Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding the rank of a particular number in a sequence of the sum of numbers and their highest prime factor This question comes from a maths contest (infer no calculators or other electronic calculating aids) for 14-16 year olds (infer no use of complicated theorems, but those accessible to high-school students).
Consider the sequence $\;a_1,a_2,a_3,a_4,... \; $ such that $\;a_1=2 \;$ and for every positive integer $n$,
$a_{n+1}=a_n+p_n \;,\;$ where $\;p_n\;$ is the largest prime factor of $\;a_n\;$
The first few terms of the sequence are $\;2,4,6,9,12,15,20.\;$ What is the largest value of $\;n\;$ such that $\;a_n\;$ is a four-digit number ?
The road so far:
Continuing the sequence:
$\underbrace{2,4,\color{red}{6}}_{\text{add}\; 2},\underbrace{9,12,\color{red}{15}}_{\text{add}\; 3},\underbrace{20,25,30,\color{red}{35}}_{\text{add}\; 5},\underbrace{42,49,56,63,70,\color{red}{77}}_{\text{add}\; 7},\underbrace{88,99,110,121,132,\color{red}{143}}_{\text{add}\; 11},\underbrace{156,\ldots}_{\text{add}\; 13} $
Inferring from the above:
*
*The numbers to be added to get the next number in the sequence are the prime numbers: $\qquad \qquad \qquad \qquad \qquad2,3,5,7,11,13,17,\ldots$
*The numbers in red are the product of two successive prime numbers:$$\begin{align}
2 &\times 3=6\\
3 &\times 5=15\\
5 &\times 7=35\\
&\vdots\\
\text{thence}&\\
&\vdots\\
91 &\times 97=8827\\
97 &\times 101=\color{blue}{9797} \longleftarrow \\
101 &\times 103=10403\\
\end{align}$$ and $9797$ is the largest $4$-digit number in the sequence.
*$$
\begin{array}{c|ccccc}
n & 3 & 6 & 10 &16& 22 \\
\hline \hline
\text{current}\; n-\text{previous}\; n & - &\color{DarkOrange}{ 3 }& \color{DarkOrange}{4} &\color{DarkOrange}{6}& \color{DarkOrange}{6}\\
\hline
a_n & 6&15&35&77&143 \\
\hline
\text{highest prime factor of}\; a_n & 3&5&7&11&13\\
\text{lowest prime factor of}\; a_{n-1} & -&2&3&5&7&\\
\text{difference} & -&\color{DarkOrange}{ 3 }& \color{DarkOrange}{4} &\color{DarkOrange}{6}& \color{DarkOrange}{6}\\
\end{array}$$
My problem is how do I now correlate the numbers in $\;\color{DarkOrange}{\text{orange}}\;$ to find the $\;n\;$ which results in $\;a_n=9797\;$.
| Just observe that to get a product of the form $p_np_{n+1}$ from $p_np_{n-1}$ you have to wait for $p_{n+1}-p_{n-1}$ terms. So the required $n$ is, to get to a prime product $p_kp_{k+1}$ is $\sum_{j=1}^k(p_{j+1}-p_{j-1})+1=p_{k+1}+p_k-2$ where I have defined $p_0=1$. So for your case the answer is $103+101-2=202$.
| {
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Other Diophantine problems that use a Pell equation What Diophantine equations employ Pell equations in their solutions? A well-known example is the case of Pythagorean triples where the legs differ by 1, like,
$$20^2+21^2 = 29^2$$
These are completely parameterized as,
$$\big(\tfrac{x-1}{2}\big)^2+\big(\tfrac{x+1}{2}\big)^2=y^2$$
where $x^2-2y^2 =-1$. These $x,y$ also imply,
$$x^4+(y^2-1)^2 = y^4$$
$$x^4+(2y)^2 = (2y^2)^2+1$$
$$1^3+2^3+3^3+\dots+(x^2)^3 = (xy)^4$$
$$1^3+3^3+5^3+\dots+(2y-1)^3 = (xy)^2$$
Other equations that are to be solved in the integers and use Pell equations (at the least as a partial solution) are in the list below. (It is to be understood that $a,b,c,d$ are constants, while $x_i,\,y_i,\,z_i$ are unknowns.)
I. Simultaneous equations
*
*$x^2+y^2-1,\;x^2-y^2-1,\;\text{both squares}$
*$xy+1,\;xz+1,\;yz+1,\;\text{all squares}$
*$xy-1,\;xz-1,\;yz-1,\;\text{all squares}$
II. Quadratics
*
*$ax^2+bx+c = dy^2$
*$ax^2+bxy+cy^2 = d$
*$ax^2+bxy+cy^2 = dz^2+e$
*$x^2-x^2y^2+y^2=1$
*$x_1^2+x_2^2\pm1 = y^2 $
*$x_1^2+x_2^2+\dots+x_m^2\pm k = y_1^2+y_2^2+\dots+y_n^2$
III. Higher Powers
*
*$x^2+y^3 = z^4$
*$x^4+y^3 = z^2$
*$x^3+y^3+z^3+x+y+z = 0$
*$x_1^3+x_2^3+x_3^3 = 1$
*$x_1^5+x_2^5+\dots+x_7^5 = 1$
*$x_1^7+x_2^7+\dots+x_9^7 = 1$
*$x_1^4+x_2^4 = y^2+1$
*$x_1^4+x_2^4+x_3^4 = y_1^4+y_2^4+1$
*$x_1^6+x_2^6+x_3^6+x_4^6 = y_1^6+y_2^6+y_3^6+1$
Q: What other examples are not covered in the list above?
| What two unequal numbers $m$ and $n$, letting $m<n$ are such that the sum from $m$ to $n$ is equal the product of $m$ and $n$?
$$m +(m+1) + (m+2) + ... + (n-2) + (n-1) +n = mn$$
This can actually be manipulated into
$$2(2m-1)^2-(2n-2m+1)^2=1$$
Which of course is a pell equation :
$$2\alpha^2 - \beta^2=1$$
We see that
$$\begin{cases} \alpha=2m-1 \\ \beta=2n-(2m-1) \end{cases}$$
implying our solution set:
$$\begin{cases} m=\frac{\alpha+1}{2} \\ n=\frac{\alpha+\beta}{2} \end{cases}$$
In terms of $\alpha$ and $\beta$ as solutions of $2\alpha^2 - \beta^2=1$
$\underline{(\alpha,\beta)\to(m,n)}$
$(1,1)\to(1,1)$
$(5,7)\to(3,6)$
$(29,41)\to(15,35)$
$(169,239)\to(85,204)$
$\vdots$
and so on...
as a check verify that
$$\begin{align}
3+4+5+6 &= 3\cdot6 \\
15+16+\cdots+34+35&=15\cdot35 \\
85+86+\cdots+203+204&=85\cdot204 \\
&\vdots
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $3*5*17*257-3^{16}$ using factorization formulas? look at this:
$2*4*10*82*6562-3^{16}$
It's easy to calculate it with elementary arithmetic.
but how to calculate it using factorization formulas?
| $$2\times 4\times 10\times 82\times 6562-3^{16} $$ $$=(3^1-1)(3^1+1)(3^2+1)(3^4+1)(3^8+1)-3^{16}$$ $$=(3^2-1)(3^2+1)(3^4+1)(3^8+1)-3^{16}$$ $$=(3^4-1)(3^4+1)(3^8+1)-3^{16}$$ $$=(3^8-1)(3^8+1)-3^{16}$$ $$=(3^{16}-1)-3^{16}$$ $$=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Memorizing the identities $\cos {\pi \over 3}=\sin {\pi \over 6} = {1 \over 2}$ I memorized $\sin {\pi \over 4} = \cos {\pi \over 4}= {1\over \sqrt{2}}$ easily by using the diagonal inside the unit square.
I am having great trouble memorizing the identities $\cos {\pi \over 3}=\sin {\pi \over 6} = {1 \over 2}$ because I keep confusing whether it is $\cos {\pi \over 3}$ or $\cos {\pi \over 6}$ that equals ${1\over 2}$.
Is there a picture similar to the unit square picture or something
like it to memorize this identity?
| There is an old trick, just memorize that
$$
\sin 0 =\frac{\sqrt{0}}{2}\qquad
\sin \frac{\pi}{6}=\frac{\sqrt{1}}{2} \qquad
\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2} \qquad
\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \qquad
\sin \frac{\pi}{2}=\frac{\sqrt{4}}{2} \qquad
$$
and $\cos x$ goes the other way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$ I cannot seem to figure out how to solve the following problem:
$$\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$$
I have tried by using integration by parts.
I set $u = \sin(t)$, $v = \tan^{-1}(x)$ and $dv = \dfrac{1}{1+t^2}dt$. Thus,
$$\int_0^x \dfrac{\sin(t)}{1+t^2}dt = \left[\sin(t)\cdot \tan^{-1}(t)\right]_0^x - \int_0^x \cos(t)\cdot \tan^{-1}(t)dt$$
Is this correct or not? I would like to get some hints, I am currently stuck.
Best regards
| i will use series expansion and find a recurrence relation. first define
$$ J_k = \int_0^x t^{2k}\sin t \,dt, k = 0, 1, \dots $$
we have $J_0 = \int_0^x \sin \, dt = 1 - \cos x.$ now,
$\begin{align}
J_k &= \int_0^x t^{2k} \sin t \, dt = -t^{2k}\cos t|_0^x + \int_0^x 2k t^{2k-1} \cos t\, dt\\
&= 2kt^{2k-1}\sin t-t^{2k}\cos t|_0^x - 2k(2k-1)\int_0^x t^{2k-2} \sin t\, dt\\
&= 2kx^{2k-1}\sin x - x^{2k}\cos x -2k(2k-1)J_{k-1} = x^{2k-1}(2k\sin x - x\cos x) -2k(2k-1)J_{k-1}
\end{align}$
the recurrence relation is $$J_k = x^{2k-1}(2k\sin x - x\cos x) -2k(2k-1)J_{k-1}\, , \, J_0 = 1 - \cos x\tag 1 $$
we can compute $$J_1 = x(2\sin x - x\cos x) -2(1-\cos x),\\ J_2 = x^3(4\sin x - x \cos x )-12x(2\sin x - x \cos x)+24(1-\cos x), \dots$$
$$\int_0^x \frac{\sin t}{1 + t^2}\, dt = \int_0^x \sin t\left(1 -t^2 + t^4 + \dots\right) = J_0- J_1 + J_2 + \dots, \text{ for } -1 < x < 1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Give a Sequence, Find Its Generating Function The Full Question
Find the generating function(closed form) of the following sequence:
$\binom{8}{1},2\binom{8}{2},3\binom{8}{3},\dots , 8\binom{8}{8}$
My Work
The open form of this generating function is:
$1\binom{8}{1}x^0+2\binom{8}{2}x^2 +3\binom{8}{3}x^3 + \cdots +8\binom{8}{8}x^8$
This can be re-written as:
$0\binom{8}{0}x^0+1\binom{8}{1}x^0+2\binom{8}{2}x^2 +3\binom{8}{3}x^3 + \cdots +8\binom{8}{8}x^8$
Which in Sigma Notation is:
$\sum_{k=0}^{8}k\binom{8}{k}x^k$
Part of this series is obviously binomial, which has a closed form of $(1+x)^8$ I don't know how to deal with the $k$ that is always being multiplied in. I thought I could maybe use the multiplication rule for sums, but it doesn't seem to apply here because we have $a_1b_1$ instead of $a_1b_n$
My Problem
How do we deal with the k that is being multiplied in there constantly? How can I get a nice closed form generating function with that $k$ getting in my way?
| Hint: $i { 8 \choose i } = 8 { 7 \choose i-1 } $
You should be able to quickly conclude that the generating function is
$ 8 ( 1 + x) ^ 7 x $
| {
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If $n$ is an odd integer, does $x+y$ divide $x^n + y^n$? I believe that the answer is yes.
Here's my thinking:
*
*$x^n + y^n -(x+y)(x^{n-1} + y^{n-1}) = -x^{n-1}y -xy^{n-1}$
*$-x^{n-1}y -xy^{n-1} - (x+y)(-x^{n-2}y -xy^{n-2}) = x^{n-2}y^2 + x^2y^{n-2}$
*So, at each step, we get $(-1)^{c}(x^{n-c}y^{c} + x^{c}y^{n-c})$ with $c$ increasing by $1$ at each step and
*Since $n$ is odd, eventually, we get: $(-1)^c(x^{c}y^{c+1} + x^{c}y^{c+1}) = (x+y)(-1)^c(x^{c}y^{c})$
Am I correct? Am I making any mistakes in my thinking?
If this is correct, is the standard formula:
$$x^n + y^n = (x+y)(x^{n-1} -x^{n-2}y -xy^{n-2} + \dots + (-1)^{\frac{n-1}{2}}x^{\frac{n-1}{2}}y^{\frac{n-1}{2}} + y^{n-1})$$
Thanks,
-Larry
Edit: Fixed the equation based on comment.
| $\text{ We have } P(x) = x^n + y^n \to P(-y) = (-y)^n + y^n = -y^n +y^n = 0 \to x+y \text{ is a factor of P(x) }$.
| {
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How does $\sqrt{n+1}-\sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1}+\sqrt{n}}$? How does $\sqrt{n+1}-\sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1}+\sqrt{n}}$ ?
What are the exact steps to get to the right side from the left?
| It's the hint with almost solution and quite good technic for such problems.
$$ \bigg( \sqrt{n+1}-\sqrt{n} \bigg) \cdot 1 = \bigg( \sqrt{n+1}-\sqrt{n} \bigg) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$$
Useful formula for this task:
$$ \bigg( \sqrt{a} - \sqrt{b} \bigg) \cdot \bigg( \sqrt{a} + \sqrt{b} \bigg) = a - b $$
| {
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Sherman-Morrison Formula to determine the inverse of a $3 \times 3$ matrix Given $$A = \pmatrix{2 & 0 & -1 & \\ -1 & 1 & 1 \\ -1 & 0 & 1}$$ and $$A^{-1} = \pmatrix{1 & 0 & 1 & \\ 0 & 1 & -1 \\ 1 & 0 & 2}$$ I want to use the Sherman-Morrison Formula to determine $A^{-1}$ if we update $A$ so that the $a_{32}$ element is $2$. In other words, now $$A = \pmatrix{2 & 0 & -1 & \\ -1 & 1 & 1 \\ -1 & 2 & 1}$$
I've seen how to do this for $2\times2$ matrices, but I cannot seem to figure out how to do this for $3 \times 3$ matrices.
Note: I believe the answer is the following: $$\pmatrix{1 & 2 & -1 & \\ 0 & -1 & 1 \\ 1 & 4 & -2}$$
| you want to find the inverse of $A + 2e_3e_2^T?$ we will use the facts that
$$(I + ab^T)^{-1} = I - \frac{ab^T}{1 + a^Tb}, (A+ab^T)^{-1} = A^{-1} - \frac{A^{-1}ab^TA^{-1}}{1+b^TA^{-1}a.}$$
we will take $$a = 2e_3, b =e_2, A^{-1}a = 2(1,-1,2)^T, b^TA^{-1} = (0,1,-1), b^TA^{-1}a=-2$$ putting all these together gets you
$$(A + ab^T)^{-1} = A^{-1} +2 \pmatrix{1\\-1\\2}\pmatrix{0&1&-1}
=\pmatrix{1&0&1\\0&1&-1\\1&0&2} +2\pmatrix{0&1&-1\\0&-1&1\\0&2&-2} =
\pmatrix{1&2&-1\\0&-1&1\\1&4&-2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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conjugate complex numbers (ahlfors) I am solving the ahlfors for my course. I have some questions:
1.Verify the calculation that the values of $\dfrac{z}{z^{2}+1}$ for $z=x+iy$ and $z=x-iy$ are conjugate.
The first step I calculated what the value is $\dfrac{2}{z^{2}+1}$ for $z=x+iy$
Then I did the same thing for $z=x+iy$
But the solutions are different. Might I do something wrong?
*The question is: "what is the solution of $z^{2}+\left( \alpha +i{\beta }\right) z+\gamma+i\delta =0$"
First I tried to solve this equation for putting $z=x+iy$. But in so many numbers I couldn't get what to find. I think I need to find x and y but I got lost.
| For $z$, we have that $z^2 = x^2 - y^2 + 2xyi$.
\begin{align*}
\frac{z}{z^2 + 1}
& = \frac{x + iy}{x^2 - y^2 + 1 + 2xyi}\\
& = \frac{x + iy}{x^2 - y^2 + 1 + 2xyi}
\frac{x^2 - y^2 + 1 - 2xyi}{x^2 - y^2 + 1 - 2xyi}\\
& = \frac{x(x^2 - y^2 + 1) + 2xy^2 + iy(x^2 - y^2 + 1 - 2x^2)}
{(x^2 - y^2 + 1)^2 + 4x^2y^2}\tag{1}
\end{align*}
For $\bar{z}$, we have that $\bar{z}^2 = x^2 - y^2 - 2xyi$.
\begin{align*}
\frac{\bar{z}}{\bar{z}^2 + 1}
& = \frac{x - iy}{x^2 - y^2 + 1 - 2xyi}\\
& = \frac{x - iy}{x^2 - y^2 + 1 - 2xyi}
\frac{x^2 - y^2 + 1 + 2xyi}{x^2 - y^2 + 1 + 2xyi}\\
& = \frac{x(x^2 - y^2 + 1) + 2xy^2 - iy(x^2 - y^2 + 1 - 2x^2)}
{(x^2 - y^2 + 1)^2 + 4x^2y^2}\tag{2}
\end{align*}
Therefore, we have that equations $(1)$ and $(2)$ are conjugates.
I asked this problem here on Dec 18, 2014 wondering if it could be simplified more but I was told it couldn't be reduced in further.
The quadratic equation is $x = \frac{-b\pm\sqrt{b^2 - ac}}{2}$.
For the complex polynomial, we have
$$
z = \frac{-\alpha - \beta i\pm
\sqrt{\alpha^2 - \beta^2 - 4\gamma + i(2\alpha\beta - 4\delta)}}{2}
$$
Let
$a + bi = \sqrt{\alpha^2 - \beta^2 - 4\gamma + i(2\alpha\beta - 4\delta)}$.
Then
$$
z = \frac{-\alpha - \beta\pm (a + bi)}{2}
$$
| {
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Guessing a numerical middle bound If $x \geq 5$ and $x$ is a prime number, then what number lies between
$$\frac{x + 2}{x}$$
and
$$\frac{x + 3}{x}?$$
Here is my attempt:
Let $\theta \in \mathbb{R}$ such that
$$\frac{x + 2}{x} < \theta < \frac{x + 3}{x}.$$
Then,
$$1 < \theta < \frac{8}{5}.$$
Alas, this is where I get stuck. I was thinking of getting the average
$$\dfrac{\frac{x + 2}{x} + \frac{x + 3}{x}}{2} = \dfrac{2x + 5}{2x} \leq \frac{3}{2}$$
but how would I know whether this lies to the left or right of the quantity $\theta$?
Thanks!
| No fixed number independent of $x$ will be between $\dfrac{x + 2}{x}$ and $\dfrac{x+3}{x}$.
To see this, note that it's equivalent to finding a number that is between $\dfrac{2}{x}$ and $\dfrac{3}{x}$, and then adding $1$. Then for any number $\theta > 0$, choosing $x$ larger than $\dfrac{3}{\theta}$ will make $\dfrac{3}{x} < \theta$.
For instance, if $\theta = \frac{1}{2}$, we might choose $x > \dfrac{3}{\frac{1}{2}} = 6$. To choose a prime, we might choose $x = 7$. Then $1 + \frac{1}{2}$ is not between $\dfrac{7 + 2}{7}$ and $\dfrac{7 + 3}{7}$.
| {
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Solve $2^x+3^y=z^2$ in nonnegative integers. So, we are trying to find all the solutions to $2^x + 3^y = z^2$ in nonnegative integers. Here are my insights:
First of all, $z^2$ can be either $0$ or $1$ modulo $3$. If $z^2 = 3k$, then
LHS cannot be divisible by $3$ unless $y = 0$ and $x=2k+1$, and so we have $2^x+1 = z^2$, so one of the solutions is $(3,0,3)$ (but we still need to show that this is the only solution in this case).
Now let $y > 0$, $x=2k$. In this case $2^x \equiv 1\ (\mod\ 3)$ and $z^2 \equiv 2^x (\mod 3)$. We have
$$
3^y = (z-2^k)(z+2^k),
$$
so $(z-2^k)=3^a$ and $(z+2^k)=3^b$, so $3^b-3^a = 2^{k+1}$. If $a > 0$, then
$2^{k+1}$ must be divisible by $3$, but it is not, so suppose $a = 0$.
Now we have
$$
2^{k+1} = 3^b - 1,
$$
Аnd this is where I stuck. It is easy to find two more solutions just iterating over $k$, but I do not know how to prove that there are no others. As far as I understand, we need to find some kind of an upper bound for $k$. Note: I am trying to avoid the use of Catalan's conjecture here.
Can you please help me to show that there are no other solutions to this equation and that $2^x+1=z^2$ has a unique solution?
| The second question answer:
$2^x = z^2 - 1 \Rightarrow 2^x = (z+1)(z-1)$. If $x$ and $z \in \mathbb N$, then this can be true only for $z=3$ as difference between $z+1$ and $z-1$ equals $2$ and we have no other degrees of $2$ which are different by $2$.
| {
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Limit of the form $0/0$ How do I find $$\lim_{x \to 0}\frac{\cos(\sin x)-\cos x}{x^4}$$ I tried rewriting $\cos$ in the form of $\sin$ and then applying $\sin x/x$ form, but it doesn't work out. Does it?
| First use Taylor series $$\sin x=x-\frac{x^3}{6}+o(x^3)$$ to get $$\lim_{x\to 0}\frac{x-\sin x}{x^3}=\frac{1}{6}$$ Next we can proceed as follows
\begin{align}
L&=\lim_{x\to 0}\frac{\cos \sin x - \cos x}{x^4}\notag\\
&=\lim_{x\to 0}\dfrac{2\sin\left(\dfrac{x+\sin x}{2}\right)\sin\left(\dfrac{x-\sin x}{2}\right)}{x^4}\notag\\
&=\lim_{x\to 0}2\cdot\dfrac{\sin\left(\dfrac{x+\sin x}{2}\right)}{\dfrac{x+\sin x}{2}}\cdot\dfrac{x+\sin x}{2x}\notag\\
&\,\,\,\,\dfrac{\sin\left(\dfrac{x-\sin x}{2}\right)}{\dfrac{x-\sin x}{2}}\cdot\dfrac{x-\sin x}{2x^3}\notag\\
&=2\cdot 1\left(\frac{1}{2}+\frac{1}{2}\right)1\cdot\frac{1}{12}=\frac{1}{6}\notag
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1164544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving equation with complex roots. I have the following question. My problem lies in (c).
