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$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational roots: $+-1$; number of possible real roots - positive: four or two or zero, negative: zero; actual roots: $x = 1, 1, 1, 1$ (a quadruple root).
Using the rational root theorem, you divide the factors of the constant, $1$, by the factors of the lead coefficient, also a 1. That step gives you only two different possibilities for rational roots: $1$ and $-1$.
The signs change four times in the original polynomial, indicating $4$ or $2$ or $0$ positive real roots. Replacing each $x$ with $-x$, you get $x^4 + 4x^3 + 6x^2 + 4x + 1 = 0$. The signs never change. The polynomial is the fourth power of the binomial $(x - 1)$, so it factors into $(x - 1)^4 = 0$, and the roots are $1, 1, 1, 1$. There are four positive roots (all the same number, of course).
Can someone explain, the factorization of the polynomial? I do not understand, how it factors into $(x - 1)^4$.
|
You can factor it as
$$x^{ 4 }-4x^{ 3 }+6x^{ 2 }-4x+1={ x }^{ 4 }-2{ x }^{ 3 }+{ x }^{ 2 }-2{ x }^{ 3 }+4{ x }^{ 2 }-2x+{ x }^{ 2 }-2x+1={ x }^{ 2 }\left( { x }^{ 2 }-2x+1 \right) -2x\left( { x }^{ 2 }-2x+1 \right) +{ x }^{ 2 }-2x+1=\\ =\left( { x }^{ 2 }-2x+1 \right) \left( { x }^{ 2 }-2x+1 \right) ={ \left( x-1 \right) }^{ 2 }{ \left( x-1 \right) }^{ 2 }={ \left( x-1 \right) }^{ 4 }\\ $$
|
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|
Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know that there is the possibility to use $\det(A^5)=\det(A)^5)$, but this is too long for given time to resolve the problem.
|
You can much easier get your expression by subtracting the third line from the fourth, then the second from the third and finally, the first from the second:
$$
\det A=
\det \begin{pmatrix}
a&a&a&a\\
a&b&b&b\\
a&b&c&c\\
0&0&0&d-c
\end{pmatrix}=
\det \begin{pmatrix}
a&a&a&a\\
a&b&b&b\\
0&0&c-b&c-b\\
0&0&0&d-c
\end{pmatrix}=
\det \begin{pmatrix}
a&a&a&a\\
0&b-a&b-a&b-a\\
0&0&c-b&c-b\\
0&0&0&d-c
\end{pmatrix}=a(b-a)(c-b)(d-c)
$$
You can easily find $\det (A^5)$ now and even say that
$$
\det A^n = \det \begin{pmatrix}
a^n&*&*&*\\
0&(b-a)^n&*&*\\
0&0&(c-b)^n&*\\
0&0&0&(d-c)^n
\end{pmatrix}
$$
|
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|
Prove the Inequality proves true, three variables Prove that for all positive real numbers $a,b,c$ we have:
$$\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\geq\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$$
So far I have solved for $a=b=c=1$ and $a=1$, $b=2$, $c=3$ to show that the inequality holds true but I am needing the answer written more like a proof, I'm just not sure what theorems to apply actually prove the inequality holds true.
|
For all real numbers it's wrong, of course.
For positives $a$, $b$ and $c$, we can use an Integral method:
We see that for all positives $x$, $y$ and $z$ the following inequality holds.
$$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)\geq0$$
Indeed, by Schur
$$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)=\sum\limits_{cyc}(x^4-x^3y-x^3z+x^2yz)+\sum\limits_{cyc}xy(x-y)^2\geq0$$
Hence, $\sum\limits_{cyc}\left(t^{4a}-2t^{2a+2b}+t^{2a+b+c}\right)\geq0$, where $t>0$ or
$\sum\limits_{cyc}\left(t^{4a-1}-2t^{2a+2b-1}+t^{2a+b+c-1}\right)\geq0$.
Thus, $\int\limits_{0}^1\sum\limits_{cyc}\left(t^{4a-1}-2t^{2a+2b-1}+t^{2a+b+c-1}\right)dt\geq0$,
which gives your inequality. Done!
|
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|
Computing $\int \frac{1}{(1+x^3)^3}dx$ I tried various methods like trying to break it into partial fractions after factorizing, applying substitutions but couldn't think of any. How will we integrate this?
|
Hint. By writing
$$
1+x^3=(1+x)(1-x+x^2), \qquad (1+x^3)^3=(1+x)^3(1-x+x^2)^3,
$$ one may observe that there is a partial fraction decomposition of the form
$$
\begin{align}
\frac{1}{(1+x^3)^3}&=\frac{a_1}{(1+x)}+\frac{a_2}{(1+x)^2}+\frac{a_3}{(1+x)^3}
\\\\&+\frac{\alpha_1x+\beta_1}{1-x+x^2}+\frac{\alpha_2x+\beta_2}{(1-x+x^2)^2}+\frac{\alpha_3x+\beta_3}{(1-x+x^2)^3}
\end{align}
$$ then each term can be classically integrated. One may notice that
$$
\frac{\alpha_3x+\beta_3}{(1-x+x^2)^3}=\frac12\frac{\alpha_3(2x-1)+2\beta_3}{(1-x+x^2)^3}+\frac12\frac{\alpha_3+2\beta_3}{(1-x+x^2)^3}
$$ which is easier to integrate and that
$$
(1-x+x^2)^3=((x-1/2)^2+3/4)^3=(3/4)^3(u^2+1)^3,
$$ the changes of variable $ x-1/2=\sqrt{3}u/2$, $u=\tan t$, will turn out useful.
|
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|
A nice problem on divisor and sum of divisor function. Find all $n$ such that $\sigma(n)+d(n)=n+100$.
My Try : Let $n=\prod\limits_{k=1}^{m}p_k^{\alpha_k};\quad\sigma(n)=\prod\limits_{k=1}^{m}\tfrac{p_k^{\alpha_k+1}-1}{p_k-1}=n\cdot\prod\limits_{k=1}^{m}\tfrac{p_k-p_k^{-\alpha_k}}{p_k-1}>n;\quad d(n)=\prod\limits_{k=1}^{m}(1+\alpha_k)\ge2^m$
Then $2^m<100\implies m\le 6$ thus $n$ atmost could have 6 distinct prime factors. I checked that there are no such $n$ satisfying the equation for cases $m=1,6$ but for the rest mere brute checking becomes very troublesome. Any idea or may be a trick would be great.
|
If $\tau(n)+\sigma(n)=n+100$ then the sum of the proper divisors of $n$ must be less than $100$. Notice that if $n$ has $s$ prime divisors then we can consider all the products of $s-1$ primes, and these are proper divisors.
If $n\geq 4$ then the sum of these divisors is at least $(2\times3\times 5)+ (2\times 5\times 7) + (2\times3\times 7)+(3\times 5\times 7) >100$.
We conclude $n$ has at most three prime divisors.
If $n$ has exactly one prime divisor then let $n=p^a$, then we need $1+p+p^2+\dots+p^{a-1}+a+1=100$. If $a=1$ it won't work as we need $1+2=100$. If $a=2$ we need $1+p+3=100\implies p=96$, which does not work, since $96$ is not prime. If $a=3$ we need $1+p+p^2+3=100\implies p(p+1)=95$, which clearly has no solutions. For $a\geq 4$ is is clear only $p=2$ could work, so we need $2^a-1+a+1=100\iff 2^a+a=100$, which is not possible (check $a=4,5,6$).
If $n$ has exactly two prime divisors is missing
Suppose $n$ has exactly three prime divisors, say $p<q<r$ and let $n=p^aq^br^c$. Suppose $(p,q,r)\neq(2,3,5)$ and suppose at least one of $a,b,c>1$.
Notice $pqr+pq+qr+qr+p+q+r+1+d(n)\geq (2\times3\times 7)+(2\times 3)+(2\times 7)+(3\times 7)+2+3+7+1+12>100$.
So the only cases that remain are $n=pqr$ and $n=2^a3^b5^c$
If $n=pqr$ then we need $pq+pr+qr+p+q+r+1=92$. Clearly $p=2$ as otherwise the sum would be odd, so $3(q+r)+qr=89$. So $q\geq5$ or the left is a multiple of $3$. Notice $q=5,r=7$ doesn't work, and $q=5,r=11$ is too big, so there are no solutions.
If $n=2^a3^b5^c$ then clearly $c=1$, otherwise $2\times5^2+3\times5^2>100$. We also have $b=1$, otherwise $(9\times 2)+(9\times 5)+(2\times 3 \times 5)+9>100$.
So the only remaining option is $2^a3^15^1$. If $a\geq 3$ then $2^a\times 15>100$. So we only have to check $n=30$ and $n=60$ and neither works.
|
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|
$3\times3$ Magic Square and hidden formula [Solved] I was doing the $3 \times3$ Magic Square of Squares problem -- found here: http://www.multimagie.com/English/SquaresOfSquaresSearch.htm -- and I figured out that if such Magic square exists with $9$ distinct numbers, then it must satisfy the following equation:
$$\sqrt{a} + \sqrt{b} - \sqrt{c} = \sqrt{d}$$
where the numbers $a,b,c$ and $d$ are distinct entries in the square.
Crucial notes:
1) $a+b$ must also be a square number
2) The numbers $a,b,c,d$ can be any distinct integers greater than 0.
My attempt was
$$\sqrt{36} + \sqrt{64} - \sqrt{75} = \sqrt{25}$$
which isn't quite equal and doesn't work because 75 isn't a perfect square.
In my amateur opinion there isn't a solution, but would like someone good in mathematics to prove or disprove above equation.
In advance, I'm very thankful for any answers.
|
From what I understand in your post and the comments, you would like to find -- as JMoravitz said -- $a,b,c,d,e$ such that
$$a^2 + b^2 = c^2 + d^2=e^2$$
We necessarily have
$$a^2+b^2-d^2=c^2$$
And if I may explain how this relates to magic squares problems, as most people won't know. The existence of a 3x3 magic square of square numbers is currently unknown. There is a prize for finding one (Read here: http://www.multimagie.com/English/SquaresOfSquaresSearch.htm) But if a solution did exist it would have the following form
$$\begin{bmatrix}
a^2 & b^2 & c^2 \\
d^2 & e^2 & f^2 \\
g^2 & h^2 & i^2
\end{bmatrix}$$
Being a magic square means the sum of the entries of each column, row, and the 2 diagonals are equal. From this, one can show that
$$a^2 + i^2 = b^2 + h^2 = c^2 + g^2=2e^2$$
Not exactly what you wanted though -- I think. However from this we can derive other interesting equalities. For example
$$(ai)^2 + \Big(\frac{a^2-i^2}{2}\Big)^2=
(bh)^2 + \Big(\frac{b^2-h^2}{2}\Big)^2=
(cg)^2 + \Big(\frac{c^2-g^2}{2}\Big)^2 = e^4$$
Maybe this is what you were tackling?
Anyways, to solve your question -- and the magic square of squares problem -- it is very important to understand the two square identity
$$(a^2 + b^2)(c^2 + d^2) = (ac \mp bd)^2 + (ad \pm bc)^2$$
It's closely related to the Gaussian integers (integer complex numbers. read here: https://en.wikipedia.org/wiki/Complex_number#Multiplication_and_division)
To solve the first equation, lets break it up into to 2
$$a^2 + b^2 = e^2$$
$$c^2 + d^2 = e^2$$
Each equation is solved if we substitute
$$a=2mn \quad b=m^2-n^2 \quad c = 2pq \quad d=p^2-q^2$$
$$e=m^2 + n^2 = p^2 + q^2$$
I mentioned the two square identity because it can be used to find this substitution. Now we must also find $m,n,p,$ and $q$ such that $m^2 + n^2=p^2+q^2$, which again can be done with the two square identity. The smallest example that gives a distinct solution without zeros is
$$(2^2 + 1^2)(3^2 + 2^2) \ \ =\ \ 8^2 + 1^2 \ \ =\ \ 7^2 + 4^2 \ \ =\ \ 65$$
So we can use $m=8$, $n=1$, $p=7$, and $q=4$ to get our first -- and the smallest -- solution
$$16^2 + 63^2 = 56^2 + 33^2 = 65^2$$
It would be fitting here to also explain why there are guaranteed to be infinitely many solutions. Any prime number that leaves a remainder of 1 when divided by 4, can be expressed as the sum of two squares (read more here: https://en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares). For the previous solution I used the first two such primes: $5=2^2 + 1^2$ and $13 = 3^2 + 2^2$.
$\quad$ There are indeed infinitely many of this kind of prime just like there are infinitely many primes. The first few are $5,13,17,29,37,41,...$ (https://oeis.org/A002144). You can pick any two distinct products of these primes to make a new solutions.
For example, starting with $m=7$, $n=4$, $p=5$, and $q=2$ will yield this solution
$$1836^2 + 427^2 = 1813^2 + 516^2 = 1885^2$$
That choice of $m,n,p,$ and $q$ works because $7^2+4^2=65=5\cdot 13$ and $5^2 + 2^2 = 29$. The numbers $5,13,$ and $29$ are primes of the form $4k+1$.
All solutions to $a^2 + b^2 = c^2 + d^2=e^2$ can either be generated with this method or by multiplying the $a,b,c,d$ and $e$ of some such solution by an arbitrary integer.
Hope this helps. Good luck getting that prize :)
|
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If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$. Question:
If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$.
My attempt:-
$$\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx$$
$$\frac{dy}{1+y}=-\frac{cos(x)}{2+sin(x)}dx$$
Integrating both sides
$$\int\frac{dy}{1+y}=-\int\frac{cos(x)}{2+sin(x)}dx$$
$$log(1+y)=-log(2+sin(x))+C$$
$$log(1+y)=log(\frac{1}{2+sin(x)})+C$$
Antilog on both sides
$$1+y=\frac{1}{2+sin(x)}+C$$
$$y=\frac{-1-sin(x)}{2+sin(x)}+C$$
Now it is given that $y(0)=1$
So $$y(0)=1=\frac{-1-sin(0)}{2+sin(0)}+C$$
$$1=\frac{-1}{2}+C$$
$$C=\frac{3}{2}$$
$$y(\frac{\pi}{2})=\frac{-1-sin(\frac{\pi}{2})}{2+sin(\frac{\pi}{2})}+C$$
$$y(\frac{\pi}{2})=\frac{-2}{3}+\frac{3}{2}$$
$$y(\frac{\pi}{2})=\frac{5}{6}$$
But the answer happens to be $\frac{1}{3}$.
|
$$
\frac{2+\sin(x)}{y+1}\frac{\mathrm{d}y}{\mathrm{d}x}=-\cos(x)
$$
Separating variables gives
$$
\int\frac1{y+1}\mathrm{d}y=-\int\frac{\cos(x)}{2+\sin(x)}\,\mathrm{d}x
$$
Integrating yields
$$
\log(y+1)=C_1-\log(2+\sin(x))
$$
or
$$
y=\frac{C}{2+\sin(x)}-1
$$
Since $y(0)=1$, we get $C=4$. Thus, $y\,\left(\frac\pi2\right)=\frac13$.
|
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|
Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
|
Hint
Let
$$ax^2+bxy+cy^2+dx+ey+f=0$$
Note
$$\tan 2\theta=\frac{b}{a-c}$$
or
$$\cot \theta=\frac{a-c}{b}+\sqrt{1+\left(\frac{a-c}{b}\right)^2}$$
we have
$$\cot \theta=\frac{1+\sqrt{5}}{2}$$
find $\sin\theta$ and $\cos\theta$ and apply
\begin{cases}
x=X\cos\theta-Y\sin\theta\\
y=X\sin\theta+Y\cos\theta
\end{cases}
|
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How to evaluate $\int \frac{1+x}{1+\sqrt x} dx$ I picked this exercice from B. Demidovitch and i started solving, but when it's complete, i get only the half of the solution (the first two fractions and the ln, doesn't show up).
$$\int \frac{1+x}{1+\sqrt x} dx$$
The Solution is:
$$2 \left[\frac{\sqrt{x^3}}{3}-\frac{x}{2}+2\sqrt{x}-2ln(1+\sqrt{x})\right] + C$$
|
Let $u=\sqrt x+1$, so that $\mathrm{d}x=2(u-1)\,\mathrm{d}u$.
$$\int\frac{1+x}{1+\sqrt x}\,\mathrm{d}x=\int\frac{1+(u-1)^2}{u}(2(u-1))\,\mathrm{d}u$$
Expanding makes the integral trivial. (added some more details below)
$$\begin{align*}2\int\frac{u-1+(u-1)^3}{u}\,\mathrm{d}u&=2\int\frac{u^3-3u^2+4u-2}{u}\,\mathrm{d}u\\[1ex]&=2\int\left(u^2-3u+4-\frac{2}{u}\right)\,\mathrm{d}u\\[1ex]&=\frac{2}{3}u^3-3u^2+8u-4\ln|u|+C\\[1ex]&=\frac{2}{3}\left(\sqrt x+1\right)^3-3\left(\sqrt x+1\right)^2+8\left(\sqrt x+1\right)-4\ln\left|\sqrt x+1\right|+C\\[1ex]&=\frac{2}{3}x^{3/2}-x+4x^{1/2}-4\ln\left(\sqrt x+1\right)+C\end{align*}$$
|
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Let a, b, c be positive real numbers. Prove that Let a,b,c be positive real numbers. Prove that
$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$
I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in finding my mistakes :]
|
The right inequality.
By AM-GM we obtain:
$$\prod\limits_{cyc}(a^2+bc)=\sqrt[3]{\frac{\prod\limits_{cyc}((a^2+bc)(b^2+ac))^2}{\prod\limits_{cyc}(a^2+bc)}}=\sqrt[3]{\frac{\prod\limits_{cyc}(ab(c^2+ab)+c(a^3+b^3))^2}{\prod\limits_{cyc}(a^2+bc)}}\geq$$
$$\geq\sqrt[3]{\frac{\prod\limits_{cyc}(4(a^3+b^3)(c^2+ab)abc)}{\prod\limits_{cyc}(a^2+bc)}}\geq\sqrt[3]{\frac{\prod\limits_{cyc}((a+b)^3(c^2+ab)abc)}{\prod\limits_{cyc}(a^2+bc)}}=abc\prod\limits_{cyc}(a+b)$$
|
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|
If $a +b $,= 5 and $ab = 6$ . Find $\frac{1}{a}$ + $\frac{1}{b}$ $a +b $ = 5 and $ab = 6$ . Find $\cfrac{1}{a}$ + $\cfrac{1}{b}$
Any Ideas on how to begin?
Many Thanks
|
Since $a+b = 5$ and $ab = 6$, then $a,b$ can be the roots of polynomial $x^2-5x+6=0$
By fraction, $(x-2)(x-3) = 0$.
Out of symmetry, let $a = 2,b=3$, so $\frac{1}{a}+\frac{1}{b}= \frac{5}{6}$
|
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Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might argue:
$$ \sum_{n \geq 1} (-1)^{n+1} \sqrt{n} = \frac{1}{2}\sum_{m \geq 1} \frac{1}{\sqrt{2m}} = \infty $$
Are these Cesaro summable? For an even number of terms:
$$\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots - \sqrt{2n}
\approx - \frac{1}{2\sqrt{2}}\left( \frac{1}{\sqrt{1}} +
\frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \right)
\approx \sqrt{\frac{n}{2}}$$
so the Cesaro means tend to infinity. Does any more creative summation method work?
The result is from paper called "The Second Theorem of Consistency for Summable Series" in Vol 6 of the Collected Works of GH Hardy
the series $1 - 1 +1 - 1 \dots$ is summable $(1,k)$ for any $k$ but not summable $(e^n, k)$ for any value of $k$.
The series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ is summable $(n,1)$ but not $(e^{\sqrt{n}},1)$ and so on...
Here things like $(1,k), (n,1)$ refer to certain averaging procedures, IDK
|
Hopefully I'm not copying Gottfried Helms' Euler sum, but the Euler summation formula allows you to sum this divergent series.
$$\sum_{k=1}^\infty(-1)^{k+1}\sqrt k=\sum_{n=0}^\infty\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^k\sqrt{k+1}$$
This converges extremely quickly to the Riemann zeta function,
$$\zeta(-1/2)=\frac1{1-\sqrt8}\sum_{n=0}^\infty\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^k\sqrt{k+1}$$
|
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|
Infimum of $\frac{(1+2a)(1+2b)(1+3a-b-ab)}{(1-b)(1+2b)(1+2a-a^2)+3a(1+2a)(1+2b-b^2)}$ Let $0<a,b<1$. What is the infimum of $$f(a,b)=\frac{(1+2a)(1+2b)(1+3a-b-ab)}{(1-b)(1+2b)(1+2a-a^2)+3a(1+2a)(1+2b-b^2)}?$$
It is possible to make the term $1+3a-b-ab$ in the numerator arbitrarily close to $0$ by letting $a\rightarrow 0$ and $b\rightarrow 1$. But this also makes the denominator go to zero, so it doesn't show that $f(a,b)$ can be close to zero.
|
To find the extrema inside the region $]0,1[ \times ]0,1[$, we should solve
$$\frac{\partial f(a,b)}{\partial a}=0$$
$$\frac{\partial f(a,b)}{\partial b}=0$$
After many algebraic operators, we obtain
$$(1+2b)(b-1)(2a^2b^2-b^2+b+2ab-3a^2b-2a-5a^2)=0$$
$$a(1+2a)(5a-2b+1-10ab+b^2-ab^2-4a^2b+2a^2b^2+8a^2)=0$$
We are only interested in solutions for which $0<a,b<1$, so
$$2a^2b^2-b^2+b+2ab-3a^2b-2a-5a^2=0$$
$$5a-2b+1-10ab+b^2-ab^2-4a^2b+2a^2b^2+8a^2=0$$
Now, I'm not sure how to continue with pen and paper, as the set of equations are quite lengthy and not easy to simplify. Using mathematical software, I received multiple solutions, but only one for which $0<a,b<1$
$$a_{min} = 0.1440863966$$
$$b_{min} = 0.6394981637$$
$$f(a_{min},b_{min})=0.9876434319$$
We have found the minimum inside $]0,1[ \times ]0,1[$, but it is still possible that the global minimum is located at the edges
$$f(a=0,b)=1$$
$$f(a=1,b)=\frac{6(1+2b)(2-b)}{11+20b-13b^2}$$
$$f(a,b=0)=\frac{(1+2a)(1+3a)}{5a^2+5a+1}$$
$$f(a,b=1)=1$$
For $a=1$ and $b=0$, the minimum values are
$$\frac{\partial f(a=1,b)}{\partial b}=0 \Rightarrow b=1, f(a=1,b=1)=1$$
$$\frac{\partial f(a,b=0)}{\partial a}=0 \Rightarrow a=0, f(a=0,b=0)=1$$
The values of $f(a,b)$ at the edges are larger than the minimum value of $f(a,b)$ in $]0,1[ \times ]0,1[$, so as the final answer we have
$$a_{min} = 0.1440863966$$
$$b_{min} = 0.6394981637$$
$$f(a_{min},b_{min})=0.9876434319$$
|
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|
A Trig Identity There are many trig identities that can be proven with known identities, but often the identities are tricky in finding the right methods to use. This is one of those identities that seems easy, but elusive. Consider the cubic equation
$$ x^3 - 14 x^2 + 56 x - 56 = 0$$
for which one solution is $8 \sin^2(2\pi/7)$ and is equal to $4.89008...$. Now, another solution can be found to be
$$7 - 6 \cos(\pi/7) + 6 \cos(2\pi/7) - 2 \cos(3\pi/7)$$
which is also found to have the numerical value $4.89008...$. Since both forms have the same numerical value and satisfy the same equation they must be equal. This leads to the identity
$$8 \sin^2(2\pi/7) = 7 - 6 \cos(\pi/7) + 6 \cos(2\pi/7) - 2 \cos(3\pi/7).$$
The purpose of this question is to ask what methods are best suited to prove this equation by use of trig identities, or a proof of this identity.
|
Writing $7x=\pi$
We need to prove $$7-6\cos x+6\cos2x-2\cos3x=8\sin^22x=4(1-\cos4x)\ \ \ \ (1)$$
Now as $\cos2x=\cos(\pi-5x)=-\cos5x,\cos4x=\cdots=-\cos3x$
$(1)$ reduces to $$\cos x+\cos3x+\cos5x=\dfrac12$$
Can you use Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ to prove this?
