Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Let $A = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)$. Prove for $n \geq 1$ using induction that $A^n =$ ... Can someone check to see if my proof is correct? Feel free to nitpick, trying to get better at writing proofs. Here's the problem:
Let $A = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smal... | The proof looks perfect mathematically. Stylistically, it looks pretty good as well, although since you mentioned that you'd like some nitpicking, here goes:
Depending on the context in which you need to write this proof, the phrasing of the induction could be improved a little bit. (If this is for a class, and you've ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all polynomials $P, Q$ such that $(x^2+ax+b)P(x^2+cx+d)=(x^2+cx+d)Q(x^2+ax+b)$
Find all polynomials $P(x), Q(x)$ such that
$(x^2+ax+b)P(x^2+cx+d)=(x^2+cx+d)Q(x^2+ax+b)$
where $a, b, c, d$ are all different real numbers.
I found this problem while solving functional equations in polynomials. Here's what I've foun... | As Ng Chung Tak said in the comments above, the only solutions are $P = Q = \lambda X$, $\lambda\in \mathbb{R}$, if $a \neq c$ (though the answer seems more complicated for $a = c$, but this is not the object of this question).
Let $(P,Q) \in \mathbb{R}[X]^2$ be a solution of our polynomial equation, equation which we'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A better lower bound for $f(t):=((t-a)^2+b)^2+d(t-c)^2$ on $[0,\infty[$ where $a,b,c,d>0$ and $c>a$ Let
$f(t):=((t-a)^2+b)^2+d(t-c)^2$ where $a,b,c,d>0$ and $c>a$, and $t\geq 0$. I am trying to find a better lower bound for $f$ than $b^2$, if it exists.
The parabola $t \mapsto (t-a)^2+b$ has its unique minimum $b$ at ... | For a real-valued function, the maximum or minimum is taken where the first derivative is zero. To check whether there is a maximum or a minimum there is to test the sign of the second derivative.
For the set of four parameters as in the given function, the condition that there is an extremum is given by the following ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show $a^3+b^3+c^3+d^3 \le 27$ Given $a, b, c, d$ are real numbers and $a^2+b^2+c^2+d^2 = 9$ show that
$$a^3+b^3+c^3+d^3\le27$$
I have tried to do the following:
$$(a^2+b^2+c^2+d^2)^{\frac{3}{2}} = 27 \\ \text{after simplifying} \\ (a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=27$$
By setting the fol... | Let $A=a^2, B=b^2, C=c^2, D=d^2$. You want to find the maximum of $\sum_{cyc}A^{3/2}$ on the set $\{A+B+C+D=9, A,B,C,D\geq 0\}$, i.e. the maximum of a convex function on a convex, closed and bounded set, namely a tetrahedron. By the properties of convex functions such maximum is attained at the vertices of the tetrahed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim\limits_{(x,y)\to(0,0)}\frac{x^6+y^6}{x^3+y^3}$ using the $\epsilon-\delta$ definition? My textbook asks the question
$$f(x,y) = \frac{x^6+y^6}{x^3+y^3}$$
Does $f(x,y)$ have a limit as $(x,y) \rightarrow (0,0)$?
I used polar coordinates instead of solving explicitly in $\mathbb R^2 $, and it went as the fol... | Top tip: don't trust wolfram! It gets the answer wrong sometimes!
Suppose we set $y = x$ and take $x \to 0$. Then
\begin{align*}
\lim_{(x,y(x)) \to (0,0)}\frac{x^6+y^6}{x^3+y^3} &= \lim_{x \to 0}\frac{2x^6}{2x^3}\\
&= 0
\end{align*}
Now let's suppose we take $y = -x$ and take $x \to 0$. Then
\begin{align*}
\lim_{(x,y(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{... | As the coefficients are real thus the complex roots would appear in conjugate pair. Thus the second root would be $1-i$ and the third root (that is $-1$) can be calculated by using the fact that the sum of the roots of a cubic equation $ax^3+bx^2+cx+d$ is $-b/a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Is difference equation $t_n = (a+b) t_{n-1} - ab \cdot t_{n-2}$ possible ($t_1 = a + b$ and $t_2 = a^2 + ab + b^2$)? I had problem solving difference equation $t_n = (a+b) t_{n-1} - ab \cdot t_{n-2}$ where $t_1 = a + b$ and $t_2 = a^2 + ab + b^2$.
Here is what I have done:
First, I've written down the equation as $$t^2... | Solving by characteristics method:
$t^2-(a+b) t+ab=0$
Solution $t=a, b$
Case 1 : $a\neq b$
Then $t_n=c_1a^n+c_2b^n$ for $c_1, c_2$ constants.
$t_1=a+b$ implies $c_1 a+c_2b=a+b \space \space ......(1) $
$t_2=a^2+ab+b^2$ implies $c_1 a^2+c_2b^2=a^2+ab+b^2..... (2) $
$(2) -a×(1) $ implies $c_2=\frac{b}{b-a}$ and using $c_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Systems of cubic equations I am looking for a general yet tailor-made methods to solve system of equations like that
$a^3 + 6ab^2 = 7$ and $3a^2b +2b^3 = 5$
that involve terms of $(a +b)^3$.
Does this have a name?
| Let $\,u = a + b\sqrt{2}\,$ then $\,u^3 = a^3 + 6ab^2 + (3a^2b+2b^3) \sqrt{2}=7 + 5\sqrt{2}\,$, so the problem is equivalent to denesting $\,\sqrt[3]{5\sqrt{2} \pm 7} = b\sqrt{2} \pm a\,$.
The following steps are similar to my more general answer here.
*
*Let $\,v=\sqrt[3]{5 \sqrt{2} - 7}\,$, then $\,u^3-v^3=14\,$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4466070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Determine the image of the unit circle $S^1$ by the action of the matrix $e^A$. We have:
$$e^{ \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} }$$
I need to determine the image of the unit circle $S^1$ by the action of the matrix $e^A$.
I think that I know how to calculate $e^A$:
I get the Jordan decomposition:
$$A = \be... | The matrix exponentiation is much simpler than it looks. When you find that $1$ is the only eigenvalue, render
$\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}=\begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}+\begin{pmatrix}-6 & 9\\-4 & 6\end{pmatrix}.$
The first matrix on the right just gives a factor of $e$ to the overall expon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to find $40! \mod 5^{10}$? I'm trying to solve this congruence. I found that in $40!$ there are $9$ powers of $5$, but I'll need one more for $40!=0 \mod 5^{10}$. I thought about Euler's theorem or Wilson's, but couldn't find any correlation.
| Because $5^9$ divides $40!$, we can use $ka \bmod kb =k(a \bmod b)\,$ [mod distributive law] to get $40! \bmod 5^{10} = 5^9\cdot(\frac{40!}{5^9} \bmod 5)$. Hence we really only need to find the residue of $\frac {40!}{5^9} \bmod 5$. So separate $n \in 1 \cdots 40$ into two sets, those divisible by $5$ and those coprim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnece... | A more efficient answer than my previous approach:
Starting in the same way:
Since $\sin\left(\frac{\pi}{2}-x\right)=\cos(x)$, $f(x)$ is symmetric around $x=\frac{\pi}{4}$. This implies that $x=\frac{\pi}{4}$ is a critical point.
Writing $f(x)=(1+\sin(x)^{-6})(1+\cos(x)^{-6})$, we find that
$$
f'(x)=-6\sin(x)^{-7}\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
How to solve $\sin(x) = \pm a$ for $a \not = 0$? I was solving the below equation: $\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1$ for $x \in [0, 2\pi]$.
My attempt:
$$\begin{align}&\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1\\\implies& \left|\sqrt{2... | Use abbreviation $s = \sin x$ and $1-s^2 = \cos^2x$:
$$\begin{align}
1
&= \left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| \\
&= \Big |\sqrt{2s^2+18(1-s^2)} - \sqrt{2(1-s^2) + 18 s^2} \Big| \\
&= \Big |\sqrt{18-16s^2} - \sqrt{2+16s^2} \Big| \\
&= \Big (\sqrt{18-16s^2} - \sqrt{2+16s^2} \Big)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$ I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$.
