Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all values of a for which the following system has a solution, no solution and infinitely many solution. Find all values of a for which the following system
$$\begin{cases}x+2y+z=a^2
\\x+y+3z=a
\\3x+4y+8z=8\end{cases}$$
has a solution, no solution and infinitely many solution.
I found the reduced row echelon form... | Yes since the RREF is the following
$$\begin{pmatrix}
1 & 2 & 1 &a^2\\
1 & 1 & 3 & a\\
3 & 4 & 8 & 8
\end{pmatrix} \to \begin{pmatrix}
1 & 2 & 1 &a^2\\
0 & 1 & -2 & a^2-a\\
0 & 0 & 1 & 8-2a-a^2
\end{pmatrix}$$
by Rouché–Capelli theorem, we always have exacty one soution for the system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computation of 10th power of a matrix I have been given the following matrix $A$. I am required to find 10th power of $A$.
$$
A = \begin{pmatrix}
3 & 1 & 1 \\
2 & 4 & 2 \\
-1 & -1 & 1 \\
\end{pmatrix}
$$
I tried using Cayley Hamilton Theorem and found the characteristic polynomial as:
$t^3-8t^2+20t-16=0$.From here I... | The eigenvectors for this matrix come out to be
$$\left( \begin{array}{c} -1\\1\\0\end{array} \right),\left( \begin{array}{c} -1\\0\\1\end{array} \right), \left( \begin{array}{c} -1\\-2\\1\end{array} \right)$$
To diagonalize the matrix, if $A = SDS^{-1}$, then
$$S = \left( \begin{array}{c c c} -1&-1&-1\\1&0&-2\\0&1&1\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let f(x) = $x^2+ax+b,a,b \in R$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is ..... Let $f(x) = x^2+ax+b$ for $a,b \in \mathbb{R}$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is
(A) real
(B) imaginary
(C) real and distinct
(D) equal roots
My attemp... | We have three real terms summing up to $0$.
They can't be all zero as a quadratic has at most two zeroes.
Hence at least one term is positive and at least one term is negative, hence the roots must be distinct real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would... | Use QM-AM to obtain \begin{align*}\sum_{cyc}\sqrt{a^2+ab+b^2}\geqslant \sum_{cyc}\frac{a+b+\sqrt{ab}}{\sqrt{3}}&=\frac{2(a+b+c)}{\sqrt{3}}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&=2\sqrt{3}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&>\sqrt{3}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Given the two functions $f(x)=x^2+3x+5$ and $g(x)=\sqrt{2-|x-4|}$ work out $g^{-1}$ (Sweden 1961) Given the two functions $f(x)=x^2+3x+5$ and $g(x)=\sqrt{2-|x-4|}$ work out $g^{-1}$ for some subdomain.
I have managed to prove that $g^{-1}$ exists in the following way:
$g(x)=\sqrt{6-x}$ with $x\in[4,6]$ we have that g i... | Why did you restrict to the domain $[4,6]$? It is not the complete domain of $g$.
the domain is all $x$ where $|x-4| \le 2$ so $0\le x-4 \le 2$ or $-2 \le x-4 \le 0$ so $4\le x \le 6$ or $2\le x \le 4$. So the domain is $[2,6]$
And we can easily see that $g$ is not one to one because for any $x \in (4,6]$ where $g(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Use proof by induction to show that for any positive integer $n\geq 2$ the following holds Use proof by induction to show that for any positive integer $n\geq 2$ the following holds:
$(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})>\frac{\sqrt{2n+1}}{3}$
Proof: Base case: $n=2$. $LHS:\frac{4}{3}$; $RHS:\frac{\sqrt{... | Assuming that $(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})>\frac{\sqrt{2n+1}}{3}$ consider $(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})(1+\frac1 {2n+1})$. This quantity is greater than $\frac{\sqrt{2n+1}}{3} (1+\frac1 {2n+1})$. We want to show that $\frac{\sqrt{2n+1}}{3} (1+\frac1 {2n+1}) >\frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Logaritmic equation with floor brackets $ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor=\log_{10}x$ Solve equation
$ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor= \log_{10}x$
I dont understand solution explained in the book.
$ \left \lfloor ... | $\{x\}$, the fractional part of x, is defined as $\{x\} = x - \lfloor x\rfloor$. (For example, $\{3.14\} = 0.14$, $\{-0.6\} = 0.4$, $\{1\} = 0$, etc. Thus, $\boxed{\frac{x}{2}-\left\lfloor\frac{x}{2}\right\rfloor = \bigg\{\frac{x}{2}\bigg\}.}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3868711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.... | Let $F$ denote the function $xy + yz + xz - xyz$. Since $x + y + z = 2$, the area of the triangle by Heron's formula is
$$A = \sqrt{(1-x)(1-y)(1-z)} = \sqrt{F - 1}.$$
Hence the minimum value of $F$ is $1$, which is attained iff we allow triangles of zero area.
The maximum $F$ occurs for maximum $A$. If we're allowed to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to find $AL$ in a triangle with one side trisected? Here is the original diagram of the question:
Let $D$ and $E$ be the trisection points of $BC$. And let $K$ and $L$ be points on $AB$ and $AC$ such that $\angle BKE= \angle DLC = \alpha$. If $KB=16$, $AK=12$ and $LC=6$. Then find the length of $AL$.
The only thin... | Extends $LD$ to meet $AB$ at $F$, extends $KE$ meets $AC$ at $G$.
Let $L'$ be on $FL$ such that $BL'\mathbin{\!/\mkern-5mu/\!} AL$. Similarly $CK'\mathbin{\!/\mkern-5mu/\!}AK$.
Then $BL'=\frac 12CL=3$, $CK'=\frac 12BK=8$.
Now since triangles $CK'G$ and $AKG$ are similar, we have $AG=AC\cdot \frac{AK}{AK-CK'}=3(x+6)$. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}... | $$\lim_{x\to1}\frac{\sqrt{1-x^2}-0}{x-1}=\lim_{x\to1}\sqrt{\frac{1+x}{1-x}}.$$
You can conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Proof: For all real numbers $x$, If $x^2−5x+4≥0$, then either $x≤1$ or $x≥4$. Is this a valid technique? For all real numbers $x$, If $x^2−5x+4≥0$, then either $x≤1$ or $x≥4$. Is this a valid technique?
Instead of showing that either (x-4) and (x-1) are both positive or both negative, can I just do what I have written ... | Yes, it is valid. Well done!
Here I propose an alternative way to solve it:
\begin{align*}
x^{2} - 5x + 4 = \left(x^{2} - 5x + \frac{25}{4}\right) - \frac{25}{4} + 4 = \left(x - \frac{5}{2}\right)^{2} - \frac{9}{4} \geq 0
\end{align*}
which is equivalent to
\begin{align*}
\left|x - \frac{5}{2}\right|^{2} \geq \frac{9}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\cos x-\sin 3x=\cos 2x$? My attempt:
$$\begin{align} (\cos x- \cos 2x) - \sin 3x &= 0 \\
2\sin \frac{3x}{2}\sin \frac{x}{2} - 2 \sin \frac{3x}{2}\cos \frac{3x}{2}&=0\\
\sin\frac{3x} {2} \left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)&=0 \end{align}$$
Now either $\sin\frac{3x} {2} = 0$ or $\left(\sin\fra... | This is an elementary equation $\cos a=\cos b$ under cover: just consider that $\sin a=\cos(\pi/2-a)$. Thus you get
$$
\cos\Bigr(\frac{\pi}{2}-\frac{x}{2}\Bigr)=\cos\frac{3x}{2}
$$
and you get the two families of solutions
$$
\frac{3x}{2}=\frac{\pi}{2}-\frac{x}{2}+2n\pi
\qquad\text{or}\qquad
\frac{3x}{2}=-\frac{\pi}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $3\left(9-5\sqrt{3}\right) \sum \frac{1}{a} \geqslant \sum a^2+\frac32\cdot\frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}$ Let $a,\,b,\,c$ are positive real numbers satisfy $a+b+c=3.$ Prove that
$$3\left(9-5\sqrt{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a^2+b^2+c^2 + \frac32 \cdot ... | $uvw$ kills this problem!
