Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Set of linear equations Find eigenvalues and eigenvectors of the matrix:
$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$
$\begin{pmatrix} 1-\lambda & 0 & -2 \\ 1 & 3-\lambda & -1 \\ -1 & 0 & 2-\lambda \end{pmatrix}$
I have found the eigenvalues, which are: 0 and 3.
For the 0-eig I have found thei... | First thing first. To find the eignvalues and the eigenvectors of a matrix you need to compute characteristic polynomial, which is given by the evaluation of
$$\mathcal{X}_A(\lambda) =det(\lambda - A) = det (A - \lambda) = 0$$
where $$ A=\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$$
Once you h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limits of $ \mathop {\lim }\limits_{n \to + \infty } \left( {1 + n + n\cos n} \right)^{\frac{1}{{2n + n\sin n}}} $ To calculate the limts
the First
$\mathop {\lim }\limits_{x \to 0^ + } \left( {2\sin \sqrt x + \sqrt x \sin \frac{1}{x}} \right)^x$
Suppose:
$\frac{1}{x} = n$
We find
$
\mathop {\lim }\limits_{x \to 0^ +... | For $n$ sufficiently large:
$$0\le\frac{1}{2n+n\sin n}\log(1+n+n\cos n)\le\frac{\log(3n)}{n}=\frac{\log3+\log n}{n}\xrightarrow{n\to\infty}0$$
hence the desired limit is $e^0=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simple algebra formula for which I can't find the right answer I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.
Somebody showed me how it's done:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cd... | Your step $y=\frac12 z^2+3z+2-z-1$ is wrong because $k(a+b)=ka+kb\neq ka+b$. A faster way to tackle the question is this.
$y+z+1=\frac12 (z+1)(z+2)$
$y=\frac12 (z+1)(z+2)-(z+1)$
$y=(z+1)(\frac 12 z+1-1)$
$y=\frac12 z(z+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Linear algebra, power of matrices $P^{-1}AP =
\begin{pmatrix}
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 2 \\
\end{pmatrix}
$
with
$P=
\begin{pmatrix}
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$
and $P^{-1}$ is the inverse of $P$
Find $... | Note that $B=P^{-1}AP=(-I_3+N)\oplus2$. Therefore $B^{100}=P^{-1}A^{100}P=(-I_3+N)^{100}\oplus2^{100}$. Since $N^3=0$, by binomial theorem,
$$(-I+N)^{100}=(-I)^{100}+\binom{100}1(-I)^{99}N+\binom{100}2(-I)^{98}N^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?
| You know that $(a-b)^2=30$. The same strategy tells you that $(a+b)^2=42$.
Thus
$$a+b=\pm \sqrt{42}\quad\text{and}\quad a-b=\pm\sqrt{30}.$$
Now by adding and subtracting, we can find $2a$ and $2b$. and hence $a$ and $b$. Note that there are $4$ combinations, though if we have found one solution $a=p$, $b=q$, the other... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Improper integral comparison test Having this integral
$$\int_1^{\infty}\frac{3x^2+2x +1}{x^3+6x^2+x+4}$$
In order to do the comparison test at some point it gets like
$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\geq \frac{1}{4x}$$
How is $\frac{1}{4x}$ found ? It doesnt seem obvious to me.
EDIT
Also how is this found
$$0\leq\fra... | For example, and taking into account the fact that for large values $\;x\;$ only the summands with the largest exponent of $\;x\;$ are important (in this sense), we get
$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\ge\frac{3x^2}{x^3+6x^3+x^3+4x^3}=\frac1{4x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions of $x^2-6x+\lfloor x \rfloor+7=0$
What are the roots of $x^2-6x+\lfloor x \rfloor+7=0$, where $\lfloor x \rfloor$ is the greatest integer function?
Is there some way to solve the equation without graphing?
| $x-1 < [x] \le x$, so $x^2-5x+6<0$ while $x^2-5x+7\ge 0$, which furthur reduces to $2<x<3$. We know $[x]=2$, so $x^2-6x+9=0$ and $x=3$. But this contradicts $x<3$ and hence no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Two ways to show that $\sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ Show that: $\large \sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ on: $0<x<\frac {\pi}2$
I tried to solve it in two ways and got a little stuck:
One way is to use Cauchy's MVT, define $f,g$ such that $f(x)=\sin x -x +\frac {x^3}{3!}$ and $g(x)=\frac ... | Repeat use of the Cauchy Mean Value theorem $3$ times we have to prove:
$\dfrac{sinx}{6x} < 1$. But since $x > 0$, we need to prove: $h(x) = sinx - 6x < 0$.
$h'(x) = cosx - 6 < 0$, so $h(x) < h(0) = 0$. Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/826015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Divergence theorem integrating over a cylinder Problem: Calculate $\int \int_S \langle F,n \rangle dS$ where $S$ is the half cylinder $y^2+z^2=9$ above the $xy$-plane, and $F(x,y,z) = (x,y,z).$
My working: I did this using a surface integral and the divergence theorem and got different results. First, using a surface i... | For the surface $z=h(x,y)=(9-y^2)^{\frac1{2}}$ the outward unit normal vector is
$$n = \Big(0,\frac{y}{3},\frac{(9-y^2)^{\frac1{2}}}{3}\Big)$$
and the differential surface area element is
$$dS = \sqrt{1+h_x^2+h_y^2}dxdy= 3(9-y^2)^{\frac{-1}{2}}dxdy$$
so the surface integral over the top is
$$\int F \cdot n dS= \int_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the indefinite integral $ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $ I have been having extreme difficulties with this integral. I would appreciate any and all help.
$$
\int \sqrt{\tan x} ~ \mathrm{d}{x}.
$$
| Let $I = \int\sqrt{\tan x}\;\mathrm{d}x$ and $J = \int\sqrt{\cot x}\;\mathrm{d}x$.