I tried using the following:
\begin{align*}
\ln(1+x)&\approx x,\\
\sin(x)&\approx x-\frac{x^3}{2},\\
\cos(x)&\approx 1-\frac{x... | Write: $Q(x) =\dfrac{\cos x -1 +\dfrac{x}{2}\cdot\sin x}{\ln^4(1+x)} = \dfrac{-2\sin^2\left(\frac{x}{2}\right)+2\cdot\left(\dfrac{x}{2}\right)^2\cdot\dfrac{\sin x}{x}}{\ln^4(1+x)} = \dfrac{\dfrac{x^2}{2}}{\ln^2(1+x)}\cdot\dfrac{\left(\dfrac{\sin x}{x}-\dfrac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
help me to evaluate these integrals Find the value of $$\int\frac{x}{x^2-x+1} dx$$ and $$\int\frac{1}{x^2-x+1} dx$$ This was not the original question. Original question was tougher and I have simplified that to these two integrals. I am having a hard time evaluating these two integrals but what I know is that, in thes... | The real problem is the second integral since the first one has the form
\begin{align}
\int{{x \over x^2-x+1}dx} & = \int{{x-{1 \over 2} \over x^2-x+1}dx}+{1 \over 2}\int{{1 \over x^2-x+1}dx} \\
& = {1 \over 2}\int{{1 \over x^2-x+1}d(x^2-x+1)}+{1 \over 2}\int{{1 \over x^2-x+1}dx} \\
& = {1 \over 2}\bigg(\log{(x^2-x+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the first three Laurent expansion terms of $\frac{1-z}{z^2} e^z$ I've this function $f(z)=\frac{1-z}{z^2} e^z$ and I've to find the first three Laurent expansion terms in $z=0$.
I've proceeded in this way:
First of all I've considered the expansion series of $e^z = \sum_{n=0}\frac{z^n}{n!}$ and I found the first t... | $$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$$
So:
$$\frac{e^z}{z^2}=\sum_{n=0}^{\infty}\frac{z^{n-2}}{n!}=\sum_{n=-2}^{\infty}\frac{z^n}{(n+2)!}=\frac{1}{z^2}+\sum_{n=-1}^{\infty}\frac{z^n}{(n+2)!}$$
and
$$\frac{e^z}{z}=\sum_{n=0}^{\infty}\frac{z^{n-1}}{n!}=\sum_{n=-1}^{\infty}\frac{z^n}{(n+1)!}$$
Thus, since both converge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
1 & 0 & 0 &\cdots & 0 \\
0 & 1 & 0 &\cdots & 0 \\
0 & 0 & 1 &\cdots & 0 \\
\vdot... | Here's a geometric proof: whether it's simpler/more intuitive than the row manipulations proof is a matter of taste.
Let $e_i$ be the elementary vector with a $1$ in the $i$th entry and $0$ everywhere else, and let $u$ denote the all-ones vector. We seek to find the determinant of the matrix $M$ whose $i$th row is $v_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
trying to solve the nested sum $\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2} $ This sum grabbed my curiousity
$$
\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2}
$$
I want to solve it with Calc II methods (that's the math I have).
By getting a closed form for the expression in the denominator, it becomes this:
$$
6 ... | We have
\begin{align}
\sum_{n=1}^\infty\left(\frac{1}{n}+\frac{1}{n+1}-\frac{4}{2n+1}\right) & =
\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{2}{2n+1}+\frac{1}{n+1}-\frac{2}{2n+1}\right)\\
&= \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{2}{2n+1}\right)+\sum_{n=1}^\infty\left(\frac{1}{n+1}-\frac{2}{2n+1}\right)\\
&= 2\sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ For $n \in \mathbb{N}$, evaluate
$$\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
I could not use wolframalpha, I do not know the reason.
For $... | Let
$$a = \int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
Modified solution (now the main solution)
We have two choices: integral first then sum or vice versa.
Taking the integral first gives the following for the (finite) sum which we split immediately into even and odd t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.
Now what I thought is to manipulate given result somehow to get something in the form of $a + b$:
\begin{align... | Rearrange $a^3-a^2=b^3-b^2$ to get
$$
b^3-a^3=b^2-a^2.
$$
Since $a\neq b$ we can factor out $(b-a)$ from both sides to get
$$
a^2+ab+b^2=a+b
$$
which is equivalent to
$$
(a+b)^2-ab=a+b
$$
Let $Q\equiv a+b$ then by the AM-GM inequality $\sqrt ab\leq(a+b)/2=Q/2$, and since $a\neq b$ the inequality is strict ($\sqrt ab < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Question on calculating a Fibonacci Number using Matrix Exponentiation We know that $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^k = \begin{pmatrix}F_{k + 1} & F_k \\ F_k & F_{k - 1}\end{pmatrix},$ of which there is a simple proof by induction. However, since matrix multiplication is associative, we should be able to ... | I can't honestly call this proof "more elementary" than the proof you've given in the question because I find that proof to be quite elementary, but perhaps you'll find it more palatable.
Decomposing sequences defined by linear recurrences:
Let $A_k, B_k$ denote two sequences that satisfy the same recursive relation as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left... | A complete solution, building on your last identity.
Multiply your last equation by $x-\sqrt{x^2-1}\neq0$ to obtain:
$$\begin{align}x^3-3x+1&=\sqrt{x^2-1}-x\\x^3-2x+1&=\sqrt{x^2-1}\\x(x^2-2)+1&=\sqrt{(x^2-2)+1}\end{align}$$
$1$ is a solution by inspection of the second equation, and $\pm\sqrt{2}$ are solutions by inspe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
How to solve this function $f(x)$ for degree $3$ or $4$
If $f(x)=0$ is a polynomial whose coefficients are either $1$ or $-1$ and whose roots are all real, then the degree of $f(x)$ can be equal to$:$
$A$. $1$
$B$. $2$
$C$. $3$
$D$. $4$
My work$:$
For linear only four polynomials are possible which are $x+1$ ... | Let's analyze the cubic case.
Note that the general cubic equation
$$x^3+bx^2+cx+d$$ may be reduced to the so-called depressed cubic with the substitution $x = y-b/3$. After the algebra, we get the equivalent equation
$$y^3+py+q = y^3+(c-\frac{b^2}{3})y+(\frac{2}{27}b^3-\frac{bc}{3}+d) = 0.$$
The discriminant $\Delta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Parameterizing the intersection between $4x^2+y^2=z^2$ and $xy-1=z$ I would like to parameterize the intersection between the surfaces
\begin{align}
4x^2+y^2&=z^2\\
xy&=z+1
\end{align}
I started by noting that $z=xy-1$ and then substituting that into the first equation to get \begin{align*}
4x^2+y^2&=(xy-1)^2\\
&=x^2y^... | You have that
$$
4x^2 + y^2 = \left( {xy - 1} \right)^2
$$
thus
$$
\left( {2x + y} \right)^2 = \left( {xy + 1} \right)^2
$$
This means
$$
2x + y = \pm \left( {xy + 1} \right)
$$
So you have two cases: the first
$$
\left\{ \begin{array}{l}
2x + y = xy + 1 \\
z = xy - 1 \\
\end{array} \right.
$$
and the second... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+... | Maybe it is not the fastest way to solve it, but it is faster than computing the original integral:
\begin{align*}
I = \int_{0}^{1}x^{4}(1 - x)^{5}\mathrm{d}x = \int_{0}^{1}(1 - x)^{4}x^{5}\mathrm{d}x & \Rightarrow 2I = \int_{0}^{1}x^{4}(1-x)^{4}(x + (1 - x))\mathrm{d}x\\\\
& \Rightarrow 2I = \int_{0}^{1}x^{4}(1-x)^{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{... | $$\lim_{x\to0}\frac {12-6x^2-12\cos x}{x^4}=\lim_{x\to0}\frac{-12x+12\sin x}{4x^3}=\lim_{x\to0}\frac{-12+12\cos x}{12x^2}=\lim_{x\to0}\frac{-12\sin x}{24x}=\lim_{x\to0}-1/2\cos x=-1/2$$, by repeated application of L'hopital.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
If $\{v_1+v_2, v_2+v_3, v_1+v_3\}$ are linearly independent then $\{v_1, v_2, v_3\}$ are linearly independent Problem. Prove that
for $v_1, v_2, v_3 \in \mathbb{R}^3$, if $\{v_1+v_2, v_2+v_3, v_1+v_3\}$ are linearly independent then $\{v_1, v_2, v_3\}$ are linearly independent.
What I tried:
Let $m,n,p \in \mathbb{... | You are correct. But we can solve this in a general setting for any vector space. Let's rename the vectors in this way: $\alpha:=v_1+v_2, \beta:=v_2+v_3, \gamma:=v_3+v_1$, then we have $v_1=\frac{1}{2}(\alpha-\beta+\gamma), v_2=\frac{1}{2}(\beta-\gamma +\alpha), v_3=\frac{1}{2}(\gamma-\alpha+\beta)$. In other words, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Calculating the center of mass of a lemniscate rotated around the x-axis? This is a problem I have been stuck on a while, it goes as follows:
A lemniscate has the equation $(x^2+y^2)^2 = 4(x^2-y^2)$. Let the part of the curve that lies in the first quadrant rotate around the $x$-axis to create an object. This object ha... | Because you are rotating something around the $x$-axis, we will have $y=z=0$.
