Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding a new basis for the null space of a matrix Given $$
A=\begin{pmatrix}
-1 & -2 & 3 & -4 & -5 \\
3 & 6 & -1 & 4 & 2 \\
-2 & -4 & 0 & -2 & 0 \\
-2 & -4 & 1 & -3 & 1 \\
\end{pmatrix}
$$
and in solving for a basis for the null space of $A$, I found that:
$$ \begin{pmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\
\end{pmatrix}=x_2 \begin{pmatrix}
-2 \\ 1 \\ 0 \\ 0 \\ 0 \\
\end{pmatrix} + x_4\begin{pmatrix}
-1 \\ 0 \\ 1 \\ 1 \\ 0 \\
\end{pmatrix}$$
Thus, a basis for the null space of A is $$B_A=\begin{Bmatrix} \begin{pmatrix}
-2 \\ 1 \\ 0 \\ 0 \\ 0 \\
\end{pmatrix},\begin{pmatrix}
-1 \\ 0 \\ 1 \\ 1 \\ 0 \\
\end{pmatrix} \end{Bmatrix} $$
My question is, how would I find a new basis for the null space of $A$, whose vectors are NOT multiples? Are there any fast ways of doing this?
I am aware that you can use a change of basis matrix, but if a question asks you to find this new basis for null space and then find the change of basis matrix (not the other way around), how would this be done?
| You probably want to find an orthonormal basis. Look into the Gram-Schmidt process.
| {
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"url": "https://math.stackexchange.com/questions/2748358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a derivative with $\log$ $f(x)=\log{\frac{x}{1+\sqrt{5-x^2}}}$
I have to find the derivative of $f(x)$. Please tell me if my steps are correct.
$\frac{1+\sqrt{5-x^2}}{x}×\frac{1}{\ln 10}×\frac{1+\sqrt{5-x^2}-x×\frac{1}{2\sqrt{5-x^2}}}{(1+\sqrt{5-x^2})^2}×\frac{1}{2\sqrt{5-x^2}}×(-2x)$
| I would write $$f(x)=\log(x)-\log(1+\sqrt{5-x^2})$$ then we get
$$f'(x)=\frac{1}{x}-\frac{1}{1+\sqrt{5-x^2}}\cdot \frac{1}{2}(5-x^2)^{-1/2}\cdot (-2x)$$ if you mean $$\log(x)=\ln(x)$$ the logarithm to the base $e$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
$$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?
| Yet another argument: It is easy to check that
\begin{equation}
\left(\begin{array}{ccc}
b+c & a & a \\ b & c+a & b \\ c & c & a+b
\end{array}\right)
= L N ,
\end{equation}
where the matrices $L$ and $N$ are defined by
\begin{equation}
L = \left(\begin{array}{ccc}
0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0
\end{array}\right)
\qquad \text{and} \qquad
N = \left(\begin{array}{ccc}
0 & c & b \\ c & 0 & a \\ b & a & 0
\end{array}\right)
.
\end{equation}
Since the determinant is multiplicative, it thus suffices to prove that $\det L = 2$ and $\det N = 2abc$. This is easy.
| {
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"url": "https://math.stackexchange.com/questions/2751439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum to $n$ terms of the given series Find the sum to $n$ terms of the given series:
$$0.3+0.33+0.333+0.3333+\cdots$$
My Attempt:
Let
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$
$$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots \text{ to $n$ terms}\right]$$
How do I continue from here?
| $$\dfrac {3}{10} \left(1+\dfrac {11}{10}+\dfrac {111}{100}+\dfrac {1111}{1000}+\ldots\right) \\
=\dfrac {3}{10} \left(\dfrac{1}{10^0}+\dfrac {10 +1}{10^1}+\dfrac {10^2 +10 +1}{10^2}+\dfrac {10^3+10^2+10+1}{10^3}+\ldots\right)
$$
Each fraction has a (finite) geometric series in the denominator (there is a formula for those) and a power of ten in the denominator.
$$=\frac3{10}\sum_{k=0}^{n-1} \frac1{10^k} \sum_{j=0}^k 10^j .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$? How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$?
Note: $I$ denotes the identity matrix of size $3 \times3$ and $A^{T}$ represents the transpose of the matrix $A$.
I took transpose on both sides in given equation to get $A=(A^T)^2-I$ and then I put value of $A^T$ in this equation using $A^{T}=A^2-I$ to get $A^3-2A-I=0$ which ultimately gave $(A^T-A)(A+I)=0$ . How to deal the problem from here? And is there any better approach to tackle this problem? The given answer is $0$.
| Since $A^3-2A-I=0$ , any eigenvalue of $A$ satisfies
$$\lambda^3-2\lambda-1=0 \Rightarrow \lambda=-1, \frac{1 \pm \sqrt{5}}{2}$$
Let $\lambda_{1,2,3}$ be the eigenvalues of $A$. Then
$$tr(A^T)=tr(A^2-I) \Rightarrow
\lambda_1+\lambda_2+\lambda_3 = \lambda_1^2+\lambda_2^2+\lambda_3^2-3 \\
\Rightarrow \sum_{j=1}^3(\lambda_j^2 -\lambda_j -1) =0 \qquad (*)$$
Note that $\lambda= \frac{1 \pm \sqrt{5}}{2}$ satisfy $\lambda^2 -\lambda-1=0$, while $\lambda=-1$ satisfies $\lambda^2 -\lambda-1>0$
It follows from $(*)$ that $-1$ is NOT an eigenvalue of $A$.
Therefore, the minimal polynomial of $A$ is
$$\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2})^\alpha (x- \frac{1 -\sqrt{5}}{2})^\beta$$
Moreover, since $A^3-2A-I=0$ we have $\mu_A(x) | X^3-2X-1$ and hence $\alpha, \beta \leq 1$.
Therefore, we have three choices left:
$$\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2})\\
\mu_A(x)= (x- \frac{1 -\sqrt{5}}{2})\\
\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2}) (x- \frac{1 -\sqrt{5}}{2})$$
The first two lead to the solutions
$$A=\lambda I \qquad \lambda= \frac{1 \pm +\sqrt{5}}{2}$$
While the last gives
$$A^2-A-I =0$$
and hence
$$A=A^T$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
} |
Vandermonde matrix unique solution to polynomial equation
I want to verify if I am thinking of parts b) and c) correctly.
To find its kernel, we must find the nullspace of the matrix equation
$
\begin{bmatrix}
5^5 & 5^4 & 5^3 & 5^2 & 5 & 1 \\
4^5 & 4^4 & 4^3 & 4^2 & 4 & 1 \\
3^5 & 3^4 & 3^3 & 3^2 & 3 & 1 \\
2^5 & 2^4 & 2^3 & 2^2 & 2 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
a_5 \\
a_4 \\
a_3 \\
a_2 \\
a_1 \\
a_1
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0 \\
0 \\
0 \\
0 \\
\end{bmatrix}
$
Due to the fact that this is a vandermonde matrix with distinct geometric ratios for its rows, this matrix has full rank and thus has a trivial nullspace I think?
Then, to show part c that there is a unique polynomial such that $f(j) = a_j$, can we then just say that
$
\begin{bmatrix}
5^5 & 5^4 & 5^3 & 5^2 & 5 & 1 \\
4^5 & 4^4 & 4^3 & 4^2 & 4 & 1 \\
3^5 & 3^4 & 3^3 & 3^2 & 3 & 1 \\
2^5 & 2^4 & 2^3 & 2^2 & 2 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
a_5 \\
a_4 \\
a_3 \\
a_2 \\
a_1 \\
a_1
\end{bmatrix}
=
\begin{bmatrix}
a_5 \\
a_4 \\
a_3 \\
a_2 \\
a_1 \\
a_0 \\
\end{bmatrix} $
has a unique solution due to nontrivial nullspace?
| Well, you are right saying that this is a Vandermonde matrix and hence is invertible, however, this exercise is basically how one proves this fact. You can use the formula to compute such a determinant (which does not use this exercise), but I guess this is not the spirit of the assignment.
How about saying instead that $T$ is invertible, since if $f\in V$ is such that $T(f)=0$, then $f$ has $6>5$ distinct roots, so that $f=0$.
For the rest of the exercise, your solution is fine.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is this strange notation in the output of Wolfram Alpha? For some reason I need to find the values of two positive integers $k$ and $x$ such that:
$$
k x^3 (2 x + 1)^3>2 (k + 1) (x + 1)^2 (2 x^4 + 2 x^3 + 3 x^2 + 2 x + 1).
$$
Inputting this in Wolfram Alpha gives the following strange result (among others):
$$ k>1, x>Root[\#1^6 (4 k - 4) - 12 \#1^5 + \#1^4 (-12 k - 18) + \#1^3
> (-19 k - 20) + \#1^2 (-16 k - 16) + \#1 (-8 k - 8) - 2 k - 2\&, 2] $$
What does this mean? What is the role of the symbols $\#$ and $\&$?
| Using a command like
RegionPlot[
$k x^3 (2 x + 1)^3 -
2 (k + 1) (x + 1)^2 (2 x^4 + 2 x^3 + 3 x^2 + 2 x + 1) > 0$, {$k$, 0,
10}, {$x$, 0, 10}]
You will obtain an useful plot
| {
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"url": "https://math.stackexchange.com/questions/2753044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?
| Squaring both sides
$$
25f_{n+1}^2-80f_{n}f_{n+1}+100f_n^2 = 36 \times 4^n
$$
or
$$
(5 f_{n+1}-\lambda_1f_n)(5f_{n+1}+\lambda_2f_n) = 36\times 4^n
$$
with $\lambda = 8\pm i 6$ This can be handled as
$$
\left\{
\begin{array}{}
5 f_{n+1}-\lambda_1f_n & = & a\\
5 f_{n+1}-\lambda_2f_n & = & b
\end{array}
\right.
$$
with $a \times b = 4^n$
then
$$
f_{n+1} = \frac{ ((3 + 4 i) a + (3 - 4 i) b)}{30}\\
f_{n} = \frac{i(a-b)}{12}
$$
etc.
For real solutions $a=b \Rightarrow f_{n+1} = \frac{2^{n+1}}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Calculating $3^{m-n}=?$ $$9^m + 9^n = 52$$
$$9^m -4 = 2 \cdot 9^n$$
$$3^{m-n}=?$$
Let me show what I've tried
Simpifyling the both equalities.
$$3^{2^m} + 3^{2n} = 2 \cdot 13 \tag{1}$$
$$3^{2m} -2^2 = 2 \cdot 3^{2n} \tag{2}$$
Diving the second equality by $2$ and we have
$$\frac{3^{2m} -2^2}{2} =3^{2n} \tag{3}$$
Here is where I'm stuck.
My Kindest Regards!
| $$9^m + 9^n = 52$$
$$9^m -4 = 2 \cdot 9^n$$
By substituion, we have
$$(52-9^n)-4=2\cdot 9^n$$
$$48-9^n = 2 \cdot 9^n$$
$$9^n = 16$$
$$3^n=4$$
Substitute $9^n$ inside the first equation and do the same trick to complete the task.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a real matrix A such that (Exponential of matrices) Is there a real matrix A such that
$$\exp(A) = \begin{bmatrix} -\alpha & 0 \\ 0 & -\beta \end{bmatrix}, \text{ where }\alpha,\beta>0?$$
(Hint): In two dimensions the exponential matrix of
$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
is given by:
$$\exp(A) = e^{\delta}\cos(\Delta)\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + e^{\delta}\frac{\sin(\Delta)}{\Delta}\begin{bmatrix} \phi & b \\ c & -\phi \end{bmatrix},$$
where $\delta=\dfrac{a+d}{2}$, $\phi=\dfrac{a-d}{2}$, $\Delta=\sqrt{\phi^2+bc}$.
Does this exercise only reduce the multiplication of the matrices given in the tip?
| $$
A =
\left(
\begin{array}{cc}
0 & \pi \\
- \pi & 0
\end{array}
\right)
$$
The point is this: by straightforward summing, we can find, for real number $t,$ given
$$
B =
\left(
\begin{array}{cc}
0 & t \\
- t & 0
\end{array}
\right) \; \; ,
$$
we get
$$
e^B =
\left(
\begin{array}{cc}
\cos t & \sin t \\
- \sin t & \cos t
\end{array}
\right) \; \; .
$$
That is, even powers of $B$ are diagonal, and odd powers of $B$ are off-diagonal, in fact skew symmetric. We are looking for
$$ e^B = I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \frac{1}{24} B^4+ \frac{1}{120} B^5+ \frac{1}{720} B^6 + \cdots $$
We get the cosines on the diagonal from
$$ I + \frac{1}{2} B^2 + \frac{1}{24} B^4+ \frac{1}{720} B^6 + \cdots $$
and the sines skew-symmetric from
$$ B + \frac{1}{6} B^3 + \frac{1}{120} B^5 + \cdots $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exhibiting $f$ such that a vector field $\overline{v} = \nabla f$? The first is suppose $\overline{v} = (x, 2yz^3, 3y^2z^2)$ and we want to exhibit an $f$ such that $\overline{v} = \nabla f$. Since $\nabla \times \overline{v}=0$, $f$ exists and we can integrate:
$$\displaystyle \int f \,dx = \frac{1}{2}x^2+c(y,z), ~ \int f \,dy = y^2z^3+c(x,z), ~ \int f \,dz = y^2z^3+c(y,z)$$
So $\displaystyle f = \frac{1}{2} x+c(y, z) = y^2z^3+c(x,z) = y^2z^3+c(x,z)$. How do you determine $c(x,y), c(y, z), c(x,z)$?
| $$ f_x = x \implies f = \frac{1}{2}x^2 + g(y,z) $$
We know $f_y = 2yz^3$ so
$$ f_y = 0 + g_y = 2yz^3 \implies g(y, z) = y^2z^3 + h(z) $$
So now we have
$$ f = \frac{1}{2}x^2 + y^2z^3 + h(z) $$
We know $f_z = 3y^2z^2$ so
$$ f_z = 0 + 3yz^2 + h'(z) = 3y^2z^2 \implies h(z) =c, c\in\mathbb{R} $$
And we have that
$$ f = \frac{1}{2}x^2 + y^2z^3 + c $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757201",
"timestamp": "2023-03-29T00:00:00",
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Recursive sequence for the Koch Snowflake
Let $C_0$ an equilateral triangle of side 1 and $A_{n}$ the area of
the figure $C_{n+1}, n\in\mathbb{N}\cup {0}$
Write $A_{n+1}$ based on $A_n$ and the area of $C_0$
I noted that $C_1$ has been built by adding a smaller equilateral triangle on each side of $C_0$. Then $C_2$ is built by adding an even smaller equilateral triangle on each side of $C_1$. Figure $C_3$ is constructed by adding a smaller equilateral triangle on each side of $C_2$. So each new figure is formed by adding small equilateral triangles on the sides of the previous figure.
The area of one smaller triangle added in each figure can be calculated using the formula:
$$a=\frac{s^2\sqrt{3}}{4}$$
s is the length of the side, and knowing that the side of $C_0$ is $1$, I noted that the length of each segment of $C_{n+1}$ is given by $\left(\frac{1}{3}\right)^{n+1}$ so
$$a=\frac{\left(\left(\frac{1}{3}\right)^{n+1}\right)^2\sqrt{3}}{4}$$
I also noted that the number of sides in figure $C_{n+1}$ is $3 \times 4 ^ n$ so:
$$A_{n+1}=A_n+\frac{\left(\left(\frac{1}{3}\right)^{n+1}\right)^2\sqrt{3}}{4}\times 3\times 4^n=A_n+\left(\frac{1}{9}\right)^{n+1}\times \frac{\sqrt{3}}{4}\times3\times4^n$$
$$\boldsymbol{A_{n+1}=A_n+\frac{3\sqrt{3}}{16}\left(\frac{4}{9}\right)^{n+1}}$$
Is there another way or reasoning to get the same recursive sequence?
| One alternative reasoning might follow from:
$$A_{n+2} - A_{n+1} = \frac{4}{9} \left(A_{n + 1} - A_{n}\right) \quad;\quad A_1 = \frac{4}{3} A_0 \quad;\quad A_0 = \frac{\sqrt{3}}{4}\\
A_{n+2} - A_0 = \left(A_{n+2} - A_{n+1}\right) + \left(A_{n+1} - A_n\right) + \cdots + \left(A_1 - A_0\right)$$
I'll leave you to flesh out the details.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the following inequality is true: $(\frac{1}{a} + \frac{1}{bc}) (\frac{1}{b} + \frac{1}{ca})(\frac{1}{c} + \frac{1}{ab}) \geq 1728$ This is a question from a past Olympiad paper:
Three positive real numbers $a, b, c$ satisfy the following constraint:
$a+b+c = 1$. Show that the following inequality is true:
$\left(\dfrac{1}{a} + \dfrac{1}{bc}\right) \left(\dfrac{1}{b} + \dfrac{1}{ca}\right)\left(\dfrac{1}{c} + \dfrac{1}{ab}\right) \geq 1728$.
Starting from the L.H.S, I end up with:
$\dfrac{abc(abc+a^2 + b^2 + c^2 + 1) + a^2b^2 + a^2c^2+b^2c^2}{(abc)^3}$.
From the numerator I suspect $a+b+c$ will be factorised, but this is the furthest I have got to.
Using the hint given by timon92:
$\dfrac{1}{a} + \dfrac{1}{bc} = \dfrac{(a+b)(a+c)}{abc}$
and likewise for other two, I end up with:
$(a+b)^2(b+c)^2(a+c)^2 \geq 1728(abc)^3$
$\left((a+b)(b+c)(a+c)\right)^\frac{2}{3} \geq 8abc$
Many thanks in advance.
| Hint: $$\frac 1a + \frac{1}{bc} = \frac{bc+a}{abc} = \frac{bc+a(a+b+c)}{abc} = \frac{(a+b)(a+c)}{abc}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \frac {x}{1+x^2}dx+2\int_{0}^{1}\frac 1{1+x^2}dx-\int_{0}^1\frac {(x-1)^2}{(x^2+1)^2+(\frac {\sqrt{3}}2)^2}dx$$
But I don't know how to solve the last term.
I answered the question... I realized my silly mistake, sorry for bothering...
| I'm sorry but I had a really stupid mistake and now I solved the integral really easily...
So this is what I did:
$$I=\int_0^1 \frac {x^4+1}{(x^2)^3+1}dx=\int_0^1\frac{(x^2+1)^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx$$
then we divide top and bottom by $x^2$ and separate the integrals...
we get:
$$I=\int_0^1\frac{1+\frac 1{x^2}}{(x-\frac 1x)^2+1}dx-2\int_0^1\frac{x^2}{(x^3)^2+1}dx$$
and this is:
$$I=arctan(x-\frac 1x)-\frac 23arctan(x^3)$$
which from $0$ to $1$, we get: $I=\frac \pi3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve for x in the equation: $3^x + 9^x = 27^x$ Solve for x in the equation: $3^x + 9^x = 27^x$
Via calculator, the x is = 0.438 which is correct as it is in the choices. However I want to know how to solve it manually, hence the question.
I'm stuck at : $3^x + 3^{2x} = 3^{3x}$
I don't know what to do next. Any help would be appreciated.
| $$3^x + 9^x = 27^x$$
$$3^x+3^{2x} = 3^{3x}$$
let $y= 3^x$
$$y+y^2=y^3$$
$$y^3-y^2-y = 0 $$
$$y(y^2-y-1)=0$$
so either $y=0$ or $(y^2-y-1)=0$
since $3^x$ can never be $0$;
$(y^2-y-1)=0$
$\implies y = \dfrac{1\pm\sqrt{5}}{2}$
$\implies 3^x = \dfrac{1\pm\sqrt5}{2}$
$\implies x =\log_3\left(\dfrac{1\pm\sqrt5}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculating P(X>6|X<8) from pdf I have a pdf $\frac{k}{x^2}$ for $4<x<\infty$ ($0$ otherwise) and need to work out the $P(X>6 | X<8)$ I already worked out k to equal 4.
To work out this probability I did $\frac{P(X>6)}{P(X<8)}$ but I get a weird answer/
For $P(X>6)$ i did $\int_{6}^{\infty} \frac{4}{x^2}dx = \frac{2}{3} $
and for $P(X<8)$ i did $\int_{4}^{8} \frac{4}{x^2}dx = \frac{1}{2} $
My answer of $\frac{3}{4} $ is obviously wrong, but I am not sure where I went wrong.
| $$P(X>6 | X < 8) = \frac{P((6<X)\cap (X<8))}{P(X<8)} = \frac{P(6 < X<8)}{P(X<8)} = \frac{k\int_6^8\frac{dx}{x^2}}{k\int_4^8\frac{dx}{x^2}} = \frac{\int_6^8\frac{dx}{x^2}}{\int_4^8\frac{dx}{x^2}}$$
Note that the probability density function lives on $(4,\infty$). That's why the lower integral starts at 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $F(G(x))=G(F(x)).$
Let $V$ be a finitely dimensional linear space, and $F:V\rightarrow V$
and $G:V\rightarrow V$ linear transformations that are diagonalizable,
that is: there exists a base $\mathbf{e}$ for $V$ such that the matrix
to $F$ in the basis $\mathbf{e}$ is diagonal and a base $\mathbf{f}$
for $V$ such that the matrix to $G$ is diagonal.
i) Show that if $F$ and $G$ are simultaneously diagonalizable, that is if $\mathbf{e}=\mathbf{f},$ then it follows that
$$F(G(x))=G(F(x)), \quad \forall x\in V. \tag{1}$$
ii) Show that if $(1)$ holds, then there exists a basis $\mathbf{g}$ (possibly different from both $\mathbf{e}$ and $\mathbf{f}$) such that
the matrices to both $F$ and $G$ in the basis $\mathbf{g}$ are
diagonal.
