2}}=3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2922166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to solve Recurrence relation: $T(n) = T(n/3) + T(n/5) + n$ Given the following: $T(n) = T(n/3) + T(n/5) + n$
The aim is to find the upper bound of this solution.
I have tried to solve this using Master's Theorem, but this does not apply for this problem.
My next best options would be Substitution or Iteration, but I am having troubles finding examples similar enough to follow these methods to solve this relation.
If I were to make a "guess" to use for substitution method I would use $T(n) \leq cn$ but after working through the problem I am unable to reach any sort of a solution
|
Let's check if
\begin{equation}
T(n) = \frac{15}{7}n
\end{equation}
is a solution for our recursion.
\begin{align}
T(n) &= \frac{15}{7}n \\
T(\frac{n}{3}) &=\frac{1}{3} \frac{15}{7}n\\
T(\frac{n}{5}) &=\frac{1}{5} \frac{15}{7}n
\end{align}
So
\begin{equation}
T(\frac{n}{3}) + T(\frac{n}{5}) + n
=
\frac{1}{3} \frac{15}{7}n +
\frac{1}{5} \frac{15}{7}n
+
n
\end{equation}
which is
\begin{equation}
T(\frac{n}{3}) + T(\frac{n}{5}) + n
=
\frac{3}{7}n + \frac{5}{7}n + n
=
\frac{8+7}{7}n
=
\frac{15}{7}n
=
T(n)
\end{equation}
So
\begin{equation}
T(n) = \frac{15}{7}n
\end{equation}
is indeed a solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2923492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solution of a quadratic equation to satisfy a constraint I have the roots of a quadratic equation as
$$x = \frac{1 \pm \sqrt{1-4\theta^2}}{2\theta}$$ I know that $|\theta| < 1/2$. Among these two roots, I want to find the one which has a value $|x| < 1$.
My attempt is:
It can't be $\frac{1 + \sqrt{1-4\theta^2}}{2\theta}$ as numerator exceeds one and denominator $2|\theta| < 1$, hence the modulus of the root is greater than one.
If $x = \frac{1 - \sqrt{1-4\theta^2}}{2\theta}$:
\begin{equation}
0 < 1 - 4\theta^2 < 1 \implies 0 < \sqrt{1 - 4\theta^2} < 1.
\end{equation}
\begin{equation}
\left[(1 - \sqrt{1 - 4\theta^2}) - 2|\theta|\right]^2 > 0
\end{equation}
\begin{equation}
1 - \sqrt{1-4\theta^2} - 2|\theta|\sqrt{1-4\theta^2} > 0
\end{equation}
\begin{equation}
\frac{1 - \sqrt{1-4\theta^2}}{2|\theta|} > \sqrt{1-4\theta^2}
\end{equation}
Using the fact that $0 < \sqrt{1 - 4\theta^2} < 1$, left hand side of the inequality has to be less than one.
Is there a more elegant way of finding this?
|
Alt. hint: $\;x_1+x_2=1/\theta \gt 0\,$ and $\,x_1x_2=1 \gt 0\,$, so both $\,x_1,x_2 \gt 0\,$. Since $\,x_1x_2=1\,$ it follows that $\,0 \lt x_1 \lt 1 \lt x_2\,$, so the smaller root, corresponding to the "$-\sqrt{\Delta}$" sign, is within $\,(0,1)\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2927419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Partial decomposition of $\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$ Recently I have come arcross the following fraction
$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$$
Hence I have encountered this fraction within a task of integration I want to do a partial decomposition. First of all I rewrote it as following
$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}=\dfrac{Az+B}{2z^2+3}+\dfrac{Cz+D}{(s^2+1)z^2+1} $$
which yields to a system of equations for $A,B,C$ and $D$
$$\begin{align}
0&=A(1+s^2)+3C\\
0&=B(1+s^2)+3D\\
0&=s^2A+2C\\
1&=s^2B+2D
\end{align}$$
Solving this system gives $A=C=0$ and $B=\dfrac{3}{s^2-2}$ and $D=-\dfrac{s^2-1}{s^2-2}$. By plugging in these values we arrive at
$$\dfrac{3}{(s^2-2)(2z^2+3)}-\dfrac{s^2-1}{(s^2-2)((s^2+1)z^2+1)}=\dfrac{s^2(z^2-3)+5z^2+6}{(s^2-2)(2z^2+3)((s^2+1)z^2+1)}$$
which is in fact not the wanted fraction. From the original task(solution to Problem 2 by ysharifi) I am going through I know, that the right partial decomposition should look like
$$\dfrac1{3s^2+1}\left(\frac3{2z^2+3}-\frac1{(s^2+1)z^2+1}\right)$$
but honestly speaking I have no clue how to get to this. Could someone please go through the whole process with me. I am quite confused right now.
Thanks in advance.
|
Even though we could solve this using the method of residues, I'll follow your method
\begin{equation}
\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
=
\frac{Az + B}{2z^2 + 3}
+
\frac{Cz + D}{(s^2+1)z^2 + 1}
\tag{1}
\end{equation}
which is
\begin{equation}
\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
=
\frac{(Az + B)((s^2+1)z^2 + 1) + (Cz + D)(2z^2 + 3)}{(2z^2 + 3)((s^2+1)z^2+1)}
\end{equation}
Expanding
\begin{equation}
\frac{ B+3D + (A+3C)z + (B+2D+Bs^2)z^2 + (A+2C+As^2)z^3}{(2z^2 + 3)((s^2+1)z^2+1)}
\end{equation}
By identification of coefficients, we have
\begin{align}
B+3D &= 0\\
A+3C &= 0 \\
B+2D+Bs^2 &= 1 \\
A+2C+As^2 &= 0
\end{align}
From the first two, we get $A = -3C$ and $B= - 3D$, so replace in the last two,
\begin{align}
-D -3s^2D &= 1\\
-C -3s^2C &= 0
\end{align}
which means that $C =0$ (also $A=0$) and
\begin{equation}
D = -\frac{1}{1+3s^2}
\end{equation}
So replacing in $(1)$, we get
\begin{equation}
\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
=
\frac{1}{1+3s^2}
\Bigg(
\frac{ 3}{2z^2 + 3}
-
\frac{ 1}{(s^2+1)z^2 + 1}
\Bigg)
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2929238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq125 abc$
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that
$$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$
My try:
First I wrote the inequality as
$$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125. $$
Then I noted that
$$a^2+\frac{b}{a}\geq a^2+\frac{2}{a}, \\ b^2+\frac{c}{b}\geq b^2+\frac{2}{b}, \\ c^2+\frac{a}{c}\geq c^2+\frac{2}{c}. $$
But I don't know how this can help me.
|
You did 90% of the work. You can finish it by using a little bit of Calculus.
Let
$$f(x)=x^2+\frac{2}{x}.$$
Since the derivate of $f(x)$ is positive, the function $f(x)$ is increasing for $x\geq2$, and then
$$f(x)\geq f(2)=5.$$
Thereby,
$$a^2+\frac{b}{a}\geq 5, \\ b^2+\frac{c}{b}\geq 5,\\ c^2+\frac{a}{c}\geq 5,$$
which implies that
$$ \left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2930755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
}
|
Finding $\text{Res}(f,0)$ where $f(z)=\frac{1}{z^2\sin(z)}$ I am trying to determine the residue of $z=0$ where $f(z)=\frac{1}{z^2\sin(z)}$.
