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Compute $\int^{\pi/2}_0 \frac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$ I have tried solving this for about an hour and will probably resort to head banging in some time:
$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$
I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:
$$\int ^{\infty}_{0} \dfrac{1+t^2}{(a^2 + b^2 t^2)^2}dt$$
This has become unmanageable. Neither splitting the numerator, nor Partial fraction (taking t^2 = z and applying partial fraction) seems to work.
|
$$
\begin{align}
\int_0^\infty\frac{1+t^2}{(a^2+b^2t^2)^2}\,\mathrm{d}t
&=\frac1{a^3b}\int_0^\infty\frac{1+\frac{a^2}{b^2}t^2}{(1+t^2)^2}\,\mathrm{d}t\tag{1}\\
&=\frac1{2a^3b}\int_0^\infty\frac{t^{-1/2}+\frac{a^2}{b^2}t^{1/2}}{(1+t)^2}\,\mathrm{d}t\tag{2}\\
&=\frac1{2a^3b}\frac{\Gamma\left(\frac12\right)\Gamma\left(\frac32\right)}{\Gamma(2)}+\frac1{2ab^3}\frac{\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma(2)}\tag{3}\\[3pt]
&=\frac\pi4\frac{a^2+b^2}{a^3b^3}\tag{4}
\end{align}
$$
Explanation:
$(1)$: substitute $t\mapsto\frac abt$
$(2)$: substitute $t\mapsto t^{1/2}$
$(3)$: use the Beta Function integral
$(4)$: simplify
|
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"url": "https://math.stackexchange.com/questions/1939026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
When does the function $a\cdot \sin(x)+b\cdot \cos(x)-x$ have exactly one real root with multiplicity $1$?
*
*When does the function $$f(x)=a\cdot \sin(x)+b\cdot \cos(x)-x$$ have exactly one real root ($a$ and $b$ are real numbers) ?
*When does $f(x)$ have a multiple root , in other words for which real numbers $a$ and $b$ does the system $$a\cdot \sin(x)+b\cdot \cos(x)=x$$ $$a\cdot \cos(x)-b\cdot\sin(x)=1$$ have a real solution $x$ ?
A root with multiplicity $3$ occurs, if we have $a=1$ and $b=0$. The function $f(x)=\sin(x)-x$ and the first two derivates are $0$ for $x=0$.
If we choose $a=4$ and $b=-6$, we have $5$ real simple roots. The choice $a=1$ and $b=1$ leads to a unique simple root.
$f(x)$ has always a real root because of $$\lim_{x\rightarrow \pm \infty} f(x)=\mp \infty$$
Any ideas ?
|
Okay, so we have:
$$a\cdot\sin(x)+b\cdot\cos(x)=x$$
$$a\cdot\cos(x)-b\cdot\sin(x)=1$$
multiply bottom one by $\dfrac ab$
$$\dfrac{a^2}{b}\cdot\cos(x)-b\cdot\sin(x)=\dfrac ab$$
Add it to the top one:
$$(\dfrac{a^2+b^2}{b})\cos(x)=x+\dfrac{a}{b}$$
So:
$$\cos(x)=\dfrac{bx+a}{a^2+b^2}$$
Now multiply the bottom one by $-\dfrac{b}{a}$:
$$-b\cdot\cos(x)+\dfrac{b^2}{a}\cdot\sin(x)$$
Add it to the top one:
$$(\dfrac{a^2+b^2}{a})\cdot\sin(x)=x-\dfrac{b}{a}$$
So:
$$\sin(x)=\dfrac{ax-b}{a^2+b^2}$$
We now take:
$$\tan(x)=\dfrac{\sin(x)}{\cos(x)}=\dfrac{ax-b}{bx+a}$$
Now, we use $\tan(\dfrac{\pi}{4})=1$ to obtain:
$$\dfrac{a\pi+4b}{b\pi-4a}=1$$
Use that equation to get $b$ in terms of $a$:
$$a\pi+4b=b\pi-4a$$
$$a\pi+4a=b\pi-4b$$
$$b=a\dfrac{\pi+4}{\pi-4}$$
Now we have:
$$a(\sin(x)+\dfrac{\pi+4}{\pi-4}\cdot\cos(x))=x$$
$$a(\cos(x)-\dfrac{\pi+4}{\pi-4}\cdot\sin(x))=1$$
Divide the top one by the bottom one:
$$\dfrac{\sin(x)+\dfrac{\pi+4}{\pi-4}\cdot\cos(x)}{\cos(x)-\dfrac{\pi+4}{\pi-4}\sin(x)}=\dfrac{(\pi-4)\sin(x)+(\pi+4)\cos(x)}{(\pi-4)\cos(x)-(\pi+4)\sin(x)}=x$$
For every $x$ that is s solution to this equation, you can find $a$ and $b$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1939250",
"timestamp": "2023-03-29T00:00:00",
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|
Elliptical Integral: Finding $I= \int \sqrt{1-k^2\sin^2(x)} \cos^2(x) dx $ Just a sidenote: The part where I am stuck on is at the very end , this is a long question so bear with me until I get to the last part!
Question 1: Two famous integrals are $F= \int \frac{dx}{\sqrt{1-k^2\sin^2(x)}}$ and $E= \int \sqrt{1-k^2\sin^2(x)} dx$ , $k<1$ They are called elliptic integrals of the first
and second kind respectively. (There is a third type we will not consider.) If an integral can be reduced to a function together with a combination of integrals of this type it is called an
elliptic integral. It is know that elliptic integrals cannot be integrated exactly.
Defining $\zeta$ by $\zeta (x)= \sqrt{1-k^2\sin^2(x)}$, we have $F = \int \frac{dx}{\zeta}$ , $E = \int \zeta dx $
The problem is to show that $I= \int \sqrt{1-k^2\sin^2(x)} \cos^2(x) dx $ is an elliptic function by showing it can be expressed in terms of elliptic integrals of the first and second kind.
(i) Find $\frac{d \zeta}{dx}$ expressing your answer in terms of $\sin(x)$. $\cos(x)$ in the numerator and $\zeta$ in the denominator.
Let $\zeta(x) = \sqrt{1-k^2\sin^2(x)} = (1-k^2\sin^2(x))^{\frac{1}{2}}$
Then $\zeta'(x) = \frac{1}{2} (1-k^2\sin^2(x))^{\frac{-1}{2}} \cdot -2k^2 \sin(x)\cos(x) \implies \frac{-k^2\sin(x)\cos(x)}{\zeta}$
(ii) Differentiate $\sin(x)\cos(x) \zeta$, bringing your answer to a common denominator of $\zeta$.
Let $p(x) = \sin(x)\cos(x) \sqrt{1-k^2\sin^2(x)} = 0.5\sin(2x)(1-k^2\sin^2(x))^{\frac{1}{2}}$
Then $p'(x) = \cos(2x)(1-k^2\sin^2(x))^{\frac{1}{2}} + 0.5\sin(2x)(\frac{1}{2} (1-k^2\sin^2(x))^{\frac{-1}{2}} \cdot -2k^2 \sin(x)\cos(x))$
$ \Leftrightarrow (\cos^2(x)-\sin^2(x))\zeta + \frac{(\sin(x)\cos(x))-k^2\sin(x)\cos(x)}{\zeta} $
$ \Leftrightarrow \frac{(\cos^2(x)-\sin^2(x))\zeta^2-k^2\sin^2(x)\cos^2(x)}{\zeta} $
(iii) By setting $v=\sin^2(x)$ and using this to replace all occurrences of $\zeta^2$ and $\cos^2(x)$ as well show that
$$ \frac{d(\sin(x)\cos(x)\zeta)}{dx} = \frac{1}{\zeta} \left[1-2v(k^2+1)v+3k^2v^2 \right]$$
Hence $\sin(x)\cos(x)\zeta = \int \frac{dx}{\zeta} - 2(k^2+1)\int \frac{vdx}{\zeta} + 3k^2 \int \frac{v^2dx}{\zeta}$
From part(ii) we have that $\frac{d(\sin(x)\cos(x)\zeta)}{dx} = \frac{(\cos^2(x)-\sin^2(x))\zeta^2-k^2\sin^2(x)\cos^2(x)}{\zeta} $
Since $v=\sin^2(x)$ it implies that $1-v=\cos^2(x)$ so
$$\frac{d(\sin(x)\cos(x)\zeta)}{dx} = \frac{(1-2v)(1-k^2v)-k^2v(1-v)}{\zeta} $$
$$ \Leftrightarrow \frac{d(\sin(x)\cos(x)\zeta)}{dx} = \frac{1}{\zeta} (1+3k^2v^2-2k^2v-2v) $$
$$ \therefore \frac{d(\sin(x)\cos(x)\zeta)}{dx} = \frac{1}{\zeta} \left[1-2v(k^2+1)+3k^2v^2 \right] $$
and if we integrate bothsides we get
$$ \sin(x)\cos(x)\zeta = \int \frac{dx}{\zeta} - 2(k^2+1)\int \frac{vdx}{\zeta} + 3k^2 \int \frac{v^2dx}{\zeta}$$
Now this part of the question is where I get stuck from
(iv) Show that $$I= \frac{1}{3} \zeta \sin(x)\cos(x) + \frac{1+k^2}{3k^2}E - \frac{1-k^2}{3k^2}F$$
How would I do this?
|
$$A:=k^2\int\frac{v^2}{\zeta}\;dx$$
$$
\zeta^2=1-k^2v\implies k^4v^2=(1-\zeta^2)^2\\
\begin{align}
\therefore k^2A&=\int\frac{1}{\zeta}\;dx-2\int{\zeta}\;dx+\int\zeta^3\;dx\\
&=F-2E+\int\zeta(1-k^2v)dx\\
&=F-E-k^2\int\zeta v\;dx\\
&=F-E-k^2\int\zeta(1-\cos^2x)\;dx\\
&=F-(1+k^2)E+k^2\int\zeta\cos^2x\;dx\\
&=F-(1+k^2)E+k^2I
\end{align}
$$
$$
B:=\int\frac{v}{\zeta}\;dx
$$
$$\begin{align}
k^2B=\int\frac{1-\zeta^2}{\zeta}\;dx=F-E
\end{align}$$
From $(iii)$ we have
$$\begin{align}
\sin(x)\cos(x)\zeta&=F-2(1+k^2)B+3A\\
&=F-2\frac{1+k^2}{k^2}(F-E)+\frac{3}{k^2}[F-(1+k^2)E+k^2I]
\end{align}$$
which after a little algebraic manipulation yields
$$
I={1\over3}\sin(x)\cos(x)\zeta+\frac{1+k^2}{3k^2}E-\frac{1-k^2}{3k^2}F
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form solution for infinite summation Thomas, Bruckner & Bruckner, Elementary Real Analysis.
Prove that for all r > 1,
$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1} + \cdots$$
So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty \frac{2^n}{r^{2^n} + 1} = \sum_{n=1}^\infty \left(\frac{2^n}{r^{2^n}} - \frac{2^n}{r^{4^n} + r^{2^n}}\right)$$
|
Rinse and repeat the first line below
$$\begin{align}
{1\over \color{blue}{r}-1}&={1\over \color{blue}{r}+1}+{2\over \color{blue}{r}^2-1}\\
&={1\over r+1}+2\left({1\over\color{blue}{r^2}+1}+{2\over (\color{blue}{r^2})^2-1}\right)\\
&={1\over r+1}+\frac{2}{r^2+1}+\frac{4}{\color{blue}{r^4}-1}\\
&={1\over r+1}+\frac{2}{r^2+1}+4\left(\frac{1}{\color{blue}{r^4}+1}+\frac{2}{(\color{blue}{r^4})^2-1}\right)\\
&=\cdots
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1942285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Given an $n \times n $ grid, how many squares exist? How many squares exist in an $n \times n$ grid? There are obviously $n^2$ small squares, and $4$ squares of size $(n-1) \times (n-1)$.
How can I go about counting the number of squares of each size?
|
Let the vertices of the $n \times n$ grid by $\{(x,y)| 0\le x \le n; 0 \le y \le n\}$.
(Is that what an $n \times n$ grid is? A $1 \times 1$ has $4$ vertices and an $n \times n$ grid has $(n+1)^2$ vertices? Or is a $1 \times 1$ grid a single point? I'm assuming the former.)
A $k \times k$ square will have an lower left hand vertex as $(x,y)$ and a upper right hand vertex at $(x+k, y+k)$ with the stipulation $0 \le x; 0 \le y; x+k \le n; y+k \le n$ or in other words: $0 \le x \le n-k; 0 \le y \le n-k$.
There are $n-k+1$ possible options for $x$ and $n-k+1$ options for $y$ so there $(n-k+1)^2$ $k\times k$ squares.
So the total number of squares is $\sum{k=1}^n(n-k+1)^2$. Let $j = n-k+1$ and we have #squares = $\sum_{j = n;j--}^1j^2 = \sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}6$
Or to put it more simply.... A $k \times k$ square has a side of length $k$. There are $n-(k-1)$ ways to choose this side from the grid that is $n$ long so for squares of lengths $1,2, ...., n$ there are $n, n-1....., 1$ way to choose the horizontal side and $n, n-1,...., 1$ ways to choose the vertical sides. So there are $n^2, (n-1)^2,....., 1^2$ possible $1\times 1, 2\times 2, ...,n \times n$ squares. So there are $\sum k^2 = \frac{n(n+1)(2n+1)}6$ total squares.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1942364",
"timestamp": "2023-03-29T00:00:00",
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|
Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$ without using Calculus techniques
Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$.
It is easy if we apply Calculus techniques (e.g., derivatives, Lagrange multipliers, etc). However, the problem is assigned for students who have not yet learned Calculus. Is there a way to find the maximum without using Calculus techniques?
I have a few ideas:
*
*Apply geometric means: $xy \leq \frac{x^2+y^2}{2} = -x + 3/2 $.
*Change object function to $(xy)^2$, then we have $(xy)^2 = x^2(4-(x+1)^2)=x^2(-x^2 - 2x +3)$. That is, maximize $x^2(-x^2 - 2x +3)$ on $(-3,1)$.
|
Your constraint is a circle centered at $(-1,0)$ with radius $2$. And the level curves of $xy$ with positive level are hyperbolae in the first and third quadrant. Since the circle is centered left of the origin, visualizing this it should be clear that the maximum value of $xy$ happens when both $x$ and $y$ are negative.
Furthermore, the curve $xy=M$ (where $M$ is the maximal value subject to the constraint) must intersect the circle in a tangency instead of at crossings, or else a slightly larger $M$ would be obtainable. For these two curves to intersect with tangency, there must be a repeated root to the equation that you get by eliminating one variable.
That is, if you eliminate $y$ to get $$(x+1)^2+\left(\frac{M}{x}\right)^2=4$$ and then expand and clear $x$ from the denominator to get $$x^4+2x^3-3x^2+M^2=0$$ we need to have a polynomial equation with a doubled root. So for some $a,b,c$:
$$\begin{align}
&x^4+2x^3-3x^2+M^2\\
&=(x-a)^2(x-b)(x-c)\\
&=(x^2-2ax+a^2)(x^2-(b+c)x+bc)\\
&=x^4-(2a+b+c)x^3+(a^2+2a(b+c)+bc)x^2-(a^2(b+c)+2abc)x+a^2bc\\
\end{align}$$
which implies the system
$$\begin{cases}
2a+b+c&=-2\\
a^2+2a(b+c)+bc&=-3\\
a^2b+a^2c+2abc&=0
\end{cases}$$
In the last equation we know $a\neq0$ so we can divide by $a$. Then we can eliminate $c$ by solving for it in the first equation and substituting it into the other two. All this gives:
$$\begin{cases}
-4a-3a^2-2b-2ab-b^2&=-3\\
-2a-2a^2-4b-4ab-2b^2&=0
\end{cases}$$
Double the top and subtract from the second to get:
$$6a+4a^2=6$$ from which we can solve for $a$ using the quadratic formula. Remember that $a$ is the doubled root, and so it is the same as the $x$-value where $xy$ is maximal. Also we know this number is negative, which helps to discern between the two quadratic solutions. We find that $x=a=-\frac14\left(3+\sqrt{33}\right)$ at the point where $xy$ is maximal.
Now we can find:
$$\begin{align}
y&=-\sqrt{4-(x+1)^2}\\
&=-\sqrt{4-\left(-\frac14\left(3+\sqrt{33}\right)+1\right)^2}\\
&=-\sqrt{4-\left(\frac14\left(1-\sqrt{33}\right)\right)^2}\\
&=-\sqrt{4-\frac1{16}\left(34-2\sqrt{33}\right)}\\
&=-\frac14\sqrt{30+2\sqrt{33}}
\end{align}$$
And so the maximal value is $$M=xy=\frac{1}{16}\left(3+\sqrt{33}\right)\sqrt{30+2\sqrt{33}}$$
I submit visual evidence that this is the maximal value. A plot of the circle and the curve $xy=M$. A larger $M$, and the two curves would not intersect.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1942447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
|
Easy, download Photmath!
Answer = $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$
$$(x^2-3x+1)^2-(2x-3)^2=0$$ $$x^2-5x+4=0$$ $$x^2-x-2=0$$
$$x=4$$
$$x=1$$
$$x=2$$
$$x=-1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1943179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
}
|
Find the base numeric system Find the numeric base we are using if $ x = 4 $ and $ x = 7 $ are zeros of $ 5x^2 - 50x + 118. $
When I plug in $ x = 4 $ and $ x = 7, $ I receive $ -2 $ and $ 13, $ respectively, but how do I proceed from that?
|
$$5_bx^2-50_bx+118_b=5_b(x-4_b)(x-7_b)$$
From $118_b$, we can assume that $b > 8$.
This implies $5_b = 5,\; 4_b = 4,\; 7_b = 7,\;$ and $\; 118_b = b^2 + b + 8$.
Hence
\begin{align}
b^2 + b + 8 &= 5 \times 4 \times 7 \\
b^2 + b - 132 &= 0 \\
(b - 11)(b + 12) &= 0 \\
b &= 11
\end{align}
CHECK:
\begin{align}
5_{11}(x-4_{11})(x-7_{11})
&= 5_{11}(x^2 - 10_{11}x + 26_{11}) \\
&= 5_{11}x^2 - 50_{11}x +118_{11} \checkmark
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1944566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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|
Show that there are infinitely many pairs (a,b) such that both $x^2+ax+b$ and $x^2+2ax+b$ have integer roots The question is :
Show that there are infinitely many pairs (a,b) of relatively prime integers (not necessarily positive ) such that both quadratic equations $x^2+ax+b$ =0
and $x^2+2ax+b$ =0 have integer roots.
This is the problem .I tried it to do in many way.
At first we let that there are finely many pairs then we shall show it is a contradiction. It is an INMO problem. Somebody please help me.
Thank you
|
Wanting $a^2-b=c^2$ and $a^2-4b=d^2$, we can see that we want $4c^2-d^2=3a^2$. If $a$ is odd, we can solve $2c-d=3$, $2c+d=a^2$ getting:
$$c=\frac{a^2+3}{4}, d=\frac{a^2-3}{2}$$
Then use $3b=c^2-d^2$ to determine $b$ in terms of $a$, and then do a lot of arithmetic to get an explicit infinite family. There is another (simple) condition required on $a$ to make $a,b$ relatively prime.
You get, after some work, that if $a=2k-3$ then $b=-k(k-1)(k-2)(k-3)$, then $$x^2+ax+b=0$$ has two integer solutions, $k(k-2)$ and $-(k-1)(k-3)$, and $$x^2+2ax+b=0$$ has two integer solutions $k^2-k$ and $-(k-2)(k-3)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1945233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Probability of 4 consecutive heads in 10 coin tosses I am trying to compute the probability of having 4 (or more) consecutive heads in 10 coin tosses.
I tried using recursion but it led to a complicated expression so i think i did not quite manage.
I saw similar questions asked here that were solved with difficult approaches, but this problem looks like it could be solved in a couple of lines so I must be doing something wrong.
If anyone could help me understand it or propose a different approach for solving the problem i would be very grateful.
Have a nice day!
|
Here we are looking for the number $f_N$ of binary strings of length $N$ which do not contain the substring $HHHH$. The probability $p$ is then $$p=1-\frac{f_{10}}{2^{10}}$$
The so-called Goulden-Jackson Cluster Method is a convenient technique to derive a generating function for problems of this kind.
We consider words of length $N\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the set $\mathcal{B}=\{HHHH\}$ of bad words which are not allowed to be part of the words we are looking for.
We derive a function $F(x)$ with the coefficient of $x^N$ being the number of wanted words of length $N$.