Okay, I usually post good questions where I've done my own work before asking for help. But this one has thrown me away. I seriously don't even know how to begin or what I should proove.
This is no homework or assignment, I'm just doing practice problems for my exam at the end of this month. Any tips/tricks/translation of the problem statement is welcome!
| Regarding part 1: If $F$ and $G$ are simultaneously diagonalizable, then with respect to some basis of $V$ we can represent $F$ and $G$ as
\begin{align*}
F = \left( \hspace{-0.4cm}
\begin{matrix}
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right), \quad G = \left( \hspace{-0.4cm}
\begin{matrix}
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right),
\end{align*}
the point being that both transformations can be represented as diagonal matrices with respect to the same basis. But then, since $F$ and $G$ are diagonal matrices, they commute:
\begin{align*}
FG &= \left( \hspace{-0.4cm}
\begin{matrix}
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right) \left( \hspace{-0.4cm}
\begin{matrix}
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right)\\
&= \left( \hspace{-0.4cm}
\begin{matrix}
&\lambda_1 \kappa_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 \kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n \kappa_n\\
\end{matrix} \right) \\
&= \left( \hspace{-0.4cm}
\begin{matrix}
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right) \left( \hspace{-0.4cm}
\begin{matrix}
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right) = GF.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How can I solve this Combination with indistinguishable-objects problem? This question is not about how to solve the problem, but is about why doesn't my solution work.
A bowl has $2$ red, $2$ green, and $2$ blue marbles. How many combinations are possible if we take $3$ random marbles at a time?
I know that the answer should be, $7$.
Coz,
*
*BBG
*BBR
*BGG
*BGR
*BRR
*GGR
*GRR
But, when I want to calculate this using a combination formula, nothing comes out which is near $7$:
$\frac{^6C_3}{2! \cdot 2! \cdot 2!}$
$ = \frac{6!}{3! \cdot (6-3)!} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6!}{3! \cdot 3!} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6 \cdot 5 \cdot 4 }{3 \cdot 2 \cdot 1} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{5 \cdot 4 }{2! \cdot 2! \cdot 2!}$
$ = \frac{5}{2!} $
$ = \frac{5}{2}$
So, what am I doing wrong?
| use generating function to count :
you can select either $0$ ball ; $1$ ball ; $2$ ball of same colour
so $f(x)= (x^0+x+x^2)(x^0+x+x^2)(x^0+x+x^2)=(1+x+x^2)^3$
so, to find number of ways to pick $3$ balls out of the bowl, just find coefficient of $x^3$ in expansion of $f(x)$
now , coefficient of $x^3$ in expansion of $f(x)=(1+x)^3+3.(1+x)^2.x^2+3(1+x)x^4+x^6$ is $7$ so there will be 7 combinations of drawing a ball from the bowl.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $$
From here, I got stuck. I have to remove y from here to solve it.
The answer in the textbook gave $ y= (k(1+x^2) - 1)/{2}$ I believe $k$ is the integration constant. How do I remove the $\ln$ from both sides?
| To get rid of the ln on both sides just do $e^{ln|1+2y|}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Expand Expression in Tensor Notation I have an expression given to me in tensor/index notation and I need help expanding out into something more readable. The expression is
$$\frac{b_{ij}b_{jk}b_{ki}}{6}$$
and $b$ is a symmetric second order tensor, and the indices range from 1 to 3.
Can somebody please expand this expression for me?
I'm trying to code this expression up and I keep getting complex numbers, so I know I'm doing something wrong.
This is my attempt:
$(b_{11}b_{11}b_{11} + b_{11}b_{12}b_{21} + b_{11}b_{13}b_{31} + b_{12}b_{21}b_{11} + b_{12}b_{22}b_{21} + b_{12}b_{23}b_{31} + b_{13}b_{31}b_{11} + b_{13}b_{32}b_{21} + b_{13}b_{33}b_{31} + b_{21}b_{11}b_{12} + b_{21}b_{12}b_{22} + b_{21}b_{13}b_{32} + b_{22}b_{21}b_{12} + b_{22}b_{22}b_{22} + b_{22}b_{23}b_{32} + b_{23}b_{31}b_{12} + b_{23}b_{32}b_{22} + b_{23}b_{33}b_{32} + b_{31}b_{11}b_{13} + b_{31}b_{12}b_{23} + b_{31}b_{13}b_{33} + b_{32}b_{21}b_{13} + b_{32}b_{22}b_{23} + b_{32}b_{23}b_{33} + b_{33}b_{31}b_{13} + b_{33}b_{32}b_{23} + b_{33}b_{33}b_{33})/ 6$
EDIT: I was computing the tensor $b$ incorrectly
| Let us define a second-order symmetric tensor $\boldsymbol{B} = [b_{i j}]$ given by its coordinates $b_{\bullet\bullet}$ over an orthonormal basis (the bullets could be any distinct symbols). Using the dot product and Einstein notation, one can write
$$
[(b^3)_{i \ell}] = \boldsymbol{B}^3 = \boldsymbol{B}\cdot\boldsymbol{B}\cdot\boldsymbol{B} = [ b_{i j}b_{jk}b_{k\ell} ] \, ,
$$
where $(b^3)_{\bullet \bullet}$ denotes the coordinates of $\boldsymbol{B}^3$.
Taking the trace, the result in @Chappers comment follows:
$$
\frac{1}{6}(b^3)_{i i} = \frac{b_{i j}b_{jk}b_{k i}}{6} = \frac{1}{6}\text {tr} (\boldsymbol{B}^3)\, .
$$
Note that the symmetry hypothesis was not used here.
Let us use the tensor's symmetry to obtain an alternative formula. We apply the spectral theorem and the Cayley–Hamilton theorem, so that
$$
\boldsymbol{B}^3 - {B}_\text{I}\, \boldsymbol{B}^2 + {B}_\text{II}\, \boldsymbol{B}^1 - B_\text{III} \boldsymbol{B}^0 = \boldsymbol{0} \, .
$$
The coefficients -- principal invariants of $\boldsymbol{B}$ -- in the characteristic polynomial above are computed separately:
\begin{aligned}
{B}_\text{I} &= \text{tr}(\boldsymbol{B}) &&= b_{ii}\, ,\\
{B}_\text{II} &= \tfrac{1}{2} \left(\text{tr}(\boldsymbol{B})^2 - \text{tr}(\boldsymbol{B}^2)\right) &&= \tfrac{1}{2} \left((b_{ii})^2 - b_{ij}b_{ij}\right) ,\\
{B}_\text{III} &= \det \boldsymbol{B} &&= \varepsilon_{ijk}b_{1i}b_{2j}b_{3k} \, ,
\end{aligned}
where $\varepsilon_{ijk}$ is the Levi-Civita permutation symbol.
Taking the trace, one obtains
\begin{aligned}
\text {tr} (\boldsymbol{B}^3) &= 3 {B}_\text{III} - {B}_\text{II}{B}_\text{I} + {{B}_\text{I}}\, \text {tr} (\boldsymbol{B}^2)\\
&= 3 {B}_\text{III} - 3 {B}_\text{II}{B}_\text{I} + {{B}_\text{I}}^3 \, .
\end{aligned}
Note that the invariants are related to the eigenvalues $\lambda_1, \lambda_2, \lambda_3$ of $\boldsymbol{B}$ through
\begin{aligned}
{B}_\text{I} &= \lambda_1 + \lambda_2 + \lambda_3\, ,\\
{B}_\text{II} &= \lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \lambda_2 \lambda_3 \, ,\\
{B}_\text{III} &= \lambda_1 \lambda_2 \lambda_3 \, .
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$x_1=1,x_n=x_{n+1}+\ln (1+x_{n+1})$, prove $x_n\leq\frac{1}{2^{n-2}}$. $x_1=1,x_n=x_{n+1}+\ln (1+x_{n+1})$, prove
$x_n\leq\frac{1}{2^{n-2}}$.
I proved $0<x_{n+1}<x_n$ by contradiction, and I also get $x_n\geq\frac{1}{2^{n-1}}$ by induction. I tried $\ln(1+x)\geq x-\frac{1}{2}x^2$, but it did not work. Could anyone help me? Thanks!
I tried a new way
$\forall \epsilon>0, \ln(1+x)>(1-\epsilon)x$, then $x_n>(2-\epsilon)x_{n+1}$, thus $x_{n+1}<\frac{1}{(2-\epsilon)2^{n-2}}\to \frac{1}{2^{n-1}}$.
Is it right?
| well, I think I misunderstand this question, sorry.
as you said
$x_{n+1}\ge ln(1+x_{n+1})\geq x-\frac{1}{2}x^2$
we can get $x_n\in(\frac{1}{2^{n-1}}, 2-\sqrt{4-2x_{n-1}})$
Thus we need to show $2-\sqrt{4-2x_{n-1}}\le \frac{1}{2^{n-2}}$
in fact, the above $2-\sqrt{4-2x_{n-1}}$ is determined by $ 2-\sqrt{2\sqrt{2\sqrt{2...}}}$ which you can check with $\frac{1}{2^{n-2}} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Shortest distance between ellipsoid and plane Find the shortest distance between the points $A = (x_1,y_1,z_1)$ and $B = (x_2,y_2,z_2)$ if $A$ lies on the plane $x+y+z=2a$ and $B$ lies on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
| The prameterization of an ellipsoid is
$$x=a\cos\theta\sin\phi \\
y=b\cos\theta\cos \phi \\
z=c\sin\theta$$
distance of a point from a $x+y+z-2a=0$ is given by $$l=\frac{x+y+z-2a}{\sqrt3} \\ \implies l=\frac{a\cos\theta\sin\phi+b\cos\theta\cos\phi+c\sin\theta-2a}{\sqrt 3}$$
For minimum distance $$\frac{\partial l}{\partial\theta}=0;\quad\frac{\partial l}{\partial\phi}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c} $
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c}, c > 0 $
It seems to me that the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ of the maximum area when $ a = c; b = \frac{1}{c} $
That is, it looks like this:
The problem is that I'm not sure that an ellipse here can be a finite area
I will be glad to any hint or solution
| Expanding in Taylor series around $x=0$
$$
y = b\sqrt{1-\left(\frac{x}{a}\right)^2} = b-\frac{b x^2}{2 a^2}-\frac{b x^4}{8 a^4}+O\left(x^5\right)
$$
and
$$
y = \frac{1}{(x^2+c)} = \frac{1}{c}-\frac{x^2}{c^2}+\frac{x^4}{c^3}+O\left(x^5\right)
$$
Now solving
$$
\left\{
\begin{array}{rcl}
b & = & \displaystyle\frac{1}{c}\\
\displaystyle\frac{b}{2 a^2} & = & \displaystyle\frac{1}{c^2}
\end{array}
\right.
$$
we get
$$
a = \sqrt{\frac{c}{2}}, b = \frac{1}{c}
$$
and the area is
$$
S = \pi a b = \frac{\pi}{\sqrt{2 c}{}}
$$
NOTE
In the series expansion the third term is negative for the ellipse.This guarantees the inclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Computing the coefficient of the term of a certain degree in a polynomial Given the polynomial
${1\over8}((1+z)^9 + 3(1-z)^4(1+z)^5 + (1-z)^6(1+z)^3)$
(which is the weight enumerator of a code)
how do I find out the coefficient of $z^2$?
The solution given is ${1 \over 8}(36-12+0) = 3$.
I got $36$ for the $z^2$ coefficient of $(1+z)^9$ using the Binomial Theorem, but I don't know how to get $-12$ for the $z^2$ coefficient of $3(1-z)^4(1+z)^5$. By using the Binomial Theorem separately on $(1-z)^4$ and $(1+z)^5$ I get the following two polynomials, repsectively:
$z^4-4z^3+6z^2-4z+1$
$z^5+5z^4+10z^3+10z^2+5z+1$
I am unsure what to do next, or even if this is going in the right direction.
| A short cut or two:
$$(1-z)^4(1+z)^5=(1-z^2)^4(1+z)=(1-4z^2+\cdots)(1+z)=1+z-4z^2+\cdots$$
and
$$(1-z)^6(1+z)^3=(1-z^2)^3(1-z)^3=(1-3z^2+\cdots)(1-3z+3z^2-z^3)
=1-3z+0z^2+\cdots$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Prove that $x^{2/3}+ y^{2/3}= a^{2/3}$ $BE=x, FC=y, BC=a$
Then prove that $x^{2/3}+ y^{2/3}= a^{2/3}$
| By similar triangles $\triangle CDF,\triangle ABC$
$$\dfrac{CF}{BC}=\dfrac{AE}{AB}\implies\dfrac y{a\sin B}=\dfrac{a\cos B-x}{a\cos B} \iff x\sin B+y\cos B=a\cos B\sin B\ \ \ \ (1)$$
Similarly, by similar triangles $\triangle DAF,\triangle ABC$
$$\dfrac{AF}{CA}=\dfrac{DF}{AB}\implies\dfrac{a\cos B-x}{a\sin B}=\dfrac{a\sin B-y}{a\cos B}\iff x\cos B-y\sin B=a\cos2B\ \ \ \ (2)$$
Solve $(1),(2)$ for $x,y$ to find $x=a\cos^3B,y=a\sin^3B$
Use Prove $\sin^2\theta + \cos^2\theta = 1$
Alternatively, let $AB=c,CA=b\implies a^2=b^2+c^2\ \ \ \ (3)$
by similarities of the triangles we have $$\dfrac y{c-x}=\dfrac bc\ \ \ \ (4)$$ and $$\dfrac{c-x}{b-y}=\dfrac bc\ \ \ \ (5)$$
Solve $(4),(5)$ for $x,y$ and use $(3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of numbers between 250 and 350 which are divisible by 7?
What is the sum of numbers between $250$ and $350$ which are divisible by $7$?
My attempted solution:
7|250|35
|21 |
-----
40
35
----
5
The 1st number divisible by $7$ is: $(250-5)+7 = 252$.
7|350|50
|35 |
-----
0
The last number divisible by $7$ is: $350$.
Total numbers between 250 and 350 divisible by $7$ is:
$((350-252)/7) + 1$
$ = (98/7)+1$
$ = 14+1$
$ = 15$
$ Sum = \frac{15}{2} [2 \cdot 252 + (15 - 1) \cdot 7 ]$
$ = \frac{15}{2} [2 \cdot 252 + 14 \cdot 7 ]$
$ = \frac{15}{2} [504 + 98 ]$
$ = \frac{15}{2} \cdot 602$
$ = 15 \cdot 301$
$ = 4515 $
(Ans.) $4515$
Is there a more efficient way to do this?
| HINT
The first divisible number is $252$ the last $350$ then
$$252+\dots+350=7(36+\dots+50)=7\left(\sum_{k=1}^{50}k-\sum_{k=1}^{35}k\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the limit of this series I was trying to figure out the limit of the function below
$a_n = \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}$
The answer to the question is $\frac{1}{8}$ but when I divide through by $3^{n+3}$ (because it has the highest power) I cant figure out how they got that answer, Im really stuck has anyone go any hints
| Look for hidden powers:
$a_n
= \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}
= \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+2^{2n+2}}
$.
The $3^n$ terms are negligible
for large $n$,
so
$a_n
\approx \frac{2^{2n-1}}{2^{2n+2}}
=2^{(2n-1)-(2n+2)}
=2^{-3}
=\frac18
$.
To check,
$\begin{array}\\
a_n-\dfrac18
&= \dfrac{2^{2n-1}+3^{n+3}}{3^{n+2}+2^{2n+2}}-\dfrac18\\
&= \dfrac{8(2^{2n-1}+3^{n+3})-(3^{n+2}+2^{2n+2})}{8(3^{n+2}+2^{2n+2})}\\
&= \dfrac{2^{2n+2}+8\cdot 3^{n+3})-3^{n+2}-2^{2n+2}}{8(3^{n+2}+2^{2n+2})}\\
&= \dfrac{8\cdot 3^{n+3}-3^{n+2}}{8(3^{n+2}+2^{2n+2})}
\qquad\text{The } 2^{2n+2} \text{ is cancelled out}\\
\text{so}\\
|a_n-\dfrac18|
&= \dfrac{|8\cdot 3^{n+3}-3^{n+2}|}{|8(3^{n+2}+2^{2n+2})|}\\
&\lt \dfrac{|8\cdot 3^{n+3}|}{2^{2n}|8(3^3(3/4)^{n}+2^{2})|}\\
&= \left(\dfrac{3}{4}\right)^n\dfrac{|8\cdot 3^{3}|}{|8(3^3+4)|}\\
&\lt \left(\dfrac{3}{4}\right)^n\dfrac{216}{248}\\
&\to 0\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Finding conditional expectation of two random variables Let $X$ and $Y$ be independent and uniform on $[0,3]$. I want to calculate $E(Y| X<1 \cup Y<1 )$.
Attempt.
First, we calculate the distribution via the cdf:
$$ P(Y \leq y | X<1 \cup Y<1 ) = \frac{ P( \{Y \leq y \} \cap \{X < 1 \cup Y < 1 \} ) }{P(X<1 \cup Y<1) } = $$
$$ \frac{ P( [ \{Y \leq y \} \cap \{ X < 1 \} ] + [ \{Y \leq y \} \cap \{ Y < 1 \} ] - P(\{Y \leq y \} \cap \{X <1\} \cap \{Y < 1\}) }{P(X<1) + P(Y<1) - P(X<1)P(Y<1) } =$$
$$ \frac{ P( [ \{Y \leq y \} \cap \{ X < 1 \} ] \cup [ \{Y \leq y \} \cap \{ Y < 1 \} ] }{P(X<1) + P(Y<1) - P(X<1)P(Y<1) } $$
$$ \frac{ P(Y \leq y)P(X<1) +P(Y \leq \min(y,1) ) - P(Y \leq \min(y,1))P(X<1) }{P(X<1) + P(Y<1) - P(X<1)P(Y<1) } $$
Now, Notice $P(X<1) = P(Y<1) = \frac{1}{3}$ and $P(Y \leq y ) = \frac{ y }{3}$. Thus, the conditional cdf is
$$ \begin{cases} \frac{y}{5} \; \; \; 0<y<1 \\ \frac{y+2}{5} \; \; \; 1<y<3 \end{cases} $$
now, by taking the derivative we see that the density function is just $\frac{1}{5}$ over $y \in [0,3]$, now thus
$$ E(Y| X<1 \cup Y<1 ) = \int_0^3 y \frac{1}{5} = \frac{9}{10} = 0.9 $$
Now, this is incorrect according to my answer sheet which gives $\boxed{1.1}$ as the answer. What is my mistake?
| Here another short solution:
*
*$U = X<1 \cup Y <1 = [0,3]\times [0,1) \cup [0,1)\times [1,3]$ (disjoint union of $5$ squares with probability of $\frac{1}{9}$)
*$P([0,3]\times [0,1)| U) = \frac{3}{5}$, $P([0,1)\times [1,3] |U) = \frac{2}{5}$
*$\Rightarrow$ The mean of $Y$ on $[0,3]\times [0,1)$ is $\frac{1}{2}$ with a weight of $\frac{3}{5}$.
*$\Rightarrow$ The mean of $Y$ on $[0,1)\times [1,3]$ is $2$ with a weight of $\frac{2}{5}$.
*$\Rightarrow$ $E(Y|U)= \frac{1}{2}\cdot \frac{3}{5} + 2 \cdot\frac{2}{5} = \frac{11}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
$x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational.
If $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational numbers. What does $x^3 - 5x$ equals to?
| Note that $(x-1)^2=y$ is a rational number. So $x = 1 \pm \sqrt{y}$. Also,
$x^3-5x=3y-4 \pm\sqrt{y}(y-2)$ is rational, which is possible only when
$$y=2.$$
Therefore, $x = 1\pm \sqrt{2}$ and $x^3-5x = 3y-4 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\lim_{x \to 0}\frac{2\mu(e^x(x^2-x+1)-1)-e^x(x^2-2x+2)+2}{x(2\mu(e^x(x-1)+1)-e^x(x-2)-2)}=\mu$ I would like to ask kindly for any help to show below limit:
$$\lim_{x \to 0}\frac{2\mu(e^x(x^2-x+1)-1)-e^x(x^2-2x+2)+2}{x(2\mu(e^x(x-1)+1)-e^x(x-2)-2)}=\mu$$
I have tried to use the expansion of $e^x=1+x+x^2/2+x^3/6+\mathcal{O}(x^4)$ but still I can't show it.
| Note that
$$\frac{2\mu(e^x(x^2-x+1)-1)-e^x(x^2-2x+2)+2}{x(2\mu(e^x(x-1)+1)-e^x(x-2)-2)}=\frac{[(2\mu-1)x^2-(2\mu-2)x+(2\mu-2)]e^x-(2\mu-2) }{[(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x}=\frac{ [(2\mu-1)x^2-(2\mu-2)x]e^x +(2\mu-2)x +(2\mu-2)e^x-(2\mu-2)x-(2\mu-2)}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }=\\=1+(2\mu-2)\frac{ e^x-x-1}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }=\\=1+(2\mu-2)\frac{ {e^x-x-1}}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }\to 1+(2\mu-2)\cdot \frac12=\mu$$
indeed
$$\frac{ {e^x-x-1}}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }=\\=\frac{ {e^x-x-1}}{x^2}\frac{x}{ [(2\mu-1)x-(2\mu-2)]e^x+(2\mu-2) }\to \frac12\cdot 1=\frac12$$
indeed by $e^x=1+x+\frac12x^2+o(x^2)$
$$\frac{ {e^x-x-1}}{x^2}=\frac12 + o(1) \to \frac 12$$
and by $e^x=1+x+o(x)$
$$\frac{x}{ [(2\mu-1)x-(2\mu-2)]e^x+(2\mu-2) }=\frac{x}{ (2\mu-1)x+(2\mu-1)x^2-(2\mu-2)x+o(x)}=\frac{1}{1+(2\mu-1)x+o(1)}\to 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$ How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$?