I have determined that $z=0$ is a pole of order $3$. Hence to compute the residue, I use $$\text{Res}(f,0)=\frac{1}{2}\lim_{z\to 0}\frac{\partial^2}{\partial z^2}\left(\frac{z}{\sin(z)}\right).$$
I am convinced I am doing something incorrect, as the calculations become very convoluted. I have computed that $$\frac{\partial^2}{\partial z^2}=\frac{z\sin^3(z)-2\sin^2(z)\cos(z)+2z\sin(z)\cos^2(z)}{\sin^4(z)},$$ but am unsure of how to further compute the residue, as I'm unable to solve this limit.
edit
I have solved the problem via the repeated use of L'Hopitals rule, but this took the work of computing some ugly limits. Is there another way?
|
$$\sin z=z-\dfrac{1}{3!}z^3+\dfrac{1}{5!}z^5-\dfrac{1}{7!}z^7+\cdots$$
then
\begin{align}
\dfrac{1}{z^2\sin z}
&= \dfrac{1}{z^3\left(1-\dfrac{1}{3!}z^2+\dfrac{1}{5!}z^4-\dfrac{1}{7!}z^6+\cdots\right)} \\
&= \dfrac{1}{z^3}\left(1+\dfrac{1}{6}z^2+(\dfrac{1}{36}-\dfrac{1}{120})z^4+\cdots\right) \\
&= \dfrac{1}{z^3}+\dfrac{1}{6z}+\dfrac{7z}{120}+\cdots
\end{align}
s0 $a_{-1}=\dfrac16$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2931092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How many words from $5$ different letters can be formed out of $10$ consonants and $5$ vowels if... How many words from $5$ different letters can be formed out of $10$ consonants and $5$ vowels if 'a' is always one of the vowels and the words have at least $2$ consonants?
My approach to this has been like this:
$1$ way of choosing "a".
${10\choose 2}$ ways of choosing the two mandatory consonant.
${14\choose 2}$ ways of choosing the letters for the other two letters of the word.
$5!$ ways of rearranging everything.
So the answer is: $1{10\choose 2}{14\choose 2}5!$
Is this correct?
I think another approach could be calculate all the possible words and subtract the combinations for one vowel and the combinations from $2$ consonants, but I don't get the same result.
|
How many words with five different letters can be formed from ten consonants and five vowels if $a$ is always one of the vowels and the word contains at least two consonants?
Method 1: Consider cases, depending on the number of consonants. Since $a$ must be included, if you have at least two consonants, then the word contains either two, three, or four consonants, with the rest of the letters being vowels. A five-letter word with exactly $k$ of the ten consonants and an $a$ must contain $4 - k$ of the other four vowels. Hence, the number of words with exactly $k$ consonants is
$$\binom{1}{1}\binom{10}{k}\binom{4}{4 - k}5! = \binom{10}{k}\binom{4}{4 - k}5!$$
Therefore, the number of admissible words is
$$5!\sum_{k = 2}^{4} \binom{10}{k}\binom{4}{4 - k} = 5!\left[\binom{10}{2}\binom{4}{2} + \binom{10}{3}\binom{4}{1} + \binom{10}{4}\binom{4}{0}\right]$$
What did you do wrong?
In your attempt you chose $a$, two consonants, and two additional letters, then arranged them. Since those letters must be different from those you have already selected, you only have $15 - 1 - 2 = 12$ (not $14$) letters left from which to choose. Had you noticed that your approach would have produced the answer
$$\binom{1}{1}\binom{10}{2}\binom{12}{2}5!$$
However, that is still wrong. The reason is that by designating two of the consonants as the two consonants that must be included in the word, you count each arrangement with three consonants three times, once for each of the $\binom{3}{2}$ ways you could designate two of those three consonants as the designated consonants, and each arrangement with four consonants six times, once for each of the $\binom{4}{2}$ ways you could designate two of those four consonants as the designated consonants. Notice that
$$5!\left[\color{red}{\binom{2}{2}}\binom{10}{2}\binom{4}{2} + \color{red}{\binom{3}{2}}\binom{10}{3}\binom{4}{1} + \color{red}{\binom{4}{2}}\binom{10}{4}\binom{4}{0}\right] = \binom{10}{2}\binom{12}{2}5!$$
Method 2: There are $\binom{15}{5}$ ways to select five of the fifteen available letters. From these, we subtract those that do not contain an $a$, of which there are $\binom{14}{5}$, those that do not contain a consonant, of which there are $\binom{5}{5}$, and those that contain a $a$ and only one consonant, of which there are $\binom{1}{1}\binom{10}{1}\binom{4}{3}$. We then multiply by the $5!$ ways of arranging five distinct letters, which yields
$$5!\left[\binom{15}{5} - \binom{14}{5} - \binom{5}{5} - \binom{1}{1}\binom{10}{1}\binom{4}{3}\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2931662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find $a$ so that a line is tangent, secant or external from a sphere I am given the following problem:
Given the line $$r \{ R = (1,0,a) + \lambda [a \quad a \quad 0]$$ and the sphere $$S \{ 8x^2 + 8y^2 +8z^2 - 16x +24y -8z + 19 = 0$$ find, relating to values of $a$, when the line is external, tangent and secant to the sphere.
I completed the square on the sphere's equation for some clarity
$$(x-1)^2 + \left( y + \frac{3}{2} \right)^2 + \left( z - \frac{1}{2} \right)^2 = \frac{9}{8}$$
which gave me a center $C = \left( 1 , - \frac{3}{2}, \frac{1}{2} \right)$ and a radius $r = \frac{3}{2 \sqrt{2}}$.
To evaluate when a line is external or not to a sphere one must relate the distance of the center of the sphere to the line and check if it is greater (or not) than the radius. And my question is: how can I do that given the fact that the line has two variables?
Is the problem not well-made?
|
let $u$ be the vector from $(1,0,a)$ to the center of the circle (1,-\frac 32, \frac 12)
$u = (0,-\frac 32, \frac 12-a)$
let $v$ be the direction vector of the line. $(a,a,0)$
$\frac {u\cdot v}{v\cdot v} v$ will give the projection of $u$ onto $v$
$u - \frac {u\cdot v}{v\cdot v} v$ will be orthogonal to $v$ and be the shortest vector from the line to the center of the circle
If the magnitude of this vector equals the radius of the circle:
$\|u - \frac {u\cdot v}{v\cdot v} v\| = \sqrt{\frac 98}$
Then our line will be tangent.
$\|(0,-\frac 32, \frac 12-a) - \frac {-\frac 32 a}{2a^2} (a,a,0)\|\\
\|(0,-\frac 32, \frac 12-a) + (\frac 34,\frac 34,0)\|\\
\|(\frac 34,-\frac 34, \frac 12-a)\|\\
\frac {9}{16} + \frac {9}{16} + (\frac 12 - a)^2 = \frac 98\\
(\frac 12 - a)^2 = 0\\
a = \frac 12$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2932095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to solve following integral I am trying to solve following integral.
$$=\int_0^\infty\frac{x}{x^n+(\epsilon+\sigma)}dx$$
I started by assuming that $a = \epsilon + \sigma$ and $$\implies x^n = A\tan^2(\theta)$$
Then,
$$\implies x = a^{1/n}\tan^{2/n}(\theta)$$
$$\implies dx = \frac{2a^{1/n}}{n} (\tan(\theta))^{2/n-1}\sec^2(\theta)\text{ d}\theta$$
Using $\tan^2(\theta)+1 = \sec^2(\theta)$ and substituing above equations, we can write Eq. 1 as:
$$=\frac{\frac{2a^{2/n}}{n}(\tan(\theta))^{4/n-1}\sec^2(\theta)\text{ d}\theta}{\sec^2(\theta)}$$
$$=\int_0^{\pi/2}\frac{2a^{2/n}}{n}(\tan(\theta))^{4/n-1} \text{ d}\theta$$
let $B = \frac{2a^{2/n}}{n}$, then
$$=\int_0^{\pi/2}B(\tan(\theta))^{4/n-1} \text{ d}\theta$$
Now, I don't know how to proceed forward
The Final answer by authors is:
$\large \frac{\pi\sigma(\epsilon+\sigma)^{2/n-1}}{n\sin(\frac{2\pi}{n})}$
|
Observe we have
\begin{align}
\int^\infty_0 \frac{x}{x^n+a}\ dx = a^{(2-n)/n}\int^\infty_0 \frac{y}{1+y^n}\ dy.