Question
a) Find the three roots of the equation $(w+5)(w+8)(w+9)=360$.
b) Let $z_0=\sqrt{-2+6i}$, where $i^2={-1}$. Show that the solutions to the equation $z^2+6z=-11+6i$ can be expressed in the form $z=-3\pm z_0$.
c) Hence, solve the equation $(z+1)(z+2)(z+3)^2(z+4)(z+5)=360$, by reducing it to the equation in (a). Give your answer in terms of $z_0$.
Solution
a) Expanding $(w+5)(w+8)(w+9)=360$ gives, $w^3+22w^2+157w+360=360$.
Cancelling the zeros and taking out the common leaves, $w(w^2+22w+157)=0$ and so, $w=0$ and $w^2+2w+157=0$. The other two roots are
\begin{align*}
w=\dfrac{-22\pm\sqrt{22^2-4(1)(157)}}{2}=-11\pm6i.
\end{align*}
b) Rearranging the equation gives the quadratic $z^2+6z+(11-6i)=0$. Hence,
\begin{align*}
z=\dfrac{-6\pm\sqrt{4(1)(11-6i)}}{2}=-3\pm\sqrt{-2+6i}=-3\pm z_0.
\end{align*}
c)For what it's worth it is clear that $z=0$ is a solution. I tried substituting $-3+z_0$ into the given equation. This produces
\begin{align*}
(z_0^2-4)(z_0^2)(z_0^2-1)=360.
\end{align*}
I do not know what else I can try.
| Note that:
$$(z+1)(z+5)=z^2+6z+5$$
$$(z+2)(z+4)=z^2+6z+8$$
$$(z+3)^2=z^2+6z+9$$
So if you put $w=z^2+6z$ you get $(w+5)(w+8)(w+9)=360$ from (a).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a first order non-linear homogenous ODE Original problem is:
$$
\frac{dy}{dx} = \frac{3y + x}{3x + y}
$$
After substituting multiplying the top and bottom by $\frac{1}{x}$ and substituting $v = \frac{y}{x}$ I arrived at
$$
\frac{1}{x}dx = \frac{3+v}{1-v^2}dv
$$
After integrating initially using partial fractions and getting:
$$
\ln |x| = \frac{3}{2}\ln(1+v) - \frac{3}{2}\ln(1-u) - \frac{1}{2}\ln(1-v^2)
$$
I went to wolfram alpha and found the simplified version of the integral was
$$
\ln |x| = \ln(1+v) - 2\ln(1-v)
$$
Which made my final answer without solving for y
$$
\ln|x| = \ln\frac{1+\frac{y}{x}}{\left(1-\frac{y}{x}\right)^2}
$$
The answer in the textbook, however, was:
$$
\ln|x| = \ln \left(\left(1+\frac{y}{x}\right) \left(1-\frac{y}{x}\right)^2\right)
$$
Where did I go wrong?
| As:
$$\frac{dy}{dx} = \frac{3y + x}{3x + y}\stackrel{v=y/x}\implies x\frac{dv}{dx}+v=\frac{3v+1}{3+v}\implies x\frac{dv}{dx}=\frac{1-v^2}{3+v}\implies \frac{dx}{x}=\frac{v+3}{1-v^2}dv$$
Then partial fractions:
$$\frac{dx}{x}=\left(\frac{2}{1-v}+\frac{1}{1+v}\right )dv$$
So:
$$\ln|x|=-2\ln|1-v|+\ln|1+v|+c$$
So:
$$x=k\frac{1+v}{(1-v)^2}=k\frac{1+y/x}{(1-y/x)^2}$$
So your answer is correct, as a proof I can present a maxima output:
Converting last output:
$$k'x=\frac{1+y/x}{(1-y/x)^2}$$
| {
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Show that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$ Problem
Assume that $a,b\in\mathbb{R}-\{0\}$ and that $a+b\not=0$. Prove that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.
My Proof
Let's assume that $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$, then it follows that
$$
\begin{equation}
(a+b)^2-ab=0
\end{equation}
$$
Let $x=a+b$ and $y=ab$. Now $b=x-a$ and so $y=a(x-a)=ax-a^2$. The previous equation can be written as
$$
x^2-y=0
$$
Substituting $y=ax-a^2$ in this equation gives
$$
x^2-ax+a^2=0
$$
The discriminant of this quadratic equation (in $x$) is $-3a^2<0$ and therefore $x=a+b$ has no real solution. This means $a+b\in\mathbb{C}$ and therefore either of $a$ or $b$ or both are not real but this contradicts our assumption that $a,b$ are real numbers. Therefore by contradiction $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.
My Question
Is my proof correct? are there any alternative proofs?
| Here is another approach:
You have $(a+b)^2 = ab$, so $ab>0$. Expanding the equation gives $a^2+b^2+ab = 0$. Hence we must have $ab<0$, a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Circles question on proof It is given that a, b, and c are the sides of a triangle and c is the hypotenuse. There is an incircle inside the triangle with radius = r.
We need to prove that $r=\dfrac{a+b-c}{2}$
Image:
My attempt
Area of circle = $\dfrac{ab}{2} = \dfrac{ar+br+cr}{2}$ (area of individual triangles)
$r=\dfrac{ba}{a+b+c}$
but we have to prove that $r=\dfrac{a+b-c}{2}$ . How can that be shown? Thank you very much
| Continuing from where you had left, $$r=\frac{ab}{a+b+c}=\frac{ab(a+b-c)}{(a+b)^2-c^2}$$
$$\rightarrow r=\frac{ab(a+b-c)}{(a^2+b^2-c^2)+2ab}=\frac{ab(a+b-c)}{2ab}=\frac{a+b-c}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integral solutions of $y^2=x^3+7$ Find all integral solutions of $y^2 = x^3+7$.
I tried to use many different moduli, but it never works. With modulo $9$, you can get $x$ is divisible by $3$.
| Lemma. If $p=4n+3$ is a prime then $x^2+y^2\equiv 0\pmod p$ implies $x,y\equiv 0\pmod p$.
Proof. Suppose that $x\not\equiv \pm y\pmod p$. (Because if this is the case, then $(x,y)\equiv (\ell,\pm \ell)\pmod p$, and then $x^2+y^2\equiv 0\pmod p$ would imply $2\ell^2\equiv 0\pmod p$, i.e. $\ell\equiv 0\pmod p$, so the lemma holds in this case.)
Multiply $x^2+y^2\equiv 0\pmod p$ by $x^2-y^2$ (which is not $\equiv 0$, by what we said above) to get $x^4\equiv y^4\pmod p$. Raise this to the $n$th power to get $x^{4n}\equiv y^{4n}$. Multiply both sides by $x^2y^2$ to get $x^{4n+2} y^2\equiv y^{4n+2} x^2\pmod p$. Now $p=4n+3$ implies $p-1=4n+2$, hence the previous equation re-writes as $x^{p-1} y^2\equiv y^{p-1} x^2\pmod p$, or by Fermat's Little Theorem, $x^2\equiv y^2\pmod p$. But we also have $x^2+y^2\equiv 0\pmod p$. Thus subtracting yields $2y^2\equiv 0\pmod p$ and so $y\equiv 0\pmod p$; and this gives $x\equiv 0\pmod p$. Q.E.D.
Solution. Obviously $x$ cannot be even, or else $y^2=x^3+7\equiv 7\pmod 4$, i.e. $y^2\equiv 3\pmod 4$, which is absurd (quadratic residues). So $x$ must be odd.
Now then, add $1$ to both sides and factor as $$y^2+1=(x+2)((x-1)^2+3).$$ Since $x$ is odd, $(x-1)^2+3\equiv 3\pmod 4$. Thus there exists a prime $p$ of the form $4n+3$ dividing $(x-1)^2+3$. Then $p$ divides the $\text{LHS}$ $y^2+1$, and so by the lemma it divides $y$ and $1$. But a prime cannot divide $1$. We conclude that our supposition cannot be true. Therefore, there exist no integral solutions to the equation. Q.E.D.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the next term in the sequence. $\frac{7}{3},\frac{35}{6},\frac{121}{12},\frac{335}{36},\ldots $ $\dfrac{7}{3},\dfrac{35}{6},\dfrac{121}{12},\dfrac{335}{36},\ldots $
$\bf\text{Answer}$ given is $\dfrac{865}{48}$
I found that $4^{th}$ differencess of the numbers $7,35,121,335\cdots$ are not constant .
and the second differences of the denominator drastically changes,
$3\quad 6\quad 12\quad 36\quad 48\\~\\
\quad 3\quad 6\quad 24\quad \color{red}{12}$
decimal value is also not showing any pattern.
$\frac{7}{3},\ \frac{35}{6},\ \frac{121}{12},\ \frac{335}{36},\ldots $
$2.33,\ 5.83,\ 10.08,\ 9.33,\ldots $
| If your fourth term is $\dfrac{335}{\color{red}{24}}$ instead of $~\dfrac{335}{\color{red}{36}}~,~$ then the pattern you're looking for is
a recurrence relation of the form $~a_{n+1}~=~6n+1-\dfrac{a_n}2~,$ with $~a_1=\dfrac73~.$ This idea
came to me while trying to approximate each term with its nearest integer; in particular,
by noticing that $~6^2=36\simeq35$, and $(12-1)^2=121$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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About the Polya-Knopp-like inequality $\sum_{k=1}^{n}\frac{k^2}{a^2_{1}+\cdots+a^2_{k}}\le\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)^2$ I was inspired by other question I came out with the inequality:
let $a_{i}>0,i=1,2,\cdots,n$
Prove that
$$\sum_{k=1}^{n}\dfrac{k^2}{a^2_{1}+\cdots+a^2_{k}}\le\left(\dfrac{1}{a_{1}}+\cdots+\dfrac{1}{a_{n}}\right)^2\tag{1}$$
and I believe that (1) follows, by some way, from Carleman's inequality and Hardy's inequality but I did not manage to prove it.
case $n=2$,$$\Longleftrightarrow \dfrac{1}{x^2}+\dfrac{4}{x^2+y^2}\le\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\Longleftrightarrow \dfrac{4}{x^2+y^2}\le\dfrac{1}{y^2}+\dfrac{2}{xy}\Longleftrightarrow (x+2y)(x^2+y^2)\ge 4xy^2$$ This is clear hold,because $x^2+y^2\ge 2xy,x+2y\ge 2y$
| Following Omran Kouba's approach in a linked question, from the Holder inequality we have:
$$\sum_{j=1}^{k} j^2 \leq \sqrt[3]{\sum_{j=1}^{j}j^3 a_j}\cdot \sqrt[3]{\sum_{j=1}^{j}j^3 a_j}\cdot\sqrt[3]{\sum_{j=1}^{k}\frac{1}{a_j^2}}\tag{0}$$
from which it follows that:
$$\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq \frac{27}{k(k+1)^3(k+1/2)^3}\left(\sum_{j=1}^{k}j^3 a_j\right)^2\leq\frac{9}{2}\left(\frac{1}{k^6}-\frac{1}{(k+1)^6}\right)\left(\sum_{j=1}^{k}j^3 a_j\right)^2 $$
and by Cauchy-Schwarz inequality:
$$\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq\frac{9}{2}\left(\frac{1}{k^6}-\frac{1}{(k+1)^6}\right)\left(\sum_{j=1}^{k}a_j\right)\left(\sum_{j=1}^{k}j^6 a_j\right).\tag{1}$$
If we set $S_k\triangleq \frac{1}{k^6}\sum_{j=1}^{k-1}j^6 a_j$ and $S_1=0$ we have:
$$\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq \frac{9}{2}(S_k-S_{k+1}+a_k)\sum_{j=1}^{k}a_j\leq \frac{9}{2} a_k \sum_{j=1}^{k}a_j.\tag{2}$$
Now we set $A_k=\sum_{j=1}^{k}a_j.$ From the previous line:
$$\sum_{k=1}^{n}\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq\frac{9}{2}\sum_{k=1}^{n} a_k A_k \tag{3}$$
and from summation by parts:
$$ \sum_{k=1}^{n}a_k A_k = A_n^2 - \sum_{k=2}^{n}A_{k-1} a_{k}=A_n^2-\sum_{k=1}^{n}A_k a_k+\sum_{k=1}^{n}a_k^2 $$
hence:
$$ 2\sum_{k=1}^{n}a_k A_k = \left(\sum_{j=1}^{n}a_j\right)^2+\sum_{j=1}^{n}a_j^2 $$
proves your inequality up to a multiplicative factor $\color{red}{\frac{9}{2}}$.
Edit: If in line $(0)$ we replace the LHS with $\sum_{j=1}^{n}\sqrt{j}$ and follow the same approach, we end with a multiplicative factor equal to $\color{red}{\frac{9}{4}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$ Proving $$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$$
| $$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}\\=
\sum_{n=1}^{\infty }0.5^n\left(\frac{1/6}n+\frac{-1/2}{n+1}+\frac{1/2}{n+2}+\frac{-1/6}{n+3}\right)\\
=\frac16\sum_{n=1}^{\infty }\frac{0.5^n}n-\frac1{2\cdot0.5}\sum_{n=2}^{\infty }\frac{0.5^n}n+\frac1{2\cdot0.5^2}\sum_{n=3}^{\infty }\frac{0.5^n}n-\frac1{6\cdot0.5^3}\sum_{n=4}^{\infty }\frac{0.5^n}n\\
=\frac16(\ln2)-(\ln2-0.5)+2(\ln2-0.5-0.5^2/2)-\frac43(\ln2-0.5-0.5^2/2-0.5^3/3)\\
=\frac5{36}-\frac16\ln2$$
| {
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Some help with basic probability 1) A four-digit number is formed by
randomly selecting four digits, without replacement,
from the set D = { 1, 2, 3, 4, 5, 6, 7 }.
What is the probability that the resulting number is
greater than 4200 ?
2) A fair six-sided die is tossed four times,
each time noting which numbered face lands up.
What is the probability that at least two different numbered
faces land up during the four tosses?
I'm not really sure how to approach the first one. And in regards to the second question, I know that I must subtract the complement from 1. Any hints are appreciated.
| For the first question:
Let's think about just selecting the first digit. There's a probability of $\frac 1 7$ of selecting $7$, $\frac 1 7$ of selecting $6$, and $\frac 1 7$ of selecting $5$. In any of these cases, the resulting number is greater than $4200$. Add up these probabilities: $\frac 1 7 + \frac 1 7 + \frac 1 7 = \frac 3 7$.
Then there's also a $\frac 1 7$ probability of selecting a $4$. Then we need to pick the second digit. If we pick a $2$, then we note that our number will still be greater than $4200$, because picking zeroes for the last two digits is not an option. So the probability of picking one of $\{2,3,5,6,7\}$ is $\frac 5 6$, and any of these will yield a number greater than $4200$. Note that we can't pick $4$ as the second digit because we're selecting from $D$ without replacement.
What's the probability of picking a $4$ and then picking one of $\{2,3,5,6,7\}$? They are independent, so by the multiplication rule we can just multiply them: $\frac 1 7 \cdot \frac 5 6 = \frac 5 {42}$.
So the total probability is $\frac 3 7 + \frac 5 {42} = \frac {18} {42} + \frac {5}{42} = \frac {23}{42}$.
For the second question:
Probability of seeing at least two differently numbered faces $= 1-$
Probability of all four tosses showing the same face.
How do we calculate the probability of all four tosses showing the same face?
Well, suppose we said that all four tosses had to show $6$. The probability of that for a single toss is $\frac 1 6$, and by the rule of multiplication for probabilities of independent events, the chance of tossing four $6$s is $\left(\frac 1 6\right)^4$.
And that's the exact same for all other faces $1,2,3,4,5$. There are six such faces in total, so the probability of all four tosses showing the same face is $6 \cdot \left(\frac 1 6 \right)^4 = 6 \cdot \frac 1 {6^4} = \frac 6 {6^4} = \frac 1 {6^3}$.
(If you want more intuition on this -- draw out the table of probabilities for two rolls of the dice, and note that the diagonal is exactly six events.)
So the probability of seeing at least two numbered faces is $1 - \frac 1 {6^3} = 1 - \frac 1 {216} = \frac {215}{216}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the conditions on a, b, and c so that the system has a solution? (Matrices) Determine the values of a, b, and c so that the system has a solution:
x + 2y + 4z = a
2x + 3y - z = b
3x + y + 2z = c
Now, all I did was reduce the system to row-echelon form using gauss elimination as such:
$$ \left[
\begin{array}{ccc|c}
1 & 2 & 4 & (a) \\
2 & 3 & -1 & (b) \\
3 & 1 & 2 & (c) \\
\end{array}
\right] $$
~
$$ \left[
\begin{array}{ccc|c}
1 & 2 & 4 & (a) \\
0 & -1 & -9 & (b - 2a) \\
0 & -5 & -10 & (c - 3a) \\
\end{array}
\right] $$
~
$$ \left[
\begin{array}{ccc|c}
1 & 2 & 4 & (a) \\
0 & -1 & -9 & (b - 2a) \\
0 & 0 & 35 & (c +7a -5b) \\
\end{array}
\right] $$
so the last equation of the system becomes 35 = c + 7a - 5b but from here I'm not sure how to choose the values of a, b, and c for the system to have a solution.
| You can further row reduce and end up with:
$$ \left[
\begin{array}{ccc|c}
1 & 0 & 0 & \dfrac{1}{5} (2 c-a) \\
0 & 1 & 0 & \dfrac{1}{35} (7 a+10 b-9 c) \\
0 & 0 & 1 & \dfrac{1}{35} (7 a-5 b+c) \\
\end{array}
\right] $$
Now, notice that you have three free variables and they can be chosen any way you like for an infinite number of unique solutions.
| {
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A cubic factor of a nine degree polynomial The polynomial $x^3−3x^2+4x−1$ is a factor of $x^9+px^6+qx^3+r$. Find $p+q+r$. How do I start with the problem?
| Let $f(x) = x^3-3x^2+4x-1 $ Because the $ 9^\text{th}\ $ degree polynomial only have terms of multiples of $ 3\ $. I consider that it is the root to $ f \left ( \sqrt[3]{x} \right ) = 0 $
We have
$ \begin{eqnarray}
f \left ( \sqrt[3]{x} \right ) & = & 0 \\
x - 3x^{2/3} + 4x^{1/3} - 1 & = & 0 \\
(x-1) & = & x^{1/3} \cdot (3x^{1/3} - 4) \\
(x-1)^3 & = & x( 27x - 64 - 36x^{1/3} (3x^{1/3} - 4)) \\
& = & x(27x- 64 - 36(x-1)) \\
x^3 + 6x^2 + 31x - 1 & = & 0 \\
\end{eqnarray} $
Replace $ x $ with $ x^3 $ and compare, we have
$ p = 6, q = 31, r = -1 $
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometry problem on product of trig functions. Questions:
*
*$\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=?$
*$\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}=?$
*$\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}=?$
Approach:
If $x=\dfrac{\pi}{7}\implies 7x=\pi\implies 4x=\pi-3x\implies \sin 4x = \sin 3x$.Upon expansion, $$8cos^3x-4cos²x-4cosx+1=0$$
"$\cos \dfrac{\pi}{7}, \cos \dfrac{3\pi}{7}, \cos \dfrac{5\pi}{7}$will satisfy this cubic."
Note: I'm not able to understand the above line. Why do these values satisfy this cubic and how did we get these values?
These are therefore the roots of the cubic, and their product is:
$$\cos\dfrac{\pi}{7}\cos\dfrac{3\pi}{7}\cos\dfrac{5\pi}{7}=\dfrac{-1}{8}$$
Finally,
$\cos\dfrac{5\pi}{7} = \cos\left(\pi-\dfrac{2\pi}{7}\right)=-\cos \dfrac{2\pi}{7}$. Replacing this in (1) gives $$\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac{1}{8}$$
How can we apply this concept to find the values of
*$\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}?$
*$\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}?$
I'm not able to form the equations for these two problems. Please Help. .
| Let $a=\pi/7$.
We know $x=\cos a$ is a root of $8x^3-4x^2-4x+1=0$. This is because $$3a+4a=\pi\Rightarrow \sin(3a)=\sin(4a)$$ as you wrote.
Now, note that $3\cdot 3a+4\cdot 3a=3\pi\Rightarrow \sin(3\cdot 3a)=\sin(4\cdot 3a)$. This can be done for $5a$ as well. This is why $\cos (3a),\cos (5a)$ satisfy the cubic.
For 3, since $3a+4a=\pi\Rightarrow \tan(3a)+\tan(4a)=0$, we have
$$\frac{\tan a+\tan(2a)}{1-\tan a\tan(2a)}+\frac{2\tan(2a)}{1-\tan^2(2a)}=0,$$
i.e.
$$\tan a+3\tan(2a)-3\tan a\tan^2(2a)-\tan^3(2a)=0.$$
Setting $x=\tan a$ gives us
$$x+\frac{6x}{1-x^2}-\frac{12x^3}{(1-x^2)^2}-\frac{8x^3}{(1-x^2)^3}=0,$$
i.e.
$$x^6-21x^4+35x^2-7=0.$$
Here, note that $x=\tan a,\tan(2a),\tan(3a)$ are roots of this equation by the similar argument above. Hence, $\tan^2(ka)\ (k=1,2,3)$ are roots of
$$x^3-21x^2+35x-7=0.$$
So, we have
$$\tan^2a\tan^2(2a)\tan^2(3a)=7\Rightarrow \tan a\tan(2a)\tan(3a)=\sqrt 7.$$
Finally, we have
$$\sin a\sin(2a)\sin(3a)=\cos a\cos(2a)\cos (3a)\times \tan a\tan(2a)\tan(3a)=\frac{\sqrt 7}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Show $\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$ for $a,b,c>0$ If $a$, $b$, and $c$ are positive real numbers, show that
$$
\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \geq \sqrt{b^2 + bc + c^2}.
$$
What values of $a$, $b$, and $c$ ( or relation among them) make both sides equal ( makes it an equality)?
| We note that
$$
\text{LHS}=\sqrt{(-(b/2)+a)^2+(\sqrt{3}b/2)^2} + \sqrt{((c/2)-a)^2+(\sqrt{3}c/2)^2}\tag{$*$}
$$
so the Minkowski inequality applies to give
$$
\text{LHS}\geq\sqrt{\frac{1}{4}(c-b)^2+\frac{3}{4}(b+c)^2}=\text{RHS}.
$$
Equality is if and only if there is some $\lambda\geq 0$ such that
$$
(a-b/2,\sqrt{3}b/2)=\lambda(-a+c/2,\sqrt{3}c/2).
$$
p.s.: The version of the Minkowski inequality that we have used is
$$
(|x_1|^p+|x_2|^p)^{1/p}+(|y_1|^p+|y_2|^p)^{1/p}\geq(|x_1+y_1|^p+|x_2+y_2|^p)^{1/p}
$$
with $p=2$ and $x_1,x_2,y_1$ and $y_2$ are as in ($*$) (in that order). (e.g. $x_1=a-b/2$, etc.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Relation between units in $\mathbb{Q}(\sqrt{13})$ and integral solutions to $x^2 - 13y^2 = \pm 1$ I have shown that in the number ring of $\mathbb{Q}(\sqrt{13})$, the units are precisely $\pm \left(\frac{3+\sqrt{13}}{2} \right)^n$.
