Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
| Put $a = \sqrt[3]{5x+7}, b = \sqrt[3]{5x-12}\implies a - b = 1, a^3 - b^3 =19\implies a^2+ab+b^2 = \dfrac{a^3-b^3}{a-b} = \dfrac{19}{1} = 19\implies (a-b)^2+3ab = 19\implies 1^2+3ab = 19 \implies ab = 6\implies (b+1)b = 6\implies b^2+b-6 = 0\implies (b+3)(b-2)=0\implies b = 2, -3\implies 5x-12 = 2^3, (-3)^3 = 8,-27\implies 5x = 20, -15\implies x = 4, -3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
How i can prove that $\sum_{i=0}^{r+1}\binom{n-i}{r-i+1} = \binom{n}{r}$ I'm trying to prove this binomial coefficient identity.
$$\sum_{i=0}^{r+1}\binom{n-i}{r-i+1} = \binom{n}{r} $$
The textbook's hint is about using the identity
$$\binom{n}{r} =\binom{n}{n-r}$$
I've only reached this
$$\frac{1}{(n-1-r)!}\left(\frac{(n-2)!}{(r-1)!}+\frac{(n-3)!}{(r-2)!}+\dots+\frac{(n-r-1)!}{0!}\right)$$
| You can use the Identity:
$$ \begin{pmatrix}
n \\
r
\end{pmatrix} =
\begin{pmatrix}
n - 1 \\
r
\end{pmatrix} + \begin{pmatrix}
n - 1 \\
r - 1
\end{pmatrix}$$
Recursively in the last term { $C(n - 1, r - 1)$ }. Doing this you will have:
$$ \begin{pmatrix} n \\r \end{pmatrix} = \begin{pmatrix} n - 1 \\r \end{pmatrix} + \begin{pmatrix} n - 1 \\r -1 \end{pmatrix} $$
for the first iteration, then:
$$\begin{pmatrix} n \\r \end{pmatrix} = \begin{pmatrix} n - 1 \\r \end{pmatrix} + \begin{pmatrix} n - 2 \\r -1 \end{pmatrix} + \begin{pmatrix} n - 2 \\r - 2 \end{pmatrix}$$
Recursively this leads to:
$$\begin{pmatrix} n \\r \end{pmatrix} = \begin{pmatrix} n - 1 \\r \end{pmatrix} + \begin{pmatrix} n - 2 \\r -1 \end{pmatrix} + \begin{pmatrix} n - 3 \\r - 2 \end{pmatrix} + \begin{pmatrix} n - 4 \\r - 3 \end{pmatrix} + ... + \begin{pmatrix} n - r + 1 \\r - r \end{pmatrix} $$
For all other coming values the terms vanishes to zero ($ C(n, r) = 0 \hspace{2mm} \forall \hspace{2mm} r > n $)
this can be written is sigma notation as:
$$ \begin{pmatrix} n \\r \end{pmatrix} = \sum_{i=0}^{r} \begin{pmatrix} n - 1 - i \\r - i \end{pmatrix} $$
Changing the index
$$ j = i + 1 $$
We can write the summation as fallows:
$$ \sum_{i=0}^{r} \begin{pmatrix} n - 1 - i \\r - i \end{pmatrix} =
\sum_{j = 1}^{r + 1} \begin{pmatrix} n - 1 - (j - 1) \\r - (j - 1) \end{pmatrix}
$$
This is:
$$
\sum_{j = 1}^{r + 1} \begin{pmatrix} n - j \\r - j + 1 \end{pmatrix}
$$
And obviously we can change the dummie index as $ j \longrightarrow i $ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An object travelling on $x^2$ I have an object traveling along the curve $y=x^2$. $z$ is the distance from the origin and $dz/{dt}=1$ is the rate it's increasing per unit time. At what rate are my $x$ and $y$'s moving at the point (2,4)? In other words what is $dy/dt$ and $dx/dt$.
I've seen the geometry. Found $z$ as the hypotenuse of $x$ and $y$. Also, I infer that $dz=dt $ so they are interchangable. How do I solve this?
| I think this is simpler to understand. By definition
$$z=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$$
Since $y=x^2$
$$z=\sqrt{x^2+x^4}$$
Taking the derivative of this equation using the chain rule gives
$$z^{'}=\left({(x^2+x^4)}^{1/2}\right)^{'}\implies\frac{dz}{dt}=1/2(x^2+x^4)^{1/2-1}(x^2+x^4)^{'}\implies\frac{dz}{dt}=(1/2)(x^2+x^4)^{-1/2}(2x+4x^3)\frac{dx}{dt}$$
Then substitue $\frac{dz}{dt}=1$ and $x=2$ into the previous equation
This gives $1=\frac{1}{2}\frac{1}{\sqrt{20}}(36)\frac{dx}{dt}\implies1=\frac{18}{\sqrt{20}}\frac{dx}{dt}\implies\frac{dx}{dt}=\frac{\sqrt{20}}{18}$
Then take the derivative of $y=x^2$ on both sides to get
$$\frac{dy}{dt}=2x\frac{dx}{dt}$$
Then substitue $x=2$ and $\frac{dx}{dt}=\frac{\sqrt{20}}{18}$ to get your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a 3 x 3 orthogonal matrix Find a 3x3 orthogonal matrix whose first column is $[\frac1{\sqrt{3}},{- \frac 1{\sqrt{3}}}, \frac1{\sqrt{3}}]$.
I was thinking this has to do with: ${Q}^T•Q=I_{n}$, but I still don't understand how to go about this.
| In this case we can also guess $3$ vectors without calling for Gram-Schmidt by noticing
$\begin{pmatrix} 1 \\ -1 \\ 1\end{pmatrix}\text{ , }\begin{pmatrix} 1 \\ 2 \\ 1\end{pmatrix}\text{ and }\begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix}\text{ are orthogonal.}$
By keeping first and third coordinate of $v_1,v_2$ equal, we were able to find an easy $v_3$.
Normalizing this gives $Q=\frac 1{\sqrt{6}}\begin{pmatrix} \sqrt{2} & 1 & \sqrt{3} \\ -\sqrt{2} & 2 & 0 \\ \sqrt{2} & 1 & -\sqrt{3} \end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability of picking at least x blue balls out of red and blue balls Given $n$ balls in total.
Each ball is either red or blue.
There are $ r $ red balls and $ b $ blue balls.
At random, and without replacement, we pick $ t $ balls out of the $ n $ balls.
How to calculate the probability for the event that at least $ x $ of the picked balls are blue?
For an example consider: $n=5000$, $r=1500$, $b=3500$, $t=500$, $x=350$
| The probability of first getting $x$ blue balls, followed by $t-x$ red balls, equals:
$$\frac{b}{n} \cdot \frac{b-1}{n-1} \cdot \ldots \cdot \frac{b-x+1}{n-x+1} \cdot \frac{r}{n-x} \cdot \frac{r-1}{n-x-1} \cdot \ldots \cdot \frac{r-t+x+1}{n-t+1} = \frac{b!}{(b-x)!} \frac{r!}{(r-t+x)!} \frac{(n-t)!}{n!}$$
There are ${{t}\choose{x}}$ ways to select the turns in which we pick a blue ball, so the probability of selecting exactly $x$ blue balls equals:
$${{t}\choose{x}} \frac{b!}{(b-x)!} \frac{r!}{(r-t+x)!} \frac{(n-t)!}{n!}$$
The probability of selecting at least $x$ balls then becomes:
$$\sum_{x=350}^{500}{{t}\choose{x}} \frac{b!}{(b-x)!} \frac{r!}{(r-t+x)!} \frac{(n-t)!}{n!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the angle between 2 intersection lines of four planes?
Given $L_1=\begin{cases}P_1: x-y+z=4\\ P_2: 2x+y-z=6 \end{cases}$
and $L_2=\begin{cases}P_1:x+y+z=4\\ P_2:2x+3y-z=6 \end{cases}$
$L_1$ and $L_2$ are the intersection lines of the given planes. Find the angle between the lines $L_1,L_2$
First in order to find the intersection lines between the respective planes it is enough to find to dots that belong to the intersection line, so for $L_1$:
$$
P_1+P_2 \Rightarrow 3x=10 \Rightarrow x={10 \over 3}
$$
Then:
$$
2 \cdot \frac{10}{3}+y-z=6 \Rightarrow y-z=-\frac{2}{3}
$$
For $z=0$ and $z=1$ we get $y=-\frac{2}{3}$ and $y=\frac{1}{3}$ respectively so now can represent $L_1$ parametrically: $(\frac{10}{3},\frac{1}{3}, 10)+t(0,-1,0)$ where $t(0,-1,0)$ is the direction vector.
After identical calculations we can find $L_2: (0,\frac{10}{4}, \frac{3}{2})+t(\frac{10}{3}, 0,-\frac{2}{3})$.
Now we can calculate the angle of $L_1, L_2$ via scalar product:
$$
\cos \theta=\frac{L_1 \cdot L_2}{\Vert{L_1}\Vert \cdot \Vert L_2 \Vert}=0 \Rightarrow \theta=90^{\circ}
$$
I'm not sure that my answer is correct and I feel I'm missing something.
| Let $\vec{l_1}(a,b,c)$.
Thus, $a-b+c=0$ and $2a+b-c=0$, which gives $a=0$ and $b=c$ and $\vec{l_1}(0,1,1).$
Let $\vec{l_2}(a,b,c)$.
Thus, $a+b+c=0$ and $2a+3b-c=0$, which gives $b=-\frac{3}{4}a$, $c=-\frac{1}{4}a$
and $\vec{l_1}(4,-3,-1).$
Now,
$$\measuredangle\left(L_1,L_2\right)=\arccos\frac{|0\cdot4+1\cdot(-3)+1\cdot(-1)|}{\sqrt{4^2+(-3)^2+(-1)^2}\sqrt{0^2+1^2+1^2}}=\arccos\frac{2}{\sqrt{13}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find in a more formal way $\lfloor{(2+\sqrt3)^4}\rfloor$
Problem Statement:-
The largest integer which is less than or equal to $(2+\sqrt3)^4$ is
$\text{(A)}192\qquad\qquad \text{(B)}193\qquad\qquad \text{(C)}194\qquad\qquad\text{(C)}195\qquad\qquad$
My Solution:-
I started with the binomial expansion but as I was not able to think of any manipulation which would make my life easier so I just did the following.
$$(2+\sqrt3)^2=\binom{4}{0}2^4+\binom{4}{1}(2^3)(\sqrt3)+\binom{4}{2}(2)^2(\sqrt3)^2+\binom{4}{3}(2)(\sqrt3)^3+\binom{4}{4}(\sqrt3)^4$$
On arranging the integers and the irrational terms separately I ended up with $$(2+\sqrt3)^4=97+56\sqrt3=97+56+56(\sqrt3-1)=153+56(\sqrt3-1)$$
At this point still no manipulation was obvious to me that I can do that would give me answer to $\lfloor{(2+\sqrt3)^4}\rfloor$, so I just used a rough estimate of $(\sqrt3-1)\approx0.732$ to get $56(\sqrt3-1)=40.992$.
The rough estimate of $56(\sqrt3-1)\approx40.992$ is dangerously close to $41$ so at this point I was pretty confused as to how to ascertain that $56(\sqrt3-1)\lt 41$ or $56(\sqrt3-1)\gt 41$, which had left me with the answer as one of the options out of (C) and (D).
Your help is needed for resolving this doubt of mine.
Also, what would have been a more subjective approach.
| Another way . . .
Let $a = 2 + \sqrt{3},\;\;b = 2 - \sqrt{3}$.
Then
$$a + b = 4$$
$$ab = 1$$
so
$$a^2 + b^2 = (a+b)^2 - 2ab = 4^2-2(1) = 14$$
hence
$$a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = 14^2 - 2(1)^2 = 194\\[0pt]$$
Then
\begin{align*}
&a^4 + b^4 = 194\\[6pt]
\implies\; &193 < a^4 < 194\qquad\text{[since $0 < b < 1$]}\\[6pt]
\implies\; &\lfloor{a^4}\rfloor=193
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given the square, calculate $\tan{\alpha}.$ Problem: Given the square $ABCD$, let $M$ be the midpoint on the side $|CD|$ and designate $\alpha=\angle AMB$. Calculate $\tan{\alpha}.$
Attempt: We can, without compromising generality, assume that the side of the square is equal to 1. So drawing a figure we get
I know the following:
1) That $|AM|=|BM|=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}.$
2) The area of $ABM$ can be expressed in two ways. One way with normal geometry for triangle and another way is by using the areakit involving $\sin{\alpha}$. So: $$\begin{array}{lcl}
A_1 & = & \frac{1\cdot 1}{2} = \frac{1}{2} \\
A_2 & = & \frac{|AM|\cdot|BM|\cdot\sin{\alpha}}{2} = \frac{5}{2}\cdot\sin{\alpha} \\
\end{array}$$
3) I know that $\tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}},$ so finding $\sin{\alpha}$ and $\cos{\alpha}$ and dividing these two will solve this problem.
Setting $A_1=A_2$ yields the equation $$\frac{5}{2}\sin{\alpha}=\frac{1}{2} \ \Longleftrightarrow \ \sin{\alpha} = \frac{1}{5}.$$
Using the law of cosines in the triangle $ABM$ I get
$$\begin{array}{lcl}
|AB|^2 & = & |AM|^2+|BM|^2 -2|AM||BM|\cos{\alpha} \\
1 & = & \sqrt{5}-\sqrt{5}\cos{\alpha} \ \Leftrightarrow \ \cos{\alpha} = \frac{\sqrt{5}-1}{\sqrt{5}} \\
\end{array}$$
And finally we have $$\tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}}=\frac{\frac{1}{5}}{\frac{\sqrt{5}-1}{\sqrt{5}}} = \frac{5+\sqrt{5}}{20}.$$
But it's not correct. Any idea where I'm making the mistake in my attempt, and is there an easier way of solving this problem?
| You've already found the lengths shown below.
Hence
\begin{align}
u^2 &= \left(\frac{\sqrt 5}{2}\right)^2 -\left(\frac{2}{\sqrt 5}\right)^2 \\
&= \frac 54 - \frac 45 \\
&= \frac{9}{20} \\
u &= \frac{3}{2\sqrt 5}
\end{align}
It follows that
$ \tan \alpha = \dfrac{\frac{2}{\sqrt 5}}{\frac{3}{2\sqrt 5}} = \frac 43$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{5}$ and $abc = 5$, solve for $a^3 + b^3 + c^3$ I recently encountered this question and have been stuck for a while. Any help would be appreciated!
Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \label{eq:2}$$
Find $a^3 + b^3 + c^3$. It wasn't specified in the question but I think it can be assumed that $a, b, c$ are real numbers.
My approach:
$$ ab + ac + bc = \frac{1}{5} abc = 1 $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3[(a + b + c)(ab + ac + bc) - abc] $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c) + 15 $$
From there, I'm not sure how to go about solving for $a + b + c$.
Something else I tried was letting $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$, so we get $$ xyz = x + y + z = \frac{1}{5} $$Similarly, I'm not sure how to continue from there.
| Let $k$ be an arbitrary real constant and $a$, $b$, $c$ be the roots of the cubic equation $x^3-kx^2+x-5=0$. Then we have $ab+bc+ca=1$ and $abc=5$. Also, we have $a+b+c=k$.
$$a^3+b^3+c^3=k(a^2+b^2+c^2)-(a+b+c)+15=k(a+b+c)^2-2k(ab+bc+ca)-(a+b+c)+15=k^3-2k-k+15=k^3-3k+15$$
Since there are infinitely many possible $k$ such that $x^3-kx^2+x-5=0$ has 3 real roots, the problem has infinitely many possible answers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$ How can I solve the following integral?
$$\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$$
$$\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\end{align}$$
$$u = x+\frac{7}{2}, \quad a = \frac{7}{2}$$
$$\int \frac{1}{\sqrt{u^2 - a^2} +3}\, \,du$$
Attempt I - By Trigonometric substitution
$$\sqrt{u^2 - a^2} = \sqrt{a\sec^2\varTheta - a^2} = \sqrt{a^2(\sec^2\varTheta - 1)} = \sqrt{a^2\tan^2\varTheta} = a\tan\varTheta $$
$$u = a\sec\varTheta \implies du = a\sec\varTheta \tan\varTheta d\varTheta$$
$$\int \frac{1}{a\tan\varTheta + 3}\, a\sec\varTheta \tan\varTheta d\varTheta$$
$$a\int \frac{\sec\varTheta \tan\varTheta}{a\tan\varTheta + 3}\, d\varTheta$$
How can I continue here ?
Attempt II - By First Euler Substitution
$$\sqrt{u^2 - a^2} = u - t$$
$\sqrt{u^2 - a^2} = u - t$
$ u^2 - a^2 = (u - t)^2$
$u^2 - a^2 = u^2 -2ut + t^2$
$- a^2 = -2ut + t^2$
$2ut = t^2 + a^2$
$$u = \frac{t^2 + a^2}{2t} \implies du = \frac{1}{2} - \frac{a^2}{2t^2}\, dt$$
Thus the integral takes the form
$$\int \frac{1}{u - t +3}\, \left( \frac{1}{2} - \frac{a^2}{2t^2}\right) \,du$$
Since $u = \frac{t^2 + a^2}{2t}$ and $\left( \frac{1}{2} - \frac{a^2}{2t^2}\right) = -\frac{a^2 - t^2}{2t^2}$ then
$$I = \int \frac{1}{ \frac{t^2 + a^2}{2t} - t +3} \left( -\frac{a^2 - t^2}{2t^2} \right) \, dt$$
$$-\int \frac{a^2 - t^2}{ ta^2 - t^3 +6t^2}\, dt$$
$$-\int \frac{a^2}{ ta^2 - t^3 +6t^2}\, dt + \int \frac{t^2}{ ta^2 - t^3 +6t^2}\, dt$$
How can I continue?
Attempt III -By First Euler Substitution
$$\sqrt{x^2 + 5x} = x + t$$
$\sqrt{x^2 + 5x} = x + t$
$ x^2 + 5x = (x + t)^2$
$x^2 + 5x = x^2 + 2ut + t^2$
$x^2 + 5x = x^2 + 2xt + t^2$
$5x = 2xt + t^2$
$5x -2xt = t^2$
$x(5 -2t) = t^2$
$$x = \frac{t^2}{(5 -2t)} \implies \frac{dt}{dx} = -\frac{2(t - 5)t}{(5 - 2t)^2 } \iff dx = -\frac{(5 - 2t)^2 }{2(t - 5)t}\,dt $$
$$\int \frac{1}{x + t + 3}\, \left( -\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$
$$-\int \frac{1}{\frac{t^2}{(5 -2t)} + t + 3}\, \left(\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$
$$-\int\frac{(5 - 2t)^2}{\frac{2(t - 5)t^3}{5 -2t} + 2(t - 5)t^2 + 6(t - 5)t}\,dt$$
How can I continue here?
| Hint. By the change of variable
$$
x+\frac72=\frac72 \cdot \cosh u,\qquad dx=\frac72 \cdot \sinh u\:du,
$$ one gets
$$
\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\\\\&=\int \frac{\frac72\sinh u}{\frac72\sinh u + 3}\, du
\end{align}
$$ then one may conclude with the change of variable
$$
t=\tanh \frac u2, \quad \sinh u= \frac{2t}{1-t^2},\quad du=\frac{2}{1-t^2}\:dt.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$? The function
$$
f(z) = \frac{z}{z^2 - 1}
$$
has singularities at $z = \pm 1$. I will expand about the point $z = 1$. Then, with the substitution $w = z+1$,
\begin{align*}
f(x) =& \frac{z}{z^2 - 1} \\[3mm]
=& \frac{z}{(z+1)(z-1)} \\[3mm]
=& \frac{w-1}{w(w-2)} \\[3mm]
=& \frac{1-w}{2w} \cdot \frac{1}{1 - \frac{w}{2}} \\[3mm]
=& \frac{1-w}{2w} \sum_{n=0}^\infty \frac{w^n}{2^n} \\[3mm]
=& \sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{w^n}{2^{n+1}} \\[3mm]
=& \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}
\end{align*}
This is almost in the form of a Laurent expansion, but not quite. How can I continue this to get the correct final result?
| At the penultimate line you have
$$\sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}}
-\sum_{n=0}^\infty \frac{w^n}{2^{n+1}}.$$
This is
$$\frac1{2w}+\frac14+\frac w8+\frac{w^2}{16}+\cdots-\frac12-\frac w4-\frac{w^2} 8-\cdots=\frac1{2w}-\sum_{n=0}^\infty\frac{w^n}{2^{n+2}}.$$
Does this look more like a Laurent series?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A$, $B$, and $C$.
If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A, B$, and $C$.
Would I use half-angle identities for this problem?
| No, just take $x=0,\pi/4,\pi/2$, and you'll get three equations on $A,B,C$:
\begin{align}
0&=A+B+C\\ \ \\
\frac14&=A+0-C\\ \\
1&=A-B+C
\end{align}
Solving, we get $A=3/8$, $B=-1/2$, $C=1/8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Solving simultaneous equations with complex numbers and moduli Would anyone be able to give me a starting point as to how to approach this? A friend suggested squaring both sides, but I read something that said you can't square both sides of an equation(?).