My attempt
$$(3x+5y)^{12}=\left(3x\left(1+\frac{5y}{3x}\right)\right)^{12}$$
$$=3^{12}x^{12}\left(1+\frac{5y}{3x}\right)^{12}$$
I then compared $\left(1+\frac{5y}{3x}\right)^{12}$ with $(1+x)^n$ and got $n=17,\space x=(\frac53)(\frac yx)=(\frac53)\frac {\frac43}{\frac12} =\frac{40}{9}$.
Then I used $$\frac{(n+1)|x|}{1+|x|}=\frac{(12+1)\left(\frac{40}{9}\right)}{1+\frac {40}{9}}$$
After solving, I got
$$=10\frac{30}{49}$$ which is not an integer.
I am stuck here, can anyone explain how to solve this problem.
| How to find the numerically greatest term (NGT) in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$?
$$(3x+5y)^{12}=(3x)^{12}\left(1+\frac{5y}{3x}\right)^{12}$$
When compared $\left(1+\frac{5y}{3x}\right)^{12}$ with $(1+x)^n$, we got, $n=12$, a positive integer, and $x=\left(\frac53\right)\left(\frac yx\right)=\left(\frac53\right)\left(\frac{4/3}{1/2}\right) =\frac{40}{9}$. Thus, if rth-term $T_r$ is the numerically greatest term, $\left(T_{r+1}/T_r\right) \lt 1$.
$$T_{r+1}=~^nC_r(|x|)^r=\frac{n!}{(n-r)!r!}(|x|)^r ~\text{and} ~T_{r}=~^nC_{r-1}(|x|)^{r-1}=\frac{n!}{(n-r+1)!r-1!}(|x|)^{r-1}$$
$$\text{Thus,}~\left(\frac{T_{r+1}}{T_r}\right)=\left(\frac{n-r+1}{r}\right)|x| \lt 1 ~\text{only when,} $$
$$\frac{(n+1)|x|}{1+|x|}\lt r$$
$$\frac{(n+1)|x|}{1+|x|}=\frac{(12+1)\left(\frac{40}{9}\right)}{1+\frac {40}{9}}=\frac {13\cdot 40}{49}=10\frac {30}{49} \lt r$$
Note that if $\frac{(n+1)|x|}{1+|x|}$ is an integer, then both $T_r$ and $T_{r+1}$ are numerically greatest terms (that's the reason I derive commonly used unequality to show you).
Thus, considering $(1+x)^n$ expression,
$$T_r=T_{11}=~^{12}C_{10}\left(\frac{40}{9}\right)^{10}=\frac{12!}{(13-11)!10!}\left(\frac{40}{9}\right)^{10}=66\cdot \left(\frac{40}{9}\right)^{10}$$
Therefore the complete NGT is:
$$\text{NGT}=(3x)^{12}\cdot 66\cdot \left(\frac{40}{9}\right)^{10}= \left(\frac{3}{2}\right)^{12}\cdot 66\cdot \left(\frac{40}{9}\right)^{10}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du
= \int \cos^2{u} \, du \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\
\end{eqnarray*}
Now, I think I am right so far but I do not know have to get rid of the $u$ in
the $\cos^2(u)$ term. Please help.
Thanks
Bob
| Since you have $x=\tan u$, think of $u$ as an angle. Then
$$\tan u = \frac{x}{1}.$$
Draw a right triangle with legs $x$ and $1$ to demonstrate this fact (with $u$ as the angle opposite the $x$.) The hypotenuse is $\sqrt{1+x^2}.$ Now you can evaluate any trig function of $u$ that you please. E.g.,
$$\cos u = \frac{1}{\sqrt{1+x^2}}.$$
So to evaluate
$$\sin 2u = 2\sin u \cos u = 2\frac{x}{\sqrt{1+x^2}}\frac{1}{\sqrt{1+x^2}} = \frac{2x}{1+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding value of $\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$
Finding value of $$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$
Solution I tried:
$$(1+x)^{10}=1+\binom{10}{1}x+\binom{10}{2}x^2+\cdots +\binom{10}{10}x^{10}$$
$$(x+1)^{10}=x^{10}+\binom{10}{1}x^9+\binom{10}{2}x^7+\cdots +\binom{10}{10}$$
I did not find how do multiply terms such that i get my result. Help me
| It is Chu–Vandermonde identity
$$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}=\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{4-2}={12+1\choose 4+1}=1287.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x}{2} \right) \right) ^2}}\,dx\end{align}$$
| Hint:
$$\dfrac1{1+\sin x}=\dfrac1{1+\cos\left(\dfrac\pi2-x\right)}=\dfrac{\sec^2\left(\dfrac\pi4-\dfrac x2\right)}2$$
Alternatively,
$$\dfrac1{1+\sin x}=\dfrac{\sec^2\dfrac x2}{\left(1+\tan\dfrac x2\right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.
$$
How do I proceed further and find the solution or is there any easier way ?
| I would proceed as follows:\begin{align}\int\frac{y^2}{\sqrt{1-y^2}}\,\mathrm dy&=\int y\frac y{\sqrt{1-y^2}}\,\mathrm dy\\&=-y\sqrt{1-y^2}+\int\sqrt{1-y^2}\,\mathrm dy.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Integrate $\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx$
Integrate $\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx$
My Attempt
Put $x=\cos2a\implies dx=-2\sin2a.da$
$$
\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx=\int\tan^{-1}\sqrt{\frac{1-\cos2a}{1+\cos2a}}.-2\sin2a.da\\
=-2\int\tan^{-1}\sqrt{\frac{2\sin^2a}{2\cos^2a}}.\sin2a.da=-2\int\tan^{-1}(\tan a)\sin2a.da
$$
We have $y=\tan^{-1}(\tan a)\implies\tan y=\tan a\implies y=n\pi+a$
$$
\begin{align}
&\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx=-2\int(n\pi+a)\sin2a.da=-2n\pi a-2\int a.\sin2a.da\\
&=-2n\pi a-2\bigg[a\frac{-\cos2a}{2}-\int\frac{-\cos2a}{2}da\bigg]\\
&=-2n\pi a+a.\cos2a+\frac{\sin2a}{2}+C\\
&=-2n\pi.\frac{1}{2}\cos^{-1}x+\frac{1}{2}\cos^{-1}x.x+\frac{\sqrt{1-x^2}}{2}+C\\
&\color{red}{=\frac{1}{2}\bigg[-2n\pi\cos^{-1}x+x\cos^{-1}x-\sqrt{1-x^2}\bigg]}
\end{align}
$$
My reference has the solution $\frac{1}{2}\bigg[x\cos^{-1}x-\sqrt{1-x^2}\bigg]$. But, why am I getting the solution as above ?
| You need to notice that you just substituted $$a=\frac {\arccos x}{2}$$ whose range is itself $$\left[0,\frac {\pi}{2 }\right]$$.
And as you might know that for $\alpha \in \left(\frac {-\pi}{2},\frac {\pi}{2}\right)$ , $$\arctan (\tan \alpha) =\alpha$$
Hence $$\arctan (\tan a) =a$$ Therefore you won't get an $n\pi$ term
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using identity Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using the identity $$\frac{1}{(1-x)(1-x^2)}=\frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right)$$
where $p_k(n)$ is the number of partitions of an integer $n$ into a most $k$ parts. The generating function $P_k(x)$ of $\{p_k(n)\}$ is $$P_k(x) = \frac{1}{\prod_{r=0}^{k}(1-x^k)}$$
Therefore, the generating function $P_2(x)$ of ${p_2(n)}$ is
$$P_2(x) = \frac{1}{(1-x)(1-x^2)}$$
Now I see here that we can use the identity given above, but I am confused of how to apply it to prove the desired statement.
| Note that
$$\begin{align}\frac{4}{(1-x)(1-x^2)}&=\frac{1}{1+x}+\frac{1}{1-x}+\frac{2}{(1-x)^2}\\
&=\frac{1}{1+x}+\frac{1}{1-x}+\frac{d}{dx}\left(\frac{2}{1-x}\right)\\
&=\sum_{n=0}^{\infty}(-x)^n+\sum_{n=0}^{\infty}x^n
+2\sum_{n=1}^{\infty}nx^{n-1}\\
&=\sum_{n=0}^{\infty}((-1)^n+1+2(n+1))x^n
\end{align}$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which one of the following are correct?
Let, $A= \left[ {\begin{array}{cc} -1 & 2 \\ 0 & -1 \\
\end{array} } \right]$ , and $B = A + A^2 + A^3 +···+ A^{50}$. Then
$(A) B^2 = I $
$(B) B^2 = 0$
$(C) B^2 = A$
$(D) B^2 = B$
Eigenvalues of $A$ are $-1,-1$. So, Eigenvalues of $B$ are $0,0$. So, $\det(B)=0$. So, options (A) and (C) can be eliminated. How do I eliminate further?. By Cayley-Hamilton Theorem $A^2+2A+I=0$. I found $A^3,A^4,...$ using Cayley-Hamilton Theorem. It was time-consuming. Given matrix is not diagonalizable. Can you please help me?
| Let us begin by finding the remainder when the polynomial $x + \cdots + x^{50}$ is divided by $(x+1)^2$. So, suppose
$$x + \cdots + x^{50} = (x+1)^2 q(x) + (ax + b).$$
Then by substituting $x := -1$, we see that $-a + b = 0$. Now, if we take the derivative on both sides of the above equation, we see
$$1 + 2x + \cdots + 50 x^{49} = 2 (x+1) q(x) + (x+1)^2 q'(x) + a.$$
Substituting $x := -1$ in this equation gives $a = 1 - 2 + \cdots - 50 = -25$. Thus, $a = b = -25$.
Now, if we substitute $A$ for $x$ in the original equation, we conclude
$$B = A + \cdots + A^{50} = (A+I)^2 q(A) - 25 (A + I) = \begin{bmatrix} 0 & -50 \\ 0 & 0 \end{bmatrix}.$$
Here, we use that $(A + I)^2 = 0$ to eliminate the $(A + I)^2 q(A)$ term. From here, it should be easy to answer the original question.
(Or, if you want a briefer way to eliminate the possibility that $B = 0$: essentially, what we are doing here is showing that $x + \cdots + x^{50}$ is not divisible by $(x+1)^2$ by showing that the polynomial's derivative at $x := -1$ is not zero. Therefore, since $(\lambda+1)^2$ is the minimal polynomial of $A$, it follows that applying this polynomial to $A$ gives a nonzero matrix. On the other hand, since the polynomial is divisible by $x+1$, its square is divisible by $(x+1)^2$, which implies that $B^2 = 0$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I start this integral problem: $\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $? How do I prove this?
$$\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $$
I'm guessing using Riemann zeta function? But then how do I start?
| $$
\begin{align*}
\int_0^1 \frac{\ln^3(x)}{1-x}dx &= \int_0^1\ln^3(x)\sum_{n=0}^\infty x^n \; dx\\
&= \sum_{n=0}^\infty \int_0^1 x^n \ln^3(x) \; dx \\
&= -6\sum_{n=0}^\infty \frac{1}{(n+1)^4} \\
&= -6 \zeta(4) \\
&= -6 \times \frac{\pi^4}{90} \\
&= -\frac{\pi^4}{15}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve this system of equations for the value of $a,b$ and $c$ such that $a^3+3ab^2+3ac^2-6abc=1$,$b^3+3ba^2+3bc^2-6abc=1$,$c^3+3cb^2+3ca^2-6abc=1$ Solve this system of equations for the value of $a,b$ and $c$.
$$a^3+3ab^2+3ac^2-6abc=1$$
$$b^3+3ba^2+3bc^2-6abc=1$$
$$c^3+3cb^2+3ca^2-6abc=1$$
This is symmetric equation.
My first attempt was,
$a^3+3ab^2+3ac^2=1+6abc$,
$b^3+3ba^2+3bc^2=1+6abc$,
$c^3+3cb^2+3ca^2=1+6abc$,
then, from above three equations,
$a^3+3ab^2+3ac^2=b^3+3ba^2+3bc^2=c^3+3cb^2+3ca^2$
Then, what will I continue?
| If we substrac 1. equation -2. equation we get:
$$(a-b)\Big((a^2+ab+b^2) -3ab+3c^2\Big)=0$$
so if $a\ne b$ we get $$(a-b)^2+3c^2=0 \implies a=b, c=0 $$
a contradiction.
So $a=b$. The same way (2.equation-3.equation) we get $b=c$. S0 $a=b=c =:x$ and now you have to solve: $x^3=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluate $\int e^{-3x} \cos^3x\,dx$
Evaluate $\int e^{-3x}\cos^3x\,dx$
My Attempt
\begin{align}
& \int e^{-3x} \cos^3x\,dx=\cos^3x \cdot \frac{e^{-3x}}{-3}-\int3\cos^2x\sin x \cdot \frac{e^{-3x}}{-3} \, dx \\
= {} &\frac{-1}{3}\cos^3x \cdot e^{-3x}+\int\cos^2x\sin x \cdot e^{-3x} \, dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x} - \int\bigg[-2\cos x\sin x\cdot e^{-3x}-3\cos^2x\cdot e^{-3x}\bigg](-\cos x)\,dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x}+2\int \cos^2x\sin x \cdot e^{-3x} \, dx-3\int\cos^3x\cdot e^{-3x}\,dx
\end{align}
How do I solve this integral ?
| Hint:
$$\cos(x)\cos(y) = \dfrac{\cos(y + x) + \cos(y - x)}2\tag{1}$$
Now, using $(1)$,
$$\begin{align}
\int e^{-3x}\cdot\cos^3x\,\mathrm dx &= \int e^{-3x}\cdot\cos x\cdot\cos^2x\,\mathrm dx \\
&= \int e^{-3x}\cdot\cos x\cdot\dfrac{1 + \cos 2x}2\,\mathrm dx \\
&= \dfrac12\int e^{-3x}\cdot\cos x\,\mathrm dx + \dfrac12\int e^{-3x}\cos x\cos 2x\,\mathrm dx\\
&= \dfrac12\int e^{-3x}\cdot\cos x\,\mathrm dx + \dfrac12\int e^{-3x}\cdot\dfrac{\cos3x + \cos x}2\,\mathrm dx \\
&= \dfrac34\int e^{-3x}\cdot\cos x\,\mathrm dx + \dfrac14\int e^{-3x}\cos3x\,\mathrm dx
\end{align}$$
From here onward, solve both integrals by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Solving $\frac{\sqrt{r+1}-\sqrt{r-1}}{\sqrt{r+1}+\sqrt{r-1}}=\log_2\left(|x-2|+|x+2|\right)-\frac{11}{9}$, where $r=\frac{1+x^2}{2x}$ The problems in my sophomore workbook are marked with three colors that signify how hard is the marked problem: green for D and C, yellow for B and A and red for advanced students. All problems are also ordered from the easiest to the hardest.
This problem is marked red and appears last in the section of logarithmic problems, which means that it's the hardest logarithmic problem in the whole workbook. I know what its solution is (there are solutions at the end of the workbook), but I'd like to see how do we arrive at it.
I wrote the solution here, but it's hidden until it's hovered over. I don't want to spoil fun to those who want to find it by themselves.
Find the value(s) of $x$ when the left-hand side is equal to the right-hand side.
$$\frac{\sqrt{\dfrac{1+x^2}{2x}+1}\;-\;\sqrt{\dfrac{1+x^2}{2x}-1}}{\sqrt{\dfrac{1+x^2}{2x}+1}\;+\;\sqrt{\dfrac{1+x^2}{2x}-1}}=\log_2(|x-2|+|x+2|)-\frac{11}{9}$$
The solution is
$$x_1=\frac{7}{9},x_2=\frac{9}{7}$$
| Hint:
Since
$$\sqrt{\frac{1+x^2}{2x}\pm1}= \sqrt{(x\pm1)^2\over 2x} = {|x\pm1|\over \sqrt{2x}}$$
we have
$$\frac{\sqrt{\frac{1+x^2}{2x}+1}-\sqrt{\frac{1+x^2}{2x}-1}}{\sqrt{\frac{1+x^2}{2x}+1}+\sqrt{\frac{1+x^2}{2x}-1}}={|x+1|-|x-1|\over |x+1|+|x-1|}$$
Let $f(x)=|x+2|+|x-2| $, then $f(x)=4$ for $x\in[-2,2]$ and for $|x|>2$ we have $f(x)>4$.
So, for $|x|>2$ we have $\log_2f(x)-11/9>7/9$ so we have $|x+1|>8|x-1|$
a) if $x>2$ we get $x<9/7$ a contradiction and
b) if $x<-2$ we get $x>7/9$ a contradiction again.
So we are left with the $|x|\leq 2$ so we have $$|x+1|=8|x-1|$$
Can you finish?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Integral $\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx$ I am trying to evaluate $$I=\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx=\int_0^1 \frac{\sqrt x (1+x)\ln x} {1+x^3}dx$$ Now if we expand into geometric series: $$I=\sum_{n=0}^{\infty} (-1)^n \int_0^1 (x^{3/2}+x^{1 /2})x^{3n}\ln x dx$$ Also since $$I(k) =\int_0^1 x^kdx=\frac{1} {k+1}$$ Giving: $$I'(k) =\int_0^1 x^k\ln x dx=-\frac{1} {(k+1)^2 }$$ so using this we get $$I=\sum_{n=0}^{\infty} (-1)^{n+1}\left(\frac{1} {(6n+3)^2 }+\frac{1} {(6n+1) ^2 }\right)$$ Now when I plug this into wolfram-alpha the result differs from the value of the integral, also if I multiply by a half it is really close to it. Where did I go wrong?
Edit: Looks like I forgot a 2 in the denominator and to add $+1$ from $I'(k) $ and the correct series should be:$$I=\frac{4}{36}\sum_{n=0}^{\infty} (-1)^{n+1}\left(\frac{1} {(n+5/6)^2 }+\frac{1} {(n+1/2) ^2 }\right)$$ The second one is just $-4G$ where $G$ is the Catalan constant and can you show me how to transform the sum into a closed form? Trigamma or hurwitz zeta function as wolfram alpha gives as a solution. Many thanks in advance!
| $$ \frac{\sqrt{x}}{x^2-x+1} = \sum_{k=0}^\infty (-1)^k \left(x^{3k+1/2} + x^{3k + 3/2}\right) $$ so
$$ \frac{\sqrt{x} \ln(x)}{x^2-x+1} = \left.\sum_{k=0}^\infty (-1)^k \dfrac{d}{dp}\left(x^{3k+1/2+p} + x^{3k+3/2+p}\right)\right|_{p=0}$$
$$ \eqalign{\int_0^1 \frac{\sqrt{x} \ln(x)}{x^2-x+1} \; dx &= \sum_{k=0}^\infty (-1)^k \left.\dfrac{d}{dp} \left( \frac{1}{3k+3/2+p} + \frac{1}{3k+5/2+p} \right)\right|_{p=0}\cr &= \sum_{k=0}^\infty (-1)^{k+1} \left( \frac{1}{(3k+3/2)^2} + \frac{1}{(3k+5/2)^2}\right)\cr}$$
EDIT: In terms of $\zeta(2,v) = \sum_{j=0}^\infty 1/(j+v)^2$, we can write this as
$$ \frac{1}{36} \left(-\zeta(2,1/4) -\zeta(2,5/12) + \zeta(2,3/4)+\zeta(2,11/12)\right) $$
since for $k=2j$, $3k+3/2 = 6 (j+1/4)$, for $k=2j+1$, $3k+3/2 = 6 (j+3/4)$, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Evaluate $\,\lim_{n\to\infty}\frac{n^2}{({n^2+1^2})^{3/2}} + \frac{n^2}{({n^2+3^2})^{3/2}} + \cdots + \frac{n^2}{({n^2+(n-1)^2})^{3/2}}$ $$= \sum_{r=1,3,5 \ldots}^{n-1} \frac{n^2}{({n^2 + r^2})^{\frac32}} = \sum_{r=1,3,5 \ldots}^{n-1} \frac{1}{n({1 + (r/n)^2})^{\frac32}} = \int_{0}^{1}\frac{dx}{({1+x^2})^{\frac32}} = \frac1{\sqrt2}$$ (obtained via subsitution $x=\tan\vartheta$)
Does this answer seems fine?
| Minor corrections
$$
\sum_{r=1,3,5,\ldots}^{n-1} \frac{n^2}{({n^2 + r^2})^{\frac32}} = \frac{1}{n}\sum_{k=1}^{n/2} \frac{1}{\big({1 + \big(\frac{2k-1}{n}\big)^2}\big)^{\frac32}} \to\frac{1}{2}\int_{0}^{1}\frac{dx}{({1+x^2})^{\frac32}} = \frac1{2\sqrt2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a cleaner way of solving this number theory problem? The problem is
If $a$ has order $3\pmod{p}$ where $p$ is an odd prime, show that $(a+1)$ has order $6\pmod{p}$.
