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Can Angles A and B In A Trapezium Be Solved Using Basic Geometry?
Can angles A and B be solved? Neither the area nor the perimeter was given. Thank you very much if you can help! :)
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If I'm not mistaken, I think the followings tell us that the angles $A,B$ are not determined. Set
$$A(0,0),B(9,0),C(9+7\cos(180^\circ-B),7\sin(180^\circ-B)),D(4\cos A,4\sin A).$$
Then, we have
$$CD^2=5.9^2=\{9+7\cos(180^\circ-B)-4\cos A\}^2+\{7\sin(180^\circ-B)-4\sin A\}^2$$
$$\iff 5.9^2=9^2+7^2+4^2-126\cos B-72\cos A+56\cos A\cos B-56\sin B\sin A.$$
Here, even if we suppose that $0\le A,B\le 90^\circ,$ we have
$$5.9^2=9^2+7^2+4^2-126y-72x+56xy-56\sqrt{(1-x^2)(1-y^2)}$$
where $x=\cos A, y=\cos B, \sin A=\sqrt{1-x^2}, \sin B=\sqrt{1-y^2}.$
For example, if $A=90^\circ$, then $\cos B=y=39.19/70\approx 0.559857.$
If $A=45^\circ$, then $\cos B=y\approx 0.254432.$
Since we have at least two concrete examples, it's obvious that there are infinitely many pairs $(A,B)$ because $C$ is on the circle whose center is $B$ with radius $7$ and $D$ is on the circle whose center is $A$ with radius $4$.
|
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|
Given any positive real numbers $a,b,c$, we have $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$ I have a beautiful inequality, but I can only prove part of cases. Given any positive real numbers $a,b,c$, we have
$$(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$$
How can we prove this statement?
|
Hint: you can use Cauchy Schwarz inequality to prove stronger inequality
$$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\ge 9(ab+bc+ac)$$
Use Cauchy-Schwarz inequality we have
$$(b^2+1)(1+c^2)\ge (b+c)^2, (b^2+c^2)(1+1)\ge (b+c)^2$$
so
$$(b^2+2)(c^2+2)=(b^2+1)(1+c^2)+b^2+c^2+3\ge (b+c)^2+\dfrac{1}{2}(b+c)^2+3=3[1+\dfrac{(b+c)^2}{2}]$$
and use Cauchy-Schwarz inequality
$$(a^2+2)[1+\dfrac{(b+c)^2}{2}]\ge (a+b+c)^2$$
|
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|
Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
|
After two applications of L'Hôpital's rule: $$ \displaystyle \begin{aligned} \lim_{x \to 0} \left( \dfrac{1-\cos^3 (x)}{x\sin (2x)} \right) & \ = \ \lim_{x \to 0} \left( \dfrac{3\sin (x) \cos^2 (x)}{\sin (2x) + 2x\cos (2x)} \right) \\ & \ = \ \lim_{x \to 0} \left( \dfrac{3\cos^3 (x) - 6\cos (x) \sin (x)}{4\cos (2x) - 4x \sin (2x)} \right) = \ ? \end{aligned} $$
|
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|
Solve trigonometric function $x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0$ I need to solve a trigonometric function similar to the following one for $\alpha$.
$$
x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0
$$
I found a solution to a very similar problem
$$\frac{1}{2} \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0. $$
here, but the $\cos(2\alpha)$ term causes me some trouble, to adapt this solution.
TIA for your time!
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Using the complex representation, we have $z:=e^{i\alpha}$, $\cos\alpha=\frac{z+z^{-1}}2$, $\sin\alpha=\frac{z-z^{-1}}{2i}$, and $\cos2\alpha=\frac{z^2+z^{-2}}2$, $\sin2\alpha=\frac{z^2-z^{-2}}{2i}$.
$$x_1\frac{z^2-z^{-2}}{2i}+x_2\frac{z^2+z^{-2}}2-x_3\frac{z-z^{-1}}{2i}-x_4\frac{z+z^{-1}}2=0,$$
or
$$x_1\frac{z^4-1}{2i}+x_2\frac{z^4+1}2-x_3\frac{z^3-z}{2i}-x_4\frac{z^3+z}2=0.$$
This is a general quartic equation in $z$ and unless the coefficients have special values, there is no easy solution.
|
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|
Find a closed form for this infinite sum: $ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+ \dots$ How to find a closed form for the expression??
$$ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+\frac{1\times 2\times 3\times 4}{2 \times 5\times 8\times 11}+ \cdots$$
Wolfram alpha gives,
$$\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{4\sqrt[3] {2}}+\frac{\sqrt{3}}{2\sqrt[3]{2}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}$$
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First of all, let us note that
$$\int_0^1(1-x)^{n-1}x^{-1/3}\,dx=\mathrm{B}\left(n,\frac23\right)=\frac{\Gamma(n)\Gamma(\frac23)}{\Gamma(n+\frac23)}=\frac{(n-1)!\,\Gamma(\frac23)}{(n-\frac13)(n-\frac43)\ldots\cdot \frac23 \Gamma(\frac23)}=\frac{3^n (n-1)!}{2\cdot 5\cdot\ldots\cdot(3n-1)},$$
which is almost the main term of our series (becomes it after multiplicating by $n/3^n$). Then the series in question without initial term $1$ equals to
$$S=\sum_{n=1}^\infty \frac{n!}{2\cdot 5\cdot\ldots\cdot(3n-1)}=\sum_{n=1}^\infty \frac{n}{3^n}\int_0^1(1-x)^{n-1}x^{-1/3}\,dx=$$
$$=\sum_{n=1}^\infty \frac{n}{3^n}\int_0^1(1-u^{3/2})^{n-1}\frac32\,du=\frac12\int_0^1 \sum_{n=1}^\infty n\left(\frac{1-u^{3/2}}{3}\right)^{n-1}\,du.$$
Using the relation $$\sum_{n=1}^\infty nq^{n-1}=\frac{1}{(1-q)^2}$$ (which is just the derivative of $\sum\limits_{n=1}^\infty q^{n}=\frac{1}{1-q}$) with $q=\frac{1-u^{3/2}}{3}$ we get
$$S=\frac12\int_0^1\frac{du}{(1-\frac{1-u^{3/2}}{3})^2}=\frac92\int_0^1\frac{du}{(2+u^{3/2})^2}=9\int_0^1 \frac{t\,dt}{(t^3+2)^2}.$$
Since we have antiderivative for this function of the form
$$\int \frac{9t\,dt}{(t^3+2)^2}=\frac{3t^2}{2(t^3+2)}+\frac{1}{4\sqrt[3]{2}}\ln(t^3+2)-\frac{3}{4\sqrt[3]{2}}\ln(t+\sqrt[3]{2})+\frac{\sqrt3}{2\sqrt[3]{2}}\arctan\frac{\sqrt[3]{4}t-1}{\sqrt3}+C,$$ by the fundamental theorem of calculus $$S=\frac12+\frac{1}{4\sqrt[3]{2}}\ln3-\frac{3}{4\sqrt[3]{2}}\ln(1+\sqrt[3]{2})+\frac{\sqrt3}{2\sqrt[3]{2}}\arctan\frac{\sqrt[3]{4}-1}{\sqrt3}+\frac{\sqrt3}{2\sqrt[3]{2}}\arctan\frac{1}{\sqrt3}=$$
$$=\frac12+\frac{1}{4\sqrt[3]{2}}\ln(\sqrt[3]{2}-1)+\frac{\sqrt3}{2\sqrt[3]{2}}\left(\frac{\pi}{6}+\arctan\frac{\sqrt[3]{4}-1}{\sqrt3}\right).$$
Adding again omitted earlier first term $1$, we obtain an expression for required sum which is equivalent to one given by Mathematica.
|
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Evaluate a limit using Taylor series Let $$\lim\limits_{x\to 0}\frac{({\ln(1+x) -x +\frac{x^2}{2})^4}}{(\cos(x)-1+\frac{x^2}{2})^3}$$
Now, I know that I should utilize Taylor polynomial.
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...$
$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ...$
Plugin it into the limit:
$$\lim\limits_{x\to 0}\frac{({x-\frac{x^2}{2} + \frac{x^3}{3} + R_3(x) -x +\frac{x^2}{2})^4}}{(1-\frac{x^2}{2} + \frac{x^4}{4!} + S_3(x) -1+\frac{x^2}{2})^3}$$
Simplyfing:
$$\lim\limits_{x\to 0}\frac{({\frac{x^3}{3} + R_3(x))^4}}{(\frac{x^4}{4!} + S_3(x))^3}$$
$R_3(x)$ and $T_3(x)$ are the remainders (with order of $3$).
We've learned in class that if you divide the remainder by the same order the limit is still approaces $0$ and I think that's the case here.
I'd be glad if you could show me how to end this exercise and explaining more about the remainder (Which approaches $0$ even if divided by a polynomial with the same order).
Thanks!
|
You might rearrange and factor some parts of the fraction appropriately to use $$\frac{\ln(1+x)}{x}\to 1,~~~\frac{\cos x-1}{x}\to 0$$ while $x\to 0$. Note that when $x\to 0$ then, we can simplify something like $\frac{x^4}{x^2}$ and get $x^2$.
|
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|
How do you find the modular inverse of $5\pmod{\!11}$ I need to find out the modular inverse of 5(mod 11), I know the answer is 9 and got the following so far and don't understand how to than get the answer. I know how to get the answer for a larger one such as 27(mod 392) but am stuck because they are both low numbers.
11=5 (2)+1
5=1 (5)
|
In finding a modular inverse, you are trying to solve the modular equation
$$
ax\equiv 1\pmod n.
$$
Ordinarily, you use the Extended Euclidean Algorithm for this to solve the equation $ax+ny=1$. If the numbers $a$, and $n$ are small, then simple trial and error is probably just as fast or faster.
For your example, we have $a=5$ and $n=11$, which means would could just use trial and error.
\begin{align}
1\cdot 5 &\equiv 5\\
2\cdot 5 &\equiv 10\\
3\cdot 5 &\equiv 15\equiv 4 \\
4\cdot 5 &\equiv 20\equiv -2\\
5\cdot 5 &\equiv 25\equiv 3 \\
6\cdot 5 &\equiv 30\equiv -3 \\
7\cdot 5 &\equiv 35\equiv 2 \\
8\cdot 5 &\equiv 40\equiv -4 \\
9\cdot 5 &\equiv 45\equiv 1 \\
10\cdot 5 &\equiv 50\equiv -5 \\
\end{align}
Since $9\cdot 5 \equiv 1$ then we have found the modular inverse to be 9.
When looking at those numbers on the far right side, keep in mind that any multiple of 11 made be added or subtracted to the modulus and it is still equivalent. That is, $-3\equiv 30\equiv 8$ since $-3+3(11)=30$ and $8+2(11)=30$.
|
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|
How to solve recursion? I have tried to solve some recursion:
$$f_n = \frac{2n-1}{n}f_{n-1} - \frac{n-1}{n} f_{n-2} + 1, \quad f_0 = 0, f_1 = 1$$
I would like to use a generating function:
$$F(x) = \sum_{n=0}^{\infty}f_nx^n$$
Then:
$F(x) = f_0 + f_1x + \sum_{n=2}^{\infty} \frac{2n-1}{n}f_{n-1}x^n - \sum_{n=2}^{\infty}\frac{n-1}{n} f_{n-2}x^n + \sum_{n=2}^{\infty}1x^n$
But As You see I have a problem. Could you help me ?
|
You have:
$$ x\, F'(x) = \sum_{n=1}^{+\infty} n\,f_n\, x^n,$$
$$ x\, F(x) = \sum_{n=1}^{+\infty} f_{n-1} x^n,$$
$$ x^2\, F'(x) + x F(x) = \sum_{n=1}^{+\infty}n\, f_{n-1}\, x^n,$$
$$ x^2 F(x) = \sum_{n=2}^{+\infty} f_{n-2} x^n,$$
$$ 2x^2 F(x) + x^3 F'(x) = \sum_{n=2}^{+\infty} n f_{n-2} x^n.$$
The recursion now gives:
$$ xF'(x)-x = 2(x^2 F'(x)+xF(x))-xF(x)-(2x^2 F(x)+x^3 F'(x))+x^2F(x)+\sum_{n=2}^{+\infty}n x^n$$
or just:
$$ (1-x)^2 F'(x) = (1-x)F(x)+\sum_{n=1}^{+\infty}(n+1)x^n$$
$$ F'(x) = \frac{F(x)}{(1-x)}-\frac{1}{(1-x)^2}+\frac{1}{(1-x)^4}.$$
By setting $F(x)=G(x)-\frac{1}{2(1-x)}+\frac{1}{2(1-x)^3}$ we get the ODE:
$$ G'(x) = \frac{G(x)}{1-x},$$
so:
$$ F(x) = \frac{K}{1-x}+\frac{1}{2(1-x)^3}.$$
Since $F(x)$ has an only pole of order three in $x=1$, we have that $f_n$ is a second degree-polynomial in $n$. By interpolating the first three values we further get:
$$ f_n = \frac{n(n+3)}{4}.$$
|
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|
Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\left(n-1\right)+8\right)\right)}{k\left(\left(7n+15\right)-\left(7\left(n-1\right)+15\right)\right)}=\frac{3}{7}$
As this applies for any term:
$\frac{T_{12}}{t_{12}}=\frac{3}{7}$
But this is not the answer. The actual answer is $\frac7{16}$. I know how to obtain that answer.
But why is my solution wrong? Its probably the concept I guess.
|
A short and direct answer to your question, distilling the wisdom in answers given by others here earlier:
Both $S_n$ and $s_n$ are of the form $n(An+B)$.
However when you take the ratio, $n$ gets cancelled out.
Therefore when you take the ratio of the $\underline{\text{difference}}$, $n$ (and also $n-1$) should be reinstated accordingly to arrive at the correct answer.
In summary,
$$\begin{align}
\dfrac{S_n}{s_n}&=\dfrac{3n+8}{7n+15}=\dfrac{n(3n+8)}{n(7n+15)}\\
\dfrac{S_{12}}{s_{12}}&=\dfrac {44}{99}\qquad =\dfrac {12\cdot 44}{12\cdot 49}\\
\dfrac{S_{12}-S_{11}}{s_{12}-s_{11}}&\quad\qquad\;\quad=\dfrac{12\cdot44-11\cdot41}{12\cdot 99-11\cdot 92}=\dfrac{48-41}{108-92}=\dfrac7{16}\\
\end{align}$$
|
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Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
|
This is another way of solving this problem using Completing the Square method.
$x^2+2xy-y^2=??$
First we have,
$x^2+2xy-y^2=0$
$x^2+2xy=y^2,$
By adding both sides by $y^2$ to make it perfect square then,
$x^2+2xy+y^2=y^2+y^2$
$(x+y)^2=2y^2$
$x+y=[2^(1/2)]y$
$x=(+ or -)[2^(1/2)]y-y$
$x=[2^(1/2)]y-y ; x=-[2^(1/2)]y-y $
$x=[2^(1/2)-1]y ; x=-[2^(1/2)-1]y$
therefore we have,
${x-[2^(1/2)-1]y}{x+[2^(1/2)-1]y}$
By solving
$x^2-2xy-y^2=??$
Try to solve this by relying with this pattern. Thank you.
|
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|
Ranges in trigonometry How to find the range of the sum or difference of two trigonometric functions?
$2\sin x-3\cos x$
Before this whenever the question of range i have solved they were either single trigonometric function or if they were in pair then they were in form of squires (eg. $2\cos^2x + \sin^2x$) so i used to convert them into one function.
I don't know how to solve this type of problem, i think we here also we need to change the it into single function?
|
Let $\theta$ be an angle which satisfies $\cos \theta = \dfrac{2}{\sqrt{13}}$ and $\sin \theta = -\dfrac{3}{\sqrt{13}}$. Then we have:
$2\sin x - 3\cos x$
$= \sqrt{13}\left(\dfrac{2}{\sqrt{13}}\sin x - \dfrac{3}{\sqrt{13}}\cos x\right)$
$= \sqrt{13}\left(cos\theta\sin x + \sin\theta\cos x\right)$
$= \sqrt{13}\sin(x+\theta)$.
Do you know what the range of $\sqrt{13}\sin(x+\theta)$ is?
Using the same method, we can get the general result $A\sin x + B\cos x = \sqrt{A^2+B^2}\sin(x+\theta)$ where $\theta$ satisfies $\cos\theta = \dfrac{A}{\sqrt{A^2+B^2}}$ and $\sin\theta = \dfrac{B}{\sqrt{A^2+B^2}}$. See this pdf for details.
Thanks Freddy for suggesting to include that link in my answer.
|
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|
Distinct balls into distinct boxes with a minimal number of balls in each box
Find the number of ways to distribute $8$ distinct balls into $3$ distinct boxes if each box must hold at least $2$ balls.
The stars and bars approach would not work because the balls are non-identical. Stirling Numbers of the second kind would also not work directly, because each box doesn't just have to be non-empty; they must contain at least $2$ balls each.
I tried the following approach: Since each box must contain at least $3$ balls each, then the $8$ distinct balls must be divided into distinct partitions of sizes $2,2,4$ or $2,3,3$.
For the first case, we choose $2$ of the $8$ balls for the first partition, $2$ of the $6$ remaining balls for the second partition, and $4$ of the $4$ remaining balls for the third partition. Since the partitions are distinct, we multiply by $3!$ to give:
$$\binom{8}{2}\binom{6}{2}\binom{4}{4}3!$$
Apply the same approach, the number of ways for the second case is:
$$\binom{8}{2}\binom{6}{3}\binom{3}{3}3!$$
Hence, the total number of ways is
$$\binom{8}{2}\binom{6}{2}\binom{4}{4}3! +\binom{8}{2}\binom{6}{3}\binom{3}{3}3!$$
Is my approach valid? Also, is it possible to transform the question to make use of Stirling numbers of the second kind?
|
Associate a different colour to each of the three boxes; then you are looking for the number of ways to colour the balls such that the frequency vectors of colours used is one of
$$
(2,2,4), (2,3,3), (2,4,2), (3,2,3), (3,3,2), (4,2,2)
$$
The number for each of these colourings is given by a trinomial coefficient, and by symmtry one can simply compute
$$
3\binom8{2,2,4}+3\binom8{2,3,3} = 3\times(420+560) = 2940.
$$
So basically your approach is correct, but your rather unclearly motivated multiplication by $3!=6$ is not right; rather, each pattern should be multiplied by the number of its distinct permutations, which happens to be$~3$ here both for $(2,2,4)$ and for $(2,3,3)$.
|
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Sum of an unorthodox infinite series $ \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots $
This is a pretty unorthodox problem, and I'm not quite sure how to simplify it. Could I get a solution? Thanks.
|
Another approach is to use summation by parts. This method works as follows. If
$$S_N = \sum_{n=1}^{N} a_n b_n$$
then we can define
$$B_n = \sum_{k=1}^{n} b_k$$
Then
$$S_N = a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n)$$
In this problem, we can set $a_n = 2n-1$ and $b_n = 1/2^n$. Then
$$B_n = \sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n}$$
and $a_{n+1} - a_n = 2$ for all $n$. Therefore,
$$\begin{align}S_N &= a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n) \\
&= (2N-1)\left(1 - \frac{1}{2^N}\right) - \sum_{n=1}^{N-1} 2\left(1 - \frac{1}{2^n}\right) \\
&= 2N - 1 - \frac{2N - 1}{2^N} - 2(N-1) + 2\sum_{n=1}^{N-1} \frac{1}{2^n} \\
&= 1 - \frac{2N - 1}{2^N} + 2\sum_{n=1}^{N-1} \frac{1}{2^n} \\
\end{align}$$
The middle term goes to zero as $N \rightarrow \infty$, and the rightmost term goes to $2$. So in the limit, we get $1 + 2 = 3$.
|
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How prove this triangulation with indentity
let $x,y,z\in (0,\pi)$, prove or disprove
$$\sin{(x+y)}\cdot\sin{(y+z)}\cdot\sin{(x+z)}\cdot\sin{(x+y+z)}
=[\sin{(x+y+z)}\cdot\sin{x}+\sin{y}\cdot\sin{z}]\times[\sin{(x+2y+z)}\cdot\sin{z}+\sin{(x+y)}\cdot\sin{y}]$$
if this is ture, we can use this
$$2\sin{x}\sin{y}=\cos{(x-y)}-\cos{(x+y)}$$
But I fell very ugly.
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HINT:
From the RHS,
Using Werner Formula
$$2\sin(x+2y+z)\sin z=\cos(x+2y)-\cos(x+2y+2z)$$
and $$2\sin(x+y)\sin y=\cos x-\cos(x+2y)$$
Using Prosthaphaeresis formula,
$$2\sin(x+2y+z)\sin z+2\sin(x+y)\sin y=\cos x-\cos(x+2y+2z)=2\sin(y+z)\sin(x+y+z)$$
Can you apply the same method on $$[\sin(x+y+z)\sin x]+[\sin y\sin z]?$$
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|
Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$ I am trying to find out the sum of this
$$1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\frac{1\cdot 2\cdot 3\cdot 4}{2\cdot 5\cdot 8\cdot 11}+\cdots$$.