Complex numbers and simultaneous equations question
If $x$ and $y$ are both real numbers, find all the solutions $(x,y)$ of the simultaneous equations
$$
\begin{cases}
\lvert x + i y \rvert &= 1\\[3pt]
\big\lvert x + i y - \frac{3}{2} \big\rvert &= 2
\end{cases}
$$
Thank you in advance!
Edit: So, would you square the real and imaginary parts separately? Giving: $x^2+y^2=1$ for the first one? Then, $(x-3/2)^2+y^2=4$ for the second one?
| Hint: let $z=x+iy$ and write the system as:
$$
\begin{cases}
\begin{align}
\lvert z \rvert &= 1 \\
\left| z - \frac{3}{2} \right| &= 2
\end{align}
\end{cases}
$$
Since $z \bar z = |z|^2=1$ the second equation is equivalent to:
$$
2^2 = \left| z - \frac{3}{2} \right|^2 = \left( z - \frac{3}{2} \right) \left( \bar z - \frac{3}{2} \right) = z \bar z - \frac{3}{2}(z+\bar z)+\frac{9}{4} = -3 \operatorname{Re}(z)+\frac{13}{4}
$$
Therefore $\operatorname{Re}(z)=x=-\cfrac{1}{4}\,$, then $\operatorname{Im}(z) = y = \pm \sqrt{1 - \operatorname{Re}^2(z)}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$
The expression $\sin(3x)$ is equivalent to:
A. ...
My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x)\cos(x)-\sin(x)+\cos(x)2\sin(x)\cos(x) = \\
\sin(x)(2\cos^2(x)-1+2\cos^2(x)) = \\
\sin(x)(4\cos^2(x)-1) = \\
4\cos^2(x)\sin(x)-\sin(x) = ???$$
What do I do next?
| $$\sin(3x) = \sin(x+2x)$$
$$= \sin(x) \cos(2x) + \sin(2x) \cos(x)$$
$$ = \sin(x) ( \cos^2(x) - \sin^2 (x)) + \cos(x) (2 \sin(x) \cos(x)) $$
$$= 3 \sin(x) \cos^2 (x) - \sin^3 (x )$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the value of $f(1)$ from a given functional equation
A function $f: \mathbb{Q}^+ \cup \{0\} \to \mathbb{Q}^+ \cup \{0\}$ is defined such that $$ f(x) + f(y) + 2xyf(xy) = \frac{f(xy)}{f(x+y)}$$
Then what is the value of $\left[f(1)\right]$ (where $[.]$ denotes the greatest integer function)?
I proceeded this way:
Putting $x=y=0$ I got $f(0) = \frac{1}{2}$ (assuming $f(0) \neq 0$)
Again putting $y=0$ I got $f(x) + f(0) = \frac{f(0}{f(x)}$ which gave 2 values of $f(x)$ as $-1$ and $\frac{1}{2}$.
As $f(0)$ was equal to $\frac{1}{2}$ so I assumed $f(x)$ as a constant function having value $\frac{1}{2}$ for all $x$.
When $f(0) = 0$ then I got $f(x) = 0$ for all $x$.
But the answer was given to be equal to $1$ which means $f(1) \in [1,2)$. Where am I wrong?
| *
*Assume $f(0) \neq 0$.
Then $f(0) + f(0) = \frac{f(0)}{f(0)} \rightarrow f(0) = \frac{1}{2}$,
and $f(x) + f(0) = \frac{f(0)}{f(x)} \rightarrow f(x)^2 + \frac{1}{2}f(x) -\frac{1}{2} = 0.$
*Assume $f(1) \neq 0$.
We also know that $f(1) + f(1) + 2f(1) = \frac{f(1)}{f(2)}$. Or $f(2) = \frac{1}{4}$.
However, $(\frac{1}{4})^2 + \frac{1}{2}\cdot\frac{1}{4} - \frac{1}{2} \neq 0$. Contradiction.
*Assume $f(1) = 0$.
$0^2 + \frac{1}{2}\cdot 0 - \frac{1}{2} \neq 0$. Contradiction.
Therefore we have $f(0) = 0$. However then we have for any $x$:
$$f(x) = \frac{0}{f(x)}$$
If $f(x) \neq 0$ then the above formula implies $f(x)^2 = 0$ which is a contradiction.
So the function must be $f(x) = 0$ everywhere however if this is the case then the 'definition' is ill-defined (division by zero) everywhere, therefore I argue that $f$ itself is ill-defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trouble calculating Fourier series of a function Calculate the Fourier series
$$
f(x) = \left\{\begin{aligned}
& 7\sin(x), && 0 \le x \le \pi\\
& 0, && \pi \le x \le 2\pi
\end{aligned}
\right.$$
I know that when $f(x+L) = f(x)$, for all real $x$, the Fourier series expansion is
$$f(x) = \dfrac{a_0}{2} + \sum_{n = 1}^{\infty} a_n \cos\left(n \tfrac{2\pi}{L}x \right) + b_n\sin\left(n \dfrac{2\pi}{L}x \right)$$
where
$$a_n = \dfrac{2}{L}\int\limits_{x_0}^{x_0+L}f(x)\cos\left( n \tfrac{2\pi}{L} x \right) dx \qquad x_0 \in \mathbb{R}$$
and $$b_n = \dfrac{2}{L}\int\limits_{x_0}^{x_0+L}f(x)\sin\left( n \tfrac{2\pi}{L} x \right) dx \qquad x_0 \in \mathbb{R}$$
But if we attempt to calculate $a_n$, we get $a_n = \dfrac{7}{\pi}\int^{\pi}_0 \sin(x)\cos(nx) dx$. For $n \ge 1$, I do not understand how to calculate this integral. Even If I use integration by parts, I still get an integral that is not solvable.
EDIT: To be honest, I am not even sure I got this part correct. We are supposed to have $-L \le x \le L$, but for this problem, we do not have a $-L$? Instead, we have $0 \le x \le 2\pi$. What is going on here? (This was corrected in the equations above.)
I would greatly appreciate it if people could please take the time to explain how to solve such a problem.
| Hint: The Fourier series of a function $f$ in $[a,a+2\ell]$ with period $T=2\ell$ is
$$f(x) = \dfrac{a_0}{2} + \sum_{i = 1}^{\infty} a_n \cos\left( \dfrac{n\pi x}{\ell} \right) + b_n\sin\left( \dfrac{n\pi x}{\ell} \right)$$
where
\begin{eqnarray}
a_0 &=& \dfrac{1}{\ell}\int_a^{a+2\ell}f(x) dx \\
a_n &=& \dfrac{1}{\ell}\int_a^{a+2\ell}f(x)\cos\left( \dfrac{n\pi x}{\ell} \right) dx \\
b_n &=& \dfrac{1}{\ell}\int_a^{a+2\ell}f(x)\sin\left( \dfrac{n\pi x}{\ell} \right) dx
\end{eqnarray}
Edit: Here $a=0$ and $\ell=\pi$ then
\begin{eqnarray}
a_0 &=& \dfrac{1}{\pi}\int_0^\pi7\sin x\,dx\\
&=&\dfrac{7}{\pi}\int_0^\pi\sin x\,dx\\
&=&\color{blue}{\dfrac{14}{\pi}}\\
a_n&=&\dfrac{1}{\pi}\int_0^\pi7\sin x\cos nx\,dx\\
&=&\dfrac{7}{2\pi}\left(-\dfrac{\cos(1-n)x}{1-n}-\dfrac{\cos(1+n)x}{1+n}\right)_0^\pi\\
&=&\color{blue}{\dfrac{7}{\pi}\dfrac{(-1)^n+1}{1-n^2}}
\end{eqnarray}
for $n\neq1$ and when $n=1$
$$a_1=\dfrac{1}{\pi}\int_0^\pi7\sin x\cos x\,dx=\dfrac{7}{2\pi}\int_0^\pi\sin2x\,dx=\color{blue}{0}$$
also
\begin{eqnarray}
b_n
&=&\dfrac{1}{\pi}\int_0^\pi7\sin x\sin nx\,dx\\
&=&\dfrac{-7}{2\pi}\int_0^\pi\cos(1+n)x-\cos(1-n)x\,dx\\
&=&\dfrac{-7}{2\pi}\left(\dfrac{\sin(1+n)x}{1+n}-\dfrac{\sin(1-n)x}{1-n}\right)_0^\pi\\
&=&\color{blue}{0}~,~(n>1)\\
b_1&=&\dfrac{1}{\pi}\int_0^\pi7\sin x\sin x\,dx\\
&=&\dfrac{7}{2\pi}\int_0^\pi1-\cos2 x\,dx\\
&=&\color{blue}{\dfrac{7}{2}}
\end{eqnarray}
thus
$$\color{blue}{\boxed{f(x)=\dfrac{7}{\pi}+\dfrac{7}{2}\sin x+\dfrac{7}{\pi}\sum_{n=2}^\infty\dfrac{(-1)^n+1}{1-n^2}\cos nx}}$$
or
$$\color{blue}{\boxed{f(x)=\dfrac{7}{\pi}+\dfrac{7}{2}\sin x-\dfrac{14}{\pi}\left(\dfrac{\cos 2x}{3}+\dfrac{\cos 4x}{15}+\dfrac{\cos 6x}{35}+\cdots\right)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the $2\times 2$ matrix $D$ such that $P^{-1}DP=\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$ Let $P = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Find the $2 \times 2$ matrix $D$ such that
$$P^{-1} DP = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$$
I think matrix multiplication is not associative, so what else can I do?
| Hint: The matrix $P$ is diagonalizable with distinct eigenvalues $1$ and $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\int ^{\infty}_{-\infty} \frac{\cos x}{e^x+e^{-x}} dx$ Show that $$\int ^{\infty}_{-\infty} \frac{\cos x}{e^x+e^{-x}}dx=\frac{\pi}{e^\frac{\pi}{2}+e^{-\frac{\pi}{2}}}.$$
How do we solve this? Since given function is even, it follows that
$$ \int ^{\infty}_{-\infty} \frac{\cos x}{e^x+e^{-x}}dx= 2\int ^{\infty}_{0} \frac{\cos x}{e^x+e^{-x}}dx,$$
but I can't go for further.
| $$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{\cos x}{e^x+e^{-x}} d x
= & 2 \operatorname{Re} \int_0^{\infty} \frac{e^{-x} \cdot e^{i x}}{1+e^{-2 x}} d x \\
= & 2 \operatorname{Re} \sum_{n=0}^{\infty} \int_0^{\infty} e^{(i-1-2 n) x} d x \\
= & 2 \operatorname{Re} \sum_{n=0}^{\infty} \left(-1\right)^n\left[\frac{e^{(i-1-2 n) x}}{i-1-2 n}\right]_0^{\infty} \\
= & 2 \operatorname{Re} \sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+1-i} \\
= & 2 \operatorname{Re}\sum_{n=0}^{\infty} \frac{(-1)^n(2 n+1+i)}{(2 n+1)^2+1} \\
= & 2 \sum_{n=0}^{\infty} \frac{(-1)^n(2 n+1)}{(2 n+1)^2+1} \\
= & \frac{\pi}{2} \operatorname{sech}\left(\frac{\pi}{2}\right)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rational numbers inequality proof - is this proof valid?
Knowing that $x^2+y^2 = 2 $ prove that $x+y \le 2$
I rewrite this as
$$x^2+y^2 \ge x+y$$
Now, multiply both sides by 2
$$x^2 -2x + 1 + y^2-2y + 1 -2 +x^2 + y^2 \ge 0$$
I substitue $2$ for $x^2+y^2$
$$(x-1)^2+(y-1)^2\ge0$$
Which is true
| Or (shortened, once we know how to poceed):
$$x+y\le x+y+\tfrac{ (x-1)^2+(y-1)^2}2=x+y+\tfrac{x^2-2x+1+y^2-2y+1}2=\frac{x^2+y^2+2}2=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solve this trigonometric inequality The values of $\lambda$ for which the equation $2\sin (x)-\sqrt \lambda \cos(x)=\sqrt2+\sqrt{2-\lambda}$ has solutions.
MY APPROOACH:
Now I know that max value of LHS of above expression can be $\sqrt{4+\lambda}$ and min value can be $\sqrt{4-\lambda}$. So $\sqrt2+\sqrt{2-\lambda}\in [\sqrt{4-\lambda},\sqrt{4+\lambda}]$. So I get $\lambda \ge \frac{1}{2}$. Now how do I find the max value because I am unable to get a conclusion from $\sqrt{4+\lambda}\ge \sqrt2+\sqrt{2-\lambda}$
| Obviously, we need $-4\le\lambda\le2$.
\begin{align*}
\sqrt{4+\lambda}&\ge\sqrt{2}+\sqrt{2-\lambda}\\
4+\lambda&\ge2+2\sqrt{2(2-\lambda)}+2-\lambda\\
\lambda&\ge \sqrt{4-2\lambda} \qquad\qquad (\text{Note: this implies that }\lambda\ge0)\\
\lambda^2&\ge4-2\lambda\\
(\lambda+1)^2&\ge5\\
\lambda+1&\ge\sqrt{5}\\
\lambda&\ge\sqrt{5}-1
\end{align*}
So $\sqrt{5}-1\le \lambda\le2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Multiple of $5$ If two natural nos. $x,y$ are selected at random then the probability that $x^2+y^2$ is a multiple of 5?
I think they should give a finite set for selecting natural nos. otherwise there are infinite ordered pairs of $x$ and $y$.
Is the question incomplete or am I missing some trick?
| Without loss of generality, you can assume that $x$ and $y$ are restricted to the set $\{0, 1, 2, 3, 4\}$, as pointed out by Lord Shark the Unknown in a comment. Now,
$$z \equiv 0 \pmod 5 \implies z^2 \equiv 0 \pmod 5$$
$$z \equiv 1 \pmod 5 \implies z^2 \equiv 1 \pmod 5$$
$$z \equiv 2 \pmod 5 \implies z^2 \equiv 4 \pmod 5$$
$$z \equiv 3 \pmod 5 \implies z^2 \equiv 4 \pmod 5$$
$$z \equiv 4 \pmod 5 \implies z^2 \equiv 1 \pmod 5$$
Thus, $x^2 + y^2 \equiv 0 \pmod 5$ can only happen if
$$x \equiv y \equiv 0 \pmod 5$$
or $x \equiv 1 \pmod 5$ and $y \equiv 2 \pmod 5$, or $x \equiv 4 \pmod 5$ and $y \equiv 2 \pmod 5$, or $x \equiv 1 \pmod 5$ and $y \equiv 3 \pmod 5$, or $x \equiv 4 \pmod 5$ and $y \equiv 3 \pmod 5$, or with the roles of $x$ and $y$ interchanged in the last $4$ cases.
Therefore, this gives a probability of
$$\frac{1}{25}+\frac{4}{25}+\frac{4}{25}=\frac{9}{25}=36\%.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Logarithmic-trigonomic inequality $\log_5 \sin(x) > \log_{25} \cos(\frac{x}{2}) $ $$\log_5 \sin(x) > \log_{25} \cos(\frac{x}{2})$$
I have simplified it to
$$\sin(\frac{x}{2}) \cdot \sin(\frac{x}{2}) \cdot \cos(\frac{x}{2}) > \frac{1}{4}$$
But what should I do next?
I have no idea about next steps!
Thanks!
| Note that
$$\log_5 \sin(x) = \frac{\log\sin(x)}{\log 5}$$
and
$$\log_{25} \cos(x/2) = \frac{\log\cos(x/2)}{\log 25}$$
so that your inequality reduces to
$$2\log\sin(x) > \log\cos(x/2).$$
This becomes
$$\log\sin^2(x) > \log\cos(x/2).$$
Since the logarithm is a monotonically increasing function, then
$$\sin^2(x) > \cos(x/2).$$
But we have
$$\sin^2(x) = 1 - \cos^2(x) = 1 - \bigg(2\cos^2(x/2) - 1\bigg)^2 = 1 - 4\cos^4(x/2) + 4\cos^2(x/2) - 1 = -4\cos^4(x/2) + 4\cos^2(x/2).$$
We obtain
$$-4\cos^4(x/2) + 4\cos^2(x/2) > \cos(x/2).$$
Now, set $u = \cos(x/2)$, and get
$$4u^4 - 4u^2 + u < 0$$
$$u(4u^3 - 4u + 1) < 0$$
Using WolframAlpha, we obtain
$$u > 0.837565$$
or
$$0 < u < 0.269594$$
or
$$u < -1.10716.$$
Therefore,
$$\cos(x/2) > 0.837565$$
or
$$0 < \cos(x/2) < 0.269594$$
or
$$\cos(x/2) < -1.10716.$$
The last inequality is extraneous.
Since $\cos(x/2)$ is decreasing, this means that
$$x/2 < \cos^{-1}(0.837565)$$
or
$$\cos^{-1}(0.269594) < x/2 < \cos^{-1}(0)$$
from which we get
$$x/2 < 0.577985$$
or
$$1.297825 < x/2 < \frac{\pi}{2}$$
so that the solution set for the original inequality is
$$\{x < 1.15597\} \cup \{2.59565 < x < \pi\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_0^1{\frac{u^{n+2}+u^n}{u+1}}\,du$ for $n\in\mathbb{N}$, $n\neq-1$ Background:
I got stuck in the question below because of a silly error.
Evaluate the integral using the substitution $u=\tan x$
$$\int^{\frac{1}{4}\pi}_0{(\tan^{n+2}x+\tan^{n}x)}\,dx$$
I realized that $du = \sec^2x\,dx$ and hence $du=\mathbf{(u+1)}\,dx$. The error being that $\sec^2x = u^2 +1$, NOT $u+1$.
If we proceed and use the erroneous expression $du = (u+1)\,dx$, we get:
$$\int_0^1{\frac{u^n(u^2+1)}{u+1}}\,du$$
Which is the question that I'm asking: how do you evaluate this integral?
And more generally, how can you evaluate an integral of the form:
$$\int{\frac{u^n}{u+a}\,du}$$
where $a\in\mathbb{Z}$
| $$\frac{u^n}{u+a} = u^{n-1} - au^{n-2} + a^2u^{n-3} - \cdots + a^{n-2}u - a^{n-1} -\frac{a^n}{u+a} $$
So
\begin{align}
\int \frac{u^n}{u+a} du &= \int \left(u^{n-1} - au^{n-2} + a^2u^{n-3} - \cdots + a^{n-2}u - a^{n-1} -\frac{a^n}{u+a}\right) du\\
&= \frac{u^n}{n} - a \frac{u^{n-1}}{n-1} + a^2 \frac{u^{n-2}}{n-2} - \cdots + a^{n-2} \frac{u^2}{2} - a^{n-1}u - a^n \ln{|u+a|}\\
&= \sum_{m=0}^{n-1} (-1)^ma^m \frac{u^{n-m}}{n-m} - a^n\ln{|u+a|}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding a generator of $\mathbb{F}_{49}^*$ I have a little doubt.
Let $\alpha\in\mathbb{F}_{49}$ such that $\alpha^2=3$ and $\mathbb{F}_{49}=\mathbb{F}_7(\alpha)$. Find a generator for the cyclic group $\mathbb{F}_{49}^*$ and find the minimal polynomial over $\mathbb{F}_7$
For the first part, it's enough to find a primitive root modulo 49?
Thanks!
| This are the powers of $1 + \alpha$ in $\mathbb{F}_{49}$:
$$\begin{array}{cccccc}
\color{blue}{1 + 1\alpha} &
4 + 2\alpha &
3 + 6\alpha &
0 + 2\alpha &
\color{blue}{6 + 2\alpha} &
5 + 1\alpha \\
\color{blue}{1 + 6\alpha} &
\color{red}{5 + 0\alpha} &
5 + 5\alpha &
6 + 3\alpha &
\color{blue}{1 + 2\alpha} &
0 + 3\alpha \\
\color{blue}{2 + 3\alpha} &
4 + 5\alpha &
5 + 2\alpha &
\color{red}{4 + 0\alpha} &
\color{blue}{4 + 4\alpha} &
2 + 1\alpha \\
\color{blue}{5 + 3\alpha} &
0 + 1\alpha &
3 + 1\alpha &
6 + 4\alpha &
\color{blue}{4 + 3\alpha} &
\color{red}{6 + 0\alpha} \\
\color{blue}{6 + 6\alpha} &
3 + 5\alpha &
4 + 1\alpha &
0 + 5\alpha &
\color{blue}{1 + 5\alpha} &
2 + 6\alpha \\
\color{blue}{6 + 1\alpha} &
\color{red}{2 + 0\alpha} &
2 + 2\alpha &
1 + 4\alpha &
\color{blue}{6 + 5\alpha} &
0 + 4\alpha \\
\color{blue}{5 + 4\alpha} &
3 + 2\alpha &
2 + 5\alpha &
\color{red}{3 + 0\alpha} &
\color{blue}{3 + 3\alpha} &
5 + 6\alpha \\
\color{blue}{2 + 4\alpha} &
0 + 6\alpha &
4 + 6\alpha &
1 + 3\alpha &
\color{blue}{3 + 4\alpha} &
\color{red}{1 + 0\alpha}
\end{array}$$
(in $\color{red}{\text{red}}$ all the elements in $\mathbb{F}_7$ and in $\color{blue}{\text{blue}}$ all the generators).
