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Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies
$$\sin A+1=4\cos A.$$
Then what to do?
|
Let $t=\tan \frac{A}{2}$ then $\tan A = \frac{2t}{1-t^2}$ and $\cos A = \frac{1-t^2}{1+t^2}$ so $$\begin{align*}\frac{1+t^2}{1-t^2} + \frac{2t}{1-t^2}&=\frac{(1+t)^2}{1-t^2} \\ & = \frac{(1+t)(1+t)}{(1-t)(1+t)} \\ & = \frac{1+t}{1-t} = 4\end{align*}$$
So, assuming $t\neq 1$ we get $t = \frac{3}{5}$. From this, you can find $\cos A$ and $\sin A$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2243571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Subtraction of fractional exponents with different bases I wanna know if $$\left[(x^2+h)^3-(x+h)\right]^{1/2}-[x^3-x]^{1/2}=\left[(x+h)^3-(x+h)-x^3+x\right]^{1/2}.$$
|
Does $a^2 - b^2 = (a-b)^2\;$?
No, Because $$(a-b)^2 = a^2 -2ab +b^2 \neq a^2 - b^2$$
Alternatively $$a^{1/2} - b^{1/2} = \sqrt a -\sqrt b \neq \sqrt{a-b}$$
Simply take $a = 4, b = 9$. Then we have $2 - 3 = -1$. Whereas $\sqrt{2-3}$ is undefined in the reals, and which will never be equal to a negative real number.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
|
One more way to prove it is by substituting out for $z$:
$$
x^2+y^2+z^2 = x^2+y^2+(1-x-y)^2=2x^2+2y^2+2xy-2x-2y+1
$$
Now, substitute $x=\hat x+a$ and $y=\hat y+b$. We get
$$
2\hat x^2+2\hat y^2+2\hat x\hat y+(4a+2b-2)\hat x+(2a+4b-2)\hat y+(2(a^2+ab+b^2-a-b)+1)
$$
We can eliminate the order 1 terms by letting $4a+2b-2=2a+4b-2=0$, which gives $a=b=\frac13$. With this substitution, we have
$$
2(\hat x^2+\hat x\hat y+\hat y^2)+\frac13
$$
We can express the bracketed term as a sum of squares, giving us
$$
3(\hat x+\hat y)^2+(\hat x-\hat y)^2 + \frac13
$$
and we can see that the smallest value this can take is $\frac13$. Indeed, it takes this value when $\hat x=\hat y=0$ - that is, when $x=y=\frac13$.
|
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"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
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|
Reverse engineer the gamma question from a typo $\int_0^1(\ln1/x)^{1/3}$dx and the answer $\Gamma(4/3)$ Background: question 9.1.2.d in a newly published book Mathematics for Physical Science by Harris.
My question is what was the question supposed to be? Taking ln(x) for ln1 I'm doing something wrong or not guessing the right typo:
$$\int_0^1(\ln(x)/x)^{1/3}\mathrm{dx}$$
$$x=e^{-u}$$
$$-\int_0^\infty (-u)^{1/3}*e^{-2u/3}(-\mathrm{du})$$
$$\int_0^\infty (-3/2u)^{1/3}*e^{-u}*(\frac{3}{2}\mathrm{du})$$
|
Notice that:
$$\int_0^\infty \frac{3}{2}\left(-3/2u\right)^{\frac{1}{3}}e^{-u}\mathrm{du} = \int_0^\infty \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}u^{\frac{1}{3}}e^{-u}\mathrm{du} = \\
\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\int_0^\infty u^{\frac{4}{3}-1}e^{-u}\mathrm{du} =\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{4}{3}\right). $$
Recall also that $$\Gamma(z) = (z-1)\Gamma(z-1).$$
Then:
$$\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{4}{3}\right) = \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right) = \frac{1}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{1}{3}\right).$$
The results is confirmed also by Wolframalpha. Actually, you can't get rid of the multiplied term in front of the final expression.
|
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|
Let $x, y, z$ be different prime numbers with $x, y, z > 3$. Prove that if $x + z = 2y$, then $6 | (y - x)$. I have troubles to prove the following task:
Let $x, y, z$ be different prime numbers with $x, y, z > 3$. Prove that if $x + z = 2y$, then $6 | (y - x)$.
The only idea I have is that every prime number $> 3$ divided by $6$ has remainder $1$ or $5$.
But I do not have any idea how to prove this statement?!
Thank you for any help!
|
$6| y-x$ if and only if $3|y-x$ and $2|y-x$.
1) $2|y-x$.
$y$ and $x$ are both primes larger than $2$ so $x$ and $y$ or both odd. So $y-x$ is even.
2) $3|y-x$.
$y > 3$ and $y$ is prime so $y \equiv \pm 1 \mod 3$. $x > 3$ and $x$ is prime so $x \equiv \pm 1 \mod 3$. and similarly $z \equiv \pm 1 \mod 3$.
Either $x \equiv y \mod 3$ or $x \equiv -y \mod 3$.
If $x \equiv -y \mod 3$ then $2z \equiv x+y \equiv -y+y \equiv 0 \mod 3$. But $2z \equiv 2*\pm 1 \equiv \mp 1 \not \equiv 0 \mod 3$.
So $x \equiv y \mod 3$. So $y-x \equiv 0 \mod 3$ and $3|y-x$.
|
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|
Use relations to show the value of $ C $ I am asked to show that $$ \lim_{n \to \infty} \frac{n!}{a_n} = \sqrt{2 \pi}, $$ where $$ a_n = \sqrt{n} \cdot \left (\frac{n}{e} \right)^n. $$
And I am only allowed to use the following relations $$ \sum_{k = 0}^{2n} \dbinom{2n}{k} \cdot 2^{-2n} = 1, \qquad \int_{- \infty}^{\infty} e^{-x^2} \,dx = \sqrt{\pi}. $$
I know that this is the Sterling's approximation formula for $ n! $, but I'm not sure what I'm supposed to do. Any help is appreciated.
|
The following proof has two parts: in the first part I will prove that $n!\sim C\sqrt{n}\left(\frac{n}{e}\right)^n$, in the second part I will show that $C=\sqrt{2\pi}$. It is crucial to notice that
$$ m = \prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)\tag{1}$$
from which:
$$ n! = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right) = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}\tag{2}$$
and
$$ \log\left(\frac{n^n}{n!}\right) = (n-1)-\sum_{k=1}^{n-1}\left[1-k\log\left(1+\frac{1}{k}\right)\right] \tag{3} $$
where $1-k\log\left(1+\frac{1}{k}\right)$ is bounded between $\frac{1}{2k}-\frac{1}{3k^2}$ and $\frac{1}{2k}$ for any $k\geq 1$. This leads to
$$ \log\left(\frac{n^n}{n!}\right) = n-\frac{1}{2}\log(n)+K+O\left(\frac{1}{n}\right)\tag{4} $$
and the first part is granted, $n!\sim C\sqrt{n}\left(\frac{n}{e}\right)^n$. As a consequence,
$$ \lim_{n\to +\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n} = \lim_{n\to +\infty}\frac{C\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}\sqrt{n}}{4^n C^2 n \left(\frac{n}{e}\right)^{2n}}=\frac{\sqrt{2}}{C}\tag{5}$$
and it is enough to show that $ \lim_{n\to +\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n} = \frac{1}{\sqrt{\pi}}$. On the other hand
$$ \frac{(2n+1)!!}{(2n)!!}=\prod_{k=1}^{n}\left(1+\frac{1}{2k}\right) = \sqrt{\prod_{k=1}^{n}\left(1+\frac{1}{k}\right)\prod_{k=1}^{n}\left(1-\frac{1}{(2k+1)^2}\right)^{-1}} \tag{6} $$
leads to the identity
$$ \frac{2n+1}{4^n}\binom{2n}{n} = \sqrt{n+1}\sqrt{\prod_{k=1}^{n}\left(1-\frac{1}{(2k+1)^2}\right)^{-1}} \tag{7} $$
and
$$ \prod_{k\geq 1}\left(1-\frac{1}{(2k+1)^2}\right)=\frac{\pi}{4}\tag{8} $$
by the Weierstrass product for the cosine function. This finishes the proof.
As an alternative,
$$ \frac{(2n+1)!!}{(2n)!!}=\prod_{k=1}^{n}\left(1+\frac{1}{k}\right) = \frac{2\,\Gamma\left(\frac{3}{2}+n\right)}{\Gamma\left(\frac{1}{2}\right)\,\Gamma(n+1)}\sim \frac{2}{\Gamma\left(\frac{1}{2}\right)}\sqrt{n}\tag{9} $$
for large $n$s by Gautschi's inequality ($\log\Gamma$ is a convex function on $\mathbb{R}^+$, by the Bohr-Mollerup characterization of just by applying the Cauchy-Schwarz inequality to its integral representation), and
$$ \Gamma\left(\frac{1}{2}\right) = \int_{0}^{+\infty}z^{-1/2}e^{-z}\,dz = 2 \int_{0}^{+\infty}e^{-x^2}\,dx = \int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}.\tag{10}$$
|
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|
If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \sqrt{ab}$.
Because $\gcd(a,b) = 1$, there exists integers $k$ and $l$ such that $ak + bl = 1$. This means that $\sqrt{a}(k\sqrt{}) + \sqrt{b}(l\sqrt{b}) = 1$.
Hence $\sqrt{a}$ and $\sqrt{b}$ are both positive integers and we can set $\sqrt{a} = c$ for some arbitrary integer $c$ and $\sqrt{b} = d$ for some arbitrary integer $d$. Therefore, $a = c^2$ and $b = d^2$.
|
Because $n$ is a positive integer, $n^2$ has an even number of factors, and each factor appears an even number of times in its factorization. The prime factorizations of $a$ and $b$ must together contain $2$ of each of all elements of the prime factorization of $n$. However, since $\gcd(a,b)=1$, $a$ and $b$ share no factors. Then each factor $f_k$ of $n$ must occur twice as many times in either $a$ or $b$ (not both) than it does in $n$, so each factor of $a$ and $b$ occurs an even number of times. Since each factor occurs an even number of times in both $a$ and $b$, they are both perfect squares.
|
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if $2^n+1$ is prime, then n = $2^r$ and $(x - y) \mid(x^k - y^k$) only with natural numbers I need to prove that if $2^n + 1$ is prime, then $n = 2^r$ for a natural number $r$.
I do know how to prove it with the lemma $(x - y) \mid (x^k - y^k)$, but in order to prove it with this lemma, I need to substitute $y$ with $-1$ and my problem is, that in my case, this lemma is only given if $x$, $y$, $k$ are natural numbers.
Could you please tell me how to prove it without needing a number from $Z$ to prove this theorem?!
Thank you!
|
Suppose that $p$ is odd prime divisior of $n$. Then $n=p\cdot m$,
$$ 2^n +1 = (2^m)^p + 1^p = (2^m + 1)\cdot ((2^m)^{p-1}-(2^m)^{p-2}+ \cdots + 1)$$
and each factor greater than $1$. So $2^n +1 $ isn't prime which gives a contradiction.
Some Notes:
1) We use the identity $x^{2k+1}+y^{2k+1}=(x+y)\cdot (x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k})$. Proof of it is not difficult. By distribution property
$(x+y)\cdot (x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k}) \\ =\left(x^{2k+1}-x^{2k}y + x^{2k-1}y^2 - \cdots -x^2y^{2k-1}+xy^{2k}\right)+\left(x^{2k}y-x^{2k-1}y^2 + x^{2k-2}y^3 - \cdots -xy^{2k}+y^{2k+1}\right) \\ =x^{2k+1}+y^{2k+1}$.
2) We can see that $x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k} >1$. Otherwise,if $$x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k} =1$$
then
$$2^n +1=2^m +1 $$
and $n=m$ which gives a contradiction.
|
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normal equations and projections (linear algebra) I have three points $(-1,7)$, $(1,7)$ and $(2,21)$ and a linear model $b = C + Dt$. It is asking to give and solve the normal equations in $\hat{x}$ to find the projection $p = A\hat{x}$ of $b$ onto the column space of $A$ (i.e. find $\hat{x}$ with least $||A\hat{x} - b||^2$).
I think I've found the normal equations to be $3C + 2D = 35$ and $2C + 6D = 42$ and the $b = 9 + 4t$, but I'm not sure where to go from there.
|
Problem statement
Given a sequence of $m=3$ data points of the form $\left\{ x_{k}, y_{k} \right\}$, and a model function
$$
y(x) = a_{0} + a_{1} x
$$
find the least squares solution
$$
a_{LS} = \left\{
a\in\mathbb{C}^{n} \colon
\lVert
\mathbf{A} a - y
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Linear system
$$
%
\begin{align}
%
\mathbf{A} a & = y \\
%
\left[
\begin{array}{cr}
1 & -1 \\
1 & 1 \\
1 & 2 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
a_{0} \\
a_{1}
\end{array}
\right]
%
&=
\left[
\begin{array}{c}
7 \\
7 \\
21 \\
\end{array}
\right]
\end{align}
%
$$
Normal equations
$$
%
\begin{align}
%
\mathbf{A}^{*} \mathbf{A} a &= \mathbf{A}^{*} y \\
%
\left[
\begin{array}{cc}
3 & 2 \\
2 & 6 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
a_{0} \\
a_{1}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
35 \\
42
\end{array}
\right]
%
\end{align}
%
$$
Solution via normal equations
$$
%
\begin{align}
%
a_{LS} &= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}y \\
%
&=\frac{1}{14}
%
\left[
\begin{array}{rr}
6 & -2 \\
-2 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
35 \\
42
\end{array}
\right] \\
%
&=
%
\left[
\begin{array}{c}
9 \\
4
\end{array}
\right]
%
\end{align}
%
$$
Solution function
$$ y =9 + 4 x $$
Is the best fit a good fit?
The residual error vector is
$$
r =\mathbf{A} a_{LS} - y =
\left[
\begin{array}{r}
-2 \\
6 \\
-4
\end{array}
\right]
$$
The function which was minimized is the total error
$$
r^{2} = r \cdot r = 56.
$$
Solution plotted against the data points:
Projections
What is the projection on the column space?
$$
\mathbf{A} a = a_{0} \mathbf{A}_{1} + a_{1} \mathbf{A}_{2} = 9
\left[
\begin{array}{r}
1 \\
1 \\
1
\end{array}
\right]
+ 4
\left[
\begin{array}{r}
-1 \\
1 \\
2
\end{array}
\right] =
%
\left[
\begin{array}{r}
5 \\
13 \\
17
\end{array}
\right]
\in
\mathcal{R}\left( \mathbf{A} \right)
$$
Errata
Given an invertible matrix $\mathbf{A}$, the inverse matrix can be computed using
$$
\mathbf{A}^{-1} = \frac{\text{adj }\mathbf{A}} {\det \mathbf{A}}
$$
where the adjugate matrix adj $\mathbf{A}$ is the matrix of cofactors and $\det \mathbf{A}$ is the determinant.
For the product matrix in the example, the adjugate matrix is
$$
\text{adj }\mathbf{A}^{*}\mathbf{A} =
%
\left[
\begin{array}{rr}
6 & -2 \\
-2 & 3 \\
\end{array}
\right]
%
$$
and the determinant is
$$
\det \mathbf{A}^{*}\mathbf{A} = 18 - 4 = 14.
$$
The inverse of the product matrix is
$$
\left( \mathbf{A}^{*}\mathbf{A} \right) ^{-1}
= \frac{1}{14}
%
\left[
\begin{array}{rr}
6 & -2 \\
-2 & 3 \\
\end{array}
\right].
$$
Thanks to @spacedustpi for clarification.
|
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|
Maximum value of $4x-9y$ Suppose xand y are real numbers and that $x^2 +9y^2 -4x +6y+4=0$ then we have to find the maximum vale of 4x-9y
I found that the equation given is that of ellipse .
I can consider a line $4x-9y=0$ which cut the ellipe .
bt by doing this its getting very comlicated .
|
We'll use Lagrange's multiples:
$$f(x,y) = 4x-9y , \phi(x,y) = (x-2)^2 + (3y+1)^2 $$
$$F(x,y) = 4x-9y - \lambda(x-2)^2 - \lambda (3y+1)^2 $$
And so:
$$\frac{\partial F}{\partial x} = 4 - 2\lambda (x-2) = 0 \Rightarrow x-2 = \frac{2}{\lambda} $$
$$ \frac{\partial F}{\partial y} = -9 - 6\lambda (3y+1) = 0 \Rightarrow 3y+1=-\frac{3}{2\lambda}$$
Now substitute in $\phi$:
$$(x-2)^2+(3y+1)^2=1 \Rightarrow \frac{4}{\lambda^2} + \frac{9}{4\lambda^2}=1 $$
$$ \lambda^2 = 4 + \frac94 = \frac{25}4 \Rightarrow \lambda = \pm \frac{5}{2} $$
Substitute $\lambda$ in $x,y$ and get:
$$ x-2 = \frac{2}{\lambda} \Rightarrow x = \frac{14}{5} , \frac{6}{5} $$
$$ 3y + 1 = -\frac{3}{2\lambda} \Rightarrow y = -\frac{8}{15} , -\frac{2}{15} $$
Now we need to solve for $4x-9y$. For $\lambda = \frac{5}{2}$ we get $16$ and for $\lambda = -\frac{5}{2} $ we get $6$. So the maximum is 16!
|
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|
Find all positive integer solutions of the equation $n^5+n^4 = 7^m-1$
Find all positive integer solutions of the equation $n^5+n^4 = 7^m-1$.
Suppose $n$ is odd. Then we have $$n^5+n^4 \equiv \pm 2 \not \equiv 7^m-1 \pmod{7}.$$ Thus $n$ is even. Let $n = 2k$ for some positive integer $k$, and we have $$n^4(n+1) = 16k^4(2k+1) = 7^m-1,$$ so since $7^m \equiv 7 \pmod{16}$ for odd $m$ and $7^m \equiv 1 \pmod{16}$ for even $m$, it follows that $m$ is even. Let $m = 2j$ for some positive integer $j$. Then the equation is $$16k^4(2k+1) = 7^{2j}-1.$$ Thus, $7^{2j} = 16k^4(2k+1)+1$.
I didn't see how to continue.
|
Let's start by observing that
$$
n^5+n^4 + 1 = (n^2 + n + 1)(n^3-n+1).
$$
Now, let $d=gcd(n^2 + n + 1,n^3-n+1)$. Note that, $d\mid n^2 + n + 1 \implies d\mid n^3 - 1$. Hence, $n^3 \equiv 1\pmod{d}.$ But since $d|n^3-n+1$, we must have, $n\equiv 2\pmod{d}$. Hence, $d|7$, therefore $d$ is either $1$ or $7$.
Note that $n=2$ and $m=2$ is a solution. Suppose $n \geq 3$. $n^3-n+1>n^2+n+1$. From the condition on $d$, we must either have $n^2+n+1 = 1$ or $7$. But clearly, the only $n$ that fulfills this is $n=2$.
Hence, $(2,2)$ is the only solution.
|
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|
Probability of drawing a spade on the first draw, a heart on the second draw, and an ace on the third draw My answer is
$(\frac{1}{52})(\frac{1}{51})(\frac{2}{50})$ + $2(\frac{1}{52})(\frac{12}{51})(\frac{3}{50})$ + $(\frac{12}{52})(\frac{12}{51})(\frac{4}{50})$
= $\frac{1}{204}$
But the answer in the textbook is
$(\frac{13}{52})(\frac{13}{51})(\frac{2}{50})$ + $2(\frac{13}{52})(\frac{13}{51})(\frac{3}{50})$ + $(\frac{13}{52})(\frac{13}{51})(\frac{4}{50})$
= $\frac{13}{850}$
Which answer is correct?
|
Using the probability tree diagram:
|
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|
How to prove this inequality? Thanks
Let $x_{1},x_{2},\cdots,x_{n}\in [0,1]$, show that
$$\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^{\frac{n}{2}}\cdot\sqrt{1-x_{i}}\right]\le 1$$
Maybe it can use McLaughlin inequality to solve it?I found this inequality simaler this problem
|
Claim: For all integer $n$, positive $c$ satisfying $n/2 \le c$ and $x_{1},x_{2},\cdots,x_{n}\in [0,1]$,$$\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]\le 1$$
Proof:
Let $f(x_1, x_2, \cdots, x_n, c)=\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]$.
Let's proceed by mathematical induction.
Base case is trivial because $\sqrt{1-x_1} \le 1$.
For induction case, we need to check the critical points and when any of $x_i$ is $0$ or $1$.
Firstly, the cases where any of $x_i$ is $1$ is covered by induction hypothesis. Also, when $x_i=0$, $f=\left(\prod_{j\neq i }x_{j}\right)^c\le1$.