Now $$\begin{align}I + J
&= \int\left(\sqrt{\tan x} + \sqrt{\cot x}\right) \;\mathrm{d}x \\
&= \sqrt{2} \int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \;\mathrm{d}x \\[5pt]
&= \sqrt{2} \int\frac{(\sin x - \cos x)'}{\sqrt{1-(\sin x - \cos x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 7,
"answer_id": 1
} |
Applying Rouche's Theorem for an Annulus We hadn't really covered much of Rouche's Theorem within class, so I'm kind of asking a trial question to see if I understand it.
I've been asked to find out how many zeros of the function $z^5 + \frac{1}{8}\cdot e^z + 1$, counting multiplicity, that lie inside the annulus $\{ z... | Yes, this is a correct way to apply Rouché's theorem here.
However, the first of two applications could just as well be replaced by one-line argument with the reverse triangle inequality. Namely: when $|z|\le 1/2$,
$$\left|z^5 + \frac{1}{8}\cdot e^z + 1\right|\ge |1|-\left|z^5\right| - \left|\frac{1}{8}\cdot e^z \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix Power Formula
Prove that for a fixed $a \in \mathbb{R}$ we have the matrix power formula for all $n \in \mathbb{Z}_+$:
$$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix}$$
How would we prove this? Right now we are doing work with proofs by induction... but how ... | For $n = 1$ we should have (according to the rule)
$$
\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^1 = \begin{pmatrix}a^1 & 1\cdot a^0\\0 & a^1\end{pmatrix}
$$
but of course
$$
\begin{pmatrix}a^1 & 1\cdot a^0\\0 & a^1\end{pmatrix} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix}
$$
so it's correct (this is called the base case ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Show $ f(x,y)= 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - x\cos(\frac{1}{\sqrt{x^2 + y^2}}) \sqrt{x^2+y^2}^{-1}$ not continuous at $(0,0)$ $f(x,y) = 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}}
$
I'm a bit puzzled. The statement is obviously true if you plot the function.
For... | Let $x =r \cos(\phi), y = r\sin(\phi)$ then you have
\begin{align*}
\lim_{(x,y)\to(0,0)}f(x,y) &= \lim_{(x,y)\to(0,0)}2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}} \\
&= \lim_{r\to 0}2 r\cos(\phi) \sin\left(\dfrac{1}{r}\right)- \dfrac{r\cos(\phi)\cos\left(\dfrac{1}{r}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Problem about functional equations: $ f \left( x ^ 2 + y \right) = f ( x ) ^ 2 + \frac { f ( x y ) } { f ( x ) } $ This problem was taken from the Bulgarian selection team test for the 47th IMO and appeared in a Chinese magazine, I came across it in my own training.
http://www.math.ust.hk/excalibur/v10_n4.pdf
The probl... | This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).
Determine the value of $f(1)$
Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Equations of lines tangent to an ellipse Determine the equations of the lines that are tangent to the ellipse $\displaystyle{\frac{1}{16}x^2 + \frac{1}{4}y^2 = 1}$
and pass through $(4,6)$.
I know one tangent should be $x = 4$ because it goes through $(4,6)$ and is tangent to the ellipse but I don't know how to find th... | We have that $x^2+4y^2=16$, so using implicit differentiation gives $2x+8yy^{\prime}=0$ and therefore $\;\;\;\displaystyle y ^{\prime}=-\frac{x}{4y}$.
Since the slope of the line between $(x,y)$ and $(4,6)$ is given by $\frac{y-6}{x-4}$, you have that $\displaystyle -\frac{x}{4y}=\frac{y-6}{x-4}$.
This gives $-x^2+4x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Solve $y^{2/3}+(y')^{2/3}=1$ other than the direct method? Is there any way to solve
$$y^{2/3}+(y')^{2/3}=1$$
other than just solving for $y'$ and then integrate?
| $\cos^2 u + \sin^2 u = 1 \Rightarrow (\cos^3 u)^{2/3} + (\sin^3 u)^{2/3} = 1$
Let $y = \cos^3 u$ such that $y' = \sin^3 u$, if possible.
$y = \cos^3 u \Rightarrow\\
y' = -3 (\cos^2 u \sin u) u' \Rightarrow\\
-3 u'\cos^2 u \sin u = \sin^3 u \Rightarrow\\
3u' = -\tan^2 u \Rightarrow\\
\displaystyle \int -3 \cot^2 u\, du ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How does Wolfram get from the first form to the second alternate form? So, I was trying to compute an integral but I couldn't actually manage getting anywhere with it in its initial form. So, I inserted the function in Wolfram Alpha and I really got a nicer form (second alternate form). But I want to understand how tha... | One trick to help deriving the $2^{nd}$ alternate form is following identity:
$$\frac{1}{(z+a)(z+b)} = \frac{1}{b-a}\left(\frac{1}{z+a} - \frac{1}{z+b}\right)$$
In particular, when $a,b$ differs by an integer, it reduces to a pattern relatively easy
to detect and apply this trick. e.g.
$$\begin{align}
\frac{1}{(z+1)(z+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
finding the explicit function of a recursive sequence So I have the recursive sequence $f(0) = 0, f(n+1) = 2f(n)+ (n+1)^2$, and I'm not quite sure how to make it explicit. Substituting $n$ for $n+1$ cleans it up a little, yielding $f(n) = 2f(n-1) + n^2$, but I'm not quite sure what to do after that. It seems to be a lo... | Here is a generating functions solution.