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\frac{9(9-5\sqrt3)v^2}{w^3}\geq9u^2-6v^2+\frac{3(w^3-3(2-\sqrt3)v^2)^2}{2w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=6(9-5\sqrt3)u^4v^2-2(3u^2-2v^2)uw^3-3(w^3-3(2-\sqrt3)uv^2)^2.$$
But $$f''(w^3)=-6<0,$$ which sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that... | The tens digit is $7$, not the units and you want to find the unit digit.
So if the unit digit is $x$ then the number ends with $7x$ and $\pmod{100}$ what you are trying to say is $n^2 \equiv 70 + x\pmod {100}$.
The way I would do this is let $n= 10k + a$ where $a,k$ are single digits. And the hundreds place don't aff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Volume with spherical polar coordinates Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.
I convert to spherical polar coordinates.
$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$
$$\iiint_{K} (\sqrt {4-r^2\s... | Area on XY plane is bound by $x^2 + y^2 \leq 1, y \geq 0, x + y \geq 0$
This is a sector of the circle $x^2 + y^2 \leq 1$ bound between positive $X$-axis and line $y = -x$ in the second quadrant. This comes from the fact that $y \geq 0$ so part of the circle in third and fourth quadrant of $XY$ plane is not included. $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How many 4 digit number can be formed by using 1,2,3,4,5,6,7 if at least one digit is repeated My approach for this was selecting 4 numbers from which one is being repeated
$$\boxed{A}\boxed{B}\boxed{C}\boxed{A} $$
$$\boxed{A}\boxed{B}\boxed{A}\boxed{A} $$
$$\boxed{A}\boxed{A}\boxed{A}\boxed{A} $$
and permitting them... | The quick way
$$7^4 - \left[\binom{7}{4} \times 4!\right].$$
The hard way:
$$S_1 + S_2 + S_3 + S_4.$$
$S_1 = 7 =$ # ways all 4 numbers the same.
$S_2 = 7 \times 6 \times 4 =$ # ways 3 numbers the same.
$S_3 = \binom{7}{2}\times\binom{4}{2} =$ # ways of having two pairs of numbers.
$S_4 = 7 \times \binom{6}{2} \times 4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b... | I apologize for the first non-obvious proof and I give you a refinement
Hint :$a\geq b \geq c$
First prove that :
$$\frac{2(x-y)}{x+y+y^2}\leq \frac{x}{y}-\frac{x+2}{y+2}\quad (1)$$
For that multiply by $y(y+2)(x+y+y^2)$ , put in factor and it becomes :
$$2(x-y)^2\geq 0$$
Apply $(1)$ for $(a,b)$,$(b,c)$,$(c,a)$
Now we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Irreducible factors of $x^q-1$ over $\mathbb{Z}_p$ The polynomial $x^3-1$ over $\mathbb{Z}_p$, where $p$ is a prime, factors as $(x-1)(x^2+x+1)$.
The polynomial $x^7-1$ over $\mathbb{Z}_{13}$ factors as, $(x-1)(x^2+3x+1)(x^2+5x+1)(x^2+6x+1)$.
All these $(x-1),(x^2+x+1),(x^2+3x+1),(x^2+5x+1),(x^2+6x+1)$ are irreducible ... | No. For example, $x^7 - 1 = (x-1)(x^6+x^5+ x^4+x^3+x^2+x+1)$ modulo $19$ as a product of irreducible factors. It is easy to see that $x-1$ is always a factor, but from that point on I think more specific conditions may be required on $p$ and $q$ ($q<p$ seems arbitrary in this respect) for this breakup to occur.
Note : ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that... | Let us find the general term of $1,3,7,13,\cdots$
As the next consecutive differences are $2,4,6,\cdots$
$$T_n=an(n-1)+b(n-1)(n-2)+cn(n-2)$$
$n=1\implies1=c(1-2),c =-1$
$n=2\implies3=2a$
$n=3,7=6a+2b+3c=3(3)+2b+3(-1),2b=1$
$$\implies T_n=\cdots=n^2-n+1$$
We need the sum of $n$ terms from $k=n^2-n+1$ to $(n+1)^2-(n+1)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer? Which positive integers
$a$ and $b$ make
$(ab)^2-4(a+b)
$
a square of an integer?
I saw this in quora,
and found that
the only solutions with
$a \ge b > 0$
are
$(a, b, (ab)^2–4(a+b)) = (5, 1, 1)$
and $(3, 2, 16)$.
Another “solution” is
$a... | If $(ab)^2-4(a+b)$ is greater than $(ab-1)^2$ then it cannot be a square, since it is strictly between two consecutive squares. Hence
$$(ab)^2-4(a+b) \le (ab-1)^2=(ab)^2-2ab+1$$
$$2ab-4a-4b-1\le0$$
$$2(a-2)(b-2)=2ab-4a-4b+8\le 9$$
which, again, gives a finite set of possibilities to be checked.
WLOG suppose $a \ge b$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A unique question in combinatorics and non-commutative variables. Let $x,y$ be two variables that satisfy:
$xy=yx+1$ (they are not commutative).
Find $(xy)^2 ,(xy)^3, (yx)^2, (yx)^3$ as a linear combination in terms of $y^jx^j$.
Then find a formula for $(xy)^n$ and $(yx)^n$.
After some calculations we get:
$(xy)^2=y^2... | The following is valid for $n\geq 0$:
\begin{align*}
\color{blue}{(xy)^n=\sum_{k=0}^n{n+1\brace k+1}y^kx^k}\tag{1}
\end{align*}
where ${n\brace k}$ are the Stirling numbers of the second kind. We show (1) by deriving the recurrence relation
\begin{align*}
&{n+1\brace k}=k{n\brace k}+{n\brace k-1}\qquad\qquad n, k\geq 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that ... | I continue from Thomas Browning's (the author of the question) answer. We desire to show that
$$nx^2-4(n-1)y^2=1$$
has no solutions. Note that any solution must satisfy $\gcd(nx,y)=1$. We can rewrite the equation as
$$(nx)^2-4n(n-1)y^2=n,$$
so if
$$x^2-4n(n-1)y^2=n$$
has no solutions with $\gcd(x,y)=1$ then we're done.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 0
} |
$\int_{-1}^1{x^4}dx$ by substitution My student asked me: While solving $\int_{-1}^1{x^4}dx$, why can't we substitute as $t=x^2$
$$x^2 =t, 2xdx = dt, x=\pm \sqrt t, \int_{-1}^{1}(x^2 \cdot x \cdot x) dx = \int_1^1{\pm t
\sqrt{t}dt} =0 \ne \frac{1}{5}$$
I think the problem comes from $x = \pm \sqrt t$, but I can't prop... | Since for $r>0$, $\dfrac{\partial x^r}{\partial x}=r\times x^{r-1}$ and if $r-1+r=4$ then $r=\dfrac{5}{2}$.
Thus,
\begin{align}\int_{-1}^1 x^4\,dx&\overset{x\rightarrow x^4 \text{is even function}}=2\int_0^1 x^4 dx\\
&\overset{y=x^{\frac{5}{2}}}=2\times \frac{2}{5}\int_0^1 ydy\\
&=\frac{4}{5}\int_0^1 y dy\\
&=\frac{4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3904439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a +... | You made a mistake. From here
$$ a^2 - (1 + b)a + b \geq 0$$
you went here
$$ a=\frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
While the correct solution is
$$a=\frac{-(1 + b) + \sqrt{ (1 + b)^2-4b}}{2} =\frac{-(1 + b) + \sqrt{ (1 - b)^2}}{2}$$
$$a_1=1;\;a_2=b$$
$$a\le 1\lor a\ge b$$
that is
$$x^2\le 1 \lor x^2\ge (m+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3904731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
About the divergence of an infinite series. Why is the series $\sum \frac{1}{n}$ not convergent, even though $$
\lim_{n \to \infty}\frac{1}{n}=0?$$
| $$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n}=\\
=1+\frac{1}2+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\cdots+\frac{1}{8}\right)+\cdots+\left(\frac{1}{2^{n-1}+1}+\cdots+\frac{1}{2^n}\right)
$$
But
$$
\frac{1}{3}+\frac{1}{4}\ge \frac{2}{4}=\frac{1}{2}, \\
\frac{1}{5}+\cdots+\frac{1}{8} \ge \frac{4}{8}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
System of equations involving 4 variables
If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$
I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1... | Define $A_i$ as follows
\begin{eqnarray*}
a+b=A_0 \\
ax+by=A_1 \\
ax^2+by^2=A_2 \\
ax^3+by^3=A_3 \\
ax^4+by^4=A_4 .