Now using Trigonometry Sine Multiple Angle Formulae,
$$\sin7x=7\sin x-56\sin^3x+112\sin^5x-64\sin^7x$$
Now $\sin7x=0,7x=m\pi$ where $m$ is any integer, $x=\dfrac{m\pi}7$ where $m\equiv0,\pm1,\pm2,\pm3\pmod7$
So, the roots of $7s-56s^3+112s^5-64s^7=0$ are $\sin\dfrac{m\pi}7$ where $m\equiv0,\pm1,\pm2,\pm3\pmod7$
As $\sin0=0,$ the roots of $7-56(s^2)+112(s^2)^2-64(s^2)^3=0\ \ \ \ (1)$ are $\sin\dfrac{m\pi}7$ where $m\equiv\pm1,\pm2,\pm3\pmod7$
Let $y=8\sin^2\dfrac{2\pi}7,$ as $\sin\dfrac{2\pi}7$ is a root of $(1)$
$$7-56\cdot\dfrac y8+112\left(\dfrac y8\right)^2-64\left(\dfrac y8\right)^3=0$$
Can you take it from here?
|
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|
Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$ If $\frac{x}{y}$ + $\frac{y}{x}$ = 3 Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$
Any Ideas on how to begin ?
|
\begin{align*}
\frac{x}{y} + \frac{y}{x} = 3 &\Longrightarrow \left(\frac{x}{y} + \frac{y}{x}\right)^2 = 9 \\
&\Longrightarrow \frac{x^2}{y^2} + \frac{y^2}{x^2} = 9-2=7
\end{align*}
|
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|
Is there a neater proof for $\lim_{x \rightarrow 1} 1/x = 1$? What I need to prove:
For all $\epsilon>0$, there exist a $\delta>0$, such that for all $x$ satisfying $0<|x-1|<\delta$, we have $|1/x-1|<\epsilon$.
My proof:
Let $\delta=\min \{ 1/2 ,\epsilon/2 \}$. Then when $\epsilon\geq 1$,
$$
|x-1| < \frac{1}{2} \\
(*) \frac{1}{2} < x < \frac{3}{2} \\
\frac{2}{3} < \frac{1}{x} < 2 \\
-\frac{1}{3} < \ \frac{1}{x}-1 < 1
$$
and since $-1<-1/3$, we have $|1/x-1|<1\leq \epsilon$.
When $\epsilon<1$,
$$
|x-1| < \frac{\epsilon}{2} \\
\frac{2-\epsilon}{2} < x < \frac{2+\epsilon}{2} \\
-\frac{\epsilon}{2+\epsilon} < \frac{1}{x}-1 < \frac{\epsilon}{2-\epsilon}
$$
and since $\epsilon>0$, we have $2+\epsilon > 2-\epsilon$. So
$$
-\frac{\epsilon}{2-\epsilon} < -\frac{\epsilon}{2+\epsilon} \implies \left| \frac{1}{x}-1 \right|< \frac{\epsilon}{2-\epsilon} < \epsilon.
$$
What I'm looking for is a proof that doesn't "deconstruct" the absolute values as in the starred inequality, ie uses only absolute value identities such as the triangle inequality and its variations.
|
If $|x - 1| \leqslant 1/2$, then $1 - x \leqslant |1 - x| = |x - 1| \leqslant 1/2$, therefore $x \geqslant 1/2$.
Since $x > 0$, the expression $1/x$ is well-defined.
Also because $x > 0$, we have $|x| = x \geqslant 1/2$.
If you must use the Triangle Inequality, you can argue (using it in the form $|u - v| \geqslant ||u| - |v||$):
$$
|x| = |1 - (1 - x)| \geqslant |1 - |1 - x|| = |1 - |x - 1|| \geqslant 1/2 \quad
(\text{because } 1 - |x - 1| \geqslant 1/2 > 0),
$$
but that only seems to make the argument more complicated, not less.
(It looks a bit less unnatural, however, if you start with the substitution $x = 1 - h$, or $x = 1 + h$.)
Therefore, if $|x - 1| \leqslant 1/2$,
$$
\left\lvert\frac{1}{x} - 1\right\rvert =
\left\lvert\frac{1 - x}{x}\right\rvert =
\frac{|1 - x|}{|x|} =
\frac{|x - 1|}{|x|}
\leqslant 2|x - 1|.
$$
Therefore, for all $\epsilon > 0$,
$$
\text{if } 0 < |x - 1| < \frac{\min\left\{1, \epsilon\right\}}{2},
\text{ then } \left\lvert\frac{1}{x} - 1\right\rvert < \epsilon.
$$
So, just as you found, we can take:
$$
\delta = \frac{\min\left\{1, \epsilon\right\}}{2}.
$$
If you were asked to prove, more generally,
$$
a \ne 0 \implies \lim_{x \to a} \frac{1}{x} = \frac{1}{a}
$$
(directly, that is, rather than deducing it from the case $a = 1$, via a result about limits of constant multiples of functions), there would be a bit more scope for manipulating absolute values using the Triangle Inequality:
If $|x - a| \leqslant |a|/2$, then
$$
|a| - |x| \leqslant ||a| - |x|| \leqslant |a - x| = |x - a| \leqslant |a|/2,
$$
therefore $|x| \geqslant |a|/2$, therefore
$$
\left\lvert\frac{1}{x} - \frac{1}{a}\right\rvert =
\left\lvert\frac{a - x}{xa}\right\rvert =
\frac{|a - x|}{|xa|} =
\frac{|x - a|}{|x|\cdot|a|}
\leqslant 2\frac{|x - a|}{|a|^2}.
$$
Therefore, for all $\epsilon > 0$,
$$
\text{if } 0 < |x - a| < \frac{|a|\cdot\min\left\{1,\, |a|\cdot\epsilon\right\}}{2},
\text{ then } \left\lvert\frac{1}{x} - \frac{1}{a}\right\rvert < \epsilon,
$$
which isn't much more complicated than the special case $a = 1$.
I see that Michael Spivak, Calculus, 3rd ed. (1994), p.101, does it this way - so at least it's not wrong!
|
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|
If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$?
Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$. What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?
After clearing the denominators, we have $$a(c+a)(a+b)+b(b+c)(a+b)+c(b+c)(c+a)=(a+b)(b+c)(a+c)\,.$$
That is,
$$a^3+b^3+c^3+abc=0\,.$$
But then I'm stuck. This question is related, but a bit different.
Thank you for your help!
|
Hint. You have
$$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - 1 = \frac{a^3+b^3+c^3+abc}{(a+b)(b+c)(c+a)} $$
as well as
$$ \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = \frac{(a^3+b^3+c^3+abc)(a+b+c)}{(a+b)(b+c)(c+a)}. $$
|
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|
Prove two vectors are parallel Let $ABCD$ be a parallelogram and $P$, $Q$ two points so that $\overrightarrow{PC} = \frac{1}{3}\overrightarrow{AC}$ and $\overrightarrow{BQ} = 2\overrightarrow{QD}$.
Find $\alpha$, $\beta$ $\in \mathbb{R}$ with the following property: $\overrightarrow{AP} + \overrightarrow{BQ} = \alpha\overrightarrow{AB} + \beta\overrightarrow{AD}$ and prove that $PQ$ and $BA$ are parallel lines.
For the first part of this problem I found $\alpha = \frac{4}{3}$ and $\beta = \frac{8}{3}$. However, I can't solve the last part. All I know is that I have to find a $\lambda \in \mathbb{R}$ so that $\overrightarrow{PQ} = \lambda\overrightarrow{AB}$, but I can't figure out how to find it.
|
$\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {AD}\\
\overrightarrow {AP} = \frac 23 \overrightarrow {AC}\\
\overrightarrow {BD} = -\overrightarrow {AB} + \overrightarrow {AD}\\
\overrightarrow {BQ} = \frac 23 \overrightarrow {BD}$
$\overrightarrow {AP}+\overrightarrow {BQ} =$$ \frac 23 \overrightarrow {AC} + \frac 23 \overrightarrow {BD}\\
\frac 23 \overrightarrow {AB} + \frac 23\overrightarrow {AD} -\frac 23\overrightarrow {AB} + \frac 23\overrightarrow {AD}\\
\frac 43 \overrightarrow {AD}$
$\overrightarrow {PQ} = $$\overrightarrow {AB} + \overrightarrow {BQ} - \overrightarrow {AP}\\\overrightarrow {AB} -\frac 23 \overrightarrow {AB}+\frac 23\overrightarrow {AD}-\frac 23 \overrightarrow {AB} - \frac 23\overrightarrow {AD}\\- \frac 13\overrightarrow {AB}$
|
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|
Number of digits in row of Pascal's triangle is $O\left(n^2\right)$
Let $a_n$ be the number of decimal digits in the $n$-th row of Pascal's Triangle
(so $a_0=1, a_1=2, a_2=3, a_3=4, a_4=5, a_5=8,\dots$).
Prove that $\frac{a_n}{n^2}$ converges and find the limit.
It's very easy to see that it converges. Indeed, it's well known that $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$ as $n\to\infty$. Letting $b_n=\log_{10}\binom{2n}{n}$, clearly $a_n\leq(n-2)b_n+2$. Then \begin{align*}\lim_{n\to\infty}\frac{a_n}{n^2}&\leq\lim_{n\to\infty}\frac{\log_{10}\frac{4^n}{\sqrt{\pi n}}}{2n}\\&=\lim_{n\to\infty}\frac{4n\ln(2)-1}{4n\ln(10)}\\&=\frac{\ln(2)}{\ln(10)}\approx0.301\dots\end{align*} However, empirical testing up to $2000$ suggests that this is a bad bound, and indeed, I'm sure that we can do a lot better, especially as I don't think $a_n\to(n-2)b_n+2$ as $n\to\infty$.
|
The quantity you're investigating is $\displaystyle \sum_{k=0}^n \left( \lfloor \frac{\ln \binom{n}{k} }{\ln 10}\rfloor +1\right) = n + O(n) + \frac{1}{\ln 10}\sum_{k=0}^n \ln \binom{n}{k} $
A simple computation yields $\displaystyle \frac{1}{n^2}\sum_{k=0}^n \ln \binom{n}{k} = \frac 2n \sum_{k=1}^n \frac kn \ln\left( \frac kn\right) + \frac{n+1}{n^2}\ln\left(\frac{n^n}{n!}\right)$
First term is a Riemann sum that goes to $-\frac 12$, and the second term goes to $1$ (with Stirling).
Hence $$\begin{align}\displaystyle \sum_{k=0}^n \left( \lfloor \frac{\ln \binom{n}{k} }{\ln 10}\rfloor +1\right) &= n + O(n) + \frac 1{2\ln 10} n^2 + o(n^2)\\ &= \frac 1{2\ln 10} n^2 + o(n^2)\end{align}$$
Telescopic estimates for $\sum_{k=1}^n k\ln k$ yield $\displaystyle \sum_{k=1}^n k\ln k=\frac{n^2\ln n}2 - \frac{n^2}4 +\frac{n\ln n}{2} + o(n\ln n)$
Since $\displaystyle\sum_{k=0}^n \ln \binom{n}{k} = -(n+1)\ln(n!)+2\sum_{k=1}^n k\ln k$, we get the sharper estimate $$\displaystyle \sum_{k=0}^n \left( \lfloor \frac{\ln \binom{n}{k} }{\ln 10}\rfloor +1\right) = \frac 1{2\ln 10} n^2 - \frac{1}{2\ln(10)}\frac{\ln n}{n} + o\left(\frac{\ln n}{n} \right)$$
|
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|
Diophantine equation: $x^2 + 4y^2- 2xy -2x -4y -8 = 0$ How many integer pairs $ (x, y) $ satisfy $$x^2 + 4y^2- 2xy -2x -4y -8 = 0 ?$$
As it seems at first glance, and hence, my obvious attempt was to make perfect squares but the $2xy$ term is causing much problem.
|
Too long for a comment.
$$x^2 + 4y^2- 2xy -2x -4y -8 = (x-y)^2 + 3(y-1)^2 -2x + 2y - 11\\ = (x-y-1)^2 + 3(y-1)^2 - 12 = 0,$$
$$(x-y-1)^2+3(y-1)^2=12,$$
$$(x-y-1,y-1)\in\{(0,2), (0,-2), (3,1), (3,-1), (-3,1), (-3,-1)\},$$
$$(x,y)\in\{(4,3), (0,-1), (6,2), (4,0), (0,2), (-2,0)\}.$$
|
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|
Show that $\frac{1}{2}\le \sum\limits_{k=0}^n\frac{1}{n+k}\le1$ for $n\in \mathbb N^+$
$$\frac{1}{2n}\le \frac{\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{n+n}}{n}\le\frac{1}{n}$$
I tried math induction and I tried take integral but I want to solve this with most elementary methods please give me hint or just show that. Thanks....
|
We show the inequality chain
\begin{align*}
\frac{1}{2}\leq\sum_{k=0}^n\frac{1}{k+n}\leq 1\qquad\qquad\qquad n\geq 1\tag{1}
\end{align*}
is not valid for $n=1,2$ and valid for $n\geq 3$.
We denote the sum with $A(n):=\sum_{k=0}^n\frac{1}{k+n}$.
Case $n=1,2,3$ :
\begin{align*}
A(1)&=\sum_{k=0}^1\frac{1}{k+1}=1+\frac{1}{2}=\frac{3}{2}>1\\
A(2)&=\sum_{k=0}^2\frac{1}{k+2}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\\
A(3)&=\sum_{k=0}^3\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{19}{20}<1\\
\end{align*}
We observe $A(1)$ and $A(2)$ are greater than $1$, while $\frac{1}{2}\leq A(3)\leq 1$.
Conclusion:
*
*The inequality chain (1) is not valid for $n=1,2$.
*Since $\frac{1}{2}\leq A(3)=\frac{19}{20}\leq 1$ the inequality chain (1) is valid for $n=3$.
$$ $$
Monotonicity of $A(n)$:
We want to compare $A(n)$ with $A(n+1)$. We obtain for $n\geq 1$
\begin{align*}
A(n+1)&=\sum_{k=0}^{n+1}\frac{1}{k+n+1}\\
&=\sum_{k=1}^{n+2}\frac{1}{n+k}\\
&=\sum_{k=0}^{n}\frac{1}{k+n}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}\\
&=A(n)+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}
\end{align*}
When we consider with some help of Wolfram Alpha the function
$$f(x)=\frac{1}{2x+1}+\frac{1}{2x+2}-\frac{1}{x}$$
with $x$ real, we see there is just one zero at $x=-\frac{2}{3}$. Since $f(1)=-\frac{5}{12}$, the function is negative for $x\geq 1$ and so
\begin{align*}
A(n+1)-A(n)=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}<0\qquad\qquad n\geq 1
\end{align*}
Conclusion:
*
*$A(n)$ is monotonically decreasing with increasing $n$.
*Since $A(3)\leq 1$ we see that $1$ is an upper limit of $A(n)$ for $n\geq 3$.
Finally we show $\frac{1}{2}$ is a lower limit of $A(n)$.
Harmonic numbers $H_n$:
Note that $A(n)$ is closely related with harmonic numbers $H_n=\sum_{k=1}^n\frac{1}{k}$.
We obtain
\begin{align*}
A(n)=\sum_{k=0}^n\frac{1}{k+n}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n-1}\frac{1}{k}=H_{2n}-H_{n-1}\qquad\qquad n\geq 1
\end{align*}
The harmonic numbers are asymptotically equal to
\begin{align*}
H_n\sim \ln n+\gamma
\end{align*}
with $\gamma$ the Euler constant. We obtain
\begin{align*}
\lim_{n\rightarrow\infty}A(n)&=\lim_{n\rightarrow \infty}\left(H_{2n}-H_{n-1}\right)\\
&\sim
\ln(2n)+\gamma-\ln(n-1)-\gamma\\
&\sim\ln 2
\end{align*}
Conclusion:
*
*Since $\ln 2\doteq 0.69314>\frac{1}{2}$ we see $A(n)\geq\frac{1}{2}$ for all $n\geq 3$.
$$ $$
Summary:
*
*The inequality chain (1) is not valid for $n=1,2$ and valid for all $n\geq 3$.
*The sum is monotonically decreasing with increasing $n$.
$$\sum_{k=0}^n\frac{1}{k+n}\searrow$$
*The limit of the sum is $\ln 2$.
$$\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{1}{k+n}=\ln 2$$
|
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|
Closed form of the following Recurrence Relation Let $L\colon\mathbb{N}^3 \to \mathbb{N}$ satisfy the following recurrence relationship,
$$
L(a,b,c) = 1 + \sum_{i=0}^{a-1} \sum_{j=0}^{b-1} \sum_{k=0}^{c-1} L(i,j,k),
$$
With "initial conditions" $L(0,a,b) = L(c,0,d) = L(0,e,f) = 0$. I am interested in knowing a closed form of $L$.
Work I have done:
I have investigated a simpler case
$
G\colon \mathbb{N}^2\to \mathbb{N}
$, satisfying
$$G(a,b) = 1 + \sum_{i=0}^{a-1} \sum_{j=0}^{b-1} G(i,j)$$
with similar "initial conditions" and can obtain $$G(a,b) = \binom{a+b-2}{a-1}= \binom{a+b-2}{b-1}= \frac{(a-b-2)!}{(a-1)!(b-1)!},$$ so I would have guessed that $$L(a,b,c) = \frac{(a+b+c-3)!}{(a-1)!(b-1)!(c-1)!}, $$ but this isn't correct. I would appreciate any hints on how to find a closed form solution for this.
|
Continuing according to Fred's very clever deduction of the multiple z-tranform
( the credit should go to him, so I invite him to post it, and votes to be casted on it as well)
$$
F(x,y,z) = \frac{{\frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}}}
{{1 - \frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}}} = \sum\limits_{1\, \leqslant \,n} {\left( {\frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}} \right)^{\,n} }
$$
and considering that
$$
\frac{{z^n }}
{{\left( {1 - z} \right)^n }} = \sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
n - 1 + k \\
k \\
\end{gathered} \right)\,\,z^{\,n + k} } = \sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered}
j - 1 \\
j - n \\
\end{gathered} \right)\,\,z^{\,j} }
$$
then we have
$$
\begin{gathered}
\sum\limits_{1\, \leqslant \,n} {\left( {\frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}} \right)^n } = \hfill \\
= \sum\limits_{0\, \leqslant \,a,\;b,\;c\;} {\left( {\sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
a - 1 \\
a - n \\
\end{gathered} \right)\left( \begin{gathered}
b - 1 \\
b - n \\
\end{gathered} \right)\left( \begin{gathered}
c - 1 \\
c - n \\
\end{gathered} \right)} } \right)x^{\,a} \,y^{\,b} \,z^{\,c} } \hfill \\
\end{gathered}
$$
i.e.
$$
\begin{gathered}
L(a,b,c) = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
a - 1 \\
a - n \\
\end{gathered} \right)\left( \begin{gathered}
b - 1 \\
b - n \\
\end{gathered} \right)\left( \begin{gathered}
c - 1 \\
c - n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
a - 1 \\
n - 1 \\
\end{gathered} \right)\left( \begin{gathered}
b - 1 \\
n - 1 \\
\end{gathered} \right)\left( \begin{gathered}
c - 1 \\
n - 1 \\
\end{gathered} \right)} \hfill \\
\end{gathered}
$$
Note that:
*
*in 1D it becomes $L(a,b,c) = 2^{\,a - 1} $
*in 2D
$$
\begin{gathered}
L(a,b) = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b} \right)} \right)} {\left( \begin{gathered}
a - 1 \\
a - n \\
\end{gathered} \right)\left( \begin{gathered}
b - 1 \\
b - n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b} \right)} \right)} {\left( \begin{gathered}
a - 1 \\
n - 1 \\
\end{gathered} \right)\left( \begin{gathered}
b - 1 \\
b - n \\
\end{gathered} \right)} = \left( \begin{gathered}
a + b - 2 \\
b - 1 \\
\end{gathered} \right) \hfill \\
\end{gathered}
$$
*but in 3D I do not know if there is a closed form.
Addendum
Again thanks to Fred's hint, actually $L(a,b,c)$ can also be expressed in terms of Hypergeometric Function, as
$$
\begin{gathered}
L(a,b,c) = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
a - 1 \\
a - n \\
\end{gathered} \right)\left( \begin{gathered}
b - 1 \\
b - n \\
\end{gathered} \right)\left( \begin{gathered}
c - 1 \\
c - n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{0\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right) - 1} \right)} {\left( \begin{gathered}
a - 1 \\
n \\
\end{gathered} \right)\left( \begin{gathered}
b - 1 \\
n \\
\end{gathered} \right)\left( \begin{gathered}
c - 1 \\
n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{0\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right) - 1} \right)} {\left( { - 1} \right)^{\,n} \left( \begin{gathered}
n - a \\
n \\
\end{gathered} \right)\left( { - 1} \right)^{\,n} \left( \begin{gathered}
n - b \\
n \\
\end{gathered} \right)\left( { - 1} \right)^{\,n} \left( \begin{gathered}
n - c \\
n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{0\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right) - 1} \right)} {\frac{{\left( {1 - a} \right)^{\,\overline {\,n\,} } \left( {1 - b} \right)^{\,\overline {\,n\,} } \left( {1 - c} \right)^{\,\overline {\,n\,} } }}
{{1^{\,\overline {\,n\,} } \;1^{\,\overline {\,n\,} } }}\frac{{\left( { - 1} \right)^{\,n} }}
{{n!}}} = \hfill \\
= {}_3F_{\,2} \left( {\left( {1 - a} \right),\left( {1 - b} \right),\left( {1 - c} \right);\;\;1,1;\;\; - 1} \right) \hfill \\
\end{gathered}
$$
although, having the variable $z$ fixed at $-1$, we are not much exploiting that function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1913600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Find m, n, p of given expression with some conditions Find $m, n, p$ such that in the expansion of the expression $ \left( x^m + \frac{1}{x^p} \right)^n$ the 12th and the 24th terms contain $x$, respectively $x^5$, and furthermore, this expansion also contains one free term (without $x$).
By using what we are given, I got the following:
$$mn - k(m + p) = 0$$
$$mn - 11(m + p) = 1$$
$$mn - 23(m + p) = 5$$
From these expressions you can easily conclude that $mn = -\frac{8}{3}$ and $m + p = -\frac{1}{3}$. By using these in the first equation, we get $k = 8$.
From this point we have three equations with 3 unknown variables. I have no idea what to do next.
Thank you in advance!
|
Expanding the binomial we obtain
\begin{align*}
\left( x^m + \frac{1}{x^p} \right)^n&=\sum_{k=0}^n\binom{n}{k}x^{mk}x^{-p(n-k)}\\
&=\sum_{k=0}^n\binom{n}{k}x^{(m+p)k-pn}
\end{align*}
According to the requirements we derive three equations:
\begin{align*}
11(m+p)-pn&=1\tag{1}\\
23(m+p)-pn&=5\tag{2}\\
k(m+p)-pn&=0\tag{3}\\
\end{align*}
Subtracting (2) from (1) and (3) from (2) gives
\begin{align*}
12(m+p)&=4\\
(k-23)(m+p)&=-5
\end{align*}
From the first we derive $m+p=\frac{1}{3}$ and obtain by putting this value in the second equation according to OPs result
\begin{align*}
k&=8
\end{align*}
With $k=8$ we obtain from (3)
\begin{align*}
p\cdot n=k(m+p)=\frac{8}{3}\tag{4}
\end{align*}
Since we have only to find a proper tripel $(n,m,p)$ we are free to specify appropriate values of $p$ and $n$ in (4). According to the requirement that the exponent of $x$ of the $24$-th term is $5$, we have to set $n$ at least to $24$ and we just take this value for $n$.
We obtain with $n=24$ from (4) and since $m+p=\frac{1}{3}$
\begin{align*}
p=\frac{1}{9},
m=\frac{2}{9}
\end{align*}
We finally get a solution
\begin{align*}
\left(x^{\frac{2}{9}}+x^{-\frac{1}{9}}\right)^{24}=x^{-\frac{8}{3}}(1+x^{\frac{1}{3}})^{24}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1916312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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|
For what value $k$ is $f(x) = \begin{cases} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}$ continuous at $x=2$? For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$
All those square roots are weighing me down! And $k$? My mind's not where it's supposed to be today. Thanks in advance for posting a solution!
|
Note that $$\frac { \sqrt { 2x+5 } -\sqrt { x+7 } }{ x-2 } =\frac { \left( \sqrt { 2x+5 } -\sqrt { x+7 } \right) \left( \sqrt { 2x+5 } +\sqrt { x+7 } \right) }{ \left( x-2 \right) \left( \sqrt { 2x+5 } +\sqrt { x+7 } \right) } =\\ =\frac { 1 }{ \sqrt { 2x+5 } +\sqrt { x+7 } } $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1920906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge {9+3\sqrt{3}\over2\sqrt{a+b+c}}$$
I tried AM-GM, which only gives large terms without any answer.
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge 3\sqrt[3]{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)}$$
Now this will never simplify to anything.
using a different approach i got,
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge \sqrt{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right)}$$
which also does not simplify further.
breaking the individual terms on the LHS also does not help nor does multiplying each term on LHS by its conjugate. :<
This was a introductory problem, so i guess there must exist a easy solution which i can't find.