Let's compute the critical points of $f$. First note that
$\frac{\partial f}{\partial x_i}=-\frac{1}{2\sqrt{1-x_i}}\left(\prod_{j\neq i }x_{j}\right)^c+\frac{c}{x_i}\sum_{k\neq i}\left[\left(\prod_{j\neq k }x_{j}\right)^c\cdot\sqrt{1-x_{k}}\right]$ and assuming $\prod_ix_i\neq 0$, $\frac{\partial f}{\partial x_i}=0$ is equivalent to$$\frac{2c-(2c-1)x_i}{2x_i^{c}\sqrt{1-x_i}}=c\sum_{k}\left[\frac{\sqrt{1-x_{k}}}{x_k^c}\right]$$and this means $\frac{2c-(2c-1)x_i}{2x_i^{c}\sqrt{1-x_i}}$ is same for all $x_i$. One can check that the function $g(x)=\frac{2c-(2c-1)x}{2x^{c}\sqrt{1-x}}$ has one local minimum, therefore on a critical point, there are only $2$ values that $x_i$s can get. Let $y, z$ be numbers satisfying $y \le z, g(y)=g(z)$ and suppose there are $l$ $x_i$s satisfying $x_i=y$, and other $n-l$ $x_i$s satisfies $x_i=z$. Then,
$f=\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]=y^{cl}z^{c(n-l)}\left(l\frac{\sqrt{1-y}}{y^c}+(n-l)\frac{\sqrt{1-z}}{z^c}\right)$
From $g(y)=g(z)$, $\frac{\sqrt{1-y}}{y^c}=\frac{\sqrt{1-z}}{z^c}\left(\frac{2c-(2c-1)z}{1-z}\right)\left(\frac{1-y}{2c-(2c-1)y}\right)$. Applying this,$$f=y^{cl}z^{c(n-l-1)}\sqrt{1-z}\left(l\left(\frac{2c-(2c-1)z}{1-z}\right)\left(\frac{1-y}{2c-(2c-1)y}\right)+n-l\right)$$
Note that $l\left(\frac{2c-(2c-1)z}{1-z}\right)\left(\frac{1-y}{2c-(2c-1)y}\right)+n-l\le n$ and $y^{cl}z^{c(n-l-1)}\sqrt{1-z} \le z^{c(n-1)}\sqrt{1-z} \le z^{n(n-1)/2}\sqrt{1-z}$. $z^{n(n-1)/2}\sqrt{1-z}$ is maximized at $z=1-\frac{1}{n^2-n+1}$. Then we know$$z^{c(n-1)}\sqrt{1-z} \le \left(1-\frac{1}{n^2-n+1}\right)^{n(n-1)/2}\sqrt{\frac{1}{n^2-n+1}}$$
and $\left(1-\frac{1}{n^2-n+1}\right)^{n(n-1)/2}\approx\exp(-0.5)$, in fact, it is less than $2/3$ for all integer $n>1$. Also, $\frac{2n}{3(n-0.5)}<1$ if $n \ge 2$. Therefore, we finally get$$f \le (z^{c(n-1)}\sqrt{1-z})n\le \left(1-\frac{1}{n^2-n+1}\right)^{n(n-1)/2}\sqrt{\frac{n^2}{n^2-n+1}}\le\frac{2n}{3(n-0.5)}<1$$
And we proved the claim.
|
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|
Volume of the revolution solid Let $K \subset \mathbb R^2$ be area bounded by curves $x=2$, $y=3$, $xy=2$ and $xy=4$.
What is the volume of the solid we get after rotating $K$ around $y$-axis?
I tried to express the curves in terms of $y$ and solved the integral $\displaystyle \pi\left(\int_2^3 \frac{16}{y^2}dy - \int_1^3\frac{4}{y^2}dy \right)$ but it came out zero. I got the integral by subtracting the volume of the inner shape from the volume of the outer shape, if that makes any sense.
|
Do this in two parts. First find the section of the volume between $y=3$ and $y=2$, then the section of the volume between $y=2$ and $y=1$. For the first volume, use the integral
$$\pi\int_{2}^{3} \frac{16}{x^2}dx-\pi\int_{2}^{3} \frac{4}{x^2}dx$$
This is the difference of the volumes formed by rotating the regions between $xy=4$ and the y-axis and between $xy=2$ and the y-axis between $y=2$ and $y=3$. When you evaluate these integrals, you get
$$\pi\int_{2}^{3} \frac{12}{x^2}dx$$
$$\pi(-\frac{12}{3}+\frac{12}{2})$$
$$\pi(6-4)$$
$$2\pi$$
Now for the second region. For this I use the integrals
$$4\pi-\pi\int_{1}^{2} \frac{4}{x^2}dx$$
Which is the difference of the volumes of the cylinder made by rotating a rectangle about the x-axis and the volume of the area under $xy=2$ rotated about the x-axis. This gives us
$$4\pi-\pi(-\frac{4}{2}+\frac{4}{1})$$
$$4\pi-2\pi$$
$$2\pi$$
The total volume is the sum of the volumes, which is
$$4\pi$$
Is this the correct answer?
|
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|
Deriving left nullspace of matrix from $EA=R$ Let $A$ be an $m\times n$ matrix, $R$ be its row-reduced echelon form, and $E$ be the sequence of matrices $E_k\dots E_1$ used to bring $A$ to $R$, such that $EA = R$. In one of the books on linear algebra it is said that we can use the fact that $EA=R$ to find the basis for the left nullspace of $A$, without the need to bring $A^T$ to the row-reduced echelon form. Is my understanding correct that all we need to do is, since $(EA)^T=A^T$, just row-reduce $R^T$? Or, even better, just read the basis off from $R^T$ without even row-reducing it?
|
Here are two detailed examples:
Find base and dimension of given subspace
Given a matrix and its reduced row echelon form, resolve the image and the kernel.
Fundamental Theorem of Linear Algebra
Given $\mathbf{A}\in\mathbb{C}^{m\times n}$, the four fundamental subspaces are
$$
\begin{align}
%
\mathbf{C}^{n} =
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
\mathbf{C}^{m} =
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
\color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
\end{align}
$$
Column space
$$
\begin{align}
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}_{4} \\
\end{array}
\right]
& \mapsto
\left[
\begin{array}{c|c}
\mathbf{E_{A}} & \mathbf{R} \\
\end{array}
\right] \\
%
\left[
\begin{array}{rrr|cccc}
1 & 1 & -1 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 & 0 \\
-1 & 0 & 1 & 0 & 0 & 1 & 0 \\
-2 & 1 & 2 & 0 & 0 & 0 & 1 \\
\end{array}
\right]
& \mapsto
\left[
\begin{array}{ccr|rrcc}
\boxed{1} & 0 & -1 & 1 & -1 & 0 & 0 \\
0 & \boxed{1} & 0 & 0 & 1 & 0 & 0 \\\hline
0 & 0 & 0 & \color{red}{1} & \color{red}{-1} & \color{red}{1} & \color{red}{0} \\
0 & 0 & 0 & \color{red}{2} & \color{red}{-3} & \color{red}{0} & \color{red}{1} \\
\end{array}
\right]
\tag{1}
\end{align}
$$
The boxed pivot entries identify the fundamental columns of the $\color{blue}{range}$ space. The $\color{red}{red}$ row vectors form a span for the $\color{red}{null}$ space.
The column space is now resolved.
$$
\begin{align}
%
\mathbf{C}^{m} =
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
\color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
% range
&=
\text{span } \left\{ \,
\color{blue}{\left[
\begin{array}{r}
1 \\
0 \\
-1 \\
-2 \\
\end{array}
\right]}, \,
\color{blue}{\left[
\begin{array}{r}
1 \\
1 \\
0 \\
1 \\
\end{array}
\right]},
\, \right\}
% null
\oplus
\text{span } \left\{ \,
\color{red}{
\left[
\begin{array}{r}
1 \\
-1 \\
1 \\
0 \\
\end{array}
\right],
\,
\left[
\begin{array}{r}
2 \\
-3 \\
0 \\
1 \\
\end{array}
\right]}
\, \right\}%
%
\end{align}
$$
Row space
$$
\begin{align}
%
\left[
\begin{array}{c|c}
\mathbf{A}^{T} & \mathbf{I}_{4} \\
\end{array}
\right]
& \mapsto
\left[
\begin{array}{c|c}
\mathbf{E_{A^{T}}} & \mathbf{R} \\
\end{array}
\right] \\
%
\left[
\begin{array}{rcrr|ccc}
1 & 0 & -1 & -2 & 1 & 0 & 0 \\
1 & 1 & 0 & 1 & 0 & 1 & 0 \\
-1 & 0 & 1 & 2 & 0 & 0 & 1 \\
\end{array}
\right]
&\mapsto
\left[
\begin{array}{ccrr|rcc}
\boxed{1} & 0 & -1 & -2 & 1 & 0 & 0 \\
0 & \boxed{1} & 1 & 3 & -1 & 1 & 0 \\\hline
0 & 0 & 0 & 0 & \color{red}{1} & \color{red}{0} & \color{red}{1} \\
\end{array}
\right]
\tag{2}
%
\end{align}
$$
The row space is now resolved:
$$
\begin{align}
%
\mathbf{C}^{n} =
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
\color{red} {\mathcal{N} \left( \mathbf{A} \right)}
%
% range
&=
\text{span } \left\{ \,
\color{blue}{\left[
\begin{array}{r}
1 \\
1 \\
-1 \\
\end{array}
\right]}, \,
\color{blue}{\left[
\begin{array}{r}
0 \\
1 \\
0 \\
\end{array}
\right]},
\right\}
% null
\oplus
\text{span } \left\{ \,
\color{red}{
\left[
\begin{array}{r}
1 \\
0 \\
1 \\
\end{array}
\right]}
\, \right\}
%
\end{align}
$$
Challenge
As an example for this question, the task is to use the blue row vectors in (1)
$$
\left[
\begin{array}{ccr|rrcc}
\color{blue}{1} & \color{blue}{0} & \color{blue}{-1} & 1 & -1 & 0 & 0 \\
\color{blue}{0} & \color{blue}{1} & \color{blue}{0} & 0 & 1 & 0 & 0 \\\hline
0 & 0 & 0 & \color{red}{1} & \color{red}{-1} & \color{red}{1} & \color{red}{0} \\
0 & 0 & 0 & \color{red}{2} & \color{red}{-3} & \color{red}{0} & \color{red}{1} \\
\end{array}
\right],
\tag{3}
$$
to find a vector in
$$
\text{span } \left\{ \,
\color{red}{
\left[
\begin{array}{r}
1 \\
0 \\
1 \\
\end{array}
\right]}
\, \right\}.
\tag{4}
$$
What the reduction in (1) provides is another span
$$
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}
=
\text{span } \left\{ \,
\color{blue}{\left[
\begin{array}{r}
1 \\
1 \\
-1 \\
\end{array}
\right]}, \,
\color{blue}{\left[
\begin{array}{r}
0 \\
1 \\
0 \\
\end{array}
\right]}
\, \right\}.
$$
And yes, you could look at that span and conclude (4). But this is not "reading" the vectors directly as in the red terms in (1) and (2). For example, can you look at (3) and guess the spans for the column space?
Conclusion
To summarize: There is no general method to resolve the row space from the reduction matrix $\left[
\begin{array}{c|c}
\mathbf{E_{A}} & \mathbf{R} \\
\end{array}
\right]$.
The row vectors in the upper left quadrant of $\mathbf{E_{A}}$ are in the span of $\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}$ and in some simple cases you will be able to deduce a span for $\color{red}{\mathcal{N} \left( \mathbf{A} \right)}$.
|
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|
Uniform convergence of $f_n(x) =\sqrt[n]{1+x^n}$ I need to decide if the following function uniformly converges:
$$f_n(x) =\sqrt[n]{1+x^n} \quad, \quad x\in[0,\infty)$$
I found the sequence pointwise converges to
$$f(x) =
\begin{cases}
1 \; , & \text{$x\in[0,1]$} \\
x \; ,& \text{$x \in(1,\infty)$}
\end{cases}$$
I tried to find the supremum of $\;|f_n-f|\;$ but got nothing. Any ideas how I can get forward?
|
Hint. For $x>1$,
$$ \sqrt[n]{1+x^n} - x = \frac{1}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}(1+x^n)^{\frac{1}{n}}+\cdots +x^{\frac{1}{n}}{(1+x^n)^{\frac{n-1}{n}}}} \le \frac{1}{n}$$
and for $x\le 1$,
$$ \sqrt[n]{1+x^n} - 1 = \frac{x^n}{1+(1+x^n)^{\frac{1}{n}}+\cdots +{(1+x^n)^{\frac{n-1}{n}}}} \le \frac{1}{n}$$
|
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|
Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it's maximum whenever $z=0$. (Is this conclusion correct? I have doubt here).
So the problem reduces to maximise $f(x, y, 0)$ which again can be shown that $f(x, y, 0)\le f(x, 2x,0)$ and this completes the proof with maximum of $9$ and equality at $(1,2,0)$ and it's permutations.
Is it correct? I strongly believe even it might have faults there must be a similar way and I might have made mistakes. Every help is appreciated
|
You have correctly established that $z=0$. From there you have $y=3-x$ so substitute that into $f$. As $y\le2$ then $1\le x\le2$.
$$f(x,3-x,0)=x^3+(3-x)^3=9x^2+27x+27=9(x^2+3x+3)$$
$$=9\left(x-\frac{3}{2}\right)^2+\frac{27}{4}$$
This quadratic has minimum at $x=\frac{3}{2}$ and the maximum is only limited by the domain of $x$ which leads to the answer of $x=1$ or $x=2$ so the three numbers are $0,1,2$ and the maximum is $9$.
|
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|
Infinite Sum Calculation: $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$ Problem
Show the following equivalence: $$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Good afternoon, dear StackExchange community. I'm studying real analysis (the topic right now is exchanging limits) and I can't wrap my head around this problem. It might be a duplicate, but I couldn't find a thread that helped me to solve this problem.
Also, I know that both terms equals $\frac{\pi^2}{8}$ (courtesy of Wolframalpha), but I think I should show the original problems via exchanging limits of double series or with analyising the summands, because we didn't introduce the identity $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi}{6}$ yet.
My Attempts
If we consider the first summands of both sums, we have:
\begin{array}{|c|c|c|c|}
\hline
k & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{(2k+1)^2} & 1 & \frac{1}{9} & \frac{1}{25} & \frac{1}{49} & \frac{1}{81} & \frac{1}{121} & \frac{1}{169} \\ \hline
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{n^2} & / & 1 & \frac{1}{4} & \frac{1}{9} & \frac{1}{16} & \frac{1}{25} & \frac{1}{36} \\ \hline
\end{array}
We immediately see that in $\frac{1}{n^2}$ is every summand of $\frac{1}{(2k+1)^2}$ is included. Additionally, we see that the extra summands in $\frac{1}{n^2}$ have the form $\frac{1}{(2x)^2}$.
Therefore, we could write,
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}$$
and the original equation would be
$$\Rightarrow 1 + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \left( 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\right)$$.
But I don't know how this could help me.
Anyways, series and infinite sums give me headaches and I would appreciate any help or hints. Thank you in advance.
Wow! Thank you for the really quick replies!
|
$\displaystyle \begin{align}\sum_{n\ge 0} \dfrac{1}{(2n+1)^s}&= \sum_{n=1}^\infty \dfrac{1}{n^s}-\sum_{n=1}^\infty \dfrac{1}{(2n)^s} \\ &= \sum_{n=1}^\infty \dfrac{1}{n^s}(1-2^{-s}) \\ &= (1-2^{-s})\sum_{n=1}^{\infty} \dfrac{1}{n^s}\end{align}$
Putting $s=2$ gives you the result and obviously the above is true for $s\in\mathbb{N}$
|
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|
If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^2 x$$
$$=\sin^4 x + \sin^3 x + \sin^3 x + \sin^2 x$$
$$=\sin^3 x(\sin x +1) +\sin^2 x(\sin x +1)$$
$$=(\sin x +1) (\sin^3 x +\sin^2 x)$$
How do I proceed further?
|
Almost finished from there:
$$(\sin x + 1)(\sin^3 x + \sin^2 x)\\ = (\sin x + 1)(\sin x + 1) \sin^2 x\\ = (\sin^2 x + \sin x) (\sin^2 x + \sin x) \\= 1$$
|
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|
Different ways to solve $\int_0^1 \sqrt{1+x^2} dx $ So I have the task to solve this integral in $4$ different ways, but have solved it only with substitution. ($x=tg(t)$, $dx=\sec^2(t)dt$) and so on. Any advice on the other $3$ ways ? Thank you :)
|
IBP? $u = \sqrt{1+x^2}$ and $\mathrm{d}v = 1$ so $v = x$. Hence $$I = \int_0^1 \sqrt{1+x^2} \, \mathrm{d}x = \bigg[x\sqrt{1+x^2}\bigg]_0^1 - \int_0^1 \frac{x^2}{\sqrt{1+x^2}} \, \mathrm{d}x$$
Then $\int_0^1 \frac{x^2}{\sqrt{1+x^2}} \, \mathrm{d}x = \int_0^1 \frac{1+x^2}{\sqrt{1+x^2}} \, \mathrm{d}x - \int_0^1 \frac{1}{\sqrt{1+x^2}} \, \mathrm{d}x$. So $$I = \sqrt{2} - I + \int_0^1 \frac{1}{\sqrt{1+x^2}} \, \mathrm{d}x$$ or equivalently $$I = \frac{\sqrt{2}}{2} + \frac{1}{2}\text{arsinh} \,1$$
Alternatively, if you accept other substitutions $\sqrt{x^2 + 1} + x = t$ should work; this is known as Euler's substitution.
|
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|
How many real roots of $f(x)=x^{12}-x^9+x^4-x+1$ between $0$ and $1$? How can we know if $f(x) = x^{12} - x^9 + x^4 - x + 1$ has How many real roots between 0 and 1. ?
Okay so at both 1 and 0 the value of function is 1. So there can be no or even roots between
them so how can we tell can we use calculus ?
|
$\begin{cases}
f(x)=x^{12}-x^9+x^4-x+1 \\
f'(x)=12x^{11}-9x^8+4x^3-1 \\
f''(x)=132x^{10}-72x^7+12x^2=12x^2(11x^8-6x^5+1)=12x^2g(x) \\
g'(x)=88x^7-30x^4=x^4(88x^3-30)
\end{cases}$
And we can stop here, because we are only interested in the signs of the derivatives.
$a=\sqrt[3]{\frac{30}{88}}$
$\begin{array}{|c|ccccc|}
\hline
x & 0 && a && 1\\
\hline
g'& 0 & - & 0 & + & 58 \\
\hline
g & 1 & \searrow & \simeq 0.62 & \nearrow & 6 \\
\hline
f'' & 0 && + && 72\\
\hline
f' & -1 && \nearrow && 6 \\
\hline\end{array}$
So there exists $b\in]0,a[$ such that $f'(b)=0$.
$\begin{array}{|c|ccccc|}
\hline
x & 0 && b && 1\\
\hline
f'& -1 & - & 0 & + & 6 \\
\hline
f & 1 & \searrow & f(b) & \nearrow & 1 \\
\hline
\hline\end{array}$
From there you can calculate $b$ by dichotomy $b\simeq 0.6746$ and then compute $f(b)\simeq 0.51245>0$.
We are dealing with polynomial here, so continuity is assured and abrupt variations excluded so we can be quite confident that $f>0$ on $[0,1]$.
Now, if you want to pursue the algebraic proof, you'll have to substitute $b$ : we know that $bf'(b)=12b^{12}-9b^9+4b^4-b=0$
So $12f(b)=(9b^9-4b^4+b)+12(-b^9+b^4-b+1)=-3b^9+8b^4-11b+12$
And then carry on with this lesser degree equation until you get $f(b)>0$.
You wanted a general method without tricks, you got one.
|
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"source": "stackexchange",
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|
$\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$ In MAO 1991,
Find $2x+5$ if $x$ satisfies $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$
My attempt,
I squared the the equation then I got $144x^2+1648x+4480=144x^2-1632x+4624$, which results $x=-9$, and $2x+5=-13$.
I want to ask is there another way to solve this question as my method is very tedious. Thanks in advance.
|
An $x$ that solves your equation also solves the system
\begin{align}
40&&-9x&-a^2&=0\\
7&&-x&-b^2&=0\\
&&-x&-c^2&=0\\
&&a-2b&-c&=0\\
\end{align}
If we subtract the second and the third equation we get
$$-b^2+c^2+7=0.$$
It turns out that
$$-9(-b^2+c^2+7)=-(3b-4c)(-3b-4c)-(-7c^2+63).$$
Therefor one solution is $c^2-9=0$ and $3b-4c=0$. Both solutions to $c$ work. If we pick $c=3$ we get $x=-9$ from the third equation and $b=4$ which makes $x=-9$ consistent with the second equation.
|
{
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|
Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$ Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$
Hints are appreciated. Thanks in advance.
|
Another way:
$$x^{18} + x^{24} = x^{21} (x^3 + \frac{1}{x^3})\\ = x^{21} ((x+\frac{1}{x})^3 - 3(x+\frac{1}{x}))\\ = x^{21}(3\sqrt{3} - 3\sqrt{3}) = 0. $$
|
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"source": "stackexchange",
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|
For integer m greather than 2, $\frac{1}{m} + \frac{1}{m+2}$, the numerators and denomitors are primitive pythagorean triples $a$ and $b$ For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of $\frac{1}{m} + \frac{1}{m+2}$ show primitive pythagorean triples a and b.
|
$$\frac{1}{m} + \frac{1}{m+2} = \frac{2m+2}{m^2 + 2m}$$
Now notice that
$$(2m+2)^2 + (m^2 + 2m)^2=8m^2+4m+4+m^4+4m^3=(m^2+2m+2)^2$$
|
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|
Proving Ramanujan's Nested Cube Root Ramanujan's Nested Cube:
If $\alpha,\beta$ and $\gamma$ are the roots of the cubic equation$$x^3-ax^2+bx-1=0\tag{1}$$then, they satisfy$$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=(a+6+3t)^{1/3}\tag{2.1}$$
$$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=(b+6+3t)^{1/3}\tag{2.2}$$where$$t^3-3(a+b+3)t-(ab+6(a+b)+9)=0\tag3$$
The formula (2.1) is what Ramanujan used to get
$$\sqrt[3]{\cos\tfrac {2\pi}7}+\sqrt[3]{\cos\tfrac {4\pi}7}+\sqrt[3]{\cos\tfrac {8\pi}7}=\sqrt[3]{\tfrac 12\left(5-3\sqrt[3]7\right)}$$
by starting with $x^3+x^2-2x-1=0$ along with its trigonometric roots $\cos\frac {2\pi}7,\>\cos\frac {4\pi}7, \>\cos\frac {8\pi}7$ on LHS, and then getting RHS with $a=-1,b=-2$.
Question:
*
*How to prove the formulas (2.1) and (2.2)?
*Is there a standard procedure to find trigonometric roots of a polynomial?