We know that $\mathbb{F}_{49}^*$ has $48$ elements and if $g$ is a generator, then $g^k$ is also a generator for every positive integer $k < 48$ coprime with $48$. Thus, there are $16$ generators. We can discard the elements in $\mathbb{F}_7^*$ and the elements in $\alpha\mathbb{F}_7^*$ since those cannot have degree larger than $6$ and $12$, respectively. Therefore we have $16$ generators out of $36$ candidates. I suppose one could use some more powerful tools to narrow even more the search, but I was lucky with the first pick.
Its minimal polynomial over $\mathbb{F}_7$ is $X^2 - 2X - 2$.
Indeed it has to be of degree $2$ since $\mathbb{F}_7(\alpha)$ is a quadratric extension over $\mathbb{F}_7$. And $$(1+\alpha)^2 - 2(1+\alpha) - 2 = 0.$$
| {
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"url": "https://math.stackexchange.com/questions/2282918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove an inequality involving powers of two Let $a,b,c$ be mutually distinct positive integers.
Prove
$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$
Any hints on how to start proving it?
| I assume the set of Natural Numbers to be $\mathbb N = \{1,2,3 \ldots \}$
Let $$\alpha =2^{-a} + 2^{-b} + 2^{-c}$$
And
$$\beta =1+2^{-(a+b+c)}$$
We can easily see that the maximum value of $\alpha$ will occur for minimum values of $a,b$, and $c$.
The minimum values of $a,b$ and $c$ , are $1,2$, and $3$ (All are distinct).
Hence $$\alpha =2^{-a} + 2^{-b} + 2^{-c} \le 2^{-1} + 2^{-2} + 2^{-3}=\frac 78$$
And we also have $\beta$ :
$$2^{-(a+b+c)} \gt 0 \implies \beta= 1+2^{-(a+b+c)} \gt 1$$
Thus we have :
$$\alpha \le \frac78 ~~{and}~~ \beta \gt 1$$
Hence, obviously $\alpha \lt \beta$
Now multiplying both the sides by any positive quantity won't change the inequality, so we multiply both the sides by $2^{a+b+c}$
Hence, finally we get :
$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$
| {
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"url": "https://math.stackexchange.com/questions/2283150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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expanding $\frac 1{\log\left( \frac{z+1}{z-1} \right) }$ at infinity Just as the title says, how do we expand such a function at infinity? I am told that the first term is $z/2$. My knowledge in complex analysis is very rusty, so if anyone can help me I'd be really thankful.
| The "quick and dirty" way to do this is to view this as a composition of series with well-known forms:
\begin{align*}
\left[ \ln \left( \frac{z + 1}{z - 1} \right) \right]^{-1} &= \left[ \ln \left( 1 + \frac{1}{z} \right) - \ln \left(1 - \frac{1}{z} \right) \right]^{-1} \\
&= \left[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n z^n} + \sum_{n=1}^{\infty} \frac{1}{n z^n} \right]^{-1} \\
&= \left[ \sum_{\text{odd $n$}} \frac{2}{n} z^{-n}\right]^{-1} \\
&= \left( \frac{2}{z} \right)^{-1} \left( 1 + \frac{1}{3} z^{-2} + \frac{1}{5} z^{-4} + \dots \right)^{-1} \\
&= \frac{z}{2} \left[ 1 - \left( \frac{1}{3} z^{-2} + \frac{1}{5} z^{-4} + \dots \right) + \left( \frac{1}{3} z^{-2} + \frac{1}{5} z^{-4} + \dots \right)^2 + \dots \right]
\end{align*}
At this point you can expand as many terms in the "series of series" in the last expression to get the series to the order that you need. This is likely to be tedious for anything but the lowest few orders in the series, but in principle it can be done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this partial fraction decomposition? Please help me to solve the following partial fraction decomposition:
$$\frac{1-v^2}{v+v^3} = \frac{A}{v}+\frac{Bv+C}{1+v^2}$$
| The usual method would be to multiply both sides of this equation by $v+v^3$:
$$\begin{align}\frac{1-v^2}{v+v^3} &= \frac{A}{v}+\frac{Bv+C}{1+v^2}\\
1-v^2 &= A(1+v^2) + (Bv+C)(v)\\
1-v^2 &= Av^2 + A + Bv^2 + Cv\\
-v^2 + 0v + 1 &= (A+B)v^2 + Cv + A
\end{align}$$
At this point, you can equate coefficients on the two sides. Since the $v^2$ coefficients must match, we have $-1=A+B$. Since the $v$ coefficients must match, we have $0=C$, and since the constant terms must match, we have $1=A$.
Do you see how that's working?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all postive integer $n$,such $(x+y+z)|(x^{2n+1}+y^{2n+1}+z^{2n+1})$
Let $0<x<y<z<p$,where $x,y,z$ are postive intgers,$p$ is prime number, and such
$$x^3\equiv y^3\equiv z^3\pmod p$$
Find all postive integer $n$,such $$(x+y+z)|(x^{n}+y^{n}+z^{n})$$
$n=1$ is clear
$n=2$
I have show that
$$p|x^2+xy+y^2, p|y^2+yz+z^2, p|x^2+xz+z^2$$
and
$$p|x+y+z$$
so use this
$$2\sum_{cyc}(x^2+xy+y^2)-(x+y+z)^2=3(x^2+y^2+z^2)$$
so $n=2$ is right.
also I have show $n=5$ also is right
| For the congruence $x^3\equiv a\pmod p$ to have three non-congruent solutions $x,y,z$ it is necessary that there exists a primitive cubic root $\omega\in\Bbb{Z}_p$. This is the case if and only if $p\equiv1\pmod3$. Otherwise cubing will be a bijection from $\Bbb{Z}_p$ to itself.
In that case we have $y\equiv\omega x$ and $z\equiv \omega^2x$ or the other way around. This is because in $\Bbb{Z}_p^*$ we have $(y/x)^3=1$ and $(z/x)^3=1$, and the residue classes $1,\omega,\omega^2$ are the only classes
with cubes $\equiv1$.
Without loss of generality we can assume the first case by replacing $\omega$ with $\omega^2$, if necessary. Anyway, because $\omega\not\equiv 1$ and $\omega^3-1=(\omega-1)(\omega^2+\omega+1)\equiv0\pmod p$, we can conclude that $p\mid 1+\omega+\omega^2$ and consequently
$$
p\mid x(1+\omega+\omega^2)\equiv x+y+z.
$$
The inequalities imply that $0<x+y+z<3p$, so we also know that either $x+y+z=p$ or $x+y+z=2p$.
Claim. $(x+y+z)\mid (x^n+y^n+z^n)$ if and only if $3\nmid n$.
Proof. Let's first show that $p\mid (x^n+y^n+z^n)$ if and only if $3\nmid n$.
*
*If $n\equiv1\pmod 3$, then $y^n\equiv \omega^nx^n\equiv\omega x^n$ and similarly $z^n\equiv\omega^{2n}x^n\equiv \omega^2x^n$. So in this case
$$x^n+z^n+y^n=x^n(1+\omega+\omega^2)$$ is divisible by $p$.
*If $n\equiv2\pmod 3$, then we proceed similarly. Only this time $y^n\equiv \omega^2x^n$ and $z^n\equiv \omega x^n$. Again $p\mid(x^n+y^n+z^n)$.
*If $3\mid n$, then, as a consequence of $\omega^3\equiv1$ we get that $x^n,y^n,z^n$ are all congruent modulo $p$. Because $p>3$ and $p\nmid x^n$, the sum $x^n+y^n+z^n$ won't be divisible by $p$.
It remains to show that if $x+y+z=2p$ then $x^n+y^n+z^n$ is also an even number. But this is easy. Among $x,y,z$ there are either $0$ or $2$ odd numbers, so the same holds for the powers $x^n,y^n,z^n$. So $x^n+y^n+z^n$ is divisible by $2p=x+y+z$ if and only if it is divisible by $p$.
The claim follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$$
how should I approach this?
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)}=0$$ but I still have $$\lim_{(x,y)\to(0,0)}\frac{1}{x^2y^2}$$
The answer says there is no limit
| Note that
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}=\frac{1}{2}.$$
So in there is a neighborhood $U$ of $(0,0)$ where $\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}\geq \frac{1}{4}.$
Hence, for $x\not=0$, $y\not=0$ and $(x,y)\in U$,
$$\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}=\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}\cdot \frac{x^2+y^2}{x^2y^2}\geq \frac{1}{4}\cdot \left(\frac{1}{x^2}+\frac{1}{y^2}\right)\to +\infty$$
as $(x,y)\to(0,0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number reduction before mod $$15\cdot25 \pmod {11}\equiv 4\cdot3 \pmod {11}$$
How does it work?
Full example
$$3^{(11-1)} \pmod{11} = 3\cdot27\cdot27\cdot27 \pmod{11}= 3\cdot5\cdot5\cdot5 \pmod{11} = 15\cdot25 \pmod{11}= 4\cdot3 \pmod{11} =1 \pmod{11}$$
| $a\equiv b \pmod{11}$ means that $11$ divides $a-b$. So if we want to see that $$(a+11)(b+11)\equiv ab \pmod{11}$$ we should check that $11$ is a divisor of $$(a+11)(b+11)-ab=ab+11a+11b+11*11-ab=11a+11b+121$$ This is clearly the case (in your example we have that $a=4$ and $b=14$). You can do this again for $b$ and you see that $15\cdot 25 \equiv 3 \cdot 4 \pmod{11}$.
| {
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Finding coefficient of $x^{15}$ $$(x+x^2+x^3+x^4+x^5)\cdot (x^2+x^3+x^4+…)^5$$
I have done a little :
$$x(1 + x+x^2+x^3+x^4)\cdot x^{10}(1 + x^2+x^3+…)^5$$
By generating functions:
$$\begin{align}&x^{11}\cdot\frac{1 - x^5}{1-x}\cdot\frac{1}{(1-x)^5}\\[1ex]
\implies &x^{11}(1 - x^5)\cdot\frac{1}{(1-x)^6}\\[1.5ex]
\implies &x^{11}(1 - x^5) \cdot\sum_{n=0}^{\infty}\binom{n+5}{5} x^n\\
\end{align}$$
How am I supposed to find $x^{15}$or any other one like $x^{18} , x^{19}$
Any clues? Thanks.
| You are most of the way home with what you have done already, the last step involves manipulating the summation by multiplying through by $x^{11}(1-x^5)=x^{11}-x^{16}$ like so
$$\begin{align}(x^{11}-x^{16})\sum_{n=0}^{\infty}\binom{n+5}{5}x^n&=\sum_{n=0}^{\infty}\binom{n+5}{5}(x^{n+11}-x^{n+16})\\&=\sum_{n=0}^{\infty}\binom{n+5}{5}x^{n+11}-\sum_{n=0}^{\infty}\binom{n+5}{5}x^{n+16}\end{align}$$
relabelling summation indices $n\rightarrow n-11$ and $n\rightarrow n-16$ for the first and second summations respectively gives
$$\sum_{n=11}^{\infty}\binom{n-6}{5}x^{n}-\sum_{n=16}^{\infty}\binom{n-11}{5}x^{n}$$
but if we define $\binom{a}{b}=0$ for $a\lt b$ such that $a,b\in \mathbb{Z}$ then we can write this as
$$\sum_{n=0}^{\infty}\binom{n-6}{5}x^{n}-\sum_{n=0}^{\infty}\binom{n-11}{5}x^{n}$$
or simply
$$\sum_{n=0}^{\infty}\left(\binom{n-6}{5}-\binom{n-11}{5}\right)x^n$$
hence your coefficients $c_n$ in
$$\frac{x^{11}-x^{16}}{(1-x)^6}=\sum_{n=0}^{\infty}c_nx^n$$
are
$$c_n=\binom{n-6}{5}-\binom{n-11}{5}\tag{Answer}$$
e.g. To find the coefficient $x^{15}$ plug $n=15$ into that formula.
| {
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If $a,b,c \geq 1$ then prove $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$ If $a,b,c \geq 1$, then prove:
$$\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$$
My try:
$$A=\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \\ A^2 =(a-1)+(b-1)+(c-1)+2\sqrt{(a-1)(b-1)}+2\sqrt{(a-1)(c-1)}+2\sqrt{(b-1)(c-1)} $$
$$B=\sqrt{abc+c} \\B^2=abc+c=c(ab+1) $$
Now what?
| $$\sqrt{a-1}+\sqrt{b-1}=\sqrt{a+b-2+2\sqrt{(a-1)(b-1)}}=$$
$$=\sqrt{ab-(a-1)(b-1)+2\sqrt{(a-1)(b-1)}-1}=$$
$$=\sqrt{ab-\left(\sqrt{(a-1)(b-1)}-1\right)^2}\leq\sqrt{ab}.$$
Thus,
$$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq\sqrt{ab}+\sqrt{c-1}=$$
$$=\sqrt{ab+1-1}+\sqrt{c-1}\leq\sqrt{(ab+1)c}=\sqrt{abc+c}.$$
Done!
| {
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Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ where $n$ is a positive integer.
In order for a positive integer $k$ to begin with a $1$, we need $10^m \leq k < 2 \cdot 10^m$ for some positive integer $m$. Thus since $5^n$ begins with a $1$, we have $10^m \leq 5^n < 2 \cdot 10^m$ for some positive integer $m$. Then multiplying by $\dfrac{2^{n+1}}{5^n}$ we get $10^m \cdot \dfrac{2^{n+1}}{5^n} \leq 2^{n+1} < 10^m \cdot \dfrac{2^{n+2}}{5^n}$, but I didn't see how this helped.
| $10^m < 5^n < 2*10^m$ [Note: equality can not hold as $5^n$ has no factor of $2$.]
$10^m < \frac {10^n}{2^n} < 2*10^m$
$1 < \frac {10^{n-m}}{2^n} < 2$
$2^n < 10^{n-m} < 2^{n+ 1}$
So $10^{k} < 2^{n+1}$ for $k = n-m$.
...
And $2^m5^m < 5^n < 2^{m+1} 5^m$
$2^{n+1}5^m < 2^{n-m + 1}5^n$
$2^{n+ 1} < 2^{n-m+1}5^{n-m}$
$2^{n+1} < 2*10^{n-m}$
So $2^{n+1} < 2*10^k$ for $k = n-m$
So $10^k < 2^{n+1}< 2* 10^k$.
So $2^{n+1}$ has a first digit of $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which occur at $x = \frac{(4n+1)\pi}{2} $ and $\frac{-(4n+3) \pi}{2}$
Equating this to its Taylor series, we get
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x)(1 + \frac{2}{3\pi}x)(1 - \frac{2}{5\pi}x)(1 + \frac{2}{7\pi}x)... $$
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x(1-\frac13)+O(x^2))(1 - \frac{2}{\pi}x(\frac15-\frac17)+O(x^2))... $$
$$1 - x + O(x^3) = 1 - \frac{2}{\pi}x(1-\frac13+\frac15-\frac17+...)+O(x^2)$$
$$x - O(x^3) = \frac{2}{\pi}x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}-O(x^2)$$
Equating $x$ terms,
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \frac{\pi}{2}$$
However, I seem to be off by a factor of $2$ (actual answer is $\frac{\pi}{4} $). Does anyone see where this went wrong?
| While $f(x) = 1 - \sin x$ is indeed an entire function, the roots at $x = \frac{(4n+1)\pi}{2}$ and $\frac{-(4n+3)\pi}{2}$ are in fact double roots, which you can check since they are also roots of $f'(x)$. So, the factorisation you give isn't valid.
| {
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How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}
At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$
This is not the answer. Could someone please help me figure out where I've gone wrong?
|
$$\frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)} = \frac{\color{red}{8-x}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}$$
Careful in the numerator:
$$\frac{1}{x+1}-\frac{1}{9} \ne 8-x$$
but rather:
$$\frac{1}{x+1}-\frac{1}{9}= \frac{9}{9(x+1)}-\frac{x+1}{9(x+1)} = \frac{8-x}{\color{blue}{9(x+1)}}$$
So then after cancelling/simplifying:
$$\frac {\frac{-1}{\color{blue}{9(x+1)}}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \xrightarrow{x \to 8} -\frac{1}{54}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Theory of Equations and co prime
If the range of quadratic polynomial $f(x)= 2x^{2}+3x+\frac{a}{b}$ is $[4, \infty)$ and $a$ and $b$ are co prime. Find $a+b$ .
Options are $50,46,49,58$.
What I did was I made a rough graph and the roots are complex. So I saw Discriminant is less than zero. Then another method I tried was I made some cases like if $a$ and $b$ are co prime then either both are odd which lead to both will be prime or one is even and one is odd. Any method I try I get stuck. Please help me . Thank you for your time .
| $$f(x)=2\left(x^2+\frac{3}{2}x\right)+\frac{a}{b}=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}+\frac{a}{b}$$
The minimum value is
$$\frac{-9}{8}+\frac{a}{b}=4$$
$$\frac{a}{b}=\frac{41}{8}$$
$a=41$ and $b=8$.
$a+b=49$.
Another method:
The minimum value of $f(x)$ is $4$. So the graph has the vertex with $y$-coordinate equals to $4$ and hence the equation $f(x)-4=0$ has a double root.
$f(x)-4=2x^2+3x+\frac{a}{b}-4$.
$$3^2-4(2)\left(\frac{a}{b}-4\right)=0$$
$$\frac{a}{b}=\frac{41}{8}$$
| {
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How to prove the given function is not differentiable analytically? Well the question presented to me is this. The given function is,
$$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{2}x + 2,\,\;\;\;x < 2}\\{\sqrt {2x} ,\;\;\;\;\;\;x \ge 2}\end{array}} \right.
$$
Now have to check whether the given function is differenciable at $x=2$ ?
My approach:
For this function to be differentiable, the left hand derivative and the right hand derivative must exits, and both be equal.
Left hand derivative:
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {\frac{1}{2}x + 2} \right) - \left( {\frac{1}{2}\left( 2 \right) + 2} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {\frac{1}{2}x + 2} \right) - 3}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\frac{1}{2}x - 1}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x - 2}}{{2\left( {x - 2} \right)}}\end{array}
$$
This gives,
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{}}{{2}}\\ = \frac{1}{2}\end{array}
$$
Right hand derivative:
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left( {\sqrt {2x} } \right) - \left( {\sqrt {2\left( 2 \right)} } \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - \sqrt 4 }}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - 2}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - 2}}{{x - 2}} \times \frac{{\sqrt {2x} + 2}}{{\sqrt {2x} + 2}}\end{array}
$$
This gives,
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{2x - 4}}{{\left( {x - 2} \right)\left( {\sqrt {2x} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{2}{{\left( {\sqrt {2x} + 2} \right)}}\\ = \frac{2}{{2 + 2}}\\ = \frac{1}{2}\end{array}
$$
Problem is right hand derivative and left hand derivative are coming same. Fooling me that its differentiable at $x=2$.
The graph of the function is,
$f(x)$ Graph
The function is discontinues at $x=2$. So, its not differentiable at $x=2$.
Where have I gone wrong ? How do I prove analytically without taking help of graph that the function is not differentiable at $x=2$?
| $$f (2^-)=\lim_{x\to 2^-} (\frac {x}{2}+2)=1+2=3$$
$$ f (2^+)=\lim_{2^+}\sqrt {2x}=2$$
$f $ is not continuous at $x=2$ thus it is not differentiable at $x=2$.
By definition, differentiable at $x=x_0$ means
$$\exists L\in \mathbb R \;\exists \eta>0 :\forall x\in (2-\eta,2+\eta) $$
$$f (x)=f (2)+(x-2)\Bigl (L+\epsilon (x)\Bigr) $$
with $$\lim_{x\to 2}\epsilon (x)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
I'm stuck at integrating an arc length over a circle + angle between curves Let $(\phi, \theta)$ be the usual spherical coordinates. And let $\gamma (t) = (\phi (t), \theta (t))$ be the curve given by
$$ \phi (t) = \ln\left(\cot\left(\frac{\pi}{4} - \frac{t}{2}\right)\right),$$ $$ \theta (t) = \frac{\pi}{2} - t.$$
$t \in (0, \frac{\pi}{2})$
a) Compute the arc length of $ \gamma (t). $
b) Show that the curve cuts each "parallel" in a constant angle, by computing $(\gamma (t))'$ and $\psi_{\phi}$
c) Make a diagram of the curves
So the formula for arc length would be $$ \begin{equation}
\ell_\gamma = \int_0^1 \sqrt{ \left ( \frac{dr}{dt} \right )^2 +
r^2\sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2 +
r^2 \left ( \frac{d \theta}{dt} \right )^2} \, dt.
\end{equation}$$
In my case, as $$\phi'(t) = \frac 1 {2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right) \sin\left(\frac{\pi}{4} - \frac{t}{2}\right)}, $$ and $$ \theta ' (t) = -1 ,$$ then
\begin{align*}
\ell_\gamma &= \int_0^1 \sqrt{
\sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2 +
\left ( \frac{d \theta}{dt} \right )^2} \, dt \\
&= \int_0^1 \sqrt{ \sin^2 \theta \left (\frac 1 {2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2 + 1} \, dt\\
&= \int_0^1 \sqrt{
\left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2 +
1} \, dt.
\end{align*}
I am stuck here. Before I go further, would this integral be correct? That $\sin^2(\theta)$ confuses me maybe, I don't know if I should change the argument to something else.
For exercise b) Could someone verify this reasoning?