According to the paper (p.7) the generating function $F(x)$ is
\begin{align*}
F(x)=\frac{1}{1-dx-\text{weight}(\mathcal{C})}
\end{align*}
with $d=|\mathcal{V}|=2$, the size of the alphabet and with the weight-numerator $\mathcal{C}$ with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[HHHH])
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C}[HHHH])&=-x^4-\text{weight}(\mathcal{C}[HHHH])\left(x+x^2+x^3\right)
\end{align*}
It follows:
A generating function $F(x)$ for the number of words built from $\{H,T\}$ which do not contain the subword $HHHH$ is
\begin{align*}
F(x)&=\frac{1}{1-dx-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-2x+\frac{x^4}{1+x+x^2+x^3}}\\
&=\frac{1+x+x^2+x^3}{1-x-x^2-x^3-x^4}
\end{align*}
Since the generating function counting the number $2^N$ of all binary strings of length $N$ is
\begin{align*}
\frac{1}{1-2x}=1+2x+4x^2+\cdots
\end{align*}
A generating function for the number binary strings of length $N$ which contains the string $HHHH$ is
\begin{align*}
\frac{1}{1-2x}-F(x)&=\frac{1}{1-2x}-\frac{1+x+x^2+x^3}{1-x-x^2-x^3-x^4}\\
&=\frac{x^4}{(1-2x)(1-x-x^2-x^3-x^4)}\\
&=x^4+3x^5+8x^6+\color{green}{20}x^7+48x^8+111x^9+\color{blue}{251}x^{10}\\
&\qquad 558x^{11}+1224x^{12}+2656x^{13}+5713x^{14}+12199x^{15}+\cdots\tag{1}
\end{align*}
The last line was calculated with the help of Wolfram Alpha and we see there are $\color{blue}{251}$ strings of length $10$ which contain the subword $HHHH$.
For example the $20$ strings of length $7$ containing the substring $HHHH$ are
\begin{array}{lllll}
\color{green}{HHHH}HHH\quad&\quad \color{green}{HHHH}HHT\quad&\quad \color{green}{HHHH}HTH\quad&\quad \color{green}{HHHH}HTT\\
\color{green}{HHHH}THH\quad&\quad \color{green}{HHHH}THT\quad&\quad \color{green}{HHHH}TTH\quad&\quad \color{green}{HHHH}TTT\\
HHT\color{green}{HHHH}\quad&\quad HT\color{green}{HHHH}H\quad&\quad HT\color{green}{HHHH}T\quad&\quad HTT\color{green}{HHHH}\\
T\color{green}{HHHH}H\quad&\quad T\color{green}{HHHH}HT\quad&\quad T\color{green}{HHHH}TH\quad&\quad T\color{green}{HHHH}TT\\
THT\color{green}{HHHH}\quad&\quad TTH\color{green}{HHHH}\quad&\quad TT\color{green}{HHHH}T\quad&\quad TTT\color{green}{HHHH}\\
\end{array}
We finally conclude from (1): The probability of $4$ heads in $10$ coin tosses is $$\frac{251}{2^{10}}\doteq 0.2451$$
|
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|
$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$ I am trying to formally evaluate the following limit:
$$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$$
Empirically, the limit seems to be converging to $1.5$, although I am not sure how to formally prove this. I had one idea thus far: it appears the constant terms within the square roots do not matter, so I rewrote the limit as
$$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x-3}-\sqrt[]{x^2-x-6}
=\lim \limits_{x \rightarrow \infty} \sqrt[]{(x+3)(x-1)}-\sqrt[]{(x+2)(x-3)}.$$
In each square root, there are two factors. I assumed that in the limit the geometric mean of the two factors (i.e. square root of the product) is equal to their arithmetic mean. This is a step I am not quite certain of, and if it is true I would like to prove it. However, as I found it led to the correct answer, since
$$\lim \limits_{x \rightarrow \infty} \frac{(x+3)+(x-1)}{2}-\frac{(x+2)+(x-3)}{2}=\lim \limits_{x \rightarrow \infty} (x+1)-(x-0.5)=1.5.$$
I am not sure if the result was coincidental, but if not, I would like some help formalizing each of my steps. I would also like to hear of alternate approaches that do not alter the original problem in the way I did.
|
$$\sqrt{x^2+2x+3}-(x+1) = \frac{2}{x+1+\sqrt{x^2+2x+3}}=O\left(\frac{1}{x}\right)\tag{1}$$
$$\sqrt{x^2-x-5}-\left(x-\frac{1}{2}\right) = -\frac{5+\frac{1}{4}}{\left(x-\frac{1}{2}\right)+\sqrt{x^2-x-5}}=O\left(\frac{1}{x}\right)\tag{2} $$
$$\lim_{x\to +\infty}\sqrt{x^2+2x+3}-\sqrt{x^2-x-5}=\lim_{x\to +\infty}(x+1)-\left(x-\frac{1}{2}\right) = \color{red}{\frac{3}{2}}.\tag{3}$$
|
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|
Simplifying a radical with complex fractions So I understand to simplify this:
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}}
$$
I can just multiply
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}} \cdot\frac{2t^4}{2t^4}
$$
and get
$$
\frac{-3}{t\sqrt{\frac{1}{4t^6} - 1}}
$$
But how do you simplify further getting rid of the complex fraction inside the radical?
|
$\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}} \frac{\sqrt {4t^6}}{\sqrt{4t^6}}$ first step will be to get rid of the fraction inside the radical. Note that ${\sqrt{4t^6}} = |2t^3|$
$\frac{\frac{-3}{2t^4}|2t^3|}{|\frac{1}{2t^3}|\sqrt{1 - 4t^6}} \frac{\sqrt {1-4t^6}}{\sqrt{1-4t^6}}$ Then we get the radical outside of the numerator.
$
\frac{\frac{-3}{2t^4}|2t^3|\sqrt {1-4t^6}}{|\frac{1}{2t^3}|(1 - 4t^6)} \frac{|2t^3|}{|2t^3|}$ Now I am taking on what you did in the first step. I think it is best to attack the messiest parts first.
$
\frac{\frac{-3}{2t^4}(4t^6)\sqrt {1-4t^6}}{(1 - 4t^6)}\\
\frac{-6t^2\sqrt {1-4t^6}}{(1 - 4t^6)}$
|
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|
Integrate $\frac{x^3-3xy^2}{(x^2+y^2)^2}$ over $1\leq x^3-3xy^2\leq2,2\leq3x^2y-y^3\leq4$ and $x,y\geq0.$
Integrate $\frac{x^3-3xy^2}{(x^2+y^2)^2}$ over $1\leq x^3-3xy^2\leq2,2\leq3x^2y-y^3\leq4$ and $x,y\geq0.$
Let $R$ be the region. I substitute $u=x^3-3xy^2,v=3x^2y-y^3$. $\det J(x,y)=9(x^2+y^2)^2.$ Also notice that $(x+yi)^3=u+vi$, so $(x^2+y^2)^3=u^2+v^2$.
Then the rquired integral
$$=\int_R\frac{x^3-3xy^2}{9(x^2+y^2)^4}9(x^2+y^2)^2dxdy$$
$$=\int^4_2\int^2_1\frac u{9(u^2+v^2)^{4/3}}dudv$$
$$=\int^4_2\frac 1 6[\frac 1 {(1+v^2)^{1/3}}-\frac 1{(4+v^2)^{1/3}}]dv$$
This has no closed form solution. I wonder where I did wrong... Thank you.
|
I think that your calculations are correct. The final result is given in terms of an hypergeometric function.
You may make use of
$$
\int \frac{\mathrm{d} u}{(a^2+u^2)^{1/3}} = \frac{u}{a^{2/3}} \, {}_2F_1 \left( \frac{1}{3}, \frac{1}{2}; \frac{3}{2}; -\frac{u^2}{a^2} \right) \, ,
$$
to find a closed form of your integral.
Cheers,
R
|
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|
What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
|
*
*Take any sequence of consecutive square numbers. What do you know (or, if you don't know, try it: what do you notice) about the second differences?
*Each row of the matrix has four consecutive square numbers.* Note that column operations preserve the determinant in the same way that row operations do (this is obvious if we recall that the determinant of a matrix equals the determinant of its transpose, and column operations on your matrix are row operations on its transpose). One can fill the final three columns of your matrix with the first differences of your four consecutive square numbers, by applying the following column operations: subtract column 3 from column 4; subtract column 2 from column 3; subtract column 1 from column 2. (Why did we have to do it in this order?)
*Can you perform another set of column operations to difference these three first differences? The final two columns of your matrix will then be filled with the second differences of your original sequences of four consecutive square numbers. Recall the result from (1). What does this tell you about these two columns and hence the determinant?
$(*)$ Perhaps $a^2, \dots, (a+3)^2$ may not be "square numbers" in the sense that $a$ may not be a natural number. But this doesn't matter very much; the reason the second differences of the sequence $n^2, \, n\in\mathbb{N}$ are so nice is because of algebra that works just as well on $x^2, \, x\in \mathbb{R}$. In particular, what is $\left((x+2)^2-(x+1)^2\right) - \left((x+1)^2-x^2\right)$?
If you brush up a little on finite differences of higher polynomials you might want to have a think about how you could determine the following determinant, where each row has six consecutive fourth powers:
$$\begin{vmatrix}a^4 & (a+1)^4 & (a+2)^4 & (a+3)^4 & (a+4)^4 & (a+5)^4 \\ (b+6)^4 & (b+7)^4 & (b+8)^4 & (b+9)^4 & (b+10)^4 & (b+11)^4 \\ (c-3)^4 & (c-2)^4 & (c-1)^4 & c^4 & (c+1)^4 & (c+2)^4 \\ (d+20)^4 & (d+21)^4 & (d+22)^4 & (d+23)^4 & (d+24)^4 & (d+25)^4 \\ (e-8)^4 & (e-7)^4 & (e-6)^4 & (e-5)^4 & (e-4)^4 & (e-3)^4 \\ (f + 2016)^4 & (f+2017)^4 & (f+2018)^4 & (f+2019)^4 & (f+2020)^4 & (f+2021)^4 \end{vmatrix} $$
Expanding this out in full may not necessarily develop the situation to your advantage. But this time, with a fourth power polynomial, it isn't the second differences that come out the same...
|
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|
Finding that a two variable limit does not exist trouble: $\lim_{(x, y) \to (0, 0)} \frac{y^2 \sin^2 x}{x^4 + y^4}$ I have
$$ \lim_{(x, y) \to (0, 0)} \frac{y^2 \sin^2 x}{x^4 + y^4} $$
I've tried setting $y= 0, x=0, y=mx, y=x$, and lots of other things but because of the $\sin$ I keep getting a limit of zero. How do you show this doesn't exist?
|
As you say: if you let $x = 0$ or $y = 0$ your answer will be 0. If you let $y = x$, however, you get
\begin{equation*}
\lim_{x \rightarrow 0} \frac{x^2 \sin^2(x)}{x^4 + x^4}
= \lim_{x \rightarrow 0} \frac{\sin^2(x)}{2x^2}
= \lim_{x \rightarrow 0} \frac{\sin(x)}{x} \cdot \frac{\sin(x)}{x} \cdot \frac{1}{2} = \frac{1}{2}.
\end{equation*}
|
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|
Prove that $\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$
Prove that $$\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$$
$\bf{My\; Try::}$ Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+........+\binom{n}{n}x^n$$
and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-1}+........+\binom{n}{n}$$
Now Coefficients of $x^n$ in $\displaystyle (1+x)^{2n} = \binom{2n}{n} = \binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2$
Now Using $\bf{Cauchy\; Schwarz}$ Inequality
$$\left[\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2\right]\cdot \left[1^2+1^2+....+1^2\right]\geq \left[\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+.....+\binom{n}{n}\right]^2=2^{2n}=4^n>4n\forall n\geq 2,n\in \mathbb{N}$$
So $$\binom{2n}{n} = \binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2>\frac{4n}{n+1}\forall n\geq 2,n\in \mathbb{N}$$
My question is , Is my proof is right, If not then how can i solve it.
Also plz explain any shorter way, Thanks
|
We prove the stronger bound $\binom{2n}{n} \geq \frac{4^n}{2n+1}$, which is bigger than $\frac{4}{n+1}$ for all $n \geq 2$.
Note that the $2n+1$ binomial coefficients $\binom{2n}{0}$, $\binom{2n}{1}$, ..., $\binom{2n}{2n}$ have average
$$
\frac{\binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{2n}}{2n+1} = \frac{2^{2n}}{2n+1} = \frac{4^n}{2n+1}.
$$
Since $\binom{2n}{n}$ is the largest of these binomial coefficients, it is surely "above average": $\binom{2n}{n} \geq \frac{4^n}{2n+1}$.
|
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|
Nonlinear equations in 3 variables Solve for $x,y\ \mbox{and}\ z$:
$$
\left\{\begin{array}{rcr}
x + y + z & = & 2
\\[1mm]
\left(x + y\right)\left(y + z\right) + \left(y+z\right)\left(z+x\right) +
\left(z + x\right)\left(x + y\right) & = & 1
\\[1mm]
x^{2}\left(y + z\right) + y^{2}\left(x + z\right) + z^{2}\left(x + y\right)
& = & -6
\end{array}\right.
$$
I tried this by adding the second and third equations, and then using the first equation to substitute values. However, this is getting nowhere.
|
Considering the equations$$x+y+z-2=0\tag 1$$ $$(x+y) (x+z)+(x+z) (y+z)+(x+y) (y+z)-1=0 \tag 2$$ $$x^2 (y+z)+y^2 (x+z)+z^2 (x+y)+6=0\tag 3$$ first eliminate $z$ from $(1)$ $$z=-x-y+2\tag 4$$ Replace in $(2)$ to get $$\left(-x^2+2 x+3\right)+(2-x) y-y^2=0\tag 5$$ Solve the quadratic in $y$ and keep the largest root; this gives $$y=\frac{1}{2} \left(\sqrt{-3 x^2+4 x+16}-x+2\right)\tag6$$ Plug $z$ and $y$ in $(3)$ (this is the tedious part) and get $$-3 x^3+6 x^2+9 x=0 \tag 7$$ So the possible roots $$x= -1\qquad ,\qquad x= 0\qquad ,\qquad x= 3$$
|
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|
Sum of three numbers and the sum of their squares: $x^2+y^2+z^2 \geq \frac{1}{3}$ for $x+y+z=1$ We have $x, y, z \in \mathbb R$ such that $x+y+z=1$. Prove that $x^2+y^2+z^2 \geq \frac{1}{3}$. I am able to do this using the relationship between the power and arithmetic means. Is there a way to not use this relationship?
|
We have
$$
x+y+z=1\\
(x+y+z)^2=1\\
x^2+y^2+z^2+2(xy+xz+yz)=1
$$
By the rearrangement inequality, $x^2+y^2+z^2\geq xy+xz+yz$. Inserting that gives you
$$
3(x^2+y^2+z^2)\geq1
$$
|
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|
Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$ Let $a,b,c>0$ prove or disprove
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
since
$$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+a^2c^2)\tag{1}$$
other
$$(a+b+c)^2\le 3(a^2+b^2+c^2)\tag{2}$$
Because of the inequality sign in a different direction so we can't $(1)\times (2)$
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\sqrt[3]{u(3u^2-2v^2)}\geq\sqrt[4]{3v^4-2uw^3}$$ or
$f(w^3)\geq0$, where $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $w^3$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or
$x^3-3ux^2+3v^2x=w^3$.
Hence, a line $y=w^3$ and a graph of $y=x^3-3ux^2+3v^2x$ have three common points.
Thus, $w^3$ gets a minimal value, when a line $y=w^3$ is a tangent line
to the graph of $y=x^3-3ux^2+3v^2x$, which happens for equality case of two variables.
Also we must check the case $w^3\rightarrow0^+$.
Id est, it's enough to prove our inequality in the following cases.
*
*$b=c=1$, which gives
$$(a-1)^2(a^{10}+10a^9+51a^8+188a^7+557a^6+1374a^5+983a^4+1616a^3+501a^2+538a+13)\geq0$$
2. $w^3\rightarrow0^+$.
Let $c\rightarrow0^+$ and $b=1$.
We obtain $(a^3+a^2+a+1)^4\geq243a^6$, which is obvious by AM-GM.
Done!
|
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|
cubic equation has $3$ distinct roots $\alpha,\beta,\gamma$ equation $x^3-9x^2+24x+c=0$ has $3$ distinct roots $\alpha,\beta,\gamma$, then $\lfloor \alpha \rfloor+\lfloor \beta \rfloor +\lfloor \gamma \rfloor =$
and $\lfloor \alpha \rfloor = \alpha-\{\alpha\},\lfloor \beta \rfloor = \beta-\{\beta\},\lfloor \gamma \rfloor = \gamma-\{\gamma\},0\leq \{\alpha\},\{\beta\},\{\gamma\}<1$
we assume $f(x) = x^3-9x^2+24x+c$, using $f'(x) = 3x^2-18x+24$
put maximum , minimum $f'(x)=0$
$3x^2-18x+24=0$
$x^2-6x+8=(x-4)(x-2)=0$
$x=2,x=4$
put $x=2,3$ in $f''(x)=6x-18=6(x-3)$
$f''(2) = -6<0$ means $x=2$ is a point of local maximum
and $f''(4)=6>0$ means $x=4$ is a point of local minimum
i can not go further
|
We may assume that $\alpha\lt \beta\lt \gamma$.
We have to have
$$f(2)=20+c\gt 0\quad\text{and}\quad f(4)=16+c\lt 0$$
Note that we have
$$f(1)=16+c\lt 0\quad\text{and}\quad f(5)=20+c\gt 0$$
from which
$$1\lt\alpha\lt 2\implies 0\lt\{\alpha\}\lt 1$$
$$4\lt\gamma\lt 5\implies 0\lt\{\gamma\}\lt 1$$
follow.
Thus, since
$$0\lt \{\alpha\}+\{\beta\}+\{\gamma\}\lt 3\implies \{\alpha\}+\{\beta\}+\{\gamma\}=1,2$$
and
$$\alpha+\beta+\gamma=9$$
we get
$$\lfloor\alpha\rfloor+\lfloor\beta\rfloor+\lfloor\gamma\rfloor=9-(\{\alpha\}+\{\beta\}+\{\gamma\})=\color{red}{7,8}$$
(Since $f(3)=18+c$, we have that $\lfloor\alpha\rfloor+\lfloor\beta\rfloor+\lfloor\gamma\rfloor=7$ for $-20\lt c\lt -18$ and that $\lfloor\alpha\rfloor+\lfloor\beta\rfloor+\lfloor\gamma\rfloor=8$ for $-18\le c\lt -16$.)
|
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|
The ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime
I tried to show that the ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime in $\mathbb{C}[X,Y]$.
I noticed that $X^{2}-Y^{3}-(Y^{2}-X^{3})=(X-Y)(X^{2}+XY+Y^{2}+X+Y)$. Now, if $X-Y$ is in $(X^{2}-Y^{3},Y^{2}-X^{3})$ there exists $f(X,Y),\ g(X,Y)\in \mathbb{C}[X,Y] $ s.t. $$X-Y=f(X,Y)(X^{2}-Y^{3}) + g(X,Y)(Y^{2}-X^{3}).$$ For $Y=0$ it results $X=f(X,0)X^{2} - g(X,0)X^{3}\Rightarrow 1=f(X,0)X-g(X,0)X^{2}$ (False).
A similar argument goes for$X^{2}+XY+Y^{2}+X+Y$. Is this a right solution?
|
Here is a more geometric solution.
To show that $I=(x^2-y^3,y^2-x^3)$ is not prime, one way is to show that their set of common zeroes has cardinality $>1$.
This works because the irreducible components of their common zero set correspond to the radicals of the primary components in the decomposition of $I$.
So suppose $x^2-y^3=y^2-x^3=0$. From the first equality, you get $$x^3-y^2=x^2 x - y^2=y^3x-y^2 = y^2(x-1)=0.$$
But then either $y=0$ or $x=1$. If $x=1$, then we must have $y=1$. If $y=0$, we must have $x=0$. Hence we have found two distinct points solving the equation, hence it cannot be prime.
|
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|
To find number of distinct terms of binomial expansion I have binomial expansion as
$(x+\frac{1}{x} + x^2 + \frac{1}{x^2})^{15}$.
How do i find number of distinct terms in it. Distinct in sense means terms having different powers of $x$?