I first started off with a function $x^3-px^2+qx-1=0$ and assumed that the roots were $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$. That way, by Vieta's formula, we have$$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=p$$$$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=-q$$However, I'm not sure how to represent the RHS in $(2.1)$ or $(2.2)$
EDIT: I found a proof, but something doesn't match up. I have posted another question here
|
We have $\alpha$ , $\beta$ and $\gamma$ are roots of the equation $x^3-ax^2+bx-1=0$ so
\begin{eqnarray*}
a= \sum \alpha \\
b= \sum \alpha \beta \\
\alpha \beta \gamma =1
\end{eqnarray*}
so $\sqrt[3]{\alpha \beta \gamma} =1$ and let
\begin{eqnarray*}
A= \sum \sqrt[3]{\alpha} \\
B= \sum \sqrt[3]{\alpha \beta}
\end{eqnarray*}
Cube these equations and multiply them
\begin{eqnarray*}
A^3= a+3\sum \sqrt[3]{\alpha^2 \beta} +6\\
B^3= b+3\sum \sqrt[3]{\alpha^2 \beta} +6\\
AB= \sum \sqrt[3]{\alpha^2 \beta}+3
\end{eqnarray*}
let $t=\sum \sqrt[3]{\alpha^2 \beta}$ and cube $AB=t+3$ We have
\begin{eqnarray*}
t^3+9t^2+27t+27 = A^3B^3=(a+3t +6)(b+3t +6)\\
t^3=3(a+b+3)t+(ab+6(a+b)+9).
\end{eqnarray*}
Thus the equation is shown.
|
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|
Find all solutions to $|x^2-2|x||=1$ Firstly, we have that $$\left\{
\begin{array}{rcr}
|x| & = & x, \ \text{if} \ x\geq 0 \\
|x| & = & -x, \ \text{if} \ x<0 \\
\end{array}
\right.$$
So, this means that
$$\left\{
\begin{array}{rcr}
|x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\
|x^2+2x| & = & 1, \ \text{if} \ x<0 \\
\end{array}
\right.$$
For the first equation, we have
$$|x^2-2x|\Rightarrow\left\{\begin{array}{rcr}
x^2-2x & = & 1, \ \text{if} \ x^2\geq 2x \\
x^2-2x & = & -1, \ \text{if} \ x^2<2x \\
\end{array}
\right.$$
and for the second equation, we have
$$|x^2+2x|\Rightarrow\left\{\begin{array}{rcr}
x^2+2x & = & 1, \ \text{if} \ x^2+2x\geq 0 \\
x^2+2x & = & -1, \ \text{if} \ x^2+2x<0 \\
\end{array}
\right.$$
Solving for all of these equations, we get
$$\left\{\begin{array}{rcr}
x^2-2x & = 1 \Rightarrow& x_1=1+\sqrt{2} \ \ \text{and} \ \ x_2=1-\sqrt{2}\\
x^2-2x & =-1 \Rightarrow& x_3=1 \ \ \text{and} \ \ x_4=1\\
x^2+2x & = 1 \Rightarrow& x_5=-1-\sqrt{2} \ \ \text{and} \ \ x_6=-1+\sqrt{2}\\
x^2+2x & =-1 \Rightarrow& x_7=-1 \ \ \text{and} \ \ x_8=-1
\end{array}
\right.$$
So we have the roots $$\begin{array}{lcl}
x_1 = & 1+\sqrt{2} \\
x_2 = & 1-\sqrt{2} \\
x_3 = & -1+\sqrt{2} \\
x_4 = & -1-\sqrt{2} \\
x_5 = & 1 \\
x_6 = & -1
\end{array}$$
But according to the book, the answer is
\begin{array}{lcl}
x_1 & = & 1+\sqrt{2} \\
x_4 & = & -1-\sqrt{2} \\
x_5 & = & 1 \\
x_6 & = & -1
\end{array}
What happened to $x_2$ and $x_3$? Any other way to solve this equation quicker?
|
The most general way is to grow a tree and then for each leaf of the tree a table. If more than a couple of layers of $|.|$ you will probably be starting to confuse yourself if you don't stick to a systematic approach.
Each branch in the tree reduce the set the variable is valid for. We need to store a pair $(expression,set)$ at each node and at the leafs of the tree, we will have an expression with no $|.|$ left, just a polynomial and a set. That is when we can make a table splitting the real number line.
*
*First tree branch is due to $|x|$: $x\in[0,\infty]$ left $x\in[-\infty,0]$ right.
*left does $|x|\to x$, right does $|x|\to -x$
*Now store pairs sets and expressions $|x^2-2x|$ left , $|x^2+2x|$ right
*In our new nodes we need to factor polynomials to find how to split the tree up in subsets > and <0. But hopefully the systemacy of the approach is clear enough by now.
|
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|
$a/b < (1 + \sqrt{5})/2 \iff a^2 - ab - b^2 < 0$? For positive integers $a$ and $b$, I want to show that $a/b < (1 + \sqrt{5})/2$ if and only if $a^2 - ab - b^2 < 0$.
I had a loose proof ready to go, but I noticed a fatal flaw. Perhaps there is a way to work around this though.
My tactic was to start from $a^2 - ab - b^2 < 0$ and complete the square on the LHS for $a$. I ended up with
$$
\left(a-\frac{b}{2}\right)^2 - \frac{5b^2}{4} < 0 \iff \left(a - \frac{b}{2} \right)^2 < \left( \frac{\sqrt{5} \, b}{2} \right)^2.
$$
Now the tempting thing to do is to show this is equivalent to saying
$$
\quad \quad \qquad \quad \, \, \iff a-\frac{b}{2} < \frac{\sqrt{5} \, b}{2}
$$
but obviously it is necessary for $a > b / 2$ for this to work.
This is not necessarily the case because if $a = 1$ and $b = 5$, then $a^2 - ba - b^2 = -29 < 0$ and $a/b = 1/5 < (1+\sqrt{5})/2$ so the initial claim is true but $a \leq b/2$.
So is there some kind of assumption I can make to get around this, and without loss of generality? Or should I rethink the entire structure of the proof? Cheers!
|
If $b=0, a^2<0$ which is impossible
So, $b\ne0, b^2>0$ consequently, $$a^2-ab-b^2<0\iff\left(\dfrac ab\right)^2-\left(\dfrac ab\right)-1<0$$
Now the roots of $x^2-x-1=0$ are $$x=\dfrac{1\pm\sqrt5}2$$
We can prove if $(x-a)(x-b)<0$ with $a<b;$
$$a<x<b$$
|
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|
Is there name for Ring, with the following Cayley tables I'm solving a Cayley table for Ring with four elements and noticed that the resulting multiplication table looks pretty interesting. It has two left multiplication 1s and two left multiplications zeros and one of those zeros is ofcourse the zero in the Additive group of the Ring. Is there any special name for this Ring?
Let R = {a, b, c, d}
$$
\begin{array}{ l | c c c c }
+ & a & b & c & d \\ \hline
a & d & c & b & a \\
b & c & d & a & b \\
c & b & a & d & c \\
d & a & b & c & d
\end{array} \quad \quad \quad \quad
\begin{array}{ l | c c c c }
* & a & b & c & d \\ \hline
a & ~ & d & d & d \\
b & ~ & ~ & ~ & d \\
c & a & b & ~ & d \\
d & d & d & d & d
\end{array} \\
cc = c(a + b) = ca + cb = a + b = c \\
aa = a(c + b) = ac + ab = d + d = d \\
ba = (c + a)a = ca + aa = a + d = a \\
bb = (c + a)b = cb + ab = b + d = b \\
bc = (c + a)c = cc + ac = c + d = c \\
\begin{array}{ l | c c c c }
+ & a & b & c & d \\ \hline
a & d & c & b & a \\
b & c & d & a & b \\
c & b & a & d & c \\
d & a & b & c & d
\end{array} \quad \quad \quad
\begin{array}{ l | c c c c }
* & a & b & c & d \\ \hline
a & d & d & d & d \\
b & a & b & c & d \\
c & a & b & c & d \\
d & d & d & d & d
\end{array}
$$
|
I've never heard a name for this rng, but I have seen it expressed this way:
Given the two-element semigroup $S=\langle b,c\mid b^2=cb=b, c^2=bc=c\rangle$, your rng is the semigroup rng $F_2[S]$.
|
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|
Show without multiplying out Show without multiplying out,
$$\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=\frac{(a-b)(b-c)(a-c)}{abc}$$
So, I'm given a solution which is
\begin{align}\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}&=\frac{bc(b-c)+ac(c-a)+ab(a-b)}{abc}\\
&=\frac{bc(b-c)+a^2(b-c)-a(b^2-c^2)}{abc}\\
&=\frac{(b-c)(bc+a^2-ab-ac)}{abc}\\
&=\frac{(a-b)(b-c)(a-c)}{abc}\end{align}
My question is, I don't know how to go from 1st line to 2nd line. I've no idea about it. Can anyone mind to explain for it? Thanks in advance.
My another question, is there any other way to solve this? Thanks.
|
Between line $1$ and line $2$ you've got $ac(c-a)+ab(a-b)$ as terms in the numerator on the right-hand-side. These terms expand out as $ac^2-a^2c+a^2b-ab^2$ (we're not breaking the rules since we're not multiplying out the whole thing). Notice the two terms with $a^2$. We can rearrange these terms to $a^2(b-c)-a(b^2-c^2)$ (as your solution has done).
We then know by the difference of squares that $b^2-c^2=(b+c)(b-c)$ so then the numerator is $bc(b-c)+a^2(b-c)-a(b+c)(b-c)$, between lines $2$ and $3$. This means every term in the numerator has $(b-c)$ as a factor so $(b-c)$ can be factored out. We then end up with $(b-c)(bc+a^2-a(b+c))$, which simplifies to what you've got in line $3$.
So the question is how to factor $(a^2-ab-ac+bc)$. We can do this by inspection by noticing the $a^2$ term which tells us we've got $(a+?)(a+?)$. But what two numbers have a sum of $(-b-c)$ but a product of $bc$? The answer is $-b$ and $-c$. This tells us that $(a^2-ab-ac+bc)=(a-b)(a-c)$
And voila, out falls $bc(b-c)+ac(c-a)+ab(a-b)=(b-c)(a-b)(a-c)$.
|
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|
fine the limits :$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$ fine the limits-without-lhopital rule and Taylor series :
$$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$$
i know that :
$$\lim_{x \to 0} \frac{\sin x}{x}=1=\lim_{x \to 0}\frac{\tan x}{x}$$
But I can not answer please help .
|
If you know, that $\enspace\displaystyle \lim\limits_{x\to 0}\frac{1}{x^2}(1-\frac{\sin x}{x})=\frac{1}{3!}\enspace$ then you can answer your question easily:
$\displaystyle \frac{(\sin(2x)-2x\cos x)(\tan(6x)+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x\tan x\sin(2x)}=$
$\displaystyle =\frac{(\sin(2x)-2x\cos x)(\frac{\sin(6x)}{\cos(6x)}-\frac{\sin(6x)}{\cos(\frac{\pi}{3}-2x)\cos(\frac{\pi}{3}+4x)})}{x\sin x\tan x\sin(2x)}$
$\displaystyle =\frac{2\sin x\cos x -2x\cos x}{\sin x\tan x\sin(2x)}6\frac{\sin(6x)}{6x}(\frac{1}{\cos(6x)}-\frac{1}{\cos(\frac{\pi}{3}-2x)\cos(\frac{\pi}{3}+4x)})$
$\displaystyle =-\frac{1}{x^2}(1-\frac{\sin x}{x}) (\frac{x}{\sin x}\cos x)^2 \frac{2x}{\sin(2x)} 6\frac{\sin(6x)}{6x}(\frac{1}{\cos(6x)}-\frac{1}{\cos(\frac{\pi}{3}-2x)\cos(\frac{\pi}{3}+4x)})$
$\displaystyle \to -\frac{1}{3!}6(1-4)=3\enspace$ for $\enspace x\to 0$
A note about what I have used:
$\displaystyle \tan x=\frac{\sin x}{\cos x}$
$\sin(2x)=2\sin x\cos x$
$\displaystyle \tan x-\tan y=\frac{\sin(x-y)}{\cos x\cos y}$
|
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|
Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
Let $x^3+ax^2+bx+c=0$ are $\alpha, \beta, \gamma$. Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
My attempt,
As I know from the original equation,
$\alpha+\beta+\gamma=-a$
$\alpha\beta+\beta\gamma+\alpha\gamma=b$
$\alpha\beta\gamma=-c$
I've tried to expand $(\alpha+\beta+\gamma)^3$ which is equal to $\alpha^3+\beta^3+\gamma^3+3\alpha^2\beta+3\alpha\beta^2+3\alpha^2\gamma+6\alpha\beta\gamma+3\beta^2\gamma+3\gamma^2\alpha+3\gamma\beta$
Basically, I know I've to find what's the value of $\alpha^3+\beta^3+\gamma^3$, $\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3$ and $\alpha^3\beta^3\gamma^3$. But I
m stuck at it. I would appreciate can someone explain and guide me to it. Thanks a lot.
By the way, I would appreciate if someone provides another tactics to solve this kind of routine question. Thanks a lot.
|
Using basic algebra:
*
*Calculating $\alpha^3 + \beta^3 + \gamma^3$ :
As, \begin{aligned}
(&\alpha + \beta + \gamma)^3 = \alpha^3 + \beta^3 + \gamma^3 + 3\alpha^2\beta + 3\alpha^2\gamma + 3\alpha\beta^2 + 3\beta^2\gamma + 3\alpha\gamma^2 + 3\beta\gamma^2 + 6\alpha\beta\gamma
\end{aligned}
We can factor this, \begin{aligned}
(&\alpha + \beta + \gamma)^3 = (\alpha^3 + \beta^3 + \gamma^3) + 3(\alpha^2\beta + \alpha^2\gamma + \alpha\beta^2 + \beta^2\gamma + \alpha\gamma^2 + \beta\gamma^2) + 6\alpha\beta\gamma
\end{aligned}
\begin{aligned}\qquad = (\alpha^3 + \beta^3 + \gamma^3) + 3[\,(\alpha + \beta + \gamma)(\alpha\beta + \alpha\gamma + \beta\gamma) - 3\alpha\beta\gamma \,] + 6\alpha\beta\gamma
\end{aligned}
\begin{aligned} = (\alpha^3 + \beta^3 + \gamma^3) + 3(\alpha + \beta + \gamma)(\alpha\beta + \alpha\gamma + \beta\gamma) - 3\alpha\beta\gamma
\end{aligned}
We can rearrange this equation to get,
\begin{aligned} \boldsymbol\alpha^\boldsymbol3 \boldsymbol+ \boldsymbol\beta^\boldsymbol3 \boldsymbol+ \boldsymbol\gamma^\boldsymbol3 \boldsymbol= \boldsymbol(\boldsymbol\alpha \boldsymbol+ \boldsymbol\beta \boldsymbol+ \boldsymbol\gamma\boldsymbol)^\boldsymbol3 \boldsymbol- \boldsymbol3\boldsymbol(\boldsymbol\alpha \boldsymbol+ \boldsymbol\beta \boldsymbol+ \boldsymbol\gamma\boldsymbol)\boldsymbol(\boldsymbol\alpha\boldsymbol\beta \boldsymbol+ \boldsymbol\alpha\boldsymbol\gamma \boldsymbol+ \boldsymbol\beta\boldsymbol\gamma\boldsymbol) \boldsymbol+ \boldsymbol3\boldsymbol\alpha\boldsymbol\beta\boldsymbol\gamma
\end{aligned}
*Calculating $\alpha^3\beta^3 + \alpha^3\gamma^3 + \beta^3\gamma^3$
Since,
\begin{aligned} (\alpha\beta + \alpha\gamma + \beta\gamma)^3= \alpha^3\beta^3 + \alpha^3\gamma^3 + \beta^3\gamma^3 + 3\alpha^3\beta\gamma^2 + 3\alpha\beta^3\gamma^2 + 3\alpha^3\beta^2\gamma+ 3\alpha^2\beta^3\gamma + 3\alpha\beta^2\gamma^3 + 3\alpha^2\beta\gamma^3 + 6\alpha^2\beta^2\gamma^2 \end{aligned}
Factorizing this gives,
\begin{aligned} (\alpha\beta + \alpha\gamma + \beta\gamma)^3= (\alpha^3\beta^3 + \alpha^3\gamma^3 + \beta^3\gamma^3) + 3(\alpha^3\beta\gamma^2 + \alpha\beta^3\gamma^2 + \alpha^3\beta^2\gamma+ \alpha^2\beta^3\gamma + \alpha\beta^2\gamma^3 + \alpha^2\beta\gamma^3) + 6(\alpha\beta\gamma)^2 \end{aligned}
\begin{aligned} = (\alpha^3\beta^3 + \alpha^3\gamma^3 + \beta^3\gamma^3) + 3[\,(\alpha + \beta + \gamma)(\alpha\beta + \alpha\gamma + \beta\gamma)(\alpha\beta\gamma) - 3\alpha^2\beta^2\gamma^2\,] + 6(\alpha\beta\gamma)^2 \end{aligned}
\begin{aligned} = (\alpha^3\beta^3 + \alpha^3\gamma^3 + \beta^3\gamma^3) + 3(\alpha + \beta + \gamma)(\alpha\beta + \alpha\gamma + \beta\gamma)(\alpha\beta\gamma) - 3(\alpha\beta\gamma)^2 \end{aligned}
Rearranging this gives,
\begin{aligned} \boldsymbol\alpha^\boldsymbol3\boldsymbol\beta^\boldsymbol3 \boldsymbol+ \boldsymbol\alpha^\boldsymbol3\boldsymbol\gamma^\boldsymbol3 \boldsymbol+ \boldsymbol\beta^\boldsymbol3\boldsymbol\gamma^\boldsymbol3 \boldsymbol= \boldsymbol(\boldsymbol\alpha\boldsymbol\beta \boldsymbol+ \boldsymbol\alpha\boldsymbol\gamma \boldsymbol+ \boldsymbol\beta\boldsymbol\gamma\boldsymbol)^\boldsymbol3 \boldsymbol- \boldsymbol3\boldsymbol(\boldsymbol\alpha \boldsymbol+ \boldsymbol\beta \boldsymbol+ \boldsymbol\gamma\boldsymbol)\boldsymbol(\boldsymbol\alpha\boldsymbol\beta \boldsymbol+ \boldsymbol\alpha\boldsymbol\gamma \boldsymbol+ \boldsymbol\beta\boldsymbol\gamma\boldsymbol)\boldsymbol(\boldsymbol\alpha\boldsymbol\beta\boldsymbol\gamma\boldsymbol) \boldsymbol+ \boldsymbol3\boldsymbol(\boldsymbol\alpha\boldsymbol\beta\boldsymbol\gamma\boldsymbol)^\boldsymbol2 \end{aligned}
p.s.I know it's too late to reply now but, hope it's useful for somebody :)
|
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|
How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers. Problem: How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers.
I tried a similar approach to the ones described here, but to no avail. I can't split up the RHS so that all the factors have the same base.
|
The product $n=3^8\cdot 5^6\cdot13^4$ has a total of $(8+1)(6+1)(4+1)=315$ factors $d$. Those can be paired up as $(d,n/d)$. For the choice $d=n/d=\sqrt n$ the two factors are equal. For the remaining $157=(315-1)/2$ pairs we must use the bigger one as $x+y$ and the smaller as $x-y$. Because all the factors $d, n/d$ are odd, the resulting system has a solution $(x,y)$ in integers.
The answer is thus $157$. They come from solutions of the system $x+y=n/d, x-y=d$, that is,
$$
x=\frac{(n/d)+d}2,\quad y=\frac{(n/d)-d}2,
$$
with $d$ ranging over the set of factors $<\sqrt n$.
|
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|
Values of $m$ that satisfy the quartic equation $x^4-(3m+2)x^2+m^2 = 0$ which has 4 real roots in AP The question says find all values for $m$, such that the below equation has 4 real roots in Arithmetic Progression.
My approach:
\begin{gather}
x^4-(3m+2)x^2+m^2 = 0 \\
\\
\text{Let roots be }\\
\beta+d \\
\beta-d \\
\beta+2d \\
\beta-2d \\
\\
x^4+0x^2-(3m+2)x^2+0x+m^2 = 0\\
\implies 0 = (\beta-2d) + (\beta-d) + (\beta+2d) + (\beta+d) = 4\beta \\
\implies \beta = 0\\
\\
-(3m+2) = (-d)(d)+(-d)(2d)+(-d)(-2d)+(d)(2d)+(d)(-2d)+(2d)(-2d)\\
\implies -(3m+2) = -d^2 -2d^2 + 2d^2 +2d^2 -2d^2 -4d^2 = -5d^2\\
\implies \frac{3m+2}{5} = d^2 \\
\\
m^2 = (-2d)(2d)(-d)(d) = 4d^4\\
\implies m^2 = 4\cdot\frac{(3m+2)^2}{25}\\
\implies 25m^2 = 4\cdot(9m^2 +12m + 4) \\
\implies 0 = 36m^2 -25m^2 +48m + 16 \\
\implies 0 = 11m^2+48m+16 \\
\implies m = -4 \text{ or } m = -\frac{4}{11} \\
\end{gather}
However, when you plug the values back into the equation, you get complex roots for $m = -4$ and only 3 real roots for $m = -\frac{4}{11}$.
I am stuck here, I cannot see how to move forward. I also do not know if what I did was correct or not.