How can one deduce the integral solutions to the related Pell's equation in this manner?
| The connection is the norm function: $$N\left(\frac{a}{2} + \frac{b \sqrt{13}}{2}\right) = \left(\frac{a}{2} - \frac{b \sqrt{13}}{2}\right)\left(\frac{a}{2} + \frac{b \sqrt{13}}{2}\right) = \frac{a^2}{4} - \frac{13b^2}{4}.$$ It's important that $a$ and $b$ be of the same parity. But if $a$ and $b$ are both even, you can just halve them both without regard for the new parity and do $$N(a + b \sqrt{13}) = a^2 - 13b^2.$$ This should remind you of the $x^2 - 13y^2 = \pm 1$ equation. So what you need here are those powers of $\frac{3}{2} + \frac{\sqrt{13}}{2}$ that are of the form $a + b \sqrt{13}$ with $a, b \in \mathbb{Z}$. This turns out to be $\left(\frac{3}{2} + \frac{\sqrt{13}}{2}\right)^{3n}$. Indeed:
*
*$N(18 + 5 \sqrt{13}) = -1$ and $18^2 - 13 \times 5^2 = -1$.
*$N(649 + 180 \sqrt{13}) = 1$ and $649^2 - 13 \times 180^2 = 1$.
*$N(23382 + 6485 \sqrt{13}) = -1$ and $23382^2 - 13 \times 6485^2 = -1$.
*etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Differentiate. $y = (1 − \sec x) / \tan x$ I have no idea what I have done wrong. Please criticise.
Differentiate $y = \frac{1 - \sec x}{\tan x}$.
\begin{align*}
\frac{d}{dx} \left[ \frac{1 - \sec x}{\tan x} \right]
&= \frac{\tan x \frac{d}{dx} [ 1 - \sec x ] - (1 - \sec x) \frac{d}{dx} \tan x }{(\tan x)^2} \\
&= \frac{\tan x (- \sec x \tan x) - [(1 - \sec x) (-\csc^2 x)]}{(\tan x)^2} \\
&= \frac{-\tan^2 x \sec x + \csc^2 x - \sec x \csc^2 x}{(\tan x)^2}
\end{align*}
| This is equivalent to
$1-\csc x$
$y'=\cot x \csc x$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Find the point $T(a, b)$ on the curve $y = x^2 + 1$ whose tangent passes through the point $P(1, 0)$ from the left. Find the point $T(a, b)$ on the curve $y = x^2 + 1$ whose tangent passes through the point $P(1, 0)$ from the left.
$$
\\ \begin{align}
\\
\\ f(x) &= x^2 + 1
\\ f(a) &= a^2 + 1 = b
\\
\\ T(a, f(a))
\\
\\ f'(x) &= 2x
\\ f'(a) &= 2a = m_{tan}
\\
\\ x_1 &= 1
\\ y_1 &= 0
\\
\\ P(x_1, y_1)
\\
\\ y_1 - b &= m_{tan}(x_1 - a)
\\ 0 - a^2 - 1 &= 2a(1 - a)
\\ -a^2 - 1 &= 2a - 2a^2
\\ a^2 - 2a - 1 &= 0
\\ (a - 1)(a - 1) &= 0
\\ a &= 1
\\ b &= 1^2 + 1 = 2
\\
\\ T(1, 2)
\\ P(1, 0)
\\ \end{align}
$$
At this point, I see that the line that goes through the points $T$ and $P$ would have to have a vertical slope.
I've made an error somewhere along the line, but I don't quite see it. Perhaps I'm going about this all wrong?
|
Find the point $T(a,b)$ on the curve $y=x^2+1$ whose tangent passes through the point $P(1,0)$ from the left.
Parametrize $(t,t^2+1)$, $y'=2x=2t$, Eqn. of tangent, $y-(t^2+1)=2t(x-t)\implies y=1+2tx-t^2$, at $y=0$, $x=\frac{t^2-1}{2t}<1\implies t\in(-\infty,1-\sqrt2)\cup(0,1+\sqrt2)$
So, $$x\in(\infty,1-\sqrt2)\cup(0,1+\sqrt2)$$
More shorter method:
Write Chord of Contact (while keeping the graph of $x^2+1$ in mind) and find intersection with curve:
$$\frac12(y+0)=x.1+1\implies y=2x+2=x^2+1\implies x=1-\sqrt2,1+\sqrt2$$
It's fairly easy to proceed now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How do I calculate the number of permutations of the list $(6, 6 ,5, 4)$? I have the list $l = (6, 6, 5, 4)$ and want to how to calculate the possible number of permutations.
By using brute force I know that there are 12 possible permutations:
$$\{(6, 5, 6, 4),
(6, 6, 5, 4),
(5, 6, 6, 4),
(6, 4, 5, 6),
(6, 5, 4, 6),
(4, 6, 6, 5),
(4, 5, 6, 6),
(4, 6, 5, 6),
(6, 4, 6, 5),
(6, 6, 4, 5),
(5, 4, 6, 6),
(5, 6, 4, 6)\}$$
But how would I calculate this?
I know we talk about permutation as order matters also I know that repetition is allowed (at least for $x = 6$).
| Here is another way of thinking about the problem: There are four positions to fill. We can fill two of them with a $6$ in $\binom{4}{2}$ ways. Once those two positions have been filled, we fill one of the remaining two positions with a $5$ in $\binom{2}{1}$ ways. Once those three positions have been filled, we can fill the remaining position with a $4$ in $\binom{1}{1}$ way. Thus, the number of ways we can fill four positions with two sixes, one five, and a four is
$$\binom{4}{2}\binom{2}{1}\binom{1}{1} = \frac{4!}{2!2!} \cdot \frac{2!}{1!1!} \cdot \frac{1!}{0!1!} = \frac{4!}{2!1!1!} = 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $f$ is differentiable at $4$ and $f^\prime(4) = 5$, evaluate the following limit: $\lim_{x \to 4} \frac{f(x)-f(4)}{\sqrt{x} - 2}$. If $f$ is differentiable at $4$ and $f^\prime(4) = 5$, evaluate the following limit: $\lim_{x \to 4} \frac{f(x)-f(4)}{\sqrt{x} - 2}$.
\begin{align*}
\\ h &= \sqrt{x} - 2
\\
\\ h \to 0 &\text{ as } x \to 4
\\
\\ \lim_{x \to 4} \frac{f(x)-f(4)}{\sqrt{x} - 2} &= \lim_{h \to 0} \frac{f(x)-f(4)}{h}
\\ \lim_{h \to 0} \frac{f(4-h)-f(x)}{-h} &\implies \lim_{h \to 0} \frac{f(4-h)-f(4)}{-h} = f^\prime(4) = 5
\end{align*}
| $$\lim_{x \to 4} \frac{f(x)-f(4)}{\sqrt{x} - 2}$$
$$=\lim_{x \to 4} \frac{f(x)-f(4)}{x-4}(\sqrt x+2)$$
$$=f'(4)\cdot\lim_{x \to 4}(\sqrt x+2)=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a general formula for $c_{n+1}=c_n+\frac1{c_n}$ (Recursive) Find a general formula for $c_{n+1}$
If $c_1 = 1$ and $$c_{n+1} = c_n + \frac{1}{c_n}$$
$$c_1 = 1$$
$$c_2 = 1 + 1 = \frac{2}{1}$$
$$c_3 = 2 + 1/2 = \frac{5}{2}$$
$$c_4 = \frac{5}{2} + \frac{2}{5} = \frac{29}{10}$$
$$c_5 = \frac{29}{10} + \frac{10}{29} = \frac{941}{290}$$
| Searching the Online Encyclopedia of Integer Sequences is always a good idea.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation over a product of binomial coefficients Question: I can't figure out why the following equality is true
$\sum_\limits{k=a-b-c}^{d} (-1)^k \binom{d}{k}\binom{k+b+c}{a} = (-1)^d \binom{b+c}{a-d} $
How can this be shown? (In the book it just says that the summation is accomplished by elementary means.)
Attempt:
I tried setting $d=a-b-c+n$ so there will be $n$ terms in the summation. Using the briefer notation $x=a-b-c$, I get the LHS
\begin{align}
\sum_\limits{k=x}^{x+n} (-1)^k \binom{d}{k}\binom{k+b+c}{a} = &
(-1)^{x}\binom{x+n}{x}\binom{a}{a} + (-1)^{x+1}\binom{x+n}{x+1}\binom{a+1}{a} +
\\ & \dots + (-1)^{x+i}\binom{x+n}{x+i}\binom{a+i}{a} + \dots + \\ & (-1)^{x+n-1}\binom{x+n-1}{x+n}\binom{a+n-1}{a} \\
= & \sum_\limits{i=0}^{n-1} (-1)^{x+i} \binom{x+n}{x+i}\binom{a+i}{a}
\end{align}
and then I get stuck. I can't recognize it as being of the form of some of the normal products or summations of binomial coefficents.
I can verify the equation if I put in numbers or just make n=1 or n=2 or similar. So I thought of using induction to prove it. However with the $n$ in $\binom{x+n}{x+i}$ part, I am not sure that if it is possible. I thought of maybe using the relation
$\binom{n}{k} k = n \binom{n-1}{k-1}$ but I can't see that making sense.
| Consider the series
\begin{align}
S = \sum_{k=a-b-c}^{d} (-1)^k \binom{d}{k}\binom{k+b+c}{a}
\end{align}
The following is the reduction of this series.
\begin{align}
S &= \sum_{k=0}^{d - a + b + c} (-1)^{k+a - b - c} \binom{d}{k+a-b-c} \binom{k+a}{a} \\
&= (-1)^{a - b -c} \, \frac{d! (a-b-c)!}{(d-a+b+c)!} \, \sum_{k=0} \frac{(-1)^{k} (a+1)_{k} }{ k! \, (a-b-c+1)_{k} (d-a+b+c+1)_{-k}} \\
&= (-1)^{a - b -c} \, \frac{d! (a-b-c)!}{(d-a+b+c)!} \, \sum_{k=0} \frac{(a+1)_{k} (a-d-b-c)_{k}}{k! \, (a-b-c+1)_{k} } \\
&= (-1)^{a - b -c} \, \frac{d! (a-b-c)!}{(d-a+b+c)!} \, {}_{2}F_{1}(a+1, a-b-c-d; a-b-c+1; 1) \\
&= (-1)^{a - b -c} \, \frac{d! (a-b-c)!}{(d-a+b+c)!} \, \frac{ (a-b-c)! (a-b-c)! (d-a-1)!}{d! \, (-b-c-1)! (d-a+b+c)!} \\
&= (-1)^{d} \binom{b+c}{a-d}.
\end{align}
The term $(a)_{k}$ is called the Pochhammer symbol and ${}_{2}F_{1}$ is the hypergeometric function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Bernoulli Number analog using Cosine I know that Bernoulli Numbers can be found with the generating function
$$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$$
I was wondering if any work has been done using a similar equation
$$\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_n}{n!}x^{2n}$$
I'm particularly interested in the $C_n$. Can anyone help me with a reference to work with this particular generating function or to the coefficients $C_n$?
| Here is a more natural derivation of the formula:
$$
\frac{z^2}{1 - \cos z} =
2\sum_{n = 0}^\infty (-1)^{n - 1}\frac{(2n - 1)B_{2n}}{(2n)!}z^{2n}
\ \ \ (\left\lvert{z}\right\rvert < 2\pi).
$$
We start with the defining result that
for all $z \in \mathbb{C}$ such that $\left\lvert{z}\right\rvert < 2\pi$,
$$
\frac{z}{e^z - 1} =
\sum_{m = 0}^\infty \frac{B_m}{m!}z^m.
$$
Then:
\begin{gather*}
\coth\frac{z}{2} =
\frac{\cosh\frac{z}{2}}{\sinh\frac{z}{2}} =
\frac{e^z + 1}{e^z - 1}
= \frac{2}{e^z - 1} + 1, \\
\therefore\
\frac{z}{2}\coth\frac{z}{2} =
\frac{z}{e^z - 1} + \frac{z}{2}, \\
\therefore\
\frac{z}{2}\cot\frac{z}{2} =
\frac{iz}{2}\coth\frac{iz}{2} =
\frac{iz}{e^{iz} - 1} + \frac{iz}{2} \\
= \sum_{n = 0}^\infty (-1)^n\frac{B_{2n}}{(2n)!}z^{2n} +
iz\sum_{n = 0}^\infty (-1)^n\frac{B_{2n + 1}}{(2n + 1)!}z^{2n} +
\frac{iz}{2}
\end{gather*}
But this is a real-valued function of $z$ when $z$ is real,
therefore:
\begin{gather*}
B_1 = -\frac{1}{2}; \ \ \
B_{2n + 1} = 0 \ \ \ (n \geqslant 1); \\
\frac{z}{2}\cot\frac{z}{2} =
\sum_{n = 0}^\infty (-1)^n\frac{B_{2n}}{(2n)!}z^{2n}
\ \ \ (\left\lvert{z}\right\rvert < 2\pi).
\end{gather*}
(We could also have argued, earlier, that
$\frac{z}{2}\coth\frac{z}{2}$ is an even function of $z$.)
Differentiating the series term-by-term, and multiplying by $z$:
\begin{gather*}
\sum_{n = 0}^\infty (-1)^n\frac{2nB_{2n}}{(2n)!}z^{2n} =
z\frac{d}{dz}\left(\frac{z}{2}\cot\frac{z}{2}\right) \\
= \frac{z}{2}\cot\frac{z}{2} - \frac{z^2}{4}\mathrm{cosec}^2\frac{z}{2}
= \frac{z}{2}\cot\frac{z}{2} - \frac{1}{2}\cdot\frac{z^2}{1 - \cos z},
\end{gather*}
which gives the required result.
| {
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"url": "https://math.stackexchange.com/questions/1195528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
reduction formulae for $\int_0^1 (x^{\frac{n+2}2}\sqrt{1-x})dx\,\!$ how do I prove that
$\int_0^1 (x^{\frac{n+2}2}\sqrt{1-x})dx\,\!={\frac{n+2}{n+5}}\int_0^1 (x^{\frac{n}2}\sqrt{1-x})dx\,\!$
I couldn't find which functions u and v' to use for the intgeration by parts.
Help please.
| Let $$I_n = \int_0^1 x^{\frac{n+2}2}\sqrt{1-x}dx\,$$
Noticing $\frac{n+2}{2} = \frac{n}{2}+1$, you want to set
$$u = x^{\frac{n+2}{2}} \text{ and } dv = \sqrt{1-x}dx$$
That gives $$du = \frac{n+2}{2}x^{\frac{n}{2}} dx\text{ and } v = -\frac{2}{3}(1-x)\sqrt{1-x}$$
Integrating by parts yields
$$\begin{align}I_n &= -x^{\frac{n+2}{2}} \frac{2}{3}(1-x)^{3/2}\Bigg|_0^1 + \int_0^1 \frac{n+2}{2}x^{\frac{n}{2}}\frac{2}{3}(1-x)\sqrt{1-x} dx\\~\\&= 0 + \frac{n+2}{3}\int_0^1 x^{\frac{n}{2}}(1-x)\sqrt{1-x}dx\\~\\&=\frac{n+2}{3}\left(\int_0^1x^{\frac{n}{2}} \sqrt{1-x}dx -\int_0^1x^{\frac{n}{2}+1} \sqrt{1-x}dx\right)\\~\\&=\frac{n+2}{3}\left(I_{n-1} - I_n\right) \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A second-order non-constant coefficient differential equation I encountered the following second-order differential equation in my research:
$$x^2(1-x)^2y''+Ax(1-x)y'+By=0$$
where $y$ is a function of $x$ we are looking for, and $A$ and $B$ are constants. It looks kind of regular, but many ways failed at some subtle points. Can someone give me any suggestion on solving this equation? Thanks a lot!
| For the differential equation
\begin{align}
x^2(1-x)^2y''+Ax(1-x)y'+By=0
\end{align}
make the substitution $y(x) = x^{\mu} (1-x)^{\nu} g(x)$ to obtain the equation
\begin{align}
x(1-x) g'' + (A + 2 \mu - (2 \mu + 2 \nu) x ) g' + \left[ \frac{\nu (\nu - A - 1) + B}{1-x} + \frac{\mu (\mu + A - 1) + B}{x} - (\mu + \nu)(\mu + \nu -1) \right] g = 0.
\end{align}
Let $\nu^{2} - (A+1) \nu + B=0$ and $\mu^{2} - (1-A)\mu + B =0$ to obtain
\begin{align}
\nu &= \frac{1}{2} \left( 1 + A + \sqrt{(1 + A)^{2} - 4 B} \right) \\
\mu &= \frac{1}{2} \left( 1 - A + \sqrt{(1-A)^{2} - 4 B} \right)
\end{align}
and
\begin{align}
x(1-x) g'' + (A + 2\mu - (2 \nu + 2 \mu)x)g' - (\mu + \nu)(\mu + \nu -1) g=0
\end{align}
which is the hypergeometric differential equation and has the first solution
\begin{align}
g(x) = {}_{2}F_{1}(\mu + \nu, \mu + \nu -1; A+2\mu ; x).
\end{align}
With this it is seen that
\begin{align}
y(x) = C_{0} \, x^{\mu} (1-x)^{\nu} \, {}_{2}F_{1}(\mu + \nu, \mu + \nu -1; A+2\mu ; x).
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that if $p'=2p+1$ and $p\equiv 3\pmod{4}$, then $2^p\equiv 1\pmod{p'}$ Unsure if this is relevant: In the previous part of the question I deduced that if a number of the form $2^n-1$ is prime (a Mersenne prime), $n$ must also be prime.
Question: For primes $p$ and $p'$, where $p'=2p+1$, does $p\equiv 3\pmod{4}$ imply that $2^p\equiv 1\pmod{p'}$?
Attempt: By Euler's theorem, since $(2,p)=(2,p')=1$ we have $2^{\phi(p')}\equiv 2^{p'-1}\equiv 1\pmod{p'}$. Substituting $p'=2p+1$, we obtain $2^{p'-1}\equiv2^{2p}\equiv 1\pmod{p'}$, and thus $2^p\equiv\pm 1\pmod{p'}$.
Letting $p\equiv 4p_0+3$, then $p'\equiv 2(4p_0+3)+1 \equiv 8p_0+7$. So $p'\equiv 7\pmod{8}$.
I am stuck as to how to link this last bit to show that $2^p\equiv 1\pmod{p'}$, not $2^p\equiv -1\pmod{p'}$.
| As you pointed out, $p'$ is of the form $8k+7$. You have probably already seen the proof of the fact that $2$ is a quadratic residue of the odd prime $q$ if and only if $q\equiv \pm 1\pmod{8}$.
So $2$ is a quadratic residue of $p'$, say $2\equiv x^2\pmod{p}$. But $x^{p'-1}=x^{2p}\equiv 1\pmod{p'}$ and therefore $2^{p}\equiv 1\pmod{p'}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the exponential equation Find the number of roots to $e^x = ax^2$ for all values of $a$. (x is real and so is $a$). I have tried some things but I am stuck.
| For $a>0$ the roots can be expressed in terms of Lambert $W$ function:
\begin{align}
a x^2&=\exp(x) \\
x^2\exp(-x)&=\frac{1}{a} \\
x\exp(-\frac{x}{2})&=\pm\frac{1}{\sqrt{a}} \\
-\frac{x}{2}\exp(-\frac{x}{2})&=\mp\frac{1}{2\sqrt{a}} \\
-\frac{x}{2}&=\mathrm{W}\left( \mp\frac{1}{2\sqrt{a}} \right)\\
x&=-2\mathrm{W}\left( \mp\frac{1}{2\sqrt{a}} \right)
\end{align}
So, $\forall a>0$ there is always one negative real root
$x_1=-2\mathrm{W_0}\left(\frac{1}{2\sqrt{a}} \right)$.
And since the Lambert $W(u)$ function has two real branches
for $-\exp(-1)<u<0$, it would be two more (positive) real roots (three total):
\begin{align}
x_2&=-2\mathrm{W_0}\left(-\frac{1}{2\sqrt{a}} \right), \\
x_3&=-2\mathrm{W_{-1}}\left(-\frac{1}{2\sqrt{a}} \right)
\end{align}
for $a>\exp(2)/4$.
Also, when $a=\exp(2)/4$ then $x_2=x_3$ and there are two real roots total.
Summarizing, the number $n$ of real roots to $\exp(x) = a x^2$:
\begin{align}
n&=
\begin{cases}
0,\quad a\le0 \\
1,\quad 0<a<\exp(2)/4 \\
2,\quad a=\exp(2)/4 \\
3,\quad a>\exp(2)/4 \\
\end{cases}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Verification of a certain identity in wavelet basis lemma. This is from Lemma 7.1 in Mallat's Wavelet Tour 2nd edition.
I am trying to show that
$$
b(2x)h(x) + c(2x)g(x) = a(x)
$$
when
\begin{align*}
b(2x) &= \frac{1}{2}\left[ a(x)h(x)^* + a(x+\pi)h(x+\pi)^* \right] \\
c(2x) &= \frac{1}{2}\left[ a(x)g(x)^* + a(x+\pi)g(x+\pi)^* \right]
\end{align*}
where $z^*$ denotes complex conjugation.