I want the angle between two curves on the sphere, which is defined as the angle between the tangent vectors of this curves at the point of intersection. So I want to find $\alpha$ such that $\alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) $, because that is what the angle between two vectors in a vector space is.
So if $\beta (t) = (t, \theta_0)$ is a curve that describes the parallel curve in a sphere, with $\theta_0$ constant, Its tangents are given by $(1, 0),$. I already have the derivatives of $\gamma$ so $$ \alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) = cos^{-1}(\frac{<(1, 0) \dot (\frac{1}{sin{\frac{\pi}{2}- t}}, -1)>}{(1) \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$
$$ = cos^{-1}(\frac{1}{sin{\frac{\pi}{2}- t}}{ \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$.
Thanks in advance.
| \begin{align}
& \int_0^1 \sqrt{ \left(\frac{\sin (\frac{\pi}{2} - t) }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt]
\end{align}
As $$\sin^2(\frac \pi 4 - \frac t 2)cos^2(\frac \pi 4 - \frac t 2) = \frac{1 - \cos(\pi - 2t)}{8},$$ then
\begin{align}
& \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{1 -\cos(2(\frac{\pi}{2} - t ))}\right) + 1} \, dt & = \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{1 -[\cos^2(\frac{\pi}{2} - t )- \sin^2(\frac{\pi}{2} - t ))]}\right) + 1} \, dt \end{align}
\begin{align} = \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{2\sin^2(\frac{\pi}{2} - t)}\right) + 1} \, dt = \sqrt2 \\[10pt]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated.
| On your last equation, putting $n=-2,-3,-4$ leads the equation equal to zero. Then $n=1$ and the value of $\frac{1*2*3*4*5}5$ leads $\frac{n(n+1)(n+2)(n+3)(n+4)}5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Find the extrema of $f(x,y) = \left( \frac{1}{2} - x^2 + y^2 \right) \cdot e^{1-x^2-y^2}$ I am asked to find the extrema of $$f(x,y) = \left( \frac{1}{2} - x^2 + y^2 \right) \cdot e^{1-x^2-y^2}$$
The partial derivatives could be easier, but that's not the issue:
\begin{align*}
f_x &= \left( e^{1-x^2-y^2} \right) \left( 2x^3-2xy^2 -3x \right) = 0\\
f_y &= \left( e^{1-x^2-y^2} \right) \left( -2yx^2+2y^3-y\right) = 0
\end{align*}
Exponentials are never null so
\begin{align*}
x \left( 2x^2-2y^2 -3 \right) &= 0\\
y \left( -2x^2+2y^2-1\right) &= 0
\end{align*}
The extrema $(0,0)$ (saddle point) is pretty obvious but what about the other extrema: how could I find those given this system of equations?
Any help is appreciated! Thank you.
Link for the complete answer (Wolfram)
| For $x=0$ but $y\ne 0$, we have
$$2y^2-1=0$$
$$y=\pm\sqrt{\frac{1}{2}}$$
For $y=0$ but $x\ne 0$, we have
$$2x^2-3=0$$
$$x=\pm\sqrt{\frac{3}{2}}$$
For $x,y\ne 0$, we have
$$2x^2=2y^2+3$$
$$-2y^2 -3+2y^2-1=0$$
$$-4=0$$
So no solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$? I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
However, this strikes me as not being simplified. Isn't it more proper to write it like this?
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$
$$x=\frac{-b\pm((b-2)\sqrt{ac})}{2a}$$
$$x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$$
Why isn't $x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$ more commonly used as the quadratic formula??
I'm sorry for my typo, I have edited it.
I have now edited in my steps, per request of commenters.
| Because $\sqrt{b^2 - 4ac}\neq 2b\sqrt{ac}$. Say for example that $c = 0$ and $b\neq 0$. Then you have
\begin{align*}
\sqrt{b^2 - 4ac} &= \sqrt{b^2}\\
&= \left|b\right|\\
&\neq 0,
\end{align*}
so you can see that this simplification cannot be correct.
It seems that in your proposed simplification, you have completely disregarded the subtraction occurring in the radical. Moreover, $\sqrt{b^2} = \left|b\right|$, not just $b$. To see this with an example, take $b = -1$. Then $\sqrt{(-1)^2} = \sqrt{1} = 1 = \left|-1\right|\neq -1$.
EDIT:
Again, the simplification is incorrect. While it is true in general that $\sqrt{a^2b} = \left|a\right|\sqrt{b}$, this is not the situation you are in here:
$$
b^2 - 4ac\neq (b - 2)^2 ac = (b^2 - 4b + 4)ac = b^2 ac - 4abc + 4ac.
$$
You seem to have made a few mistakes here (if I'm to take a stab at the reasoning behind the simplification): first you've incorrectly simplified $b^2 - 4ac$ as $(b^2 - 4)ac$ (which is not true, because the first term in the former has no $ac$), and then you've simplified $b^2 - 4$ as $(b - 2)^2$, which is also not true (take $b = 0$ to see why). In general, $(x + y)^n\neq x^n + y^n$: this is a common mistake algebra learners make! Remember that when expanding $(x + y)^2$, we need to use the distributive property, and not simply regard squaring as linear:
\begin{align*}
(x + y)^2 &= (x + y)(x + y)\\
&= x^2 + yx + xy + y^2\\
&= x^2 + 2xy + y^2.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$
My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor=
\lim_{x\to 0}\left\lfloor\dfrac{2\cos x}{\cos 2x}\right\rfloor $$
Now what?
| Let's consider
$$
f(x)=\frac{2\cos x}{\cos2x}
$$
over $(-\pi/2,\pi/2)$. Then
$$
f'(x)=2\frac{-\sin x\cos2x+2\cos x\sin2x}{\cos^22x}=
2\frac{\sin x(2\cos^2x+1)}{\cos^22x}
$$
Therefore $f$ has a local minimum at $0$ and $f(0)=2$.
In a suitable neighborhood of $0$ we have $2\le f(x)<3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the absolute max and min of f subject to the given constraint $f(x)=4x^4+y^4$ subject to the constraint $x^2+y^2=1$
My attempt at it was by finding partial derivitives:
$$f_x=16x^3 =0$$
$$f_y=4y^3=0$$
this means that we have an interior point $(0,0)$
Now Let $x=\cos(t)$, $y=\sin(t)$
$$f(\cos(t),\sin(t))=4\cos^4t+\sin^4t$$
$$<=> 3\cos^4t+1$$
we can now note that $f$ has max when $t=0,\pi$, or $2\pi $ and min when $t= \pi/t +k\pi)$
this means that we have the following potential points
$f(1,0),f(-1,0), f(0,1), f(0,-1), f(0,0)$
$$f(0,0)=0$$
$$f(1,0)=4$$
$$f(0,1)=1$$
Max =4
Min = 0
Anyone can help verifying my answer, because i tried the lagrange way and it didnt work
| Express $(x,y)$ in polar coordinates as $(\cos\theta,\sin\theta)$, and then let
$c= \cos^2 \theta$. Note that $\sin^2 \theta = 1-\cos^2 \theta$ so $x^4=c^2$ and $y = (1-c)^2$.
Then $$4x^4+y^4 = 4c^2 + (1-c)^2 = 5c^2 -2c+1
$$
Since $c$, being $\cos^2\theta$, is restricted by $0 \leq |c|\leq 1$,
this is maximized at $c=1 = \cos^2\theta$, at which point $x=\pm1, y=0, f(x,y) =4 $.
Trivial differentiation locates the minimum at $c = 1/5$, at which point
$x=\pm \frac{\sqrt{5}}{5}, y=\pm 2\frac{\sqrt{5}}{5} , f(x,y) =\frac4{5} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Having trouble integrating $ \ \ \int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $. I'm having trouble integrating $$\int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $$
My approach
*
*$ \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} =\frac{Ax+B}{x^{2}}+\frac{Cx+D}{x^{2}+2x+2} $ ,
*or, $ x^{3}+3x^{2}+2x+4=(Ax+B)(x^{2}+2x+2)+(Cx+D)x^{2} $ .
But is the process right ? If not, Any suggestions are welcome.
| HINT: we have $$\frac{x^3+3x^2+2x+4}{x^2(x^2+2x+2)}=-\frac{1}{x}+\frac{2}{x^2}+\frac{2x+3}{x^2+2x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove inequality $a<1Given $a,b,c$ are positive number satisfy $a<b<c;a+b+c=6;ab+bc+ca=9$. Prove that $a<1<b<3<c<4$
i think use quality:"The solutions of function $f(x)=(x-a)(x-b)(x-c)$ satisfy $a\le x_1\le b\le x_2\le c$ with $x_1,x_2\ (x_1<x_2)$ are extreme point of $f(x)$ ".
Help me
| Your idea is correct. $a, b, c$ are the zeros of
$$
f(x)=(x-a)(x-b)(x-c) = x^3 - 6 x^2 + 9x - abc
$$
Rolles's theorem states that $f'$ has a zero in each interval
$(a, b)$ and $(b, c)$. But
$$
f'(x) = 3x^2 - 12 x + 9 = 3(x-1)(x-3)
$$
has zeros $x_1=1$ and $x_2=3$. It follows that
$$
a < x_1 = 1 < b < x_2 = 3 < c \, .
$$
It remains to show that $c < 4$: $f$ changes signs exactly at the
zeros $a, b, c$, and $f(4) = 4 - abc = f(1)$. Therefore $4$ must lie in the interval $(c, \infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Confusing differentials problem - Electrical resistance $ R = \frac{k}{r^2} $ with $ dr = 5\% $ I am struggling with a confusing differentials' problem. It seems like there is a key piece of information missing:
The problem:
The electrical resistance $ R $ of a copper wire is given by $ R = \frac{k}{r^2} $ where $ k $ is a constant and $ r $ is the radius of the wire. Suppose that the radius has an error of $ \pm 5\% $, find the $\%$ error of $ R $.
My solution:
\begin{align*}
R &= \frac{k}{r^2}\\
\frac{dR}{dr} &= k \cdot (-2) \cdot r^{-3} \quad \therefore \quad dR = \frac{-2k \cdot 0.05}{r^3} = \frac{-0.1k}{r^3}\\
\end{align*}
So the percentage error is given by
\begin{align*}
E_\% = \frac{\frac{-0.1k}{r^3}}{\frac{k}{r^2}} = - \frac{0.1}{r}
\end{align*}
My question: Am I missing something? Should I have arrived in a real value (not a function of $ r $ )? Is there information missing on the problem?
Thank you.
| Just doing this by brute force:
$$R=\frac{k}{r^2}\quad\text{percentage error in $r$ is $\pm 5$ percent}$$
So $R_{min} =\frac{k}{(r+0.05r)^2} =\frac{k}{1.05^2r^2}=\frac{400}{441}\cdot\frac{k}{r^2}$
$R_{max} = \frac{k}{(0.95r)^2} =\frac{400}{361}\cdot \frac{k}{r^2}$
Then the percentage error is given by:
$$\frac{R_{max}-R_{min}}{2R}\cdot 100 =\frac{\frac{k}{r^2}\left(\frac{400}{361}-\frac{400}{441}\right)}{2\cdot \frac{k}{r^2}}\cdot 100 \approx 10.1\%\quad\text{3s.f.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
minimizing vectors Could please check whether my solution is right?
Q. $$A=\begin{pmatrix}
1 & 1 \\
1 & 1 \\
0 & 0 \\
\end{pmatrix}$$Find the set of vectors $x$ that minimize the value $$||
A
x-
\begin{pmatrix}
1 \\
2 \\
3 \\
\end{pmatrix}||$$
My solution.
$Ax-(1,2,3)^T=0$ (inconsistent) → least square method
$A^TAx=A^Tb$, $b=(1,2,3)^T$
$$
\begin{pmatrix}
2 & 2 \\
2 & 2 \
\end{pmatrix}\begin{pmatrix}
x_1 \\
x_2 \\
\end{pmatrix}=\begin{pmatrix}
3 \\
3 \\
\end{pmatrix}$$
$2x_1+2x_2=3$ → $x_2=-x_1+1.5$
Therefore the set of $x$ is {$
\begin{pmatrix}
k \\
-k+1.5 \\
\end{pmatrix}$| k∈ℝ }
| I would use a more "basic" method.
Writing the vector x as $\begin{pmatrix}x \\ y \end{pmatrix}$, $Ax= \begin{pmatrix}1 & 1 \\ 1 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ y \\ x+ y \\ 0 \end{pmatrix}$ so that $Ax- \begin{pmatrix}1 \\ 2 \\ 3 \end{pmatrix}= \begin{pmatrix}x+ y- 1 \\ x+ y- 2 \\ -3\end{pmatrix}$.
The norm of that is $\sqrt{(x+ y- 1)^2+ (x+ y- 2)^2+ 9}$.
Writing that as f(x,y), we have $f_x= (1/2)((x+ y- 1)^2+ (x+ y- 2)^2+ 9)^{-1/2}[(2(x+y-1)+ 2(x+ y- 2)]= 0$ and $f_y= (1/2)((x+ y- 1)^2+ (x+ y- 2)^2+ 9)^{-1/2}[(2(x+y-1)+ 2(x+ y- 2)]= 0$. Those both reduce to $2[(x+ y- 1)+ 2(x+ y- 2)]= 4x+ 4y- 6= 0$. The function will be minimized at any point along the line $x+ y= \frac{3}{2}$. That is, $y= \frac{3}{2}- x$ which is exactly what you have.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How many numbers in between $1$ and $250$ are not divisible by $2, 3, 5$ or $7$? How many numbers in between $1$ and $250$ are not divisible by $2, 3, 5$ or $7$? Using the Inclusion Exclusion Principle, I got the answer to be $57$ (i.e. $250-193$), but I'm worried I might have screwed up the arithmetic. Is there a quick and easy way to check whether long calculations like these are right?
| I will first consider all integers up to $210$, as it is the L.C.M. of $2$, $3$, $5$ and $7$. The number of integers from $1$ to $210$ which are not divisible by $2$, $3$, $5$ or $7$ is
$$210-\frac{210}{2}-\frac{210}{3}-\frac{210}{5}-\frac{210}{7}+\frac{210}{6}+\frac{210}{10}+\frac{210}{14}+\frac{210}{15}+\frac{210}{21}+\frac{210}{35}-\frac{210}{30}-\frac{210}{42}-\frac{210}{70}-\frac{210}{105}+\frac{210}{210}$$
which is equal to
$$210\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)=48$$
The integers from $211$ to $250$ can be replaced by $1$ to $40$. I prefer to simply list out all integers satisfying the conditions: $1$, $11$, $13$, $17$, $19$, $23$, $29$, $31$, $37$.
So, there are $48+9=57$ such integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Find the value of $\frac{\sin 24x}{\cos18x \cos 6x}$. Given $\tan x = \sqrt{5} - 2 $, find the value of $$\frac{\sin 24x}{\cos 18x\cos 6x}.$$
Now I wrote numerator as $\sin (18 + 6)x$ and I used identity $\sin(A+B)$, after that I reduced problem to $\tan 18x + \tan 6x$
How do I proceed?
Thanks
| We have
\begin{align}
\tan 2x&=\frac{2\tan x}{1-\tan^2x}\\
&=\frac{2(\sqrt{5}-2)}{1-(\sqrt{5}-2)^2}\\
&=\frac{2(\sqrt{5}-2)}{1-5+4\sqrt{5}-4}\\
&=\frac{2(\sqrt{5}-2)}{4(\sqrt{5}-2)}\\
&=\frac{1}{2}
\end{align}
Then use
$$\tan 3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A}$$
to deduce that
$$\tan 6x=\frac{11}{2}$$
Use it once again to deduce that
$$\tan18x=\frac{1199}{718}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The value of k is given to be 4, but I am unable to derive that result.
Kindly provide me with some insight, or with a step-by-step solution.
Thanks in advance,
Abhigyan
| Set $z=e^{i\pi/7}$, so $z^7=e^{i\pi}=-1$ and $z^{-1}=\bar{z}$.
Then
$$
\cos\frac{\pi}{7}=\frac{z+\bar{z}}{2}=\frac{z^2+1}{2z},
\qquad
\sin\frac{\pi}{7}=\frac{z-\bar{z}}{2i}=\frac{z^2-1}{2iz}
$$
Therefore
$$
\tan\frac{\pi}{7}=-i\frac{z^2-1}{z^2+1}
$$
Plugging in the left hand side, we get
$$
\frac{3(z^2+1)^2+(z^2-1)^2}{(z^2+1)^2+(z^2-1)^2}=
\frac{4z^4+4z^2+4}{2z^4+2}=2\frac{z^4+z^2+1}{z^4+1}
$$
and finally
$$
k=2\frac{z^4+z^2+1}{z^4+1}\frac{2z}{z^2+1}=
\frac{4z(z^4+z^2+1)}{(z^4+1)(z^2+1)}
$$
The denominator reads $z^6+z^4+z^2+1$, but from $z^7+1=0$ we can deduce $z^6-z^5+z^4-z^3+z^2-z+1=0$ and so
$$
z^6+z^4+z^2+1=z^5+z^3+z=z(z^4+z^2+1)
$$
and you get
$$
k=4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
| $S=\sum_{n=1}^{+\infty}(\frac {3}{(n+1)(n+2)}+\frac {7}{n (n+1)(n+2)}) $
$$=3\sum_{n=1}^{+\infty}(\frac {1}{n+1}-\frac {1}{n+2}) $$
$$+7\sum_{n=1}^{+\infty}\frac{1}{n (n+1)(n+2)} $$
$$\implies S=\frac {3}{2}+7A$$
and
$$S=\sum_{n=1}^{+\infty}\frac {3 (n+2)+1}{n (n+1)(n+2)} $$
$$=3\sum_{n=1}^{+\infty}(\frac {1}{n}-\frac {1}{n+1})+A$$
$$\implies S=3+A $$
thus after eliminating $A $, we get
$$6S=21-\frac {3}{2}\implies S=\frac {13}{4} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
What are the area of a triangle with side lengths $\tan(x)$, $\cos(x)$ and $\sin(x)$?
Consider a nondegenerated right triangle with sides of length $\sin x$, $\cos x$, and $\tan x$ where $x$ is a real number.
Compute the possible values of the area of this triangle.
*
*I was thinking of more along the lines of Heron's formula but that was very nasty indeed, rather I have attempted to find out what are the lengths of the triangle and I have done the Pythagorean theorem which was in vain.
*Another thing that I have done was to graph all three and to see which one was the largest in the y-value but it turned out that at times one graph was larger than the other and at other times it wasn't.
*And another thing that I have done was to plug and chug in values such as the number 3 into all of the trig functions and do Pythagorean theorem to see if it satisfied it but none of them didn't
The reason why that I have done those steps was so that I can multiply the legs and divide by two to find the area of the triangle but in order to do that I must know what are the legs.
I was wondering if there was any other way?
| Since $\sin x$, $\cos x$ and $\tan x$ are all positive, we may assume that $\displaystyle x\in\left(0,\frac{\pi}{2}\right)$. Therefore, $\tan x>\sin x$.
If $\cos x>\tan x$, then $\cos x$ is the hypotenuse and
\begin{align}
\sin^2x+\tan^2x&=\cos^2x\\
\sin^2x(\cos^2x+1)&=\cos^4x\\
1-\cos^4x&=\cos^4x\\
\cos x&=\frac{1}{\sqrt[4]{2}}
\end{align}
The area of the triangle is
\begin{align}
\frac{1}{2}\sin x\tan x&=\frac{\sin^2 x}{2\cos x}\\
&=\frac{1-\cos^2 x}{2\cos x}\\
&=\frac{\displaystyle1-\frac{1}{\sqrt{2}}}{\displaystyle\frac{2}{\sqrt[4]{2}}}\\
&=\frac{\sqrt[4]{2}(2-\sqrt{2})}{4}\\
\end{align}
If $\tan x>\cos x$, then $\tan x$ is the hypotenuse and
\begin{align}
\sin^2x+\cos^2x&=\tan^2x\\
x&=\frac{\pi}{4}
\end{align}
The area of the triangle is
$$\frac{1}{2}\sin x\cos x=\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$ Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$
The root of equation very bad
My try 1:
Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$
Have: $a+b=4-x$
=>Root of a system of equation bad, too
My try 2:
Use quality $\sqrt[3]{a}\pm \sqrt[3]{b}=\frac{a\pm b}{\sqrt[3]{a^2}\mp \sqrt[3]{ab}+\sqrt[3]{b^2}}$ :
$\sqrt[3]{x+2}-ax+\sqrt[3]{2x-1}-bx=4-x$
Need find $ax$, $bx$ but it's very bad, too
| If we set $a = \sqrt[3]{x+2}$ and $b = \sqrt[3]{2x-1}$, since $a + b = 4 - x = 1 + (3 - x)$, we obtain that
\begin{cases}
a^{3} - b^{3} = a + b - 1\\
b^{3} = 2a^{3} - 5
\end{cases}
Hence we conclude that $b = 6 - a - a^{3}$, from whence we have the equation
\begin{align*}
(6 - a - a^{3})^{3} = 2a^{3} - 5
\end{align*}
Which can be solved numerically. Wolfram, for example, gives $a \approx 1.49263$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a^3+b^3+c^3=3$ so $2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$ I was looking at this question, and I derived this inequality from that
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3.$ Prove that:
$$2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$
I couldn't solve Rozenberg's inequality but I assumed it's true. So if $a^3+b^3+c^3=3$, then
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}\tag {1}$$
and also by using this, if $b^3+a^3+c^3=3$, then
$$\frac{b^3}{b+a}+\frac{a^3}{a+c}+\frac{c^3}{c+b}\geq\frac{3}{2}\tag {2}$$
using $(1)$ and $(2)$, if $a^3+b^3+c^3=3$, then
$$\frac{a^3+b^3}{a+b}+\frac{b^3+c^3}{b+c}+\frac{c^3+a^3}{c+a}\geq3$$
$$\Leftrightarrow \sum\limits_{cyc}(a^2-ab+b^2)\geq 3$$
$$\Leftrightarrow 2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$
If I'm not making any silly mistake, this inequality should be true. I couldn't solve yet, but I wanted to share with you.