Any hints will be helpful.
|
Hint:
If you multiply $a$, $b$ and $C$ by a same positive constant, $k^2$, then both LHS and RHS get multiplied by $k^{-1}$ (i.e. the inequality is homogeneous). Therefore you can assume WLOG that $a+b+c= 1$.
Simplify the inequality and you get:
$$\dfrac{1}{1-\sqrt{a}} + \dfrac{1}{1-\sqrt{b}} + \dfrac{1}{1-\sqrt{c}} \geq \dfrac{9 - 3\sqrt{3}}{2}.$$
Perhaps this is easier to prove.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1921833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Orthogonal change of variables - elimination of cross terms
Eliminate cross terms of $Q$ and express $Q$ in terms of the new variables.
$$\textit{Q}=2x^2+5y^2+5z^2+4xy-4xz-8yz$$
$$\begin{pmatrix}
2 &2 &-2 \\
2 &5 &-4 \\
-2 &-4 &5
\end{pmatrix}$$
$$-\lambda^3+12\lambda^2-21\lambda +10$$
$$\lambda _1=1, v_1=\begin{pmatrix}
-2 &1 &0
\end{pmatrix}$$
$$\lambda _2=1, v_2=\begin{pmatrix}
2 &0 &1
\end{pmatrix}$$
$$\lambda _3=10, v_3=\begin{pmatrix}
-1 &-2 &2
\end{pmatrix}$$
Do I have enough information to make a solution?
$${\textit{Q}}'={x}'+{y}'+10{z}'$$
or do I need to normalize and find the orthonormal bases -if so where to start from here?
|
Updated:
Note that $\lambda_{1}=\lambda_{2}$, $\boldsymbol{v}_{1} \perp \boldsymbol{v}_{3}$ and $\boldsymbol{v}_{2} \perp \boldsymbol{v}_{3}$, $\boldsymbol{v}_{1}$ and $\boldsymbol{v}_{2}$ can be resolved into linear combination of $\boldsymbol{e}_{1}$ and $\boldsymbol{e}_{2}$.
Let $\boldsymbol{e}_{1}=\displaystyle \frac{1}{\sqrt{5}}
\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$
and $\boldsymbol{e}_{3}=\displaystyle \frac{1}{3}
\begin{pmatrix} -1 \\ -2 \\ 2 \end{pmatrix}$, then
$$\boldsymbol{e}_{2}=\boldsymbol{e}_{3} \times \boldsymbol{e}_{1}=
\begin{pmatrix} -\frac{2}{3\sqrt{5}} \\ \frac{\sqrt{5}}{3} \\ \frac{4}{3\sqrt{5}} \end{pmatrix}$$
Therefore
\begin{align*}
\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} &=
\begin{pmatrix}
\frac{2}{\sqrt{5}} & -\frac{2}{3\sqrt{5}} & -\frac{1}{3} \\
0 & \frac{\sqrt{5}}{3} & -\frac{2}{3} \\
\frac{1}{\sqrt{5}} & \frac{4}{3\sqrt{5}} & \frac{2}{3} \\
\end{pmatrix}^{-1}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} \\
&=
\begin{pmatrix}
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
-\frac{2}{3\sqrt{5}} & \frac{\sqrt{5}}{3} & \frac{4}{3\sqrt{5}} \\
-\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
\end{align*}
$$
\begin{pmatrix}
\frac{2}{\sqrt{5}} & -\frac{2}{3\sqrt{5}} & -\frac{1}{3} \\
0 & \frac{\sqrt{5}}{3} & -\frac{2}{3} \\
\frac{1}{\sqrt{5}} & \frac{4}{3\sqrt{5}} & \frac{2}{3} \\
\end{pmatrix}
\begin{pmatrix} 2 & 2 & -2 \\ 2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}
\begin{pmatrix}
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
-\frac{2}{3\sqrt{5}} & \frac{\sqrt{5}}{3} & \frac{4}{3\sqrt{5}} \\
-\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
\end{pmatrix} =
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 10 \end{pmatrix}$$
$$Q = x'^2+y'^2+10z'^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1922198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$
|
$$\frac{s^2}{c^2}+\frac{c^2}{s^2}=\frac{s^4+c^4}{s^2c^2}=\frac{(s^2+c^2)^2-2s^2c^2}{s^2c^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1923555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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|
$x^2 dy/dx + 2xy=y^{-3}$, find $y(x)$ The question states to start by substituting $u=y^{-2}$, however I am not sure how to deal with this as-as far as I can see there is no easy substitution.
Are you meant to find that $y=(1/u)^{1/2}$ and then work with that(seems a little messy)?
Thanks,
|
Rewrite it as follows
$$
(2xy-y^{-3})dx+x^2dy=0
$$
Multiply it by $x^6y^3$, then
$$
(2x^7y^4-x^6)dx+x^8y^3dy=0
$$
Which is full differential. Hence:
$$
F=\int{x^8y^3dy}=\frac{x^8y^4}{4}+f(x)\\
\frac{dF}{dx}=2x^7y^4+\frac{df}{dx}=2x^7y^4-x^6\to\\
f=-\frac{x^7}{7}\to F=\frac{x^8y^4}{4}-\frac{x^7}{7}=C^*
$$
Which represent family of soutions:
$$
y^4=\frac{C}{x^8}+\frac{4}{7x}
$$
Edit
Consider substitution $u=y^{-2}\to du=-2y^{-3}dy$, then we have following equation:
$$
-\frac{x^2y^3}{2}\frac{du}{dx}+2xy=y^{-3}
$$
Divide both sides of equation by $y^3$
$$
-\frac{x^2}{2}\frac{du}{dx}+2xy^{-2}=y^{-6}\to
-\frac{x^2}{2}\frac{du}{dx}+2xu=u^3\to\\
(u^3-2xu)dx+\frac{x^2}{2}du=0
$$
Multiply equation by $u^{-3}x^6$, then we have full differential
$$
(x^6-2x^7u^{-2})dx+\frac{x^8u^{-3}}{2}du=0
$$
From which
$$
F=\int\frac{x^8u^{-3}}{2}du=-\frac{x^8u^{-2}}{4}+f(x)\\
\frac{dF}{dx}=-2x^7u^{-2}+\frac{df}{dx}=x^6-2x^7u^{-2}\to\\
f=\frac{x^7}{7}\to F=\frac{x^7}{7}-\frac{x^8u^{-2}}{4}=C^*
$$
Which after inverse substitution leads to the same family of solutions
$$
y^4=\frac{C}{x^8}+\frac{4}{7x}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1925305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Simplify the subtraction of two summations. $$\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) $$
I thought I would use the distributive rule but the question stipulates that my answer should not include a summation sign...
|
Method 1:
$$\begin{align}
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =\sum_{i=1}^n (3i^2 +4) - \sum_{j=1}^n (3(j+1)^2 +1) \\
& = \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2+6j+4) \\
& = \sum_{i=1}^n \left[(3i^2 +4) - (3i^2+6i+4)\right] \\
& = \sum_{i=1}^n -6i \\
& = -6\cdot\frac{n(n+1)}2 \\
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =-3n(n+1) \\
\end{align}$$
Method 2:
$$\begin{align}
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & = 7+\sum_{i=2}^n (3i^2 +4) - \sum_{j=2}^n (3j^2 +1)-3(n+1)^2-1 \\
& =6-3(n+1)^2+\sum_{i=2}^n \left[(3i^2 +4) - (3i^2 +1)\right] \\
& =6-3(n+1)^2+\sum_{i=2}^n3 \\
& =6-3(n+1)^2+\underbrace{3+3+3+\dots+3}_{n-1} \\
& =6-3(n+1)^2+3(n-1) \\
& =-3n^2-3n \\
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =-3n(n+1) \\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
General term of recurrence relation Find the general term of the following recurrence relation:
$$a_{1} = 2$$
$$a_{n+1} = \frac{2a_{n} - 1}{3}$$
I've tried to find the first few terms:
$$a_{1} = 2$$
$$a_{2} = \frac{2 \cdot 2 - 1}{3} = 1$$
$$a_{3} = \frac{2 \cdot 1 - 1}{3} = \frac{1}{3}$$
$$a_{4} = -\frac{1}{9}$$
$$a_{5} = -\frac{1}{27}$$
but can't see any pattern, especially considering the first 2 terms.
|
We have
$$ 3a_{n+1}=2a_n-1 \tag{1}$$
and we may try to get rid of the inhomogeneous term $-1$ by a suitable substitution.
For instance, by setting $a_n=b_n-1$, we get $b_1=3$ and
$$ 3b_{n+1}=2b_n \tag{2} $$
from which $b_n = 3\cdot\frac{2^{n-1}}{3^{n-1}}$ readily follows. The general term of the original sequence is so:
$$ a_n = \color{red}{3\cdot\frac{2^{n-1}}{3^{n-1}}-1}.\tag{3}$$
|
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"url": "https://math.stackexchange.com/questions/1927862",
"timestamp": "2023-03-29T00:00:00",
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|
How to sum up these fractions? Found this question in a competitive math test for elementary students. The long way is to add all the decimal values but is there a pattern/trick to solve this question (or these types)? I don't know how to solve this except by the long method of adding all the decimal equivalents.
The Answer is $1$
Compute:
$$\frac17 + \frac18 + \frac19 + \frac1{10} + \frac1{11} + \frac1{12} + \frac1{14} + \frac1{15} + \frac1{18} + \frac1{22} + \frac1{24} + \frac1{28} + \frac1{33} = ? $$
|
Not sure if this is what you are looking for but this is what I did:
$$S = \frac17 + \frac18 + \frac19 + \frac1{10} + \frac1{11} + \frac1{12} + \frac1{14} + \frac1{15} + \frac1{18} + \frac1{22} + \frac1{24} + \frac1{28} + \frac1{33}$$
I observed the following groupings:
*
*7, 14, 28
*8, 12, 24
*11, 22, 33
*I did grunt work on the remaining four numbers and found that the LCM is $90$
$$S = \underbrace{\frac17 + \frac1{14} + \frac1{28}}_{\frac14} + \underbrace{\frac18 + \frac1{12} + \frac1{24}}_{\frac14} + \underbrace{\frac1{11} + \frac1{22} + \frac1{33}}_{\frac16} + \underbrace{\frac19 + \frac1{10} + \frac1{15} + \frac1{18}}_{\frac13}$$
$$S = \frac14 + \frac14 + \frac16 + \frac13$$
$$ = \frac12 + \frac12 = 1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
|
Let $f(x)=ax^3+8x^2+bx+6$, and let $g(x)=x^2-2x-3$. Since the question says that $g(x)|f(x)$, then write $f(x)=g(x)(px+q)$. Note the linear polynomial, we did that because dividing $f(x)$ by $g(x)$ will result in degree $1$. Now, expand the LHS and RHS, and you get something like this:
$$ax^3+8x^2+bx+6=(x^2-2x-3)(px+q)=px^3+x^2(q-2p)-x(2q+3p)-3q$$
Clearly, $q=-2,$ and $p=-5$. Now get the values of $a$ and $b$ by comparing the co-efficients.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 3
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|
Finding a Power Series representation for the function $f(x) = \frac{2}{3-x}$ Let's say I want to find a Power Series representation of the function $f(x) = \frac{2}{3-x}$
Now I know we can write this as a geometric series
$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$
But I see two possible ways two write it as a geometric series. $(1)$ and $(2)$ below
Method $(1)$:
We let $a=2$, and $r = x-2$, thus we transform $\frac{2}{3-x}$ into the form $\frac{a}{1-r}$ and we write the Power Series for $f$ as follows:
$$f(x) = \sum_{n=0}^{\infty}\ 2(x-2)^n$$
Method $(2)$:
We divide both numerator and denominator by a factor of $3$ to go from $\frac{2}{3-x}$ to $$\frac{\frac{2}{3}}{1-\frac{x}{3}}$$
and we can then write $f$ as follows:
$$\begin{aligned}f(x) &= \sum_{n=0}^{\infty}\ \frac{2}{3}\left(\frac{x}{3}\right)^n
\\
&= \sum_{n=0}^{\infty}\left(\frac{2}{3^{n+1}}\right)x^n
\end{aligned}$$
But only $(2)$ is correct, and $(1)$ is incorrect, but I can't seem to see why. What I did in $(1)$ seemed like perfectly valid algebraic manipulations, so why does $(1)$ result in an erroneous answer?
|
Both methods are quite ok and valid and there are even many more possibilities to expand the function into a power series.
In order to better see what's going on, we should at first look somewhat closer at the function $f$. We consider the full specification of $f$ and choose as domain and codomain
\begin{align*}
&f:\mathbb{R}\setminus\{3\}\longrightarrow\mathbb{R}\\
&f(x)=\frac{2}{3-x}
\end{align*}
Note that we have a singularity, a simple pole at $x=3$. This is crucial when expanding a function as power series.
As important as domain and codomain is the radius of convergence when expanding a function as power series.
\begin{align*}
\sum_{n=0}^\infty a x^n=\frac{a}{1-x}\qquad\qquad |x|<1
\end{align*}
It is the radius of convergence which determines where the representation of $f$ as power series is valid.
If we look at the series expansion of the first method
\begin{align*}
f(x)=\sum_{n=0}^\infty2(x-2)^n\qquad\qquad\qquad |x-2|<1\tag{1}
\end{align*}
and the second method gives
\begin{align*}
f(x)=\sum_{n=0}^\infty\left(\frac{2}{3^{n+1}}\right)x^{n}=\frac{\frac{2}{3}}{1-\frac{x}{3}}\qquad\qquad \left|\frac{x}{3}\right|<1\tag{2}
\end{align*}
Attention: We have to explicitely state in (1) and (2) the range of validity since the power series is not defined outside this range. On the other hand $f$ can be defined on a much larger domain $\mathbb{R}\setminus\{3\}$.
Here's a graphic which illustrates both methods. We see the graph of $f$ with the asymptote at $x=3$.
*
*Method 1: The point $A=(2,2)$ is the center of an interval with length $2$ showing the validity of the power expansion. We clearly see the interval is bounded by the asymptote.
*Method 2: The point $B=\left(0,\frac{2}{3}\right)$ is the center of an interval with length $6$ showing the validity of the power expansion.
More expansions:
The graphic indicates what's going on, when we expand $f$ in a power series at a point $x=x_0$. The point $x_0$ is the center of an interval and the length is determined by the distance from $x_0$ to the asymptote $x=3$.
We can now expand $f$ at any point $x_0\in \mathbb{R}\setminus \{3\}$ as follows:
\begin{align*}
\frac{2}{3-x}&=\frac{2}{(3-x_0)-(x-x_0)}\\
&=\frac{2}{3-x_0}\cdot\frac{1}{1-\frac{x-x_0}{3-x_0}}\\
&=\frac{2}{3-x_0}\sum_{n=0}^\infty\left(\frac{x-x_0}{3-x_0}\right)^n\\
&=\sum_{n=0}^\infty\frac{2}{(3-x_0)^{n+1}}(x-x_0)^n\qquad\qquad \left|\frac{x-x_0}{3-x_0}\right|<1
\end{align*}
Of course, setting $x_0=2$ we obtain $\sum_{n=0}2(x-2)^n$ and setting $x_0=0$ we obtain $\sum_{n=0}^\infty\frac{2}{3^{n+1}}x^n$ and get back both methods as special cases.
|
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"url": "https://math.stackexchange.com/questions/1930354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Dice roll game probability Suppose we play a 1 vs 1 game. I roll a 6 sided fair die, if I get a 1 I win, if not you roll. If you get a 6 you win, otherwise I go again.... What are my chances of winning and what are yours?
|
The probability of you winning immediately is $\frac{1}{6}$. The probability of you winning on the third roll (after your opponent rolls for the first time) is $\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}$. This represents the probability that you do not win on the first roll, your opponent does not win on the second roll (given you did not win on the second roll), and you win on the third roll (given no one won on the first two rolls).
We can repeat this process to calculate the probability of you winning on the fifth roll as: $\frac{5}{6}\times\frac{5}{5}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}$.
We note that this is a geometric sequence with initial value $\frac{1}{6}$ and ratio $(\frac{5}{6})^2$. To find the probability that you will win the game, we want to sum:
$$\frac{1}{6} + (\frac{5}{6})^2\times\frac{1}{6}+(\frac{5}{6})^4\times\frac{1}{6}+...$$
$$=\frac{1}{6} \times(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)$$
$$=\frac{1}{6}\times\frac{1}{1-(\frac{5}{6})^2}$$
$$=\frac{1}{6}\times\frac{1}{1-\frac{25}{36}}$$
$$=\frac{1}{6}\times\frac{36}{11}$$
$$=\frac{6}{11}$$
The probability that you win is $\frac{6}{11}$. We notice that someone must eventually win the game, so the probability that your opponent wins will be $1-\frac{6}{11}=\frac{5}{11}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1930770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
A different way to prove that $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$ From the fact that $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \int_{0}^{\pi/8} \ln(1+ \cot u) \, du = \frac{3G}{4} + \frac{\pi}{16} \, \ln 2\tag{1}.$$
What is another way to prove $(1)$ that preferably doesn't involve the dilogarithm function?
EDIT:
In response to Dr. MV's comment, the following is how I deduced $(1)$ from that integral representation of Catalan's constant.
$$ \begin{align} G&= \int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{2-y^{2}} \int_{0}^{1} \frac{1}{1-\frac{x^{2}}{2-y^{2}}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{\sqrt{2-y^{2}}} \, \text{artanh} \left(\frac{1}{\sqrt{2-y^{2}}} \right) \, dy \\ &= \int_{0}^{\pi/4} \text{artanh} \left(\frac{1}{\sqrt{2} \cos \theta} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{\sqrt{2} \cos \theta +1}{\sqrt{2} \cos \theta -1} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{(\sqrt{2} \cos \theta+1)^{2}}{2 \cos^{2} \theta -1} \right) \, d \theta \\ &=\int_{0}^{\pi/4} \ln (\sqrt{2} \cos \theta +1) \, d \theta - \frac{1}{2} \int_{0}^{\pi/4} \ln(\cos 2 \theta) \, d \theta \\ &= \int_{0}^{\pi/4} \ln\left(\sqrt{2} \cos \left(\frac{\pi}{4} - \phi\right)+1\right) \, d \phi - \frac{1}{4} \int_{0}^{\pi/2} \ln( \cos \tau) \, d \tau \\ &= \int_{0}^{\pi/4} \ln \left(\sin(\phi) + \cos(\phi)+1\right) \, d \phi - \frac{1}{4} \left(- \frac{\pi}{2} \, \ln 2 \right) \\ &= \int_{0}^{\pi/4} \ln (\sin \phi) \, d \phi + \int_{0}^{\pi/4} \ln \left(1+ \frac{1+ \cos \phi}{\sin \phi} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{4} \, \ln 2 + \int_{0}^{\pi/4} \ln \left(1+ \cot \frac{\phi}{2} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{8} \, \ln 2 + 2 \int_{0}^{\pi/8} \ln (1 + \cot u) \, du \end{align}$$
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Perform the change of variable $y=\arctan x$,
$\displaystyle I=\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln(1+\tan x)dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}} \ln\left(\sin x+\cos x\right)dx-\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos x\right)dx$
$\begin{align}\displaystyle \int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}} \ln\left(\sin x+\cos x\right)dx&=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\sqrt{2}\cos\left(\dfrac{\pi}{4}-x\right)\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos\left(\dfrac{\pi}{4}-x\right)\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{\tfrac{\pi}{8}}^{\tfrac{\pi}{4}}\ln(\cos x)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x)dx
\end{align}$
Perform the change of variable $y=\dfrac{\pi}{2}-x$,
$\displaystyle \int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos x\right)dx=\int_{0}^{\tfrac{\pi}{8}}\ln\left(\sin x\right)dx$
Therefore,
$\begin{align}
\displaystyle I&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln\left(\sin x\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x\sin x)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln\left(\dfrac{\sin(2x)}{2}\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln\left(\dfrac{\sin x}{2}\right)dx\\
&=\dfrac{3\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln\left(\sin x\right)dx\\
&=\dfrac{3\pi}{16}\ln 2+\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx+\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln(\cot x)dx
\end{align}$
Since,
$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2} G-\dfrac{1}{4}\pi\ln 2$
and,
$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cot x)dx=G$
then,
$\begin{align}
I&=\dfrac{3\pi}{16}\ln 2+\dfrac{1}{4} G-\dfrac{1}{8}\pi\ln 2+\dfrac{1}{2}G\\
&=\dfrac{\pi}{16}\ln 2+\dfrac{3}{4}G
\end{align}$
PS:
Proof of
$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2} G-\dfrac{1}{4}\pi\ln 2$
$\begin{align}
\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx&=\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ln(\sin x)dx\\
&=\int_0^{\tfrac{\pi}{2}}\ln(\sin x)dx\\
&=2\int_0^{\tfrac{\pi}{4}}\ln\left(\sin(2x)\right)dx\\
&=2\int_0^{\tfrac{\pi}{4}}\ln\left(2\sin x\cos x\right)dx\\
&=\dfrac{1}{2}\pi\ln 2+2\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+2\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx
\end{align}$
Therefore,
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx=-\dfrac{1}{2}\pi\ln 2$
and,
$\displaystyle G=\int_0^{\tfrac{\pi}{4}}\ln(\cot x)dx=\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx$
therefore,
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2}G-\dfrac{1}{4}\pi\ln 2$
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx=-\dfrac{1}{2}G-\dfrac{1}{4}\pi\ln 2$
|
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"url": "https://math.stackexchange.com/questions/1931724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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|
Do the Taylor series of $\sin x$ and $\cos x$ depend on the identity $\sin^2 x + \cos^2 x =1$? I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatorics identitie.
The proof:
$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!}+\frac{1}{5!1!}\right )-...\\\\\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\\\\\cos^2x\!=\!1\!-\!x^2\left(\!\frac{1}{0!2!}\!+\!\frac{1}{2!0!}\!\right)\!+\!x^4\left(\!\frac{1}{0!4!}\!+\!\frac{1}{2!2!}\!+\!\frac{1}{4!0!}\!\right)\!-\!x^6\left(\!\frac{1}{0!6!}\!+\!\frac{1}{2!4!}\!+\!\frac{1}{4!2!}\!+\!\frac{1}{6!0!}\!\right)\!+...$$We should have shown that the series for both $\sin x$ and $\cos x$ converge absolutely (since we changed the arrangement), but it's obvious since the absolute value of all terms of $\sin x+\cos x$ add up to $e^x$.$$\sin^2x+\cos^2x=\\=1-x^2\left(\frac{1}{0!2!}-\frac{1}{1!1!}+\frac{1}{2!0!}\right)+x^4\left(\frac{1}{0!4!}-\frac{1}{1!3!}+\frac{1}{2!2!}-\frac{1}{3!1!}+\frac{1}{4!0!}\right)-x^6\left(\frac{1}{0!6!}-\frac{1}{1!5!}+\frac{1}{2!4!}-\frac{1}{3!3!}+\frac{1}{4!2!}-\frac{1}{5!1!}+\frac{1}{6!0!}\right)+...=\\\\=1+\sum_{n=1}^{\infty}(-1)^nx^{2n}\sum_{k=0}^{2n}\frac{(-1)^k\binom{2n}{k}}{(2n)!}$$
Since we can show easily that $\sum_{i=0}^n(-1)^i\binom{n}{i}=0$ by expanding $(1-1)^n$ using Binom's formula. So:$$\sin^2x+\cos^2x=1-0+0-0+...=1$$
I think it's beautiful. I just wanted to ask, do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.
|
No, they do not depend on the Pythagoren identity. Using the differential equations that define $\sin$ and $\cos$
*
*$\cos(0) = 1$
*$\sin(0) = 0$
*$\cos' = -\sin$
*$\sin' = \cos$
it's very easy to show that $\forall x \in \mathbb{R}\cos(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i}}{(2x)!}$ and $\forall x \in \mathbb{R}\sin(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i + 1}}{(2x + 1)!}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1934790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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|
If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$$\frac {1}{1 + x} = y$$
$$(1 + x)y = 1\tag{1}$$
Similarly,
$$(1 + y)z = 1\tag{2}$$
$$(1 + z)x = 1\tag{3}$$
Please help me to continue from here.
|
From
$\cos A=\tan B,
\cos B=\tan C,
\cos C=\tan A
$,
$\begin{array}\\
\cos A
&= \sin B/\cos B\\
&= \sin B/\tan C\\
&= \sin B/(\sin C/\cos C)\\
&= \sin B \cos C/\sin C\\
&= \sin B (\sin A/\cos A)/\sin C\\
\end{array}
$
or
$\cos^2 A = \sin A\sin B/\sin C
$.
