Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
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:
$$ \begin{pmatrix}
x_1 \\ x_2 \\ x_3... | You probably want to find an orthonormal basis. Look into the Gram-Schmidt process.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a derivative with $\log$ $f(x)=\log{\frac{x}{1+\sqrt{5-x^2}}}$
I have to find the derivative of $f(x)$. Please tell me if my steps are correct.
$\frac{1+\sqrt{5-x^2}}{x}×\frac{1}{\ln 10}×\frac{1+\sqrt{5-x^2}-x×\frac{1}{2\sqrt{5-x^2}}}{(1+\sqrt{5-x^2})^2}×\frac{1}{2\sqrt{5-x^2}}×(-2x)$
| I would write $$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)$$ the logarithm to the base $e$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove the following determinant Prove the following:
$$\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... | 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)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
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
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$
$$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {... | $$\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)
$$
Each fraction has a (finite) geometric series in the denominator (there is a formula for th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$? How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$?
Note: $I$ denotes the identity matrix of size $3 \times3$ and $A^{T}$ represents the transpose of the matrix $A$.
I to... | Since $A^3-2A-I=0$ , any eigenvalue of $A$ satisfies
$$\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(\lambd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"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^... | 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 th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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):
$$ k>... | Using a command like
RegionPlot[
$k x^3 (2 x + 1)^3 -
2 (k + 1) (x + 1)^2 (2 x^4 + 2 x^3 + 3 x^2 + 2 x + 1) > 0$, {$k$, 0,
10}, {$x$, 0, 10}]
You will obtain an useful plot
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$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}$$
Now letting $b_n=\... | Squaring both sides
$$
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.
$$
with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Calculating $3^{m-n}=?$ $$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}$$
H... | $$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$$
Substitute $9^n$ inside the first equation and do the same trick to complete the task.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a real matrix A such that (Exponential of matrices) Is there a real matrix A such that
$$\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}$$
is given by:
$$\ex... | $$
A =
\left(
\begin{array}{cc}
0 & \pi \\
- \pi & 0
\end{array}
\right)
$$
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
$$
e^B =
\left(
\begin{array}{cc}
\cos t & \sin t \\
- \sin t & \cos t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Exhibiting $f$ such that a vector field $\overline{v} = \nabla f$? The first is suppose $\overline{v} = (x, 2yz^3, 3y^2z^2)$ and we want to exhibit an $f$ such that $\overline{v} = \nabla f$. Since $\nabla \times \overline{v}=0$, $f$ exists and we can integrate:
$$\displaystyle \int f \,dx = \frac{1}{2}x^2+c(y,z), ~... | $$ f_x = x \implies f = \frac{1}{2}x^2 + g(y,z) $$
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
$$ f = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Recursive sequence for the Koch Snowflake
Let $C_0$ an equilateral triangle of side 1 and $A_{n}$ the area of
the figure $C_{n+1}, n\in\mathbb{N}\cup {0}$
Write $A_{n+1}$ based on $A_n$ and the area of $C_0$
I noted that $C_1$ has been built by adding a smaller equilateral triangle on each side of $C_0$. Then $C_2$ ... | One alternative reasoning might follow from:
$$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)$$
I'll leave you to flesh o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that the following inequality is true: $(\frac{1}{a} + \frac{1}{bc}) (\frac{1}{b} + \frac{1}{ca})(\frac{1}{c} + \frac{1}{ab}) \geq 1728$ This is a question from a past Olympiad paper:
Three positive real numbers $a, b, c$ satisfy the following constraint:
$a+b+c = 1$. Show that the following inequality is true:... | Hint: $$\frac 1a + \frac{1}{bc} = \frac{bc+a}{abc} = \frac{bc+a(a+b+c)}{abc} = \frac{(a+b)(a+c)}{abc}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \... | 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:
$$I=\int_0^1\frac{1+\frac 1{x^2}}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve for x in the equation: $3^x + 9^x = 27^x$ Solve for x in the equation: $3^x + 9^x = 27^x$
Via calculator, the x is = 0.438 which is correct as it is in the choices. However I want to know how to solve it manually, hence the question.