Now, for this case, I prefer to use the traditional coordinates. First we have ro solve for $y$ the equation:
$$(x^2+y^2)^2-4(x^2-y^2)=0\leftrightarrow \left[t=x^2\right] \leftrightarrow y=\pm\sqrt{\frac{-2x^2-4\pm\sqrt{(-2x^2-4)^2-4(x^4-4x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the u... | $$
\begin{align*}
\int_0^1 \frac{\arcsin(x)\arccos(x)}{x}\,dx& = \int_0^1 \frac{\arcsin(x)\left(\frac{\pi}{2}-\arcsin(x)\right)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^1 \frac{\arcsin(x)}{x}\,dx-\int_0^1 \frac{\arcsin^2(x)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^{\pi/2} \frac{x}{\sin(x)}\cos(x)\,dx-\int_0^{\pi/2} \frac{x^2}{\sin(x)}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Limit of $n^2$ and a recurrence relation with ceiling function For all positive integer $n$ we define a finite sequence in the following way:
$n_0 = n$, then $n_1\geq n_0$ and has the property that $n_1$ is a multiple of $n_0-1$ such that the difference $n_1 - n_0$ is minimal among all multiple of $n_0 -1 $ that are bi... | Not sure but I think I did one direction in another way:
Claim: For $ n/2 < k < n$ and $n$ big enough we have that
$$ \left \lceil \frac{n_{k-1}}{n-k} \right \rceil \leq \frac{(k+1)^2}{\pi} \frac{ \Gamma(k+1/2)\Gamma(k+3/2)}{\Gamma(k+1)^2} $$
hence
$$ n_k = (n-k) \left \lceil \frac{n_{k-1}}{n-k} \right \rceil \leq (n-k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
$\varepsilon-\delta$ proof of this multivariable limit We've to prove that
$$
\lim_{(x,y)\to(0,0)} \frac{x^3+y^4}{x^2+y^2} =0
$$
Kindly check if my proof below is correct.
Proof
We need to show there exists $\delta>0$ for an $\varepsilon>0$ such that
$$
\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon \implies \sqr... | There is a simple strategy for dealing with such limits, which is just switching to polar coordinates. By setting $x=\rho\cos\theta, y=\rho\sin\theta$ we have $x^2+y^2=\rho^2$ and
$$ \left|\frac{x^3+y^4}{x^2+y^2}\right| = \rho\left|\cos^3\theta+\rho \sin^4\theta\right|. $$
The RHS is trivially bounded by $\rho(1+\rho)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right)$ Recently, I stumbled upon a summation
$$S=\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right)$$
which can luckily be summed to a good number.
Use $\psi^1(z)=\sum_{n=0}^{\infty}\frac{1}{(n+z)^2}$ to write
$$ S... | Combining consecutive terms, we have $$S = \sum_{m=0}^\infty \left(-\frac{1}{(6m+1)^2} + \frac{1}{(6m+2)^2} + \frac{1}{(6m+4)^2} - \frac{1}{(6m+5)^2}\right). \tag{1}$$
Now recall
$$\zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}, \tag{2}$$ hence
$$\sum_{m=1}^\infty \frac{1}{(3m)^2} = \frac{\zeta(2)}{9} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{\sin^3 \beta}{\sin \alpha} + \frac{\cos^3 \beta}{\cos \alpha} = 1$ with certain given $\alpha, \beta$ Let $$\frac{\sin (\alpha)}{\sin (\beta)} + \frac{\cos (\alpha)}{\cos (\beta)} = -1 \tag{$1$}$$ where $\alpha, \beta$ are not multiples of $\pi / 2$. Show that
$$\frac{\sin^3 (\beta)}{\sin (\alpha)} + \... | Eq.$(1)$ is equivalent to
$$\sin(a+b)=-\frac{1}2\sin(2b)\tag{3}$$
Now start from the LHS of Eq.$(2)$, and we will show it equals $1$
$$\begin{align}
\text{LHS}=\frac{\sin^3(b)\cos(a)+\cos^3(b)\sin(a)}{\sin(a)\cos(a)}\end{align}$$
Deal with the numerator:
$$\begin{align}
\text{Numerator}&=(1-\cos^2(b))\sin(b)\cos(a)+(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdo... | $\sum_{k=1}^{N} \frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)\cdot\cdot\cdot(N+i)}$ is easy to prove by induction.
It is true for $N=1$ since
$\sum_{k=1}^{1} \frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{(1+i)!}$ equals $\frac{1}{i\cdot i!}-\frac{1}{i\cdot 2\cdot 3\cdot\cdot\cdot(1+i)}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Rectangle of maximum area under the curve $y=x(x-1)^2$ A rectangle's bottom is $y=0$. While the top corners are on the curve $y=x(x-1)^2$ between $x=0$ and $x=1$. Find the maximum area of this rectangle.
My Progress
Defining $f(x) =x(x-1) ^2$
I first starting by assuming that there exist $a$ and $b$ such that $0<a, b<1... | There is really only one free variable, which is without loss of generality, the $x$ coordinate of the vertex of the rectangle that is closest to the origin. If we call this value $a$, then the other vertex coordinates are uniquely defined: in total, we have
$$(a,0), (a, a(1-a)^2), (b,a(1-a)^2), (b, 0)$$ where $$b =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many method are there to handle the integral $\int \frac{\sin x}{1-\sin x \cos x} d x$ $$
\begin{aligned}
\int \frac{\sin x}{1-\sin x \cos x} d x =& \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{1-\sin x \cos x} d x \\
=& \int \frac{d(\sin x-\cos x)}{2-2 \sin x \cos x}-\int \frac{d(\sin x+\cos x)}{2-2 \si... | Multiplying both numerator and denominator of the integrand by $1+\sin x \cos x$ transforms the integral into
$$
\begin{aligned}
I &=\int \frac{\sin x(1+\sin x \cos x)}{1-\sin ^{2} x \cos ^{2} x} d x \\
&=\int \frac{\sin x d x}{1-\sin ^{2} x \cos ^{2} x}+\int \frac{\sin ^{2} x \cos x}{1-\sin ^{2} x \cos ^{2} x} d x \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Given that $a>b>0$, can we find $\int_{0}^{2 \pi} \frac{\cos n x}{a+b \cos x} d x$ and $\int_{0}^{2 \pi} \frac{\sin n x}{a+b \cos x} d x?$ Let’s first consider
$$
\int_{0}^{2 \pi} \frac{\cos n x}{a+b \cos x} d x+i \int_{0}^{2 \pi} \frac{\sin n x}{a+b \cos x} d x = \int_{0}^{2 \pi} \frac{e^{n x i}}{a+b \cos x} dx=I \tag... | Utilize the Fourier series
$$\eqalign{
\frac{1-r^2}{1+r^2+2r\cos x}=1+2\sum_{k>0}(-r)^k\cos k x
}
$$
and let $r=\frac {a-\sqrt{a^2-b^2}}b$ to integrate
\begin{align}
&\int_{0}^{2 \pi} \frac{e^{i n x}}{a+b \cos x} dx\\
=&\ \frac1{\sqrt{a^2-b^2}}\int_{0}^{2 \pi} \frac{(1-r^2) e^{i n x}}{1+r^2+2r\cos x} dx\\
=&\ \frac1{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving recurrence relation $a_n = a_{n-1} - a_{n-2}$ I am given a sequence of determinants of matrices $M_n$, where the matrix elements $(M_n)_{ij}$ of $M_n$ are $0$ whenever $|i-j|>1$ and $1$ whenever $|i-j| ≤ 1$. Writing out the first five matrices, it becomes apparent that $\det(M_n) = \det(M_{n-1}) - \det(M_{n-2})... | You are asking for the recurrence, I think, $a_0=a_1=1$, $a_n=a_{n-1}-a_{n-2}$. Let $T:\Bbb R^3\to\Bbb R^3$ be the linear operator with the following matrix (on the standard basis): $$\begin{pmatrix}1&-1&0\\1&0&0\\0&1&0\end{pmatrix}$$And let, $n\ge 0$, the vectors $v_n$ be given by: $$v_n=\begin{pmatrix}a_{n+2}\\a_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove the following algebraic asymmetric inequality: $\sqrt{3abc(a + b + c)} + 2(a - c)^2 \geq a^2 + b^2 + c^2$
Consider $a \geq b \geq c \geq 0$ real numbers. Prove that:
$$\sqrt{3abc(a + b + c)} + 2(a - c)^2 \geq a^2 + b^2 + c^2$$
Source: RMM 2022
Comments & discussion:
A weaker form of the inequality may be obta... | Using $ab + bc + ca \ge \sqrt{3abc(a + b + c)}$, we have
$$\sqrt{3abc(a+b+c)}
\ge \frac{3abc(a+b+c)}{ab + bc + ca}.$$
Using $a \ge b \ge c$, we have
$$2(a-c)^2 = (a - c)^2 + (a - b + b - c)^2 \ge (a - c)^2 + (a - b)^2 + (b - c)^2.$$
It suffices to prove that
$$ \frac{3abc(a+b+c)}{ab + bc + ca} + (a - c)^2 + (a - b)^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
S... | Write,
$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a\ge b$$
and we obtain
$$A=\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}=2\sqrt a$$
Then, we want to find ratinal $a,b$ such that:
$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a>b$$ holds.
