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Limit proof, finding upper bounds I need to prove that $\frac{1}{x-3} \to 1$ as $x \to 4$
So I did $\left | \frac{1}{x-3} - 1 \right | = \left | \frac{4 - x}{x-3} \right | = \left | \frac{x-4}{x-3} \right | = \frac{|x-4|}{|x-3|} < K|x-4|$. So I need to pick $| x - 4| < \delta = \frac{\epsilon}{K}$
Now the problem is that I can't bound my $\frac{1}{|x-3|}$
|
Let $\epsilon>0$. Suppose $|x-4|<1/2$. Then $-1/2<x-4<1/2$ which implies that $1/2<x-3<3/2$. Hence, $|x-3|\ge x-3>1/2$, that is, if $x\ne 3$ then $\frac{1}{|x-3|}<2$. Define $\delta=\min\left\{\frac{1}{2},\frac{\epsilon}{2}\right\}$. Let $0<|x-4|<\delta$. Then certainly $x\ne 3$. Hence,
$\left| \frac{1}{x-3}-1\right|=\left| \frac{x-4}{x-3}\right|<|x-4|\cdot 2 < \frac{\epsilon}{2}\cdot 2=\epsilon$.
The result follows.
|
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|
Using formula of sum of two squares ... create Pythagorean triplets from pairs of primitive Pythagorean triplets? Can you use a formula for writing products of two sums of squares as a sum of squares to create Pythagorean triplets from pairs of primitive Pythagorean triplets?
First, the formula for writing products of two sums of squares as a sum of squares is:
$$(a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2$$
Attempt:
A Pythagorean triplet is: $a^2+b^2=c^2$.
Two examples of a Pythagorean triplet is $(3,4,5)$ and $(5,12,13)$.
Where $3^2+4^2=5^2$ and $5^2+12^2=13^2$.
Now, using the previously mentioned formula:
$$(3^2+4^2)(5^2+12^2)=(3 \cdot 12 - 4 \cdot 5)^2+(3 \cdot 5+4 \cdot 12)^2=5^2\cdot 13^2=4225$$
So, $16^2+63^2=65^2$. Therefore the answer is yes. Is this sufficient? I can't help but feel I have overlooked something. The question thankfully doesn't say "prove", but maybe a "proof" is what I am looking for. Can anyone help me with this?
|
Let the following be your Pythagorean triplets.
$$a_1^2 +b_1^2 = c_1^2$$
$$a_2^2 +b_2^2 = c_2^2$$
We have (from the formula):
$$(a_1^2 +b_1^2)\cdot(a_2^2 +b_2^2)=(a_1\cdot b_2-a_2\cdot b_1)^2+(a_1\cdot a_2-b_1\cdot b_2)^2$$
Obviously the RHS is the sum of two squares. To prove this is a Pythagorean triplet, we need to show the LHS is a perfect square. We know that the sum $a_1^2 +b_1^2 = c_1^2$, and similar for the other factor. Thus, we have:
$$(c_1^2)\cdot(c_2^2)=(a_1\cdot b_2-a_2\cdot b_1)^2+(a_1\cdot a_2-b_1\cdot b_2)^2$$
$$(c_1\cdot c_2)^2=(a_1\cdot b_2-a_2\cdot b_1)^2+(a_1\cdot a_2-b_1\cdot b_2)^2$$
Which is a Pythagorean triple.
|
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|
Sum of special series: $1/(1\cdot2) + 1/(2\cdot3 )+ 1/(3\cdot4)+\cdots$ The question is:
Find the sum of the series $$ 1/(1\cdot 2) + 1/(2\cdot3)+ 1/(3\cdot4)+\cdots$$
I tried to solve the answer and got the $n$-th term as $1/n(n+1)$. Then I tried to calculate $\sum 1/(n^2+n)$. Can you help me?
|
Find the sum of the series $ \frac{1}{1\cdot 2} + \frac{1}{2\cdot3}+ \frac{1}{3\cdot4}+\cdots +\frac{1}{n(n+1)}$
You have a series called Telescoping series as M. Strochyk pointed out in the comment. Its $i^{th}$ term is $\frac{1}{i(i+1)}$ and $n^{th}$ term is $\frac{1}{n(n+1)}$ as shown in the series. Thus,
$$
\sum_{i=1}^n \frac 1{i(i+1)} = \sum_{i=1}^n \left( \frac 1i - \frac 1{i+1} \right) \\
= \sum_{i=1}^n \frac 1i - \sum_{i=1}^n \frac 1{i+1} \\
=\frac{1}{1} + \frac{1}{2}+ \frac{1}{3}+\cdots +\frac{1}{n} -\left(\frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+\cdots +\frac{1}{(n+1)}\right)\\
= 1 + \left(\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{1}{3} - \frac{1}{3}\right) + \left(\frac{1}{4} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} -\frac{1}{n}\right) - \frac{1}{(n+1)}$$
$$\therefore \ \sum_{i=1}^n \frac 1{i(i+1)} = 1 - \frac 1{n+1}$$
Therefore, if $n \to \infty$, the series converges to $1$ as
$$
\lim_{n \to \infty} \left(1 - \frac 1{n+1}\right) \to 1.
$$
|
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|
A question on complex numbers We are given
If $\cos(a+ib)$=$r (\cos\theta +i\sin\theta)$
then prove that $e^{2b} = \sin(a-\theta)/\sin(a+\theta)$
I just tried and got $b = 0$ such that $\cos(a) = ra$. Will there be other solutions?
|
$\displaystyle \cos (a+ib) = \cos a \cos ib - \sin a \sin ib = \cos a \cosh b - i \sin a \sinh b = r (\cos \theta + i \sin \theta) \\
\displaystyle r \cos \theta = \cos a \cosh b = \cos a \frac {e^{2b} + 1}{2e^b}\\
\displaystyle r \sin \theta = -\sin a \sinh b = -\sin a \frac {e^{2b} - 1}{2e^b} \\
\displaystyle \tan \theta = -\tan a \frac {e^{2b}-1}{e^{2b}+1} \\
\displaystyle \tan \theta (e^{2b}+1) = -\tan a(e^{2b}-1)\\
\displaystyle e^{2b}(\tan a + \tan \theta) = \tan a - \tan \theta \\
\displaystyle e^{2b} = \frac {\tan a - \tan \theta}{\tan a + \tan \theta} = \frac{\frac {\sin (a - \theta)}{\cos a \cos \theta}}{\frac{\sin(a + \theta)}{\cos a \cos \theta}} = \frac{\sin(a - \theta)}{\sin (a + \theta)}$
|
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|
If $(a,b)=1$ then prove $(a+b, ab)=1$.
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.
$(a,b)=1$ means $a$ and $b$ have no prime factors in common
$ab$ is simply the product of factors of $a$ and factors of $b$.
Let's say $k\mid a+b$ where $k$ is some factor of $a$.
Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.
So $a(k-l)=b, \ a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].
But $(a,b)=1$ so $a$ cannot divide $b$.
We have a similar argument for $b$.
So $a+b$ is not divisible by any factors of $ab$.
Therefore, $(a+b, ab)=1$.
Would this be correct? Am I missing anything?
|
HINT: Suppose that $p$ is a prime that divides $ab$ and $a^2+b^2$. Then $p$ divides both $a^2+2ab+b^2=(a+b)^2$ and $a^2-2ab+b^2=(a-b)^2$. This in turn implies that $p$ divides both $a+b$ and $a-b$. (Why?) Use this to show that $p$ divides both $a$ and $b$.
|
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Proving a number defined by a sequence is a square number I found this problem in a math magazine:
Given the sequence $(x_n)_{n \in \mathbb{N}}$ defined by:
$$
x_0 = 0\\
x_1 = 1\\
x_{n+2}+x_{n+1}+2x_{n}=0
$$
Prove that $s_n = 2^{n+1}-7x_{n-1}^2, n > 0$ is a square number.
I tried searching a rule between the numbers of the sequence and the square numbers ($s_n$) formed:
$$
x_2=-1\\
x_3=-1\\
x_4=3\\
x_5=-1\\
x_6=-5\\
x_7=7\\
x_8=3\\
x_9=-17\\
$$
$$
s_1 = 2^2\\
s_2 = 1^2\\
s_3 = 3^2\\
s_4 = 5^2\\
s_5 = 1^2\\
s_6 = 11^2\\
s_7 = 9^2\\
$$
If I rewrite the rule and square it:
$$
x_{n-1} = -x_{n-2}-2x_{n-3}, n > 3\\
x_{n-1}^2=x_{n-2}^2 + 4x_{n-3}^2 + 4x_{n-2}x_{n-3}\\
$$
apply for the next two, $x_{n-2}$, $x_{n-3}$ the same rule:
$$
x_{n-2}^2=x_{n-3}^2 + 4x_{n-4}^2 + 4x_{n-3}x_{n-4}\\
x_{n-3}^2=x_{n-4}^2 + 4x_{n-5}^2 + 4x_{n-4}x_{n-5}\\
$$
substitute:
$$
x_{n-1}^2=x_{n-3}^2 + 4x_{n-4}^2 + 4x_{n-3}x_{n-4}+4x_{n-3}^2 + 4x_{n-2}x_{n-3}\\
x_{n-1}^2=5x_{n-3}^2 + 4x_{n-4}^2 + 4x_{n-2}x_{n-3} + 4x_{n-3}x_{n-4} \\
x_{n-1}^2=5(x_{n-4}^2 + 4x_{n-5}^2 + 4x_{n-4}x_{n-5}) + 4x_{n-4}^2 + 4x_{n-2}x_{n-3} + 4x_{n-3}x_{n-4}\\
x_{n-1}^2=9x_{n-4}^2 + 20x_{n-5}^2 + 4x_{n-2}x_{n-3} + 4x_{n-3}x_{n-4} + 20x_{n-4}x_{n-5}\\
$$
It seems to lead to a dead end.
A help would be really appreciated.
|
This may be a little brute force, but it's a solution:
If you're familiar with solving linear recurrences, you'll know that you can write a formula for the $x_n$ of the form $$x_n = A\lambda_+^n + B\lambda_-^n,$$ where $A$ and $B$ are constants and the $\lambda_\pm$ are the roots of the polynomial $x^2 + x + 2$, that is, $$\lambda_\pm = \frac{-1 \pm i\sqrt{7}}{2}.$$ Using the initial conditions $x_0 = 0$ and $x_1 = 1$, you can solve for the constants $A$ and $B$:$$ A = -B = \frac{-i}{\sqrt{7}}.$$ This then gives the explicit formula for the $x_n$:$$ x_n = \frac{-i}{\sqrt{7}}(\lambda_+^n - \lambda_-^n).$$ Now one has $$s_n = 2^{n+1} - 7x_{n-1}^2 = 2^{n+1} + (\lambda_+^{n-1} - \lambda_-^{n-1})^2 = 2^{n+1} + \lambda_+^{2(n-1)} + \lambda_-^{2(n-1)} - 2(\lambda_+\lambda_-)^{n-1}.$$ Using the fact that $\lambda_+\lambda_- = 2$, this becomes $$s_n = \lambda_+^{2(n-1)} + \lambda_-^{2(n-1)} + 2^n.$$ But then we can use $\lambda_+\lambda_- = 2$ again to write $$s_n = \lambda_+^{2(n-1)} + \lambda_-^{2(n-1)} + 2(\lambda_+\lambda_-)^{n-1} = (\lambda_+^{n-1} + \lambda_-^{n-1})^2.$$ Now it only remains to show that $\lambda_+^{n-1} + \lambda_-^{n-1}$ is an integer. Define $\alpha_n = \lambda_+^n + \lambda_-^n$ for each $n$. Then $\alpha_0 = 2$, $\alpha_1 = -1$, and $\alpha_{n+2} + \alpha_{n+1} + 2\alpha_n = 0$ for each $n\geq 2$, so $\alpha_n$ is an integer for all $n$. This completes the proof.
|
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|
show that if $a,b,c \in \mathbb{R}^+$ show that if $a,b,c \in \mathbb{R}^+$ different from zero, then:
$$(a^2+b^2+c^2)\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq(a+b+c)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
I had no success in my attempts
|
$a=2012,b=c=1$.
Any other large number should work instead of $2012$.
If the inequality is reversed, just multiply and prove the following simple Lemma:
Lemma $f(x) =x+ \frac{1}{x}$ is increasing on $[1, \infty)$. In particular, for all $x\geq 1$ we have
$$f(x^2) \geq f(x) \,.$$
P.S. Maybe even simpler
$$\frac{a}{b}+\frac{b}{a} \geq 2 \Rightarrow \frac{a^2}{b^2}+\frac{b^2}{a^2}-\frac{a}{b}+\frac{b}{a}=(\frac{a}{b}+\frac{b}{a})^2-2-(\frac{a}{b}+\frac{b}{a}) \geq 2(\frac{a}{b}+\frac{b}{a} )-2-(\frac{a}{b}+\frac{b}{a}) \geq 0 $$
|
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How can I solve quadratic equations using modular arithmetic How can I solve quadratic equations using modular arithmetic? E.g.
$$2x^2 + 8x + 2 = 0 \pmod{23}$$
N.b. I have changed the figures from those in my homework question because I don't want a solution I want to understand the process. Consequently the example I gave might not have solutions. For the example I am working from divide the LHS by 2.
|
We have $2x^2+8x+2\equiv 0\pmod{23}$ if and only if $x^2+4x+1\equiv 0\pmod{23}$.
Now complete the square. We have $x^2+4x+1=(x+2)^2-3$. So we want to solve the congruence $(x+2)^2\equiv 3\pmod{23}$.
Let $y=x+2$. We want to solve the congruence $y^2\equiv 3\pmod{23}$.
There is general theory that, for large $p$, helps us determine whether a congruence $y^2\equiv a \pmod{p}$ has a solution, and even to compute a solution. But at this stage you are probably expected to solve such things by inspection.
Note that $y\equiv 7\pmod{23}$ works, and therefore $y\equiv -7\equiv 16\pmod{23}$ also works. We have found two solutions, and by general theory if $p\gt 2$ there are either $2$ solutions or none, so we have found all the solutions.
From $y\equiv 7\pmod{3}$ we conclude that $x+2\equiv 7\pmod{23}$, and therefore $x\equiv 5\pmod{23}$.
From $y\equiv 16\pmod{23}$ we conclude that $x\equiv 14\pmod{23}$.
Remarks: If our congruence had been (for example) $x^2+7x-8\equiv 0\pmod{23}$, there would be a bit of unpleasantness in completing the square, since $7$ is odd. But we could replace $7$ by, say, $30$, and complete the square to get $(x+15)^2-225-8$. So our congruence would become $(x+15)^2\equiv 233\pmod {23}$, or equivalently $(x+15)^2\equiv 3\pmod {23}$.
In general, if we have $ax^2+bx+c\equiv 0\pmod{p}$, it is useful to multiply through by the inverse of $a$ modulo $p$ to make the lead coefficient equal to $1$. There are a number of other helpful "tricks."
|
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|
Fourier transform of $f(x)=\frac{1}{x^2+6x+13}$ How to find the Fourier transform of the following function:
$$f(x)=\frac{1}{x^2+6x+13}$$
|
This begs for completing the square:
$$
x^2 + 6x + 13 = \Big( x^2 + 6x + 9 \Big) + 4 = (x+3)^2 + 4.
$$
Then
$$
\frac{1}{(x+3)^2 + 4} = \frac 1 4 \cdot \frac{1}{\left(\dfrac{(x+3)^2}{4}\right) + 1} =\frac 1 4 \cdot \frac{1}{\left(\frac{x+3}{2}\right)^2+1} = \frac 1 4 \cdot \frac{1}{w^2 + 1}
$$
The Laplace transform involves an integral with respect to $x$. So
$$
w = \frac{x+3}{2}
$$
$$
dw = \frac{dx}{2}
$$
$$
dx = 2\,dw
$$
If you can find $\displaystyle \int_0^\infty \frac{1}{1+w^2} e^{-tw}\,dw$, you should be able to get what you're looking for.
If you'd had a number smaller than $9$ where $13$ appeared, then things would be different. For example:
$$
x^2 + 6x + 5 = \Big(x^2+6x+0\Big) - 4 = (x+3)^2 - 2^2 = (x+3+2)(x+3-2)
$$
$$
=(x+5)(x+1).
$$
Then
$$
\frac{1}{x^2+6x+5} = \frac{1}{(x+5)(x+1)} = \frac{A}{x+5}+\frac{B}{x+1},
$$
and you'd have to find $A$ and $B$.
|
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How to solve system of equations? I want to sove the system of equations $$\begin{cases}
x^3 y-y^4=7,\\
x^2 y+2 xy^2+y^3=9.
\end{cases}
$$
I tried divide these two equations we obtain
$$\dfrac{x^3 - y^3}{(x+y)^2 } = \dfrac{7}{9}$$
From here, I don't know how to solve.
|
We'll prove that $(2,1)$ is the only real solution. First note that if $y = 0$, the second equation is unsolvable, so $y\neq 0$.
Solving the second equation for $x$ via the quadratic formula gives \begin{align*} x &= \frac{-2y^2 \pm \sqrt{4y^4-4y(y^3-9)}}{2y}\\ &= \frac{-y^2\pm 3\sqrt{y}}{y} \\ &= -y \pm y^{-\frac{1}{2}}.\end{align*}
In particular, $y \geq 0$. Subbing this into the first equation gives \begin{align*}7 &= \left(-y \pm 3y^{-\frac{1}{2}} \right)^3 y - y^4 \\ &= -2y^4\pm 9y^{\frac{5}{2}}-27y\pm 27y^{-\frac{1}{2}}.\end{align*}
This is equivalent to $$ 0 = -2y^{\frac{9}{2}} \pm 9y^{\frac{6}{2}}-27y^{\frac{3}{2}} -7y^\frac{1}{2} \pm 27.$$
Now, substitute in $z = \sqrt{y}$, which is possible since we already know $y\geq 0$. This gives $$0 = -2z^9 \pm 9z^6 - 27z^3 -7z \pm 27$$
If we choose the $-$ sign, then clearly there is no solution with $z$ real, so we must have $$0 = -2z^9 + 9z^6 - 27z^3-7z+27.$$
Now, using the rational roots theorem, maple, direct inspection, etc, we see $z=1$ is a solution.
I claim that $f(z)-2z^9 + 9z^6 - 27z^3 - 7z + 27$ is always decreasing, so $z=1$ is the only solution.
To see this, compute the derivative, getting $f'(z)=-18z^8 + 54z^5-81z^2 - 7 = 9(-2z^8 + 6z^5-9z^2 - \frac{7}{9})$. I claim this is always negative. The $9$ out in front doesn't effect this, so we'll ignore it in intermediate computations.
Why is $f'(z)$ always negative? Lets find its maximum. So, take another derivative, getting $f''(z)-16z^7+30z^4-18z = 2z(-8z^6 + 15z^3 - 9)$. Of course, this is $0$ when $z=0$, but the other part is quadratic in $z^3$ and so we see that it's never $0$. Since $f'(z)$ goes to $-\infty$ as $x\rightarrow \pm \infty$, it follows that if it only has one critical point, this must be a max. So, the max of $f'(z)$ occurs when $z = 0$, where $f'(0) = -7 < 0$.
Since $f'(z) < 0$, $f(z)$ is always decreasing, so $z= 1$ must be its only solution.
When $z=1$, $\sqrt{y} = 1$ so $y=1$, and subbing back in shows that $x=2$.
|
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|
Four digit reversal numbers How to prove without an exhaustive checking that there are only 2 (nontrivial) four digit reversal numbers?
|
Let $A = \{2,3,4,5,6,7,8,9\}$.
Let the $4$ digit reversal number be $abcd$ i.e. $1000a + 100b + 10 c + d$, where $a \in \{1\} \cup A$, $d \in \{1,2,3,\ldots,a-1\}$ and $b,c \in \{0,1\} \cup A$. Its reversed number is $$dcba = 1000d + 100c + 10b + a$$ We want $abcd = k \times (dcba)$ where $k \in A$. This gives us
$$(1000-k)a + (100-10k)b + (10-100k)c + (1-1000k) d = 0$$
$$\color{red}{\text{First note that $10 \vert (ka-d)$. This also means that if $a$ is even, then $d$ also has to be even.}}$$
Let us call the above necessary condition in red as $\star$. As explained below, this necessary condition filters out almost all the solutions that are not possible.
*
*If $a=2$. Then $d=1$. Violates $\star$.
*If $a=3$. Then $d=1$. Hence, $k=1$ or $2$. $ka - d = 1 or 5$. Violates $\star$.
*If $a=4$. Then $d=2$. If $d=2$, then $k=2$, then $ka-d = 6$. Violates $\star$.
*If $a=5$. Then $d=1$ or $2$. If $d=1$, $k = 3$ or $4$ or $5$. But $10$ doesn't divide $ka - d$ if $d=1$. Violates $\star$. If $d=2$, $k=2$. Violates $\star$.
*If $a=6$, then $d=2$. If $d=2$, then $k=3$. Violates $\star$.