\end{align}
Hence it suffices to evaluate
\begin{align}
\int^\infty_0 \frac{y}{1+y^n}\ dy.
\end{align}
The easiest way is to use contour integration. Observe
\begin{align}
\int^R_0 \frac{y}{1+y^n}\ dy + \int_{C_R} \frac{z}{1+z^n}\ dz - \int_0^R \frac{(R-t)e^{i4\pi/n}}{1+(R-t)^n}dt = 2\pi i \operatorname{Res}\left(\frac{z}{1+z^n}, z_0=e^{i\pi/n} \right)
\end{align}
where $C_R$ is the arc of the circle of radius $R$ and angle $2\pi/n$. Hence we see that
\begin{align}
(1-e^{i4\pi/n}) \int^R_0 \frac{x}{1+x^n}\ dx +\int_{C_R} \frac{z}{1+z^n}\ dz= 2\pi i\frac{z_0}{nz_0^{n-1}} = \frac{2\pi i}{ne^{i\pi (n-2)/n}}
\end{align}
which means
\begin{align}
\int^\infty_0 \frac{x}{1+x^n}\ dx =& \frac{2\pi i}{ne^{i\pi(n-2)/n}(1-e^{4i\pi/n})}= \frac{2\pi i}{ne^{i\pi }(e^{-2\pi i/n}-e^{2\pi i/n})} \\
=& \frac{\pi}{n\sin\left(\frac{2\pi}{n} \right)}.
\end{align}
Finally, combining everything yields
\begin{align}
\int^\infty_0 \frac{x}{x^n+(\sigma+\varepsilon)}\ dx = \frac{\pi (\sigma+\varepsilon)^{2/n-1}}{n\sin\left(\frac{2\pi}{n} \right)}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2932746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Prove: $x^4 + y^4 + z^4 - 4(x^3 + y^3 + z^3) + 5(x^2 + y^2 + z^2) \leqslant 4 $ Prove that $$x^4 + y^4 + z^4 - 4(x^3 + y^3 + z^3) + 5(x^2 + y^2 + z^2) \leqslant 4$$ for all $x, y, z \geqslant 0$ satisfying $x + y + z = 2$. When does equality occur?
Here is my aproach:
For each $x\in [0,2]$ we have: $$x^4-4x^3+5x^2-2x\leq 0$$
since $$(x-1)^2(x-2)x\leq 0$$
Equality holds for any permutation of $0,0,2$ or $0,1,1$.
Any idea how to solve this with Muirhead or Am-Gm or Cauchy?
|
The homogenization gives
$$\frac{1}{4}(x+y+z)^4-\frac{5}{4}(x^2+y^2+z^2)(x+y+z)^2+2(x+y+z)(x^3+y^3+z^3)-x^4-y^4-z^4\geq0$$ or
$$\sum_{cyc}(x^3y+x^3z-2x^2y^2+x^2yz)\geq0,$$ which is true by AM-GM or by Muirhead because $(3,1,0)\succ(2,2,0).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2933273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Finding $ \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$
Finding $\displaystyle \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$
Try: Let $$\displaystyle I = \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$$ (Function is even )
$$I = 2\int^{1}_{0}\bigg[\ln(1+x)-\ln(1-x)\bigg]\frac{x^3}{\sqrt{1-x^2}}dx$$
Using By parts
$$I = 2\bigg[\ln\bigg(\frac{1+x}{1-x}\bigg)\bigg(\sqrt{1-x^2}-\sqrt[3]{1-x^2}\bigg)\bigg]\bigg|^{1}_{0}+4\int^{1}_{0}\frac{1}{1-x^2}\cdot \bigg(\sqrt{1-x^2}-\sqrt[3]{1-x^2}\bigg)dx$$
Could some help me to solve second Integration, Thanks
|
Let us consider your integral:
$$I = \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$$
Separate the integral:
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx-\int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1-x)\ dx$$
Consider the right-hand integral. Let $x=-t$ such that $dx=-dt$, where $t\in (1,-1)$:
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx-\int^{-1}_{1}\frac{(-t)^3}{\sqrt{1-(-t)^2}}\ln (1-(-t))\ (-dt)$$
Incorporate a minus sign into the integral in order to switch to interval to $t\in (-1,1)$:
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx+\int^{1}_{-1}\frac{t^3}{\sqrt{1-t^2}}\ln (1+t)\ dt$$
Recognize that $t$ is a dummy variable. Let $t=x$ such that $dt=dx$, where $x\in (-1,1)$:
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx+\int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx=2\int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx$$
Apply integration by parts:
$$I=-\frac{2}{3}\sqrt{1-x^2}(x^2+2)\ln(1+x)\bigg|^{1}_{-1}+\frac{2}{3}\int^{1}_{-1}\sqrt{1-x^2}\frac{x^2+2}{1+x}\ dx=\frac{2}{3}\int^{1}_{-1}\sqrt{1-x^2}\frac{x^2+2}{1+x}\ dx$$
Let $x=\sin(u)$ such that $dx=\cos(u)\ du$, where $u\in (-\pi/2,\pi/2]$:
$$I=\frac{2}{3}\int^{\pi/2}_{-\pi/2}\cos^2(u)\frac{\sin^2(u)+2}{1+\sin(u)}\ du=\frac{2}{3}\int^{\pi/2}_{-\pi/2}(\sin^2(u)+2)\frac{1-\sin^2(u)}{1+\sin(u)}\ du$$
Perform long-division on the integrand:
$$I=\frac{2}{3}\int^{\pi/2}_{-\pi/2}(\sin^2(u)+2)(1-\sin(u))\ du=\frac{2}{3}\int^{\pi/2}_{-\pi/2}\sin^2(u)\ du-\frac{2}{3}\int^{\pi/2}_{-\pi/2}\sin^3(u)\ du+\frac{4}{3}\int^{\pi/2}_{-\pi/2}du-\frac{4}{3}\int^{\pi/2}_{-\pi/2}\sin(u)\ du$$
Since $\sin^3(u)$ and $\sin(u)$ are odd, their integrals evaluate to zero. Rewrite $\sin^2(u)$ by making use of a double-angle identity:
$$I=\frac{1}{3}\int^{\pi/2}_{-\pi/2}(1-\cos(2u))\ du+\frac{4}{3}\int^{\pi/2}_{-\pi/2}du=\frac{5}{3}u-\frac{1}{6}\sin(2u)\bigg|^{1}_{-1}$$
Thus,
$$I=\frac{5}{3}\pi$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2935264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How can you simplify $\sqrt{9-6\sqrt{2}}$? How do you simplify: $$\sqrt{9-6\sqrt{2}}$$
A classmate of mine changed it to $$\sqrt{9-6\sqrt{2}}=\sqrt{a^2-2ab+b^2}$$ but I'm not sure how that helps or why it helps.
This questions probably too easy to be on the Math Stack Exchange but I'm not sure where else to post it.
|
Your class mate is being.... clever.
If $\sqrt {9-6\sqrt 2}=a-b $ then $9-6\sqrt 2=a^2-2ab+c^3$
Let $2ab=6\sqrt 2$ and $a^2+b^2=9$.
Can we do that?