We have
$$3\pm \sqrt 5=a+b+2\sqrt {ab}\\
\implies \begin{cases}a+b=3\\ ab=\frac 54\end{cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
} |
Ideals having the same norm as a prime ideal in ring of integers of a number field Let $K$ be a number field and $\mathcal{O}_K$ its ring of integers. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$, and $I$ be an ideal of $\mathcal{O}_K$ such that $N(I) = N(\mathfrak{p})$, where $N(\cdot)$ is the ideal norm (i.... | The following is perhaps too long for a comment, so I'll put it in the answer. By looking at LMFDB I found that the cubic field $\mathbb{Q}(\sqrt[3]{11})$ has class number $2$. The Minkowski bound is $M = \dfrac{4}{\pi}\cdot \dfrac{6}{27} \sqrt{3267} = \dfrac{88}{\sqrt{3}\pi}<17$, so we need to consider the factorizati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving floor function system of equations given $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144$
$x$ and $y$ are real numbers satisfying $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144.$ Find $x$ and $y.$
I first a... | By adding the given equations,
$$\begin{align*}
(x+y)(\lfloor x\rfloor + \lfloor y\rfloor) &= 210\\
(\lfloor x\rfloor+\lfloor y\rfloor + \{x\}+\{y\})(\lfloor x\rfloor +\lfloor y\rfloor) &= 210\\
\left(\lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2\right)^2 - \left(\frac{\{x\}+\{y\}}2\right)^2 &= 210\\
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find... | You can also use standard solution which is the equation of a plane passing through a given point and a line. Suppose a plane passes through point M(x_0, y_0, z_0) and a line with following equation:
$\frac {x-x_1}{l}=\frac {y-y_1}{m}=\frac {z-z_1}{n}$
Then the equation of plane is:
$\begin {vmatrix}x-x_0&y-y_0& z-z_0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find this indefinite integral? $\int\frac{1+x^4}{(1-x^4)\cdot \sqrt{1+x^4}}dx$ I am thinking of a trig sub of $x^2 = \tan{t}$ but its not leading to a nice trigonmetric form, which i can integrate. Our teacher said that it can be computed using elementary methods, but I'm unable to think of the manipualtion.
| \begin{aligned}\int{\frac{1+x^{4}}{\left(1-x^{4}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}&=\int{\frac{1+x^{4}}{\left(1-x^{2}\right)\left(1+x^{2}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}\\ &=\int{\frac{x^{2}+\frac{1}{x^{2}}}{\left(1-x^{2}\right)\left(x+\frac{1}{x}\right)\sqrt{x^{2}+\frac{1}{x^{2}}}}\,\mathrm{d}x}\\ &=\int{\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Sum of squares of ratios of diagonal of a regular heptagon A problem from the 1998 Lower Michigan Math Competition...
A regular heptagon has diagonals of two different lengths. Let $a$ be the length of a side, $b$ the length of a shorter diagonal, and $c$ the length of a longer diagonal. Prove that
$$
\frac{a^2}{b^2}... | Use $\displaystyle e^{\tfrac{2\pi i}{7}}=\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7} $ to get $$|1+\omega|=\left|1+\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}\right|$$$$=\left|2\cos^2 \frac{\pi }{7}+2i\sin \frac{\pi }{7}\cos \frac{\pi }{7}\right|=2\cos \frac{\pi }{7}\left|\cos \frac{\pi }{7}+i \sin\frac{\pi }{7} \right|$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$ Question:
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.
So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to... | One of $a$ or $b$ is non zero, or else the question is not well defined.
If $a=b$, then we get left hand side of inequality to be $2/3$, which is $\geq 1/2$ as desired.
So assume $a>b$. Then
$$ \frac{a}{a+2b} + \frac{b}{b + 2a}
\\ = \frac{a(b+2a)+b(a+2b)}{(a+2b)(b+2a)}
\\ = \frac{2ab+2a^2+2b^2}{3ab + (2ab+2a^2 + 2b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving system of logarithmic equations. I have two equations
\begin{align}
\frac{1}{2} \ln(w+b-0.4)+\frac{1}{2}\ln(w-0.4) & =\ln(w)\\
\frac{1}{2} \ln(w+b-0.4)+\frac{1}{2}\ln(w-0.4) & =1.2
\end{align}
I want to find $b$ and $w$.
—-
I find $w=e^{1.2}$
But when I derive $b$, I obtain that $b= \frac{e^{2.4}}{e^{1.2}-0.4}-... | From the first equation, we have
\begin{align}
\frac12\log(w+b-0.4)+\frac12\log(w-0.4)&=\log w\\
\implies\log\left(\sqrt{w+b-0.4}\sqrt{w-0.4}\right)&=\log w\\
\implies\sqrt{(w+b-0.4)(w-0.4)}&=w\\
\implies(w+b-0.4)(w-0.4)&=w^2\\
\implies w&=0.4\left(\frac{5b-2}{5b-4}\right)\\
\implies w&=0.4\left(1+\frac2{5b-4}\right)\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Methods of deriving $\sum^n_{x=0} xr^{x-1}$ To calculate this sum, we can differentiate $\sum^n_{x=0} r^{x}$:
$$ \sum^n_{x=0} xr^{x-1} = \frac{d}{dr} \sum^n_{x=0} r^{x} = \frac{d}{dr} \left(\frac{r^{n+1}-1}{r-1}\right) = \frac{(n+1)r^{n}}{r-1} - \frac{r^{n+1}-1}{(r-1)^2}
$$
However, this disagrees with the method of... | The two expressions are the same:
$$\begin{align} \frac{(n+1)r^{n}}{r-1} - \frac{r^{n+1}-1}{(r-1)^2} &= \frac{nr^{n}}{r-1} + \frac{r^n}{r-1}- \frac{r^{n+1}-1}{(r-1)^2}\\ &= \frac{nr^{n}}{r-1} + \frac{r^n(r-1)}{(r-1)^2}- \frac{r^{n+1}-1}{(r-1)^2}\\ &= \frac{nr^n}{r-1} - \frac{r^n-1}{(r-1)^2}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k... | The original question looks wrong or at least badly worded. From continuity and differentiability at $k$, one indeed gets $g(k)=g'(k)=g''(k)$.
First assume that $a\neq 0$ and therefore $b+c\neq 0$.
Solving the differentiability condition gives $k=1-\frac{b}{2a}=\frac{b+2c}{2b+2c}$, using $b+c=a$.
The continuity conditi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials... | Alternative approach:
Stealing the insight from the comment of Ethan Bolker.
Mathematicians discovered that
$$(1 + x + x^2 + \cdots + x^n) \times (1 - x) = 1 - x^{n+1}. \tag1 $$
For $0 \neq a,b$,
$a^3 + b^3$ can be rewritten as
$$a^3 \times \left[1 - \left(\frac{-b}{a}\right)^3\right]. \tag2 $$
Setting $~\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).... | $$\dfrac1{2\sqrt2}=\sin\left(\alpha-\dfrac\pi4\right)=\sin x$$ where $\alpha-\dfrac\pi4=x$
$$\sin^5\alpha-\cos^5\alpha=\sin^5\left(x+\dfrac\pi4\right)-\cos^5\left(x+\dfrac\pi4\right)=\dfrac{(\cos x+\sin x)^5-(\cos x-\sin x)^5}{(\sqrt2)^5}$$
Now,
$$(\cos x+\sin x)^5-(\cos x-\sin x)^5=2\sin^5x+2\binom53\sin^3x\cos^2x+2\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | Yes, (though strangely written) your method is correct. It can be reformulated noticing more explicitely that $f(x):=4x^2-2ax+a^2-5a+4$ is decreasing for $x\le a/4$ and increasing for $x\ge a/4$, since $f'(x)=8x-2a.$ We thus recover your 3 cases (and the same numerical results as Andreas, but more directly):
*
*Case ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 1
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In Tom Apostol Calculus: method of exhaustion for the area of a parabolic segment of $x^2$, proving $b^3/3$ is the only number between $s_n$ and $S_n$ In the method of exhaustion for the area of a parabolic segment of $x^2$, the part where the author proves that $b^3/3$ is the only number which satisfies $s_n <A<S_n$
T... | You have showed that, if we assume that $A > \frac{b^3}{3}$, then equation (1) is true only when $n < \frac{b^2}{A - \frac{b^3}{3}}$. However, the equation (1) must be true for all integers $n \geq 1$.