\end{eqnarray*}
Multiply the first & third equations & subtract the square of the second equation
\begin{eqnarray*}
A_0 A_2 -A_1^2 = ab(x-y)^2.
\end{eqnarray*}
Multiply the first & fourth equations & su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve the inequality $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$ So am trying to solve this inequality, $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$, the problem is of course for values $0<a<1$, I tried working through it multiple times but the solutions i found, which... | You obviously need $x > 0$ and further have
$$\sqrt[3]{(1+\sqrt x)^3} - \sqrt[3]{(1-\sqrt x)^3} > x+a$$
$\iff 1-a > (\sqrt x - 1)^2 \iff \sqrt{1-a} > |\sqrt x - 1|$
For $x \in (0, 1)$, we have this as
$ \sqrt{1-a} > 1-\sqrt x \implies x > (1-\sqrt{1-a})^2 $
For $x \geqslant 1$, we have this as
$ \sqrt{1-a} > \sqrt x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove a hard inverse trigonometry mathematical induction $\arctan\left(\frac{1}{2\left(1\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(2\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(3\right)^{2}}\right)+...+\arctan\left(\frac{1}{2\left(n\right)^{2}}\right)=\frac{\pi}{4}-\arctan\left(\frac{1}{2n+1}\right)$
I... | I'll assume you were able to prove the base case $n = 1$.
The inductive hypothesis is to assume $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+1}. \quad (1)$$
Then, you need to prove $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} +\arctan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help solving the limit of the sequence: $\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$ Given the following sequence,
$$\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$$
I am asked to determine to which $l \in \mathbb{R}$ the sequence converges... | Hint
$$=\left(\left(1-\dfrac2{2n^2+1}\right)^{-(2n^2+1)/2}\right)^{-\frac{2(2n^3-n)}{(2n^2+1)(n+3)}}$$
Alternatively
$$=\left(\dfrac{(1-1/2n^2)^{2n^2}}{(1+1/2n^2)^{2n^2}}\right)^{\dfrac{2n^3-n}{2n^2(n+3)}}$$
Now the numerator converges to $e^{-1}$
What about the denominator & the exponent?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is | Start with the general form of a circle:
$$x^2+y^2 + ax+by+c = 0$$
It passes through $(1, 1)$ and $(2,3)$, so $a+b+c = -2$ and $2a+3b+c = -13$. Therefore $(2a + 3b + c) - 2(a+b+c) = b-c = -9 \Rightarrow b = c - 9$.
Now, since the line $x-y+1 =0 \Rightarrow y = x+1$ is tangent to the circle, substitute in $y = x+1$.
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Proving identity in infinite series I am trying to prove the equation
\begin{align*}
\sum_{n=0}^{\infty} a^n \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d} = \prod_{i=1}^d \frac{1}{1 - a b_i},
\end{align*}
where we have that $n = c_1 + c_2 + ... c_d$. This is what I have tried so far:
\begin{align*}
\sum_{n=0}^{\inf... | By induction on $d$.
$$\begin{aligned}
\frac{1}{1-ab_1}
\frac{1}{1-ab_2} &= \left(\sum_{n=0}^\infty (ab_1)^n\right)\left(\sum_{n=0}^\infty (ab_2)^n\right)\\
&=\sum_{n=0}^\infty \left(\sum_{l=0}^nb_1^lb_2^{n-l}\right) a^n\\
&=\sum_{n=0}^\infty \left(\sum_{c_1+c_2=n}b_1^{c_1} b_2^{c_2}\right) a^n
\end{aligned}$$
accordin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove Inequalities $\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$ Using Calculus I am often asked to prove inequalities with natural logarithms. An example question I am stuck on is this:
Let $f(x) = \ln(\frac{1+x}{1-x})$ for $-1 < x < 1$. Prove that
$$\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$$ with... | Let
$$g(x)=\frac{x(2-x)}{1-x}- \ln\frac{1+x}{1-x}
$$
$$g’(x)= \frac{x(x^2-x+2)}{(1-x)^2(1+x)},\>\>\>
g’’(x) =\frac{2(3x^2+1)}{(1-x)^3(1+x)^2}
$$
Note $g(0)=g’(0)=0$ and $g’’(x)>0$, which means that $g(x)$ is convex with the minimum at $g(0)=0$. Thus, $g(x)\ge 0$, or
$$\ln\frac{1+x}{1-x} \le \frac{x(2-x)}{1-x}$$
Simila... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$ Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
I couldn't do this question and hence I looked at the solution which goes as follows:
if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$
I state that $y=lx$. Hence the system becomes:... | Alternative approach.
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
$3x^2 + 6xy = 15~$ and $~2y^2 - 6xy = -4.$
Adding gives $3x^2 + 2y^2 = 11.$
Then, you have that $5 - x^2 = 2xy.$
Therefore, $25 - 10x^2 + x^4 = 4x^2y^2 = (2x^2)(11 - 3x^2) = 22x^2 - 6x^4.$
Therefore, $7x^4 - 32x^2 + 25 = 0.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find centroid of shape given by a circle and a cardoid I'm stuck on finding the centroid of this shape (the shaded area):
With the sphere having a radius of $r=2$, and the cardioide being given by $r=1+\cos\phi$
I found the Area, which is
$A=\frac12(2\pi r^2-\frac32 \pi)=\frac{13}{4} \pi$
Now I don't know how to proce... | There is a mistake in your area calculation -
$A = \frac12(\pi r^2-\frac32 \pi)=\frac{5}{4} \pi$ (you wrote $2\pi r^2 $ instead of $\pi r^2$ by mistake)
Now coming to finding centroid,
We know in polar coordinates, $x = r\cos \theta, y = r \sin \theta$ so the centroid ($\overline{x}, \overline {y}$) of the polar region... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$? Consider the integral domain $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?
I know the following elementary facts. We have
\begin{equation}
\math... | Yes, $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ is a UFD because it is norm-Euclidean.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Does anyone know of a much faster way to solve this polynomial question? A monic cubic $p(x) $ is divided by $(x^2 +x +1)$ and the remainder is $(2x+3)$. When $p(x)$ is divided by $x(x+3)$, the remainder is $5(x+1)$. Find $p(x)$
So the way I solved it was through using long division - as you can imagine very lengthy, a... | This may not be a quicker way, but it is systematic and will work in all cases where the two moduli have no common factors. Hopefully, it also might make the use of the Extended Euclidean Algorithm with the Chinese Remainder Theorem more transparent.
We want
$$
\begin{align}
p(x)&\equiv2x+3&&\pmod{x^2+x+1}\tag{1a}\\
p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$ Number Theory : Solve the system of congruence
(1) $28x+17y\equiv 18 \pmod{41}$
(2) $31x+11y\equiv 35\pmod{41}$
Attempt :
we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular a... | Let $p \in \Bbb N$ be a prime number.
Recall the following two facts from elementary number theory:
Let $p$ be a prime integer.