*If $a=7$, then $d=1,2,3$. If $d=1$, $k=4,5,6,7$. Violates $\star$. If $d=2$, then $k=3$. Violates $\star$.
*If $a=8$, then $d=2,4$. If $d=4$, $k=2$. Violates $\star$. If $d=2$, $k=3,4$. $k=3$ violates $\star$. However, $k=4$ satisfies $\star$. Hence, $a=8$, $d=2$ and $k=4$ is a possible candidate. Let us return to this candidate later.
*If $a=9$, then $d=1,2,3,4$. If $d=1$, $k=5,6,7,8,9$. Except $k=9$, the rest violate $\star$. We will return to the case $a=9$, $d = 1$ and $k=9$ later. If $d=2$, then $k=4$. Violates $\star$. If $d=3$, $k=3$. Violates $\star$. If $d=4$, $k=2$. Violates $\star$.
Hence, after this elimination using the necessary condition $\star$, we get only two possibilities.
*
*$a = 8$, $d=2$ and $k=4$. This means $60b - 390c + 996 \times 8 - 3999 \times 2 = 0$. Hence, $2b - 13c = 1$. This gives $b=7$ and $c=1$.
*$a = 9$, $d=1$ and $k=9$. This means $10b - 890c + 991 \times 9 - 8999 \times 1$. Hence, $b-89c = 8$. This gives us $b=8$ and $c=0$.
Hence, the only two numbers are $8712$ and $9801$.
|
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}
|
Let $\{a_{n}\}$ be any sequence of reals such that $\lim \limits_{n\rightarrow \infty} na_{n} =0$... Let $\{a_{n}\}$ be any sequence of reals such that $\lim \limits_{n\rightarrow \infty } na_{n} =0$. Prove that $$\lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} + a_{n}\right)^{n} =e$$
I think I have shown that $\lim \limits_{ n\rightarrow \infty} \{ a_{n} \}$ must be zero
and. I am thinking maybe showing the sequence above is bounded and decreasing might help. Basically I want to show it has the same limit as $\lim \limits_{n \rightarrow \infty } ( 1 + \frac{1}{n}) $ which is defined as e in my book.
|
First, let's factor out the part we know, then simplify
$$ \begin{align}
\lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} + a_{n}\right)^{n}
&= \lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} \right)^{n}
\left( \frac{1 + \frac{1}{n} + a_{n}}{1 + \frac{1}{n}} \right)^{n}
\\ &= \lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} \right)^{n}
\lim_{n \rightarrow \infty }
\left( \frac{1 + \frac{1}{n} + a_{n}}{1 + \frac{1}{n}} \right)^{n}
\\ &=e \lim_{n \rightarrow \infty } \left(1 + \frac{n a_{n}}{n + 1} \right)^{n}
\end{align}$$
What's left looks sort of like the usual limit for $e$, so let's fiddle with it
$$ \begin{align}
\cdots &=e \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n-1}
\\ &= e \lim_{n \rightarrow \infty } \left(\left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{\frac{n-1}{n}}
\\ &= e \left( \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{\lim_{n \rightarrow \infty } \frac{n-1}{n}}
\\ &= e \left( \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{1}
\\ &= e \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n}
\end{align}
$$
Better. This looks even more like the limit for $e^x$; but the part that should be $x$ is going to $0$. So let's call it $L$ and bound it:
$$ \begin{align}
L &\leq e \lim_{n \rightarrow \infty } \left(1 + \frac{x}{n} \right)^{n}
\\ &= e^{1+x}
\end{align}
$$
$$ \begin{align}
L &\geq e \lim_{n \rightarrow \infty } \left(1 - \frac{x}{n} \right)^{n}
\\ &= e^{1-x}
\end{align}
$$
for all $x > 0$. That is, if $L$ is the limit, then
$$ e^{1-x} \leq L \leq e^{1+x} $$
Now we can use the method of exhaustion (i.e. squeeze theorem):
$$ e = \lim_{x \to 0^+} e^{1-x} \leq L \leq \lim_{x \to 0^+} e^{1+x} = e$$
and therefore $L = e$.
|
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|
product of two distinct squares Is there any shorter and efficient way to find
if a number can be formed by the product of two distinct square numbers
for example
*
*36=4*9
*144=16*9
help me with an algorithm or the logic
|
You are looking for a number formed by product of two squares, so you have it written in this form: $n = a^2 b^2$, apply product of powers property, so you have: $n = a^2 b^2 = (ab)^2$, so at the end you have to check if this number is a square. If you want to control if the number is made of two factors you have to find its square root, then check if it's a prime number (as in comment $49 = 7\cdot 7$, $7$ is prime, so you will get only $1 \cdot 49$).
So taking your examples: 36 is a square because $36 = 6 \cdot 6 $, now take factors: $6 = 1 \cdot 2 \cdot 3$, so $36 = 1 \cdot 4 \cdot 9 $, at the end you multiply some of them: $36 = 4 \cdot 9$ or $36 = 1 \cdot 36$.
144 is a square because $144 = 12 \cdot 12 $, now take factors: $12 = 1 \cdot 2 \cdot 2 \cdot 3$, so $144 = 1 \cdot 4 \cdot 4 \cdot 9 $, at the end you multiply some of them: $144 = 1 \cdot 144$ or $144 = 4 \cdot 36$ or $144 = 9 \cdot 16$.
If you try with $15$ you can find any product of squars because $15$ isn't a square.
|
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|
Calculating that 13 is a Wilson Prime I'm confused by the idea of a Wilson Prime. The theorem states that $$p^2=(p-1)!+1$$
This makes sense for $5$: $$5^2=(4\times3\times2)+1$$ so $5^2=25$
But it makes no sense to me for $13$: $$13^2=4790016001$$
Clearly I am as far from a mathematician as possible. If you can help in very simple terms...
|
To prove that $13$ is a Wilson prime, you need to show that
$$12! \equiv -1 \pmod {13^2} \,.$$
To make the computations faster, observe that
$$12!= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot (2 \cdot 5)\cdot 11 \cdot (4 \cdot 3)$$
We split the product in three:
$$2 \cdot 3 \cdot 4 \cdot 7 =(2 \cdot 7)(3 \cdot 4)=(13+1)(13-1)\equiv -1 \pmod {13^2} $$
Also
$$6 \cdot 11 \cdot 8 \cdot 2 \cdot 4 = (6 \cdot 11)(8 \cdot 8)=(5 \cdot 13+1)(5 \cdot 13-1) \equiv -1 \pmod{13^2}$$
and
$$ 5 \cdot 9 \cdot \cdot 5 \cdot 3=25 \cdot 27=(2 \cdot 13-1)(2 \cdot 13+1) \equiv -1 \pmod {13^2} \,.$$
Multiplying we get
$$12! \equiv -1 \pmod{13^2} \,.$$
|
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|
Prime sum identity Let $ \Lambda(k) $ denote the von Mangoldt function:
$$
\Lambda(k) \stackrel{\text{def}}{=}
\begin{cases}
0 & \text{if $ k $ is not a prime power}, \\
\ln(p) & \text{if $ k = p^{j} $}.
\end{cases}
$$
Also, let $ \lfloor x \rfloor $ be the floor function.
Can anyone prove any of the following identities:
$$
\sum_{k=0}^{\left\lfloor (n - 1)/3 \right\rfloor} \ln(3k+1) = \sum_{k=0}^{\left\lfloor (n - 1)/3 \right\rfloor} \Lambda(3k + 1) \left\lfloor \frac{n}{9k + 3} + \frac{2}{3} \right\rfloor + \sum_{k=0}^{\left\lfloor (n - 2)/3 \right\rfloor} \Lambda(3k + 2) \left\lfloor \frac{n}{9k + 6} + \frac{1}{3} \right\rfloor.
$$
$$\sum_{k=0}^{\left\lfloor (n - 3)/4 \right\rfloor} \ln(4k+1)=\sum_{k=0}^{\left\lfloor (n - 1)/4 \right\rfloor} \Lambda(4k + 1) \left\lfloor \frac{n}{16k + 4} + \frac{1}{4} \right\rfloor + \sum_{k=0}^{\left\lfloor (n - 3)/4 \right\rfloor} \Lambda(4k + 3) \left\lfloor \frac{n}{16k +12} + \frac{3}{4} \right\rfloor.$$
$$\sum_{k=1}^{n} \ln(k)= \sum_{k=1}^{n} \Lambda(k) \left\lfloor \frac{n}{k} \right\rfloor$$
The latter ones obvious, I just thought id post it for comparison.
|
A perhaps simpler proof, using the $p$-adic expansion of $n!$:
$$\log(n!)=\sum_{p \text{ prime}}\log(p)\left(\left \lfloor \frac{n}{p} \right \rfloor+\left \lfloor \frac{n}{p^2} \right \rfloor+...\right)=\sum_{k=1}^n \Lambda(k)\left \lfloor \frac{n}{k} \right \rfloor. $$
|
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|
How to prove this inequality with the following condition? Let $x$, $y$, $z$ be three positive real numbers satisfying
\begin{equation}
x + y +z + 1 =4xyz.\tag{1}
\end{equation}
Prove that
\begin{equation}
xy + yz + zx \geqslant x + y + z.\tag{2}
\end{equation}
I don't know how to start?
|
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, the condition gives $3u+1=4w^3$, which does not depend on $v^2$.
We need to prove that $v^2\geq u$, for which it's enough to prove it for a minimal value of $v^2$.
In another hand $x$, $y$ and $z$ are positive roots of the following equation.
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$3v^2X=-X^3+3uX^2+w^3.$$
Let $f(X)=-X^3+3uX^2+w^3$.
Hence, the line $Y=3v^2X$ and graph of $f$ have three common points and $v^2$ gets a minimal value,
when the line $Y=3v^2X$ is a tangent line to the graph of $f$,
which happens for equality case of two variables.
Let $y=x$.
Hence, the condition gives $z=\frac{1}{2x-1}$, where $x>\frac{1}{2}$ and we need to prove that
$$x^2+\frac{2x}{2x-1}\geq2x+\frac{1}{2x-1}$$ or
$$x^2-2x+\frac{2x-1}{2x-1}\geq0$$ or
$$(x-1)^2\geq0.$$
Done!
|
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|
Find the standard form of the conic section $x^2-3x+4xy+y^2+21y-15=0$ Find the standard form of the conic section $x^2-3x+4xy+y^2+21y-15=0$.
I understand the approach in trying to solve these problems. But the $4xy$ is confusing me. I am not sure of where to start on this one.
UPDATE:
i have found a way of doing this with reference to Canonical form of conic section
Solution:
$x^2-3x+4xy+y^2+21y-15=(x+2y)^2-3y^2-3x+21y$
Let:
$X = x+2y$
$Y=y$
$x = X - 2Y$
Now we substitute:
$X^2-3Y^2-3X+27Y-15=0$
so it is a hyperbola?
Please check.
|
Consider the second degree terms: $x^2 + 4xy + y^2$. In general, if the second degree term is $Ax^2 + Bxy + Cy^2$, then the discriminant is $D = B^2 - 4AC$. If the conic section is non-degenerate, then
\begin{align}
D > 0 & \implies \text{Hyperbola}\\
D = 0 & \implies \text{Parabola}\\
D < 0 & \implies \text{Ellipse}
\end{align}
Hence, in your case, it can be a hyperbola or a pair of straight lines.
Now factorize the second degree term as follows.
$$x^2 + 4xy + y^2 = \underbrace{(x+(2+\sqrt{3}) y)}_{X}\underbrace{(x+(2-\sqrt{3}) y)}_{Y}$$
$$x^2 + 4xy + y^2 -3x + 21 y - 15 = XY + aX + bY - 15 =0$$
Hence, we get that $a+ b= -3$ and $(2+\sqrt{3})a+ (2-\sqrt{3})b= 21$.
This gives us $a = \dfrac32 (-1 + 3 \sqrt{3})$ and $b = \dfrac32(-1 - 3 \sqrt{3})$.
Hence, $$x^2 + 4xy + y^2 -3x + 21 y - 15 = XY + aX + bY - 15 = (X+b)(Y+a) - ab - 15 = 0$$
Hence, $$(X+b)(Y+a) = 15 - \dfrac{117}2 = -\dfrac{87}2$$
in the canonical hyperbola form.
|
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|
Inspecting the function $f(x)=-x\sqrt{1-x^2}$ We are just wrapping up the first semester calculus with drawing graphs of functions. I sometimes feel like my reasoning is a bit shady when I am doing that, so I decided to ask you people from Math.SE.
I am supposed to draw a graph (and show my working) of the function $f(x)=-x\sqrt{1-x^2}$: Below is my work, I'd be very grateful for any comments on possible loopholes in my reasoning (the results should be correct), thanks!
$$\text{1. } f(x)=-x\sqrt{1-x^2}$$
Domain: We have a square-root function, therefore we need $1-x^2\geq0$. From that we get$x\in[-1,1]$. As this is the only necessary condition, $D(f)=[-1,1]$. The function is also continuous on this interval as there is nothing that would produce a discontinuity.
Symmetry: $f(x)$ is an odd function, as $-f(x)=f(-x)$, as demonstrated here:
$$-f(x)=f(-x)$$
$$-(-x\sqrt{1-x^2})=-(-x)\sqrt{1-(-x)^2}$$
$$x\sqrt{1-x^2}=x\sqrt{1-x^2}$$
Therefore we are only interested in the interval $[0,1]$, as the function will behave symmetrically on $[-1,0]$.
x and y intercepts: $f(x)=0$ holds when $x=0,1$, those are then the x-intercepts. Thus also the y-intercept is at $(0,0)$
First derivative
$$\begin{align}
\ f'(x) &=(-x\sqrt{1-x^2})' \\
& = (-x)'\sqrt{1-x^2}+(-x)(\sqrt{1-x^2})' \\
& = -\sqrt{1-x^2}+(-x)\frac{1}{2\sqrt{1-x^2}}(-2x) \\
& = -\frac{1-x^2}{\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}} \\
& = \frac{2x^2-1}{\sqrt{1-x^2}} \\
\end{align}$$
Local minima and maxima: The equality $\frac{2x^2-1}{\sqrt{1-x^2}}=0$ yields $x=\frac{1}{\sqrt2}$. As $f(\frac{1}{\sqrt2})=-\frac{1}{2}$ and because we know the x-intercepts, we can conclude that this is the local minimum and the function is decreasing at $x\in [0,\frac{1}{\sqrt2})$ and increasing at $x\in(\frac{1}{\sqrt2},1]$. That can also be concluded from the fact that $f′≤0$ on the interval $[0,\frac{1}{\sqrt2}]$ and $f′≥0$ on the interval $[\frac{1}{\sqrt2},1]$.
Second derivative
$$\begin{align}
\ f''(x) &=(\frac{2x^2-1}{\sqrt{1-x^2}})' \\
& = \frac{(2x^2-1)'\sqrt{1-x^2}-(2x^2-1)(\sqrt{1-x^2})'}{1-x^2} \\
& = \frac{4x\sqrt{1-x^2}-(2x^2-1)\frac{-x}{\sqrt{1-x^2}}}{1-x^2} \\
& = \frac{4x\frac{1-x^2}{\sqrt{1-x^2}}-(2x^2-1)\frac{-x}{\sqrt{1-x^2}}}{1-x^2} \\
& = \frac{\frac{-2x^3+3x}{\sqrt{1-x^2}}}{1-x^2} \\
& = \frac{-2x^3+3x}{\sqrt{(1-x^2)^3}} \\
\end{align}$$
Inflection point and concavity: $f''(x)=0$ has only one result and that is $x=0$. If we look at the inequality for $x\in(0,1]$.
$$\begin{align}
\ 0&<f''(x) \\
0&<\frac{-2x^3+3x}{\sqrt{(1-x^2)^3}} \\
0&<-2x^3+3x \\
2x^3&<3x \\
2x^2&<3 \\
x^2&<3/2 \\
\end{align}$$
We can see that $x^2<3/2$ is satisfied on $x\in(0,1]$, thus the function is convex on this interval.
Conclusion: Using the symmetricity of the function, we can conclude that the function is increasing on $[-1,-\frac{1}{\sqrt2}]$ and $[\frac{1}{\sqrt2},1]$ and decreasing on $[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}]$. It has local minimum at $x=\frac{1}{\sqrt2}$ and local maximum at $x=-\frac{1}{\sqrt2}$. It is concave on $[-1,0]$ and convex on $[0,1]$, with point of inflection at $x=0$.
|
Your answer and/or analysis of the function $$f(x)=-x\sqrt{1-x^2}$$ is accurate, thorough, well-justified, and consistent with its graph: $\quad \quad f(x)=-x\sqrt{1-x^2}$.
$\quad$Source: Wolfram Alpha.
Kudos for the effort you've shown and your accurate and detailed analysis! (Just don't forget to graph the function too!)
|
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|
Causal Inverse Z-Transform of Fibonacci Say the Fibonacci sequence is defined by:
$y(n) = y(n-1) + y(n-2)$
initial conditions: $y(0)=0, y(1)=1$
I incorporate those initial conditions as:
$y(n) = y(n-1) + y(n-2) + \delta(n-1)$
I compute the z-transform as:
$Y(z) = z^{-1}Y(z) + z^{-2}Y(z) + z^{-1}$
solve for $Y(z)$:
$Y(z) = \frac{z^{-1}}{1-z^{-1}-z^{-2}}$
Now I am having the hardest time taking the causal inverse z-transform. Any tips?
|
Using partial fraction techniques and denoting the zeroes of $z^2-z-1$ by $\alpha_1$ and $\alpha_2$, we have
$$ Y(z) = \frac{z}{z^2-z-1} \implies \frac{Y(z)}{z} = \left(-\frac{1}{5}+\frac{2}{5}\alpha_1\right) \frac{1}{z-\alpha_1}+ \left(-\frac{1}{5}+\frac{2}{5}\alpha_2\right) \frac{1}{z-\alpha_2} $$
$$ \implies Y(z) = \beta_1 \frac{z}{z-\alpha_1}+\beta_2 \frac{z}{z-\alpha_2} $$
$$ \implies Y(z) = \beta_1\left(1-\frac{\alpha_1}{z}\right)^{-1}+\beta_2\left(1-\frac{\alpha_2}{z}\right)^{-1} $$
where $\beta_1 =\left(-\frac{1}{5}+\frac{2}{5}\alpha_1\right)$ and $\beta_2=\left(-\frac{1}{5}+\frac{2}{5}\alpha_2\right)$.
$$ \implies Y(z) = \beta_1 \sum_{n=0}^{\infty}\frac{\alpha_1^n}{z^n}+\beta_2 \sum_{n=0}^{\infty}\frac{\alpha_2^n}{z^n}$$
$$ \implies y(n)= \beta_1\alpha_1^n + \beta_2\alpha_2^n . $$
Note: You can use the inverse Z-transform formula which uses the contour integration
$$ y(n)=\frac{1}{2\pi i}\int_{C} z^{k-1}Y(z) dz .$$
|
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|
What is the Maximum Value? If $a, b, c, d, e$ and $f$ are non negative real numbers such that $a + b + c + d + e + f = 1$, then what is the maximum value of $ab + bc + cd + de + ef$?
|
Here's the full solution in case the comments weren't enough.
Note that $(a+c+e)(b+d+f) = (a+c+e)(1-(a+c+e)) \le \frac{1}{4},$ with equality iff $a + c + e = \frac{1}{2}.$ Expanding the first expression above gives $(ab+bc+cd+de+ef)+(ad+af+be+cf) \le \frac{1}{4}.$ Since all of our variables are non-negative, $ad + af + be + cf \ge 0,$ or $-(ad+af+be+cf) \le 0,$ which gives $ab+bc+cd+de+ef \le \frac{1}{4} - (ad+af+be+cf) \le \frac{1}{4}.$ To achieve equality, we need the following:
1) $a + c + e = \frac{1}{2},$ our original condition.
2) $\frac{1}{4} - (ad + af + be + cf) = \frac{1}{4},\, \textrm{ i.e. } ad + af + be + cf = 0.$
So we have shown that the upper bound is $\frac{1}{4},$ and that we can achieve this upper bound. For example, let $a = \frac{1}{2} = b$ and $c = d = e = f = 0.$ Another example is $a = 0,\, b = \frac{1}{3},\, c = \frac{1}{2},\, d = \frac{1}{6}, e = 0,\, f = 0.$
|
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|
show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion I am trying to show that
$$
\ln{\Big|\frac{1-x}{1+x}\Big|}
$$
belongs to $L^2({\Bbb{R}})$ but not to $L^1({\Bbb{R}})$ by using it's taylor expansion (this is the entire statement of the problem). More important to me than the solution to this particular problem is to understand the general technique I may use to do this.
Since there are singularities at $\pm1$ I am inclined to think that this needs to be done in three parts. Expanding around $0$ I find:
$$
ln\frac{1-x}{1+x} \approx -2x - \frac{2x^3}{3} - \frac{2x^5}{5} - \frac{2x^7}{7} ...
= \sum_0^{\infty}{\frac{-2}{2n-1}x^{2n-1}}
$$
Which is (please correct me if I am wrong) valid between $\pm1$ because of the singularities at $\pm1$.