We have reached a contradiction since our assumption implies an upper bound for $n$ in equation (1), but the equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is this proof of $\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$ correct? I've been practicing proving things about floor and ceiling functions, so I thought I'd try to prove this well-known identity:
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil... | Your proof looks okay (although the second equality in the $n/m$ not being integer might need some clarification). Notice that this identity can be proven essentially the same way as Prove that $\left\lceil \frac{n}{m} \right\rceil =\left \lfloor \frac{n+m-1}{m} \right\rfloor$ . Here is one variant.
By division with re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Deriving $\gamma \approx H(n)-\ln(n+1)+\frac{1}{2(n+1)}+\frac{1}{12(n+1)^2}$ The Euler-Mascheroni constant can be represented geometrically by the infinite sum of the areas in blue in the following picture, which is the area between the curve $y=1/x$ and the harmonic numbers.
Thus, the total area can be approximated b... | Consider one of the intervals from $n$ to $n+1$
After your $\frac{1}{2(n+1)}$ approximation, there is still some area between the $y=1/x$ curve and the hypotenuse of each triangle.
It's possible to approximate the function $y=1/x$ locally by a parabola with the Taylor series at $x=n+1/2$. Let this approximation be $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum of Consecutive Powers I am trying to understand a cross between steps when finding the formula for the sum of consecutive powers.
I am following the steps from a webpage here http://mikestoolbox.com/powersum.html but will also provide them below:
Step 1: Sum = x^5 + x^4 + x^3 + x^2 + x + 1
Step 2: Sum · x + 1 = x... | You do not need to add 1 after multiplying with $x$.
$$S=1+x+x^2+\ldots+x^n$$
$$xS=x+x^2+\ldots+x^n+x^{n+1}$$
Subtract both equations:
$$S(1-x)=1-x^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Find value of $2x$
I worked it out as such:
$\sin^2x = \frac{2+ \sqrt{3}}{4} \implies \frac{1- \cos 2x}{2} = \frac{2+ \sqrt{3}}{4}$
$2 - 2 \cos 2x = 2 + \sqrt{3}$
$- 2 \cos 2x = + \sqrt{3}$
$ \cos 2x = \frac{- \sqrt{3}}{2}$
And $ 2x... | It's because there are always two solutions for the equation $\cos x = a$ on $[0,2\pi)$ when $|a|<1$. If $x_1$ is one solution, then $x_2 = 2\pi - x_1$ is another solution. Indeed
$$\cos x_2 = \cos(2\pi - x_1) = \cos(-x_1) = \cos x_1 $$
where the second equality holds since $\cos$ is periodic with period $2\pi$ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Representing $\frac{1}{x^2}$ in powers of $(x+2)$ I was asked to represent $f(x)=\frac{1}{x^2}$ in powers of $(x+2)$ using the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I am able to represent $\frac{1}{x^2}$ as a power series, but I am struggling withdoing it in powers of $(x+2)$. This is what I attempted.... | The standard riff here is to first notice that $1/x^2$ is ($-1$ times) the derivative of $1/x$... and that differentiation preserves power series with a given center. Then some algebra gives
$$ 1/x \;=\; 1/(x+2-2) \;=\; {-1\over 2}{1\over 1-{x+2\over 2}}
\;=\; {-1\over 2} \sum_{n\ge 0} ({x+2\over 2})^n
$$
Differentiati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\epsilon-N$ proof for the limit of $a_n=n\left(\sqrt{1+\frac{1}{n}}-1\right)$ We are told the limit is 1/2, and I know the definition for $\epsilon-N$ convergence, and generally what I am looking for, but just cannot find the right algebra as scratchwork to find how to choose $N$ at the actual top of the proof. So far... | We have that $\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} <0$ then we can consider, for $0<\varepsilon <\frac12$
$$\frac12-\frac{1}{\sqrt{1+\frac{1}{n}}+1} <\varepsilon \iff \frac{1}{\sqrt{1+\frac{1}{n}}+1} >\frac{1-2\varepsilon}{2}$$
$$\iff \sqrt{1+\frac{1}{n}}+1<\frac{2}{1-2\varepsilon} \iff \sqrt{1+\frac{1}{n}} <\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Proving $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ I hope I'm allowed to ask this question here, but I have to prove that $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ using the following Fourier series:
$$
1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n... | $\sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}
=\frac{i}{\pi}\sum_{n=1}^{\infty}(\frac{1}{n+i/\pi}-\frac{1}{n-i/\pi})
=\frac{i(-\phi^{(0)}(\infty-i/\pi+1)+\phi^{(0)}(\infty+i/\pi+1)-\phi^{(0)}(1+i/\pi)+\phi^{(0)}(1-i/\pi))}{2\pi}=\frac{1}{2(e-1)}-\frac{1}{2(e+1)}=\frac{1}{e^2-1}$
where $\phi^{(0)}$ is the poly-gamma function... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Inverse of the function $f(x)=\sqrt{x-3}-\sqrt{4-x}+1$ I am trying to find the inverse of the function $$f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$
First of all its domain is $[3,4]$
As far as my knowledge is concerned, since $f'(x)>0$, it is monotone increasing in $[3,4]$, so it is injective. Also the Range is $[0,2]$. So $$f:[3,... | Complete the square from where you left off:
$$\begin{align}
4x^2-28x&=-y^4+4y^3-4y^2-48\\
4x^2-28x+49&=49-y^4+4y^3-4y^2-48\\
(2x-7)^2&=49-y^4+4y^3-4y^2-48\\
2x-7&=\pm\sqrt{1-y^4+4y^3-4y^2}\\
\end{align}$$
Since $x$ comes from $[3,4]$, then $2x-7$ comes from $[-1,1]$, sometimes positive, sometimes negative. So it is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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An estimate for $x/|x|$ I want to show that $$\left|\frac{x-y}{|x-y|}-\frac{x}{|x|}\right|\leq 4\frac{|y|}{|x|}$$ whenever $|x|>2|y|$ and $x,y\in\mathbb R^n$. This is at many places in the literature and related to singular integrals. Can anyone tell me how to prove this? I tried to prove this using mean value theorem ... | You can notice that
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| = \left|\frac{|x| - |x-y|}{|x-y|}\,\frac{x}{|x|} - \frac{y}{|x-y|}\right|
$$
so that by the triangle inequality
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{\left||x| - |x-y|\right|}{|x-y|} + \frac{|y|}{|x-y|} \leq \frac{2\left|y\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is $\sum_i c_i (x^{n_i} + x^{-n_i} ) \ge 0$ for $\sum_i c_i =0$ and $\sum_i c_i n_i^2 > 0$? Let $ f(x) = \sum_i c_i (x^{n_i} + x^{-n_i} )$ for an index set $\{i\}$ with $n_i \in \cal{{N}}_0$; $\sum_i c_i = 0$, and $c_i >0 $ for the maximum $n_i$.
Further, it is required that $\sum_i c_i n_i^2 > 0$, which ensures tha... | A counterexample:
$$
\begin{align}
f(x) &= (x^3+x^{-3}) - 50 \cdot(x^2+x^{-2}) + 200 \cdot(x + x^{-1}) - 151\cdot 2 \\
&= \frac{(x-1)^2(x^4-48 x^3+103 x^2-48 x+1)}{x^3}
\end{align}
$$
satisfies the given conditions, but $f(2) = -51/8 < 0$.
Remark: This was inspired by the representation
$$\sum_ic_i\cosh tn_i=\frac{t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Finding the value of $\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta$. It is given that $$
I=\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta=\pi\ln k-\frac{\pi^2}{w}
$$ and was asked to find the value $kw$. Here is my try on it:
Substituting $\theta = \dfrac{\pi}{4... | Note that
\begin{align}
\int_0^{\pi/2}\frac x{\tan x}dx
=&\int_0^{\pi/2}\int_0^1 \frac1{1+y^2\tan^2 x}dy \ dx
=\frac\pi2\int_0^1\frac1{1+y}dy=\frac\pi2\ln2
\end{align}
and
\begin{split}
I= &\ 2 \int_{-\frac{\pi}{2}}^0\dfrac{x(1+\tan x)}{\tan x} \, dx
= 2 \int^{\frac{\pi}{2}}_0\left(\frac x{\tan x}- x\right) dx
= \pi \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
The value of $(a^2+b^2)(c^2+1)$. The question is:
Given real numbers $a,b,c$ that satisfy
$$ab(c^2-1)+c(a^2-b^2)=12$$
$$(a+b)c+(a-b)=7$$
Find the value of $(a^2+b^2)(c^2+1)$
From what I've done, I got $7(3ac+3a+3bc-b)-2ab(c+1)(c-1)=(a^2+b^2)(c^2+1)$. I think I'm inching further from the actual solution. Can anyone gi... | Since you are asking for the value of $(a^2+b^2)(c^2+1)$, we may assume that the value is equal for all $b$, so we can just set $b=0$, so that $(a-3)(a-4)=0$ and $c=12/a^2$. For $a=3$ we obtain $c=4/3$, so that $(a^2+b^2)(c^2+1)=16+9=25$. The other case, $a=4$ gives $c=3/4$ and again $(a^2+b^2)(c^2+1)=9+16=25$.