I'm stuck at : $3^x + 3^{2x} = 3^{3x}$
I don't know what to do next. Any help w... | $$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$$
so either $y=0$ or $(y^2-y-1)=0$
since $3^x$ can never be $0$;
$(y^2-y-1)=0$
$\implies y = \dfrac{1\pm\sqrt{5}}{2}$
$\implies 3^x = \dfrac{1\pm\sqrt5}{2}$
$\implies x =\log_3\left(\dfrac{1\pm\sqrt5}{2}\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculating P(X>6|X<8) from pdf I have a pdf $\frac{k}{x^2}$ for $4<x<\infty$ ($0$ otherwise) and need to work out the $P(X>6 | X<8)$ I already worked out k to equal 4.
To work out this probability I did $\frac{P(X>6)}{P(X<8)}$ but I get a weird answer/
For $P(X>6)$ i did $\int_{6}^{\infty} \frac{4}{x^2}dx = \frac{2}... | $$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}$
... | 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... | {
"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 answ... | 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... | {
"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$$
... | To get rid of the ln on both sides just do $e^{ln|1+2y|}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Expand Expression in Tensor Notation I have an expression given to me in tensor/index notation and I need help expanding out into something more readable. The expression is
$$\frac{b_{ij}b_{jk}b_{ki}}{6}$$
and $b$ is a symmetric second order tensor, and the indices range from 1 to 3.
Can somebody please expand this ex... | Let us define a second-order symmetric tensor $\boldsymbol{B} = [b_{i j}]$ given by its coordinates $b_{\bullet\bullet}$ over an orthonormal basis (the bullets could be any distinct symbols). Using the dot product and Einstein notation, one can write
$$
[(b^3)_{i \ell}] = \boldsymbol{B}^3 = \boldsymbol{B}\cdot\boldsymb... | {
"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 ... | 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.... | {
"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 $$\frac{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c} $
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c}, c > 0 $
It seems to me that the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ of the maximum area when $ a = c; b = \frac{1}{c} $
That is, it lo... | Expanding in Taylor series around $x=0$
$$
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}\... | {
"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 $z^2$ coeffic... | 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 $\triangle DAF,\triangle ABC$
$$\dfrac{AF}{CA}=\dfrac{DF}{AB}\implies\dfrac{a\cos B-x}{a\sin B}=\dfrac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is the sum of numbers between 250 and 350 which are divisible by 7?
What is the sum of numbers between $250$ and $350$ which are divisible by $7$?
My attempted solution:
7|250|35
|21 |
-----
40
35
----
5
The 1st number divisible by $7$ is: $(250-5)+7 = 252$.
7|350|50
|35 |
-----
0
The last... | HINT
The first divisible number is $252$ the last $350$ then
$$252+\dots+350=7(36+\dots+50)=7\left(\sum_{k=1}^{50}k-\sum_{k=1}^{35}k\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to find the limit of this series I was trying to figure out the limit of the function below
$a_n = \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}$
The answer to the question is $\frac{1}{8}$ but when I divide through by $3^{n+3}$ (because it has the highest power) I cant figure out how they got that answer, Im really stu... | Look for hidden powers:
$a_n
= \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}
= \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+2^{2n+2}}
$.
The $3^n$ terms are negligible
for large $n$,
so
$a_n
\approx \frac{2^{2n-1}}{2^{2n+2}}
=2^{(2n-1)-(2n+2)}
=2^{-3}
=\frac18
$.
To check,
$\begin{array}\\
a_n-\dfrac18
&= \dfrac{2^{2n-1}+3^{n+3}}{3^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding conditional expectation of two random variables Let $X$ and $Y$ be independent and uniform on $[0,3]$. I want to calculate $E(Y| X<1 \cup Y<1 )$.
Attempt.