I have simplified this as $\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$. How do i proceed
Thanks
|
$$\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}=\frac{1}{x^{30}}(x^3(x+1)+x+1)^{15}=$$
$$=\frac{1}{x^{30}}(x^3+1)^{15}(x+1)^{15}=\frac{1}{x^{30}}\sum_{i=0}^{15}\binom{15}{i}x^{3i}\sum_{j=0}^{15}\binom{15}{j}x^{j}=$$
$$=\sum_{i=0}^{15}\binom{15}{i}\sum_{j=0}^{15}\binom{15}{j}x^{3i+j-30}$$
Number of elements in set
$$\{3i+j-30:i,j\in\{0,1,...,15\}\}=\{-30,-29,...,0,1,...,30\}$$ show the distinct powers of $x$
|
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|
Hints to find solution of : $\left\lfloor \frac{x}{100} \left\lfloor \frac{x}{100}\right\rfloor \right\rfloor=5$ Find the number of integer solution of $$\left\lfloor \dfrac{x}{100} \left\lfloor \dfrac{x}{100}\right\rfloor \right\rfloor=5$$
$\lfloor . \rfloor$ is Greatest Integer Function or floor function.
|
$\lfloor\frac{{x}}{\mathrm{100}}\lfloor\frac{{x}}{\mathrm{100}}\rfloor\rfloor=\mathrm{5}\:\Rightarrow\lfloor\frac{{x}}{\mathrm{100}}\rfloor=\lfloor\sqrt{\mathrm{5}}\rfloor=\mathrm{2} \\ $
$\lfloor\frac{{x}}{\mathrm{100}}.\mathrm{2}\rfloor=\mathrm{5}\Rightarrow\frac{{x}}{\mathrm{50}}\in\left[\mathrm{5},\mathrm{6}\right) \\ $
${x}\in\left[\mathrm{250},\mathrm{300}\right) \\ $
|
{
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"url": "https://math.stackexchange.com/questions/1962599",
"timestamp": "2023-03-29T00:00:00",
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|
Angle of a triangle using trigonometry. If in a triangle ABC , cos A+ cos C= sin B then what is the measurement of angle A? I have tried to solve it using trigonometric identities but failed to solve it.
|
Using Prosthaphaeresis Formula, $$\cos A+\cos C=2\cos\dfrac{A+C}2\cos\dfrac{A-C}2$$
Now $\cos\dfrac{A+C}2=\cos\dfrac{\pi- B}2=\sin\dfrac B2$
and $\sin B=\sin\left(2\cdot\dfrac B2\right)2=\sin\dfrac B2\cos\dfrac B2$
As $0<B<\pi,0<\dfrac B2\le\dfrac\pi2\implies\sin\dfrac B2>0$
So we have $$\cos\dfrac{A-C}2=\cos\dfrac B2$$
$\implies\dfrac{A-C}2=2m\pi\pm\dfrac B2\iff A-C=4m\pi\pm B$ where $m$ is any integer.
As $A+B+C=\pi,0<A,B,C<\pi$ we must have $m=0$
$\implies A-C=\pm B$
If $A-C=+B\iff A=B+C=\dfrac{A+B+C}2=\dfrac\pi2$
Else $A-C=-B\iff C=A+B\implies C=\dfrac\pi2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1962983",
"timestamp": "2023-03-29T00:00:00",
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|
Find the coefficient of $x^{100}$ in the power series representing the function $\dots$ Find the coefficient of $x^{100}$ in the power series representing the function:
$$f(x)=(x+x^2+x^3+ \cdots) \cdot (x^2+x^3+x^4 \cdots) \cdot (x^3+x^4+x^5 \cdots)$$
Here is what I have so far:
$x+x^2+x^3+ \cdots = x(1+x+x^2+x^3+ \cdots)$ and this equals to the geometric series $\frac{x}{1-x}$
$x^2+x^3+x^4 \cdots = x^2(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^2}{1-x}$
$x^3+x^4+x^5 \cdots = x^3(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^3}{1-x}$
So now we have $F(x)=\left(\frac{x}{1-x}\right)\left(\frac{x^2}{1-x}\right)\left(\frac{x^3}{1-x}\right)$
This is the same as $F(x)=x^6(1-x)^{-1}$
Now we can use the following way to expand and find the coefficient in front of $x^{100}$, $(1+x)^{\alpha}=1+\alpha x + {\alpha \choose 2}x^2 + {\alpha \choose 3}x^3 + \dots$
For our function we get $x^6(1+(-1)(-x) + {-1 \choose 2}(-x)^2 + {-1 \choose 3}(-x)^3 + \dots)$, but $-1 \choose \beta$ where $\beta \ge 2$ will always equal $1$ or $-1$ and I do not think I am doing this correct because according to this the coefficient of $x^{100}$ will be either $1$ or $-1$?
What am I doing wrong?
|
A different conceptual interpretation: We have
$$
F(x) = x^6 (1+x+x^2+\cdots)^3
$$
We can interpret this as a generating function representing a fixed prefix of six balls, plus three (distinguishable) sacks of zero or more balls each. The coefficient of $x^{100}$ is then the number of ways to total $100$ balls. Factoring out the six-ball prefix, that leaves the number of ways to obtain $94$ balls in the three sacks, which we can determine using stars-and-bars as
$$
\binom{96}{2} = 4560
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show $\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$ I want to verify the following inequality:
Let $a, b$ be non negative number
$$\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$$
I decided to analyse the sign of $$\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 - 2\left(\frac{a}{b} + \frac{b}{a}\right)$$
But I'm not getting anywhere.
I'm having trouble deciding whether $\dfrac{a^2(a-2)+b^2(b-2)}{ab} +3$ is positive or negative.
|
You actually can prove for +2 also like $\frac{a^2}{b^2}+1\geq 2\frac{a}{b}$ and
$\frac{b^2}{a^2}+1\geq 2\frac{b}{a}$ by AM-GM and by adding you get the desire result.
|
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|
Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
|
Let $f(x) = x^2 -ax + 1$ and let $g(x) = f(x+2) = x^2 - (a-4)x + (5-2a)$
Find all $x$ such that $|f(x)|\le\frac{1}{4}$ and $|g(x)|\le\frac{1}{4}$
If $f(x)$ and $g(x)$ are to share that property at the same point, that point must be the one point that they have in common.
\begin{align}
f(x) &= g(x) \\
x^2 -ax + 1 &= x^2 - (a-4)x + (5-2a) \\
-4x &= 4-2a \\
x &= -1 + \dfrac{a}{2} \\
y &= 2-\dfrac{a^2}{4}
\end{align}
The point that they have in common is the point
$\left(-1 + \dfrac a2,2-\dfrac{a^2}{4} \right)$.
So we need to solve
\begin{array}{c}
\left| 2-\dfrac{a^2}{4} \right| \le \dfrac 14 \\
|8 - a^2| \le 1 \\
-1 \le a^2 - 8 \le 1 \\
7 \le a^2 \le 9 \\
a \in (-3, -\sqrt 7) \cup (\sqrt 7, 3)
\end{array}
And finally, $a = 2x + 2 \implies$
\begin{array}{c}
-3 \le 2x+2 \le -\sqrt 7 \\
-\dfrac 52 \le x \le -\dfrac{\sqrt 7 + 2}{2} \\
\hline
\sqrt 7 \le 2x+2 \le 3 \\
\dfrac{\sqrt 7 - 2}{2} \le x \le \dfrac 12 \\
\hline
x \in \left[-\dfrac 52, -\dfrac{\sqrt 7 + 2}{2} \right]
\cup \left[\dfrac{\sqrt 7 - 2}{2} ,\dfrac 12 \right]
\end{array}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\frac{n(n+1)}{2}= 45$ I have this equation:
$\frac{n(n+1)}{2}= 45$
I started:
$$n^2+n= 90$$
What do I do next?
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$n^2+n=90$
We can write this as-
$n^2+n-90=0$
Now, it is in the quadratic equation form, .i.e., $ax^2+bx+c=0$
Factor out the resulting equation, break $n$ into two terms $10n-9n$
$$n^2+10n-9n-90=0$$
Take $n$ common from first two terms and $-9$ from last two terms-
$$n(n+10)-9(n+10)=0$$
$$(n-9)(9+10)=0$$
$n-9=0\;\implies n=9$
$n+10=0\;\implies n=-10$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integration of $\frac{2x^2}{\sqrt{x^2-1}}\, dx$ Integration of $\dfrac{2x^2}{\sqrt{x^2-1}}\,\mathrm dx$
My try:
Let $u=\sqrt{x^2-1}$
Then $x=\sqrt{u^2+1}$
$x \,\mathrm dx =u\,\mathrm du$
$$\int \frac{2u(u^2+1)}{u\sqrt{u^2+1}} \, \mathrm du$$
$$=2\int \sqrt {u^2+1} \, \mathrm du$$
True ? and what about the last integration?
|
\begin{align}
x & = \sec\theta \\[5pt]
dx & = \sec\theta\tan\theta\, d\theta \\[5pt]
\sqrt{x^2-1} & = \tan\theta \\[5pt]
\frac{2x^2}{\sqrt{x^2-1}}\, dx & = \frac{2\sec^2\theta}{\tan\theta} \sec\theta\tan\theta\,d\theta = 2\sec^3\theta\,d\theta
\end{align}
The integral of secant cubed is well known to be challenging to those first encountering it, but it does not require advanced methods.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
To find the range of $\sqrt{x-1} + \sqrt{5-x}$ To find the range of $\sqrt{x-1} + \sqrt{5-x}$.
I do not know how to start?
Thanks
|
Note that
$$
\left(\sqrt{x-1}+\sqrt{5-x}\right)^2=4+2\sqrt{4-(x-3)^2}
$$
Since $4-(x-3)^2\le4$ and $4+2\sqrt{4-(x-3)^2}\ge4$, we get that
$$
2\le\sqrt{x-1}+\sqrt{5-x}\le2\sqrt2
$$
|
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|
Find the equation solution $\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$ solve this following equation
$$\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$$
the equation obvious have a root $x=3$. But how to solve other roots?
|
Obviously $x>0$, suppose that $\lfloor x \rfloor =k$ (clearly $k$ must be between $0$ and $5$), then the solutions are $\lfloor \frac{9}{x} \rfloor = 6-k\iff 6-k\leq \frac{9}{x} < 6-k+1\iff \frac{9}{6-k+1}< x \leq \frac{9}{6-k}$.
So the solution set is $\bigcup\limits_{k=0}^{5}( (\frac{9}{6-k+1},\frac{9}{6-k}]\cap [k,k+1))$.
This is just a simple and fast calculation now.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
How to put $\frac{1}{\cos\theta - j\sin \theta}$ in the form $a+jb$? ($j^2=-1$) I am new to complex numbers and am having trouble putting them in the form $a+jb$ (or $a+bi$) How would I go about putting this expression in the form $a+jb$?
$$\frac{1}{\cos\theta - j\sin \theta}$$
|
$$\frac{1}{\cos \theta - j \sin \theta}=\frac{\cos \theta + j \sin \theta}{(\cos \theta - j \sin \theta)(\cos \theta + j \sin \theta)}=\frac{\cos \theta+j \sin \theta}{\cos^{2}\theta-j^2 \sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{\cos^2 \theta - (-1)\sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{\cos^2 \theta + \sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{1}=\cos \theta + j \sin \theta$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1979977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving $xy''-y'+4x^3y=0$ I need to solve
$$xy''-y'+4x^3y=0$$
by reduction or order, knowing one solution is $y=\sin x^2$. That's what I did:
$y_2 = v(x)\sin x^2\implies \\y_2' = v'\sin x^2+2xv\cos x^2\implies\\y_2'' = v''\sin x^2+2xv'\cos x^2+2(-2vx^2\sin x^2+v'x\cos x^2+v\cos x^2)$
Then,
$$xy_2''-y_2'+4x^3y_2 = \\xv''\sin x^2+2x^2v'\cos x^2-4vx^3\sin x^2+2x^2v'\cos x^2+2xv\cos x^2-v'\sin x^2-2vx\cos x^2+4x^3v\sin x^2 = \\$$
$$xv''\sin x^2+4v'x^2\cos x^2-v'\sin x^2$$
which seems like a difficult equation to solve, so I guess it's wrong. Also, wolfram alpha says the solution for $v$ is $c_1\cot x^2+c_2$, but my book says that $y_2$ is $\cos x^2$. What did I do wrong?
|
$y=\sin x^2 \int v(x) dx $
$y' =2x \cos x^2 \int v(x) dx +v(x)\sin x^2 $
$y''=2\cos x^2 \int v(x) dx -4x^2 \sin x^2 \int v(x) dx +2x v(x)\cos x^2 +v'(x)\sin x^2 +2xv(x) \cos x^2 $
so
$xy'' -y' +4x^3 y =x\left(2\cos x^2 \int v(x) dx -4x^2 \sin x^2 \int v(x) dx +2x v(x)\cos x^2 +v'(x)\sin x^2 +2xv(x) \cos x^2\right) - \left(2x \cos x^2 \int v(x) dx +v(x)\sin x^2\right) +4x^3 \sin x^2 \int v(x) dx =4x^2 v(x)\cos x^2 +xv'(x) \sin x^2 -v(x) \sin x^2$
hence
$\frac{v'}{v} =\frac{1}{x} +4x\cot x^2 $
thus
$\ln v = \ln x +2\ln |\sin x^2 |+ \ln C$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Mediant Inequality Proof: $\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$ If $\frac{a}{b}$ < $\frac{c}{d}$ then
$$\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$$
I have been searching and trying and couldnt find a reliable proof.
One might do this like here which I think it is very wrong:
$\frac{a}{b}$ < $\frac{c}{d}$ then $a.d$ < $c.b$
Because $\frac{2}{-3}$ < $\frac{4}{5}$ doesn't mean $10 < -12$
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Hint. As regards $\frac{a}{b} < \frac{a+c}{b+d}$, show that
$$\frac{a+c}{b+d}-\frac{a}{b}=\frac{\frac{c}{d}-\frac{a}{b}}{\frac{b}{d}+1}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1989104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating the improper integral $\int_0^\infty \frac{x\cos x-\sin x}{x^3} \cos(\frac{x}{2}) \mathrm dx $ I've been working through the following integral and am stumped:
$$\int_0^\infty \frac{x\cos x-\sin x}{x^3}\cos\left(\frac{x}{2}\right)\mathrm dx$$
Given the questions in my class that have proceeded and followed this integral, I believe that this is some form of Fourier transform/integral. However, it doesn't look like any of the content surrounding it. That is, there is no $e^{-ikx}$ or $g(k)$ or anything else that I'm familiar with.
I know that it's an even function, but that's about as far as I can get. If I try to split it over the subtraction, I get two non-converging integrals, so that wasn't much help either. I've been throwing lots of trig identities at it but nothing familiar has appeared yet.
Any help would be greatly appreciated.
Thank you.
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Integration by parts can be performed for the indefinite integral, using relations
$$\dfrac{x\cos x-\sin x}{x^2} = \left(\dfrac{\sin x}{x}\right)',\quad
\sin^3 z =\frac{3\sin z-\sin3z}4,\quad \cos^3z=\frac{3\cos
z+\cos3z}4.$$
One can get
\begin{align}
&\int \dfrac{x\cos x-\sin x}{x^3}\,\cos\dfrac x2\, \mathrm dx
= \int\dfrac1{4\sin \frac x2}\,\mathrm d\left(\dfrac{\sin x}{x}\right)^2 \\[4pt]
&=\dfrac1{4\sin \frac x2}\left(\dfrac{\sin x}{x}\right)^2
+\int\left(\dfrac{\sin x}{x}\right)^2\dfrac{\cos\frac x2}{8\sin^2 \frac x2}\,\mathrm dx
=\dfrac1{x^2}\sin\frac x2\,\cos^2 \frac x2
+\dfrac12\int\dfrac{\cos^3\frac x2}{x^2}\,\mathrm dx\\[4pt]
&=\dfrac1{x^2}\sin\frac x2
-\dfrac1{4x^2}\left(3\sin\frac x2-\sin \frac {3x}2\right)
+\dfrac18\int\dfrac{3\cos\frac x2+\cos\frac{3x}2}{x^2}\,\mathrm dx\\[4pt]
&=\dfrac1{4x^2}\left(\sin\frac x2+\sin \frac {3x}2\right)
-\dfrac18\int\left(3\cos\frac x2+\cos\frac{3x}2\right)\,\mathrm d\frac1x\\[4pt]
&=\dfrac1{8x^2}\left(2\sin\frac x2+2\sin\frac{3x}2-3x\cos\frac {x}2-x\cos \frac {3x}2\right)
-\dfrac3{16}\int\frac1x\left(\sin\frac x2+\sin\frac{3x}2\right)\,\mathrm dx.
\end{align}
Since
$$\lim\limits_{x\to0}\dfrac{2\sin\frac x2+2\sin\frac{3x}2-3x\cos\frac {x}2-x\cos\frac {3x}2}{8x^2}
= \lim\limits_{x\to0}\dfrac{2\frac x2+2\frac{3x}2-3x-x+O(x^3)}{8x^2}=0,$$
$$\int\limits_{0}^{\infty} \dfrac{\sin x}x\,\mathrm dx =\frac\pi2,$$
then
$$\int\limits_{0}^{\infty} \dfrac{x\cos x-\sin x}{x^3}\,\cos\dfrac x2\, \mathrm dx
=-\frac3{16}\left(\int\limits_{0}^{\infty} \dfrac{\sin\frac x2}{\frac x2}\,\mathrm d\frac x2+\int\limits_{0}^{\infty} \dfrac{\sin \frac{3x}2}{\frac{3x}2}\,\mathrm d\frac{3x}2\right) = \color{green}{\mathbf{-\frac{3\pi}{16}}}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why $\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim\limits_{t\to 0}\frac{\sin t}{t}$? Why $\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{t\to 0}\frac{\sin t}{t}$( and hence equals to $1$)?
Any rigorous reason? (i.e. not just say by letting $t=x^2+y^2$.)
|
Using $x = t \cos \alpha $ and $y = t \sin \alpha$, then:
$$ \frac{ \sin (x^2 + y^2 ) }{x^2+y^2} = \frac{ \sin (t^2(\sin^2 \alpha + \cos^2 \alpha)) }{t^2(\sin^2 \alpha + \cos^2 \alpha)} = \frac{ \sin (t^2) }{t^2}$$
.