Furthermore, in addition to the fact that these are all $2\pi$ periodic functions, we have the following identities to make use of:
\begin{align}
|g(x)|^2 + |g(x+\pi)|^2 &= 2 \\
|h(x)|^2 + |h(x+\pi)|^2 &= 2 \\
g(x)h(x)^* + g(x+\pi)h(x+\pi)^* &= 0
\end{align}
I thought this was supposed to be a straightforward calculation, but apparently I'm missing something.
My attempt:
In the interest of things not spilling over margins, let me denote $g(x) = g$ and $g(x+\pi) = g_{\pi}$ for all functions. Now, I just plug away...
\begin{align*}
& b(2x)h(x) + c(2x)g(x) \\
&= \frac{1}{2}a\left[|g|^2 + |h|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a\left[4-(|g_{\pi}|^2 + |h_{\pi}|^2)\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= 2a-\frac{1}{2}a\left[|g_{\pi}|^2 + |h_{\pi}|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= a + \left\{a - \frac{1}{2}a\left[|g_{\pi}|^2 + |h_{\pi}|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right]\right\}
\end{align*}
I've used the first and second identities in the second line. The last line implies necessarily that
\begin{align*}
0 &= a - \frac{1}{2}a\left[|g_{\pi}|^2 + |h_{\pi}|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a \left(2 - \left[|g_{\pi}|^2 + |h_{\pi}|^2\right] \right) + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a \left(|h|^2 - |g_{\pi}|^2 \right) + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a \left(|g|^2 - |h_{\pi}|^2 \right) + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
\end{align*}
I'm not really sure what to do with the $a_{\pi}$ term, or what do at all from here except run in circles. I appreciate any help.
| I figured it out. I'll provide the verification (for future googlers) in a way that elucidates more clearly where the definition of $b(2x)$ and $c(2x)$ arise from.
We have the equation:
$$
a(x) = b(2x)h(x) + c(2x)g(x)
$$
Since everyone's $2\pi$ periodic, we also have the equation:
$$
a(x+\pi) = b(2x)h(x+\pi) + c(2x)g(x+\pi)
$$
Thus, using the notation introduced in the question, we have a 2D linear system:
$$
\begin{bmatrix}
a \\
a_{\pi}
\end{bmatrix}
=
\begin{bmatrix}
h & g \\
h_{\pi} & g_{\pi}
\end{bmatrix}
\begin{bmatrix}
b \\
c
\end{bmatrix}
$$
So that:
$$
\begin{bmatrix}
b \\
c
\end{bmatrix}
=
\frac{1}{hg_{\pi} - a_{\pi}g}
\begin{bmatrix}
g_{\pi} & -g \\
-h_{\pi} & h
\end{bmatrix}
\begin{bmatrix}
a \\
a_{\pi}
\end{bmatrix}
$$
Focusing on $b$ (c follows similarly):
$$
b = \frac{ag_{\pi} - a_{\pi}g}{hg_{\pi} - h_{\pi}g} = \frac{ag_{\pi}}{hg_{\pi} - h_{\pi}g} - \frac{a_{\pi}g}{hg_{\pi} - h_{\pi}g}
$$
where
\begin{align*}
\frac{g_{\pi}}{hg_{\pi} - h_{\pi}g} &= \frac{h^*g_{\pi}}{|h|^2g_{\pi} - h_{\pi}gh^*} \\
&= \frac{h^*g_{\pi}}{|h|^2g_{\pi} + h_{\pi}g_{\pi}h_{\pi}^*} \tag{1} \\
&= \frac{h^*g_{\pi}}{g_{\pi}\left(|h|^2 + |h_{\pi}|^2\right)} \\
&= \frac{h^*}{2} \tag{2} \\
\end{align*}
with $(1)$ using $gh^* + g_{\pi}h_{\pi}^* = 0$ and $(2)$ using $|h|^2 + |h_{\pi}|^2 = 2$.
Similarly,
\begin{align*}
\frac{-g}{hg_{\pi} - h_{\pi}g} &= \frac{-h_{\pi}^*g}{hg_{\pi} h_{\pi}^* - |h_{\pi}|^2g} \\
&= \frac{-h_{\pi}^*g}{-hg h^* - |h_{\pi}|^2g} \tag{3} \\
&= \frac{-h_{\pi}^*g}{-g \left( |h|^2 + |h_{\pi}|^2 \right)} \\
&= \frac{h_{\pi}^*}{2} \tag{4}
\end{align*}
where $(3), (4)$ use identities as with $(1),(2)$.
SO FINALLY...
$$
b(2x) = \frac{1}{2} \left( a(x)h(x)^* + a(x+\pi)h(x+\pi)^* \right)
$$
as defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Geometry: Find angle x in triangle I have not been able to find a euclidean geometry solution to this, but any other solutions are also appreciated.
Let ABC be a triangle with AB=CD and angles as marked in the diagram. Find the measure in degrees of angle $x$.
Answer: $15^o$.
| let $AB = DC = 1, AD = a.$ the by the rule sine applied to the triangle $ABD,$ we have $$\frac a{\sin x} = \frac 1{\sin 3x}\to a = \frac{\sin x}{\sin 3x} = \frac{\sin x}{\sin x \cos 2x + \sin 2x \cos x} = \frac{1}{\cos 2x + 2\cos^2x} $$
apply the rule of sine to the triangle $ABC,$ then $$\frac{1+a}{\sin 7x} =\frac1{\sin 9x}$$ so we need to solve $$\sin 9x =(\sin 7x - \sin 9x)(\cos 2x + 2\cos^2 x),\, 0 < x < 20^\circ.$$
this has one positive solution in the $x = 15^\circ$ in the interval $0 < x < 20^\circ.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The limit of sums of the form $ \frac{1}{\sqrt{2n}}- \frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+2}}-\dotsb-\frac{1}{\sqrt{4n}}$ I need to calculate limit:
$$
\lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{2n}}- \frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+2}}-\dotsb+\frac{1}{\sqrt{4n}}\right )
$$
Any hints how to do that would be appreciated.
| The quantity inside the limit is between $\frac{1}{\sqrt{2n}}$ and $\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}$, hence the limit is zero by squeezing. To notice it, it is sufficient to couple consecutive terms: let $A_n=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$.
Then $A_n>0$ and:
$$ \frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\ldots+\frac{1}{\sqrt{4n}}=\frac{1}{\sqrt{2n}}-\sum_{k=n}^{2n-1}A_{2k+1}<\frac{1}{\sqrt{2n}} $$
as well as:
$$ \frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\ldots+\frac{1}{\sqrt{4n}}=A_{2n}+\sum_{k=n+1}^{2n-1}A_{2k}+\frac{1}{\sqrt{4n}}>\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find Matrix $A^{50}$? Find the matrix $A^{50}$ given
$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$
I was practicing some questions for my exam and I found questions of this form in a previous year's paper.
I don't know how to do such questions.
Please assist over this question.
Thank You
| This is for the updated version of question where $A = \begin{bmatrix}2 & -1\\0 & 1\end{bmatrix}$. For the original version of $A$, the derivation is similar.
The characteristic polynomial for matrix $A$ is
$$\chi_A(\lambda) = \det\begin{bmatrix}\lambda-2 & 1\\0 & \lambda - 1\end{bmatrix}
= (\lambda - 2)(\lambda - 1) = \lambda^2 - 3\lambda + 2$$
By Cayley Hamilton theorem, we have
$$A^2 - 3A + 2I_2 = 0$$
If we divide the polynomial $x^{50}$ by $x^2 - 3x + 2$ with long division,
we know there are polynomial $p(x)$ and coefficients $\alpha, \beta$ such that
$$x^{50} = p(x)(x^2 - 3x + 2) + \alpha x + \beta$$
To determine $\alpha, \beta$, substitute $x$ by $1$ and $2$ in above expression. We get
$$\begin{cases}
1 &= \alpha + \beta\\
2^{50} &= 2\alpha + \beta
\end{cases}
\quad\implies\quad
\begin{cases}
\alpha &= 2^{50} - 1\\
\beta &= 2 - 2^{50}
\end{cases}$$
Form this, we get
$$\require{cancel}
\begin{align}
A^{50} &= p(A)\color{red}{\cancelto{0}{\color{gray}{(A^2 - 3A + 2I_2)}}} + \alpha A + \beta = \alpha A + \beta\\
&= (2^{50}-1)\begin{bmatrix}2 & -1\\0 & 1\end{bmatrix}
+(2 - 2^{50})\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}
= \begin{bmatrix}2^{50} & 1 - 2^{50}\\0 & 1\end{bmatrix}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:
Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.
I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.
I also figured maybe I should play with it this way:
$2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$
But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.
So what am I missing (other than sleep and food)? What route should I take to
prove this?
Thanks in advance!
| Note
$$1 + \sqrt[3]{2} + \sqrt[3]{4} = \frac{(\sqrt[3]{2})^3 - 1}{\sqrt[3]{2} - 1} = \frac{1}{\sqrt[3]{2} - 1}.$$
So if $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, then $1/(\sqrt[3]{2} - 1)$ is rational, which implies $\sqrt[3]{2} - 1$ is rational. Then $\sqrt[3]{2}$ is rational, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 6,
"answer_id": 0
} |
How is $x^\frac{9}{4} \cdot x^\frac{1}{2} = x^\frac{11}{4}$ How is $x^\frac{9}{4} \cdot x^\frac{1}{2} = x^\frac{11}{4}$?
I don't really understand it.
| Use the exponent rule $x^a\cdot x^b = x^{a + b}$ with $a = 9/4$ and $b = 1/2$. Since $b = 2/4$, $a + b = 9/4 + 2/4 = 11/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find $\log (2-x)$ in powers of x Find $\log (2-x)$ in powers of $x$.
I know this is how I've gotten so fare,
$$\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n$$
given that $\left| x\right| <1$.
Now
$$\begin{align}\frac{1}{2-x}&=\frac{1}{2} \sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n \end{align}$$
I therefore try integrating
$$\begin{align} \int \frac{1}{2-x} \, dx&=-\log(2 - x)\\
\int \frac{1}{2} \left(\frac{x}{2}\right)^n dx&=
\frac{2^{-n-1} x^{n+1}}{n+1}\end{align}$$
Now multiplying both sides with $-1$ i still not get the answer which is
$$\begin{align}\ln(2-x)&=\ln(2)-\sum_{n=1}^{+\infty}\frac{x^n}{2^nn} & -2\leq x<2\end{align}$$
Could someone please show me the steps?
| Consider the series
\begin{align}
\frac{1}{2-x} = \sum_{n=0}^{\infty} \frac{x^{n}}{2^{n}} = 1 + \sum_{n=1}^{\infty} \frac{x^{n}}{2^{n}}
\end{align}
Now this series can be integrated from 0 to $x$ to obtain
\begin{align}
\int_{0}^{x} \frac{dt}{2-t} &= \int_{0}^{x} \left( 1 + \sum_{n=1}^{\infty} \frac{ t^{n}}{2^{n}} \right) \, dt \\
\left[ - \ln(2-t) \right]_{0}^{x} &= \left[ t + \sum_{n=1}^{\infty} \frac{t^{n+1}}{2^{n} (n+1)} \right]_{0}^{x} \\
- \ln(2-x) + \ln(2) &= x + \sum_{n=1}^{\infty} \frac{x^{n+1}}{2^{n}(n+1)}
\end{align}
The series can now be changed slightly and is seen as:
\begin{align}
\ln(2-x) &= \ln(2) - x - \sum_{n=2}^{\infty} \frac{x^{n}}{2^{n-1} \, n} \\
&= \ln(2) - \sum_{n=1}^{\infty} \frac{x^{n}}{2^{n} \, n}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to find periodic continued fraction expansion of $\frac{\sqrt{7}}3$
How to find periodic continued fraction expansion of $\frac{\sqrt{7}}3$
Using this formula here (it begins in the middle of the page),
I obtained $\frac{\sqrt{7}}3=[0;1,\overline{7,2}]$
but how can I justify rigorously that the period starts at $a_2$, is it enough to say that $\bar{\zeta_2}=\frac{7-\sqrt{63}}{2}=-0.133\dots$, so it is between $0$ and $-1$ and this is not true for $\bar{\zeta_1}$
| *
*We have that $\frac{\sqrt{7}}{3}$ is positive but less than one, hence its continued fraction starts with a $\color{red}{0}$;
*$\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7}=1+\frac{3\sqrt{7}-7}{7}$, hence the next element of the continued fraction is $\color{red}{1}$;
*$\frac{7}{3\sqrt{7}-7}=\frac{3\sqrt{7}+7}{2}=7+\frac{3\sqrt{7}-7}{2}$, hence the next element of the continued fraction is $\color{red}{7}$;
*$\frac{2}{3\sqrt{7}-7}=\frac{3\sqrt{7}+7}{7}=2+\frac{3\sqrt{7}-7}{7}$, hence the next element of the continued fraction is $\color{red}{2}$ and we are back to step 3.
This gives:
$$ \alpha = \frac{\sqrt{7}}{3} = [0;1,\overline{7,2}].$$
Now a double-check. If $\beta = [\overline{7,2}]$, then $\beta=[7;2,\beta]$ hence $\beta$ is a root of:
$$ \beta=7+\frac{1}{2+\frac{1}{\beta}}=\frac{15\beta+7}{2\beta+1}$$
hence:
$$ 2\beta^2+\beta = 15\beta + 7 $$
or $2\beta^2-14\beta-7=0$, from which:
$$ \beta = \frac{7+3\sqrt{7}}{2} $$
follows, and since $[0,1,\beta]=\frac{1}{1+\frac{1}{\beta}}=\frac{\beta}{\beta+1}$,
$$ [0,1,\overline{7,2}] = [0,1,\beta] = \frac{\sqrt{7}}{3}, $$
ok.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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divisible large degree polynomial Let $n$ be an even positive integer and $a$, $b$ real numbers such that $b^n=3a+1$. Prove that if $(X^2+X+1)^n-X^n-a$ is divisible by $X^3+X^2+X+b$, then $a=0$ and $b=1$.
I am thinking of using the fact that $X^2+X+1=-b/X$, but I don't see it really helping
| First
If $b = 1$, then $q(x) = x^3 + x^2 + x + 1 = (x + 1)(x + i)(x − i)$. Also, $a = 0$, so $p(x) = (x^2 +x+1)^n−x^n$, and it is easy to verify that $−1$ and $\mp i$ are indeed roots of $p(x)$. Thus $q(x)|p(x)$.
Second
The other direction, assume that $q(x) | p(x)$. Then $q(x)$ divides $$x^np(x) = (x^3+x^2+x)^n−x^{2n}−ax^n\tag1$$ But also $q(x)$ divides
$$(x^3+x^2+x)^n−(−b)^n = (x^3+x^2+x+b)\sum_{i=0}^{n-1}(x^3 +x^2 +x)^i(−b)^{n−1−i}\tag2$$
A subtraction operation between $(1)$ and $(2)$ noticing that $n$ is even, so that $(−b)^n = b^n$ gives us
$$ q(x)|(x^{2n} +ax^n −b^n)=f(x) \tag3 $$
We now consider the roots of $f(x)$ Substituting $y = x^n$, we have $f(x) = y^2 +ay−b^n=g(y)$
Because $n$ is even, $b^n$ is nonnegative, and it follows that the discriminant of $g(y)$ is nonnegative. Thus, $g(y)$ has two roots $u$ and $v$ in $\mathbb{R}$, with $$u+v =−a, uv =−b^n \tag{4}$$
Finally every root of $f(x)$ verify $$x^n=u \text{ or } x^n=v\tag{5}$$
Let $\alpha,\beta,\gamma$ be the roots of $p(x)$ then :
$$\alpha\beta\gamma=-b \tag{6}$$
and we know that $q(x)|f(x)$ then $\alpha,\beta,\gamma$ verifies $(5)$, now we have $4$ cases:
*
*Two roots verify $x^n=u$ and one root verify $x^n=v$, we can assume WLOG that:
$$\alpha^n=\beta^n=u \text{ and } \gamma^n=v \tag{7}$$
now using $(6)$ we have $|\alpha||\beta||\gamma|=|b|$ and using $(4)$ and $(7)$ we have $|\alpha||\gamma|=|\beta||\gamma|=|b|$ hence : $|\alpha|=|\beta|=1$ and $|\gamma|=b$, and because $u,v$ are reals then using $(7)$ we have $u=1$ and $v=-b^n$ or $u=-1$ or $v=b^n$ (see $(4)$), in this two cases we have $u+v=\mp(1-b^n)=\mp(1-(3a+1))=\mp 3a$ and returning to $4$ again we have $a=0$ and hence $b=\mp1$, if $b=-1$ then $q(x)=x^3+x^2+x-1$ and $f(x)=x^{2n}-1$ it's easy to verify that $f(x)$ does not divide $f(x)$ because $f$ has only the two real roots $\mp1$ and $q$ has another different real root. Here we conclude that $a=0$ and $b=1$
*Two roots verify $x^n=v$ and one root verify $x^n=u$ (the same approach as in the case $(1)$
*The three roots verify $x^n=u$ , then because of $(6)$ each root has absolute value $|b|^{1/3}$, On the other hand, at least one root is real because $q(x)$ has odd degree. Thus, either $c=b^{1/3}$ or $-c=−b^{1/3}$ must be a root of $q(x)$ and hence in the first case $q(c)=2c^3 +c^2 +c = 0$ hence $c=0$(impossible) or $0<2c^2 +c+1 = 0$ (impossible),so $q(-c)=0$ and $c^2-c=0$ and because $c\neq 0$ we have $c=1$ thus $b=1$.
*The three roots verify $x^n=v$ (the same approach as in $(3)$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive rational numbers $a$,$b$ satisfy $a^3+4a^2b=4a^2+b^4$. Prove that $\sqrt{\sqrt{a}-1}$ is also rational.
Positive rational numbers $a$,$b$ satisfy $a^3+4a^2b=4a^2+b^4$. Prove that $\sqrt{\sqrt{a}-1}$ is also rational.
My try: $a(a+2b)^2=a^3+4a^2b+4b^2a=4a^2+b^4+4b^2a=(2a+b^2)^2$, so $\sqrt{a}=\frac{2a+b^2}{a+2b}$.
| Your approach is correct you can just continue, by proving that
$$\sqrt a-1=\left(\frac b{\sqrt a}-1\right)^2$$
and remember that $\sqrt a$ is proved to be rational!
| {
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"url": "https://math.stackexchange.com/questions/1213400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Is $\frac{n}{3}! = (\frac{1}{3})^n n!$ Is $$\frac{n}{3}! = (\frac{1}{3})^n n!$$
I thought I could take all the (1/3) out of the factorial, but wolfram alpha says this is false.
| No, compare
$$\frac{n!}{3^n}=\left(\frac{1\cdot2\cdot3}{3\cdot3\cdot3}\right)\left(\frac{4\cdot5\cdot6}{3\cdot3\cdot3}\right)\left(\frac{7\cdot8\cdot9}{3\cdot3\cdot3}\right)\cdots\left(\frac{(n-2)\cdot(n-1)\cdot n}{3\cdot3\cdot3}\right)$$
and
$$\left(\left(\frac n3\right)!\right)^3=\left(\frac{3\cdot3\cdot3}{3\cdot3\cdot3}\right)\left(\frac{6\cdot6\cdot6}{3\cdot3\cdot3}\right)\left(\frac{9\cdot9\cdot9}{3\cdot3\cdot3}\right)\cdots\left(\frac{n\cdot n\cdot n}{3\cdot3\cdot3}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| It is a pity you did not have a minus-sign in the numerator, since
$$
D\ln(3\cos x+2\sin x)=\frac{2\cos x-3\sin x}{3\cos x+2\sin x},
$$
but let us see how we can use this fact anyways.
Let us aim at writing
$$
\frac{2\cos x+3\sin x}{3\cos x+2\sin x}=c_1 \frac{2\cos x-3\sin x}{3\cos x+2\sin x}+c_2 \frac{3\cos x+2\sin x}{3\cos x+2\sin x}
$$
since both those terms are easy to integrate. This leads us to the linear equations $2=2c_1+3c_2$ and $3=-3c_1+2c_2$. The solution to this system is
$c_1=-5/13$ and $c_2=12/13$.
Thus
$$
\int\frac{2\cos x+3\sin x}{3\cos x+2\sin x}\,dx = -\frac{5}{13}\int \frac{2\cos x-3\sin x}{3\cos x+2\sin x}\,dx +\frac{12}{13}\int \frac{3\cos x+2\sin x}{3\cos x+2\sin x}\,dx.
$$
I guess you can take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Prove that $\frac{a^n-1}{b^n-1}$ and $\frac{a^{n+1}-1}{b^{n+1}-1}$ can't both be prime.
Prove that $$\frac{a^n-1}{b^n-1} \ \text{and} \ \frac{a^{n+1}-1}{b^{n+1}-1}$$ cannot both be prime ($a>b>1,n\ge 2$).
Clearly $(a^n-1,a^{n+1}-1)=a-1$ and $(b^n-1,b^{n+1}-1)=b-1$.
Clearly $b-1\mid a-1$ (since $b-1\mid a^n-1,a^{n+1}-1$), so $a=bk-k+1$ for some $k\in\mathbb Z_{\ge 2}$.
A hint would probably be enough. I haven't found an approach I could use.
| Let $\let\v\nu\let\geq\geqslant\v_p(n)$ denote the exponent of the prime $p$ in the factorisation of $n\in\mathbb N$.
Lemma (Lifting The Exponent) If $p$ is odd, $p\nmid a$ and $p\mid a-b$, then $$\v_p(a^n-b^n)=\v_p(a-b)+\v_p(n).$$
Proof. See ProofWiki.
We already have $b-1\mid a-1$. Let $r$ be any prime divisor of $b-1$ and thus of $a-1$. If $r$ is odd, then (by the lemma)
$$\v_r\left(\frac{a^n-1}{b^n-1}\right)=\v_r(a-1)-\v_r(b-1).$$
Because $\frac{a^n-1}{b^n-1}=p$ is prime, this shows that $\v_r(a-1)=\v_r(b-1)$ for all primes $r\mid b-1$ with $r\neq2$ and $r\neq p$. Hence $a-1=2^kp^xg(b-1)$ for some $k,x$ and $g$, $\gcd(g,b-1)=1$.
We can do the same for $\frac{a^{n+1}-1}{b^{n+1}-1}=q$, giving $a-1=2^lq^yh(b-1)$ for some $l,y$ and $h$, $\gcd(h,b-1)=1$.