Similarly,
$\begin{array}\\
\cos B
&= \sin C/\cos C\\
&= \sin C/\tan A\\
&= \sin C/(\sin A/\cos A)\\
&= \sin C \cos A/\sin A\\
&= \sin C (\sin B/\cos B)/\sin A\\
\end{array}
$
or
$\cos^2 B = \sin B\sin C/\sin A
$.
Also,
$\begin{array}\\
\cos C
&= \sin A/\cos A\\
&= \sin A/\tan B\\
&= \sin A/(\sin B/\cos B)\\
&= \sin A \cos B/\sin B\\
&= \sin A (\sin C/\cos C)/\sin B\\
\end{array}
$
or
$\cos^2 C = \sin A\sin C/\sin B
$.
Letting
$x=\sin A,
y = \sin B,
z = \sin C
$,
these become
$1-x^2 = xy/z,
1-y^2 = yz/x,
1-z^2 = zx/y
$.
Dividing the first two,
$\frac{1-x^2}{1-y^2}
=\frac{x^2}{z^2}
$.
From the last one,
$x = \frac{y(1-z^2)}{z}
$
so
$1-y^2
=\frac{zy}{\frac{y(1-z^2)}{z}}
=\frac{z^2}{1-z^2}
$
or
$y^2
=1-\frac{z^2}{1-z^2}
=\frac{1-2z^2}{1-z^2}
$.
Similarly,
$\frac{1-x^2}{\frac{z^2}{1-z^2}}
=\frac{x^2}{z^2}
$
or
$(1-x^2)(1-z^2)
=x^2
$
or
$1-z^2
=\frac{x^2}{1-x^2}
$
or
$z^2
= 1-\frac{x^2}{1-x^2}
= \frac{1-2x^2}{1-x^2}
$.
Similarly,
again,
we get
$\begin{array}\\
x^2
&= \frac{1-2y^2}{1-y^2}\\
&= \frac{1-2\frac{1-2z^2}{1-z^2}}{1-\frac{1-2z^2}{1-z^2}}\\
&= \frac{1-z^2-2(1-2z^2)}{1-z^2-(1-2z^2)}\\
&= \frac{3z^2-1}{z^2}\\
&= \frac{3\frac{1-2x^2}{1-x^2}-1}{\frac{1-2x^2}{1-x^2}}\\
&= \frac{3(1-2x^2)-(1-x^2)}{1-2x^2}\\
&= \frac{2-5x^2}{1-2x^2}\\
\text{or}\\
x^2-2x^4
&=2-5x^2\\
\text{or}\\
0
&=2x^4-6x^2+2\\
\text{or}\\
0
&=x^4-3x^2+1\\
\text{or}\\
x^2
&=\dfrac{3\pm\sqrt{9-4}}{2}\\
&=\dfrac{3\pm\sqrt{5}}{2}\\
\end{array}
$
Since
$0 \le x^2 \le 1$,
we must have
$x^2
=\dfrac{3-\sqrt{5}}{2}
$,
so
$x
=\pm\dfrac{\sqrt{5}-1}{2}
$.
Going through the same thing
for $y$ and $z$,
we get
$x = y = z
=\pm\dfrac{\sqrt{5}-1}{2}
$.
To determine the signs,
multiply
$1-x^2 = xy/z,
1-y^2 = yz/x,
1-z^2 = zx/y
$
together.
We get
$(1-x^2)(1-y^2)(1-z^2)
=xyz
$.
Therefore
$xyz > 0$
so that
0 or 2 of them
are negative
and the other(s)
are positive.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$
Let $a$, $b$, and $c$ be positive real numbers with $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that:
$$
ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc.
$$
Let us consider the following proofs.
$$
a^{2}+b^{2}+c^{2} \geq ab+bc+ca
$$
By the Arithmetic Mean-Geometric Mean Inequality we have
$$
a^{2}+b^{2} \geq 2ab,\ \ b^{2}+c^{2} \geq 2bc,\ \ c^{2}+a^{2} \geq 2ca \tag{1}
$$
If we add together all the inequalities $(1)$, we obtain
$$
2a^{2}+2b^{2}+2c^{2} \geq 2ab+2bc+2ca
$$
By dividing both side by $2$, the result follows.
Now let us consider,
$$
ab(a+b) + bc(b+c) + ac(a+c) \geq 6abc \tag{2}
$$
I already have proved $(2)$
Then,
We are given,
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3 \implies bc+ac+ab=3abc \tag{3}
$$
Notice that we have
$$
a^{2}+b^{2}+c^{2} \geq bc+ac+ab=3abc
$$
So,
$$
a^{2}+b^{2}+c^{2} \geq 3abc \tag{4}
$$
Let us multiply both side of $(4)$ by $\displaystyle\frac{2}{3}$, yield
$$
\frac{2}{3}(a^{2}+b^{2}+c^{2}) \geq 2abc
$$
Here where I stopped. Would someone help me out ! Thank you so much
|
We need to prove that
$$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2(a^2+b^2+c^2)}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+4abc$$ or
$$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2abc(a^2+b^2+c^2)}{ab+ac+bc}+4abc$$ or
$$\sum\limits_{cyc}ab\sum\limits_{cyc}(a^2b+a^2c)\geq2abc(a^2+b^2+c^2)+4abc(ab+ac+bc)$$ or
$$\sum_{cyc}(a^3b^2+a^3c^2+2a^3bc+2a^2b^2c)\geq\sum\limits_{cyc}
(2a^3bc+4a^2b^2c)$$
$$\sum\limits_{cyc}(a^3b^2+a^3c^2-2a^2b^2c)\geq0$$ or
$$\sum\limits_{cyc}c^2(a+b)(a-b)^2\geq0$$
Done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding solutions to the diophantine equation $7^a=3^b+100$
Find the positive integer solutions of the diophantine equation $$7^a-3^b=100.$$
So far, I only found this group $7^3-3^5=100$.
|
note: this answer has been found to be -at least- incomplete, see comments of Gottfried Helms and piquito
$100$ has no primitive root therefore $7^m$ and $3^n$ are congruent with $1$ modulo $100$ for some integers $m,n$ smaller than $100$ and these numbers should divide $100$.
We have $7^4\equiv 3^{20}\equiv 1\pmod{100}\Rightarrow 7^{4m}\equiv 3^{20n}\equiv 1\pmod {100}$ which could be a starting point to calculate possible solutions. However we notice in this search that (in the ring $\Bbb Z/100\Bbb Z$ for short) the solution $7^3=3^5+100$ so we have
$$\begin{cases}7^a=3^b+100\\7^3=3^5+100\end{cases}\Rightarrow 7^a-7^3=3^b-3^5$$
For the values $a=1,2,3$ and $b=1,2,3,4,5$ it is not verified except for $(a,b)=(3,5)$ which does not provide another solution. Hence $a\gt 3$ and $b\gt 5$. It follows
$$7^3(7^{a-3}-1)=3^5(3^{b-5}-1)\Rightarrow 7^{a-3}=244=2^2\cdot61\text{ and } 3^{b-5}=344=2^3\cdot43$$ which is absurde. Consequently $(a,b)=(3,5)$ is the only solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inverse of a $2 \times 2$ block matrix Let
$$S := \pmatrix{A&B\\C&D}$$
If $A^{-1}$ or $D^{-1}$ exist, we know that matrix $S$ can be inverted.
$$S^{-1} = \pmatrix{A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}}$$
But, what if $A^{-1}$ and $D^{-1}$ do not exist? Can we invert matrix $S$?
For example,
$$S = \pmatrix{0&1\\1&0}$$
or
$$S = \pmatrix{2&3&1&1\\4&6&1&2\\1&1&3&1\\4&1&12&4}$$
both their $A^{-1}$ and $D^{-1}$ don't exist, but $S^{-1}$ exists.
|
One good reference is the Handbook Matrix Mathematics Theory, Facts and Formulas by Dennis S. Bernstein. Let $\mathbb{F}$ equal to $\mathbb{R}$ or $\mathbb{C}$.
Proposition 3.9.7. Let $A \in \mathbb{F}^{n \times n}, B \in \mathbb{F}^{n \times m}, C \in \mathbb{F}^{m \times n}$, and $D \in \mathbb{P}^{m \times m}$. If $A$ and $D-C A^{-1} B$ are nonsingular, then
$$
\left[\begin{array}{ll}
A & B \\
C & D
\end{array}\right]^{-1}=\left[\begin{array}{rr}
A^{-1}+A^{-1} B\left(D-C A^{-1} B\right)^{-1} C A^{-1} & -A^{-1} B\left(D-C A^{-1} B\right)^{-1} \\
-\left(D-C A^{-1} B\right)^{-1} C A^{-1} & \left(D-C A^{-1} B\right)^{-1}
\end{array}\right]
$$
There are three more cases to consider in addition to the above proposition.
*
*If $D$ and $A-B D^{-1} C$ are nonsingular;
*If $B$ and $C-D B^{-1} A$ are nonsingular;
*If $C$ and $B-A C^{-1} D$ are nonsingular;
For the sake of illustration we will prove only the case where $C$ and $B-A C^{-1} D$ are nonsingular;
To prove these cases it is enough to use the matrix $J=\left[\begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]$ to reduce them to the same case as the proposition above. Observe that
$
\left[\begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]^{-1}
=
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\mbox{ once }
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
=
\left[ \begin{array}{cc} I & 0 \\ 0 & I \end{array} \right].
$
Another important thing to note is that
$$
\;(\ast)\;\;\qquad \left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right]
=
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\cdot
\left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right]
=
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\cdot
\left[ \begin{array}{cc} X & Y \\ U & V \end{array} \right].
$$
and
$$
(\ast\ast)\qquad \left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right]
=
\left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right]
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
=
\left[ \begin{array}{cc} V & U \\ Y & X \end{array} \right]
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
$$
Suppose there exists the inverse $C^{-1}$ of $C$ and there exists the inverse $(B -AC^{-1}D)^{-1}$ of $(B -AC^{-1}D)$. What can we say about the inverse of $\left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]$? By proposition above we have
$$
\left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]^{-1}
=\left[\begin{array}{rr}
C^{-1}+C^{-1} D\left(B-A C^{-1} D\right)^{-1} A C^{-1} & -C^{-1} D\left(B-A C^{-1} D\right)^{-1} \\
-\left(B-A C^{-1} D\right)^{-1} A C^{-1} & \left(B-A C^{-1} D\right)^{-1}
\end{array}\right]
$$
And more, by $(\ast)$ and $(\ast\ast)$ we have
\begin{align}
\left[ \begin{array}{cc} A & B \\ C & D \end{array} \right]^{-1}
&=
\left(
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\cdot
\left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]
\right)^{-1}
\\
&=
\left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]^{-1}
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]^{-1}
\\
&=
\left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]^{-1}
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\\
&=
\left[\begin{array}{rr}
C^{-1}+C^{-1} D\left(B-A C^{-1} D\right)^{-1} A C^{-1} & -C^{-1} D\left(B-A C^{-1} D\right)^{-1} \\
-\left(B-A C^{-1} D\right)^{-1} A C^{-1} & \left(B-A C^{-1} D\right)^{-1}
\end{array}\right]
\cdot
\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]
\\
&=
\left[\begin{array}{rr}
-C^{-1} D\left(B-A C^{-1} D\right)^{-1} & C^{-1}+C^{-1} D\left(B-A C^{-1} D\right)^{-1} A C^{-1} \\
\left(B-A C^{-1} D\right)^{-1} & -\left(B-A C^{-1} D\right)^{-1} A C^{-1}
\end{array}\right]
\end{align}
|
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|
If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{a+b+c+15}}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$
It seems nice enough.
I proved this inequality by Holder, but it quits very ugly.
Maybe there is something nice? Thank you!
|
Put the following substitution :
$\frac{a}{b+c}=x$$\quad$$\frac{b}{a+c}=y$$\quad$$\frac{c}{b+a}=z$
Remark that :
$$a+b+c=((\frac{1}{x}+1)(\frac{1}{y}+1)(\frac{1}{z}+1))^{\frac{1}{3}}$$
And :
$$abc=1 \iff -2=-\frac{1}{xyz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
So we get the inequality related to this Post that you have proved by your own with brio .
|
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|
Find the set of $n\in\Bbb Z^+$ with $M=\{n,n+1,n+2,n+3,n+4,n+5\}$ partitionable into two sets
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.
If $n=1$ them $M=\{1,2,3,4,5,6\}$ and there is no such partition, so $n \ge 2$.
If the prime $p|n$ then either $p|2$ or $p|3$ or $p|5$ which means either $p=2$ or $p=3$ or $p=5$.
Suppose $n=2k$. Then $M=\{2k, 2k + 1,2k + 2,2k + 3,2k + 4,2k + 5\}$. I have no idea how to proceed.
|
Partition $M$ into $A$ and $B.$ For brevity let $A^*=\prod_{x\in A}x$ and $B^*=\prod_{y\in B}y.$ Assume that $A^*=B^*$.
We may assume $n+5\in B.$
Observe that $5|n$ otherwise exactly one member of $M$ is divisible by $5,$ implying that $5$ divides exactly one of $A^*, B^*.$ So let $M=\{5m,5m+1,5m+2,5m+3,5m+4,5m+5\}.$
Now let $p$ be any prime divisor of $m+1.$
(1).If $p=5$ then $5^2 | 5(m+1),$ so $5^2 | B^*, $ but then the only member of $A$ that is divisible by $5$ is $5m,$ which is not divisible by $5^2.$ Therefore $p\ne 5.$
(2). So $p| (m+1)$, and $p|B^*$ and $p\not |\; 5m,$ so $p$ divides at least one member of $\{5m+j:1\leq j\leq 4\} $ (Otherwise $p\not |\; A^*.$) Now $p|(5m +5)$ and $p|(5m+j)$ for some $j\in \{1,2,3,4\},$ implying $p|(5m+5)-(5m+j)=5-j$ for some $j\in \{4,3,2,1\}.$ So the only possible prime divisors of $m+1$ are $2$ and $3.$
(3). If $3|(m+1)$ then the only other member of $M$ that is divisible by $3$ is $5m+2,$ so $5m+2\in A$ and $5m+5\in B.$ And $3^2$ cannot divide either $5m+2$ or $5m+5,$ otherwise exactly one of $A^*,B^*$ is divisible by $3^2.$ Therefore if $3|(m+1)$ then $3^2\not |\; (m+1).$
(4). If $2|(m+1)$ then $5m+1,5m+3, 5m+5$ are even while $5m, 5m+2,5m+5$ are odd. Now if $2^4|(m+1)$ then $2^4|B^*, $ but also $2^4\not |\; (5m+3)(5m+1)$, implying $2^4\not |\; A^*.$ Therefore if $2|(m+1)$ then $2^4\not |\; (m+1).$
(5).Therefore $m+1\in \{2^a3^b: a\leq 3\land b\leq 1\}$ \ $\{1\}=\{2,4,8, 3, 6, 9, 12\}.$ So $n+5=5m+5\in\{10,20,40,15,30,45,60\}.$ But for each of these potential values for $n+5,$ there is a prime $q\geq 7$ that belongs to $M,$ and does not divide any other member of $M, $ so $q$ divides exactly one of $A^*,B^*.$
So $A^*\ne B^*$ and there are no solutions.
|
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|
Proving $a^n - b^n \le na^{n - 1}(a-b)$ with induction The question is to prove $a^n - b^n \le na^{n - 1}(a-b)$ with induction where $0 < b < a$ and $ \forall n \ge 1$, $n \in \mathbb{Z}$
I finished the different parts of the proof, including the predicate, base case, and inductive step, but I want to make sure that I'm not missing anything - especially in the inductive step.
I'm also a little confused about moving from Step 1 to the rest of the steps because it seems that the inductive hypothesis applies to $b(a^n-b^n$, but I'm not sure how the other part of the expression, namely $a^n(a-b)$ is used.
Let $P(n) = a^n - b^n \le na^{n - 1}(a-b)$
Base case: $P(0) = a - b \le a - b$
Inductive step - need to prove $P(n + 1)$:
*
*$a^{n + 1} - b^{n + 1} = a^{n + 1} - a^nb + a^nb - b^{n + 1} = a^n(a-b) + b(a^n - b^n)$
*Because $0 < b < a$, $b(a^n-b^n) \le a(a^n - b^n)$
*Using the induction hypothesis, $a(a^n - b^n) \le a \cdot n \cdot a^{n - 1}(a-b)$
*$\rightarrow a(a^n - b^n) \le na^{n}(a-b) \le (n + 1)a^{n}(a-b)$
*$\rightarrow a(a^n - b^n) \le (n + 1)a^{n}(a-b)$
|
\begin{align}
&a^{n+1} - b^{n+1} \\
=\ &a^{n+1} - a^nb + a^nb - b^{n+1} \tag{your step $1$} \\
=\ &a^n(a - b) + \color{red}{b(a^n - b^n)} \\
\leq\ &a^n(a - b) + \color{red}{a(a^n - b^n)} \\
\leq\ &a^n (a - b) + \color{red}{a \cdot na^{n-1}(a-b)} \\
\leq \ &a^n(a - b) + \color{red}{n\cdot a^n(a-b)} \tag{your step $4$}\\
=\ &(1 + n)a^n(a - b)
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all continuous functions $f(x+1)=f(x)+1$ and $f\left(\frac1x\right)=\frac1{x^2}f(x)$
Find all continuous functions $f:\mathbb R \rightarrow \mathbb R$ such that
1) $\forall x \in \mathbb R$
$$f(-x)=-f(x)$$
2) $\forall x \in \mathbb R$
$$f(x+1)=f(x)+1$$
3$\forall x \in \mathbb R/ \{0\}$
$$f\left(\frac1x\right)=\frac1{x^2}f(x)$$
My work so far:
1) $f(0)=0$
2) $\forall x \in \mathbb Z$ $$f(x+m)=f(x)+m$$
3) I see that the answer $f(x)=x$, but can not rove.
|
Get $y = f(x)$ , then i have
$$f(x+1)=y+1,f(\frac{1}{x+1})=\frac{y+1}{(x+1)^2},f(\frac{-1}{x+1})=\frac{-(y+1)}{(x+1)^2}$$
$$f(\frac{x}{x+1})=f(-\frac{1}{x+1}+1)=\frac{-(y+1)}{(x+1)^2}+1=\frac{x^{2}+2x-y}{(x+1)^2}$$
$$f(\frac{x}{x+1})=f(\frac{1}{\frac{x+1}{x}})=\frac{f(\frac{x+1}{x})}{(\frac{x+1}{x})^2}=\frac{\frac{x^{2}+y}{x^2}}{(\frac{x+1}{x})^2}=\frac{x^2+y}{(x+1)^2}$$
So i have
$$\frac{x^{2}+2x-y}{(x+1)^2}=\frac{x^2+y}{(x+1)^2}$$
$$x=y$$
And easy to check $f(0)=0,f(-1)=-1$ so
$$f(x) \equiv x$$
|
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|
Show that as $n\rightarrow \infty,$ $\frac{n^4}{3^n}\rightarrow 0$ linearly with rate $\frac{1}{3}$ Show that as $n\rightarrow \infty,$ $\frac{n^4}{3^n}\rightarrow 0$ linearly with rate $\frac{1}{3}$
linearly means take $a_n=\frac{n^4}{3^n}$
$\lim_{n\rightarrow \infty} (a_n)^\frac{1}{n}$
$\lim_{n\rightarrow\infty } (\frac{n^4}{3^n})^\frac{1}{n}=\frac{1}{3}$
how to solve this problem
|
Since
$$\lim_{n\to\infty}\left(\frac{n^4}{3^n}\right)^\frac{1}{n} = \lim_{n\to\infty}\frac{\left(n^4\right)^\frac{1}{n}}{\left(3^n\right)^\frac{1}{n}}=\lim_{n\to\infty}\frac{n^\frac{4}{n}}{3},$$
it will suffice to show that
$$\lim_{n\to\infty}n^\frac{4}{n} = \left(\lim_{n\to\infty} n^\frac{1}{n}\right)^4 = 1.$$
Now let $n^{1/n} = (1+a)$. Then by the binomial theorem,
\begin{eqnarray}
n = (1+a)^n &=& \sum_{k=0}^n \binom{n}{k}a^k\\
&=& 1+ na + \frac{n(n-1)}{2}a^2 + \cdots + a^n\\
&>& \frac{n(n-1)}{2}a^2.
\end{eqnarray}
Thus $a^2 < \dfrac{2n}{n(n-1)} = \dfrac{2}{n-1}$, which implies $a < \sqrt{\frac{2}{n-1}}$. Therefore,
$$1 \le \lim_{n\to\infty} n^{1/n} < \lim_{n\to\infty}\left(1 + \sqrt{\frac{2}{n-1}}\right) = 1.$$
|
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|
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$
Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$
Now, how can I solve it after that, Help Required, Thanks
|
Hint:
$$\binom{n}{0}= \mbox{ coefficient of } x^n \mbox{ in } (1+x)^n \\
\binom{n-1}{1}= \mbox{ coefficient of } x^n \mbox{ in } x^2(1+x)^{n-1} \\
\binom{n-2}{2}= \mbox{ coefficient of } x^n \mbox{ in } x^4(1+x)^{n-2} \\
...$$
Hint 2:
$$(1+x)^n-x^2(1+x)^{n-1}+x^4(1+x)^{n-2}-..=(1+x)^n \left(1-(\frac{x^2}{1+x})+ (\frac{x^2}{1+x})^2-(\frac{x^2}{1+x})^3 ...\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving an ordinary differential equation nonlinear I have the following
$$ y' = \frac{ xy }{x^2 + y^2} $$
My approach would be to rewrite this in terms of $x$
$$ x' = \frac{x}{y} + \frac{y}{x} $$
And then let $u = \frac{y}{x} \implies u' = \frac{x + y x'}{x^2} = \frac{1}{x} + \frac{u}{x} x' = \frac{1}{x} + \frac{u}{x} \left( \frac{1}{u} + u\right)$. Therfore,
$$ u' = \frac{1}{x} + \frac{1}{x} + \frac{u^2}{x} = \frac{2 + u^2}{x}$$
Thus,
$$ \int \frac{ d u}{2 + u^2} = \int \frac{ dx}{x} \implies \frac{1}{\sqrt{2}} \arctan (u/ \sqrt{2}) = \ln |x| + C$$
So general solution ( with $u = y/x$) is
$$ \boxed{ \arctan \left( \frac{y}{x \sqrt{2}} \right) = \sqrt{2} \ln |x| + \sqrt{2} +C }$$
is this sufficient? is there an easier way to solve this ODE?
|
This belongs to the class of homogeneous differential equations of order 0, that is, it satisfies $f(\lambda x, \lambda y) = f(x,y)$. One way to solve such a equation is to write the differential equation in the form: $y' = f(\frac{y}{x})$ and make the substitution $v=\frac{y}{x}$.
|
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|
Find $a,b,c$ where $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ Find integers $a,b,c$ such that $1<a<b<c$ and $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ .I tried it solve using elementary number theory but I can't proceed . Somebody help me.
|
Denote $\bullet = \dfrac{abc-1}{(a-1)(b-1)(c-1)}$
Note that $$
\bullet = \dfrac{a}{a-1}\cdot \dfrac{bc-1}{(b-1)(c-1)} + \dfrac{1}{(b-1)(c-1)} \\
=\dfrac{a}{a-1} \cdot \left(1 + \dfrac{1}{b-1} + \dfrac{1}{c-1}\right) + \dfrac{1}{(b-1)(c-1)}.\tag{1}
$$
$(1)\Rightarrow$ (lower bound)
$$
\dfrac{a}{a-1}<\bullet.\tag{2}
$$
If $a=2$, then (since $c-1\ge b$) from $(1), (2)$ we get
$$2<\bullet \leq 2\cdot \left(1+\dfrac{1}{b-1}+\dfrac{1}{b}\right) + \dfrac{1}{(b-1)b};\tag{3}$$
and if (in addition) $b\ge 5$, then $(3)\Rightarrow$
$$2<\bullet \leq 2\cdot \left(1+\dfrac{1}{4}+\dfrac{1}{5}\right) + \dfrac{1}{20}=\dfrac{59}{20}<3,\tag{4}$$
so expression $\bullet$ cannot be integer in this case.
Considering cases $(a,b,c)=(2,3,c)$ and $(a,b,c)=(2,4,c)$, we find first solution$$(a,b,c)=(2,4,8).\tag{*}$$
If $a=3$, then from $(1), (2)$ we get
$$\dfrac{3}{2}<\bullet \leq \dfrac{3}{2}\cdot \left(1+\dfrac{1}{b-1}+\dfrac{1}{b}\right) + \dfrac{1}{(b-1)b};\tag{5}$$
and if (in addition) $b\ge 7$, then $(5)\Rightarrow$
$$\dfrac{3}{2}<\bullet \leq \dfrac{3}{2}\cdot \left(1+\dfrac{1}{6}+\dfrac{1}{7}\right) + \dfrac{1}{42}=\dfrac{167}{84}<2,\tag{6}$$
so expression $\bullet$ cannot be integer in this case.