Note as $x,y \to 0,0$, then $t \to 0 $.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking $$x=\sqrt{2}y$$ then using integration by parts, we have
$$I=\frac{\pi^5}{2048}-\frac{1}{4}\int_0^1{\arcsin^4\left( \frac{z}{\sqrt{2}}\right) }\frac{dz}{\sqrt{1-x^2}}$$
But how to calculate this integral? I would appreciate your help
|
A slight detour from Ron Gordon's answer, using the square of the $\arcsin$ series:
$$\begin{align*}
I &= \int_0^1 \frac{\arcsin(x) \arcsin\left(\frac x{\sqrt2}\right)}{\sqrt{2-x^2}} \, dx \\[1ex]
&= \frac{\pi^3}{64} - \frac12 \int_0^1 \arcsin^2\left(\frac x{\sqrt2}\right) \, \frac{dx}{\sqrt{1-x^2}} \tag{1} \\[1ex]
&= \frac{\pi^3}{64} - \frac14 \sum_{n=1}^\infty \frac{2^n}{n^{2} \binom{2n}n} \int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}} \, dx \tag{2} \\[1ex]
&= \frac{\pi^3}{64} - \frac18 \sum_{n=1}^\infty \frac{2^n}{n^2 \binom{2n}n} \int_0^1 x^{n-\frac12} (1-x)^{-\frac12} \, dx \tag{3} \\[1ex]
&= \frac{\pi^3}{64} - \frac18 \sum_{n=1}^\infty \frac{2^n}{n^2 \binom{2n}n} \frac{\Gamma\left(n+\frac12\right)\Gamma\left(\frac12\right)}{\Gamma(n+1)} \tag{4} \\[1ex]
&= \frac{\pi^3}{64} - \frac\pi8 \sum_{n=1}^\infty \frac1{2^nn^2} \tag{5} \\[1ex]
&= \frac{\pi^3}{64} - \frac\pi8 \operatorname{Li}_2\left(\frac12\right) \tag{6} \\[1ex]
&= \frac{\pi^3}{64} - \frac\pi8 \left(\frac{\pi^2}{12} - \frac{\log^2(2)}2\right) \\[1ex]
&= \boxed{\frac{\pi^3}{192} + \frac{\pi\log^2(2)}{16}}
\end{align*}$$
*
*$(1)$ : integrate by parts
*$(2)$ : power series of $\arcsin^2(x)$
*$(3)$ : substitute $x\mapsto\sqrt x$
*$(4)$ : beta function
*$(5)$ : simplify gammas and central binomial coefficient; in particular, we have the identity $\dfrac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)\Gamma(n+1)} = \dfrac1{4^n} \dbinom{2n}n$
*$(6)$ : dilogarithm
I suspect one could adapt Jack D'Aurizio's general solution for a related integral... an exercise for the reader.
|
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}
|
Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
|
Use this simple lemma
$(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$
|
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|
If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
$\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)$$
$$ = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)$$
$$ = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c$$
Now how can i solve it after that , Help required, Thanks
|
Let $\sin (a)=x,\sin (b)=y,\sin (c)=z $ thus continuing from your simplified version we have $$\frac {\sum ^{cyc} x\sqrt {(1-y^2)(1-z^2)}}{x+y+z}$$ then using Am-Gm for each sqyare root we have $$\frac {\sum^{cyc} x\sqrt{(1-y^2)(1-z^2)}}{x+y+z }\leq \frac {2 (x+y+z)-2 (xy+yz+xz)}{2 (x+y+z)} $$ thus its $1-\bf{something} $ we can easily see that the second bracket has maximum value $1$
but in the problem as $90^{0}$ is not in domain so its always less than $1$
Thus the original expression is less than $1$
|
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|
Show series $1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2$ ... does not converge I was wondering if my proof is correct, and if there are any better alternative proofs. Or maybe proof that use nice tricks i might need in the future.
$$1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} \ldots = \sum_{n =1}^\infty \left(\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2}\right) + 1 - \frac{1}{2^2}$$
Now we know that $$\sum_{n = 1}^\infty \frac{1}{2n + 1}$$ diverges and $$\sum_{n = 1}^\infty -\frac{1}{(2n + 2)^2}$$ converges. Hence their sum diverges (I proved this fact). Hence, the series diverges. Any obvious mistake or better way of tackling it? Maybe using partial sums since i am clueless how to use them.
|
Call $s_k$ the partial sums of the series. That is $s_k=\sum_{n=1}^ka_n$, where $a_n=\frac{1}{n}$ if $n$ is odd, and $a_n=-\frac{1}{n^2}$ if $n$ is even. As $\sum_{j=1}^\infty \frac{1}{j^2}=\frac{\pi^2}{6}$, we have the estimate
$$
s_k\geq-\frac{\pi^2}{6}+\sum_{n=1}^ka_{2n-1}=-\frac{\pi^2}{6}+\sum_{n=1}^k\frac{1}{2n-1}.
$$
By comparison with the harmonic series the last sum is divergent (when $k\to\infty$) so $s_k$ must be too, proving that your series diverges.
|
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|
Evaluate $\int \frac{1}{1+x^4} dx$
Evaluate
$$\int_0^\infty \frac{1}{1+x^4} dx$$
It seems to me that the most natural way would be to use contour integral. The integral on the semi circle will disappear. The only thing I'm not sure about is how to find the residue. Four roots of $1+z^4 = 0$ can be found easily and they are
$$e^{\theta i} \text{ where } \theta = \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$$
Let four solutions be $z_1,z_2,z_3,z_4$ respectively, then
$$\frac{1}{1+x^4}=\frac{1}{(z - z_1)(z-z_2)(z-z_3)(z-z_4)}$$
What would be the best way to find the residue at two points in the upper half plane, except just plug in and multiply a lot?
|
Let $x^{4}=z$
\begin{align}
\int\limits_{0}^{\infty} \frac{1}{1+x^{4}} dx
&= \frac{1}{4} \int\limits_{0}^{\infty} \frac{z^{-3/4}}{1+z} dz \\
&= \frac{1}{4} \mathrm{B}\left(\frac{1}{4}, \frac{3}{4}\right) \\
&= \frac{\Gamma(1/4)\Gamma(3/4)}{4\Gamma(1)} \\
&= \frac{\pi}{2\sqrt{2}}
\end{align}
|
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|
If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \cos^{2} a)$$ But now, I'm stuck. Solutions are greatly appreciated.
|
Hint: if $\sin\alpha+\cos\alpha=1.2$ then
$$
(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=1+2\sin\alpha\cos\alpha=1.44.
$$
|
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|
A problem related to arithmetico-geometric sequence Question:
Find the sum of the series: $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ... = 1+\sum_{n=1}^\infty \frac{(4n-2)}{3^n}$
My doubt:
I have taken $\frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ...$ to be in A.G.P. and calculated the sum using the formula $$S_\infty=\frac{a}{1-r}+\frac{d \cdot r}{(1-r)^2},$$ where $a=\frac23, r=\frac13$ and $d=4$.
I get $S_\infty=4$ and then I add $1$ to it to get the answer $5$.
But my book shows a different method with a different answer ($3$).
Please tell what is wrong with this method.
|
We can use Telescoping series as well
$$\dfrac{4r-2}{3^r}=f(r+1)-f(r)$$ where $f(n)=\dfrac{a_0+a_1n+a_2n^2+\cdots}{3^n}$
So that $$\sum_{r=1}^n\dfrac{4r-2}{3^r}=\sum_{r=1}^n(f(r+1)-f(r))=f(n+1)-f(1)$$
$$4r-2=a_0+a_1(r+1)+a_2(r+1)^2+\cdots-3\left(a_0+a_1r+a_2r^2+\cdots\right)$$
As the coefficients of $r^m$ is zero for $m\ge2\implies a_m=0\forall m\ge2$
$$4r-2=a_0+a_1(r+1)-3\left(a_0+a_1r\right)=-2a_1r+a_1-2a_0$$
Comparing the coefficients of $r,4=-2a_1\iff a_1=?$
Comparing the constants, $a_1-2a_0=-2\implies a_0=?$
For infinite sum, establish $$\lim_{n\to\infty}f(n)=0$$
|
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|
Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
$$41-4\cdot17=1$$
So the inverse of $17$ is $-4$.
That is, the inverse of $17$ is $37$
Am I right?
|
For the first Q.
Let $17=x_1.$ For each $n,$ if $|x_n|>1,$ take positive integer $K_n$ such that $$|K_nx_n|<41<|(1+K_n)x_n| .$$ Take $x_{n+1}$ with $41>|x_{n+1}|$ where: ...
(i). If $41-|K_nx_n|<|(1+K_n)x_n|-41$ then $x_{n+1}\equiv K_nx_n \pmod {41}$...
(ii). Otherwise $x_{n+1}\equiv (1+K_n)x_n \pmod {41}.$
(This is faster than simply taking $x_{n+1}\equiv 41- K_nx_n \pmod {41},$ as it ensures that if $|x_n|>1$ then $|x_{n+1}|\leq |x_n|/2.)$
If $|x_n|=1$ then stop.
We have $17=x_1$ and $34=17\cdot 2<41<17\cdot 3=51$ so $K_1=2.$ We have $41-34<51-41$ so we take $x_2\equiv 17\cdot 2 \pmod {41}$ . That is, $x_2=-7$. Now we have $35=5|x_2|<41<6|x_2|=42$ so $K_2=5$. And $41-35>42-41$ so we take $x_3\equiv (1+K_2)x_2 \pmod {41}.$ That is, $x_3= -1.$
Now modulo 41 we have $-1\equiv (1+k_2)\cdot k_1\cdot 17 \equiv 6\cdot 2\cdot 17 \equiv 12\cdot 17.$ Therefore $1\equiv (-12)17\equiv 29\cdot 17.$
|
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|
points on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is minimum
Find point on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is, minimum
$\bf{My\; Try::}$ Let $(x,y)$ be any point on the curve $x^2+2y^2=6\;,$ Then we have to
minimize $\displaystyle \left|\frac{x+y-7}{\sqrt{2}}\right|$
Using $\bf{Cauchy\; Schwarz}$ Inequality
$$\left[x^2+\left(\sqrt{2}y\right)^2\right]\cdot \left[1^2+\left(\frac{1}{\sqrt{2}}\right)^2\right]\geq (x+y)^2$$
So $$6\times \frac{3}{2}\geq (x+y)^2\Rightarrow (x+y)^2\leq 3^2\Rightarrow-3 \leq (x+y)\leq 3$$
So We get $$\frac{-3-7}{\sqrt{2}}\leq \frac{x+y-7}{\sqrt{2}}\leq \frac{3-7}{\sqrt{2}}\Rightarrow -5\sqrt{2}\leq \frac{x+y-7}{\sqrt{2}}\leq -2\sqrt{2}$$
and equality hold when $\displaystyle \frac{x}{1} = \frac{2y}{1}$
Now i did not understand how can i calculate $\displaystyle \min \left|\frac{x+y-7}{\sqrt{2}}\right|,$ Help me thanks
|
WLOG any point on $x^2+2y^2=6$ can be P$(\sqrt6\cos t,\sqrt3\sin t)$
Now the distance of $P$ from $x+y-7=0$ is $$\dfrac{|\sqrt6\cos t+\sqrt3\sin t-7|}{\sqrt{1^2+1^2}}=\dfrac{\left|3\cos\left(t-\arccos\dfrac{\sqrt6}3\right)-7\right|}{\sqrt2}$$
Now $-1\le\cos\left(t-\arccos\dfrac{\sqrt6}3\right)\le1$
$\implies-3-7\le\sqrt6\cos t+\sqrt3\sin t-7\le3-7$
$\implies4\le|\sqrt6\cos t+\sqrt3\sin t-7|\le10$
|
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|
Integral of $x^2/(x^2 + \alpha^2)^2$ I want to find the indefinite integral
$$\int\frac{x^2}{(x^2 + \alpha^2)^2}\,dx.$$
The answer (from Mathematica) is
$$-\frac{x}{2(x^2+\alpha^2)} + \frac{1}{2\alpha}\tan^{-1}\left(\frac{x}{\alpha}\right).$$
I can tell from the answer that there must be some integration by parts and some sort of trig substitution, but I can't seem to get it to work. Any ideas?
|
The usual method of trigonometrical substitution would suggest to put
\begin{align*}
x&=a\cdot \tan\theta\\
dx&=a\cdot \sec^2\theta\,d\theta
\end{align*}
and go on. We obtain
\begin{align*}
\int \dfrac{x^2}{(x^2+a^2)^2}\,dx&=\int \dfrac{a^2\tan^2\theta}{(a^2\tan^2\theta+a^2)^2}\,a\sec^2\theta\,d\theta\\
&=\int \dfrac{a^2\tan^2\theta}{(a^2(1+\tan^2\theta))^2}\,a\,\sec^2\theta\,d\theta\\
&=\int \dfrac{a^3\tan^2\theta \sec^2\theta}{a^4(\sec^2\theta)^2}\,d\theta\\
&=\int \dfrac{\tan^2\theta}{a\sec^2\theta}\,d\theta\\
&=\dfrac{1}{a}\int \dfrac{\dfrac{\sin^2\theta}{\cos^2\theta}}{\dfrac{1}{\cos^2\theta}}d\theta\\
&=\dfrac{1}{a}\int \sin^2\theta\,d\theta\\
&=\dfrac{1}{a}\int \dfrac{1-cos(2\theta)}{2},d\theta\\
&=\dfrac{1}{2a}\theta - \dfrac{1}{4a}\sin(2\theta)+C.
\end{align*}
To recover the original variables we do:
\begin{align*}
\tan\theta=\dfrac{x}{a} &\Rightarrow \sin\theta=\dfrac{x}{\sqrt{x^2+a^2}} \text{ and } \cos\theta=\dfrac{a}{\sqrt{x^2+a^2}}\\
&\Rightarrow \sin(2\theta)=2\sin\theta\cos\theta=\dfrac{2ax}{x^2+a^2}\\
&\\
\tan\theta=\dfrac{x}{a} &\Rightarrow \theta=arctan\left(\dfrac{x}{a}\right)
\end{align*}
Obtaining the following:
\begin{align*}
\int \dfrac{x^2}{(x^2+a^2)}\,dx&=-\dfrac{1}{4a}\sin(2\theta) + \dfrac{1}{2a}\theta + C\\
&=-\dfrac{1}{4a}\dfrac{2ax}{x^2+a^2}+\dfrac{1}{2a}\arctan\left(\dfrac{x}{a}\right)+C\\
&=-\dfrac{x}{2(x^2+a^2)}+\dfrac{1}{2a}\arctan\left(\dfrac{x}{a}\right)+C
\end{align*}
|
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|
Finding the nullspace of a square matrix I wish to find the nullspace of $(A-2I)$, where $A$ is given below, as part of finding the kernel of a linear transformation.
And the answer provided is
I am able to obtain the vector on the left, but I am not too sure how the vector on the right was acquired. I do not require an answer containing the exact computation. A brief explanation of the main steps involved is sufficient.
|
You're computing the eigenspaces, the characteristic polynomial of the matrix is
$$
(3-X)(2-X)^2
$$
so the eigenvalues are $3$ and $2$. The eigenspace relative to $2$ is the null space of
$$
\begin{pmatrix}
3-2 & 0 & -1 \\
0 & 1-2 & 1 \\
0 & -1 & 3-2
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & -1 \\
0 & -1 & 1 \\
0 & -1 & 1
\end{pmatrix}
$$
and a Gaussian elimination gives
$$
\begin{pmatrix}
1 & 0 & -1 \\
0 & -1 & 1 \\
0 & -1 & 1
\end{pmatrix}
\to
\begin{pmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{pmatrix}
$$
so the equations for the eigenvectors are
$$
\begin{cases}
x_1=x_3\\[4px]
x_2=x_3
\end{cases}
$$
so the eigenspace has dimension $1$ and it is generated by
$$
\begin{pmatrix}1\\1\\1\end{pmatrix}
$$
The eigenspace relative to $3$ is the null space of
$$
\begin{pmatrix}
3-3 & 0 & -1 \\
0 & 1-3 & 1 \\
0 & -1 & 3-3
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 & -1 \\
0 & -2 & 1 \\
0 & -1 & 0
\end{pmatrix}
$$
A Gaussian elimination gives the equation $x_2=x_3=0$, so the eigenspace is generated by
$$
\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}
$$
|
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|
Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.
Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.
Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2<n^7-n^3=\text{r.h.s}.$ This means that either $p>n^2$ or $p<n^2$ must hold.
Assuming $p>n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p>n^2)$.
Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.
Finally, we get $(5,2)$ as the solution to the given equation.
But how do I conclude this is the only solution possible? Also, why does'nt $p<n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem.
|
We have $$p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$$ so, clearly, $n=1$ and $n\ge p$ are discarded hence one has $1\lt n\lt p$ and $(n,p)=1$ and $p$ neither divides $n^3$ nor $n-1$. Besides the possibility $p=n+1$ is easily discarded.
It follows $p$ divides $n^2+1$ and since $n^2+1$ neither divides $p-1$ nor $p+1$ then we get $$p=n^2+1$$
which gives immediately the solution $(p,n)=(5,2)$.
That this solution is the only one is deduced putting the value of $p$ in the given equality so we have
$$n^2(n^2+1)(n^2+2)=n^7-n^3\iff n^5-n^4-3n^3-n-2=0$$ this last equation has as only real root $n=2$ (the other roots are $\pm i$ and $\pm \sqrt[3]{-1}$).
Thus $(p,n)=(5,2)$ is the only solution.
|
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|
To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to directly expand the determinant before greatly simplifying it).
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}= (ab+bc+ca)^3$$
I tried everything:
$R_1\to R_1+R_2+R_3$ and similar transformations to extract that $ab+bc+ca$ term, but to no avail. $C_2\to C_1+C_2$ and $C_3\to C_3+C_1$ seemed to be a good lead, but I couldn't follow up.
How can I solve this question?
|
It is easy to check that the equality holds if $bc=0$, $ca=0$ or $ab=0$.
Let $bc=x$, $ca=y$ and $ab=z$. Then the determinant in L.H.S. can be written as
\begin{align*}
\begin{vmatrix} -x & x+zx/y & x+xy/z \\ y+yz/x & -y & y+xy/z \\ z+yz/x & z+zx/y & -z \end{vmatrix}&=\frac{1}{xyz}\begin{vmatrix}-x^2 & zx+xy & zx+xy \\ yz+xy & -y^2 & yz+xy\\ zx+yz & zx+yz & -z^2\end{vmatrix}\\
&=\begin{vmatrix}-x & y+z & y+z\\z+x&-y&z+x\\x+y&x+y&-z\end{vmatrix}\\
&=\begin{vmatrix}x+y+z & x+y+z & x+y+z\\z+x&-y&z+x\\x+y&x+y&-z\end{vmatrix}\\
&=(x+y+z)\begin{vmatrix}1 & 1 & 1\\z+x&-y&z+x\\x+y&x+y&-z\end{vmatrix}\\
&=(x+y+z)\begin{vmatrix}0 & 1 & 0\\x+y+z&-y&x+y+z\\0&x+y&-(x+y+z)\end{vmatrix}\\
&=(x+y+z)^3
\end{align*}
|
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|
what kind of test should I use to determine if this sequence converge or diverge? what kind of test should I use to determine if this sequence converge or diverge?
$$\sum_{n=1}^\infty \left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right)$$
|
$$\sum_{n=1}^\infty \left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right) = \left(\sin1-\sin \frac{1}{2}\right)+ \left(\sin\frac{1}{2}-\sin \frac{1}{3}\right)+\cdots+\left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right) = \sin1+\left(-\sin \frac{1}{2}+ \sin\frac{1}{2}\right)+\left(-\sin \frac{1}{3}+\cdots+\sin\frac{1}{n}\right)-\sin \frac{1}{n+1}=\lim_{n\to \infty} \left(\sin1-\sin \frac 1 n\right) = \sin 1$$
|
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|
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$ x^2+4x+3 $$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|}
\hline
x^2 & \\
\hline
& 3 \\
\hline
\end{array}
Then find HCF of 3:
$$3\\
1 | 3
$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|}
\hline
x^2 & 1x \\
\hline
3x& 3 \\
\hline
\end{array}
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=3(x+3)
$$
Therefore:
$$x^2+4x+3=(x+1)(x+3)$$
It follows that:
$$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
|
Well done. And for those of us unaware of the box method, the following would have also worked: $$x^2+4x+3=x^2+4x+4-1=(x+2)^2-1=(x+2-1)(x+2+1)=(x+1)(x+3)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $ This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$
having only the left-hand side?
|
To denest the nested radical $\sqrt{3+2\sqrt{2}}$, we have a useful formula. Namely
Given a radical of the form $\sqrt{X\pm Y}$, we can rewrite it into$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}\tag1$$
With $X>Y$ and $X,Y\in\mathbb{R}$.
In $\sqrt{3+2\sqrt{2}}$, we have $X=3,Y=\sqrt8$. Using $(1)$, we get$$\sqrt{3+2\sqrt{2}}=\left(\frac {3+1}2\right)^{1/2}+\left(\frac {3-1}2\right)^{1/2}=\sqrt2+1\tag2$$
From here, can you continue?
|
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|
Equation of a line making equal and positive intercepts Q. Find the equation of line passing through the pt of intersection of the lines- $3x-y=5$ and $x+3y=1$ and making equal and positive intercepts on both the axes.
A. To find the pt of intersection substitute $x=1-3y$ in $3x-y=5$. Thus, $y= \frac{-1}{5}$ and $x= \frac{8}{5}$.
Thus, $b(x-h)-a(y-k)=0$ where a and b are the x and y intercepts respectively. And h and k the pt of intersection the x and y axis respectively.
Since, $a=b=z(say)$-
$$z[x-h-y+k]=0$$
$$x- \frac{8}{5} -y - \frac{1}{5}=0$$
$$5x-5y-7=0$$
But the answer key says it to be $5x+5y-7=0$
Also, I would like the derivation of the formula
$b(x-h)-a(y-k)=0$ which was simply stated in my book.
|
Hint:
A line passing thorough the point $P=(\frac{8}{5},\frac{-1}{5})$ (and not parallel to the $y$ axis) has equation:
$$
y+\frac{1}{5}=m\left(x-\frac{8}{5}\right)
$$
and, if this line has equals positive intercepts with the $x$ and $y$ axis, than $m=-1$, so it has equation:
$$
y+\frac{1}{5}+\left(x-\frac{8}{5}\right)=0
$$
|
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|
Limit of a hyperbolic trig function inside a square root I am asked to find this limit here:
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
I combined the terms to get
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
But if I try and factor out terms, I get
$$\lim_{x\to\infty} \frac{x^2\sqrt{1+\frac{\tanh(x)}{x}+\frac{1}{x^{2}}}-x^2(1+\frac{1}{x})}{x(1+\frac{1}{x})}$$
and that won't cancel with the x which I have on the bottom so the limit just blows up. Did I make a mistake somewhere?