Edit: I am stupid and it is $\beta -3d$ and $\beta + 3d$.
|
In case others need help in a similar question, I am putting up the solution.
\begin{gather}
x^4-(3m+2)x^2+m^2 = 0 \\
\\
\text{Let roots be }\\
\beta+d \\
\beta-d \\
\beta+3d \\
\beta-3d \\
\\
x^4+0x^2-(3m+2)x^2+0x+m^2 = 0\\
\implies 0 = (\beta-3d) + (\beta-d) + (\beta+3d) + (\beta+d) = 4\beta \\
\implies \beta = 0\\
\\
-(3m+2) = (-d)(d)+(-d)(3d)+(-d)(-3d)+(d)(3d)+(d)(-3d)+(3d)(-3d)\\
\implies -(3m+2) = -d^2 -3d^2 + 3d^2 +3d^2 -3d^2 -9d^2 = -10d^2\\
\implies \frac{3m+2}{10} = d^2 \\
\\
m^2 = (-3d)(3d)(-d)(d) = 9d^4\\
\implies m^2 = 9\cdot\frac{(3m+2)^2}{100}\\
\implies 100m^2 = 9\cdot(9m^2 +12m + 4) \\
\implies 0 = -19m^2 + 108m + 36 \\
\implies m = 6 \text{ or } m = -\frac{6}{19} \\
\end{gather}
|
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|
Logarithmic inequality, can't define the scope of $x$ I'm solving and getting answer that $x >1$.
\begin{align*}
\ln(x)-\ln(2-x)&>0\\
\implies \ln\left(\frac{x}{2-x}\right)&>0\\
\implies e^{\ln(x/(2-x))}&>e^0\\
\implies \frac{x}{2-x}&>1\\
\implies x &> 2-x\\
\implies 2x &> 2\\
\implies x &> 1
\end{align*}
But when I assign a value of $e$ to $x$, which is greater than $1$, I get an error, because I get $\ln(2-e)$.
|
You have $\dfrac x {2-x} > 1.$
You cannot go from there to $x>2-x$ unless you somehow know that $2-x$ is positive.
There are at least two ways to deal with that:
*
*First of course you rule out $x=2$ as a possible solution because it makes the denominator zero. Then consider what happens when $x>2.$ Then $2-x$ is negative, so from $\dfrac x {2-x}>1$ you conclude $x< 2-x,$ so $2x<2$, so $x<1,$ but $x$ cannot be $<1$ if $x>2.$ Therefore there are no solutions greater than $2.$
*From $\dfrac 2 {2-x}>1$ deduce that $\dfrac 2 {2-x} - 1 > 0.$ The common denominator is $2-x$, so you have $$ \frac 2 {2-x} - \frac{2-x}{2-x}>0 $$ $$ \frac{x}{2-x}>0 $$ The fraction $x/(2-x)$ changes signs at $0$ and and $2$. So consider the three intervals $(-\infty,0),$ $(0,2),$ and $(2,\infty).$ You will find that the fraction is positive only if $0<x<2.$
|
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|
$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $ fine limit :
$$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $$
such that : $$ a \in (0,2)$$
and :
$[x]: \ \ $ floor function
My Try :
$$f(x):=( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]}$$
$$\ln f(x)=(x-[x])\ln(\sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )$$
Now ?please help
|
$\begin{array}\\
( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]}
&=(x^{2/3} \sqrt[3]{1+4 x^{a-2}} - x^{2/3}\sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\
&=x^{2(x-[x])/3} (\sqrt[3]{1+4 x^{a-2}} - \sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\
&=x^{2(x-[x])/3} (1+4 x^{a-2}/3+O(x^{a-3}) - (1+ x^{a-2}/3+O(x^{a-3})) )^{x-[x]}\\
&=x^{2(x-[x])/3} ( x^{a-2}+O(x^{a-3} ))^{x-[x]}\\
&=x^{(x-[x])(2/3+a-2)} ( 1+O(x^{-1} ))^{x-[x]}\\
&=x^{(x-[x])(a-4/3)} ( 1+O(x^{-1} ))^{x-[x]}\\
\end{array}
$
As Paramanand Singh's answer states,
if $a = 4/3$,
this is
$( 1+O(x^{-1} ))^{x-[x]}
$
which goes to $1$.
If $a \ne 4/3$,
this goes to
$x^{(x-[x])(a-4/3)}
$.
This goes from
$1$
(when $x = [x]$)
to
$x^{a-4/3}$
(when
$x-[x] \approx 1$).
Therefore,
when $a \ne 4/3$,
the limit does not exist.
|
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|
how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f,g,h,i\in \{0,1,2\}$
$$M^{T}M= \begin{pmatrix}
a & d & g\\
b & e & h\\
c & f & i
\end{pmatrix}\begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix} $$.
sum of diagonal entries $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2 = 5$$
How can I form different cases?
|
Following Lord Shark the Unknown's answer.
For five 1s and the rest zero, there are $C^9_5$ matrices.
For one 1 and one 2, there are $9\times 8$ matrices.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A summation question ,Is $S \to \ln 2$? I am in doubt with this question .
let $S=\dfrac{\dfrac12}{1} +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+\dfrac{(\dfrac12)^5}{5}+...$ Is it converge to $\ln 2$ ?
I tried this
$$x=\dfrac12 \to 1+x+x^2+x^3+x^4+...\sim\dfrac{1}{1-x}\to 2$$ by integration wrt x we have $$\int (1+x+x^2+x^3+x^4+...)dx=\int (\dfrac{1}{1-x})dx \to\\
x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+...=-\ln(1-x)$$then put $x=0.5$ $$\dfrac{\dfrac12}{1} +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+\dfrac{(\dfrac12)^5}{5}+..\sim -\ln(0.5)=\ln 2$$ now my question is : Is my work true ?
I am thankful for you hint,guide,idea or solutions. (I forgot some technics of calculus)
|
Yes.
This series was already known to Jacob Bernoulli (Gourdon and Sebah http://plouffe.fr/simon/articles/log2.pdf, formula 14) and can be written
$$S=\sum_{k=0}^\infty \frac{1}{k+1}\left(\frac{1}{2}\right)^{k+1}$$
To evaluate it, we can change the identity
$$\int_0^1 x^n dx = \frac{1}{n+1}$$
into
$$\int_0^\frac{1}{2} x^n dx = \frac{1}{n+1}\left(\frac{1}{2}\right)^{n+1}$$
and then
$$\begin{align}
S&=\sum_{k=0}^\infty \frac{1}{k+1}\left(\frac{1}{2}\right)^{k+1}\\
&=\sum_{k=0}^\infty \int_0^\frac{1}{2} x^k dx \\
&=\int_0^\frac{1}{2}\left(\sum_{k=0}^\infty x^k\right) dx\\
&=\int_0^\frac{1}{2} \frac{1}{1-x} dx\\
&=-\log(1-x)|_0^\frac{1}{2}\\
&=\log(2) \\
\end{align}$$
A nice way to encode formulas like
$$\log(2)=\sum_{n=1}^\infty \frac{1}{n2^n}$$
is noting that the numerator is one so the sequence of integer denominators can represent the series. When we search the OEIS for $2,8,24,64$ (http://oeis.org/A036289) we find that
$$\sum_{n=1}^\infty \frac{1}{a(n)} = \log(2)$$
is one of the formulas given.
Your series is a base-2 BBP-type formula for $\log(2)$. The base-3 version is
$$\log(2)=\frac{2}{3} \sum_{k=0}^\infty \frac{1}{(2k+1)9^k} $$
and the sequence of denominators is OEIS http://oeis.org/A155988.
|
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|
completing square for a circle In the following question:
I don't understand how we can get from the original equation to the final equation using completing the square.
Any thoughts as how to get to the final equation?
|
$x + y = c(x^2 + y^2 + 1)$
$x^2 + y^2 - \frac xc - \frac yc + 1 = 0$
$x^2 - \frac xc + (\frac 1{2c})^2- (\frac 1{2c})^2 + y^2 - \frac yc + (\frac y{2c})^2- (\frac 1{2c})^2+ 1 = 0$
$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 + 1 - \frac 1{4c^2} - \frac 1{4c^2} = 0$
$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 = \frac 1{2c^2} -1$.
It's.... completing the square and exactly what it says.
|
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|
Show that $g(x) = \sqrt{1 + x^2}$ is continuous Show that $g(x) = \sqrt{1 + x^2}$ is continuous on $\mathbb{R}$ using the $\epsilon - \delta$ argument.
This is what I was thinking...
Let $p \in \mathbb{R}$. Then $g(x)-g(p) = \sqrt{1 + x^2} - \sqrt{1 + p^2}$
$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{x^2-p^2}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} = \frac{(x+p)(x-P)}{\sqrt{1 + x^2} + \sqrt{1 + p^2}}$.
Thus, $g(x) - g(p)| = \frac{|x+p||x-p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}}$
$ \quad \quad \quad \quad \quad \quad \leq \frac{|x| + |p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} |x -p|$ $= (\frac{|x|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} + \frac{|p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} ) |x-p| $
$ \quad \quad \quad \quad \quad \quad \leq (\frac{|x|}{\sqrt{1 + x^2}} + \frac{|p|}{\sqrt{1 + p^2}} ) |x-p|$
$ \quad \quad \quad \quad \quad \quad \leq (1+1) |x-p| = 2|x-p|$
Choose $\epsilon > 0$. Then $|g(x)-g(p)| < \epsilon$ whenever $|x-p| < \frac{\epsilon}{2}$. Let $\delta = \frac{\epsilon}{2}> 0$. Then for $\epsilon > 0$ there exists $\delta > 0$ such that $|g(x)-g(p)| < \epsilon$ whenever $|x-p|< \delta$. Hence, $g$ is continuous at $p \in \mathbb{R}$. Since $p$ is an arbitrary real number, $g$ is continuous on $\mathbb{R}$.
Is this correct?
|
A slightly "different" approach is to use ( to abuse to be correct ) the big brother inequality called Minkowski inequality of the form: $\sqrt{(a+b)^2 + (c+d)^2} \le \sqrt{a^2+c^2} +\sqrt{b^2+d^2}$. Use this inequality with $a = b = 1, c = x, d = p$ as in your answer, the follows it with the popular trigonometric substitution $\dfrac{|x|}{\sqrt{1+x^2}} = \sin \theta\le 1$, and the other part is the same which is $|x-p|$.
|
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|
How to find the value of this integral? $\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$ $$I=\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$$
Where G is the Catalan's contant
My try:
Integrating by parts, we have:
$$3I=\int_0^1x^4\ln(1-x)\frac{dx}{\sqrt{1-x^2}}+\int_0^1x^3\sqrt{1-x^2}\frac{dx}{1-x}=S+T$$
$$\frac{x^3}{1-x}=-x^2-x-1+\frac{1}{1-x}$$
It is easy to deduce: $$\int_0^1x^2\sqrt{1-x^2}dx=\frac{\pi}{16},\int_0^1x\sqrt{1-x^2}dx=\frac{1}{3},\int_0^1\sqrt{1-x^2}dx=\frac{\pi}{4},\int_0^1\sqrt{1-x^2}\frac{dx}{1-x}=\frac{\pi}{2}+1$$
$$3I=S+\frac{3\pi}{16}+\frac{2}{3}$$
But how to deduct S?
|
It is easy to check that $\int_{0}^{1}x^2\sqrt{1-x^2}\,dx=\frac{\pi}{16}$. By setting $x=\cos\theta$ the problem boils down to computing
$$ \int_{0}^{\pi/2}\cos^2(\theta)\sin^2(\theta)\log\sin^2\frac{\theta}{2}\,d\theta$$
that is simple through the Fourier series of $\log\sin$ and $\log\cos$.
|
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|
Differentiate function: $\ y=\sin(x)\sqrt{x+3}$ Differentiate function:
$\ y=\sin x \sqrt{x+3}$
$\frac{1}{y}dy= \frac{1}{\sqrt{x+3}}dx + \frac{1}{\sin x}\cos xdx$
$\ln(y)=\ln(\sqrt{x+3})+\ln(\sin x)$
$\frac{dy}{dx}=y \left(\frac{1}{\sqrt{x+3}}+\frac{\cos x}{\sin x} \right)$
$\frac{dy}{dx}=\sqrt{x+3}\sin x \left(\frac{1}{\sqrt{x+3}}+\frac{\cos x}{\sin x} \right)$
$\frac{dy}{dx}=\sin x+\cos x\sqrt{x+3}$
But when I check with a product rule I get...
$\large \frac{dy}{dx}=\frac{\sin x}{2\sqrt{x+3}}+\cos x\sqrt{x+3}$
Where is a mistake?
|
I assume the function is $y=\sqrt{x+3}\sin x$. With the (formal) logarithmic derivative you have
$$
\log y=\frac{1}{2}\log(x+3)+\log\sin x
$$
so
$$
\frac{y'}{y}=\frac{1}{2}\frac{1}{x+3}+\frac{\cos x}{\sin x}
$$
and therefore
$$
y'=\sqrt{x+3}\,\sin x\left(\frac{1}{2}\frac{1}{x+3}+\frac{\cos x}{\sin x}\right)=
\frac{\sin x}{2\sqrt{x+3}}+\sqrt{x+3}\,\cos x
$$
Note that
$$
D(\log(\sqrt{x+3})=\frac{1}{\sqrt{x+3}}\frac{1}{2\sqrt{x+3}}
$$
and you forgot the second factor (chain rule).
|
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|
Prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$
How to prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$?
I tried Eisenstein criteria on $f(x+n)$ with $n$ ranging from $-10$ to $10$. None can be applied. I tried factoring over mod $p$ for primes up to $1223$. $f(x)$ is always reducible over these.
$f(x)$ has roots $\pm\sqrt{\left(2\pm\sqrt{2}\right) \left(3\pm\sqrt{3}\right)}$, and according to computation by PARI, should have Galois group isomorphic to the quaternion group. The splitting field of $f(x)$ is $\mathbb{Q}(\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)})$, and it contains $\sqrt{2}, \sqrt{3}, \sqrt{6}$, so we know $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is in the splitting field, so $\mathbb{Q}(\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)})$ has degree $4$ or $8$.
I tried showing the degree is $8$ by showing that $(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})^2=(2+\sqrt{2})(3+\sqrt{3})$ cannot have a solution with $a, b, c, d \in \mathbb{Q}$, and got these equations:
$$a^2+2b^2+3c^2+6d^2=6$$
$$2ab + 6cd = 3$$
$$ac+2bd = 1$$
$$2ad+2bc = 1$$
which I'm unable to handle.
Addendum: now that I've solved this problem thanks to the answers, I found some additional related information:
In A Rational Polynomial whose Group is the Quaternions, a very similar polynomial, $$f(x)=x^8 - 72 x^6 + 180 x^4 - 144 x^2 + 36$$ is studied and its Galois group is proven to be the quaternion group. I subjected this polynomial, as well as two related ones: $f(\sqrt{x})$, $f(6\sqrt{x})/36$, to the same battery of tests (Eisenstein; mod p) and these tests also failed to show them to be irreducible. Maybe there's something common about these polynomials.
So I subjected $f(x)$ to the prime numbers test demonstrated by Robert Israel, and found that it is prime at $\pm\{7, 13, 23, 25, 49, 53, 55, 79, 91, 127, 139, 145, 151, 181, 239, 251, 277, 283, 319, 355, 379, 403, 413, 425, 473, 485, 595, 607, 623, 679, 733, 743, 779, 827, 851, 923, 965, ...\}$ and thus $f(x)$ is irreducible.
|
Assume that $f(x) = f_1(x) f_2(x)$.
Note that $f(1) = - 23$, so one of $f_1(1)$, $f_2(1) = \pm 1$, and the other $\mp 23$.
The polynomial factors $\mod 5$ as $(x^4+2)(x^4 + x^2 + 2)$ ( see Jyrki's answer). Say we have $f_1(x)\equiv x^4 + 2$ and $f_2(x) \equiv x^4 + x^2 + 2 \mod 5$. Now, $f_1(1) \equiv 1^4 + 2 = 3 \mod 5$, and $f_2(1) \equiv 1^4 + 1^2 + 2 \equiv -1 \mod 5$. We conclude that $f_1(1) = 23$, and $f_2(1) = -1$.
Now, $f(x)$ factors $\mod 3$ as $x^8$. We have $f_1(x) \equiv x^4$, and $f_2(x) \equiv x^4 \mod 3$. This implies $f_1(1), f_2(1) \equiv 1 \mod 3$. We got a contradiction
$\bf{Added:}$ Another proof: Assume that $(2+\sqrt{2})(3+ \sqrt{3})$ is a square in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. Taking the norm to $\mathbb{Q}(\sqrt{2})$ we get
$(2+\sqrt{2})^2 (3+ \sqrt{3})(3-\sqrt{3})$ is a square in $\mathbb{Q}(\sqrt{2})$, so $6$ is, contradiction.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find all prime $a, b, c$ such that $ab+bc+ac > abc$ Find all primes $a, b, c$ such that $ab + bc + ac > abc$
|
let $a\geq b \geq c$
if $c\geq 3$ then $ab+bc+ac\leq 3ab\leq cab\implies c=2$.
We now need $ab+2(a+b)> 2ab\iff 2(a+b)>ab$
if $b\geq 4$ then $2(a+b)\leq4a \leq ab$, so $b\in \{2,3\}$
if $b=2$ then $2(a+2)>2a$.
if $b=3$ then $2a+6>3a\iff a<6$.
Hence ordered solutions are $(p,2,2)$ and $(3,3,2),(5,3,2)$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\int_{-1}^1\frac{y}{\pi\sqrt{1-y^2}}\,dy$ The solution says that we will get this $$\left[-\frac{\sqrt{1-y^2}}{\pi} \right]^1_{-1} $$.
How do we get this definite integral. I know we are integrating an odd function over symmetric domain so the answer will be 0.
|
Let $u = 1-y^2$, then $du = -2y ~dy$ so that
\begin{align}
\int_{-1}^{1} \frac{y}{\pi\sqrt{1-y^2}} dy &= \int_{y=-1}^{y=1} \frac{\frac{-1}{2}du }{\pi\sqrt{u}} = -\frac{1}{2\pi}\int_{y=-1}^{y=1} u^{\frac{-1}{2}}du = -\frac{1}{2\pi}\left(2u^{\frac{1}{2}}\right)_{y=-1}^{y=1}\\
&=-\left(\frac{\sqrt{1-y^2}}{\pi}\right)_{y=-1}^{y=1}
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Doubts about derivative of $|x|$ We know that the derivative of $f(x)=|x|$ is $-1$, if $x\lt 0$ and $1$, if $x\ge 0$.
However for every $x\in \mathbb R$, the following holds
$f(x)=|x|=\sqrt{x^2}\implies f'(x)=\frac{1}{2}(x^2)^{-1/2}2x=1$.
So where is my mistake?
|
As you said
$$ f(x) = \sqrt{x^2} = |x| = \begin{cases} x, & \text{if } x\ge0\\ -x, & \text{if } x < 0, \end{cases} $$
then
$$\frac{1}{\sqrt{x^2}} = \frac{1}{|x|} = \begin{cases} \frac{1}{x}, & \text{if } x > 0, \\ -\frac{1}{x} , & \text{if } x < 0. \end{cases}$$
Hence
$$ f′(x)=\frac{1}{2}(x^2)^{−1/2}2x=\frac{x}{|x|} = \begin{cases}1, & \text{if } x > 0, \\ -1, & \text{if } x < 0. \end{cases} $$
|
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|
Show that $x^3-x^2+8=y^2$ has no integer solution
Show that $x^3-x^2+8=y^2$ has no integer solution.
I spent several hours on this problem but I couldn't figure out how to solve it. A hint from the professor was to find a proper field where to find the solutions, or use Gaussian integers, but I still can't find the proof. I only showed that $y$ must be even, and I tried to factorize $(x^3+8)$ and equalize the two members with $x^2+y^2$, but nothing came out of it.
This problem was given at the Competencia Interuniversitaria Matemática Argentina (CIMA).
|
Here is an official solution: https://drive.google.com/file/d/1qOPYHK3p_S8tlwlewQ91X6JbDrZ-oKrc/view. Following is an English version (not literal translation, also little reordered):
If $x$ is even, say $x=2a$, we have $y^2=x^3-x^2+8=8a^3-4a^2+8$, implying $2 \mid y^2$, and so $2 \mid y$. Let $y=2b$, giving $8a^3-4a^2+8=4b^2$, and after simplification $2a^3-a^2+2=b^2$. Now if $a$ is even, then we have $b^2 \equiv 2 \pmod 4$, and if $a$ is odd, then $b^2 \equiv 3 \pmod 4$, either way impossible.
If $x \equiv 3 \pmod{4}$, then left side $x^3-x^2-8\equiv 2 \pmod{4}$. This implies $y^2 \equiv 2 \pmod {4}$, impossible.
For final case $x \equiv 1 \pmod {4}$, consider
$$(x+2)(x^2-2x+4)=x^2+y^2.$$
Since we have $x+2 \equiv 3 \pmod 4$, there must be prime $p \equiv 3 \pmod {4}$ such that $p \mid x+2$. From equation above $p \mid x^2+y^2$, or $x^2 \equiv -y^2 \pmod {p}$. So $p \mid x$ iff $p \mid y$. We also have
$$
x^{p-1} \equiv (x^2)^{\frac{p-1}{2}} \equiv (-y^2)^{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}y^{p-1} \pmod{p}.
$$
Now clearly $p \nmid x$ (otherwise we would have $p=2$, a contradiction). So we have $p\nmid x, p \nmid y$. From Fermat's Little Theorem we have $x^{p-1}\equiv y^{p-1} \equiv 1 \pmod p$, and so from above equation $(-1)^{\frac{p-1}{2}}\equiv 1 \pmod{p}$. This implies $x\equiv 3 \pmod{p}$, a contradiction.
So there is no integer solution to the equation.
|
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|
Solving $(a-1)^{p+1}+a^p=(a+1)^{p-1}$
Suppose that $a,p$ are nonnegative integers such that $p$ is prime and $p\nmid (a-1)$. If $(a-1)^{p+1}+a^p=(a+1)^{p-1}$, find the sum of all possible values of $a$.