If we let $b^2=k $ and $a^2=9-k$ then $ab=\sqrt {k (9-k)}=3\sqrt 2=\sqrt {18} $. Solving $k (9-k)=18$ for $k $ (if it isn't visiblely obvious that we can do it in our heads, it is just a quadratic that we can solve by quadratic formula) and, for covenience, choicing the smaller solution (because we want $a>b $), we get $k =3$ is a good solution..
So $a=\sqrt 6$ and $b=\sqrt 3$.
I.e. in other words
$\sqrt {9-6\sqrt 2}=$
$\sqrt {6-2\sqrt 6\sqrt 3 +3}=$
$\sqrt {(\sqrt 6- \sqrt 3)^2}=$
$|\sqrt 6 - \sqrt 3|=$
$\sqrt 6 -\sqrt 3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2936269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
Laurent expansion for $1/\cos(z)$ I have a quick question.
How to find the Laurent expansion for $1/\cos(z)$
In the link above the person asks how to find the Laurent expansion for $\frac{1}{cos(z)}$.
The accepted answer utilizes the fact that \begin{align}
\frac{1}{\sin t}&=\frac{1}{ t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots } \\
&=\frac{1}{t}\frac{1}{1-\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)}\\
&=\frac{1}{t}\left[1+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)^2+\dots\right]\\
&=t^{-1}+\frac{1}{3!}t+\left[\left(\frac{1}{3!}\right)^2-\frac{1}{5!}\right]t^3+\dots
\end{align}
Fair enough. However, I'm not sure how the person motivates that $|\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots| < 1$ (which is necessary for the above calculation to be valid, i.e. the part of writing $\frac{1}{1-\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots}$ as a geometric series).
This may be obvious but I cant seem to figure it out. It feels valid to say that it should be less than one, but how do you show this?
Thanks!
|
Here is a simple justification:
$$1-\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}-\dotsm\right)=\frac{\sin t}t,$$
so that $\enspace\dfrac{t^2}{3!}-\dfrac{t^4}{5!}+\dfrac{t^6}{7!}-\dotsm=1-\dfrac{\sin t}t$, and it is well-known the cardinal sine function
$$\operatorname{sinc}(t)=\begin{cases}\dfrac{\sin t}t&\text{if }t\ne 0\\1&\text{if }t=0\end{cases}$$
has values in $[0,1]\:$ for all $|t|\le\frac\pi2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2939691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find height of a wall if at the beginning it exceeds $10$ meters and then $8$ meters When the foot of a staircase is $5$ meters from the base of a wall, it protrudes $10$ meters above the wall; and if it is $9$ meters from the base, it stands $8$ meters.
Find the height of the wall.
Using the Pythagorean theorem for the two situations I have
$$\begin{cases}(10-x)^2=5^2+y^2\\(8-x)^2=9^2+y^2,\end{cases}$$ where $x$ is the hypotenuse and $y$ is the height of the wall.
Solving that system of equations I have that $x=23$ and $y=\pm12$, but since a height is always positive, the solution is $\boxed{12~\text{meters}}$.
Is it correct?
I am not sure of the $x$ value because first it has a value but then it has another, so maybe we have to use $x_1$ and $x_2$, but then we have $3$ equations with $2$ variables, so it is not possible.
Thanks!
|
Let $z$ be the length of the ladders, then
$$(z-10)^2=5^2+y^2$$
$$(z-8)^2=9^2+y^2$$
subtracting the two equations, we have
$$-2(2z-18) = (5-9)(5+9)$$
$$4(z-9) = (9-5)(5+9)=4(14)$$
$$z=9+14=23$$
$$(23-10)^2-5^2=y^2$$
$$y^2=13^2-5^2=12^2$$
Hence $y=12$.
Note my definition of $z$ is not the hypothenus but the length of ladder.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2940122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that of a hyperbola.
|
Without derivatives,
Considering the functions
$$
f(x,y)=y-x-2\sqrt{x^2-\sqrt 2 x-1} = 0\\
y = \lambda
$$
Their intersection is at the solution for
$$
f(x,\lambda)=\lambda-x-2\sqrt{x^2-\sqrt 2 x-1} = 0
$$
or squaring
$$
(x-\lambda)^2-2(x^2-\sqrt 2 x-1)=0
$$
Solving for $x$ we have
$$
x = \frac{1}{3} \left(2 \sqrt{2}\pm 2 \sqrt{\lambda ^2-\sqrt{2} \lambda -1}-\lambda\right)
$$
but at tangency between $f(x,y)=0, y = \lambda$ we have
$$
\lambda ^2-\sqrt{2} \lambda -1 = 0\Rightarrow \lambda = \frac{1}{2} \left(\sqrt{2}+\sqrt{6}\right)
$$
as the feasible minimum.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2942263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
}
|
Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4\sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
|
Note that$$z=7\left(-\frac17+\frac{4\sqrt3}7i\right).\tag1$$Now, since $\left(-\frac17\right)^2+\left(\frac{4\sqrt3}7\right)^2=1$, the expression $(1)$ expresses $z$ as $7\bigl(\cos(\alpha)+\sin(\alpha)i\bigr)$, for some $\alpha$. So, a square root of $z$ is $\sqrt7\left(\cos\left(\frac\alpha2\right)+\sin\left(\frac\alpha2\right)i\right)$. Now, note that if $c=\cos\left(\frac\alpha2\right)$ and $s=\sin\left(\frac\alpha2\right)$, then $c^2+s^2=1$ and $c^2-s^2=\cos(\alpha)=-\frac17$. This allows you to compute the square roots of $z$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2943851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
}
|
Determine linear map so that equation is correct Let $f:\mathbb R^3→\mathbb R^2$, $(x,y,z)^T\mapsto(x^2y-5z,2x+4yz-3z^3)^T$,
and $(x_0, y_0, z_0)^T=(-1,0,1)^T$.
I need to determine the linear map $A∈L(\mathbb R^3,\mathbb R^2)\cong\mathbb R^{2×3}$ so that$$
f \begin{pmatrix} x \\ y \\ z \end{pmatrix}=f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}+A\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}+\left\|\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}\right\| ε\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}
$$
with $\lim\limits_{\vec h \to 0} ε(\vec h)=0$.
How can this be done? I don't know how to start (except for calculating $f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}$).
$f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}=\begin{pmatrix} -5 \\ -5 \end{pmatrix}$
$A=\begin{pmatrix} 0 && 1 && -5\\ 2 && 4 && -9 \end{pmatrix}$
$\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}=\begin{pmatrix} x+1 \\ y \\ z-1 \end{pmatrix}$
Therefore:
$f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}+A\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}=\begin{pmatrix}y-5z \\ 6+2x+4y-9z\end{pmatrix}$
|
Guide:
Use Taylor series expansion.
Let $(x,y,z) = (x_1, x_2, x_3)$ and $f(x,y,z)=(w_1,w_2)$.
Compute $A_{ij} = \frac{\partial w_i}{\partial x_j}$ and evaluate at $(x_0, y_0, z_0)$.
Edit:
For checking, I just drop higher order terms.
\begin{align}
x^2y-5z &= [(x-1)+1]^2y-5z \\
&\approx [2(x-1)+1]y - 5z \\
&\approx y-5z\\
2x+4yz-3z^3&= 2x+4y[(z-1)+1]-3[(z-1)+1]^3\\
&\approx2x+4y-3[3(z-1)+1]\\
&\approx 2x+4y-9z+6
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2944182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to prove divisibility of a number using the binomial expansion? I have the following problem:
Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.