The limit is apparently $\frac{-1}{2}$ but i'm not sure how that's the case. Thanks.
|
I do not knwo how much this is valid or not.
When $x\to\infty$, $\tanh(x)\to 1$. So, it seems to me that $$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x=\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}+x^2}}{x+1}-x$$ Now, $$\sqrt{x^4+x^{3}+x^2}=x^2\sqrt{1+\frac 1x+\frac 1 {x^2}}$$ Now, using $$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ and replacing $\epsilon$ by $\left(\frac 1x+\frac 1 {x^2}\right)$, $$\sqrt{1+\frac 1x+\frac 1 {x^2}}=1+\frac{1}{2 x}+\frac{3}{8 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\sqrt{x^4+x^{3}+x^2}=x^2\sqrt{1+\frac 1x+\frac 1 {x^2}}=x^2+\frac{x}{2}+\frac{3}{8}+O\left(\frac{1}{x}\right)$$ Now, using long division to get $$\frac{\sqrt{x^4+x^{3}+x^2}}{x+1}=x-\frac{1}{2}+\frac{7}{8 x}+O\left(\frac{1}{x}\right)$$ $$\frac{\sqrt{x^4+x^{3}+x^2}}{x+1}-x=-\frac{1}{2}+\frac{7}{8 x}+O\left(\frac{1}{x}\right)$$ Using $x=10$, the "exact" value is $\approx -0.422133$ while the above asymptotics gives $-\frac{33}{80}=-0.4125$
|
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|
Quadratic equation 4 need some guidance with a quadratic equation.
Suppose $x^2+20x-4000=0$
Here is what I have done so far;
Using the Quadratic Equation $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where from the above equation, $a=1$, $b=20$, $c=-4000$, we find
$$\begin{align*}x&=\frac{-20+\sqrt{(20)^2-(4)(1)(-4000)}}{2\times 1}\\
&=\frac{-20+\sqrt{16400}}{1}\\
&=108.0624847 \end{align*}$$
and
$$\begin{align*} x&=\frac{-20-\sqrt{20^2-(4)(1)(-4000)}}{2\times1}\\
&=\frac{-20-\sqrt{16400}}{1} \\
&=-148.0624847 \end{align*}$$
Neither of these values for $x$ prove correct when applied to $x^2+20x-4000=0$.
I know something is wrong but I can’t figure out what. Any guidance would be appreciated.
|
$2\times 1=2$, not $1$. In the denominator.
|
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|
Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
|
$x^8+x^6−4x^4+x^2+1$
$=(x^8−2x^4+1)+(x^4-2x^2+1)x^2$
$=(x^4-1)^2+(x^2-1)^2x^2$
$≥0 $
|
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|
Primitive $6^{th}$ root of unity Why does any extension of $\mathbb{Q}$ that contains a root of $x^6+3$ have to contain a primitive $6^{th}$ root of unity? The roots of this equation are $\{\sqrt[6]{3}e^{\pi i/2},\sqrt[6]{3}e^{5\pi i/6},\sqrt[6]{3}e^{7\pi i/6},\sqrt[6]{3}e^{3\pi i/2},\sqrt[6]{3}e^{11\pi i/6},e^{13\pi i/6}\}$. A primitive $6^{th}$ root of unity has the form $\frac{1}{2}+\frac{\sqrt{3}}{2}i$...
I don't see how to get $\frac{1}{2}+\frac{\sqrt{3}}{2}i$ from $\sqrt[6]{3}e^{\pi i/2}$, for example.
|
Let $\alpha$ be such a root, then $\beta=\alpha^3$ satisfies $\beta^2+3=0$, hence ${1+\beta\over 2}={1+\alpha^3\over 2}$ is a $6^{th}$ root of $1$.
You can even verify this directly:
$$(1+\beta)^6 = 1+6\beta+15\beta^2+20\beta^3+15\beta^4+6\beta^5+\beta^6$$
Now use that $\beta^2=-3$ to reduce this to:
$$(1+5\cdot(-3)+15\cdot(-3)^2+(-3)^3)+\beta(6+20\cdot(-3)+6\cdot(-3)^2)$$
$$=(1-45+135-27)+\beta(6-60+54)= 64+0\beta $$
$$= 2^6.$$
And this is primitive as if we only cube things we get
$$(1+\beta)^3 = 1+3\beta + 3\beta^2+\beta^3 = 1-9 + \beta(3-3) = -8=-2^3.$$
|
{
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|
Choosing a basis for a subspace in $\mathbb{R}^{3\times 3}$ and finding a coordinate vector with respect to it We observe the subspace known as $U\subset \mathbb{R}^{3\times 3}$ that is spanned by the following vectors:
$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
$, $
\begin{bmatrix}
0 & -3 & 0 \\
0 & 2 & 0 \\
0 & -1 & 0 \\
\end{bmatrix}
$, $
\begin{bmatrix}
0 & 0 & 1 \\
0 & -2 & 0 \\
3 & 0 & 0 \\
\end{bmatrix}
$, $
\begin{bmatrix}
0 & 0 & 0 \\
-1 & 2 & -3 \\
0 & 0 & 0 \\
\end{bmatrix}
$.
Choose a basis for $U$ and show that $
\begin{bmatrix}
2 & -3 & -2 \\
-3 & 8 & -9 \\
-6 & -1 & 6 \\
\end{bmatrix}
\in U$.
Determine the coordinate vector for this vector with respect to the chosen basis for $U$.
Hints to how I solve this?
|
We have:
$$ (1) \qquad a \begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
+b \begin{bmatrix}
0 & -3 & 0 \\
0 & 2 & 0 \\
0 & -1 & 0 \\
\end{bmatrix}
+c \begin{bmatrix}
0 & 0 & 1 \\
0 & -2 & 0 \\
3 & 0 & 0 \\
\end{bmatrix}
+d \begin{bmatrix}
0 & 0 & 0 \\
-1 & 2 & -3 \\
0 & 0 & 0 \\
\end{bmatrix}
=\begin{bmatrix}
a & -3b & c \\
-d & -2(a-b+c-d) & -3d \\
3c & -b & 3a \\
\end{bmatrix} =
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
$$
iff $ a=b=c=d=0$. So these matrices are linearly independent and they are a basis $\mathcal{U}$ fo $U$.
And, for $(a,b,c,d)=(2,1,-2,3)$ the linear combination $(1)$ gives the matrix
$$M=
\begin{bmatrix}
2 & -3 & -2 \\
-3 & 8 & -9 \\
-6 & -1 & 6 \\
\end{bmatrix}
$$
so $(2,1,-2,3)$ are the components of this matrix in the basis $\mathcal{U}$.
|
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|
Axes of Symmetry for a General Ellipse
Given a general ellipse
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
where $B^2<4AC$, what are the major and minor axes of symmetry in the form $ax+by+c=0$?
It is possible of course to first work out the angle of rotation such that $xy,x,y$ terms disappear, in order to get an upright ellipse of the form $x^2/p^2+y^2/q^2=1$ and proceed from there. This may involve some messy trigonometric manipulations.
Could there be another approach perhaps, considering only quadratic/diophantine and linear equations?
Addendum
Here's a graphical implementation based on the answer by Ng Chung Tak.
Addendum 2
Based on the answers by amd and by Ng Chung Tak, the equations for the axes are
$$\color{red}{\left(y-\frac {2AE-BD}{B^2-4AC}\right)=\frac {C-A\pm \sqrt{(A-C)^2+B^2}}B\left(x-\frac {2CD-BE}{B^2-4AC}\right)}$$
Note that
$$\frac{C-A\pm \sqrt{(A-C)^2+B^2}}B\cdot \color{lightgrey}{\frac {C-A\mp\sqrt{(A-C)^2+B^2}}{C-A\mp\sqrt{(A-C)^2+B^2}}}=-\frac B{C-A\mp\sqrt{(A-C)^2+B^2}}$$
i.e. it is equal to the negative of its own reciprocal. Hence the equations for the axes can also be written as
$$\color{red}{\left(x-\frac {2CD-BE}{B^2-4AC}\right)=-\frac {C-A\mp \sqrt{(A-C)^2+B^2}}B\left(y-\frac {2AE-BD}{B^2-4AC}\right)}$$
hence the two symmetrical anti-symmetrical forms for the axes. Here's the graphical implementation.
|
The trigonometric manipulations are not messy.
After centering, the equation is, in polar coordinates,
$$\rho^2=-\frac{F'}{A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta}.$$
The denominator
$$A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta=\frac12\left((A-C)\cos\frac\theta2+B\sin\frac\theta2\right)$$ achieves its extrema when
$$(A-C)\sin\frac\theta2=B\cos\frac\theta2,$$ which is elementary.
Update:
Read $\color{green}{2\theta}$, not $\color{red}{\dfrac\theta2}$ !
$$A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta=\frac12\left((A-C)\cos2\theta+B\sin2\theta+(A+C)\right)$$
and
$$(A-C)\sin2\theta=B\cos2\theta.$$
|
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|
Solve the equation $x^7- 2ix^4 - ix^3 -2 = 0$ for $x$ I am having difficulty factorising the equation.
|
$$x^7-ix^3-2ix^4-2=0\\(x^7-ix^3)-2(ix^4+1)=0\\
x^3(x^4-i)-2(ix^4+1)=0\\\frac{i}{i}x^3(x^4-i)-2(ix^4+1)=0\\
\frac{1}{i}x^3(ix^4-i^2)-2(ix^4+1)=0\\
\frac{1}{i}x^3(ix^4+1)-2(ix^4+1)=0\\
(\frac{x^3}{i}-2)(ix^4+1)=0\\
(\frac{x^3-2i}{i})(ix^4+1)=0\\ \to
(x^3-2i)(ix^4+1)=0\\$$
|
{
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|
Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
|
After proceeding with the polar coordinate, it suffices to evaluate the following: $$I = 4\int_{0}^{\frac{\pi}{2}}\dfrac{d\theta}{2-\sin^2(2\theta)}= 2\int_0^{\pi}\dfrac{dx}{2-\sin^2x}$$.
Now do a tangent substitution: $$\tan x = t\Rightarrow \sin^2x = \dfrac{t^2}{1+t^2},\,\, dx = \dfrac{dt}{1+t^2}$$
Finally, $$I = 4\int_0^{\infty}\dfrac{dt}{2+t^2} =2\sqrt{2}\cdot(\arctan(\infty) - \arctan(0)) = \sqrt{2}\pi = 4.4429...$$
Or this is also a routine exercise on complex contour integration using the Euler identity $\sin x = \dfrac{e^{ix}-e^{-ix}}{2i}$.
|
{
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|
Finding volume of a solid over region E, bounded below by cone $z = \sqrt{x^2 + y^2}$ and above by sphere $ z^2 = 1- x^2 - y^2$ I set up the triple integral as follows:
$v=\int^1_0\int^{\sqrt{x-1}}_{-\sqrt{x}}\int^{\sqrt{1-x^2-y^2}}_{\sqrt{x^2+y^2}}(1)dzdydx$ Then went to polar coords getting $$\int^{2\pi}_0\int_0^1(\sqrt{1-r^2}r=r^2)drd\theta$$$$2\pi(\frac{-1}{2})[\frac{2}{3}(1-r^2)^{3/2}-1]|^1_0=-\pi[(0-1)-(\frac{2}{3}-1)]=\pi+\frac{1}{3}$$
The answer should be $\frac{\pi}{3}(2-\sqrt{2})$
|
The region $E$ in cylindrical coordinate is given by
$$ E = \left\{(r,\theta,z)\colon 0\le r\le 1/\sqrt{2}, 0\le\theta\le 2\pi, r\le z\le \sqrt{1-r^2}\right\}. $$
To see this, observe that $z$ is bounded below by the given cone means that
$$ z\ge \sqrt{x^2+y^2} = r,$$
and $z$ is bounded above by the sphere of radius 1 means that
$$ z\le\sqrt{1-x^2-y^2} = \sqrt{1-r^2}.$$
Moreover, the cone intersects the sphere when $x^2+y^2=z^2=1-x^2+y^2$, i.e.
$$r^2=1-r^2\implies 2r^2=1\implies r=\pm\frac{1}{\sqrt{2}}.$$
Thus, the volume of solid over $E$ is given by
\begin{align*}
V = \int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\int_r^{\sqrt{1-r^2}} rdzdrd\theta & = 2\pi \int_0^{\frac{1}{\sqrt{2}}}\int_r^{\sqrt{1-r^2}} rdzdr\\
& = 2\pi\int_0^{\frac{1}{\sqrt{2}}} \left(r\sqrt{1-r^2} - r^2\right)\, dr.
\end{align*}
The second integral is
$$ \int_0^{\frac{1}{\sqrt{2}}} r^2\, dr = \frac{r^3}{3}\bigg|_0^{\frac{1}{\sqrt{2}}} = \frac{1}{6\sqrt{2}}$$
while the first integral is (upon integration by substitution)
\begin{align*}
\int_0^{\frac{1}{\sqrt{2}}} r\sqrt{1-r^2}\, dr & = -\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)(1-r^2)^{3/2}\bigg|_0^{\frac{1}{\sqrt{2}}}\\
& = -\frac{1}{3}\left(\frac{1}{2\sqrt{2}}-1\right)\\
& = \frac{2\sqrt{2}-1}{6\sqrt{2}}.
\end{align*}
Hence,
\begin{align*}
V=2\pi\left(\frac{2\sqrt{2}-1}{6\sqrt{2}}-\frac{1}{6\sqrt{2}}\right) & = 2\pi\left(\frac{2\sqrt{2}-2}{6\sqrt{2}}\right) \\
& = 2\pi\left(\frac{\sqrt{2}-1}{3\sqrt{2}}\right) \\
& = \frac{2\pi}{3}\left(1-\frac{1}{\sqrt{2}}\right) \\
& = \frac{\pi}{3}(2-\sqrt{2}).
\end{align*}
|
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|
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
|
$\frac{5}{x+3}$+$\frac{2}{2(x+3)}$=4
$\frac{(5*2+2)}{2(x+3)}$=4
12=8(x+3)
x=$\frac{-12}{8}$
x=$\frac{-3}{2}$
|
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|
Calculating $\int_0^{\pi/3}\cos^2x+\dfrac{1}{\cos^2x}\mathrm{d}\,x$ I've been given the following exercise:
Show that the exact value of $$\int_0^{\pi/3}\cos^2x+\frac{1}{\cos^2x}\,\mathrm{d}x = \frac{\pi}{6}+\frac{9}{8}\sqrt{3}$$
Can someone help me with this?
|
$\displaystyle \int \cos^2 x+\dfrac{1}{\cos^2 x}\,dx$
$=\displaystyle \int \dfrac{1}{2}(1+\cos 2x)+\sec^2 x\,dx$
$=\dfrac{1}{2}x+\dfrac{1}{4}\sin 2x+\tan x+C$
So integrating between the limits
$\left[0,\dfrac{\pi}{3}\right]$
$=\dfrac{\pi}{6}+\dfrac{1}{4}\sin \dfrac{2\pi}{3}+\tan \dfrac{\pi}{3}-0$
$=\dfrac{\pi}{6}+\dfrac{1}{4}\times \dfrac{\sqrt{3}}{2}+\sqrt{3}$
$=\dfrac{\pi}{6}+\sqrt{3}\left(\dfrac{1}{8}+1\right)$
$=\dfrac{\pi}{6}+\dfrac{9}{8}\sqrt{3}$
|
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|
Prove that $ \lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2} $ $$\lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2}$$
Attempt:
According to me, it has to be $1$, since the outermost exponent tends to $0$.
But anyways, would taking the limit as equal to $L$. And taking ($\log$) on both sides help?
|
Let us consider $n=2$ first, for simplicity, and use $x^2$ instead of $\sin^2x$, as $x\to0$. As you suggested, since the function has the indeterminate form $\infty^0$ as $x\to0$, let us consider the logarithm of the original function
$$
x^2\log\left(1+2^{1/x^2}\right) = \frac{\log\left(1+2^{1/x^2}\right)}{1/x^2},
$$
which, as $x\to 0$, is of the form $\infty/\infty$. We can apply l'Hospital's rule getting
$$
\frac{\log 2}{1+2^{-1/x^2}},
$$
which tends to $\log 2$ as $x\to0$.
Therefore
$$
\lim_{x\to 0} \left(1+2^{1/x^2}\right)^{x^2}=2.
$$
In general, we may apply l'Hospital's rule to
$$
\frac{\log\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)}{1/x^2}
$$
getting
$$
\frac{\left(\frac{2}{n}\right)^{1/x^2}\log 2+\left(\frac{3}{n}\right)^{1/x^2}\log 3+\ldots+\log n}{1+\ldots + \left(\frac{3}{n}\right)^{1/x^2}+\left(\frac{2}{n}\right)^{1/x^2}+n^{-1/x^2}},
$$
which tends to $\log n$ as $x\to0$, proiving
$$
\lim_{x\to0}\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)^{x^2}=n.
$$
|
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|
Let $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$. Why does $a=b=c$? Let $a,b,c\in \mathbb{R}$ and $a,b,c \ne 0$ and $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$.
Why does $a=b=c$?
|
Let $a=cx$ and $b=cy$. The equation becomes $x^2y^2+y^2+x^2=xy(x+y+1)$, or
$$(y^2-y+1)x^2-(y^2+y)x+y^2=0$$
As a quadratic in $x$, the discriminant, which must be non-negative in order for $x$ to be real, is
$$(y^2+y)^2-4(y^2-y+1)y^2=-3y^2(y-1)^2$$
So if $y\not=0$, we must have $y=1$, which in turn implies $x=1$.
|
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|
Probability that the sum 3 integers is divisible by 3
I found the sample space, but i was having difficulty finding the favorable outcomes.
But how do i find numbers that add up to be divisible by 3?
|
Al integers are of the form $m=3i_m + k_m$ where $i_m$ is an integer and $k_m$ is either $0,1$ or $2$. As the the $3n$ numbers are consecutive the number of integers where $k_m$ is $0, 1$ or $2$ are equal.
The ways in which $a+b+c$ is divisible by three is that: i) all three have $k_a = k_b = k_c$ or that ii) $\{k_a,k_b,k_c\} = \{0,1,2\}$. (If $k_i = k_j=0,1,2$ then $k_i + k_j = 0,2,4$ and for $0,2,4 + k_l$ to be divisible by $3$ then $k_l = 0, 1, 2 = k+i=k_j$. So either they are all the same, or all different.)
The probability of i) $1*\frac{n-1}{3n-1}*\frac{n-2}{3n-2}$ (because $k_a$ may be anything. The probability of $k_b = k_a$ is $\frac{n-1}{3n-1}$ because there are $n-1$ numbers with the $k$ value of $k_a$ and there are $3n-1$ numbers left. And so on.)
The probability of ii) is $\frac{n}{3n}\frac{n}{3n-1}\frac{n}{3n-2}*3!$. There maybe be other ways to calculate this but $\frac{n}{3n}$ is the probability $k_a = 0$ and $\frac{n}{3n-1}$ is the probability $k_b = 1$ given that $k_a = 0$ and $\frac{n}{3n-2}$ is the probability that $k_c =2$ given that $k_a=0$ and $k_b= 1$ and $3!$ is the number of ways we can order $\{k_a,k_b,k_c\}$ to be equal to $\{0,1,2\}$.
As the events are distinct/independant the probability is:$1*\frac{n-1}{3n-1}*\frac{n-2}{3n-2} + \frac{n}{3n}\frac{n}{3n-1}\frac{n}{3n-2}*3!=\frac{(n-1)(n-2)}{(3n-1)(3n-2)} + \frac{2n^2}{(3n-1)(3n-2)}=\frac{3n^2 - 3n +2}{9n^2 - 9n + 2}= \frac{3n^2 - 3n +\frac 23}{9n^2 - 9n + 2}+ \frac 43*\frac1{(3n-1)(3n-2)}=\frac 13 + \frac{4}{3(3n-1)(3n-2)}=1 - \frac{6n(n-1)}{(3n-1)(3n-2)}$ .
For $n=1$ then probability is $1$. And as $n$ increases the probability decreases and gets closer to $1/3$.
|
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|
Reflexion of an element on $P_1(\Bbb R)$ Let $p, q \in P_3(\Bbb R)$. Define the inner product by
$$\langle p(x), q(x)\rangle = \int_0^1 p(x) q(x) \, dx $$
What is the reflection of $q(x) = x ^ 2$ on the subspace $P_1(\Bbb R)$?