$\mathbf {EDIT:}$ To be clear, I think that I derived an inequality weaker than Rozenberg's inequality by using his inequality, and I presented it for those of you who might interest trying to solve inequalities by ownselves
| Alternative proof: the pqr method
Let $p = a + b + c$ and $q = ab + bc + ca$ and $r = abc$.
We will use $p^2 \ge 3q$ and $q^2 \ge 3pr$ which are well-known. See the remarks at the end.
Using $a^3 + b^3 + c^3 = p^3 - 3pq + 3r$, the condition is written as
$$p^3 - 3pq + 3r = 3. \tag{1}$$
Using (1) and $q^2 \ge 3pr$, we have
$$p^3 - 3pq + 3\cdot \frac{q^2}{3p} \ge 3$$
or
$$\frac{1}{p}\left(\frac32p^2 - q\right)^2 \ge \frac54 p^3 + 3$$
which results in (using $p^2 \ge 3q$)
$$\frac32p^2 - q \ge \frac12\sqrt{5p^4 + 12p}$$
or
$$q \le \frac{3}{2}p^2 - \frac12\sqrt{5p^4 + 12p}. \tag{2}$$
Using (2), we have
\begin{align*}
&2(a^2 + b^2 + c^2) - (ab + bc + ca) - 3\\
=\,& 2(p^2 - 2q) - q - 3\\
=\,& 2p^2 - 5q - 3\\
\ge\,& 2p^2 - 5\cdot \left(\frac{3}{2}p^2 - \frac12\sqrt{5p^4 + 12p}\right) - 3 \\
=\,& \frac52\sqrt{5p^4 + 12p} - \frac{11}{2}p^2 - 3\\
\ge\,& 0
\end{align*}
where we have used
$$\left(\frac52\sqrt{5p^4 + 12p} \right)^2 - \left( \frac{11}{2}p^2 + 3\right)^2
= (3 - p)(-p^3 - 3p^2 + 24p - 3) \ge 0.$$
(Note: It is easy to prove that $1 \le p \le 3$. Also,
$-p^3 - 3p^2 + 24p - 3 \ge 0$ for all $1 \le p \le 3$. See the remarks at the end.)
We are done.
Remarks:
*
*$p^2 \ge 3q$ is just $a^2 + b^2 + c^2 \ge ab + bc + ca$.
*$q^2 \ge 3pr$ is just $(ab)^2 + (bc)^2 + (ca)^2 \ge ab \cdot bc + bc \cdot ca + ca \cdot ab$.
*We have $p \le 3$ which follows from the power mean inequality $\sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} \ge \frac{a + b + c}{3}$.
*We have $p \ge 1$.
Indeed, if $p < 1$, then $a, b, c \le 1$ and
$1 > a + b + c \ge a^3 + b^3 + c^3$ which contradicts the condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\gcd (3 + \sqrt{13},\ 2 + 5\sqrt{13})$ in $\mathbb Z[(1+\sqrt{13})/2]$ My task is find $\gcd (3 + \sqrt{13},\ 2 + 5\sqrt{13})$ in $\mathbb Z[(1+\sqrt{13})/2]$.
Can you give me some advice?
Any help is highly appreciated.
| \begin{align*}
\text{Let}\;\;R &= \mathbb{Z}[(1+\sqrt{13})/2]\\[4pt]
a &=3 + \sqrt{13}\\[4pt]
b &=2 + 5\sqrt{13}\\[8pt]
\end{align*}
and suppose that $d \in R$ is a common divisor of $a,b$.
By definition,
\begin{align*}
& N(a) = a\bar{a} = \left(3 + \sqrt{13}\right)\left(3 - \sqrt{13}\right) = -4\\[4pt]
& N(b) = b\bar{b} = \left(2 + 5\sqrt{13}\right)\left(2 - 5\sqrt{13}\right)= -321\\[4pt]
\end{align*}
so in the ring $R$, we have
\begin{align*}
&d\,{\mid}\,a\;\;\text{and}\;\;a\,{\mid}\,(-4)\implies d\,{\mid}\,(-4)\\[4pt]
&d\,{\mid}\,b\;\;\text{and}\;\;b\,{\mid}\,(-321)\implies d\,{\mid}\,(-321)\\[4pt]
\end{align*}
Since $-4$ and $321$ are relatively prime in $\mathbb{Z}$, there is an ordinary integer linear combination of $-4$ and $-321$ which equals $1$. For example
$$(80)(-4) + (-1)(-321) = 1$$
Since in the ring $R$, $d\,{\mid}\,(-4)$ and $d\,{\mid}\,(-321)$, it follows that $d$ divides any integer linear combination of $-4$ and $-321$.
In particular, $d\,{\mid}\,1$, so $d$ is a unit in $R$.
Therefore $\gcd(a,b)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:
This then gave me the following equation to solve:
\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}
And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}
I solved these to obtain:
$$a=2,
b=2,
c=3,
d=1,
y=4$$
And thus $$x=4^2-9=7$$
My question:
Is there another way I could have solved this, using the hint to think outside the box?
| The hint may assume two things: to draw extra lines 1) outside; 2) inside the box so that the solution is simple to understand.
Here is yet another method: draw the lines to divide the square into four rectangles indicated by $A, B, C$ and $3+3$:
$$\begin{cases}
A+B=8 \\
A+C=4 \\
CB=6A
\end{cases} \stackrel{(1)-(2)}\Rightarrow \begin{cases}
B=C+4 \\
A=4-C \\
C(C+4)=6(4-C)
\end{cases} \Rightarrow $$
$$C^2+10C-24=0 \Rightarrow C=2, A=2, B=6$$
Hence: Blue area = $(A+B+C+6)-(2+3+4)=16-9=7.$
P.S. I wonder what brilliant method the proposer (on the brilliant.org) has.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 0
} |
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
| We work with the general form of the equation $f(x)$, assuming it to be a cubic equation: $$f(x)=ax^3+bx^2+cx+d$$
We differentiate this to see that $$f'(x)=3ax^2+2bx+c$$
Now we can say that \begin{align}f(x)+f'(x)&=(ax^3+bx^2+cx+d)+(3ax^2+2bx+c)\\
x^3+5x^2+x+2&=ax^3+(b+3a)x^2+(c+2b)x+(c+d)\end{align}
We can compare coefficients to solve this equation:
\begin{align}a&=1,\tag{$x^3$ term}\\\\
5&=b+3a\tag{$x^2$ term}\\
5&=b+3\times 1\\
&\Downarrow\\
b&=2,\\\\
1&=c+2b\tag{$x$ term}\\
1&=c+2\times 2\\
&\Downarrow\\
c&=-3,\\\\
2&=c+d\tag{constant term}\\
2&=(-3)+d\\
&\Downarrow\\
d&=5\end{align}
Therefore, we have that $$f(x)=x^3+2x^2-3x+5$$
We can check that this is a solution to the equation, by first computing $f'(x)$:
$$f'(x)=3x^2+4x-3$$
and then checking the value of $f(x)+f'(x)$:
\begin{align}f(x)+f'(x)&=(x^3+2x^2-3x+5)+(3x^2+4x-3)\\
&=x^3+(2+3)x^2+(-3+4)x+(5+(-3))\\
&=x^3+5x^2+x+2\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
How to derive the Rotation Matrix from the Euler Formula I'm trying to understand how the two dimensional rotation matrix (i.e. $R \in \mathbb{R}^2$) can be derived from the Euler Formula ($e^{i\theta} = \cos \theta + i \sin \theta$). $R$ is given as:
$$
R(\theta) =
\begin{bmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{bmatrix}
$$
$$
\begin{bmatrix}
x' \\
y'
\end{bmatrix} =
\begin{bmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
$$
$$
x' = x \cos \theta - y \sin \theta
$$
$$
y' = x \sin \theta + y \cos \theta
$$
My questions are:
*
*Why can be $i$ omitted from the rotation matrix? (I tried to look for explanations 1,2 but none of these explanations goes beyond that $i$ is omitted)
*Why can we derive a rotation matrix for $\mathbb{R}^2$ from a form that is defined in $\mathbb{C}^2$? How comes we don't get complex numbers as a result after some rotations?
| The complex number $a+bi$ can be represented by the matrix $\displaystyle \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$.
Note that $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$ and
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \pm\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} a\pm c & -(b\pm d) \\ b\pm d & a\pm c \end{pmatrix}$$
Also we have $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ and
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} ac-bd & -(ad+bc) \\ ad+bc & ac-bd \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
probability or rolling 1 1 2 with six dice What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?
Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.
There are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478
I'm not sure how to get this without the help of a computer though.
| You can consider the possibility to find probability directly without inclusion-exclusion principle. This is a longer way, but there are those who like it. Consider all possibilities to roll at least two ones and at least one two with six 6-sided die. We can have exactly two $1$ and one $2$. Denote this event by $A_{21}$. Or we can have exactly two $1$ and two $2$: event $A_{22}$. There are $10$ disjoint events total: $A_{21}$, $A_{22}$, $A_{23}$, $A_{24}$, $A_{31}$, $A_{32}$, $A_{33}$, $A_{41}$, $A_{42}$, $A_{51}$.
Find and then add their probabilities. The probability of $A_{21}$ is
$$
\mathbb P(A_{21})=\frac{\binom{6}{2}\binom{6-2}{1}4^{6-2-1}}{6^6}=\frac{15\cdot 4\cdot 4^3}{6^6}.
$$
By first multiplier $\binom{6}{2}=15$ we choose the rolls which result in $1$, next multiplier $\binom{6-2}{1}=4$ we choose the roll among the rest which results in $2$, and the last multiplier $4^3$ is number of possible outcomes in the rest $6-2-1=3$ rolls.
Similar,
$$\mathbb P(A_{22})=\frac{\binom{6}{2}\binom{4}{2}4^{2}}{6^6}=\frac{15\cdot 6\cdot 16}{6^6},$$
$$\mathbb P(A_{23})=\frac{15\cdot 4\cdot 4}{6^6},\quad \mathbb P(A_{24})=\frac{15}{6^6},\quad \mathbb P(A_{31})=\frac{20\cdot 3\cdot 16}{6^6},$$
$$\mathbb P(A_{32})=\frac{20\cdot 3\cdot 4}{6^6}, \quad \mathbb P(A_{33})=\frac{20}{6^6}, \quad \mathbb P(A_{41})=\frac{15\cdot 2\cdot 4}{6^6},$$
$$\mathbb P(A_{42})=\frac{15}{6^6}, \quad \mathbb P(A_{51})=\frac{6}{6^6}.$$
The total probability of rolling at least two ones and at least one two with six 6-sided die is equal to the sum of probabilities above:
$$
\frac{6896}{6^6}=0,147805213.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding integer part of $(3+\sqrt{3})^4$ I need to find the integer part of $(3+\sqrt{3})^4$, ie the floor of it and I am unable to proceed.
I can find integer part of $(2+\sqrt{3})^4$ using the binomial theorem as $2-\sqrt{3}<1$.
Please give a hint. Thanks!
| One possibility is to look at $f(n) = (3+\sqrt{3})^n + (3-\sqrt{3})^n$. It's easy to see that $f(4) = 504$. You have
$$(3 + \sqrt{3})^4 = 3^4 + 4 \cdot 3^3 \cdot \sqrt{3} + 6 \cdot 3^2 \cdot 3 + 4 \cdot 3 \cdot 3 (\sqrt{3}) + 9 = 252 + 144 \sqrt{3}$$
and when you expand out $(3-\sqrt{3})^4$ similarly the terms with $\sqrt{3}$ come out negated, so you get $(3 - \sqrt{3})^4 = 252 - 144 \sqrt{3}$.
Now what else can you say about $(3-\sqrt{3})^4$? In particular, can you find its integer part? (This might require a bit of hand calculation.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Finding polynomial with roots $1/\alpha$, $1/\beta$, $1/\gamma$ for the roots of $x^3-4x^2+x+6=0$ If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3-4x^2+x+6=0$, find the equation whose roots are
1) $1/\alpha$, $1/\beta$, $1/\gamma$
2) $\alpha^2$, $\beta^2$, $\gamma^2$
How to do these kind of questions. I am stuck.
I know,
$\alpha+\beta+\gamma=-b/a$
$\alpha\beta+\beta \gamma+\alpha \gamma=c/a$
$\alpha \beta\gamma=-d/a$
| For the first one, set $x=1/y\implies$ $$\left(\dfrac1y\right)^3-4\left(\dfrac1y\right)^2+\left(\dfrac1y\right)+6=0$$
Multiply by $y^3$
For the second, $x^2=y,$
$$x(x^2+1)=4x^2-6\implies x(y+1)=4y-6$$
Square both sides and replace $x^2$ with $y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the functional equation $ f \left( \sqrt { x ^ 2 + y ^ 2 } \right) = f ( x ) f ( y ) $ I happened to run across this problem in Larson's Problem Solving through Problems and I wanted to ask how you would approach it.
A real-valued continuous function satisfies for all real $ x $ and $ y $ the functional equation
$$ f \left( \sqrt { x ^ 2+ y ^ 2 } \right) = f ( x ) f ( y ) \text . $$
Prove that $ f ( x ) = { [ f ( 1 ) ] } ^ { x ^ 2 } $.
Attempt:
I know that if we define $ f ( n ) = 2 ^ { \frac n 2 } $, $ f ( 1 ) = 2 ^ { \frac 1 2 } $ resulting in $ f ( n ) = 2 ^ { \frac n 2 } = 2 ^ { \frac { n ^ 2 } 2 } $. But this doesn't seem to make sense for me. Any clues?
EDIT:
If I were to use the hint given in the book where I am asked to first prove the theorem for all numbers of the form $ 2 ^ { \frac n 2 } $, where $ n $ is an integer and then prove the theorem for all numbers of the form $ \frac m { 2 ^ n } $, with $ m $ an integer and $ n $ a nonnegative integer, how would that help me?
| Note that the constant zero function satisfies the equation. So here you must accept that $ 0 ^ x = 0 $, for all real numbers $ x $ ( which is problematic when $ x $ is close to $ 0 $). Otherwise your claim that $ f ( x ) = f ( 1 ) ^ { x ^ 2 } $ is not correct.
Letting $ x = y = 0 $ in
$$ f \bigg( \sqrt { x ^ 2 + y ^ 2 } \bigg) = f ( x ) f ( y ) \tag 0 \label 0 $$
we have $ f ( 0 ) \big( f ( 0 ) - 1 \big) = 0 $ and thus $ f ( 0 ) = 0 $ or $ f ( 0 ) = 1 $. If $ f ( 0 ) = 0 $, then letting $ y = 0 $ in \eqref{0} we get $ f ( | x | ) = 0 $. Now letting $ y = x $ in \eqref{0}, we have $ f ( x ) ^ 2 = f \big( \big| \sqrt 2 x \big| \big) $ and hence in this case $ f $ is identically zero. So, from now on, we assume that $ f ( 0 ) = 1 $. In this case, letting $ y = 0 $ in \eqref{0} we get $ f ( | x | ) = f ( x ) $ which means $ f $ is an even function. Also since $ f ( | x | ) = f \Big( \frac x { \sqrt 2 } \Big) ^ 2 $, we have $ f ( x ) \ge 0 $. For convenience, we define $ g : [ 0 , + \infty ) \to [ 0 , + \infty ) $ with the equation $ g ( x ) = f \big( \sqrt x \big) $. Substituting $ \sqrt x $ for $ x $ and $ \sqrt y $ for $ y $ in \eqref{0}, we have
$$ g ( x + y ) = g ( x ) g ( y ) $$
which yields
$$ g ( x ) ^ m = g \Big( \frac m n x \Big) ^ n \tag 1 \label 1 $$
for every positive integers $ m $ and $ n $, using a simple induction. Using continuity of $ f $ at $ 0 $, one can show that for large enough $ n $, $ g \big( \frac x n \big) > 0 $, which by \eqref{1} gives us $ g ( x ) > 0 $. By continuity, this lets us generalize $ g \big( \frac m n \big) = g ( 1 ) ^ { \frac m n } $ to $ g ( x ) = g ( 1 ) ^ x $ for every non-negative real number $ x $. Finally we have $ f ( x ) = f ( | x | ) = f \Big( \sqrt { x ^ 2 } \Big) = g \big( x ^ 2 \big) = g ( 1 ) ^ { x ^ 2 } = f ( 1 ) ^ { x ^ 2 } $.
| {
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How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0
How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0
I thought that this was going to be a simple problem but then it got more complicated than I have expected
So I first multiplied the expression with its conjugate on both the denominator and the numerator
the conjugate was the following:
$$\sqrt{x^2+9}+3$$
but when I multiplied by the conjugate I actually didn't even get close to the answer because the answer was supposed to be $\frac{1}{6}$
instead I have gotten the following by multiplying by its conjugate
$\frac{x^2+9-9}{x^2\sqrt{x^2+9}-3x^2}$ which obviously doesn't return 1/6
I was wondering if I was just simply making a mistake or if I have to choose a different option in solving this?
| Note:
\begin{align*}
\frac{\sqrt{x^2+9}-3}{x^2}
= & \; \frac{\big(\sqrt{x^2+9}-3 \big) \big(\sqrt{x^2+9}+3 \big)}{x^2 \big( \sqrt{x^2+9}+3 \big)} \\
= & \; \frac{x^2+9-3^2}{x^2\big(\sqrt{x^2+9}+3 \big)} \\
= & \; \frac{1}{\sqrt{x^2+9} + 3}.
\end{align*}
So we have
\begin{align*}
\lim_{x \to 0} \frac{\sqrt{x^2+9}-3}{x^2} = \lim_{x \to 0} \frac{1}{\sqrt{x^2+9} + 3} = \frac{1}{\sqrt{9}+3}=\frac{1}{6}.
\end{align*}
I guess you made some mistake when you multiplied the conjugate.
| {
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If $a,b,c,d>0$ prove $\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$ If $a,b,c,d>0$ prove
$$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$$
I tried cauchy: $$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b}$$ with $$(a+b+c + b+c+d +c+d+a +d+a+b)$$ but I got $$\sqrt{2a^2+bc} + \sqrt{2b^2+cd}+\sqrt{2c^2+da}+\sqrt{2d^2+ab}$$. If without square root, I can get $$(a+b)^2+(b+c)^2+(c+d)^2+(d+a)^2$$ and I can apply p-norm.
Thanks
| We need to prove that
$$\sum_{cyc}\frac{2a^2+bc}{a+b+c}\geq\sum_{cyc}a$$ or
$$\sum_{cyc}\left(\frac{2a^2+bc}{a+b+c}-a\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a-c)}{a+b+c}\geq0.$$
Now, easy to use BW.
Indeed, let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$ and the rest for you.
| {
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Chinese Remainder theorem polynomials Is there a program, small script, or easy method I can use to find the polynomial P(x) such that:
$P(x) = x+1$ $\pmod {x^2+1}$
$P(x) = x+2$ $\pmod {x^2+2}$
$P(x) = x+3$ $\pmod {x^2+3}$
$P(x) = x+4$ $\pmod {x^2+4}$
.........
This is easy to do with integers, but I can't get across this with polynomials. Thanks for help!
| It didn't take too long to compute the following table of moduli.
\begin{array}{r|cccc}
& x^2 + 1 & x^2 + 2 & x^2 + 3 & x^2 + 4 \\
\hline
\dfrac 16(x^2+2)(x^2+3)(x^2+4) & 1 & 0 & 0 & 0 \\
-\dfrac 12(x^2+1)(x^2+3)(x^2+4) & 0 & 1 & 0 & 0 \\
\dfrac 12(x^2+1)(x^2+2)(x^2+4) & 0 & 0 & 1 & 0 \\
-\dfrac 16(x^2+1)(x^2+2)(x^2+3) & 0 & 0 & 0 & 1 \\
\hline
\end{array}
It follows that
\begin{align}P(x) =
& \dfrac 16(x^2+2)(x^2+3)(x^2+4)(x+1)- \\
& \dfrac 12(x^2+1)(x^2+3)(x^2+4)(x+2)+ \\
& \dfrac 12(x^2+1)(x^2+2)(x^2+4)(x+3)- \\
& \dfrac 16(x^2+1)(x^2+2)(x^2+3)(x+4)
\end{align}
With the help of Wolfram alpha, this expands to $f(x) = x-x^2$
| {
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Find $[T_{c}]^{\alpha}_{\alpha}$ For $c \in R$, let $T_{c}: M_{2x2}(R) \rightarrow M_{2x2}(R)$ be the linear transformation defined by $T_{c}(A) = A + cA^{T}$ for every $A \in M_{2x2}$.