Considering cases $(a,b,c)=(3,4,c)$, $(a,b,c)=(3,5,c)$ and $(a,b,c)=(3,6,c)$, we find second solution $$(a,b,c)=(3,5,15).\tag{**}$$
If $a\ge 4$, then $b\ge 5$, $c\ge 6$, and $(1), (2) \Rightarrow$
$$1<\bullet \le \dfrac{4}{3} \cdot \left(1 + \dfrac{1}{4} + \dfrac{1}{5}\right) + \dfrac{1}{20} = \dfrac{119}{60}<2.\tag{7}$$
Therefore there are no solutions for $a\ge 4$ (since value of expression $\bullet$ is between $1$ and $2$).
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Find the sum $\sum_{n=1}^{\infty} \frac{4n}{n^4+2n^2+9}$ Find the sum
$$\sum_{n=1}^{\infty} \dfrac{4n}{n^4+2n^2+9}.$$
By calculator, we can predict that its sum is equal to $\dfrac{5}{6}$ so I think we should use inequalities to prove it. And I found that
$\dfrac{5}{6(n^4+n^2)} < \dfrac{4n}{n^4+2n^2+9}< \dfrac{5}{6(n^2+n)}$ for all $n\ge n_0$, $n_0$ is large enough.
And $\sum_{n=1}^{\infty} \dfrac{5}{6(n^4+n^2)}= \sum_{n=1}^{\infty}\dfrac{5}{6(n^2+n)}=\dfrac{5}{6}$.
But it is not enough to confirm that the given series converges to $\dfrac{5}{6}$. Can someone help me, please? Thanks in advanced.
|
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\sum_{n = 1}^{\infty}{4n \over n^{4} + 2n^{2} + 9}}$.
\begin{align}
&\sum_{n = 1}^{\infty}{4n \over n^{4} + 2n^{2} + 9} =
\sum_{n = 0}^{\infty}{4n \over \pars{n^{2} - 2n + 3}\pars{n^{2} + 2n + 3}}
=
\sum_{n = 0}^{\infty}\pars{{1 \over n^{2} - 2n + 3} - {1 \over n^{2} + 2n + 3}}
\\[5mm] = &\
\lim_{N \to \infty}\pars{\sum_{n = 0}^{N}{1 \over \pars{n - 1}^{2} + 2} -
\sum_{n = 0}^{N}{1 \over \pars{n + 1}^{2} + 2}} =
\lim_{N \to \infty}\pars{\sum_{n = -1}^{N - 1}{1 \over n^{2} + 2} -
\sum_{n = 1}^{N + 1}{1 \over n^{2} + 2}}
\\[5mm] = &\
\lim_{N \to \infty}\pars{{1 \over 3} + {1 \over 2} +
\sum_{n = 1}^{N - 1}{1 \over n^{2} + 2} -
\sum_{n = 1}^{N - 1}{1 \over n^{2} + 2} - {1 \over N^{2} + 2} -
{1 \over \pars{N + 1}^{2} + 2}}
\\[5mm] = &\
\lim_{N \to \infty}\pars{{5 \over 6} - {1 \over N^{2} + 2} -
{1 \over \pars{N + 1}^{2} + 2}} = \bbx{\ds{5 \over 6}}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find Min of function $x+\frac{2}{x}$ for x>0 with following solution? I want to find Minimum of function $x+\frac{2}{x}$ for $x>0$ with following solution. My problem is how can we say we found the answer with following solution.
Solution:
Let Minimum be something like c. We have:
$x+\frac{2}{x} \ge c \to \frac{x^2+2}{x} \ge c \to x^2 + 2 \ge cx \to x^2 -cx +2 \ge 0 $
So $\Delta \le 0$ because the quadratic must not have two solutions so it will be only positive or zero. So:
$\Delta = c^2 - 8 \le 0 \to c^2 \le 8 \to -2\sqrt{2} \le c \le +2\sqrt{2}$
My question is how can we say with following solution that the minimum of $x+\frac{x}{2}$ must be $+2\sqrt{2}$? We find the maximum of $c$, How can we say that maximum is minimum of function? Is it a valid solution?
Thanks and sorry for my English.
|
Using $AM-GM$ inequality, we have(since $x>0$)
$$ x + \frac{2}{x} \geq 2 \sqrt{ x \cdot \frac{2}{x} } = 2 \sqrt{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1964011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the minimum value of $\sqrt{x+y}+\sqrt{(1-x)+y}+\sqrt{x+(1-y)}+\sqrt{(1-x)+(1-y)}$ Given that $x,y\in[0,1]$, find the minimum of the following
$$f(x,y)=\sqrt{x+y}+\sqrt{(1-x)+y}+\sqrt{x+(1-y)}+\sqrt{(1-x)+(1-y)}$$
I have found $$f(0,0)=f(0,1)=f(1,0)=f(1,1)=2+\sqrt{2}, f(\dfrac{1}{2},\dfrac{1}{2})=2+\sqrt{6}$$
so is $2+\sqrt{2}$ the minimum? How to prove this?
|
$\frac{\partial^2f}{\partial x^2}=\frac{\partial^2f}{\partial y^2}=-\frac{1}{4}\left((x+y)^{-\frac{3}{2}}+(1-x+y)^{-\frac{3}{2}}+(1-y+x)^{-\frac{3}{2}}+(2-x-y)^{-\frac{3}{2}}\right)<0$,
which says that $f$ is a concave function of $x$ and of $y$.
Id est, $\min\limits_{\{x,y\}\subset[0,1]}f=\min\limits_{\{x,y\}\subset\{0,1\}}f=f(0,0)=2+\sqrt2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1964851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$q^{3}\leq aq^{2}+bq+c$ finite amount of q problem Im a bit confused why this is true it's simple but i seem to have some problems with it $q^{3}\leq aq^{2}+bq+c$ is only true for a finite amount of $\forall q \epsilon \mathbb{N}$
|
First, you can assume that $a, b, c \ge 0$, for if they're negative, then $q^3 \le aq^2 + bq + c \le |a|q^2 + |b|q | c$, and if there are at most finitely many solutions the the latter, then there are to the former as well. For the same reason, we can assume $c \ge 1$, so that $a+b+c \ge 1$.
Claim: For $q \ge a+b+c$, the left hand side is larger than the right hand side. That shows that there are at most $a+b+c$ solutions.
Why is the claim true? Because
\begin{align}
q^3 &\ge (a+b+c)^3 \\
& \ge a(a+b+c)^2 + b(a + b + c)^2 + c(a+b+c)^2 \\
& \ge a(a+b+c)^2 + b(a + b + c)^1 + c(a+b+c)^0 \text{, because $a+b+c > 1$} \\
& \ge aq^2 + bq + c \text{, by substitution, $q=a+b+c$} \\
\end{align}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
$x^2$ is congruent to $-1 \bmod p$ In general, how do you solve $x^2$ is congruent to $-1 \bmod p$, where $p$ is an odd prime and $x$ is an integer.
Specifically, I need to solve $x^2\equiv-1 \pmod{29}$
|
Since you can figure out
$$ 29 = 25 + 4 = 5^2 + 2^2 $$
in your head, you see
$$ 5^2 + 2^2 \equiv 0 \pmod {29}, $$
$$ 5^2 \equiv - 2^2 \pmod {29}. $$
Bothe 2 and 5 are relatively prime to $29,$ either one has a multiplicative inverse mod 29.
$$ \frac{5^2}{2^2} \equiv -1 \pmod {29}, $$
$$ \left( \frac{5}{2} \right)^2 \equiv -1 \pmod {29}. $$
So, you need to figure out the multiplicative inverse of $2 \pmod {29}$ and multiply that by $5.$ whatever that becomes $\pmod {29}$ is what you need.
Not a coincidence: the other square root is
$$ - \frac{5}{2} \equiv \frac{2}{5} \pmod{29}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1967582",
"timestamp": "2023-03-29T00:00:00",
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|
Solving indefinite integral $\int \frac{dx}{(x^4-1)^3}$ I'm trying to solve next integral, but I can't start. WolframAlpha gives me really terrible answer and can't give any step-by-step instructions, so I simply does not know how to start.
$$\int \frac{dx}{(x^4-1)^3}$$
Please give any hint or start point of solving this integral? Maybe there any way to simplify it?
|
Let
$$I:=\int\frac{dx}{x^4-1}, J:=\int\frac{dx}{(x^4-1)^2}, K:=\int\frac{dx}{(x^4-1)^3}.$$
By parts,
$$I=\frac x{x^4-1}+4\int\frac{x^4\,dx}{(x^4-1)^2}=\frac x{x^4-1}+4(I+J)$$
and
$$J=\frac x{(x^4-1)^2}+8\int\frac{x^4\,dx}{(x^4-1)^3}=\frac x{(x^4-1)^2}+8(J+K).$$
On another hand,
$$I=\int\frac{dx}{x^4-1}=\frac12\int\frac{dx}{x^2-1}-\frac12\int\frac{dx}{x^2+1}=-\frac12(\text{argtanh }x+\arctan x).$$
Then
$$J=-\frac14\left(3I+\frac x{x^4-1}\right)$$
and
$$K=-\frac18\left(7J+\frac x{(x^4-1)^2}\right).$$
The generalization is immediate, and
$$L:=\int\frac{dx}{(x^4-1)^4}=-\frac1{12}\left(11K+\frac x{(x^4-1)^3}\right),$$
$$M:=\int\frac{dx}{(x^4-1)^5}=-\frac1{16}\left(15L+\frac x{(x^4-1)^4}\right)$$
$$\vdots$$
We get linear combinations of $\dfrac x{(x^4-1)^k}$, and $I$, with simple coefficients involving the product of fractions $\dfrac{4k-1}{4k}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$ I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-1})$$
The solution to the problem, contained in the answer book, is as follows:
\begin{align*}
x^n-y^n &= (x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&= x(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&\qquad -[y(x^{n-1}+x^{n-2}y+{...}+xy^{n-2}+y^{n-1})]&\Rightarrow \mathbf{Equation1}\\
&=x^n+x^{n-1}y+\dotsb+x^2y^{n-2}+xy^{n-1}\\
&\qquad -[x^{n-1}y+x^{n-2}y^2+xy^{n-1}+y^n]&\Rightarrow \mathbf{Equation2}\\
&=x^n-y^n
\end{align*}
While I believe that the distributive law was used to arrive at Equation 1, I do not understand how Equation 2 was arrived at.
I have tried to solve this independently to no avail. I cannot seem to understand how $x^n$ and $y^n$ came about in Equation 2 for example.
To summarise the question, what principle was Equation 2 based upon? And how was this applied in the above problem.
|
The step from equation 1 to equation 2 uses both the distributive law as well as the commutative law.
In order to better see what's going on, we look at small special cases $n=2,3$. We also use somewhat more detailed transformations.
n=2:
\begin{align*}
(x-y)\left(x^{1}+y^1\right)&=x\left(x^{1}+y^1\right)-y\left(x^1+y^1\right)\\
&=\left(x\cdot x^1+x\cdot y^1\right)-\left(y\cdot x^1+y\cdot y^1\right)\\
&=\left(x^2+xy\right)-\left(yx+y^2\right)\\
&=x^2+xy-yx-y^2\\
&=x^2-y^2
\end{align*}
n=3:
\begin{align*}
&(x-y)\left(x^{2}+x^1y^1+y^2\right)\\
&\qquad=x\left(x^{2}+x^1y^1+y^2\right)-y\left(x^{2}+x^1y^1+y^2\right)\\
&\qquad=\left(x\cdot x^2+x\cdot x^1y^1+x\cdot y^2\right)-\left(y\cdot x^2+y\cdot x^1y^1+y\cdot y^2\right)\\
&\qquad=\left(x^3+x^2y+xy^2\right)-\left(yx^2+y^2x+y^3\right)\\
&\qquad=x^3+x^2y+xy^2-yx^2-y^2x-y^3\\
&\qquad=x^3-y^3
\end{align*}
general n:
\begin{align*}
&(x-y)\left(x^{n-1}+x^{n-2}y^1+\cdots+xy^{n-1}+y^{n-1}\right)\\
&\qquad=x\left(x^{n-1}+x^{n-2}y^1+\cdots\cdots\cdot\cdot+xy^{n-1}+y^{n-1}\right)\\
&\qquad\qquad\qquad\ \ -y\left(x^{n-1}+x^{n-2}y^1+\cdots\cdots+xy^{n-1}+y^{n-1}\right)\\
&\qquad=\left(x\cdot x^{n-1}+\color{blue}{x\cdot x^{n-2}y^1}+\cdots\cdots+x\cdot xy^{n-1}+\color{blue}{x\cdot y^{n-1}}\right)\\
&\qquad\qquad\qquad\quad-\left(\color{red}{y\cdot x^{n-1}}+y\cdot x^{n-2}y^1+\cdots\cdots\cdot+\color{red}{y\cdot xy^{n-1}}+y\cdot y^{n-1}\right)\\
&\qquad=\left(x^{n}+ \color{blue}{x^{n-1}y}+\cdots\cdots\cdots\cdots+x^2y^{n-1}+\color{blue}{xy^{n-1}}\right)\\
&\qquad\qquad\ -\left( \color{red}{y x^{n-1}}+x^{n-2}y^2+\cdots\cdots\cdots\cdots\cdot+\color{red}{xy^{n-1}}+y^{n}\right)\\
&\qquad=x^{n}\;+ \color{blue}{x^{n-1}y}+\cdots\cdots\cdots\cdots+x^2y^{n-1}+\color{blue}{xy^{n-1}}\\
&\qquad\qquad\ \ -\color{red}{x^{n-1}y}-x^{n-2}y^2-\cdots\cdots\cdots\cdots-\color{red}{xy^{n-1}}-y^{n}\\
&\qquad=x^n-y^n
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form of a sum $ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$ Consider a sum:
$$ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$$
with $x$ and $y$ being (non-zero) constants. Is it possible to obtain a nice closed form of this expression?
|
As already wrote, there is no nice closed form. A way to see an approximation is the following. Using Abel's summation we have $$S_{N}\left(z\right)=\sum_{n\leq N}\frac{1}{1+n^{2}z^{2}}=\frac{N}{1+N^{2}z^{2}}+2z^{2}\int_{1}^{N}\frac{\left\lfloor t\right\rfloor t}{\left(1+t^{2}z^{2}\right)^{2}}dt
$$ where $\left\lfloor t\right\rfloor
$ is the floor function and since $t-1<\left\lfloor t\right\rfloor \leq t
$ we get $$S_{N}\left(z\right)<\frac{N}{1+N^{2}z^{2}}+2z^{2}\left(\int_{1}^{N}\frac{1}{1+t^{2}z^{2}}-\frac{1}{\left(1+z^{2}t^{2}\right)^{2}}dt\right)
$$ $$=\frac{\arctan\left(Nz\right)}{z}-\frac{\arctan\left(z\right)}{z}+\frac{1}{z^{2}+1}
$$ and in the same way we get $$S_{N}\left(z\right)>\frac{N}{1+N^{2}z^{2}}+2z^{2}\int_{1}^{N}\frac{t^{2}-t}{\left(1+t^{2}z^{2}\right)^{2}}dt
$$ $$=\frac{\arctan\left(Nz\right)}{z}-\frac{\arctan\left(z\right)}{z}+\frac{1}{1+N^{2}z^{2}}
$$ and $$\sum_{n=0}^{N-1}\frac{1}{y^{2}+n^{2}x^{2}}=\frac{1}{y^{2}}+\frac{1}{y^{2}}\sum_{n=1}^{N-1}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}
$$ $$=\frac{1}{y^{2}}+\frac{S_{N-1}\left(\frac{x}{y}\right)}{y^{2}}.$$
For the complete series another interesting method is the following $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\}
$$ and observing that $$\frac{1}{y^{2}}\sum_{n=1}^{\infty}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}=\frac{1}{2y^{2}}+\frac{1}{2y^{2}}\sum_{n\in\mathbb{Z}}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}
$$ and since we have poles at $\pm\frac{iy}{x}
$ we get $$\sum_{n\in\mathbb{Z}}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}=\frac{y\pi\coth\left(\pi y/x\right)}{x}
$$ and so $$\frac{1}{y^{2}}\sum_{n=1}^{\infty}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}=\color{red}{\frac{1}{2y^{2}}+\frac{\pi\coth\left(\pi y/x\right)}{2yx}}.$$
|
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"url": "https://math.stackexchange.com/questions/1969295",
"timestamp": "2023-03-29T00:00:00",
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|
How to show $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$ I am new to the subject of rings of formal power series. The problem is to show that $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$, where $\mathbb{Z_6}[[x]]$ is the ring of formal power series with coefficients in $\mathbb{Z}_6$, also known as $ \mathbb{Z }/6\mathbb{Z}$.
My work is as follows:
$ \begin{align} (1+3x)^{-1} &= \frac{1}{1+3x}
\\&= \frac{1}{1-(-3x)}
\\ &= 1 +(-3x) + (-3x)^2 + (-3x)^3 ...
\\ &= 1 - 3x + 3^2x^2 -3^3x^3 ...
\\ &=_6 1 + 3x +3x^2 + 3x^3 + ... \end{align}$
This is supposed to show that the inverse of $1 + 3x$ is $ 1 + 3x + 3x^2 + ...$ in the ring $\mod 6$. I am confused what this actually means. When I plug in $x =3$ into $1+3x$, this produces $4$ in mod 6 and there is no modular inverse of $4$. So why is $1+3x$ called the inverse?
I notice that we can check the calculation manually.
$\begin{align}( 1 + 3x) ( 1 + 3x + 3x^2 + 3x^3 + ...) &= 1 + 3x + 3x + 3^2x^2 + 3x^2 +...
\\ &=_6 1 + 3x + 3x + 3x^2 + 3x^2 +...
\\ &=_6 1 + 6x + 6x^2 + ...
\\ &=_6 1 + 0 + 0 + ...
\\ &=_6 1
\end{align}$
What bothers me is that some values you cannot plug into x, the modulo inverse is not defined.
|
You need to show that
$$(1 + 3x + 3x^2 + 3x^3 + \cdots)(1 + 3x) = 1$$
using the addition and multiplication rules in $\Bbb Z_6[[x]]$. The element $1 + 3x + 3x^2 + 3x^3 + \cdots$ is $\sum\limits_{j = 0}^\infty 3x^j$, so over $\Bbb Z_6[[x]]$ we write
$$\left(\sum_{j = 0}^\infty 3^jx^j\right)(1 + 3x) = \sum_{j = 0}^\infty (3x^j + 3x^{j+1}) = 1 + \sum_{j = 1}^\infty 3x^j + \sum_{j = 0}^\infty 3x^{j+1} = 1 + 2\sum_{j = 1}^\infty 3x^j$$
Now since $3^j \equiv 3\pmod 6$ for all $j \ge 1$,
$$2\sum_{j = 1}^\infty 3^jx^j = 2\sum_{j = 1}^\infty 3x^j = \sum_{j = 1}^\infty 6x^j = 0$$
and so $$\left(\sum_{j = 0}^\infty 3^jx^j\right)(1 + 3x) = 1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How should I calculate the determinant? $\left|\begin{array}{cccc}1&a&b&c+d\\1&b&c&a+d\\1&c&d&a+b\\1&d&a&b+c\end{array}\right|=
\left|\begin{array}{cccc}1&a&b&c\\1&b&c&a\\1&c&d&a\\1&d&a&b\end{array}\right|+
\left|\begin{array}{cccc}1&a&b&d\\1&b&c&d\\1&c&d&b\\1&d&a&c\end{array}\right|$
I tried to calculate the determinant but I couldn't do it after separating the determinant by the property. How should I calculate it?
|
Compute
$$
\begin{bmatrix}
1 & a & b & c+d \\
1 & b & c & d+a \\
1 & c & d & a+b \\
1 & d & a & b+c
\end{bmatrix}
\begin{bmatrix}
a+b+c+d \\ -1 \\ -1 \\ -1
\end{bmatrix}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ where $x, y, z$ are real numbers satisfying the condition $x + y + z = 1 $.
My Attempt
$$
\begin{align}
\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}&=\frac{1}{x+\frac{1}{x}}+\frac{1}{y+\frac{1}{y}}+\frac{1}{z+\frac{1}{z}}\\&\geq \frac{(1+1+1)^2}{x+\frac{1}{x}+y+\frac{1}{y} + z+\frac{1}{z}}\\&=\frac{9}{x+y+z +(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}
\end{align}
$$
From AM-HM
$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 3^2 $$
$$\implies (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9 $$
$$\implies (x+y+z)+ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 10$$
From this point i am not being able o proceed as inequality signs are getting mixed. Can anyone help out? Thanks in advance.
|
You've already been given a hint to try Jensens inequality, which works fine in this case. Alternatively by AM-GM
$$9\frac{x}{9+(3x)^2}\leqslant 9\frac{x}{10(3x)^{2/10}}=\frac3{10}(3x)^{4/5}$$
Sum that up and observe by power means:
$$\sum (3x)^{4/5} \leqslant 3\left(\frac{3x+3y+3z}3\right)^{4/5}=3$$
Equality is possible when $x=y=z=\frac13$, so these inequalities give you the maximum.
|
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|
Show that $\sum_{k=1}^{\infty}\frac{k}{2^k}=2$
Show that $\sum_{k=1}^{\infty}\frac{k}{2^k}=2$
I have no idea to calculate this sum. I try shift index since $k=0$ gives 0, tis won't change the sum. But I don't know how to keep going. Can someone give me a hint or suggestion to calculate this sum? Thanks
|
Here's another solution.
Let
\begin{align}
S=\sum^\infty_{k=1}\frac{k}{2^k} =&\ \sum^\infty_{k=1} \frac{k-1+1}{2^k} = \frac{1}{2}\sum^\infty_{k=1} \frac{k-1}{2^{k-1}} + \sum^\infty_{k=1} \frac{1}{2^k}\\
=&\ \frac{1}{2}\sum^\infty_{k=2} \frac{k-1}{2^{k-1}}+\frac{1}{2} = \frac{1}{2}\sum^\infty_{k=1}\frac{k}{2^k}+1 \\
=& \frac{1}{2}S+1.
\end{align}
Solving for $S$ yields $S=2$.
|
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|
What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ I've tried this but didn't get the answer :
Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$,
\begin{align}
S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1
\\&= 3\left ( 2014(2013)+2012(2011)+2010(2009)+ \ldots+2(1) \right ) + 1(1007)
\\&= 3\left ( \sum_{n=1}^{1007}2n(2n-1) \right )+1007\\
=& \left ( \sum_{n=1}^{1007}4n^2-\sum_{n=1}^{1007}2n \right )+1007
\\=&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007
\end{align}
This is divisible by $1007$ but not by $1007^2$ which is the correct answer. Where have I gone wrong ?
|
In general you obtain, for $n^3-(n-1)^3+\cdots + 2^3-1^3$ and $n$ even the formula
$$
3\left(\sum_{k=1}^{n/2} 2k(2k-1)\right)+\frac{n}{2}.
$$
Now it is easier to see that this is divisible by $(\frac{n}{2})^2$; and you can test this first for $n=2,4,6,\ldots$ before dealing with $n=2014$. In fact, your computation is correct, except for the last step, where you did not realize how to split of the factor $(\frac{n}{2})^2$. Working with general $n$, you do not need a calculator.
|
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|
maximum value of $\int^{\frac{3\pi}{2}}_{-\frac{\pi}{2}}\sin xf(x)dx$ subjected to the condition$|f(x)|\leq 5$ maximum value of $\displaystyle \int^{\frac{3\pi}{2}}_{-\frac{\pi}{2}}\sin xf(x)dx$ subjected to the condition$|f(x)|\leq 5$
could some help me with this, thanks
|
If $|f(x)|\le 5$ for all $x\in \left[-\frac{\pi}{2},\frac{3\pi}{2}\right]$, we have $\sin(x) f(x)\le|\sin(x) f(x)|\le 5|\sin(x)|$ for all $x\in\left[-\frac{\pi}{2},\frac{3\pi}{2}\right]$.
We thus have $$\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin(x)f(x)\mathrm{dx}\le\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}5|\sin(x)|\mathrm{dx}$$
But we have $$\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}5|\sin(x)|\mathrm{dx}=20$$
Hence $$\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin(x)f(x)\mathrm{dx}\le 20$$
Thus the upper bound for the maximum value of $\displaystyle\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin(x)f(x)\mathrm{dx}$ is $20$.