Let
$$
F(x) = \sum_{n \ge 0} f(n) x^n.
$$
Then the recurrence $f(n) = 2 f(n-1) + n^2 \;\; (n \ge 1)$ becomes
$$
F(x) = 2xF(x) + \sum_{n \ge 1} n^2 x^n + f(0)
$$
or
$$
F(x) = 2xF(x) + \frac{x(1 + x)}{(1 - x)^3}.
$$
Solving,
$$
F(x) = \frac{x + x^2}{(1 - x)^3(1 - 2x)}
$$
The parti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Area of a triangle in terms of areas of certain subtriangles In triangle $ABC$ , $X$ and $Y$ are points on sides $AC$ and $BC$ respectively . If $Z$ is on the segment $XY$ such that $\frac{AX}{XC} = \frac{CY}{YB} = \frac{XZ}{ZY}$ , then how to prove that the area of triangle $ABC$ is given by $[ABC]=([AXZ] ^{1/3} + ... | We can say that points $X$, $Y$, $Z$ separate respective segments $AC$, $CB$, $XY$ into sub-segments with ratios $b:r$ (think "blue : red"), where $b+r=1$.
Let's add a few perpendiculars, transfer our ratio-colors accordingly (as the reader may verify, using similar triangles), and compute the area of $\triangle AXZ$:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/836875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit Set of All Real Numbers Not even sure how to start this one. Does anyone know how to do this?
Prove that there exists a sequence such that its limit set is the set
of all real numbers
Limit set is the set of all subsequence limits btw
| Start the sequence with $0$.
Then write $-1,1$.
Then write $-\frac{4}{2},-\frac{3}{2},-\frac{1}{2}, \frac{1}{2},\frac{2}{2},\frac{3}{2},\frac{4}{2}$.
Then write $-\frac{16}{4},-\frac{15}{4}, -\frac{13}{4},\dots, -\frac{1}{4},\frac{1}{4},\dots, \frac{16}{4}$.
Then write $-\frac{64}{8}, \frac{63}{8},\dots, \frac{63}{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/839227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find length of $CD$ where $\measuredangle BCA=120^\circ$ and $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$ $ABC$ is a triangle with $BC=a,CA=b$ and $\measuredangle BCA=120^\circ$. $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$. Then the length of $CD$ is ____ ?
A)$\frac{a+b}{4}$
B)$... | Use the sine law. Let $x=BD$ and $y=AD$ and $z=CD$ which is what we want to find.Then we have in triangle $ABC$,
$$\frac{\sin 120}{x+y}=\frac{\sin A}{a}=\frac{\sin B}{b}$$ and so
$$\sin A=\frac{\sqrt{3}}{2}\frac{a}{x+y}$$
$$\sin B=\frac{\sqrt{3}}{2}\frac{b}{x+y}$$
Next in triangles $ACD$ and $BCD$ you have
$$\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For an odd prime $p$, $\;p =1 \pmod 4,\; \;x^2+1 = 0 \pmod p$ $$\text{ If } p \neq 2 \text{ is a prime, we know that: }$$
$$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$$
According to this,prove that:
$$p \equiv 1 \pmod 4 \Rightarrow x^2+1\equiv 0 \pmod p \text{ has a solution}$$
$$\te... | That seems correct to me. The last part comes from the fact that the equation $x^2 +1 \equiv 0$ (mod $p$) has at most 2 distinct solutions because it is a degree 2 polynomial and $p$ is prime. And if $x$ is a solution, then so is $-x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/843519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Infinite Sum of algebraic expression Prove that $$\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} = \frac89 -\frac23\ln2$$
I tried using integration but failed miserably. Hints please.
| You can do it by integration too.
$$\begin{aligned}
\frac{2}{3}\sum_{i=1}^{\infty} \left(\frac{1}{2i}-\frac{1}{2i+3}\right) &=\frac{2}{3}\int_0^1 \sum_{i=1}^{\infty} \left(x^{2i-1}-x^{2i+2}\right)\,dx\\
&=\frac{2}{3}\int_0^1 \left(\frac{1}{x}-x^2\right)\frac{x^2}{1-x^2}\,dx=\frac{2}{3}\int_0^1 \frac{(1-x)(1+x+x^2)x}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/843669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Square roots modulo powers of $2$ Experimentally, it seems like every $a\equiv1 \pmod 8$ has $4$ square roots mod $2^n$ for all $n \ge 3$ (i.e. solutions to $x^2\equiv a \pmod {2^n}$)
Is this true? If so, how can I prove it? If not, is it at least true that the maximum (over $a$) number of square roots of $a$ mod $2^n... | If $n\ge3$ and $a=8k+1$ then $a$ has exactly four distinct square roots modulo $2^n$.
Proof that there is at least one square root: use induction. The case $n=3$ is easy to check; if $x^2\equiv a\pmod{2^n}$ then $x^2=a+2^nk$ for some integer $k$; it is easy to see that $x$ is odd and so
$$(x+2^{n-1}k)^2=a+2^nk(1+x)+2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
How to use mathematical induction with inequality? I am stuck with this question.
Given that $n$ is a positive integer where $n≥2$, prove by the method of mathematical induction that
(a) $$ \sum_{r=1}^{n-1} r^3 < \frac{n^4}{4} $$
(b) $$ \sum_{r=1}^{n} r^3 > \frac{n^4}{4} $$
| Hint for the first question:
*
*Assume $\displaystyle\sum\limits_{r=1}^{n-1}r^3<\frac{n^4}{4}$
*Prove $\displaystyle\sum\limits_{r=1}^{n}r^3<\frac{(n+1)^4}{4}$
$\displaystyle\sum\limits_{r=1}^{n}r^3=$
$\displaystyle\sum\limits_{r=1}^{n-1}r^3+n^3<$
$\displaystyle\frac{n^4}{4}+n^3=$
$\displaystyle\frac{n^4+4n^3}{4}<$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
When is the sum of two squares the sum of two cubes When does $a^2+b^2 = c^3 +d^3$ for all integer values $(a, b, c, d) \ge 0$. I believe this only happens when: $a^2 = c^3 = e^6$ and $b^2 = d^3 = f^6$. With the following exception:
*
*$1^3+2^3 = 3^2 + 0^2$
Would that statement be correct?