First, we calculate the distribution via the cdf:
$$ 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) } =... | 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 $\frac{1}{2}$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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$x$ 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",
"answer_count": 6,
"answer_id": 0
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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+\mathca... | Note that
$$\frac{2\mu(e^x(x^2-x+1)-1)-e^x(x^2-2x+2)+2}{x(2\mu(e^x(x-1)+1)-e^x(x-2)-2)}=\frac{[(2\mu-1)x^2-(2\mu-2)x+(2\mu-2)]e^x-(2\mu-2) }{[(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x}=\frac{ [(2\mu-1)x^2-(2\mu-2)x]e^x +(2\mu-2)x +(2\mu-2)e^x-(2\mu-2)x-(2\mu-2)}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }=\\=1+(2\mu-2)\frac{ e^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$ How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$?
My attempt
$$(3x+5y)^{12}=\left(3x\left(1+\frac{5y}{3x}\right)\right)^{12}$$
$$=3^{12}x^{12}\left(1+\frac{5y}{3... | 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... | {
"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
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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{\s... | 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.,
$$\cos u = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding value of $\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$
Finding value of $$\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}$$
I did not find h... | It is Chu–Vandermonde identity
$$\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... | 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 $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqr... | I would proceed as follows:\begin{align}\int\frac{y^2}{\sqrt{1-y^2}}\,\mathrm dy&=\int y\frac y{\sqrt{1-y^2}}\,\mathrm dy\\&=-y\sqrt{1-y^2}+\int\sqrt{1-y^2}\,\mathrm dy.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Integrate $\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx$
Integrate $\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx$
My Attempt
Put $x=\cos2a\implies dx=-2\sin2a.da$
$$
\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\ta... | 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 $n\pi$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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 a... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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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... | 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
$$1 + 2x + \cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How do I start this integral problem: $\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $? How do I prove this?
$$\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*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Solve this system of equations for the value of $a,b$ and $c$ such that $a^3+3ab^2+3ac^2-6abc=1$,$b^3+3ba^2+3bc^2-6abc=1$,$c^3+3cb^2+3ca^2-6abc=1$ Solve this system of equations for the value of $a,b$ and $c$.
$$a^3+3ab^2+3ac^2-6abc=1$$
$$b^3+3ba^2+3bc^2-6abc=1$$
$$c^3+3cb^2+3ca^2-6abc=1$$
This is symmetric equation.... | If we substrac 1. equation -2. equation we get:
$$(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 $$
a contradiction.
So $a=b$. The same way (2.equation-3.equation) we get $b=c$. S0 $a=b=c =:x$ and now you have to solve: $x^3=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluate $\int e^{-3x} \cos^3x\,dx$
Evaluate $\int e^{-3x}\cos^3x\,dx$
My Attempt
\begin{align}
& \int e^{-3x} \cos^3x\,dx=\cos^3x \cdot \frac{e^{-3x}}{-3}-\int3\cos^2x\sin x \cdot \frac{e^{-3x}}{-3} \, dx \\
= {} &\frac{-1}{3}\cos^3x \cdot e^{-3x}+\int\cos^2x\sin x \cdot e^{-3x} \, dx\\
= {} & \frac{-1}{3}\cos^3x\cd... | Hint:
$$\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\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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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 p... | 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 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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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 $$I(k) =\int_0^1 x^... | $$ \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 (-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Evaluate $\,\lim_{n\to\infty}\frac{n^2}{({n^2+1^2})^{3/2}} + \frac{n^2}{({n^2+3^2})^{3/2}} + \cdots + \frac{n^2}{({n^2+(n-1)^2})^{3/2}}$ $$= \sum_{r=1,3,5 \ldots}^{n-1} \frac{n^2}{({n^2 + r^2})^{\frac32}} = \sum_{r=1,3,5 \ldots}^{n-1} \frac{1}{n({1 + (r/n)^2})^{\frac32}} = \int_{0}^{1}\frac{dx}{({1+x^2})^{\frac32}} = \... | Minor corrections
$$
\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}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a cleaner way of solving this number theory problem? The problem is
If $a$ has order $3\pmod{p}$ where $p$ is an odd prime, show that $(a+1)$ has order $6\pmod{p}$.