My solution:
$$a^3-1\equiv 0\pmod{p}$$
$$(a-1)(a^2+a+1)\equiv 0 \pmod{p}$$
$(a-1)$ cannot be divisible by $p$ because the order of $a$ is $3$, so,
$$a^2+a+1\equiv 0\pmod{p}$$
$$(1)\qquad -a(a+1)\equiv 1\pmod{p}$$
Raising both sides to $6$, we get
$$a^6(a+1)^6\equiv 1\pmod{p}\qquad (a+1)^6\equiv 1\pmod{p}$$
To show that $6$ is the smallest integer $s$ such that $$(2)\quad (a+1)^s\equiv 1\pmod{p}$$ we use contradiciton.
From $(1)$, $(a+1)^3\equiv -1\pmod{p}$ so $s=3$ doesn't satisfy $(2)$. If $s\in\{1,2,4\}$ and $s$ satisfies $(2)$, then that would imply
$a^2\equiv 1\pmod{p}$ but that is a contradiction. If $s=5$, then
$$(a+1)^6\equiv a+1\equiv 1\pmod{p}$$
but this is impossible since $a$ is a least residue.
Is my solution correct? If so, is there a better way of solving this problem ?
| I'll take all congruences modulo $p$.
As you say, $a^2\equiv -a-1$. Therefore $(a+1)^2\equiv -a-1+2a+1\equiv a$
and $(a+1)^3\equiv a^2+a\equiv-1$. From this, we have $(a+1)^6\equiv1$,
$(a+1)^2\not\equiv1$ and $(a+1)^3\not\equiv1$. Thus $a+1$ must have order $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Trigonometric integral using residues I am trying to evaluate following integral by using the Residue Theorem:
$$\int_0^{2\pi}\frac{\cos^2x}{13+12\cos x}dx.$$
Then $\cos x=\frac{z^2+1}{2z}$, $dx=\frac{dz}{iz}$ and we have $f(z)=\frac{(z^2+1)^2}{2z+3}$, $$\int_\gamma\frac{f(z)}{z^2(3z+2)}dz$$
The poles $z=0,\frac{-2}{3}$ are inside $\gamma=\{z\in C:|z|=1\}$, but I can't partition off simple fractions.
| Your reduction of the real integral to the complex one is correct
$$\int_0^{2\pi}\frac{\cos^2 x}{13+12\cos x}dx=\frac{1}{4i}\int_{|z|=1}f(z)dz
=\frac{\pi}{2}\left(\text{Res}(f,0)+\text{Res}\left(f,-\frac{2}{3}\right)\right)$$
where $\displaystyle f(z)=\frac{(z^2+1)^2}{z^2(2z+3)(3z+2)}$.
Now by the partial fraction decomposition there are constant $A,B,C,D,E$ such that $$f(3z+2z)=\frac{A}{z}+\frac{B}{z^2}+\frac{C}{2z+3}+\frac{D}{3z+2}+E.$$ You need to find $A$ and $D$.
Instead of looking for the partial fraction decomposition of $f$ why don't you compute the residues directly?
Take a look at Calculating residues. Then
1) For the residue at $-2/3$, which is a simple pole, consider the limit
$\lim_{z\to -2/3}(z+2/3)f(z)$.
2) For the residue at $z=0$, whose order is two, you may use the limit formula for higher order poles or just note that
$$\frac{(z^2+1)^2}{(2z+3)(3z+2)}=\frac{1+2z^2+z^4}{6(1+\frac{2z}{3})(1+\frac{3z}{2})}=\frac{1}{6}\left(1-\frac{2z}{3}-\frac{3z}{2}+o(z)\right).$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another way to find the sum $S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}$ I want to know the sum
$$S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}.$$ This is my way.
First, we find the sum
$$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}=\sum _{k=1}^n (2 k-1) x^{k-1}.$$
We have
$$\sum _{k=1}^n (2 k-1) x^{k-1} = 2\sum _{k=1}^n k\cdot x^{k-1}-\sum _{k=1}^n x^{k-1}. $$
Note that
$$\sum _{k=1}^n k\cdot x^{k-1} = \left (\sum _{k=1}^n x^k\right)' = \left (\dfrac{x\left(x^n-1\right)}{x-1}\right)'=\dfrac{n x^{n+1}-(n+1)x^n+1}{(x-1)^2}.$$
Another way
$$\sum _{k=1}^n x^{k-1} = \dfrac{x^n-1}{x-1}.$$
From the above results, we have
$$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}= \dfrac{(2 n-1) x^{n+1}-(2 n+1)x^n+x+1}{(x-1)^2}.$$
With $x=-\dfrac{1}{2}$,we get
$$S=\dfrac{2^n + (-1)^{n+1} \cdot(6n+1)}{9 \cdot 2^{n-1}}.$$
How to find the result with another way?
| I would have done it in a similar but different way. Consider the polynomial$$1-3x^2+5x^4-7x^6+\cdots+(-1)^{n-1}(2n-1)x^{2n-2}.$$If its sum is denoted by $f(x)$ and if$$F(x)=x-x^3+x^5-x^7+\cdots+(-1)^{n-1}x^{2n-1},$$then $F'=f$. On the other hand$$F(x)=x\left(1-x^2+x^4-x^6+\cdots+(-1)^{n-1}x^{2n-2}\right)=\frac{x-(-1)^nx^{2n}}{1+x^2}.$$Therefore,$$f(x)=\frac{-(-1)^n (2 n+1) x^{2 n}+(-1)^n (1-2 n) x^{2n+2}-x^2+1}{(1+x^2)^2}$$and, in particular, your sum is$$f\left(\sqrt{\frac12}\right)=\frac{1}{9} 2^{1-n}\bigl(2^n-6 (-1)^n n-(-1)^n\bigr).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find an element of order $5$ of the field $\mathbb{Z}_3[x]/\langle x^4 + x + 2 \rangle$. Find an element of order $5$ of the field $\mathbb{Z}_3[x]/\langle x^4 + x + 2 \rangle$.
I tried the systematic method of considering
$$1=(ax^3 + bx^2 + cx + d)^5$$
and using the relation $x^4 = 2x+1$ but the algebra got messy. Any idea?
| It is maybe good to know that the one or the other computer algebra system can deal systematically with computations in finite fields.
For instance, using sage, a small dialog with the sage interpreter detects all elements of multiplicative order $=5$, the code has a digestible mathematical syntax, that will be explained in the sequel:
sage: R.<X> = PolynomialRing( GF(3) )
sage: F.<x> = GF( 3^4, modulus=X^4+X+2 )
sage: F.multiplicative_generator()
x
sage: F.multiplicative_generator().multiplicative_order()
80
sage: for k in [ 16, 32, 48, 64 ]:
....: print ( "x^%s = %s is an element of multiplicative order %s"
....: % (k, x^k, (x^k).multiplicative_order() ) )
....:
....:
x^16 = 2*x^3 + x + 2 is an element of multiplicative order 5
x^32 = x^3 + 2*x^2 + 2 is an element of multiplicative order 5
x^48 = 2*x^2 + 2*x + 2 is an element of multiplicative order 5
x^64 = 2*x^2 + 2 is an element of multiplicative order 5
sage: [ y for y in F if not y.is_zero() and y.multiplicative_order() == 5 ]
[2*x^3 + x + 2, x^3 + 2*x^2 + 2, 2*x^2 + 2*x + 2, 2*x^2 + 2]
We have solved twice, without complications...
Comments for the code. Above, we have first initialized the ring $R=\Bbb F_3[X]$, since we need the "modulus" $X^4+X+2$, we construct $F=\Bbb F_3[X]/(X^4+X+2)$, this is a field with $3^4=81$ elements, then the group of its non-zero elements (with the multiplication) is a group of order $80$, it is cyclic, a generator is $x$, given by sage. (Of course, $y^{80}=1 for all $y\in F$, $y\ne 0$.)
Knowing it, we expect that all elements of order $5$ are of the shape
$$
x^{16k}\ ,\qquad k=1,2,3,4\ .
$$
The code asks for the explicit representation of these elements, checks also their multiplicative order.
One can also use the one-liner [ y for y in F if not y.is_zero() and y.multiplicative_order() == 5 ] to get them in an unstructural way.
Not needed, but to do something concrete in the direction of a human effort, let us check for instance (humanly) that the last element, $2(x^2+1)=-(x^2+1)$ has order $5$, computations get simpler if we use the Frobenius, so...
$$
\begin{aligned}
(-(x^2+1))^5
&=-(x^2+1)^5
=(?)\ ,
\\
&\qquad\text{ so let us start with a related expression}\dots
\\[2mm]
(x^2+1)^6
&=(x^2+1)^3(x^2+1)^3
\\
&=(x^6+1)(x^6+1)
\\
&=x^{12}-x^6+1
\\
&=(x^4)^3-x^4\cdot x^2+1
\\
&=(1-x)^3-(1-x)\cdot x^2+1
\\
&=(1-x^3)-(x^2-x^3)+1
\\
&=-(x^2+1)\ ,\qquad\text{ so}
\\[2mm]
-(x^2+1)^5
&=
1\ .
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Differentiating $x = \sec^2{3y}$ for $d^2y/dx^2$ I want to ask a question about differentiating trigonometric functions.
I am trying to find $\frac{d^2y}{dx^2}$ for $x = \sec^2{3y}$
Now, this is what I did.
$$\frac{dx}{dy} = 6\sec^2{3y}\tan{3y} = 6x(x-1)^{\frac{1}{2}}$$
I got the $(x-1)^{\frac{1}{2}}$ term as follows:
$$\sin^2{3y} + \cos^2{3y} = 1$$
$$\tan^2{3y} + 1 = \sec^2{3y} $$
$$\tan^2{3y} + 1 = x $$
$$\tan^2{3y} = x - 1$$
$$\tan{3y} = (x - 1)^{\frac{1}{2}}$$
hence $$\frac{dy}{dx} = \frac{1}{6x(x-1)^{\frac{1}{2}}}$$
Now, to differentiate again for $\frac{d^2y}{dx^2}$, I use the quotient property of differentiation:
$$\frac{d^2y}{dx^2} = \frac{-\left(6(x-1)^\frac{1}{2} + 3x(x-1)^{-\frac{1}{2}} \right)}{36x^2(x-1)}$$
$$= \frac{-\left(6(x-1) + 3x \right)}{36x^2(x-1)^{\frac{3}{2}}}$$
$$= \frac{6-9x}{36x^2(x-1)^{\frac{3}{2}}}$$
$$= \frac{3-2x}{12x^2(x-1)^{\frac{3}{2}}}$$
But what I did originally was find the second derivative for $\frac{d^2x}{dy^2}$ as follows:
$$\frac{dx}{dy} = 6x(x-1)^{\frac{1}{2}}$$
$$\frac{d^2x}{dy^2} = 6(x-1)^\frac{1}{2} + 3x(x-1)^{-\frac{1}{2}}$$
so why is it that $\frac{1}{\frac{d^2x}{dy^2}}$ is not the same as $\frac{d^2y}{dx^2}$ when in the first part, $\frac{1}{\frac{dx}{dy}} = \frac{dy}{dx}$?
| Because in general, $\dfrac{d^2 x}{dy^2}\dfrac{d^2 y}{dx^2}\ne 1$. To take a much simpler example, if $y=x^3$ then $x=y^{1/3}$ and $$\dfrac{d^2 x}{dy^2}=-\dfrac{2}{9}y^{-5/3}=-\dfrac{2}{9}x^{-5},\,\dfrac{d^2 y}{dx^2}=6x.$$In general, $$\frac{d^2 x}{dy^2}=\frac{d}{dy}((\frac{dy}{dx})^{-1})=-(\frac{dy}{dx})^{-2}\frac{d}{dy}\frac{dy}{dx}=-(\frac{dy}{dx})^{-3}\frac{d}{dx}\frac{dy}{dx}=-(\frac{dy}{dx})^{-3}\frac{d^2 y}{dx^2}.$$Going back to my example, both sides are $-\frac{2}{9}x^{-5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Catenoid is a minimal surface i want to show that the catenoid is a minimal surface. I have given
$f:I \times (0,2\pi)\longrightarrow \mathbb{R}^3$ with $f(r,\phi)=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\r\end{array} \right)$.
I know that:
$f$ is minimal surface $\Longleftrightarrow$ $\Delta f=0$.
$f$ is given in polar coordinates so i have to calculate the following:
$\Delta f= \frac{\partial^2f}{\partial r^2}+\frac{1}{r}\frac{\partial f}{\partial r}+\frac{1}{r^2}\frac{\partial^2f}{\partial \phi^2}$
$\frac{\partial f}{\partial r}= \left( \begin{array}{c}\sinh(r) \;\cos(\phi)\\\sinh(r) \;\sin(\phi)\\1\end{array} \right)$ , $\frac{\partial^2f}{\partial r^2}=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\0\end{array} \right)$
$\frac{\partial f}{\partial \phi}=\left( \begin{array}{c}-\cosh(r) \;\sin(\phi)\\\cosh(r) \;\cos(\phi)\\0\end{array} \right)$ , $\frac{\partial^2 f}{\partial \phi^2}=\left( \begin{array}{c}-\cosh(r) \;\cos(\phi)\\-\cosh(r) \;\sin(\phi)\\0\end{array} \right)$.
But when I put all together I can not show that $\Delta f$ is 0. What did I do wrong?
Can someone help me please?
Thanks in advance.
|
This not a complete answer but that's too long for me to post it in comment.
Preliminary for differential geometry of surfaces
\begin{align*}
\mathbf{x}(u,v)
&= \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} \\
\mathbf{x}_u &= \frac{\partial \mathbf{x}}{\partial u} \\
\mathbf{x}_v &= \frac{\partial \mathbf{x}}{\partial v} \\
\mathbf{N} &=
\frac{\mathbf{x}_u \times \mathbf{x}_v}{|\mathbf{x}_u \times \mathbf{x}_v|}
\tag{unit normal vector} \\
\end{align*}
First fundamental form
$$\mathbb{I}=
\begin{pmatrix} E & F \\ F & G \end{pmatrix}=
\begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} $$
Second fundamental form
$$\mathbb{II}=
\begin{pmatrix} e & f \\ f & g \end{pmatrix}=
-\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix}$$
Metric
$$ds^2=E\, du^2+2F\, du\, dv+G\, dv^2$$
Element of area
$$dA=|\det \mathbb{I}| \, du \, dv
=|\mathbf{x}_u \times \mathbf{x}_v| \, du\, dv
=\sqrt{EG-F^2} \, du\, dv$$
Principal curvatures
Let $\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}=
\mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}$
where $\mathbb{A}=
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$.
Now
\begin{align*}
\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} &=
\mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} \\
-\begin{pmatrix} e & f \\ f & g \end{pmatrix} &=
\mathbb{A} \begin{pmatrix} E & F \\ F & G \end{pmatrix} \\
\mathbb{A} &=
-\begin{pmatrix} e & f \\ f & g \end{pmatrix}
\begin{pmatrix} E & F \\ F & G \end{pmatrix}^{-1}
\end{align*}
The principal curvatures $k_{1}, k_{2}$ are the eigenvalues of $-\mathbb{A}$.
Mean curvature
$$H=\frac{k_{1}+k_{2}}{2}
=-\frac{1}{2} \operatorname{tr} \mathbb{A}
=\frac{eG-2fF+gE}{2(EG-F^2)}$$
Gaussian curvature
$$K=k_{1} k_{2}
=(-1)^{2} \det \mathbb{A}
=\frac{eg-f^2}{EG-F^2}$$
For minimal surface, $$H=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of given sequence Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1$$
My try:
$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$.
Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$
$$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$
$\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$
$\Rightarrow x^3-2x+1=0$
and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$
So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.
how can i say that the limit is
$\dfrac {-1 + \sqrt {5}}{2}$?
| We will prove that all $a_n$ are smaller than ${2 \over 3}=0.6666...$.
By induction, suppose that $0, 1/2, ... a_{n-1}, a_n < 2/3$.
then $a_{n+1} < {1 + 2/3 + 8/27 \over 3 }= {53 \over 81} < {54 \over 81} = {2\over 3}$
since $a_1 =0<{2 \over 3}$, for all n , $0 \leqslant a_n < {2 \over 3}$.
To prove the convergence we will show that a_n is increasing. Again, by induction,
$a_{n+1} - a_n = {a_n - a_{n-1} \over 3} + { (a_{n-1} - a_{n-2}) ( a_{n-1}^2 + a_{n-1} a_{n-2} + a_{n-2}^2 ) \over 3} > 0$ where
$a_4-a_3 = {1 \over 24} > 0 $ and $a_5- a_4 = {1 \over 72} > 0 $ are the two first consecutive positive differences.
We have here a strictly increasing bounded sequence i.e. a convergent one.
So the limit calculus in the question is valid. The limit is $\dfrac {-1 + \sqrt {5}}{2} \approx 0.618$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How to simplify a polynomial expression? Let's say I have a polynomial expression as:
$2 - 3x + x^3 $
How do I simplify the above expression to:
$(1 - x)^2(2 + x)$
| In this case where you know the ending expression, you can add-and-subtract in such a way that forces terms of the final expression to appear. For example, to force $2+x$ to appear, we have:
\begin{align*}
2-3x+x^3&=2+\color{red}{x}-\color{red}{x}-3x+x^3\\
&=2+x-4x+x^3\\
&=2+x-4x-\color{red}{2x^2}+\color{red}{2x^2}+x^3\\
&=(2+x)-2x(2+x)+x^2(2+x)\\
&=(2+x)(1-2x+x^2).
\end{align*}
Each time $\color{red}{\text{add}}$-and-$\color{red}{\text{subtract}}$ happens, it does so in a way to induce $x+2$ to happen. Now you can do the same to $1-2x+x^2$ to make $1-x$ appear:
$$
1-2x+x^2=1-\color{red}{x}+\color{red}{x}-2x+x^2=1-x-x+x^2=(1-x)-x(1-x)=\ldots
$$
How does one know $2+x$ is a factor of $2-3x+x^3$? In general, $px+q$ with $p\neq 0$ is factor of a polynomial $P(x)$ iff $-\frac{q}{p}$ is a root of $P(x)$. And to find such $p$ and $q$, you can refer to the Rational root theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$3^n-1$ is divisible by $4 \implies n $ is even What is the easiest ways to prove this: $3^n-1$ is divisible by $4 \implies n $ is even? Moreover, how would I figure out that $n$ must be even if I didn't know the result?
My approach is this: suppose $n=2k+1$. Consider $3^{2k+1}-3$. If we show that it is divisible by $4$, it will follow that $3^n-1$ has remainder $2$ in the division by $4$ if $n$ is odd. We have $3^{2k+1}-3=3(3^k-1)(3^k+1)$. Each of the two brackets is divisible by $2$, so the whole thing is divisible by $4$. Is this correct?
| If you know modulo arithmetic $3^n \equiv (-1)^n \equiv 1\mod 4 \iff n$ is even
So $3^n - 1\equiv 1-1\equiv 0 \mod 4 \iff n$ is even.
If you don't know moulo arithmetic then just... do it.
Let $n = 2k$ and notice $3^{2k} - 1 = 9^k - 1= (8 + 1)^k - 1 =8^k + 8^{k-1}k + {k\choose 2}8^{k-2} + ...... + 8k + 1 - 1= 8^k + 8^{k-1}k + {k\choose 2}8^{k-2} + ...... + 8k$ is divisible by $4$.
But If $n = 2k + 1$ then $3^{2k + 1} - 1 = 9^k*3 - 1 =(8+1)^k*3 - 1= 3((8 + 1)^k - 1 =8^k + 8^{k-1}k + {k\choose 2}8^{k-2} + ...... + 8k + 1)-1 = 3((8 + 1)^k - 1 =8^k + 8^{k-1}k + {k\choose 2}8^{k-2} + ...... + 8k) + 3-1 = 3((8 + 1)^k - 1 =8^k + 8^{k-1}k + {k\choose 2}8^{k-2} + ...... + 8k) + 2$.
$3((8 + 1)^k - 1 =8^k + 8^{k-1}k + {k\choose 2}8^{k-2} + ...... + 8k)$ is divisible by $4$ so $3((8 + 1)^k - 1 =8^k + 8^{k-1}k + {k\choose 2}8^{k-2} + ...... + 8k)+2$ will have remainder $2$ when divided by $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $y$ in the equation $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$ As part of a larger calculation, I came across $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$, which I now have to solve for $y$. My initial idea, besides a failed attempt to use the general quadratic formula, was, incorrectly:
\begin{align}
\dots \Leftrightarrow 6xy^2 + 24xy + 24x &= 3y^2 + 14y + 16 \\
\Leftrightarrow \frac{d}{dy} 6xy^2 + 24xy + 24x &= 3y^2 + 14y + 16 \frac{d}{dy} \\
\Leftrightarrow 12xy + 24x &= 6y + 14 \\
\Leftrightarrow (12x-6)y &= 14 - 24x \\
\Leftrightarrow y &= \frac{14 - 24x}{12x-6} \\
\Leftrightarrow y &= \frac{7 - 12x}{6x-3}
\end{align}
This is very "close" to the correct solution as given by WolframAlpha: $y = \frac{8 - 12x}{6x - 3}$. However, the derivation step is incorrect, because $x$ is not independent of $y$ and thus can't be regarded as a constant.