Note that $a-1=2^lq^yh(b-1)$ uniquely determines $q$ if $q\mid b-1$ and $\gcd(n,b-1)=1$:
*
*If $q$ is odd, it is the unique odd prime $r\mid b-1$ with $\v_r(a-1)>\v_r(b-1)$
*$q=2$ iff there is no odd prime $r\mid b-1$ with $\v_r(a-1)>\v_r(b-1)$
So we actually have that $p=q$:
$$\frac{a^{n+1}-1}{b^{n+1}-1}=\frac{a^n-1}{b^n-1}$$
hence
$$a^{n+1}b^n-a^{n+1}-b^n=b^{n+1}a^n-b^{n+1}-a^n$$
$$\color{darkgreen}{(a-b)}\color{blue}{(a^nb^n+\frac{a^n-b^n}{a-b}-\frac{a^{n+1}-b^{n+1}}{a-b})}=0$$
but, because $b^n\geq2^n\geq n+1$,
$$\color{blue}{a^nb^n+\frac{a^n-b^n}{a-b}}\geq (n+1)a^n\color{blue}>a^n+a^{n-1}b+\cdots+ab^{n-1}+b^n=\color{blue}{\frac{a^{n+1}-b^{n+1}}{a-b}}$$
contradicting $\color{darkgreen}{a\neq b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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"answer_id": 0
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How to verify $\frac{\sin x\pm\sin y}{\cos x+\cos y}=\tan\frac{x\pm y}{2}$ Can anyone help me verify the following identity?
$$
\frac{\sin x\pm\sin y}{\cos x+\cos y}=\tan\frac{x\pm y}{2}
$$
I could really use some guidance.
| $$\sin x+\sin y=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
$$\sin x -\sin y=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$$
$$\cos x+\cos y=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
$$\frac{\sin x+\sin y}{\cos x+\cos y}=\frac{2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})}{2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})}=\frac{\sin(\frac{x+y}{2})}{\cos(\frac{x+y}{2})}=\tan\left(\frac{x+y}{2}\right)$$
and the same method for the other!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The maximum value of expression $ \sqrt{\sin^2x+ 2a^2} - \sqrt{-1 -\cos^2x+ 2a^2} $ If $a,x\in\Bbb R$, what is the maximum value of the expression
$ \sqrt{\sin^2x+ 2a^2} - \sqrt{-1 -\cos^2x+ 2a^2} $?
I tried to differentiate but it became messy.
| With a difference of square roots, there are many potential approaches to consider for basic evaluation. Of note in this case is that we have some nice trigonometry happening that gives the problem an "easy" out: $-\cos^2x = -1+\sin^2x$:
$$\sqrt{\sin^2x+ 2a^2} - \sqrt{-1 -\cos^2x+ 2a^2}=\sqrt{\sin^2x+ 2a^2} - \sqrt{-2+\sin^2x+ 2a^2}$$
We can immediately see that we have $u=\sin^2x+2a^2$ transforms the above to
$$\sqrt{u}-\sqrt{u-2}$$
Now what is left is to find the limits on $u$, including the minimum and maximum possibilities. The most obvious limitation is that $a\ge \frac{\sqrt 2}2$, since the maximum value available from $\sin^2x$ is $1$, and $u-2\ge0$ must be maintained. After this, it looks like taking derivatives should be fairly straightforward to approaching the result.
It is also worth noting that another valuable approach to a difference of square roots is as follows:
$$\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b}$$
where we have treated $a-b$ as a "difference of squares".
Putting this together, we complete the result as follows:
$$\sqrt u-\sqrt{u-2}=\frac {u-(u-2)}{\sqrt u+\sqrt{u-2}}=\frac{2}{\sqrt u+\sqrt{u-2}}\\u\ge 2\implies \sqrt u+\sqrt{u-2}\ge \sqrt 2\\ \implies\frac2{\sqrt u+\sqrt{u-2}}\le \frac 2{\sqrt 2}=\sqrt 2$$
Therefore our maximum value is obtained by minimizing $u$, and knowing that we have a minimum possible value for $u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Elliptic Function Let $y$ be the function defined by
$$y(\theta)=2sin\frac{\theta}{2}\prod_{k=1}^{\infty}\frac{(1-e^{i\theta}q^k)(1-e^{-i\theta}q^k)}{(1-q^k)^{2}}$$
where $q = e^{2\pi i\tau}$
Show that $y$ has simple zeroes at each $\theta=2\pi m+2\pi n \tau$ for all integers $m$ and $n$.
Prove that
$y(\theta+2\pi \tau)=-e^{i \theta}q^{-1/2}y(\theta)$
$y(2\theta+4\pi \tau)=e^{4i \theta}q^{-2}y(2\theta)$
| We have
\begin{align}\frac{y(\theta + 2\pi\tau)}{y(\theta)} &= \frac{\sin\left(\frac{\theta}{2} + \pi \tau\right)}{\sin \frac{\theta}{2}} \prod_{k = 1}^\infty \frac{(1 - e^{i\theta}q^{k+1})(1 - e^{-i\theta} q^{k-1})}{(1 - e^{i\theta}q^k)(1 - e^{-i\theta}q^k)}\\
&= \frac{\sin\left(\frac{\theta}{2} + 2\pi \tau\right)}{\sin \frac{\theta}{2}}\frac{1 - e^{-i\theta}}{1 - e^{i\theta}q}\\
&= \frac{\sin\left(\frac{\theta}{2} + 2\pi \tau\right)}{\sin \frac{\theta}{2}}\frac{e^{-i\theta/2}}{e^{i\theta/2}q^{1/2}} \frac{e^{i\theta/2} - e^{-i\theta/2}}{e^{-i\theta/2+\pi\tau} - e^{i\theta/2 + \pi \tau}}\\
&= \frac{\sin\left(\frac{\theta}{2} + 2\pi \tau\right)}{\sin \frac{\theta}{2}} e^{-i\theta}q^{-1/2}\left[-\frac{\sin \frac{\theta}{2}}{\sin\left(\frac{\theta}{2} + \pi \tau\right)}\right]\\
&= -e^{-i\theta}q^{-1/2}.
\end{align}
Thus $y(\theta + 2\pi \tau) = -e^{-i\theta}q^{-1/2}y(\theta)$.
Using the above identity, we obtain
\begin{align}y(2\theta + 4\pi \tau) &= y((2\theta + 2\pi \tau) + 2\pi \tau)\\
&= -e^{-i(2\theta + 2\pi \tau)}q^{-1/2}y(2\theta + 2\pi \tau)\\
&= -e^{-i(2\theta + 2\pi \tau)}q^{-1/2}[-e^{-i(2\theta)}q^{-1/2}]y(2\theta)\\
&= e^{-4i\theta}q^{-1 - 1/2 - 1/2}y(2\theta)\\
&= e^{-4i\theta}q^{-2}y(2\theta).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is $\tan \alpha$ if $\sin \alpha + \cos \alpha = \frac{\sqrt{3}-1}{2}$ and $\alpha \in (90^\circ,135^\circ)$ I am stuck with this problem. Any ideas on how to solve this?
| $$\sin\alpha+\cos\alpha=\sqrt2\sin\left(\alpha+\frac{\pi}4\right)=\frac{\sqrt3-1}2\implies \alpha=\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)-\frac{\pi}4$$
Now:
$$\tan\alpha=\tan\left(\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)-\frac{\pi}4\right)=\frac{\tan\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)-\tan\frac{\pi}4}{1+\tan\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)\tan\frac{\pi}4}$$
Now:
$$\tan\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)=\frac{\sin\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}{\cos \arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}=\frac{\sin\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}{\sqrt{1-\sin^2 \arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}}=\frac{\frac{\sqrt3-1}{2\sqrt2}}{\sqrt{1-\left(\frac{\sqrt3-1}{2\sqrt2}\right)^2}}=2-\sqrt3$$
Now answer is:
$$\tan\alpha=\frac{2-\sqrt3-1}{1+2-\sqrt3}=\frac{1-\sqrt3}{3-\sqrt3}=-\frac1{\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Evaluating $\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}$ How do I evaluate
$$\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{\dfrac{1}{1-\cos x}}\ ?$$
I tried using the fact that $\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}} = \exp\left(\ln\bigg(\frac{\sin x}{x}\right)\frac{1}{1-\cos x}\bigg)$ and then I am already stuck.
| With l'Hospital directly:
\begin{align*}
\lim_{x\to 0}\frac{\log\frac{\sin x}x}{1-\cos x} & \stackrel{\text{l'H}}=\lim_{x\to 0}\frac{\frac 1{\sin x}\cdot\frac{x\cos x-\sin x}x}{\sin x} \\
& =\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin^2x}\\
& \stackrel{\text{l'H}}=\lim_{x\to 0}\frac{-x\sin x}{\sin^2x+2x\sin x\cos x}\\
&=-\lim_{x\to 0}\frac{x}{\sin x+2x\cos x}\\&=-\lim_{x\to 0}\frac1{\frac{\sin x}x+2\cos x}\\
&=-\frac13
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Given several integrals calculate $\int\limits_5^6 f(x) dx$ Let
$\int\limits_4^7 f(x)\,dx = 2$, $\int\limits_6^7 f(x)\,dx = 17$, and $\int\limits_4^5 f(x)\,dx = 3$
Calculate
$$\int\limits_5^6 f(x)\,dx$$
I guess I am to assume that $f(x)$ is the same in all integrals.
Should I figure out what $\Delta x$ is? Should I try to use identities of integrals?
| Note: Although this problem is older the method displayed is useful as a secondary demonstration.
The problem presents three equations and asks to evaluate a fourth. Since the results are simple the form of the integrand can be considered in a simple form, ie $f(x) = a + b \, x + c \, x^2$. In this view it can be seen that:
$$\int f(x) \, dx = a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3}$$
for which
\begin{align}
\int_{4}^{7} f(x) \, dx = 2 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3} \\
\int_{6}^{7} f(x) \, dx = 17 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3} \\
\int_{4}^{5} f(x) \, dx = 3 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3}.
\end{align}
Solving this system f equations yields
$$f(x) = \frac{1}{6} \, ( 4729 - 1806 \, x + 168 \, x^2)$$
and leads to
$$\int_{5}^{6} f(x) \, dx = \frac{1}{6} \, [ 4729 \, x - 1403 \, x^2 + 56 \, x^3]_{5}^{6} = -18.$$
In comparison it is seen that
$$\int_{5}^{6} f(x) \, dx = \left(\int_{4}^{7} - \int_{6}^{7} - \int_{4}^{5} \right) \, f(x) \, dx = (2 - 17 - 3) = -18.$$
An advantage for finding and solving the system of equations is by considering if the proposed question had asked for the value of the integral for the range (0,10), or (0,a), etc..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Question about a solution of a system of three non linear equations in three unknowns Let $a$, $b$ and $c$ be positive real numbers such that $$ a + \frac{1}{b} = 3$$ $$b + \frac{1}{c} = 4$$ $$ c + \frac{1}{a} = \frac{9}{11} $$ then $$ a \times b \times c =?$$
I tried doing this problem but I was unsuccessful. Tried a lot but couldn't get the answer! The answer is a numerical value...
| Here is a way
we have :
$$b=\dfrac{1}{3-a}\ \ \ c=\dfrac{1}{4-b}\tag1$$
hence :
$$\dfrac{1}{4-\dfrac{1}{3-a}}+\frac{1}{a}=\frac{3-a}{11-4a}+\frac{1}{a}=\frac{9}{11}$$
so $11((3-a)a-11+4a)=9a(11-4a)$ or $25a^2-110a+121=0$ so $a=\frac{11}{5}$ and here you can find $a$ and repalce in $(1)$ to find $b$ and $c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rewrite also as: $p^2+pq\equiv 3 \pmod5$.
Now I analyzed five possible cases: $p^2\equiv 0$, $p^2\equiv 1$, $p^2\equiv 2$, $p^2\equiv 3$ and $p^2\equiv 4$.
*
*We can not that if $p^2\equiv 0$ also $pq\equiv 0$ therefore the equation is impossible
*$p^2\equiv 2$ is impossible because $2$ is a non-quadratic residue $\pmod5$
*also $p^2\equiv 3$ is impossible because $3$ is a non-quadratic residue $\pmod5$.
*If $p^2\equiv 1 \pmod5$, $p\equiv \pm 1$, therefore $p\equiv -1\equiv 4 \pmod5$ and $p2\equiv 1 \pmod5$. These values are not acceptable because if $p=9$ ($9\equiv 4 \pmod5$) $275\cdot 9>2008$, if $p=11$ ($11\equiv 1\pmod5$) $275\cdot 9>2008$ therefore two values aren't acceptable.
*If $p^2\equiv 4 \pmod5$, $p\equiv \pm2$ and if $p\equiv -2\equiv 3 \pmod5$ there not exist solutions for the same reason of previous example. If $p\equiv 2 \pmod5$ possible values of $p$ are $2$ or $7$ because $275p<2008$. Replacing $p=2$ we obtain a valor not integer of $q$ while replacing $p=7$ we obtain $q=2$ therefore this is the only solution of equation.
Is correct my process or are there other solutions?
Thanks:)
| $p^2+pq+275p$ is even. If $p=2$, then $q\not\in\mathbb Z$. Otherwise $p+q+275$ is even. This implies $q=2$. Then $p=7$. So the only solution is $(p,q)=(7,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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What is my error in applying this Laplace Transform? So, our book has the seemingly innocuous problem: $y''-y'-6y=0$. I was able to solve by hand, and come up with $${\scr L}(y)=\frac{s-2}{s^{2}-s+6}$$.That completed, I factored the bottom to $(s-3)(s+2)$. Our professor gave us a sheet that said ${\scr L}\mathrm{{}f(t)\mathrm{}}=F(s)$ which corresponds along the rows to a column $f(t)$. I broke the equation down to $$\frac{s}{(s-3)(s+2)}-\frac{2}{(s-3)(s+2)}$$ and the corresponding $f(t)$ were $$\frac{ae^{at}-be^{bt}}{a-b}$$ and $$\frac{e^{at}-e^{bt}}{a-b}$$, respectively. The summation of these leads to:
$$\frac{1}{5}(-e^{2t}+5e^{-3t})$$. However, the answer in the book is: $$\frac{1}{5}(e^{3t}+4e^{-2t})$$.
I have literally no idea why my method is working. Would someone be so kind as to enlighten me?
| For the differential equation $y'' - y' -6y = 0$ the Laplace transform is
\begin{align}
( s^2 y(s) - y'(0) - s y(0) ) - (s y(s) - y(0)) - 6 y(s) = 0
\end{align}
which leads to $(s^2 - s - 6) y(s) = y'(0) + y(0) (s-1)$ and
\begin{align}
y(s) &= \frac{y'(0) - y(0)}{s^2 - s -6} + \frac{y(0) \, s}{s^2 - s - 6} \\
&= \frac{y'(0) - y(0)}{5} \left( \frac{1}{s-3} - \frac{1}{s+2} \right) + \frac{y(0)}{5} \left( \frac{2}{s+2} + \frac{3}{s-3} \right) \\
&= \frac{y'(0) + 2 y(0)}{5} \frac{1}{s-3} + \frac{3 y(0) - y'(0)}{5} \frac{1}{s+2}.
\end{align}
Inversion leads to
\begin{align}
y(t) = \frac{y'(0) + 2 y(0)}{5} \, e^{3t} + \frac{3y(0) - y'(0)}{5} \, e^{-2t}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Working out $\tan x$ using sin and cos expansion
Using only the series expansions $\sin x = x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...$ and $\cos x = 1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...$
Find the series expansions of the $\tan x$ function up to the $x^5$ term.
So it is:
$$
\frac {x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...} {1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...}
$$
However many times I retry, trying long division results :
$$
\require{enclose}
\begin{array}{rll}
Q(x) \\[-3pt]
1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ... \enclose{longdiv}{x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...}\kern-.2ex \\[-3pt]
\underline{x-\dfrac{x^3} {2!} + \dfrac{x^5} {4!}\phantom{0000}} \\[-3pt]
\dfrac{-x^3} {3!} + \dfrac{x^3} {2!}+ \dfrac{x^5} {5!}-\dfrac{x^5} {4!} \phantom{00}&& \\[-3pt]
\underline{\phantom{0}\dfrac{-x^3} {3!} + \dfrac{x^3} {2!}- \dfrac{x^5} {2!2!}+ \dfrac{x^5} {3!2!}} && \\[-3pt]
\dfrac{x^5} {5!} - \dfrac{x^5} {4!} + \dfrac{x^5} {2!2!}-\dfrac{x^5} {3!2!}&& \\[-3pt]
\end{array}
$$
Sorry for the horrible format. but this is the best i can do..
$$
Q(x) = x + (\frac {x^3}{2!} - \frac {x^3}{3!}) + (\dfrac {x^5}{5!} - \dfrac {x^5}{4!} + \dfrac {x^5}{2!2!} - \dfrac {x^5}{3!2!}) + ...
$$
Which results the coefficients of the $x^5$ term $\dfrac 1 {20}$.
This is clearly a wrong value.
I can't seem to get $\dfrac 2 {15}$ instead.
Where did I get wrong?
Many thanks in advance.
| You may exploit the fact that $\tan x$ is an odd function, hence in a neighbourhood of the origin:
$$ \tan x=\sum_{n\geq 0} a_{2n+1}\, x^{2n+1}\tag{1} $$
as well as:
$$ \frac{d}{dx}\tan x=\frac{1}{\cos^2 x}=1+\color{purple}{\tan^2 x} = \sum_{n\geq 0}(2n+1)\,a_{2n+1}\,x^{2n} \tag{2}$$
from which it follows that $a_1=1$ and:
$$ a_{2n+1} = \frac{1}{2n+1}\sum_{k=1}^{n}a_{2k-1}a_{2n-2k+1}\tag{3}$$
so the Taylor coefficients of the tangent function can be computed with a rather simple recursion. $(3)$ leads to:
$$ a_3 = \frac{1}{3}a_1^2 = \frac{1}{3},\quad a_5=\frac{2}{5}a_1 a_3=\frac{2}{15}.\tag{4}$$
No polynomial divisions involved. Pretty nice, don't you think?
| {
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"url": "https://math.stackexchange.com/questions/1238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"$111 \dots$ upto $3^n$ digits" is divisible by $3^n$
Prove that an integer of the form "$111 \dots$ upto $3^n$ digits" is divisible by $3^n$
My attempt
For $n=1,$ $111$ is divisible by 3.
Let $T_n=111...$ upto $3^n$ digits is divisible by $3^n$.
Then $T_{n+1}=111...$ upto $3^{n+1}$ digits. I am unable to prove that Then $T_{n+1}$ is divisible by $3^n$.
| $n=1$ case: 111 is divisible by 3.
$n \Rightarrow n+1$ case:
To demontrate that $T_{n+1}$ can be divided by $3^{n+1}$ I write $T_{n+1}$ as:
$$
T_{n+1} = T_n \cdot (10^{2 \cdot 3^n}+10^{3^n}+1)
$$
$T_n$ is divisibile by $3^n$ by induction hypotesis, $10^{2 \cdot 3^n}+10^{3^n}+1$ is divisible by 3 because the sum of digits is 3.
So $T_{n+1}$ is divisible by $3^n \cdot 3 = 3^{n+1}$ which is what we wanted.
Why $T_n$ is that one?
What we want to do is splitting $T_{n+1}$ to obtain an expression of $T_n$. We can do it by splitting the number in three groups of ones.
So $\underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{111 \dots 111}_{3^n \text{ digits}} = \underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{000 \dots 000}_{3^n \text{ digits}} \underbrace{000 \dots 000}_{3^n \text{ digits}} + \underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{000 \dots 000}_{3^n \text{ digits}} + \underbrace{111 \dots 111}_{3^n \text{ digits}}$
Formally it's done by writing the expression $T_{n+1} = T_n \cdot (10^{2 \cdot 3^n}+10^{3^n}+1)$
For example we can observe that $$T_2 = 111\,111\,111 = 111 \cdot 10^6 + 111 \cdot 10^3 + 111 = 111 \cdot (10^6+10^3+1) = \\ =T_1 \cdot (10^6+10^3+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find x,y & z (xyz+xyz=zyx) I saw this problem the other day at work and found it pretty interesting:
$$xyz + xyz = zyx$$
Find $x, y, z$ and the base(s) which this is true.
Note that $x,y,z$ are simply digits concatenated, they are not multiplied against each other. I ended up writing a program to brute force all solutions with base 2 - base 36. I left out $x=0,y=0,z=0$ since that seems trivial to me.
The solutions I found are below (I included the b10 value of xyz for reference):
x=0,y=1,z=1,b=2, ->b10 value=3
x=1,y=4,z=3,b=5, ->b10 value=48
x=2,y=7,z=5,b=8, ->b10 value=189
x=3,y=a,z=7,b=11, ->b10 value=480
x=4,y=d,z=9,b=14, ->b10 value=975
x=5,y=g,z=b,b=17, ->b10 value=1728
x=6,y=j,z=d,b=20, ->b10 value=2793
x=7,y=m,z=f,b=23, ->b10 value=4224
x=8,y=p,z=h,b=26, ->b10 value=6075
x=9,y=s,z=j,b=29, ->b10 value=8400
x=a,y=v,z=l,b=32, ->b10 value=11253
x=b,y=y,z=n,b=35, ->b10 value=14688
Looking at the solution set, I noticed a few relationships which surprised me:
*
*$x$ is valid for integers [0-11]
*$y = 3x+1$
*$z = 2x+1$
*$R\text{ ($xyz$ in base 10)} = 9x^3 + 21x^2 + 15x + 3$
Are these formula's just chance? Or can they actually be derived without knowing the solution set ahead of time?
I started trying to work out a general solution and got stuck pretty fast:
$$2xb^2 + 2by + 2z = zb^2 + by + x$$
And then I stopped because I didn't understand how to find four unknowns with one equation. Thoughts?
| Th e equation you described can be written:
$$2xb^2 + 2by + 2z = zb^2 + by + x$$
with the constraint $0\leq x,y,z \leq b-1$ now the first thing to do is to find $z$, we have $2z=x\mod b$ so we have two cases :
*
*$z\leq \frac{b-1}{2}$ and hence $x=2z$ and in this case we have:
$2xb+2y=zb+y$ and from here we have again $y\equiv 2y\mod b$ hence $y=0$ and $4z=z$ so $z=0$ there is no solution other then $xyz=000$
*$z>\frac{b-1}{2}$ and hence $x=2z-b$ here we replace in the equation:
$2xb+2y=zb+y-1$ but here again we have $2y\equiv y-1\mod b$ hence $y=b-1$ and $2x+1=z$ so $4z-2b+1=z$ hence $3z=2b+1$ so here we have two cases:
*
*If $3$ does not divide $2b-1$ no solution
*If $3$ divides $2b-1=3k$ then $z=k$ and $x=2k-b$ and hence:$xyz=(2k-b)b^2+(b-1)b+k=\frac{b^3+b^2-b-1}{3}$
Conclusion:
*
*$2b-1$ divisible by $3$ the only solution $\overline{xyz}^{(b)}$ is the number $\frac{b^3+b^2-b-1}{3}$ with the values ($x=\frac{b-2}{3},y=b-1,z=\frac{2b-1}{3}$)
*Otherwise no solution.