I tried with binomial theorem with rational index. But in vain. What shall I do to solve it?
If it has been solved already, kindly provide me the link. I am unable to get that.
Thanks for your help
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Since $$S=1+\sum_{k=0}^{+\infty}\prod_{j=0}^k\frac{j+1}{3j+2}=1+\sum_{k=1}^{+\infty}k\, 3^{-k} B(2/3,k),$$
where $B(2/3,k)$ is the Euler Beta function:
$$ B(2/3,k)=\int_{0}^{1}x^{-1/3}(1-x)^{k-1}$$
we have: $$S=1+3\int_{0}^{1}\frac{dx}{(1-x)^{1/3}(3-x)^2}.$$
Now the last integral can be computed explicitly, and leads to:
$$S=\frac{3}{2}-2^{-4/3}\sqrt{3}\operatorname{arccot}\left(\frac{1-2^{4/3}}{\sqrt{3}}\right)+2^{-7/3}\log\left(4+2^{4/3}-2^{5/3}\right)-2^{-4/3}\log\left(2+2^{2/3}\right).$$
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Squared binomial paradox? When you square this $$(5-2)^2$$ you will get 49 $$ 5^2 - 2 * 5 * (-2) + (-2)^2$$
$$25 + 20 + 4 = 49$$ but if you do it like this (5-2) * (5-2) you will get 9 $$ 5(5-2) - 2(5-2)$$
$$25-10-10+4$$
$$25-20+4$$
$$5+4$$
$$9$$
Why do I always get different results if I'm doing the same thing?Am I doing something wrong?
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The binomial formular is $(a + b)^2 = a^2 + 2ab + b^2$
If you apply it:
$$(5 - 2)^2 = 5^2 + 2\cdot5\cdot(-2) + 2^2 = 25-20 + 4 = 9$$
The second binomial formular is $(a - b)^2 = a^2 - 2ab + b^2$
If you apply it:
$$(5 - 2)^2 = 5^2 - 2\cdot5\cdot 2 + 2^2 = 25-20 + 4 = 9$$
Your mistake was, that you used the second binomial formular but took $a$ as $-2$ instead as $2$.
|
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Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
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I find this easier to visualize if I write this problem in terms of powers:
$$\dfrac{n^{1/2}}{\left[n + \left(n + n^{1/2}\right)^{1/2}\right]^{1/2}}\text{.} $$
Division by $n^{1/2}$ for both the numerator and denominator turns the numerator into $1$ and the denominator into
$$\begin{align*}
\dfrac{\left[n + \left(n + n^{1/2}\right)^{1/2}\right]^{1/2}}{n^{1/2}} &= \left[\dfrac{n + \left(n + n^{1/2}\right)^{1/2}}{n}\right]^{1/2}\quad \text{ since }\dfrac{a^c}{b^c} = \left(\dfrac{a}{b}\right)^{c} \\
&= \left[\dfrac{n}{n} + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{n}\right]^{1/2} \quad \text{ since }\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c} \\
&= \left[1 + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{n}\right]^{1/2} \\
&= \left[1 + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{\left(n^{2}\right)^{1/2}}\right]^{1/2} \\
&= \left[1 + \left(\dfrac{n+n^{1/2}}{n^{2}}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a^c}{b^c} = \left(\dfrac{a}{b}\right)^{c} \\
&= \left[1 + \left(\dfrac{n}{n^2}+\dfrac{n^{1/2}}{n^2}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c} \\
&= \left[1 + \left(\dfrac{1}{n}+\dfrac{1}{n^{3/2}}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a^b}{a^c} = a^{b-c} \\
&= \left\{1 + \left[\dfrac{1}{n}+\dfrac{1^{1/2}}{(n^{3})^{1/2}}\right]^{1/2}\right\}^{1/2} \\
&= \left\{1 + \left[\dfrac{1}{n}+\left(\dfrac{1}{n^3}\right)^{1/2}\right]^{1/2}\right\}^{1/2} \\
&= \sqrt{1+\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}}}}\text{.}
\end{align*}$$
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A hard square root question This is my first question on StackExchange. So my question is:
If $$x = \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{\sqrt{\sqrt5 + 1}} + \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{2\sqrt{\sqrt5 + 1}} - \sqrt{3-2\sqrt2}$$
What is the value of $x^2$?
If someone can also tell me how to input mathematical functions on the StackExchange it would be appreciated.
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Letting
$$\frac{\sqrt{\sqrt 5+2}+\sqrt{\sqrt 5-2}}{\sqrt{\sqrt 5+1}}=\alpha,$$
we have
$$\alpha^2=\frac{(\sqrt 5+2)+(\sqrt 5-2)+2\sqrt{(\sqrt 5+2)(\sqrt 5-2)}}{\sqrt 5+1}=\frac{2(\sqrt 5+1)}{\sqrt 5+1}=2\Rightarrow \alpha=\sqrt 2.$$
Hence, we have
$$x=\alpha+\frac{\alpha}{2}-\sqrt{(\sqrt 2-1)^2}=\frac{3}{2}\alpha-(\sqrt 2-1)=\frac{\sqrt 2}{2}+1\Rightarrow x^2=\frac{3}{2}+\sqrt 2.$$
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Minimum eigenvalue of product of two matrices Abstract description:
Let $\mathbf{A}$ and $\mathbf{B}$ be two $n \times n$ real matrices. Let $\sigma( \mathbf{A B} )$ denote the spectrum of $\mathbf{A B}$.
Assume that
(A1) $\mathbf{A}$ is symmetric and positive definite but not diagonal.
(A2) $\mathbf{B}$ is not symmetric but similar to a symmetric and positive semidefinite (but not positive definite) matrix.
(A3) $\sigma( \mathbf{A B} ) \subset \mathbb{R}$.
Question 1: Do (A1) and (A2) imply (A3)?
Question 2: Can we say something about $\min\sigma( \mathbf{A B} )$?
Detailed description and background:
Let $G$ be a simple graph of order $m$. That is, $G$ is undirected and without loops. $G$ can have vertices with zero degree. Let $H$ denote the graph resulting from G by adding loops to all vertices with zero degree. Let $\mathbf{I}_m$ denote the identity matrix of order $m$ and $\mathbf{O}_m$ the zero matrix of order $m$. Let $\mathbf{S}$ denote the adjacency matrix of $H$. Hence, $\mathbf{S}$ is an $m \times m$ symmetric matrix with nonnegative entries. Let $\mathbf{D}$ be the degree matrix of $H$. Hence, $\mathbf{D}$ is an $m \times m$ diagonal and positive definite matrix. Let $c \in ( -1, 1)$. Note that $\mathbf{I}_m - c \mathbf{D}^{-1} \mathbf{S}$ is nonsingular because $\mathbf{D}^{-1} \mathbf{S}$ is a right stochastic matrix that has spectral radius equal to $1$. Define $\mathbf{M} := ( \mathbf{I}_m - c \mathbf{D}^{-1} \mathbf{S} )^{-1}$.
Conjecture: $\min\sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m ) > 0$ or, equivalently, $\mathbf{M} + \mathbf{M}' - \mathbf{I}_m$ is positive definite.
So far, my thoughts on a proof of this conjecture are as follows:
(P1) $\mathbf{D}^{-1} = \mathbf{P}^{2}$ with $\mathbf{P}$ diagonal and positive definite. That is, $\mathbf{P}$ is the unique square root of $\mathbf{D}^{-1}$.
(P2) $\mathbf{D}^{-1} \mathbf{S}$ is similar to the symmetric matrix $\mathbf{P} \mathbf{S} \mathbf{P}$. Indeed, $\mathbf{P}^{-1} \mathbf{D}^{-1} \mathbf{S} \mathbf{P} = \mathbf{P} \mathbf{S} \mathbf{P}$.
(P3) $\sigma( \mathbf{M} - (1/2) \mathbf{I}_m ) \subset \mathbb{R}$ because of (P2).
(P4) It can be shown that $\sigma( \mathbf{M} - (1/2) \mathbf{I}_m ) \subset ( 0, +\infty )$.
(P5) We have $$\begin{align*}
& \sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m )\\
& = \sigma\bigl( \mathbf{M} - (1/2) \mathbf{I}_m + \mathbf{M}' - (1/2) \mathbf{I}_m \bigr)\\
& = \sigma\bigr( \mathbf{P}^{-1} \bigl( \mathbf{M} - (1/2) \mathbf{I}_m + \mathbf{M}' - (1/2) \mathbf{I}_m \bigr) \mathbf{P} \bigr)\\
& = \sigma\bigr( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m + \mathbf{P}^{-2} ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} \mathbf{P}^{2}- (1/2) \mathbf{I}_m \bigr)\\
& = \sigma\Biggl( \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr)' \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr)^{-1} \biggl( \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr) \biggr) \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr) \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr) \Biggr)\\
& \subset \sigma\Biggl( \biggl( \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr) \biggr) \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr) \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr) \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr)' \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr)^{-1} \Biggr)\\
& = \sigma\Biggl( \biggl( \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr) \biggr) \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{D}\\\mathbf{D}^{-1} & \mathbf{I}_m\end{array} \biggr) \Biggr)\\
& = \sigma( \mathbf{A} \mathbf{B} ),
\end{align*}$$ where $\mathbf{A} := \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr)$ is symmetric and positive definite because of (P4) and $\sigma( \mathbf{M} - (1/2) \mathbf{I}_m ) = \sigma( \mathbf{P}^{-1} ( \mathbf{M} - (1/2) \mathbf{I}_m ) \mathbf{P} ) = \sigma( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m )$ and $$\mathbf{B} := \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{D}\\\mathbf{D}^{-1} & \mathbf{I}_m\end{array} \biggr) = \dfrac{1}{\sqrt{2}} \biggl( \begin{array}{cc}\mathbf{I}_m & -\mathbf{I}_m\\\mathbf{D}^{-1} & \mathbf{D}^{-1}\end{array} \biggr) \biggl( \begin{array}{cc}2\mathbf{I}_m & \mathbf{O}_m\\\mathbf{O}_m & \mathbf{O}_{m}\end{array} \biggr) \biggl( \dfrac{1}{\sqrt{2}} \biggl( \begin{array}{cc}\mathbf{I}_m & -\mathbf{I}_m\\\mathbf{D}^{-1} & \mathbf{D}^{-1}\end{array} \biggr) \biggr)^{-1}$$ is (in general) not symmetric but similar to the symmetric and positive semidefinite matrix $$\biggl( \begin{array}{cc}2\mathbf{I}_m & \mathbf{O}_m\\\mathbf{O}_m & \mathbf{O}_{m}\end{array} \biggr) = \biggl( \begin{array}{cc} 2 & 0\\ 0 & 0 \end{array} \biggr) \otimes \mathbf{I}_m.$$ Note that $\sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m ) \cup \{ 0 \} = \sigma( \mathbf{A} \mathbf{B} )$, so that $\sigma( \mathbf{A} \mathbf{B} ) \subset \mathbb{R}$ because $\sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m ) \subset \mathbb{R}$.
Any ideas as to how to finish the proof? So far, I did not come up with a counter example.
EDIT:
I have reformulated assumption (A2) because it was stated inconsistently. It read as follows: $\mathbf{B}$ is not symmetric but similar to a nondiagonal, symmetric, and positive semidefinite matrix. In addition, I added the condition that $\mathbf{B}$ is not positive definite. I have also made some minor changes in Detailed description and background, which are related to $\mathbf{B}$.
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According to the new statement, $A=\begin{pmatrix}20&-4&0\\-4&4&0\\0&0&1\end{pmatrix},B=\begin{pmatrix}1&0&0\\2&2&0\\0&0&0\end{pmatrix}$.
When the next change ?
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Exotic proofs of $\sum_{j=0}^{n-1}\binom{p+j}{p}=\binom{p+n}{p+1}$ Let $p,n$ be positive integers.
The following identity $\displaystyle \sum_{j=0}^{n-1}\binom{p+j}{p}=\binom{p+n}{p+1}$ may be proved by induction or by successive uses of Pascal's rule (both techniques are demonstrated here).
Question
Is there a combinatorial proof for this identity ? Do you know other proofs ?
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First Proof:
\begin{align}
\sum^{n-1}_{j=0}\binom{p+j}{p}
&=\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^p}{z^{p+1}}\sum^{n-1}_{j=0}(1+z)^j{\rm d}z\\
&=\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^{n+p}-(1+z)^p}{z^{p+2}}{\rm d}z\\
&=\frac{1}{(p+1)!}\lim_{z \to 0}\frac{{\rm d}^{p+1}}{{\rm d}z^{p+1}}\left[(1+z)^{n+p}-(1+z)^p\right]\\
&=\frac{(n+p)(n+p-1)\cdots(n+1)(n)}{(p+1)!}\\
&=\binom{n+p}{p+1}
\end{align}
Second Proof:
\begin{align}
\sum^{n-1}_{j=0}\binom{p+j}{p}
&=[x^N]\sum^\infty_{N=0}\sum^{N=n-1}_{j=0}\binom{p+j}{j}x^N\\
&=[x^N]\sum^\infty_{j=0}\binom{p+j}{j}\sum^\infty_{N=j}x^N\\
&=[x^N]\sum^\infty_{j=0}\binom{p+j}{j}\frac{x^j}{1-x}\\
&=[x^{n-1}]\frac{1}{(1-x)^{p+2}}\\
&=\binom{p+2+n-1-1}{n-1}\\
&=\binom{p+n}{p+1}
\end{align}
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|
Is my solution to this differential equation correct?
My answer is: $[(1+x^2)^3]y = \dfrac{(1+x^2)^3}3+C$
But this option is not given, so is it correct?
Thanks
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$$(1+x^2)\frac{dy}{dx}+6xy=2x \Rightarrow (1+x^2)\frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2)\frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{1}{1-3y}dy=\frac{2x}{1+x^2}dx \Rightarrow \int \frac{1}{1-3y}dy=\int \frac{2x}{1+x^2}dx \\ \Rightarrow \frac{-1}{3}\ln{|1-3y|}=\ln{|1+x^2|}+c\Rightarrow \ln{|1-3y|}^{\frac{-1}{3}}=\ln{|1+x^2|}+c \\ \Rightarrow e^{\ln{|1-3y|}^{\frac{-1}{3}}}=e^{\ln{|1+x^2|}+c } \\ \Rightarrow |1-3y|^{\frac{-1}{3}}=C|1+x^2|, C=e^c$$
Solve now for $y$.
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How to simplify $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1}$? This is the original problem:
$\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = x$.
I'm really confused about how to solve this problem, I come as far as saying this: $\sqrt[4]{5} + \sqrt{1}\cdot \sqrt[4]{5}-\sqrt{1}$.
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Hint: Start with $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = \sqrt{(\sqrt{5}+1) \cdot (\sqrt{5}-1)}$, and then use the difference of squares identity, i.e. $(a+b)(a-b) = a^2-b^2$.
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If the equation $|x^2+bx+c|=k$ has four real roots .....
If the equation $|x^2+bx+c|=k$ has four real roots then which of the following option is correct :
(a) $b^2-4c >0$ and $0<k<\frac{4c-b^2}{4}$
(b) $b^2-4c <0$ and $0<k<\frac{4c-b^2}{4}$
(c) $b^2-4c >0$ and $0<k>\frac{4c-b^2}{4}$
Please suggest how to proceed in this problem , I am getting no clue on this.. please help thanks.
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Hint: Clearly $k$ is positive. Also, both
$$ x^2+bx + (c-k) = 0 $$
and $$ x^2+bx + (c+k) = 0 $$
have real roots (that is, both equation discriminants are positive).
(I have been assuming everywhere that there are four distinct real roots.)
Note that $k>0$ and positivity of the second discriminant, $b^2-4(c+k)>0$ imply $b^2-4c>4k>0$, so $b^2-4c>0$. The third inequality comes directly from the positivity of the first discriminant, $b^2-4(c-k)>0$.
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Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$ if $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$
if $a,b,c,d$ are positive real numbers and $$a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$$
Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$
Things i have done: from assumption $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$ I can conclude that $$a^2+b^2-ab=c^2+d^2-cd$$
Powering both sides to two gives $$a^4+b^4+a^2b^2-2a^3b-2ab^3=c^4+d^4+c^2d^2-2c^3d-2cd^3$$
And $$a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4 \rightarrow 2a^4+2b^4+6a^2b^2-2ab^3-2a^3b=2c^4+2d^4+6c^2d^2-2cd^3-2c^3d$$
I can't continue any more.
|
Observe that:
$$
a^4+b^4+(a-b)^4-c^4-d^4-(c-d)^4\\
=2(a^2-ab+b^2-c^2+cd-d^2)(a^2-ab+b^2+c^2-cd+d^2)
$$
so the result follows because you have proved the first term on the RHS is $0$.
|
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|
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
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$(x+y)^2 \geq xy$
If one of $x, y$ is negative, this is obvious due to sign. Otherwise, WLOG, assume both $x$ and $y$ are positive. Since $x+y \geq x$ and $x+y \geq y$, then $(x+y)^2 \geq xy$.
|
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|
Complex analysis: Rewrite $\cos^{-1}{i}$ in algebraic form I'm stuck in this problem (complex analysis), my answer is not the one reported in the book:
Rewrite $\cos^{-1}{i}$ in the algebraic form. A: $k\pi + i \frac{\ln{2}}{2}\ \forall\ k \in \mathbb{Z}$
So I tried this approach in order to solve it:
*
*As $\cos^{-1}{z} = -i \ln{\left( z \pm \sqrt{z^2 - 1} \right)}$, doing $z = i$ and $i^2 = -1$ where necessary, we have
$$\cos^{-1}{i} = -i \ln{\left( i \pm \sqrt{i^2 - 1} \right)} = -i \ln{\left( i \pm \sqrt{-2} \right)} = -i \ln{\left( i \pm i \sqrt{2} \right)}$$
*
*Factoring $i$ and separating the obtained logarithm of product into sum of logarithms:
$$\cos^{-1}{i} = -i \ln{\left[ i \left( 1 \pm \sqrt{2} \right) \right]} = -i \big[ \ln{i} + \ln{\left( 1 \pm \sqrt{2} \right)} \big]$$
*
*Solving $\ln{i}$ separately, we obtain $\ln{i} = \pi i \left( 2k + \frac{1}{2} \right)$
*Solving $\ln{\left( 1 \pm \sqrt{2} \right)}$, on the other hand, yields
$$\ln{\left( 1 \pm \sqrt{2} \right)} = \begin{cases} \ln{\left( \sqrt{2} + 1 \right)} + \pi i \cdot 2k & (+) \\ \ln{\left( \sqrt{2} - 1 \right)} + \pi i \left( 2k + 1 \right) & (-) \end{cases}$$
*
*Substituting these expressions into the original one, we have
$$\cos^{-1}{i} = \begin{cases} \pi \left( 4k + \frac{1}{2} \right) -i \ln{\left( \sqrt{2} + 1 \right)} & (+) \\ \pi \left( 4k + \frac{3}{2} \right) -i \ln{\left( \sqrt{2} - 1 \right)} & (-) \end{cases}$$
which obviously doesn't correspond with the book. Is something wrong?
Thanks in advance :)
|
Let me obtain an answer similar to yours by different means. Rather than use the algebraic form of the inverse cosine, I'll invert to solve for $x=e^{iz}$ and then invert again to get $z$. Carrying this out gives
\begin{align}
z=\cos^{-1} i
&\underset{\text{invert}}{\implies} i=\cos z=\frac{1}{2}(x+x^{-1})\implies x^2-2i x+1=0\\
&\underset{\text{solve}}{\implies} x=\dfrac{1}{2}\left(2i\pm \sqrt{-8}\right)=\pm i(\sqrt{2}\pm 1)\\
&\underset{\text{invert}}{\implies} z=-i\log x=\pm \frac{\pi}{2}+2\pi n-i\log(\sqrt{2}\pm 1)
\end{align}
This does not appear to agree with the book either...
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Proving a trigonometric inequality $(1-\sin a)x^2 -2x\cos a + 1+ \sin a \ge 0$ $(1-\sin (a))x^2 -2x\cos(a) + 1+ \sin( a) \ge 0$, where $a,x$ are any two real constants.
Any suggestions on how to prove this ? I tried playing with it, but nothing useful came out.
|
$\cos^2 a = 1-\sin^2a=(1+\sin a)(1-\sin a)$
$1+\sin a = \frac {\cos^2 a}{1-\sin a}$
Substituting this value in the equation --
$(1-\sin a)x^2 -2x\cos(a) + [1+ \sin( a)] $
=$(1-\sin a)x^2 -2x\cos(a) + \frac {\cos^2 a}{1-\sin a} $
=$[x \root \of {1-\sin a} - \frac {\cos a}{\root \of {1-\sin a}} ]^2$
$\ge 0$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Linear algebra system of equations Given the following linear system:
$x - y + 2z + 3w = 12$
$ x - 2y + z + w = 8$
$4x + y + 2z - w = 3$
1) find all solutions of the system. Write the solution in Vector Form
2) find all solutions of the corresponding homogeneous system
I have turned this into a matrix and worked it out two ways resulting in different outcomes, neither of which is solvable...