Is there a general formu... | Let $c=x^2,d=y^2,i=\sqrt{-1}$, then $$c^3+d^3=x^6+y^6=(x^3-(yi)^3)(x^3+(yi)^3)\\
=(x-yi)(x^2+xyi-y^2)(x+yi)(x^2-xyi-y^2)$$
Let $a+bi=(x-yi)(x^2-xyi-y^2),a-bi=(x+yi)(x^2+xyi-y^2)$, then $a^2+b^2=c^3+d^3$.
We get $a=x^3-2xy^2,b=y^3-2x^2y,c=x^2,d=y^2.$
If $a<0$ or $b<0$, we can take the absolute value.
For example, let $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Limit with cube root conjugate Calculate the limit as $n -> \infty$ of $$n(\sqrt[3]{(1+1/n)} -1)$$
Summary of my work:
*
*Put n into the denominator in the form of $1/n$
*Multiplied by $n/n$
*Multiplied by conjugate of numerator: $[\sqrt[3]{(n^3 + n^2)^2} + 2\sqrt[3]{n^3 +n^2} + n^2]/[\sqrt[3]{(n^3 + n^2)^2} + 2\s... | $$n\left(\sqrt[3]{1+\frac{1}{n}}-1\right)=n\left(\frac{\frac{1}{n}}{\sqrt[3]{(1+\frac{1}{n})^{2}}+\sqrt[3]{1+\frac{1}{n}}+1}\right)$$
which tends to $\frac{1}{3}$ as $n\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple Differentiation Problem Involving Area Radius and Circumference A stone is dropped into a pool of water, and the area covered by the spreading ripple increases at a rate of $4 m^2 s^{-1} $.
Calculate the rate at which the circumference of the circle formed is increasing 3 seconds after the stone is dropped.
My m... | $$\begin{align} & \cfrac{dA}{dr} = 2 \pi r \\ & \cfrac{dC}{dt} = 2\pi r \\ & \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\ & 4 = 2 \pi r \times \cfrac{dr}{dt} \\ & \cfrac{dr}{dt} = \cfrac{4}{2\pi r} \\ & \cfrac{dr}{dt} = \cfrac{2} {\pi r} \\ & r dr = \cfrac{2}{\pi } dt \\ & \int r dr = \int \cfrac{2}{\pi } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $ x^2+y^2+z^2 =1$ for $x,y,z \in \mathbb{R}$, then find maximum value of $ x^3+y^3+z^3-3xyz $. If $ x^2+y^2+z^2 =1$, for $x,y,z \in \mathbb{R}$, what is the maximum of
$ x^3+y^3+z^3-3xyz $ ?
I factorize it... Then put the maximum values of $x+y+z$ and min value of $xy+yz+zx$...
But it is wrong as they don't hold si... | Let $f(t)=\frac{t}{2}(3-t^2)$. It is straightforward to check that $f$ attains its maximum on the interval $[-\sqrt{3},\sqrt{3}]$ at $t=1$, with $f(1)=1$.
Now, if $x^2+y^2+z^2=1$ then
$$\eqalign{
x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\cr
&=\frac{ x+y+z }{2}\left(3(x^2+y^2+z^2)-(x+y+z)^2\right)\cr
&=f(x+y+z)
}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ et... | By Cauchy-Schwarz $$\begin{align}\sum_{cyc}\frac{a}{b+c}&=\sum_{cyc}\left(\frac{a}{b+c}+1\right)-3\\
&=\sum_{cyc}a\sum_{cyc}\frac{1}{b+c}-3\\
&=\frac{1}{2}\sum_{cyc}(b+c)\sum_{cyc}\frac{1}{b+c}-3\\
&\geq\frac{1}{2}\cdot9-3=\frac{3}{2}.\end{align}$$
Done!
Another way.
By C-S $$\sum\limits_{cyc}\frac{a}{b+c}=\sum\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
} |
Proof of $\sin2x+x\sin^2x \lt\dfrac{1}{4}x^2+2$ How can be proven the following inequality?
$$\forall{x\in\mathbb{R}},\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$$
Thanks
| For $x<0$ inequality is obvious.
$\displaystyle(\frac{x}{2}-1)^2=\frac{x^2}{4}-x+1 \geq 0$, so
$\displaystyle \frac{x^2}{4}+2 \geq x+1$, but for $x \geq 0$
$x \geq x\ (\sin x)^2$ and $1 \geq \sin 2x$
If you want to have $\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$ instead of $\left[\sin(2x)+x\sin(x)^2\right]\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/856237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
System of quadratic equations with three variables The problem is as follows:
For $x,y,z \in R$,
$$
\left\{
\begin{array}{l}
x^{2} -yz-8x+7=0 \\
y^{2}+z^{2}+yz-6x+6=0
\end{array}
\right.
$$
What is the domain of $x$?
One way to solve this is to use another variable. This is shown in this answer. What other ways are ... | We have that $yz=x^2-8x+7$ and, substituting into the 2nd equation, $y^2+z^2=6x-6-yz=6x-6-(x^2-8x+7)=-x^2+14x-13$.