For the basis $\alpha = \left\{\begin{bmatrix} 1 & 0\\ 0 & 0\\\end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 1\\\end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0\\\end{bmatrix} \begin{bmatrix} 0 & 1\\ -1 & 0\\\end{bmatrix}\right\}$ of $M_{2x2}(R)$, find $[T_{c}]^{\alpha}_{\alpha}$ and for what values of c is $T_{c}$ an isomorphism?
My working:
$T_{c}\{\begin{bmatrix} 1 & 0\\ 0 & 0\\\end{bmatrix}\} = \begin{bmatrix} 1+c & 0\\ 0 & 0\\\end{bmatrix}$
$T_{c}\{\begin{bmatrix} 0 & 0\\ 0 & 1\\\end{bmatrix}\} = \begin{bmatrix} 0 & 0\\ 0 & 1+c\\\end{bmatrix}$
$T_{c}\{\begin{bmatrix} 0 & 1\\ 1 & 0\\\end{bmatrix}\} = \begin{bmatrix} 0 & 1+c\\ 1+c & 0\\\end{bmatrix}$
$T_{c}\{\begin{bmatrix} 0 & 1\\ -1 & 0\\\end{bmatrix}\} = \begin{bmatrix} 0 & c-1\\ c-1 & 0\\\end{bmatrix}$
So $[T_{c}]^{\alpha}_{\alpha} = \begin{bmatrix} 1+c & 0 & 0 & 0\\ 0 & 0 & 0 & 1+c\\ 0 & 1+c & 1+c & 0\\ 0 & c-1 & c-1 & 0\\ \end{bmatrix}$
$T_{c}$ is an iso-morphism when $c \neq 1$ and $c \neq -1$
Would you say what I did is correct or am I going in the wrong direction?
| You are on the right way, but you made mistakes in your computation.
First:
$$
T_c\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}+c\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&1-c\\c-1&0\end{pmatrix}.
$$
Second: Since
$$
T_c\begin{pmatrix}1&0\\0&0\end{pmatrix}=(1+c)\begin{pmatrix}1&0\\0&0\end{pmatrix}\\
T_c\begin{pmatrix}0&0\\0&1\end{pmatrix}=(1+c)\begin{pmatrix}0&0\\0&1\end{pmatrix}\\
T_c\begin{pmatrix}1&0\\0&1\end{pmatrix}=(1+c)\begin{pmatrix}1&0\\0&1\end{pmatrix}\\
T_c\begin{pmatrix}0&1\\-1&0\end{pmatrix}=(1-c)\begin{pmatrix}0&1\\-1&0\end{pmatrix}
$$
you get
$$
[T_c]^\alpha_\alpha=\begin{pmatrix}1+c&0&0&0\\0&1+c&0&0\\0&0&1+c&0\\0&0&0&1-c\end{pmatrix}.
$$
But your statement that $T_c$ is an isomorphism if $c\neq 1$ and $c\neq -1$ holds.
If we write $\alpha=\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$
it seems, you wrote the elements of $T_c\alpha_i$ in the $i$-th row of $[T_c]_\alpha^\alpha$.
But you have to write $T_c\alpha_i=a_1\alpha_1+a_2\alpha_2+a_2\alpha_3+a_4\alpha_4$ where $a_1,a_2,a_3,a_4\in\mathbb{R}$. Then your $i$-th column of $[T_c]_\alpha^\alpha$ is $\begin{pmatrix}a_1\\a_2\\a_3\\a_4\end{pmatrix}$
| {
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Rational values of trigonometric functions I am using extensively trigonometric functions when an angle is given in degrees.
Some of these functions like sine or cosine have rational values, for example, the well known example is that $\cos(\theta) =0.6 $ and $\sin(\theta) =0.8 $.
However besides the case of multiplicity of $90^\circ$ it seems there are no rational numbers $\theta$ with simultaneously rational values of sine and cosine.
*
*Is it possible somehow to prove that for rational values of an angle given in degrees there are no values simultaneously rational of sine and cosine functions, beside obvious case of multiplicities of $90^\circ$?
| Let us take the Pythagorean theorem $a^2 + b^2 = c^2$ and divide both sides by $c^2$ to get $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$. Now, we can use the well-known relationship between sine and cosine: $\sin^2(x) + \cos^2 (x) = 1$, and let $\sin(x) = \frac{a^2}{c^2}$, and $\cos(x) = \frac{b^2}{c^2}$.
Since the hypotenuse of a triangle is always longer than the two legs, $\frac{a^2}{c^2}, \frac{b^2}{c^2} < 1$. Therefore, there exists a bijection between one triplet of $a,b,c$ and $\frac{a^2}{c^2}, \frac{b^2}{c^2}$, and at least one value of $\frac{a^2}{c^2}, \frac{b^2}{c^2}$ for $\sin(x)$.
One example using a $3,4,5$ triangle shows that $\cos(x) = \frac{9}{25}$, and using the second equation $\sin(x) = \frac{16}{25}$. As a bonus, since there are infinitely many Pythagorean triples, this shows that there are infinitely many values of $\sin(x)$ and $\cos(x)$ that are rational.
| {
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How to find this minimum of the value
Let $x_{i}$, where $i\in\{1,2,\cdots,n\}$ be distinct real numbers. Find the minimum of the value of
$$\sum_{1\le i<j\le n}\left(\dfrac{1-x_{i}x_{j}}{x_{i}-x_{j}}\right)^2$$
It is clear when $n=2$ minimum of the value is $0$, when $x_{1}x_{2}=1$. But for $n\ge 3$,I can't find it.
Thanks.
| We begin with a short proof for the case $n=3$ :
We make a substitution so we put :
$\frac{\lambda_1}{2}(\frac{1+(a-b)^2}{a-b})^2=(\frac{1-xy}{x-y})^2$
$\frac{\lambda_2}{2}(\frac{1+(b-c)^2}{c-b})^2=(\frac{1-zy}{z-y})^2$
$\frac{\lambda_3}{2}(\frac{1+(a-c)^2}{a-c})^2=(\frac{1-xz}{x-z})^2$
With $a,b,c$ and the condition $a\geq b \geq c$ real numbers and $\lambda_1,\lambda_2,\lambda_3$ real positive numbers with the condition $\lambda_1+\lambda_2+\lambda_3=1$
So we have to find the minimum value of :
$$\frac{\lambda_1}{2}(\frac{1+(a-b)^2}{a-b})^2+\frac{\lambda_2}{2}(\frac{1+(b-c)^2}{c-b})^2+\frac{\lambda_3}{2}(\frac{1+(a-c)^2}{a-c})^2$$
We could apply Jensen's inequality for $f(x)=\frac{1}{2}(\frac{1+x^2}{x})^2$ it becomes :
$$\frac{1}{2}[\lambda_1(\frac{1+(a-b)^2}{a-b})^2+\lambda_2(\frac{1+(b-c)^2}{c-b})^2+\lambda_3(\frac{1+(a-c)^2}{a-c})^2]\geq \frac{1}{2}[(\frac{1+(\lambda_1(a-b)+\lambda_2(b-c)+\lambda_3(a-c))^2}{((\lambda_1(a-b)+\lambda_2(b-c)+\lambda_3(a-c))})^2]$$
But $\lambda_1(a-b)+\lambda_2(b-c)+\lambda_3(a-c)>0$
So the minimum of the function $f(x)$ is 2 .
| {
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Integration of secant $$\begin{align}
\int \sec x \, dx
&= \int \cos x \left( \frac{1}{\cos^2x} \right) \, dx \\
&= \int \cos x \left( \frac{1}{1-\sin^2x} \right) \, dx \\
& = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\
&= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx
\end{align}$$
I am stuck in here. Any help to integrate secant?
| \begin{align*}\int\sec x\,\mathrm dx&=\int\frac1{\cos x}\,\mathrm dx\\&=\int\frac{\cos x}{\cos^2x}\,\mathrm dx\\&=\int\frac{\cos x}{1-\sin^2x}\,\mathrm dx.\end{align*}
Now, doing $\sin x=t$ and $\cos x\,\mathrm dx=\mathrm dt$, you get $\displaystyle\int\frac{\mathrm dt}{1-t^2}$. But\begin{align*}\int\frac{\mathrm dt}{1-t^2}&=\frac12\int\frac1{1-t}+\frac1{1+t}\,\mathrm dt\\&=\frac12\left(-\log|1-t|+\log|1+t|\right)\\&=\frac12\log\left|\frac{1+t}{1-t}\right|\\&=\frac12\log\left|\frac{(1+t)^2}{1-t^2}\right|\\&=\log\left|\frac{1+t}{\sqrt{1-t^2}}\right|\\&=\log\left|\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right|\\&=\log\left|\frac1{\cos x}+\frac{\sin x}{\cos x}\right|\\&=\log|\sec x+\tan x|.\end{align*}
| {
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Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$, find the value of $P(n+1)$ Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
I tried by making a $f(x) = (x^3+1) P(x) - x^2$. But this equation will have $n+3$ roots, out of which $n+1$ will be $0,1,2,...,n$. What about the other two roots? Can we simplify $x^2 \over x^3+1$ more to help?
| If $p(x)$ is a polynomial with degree $\leq n$ and $\delta$ is the forward difference operator, $(\delta p)(x)=p(x+1)-p(x)$, we have that
$$ \left(\delta^{n+1}p\right)(x) = \sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k\,p(x+k) = 0\tag{1} $$
hence in our case
$$ P(n+1)=(-1)^n\sum_{k=0}^{n}\binom{n+1}{k}(-1)^k \frac{k^2}{k^3+1}.\tag{2} $$
Addendum. Can we compute a closed form for the RHS of (2)? With a bit of patience, yes.
Let $\omega=\exp\left(\frac{\pi i}{3}\right)=\frac{1+i\sqrt{3}}{2}$. By partial fraction decomposition we have
$$ \frac{k^2}{k+1} = \frac{1}{3}\left(\frac{1}{k+1}+\frac{1}{k-\omega}+\frac{1}{k-\overline{\omega}}\right)\tag{3} $$
and since $\frac{1}{k+1}=\int_{0}^{1}x^k\,dx$,
$$ \sum_{k=0}^{n}\binom{n+1}{k}\frac{(-1)^k}{k+1} = \frac{1+(-1)^n}{n+2}\tag{4} $$
$$ \sum_{k=0}^{n}\binom{n+1}{k}(-1)^k\left(\frac{1}{k-\omega}+\frac{1}{k-\overline{\omega}}\right)\\ = \frac{(-1)^n(2n+1)}{n^2+n+1}+\underbrace{2\,\text{Re}\,B\left(n+2,\frac{-1+i\sqrt{3}}{2}\right)}_{\text{wildly oscillating term.}}\tag{5} $$
| {
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How to solve these functional equations $f(x) = 1 + f(x) + f(x^2) + \ldots $ and $f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \ldots $? The identity $\frac{1}{1-f(x)}=1+f(x)+\big(f(x)\big)^2+\big(f(x)\big)^3 + \ldots$ is well known.
Is there somewhere in the literature a method allowing us to solve the following functional equations?
First equation:
$$f(x) = 1 + f(x) + f(x^2) + f(x^3) + \ldots $$
Second equation:
$$f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \big(f(x^3)\big)^3 + \ldots $$
| I think that there also isn't a solution for the second equation.
Plug in $x=0$. Then we get:
$
f(0) = 1 + f(0) + f(0)^2 + \cdots
$
So $f(0) = \frac{1}{1-f(0)}$ and $f(0)^2 - f(0) + 1 = 0$, and this has no roots over the reals. (Note that $f(0) \neq 1$ as then the equation for $f(0)$ won't converge).
Also note that for both equation you could do the exact same thing for $f(1)$.
| {
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If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to sum and product of roots form, but this way is too complicated!
Please suggest a simpler process.
| The answer is $-10$.
Find the common denominator:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\frac{(1-ab-(a-b))^3+(1-ab+(a-b))^3}{(1+ab+(a+b))^3}=$$
$$\frac{2\cdot\left(\frac23\right)^3+2\cdot 3 \cdot\left(\frac23\right)\cdot (a-b)^2}{(\frac{2}{3})^3}=\frac{2\cdot\left(\frac23\right)^3+4\cdot ((a+b)^2-4ab)}{(\frac{2}{3})^3}=\frac{-10(\frac{2}{3})^3}{(\frac23)^3}=-10.$$
| {
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Trig equation $\sin(x)+\sin(3x)=0$, the answers are given in a factored form $\sin(x)+\sin(3x)=0$
So to solve this I tried the following:
-First I transformed this expression from sum to product because it equals zero
$\sin(x)+\sin(3x)=0 => 2\sin(\frac{x+3x}{2})\cdot\cos(\frac{x-3x}{2})=0$
-Second part I got to
$2\sin(2x)\cos(x)=0$
-Third part:
I have separated the equation into two parts:
$2\sin(2x)=0$ and $\cos(x)=0$
From now on I don't know how to get to the correct answers, which option should i circle and why:
a) $x=(2k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$
b) $x=(k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$
c) $x=(3k+1)\frac{\pi}{3}$ or $x=\frac{3k\pi}{2}$
d) $x=(k-1)\frac{\pi}{2}$ or $x=\frac{3k\pi}{2}$
where $k$ is any integer.
| \begin{align*}\sin (2x)=0 &\Longrightarrow 2x=k\pi\to x=\frac{k\pi}{2}\\
\cos x=0&\Longrightarrow x= \frac{\pi}{2}+k\pi=\frac{(2k+1)\pi}{2}\end{align*}
| {
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Prove that $(k^3)!$ is divisible by $(k!)^{k^2 + k + 1}$. Proof using properties of greatest integer function? I came across this problem in a regional Olympiad Exam. I'm giving the outline of my solution and a question related to it.
Problem
Let $k$ be any positive integer. Prove that $(k^3)!$ is divisible by $(k!)^{k^2 + k + 1}$.
Question
My first approach was to use the fact that for any prime $p \leq k$, the highest power of $p$ contained in $k!$ is given by $\sum _{j=1}^{\infty} \big[ \frac{k}{p^j}\big]$, where $[.]$ is the greatest integer (floor) function, then somehow use the properties of this function to show that for every prime $p$ contained in $k!$, it's power in $k^3!$ is at least $k^2+k+1$ times that in $k!$.
However, this approach did not lead me anywhere beyond a couple of steps. The reason might be that I'm missing some properties of the greatest integer function. Please post your proof if you can make this approach work.
I finally gave up this approach and came up with the following solution.
Solution
I somehow managed to prove a generalized statement using induction.
We will prove that $\forall k,n \in \mathbb{Z}^+$, $(k^n)!$ is divisible by $(k!)^{k^{n-1} + k^{n-2} + \cdots + k^2 + k + 1}$.
Let $P(k,n)$ be the statement that $(k!)^{k^{n-1} + k^{n-2} + \cdots + k^2 + k + 1} \mid (k^n)!$.
$P(k,1)$ is obviously true for all $k \in \mathbb{Z}^+$.
Chose any $k$ and assume that $P(k,n)$ is true for some positive integer $n$.
Now, $(k^{n+1})!$ can be written as follows:
$$\begin{eqnarray*}
(k^{n+1})! &=& \Big(1 \cdot 2 \cdot 3 \cdots k^n \Big) \cdot %
\Big((k^n + 1) \cdot (k^n + 2) \cdots (2k^n) \Big) \cdot %
\Big((2k^n + 1) \cdot (2k^n + 2) \cdots (3k^n) \Big) \cdots %
\Big(((k - 1)k^n + 1) \cdot ((k-1)k^n + 2) \cdots (k\cdot k^n) \Big) \\
&=& \alpha_1 \cdot \alpha_2 \cdot \alpha_3 \cdots \alpha_k
\end{eqnarray*}$$
where
$$\begin{eqnarray*}
\alpha_j &=& \big((j - 1)k^n + 1\big) \cdot \big((j-1)k^n + 2\big) \cdots \big(j\cdot k^n \big) \\
&=& \prod \limits_{m = 1}^{k^n} \big((j - 1)k^n + m\big) \\
&=& j \cdot k^n \cdot \prod \limits_{m = 1}^{k^n -1} \big((j - 1)k^n + m\big)
\end{eqnarray*}$$
Now, $\; \prod \limits_{m = 1}^{k^n -1} \big((j - 1)k^n + m\big) \;$ is a product of $(k^n - 1)$ consecutive integers. Hence $(k^n - 1)! \;$ divides $\; \prod \limits_{m = 1}^{k^n -1} \big((j - 1)k^n + m\big) \;$. Therefore, $(k^n)! \quad$ divides $\quad k^n \cdot \prod \limits_{m = 1}^{k^n -1} \big((j - 1)k^n + m\big)$. Therefore, $\alpha_j$ can be written as
$$\alpha_j = j \cdot \beta_j \cdot (k^n)!$$ for some $\beta_j \in \mathbb{Z}^+$.
By the inductive hypothesis, $(k^n)! = \nu \cdot (k!)^{k^{n-1} + k^{n-2} + \cdots + k^2 + k + 1}$ for some $\nu \in \mathbb{Z}^+$. Thus,
$$\alpha_j = j \cdot \beta_j \cdot \nu \cdot (k!)^{k^{n-1} + k^{n-2} + \cdots + k^2 + k + 1}$$
Thus,
$$\begin{eqnarray*}
(k^{n+1})! &=& \alpha_1 \cdot \alpha_2 \cdots \alpha_k \\
&=& \nu^k \cdot \Big(\prod \limits_{j = 1}^{k} \beta_j \Big) %
\cdot \Big(\prod \limits_{j = 1}^{k} j \Big) \cdot %
\Big( (k!)^{k^{n-1} + k^{n-2} + \cdots + k^2 + k + 1} \Big)^k \\
&=& \nu^k \cdot \Big(\prod \limits_{j = 1}^{k} \beta_j \Big) %
\cdot (k!) \cdot %
\Big( (k!)^{k^{n} + k^{n-1} + \cdots + k^3 + k^2 + k} \Big) \\
&=& \nu^k \cdot \Big(\prod \limits_{j = 1}^{k} \beta_j \Big) %
\cdot %
\Big( (k!)^{k^{n} + k^{n-1} + \cdots + k^3 + k^2 + k + 1} \Big)
\end{eqnarray*}$$
Thus $P(k,n+1)$ is true if $P(k,n)$ is true.
| In the interval $[1,k^3]$ there are $k^3-k^2$ numbers $\not\equiv 0\pmod{k}$, $k^2-k$ numbers that are $\equiv 0\pmod{k}$ but $\not\equiv 0\pmod{k^2}$, $k-1$ numbers that are $\equiv 0\pmod{k^2}$ but $\not\equiv 0\pmod{k^3}$ and just $1$ element that is a multiple of $k^3$. In particular $(k^3)!$ is for sure a multiple of
$$ k^3 \cdot k^{2(k-1)} \cdot k^{k^2-k} =k^{k^2+k+1}.$$
The product of $(k-1)$ consecutive numbers $\neq 0\pmod{k}$ is always a multiple of $(k-1)!$, since
$$ \frac{1}{(k-1)!}\prod_{j=1}^{k-1}(a+j) = \binom{a+k-1}{k-1} = \binom{a+k-1}{a}\in\mathbb{Z}. $$
It follows that the product of the elements of $[1,k^3]$ that are not $\equiv 0\pmod{k}$ is a multiple of $(k-1)!^{k^2}$ and the claim simply follows by rearranging the remaining terms and applying the considerations above. It is enough to consider that
$$(k!)^{k^2+k+1} = k^{k^2+k+1}\cdot (k-1)!^{k^2+k+1}. $$
| {
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"source": "stackexchange",
"question_score": "3",
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} |
How to evaluate $\int \cos^2x \ dx$ I am new to integration and I want to evaluate
$ \int \cos^2x \, dx $
What I done
Using simple chain rule,
$ \int(\cos x)^2 \\
= \int t^2 \,dt \ (t = \cos x) \\
= \frac{t^3}{3} \ + C $
By substitution of $t = \cos(x)$ I got with answer.
| You can make use of the double angle formula for cosine:
$$\begin{eqnarray}\cos(2x) & = & 2 \cos^2 (x) - 1 \\
\cos^2 (x) & = & \frac{1}{2}\left(\cos (2x) + 1\right)\\
\int \cos^2 x\ dx & = & \frac{1}{2} \int \left(\cos(2x) + 1 \right) dx \\
& = & \frac{1}{2}\left[\frac{1}{2}\sin(2x) + x + C \right] \\
& = & \frac{1}{4}\sin(2x) + \frac{1}{2}x + C \end{eqnarray}$$
| {
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For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is?
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
$\frac{a+b} {c}$ is how much?
Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3abc$$
It would e helpful if someone could tell me how should I proceed.
| $a^2 + b^2 +c^2 = \frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}= ab + bc + ac\\
\frac {a^2 -2ab +b^2}{2} + \frac {b^2 -2bc +c^2}{2} + \frac {a^2 -2ac +c^2}{2}= 0\\
(a-b)^2 + (b-c)^2 + (c-a)^2 = 0\\
a = b = c$
$\frac {a+b}{c} = 2$
| {
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How many times does $k$ occur in the composition of $n$? How many times does the number $k$ occur in the composition of $n$?
Composition of Integer
In short, the difference between the partition of an integer and composition is the order of numbers. In partition, the order doesn't matter whereas, in composition, it does. That's why Partitions are sometimes called as ordered Compositions.