With robjohn's suggestion an expansion for all x such that |x| > 1 is given by
$$
\ln{\Big|\frac{1-x}{1+x}\Big|} = \ln{\Big|\frac{\frac{1}{x}-1}{\frac{1}{x}+1}\Big|} = \ln{\Big|\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\Big|}
$$
And since $|x|>1 => |\frac{1}{x}|<1$ we are in the domain of our previous expansion therefore:
$$
ln\frac{1-x}{1+x} \approx -\frac{2}{x} - \frac{2}{3x^3} - \frac{2}{5x^5} - \frac{2}{7x^7} ...
= \sum_0^{\infty}{\frac{-2}{2n-1x^{2n-1}}}
$$
Thank you.
|
Let's begin with
$$\int_{-\infty}^{\infty} \left ( \log{\Big|\frac{1-x}{1+x}\Big|} \right )^2 dx $$
Note that the integrand is an even function. You can then split this integral up into two pieces:
$$ 2 \int_{0}^{1} \left ( \log{\Big|\frac{1-x}{1+x}\Big|} \right )^2 dx + 2 \int_{1}^{\infty} \left ( \log{\Big|\frac{x-1}{x+1}\Big|} \right )^2 dx $$
Taylor expand the integrand in $x$ in the first integral, and in $1/x$ in the second integral. The series in the first integral goes as
$$ \left ( \log{\Big|\frac{1-x}{1+x}\Big|} \right )^2 = 4 x^2 + O(x^4) $$
so we have convergence over $[0,1]$. For the second integral:
$$ \left ( \log{\Big|\frac{x-1}{x+1}\Big|} \right )^2 = \frac{4}{x^2} + O \left ( \frac{1}{x^4} \right )$$
For
$$\int_{-\infty}^{\infty} \log{\Big|\frac{1-x}{1+x}\Big|} dx $$
The series for the analogous second integral over $[1,\infty)$ goes as
$$ \log{\Big|\frac{x-1}{x+1}\Big|} = -\frac{2}{x} + O \left ( \frac{1}{x^2} \right )$$
The integral of this is divergent at $\infty$. Thus, $\log{\Big|\frac{1-x}{1+x}\Big|}$ is $L^2(\mathbb{R})$ but not $L^1(\mathbb{R})$.
|
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|
Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
|
Remember, that for any nonzero $x$, one has
$$
x+\frac{1}{x} \geq 2 \tag{1}
$$
We apply this inequality repeatedly. Put
$$
T=\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab}
$$
We have
$$
T=\frac{a^2+b^2}{abc}+\frac{c}{ab} =\frac{\sqrt{a^2+b^2}}{ab} \bigg(\frac{\sqrt{a^2+b^2}}{c}+\frac{c}{\sqrt{a^2+b^2}
}\bigg) \geq \frac{2(\sqrt{a^2+b^2})}{ab},
$$
by using (1) with $x=\frac{\sqrt{a^2+b^2}}{c}$. Then,
$$
T \geq \frac{2}{\sqrt{ab}} \sqrt{\frac{a}{b}+\frac{b}{a}} \geq \frac{2\sqrt{2}}{\sqrt{ab}}
$$
by using (1) with $x=\frac{a}{b}$. So $T \geq \sqrt{\frac{8}{ab}}$.
We see that if $ab \leq 8$, we are done. So we may assume $ab \geq 8$. By symmetry, we may also assume
$ac \geq 8, bc \geq 8$. But then
$$
1 \leq \dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \leq \frac{1}{8}+\frac{1}{8}+\frac{1}{8} =\frac{3}{8},
$$
which is impossible.
|
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|
Integral help here please? I have to solve this
$$\int \frac{dx}{(x+1)\sqrt{x^2+2x}}$$
First make the substitution $u=x+1$ and get
$$\int \frac{ds}{u\sqrt{u^2-1}}$$
Next I substitute $s=\sqrt{u^2-1}$ and I get $ds=u\,du/\sqrt{u^2-1}$.
I have integral of $1/s^2+1\dots$ I replace for $s$ and I get the integral as
$$\arctan\sqrt{u^2-1} \, .$$
The answer on my textbook is $-\arcsin[1/(x+1)]$
|
You wanted $\;s=\sqrt{u^2-1}$, so $\;s^2=u^2-1$, $\;2s\,ds=2u\,du$, and $\;s\,ds=u\,du$.
Substituting gives us:
$$
\begin{align}\int\frac{du}{u\sqrt{u^2-1}} &= \int\frac{u\,du}{u^2 \sqrt{u^2-1}} \\&= \int\frac{s\,ds}{(s^2+1)s}\\&= \int\frac{ds}{s^2+1} \\ \\&= \arctan s + C.
\end{align}
$$
Now don't forget to "back substitute": $s = \sqrt{u^2-1} = \sqrt{(x+1)^2 - 1} = \sqrt{x^2 + 2x}$
So our answer is: $$\quad\arctan(\sqrt{x^2 + 2x}) + C$$
Note that with trig substitution, and trigonometric solutions, there can be any number of solutions (constants may vary), as is evident by the number of trigonometric identities! E.g.:
$$\arctan\sqrt{x^2+2x}-\left(\arcsin\left(\frac1{x+1}\right)-\frac\pi2\right) = 0$$
$$-\arcsin\frac1{x+1}=\arctan\sqrt{x^2+2x}-\text{ constant} = \arctan \sqrt{x^2 + 2x} + C$$
To check your integration answers, you can always take the the derivative of your solutions, and if you can derive the original integrand, you are "good to go." Try differentiating each solution (your's and the text's) and both will give you the equivalent of your original integrand.
|
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|
Fibonacci Proof
Prove that: $$F_1F_2+F_2F_3+F_3F_4+\cdots+F_{2n-1}F_{2n}=F_{2n}^2$$
I set it up so:
$$F^2(2k) + F(2k+1)F(2k+2) = F^2(2k+2)$$
I've tried:
$$F(2k)^2 + F(2k+1)*F(2k+2) = F(2k+2)*F(2k)+F(2k+2)*F(2k+1)$$
$$F(2k)^2 = F(2k+2)*F(2k)$$
$$F(2k+1)*F(2k) - F(2k)*F(2k-1) = F(2k)*F(2k)+F(2k)*F(2k+1)$$
$$-F(2k)*F(2k-1) = F(2k)^2$$
??? Is there something wrong with my initial induction hypothesis ?
|
We can avoid induction using Binet's Fibonacci Number Formula, $$F_n=\frac{a^n-b^n}{a-b}$$ where $a,b$ are the roots of $x^2-x-1=0\iff a+b=1,ab=-1$ also, $a^2-a-1=0$
$$F_rF_{r+1}=\frac{(a^r-b^r)}{(a-b)}\frac{(a^{r+1}-b^{r+1})}{(a-b)}=\frac{a^{2r+1}+b^{2r+1}-(ab)^r(a+b)}{(a-b)^2}$$
$$=\frac{a^{2r+1}+b^{2r+1}-(-1)^r}{(a-b)^2} \text{ as } ab=-1$$
So, $$(a-b)^2\sum_{m\le r\le m+n}F_rF_{r+1}$$
$$=\sum_{m\le r\le m+n}a^{2r+1}+\sum_{m\le r\le m+n}b^{2r+1}-\sum_{m\le r\le m+n}(-1)^r$$
$$=\frac{a^{2m+1}\{(a^2)^{n+1}-1\}}{a^2-1}+\frac{b^{2m+1}\{(b^2)^{n+1}-1\}}{b^2-1}-(-1)^m\sum_{0\le r\le n}(-1)^r$$
$$=a^{2(m+n+1)}+b^{2(m+n+1)}-(a^{2m}+b^{2m})-(-1)^m\sum_{0\le r\le n}(-1)^r$$
as $a^2-1=a$ as $a^2-a-1=0$
$$=(a^{m+n+1}-b^{m+n+1})^2+2(ab)^{m+n+1}-\{(a^m-b^m)^2+2(ab)^m\}-(-1)^m\sum_{0\le r\le n}(-1)^r$$
$$=(a^{m+n+1}-b^{m+n+1})^2-(a^m-b^m)^2-2(-1)^m\{(-1)^{n+1}-1\}-(-1)^m\sum_{0\le r\le n}(-1)^r$$
As the sum of even number of alternative positive negative real terms with same absolute value is $0,$ if $n$ is odd $=2k-1$(say),
$$(a-b)^2\sum_{m\le r\le m+2k-1}F_rF_{r+1}=(a^{m+2k}-b^{m+2k})^2-(a^m-b^m)^2$$
$$\implies \sum_{m\le r\le m+2k-1}F_rF_{r+1}=F_{m+2k}^2-F_m^2$$
$$m=0\implies \sum_{0\le r\le 2k-1}F_rF_{r+1}=F_{2k}^2-F_0^2=F_{2k}^2 \text{ as } F_0=0$$
$$\implies \sum_{1\le r\le 2k-1}F_rF_{r+1}=F_{2k}^2-F_0F_1=F_{2k}^2 \text{ as } F_0=0$$
|
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|
Working with exponent on series Hi have this sequence:
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$
I understand that this is a Geometric series so this is what I've made to get the sum.
$$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$
$$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4})}^n$$
So $a= (-1)^n\cdot 3^{-2}$ and $r=\frac{3}{4}$ and the sum is given by
$$(-1)^n\cdot 3^{-2}\cdot \frac{1}{1-\frac{3}{4}}$$
Solving this I'm getting the result as $\frac{4}{9}$ witch I know Is incorrect because WolframAlpha is giving me another result.
So were am I making the mistake?
|
You have
$$\begin{align}
\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}
&= \sum_{n=1}^{\infty} \frac{(-1)(-1)^{n-1}\frac{1}{3}3^{n-1}}{4\cdot 4^{n-1}} \\
&= \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-1)^{n-1}3^{n-1}}{4^{n-1}}\\
&= \sum_{n=1}^{\infty} \frac{-1}{12}\left(\frac{-3}{4}\right)^{n-1}
\end{align}$$
So you have $a = \frac{-1}{12}$ and $r = \frac{-3}{4}$.
Note the key thing here that both $a$ and $r$ are constants/numbers. They do not depend on $n$. The idea is that you rewrite your series so that it is of exactly the form
$$
\sum_{n=1}^{\infty} a r^{n-1}
$$
where again $a$ and $r$ are constants/numbers.
|
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|
Ratio of areas of two triangles Based on the figure below, what is the ratio of the area of triangle $CGI$ to the area of triangle $ABC$, in terms of $\theta$?
|
Mathematica isn’t needed for the simplification, which is quite straightforward:
$$\begin{align*}
\frac{\left({\sin\theta \over \cos\theta} - \sin\theta\tan\theta\right)^2 \tan\theta}{\sin\theta\big(\cos\theta+\sin\theta\tan\theta\big)}&=\frac{\left(\tan\theta-\sin\theta\tan\theta\right)^2\tan\theta}{\frac{\sin\theta}{\cos\theta}\left(\cos^2\theta+\sin^2\theta\right)}\\\\
&=\frac{\tan^3\theta(1-\sin\theta)^2}{\tan\theta}\\\\
&=\tan^2\theta(1-\sin\theta)^2\;.
\end{align*}$$
|
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|
How to calculate $\sum_{n = 2}^{+\infty} \frac{5 + 2^n}{3^n}$? I need to calculate:
$$\sum_{n = 2}^{+\infty} \frac{5 + 2^n}{3^n}$$
Actually, I'm not particularly interested in knowing the result. What I'm really interested in is:
*
*What kind of series is it?
*How should I start calculating series like that?
|
Your sum is the sum of two geometric series:
$$\sum_{n=2}^\infty \frac{5 + 2^n}{3^n} = 5\sum_{n=2}^\infty \left(\frac{1}{3}\right)^n + \sum_{n=2}^\infty \left(\frac{2}{3}\right)^n$$
In general, $$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$$ which converges when the magnitude of $r < 1$.
In your sum of two geometric series, for the first, we have $a = 5$, $r = \dfrac 13$,
so we start first with $$5\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n = \frac{5}{1-1/3} = \frac{15}{2}$$
But your sum starts from $n=2$, not $n = 0$, so we start from the basic formula for evaluating a convergent geometric sum, and then subtract from that the value of the terms at $n = 0, n=1$ $$5\sum_{n=2}^\infty \left(\frac{1}{3}\right)^n\,.$$
Now we need to subtract from $\dfrac{15}{2}$ the evaluation of the terms when $n = 0, n = 1$:
$n = 0 \to \dfrac{5}{3^0} = 5$
$n = 1 \to \dfrac{5}{3}$
Hence, the sum of the first of the split sums evaluates to $\dfrac{15}{2} - 5 - \dfrac{5}{3} = \dfrac{5}{2} - \dfrac{5}{3} = \dfrac{5}{6}$
Similarly, you can compute the sum of the second of the split sums, using this strategy.
|
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|
Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2} $$
Change of variable
$$ t=\frac{x}{n}$$
$$ I_n = \frac{1}{2n}\int_0^n \frac{\left(1-\frac{x}{n}\right)^n \mathrm dx}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}$$
Taylor expansions:
$$ \left(1-\frac{x}{n}\right)^n = e^{-x} \left(1-\frac{x^2}{2n}+\frac{3x^4-8x^3}{24n^2}+\mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
$$ \frac{1}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)} = 1+\frac{x}{n}+\frac{x^2}{2n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) $$
$$ \frac{\left(1-\frac{x}{n}\right)^n }{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}=e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
So
$$ I_n = \frac{1}{2n} \int_0^n e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right) \mathrm dx $$
$$ I_n = \frac{1}{2n} \left(1+\frac{1}{n}+\frac{1}{2n^2}\times 2-\frac{1}{2n}\times 2 - \frac{1}{2n^2} \times 6 + \frac{1}{8n^2} \times 24 - \frac{1}{3n^2} \times 6+ \mathcal{O} \left(\frac{1}{n^3} \right) \right) $$
$$ I_n = \frac{1}{2n}-\frac{1}{2n^3}+\mathcal{O} \left(\frac{1}{n^4} \right)$$
For example Wolfram gives:
$$ 1-1000^2+ 2\times1000^3\int_0^{\pi/4} \tan(x)^{1000} \mathrm dx \approx 4.9\times 10^{-6}$$
I'm quite sure of my work, I would just like to know if everything is correct!
|
I figured I'd throw my hat into the ring as well.
We can appeal to Watson's lemma to find a full asymptotic expansion of the integral.
After your substitution $t=\tan x$, make another substitution $t = e^{-s}$. This gives
$$
\int_0^1 t^n \frac{dt}{1+t^2} = \int_0^\infty e^{-ns} \frac{ds}{e^s+e^{-s}}.
$$
By Watson's lemma we have
$$
\int_0^\infty e^{-ns} \frac{ds}{e^s+e^{-s}} \approx \frac{1}{2} \sum_{k=0}^{\infty} \frac{E_k}{n^{k+1}}
$$
as $n \to \infty$, where $E_k$ is the $k^{\text{th}}$ Euler number.
|
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|
Floor function inequality I am racking my brain trying to get this problem solved and I can't seem to break it...
Let $m, n$ be positive integers, with $m > 1$. Prove
$$\left\lfloor\frac{n}m\right\rfloor+\left\lfloor\frac{n+1}m\right\rfloor\le\left\lfloor\frac{2n}m\right\rfloor$$
I started trying to use the inequalities that say
$$\lfloor x\rfloor + \lfloor y\rfloor + \lfloor x+y\rfloor \le \lfloor 2x\rfloor + \lfloor 2y\rfloor\;.$$
From there I rewrote both sides with the notion that for a real number $y$,
$y = \lfloor y\rfloor + \{y\}$, where $\{y\}$ is the fractional part, but I keep going in circles and can not get to the desired result...
|
If you want to use you that inequality of yours let $x=\dfrac{n}{m}$ and $y=\dfrac{1}{m}:$
$$\bigg\lfloor \frac{n}{m}\bigg\rfloor+\bigg\lfloor \frac{1}{m}\bigg\rfloor+\bigg\lfloor \frac{n+1}{m}\bigg\rfloor\leq\bigg\lfloor \frac{2n}{m}\bigg\rfloor+\bigg\lfloor \frac{2}{m}\bigg\rfloor$$
Since $m>1\Rightarrow \bigg\lfloor\dfrac{1}{m}\bigg\rfloor=0$. If additionally $m>2$ then also $\bigg\lfloor\dfrac{2}{m}\bigg\rfloor=0$ so all we need to do is check for $m=2:$
$$\bigg\lfloor \frac{n}{2}\bigg\rfloor+\bigg\lfloor \frac{n+1}{2}\bigg\rfloor\leq n+1$$
Since $\displaystyle\bigg\lfloor \frac{n}{2}\bigg\rfloor+\bigg\lfloor \frac{n+1}{2}\bigg\rfloor\leq \dfrac{n}{2}+\dfrac{n+1}{2}=n+\dfrac{1}{2}<n+1$ this case is verified.
|
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|
Another Fibonacci identity: $F_{2n-1} = F_{n}^2 + F_{n-1}^2$ Here's a problem that is leading me in circles.
Consider the Fibonacci number $F_n$ defined by $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n \geq 2$.
Prove that $F_{2n-1} = F_{n}^2 + F_{n-1}^2$.
Tried an induction proof that lead me nowhere fast. So far we have only been given limited identities, and they are all basically summation formulas for the odd terms, or even terms, or all terms up to n. I also though that maybe if it was a difference, the difference of squares could factor to something, but I'm just not getting it.
|
The identity may be derived from the interesting fact that
$$\left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^k = \left ( \begin{array} \\ F_{k+1} & F_k\\F_k & F_{k-1} \\ \end{array} \right ).$$
From this, we may observe that
$$\begin{align} \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^m \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^n &= \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^{m+n} \\ &= \left ( \begin{array} \\ F_{m+1} & F_m\\F_m & F_{m-1} \\ \end{array} \right ) \left ( \begin{array} \\ F_{n+1} & F_n\\F_n & F_{n-1} \\ \end{array} \right )\\ &= \left ( \begin{array} \\ F_{m+n+1} & F_{m+n}\\F_{m+n} & F_{m+n-1} \\ \end{array} \right ) \end{align} $$
Therefore we can say that, for example,
$$F_{m+n-1} = F_m F_n + F_{m-1} F_{n-1}$$
Plug in $m=n$ and the desired identity follows.
EDIT
The first identity quoted above follows from putting the Fibonacci recurrence into matrix form:
$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) \left ( \begin{array} \\F_{k+1} \\ F_k \\ \end{array} \right )$$
which may be immediately verified. We may repeat this matrix multiplication $k$ times to get
$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\F_{2} \\ F_1 \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\1 \\ 1 \\ \end{array} \right )$$
Noting that $F_{k+2}= F_{k+1}+ F_k$ and $ F_{k+1} = F_k + F_{k-1}$, the stated identity follows.
|
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|
Baricenter of $4$ intersection points of parabola with circle lies on axis of parabola Show that the baricenter of the $4$ intersection points of a parabola with a circle is on the axis of the parabola.
Let $p$ be a parabola, $c$ a circle and $p\cap c=\{P_1,P_2,P_3,P_4\} \Rightarrow B= \frac{P_1+P_2+P_3+P_4}{4} \in$ axis of $p$.
|
If we expand a fourth-degree polynomial function $x\mapsto(x-p)(x-q)(x-r)(x-s)$, we get
$$
x^4 - (p+q+r+s)x^3 + \cdots.
$$
The sum of the roots is minus the coefficient of $x^3$.
The $x$-coordinates of the intersection of the parabola $y=x^2$ with the circle $(x-h)^2+(y-k)^2 = r^2$ are the roots of
$$
(x-h)^2 + (x^2-k)^2 = r^2.
$$
That is a fourth-degree polynomial in $x$. The coefficient of $x^3$ is $0$. Therefore the sum of the roots is $0$.
|
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Define $x_{2n}=\frac{x_{2n-1}+2x_{2x-2}}{3}\,\,\,\,\,,x_{2n+1}=\frac{2x_{2n}+x_{2n-1}}{3}$ for $n \in \mathbb{N^+}$. Let $x_0=a$ and $x_1=b$ with $b>a$. Define $$x_{2n}=\frac{x_{2n-1}+2x_{2n-2}}{3}\,\,\,\,\,,x_{2n+1}=\frac{2x_{2n}+x_{2n-1}}{3}$$ for $n \in \mathbb{N^+}$. Show that the sequence $(x_n)$ converges. my attempt is to prove that $(x_{2n})$ is increasing and $(x_{2n+1})$ is decreasing. But I don know how to prove it. Anyone can help ?
|
Closed forms for linear recurrence relations can be found using matrices and diagonalization.
Let $v_i = \begin{bmatrix}x_i & x_{i - 1}\end{bmatrix}^\intercal$. We have $v_1 = \begin{bmatrix}b & a\end{bmatrix}^\intercal$ and for any positive integer $n$,
$$\begin{align}
v_{2n} &= \begin{bmatrix}1/3 & 2/3\\1 & 0\end{bmatrix}v_{2n - 1}\\
v_{2n + 1} &= \begin{bmatrix}2/3 & 1/3\\1 & 0\end{bmatrix}v_{2n}\\
\end{align}$$
It is interesting to note that both matrices are right-stochastic (rows sum to 1).