If we let $f(x):=-5$ for $x\in\left[-\frac{\pi}{2},0\right]\cap\left[\pi,\frac{3\pi}{2}\right]$ and $f(x):=5$ for $x\in (0,\pi)$, then one has $\displaystyle\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin(x)f(x)\mathrm{dx}=20$. Since this is the same as the upper bound for maximum value, the maximum value of $\displaystyle\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin(x)f(x)\mathrm{dx}$ is $20$.
|
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|
How to calculate the limit of $\frac{a_n}{n^2}$ for the sequence $a_{n+1}=a_n+\frac{2 a_{n-1}}{n+1}$? Assume $\{a_n\}$is a sequence which satisfies the following recursion formula:
$$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}~(n\geqslant 1).$$
and $a_0=\pi,a_1=\pi^2 $.
How to compute $\displaystyle\lim_{n\to \infty}\frac{a_n}{n^2}$?
|
Let we assume that
$$ f(x) = \pi+\pi^2 x +a_2 x^2+\ldots = \sum_{n\geq 0}a_n x^n \tag{1}$$
The recursion turns into the differential equation
$$\frac{f(x)-\pi-\pi^2 x}{x}=f(x)-\pi+\frac{2}{x}\int_{0}^{x}t\,f(t)\,dt \tag{2} $$
and assuming $G'(x)=x\,f(x)$, that leads to
$$ f(x) = \frac{\pi(9-5\pi)}{4}\cdot\frac{e^{-2x}}{(1-x)^3}+\frac{\pi(\pi-1)}{4}\cdot\frac{2x^2-6x+5}{(1-x)^3} \tag{3}$$
so $f$ is a meromorphic function with a triple pole at $x=1$. By computing the MacLaurin expansion of $f(x)(1-x)^3$ at $x=1$, it follows that the wanted limit equals
$$ \frac{9\pi-5\pi^2}{8e^2}+\frac{\pi^2-\pi}{8}\approx \color{red}{0.4845}.\tag{4}$$
|
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|
How to prove $\sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}$ where $x=\frac{1}{3}e^{-1/3}$ How to prove that
$$
\sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}, ~\text{where}~~ x=\frac{1}{3}e^{-1/3}~?
$$
I found this sum in my notes, but I don't remember where I got it. Any hints or references would be nice.
|
Let's consider $\sum_{k=1}^\infty\frac{k^k}{k!}\big(\frac{1}{ae^{1/a}}\big)^k=\frac{1}{a-1}$ for $a > 1$ (as in the question).
Then $\frac{1}{a-1} = - 1+ \frac{1}{1-1/a} = \sum_{k=1}^\infty (\frac1a)^k$ by the geometric series. So one has to show
$\sum_{k=1}^\infty\Big[-1 + \frac{k^k}{k!}{e^{-k/a}}\Big](\frac{1}{a}\big)^k=0$
Writing some terms
$0 = \Big[-1 + {e^{-1/a}}\Big](\frac{1}{a}\big) + \Big[-1 + \frac{2^2}{2!}{e^{-2/a}}\Big](\frac{1}{a}\big)^2 + \Big[-1 + \frac{3^3}{3!}{e^{-3/a}}\Big](\frac{1}{a}\big)^3 + \cdots$
Expanding the exponentials and writing the terms up to $(1/a)^3$, say, requires to take the terms
$0 = \Big[-\frac{1}{a} + \frac12 (\frac{1}{a})^2 + \cdots \Big](\frac{1}{a}\big) + \Big[-1 + \frac{2^2}{2!}(1-\frac{2}{a} + \cdots)\Big](\frac{1}{a}\big)^2 + \Big[-1 + \frac{3^3}{3!}(1 + \cdots)\Big](\frac{1}{a}\big)^3 + \cdots$
Sorting for powers of $1/a$ gives
$0 = \Big[0 \Big](\frac{1}{a}\big) + \Big[-1 -1 + \frac{2^2}{2!}\Big](\frac{1}{a}\big)^2 + \Big[\frac12 - 2 \frac{2^2}{2!} -1 + \frac{3^3}{3!}\Big](\frac{1}{a}\big)^3 + \cdots$
and all the prefactors are zero, as required. The full solution then comes upon expanding all exponentials. Clearly this requires some more formal work.
|
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|
Prove the following Inequality. Given that $\alpha,\beta$ and $\gamma\in (0,\pi)$ and $\alpha+\gamma+\beta=\pi$, show that $$\cos\alpha+\cos\gamma+\sin\beta\leq\frac{3\sqrt3}{2}.$$
Now I am aware of the following Inequality: $\cos\alpha+\cos\gamma+\cos\beta\leq\frac{3}{2}$, but notice that the term $\cos\beta$ has been replaced with $\sin\beta$. I also tried substituting $\sin\beta=\sin(\pi-(\alpha+\gamma))=\sin(\alpha+\gamma)$ and expanding the resulting expression but I was unable to deduce anything meaningful.
|
$\cos(\alpha)+\cos(\gamma)+\sin(\beta)=\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)+\sin(\beta)$
1) Suppose $\beta \leq \frac{\pi}{2}$
Using Jensen inequality we get: $\frac{\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)+\sin(\beta)}{3} \leq \sin(\frac{\frac{\pi}{2}-\alpha+\frac{\pi}{2}-\gamma+\beta}{3})=\sin(\frac{2\beta}{3})$
Since $\frac{2\beta}{3}\leq \frac{\pi}{3}$, we get : $\cos(\alpha)+\cos(\gamma)+\sin(\beta) \leq 3 \sin(\frac{\pi}{3})=\frac{3\sqrt{3}}{2}$
1) Suppose $\beta \geq \frac{\pi}{2}$
Using Jensen inequality we get: $\frac{\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)}{2} \leq \sin(\frac{\frac{\pi}{2}-\alpha+\frac{\pi}{2}-\gamma}{2})=\sin(\frac{\pi-\beta}{2})$
Since $\frac{\pi-\beta}{2} \leq \frac{\pi}{4}$, we get : $\cos(\alpha)+\cos(\gamma)+\sin(\beta) \leq 2 \sin(\frac{\pi}{4})+\sin(\beta)=1+\sqrt{2}\leq \frac{3\sqrt{3}}{2}$
|
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|
To find the equation of curve. We have given a differential equation we have to find the equation of curve. I tried it a lot but not able to proceed. Can someone give me some hints.
$(1-xy+x^2y^2)~dx=x^2~dy$
I tried to substitute $xy=t$ but got no result. Please help.
|
$(1-xy+x^2y^2)~dx=x^2~dy$
$\dfrac{dy}{dx}=\dfrac{1}{x^2}-\dfrac{y}{x}+y^2$
Let $y=-\dfrac{1}{u}$ ,
Then $\dfrac{dy}{dx}=\dfrac{1}{u^2}\dfrac{du}{dx}$
$\therefore\dfrac{1}{u^2}\dfrac{du}{dx}=\dfrac{1}{x^2}+\dfrac{1}{xu}+\dfrac{1}{u^2}$
$\dfrac{du}{dx}=\dfrac{u^2}{x^2}+\dfrac{u}{x}+1$
Let $v=\dfrac{u}{x}$ ,
Then $u=xv$
$\dfrac{du}{dx}=x\dfrac{dv}{dx}+v$
$\therefore x\dfrac{dv}{dx}+v=v^2+v+1$
$x\dfrac{dv}{dx}=v^2+1$
$\dfrac{dv}{v^2+1}=\dfrac{dx}{x}$
$\int\dfrac{dv}{v^2+1}=\int\dfrac{dx}{x}$
$\tan^{-1}v=\ln x+c$
$v=\tan(\ln x+c)$
$\dfrac{u}{x}=\tan(\ln x+c)$
$-\dfrac{1}{xy}=\tan(\ln x+c)$
$y=-\dfrac{\cot(\ln x+c)}{x}$
$y=\dfrac{\cot(-\ln x-c)}{x}$
$y=\dfrac{\tan(\ln x+C)}{x}$
|
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|
Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that
cos$\alpha$ + cos$\beta$ + cos$\gamma$ - 1 $\geq$ $2\sqrt{\cos\alpha\times\cos\beta\times\cos\gamma}$.
Here's what I've tried. I know that $0 < \alpha, \beta, \gamma < \pi/2$ since they are acute angles. So I substituted $\alpha+\beta$ for $\gamma$
$\cos\alpha$ + $\cos\beta$ + $\cos(\alpha+\beta)$- 1 $\geq$ $2\sqrt{\cos\alpha\cos\beta\cos(\alpha+\beta)}$.
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)]\cos(\alpha+\beta)}.$
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{2}[\cos(\alpha-\beta)\cos(\alpha+\beta)+\cos^{2}(\alpha+\beta)]}$
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{4}[\cos(-2\beta)+\cos(2\alpha)]+\cos^{2}(\alpha+\beta)}$
I got stuck from there and I'm not sure where to go from here. Any help would be appreciated.
|
Let $\alpha\geq\beta$
Since $\cos$ is a decreasing function on $\left(0,\frac{\pi}{2}\right)$, $\frac{\alpha+\beta}{2}<\frac{\pi}{4}$ and $\frac{\alpha-\beta}{2}<\frac{\pi}{4}$, we obtain:
$$\cos\alpha+\cos\beta+\cos(\alpha+\beta)-1=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos^2\frac{\alpha+\beta}{2}-2>$$
$$>2\cdot\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}+2\cdot\left(\frac{1}{\sqrt2}\right)^2-2=0$$
Let $\cos\alpha=x$, $\cos\beta=y$ and $\cos(\alpha+\beta)=z$.
Hence, we need to prove that $$(x+y-1+z)^2\geq4xyz$$ or
$$z^2+2(x+y-1-2xy)z+(x+y-1)^2\geq0.$$
$$\frac{\Delta_{z}}{4}=(x+y-1-2xy)^2-(x+y-1)^2=$$
$$=(x+y-1-2xy-(x+y-1))(x+y-1-2xy+x+y-1)=$$
$$=4xy(1-x)(1-y).$$
Thus, it remains to prove that
$$z\leq-(x+y-1-2xy)-\sqrt{4xy(1-x)(1-y)}$$ or
$$z\leq(1-x)(1-y)+xy-\sqrt{4xy(1-x)(1-y)}$$ or
$$\cos(\alpha+\beta)\leq\cos\alpha\cos\beta+(1-\cos\alpha)(1-\cos\beta)-2\sqrt{\cos\alpha\cos\beta(1-\cos\alpha)(1-\cos\beta)}$$ or
$$-\sin\alpha\sin\beta\leq4\sin^2\frac{\alpha}{2}\sin^2\frac{\beta}{2}-4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sqrt{\cos\alpha\cos\beta}$$ or
$$-\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\leq\sin\frac{\alpha}{2}\sin\frac{\beta}{2}-\sqrt{\cos\alpha\cos\beta}$$ or
$$\cos\frac{\alpha-\beta}{2}\geq\sqrt{\cos\alpha\cos\beta}$$ or
$$1+\cos(\alpha-\beta)\geq2\cos\alpha\cos\beta$$ or
$$1\geq\cos(\alpha+\beta)$$
Done!
|
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|
Is this valid? Show that $(x-2)\mid x^{3} - 4x$ Is this valid? Here $\textsf{R}(x)$ stands for the remainder.
Show that $(x-2)\mid x^{3} - 4x$
I know that I can simply do this: Let $p(x) = x^{3} - 4x$ then if $(x-2)\mid x^{3} - 4x$ it follows that $p(2) = 0$. So we have $2^{3} - 4(2) = 0$. However I am wondering about the following.
The process (in lack of a better word):
$\textsf{R}(x)_{1} = x^{3} - 4x -x^{2}(x-2) = 2x^{2} - 4x$
$\textsf{R}(x)_{2} = 2x^{2} - 4x - 2x(x-2) = 0$
so
$ x^{3} - 4x = \textsf{R}(x)_{1} + x^{2}(x-2) \Rightarrow 2x^{2} - 4x + x^{2}(x-2) \iff 2x(x-2) + x^{2}(x-2) = (x-2)(2x + x^{2})$
My question is this: Does $\text{deg} \space \textsf{R}(x)$ have to be less than $\text{deg}(x-2)$ in the process? I get the correct answer.
Any answer is appreciated.
|
$x^3-4x=x(x^2-4)=x(x+2)(x-2)$. So, you can clearly see that $x-2$ is a factor.
Or you can do it this way,
Let $f(x)=x^3-4x$. See that $f(2)=0$. Hence, by factor theorem, $x-2|f(x).$
|
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|
Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\
&=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right)
\end{align}$$
and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.
Once the solution
$$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$
is known, the telescoping sum can be easily derived, i.e.
$$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$
However, without knowing this a priori, how would we have approached this problem?
|
Another approach. By Euler/De Moivre's formula we have
$$A_n=\frac{1}{4^n}\binom{2n}{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\cos^{2n}(x)\,dx \tag{1}$$
since $\cos(x)^{2n} = \frac{1}{4^n}\sum_{j=0}^{2n}\binom{2n}{j}e^{(2n-j)ix}e^{-jix}$ and $\int_{-\pi}^{\pi}e^{kix}\,dx =2\pi \delta(k)$, hence only the contribute given by $j=n$ survives. It follows that
$$ \color{red}{\sum_{n=0}^{N}\frac{1}{4^n}\binom{2n}{n}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1-\cos^{2N+2}(x)}{\sin^2(x)}\,dx =(2N+2)\,A_{N+1}=\color{red}{\frac{N+1}{2^{2N+1}}\binom{2N+2}{N+1}}\tag{2}$$
by integration by parts ($\int\frac{dx}{\sin^2 x}=-\cot x$).
|
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|
Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$?
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$
$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left side, $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = \sqrt{9\sin^250^\circ-8\sin^350^{\circ}}$$
Now How can i solve it after that , Help required, Thanks
|
There is not a unique solution; we will be able to provide a solution as a function of $a$. Note that if $a = 0$, then $ab=0$. Henceforth, assume $a \neq 0$.
Let $x = \sin(50^\circ) \neq 0$ and $b = c/a$ so that we wish to solve for $c$. Then the given equation is
$$ \sqrt{9 - 8 x} = a + \frac{c}{a x} \text{.} $$
Squaring, we get
$$ 9 - 8 x = a^2 + \frac{2 c}{x} + \frac{c^2}{a^2 x^2} \text{.} $$
(And we will remember to check all solutions back in the original equation since this step may have introduced spurious solutions.) Multiply through by $a^2 x^2$ and collect everything on the right (then swap sides):
$$ c^2 + 2a^2 x c + a^2 x^2( a^2 + 8 x - 9) = 0 \text{.} $$
Applying the quadratic formula, \begin{align*}
c &= \frac{ -2 a^2 x \pm \sqrt{4 a^4 x^2 - 4a^2 x^2(a^2 + 8 x - 9)}}{2} \\
&= -a^2 x \pm a x \sqrt{a^2 - a^2 - (8 x - 9)} \\
&= -a^2 x \pm a x \sqrt{9 - 8 x} \text{.}
\end{align*}
Plugging back in to the first display above, we quickly see that the $+$ choice works and the $-$ choice does not. Converting back to the original symbols, the solution is
$$ ab = -a \sin(50^\circ) \left(a - \sqrt{9 - 8 \sin(50^\circ)}\right) \text{.} $$
Happily, when $a=0$, this is also $0$, so we do not need to report the solution piecewise.
|
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|
Finding maximum value from a statement/equation The statement has values of $x$ and $y$ as positive integers: $$\sqrt{x} - \sqrt{11} = \sqrt{y}$$
I have to find the maximum possible values of $\frac{x}{y}$, this what I have done so far:
$$x = (\sqrt{y} + \sqrt{11})^2$$
$$y = (\sqrt{x} - \sqrt{11})^2$$
therefore: $$\frac{x}{y} = \frac{(\sqrt{y} + \sqrt{11})^2}{(\sqrt{x} - \sqrt{11})^2}$$
...
$$\frac{x}{y} = \frac{y + 11 + 2\sqrt{y}\sqrt{11}}{x + 11 - 2\sqrt{x}\sqrt{11}} = \frac{y+\sqrt{y}}{x-\sqrt{x}} + 1$$
and from here I'm not sure what to do...
(Again better title suggestions are also welcome)
|
Hint: $\frac{x}{y}=\frac{(\sqrt{y}+\sqrt{11})^2}{y}=\frac{y+2\sqrt{y}\sqrt{11}+11}{y}=1+\frac{2\sqrt{y}\sqrt{11}+11}{y}=f(y)$. Find zeros of $f'(y)$ and determine if they are minima or maxima by the second derivative test.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Circle incribed within a triangle percentage
I have worked out the areas as $\pi/3$ for the circle and $2/\sqrt3$ for the triangle but don't know how to convert into a percentage without a calculator.
|
$$
\begin{gathered}
side_{\,tr} = 1\quad \quad h_{\,tr} = \frac{{\sqrt 3 }}
{2}\quad \quad r_{\,circ} = \frac{{\sqrt 3 }}
{6} \hfill \\
A_{\,circ} = \pi \frac{3}
{{36}} = \frac{\pi }
{{12}}\quad \quad A_{\,tr} = \frac{{\sqrt 3 }}
{4} \hfill \\
\rho = \frac{{A_{\,circ} }}
{{A_{\,tr} }} = \frac{\pi }
{{12}}\frac{4}
{{\sqrt 3 }} = \frac{\pi }
{3}\frac{1}
{{\sqrt 3 }} \approx \frac{{3.15}}
{3}\frac{1}
{{\sqrt 3 }} = \frac{{1.05}}
{{\sqrt 3 }} \hfill \\
\rho ^{\,2} \approx \frac{{\left( {1.05} \right)^{\,2} }}
{3} = \frac{1}
{3}\left( {1 + \frac{5}
{{100}}} \right)^{\,2} \approx \frac{1}
{3}\left( {1 + 2\frac{5}
{{100}}} \right) = \frac{{110}}
{{300}} \approx \frac{{36}}
{{100}} \hfill \\
\rho \approx \sqrt {\frac{{36}}
{{100}}} = \frac{6}
{{10}} = \frac{{60}}
{{100}} = 60\% \hfill \\
\end{gathered}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1991600",
"timestamp": "2023-03-29T00:00:00",
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|
Surface area of sphere within a cylinder I have to
Compute the surface area of that portion of the sphere $x^2+y^2+z^2=a^2$ lying within the cylinder $\Bbb{T}:=\ \ x^2+y^2=by.$
My work:
I start with only the $\Bbb{S}:=\ \ z=\sqrt{a^2-x^2-y^2}$ part and will later multiply it by $2$.
$${\delta z\over \delta x}={-x\over \sqrt{a^2-x^2-y^2}}\ ;\ {\delta z\over \delta y}={-y\over \sqrt{a^2-x^2-y^2}}$$
Using the formula $$a(\Bbb{S})=\iint\limits_\Bbb{T}\sqrt{1+\left({\delta z\over \delta x}\right)^2+\left({\delta z\over \delta y}\right)^2}dx \ dy$$
I get $$a(\Bbb{S})=\iint\limits_{x^2+y^2=by}\sqrt{1+{by\over a^2-by}}\ dy\ dx\\=\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\sqrt{1+{by\over a^2-by}}\ dy\ dx\\=a\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}{1\over\sqrt{a^2-by}}dy \ dx\\=a\int_0^{b/2}\left\{\left[{-2\sqrt{a^2-by}\over b}\right]_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\right\}\ dx$$
How to proceed now?The integral seems too bad.
OR
Is there a simpler parametrization ?
|
Your derivatives are wrong by a factor of $2$. This causes your integrand to be
$$
\sqrt{1-{x^2+y^2\over a^2-x^2-y^2}}=\frac a{\sqrt{a^2-x^2-y^2}}
$$
If we represent the cylinder in polar coordinates, we have $$r^2=b r \sin\theta.$$After cancelling $r$, we get $r=b\sin\theta$. So the interior of the cylinder is the region
$$
0\leq\theta\leq\pi,\ \ \ 0\leq r\leq b\sin\theta
$$
(the restriction on $\theta$ comes from $\sin\theta\geq r/b\geq0$). So the area is
\begin{align}
a(\Bbb{S})&=\iint\limits_\Bbb{T}\sqrt{1+\left({\partial z\over \partial x}\right)^2+\left({\partial z\over \partial y}\right)^2}dx \ dy\\ \ \\
&=\int_0^\pi\int_0^{b\sin\theta}\frac {ar}{\sqrt{a^2-r^2}}\,dr\,d\theta\\ \ \\
&=-a\,\int_0^\pi \left.\vphantom{\int}{\sqrt{a^2-r^2}}\,\right|_0^{b\sin\theta}\,d\theta\\ \ \\
&=a\,\int_0^\pi \left(a-{\sqrt{a^2-b^2\sin^2\theta}}\right)\,d\theta\\ \ \\
&=a^2\pi-a\int_0^\pi {\sqrt{a^2-b^2\sin^2\theta}}\,d\theta.\\ \ \\
\end{align}
This last integral is an elliptic integral (unless $b=a$). It looks to me like all approaches to this problem will lead to this integral (I will be happy to be proven wrong, though).
|
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|
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing.
|
Here is mine...
$$\log_2 3=\frac{\log_2 243}5<\frac{\log_2 256}5=8/5$$
$$\log_3 2=\frac{\log_3 32}5>\frac{\log_3 27}5=3/5$$
$$\log_3 6=\log_3 2+\log_3 3>1+3/5=8/5>\log_2 3$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $A$ is a real $n \times n$ matrix satisfying $A^3 = I$ then Trace of $A$ is always If $A$ is a real $3 \times 3$ matrix satisfying $A^3$ = I such that $ A \neq I $ .Then, Trace of A is always
*
*$0$
*$1$
*$-1$
*$3$
I proceed as follows: from given,
$\min(x)=x-1$ or $x^2+x+1$ or $(x-1)(x^2+x+1)$
$\min(x)\ne x-1$ as $A\ne I$
so, $\min(x)=x^2+x+1$ or $(x-1)(x^2+x+1)$
now, how to proceed after this step
any help would be appreciated
|
$$A^3=I\iff A^2=A^{-1}$$ If the question is valid for all matrix then with a single example one can answer. Consider the matrix
$$A = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}$$
Calculation gives $$A^{-1}=\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\text { and } A^2=\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}$$ We see then that $A^3=I$ and here the trace is obviously equal to $0$. Hence the answer is $0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the subinterval for which $f_{n}(x) = \frac{n\sin(x)}{1+n^2\sin^2(x)}$ is Uniformly Convergent in Let $f_{n}(x) = \frac{n\sin(x)}{1+n^2\sin^2(x)}$ for $x \in [0,\pi]$. For $\delta > 0$, I want to find $E_{\delta} \subset [0, \pi]$ where $f_{n}(x)$ converges uniformly and $\mu([0,\pi] \setminus E_{\delta}) < \delta$ where $\mu$ denotes measure.
I've shown that $\lim_{n \to \infty} \sup_{x \in [0,\pi]} \left \{ \frac{n\sin(x)}{1+n^2\sin^2(x)} \right \} \neq 0$, but am not sure how to find this $E_{\delta}$. Ideas?
|
Let's look on the interval $[0,\pi/2]$ and choose an integer $m > 2$.
Then for $n > 2m^2$ and $x > \frac{1}{m}$, we have
\begin{eqnarray}
2m^2 < n &\iff& 2m < \frac{n}{m}\\
&\implies& 2m < nx \tag{since $x> \tfrac{1}{m}$}\\
&\implies& \frac{2}{nx} < \frac{1}{m}
\end{eqnarray}
Now because $\sin (x) > \frac{x}{2}$ in $[0,\pi/2]$, we have $\frac{1}{\sin(x)} < \frac{2}{x}$ so that
$$\frac{n \sin(x)}{1+n^2 \sin^2(x)} < \frac{n \sin(x)}{n^2 \sin^2(x)} = \frac{1}{n \sin(x)} < \frac{1}{n \frac{x}{2}} = \frac{2}{nx} < \frac{1}{m}$$
Thus for large enough $n$, $f_n(x) < \frac{1}{m}$ on the interval $\left(\frac{1}{m},\frac{\pi}{2}\right)$.
You can apply a symmetry argument to get a similar result on the interval $[\pi/2,\pi]$.
Finally, for $\delta > 0$, pick $m > \frac{1}{3\delta}$ and $E_\delta = \left(\frac{1}{m},\pi-\frac{1}{m}\right)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Mathematical Induction for Recurrence Relation I have solved the following recurrence relationship:
$T(1) = 1$
$T(n) = T(n-1) + n + 2$
so
$T(n) = \frac{1}{2}n^2+\frac{5}{2}n -2$
I am now trying to perform mathematical induction to prove this.