My solution:
$$a^3-1\equiv 0\pmod{p}$$
$$(a-1)(a^2+a+1)\equiv 0 \pmod{p}$$
$(a-1)$ cannot be divisible by $p$ because the order of $a$ is $3$, ... | I'll take all congruences modulo $p$.
As you say, $a^2\equiv -a-1$. Therefore $(a+1)^2\equiv -a-1+2a+1\equiv a$
and $(a+1)^3\equiv a^2+a\equiv-1$. From this, we have $(a+1)^6\equiv1$,
$(a+1)^2\not\equiv1$ and $(a+1)^3\not\equiv1$. Thus $a+1$ must have order $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric integral using residues I am trying to evaluate following integral by using the Residue Theorem:
$$\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$$
The poles $z=0,\frac{-2}{... | Your reduction of the real integral to the complex one is correct
$$\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)$$
where $\displaystyle f(z)=\frac{(z^2+1)^2}{z^2(2z+3)(3z+2)}$.
Now by the partial fraction deco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Another way to find the sum $S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}$ I want to know the sum
$$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}=... | 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$$F(x)=x\left(1-x^2+x^4-x^6+\cdots+(-1)^{n-1}x^{2n-2}\right)=\frac{x-(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find an element of order $5$ of the field $\mathbb{Z}_3[x]/\langle x^4 + x + 2 \rangle$. Find an element of order $5$ of the field $\mathbb{Z}_3[x]/\langle x^4 + x + 2 \rangle$.
I tried the systematic method of considering
$$1=(ax^3 + bx^2 + cx + d)^5$$
and using the relation $x^4 = 2x+1$ but the algebra got messy. Any... | It is maybe good to know that the one or the other computer algebra system can deal systematically with computations in finite fields.
For instance, using sage, a small dialog with the sage interpreter detects all elements of multiplicative order $=5$, the code has a digestible mathematical syntax, that will be explain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Differentiating $x = \sec^2{3y}$ for $d^2y/dx^2$ I want to ask a question about differentiating trigonometric functions.
I am trying to find $\frac{d^2y}{dx^2}$ for $x = \sec^2{3y}$
Now, this is what I did.
$$\frac{dx}{dy} = 6\sec^2{3y}\tan{3y} = 6x(x-1)^{\frac{1}{2}}$$
I got the $(x-1)^{\frac{1}{2}}$ term as follows... | Because in general, $\dfrac{d^2 x}{dy^2}\dfrac{d^2 y}{dx^2}\ne 1$. To take a much simpler example, if $y=x^3$ then $x=y^{1/3}$ and $$\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}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $\Longleftrightarrow$ $\Del... |
This not a complete answer but that's too long for me to post it in comment.
Preliminary for differential geometry of surfaces
\begin{align*}
\mathbf{x}(u,v)
&= \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} \\
\mathbf{x}_u &= \frac{\partial \mathbf{x}}{\partial u} \\
\mathbf{x}_v &= \frac{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of given sequence Find the limit of the sequence {$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 lim... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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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... | 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^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
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}
\do... | 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:
$$x=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving $a_{n+1}^2=a_n·a_{n+2}+(-1)^n$ for Fibonacci sequence $\{a_n\}$ Let $a_1,a_2,a_3,...,a_n$ be the sequence of Fibonacci numbers.
Then we are required to prove by induction that $$a_{n+1}^2=a_n \cdot a_{n+2} + (-1)^n$$
My Attempt:
Initially putting a few values like $n=1,2 \: \text{or}\ 3$ makes it certain that w... | A sneaky linear algebra way. Let $$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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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^... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
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... | 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. $0;1;2;-1,i,-i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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 ... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
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{-... | 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
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prime number and golden ratio I think the following is true, but can't show it.
Let $p$ be a prime number, and let $f(x)$ be a polynomial which satisfies
$${\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$$
which can be useful in order to solve the main problem.
If $p\equiv 1\pmod 5$ or $p\equiv 4\pmod 5$, then the equation $x^2-x-1$ has two roots in $\Bbb Z/p\Bbb Z$, say $u,v$.