How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.
|
$$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1+{n \choose n}1$$
$$(5+1)^n-1 = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1+{n \choose n}1-1$$
${n \choose n}1 = 1$ so $1$ and $-1$ cancel out.
$$\implies (5+1)^n-1 = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2944723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?
|
In general, if you want to complete the square on $ax^2+bx+c$, here's the formula and derivation
Suppose that
$$ax^2+bx+c=A(x+B)^2+C$$
Then:
$$ax^2+bx+c=Ax^2+2ABx+AB^2+C$$
Therefore:
$$A=a$$
$$b=2aB$$
$$c=aB^2+C$$
Therefore:
$$B=\frac{b}{2a}$$
$$C=-a(\frac{b}{2a})^2+c=-\frac{b^2}{4a}+c$$
Plug it all in:
$$ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$$
Which is the complete square.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2944950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
}
|
Why does Wolfram|Alpha make a mistake here? We want to evaluate $$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}.$$The solving process can be written as follows:\begin{align*}\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x \to -8}\left[\frac{(\sqrt{1-x}-3)(\sqrt{1-x}+3)}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+\sqrt[3]{x^2})}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=\lim_{x \to -8}\left[\frac{-(x+8)}{x+8}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=-\lim_{x \to -8} \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\\&=-2.\end{align*}
But when I input this
lim\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} as x to -8
into Wolfram|Alpha, it gives the limit $0$.
Why is Wolfram|Alpha making a mistake here?
|
In Mathematica 11.3 I get
In[1]:= Limit[(Sqrt[1 - x] - 3)/(2 + CubeRoot[x]), x -> -8]
Out[1]= -2
Mathematica Documentation says CubeRoot[x] gives the real-valued cube root of $x$.
Even
In[4]:= -8^(1/3)
Out[4]= -2
Mathematica gives me the correct answers
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2946885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 1
}
|
$\frac{a^{3}+1}{b+1}+\frac{b^{3}+1}{a+1}$ an integer $\Rightarrow \frac{a^{3}+1}{b+1}$ and $\frac{b^{3}+1}{a+1}$ are integers. I want to show that if the natural numbers $a,b \in \mathbb{N}$ are such that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, then, necessarily, $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$.
I have thought the following.
We are given that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$.
This means that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1}=k, \text{ for some } k\in \mathbb{N}$.
We also have that $b+1 \mid a^3+1$ and $a+1 \mid b^3+1$, right?
So does it suffice to reject the cases that $a^3+1=k_1(b+1)$ and $b^3+1=k_2 (a+1)$ for negative $k_1, k_2$ ? If so, then we pick all the possible combinations and want to get a contradiction from the fact that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, or not?
Or do we show somehow else that $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$ ?
|
Hint: You can prove this more general statement:
Let $w,x,y,z$ be positive integers so that
$$\frac{w}{x}+\frac{y}{z}\in\mathbb{N}$$
and $z|w,x|y$. Then
$$\frac{w}{x},\frac{y}{z}\in\mathbb{N}.$$
To do so, consider $d=\gcd(x,z)$, which must divide each of $w$ and $y$, so by dividing each variable by $d$ it suffices to consider the case where $\gcd(x,z)=1$. Can you finish from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2947342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\dfrac{a}{4}+i\left(b+\dfrac{b}{4}\right)$
So
$$
\begin{split}
\left|z-\frac{1}{z}\right|
&=\sqrt{\left(a-\dfrac{a}{4}\right)^2+\left(b+\dfrac{b}{4}\right)^2}\\
&=\sqrt{4+\dfrac{1}{4}-\dfrac{a^2}{2}+\dfrac{b^2}{2}}\\
&=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}
\end{split}
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $\sqrt{2+\dfrac{1}{4}}=\dfrac{3}{2}$
Now, comes the maximum value.
We can write $$\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(4-2a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+2-a^2}$$
$$=\sqrt{6+\dfrac{1}{4}-a^2}$$
Setting $a=0$ gives the maximum value $\sqrt{6+\dfrac{1}{4}}=\dfrac{5}{2}$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
|
Your idea is good, but you lose yourself in some computations (the maximum is correct, though).
Consider the square of the modulus:
$$
f(z)=\left|z-\frac{1}{z}\right|^2=\frac{|z^2-1|^2}{|z|^2}
$$
Since $|z|=2$ by assumption, we can as well consider
$$
g(z)=|z^2-1|^2=(z^2-1)(\bar{z}^2-1)=z^2\bar{z}^2-z^2-\bar{z}^2+1=5-z^2-\bar{z}^2
$$
If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $\bar{z}^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then
$$
g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2
$$
The maximum is for $a=0$, the minimum for $a=\pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.
We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2948812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
}
|
How to integrate $1/(\sin x + a\sec x)^2$? How to integrate $1/(\sin x + a\sec x)^2$?
I tried by squaring the contents and trying to make use of some identities, but it became very messy and difficult to solve.
|
$\displaystyle I(a) = \int \dfrac{\mathrm{d}x}{(\sin x + a\sec x)^2}$
Multiplying the numerator and denominator by $\cos^2x,$
$\displaystyle I(a) = \int \dfrac{\cos^2x\,\mathrm{d}x}{(\sin x\cos x + a)^2}$
$\displaystyle I(a) = \int \dfrac{\cos^2x\,\mathrm{d}x}{\sin^2x\cos^2x + 2a\sin x\cos x + a^2}$
$\displaystyle I(a)= \int \dfrac{\cos^2x\,\mathrm{d}x}{a^2 + a\sin 2x + \frac{\sin^22x}{4}}$
$\displaystyle I(a)= \int \dfrac{4\cos^2x\,\mathrm{d}x}{(4a^2 + 4a\sin2x + \sin^22x)} = 2\int\dfrac{(1+\cos 2x)\,\mathrm{d}x}{(2a + \sin2x)^2}$
$\displaystyle I(a) = 2\underbrace{\int\dfrac{\mathrm{d}x}{(2a + \sin2x)^2}}_{I_1} + 2\underbrace{\int\dfrac{\cos 2x\,\mathrm{d}x}{(2a + \sin2x)^2}}_{I_2}$
Evaluating $I_2$ should be easy, we let $(2a + \sin 2x) = t \implies 2\cos 2x\,\mathrm{d}x = \mathrm{d}t$
$\displaystyle I_2 = \dfrac{1}{2}\int \dfrac{\mathrm{d}t}{t^2} = -\dfrac{1}{2}\cdot\dfrac{1}{(2a + \sin 2x)} + C$
Solving $I_1$ is a bit challenging,
Assume $Z = \dfrac{\cos2x}{2a + \sin 2x}$
$\dfrac{dZ}{dx} = \dfrac{(2a + \sin2x)(-2\sin2x) - \cos 2x(2\cos2x)}{(2a + \sin 2x)^2}$
$\implies \dfrac{dZ}{dx} = \dfrac{-4a\sin 2x -2}{(2a + \sin 2x)^2}$
$\implies \dfrac{dZ}{dx} = \dfrac{-4a(\sin 2x + 2a) - 2 + 8a^2}{(2a + \sin 2x)^2}$
$\implies \dfrac{dZ}{dx} = -\dfrac{4a}{(2a + \sin 2x)} + \dfrac{(8a^2 - 2)}{(2a + \sin 2x)^2}$
Now integrate both sides w.r.t $x,$
$\displaystyle Z = -4a\int\dfrac{\mathrm{d}x}{(2a + \sin2x)} + (8a^2 - 2)I_1$
$\displaystyle\implies (8a^2 - 2)I_1 = Z + 4a\int\dfrac{\sec^2x\mathrm{d}x}{2a + 2\tan x + 2a \tan^2x}$
Substitute $\tan x = p,$
$\displaystyle (8a^2 - 2)I_1 = Z + \dfrac{4a}{2a}\int\dfrac{dp}{p^2 + \frac{p}{a} + 1} = Z + 2\int\dfrac{dp}{\left(p + \frac{1}{2a}\right)^2 + \left(1 - \frac{1}{4a^2}\right)}$
$\implies (8a^2 - 2)I_1 = Z + \dfrac{4a}{\sqrt{4a^2 - 1}}\tan^{-1}\left(\dfrac{2ap + 1}{\sqrt{4a^2 - 1}}\right)$
$\implies (8a^2 - 2)I_1 = \dfrac{\cos 2x}{2a + \sin 2x} + \dfrac{4a}{\sqrt{4a^2 - 1}}\tan^{-1}\left(\dfrac{2ap + 1}{\sqrt{4a^2 - 1}}\right)$
$\implies I_1 = \dfrac{1}{(8a^2 - 2)}\cdot\dfrac{\cos 2x}{2a + \sin 2x} + \dfrac{2a}{(4a^2 - 1)^{3/2}}\tan^{-1}\left(\dfrac{2ap + 1}{\sqrt{4a^2 - 1}}\right)$
Hence we finally obtain $I$ as,
$\boxed{I(a) = \dfrac{1}{(4a^2 - 1)}\dfrac{\cos 2x}{(2a + \sin 2x)} + \dfrac{4a}{(4a^2 - 1)^{3/2}}\tan^{-1}\left(\dfrac{2a\tan x + 1}{\sqrt{4a^2 - 1}}\right) - \dfrac{1}{2a + \sin 2x} + C}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2949785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Let $f$ be a function $f : \mathbb{N} \to \mathbb{N}$ such that $f(2x+3y)=f(x)f(y)$, determine $f$ here what I did .