But eve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\... | The “brute-force” way is to use the Binomial Theorem:
$$(x+a)^5 = x^5 + 5ax^4 + 10a^2x^3 + 10a^3x^2 + 5a^4x + a^5$$
Which gives us:
$$(x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1) + (x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243) = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + 260x^2 + 410x + 242 = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Describe the Jordan form of a linear operator $T:\mathbb{R}^7\to\mathbb{R}^7$ with characteristic polynomial $p_T(t)=(t-1)^2(t-2)^4(t-3)$
Describe the Jordan form of a linear operator $T:\mathbb{R}^7\to\mathbb{R}^7$ with characteristic polynomial $$p_T(t)=(t-1)^2(t-2)^4(t-3)$$ and such as $\dim(\ker(T-2I))=2, \dim(\ke... | The Jordan blocks associated with eigenvalues $1$ and $3$ are correct. To deduce the correct Jordan block associated with eigenvalue $2$, first note that $\dim(\ker(T - 2I)) = 2$ implies that there are two possibilities below:
\begin{align*}
& \text{(a).}\; J_a = \begin{bmatrix}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the line $l$ that goes through $P$ and intersects line $l_1$ and $l_2$. We have line $l_1:\begin{cases} x=1+t_1 \\ y=t_1 \\ z=-1+t_1\end{cases}$ and $l_2:\begin{cases} x=10+5t_2 \\ y=5+t_2 \\ z=2+2t_2\end{cases}$.
Find the line $l$ that goes through $P:(3, 2, −1)$ and intersects line $l_1$ and $l_2$.
I tried findi... | First, find the plane that contains the point $Q=(3, 2, -1)$ and the line $\ell_1$. To do that, find a point on $\ell_1$, for example, by substituting $t_1 = 0 $, then $P_1 = (1, 0, -1)$ is on this plane as well as $Q=(3, 2, -1)$.
The direction vector of $\ell_1$ is $v_1 = (1, 1, 1)$. Now define the vector
$v_2 = P_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Does $S_b$ become eventually greater than $S_{b+1}$? For a base $b \geq 2$, let $S_b$ be the sequence the $n$−th term of which is found by concatenating $n$ copies of the number $n$ written in base $b$, and then seeing the resulting string as a number in base $b$.
For example, given $b=2$, we get the sequence $1$, $101... | The proposition is false. Consider $S_2(n)$ and $S_3(n)$, where $n=3^k$, $ k\in\mathbb{N}.$
$ n = {1\underbrace{00\ldots 00}_{k \text{ zeros}} } (\text{ base } 3). $ This pattern ${1\underbrace{00\ldots 00}_{k \text{ zeros}} }$ repeated i.e. concatenated onto itself $\ 3^k$ times, $ \underbrace{100\ldots 00}_{1}\ \unde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to get the theta order of a non-homogeneous recurrence of Fibonacci sequence I'm recently learning recurrence and I'm stuck with how to find the theta order of a non-homogeneous Fibonacci sequence.
Given a Fibonacci sequence $F(N) = F(N-1)+F(N-2)+f(N)$. How can I determine the theta-order of this Fibonacci sequence... | Case $f(n) = n$
We have
$$F(n) = F(n-1)+F(n-2)+n$$
$$\Longleftrightarrow F(n)+ n + 3 = \left(F(n-1) + (n-1)+3\right)+\left(F(n-2) + (n-2)+3\right)$$
Denote $G(n) = F(n)+n+3$, then $G(n)$ is a Fibonacci sequence with $G(n) = G(n-1)+G(n-2)$. By consequence,
$$F(n) = G(n) - n-3 = A\varphi^n + B\varphi^{-n} - n-3$$
with th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining the cartesian equation of an ellipse ( be it neither in standard position nor orientation) given center, one vertex and one semi -axis Desmos construction, with sliders waiting for being launched : https://www.desmos.com/calculator/vuou1gnese
Question : I obtained the final formula while thinking of cent... | Rotate then Shift.
Starting from the algebraic equation of an ellipse in standard format
$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $
If you define the position vector $\mathbf{p} = [x, y]^T $ , then the above equation can be written concisely as follows
$ \mathbf{p}^T \ D \ \mathbf{p} = 1\hspace{50pt}(*)$
where
$ D = \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area between line and curve Point $A$ lies on the curve $g(x) = x$ And lies on a value $x > 0$. From point $A$ a line is drawn parallel to the $y$-axis. The second drawn graph is $f(x)=x^3$.
Calculate the coordinates of point $A$ so that the area of $B$ and $C$ will be equal.
x must be less than 1, accordin... | Let abscissa of $A$ be $A_x$.
We need to calculate coordinates of $A$ such that area of $B$ = area of $C$
$$\begin{align}\implies& \int_0^{A_x} x^3 \ dx = \int_0^{A_x} x - x^3\ dx\\\implies &\frac{(A_x)^4}{4 } = \frac{(A_x)^2}{2} - \frac{(A_x)^4}{4}\\\implies & \frac{(A_x)^4}{2} = \frac{(A_x)^2}{2}\\\implies & (A_x)^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)... | $$\begin{align}\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+1&=\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+\ln(e)=\\&=\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)\;.\end{align}$$ $$\begin{align}\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)&=\sum^\infty_{n=2}\ln\left(\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find random variable $X$ with $\text{E}(X) = 0$, $\text{Var}(X) = 4$, and $\text{P}( \vert X \vert \geq 4 ) = 0.25$ As the title states, my task is to find a random variable $X$ with $\text{E}(X) = 0$, $\text{Var}(X) = 4$, and $\text{P}( \vert X \vert \geq 4 ) = 0.25$.
My first attempt was to consider $X \sim \text{Nor... | Consider the discrete random variable $X$ with PMF given by
$$\text{P}(X = x) = \begin{cases}
\frac{3}{4} & \text{ for } x = 0 \\
0 & \text{ for } \vert x \vert = 1 \\
0 & \text{ for } \vert x \vert = 2 \\
0 & \text{ for } \vert x \vert = 3 \\
\frac{1}{8} & \text{ for } \vert x \vert = 4 \\
0 & \text{ for all other $x ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express the following permutation as a product of disjoint $2$ cycles :$(4,2,5,3)(1,5,2,4)$.
Express the following permutation as a product of disjoint $2$ cycles:
$$(4,2,5,3)(1,5,2,4).$$
I can find out the permutation as $(1,5,3)(2)(4)$. We can write $$(1,5,3)(2)(4)=(1,5)(1,3)(2)(4).$$
But how to express them as dis... | Given are two permutations $\pi,\sigma$ with
\begin{align*}
\color{blue}{\pi}&\color{blue}{=(4,2,5,3)}\\
\color{blue}{\sigma}&\color{blue}{=(1,5,2,4)}\\
\end{align*}
Composition of $\pi\sigma$ (order right to left) gives using two notations
\begin{align*}
\color{blue}{\pi\sigma}&=(4,2,5,3)(1,5,2,4)=(1,3,4)(2)(5)\color{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis ,$y$-axis
Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis, $y$-axis.
My attempt: I tried using the change of variable concept:
Let $u=x-y$ and $v=x+y$ ,$|J|= \frac{1}{2}$
Then ... | No change of variable seems necessary.I'll use Sage to explain.
Define the function :
$$f(x,y)=\left(\frac{x-y}{x+y}\right)^4$$
sage: f(x,y)=((x-y)/(x+y))^4
Find an antiderivative of this function with respect to $y$ :
sage: A1=integrate(f(x, y), y) ; A1
-8*x*log(x + y) + y - 8/3*(5*x^4 + 12*x^3*y + 9*x^2*y^2)/(x^3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k$ by Stolz–Cesàro $$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k, k \in \mathbb{N}$$
Putting $\sqrt{n^k}$ in denominator, we ge... | By Cesaro-Stolz theorem
$$\lim_{n \to \infty} \frac{\sum_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\sum_{i = 1}^{n+1} \frac{1}{\sqrt{i}} - \sum_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n+1} - \sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof verification: $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ The question states:
Let $a, b, c$ be real numbers such that $$\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}=0$$ Prove that $$\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$$
Now this can be solved with... | C-S it's the following:
$$\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2\geq\left(\sum_{i=1}^na_ib_i\right)^2$$ for any reals $a_i$ and $b_i$,
which says that your solution is wrong because $bc-a^2$ may be negative and can not be square of real number. You can not use C-S so.