Example: $k$ = $1$ & $n$ = $5$
The composition of 5 are:
$5$
$4 + 1$
$3 + 2$
$3 + 1 + 1$
$2 + 3$
$2 + 2 + 1$
$2 + 1 + 2$
$2 + 1 + 1 + 1$
$1 + 4$
$1 + 3 + 1$
$1 + 2 + 2$
$1 + 2 + 1 + 1$
$1 + 1 + 3$
$1 + 1 + 2 + 1$
$1 + 1 + 1 + 2$
$1 + 1 + 1 + 1 + 1$
In all $1$ occurs $28$ times in the composition of $5$
Similarly is there any relation between $k$ and $n$ for all $n \geq 0$ & $k \leq n$
| Suppose that $k$ (such that $0<k<n$) occurs at "position $p$" in a composition of $n$: $$\ldots+k+\ldots$$
by which I mean that everything to its left is a composition of $p$, and everything to its right is a composition of $n-p-k$.
If $p>0$ and $n-p-k>0$, there are $2^{p-1}$ compositions of $p$ and $2^{n-p-k-1}$ compositions of $n-p-k$, thus there are $2^{p-1}\cdot2^{n-p-k-1}=2^{n-k-2}$ possibilities for the rest of the composition of $n$. If $p=0$ or $n-p-k=0$, there are $2^{n-k-1}$ possibilities for the rest of the composition of $n$.
Summing over $p$, you will find that for $k<n$ the answer is: $$2\cdot2^{n-k-1}+\sum_{p=1}^{n-k-1}2^{n-k-2}=(n-k+3)\cdot2^{n-k-2}$$
| {
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"source": "stackexchange",
"question_score": "6",
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Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ cancel eachother out outside the integral, so we have:
$$-\int \frac{sec\ \theta\ tan\ \theta\ d\theta}{tan\ \theta}$$
The $tan\ \theta$ terms cancel, so we're left to integrate $sec\ \theta$ which is equal to:
$$- \ln\ (tan\ \theta + \ sec\ \theta) + C$$
Since $tan\ \theta = \sqrt{x^2-9}\ $ since it replaced it in the integral, and $sec\ \theta = \frac{x}{3}$, the answer is:
$$-\ ln\ (\sqrt{x^2-9}\ + \frac{x}{3})+C$$
However, using an online calculator, the answer turned out to be:
$$-\ ln\ (\sqrt{x^2-9}\ + x)+C$$
It seemed to come about due to their substitution of $u = \frac{x}{3}$ we led them to get the standard integral of $sec^{-1}x$, which would imply that
$$\sqrt{\frac{x^2}{9} - 9}$$
becomes $$\sqrt{u^2 - 1}$$
And I just don't see how. Can someone explain why what they did is valid and/or where my mistake was?
| $$\int\frac{dx}{\sqrt {x^2-9}}$$
By substitution
$$x=3\sec\theta $$
$$dx=3\sec\theta \tan \theta d\theta$$
$$\int \frac{3\sec\theta \tan \theta d\theta}{\sqrt{9\sec^2\theta-9}} $$
$$\int \frac{3\sec\theta \tan \theta d\theta}{3\tan\theta} $$
$$\int \sec\theta d\theta $$
$$ ln|\sec\theta+\tan\theta|$$
$$ ln| \frac{1+\sin \theta}{cos \theta}|$$
$$ \cos \theta=\frac{3}{x}$$
Recall that $\sin \theta= \sqrt{1-\cos^2\theta}$
$$ln|\frac{x(1+\sqrt{1-\frac{9}{x^2})}}{3}|$$
$$ ln|x+\sqrt{x^2-9}|-ln3$$
| {
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Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram. Problem
Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.
My Solution
B (Conclusion): The midpoints of the sides of a space quadrilateral form a parallelogram.
A (Hypothesis): Let $A$, $B$, $C$, $D$ be four points such that they form a space quadrilateral.
B1: $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ where $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B}$ and $\dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ are congruent sides. The same can be said for the other two sides.
A1: $\mathbf{A} + \mathbf{B} = \mathbf{C} + \mathbf{D}$ by the definition of quadrilaterals.
$\implies \dfrac{1}{2} \left( \mathbf{A} + \mathbf{B} \right) = \dfrac{1}{2} \left( \mathbf{C} + \mathbf{D} \right)$
$\implies \dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$
$Q.E.D.$
I would greatly appreciate it if people could please review my proof for correctness.
| Hint: If your four points are $a, b, c, d$, then the midpoints, in order around the quad, are
$$
p = \frac{1}{2}(a+b), q = \frac{1}{2}(b+c), r = \frac{1}{2}(c+d), s = \frac{1}{2}(d+a).
$$
For $pqrs$ to be a parallelogram, you need the edge from $p$ to $q$ to have the same direction vector as the edge from $s$ to $r$; you need a similar thing to hold for the edges from $q$ to $r$ and $p$ to $s$.
What's the direction vector of the edge from $p$ to $q$? Can you express it in terms of $a, b, c, d$?
| {
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How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me in finding the value.
Any good hint is welcomed.
| The roots of the equation $ax^2+bx+c=0$ are:
$$\alpha=\frac{-b-\sqrt{b^2-4ac}}{2a}; \beta=\frac{-b+\sqrt{b^2-4ac}}{2a}.$$
Now substitute these into the expression:
$$\frac{1}{(a\alpha+b)^2} + \frac{1}{(a\beta+b)^2}=$$
$$\frac{1}{(\frac{-b-\sqrt{b^2-4ac}}{2}+b)^2} + \frac{1}{(\frac{-b+\sqrt{b^2-4ac}}{2}+b)^2}=$$
$$\frac{4}{(b-\sqrt{b^2-4ac})^2}+\frac{4}{(b+\sqrt{b^2-4ac})^2}=$$
$$\frac{4(2b^2+2(b^2-4ac))}{(4ac)^2}=\frac{b^2-2ac}{a^2c^2}.$$
| {
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prove $\sum_{cyc}\frac{1}{a(a+b)}\ge\frac{4}{ac+bd}$ Let $a,b,c,d$ be positives. Show that
$$\sum_{cyc}\dfrac{1}{a(a+b)}\ge\dfrac{4}{ac+bd}$$
Use Cauchy-Schwarz inequality we have
$$\sum_{cyc}\dfrac{1}{a(a+b)} \cdot\sum_{cyc}a(a+b)\ge (1+1+1+1)^2$$
it suffient
$$\dfrac{16}{\sum\limits_{cyc}(a^2+ab)}\ge\dfrac{4}{ac+bd}$$
However this is wrong inequality
| By AM-GM $$\frac{2}{\sqrt{abcd}}\geq\frac{4}{ac+bd}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{1}{a^2+ab}\geq\frac{2}{\sqrt{abcd}}.$$
Since the last inequality is homogeneous, we can assume that $abcd=1$.
Now, let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{t}{z}$, where $x$, $y$, $z$ and $t$ are positives.
Hence, $d=\frac{x}{t}$ and we need to prove that
$$\sum_{cyc}\frac{1}{\frac{y}{x}\left(\frac{y}{x}+\frac{z}{y}\right)}\geq2$$ or
$$\sum_{cyc}\frac{x^2}{y^2+xz}\geq2.$$
But by C-S we have
$$\sum_{cyc}\frac{x^2}{y^2+xz}=\sum_{cyc}\frac{x^4}{x^2y^2+x^3z}\geq\frac{(x^2+y^2+z^2+t^2)^2}{\sum\limits_{cyc}(x^2y^2+x^3z)}.$$
Thus, it remains to prove that
$$(x^2+y^2+z^2+t^2)^2\geq2\sum\limits_{cyc}(x^2y^2+x^3z)$$ or
$$(x^2+z^2)(x-z)^2+(y^2+t^2)(y-t)^2\geq0.$$
Done!
| {
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Find out |a+p+b+q| for the following question. Let$$f(x)=\begin{cases}
ax(x-1)+b&;x\lt1\\
x+2&;1\le x\le3\\
px^2+qx+2&;x\gt3
\end{cases}$$ is continuous for all x belongs to R except $x=1$ but $|f(x)|$ is differentiable everywhere and $f'(x)$ is continuous at x=3 and $|a+p+b+q|=k$, then k=?
|
$f′(x)$ is continuous at $x=3$
So derivatives of $x+2$ and $px^2 + qx + 2$ must be equal.
$1=2px + q$; $x=3$
$\boxed{1=6p + q}$
continuous for all x belongs to R except x=1
but |f(x)|
is differentiable everywhere
So it's graph should be like this (drawed very roughly, slopes should be equal at $x=1$)
If $x+2$ is positive at $x=1$, then $ax(x−1)+b$ must be negative.
If $x+2$ is negative at $x=1$, then $ax(x−1)+b$ must be positive.
So $ax(x−1)+b = -x-2$ for $x=1$
$a(0) + b = -3$
$\boxed{b=-3}$
Also the derivatives of $ax(x−1)+b$ and $-x-2$ must be equal too. (at $x=1$)
$2ax-a= -1$
$\boxed{a= -1}$
Jump to
$f′(x)$ is continuous at $x=3$
$x+2$ and $px^2 + qx + 2$ must be equal for $x=3$
$9p + 3q + 2 = 5$
$\boxed{3p + q = 1}$
Remember, we found that $1=6p + q$
Solving system of equations $1=6p + q$ and $3p + q = 1$
$\boxed{p=0,q=1}$
Now, $|0+1-1-3| = 3$, the answer should be $3$.
And our function is:
$$f(x)=\begin{cases}
-x(x-1)-3&;x\lt1\\
x+2&;1\le x\le3\\
x+2&;x\gt3
\end{cases}$$
| {
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
|
For what values of $a$ does the equation
$$
\frac{2x}{1+x^2}=a
$$
admit a solution?
The equation can be rewritten
$$
ax^2-2x+a=0
$$
and its discriminant is
$$
4(1-a^2)
$$
which is $\ge0$ if and only if $-1\le a\le 1$. For $a=0$ it's not a degree $2$ equation, but of course $f(0)=0$, so also $a=0$ is a value attained by the function.
| {
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Integral $\int \frac{x^2}{x^4+x^2+1}\ dx$ I have taken $(x^4+x^2+1)=u$ to differentiate it w.r.t. $x$ to get
$$\int \frac{x}{2u(2x^2+1)}\ du$$
| $$\dfrac{2x^2}{x^4+x^2+1}=\dfrac{x^2+1}{x^4+x^2+1}+\dfrac{x^2-1}{x^4+x^2+1}$$
$$=\dfrac{1+\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}+\dfrac{1-\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}$$
For the first integral:
As $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x,$
write $x^2+1+\dfrac1{x^2}=1+\left(x-\dfrac1x\right)^2+2$
Can you handle the second integral?
| {
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Value of a definite integral with irrational terms in the denominator Can someone help me to find the value of the following integration?
$$ \int_0^3\frac{dx}{\sqrt{x+1} + \sqrt{5x+1}} $$
I tried to rationalize the given expression, but that was not of much help.
I also tried to apply the following formula,
$$ \int_0^a{f(x)dx} = \int_0^a{f(a-x)dx} $$ and got the following form
$$ \int_0^3\frac{dx}{\sqrt{4-x} + \sqrt{16-5x}} $$
How do I reach to the final value?
| Rationalizing is probably the way to go. After rationalizing you can rewrite the integrand as,
$$ L =- \int_0^3 \frac{\sqrt{x+1} - \sqrt{5x+1}}{4x}\ dx $$
which can further be expanded into,
$$ L = -\frac 1 4 \int_0^3 \left( \frac{\sqrt{x+1}}{x} - \frac{\sqrt{5x+1}}{x} \right) \ dx \quad\quad \text{(1)}$$
which has an antiderivative at limits $0, 3$,
$$L = -\frac 1 4 \left( -2\sqrt{x+1} + 2\sqrt{5x+1} + 2\tanh^{-1}\left(\sqrt{x+1}\right)-2\tanh^{-1}\left(\sqrt{5x+1}\right) \right)\big|_0^3 \quad \text{(2)}$$
you'll have to take the limit at $x=0$, but after evaluation you'll get
$$ L = 1 - \frac 1 2 \log\left( \frac 5 3 \right) $$
Edit: We make the step from $\text{(1)}$ to $\text{(2)}$ due to the relation,
$$ \int\frac {\sqrt {a x + b}} {x}\, dx =
2\sqrt {a x + b} - 2\sqrt {b} \tanh^{-1}\left(\frac {\sqrt {a x + b}} {\sqrt {b}}\right) + c $$
| {
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} |
Polynomial having rational coefficients and one root: $\sqrt{2}+\sqrt{3}-\sqrt{5}$
Form a polynomial of smallest degree having rational coefficients and one root as $\sqrt{2}+\sqrt{3}-\sqrt{5}$
Idea 1:
I thought that other roots would be just different combination of signs on the surds, ie
*
*$\sqrt{2}+\sqrt{3}+\sqrt{5}$
*$\sqrt{2}-\sqrt{3}+\sqrt{5}$
so least degree would be $2^3 = 8$.
Polynomial then could be formed using viete's formulas.
Idea 2:
We let $x = \sqrt{2}+\sqrt{3}-\sqrt{5}$. Then rearranging and squaring repeatedly gives us the polynomial.
Questions
*
*This method seems unsatisfactory and is just a thought. Please help me with a proper method.
*Is the polynomial i found unique? or there are more polynomials with rational coefficients with this root ($\sqrt{2}+\sqrt{3}-\sqrt{5}$)?
*Also can we generalise this result: that the least degree of a polynomial whose root is a sum of $n$ distinct surds is $\sum \binom{n}{k} = 2^n$ ?
Edit
As stated by Hagen Von Elitzen, the result in third question is correct only for square roots of numbers which are pairwise coprime. Eg. ($\sqrt{2}, \sqrt{3}, \sqrt{5}$) and not ($\sqrt{2}, \sqrt{5}, \sqrt{10}$)
|
If $f(x)\in \mathbb{Q}[x]$ has $\sqrt{2}+\sqrt{3}-\sqrt{5}$ as a root,
then it has all eight numbers $\pm \sqrt{2} \pm \sqrt{3} \pm \sqrt{5}$
as roots. So the smallest degree of your polynomial is $8$.
The reason is that we have some automorphisms over $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. One is
$$\sigma(\sqrt{2}) = \sqrt{2} \quad \sigma(\sqrt{3}) = \sqrt{3} \quad \sigma(\sqrt{5}) = -\sqrt{5}$$ This automorphism also fixes $\mathbb{Q}$
Assume $f(x)\in \mathbb{Q}[x]$ has $\sqrt{2}+\sqrt{3}-\sqrt{5}$ as root, write $f(x) = a_0 + a_1 x + ... + a_n x^n$, $\alpha = \sqrt{2}+\sqrt{3}-\sqrt{5}, \beta = \sqrt{2}+\sqrt{3}+\sqrt{5}$. Then we have $\sigma(\alpha) = \beta$.
We have $a_0 + a_1 \alpha + ... +a_n \alpha^n = 0$, taking $\sigma$ on both sides give
$$\sigma(a_0)+\sigma(a_1)\sigma(\alpha) + ... + \sigma(a_n)\sigma(\alpha)^n = 0 \quad \Rightarrow \quad a_0 + a_1 \beta + ... + a_n \beta^n = 0$$
Hence $f(\alpha) = 0$ implies $f(\beta) = 0$.
Using the same argument, we can show that all remaining 7 numbers are roots. Proving the existence of these automorphisms has to resort to Galois theory.
| {
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} |
$\int_\gamma{(x-y)dx + (x+y)dy}, \quad \gamma : x^2 + 2y^2 = 1 , \quad 0 \leq y $ I'm asked to find
$$\int_\gamma{(x-y)dx + (x+y)dy}$$
where
$$\gamma : x^2 + 2y^2 = 1 , \quad 0 \leq y$$
(with positive direction) i.e the upper half of the ellipse $x^2 + 2y^2 = 1$.
My attempt
Let $\sigma = \gamma + \gamma_1$ where $\gamma_1 = (t,0) \quad , \quad 0\leq t\leq 1.$
Since $\sigma$ is both positive and closed, Greens Formula can be used with
$\frac{dQ}{dx} - \frac{dP}{dy} = 2.$
$$\int_\gamma(x-y)dx + (x+y)dy = \int\int_D2dxdy - \int_{-1}^1t\cdot dt$$
$$= 2\cdot\frac{1}{2}\cdot (\frac{1}{\sqrt{2}}\pi) - 2 = \frac{\pi}{\sqrt{2}} - 2$$
where I evaluate the double integral simply by getting half of the area of the ellipsoid $ = \frac{1\cdot\frac{1}{\sqrt{2}}\cdot\pi}{2}.$
However, the answer is supposed to be $\frac{\pi}{\sqrt{2}}.$
What am I doing wrong?
| Here is an alternate approach using straight forward substitution.
Let $x=\cos\theta,\,y=\frac{\sqrt{2}}{2}\sin\theta,\,0\le\theta\le\pi$
\begin{eqnarray}
\int_\gamma{(x-y)dx + (x+y)dy}&=&\int_0^\pi-\left(\cos\theta+\frac{\sqrt{2}}{2}\sin\theta\right)\sin\theta+\left(\cos\theta+\frac{\sqrt{2}}{2}\sin\theta\right)\cdot\frac{\sqrt{2}}{2}\cos\theta\,d\theta\\
&=&\int_0^\pi-\frac{1}{4}\sin2\theta+\frac{\sqrt{2}}{2}\,d\theta\\
&=&\left[\frac{1}{8}\cos2\theta+\frac{\theta}{\sqrt{2}}\right]_0^\pi\\
&=&\frac{\pi}{\sqrt{2}}
\end{eqnarray}
| {
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Prove this trigonometric inequality about the angles of $\triangle ABC$
In $\Delta ABC$ show that
$$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$
since
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\dfrac{A-B}{2}\right)=\frac{\sqrt{3}}{2}\sum\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{B}{2} \sin\frac{C}{2}$$
| We can make also the following thing.
We'll prove that for all triangle the following inequality is true.
$$\sin\alpha+\sin\beta+\sin\gamma\geq\frac{\sqrt{3}}{2}(\cos(\alpha-\beta)+\cos(\alpha-\gamma)+\cos(\beta-\gamma)).$$
Indeed, let $R$ be a radius of circumcircle, $r$ be a radius of inscribed circle and $p$ be a semiperimeter of the triangle.
Hence, $$\sum_{cyc}\sin\alpha=\sum_{cyc}\frac{a}{2R}=\frac{p}{R}.$$
In another hand, easy to show that
$$\sum_{cyc}\cos(\alpha-\beta)=\frac{p^2-2R^2+2Rr+r^2}{2R^2}.$$
Thus, we need to prove that
$$\frac{p}{R}\geq\frac{\sqrt3(p^2-2R^2+2Rr+r^2)}{4R^2}$$ or
$$\sqrt3p^2-4Rp-2\sqrt3R^2+2\sqrt3Rr+\sqrt3r^2\leq0$$ and since
$$-2\sqrt3R^2+2\sqrt3Rr+\sqrt3r^2=-2\sqrt3R(R-r)+\sqrt3r^2\leq-2\sqrt3\cdot2r(2r-r)+\sqrt3r^2<0,$$
we need to prove that
$$p\leq\frac{2R+\sqrt{10R^2-6Rr-3r^2}}{\sqrt3}.$$
Now, by Gerretsen inequality $p\leq\sqrt{4R^2+4Rr+3r^2}$.
Id est, it's enough to prove that
$$\sqrt{4R^2+4Rr+3r^2}\leq\frac{2R+\sqrt{10R^2-6Rr-3r^2}}{\sqrt3}.$$
Let $R=2rx$.
Hence, $x\geq1$ and we need to prove that:
$$\sqrt{16x^2+8x+3}\leq\frac{4x+\sqrt{40x^2-12x-3}}{\sqrt3}$$ or
$$2x^2-9x-3+2x\sqrt{40x^2-12x-3}\geq0$$ or
$$2x^2+x-3+2x\left(\sqrt{40x^2-12x-3}-5\right)\geq0$$ or
$$(x-1)\left(2x+3+\frac{8x(10x+7)}{\sqrt{40x^2-12x-3}+5}\right)\geq0,$$
which is obvious.
Now, since the inequality
$$\sin\alpha+\sin\beta+\sin\gamma\geq\frac{\sqrt{3}}{2}(\cos(\alpha-\beta)+\cos(\alpha-\gamma)+\cos(\beta-\gamma))$$
is true for all triangle, this inequality is true also for any acute-angled triangle
and after replacing in the last inequality
$\alpha$ on $90^{\circ}-\frac{\alpha}{2}$, $\beta$ on $90^{\circ}-\frac{\beta}{2}$ and $\gamma$ on $90^{\circ}-\frac{\gamma}{2}$ we'll get the starting inequality.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Quartic with 3 distinct roots Consider a quartic equation
$$x^4 – kx^3 + 11x^2 – kx + 1 = 0$$
The value of $k$ so that given equation has three real and distinct solutions can be equal to - i think that for any four degree to have 3 roots, its $f'(x)$ should have two zero and by the same way its $f''(x)$ should have one zero which means that the determinant of $f''(x)$ should be zero but this is wrong since the correct answer to this question is $\frac{13}{2}$.
| You already know a polynomial having one distinct root whose derivative has one root: $f(x) = x^2$. It is not automatic that the number of distinct roots decreases when you take the derivative of a polynomial -- the total number of roots (including multiplicities) decreases by one. So if it is one of the multiple roots that is removed, the number of distinct roots is unchanged.