$$f(0)=f(0)^2$$
so $f(0)=1$ or $f(0)=0$
IF $f(0)=1$
we have $f(2y)=f(y)$
$$f(1)=f(2)=\ldots=f(2^n)=a$$
the equation $f(x)-a=0$ has infinitly many solutions , so $f(x)=a$
since f(0)=1 , $f(x)=1$.
i don't know how to handle the other case , i just found that $f(2x)=f(3x)=0$
|
A priori, there is no reason to expect $f$ to be a polynomial. So knowing that $f(x) = 0$ for infinitely many $x$ doesn't mean that $f$ is identically zero.
I consider the case when $f(0) = 1$. Then $f(2x + 0y) = f(x)$, so $f(x) = f(2x)$ for all $x$. Similarly, $f(x) = f(3x)$ for all $x$.
Let's evaluate $f(1)$.
Note that $f(5) = f(1\cdot 2 + 1\cdot 3) = f(1)^2$.
Also note that $13 = 5 \cdot 2 + 1 \cdot 3$ and $13 = 2 \cdot 2 + 3 \cdot 3$. Thus $f(13) = f(5)f(1) = f(1)^3$ and $f(13) = f(2)f(3) = f(1)^2$.
Thus $f(1)^2 = f(1)^3$, forcing $f(1) = 0$ or $f(1) = 1$.
It is a straightforward induction argument to show that $f(n)$ is a power of $f(1)$ for all $n \geq 1$. In either case, it is now apparent that $f(0) = 1$ and $f(n) = 1$ or $f(n) = 0$ for $n \geq 1$ are two possible solutions.
If $f(0) = 0$, then $f(2x + 0y) = f(x)f(0) = 0$, and similarly $f(3x) = 0$ for all $x$.
Let's evaluate $f(1)$ again.
Now note that $f(13) = f(5 \cdot 2 + 1 \cdot 3) = f(5)f(1)$, and $f(5) = f(1 \cdot 3 + 1 \cdot 2) = f(1)f(1)$, so taht $f(13) = f(1)^3$ (as above). But $f(13) = f(2 \cdot 2 + 3 \cdot 3) = f(2)f(3) = 0$, so that $f(1) = 0$.
Given any $n$ odd, we can write $n = 2x + 3y$ with $y = 1$. Then $f(n) = f(?)f(1) = 0$. Thus $f(n) = 0$ if $n$ is odd.
Given any $n$ even, as $f(2x) = 0$ for all $x$, we see that $f(n) = 0$. Thus we have shown that $f(x)$ is identically $0$ if $f(0) = 0$.
Thus in total, there are three such functions. They are
*
*$f(n) = 0$ for all $n$.
*$f(n) = 1$ for all $n$.
*$f(0) = 1$, and $f(n) = 0$ for all $n \geq 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2951724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Roots of polynomial equation $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$
If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_j|=2\space\space\space\forall j\in\{1,2,3,4,5,6\}$
I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=\sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?
|
Hint:\begin{multline}x^6+2x^5+4x^4+8x^3+16x^2+32x+64=\\=64\left(\left(\frac x2\right)^6+\left(\frac x2\right)^5+\left(\frac x2\right)^4+\left(\frac x2\right)^3+\left(\frac x2\right)^2+\frac x2+1\right).\end{multline}Also, use the fact that $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2952564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Sum of $\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - ...$ I have to find the sum of :
$$\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - \frac{x^5}{5*4} +\cdots$$
So far I have :
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)}$$
which is very close to $\ln(1+x)$... but I just can't figure out what I have to do from there.
|
Hint:
On its domain of convergence, the derivative of the sum of this power series is
$$x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dotsm=\ln(1+x).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2954240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Problems with an exact differential eqution Consider the following differential equation
$$
\left(\frac{1}{x}-\frac{y^2}{(x-y)^2}\right)dx=\left(\frac{1}{y}-\frac{x^2}{(x-y)^2}\right)dy
$$
I want to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not work. Can someone give me a hit?
|
The differential is indeed exact
$$\left(\frac{1}{x}-\frac{y^2}{(x-y)^2}\right)dx-\left(\frac{1}{y}-\frac{x^2}{(x-y)^2}\right)dy=0$$
Note that
$$\frac {dx}x=d\ln (x)$$
$$\frac {dy}y=d\ln (y)$$
And also that
$$
\begin{align}
E=&-\frac{y^2}{(x-y)^2}dx+\frac{x^2}{(x-y)^2}dy \\
E=&\frac{-y^2dx+x^2dy}{(x-y)^2}\\
E=&\frac{-y^2dx+x^2dy}{x^2y^2}\frac {(xy)^2}{(x-y)^2}\\
E=&(d(\frac 1x- \frac 1y))\frac {(xy)^2}{(x-y)^2}\\
E=&(\frac {xy}{y-x})^2d(\frac {y-x}{xy}) \\
\end{align}
$$
$$ \text {Since we have } \frac {dv}{v^2}=-d\left(\frac 1v \right ) \implies E=-d(\frac {xy}{y-x})$$
Therefore we have
$$d \ln x -d \ln y +d(\frac {xy}{x-y})=0$$
$$\boxed{\ln (\frac xy)+\frac {xy}{x-y}=K}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2955435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
A theorem for calculating the perpendicular diagonals' slopes of a quadrilateral
If the slopes of side $AB$, $BC$, $CD$, $DA$ of a quadrilateral $ABCD$ are respectively $m_1, m_2, m_3, m_4$, and the diagonals (or the extended lines of them) are perpendicular, then the equation in $p$:
$$(m_1m_3-m_2m_4)p^2-((m_2+m_4)(1+m_1m_3)\\-(m_1+m_3)(1+m_2m_4))p-(m_1m_3-m_2m_4)=0$$
is satisfied by both slopes of the diagonals (in case both exist) or by the only existing one.
This theorem can be proved by means of the theory of circumscribing conics (see my dummy answer below).
Does anyone know a different, more simple proof?
|
$\def\peq{\mathrel{\phantom{=}}{}}$Since the equation remains the same under translation, without loss of generality assume that line $AC$ and $BD$ intersect at the origin $O$. Denote by $p$ the slope of $AC$ (Assume that it exists).