Obviously, the result from WA doesn't help much without a derivation, so that's what I'm looking for here (and maybe some useful tips for dealing with such expressions in general).
Thanks in advance!
| I do not get, why you try to take the derivative. Why should that help in solving this equation for y?
Anyways:
$$x=\frac{3y^2+14y+16}{6y^2+24+24}=\frac{3y^2+14y+16}{6(y^2+4y+4)}=\frac{3y^2+6y+8y+16}{6(y+2)^2}=\frac{3y(y+2)+8(y+2)}{6(y+2)^2}=\frac{(y+2)(3y+8)}{6(y+2)^2}=\frac{3y+8}{6(y+2)}$$
Now you can proceed:
$$x=\frac{3y+8}{6(y+2)}$$
$$6(y+2)x=3y+8$$
Bring everything with $y$ on one side:
$$6yx-3y=8-12x$$
Factor out $y$:
$$y(6x-3)=8-12x$$
Divide by $6x-3$:
$$y=\frac{8-12x}{6x-3}$$
Note, that it is "lucky" that we can reduce the fraction to linear numerator and denominator and we do not have to deal with quadratic expression, which would lead into a more complicated calculation involving the quadratic formula.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $a_{n+1}^2=a_n·a_{n+2}+(-1)^n$ for Fibonacci sequence $\{a_n\}$ Let $a_1,a_2,a_3,...,a_n$ be the sequence of Fibonacci numbers.
Then we are required to prove by induction that $$a_{n+1}^2=a_n \cdot a_{n+2} + (-1)^n$$
My Attempt:
Initially putting a few values like $n=1,2 \: \text{or}\ 3$ makes it certain that we can move ahead.
Following the conventional way we can assume that the proposition holds true for some $k$ s.t.$$a_{k+1}^2=a_k \cdot a_{k+2} + (-1)^k \tag1$$
To prove that the proposition holds for all values of $n$, we need to prove for the same for $n=k+1$ too.
In short we need to prove that:
$$a_{k+2}^2=a_{k+1} \cdot a_{k+3} + (-1)^{k+1} \tag2$$
We know that
$$ a_{k+2}^2=(a_{k+3}-a_{k+1})^2$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_{k+1}^2-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_k \cdot a_{k+2} + (-1)^k-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}[a_{k+3}- 2\cdot a_{k+1}]+a_k \cdot a_{k+2} + (-1)^k \tag{*}$$
It is easy to see that $$a_{k+3}-a_{k+1}=a_{k+2} \tag3$$
And $$a_{k+2}-a_{k+1}=a_{k} \tag4$$
Adding equation $3$ and $4$, we get$$a_{k+3}-2a_{k+1}=a_{k} \tag5$$
Putting eq$(5)$ in eq$(*)$
$$a_{k+2}^2=a_{k+3}[a_k]+a_k \cdot a_{k+2} + (-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k[a_{k+3}+a_{k+2}]+(-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k(a_{k+4})+(-1)^k$$
This I indeed completely different from something that I wanted to prove. Is there any way to get ahead of this or is there something wrong with what I have done?
| A sneaky linear algebra way. Let $$A_{n}=\begin{pmatrix}a_{n+1}&a_{n}\\a_{n+2}&a_{n+1}\end{pmatrix}$$
Then show that $$A_{n+1}=\begin{pmatrix}0&1\\ 1&1\end{pmatrix}A_n$$
using the recurrence relation.
Then show $$\det A_{n+1} = -\det A_n\tag{1}$$
Finally, the only induction you need is to show, using (1), that $$\det A_{n}=(-1)^{n-1}\det A_1$$
which makes the induction simple.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Where is the mistake in my reasoning? I have a statement that says:
If $a^2 + b^2 + c^2 = 2$ and $(a + b + c)(1 + ab + bc + ac) = 3^2$
What is the value of $( a + b + c )$ ?
My reasoning was:
$a^2 + b^2 + c^2 = 2$, rewritten as:
*
*$(a + b + c)^2 = 2 + 2ab + 2ac + 2bc$
Since, $(a + b + c)(1 + ab + bc + ac) = 3^2$
*$(1 + ab + bc + ac) = \frac{9}{(a + b + c)}$
Now, replacing in the 1.
*Factorize $(a + b + c)^2 = 2(1 + ab + ac + bc)$
*Replacing $(a + b + c)^2 = \frac{18}{a + b + c}$
*Multiplying by $(a + b + c)$
*$(a + b + c)^3 = 18$. Thus $a + b + c = \sqrt[3]{18}$. That is my result.
But, the correct result should be $4$, then where is my mistake ?
| Your work is perfectly correct. But I'm pretty sure you misinterpreted the original question: there should be $\color{green}{32}$, not $\color{red}{3^2}$, in the given data. With that value, the same work as you did will indeed yield $4$ as the final answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
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What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.
My attempt
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
$$\left(\sqrt{x+4+2\sqrt{x+3}}\right)^2=\left(\frac{x+8}{3}\right)^2$$
$$x+4+2\sqrt{x+3}=\frac{x^2+16x+64}{9}$$
$$9x+36+18\sqrt{x+3}=x^2+16x+64$$
$$-x^2-7x-28=-18\sqrt{x+3}$$
$$x^2+7x+28=18\sqrt{x+3}$$
$$(x^2+7x+28)^2=(18\sqrt{x+3})^2$$
$$x^4+7x^3+28x^2+7x^3+49x^2+196x+28x^2+196x+784=324x+972$$
$$x^4+14x^3+105x^2+68x-188=0$$
Where is my mistake? Even if this were true, I still wouldn't be able to solve it without a calculator (I can't use Rational Root Theorem on such a big numbers!).
By the way, the solution should be (again, according to Symbolab) $x \in \{1,-2\}$.
| You can brutally use Ferrari formula but no need here.
Search for obvious Roots , here 1 is clearly an obvious Roots then factorise by :
$$X-1$$
And try again with obvious roots. $0;1;2;-1,i,-i$
| {
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What is the relative maximum or minimum and the point of inflection?
$f(x) = ax^3+bx^2+cx +d$, determine a, b, c, and d such that the graph of $f$ has a extreme in $(0,3)$ and a point of inflection in $(-1,1)$.
When is a quadratic I know that the formula $V=(\frac{-b}{2a},\frac{-\triangle}{4a})$ gives the maximum or minimum of the function, but for $f(x) = ax^3+bx^2+cx +d$, not seems the same. I tried to make $f(x) = x(ax^2+bx+c) +d$ but doesn't work and for the point of inflection I know that is the point of the second derivative is equal zero and change of the signal, but how I can make the function change in (-1,1)?
| Extremum at $(0,3)$:
We need $f'(0) = 0$ and $f''(0) \ne 0$ for a local minimum or maximum at $x=0$, so
$$
f'(x) = 3ax^2 + 2bx + c \Rightarrow
f'(0) = c = 0 \\
f''(x) = 6ax + 2b \Rightarrow
f''(0) = 2b \ne 0
$$
then $f(0) = 3$ was required:
$$
f(0) = d = 3
$$
Inflection point at $(-1, 1)$:
Further we need $f''(-1) = 0$, because the curvature
$$
k(x)= \frac{f''(x)}{(1+f'(x)^2)^{3/2}}
$$
is required to change sign for a point of inflection, and it has the same sign as $f''$, so
$$
f''(x) = 6ax + 2b \Rightarrow
f''(-1) = -6a + 2b = 0
$$
and $f(-1) = 1$ was required:
$$
f(-1) = -a + b -c + d = -a + b+3 = 1
$$
so we need to solve
$$
-6a + 2b = 0 \\
-a + b + 3 = 1
$$
which is equivalent to
$$
b = 3a \\
-a + 3a = -2
$$
which is equivalent to
$$
a = -1 \\
b = -3
$$
Checking the solution:
This gives
$$
(a, b, c, d) = (-1, -3, 0, 3)
$$
which is the function
$$
f(x) = -x^3 - 3x^2 + 3
$$
It has the derivatives
$$
f'(x) = -3x^2 -6x \\
f''(x) = -6x - 6
$$
with $f'(0) = 0$, $f''(0) = -6 \ne 0$, $f(0) = 3$, so we have a local maximum at $(0,3)$.
Further it has $f''(-1) = 0$, $f(-1) = 1$. Then we have for $\epsilon > 0$
$$
f''(-1 - \epsilon) = -6(-1-\epsilon) - 6 = 6 \epsilon > 0 \\
f''(-1 + \epsilon) = -6(-1+\epsilon) - 6 = -6 \epsilon < 0
$$
so we have a change of sign of $f''$ and thus also for the curvature $k$ around $x=-1$, which makes $(-1,1)$ a point of inflection.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the recurrence $T(n) = T(n-1) + 2T(n-2) + 2$ $T(0) = 1,
T(1) = 0$
I ain't able to get answer from any of the methods.
Substitution:
$T(n) = x^n $
\begin{align}
& x^n = x^{n-1} + 2x^{n-2} + 2 \\
& x^2 = x + 2 + 2x^2\\
& x^2 + x + 2 =0
\end{align}
solving this I will get a complex root.
\begin{align}
x & = \frac{-1 \pm \sqrt{1-8}}{2}
x & = \frac{-1 \pm 7i}{2}
\end{align}
Now how to go further.
General:
\begin{align}
T(2) & = 4\\
T(3) & = 6\\
T(4) & = 16\\
T(5) & = 30\\
T(6) & = 64\\
T(7) & = 126\\
T(8) & = 256
\end{align}
From this i can deduce if $n$ is even then $2^n$.
I cant deduce for if $n$ is odd.
| Put $U(n) - 1 = T(n)$ and the equation is
$$
U(n+2) = U(n-1) + 2U(n)
$$
let as you do $U(n) = x^n$ and we get
$$
x^{n+2} = x^{n-1}+2x^n
$$ or we need to solve
$$
x^2-x-2 = 0
$$
hence
$$
x = \frac{1}{2} \pm \sqrt{1/4+8/4} = \frac{1 \pm 3}{2}
$$
So
$$
U(n) = A 2^n + B (-1)^n
$$ and
$$
T(n) = A 2^n + B (-1)^n - 1
$$
So
$$
T(0) = A+B-1 = 1
$$ and
$$
T(1) = 2A-B -1 = 0
$$ gives
$$
A = 1, \,B=1
$$
And we conclude
$$
T(n) = 2^n+(-1)^n-1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prime number and golden ratio I think the following is true, but can't show it.
Let $p$ be a prime number, and let $f(x)$ be a polynomial which satisfies
$${\Bigl(1-x+\frac{1}{x}\Bigr)}^p-1=f(x)+f\Bigl(-\frac{1}{x}\Bigr).$$
Then
$$f(0)\neq0\pmod{p^2}.$$
This question is equivalent to my previous question.
| In this answer I'll prove the identity
$$4f(0)\equiv -p\sum_{k=1}^{(p-1)/2}\frac{5^k-1}k\pmod{p^2}\tag 1$$
which can be useful in order to solve the main problem.
If $p\equiv 1\pmod 5$ or $p\equiv 4\pmod 5$, then the equation $x^2-x-1$ has two roots in $\Bbb Z/p\Bbb Z$, say $u,v$.
If $\delta$ denote the $p$-derivation $x\mapsto\frac{x-x^p}p$, then from (1) we get:
$$4f(0)\equiv-2p(u-3)\delta(u)\pmod{p^2}$$
Since $u\not\equiv 3\pmod p$, we get $f(0)\equiv 0\pmod{p^2}$ if and only if $\delta(u)\equiv 0\pmod p$ which is equivalent to $u^{p-1}\equiv 1\pmod{p^2}$.
proof of (1): Starting from
\begin{align}
\left(x-\frac 1x+1\right)^p
&=\sum_{a+b+c=p}\frac{p!}{a!b!c!}x^a\left(-\frac 1x\right)^b\\
&=\sum_{a+b+c=p}\frac{p!}{a!b!c!}(-1)^bx^{a-b}
\end{align}
we get for odd $p$
\begin{align}
2f(0)
&=-1+\sum_{2a+c=p}\frac{p!}{(a!)^2c!}(-1)^b\\
&=-1+\sum_{a=0}^{(p-1)/2}\frac{p!}{(a!)^2(p-2a)!}(-1)^a\\
&=\sum_{a=1}^{(p-1)/2}\frac{p!}{(a!)^2(p-2a)!}(-1)^a\\
&=p\sum_{a=1}^{(p-1)/2}\frac{(p-1)!}{(a!)^2(p-2a)!}(-1)^a
\end{align}
Now
\begin{align}
\frac{(p-1)!}{(p-2a)!}
&=\prod_{n=1}^{2a-1}(p-n)\\
&\equiv-\prod_{n=1}^{2a-1}n\\
&\equiv -(2a-1)!\pmod p
\end{align}
hence
$$2f(0)\equiv -p\sum_{a=1}^{(p-1)/2}(-1)^a\frac{(2a-1)!}{(a!)^2}\pmod{p^2}$$
Recall that
$$(-1)^a\binom{2a}a=4^a\binom{-1/2}a\equiv 4^a\binom{(p-1)/2}a\pmod p$$
Starting from
$$\sum_{a=0}^{(p-1)/2}\binom{(p-1)/2}ax^a=(1+x)^{(p-1)/2}$$
we get
\begin{align}
\sum_{a=1}^{(p-1)/2}\binom{(p-1)/2}a\frac{x^a}a
&=\sum_{a=1}^{(p-1)/2}\binom{(p-1)/2}a\int_0^x t^{a-1}\mathrm dt\\
&=\int_0^x\frac{(1+t)^{(p-1)/2}-1}t\mathrm dt\\
&=\int_1^{1+x}\frac{u^{(p-1)/2}-1}{u-1}\mathrm du\\
&=\sum_{k=1}^{(p-1)/2}\int_1^{1+x}u^{k-1}\mathrm du\\
&=\sum_{k=1}^{(p-1)/2}\frac{(1+x)^k-1}k
\end{align}
Finally, for $x=4$, we get
\begin{align}
\sum_{k=1}^{(p-1)/2}\frac{5^k-1}k
&=\sum_{a=1}^{(p-1)/2}\binom{(p-1)/2}a\frac{4^a}a\\
&\equiv\sum_{a=1}^{(p-1)/2}\frac{(-1)^a}a\binom{2a}a\\
&\equiv 2\sum_{a=1}^{(p-1)/2}(-1)^a\frac{(2a-1)!}{(a!)^2}\pmod p
\end{align}
from which the assertion follows.
| {
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} |
General solution of the ordinary differential equation $(D^4+D^2+1)y=0$ $(D^4+D^2+1)y=0$ where $D=\frac{d}{dx}$
$$D^4+D^2+1 =0 \Rightarrow D^4 + D^2 + \frac14 +\frac{{3}}{4} = 0 \Rightarrow (D^2 + \frac12) = \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D^2 = \frac{-1}{2} \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D= \pm \sqrt{\frac{-1}{2} \pm \frac{\sqrt{3\iota}}{2}} = \pm \sqrt{e^{\pm(2/3)\pi\iota}} = \pm e^{\pm(1/3)\pi\iota} $$
Is there a way to simplify this?
| Your equation is
$$x^4+x^2+1=0$$
Let $y=x^2$, then we have that:
$$y^2+y+1=0$$
And the solution is $y=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$, or in exponential form: $y_1=e^{\frac{2}{3}i\pi}$ and $y_2=e^{-\frac{2}{3}i\pi}$. And from $x^2=y$, we get that $x_{12}=\exp\left(\frac{\frac{2}{3}i\pi+2ni \pi}{2}\right)$ and $x_{34}=\exp\left(\frac{-\frac{2}{3}i\pi+2ni \pi}{2}\right)$ for $n=0$ and $n=1$. So the roots are:
$$x_1=\exp\left(\frac{1}{3}i \pi\right)$$
$$x_2=\exp\left(\frac{4}{3}i \pi\right)$$
$$x_3=\exp\left(-\frac{1}{3}i \pi\right)=\exp\left(\frac{5}{3}i \pi\right)$$
$$x_4=\exp\left(\frac{2}{3}i \pi\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
| Suppose that you have 4 consecutive numbers $a, b, c, d$. They can be expressed as $a=t-\frac{3}{2}$, $b=t-\frac{1}{2}$, $c=t+\frac{1}{2}$ and $d=t+\frac{3}{2}$ for some number $t$.
Now,
$$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = t^2 - \left(\frac{3}{2}\right)^2 = t^2 - \frac{9}{4}$$
and
$$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = t^2 - \left(\frac{1}{2}\right)^2 = t^2 - \frac{1}{4}$$
If we define $y = t^2 - \frac{5}{4}$, we have
$$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = y - 1$$
and
$$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = y + 1$$
Furthermore, since the LHS in both cases is an integer, it is clear that $y$ is an integer.
So the product of all four numbers is
$$ abcd = (y - 1)(y+1) = y^2 - 1 $$
one less than the square of an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 11
} |
Calculating volume using cylindrical coordinates.
A solid $D$ is defined by the following inequalities: $$\begin{align}x^2+y^2+(z-1)^2 &\le 1\\ z^2 &\le x^2+y^2\end{align}$$ Calculate the volume of $D$.
Attempt to solve:
$$\begin{align}x&=r\cos\theta \\ y&=r\sin\theta\end{align}$$
Plugging in these values into the first equation we get :
$$\begin{align}
r^2+(z-1)^2&=1 \\
(z-1)^2&=1-r^2 \\
z&=1\pm\sqrt{1-r^2}
\end{align}$$
Since $z^2=r^2$ from the 2nd inequality, we'll have $z=\pm r$.
Solving for $r$:
$$r=1+\sqrt{1-r^2} \implies r=1$$
and $0<\theta<2\pi$.
However, I'm confused about how to define the limits of $z$ since it could be equal to $1+\sqrt{1-r^2}$ or $1+\sqrt{1-r^2}$.
Thanks in advance.
| For limits of $z$ your lower bound is the lower hemisphere $z=1-\sqrt {1-r^2}$ and the upper limit is the cone $ z=r $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find smallest set of natural numbers whose pairwise sums include 0..n Given a positive integer $n$, how do you find the smallest set of nonnegative integers $S$ such that for each integer $m$, where $0\leq m<n$, there exist two (not necessarily distinct) members of set $S$, say $x$ and $y$ such that $x+y=m$.
For example, consider the case $n=50$. Suppose the length of $S$ is $L$.
For a lower bound, if the elements of $S$ have pairwise distinct sums, then there are $\dbinom{L+1}{2}$ sums (the plus 1 is because numbers can be added to themselves). Thus, $$\binom{L+1}{2}\geq50\implies L\geq10$$.
I can acheive $L=12$ with the set {0, 1, 2, 3, 7, 10, 15, 18, 22, 23, 24, 25} (done with very inefficient program which searches randomly among all sets). For $L=10$, I feel like it should be impossible; we only have to show that more than 5 numbers can be expressed as a sum in more than 1 way, which should be able to be done through some casework. However, is $L=11$ possible? I think so.
Similarly, for $n=100$, I have $L=17$ from my program: {0, 1, 3, 4, 9, 11, 16, 20, 25, 30, 34, 39, 41, 46, 47, 49, 50}. But the lower bound only gives $L\geq 14$, so at least $L=15$ or $L=16$ should be possible.
In general, how do you do it efficiently for any given $n$?
| Let $n+4=s^2+r$ with $r,s\in\Bbb N$ and $0\leq r\leq 2s$.
Then an upper bound is given by
$$L\leq
\begin{cases}
\lceil{\frac rs}\rceil+2s-3&2\mid s\\
\lceil{\frac{r+1}{s+1}}\rceil+2s-3&2\nmid s
\end{cases}$$
which gives $L\leq 12$ for $n=50$ and $L\leq 18$ for $n=100$.
In general, for large $n$ this gives $L=O(\sqrt n)$.
A set $S$ corresponding to this upper bound is given by
\begin{align}
S
&=\{i\in\Bbb N:0\leq i<q-1\}\\
&\cup\{(q-1)+jq:0\leq j\leq k-1\}\\
&\cup\{i\in\Bbb N:(q-1)+(k-1)q<i\leq 2(q-1)+(k-1)q\}
\end{align}
for suitable values of $q$ and $k$.
Then $|S|=k+2(q-1)$ and then summing each pair of its elements we get all the natural number less or equals to $2(2(q-1)+(k-1)q)=2\max S$.
Consequently, we choose
$$k=\left\lceil\frac{n+4}{2q}\right\rceil-1$$
The function
$$\frac{n+4}{2q}-1+2(q-1)$$
attains a minimum at $q=\frac{\sqrt{n+4}}2$.