Since the system of equations $y^2+z^2=a$ and $yz=b$ has a solution iff $a\ge2b$ and $a\ge-2b$,
as shown below,
the given system has a solution iff
1) $-x^2+14x-13\ge2(x^2-8x+7)$ and 2) $-x^2+14x-13\ge -2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Is $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ integrally closed?
Is $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ integrally closed? (Or could it have a relation to another domain like $\mathbb{Z}[\sqrt{-3}]$ does with $\mathbb{Z}[\omega]$?) Also, is it UFD? What are its units?
I have never read about this domain in any book, though i di... | Interestingly enough this is not integrally closed. (spelling not corrected)
We determine the rings of integers of
$\mathbb{Z}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$
We prove first that
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \in \mathbb{Z}[\sqrt{2}+\sqrt{3}]$$ if and only if $b$ and $c$ have the same pari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/859448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Finding the conditions of (x,y,z,t) for them to belong to the span of a set of vectors So I got this math exercise, and I don't know how to go about it:
In $\mathbb{R}^4$, $S$ is the subspace spanned by the following set of vectors: $(1, 1, 1, 0) , (1, 2, 1, 1) , (2, 0, 1, 1) , (3, 0, 4, 2)$
Find the condition (or a s... | If a vector is in the subspace spanned by those four vectors, then we have:
$$\begin{pmatrix} x\\ y\\ z\\ t\\
\end{pmatrix} =\alpha \begin{pmatrix}1\\1\\1\\0\\ \end{pmatrix}+\beta \begin{pmatrix}1\\2\\1\\1\\ \end{pmatrix} + \gamma \begin{pmatrix}2\\0\\1\\1\\\end{pmatrix} + \delta \begin{pmatrix}3\\0\\4\\2\\\end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/863266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the derivative of sinus and cosinus. Trigonometric identities How can we see that $$\sin(x+h)-\sin(x)=2\sin\left(\frac h2\right)\cos\left(x+\frac h2\right)$$
How can we see that $$\cos(x+h)-\cos(x)=-2\sin\left(\frac h2\right)\sin\left(x+\frac h2\right)$$
Do these identities have a name?
| For any $a,b$, we have the well known "addition theorems"
$$ \sin(a+b) = \sin a \cos b + \sin b \cos a$$
and $$ \sin(a-b) = \sin a \cos b - \sin b \cos a $$
Subtracting these two equations, we get
$$ \sin(a+b) - \sin(a-b) = 2\sin b \cos a $$
For the cosine, we have
$$ \cos(a+b) = \cos a \cos b - \sin a \sin b $$
and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\alpha, \beta \in [0,\pi]$ then the minimum value of $\sin(\frac{\alpha +\beta}{2})$ is... Problem : If $\alpha, \beta \in [0,\pi]$ then the minimum value of $\sin(\frac{\alpha +\beta}{2})$ is
a) $\frac{\sin\alpha +\sin\beta}{2}$
b) $|\sin\alpha -\sin\beta|$
c) $\frac{\cos\alpha +\cos\beta}{2}$
d) $|\cos\al... | Using Werner Formula,
$$2\sin\frac{A+B}2\cos\frac{A-B}2=\sin A+\sin B$$
But,
$$2\sin\frac{A+B}2\cos\frac{A-B}2\le2\sin\frac{A+B}2$$ as for $\displaystyle A,B\in[0,\pi]; 0\le \cos\frac{A-B}2\le1$(why?)
$$\implies2\sin\frac{A+B}2\ge\sin A+\sin B$$
Please try Werner formula, with $\displaystyle2\sin\frac{A+B}2\sin\frac{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving 7n+5 is never a cubic number? This is from a question that starts with:
An arithmetic progression of integers an is one in which $a_n=a_0+nd$, where $a_0$ and $d$ are integers and n takes successive values $0, 1, 2, \cdots$ Prove that if one term in the progression is the cube of an integer there will be an inf... | This is quite a quick thing to prove with modular arithmetic, but I will avoid that and use first principles. Suppose $m^3 = 7n+5$. There are seven cases:
*
*$m=7k$
*$m=7k+1$
*$m=7k+2$
*$m=7k+3$
*$m=7k+4$
*$m=7k+5$
*$m=7k+6$
So basically, $m=7k+i$ for $i$ in $\{0,1,\ldots,6\}$.
Now
$$\begin{align}
(7k+i)^3 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Closed form for the sum: $\sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)}$ I tried using partial fractions to compute the sum of the series
$$
\sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)}
$$
Another technique is to turn this series into a definite integral of 0 to 1. but do not know how to do.
Thanks for any help.
| Note that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n(n+1/3)} = \sum_{n=1}^{\infty}\dfrac{9}{3n(3n+1)} = 9\sum_{n=1}^{\infty}\left(\dfrac{1}{3n}-\dfrac{1}{3n+1}\right)$.
Now, let $f(x) = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{x^{3n}}{3n}-\dfrac{x^{3n+1}}{3n+1}\right)$. Clearly, $f(0) = 0$ and our sum is $9f(1)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Perfect squares and modularly congruency in mod 5 There is not any perfect square $k$ such that $k \equiv 3\ (\textrm{mod}\ 5)$? Why? How can I prove it?
| You can check the 5 cases:
If $n \equiv 0 \pmod{5} \Rightarrow n^2 \equiv 0 \pmod{5}$
If $n \equiv 1 \pmod{5} \Rightarrow n^2 \equiv 1 \pmod{5}$
If $n \equiv 2 \pmod{5} \Rightarrow n^2 \equiv 4 \pmod{5}$
If $n \equiv 3 \pmod{5} \Rightarrow n^2 \equiv 4 \pmod{5}$
If $n \equiv 4 \pmod{5} \Rightarrow n^2 \equiv 1 \pmod{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/870314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
| $n^{4}-1 = (n^{2}-1)(n^{2}+1)$ and (mod $5$) we have $n^{2}+1 \equiv n^{2} - 5n +6 = (n-2)(n-3),$
while also (mod $5$), we have $n^{2}-1 = (n-1)(n+1) \equiv (n-1)(n-4).$ Hence we have $(n^{4}-1) \equiv (n-1)(n-2)(n-3)(n-4) $ (mod $5$), and as long as $n$ is not divisible by $5$, one of $n-1,n-2,n-3,n-4$ is divisible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 15,
"answer_id": 9
} |
How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?
| By induction.