My result is $R(q)(x) = \frac{2}{3} + \frac{3x}{2} - x^2 $
But the answer in the book is $ R(q)(x) = -\frac{1}{3} + 2x - x^2$
I don't what is the right procedure to this problem.
|
You have to start with an orthogonal base for $P_1(\Bbb R)$ in order to calculate the reflection.
*
*We take $1$ as the first element of our orthogonal base for $P_1(\Bbb R)$.
*
*$\lVert1\rVert^2 = \langle1,1\rangle = 1$ is obvious.
*We apply the Gram-Schmidt process on $x$ and $x^2$ to obtain the second and third element in this basis. At this stage, we'll have the projection of $q(x)=x^2$ on $P_1(\Bbb R)$, denoted by $\pi(q(x))=\pi(x^2)$.
*
*$x-\dfrac{\langle x,1 \rangle}{\lVert1\rVert^2}\cdot 1 = x - \dfrac{\int_0^1 x\,\mathrm{d}x}{1} = x - \dfrac12$, which is the second element in our orthogonal basis.
*Find $\pi(q(x))$.
\begin{align}
& \pi(q(x)=\pi(x^2) \\
=& \frac{\langle x^2, 1\rangle}{\langle 1, 1\rangle} \cdot 1 + \frac{\langle x^2, x - \frac12 \rangle}{\langle x - \frac12, x - \frac12 \rangle} \cdot \left(x - \frac12\right) \\
=& \frac{\int_0^1 x^2\,\mathrm{d}x}{1} \cdot 1 + \frac{\int_0^1\left(x^3-\frac12 x^2\right)\,\mathrm{d}x}{\int_0^1 \left(x - \frac12\right)^2\,\mathrm{d}x} \cdot \left(x - \frac12\right) \\
=& \frac{\frac13}{1}\cdot 1 + \frac{\left.\frac14 x^4 - \frac16 x^3 \right\rvert_0^1}{\left.\frac13\left(x-\frac12\right)^3\right\rvert_0^1}\cdot \left(x - \frac12\right) \\
=& \frac13 + \frac{\frac14-\frac16}{\frac23 \left(\frac12\right)^3} \cdot \left(x - \frac12\right) \\
=& x - \frac16
\end{align}
*We calculate
\begin{align}
R(q)(x) =& 2\pi(q(x))-x^2 \\
=& 2\pi(x^2)-x^2 \\
=& 2 \left(x - \frac16\right)-x^2 \\
=& -x^2 + 2x -\frac13.
\end{align}
|
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|
Mathematical Induction with Exponents: $1 + \frac12 + \frac14 + \dots + \frac1{2^{n}} = 2 - \frac1{2^{n}}$ Prove $1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}} = 2 - \frac{1}{2^{n}}$ for all positive integers $n$.
My approach was to add $\frac{1}{2^{n + 1}}$ to both sides for the induction step. However, I got lost in the algebra and could not figure out the rest of the proof. Any help would be greatly appreciated.
Thank you in advance.
|
Hint
if we add $\frac{1}{2^{n+1}}$ to the right side, we get
$$2-\frac{1}{2^n}+\frac{1}{2^{n+1}}$$
$$=2-\frac{2}{2^{n+1}}+\frac{1}{2^{n+1}}$$
$$2-\frac{1}{2^{n+1}}$$
qed.
|
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|
How to prove $(n+2)(n+1)k > (n-k+3)(n-k+2)(n-k+1)$ for $k\le n$ and $3k>n+1$ How to prove the following inequality:
$$(n+2)(n+1)k > (n-k+3)(n-k+2)(n-k+1)$$
where $k\le n$ and $3k>n+1$ for integers $n,k$.
|
Since $3k-1>k$, we obtain $k\geq1$.
In another hand, $k\leq n\leq3k-2$.
Let $f(n)=(n-k+3)(n-k+2)(n-k+1)-k(n+1)(n+2)$.
$f(3k-2)=-k(k-1)(k-2)\leq0$
and $f(k)=-k^3-3k^2-2k+6\leq0$.
$f'(x)=3x^2-(8k-12)x+3k^2-15k+11$,
which says that our inequality is proven for $3k^2-15k+11\geq0$
because there is also $x_1<0$ for which $f(x_1)=0$. (Draw a cubic parabola $y=f(x)$).
Thus, it remains to prove our inequality for $3k^2-15k+11<0$ or $k\in\{1,2,3,4\}$,
which gives that for $k=1$ or $k=2$ your inequality is wrong and for $k\in\{3,4\}$ it's true.
Done!
|
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|
Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$. The cosecant of $\frac{\pi}{3}$ is $\frac{2\sqrt{3}}{3}$. Using the negative, the answer is $-\frac{2\sqrt{3}}{3}$.
My questions:
is how do one tell how many revolutions it takes, based off the fraction they give?
And how does one find the cosecant of $\frac{\pi}{3}$? On the calculator I use sine and then take the reciprocal of sine, but sine come out to $.018276028$. And I'm not sure on how to get the reciprocal from that or even find the cosecant of that number.
|
Notice this. Quite simple approach
|
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|
Residue ${{(z+1)^2}\sin(1/z)}\over{(3z+1)^2}$
Calculate residue at $z=0$ of $\dfrac{{(z+1)^2\sin(1/z)}}{(3z+1)^2}$
It is essential singularity.So I am trying to get Laurent series expansion.Which in this case is hard because requires combing two expansions.
|
If you are familiar with the residue at infinity, you can reduce your problem to a simpler one. Set
$$ f(z) = \left( \frac{z+1}{3z+1} \right)^2 \sin \left( \frac{1}{z} \right). $$
The function $f$ has isolated singularities at $z = 0, -\frac{1}{3}, \infty$. Let us calculate the residue at infinity first. We have
$$ -\frac{1}{z^2} f \left( \frac{1}{z} \right) = -\frac{1}{z^2} \left( \frac{\frac{1}{z} + 1}{\frac{3}{z} + 1} \right)^2 \sin(z) = -\frac{1}{z^2} \left( \frac{1+z}{3 + z} \right)^2 \sin(z). $$
Since $\left( \frac{1+z}{3+z} \right)^2 \sin(z)$ is holomorphic at $z = 0$, we see that the residue at infinity is zero. Since the total sum of residues is always zero, we have
$$ \operatorname{Res}(f, 0) = -\operatorname{Res} \left( f, -\frac{1}{3}
\right) $$
so it is enough to calculate $\operatorname{Res} \left( f, -\frac{1}{3}
\right)$. We have
$$ f(z) = \frac{(z+1)^2}{9} \sin \left( \frac{1}{z} \right) \frac{1}{ \left( z + \frac{1}{3} \right)^2} = \frac{g(z)}{\left( z + \frac{1}{3} \right)^2}$$
for $g$ which is holomorphic at $z = -\frac{1}{3}$. Hence, the residue $\operatorname{Res} \left( f, -\frac{1}{3}
\right)$ will be $g' \left( -\frac{1}{3} \right)$ which can be calculated directly.
|
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|
How do you find the derivative of $\sqrt{x^2+9}$ using the definition of a derivative? I had this question on a not-so-recent math test. It asked to find the derivative of $\sqrt{x^2+9}$ with the definition of a derivative. Using the chain rule, I can figure out that the derivative is $\frac x{\sqrt{x^2+9}}$, but how can it be done with only the definition of a derivative? I tried multiplying both sides of the fraction by the square roots, but that just makes a mess of everything.
|
So we have $f(x) = \sqrt{x^2 +9}$.
Then using the definition of the derivative, we have $f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)^2 +9} - \sqrt{x^2+9}}{h} =\lim_{h \to 0} \frac{\sqrt{(x+h)^2 +9} - \sqrt{x^2+9}}{h} * \frac{\sqrt{(x+h)^2 +9}+ \sqrt{x^2 +9}}{\sqrt{(x+h)^2 +9} + \sqrt{x^2+9}} = \lim_{h \to 0} \frac{2xh + h^2}{h(\sqrt{(x+h)^2 +9} + \sqrt{x^2 +9})}= \lim_{h \to 0} \frac{2x + h}{\sqrt{(x+h)^2 +9} + \sqrt{x^2 +9}}$.
As $h \to 0$, the limit tends to $\frac{2x}{2\sqrt{x^2 +9}}= \frac{x}{\sqrt{x^2 +9}}$.
|
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|
Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be observed that for $x=-1$ the condition of $n$ being a positive integer is satisfied. But I could not get others. Moreover to satisfy that $n$ is positive we must have $x<-3/4$.
Please help me in this regard. Thanks.
|
$x = 0$ cannot be a root so suppose $x > 0$.
The equation is equivalently written as $nx^3+3/x = -4$.
But by AM GM, $nx^3+3/x \ge 4 \sqrt[4]{n} > -4$ so no positive roots.
Likewise, let $x = -r$ where $r$ is a positive number.
Then we have $nr^4-4r+3 = 0$ or $nr^3+3/r = 4$.
Again by AM GM, we have $nr^3+3/r \ge 4\sqrt[4]{n} \ge 4$ with equality at $n = 1$
which is the unique positive integer.
|
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|
Find all complex numbers satisfying the equation $\bar{z}+1=iz^2+|z|^2$ Find all complex numbers z satisfying $$\bar{z}+1=iz^2+|z|^2$$
where $i=\sqrt{-1}$
I only know one way i.e. assuming $z=x+iy$ but that process is very cumbersome. I don't know how to proceed otherwise with a shorter approach.
|
I do not find it so cumbersome to replace $z=x+iy$ if you write out the details (it looks like it might be a mess, and, without trying it out, it's hard to see where the problem simplifies). Substituting $z=x+iy$ into the equation gives
$$
(x+1)+(-y)i=(x^2+y^2-2xy)+(x^2-y^2)i=(x-y)^2+(x^2-y^2)i.
$$
Therefore
\begin{align*}
x+1&=(x-y)^2\\
-y&=x^2-y^2
\end{align*}
From here, you have that $x^2=y^2-y$. You could, at this point, take the first equation and rewrite it as
$$
x+2xy=x^2+y^2-1.
$$
Factoring the LHS gives $x(1+2y)$ and substituting the formula for $x^2$ into the RHS gives $2y^2-y-1=(2y+1)(y-1)$. So, this equation simplifies to:
$$
x(1+2y)=(1+2y)(y-1).
$$
Moving all the terms to one side, we get
$$
(1+2y)(y-x-1)=0.
$$
Therefore, either $1+2y=0$ or $y=1+x$. In the first case, $y=-\frac{1}{2}$, we can find $x$ via the equation $x^2=y^2-y$, and then check our answers are right in the original expression. If $y=1+x$, you can, again, substitute into $x^2=y^2-y$ to get a formula for $x$ (and then $y$). Whichever of these solutions satisfies the original equation are the answers that you're looking for.
|
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|
How to show $y^2,z^2,u^2$ are roots of $(1)$ Euler's solution for the general quartic are as follows:
From the depressed quartic, $x^4+px^2+qx+r=0$, assume $x=y+z+u$ and it may be shown that $u^2,y^2,z^2$ are roots of the cubic$$t^3+\dfrac q2t^2+\dfrac {q^2-4s}{16}t-\dfrac {r^2}{64}=0\tag1$$
$$\vdots$$
However, how would you go about showing that $u^2,z^2,y^2$ are roots of $(1)$? I originally started by writing $(1)$ as$$(t-u^2)(t-y^2)(t-z^2)\tag2$$
And expanding to get $t^3-(u^2+y^2+z^2)t^2+(u^2y^2+u^2z^2+y^2z^2)t-u^2y^2z^2$. However, now I don't know what to do. Any pointers on what to do now?
|
COMMENT.-Euler himself used the following equation to apply his method:
$$x^4-8x^3+14x^2+4x-8=0\tag1$$
I reproduce this solution because it seems to me illustrative for who likes the subject and of certain historical interest.
First a synthesis of the Euler method to solve the quartic.
$$***$$
*
*$Ax^4+Bx^3+Cx^2+Dx+E=0$ is reduced to $x^4-ax^2-bx-c=0$ through the change of variables $y=x-\dfrac{B}{4A}$ similar to the known change for the cubic equation.
*Euler assumes that a quartic of the form $x^4-ax^2-bx-c=0$ has a solution of the form $$x=\sqrt p+\sqrt q+\sqrt r\tag{2}$$
*Taking twice the square in $(2)$ he got
$$x^4-2fx^2-8\sqrt h x-(4g-f^2)=0$$ where $f=p+q+r$, $g=pq+pr+qr$ and $h=pqr$. (It is clear that $p,q,r$ are roots of $t^3-fx^2+gx-h=0$)
*Comparing (3) with (2) he got $f=\dfrac a2$, $h=\dfrac{b^2}{64}$ and $g=\dfrac{4c+a^2}{16}$
$$***$$
$$\color{blue}{\text{ Solve the equation } x^4-8x^3+14x^2+4x-8=0}$$
The reduced equation through $y=x+2$ is $$x^4+10x^2-4x+8=0$$
Since $f=\dfrac{10}{2}=5,h=\dfrac 14$ and $g=\dfrac{17}{4}$, $p,q,r$ are roots of the equation $$t^3-5t^2+\dfrac{17}{4}t-\dfrac 14=0\tag3$$ whose solutions are
$$p=1,\space q=\frac{4+\sqrt{15}}{2},\space r= \frac{4-\sqrt{15}}{2}\tag4$$
Hence
$$\sqrt p=\pm1,\space \sqrt q=\sqrt{\frac{4+\sqrt{15}}{2}},\space \sqrt r=\sqrt{\frac{4-\sqrt{15}}{2}}\tag5$$ Euler noticed that $(\sqrt5\pm\sqrt3)^2=\sqrt{8\pm 2\sqrt{15}}$ so he simplified
$$\sqrt p=\pm1,\space \sqrt q=\pm\frac{\sqrt5+\sqrt3}{2},\space \sqrt r=\pm\frac{\sqrt5-\sqrt3}{2}\tag6$$ Furthermore, because of $\dfrac q8=\sqrt h=\sqrt p\sqrt q\sqrt r\gt 0$, Euler had to give signs to the three square roots so that his product be positive which gives
$$\begin{align*} & \sqrt p+\sqrt q+\sqrt r=1+\sqrt5\\ & \sqrt p-\sqrt q-\sqrt r=1-\sqrt5\\ & -\sqrt p+\sqrt q-\sqrt r=-1+\sqrt5\\ & -\sqrt p-\sqrt q+\sqrt r=-1-\sqrt5\end{align*}\tag7$$ Thus, taking into account $y=x+2$ Euler got the four roots of his equation $$x_{1,2}=3\pm\sqrt5\\x_{3,4}=1\pm\sqrt3$$
|
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|
Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and should be added to fraction, while the book treats it as part of fraction itself, thus multiplying it with $n^2+5$.
So, I just want to understand which convention is correct.
This is from problem 6 in exercise 9.1 on page 180 of the book Sequences and Series.
Here is the answer sheet from the book (answer 6, 3rd element):
*
*$3,8,15,24,35$
*$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$
*$2, 4, 8, 16 \text{ and } 32$
*$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$
*$25,-125,625,-3125,15625$
*$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$
*$65, 93$
*$\dfrac{49}{128}$
*$729$
*$\dfrac{360}{23}$
*$3, 11, 35, 107, 323$; $3+11+35+107+323+...$
*$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$
*$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$
*$1,2,\dfrac{3}{5},\dfrac{8}{5}$
|
It just depends on context.
In some rare cases
$$
a\frac{c}{d}:=a+\frac{c}{d}
$$
which is the interpretation in your answer, but mostly
$$
a\frac{c}{d}:=a\cdot\frac{c}{d}
$$
|
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Find the pseudo-inverse of the matrix $A$ without computing singular values of $A$. Consider the following Least Square Minimization problem:$min_{x \in \mathbb C^3} |x_1 +x_3-3|^2+|x_2 -x_3|^2+|x_1+x_3-4|^2$
Find the pseudo-inverse of the matrix $A=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & -1 \\
1 & 0 & 1
\end{bmatrix}$
without computing singular values of $A$.
It seems we cannot use normal equations to denote $A^{+}=(A^{T}A)^{-1}A^{T}$
since $A^{T}A^{-1}$ is not invertible. Any hints would be appreciated.
|
There are several methods to calculate $A^+$ without calculating singular values. Here are two of them.
Method 1.
Use the formula $A^+=\lim_{t\to0}A^T(AA^T+tI)^{-1}$. In your case, you should get something like
\begin{align*}
A^+&=\lim_{t\to0}A^T(AA^T+tI)^{-1}\\
&=\lim_{t\to0}\pmatrix{1&0&1\\ 0&1&0\\ 1&-1&1}\pmatrix{t+2&-1&2\\ -1&t+2&-1\\ 2&-1&t+2}^{-1}\\
&=\lim_{t\to0}\frac1{t^2+6t+6}
\pmatrix{t+2&2&t+2\\ 1&t+4&1\\ t+1&-t-2&t+1}\\
&=\frac16\pmatrix{2&2&2\\ 1&4&1\\ 1&-2&1}.
\end{align*}
Method 2.
Use the property that if $X$ has orthonormal columns or $Y$ has orthonormal rows, then $(XY)^+=Y^+X^+$. Decompose $A$ as the product $\pmatrix{1&0\\ 0&1\\ 1&0}\pmatrix{1&0&1\\ 0&1&-1}$. Gram-Schmidt orthogonalisation gives
\begin{align*}
A
&=\left[
\pmatrix{\frac1{\sqrt{2}}&0\\ 0&1\\ \frac1{\sqrt{2}}&0}
\pmatrix{\sqrt{2}\\ &1}
\right]
\left[
\pmatrix{\sqrt{2}&0\\ -\frac1{\sqrt{2}}&\sqrt{\frac32}}
\pmatrix{\frac1{\sqrt{2}}&0&\frac1{\sqrt{2}}\\ \frac1{\sqrt{6}}&\sqrt{\frac23}&-\frac1{\sqrt{6}}}
\right]\\
&=\underbrace{\pmatrix{\frac1{\sqrt{2}}&0\\ 0&1\\ \frac1{\sqrt{2}}&0}}_{X}
\underbrace{\pmatrix{2&0\\ -\frac1{\sqrt{2}}&\sqrt{\frac32}}}_{Y}
\underbrace{\pmatrix{\frac1{\sqrt{2}}&0&\frac1{\sqrt{2}}\\ \frac1{\sqrt{6}}&\sqrt{\frac23}&-\frac1{\sqrt{6}}}}_{Z}.
\end{align*}
Since $X$ has orthonormal columns, $Y$ is invertible and $Z$ has orthonormal rows, we have
$$
A^+=(XYZ)^+=Z^+Y^+X^+=Z^TY^{-1}X^T.
$$
|
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|
in a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is In a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is
cosine formula $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{c^2}{2ab}<\frac{(a+b)^2}{2ab}=\frac{a^2+b^2}{2ab}+1$
wan,t be able to process after that, could some help me
|
HINT:
Let $a=p-q,b=p+q,c^2=p^2+q^2$
$\cos C=\cdots=\dfrac{c^2}{2ab}=\dfrac12+\dfrac{p^2+q^2}{2(p^2-q^2)}-\dfrac12\ge\dfrac12$
|
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|
Maximum value of $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$ If $a,b,x$ are real and $$f(x)=2(a-x)(x+\sqrt{x^2+b^2}),$$ then find the maximum value of $f(x)$.
Is there any method to solve this question without differentiation because using differentiation I am getting an ugly expression.
|
let $\sqrt{x^2+b^2}+x=y,$ then $\displaystyle \sqrt{x^2+b^2}-x = \frac{b^2}{y}$ and $\displaystyle 2x=y+\frac{b^2}{y}$
$\displaystyle f(x) = (2a-2x)(x+\sqrt{b^2+y^2}) = (2a-y-\frac{b^2}{y})\cdot y = (2ay-y^2-b^2) $
$f(y)= (a^2+b^2)-(y-b)^2\leq (a^2+b^2)$ equality hold when $y=b$
|
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|
Diophantine system of equations I'm trying to know if there is an efficient way to find the smallest (i.e lexicographically) triplet $(a,b,c)$ of integers verifying $$a^2+b^2+c^2 = x$$ $$a^3+b^3+c^3=y$$ $$a^4+b^4 + c^4 = z$$ if $(x,y,z)$ is known.
We assume that a solution exists for that triplet $(x,y,z)$.
Originally, this question is part of an algorithmic problem. So what I want is a fast way to find the triplet without having to brute-force.
What I've tried (the brute-force approach): going through all the possible values of $a$ since we can deduce an upper bound for its value, we remain with three equations where it's easy to find $b$ and $c$ and see if they are integers.