This formula works for your given values, and if you want to find the value of $x$ you can replace in the equations $b=3x+2$ so that you obtain:
$$b=3x+2,y=3x+1,z=2x+1, R=9x^3+21x+15x+3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometry with complex numbers. Let $a$, $b$, $c$, and $d$ be four complex numbers on the unit circle, such that the line joining $a$ and $b$ is perpendicular to the line joining $c$ and $d$.
Find a simple expression for $d$ in terms of $a$, $b$, and $c$.
I have thought of using power of a point, but that has gotten me nowhere so far. Can anyone help me?
| Since $|a| = 1$, we have $a \overline{a} = |a|^2 = 1$, so $\overline{a} = 1/a$. Similarly, $\overline{b} = 1/b$, $\overline{c} = 1/c$, and $\overline{d} = 1/d$.
The line joining $a$ and $b$ is perpendicular to the line joining $c$ and $d$, so
$(b - a)(\overline{d} - \overline{c}) + (\overline{b} - \overline{a})(d - c) = 0,$
which becomes
\begin{align*}
(b - a) \left( \frac{1}{d} - \frac{1}{c} \right) + \left( \frac{1}{b} - \frac{1}{a} \right) (d - c) &= 0 \\
\Rightarrow \quad \frac{(b - a)(c - d)}{cd} + \frac{(a - b)(d - c)}{ab} &= 0 \\
\Rightarrow \quad (b - a)(c - d) \left( \frac{1}{cd} + \frac{1}{ab} \right) &= 0.
\end{align*}
Since $b \neq a$ and $c \neq d$, we may divide both sides by $(b - a)(c - d)$, to get
$\frac{1}{cd} + \frac{1}{ab} = 0.$
Solving for $d$, we find $d = \boxed{-ab/c}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finite field, how to satisfy equation? Suppose we have $\mathbb{F}_{2^5}$ defined by polynomial $x^5+x^2+1$, and (this is homework exercise, which I kinda solved) it is required to find suitable elements $b$, so that it satisfies equation $b^2+b=x^3+1$
In this field, only 32 items and it is not a big deal to reiterate them all, here is cloud.sage script which does that.
Suitable are $b=x^7=x^4 + x^2$ with $b^2=x^{14}=x^4 + x^3 + x^2 + 1$
And $b=x^{22}=x^4 + x^2 + 1$ with $b^2=x^{44}=x^4 + x^3 + x^2$
(It is obvious that in both cases $b+b^2$ gives $x^3+1$)
It is dumb computational iteration, however. Is there any neat ways to get to this result without going through all $32$ elements?
| Write $b = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$, $a_i \in \{0,1\}$. Then compute
$$b^2 + b = (a_2 + a_4) x^4 + a_4 x^3 + (a_1 + a_2 + a_4) x^2 + (a_1 + a_3) x + a_4$$
so if this is to be $x^3 + 1$, we must have
$$ \eqalign{ a_2 + a_4 &= 0\cr
a_4 &= 1\cr
a_1 + a_2 + a_4 &= 0\cr
a_1 + a_3 &= 0\cr
a_4 &= 1\cr}$$
(all mod 2). It's not hard to find the solution: $a_1 = 0$, $a_2 =1$, $a_3 = 0$, $a_4 = 1$, $a_0$ arbitrary. So $b = x^2 + x^4$ or $1 + x^2 + x^4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating the value of $\cos \left(\frac{1}{2} \arccos \frac{3}{5}\right)$ I need help finding the value of $$\cos \left(\frac{1}{2} \arccos \frac{3}{5}\right)$$
My try:
let $\theta = \frac{1}{2} \arccos \frac{3}{5}$, then $2\theta = \arccos \frac{3}{5}$ and $\theta \in [0,\frac{\pi}{2}]$
Now let $A = 2\theta$, and $A \in [0,\frac{\pi}{2}]$ So $\cos A = \frac{3}{5} $ and $\sin A = \frac{4}{5} $. I noticed that I got stuck and wasn't getting closer to the answer.
Can someone please give me a HINT AND NOT THE ANSWER? Thanks.
| Since:
$$\cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}$$
it follows that:
$$\cos\left(2\arccos\frac{3}{5}\right)=\sqrt{\frac{1+\frac{3}{5}}{2}}=\frac{2}{\sqrt{5}}.$$
| {
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"url": "https://math.stackexchange.com/questions/1245197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Integrating $\int \frac{\sin^3x}{(\cos x)^\frac 4 3} dx$
Find: $\int \frac{\sin^3x}{(\cos x)^\frac 4 3} dx$
My attempt:
Set $u=(\cos x)^\frac 4 3 $ so $du= \frac 4 3 (\cos x)^\frac 1 3 \sin x dx \Rightarrow dx= \frac 3 {4 (\cos x)^\frac 1 3 \sin x}du$
\begin{align}
\int \frac{\sin^3x}{(\cos x)^\frac 4 3} dx&=\frac 3 4 \int \frac{\sin ^2 x}{(\cos x)^\frac 5 3} du \\
&=\frac 3 4 \int \frac{1- \cos ^2 x}{(\cos x)^\frac 5 3} du\\
&=\frac 3 4 \int u^{-\frac 5 4}-u^\frac 1 4 du\\
&=-3(\cos^{-\frac 1 3} x +\frac 1 5 \cos ^{\frac 5 3 }x) +c
\end{align}
It's very close to the right answer, but not quite, what did I do wrong? Alternatively, is there a better way?
http://www.integral-calculator.com/#expr=%28sinx%29%5E3%2F%28cosx%29%5E%284%2F3%29
| Let $u=\cos x$. Then $\mathrm du=-\sin x\mathrm dx$. Therefore
$$\begin{split}\int\frac{\sin^3x}{(\cos x)^{4/3}}\mathrm dx&=\int\frac{u^2-1}{u^{4/3}}\mathrm du\\&=\int \left(u^{2/3}-u^{-4/3}\right)\mathrm du\\&=\frac35u^{5/3}+3u^{-1/3}+c\\&=\frac35(\cos x)^{5/3}+3(\cos x)^{-1/3}
+c.\end{split}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Curvature of plane parametric curves What is the neatest way to derive the following formula for the curvature of a parametric curve? $$\kappa =\frac{\|y'x''-y''x'\|}{(x'^2+y'^2)^{\frac{3}{2}}} $$
| The definition of curvature $\kappa$ is
$$
\kappa = \frac{d\alpha}{ds}
$$
where $\alpha = \arctan(\frac{dy}{dx})$ and $s$ is distance along the curve. $ds = dx\sqrt{(1 + (\frac{dy}{dx})^2)}\ $. Curvature has the units angle per length.
Start with the simplest parametric curve in two dimensions: $y = y(x)$. We then have:
$$
\begin{align*}
\kappa &= \frac{d\alpha}{ds} \\
&= \frac{d\alpha/dx}{ds/dx} \\
&= \frac{\frac{1}{1+(\frac{dy}{dx})^2} \frac{d^2y}{dx^2} }{\sqrt{(1 + (\frac{dy}{dx})^2)}} \\
&= \frac{\frac{d^2y}{dx^2}}{\left(1 + (\frac{dy}{dx})^2\right)^{3/2}}
\end{align*}
$$
Now consider a two-dimensional parameterized curve, given by $x = x(t)$ and $y = y(t)$. Denote $\frac{dx}{dt} = x'$ and $\frac{dy}{dt} = y'$ Then
\begin{align*}
\frac{dy}{dx} &= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'}{x'}\\
\end{align*}
\begin{align*}
\frac{d^2y}{dx^2} &= \frac{\frac{d}{dt}\left( \frac{dy}{dt} / \frac{dx}{dt} \right) }{\frac{dx}{dt}} \\
&= \frac{1}{\left( \frac{dx}{dt} \right)^3} \left[ \frac{dx}{dt} \frac{d^2y}{dt^2} - \frac{dy}{dt} \frac{d^2x}{dt^2} \right] \\
&= \frac{x'y'' - x''y'}{x'^3}
\end{align*}
Substituting $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the definition of $\kappa$, we get
$$
\kappa = \frac{x'y'' - x''y'}{\left( x'^2 + y'^2 \right)^{3/2}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Diophantine Equations : Solving $a^2+ b^2=2c^2$ I was working through some number theory problems , when I came across the following question :
Find all solutions of $a^2+b^2=2c^2$
My Solution (Partial) :
*
*We can rewrite the above equation as : $c^2 = (a^2 + b^2)/2 $
*So $\Rightarrow$ $a^2 , c^2 , b^2$ are in an Arithmetic Progression $\Rightarrow$ there exist infinite solutions
*WLOG , let $a^2$ be $t$ , $b^2$ be $s$ and $c^2$ be $m$
*The equation can be re-written as $ t + s = 2m$
*Now , what I was thinking was that should I solve this Diophantine equation for $t$ & $s$ in terms of $m$ ; substitute their values back into the equation and find out $m$ ; and then finally put this value of $m$ back into the values of $t$ and $s$ to derive a general solution
Can someone help me out ? Maybe a hint ...
| Assume you have the Pythagorean relation $u^2 + v^2 = c^2$
Then
$$
\begin{align}
(u^2 + v^2) + (u^2 + v^2) & = 2c^2\\
(u^2 + v^2 + 2uv) + (u^2 + v^2 - 2uv) & = 2c^2\\
(u + v)^2 + (u - v)^2 & = 2c^2\\
\end{align}
$$
Thus if $a = u + v$ and $b = |u - v|$
$a^2 + b^2 = 2c^2$
Update
Given any triple of integers $a, b, c : a^2 + b^2 = 2c^2$
$a^2 + b^2 \equiv 0 \mod 2$
Which implies $a + b \equiv 0 \mod 2$,
So $a \equiv b \mod 2$ and
$a - b \equiv 0 \mod 2$
WLOG, assume $a \ge b$
Both $a + b$ and $a - b$ are even.
Let $2u = a + b$ and $2v = a - b$
$4u^2 + 4v^2 = (a + b)^2 + (a - b)^2 = 2(a^2 + b^2) = 4c^2$
$u^2 + v^2 = c^2$
i.e., $u, v, c$ is a Pythagorean triple.
So every triple of integers $a, b, c : a^2 + b^2 = 2c^2$ corresponds to a Pythagorean triple.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "16",
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Find the value of $a$, $b$ and $c$ for the given limit. Question -
Find the values of $a$, $b$ and $c$ so that $$ \lim_{x\to 0} \cfrac{ae^x - b\cos x +c e^{-x} }{x\sin x} = 2 $$
This is what I've tried yet :
For $ x\to 0 $ the numerator must also tend to zero as : $ e^x , \ \cos x, e^{-x} $ all $\to 1$ for $x \to 0$ .
Therefore, we have:
$$
a - b + c = 0 \\
\color{blue}{\text{OR}} \\ b = a + c \\
$$
Now, plugging this in the original equation:
$$\lim_{x\to 0} \cfrac{ae^x -(a+c)\cos x + ce^{-x} }{x^2 \times \frac{\sin x}{x} } = 2 $$
Which implies -
$$\lim_{x\to 0} \cfrac{a(e^x - \cos x) + c(e^{-x} - \cos x) }{x^2} =2 $$
I do get a positive wave that there is some application of expansion series of $e^x \ \& \ \cos x $ . I tried that too but that just didn't work. May be I'm not doing right simplification. Any help will be greatly appreciated.
| If $b\neq a+c$ the limit would not be finite. Therefore we need $b=a+c$. Using approximation $e^x=1+x+\frac{x^2}{2}+O(x^3)$, $\cos x=1-\frac{x^2}{2}+O(x^3)$, and $\sin x=x+O(x^2)$ for $x\to 0$ we get the expression
$$
\frac{a(1+x+\frac{x^2}{2})-b(1-\frac{x^2}{2})+c(1-x+\frac{x^2}{2})+O(x^3)}{x(x+O(x^2))},
$$
which simplifies [considering $b=a+c$] to
$$
\frac{(a-c)x+\frac{1}{2}(a+b+c)x^2+O(x^3)}{x^2+O(x^3)}.
$$
This implies that if $a\neq c$ then the limit would be not finite. Therefore
$$
a=c=\frac{b}{2}.
$$
In such case we would obtain that the limit is finite and it is exactly
$$
\frac{1}{2}(a+b+c)=2a.
$$
Hence, the unique triple $(a,b,c)$ which satisfies the relation is $(1,2,1)$.
| {
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Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$ Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$
We have to show that :
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$
I made three equations using cross multiplication :
$$1.~~x^{c-a}=y^{b-c}$$
$$2.~~y^{a-b}=z^{c-a}$$
$$3.~~z^{b-c}=x^{a-b}$$
How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents.
Thank you.
| We have
$$\dfrac{\log(x)}{b-c} = \dfrac{\log(y)}{c-a} = \dfrac{\log(z)}{a-b} = t$$
This gives us
$$x=e^{t(b-c)}, y = e^{t(c-a)} \text{ and }z = e^{t(a-b)}$$
Hence,
\begin{align}
x^{b+c-a}\cdot y^{c+a-b} \cdot z^{a+b-c} & = e^{t\left((b-c)(b+c-a) + (c-a)(c+a-b) + (a-b)(a+b-c)\right)}\\
& = e^{t(b^2-c^2-ab+ac + c^2 - a^2 -bc + ba + a^2 - b^2 - ac + bc)} = e^0 = 1
\end{align}
| {
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Prove that $a^2+b^2+c^2\geq [2(a-b)^2(b-c)^2(a-c)^2]^{1/3}$ Mathematica seems to know that this statement is true, yet I am struggling to prove it. Possible useful inequalities are Minkowski and the geometric mean. Using the geometric mean inequality I can prove a less strict bound:
$$(a-b)^2+(b-c)^2+(a-c)^2 \geq 0 \implies 4 a^2+4 b^2+4 c^2 \geq (a-b)^2+(b-c)^2+(a-c)^2,$$
then the geometric mean inequality gives
$$ (a-b)^2+(b-c)^2+(a-c)^2 \geq 3[(a-b)^2(b-c)^2(a-c)^2]^{1/3} .$$
Combining the two we get
$$ a^2+b^2+c^2 \geq \frac{3}{4}[(a-b)^2(b-c)^2(a-c)^2]^{1/3},$$
which is not strong enough.
Can anyone prove this for $2^{1/3}$ in place of $3/4$?
| Here is another approach that might shed a bit more light on the essence of the inequality.
The AM-GM inequality gives
$$
\begin{align}
a^2+b^2+c^2
&=\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a+b+c)^2}3\\
&\ge\frac{(a-b)^2+(b-c)^2+(c-a)^2}3\\
&\ge\left[(a-b)^2(b-c)^2(c-a)^2\right]^{1/3}\tag{1}
\end{align}
$$
Equality is attained in the second inequality of $(2)$ only when $(a-b)^2=(b-c)^2=(c-a)^2$, which cannot happen since $(a-b)+(b-c)+(c-a)=0$. This indicates that a better inequality may be possible.
If we set $x=a-b$ and $y=b-c$, we can look for the maximum of
$$
\frac{x^2y^2(x+y)^2}{\left(x^2+y^2+(x+y)^2\right)^3}=\frac{z^2(z+1)^2}{\left(z^2+1+(z+1)^2\right)^3}\tag{2}
$$
where $z=x/y$ ($y=0$ is certainly not the maximum). The derivative of the log of $(2)$ is
$$
\frac{2(2z+1)}{z(z+1)}-\frac{6(2z+1)}{z^2+1+(z+1)^2}=\frac{(2+z)(1-z)(1+2z)}{z(1+z)(1+z+z^2)}\tag{3}
$$
which vanishes where $z=-\frac12$, $z=1$, and $z=-2$. Each gives a value of $\frac1{54}$ in $(2)$. Therefore,
$$
\left(x^2+y^2+(x+y)^2\right)^3\ge54x^2y^2(x+y)^2\tag{4}
$$
Substituting back for $x$ and $y$ in $(4)$ gives
$$
\frac{(a-b)^2+(b-c)^2+(c-a)^2}3\ge\left[2(a-b)^2(b-c)^2(c-a)^2\right]^{1/3}\tag{5}
$$
To get the best inequality, we can use $(5)$ instead of the AM-GM:
$$
\begin{align}
a^2+b^2+c^2
&=\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a+b+c)^2}3\\
&\ge\frac{(a-b)^2+(b-c)^2+(c-a)^2}3\\[6pt]
&\ge\left[2(a-b)^2(b-c)^2(c-a)^2\right]^{1/3}\tag{6}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does matrix addition give you a matrix or a number? I am very confused by something our lecturer said today:
We were given two matrices:
$B=\begin{pmatrix}2 & 3\\ 2 &0 \\ 0&3\end{pmatrix}$
C=$\begin{pmatrix}6 &3&4\\6&6&0\end{pmatrix}$
$\space$
And we were supposed to find: $B_{ij}+C_{ji}$
To me, this seemed like another way of writing $B+C^T \implies \begin{pmatrix}2 & 3\\ 2 &0 \\ 0&3\end{pmatrix}+ \begin{pmatrix}6 & 6\\ 3 &6 \\ 4&0\end{pmatrix}=\begin{pmatrix}8 & 9\\ 5 &6 \\ 4&3\end{pmatrix}$
Our lecturer said that when adding the two matrices I don't get a matrix, I actually get an entry. She also added that the correct way of writing the answer would be: $$\begin{pmatrix}8 & 9\\ 5 &6 \\ 4&3\end{pmatrix}_{ij}$$
How is that not a matrix? Can anyone explain what she means by this?
| Usually when one writes $A_{ij}$, one means the entry of $A$ at position $(i,j)$. For example
$$\begin{pmatrix}6 &3&4\\6&6&0\end{pmatrix}_{2,2} = 6$$
| {
"language": "en",
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How to simplify the integration? If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows the $\tan^{-1}x$ but How to simplify if we have
$$\int \frac{1}{(1+x^2)^2}dx$$
| An idea that works for $\frac{1}{(1+x^2)^2}$ and for all other exponents $\frac{1}{(1+x^2)^n}$ is to get the base of the denominator in the numerator
$$\frac{1}{(1+x^2)^2}=\frac{1+x^2}{(1+x^2)^2}-\frac{x^2}{(1+x^2)}$$
The integral of the first term we know because it has a lower exponent and for the second we can write it as $$\frac{1}{2}x\frac{2x}{(1+x)^2}$$ and integrate by parts
$$\frac{1}{2}\int x\frac{2x}{(1+x^2)^2}=\frac{1}{2}x\frac{1}{(1+x^2)}+\frac{1}{2}\int\frac{1}{(1+x^2)}$$
which makes it boil down again to the integral with the smaller exponent, which we know.
Adding the two parts we get the answer $$\arctan(x)-\frac{1}{2}x\frac{1}{(1+x^2)}-\frac{1}{2}\arctan(x)=\frac{1}{2}\arctan(x)-\frac{1}{2}x\frac{1}{(1+x^2)}$$
| {
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Prove that $p\mid a^2+b^2\,\Rightarrow\, p\equiv 1\pmod{\! 4}$ Let a prime number $p$ divide $a^2+b^2$ with some $a,b \in \left\{ 1,2, \ldots , p-1 \right\}$ Prove that $p\equiv 1 \pmod{4}$. Is the converse true?
I know that $a^2+b^2\equiv 0 \pmod{p}$ and I don't know.
| Proof by exhaustion.
$$
p = a^2 + b^2 = 3 \pmod 4 \implies
$$
If
$$
p = 1^2 + 1^2 = 2, \\
p = 1^2 + 2^2 = 1, \\
p = 1^2 + 3^2 = 2, \\
p = 2^2 + 2^2 = 0, \\
p = 2^2 + 3^2 = 1, \\
p = 3^2 + 3^2 = 2, \\
\pmod 4
$$
Then we have a contradiciton. By symmetry of variables $a,b$ (namely commutativity of $+$) we have covered all possibilities.
$\blacksquare$
The other proofs are impossible to understand easily. This one is plain and simple.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help understanding how to factor completely $x^3-x^2-x+1$ I need someone to help explain the steps to completely factor the problem $x^3-x^2-x+1$.
Here is what I have done so far:
$x^3-x^2-x+1$
to
$x^3-x^2+-1(x+1)$
Since there is a negative I changed $-x+1$ to $-1(x+1)$
to
$x^2(x-1)+-1(x+1)$
to
$(x^2-1)(x-1)(x+1)$.
Is this correct? If not please explain what I did wrong and what I should do. If it is then thank you for saying so, if not, thank you for helping me to understand.
| Let $f(x)=x^3-x^2-x+1$. By Rational Root Theorem, any rational root $\frac{p}{q}$ of $g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ has $p\mid a_0$ and $q\mid a_n$ (i.e. $q$ divides $a_n$). Proof is simple: substitute $x=\frac{p}{q}$, multiply both sides by $q^n$, observe divisibility.