Assume without loss of generalization that $A \leq B$
Note that $A^{n-1}(B-A) \leq B^{n-1}(B-A) \Rightarrow A^{n-1}B - A^n \leq B^n - AB^{n-1} \Rightarrow AB^{n-1} + A^{n-1}B) \leq A^n + B^n$
For $p = 1$ this is obvious. Supose true for all $p < n$.
$(A+B)^n = (A+B)^{n-1}(A+B) \leq 2^{n-1}(A^{n-1}+B^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ the... Problem :
If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ then c +d equals
(a) 60
(b) 50
(c) 40
(d) 30
Solution :
Equation of common chord of the circles is given ... | The Common chord has equation $6x+14y+c+d=0$ and it must pass through the center $(1,-4)$ of $S'$. Thus $c+d=50$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/873506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Can you show me where the "1/2" comes from? I'm struggling with a math assignment:
$$\frac12 \cos(x)·(3+2\sin(2x))−\cos(x)=0 ⇔ \cos(x)\left(\color{red}{\frac12}+\sin(2x)\right)=0$$
According to my knowledge it needs to be:
$$\frac12 \cos(x)·(3+2\sin(2x))−\cos(x)=0 ⇔ \cos(x)\left(\color{red}{\frac32}+\sin(2x)\right)=0$... | \begin{align*}
\frac{1}{2} \cos(x)( 3 + 2 \sin(2x)) - \cos(x)
&= \color{blue}{\frac{3}{2} \cos(x)} + \cos(x) \sin (2x) \color{blue}{-\cos(x)} \\
&= \color{blue}{\frac{1}{2} \cos(x)} + \cos(x) \sin (2x) \\
&= \cos(x)(\frac{1}{2} + \sin(2x))
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/873972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove this $3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}}$ Question:
show that:
$$3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}},n\geq 1$$
My idea: since
I have prove
$$5^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
$$3^{2^n}-1\equiv 0\pmod {2^{n+2}}$... | This is not even true. Just try with n=0
$3^1 = (-5)^{1/2}$ (mod 16) makes no sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
| $\sin x+\cos x=\frac12$ $\Rightarrow$
$(\sin x+\cos x)^2=\frac14$ $\Rightarrow$
$\sin^2 x+2\sin x\cos x+\cos^2 x=\frac14$ $\Rightarrow$
$1+2\sin x\cos x=\frac14$ $\Rightarrow$
$2\sin x\cos x = -\frac34$ $\Rightarrow$
$\sin x\cos x=-\frac38$.
If $a=\sin x$, $b=\cos x$, we have the system of equations
$$
\begin{align}
a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 1
} |
taylor series of $\ln(1+x)$? Compute the taylor series of $\ln(1+x)$
I've first computed derivatives (up to the 4th) of ln(1+x)
$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{... | Start with the Taylor series of the derivative
$$\frac 1{1+x}=\frac 1{(1+a)+(x-a)}=\frac 1{1+a}\,\,\frac 1{ 1+\frac {x-a}{1+a}}$$ Let $t=\frac {x-a}{1+a}$
$$\frac 1{1+t}=\sum_{n=0}^\infty (-1)^n t^n$$ So
$$\frac 1{1+x}=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \left(\frac{x-a}{a+1}\right)^n$$
Now, integrate
$$\log(1+x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sin(12^\circ)\sin(48^\circ)\sin(54^\circ)=\frac18$ Prove that $$\sin(12^\circ)\sin(48^\circ)\sin(54^\circ)=\frac18.$$ Without using a calculator. I tried all identities I know but I have no idea how to proceed. I always get stuck on finding $\sin36^\circ$.
| $2\sin 12\cdot \sin 48 = \cos (48 - 12) - \cos (48 + 12) = \cos 36 - \dfrac{1}{2}$. So the problem is to find $\cos 36$. Use $1 - 2x^2 = 3x - 4x^3$ to solve for $\sin 18$ ( not hard ), then find $\cos 36$, and $\sin 54$. In fact, the equation is: $4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$, so $\sin 18 = x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How prove $x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$
let $x,y,z\ge 0$,and such
$$x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$$
Find the maximum of the $C$
witout loss of we assume that $$x+y+z=1$$
I think
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)=1-3(yz+xz+xy)$$
then $$(x-y)(y-z)(x-z)... | this might be helpful
$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? I want to show that:
$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$
That's what I have tried:
*
*$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \i... | Since the function $f(x)=\{x\}=x-\lfloor x\rfloor$ is periodic with period one, the function:
$$ g(n) = \left\lfloor\frac{n}{2}\right\rfloor+\left\lceil\frac{n}{2}\right\rceil -n$$
is periodic with period $2$. Since $g(0)=g(1)=0$, $g(n)=0$ for every $n\in\mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
The limit of $((1+x)^{1/x} - e+ ex/2)/x^2$ as $x\to 0$ $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
*
*by directly substituting $x=0$ i got $\infty$
*by using L-H's rule, i got $-1/8$
the given options are
$a)\frac{24e}{11}$
$b)\frac{11e}{24}$
$c)\frac{e}{11}$
$d)\frac{e}{24}$
may b... | We will use three limits derived using L'Hospital:
$$
\lim_{x\to0}\frac{1-e^x}{x}=-1\tag1
$$
and
$$
\lim_{x\to0}\frac{1+x-e^x}{x^2}=-\frac12\tag2
$$
and
$$
\lim_{x\to0}\frac{1+x+\frac12x^2-e^x}{x^3}=-\frac16\tag3
$$
Now
$$
\begin{align}
&\frac{(1+x)^{1/x}-e+\frac{ex}2}{x^2}\\[9pt]
&=\frac{e\left(1+\frac{1+x-e^x}{e^x}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Coordinate Geometry of circles; Radical Axis question
If one of the diameters of the circle $x^2+y^2-2x-6y+6=0$ is a chord to the circle with center at $(2, 1)$, then the radius of the second circle is?