$Basis:$
$T(1)=1=3-2=\frac{1}{2} + \frac{5}{2} - 2 $
$Induction:$
$T(k+1) = T(k) + k+1 + 2$
$= \frac{1}{2}k^2 + \frac{5}{2}k -2 + k+1 +2$
$= \frac{1}{2}k^2 + \frac{7}{2}k +1$
What can I do next?
|
HINT: Try to express the last formula in terms of $(k+1) $, rather than a function of $k $.
|
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|
Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
Given, for every $x>1$,
$$f(x)=4\arctan\frac{1}{\sqrt{x-1}+\sqrt{x}}$$
Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
I have tried to use the fact that $\arctan(x)+\arctan(1/x)=\frac{\pi}{2
}$
So I obtain: $f(x)=4(\frac{\pi}{2}-\arctan(\sqrt{x-1}+\sqrt{x})$
I am stuck here !
|
Your statement is equivalent to proving that, for
$$
g(x)=2\arctan\frac{1}{\sqrt{x}+\sqrt{x-1}}
$$
we also have
$$
g(x)=\frac{\pi}{2}-\arctan\sqrt{x-1}
$$
Note that
$$
\frac{1}{\sqrt{x}+\sqrt{x-1}}=\sqrt{x}-\sqrt{x-1}
$$
Set $h(x)=\sqrt{x}-\sqrt{x-1}$ and prove that, for $x>1$, $0<h(x)<1$.
It follows that $0<g(x)<\pi/2$.
Now
\begin{align}
\tan g(x)
&=\tan(2\arctan h(x))\\[6px]
&=\frac{2\tan\arctan(h(x))}{1-\tan^2(\arctan h(x))}\\[6px]
&=\frac{2h(x)}{1-(h(x))^2}\\[6px]
&=2\frac{\sqrt{x}-\sqrt{x-1}}{1-x-(x-1)+2\sqrt{x(x-1)}}\\[6px]
&=\frac{\sqrt{x}-\sqrt{x-1}}{\sqrt{x-1}(\sqrt{x}-\sqrt{x-1})}\\[6px]
&=\frac{1}{\sqrt{x-1}}
\end{align}
Therefore
$$
\tan\left(\frac{\pi}{2}-g(x)\right)=
\cot g(x)=\sqrt{x-1}
$$
and so
$$
\frac{\pi}{2}-g(x)=\arctan\sqrt{x-1}
$$
|
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"url": "https://math.stackexchange.com/questions/1996649",
"timestamp": "2023-03-29T00:00:00",
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|
How many five-digit numbers can be formed using the digits 1-9 which have at least three identical digits? How many five-digit nos. can be formed using the digits 1-9 having at least three identical digits?
My attempt:
Total no. of possible nos. with no restrictions $=9^5$
No. of numbers having two identical digits$=9\times 8^4$
Hence the answer should be
$( 9^5)-(9\times 8^4) = 22185$
Is this correct? If not then please suggest alternate method.....Thanks in advance!!
|
First approach: Pick which three spots host the identical digits; there are $\binom53$ ways to do this. Fill those three spots in one of $9$ ways, and then fill the other two spots in any of $9$ ways. Thus, $\binom53\times 9^3$
Problem with this approach: the number $12222$ has been counted $4$ times, once as $1\color{red}{222}2$, once as $1\color{red}{22}2\color{red}{2}$, once as $1\color{red}{2}2\color{red}{22}$, and once as $12\color{red}{222}$. Similarly, a number such as $44444$ has been overcounted by a factor of $\binom53$.
This can be fixed by subtracting out the repetitions. There are $5\times 9\times 8$ numbers with exactly $4$ digits identical, and $9$ numbers with exactly $5$ digits identical.
Thus: $\binom53\times 9^3 - (5\times 9\times 8)\times(4-1) - 9\times\left(\binom53 - 1\right)$
This comes out to $7290 - 1080 - 81 = 6129$
Doublecheck: Count the numbers with exactly $3$ identical digits, then the ones with exactly $4$, then the ones with exactly $5$:
$\binom53\times 9\times 8\times 8 + \binom54\times 9\times 8 + 9$
which comes out to: $5760 + 360 + 9 = 6129$
|
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|
find the range of the $b$, if $a^2+b^2+c^2=21~~~~2b=a+c$ Let $\Delta ABC$ such $|AC|=b,|BC|=a,|AB|=c$,and such $$\begin{cases}
2b=a+c\\
a^2+b^2+c^2=21
\end{cases}$$
find the range of the $b$
since
$$(a+c)^2-2ac+b^2=21$$
so we have
$$2ac=5b^2-21$$
then $a,c$ is a equation
$$x^2-2bx+\dfrac{5b^2-21}{2}=0$$ roots.so we have
$$\Delta =4b^2-2(5b^2-21)=42-6b^2\ge 0\Longrightarrow -\sqrt{7}\le b\le \sqrt{7}$$
the book aswer: $\sqrt{6}<b\le\sqrt{7}$
|
For the lower bound we have: $b > |a-c| \implies b^2 > (a-c)^2, (2b)^2 = (a+c)^2\implies 5b^2 > (a-c)^2 + (a+c)^2 = 2(a^2+c^2) = 2(21- b^2) \implies 7b^2 > 42 \implies b^2 > 6 \implies b > \sqrt{6}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Help with tricky double integral
Consider the region $R$ bounded by the circles
$$x^{2}+y^{2}=Ax$$$$x^{2}+y^{2}=Bx$$$$x^{2}+y^{2}=Cy$$$$x^{2}+y^{2}=Dy$$
where $B>A$ and $D>C$.
Use the change of variables
\begin{cases} u=\frac{x}{x^2+y^2}\\
v=\frac{y}{x^2+y^2} \end{cases}
to evaluate the integral $$\int\int_R \frac{dxdy}{(x^2+y^2)^3}\quad
(1)$$
My work so far: the Jacobian is
$$\begin{bmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}
\end{bmatrix}$$
With $\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}=\frac{-2xy}{(x^2+y^2)^2} $ and $\frac{\partial u}{\partial x}=\frac{y^2-x^2}{(x^2+y^2)^2}$
and $\frac{\partial v}{\partial y}=\frac{x^2-y^2}{(x^2+y^2)^2}$
: $$J(u,v)=\frac{\partial (x,y)}{\partial (u,v)}=\frac{-1}{(x^2+y^2)^2} \quad(2)$$
My first approach to get the limits is rearranging $x^{2}+y^{2}=Ax$ and hence $\frac{1}{A}=\frac{x}{x^2+y^2}$ hence $1/A$ is a limit and the rest is $1/B$, $1/C$ and $1/D$.
My questions:
*
*How do I find the limits of integration?
*How do I express the integral is terms of $u$ and $v$ ?
Any tips will be appreciated thanks!!
|
*
*You pretty much got it:
$$
R(u,v)=\left\{(u,v)\;|\; \frac{1}{B}\le u\le\frac{1}{A} , \frac{1}{D}\le u\le \frac{1}{C}\right\}
$$
The Figures below show the case where $A=C=1$, $B=D=2$:
So the region we are interested in is
*We are given \begin{cases} u=\frac{x}{x^2+y^2}\\
v=\frac{y}{x^2+y^2} \end{cases}
Solving for $x$ and $y$ yields:
\begin{cases} x=\frac{u}{u^2+v^2}\\
y=\frac{v}{u^2+v^2} \end{cases}
The Jacobian of such a transformation equals $\frac{1}{(u^2+v^2)^2}$, and $\frac{1}{(x^2+y^2)^3}=(u^2+v^2)^3$ therefore the integral equals
$$
\int_{1/B}^{1/A}\int_{1/D}^{1/C}(u^2+v^2)^3\frac{dvdu}{(u^2+v^2)^2}=\int_{1/B}^{1/A}\int_{1/D}^{1/C}u^2+v^2\;dvdu
$$
This is very easy to integrate, although the final result does not simplify too well. The case with $A=C=1$, $B=D=2$ gives $\frac{7}{24}$.
|
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|
A proof related to ellipse. A tangent is drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to cut the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ at the points $P$ and $Q$. If tangents at $P$ and $Q$ to the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ intersect at right angle, then prove that $\frac{a^2}{c^2}+\frac{b^2}{d^2}=1$.
My attempt at a solution:
Since it is told that the tangents at $P$ and $Q$ intersect at right angle, therefore the point of intersection of the tangents must lie on the director circle of the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ , whose equation is $x^2 + y^2 = c^2 + d^2$.
Let $(a\space \cos\theta \space, \space b \space \sin\theta)$ be a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at which a tangent is drawn. Therefore equation of the tangent would be $\frac{x\space \cos\theta}{a}+\frac{y\space \sin\theta}{b}=1$. Let this tangent be called $PQ$.
Now, let $(c\space \cos\theta_1 \space, \space d \space \sin\theta_1)$ and $(c\space \cos\theta_2 \space, \space d \space \sin\theta_2)$ be points $P$ and $Q$ on the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ where the tangent intersects it.
Let $R$ be the point of intersection of the tangents drawn at $P$ and $Q$.
The coordinates of $R$ is given by $(\frac{a \space \cos(\frac{\theta_1 + \theta_2}{2})}{\cos(\frac{\theta_1 - \theta_2}{2})},\frac{b \space \sin(\frac{\theta_1 + \theta_2}{2})}{\cos(\frac{\theta_1 - \theta_2}{2})})$
Now I get 3 equations, as follows:
*
*The point $R$ satisfies the equation of the director circle.
*The point $P$ satisfies the equation of the tangent PQ.
*The point $Q$ satisfies the equation of the tangent PQ.
Solving these equations, I get something like:
$(c^2-d^2)\sin^4(\theta)+(a^2+b^2+d^2-c^2) \sin^2(\theta)-b^2=0$
Now since it is told in the problem only "A tangent is drawn..." I tried to prove the given equation by making the discriminant of the above equation to 0, but I am unable to get the condition. Is there a simpler approach to this? Thanks for any help.
|
1) The first ellipse $E_1$ has equation $$F(x,y)=b^2x^2+a^2y^2-a^2b^2=0$$ and its derivative at the point $(x,y)$ is equal to $$-\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial y}}=\frac{-b^2x}{a^2y}$$
2) Clearly for an ellipse of equation $\dfrac{X^2}{A^2}+\dfrac{Y^2}{B^2}=1$ the vertical $X=A$ and the horizontal $Y=B$ are tangents to the ellipse which intersects making a right angle at the point $(X,Y)=(A,B)$ (we clearly get a rectangle).
3) Take the tangent line at a point $(x_0,y_0)\in E_1$. You get $$y=-\frac{b^2x_0}{a^2y_0}x+\frac{b^2x_0^2+a^2y_0^2}{a^2y_0}=-\frac{b^2x_0}{a^2y_0}x+\frac{b^2}{y_0}$$
4) The crossing points of this line with the axis are $\left(\dfrac{a^2}{x_0},\space0\right)$ and $\left(0,\space\dfrac{b^2}{y_0}\right)$ and these points are the axes of the given second ellipse $E_2$ of equation (See paragraph 2)) $$\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$$ so this equation becomes $$\frac{x_o^2x^2}{a^4}+\frac{y_0^2y^2}{b^4}=1$$
It is clear now that $(x,y)=(a,b)$ is a point of the ellipse $E_2$ (puting this point in $E_2$ we simply verify that $(a,b)\in E_1$
|
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|
Finding a simpler representation for the set $\{z\in\mathbb{C}|\|z+i\|+\|z-i\|=4\}$
Sketch the following set: $\{z\in\mathbb{C}|\|z+i\|+\|z-i\|=4\}$
I tried thinking about it geometrically, where we are interested in all triangles of the form (x,y-1), (x,y+1), (0,0) and we want the sum of the sides next to (0,0) be 2. Intuitively this makes me think the solution will be an ellipse of sort (where $\pm i$, and $\pm3$ being in the set helps this feeling), but I wasn't able to prove it.
Algebraically, I tried squaring both sides, to get
$$ \left|z-i\right|^{2}+\left|z+i\right|^{2}+2\left|z-i\right|\left|z+i\right|=16 $$
and while the first two summands behave nicely
$$\begin{aligned}\left|z-i\right|^{2} & = & \left(z-i\right)\overline{\left(z-i\right)}\\
& = & z\overline z-i\overline z+iz+1\\
& = & \left|z\right|^{2}+i\left(z-\overline{z}\right)+1\\
& = & \left|z\right|^{2}+i\left(2i{\rm Im}z\right)+1\\
& = & \left|z\right|^{2}-2{\rm Im}z+1
\end{aligned}$$
and $$\begin{aligned}\left|z+i\right|^{2} & = & \left(z+i\right)\overline{\left(z+i\right)}\\
& = & z\overline z-iz+i\overline{z}+1\\
& = & \left|z\right|^{2}-i\left(z-\overline z\right)+1\\
& = & \left|z\right|^{2}+2{\rm Im}z+1
\end{aligned}$$
the third one leaves me a bit stuck. Any help on proceeding?
|
This is the "two nails and fixed cord" construction of an ellipse.
Computing the minor axis:
For $z = a + 0\cdot i$ we have
$$
\lVert z+i \rVert + \lVert z-i \rVert
= \sqrt{a^2 + 1^2} + \sqrt{a^2 + (-1)^2}
= 2 \sqrt{1 + a^2} = 4 \iff \\
1 + a^2 = 4 \iff \\
a = \sqrt{3}
$$
For $z = 0 + b\cdot i$ we have
$$
\lVert z+i \rVert + \lVert z-i \rVert
= \sqrt{0^2 + (b+1)^2} + \sqrt{0^2 + (b-1)^2}
= b + 1 + b - 1 \\
= 2b = 4 \iff \\
b = 2
$$
We have $a = \sqrt{3}$ and $b=2$, so it should be
$$
\left( \frac{x}{\sqrt{3}} \right)^2 + \left( \frac{y}{2} \right)^2 = 1
$$
Now lets try this from scratch:
$$
4 = \lVert z+i \rVert + \lVert z-i \rVert \\
= \sqrt{x^2 + (y+1)^2} + \sqrt{x^2 + (y-1)^2} \iff \\
= \sqrt{x^2 + y^2 + 1 + 2y} + \sqrt{x^2 + y^2 + 1 - 2y} \iff \\
16 = 2(x^2 + y^2 + 1)+2\sqrt{(x^2 + y^2 + 1)^2 - 4y^2} \iff \\
8 = x^2 + y^2 + 1 + \sqrt{(x^2 + y^2 + 1)^2 - 4y^2} \iff \\
(8 - (x^2 + y^2 + 1))^2 = (x^2 + y^2 + 1)^2 - 4y^2 \iff \\
64 + (x^2 + y^2 + 1)^2 - 16 (x^2 + y^2 + 1) = (x^2 + y^2 + 1)^2 - 4y^2 \iff \\
64 = 16 (x^2 + y^2 + 1) - 4y^2 = 16 x^2 + 12 y^2 + 16 \iff \\
48 = 16 x^2 + 12 y^2 \iff \\
12 = 4 x^2 + 3 y^2 \iff \\
1 =x^2 / 3 + y^2 / 4 \iff \\
1 = \left(\frac{x}{\sqrt{3}}\right)^2 + \left(\frac{y}{2}\right)^2
$$
|
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|
Where's this 1/2 coming from squaring the derivative for arc length? I am currently studying arc length in calculus 2 and I am having trouble understanding one of the solutions. I understand the concept and how the procedure works, but it seems like in a lot of problems you're adding on a 1/2 and bringing it in front of the integral, and I am not sure how or why.
Above is the image that shows the calculation for finding the derivative and then squaring that derivative. $1/2y^3 - 1/2y^{-3}$ squared should be $1/4y^6 + 1/4y^{-6}$.
Where is that $-1/2$ coming from in the 4th line of this solution? Is it so we can get a perfect square? And if so how do we just add that without changing the value of the equation?
Thanks!
|
You've forgotten the middle term in $(a-b)^2 = a^2 - 2 a b + b^2$ and that $y^3 \cdot y^{-3} = 1 \neq 0$. Distibuting twice, then collecting...
\begin{align*}
&\left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right)^2 \\
&\qquad = \left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right) \left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right) \\
&\qquad = \left( \frac{1}{2} \right) y^3\left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right) - \left( \frac{1}{2} \right) y^{-3}\left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right) \\
&\qquad = \left( \left( \frac{1}{4} \right) y^6 - \left( \frac{1}{4} \right) y^{0} \right) - \left( \left( \frac{1}{4} \right) y^0 + \left( \frac{1}{4} \right) y^{-6} \right) \\
&\qquad = \left( \frac{1}{4} \right) y^6 - \left( \frac{2}{4} \right) y^{0} + \left( \frac{1}{4} \right) y^{-6} \\
&\qquad = \left( \frac{1}{4} \right) y^6 - \left( \frac{1}{2} \right) + \left( \frac{1}{4} \right) y^{-6} \text{.}
\end{align*}
|
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|
Locus of focus of variable parabola
A variable parabola is drawn to pass through A & B, the ends of a
diameter of a given circle with centre at the origin and radius c & to
have as directrix a tangent to a concentric circle of radius 'a' (a>c) ; the axes being AB & a perpendicular diameter, prove that the locus of the focus of the parabola is the standard ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ where $b^2= a^2– c^2$.
Doubt:
I am not being able to frame to equation of the variable parabola. I feel that any parabola can have a double ordinate whose length is equal to $2c$, which can pass through the ends of diameter of the smaller circle. However, one catch is that the directrix must be a tangent to the bigger circle. But how to write the equation of such a variable parabola, whose locus of focus I need to find ?
|
We don't need to know the equation of the parabola.
We may assume that $A(c,0),B(-c,0)$.
Let $F(X,Y)$ be the coordinates of the focus of the parabola.
The equation of the directrix is given by $mx+ny=a^2$ where $P(m,n)$ is the tangent point such that $m^2+n^2=a^2$.
Now, by the definition of the parabola, the distance from the focus to $(\pm c,0)$ is equal to the distance from $(\pm c,0)$ to the directrix.
So, we get
$$\sqrt{(X-c)^2+(Y-0)^2}=\frac{|mc+n\cdot 0-a^2|}{\sqrt{m^2+n^2}},$$
i.e.
$$a^2(X-c)^2+a^2Y^2=m^2c^2-2mca^2+a^4\tag1$$
Also,
$$\sqrt{(X+c)^2+(Y-0)^2}=\frac{|m(-c)+n\cdot 0-a^2|}{\sqrt{m^2+n^2}},$$
i.e.
$$a^2(X+c)^2+a^2Y^2=m^2c^2+2mca^2+a^4\tag2$$
From $(1)-(2)$,
$$a^2(-4cX)=-4mca^2\implies X=m$$
From $(1)+(2)$,
$$a^2X^2+a^2c^2+a^2Y^2=m^2c^2+a^4\implies a^2X^2+a^2c^2+a^2Y^2=X^2c^2+a^4$$
which can be written as
$$\frac{X^2}{a^2}+\frac{Y^2}{a^2-c^2}=1$$
|
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|
The sum of 50 terms of the series $\frac{3}{1} $+ $\frac{5}{1+2^2} $+ $\frac{7}{1+2^2+3^2}$+ $\frac{11}{1+2^2+3^2+4^2}$... The numerator is increasing by 2.
The number of terms in denominator is increasing, and they are squares of natural numbers (i.e. $1^2, 2^2, 3^2, 4^2...$)
Options are:
(a)$\frac {100}{17}$
(b)$\frac {150}{17}$
(c)$\frac {200}{51}$
(d)$\frac {50}{17}$
|
Start by simplifying the denominator of the $n^{th}$ term.
$$U_n=\frac{2n+1}{\sum_{i=1}^ni^2}$$
$$=\frac{6(2n+1)}{n(n+1)(2n+1)}$$
$$=\frac{6}{n(n+1)}$$
$$=\frac{6}{n}-\frac{6}{n+1}$$
So the sum of the first 50 terms will be:
$$S_{50}=U_1+U_2+U_3+\cdots+U_{48}+U_{49}+U_{50}$$
$$=\frac{6}{1}-\frac{6}{2}+\frac{6}{2}-\frac{6}{3}+\frac{6}{3}-\frac{6}{4}+\cdots+\frac{6}{48}-\frac{6}{49}+\frac{6}{49}-\frac{6}{50}+\frac{6}{50}-\frac{6}{51}$$
$$=6-\frac{6}{51}$$
$$=\frac{100}{17}$$
|
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|
How to find condition on the constant term of a cubic equation with no square term such that the cubic equation has atleast two integer roots? The number of integers $k$ for which the equation $x^3-27x+k=0$ has at least two distict integer roots is ?
a. 1 b. 2 c. 3 d. 4
My attempt is to differentiate this cubic, I get a quadratic with real and distinct roots that is this cubic has a maximum and minimum, hence it cuts the $x$ axis always at three distinct points for any $k$.
Now using Vieta's formulas I conclude that if two roots of the cubic have to be integers, k as well as the third root should be an integer ?
Also there will be one root between $-3$ and $3$ i.e. points of maximum and minimum. Hence taking integer roots between them i.e. $1,2,-1,-2,0$, I get that for none of it I get integer roots for the remaining cubic.
Hence no $k$ exists, i.e. $No.\; of\; k = 0$ which is incorrect.
Where am I wrong or is there a simpler way to find ? Please use basic techniques of algebra to solve not with the help of very advanced topics.
|
In another way,
you already found that
$$
\begin{gathered}
x_{\,1} ,\,x_{\,2} ,\,x_{\,3} ,k\;\text{integers} \hfill \\
x_{\,1} \leqslant - 3,\quad - 3 \leqslant x_{\,2} \, \leqslant 3,\quad 3 \leqslant x_{\,3} \hfill \\
\end{gathered}
$$
where we include the $=$ sign since the roots may coincide at the abscissa of max or min.
Besides that, for the values of the max and min we shall have that
$$
\left\{ \begin{gathered}
0 \leqslant p( - 3) = \max = 54 + k \hfill \\
p(3) = \min = - 54 + k \leqslant 0 \hfill \\
\end{gathered} \right.\quad \Rightarrow - 54 \leqslant k \leqslant 54
$$
Then, Vieta's formulas tell us that
$$
\left\{ \begin{gathered}
- x_{\,1} x_{\,2} x_{\,3} = k \hfill \\
x_{\,1} + x_{\,2} + x_{\,3} = 0 \hfill \\
x_{\,1} x_{\,2} + x_{\,1} x_{\,3} + x_{\,2} x_{\,3} = - k\left( {\frac{1}
{{x_{\,1} }} + \frac{1}
{{x_{\,2} }} + \frac{1}
{{x_{\,3} }}} \right) = - 27 \hfill \\
\end{gathered} \right.
$$
wherefrom
$$
0 = \left( {x_{\,1} + x_{\,2} + x_{\,3} } \right)^{\,2} = x_{\,1} ^{\,2} + x_{\,2} ^{\,2} + x_{\,3} ^{\,2} + 2\left( {x_{\,1} x_{\,2} + x_{\,1} x_{\,3} + x_{\,2} x_{\,3} } \right)
$$
which implies:
$$
x_{\,1} ^{\,2} + x_{\,2} ^{\,2} + x_{\,3} ^{\,2} = 54\quad \Rightarrow \quad x_{\,3} ^{\,2} ,x_{\,1} ^{\,2} \leqslant 54
$$
Combining the above with previous equality for the sum, we get the ellipse
$$
\begin{gathered}
\left\{ \begin{gathered}
x_{\,1} + x_{\,2} + x_{\,3} = 0 \hfill \\
x_{\,1} ^{\,2} + x_{\,2} ^{\,2} + x_{\,3} ^{\,2} = 54 \hfill \\
\end{gathered} \right.\quad \Rightarrow \quad x_{\,1} ^{\,2} + \left( { - \left( {x_{\,1} + x_{\,3} } \right)} \right)^{\,2} + x_{\,3} ^{\,2} = 54\quad \Rightarrow \hfill \\
\Rightarrow \quad x_{\,1} ^{\,2} + x_{\,3} ^{\,2} + x_{\,1} x_{\,3} = 27\quad \Rightarrow \quad \frac{{\left( {x_{\,3} + x_{\,1} } \right)}}
{{6^{\,2} }}^{\,2} + \frac{{\left( {x_{\,3} - x_{\,1} } \right)}}
{{\left( {6\sqrt 3 } \right)^{\,2} }}^{\,2} = 1\quad \Rightarrow \hfill \\
\Rightarrow \quad \frac{{x_{\,2} ^{\,2} }}
{{6^{\,2} }} + \frac{{\left( {x_{\,3} - x_{\,1} } \right)}}
{{\left( {6\sqrt 3 } \right)^{\,2} }}^{\,2} = 1 \hfill \\
\end{gathered}
$$
finally reaching to:
$$
\left( {x_{\,3} - x_{\,1} } \right) = 6\sqrt 3 \sqrt {1 - \frac{{x_{\,2} ^{\,2} }}
{{6^{\,2} }}} \quad \left| {\; - 3 \leqslant x_{\,2} \, \leqslant 3} \right.
$$
and we get that the only possible solutions are:
$$
\left\{ \begin{gathered}
x_{\,2} = \pm 3 \hfill \\
\left( {x_{\,3} - x_{\,1} } \right) = 9 \hfill \\
k = \pm 54 \hfill \\
\end{gathered} \right.
$$
|
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|
Evaluate the integral $\int \frac{dx}{1+x^5}$
$$\int \frac{dx}{1+x^5}$$
I have tried to add a $x^5$ and subtract $x^5$, but got nothing.
|
Hint. One may start with
$$
x^5+1=(x+1)\left(x^2-\frac{\sqrt{5}+1}{2} x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2} x+1\right)
$$ then one may obtain a partial fraction decomposition,
$$
\frac1{x^5+1}=\frac{a_0}{x+1}+\frac{a_1x+b_1}{x^2-\frac{\sqrt{5}+1}{2} x+1}+\frac{a_2x+b_2}{x^2+\frac{\sqrt{5}-1}{2} x+1}
$$ and integrate classically each term.
|
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|
solve $x^2 - 25 xy + y^2 = 1$ does it have a solution? Usually the Pell equation is written $x^2 - dy^2 = 1$ but here I am looking for solutions to an equation of the type:
$$ x^2 - k xy + y^2 = 1 $$
and In particular, $k$ is a perfect square. So I am picking $k = 25$ an example.