For all $u,v,w \in \Bbb Z$ satisfying $u\not\equiv 0 \pmod{p}$, $v\not\equiv 0 \pmod{p}$ and $w\not\equiv 0 \pmod{p}$,
$\tag 1 u \equiv v \pmod{p} \; \text{ iff } \; uw \equiv vw \pmod{p}$
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Justifying a solution for a system of equations in linear algebra I'm trying to understand how to solve systems of linear equations using matrices, however I'm faced with the following problem. Given the following system:
$$
\begin{cases}
x+2y+3z-3w=a \\
2x-5y-3x+12w=b \\
7x+y+8x+5w=c
\end{cases}
$$
What steps do I ne... | You can get your answer using simple elementary row operations:
$$\left(
\begin{matrix}
1 & 2 & 3 & -3 & | & a\\
2 & -5 & -3 & 12 & | & b \\
7 & 1 & 8 & 5 & | & c \\
\end{matrix}
\right) >>{\text{$R2=R2-2R1, R3=R3-7R1$}}>>$$ $$\left(
\begin{matrix}
1 & 2 & 3 & -3 & | & a\\
0 & -9 & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lagrange's Multiplier for $f(x,y,z) = (xyz)^2$ Given a function:
$$f(x,y,z) = (xyz)^2$$
And the set: $$S = \{(x,y,z) : x^2 + y^2 + z^2 = a^2\} $$
Does $f$ have a maxima on $S$?
(Is this argument correct? : Yes, because f has a maxima and minima in S if S is compact and f is continuous. But here S is compact so we have... | Define : $\mathcal{L}(x,y,z,\lambda) = (xyz)^2 - \lambda\cdot(x^2 + y^2 + z^2 - a^2)$ for any $\lambda \in \mathbb{R}$ , and for any $x, y, z \in \mathbb{R}$ .
So, $~\mathcal{L}_x = 2xy^2z^2 - 2\lambda x~$ , $~\mathcal{L}_x = 2x^2yz^2 - 2\lambda y~$ , and $~\mathcal{L}_x = 2x^2y^2z - 2\lambda z~$ . Also, we have : $~\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Range of quadratic function using discriminant
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.
Let $2x^2-2xy+y^2=a$.
Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.
We divide both side by $y^2$ and let $t=\frac{x}{y}$.
Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.
Since its ... | Others have pointed out where you went wrong. I just want to provide an alternative proof.
$$x^2-2xy-3y^2=4=(x+y)(x-3y)$$
Denote $z=x+y, w=x-3y$, then $zw=4, x=(3z+w)/4, y=(z-w)/4$.
We want to find the range of $$2x^2-2xy+y^2=\frac{1}{16} (5w^2+14wz+13z^2)=\frac 72+\frac{1}{16} (5w^2+13z^2) \tag 1$$ under the constrain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Am I allowed to use this method? Question: solve
$\frac{d^2y}{dx^2}+\frac{dy}{dx}=2x+3$.
My attempt:
The general solution to the corresponding homogeneous DE is $y_g=c_1+c_2e^{-x}$.
Now to find the particular integral, rewrite the DE as $$(D^2+D)y=2x+3$$
We have $y_p=\frac{1}{D(D+1)} (2x+3)$.
Then $y_p=\frac{1}{D+1}(x^... | $$(D^2+D)y=2x+3$$
It means:
$$y''+y'=2x+3$$
Not the equation you write in the first line.
$$\frac{d^2y}{dx^2}y+\frac{dy}{dx}y=2x+3$$
$$y''y+y'y=2x+3$$
And yes you have that:
$$\dfrac 1{1+D}=1-D+D^2-D^3.......$$
But you have:
$$y_p=\frac{1}{D(D+1)} (2x+3)$$
$$y_p=\left (\dfrac{1}{D}-\dfrac 1 {D+1} \right)(2x+3)$$
$$y_p=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan... | I think the mistake is from the first step.Emm...in fact,$$\lim \frac{{\ln (1 + x\arctan x)}}{{\sqrt {1 + 2{x^4}} - 1}} = \frac{{\ln (1 + x\arctan x)(\sqrt {1 + 2{x^4}} + 1)}}{{2{x^4}}} = \infty , \\ \lim \frac{{{e^{{x^2}}} - 1}}{{\sqrt {1 + 2{x^4}} - 1}} = \frac{{({e^{{x^2}}} - 1)(\sqrt {1 + 2{x^4}} + 1)}}{{2{x^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{x \to + \frac{1}{2}} [ \tan(\pi x^2)+(2x-1)4^x+6x-4 ] \cot(x-\frac{1}{2})$ I've tried to solve this limit:
$\lim_{x \to + \frac{1}{2}} \left [ \tan(\pi x^2)+(2x-1)4^x+6x-4 \right ] \cot(x-\frac{1}{2})$
the first parentesis should tend to $\frac{\sqrt{2}}{2}-1$ but I don't know how to procede.
The final result s... | To make life easier, let $x=y+\frac 12$ to make the expression
$$\left(\left(4^{y+1}+6\right) y+\tan \left(\frac{1}{4} \pi (2
y+1)^2\right)-1\right) \cot (y)$$ and now, compose series one piece at the time
$$4^{y+1}=4 e^{y \log(4)}=4+4 y \log (4)+2 y^2 \log ^2(4)++O\left(y^3\right)$$
Now, after expansion of the a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determinant and inverse matrix calculation of a special matrix Is there any smart way of calculating the determinant of this kind matrix?
\begin{pmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmat... | Call your matrix $A$. Let $\{e_1,e_2,\ldots,e_{n-1}\}$ be the standard basis of $\mathbb R^{n-1}$. Let also $e=\sum_ie_i=(1,1,\ldots,1)^T$ and
$$
L=\pmatrix{1\\ -1&1\\ &\ddots&\ddots\\ &&-1&1}.
$$
Then $B:=LAL^T=\pmatrix{1&e^T\\ e&-2I_{n-1}}$. Using Schur complement, we obtain
$$
\det(A)=\det(B)=\det(-2I_{n-1})\left(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ without using antiderivative? Someone gives a solution here:
\begin{align*} \int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}&=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} (-\cos\alpha\cos\beta)^n{\rm d}\alpha\\ &=\sum_{... | For any $\beta \textrm{ is not a multiple of }\pi$, we multiply both the numerator and denominator by $1-\cos \alpha \cos \beta.$
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{d \alpha}{1+\cos \alpha \cos \beta}
=& \int_0^{\frac{\pi}{2}} \frac{1-\cos \alpha \cos \beta}{1-\cos ^2 \alpha \cos ^2 \beta} d \alpha \\
=& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve integral $\int^1_0 \frac{1-x^2}{{(1+x^2)}\sqrt{1+x^4}}dx$ using subsituition $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ Hi this question has been posed in my integration book where it has been asked to solve it using the given substitution or any other substitution, I've found identical question posted here {1} with ... | Divide top and bottom by $x^2$:
$$\int_0^1\frac{\frac{1}{x^2}-1}{\left(\frac{1}{x}+x\right)\sqrt{\frac{1}{x^2}+x^2}}dx$$
which suggests using the substitution $t = \frac{1}{x}+x$:
$$\int_2^\infty \frac{dt}{t\sqrt{t^2-2}} = \frac{1}{\sqrt{2}}\sec^{-1}\left(\frac{t}{\sqrt{2}}\right)\Biggr|_2^\infty = \frac{\pi}{4\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
A question related to $f(n)=n^4+n^2+1$ If $$f(n)=n^4+n^2+1$$ then we have to evaluate $$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}$$ which, when run in Desmos, returns 6... | We note that $n^4+n^2+1 = (n^4+2n^2+1)-n^2 = (n^2+1)^2-n^2 = (n^2+n+1)(n^2-n+1)$
Writing the value of $(n^2+n+1) = g(n)$, the value of $g(n-1) = n^2-n+1$
So we get eg $f(n)=g(n)g(n-1)$
And $\frac{f(3)f(5)f(7)...f(13)}{f(2)f(4)f(6)...f(12)}$
is equal to $\frac{g(2)g(3)g(4)...g(12)g(13)}{g(1)g(2)g(3)g(4)...g(12)} = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Algebra problem with isomorphism Let $G=(-1,1)$ and the binary operation such that
$x*y=\frac{x\sqrt{1-y^2}+y\sqrt{1-x^2}}{\sqrt{1-(xy)^2+2xy\sqrt{(1-x^2)(1-y^2)}}}$.
Prove that (G,*) it is a isomorph group with the group (R,+).