I'll try to show my work, but I am new to this site and the formatting for matrices is fairly hard to figure out...
1 -1 2 3 12
1 -2 1 1 8
4 1 2 -1 3 (-4 row1)
1 -1 2 3 12
1 -2 1 1 8 (-row1)
0 5 -6 -13 -45
1 -1 2 3 12
0 -1 -1 -2 -4
0 5 -6 -13 -45(+5row2)
1 -1 2 3 12
0 -1 -1 -2 -4
0 0 -11 -23 -65
from here I don't think there's anything else I can do, I'm fairly sure there are infinitely many solutions.
|
Oh of course there is more that can be done!
So starting from where you left, I would first multiple $R_2$ with the scalar $(-1)$ and then add the new $R_2$ to $R_1$ and replace it. Then I get,
$$
\left({\begin{matrix}
1 & 0 & 3 & 5 & 16 \\
0 & 1 & 1 & 2 & 4 \\
0 & 0 & -11 & -23 & -65 \\
\end{matrix}}\right)
$$
$$\underrightarrow{\left({\frac{-1}{11}}\right) R_3 \ \text{and} -1 R_3 + R_2 \ \text{and} -3R_3 + R_1 } $$
$$ \left({\begin{matrix}
1 & 0 & 0 & \frac{-14}{11} & \frac{-19}{11}\\
0 & 1 & 0 & \frac{-1}{11} & \frac{-21}{11} \\
0 & 0 & 1 & \frac{23}{11} & \frac{65}{11} \\
\end{matrix}}\right) $$
Now translating back to equations you have,
$$ x = \frac{14}{11}w + \frac{19}{11}$$
$$ y = \frac{1}{11}w + \frac{21}{11}$$
$$ z = \frac{-23}{11}w + \frac{-65}{11}$$
Now any value you take for $w$ will solve the system. The method underlined here is made formal in Linear Algebra by Hoffman, Kunze which you should definitely read.
Now the solution set in vector form would be,
$$ \left({\begin{matrix}
x \\
y \\
z \\
w \\
\end{matrix}}\right) = w\left({\begin{matrix}
\frac{14}{11} \\
\frac{1}{11} \\
\frac{-23}{11} \\
1 \\
\end{matrix}}\right) + \left({\begin{matrix}
\frac{19}{11} \\
\frac{21}{11} \\
\frac{-65}{11} \\
0 \\
\end{matrix}}\right)$$
where $w$ is any scalar in $\Bbb R$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the generating function of the sequence $a_n = \sum\limits_{k=0}^n k(k-1)$
Find the generating function of the sequence $ a_n =\sum\limits_{k=0}^n k(k-1)$
My try:
Let's assume $k(k-1)$ is genereated by $F(x)$ then $a_n$ is generated by $\frac{F(x)}{1-x}$ (that's a common trick).
So have we reduced the problem to: Find the generating function of $a_k = k(k-1)$.
Now, there's another trick: If $G(x)$ generates $b_k$ then $x\cdot G'(x)$ generates $c_k = n\cdot b_k$
If so, we need to find the generating function of $c_k = k-1$.
I'm not sure I'm on the right path, can you help me with it?
Thanks!
|
from your initial conditions, we have
$$a_n=\sum_{k=0}^nk^2-\sum_{k=0}^nk\\
=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\\
=\frac{n(n+1)(n-1)}{3}$$
thus the generating function is
$$G(a_n,x)=\frac{1}{3}\sum_{n=0}^\infty(n^3-n)x^n\\
=\frac{1}{3}\left[\sum_{n=0}^\infty n^3x^n-\sum_{n=0}^\infty nx^n\right]$$
other hand, we have
$$<1,1,1,1...>=\frac{1}{1-x}\\
\Rightarrow<1,2,3,4...>=\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}\\
\Rightarrow<0,1,2,3...>=x\cdot\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}$$
so $\sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}$.
Then
$$<1,4,9,16...>=\frac{d}{dx}\frac{x}{(1-x)^2}=\frac{1+x}{(1-x)^3}\\
\Rightarrow<0,1,4,9...>=x\cdot\frac{1+x}{(1-x)^3}=\frac{x(1+x)}{(1-x)^3}\\
\Rightarrow<1,2^3,3^3...>=\frac{d}{dx}\frac{x(1+x)}{(1-x)^3}=\frac{1+4x+x^2}{(1-x)^4}\\
\Rightarrow<0,1,2^3,3^3...>=x\cdot\frac{1+4x+x^2}{(1-x)^4}=\frac{x+4x^2+x^3}{(1-x)^4}$$
which means $\sum_{n=0}^\infty n^3x^n=\frac{x+4x^2+x^3}{(1-x)^4}$.
So the result is
$$G(a_n,x)=\frac{1}{3}\left[\frac{x+4x^2+x^3}{(1-x)^4}-\frac{x}{(1-x)^2}\right]\\
=\frac{1}{3}\frac{6x^2}{(1-x)^4}\\
=\frac{2x^2}{(1-x)^4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $m_1 , m_2,m_3,m_4\in\mathbb{Q}$ s.t. $\forall a_k,b_k\in\mathbb Z,\,m_1(a_1^2+a_2^2)+m_2(a_3^2+a_4^2)\neq m_3(b_1^2+b_2^2)+m_4(b_3^2+b_4^2)$ Let us assume that $a_1 , a_2 , a_3 ,a_4,b_1,b_2,b_3,b_4\in\mathbb{Z}$.
If $m_1 , m_2,m_3,m_4\in\mathbb{Q}$, then how can I choose $m_1,m_2,m_3,m_4$, such that the following equation is $never$ satisfied? (all $a_i$'s and $b_i$'s can not be all zero at the same time) $$m_1(a_1^2+a_2^2)+m_2(a_3^2+a_4^2)=m_3(b_1^2+b_2^2)+m_4(b_3^2+b_4^2)$$
Note that the $m_i$'s are all $positive$ numbers and can not be varying with $a_i$'s and $b_i$. Thank you.
|
As I said, for 8 unknown parameters and the formula goes bulky.
$$a(z_1^2+z_2^2)+b(z_3^2+z_4^2)=c(z_5^2+z_6^2)+j(z_7^2+z_8^2)$$
3 - the formula looks like this: Solutions to $ax^2 + by^2 = cz^2$
Will consider here the special case when: $a+b=c+j$ $(1)$
Then the solutions are of the form:
$$z_1=js^2+jt^2+ck^2+cp^2-bq^2-bx^2-ay^2$$
$$z_2=js^2+jt^2+ck^2+cp^2-bq^2-bx^2+ay^2+2(bx+bq-js-jt-ck-cp)y$$
$$z_3=js^2+jt^2+ck^2+cp^2-bq^2+bx^2-ay^2+2(ay+bq-js-jt-ck-cp)x$$
$$z_4=js^2+jt^2+ck^2+cp^2+bq^2-bx^2-ay^2+2(ay+bx-js-jt-ck-cp)q$$
$$z_5=js^2+jt^2+ck^2-cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-js-jt-ck)p$$
$$z_6=js^2+jt^2-ck^2+cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-js-jt-cp)k$$
$$z_7=js^2-jt^2+ck^2+cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-js-ck-cp)t$$
$$z_8=jt^2-js^2+ck^2+cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-jt-ck-cp)s$$
$s,t,k,p,q,x,y$ - integers asked us.
It is clear that if you will satisfy the condition $(1)$ we can always write such a simple solution. It is easy enough to see how it turns out.
|
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|
Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$ I have a problem with the following integral:
$$
\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\,
{{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}
$$
The first idea was to use the integration by parts because
$$
\int{{\rm d}x \over \,\sqrt{x\left(\,1 - x\,\right)\,}\,}
=\arcsin\left(\,2x - 1\,\right) + C
$$
but what would be the next step is not clear. Another idea would be expand
$\ln\left(\,\cdot\right)$ into Taylor series but it seems to be even worse option.
So, what are the other options?
|
We have:
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx = 2\pi\log\frac{2+\sqrt{3}}{2}.\tag{1}$$
This happens because:
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=2\int_{0}^{1}\frac{\log(3+x^2)}{\sqrt{1-x^2}}\,dx =\int_{-\pi/2}^{\pi/2}\log(3+\cos^2\theta)\,d\theta,$$
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=\frac{1}{2}\int_{-\pi}^{\pi}\log\left(\frac{7+\cos\theta}{2}\right)d\theta=-\pi\log 2+\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta.$$
Now comes an interesting technique - we have:
$$\begin{eqnarray*}\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta &=& \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n}\log\left(7+\cos\frac{k\pi}{n}\right)\\&=&\lim_{n\to +\infty}\frac{\pi}{n}\log\prod_{k=1}^{n}\left(7+\cos\frac{k\pi}{n}\right)\end{eqnarray*}$$
but since:
$$z^{2n}-1 = \prod_{k=1}^{2n}\left(z-e^{\frac{\pi i k }{n}}\right)=(z^2-1)\prod_{k=1}^{n-1}\left(z^2+1-2z\cos\frac{k\pi}{n}\right)$$
and the solutions of
$$\frac{z^2+1}{2z}=-7$$
are $z=\pm 4\sqrt{3}-7$, it follows that:
$$\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta=\pi\log\left(\frac{7}{2}+2\sqrt{3}\right).$$
With the same technique we can prove:
$$\int_{0}^{1}\frac{\log(3-x)}{\sqrt{x(1-x)}}\,dx = \pi\log\left(\frac{5}{4}+\sqrt{\frac{3}{2}}\right),\tag{2}$$
hence we have:
$$\int_{0}^{1}\frac{\log\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\,dx = \pi\log\left((7+4\sqrt{3})(5-2\sqrt{6})\right).\tag{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $ \int \frac{dx}{x\sqrt{1-x^4}}$ Find $\displaystyle \int \dfrac{dx}{x\sqrt{1-x^4}}$
I cannot figure out how start this problem, can anyone explain
|
The substitution $u=\sqrt{1-x^{4}}$ converts the integral into a rational
function of $u$. Since $x^{4}=1-u^{2}$ and $du=-\frac{2x^{3}}{u}dx$, we get
\begin{equation*}
I=\int \frac{dx}{x\sqrt{1-x^{4}}}=\int \frac{1}{xu}\left( -\frac{u}{2x^{3}}%
\right) du=\frac{1}{2}\int \frac{1}{u^{2}-1}\,du.
\end{equation*}
Expanding the integrand into partial fractions
\begin{equation*}
\frac{1}{u^{2}-1}=\frac{1}{2\left( u-1\right) }-\frac{1}{2\left( u+1\right) }
\end{equation*}
the integral is expressable in terms of logarithms
\begin{equation*}
I=\frac{1}{4}\int \frac{du}{u-1}-\frac{1}{4}\int \frac{du}{u+1}=\frac{1}{4}
\ln \left( \left\vert \frac{\sqrt{1-x^{4}}-1}{\sqrt{1-x^{4}}+1}\right\vert
\right) +C.
\end{equation*}
Alternatively, using the standard integral
\begin{equation*}
\int \frac{1}{1-u^{2}}\,du=\operatorname{artanh}u+C
\end{equation*}
we obtain an inverse hyperbolic function
\begin{equation*}
I=-\frac{1}{2}\operatorname{artanh}\sqrt{1-x^{4}}+C.
\end{equation*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
find the slope of the tangent line Sorry I couldn't figure out how to use the editor right now and I really need help with this question.
Let $f(x)=\frac{1}{\sqrt[3]{x}}$. find the slope of the tangent line at $(8,\frac{1}{2})$
So far this is what I have I just don't know what to do from here
$$
\begin{array}{lll}
m&=&\lim_{h \to 0}\frac{f(8+h)-f(8)}{h}\\
&=&\lim_{h \to 0} \frac{\frac{1}{\sqrt[3]{8+h}}-1/2}h \\
&=&\lim_{h \to 0} \ 2-\frac{\sqrt[3]{8+h}}{2\sqrt[3]{8+h}}\\
&=&\lim_{h \to 0} \frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}\\
\end{array}
$$
Can someone please help me out please
|
From where you're at, do this trick:
\begin{align*}
&\lim_{h \to 0} \frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}=\lim_{h \to 0} \frac{\left(2-\sqrt[3]{8+h}\right)}{2h\sqrt[3]{8+h}}\frac{\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}{\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}=\\
&\lim_{h\to 0}\frac{-h}{2h(8+h)^{1/3}\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}= \\
&\lim_{h\to 0} -\frac{1}{2(8)^{1/3}(8^{2/3}+2\cdot8^{1/3}+4)}=-\frac{1}{48}
\end{align*}
Here's an equivalent limit proof.
\begin{align*}
&f'(8)=\lim_{x\to 8}\frac{f(x)-f(8)}{x-8} = \lim_{x\to 8}\frac{\frac{1}{x^{1/3}}-\frac{1}{2}}{(x^{1/3}-2)(x^{2/3}+2x^{\frac{1}{3}}+4)}= \\
&\lim_{x\to 8} \frac{\frac{2-x^{1/3}}{2x^{1/3}}}{(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)}=-\frac{1}{2}\lim_{x\to 8}\frac{1}{x^{1/3}(x^{2/3}+2x^{1/3}+4)}=\\
&-\frac{1}{2}\left(\frac{1}{2(4+4+4)}\right)=-\frac{1}{48}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that the square root of 3 is irrational I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if $x$ is divisible by $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of $6$?
|
suppose $\sqrt{3}$ is rational, then $\sqrt{3}=\frac{a}{b} $ for some $(a,b)
$
suppose we have $a/b$ in simplest form.
\begin{align}
\sqrt{3}&=\frac{a}{b}\\
a^2&=3b^2
\end{align}
if $b$ is even, then a is also even in which case $a/b$ is not in simplest form.
if $b$ is odd then $a$ is also odd.
Therefore:
\begin{align}
a&=2n+1\\
b&=2m+1\\
(2n+1)^2&=3(2m+1)^2\\
4n^2+4n+1&=12m^2+12m+3\\
4n^2+4n&=12m^2+12m+2\\
2n^2+2n&=6m^2+6m+1\\
2(n^2+n)&=2(3m^2+3m)+1
\end{align}
Since $(n^2+n)$ is an integer, the left hand side is even. Since $(3m^2+3m)$ is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int\frac{\cos^2x}{1+\tan x}dx$ Integrate $$I=\int\frac{\cos^2x}{1+\tan x}dx$$
$$I=\int\frac{\cos^3xdx}{\cos x+\sin x}=\int\frac{\cos^3x(\cos x-\sin x)dx}{\cos^2x-\sin^2x}=\int\frac{\cos^4xdx}{1-2\sin^2x}-\int\frac{\cos^3x\sin xdx}{2\cos^2x-1}$$
Let $t=\sin x,u=\cos x,dt=\cos xdx,du=-\sin xdx$
$$I=\underbrace{\int\frac{-u^4du}{(2u^2-1)\sqrt{1-u^2}}}_{I_1}+\underbrace{\int\frac{u^3du}{2u^2-1}}_{I_2}$$
I have found(using long division): $$I_2=\frac{u^2}2+\frac18\ln|2u^2-1|+c=\frac12\cos^2x+\frac18\ln|\cos2x|$$
I have converted $I_1$ into this:
$$I_1=\frac12\left(\int(-2)\sqrt{1-u^2}du+\int\frac{du}{\sqrt{1-u^2}}\right)+\frac14\underbrace{\int\frac{du}{(2u^2-1)\sqrt{1-u^2}}}_{I_3}$$
Now I have took $v=1/u$ in $I_3$ so that $du=-(1/v^2)dv$:
$$I_3=\int\frac{vdv}{(v^2-2)\sqrt{v^2-1}}$$
Now I took $w^2=v^2-1$ or $wdw=vdv$ to get:
$$I_3=\int\frac{dw}{w^2-1}=\frac12\ln\left|\frac{w-1}{w+1}\right|$$
I have not yet formulated the entire thing;
*
*Is this correct?
*This is very long, do you have any "shorter" method?
|
The integral is equivalent to $ \displaystyle{\int\frac{\cos^3 x}{\sin x + \cos x}\,\mathrm{d}x}. $ Now, consider
$$ \mathcal{I}_{1} = \int\frac{\cos^3 x}{\sin x + \cos x}\,\mathrm{d}x \qquad\text{and}\qquad\mathcal{I}_{2}=\int\frac{\sin^3 x}{\sin x + \cos x}\,\mathrm{d}x. $$
Observe that
$$
\begin{aligned}
\mathcal{I}_{1}+\mathcal{I}_{2} &= \int\frac{\cos^3 x + \sin^3 x}{\sin x + \cos x}\,\mathrm{d}x\\
&=\int\frac{(\cos x + \sin x)(\cos^2 x - \sin x\cos x + \sin^2 x)}{\cos x + \sin x}\,\mathrm{d}x\\
&=x - \frac{\sin^2 x}{2}+C_{1},
\end{aligned}
$$
and
$$
\begin{aligned}
\mathcal{I}_{1}-\mathcal{I}_{2}&=\int\frac{(\cos x - \sin x)(1 + \sin x \cos x)}{\cos x + \sin x}\,\mathrm{d}x\\
&=\int\frac{(\cos x - \sin x)(\cos x + \sin x)(1 + \sin x\cos x)}{(\cos x + \sin x)^2}\,\mathrm{d}x\\
&=\int\frac{1 +\frac{\sin 2x}{2}}{1 + \sin 2x}\cos 2x\,\mathrm{d}x\\
&=\underbrace{\frac{1}{4}\int\left(1 + \frac{1}{1+\sin 2x}\right)2\cos 2x\,\mathrm{d}x}_{t = \sin 2x\implies\mathrm{d}t=2\cos 2x\,\mathrm{d}x}\\
&=\frac{1}{4}\int1+\frac{1}{1+t}\,\mathrm{d}t\\
&=\frac{t}{4}+\frac{1}{4}\log|t+1|+C_{2}\\
&=\frac{\sin2x}{4}+\frac{1}{4}\log(1+\sin2x)+C_{2}
\end{aligned}
$$.
Thus,
$$
\begin{aligned}
(\mathcal{I}_{1}+\mathcal{I}_{2})+(\mathcal{I}_{1}-\mathcal{I}_{2})&=2\mathcal{I}_{1}\\
&=x+\frac{\sin2x}{4}+\frac{1}{4}\log(1+\sin2x)-\frac{\sin^2 x}{2} + C_{3},
\end{aligned}
$$
which yields
$$
\begin{aligned}
\mathcal{I}_{1}&=\int\frac{\cos^3x}{\sin x + \cos x}\,\mathrm{d}x\\
&=\int\frac{\cos^2 x}{1+\tan x}\,\mathrm{d}x\\
&=\frac{x}{2}+\frac{\sin 2x}{8}+\frac{1}{8}\log(1+\sin2x)-\frac{\sin^2 x}{4} + C,
\end{aligned}
$$
or, in an equivalent manner,
$$
\int\frac{\cos^2 x}{1+\tan x}\,\mathrm{d}x=\frac{x}{2}+\frac{\sin x\cos x}{4}+\frac{1}{4}\log\left|\cos\!\left(x-\frac{\pi}{4}\right)\right|-\frac{\sin^2 x}{4}+\mathcal{C}.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{\partial \coth ^{-1}(x)}{\partial x}$ I am asked to find
$\dfrac{d\coth ^{-1}(x)}{dx}$
I rewrite it to become
$x=\dfrac{1}{\tan(y)}$
$\dfrac{\text{dx}}{\text{dy}}=-\dfrac{1}{\frac{\sec ^2(x)}{\tan ^2(x)}}=-\sin^2(x)$
However the answer should be
$\dfrac{1}{1-x^2}$
What am I doing wrong?
|
First note that
$$ \coth^{-1} x=\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right), \mbox{for}\ |x|\gt 1$$
Then
$$ \frac{d}{dx}[\coth^{-1} x]=\frac{1}{2}\frac{d}{dx}\left[\ln\left(\frac{x+1}{x-1}\right)\right]$$
$$ =\frac{1}{2}\left[\frac{d}{dx}[\ln(x+1)]-\frac{d}{dx}[\ln(x-1)]\right] $$
$$ =\frac{1}{2}\left[\frac{1}{x+1}-\frac{1}{x-1}\right]=\frac{1}{2}\left[\frac{x-1-x-1}{(x+1)(x-1)}\right] $$
$$ =\frac{1}{2}\left[\frac{-2}{x^2-1}\right]=-\frac{1}{x^2-1} =\frac{1}{1-x^2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$
I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$
What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...
|
The expression under limit can be written as $$\frac{y^{1/3}-1}{x}=\frac{y^{1/3}-1}{y-1}\cdot\frac{y-1}{x}\tag{1}$$ where $$y=\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}=\frac{x}{1+\sqrt{1-x}}\cdot\frac{\sqrt{1+x}+1}{x}=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}\tag{2}$$ Since $y\to 1$ as $x\to 0$ we can see that the first fraction in $(1)$ tends to $1/3 $ via the standard formula $$\lim_{t\to a} \frac{t^n-a^n}{t-a}=na^{n-1}\tag{3}$$ Hence the desired limit is equal to the limit of $$\frac{1}{3}\cdot\frac{y-1}{x}=\frac{\sqrt {1+x}-\sqrt{1-x}}{3x(1+\sqrt{1-x})}=\frac{2x}{3x}\cdot\frac{1}{(\sqrt{1+x}+\sqrt{1-x})(1+\sqrt{1-x})}$$ It follows that the desired limit is $$\frac{2}{3}\cdot\frac{1}{2\cdot 2}=\frac{1}{6}$$ In general avoid rationalization of expressions containing radicals beyond square roots and instead use the limit formula $(3)$. This saves a lot of typing effort and reduces visual clutter.
|
{
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|
Multiplying both sides of an equation when there's a limit on one side?