Since $f(1)=0$, we know $x^3-x^2-x+1=(x-1)P(x)$ for some $P(x)$, which of course must be quadratic, so $P(x)=x^2+ax+b$.
$$(x-1)(x^2+ax+b)=x^3+(a-1) x^2+(b-a)x-b=x^3-x^2-x+1$$
So $a-1=-1, b-a=-1, -b=1$, so $b=-1, a=0$.
$$x^3-x^2-x+1=(x-1)(x^2-1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\int\limits_{0}^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$ I have to find $$\int_0^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$$
I tried to do it by substitution $t = \tan(\frac{x}{2})$
Then we have that $$\cos(x) = \frac{1-t^2}{1+t^2} \quad dx = \frac{2\,dt}{1+t^2}$$ but then also limits of integration are changing so we have
$$\int\limits_{t(0)}^{t(2\pi)} \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2dt}{1+t^2} = \int\limits_0^0 \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2\,dt}{1+t^2} = 0$$
I figured out that it is not correct because $\tan(\frac{\pi}{2})$ is not defined and $t(\pi) = \tan(\frac{\pi}{2})$ and $\pi \in [0, 2\pi]$. How can I "repair" that and do it right?
| We have
$$\dfrac1{5-3\cos(x)} = \dfrac15 \cdot \dfrac1{1-\dfrac{3\cos(x)}5} = \dfrac15 \sum_{k=0}^{\infty} \left(\dfrac35\right)^k \cos^k(x)$$
Hence, the integral becomes
$$I = \underbrace{\sum_{k=0}^{\infty} \dfrac15 \cdot \left(\dfrac35\right)^k \int_0^{2\pi} \cos^k(x)dx = \sum_{k=0}^{\infty} \dfrac15 \cdot \left(\dfrac35\right)^{2k} \int_0^{2\pi} \cos^{2k}(x)dx}_{\text{Integral of odd powers of $\cos$ over $[0,2\pi]$ vanish}}$$
From here, we have that
$$\int_0^{2\pi} \cos^{2k}(t) = 4\int_0^{\pi/2} \cos^{2k}(t) = \dfrac{\pi}{2^{2k-1}} \dbinom{2k}k$$
Further, we have
$$\sum_{k=0}^{\infty} \dbinom{2k}k x^k = \dfrac1{\sqrt{1-4x}}$$
Hence, we obtain
$$I = \dfrac{2\pi}5 \sum_{k=0}^{\infty} \dbinom{2k}k \left(\dfrac3{10}\right)^{2k} = \dfrac{2\pi}5 \cdot \dfrac1{\sqrt{1-4 \cdot 9/100}} = \dfrac{2\pi}5 \cdot \dfrac54 = \dfrac{\pi}2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluating $\lim_{x\to -\infty}\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2} \right )$ For some reason I fail to evaluate this (apparently) simple limit:
$$\lim_{x\to -\infty}\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2} \right )$$
I tried conjugate multiplication* however it didn't work for me. I thought about sandwiching, but I don't see how to do it here. I was also trying to evaluate it as a composition of functions but with no luck. Any suggestions?
*
Conjugate multiplication gives
$$\frac{2x}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$
I tried working with this by factoring and canceling things out but it didn't work.
| Conjugate multiplication will help and what you got is correct.
$$\lim_{x\to -\infty}\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$$$$=\lim_{x\to -\infty}(\sqrt{1+x+x^2}-\sqrt{1-x+x^2})\cdot\frac{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$
$$=\lim_{x\to -\infty}\frac{2x}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$
(here setting $t=-x$ gives you)
$$=\lim_{t\to \color{red}{+}\infty}\frac{-2t}{\sqrt{1-t+t^2}+\sqrt{1+t+t^2}}$$
$$=\lim_{t\to\infty}\frac{-2}{\sqrt{\frac{1}{t^2}-\frac{1}{t}+1}+\sqrt{\frac{1}{t^2}+\frac 1t+1}}$$
$$=\frac{-2}{1+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Polynomial satisfying $f(x) = f'(x) \cdot f''(x)$ If a polynomial of degree $n$ satisfies $f(x) = f'(x)\cdot f''(x)$ (such that $n$ belongs to $\mathbb R$) then $f(x)$ is?
A) an onto function
B) an into function
C) no such function possible
D) even function
I tried this question by letting a polynomial $f(x) = ax^n + bx^{n-1} \cdots$ and then derivated it but it did not help much , then i assumed a quadratic expression $ax^2 + bx + c$ and got $ax^2 + x(b-4a^2)$.
| Suppose that $n\geq2$. Then we have that the degree of the left side is $n$, but that of the right side is $(n-1)+(n-2)=2n-3$.
So we want to solve $2n-3=n$, or $n=3$.
Then write $f(x)=ax^3+bx^2+cx+d$ with $a \neq 0$, then we have $f'(x)= 3ax^2+2bx+c$ and $f''(x)= 6ax+2b$. Thus $$f'(x)f''(x) = (3ax^2+2bx+c)(6ax+2b) = 18a^2x^3+18abx^2+(6ac+4b^2)x+2bc$$
Now we have $18a^2=a$, thus $a = \frac{1}{18}$. The second coefficient doens't give us anything. The third gives $\frac{1}{3}c+4b^2=c$, and thus $b^2=\frac{1}{6}c$. Furthermore we have $2bc=d$, so $12b^3=d$. This gives the following equation for $f(x)$:
$$f(x)=\frac{1}{18}x^3 + bx^2+ 6b^2x+12b^3$$
for some $b \in \mathbb R$.
If $n$ is 1, the LHS is a linear function and the RHS a constant.
If $f$ is a constant unequal to zero, the RHS is a constant equal to zero.
This leaves only the zero polynomial $f(x)=0$, that indeed statistifies the equation.
Conclusion:
The only polynomials that statistify the given equations are $f(x)=0$ and
$$f(x)=\frac{1}{18}x^3 + bx^2+ 6b^2x+12b^3$$
for some $b \in \mathbb R$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the eccentricity of the elliptical section of a right circular cone A right circular cone, with the apex angle $\alpha=60^{o}$, is thoroughly cut with a smooth plane inclined at an acute angle $\theta=70^{o}$ with its geometrical axis to generate an elliptical section (As shown in the diagram) .
How to calculate the eccentricity of the elliptical section generated?
Is there any set formula to calculate eccentricity for the similar case i.e. eccentricity, $e$ as the function $\alpha$ & $\theta$ i.e. $e=f(\alpha, \theta)$?
| Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $\theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $\beta$ so that $\beta = \alpha/2$ (this will help simplify the calculations later).
Clearly in $\Delta AOC$ we have $$\frac{OA}{\sin \angle OCA} = \frac{AC}{\sin \angle AOC}$$ so that $$b = AC = \dfrac{\sin \theta}{\sin(\theta + \beta)}\tag{1}$$ Similarly $$c = AB = \frac{\sin \theta}{\sin(\theta - \beta)}\tag{2}$$ and $$a = BC = \frac{b\sin 2\beta}{\sin(\theta - \beta)} = \frac{\sin\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}\tag{3}$$ Let the incircle of $\Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $\Delta ABC$ then we know that $$\tan \frac{C}{2} = \frac{r}{CD}$$ so that $$CD = \frac{r}{\tan(C/2)} = \dfrac{r}{\tan\left(\dfrac{\pi - \theta - \beta}{2}\right)} = r\tan\left(\frac{\theta + \beta}{2}\right)\tag{4}$$ Note further that the inradius $r$ is given by the formula
\begin{align}
r &= \frac{\Delta}{s}\notag\\
&= \frac{bc\sin 2\beta}{a + b + c}\notag\\
&= \dfrac{\dfrac{\sin^{2}\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}}{\dfrac{\sin\theta\{\sin(\theta + \beta) + \sin(\theta - \beta) + \sin 2\beta\}}{\sin(\theta + \beta)\sin(\theta - \beta)}}\notag\\
&= \frac{\sin\theta\sin 2\beta}{2\sin\theta\cos\beta + 2\sin\beta\cos\beta}\notag\\
&= \frac{2\sin\theta\sin\beta\cos\beta}{2\cos\beta\{\sin\theta + \sin\beta\}}\notag\\
&= \frac{\sin\theta\sin\beta}{\sin\theta + \sin\beta}\tag{5}
\end{align} If $e$ is the eccentricity then we know that $$e = \frac{\text{distance of focus from center}}{\text{length of semi major axis}} = \frac{(a/2) - CD}{a/2} = \frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as
\begin{align}
e &= \dfrac{a - 2r\tan\left(\dfrac{\theta + \beta}{2}\right)}{a}\notag\\
&= 1 - \frac{2r}{a}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\
&= 1 - \frac{2\sin\theta\sin\beta}{\sin\theta + \sin\beta}\cdot\frac{\sin^{2}\theta - \sin^{2}\beta}{2\sin\theta\sin\beta\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\
&= 1 - \frac{\sin\theta - \sin\beta}{\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\
&= 1 - \dfrac{2\cos\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\
&= 1 - \dfrac{2\sin\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\notag\\
&= 1 - \frac{\cos \beta - \cos \theta}{\cos \beta}\notag\\
&= \frac{\cos \theta}{\cos \beta} = \frac{\cos\theta}{\cos(\alpha/2)}\tag{6}
\end{align}
Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < \beta = \alpha / 2 < \theta \leq \pi/2$. If $\beta = \alpha/2 = \theta$ then the section becomes a parabola and if $\beta = \alpha/2 > \theta$ then the section becomes a hyperbola. For the angles given in question we have $e = \cos 70^{\circ}/\cos 30^{\circ} = 0.39493084\ldots$.
Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = \cos\theta/\cos\beta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is indefinite $ \int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - x^{4}}}{\rm d}x . $ $$
\int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - x^{4}}}{\rm d}x .
$$
Hello.
Today, in exam, we had to evaluate this integral. Noone was able to do it. Appreciate any help.
| Hint:
$\int\dfrac{\sin \cos x}{\sqrt{3-x^4}}dx$
$=\int\dfrac{\sin2x}{2\sqrt{3-x^4}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nx^{2n+1}}{(2n+1)!\sqrt{3-x^4}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n-1}x^{2n}}{(2n+1)!\sqrt{3-x^4}}d(x^2)$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n-1}u^n}{(2n+1)!\sqrt{3-u^2}}du$ $(\text{Let}~u=x^2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality - L'Hôpital's rule, logarithms Let's consider a function: (EDIT here: the number "2" was lacking)
$\mathcal{B}(x) = \frac{2}{x^2} \big( (1+x) log(1+x) - x \big)$
I need to prove inequality:
$ (1 + \frac{1}{3}x)\, \mathcal{B}(x) \geq 1 \, , \qquad x>0 \, ,$
using a hint: "Apply L'Hôpital's rule twice to reduce left-hand side to
$ \frac{1 + \frac{1}{3}x ^\ast}{1 + x^\ast} + \frac{2}{3} log (1 + x^\ast)$
for some $x^\ast$ less that $x$. Then use: $ \quad log(1+x^\ast) \geq \frac{x^\ast}{1 + x^\ast} \,$".
To be honest, I need a hint how to make use of this hint.
(EDIT:)
What I can do is compute:
$f(x) = (1 + \frac{1}{3}x) \big( (1+x) log(1+x) - x \big) \\
f'(x) = \frac{1}{3} \big( (1+x) log(1+x) - x \big) + (1 + \frac{1}{3}x)log(1+x) \\
f''(x) = \frac{(1 + \frac{1}{3}x)}{1+x} + \frac{2}{3}\log(1+\frac{1}{3}) \\
g(x) = \frac{1}{2}x^2\\
g'(x) = x \\
g''(x) = 1\\
\underset{x \rightarrow 0^+}{\lim}f(x)=\underset{x \rightarrow 0^+}{\lim}f'(x)=\underset{x \rightarrow 0^+}{\lim}g(x)=\underset{x \rightarrow 0^+}{\lim}g'(x)=0 \\
\underset{x \rightarrow 0^+}{\lim}f''(x) = 1 = \underset{x \rightarrow 0^+}{\lim}g''(x)$
Since all of functions above are continuous and differentiable on $(0,\infty)$, limits of $f$, $f$, $g$, $g'$ when $x \rightarrow 0^+$ are equal to $0$ and limits of $f''$ and $g''$ are finite and $g''(x) \ne 0$, we obtain:
$$ \underset{x \rightarrow 0^+}{\lim}(1 + \frac{1}{3}x)\, \mathcal{B}(x) = 1 $$
But I don't understand the part of hint about "reducing the left hand side" to prove the inequality. Taking what I computed above and showing that the function is non-decreasing would be sufficient. But I want to understand method suggested by hint.
| There is something strange in this problem, since, built at $x=0$, a Taylor expansion gives $$\mathcal{B}(x) = \frac{1}{x^2} \big( (1+x) \log(1+x) - x \big)=\frac{1}{2}-\frac{x}{6}+\frac{x^2}{12}-\frac{x^3}{20}+O\left(x^4\right)$$ Applying L'Hopital twice, you end with $$\frac{1}{2 (x+1)}$$ which just confirms the limit at $x=0$.
Then $$f(x)=(1 + \frac{x}{3})\, \mathcal{B}(x)=\frac{1}{2}+\frac{x^2}{36}-\frac{x^3}{45}+O\left(x^4\right)$$ Function $f(x)$ is continuously increasing and $f(x) >1$ only if $x>39.1242$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac {\sin^4\theta}a +\frac {\cos^4\theta}b=\frac 1 {a+b}$ P.T $\frac {\sin^8\theta}{a^3} + \frac {\cos^8\theta} {b^3} = \frac1{(a+b)^3}$ The question reads if
$$\frac {\sin^4\theta} {a} + \frac {\cos^4\theta} {b} = \frac 1 {a+b}$$
Then prove that
$$\frac {\sin^8\theta} {a^3} + \frac {\cos^8\theta} {b^3} = \frac 1 {(a+b)^3}$$
I know simply cubing the first equation would not work. I have also tried arriving at the solution using various identities like $ \sin^2\theta + \cos^2\theta = 1$ but to no avail. I would also like to know the approach one should use while solving such trigonometry questions.
I am an eleventh grader so sorry if this question is too silly.
| Let $\sin^2 \theta = k$
Then our equation is.
$$\frac{k^2}{a}+\frac{(1-k^2)^2}{b}=\frac{1}{a+b}$$
I won't solve the whole equation but i will put in some steps.
We will be getting something messy like:
$$k^2ab+k^2b^2+a^2+a^2k^2-2a^2k^2+ab+abk^2-2abk=ab$$
And the best part is this all simplifies to:
$$(k\times (a+b)-a)^2=0$$
Hence $k=\frac{a}{a+b}=\sin^2 \theta$
$(1-k)=\frac{b}{a+b}=\cos^2 \theta$
Let us prove the general case. $$\frac{\sin^{4n}\theta}{a^{2n-1}}+\frac{\cos^{4n} \theta}{b^{2n-1}}=\frac{1}{(a+b)^{2n-1}}$$
Substituting our values on LHS.
$$\frac{1}{a^{2n-1}}\times \left(\frac{a}{a+b}\right)^{2n}+\frac{1}{b^{2n-1}}\times \left(\frac{b}{a+b}\right)^{2n}$$
$$\frac{a}{(a+b)^{2n}}+\frac{b}{(a+b)^{2n}}=\frac{1}{(a+b)^{2n-1}}=RHS$$ Hence result is true $\forall$ $n \in N$
| {
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How to prove $\lim\limits_{(x,y)\to(0,0)}\frac{{x^3{y^2}}}{{{x^4} + {3y^4}}} = 0$? To prove that $$\lim\limits_{(x,y)\to(0,0)}\frac{{x^3{y^2}}}{{{x^4} + {3y^4}}} = 0$$
I start with
$$\left| {\frac{{{x^3}{y^2}}}{{{x^4} + 3{y^4}}}} \right| \leqslant \left| {\frac{{{x^3}{y^2}}}{{{x^4}}}} \right| = \left| {\frac{{{y^2}}}{x}} \right|.$$
But I do not know how to show $| {\frac{{{y^2}}}{x}}|$ is bounded using the hypothesis that $0<|x|<\delta$, $0<|y|<\delta$ and $0<\sqrt{x^2+y^2}<\delta$ since the quarters powers of $x$ and $y$ are very difficult to manage. Even setting $\delta=1$ gets me nowhere.
| Another idea is using polar coordinates. Then
\begin{align}
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^4+3y^4} &= \lim_{r \to 0^+} \frac{r^5\cos^3x\sin^2x}{r^4\cos^4x+3r^4\sin^4x} \\
&=\lim_{r \to 0^+}r \frac{\cos^3x\sin^2x}{\cos^4x + 3\sin^4x}
\end{align}
Now, since
$$0 \leq \cos^3x\sin^2x \leq 1$$
and
$$\frac{3}{4} \leq \cos^4x+3\sin^4x \leq 3$$
We have
$$\lim_{r \to 0^+}0 \cdot r \leq \lim_{r \to 0^+}r \frac{\cos^3x\sin^2x}{\cos^4x + 3\sin^4x} \leq \lim_{r \to 0^+} \frac{4r}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $\mathbb Q[x]$ Trying to prove that the following polynomial is irreducible in $\mathbb Q[x]$:
$x^4-16x^3+20x^2+12$
What I have tried:
1.) Eisenstein's Criterion, but there exists no suitable prime.
2.) reducing to modulo 2, 3, 5 ,7, 11, but by my calculations, reduction to mod 2, mod 3, mod 5, mod 7, yields a reducible polynomial. Mod 11 seems like it could potentially work, but I can't believe that would be the correct approach, given the sheer number of potential quadratic factors one would have to check.
3.) This polynomial fails the rational roots test, so I know that the only possible factors would involve second degree polynomials. Guided by some of the previous posts on related questions, I have attempted to work out some type of contradiction by assuming the polynomial can be factored like $(x^2+ax+b)(x^2+cx+d)$, but I haven't had much luck with this approach.
I imagine I'm staring at something obvious but can not see it. Any help would be appreciated.
| I am here to prove by Contradiction.
Assume $x^4-16x^3+20x^2+12$ can be factorized, so we have:
\begin{align}
x^4-16x^3+20x^2+12&=(x^2+ax+b)(x^2+cx+d)\\&=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd
\end{align}
By comparing similar terms, we have:
$$ad+bc=0------(1)$$
$$ac+b+d=20----(2)$$
$$a+c=-16-----(3)$$
$$bd=12-------(4)$$
$d(1),(4)$:
$$ad^2+bdc=0$$
$$d=\pm{\sqrt{\frac{-12c}a}}$$
For $d\in{\Bbb{Q}}, \exists{k\in{\Bbb{Q}}\text{\0}}$, $\frac{c}a=-3k^2----(5)$,
so $d=\pm{6k}-----(6)$
Sub (5),(6) into $(1)/a$,
$$\pm{6k}+b(-3k^2)=0$$
$$b=\pm\frac2k$$
Sub (5) into (3),
$$a-3ak^2=-16$$
$$a=\frac{16}{3k^2-1}$$
$$c=-3ak^2=\frac{-48k^2}{3k^2-1}$$
Sub $a,b,c,d$ to original expression,
\begin{align}
(x^2+ax+b)(x^2+cx+d)&=(x^2+\frac{16}{3k^2-1}x\pm\frac2k)(x^2+\frac{-48k^2}{3k^2-1}x\pm{6k})\\&=((3k^2-1)x^2+16x\pm{\frac{2(3k^2-1)}{k}})((3k^2-1)x^2-48k^2\pm{6k(3k^2-1)})
\end{align}
So
$$12(3k^2-1)^2=12$$
$$3k^2=\pm{1}+1$$
$$k^2=\frac23\text{ or } 0 \text{ (rej.) }$$
$$k=\pm{\sqrt{\frac23}}\notin{\Bbb{Q}}$$
Recall
For $d\in{\Bbb{Q}\text{\0}}, \exists{k\in{\Bbb{Q}}}$
Contradiction!
$x^4-16x^3+20x^2+12$ cannot be factorized into expression in which $k\in{\Bbb{Q}}$.
Therefore the irreducibility of $x^4-16x^3+20x^2+12$ is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Jaccard dissimilarity and the triangle inequality Suppose that $\delta(A, B) = \dfrac{|A \Delta B|}{|A \cup B|}$, where $\Delta$ represents symmetric difference. Then how does one prove the triangle inequality, viz that $\delta(A, B) + \delta(B, C) \ge \delta(A, C)$?
| If I understand you correctly, the objects you are considering are finite sets and your dissimilarity function is defined as $d(A, B) = \frac{|A \ominus B|}{|A \cup B|}$, where $\ominus$ denotes the symmetric difference.
Consider the Venn diagram for three sets $A$, $B$, and $C$.
Then the triangle inequality boils down to
$$
\frac{a+b+e+f}{a+b+d+e+f+g} + \frac{b+c+d+f}{b+c+d+e+f+g} \geq \frac{a+c+d+e}{a+c+d+e+f+g}
$$
Plugging this into Wolfram Alpha, we see that it "simplifies" to:
$$
\frac{a^2 b + a b^2 + a^2 c + 2 a b c + b^2 c + a c^2 + b c^2 + a^2 d + 2 a b d + b^2 d + 2 a c d + 2 b c d + a d^2 + b d^2 + 2 a b e + b^2 e + 2 a c e + 2 b c e + c^2 e + a d e + 2 b d e + c d e + b e^2 + c e^2 + a^2 f + 4 a b f + 2 b^2 f + 4 a c f + 4 b c f + c^2 f + 4 a d f + 5 b d f + 3 c d f + 2 d^2 f + 3 a e f + 5 b e f + 4 c e f + 4 d e f + 2 e^2 f + 3 a f^2 + 4 b f^2 + 3 c f^2 + 4 d f^2 + 4 e f^2 + 2 f^3 + 2 a b g + 2 b^2 g + 2 a c g + 2 b c g + a d g + 3 b d g + 3 b e g + c e g + 3 a f g + 6 b f g + 3 c f g + 4 d f g + 4 e f g + 4 f^2 g + 2 b g^2 + 2 f g^2}{(a + b + d + e + f + g) (a + c + d + e + f + g) (b + c + d + e + f + g)} \geq 0
$$
Since all the coefficients in the above expression are positive, and $a, b, c, d, e, f, g \geq 0$, we conclude that the inequality always holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Partial fractions for $\left (\frac{z^2+1}{z^2-1} \right )^2$ I am trying to decompose this function: $$\left (\frac{z^2+1}{z^2-1} \right )^2$$ into partial fractions.
What I've done up to now is this:
$$\frac{z^2+1}{(z-1)(z+1)}=\frac{A}{z-1}+\frac{B}{z+1}$$
Which doesn't give me the correct $A$ and $B$.
What is wrong with the not-squared experssion?
| $$\frac{z^2+1}{z^2-1}=1+\frac{2}{z^2-1}=1+\frac{1}{z-1}-\frac{1}{z+1}\tag{1}$$
hence by squaring the previous line:
$$\left(\frac{z^2+1}{z^2-1}\right)^2 = 1+\frac{1}{(z-1)^2}+\frac{1}{(z+1)^2}+\frac{2}{z-1}-\frac{2}{z+1}-\frac{2}{z^2-1}\tag{2}$$
so, using $(1)$ again:
$$\left(\frac{z^2+1}{z^2-1}\right)^2 = 1+\frac{1}{(z-1)^2}+\frac{1}{(z+1)^2}+\frac{1}{z-1}-\frac{1}{z+1}.\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Integral solutions to $56u^2 + 12 u + 1 = w^3$ I would like to find all integer solutions to $$56u^2 + 12 u + 1 = w^3.$$ My computer thinks the only integral point is $(0,1).$
This problem arises from Integer solutions of $x^3 = 7y^3 + 6 y^2+2 y$? and is likely to be easier.