Let $M = \begin{bmatrix}2/3 & 1/3\\1 & 0\end{bmatrix}\begin{bmatrix}1/3 & 2/3\\1 & 0\end{bmatrix} = \begin{bmatrix}5/9 & 4/9\\1/3 & 2/3\end{bmatrix}$;
$$\begin{align}
v_{2n} &= \begin{bmatrix}1/3 & 2/3\\1 & 0\end{bmatrix}M^{n - 1}\begin{bmatrix}b\\a\end{bmatrix}\\
v_{2n + 1} &= M^n\begin{bmatrix}b\\a\end{bmatrix}\\
\end{align}$$
We may diagonalize $M = \begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}2/9 & 0\\0 & 1\end{bmatrix}\begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}^{-1}$
Substitute $u_i = \begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}^{-1}v_i$ to get
$$\begin{align}
u_{2n + 1} &= \begin{bmatrix}(2/9)^n & 0\\0 & 1\end{bmatrix}u_1 = \frac{1}{7}\begin{bmatrix}3(2/9)^n(a - b)\\4a - 3b\end{bmatrix}\\
\implies \begin{bmatrix}x_{2n + 1}\\x_{2n}\end{bmatrix} = v_{2n + 1} &= \begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}3(2/9)^n(a - b)\\4a - 3b\end{bmatrix}
\end{align}$$
This should answer your questions.
|
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|
A log improper integral Evaluate :
$$\int_0^{\frac{\pi}{2}}\ln ^2\left(\cos ^2x\right)\text{d}x$$
I found it can be simplified to
$$\int_0^{\frac{\pi}{2}}4\ln ^2\left(\cos x\right)\text{d}x$$
I found the exact value in the table of integrals:
$$2\pi\left(\ln ^22+\frac{\pi ^2}{12}\right)$$
Anyone knows how to evaluate this?
|
$$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \sin }^{ 2m-1 }\left( x \right) { \cos }^{ 2n-1 }\left( x \right) dx } =B\left( m,n \right) \\ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \sin }^{ 2m-1 }\left( x \right) { \left( { \cos }^{ 2 }x \right) }^{ \frac { 2n-1 }{ 2 } }dx } =B\left( m,n \right) $$
On differentiating it twice w.r.t. to n and taking $m=\frac{1}{2}$ and $n=\frac{1}{2}$, we get
$$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\ln { \left( { \cos }^{ 2 }x \right) dx })^2 } =\frac { 1 }{ 2 } \frac { { \left( \Gamma \left( \frac { 1 }{ 2 } \right) \right) }^{ 2 } }{ 1 } \left\{ { \left( \psi \left( \frac { 1 }{ 2 } \right) -\psi \left( 1 \right) \right) }^{ 2 }+\psi '\left( \frac { 1 }{ 2 } \right) -\psi '\left( 1 \right) \right\} $$
$$\therefore \int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\ln { \left( { \cos }^{ 2 }x \right) dx })^2 } =\frac { { \pi }^{ 3 } }{ 6 } +2\pi { \left( \ln { 2 } \right) }^{ 2 }$$
|
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|
Rolling three dice...am I doing this correctly? Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have $\frac{6}{6}$. The second die needs to be equal to the first, so you have probability of $\frac{1}{6}$. Then the third die can't be equal to the first and second dice, so it's $\frac{5}{6}$.
All together you get $1 \cdot \frac{1}{6} \cdot \frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3 \cdot \left(1 \cdot\frac{1}{6} \cdot \frac{5}{6}\right)=\frac{5}{12}$$
Did I do this correctly?
Thank you.
|
One more way you can solve this is using complimentary counting. There are 6*$\dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $\frac{90}{216}=\frac{5}{12}$
|
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|
Diophantine equations - Perfect square and Perfect cube related Solve following Diophantine equations:
$1) \ a^3-a^2+8=b^2$
2) $a, \ b,\ c \in \mathbb{Z^+}$$$\frac{a^3}{(b+3)(c+3)} + \frac{b^3}{(c+3)(a+3)} + \frac{c^3}{(a+3)(b+3)} = 7$$
3) $a^3-8=b^2$
In Problem 2 I tried to use inequality, then I can 'limit' that: $25 \ge a+b+c$ and $a^3 + b^3 + c^3 \ge 112$
Please use elementary way to solve it, I haven't studied elliptic curve yet, thanks.
|
(2) Extremely ugly solution.
You have
$$a^3(a+3)+b^3(b+3)+c^3(c+3)=7(a+3)(b+3)(c+3)$$
or
$$(x-3)^3x+(y-3)^3y+(z-3)^3z=7xyz$$
Since it is symmetric, we can look for the solutions where $x \geq y \geq z$.
It is easy to check that for $x>15$ we have $(x-3)^3>7x^2$. This shows that $4 \leq x \leq 14$. For each particular $x$ you get a simpler solution which can be solved the same way.
P.S. It probably also helps observing that modulo 3 you have
$$(w-3)^3w \equiv 0,1 \pmod 3$$
Then, if none of $x,y,z$ is divisible by $3$ the LHS is 0 $\pmod 3$ which is not possible.
If one of $x,y,z$ is divisible by $3$, then all of them must be divisible by 3, and looking at the equation it follows that one of them is divisible by 9.
This should solve the equation.
|
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|
Assume $n$ is even. Prove that $323$ divides $20^n+16^n-3^n-1$. I'm unclear what is the best method to teach this with minimum math experience.
|
We have $323 = 17 \cdot 19$ with $\gcd(17, 19) = 1$. Also, $19 = 20-1 \mid 20^n - 1, 19 = 16 + 3 \mid 16^n - 3^n \Rightarrow 19 \mid 20^n+16^n-3^n-1$
and $17 = 20 - 3 \mid 20^n - 3^n$, $17 = 16+1 \mid 16^n +1 \Rightarrow 17 \mid 20^n+16^n-3^n-1$
$\gcd(17,19)=1 \Rightarrow 323 = 17 \cdot 19 \mid 20^n+16^n-3^n-1$
|
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|
2nd order implicit derivative What would the 2nd order implicit derivative of $y^2=12x$ be?
I get the first derivative is $2yy'=12$ but Wolfram gives the second as $y''=\frac{-3}{xy}$ and I don't understand how they get that.
|
$$ y^2 = 12x \Rightarrow \\ 2\cdot y \frac{dy}{dx} = 12 \Rightarrow \\ \frac{dy}{dx} = \frac{12}{2y} = \frac{6}{y} \Rightarrow \\ \frac{d^2y}{dx^2} = -\frac{6}{1}\cdot \frac{1}{y^2}\cdot \frac{dy}{dx} = -\frac{6}{1}\cdot \frac{1}{y^2}\cdot \frac{6}{y} = -\frac{36}{y^3} = -36\cdot \frac{1}{y^2\cdot y} = -36\cdot \frac{1}{12xy} = \frac{-3}{xy} $$
|
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|
Integer solution $(x,y,z)$ to $xyz=24$
Find the total number of integer solutions $(x,y,z)\in\mathbb{Z}$ of the equation $xyz=24$.
I have tried $xyz = 2^3 \cdot 3$
My Process:
Factor $x$, $y$, and $z$ as
$$
\begin{cases}
x = 2^{x_1} \cdot 3^{y_1}\\
y = 2^{x_2} \cdot 3^{y_2}\\
z = 2^{x_3} \cdot 3^{y_3}
\end{cases}
$$
Now $2^{x_1+x_2+x_3} \cdot 3^{y_1+y_2+y_3} = 2^3 \cdot 3$, so
$$
\begin{cases}
x_1+x_2+x_3 = 3\\
y_1+y_2+y_3 = 1
\end{cases}
$$
where $x_1,x_2,x_3,y_1,y_2,y_3\geq 0$. Thus the number of solutions for the first is $\binom{5}{2}=10$ and second is $\binom{3}{2}=3$. This gives a total number of positive solutions of $10 \cdot 3 = 30$
How can I calculate the total number of solutions? Is this process correct?
|
Yes, this looks right for finding the number of positive integer solutions.
Since you originally asked for all integer solutions, you'll also have to take into account the signs of the factors. There are $\binom{3}{0} + \binom{3}{2} = 4$ ways of adding zero or two minus signs, for a total of 120 integer solutions.
|
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|
Finding the value of the integral $\int_{1/2}^{1}[1/x^2]dx$ I came across the following problem that says:
For any $x \in \mathbb R$,let $[x]$ denote the greatest integer smaller than or equal to $x$. Then the value of the integral $\int_{1/2}^{1}[1/x^2]dx$ is which of the following:
1.$\frac {1}{\sqrt 3}+\frac {1}{\sqrt 2}-\frac {1}{2}$,
2.$\frac {1}{\sqrt 3}+\frac {1}{\sqrt 2}$,
3.$\frac {1}{\sqrt 3}+\frac {1}{\sqrt 2}+\frac {1}{2}$,
4.$\frac {1}{\sqrt 3}-\frac {1}{\sqrt 2}$.
I do not know how to progress with the problem. Can someone point me in the right direction? Thanks in advance for your time.
|
When $x=1/2$, $[1/x^2] = 4$; when $x=1/\sqrt{3}$, it is equal to $3$, etc. The integral is then
$$\int_{1/2}^1 dx \: [1/x^2] = -4 \left ( \frac{1}{2} - \frac{1}{\sqrt{3}} \right ) - 3 \left ( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{2}} \right ) - 2 \left ( \frac{1}{\sqrt{2}} - 1 \right )$$
or,
$$\int_{1/2}^1 dx \: [1/x^2] = \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}} $$
|
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|
Algebraically Solve Limit $$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2}$$
I can solve it using l'Hôpital but just cannot find a way to do it algebraically.
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We have $\sqrt{1+x}=1+\frac{1}{2}x+(\frac{1}{2})(\frac{-1}{2})\frac{x^2}{2}+o(x^2)$, so
$$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2}=\lim_{x \to 0}\dfrac{-\frac{1}{4}x^2+o(x^2)}{x^2}=-\frac{1}{4}$$
|
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|
Show that the difference of two consecutive cubes is never divisible by $3$. Here is my proof:
Let $n \in \Bbb Z$. Then, $n$ is of the form $2k$(even) or $2k + 1$(odd), for some $k \in \Bbb Z$.
Without loss of generality (not sure if I can use this), let $n = 2k$.
Then, $n + 1 = 2k + 1$.
$$\begin{align}
(n + 1)^3 - n^3 & = (2k + 1)^3 - (2k)^3 \\
& = 8k^3 + 12k^2 + 6k + 1 - 8k^3 \\
& = 3(4k^2 + 2k) + 1
\end{align}$$
Let $m = (n + 1)^3 - n^3$.
Then, $m = 1 + 3(4k^2 + 2k) \Rightarrow m \equiv 1 \pmod 3$.
Therefore, $\forall m$, when divided by $3$, there remains a remainder $1$. So $(n + 1)^3 - n^3$ is not divisible by $3$.
Do I also have to prove this for the other case where $n$ is odd, or is this proof sufficient?
|
$(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3 = 3(n^2 + n) + 1 \equiv 1$ (mod $3$)
|
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|
Proof by induction help. I seem to be stuck and my algebra is a little rusty Stuck on a homework question with mathematical induction, I just need some help factoring and am getting stuck.
$\displaystyle \sum_{1 \le j \le n} j^3 = \left[\frac{k(k+1)}{2}\right]^2$
The induction part is: $\displaystyle \left[\frac{k(k+1)}{2}\right]^2
+(k+1)^3$ is where I am having a problem.
If you could give me some hints as to where to go since I keep getting stuck or writing the wrong equation.
I'll get to $\displaystyle \left[{k^2+2k\over2}\right]^2 + 2{(k+1)^3\over2}$
Any push in the right direction will be appreciated.
|
You have:
$$
\begin{align*}
\left[ \frac{n (n + 1)}{2} \right]^2 + (n + 1)^3
&= \frac{n^2 (n + 1)^2 + 4 (n + 1)^2 (n + 1)}{4} \\
&= \frac{(n^2 + 4 (n + 1)) (n + 1)^2}{4} \\
&= \frac{(n^2 + 4 n + 4) (n + 1)^2}{4} \\
&= \frac{(n + 2)^2 (n + 1)^2}{4} \\
&= \left[\frac{(n + 1) (n + 2)}{2} \right]^2
\end{align*}
$$
|
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|
Solving an equation with fractional powers I was trying to find the maximum value for a function. I took the first derivative and arrived at this horrible expression:
$$ (x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
How can I find the extrema by hand?
|
$$ (x^2 + y^2)^{3/2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
$$\iff (x^2 + y^2)^{1/2}\left((x^2 + y^2 - 3y^2)\right) = 0$$
$$\iff (x^2 + y^2)^{1/2}\left(x^2 - 2y^2\right) = 0$$
$$\iff (x^2 + y^2) = 0 \;\;\text{ or }\;\;x^2 = 2y^2$$
$$\iff \quad ?$$
|
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|
A improper integral with complex parameter For a complex number $\displaystyle z$, How to evaluate
$$\int_0^\infty\frac{\text{d}x}{x^2+(1-z^2x^2)^2}$$
|
By the quadratic formula, the denominator of the integrand has roots
$$\begin{align}
r_1,r_2,r_3,r_4 &= \pm\sqrt{\frac{2z^2-1\pm\sqrt{1-4z^2}}{2z^4}}\\
&=\frac{\pm1}{2z^2}\left(i\pm\sqrt{4z^2-1}\right)
\end{align}$$
and since the integrand is an even function,
$$\begin{align}
\int_0^\infty \frac{dx}{x^2+(1-z^2x^2)^2} &= \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{x^2+(1-z^2x^2)^2} \\
&= \frac{1}{2}\int_{-\infty}^\infty \frac{1/z^4}{x^4+(1-2z^2)x^2/z^4+1/z^4} dx \\
&= \frac{1}{2z^4}\int_{-\infty}^\infty f(x) dx
\end{align}$$
where $$f(x) = \frac{1}{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}.$$ Since $f(z)$ is analytic in the upper half complex plane (except for a finite number of poles), and since $f(z)$ vanishes faster than $1/z^2$ for $|z|\rightarrow\infty$, the residue theorem gives $$\frac{1}{2z^4}\int_{-\infty}^\infty f(x)dx = \frac{\pi i}{z^4}\sum\mathrm{res\,\,}f$$ where $\sum\mathrm{res\,\,}f$ is the sum of the residues in the upper-half plane. The trick then is to determine which of the four poles are on the upper half plane. We define the roots as
$$\begin{align}
r_1 &= \frac{1}{2z^2}\left(i+\sqrt{4z^2-1}\right) \\
r_2 &= -r_1 \\
r_3 &= \frac{1}{2z^2}\left(i-\sqrt{4z^2-1}\right) \\
r_4 &= -r_3
\end{align}$$
If $z$ is purely real, then $r_1$ and $r_3$ are in the upper half plane. Thus
$$\begin{align}
\sum\mathrm{res\,\,}f &= \frac{1}{(r_1-r_2)(r_1-r_3)(r_1-r_4)} + \frac{1}{(r_3-r_1)(r_3-r_2)(r_3-r_4)} \\
&= -i\frac{z^4}{2}
\end{align}$$
after much algebra. Therefore we have
$$\begin{align}
\int_0^\infty \frac{dx}{x^2+(1-z^2x^2)^2} &= \frac{\pi i}{z^4}\frac{-iz^4}{2} \\
&= \frac{\pi}{2}
\end{align}$$
which is independent of $z$ (if $z$ is purely real)! In the general case, however, compute the residues of the poles that are in the upper half plane, to be determined by the value of $z$.
|
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|
What is the remainder when $x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$ is divided by: $x^2-8x+12$. What is the remainder when: $$x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$$ is divided by: $$x^2-8x+12$$
$x^2-8x+12$ $\leftrightarrow (x-2)(x-6)$
This gives me the polynomial: $(x-2)(x-6)k(x) + r(x)$, where $r(x)$ is the remainder.
The expression gives me the remainder when $x=2$ or $x = 6$
By replacing $x$ with $2$ into the original polynomial you get: $$x^{100} - 4x^{100}+3x^{100}-3x^{10}+12x^{10}-9x^{10}+3x^{2}-29x + 41 = 3x^2-29x + 41$$
By plugging in $2$ I get: $12-58 + 41 = -5$.
Hence $r(x)$ should be: $-5$.
This is apparently the wrong answer. What did I do wrong?
Thank you kindly for your help!
|
First note that $r(x)=ax+b$ for some $a,b\in\mathbb Z$
For $x=2$ you get $r(2)=-5$ (not $r(x)=-5$) $ \Rightarrow 2a+b=-5$. Do the same for $x=6$ to find r(6).Then find $a,b$.
|
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|
Is this function decreasing? I have to prove that the following function is decreasing when $x>0$
$$f(x)=\frac{(x+1)^2}{\sqrt{x^2+2x}}-(x+1)^2\arcsin\Big(\frac{1}{x+1}\Big)$$
but calculating its derivative I don't succed in proving this result.
|
$f'(x) = \dfrac{3(x+1)-\dfrac{(x+1)^3}{(x+2)x}}{\sqrt{x(x+2)}}-2(x+1)\arcsin(\dfrac{1}{1+x})$
Since $1+x>0$,thus we just prove
$\dfrac{3-\dfrac{(1+x)^2}{(x+2)x}}{\sqrt{x(x+2}}-2\arcsin(\dfrac{1}{x+1})\le 0$
Since $\arcsin(\dfrac{1}{1+x})\ge \dfrac{1}{1+x}$, thus
we just prove
$\dfrac{3-\dfrac{(1+x)^2}{(x+2)x}}{\sqrt{x(x+2)}}-\dfrac{2}{x+1}\le 0$
which is equivalent to
$3-\dfrac{(x+1)^2}{x(x+2)}-\dfrac{2\sqrt{x(x+2)}}{x+1}\le 0$
take $u = \dfrac{1+x}{\sqrt{x(x+2)}}$, then
we just have to show
$3-u^2-\dfrac{2}{u}\le 0$.
AM-GM Inequality
$u^2+\dfrac{1}{u}+\dfrac{1}{u}\ge 3\sqrt[3]{1}=3$
And we know $u\neq 1$, thus the equality cannot be achieved.
|
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|
Find parametric expression of an arc given its start point, end point and central angle in 3D cartesian coordinate system In a 3D cartesian coordinate system, the coordinates of start point and end point have been given as $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. If the central angle of the two points (the one smaller than 180 degrees) is known as $\theta$, and the plane on which the arc lies is perpendicular to $XY$ plane, how to find the parametric expression $(x, y, z)$ of the arc?