If we complete the square then $25/2$ is not an integer.
$$ xy + x^2 - 26xy + y^2 = x^2 + xy + \big(x - 13y\big)^2 = 170$$
I think I am better of solving the orginal problem. Can any variant on the Pell equation work?
|
Your discriminant, for $x^2 - k x y + y^2,$ is
$$ \Delta = k^2 - 4. $$ The smallest solution, in positive integers, to
$$ \tau^2 - \Delta \sigma^2 = 4 $$
is $\tau = k, \sigma = 1.$ You have the obvious $-1$ automorphism given by interchanging $x,y.$ The matrix generating the oriented automorphisms is
$$
\left(
\begin{array}{rr}
\frac{\tau - B \sigma}{2} & - C \sigma \\
A \sigma & \frac{\tau + B \sigma}{2}
\end{array}
\right).
$$
With $A = 1, B = -k, C = 1$ this becomes
$$
P =
\left(
\begin{array}{rr}
k & - 1 \\
1 & 0
\end{array}
\right).
$$
$$
P^{-1} =
\left(
\begin{array}{rr}
0 & 1 \\
-1 & k
\end{array}
\right).
$$
Beginning with a column vector with entries $(1,0),$ multiplying $P$ times the column repeatedly gives (columns) $(1,0),$ $(k,1),$ $(k^2 - 1,k),$ $(k^3 - 2k,k^2 - 1),$ and so on forever.
as in
$$
\left(
\begin{array}{r}
x_{n+1} \\
y_{n+1}
\end{array}
\right)
=
\left(
\begin{array}{rr}
k & - 1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{r}
x_n \\
y_n
\end{array}
\right)
$$
From Cayley-Hamilton on the matrix $P$, we get that both $x_n$ and $y_n$ sequences obey linear recursions
$$ x_{n+2} = k x_{n+1} - x_n, $$
$$ y_{n+2} = k y_{n+1} - y_n. $$
For this particular problem, this is also evident from the matrix $P$ since the $x$ and $y$ sequences are the same, only $y_{n+1}= x_n.$
|
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|
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
$\bf{My\; Try::}$ Given $$ \alpha = \arctan \bigg(\frac{2(2\sqrt{2}-1)}{1-(2\sqrt{2}-1)^2}\bigg)=\arctan \bigg(\frac{(2\sqrt{2}-1)}{2\sqrt{2}-4}\bigg)$$
and $$\beta = \arcsin\bigg(1-\frac{4}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg) = \arcsin\bigg(\frac{23}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg)$$
So $$\arcsin\bigg[\frac{23}{27}\cdot \frac{4}{5}+\frac{3}{5}\cdot \frac{10\sqrt{2}}{23}\bigg]$$
Now how can i solve it, Help required, Thanks
|
\begin{align}
\beta&=\sqrt{10}\left(\frac{3}{\sqrt{10}}\arcsin\frac13+\frac{1}{\sqrt{10}}\arcsin\frac35\right)\\
&<\sqrt{10}\arcsin\left(\frac{4\sqrt{10}}{25}\right)
\end{align}
where I use the convexity of $\arcsin$. On the other hand using the definition of $\arctan$ we obtain $$\arcsin\left(\frac{4\sqrt{10}}{25}\right)=\arctan\left(\frac{4\sqrt{2}}{\sqrt{93}}\right).$$
Now further note that $\sqrt{10}<4$ hence we should check wether $$2\arctan\left(\frac{2\sqrt{2}}{\sqrt{93}}\right)<\arctan(2\sqrt2-1)$$
We hence (with $\tan2x=\frac{2\tan x}{1-\tan^2x}$) should prove that $$\frac{8\sqrt{186}}{61}<2\sqrt2-1.$$
|
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|
factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, as he was rushing) ended up with:
$(x^2 - 4)(x^4 + 4x^2 + 16)$
I pointed out two things:
1:
$(x^6 - 64)$ was also a difference of two perfect squares, and we could factor it into this:
$(x^3 + 8)(x^3 - 8)$
Which could then be factored further into:
$(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$
and 2:
the $(x^2 - 4)$ was also a difference of two perfect squares, and we could factor further:
$(x + 2)(x - 2)(x^4 + 4x^2 + 16)$
Now for my question:
How are $(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$ and $(x + 2)(x - 2)(x^4 + 4x^2 + 16)$ equivalent?
The teacher didn't seem to know off the top of his head, and I can't figure it out after trying for a half an hour. I've heard that you can factor the sum of two perfect squares with imaginary numbers, so maybe i can do something there to help explain this?
edit: fixed instances of "- 16" to "+ 16", and "4x" to "4x^2" (thanks for pointing that out)
|
It's a very common mistake (*) to think that, since $t^2+4t+16$ is irreducible over the reals, also the polynomial obtained with $t=x^2$ is irreducible.
Over the reals, every polynomial of degree $>2$ is reducible, so also $x^4+4x^2+16$ is reducible.
The trick here is to push in a difference of squares: if you add and subtract $4x^2$, you get
$$
x^4+8x^2+16-4x^2=(x^2+4)^2-(2x)^2=(x^2+4-2x)(x^2+4+2x)
$$
so you get the same factorization as with the other method.
Another funny example is $x^4+1$:
$$
x^4+1=x^4+2x^2+1-2x^2=
(x^2+1)^2-(\sqrt{2}\,x)^2=
(x^2+1-\sqrt{2}\,x)(x^2+1+\sqrt{2}\,x)
$$
which is the full factorization over the reals, since those degree $2$ polynomials have no real roots.
(*) Even among teachers, unfortunately.
|
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|
Expansion formula for $\tan$ We know that we can easily write the expansion of $\sin$ and $\cos$. But writing the expansion of $\tan$ becomes very difficult as coefficients of the powers of $x$ can't be found out easily. Is there a simple way of finding these coefficients, say a general formula for writing up to any number of terms.
|
If you are only concerned about the answer:
$$tanx=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\frac{62}{2835}x^9+... $$
As to the method of deriving using the expansions of $sin(x)$ and $cos(x)$, use $$cos(x) \times tan(x) = sin(x)$$
$$(1-\frac{x^2}{2!}+\frac{x^4}{4!}...) \times (a+bx+cx^2+dx^3+ex^4+fx^5...) = (x-\frac{x^3}{3!}+\frac{x^5}{5!}...)$$
You can equate coefficients from both the sides to get some initial constants.
Firstly there are no constants on the right, so $a=0$ (because $a \times 1$ in the expansion of the $LHS$ would give you a constant).
Next, $$bx\times1 = x$$ therefore $b=1$.
Further, $$cx^2\times1 + a\times(\frac{-x^2}{2!}) = 0$$ therefore $c=0$ (because we earlier found $a=0$).
And one more, $$dx^3\times1 + (bx)\times(\frac{-x^2}{2!}) = \frac{-x^3}{3!}$$ Solving this gives you $d = \frac{1}{3}$. (We found $b=1$).
And so on. Do it for as many terms as you feel necessary.
|
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|
$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.
|
For a real $x$, we have
$$f(x) = \frac{9x}{16x^2+1} + \frac{2(2x^2-1)}{16x^2+1}i $$
When $x=0$, we have $f(0) = -2i$ and when $x \rightarrow \infty$, $f(x) \rightarrow \frac{1}{4}i$. Clearly, changing $x$ to $-x$ does not change the imaginary part of $f(x)$ and hence $f(\mathbb{R})$ is symmetrical about the imaginary axis. Thus $(0,-2)$ and $(0,\frac{1}{4})$ are ends of diameters of the image circle. Hence center is $\left(0, -\frac{7}{8}\right)$ and radius is $\frac{9}{8}$.
It is easy to see that $\frac{i}{4}$ maps to the center of the circle.
A simple computation shows that
$$\left(\frac{9x}{16x^2+1}\right)^2 + \left(\frac{2(2x^2-1)}{16x^2+1}+\frac{7}{8}\right)^2 = \left(\frac{9}{8}\right)^2$$
|
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|
Proving this sequence ( probably using induction? ) Question:
For $n>0$ and $\theta\in\Re$, Let $$t_n = \sum_{k=1}^n (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
Using the identity $cos^3\phi=\frac{1}{4}cos3\phi+\frac{3}{4}cos\phi$,
Show that $$t_n = \frac{(-1)^n 3^n}{4}cos\frac{\theta}{3^n} - \frac{1}{4}cos\theta$$ and evaluate$ \lim_\limits{n \to \infty} t_n$
My attempt(using induction):
Base case(n=1) : $$t_1 = \sum_{k=1}^1 (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
$$=(-1)3^{1-1}cos^3\frac{\theta}{3}$$
$$=-(\frac{1}{4}cos\theta + \frac{3}{4}cos\frac{\theta}{3})$$
$$=\frac{(-1)^1 3^1}{4}cos\frac{\theta}{3^1}-\frac{1}{4}cos\theta$$
Assumption:
$$t_m = \frac{(-1)^m 3^m}{4}cos\frac{\theta}{3^m} - \frac{1}{4}cos\theta$$
$$t_{m+1} = \sum_{k=1}^{m+1} (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
$$=(-1)^{m+1}3^mcos^3\frac{\theta}{3^{m+1}}+ \sum_{k=1}^m (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
$$=(-1)^{m+1}3^mcos^3\frac{\theta}{3^{m+1}}+ t_m$$
Then using assumption , am I correct?
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Hint. One may observe that, for $k=1,2,3\cdots,$ we have (why?)
$$
\frac{(-1)^k 3^k}{4}\cdot\cos\frac{\theta}{3^k}-\frac{(-1)^{k-1} 3^{k-1}}{4}\cdot\cos\frac{\theta}{3^{k-1}}=(-1)^k 3^{k-1} \cdot\cos^3{\frac{\theta}{3^k}}
$$ then summing from $k=1$ to $k=n$ one may recognize a telescoping sum.
|
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|
Linear differential equations doubt There is solution below with the question, I have doubt in the last step.
Solve $$\frac{d^2 y}{dx^2} + 2y = x^2 e^{3x} - \cos 2x + $e^x$$
Solution: Auxillary equation is $D^2 + 2 = 0$
$$D=\pm i\sqrt{2}$$
Complementary function is $e^{0x}(C_1 \cos \sqrt{2} x + C_2 \sin \sqrt{2}x)$
Particular Integral is $\frac{1}{f(D)}(X)$
\begin{align}
&= \frac{1}{(D^2+2)} e^x + \frac{1}{(D^2+2)} x^2 e^{3x} - \frac{1}{(D^2+2)} \cos2x\\
&= \frac{1}{(1^2+2)} e^x + \frac{1}{(D+3)^2 + 2} x^2 e^{3x} - \frac{1}{(-2)^2+2} \cos2x
\end{align}
Value of $D$ in the last step is $D=D+3$, $D=1$, $D=-2$.
Why is it that the value of $D$ keeps on changing for every part in the equation?
Kindly help.
|
$$
u=(D^2+2)^{-1}(x^2·e^{ 3x})
$$
is just a fancy way to say that $u$ is a solution of
$$
u''+2u=x^2·e^{ 3x}
$$
which does not tell you much. You could then use a trial solution with undetermined coefficients $$u=(A+Bx+Cx^2)e^{3x}$$ to get
$$
u''+2u=e^{3x}(9(A+Bx+Cx^2)+6(B+2Cx)+2C+2(A+Bx+Cx^2))\\
=(11A+6B+2C+(11B+12C)x+11Cx^2)e^{3x}
$$
and comparing coefficients of like expressions
one gets backwards $C=\frac1{11}$, $B=-\frac{12}{11^2}$, $A=\frac{50}{11^3}$.
If one wants to get a more complicated way, write $x^2e^{3x}=\frac{∂^2}{∂^2α}e^{(3+α)x}{}_{\Big|α=0}$ so that the inversion becomes
$$
\frac{∂^2}{∂^2α}\frac{e^{(3+α)x}}{(3+α)^2+2}{}_{\Big|α=0}\\
=\frac{∂^2}{∂^2α}\left(e^{3x}(1+αx+\tfrac12(αx)^2)·\frac1{11}·(1-\tfrac{α(6+α)}{11}+\tfrac{α^2(6+α)^2}{11^2})+O(α^3)\right)_{\Big|α=0}
$$
which comes close to the indicated formalism.
|
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|
Minimal perimeter The problem is: Find the angle to OX axis, at
which a line should be drawn through the point A
(a,b) (a>0, b>0), so, that triangle, formed by this
line and positive coordinate semi-axes had the
minimal perimeter. I.e. the triangle vertices are (0,0), and two intersection points of line passing through A with OX and OY axes: (x,0), x>0 and (0,y),y>0.
I found the function appearance: the function to minimize is
$$
f(\varphi)=\left({a\over\cos \varphi}+{b\over\sin \varphi}\right)(1+\cos \varphi+\sin \varphi)
$$
and found its derivative,but I failed to solve equation derivative=0: there is 4th degree equation with respect to $\sin\varphi$, $\cos\varphi$ which I failed to solve. The answer is known, it was in the same book, but I can't come to that answer. Could somebody, please, help me?
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I got the same function $f(\varphi)$ with $0<\varphi<\frac{\pi}{2}$, so I tried to derivate and to solve the equation:
\begin{equation}
f'(\varphi)=\left(a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}\right)(1+\cos\varphi+\sin\varphi)+\left(\frac{a}{\cos\varphi}+\frac{b}{\sin\varphi}\right)(-\sin\varphi+\cos\varphi)=\\=\left(a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}\right)(1+\cos\varphi+\sin\varphi)+\left(\frac{a}{\cos\varphi}+\frac{b}{\sin\varphi}\right)(-\sin\varphi+\cos\varphi)=\\=a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}+a\frac{\sin\varphi}{\cos\varphi}-b\frac{\cos^2\varphi}{\sin^2\varphi}+a\frac{\sin^2\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin\varphi}-a\frac{\sin\varphi}{\cos\varphi}+a-b+b\frac{\cos\varphi}{\sin\varphi}\\=a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}-b\frac{\cos^2\varphi}{\sin^2\varphi}+a\frac{\sin^2\varphi}{\cos^2\varphi}+a-b=\\=\frac{a\sin^3\varphi-b\cos^3\varphi-b\cos^4\varphi+a\sin^4\varphi+a\sin^2\varphi\cos^2\varphi-b\sin^2\varphi\cos^2\varphi}{\sin^2\varphi\cos^2\varphi}=\\=\frac{a\sin^2\varphi(\sin^2\varphi+\sin\varphi+\cos^2\varphi)-b\cos^2\varphi(\cos^2\varphi+\cos\varphi+\sin^2\varphi)}{\sin^2\varphi\cos^2\varphi}=\\=\frac{a\sin^2\varphi(1+\sin\varphi)-b\cos^2\varphi(1+\cos\varphi)}{\sin^2\varphi\cos^2\varphi}
\end{equation}
Putting it equal zero:
\begin{equation}
\frac{a\sin^2\varphi(1+\sin\varphi)-b\cos^2\varphi(1+\cos\varphi)}{\sin^2\varphi\cos^2\varphi}=0\\a\sin^2\varphi(1+\sin\varphi)-b\cos^2\varphi(1+\cos\varphi)=0
\end{equation}
Now we can use $\sin\varphi=\frac{2t}{1+t^2}$ and $\cos\varphi=\frac{1-t^2}{1+t^2}$ with $t=\tan\frac{\varphi}{2}, \varphi\ne\pi+2k\pi$:
\begin{equation}
a\frac{4t^2}{(1+t^2)^2}\left(1+\frac{2t}{1+t^2}\right)-b\frac{(1-t^2)^2}{(1+t^2)^2}\left(1+\frac{1-t^2}{1+t^2}\right)=0\\4at^2\left(\frac{1+t^2+2t}{1+t^2}\right)-b(1-t^2)^2\left(\frac{1+t^2+1-t^2}{1+t^2}\right)=0\\4at^2\frac{(t+1)^2}{1+t^2}-b(1-t^2)^2\left(\frac{2}{1+t^2}\right)=0\\4at^2(t+1)^2-2b(1-t^2)^2=0\\2at^2(t+1)^2-b(1-t)^2(1+t)^2=0\\2at^2-b(1-t)^2=0\\2at^2-b(1+t^2-2t)=0\\(2a-b)t^2+2bt-b=0
\end{equation}
Solving this equation I got that the minimum is:
\begin{equation}
t=\frac{-b+\sqrt{2ab}}{2a-b}\\\tan\frac{\varphi}{2}=\frac{-b+\sqrt{2ab}}{2a-b}
\end{equation}
so $\varphi=2\arctan\frac{-b+\sqrt{2ab}}{2a-b}$.
This is my solution but if someone checked it I would be grateful; I hope it is helpful anyway.
|
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|
Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$
Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$ subject to $x,y,z>0$ and $x+y+z=1$
This inequality is equivalent to;
$\left(\frac{x+1}{4}\cdot\frac1x\right)\left(\frac{y+1}{4}\cdot\frac1y\right)\left(\frac{z+1}{4}\cdot\frac1z\right)\ge1$ taking logarithm of both sides we have to show that LHS is greater or equal to $0$, but by convexity of $\ln\left(\frac{x+1}{4}\right)-\ln(x)$, i.e. $\frac{d^2}{dx^2}\left[\ln\left(\frac{x+1}{4}\right)-\ln(x)\right]= \frac{1}{x^2}-\frac{1}{(x+1)^2}>0$ we have
$\sum\limits_{cyc}\ln\left(\frac{x+1}{4}\right)-\ln(x)\ge3\ln\left(\sum\limits_{cyc}\frac{x+1}{4}\right)-3\ln\left(\sum\limits_{cyc}x\right)=3\ln(1)-3\ln(1)=0$
Is this OK, or is there a much simpler way ? (I did the same as in example $4.1.2.$ here)
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By Holder and AM-GM $\prod\limits_{cyc}\left(1+\frac{1}{x}\right)\geq\left(1+\frac{1}{\sqrt[3]{xyz}}\right)^3\geq\left(1+\frac{1}{\frac{1}{3}}\right)^3=64$
|
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|
Please help me with this limit $$
\lim_{h\to 0} {\frac {\sqrt[3]{x+h}-\sqrt[3]{x}} h}
$$
I tried with the conjugation in the top and under but i still have the determiantion the result must be 1/3 i did it with Matlab, you CAN'T use Lhopital
|
Just multiply both side to $ \left( \sqrt [ 3 ]{ { \left( x+h \right) }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right) } +\sqrt [ 3 ]{ { x }^{ 2 } } \right) $ to get announced result $$\lim _{ h\to 0 }{ \frac { \sqrt [ 3 ]{ x+h } -\sqrt [ 3 ]{ x } }{ h } } =\lim _{ h\to 0 }{ \frac { \left( \sqrt [ 3 ]{ x+h } -\sqrt [ 3 ]{ x } \right) \left( \sqrt [ 3 ]{ { \left( x+h \right) }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right) } +\sqrt [ 3 ]{ { x }^{ 2 } } \right) }{ h\left( \sqrt [ 3 ]{ { \left( x+h \right) }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right) } +\sqrt [ 3 ]{ { x }^{ 2 } } \right) } } =\\ =\lim _{ h\to 0 }{ \frac { h }{ h\left( \sqrt [ 3 ]{ { \left( x+h \right) }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right) } +\sqrt [ 3 ]{ { x }^{ 2 } } \right) } } =\frac { 1 }{ 3\sqrt [ 3 ]{ { x }^{ 2 } } } $$
|
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|
Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e^{2x} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}
\end{align*}
I don't know how to proceed with the rightmost term. So far I've been applying the Leibniz Rule whenever I've had to find the $n$ derivative of a function of the form $f(x)g(x)$, because is clear that either $f(x)$ or $g(x)$ has a derivative a $k$ derivative ($1<k<n$) equal to zero, which simplifies the expression nicely.
But here
$$\frac{e^{2x}\cos 2x}{2}$$
both functions are infinitely differentiable on $\mathbb{R}$ which makes things a bit different.
My only attempt was to write its n derivative in this form
\begin{align*}
&=\frac{1}{2}\sum_{k=0}^n{n \choose k} \big(2^k e^{2x}\big)\bigg(2^{n-k}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\bigg)\\
&=\sum_{k=0}^n {n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]
\end{align*}
So
\begin{align*}
y^{(n)} &= 2^{n-1}e^{2x} -\sum_{k=0}^n{n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\\
&= 2^{n-1}e^{2x}\left(1 -\sum_{k=0}^n{n \choose k} \cos \left[2x + \frac{\pi(n-k)}{2}\right]\right)
\end{align*}
But the textbook's answer is
$$2^{n-1}e^{2x}\left(1 -2^{n/2}\cos \left[2x + \frac{\pi n}{4}\right]\right)$$
For some reason I have the feeling that a little of modular arithmetic has to be applied on $\frac{\pi(n-k)}{2}$
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Calculate the fist few derivatives and see if a recognizable pattern shows up.
$f(x) = \frac {e^{2x}}{2} + \frac {e^{2x}\cos2x}{2}\\
f'(x) = e^{2x} + e^{2x} \cos 2x - e^{2x}\sin 2x\\
f''(x) = 2e^{2x} - 2e^{2x}\cos 2x - 2e^{2x}\sin 2x - 2e^{2x}\sin 2x - 2e^{2x} \cos 2x = 2e^{2x} - 4e^{2x}\sin 2x\\
f'''(x) = 4e^{2x} - 8 e^{2x}\sin 2x - 8 e^{2x}\cos 2x\\
f^{(4)} = 8e^{2x} - 16\cos 2x$
Do you see a pattern? Can you prove that the pattern will continue?
If you know a little bit about complex exponential you could say:
$f(x) = \frac {e^{2x}}{2} + Re[\frac {e^{(2+2i)x}}{2}]\\
f^{(n)}(x) = 2^n\frac {e^{2x}}{2} + \frac {(2\sqrt2)^n}{2} Re[\frac {e^{(2+2i)x + \frac {n\pi}{4}i}}{4}]\\
f^{(n)}(x) = 2^n\frac {e^{2x}}{2} + \frac {(2\sqrt2)^n}{2} e^{2x} \cos(2x + \frac {n\pi}{4})\\$
|
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|
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
|
You can write
$$
f(x) = \frac{1}{2}\cdot\frac{1}{1+(x+\frac{x^2}{2})}.
$$
Setting $\stackrel{\rm def}{=} x+\frac{x^2}{2}\xrightarrow[x\to0]{}0$, you can use the Taylor expansion of $\frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^n u^n$ around $0$ to get
$$\begin{align}
f(x) &= \frac{1}{2}\left(1-u+u^2-u^3+u^4+o(u^4)\right)\\
&= \frac{1}{2}\left(1-(x+\frac{x^2}{2})+(x+\frac{x^2}{2})^2-(x+\frac{x^2}{2})^3+(x+\frac{x^2}{2})^4+o(x^4)\right)\\
&= \frac{1}{2}\left(1-x-\frac{x^2}{2}+(x^2+x^3+\frac{x^4}{4})-(x^3+\frac{3}{2}x^4)+x^4+o(x^4)\right)
\end{align}$$
(where when we expand we neglect all terms $x^k$ for $k>4$), which gives
$$\begin{align}
f(x) &= \frac{1}{2}\left(1-x+\frac{x^2}{2}-\frac{1}{4}x^4+o(x^4)\right)
\end{align}$$
i.e., the result you found. Note that with this technique, going to order say $x^9$ is much simpler than by differentiating repeatedly 9 times.
|
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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.