We can't have $p = 2$ since the equation $$(a-1)^3+a^2 = (a+1)$$ has no integer solutions. We now take two cases:
Case 1: $p \mid (a+1)$
In this case, $a \equiv -1 \pmod{p}$. Then $(a-1)^{p+1} \equiv (-2)^{p+1} \equiv 4 \equiv -a^p \equiv -a \pmod{p}$. Thus $a \equiv 4 \pmod{p}$ and therefore $p \mid 3$, so that $p = 3$. Thus $a = 2$ in this case.
Case 2: $p \nmid (a+1)$
We have $$(a-1)^{p+1}+a^p \equiv (a-1)^2+a^p \equiv 1 \pmod{p}.$$ If $a \not \equiv 0 \pmod{p}$, then we have $(a-1)^2+a \equiv 1 \pmod{p}$, which gives $a(a-1) \equiv 0 \pmod{p}$. Thus $p \mid (a-1)$, a contradiction. Thus $a \equiv 0 \pmod{p}$.
How do I continue from here?
|
I will assume $p$ is odd.
Consider the monic polynomial $f(X) = (X-1)^{p+1} + X^p - (X+1)^{p-1}$.
It has $0$ as a root, so dividing by $X$ we find $g(X) = f(X)/X$ is another monic polynomial.
The constant term of $g$ is ${p+1 \choose p}(-1)^p-{p-1 \choose p-2} = (p+1)(-1)^p-(p-1) = -2p$.
The rational root theorem gives us the possible integer roots of $f$ as $0, \pm 1, \pm 2, \pm p, \pm 2p.$
We should be able to test each of these candidates and thus find all integer roots, and their sum.
Since we are interested only in positive integers (and $0$ would not affect the sum) we only need to check $1, 2, p, 2p$.
$a = 1$ is impossible because $f(1) = 1 - 2^{p-1} = 0$ requires $p = 1$ which is not prime.
$a = 2$ gives the equation $2^p + 1 = 3^{p-1}$ which is true only when $p = 3$.
$a = p$ leads to $f(p) = (p-1)^{p+1} + p^p - (p+1)^{p-1}=0$.
But $f(p) \geq (p-1)^{p-1}\left((p-1)^2+(p-1)-\left(1+\frac{2}{p-1}\right)^{p-1}\right)\geq 4^4(4^2+4-e^2) \gt 0$ for $p \geq 5.$ We can check $p = 3$ individually to find $f(3) = 27 \gt 0$ so $p$ is not a root.
[A much simpler way occurred to me, since $p$ is odd this is impossible because $(p-1)^{p+1} + p^p - (p+1)^{p-1}$ is odd too].
$a = 2p$ should work similarly as well.
|
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|
Find all positive integers $n$ such that $3^n+5^n = x^3$
Find all positive integers $n$ such that $3^n+5^n = x^3$ for some positive integer $x$.
One solution is $n = 1, x = 2$.
We have $1 < 3^n+5^n \leq 8^n$, so $1 < 3^n+5^n \leq 2^{3n}$. Thus $1 < x \leq 2^n$. How can we continue from here?
|
Modulo $9$, if $n>1$, the equation simplifies to $5^n\equiv x^3 \pmod 9$. Because the cubes mod 9 are $0,1,8$, and $5^n \pmod 9$ repeats with period $\phi(9)=6$ with the pattern (starting at $n=0$) $1,5,7,8,4,2,\ldots$, we have that $n$ is a multiple of $3$ (if $n>1$).
However, working modulo $7$, where $3^n+5^n$ also has period $\phi(7)=6$, we see that $3^{3k}+5^{3k}$ is never a cube mod 7. (The cubes are 0,1,6, and $3^n+5^n$ repeats $2,1,6,5,6,1,\ldots,$ but neither $2$ nor $5$ are cubes).
Therefore, if $n>1$, combining both results, we see there are no solutions. However, $n=1, x=2$ is a solution. Hence, it is the only solution.
|
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|
If $f(x) =mx$, then $f(a + b) = f(a) + f(b)$ for all $a$ and $b$?
If $f(x) =mx$, then $f(a + b) = f(a) + f(b)$ for all $a$ and $b$. True or False.
Verification of work: I found a similar problem where $f(x)=y-mx+b$ and test values for $m$ and $b$ were used. $f(x)=y-mx+b$
My problem has $m$ and $x$ as the values so I worked it as such and came to the conclusion that the question is True.
Give $m$ and $x$ the values of $3$ and $1$, respectively. So that, $f(x)=mx$ becomes $f(x)=(3)(1)$.
Give $a$ and $b$ the values of $2$ and $4$. Now we have:
$$f(2)+f(4)=(3)(2)+(3)(4)=18$$
$$f(2+4)=f(6)=(3)(6)=18$$
$18=18,$ so that $f(x) =mx$, then $f(a + b) = f(a) + f(b)$
|
So if $f(x) = mx$, $f(a+b) = m(a+b) = ma + mb = f(a) + f(b)$
|
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|
Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1)4^2+\dots+(4-1)4^{z-2}+(4-1)4^{z-1}$$
Which becomes:
$$4^z=1+4^1-4^0+4^2-4^1+4^3-4^2+\dots+4^{z-1}-4^{z-2}+4^z-4^{z-1}$$
After canceling everything out you are left with:
$$4^z=4^z$$
More generally:
$$a^z =1 + (a-1)\sum_{k=0}^{z-1} a^k$$
Therefore: $(2^x-1)/3$ will always be an integer when $x$ is even.
My question is: will there ever be an integer solution to $(2^x-1)/3$ when $x$ is odd?
|
If x is odd we can write x= k+ 1 for k an even integer. Then $\frac{2^x- 1}{3}= \frac{2^{k+1}- 1}{3}= \frac{2(2^k)- 1}{3}= \frac{2(2^k)- 2+ 1}{3}= \frac{2(2^k- 1)+ 1}{3}= 2\frac{2^k- 1}{3}+ \frac{1}{3}$.
You have already shown that $\frac{2^k-1}{3}$ is an integer so this is an integer plus 1/3, not an integer.
|
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|
Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}{4}} &=
\frac{\cos(2x)}{\frac{4x-\pi}{4}} \\&=
\frac{4\cos(2x)}{4x-\pi} = \,\,???\end{align}$$
What do I do next?
|
You are trying this the hard way.
Let $x = \frac\pi 4 -y$. Then
$$
\lim_{y\to 0}\frac{\cos(2(\frac\pi 4 -y))}{\sin(- y)}
=\lim_{y\to 0}\frac{\cos(\frac\pi 2-2y)}{-\sin y}
=-\lim_{y\to 0}\frac{\sin(2y)}{\sin y}
=-\lim_{y\to 0}\frac{2\sin y \cos y}{\sin y}
=-\lim_{y\to 0}2 \cos y=-2
$$
|
{
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|
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
I've tried different substitutions like put $Q(x)$ $=$ $k$ for some $k$ and getting the equation $k(x^2-6x+8)$ $=$ $(k-2)(x^2-6x)$ $<=>$ $k$=$(6x-x^2)/4$, but that doesn't give equlity. Thank you for your help
|
maybe Hint:Some information come out from $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$
$$x=0 \to Q(0)(8)=Q(-2)0 \to Q(0)=0\\
x=6 \to Q(6)(8)=Q(6-2)0 \to Q(6)=0\\
x=4 \to Q(4)(0)=Q(2)(16-24) \to Q(4)=0$$ so $Q(x)$ at least contain $$Q(x)=(x-0)(x-4)(x-6)$$ or
$$Q(x)=(x-0)(x-6)(x-4)q(x)$$
|
{
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|
Is $e$ a coincidence? $e$ has many definitions and properties. The one I'm most used to is
$$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n $$
If someone asked me (and I didn't know about $e$):
Is there a constant $c$ such that the equation $\frac{d}{dx}c^x=c^x $ is true for all $x$?
Then I'd likely answer that:
I doubt it! That would be a crazy coincidence.
I'm curious, is it a coincidence that there is a constant that makes this true?
|
Just to add how these definitions of $e$ are matched together. You can call this an overkill. I use Taylor polynomials around $x=0$ and then Taylor series to show that $e$ satisfies the equation.
We first note that $c=0$ will not work (why?). It is obvious that the function $f(x)=c^x$ is infinitely many times differentiable. So, $f(0)=f'(0)=...=f^{(n)}(0)=1$. So the Taylor polynomial around $x=0$ is given as:
\begin{align}
T_n(x)=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}
\end{align}
With remainder term:
\begin{align}
R_n(x) = \frac{c^A}{(n+1)!}x^{n+1}
\end{align}
for $A$ between $x$ and $0$. It is obvious that $\vert R_n(x) \vert \to 0$ as $n \to \infty$ for all $x\in \mathbb{R}$. So the function can be expressed in a Taylor series and it is equal to the function on all of $x\in\mathbb{R}$:
\begin{align}
f(x)=1+\sum_{k=1}^{\infty} \frac{x^k}{k!}
\end{align}
So:
\begin{align}
f(1)=c=1+\sum_{k=1}^{\infty} \frac{1}{k!}
\hspace{25pt} (1)
\end{align}
If your definition of $e$ was as $(1)$, then we would be finished. But if it was:
\begin{align}
e=\lim\limits_{n\to \infty}\left(1+\frac 1 n\right)^n \hspace{25pt} (2)
\end{align}
Then we should do more work. In case you wonder how to do that, define:
\begin{align}
e_n = \left( 1 +\frac 1 n\right)^n
\end{align}
Now we have:
\begin{align}
e_n &= \left( 1 +\frac 1 n\right)^n\\
&= \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} \\
&= 1+ \sum_{k=1}^n \binom{n}{k}\frac{1}{n^k} \\
&= 1+ \sum_{k=1}^n \frac{1}{k!}\frac{n \cdot (n-1) \cdot ... \cdot (n-k+1)}{n\cdot n ... \cdot n}\\
&\leq 1+ \sum_{k=1}^n \frac{1}{k!}\frac{n \cdot n \cdot ... \cdot n}{n\cdot n ... \cdot n}\\
& = 1+ \sum_{k=1}^n \frac{1}{k!}
\end{align}
Taking $n \to \infty$ we get: $e\leq f(1)=c$. We can do something similar to get: $e\geq f(1)=c$. And that implies $c=e$ as desired.
You see now how some of the definitions of $e$ imply the other.
Edit:
@Alex M. asked me to add an elaboration on the part $e\geq f(1)=c$.
Note that for $n\geq m$ we have:
\begin{align}
e_n =& 1+\sum\limits_{k=1}^n \binom{n}{k} \frac{1}{n^k}\\
=& 1+ \sum_{k=1}^n \frac{1}{k!}\frac{n \cdot (n-1) \cdot ... \cdot (n-k+1)}{n\cdot n ... \cdot n}\\
\geq& 1+ \sum_{k=1}^m \frac{1}{k!}\frac{n \cdot (n-1) \cdot ... \cdot (n-k+1)}{n\cdot n ... \cdot n} \hspace{15pt}\text{ (since } n\geq m)\\
=& 1+\sum_{k=1}^m \frac{1}{k!} \cdot \frac{n}{n} \cdot \frac{n-1}{n} \cdot ... \cdot \frac{n-k+1}{n}
\end{align}
Now we take $n \to \infty$ to obtain:
\begin{align}
e\geq 1+\sum_{k=1}^m \frac{1}{k!}
\end{align}
Now we take $m \to \infty$ to get:
\begin{align}
e\geq 1+\sum_{k=1}^\infty \frac{1}{k!} = f(1)=c
\end{align}
|
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|
Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$ I know this can be done by evaluating $f$ at the points $0,1,...10$ to check if $f$ has a linear factor.
Is there any other shorter way?
|
We start with $$x^2+x+4=0$$ multiplying with $4$ we get $$4x^2+4x+16=0$$ This can be written as $$(2x+1)^2+15=0$$ modulo $11$ this is equivalent to $$(2x+1)^2-7=0$$
So we have to solve $$(2x+1)^2\equiv 7\mod 11$$
Using the quadratic reciprocity law we can calculate the legendre symbol
$$(\frac{7}{11})=-(\frac{11}{7})=-(\frac{4}{7})=-1$$
Hence $u^2\equiv 7\mod 11$ is not solveable because $7$ is not a quadratic residue modulo $11$. Since $x^2+x+4$ has no root in $\mathbb Z_{11}$, it follows that $x^2+x+4$ is irreducible in $\mathbb Z_{11}$
|
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|
What is the set of solutions to $2\log_{\cos x}\sin x\le \log_{\sin x}\cot x$ where $0
$$2\log_{\cos x}\sin x\le \log_{\sin x}\cot x$$ where $0<x<\pi$
My process:
Since both $\sin x$ and $\cos x$ must be greater than $0$ and less than $1$, the initial set shrinks to $(0,\pi/2)$
Then after transformations I get
$2\log_{\cos x}\sin x+1-\log_{\sin x}\cos x\le 0$
.
.
.
$2t^2+t-1\le 0$
. . .
$\log_{\cos x}\sin x\in [-1,1/2] $
Can't be $[-1,0)$
But I'm not sure what the interval is further
I'd venture a guess as to say that $x\in (0,\frac{\sqrt5-1}{2}]$ but that doesn't fit with the solution on the exam, which doesn't explicitly give the solution but says it is in the form of $(a,b)$
Any help appreciated.
|
We need $\displaystyle 0<x<\frac{\pi}{2}$. Note that $\ln\cos x<0$ and $\ln \sin x<0$.
\begin{align}
\frac{2\ln \sin x}{\ln \cos x}&\le\frac{\ln \cos x-\ln\sin x}{\ln\sin x}\\
2(\ln\sin x)^2&\le(\ln\cos x)^2-\ln\sin x\ln \cos x\\
(\ln\cos x)^2-\ln\sin x\ln \cos x-2(\ln\sin x)^2&\ge0\\
(\ln \cos x+\ln \sin x)(\ln\cos x-2\ln \sin x)&\ge0\\
\ln\cos x-2\ln\sin x&\le0\\
\ln\cos x&\le\ln\sin^2x\\
\cos x&\le1-\cos^2x\\
\left(\cos x+\frac{1}{2}\right)^2&\le\frac{5}{4}\\
0<\cos x&\le\frac{-1+\sqrt{5}}{2}\\
\cos^{-1}\left(\frac{-1+\sqrt{5}}{2}\right)\le x&<\frac{\pi}{2}
\end{align}
|
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|
Injectivity and Range of a function Let, $A=\{(x,y)\in \Bbb R^2:x+y\neq -1\}$
Define, $f:A\to \Bbb R^2$ by $f(x,y)=\left(\displaystyle \frac{x}{1+x+y},\frac{y}{1+x+y}\right)$
$(1)$Is $f$ injective on $A$?
$(2)$What is the Range of $f$?
I started with $f(x,y)=f(u,v)$, then tried to show $(x,y)=(u,v)$. But I could not do it. Can someone help me out for this.
Secondly for computing the range,
$$\begin{align}
f(x,y)=(a,b)\\
\implies \displaystyle \frac{x}{1+x+y}=a,\displaystyle \frac{y}{1+x+y}=b\\
\implies (1-a)x-ay=a\\
-bx+(1-b)y=b\\
\implies x=\displaystyle \frac{a}{1-(a+b)},y=\displaystyle \frac{b}{1-(a+b)}
\end{align}$$
So, $f(A)=\Bbb R^2-\{(a,b)\in \Bbb R^2:a+b=1\}$
Is the range correct? If yec can it be obtained in some other way?
|
To continue our discussion in the comments, pick any $(x,y), (u,v) \in A$.
Suppose $f(x,y) = f(u,v)$. Further suppose $x, y, v \neq 0$. (You have to prove the cases (i) $x = 0$, (ii) $y = 0$ and (iii) $v = 0$ later on)
Then $$\frac{x}{1+x+y} = \frac{u}{1+u+v}\quad \text{and} \quad \frac{y}{1+x+y} = \frac{v}{1+u+v}$$
For $\frac{y}{1+x+y} \neq 0$, we can divide $\frac{x}{1+x+y}$ by $\frac{y}{1+x+y}$, which is $\frac{v}{1+u+v}$. Then
$$\frac{x}{1+x+y} \frac{1+x+y}{y} = \frac{u}{1+u+v}\frac{1+x+y}{y}=\frac{u}{1+u+v}\frac{1+u+v}{v}$$
$$\frac{x}{y} = \frac{u}{v}$$
For $y, v \neq 0$, we can divide the numerator and divisor by $y$ and $v$.
$$\frac{\frac{x}{y}}{\frac{1}{y}+\frac{x}{y}+1} = \frac{\frac{u}{v}}{\frac{1}{v}+\frac{u}{v}+1}= \frac{\frac{x}{y}}{\frac{1}{v}+\frac{x}{y}+1} $$
And we can cancel $\frac{x}{y}$ on both sides, then
$$\frac{1}{\frac{1}{y}+\frac{x}{y}+1}= \frac{1}{\frac{1}{v}+\frac{x}{y}+1} $$
then
$$\frac{1}{y}+\frac{x}{y}+1 = \frac{1}{v}+\frac{x}{y}+1$$
then $v = y$. Then I think you can show that $x =u$ from now.
|
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|
How to solve $\frac{2}{3}tx+\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x}=t+\sqrt{2}+\sqrt{6}$ where $t=\sqrt{2}-\sqrt{3}-\sqrt{5}+\sqrt{6}$?
Let $t=\sqrt{2}-\sqrt{3}-\sqrt{5}+\sqrt{6}$
then fine the $x$:
$$\frac{2}{3}tx+\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x}=t+\sqrt{2}+\sqrt{6}$$
since : $$\sqrt{2x-3} \to 2x-3 \geq 0 \to x\geq\dfrac{3}{2}$$
$$\sqrt{2x-1} \to 2x-1 \geq 0 \to x\geq\dfrac{1}{2}$$
$$\sqrt{3-x} \to 3-x \geq 0 \to x\leq 3$$
So :
$$ \dfrac{3}{2} \leq x \leq 3$$
Now what ?
|
if you see $d(d(\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x})/dx)/dx$, you will see that is less than zero from $\frac{3}{2}$ to $3$. So $f = \frac{2}{3}tx + \sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x} - t - \sqrt{2} - \sqrt{6}$ is convex upward. At $x = \frac{3}{2}$ is zero, at $x = 3$ is zero too. So we have no other roots.
|
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|
Completing the square but in different situation
Solve the equation $$x^2+4\left(\frac{x}{x-2}\right)^2=45$$
My attempt,
I decided to use completing the square method, so I change it to $$x^2+\left(\frac{2x}{x-2}\right)^2=45$$
But I never encounter this before. Normally, for example $x^2+4x=5$, we can change it to $x^2+4x+(\frac{4}{2})^2=45+(\frac{4}{2})^2$. But in this question is different. Could someone give me some hints for it? Thanks in advance.
|
The suggested solution posted by OP provide a clever method.
\begin{align}
x^2+4\left(\frac{x}{x-2}\right)^2&=45\\
x^2+2(x)\left(\frac{2x}{x-2}\right)+\left(\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(x+\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(\frac{x^2}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(\frac{x^2}{x-2}\right)^2-4\left(\frac{x^2}{x-2}\right)+4&=49\\
\left(\frac{x^2}{x-2}-2\right)^2&=49\\
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that $\log_a b \cdot \log_b a = 1$
Prove that $\log_a b \cdot \log_b a = 1$
I could be totally off here but feel that I have at least a clue.
My proof is:
Suppose that $a = b$, then $a^{1} = b$ and $b^{1} = a$ and we are done.
Suppose now that $a \neq b$. We wish to show that: $\log_b a = \frac{1}{\log_a b}$.
\begin{align*}
1 &= \frac{1}{\log_a b} \cdot \frac{1}{\log_b a} \cdot \left(\log_a{b} \cdot \log_b a \right) \\
&= \left(\log_a{b} \cdot \frac{1}{\log_a b} \right) \cdot \left(\log_b a \cdot \frac{1}{\log_b a}\right) \\
&= \left(\log_b a \cdot \frac{1}{\log_b a}\right) \cdot 1 \\
&=1 \cdot 1 \\
&= 1
\end{align*}
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Method I
Let $\log_ab=x$ and $\log_ba=y$. Then we have
$$a^x=b \quad\text{and}\quad b^y=a$$
So,
\begin{align}
(b^y)^x&=b\\
b^{xy}&=b\\
xy&=1
\end{align}
Method II
Let $\log_ab=x$. Then we have
\begin{align}
a^x&=b\\
\log_b(a^x)&=\log_bb\\
x\log_ba&=1
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving equation of type $a\cos x+b\cos y-c=0$ and $a\sin x+b\sin y-d=0$ Here's the questions
There are two equations:
$a\cos x+b\cos y-c=0 $ and $ a\sin x+b\sin y-d=0$ .
For instance
What is the value of $x$ and $y$ in following question?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$
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The equations can be rewritten as
$$a\cos x+b\cos y=c$$
$$a\sin x+b\sin y = d$$
Squaring both gives us
$$a^2\cos^2x+b^2\cos^2 y+2ab\cos x\cos y=c^2$$
$$a^2\sin^2 x+b^2\sin^2 y+2ab\sin x\sin y=d^2$$
Adding these together gives
$$a^2+b^2+2ab\cos(x-y)=c^2+d^2$$
Which allows us to solve for $\cos(x-y)$ in terms of given constants. Multiplying the first two equations by $\cos y$ and $\sin y$ respectively and adding them gives us
$$a\cos(x-y)+b=c\cos y+d\sin y$$
which allows you to solve for $y$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{4^2+7^2+3^2}}\right) \\
&= \cos^{-1}(0)=90° \\
\alpha_2 &= \cos^{-1}\left(\frac{(4,7,3)(2,6,8)}{\sqrt{4^2+7^2+3^2}\sqrt{2^2+6^2+8^2}}\right) \\
&= \cos^{-1}\left(\frac{74}{\sqrt{74}\sqrt{104}}\right)=32.49\\
\alpha_3 &= \cos^{-1}\left(\frac{(2,6,8)(2,1,-5)}{\sqrt{2^2+6^2+8^2}\sqrt{2^2+1^2+(-5)^2}}\right) \\
&= \cos^{-1}\left(\frac{-30}{\sqrt{104}\sqrt{30}}\right)=122.5°
\end {aligned}$$
As you can see, these angles don't even form a triangle, what am I doing wrong, any thoughts?