I tried to find the bijective function $f:G\to R$ such that $f(x*y)=f(x)+f(y)$ but I co... | Let $x=\sin\alpha$, $y=\sin\beta$, so $\alpha,\beta\in(-\pi/2,\pi/2)$. (always do such substitution when there is a $\sqrt{1-x^2}$)
Then $x*y = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sqrt{1 - \sin^2\alpha\sin^2\beta + 2\sin\alpha\sin\beta\cos\alpha\cos\beta}} = \frac{\sin(\alpha+\beta)}{\sqrt{\cos^2\alpha\cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Markov's Chain stationary matrix determination I have the following transition matrix $P$
\begin{equation}
P=
\begin{pmatrix}
1/2 & 1/3 & 1/6\\
1/4 & 3/4 & 0\\
1/5 & 2/5 & 2/5
\end{pmatrix}
\end{equation}
I have tried to calculate the stationary matrix as
\begin{equation}
P^n= C D^n C^{-1}
\end{equation}
With $D=C^{-1... | Since $1$ is the dominant eigenvalue of the Markov Chain (the derivation of that fact can be found elsewhere, thus I leave that up to the reader to investigate if desired), all we need to solve is
$$ \begin{bmatrix} 1/2 & 1/4 & 1/5 \\1/3 & 3/4 & 2/5 \\ 1/6 & 0 & 2/5 \end{bmatrix} \begin{bmatrix} x \\y \\ z \end{bmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric problem (problem from a Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two angles are $135^\circ$ and the sum of their tangents are $5$. Calculate the angles.
Is there a shorter/simpler solution than the one pres... | Recall that $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. Now, given $\alpha+\beta=135^{\circ}$ and $\tan(\alpha+\beta)=\tan 135^\circ=-1$, you get $\frac{5}{1-\tan\alpha\tan\beta}=-1$, i.e. $\tan\alpha\tan\beta=6$. Now, knowing the sum ($5$) and the product ($6$) of $\tan\alpha$ and $\tan\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number? Is there a way to show that
$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$
is a rational number?
I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.
| Anay's idea of cubing both sides looks promising, but there is a little subtlety in the method. Let's explore.
Start with
$\alpha=\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$
Cube with the Binomial Theorem:
$\alpha^3=\color{blue}{7+5\sqrt2}+\color{brown}{3(\sqrt[3]{7+5\sqrt2})^2\sqrt[3]{20-14\sqrt2}+3\sqrt[3]{7+5\sqrt2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've fou... | Here a relatively "natural" way to show the inequality. It avoids a time consuming (or rather time wasting) analysis of a quite specific 6th degree polynomial where the amount of learnings and insights to be expected doesn't seem to justify the effort.
To show is equivalently
$$\frac{x^2}y + \frac{y^2}x \geq \sqrt 2$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 10,
"answer_id": 0
} |
Find a basis of the generated subspace Consider the vectors
$$
v_1 =\begin{pmatrix}
1\\ 0\\ 0\\ 1
\end{pmatrix}, \quad v_2=\begin{pmatrix}
\sqrt{2}\\ 1\\ 2\\0
\end{pmatrix}, \quad v_3 =\begin{pmatrix}
\sqrt{2}+2\\ 1\\ 2\\ 2
\end{pmatrix}
$$
and let $S$ be the subspace generated by $v_1, v_2, v_3$. Find the dimension of... | It's correct, however an overkill.
You observed correctly that $v_3=2v_1+v_2$, so they are linearly dependent, but clearly $v_1,v_2$ are not parallel, hence independent, so the dimension they span is indeed $2$.
But then any $2$ independent elements of the span form a basis, so just take $v_1$ and $v_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expressing $ \frac{\cos x + \sin x}{\cos x - \sin x} $ as a single trigonometric function For a while now, I have been trying to get this expression into 1 term instead of the 2 term fraction.
$$
\frac{\cos x + \sin x}{\cos x - \sin x}
$$
The most I have gotten down to is get it to a single ratio, ie,
$$
\frac{1 + \tan... | $
\dfrac{\cos x + \sin x}{\cos x - \sin x} = \dfrac{1+ \tan x}{1 - \tan x}
$
Now, $\tan (x+y) = \dfrac{\tan x+ \tan y}{1 - \tan x \tan y}$
Therefore, $\tan (x+ \frac{\pi}{4}) = \dfrac{\tan x+ 1}{1 - \tan x}$
Therefore, $\dfrac{\cos x + \sin x}{\cos x - \sin x} = \tan(x+ \frac{\pi}{4})$
Is this what you meant by 1 term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
simplify fraction (stone drops off a cliff) How do I get from
$$ \frac {dt} {T} = \frac {dx} {gt} \sqrt { \frac {g} {2h} } $$
to
$$ \frac {dt} {T} = \frac {1} {2 \sqrt {hx} } dx $$
where $x(t) = \frac {1} {2} gt^2 $ and $ T = \sqrt { \frac {2h} {g}}$.
I'm currently struggling with Griffiths' Introduction to Quantum Mec... | We have
$$
x=\frac{1}{2}gt^2
\quad\Leftrightarrow\quad
gt^2=2x
$$
so
\begin{align*}
\frac{dt}{T}
&
=\frac{dx}{gt}\sqrt{\frac{g}{2h}}
=\sqrt{\frac{g}{g^2t^2\cdot2h}}\,dx
=\sqrt{\frac{1}{gt^2\cdot2h}}\,dx
\\&
=\sqrt{\frac{1}{2x\cdot2h}}\,dx
=\sqrt{\frac{1}{4hx}}\,dx
=\frac{1}{\sqrt{4hx}}\,dx
=\frac{1}{2\sqrt{hx}}\,dx.
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A simple but nonsense infinite sums rule Assume $a>0$ and $0<r<1$ are given. The geometric series is known to be $\sum_0^{\infty} ar^n = \frac{a}{1-r}$. Now differentiate w.r.t $r$ to obtain $\sum_0^{\infty}anr^n = \frac{ar}{(1-r)^2}$. So far we're cool.
Now take the two series $$S_1 = \sum_n \frac{1}{2^n}$$ and $$S_2... | Yes, you are right $S_1 = S_2 = 2$ and it might be somewhat astonishing the first time. But first let's have a look at your derivation which needs to be somewhat revised. We can take $a=1$ and consider
\begin{align*}
S_1(r)=\sum_{n=0}^\infty r^n=\frac{1}{1-r}\tag{1}
\end{align*}
Derivation with respect to $r$ gives
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Problem proving $\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx$ I am trying to show that the value of the following integral is:
$$\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=2^{n-1}\pi^{\frac{1}{2}}(2n+1)n!\delta_{mn}+2^n\pi^{\frac{1}{2}}(n+2)!\delta_{(n+2)m}+2^{n-2}\pi^{\frac{1}{2}}n!\delta_{(n-2)m},$$
where ... | Starting from here
$$
\cfrac{\sqrt{\pi}}{4}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}
$$
First, rewrite the $\delta$s so that they all have only $m$. We do this by noting that, for example, if $m-1 = n+1$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the integer solutions to $4^x - 9^y = 55$ I want to find the integer solutions of:
$$ 4^x - 9^y = 55$$
For now, I see that $x = 3, y = 1$ is an integer solution to the equation. How can I rigorously prove there are no other solutions for $x, y$ integers?
I tried to solve for $y$, but to no avail. WolframAlpha tell... | I like the explanations given in @MishaLavrov's answer, and I understand this only as a comment to his answer.
There, the second sentence introduces a bit obfuscation, I think: the bare idea of needing a crosscheck of $\{0,1,2,3,4,5\} \times
\{0,1,2,3\}$ insertions is somehow surprising (and misleading) after the intr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Some horrid integrals I am confused with $$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$
Speaking as an example on (1).
The only thing I could do from here is only to do a u-sub:
$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$
And so this be... | Concerning $(3)$ $$\int_{\frac 1e}^{\infty} \sqrt{\frac{\log{(x)}}{x^3 + 1}}\,dx $$ it is normal to get a complex result.