EDIT: We may assume as well that $a<b<c$.
|
If you have software to find real roots, the following result provides another way to find qualifying triples $a,b,c$ ...
Proposition: If $a,b,c$ satisfy $a^2 + b^2 + c^2 = x$, $a^3 + b^3 + c^3 = y$, $a^4 + b^4 + c^4 = z$, then $a,b,c$ are roots of the following 12'th degree polynomial:
$$
p(t) = \left(12\right)\left(t^{12}\right)
-\left(24x\right)\left(t^{10}\right)
-\left(16y\right)\left(t^9\right)
+\left(24x^2-12z\right)\left(t^8\right)
+\left(48xy\right)\left(t^7\right)
$$
$$
-\left(24x^3+8y^2\right)\left(t^6\right)
-\left(24x^2y-24yz\right)\left(t^5\right)
$$
$$
+\left(15x^4+6x^2z-24xy^2+3z^2\right)\left(t^4\right)
+\left(8x^3y-24xyz+16y^3\right)\left(t^3\right)
$$
$$
-\left(6x^5-12x^2y^2-6xz^2+12y^2z\right)\left(t^2\right)
+\left(x^6-4x^3y^2-3x^2z^2+12xy^2z-4y^4-2z^3\right)
$$
|
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|
How to put numbers $1$ to $8$ in a row that every number shouldn't be bigger than the sum of numbers next to it? How to put numbers $1$ to $8$ in a row that every number shouldn't be bigger than the sum of numbers next to it?
My attempt: I tried to put numbers from left to right and count the ways but it was to hard and didn't give the answer.Any hints(not answers).
|
Suppose that the first three numbers satisfy the condition, but the whole seqence does not.
Since $1+2+3+4>8$ we must necessarily have $a_4>a_1+a_2+a_3$.
Case $1: a_4=7$, in this case we must have that $a_1,a_2,a_3$ are a permutation of $1,2,3$. So there are $3!\times 4!$ counterexamples of this form.
Case $2: a_4=8$, In this case we can have that $a_1,a_2,a_3$ are a permutation of $1,2,3$ or $1,2,4$. In total there are $2\times 3!\times 4!$ counterexamples.
So how many permutations satisfy that $a_1>a_2$ and $a_1+a_2>a_3$. Since clearly the two conditions are independent we must only count the triples $(a_1,a_2,a_3)$ with $a_1>a_2$ with $a_1+a_2<a_3$.
They can be enumerated rather quickly:
$(1,a,k)$ gives $5+4+3+2+1=15$ solutions.
$(2,a,k)$ gives $3+2+1=6$ solutions.
$(3,a,k)$ gives $1$ solution.
So there are $22$ solutions.
This means that there are $\frac{8\times 7 \times 6 - 2\times 22}{2}$ triples that satisfy $a_1<a_2$ and $a_1+a_2\geq a_3$. So our final answer is:
$$\frac{8\times 7 \times 6 - 2\times 22}{2}\times 5!-3\times 3!\times 4!$$
|
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|
A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alternative proof, or at least a hint to it. My proof as follows:
Bernoulli inequality states that for $-1<x, x\neq 0, n\in \mathbb{N},n>1$ the following is true:$(1+x)^n>1+nx$.
Thus, $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ is also true.
Then I need to show that $(1-x)^n<\frac{1}{(1+x)^n}$, which is equivalent to $\frac{1}{(1-x)^n}>(1+x)^n$, which I prove by induction:
Basecase: $n=1$ $1=\frac{1-x}{1-x}\Leftrightarrow 1=\frac{1}{1-x}-\frac{x}{1-x}\Leftrightarrow 1+\frac{x}{1-x}=\frac{1}{1-x}$
Let $a,b\in \mathbb{R_{>0}}$ and $0>b>1$ Then $\left[ a>ab \right]\Leftrightarrow \left[a<\frac{a}{b}\right]$. Thus $\left[ 0<x<1\right] \Rightarrow \left[ x<\frac{x}{1-x}\right]$
Thus $1+\frac{x}{1-x}=\frac{1}{1-x} \Rightarrow 1+x<\frac{1}{1-x} \square$
Inductive step: Assume $(1+x)^n<\frac{1}{(1-x)^n}$. Need to show $(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}$.
$(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}\Leftrightarrow (1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x}$
Let $a,b,c,d \in \mathbb{R_{>0}}$. Then $[a>c]\wedge[b>d] \Rightarrow [ab>cd]$. $(1+x)^n<\frac{1}{(1-x)^n}$ was the assumption and $1+x>\frac{1}{1-x}$ was the basecase, therefore $(1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x} \Leftrightarrow (1+x)^{n+1}<\frac{1}{(1-x)^{n+1}} \square$
Thus $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ and $(1-x)^n<\frac{1}{(1+x)^n}$ are both true, which implies the original statement $(1-x)^n<\frac{1}{1+nx} \square$ If I were to count the proof of the Bernoulli inequality by induction, it would mean that I used induction twice in order to prove something that basic, which to me doesn't seem to be a sensible thing to do.
|
What about AM-GM?
$$(1-x)^n(1+nx) = \text{GM}(1-x,1-x,\ldots,1-x,1+nx)^{n+1} \color{red}{\leq} \text{AM}(1-x,1-x,\ldots,1-x,1+nx)^{n+1}=1$$
and the inequality holds tight since $(1-x)\neq(1+nx)$. Done.
|
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|
Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$ Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots are possible. Thus, $y=\ln\left(x\pm\sqrt{x^2-1}\right)$. Why did we fail to eliminate the minus?
|
since the definition of the $$\sqrt{x^2-1}$$ we get $$|x|>1$$ which is for $$x-\sqrt{x^2-1}$$ impossible
|
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|
Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation? Here's Prob. 17, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Fix $\alpha > 1$. Take $x_1 > \sqrt{\alpha}$, and define $$x_{n+1} = \frac{\alpha + x_n}{1+x_n} = x_n + \frac{\alpha - x_n^2}{1+x_n}.$$
(a) Prove that $x_1 > x_3 > x_5 > \cdots$.
(b) Prove that $x_2 < x_4 < x_6 < \cdots$.
(c) Prove that $\lim x_n = \sqrt{\alpha}$.
My effort:
From the recursion formula, we can obtain
$$
\begin{align}
x_{n+1} &= \frac{ \alpha + x_n}{1+ x_n} \\
&= \frac{ \alpha + \frac{\alpha + x_{n-1}}{1+x_{n-1}} }{ 1 + \frac{\alpha + x_{n-1}}{1+x_{n-1}} } \\
&= \frac{ (\alpha + 1) x_{n-1} + 2 \alpha }{ 2x_{n-1} + ( 1 + \alpha ) } \\
&= \frac{\alpha+1}{2} + \frac{2 \alpha - \frac{(\alpha+1)^2}{2} }{2x_{n-1} + ( 1 + \alpha ) } \\
&= \frac{\alpha+1}{2} + \frac{ \alpha - \frac{\alpha^2+1 }{2} }{2x_{n-1} + ( 1 + \alpha ) }.
\end{align}
$$
What next?
|
Notice that
$$ x_{n+1} = \frac{\alpha+x_n}{1+x_n} = 1 + \frac{\alpha-1}{1+x_n}. $$
Since $\alpha-1>0$, we have that if $x_n<\sqrt{\alpha}$, then
$$ x_{n+1} > 1 + \frac{\alpha-1}{1+\sqrt{\alpha}} = \frac{1+\sqrt{\alpha}+\alpha-1}{1+\sqrt{\alpha}} = \frac{\sqrt{\alpha}+\alpha}{1+\sqrt{\alpha}} = \sqrt{\alpha}$$
and similarly, if $x_n > \sqrt{\alpha}$, then $x_{n+1} < \sqrt{\alpha}$. Since $x_1>\sqrt{\alpha}$, it follows from induction that $x_n<\sqrt{\alpha}$ for $n$ even and $x_n>\sqrt{\alpha}$ for $n$ odd. In particular, $x_{n+1}-x_n > 0 $ if $n$ is even, and $x_{n+1}-x_n < 0 $ if $n$ is odd.
Notice that
$$ x_{n+1} = \frac{\alpha+x_n}{1+x_n}\implies x_{n+1}(1+x_n) = \alpha+x_n \implies x_nx_{n+1} = \alpha - (x_{n+1}-x_n) $$
and hence
$$ x_n(x_{n+1}-x_{n-1}) = (x_n-x_{n-1}) - (x_{n+1}-x_n). $$
It is clear that $x_n>0$ for all $n$, so we see that if $n$ is odd, then $x_n-x_{n-1}>0$ and $x_{n+1}-x_n<0$, so $x_{n+1}-x_{n-1} > 0$, while if $n$ is even, then $x_n-x_{n-1}<0$ and $x_{n+1}-x_n>0$, so $x_{n+1}-x_{n-1} <0$.
Thus, $x_3-x_1<0$, $x_5-x_3<0$, and so on, so $x_1 > x_3 > x_5 > \dots$, while $x_4-x_2>0$, $x_6-x_4>0$, and so on, so $x_2 < x_4 < x_6 < \dots$.
This proves (a) and (b).
Now, since $x_n>\sqrt{\alpha}$ for $n$ odd, and $x_1>x_3>x_5>\dots$, it follows that the subsequence of odd terms $\{x_{2n-1}\}$ is monotonically decreasing and bounded from below, and hence has a limit (say $L$). Similarly, the subsequence of even terms $\{x_{2n}\}$ is monotonically increasing and bounded from above (namely by $\sqrt{a}$), and so it has a limit as well (say $M$). These limits must satisfy $L\ge\sqrt{\alpha}$ and $M\le\sqrt{\alpha}$. From the equation
$$ x_{n+1} = \frac{\alpha+x_n}{1+x_n} $$
if we consider $n$ odd and take limits on both sides, we obtain
$$ M = \frac{\alpha+L}{1+L} $$
while if we consider $n$ even and take limits, we obtain
$$ L = \frac{\alpha+M}{1+M}. $$
Thus, if we define the sequence $\{y_n\}$ by $y_1 = L$ and $y_{n+1} = \frac{\alpha+y_n}{1+y_n}$, then the sequence $\{y_n\}$ is just $\{L,M,L,M,\dots\}$. If $L>\sqrt{\alpha}$, then we can actually apply what we proved in part (a) to conclude that $y_1 > y_3 > y_5 > \dots$, which is impossible since all odd terms are $L$. Since $L\ge\sqrt{\alpha}$, it must follow that $L = \sqrt{\alpha}$, and you can easily check that this forces $M = \sqrt{\alpha}$ as well. Thus, the odd and even subsequential limits are both $\sqrt{\alpha}$, so the limit of the sequence is $\sqrt{\alpha}$ as well.
|
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|
Infinitely nested radical formulas for $\pi$ Has anyone come by the infinitely nested radical formula:
$$\Bigg\{\pi=\frac{12}{5}\cdot \lim\limits_{n \to \infty} 2^{n}\cdot\frac{1}{2^{\frac{2^{n}+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}} -\cdot\cdot\cdot\cdot\sqrt{ 2^{\frac{3}{2}} +\sqrt{ 2^2 + (\sqrt{ 6}- \sqrt{ 2})}}}\Bigg\}$$ And can anyone prove it?
|
I appreciate your work and finding. We may derive $\pi$ In infinite number of ways with the help of nested square roots
For explaining your derivation we have to think infinite nested radicals from the end.
Let us start from chord inside unit circle and value of lengths bisecting the chord
The length of chord is $2\sin\frac{75^\circ}{2}$ or 2 $\sin\frac{5\pi}{12}$$\frac{5\pi}{12}$ or $75^\circ$" />
The length of centre line bisecting the arc till the chord from centre is $\cos\frac{75^\circ}{2}$ as it bisects the angle, is equal to $\frac{(\sqrt6-\sqrt2)}{4}$ which can be simplified to $\frac{\sqrt{2-\sqrt3}}{2}$.
Length of bisected chord will be $2\sin\frac{75^\circ}{2}$ = $2\cdot\frac{\sqrt{2+\sqrt3}}{2}$
Now by bisecting the chord into halves leads to calculation of length from centre to midpoint of the chord which becomes smaller and smaller approaching towards unit of radius
Here you have applied 1/2 angle formula leading to nested square roots as follows
$\cos(\frac{75^\circ}{2^2})$ as $\frac{{\sqrt{2+{\sqrt{2-{\sqrt{3}}}}}}}{2}$, number of chords is 4
Now the number of chords within $75^\circ$ becomes 2^2
Sum of lengths of 4 chords will be $4\sin\frac{75^\circ}{2^2}$ $4\cdot\frac{{\sqrt{2-{\sqrt{2-{\sqrt{3}}}}}}}{2}$
Like this next iteration will be
$2^3\cdot \frac{{\sqrt{2-{\sqrt{2+{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$
When we do this repeatedly we get $\cos(\frac{75^\circ}{2^n})$ as $\frac{{\sqrt{2+{\sqrt{2+...(n-1) times{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$
Finally when we take sine value of the infinitesimal small angle $\frac{75^\circ}{2^n}$ it will be $$\frac{{\sqrt{2-{\sqrt{2+...(n-2) times{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$$
Number of chords will be $2^n$ times
Chord length in radians is $r\theta$ = $\frac{5\pi}{12}$ which is nothing but $2^n\cdot \frac{{\sqrt{2-{\sqrt{2+...(n-1) times{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$
Therefore $$\pi = \frac{12}{5}\cdot\lim_{n\to\infty} 2^n\cdot {\sqrt{2-{\sqrt{2+...(n-1) times{\sqrt{2-{\sqrt{3}}}}}}}}$$
Please forgive me for such a lengthy answer.
Please forgive for mistakes if you find any.
Thank you
My work
|
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How to solve these general $n \times n$ determinants? How would you solve these two general determinants?
$$
\begin{vmatrix}
2 & 1 & 0 & \cdots & 0 & 0 \\
1 & 2 & 1 & \cdots & 0 & 0\\
0 & 1 & 2 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 2 & 1\\
0 & 0 & 0 & \cdots & 1 & 2\\
\end{vmatrix}
$$
$$
\begin{vmatrix}
3 & 2 & 2 & \cdots & 2 \\
2 & 3 & 2 & \cdots & 2 \\
2 & 2 & 3 & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & 3 \\
\end{vmatrix}
$$
And if there are any tips for counting determinants of this type, please let me know :-)
|
For the second one, in general, if $$\Delta_n=\begin{vmatrix}
a & b & b & \ldots & b\\
b & a & b & \ldots & b \\
b & b & a & \ldots & b \\
\vdots&&&&\vdots\\
b & b & b & \ldots & a
\end{vmatrix}.$$ First step $R_i\to R_i-R_n\;(i\ne n)$ implies $$\Delta_n=
\begin{vmatrix}
a & b & b & \ldots & b\\
b -a& a-b & 0 & \ldots & 0 \\
b-a & 0 & a-b & \ldots & 0 \\
\vdots&&&&\vdots\\
b -a& 0 & 0 & \ldots & a
-b\end{vmatrix}.$$
Second step $C_1\to C_1+C_2+\cdots +C_n$ implies
$$\Delta_n=
\begin{vmatrix}
a +(n-1)b& b & b & \ldots & b\\
0 & a-b & 0 & \ldots & 0 \\
0 & 0 & a-b & \ldots & 0 \\
\vdots&&&&\vdots\\
0 & 0 & 0 & \ldots & a
-b\end{vmatrix}=[a+(n-1)b](a-b)^{n-1}.$$
|
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Why is $\lim\limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ equal to $0$? So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:
$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $
$\lim \limits_{x\to +\infty} \frac{\sqrt{x^2}\sqrt{1-\frac{2}{x}}+x}{-2x}$
$\lim \limits_{x\to +\infty} \frac{\sqrt{1-\frac{2}{x}}+1}{-2} = -1 $
The next exercise is $\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$
I thought that I could solve this one in the same way by:
$\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} $
$\lim \limits_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x}$
$\lim \limits_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} = 1 $
But apparently, the answer is $0$...
Why can't the second one be solved in the same way as the first?
And why is the answer $0$?
|
We already have answers showing correct ways to solve the problem.
What I found interesting was the implied question:
where is the error in the steps taken in the body of the question,
which appeared to show that the limit is $1$?
The steps in question are
\begin{align}
\newcommand{\?}{\stackrel{?}{=}}
\lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}
&\? \lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}
\frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} \tag1\\
&\? \lim_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x} \tag2\\
&\? \lim_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} \tag3\\
&\? 1 \tag4
\end{align}
The error is in the third step (from line $2$ to line $3$).
In fact, if $x < 0,$ then $\sqrt{x^2} = -x,$ so
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x}
&= \lim_{x\to -\infty} \frac{-x\sqrt{1+\frac{2}{x}}+x}{2x} \\
&= \lim_{x\to -\infty} \frac{-\sqrt{1+\frac{2}{x}}+1}{2} \\
&\neq \lim_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2}
\end{align}
One way to avoid errors of this sort (other than constant vigilance
when manipulating square roots of expressions over negative-valued variables)
is to begin by observing that
$$
\lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x} - x}
= \lim_{y\to +\infty} \frac{1}{\sqrt{y^2-2y} + y}.
$$
One can then work on the limit over $y,$ confident in the knowledge that
$\sqrt{y^2}/y = 1.$
|
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|
Find all real solutions to the equation$4^x+6^{x^2}=5^x+5^{x^2}$ Find all real solutions to the equation:$$4^x+6^{x^2}=5^x+5^{x^2}$$
Obviously $0$ and $1$ satisfy this equation, but here $x$ is a real number, and I do not know how to do the next step, in fact, I guess this equation is not in addition to $0$ and $1$ other than the root. But you know, I need a proof! Could anyone help me? Thanks a lot!
|
For any $x \in \mathbb{R}$, apply MVT to $t^{x^2}$ on $[5,6]$ and $t^x$ on $[4,5]$,
we find a pair of numbers $\xi \in (5,6)$ and $\eta \in (4,5)$ (both dependent on $x$) such that
$$6^{x^2} - 5^{x^2} = x^2 \xi^{x^2-1}\quad\text{ and }\quad
5^x - 4^x = x\eta^{x-1}$$
This implies
$$
{\tt LHS}-{\tt RHS} = (4^{x} + 6^{x^2}) - (5^x + 5^{x^2}) = x^2\xi^{x^2-1} - x\eta^{x-1}
$$
There are 3 cases we need to study:
*
*$x \in (1,\infty)$ - Both $x^2 - 1$ and $x - 1$ are positive, we have
$$\begin{align}
x^2\xi^{x^2-1} & \ge x^2\inf\{ t^{x^2-1} : t \in (5,6) \} = x^2 \left(\inf\{ t : t \in (5,6)\}\right)^{x^2-1} = x^2 5^{x^2-1}\\
x\eta^{x-1} & \le x\sup\{ t^{x-1} : t \in (4,5)\} = x\left(\sup\{ t : t \in (4,5)\}\right)^{x-1} = x 5^{x-1}
\end{align}\\
\implies
{\tt LHS} - {\tt RHS} \ge x^2 5^{x^2-1} - x5^{x-1} = (\underbrace{x}_{> 1}\underbrace{5^{x^2-x}}_{>1} - 1)\underbrace{x5^{x-1}}_{>0} > 0$$
*$x \in (0,1)$ - Both $x^2-1$ and $x-1$ are negative, we have
$$\begin{align}
x^2\xi^{x^2-1} & \le x^2\sup\{ t^{x^2-1} : t \in (5,6) \} = x^2 \left(\inf\{ t : t \in (5,6)\}\right)^{x^2-1} = x^2 5^{x^2-1}\\
x\eta^{x-1} & \ge x\inf\{ t^{x-1} : t \in (4,5)\} = x\left(\sup\{ t : t \in (4,5)\}\right)^{x-1} = x 5^{x-1}
\end{align}\\
\implies
{\tt LHS} - {\tt RHS} \le x^2 5^{x^2-1} - x5^{x-1} = (\underbrace{x}_{< 1}\underbrace{5^{x^2-x}}_{<1} - 1)\underbrace{x5^{x-1}}_{>0} < 0$$
*
*$x \in (-\infty,0)$ - We have
$$x^2 \xi^{x^2-1} > 0 \land x \eta^{x-1} < 0
\quad\implies\quad
{\tt LHS} - {\tt RHS} = x^2\xi^{x^2-1} - x\eta^{x-1} > 0$$
Combine these 3 cases, we have ${\tt LHS} \ne {\tt RHS}$ for $x \in \mathbb{R} \setminus \{ 0, 1 \}$.