I mean, each coordinate of $(x, y, z)$ needs to be represented by variables given above. Can anyone help? Thanks!
|
The expression for the locus of points of the circle containing the arc is
$$ {\rm RZ}(\psi) \begin{pmatrix}
x_c + r \cos \varphi \\
y_c \\
z_c + r \sin \varphi
\end{pmatrix} = \begin{pmatrix}
r \cos\varphi \cos\psi + x_c \cos\psi - y_c \sin\psi \\
r \cos\varphi \sin\psi + y_c \cos\psi + x_c \sin\psi \\
r \sin\varphi + z_c
\end{pmatrix} $$
where $(x_c,y_c,z_c)$ is the center of the circle, $r$ is the radius and $\varphi$ is the azimuthal position along the circle. Also $\psi$ is the orientation of the plane of the arc relative to the XZ plane, as rotated by the z axis. For the end points of the arc, one can use an unknown center azimum $\varphi_c$ and the range $$\varphi = \varphi_c - \frac{\theta}{2} \ldots \varphi_c + \frac{\theta}{2}$$
Now we have the equations
$$ \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} = \begin{pmatrix}
r \cos(\varphi-\frac{\theta}{2}) \cos\psi + x_c \cos\psi - y_c \sin\psi \\
r \cos(\varphi-\frac{\theta}{2}) \sin\psi + y_c \cos\psi + x_c \sin\psi \\
r \sin(\varphi-\frac{\theta}{2}) + z_c
\end{pmatrix} $$
$$ \begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} = \begin{pmatrix}
r \cos(\varphi+\frac{\theta}{2}) \cos\psi + x_c \cos\psi - y_c \sin\psi \\
r \cos(\varphi+\frac{\theta}{2}) \sin\psi + y_c \cos\psi + x_c \sin\psi \\
r \sin(\varphi+\frac{\theta}{2}) + z_c
\end{pmatrix} $$
By subtracting the 2nd point from the first point and dividing the y coordinate with the x coordinate the orientation of the plane $\psi$ is found
$$ \tan \psi = \frac{y_2-y_1}{x_2-x_1} $$
Further manipulation of the subtraction of the two points yields the center angle
$$ \tan \varphi_c = \frac{\sqrt{2} (\sin \frac{\theta}{2}) \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}{(z_2-z_1)\sqrt{1-\cos\theta}} $$
and the radius of the circle
$$ r = \frac{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}{\sqrt{2}\sqrt{1-\cos\theta}} $$
Now by adding the end points, one finds the center of the circle
$$ \begin{pmatrix} x_c \\ y_c \\ z_c \end{pmatrix} = \begin{pmatrix}
\frac{x_1+x_2}{2} \cos\psi + \frac{y_1+y_2}{2} \sin\psi - r \cos\varphi_c \cos \frac{\theta}{2} \\
-\frac{x_1+x_2}{2} \sin\psi + \frac{y_1+y_2}{2} \cos\psi \\
\frac{z_1+z_2}{2} - r \sin\varphi_c \cos \frac{\theta}{2}
\end{pmatrix} $$
The arc is now fully defined.
|
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$(y^2+2iy)^{m-\frac{1}{2}}=y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{m+\frac{1}{2}})(0\leq y \leq 1)?$ This question comes from stein's book Introduction to fourier analysis on euclidean space,page 159.
when $m > 1/2$,
$$(y^2+2iy)^{m-\frac{1}{2}}=y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{m+\frac{1}{2}})(0\leq y \leq 1)$$
$$(y^2+2iy)^{m-\frac{1}{2}}=y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{2m-1})(1\leq y \leq \infty)$$
why $(y^2+2iy)^{m-\frac{1}{2}}$ can be written in this way? It is not obviously at all.
|
The first equality comes from the fact $\forall \alpha>0, (1+x)=_01+O(x)$ where $=_0$ means that the equality is true for $x$ in a neighborhood of $0$. Then we have:
$$(y^2+2iy)^{m-\frac{1}{2}}=(2iy)^{m-\frac{1}{2}}\left((\frac{1}{2i})^{m-\frac{1}{2}}y+1\right)=_0 (2iy)^{m-\frac{1}{2}}(1+O(y))\\=_0y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{m+\frac{1}{2}}).$$
The second equality is a little weird, I find this result:
$$(y^2+2iy)^{m-\frac{1}{2}}=(y^2)^{m-\frac{1}{2}}(1+\frac{2i}{y})^{m-\frac{1}{2}}=_{+\infty}(y^2)^{m-\frac{1}{2}}(1+O(\frac{1}{y}))\\=_{+\infty}y^{2m-1}+O(y^{2m-2}).$$
If there is a correction, it is welcome.
|
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|
>Prove that $\frac d {dx} x^n=nx^{n-1}$ for all $n \in \mathbb R$.
Prove that $\frac d {dx} x^n=nx^{n-1}$ for all $n \in \mathbb R$.
I saw some proof of $\frac d {dx} x^n=nx^{n-1}$ using binomial theorem, which is only available for $n \in\mathbb N$. Do anyone have the proof of $\frac d {dx} x^n=nx^{n-1}$ for all real $n$? Thank you.
|
From the definition of derivatives, we have
$$ f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} $$
Let's assume here that $f(x) = x^n$. Then,
*
*Case 1, $n \in \mathbb{N}$
$$
\begin{align}
f'(x) &= \lim_{h \to 0} \frac{(x + h)^n - x^n}{h} \\
&= \lim_{h \to 0} \frac{x^n + nhx^{n-1} + \frac{n(n-1)}{2!}h^2x^{n-2} + \cdots - x^n}{h}\\
&= \lim_{h \to 0} \text{ } h \text{ } \frac{nx^{n-1} + \frac{n(n-1)}{2!}hx^{n-2} + \cdots \text{ higher powers in h}}{h} \\
&= nx^{n-1} + \lim_{h \to 0} \text{ } h \cdot \left( \frac{n(n-1)}{2!}x^{n-2} + \cdots \text{multiples of h}\right)\\
&= nx^n + 0\\
&= nx^n
\end{align}
$$
*
*Case 2, $r \in \mathbb{R}$: A similar expansion of binomials exists for real powers, as given by Issac Newton.
$$
\begin{align}
f'(x) &= \lim_{h \to 0} \frac{(x + h)^r - x^r}{h} \\
&= \lim_{h \to 0} \frac{x^r + hrx^{r-1} + \frac{r(r-1)}{2!}h^2x^{r-2} + \cdots - x^r}{h}\\
&= \lim_{h \to 0} \text{ } h \text{ } \frac{rx^{r-1} + \frac{r(r-1)}{2!}hx^{r-2} + \cdots \text{ higher powers in h}}{h} \\
&= rx^{r-1} + \lim_{h \to 0} \text{ } h \cdot \left( \frac{r(r-1)}{2!}x^{r-2} + \cdots \text{multiples of h}\right)\\
&= rx^r + 0\\
&= rx^r
\end{align}
$$
|
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|
Show that the curve $x^2+y^2-3=0$ has no rational points
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.
Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.
Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.
Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.
In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.
Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.
Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.
Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.
We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.
|
Suppose to the contrary that there is a rational solution of the equation. Then there exist integers $a$, $b$, and $q$, with $q\ne 0$, such that $a^2+b^2=3q^2$, and $a$, $b$, and $q$ have no common factor greater than $1$.
Note that $a$ and $b$ must both be divisible by $3$. For if an integer $m$ is not divisible by $3$, then $m^2$ has remainder $1$ on division by $3$. So if one or both of $a$ and $b$ is not divisible by $3$, then $a^2+b^2$ has remainder $1$ or $2$ on division by $3$, and therefore cannot be of the shape $3q^2$.
Thus both $a$ and $b$ are divisible by $3$. It follows that $q$ is divisible by $3$, contradicting our assumption that $a$, $b$, and $q$ have no common divisor greater than $1$.
|
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|
$y''=y$, $x''=-x$. Write this equation in terms of the first-order system I'm having trouble with my homework on higher-order equations and their equivalent systems. :(
This is the problem:
Write this equation in terms of the first-order system.
$$\left\{ \begin{array}{l} \frac{d^2x}{dt^2}=-x \\ \frac{d^2y}{dt^2}=y \end{array} \right.$$
(Hint: Write each second-order equation as two first-order equations.
|
If $X= \left( \begin{array}{c} x' \\ x \end{array} \right)$, then $X'= \left( \begin{array}{c} x'' \\ x' \end{array} \right) = \left( \begin{array}{c} -x \\ x' \end{array} \right)= \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{c} x' \\ x \end{array} \right)=AX$. You can do the same thing for $y$.
|
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|
Solutions of the Diophantine equation $x^2(x^2+10)=3y^2(y^2+10)$ I am looking for the solutions of the Diophantine equation
$$x^2(x^2+10)=3y^2(y^2+10).$$
Is there any solution of this equation except when $(x,y)=(0,0)$?
Or
Any computer programme such as MAGMA could solve this problem?
Thank you very much.
|
If $x$ and $y$ are not both $0$, let $3^k$ be the highest power of $3$ that divides both $x$ and $y$. Let $x=3^ks$ and $y=3^k t$. Then $3$ cannot be a common divisor of $s$ and $t$.
Substitute and cancel. We get $s^2(x^2+10)=3t^2(y^2+10)$. Since $3$ cannot divide $x^2+10$, it must divide $s$. Say $s=3u$. Then $9u^2(x^2+10)=3t^2(y^2+10)$, and therefore $3u^2(x^2+10)=t^2(y^2+10)$. But then $3$ divides $t$, contradicting the fact that $3$ is not a common divisor of $s$ and $t$.
|
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|
Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ Could someone help with the following integration:
$$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$
So far I have done the following, but I am stuck:
I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$
Then I am really stuck. Could someone help me?
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Let $$I = \int_0^{\pi} \dfrac{x \sin(x)}{1+\cos^2(x)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \sin(x+\pi/2)}{1 + \cos^2(x+\pi/2)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx $$
Now
$$\int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx = \int_{-\pi/2}^{\pi/2} \underbrace{\dfrac{x \cos(x)}{1 + \sin^2(x)}}_{\text{Odd function}} dx + \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx $$
Hence, we get that
$$I = \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx = \dfrac{\pi}2 \cdot \int_{-1}^1 \dfrac{dt}{1+t^2} = \dfrac{\pi}2 \cdot \left( \dfrac{\pi}4 - \dfrac{-\pi}4\right) = \dfrac{\pi^2}4$$
The integral you are after is $-I$ and hence the answer is $-\dfrac{\pi^2}4$.
|
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|
Fourier series of function $f(x)=0$ if $-\pi$$f(x) = \begin{cases}0 & \text{if }-\pi<x<0, \\
\sin(x) & \text{if }0<x<\pi.
\end{cases}$$
My attempt:
I went the route of expanding this function with a complex Fourier series.
$$f(x) = \sum_{n=-\infty}^{+\infty} C_{n}e^{inx}$$
$$C_{n} = \frac {1}{2\pi} \int_{0}^{\pi} \frac {e^{ix}-e^{-ix}}{2i} e^{-inx} \,\mathrm dx = \frac {1}{\pi}\left(\frac {1}{1-n^2}\right)$$
because only even $n$ terms survive, odd $n$ are 0
$$
C_0 = \frac {1}{2\pi} \int_{0}^{\pi} \sin(x)\, \mathrm dx = \frac {1}{\pi}
$$
so
$$
f(x) = \frac{1}{\pi} + \frac {1}{\pi} \left(\frac {e^{i2x}}{1-2^2} + \frac {e^{i4x}}{1-4^2}+\frac {e^{i6x}}{1-6^2}+\cdots\right) + \frac {1}{\pi} \left(\frac {e^{-i2x}}{1-2^2} + \frac {e^{-i4x}}{1-4^2}+\frac {e^{-i6x}}{1-6^2}+\cdots\right)
$$
In sine and cosine terms,
$$
f(x) = \frac{1}{\pi} + \frac {2}{\pi} \left(\frac {\cos(2x)}{1-2^2} + \frac {\cos(4x)}{1-4^2}+\frac {\cos(6x)}{1-6^2}+\cdots\right)
$$
But the answer in my book is given as
$$
f(x) = \frac{1}{\pi} + \frac{1}{2} \sin(x)+ \frac {2}{\pi} \left(\frac {\cos(2x)}{2^2-1} + \frac {\cos(4x)}{4^2-1}+\frac {\cos(6x)}{6^2-1}+\dotsb\right)$$
I don't understand how there is a sine term and the denominator of the cosines has $-1$.
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The $\sin$ term comes form $n=1$ you can't devide by zero.
mh I calculated again, your $\cos(x)$ terms are right, there shouldn't be a $-$ in the denominator
For $$\int_0^\pi \sin(x) e^{inx}\, \mathrm{d} x = \frac{1+ e^{i \pi n}}{1-n^2}$$
we have to check the case $n=1$ seperate as we can't devide by zero.
The case $n=1$ give
$$\int_0^\pi \sin(x) \exp(x)\, \mathrm{d}x=\frac{i \pi }{2}$$
|
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|
Partial Derivative of $f(x,y) =\ln(x^{2} + y^{2}) + \sqrt{x^{2}\cdot y^{3}}$ $$f(x,y) = \ln(x^{2} + y^{2}) + \sqrt{x^{2}\cdot y^{3}}$$
What is the value of $f_{x}\left ( 0,1 \right )$ and $f_{y}\left ( 0,1 \right )$?
I tried but I found the denominator as zero.
|
$$f(x,y) = \ln(x^{2} + y^{2}) + \sqrt{x^{2}\cdot y^{3}}$$
Using chain rule of differentiation:
$$
\begin{align}
f_x(x, y) &= \dfrac{2x}{x^2 + y^2} + \dfrac{1}{2} 2xy^3 \left(x^2 \cdot y^3\right)^{\frac{-1}{2}} \\
&= \dfrac{2x}{x^2 + y^2} + \dfrac{xy^3}{\sqrt{x^2 \cdot y^3}} \\
&= \dfrac{2x}{x^2 + y^2} + y^{\frac{3}{2}}
\end{align}
$$
Similarly,
$$
\begin{align}
f_y(x, y) &= \dfrac{2y}{x^2 + y^2} + \dfrac{1}{2} 3x^2y^2 \left(x^2 \cdot y^3\right)^{\frac{-1}{2}} \\
&= \dfrac{2y}{x^2 + y^2} + 3xy^{\frac{1}{2}}
\end{align}
$$
|
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|
Variable Exponents Question If $a,b$ and $c$ are different positive integers and $2^a\cdot2^b\cdot2^c =64$ then $2^a+2^b+2^c$=?
This is so far my work: I got $2^a\cdot2^b\cdot2^c=2^6$ then $abc=6$ is this so far in the right track?
|
$$2^a \cdot 2^b\cdot 2^c = 2^{a + b + c} = 2^6\;\;\iff\;\; a + b + c = 6$$
(Recall that $2^x\cdot 2^y = 2^{x+y}.$)
The only possible combinations of distinct $a, b, c$ which sum to $6$ is $\;(a, b, c) = (1, 2, 3),\;$ or any permutation thereof.*
That is, $$a \neq b \neq c \implies 2^a + 2^b + 2^c = 2^1 + 2^2 + 2^3 = 2 + 4 + 8 = 14.$$
*(Because of symmetry, we do not need to consider $a = 2, b = 1, c = 3,$ etc. All we need to know is if a, b, c are distinct, then we need one of a, b, c to be $1$, one to be $2$, and one to be $3$.)
|
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|
How to solve $\int\cot^5x\sin^2x\ dx$? I'm not quite sure how to approach this without it getting extremely messy... and even then, I don't know if it will come out right.
The best I can think of is to use IBP, but neither of those functions are easy to integrate.
I did integrate $\sin^2x$ to get $$\int\sin^2x\ dx=\frac{2x+\sin{2x}}{4}+C$$
but I dare not try to go further as it looks like hell to apply IBP after this point.
I did consider converting $\cot^5x$ to $\displaystyle\frac{\cos^5x}{\sin^5x}$, but then I wasn't any better off...
$$\cot^5x\sin^2x=\frac{\cos^5x}{\sin^5x}\sin^2x=\cos^5x\csc^3x=\cos^2x\cot^3x$$
|
We have:
\begin{align}
I = \int \cot^5 x \sin^2 x \,dx &= \int \frac{\cos^5 x}{\sin^5 x} \sin^2 x \,dx \\
&= \int \frac{\cos^5 x}{\sin^3 x} \,dx \\
&= \int \frac{\left(1 - \sin^2 x\right)^2}{\sin^3 x} \cos x \,dx
\end{align}
Now put $u = \sin x$ to get:
\begin{align}
I &= \int \frac{\left(1 - u^2\right)^2}{u^3} \,du \\
&= \int \left(u^{-3} -2 u^{-1} + u\right) \,du \\
&= -\frac{u^{-2}}{2} -2 \log\left|u\right| + \frac{u^2}{2} + C
\end{align}
Put $u = \sin x$ to get the final result:
$$
I = -\frac{\sin^{-2} x}{2} -2 \log\left|\sin x\right| + \frac{\sin^2 x}{2} + C
$$
|
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|
$\overline{xyz} =x!+y!+z!$ how to find all of 3-digit integer number $\overline{xyz}$ such that:
$\overline{xyz} =x!+y!+z!$
|
As I was partway through typing this answer, Dan Shved's answer appeared which is much better than mine, but I'll include what I have so far as I use some different ideas. I reduce the number of possibilities to $13$ (though of course, I could reduce this number even further using the reasons in Dan's answer).
\begin{align*}
0! &= 1\\
1! &= 1\\
2! &= 2\\
3! &= 6\\
4! &= 24\\
5! &= 120\\
6! &= 720
\end{align*}
As $7! = 5040$, each of $x, y,$ and $z$ is between $0$ and $6$. Furthermore, at most one of $x, y,$ and $z$ can be $6$ otherwise $x! + y! + z!$ is not a three digit number (too big). Similarly, at least one of $x, y,$ and $z$ is a $5$ or $6$ otherwise $x! + y! + z!$ is not a three digit number (too small). Also, $x$ can't be $0$. We are left with $229$ possibilities.
Note that $\overline{xyz}$ can't end with a $0$ or a $1$ because there is no possible $x$ and $y$ such that $x! + y! \equiv 9\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $2$ or a $4$, we must have $\overline{xy} \in \{34, 43, 56, 65\}$
as these are the only choices with $x! + y! \equiv 0\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $3$ we must have $\overline{xy} \in \{13, 31\}$ as these are the only choices with $x! + y! \equiv 7\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $5$ we must have $\overline{xy} \in \{14, 40, 41\}$ as these are the only choices with $x! + y! \equiv 5\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $6$ we must have $\overline{xy} \in \{24, 42\}$ as these are the only choices with $x! + y! \equiv 6\ \textrm{mod} 10$.
So we are left with $15$ possibilities:
$$342, 432, 562, 652, 344, 434, 564, 654, 133, 313, 145, 405, 415, 246, 426.$$
Note that $\overline{xyz}$ is divisible by $\min\{x!, y!, z!\} = (\min\{x, y, z\})!$. This condition is automatically satisfied if the smallest digit is $0, 1,$ or $2$. However, this condition rules out both $564$ and $654$ as they're not divisible by $4!$.
|
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|
Solving: $\frac{3x-1}{2} =\frac{-2}{x+2}$ How to solve :
$$\frac{3x-1}{2} =\frac{-2}{x+2} $$
|
$$\frac{3x-1}{2} =\frac{-2}{x+2} $$
$$(x+2)(3x-1)=2\cdot(-2) $$
$$3x^2+5x+2=0$$
$$a=3,b=5,c=2$$
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
|
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|
System of equations wrt self-adjoint operators $X = \left( \begin{matrix} 2&s\\ 8&2 \end{matrix} \right)$ and $Y = \left( \begin{matrix} 2&-1\\ 2&2 \end{matrix} \right)$ are two operators wrt the same orthonormal basis $B$ in a 2D scalar product space. Furthermore, $s$ is a fixed, complex-valued parameter.
The associated system of equations is $P^2Q^{-1} = X$ and $PQ = Y$, where $P = P^*$ and $Q = Q^*.$
1) Assume that this system of equations is guaranteed to have a solution. Determine the value of $s$ and show that the solution for $P$ and $Q$ is unique.
2) Using the value of $s$ found in part (1), compute $P$ and $Q$ wrt $B$.
|
We have $XY^\ast=P^3$. Therefore $XY^\ast=\begin{pmatrix}4-s&2s+4\\14&20\end{pmatrix}$ is self-adjoint and hence $s=5$. Note that $P,Q$ are invertible because $X,Y$ are invertible. Also, since $XQ=P^2$ and $QY^{-1}=P^{-1}$ are self-adjoint, we have
\begin{align}
XQ&=QX^T,\tag{1}\\
QY^{-1}&=Y^{-T}Q,\tag{2}\\
XQ &= (YQ^{-1})^2.\tag{3}
\end{align}
Let $Q=\begin{pmatrix}a&b+yi\\b-yi&d\end{pmatrix}$. Using the above two equations, you will find that $Q$ and hence $P=YQ^{-1}$ are uniquely determined:
$$
Q=\begin{pmatrix}0&1\\1&0\end{pmatrix},
\ P=\begin{pmatrix}-1&2\\2&2\end{pmatrix}.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $y^2= x^3 − 33$ in integers This is not homework, could someone provide a nice clear proof as I have been struggling with this for some time.
Solve the equation $y^2= x^3 − 33$; $x, y \in \mathbb{Z}$
|
$$ y^2 + 25 = x^3 - 8. $$
Note $y^2 + 25$ can not be divisible by any prime $q \equiv 3 \pmod 4,$ therefore not by any number $m \equiv 3 \pmod 4.$
If $y$ were odd, $y^2 + 25 \equiv 2 \pmod 4,$ impossible for $x^3 - 8.$ Therefore $y$ is even and $x$ is odd.
$$ y^2 + 25 = (x - 2)(x^2 + 2 x + 4). $$
As we need $x-2 \equiv 1 \pmod 4,$ we know $x \equiv 3 \pmod 4.$ However, then $(x^2 + 2 x + 4) \equiv 1 + 2 + 4 \equiv 3 \pmod 4. $ This is prohibited from dividing $y^2 + 25,$ so we have arrived at a contradiction of the equation in integers.