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To find $\alpha_3=\angle ABC$, you should consider $\overrightarrow{BA}\cdot \overrightarrow{BC}$ instead of $\overrightarrow{AB}\cdot \overrightarrow{BC}$. That's why you have a negative cosine and obtained the supplementary angle of the correct answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT I used the identity $\cos{2x}=2\cos^{2}x-1$ to solve it in a simpler way. viz.
LHS $= 2\cos^{2}3x-1$
$=2{(4\cos^{3}x-3\cos{x})}^2-1$
$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$
$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$
Still thank you for the answers!
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Expand $(\cos(x)+i\sin(x))^6=\cos(6x)+i\sin(6x) $ and take raeal part of both sides
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|
Function question $f(x+1)$ and $f(x)$
Given that $f(x+1)-f(x)=4x+5$, $f(0)=6$. Find $f(x)$.
My attempt,
$f(1)-6=5$,
$f(1)=11$
How to proceed then? I've never solve this kind of question before.
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This type of question is called a functional equation. First, suppose $f(x)$ is in the form
$$f(x)=ax+b$$
Then you can use this assumption to solve for $a$ and $b$. So
$$a(x+1)+b-ax-b=4x+5$$
$$a=4x+5$$
Which cannot be, since $a$ is a constant. So $f(x)$ must not be of that form. So, instead consider the quadratic form
$$f(x)=ax^2+bx+c$$
Then we have
$$a(x+1)^2+b(x+1)+c-ax^2-bx-c=4x+5$$
$$a(x^2+2x+1)+b(x+1)-ax^2-bx=4x+5$$
$$ax^2+2ax+a+bx+b-ax^2-bx=4x+5$$
$$2ax+a+b=4x+5$$
So now we have a system that we can use to solve for the constants $a$ and $b$:
$$2a=4$$
$$a+b=5$$
Which gives us $a=2$, $b=3$. Now you need to solve for $c$ given that $f(0)=6$, giving us $c=6$, so
$$f(x)=2x^2+3x+6$$
If you have any questions, just ask!
|
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|
$x^4 -ax^3 +2x^2 -bx +1$ has real root $\implies$ $a^2+b^2 \ge 8$ it is requested to show that if the quartic polynomial $f(x) \in \mathbb{R}[x]$, defined by:
$$
f(x) = x^4 -ax^3 +2x^2 -bx +1,
$$
has a real root, then
$$
a^2 +b^2 \ge 8
$$
this question was asked by @medo, then deleted a few minutes ago. however having spent a little time on it, i think the problem seems sufficiently instructive to be worth resuscitating. it is not deep or difficult, but to find the right way of rewriting the polynomial to demonstrate the result is an interesting coffee-break challenge.
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Let $x$ be a root. Thus, $x\neq0$ and $b=\frac{x^4-ax^3+2x^2+1}{x}$ and we need to prove that
$$a^2+\frac{(x^4-ax^3+2x^2+1)^2}{x^2}\geq8$$ or
$$(x^6+x^2)a^2-2(x^7+2x^5+x^3)a+x^8+4x^6+6x^4-4x^2+1\geq0,$$
for which it's enough to prove that
$$(x^7+2x^5+x^3)^2-(x^6+x^2)(x^8+4x^6+6x^4-4x^2+1)\leq0$$ or
$$(x^2-1)^4\geq0.$$
Done!
|
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|
Give me some hints in calculation this limit. $$\lim_{x\to2\ \\ y\to2}\frac{x^{6}+ \tan (x^{2}-y^{2}) - y^{6}}{\sin(x^{6}-y^{6}) - x^{5}y +xy^{5} + \arctan(x^{2}y -xy^{2})}$$
I used a fact that $$\tan \alpha \sim \alpha \\ \arctan \alpha \sim \alpha \\ \sin\alpha \sim \alpha$$
Since now we have $$\lim_{x\to2\ \\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}$$
Then $$\lim_{x\to2\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}=\\ \\
=\lim_{x\to2\ \\ y\to2}\frac{(x+ y)(x-y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})+(x+y)(x-y)} {(x+ y)(x-y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})-xy(x^{2}-y^{2})(x^{2}+y^{2}) - xy(x-y)}=\\=\lim_{x\to2\ \\y\to2}\frac{(x+ y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})+(x+y)} {(x+ y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})-xy(x+y)(x^{2}+y^{2}) - xy} $$
And what to do next what multipliers to group? Help please.
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Observe that
\begin{align*}
\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}&=\frac{\left(x^{6}+ x^{2}-y^{2} - y^{6}\right)/(x-y)}{\left(x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}\right)/(x-y)}\\
\end{align*}
$$=\frac{x^5+x^4y+x^3y^3+x^2y^3+xy^4+y^5+x+y}{x^5+x^4y+x^3y^3+x^2y^3+xy^4+y^5-xy(x^3+x^2y+xy^2+y^3)+xy}$$
Then
\begin{align*}
\lim_{x\to2\\y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}&=\frac{6\cdot 2^5+2\cdot 2}{6\cdot 2^5-4(4\cdot 2^3)+4}\\
&=\frac{49}{17}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$ Question:
Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$
Attempt:
Substituting $n = -2,-1,0,1,2$into $\sum_{r=-1}^{n}{r^3}$ we get:
$$\sum_{r=-2}^{-2}{r^3} = -8 =16a-8b+4c-2d+e$$
$$\sum_{r=-2}^{-1}{r^3} = -9 = a-b+c-d+e$$
$$\sum_{r=-2}^{0}{r^3} = -9 = e$$
$$\sum_{r=-2}^{1}{r^3} = -8 = a+b+c+d+e$$
$$\sum_{r=-2}^{2}{r^3} = 0 =16a+8b+4c+2d+e$$
After some algebra we now know $a= -\frac12, b = -1, c = 1, d = \frac32, e = -9$:
$$\therefore \sum^{n}_{r=-2}{r^3}= -\frac12n^4 -n^3 + n^2 +\frac32n -9$$
We define $\sum^{n}_{r=0}{r^3}$:
$$\sum^{n}_{r=0}{r^3}=\sum^{n}_{r=-2}{r^3} - \sum^{-2}_{r=-2}{r^3}=-\frac12n^4 -n^3 + n^2 +\frac32n -9 -(-9)$$
$$-\frac12n^4 -n^3 + n^2 +\frac32n = \frac14(-2n^4-4n^3+4n^2 + 6n)$$
$$\frac14n^2\biggl(-2n^2-4n+4+\frac6n\biggl) = \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
$$ \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
My problem:
I got $d = \frac32$ and as far as i can see since it need to end up with $d = 0$ as the smallest term that can be left is a $n^2$ term as $\frac14n^2(n+1)^2$ is multiplied by $n^2$ meaning $e$ has to cancel out (which it does) but i don't know how to get rid of $d$, if you could explain it to me it would be much appreciated
|
I would propose a more efficient strategy. Given that $\sum_{r=-2}^{n}r^3$ is a fourth-degree polynomial in the variable $n$, the same holds for $\sum_{r=0}^{n}r^3=\sum_{r=1}^{n}r^3$, since the two sums differ by a constant. Two different fourth-degree polynomials cannot agree on $5$ points or more: otherwise their difference would be a non-constant polynomial with degree $\leq 4$ and with $\geq 5$ roots, impossible. So, in order to prove
$$ \sum_{r=0}^{n}r^3 = \frac{n^2(n+1)^2}{4} \tag{1}$$
for any $n\in\mathbb{N}$, it is enough to check that such identity holds at $n\in\{0,1,2,3,4\}$:
$$ \begin{array}{|c|c|c|c|c|c|}\hline n & 0 & 1 & 2 & 3 & 4 & 5\\
\hline \sum_{r=0}^{n}r^3 & 0 & 1 & 9 & 36 & 100 & 225\\
\hline \frac{1}{4}\left(n(n+1)\right)^2 & 0 & 1 & 9 & 36 & 100 & 225
\\\hline \end{array}\tag{2}$$
and we are done.
|
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|
Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me anymore:
$$\begin{align*}
&= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\
&= \frac{(k+1)^2(k+2)^2}{4}.\\
\end{align*}$$
I understand that first $(k+1)^3$ was changed to have the same denominator as the main term (which is $4$). Can someone help me break down the steps how the polynomials are added then after that, the powers confuse me a bit.
|
$\frac{k^2 (k+1)^2 + 4(k+1)^3}{4} = (k+1)^2 \cdot \frac{k^2 + 4(k+1)}{4} = (k+1)^2 \cdot \frac{k^2 + 4k + 4}{4} = (k+1)^2 \cdot \frac{(k + 2)^2}{4}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we get
$$A^2B^4=(AB^2)A=BA^2$$ hence
$$A^2B^4=BA^2 \tag{2}$$ Now post multiplying with $B^4$ and using $(2)$we get
$$A^2B^8=B(A^2B^4)=B^2A^2$$ hence
$$A^2B^8=B^2A^2 \tag{3}$$
Now Pre Multiply with $B^2$ and use $(3)$ we get
$$B^2A^2B^8=B^4A^2$$ $\implies$
$$A^2B^8B^8=B^4A^2$$
$$A^2B^{16}=B^4A^2$$
Now pre multiply with $A^2$ and use $(2)$we get
$$A^4B^{16}=A^2B^4A^2$$ $\implies$
$$B^{16}=BA^4=B$$
is there any other approach to solve this?
|
The relation $AB^2 = BA$ allows to reduce the number of $B$s by one. Using this idea, we can show that for every even power $2n$, we have that
$$AB^{2n} = B^nA.$$
Applying this and using $A^4 = I$, we can simplify
$$B^{16} = A^4B^{16} = A^3B^8A = \ldots = B.$$
When we pull the first $A$ through, we half the power of $B$. Now it should be easy to see how to continue and why we will be left with $B$ in the end.
Note that you used basically the same idea, you just hid it behind multiple steps.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2332199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculate line integral $\oint _C d \overrightarrow{r } \times \overrightarrow{a }$ Calculate line integral:
$$\oint_C d \overrightarrow{r} \times \overrightarrow{a},$$
where $\overrightarrow{a} = -yz\overrightarrow{i} + xz \overrightarrow{j} +xy \overrightarrow{k}$ , and curve $C$ is intersection of surfaces given by $ x^2+y^2+z^2 = 1$, $y=x^2$, positively oriented looked from the positive part of axis $Oy$.
I first tried to calculate the intersection vector $\overrightarrow{r }$ by substituting $y=x^2$ into $ x^2+y^2+z^2 = 1$, and completing the square with $y+1/2$ and got the following : $ \overrightarrow{r}= \pm\sqrt{ \sqrt{5/4} \cos(t) - 1/2}\overrightarrow{i}, (\sqrt{5/4}\cos(t) - 1/2 )\overrightarrow{j}, (\sqrt{5/4}\sin(t) )\overrightarrow{k}$,
$ -\cos^{-1} (1/\sqrt 5)<t< + \cos^{-1}(1/\sqrt 5)$
After that i tried to vector multiply the two vectors, after differentiating $\overrightarrow{r}$ first, but I got some difficult expressions to integrate.
Is there an easier way to do this?
|
$x^2 + y^2 + z^2 = \frac 34\\
x^2 = y\\
y^2 + y + z^2 = 1\\
(y+\frac 12)^2 + z^2 = \frac 54$
$y = \frac {\sqrt 5}{2} \cos\theta - \frac 12\\
z = \frac {\sqrt 5}{2} \cos\theta\\
x = \sqrt {\frac {\sqrt 5}{2} \cos\theta - \frac 12}$
or we could do standard spherical:
$x = \sin\phi\cos\theta\\
y = \sin\phi\sin\theta\\
z = \cos\phi$
and then incorporate the parabola $y=x^2$
$\sin\phi\sin\theta = \sin^2\phi\cos^2\theta\\
\sin\phi = \sec\theta\tan \theta\\
\phi = \arcsin\sec\theta\tan \theta$
$x = \tan \theta\\
y = \tan^2 \theta\\
z = \sqrt{1-\sec^2\theta\tan^2 \theta}$
would be another.
But since it is a closed curve, perhaps Stokes theorem is in order
$\nabla \times a = (0,2y,2z)$
Lets use the surface of the parabolic cylinder:
$x = x\\
y = x^2\\
z = z$
$dS = (-2x,1, 0)$
$\iint 4x^2 \ dz \ dx\\
\int 4 x^2z \ dz$
I need some limits of integration.
$x^2 + x^4 + z^2 = 1\\
z = \pm\sqrt{1-x^4-x^2}\\
x^4 + x^2 - 1 = 0\\
x^2 = \frac 12(1-\sqrt5)$
$\int_{-\sqrt \phi}^{\sqrt \phi} 8 x^2\sqrt{1-x^4-x^2} \ dx$
And that is as far as I can take this.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The perimeter is equal to the area The measurements on the sides of a rectangle are distinct integers. The perimeter and area of the rectangle are expressed by the same number. Determine this number.
Answer: 18
It could be $4*4$ = $4+4+4+4$ but the answer is 18.
Wait... Now that I noticed, the sides are different numbers. But I can't find a way to solve.
|
Let $x$ and $y$ be the sides of the rectangle. Then the given condition implies that $xy = 2x + 2y$. Solving for $y$ in terms of $x$, we get:
$$y = \frac{2x}{x-2}.$$
It's not immediately obvious from this equation for what integer values of $x$ this gives an integer value of $y$. However, by doing a polynomial division, we can rewrite this as:
$$y = 2 + \frac{4}{x-2}.$$
Now, it is clear that $y$ is an integer if and only if $x-2$ is a factor of 4. The integers factors of 4 are $\pm 4, \pm 2, \pm 1$; however, since $x$ must be positive, we can eliminate the cases $x-2 = -4$ and $x-2 = -2$. For the other factors, plugging into the equation we get solutions $(x,y) = (1,-2), (3, 6), (4, 4), \mathrm{or}~(6,3)$. However, $y$ must also be positive, eliminating the $(1,-2)$ solution; and we are given $x \ne y$, eliminating the $(4, 4)$ solution. Therefore, $(x,y) = (3,6)~\mathrm{or}~(6,3)$, and in both cases, $xy = 2x + 2y = 18$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\cos{\frac{\pi}{8}}$? I have to evaluate
$\cos{\frac{\pi}{8}}$
and I'm supposed to do so evaluating first
$\cos^2{\frac{\pi}{8}}$
(since it's an exercise to practice half-angle formulas). Solving this second formula I get to
$\cos^2{\frac{\pi}{8}} = \frac{1}{2} + \frac{1}{2\sqrt[]{2}}$
where I'm stuck. I'm not sure it is an useful evaluation and, worst of all, I don't think that could help me solving
$\cos{\frac{\pi}{8}}$.
I don't know how to get to
$\frac{\sqrt[]{2 + \sqrt[]{2}}}{2}$, which is given as the proper answer. Could anyone explain it to me?
Thank you very much in advance.
|
$$\cos2\theta = 2\cos^2\theta -1$$
Put $\theta = \frac{\pi}{4}$,
$$\frac{1}{\sqrt{2}} = 2\cos^2\frac{\pi}{8} - 1$$
$$\cos^2\frac{\pi}{8} = \frac{\sqrt{2}+1}{2\sqrt{2}} = \frac{2 + \sqrt{2}}{4}$$
$$\implies \cos\frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x\cos x}{2\sin x+ 2\sin x\cos x} & \text{since $\sin(2x) = 2\sin x\cos x$}&\\
& = \dfrac{2\sin x(1 - \cos x)}{2\sin x(1 + \cos x)} &&\\
& = \dfrac{(1- \cos x)}{(1+ \cos x)} &&\\
& = \dfrac{(1- \cos x)(1+ \cos x)}{(1+ \cos x)(1+ \cos x)}& \text{since $\dfrac{(1+ \cos x)}{(1+ \cos x)}=1$}&\\
& = \dfrac{(1)^2-(\cos x)^2}{(1+ \cos x)^2} & \text{since $a^2-b^2 = (a+b)(a-b)$}&\\
& = \dfrac{(\sin x)^2}{(1+ \cos x)^2} & \text{since $(\sin x)^2 + (\cos x)^2 =1$, so $(\sin x)^2 = 1- (\cos x)^2$.}&
\end{array}$$
Now, I understand that I have the $\sin x$ part on the numerator. What I have to do is get the denominator to be $\cos x$ somehow and also make the angles $\frac{x}{2}$ instead of $x$. How do I do that?
Please be through, and you can't use half-angle or triple angle or any of those formulas.
Also, we have to show left hand side is equal to right hand side, we can't do it the other way around. So please do not take $(\tan\frac{x}{2})^2$ and solve the equation.
Thank you for understanding and have a nice day :)
|
We have used here $$\sin { 2x } =2\sin { x } \cos { x } \\ \cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } \\ \cos ^{ 2 }{ \frac { x }{ 2 } } +\sin ^{ 2 }{ \frac { x }{ 2 } } =1\\ $$$$\frac { 2\sin { x } -\sin { 2x } }{ 2\sin { x+\sin { 2x } } } =\frac { 2\sin { x } \left( 1-\cos { x } \right) }{ 2\sin { x } \left( 1+\cos { x } \right) } =\frac { \cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 } -\cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 } } } } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 } +\cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 } } } } } } =\\ =\frac { 2\sin ^{ 2 }{ \frac { x }{ 2 } } }{ 2\cos ^{ 2 }{ \frac { x }{ 2 } } } =\tan ^{ 2 }{ \frac { x }{ 2 } } $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\sin(\frac{1}{2}\arccos(\frac{1}{9}))$ This question puts me in mind of the previous alike question beginning with tangent. So, tried to solve it using that method: if $\arccos(\frac{1}{9})=\alpha,$ then $\sin(\frac{\alpha}{2}).$ The formula: $$\sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}$$ doesn't work because there are signs and there wasn't said in which quarter $\frac{1}{9}$ is in. So is representing $\sin(\frac{\alpha}{2})$ via a tangent of a half angle.
|
Let $\alpha = \cos^{-1}(\frac{1}{9})$ so $\cos(\alpha)= \frac{1}{9}$, we have
\begin{eqnarray*}
\sin(\frac{1}{2} \cos^{-1}(\frac{1}{9}))= \sin(\frac{1}{2} \alpha) = \sqrt{\frac{1-\cos(\alpha)}{2}} = \sqrt{\frac{1-\frac{1}{9}}{2}}=\color{red}{\frac{2}{3}}.
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
|
Since $z^2+\frac{1}{z^2}=7$, we obtain $z+\frac{1}{z}=3$ or $z+\frac{1}{z}=-3$.
In the first case we obtain
$$z^3+\frac{1}{z^3}=3\left(z^2+\frac{1}{z^2}-1\right)=3(7-1)=18$$
in the second case we obtain
$$z^3+\frac{1}{z^3}=-3(7-1)=-18$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show that $\lim_{n\to\infty}[a_1,\cdots,a_n]$ exists if $a_k\geq 2$ for all $k$?
Consider a sequence of positive real numbers $(a_n)$. Define $[a_1]=\frac{1}{a_1}$ and recursively inductively $[a_1,\cdots,a_n]=\frac{1}{a_1+[a_2,\cdots,a_n]}$. Suppose $a_k\geq 2$ for all $k$. How to show that $\lim_{n\to\infty}[a_1,\cdots,a_n]$ exists?
I was trying to show that the sequence is monotone, which is not true. A special related case is done here, which is not very helpful to have a generalization.
[Added to answer the confusion in comments.] The definition above should be understood properly as follows. For any positive real number $a$, define $[a]:=\frac{1}{a}$. Now, given any two positive real numbers $a_1,a_2$, one can define $[a_1,a_2]:=\frac{1}{1+[a_2]}$. One can thus keep going on in this fashion to define $[a_1,a_2,\cdots,a_n]$. To write down a few terms explicitly,
$$
[a_1,a_2]=\frac{1}{a_1+\frac{1}{a_2}},\
[a_1,a_2,a_3]=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3}}},\
[a_1,a_2,a_3,a_4]=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{a_4}}}},\cdots
$$
|
You only need $a_i \ge 1$.
Define
$$\begin{bmatrix} p_{-1} & p_{0} \\ q_{-1} & q_{0} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$
Recursively, set
$$p_k = a_kp_{k-1} + p_{k - 2}, q_k = a_kq_{k-1} + q_{k - 2} \tag{1} $$
for $k \ge 1$. We will show that
$$ [a_1,\dots,a_n,a_{n+1}] = \frac{a_{n+1}p_n + p_{n - 1}}{a_{n+1}q_n + q_{n - 1}}. $$
This holds by induction:
\begin{align}
[a_1,\dots,a_n,a_{n+1}] &= \left[ a_1,\dots,a_n + \frac{1}{a_{n+1}} \right] \\
&= \frac{\left(a_n + \frac{1}{a_{n+1}} \right)p_{n-1}+p_{n-2}}{\left(a_n + \frac{1}{a_{n+1}} \right)p_{n-1}+p_{n-2}} \\
&= \frac{(a_np_{n-1}+p_{n-2)} + \frac{1}{a_{n+1}}p_{n - 1}}{(a_nq_{n-1}+q_{n-2}) + \frac{1}{a_{n+1}}q_{n - 1}} \\
&= \frac{a_{n+1}p_n + p_{n - 1}}{a_{n+1}q_n + q_{n - 1}}
\end{align}
and the base case is easy to verify.