If $n+4=s^2+r$ and $s=2t+b$ with $r\leq 2s$ and $0\leq b\leq 1$, then
$$t\leq\frac{\sqrt{n+4}}2<t+1$$
For $q=t$ we get
\begin{align}
|S_t|
&=\left\lceil\frac{n+4}{2t}\right\rceil+2t-3\\
&=\left\lceil\frac{s^2+r}{s-b}\right\rceil+s-b-3\\
&=\left\lceil\frac{r+b}{s-b}\right\rceil+2s-3
\end{align}
while for $q=t+1$
\begin{align}
|S_{t+1}|
&=\left\lceil\frac{n+4}{2t+2}\right\rceil+2t+2-3\\
&=\left\lceil\frac{s^2+r}{s-b+2}\right\rceil+s-b-3\\
&=\left\lceil\frac{r+4-3b}{s+2-b}\right\rceil+2s-3
\end{align}
Since $|S_t|\leq|S_{t+1}|$ if and only if $b(2s+1)\leq 2s-r$, the formula on the top is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?
| Using Jordan canonical form, $X$ must be similar to one of the three matrices
\begin{align}
\begin{pmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{pmatrix},
\begin{pmatrix}
\lambda & 0 \\
0 & \lambda
\end{pmatrix},
\begin{pmatrix}
\lambda & 1 \\
0 & \lambda
\end{pmatrix}
\end{align}
, where $\lambda_1 \neq \lambda_2$, which implies that $X^2 + X$ must be similar to
\begin{align}
\begin{pmatrix}
\lambda_1^2 + \lambda_1 & 0 \\
0 & \lambda_2^2 + \lambda_2
\end{pmatrix},
\begin{pmatrix}
\lambda^2 + \lambda & 0 \\
0 & \lambda^2 + \lambda
\end{pmatrix},
\begin{pmatrix}
\lambda^2 + \lambda & 2\lambda + 1 \\
0 & \lambda^2 + \lambda
\end{pmatrix}
\end{align}
, respectively. Since the latter two matrices have only a single eigenvalue $\lambda^2 + \lambda$, it is impossible for them to be similar to $J \equiv \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, which has two distinct eigenvalues. Therefore, $X$ must take the form
$$X = P^{-1}\begin{pmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{pmatrix}P, \tag{1}$$
where $P := \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is invertible. In terms of $(1)$, the equation $X^2 + X = J$ is expressed as
$$P^{-1}\begin{pmatrix}
\lambda_1^2 + \lambda_1 & 0 \\
0 & \lambda_2^2 + \lambda_2
\end{pmatrix}P = J. \tag{2}$$
Because $J$ has two distinct eigenvalues $0$ and $2$, without of loss of generality, we can set
$$\begin{cases}
\lambda_1^2 + \lambda_1 = 0 \\
\lambda_2^2 + \lambda_2 = 2,
\end{cases} \tag{3}$$
under which $(2)$ is equivalent to
$$\begin{pmatrix}
0 & 0 \\
0 & 2
\end{pmatrix}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} =
\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, $$
or
$$\begin{pmatrix}
0 & 0 \\
2c & 2d
\end{pmatrix} =
\begin{pmatrix}
a + b & a + b \\
c + d & c + d
\end{pmatrix}, $$
which requires $a = -b$ and $c = d$ (hence $a \neq 0, c \neq 0$). Therefore,
$$P = \begin{pmatrix}
a & -a \\
c & c
\end{pmatrix}, \quad
P^{-1} = \frac{1}{2ac}\begin{pmatrix}
c & a \\
-c & a
\end{pmatrix}. \tag{4}$$
Substituting $(4)$ into $(1)$ gives
$$X = \frac{1}{2}\begin{pmatrix}
\lambda_1 + \lambda_2 & \lambda_2 - \lambda_1 \\
\lambda_2 - \lambda_1 & \lambda_1 + \lambda_2
\end{pmatrix}. \tag{5}$$
While $(3)$ admits the following four combinations of $(\lambda_1, \lambda_2)$:
$$(0, -2), (0, 1), (-1, -2), (-1, 1),$$
$(5)$ entails the following four solutions of $X$:
\begin{align}
\begin{pmatrix}
-1 & -1 \\
-1 & -1
\end{pmatrix},
\begin{pmatrix}
1/2 & 1/2 \\
1/2 & 1/2
\end{pmatrix},
\begin{pmatrix}
-3/2 & -1/2 \\
-1/2 & -3/2
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Minimal area of a right triangle with inradius $1$ I got this question, solved it, then forgot how I solved it.
What is the minimal area of a right triangle with inradius $1$?
My attempt:
$r=\frac{a+b-c}2$, so $a+b=c+2$
$a^2+b^2=c^2$
This gives $ab=2(c+1)$
I remember using the AM-GM to prove that equality held, thus giving $a=b=2+\sqrt2$ and getting the area as $3+2\sqrt2$, but I can't do it now. Please help.
| Recall that $$1 = r = \frac{abc}{4sR} = \frac{abc}{4 \cdot \frac{a+b+c}2\cdot \frac{c}2} \implies ab = a + b+ c$$
where $R = \frac{c}2$ is the circumradius.
Now we have
$$ab = a + b + c = a + b + \sqrt{a^2 + b^2} \ge 2\sqrt{ab} + \sqrt{2ab} = (2+\sqrt{2})\sqrt{ab}$$
so $ab \ge (2+\sqrt{2})^2 = 2(3+2\sqrt{2})$.
The equality is attained if and only if $a = b = 2+\sqrt{2}$ so the minimum area is indeed $$\frac{ab}2 = 3+2\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Find a real matrix $B$ such that $B^3 = A$
Given $$A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}$$ find a real, invertible matrix $B$ such that $B^3 = A$
I think I am doing something wrong here, so let me describe my attempt:
1) So I started off with diagonalizing the matrix $A$ with finding the eigenvalues $\lambda_1 = -8$ and $\lambda_2 = 1$ and the corresponding eigenvectors $ \vec v_1 = \begin{bmatrix}1 & 1\\0 & 0\end{bmatrix} = x + y = 0 \Rightarrow -x = y \Rightarrow \begin{bmatrix}1\\-1\end{bmatrix}$ and $ \vec v_2 = \begin{bmatrix}1 & -\frac{1}{2}\\0 & 0\end{bmatrix} = x - \frac{1}{2}y = 0 \Rightarrow 2x = y \Rightarrow \begin{bmatrix}1\\2\end{bmatrix}$
2) With that being done I proceeded with computing $D = \begin{bmatrix}-8 & 0\\0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}$ and check everything with $D = PAP^{-1}$
3) Now I thought I will simple find a diagonal matrix $M = PBP^{-1}$ and $M^3 = D$ and the easiest solution I came up with was $M = \begin{bmatrix}\sqrt[3]{-8} & 0\\0& 1\end{bmatrix}$ so basically $M = D^{\frac{1}{3}}.$ So that $B = PMP^{-1}$. But now come the tricky part, if I compute $B$ it results in a complex matrix not a real. //It is real!
Have I perhaps overlooked something here or miscalculated the solution for $B$?
Edit:
As Cameron pointed out my calculator and I totally failed as it was in complex mode and computed one of the non-real cube roots instead of -2. So $M = \begin{bmatrix}-2 & 0\\0 & 1\end{bmatrix}$ and consequentially $B = \begin{bmatrix}-1 & 1\\2 & 0\end{bmatrix}$
| $$A=\begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}=PDP^{-1}$$
Where $$P=\begin{bmatrix}1& 1\\-1 & 2\end{bmatrix}$$is the matrix of eigenvectors
and $$D=\begin{bmatrix}-8& 0\\0 & 1\end{bmatrix}$$ is the matrix of eigenvalues.
Thus $$ B = PD^{1/3}P^{-1} = \begin{bmatrix}-1& 1\\2 & 0\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Evaluate the given limit: $\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$ Evaluate the given limit
$$\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
My Attempt :
$$=\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
$$=\lim_{x\to 0} \dfrac {x(\tan (2x)-2\tan (x))}{1-2\cos (2x)+\cos^2 (2x)}$$
| Using $$\tan (2x) = \frac{2\tan x}{1-\tan^2 x}$$
So $$\lim_{x\rightarrow 0}\frac{x\tan (2x)-2x\tan x}{(1-\cos 2x)^2} = \lim_{x\rightarrow 0}\frac{2x\tan x\cdot \tan^2 x}{4\sin^4 x(1-\tan^2 x)}$$
So Using $\displaystyle \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$ and $\displaystyle \lim_{x\rightarrow 0}\frac{\tan x}{x} = 1$
So we have $$\lim_{x\rightarrow 0}\frac{x\tan (2x)-2x\tan x}{(1-\cos 2x)^2} = \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?
| One easy approach to this problem is by dividing both numerator and denominator by $\dfrac{\tan^4x}{\tan^4x}$
$$\dfrac{1}{\sin^4x+\cos^4x}\left(\dfrac{\tan^4x}{\tan^4x}\right)=\dfrac{\sec^4x}{1+\tan^4x}=\dfrac{(1+\tan^2x)\sec^2x}{1+\tan^4x}$$
Now you can use $u-$ substitution $u=\tan x$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
} |
Evaluate $\int \sqrt {3 \tan^2 \theta - 1} d \theta$ Evaluate $I=\int \sqrt {3 \tan^2 \theta - 1} d \theta$
My attempt
$\tan \theta = t, $ then $I = \int \frac{\sqrt {3t^2-1}}{1+t^2} dt $
Now integrating by parts,
I = $\sqrt {3t^2-1} \tan^{-1} t- \int( \frac{6t}{2\sqrt {(3t^2-1)}} \tan^{-1} t) dt$
Now i am struck... How to proceed.
| The result of @Manthanein can be obtained without much difficulty.
The integral can also be written as $$\int \sqrt{3\sec^2 \theta -4} \, \mathrm d\theta$$
Now, substitute $$\begin{align}3\sec^2 \theta -4 &=x^2\\ \implies 3\sec^2 \theta \tan \theta \, \mathrm d\theta &= x \, \mathrm dx \\ \implies (x^2+4)\sqrt{\dfrac{x^2+1}3}\,\mathrm d\theta &=x\,\mathrm dx \\ \implies \mathrm d\theta &=\dfrac{\sqrt 3 x\,\mathrm dx}{(x^2+4)\sqrt{x^2+1}}\end{align}$$
Hence the integral now becomes $$\int\dfrac{\sqrt3 x^2 \mathrm dx}{(x^2+4)\sqrt{x^2+1}}$$
and the rest follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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An inequality relating the ratio of the areas of two triangles The conjecture below is a modified version of this question:
Prove that, given a triangle with sides $a,b,c$, there exists a triangle with sides $a+2b,b+2c,c+2a$ that has an area three times the original
Conjecture:
If $u$ is the area of a triangle with sides $a,b,c$, and $v$ is the area of a triangle with sides $a+2b,b+2c,c+2a$, then ${\large{\frac{v}{u}}}\ge 9$.
Remarks:
*
*For the equilateral case ($a=b=c$), we get ${\large{\frac{v}{u}}}=9$.
*Limited data testing seems to support the truth of the conjecture.
*Trying to prove the claim via Heron's formula appears to be a disaster.
Question:$\;$Is the conjecture true?
| Let $x=b+c-a\geq 0$, $y=a+c-b\geq 0$, $z=a+b-c\geq 0$.
Let $p=\frac{1}{2}(a+b+c)$ be the semiperimeter of the triangle with sides $a$, $b$, $c$. Similarly let $P=\frac{1}{2}(A+B+C)=3p$ be the semiperimeter of the triangle with sides $A=b+2c$, $B=c+2a$, $C=a+2b$. Then, by Heron's formula, the inequality is equivalent to
$$
\begin{align}0&\leq \frac{v^2}{u^2}-9^2=\frac{P(P-A)(P-B)(P-C)}{p(p-a)(p-b)(p-c)}-9^2\\&=
\frac{6((x^2y+y^2z+z^2x)+2(xy^2+yz^2+zx^2)-9xyz)}{xyz}
\end{align}$$
which holds because by the AGM inequality
$$x^2y+y^2z+z^2x\geq 3xyz\quad\mbox{and}\quad xy^2+yz^2+zx^2\geq 3xyz.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim_{n\to\infty}\sum_{k=1}^n \frac{\pi k}{2n}\int_0^1 x^{2n}\sin\frac{\pi x}{2}dx$ Hello I am trying to compute $$\lim_{n\to\infty}\sum_{k=1}^n \frac{\pi k}{2n}\int_0^1 x^{2n}\sin\frac{\pi x}{2}dx$$ I have rewrited it as: $$\lim_{n\to\infty}\frac1n\sum_{k=1}^n \frac{k}{n}\int_0^1\frac{n\pi}{2} x^{2n}\sin\frac{\pi x}{2}dx$$ Now the first part is just a Riemann integral $\int_0^1 x \,dx=\frac{1}{2}\,$ however I dont know if I am allowed to use it in this case. Now for the second part using $\frac{\pi x}{2}=t$ we have:$$\int_0^1\frac{n\pi}{2} x^{2n}\sin\frac{\pi x}{2}dx=\left(\frac{\pi}{2}\right)^nn\int_0^\frac{\pi}{2}t^{2n}\sin t\,dt=\left(\frac{\pi}{2}\right)^n\frac{n}{2n+1}\int_0^\frac{\pi}{2}\left(t^{2n+1}\right)'\sin t\, dt$$ So integrating by parts two times gives: $$I(2n)=\frac{1}{2n+1}\left(\left(\frac{\pi}{2}\right)^{2n+1}-\frac{1}{2n+2}I(2n+2)\right)$$ or rewritten as:$$I(2n)=2n\left(\left(\frac{\pi}{2}\right)^{2n-1}-(2n-1)I(2n-2)\right)$$ But how do I use the relation? Could you give me some help with this?
EDIT:Originally was:$$\lim_{n\to\infty}\sum_{k=1}^n \sin \frac{\pi k}{2n}\int_0^1 x^{2n}\sin\frac{\pi x}{2}dx$$
| Method 1 (Laplace's method).
We use the asumptotics of $$\int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx=\int^1_0 e^{2n(\log(x))} \sin\left( \frac{\pi x}{2}\right)\,dx $$
Via Laplace's method one gets as $n\to\infty$: $$ \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx \sim \frac{1}{2n } $$ The required limit is thus:
\begin{align}
\lim_{n\to\infty} \sum_{k=1}^n \frac{\pi k}{2n} \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx = \lim_{n\to\infty} \frac{\pi}{2n} \cdot \frac{(n+1)n}{2} \cdot \frac{1}{2n} = \frac{\pi}{8}
\end{align}
where we have used the well known result $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.
Method 2 (Elementary).
So let me derive this more elementary. Set $t=1-x$ to get: $$\int^1_0 x^{2n}\sin\left( \frac{\pi x}{2}\right)\,dx = \int^1_0 (1-t)^{2n} \cos\left( \frac{\pi t}{2}\right)\,dt $$ Recall the inequality $1-\frac{x^2}{2} \leq \cos(x) \leq 1$. Hence:
\begin{align}
\int^1_0 (1-t)^{2n}\left(1- \frac{\pi^2}{8}t^2\right)\,dt \leq \int^1_0 (1-t)^{2n} \cos\left( \frac{\pi t}{2}\right)\,dt\leq \int^1_0 (1-t)^{2n}\,dt
\end{align}
In other words:
\begin{align}
\frac{1}{2n+1} - \frac{\pi^2}{8}\int^1_0 (1-t)^{2n}t^2\,dt \leq \int^1_0 (1-t)^{2n} \cos\left( \frac{\pi t}{2}\right)\,dt\leq \frac{1}{2n+1}
\end{align}
Moreover
$$\int^1_0 (1-t)^{2n}t^2\,dt = \frac{1}{(2n+1)(2n+3)(n+1)} = O(n^{-3})$$
which can be easily gotten by integration by parts (or through Beta function). All in all we conclude:
\begin{align}
\frac{(n+1)\pi}{4} \frac{1}{(2n+1)}+O(n^{-2})\leq \sum_{k=1}^n \frac{\pi k}{2n} \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx \leq \frac{(n+1)\pi}{4} \frac{1}{(2n+1)}
\end{align}
where we (again) have used the formula for $\sum_{k=1}^n k$. With the Squeeze Theorem we conclude the same as in method 1.
Method 3 (DCT).
This method is inspired by Mark Viola's excellent answer. Integrate by parts once to get:
$$\int_0^1 x^{2n}\sin\left(\frac{\pi x}{2}\right)\,dx=\frac1{2n+1}-\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos\left(\frac{\pi x}{2}\right)\,dx$$
So: $$\sum_{k=1}^n \frac{\pi k}{2n} \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx = \frac{(n+1)\pi}{4} \frac{1}{(2n+1)}-\frac{(n+1)\pi}{4}\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos\left(\frac{\pi x}{2}\right)\,dx $$
Taking the limits as $n\to\infty$ together with DCT ( $|x^{2n+1}\cos\left(\frac{\pi x}{2}\right)|\leq 1$ and it converges a.e. to $0$) yields the same result.
| {
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Example $f$ Riemann-integrable, $g$ bounded and $g=f$ almost everywhere. I'm facing this problem,
Let $g:[a,b]\rightarrow \mathbb{R}$ be Riemann-integrable, $f:[a,b]\rightarrow \mathbb{R}$ a bounded function, $(x_n)$ a sequence of points in $[a,b]$ such that $f(x)=g(x)$ for all $x$ in $[a,b]$ other than the $x_n$. $\textbf{Give an example}$ to show that $f$ need not be Riemann-integrable.
Before this, the book ("A First Course in Real Analysis by Sterling Berberian",Page 164) says that for $f:[a,b]\rightarrow \mathbb{R}$ and $g:[a,b]\rightarrow \mathbb{R}$ Riemann integrable, and $f=g$ almost everywhere we can conclude that $f$ is Riemann-integrable.
Checking the examples they give on non riemann integrability, I found $f(x):[0,1]\rightarrow \mathbb{R}$, $f(x)=1$ for rationals and $f(x)=0$ for irrationals. From this I can build a function $g:[0,1]\rightarrow \mathbb{R}$, that can be equal to $f$ over the irrationals, and something different than $f$ for rationals, However, how could I build a sequence such as the one they ask me to, but over the rationals?
| You can list the rationals in $[0, 1]$ like so:
$$0, 1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{5}{6} \ldots$$
Essentially, I'm listing the points in the graph of the ruler function in lexicographic order. It's not a sequence with a "nice" formula, but it is a sequence.
You can make it a tad nicer if you don't care about injectivity:
$$0, 1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \ldots,$$
or indeed just look at dyadic rationals:
$$1, \frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}, \ldots.$$
| {
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How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.
| Use the fact that when $u\to 0$, $(1+u)^n \approx 1+nu$.
Using that fact,
$$\begin{align} L &= \lim_{x\to 0} \frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}} \\ &= \lim_{x\to 0} \dfrac{1+\dfrac 12 x^2 - (1+\dfrac 12 x)}{1+\dfrac 12 x^3 - (1+\dfrac 12 x)}\\ &= \lim_{x\to 0} \dfrac{x^2-x}{x^3-x} \\ &= \lim_{x\to 0}\dfrac{1}{1+x} \\ &= \color{red}{1}\end{align}$$
| {
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Proof of this integration shortcut: $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$ I came across this as one of the shortcuts in my textbook without any proof.
When $b\gt a$,
$$\int\limits_a^b \dfrac{dx}{\sqrt{(x-a)(b-x)}}=\pi$$
My attempt :
I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$\int\limits_0^1 \dfrac{dt}{\sqrt{t(1-t)}}$$
This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?
| \begin{align}
\tan^2 \theta &= \frac{x-a}{b-x} \\
2\tan \theta \sec^2 \theta \, d\theta &=
\frac{b-a}{(b-x)^2} \, dx \\
2\sqrt{\frac{x-a}{b-x}} \times \frac{(x-a)+(b-x)}{b-x} \, d\theta &=
\frac{b-a}{(b-x)^2} \, dx \\
2\, d\theta &= \frac{dx}{\sqrt{(x-a)(b-x)}} \\
\int \frac{dx}{\sqrt{(x-a)(b-x)}} &=
2\tan^{-1} \sqrt{\frac{x-a}{b-x}}
\end{align}
The singularity in Wolfram Alpha comes from the upper limit $b$.
Geometrical interpretation
Considering circular arc $(x,y)=(\sqrt{b-u},\sqrt{u-a})$
\begin{align}
ds &= \frac{\sqrt{b-a} \, du}{2\sqrt{(u-a)(b-u)}} \\
\tan \theta &= \sqrt{\frac{u-a}{b-u}} \\
\begin{pmatrix}
x \\ y
\end{pmatrix} &=
\begin{pmatrix}
\sqrt{b-a} \cos \theta \\
\sqrt{b-a} \sin \theta
\end{pmatrix} \\
ds &= \sqrt{b-a} \, d\theta
\end{align}
See also another integral here.
| {
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Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably I am mistaken in solving.
| What? So the left hand side is the AM of $1/(b+c+d), 1/(c+d+a), 1/(d+a+b)$ and $1/(a+b+c)$ i.e. $(1/(b+c+d)+1/(a+c+d)+1/(a+b+d)+1/(a+b+c))/4$. On the RHS we have $4$ over the sum of the reciprocals of these terms i.e. $4/((b+c+d)+(c+d+a)+(d+a+b)+(a+b+c))$ which does indeed come out as $4/3(a+b+c+d)$ Which then (multiplying both sides by 4) yields the result $1/(b+c+d)+1/(a+c+d)+1/(a+b+d)+1\(a+b+c) \geq 16 / 3(a+b+c+d)$. Isn't that cool?
| {
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Find the probability that atleast one valve is defective. A factory A produces $10$% defective valves and another factory $B$ produces 20% defective valves.A bag contains $4$ valves of factory $A$ and $5$ valves of factory B.If two valves are drawn at random from the bag,find the probability that at least one valve is defective.