If $\delta$ denote the $p$-derivation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
General solution of the ordinary differential equation $(D^4+D^2+1)y=0$ $(D^4+D^2+1)y=0$ where $D=\frac{d}{dx}$
$$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{... | 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_{3... | {
"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... | 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
$$bc = \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 11
} |
Calculating volume using cylindrical coordinates.
A solid $D$ is defined by the following inequalities: $$\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}$$
Plugging in these values into the fi... | For limits of $z$ your lower bound is the lower hemisphere $z=1-\sqrt {1-r^2}$ and the upper limit is the cone $ z=r $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find smallest set of natural numbers whose pairwise sums include 0..n Given a positive integer $n$, how do you find the smallest set of nonnegative integers $S$ such that for each integer $m$, where $0\leq m<n$, there exist two (not necessarily distinct) members of set $S$, say $x$ and $y$ such that $x+y=m$.
For exampl... | Let $n+4=s^2+r$ with $r,s\in\Bbb N$ and $0\leq r\leq 2s$.
Then an upper bound is given by
$$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}$$
which gives $L\leq 12$ for $n=50$ and $L\leq 18$ for $n=100$.
In general, for large $n$ this gives $L=O(\sqrt n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$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}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization... | Using Jordan canonical form, $X$ must be similar to one of the three matrices
\begin{align}
\begin{pmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{pmatrix},
\begin{pmatrix}
\lambda & 0 \\
0 & \lambda
\end{pmatrix},
\begin{pmatrix}
\lambda & 1 \\
0 & \lambda
\end{pmatrix}
\end{align}
, where $\lambda_1 \neq \lambda_2$, whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Minimal area of a right triangle with inradius $1$ I got this question, solved it, then forgot how I solved it.
What is the minimal area of a right triangle with inradius $1$?
My attempt:
$r=\frac{a+b-c}2$, so $a+b=c+2$
$a^2+b^2=c^2$
This gives $ab=2(c+1)$
I remember using the AM-GM to prove that equality held, thus ... | Recall that $$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}$$
so $ab \ge (2+\sqrt{2})^2 = 2(3+2\sqrt{2})$.
The equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find a real matrix $B$ such that $B^3 = A$
Given $$A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}$$ find a real, invertible matrix $B$ such that $B^3 = A$
I think I am doing something wrong here, so let me describe my attempt:
1) So I started off with diagonalizing the matrix $A$ with finding the eigenvalues $\lambd... | $$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... | 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} =... | {
"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
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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\... | 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?
| {
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"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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...... | 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 $$\begin{align}3\sec^2 \theta -4 &=x^2\\ \implies 3\sec^2 \theta \tan \theta \, \mathrm d\theta &= x \, \mathrm dx \\ \implies (x^2+4)\sqrt{\dfrac{x... | {
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"source": "stackexchange",
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An inequality relating the ratio of the areas of two triangles The conjecture below is a modified version of this question:
Prove that, given a triangle with sides $a,b,c$, there exists a triangle with sides $a+2b,b+2c,c+2a$ that has an area three times the original
Conjecture:
If $u$ is the area of a triangle with ... | Let $x=b+c-a\geq 0$, $y=a+c-b\geq 0$, $z=a+b-c\geq 0$.
Let $p=\frac{1}{2}(a+b+c)$ be the semiperimeter of the triangle with sides $a$, $b$, $c$. Similarly let $P=\frac{1}{2}(A+B+C)=3p$ be the semiperimeter of the triangle with sides $A=b+2c$, $B=c+2a$, $C=a+2b$. Then, by Heron's formula, the inequality is equivalent to... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{n\to\infty}\sum_{k=1}^n \frac{\pi k}{2n}\int_0^1 x^{2n}\sin\frac{\pi x}{2}dx$ Hello I am trying to compute $$\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\f... | 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 li... | {
"language": "en",
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"source": "stackexchange",
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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 $x_n$. $\t... | 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 sequenc... | {
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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 sho... | Use the fact that when $u\to 0$, $(1+u)^n \approx 1+nu$.