Done.
|
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|
How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$ let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that
$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
I find this inequality are same as Crux 3059 Problem.
|
Your constraints are $a,b,c,d > 0$ and $a^2+b^2+c^2+d^2 = 3$, by doing a change of variables, $x^2 = a^2/3, y^2 = b^2/3, z^2 = c^2/3, w^2 = d^2/3$. Then your constraints change to
$$x,y,z,w > 0, x^2+y^2+z^2+w^2 =1$$
and the problem is equivalent to proving $xy +xz +xw +yz+yw+zw \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$. Since $x^2+y^2+z^2+w^2 =1$ that first implies that $x,y,z,w \leq 1$ and in particular $x^2+y^2+z^2+w^2 \leq x+y+z+w$. Doing your standard expansion,
$$ (x+y+z+w)^2 = x^2+y^2+z^2+w^2 + 2(xy+xz+xw+yz+yw+zw) $$
The right hand side is bounded above by $x^2+y^2+z^2+w^2 + (x^2+z^2)+(x^2+w^2)+(y^2+z^2) +(y^2 + w^2) + 2(xy + wz) = 3(x^2+y^2+z^2+w^2) + 2(xy+wz)$
Balancing this with the RHS of the display we get
$$2(xy+xz+xw+yz+yw+zw) \leq 2(x^2+y^2+z^2+w^2) + 2(xy+wz)$$
Then we have that $xy \leq (x^2+y^2)/2$ and $wz \leq (w^2+z^2)/2$ and placing this in we have
$$xy+xz+xw+yz+yw+zw \leq 1.5(x^2+y^2+z^2+w^2) \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$$
since $\frac{2}{3}xyzw \geq 0$, $1.5 <\sqrt{3}$ and $x^2+y^2+z^2+w^2 \leq x+y+z+w$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Generating function for the sequence $1,1,3,3,5,5,7,7,9,9,\ldots$ The generating function for the sequence $\left\{1,1,1,1,...\right\}$ is $$1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$
What is the generating function for the sequence $\left\{1,1,3,3,5,5,7,7,9,9,\dots \right\}$?
This is my attempt:
what we want to do is after the initial $\left\{1,1,1,1,...\right\}$ function we want to add two more functions shifted two to the right such as $\left\{0,0,1,1,1,...\right\}$ and keep adding such function 2 at a time and each new two are shifted 2 more to the right then the previous one. so first would be $\left\{1,1,1,1,\dots \right\}$ then add 2 of $\left\{0,0,1,1,1,1,\dots\right\}$ then add 2 more of $\left\{0,0,0,0,1,1,1,1,\dots\right\}$ and keep doing that. this gives us the function
$$\frac{1}{1-x}+\frac{2x^2}{1-x}+\frac{2x^4}{1-x}+...$$ or $$\frac{1}{1-x}+\frac{2x^{2k}}{1-x}$$ however $i$ do not know where to go from here, how do $i$ finish this problem?
|
You already did the difficult part... You probably mean $$\frac1{1-x}+\sum_{k=1}^\infty\frac{2x^{2k}}{1-x}$$
So you just have to evaluate the geometric series $$\sum_{k=1}^\infty (x^2)^k = \frac1{1-x^2} - 1 = \frac{x^2}{1-x^2}$$
And obtain
$$\frac1{1-x} + \frac2{1-x}\sum_{k=1}^\infty x^{2k} = \frac{1+\frac{2x^2}{1-x^2}}{1-x} = \frac{x^2+1}{(1-x)(1-x^2)}$$
|
{
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|
Integral $\int\frac{6^x}{9^x-4^x}dx $ How to solve this integral:
$$\int\frac{6^x}{9^x-4^x}dx $$
(I notice that $\frac{6^x}{9^x-4^x}=\frac{2^x3^x}{(3^x-2^x)(3^x+2^x)}$)
Thank you!
|
Note that $$\frac{1}4\cdot\frac{3^x+2^x}{3^x-2^x}-\frac{1}4\cdot\frac{3^x-2^x}{3^x+2^x}$$ is the integrand. Now use what @J.H. suggested.
|
{
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|
$ \sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}. $ The series is:
$$
\sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}.
$$
I tried splitting the whole thing into simple fractions but I don't seem to get anywhere.
Any ideas?
|
$$\sum_{n=2}^\infty \frac{1}{n^3(n^3+1)} = \sum_{n=2}^\infty \frac{1}{n^3}+\frac{n-2}{3 (n^2-n+1)}-\frac{1}{3 (n+1)} = \\
\zeta{(3)} -1 + \sum_{n=2}^{\infty} \frac{n-2}{3 (n^2-n+1)}-\frac{1}{3 (n+1)} = \\
\zeta{(3)}-1 -\sum_{n=2}^{\infty} \frac{1}{n^3+1} \approx \\
0.015553560820970399435132949324595238582753593120862693333639...$$
|
{
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|
What are the generators for $SL_n(\mathbb{R})$ (Michael Artin's Algebra book) The book asks you to prove that $SL_n(\mathbb{R})$ is generated by elementary (row operation) matrices in which one nonzero off-diagonal entry is added to the identity matrix. For example,
$$
\begin{bmatrix}
1 & a \\
0 & 1
\end{bmatrix}
$$
acts by left multiplication on $2\times2$ matrices by adding $a$ times (row 2) to (row 1). Considering a simple example:
$$
M=
\begin{bmatrix}
a & 0 \\
0 & 1/a
\end{bmatrix}, a \neq 0$$
you can see that the matrix $M$ does belong to $SL_n(\mathbb{R})$. However, the elementary matrices composing $M$ are of the below type and not of the first type (nonzero off-diagonal entry).
$$
\begin{bmatrix}
c & 0 \\
0 & 1
\end{bmatrix}
$$
i.e. one nonzero diagonal entry added to the identity matrix. So
$$M =
\begin{bmatrix}
a & 0 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1/a
\end{bmatrix}
$$
So $M$ clearly isn't generated by the first type. What is going wrong?
|
The point is, you can do without using elementary row operations of the form $R_i\leftarrow \lambda R_i$. Given a matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ such that its first column is nonzero, we can always reduce its first column to $(1,0)^T$ using only elementary row operations of the form $R_i\leftarrow R_i+kR_j$. In fact, let $k$ be a scalar such that $ka+c\neq0$. Then
$$
\begin{pmatrix}1&0\\-(ka+c)&1\end{pmatrix}
\begin{pmatrix}1&\frac{1-a}{ka+c}\\0&1\end{pmatrix}
\begin{pmatrix}1&0\\k&1\end{pmatrix}
\begin{pmatrix}a&b\\c&d\end{pmatrix}
=\begin{pmatrix}1&\ast\\0&\ast\end{pmatrix}.
$$
The first elementary row operation (i.e. the rightmost one) modifies the $(2,1)$-th entry to nonzero; the second one makes the $(1,1)$-th entry becomes $1$ and the third (the leftmost one) kills off the $(2,1)$-th entry.
For your particular example, we may put $k=1$ and get
$$
\begin{pmatrix}1&0\\-a&1\end{pmatrix}
\begin{pmatrix}1&\frac{1-a}{a}\\0&1\end{pmatrix}
\begin{pmatrix}1&0\\1&1\end{pmatrix}
\begin{pmatrix}a&0\\0&\frac1a\end{pmatrix}
=\begin{pmatrix}1&\frac{1-a}{a^2}\\0&1\end{pmatrix}.
$$
Hence
$$
\begin{align*}
\begin{pmatrix}a&0\\0&\frac1a\end{pmatrix}
&=
\begin{pmatrix}1&0\\1&1\end{pmatrix}^{-1}
\begin{pmatrix}1&\frac{1-a}{a}\\0&1\end{pmatrix}^{-1}
\begin{pmatrix}1&0\\-a&1\end{pmatrix}^{-1}
\begin{pmatrix}1&\frac{1-a}{a^2}\\0&1\end{pmatrix}\\
&=
\begin{pmatrix}1&0\\-1&1\end{pmatrix}
\begin{pmatrix}1&\frac{a-1}{a}\\0&1\end{pmatrix}
\begin{pmatrix}1&0\\a&1\end{pmatrix}
\begin{pmatrix}1&\frac{1-a}{a^2}\\0&1\end{pmatrix}.
\end{align*}
$$
|
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|
Tensor and Kronecker product I have a follow exercise. Let $\left|\Psi\right\rangle = \tfrac{1}{\sqrt{2}}\left(\left|0\right\rangle + \left|1\right\rangle\right)$. Write out $\left|\Psi\right\rangle^{\otimes 2}$ xplicitly, both
in terms of tensor products like $\left|0\right\rangle \otimes \left|1\right\rangle$ , and using the Kronecker product.
Using tensor products I get:
$$\frac{1}{2}\left(\left|00\right\rangle + \left|01\right\rangle + \left|10\right\rangle + \left|11\right\rangle\right).$$
How I will be able to express this in terms of the Kronecker product?
|
Let $S : \mathbb{C}^2 \otimes \mathbb{C}^2 \to \mathbb{C}^4$ be the isomorphism that maps the tensor product $v \otimes w$ of $v$, $w \in \mathbb{C}^2$ to their Kronecker product. Then, in general, for $v = (v_1,v_2)^T$, $w = (w_1,w_2)^T \in \mathbb{C}^2$,
$$
S\left(v \otimes w\right) = \begin{pmatrix} v_1 w \\ v_2 w \end{pmatrix} = \begin{pmatrix} v_1 w_1 \\ v_1 w_2 \\ v_2 w_1 \\ v_2 w_2 \end{pmatrix}.
$$
Thus, if $\left|0\right\rangle = (1,0)^T$ and $\left|1\right\rangle = (0,1)^T$ are spin up and spin down eigenvectors for the component of spin in the $z$ direction, then $\left|\Psi\right\rangle = \left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\right)^T$, and hence
$$
S\left(\left|\Psi\right\rangle^{\otimes 2}\right) = \begin{pmatrix} \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\\ \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\\ \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\\ \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\end{pmatrix} = \frac{1}{2} \begin{pmatrix}1\\1\\1\\1\end{pmatrix}.
$$
This corresponds to your calculation above in $\mathbb{C}^2 \otimes \mathbb{C}^2$, since
$$
S\left(\left|00\right\rangle\right) = S\left(\left|0\right\rangle \otimes \left|0\right\rangle\right) = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \quad S\left(\left|01\right\rangle\right) = S\left(\left|0\right\rangle \otimes \left|1\right\rangle\right) = \begin{pmatrix}0\\0\\1\\0\end{pmatrix},\\
S\left(\left|10\right\rangle\right) = S\left(\left|1\right\rangle \otimes \left|0\right\rangle\right) = \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \quad
S\left(\left|11\right\rangle\right) = S\left(\left|1\right\rangle \otimes \left|1\right\rangle\right) = \begin{pmatrix}0\\0\\0\\1\end{pmatrix}.
$$
|
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|
How many solutions are possible to this equation? Given
$$A+2B+3C=N
$$
where $N$ is a given positive integer.
$A ,B,C\in\mathbb{N}$ vary from $0$ to $\infty$.
How many solutions will be there to this equation?
|
Let $X=C, Y=B+C, Z=A+B+C$, then this is equivalent to finding the number of non-negative integer solutions to $X+Y+Z=N$ s.t. $X \leq Y \leq Z$.
First ignore the restriction on $X, Y, Z$. In total we have $\binom{N+2}{2}$ solutions.
The number of solutions with $Y=Z$ is just the number of non-negative integer solutions to $2Y \leq N$, giving $\lfloor \frac{N}{2} \rfloor+1$. By symmetry we get the same number of solutions for $X=Y, X=Z$.
The number of solutions for $X=Y=Z$ is 1 if $3 \mid N$ and 0 otherwise. We can represent this as $1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil$.
The number of solutions with exactly 2 variables equal is thus $3[(\lfloor \frac{N}{2} \rfloor+1)-(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)]$.
Now going back to the problem, each solution where $X<Y<Z$ corresponds to 6 solutions where $X, Y, Z$ are distinct. Each solution with $X \leq Y \leq Z$ and exactly 2 of them equal corresponds to 3 solutions where exactly 2 of them equal but without the condition $X \leq Y \leq Z$.
We thus get:
\begin{align}
& \frac{\binom{N+2}{2}-3(\lfloor \frac{N}{2} \rfloor+1)+2(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)}{6} \\
& +\frac{3(\lfloor \frac{N}{2} \rfloor+1)-3(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)}{3} +(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil) \\
& =\frac{\binom{N+2}{2}+3(\lfloor \frac{N}{2} \rfloor+1)+2(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)}{6}
\end{align}
|
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|
$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$ I'm having trouble proving $$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$$ where $n\in\mathbb{Z}$ and $\theta\in\mathbb{R}$. Can anyone suggest a hint?
|
Using the sum of sines and cosines with arguments in arithmetic progression as given above: if $\theta\ne0$ and let $\varphi =0$, then we have,
\begin{align} &S =\sin{(\theta)} + \sin{(2\theta)} + \cdots + \sin{(n\theta)} = \frac{\sin{\left(\frac{(n+1) \theta}{2}\right)} \cdot \sin{(\frac{n \theta}{2})}}{\sin{\frac{\theta}{2}}} \quad\hbox{and}\\[10pt] &C =\cos{(\theta)} + \cos{(2\theta)} + \cdots+ \cos{(n\theta)} = \frac{\cos{\left(\frac{(n+1) \theta}{2}\right)} \cdot \sin{(\frac{n \theta}{2})}}{\sin{\frac{\theta}{2}}}. \end{align}
From Euler's Formula and the definition of absolute value of a complex number, we can write $$\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2 = C^2 + S^2$$
$$= \frac{\sin{\left(\frac{n \theta}{2}\right)}^2}{\sin{\frac{\theta}{2}}^2}\cdot\left(\cos{(\frac{(n+1) \theta}{2})}^2 + \sin{(\frac{(n+1) \theta}{2})}^2\right)$$
$$ = \left(\frac{\sin{\frac{n \theta}{2}}}{\sin{\frac{\theta}{2}}}\right)^2$$
|
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|
Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$
We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.
Any idea? Or any other method to establish the convergence?
|
If I'm not getting it wrong, your series is
$$1-\frac 1 2 -\frac 13+\frac 14+\frac 15+\frac 16-\frac 17-\frac 18-\frac 19-\frac 1{10}+++++------\dots$$
So you have $1$ plus, $2$ minuses, $3$ pluses, $4$ minuses, and so on.
We can write your series as $a_0+a_1+a_2+a_3+\dots$ where
$$\begin{align} a_0&=(-1)^0 1 \\
a_1&=(-1)^1\sum_{k=2}^{3}\frac 1 k\\
a_2&=(-1)^2\sum_{k=4}^{6}\frac 1 k\\
a_3&=(-1)^3\sum_{k=7}^{10}\frac 1 k\\
\cdots &=\cdots \\
a_n&=(-1)^n\sum_{k=T_n+1}^{T_{n+1}}\frac 1 k\end{align}$$
Where $T_n=\frac{n(n+1)}{2}$ so it goes $1,3,6,10,\dots$
Now, we know that $$\sum_{k=1}^n \frac 1 k =\log n+\gamma+\frac 1 {2n}+O(n^{-2})$$
Thus
$$\eqalign{
& \sum\limits_{k = 1}^{{T_n}} {{1 \over k}} = \log n + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr
& \sum\limits_{k = 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {n + 2} \right) + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr} $$
Whence, after simplification
$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {1 + {2 \over n}} \right) + O(n^{-2})$$
Recall that $$\log(1+x)=x+O(x^2)$$ so
$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \frac{2}{n} +O\left(\frac 1 {n^2}\right)$$
Since $$\sum (-1)^n \frac 1 n $$ and $$\sum n^{-2}$$ converge, so does your series.
|
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|
Find all positive integers solution of $xy+yz+xz = xyz+2$ 1.Find all positive integers solution
$xy+yz+xz = xyz+2$
2.Determine all p and q which p,q are prime number and satisfy
$p^3-q^5 = (p+q)^2$
Thx for the answer
3.Find all both positive or negative integers that satisfy
$\frac{13}{x^2} + \frac{1996}{y^2} = \frac{z}{1997}$
|
1.set $x≤y≤z,xyz<xyz+2=xy+yz+zx≤3yz,$
so $0<x<3,x=1,2$
If $x=1,y+z+yz=yz+2,y+z=2$,so $x=y=z=1$
If $x=2,2y+2z+yz=2yz+2,yz=2y+2z-2<4z,2≤y<4,y=3,z=4$
so the only solution to the fisrt equation is $x=y=z=1$ or $x=2,y=3,z=4$ or change their order.
|
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|
Write in polynomial in factored form in complex number Write the following polynomial in factored form(in complex number):
$$1+z+z^2+z^3+z^4+z^5+z^6$$
Also, is there general solution of factoring for $1+z+z^2...z^n$ types of polynomial?
|
I'd extend Adi Dani's answer a bit
$$
1 + z^2 + \ldots + z^6 = \frac {1-z^7}{1-z}
$$
Now decompose $1-z^7$ in factors. In order to do that, find all roots of
$$
1-z^7 = 0 \\
z = \sqrt [7]1 = \sqrt[7]{e^{0\cdot pi}} = \{ e^{\frac {2\pi ki}7}\}, k = 0,1,\ldots,6
$$
So
$$
1-z^7 = \prod_{k = 0}^6 \left (z - e^{\frac {2\pi ki}7}\right)
$$
To remove singularity at $z = 1$ we can cancel first $z - e^{\frac 07} = z-1$ term with denominator. So
$$
\sum_{k=0}^6 z^k = \prod_{k=1}^6 \left(z-e^{\frac {2\pi ki}7} \right)
$$
Update
I overlooked second part of the problem, but it's pretty much the same as Adi Dani's answer
$$
\sum_{k=0}^n z^k = \frac {1-z^{n+1}}{1-z} = \frac {\displaystyle \prod_{k = 0}^n \left(z-e^{\frac {2\pi ki} n} \right)}{1-z} = \prod_{k = 1}^n \left(z-e^{\frac {2\pi ki}n} \right)
$$
|
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|
solving $x^2 = a \pmod {2^n}$ , $n \ge 3$ I read that the equation $x^2 = a \pmod {2^n}$ for $n \ge 3$ has four solutions and the solutions are $x_1, -x_1, x_1 + 2^{n-1}, - x_1 + 2^{n-1}$. It is easy to prove that they are indeed the solutions and are incongruent solutions. But if I need to derive that these are the only four solutions then how do I proceed?
Thanks
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We need to assume that $a$ is odd. For example, if $a=4$, modulo $32$ we have solutions $2,6,10,14,18,22,26,30$.
Suppose that $x_1^2\equiv a \pmod{2^n}$ and $x\equiv a\pmod{2^n}$. We need to show that $x$ can take on at most $4$ values.
From the two congruences, we have $x^2\equiv x_1^2\pmod{2^n}$, meaning that
$2^n$ divides $(x-x_1)(x+x_1)$.
Note that $x$ and $x_1$ are both odd. This implies that one of $x-x_1$ or $x+x_1$ is congruent to $2$ modulo $4$. The other therefore must be divisible by $2^{n-1}$. If you need details here, either $x\equiv x_1\pmod{4}$, in which case $x+x_1\equiv 2\pmod{4}$, or $x\equiv -x_1\pmod{4}$, in which case $x-x_1\equiv 2\pmod{4}$.
A little play then gives us the desired result. Perhaps $2^n$ divides $x-x_1$, giving the solution $x\equiv x_1$. Perhaps $2^n$ divides $x+x_1$, giving solution $x\equiv -x_1\pmod{2^n}$.
If neither of these cases holds, then because one of $x_1-x_2$ and $x+1+X_2$ contributes pnly one $2$, the other must contribute exactly $n-1$.
Suppose the highest power of $2$ that divides $x-x_1$ is $2^{n-1}$. Then $x\equiv x_1+2^{n-1}\pmod{2^n}$.
Suppose that the highest power of $2$ that divides $x+x_1$ is $2^{n-1}$. Then $x\equiv -x_1+2^{n-1}\pmod{2^n}$.
The cases are not distinct if $n\lt 3$, but we do not need that to show there are at most $4$ solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $\lim_{n\to\infty}n^2\left(\sin(2\pi en!)-\frac{2\pi}{n}\right)$
How to find$$\lim_{n\to\infty}n^2\left(\sin(2\pi en!)-\frac{2\pi}{n}\right)$$
I am very confused. I don't know how to show the limit exists, or what it is.
|
Recall that $e = 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots$. Hence,
$$n! e = \text{Integer} + \dfrac1{n+1} + \dfrac1{(n+1)(n+2)} + \dfrac1{(n+1)(n+2)(n+3)} + \cdots$$
Hence,
\begin{align}
\sin(2\pi n!e) & = \sin\left(\text{Integer} \times 2 \pi + \dfrac{2\pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3) \right)\\
& = \sin\left(\dfrac{2\pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3) \right)\\
& = \dfrac{2 \pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3)
\end{align}
Hence,
$$\sin(2 \pi n!e) - \dfrac{2 \pi}{n} = -\dfrac{2 \pi}{n(n+1)} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3)$$
Hence,
$$n^2 \left(\sin(2 \pi n!e) - \dfrac{2 \pi}{n}\right) = 2 \pi \left(-\dfrac{n}{n+1} + \dfrac{n^2}{(n+1)(n+2)} + \mathcal{O}(1/n)\right)$$
Hence,
$$\lim_{n \to \infty} n^2 \left(\sin(2 \pi n!e) - \dfrac{2 \pi}{n}\right) = 2 \pi \times 0 = 0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Surface area of an elliptic paraboloid The elliptic paraboloid has height h, and two semiaxes a, b. How to find its surface area? Does it possible to use a direct formula without integrals?
|
The elliptic paraboloid is represented parametrically as follows:
$$x=a \sqrt{u} \cos{v}$$
$$y=b \sqrt{u} \sin{v}$$
$$z=u$$
The surface area of this object is given by
$$\int_0^h du \: \int_0^{2 \pi} dv \: \sqrt{E \,G - F^2}$$
where the 1st fundamental form is given by
$$E=1+\frac{a^2 \cos^2{v} + b^2 \sin^2{v}}{4 u}$$
$$F=\frac{1}{4} (b^2-a^2) \sin{2 v}$$
$$G = (a^2 \sin^2{v}+b^2 \cos^2{v}) u$$
The integral simplifies to
$$\int_0^h du \: \left [\sqrt{b^2 \left(a^2+4 u\right)} E\left(\frac{4 \left(b^2-a^2\right) u}{b^2 \left(a^2+4 u\right)}\right)+\sqrt{a^2 \left(b^2+4 u\right)} E\left(\frac{4
(a-b) (a+b) u}{a^2 \left(b^2+4 u\right)}\right)\right ]$$
where $E$ is the elliptic integral defined as
$$E(m) = \int_0^{\pi/2} dx \sqrt{1-m \sin^2{x}}$$
This is about the best you'll do as far as I can see.
|
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|
Finding determinant (using row reduction) I'm trying to calculate the following
$\det \begin{pmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{pmatrix}$
and I'm sure there must be a way to get this as an upper/lower triangular matrix. However whichever way I try looking at it, I can't see how to make it into one, if you minus the bottom row from each of the other rows you're still left with the bottom row and the same goes for columns. Is there some clever trick to note here? I've also briefly entertained the idea of proving a general form (since I'm sure one exists for a matrix of this form) but it seems a bit overkill...