A different way to go after $k$, which shows that $k = -13/2$ also works (and explains why $k = \pm 6$ does not).
A fourth degree polynomial has four roots. Non-real roots come in conjugate pairs, so if three roots are real, all four roots are real. If there are only three distinct real roots, one root is duplicated. Therefore, your polynomial factors as \begin{align*}
p(x) = (x-a)^2(x-b)(x-c) \text{.}
\end{align*} (We may take all coefficients of the "$x$"s in the factors to be $1$ because the given polynomial has leading coefficient $1$. This isn't essential, but it is a little convenient. If the given polynomial's leading coefficient were not $1$, we would merely multiply $p$ by this coefficient.) Multiplying out, we get \begin{align*}
p(x) = x^4 &+ (-2a-b-c) x^3 + (a^2 + 2ab + 2ac + bc) x^2 \\
&+(-a^2 b - a^2 c - 2 a b c)x + a^2 b c \text{.}
\end{align*} Comparing coefficients, we get \begin{align*}
1 &= 1 & [x^4] \\
-k &= -2a - b - c & [x^3] \\
11 &= a^2 + 2ab + 2ac + bc & [x^2] \\
-k &= -a^2 b - a^2 c - 2 a b c & [x^1] \\
1 &= a^2 b c & [x^0] \\
\end{align*}
We can solve this system. (Useful observations, $c[x^1]$ can eliminate its first RHS term with $[x^0]$, $b[x^1]$ eliminates the second, ans $a[x^1]$ eliminates the third. $a[x^2]$ eliminates $a^2b + a^2c$ with $[x^1]$ and with a numerical coefficient, eliminates $abc$ with it. And so on...)
We eventually find solutions with $k = \pm 6$ or $k = \pm 13/2$, but when $k = \pm 6$, $b=c$, so there are only two distinct real roots and we must discard those potential solutions. Therefore, $k = \pm 13/2$. Then $a = \pm 1$, with the same sign as $k$ and $b,c = \pm \frac{1}{4}(9 \pm \sqrt{65})$, where the leading $\pm$ matches the sign of $k$ and the inner $\pm$ distinguishes $b$ from $c$. This gives all solutions and both satisfying values of $k$.
| {
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Which area is larger, the blue area, or the white area? In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
My Solution
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$
White area = ${\pi}r^2 - $ area of the intersection of the two circles.
Let the area of the intersection of the two circles be $t$.
White area = ${\pi}r^2 - t$.
The segments that make up $t$ are identical.
$t$ = area of segment ${\times}2$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $90^{\circ}$ (The circles both have radius $r$, and the outer shape is a square.
Angle of sector $ = \frac{1}{4} * {\pi}r^2$.
Area of triangle $ = \frac{1}{2} * r^2$.
Area of segment $ = \frac{{\pi}r^2 - 2r^2}{4}$.
$t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$.
$t = \frac{{\pi}r^2 - 2r^2}{2}$.
White area $ = {\pi}r^2 - \frac{{\pi}r^2 - 2r^2}{2}$.
White area $ = \frac{2{\pi}r^2 - {\pi}r^2 + 2r^2}{2}$.
White area $ = \frac{{\pi}r^2 + 2r^2}{2}$.
Blue area = $r^2\left(4 - \frac{{\pi} + 2}{2}\right)$.
Blue area = $r^2\left(\frac{8 - ({\pi} + 2)}{2}\right)$.
Blue area = $r^2\left(\frac{6 - {\pi}}{2}\right)$.
If White area $-$ Blue area $ \gt 0$, then the White area is larger.
$$r^2\left(\frac{{\pi}+2 - (6 - {\pi}}{2}\right)$$
$$r^2\left(\frac{2{\pi} - 4}{2}\right)$$
$$r^2(\pi - 2)$$
$\therefore$ the white area is larger.
My answer was wrong.
What is the error in my solution?
The provided solution:
| The area of the segment is “quarter of circle minus triangle”:
$$
\frac{1}{4}\pi r^2-\frac{1}{2}r^2=\frac{r^2}{4}(\pi-2)
$$
Thus half of the white area is “quarter of circle plus triangle minus segment”:
$$
\frac{1}{4}\pi r^2+\frac{1}{2}r^2-\frac{r^2}{4}(\pi-2)=r^2
$$
Therefore the white area is $2r^2$.
| {
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Integrate $\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t$ I am trying to solve the following integral by hand, where $c$ is a constant:
$$\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t$$
I have found a solution using Mathematica, but I was interested in how one might solve it manually.
At first I tried the substitution $t=v^2$, which almost yields the Error function, but I wasn't sure how to proceed further.
According to Mathematica, the correct solution is
$$2e^{\frac{-c^2}{4t}}\sqrt{t}+c\sqrt{\pi} \,\mathrm{erf}\!\left(\frac{c}{2\sqrt{t}} \right) $$ (plus an arbitrary constant).
| Using the substitution suggested by Robert Israel quickly yields a solution:
Let $t=\cfrac{c^2}{4v^2}$.
Thus $\mathrm{d}t = -\cfrac{c^2}{2v^3}$ and $v=\cfrac{c}{2\sqrt{t}}$.
So,
\begin{align}
\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t &= \int \left(\frac{c^2}{4}v^{-2}\right)^{-\frac{1}{2}} e^{-\frac{c^2}{4}\left(\frac{c^2}{4}v^{-2}\right)^{-1}} \cdot \left(-\frac{c^2}{2}v^{-3}\right)\, \mathrm{d}v \\
&= -\int \left(\frac{2}{c}v\right) e^{-v^2} \cdot \left(\frac{c^2}{2}v^{-3}\right)\, \mathrm{d}v \\
&= -c \int v^{-2} e^{-v^2} \, \mathrm{d}v \\
\textrm{Now, integrate by parts:} \\
&= -c\left(-v^{-1}e^{-v^2} -\int -v^{-1} \left(-2ve^{-v^2} \right) \, \mathrm{d}v \right) + k \\
&= -c \left(-v^{-1}e^{-v^2} -2 \int e^{-v^2} \mathrm{d}v\right) + k\\
\textrm{Substitute } v=\cfrac{c}{2\sqrt{t}} \mathrm{:}\\
&= 2 \sqrt{t} e^{-\frac{c^2}{4t}} + c \sqrt{\pi} \; \mathrm{erf}\!\left(\frac{c}{2\sqrt{t}} \right) + k
\end{align}
Thus we find the desired solution.
| {
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Limit $\lim_{x \to 2^{-}} \left ( \frac{1}{\sqrt[3]{x^{2} -3x+2}} + \frac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$ $$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$$
I've tried using the $A^3-B^3$ identity, but that doesn't help. Also, I tried multiplying every fraction with $\sqrt[3]{A^2}$ to get rid of the roots in the denominator, but that doesn't help either. Can someone suggest a solution? Thanks.
| Note that
\begin{align}
x^2-3x+2&=(2-x)(1-x)=(2-x)(2-x-1)
\\[4px]
x^2-5x+6&=(2-x)(3-x)=(2-x)(2-x+1)
\end{align}
Set $2-x=t^3$, so your expression becomes
$$
\frac{1}{\sqrt[3]{t^3(t^3-1)}}+\frac{1}{\sqrt[3]{t^3(t^3+1)}}
=\frac{1}{t}\frac{\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}}{\sqrt[3]{t^6-1}}
$$
and the limit is
$$
\lim_{t\to0^+}\frac{1}{\sqrt[3]{t^6-1}}\frac{\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}}{t}
$$
The first fraction has limit $-1$, whereas the second fraction has limit $f'(0)$, where $f(t)=\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}$. Since
$$
f'(t)=3t^2(t^3+1)^{-2/3}+3t^2(t^3-1)^{-2/3}
$$
we have $f'(0)=0$.
Alternatively,
\begin{align}
\frac{1}{\sqrt[3]{t^3(t^3-1)}}+\frac{1}{\sqrt[3]{t^3(t^3+1)}}
&=\frac{(1+t^3)^{-1/3}-(1-t^3)^{-1/3}}{t}\\[6px]
&=\frac{1-\frac{1}{3}t^3-1-\frac{1}{3}t^3+o(t^3)}{t}\\[6px]
&=-\frac{2}{3}t^2+o(t^2)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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What is the probability that Cathy wins?
Alice, Bob and Cathy take turns (in that order) in rolling a six sided
die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5
he wins, and Cathy wins if she rolls a 6. They continue playing until
a player wins. What is the probability (as a fraction) that Cathy
wins?
This was I question I got when competing at a Mathematics competition earlier this week. I thought the answer was $\frac{1}{18}$ but it turns out the actual answer is $\frac{1}{13}$...
This is my working out of why I thought it was $\frac{1}{18}$:
For Cathy to win:
*
*Alice needs to get a 4, 5 or 6 and there is a $\frac{3}{6}$ chance of that happening when she rolls the die
*Bob needs to get a 1, 2, 3 or 6 and there is a $\frac{4}{6}$ probability of that happening when he rolls the die
*Cathy needs to get 6 and there is a $\frac{1}{6}$ probability of that happening when she rolls the die
Those three need to happen for Cathy to win so:
$$\frac{3}{6} \times \frac{4}{6} \times \frac{1}{6} = \frac{1}{18}$$
However, the answer was $\frac{1}{13}$ so something must be wrong! Could someone please demonstrate why it was $\frac{1}{13}$?
| C wins when:
1) A loses, B loses, C wins
or
2) A loses, B loses, C loses, A loses, B loses, C wins
or
3) A loses, B loses, C loses, A loses, B loses, C loses, A loses, B loses, C wins
or
..........
There are infinitely many possibilities, what leads to infinite series
$$
\sum_{i\ge 0}\left(\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}\right)^i\cdot\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{1}{6}
$$
where $\frac{3}{6}=P(A\ loses)$, $\frac{4}{6}=P(B\ loses)$, $\frac{5}{6}=P(C\ loses)$, $\frac{1}{6}=P(C\ wins)$.
The final answer is $\frac{18}{13}\cdot\frac{1}{18}=\frac{1}{13}$.
| {
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"question_score": "1",
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Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, can you guys help me out because I can't understand which formula to use, Do we have to use the Discriminant formula ($b^2-4ac$)?
| Hint: Multiply the numerator and denominator of
$$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}}$$
by the conjugate of the denominator to obtain
$$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}} \cdot \frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}$$
then simplify.
| {
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What did I do wrong in this proof? I have this answer marked partially correct. I'd like to get an idea of where I went wrong
Prove that for all integers $a,b \in \mathbb{Z}$, $(a+b)^5 \equiv a^5+b^5 \mod 5$.
My answer:
Suppose that $a,b \in \mathbb{Z}$. Then, by definition of congruence modulo, $5|(a+b)^5-(a^5+b^5)$ meaning that there exists an integer $k$ such that $$(a+b)^5-(a^5+b^5)=5k$$ Then, $$a^5+5a^4b+10a^3b^2+10a^2b^3+5b^4a+b^5-a^5-b^5 = 5k$$ $$= 5(a^4b+2a^3b^2+2a^2b^3+b^4)=5k$$
This shows that the deduced result is divisible by 5, hence, for all integer $a,b,\in\mathbb{Z}$, $(a+b)^5 \equiv a^5+b^5 \mod 5$.
| Your idea is correct, but your proof looks backwards. I suggest the following modifications. Observe how now the proof looks like a tale of deduction rather than one of back-engineering, if you may.
Suppose that $a,b \in \mathbb{Z}$. Then, by definition of congruence
modulo, we must show that $5|(a+b)^5-(a^5+b^5)$. This means
that there exists an integer $k$ such that $$(a+b)^5-(a^5+b^5)=5k$$
Let us show this holds. To see this, note that, $$(a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5b^4a+b^5 = a^5+b^5+5k$$
where $a^4b+2a^3b^2+2a^2b^3+b^4=k$ is obtained by collecting terms. This shows that the difference $(a+b)^5 -a^5-b^5$ is divisible by 5, hence,
for all integer $a,b,\in\mathbb{Z}$, $(a+b)^5 \equiv a^5+b^5 \mod 5$.
| {
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using Parseval's identity to estimate the value of $\Sigma_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}$ There is a problem with two parts; The first part is asking to find Fourier series for $f(x)=|\sin(x)|$ on $[-\pi,\pi]$. And the second part wants to estimate the following using Parseval's identity:
$$
\sum_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}
$$
I'm not sure it is necessary to use the answer of the first part or not. I don't know what to do!
The first part answer:
$$\forall n, b_n=0$$.
$$
a_0=\frac{1}{\pi}\int_{-\pi}^\pi |\sin(x)|dx=\frac{2}{\pi}\int_0^\pi \sin(x)dx=\frac{-2}{\pi}\cos(x)]_0^\pi=\frac{-2}{\pi}(-1-1)=\frac{4}{\pi}
$$
\begin{align*}
a_n=\frac{1}{\pi}\int_{-\pi}^\pi |\sin(x)|\cos(nx) dx
= \frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)dx = \frac{1}{\pi}\int_0^\pi [\sin(x+nx)+\sin(x-nx)]\\ =\frac{1}{\pi}[\frac{-\cos((n+1)x)}{2(n+1)}+\frac{-\cos((1-n)x)}{2(1-n)}]_0^\pi \\
a_n=\frac{-1}{\pi}(\frac{\cos(n\pi)+1}{n^2-1})
\end{align*}
| The Fourier series of the function $f(x)$ is given by
$$
f(x) = \sum_{n=-\infty}^\infty c_n e^{i n x}
$$
with
$$
c_n = \frac{1}{2 \pi} \int_{-\pi}^\pi d x~f(x) e^{-\imath n x}
$$
In the case $f(x) = |\sin x |$, this results in
$$
c_n = 2 \frac{1}{2 \pi} \int_{0}^\pi d x~\sin x e^{-\imath n x} = \left\{ \begin{array}{ll} -\frac{2}{\pi (n^2-1)} & \text{for $n$ even}\\ 0 & \text{for $n$ odd} \end{array} \right.
$$
Using Parsifal's identity we have
$$
\sum_{n=-\infty}^\infty |c_n|^2 = \frac{1}{2 \pi} \int_{-\pi}^\pi d x ~ |\sin x|^2 = \frac{1}{2}
$$
Since $c_{n} = c_{-n}$ the sum on the left hand side can be rewritten as
$$
c_0^2 + 2 \sum_{n=1}^{\infty} c_n^2 = \frac{1}{2}
$$
From which it follows that with $c_0=\frac{2}{\pi}$
$$
\sum_{n=1}^{\infty} c_n^2 = \frac{1}{2}\left(\frac{1}{2} - c_0^2\right) = \frac{1}{4} - \frac{2}{\pi^2}
$$
Using the result we found for $c_n$ we get
$$
\sum_{n=1}^{\infty} c_n^2 = \sum_{n=1}^{\infty} c_{2 n - 1}^2 + \sum_{n=1}^{\infty} c_{2 n}^2 = \sum_{n=1}^{\infty} c_{2 n}^2 = \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{(4 n^2 -1)^2}
$$
and therefore we finally obtain
$$
\sum_{n=1}^{\infty} \frac{1}{(4 n^2 -1)^2} = \frac{\pi^2}{4} \left( \frac{1}{4} - \frac{2}{\pi^2} \right) = \frac{\pi^2-8}{16}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x}+x}{x}+
\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right]
+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}
\end{align}
I could not evaluate the second limit
| Or we could've approached it differently from the beginning, if we observe that
$$2\ln(\cos x)=\ln(\cos^2 x)=\ln(1-\sin^2 x)$$
and use the series for
$$\ln(1-t)=-\sum_{n=1}^{\infty}\frac{t^n}{n}=-t-\frac{t^2}{2}-\frac{t^3}{3}-\cdots.$$
Then for the original limit:
$$\lim_{x\to0}\frac{\sin^2 x+2\ln(\cos x)}{x^4}=
\lim_{x\to0}\frac{\sin^2 x+\ln(1-\sin^2 x)}{x^4}=
\lim_{x\to0}\frac{\sin^2 x-\sin^2 x-\frac{\sin^4 x}{2}-O(\sin^6 x)}{x^4}=
\lim_{x\to0}\frac{-\frac{\sin^4 x}{2}-O(x^6)}{x^4}=
-\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary. Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.
My Attempt:
$$x^2-4abx+(a^2+2b^2)^2=0$$
Comparing above equation with $Ax^2+Bx+C=0$, we get
\begin{align}
A&=1 \\
B&=-4ab \\
C&=(a^2+2b^2)^2
\end{align}
Now,
\begin{align}
B^2-4AC
&=(-4ab)^2 - 4\cdot 1\cdot (a^2+2b^2)^2 \\
&=16a^2b^2 - 4(a^4+4a^2b^2+4b^4) \\
&=-4a^4-16b^4. \end{align}
| Suppose $ab\ne 0$.
Let $f (x) $ be your quadric.
$$\lim_{\infty}f (x)=+\infty $$
thus the roots will not be real if the minimum is $$f (c)>0$$ where $c $ is such that $$f'(c)=0.$$
$$f'(x)=2x-4ab \implies c=2ab $$
$$\implies f (c)=$$
$$4a^2b^2-8a^2b^2+(a^2+2b^2)^2=$$
$$(a^2+2b^2+2ab)(a^2+2b^2-2ab) =$$
$$((a+b)^2+b^2)((a-b)^2+b^2)>0$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $f(x)\cdot g(x)$ given that $\lim_{x\to 0}{\frac{f(x)}{\sin(2x)}}=2$ and $\lim_{x\to 0}{(\sqrt{x+4}-2)\cdot{g(x)}}=5$ Question:
If $$\lim_{x\to 0}{\left[\frac{f(x)}{\sin(2x)}\right]}=2$$ and $$\lim_{x\to 0}{\left[(\sqrt{x+4}-2)\cdot{g(x)}\right]}=5$$ then find $$\lim_{x\to 0}{[f(x)\cdot{g(x)}]}$$
But, from what I found, $\lim_{x\to 0}{[g(x)]}$ does not exist. So how can I find $\lim_{x\to 0}{[f(x)\cdot{g(x)}]}$?
| Only one theorem applied.
Theorem. If $\lim\limits_{x\rightarrow c}f(x)=L$ and $\lim\limits_{x\rightarrow c}g(x)=M$, then $\lim\limits_{x\rightarrow c}[f(x)g(x)]=LM$
$$
\because \qquad \lim_{x\rightarrow0}\frac{f(x)}{\sin(2x)}=2\quad \text{and}\quad \lim_{x\rightarrow0}\left[(\sqrt{x+4}-2）\cdot g(x)\right]=5
$$
$$
\therefore\qquad \lim_{x\rightarrow0}\left\{\frac{f(x)}{\sin(2x)}\cdot \left[(\sqrt{x+4}-2）\cdot g(x)\right]\right\}=2\times5=10
$$
Rearranging:
\begin{align}
\frac{f(x)}{\sin(2x)}\cdot \left[(\sqrt{x+4}-2）\cdot g(x)\right]
&=\left[f(x)g(x)\right]\cdot\frac{\sqrt{x+4}-2}{\sin(2x)}\\
&=\left[f(x)g(x)\right]\cdot\frac{2x}{\sin(2x)}\cdot \frac{1}{2(\sqrt{x+4}+2)}
\end{align}
$$
\Rightarrow\qquad
\lim_{x\rightarrow0}\left\{\left[f(x)g(x)\right]\cdot\frac{2x}{\sin(2x)}\cdot \frac{1}{2(\sqrt{x+4}+2)}\right\}
=10
$$
$$
\because\qquad
\lim_{x\rightarrow0}\frac{\sin(2x)}{2x}=1\quad
\text{and}
\quad \lim_{x\rightarrow0}\frac{2(\sqrt{x+4}+2)}{1}=8
$$
$$
\therefore\qquad
\lim_{x\rightarrow0}
\left\{
\left[\left[f(x)g(x)\right]\cdot\frac{\sin(2x)}{2x}\cdot \frac{1}{2(\sqrt{x+4}+2)}\right]
\cdot
\left[
\frac{2x}{\sin(2x)}
\right]
\cdot
\left[
\frac{2(\sqrt{x+4}+2)}{1}
\right]
\right\}\\=10\cdot 1\cdot8=80
$$
$$
\Rightarrow\qquad
\lim_{x\rightarrow0}\left[f(x)g(x)\right]=80
$$
Tips. Treat the block of functions as one function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
On a determinant identity
Show that
$$\det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} = (-1)^n(1+x-n)(1+x)^{n-1}$$
for all $n \in \mathbb{N}$.
I tried using induction but I didn't manage (using row/column swaps). Does anyone have any idea?
| HINT
$$\begin{align}\begin{vmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{vmatrix}&=\begin{vmatrix} -x+n-1& 1 & \dots & 1 \\ -x+n-1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\-x+n-1 & 1 & \dots & -x \end{vmatrix}\\&=(-x+n-1)\begin{vmatrix} 1& 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\1 & 1 & \dots & -x \end{vmatrix}\\&=(-x+n-1)\begin{vmatrix} 1& 1 & \dots & 1 \\ 0 & -x-1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \dots & -x-1 \end{vmatrix}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.