By the way, $$\sum_{cyc}\frac{1}{ab-c^2}=\frac{(ab+ac+... | {
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Find limit of trigonometric function with indeterminacy Find limit of the given function:
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
I tried putting 0 instead of x
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arct... | The limit of a product is the product of the limits (provided they exist). Multiply and divide by terms so that the resulting limmand is a product of known limits:
$$\frac{\left(4^{\arcsin(x^2)}-1\right)\left(\sqrt[10]{1-\arctan(3x^2)}-1\right)}{(1- \cos \tan 6x)\log(1-\sqrt{\sin x^2})}$$
$$= \frac{e^{\log 4 \arcsin x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which one is bigger? $ \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx$ or $ \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx$ Which is bigger
$$ \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx$$ or $$ \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx~?$$
I let $x=\frac{\pi}{2}-t$ in the second integral, and I obtain this
$$\int_0^{\... | Actually, this is rather crude:
$$\int_0^{\frac{\pi}{2}}\dfrac{\sin x}{1+x^2}dx<\int_0^{\frac{\pi}{2}}\dfrac{x}{1+x^2}dx = \dfrac{1}{2}\ln\dfrac{4+\pi^2}{4}\approx 0.62 < 0.72\approx \dfrac{3}{2}\arctan\frac{\pi}{2} - \frac{\pi}{4}=\int_0^{\frac{\pi}{2}}\dfrac{1-x^2/2}{1+x^2}dx <\int_0^{\frac{\pi}{2}}\dfrac{\cos x}{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Cyclic inequality. $\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$ Let $x,y,z>0$. Show that
$$\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$$
Equality case is for $x=y=z$.
A hint I have is that I have to amplify with something convenient, and then apply some sort of mean inequali... | Apply AM-GM inquality now and Cauchy-Schwarz inequalities ( at later step ) we have: $\displaystyle \sum_{\text{cyclic}} \dfrac{x+2y}{\sqrt{6z(x+2y+3z)}}\ge \displaystyle \sum_{\text{cyclic}} \dfrac{2(x+2y)}{6z+(x+2y+3z)}= \displaystyle \sum_{\text{cyclic}} \dfrac{2(x+2y)}{(x+2y)+9z}$. Next, let $a = x+2y, b = y + 2z, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Taylor's Expansion Exercise I have a few troubles coming up with the solution of this limit.
$$ \lim_{x \to 0} \frac {8x^2(e^{6x}-1)}{2x-\sin(2x)}$$
I've tried using Taylor like this but I honestly have no idea if it's even remotely close
$$\lim_{x\to 0} \frac{8x^2(1+6x+o(x)-1)}{2x-2x+o(x)}$$
Thank you so much
| Using
$$ e^{ax} - 1 = (a x) \, \left(1 + \frac{a x}{2} + \frac{a^2 x^2}{6} + \mathcal{O}(x^3) \right) $$
and
$$ b x - \sin(b x) = \frac{b^3 x^3}{3!} \ \left(1 - \frac{b^2 x^2}{20} + \mathcal{O}(x^4) \right) $$
then
\begin{align}
\frac{x \, \left(e^{a x} - 1\right)}{b x - \sin(b x)} &= \frac{ a x^3 \, \left(1 + \frac{a ... | {
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How many ways to put $10$ distinguishable balls in $5$ identical boxes so that no box has more than $3$ balls? As the title says, I was asked to count the number of ways one can put $10$ different balls inside $5$ identical boxes so that no box has more than $3$ balls. I am really confused by the restriction and I can'... | A comparison with distinct balls in distinct boxes$\;$ vs$\;$ identical boxes yields a simple transformation (although $5$ cases can't be avoided here)
For the pattern $3-3-3-1-0,\;$ for distinct balls to distinct boxes, we have the product of two factorials,
[Fill Pattern]$\times$[permute Pattern]$\;=
\large\frac{10... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the $4 \times 4$ Vandermonde determinant I'm currently doing an exercise question from the textbook. The question is:
The goal of this problem is to find the $4 \times 4$ Vandermonde determinant.
$V_4 = \begin{bmatrix}1&a&a^2&a^3\\1&b&b^2&b^3\\1&c&c^2&c^3\\1&x&x^2&x^3\end{bmatrix}$.
(a). Explain why $V_4$ is a cu... | Answering my own question in a different way from
math.stackexchange.com/a/699339/42969
because it's easier to understand (for me and maybe for some other people).
(a). The determinant is a cubic polynomial because if we calculate the determinant by cofactor expansion across the last row, we have
$V_4= -1 \begin{vmatri... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Runge-Kutta Method [ Butcher Tableau ] Show $A$-Stability. We consider the following Butcher Tableau:
\begin{array}{c|cc}
\frac{1}{3} & \frac{5}{12}& -\frac{1}{12} \\
1 & \frac{3}{4} & \frac{1}{4}\\
\hline
& \frac{3}{4} & \frac{1}{4} \\
\end{array}
What do I have to do?:
First I wrote down, everything I need:
For th... | Now examine the absolute value for $z=iy$
$$
|g(iy)|^2=\left|\frac{iy+6}{-y^2-4iy+6}\right|^2=\frac{y^2+36}{(6-y^2)+16y^2}=\frac{(y^2+1)+35}{(y^2+1)^2+35}\le 1.
$$
By the maximum modulus principle the maximum absolute value for the holomorphic domain $Re(z)\le 0$ has to be on the boundary (including here the infinite p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:
Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$
My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^... | We quickly observe that $x>0$.
Letting $1-x^2=u^2, \,u≥0$ we have:
$$x^2(1+2u+u^2)=1+x^2$$
This implies that,
$$\begin{align}&(1-u^2)(u^2+2u)=1\\
\implies &u(u+1)(u+2)(u-1)=-1\\
\implies &(u^2+u)(u^2+u-2)=-1\end{align}$$
Finally letting $t=u^2+u$, we have:
$$(t-1)^2=0\implies t=1$$
This leads to:
$$\begin{align}&u^2+u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than... | Let $a<\sqrt{b^2-bc+c^2}.$
Thus, $$b^2<ac<c\sqrt{b^2-bc+c^2},$$ which gives
$$(b-c)(b^2(b+c)+c^3)<0$$ or
$$b<c.$$
Also, since $$a(2b-c)>bc,$$ we obtain: $$2b-c>0$$ and
$$\sqrt{b^2-bc+c^2}>a>\frac{bc}{2b-c},$$ which gives $$(b-c)(4b^3-4b^2c+4bc^2-c^3)>0$$ or
$$4b^3-4b^2c+4bc^2-c^3<0.$$
But $$4b^3-4b^2c+4bc^2-c^3=b(2b-c)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2$ If $a,b,c,d \in \mathbb{N}$
Given pairwise distinct $a,b,c,d \in \mathbb{N}$, prove that $$E=2$$ if $E=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}$ is an integer.
My effort:
We have: $$\begin{aligned}
& E=\frac{a}{a+b}+\frac{b}{b+c}+\fra... | $$E>\frac a{a+b+c+d}+\frac b{a+b+c+d}+\frac c{a+b+c+d}+\frac d{a+b+c+d}=1
$$
Similarly, $F=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}>1$.
Since $E+F=(\frac a{a+b}+\frac b{a+b})+(\frac b{b+c}+\frac c{b+c})+(\frac c{c+d}+\frac d{c+d})+(\frac d{d+a}+\frac a{d+a})=4$, we have $E=4-F<3$.
Since $E$ is an integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Algebraic calculation with polynomial and complex root.
Let $f=X^{3}-7 X+7$ be in $\mathbb{Q}[X]$.
Let $\alpha \in \mathbb{C}$ be a root of $f$ and hence $1, \alpha, \alpha^{2}$ be a basis of the $\mathbb{Q}$ vector space $\mathbb{Q}(\alpha)$. Let $\beta=3 \alpha^{2}+4 \alpha-14$. Write $\beta^{2}$ and $\beta^{3}$ as ... | Your mistake appears to be in the last line evaluating $\beta^3$. The expression
\begin{eqnarray*}
15\alpha^3+18⋅(−7\alpha)+77 &=& 15(7\alpha-7)-7*18\alpha+7*11\\
& =& 7(15-18)\alpha+7(-15+11) \\
& = & -21\alpha -28
\end{eqnarray*}
whereas you got $-21\alpha+62$. The expression for $\beta^3$ becomes $-21\alpha^2-28\a... | {
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"url": "https://math.stackexchange.com/questions/4600336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove or disprove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1$. Prove or disprove that the inequality
$$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$
is valid if $x,y,z$ are positive numbers and $$xyz=1.$$
My solution is:
Let $$x=\dfrac{a}{b}, y=\dfrac{b}{c}, z=\... | This is almost the same as the other answer and just slightly different: $\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}\ge\sqrt{\dfrac{b}{a+b+c}}+\sqrt{\dfrac{c}{a+b+c}}+\sqrt{\dfrac{a}{a+b+c}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a+b+c}}> 1.$ Done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Show that $(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$ is strictly increasing Let $1/2<p< 1$. I am asked to show that
$$f(x)=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$$
is strictly increasing for $x\geq 0$ and to compute $\lim_{x\to\infty} f(x)$.