Around $x=\frac 1e$ we have
$$\sqrt{\frac{\log{(x)}}{x^3 + 1}}=i\, \sum_{n=0}^\infty a_n \left(x-\frac{1}{e}\right)^n$$ and the first coefficients are
$$a_0=\frac{ e^{3/2}}{\left(1+e^3\right)^{1/2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 0
} |
Find all positive integer solutions for the following equation $5^a + 4^b = 3^c$:
Find all positive integer solutions for the following equation:
$$5^a + 4^b = 3^c$$
My first guess would be to study the equation in mod, but I tried modulo 3, 4, 5, and 9 and I can't find anything.
| Another route: Follow the reasoning in Aqua's answer up to $5^a=(3^d-2^b)(3^d+2^b)$.
Then ask if $3^d-2^b$ and $3^d+2^b$ share any factors. If $m\mid (3^d-2^b) \land m \mid (3^d+2^b)$, then $m \mid (3^d-2^b)+(3^d+2^b)=2\cdot3^d$, which means $m\in\{1,2,3^x\}$. But $2\not \mid (3^d-2^b)$ and $3\not \mid (3^d-2^b)$, so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Unable to reconstruct a new polynomial to find a given value Suppose $f(x)= x^3+2x^2+3x+3$ and has roots $a , b ,c$.
Then find the value of
$\left(\frac{a}{a+1}\right)^{3}+\left(\frac{b}{b+1}\right)^{3}+\left(\frac{c}{c+1}\right)^{3}$.
My Approach :
I constructed a new polynomial $g(x) = f\left(\frac{x^{\frac{1}{3}}}{1... | We shall make use of the following well-known identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz).$$
Now, using Vieta's Formula, $a+b+c=-2, ab+ac=bc=3$, and $abc=-3$.
Thus, by direct expansion, we have that
$(a+1)(b+1)(c+1)=-1.$
\begin{align}
\dfrac{a}{a+1}+\dfrac{b}{b+1} + \dfrac{c}{c+1} &= 3- \left(\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A Negative binomial problem $P(X \ge 5)$ equals?
Consider a sequence of independent Bernoulli trials with the
probability of success in each trial being $\dfrac{1}{3}$. Let $X$
denote the number of trials required to get the second success. Then
$P(X \ge 5)$ equals.
$A=\dfrac{3}{7}$
$B=\dfrac{16}{27}$
$C=\dfrac{2}{3}$... | Easy is $B:\frac{16}{27}$
$$P(X\geq 5)=1-P(X\leq 4)=1-\binom{2-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^0-\binom{3-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^1-\binom{4-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^2=$$
$$=1-\frac{3}{27}-\frac{4}{27}-\frac{4}{27}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? The question:
How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? At which points do these tangent lines touch the curve?
My attempt:
To find the slope of the tangent line to the curve, we need to... | For a fraction $\frac{a}{b}$ to be equal to 0, we must have $a=0$. In this case we have $$x^2+4x+1=0$$ and the result follows.
EDIT:
Note that $$\frac{x-1}{(x+1)^2}+2-\frac{x}{x+1}=\frac{x^2+4x+1}{(x+1)^2}$$
You might wanna check your steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
No. of ways of selecting 3 squares when they do not lie in same row, column, or diagonal
Total number of ways of selecting 3 small squares on a normal chess board so that they don’t belong to the same row, column or diagonal line, is equal to:
No. of ways of selecting 3 squares when they do not lie in same row or sam... | I will count the number of ways of selecting three named positions $X,Y,Z$ such that no two are in the same row or column under various conditions. The counts differ slightly according to the parity of $n$ and so I will assume that $n$ is even.
$A$. No other conditions.
$B$. $X$ and $Y$ are on the same diagonal.
$C$. $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$ I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:
$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{... | Integrate by parts to reduce the remaining integrals
$$\int \frac{dx}{(x^2-1)^3}=\int \frac1{4x^3}d\left( \frac{-x^4}{(x^2-1)^2}\right)= -\frac14 \frac{x}{(x^2-1)^2}-\frac34 \int \frac{dx}{(x^2-1)^2}
$$
$$\int \frac{dx}{(x^2-1)^2}=\int \frac1{2x}d\left( \frac{-x^2}{x^2-1}\right)= -\frac12 \frac{x}{x^2-1}-\frac12\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find the period of this sinusoid? I'm stuck trying to find the period of this sinusoid and would really like some pointers to different ways to approach this problem.
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3})$$
I would like to turn the product into a sum and do something about that $2$ in the expon... | To get the LCM, we want the smallest $T$ which is an integer multiple of the three periods. That is, we want integers $n_1,n_2,n_3$ such that $$T=\frac{5}{2}n_1=\frac{15}{46}n_2=\frac{15}{34}n_3$$ with $T$ as small as possible.
To this end we first focus on the middle equality, which rearranges to $23n_1=3n_2$. Since $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.
As an instance, I will justify an example inequality just to show you the intended strategy. Consider th... | Hint:
Set
$$
f(x)=\frac{5x^3 + 3x^2 + x}{x^3 - x^2 - 1}.
$$
Then you have
$$
\lim_{x\to+\infty}f(x)=5\tag{1}
$$
Now use the limit definition to show that $f(x)<10$ for sufficiently large $x$. So you get
$$
\frac{f(x)}{x}<\frac{10}{x}
$$
for sufficiently large $x$.
Notes: The limit (1) implies that
$$
|f(x)-5|<1
$$
for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Maximum number of non-congruent triangles with side lengths integers and less than $9$
Find the maximum number of non-congruent triangles whose side lengths are integers and less than $9$.
I tried using brute force to count all the triangles but there are so many triangles that can be formed and I counted about $60$ ... | We do some systematic counting.
First the isosceles (including equilateral) ones. If the base is $1$, other two sides can be any from $1$ to $8$. If base is $2$, other two can be any from $2$ to $8$. If base is $3$ - $2$ to $8$. Base $4$ - $3$ to $8$. Base $5$ - $3$ to $8$. Base $6$ - $4$ to $8$. Base $7$ - $4$ to $8$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x))$ Let $cot^{-1} (1+x)=a$
Then $x =1-\cot a$
So $$\sin a =\cos (\tan^{1}(1-\cot a))$$
$$a=\frac{\pi}{2}- \tan^{-1} (1-\cot a)$$
$$\cot a =\frac 12$$
So $x=\frac 12$, but given answer is $-\frac 12$. What am I doing wrong?
| $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x)).$
Let $\cot^{-1} (1+x)=A \implies \cot A =1+x.$ So $LHS=\sin A=
\frac{1}{\sqrt{2+2x+x^2}}$ Let $\tan B=x \implies RHS=\cos B =\frac{1}{\sqrt{1+x^2}}$. Finally $LHS=RHS \implies 2+2x+x^2=1+x^2 \implies x=-1/2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that $|ba^{-1}| = |b^3a^3|$. Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that
$|ba^{-1}| = |b^3a^3|$.
attempt:
Write $a*b=ab$. Let $a,b \in G$.
Then, $ab=b^5a^3$. Note that
$ba^{-1} = a^{-1}b^5a^2$,
$ab^{-1} = b^2(b^3a^3)b^{-2}$, an... | Claim 1: For any $x,y \in G$ we have $|xyx^{-1}|=|y|$
Proof: Exercise
Claim 2: For any $x \in G$ we have $|x|=|x^{-1}|$
Proof: Exercise
Now, from your last step of actual manipulation we have, $b^3a^3=b^{-2}(ab^{-1})b^2$.
So applying Claim 1 twice, we have, $|b^3a^3|=|b^{-2}(ab^{-1})b^2|=|b^{-1}(b^{-1}(ab^{-1})b)b|=|b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Primitive roots of a finite field I understand the definition of a primitive root of integers; however, I am quite confused trying to find the primitive roots of $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$. I know that $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$ has elements of the form $ax^2+... | As Leoli1 commented, you have to find an element of order $26$,
since the multiplicative group of $\mathbb F_{27}$ has order $26$.