Determine the value of $a\in\mathbb{R}$, such that $\displaystyle\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$
My attempt:
\begin{align*}
&\lim_{x\to 1} \dfrac{x^2+(3-a)x+3a}{x-1}=7\\
&\implies\lim_{x\to 1} x^2+(3-a)x+3a=7x-7\\
&\implies\lim_{x\to 1} x^2-4x-a(x-3)=-7\\
&\implies1^2-4(1)-a(1-3)=-7\\
&\implies a=-2
\end{align*}
My question is can I multiply both sides of my first line by $x-1$, even though the equation has a $\lim_{x\to 1}$ on the left side, as opposed to a normal equation?
|
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=\lim_{x\to 1}7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}-\lim_{x\to 1}7=0$$
$$\lim_{x\to 1}\left(\dfrac{x^2+(3-a)x+3a}{x-1}-7\right)=0$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a-7x+7}{x-1}=0$$
$$\lim_{x\to 1}\dfrac{x^2+(-4-a)x+3a+7}{x-1}=0$$
To the limit exists and be $0$ the roots of the numerator must be one with multiplicity two (one to eliminate the $x-1$ and the other to make the limit zero). Therefore:
$$-4-a=-(1+1)$$
$$3a+7=1\cdot 1$$
Both equations have a unique solution $a=-2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Are all those numbers coprime? The values of $4m^2+1$ and $4m^2+4m+5$ for $m\geq{1}$ are (resp.) 5,17,37,... and 13,29,53,... Those numbers seem to be all coprime : how to prove it if it is true, please ?
|
Suppose there is a common divisor $d$. Then $d|4m^2+1$ and $d|4m^2+4m+5$ implies that $d|4m^2+4m+5-4m^2-1$, which is $d|4m+4$. Since $4m^2+1$ is an odd number, so is $d$. Therefore $d|m+1$. Also, $d|4(m+1)^2$, so $d|4(m+1)^2-(4m^2+4m+5)$ and then $d|4m-1$.
Now combine $d|4m-1$ and $d|4m+4$. This gives us $d|5$.
|
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|
If $\alpha$, $\beta$ are two values of $\theta$ satisfying the equation $\cos\theta/a+\sin\theta/b=1/c$, prove that $\cot ((\alpha+\beta)/2) = b/a$ What I did was
$$b\ \cos (\theta) + a \sin (\theta) = \dfrac{ab}{c} \\
b\ \cos (\theta) = \frac{ab}{c} - a\ \sin (\theta) $$
Square both sides and using sum of roots and product of roots as
$$\alpha + \beta = \dfrac{2a^2b}{c(a^2+b^2)}\ \ \text{and}\ \
\alpha\beta = \dfrac{a^2b^2 - b^2c^2}{c^2(a^2+b^2)}$$
Now
$$b\ \cos (\theta) - \dfrac{ab}{c} = -a\ \sin (\theta)$$
Square both sides and using sum of roots and product of roots as
$$ \alpha + \beta = \dfrac{2ab^2}{c(a^2+b^2)} \\
\alpha\beta = \dfrac{a^2b^2 - a^2c^2}{c^2(a^2+b^2)} $$
I don't know how to solve further.
|
Hint. Write
$$\theta=\frac{\alpha+\beta}{2}\ ,\quad \phi=\frac{\alpha-\beta}{2}\ .$$
You need to find $\cot\theta$, and you are given that
$$\frac{\cos(\theta+\phi)}{a}+\frac{\sin(\theta+\phi)}{b}
=\frac{\cos(\theta-\phi)}{a}+\frac{\sin(\theta-\phi)}{b}\ .$$
See if you can take it from here.
|
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|
Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots$$
|
$$\begin{align}
y &= \sum_{k=1} \frac{\prod_{j=2}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{A}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
\frac{\left(-\frac 12\right)\prod_{j=0}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{B}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
\frac{\left(-\frac 13\right)^{k+1}\left(-\frac 12\right)\prod_{j=0}^k 1/3 - j} {6^k k!}\tag{C}
\\ &= \left(-\frac 12\right) + \left(-\frac 13\right)\left(-\frac 12\right)\sum_{k=0}
\frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {3\cdot 6}\right)^k\tag{D}
\\ &= \left(-\frac 12\right) + \left(\frac 16\right) \sum_{k=0} \frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {18}\right)^k\tag{E}
\end{align}$$
Generalized binomial theorem is:
$$(1 + x)^n = \sum_{k=0} \frac{ \prod_{j=0}^k n - j } {k!} x^k\tag{F}$$
So if I don't have a type then
$$y = \left(-\frac 12\right) + \left(\frac 16\right)\left(1 + -\frac{1}{18}\right)^{1/3} = \frac{1}{18}\sqrt[3]{\frac{17}{2}} - \frac 12\tag{G}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of the series $\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\dots$ I am recently struck upon this question that asks to find the sum until infinite terms
$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+.....∞$$
I tried my best to get something telescoping or something useful, but I failed. I even made a recurrence as $t_n=t_{n-1}\frac{2n-1}{2n+2}$, but this question was expected to be done with simple logic of series (Also that the recurrence on solving gives a higher order charasteristic polynomial, which might be difficult to solve without calculator). So, thus I ended up being confused with this question. So, can anyone prode a small solution to this (might be easy) problem .
|
$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}$$
Rewrite the sum as
\begin{align}
\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots
&=\sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!}\\
&=\sum^\infty_{n=0}\frac{(2n+1)!}{2^{2n+2}n!(n+2)!}\\
&=\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}\Bigg{|}_{x=1/2}
\end{align}
Differentiate once to get
\begin{align}
\frac{{\rm d}}{{\rm d}x}\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}
&=\sum^\infty_{n=0}\binom{2n+2}{n}x^{2n+1}\\
&=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{z}\sum^\infty_{n=0}\left[\frac{(x+xz)^{2}}{z}\right]^n{\rm d}z\\
&=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{z-x^2(1+z)^2}{\rm d}z\\
&=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{-x^2z^2+(1-2x^2)z-x^2}{\rm d}z\\
&=-\frac{1}{i2\pi x^2}\oint_{|z|=1}\frac{x(1+z)^2}{(z-r_+)(z-r_-)}{\rm d}z\\
&=-\frac{1}{x^2}{\rm Res}\left[\frac{x(1+z)^2}{(z-r_+)(z-r_-)},r_+\right]\\
&=-\frac{1}{x}\frac{(1+r_+)^2}{r_+ - r_-}
\end{align}
We know that $r_+=\dfrac{1-2x^2-\sqrt{1-4x^2}}{2x^2}$. Coupled with the fact that $r_+ - r_- =\dfrac{-\sqrt{1-4x^2}}{x^2}$, this gives us
\begin{align}
\sum^\infty_{n=0}\binom{2n+2}{n}x^{2n+1}
&=\frac{\sqrt{1-4x^2}}{4x^3}-\frac{1}{2x^3}+\frac{1}{4x^3\sqrt{1-4x^2}}
\end{align}
Now, integrate this expression once
\begin{align}
\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}
&=\frac{1-\sqrt{1-4x^2}}{4x^2}+C
\end{align}
It is easy to see that
\begin{align}
C
=\lim_{x\to 0}\frac{\sqrt{1-4x^2}-1}{4x^2}
=\lim_{u\to 1}\frac{u-1}{1-u^2}
=-\lim_{u\to 1}\frac{1}{1+u}
=-\frac{1}{2}
\end{align}
Finally, letting $x=\dfrac{1}{2}$ yields
\begin{align}
\sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!}
&=\frac{1-\sqrt{1-1}}{1}-\frac{1}{2}\\
&=\frac{1}{2}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Stuck on this intergral $\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{\tan^2x}{x-\tan x} dx $ calculus I $$\int^{\pi/3}_{\pi/4} \frac{\tan^2x}{x-\tan x} dx $$
this is that I have tried
$$\int^{\pi/3}_{\pi/4} \frac{\frac{\sin^2x}{\cos^2 x}}{x-\frac{\sin x}{\cos x}} dx $$
$$\int^{\pi/3}_{\pi/4} \frac{\sin^2(x)}{x \cos^2(x)-\sin(x)\cos(x)} dx $$
but I an not making anymore progress
|
Hint: For the integral
$$\int \dfrac{\tan^2(x)}{x-\tan(x)}\, dx$$
try the substitution $u = x-\tan(x)$, $du = 1-\frac{1}{\cos(x)^2} \, dx = -\tan^2(x)\, dx$
As requested here is the elaboration:
With the substitution $u = x-\tan(x)$ we get
\begin{align*}du &= \dfrac{d}{dx}\left(x-\frac{\sin(x)}{\cos(x)}\right)dx=1-\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}dx=1-\frac{1}{\cos^2(x)}dx\\&=\frac{\cos^2(x)-1}{\cos^2(x)}dx=\frac{-\sin^2(x)}{\cos^2(x)}dx=-\tan^2(x)\, dx\end{align*}
So your integral becomes
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\dfrac{\tan^2(x)}{x-\tan(x)}\, dx=-\int_{\frac{\pi}{4}-1}^{\frac{\pi}{3}-\sqrt{3}}\frac{1}{u}\,du=\ln\left(\frac{\pi}{3}-\sqrt{3}\right)-\ln\left(\frac{\pi}{4}-1\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form of $\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2}$ I'm new in the area of the series involving Bessel function of the first kind. What are
the usual tools you would recommend me for computing such a series? Thanks.
$$\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2}$$
|
Here is a route.
Recall that
$$
J_0(2n)=\frac 1\pi \int_0^\pi \cos (2n \sin x)\:{\rm d}x \tag1
$$
and that
$$\sum_{n=1}^\infty\frac{\cos nt}{n^2}=\frac{\pi^2}{6}-\frac{\pi t}{2}+\frac{t^2}{4},\quad 0\leq t\leq 2\pi. \tag2 $$
Then, due to normal convergence of the series on $[0,2\pi] $, we may write
$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0(2n)}{n^2} & =\frac 1\pi \int_0^\pi \sum_{n=1}^\infty\frac{\cos (2n \sin x)}{n^2}{\rm d}x \\\\
& =\frac 1\pi \int_0^\pi\left(\frac{\pi^2}{6}-\pi \sin x+\sin^2 x\right){\rm d}x \\\\
& =\frac{\pi^2}{6}-\frac{3}{2}\\\\
& =0.144934066848226436...
\end{align}
$$
thus
$$
\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2} = \frac{\pi^2}{6}-\frac{3}{2}.
$$
Using a similar technique, one may obtain the following result.
$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0(2 n \alpha )}{n^2} &=\frac{\pi^2}{6}-2\alpha+\frac{\alpha^2}{2},\qquad \alpha \in [0,\pi).\tag3
\end{align}
$$
|
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|
Please explain the Quotient Rule I am currently working on an equation but I'm having a hard time understanding how to get the answer.
the answer is ${(x^2-4)(x^2+4)(2x+8)-(x^2+8x-4)(4x^3)\over (x^2-4)^2(x^2+4)^2}$
The equation is $f(x)= {x\over x^2-4}-{x-1\over x^2+4}$
When I apply the quotient rule i get $f'(x)= {(1)(x^2-4)-(2x)(1)\over (x^2-4)^2}-{(1)(x^2+4)-(2x)(1)\over (x^2+4)^2}$ but it cancels each other out. I can't figure out how they had gotten the answer.
|
$$\frac{a}{b}-\frac{a}{c} \neq \frac{a-a}{b}$$ but $$\frac{a}{b}-\frac{a}{c}=\frac{ac-ab}{bc}$$
Notice that you have different denominators.
|
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|
Calculate $\pi$ By Hand? All over the internet the only hand equation i found was
$$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$
But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?
|
I skimmed the other answers. I believe this one is different in that it gives a way to rapidly calculate the digits of $\pi$ and proves from scratch that that number in fact is $\pi$.
It's obvious that $\pi = 6 \times \sin^{-1}(\frac{1}{2})$. It turns out that the derivative of $\sin^{-1}$ is elementary so we can take its power series centered at 0 and then integrate it to get the power series of $\sin^{-1}$. In general, $\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$. So $\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-\frac{1}{2}}$. Once I get the power series of $(1 + x)^{-\frac{1}{2}}$, I can substitute $-x^2$ for $x$ to get the power series of $(1 - x^2)^{-\frac{1}{2}}$. The first derivative of $(1 + x)^{-\frac{1}{2}}$ is $-\frac{1}{2}(1 + x)^{-1\frac{1}{2}}$. The second derivative is $-\frac{1}{2}(-1\frac{1}{2})(1 + x)^{-2\frac{1}{2}}$ and so on. Now we divide by the factorials to get the coefficients of the power series so $(1 + x)^{-\frac{1}{2}} = 1 + -\frac{1}{2}x + -\frac{1}{2}(-\frac{3}{4})x^2 + -\frac{1}{2}(-\frac{3}{4})(-\frac{5}{6})x^3 ...$ So the power series of $(1 - x^2)^{-\frac{1}{2}}$ is $1 + \frac{1}{2}x^2 + \frac{1}{2}(\frac{3}{4})x^4 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})x^6 ...$ Then the power series of $\sin^{-1}$ is $x + \frac{1}{2}(\frac{1}{3})x^3 + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})x^5 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})x^7 ...$ Now finally, $\pi = 6 \times \sin^{-1}(\frac{1}{2}) = 6(2^{-1} + \frac{1}{2}(\frac{1}{3})2^{-3} + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})2^{-5} + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})2^{-7} ...)$.
|
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|
What is the closed form of $\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(n\right)} ^2}{{n}^4}}$? Using Maple I am obtaining the numerical approximation
$$0.5902373619$$
Please, let me know what is the closed form. Many thanks.
|
Hint.
Observe that
$$
J_0(n)=\frac 1\pi \int_0^\pi \cos (n \sin x)\:{\rm d}x \tag1
$$
and that
$$\sum_{n=1}^\infty\frac{\cos nt}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2 t^2}{12}+\frac{\pi t^3}{12}-\frac{t^4}{48},\quad 0\leq t\leq 2\pi. \tag2 $$
Then, due to normal convergence of the series on $[0,2\pi] $, we may write
$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} & =\frac{1}{\pi^2} \int_0^\pi \!\!\int_0^\pi \sum_{n=1}^\infty\frac{\cos (n \sin x)\cos (n \sin y)}{n^4}{\rm d}x\:{\rm d}y .\tag3
\end{align}
$$
We may plug
$$
2\cos (n \sin x)\cos (n \sin y)=\cos (n(\sin x+\sin y))+\cos (n(\sin x-\sin y)) \tag4
$$
into $(3)$ and integrate as here.
Hence we obtain
$$ \sum_{n=1}^\infty\frac{J_0^2(n)}{n^4}=\frac{\pi ^4}{90}-\frac{\pi ^2}{12}-\frac{3}{64}+\frac{32}{27 \pi }. \tag5 $$
A numerical value is
$$
\sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} =0.5902373616900361395467798486 \ldots.
$$
Some details.
Identity $(2)$ may be rewritten as
$$\sum_{n=1}^\infty\frac{\cos nt}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2 t^2}{12}+\frac{\pi |t|^3}{12}-\frac{t^4}{48},\quad -\pi \leq t\leq \pi. \tag6 $$
Then using $(3)$, $(4)$ and $(6)$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} =\frac{1}{2\pi^2}\int_0^\pi \!\!\int_0^\pi
\left[\left(\frac{\pi^4}{90}-\frac{\pi^2}{12}(\sin x+\sin y)^2+\frac{\pi }{12}(\sin x+\sin y)^3-\frac{1}{48}(\sin x+\sin y)^4\right)+\left(\frac{\pi ^4}{90}-\frac{\pi ^2}{12}(\sin x-\sin y)^2+\frac{\pi }{12}\left|\sin x-\sin y\right|^3-\frac{1}{48}(\sin x-\sin y)^4\right)\right]{\rm d}x\:{\rm d}y .\tag7
\end{align}
$$
We just have to be careful with the computation of the term ($|t|^3=\left({\rm Abs} (t)\right)^3$)
$$
\begin{align}
\frac{1}{2\pi^2}\times \frac{\pi }{12}\times \int_0^\pi \!\!\int_0^\pi \left|\sin x-\sin y\right|^3 {\rm d}x\:{\rm d}y & = \frac{1}{2\pi^2}\times\frac{\pi }{12} \times 4\int_0^{\pi/2} \!\!\int_0^{\pi/2} \left|\sin x-\sin y\right|^3 {\rm d}x\:{\rm d}y \\\\
&=-\frac{13}{36}+\frac{32}{27 \pi }.
\end{align}
$$
|
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|
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162
x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928
x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750
x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900
x^8+366120 x^6+87829920 x^4+13038019200
x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$
|
I played a little bit with Jack's answer. One can expand with the Newton formula
$$ (1 + \sqrt{3} t)^n = \sum_{k=0}^n \binom{n}{k} (\sqrt{3}t)^k $$
then the integrals can be performed using
$$
\int_0^\infty dt\, t^k e^{- t} \cos(s t) = \frac{T_{1+k}\bigl(1/\sqrt{1 + s^2}\bigr) k!}{(1+s^2)^{(k+1)/2}} $$
where $T_a(x)$ is the T-Chebyshev polynomial. Then, the given sequence can be rewritten setting $s = x/\sqrt{3}$ as
$$ S_n = \frac{1}{18(s^2 + n^2)^{n+1}} \sum_{k=0}^n \frac{n!}{(n-k)!} (s^2 + n^2)^{n -\frac{k - 1}{2}} T_{1+k}\bigl(1/\sqrt{1 + s^2/n^2}\bigr)
$$
|
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|
Solving quadratic complex equation I'm trying to solve this equation:
$z^2+2iz+11-16i=0$
where $z=x+iy$ with $ x,y \in \mathbb{R}$.
I tried it with the quadratic formula $z_{1/2}=- \frac p2 \pm \sqrt{\left(\frac p2\right)^2 -q}$ and get:
$z_{1/2}=-i \pm \sqrt{i^2-11+16i} \\ \Leftrightarrow z_{1/2}=-i \pm \sqrt{12i^2+16i} $
Now I'm stucking. Any ideas how it's going on?
|
Completing the square might be simpler:
$z^2+2iz+i^2 = -11+16i+i^2$
$(z+i)^2 = -12+16i$
$z = -i \pm \sqrt{-12+16i}$
$z = -i \pm 2\sqrt{-3+4i}$ (factor out a $4$ from under the squareroot).
To simplify $\sqrt{-3+4i} = a+bi$, we need to find reals $a,b$ such that $(a+bi)^2 = (a^2-b^2)+(2ab)i = -3+4i$.
Since the numbers are small, you can try guessing integers for $a,b$.
Since $a = 1$ and $b = 2$ work, $1+2i$ is a value for $\sqrt{-3+4i}$. Can you continue from here?
|
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|
To prove triangle is equilateral> given one equation and one combined equation. show that the lines $ x^2 - 4xy +y^2 $ and $x +y=3 $ form an equilateral trinagle . Also Find Area.
Here is what I have tried:
$ l1 =x+y=3 $
The combined equation is:
$ x^2 - 4xy + y^2 $
compare it with general formula :
$ ax^2 +2hxy+y^2 $
What to do next ?
I could find the slope but how will that help ?
|
If the angle between $y=mx$ and $x+y=3\iff y=-x+3$ is $60^\circ,$
$$\left|\frac{m-(-1)}{1+m(-1)}\right|=\tan60^\circ=\sqrt3$$
$$\iff\frac{m+1}{m-1}=\pm\sqrt3\iff(m+1)^2=3(m-1)^2\iff m^2-4m+1=0$$
Setting $m=\dfrac yx\implies y^2-4xy+x^2=0$
|
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|
Finding the integer solutions of the equation $3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14$ $
3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14
$ .
I already solved this using the Cauchy–Schwarz inequality and got $x=4$ and $y=5$. But I'm sure there is a prettier, simpler solution to this and I was wondering if anyone could suggest one.
|
This is a trivial solution so that we get answer in integer
8 - x and 6 - y are under the root so x <8 and y <6 , for making 6 - y to be perfect square y=2 or 5. and similarly x=4 or 7.