Thus we get the matrix equation
$$ \begin{bmatrix} p_n & p_{n-1} \\ q_n & q_{n - 1} \end{bmatrix} \begin{bmatrix} a_{n+1} & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} p_{n+1} & p_{n} \\ q_{n+1} & q_{n} \end{bmatrix} $$
and by induction
$$ \begin{bmatrix} p_{n+1} & p_{n} \\ q_{n+1} & q_{n} \end{bmatrix} = \begin{bmatrix} a_{n+1} & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n} & 1 \\ 1 & 0 \end{bmatrix} \cdots \begin{bmatrix} a_{1} & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$
taking determinants we have
$$ p_{n+1}q_n - p_nq_{n+1} = (-1)^{n}. $$
Hence
$$ \left\lvert \frac{p_{n+1}}{q_{n + 1}} - \frac{p_n}{q_n} \right\rvert = \frac{|p_{n+1}q_n - p_nq_{n+1}|}{q_{n+1}q_n} = \frac{1}{q_{n+1}q_n}. $$
From $(1)$ and the assumption that $a_k \ge 1$ we have $q_n \ge n$. Thus the sequence $p_n/q_n$ is Cauchy and converges to some limit.
Reference
Automatic Sequences by J-P. Allouche and J. Shallit, Cambridge University Press (2003), pages 44-46.
|
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|
find matrix $A$ in other basis Original matrix is:
$$A=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$
original basis: $$\langle \mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \rangle$$
basis where I have to find matrix: $$\langle f_1,f_2,f_3 \rangle$$
Where:
$$f_1=2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3} \\ f_2 = 3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} \\f_3=\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3}$$
Therefore:
$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1f_1+y_2f_2+y_3f_3$$
Expanding that we get:
$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1(2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3})+y_2(3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} )+y_3(\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3})$$
How should I group things at right side around unit vectors? for example for first coordinate $2\mathbf{e_1} \, !=\, 3\mathbf{e_1}$, how could I group coordinates?
${\Large \textbf {Solution:}}$
Opening brackets:
$$x\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3} = 2y_1\mathbf{e_1}+3y_1\mathbf{e_2}+y_1\mathbf{e_3}+3y_2\mathbf{e_1}+4y_2\mathbf{e_2}+y_2\mathbf{e_3}+y_3\mathbf{e_1}+2y_3\mathbf{e_2}+2y_3\mathbf{e_3}$$
Sorting by unit vectors:
$$x_1\mathbf{e_2}+x_2\mathbf{e_3}+x_3\mathbf{e_3} = (2y_1+3y_2+y_3)\mathbf{e_1}+(3y_1+4y_2+2y_3)\mathbf{e_2}+(y_1+y_2+2y_3)\mathbf{e_3} \tag{1}$$
Based on fotmula:
$$A'=T^{-1}AT$$
Matrix $T$ is based on right side of equation $(1)$, and we have:
$$T=\begin{pmatrix} 2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$
The rest is more less technical.
$$T^{-1}=\begin{pmatrix} -6&5&-2\\4&-3&1\\1&-1&1 \end{pmatrix}$$
|
To compute the matrix in the other basis We need to compute $T(f_1), T(f_2), T(f_2)$, where $T$ is the linear map defined by $A$ in the basis $e$.
By definition of matrix of a linear map, the coordinates of $T(f_i)$ in the basis $e$ will be obtained if we multiply $A$ by the coordinates of $f_i$ in the basis $e$.
For example, the coordinates of $f_1$ in the basis $e$ is $(2,3,1)^{T}$ since $f_1=2e_1+3e_2+e_3$. Multiplying $A(2,3,1)^{T}=(2,3,1)^{T}$.
Therefore $T(f_1)=2e_1+3e_2+e_1=f_1=1f_1+0f_2+0f_3$. Therefore, the first column of the matrix of $T$ in the basis $f$ is $(1,0,0)^{T}$.
Doing the same for $f_2$ and $f_3$ gives you the other two columns.
|
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|
show that $x+y+z\ge 2$
Let $x,y,z\ge 0$. If
$$x^{2016}y^{2016}+y^{2016}z^{2016}+x^{2016}z^{2016}=1$$ then show
that $$x+y+z\ge 2$$
I have tried
$$(x+y+z)^2-3(xy+yz+xz)=\dfrac{1}{2}\left[ (x-y)^2+(y-z)^2+(z-x)^2\right] \ge 0$$
so we have
$$(x+y+z)^2\ge 3(xy+yz+xz)$$
but I am stuck here. How could I continue this proof?
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For $z=0$ and $x=y=1$ we get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\left(\frac{x+y+z}{2}\right)^{4032}\geq\sum_{cyc}x^{2016}y^{2016},$$
for which we'll prove the following.
Let $x$, $y$ and $z$ be non-negative numbers and nutural $n\geq2$. Prove that:
$$\left(\frac{x+y+z}{2}\right)^{2n}\geq x^ny^n+x^nz^n+y^nz^n.$$
1. $n=2$.
We need to prove that
$$(x+y+z)^4\geq16(x^2y^2+x^2z^2+y^2z^2).$$
Indeed, by AM-GM
$$(x+y+z)^4=\left(\sum_{cyc}(x^2+2xy)\right)^2\geq8\sum_{cyc}x^2\sum_{cyc}xy=$$
$$=8\sum_{cyc}(x^3y+x^3z+x^2yz)\geq8\sum_{cyc}(x^3y+x^3z)\geq8\sum_{cyc}(2x^2y^2)=16(x^2y^2+x^2z^2+y^2z^2).$$
2. Now, $$\left(\frac{x+y+z}{2}\right)^{2k+2}=\left(\frac{x+y+z}{2}\right)^4\left(\frac{x+y+z}{2}\right)^{2k-2}\geq$$
$$\geq\sum_{cyc}x^2y^2\sum_{cyc}x^{k-1}y^{k-1}=\sum_{cyc}(x^{k+1}y^{k+1}+x^{k+1}y^{k-1}z^2+x^{k+1}z^{k-1}y^2)\geq$$
$$\geq x^{k+1}y^{k+1}+x^{k+1}z^{k+1}+y^{k+1}z^{k+1}.$$
Thus, by induction for ending of the proof it remains to prove that
$$(x+y+z)^6\geq64(x^3y^3+x^3z^3+y^3z^3).$$
We can make it by the similar way:
$$(x+y+z)^6=(x+y+z)^4\sum_{cyc}(x^2+2xy)\geq16\sum_{cyc}x^2y^2\sum_{cyc}(x^2+2xy)=$$
$$=16\sum_{cyc}(x^4y^2+x^4z^2+x^2y^2z^2+2x^3y^3+2z^3y^2z+2x^3z^2y)\geq$$
$$\geq16\sum_{cyc}(x^4y^2+x^4z^2+2x^3y^3)\geq16\sum_{cyc}(2x^3y^3+2x^3y^3)=64\sum_{cyc}x^3y^3$$
and we are done!
|
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|
Solve the equation for $4\csc (x+2)-3\cot^2(x+2) = 3$ for $ 0\leq x \leq 8$ For this question, what I did was simply change all of the trigonometric functions to $\sin , \cos, \tan$ form. So the final step before I got stuck was $3(\sin^2 (x+2) + \cos^2 (x+2))=4\sin(x+2)$.
Not quite sure how to proceed from here, any hint will be appreciated.
|
As $\csc^2A-\cot^2A=1$
$$3=4\csc(x+2)-3\cot^2(x+2)=4\csc(x+2)-3\{\csc^2(x+2)-1\}$$
$$\iff\csc(x+2)\{3\csc(x+2)-4\}=0$$
As for real $x+2,\csc(x+2)\ge1$ or $\le1$
$$3\csc(x+2)-4=0\iff\sin(x+2)=\dfrac34$$
|
{
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|
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}$
$=\boxed{2a^2 + 2b^2 +2c^2 + 2ab + 2bc + 2ac}$
$2(ab + bc +ac) = 2$
$\boxed{2a^2 + 2b^2 +2c^2 + 2}$
$a^2 + b^2 + c^2 = (a+b+c)^2 -2ab - 2bc -2ac$
$a^2 + b^2 + c^2 = (2)^2 -2$
$a^2 + b^2 + c^2 = 2$
$\bbx{\color{red} {2\times 2 + 2 =6}}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$ I figure out these thing when "playing" with numbers:
$$3^2-2.2^2+1^2=2=2!$$
$$4^3-3.3^3+3.2^3-1^3=6=3!$$
$$5^4-4.4^4+6.3^4-4.2^4+1^4=24=4!$$
So I go to the conjecture that:
$$\binom{n}{n}(n+1)^n-\binom{n}{n-1}n^n+\binom{n}{n-2}(n-1)^n-...=n!$$
or
$$\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$$
Now, how can I prove this conjecture? I've tried a lot, but still couldn't have any idea.
|
We can test the case $n=1$:
\begin{align}\sum_{x=0}^1(−1)^x{1\choose 1−x}(1+1−x)^1&=(−1)^0{1\choose 1−0}(1+1−0)^1+ (−1)^1{1\choose 1−1}(1+1−1)^1\\
&=2+(-1)\\
&=1\\
&=1!\end{align}
Now we assume it holds for $n=k$, that is to say that $$\sum_{x=0}^k(−1)^x{k\choose k-x}(k+1−x)^k=k!$$
We need to prove that it holds for $n=k+1$, that is to say that $$\sum_{x=0}^{k+1}(−1)^x{k+1\choose k+1-x}(k+2−x)^{k+1}=(k+1)!$$
Note that $$(k+1)!=k!\times (k+1)$$
Can you continue from here?
|
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Simplifying the Solution to the Cubic I am trying to solve the cubic. I currently have that, for $ax^3+bx^2+cx+d=0$, a substitution to make this monic. Dividing by $a$ gives
$$x^3+Bx^2+Cx+D=0$$
where $B=\frac{b}{a}, C=\frac{c}{a}, D=\frac{d}{a}$. Then, with the substitution $x=y-\frac{B}{3}$, I got
$$y^3+\left(C-\frac{B^2}{3}\right)y+\left(D-\frac{BC}{3}+\frac{2B^3}{27}\right)=0$$
Thus, to make things simpler, i made the substitution $p=C-\frac{B^2}{3}$ and $q=D-\frac{BC}{3}+\frac{2B^3}{27}$ we have the "depressed cubic"
$$y^3+py+q=0$$
Now, using the identity,
$$(m+n)^3=3mn(m+n)+(m^3+n^3)$$
we let $y=m+n$. This then translates to $p=-3mn,$ and $q=-(m^3+n^3)$ and gives us a system of equations in $m$ and $n$. Solving for $n$ gives $n=-\frac{p}{3m}$ and back substituting yields
$$q=-m^3+\frac{p}{3m}\qquad \Rightarrow \qquad m^6+qm^3-\frac{p^3}{27}=0$$ and now we can solve the quadratic for $m$;
$$m=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}$$
and then that means, by back substitution
$$n=-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
So, I think I am almost here, because now,
$$y=m+n=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
But how can I simplify this expression? I know I can back substitute for the original $a,b,c,d$ and solve for $x$. But this sum looks complicated and my attepts to simplify the sum have not worked.
|
For the calculation of the roots of the depressed cubic
$$
y^{\,3} + p\,y + q = 0
$$
where $p$ and $q$ are real or complex,
I personally adopt a method indicated in this work by A. Cauli, by which putting
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}\quad v = - \frac{p}
{{3\,u}}\quad \omega = e^{\,i\,\frac{{2\pi }}
{3}}
$$
where for the radicals you take one value, the real or
the first complex one (but does not matter which)
then you compute the three solutions as:
$$
y_{\,1} = u + v\quad y_{\,2} = \omega \,u + \frac{1}
{\omega }\,v\quad y_{\,3} = \frac{1}
{\omega }\,u + \omega \,v
$$
Also refer to this post and to this other one.
|
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|
Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$ Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$
a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$
$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$
$b_n = \frac{1}{2\pi}\int_0^r r\sin(nr)\,dr = \frac{(-1)^n + 1}{2n}$
$f(x) = \frac{\pi}{8} - \sum_n [\frac{((-1)^n - 1)\cos(nx)}{2\pi n^2} + \frac{((-1)^n + 1)\sin(nx)}{2n}]$
The prof asked us to use this Fourier series to prove that $\pi^2/8 = 1+1/3^2+1/5^2+1/7^2+\cdots$. How do I do this?
|
First of all your $b_k$ are wrong, they should be:
$$b_k= \frac{(-1)^{k+1}}{2k}$$
Not that it matters beacause of the following. Second of all notice that $f(x)$ is continuos at zero, which is to say $f(0^+)= f(0^-)=0$. Once you get the expansion right is not that hard, just make $x=0$, easy peasy.
|
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|
Solution to $7^{2x-2} \equiv 4 \mod13$ I need to solve the congruence
$$ 7^{2x-2} \equiv 4 \mod13 $$
I think I need to use a primitive root to transform it into $2x-2 \equiv ??? \mod (\phi(13)=12)$ but I'm stumped on how to actually do it.
|
$\mod 13$:
$$7^2 \equiv 10 \equiv -3 \\
7^4 \equiv (-3)^2 \equiv 9 \equiv -4\\
7^8 \equiv (-4)^2 \equiv 16 \equiv 3\\
7^{10} \equiv 7^2\cdot 7^8 \equiv (-3)\cdot 3\equiv -9 \equiv 4
$$
So
$$
2x-2 = 10 \\
x = 6
$$
|
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|
Two ways to calculate the primitive of $f(x) =\frac{x^2}{x^4-1}$ but two different results, why? When I search primitives of $f(x)= \dfrac{x^2}{x^4-1}$, using partial fraction decomposition the solutions are
$\boxed {F(x) =\dfrac{1}{4} \ln|x-1| + \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
But if I use an other way, I've got an another result :
$f(x)=\dfrac{x^2}{x^4-1} =\dfrac{x^2-1+1}{(x^2-1)(x^2+1)} =\dfrac{x^2-1}{(x^2-1)(x^2+1)}+\dfrac{1}{(x^2-1)(x^2+1)} = \\\dfrac{1}{x^2+1} +\dfrac{1}{x^4-1}= \dfrac{1}{x^2+1} -\dfrac{1}{1-(x^2)^2}$
Thus $\boxed {F(x)= \arctan{x} -\operatorname{argth}x^2 +C} $
I think that might be wrong, because that doesn't match with the first result if I plot both on geogebra. I noticed as well their domains are different.
My question is why this two different results? Is one of them wrong?
|
Alternate forms exist. According to Wolfram Alpha:
The first one and the second one are the same because:
$$
\tanh^{-1}(x):=\frac12\log\left(\frac{1+x}{1-x}\right)
$$
|
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|
How to get the $\phi$ from $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$? $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$,
where $c=\sqrt{a^2+b^2+2ab\cos(\theta)}$, and $\displaystyle\tan(\phi)=\frac{b\sin(\theta)}{a+b\cos(\theta)}$.
I want to know how to get to this result.
I'm able to derive $c$ by taking the derivative of the equation, then squaring both and adding them together, and back-substituting the cosine of a double angle.
But how does one get to the expression for $\tan(\phi)$?
|
Use the sin addition formula $\sin(\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$
\begin{eqnarray*}
a \sin x + \underbrace{b \sin(x+\theta)}_{ b\sin x \cos \theta+b \cos x \sin \theta}= \underbrace{c \sin(x+ \phi)}_{b\sin x \cos \phi+b \cos x \sin \phi} \\
(a + b \cos \theta) \color{red}{\sin x} + b \sin \theta \color{blue}{\cos x} = c \cos \phi \color{red}{\sin x}+c \sin \phi \color{blue}{\cos x}
\end{eqnarray*}
Equate the coefficients of $ \sin x $ and $ \cos x $
\begin{eqnarray*}
a + b \cos \theta = c \cos \phi \\
b \sin \theta =c \sin \phi
\end{eqnarray*}
Now square these equations and add to get the first equation you want. & take the ratio of these equations to get the second equation.
|
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|
For a odd prime $p, \exists a,b$ such that $a^2 + ab + b^2 \equiv 0 \pmod{p} \iff \exists x,y $ such taht $x^2 + xy + y^2 = p$? We know that $p=x^2+xy+y^2$ if and only if $p \equiv 1 \pmod {3}$.
But I need $a^2+ab+b^2 \equiv 0 \pmod{p} $ if and only if $p \equiv 1 \pmod {3}$, more generalized theorem.
I think that i should proof below :
$\exists a,b$ such that $a^2 + ab + b^2 \equiv 0 \pmod{p} \iff \exists x,y $ such taht $x^2 + xy + y^2 = p$
Easy to prove ($\Longleftarrow$), but not ($\implies$).
These are some examples.
$2^2 + 2 \cdot 4 + 4^2 = 4( 1^2 + 1 \cdot 2 + 2^2) = 4 \cdot 7$
$3^2 + 3 \cdot 9 + 9^2 = 9( 1^2 + 1 \cdot 3 + 3^2) = 9 \cdot 13$
$7^2 + 7 \cdot 11 + 11^2 = 13( 2^2 + 2 \cdot 3 + 3^2) = 13 \cdot 19$
$5^2 + 5 \cdot 25 + 25^2 = 25( 1^2 + 1 \cdot 5 + 5^2) = 25 \cdot 31$
$10^2 + 10 \cdot 26 + 26^2 = 28( 3^2 + 3 \cdot 4 + 4^2) = 28 \cdot 37$
$6^2 + 6 \cdot 36 + 36^2 = 36( 1^2 + 1 \cdot 6 + 6^2) = 36 \cdot 43$
$13^2 + 13 \cdot 47 + 47^2 = 49( 4^2 + 4 \cdot 5 + 5^2) = 49 \cdot 61$
$29^2 + 29 \cdot 37 + 37^2 = 49( 2^2 + 2 \cdot 7 + 7^2) = 49 \cdot 67$
|
Given any sum with a particular prime factor $p$ we can isolate the factor $p$ via Eisenstein integer multiplication.
Take your example with $10, 26$ having the sum $28×37$. How to isolate the factor $37$?
First get rid of the common factor $2^2$ which is totally useless:
$5^2+(5×13)+13^2=7×37$
Next we associate thus quadratic form with the Eisenstein integer $5-13\omega$ whose squared norm matches the value of the quadratic form. Generally: $|a-b\omega|^2=a^2+ab+b^2$.
Now $7$ has several sums that similarly match up with Eisenstein integers having that squared norm:
$7=1^2+(1×2)+2^2 \rightarrow 1-2\omega$
$7=2^2+(2×1)+1^2 \rightarrow 2-\omega$
$7=1^2-(1×3)+3^2 \rightarrow 1+3\omega$
$7=3^2-(3×1)+1^2 \rightarrow 3+\omega$
$7=2^2-(2×3)+3^2 \rightarrow 2+3\omega$
$7=3^2-(3×2)+2^2 \rightarrow 3+2\omega$
All we need do is multiply our Eisenstein integer for $7×37$ above, $5-13\omega$, by the above multipliers, and look for a product that has a factor of $7$ which we drop out. We will then have a combination equalling $37$ with integer inputs.
$(5-13\omega)×(1-2\omega)=(5×1-13×2)-(5×2+13×1+13×2)\omega=-21-49\omega=-7×(3+7\omega)$
So $|3+7\omega|^2=37$ meaning $3^2-(3×7)+7^2=37$.
To get a solution with positive integers we can multiply $3+7\omega$ by $\omega$ to get a product with mixed signs:
$(3+7\omega)×\omega=-7-4\omega$
$(-7-4\omega)×\omega=4-3\omega$
Then $37=4^2+(4×3)+3^2$.
All these manipulations turn on the Euclidean domain property of the Eisenstein integers so we are sure they work.
|
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What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $
Solving this equation results in:
$x_{1}+x_{2}+x_{3}+x_{4}=22$
I removed restriction of $x_{i} \geq 1$ first as follows-:
$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
$\Rightarrow \binom{18+4-1}{18}=1330$
Now i removed restriction for $x_{i} \leq 6$ , by calculating the number of bad cases and then subtracting it from $1330$:
calculating bad combination i.e $x_{i} \geq 7$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
We can distribute $7$ to $2$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{2}$
We can distribute $7$ to $1$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{1}$ and then among all others .
i.e
$$\binom{4}{1} \binom{14}{11}$$
Therefore, the number of bad combinations equals $$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$
Therefore, the solution should be:
$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$
However, I am getting a negative value. What am I doing wrong?
EDIT
I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.
|
In order for the sum to equal 22, either three dice equal $6$ and one equals $4$, or two dice equal $6$ and two dice equal $5$. The number of valid outcomes thus equals:
$${4 \choose 1} + {4 \choose 2} = 4 + 6 = 10$$
As such, the probability of the four dice having a sum of $22$ equals:
$$\frac{{4 \choose 1} + {4 \choose 2}}{6^4} = \frac{10}{1296} = \frac{5}{648} \approx 0.00772$$
|
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|
$x^4 + y^4 \mod p$ The primes $p = 5, 13, 17, 29$ have the property that there exist fewer than $p$ values for $x^4 + y^4 \mod p$. For example, $x^4 \equiv 0$ or $1 \mod 5$ so $x^4 + y^4 \equiv 0, 1$ or $2$, but never $3$ or $4$, $\mod 5$. Are there any other primes with this property?
The question arose in connection with OEIS sequence A289559
|
For a given prime $p$, let $Z_p$ denote the ring of integers, mod $p$, and let $f\colon Z_p^2 \to Z_p$ be defined by $f(x,y) = x^4+y^4$.
Partial result: If $p \equiv -1\;\,(\text{mod}\;4)$, then $f$ is surjective.
Proof:
Assume $p$ is a prime, with $p \equiv -1\;\,(\text{mod}\;4)$.
Then $-1$ is not a quadratic residue, mod $p$.
Claim every square in $Z_p$ is also a $4$-th power.