Many thanks!
|
By Gaussian elimination, using the the first entry of the fifth row as the first pivot, we get that
$$
\begin{vmatrix} -1+x & -1 & -1 & -1 & -1 \\ -1 & -1+x & -1 & -1 & -1 \\ -1 & -1 & -1+x & -1 & -1 \\ -1 & -1 & -1 & -1+x & -1 \\ -1 & -1 & -1 & -1 & -1+x \end{vmatrix}
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ -1+x & -1 & -1 & -1 & -1 \\ -1 & -1+x & -1 & -1 & -1 \\ -1 & -1 & -1+x & -1 & -1 \\ -1 & -1 & -1 & -1+x & -1 \end{vmatrix}
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & x & 0 & 0 & -x \\ 0 & 0 & x & 0 & -x \\ 0 & 0 & 0 & x & -x \end{vmatrix}
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & x & 0 & -x \\ 0 & 0 & 0 & x & -x \end{vmatrix} = -\begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & 0 & -x & x^2-4x \\ 0 & 0 & 0 & x & -x \end{vmatrix}
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & 0 & -x & x^2-4x \\ 0 & 0 & 0 & 0 & x^2-5x \end{vmatrix}
= (-1)(-x)^3(x^2-5x) = x^4(x-5).
$$
|
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|
How to prove ${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$ ?thanks. How to prove
$${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$$
$a>0$,$b>0$,
thanks.
|
Dividing by $b^{a+b}$ and taking the $b^{th}$ root, we need to prove
$$\left(\dfrac{a}{b} \right)^{a/b} \geq \left(\dfrac{a/b+1}2 \right)^{a/b+1}$$
Let $a/b = t$. We then need to prove that
$$t^t \geq \left(\dfrac{1+t}2\right)^{1+t}$$
Consider the function
$$f(x) = x \log(x) - (1+x) \log \left(\dfrac{1+x}2 \right)$$
We then have
\begin{align}
f'(x) & = x \cdot \dfrac1x + \log(x) - (1+x) \cdot \dfrac1{1+x} - \log(1+x) + \log(2)\\
& = \log(2)+ \log(x) - \log(1+x) = \log\left(\dfrac{2x}{1+x}\right)\\
f''(x) & = \dfrac1x - \dfrac1{1+x} = \dfrac1{x(1+x)}
\end{align}
Hence, we see that for $x>0$, we have $f(x)$ to attain an extremum when $$f'(x) = 0 \implies 2x = 1+x \implies x = 1$$ And this extremum is a minimum since $f''(x) = \dfrac1{x(1+x)} > 0$ for $x>0$. Hence, we have $$f(x) \geq f(1) = 0$$ Hence, $f(x)$ is non-negative i.e. $f(x) \geq 0$ for all $x \in \mathbb{R}^+$. Now take $x = \dfrac{a}b$ and do some algebraic massaging to get your answer.
|
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|
What is the first non zero digit in 50 factorial (50!)? What is the first non zero digit in 50 factorial (50!)?
Any Help or hint will be appreciated.
|
First, the exponent of $2$ in $50!$ is $47$, and the exponent of $5$ in $50!$ is $12$, so we’re looking for $\frac{50!}{10^{12}}$ mod$ 10$.
Since $\frac{50!}{10^{12}}$ is even (in fact, divisible by $2^{35}$), it suffices to compute it mod $5$:
$\frac{50!}{5^{12}}$ $\equiv$ ($4!$. $1$. $4!$. $2$. $4!$. $3$. $4!$. $4$. $4!$. $1$.($4!$. $1$. $4!$. $2$. $4!$. $3$. $4!$. $4$. $4!$). $2$
=$4!^{10}$.$4!^2$.2!
$\equiv$$(-1)^{10}$.$(-1)^2$.2
$\equiv$ $2$ mod $5$
$\frac{50}{10^{12}}$ $\equiv$ $2$.$2^{-12}$ $\equiv$ $2$ . $(2^4)^{-3}$ $\equiv$ $2$ mod $5$
So, $\frac{50!}{5^{12}}$ $\equiv$ $2$ (mod $10$ )
|
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"timestamp": "2023-03-29T00:00:00",
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|
Fibonacci and the algebraic expression $x^2-x-1$ $$\left( \left( 1/2\,{\frac {-\beta+ \sqrt{{\beta}^{2}-4\,\delta\,
\alpha}}{\alpha}} \right) ^{i}- \left( -1/2\,{\frac {\beta+ \sqrt{{
\beta}^{2}-4\,\delta\,\alpha}}{\alpha}} \right) ^{i} \right) \left(
\sqrt{{\beta}^{2}-4\,\delta\,\alpha} \right) ^{-1}$$
What exactly is the connection between the above formula and
$$x^2-x-c$$
c being a constant.
I know about linear recursion, but I don't understand why the Fibonacci Numbers are associated with $$x^2-x-1$$
I know about
$${\frac {\sqrt {5}+1}{{2}}}$$
But I don't see the connection. If I were to represent these terms algebraically it seems I would do something like this:
$${\alpha}^{-1}$$
$$-{\frac {\beta}{{\alpha}^{2}}}$$
$$-{\frac {-{\beta}^{2}+\delta\,\alpha}{{\alpha}^{3}}}$$
$${\frac {\beta\, \left( -{\beta}^{2}+2\,\delta\,\alpha \right) }{{
\alpha}^{4}}}$$
$${\frac {{\beta}^{4}-3\,{\beta}^{2}\delta\,\alpha+{\delta}^{2}{\alpha}^
{2}}{{\alpha}^{5}}}$$
$$-{\frac {\beta\, \left( {\beta}^{4}-4\,{\beta}^{2}\delta\,\alpha+3\,{
\delta}^{2}{\alpha}^{2} \right) }{{\alpha}^{6}}}
$$
There also seems to be a relationship with factorials when they are used as arrays in matrices.
|
Let $F_0=0$, $F_1=1$, $\forall n \in \Bbb N, F_{n+2}=F_{n+1}+F_n$
It's the Fibonacci sequence.
Now let $\forall n \in \Bbb N,V_n=\begin{pmatrix}
F_n\\F_{n+1}\end{pmatrix}$
Let $M=\begin{pmatrix}0 & 1\\ 1 & 1\end{pmatrix}$
Then $MV_n=\begin{pmatrix}0 & 1\\ 1 & 1\end{pmatrix}\begin{pmatrix}
F_n\\F_{n+1}\end{pmatrix}=\begin{pmatrix}
F_{n+1}\\F_n+F_{n+1}\end{pmatrix}$
By induction, $\forall n \in \Bbb N, V_n=M^nV_0$
So the only thing we really need to compute is $M^n$. In order to do this, we try do diagonalize it.
Its characteristic polynomial is $\chi_M(\lambda)=\det(M-\lambda I_2)=\det \begin{pmatrix}-\lambda & 1\\ 1 & 1-\lambda\end{pmatrix}=(-\lambda)(1-\lambda)-(1)(1)=\lambda^2-\lambda-1$
Let $\varphi=\cfrac{1+\sqrt{5}}{2}$ and $\psi=\cfrac{1-\sqrt{5}}{2}$ be the two roots of that polynomial.
Since $M\in M_2(\Bbb R)$, we can find $P\in GL_2(\Bbb R)$ so that $M=PDP^{-1}$ where $D=\begin{pmatrix}\varphi & 0\\ 0 & \psi\end{pmatrix}$
$D$ is the matrix of the canonical endomophism associated to $M$ in a basis $e=(e_1,e_2)$ where $Me_1=\varphi e_1$ and $Me_2=\psi e_2$
We know that $e_1\in Ker(M-\varphi I_2) = Ker\left(\begin{pmatrix}-\varphi & 1\\ 1 & 1-\varphi\end{pmatrix}\right)=Vect\left(\begin{pmatrix}\cfrac{1}{\varphi}\\1\end{pmatrix}\right)$ because $\cfrac{1}{\varphi}\begin{pmatrix}-\varphi\\ 1 &\end{pmatrix}+1\begin{pmatrix} 1\\ 1-\varphi\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}$ so we can take $e_1=\varphi\begin{pmatrix}\cfrac{1}{\varphi}\\ 1\end{pmatrix}=\begin{pmatrix}1\\ \varphi\end{pmatrix}$
Similarly, $e_2\in Ker(M-\psi I_2) = Ker\left(\begin{pmatrix}-\psi & 1\\ 1 & 1-\psi\end{pmatrix}\right)=Vect\left(\begin{pmatrix}\cfrac{1}{\psi}\\1\end{pmatrix}\right)$ because $\cfrac{1}{\psi}\begin{pmatrix}-\psi\\ 1 &\end{pmatrix}+1\begin{pmatrix} 1\\ 1-\psi\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}$ so we can take $e_2=\psi\begin{pmatrix}\cfrac{1}{\psi}\\ 1\end{pmatrix}=\begin{pmatrix}1\\ \psi\end{pmatrix}$
And we get $P=\begin{pmatrix}1 & 1\\ \varphi & \psi\end{pmatrix}$
Then we need $P^{-1}=\cfrac{\begin{pmatrix}\psi & -1\\ -\varphi & 1\end{pmatrix}}{\psi-\varphi}$
Then $M^n=(PDP^{-1})^n=PD^nP^{-1}$ and $D^n=\begin{pmatrix}\varphi & 0\\ 0 & \psi\end{pmatrix}^n=\begin{pmatrix}\varphi^n & 0\\ 0 & \psi^n\end{pmatrix}$
So $\begin{pmatrix}
F_n\\F_{n+1}\end{pmatrix}=V_n=PD^nP^{-1}V_0=\cfrac{1}{\psi-\varphi}\begin{pmatrix}1 & 1\\ \varphi & \psi\end{pmatrix}\begin{pmatrix}\varphi^n & 0\\ 0 & \psi^n\end{pmatrix}\begin{pmatrix}\psi & -1\\ -\varphi & 1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\cfrac{1}{\psi-\varphi}\begin{pmatrix}1 & 1\\ \varphi & \psi\end{pmatrix}\begin{pmatrix}\varphi^n & 0\\ 0 & \psi^n\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}=\cfrac{1}{\psi-\varphi}\begin{pmatrix}1 & 1\\ \varphi & \psi\end{pmatrix}\begin{pmatrix}-\varphi^n\\\psi^n\end{pmatrix}=\cfrac{1}{\psi-\varphi}\begin{pmatrix}1 & 1\\ \varphi & \psi\end{pmatrix}\begin{pmatrix}-\varphi^n\\\psi^n\end{pmatrix}=\cfrac{1}{\psi-\varphi}\begin{pmatrix}-\varphi^n+\psi^n\\-\varphi^{n+1}+\psi^{n+1}\end{pmatrix}=\begin{pmatrix}\cfrac{\psi^n-\varphi^n}{\psi-\varphi}\\ \cfrac{\psi^{n+1}-\varphi^{n+1}}{\psi-\varphi}\end{pmatrix}$
From which we can deduce $F_n = \cfrac{\psi^n-\varphi^n}{\psi-\varphi}$
It's more often written as $\cfrac{\varphi^n-\psi^n}{\varphi-\psi}$ because $\varphi-\psi=\sqrt{5}$ whereas $\psi-\varphi=-\sqrt{5}$.
And we finally get $\boxed{F_n = \cfrac{\varphi^n-\psi^n}{\sqrt{5}}}$
|
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|
$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $ Help me please to prove that:
for any $\triangle ABC$ we have the following inequality: $$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $$
It's about convexity ?
thanks :)
|
$\sin A+\sin B=\sin(\frac{A+B}{2}+\frac{A-B}{2})+\sin(\frac{A+B}{2}-\frac{A-B}{2})=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\leqslant2\sin\frac{A+B}{2}$. Similarly, $\sin B+\sin C\leqslant2\sin\frac{B+C}{2},\sin C+\sin A\leqslant2\sin\frac{A+C}{2}$. Add them together yields the result.
Note that the inequality holds is due to $\frac{A+B}{2}\in(0,\frac{\pi}{2})$ and similar for the rest.
|
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The least possible value How to find the least possible value for :$$(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$$
For every real $x$
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Hint:
$(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2 \ge 0$
$5x^2+1+4+9+16+25 \ge 15x$
$5x^2 +55 -15x \ge 0 \implies x^2+11-3x \ge 0$
Aliter: $(x-3)=k$
You get
$(k+2)^2+(k+1)^2+(k)^2+(k-1)^2+(k-2)^2 \ge 0$
$k^2+11 \ge 0$
|
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|
Show that $\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$ Show that
$$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$$
|
A more general result is in fact true.
$$\sum_{k=0}^{n} \dfrac{\sin\left(3^k x\right)}{\cos\left(3^{k+1} x\right)} = \dfrac{\tan\left(3^{n+1} x\right) - \tan(x)}2 \,\,\,\, (\spadesuit)$$
Take $n=2$ in $(\spadesuit)$, to get what you want.
To prove $(\spadesuit)$, first note that
$$\dfrac{\sin \left(3^k \cdot x \right)}{\cos \left(3^{k+1} \cdot x \right)} = \dfrac{\tan \left( 3^{k+1} \cdot x\right) - \tan \left(3^{k} \cdot x\right)}2 \,\,\,\, (\clubsuit)$$
Now telescopic summation gives you $(\spadesuit)$.
$(\clubsuit)$ is proved by noting the fact that
\begin{align}
\tan \left( 3^{k+1} \cdot x\right) - \tan \left(3^{k} \cdot x\right) & = \dfrac{\sin \left( 3^{k+1} \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)} - \dfrac{\sin \left( 3^{k} \cdot x\right)}{\cos \left( 3^{k} \cdot x\right)}\\
& = \dfrac{\sin \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right) - \sin \left( 3^{k} \cdot x\right)\cos \left( 3^{k+1} \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right)}\\
& = \dfrac{\sin \left(2 \cdot 3^k \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right)} = \dfrac{2 \sin \left(3^k \cdot x\right) \cos \left(3^k \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right)}\\
& = 2 \cdot \dfrac{\sin \left(3^k \cdot x \right)}{\cos \left(3^{k+1} \cdot x \right)}
\end{align}
|
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|
Prove that $\sum_{k=n^2+1}^{n^2+2n+1}\sqrt{k}\le 2n^2+2n+\frac{7}{6}$ prove that:
$$\displaystyle\sum_{k=n^2+1}^{n^2+2n+1}\sqrt{k}\le 2n^2+2n+\dfrac{7}{6},n\ge 1$$
I find this inequality is very strong. Thank you!
such:when $n=100$,we use the mathmatic
$$\displaystyle\sum_{k=10001}^{10201}\approx 20201.16666254125412794937\cdots $$
and $2n^2+2n+\dfrac{7}{6}|_{n=100}=20201.166666666666\cdots$
|
Note that your bound gets much tighter as $n$ increases, suggesting that some kind of limiting argument is involved.
With that insight, we have that
$$ \begin{align}
\sum_{k = n^2 + 1}^{k=n^2+2n+1} \sqrt{k} & = \sum_{k=n^2} ^{k=n^2 + 2n} \sqrt{k} + 1 \\
& = 1 + \sum_{k=n^2} ^{k=n^2+2n} \int_k^{k+1} \sqrt{k} \, dx \\
& \leq 1 + \sum_{k=n^2} ^ {k=n^2 + 2n} \int_{k}^{k+1} \sqrt{x} \, dx \\
& = 1 + \int_{n^2} ^{(n+1)^2} \sqrt{x} \, dx\\
& = 1 + [ \frac {2}{3} k^{\frac {3}{2}} ]_{n^2} ^{(n+1)^2} \\
& = 1 + [\frac {2}{3} (3n^2 + 3n+1) ] \\
& = 2n^2 + 2n + \frac {5}{3}
\end{align}
$$
Now, let's consider how much we over approximated by. Looking at the inequality, we over approximated by $\int_{k}^{k+1} \sqrt{x} - \sqrt{k}$. Since the square root function is concave, we have the extra triangle on top, which tells us that we over approximated by at least $\frac {1}{2} (\sqrt{k+1}-\sqrt{k})$. Summing this up, we get
$$\sum_{k=n^2}^{n^2+2n} \frac {1}{2} ( \sqrt{k+1}- \sqrt{k} ) = \frac {1}{2} ( k+1 - k) = \frac {1}{2}. $$
Hence, (back)accounting for this over-approximation, we know that
$$ \sum_{k = n^2 + 1}^{k=n^2+2n+1} \sqrt{k} \leq 2n^2 + 2n + \frac {5}{3} - \frac {1}{2} = 2n^2 + 2n + \frac {7}{6}. $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$f(x - 1) + f(x − 2) $ and the sum of coeficients If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$
and
$f(x)= ax^2 + bx + c$
what would be the value of $a+b+c$?
I was doing
$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$
a = 5
b = -2
c = 9
$(5-3)+(-2-3)+(9-3)$
But do not think is is correct
What would be correct approach?
|
If $f(x) = ax^2 + bx+ c$, what is $f(x-1)$ and what is $f(x-2)$?
Work those out, then add both expressions to equate to $5x^2-2x+9$.
|
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|
Multivariable limit $\lim_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1}$ $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$
According to my textbook the limit equals $2$.
What I have tried:
Using the squeeze theorem:
$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le 0 $$
I have also tried to use the squeeze theorem with two other equations and obtained different values:
$$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 - 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le -1 $$
|
Remember the difference of squares algebraic identity.
$$ A^2 - B^2 = (A - B)(A + B) $$
Why is that useful? With $A = \sqrt{x^2 + y^2 + 1}$ and $B = 1$, the denominator of your expression is $A - B$. With that in mind,
$$ \begin{align}
\frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} &= \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} \cdot \frac{\sqrt{x^2 + y^2 + 1} + 1}{\sqrt{x^2 + y^2 + 1} + 1} \\
&= \frac{(x^2 + y^2)\left(\sqrt{x^2 + y^2 + 1} + 1 \right)}{x^2 + y^2} \\
&= \sqrt{x^2 + y^2 + 1} + 1
\end{align} $$
Now, you can evaluate the limit as $(x, y) \to (0, 0)$ simply by evaluation, since this expression is continuous at the origin.
$$ \lim_{(x, y) \to (0, 0)} \sqrt{x^2 + y^2 + 1} + 1 = \sqrt{0^2 + 0^2 + 1} + 1 = 2. $$
|
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|
What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 + \frac{2}{3} (x^2-1)^{-2/3} $$
whereas according to Wolfram Alpha (see alternate form), the correct result is:
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 - \frac{2}{3} (x^2-1)^{-5/3} $$
So apparently, my calculation for $u'v$ is correct, but $uv'$ is wrong. What am I missing here?
|
Your $u'v$ is wrong. Check your calculation.
|
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|
Cubes, squares and minimal sums I have trouble solving the following task: i need to find positive integers a and b such that
1) $a \neq b$
2) $ \exists c \in \mathbb{N} : ~ a^2 + b^2 = c^3$
3) $\exists d \in \mathbb{N}: ~ a^3 + b^3 = d^2$
4) sum $a + b$ is minimal possible
Thanks in advance!
|
If $a=1250$ and $b=625$, then $a=2\cdot 5^4$ and $b=5^4$, and
$$
a^2 + b^2 = 2^2\cdot 5^8 + 5^8 = (4 + 1)5^8 = 5^9 = (5^3)^3
$$
$$
a^3 + b^3 = 2^3\cdot 5^{12} + 5^{12} = (8 + 1)5^{12} = 3^2(5^6)^2 = (3\cdot 5^6)^2
$$
This was the smallest example I found by an extremely primitive computer search.
|
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|
Prove $\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$ If $a,b$ and $c \ge 0$ and $ab + bc + ca = 1$, prove that the following inequality holds:
$$\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$$
I've tried two aproaches, but it seems like both doesn't work. Here they are:
Cauchy-Scwarz inequality
I've tried using the following formula of Cauchy-Scwarz:
$$\frac{x_1^2}{a_1} + \frac{x_2^2}{a_2} + \frac{x_3^2}{a_3} \ge \frac{(x_1 + x_2 + x_3)^2}{a_1 + a_2 + a_3}$$
And i get:
$$ LHS \ge \frac{(1+1+1)^2}{2a + 2bc + 2b + 2ca + 2c + 2ba + 3}$$
$$ LHS \ge \frac{3^2}{2(a+b+c) + 5}$$
$$ LHS \ge \frac{9}{2(a+b+c) + 5}$$
Now to prove that the RHS is bigger than or equal to 1.
$$ \frac{9}{2(a+b+c) + 5} \ge 1$$
$$ 9 \ge 2(a+b+c) + 5$$
$$ 2 \ge a+b+c$$
And her I'm stuck, what can I do now?