Now x+y is also under the root so x+y has to be perfect square. So we get two pairs of (x, y)=(7,2) and (2,5) but (7,2) will not satisfy the equation so ur answer is x=4, y=5.
|
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|
Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^2x$ we have:
$$\int_0^{\pi /4}\frac{\sec^2x}{\sec^2x-3\tan^2x}dx=\int_0^{\pi /4}\frac{\sec^2x}{1-2\tan^2x}dx=\int _0^1 \frac{dt}{1-2t^2}=\int _0^1 \frac{1}{2}\frac{dt}{\frac{1}{2}-t^2}=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}t}{1+\sqrt{2}t}\right|_0^1=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}}{1+\sqrt{2}}\right|$$
But when we do the same integration by dividing the initial term by $\sec^4x$ and solving it yields an answer $$\frac{\pi }{2}$$
Am I wrong somewhere?
|
Let $$I = \int^{\frac{\pi}{4}}_{0}\frac{1}{\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x}dx = \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x(\tan ^2x+\cot^2 x-1)}dx$$
So $$I =\int^{\frac{\pi}{4}}_{0}\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+1}dx = \left[\tan^{-1}(\tan x-\cot x)\right]^{\frac{\pi}{4}}_{0}=\frac{\pi}{2}$$
|
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|
Bivariate distribution with normal conditions Define the joint pdf of $(X,Y)$ as:
$$f(x,y)\propto \exp(-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]),$$
where $A,B,C,D$ are constants.
Show that the distribution of $X\mid Y=y$ is normal with mean $\frac{By+C}{Ay^2+1}$ and variance $\frac{1}{Ay^2+1}$. Derive a corresponding result for the distribution of $Y\mid X=x$.
Attempt:
I tried to integrate the equation w.r.t. $x$ in order to find $X\mid Y=y$. However, I'm not sure if I am correct:
$$\int_{-\infty}^{\infty}\exp(-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy])\,dx$$
$$ =\left[\frac{\exp[-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]}{-Axy^2-1/2+By+C}\right]_{-\infty}^{\infty}$$
Whatever the value of the previous integration (call it "Q"), then we would divide the original equation by "Q", i.e.:
$$ \frac{f_{X,Y}(x,y)}{Q}$$
Which would give us $f_{X\mid Y=y}(X\mid Y=y)$.
How do I go about evaluating
$$ =\left[\frac{\exp[-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]}{-Axy^2-1/2+By+C}\right]_{-\infty}^{\infty}\text{ ?}$$
|
You would need to integrate out $y$, not $x$, to get a marginal density that is a function of $x$.
What you have inside of $\exp\left(\dfrac{-1}2\left(\bullet\bullet\bullet\right)\right)$ is
$$
Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy.
$$
Think of how this behaves as a function of $x$. You want
$$
\left(\frac{x-\mu}\sigma\right)^2 + (\text{something not depending on $x$}).
$$
You need to figure out what $\mu$ and $\sigma$ are. The standard method in algebra for this sort of thing is completing the square.
\begin{align}
& Ay^2x^2+x^2+y^2-2Bxy-2Cx-Dy \\[8pt]
= {} & (Ay^2+1)x^2 -2(By+C)x + \text{terms not depending on $x$} \\[8pt]
= {} & (Ay^2+1)\left(x^2 - 2\frac{By+C}{Ay^2+1} x\right) + \text{terms not depending on $x$} \\[8pt]
= {} & (Ay^2+1)\left(x^2 - 2\frac{By+C}{Ay^2+1} x + \left(\frac{By+C}{Ay^2+1}\right)^2\right) + \text{terms not depending on $x$} \\[8pt]
= {} & (Ay^2+1)\left(x - \frac{By+C}{Ay^2+1}\right)^2 + \text{terms not depending on $x$} \\[8pt]
= {} & \left(\frac{x - \frac{By+C}{Ay^2+1}}{1/\sqrt{Ay^2+1}}\right)^2 + \text{terms not depending on $x$} \\[8pt]
= {} & \left(\frac{x-\mu}{\sigma}\right)^2 + \text{terms not depending on $x$, if }\mu=\text{ what and }\sigma=\text{ what?}
\end{align}
|
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|
Bound for $\sum_{k=1}^\infty\left(\frac{1}{2^k+k^2}\right)$ I found for the series:
$$S=\sum_{k=1}^\infty\left(\dfrac{1}{2^k+k^2}\right)$$ a bound:
$$S\le\dfrac{\pi^2}{6+\pi^2}$$
which is in good agreement with the approximate value of $S$ calculated with Maple or Mathematica $(S=0.588239...)$, while the ratio: $\dfrac{\pi^2}{\pi^2+6}=0.621918...$ Is it possible to get a sharper bound for $S$? Thanks.
|
One idea:
$$\sum_{n=1}^{\infty} \frac{1}{n^2+2^n}=\sum_{n=1}^{k} \frac{1}{n^2+2^n}+\sum_{n=k+1}^{\infty} \frac{1}{n^2+2^n}$$
For some $k$
Now $\displaystyle \sum_{n=1}^{k} \frac{1}{n^2+2^n}$ is finite sum, so it's possible to calculate it directly.
In second series use inequality $\frac{1}{n^2+2^n} \leq \frac{1}{2^n}$, so:
$$\sum_{n=k+1}^{\infty} \frac{1}{n^2+2^n} \leq \sum_{n=k+1}^{\infty}\frac{1}{2^n}=\frac{1}{2^k} $$
You can get as good aproxximation as you want (for example for $k=10$ you get $4$-digit precision).
|
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|
Divisibility property of $(a+b)^n-a^n-b^n$ Let $n$ be a natural number of the form $n=6k+1$ (while $k$ is a positive integer).
Show that $(a^2+ab+b^2)^2$ divides $(a+b)^n-a^n-b^n$ for all integer numbers $a,b$ (such that $a^2+ab+b^2\ne0$).
|
Note: This proof replaces an earlier less elementary version that used the unique factorization property of $\mathbf{Z}[x,y]$.
Let $c = a^2 + ab + b^2$. Since $(a-b)(a^2 + ab + b^2) = a^3 - b^3$, we have $a^3 \equiv b^3$ modulo $c$. It follows therefore that for any $k$ we have $a^{6k} \equiv b^{6k}$ modulo $c$, hence that $c$ divides $a^{6k} - b^{6k}$.
The fact to be proved is trivial for $k=0$. Assume by induction that $(a+b)^{6k+1} \equiv a^{6k+1} + b^{6k+1}$ modulo $c^2$. Calculating modulo $c^2$, we have
$$
\begin{align}
(a+b)^{6k+7} &= (a+b)^6(a+b)^{6k+1} \\
&\equiv \left[ (a+b)^6 - 7abc^2 \right](a^{6k+1} + b^{6k+1})\\
&= \frac{a^7 + b^7}{a+b}(a^{6k+1} + b^{6k+1}) \\
&\equiv \frac {a^7 + b^7}{a+b}(a^{6k+1} + b^{6k+1}) + ab(a-b)(a^2 - ab +b^2)c(a^{6k}-b^{6k}) \\
&= \frac {a^7 + b^7}{a+b}(a^{6k+1} + b^{6k+1}) + ab\frac{a^6-b^6}{a+b}(a^{6k}-b^{6k}) \\
&=a^{6k+7} + b^{6k+7},
\end{align}
$$
where for the first equivalence we've used the induction hypothesis, and for the second the divisibility fact mentioned above. (The desired property is easy to prove separately when $a+b = 0$. Alternatively, in order for the case $a+b = 0$ to be included above, replace the fractions with the polynomials in $a$ and $b$ to which they're equal.) This completes the proof by induction on $k$.
|
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|
Two differentiation results of $\sin^{-1}(2x\sqrt{1-x^2})$ While trying to differentiate $\sin^{-1}(2x\sqrt{1-x^2})$, if we put
$x = \sin\theta$, we get,
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})\\
&= \sin^{-1}(2\sin\theta\cos\theta)\\
&= \sin^{-1}(\sin2\theta)\\
&= 2\theta\\
&= 2\sin^{-1}x.
\end{align*}
So,
\begin{align*}
\frac{dy}{dx} &= \frac{2}{\sqrt{1-x^2}}.\\
\end{align*}
But if we put $x = \cos\theta$, we get,
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\cos\theta\sqrt{1-\cos^2\theta})\\
&= \sin^{-1}(2\cos\theta\sin\theta)\\
&= \sin^{-1}(\sin2\theta)\\
&= 2\theta\\
&= 2\cos^{-1}x.
\end{align*}
This time,
\begin{align*}
\frac{dy}{dx} &= -\frac{2}{\sqrt{1-x^2}}.\\
\end{align*}
We are perplexed about the difference in sign between the two results
and thought that you could help.
(We understand that we can differentiate $\sin^{-1}(2x\sqrt{1-x^2})$ directly, without any substitution, which gives us the first result.)
|
Remember that $\sin^{-1}$ is not a "true inverse". Here is a graph of $\sin^{-1}(\sin(2x))$.
I think it is clear that this is the most likely source for your sign error. As you note, you could have a different choice for arcsin which would give you the opposite sign when differentiating.
I think the easiest way to reason into the first choice is that because $\sin^{-1}(x)$ is increasing your derivative should be positive.
If you wanted to be more precise, remember that you are only considering values of $x$ such that $-1 \leq 2x\sqrt{1-x^2} \leq 1$. If you let $x = \cos\theta$ you will use values of theta that are greater than $\pi/2$ because you need $x$ to be negative. If you let $x = \sin(\theta)$ you can let $\theta$ take on values between $-\pi/2$ and $\pi/2$ which is where $\sin^{-1}$ is "nicely defined".
After all of this, let me correct an error in your work. This is where the "paradox" arises.
For $x = \sin(\theta)$
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})\\
&= \color{red}{\sin^{-1}(2\sin\theta|\cos\theta|)}\\
\end{align*}
For $x = \cos\theta$
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\cos\theta\sqrt{1-\cos^2\theta})\\
&= \color{red}{\sin^{-1}(2\cos\theta|\sin\theta|)}\\
\end{align*}
If we chose $x = \sin(\theta)$ then $\theta$ is such that $\cos(\theta)$ is always positive and we don't need the absolute value signs and we can continue just as we did.
If we choose $x = \cos(\theta)$ then $\theta$ is such that $\sin(\theta)$ is always negative so here is the source of the discrepancy. A negative sign should be introduced after taking the square root.
|
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|
Is this the correct procedure for Integral Partial Fraction. $∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$
The first step I did is distributed the denominator so that I can find out if I should use synthetic division. Which after doing this I discovered that the denominator's exponent is greater therefore I do not use synthetic division
= $∫ (x^3+x^2+x+3)/(x^4+4x^2+3)$ Therfore I solve.
$∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$ =∫ $(Ax +B)/(x^2+1) + (Cx+D)/(x^2+3)$ then I distributed $(x^2+1)(x^2+1)$ on both sides and got.
$∫ (x^3+x^2+x+3) $= $(Ax +B)(x^2+3)+( Cx+D)(x^2+1)$
Then I solve for B.
x=0
$0+0+0+3 = (A(0)+B)(0+3)+(0+0)(0+1) =$
3=B If this is right how does one find A,CD? Do I make x =0 and B =3 and e verything else 0?
|
You have to use what is called a coefficient matrix.
So as you said we have that:
$$∫ \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)}dx =∫ \frac{Ax +B}{x^2+1} + \frac{Cx+D}{x^2+3}dx$$
Then multiplying both sides we get that:
$$x^3+x^2+x+3 = (x^2+3)(Ax+B) + (x^2+1)(Cx+D) = Ax^3+Bx^2+3Ax+3B + Cx^3+Dx^2+Cx+D$$
Now we analyze the coefficients on the LHS and RHS for each power of $x$ and the constants.
So for example we have for $x$ coefficients:
$$3A + C = 1$$
Now, we construct our equations:
For constants:
$$3B+D = 3$$
For $x$ coefficients:
$$3A + C = 1$$
For $x^2$ coefficients:
$$B + D = 1$$
For $x^3$ coefficients:
$$A + C = 1$$
Now construct your matrix and solve for $A$, $B$, $C$, and $D$.
EDIT
You don't have to use a matrix, using simple algebra can yield your answer.
Subtracting the equations gives us $A=D=0$, $B=1$ and $C=1$
Therefore, our partial fraction decomposition is:
$$ \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)} = \frac{1}{x^2+1} + \frac{x}{x^2+3}$$
|
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|
Integration Question: Completing the Square/Trig Sub yields a different answer than integral table. After the completing the square,
$$\int \frac{dx}{x^2 + 2x - 3}$$
becomes,
$$ \int \frac{dx}{(x+1)^2 - 4}$$
The integral table in my book says the antiderivative is,
$$\frac{1}{2a} ln \, \Biggl\lvert \frac{x-a}{x+a} \Biggr\rvert + \, C$$
or,
$$ \frac{1}{4} ln \, \Biggl\lvert \frac{x-1}{x+3} \Biggr\rvert + \, C$$
Using trig sub, and $\displaystyle u=a\sec\theta$ I obtained,
$$\frac{1}{2}\int csc \, \theta \, d\theta$$
which corresponds to
$$-\frac{1}{2} ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert + \, C $$
I am having a lot of trouble putting this result in the form presented in the table. Any help would be much appreciated.
|
$$-\frac{1}{2} \ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert = -\frac12 \ln \left| \sqrt{\frac{(x+3)^2}{(x-1)(x+3)}} \right| = -\frac14 \ln \left| \frac{x+3}{x-1} \right| = \frac14 \ln \left|\frac{x-1}{x+3}\right|$$
|
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|
Suddenly a wild Z-transform appears... I am attempting to complete the z transform of the following formula:
$$x(n) = \frac{1}{2}(n^2+n) u(n-1)$$
I got it into the summation form like so:
$$X(z) = \frac{3}{2} \sum_{n=1}^{\infty} n^2(\frac{1}{3} z^{-1})^2 + \frac{3}{2} \sum_{n=1}^{\infty} n(\frac{1}{3} z^{-1})^2$$
$$X(z) = ??? + 3\frac{\frac{1}{3}z^{-1}}{(1-\frac{1}{3}z^{-1})^2}$$
I'm not sure what to do about that first term there. It doesn't seem to fit any of the forms given in my Z-Transform table.
|
The crucial idea here is to use the fact that
$$
\sum_{n=1}^\infty na(n)z^{-n}=-z\frac{d}{dz}\sum_{n=1}^\infty a(n)z^{-n}.
$$
The Z-transform of your $x(n)$ will by definition be
$$
X(z)
=\sum_{n=-\infty}^\infty \frac{1}{2}(n^2+n)u(n-1)z^{-n}
=\frac{1}{2}\sum_{n=1}^\infty n^2z^{-n}+\frac{1}{2}\sum_{n=1}^\infty nz^{-n},
$$
where we can handle each of the two series separately.
We find that
$$
\sum_{n=1}^\infty nz^{-n}
=-z\frac{d}{dz}\sum_{n=1}^\infty z^{-n}
=-z\frac{d}{dz}\Big(\frac{z^{-1}}{1-z^{-1}}\Big)
=\frac{z}{(z-1)^2}
$$
and
$$
\sum_{n=1}^\infty n^2z^{-n}
=-z\frac{d}{dz}\sum_{n=1}^\infty nz^{-n}
=-z\frac{d}{dz}\bigg(\frac{z}{(z-1)^2}\bigg)
=\frac{z^2+z}{(z-1)^3}.
$$
Finally, this gives us
$$
X(z)
=\frac{1}{2}\bigg(\frac{z}{(z-1)^2}+\frac{z^2+z}{(z-1)^3}\bigg)
=\frac{z^2}{(z-1)^3}
=\frac{z^{-1}}{(1-z^{-1})^3}.
$$
|
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"url": "https://math.stackexchange.com/questions/959664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}
\end{align}
I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
|
$$\lim_{x\rightarrow \infty} (\sqrt{x^2+2x} - \sqrt{x^2-7x}) $$$$=\lim_{x\rightarrow \infty} \bigg(\sqrt{x^2+2x+1} - \sqrt{x^2-7x+\frac{49}{4}}\bigg) $$ $$=\lim_{x\rightarrow \infty} \bigg( x+1 - \bigg(x-\frac{7}{2}\bigg)\bigg) =\frac{9}{2} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
}
|
$\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers of $x$ and $y$ Give a convincing argument that $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers $x$ and $y$. Could someone please explain how to prove this? I attempted to say that the largest values that could be added to $x$ and $y$ is $0.99$ and that doing so still made $\lfloor x\rfloor + \lfloor y\rfloor < \lfloor x+y\rfloor$. However, my answer was not accepted.
|
Let $x = \lfloor x \rfloor + \epsilon_1$ and $y = \lfloor y \rfloor + \epsilon_2$.
Then $\lfloor x + y \rfloor = \lfloor\lfloor x \rfloor + \epsilon_1 + \lfloor y \rfloor + \epsilon_2\rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor \epsilon_1 + \epsilon_2 \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/963245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Equation of line passing through point. The straight line $3x + 4y + 5 = 0 $ and $4x - 3y - 10 = 0$ intersect at point $A$. Point $B$ on line $3x + 4y + 5 = 0 $ and point C on line $4x - 3y - 10 = 0$ are such that $d(A,B)=d(A,C)$.
Find the equation of line passing through line $\overline{BC}$ if it also passes through point $(1,2)$.
I have found out slope of both the lines through $A$: $\frac{-3}{4}$ and $\frac{4}{3}$.
I can't figure out how to solve it.
|
Intersection point $A$ has coordinates $(1, -2)$.
The first line equation can be written as $y = -\frac34x - \frac54$, the second: $y = \frac43x-\frac{10}{3}$, so $k_1 = -\frac34, k_2 = \frac43$.
$AB$ would lie on the first line if $y_B - y_A = k_1(x_B - x_A)$.
$AC$ would lie on the second line if $y_C - y_A = k_2(x_C - x_A)$.
This gives us $y_B = -\frac34x_B - \frac54$, $y_C = \frac43x_C - \frac{10}{3}$. $AB = AC$ means that $\sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2}$, so $\sqrt{(x_B - 1)^2 + (y_B + 2)^2} = \sqrt{(x_C - 1)^2 + (y_C + 2)^2}$
Let $y = k_3x + b_3$ be the equation of line containing $BC$. If it passes through $D = (1,2)$, then it means that it contains segments $BD$ and $CD$, so $k_3 = \frac{y_B - y_D}{x_B - x_D} = \frac{y_D - y_C}{x_D - x_C}$ (as $D$ lies between $B$ and $C$), so his gives the last equation: $\frac{y_B - 2}{x_B - 1} = \frac{2 - y_C}{1 - x_C}$.
Finally, you have to solve the following system:
\begin{cases}
y_B = -\frac34x_B - \frac54 \\
y_C = \frac43x_C - \frac{10}{3} \\
\sqrt{(x_B - 1)^2 + (y_B + 2)^2} = \sqrt{(x_C - 1)^2 + (y_C + 2)^2} \\
\frac{y_B - 2}{x_B - 1} = \frac{2 - y_C}{1 - x_C}
\end{cases}
Its solution gives you coordinates of $B$ and $C$ from which you can obtain the line equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/965015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solving $2\cos2x-4\sin x\cos x=\sqrt{6}$ How I solve the following equation for $0 \le x \le 360$:
$$
2\cos2x-4\sin x\cos x=\sqrt{6}
$$
I tried different methods. The first was to get things in the form of $R\cos(x \mp \alpha)$:
$$
2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\
2\cos2x-2\sin2x=\sqrt{6}\\
R = \sqrt{4} = 2 \\
\alpha = \arctan \frac{2}{2} = 45\\
\therefore \cos(2x + 45) = \frac{\sqrt6}{2}
$$
which is impossible. I then tried to use t-substitution, where:
$$
t = \tan\frac{x}{2}, \sin x=\frac{2t}{1+t^2}, \cos x =\frac{1-t^2}{1+t^2}
$$
but the algebra got unreasonably complicated. What am I missing?
|
Note that
$$2\cos(2x)-2\sin (2x)=\sqrt{2^2+2^2}\cos(2x+\alpha)$$
where
$$\cos\alpha=\sin\alpha=\frac{2}{2\sqrt 2}=\frac{1}{\sqrt 2}.$$
Hence, you'll have
$$2\sqrt 2\cos(2x+45^\circ)=\sqrt 6\Rightarrow\cos(2x+45^\circ)=\frac{\sqrt 6}{2\sqrt 2}=\frac{\sqrt 3}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/966798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
how to solve $\int\frac{1}{1+x^4}dx$ i want find the answer and method of solve of $\int\frac{1}{1+x^4}dx$.
I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,
How I can use this to solve of that integration.
|
$$\frac{1}{1+x^4}=\frac{Ax+B}{2\sqrt2 (-x^2+\sqrt 2 x-1)}+\frac{Cx+D}{2\sqrt2 (x^2+\sqrt 2 x+1)}=\dots A=C=1, D=-B=\sqrt 2$$
Simplify even further
$$\frac{x-\sqrt 2}{2\sqrt2 (-x^2+\sqrt 2 x-1)}= -\frac{\sqrt 2-2x}{2-x^2+\sqrt 2 x-1)}-\frac{1}{\sqrt 2(-x^2+\sqrt 2 x-1}$$
Substitute $u=-x^2+\sqrt 2 x-1$ then it's trivial. The other is quite similar.
|
{
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"url": "https://math.stackexchange.com/questions/973822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find $\int (x^3+1)(\cos(2x)) dx$ What I've tried for this problem is expanding it to $x^3\cos(2x) + \cos(2x)$ and then evaluating the respective functions as separate integrals. The first one uses tabular and the second one is simple u substitution. Is my procedure correct?
|
what we want to do is eliminate x for the integral, thus do the integral by parts.
$$
\begin{align}
\int (x^3+1)\cos (2x))\mathrm dx
&=\frac{1}{2}(x^3+1)\sin2x - \int 3x^2\sin 2x\mathrm dx\\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{2}\int 3x\cos2x\mathrm dx\\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x +\frac{3}{4}\int\sin 2x \mathrm dx \\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x -\frac{3}{8}\cos 2x +constant
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/975884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Simplify: $\sin \frac{2\pi}{n} +\sin \frac{4\pi}{n} +\ldots +\sin \frac{2\pi(n-1)}{n}$. Can you help me solve this problem?