Case 1: $p = 0$. In this case, $y_A = y_C = 0$, and $AC ⊥ BD$ implies $x_B = x_D = 0$. Thus,$$
m_1 = -\frac{y_B}{x_A},\ m_2 = -\frac{y_B}{x_C},\ m_3 = -\frac{y_D}{x_C},\ m_4 = -\frac{y_D}{x_A} \Longrightarrow m_1 m_3 - m_2 m_4 = 0,
$$
then $p = 0$ satisfies the given equation.
Case 2: $p ≠ 0$. In this case, $AC ⊥ BD$ implies the slope of $BD$ is $-\dfrac{1}{p}$, thus$$
y_A = p x_A,\ y_C = p x_C,\ y_B = -\frac{x_B}{p},\ y_D = -\frac{x_D}{p},
$$
and\begin{gather*}
m_1 = \frac{y_B - y_A}{x_B - x_A} = -\frac{x_B + p^2 x_A}{p(x_B - x_A)},\ m_2 = \frac{y_C - y_B}{x_C - x_B} = -\frac{x_B + p^2 x_C}{p(x_B - x_C)},\\
m_3 = \frac{y_D - y_C}{x_D - x_C} = -\frac{x_D + p^2 x_C}{p(x_D - x_C)},\ m_4 = \frac{y_A - y_D}{x_A - x_D} = -\frac{x_D + p^2 x_A}{p(x_D - x_A)}.
\end{gather*}
Thus,\begin{gather*}
\begin{cases}
m_1 p(x_B - x_A) + (x_B + p^2 x_A) = 0\\
m_2 p(x_B - x_C) + (x_B + p^2 x_C) = 0\\
m_3 p(x_D - x_C) + (x_D + p^2 x_C) = 0\\
m_4 p(x_D - x_A) + (x_D + p^2 x_A) = 0
\end{cases}\\
\Longrightarrow \begin{pmatrix}
p^2 - m_1 p & m_1 p + 1 & 0 & 0\\
0 & m_2 p + 1 & p^2 - m_2 p & 0\\
0 & 0 & p^2 - m_3 p & m_3 p + 1\\
p^2 - m_4 p & 0 & 0 & m_4 p + 1
\end{pmatrix} \begin{pmatrix}
x_A \\ x_B \\ x_C \\ x_D
\end{pmatrix} = \begin{pmatrix}
0 \\ 0 \\ 0 \\ 0
\end{pmatrix}.
\end{gather*}
Because $ABCD$ is not degenerate, then $(x_A, x_B, x_C, x_D)^T ≠ (0, 0, 0, 0)^T$, which implies\begin{align*}
0 &= \begin{vmatrix}
p^2 - m_1 p & m_1 p + 1 & 0 & 0\\
0 & m_2 p + 1 & p^2 - m_2 p & 0\\
0 & 0 & p^2 - m_3 p & m_3 p + 1\\
p^2 - m_4 p & 0 & 0 & m_4 p + 1
\end{vmatrix}\\
&= (p^2 - m_1 p) \begin{vmatrix}
m_2 p + 1 & p^2 - m_2 p & 0\\
0 & p^2 - m_3 p & m_3 p + 1\\
0 & 0 & m_4 p + 1
\end{vmatrix}\\
&\peq - (p^2 - m_4 p) \begin{vmatrix}
m_1 p + 1 & 0 & 0\\
m_2 p + 1 & p^2 - m_2 p & 0\\
0 & p^2 - m_3 p & m_3 p + 1
\end{vmatrix}\\
&= (p^2 - m_1 p)(m_2 p + 1)(p^2 - m_3 p)(m_4 p + 1)\\
&\peq - (p^2 - m_4 p)(m_1 p + 1)(p^2 - m_2 p)(m_3 p + 1)\\
&= p^2 (p^2 + 1) \bigl( (m_2 m_4 - m_1 m_3)(p^2 - 1)\\
&\peq + ((m_2 + m_4)(1 + m_1 m_3) - (m_1 + m_3)(1 + m_2 m_4))p \bigr).
\end{align*}
Since $p ≠ 0$, then$$
(m_2 m_4 - m_1 m_3)(p^2 - 1) + ((m_2 + m_4)(1 + m_1 m_3) - (m_1 + m_3)(1 + m_2 m_4))p = 0,
$$
i.e.$$
(m_1 m_3 - m_2 m_4)p^2 - ((m_2 + m_4)(1 + m_1 m_3)\\
- (m_1 + m_3)(1 + m_2 m_4))p - (m_1 m_3 - m_2 m_4) = 0.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2955848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Parametric equation and distance formula Help with this assignment! For the past three weeks now, I've been battling with mathematics assignment. I've successfully solved some but these ones posed a great challenge to me
*
*Eliminate the parameter and identify the graph of the resulting equation
$x = 3v+7, y = v-1$
$x = \sin t, y = \cos t, 0\leq t \leq \pi$
$x = 3\sin t, y = 2\cos t, 0\leq t \leq \pi$
$x = 7u^2-3, y = 4-2u$
$x = 2+\cos t, y = 1+\sin t$
*Prove that the distance between line $y = mx+c$ and the origin is given by $d = \frac{|c|}{\sqrt{1+m^2}}$.
*Use the above equation again to show that the distance between the line and the point $(x,y)$ is $d = \frac{|y-mx-c|}{\sqrt{1+m^2}}$.
For question 1, I just need a hint to go about solving them since they are kinda similar. I seriously need help in question 2 and 3 because I'm lacking when its comes to proving or showing relation between one thing and another.
|
$x = \sin t, y = \cos t, 0\leq t \leq \pi$
I would really hope that you just recognize this as a circle, as this is close to the unit circle definition of cosine and sine.
You could say:
$x^2 = \sin^2 t\\ y^2= \cos^2 t\\
x^2 + y^2 = \sin^2 t + \cos^2 t = 1$
Considering the limits on $t$ once that is applied, you will get a semi-circle.
Moving on to,
$x = 3\sin t, y = 2\cos t, 0\leq t \leq \pi$
That is just a circle with a dilation applied.
And using the same logic for the algebra.
$\frac {x}{3} = \sin t, \frac {y}{2} = \cos t\\
\frac {x^2}{9} + \frac {y^2}{4} = 1$
Regarding:
$x = 7u^2-3, y = 4-2u$
Isolate $u,$ and substitute.
$u = \frac {4-y}{2}\\
x = 7\frac {(4-y)^2}{4}-3$
And simplify.
As for $2, 3$ the distance between a point and a line, is the shortest distance.
I would start at the point (the origin in problem 2) and move along a line perpendicular to the targeted line, until I found the point of intersection.
$y = -\frac {1}{m} x$ is perpendicular to $y= mx + c$ and goes through the origin.
Now use substitution or elimination.
$(1+m^2) x + mc = 0\\
x = \frac {-mc}{1+m^2}\\
y = \frac {c}{1+m^2}$
$d((0,0),(x,y)) = \sqrt {x^2 + y^2} = \sqrt {\frac {c^2 + c^2m^2}{(1+m^2)^2}} = \frac {|c|}{\sqrt{1+m^2}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2957236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solving $2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$ I wanted to know if this equation could be solved any further please.
$$2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$$
I have gone this far:
$$4 \sin x \sin(60^\circ-x)= \sqrt{2}- 1$$
Thank you
|
Hint: $2\sin x(\sqrt{3}\cos x - \sin x) = \sqrt{2} - 1\implies \sqrt{3}\sin(2x) - (1- \cos(2x)) = \sqrt{2} - 1\implies \sqrt{3}\sin(2x)+\cos(2x)=\sqrt{2}\implies \sin(2x+\frac{\pi}{6})= \sin(\frac{\pi}{4})$. At this point, can you continue to the finish line ….?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2958080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Why this $12b^2<49ac< \frac{49}{4}b^2$ inequality given ? What is the use of it? Question :
Let $\alpha$ and $\beta$ be roots of $ax^2+bx+c=0$. Show that
$(4\alpha-3\beta)(4\beta-3\alpha)=\frac{49ac-12b^2}{a^2}$. If
$12b^2<49ac< \frac{49}{4}b^2$ , then show that $\beta $ lies between
$\frac{3}{4}\alpha$ and $\frac{4}{3}\alpha $.