Since we know $0, 1$ are roots of the equation at hand,
$0$ and $1$ are all the real roots of the equation.
|
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|
Solving $ {L}_{1} $ Regularized Least Squares Over Complex Domain I would like to solve the following Regularized Least Squares Problem (Very Similar to LASSO):
$$ \arg \min_{x} \frac{1}{2} {\left\| A x - b \right\|}_{2}^{2} + \lambda {\left\| x \right\|}_{1} $$
Where $ A \in {\mathbb{R}}^{m \times n} $ and $ b \in {\mathbb{R}}^{m} $.
For simplicity one could define $ f \left( x \right) = \frac{1}{2} {\left\| A x - b \right\|}_{2}^{2} $ and $ g \left( x \right) = \lambda {\left\| x \right\|}_{1} $.
For $ x \in {\mathbb{R}}^{n} $ the solution can be achieved using Sub Gradient Method or Proximal Gradient Method.
My question is, how can it be solved for $ x \in {\mathbb{C}}^{n} $ (Assuming $ A \in {\mathbb{C}}^{m \times n} $ and $ b \in {\mathbb{C}}^{m} $)?
Namely if the problem is over the complex domain.
For instance, what is the Sub Gradient?
What is the Prox (Shrinkage of Complex Number)?
Thank You.
My Attempt for Solution 001
The Gradient of $ f \left( x \right) $ is given by:
$$ {\nabla}_{x} f \left( x \right) = {A}^{H} \left( A x - b \right) $$
The Sub Gradient of $ g \left( x \right) $ is given by:
$$ {\partial}_{x} g \left( x \right) = \lambda \operatorname{sgn} \left( x \right) = \lambda \begin{cases}
\frac{x}{ \left| x \right| } & \text{ if } x \neq 0 \\
0 & \text{ if } x = 0
\end{cases} $$
Namely it is the Complex Sign Function.
Then, the Sub Gradient Method is given by:
$$ {x}^{k + 1} = {x}^{k} - {\alpha}_{k} \left( {A}^{H} \left( A {x}^{k} - b \right) + \lambda \operatorname{sgn} \left( {x}^{k} \right) \right) $$
Where $ {\alpha}_{k} $ is the step size.
Yet it won't converge to CVX Solution for this problem.
Remark on Attempt 001
I think I understood why it doesn't work well.
The Absolute Value Function in the Complex Domain is (Quoted from Wikipedia Absolute Value Derivative Section):
|
Write $A = B + Ci$, $b=c+di$ and $x=y+zi$. The objective function is
$$
\begin{align*}f(x) &= ||(B+Ci)(y+zi)-c-di||_1^2 + \lambda ||y+zi||_1 \\
&= ||By-Cz-c+(Cy+Bz-d)i||_2^2 + \lambda ||y+zi||_1 \\
&= ||By-Cz-c||_2^2 + ||Cy+Bz-d||_2^2 + \lambda \sum_{j=1}^n \sqrt{y_j^2+z_j^2} \\
&= \left\Vert\begin{pmatrix}B \\ -C \end{pmatrix}^T \begin{pmatrix}y\\z\end{pmatrix}-c\right\Vert_2^2 + \left\Vert\begin{pmatrix}C \\ B \end{pmatrix}^T\begin{pmatrix}y\\z\end{pmatrix}-d\right\Vert_2^2 + \lambda \sum_{j=1}^n \left\Vert\begin{pmatrix}e_j^T & 0 \\ 0 & e_j^T \end{pmatrix} \begin{pmatrix}y\\z\end{pmatrix}\right\Vert_2
\end{align*}
$$
where $e_j$ is the $j^{th}$ unit vector. Finding the subgradient is now straightforward:
$$
\begin{align*}
2\begin{pmatrix}B \\ -C \end{pmatrix} \left(\begin{pmatrix}B \\ -C \end{pmatrix}^T \begin{pmatrix}y\\z\end{pmatrix}-c\right) &+ 2\begin{pmatrix}C \\ B \end{pmatrix}\left(\begin{pmatrix}C \\ B \end{pmatrix}^T\begin{pmatrix}y\\z\end{pmatrix}-d\right) \\
&+ \lambda \sum_{j=1}^n \frac{\begin{pmatrix}e_j^T & 0 \\ 0 & e_j^T \end{pmatrix}^T \begin{pmatrix}e_j^T & 0 \\ 0 & e_j^T \end{pmatrix} \begin{pmatrix}y\\z\end{pmatrix}}{\left\Vert\begin{pmatrix}e_j^T & 0 \\ 0 & e_j^T \end{pmatrix} \begin{pmatrix}y\\z\end{pmatrix}\right\Vert_2}
\end{align*}$$
|
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|
Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
|
Let $f_a(x) = x^4+2ax^3+x^2+2ax+1$. If $x$ is a negative root, then we have
$$a = -\frac{x^4+x^2+1}{2x^3+2x}$$
That is, if we let
$$g(x) = \frac{x^4+x^2+1}{2x^3+2x}$$
then $a = g(-x)$. We have that $-x > 0$; using standard techniques, $g(-x)$ reaches a maxiximum (on $(-\infty, 0)$) of $3/4$ at $x=-1$.
This shows that $a\geq 3/4$. It's easy to show that $f_{3/4}(x)$ only has one negative real root, and so $a > 3/4$.
|
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|
Solving a system of equations involving an absolute value Solve the following system:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\left|\epsilon\right|=\left|-\frac{\text{c}}{\text{d}}\right|
\end{cases}
$$
I don't know how to proceed?!
With the help of the comments and the answer I got:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\epsilon=\pm\frac{\text{c}}{\text{d}}
\end{cases}\to\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\frac{\text{c}^2}{\text{d}^2}=\pm\frac{\text{c}}{\text{d}}\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)
$$
|
We are given
$$
\left\{ \begin{gathered}
bd\varepsilon ^{\,2} - \left( {bc - ad} \right)\varepsilon + ac = 0 \hfill \\
\varepsilon = \pm \frac{c}
{d} \hfill \\
\end{gathered} \right.
$$
So we have the equation of a vertical parabola in $\varepsilon$, whose intercepts with the $x$ axis
must be symmetrical (vs. $\varepsilon=0$).
Thus the parabola must be symmetrical as well, i.e. the coefficient
of the $\varepsilon$ term must be null, which gives:
$$
\begin{gathered}
\left\{ \begin{gathered}
bd\varepsilon ^{\,2} - \left( {bc - ad} \right)\varepsilon + ac = 0 \hfill \\
\varepsilon = \pm \frac{c}
{d} \hfill \\
\end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered}
bc - ad = 0 \hfill \\
b\frac{{c^{\,2} }}
{d} + ac = 0 \hfill \\
\end{gathered} \right.\quad \Rightarrow \hfill \\
\Rightarrow \quad \left\{ \begin{gathered}
bc - ad = 0 \hfill \\
d \ne 0 \hfill \\
\left( {bc + ad} \right)c = 0 \hfill \\
\end{gathered} \right. \Rightarrow \quad \left\{ \begin{gathered}
c = 0 \hfill \\
d \ne 0 \hfill \\
a = 0 \hfill \\
\forall b \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
|
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|
How to find the given limits I now that: $$\lim_{x\to 0}\frac{\sin x}{x}=1,$$ but I didnt now how to prove that:$$\lim_{x\to 0}\frac{1-\cos x}{x^2}$$
Please help me. Thanky very much for your help.
|
First method:
$$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\left(\frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}\right)=\lim_{x\to 0}\frac{1-\cos^2 x}{x^2(1+\cos x)}$$
$$=\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)}=\lim_{x\to 0}\frac{\sin x\cdot \sin x}{x\cdot x (1+\cos x)}=\lim_{x\to 0}\left(\frac{\sin x}{x}\cdot\frac{\sin x}{x}\cdot \frac{1}{1+\cos x}\right)$$
$$=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{1}{1+\cos x}=1\cdot 1\cdot \frac{1}{1+\lim_{x\to 0} \cos x}=\frac{1}{2}$$
Second method:
Use the trigonometrics formula, $1-\cos x=2\sin^2\frac{x}{2}$ we have:
$$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{x^2}$$
$$=2\lim_{x\to 0}\frac{\sin\frac{x}{2}\cdot \sin\frac{x}{2}}{\frac{x}{2}\cdot\frac{x}{2}\cdot 4}=\frac{2}{4}=\frac{1}{2}$$
|
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|
Find the sum of the reciprocals
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{30}+\cdots.$$ What is $\lfloor A\rfloor$?
The sum is greater than $2$ since $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11} > 2$. Can we show it is less than $3$?
|
Let $A$ be the set of all natural numbers $n$ not containing $9$ in their decimal representation and let $B$ be the set of all natural numbers $n$ that contain only some of the digits $0,1,2,3,4$ in their decimal representation. It is obvious that $B\subset A$. So:
$$\displaystyle{\sum_{n\in B}}\frac{1}{n}\leq \displaystyle{\sum_{n\in A}}\frac{1}{n}$$
The $\displaystyle{\sum_{n\in B}}\frac{1}{n},\displaystyle{\sum_{n\in A}}\frac{1}{n}$ mean that we have summation over the elements of $B$ and $A$ respectively.
But $\displaystyle{\sum_{n\in A}}\frac{1}{n}$ is the known Kempner series that converges and is less than $90$. Below I give a reference for it:
https://en.wikipedia.org/wiki/Kempner_series
|
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|
Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi}{3}$ which gives the answer $x^2+y^2+z^2=\frac{8{\pi}^2}{9}$.But I want a way to find the answer using equations.ANy hints?
|
\begin{eqnarray}
\sin z &=& -\sin x-\sin y\\
\cos z &=& -\cos x-\cos y
\end{eqnarray}
so
\begin{eqnarray}
\sin^2z &=& \sin^2x+\sin^2y+2\sin x\sin y\\
\cos^2z &=& \cos^2x+\cos^2y+2\cos x\cos y
\end{eqnarray}
adding to these equations concludes
\begin{eqnarray}
1 &=& 1+1+2(\sin x\sin y+\cos x\cos y)\\
0 &=& 1+2\cos(x-y)\\
-\frac{1}{2} &=& \cos(x-y)
\end{eqnarray}
and also
$$\cos^2\frac{x-y}{2}=\frac{1+\cos(x-y)}{2}=\frac{1}{4}$$
or
$$\cos\frac{x-y}{2}=\pm\frac{1}{2}$$
In the other hand
\begin{eqnarray}
(\sin x+\sin y+\sin z)^2 + (\cos x+\cos y+\cos z)^2 &=& 0\\
\sin^2x+\sin^2y+\sin^2z+2\sin x\sin y+2\sin y\sin z+2\sin z\sin x &+&\\
\cos^2x+\cos^2y+\cos^2z+2\cos x\cos y+2\cos y\cos z+2\cos z\cos x &=&0\\
3+2\big(\cos x\cos y+\sin x\sin y\big)+2\big(\cos y\cos z+\sin y\sin z\big)+2\big(\cos z\cos x+\sin z\sin x\big) &=&0\\
3+2\cos(x-y)+2\cos(y-z)+2\cos(z-x) &=&0\\
3+2(-\frac{1}{2})+4\cos\frac{x-y}{2}\cos3\frac{x+y}{2} &=&0\\
\cos\frac{x-y}{2}\cos3\frac{x+y}{2} &=&-\frac{1}{2}\\
\cos3\frac{x+y}{2} &=&\mp1\\
\end{eqnarray}
so
\begin{eqnarray}
\cos(x-y) &=& -\frac{1}{2}\\
\cos3\frac{x+y}{2} &=&\mp1
\end{eqnarray}
We have two answer for each of them
\begin{eqnarray}
x-y &=& \frac{2\pi}{3},~\frac{-\pi}{3}\\
x+y &=& 0,~\frac{2\pi}{3}
\end{eqnarray}
We obtain 4 answer from these equations and then permutation over $(x,y,z)$ from equations symmetrically, gives us 12 answer. Finally by substitution answers in equations, $x^2+y^2+z^2$ has one value ($\displaystyle\frac{2\pi}{3},\frac{-2\pi}{3},0$), it is
$$\frac{8\pi^2}{9}$$
|
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|
Generators of a quotient ring I have a polynomial ring $\mathcal{R} = \mathbb{Z}_3[x]/(x^2+x+2)$ and I am supposed to find a generator of the multiplicative group of $\mathcal{R}$.
Well, I know that the $\vert \mathcal{R} \vert = 9$, therefore the number of elements of the multiplicative group is $9-1 = 8$ and that the elements $f(x)\in$ $\mathcal R$ are $[f(x)]=\{g(x) \in \mathbb Z_3 \mid g(x) \equiv f(x)$ mod $(x^2+x+2) \}$. Though I do not know how to find the generator instead of brute-forcing. Is there a smart way to do it? Thank you in advance.
|
The ring is the splitting field of $x^9-x$; since
$$
x^9-x=x(x^8-1)=x(x-1)(x+1)(x^2+1)(x^4+1)
$$
and $x^4+1=(x^2+x+2)(x^2+2x+2)$, the complete factorization is
$$
x^9-x=x(x^8-1)=x(x-1)(x+1)(x^2+1)(x^2+x+2)(x^2+2x+2)
$$
Note that if $\alpha$ is a root of $x^2+1$, then $\alpha^4=1$, so it is not good for the purpose. Thus we remain with the roots of the other degree two factors: indeed the group is cyclic of order $8$, so it must have $\varphi(8)=4$ elements which are generators.
|
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|
Does the law of cosines imply triangle exists? This is surely trivial, but I've never seen a triangle explicitly defined in this way before. Let $a,b,c>0$. Is it true that:
$$a,b,c\text{ form a triangle}\iff -1\leq\frac{b^2+c^2-a^2}{2bc}\leq 1.$$
My attempt:
($\implies$) Suppose $a,b,c$ form a triangle, then by the law of cosines, $$\cos \vartheta=\frac{b^2+c^2-a^2}{2bc},$$ and $-1\leq \cos \vartheta\leq 1$.
($\impliedby$) Suppose $$\frac{b^2+c^2-a^2}{2bc}=\lambda,$$ with $-1\leq\lambda\leq 1$, then there exists a $\vartheta$ such that $\cos\vartheta=\lambda$ and so
$$\frac{b^2+c^2-a^2}{2bc}=\cos\vartheta.$$
Hence, by the law of cosines $a,b,c$ form a triangle.
|
Let $a,b,c>0$ and
$$ -1\leq\frac{b^2+c^2-a^2}{2bc}\leq 1. \tag1$$
Then your proof via the Law of Cosines requires the existence of
$\vartheta$ such that $0 \leq \vartheta \leq \pi$
and $\cos\vartheta = \frac{b^2+c^2-a^2}{2bc}.$
Such a value $\vartheta$ always exists under the conditions given
in Inequality $1$.
Then you need merely construct a triangle with sides $b$ and $c$
adjacent to angle $\vartheta$, and the Law of Cosines then implies
that the opposite side will have length $a$.
Hence a triangle with sides $a,b,c$ exists.
This part of the answer (which I think is what you wanted) was already
given in comments, so I'm not taking credit for it.
Someone else can post a non-community-wiki version if desired.
Alternatively, starting with $a,b,c>0$ and Inequality $1$,
it is a fact that $-2bc < 0$,
so multiplying everything by $-2bc$ reverses the
directions of both inequalities:
$$ 2bc \geq a^2 - b^2 - c^2 \geq -2bc.$$
Add $b^2 + c^2$:
$$ (b+c)^2 = b^2 + 2bc + c^2 \geq a^2 \geq b^2 - 2bc + c^2 = (b - c)^2.$$
The inequalities are preserved by taking the positive square root, so
$$ b+c \geq a \geq \lvert b - c \rvert,$$
observing that $b+c>0$ and $a>0$ but the sign of $b-c$ has not been determined.
The left-hand inequality immediately gives us one of the necessary
triangle inequalities, $a \leq b + c$.
The right-hand inequality gives us two inequalities,
\begin{align}
b - c &\leq a,\\
c - b &\leq a,\\
\end{align}
from which we get
\begin{align}
b &\leq a + c,\\
c &\leq a + b,\\
\end{align}
which are the other two triangle inequalities.
|
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|
Definite integral of $4x(1+x)^3$ I'm trying to integrate by parts, and have done:$$\int^1_0{4x(1+x)^3}=4\int^1_0{x(1+x)^3}= 4\bigg[\frac{x(1+x)^4}4-\frac{(1+x)^4 }4\bigg]=4\bigg[\frac{(x-1)(1+x)^4}4\bigg]^1_0=
\\ \bigg[(x-1)(1+x)^4 \bigg]^1_0 = 1$$
Why is this wrong?
|
Integration by parts should give, if we take $dv=(1+x)^3 dx$ so that $v=\frac{(1+x)^4}{4}$. And $u=x$ so that $du=dx$,
$$\int {4x(1+x)^3}=4\int {x(1+x)^3}= 4\bigg[\frac{x(1+x)^4}4-\int \frac{(1+x)^4 }4 \mathrm dx\bigg]$$
As $\int u \mathrm dv=vu-\int v \mathrm du$.
|
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"url": "https://math.stackexchange.com/questions/2084271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
solving Integration of trignometry In the following integral
$$\int \frac {1}{\sec x+ \mathrm {cosec} x} dx $$
My try: Multiplied and divided by $\cos x$ and Substituting $\sin x =t$. But by this got no result.
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$\displaystyle I = \int\frac{1}{\sec x+\csc x}dx = \frac{1}{2}\int\frac{\sin 2x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{(\sin x+\cos x)^2-1}{\sin x+\cos x}dx$
$\displaystyle I = \frac{1}{2}\int (\sin x+\cos x)dx - \frac{1}{2}\int \frac{1}{\sin x+\cos x}dx$
$\displaystyle I =\frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2}}\int \frac{1}{\sin (x+45^0)}dx$
$\displaystyle I = \frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2}}\int \csc (x+45^0)dx$
Use $\displaystyle \int \csc xdx = \ln |\csc x-\cot x|+\mathcal{C} = \ln\tan \frac{x}{2}+\mathcal{C}$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2087107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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|
Need help solving this Can I have help with this? I can't think of a way to get started...Thank you
a) If $k$ and $\alpha$ are constants and the following trigonometric
equation can be solved,
$$2(\cos\alpha+\sin\alpha)\cos\theta+4(\cos\alpha-\sin\alpha)\sin\theta=k$$
Prove that $k^2\le32$.
b) If $k=2\sqrt2$, and $-1\le\cot \alpha\le 7$, show that the above
equation can be solved.
https://i.stack.imgur.com/dyv9x.jpg
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hint:
For part (a)
$$2(\cos (\alpha) +\sin(\alpha))\cos \theta + 4(\cos \alpha - \sin \alpha)\sin(\theta) = k$$
By harmonic addition theorem, http://mathworld.wolfram.com/HarmonicAdditionTheorem.html
I can express the $a \cos \theta + b \sin \theta$ in the form of $R cos(\theta + \delta)$ where $R^2=a^2+b^2$.
\begin{align}R^2 &= 4(\cos (\alpha) + \sin(\alpha))^2+16(\cos \alpha - \sin \alpha)^2\\&=4+4\sin(2\alpha)+16-16\sin(2\alpha)\\
&=20-12\sin(2\alpha)
\leq 32\end{align}
Hence $k^2 \leq 32$.
For part (b)
$$2(\cos (\alpha) +\sin(\alpha))\cos \theta + 4(\cos \alpha - \sin \alpha)\sin(\theta) = 2\sqrt{2}$$
$$sign(2(\cos(\alpha)+\sin(\alpha))\sqrt{20-12\sin(2\alpha)}\cos\left(\theta+\tan^{-1}\left(-\frac{4(\cos\alpha-\sin\alpha)}{2(\cos\alpha+\sin\alpha)}\right)\right)=2\sqrt{2}$$
$$\cos\left(\theta+\tan^{-1}\left(-\frac{4(\cos\alpha-\sin\alpha)}{2(\cos\alpha+\sin\alpha)}\right)\right)=\frac{2\sqrt{2}}{sign(2(\cos(\alpha)+\sin(\alpha))\sqrt{20-12\sin(2\alpha)}}$$
Homework: check that the denominator is non-zero. If it is zero, handle that case separately.
Hence the system is solvable if
$$\left| \frac{2\sqrt{2}}{\sqrt{20-12\sin(2\alpha)}}\right| \leq 1$$
which is equivalent to
$$\left| \frac{\sqrt{2}}{\sqrt{5-3\sin(2\alpha)}}\right| \leq 1$$
Note that $5-3\sin(2\alpha) \geq 2$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2088433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.