Simplify: $\sin \dfrac{2\pi}{n} +\sin \dfrac{4\pi}{n} +\ldots +\sin \dfrac{2\pi(n-1)}{n}$.
|
Take the terms in opposite pairs, and note the change of sign,
$$\sin \dfrac{2k\pi}{n}+\sin \dfrac{2\pi(n-k)}{n}=\sin \dfrac{2k\pi}{n} +\sin(2\pi-\dfrac{2k\pi}{n})=\sin \dfrac{2k\pi}{n} -\sin\dfrac{2k\pi}{n}=0.$$
In case that $n$ is even, the central term remains, but $$\sin \dfrac{2n\pi}{2n}=0.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/977956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Drawing a Right Triangle With Legs Not Parallel to x/y Axes? I have been presented with an interesting problem. How can I decide whether a right triangle with given side lengths can be placed (with integer coordinate vertices) on a Cartesian plane so that the legs are not parallel to the x or y axes?
I'm a little stumped as to how the lengths come into play. I figured I would just be considering the slopes of the legs, and whether they were 0 or undefined. Also, try as I might, I can't make a sketch of a triangle that obeys this non-parallel alignment specification.
Any ideas?
|
Here's an example of a right triangle with integer coordinates at every vertex
and legs that are not parallel to either coordinate axis:
The vertices of this triangle are $A=(6,4),$ $B=(-6,9),$ and $C=(0,0).$
Now to find out whether a right triangle with given side lengths
can be placed with all three vertices at integer coordinates, let's first try to
determine what right triangles can be constructed on integer coordinates and
what possible sets of lengths their sides can have.
Then we can ask whether the given set of side lengths is one of those possible sets.
First of all, the three side lengths have to satisfy the Pythagorean formula.
List the lengths in order of increasing size, so the lengths are
$AC,$ $BC,$ and $AB$ with $AC \leq BC < AB.$ Then it must be true that
$(AC)^2 + (BC)^2 = (AB)^2.$
If this is false, we do not even have a right triangle.
Now suppose we have three leg lengths as described in the previous paragraph.
Without loss of generality, we can simplify our visualization and calculations
by considering only triangles that (like the one in the figure) have a right
angle at $C=(0,0),$ vertex $A$ in the first quadrant of the plane ($x>0,y>0$),
and vertex $B$ in the second quadrant ($x<0,y>0$).
Any other right triangle with integer coordinates can be translated an integer distance
up or down and an integer distance right or left so that its right-angled vertex is
at $(0,0),$ then (if needed) rotated by $90,$ $180,$ or $270$ degrees so that
the other two vertices are in the first two quadrants, and then (if needed)
"flipped" (reflected) around the $y$ axis so that the leg in the first quadrant is
the shorter one.
So for the general case, let $A=(x_A,y_A),$ $B=(x_B,y_B),$ and $C=(0,0),$
where $x_A,$ $y_A,$ $x_B,$ and $y_B$ are all integers,
$x_A>0,$ $y_A>0,$ $x_B<0,$ and $y_B>0$ as in the figure.
The squares of the lengths of the two legs are then
$(AC)^2=x_A^2+y_A^2$ and $(BC)^2=x_B^2+y_B^2.$
Right away this tells us something about the possible lengths of sides:
the length of each leg is the square root of an integer.
But not all integers are sums of squares of integers,
so only some square roots of integers are possible side lengths.
To know if an integer $N$ is a sum of squares, find the prime factorization of $N.$
If no prime of the form $4k+3$ has an odd exponent in that factorization, $N$ is a
sum of two squares.
This is due to a theorem of Fermat as explained here.
Euler's proof of this theorem also gives some useful techniques to help find
possible pairs of squares.
For the given set of side lengths of a triangle, then,
let's determine whether $AC^2$ is the sum of two squares.
If it is not, we can stop right away; the triangle's vertices
cannot all have integer coordinates.
(For this step we just need the prime factorization of $AC^2.$
Let's examine leg $AC$ a little closer before we actually look for the two squares.)
The slope of the leg $AC$ is $y_A/x_A.$ Since $y_A$ and $x_A$ are both integers,
we can reduce this fraction to lowest terms, that is we can write
$$\frac{y_A}{x_A} = \frac qp$$
where $p$ and $q$ are integers that have no common divisor,
that is, $\gcd(p,q) = 1.$
This means $x_A=mp$ and $y_A=mq$ for some integer $m$
and $(p,q)$ is the point on the leg $AC$ with integer coordinates closest to $C.$
In the example in the figure, $(p,q)=(3,2).$
Since $(p,q)$ is on the leg $AC,$ the point $(-q,p)$ ($(-2,3)$ in the figure)
is on the leg $BC,$ perpendicular to $AC.$ Moreover, $(-q,p)$ is the point on $BC$
with integer coordinates that is closest to $(0,0).$
Any other point on the line $BC$ that has integer coordinates must have
coordinates that are just $(-q,p)$ scaled up by some integer factor.
In order for $B$ to be at integer coordinates, then, we must be able to write
$x_B=-nq$ and $y_B=np$ for some integer $n.$
So we can say this about the legs of the triangle: there exist positive integers
$m,$ $n,$ $p,$ and $q$ such that $\gcd(p,q) = 1,$
$$\begin{eqnarray}(AC)^2 &=& (mp)^2 + (mq)^2 &=& m^2(p^2 + q^2), \mbox{ and}\\
(BC)^2 &=& (np)^2 + (nq)^2 &=& n^2(p^2 + q^2).\end{eqnarray}$$
In the case of a triangle with sides $AC = 2\sqrt{13},$ $BC = 3\sqrt{13},$
and $AB = 13,$ like the one in the figure, we can then look for two squares of
positive integers whose sum is $(AC)^2 = 52.$ The only choice (apart from the
order in which we list the squares) is $52 = 6^2 + 4^2.$
This gives us $m=2,$ $p=3,$ and $q=2,$ as illustrated in the figure.
Now we need to check that we can find an integer $n$ such that
$(BC)^2 = n^2(p^2 + q^2).$ We find that $n=3,$ so we know we can place all vertices
of the triangle at integer coordinates, and moreover we have enough information
to find one such set of coordinates.
In other cases there is a pitfall we must avoid.
Suppose, for example, that $AC=5\sqrt{5},$ $BC=12\sqrt{5},$ and $AB=13\sqrt{5}.$
Then $(AC)^2 = 125 = 11^2 + 2^2,$ from which we get $m=1,$ $p=11,$ and $q=2.$
But we cannot find an integer $n$ such that $(BC)^2 = 720 = n^2(11^2 + 2^2).$
Yet we can place vertices of a triangle with these three side lengths
at integer coordinates: $A=(10,5),$ $B=(-12,24),$ and $C=(0,0).$
It matters which two squares we find whose sum is $(AC)^2.$
We could have written $(AC)^2 = 125 = 10^2 + 5^2,$
and then we would have $m=5,$ $p=2,$ and $q=1.$
We would then easily find that $(BC)^2 = 720 = 12^2(2^2 + 1^2).$
In order to find suitable pairs of squares, we first find the largest square
$m^2$ that divides $(AC)^2,$ so we can write $(AC)^2 = m^2 r$ where $r$
has no square factors.
In fact, if $(AC)^2$ is the sum of two squares, then the prime factorization
of $r$ has at most one factor of $2,$ at most one factor of
any prime of the form $4k+1,$ and no other factors, and so there are integers
$p$ and $q$ such that $r = p^2 + q^2.$
Moreover, for any values of $m',$ $p',$ and $q'$ that we could choose
so that $(AC)^2 = m'^2 (p'^2 + q'^2),$
we will find that $p^2 + q^2$ is a divisor of $p'^2 + q'^2$
and that the quotient is the square of an integer.
(This follows from facts used in counting the number of pairs of squares
with the desired sum.)
So if we can find an integer $n'$ such that $(BC)^2 = n'^2 (p'^2 + q'^2),$
we can find an integer $n$ such that $(BC)^2 = n^2 (p^2 + q^2).$
So if $(AC)^2$ is the sum of two squares, write $(AC)^2 = m^2 r$ where $m$ is an integer
and $r$ is an integer with no square factors,
and write $r = p^2 + q^2$ where $p$ and $q$ are integers.
If it is possible to place the vertices of a triangle with the desired side lengths
using only integer coordinates,
then $(BC)^2/(p^2 + q^2)$ will be the square of an integer $n,$ and we will
be able to write $(BC)^2 = n^2(p^2 + q^2)$ and find integer coordinates for all
vertices of the desired triangle.
But if $(BC)^2/(p^2 + q^2)$ is not the square of an integer,
then it is not possible to draw a triangle with the desired side lengths
using only integer coordinates for the vertices.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A problem on positive semi-definite quadratic forms/matrices Suppose $a+b+c=0$ and (without loss of generality) $a\leq b\leq 0\leq c$, $a^2+b^2+c^2=1$, is the following quadratic form positive semi-definite? Thank you very much.
\begin{equation*}
\begin{split}
I(x,y,z)=&(2a^2+5b^2+5c^2)a^2x^2+(5a^2+2b^2+5c^2)b^2y^2\\
+&(5a^2+5b^2+2c^2)c^2z^2\\
-&ab(a^2+b^2+c^2+6ab)xy-ac(a^2+b^2+c^2+6ac)xz\\
-&bc(a^2+b^2+c^2+6bc)yz\\
=&(2+3b^2+3c^2)a^2x^2+(2+3a^2+3c^2)b^2y^2+(2+3a^2+3b^2)c^2z^2\\
-&ab(1+6ab)xy-ac(1+6ac)xz-bc(1+6bc)yz.
\end{split}
\end{equation*}
We can write the matrix for $I$ as a product
\begin{equation*}
\left(\begin{array}{ccc}
a&0&0\\
0&b&0\\
0&0&c
\end{array}\right)
\left(\begin{array}{ccc}
2+3b^2+3c^2&3ab+\frac{1}{2}&3ac+\frac{1}{2}\\
3ab+\frac{1}{2}&2+3a^2+3c^2&3bc+\frac{1}{2}\\
3ac+\frac{1}{2}&3bc+\frac{1}{2}&2+3a^2+3b^2
\end{array}\right)
\left(\begin{array}{ccc}
a&0&0\\
0&b&0\\
0&0&c
\end{array}\right)
\end{equation*}
Therefore we need to show the middle matrix is positive semi-definite, is it correct?
|
A sum-of-squares decomposition of the polynomial is possible (the constraints are not needed).
Here is an implementation in the MATLAB Toolbox YALMIP (developed by me). YALMIP (together with a semidefinite programming solver) easily derives a decomposition $I(a,b,c,x,y,z) = v(a,b,c,x,y,z)^TQv(a,b,c,x,y,z)$ where $Q$ is positive definite (the fact that $Q$ is rather far away from being singular can be used to show that the numerical approach actually yields a certificate, despite any possible numerical issue).
sdpvar a b c x y z
I = (2*a^2+5*b^2+5*c^2)*a^2*x^2+(5*a^2+2*b^2+5*c^2)*b^2*y^2+(5*a^2+5*b^2+2*c^2)*c^2*z^2-a*b*(a^2+b^2+c^2+6*a*b)*x*y-a*c*(a^2+b^2+c^2+6*a*c)*x*z-b*c*(a^2+b^2+c^2+6*b*c)*y*z
[~,v,Q]=solvesos(sos(I))
However, we can do better and generate a nicer certificate. One way to do that is to search for an integer matrix $Q$. This will not work here (the problem is infeasible). We can however search for simple rational decompositions. Hence, trying the most simple rationals, we enforce all elements of $Q$ to be an integer divided by $2$. Alternatively, we search for a sum-of-squares decomposition of $2I$. This leads to an integer semidefinite program, but luckily YALMIP has a built-in solver for that. We do however have to formulate the sum-of-squares problem manually, but it is pretty easy when we already have run the module once and generated a sufficient set of monomials $v$.
Q = intvar(9,9);
Match = coefficients(I*2-v{1}'*Q*v{1},[a b c x y z])==0;
optimize([Match, Q >=0])
The integer SDP is solved in no-time, and the solution is given by $2I = v^TQv$ where
value(Q)
4 0 -1 -1 0 -1 0 0 -1
0 10 -5 -1 0 0 0 -1 0
-1 -5 10 0 0 -1 0 0 0
-1 -1 0 4 0 0 0 -1 -1
0 0 0 0 10 -5 -1 0 -1
-1 0 -1 0 -5 10 0 0 0
0 0 0 0 -1 0 10 -5 -1
0 -1 0 -1 0 0 -5 10 0
-1 0 0 -1 -1 0 -1 0 4
sdisplay(v{1})
ans =
'c^2*z'
'b*c*z'
'b*c*y'
'b^2*y'
'a*c*z'
'a*c*x'
'a*b*y'
'a*b*x'
'a^2*x'
|
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|
Show that inequality holds How would you show that the following inequality holds? Could you please write your reasoning by solving this problem too?
$a^2 + b^2 + c^2 \ge ab + bc + ca$ for all positive integers a, b, c
I have tried:
$a^2 + b^2 + c^2 - ab - ab + ab \ge bc + ca $
$(a^2 -2ab + b^2) + c^2 + ab \ge bc + ca $
$(a-b)^2 + c^2 + ab - bc - ca \ge 0 $
$(a-b)^2 + c(c - a) - b (c - a) \ge 0 $
$(a-b)^2 + (c - a)(c - b) \ge 0 $
Well, $(a-b)^2$ is positive. But how do we know if $(c-a)(c-b)$ is positive as well?
PS: The problem is from the book "How to think like a mathematician" by Kevin Housten.
|
Hint: Consider $(a-b)^2+(b-c)^2+(c-a)^2$.
Remark: Using the above hint, you can show that the desired inequality holds for all real numbers $a$, $b$, and $c$.
|
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|
Basic Math, exponents and algebra I have the equation
$$\frac{x_1^{-\frac{1}{2}}}{{x_2^{-\frac{1}{2}}}} = p_l/p_2$$
How do I get $x_2$ on its own?
I have
$$x_2^{-\frac{1}{2}} = \frac{p_2(x_1^{-\frac{1}{2}})}{p_1}$$
And if you have a useful link that reviews this info, it would be highly appreciated.
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By definition, if $x \neq 0$, then
$$x^{-n} = \frac{1}{x^n}$$
Thus,
$$x_1^{-\frac{1}{2}} = \frac{1}{x_1^{\frac{1}{2}}}$$
If we make the substitution $n = -m$ in the equation
$$x^{-n} = \frac{1}{x^n}$$
we obtain
$$x^{m} = \frac{1}{x^{-m}}$$
Hence,
$$\frac{1}{x_2^{-\frac{1}{2}}} = x_2^{\frac{1}{2}}$$
Therefore, we can rewrite the equation
$$\frac{x_1^{-\frac{1}{2}}}{x_2^{-\frac{1}{2}}} = \frac{p_1}{p_2}$$
in the form
$$\frac{x_2^{\frac{1}{2}}}{x_1^{\frac{1}{2}}} = \frac{p_1}{p_2}$$
which we can solve for $x_2$ by multiplying both sides by $x_1^{\frac{1}{2}}$ to obtain
$$x_2^{\frac{1}{2}} = \frac{p_1x_1^{\frac{1}{2}}}{p_2}$$
then squaring both sides to obtain
$$x_2 = \frac{p_1^2x_1}{p_2^2}$$
|
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"url": "https://math.stackexchange.com/questions/985121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
For the series $S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2$...... Problem :
For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series.
We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1}$
$$S_n =1+ \sum \frac{(\frac{n(n+1)}{2})^2}{(2n-1)^2}$$
$$\Rightarrow S_n =\frac{n^4+5n^2+2n^3-4n+1}{(2n-1)^2}$$
But I think this is wrong, please suggest how to proceed thanks..
|
You don't need to bother with calculating $S_n - S_{n-1}$ here as the terms are explicitly given in the summation.
Here $T_n = \frac{(1+2+...+n)^2}{(1+3+5+...+(2n-1))}$
The numerator is the square of the first $n$ integers, so is equal to $(\frac{1}{2}n(n+1))^2$.
The denominator is the sum of the first $n$ odd integers. You can calculate the sum in a couple of ways, by direct application of the arithmetic series sum formula, or by subtracting the sum of even numbers from the sum of the first $2n$ integers. Either way, you'll figure out the denominator is equal to $n^2$. Now just do the division.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I used the result to find its closed-form. The possible candidate closed-form from Wolfram Alpha is
$$\pi\sqrt{\frac{1+\sqrt{2}}{2}}-\pi$$
Is this true? If so, how to prove it?
|
\begin{align}
\int_0^{\Large\frac{\pi}{4}} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx&=\int_0^1\frac{\sqrt{y(1-y)}}{1+y^2}\,dy\quad\Rightarrow\quad y=\tan x\\
&=\int_0^\infty\frac{\sqrt{t}}{(1+t)(1+2t+2t^2)}\,dt\quad\Rightarrow\quad t=\frac{y}{1-y}\\
&=\int_0^\infty\frac{2z^2}{(1+z^2)(1+2z^2+2z^4)}\,dz\quad\Rightarrow\quad z^2=t\\
&=2\int_0^\infty\left[\frac{2z^2}{1+2z^2+2z^4}+\frac{1}{1+2z^2+2z^4}-\frac{1}{1+z^2}\right]\,dz\\
&=\int_{-\infty}^\infty\left[\frac{2z^2}{1+2z^2+2z^4}+\frac{1}{1+2z^2+2z^4}-\frac{1}{1+z^2}\right]\,dz\\
&=I_1+I_2-\pi
\end{align}
\begin{align}
I_1
&=\int_{-\infty}^\infty\frac{2z^2}{1+2z^2+2z^4}\,dz\\
&=\int_{-\infty}^\infty\frac{1}{z^2+\frac{1}{2z^2}+1}\,dz\\
&=\int_{-\infty}^\infty\frac{1}{\left(z-\frac{1}{\sqrt{2}z}\right)^2+1+\sqrt{2}}\,dz\\
&=\int_{-\infty}^\infty\frac{1}{z^2+1+\sqrt{2}}\,dz\\
&=\frac{\pi}{\sqrt{1+\sqrt{2}}}
\end{align}
where the 4th line we use identity
\begin{align}
\int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx\qquad,\qquad\text{for }\, a>0.
\end{align}
The proof can be seen in my answer here. $I_2$ can be proved in similar manner (see user111187's answer).
\begin{equation}
I_2=\frac{1}{2}\int_{-\infty}^\infty\frac{1}{z^4+z^2+\frac{1}{2}}\,dz=\pi\sqrt{\frac{\sqrt{2}-1}{2}}
\end{equation}
Combine all the results together, we finally get
\begin{equation}
\int_0^{\Large\frac{\pi}{4}} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx=\frac{\pi}{\sqrt[4]{2}}\sqrt{\frac{2+\sqrt{2}}{2}}-\pi
\end{equation}
|
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"url": "https://math.stackexchange.com/questions/989021",
"timestamp": "2023-03-29T00:00:00",
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|
How to Change Summation Expression $\sum_{i=1}^N \mathbf{X}_i^{\top}\mathbf{\Omega}^{-1}\mathbf{X}_i$ into Matrix Expression Let $\mathbf{X}_i$ be a $G \times K$ matrix, and suppose are $i=1,...,N$ of these matrices. Note that
\begin{align}
\sum_{i=1}^N \mathbf{X}_i^{\top}\mathbf{X}_i
&=
\begin{bmatrix}
\mathbf{X}_1^{\top} & \mathbf{X}_2^{\top} & \cdots & \mathbf{X}_N^{\top}
\end{bmatrix}
\begin{bmatrix}
\mathbf{X}_1 \\
\mathbf{X}_2 \\
\vdots \\
\mathbf{X}_N
\end{bmatrix} \\
&=\begin{bmatrix}
\mathbf{X}_1 \\
\mathbf{X}_2 \\
\vdots \\
\mathbf{X}_N
\end{bmatrix}^{\top}
\begin{bmatrix}
\mathbf{X}_1 \\
\mathbf{X}_2 \\
\vdots \\
\mathbf{X}_N
\end{bmatrix} \\
&=\mathbf{X}_{\bullet}^{\top}\mathbf{X}_{\bullet}
\end{align}
where $\mathbf{X}_{\bullet}$ is an $NG \times K$ matrix whose $i^{th}$ row is $\mathbf{X}_i$.