Note after getting answers: comes down to the Mordell curve $Y^2 = X^3-980$, solutions below.
E_-00980: r = 2 t = 1 #III = 1
E(Q) = <(14, 42)> x <(126, 1414)>
R = 1.4319518662
10 integral points
1. (14, 42) = 1 * (14, 42)
2. (14, -42) = -(14, 42)
3. (21, 91) = -2 * (14, 42)
4. (21, -91) = -(21, 91)
5. (326, 5886) = 3 * (14, 42)
6. (326, -5886) = -(326, 5886)
7. (29, 153) = 1 * (14, 42) - 1 * (126, 1414)
8. (29, -153) = -(29, 153)
9. (126, 1414) = 1 * (126, 1414)
10. (126, -1414) = -(126, 1414)
| Rewrite as $56u^2+12u+(1-w^3)=0$
This has roots $u=\frac {-12 \pm \sqrt{144-224(1-w^3)}} {112}$
$u=\frac {-12 \pm \sqrt{144-224+224w^3}} {112}$
$u=\frac {-12 \pm \sqrt{224w^3-80}} {112}$
$u=\frac {-12 \pm 4\sqrt{14w^3-5}} {112}$
$u=\frac {-3 \pm \sqrt{14w^3-5}} {28}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find the Range of the function $f(x) = |x-6|+x^2-1$
find the Range of $f(x) = |x-6|+x^2-1$
$$ f(x) = |x-6|+x^2-1 =\left\{
\begin{array}{c}
x^2+x-7,& x>0 .....(b) \\
5,& x=0 .....(a) \\
x^2-x+5,& x<0 ......(c)
\end{array}
\right.
$$
from eq (b) i got $$f(x)= \left(x+\frac12\right)^2-\frac{29}4 \ge-\frac{29}4$$
and from eq (c) i got $$f(x)= \left(x-\frac12\right)^2+\frac{19}4 \ge\frac{19}4$$
and eq(b) tells me that it also passes through 5 and so generalize all this and found its range is $\left[-\frac{29}4 , \infty\right)$
but the graph says its range is $(5, \infty)$
| No derivatives are necessary. For $x\ge6$ the function is
$$
f(x)=x^2+x-7
$$
The graph is an arc of a parabola with its axis at $x=-1/2$, so in this interval the function is increasing, with its minimum at $6$: $f(6)=35$.
For $x<6$ the function is
$$
f(x)=x^2-x+5
$$
whose graph is an arc of a parabola with its axis at $x=1/2$. Since $1/2<6$ the minimum of $f$ is reached at $1/2$: $f(1/2)=19/4$.
Thus the range is $(19/4,\infty)$ because, clearly, $\lim_{x\to\infty}f(x)=\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Multiplying a floor function to a number Is it correct to write:
$\cfrac{\left\lfloor{\cfrac{\pi y^2}{3\sqrt{3}x^2}}\right\rfloor}{n} \times\sqrt{3}x =\left\lfloor\cfrac{\pi y^2}{3xn}\right\rfloor$ ?
| No, it is not correct.
For $x,y$ such that $\left\lfloor\frac{\pi y^2}{3\sqrt 3 x^2}\right\rfloor\not=0$, the LHS is irrational and the RHS is rational.
Added :
when could this be true?
We need $$\left\lfloor\frac{\pi y^2}{3\sqrt 3 x^2}\right\rfloor=\left\lfloor\frac{\pi y^2}{3xn}\right\rfloor=0,$$
i.e.
$$0\le \frac{\pi y^2}{3\sqrt 3 x^2}\lt 1\ \ \ \text{and}\ \ \ 0\le \frac{\pi y^2}{3xn}\lt 1,$$
i.e.
$$\pi y^2\lt 3\sqrt 3x^2\ \ \ \text{and}\ \ \ \pi y^2\lt 3xn.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Express $w=f(z)=\frac{1}{(1-z)^2}$ in the form $w=u(x,y)+iv(x,y)$ I start by writing $f(z)$ as $$\frac{1}{(1-(x+iy))^2}$$
and then I expand the bottom to get $$\frac{1}{(1-2x+x^2-y^2) + i(2y-2xy)}$$
The answer says $$w=\frac{(1+x^2-2x-y^2)-(i(2xy-2y)}{(1+x^2-2x-y^2)^2+(2xy-2y)^2}$$
How do I get to this stage?
| You can simplify this process a bit:
$$\begin{align}
\frac1{(1-z)^2} & =\left(\frac1{1-z}\right)^2=\left(\frac{1-\overline z}{(1-z)(1-\overline z)}\right)^2\\
& = \left({1-\overline z\over 1-z-\overline z+|z|^2}\right)^2\\
& = \left({1-x+iy\over 1-2x+x^2+y^2}\right)^2,\quad\text{since }z+\overline z=2x\\
& = {1-2x+x^2-y^2+2iy(1-x)\over(1-2x+x^2+y^2)^2}\\
& ={1-2x+x^2-y^2\over(1-2x+x^2+y^2)^2}+{2y(1-x)\over(1-2x+x^2+y^2)^2}i.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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how to prove by contradiction that any distance between a curve $x^4 - x^2 + y^4 - y^2 = 0$ and the origin is less than or equal to $\sqrt{2}$ Given a closed trajectory $x^4 - x^2 + y^4 - y^2= 0$
Prove that any distance between any point on the curve and the origin does not exceed $\sqrt2$ (ie, maximum distance from the origin to the curve is $\sqrt2$)
I proved it using polar transformations but i'd rather use proof by contradiction without using polar transformation, so i tried
$$\begin{array}{l}
\left( {\exists \alpha ,\beta } \right){\rm{ }}{\alpha ^2} + {\beta ^2}>2\\
{\alpha ^4} - {\alpha ^2} + {\beta ^4} - {\beta ^2} = 0\\
\Rightarrow {\alpha ^4} + {\beta ^4} = {\alpha ^2} + {\beta ^2}\\
\Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} - 2{\left( {\alpha \beta } \right)^2} = {\alpha ^2} + {\beta ^2}{\rm{ }}\left( {{\rm{Completing - Square}}} \right)\\
\Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} = {\alpha ^2} + {\beta ^2} + 2{\left( {\alpha \beta } \right)^2}\\
\Rightarrow {\left( {a + b} \right)^2} = a + b + 2ab \le 2\left( {{a^2} + {b^2}} \right){\rm{ }}\left( {{\rm{Cauchy - Schwarz - Inequality}}} \right)\\
\Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + \left( {{a^2} + {b^2}} \right) - 2ab\\
\Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + {\left( {a - b} \right)^2}\\
\Rightarrow {\left( {a - b} \right)^2} \ge \left( {a + b} \right) - \left( {{a^2} + {b^2}} \right) \ge 0\\
\Rightarrow \left( {a + b} \right) \ge \left( {{a^2} + {b^2}} \right)\\
\Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) \ge \left( {{\alpha ^4} + {\beta ^4}} \right)\\
\Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) - \left( {{\alpha ^4} + {\beta ^4}} \right) \ge 0
\end{array}
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$$
Since the last equation includes 0, I cannot show any contradiction.
I think the problem is that I did not use my hypothesis, ${\alpha ^2} + {\beta ^2}>2.
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$
I think this claim can be proven using contradiction by I'm just not sure how to use my hypothesis.
| An easier solution. Use Lagrange multipliers for $f(x,y) = x^2+y^2$ with the constraint $$g(x,y) = x^4 - x^2 + y^4 - y^2 = 0.$$ Putting $\nabla f(x,y)$ and $\nabla g(x,y)$ in rows: $$\begin{vmatrix} x & y \\ 2x^3 - x & 2y^3-y\end{vmatrix}=0 \implies xy^2-yx^3=0.$$ If $x \neq 0$ and $y \neq 0$ (the trivial case), we get $y = \pm x$. Back to $g(x,y) = 0$. $$x^4-x^2+x^2-x^2 = 0 \implies 2x^4-2x^2=0 \implies x^2-1 = 0 \implies x = \pm 1.$$ So we get the points $(1,1)$ and $(1,-1)$. And $f(1,1) = f(1,-1)= 2$. So the square of the maximum distance is $2$, hence the maximum distance is $\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor series $\ln(1+e^x)$ about $x=0$ What's the best way to determine this up $x^3$ terms?
I thought it would be to take the series for $\ln(1+x)$ and the series for $e^x$ up to $x^3$ and sub the second series into the first.
$$\ln(1+x) = x - \frac {x^2}{2} +\frac {x^3}{3} + \cdots$$
$$e^x = 1+ x + \frac {x^2}{2} +\frac {x^3}{6} + \cdots$$
This gave me $$\frac 5{6} + x + \frac {3x^2}{2} + \frac {2x^3}{3}$$
but this is not the answer in my book. help?
| Ok, we have to compute three Taylor coefficients. So, let us compute them all.
We may notice that:
$$\frac{d}{dx}\log(1+e^x)=\frac{1}{2}+\frac{1}{2}\tanh\frac{x}{2}\tag{1} $$
and since:
$$\cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{2}$$
by considering the logarithmic derivative we have:
$$\tanh(x/2) = \sum_{n\geq 0}\frac{4x}{(2n+1)^2\pi^2+x^2}\tag{3} $$
and by expanding every term in the RHS as a geometric series we get:
$$\tanh(x/2) = \sum_{n\geq 0}(-1)^n \frac{(4-4^{-n})\cdot\zeta(2n+2)}{\pi^{2n+2}} x^{2n+1} \tag{4}$$
so:
$$\log(1+e^x)=\log 2+\frac{x}{2}+\sum_{n\geq 0}(-1)^n \frac{(4-4^{-n})\cdot\zeta(2n+2)}{(2n+2)\,\pi^{2n+2}} x^{2n+2}\tag{5}$$
from which it follows that, in a nieghbourhood of the origin:
$$\log(1+e^x)=\log 2+\frac{x}{2}+\frac{x^2}{8}+o(x^3).\tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Triangle in circumference of circle Points $A$, $B$, and $C$ are on the circumference of a circle with radius 2 such that $\angle BAC = 45^\circ$ and $\angle ACB = 60^\circ$. Find the area of $\triangle ABC$.
How would I start this problem? I think it has to do with the extended law of sines, but I'm not sure.
Thanks
| Use the sines law: $\frac{AB}{sin(C)}=\frac{AC}{sin(B)}=\frac{BC}{sin(A)}=2R=2\cdot 2=4$. So:
$$AB=4 sin(60^ {\circ})=2 \sqrt{3} $$
$$ BC=4sin(45^{\circ})=2 \sqrt{2} $$
Then: $S_{ \triangle ABC}=\frac{AB \cdot BC \cdot \sin (\widehat{ABC})}{2}=\frac{2\sqrt{2}\cdot 2\sqrt{3} \cdot \sin(75^{ \circ})}{2}=2\sqrt{6}\cdot \frac{1+\sqrt{3}}{2 \sqrt{2}}=\sqrt{3}+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coordinates of the center of the circle I am stuck on this problem:
If the lines $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ are two tangents of a
circle and $(0,\sqrt{2})$ lies on this circle then what is the equation of the circle?
I found out the distance between the two tangents $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ is $3$. The radius is $3/2$ but I don't know how to find the center. I tried forming the equations but could not succeed. Please tell me the easiest way. Thank you
| Let the given point be $A(0, \sqrt{2})$ which lines on the lies on the line: $y=x+\sqrt{2}$, Now draw the perpendicular say $AN$ from the point $A(0, \sqrt{2})$ to the parallel line: $y=x-2\sqrt{2}$ at the point N.
Thus point N is foot of perpendicular & $AN$ denotes the diameter of circle. Point $N$ is determined by a general expression derived in Reflection Formula by HCR for directly calculating the co-ordinates of foot of perpendicular say $(x', y')$ drawn from any point $(x_o, y_o)$ to the straight line $y=mx+c$ given as $$(x', y')\equiv\left(\frac{x_o+m(y_o-c)}{1+m^2}, \frac{mx_o+m^2y_o+c}{1+m^2}\right)$$ Hence, the co-ordinates of the foot of perpendicular $N$ drawn from the point $A(0, \sqrt{2})$ to the line: $y=x-2\sqrt{2}$ are calculated as follows $$N\equiv\left(\frac{0+1(\sqrt{2}-(-2\sqrt{2}))}{1+1^2}, \frac{1(0)+1^2(\sqrt{2})-2\sqrt{2}}{1+1^2}\right)\equiv\left(\frac{3}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$$ Now, the center of circle is the mid-point of the line AN joining $A(0, \sqrt{2})$ & $N\left(\frac{3}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$ which is given as $\left(\frac{3}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)$ & radius of circle is $\frac{3}{2}$ i.e. half the distance ($=3$) between parallel lines.
Hence, the equation of the circle with center $\left(\frac{3}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)$ & radius $\frac{3}{2}$ $$\left(x-\frac{3}{2\sqrt{2}}\right)^2+\left(y-\frac{1}{2\sqrt{2}}\right)^2=\left(\frac{3}{2}\right)^2 $$$$\implies \color{#0b4}{4x^2+4y^2-6x\sqrt{2}-2y\sqrt{2}-4=0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluate $\sin\left(-\frac{\pi}{6} + \frac{1}{2}\arccos\left(\frac{1}{3}\right)\right)$. My task is to evaluate $$\sin\left(-\frac{\pi}{6} + \frac{1}{2}\arccos\left(\frac{1}{3}\right)\right).$$
I think I've gotten most of the way there but I keep running into trouble... any suggestions?
| We very well know $$\cos\theta=2\cos^2\frac{\theta}{2}-1$$ $$\implies \cos\frac{\theta}{2}=\pm \sqrt{\frac{1+\cos\theta}{2}}$$ If $\theta\leq\pi$ $$\implies \cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}} $$
Now, let $\cos^{-1}\frac{1}{3}=\theta $ $ (\theta<\pi)$ or $\cos\theta=\frac{1}{3}$ then we have $$\implies \cos\frac{\theta}{2}=\sqrt{\frac{1+\frac{1}{3}}{2}}=\sqrt{\frac{2}{3}}$$ $$\frac{\theta}{2}=\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\implies \frac{1}{2}\cos^{-1}\left(\frac{1}{3}\right)=\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$$ Hence, we have $$\sin\left(-\frac{\pi}{6}+\frac{1}{2}\cos^{-1}\left(\frac{1}{3}\right) \right)=\sin\left(-\frac{\pi}{6}+\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)$$ $$=\sin\left(-\frac{\pi}{6}\right)cos\left(\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)+\cos\left(-\frac{\pi}{6}\right)\sin\left(\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)$$ $$=\left(-\frac{1}{2}\right)\left(\sqrt{\frac{2}{3}}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{3}}\right)$$ $$=\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Given $\csc\theta=-\frac53$ and $\pi<\theta<\frac32\pi$, evaluate sine ,cosine, and tangent of $2\theta$ If $\csc\theta=\frac{-5}{3}$, what is the exact value of $\tan(2\theta)$, $\sin(2\theta)$, and $\cos(2\theta)$ on the interval of $\left(\pi, \frac{3\pi}{2}\right)$?
I think I'm getting the fraction negatives wrong. I've used the sine formula($2\sin{\theta}\cos{\theta}$), the cosine formula ($2\cos^2{\theta}-1$) and the tangent formula, $\left(\frac{\sin(2\theta)}{\cos(2\theta)}\right)$
originally answering the problem I got $\sin(2\theta)=\frac{24}{25}$,
$\cos(2\theta)=\frac{7}{25}$, and $\tan(2\theta)=\frac{24}{7}$
| $\csc \theta=-\frac{5}{3}$
$\implies \sin^2\theta=\frac{9}{25}$
$\implies \cos^2\theta=\frac{16}{25}$
$\implies \cos2\theta=2*\frac{16}{25}-1=\frac{7}{25}$
Also,
$\sin^22\theta=4\sin^2\theta\cos^2\theta=\frac{4\times9\times16}{25\times25}$
$\implies \sin2\theta=\frac{24}{25}$
That gives
$\tan2\theta=\frac{24}{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Zeroes of polynomials and their sum Let $a, b$ are zeroes of the polynomial $x^2-10cx-11d$ and $c,d$ are the zeroes of the polynomial $x^2-10a x-11b $ where $a,b,c, d$ are distinct reals then $a+b+c+d=?$
| If we have a monic polynomial $x^2 + Ax + B$ for some numbers $A,B$, and this polynomial has roots $r_1, r_2$, then $A$ and $B$ are deeply related to $r_1$ and $r_2$. In particular, we must have
$$ x^2 + Ax + B = (x - r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1r_2,$$
so that $-(r_1 + r_2) = A$ and $r_1r_2 = B$.
As an aside, this sort of pattern continues for higher degree polynomials. The coefficients of the polynomials are elementary symmetric polynomials in the roots.
So we are looking at the two polynomials
$$ \begin{align}
x^2 - 10cx - 11d \\
x^2 - 10ax - 11b
\end{align},$$
such that $a,b$ are the roots of the first and $c,d$ are the roots of the second. So we know that $a + b = 10c$ and $c + d = 10a$ by looking at the degree 1 coefficients, and similarly $ab = -11d$ and $cd = -11b$ from the constant coefficients.
By adding the identities of the degree $1$ coefficients, we see that $a + b + c + d = 10(a + c)$, or rather
$$ b + d = 9(a + c) \tag{1}.$$
So to understand $a + b + c + d$, it will suffice to understand either $b + d$ or $a + c$.
By multiplying the relations for the constant coefficients, we see that $abcd = 121bd$. We can cancel the $bd$ as long as $bd \neq 0$. Suppose that one of $b$ or $d$ is $0$, and without loss of generality suppose it is $b$. Then $b = 0$ is a root of the first polynomial $x^2 - 10cx - 11d$. This forces $d = 0$, so that $b = d = 0$. Since in the problem statement we are given that $b \neq d$, we see that $bd \neq 0$. So we may cancel off the $bd$, and we get that
$$ ac = 121 \tag{2}.$$
We go after either $a + c$. We know that $a$ is a root of the first polynomial and $c$ is a root of the second polynomial. So
$$\begin{align}
a^2 - 10ac - 11d = 0 \\
c^2 - 10ac - 11b = 0.
\end{align}$$
Notice we know that in the middle coefficients, we know $ac = 121$ by $(2)$. So adding these gives
$$ a^2 + c^2 - 20(121) - 11(b + d)=0.$$
We know that $11(b+d) = 99(a + c)$ by $(1)$ above, so this is
$$ a^2 + c^2 - 20(121) - 99(a + c)=0. \tag{3}$$
Notice that this is almost of the form $(a + c)^2 + \alpha (a + c) + \beta = 0$. We're off by $-2ac = -2(121)$. More explicitly, $(a + c)^2 - 2ac = (a+c)^2 - 2(121) = a^2 + c^2$, so that $(3)$ becomes
$$ (a+c)^2 - 99(a + c) - 22(121) = 0. \tag{4}$$
This is a quadratic in $a + c$. We know how to find the roots of
$$ x^2 - 99x - 22(121) = 0.$$
The quadratic formula tells us the roots are
$$ x = \frac{99 \pm 143}{2} = -22, 121$$
(and getting integer results here is a nice indication that this is the right path, and that they carefully designed the solution).
We must now ask whether both are actually possible. I had to think quite a bit on how to do this, and I find it a bit subtle. If $x = -22$, which is to say that $a + c = -22$, then in fact we have that
$$(a-c)^2 = (a+c)^2 - 4ac = 22^2 - 4(121) = 4 \cdot 11^2 - 4\cdot 11^2 = 0,$$
so that $a = c$. This contradicts the original claim that $a \neq c$, and so in fact $a + c = 121$.
Now that we know that $a + c = 121$, and we can relate $b + d$ to $a + c$ through $(1)$, we can get the answer, $a + b + c + d = 1210$. $\diamondsuit$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?) I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1.
However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CTS algorithm, you just need to divide the coefficient of the $x$ term by twice the square root of the coefficient of the $x^2$ term.
So, if you have $ax^2 + bx + c$, your perfect square would be $(\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2$
If $a$ is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there.
For example:
In the equation $5x^2 + 6x + 5 = 0$, we could do:
$25x^2 + 30x + 25 = 0$
$(5x+3)^2 = -16$
$5x+3 = \pm4i$
$x = \pm \frac{4i}{5} - \frac{3}{5}$
My questions are:
-Is this correct?
-Is there ever an advantage to using the quadratic formula?
-Are there quadratics that are unsolvable this way?
| Every quadratic can have its square completed, which is in fact where the quadratic formula comes from.
The method I use is to factor out the leading coefficient,
$$
\begin{align}
\color{red}{a} x ^2 + bx +c &= \color{red}{a}\left( x^2 + \frac{b}{\color{red}{a}} x \right) + c \\
&= a\left( \left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} \right) + c \\
&= a\left(x+\frac{b}{2a}\right)^2 + c-\frac{b^2}{4a},
\end{align}
$$
if you solve for when this is equal to zero you will get the quadratic formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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"answer_id": 2
} |
Prove factorial problem $\forall n\in\mathbb N$ $C_n=\frac{1}{n+1}\left(\begin{matrix}2n\\n\end{matrix}\right)$.
Prove that $C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)$
I tried many ways but got nothing. Appreciate any tips.
My attempt: By defining, $C_{n+1}= \dfrac{1}{n+2}\left(\begin{matrix}2n+2\\n+1\end{matrix}\right)=\dfrac{1}{n+2}\dfrac{(2n+2)!}{(n+1)!(n+1)!}=\dfrac{1}{n+2}*\dfrac{(2n+2)(2n+1)n!}{(n+1)^2n!n!}$
$=\dfrac{1}{n+2}\dfrac{2(2n+1)n!}{(n+1)n!}$
| $$C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)=\frac{1}{n+1}[\frac{(2n)!(n+1))}{n!n!}-\frac{(2n)!}{n!(n-1)!}]=\\\frac{1}{n+1}\frac{2n!}{n!}[\frac{n+1}{n!}-\frac{1}{(n-1)!}]=\\ \frac{1}{n+1}\frac{2n!}{n!n!}(n+1-\frac{n!}{(n-1)!})=\\ \frac{1}{n+1}\frac{2n!}{n!n!}(n+1-n)=\frac{1}{n+1}\binom{2n}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.