You may consider this brute force, but using $x^3\equiv x-1$ we have
$x^4\equiv x^2-x, $ $x^5\equiv -x^2+x-1$, $x^6\equiv x^2+x+1$, $x^7\equiv x^2+2x+2$, $x^8\equiv2x^2-1$,
$x^9\equiv x+1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $1/x^2$ without power rule I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents
\begin{equation*}
\int \frac{1}{x^2} \; dx
\end{equation*}
When using integration by parts, you end up with the same integral in the rhs so it seems out of l... | $$ \frac{1}{x^2} = e^{\ln \left ( \frac{1}{x^2} \right ) } = e^{\ln \left ( x^{-2} \right ) } = e^{-2\ln (x ) } = (e^{-2})^{\ln(x)} $$
$$ \text{when} ~~a \in \mathbb{R^+}: ~~~\int a^{\ln(x)} dx = \int \frac1x x^{\ln(a) + 1} dx \\ u=\ln x \Rightarrow du = \frac1x dx \\ = \int e^{(\ln(a) + 1)u} \\v = ( \ln(a) + 1)u \Ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 11,
"answer_id": 7
} |
Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longri... | $$
f(x) = x^{f(x)}\\
\ln(f(x))=f(x) \ln(x)\\
\frac{1}{f(x)}f'(x) = \frac{f(x)}{x} + \ln(x)f'(x),\\
f'(x)=\left(\frac{1}{f(x)}-\ln(x)\right)^{-1}\frac{f(x)}{x},\\
f'(x)=\left(\frac{f(x)}{1-\ln(x)f(x)}\right)\frac{f(x)}{x}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Failing a basic integration exercise; where did I go wrong? (This is a basic calculus exercise gone wrong where I need some feedback to get forward.)
I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where st... | After the substitution $u = \sqrt{x}$, the integral becomes $2\int u^2 \sin(u) \ du$ as Adam Rubinson has said. We can proceed using tabular integration:
$$\begin{array}{c|c} u^2 & \sin(u) \\ \hline 2u & -\cos(u) \\ \hline \ 2 & -\sin(u) \\ \hline \ 0 & \cos(u)\end{array}$$
where we differentiate on the left and integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
What are the symmetric elements of $SU(3)$? Is there any simple way to know what are the matrices $M\in\operatorname{SU}(3)$ s.t $M=M^t$? For exemple, for $\operatorname{SU}(2)$ it is easy to verify that the symmetric elements are either diagonal or
$$\left(\begin{array}{cc}
0 & i \\
i & 0
\end{array}\right).$$
| I assume you are asking about the triplet representation, judging from your doublet paradigm for SU(2), which should not be "either-or", but rather
$$
i~ \begin{pmatrix}
\sin\phi & \cos\phi \\
\cos\phi & - \sin\phi
\end{pmatrix} .
$$
I gather you are looking for "seat-of-the-pants" eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
If I would start from the right expression I woul... | We'll use the following identities:
$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}{2}$
$\sin^2{\theta} = \dfrac{1-\cos(2\theta)}{2}$
$\sin \theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}$
$\cos \theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2}$
$$
(\sin(\theta) - \cos(2\theta))^2 = \sin^2(\theta) + \cos^2(2\theta) - 2 \sin(\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the limit of $a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2})$ for arbitrary $a_{0}$ and $a_{1}$. Here is the problem statement:
Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by
$$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \g... | $$
\begin{bmatrix}
a_{n}\\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
a_{n-1}\\
a_{n-2}
\end{bmatrix}
$$
$$
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Deriving what $\cos(A+2B)$ is We know that $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ , but I don't understand with the $2B$. Would it just become $\cos(A+2B) = \cos(A)\cos(2B) - \sin(A)\sin(2B)$?
Thank you for any help you can give me.
| Further expansion is possible with repeated use of the angle addition identities.
$$\begin{align}
\cos (a + 2b) &= \cos a \cos 2b - \sin a \sin 2b \\
&= \cos a (\cos b \cos b - \sin b \sin b) - \sin a (\sin b \cos b + \cos b \sin b) \\
&= \cos a (\cos^2 b - \sin^2 b) - \sin a (2 \sin b \cos b) \\
&= \cos a \cos^2 b - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
I have to show that the sum of this double series is $\frac{1}{2}$ i have to solve this double series. i tried it, but i am not sure, that it is enough.
$$\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \left(\left(\frac{1}{k+1} \cdot \left(\frac{k}{k+1}\right)^{i}\right) - \left(\frac{1}{k+2} \cdot \left(\frac{k+1}{k+2}\right... | Here is my attempt.
Since all terms are positive, we are allowed to swap the two summations.
$$\sum_{i=1}^\infty \sum_{k=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \frac{1}{k+2} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty }\sum_{k=1}^N \sum_{i=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Analyse convergence of: $\displaystyle \sum^{\infty}_{n=3} \left (\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right)$
Analyse convergence of:
$$\displaystyle \sum^{\infty}_{n=3} \left(\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right )$$
I know that all elements are negative, so I need to take $-1$ out and I will have only positive el... | If you are able to prove that both series$$\sum_{n=3}^\infty\sqrt[3]{n^3+2}-n\quad\text{and}\quad\sum_{n=3}^\infty\sqrt{n^2+7}-n\tag1$$converge, or that one of them converges and the other one diverges, then you're done.
Now, note that\begin{align}\sqrt[3]{n^3+2}-n&=\sqrt[3]{n^3+2}-\sqrt[3]{n^3}\\&=\frac2{\sqrt[3]{n^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A weird contour integral calculation I have function $\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx$.
Here to use residue thm, I rewrite the integral as $\int_{-1}^{1}\frac{\sqrt{1-z^2}}{1+z^2}dz$ with the poles $z=i$ and $-i$. However there is a given condition $-1<x<1$, so it means $z^2<1$. This is the problem because w... | Time for the dog-bone contour (dumbbell contour), which works perfectly.
Indeed, consider the contour $\mathcal{C}$ given as follows:
$\hspace{12em}$
With the principal branch cut, as the radius/band-width of $\mathcal{C}$ goes to zero,
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{i(z^2+1)} \, dz
&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine the partial derivative of $\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}$.
Determine the partial derivative of $f(x)$:
$$f(x,y)=\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}$$
Here... | The only issue is that $ \frac{\partial}{\partial x} \cos^2(x^2 y^2)$ should be equal to
$$- 4 x y^2 \cos(x^2 y^2) \sin(x^2 y^2) $$
and with that correction it is good.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$. My solution Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:
*
*If $i=0$, then
\begin{align*}
n = 5k \Rightarrow n = 5k + 4(0).
\... | We proceed by induction on $n.$ Observe that $12 = 3 \cdot 4 + 0 \cdot 5$ is a non-negative linear combination of $4$ and $5.$ We will assume inductively that any integer $n \geq 12$ can be written as a non-negative linear combination $n = 4x + 5y$ of $4$ and $5.$ Given that $x \geq 1,$ we have that $$4(x - 1) + 5(y + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
General Inequality Problem. The question goes like this :
Let $a,b,c,d$ be positive reals and given that $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$
My approach goes like this:
I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$
Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$
Although ... | Use Tangent Line Method
$$6x^3-x^2 \geqslant \frac{5x-1}{8}, \quad \forall x > 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
compute $\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}$
The question is:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}
\quad \gamma:r(t)=(t,2t^2-2), \quad -1\leq t\leq 1
$$
I have tried to solve it like this:
since $Q_x=P_y$ it's potential vector field but the singularity is at origin, first i thought that i could evaluat... | $$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int_\gamma\frac{-y^2}{x^2+y^4}\,dx+\int_\gamma\frac{2xy}{x^2+y^4}\,dy$$
$$\gamma(t)=(t,2 t^2-2);\;t\in[-1,1];\;\gamma'(t)=(1, 4 t)$$
Substitute
$$\int_{-1}^1 -\frac{\left(2 t^2-2\right)^2}{\left(2 t^2-2\right)^4+t^2}\cdot 1\,dt+\int_{-1}^1 \frac{2 t \left(2 t^2-2\right)}{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inequality Induction Proof, how should I proceed? I have been trying to prove inequalities using induction to no avail.
For example,
Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$
Base Case:
$\frac{1}{2^{1}} = 2$
$\frac{1}{2}<2\;$ Which is true.