I'd like to do something similar with this summation:
\begin{equation}
\sum_{i=1}^N \mathbf{X}_i^{\top}\mathbf{\Omega}^{-1}\mathbf{X}_i
\end{equation}
where $\mathbf{\Omega}$ is some arbitrary symmetric $G \times G$ matrix, say for illustration
\begin{equation}
\mathbf{\Omega}
=\begin{bmatrix}
g_{11} & g_{12} & \cdots & g_{1G} \\
g_{12} & g_{22} & \cdots & g_{2G} \\
\vdots & \vdots & \ddots & \vdots \\
g_{1G} & g_{2G} & \cdots & g_{GG}
\end{bmatrix}
\end{equation}
This problem is not so simple as the first one, because now I have some $\mathbf{\Omega}^{-1}$ matrix in the middle of the sum. However, I think it can still be expressed in matrix notation.
For those curious, this matrix algebra is behind the theory of Generalized Least Squares. Thanks very much for your help!
Edit I think I have figured it out. I have posted an answer below -- please let me know if you think it's correct!
|
Let's try this.
\begin{align}
\sum_{i=1}^N \mathbf{X}_i^{\top}\mathbf{\Omega}^{-1}\mathbf{X}_i&=\mathbf{X}_1\mathbf{\Omega}^{-1}\mathbf{X}_1+\mathbf{X}_2\mathbf{\Omega}^{-1}\mathbf{X}_2+\cdots+\mathbf{X}_N\mathbf{\Omega}^{-1}\mathbf{X}_N \\
&=\begin{bmatrix}
\mathbf{X}_1^{\top} & \mathbf{X}_2^{\top} & \cdots & \mathbf{X}_N^{\top}
\end{bmatrix}
\begin{bmatrix}
\mathbf{\Omega}^{-1}\mathbf{X}_1 \\
\mathbf{\Omega}^{-1}\mathbf{X}_2 \\
\vdots \\
\mathbf{\Omega}^{-1}\mathbf{X}_N
\end{bmatrix} \\
&=\begin{bmatrix}
\mathbf{X}_1^{\top} & \mathbf{X}_2^{\top} & \cdots & \mathbf{X}_N^{\top}
\end{bmatrix}
\begin{bmatrix}
\mathbf{\Omega}^{-1} & \mathbf{0} & \cdots & \mathbf{0} \\
\mathbf{0} & \mathbf{\Omega}^{-1} & \cdots & \mathbf{0} \\
\vdots & \vdots & \ddots & \vdots \\
\mathbf{0} & \mathbf{0} & \cdots & \mathbf{\Omega}^{-1}
\end{bmatrix}
\begin{bmatrix}
\mathbf{X}_1 \\
\mathbf{X}_2 \\
\vdots \\
\mathbf{X}_N
\end{bmatrix} \\
&=\begin{bmatrix}
\mathbf{X}_1 \\
\mathbf{X}_2 \\
\vdots \\
\mathbf{X}_N
\end{bmatrix}^{\top}
\begin{bmatrix}
\mathbf{\Omega}^{-1} & \mathbf{0} & \cdots & \mathbf{0} \\
\mathbf{0} & \mathbf{\Omega}^{-1} & \cdots & \mathbf{0} \\
\vdots & \vdots & \ddots & \vdots \\
\mathbf{0} & \mathbf{0} & \cdots & \mathbf{\Omega}^{-1}
\end{bmatrix}
\begin{bmatrix}
\mathbf{X}_1 \\
\mathbf{X}_2 \\
\vdots \\
\mathbf{X}_N
\end{bmatrix} \\
&=\mathbf{X}_\bullet^{\top}\left(\mathbf{I}_N \otimes \mathbf{\Omega}^{-1}\right)\mathbf{X}_\bullet
\end{align}
where "$\otimes$" denotes the Kronecker Product.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+1) -2 = \frac{k+1}{2}(3(k+1)-1)$
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$$1+4+7+…+(3n-2) = \frac{n}{2}(3n-1)$$
Given identity is true For $n=1$
$$\frac{1\cdot(3-1)}{2}=1\tag{1}$$
Assume it's true for some integer $k$
$$1+4+7+…+(3k-2) = \frac{k}{2}(3k-1)\tag{2}$$
for $k+1$ .
We have to show that
$$1+4+7+…+(3k-2)+(3(k+1)-2) = \frac{k+1}{2}(3(k+1)-1)$$
$$1+4+7+…+(3k-2)+(3k+1) = \frac{k+1}{2}(3k+2)\tag{3}$$
Adding $(3k+1)$ to LHS and RHS of $(2)$
We get
$$\begin{align}
1+4+7+…+(3k-2)+(3k+1) & = \frac{k}{2}(3k-1)+(3k+1)\\
&= \frac{k(3k-1)+2(3k+1)}{2}\\
&= \frac{3k^2+5k+2}{2}\\
&= \frac{3k^2+3k+2k+2}{2}\\
1+4+7+…+(3k-2)+(3k+1) &= \frac{(3k+1)(k+2)}{2}\tag{4}\\
\end{align}$$
So, if our Identity is true for some integer $n=k$ Then it's also true for $n=k+1$, Since our identity is true for $n=1$ , Using principle of Mathematical Induction we can say that it's true for all $n\in\mathbb N$
Alternate
$$S=1+4+7+…+(3n-5)+(3n-2)\tag{1}$$
$$S=(3n-2)+(3n-5)+\cdots+7+4+1\tag{2}$$
By adding $1$ and $2$ We get,
$$2S=\underbrace{(3n-1)+(3n-1)+\cdots+(3n-1)+(3n-1)}_{\text{ n times }}=n(3n-1)$$
$$S=1+4+7+…+(3n-5)+(3n-2)=\frac{n}{2}(3n-1)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the bounds on a triple integral Problem: Find the volume enclosed by the cone $$x^2 + y^2 = z^2$$ and the plane $$2z - y -2 = 0$$
So I know that I need to do a triple integral over this region, and the integrand will be 1.
My problem is with finding the bounds for the integral. I set the $z$s equal to each other and found the intersection is $x^2 + \frac{3}{4}(y-\frac{4}{9})^2 = \frac{13}{9}$ (although I may have made a mistake here.)
I believe that the bound for $z$ is from $\sqrt{x^2 + y^2}$ to $\frac{y+2}{2}$?
Now I am not sure what to do. How can I find the bounds for $x$ and $y$? Also, I think I should do a change of variable so that the plane lies flat across the cone instead of slanted, to make the upper bound for $z$ constant. Is this a good idea? How could I do it? I am really lost with this problem.
|
The intersection is indeed an ellipse, but you seem to have made some algebra mistakes:
\begin{align*}
x^2 + y^2 &= (\tfrac{y + 2}{2})^2 \\
x^2 + y^2 &= \tfrac{1}{4}y^2 + y + 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y) &= 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y + \tfrac{4}{9} - \tfrac{4}{9}) &= 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y + \tfrac{4}{9}) - \tfrac{1}{3} &= 1 \\
x^2 + \tfrac{3}{4}(y - \tfrac{2}{3})^2 &= \tfrac{4}{3} \\
\end{align*}
Using the shadow method, notice that the above ellipse is the desired shadow that we want to integrate our $x$ and $y$ over. Solving for $x$ (for example) in the above ellipse yields:
$$
x = \pm \sqrt{\tfrac{4}{3} - \tfrac{3}{4}(y - \tfrac{2}{3})^2}
$$
A quick plot tells us that the min/max values of $y$ occur at the $y$-intercepts, where $y = \tfrac{2}{3} \pm \tfrac{4}{3}$.
so our triple integral is:
$$
V = \int_{-2/3}^2 \int_{-\sqrt{\frac{4}{3} - \frac{3}{4}(y - \frac{2}{3})^2}}^{\sqrt{\frac{4}{3} - \frac{3}{4}(y - \frac{2}{3})^2}} \int_\sqrt{x^2 + y^2}^{\frac{y + 2}{2}} \, dz \, dx \, dy
$$
To simplify this calculation consider using polar coordinates.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$? Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$?
|
To compute this series let's rewrite it first as a polygamma function :
\begin{align}
\tag{1}S&=\sum_{k=0}^{\infty }\frac{1}{(4k+3)^3}\\
S&=\frac 1{4^3}\sum_{k=0}^{\infty }\frac{1}{\left(k+\frac 34\right)^3}\\
\tag{2}S&=-\frac 1{2!\,4^3}\psi^{(2)}\left(\frac 34\right)\\
\end{align}
Since (from the previous link) $$\tag{3} \psi^{(2)}(z)=-2!\sum_{k=0}^{\infty }\frac{1}{\left(k+z\right)^3}$$
Computing the second derivative of the logarithm derivative of Euler's reflection formula for $\Gamma$ we get (since $\psi(z):=(\ln\,\Gamma(z))'\,$ and $\,(\ln\,\sin(\pi z))'=\pi\,\cot(\pi\,z)$ ) following reflection relation :
$$\tag{4}\psi^{(2)}(1-z)-\psi^{(2)}(z)=\pi\frac {d^2}{dz^2} \cot(\pi\,z)$$
that is for $z=\frac 14$ :
$$\tag{5}\psi^{(2)}\left(\frac 34\right)-\psi^{(2)}\left(\frac 14\right)=\lim_{z\to 1/4}\left[2\pi^3\cot(\pi\,z)(\cot(\pi\,z)^2+1)\right]=4\,\pi^3$$
But from $(3)$ we have too :
\begin{align}
\tag{6}\psi^{(2)}(z)+\psi^{(2)}\left(z+\frac 12\right)&=-2\left[\sum_{k=0}^{\infty }\frac{1}{\left(k+z\right)^3}+\sum_{k=0}^{\infty }\frac{1}{\left(k+z+1/2\right)^3}\right]\\
\psi^{(2)}\left(\frac 14\right)+\psi^{(2)}\left(\frac 34\right)&=-2\left[\sum_{k=0}^{\infty }\frac{1}{\left(k+1/4\right)^3}+\sum_{k=0}^{\infty }\frac{1}{\left(k+3/4\right)^3}\right]\\
&=-2\cdot 4^3\left[\sum_{k=0}^{\infty }\frac{1}{\left(4k+1\right)^3}+\sum_{k=0}^{\infty }\frac{1}{\left(4k+3\right)^3}\right]\\
&=-2\cdot 4^3\sum_{n=1}^{\infty }\frac{1}{\left(2n-1\right)^3}\\
&=-2\cdot 4^3\left[\sum_{n=1}^{\infty}\left(\frac{1}{\left(2n-1\right)^3}+\frac{1}{\left(2n\right)^3}\right)-\sum_{n=1}^{\infty}\frac{1}{\left(2n\right)^3}\right]\\
&=-2\cdot 4^3\left[\zeta(3)-\frac {\zeta(3)}8\right]\\
\tag{7}&=-2\cdot 4^3\frac 78\zeta(3)\\
\end{align}
Adding $(5)$ and $(7)$ we obtain $\,2\,\psi^{(2)}\left(\frac 34\right)\,$ at the left so that $(2)$ becomes :
$$S=-\frac 1{4\cdot 4^3}\cdot 2\,\psi^{(2)}\left(\frac 34\right)=-\frac 4{4\cdot 4^3}\pi^3+\frac{2\cdot 4^3}{4\cdot 4^3}\frac 78\zeta(3)$$
or simply
$$\tag{8}\boxed{\displaystyle S=\frac 7{16}\zeta(3)-\frac{\pi^3}{64}}$$
For a more general proof see Kölbig's $1996$ paper "The polygamma function $\psi^{(k)}(x)$ for $x=\frac 14$ and $x=\frac 34$".
For other rational arguments see "The polygamma function and the derivatives of the cotangent function for rational arguments".
|
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|
prove that $a^3 - b(b+1) = (a-2)(c^2 + 1) + 2$ has infinitely many solutions How do I prove that $$a^3 - b(b+1) = (a-2)(c^2 + 1) + 2$$ has infinitely many solutions if a, b and c are natural numbers?
I have opening the brackets and moving all the terms to one side which gets rid of the constant 2. I have also tried substituting the value of a in b, b in c, bc in a and so on... but it all results in a big mess which I can't simplify.
|
Ad-hoc solution: note that when $a=2$, $c$ doesn't matter. Solving
$$
a^3-b(b+1)=2
$$
for $b$ (given that $a=2$) gives us $b=2$ as well. Then, any triple $(2,2,c)$ $(c\in\mathbb{N})$ is a solution. There are infinitely many such triples so the claim follows.
Better solution:
solving for $c$ gives you
$$
c=\sqrt{\frac{a-a^3+b+b^2}{2-a}}
$$
so to make $c$ a natural number, it is enough that $a$ and $b$ satisfy
$$
b+b^2=2a^2-a^3\tag{i}
$$
because this will make the numerator under the square root $(2-a)a^2$. Solve (i) for $b$, which yields
$$
b=\frac{1}{2}(-1+\sqrt{1-4a+8a^2})=\frac{1}{2}\left(-1+\sqrt{(2a)^2+(2a-1)^2}\right).
$$
So we are done if we can demonstrate that there are infinitely many natural numbers $a$ such that $(2a)^2+(2a-1)^2$ is a perfect square. But that is already done here by Hagen von Eitzen. For completeness, I partially reproduce that answer here: define
$$
a_0=0, \qquad a_1=3, \qquad a_{n}=6a_{n-1}-a_{n-2}+2\text{ for }n\ge 2.
$$
Then, $a_n+(a_n+1)^2=c_n^2$ where
$$
c_0=1, \qquad c_1=5, \qquad c_{n}=6c_{n-1}-c_{n-2}\text{ for }n\ge 2.
$$
The numbers $\{a_n\}$ alternate in parity so in our context, you want $2a=a_{2k+1}+1$ for $k\geq 1$ (when $k=0$, $a=2$ so refer to the ad-hoc solution above).
|
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|
How integrate $ \iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0I'm trying to resolve this integral
$$
\iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0<a<b
$$
I tried with polar coordinates:
$$
x = r\cos{\theta} \\
y = r\sin{\theta} \\
Jacobian = r
$$
But I'm confused about in how to calculate the domain for each integral
|
if you are integrating over an annulus, then by symmetry
$$
\iint_{D} (\frac{x^2}{x^2+y^2})dA = \iint_{D} (\frac{y^2}{x^2+y^2})dA
$$
so by adding you have
$$
\iint_{D} (\frac{x^2+y^2}{x^2+y^2})dA = \iint_{D} dA = \pi(b^2-a^2)
$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
How do I know if two vectors with $n$ components are parallel? How do I know if two vectors with $n$ components are parallel?
For example
$$\begin{pmatrix}5\\2\\1\\3\\4\end{pmatrix} \text{, and } \begin{pmatrix}4\\1\\2\\3\\6\end{pmatrix}.$$
|
Put them together like: $$\begin{pmatrix} 5 & 4 \\ 2 & 1 \\ 1 & 2 \\ 3 & 3 \\ 4 & 6\end{pmatrix},$$
or in rows. If one of the $2 \times 2$ determinants there is non-zero, they're not parallel. They are not parallel because, for example, $\left|\begin{matrix} 5 & 4 \\ 2 & 1\end{matrix}\right| = -3 \neq 0$.
|
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|
Find the volume of the solid generated by revolving the region bounded by $y=x$ and $y=x^2$ about the line $y=x$ Find the volume of the solid generated by revolving the region bounded by $y=x$ and $y=x^2$ about the line $y=x$
I am confused, how do we approach such problems, where the rotation lines are not vertical/horizontal.
|
One way to do this is to rotate everything counter-clockwise about the angle $\theta = \pi/4$, so the line $y=x$ would line up with the $y$-axis
We can parametrize the curves:
$$y = x \rightarrow x_1 = t, \quad y_1 = t $$
$$y = x^2 \rightarrow x_2 = t, \quad y_2 = t^2 $$
Applying the rotation gives new coordinates
$$ x' = x\cos\theta - y\sin\theta = \frac{x-y}{\sqrt{2}} $$
$$ y' = x\sin\theta + y\cos\theta = \frac{x+y}{\sqrt{2}} $$
So curves when rotated are
$$ x_1' = \frac{t-t}{\sqrt{2}} = 0, \quad y_1' = \frac{t+t}{\sqrt{2}} = \sqrt{2}t $$
$$ x_2' = \frac{t-t^2}{\sqrt{2}}, \quad y_2' = \frac{t+t^2}{\sqrt{2}} $$
The original bounds are $(0,0)$ and $(1,1)$, rotated become $(0, 0)$ and $\left(0, \sqrt{2}\right)$, so $t$ goes from $0$ to $1$
The solid is equivalent to rotating the second curve around the $y$-axis
$$ V = \int_0^1 \pi x_2'^2 \frac{dy_2'}{dt} dt = \int_0^1 \pi \left( \frac{t-t^2}{\sqrt{2}} \right)^2 \left( \frac{1 + 2t}{\sqrt{2}} \right)\, dt $$
|
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|
derivative of $y=x^3\sqrt{x}-\frac{1}{x^2\sqrt{x}}$ $y=x^3\sqrt{x}-\dfrac{1}{x^2\sqrt{x}}$
Both terms require the product rule, right? My try:
$x^3\dfrac{1}{2}x^{-1/2}+3x^2x^{1/2}-\dfrac{-1}{2}x^{-3/2}x^{-2}--2x^{-3}x^{-1/2}$
What am I doing wrong? The correct answer is: $y\;'=3.5x^2\sqrt{x}+\dfrac{2.5}{x^3\sqrt{x}}$ and I don't see how what I got can reduce to this.
|
$$y=x^3x^{1/2} -x^{-2}x^{-1/2}$$
$$y'=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} -\left( -2 x^{-3}x^{-1/2} -\frac{1}{2}x^{-2}x^{-3/2}\right)$$
$$=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} +2 x^{-3}x^{-1/2} +\frac{1}{2}x^{-2}x^{-3/2}$$
$$=3x^{5/2} + \frac{1}{2}x^{5/2} + 2 x^{-7/2} + \frac{1}{2}x^{-7/2},$$
which matches the correct answer. You were not wrong in general (just a sign). You just needed to simplify a bit.
|
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|
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
|
Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that
$$\sum_{cyc}\frac{\frac{x}{y}}{\sqrt{1+\frac{x}{y}}}\geq\frac{3}{\sqrt2}$$ or
$$\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\geq\frac{3}{\sqrt2}.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\right)^2\sum_{cyc}xy(x+y)(2z+x+y)^3\geq\left(\sum_{cyc}x(2z+x+y)\right)^3.$$
Thus, it remains to prove that
$$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$ or
$$\sum_{sym}(x^6+9x^5y+24x^4y^2+18x^3y^3+9x^4yz-36x^3y^2z-25x^2y^2z^2)\geq0,$$
which is obviously true.
Done!
|
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|
Proving Identity of Combination The lecturer had given two questions of proving that are
$$\binom{r}{r}+\binom{r+1}{r}+...+\binom{n}{r}=\binom{n+1}{r+1}\text{for }n\geq{r}\geq{1} $$
$$\binom{r}{0}+\binom{r+1}{1}+...+\binom{r+k}{k}=\binom{r+k+1}{k}\text{for }r,k\geq{1}$$
I tried to use the induction to prove these two identites but the lecturer said these two proving questions should be related to the identity which is
$$\binom{m+n}{r}=\binom{m}{0}\binom{n}{r}+...+\binom{m}{r}\binom{n}{0}$$
|
$\bf{My\; Solution::}$ Given
$\displaystyle \binom{r}{0}+\binom{r+1}{1}+\binom{r+2}{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot +\binom{r+k}{k} = \binom{r+k+1}{k}\;,$ where $r,k\geq 1$
Now Using The formula of $\displaystyle \binom{n}{r} = \binom{n}{n-1}$ on $\bf{L.H.S\;}$
We can Write $\bf{L.H.S}$ as
$\displaystyle \binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot +\binom{r+k}{r}$
Coeff. of $x^r$ in $\displaystyle \left\{(1+x)^r+(1+x)^{r+1}+(1+x)^{r+2}+\cdot \cdot \cdot \cdot +(1+x)^{r+k}\right\} = \frac{(1+x)^{r+k+1}-(1+x)^{r}}{(1+x)-1}$
So Coeff. of $x^{r+1}$ in $\displaystyle \left\{(1+x)^{r+k+1}-(1+x)^{r}\right\} = \binom{r+k+1}{r+1} = \binom{r+k+1}{k}$
|
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|
How to prove $n!\leq(\frac{n+1}{2})^n$ Prove that for $n\in\mathbb{N}$ $$n!\leq(\frac{n+1}{2})^n.$$
I've solved base case for $n=1$
$$1\leq(\frac{1+1}{2})^1=1$$
The second step I've made was that I assumed that $n!\leq(\frac{n+1}{2})^n$ And then I have $(\frac{n+2}{2})^{n+1}$. What do I need to do now?
|
The induction proof can be made to work; use the fact that
$$\begin{align*}
\frac{\left(\frac{n+2}2\right)^{n+1}}{\left(\frac{n+1}2\right)^n}&=\frac{n+2}2\left(\frac{n+2}{n+1}\right)^n\\
&=\frac{n+2}2\left(1+\frac1{n+1}\right)^n\\\\
&>\frac{n+2}2\left(1+\frac{n}{n+1}\right)\\\\
&=\frac{(n+2)(2n+1)}{2n+2}\\\\
&=\frac{2n^2+5n+2}{2n+2}\\\\
&=n+\frac{3n+2}{2n+2}\\\\
&>n+1\;.