AM - GM inequality
This try doesn't even stand a chance, but I'll still post something, because somebody can recieve an idea.
$$2a + 2bc \ge 2\sqrt{4abc}$$
$$2a + 2bc + 1 \ge 4\sqrt{abc} + 1$$
$$\frac{1}{2a + 2bc + 1} \le \frac{1}{4\sqrt{abc} + 1}$$
And for oher 2 fraction i get the same and if I add them I get:
$$\frac{3}{4\sqrt{abc} + 1} \ge LHS \ge 1$$
Now even if i prove that: $\frac{3}{4\sqrt{abc} + 1} \ge 1$ si true, that doesn't mean the original inequality holds.
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(w^3)\geq0$, where $f(w^3)=-8w^6+A(u,v^2)w^3+B(u,v^2)$.
But $f$ is a concave function, which says that it's enough to prove our inequality for an extremal value of $w^3$ wich happens in the following cases.
*
*$w^3=0$.
Let $c=0$. Hence, $ab=1$ and we need to prove that
$$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{3}\geq1$$ or
$$\frac{a+b+1}{2(a+b)+5}\geq\frac{1}{3}$$ or
$$a+b\geq2,$$
which is AM-GM: $a+b\geq2\sqrt{ab}=2$;
*Two variables are equal.
Let $b=a$. Hence, $a\neq0$ and $c=\frac{1-a^2}{2a}$, where $0<a<1$
and after these substitutions we need to prove that
$$a^2(1-a)(1+3a-2a^2)\geq0.$$
Done!
|
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|
Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$
This is where I get stuck. Any help appreciated.
|
$\displaystyle \frac{1}{i^2}< \frac{1}{i(i-1)}=\frac{1}{i-1}-\frac{1}{i}$,for $i\ge 2$
So we have $\displaystyle 1+\sum_{i=2}^{k}\frac{1}{i^2}\le 1+\sum_{i=2}^{k}(\frac{1}{i-1}-\frac{1}{i})=2-\frac{1}{k}$
I think this is better than induction.
|
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|
Show that $x^4-10x^2+1$ is reducible over $\mathbb{Q}(\sqrt{6}), \mathbb{Q}(\sqrt{2}),$ and $\mathbb{Q}(\sqrt{3})$. Is there an easier method? Ok, I actually worked this out as I was typing it up. But my solution seems kind of inelegant and involves a lot of tedious algebra that I've omitted here. Can anyone think of an easier method?
(1) Reducibility over $\mathbb{Q}(\sqrt{6})$
Quite straightforward, as the substitution $y = x^2$ and the quadratic formula on $y^2 - 10y + 1$ give:
$x^4-10x^2+1 = (x^2-5-2\sqrt{6})(x^2-5+2\sqrt{6})$.
(2) Reducibility over $\mathbb{Q}(\sqrt{2})$
Further reduction gives: $(x-\sqrt{5-2\sqrt{6}})(x+\sqrt{5-2\sqrt{6}})(x-\sqrt{5+2\sqrt{6}})(x+\sqrt{5+2\sqrt{6}})$.
It's also not that hard to show $\sqrt{5-2\sqrt{6}} = \sqrt{2}-\sqrt{3}$ and
$\sqrt{5+2\sqrt{6}} = \sqrt{2}+\sqrt{3}$ so the reduction becomes:
$(x-\sqrt{2}-\sqrt{3})(x+\sqrt{2}+\sqrt{3})(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}-\sqrt{3})$.
If we label these a, b, c, d left to right then:
$a.c = x^2-2\sqrt{2}x -1$ and $b.d = x^2+2\sqrt{2}x -1$ so it's reducible over $\mathbb{Q}(\sqrt{2})$.
(2) Reducibility over $\mathbb{Q}(\sqrt{3})$
$a.d = x^2-2\sqrt{3}x+1$ and $b.c = x^2+2\sqrt{3} +1$ so it's reducible over $\mathbb{Q}(\sqrt{3})$.
|
Well, a possibility is to write it as a difference of two squares and thus factors into a sum and difference. For e.g.:
$$x^4-10x^2+1 = (x^2-5)^2 - 24$$
and noting that $24=4\cdot6$ is a square in $\mathbb{Q}(\sqrt{6})$, so this is the difference of two squares.
Similarly, $x^4-10x^2+1 = (x^2-1)^2 - 8x^2$ and note $8$ is a square in $\mathbb{Q}(\sqrt{2})$.
Again, $x^4 - 10x^2 + 1 = (x^2+1)^2 - 12x^2$ and $12$ is a square in $\mathbb{Q}(\sqrt{3})$.
|
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|
Sum of all numbers x such that $(3x^2+9x-2012)^{(x^3-2012x^2-10x+1)} = 1$? What is the sum of all $x$ such that $(3x^2 + 9x - 2012)^{(x^3-2012x^2-10x+1)} = 1$?
|
Recall that $a^0 = 1$ for all real $a \neq 0$, and $1^b = 1$ for all real $b$.
$$(3x^2 + 9x - 2012)^{\large(x^3-2012x^2-10x+1)} = 1 \iff$$
$$ $$
$$x^3 \color{blue}{\bf - 2012}x^2 - 10 x + 1 = 0\tag{1}$$
or $$3x^2 + 9x - 2012 = 1 \iff 3x^2 + 9x - 2013 = 0 \iff \color{red}{\bf 1}\cdot x^2 + \color{red}{\bf 3}x - 671 = 0\quad\quad\quad\quad\tag{2}$$
There exist $3$ roots $x_1, x_2, x_3$ to $(1)$, and $2$ roots $x_4, x_5$ to (2).
The sum $S$ you want is $$S = \color{blue}{\bf x_1 + x_2 + x_3} + \color{red}{\bf x_4 + x_5}$$
To obtain this, note the colored coefficients: $${\bf Sum} = \color{blue}{\bf 2012} - \color{red}{\bf \frac 31} = {\bf 2009}.$$
|
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|
Find the Laurent expansion in powers of $\,z\,$ and $\,1/z\,$ $f(z)=\large\frac{z+2}{z^2-z-2}$ in $1<|z|<2$ and then in $2<|z|<\infty$
I am unsure how the two different regions of $z$ affect the series expansion.
Any help would be appreciated.
|
$$f(z) = \frac{z+2}{z^2-z-2} = \frac{z+2}{(z-2)(z+1)} = \frac{4}{3(z-2)} - \frac{1}{3(z+1)} = -\frac{4}6 \frac{1}{z(1-\frac{z}{2})} - \frac{1}3 \frac{1}{z(1-\frac{-1}{z})} = -\frac{2}{3z} \sum_{k=0}^{\infty}\frac{z^k}{2^k} - \frac{1}{3z} \sum_{k=0}^{\infty}\frac{(-1)^k}{z^k}$$
This should give you convergence inside the annulus $1 < |z| < 2$.
|
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|
Finding the value of $\sum\limits_{k=0}^{\infty}\frac{2^{k}}{2^{2^{k}}+1}$ Does this weighted sum of reciprocals of Fermat numbers,
$$
F=\sum_{k=0}^{\infty}\dfrac{2^{k}}{2^{2^{k}}+1}
$$
have a nice closed form? Wolfram says it's $1$.
Thanks.
|
This might be another way of looking at Jyrki's hint, but here is the way I did this:
$$
\begin{align}
\color{#C0C0C0}{1-\frac1{2-1}+}\frac1{2+1}&=1-\frac2{4-1}\\
1-\frac{2}{4-1}+\frac{2}{4+1}&=1-\frac{4}{16-1}\\
1-\frac{4}{16-1}+\frac{4}{16+1}&=1-\frac8{256-1}\\
1-\frac8{256-1}+\frac8{256+1}&=1-\frac{16}{65536-1}\\
&\vdots
\end{align}
$$
Motivation behind this approach
One trick to try is conjugates. Noting that
$$
\frac{2^k}{2^{2^k}-1}-\frac{2^k}{2^{2^k}+1}=\frac{2^{k+1}}{2^{2^{k+1}}-1}
$$
we see, using Telescoping Series, that
$$
\begin{align}
\sum_{k=0}^{n-1}\frac{2^k}{2^{2^k}+1}
&=\sum_{k=0}^{n-1}\left(\frac{2^k}{2^{2^k}-1}-\frac{2^{k+1}}{2^{2^{k+1}}-1}\right)\\
&=1-\frac{2^n}{2^{2^n}-1}
\end{align}
$$
Therefore, letting $n\to\infty$,
$$
\sum_{k=0}^\infty\frac{2^k}{2^{2^k}+1}=1
$$
|
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|
Generating Functions- Closed form of a sequence We are given the following generating function : $$G(x)=\frac{x}{1+x+x^2}$$
The question is to provide a closed formula for the sequence it determines.
I have no idea where to start. The denominator cannot be factored out as a product of two monomials with real coefficients. Any sort of help to solve this problem is welcome!
|
If you want the Binet formula for this you can solve the matrix
Set up the partial fraction
$$\frac{x}{1+x+x^2}=\frac{A}{1-ax}+\frac{B}{1-bx}$$ where $$a=\frac{-1+\sqrt{3}i}{2}$$ and $$ b=\frac{-1-\sqrt{3}i}{2}$$ so that
$$A(1-bx)+B(1-ax)$$
The matrix becomes
\begin{bmatrix}
-b&-a&1\\[0.3em]
1&1&0
\end{bmatrix}
or
\begin{bmatrix}
\frac{1+\sqrt{3}i}{2}&\frac{1-\sqrt{3}i}{2}&1\\[0.3em]
1&1&0
\end{bmatrix}
which in reduced echelon form is
\begin{bmatrix}
1&0&\frac{-\sqrt{3}i}{3} \\[0.3em]
0&1&\frac{\sqrt{3}i}{3}
\end{bmatrix}
The closed form then becomes
$$a_n=\frac{-\sqrt{3}i}{3}\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\frac{\sqrt{3}i}{3}\left(\frac{-1-\sqrt{3}i}{2}\right)^n$$
$a_0=0$ $a_1=1$, $a_2=-1$, $a_3=0$, $a_4=1$, $a_5=-1$ etc.
|
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|
How can I prove that $xy\leq x^2+y^2$? How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?
|
\begin{align}
0\leq (x-y)^2 \implies
&
0\leq x^2-2xy +y^2
\\
\implies
&
2xy\leq x^2+y^2
\\
\implies
&
xy\leq \dfrac{x^2+y^2}{2}
\\
\implies
&
xy\leq {x^2+y^2}
\end{align}
since the clearly nonnegative real $x^2+y^2$ clearly satisfies $\dfrac{x^2+y^2}{2} \leq x^2+y^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving functional equation for generating function Find the functional equation for the generating function whose coefficients satisfy $$
a_n = \sum_{i=1}^{n-1}2^ia_{n-i}, \text{ for } n\ge 2, a_0 = a_1 = 1
$$
This is what I've tried so far:
$$
\begin{align}
g(x) -1 -x &= \sum_{n\ge2} \sum_{i=1}^{n-1} 2^i a_{n-i}\\
&=\sum_{n\ge2} \sum_{i=1}^n 2^i a_{n-i}x^n\\
\end{align}
$$
at which point I'm stuck. How do I proceed?
|
Let
$$g(x) = \sum_{k=0}^{\infty} a_k x^k$$
From the defining recurrence:
$$\begin{align}a_2 x^2 &= 2 a_1 x^2\\ a_3 x^3 &= 2 a_2 x^3 + 4 a_1 x^3\\a_4 x^4 &= 2 a_3 x^4 + 4 a_2 x^4 + 8 a_1 x^4\\a_5 x^5 &= 2 a_4 x^5 + 4 a_3 x^5 + 8 a_2 x^5 + 16 a_1 x^5\\ \ldots \end{align}$$
and so on. We then form
$$g(x) - a_0 - a_1 x = \sum_{k=2}^{\infty} a_k x^k$$
by summing the above terms. By rearranging like coefficients, we get
$$\begin{align}\sum_{k=2}^{\infty} a_k x^k &= \frac{1}{2} a_1 \sum_{k=2}^{\infty}(2 x)^k + \frac{1}{4} a_2 \sum_{k=3}^{\infty}(2 x)^k + \frac{1}{8} a_3 \sum_{k=4}^{\infty}(2 x)^k+\ldots\\ &= (2 x^2 a_1 + 2 x^3 a_2 + 2 x^4 a_3 + \ldots) \sum_{k=0}^{\infty} (2 x)^k \\ &= \frac{2 x}{1-2 x}\sum_{k=1}^{\infty} a_k x^k \end{align}$$
Using the initial conditions and the definition of $g$, we get the following functional equation:
$$g(x) - (1+x) = \frac{2 x}{(1-2 x)} [g(x) - 1] $$
Doing this algebra, we get Marvis's result:
$$g(x) = \frac{1-3 x - 2 x^2}{1-4 x}$$
|
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|
If $ a+b+c = \frac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54}$
If $a+b+c = \dfrac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $$\dfrac{a}{b^3+54}+\dfrac{b}{c^3+54}+\dfrac{c}{a^3+54} \qquad ?$$
My try:
$$\begin{align*}
\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54} &= \frac{a^2}{ab^3+54a}+\frac{b^2}{bc^3+54b}+\frac{c^2}{ca^3+54c}\\
&\geq \frac{(a+b+c)^2}{ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)}
\end{align*}$$
Now, how can I proceed after that? Thanks.
|
Point to note: The minimum value is not achieved when $a=b=c=\frac{3}{2}$, which gives a value of $\frac{4}{51}$. The minimum value is achieved at $a=\frac{3}{2}, b=3, c=0$, which gives $\frac{2}{27}$. Note that $\frac{4}{51}>\frac{2}{27}$. (and cyclic permutations)
We shall smooth towards this equality case.
Note: There was an error with the smoothing approach, and it has been removed. The following only addresses $c=0$.
When $c=0$, we are left to minimise $$\frac{a}{b^3+54}+\frac{b}{54}=\frac{\frac{9}{2}-b}{b^3+54}+\frac{b}{54}$$
where $0 \leq b \leq \frac{9}{2}$. This is now a 1 variable inequality, which is easy to work with. Now,
\begin{align}
& \frac{\frac{9}{2}-b}{b^3+54}+\frac{b}{54} \geq \frac{2}{27} \\
\Leftrightarrow &54(\frac{9}{2}-b)+b(b^3+54) \geq 4(b^3+54) \\
\Leftrightarrow &b^4-4b^3+27 \geq 0 \\
\Leftrightarrow &(b-3)^2(b^2+2b+3) \geq 0
\end{align}
The last inequality is clearly true, so the minimum value is $\frac{2}{27}$, when $c=0, b=3, a=\frac{3}{2}$ (and cyclic permutations)
Note: The following shows the motivation to getting the value $\frac{2}{27}$, with equality case $b=3$. Some parts may be just based on intuition and not necessarily rigorous.
Motivation:
Let us suppose that the minimum value is $k$; expanding should give us a quartic, which should have a repeated root at the value of $b$ which gives $k$. (Intuitively)
\begin{align}
& \frac{\frac{9}{2}-b}{b^3+54}+\frac{b}{54} \geq k \\
\Leftrightarrow &54(\frac{9}{2}-b)+b(b^3+54) \geq 54k(b^3+54) \\
\Leftrightarrow &b^4-54kb^3+243-54^2k \geq 0 \\
\end{align}
Let's do the substitution $l=54k$ to simplify the coefficients, so $$b^4-lb^3+(243-54l) \geq 0$$ We perform another substitution $b=3x$ to simplify further; now $0 \leq x \leq \frac{3}{2}$, and the inequality becomes $$81x^4-27lx^3+(243-54l) \geq 0$$, or equivalently $$3x^4-lx^3+(9-2l) \geq 0$$
We want this to have a repeated root, so we differentiate, and get $12x^3-3lx^2$, which has roots $x=0$ and $x=\frac{l}{4}$. If $x=0$ were to be a repeated root, we would have $l=\frac{9}{2}$ and the quartic inequality becomes $3x^4 \geq \frac{9}{2}x^3$, which fails for $0<x<\frac{9}{2}$.
Therefore let us try to make $x=\frac{l}{4}$ as the repeated root. We would then require $$0=\left(\frac{l}{4}\right)^3(3(\frac{l}{4})-l)+(9-2l)=9-2l-\frac{l^4}{256}$$
Since I want to avoid calculations as much as possible, let us substitute $m=\frac{l}{4}$. ($m$ is also the value of $x$ where there is a repeated root) Then
$$m^4+8m-9=0$$
$$(m-1)(m^3+m^2+m+9)$$
We want non-negative $m$, so let's take $m=1$. Then $k=\frac{l}{54}=\frac{4m}{54}=\frac{2}{27}$, and there is a repeated root at $b=3m=3$. Now the quartic polynomial inequality is easy to prove; since we know now that $b=3$ is a repeated root, and thus can factor out $(b-3)^2$ to get a quadratic.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/357988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
limit with $\arctan$ I have to find the limit
and want ask about a hint:
$$\lim_{n \to \infty} n^{\frac{3}{2}}[\arctan((n+1)^{\frac{1}{2}})- \arctan(n^{\frac{1}{2}})]$$
I dont have idea what to do. Derivatives and L'Hôpital's rule are so hard
|
Write
$$\tan \left(\arctan\left( (n+1)^{1\over2}\right) - \arctan\left( n^{1\over2}\right) \right) = \frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n(n+1)}} = \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}$$
Thus
$$\arctan\left( (n+1)^{1\over2}\right) - \arctan\left( n^{1\over2}\right) = \arctan \left( \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}\right) $$
And
$$ \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}} \sim \frac{n^{1\over 2} \left( 1 + \frac{1}{2n} + \mathcal{O}\left(\frac{1}{n^2} \right) -1\right)}{n} \sim \frac{1}{2n^{3\over2}}$$
So, since $\arctan x \underset{x \rightarrow 0}{\sim} x$,
$$ \arctan \left( \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}\right) \sim \frac{1}{2n^{3\over2}} $$
And your limit is $1 \over 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/358368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
Find this $\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2}{2k+1}$ Let $m,n,k\in \mathbb{N}$, and $m,n,k\ge 1,m\neq n\neq k\neq m $, such that
$$
\dfrac{1}{2m+1}+\dfrac{1}{2n+1}=\dfrac{2}{2k+1}
$$
Is there a solution? Or does this not have any solution?
|
$\dfrac{1}{2m+1} +\dfrac{1}{2n+1}$, letting $2m+1$ to be the LCM of $2m+1$ and $2n+1$, Assume $2m+1=(2n+1) \cdot k$, your expression becomes $\dfrac{k+1}{2m+1}=\dfrac{2}{2l+1}$
$\dfrac{(k+1)}{2m+1}=\dfrac{2}{2l+1} \implies k+1=\dfrac{2(2m+1)}{2l+1}$. You just need a condition that $2l+1|2m+1$. Do you see that infinitely many solutions exists for this?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/358479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Residue of hyperbolic function How would I find the residue at $z_0=0$ of $$f(z)=\frac{\sinh(z)}{z^4(1-z^2)}$$I tried writing it as a series and reach $$\frac{1}{1-z^2}\sum_{n=0}^\infty \frac{z^{2n-3}}{(2n+1)!}$$ and then don't know where to go form there. Any help/hints appreciated.
|
$$\sinh z = z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots$$
$$\frac{1}{z^4}\sinh z = \frac{1}{z^3} + \frac{1}{z3!} + \frac{z}{5!} + \cdots$$
Now,
$$\frac{1}{1-z^2} = 1 + z^2 + z^4 + \cdots$$
So then, multiplying the series (note we only want to the the $z^{-1}$ terms, so no need to multiply every term)
$$\frac{1}{z^4(1-z^2)}\sinh z = \\
\left(1 + z^2 + z^4 + \cdots\right)\left(\frac{1}{z^3} + \frac{1}{z3!} + \frac{z}{5!} + \cdots\right) = \\
\cdots + \frac{1}{z3!} + \frac{z^2}{z^3} + \cdots =\\
\cdots + \frac{7}{6z} \cdots
$$
So the residue is $\frac{7}{6}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/358566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.