When $\alpha+\beta=-\frac{b}{a} $ and $\alpha \beta = \frac{c}{a} $, I solved a quadratic question where we are asked to show that $$(4\alpha-3\beta)(4\beta-3\alpha)=\frac{49ac-12b^2}{a^2}$$
In the question it is given that $$12b^2<49ac< \frac{49}{4}b^2$$
So by using $12b^2<49ac$ ,
I showed that $\beta $ lies between $\frac{3}{4}\alpha$ and $\frac{4}{3}\alpha $ ,
By arranging $$(\beta-\frac{4}{3}\alpha)(\beta-\frac{3}{4}\alpha)=\frac{12b^2-49ac}{a^2}$$
Why is this part of the inequality given ? What is the use of it ?
$$49ac< \frac{49}{4}b^2$$
|
$$(4\alpha-3\beta)(4\beta-3\alpha)=25\alpha\beta-12(\alpha^2+\beta^2)=49\alpha\beta-2(\alpha+\beta)^2=$$
$$=\frac{49c}{a}-\frac{12b^2}{a^2}=\frac{49ac-12b^2}{a^2}.$$
$49ac<\frac{49b^2}{4}$ says that our equation has two real roots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2960293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Domain and range of $g(x) = \frac{3x^2+4}{x-2}$
$$g(x) = \frac{3x^2+4}{x-2}$$
for the domain, I find, $\{x|x\ne 2\}$'
and for the range
$y = \dfrac{3x^2+4}{x-2}$ and when $x=2-\epsilon$, $y \to \dfrac{3(2-\epsilon)^2 + 4}{(2-\epsilon)-2} = \dfrac{K}{-\epsilon} \to -\infty$
When $x = 2+\epsilon$, $y \to \dfrac{3(2+\epsilon)^2+4}{(2+\epsilon)-2} = \dfrac{K}{\epsilon} \to \infty$
Therefore, $\text{ran}(g(x)) = (-\infty, \infty)$
have I done this right or am I missing something? For instance, I thought about the case when $x \to \infty$, but this looks like $y \to 1$ in this case, nothing remarkable. I don't even know if that is what we say when we have an $\frac{\infty^2}{\infty}$....Do we just say that is one? Or do we say that is $\infty$ still?
|
The domain is $$\mathbb R\setminus\{2\}.$$
Now, let $x>2$.
Thus, by AM-GM
$$\frac{3x^2+4}{x-2}=\frac{3(x-2)^2+12(x-2)+16}{x-2}=3(x-2)+\frac{16}{x-2}+12\geq$$
$$\geq2\sqrt{3(x-2)\cdot\frac{16}{x-2}}+12=12+8\sqrt{3}.$$
The equality occurs for $3(x-2)=\frac{16}{x-2}$ or $x=2+\frac{4}{\sqrt3}.$
Also, let $x<2$.
Thus, by AM-GM again we obtain:
$$\frac{3x^2+4}{x-2}=\frac{3(x-2)^2+12(x-2)+16}{x-2}=3(x-2)+\frac{16}{x-2}+12=$$
$$=12-\left(3(2-x)+\frac{16}{2-x}\right)\leq12-2\sqrt{3(2-x)\cdot\frac{16}{2-x}}=12-8\sqrt{3}.$$
The equality occurs for $3(2-x)=\frac{16}{2-x}$ or $x=2-\frac{4}{\sqrt3}$ and since $g$ is a continuous function on any interval of the domain and
$$\lim_{x\rightarrow+\infty}g(x)=+\infty$$ and
$$\lim_{x\rightarrow-\infty}g(x)=-\infty,$$ we got the following range:
$$\left(-\infty,12-8\sqrt3\right]\cup\left[12+8\sqrt3,+\infty\right).$$
Because $$3x^2+4=3x^2-12x+12+12x-24+16=$$
$$=3(x^2-4x+4)+12(x-2)+16=3(x-2)^2+12(x-2)+16.$$
I used AM-GM:
For non-negatives $a$ and $b$ we have:
$$\frac{a+b}{2}\geq\sqrt{ab}.$$
The equality occurs for $a=b$.
For $x>2$ we have $a=3(x-2)$ and $b=\frac{16}{x-2}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2960527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Multiple divisions I was confused a bit with a little arithmetic here. For instance $1÷1÷2$ and $2÷3÷7$. BODMAS isn't effective in this case. My question is this:
$2÷3÷7$
Am I to divide $2/3$ by $7$ or divide $2$ by $3/7$??
|
You need to carry on first division first.
$$\frac{\frac{2}{3}}{7}=\frac{\frac{2}{3}}{\frac{7}{1}}=\frac{2}{3}\cdot\frac{1}{7}=\frac{2}{21}$$
while
$$\frac{2}{\frac{3}{7}}=\frac{\frac{2}{1}}{\frac{3}{7}}=\frac{2}{3}\cdot\frac{7}{1}=\frac{14}{3}$$
BODMAS is an acronym and it stands for Bracket, Of, Division, Multiplication, Addition and Subtraction .
It gives you the order of precedence of operations:
When there are similar operations, you can proceed with any one of them;
but when you have a mixture of operations:
*
*Bracket gets the highest priority
*proceed with multiplication and division before addition and subtraction
*For similar operations, you can proceed with the order in which they appear, or in any way you feel easier.
E.g. $$(4+2)-3=4+(2-3)$$
$$4\cdot \frac{9}{5} = \frac{4 \cdot 9}{5}$$
$$(4+2)\cdot3 \neq 4+(2\cdot3)$$
$$\frac{4-2}{3} \neq 4- \frac{2}{3}$$
E.g. $$4+(6-1)\cdot7-\frac{8}{2}$$
will be done like $$4+5\cdot7-4$$
$$=4+35-4$$
$$=35$$
And not like:
$$4+6-1\cdot7-\frac{8}{2}$$
$$=10-7-4$$
$$=-1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2963227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$7^x-3^y=4$ in the integers I have to find all solutions to $7^x-3^y=4$ in the integers, I've already proven that $x$ and $y$ have the same parity and that they cannot be even. But I'm stuck in the case when $x$ and $y$ are odd. Could someone show me how to solve for all solutions? I know there is at least one, $(x,y)=(1,1)$, but I don't know if there's any other solution
|
If $y\leq 1$, the only solution is $(1,1)$, which you found. If $y\geq 2$, then consider the equation $\bmod 9$. We see
$$7^x\equiv 4\bmod 9$$
We have
$$7^0\equiv 1,\ 7^1\equiv 7,\ 7^2\equiv 4,\ 7^3\equiv 1,$$
so $7^x\equiv 4\bmod 9$ iff $x\equiv 2\bmod 3$. As a result, we have $x\equiv 5\bmod 6$ (as you have already shown $x$ is odd). So, as
$$7^{12}\equiv 1\bmod 13,$$
we have
$$7^x\equiv 7^5\mathrm{\ or\ }7^{11}\equiv \pm 2\bmod 13$$
since $x\equiv 5\bmod 6\implies x\equiv 5\mathrm{\ or\ }11\bmod 12$. This means that
$$3^y=7^x-4\equiv (\pm 2)-4\in\{7,11\}\bmod 13.$$
However
$$3^0\equiv 1,3^1\equiv 3,3^2\equiv 9,3^3\equiv 1\bmod 13,$$
so we get a contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2963683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
}
|
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.