$P(\text{at least one valve is defective})=\\=1-P(\text{none of the two valves are defective})=\\=1-\left(\frac{\binom{4}{2}}{\binom{9}{2}}(0.9)^2+\frac{\binom{5}{2}}{\binom{9}{2}}(0.8)^2+\frac{\binom{4}{1}\binom{5}{1}}{\binom{9}{2}}(0.9)(0.8)\right)=\frac{517}{1800}$,
but the answer given is $\frac{303}{1800}$ I don't know where i am wrong.
| The Probability that factory $A$ produces defective values is $\dfrac{10}{100}=\dfrac{1}{10}$
The Probability that factory $B$ produces defective values is $\dfrac{20}{100}=\dfrac{1}{5}$
Given a bag contains $4$ values of factory $A$ and $5$ values of factory $B$ and two values are drawn random.
$$P(\mbox{at least one defective})=1-P(\mbox{both are non-defective})$$
$$P(\mbox{both are non-defective})=P(\mbox{both values of factory }B)\times P(\mbox{both are non-defective})+P(\mbox{both values of factory }B)\times P(\mbox{ both are non defective})+P(\mbox{one value of $A$ and other of factory $B$})\times P(\mbox{both are nondefective})$$
$$=\dfrac{\dbinom{4}{2}}{\dbinom{9}{2}}\left(\frac{9}{10}\right)^2+\dfrac{\dbinom{5}{2}}{\dbinom{9}{2}}\left(\dfrac45\right)^2+\dfrac{\dbinom{4}{1}\cdot\binom{5}{1}}{\binom{9}{2}}\times\left(\frac{9}{10}\right)\times\left(\frac{4}{5}\right)$$
$$=\dfrac{27}{200}+\dfrac{8}{45}+\dfrac{2}{5}=\dfrac{1283}{1800}$$
Now, $P(\mbox{at least one defective})=1-\dfrac{1283}{1800}=\approx0.29$
So, the answer what you got is $\dfrac{517}{1800}\approx0.29$ which is correct.
| {
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Algebraic Inequality involving AM-GM-HM If $$a,b,c \;\epsilon \;R^+$$
Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$\frac{bc}{x-a} + \frac{ab}{x-c} + \frac{ac}{x-b} \leq \frac{1}{2} \Bigl(x\Bigl)$$
This didn't help so instead I wrote it as $$2\Biggl(\frac {1}{a} + \frac{1}{b} + \frac{1}{c}\Biggl ) \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
Not sure what to to do next.
Any help would be appreciated.
| Proof
Just by $H_n \leq A_n$, we have
\begin{align*}
\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c}&=\frac{1}{2}\left(\frac{2}{\frac{1}{b}+\frac{1}{c}}+\frac{2}{\frac{1}{a}+\frac{1}{b}}+\frac{2}{\frac{1}{a}+\frac{1}{c}}\right)\\&\leq \frac{1}{2}\left(\frac{b+c}{2}+\frac{a+b}{2}+\frac{a+c}{2}\right)\\&=\frac{a+b+c}{2}.
\end{align*}
| {
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Two different answers from integrating $\int\frac{dx}{x\sqrt{x^2-1}}$ in two ways. What did I do wrong? I want to calculate the answer of the integral $$\int\frac{dx}{x\sqrt{x^2-1}}$$ I use the substitution $x=\cosh(t)$ ($t \ge 0$) which yields $dx=\sinh(t)\,dt$. By using the fact that $\cosh^2(t)-\sinh^2(t)=1$ we can write $x^2-1=\cosh^2(t)-1=\sinh^2(t)$. Since $t\ge 0$, $\sinh(t)\ge 0$, and we have $\sqrt{x^2-1}=\sinh(t)$. Now, by substituting for $x$, $dx$, and $\sqrt{x^2-1}$ in the first integral, we have
$$\int\frac{dt}{\cosh(t)}$$
Since $\cosh(t)=(e^t+e^{-t})/2$ by substituting this in this integral we have
$$2\int\frac{dt}{e^t+e^{-t}}.$$
Now by multiplying numerator and denominator in $e^t$ one can write:$$2\int\frac{e^t\,dt}{1+e^{2t}}$$
Now by using $z=e^t$ in this integral one can write ($dz=e^t\,dt$): $$2\int\frac{dz}{1+z^2}$$
So we have $$\int\frac{dt}{\cosh(t)}=2\arctan(z)=2\arctan(e^t)$$
On the other hand $$t=\cosh^{-1}(x)=\ln\left(x+\sqrt{x^2-1}\right)$$
So we have $$\int\frac{dx}{x\sqrt{x^2-1}}=\int\frac{dt}{\cosh(t)}=2\arctan(\exp(\ln(x+\sqrt{x^2-1})))$$
which yields
$$\int\frac{dx}{x\sqrt{x^2-1}}=2\arctan(x+\sqrt{x^2-1})$$
but this answer is wrong. The true answer can be obtained by direct substitution $u=\sqrt{x^2-1}$ and is
$$\int\frac{dx}{x\sqrt{x^2-1}}=\arctan(\sqrt{x^2-1})$$
I don't want to know the answer of the integral. I want to know what I did wrong? Can somebody help?
| Verify by differentiation.
$$(2\arctan(x+\sqrt{x^2-1}))'=2\frac{1+\dfrac x{\sqrt{x^2-1}}}{(x+\sqrt{x^2-1})^2+1}=2\frac{x+\sqrt{x^2-1}}{2(x^2+x\sqrt{x^2-1)}\sqrt{x^2-1}}$$
and
$$(\arctan\sqrt{x^2-1})'=\frac{\dfrac{x}{\sqrt{x^2-1}}}{(\sqrt{x^2-1})^2+1}.$$
Hence both answers are correct.
| {
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Find number of ordered pairs Find the number of ordered pairs $(p , q)$ such that $p , q $ are both prime numbers less than 50 , and $pq$+1 is divisible by 12
Edit :
What i have done is i have written down all the primes below 50 congruent to modulo 12 .
For example : 11 $ \equiv$ -3 $(mod 12)$
| Clearly we need both $p$ and $q$ to be coprime to $12$, so we cannot have $p$ or $q$ equal to either $2$ or $3$.
This means that $\{p,q\}\in\{1,5,7,11\}\bmod 12$. We can quickly calculate the products $\bmod 12$ across this set:
\begin{array}{l|cc}
p\ \backslash\ q & 1 & 5 & 7 & 11 \\ \hline
1 & 1 & 5 & 7 & 11 \\
5 & 5 & 1 & 11 & 7 \\
7 & 7 & 11 & 1 & 5 \\
11 & 11 & 7 & 5 & 1 \\
\end{array}
We are looking for $p,q$ pairs that give $11\equiv -1 \bmod 12$ in the above table so that $12\mid pq{+}1$. So we would select the pairs from the different residue class sets appropriately, which gives a way to calculate how many options in total.
For example, the set of primes in range $\equiv 1 \bmod 12$ (call this $S_1$) consists of just $\{13,37\}$. This gives us $|S_1|=2$ and similarly $|S_5|=4,$ $|S_7|=4,$ and $|S_{11}|=3$. Thus we will have $2\cdot 3 + 4\cdot 4 + 4\cdot 4 + 3\cdot 2 = 2\cdot 6+2\cdot 16= \fbox{44 }$ ordered pair solutions.
| {
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Show that if $p_1^4 + \cdots + p_{31}^4$ is divisible by $30$, then three consecutive primes are included Let $p_1<p_2<\cdots<p_{31}$ be prime numbers. Prove that if $p_1^4+p_2^4+\cdots+p_{31}^4$ is divisible by $30$, then there are three consecutive prime numbers in the sum.
Consecutive prime numbers in the sense of $2,3,5,7,11,13,17,19,23,27,29,31, \ldots$ and so on.
What I tried:
$$S=p_1^4+p_2^4+\cdots+p_{31}^4$$
$S$ must be divisible by $3$ and $10.$
The sum of 3 arbitrary prime numbers, greater than 3, to the fourth power is divisible by 3, so the sum of 27 arbitrary prime numbers, grater than 3, to the fourth power is divisible by 3.
The sum of 3,5,7,11 raised to the fourth power is divisible by 3.
Yes but the sum of 3,5,7,11 raised to fourth power plus the sum of 27 prime numbers, grater than 11, to the fourth power is not divisible by 10!
Could someone give me an idea!
| As there are an odd number of primes appearing in the sum, and the sum is to be even, we must have that $p_1 = 2$. Considering mod $30$, we must then have that
$$ p_2^4 + \cdots + p_{31}^4 \equiv 14 \pmod {30}.$$
Working mod $3$, we must then have that
$$ p_2^4 + \cdots + p_{31}^4 \equiv 2 \pmod {3}.$$
There are $30$ terms appearing in the sum. Every prime $p > 3$ satisfies $p^4 \equiv 1 \bmod 3$, and $30 \cdot 1 \equiv 0 \bmod 3$. So $p_2 = 3$.
Similar analysis mod $5$ will show that $p_3 = 5$, and so the primes must include $2,3$, and $5$.
| {
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the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$
For any acute angled triangle ABC , find the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$ .
Attempt:
As $A+B+C=\pi$
$C=\pi -(A+B)$
After differentiating it
$dA+dB+dC=0$
Now :
$\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$
$\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin (A+B)}{\pi-(A+B)}$
$(\frac{A\cos A-\sin A}{A^2})dA + (\frac{B\cos B- \sin B}{B^2})dB + (\frac{C\cos C-\sin c}{C^2})dC =0$
But could not solve further .
| By canceling the gradient of
$$\frac{\sin A}A+\frac{\sin B}B+\frac{\sin(A+B)}{\pi-A-B},$$
we must have
$$\frac{A\cos A-\sin A}{A^2}=-\frac{(\pi-A-B)\cos(A+B)+\sin(A+B)}{(\pi-A-B)^2}=\frac{B\cos B-\sin B}{B^2}.$$
As the function
$$\frac{x\cos x-\sin x}{x^2}$$ is monotonic in the first quadrant, we have $A=B$, and we now maximize
$$2\frac{\sin A}A+\frac{\sin2A}{\pi-2A}.$$
Graphically, it is obvious that the function has a single maximum, and we can verify that $A=\dfrac\pi3$ cancels the derivative. It would be better to prove that there are no other solutions.
| {
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Verifying that $u(x, t) = \frac{1}{2}c \text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$ is a solution to $u_t + 6uu_x + u_{xxx} = 0$ I'm trying to verify that
$$u(x, t) = \frac{1}{2}c \text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
is a solution to the KdV equation
$$u_t + 6uu_x + u_{xxx} = 0$$
My calculations have lead to
$$u_t = \frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
$$u_x = -\frac{c^{3/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
$$u_{xxx} = -\frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}^3\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) + c^{5/2}\text{sech}^4\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
I have used Symbolab to verify that these are correct.
So we have
$$u_t + 6uu_x + u_{xxx} = \frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) + 6 \left[ \frac{1}{2}c \text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) \right] \left[ -\frac{c^{3/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) \right] + \left[ \left( -\frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}^3\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) + c^{5/2}\text{sech}^4\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) \right) \right]$$
Which, as far as I can tell, does not equal $0$?
I'm wondering if there's an error that I'm not noticing, or whether there is some hyperbolic trig identity that I'm supposed to use?
| Begin by setting $a=\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) $and $b=\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right).$ If $c=0,$ then the function in question is clearly a solution, hence we may factor out $c^{5/2}/2.$
The computed expression simplifies to $ab-3a^2b-ab^3+2a^2b = ab-a^2b-ab^3 = ab(1-a-b^2) = 0,$ where we used the trig identity $a+b^2=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial problem with unknown coefficients $a, b, c$ $p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.
Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-\frac{a^2}{a+1}$ and then getting $c=-\frac{a^4}{a+1}$ does not help much I guess. Will someone tell me how I should be thinking about this question? Thanks.
| Since $b=-\frac{a^2}{a+1}$, we see that
$$c=-a^3-ab=-a^3+\frac{a^3}{a+1}=-\frac{a^4}{a+1}$$ and since $gcd(a^4,a+1)=1,$ we obtain:
$a+1\in\{-1,1\}$ and since $a\neq0$, we obtain $a=-2$, $c=16$, $b=4$,
$$f(x)=x^3-2x^2+4x+16$$ and
$$f(13)=13^3-2\cdot13^2+4\cdot13+16=1927.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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General solution of $(4x^2-x)y''+2(2x-1)y'-4y=12x^2-6x$ given $y_1=\frac1x$ Verify that $y_1=\frac1x$ is a solution to d.e
$\left(4x^2-x\right)y''+2\left(2x-1\right)y'-4y=0$
Find the general solution of the d.e
$\left(4x^2-x\right)y''+2\left(2x-1\right)y'-4y=12x^2-6x$
Note: I was able to prove initial value $y_1=\frac{1}{x}$ however confused on second part.
| Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + \frac{4x-2}{x(4x-1)}y' - \frac{4}{x(4x-1)}y = \frac{12x-6}{4x-1} $$
The basic idea is to try a solution of the form $$ y(x) = y_1(x)u(x)= \frac{1}{x}u(x) $$
Plugging this in, we obtain
$$ u'' + \left(-\frac{2}{x} + \frac{4x-2}{x(4x-1)} \right)u' = \frac{12x^2-6x}{4x-1} $$
which is a standard first-order ODE in $u'$ that you can solve using the integration factor.
Here's the full working
\begin{align} u'' - \frac{4}{4x-1}u' &= \frac{12x^2-6x}{4x-1} \\ \frac{u''}{4x-1} - \frac{4u'}{(4x-1)^2} &= \frac{12x^2-6x}{(4x-1)^2} \\ \left(\frac{u'}{4x-1}\right)' &= \frac34 - \frac{3/4}{(4x-1)^2} \\ \frac{u'}{4x-1} &= \frac{3x}{4} + \frac{3/16}{4x-1} + C \\ u' &= 3x^2 + \left(C-\frac{3}{16}\right)(4x-1) \\ u(x) &= x^3 + c_1(2x^2-x) + c_2 \\ y(x) &= x^2 + c_1(2x-1) + \frac{c_2}{x} \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating over a sphere Suppose $x\in\mathbb{R}^n$ is a random unit vector distributed uniform over the $(n-1)-$sphere $S_{n-1}$ (the set of unit vectors in $\mathbb{R}^n$ ). For an arbitrary vector $y\in\mathbb{R}^n$ how does one evaluate
$$\int_{S_{n-1}}e^{-||y-x||_2^2}\mathrm{d}\mu$$
where $\mu$ denotes the uniform measure over $S_{n-1}$.
My attempt:
Expanding $||y-x||_2^2=||y||_2^2+2\langle x,y\rangle+||x||_2^2$, I can reduce the integral to
$$e^{-||y||_2^2-1}\int_{S_{n-1}}e^{2\langle x,y\rangle}\mathrm{d}\mu.$$
From here, I am not able to proceed.
Can someone provide some pointers?
| As remarked by @md2perpe, since the $(n-1)$ sphere is invariant under rotation, then you could simply apply a change of variable given by $x=Rx'$, where $R$ is an orthogonal matrix, such that $R^\ast y=e_1 = (1, 0, \ldots, 0)$. Then we see that
\begin{align}
\int_{S^{n-1}} e^{-|x-y|^2}\ d\mu(x) = \int_{S^{n-1}} e^{-|Rx'-y|^2}\ d\mu(x') = \int_{S^{n-1}} e^{-|x'-e_1|^2}\ d\mu(x').
\end{align}
Next, notice
\begin{align}
\int_{S^{n-1}} e^{-|x-e_1|^2}\ d\mu(x) =&\ \int_{S^{n-1}} e^{-(x_1-1)^2-x_2^2-\ldots -x_n^2}\ d\mu(x) = \int_{S^{n-1}} e^{2x_1} e^{-|x|^2-1}\ d\mu(x)\\
=&\ e^{-2} \int_{S^{n-1}} e^{2x_1}\ d\mu(x).
\end{align}
Finally, we could parametrize the sphere by spherical coordinates
\begin{align}
(\cos \theta_1, \sin\theta_1 \cos\theta_2, \sin\theta_1 \sin\theta_2\cos\theta_3, \ldots, \sin\theta_1\sin\theta_2\cdots\sin\theta_{n-1})
\end{align}
with $0<\theta_i<\pi$ for $i=1, \ldots, n-2$ and $0<\theta_{n-1}\leq 2\pi$. This gives us
\begin{align}
&\int^{2\pi}_0\cdots \int^{\pi}_0 e^{2\cos\theta_1} \sin^{n-2}\theta_1\sin^{n-3}\theta_2\cdots \sin\theta_{n-2}\ d\theta_1\cdots d\theta_{n-1}\\
&=2\pi \left(\int^\pi_0 e^{2\cos\theta_1}\sin^{n-2}\theta_1 d\theta_1\right)\left( \int^\pi_0 \sin^{n-3}\theta_2\ d\theta_2\right)\cdots \left(\int^\pi_0 \sin\theta_{n-2}\ d\theta_{n-2} \right)
\end{align}
Additional: As pointed out by @JohnHughes, we should have
\begin{align}
\int_{S^{n-1}}e^{-|x-y|^2} d\mu(x) = e^{-|y|^2-1} \int_{S^{n-1}}e^{2|y|x_1} d\mu(x).
\end{align}
Anyhow, to evaluate
\begin{align}
I(\alpha, n)=\int^\pi_0 e^{\alpha \cos\theta} \sin^{n-2}\theta\ d\theta
\end{align}
we will consider two cases, $n$ odd and $n$ even.
When $n>1$ is odd, we see that
\begin{align}
\int^\pi_0 e^{\alpha \cos\theta} \sin^{n-2}\theta\ d\theta= \int^1_{-1}e^{\alpha u} (1-u^2)^{(n-3)/2}\ du = \int^1_{-1}e^{\alpha u} (1-u^2)^k\ du
\end{align}
for some positive integer $k$. Hence we could compute the integral by standard integration by parts method (extremely tedious). In the special case $n=3$, we have that
\begin{align}
\int^{\pi}_0 e^{2|y| \cos\theta} \sin\theta\ d\theta = \int^1_{-1}e^{2|y|u}\ du = \frac{e^{2|y|}-e^{-2|y|}}{2|y|} = \frac{\sinh(2|y|)}{|y|}
\end{align}
which means
\begin{align}
\int_{S^2} e^{-|x-y|^2} d\mu(x) = 4\pi e^{-|y|^2-1} \frac{\sinh(2|y|)}{2|y|}.
\end{align}
In particular, when $y=0$, we see that the above answer give $\frac{4\pi}{e}$ which is precisely the surface area times the value of $e^{-|x|^2}$ on the unit $2$-sphere.
For the $n$ even case, we will only consider $n=2, 4$. We start by considering the case $n=2$ to get
\begin{align}
\int^\pi_0 e^{\alpha \cos\theta} \ d\theta
\end{align}
which requires special functions to evaluate. In particular, we have that
\begin{align}
\int^\pi_0 e^{\alpha \cos\theta} \ d\theta= \pi I_0(\alpha)
\end{align}
where $I_0$ is the $0$th modified Bessel function of the first kind. Hence we see that
\begin{align}
\int_{S^1} e^{-|x-y|^2} d\mu(x) = \pi e^{-|y|^2-1} I_0(2|y|)
\end{align}
which gives us $\frac{\pi}{e}$ when $y=0$.
For the $n=4$ case, we see that
\begin{align}
\int^\pi_0 e^{\alpha \cos\theta} \sin^2\theta\ d\theta = \int^\pi_0 e^{\alpha \cos\theta} \frac{1-\cos 2\theta}{2} d\theta = \frac{\pi}{2}I_0(\alpha)-\frac{1}{2}\int^\pi_0e^{\alpha \cos\theta}\cos 2\theta\ d\theta.
\end{align}
To evaluate the latter integral we will need to use higher modified Bessel functions
\begin{align}
I_k(\alpha) = \frac{1}{\pi} \int^\pi_0 e^{\alpha \cos\theta}\cos k\theta d\theta
\end{align}
where $k$ is a positive integer. Hence it follows
\begin{align}
\int^\pi_0 e^{\alpha \cos\theta} \sin^2\theta\ d\theta= \frac{\pi}{2}(I_0(\alpha)-I_2(\alpha)).