Using that fact,
$$\begin{align} L &= \lim_{x\to 0} \frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}} \\ &= \lim_{x\to 0} \dfrac{1+\dfrac 12 x^2 - (1+\dfrac 12 x)}{1+\dfrac 12 x^3 - (1+\dfrac 12 x)}\\ &= \lim_{x\to 0} \dfrac{x^2-x}{x^3-x} \\ &= \l... | {
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"url": "https://math.stackexchange.com/questions/2852294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of this integration shortcut: $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$ I came across this as one of the shortcuts in my textbook without any proof.
When $b\gt a$,
$$\int\limits_a^b \dfrac{dx}{\sqrt{(x-a)(b-x)}}=\pi$$
My attempt :
I notice that the the denominator is $0$ at both the bounds. I thought of su... | \begin{align}
\tan^2 \theta &= \frac{x-a}{b-x} \\
2\tan \theta \sec^2 \theta \, d\theta &=
\frac{b-a}{(b-x)^2} \, dx \\
2\sqrt{\frac{x-a}{b-x}} \times \frac{(x-a)+(b-x)}{b-x} \, d\theta &=
\frac{b-a}{(b-x)^2} \, dx \\
2\, d\theta &= \frac{dx}{\sqrt{(x-a)(b-x)}} \\
\int \frac{dx}{\sqrt{(x-a)(b-x)}} &=
2\... | {
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"url": "https://math.stackexchange.com/questions/2853673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\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)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably... | What? So the left hand side is the AM of $1/(b+c+d), 1/(c+d+a), 1/(d+a+b)$ and $1/(a+b+c)$ i.e. $(1/(b+c+d)+1/(a+c+d)+1/(a+b+d)+1/(a+b+c))/4$. On the RHS we have $4$ over the sum of the reciprocals of these terms i.e. $4/((b+c+d)+(c+d+a)+(d+a+b)+(a+b+c))$ which does indeed come out as $4/3(a+b+c+d)$ Which then (multi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the probability that atleast one valve is defective. A factory A produces $10$% defective valves and another factory $B$ produces 20% defective valves.A bag contains $4$ valves of factory $A$ and $5$ valves of factory B.If two valves are drawn at random from the bag,find the probability that at least one valve is ... | The Probability that factory $A$ produces defective values is $\dfrac{10}{100}=\dfrac{1}{10}$
The Probability that factory $B$ produces defective values is $\dfrac{20}{100}=\dfrac{1}{5}$
Given a bag contains $4$ values of factory $A$ and $5$ values of factory $B$ and two values are drawn random.
$$P(\mbox{at least one ... | {
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"url": "https://math.stackexchange.com/questions/2857533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 ... | 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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"url": "https://math.stackexchange.com/questions/2861406",
"timestamp": "2023-03-29T00:00:00",
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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 &... | {
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"url": "https://math.stackexchange.com/questions/2861490",
"timestamp": "2023-03-29T00:00:00",
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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 $... | 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 $30$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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{\s... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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
$$u_t + 6uu_x + u_{xxx}... | Begin by setting $a=\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) $and $b=\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right).$ If $c=0,$ then the function in question is clearly a solution, hence we may factor out $c^{5/2}/2.$
The computed expression simplifies to $ab-3a^2b-ab^3+2a^2b = ab-a^2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polynomial problem with unknown coefficients $a, b, c$ $p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.
Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-\frac{a^2}{a+1}$ and then getting $c=-\frac{a^4}{a+1}$ does no... | Since $b=-\frac{a^2}{a+1}$, we see that
$$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.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865053",
"timestamp": "2023-03-29T00:00:00",
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General solution of $(4x^2-x)y''+2(2x-1)y'-4y=12x^2-6x$ given $y_1=\frac1x$ Verify that $y_1=\frac1x$ is a solution to d.e
$\left(4x^2-x\right)y''+2\left(2x-1\right)y'-4y=0$
Find the general solution of the d.e
$\left(4x^2-x\right)y''+2\left(2x-1\right)y'-4y=12x^2-6x$
Note: I was able to prove initial value $y_1=\frac... | Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + \frac{4x-2}{x(4x-1)}y' - \frac{4}{x(4x-1)}y = \frac{12x-6}{4x-1} $$
The basic idea is to try a solution... | {
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"timestamp": "2023-03-29T00:00:00",
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Integrating over a sphere Suppose $x\in\mathbb{R}^n$ is a random unit vector distributed uniform over the $(n-1)-$sphere $S_{n-1}$ (the set of unit vectors in $\mathbb{R}^n$ ). For an arbitrary vector $y\in\mathbb{R}^n$ how does one evaluate
$$\int_{S_{n-1}}e^{-||y-x||_2^2}\mathrm{d}\mu$$
where $\mu$ denotes the unif... | As remarked by @md2perpe, since the $(n-1)$ sphere is invariant under rotation, then you could simply apply a change of variable given by $x=Rx'$, where $R$ is an orthogonal matrix, such that $R^\ast y=e_1 = (1, 0, \ldots, 0)$. Then we see that
\begin{align}
\int_{S^{n-1}} e^{-|x-y|^2}\ d\mu(x) = \int_{S^{n-1}} e^{-|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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First Order Separable differential Equation
Problem:
Solve the following differential equation:
\begin{eqnarray*}
6x^2y \, dx - (x^3 + 1) \, dy &=& 0 \\
\end{eqnarray*}
Answer:
This is a separable differential equation.
\begin{eqnarray*}
\frac{6x^2}{x^3+1} \, dx - \frac{dy}{y} &=& 0 \\
\int \frac{6x^2}{x^3+1} \, ... | Both answers are correct.
Your answer $$(x^3+1)^2 = c |y|$$ makes the assumption that $c\ge 0$
The book's answer $$(x^3+1)^2 = |cy|$$ is OK for all values of $c$.
Thus to make sure that you can take any value for c go with the book's answer, otherwise mention that $c\ge 0$ .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$ Solve the equation:
$$\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]{... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}... | We may consider that $f(x)=\frac{1}{\sqrt{x}}$ is a convex function on $\mathbb{R}^+$, hence for any $n\in\mathbb{N}^+$ we have
$$ 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876221",
"timestamp": "2023-03-29T00:00:00",
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"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 $\cdo... | 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 ta... | {
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"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 $-1$... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 aga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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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:
$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x... | This is not a different way, but is better I think
$$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|$$
which gives both oblique asymptotes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Range of a rational function with radicals Find the range of the function
$$\frac{6}{5\sqrt{x^2-10x+29} - 2}$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$\frac{6}{5\sqrt{x^2-10x+29} - 2} = y$
getting the inverse,
$\frac{6}{5\sqrt{y^2-10y+29} - 2} = x... | Hint
If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that:
$$(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)$$
Does that help?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $z^3+bz^2+c=0$ Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5$
We have $z_1=(-\sqrt{2}-\sqrt{2}i)^5=32e^{i\frac{15\pi}{4}}$
and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=32e^{-i\frac{15\pi}{4}}$
Can we conclude straight away something about $b,c$?
| Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=z_1^*$
$$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$$
You can conclude right away anything you want here. You coul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Monotonicity of function at a point The question says :
Let
$$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}$$. Find all possible values of b such that f(x)has the smallest value at $x=1$.
Since this question was an example que... | $f$ is decreasing on $[0,1)$ and increasing on $[1,3]$. Hence $f(1)$ is the minimum iff $f(1) \geq f(1-)$ which means $-1 \geq -1+\frac {(1+b^{2})(b-1)} {(1+b)(2+b)}$ or $\frac {(b-1)} {(1+b)(2+b)}\leq 0$ . It is easy to find values of $b$ from this. (The function is not defined for $b=-1$ and $b=-2$ so consider $b$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $|z-4i|=2|z+4|$ $$\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... | 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 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
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 f... | 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 s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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,... | 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\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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 d... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.