\end{align*}$$
The first inequality follows from the binomial theorem, for instance.
|
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|
Evaluating $\int \cos(x^2 + x)dx$ I need to evaluate following integral,
$$\int \left( x + \frac{1}{2} \right) \cos(x^2 + x)\,dx$$
but this one has got me pretty stumped! Any suggestions would be appreciated. Thanks!
|
Complete the square $$x^2+x=(x+\frac12)^2-\frac14$$ and change variable $y=x+\frac12$. So, $$\cos(x^2 + x)=\cos(y^2-\frac14)=\cos(y^2)\cos(\frac14)+\sin(y^2)\sin(\frac14)$$ and then $$I=\int cos(x^2 + x)dx=\cos(\frac14)\int \cos(y^2)dy+\sin(\frac14)\int \sin(y^2)dy$$where you recognized Fresnel integrals $$\int \cos(y^2)dy=\sqrt{\frac{\pi }{2}} C\left(\sqrt{\frac{2}{\pi }} y\right)$$ $$\int \sin(y^2)dy=\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} y\right)$$ So, back to $x$, $$\int \cos(x^2+x)dx=\sqrt{\frac{\pi }{2}} \left(\cos \left(\frac{1}{4}\right) C\left(\frac{2 x+1}{\sqrt{2
\pi }}\right)+\sin \left(\frac{1}{4}\right) S\left(\frac{2 x+1}{\sqrt{2 \pi
}}\right)\right)$$
|
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|
Can two pythagoras triplet have a common number If I have a pythagoras triplet $(a,b,c)$ such that
$$a^2+b^2=c^2$$
then is there another triplet $(a,d,e)$ possible such that
$$a^2+d^2=e^2, \; b\neq d$$
|
Each of $\quad A,B,C\quad$ will have as many primitive triples as its number of unique prime factors.
Triples are often generated with Euclid's formula
$$\text{where} \quad A=m^2-k^2 \quad B=2mk \quad C=m^2+k^2$$
$A=3\times5=15\longrightarrow
f(4,1)=(15,8,17)\quad
f(8,7)=(15,112,113)$
$B=2^2\times3=12\longrightarrow
f(3,2)=(5,12,13)\quad
f(6,1)=(35,12,37)$
$C=5\times13=65\longrightarrow
f(7,4)=(33,56,65)\quad
f(8,1)=(63,16,65)$
There are no tiple pairs where two numbers a shared so there is no such thing as
$\quad a^2+b^2=c^2\quad a^2+d^2=c^2\quad b\ne d$
|
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|
Solving Trigonometric Equation Problem im kinda stuck to solve the following problems below
Problem: 4cos²2x+sin2x=3 (0 < x <= π)
Steps:
4cos²2x+sin2x=3
2+2cos4x+sin2x=3
2cos4x+sin2x=1
2(1-2sin²2x)+sinx=1
2-4sin²2x+sinx=1
Bring every term to the right side
0= 1-2+4sin²2x-sinx
4sin²2x-sinx-1=0
Let M= sin2x, sin
4m²-m-1=0
X=+0.64 , x=-0.39
|
You have:
$$4\cos^2(2x)+\sin(2x) = 3$$
Use $\cos^2(2x) = 1-\sin^2(2x)$ then:
$$4(1-\sin^2(2x))+\sin(2x) = 3\implies4-4\sin^2(2x) + \sin(2x) = 3\implies\\1+\sin(2x)-4\sin^2(2x) = 0$$
which has roots:
$$\sin(2x) = \frac{1}{8}(1-\sqrt{17})\\\sin(2x) = \frac{1}{8}(1+\sqrt{17})$$
For the first equations:
$$\sin(2x) = \frac{1}{8}(1-\sqrt{17})\implies \sin^{-1}\sin(2x) = \sin^{-1}\frac{1}{8}(1-\sqrt{17})\implies \\2x = \sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)\implies x = \frac{\sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)}{2}$$
But there is also another solution in the other quadrant such that:
$$2x = \pi - \sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)\implies\\x = \frac{\pi}{2}-\frac{\sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)}{2}$$
You have to do the same thing for the second root:
$$\sin(2x) = \frac{1}{8}(1+\sqrt{17})$$
And then you have the answer
|
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|
Limit of $\left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- 1/ {n^4}} {4+ n^7/ {3^n}}$ as $n$ tends to infinity I already took some steps to simplifying the original question and im stuck at this point:
$$ \lim_{n \to \infty} \left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- \frac {1} {n^4}} {4+ \frac {n^7} {3^n}} $$
I know that $$\lim_{n \to \infty}\left(\frac {2}{3}\right)^n=0$$
since $$0<\frac{2}{3}<1$$
and that $$\lim_{n \to \infty}\frac {1} {n^4}=0$$
and that I have to use d'Alembert's criterion to solve what $$\frac {n^7} {3^n}$$ is
but what I get is $$\frac 13*\frac {(n+7)^7}{n^7}$$
I suppose the next step is just $$\frac 13 <1$$
"therefore the limit is zero" but my question is how do I get rid of the
$$\frac {(n+7)^7}{n^7}$$
The last step is just:
$$0*\frac {1-0}{4+0}=0$$
|
I think you can get this by doing a bit of algebra. Since $1-\frac{1}{n^4}<4+\frac{n^7}{3^n}$ for all $n \in \Bbb{N}$ then $$ \left(\frac{2}{3}\right)^n \cdot n^4 \cdot \frac{1- \frac{1}{n^4}} {4+ \frac{n^7}{3^n}}< \left(\frac{2}{3}\right)^n \cdot n^4$$ Now let's use the fact that exponents grow faster than powers. We know that there exists some $M \in \Bbb{N}$ that $$\left(\frac{4}{3} \right)^m>m^4 \quad \text{for all} \quad m>M$$ which means $$\left(\frac{2}{3}\right)^m \cdot m^4< \left(\frac{2}{3}\right)^m\left(\frac{4}{3} \right)^m \\ = \left(\frac{8}{9} \right)^m$$ and we know that $\lim_{m \to \infty} \left(\frac{8}{9} \right)^m=0$. Thus, for all $m>M$ $$\lim_{m \to \infty} \left(\frac{2}{3}\right)^m \cdot m^4 \cdot \frac{1- \frac{1}{m^4}}{4+ \frac{m^7}{3^m}} < \lim_{m \to \infty}\left(\frac{8}{9} \right)^m$$ which means by the comparison test that $$\lim_{m \to \infty} \left(\frac{2}{3}\right)^m \cdot m^4 \cdot \frac{1- \frac{1}{m^4}}{4+ \frac{m^7}{3^m}}=0$$
|
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|
If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$1681\cos^2\theta-3280\cos\theta+1600=0$$
Solving the quadratic equation gives $\cos\theta=\dfrac{40}{41}$. It is not easy to solve the quadratic equation without calculator so there must be some other method, if yes then please explain.
P.S: I've found the other method so I am self-answering the question.
|
Since $41^2 - 40^2 = (41-40)(41+40) = 81 = 9^2$, there exists an angle $\phi$ satisfying $$\sin \phi = \frac{40}{41}, \quad \cos \phi = \frac{9}{41},$$ hence $$\begin{align*} 1 &= \frac{9}{41} \sin \theta + \frac{40}{41} \cos \theta \\ &= \sin\theta \cos\phi + \cos\theta \sin\phi \\ &= \sin(\phi+\theta),\end{align*}$$ from which it follows that $\phi+\theta = \pi/2 + 2\pi k$ for some integer $k$, hence $$\cos\theta = \cos\left(\frac{\pi}{2} + 2\pi k - \phi\right) = \cos\left(\frac{\pi}{2} - \phi\right) = \sin\phi = \frac{40}{41},$$ and we are done.
|
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|
Proving $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\ldots\left(1+\frac{1}{n^3}\right)<3$ for all positive integers $n$
Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)<3$ for all positive integers $n$
This problem is copied from Math Olympiad Treasures by Titu Andreescu and Bogdan Enescu.They start by stating that induction wouldn't directly work here since the right hand side stays constant while the left increases.They get rid of this problem by strengthening the hypothesis.$$\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)\le3-\dfrac{1}{n}$$
and then proceed by induction.
The problem is that I can't find a motivation for the above change.I mean,we could have subtracted a lot of things from the RHS but what should nudge us to try $\dfrac{1}{n}$?The rest of the proof is quite standard,but I can't see how I am supposed to have thought of it.Is it just experience?Or is it a standard technique?A little guidance and motivation will be appreciated.
|
If we need to prove the strong inequality and proving the weak inequalities for the simplest to imagine values is easy...
\begin{align*}
\left(3-\frac1n\right)\left(1+\frac1{(n+1)^3}\right)
&=
3+\frac3{(n+1)^3}-\frac1n-\frac1{n(n+1)^3}\\
&=3+\frac{3n-(n+1)^3-1}{n(n+1)^3}\\
&=3+\frac{-n^3-3n^2-2}{n(n+1)^3}\\
&=3+\frac{-1}{n+1}
\frac{n^3+3n^2+2}{n^3+2n^2+n}\\
&\leq3-\frac1{n+1}
\end{align*}
|
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|
Completing Cayley table for a group The task is to complete the following Cayley table
for a given group. $e$ is of course the identity element. Together with group axioms and the fact that every Cayley table of a group must be a latin square, I arrived at
Is it correct? I know there is only one possible table and it sure looks like a latin square but I would appreciate if someone could confirm it's correct or not!
|
I agree with your answer.
Completing the Cayley table of order $6$ (if $pq=e$, it must be $qp=e$, etc.), you get a non commutative ($pr\ne rp$) magma, with identity element $e$.
Each element of it has unique inverse (each one is inverse of itself, but $p$ and $q$ mutually inverses).
Finally, if you relabel $e,p,q,r,s,t$ as $1,2,3,4,5,6$, you get the table
$$\begin{array}{c|cccccc}
\cdot & 1 & 2 & 3 & 4 & 5 & 6\\
\hline
1 & \color{green}{1} & \color{red}{2} & \color{red}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6}\\
2 & \color{red}{2} & \color{red}{3} & \color{green}{1} & \color{blue}{5} & \color{blue}{6} & \color{blue}{4}\\
3 & \color{red}{3} & \color{green}{1} & \color{red}{2} & \color{blue}{6} & \color{blue}{4} & \color{blue}{5}\\
4 & \color{blue}{4} & \color{blue}{6} & \color{blue}{5} & \color{green}{1} & \color{red}{3} & \color{red}{2}\\
5 & \color{blue}{5} & \color{blue}{4} & \color{blue}{6} & \color{red}{2} & \color{green}{1} & \color{red}{3}\\
6 & \color{blue}{6} & \color{blue}{5} & \color{blue}{4} & \color{red}{3} & \color{red}{2} & \color{green}{1}
\end{array}$$
that you can check to be associative by using this brute-force Matlab script.
On the other hand, the above is a quasigroup since the Cayley table is a latin square and an associative quasigroup has identity element too.
|
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|
probability: continuous uniform distribution mean by symmetry I am trying to show that the mean of the uniform continuous distribution is $(b+a)/2$ by symmetry. The direct method is fairly simple but, for some reason, I cant get this one.
\begin{align}
E[X] &= \int_{-\infty}^{\infty}xp_X(x)dx\\
&= \int_{-\infty}^{\frac{b+a}{2}}xp_X(x)dx + \int_{\frac{b+a}{2}}^{\infty}xp_X(x)dx
\end{align}
Now let $u = x-\frac{b+a}{2}$ for the first integral and $u = \frac{b+a}{2}-x$ for the second integral. Then
$$
\int_{-\infty}^0\Big(\frac{b+a}{2} + u\Big)p_X\Big(\frac{b+a}{2} + u\Big)du + \int_0^{\infty}\Big(\frac{b+a}{2} - u\Big)p_X\Big(\frac{b+a}{2} - u\Big)du
$$
In the uniform distribution on the interval $a$ to $b$, $p_X(x) = \frac{1}{b - a}$.
At this point, I have just been winging it. For the first integral, $\frac{a-b}{2} < u < \frac{b-a}{2}$. For the second integral, we have $\frac{a-b}{2} < -u < \frac{b-a}{2}\Rightarrow \frac{a-b}{2} < u < \frac{b-a}{2}$. Therefore, $p_X\Big(\frac{b+a}{2} + u\Big) = p_X\Big(\frac{b+a}{2} - u\Big) = \frac{1}{b - a} = p_X(x)$ Then I have
$$
\int_{-\infty}^0up_X(x)du - \int_0^{\infty}up_X(x)dx + \int_{-\infty}^{\infty}\frac{b + a}{2(b - a)}du
$$
but I need
$$
\int_{-\infty}^0up_X(x)du + \int_0^{\infty}up_X(x)dx = \int_{-\infty}^{\infty}up_X(x)du = \frac{b + a}{2}
$$
Edit 1:
I then tried integrating from $(a, (a+b)/2)$ and $((a+b)/2, b)$ but I end up with
$$
\frac{a+b}{2(b-a)}\int_a^bdu + \int_a^{(a+b)/2}\frac{u}{b-a}du - \int_{(a+b)/2}^b\frac{u}{b-a}du = \frac{a+b}{2} + \frac{ab}{b - a}
$$
so apparently I am missing something.
Edit 2:
Could I say
$$
\int_a^{(a+b)/2}\frac{u}{b-a}du
$$
and
$$
\int_{(a+b)/2}^b\frac{u}{b-a}du
$$
are both half of area so there subtraction is zero? It is very cavalier though. I would like to be able to prove this assertation not just state it and conclude the correct answer.
|
Sorry, I didn't read your question carefully enough.
The limit on the second of your integrals after doing the substitutions is wrong. Since $u=-x + (b+a)/2$ and $x$ runs from $(b+a)/2$ to $\infty$, you should have $u$ running from $0$ to $-\infty$. I think this is what has mislead you. The two integrals evaluate to $(3a+b)/8$ and $(a+3b)/8$, so their sum is $(a+b)/2$ as required.
|
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|
If $\sin A+\sin B =a,\cos A+\cos B=b$, find $\cos(A+B),\cos(A-B),\sin(A+B)$ If $\sin A+\sin B =a,\cos A+\cos B=b$,
*
*find $\cos(A+B),\cos(A-B),\sin(A+B)$
*Prove that $\tan A+\tan B= 8ab/((a^2+b^2)^2-4a^2)$
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$$\begin{align*} \tan A + \tan B &= \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} \\
&= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} \\
&= \frac{\sin(A+B)}{\frac{\cos(A+B) + \cos(A-B)}{2}} \\
&= \frac{2\sin(A+B)}{\cos(A+B) + \cos(A-B)} \end{align*}$$
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Give a combinatorial argument Give a combinatorial argument to show that $$\binom{6}{1} + 2 \binom{6}{2} + 3\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} + 6 \binom{6}{6} = 6\cdot2^5$$
Not quite where to starting proving this one. Thanks!
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$$2^5$$ is just a way to count the elements in a 5 member set. (For each member of the set you make a binary decision to include it or not)
Now let's say you have a six element set.
Let's look at $$ \binom{6}{1} $$
Fix any member of the 6 element set. Now you have 5 members left. So that is $$ \binom{5}{1} $$ But you could have choosen that fixed member in 6 ways. So that is effectively $$ 6\cdot\binom{5}{1} $$
Repeat the same argument for each binomial and we get:
$$ 6(\binom{5}{1} + \binom{5}{2} +\binom{5}{3} + \binom{5}{4} + \binom{5}{5} + \binom{5}{6}) $$
But what is inside the parenthesis is simply another way of counting the sub-sets of a set of 5 elements.
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"timestamp": "2023-03-29T00:00:00",
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|
Use limit definition to find derivative of $x+\sqrt x$ The function is $f(x) = x + \sqrt x$.
How would you use the limit definition of the derivative to find the derivative of that equation?
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We have
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
Let $$f(x) = x + \sqrt x$$
$$\begin{align}f'(x)&=\lim_{h\to0}\frac{(x+h + \sqrt {x+h})-(x + \sqrt x)}{h}\\
&=\lim_{h\to0}\frac{h + \sqrt {x+h} - \sqrt x}{h}\\
&=\lim_{h\to0}1+\frac{ \sqrt {x+h} - \sqrt x}{h}\\
&=\lim_{h\to0}1+\frac{ \sqrt {x+h} - \sqrt x}{h}\frac{\sqrt {x+h} + \sqrt x}{\sqrt {x+h} + \sqrt x}\\
&=\lim_{h\to0}1+\frac{x+h - x}{h( \sqrt {x+h} + \sqrt x)}\\
&=\lim_{h\to0}1+\frac{h}{h( \sqrt {x+h}+ \sqrt x)}\\
&=\lim_{h\to0}1+\frac{1}{ \sqrt {x+h} + \sqrt x}\\
&=1+\frac{1}{ \sqrt {x} + \sqrt x}\\
f'(x)&=1+\frac{1}{ 2\sqrt {x} }\\
\end{align}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1018890",
"timestamp": "2023-03-29T00:00:00",
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|
Find maximum/minimum for $\cos(2x) + \cos(y) + \cos(2x+y) $ I have not been able to find the critical points for $\cos(2x) + \cos(y) + \cos(2x+y) $
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When you set the partial derivatives equal to $0$, you obtain the two equations
\begin{align*}
\sin 2x + \sin (2x+y) &= 0 \\
\sin y + \sin (2x+y) &= 0\,.
\end{align*}
This means that we must have $\sin 2x = \sin y$, and so either $y=x+(2k)\pi$ or $y=(2k+1)\pi-x$ for some integer $k$. Substituting, for example, $y=2x$, you'll find that
$\sin 2x = 0$ or $\cos 2x = -1/2$. Continue :)
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"timestamp": "2023-03-29T00:00:00",
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|
How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified? I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?
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Hint: \begin{align}64x^4 + 16 x^2 + 4 &= (8x^2-2)^2 + 32x^2 + 16x^2
\\&= (8x^2-2)^2 + 48x^2
\\&= (8x^2-2)^2 - (4i\sqrt 3 x)^2
\end{align}
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Volume of the solid cut by a plane.
I'd like to find the volume of the following solid.
The solid enclosed by the paraboloid $z=4-x^2-y^2$ and the plane $x+y+z=1$.
Actually original problem is the following (I made upper problem...)
The solid enclosed by the cylinder $x^2+y^2=1$ and the plane $x+y+z=1$ and $z=-5$
In this case,
(the volume of the solid)=$$
\int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \int_{-5}^{1-x-y}dzdydx$$
Now, in case of
The solid enclosed by the paraboloid $z=4-x^2-y^2$ and the plane $x+y+z=1$.
How can I solve this?
Could you give me some hint, please?
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$$\begin{align}
z & = 4 - x^2 - y^2\\
1 - x - y & = 4 - x^2 - y^2\\
0 & = 3 - x^2 + x - y^2 - y\\
0 & = 3 - x^2 + x - \frac14 + \frac14 - y^2 + y - \frac14 + \frac14\\
0 & = 3 - (x - \frac12)^2 + \frac14 - (y - \frac12)^2 + \frac14\\
\frac72 & = (x - \frac12)^2 + (y - \frac12)^2
\end{align}$$
So the region of intersection between the plane and paraboloid lies over a circle of radius $\sqrt{\frac72}$ centered at $(\frac12, \frac12)$. Now you can set up the bounds of a 2-d integral.
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}$$
But I don't know how to evaluate this?
Many thanks for any help.
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You can use l'Hospital as many times as needed as long as the indeterminate forms conditions are fulfilled. In this case, using Taylor series can be helpful, too:
$$\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \ldots = x - \frac{x^3}6 + \mathcal O(x^5)$$
$$\implies \frac{\sin x - x + \frac{x^3}6}{x^3} = \frac{\mathcal O(x^5)}{x^3} = \mathcal O(x^2) \xrightarrow[x \to 0]{} 0$$
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"timestamp": "2023-03-29T00:00:00",
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|
Manipulating roots of a cubic Given that $A,B,C$ are the roots of the equation $x^3-5x^2+x+1$, how do I find the value of $$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B}$$ I know the Vieta's formulas but I am not able to manipulate the above expression into something known. And taking the LCM doesn't help. Please help me out. Thank you.
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Hint: $(A+B)(B+C)(C+A) = (AB+AC+B^2+BC)(C+A) = ABC+A^2B+AC^2+A^2C+B^2C+B^2A+BC^2+BCA = 2ABC + A^2(B+C) + B^2(C+A) + C^2(A+B) = 2\cdot 1 + A^2(5-A) + B^2(5-B) + C^2(5-C) = 2 + 5(A^2+B^2+C^2) - (A^3+B^3+C^3) = 2 + (A+B+C) + 3 = 2+5+3=10$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1022461",
"timestamp": "2023-03-29T00:00:00",
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|
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.