We assume, for n=k, that
$\... | I finally managed to prove it with induction, which was the goal.
I followed the suggestion to prove that $\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}} = 2-\frac{k+2}{2^{k}}\;$
for k+1 we have that
$\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}}+\frac{k+1}{2^{k+1}} = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
find the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.
Consider the number $A_{1395}=\underbrace{333\cdots3}_{1395}$. what is
the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.
$a) 2\quad\quad b)3\quad\quad c)4\quad\quad d)5\quad\quad e)6$
Here is my attempt :
We have $A_{1395}=3\times \und... | Write using the binomial theorem,
\begin{align}
999\cdots 9 &= 10^{1395} -1 \\
&= (1+9)^{1395} -1 \\
&= \left(\array{1395\\1}\right)\cdot 9 + \left(\array{1395\\2}\right)\cdot9^2 + \cdots 9^{1395}
\end{align}
But $1395 = 3^2 \cdot 5 \cdot 31$. So, the first term is divisible by $3^4$ and each term thereafter is divisi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.
I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not nega... | Hint: $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
The center of the circumcircle lies on a side of a triangle Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.
$$AB=\dfrac{144\sqrt5}{5}$$
By Triangle-Angle-Bisector Theorem $$\dfrac... | Alternative solution using neither trigonometry nor power of a point:
You already figured out that $OP \parallel AB$, so draw two perpendicular lines $OQ$ and $BR$.
Let the radii $OA = OB = OP = 10x$.
Since $\triangle ABC \sim \triangle OPC$, we have $AB = \frac{20+16}{20} \cdot OP = 18x$, so $AQ = BQ = 9x$.
Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Divergence of $\lim_{N\to +\infty}[ \int_{1}^{N+1} f(x) dx -\sum_{n=1}^{N}f(n)] $ Divergence of $\lim_{N\to +\infty}[ \int_{1}^{N+1} f(x) dx -\sum_{n=1}^{N}f(n)] $ where $f(x)= sin(log_e x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})$,$ 0<a<1/2$
My try -
$\lim_{N\to +\infty}[ \int_{1}^{N+1} sin(log_e x)(\frac{1}{x^{a}}-\frac{1... | *
*Integral $I(a)=\int_{1}^{N+1} \sin(\ln x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})dx$ can be evaluated - just make a change $x=e^t$:
$I(a)=\int_{0}^{\ln(N+1)} \sin(t)(e^{-ta}-e^{(a-1)t})e^tdt=\int_{0}^{\ln(N+1)} \sin(t)(e^{t(1-a)}-e^{at})dt$ and $\sin(t)=\frac{e^{it}-e^{-it}}{2i}= \Im({e}^{it})$
Integral with exponents... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$ I accidentally found out that the two integrals below
$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluat... | By the known formula for the difference of two inverse tangents, your last integral is
$$
\int_0^{ + \infty } {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} = \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} + \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Impossible integral?
$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$
I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:
$$\frac {d}{dx}\int_{0}^{... | You are correct that this does mean there are no continuous functions satisfying the equation. (There's a quicker way to see this: set $x=1$, then $LHS=0$, but $RHS=5$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Show that $2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$ Is there a way to show that (with integer $n>2$) $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$
I tried to figure a way with derivative or by looking how both side are growing, but I have some trouble with the ceiling function.
| With $n+k= {\lceil n \log_23\rceil}<n \log_23+1$ we can use the Rhin bound (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)
$$|(n+k) \log2 - n \log3|>\frac 1{(n+k)^{13.3}}>\frac1{(n \log_23+1)^{13.3}}>\frac 1{n^a}$$
If we choose $a=15$ the above is true for $n>41$
Now (growth of $x^a$ vs $a^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric system with different coefficients I have a trigonometric system as below and I want to solve it for $x$.
\begin{gather}
A\cos(x)+B\cos(2x)+C\sin(2x) = D \\
-C\cos(2x)-A\sin(x)+B\sin(2x) = F
\end{gather}
Can anyone suggest a solution?
Regards.
| Note $(1)$ the first equation and $(2)$ the second. Multiplying $(1)$ by $C$ and $(2)$ by $B$ the sum of two resulting equations give $$A\cos(\theta-x)+\cos(2x)=\frac{BD-CF}{B^2+C^2}=M$$ Similarly you get $$A\sin(\theta-x)+\sin(2x)=\frac{CD-BF}{B^2+C^2}=N$$ where $\sin(\theta)=\dfrac CB$. Now taking the square and addi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding minimum of value $x^2+y^2+z^2+t^2$
Suppose $x,y,z,t$ are real numbers which holds for these equations:
numbers:
${ \begin{cases}{x^2+6y+2t=2} \\ {y^2+2t+y=-14} \\
{z^2+4x+2y=-27} \\ {t^2+2z+y=5}\end{cases} }$
What is the minimum
value of $x^2+y^2+z^2+t^2$ ?
$$1)12\quad\quad\quad\quad\quad\quad2)18\quad\quad\... | I have both checked your solution, and confirmed it via WolframAlpha. No real solutions exist for this system of equations, and hence, the answer should be none.
Here is the link to the WolframAlpha computation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving multivariate quadratic equations over the integers I am looking for a method (if it exists) to solve over the integers the following sum of squares equation:
$$ x_1^2 + x_2^2+x_3^2 + \cdots + x_n^2 = m,$$ with $m \in \mathbb{N}.$
Someone has any idea about books, articles dealing with this kind of problem?
Than... | If $m\in\mathbb{N}$, then $x_n\in\mathbb{N}$ meaning that any $x$-values will work. If $m$ is a perfect square, the are also infinite solutions because the Pythagorean theorem works for any number of dimensions.
For every Pythagorean triple, there is a Pythagorean quadruple where $C$ of the triple is replaced by $(A,B)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficient of $x^{100}$ in $\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$ Find the coefficient of $x^{100}$ in $$\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$$
This problem came from finding total number of 4-partition of an integer 100.
One uses mathematica and it maybe easy, but I want to find more Mathematical proof.
f[... | Partial fractions:
$$\begin{split}
\frac{288x^4}{(1-x)(1-x^2)(1-x^3)(1-x^4)}
&=\frac{288x^4}{(1-x)^4(1+x)^2(1+x+x^2)(1+x^2)}\\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-x)^2} \\
&\quad + \frac{9}{(1+x)^2} - \frac{32}{1+x+x^2} + \frac{36}{1+x^2} \\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the whole number solutions for W+X+Y+Z = 15 where W, X, Y, Z ≤ 6 I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.
First, the number of non-negative integer solutions for $W+X+Y+Z = 15$ can be calculated using stars and bars:
... | There is a mistake in your last term. The answer should be $180$ and not $172$. Also you can simplify the working.
There are multiple ways to tackle the problem -
*
*Solve as is using P.I.E what you did
*Simplify using change of variable and solve
*Solve using generating function
a) Using first method (how you did... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.
By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 root... | Given $f(x)=x^5-5x^3+4x-1$, differentiate first:
$$ f'(x) = 5x^4 - 15x^2 +4 . $$
Setting the derivative equal to $0$, we have
$$ 0 = 5x^4 - 15x^2 +4 . $$
This is quadratic in $x^2$, so we can use the quadratic formula:
$$ x^2 = \frac{15 \pm \sqrt{145}}{10} . $$
Since $\sqrt{145}<15$, $x^2>0$, so our solutions are
$$ x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Inequality with quadratic and maximum Let $x,y$ be real numbers in $[0,1]$ such that $$x \ge y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}.$$ Is it true that $x^2+2y \ge 2x$?
In order to prove this, the term $x^2$ is currently on the "wrong" side, between the condition and the needed inequalit... | nope not true...
let $x = 0.6, y=0.4, \max\left\{(0.6)^2, \frac{2*0.6}{3}, \frac{1+0.6}{4}, \frac{1}{3}\right\}=\max\left\{0.36, 0.4, 0.4, 0.33\right\} =0.4\\$
$\therefore x \geq y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}$
holds when x = 0.6 and y = 0.4. Now check if $x^2+2y \geq 2x \implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.