I first computed the derivative, but I don't see why it must be positive:
$$\frac{d f(x)}{... | Computing the limit is rather straight forward
\begin{align}
f(x)&=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}=\frac{(x+1)^p}{x^p}x-(x+1)\frac{x^p}{(1+x)^p}\\
&=\frac{(1+x)^{2p}x-(x+1)x^{2p}}{x^p(1+x)^p}=\frac{x\Big(\big(1+\tfrac1x\big)^{2p} -1\Big)-1}{\big(1+\frac{1}{x}\big)^p}\\
&=\frac{\frac{\big(1+\tfrac1x\big)^{2p}-1}{\tfra... | {
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"url": "https://math.stackexchange.com/questions/4604904",
"timestamp": "2023-03-29T00:00:00",
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Solve $\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$ using hypergeometric functions How can we solve this integral?
$$\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$$
I used geometric series
$$=\int_0^1(x+1)^b\sum_{n=0}^\infty\left(\frac{x}{\alpha}\right)^{n+1}dx=\sum_{n=0}^\infty\int_0^1\left(\frac{x}{\alpha}\right)^{n+1}(x+1)^bdx$$
Usin... | $$\frac{x(x+1)^b}{\alpha-x}=\frac{(x-\alpha)(x+1)^b+\alpha(x+1)^b}{\alpha-x}=-(x+1)^b+\frac{\alpha(x+1)^b}{(\alpha+1)-(x+1)}$$
$$I=\int \frac{x(x+1)^b}{\alpha-x} \,dx=-\frac {(x+1)^{b+1}}{b+1}+\alpha J$$
$$J=\int \frac{(x+1)^b}{(\alpha+1)-(x+1)}\,dx$$
Let $$(x+1)=(\alpha+1) t\quad \implies \quad J=(1+\alpha)^b \int\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \i... | For all intents and purposes, this amounts to the integral
$$\int_0^{2\pi} \frac{1}{1+ a \cos x} \, \mathrm{d}x$$
for $a := \cos \theta$. This is a rational function of sine and cosine, and hence the Weierstrass substitution may apply, if be a bit messy. Then
$$t = \tan \frac x 2 \implies \cos x = \frac{1-t^2}{1+t^2} ,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Why does solving this integral using trigonometric substitution lead to the wrong answer? I was solving the integral $$\int \frac{\sqrt{x^2 - 16}}{x} \, dx,$$ and I admittedly attempted to solve it blindly using trigonometric substitution:
$$\begin{align}
&\int \frac{\sqrt{x^2 - 16}}{x} \, dx, \quad \text{let } x = 4\s... | $$
\begin{aligned}
\int \frac{\sqrt{x^2-16}}{x} d x=&\int \frac{x^2-16}{x \sqrt{x^2-16}} d x \\
= & \int \frac{x^2-16}{x^2} d\left(\sqrt{x^2-16}\right) \\
= & \int\left(1-\frac{16}{x^2}\right) d\left(\sqrt{x^2-16}\right) \\
= & \sqrt{x^2-16}-16 \int \frac{d\left(\sqrt{x^2-16}\right)}{\left(\sqrt{x^2-16}\right)^2+4^2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How To Prove $a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $ if $ a > b > c > 0\,$? How to prove this :
$$a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $$
if we know: $$ a > b > c > 0 $$
My attempt:
$$\frac {a^3b＋b^3a}{2}>a^2b^2
...(1)$$
$$\frac {b^3c＋c^3b}{2}>c^2b^2...(2)$$
$$\frac {a^3c＋c^3a}{2}>a^2c^2...(3)$$
(1) + (2)+ (3) :
$$\... | We need to prove that:
$$\sum_{cyc}(a^3b-a^2b^2)>0$$ or
$$\sum_{cyc}(2a^3b-2a^2b^2)>0$$ or
$$\sum_{cyc}(a^3b+a^3c-2a^2b^2)+\sum_{cyc}(a^3b-a^3c)>0$$ or
$$\sum_{cyc}ab(a-b)^2+(a-b)(a-c)(b-c)(a+b+c)>0,$$ which is obvious.
I think, BW does not help for the following inequality.
Let $\frac{1}{\sqrt{a}},$ $\frac{1}{\sqrt{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to maximize the expression $xy(1-2x-3y)$ when $\{x>0, y>0, 2x+3y<1\}$ Find the greatest value of the following expression $xy(1-2x-3y)$ when $x>0$,$y>0$,$2x+3y<1$ and also determine in each case the value of the variables for which the greatest value is attained.
I have tried to solve it using AM-GM inequality cons... | We have a one-line solution that can be obtained using the AM - GM inequality :
$$\begin{align}&\left(1-2x-3y\right)+2x+3y\\
&\geq 3\sqrt [3]{6xy(1-2x-3y)}\\
\implies &xy\left(1-2x-3y\right)\leq\frac {1}{162}\thinspace .\end{align}$$
The equality occurs iff, when
$$\begin{align}&2x=3y=1-(2x+3y)\\
\implies &2x=1-4x\\
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculating double integral using variables substitution $\displaystyle
D = \left\lbrace \left. \rule{0pt}{12pt} (x,y) \; \right| \; 3 x^2 + 6 y^2 \leq 1 \right\rbrace$
Calculate $\displaystyle
\iint_D \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } } \; dx dy{}$.
Attempt:
$x=\frac{r}{\sqrt3}cost,y=\frac{r}{\sqrt6}sin... | With the change of variable $$(x,y)\to \left(\frac{1}{\sqrt{3}}r\cos t,\frac{1}{\sqrt{6}}r\sin t\right),$$ for $t\in [0,2\pi[$ and since $3x^2+6y^2\leqslant 1$ then $r\in [0,1]$.
The determinant of Jacobian is given by
$$\frac{\partial (x,y)}{\partial (r,t)}=\det\begin{pmatrix}\frac{\partial x}{\partial r}& \frac{\part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
| Remark: @achille hui posted a similar proof. But we got them independently.
The desired inequality is written as
$$3^{2\sqrt 2} < (2\sqrt 2)^3$$
or
$$2\sqrt 2\, \ln 3 < 3\ln (2\sqrt 2)$$
or
$$\frac{\ln 3}{3} < \frac{\ln(2\sqrt 2)}{2\sqrt 2}. \tag{1}$$
Let
$$f(x) := \frac{\ln x}{x}.$$
We have
$$f'(x) = \frac{1 - \ln x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
} |
Question about limits at infinity: $f(x) < g(x)$ and $f(x) < \epsilon$ imply $g(x) < \epsilon$? I am going over the proof of $\lim\limits_{x \rightarrow \infty} \frac{x^2+2x}{2x^2+1} = \frac{1}{2}$. Let $f(x) = \frac{x^2+2x}{2x^2+1}$.
After setting $\epsilon >0$, we have $|f(x) - \frac{1}{2}| = |\frac{x^2+2x}{2x^2+1} -... | The proof you are quoting seems to redefine $f(x)$. Worse, it is missing some connecting words to flesh out the argument. Some of the '<' you are seeing are aspirational: we want to assert inequality, but the inequality isn't true without additional conditions on $x$. The rest of the proof seeks to establish those cond... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this logical deduction regarding some modular restrictions on odd perfect numbers valid? - Part II Let $p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
He... | Yes, it does.
Moreover, we can say that
$$k=1\implies p\not\equiv 13\pmod{16}$$
Proof :
Suppose that $p\equiv 13\pmod{16}$. Then, since we have
$$p \not\equiv {5} \pmod {16}\implies \sigma(m^2) \not\equiv 3 \pmod 4$$
and
$$p\not\equiv 1,9\pmod{16}\implies p\not\equiv 1\pmod 8\implies\sigma(m^2) \not\equiv 1 \pmod 4$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximize $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$ The equation $x^2 + y^2 = 1 - xy$ represents an ellipse. I am trying to show that its major axis is along $y=-x$ and find the vertex. To find the vertex I tried to find the vector in the ellipse with the greatest norm, which is equivalent to maximizing $x^2 + y^2$ su... | $\Delta \thinspace \thinspace \rm {method \thinspace\thinspace works\thinspace .}$
You have :
$$
\begin{cases}x^2+y^2=a,\thinspace a\in\mathbb R\\
x^2+y^2+xy=1\end{cases}
$$
This implies that,
$$
\begin{cases}y=\frac {1-a}{x}\\
x^2+\frac{(a-1)^2}{x^2}-a=0\end{cases}
$$
Then, letting $x^2=u$, we have :
$$
\begin{align}&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
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