Apparently the solution is $3$, with the cryptic reasoning that $r^2= 2^2+ (3−1)^2+ (2−1)^2 ⇒ r =3$.
I have no idea... | Explanation of cryptic solution.
First let us rewrite the given equation in standard form:
$$x^2+y^2-2x-6y+6=0\implies (x-1)^2+(y-3)^2=4$$
Now we know the original circle has radius $r=2$ and is centered at $(1,3)$
Now we can make a convenient right triangle with $r$ the radius of the original circle, $d$ the distance ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A Fixed Field of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ I was working on a review problem from Dummit and Foote and came across the following issue. It is clear that the Galois group of the splitting field for the polynomial $(x^2-2)(x^2-3)(x^2-5)$ has order $8$ and one of the automorphisms is
$$
a+b\sqrt{2}+c\sqrt{3... | Under this automorphism, $\sqrt{2}\mapsto -\sqrt{2}$ and $\sqrt{3}\mapsto -\sqrt{3}$, so that $\sqrt{6}\mapsto\sqrt{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/881036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{\Gamma(\frac 1 3)^2}{\Gamma(\frac 1 6)}=\frac{\sqrt {\pi}\sqrt[3] 2}{\sqrt 3}$. Show that $$\frac{\Gamma(\frac 1 3)^2}{\Gamma(\frac 1 6)}=\frac{\sqrt {\pi}\sqrt[3] 2}{\sqrt 3}$$
Since there's $\sqrt {\pi}$, I suspect I have to related it to $\Gamma(1/2)$. Please give me some idea.
| It's the Legendre duplication formula
$$\Gamma (2z) = \frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\tfrac{1}{2}\right),$$
and Euler's reflection formula
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin (\pi z)}.$$
Thus we have
$$\Gamma\left(\tfrac{1}{3}\right)^2 = \underbrace{\frac{2^{-2/3}}{\sqrt{\pi}}\Gamma\left(\tfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Line parallel to a plane and have 45 degrees between another I need to find a direction vector for a line parallel to a plane $x+y+z = 0$ and that have $45$ degrees with the plane $x-y = 0$
So, i've assumed the vector $\vec V_r = (a,b,c)$ and since it is parallel to the first plane, the product:
$$(a,b,c)\cdot(1,1,1) =... | After "cross multiplying" and canceling the factor of $\sqrt{2}$ on each side, we have
$$
\sqrt{a^2 + b^2 + c^2} = |a - b|.
$$
Now, square both sides and remove $a^2$ and $b^2$ terms that appear on both sides, yielding
$$
c^2 = -2ab.
$$
From the original equation, we know that $a + b + c = 0$, so
$$
b = -a - c.
$$
At t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
| Hint By the monotonicity of $x \mapsto \frac{1}{x}$, we have
$$\frac{1}{n} + \dots + \frac{1}{2n} = \sum_{k=n}^{2n} \frac{1}{k} \geq \int_n^{2n} \frac{1}{x} \, dx.$$
A direct calculation of the latter integral yields an even sharper bound:
$$\frac{1}{n} + \dots + \frac{1}{2n} \geq \ln 2 > \frac{7}{12}.$$
For the last ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 0
} |
How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear... | $$
\begin{array}{ccccccccc}
& & 1 \\ \\
x^2 + 1 & ) & x^2 & - & 3 \\
& & x^2 & + & 1 \\ \\
& & & & -4 & \longleftarrow\text{remainder}
\end{array}
$$
The quotient is $1$ and the remainder is $-4$.
So $\displaystyle \frac{x^2-3}{x^2+1} = 1 - \frac{4}{x^2+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
How to prove this $(n+1)^n < n^{n+1}$ for $\space n \ge 3$ I'm having some more trouble with induction
I know how to prove this using $\ln$, but I need to use induction only.