Let $a,b \in Z_p^{*}$, and suppose $a^4=b^4$ in $Z_p$.
\begin{align*}
\text{Then in $Z_p$,}\;\;&a^4=b^4\\[4pt]
\implies\;&a^2=\pm b^2\\[4pt]
\implies\;&a^2=b^2&&\text{[since $-1$ is not a quadratic residue, mod $p$]}\\[4pt]
\implies\;&a= \pm b\\[4pt]
\end{align*}
It follows that the map $Z_p^{*} \to Z_p^{*}$ given by $x \mapsto x^4$ is exactly two-to-one, hence the set of $4$-th powers in $Z_p^{*}$ has cardinality $\frac {p-1}{2}$. Of course every $4$-th power is also a square, hence, since the cardinalities are the same, the set of $4$-th powers in $Z_p^{*}$ is the same as the set of squares in $Z_p^{*}$, which (since $0$ is also a $4$-th power), proves the claim.
For any $k \in Z_p$ which is not a square in $Z_p$,
$$Z_p = \{x^2 \mid x \in Z_p\} \cup \{kx^2 \mid x \in Z_p^{*}\}$$
Fix $k \in \{0,1,2,...,p-1\}$ as the least positive integer such that $k$ is not a square in $Z_p$. Then $k > 1$, and $k-1$ is a square in $Z_p$.
Fix $r$ is $Z_p$. We want to show $r$ is in the image of $f$.
If $r$ is a square in $Z_p$, then $r$ is a $4$-th power in $Z_p$, hence $f(x,0)=r$, for some $x$, so $r$ is in the image of $f$.
Next, suppose $r$ is not a square in $Z_p$.
Then $r=kx^2$, for some $x \in Z_p$, and $(k-1)x^2$ is a square in $Z_p$.
Let $y$ be such that $y^2 = (k-1)x^2$ in $Z_p$.
\begin{align*}
\text{Then in $Z_p$,}\;\;&y^2=(k-1)x^2\\[4pt]
\implies\;&x^2+y^2=kx^2\\[4pt]
\implies\;&x^2+y^2=r\\[4pt]
\implies\;&\text{$r$ is the sum of two squares}\\[4pt]
\implies\;&\text{$r$ is the sum of two $4$-th powers}\\[4pt]
\end{align*}
hence $r$ is in the image of $f$, as required.
|
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|
How can the sum of squares be negative?
If $a,b,c,d$ are the roots of the equation $x^4-Kx^3+Kx^2+Lx+M=0$, where $K,L,M$ are real numbers, then the mininmum value of $a^2+b^2+c^2+d^2$ is?
My answer:
$\sum a=K,\ \sum ab=K\implies$
$a^2+b^2+c^2+d^2=K^2-2K=(K-1)^2-1$. For $K=1$, $(a^2+b^2+c^2+d^2)_{min}=-1$
This matches with the answer in fact, but how can sum of squares ever result in NEGATIVE.
What's the intuition behind this answer is it wrong or I'm going the wrong way.
|
Consider $K=1,$ that is, $$f(x)=x^4-x^3+x^2+Lx+M.$$ We have that
$$f''(x)=2(6x^2-3x+1).$$ Note that $$f''=0$$ has no real roots. Thus $f'=0$ has a real root and so $f=0$ has at most two real roots. In other word, at least two roots of the equation $$f(x)=0$$ are complex. Thus there is no contradiction with the fact that the sum of the squares of the roots is $-1.$
If $L=M=0$ then the roots are $0$ (double) and $\dfrac{1\pm\sqrt {-3}}{2}=\dfrac{1\pm i\sqrt{3}}{2}.$ We have that
$$0^2+0^2+\left(\dfrac{1+i\sqrt{3}}{2}\right)^2+\left(\dfrac{1-i\sqrt{3}}{2}\right)^2=-1.$$
|
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|
Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is evident that $-1$ is not a root of the second equation, but how can I find out whether the other two roots are common or not?
|
The roots of $x^2+x+1$ are $\omega$ and $\omega^2$ (The two complex cube roots of unity)
and $\omega^3=1$. Therefore
$$ \omega^{1990}+\omega^{200}+1=\omega+\omega^2+1=0 $$
and
$$ \omega^{2\times1990}+\omega^{2\times200}+1=\omega^2+\omega+1=0$$
So $\omega$ and $\omega^2$ are the common roots.
|
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If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this question, but not able to proceed. How do I solve this?
|
On the LHS you have $$\frac{a^5}{b^3c^3} + \frac{b^5}{a^3c^3} +
\frac{c^5}{a^3b^3}$$
Relabel this as $$x_1 + x_2 + x_3$$
The standard way to proceed with AM-GM is to fiddle around with products of $x_1, x_2, x_3$ until you find some $\sqrt[n_1 + n_2 + n_3]{x_1^{n_1}x_2^{n_2}x_3^{n_3}}$ that yields one of the RHS terms.
In our case a little fiddling gives $$x_1^3x_2^3x_3^2 = \frac{1}{c^8}$$
So we have by AM-GM
$$\frac{3x_1 + 3x_2 + 2x_3}{8} \geq \sqrt[8]{x_1^3x_2^3x_3^2} = \frac{1}{c}$$
And similarly
$$\frac{3x_1 + 2x_2 + 3x_3}{8} \geq \frac{1}{b}$$
$$\frac{2x_1 + 3x_2 + 3x_3}{8} \geq \frac{1}{a}$$
Adding these three inequalities together gives you the one you set out to prove.
|
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"url": "https://math.stackexchange.com/questions/2355748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Reflection across a line Let $g$ be a line with equation $g:ax+by+c=0$ in Hesse normal form. I want to show that the reflection across $g$ is described by \begin{equation*}\binom{x}{y}\mapsto \binom{x}{y}-2(ax+by+c)\binom{a}{b}\end{equation*}
At the reflection across $g$ it holds the following for the image $P'$ of each point $P$:
*
*$P'$ lies on the perpendicular to $g$ through $P$.
*$g$ bisects $PP'$.
So to show the desired result, do we have to find the perpendicular line to $g$ ?
|
This reflection is an affine map, so we can represent it as linear map if we embed plane in $\mathbb R^3$ with $(x,y)\mapsto (x,y,1)$.
I will use the same idea as Robert Z, $P'$ is reflection of $P$ along the line $ax + by + c=0$ if two conditions are met:
(1) $\ \displaystyle\frac{P+P'}2$ lies on the line $ax + by + c = 0$
We can describe this with linear equation $$\left\langle\frac{P+P'}2,(a,b,c) \right\rangle = 0.$$
(2) $\ P-P'$ is perpendicular to the line $ax + by + c = 0$
Direction vector of the line is given by $(-b,a,0)$, so $P'-P$ must be orthogonal to it, i.e. $$\left\langle P'-P,(-b,a,0)\right\rangle = 0.$$
If we write $P = (x,y,1),\ P'=(x',y',1)$, conditions (1) and (2) can be written as
\begin{align}
ax'+by'&= -ax - by - 2c\\
-bx'+ay' &= -bx+ay
\end{align}
or in matrix form
$$
\begin{pmatrix}
a & b & 0\\
-b & a & 0\\
0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
x'\\
y'\\
1\\
\end{pmatrix} = \begin{pmatrix}
-a & -b & -2c\\
-b & a & 0\\
0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
x\\
y\\
1\\
\end{pmatrix}.$$
Invert the matrix on the left hand side to get
\begin{align}
\begin{pmatrix}
x'\\
y'\\
1\\
\end{pmatrix} &= \frac 1{a^2+b^2}\begin{pmatrix}
a & -b & 0\\
b & a & 0\\
0 & 0 & a^2+b^2\\
\end{pmatrix}\begin{pmatrix}
-a & -b & -2c\\
-b & a & 0\\
0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
x\\
y\\
1\\
\end{pmatrix}\\
&= \frac {-1}{a^2+b^2}\begin{pmatrix}
a^2-b^2 & 2ab & 2ac\\
2ab & b^2-a^2 & 2bc\\
0 & 0 & -(a^2+b^2)\\
\end{pmatrix}\begin{pmatrix}
x\\
y\\
1\\
\end{pmatrix}
\end{align}
or if you want to get back to $\mathbb R^2$, we get affine map
$$\begin{pmatrix} x\\ y\end{pmatrix} \mapsto \frac {-1}{a^2+b^2}\begin{pmatrix}
a^2-b^2 & 2ab \\
2ab & b^2-a^2
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}-\frac {1}{a^2+b^2}\begin{pmatrix} 2ac\\ 2bc\end{pmatrix}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine if these functions are injective
Determine if the following functions are injective.
$$f(x) = \frac{x}{1+x^2}$$
$$g(x) = \frac{x^2}{1+x^2}$$
My answer:
$f(x) = f(y)$
$$\implies \frac{x}{1+x^2} = \frac{y}{1+y^2}$$
$$\implies x+xy^2 =y+yx^2$$
$$\implies x=y$$
Hence $f(x)$ is injective
$g(x) = g(y)$
$$\implies \frac{x^2}{1+x^2}=\frac{y^2}{1+y^2}$$
$$\implies x^2+x^2y^2=y^2+y^2x^2$$
$$\implies x^2=y^2$$
$$\implies \pm x=\pm y$$
So $g(x)$ is not injective
|
If $x + xy^2 = y+yx^2$, then $(x - y) = yx^2 - xy^2 = xy(x-y)$. This gives $(x-y)(xy-1) = 0$, hence either $x=y$ or $xy=1$. You can check that for example, $f(\frac 12) = f(2) = \frac 25$.
Hence, $f$ is not injective.
The answer to the second question is correct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-a} + \sqrt {bx}}$$
$$=\lim_{x\to \infty} \dfrac {x-a-bx}{\sqrt {x-a} + \sqrt {bx}}.$$
How do I proceed?
|
Clearly, for $b<1$, $-\infty$ and for $b>1$, $\infty$ (because $\sqrt{1-a/x}-\sqrt b$ tends to the constant $1-\sqrt b$).
Then
$$\sqrt{x-a}-\sqrt{x}=\frac{x-a-x}{\sqrt{x-a}+\sqrt{x}}$$ and the limit is $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2362432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Simplifying $\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$ I'm struggling trying to simplify $$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$$
Here's my attempt:
All of these coefficients can be converted to base $2$ exponentials.
$$\frac{2^{4^{(x+1)}}+20^\ (2^{2^{(2x)}})}{2^{x-3}2^{3^{(x+2)}}}$$
Doing some algebra:
$$\frac{2^{4x+4}+20^\ (2^{4x})}{2^{x-3}2^{3x+6}}$$
The denominator can be simplified.
$$\frac{2^{4x+4}+20^\ (2^{4x})}{2^{4x+3}}$$
Now, here's where I think I've made a mistake, but I'm gonna try to simplify it further by splitting this up into separate fractions.
$$\frac{2^{4x+4}}{2^{4x+3}} + \frac{5 \bullet 2^2 \ 2^{4x}}{2^{4x+3}}$$
$$2\ \frac{2^{4x+3}}{2^{4x+3}} + \frac{5\ \bullet 2^{4x+2}}{2^{4x+3}}$$
$$2 + 5\ \bullet 1/2\ \frac{2^{4x+2}}{2^{4x+2}}$$
$$2 + 5/2$$
$$4/2 + 5/2$$
$$9/2$$
Whelp, I just solved it properly while putting it in as a question. However, I still have two queries:
*
*Symbolab's answer is this: . Is there logic behind this answer or is outright not simplified as well?
*Does this algorithm to solving (changing all to the right base and trying to do algebra do them apply to ones even if they aren't connected by whole number bases? Like if I were trying to do one with base $5$ and base $7$ exponentials, could I solve it using logarithm laws just the same?
|
You just need to get more comfortable converting bases. Then any way you want to do it will work:
$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$
$=\frac {2^{4(x+1)} + 5*4*2^{2*2x}}{2^{x-3}2^{3(x+2)}}$
$=\frac {2^{4x + 4} + 5*2^22^{4x}}{2^{(x-3)+3x + 6}}$
$=\frac {2^{4x+4} + 5*2^{4x+2}}{2^{4x + 3}}$
$= \frac {2^{4x+2}(2^2 + 5)}{2^{4x + 3}}$
$=9*\frac {2^{4x+2}}{2^{4x + 3}}$
$= 9*\frac 1{2^{(4x + 3)-(4x+2)}}$
$= 9*\frac 12$
"Does this algorithm to solving (changing all to the right base and trying to do algebra do them apply to ones even if they aren't connected by whole number bases? Like if I were trying to do one with base 5 and base 7 exponentials, could I solve it using logarithm laws just the same?"
Yes, but it might not be easy. And without calculators it might not be possible.
Ex: $\frac {7^x + 3^{2x}}{2^{3x + 1}} =$
$\frac {2^{x\log_2 7} + 2^{2x\log_2 3}}{2^{3x+1}}=$
$2^{x\log_2 7 + 2x \log_2 3 - 3x - 1} =$
$2^{x(\log_2 (63) -3) - 1}$.
But is that really "simpler"?
Post-script.... Actually I guess it is.
If you need to solve $\frac {7^x + 3^{2x}}{2^{3x + 1}} = 43$ we get
$2^{x(\log_2 (63) -3) - 1} = 43$ so
$x(\log_2 (63) - 3) - 1 = \log_2 43$
$x = \frac {\log_2 43 + 1}{\log_2(63) - 3}$
$\approx 2.1584348532293047695329252628754$.
We didn't have to do this base $2$. We could have used any base.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving solution set inequalities problems I need some guidance with solving problems of the following nature:
$$|x^2-3x-1|<3$$
Instinctively, my way of solving it is splitting it into the following equations:
$$x^2-3x-1< 3$$
$$x^2-3x-1> -3$$
Solving $x^2-3x-1< 3$
$$x^2-3x-4< 0$$
$$(x+1)\ (x-4)< 0$$
$$x < -1$$
$$x < 4$$
Solving $x^2-3x-1> -3$
$$x^2-3x-1> -3$$
$$x^2-3x+2> 0$$
$$(x-2)(x-1)> 0$$
$$x > 2$$
$$x > 1$$
So, here $x < -1$ is a valid solution and $x > 1$ is also valid.
In addition, $x < 4$ is valid and $ x > 2$ is also valid.
So we have $-1 > x > 1$ and $2<x<4$.
Firstly, is this possible? Does any $x$ have to satisfy $-1 > x > 1$ and $2<x<4$ simultaneously? Or are these the ranges of allowed inputs? Like, $x$ can be $-2$ despite $2<-2<4$ is not true.
Secondly, WolframAlpha has the following numberline:
Which means I've not solved this correctly. So how do I solve these properly?
|
$$|x^2-3x-1|<3\Rightarrow -3<x^2-3x-1<3\\\Rightarrow -3<x^2-2(3/2)x+(3/2)^2-(3/2)^2-1<3\\\Rightarrow 1/4<(x-3/2)^2<25/4$$
Now you have two inequation:
*
*$1/4<(x-3/2)^2\Rightarrow x-3/2<-1/2\space \text{or}\space x-3/2>1/2\Rightarrow x<1\space \text{or}\space x>2$.
*$(x-3/2)^2<25/4\Rightarrow -5/2<x-3/2<5/2\Rightarrow x>-1\space \text{or} \space x<4$
Hence the final answer is $\{x\in\mathbb{R}: -1<x<1\space\text{and}\space 2<x<4\}\space\space\space\blacksquare$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2363590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
arctan series multisection by roots of unity I'm trying to multisect the series for $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots$ using the method of roots of unity, as described in the paper linked in one of the comments in How to express a power series in closed form. I want to take every second term of this series. But I think that I'm doing something wrong, because I get some weird results :q
Here's what I do:
First, since I'll be taking every 2nd term, I need square roots of unity:
$\omega = e^{i\frac{2\pi}{2}} = e^{i\pi} = -1 \\
\omega^0 = (-1)^0 = 1 \\
\omega^1 = (-1)^1 = -1 \\
\omega^{-1} = (-1)^{-1} = \frac{1}{-1} = -1$
I assume the step $k=2$ and the initial offset $r = 0$, and the function $f(x) = \arctan(x)$.
Now I use the formula for the closed form of the multisected series:
$$F(x) = \frac{1}{k}\sum_{n=0}^{k-1}\omega^{-nr}f(\omega^nx)$$
After substituting my numbers and the roots of unity to this formula I get:
$$F(x) = \frac{1}{2}\sum_{n=0}^{1}\omega^{-n\cdot0}\arctan(\omega^nx) \\
F(x) = \frac{1}{2}\left\{ \omega^{-0\cdot0}\arctan(\omega^0x) + \omega^{-1\cdot0}\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{ \omega^0\arctan(\omega^0x) + \omega^0\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) + \arctan(-x) \right\} $$
Now since $\arctan(-x) = -\arctan(x)$, I get:
$$F(x) = \frac{1}{2}\left\{ \arctan(x) - \arctan(x) \right\} = \frac{1}{2}\!\cdot\!0 = 0$$
:-/
I thought that I perhaps should have used $r=1$ as the offset, so I tried this one too:
$$F(x) = \frac{1}{2}\sum_{n=0}^{1}\omega^{-n\cdot1}\arctan(\omega^nx) \\
F(x) = \frac{1}{2}\left\{ \omega^{-0\cdot1}\arctan(\omega^0x) + \omega^{-1\cdot1}\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{ \omega^0\arctan(\omega^0x) + \omega^{-1}\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{1\cdot\arctan(1\!\cdot\!x) + (-1)\!\cdot\!\arctan(-1\!\cdot\!x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) - \arctan(-x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) + \arctan(x) \right\} \\
F(x) = \frac{2\!\cdot\!\arctan(x)}{2} = \arctan(x)$$
So again, something seems to be wrong here: how come I obtained the same original function from taking only every second term of its power series?
Where did I make a mistake?
|
The problem is that the arctangent series has already had this procedure applied once:
$$ \arctan{x} = \frac{1}{2} (-i\log{(1+ix)}-(-i\log{(1-ix)})). $$
If one applies the sifting procedure again, the answer comes out the same (each term in the sum is sifted and returns the same terms as the original, which add to give the same function:
$$ f(x) \mapsto \frac{1}{2}(f(x)-f(-x)) \mapsto \frac{1}{2}\left(\frac{1}{2}(f(x)-f(-x))\right)-\frac{1}{2}\left(\frac{1}{2}(f(-x)-f(-x))\right) = \frac{1}{2}(f(x)-f(-x)) $$
(this procedure produces a function that has a certain symmetry; applying the operation again does not produce more symmetry).
So the arctangent series is already missing terms: you actually want every fourth term of $x+0x^2-x^3/3+0x^4/4 + \dotsb$. Hence the correct sum is
$$ \frac{1}{4}(\arctan{x}+i\arctan{ix}-\arctan{(-x)})-i\arctan{(-ix)}) = \frac{1}{2}(\arctan{x}+\arg\tanh{x}) $$
by using some complex trigonometric identities.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \theta}{4}+\frac{ir \sin \theta}{4}+\frac{6 \cos \theta}{r}-\frac{6 i \sin \theta}{r} \right|$$
So
$$\left|\frac{z}{4}+\frac{6}{z}\right|=\sqrt{\frac{r^2}{16}+\frac{36}{r^2}+3 \cos (2\theta)}$$
any clue from here?
|
$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{z^2+24}{4z}\right|$
The function $f(z)=\frac{z^2+24}{4z}$ is analitic on $1\leq|z|\leq7$. Therefore, by the maximum modulus theorem its maximum absolute value is attained at the boundary. The boundary are the circles $|z|=1$ and $|z|=7$.
For $|z|=1$, observe that $z^2$ just travels the same circle. We have $|f(z)|=|z^2+24|/4$, which is maximum for $z=1$ or $z=-1$ (such that $z^2$ and $24$ point in the same direction).
For $|z|=7$, observe that $z^2$ travels the circle $|w|=49$. We have $|f(z)|=|z^2+24|/28$, which is maximum for $z=7$ or $z=-7$ (such that $z^2$ and 24 point in the same direction).
So, $f(7)=f(-7)$ seem to be the largest.
The minimum is zero at $f(\pm\sqrt{24}i)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\det(A^3B^2+A^2B^3)$ for $A,B$ s.t. $AB = BA$ given $\det A,\det B$. If two matrices $A$ and $B$ are such that $AB = BA$ with $\det A = 1$ and $\det B = 0$, then what is $\det(A^3B^2 + A^2B^3)$ ?
|
\begin{align*}
\text{det}(A^3B^2+A^2B^3) & = \text{det}A^2 \text{det} B^2\text{det}(A+B)\\
& = (\text{det}A)^2 (\text{det} B)^2\text{det}(A+B)\\
&=0.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understand what the author tries to say here. Can't this problem be done in another manner?
|
I was thinking of writing
\begin{align}
n^4 - n^2 + 64 &= (n^2 + A)^2 \\
n^4 - n^2 + 64 &= n^4 + 2An^2 + A^2 \\
2An^2 + n^2 + A^2 &= 64 \\
n^2(2A+1) &= 64 - A^2 \\
n^2 &= -\dfrac{A^2 - 64}{2A+1} \\
n^2 &= -\frac 12A + \frac 14 + \frac{255}{4(2A+1)} \\
4n^2 &= -2A + 1 + \frac{255}{2A+1}
& \text{$2A+1$ must be a positive divisor of $255$} \\
\hline
2A + 1 &\in \{ 1,3,5,15,17,51,85,255 \} \\
A &\in \{ 0,1,2,7,8,25,42,177 \} \\
4n^2 &\in \{256, 84, 48, 4, 0, -44, -80, -252 \} \\
n &\in \pm\{ 8, 1,0 \}
\end{align}
ALSO
\begin{align}
n^4 - n^2 + 64 &= A^2 \\
(n^2-1)^2 + (n^2-1) - (a^2-64) &= 0\\
n^2-1 &= \frac{-1 \pm \sqrt{4A^2-255}}{2}\\
\hline
4A^2 - 255 &= B^2 \\
(2A-B)(2A+B) &= 255 \\
(2A-B, 2A+B) &\in \{(1,255),(3,85),(5,51),(15,17)\} \\
(A,B) &\in \{(64,127),(22,41),(14,23),(8,1)\} \\
A &\in \{64,22,14,8\} \\
n^2-1 &\in \{63, 20, 11, 0, -1\} \\
n &\in \pm\{8,1,0\}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.