\end{align}
Then we see that
\begin{align}
\int_{S^3} e^{-|x-y|^2} d\mu(x) = 2\pi^2 e^{-|y|^2-1}\{I_0(2|y|)-I_2(2|y|)\}
\end{align}
which gives us the expected value $\frac{2\pi^2}{e}$ when $y=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
First Order Separable differential Equation
Problem:
Solve the following differential equation:
\begin{eqnarray*}
6x^2y \, dx - (x^3 + 1) \, dy &=& 0 \\
\end{eqnarray*}
Answer:
This is a separable differential equation.
\begin{eqnarray*}
\frac{6x^2}{x^3+1} \, dx - \frac{dy}{y} &=& 0 \\
\int \frac{6x^2}{x^3+1} \, dx - \int \frac{dy}{y} &=& c_1 \\
2 \ln{|x^3+1|} - \ln{|y|} &=& c_1 \\
\ln{(x^3+1)^2} - \ln{|y|} &=& c_1 \\
\ln{ \Big( \frac{(x^3+1)^2}{|y|} \Big) } &=& c_1 \\
(x^3+1)^2 &=& c|y| \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
(x^3+1)^2 &=& |cy| \\
\end{eqnarray*}
Is my answer different from the book's answer? I believe it is. What am I missing?Any idea of how to proceed?
Thanks,
Bob
| Both answers are correct.
Your answer $$(x^3+1)^2 = c |y|$$ makes the assumption that $c\ge 0$
The book's answer $$(x^3+1)^2 = |cy|$$ is OK for all values of $c$.
Thus to make sure that you can take any value for c go with the book's answer, otherwise mention that $c\ge 0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$ Solve the equation:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$
Things I have done so far:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$ Deducting 2 from both sides of equation
$$\Leftrightarrow (\sqrt[3]{x^2+4}-2)=(\sqrt{x-1}-1)+2x-4$$
$$\Leftrightarrow \frac{x^2+4-8}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}=\frac{x-2}{\sqrt{x-1}+1}+2(x-2)$$
$$\Leftrightarrow (x-2)(\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2)=0$$
+)$$x-2=0\Leftrightarrow x=2$$
+)$$\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2=0(*)$$
I don't know how to solve the equation (*).
By the way, I think there must be "smart" way to solve this equation.
| As you gave,the part is $$\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}-\dfrac{1}{\sqrt{x-1}+1}-2=0$$
$$\implies\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}-\dfrac{1}{\sqrt{x-1}+1}=2..........(1)$$
$${\text{Let,}}$$
$$f(x)=\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}-\dfrac{1}{\sqrt{x-1}+1}$$
I will break $f(x)$ in two parts
$$g(x)=\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}$$
$$h(x)=\dfrac{1}{\sqrt{x-1}+1}$$
Now if equation (1) is valid then $g(x)-h(x)$ is must be $2$.
I will check is it possible or not.I will start with h(x).
For h(x) to be valid it is sure that $x \geq 1$ and for any value of $x, 0\lt h(x)\leq 1$
Now in $g(x)$ it is clear that for $x\geq 1 ,numerator\lt denominator$.So,it is also true that,$0\lt g(x)\lt 1$
$$Hence, g(x)-h(x) \neq 2$$
$$\implies f(x)\neq 2$$
so,there is no solution for $f(x)$ and there exists only one solution and that is $x=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$
My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n-1}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}\right)<\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}$$
Please check my solution for me and give me some idea.
| We may consider that $f(x)=\frac{1}{\sqrt{x}}$ is a convex function on $\mathbb{R}^+$, hence for any $n\in\mathbb{N}^+$ we have
$$ f(n-1/2)-f(n+1/2) \geq -f'(n) $$
or
$$ \frac{1}{\sqrt{n-1/2}}-\frac{1}{\sqrt{n+1/2}} \geq \frac{1}{2n\sqrt{n}} $$
such that by creative telescoping
$$ \sum_{n=1}^{N}\frac{1}{n\sqrt{n}} \leq 2\left(\frac{1}{\sqrt{1-1/2}}-\frac{1}{\sqrt{N+1/2}}\right) < 2\sqrt{2}.$$
The inequality is pretty sharp since $\zeta\left(\frac{3}{2}\right)=2.612375\ldots$
The same approach proves $\zeta\left(\frac{3}{2}\right)\leq 1+2\sqrt{\frac{2}{3}}<2.633$.
| {
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"url": "https://math.stackexchange.com/questions/2876221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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You flip a coin $10$ times. How many ways can you get at least $7$ heads?
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$\binom{10}{10}+ \binom{10}9\cdot\binom{10}1 + \binom{10}8\cdot\binom{10}2+\binom{10}7\cdot\binom{10}3$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $\cdot$ $1$ Tail $+$ $8$ Heads $\cdot$ $2$ tails $+$ $7$ Heads $\cdot$ $3$ tails.
The answer is $176$ though.
| The number of sequences of ten tosses that contain exactly seven heads and three tails is
$$\binom{10}{7}\binom{3}{3} = \binom{10}{7}$$
since there are $\binom{10}{7}$ ways to select exactly seven of the ten positions for the heads and $\binom{3}{3}$ ways to select all three of the remaining three positions for the tails.
By similar reasoning, the number of sequences of ten tosses that contain exactly $k$ heads and $10 - k$ tails is
$$\binom{10}{k}\binom{10 - k}{10 - k} = \binom{10}{k}$$
since there are $\binom{10}{k}$ ways to select exactly $k$ of the ten positions in the sequence for the heads and $\binom{10 - k}{10 - k}$ to select all $10 - k$ of the remaining $10 - k$ positions for the tails.
Hence, the number of sequences of ten coin tosses in which at least seven heads occur is
\begin{align*}
\sum_{k = 7}^{10} \binom{10}{k}\binom{10 - k}{10 - k} & = \sum_{k = 7}^{10} \binom{10}{k}\\
& = \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10}\\
& = 120 + 45 + 10 + 1\\
& = 176
\end{align*}
In your attempt, when you calculated the number of ways of selecting seven heads and three tails, you first selected seven of the ten positions for the heads and then selected three of the ten positions for the tails without taking into account that the three tails can only occupy the $10 - 7 = 3$ positions not already occupied by heads. That means you allowed heads and tails to occupy the same positions in the sequence, a physical impossibility.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to graph sinusoidal functions I understand how to graph sinusoidal functions, but how do you decide to choose an input? For $\cos(x)$, people choose $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$, etc. but for $\cos(4x)$, choosing those same inputs would give the outputs: $1, 1, 1, 1$, etc. to get the outputs $0$ and $-1$, you need other inputs. How do I decide/find those inputs?
| Consider the graph of the cosine function $f: \mathbb{R} \to [-1, 1]$ defined by $f(x) = \cos x$.
Observe that at $0$, the function obtains its maximum value of $1$. Its value falls to $0$ at $x = \pi/2$, continues to decrease to its minimum value of $-1$ at $x = \pi$, increases to $0$ at $x = 3\pi/2$, and continues to increase to its maximum value of $1$ at $x = 2\pi$. The cycle then repeats itself. In fact, the graph repeats itself every $2\pi$ radians.
If there exists a positive number $p$ such that $f(x) = f(x + p)$ for each $x$ in its domain, then we say that the function $f$ is periodic. If there exists a smallest such $p$, the function is said to have period $p$.
The cosine function has period $2\pi$.
The function $g: \mathbb{R} \to [-1, 1]$ defined by $g(x) = \cos(4x)$ has frequency $4$, meaning that it complete four full cycles in one period of the cosine function
$f: \mathbb{R} \to [-1, 1]$ defined by $f(x) = \cos x$. Consequently, it has period
$$T = \frac{2\pi}{4} = \frac{\pi}{2}$$
Notice that $g(4 \cdot 0) = g(0) = 1$. Thus, the function $g$ also assumes it maximum value of $1$ at $x = 0$. The value of the function $g(x) = \cos(4x)$ then decreases to $0$, which it reaches when
$$4x = \frac{\pi}{2} \implies x = \frac{\pi}{8}$$
It continues to decrease to its minimum value of $-1$, which it reaches when
$$4x = \pi \implies x = \frac{\pi}{4}$$
The function then increases to $0$, which it reaches when
$$4x = \frac{3\pi}{2} \implies x = \frac{3\pi}{8}$$
The function continues to increase to its maximum value of $1$, which it reaches when
$$4x = 2\pi \implies x = \frac{\pi}{2}$$
Since $g$ has period $\pi/2$, the graph then repeats itself, as shown below.
Notice that we have divided a period of the graph into four subintervals by determining the values at which the cosine function has its maximum and minimum values and $x$-intercepts, then used periodicity to draw the graph.
Alternatively, we could solve for the maxima, minima, and $x$-intercepts of $g(x) = \cos(4x)$, then draw a smooth curve through the points we obtain.
maxima:
\begin{align*}
\cos(4x) & = 1\\
\cos(4x) & = \cos(2n\pi), n \in \mathbb{Z}\\
4x & = 2n\pi, n \in \mathbb{Z}\\
x & = \frac{n\pi}{2}, n \in \mathbb{Z}
\end{align*}
minima:
\begin{align*}
\cos(4x) & = -1\\
\cos(4x) & = \cos(\pi + 2n\pi), n \in \mathbb{Z}\\
4x & = \pi + 2n\pi, n \in \mathbb{Z}\\
x & = \frac{\pi}{4} + \frac{n\pi}{2}, n \in \mathbb{Z}
\end{align*}
$x$-intercepts:
\begin{align*}
\cos(4x) & = 0\\
\cos(4x) & = \cos\left(\frac{\pi}{2} + n\pi\right), n \in \mathbb{Z}\\
4x & = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}\\
x & = \frac{\pi}{8} + \frac{n\pi}{4}, n \in \mathbb{Z}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Real value of equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$ Find the real value of x in the equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$
I tried to square the whole term and after expansion not getting the result.
| Note for the expression $(1-1/x)^{1/2}$ to have meaning, $1$ must be greater than or equal to $1/x$. So $x\geq1$.
Squaring both sides, $$x-\frac{1}{x}+2\left(x-1-\frac{1}{x}+\frac{1}{x^2}\right)^{1/2}+1-\frac{1}{x}=x^2$$
Rearranging:
$$2\left(x-1-\frac{1}{x}+\frac{1}{x^2}\right)^{1/2}=x^2-x-1+\frac{2}{x}$$
Squaring again:
$$4\left(x-1-\frac{1}{x}+\frac{1}{x^2}\right)=x^4-2x^3-x^2+6x-3-\frac{4}{x}+\frac{4}{x^2}$$
$$0=x^4-2x^3-x^2+2x+1$$
$$0=\left(x^2-x-1\right)^2$$
$$0=x^2-x-1$$
There is one solution greater than $1$, the large Golden Ratio, $\frac{1+\sqrt{5}}{2}\approx1.613\ldots$.
| {
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"url": "https://math.stackexchange.com/questions/2883280",
"timestamp": "2023-03-29T00:00:00",
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Find the oblique asymptote of $\sqrt{x^2+3x}$ for $x\rightarrow-\infty$ I want to find the asymptote oblique of the following function for $x\rightarrow\pm\infty$
$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2\left(1+\frac{3x}{x^2}\right)}\sim\sqrt{x^2}=|x|$$
For $x\rightarrow+\infty$ we have:
$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x}{x}=1$$
which means that the function grows linearly.
$$f(x)-mx=\sqrt{x^2+3x}-x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}-1\right)\sim x\left(\frac {1}{2}\cdot\frac{3}{x}\right)=\frac{3}{2}$$
The oblique asymptote is $y=x+\frac 3 2$ which is correct. For $x\rightarrow-\infty$ we have:
$$\frac{f(x)}{x}=\frac{|x|}{x}=\frac{-x}{x}=-1$$
This means that
$$f(x)-mx=\sqrt{x^2+3x}+x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty$$
Which is not what my textbook reports ($-\frac{3}{2}$). Any hints on what I did wrong to find the $q$ for $x\rightarrow-\infty$?
| This is not a different way, but is better I think
$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2+3x+(\frac32)^2-(\frac32)^2}=\sqrt{\left(x+\frac{3}{2}\right)^2-(\frac32)^2}\sim\left|x+\frac{3}{2}\right|$$
which gives both oblique asymptotes.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Range of a rational function with radicals Find the range of the function
$$\frac{6}{5\sqrt{x^2-10x+29} - 2}$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$\frac{6}{5\sqrt{x^2-10x+29} - 2} = y$
getting the inverse,
$\frac{6}{5\sqrt{y^2-10y+29} - 2} = x$
$\frac{4x^2+24x+36}{25x^2}= y^2-10y +29$
Then it would be a quadratic function in y, but the discriminant becomes really big
$100- 116(\frac{4x^2+24x+36}{25x^2})$
| Hint
If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that:
$$(x-5)^2+4 \in [4,+\infty)$$
$$\sqrt{(x-5)^2+4} \in [2,+\infty)$$
$$5\sqrt{(x-5)^2+4}-2 \in [8,+\infty)$$
Does that help?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $z^3+bz^2+c=0$ Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5$
We have $z_1=(-\sqrt{2}-\sqrt{2}i)^5=32e^{i\frac{15\pi}{4}}$
and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=32e^{-i\frac{15\pi}{4}}$
Can we conclude straight away something about $b,c$?
| Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=z_1^*$
$$z^3+bz^2+c=(z-z_1)(z-z_1^*)(z-a)$$ where $a$ has to be real, right?
$$z^3+bz^2+c=z^3-(z_1+z_1^*+a)z^2+(a(z_1+z_1^*)+|z_1|^2)z-a|z_1|^2$$
You can conclude right away anything you want here. You could quickly compute what $a$ is and find any relation between $b,c$. You could also quickly solve for $b,c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Monotonicity of function at a point The question says :
Let
$$f(x)=\begin{cases}
-x^3+\frac{b^3-b^2+b-1}{b^2+3b+2} &:0\le x\lt1\\ 2x-3 &:1\le x\le3\end{cases}$$. Find all possible values of b such that f(x)has the smallest value at $x=1$.
Since this question was an example question, the solution said,
The Limiting value of f(x) from the left of $x=1$ should be either greater or equal to the value of the function at $x=1$.
My question is for $x=1$ to have the smallest possible value, shouldn't the limiting value be Less than or equal to the function at $x=1$?
The answer is $b\in (-2,-1)\cup (1,+\infty)$
| $f$ is decreasing on $[0,1)$ and increasing on $[1,3]$. Hence $f(1)$ is the minimum iff $f(1) \geq f(1-)$ which means $-1 \geq -1+\frac {(1+b^{2})(b-1)} {(1+b)(2+b)}$ or $\frac {(b-1)} {(1+b)(2+b)}\leq 0$ . It is easy to find values of $b$ from this. (The function is not defined for $b=-1$ and $b=-2$ so consider $b$ in $(-\infty, -2),(-2,-1)$ and $(1,\infty$)).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x+8y+48
\end{align}$$
Is it okay? Thank you
| Your argument is completely fine. Continue it with:
$$x^2+\frac{32}{3}x+y^2+\frac 83 y+16=0$$
Now complete the square on the $x$'s and $y$'s
$$(x+\frac{16}{3})^2-\frac{256}{9}+(y+\frac43)^2-\frac{16}{9}+16=0$$
$$(x+\frac{16}{3})^2+(y+\frac{4}{3})^2=\frac{128}{9}$$
Thus we have a circle, centre $(-\frac{16}{3}, -\frac 43)$, radius $\frac{8\sqrt2}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $\lim\limits_{t\to\infty}x(t)$ if $x'= (x-y)(1-x^2-y^2)$, $y' = (x+y)(1-x^2-y^2)$ Suppose $x_0,y_0$ are reals such that $x_0^2+y_0^2>0.$ Suppose $x(t)$ and $y(t)$ satisfy the following:
*
*$\frac{dx}{dt} = (x-y)(1-x^2-y^2),$
*$\frac{dy}{dt} = (x+y)(1-x^2-y^2),$
*$x(0) = x_0,$
*$y(0) = y_0.$
I am asked to find $\lim_{t\to\infty}x(t).$
Dividing the two differential equations, we have $$\frac{dy}{dx} = \frac{x+y}{x-y} = \frac{1 + \frac{y}{x}}{1-\frac{y}{x}}.$$
Let $v(x)=\frac{y}{x},$ so that $y = xv(x),$ and $$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$ This implies that $$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v}.$$ Rearranging and integrating, we get $$\log(x) = \arctan(v)-\frac{1}{2}\log(1+v^2)+c,$$ so $$\log(x) = \arctan\left(\frac{y}{x}\right)-\frac{1}{2}\log\left(1+\left(\frac{y}{x}\right)^2\right)+c.$$ By writing $1+\left(\frac{y}{x}\right)^2$ as $\frac{x^2+y^2}{x^2},$ we can simplify the above equation to $$\frac{1}{2}\log(x^2+y^2) = \arctan\left(\frac{y}{x}\right) + c.$$
However, I'm not sure how to proceed from this implicit equation. Any help would be much appreciated!
| Let $r^2 = x^2+y^2$, $x = rcos\theta$, and $y = rsin\theta$.
Then, you get the following equations:
$$\begin{align}\frac{dr}{dt} &= r(1-r^2) \\
\frac{d\theta}{dt} &= (1-r^2)
\end{align}$$
The solution to the first equation is: $r^2=\frac{ke^{2t}}{1+ke^{2t}}$, where $k=\frac{r^2_0}{1-r_0^2}$.
Then, the solution to the second equation is: $\theta-\theta_0=log\ \frac{r}{r_0}$,
where $r_0^2=x_0^2+y_0^2$ and $tan\ \theta_0= \frac{y_0}{x_0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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non-abelian group of order 27 If $K$ is a field and $A= \begin{pmatrix} a & 0 &0 \\ 0 & b &0 \\ 0& 0& c \end{pmatrix}$ , $B=\begin{pmatrix} 0 & p &0 \\ 0 & 0 &q \\ r& 0& 0 \end{pmatrix}$ , $C=\begin{pmatrix} 0 & 0 &x \\ y & 0 &0 \\ 0& z& 0 \end{pmatrix}$ ,
where $a,b,c,p,q,r,x,y,z\in K$ are such that $S:=\{a,b,c,p,q,r,x,y,z\}$ is closed under multiplication , then I can show that $A^2,B^2,C^2,AB,BA,BC,CB,AC,CA \in \Bigg\{\ \begin{pmatrix} a' & 0 &0 \\ 0 & b' &0 \\ 0& 0& c' \end{pmatrix}, \begin{pmatrix} 0 & p' &0 \\ 0 & 0 &q' \\ r'& 0& 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 &x' \\ y' & 0 &0 \\ 0& z'& 0 \end{pmatrix} : a',b',c',p',q',r',x',y',z' \in S \Bigg\}$
My question is: Can we use this fact to construct a non-abelian group of order $27$ ?
The only way I know of constructing a non-abelian group of order $p^3$ for odd prime $p$ is using upper triangular $3\times3$ matrices or special kind of $2\times 2$ matrices with entries in $\mathbb Z/(p)$ or $\mathbb Z/(p^2)$, as given here http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf .
Please help.
Thanks in advance
| Yes, provided that $K$ contains a primitive cube root $\omega$ of $1$. The two nonabelian groups of order $27$ both have irreducible representations of degree $3$ that are induced from $1$-dimensional representations of elementary abelian subgroups of order $9$.
The nonabelian group of exponent $3$ $is$
$$\left\langle\,\left(\begin{array}{ccc}1&0&0 \\ 0&\omega&0 \\ 0&0&\omega^2 \end{array}\right),\,
\left(\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right)
\,\right\rangle$$
and the one of exponent $9$ is
$$\left\langle\,\left(\begin{array}{ccc}1&0&0 \\ 0&\omega&0 \\ 0&0&\omega^2 \end{array}\right),\,
\left(\begin{array}{ccc}0&0&\omega\\1&0&0\\0&1&0\end{array}\right)
\,\right\rangle.$$
To see that the first group has order $27$, call the two generators $a$ and $b$. Then $b$ is a permutation matrix, and $b^{-1}ab = \left(\begin{array}{ccc}\omega&0&0\\0&\omega^2&0\\0&0&1\end{array}\right)$, so $b^{-1}aba^{-1}$ is the scalar matrix $\omega I_3$, which commutes with $a$ and $b$.
So, putting $c=b^{-1}aba^{-1}$, $a$, $b$ and $c$ satisfy the relations of the presentation $$\langle a,b,c \mid a^3=b^3=c^3=1, c=b^{-1}aba^{-1}, [a,c]=[b,c]=1\rangle,$$
which define a group of order $27$. It's not hard to see that the group generated by $a$ and $b$ has order greater than $9$, so it must have order $27$.
You can prove that the second group has order $27$ with a similar argument.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \frac{\sqrt{1+x^8} dx}{x^{13}}$
Evaluate: $\displaystyle\int \frac{\sqrt{1+x^8}}{x^{13}}dx$
My attempt:
I have tried substituting $1+x^8=z^2$ but that did not work. I also tried writing $x^{13}$ in the denominator as $x^{16}.x^{-3}$ hoping that it would bring the integrand into some form but that too did not work.
| Let $x^4=\tan u$ so that $dx=\frac{\sec^2 u}{4x^3}du$
so our integral becomes
$\int \frac{\sqrt{1+\tan^2 u}}{x^{13}}.\frac{\sec^2 u}{4x^3}du$
or rather
$\frac{1}{4}\int \frac{\sec^3 u}{\tan^4 u}du$
which is
$\frac{1}{4}\int \cos u\sin^{-4} u du$
which equals
$-\frac{1}{12}\sin^{-3}u +C$
now substitute back from $u$ to $x$ which I think is
$-\frac{1}{12}(\frac{x^8}{1+x^8})^\frac{-3}{2}+C$
| {
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"url": "https://math.stackexchange.com/questions/2893991",
"timestamp": "2023-03-29T00:00:00",
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