prove that:
$(n+1)^n < n^{n+1}$
for any $ n\ge 3$
| Hypothesis
\[ n^{n+1}>(n+1)^{n},\ \mbox{for}\ n\ge 3\]
Basis
\[ 3^{3+1} > (3+1)^{3} \]
\[ 3^{4} > 4^{3} \]
\[ 81 > 64\]
Induction
\[
(n+1)^{n+2}=n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right]
\]
Now via the induction hypothesis, we have
\[
n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] > (n+1)^{n}\left[\frac{(n+1)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2... | You are doing the right thing. Just be careful at the end. Remember that: $$(a^b)^c = a^{bc}$$
So $$4^{2x - 2} = (4^2)^{x-1} = (2^4)^{x - 1} = 2^{4x - 4}$$This happens because $2^4 = 4^2 = 16$. In general $a^b \neq b^a$. And you can write $\frac{1}{2^{x - 2}} = 2^{2 - x}$. This gives: $$2^{4x - 4} \cdot 2^{2 -x} = 2^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Proof: Question:
If $m\cos\alpha-n\sin\alpha=p$ then prove that $m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$
My Efforts:
$(m\cos\alpha-n\sin\alpha)^{2}=p^2$
$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$
Now i think we need to add something on both side but i can't figure out what to ... | From
$$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$
we can add $m^2\sin^2\alpha+n^2\cos^2\alpha$ to both sides (to make use of the $\sin^2x+\cos^2x=1$ identity) to achieve
$$m^2(\sin^2\alpha+\cos^2\alpha)+n^2(\sin^2\alpha+\cos^2\alpha)-2mn\cos\alpha\ \sin\alpha=p^2+m^2\sin^2\alpha+n^2\cos^2\alpha
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$ Prove that (please)
$$\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$$
I've tried using Taylor series and I ended up with
$$-\sum_{m=0}^\infty\sum_{n=1}^\infty\frac{2}{n(m+n+1)^3}$$
I am stuck there and I couldn't continue it using partial fr... | The answer by user111187 is the cleanest way to approach such problem. I will elaborate on the sums (which he doesnt show).
$$I = (-)\cdot\int_{0}^{1} \frac{\log(1-x)\log^2(x) dx}{1-x}$$
Let $u = 1-x \implies du = -dx$
$$I = -\int_{0}^{1} \frac{\log(u)\log^2(1-u) du}{u}$$
$$\sum_{n=1}^{\infty} H_n u^n = -\frac{\log(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/889076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 2
} |
condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$ As the title says, what would be the condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$? Would there be infinitely many tuples that satisfy the condition above?
| Suppose that there is a set of non-zero integers $(a,b,c,d)$.
First, since $b\not=0$, we have
$$ac=bd\Rightarrow d=\frac{ac}{b}\tag1$$
Moreover, we have
$$ad+bc=ac-bd\Rightarrow (a+b)d=c(a-b).$$
If $a+b=0$, then we have $a-b=0$ because $c\not=0$. This leads $a=b=0$. This is a contradiction. Hence, we have $a+b\not =0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\frac{1}{a}\int_0^\infty{x^2}e^{-\frac{x^2}{2a}}\,dx$ Evaluate the following integral:
$$\frac{1}{a}\int_0^\infty{x^2}e^{-\large\frac{x^2}{2a}}\,dx.$$
| $$\frac{1}{a}\int_{0}^{\infty}x^{2}e^{-\frac{x^{2}}{2a}}dx=-\int_{0}^{\infty}x(e^{-\frac{x^{2}}{2a}})'dx$$
Use integration by parts and the value of Gaussian integral to evaluate:
$$=-(xe^{-\frac{x^{2}}{2a}})\vert_{0}^{\infty}+\int_{0}^{\infty}e^{-\frac{x^{2}}{2a}}dx=\sqrt{2a}\int_{0}^{\infty}e^{-(\frac{x}{\sqrt{2a}})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerat... | I think you are correct. The Wolfram answer is only defined for $x>0$, whereas $\dfrac{\sqrt{x^2+1}}{x}$ has an antiderivative for $x<0$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Divisor function asymptotics Define $\tau_{r}(n) = \sum_{d_1...d_r = n}1$.
One exercise in a book on sieve theory asked for an elementary proof by induction of the fact that
$$\sum_{n\le x}\tau_r(n) = \frac{1}{(r - 1)!}x(\ln x)^{r - 1} + O\left(x(\ln x)^{r - 2}\right)$$
The base case $r = 2$ is easy with reversing th... | We can write
$$\tau_{r+1}(n)= \sum_{d\mid n} \tau_r(d),$$
which leads to your
$$\sum_{n\leqslant x} \tau_{r+1}(n) = \sum_{d\leqslant x} \left\lfloor \frac{x}{d}\right\rfloor \tau_r(d),$$
but we can also write
$$\tau_{r+1}(n) = \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right),$$
and that gives us
$$\begin{align}
\sum_{n\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{(xy)^7}{x^8+(xy)^7+y^8}+\frac{(yz)^7}{y^8+(yz)^7+z^8}+\frac{(zx)^7}{z^8+(zx)^7+x^8}\leq1$
If $x,y,z$ are positive real numbers that $xyz=1$ , Prove
a) $\frac{xy}{x^8+xy+y^8}+\frac{yz}{y^8+yz+z^8}+\frac{zx}{z^8+zx+x^8}\leq1$
b)$\frac{(xy)^7}{x^8+(xy)^7+y^8}+\frac{(yz)^7}{y^8+(yz)^7+z^8}+\frac{(zx)^7}{z^8+(... | After the first step it will become
$\frac{(xy)^6}{x^6+y^6+(xy)^6}=\frac{1}{\frac{1}{y^6}+\frac{1}{x^6}+1}$
After set as I said it will be
$\frac{1}{a^6+b^6+1}$
Now you just need to continue what you do in a)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
| First recall that
\[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \]
And
\[ \frac{a}{\left(\frac{c}{d}\right)} =\frac{ad}{c} \]
Then
\[
\frac{2x+1}{\frac{3}{x^{2}}+\frac{2x+1}{x}}= \frac{2x+1}{\frac{3x+x^{2}(2x+1)}{x^{3}}}= \frac{x^{3}(2x+1)}{3x+x^{2}(2x+1)}= \frac{x^{3}(2x+1)}{2x^{3}+x^{2}+3x}= \frac{x^{3}(2x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Any idea how to linearize this equation? $X^2-Y^2=aZ+bZ^2$ The intention is to linearize this equation $X^2-Y^2=aZ+bZ^2$ into something which looks like $Z=mX+nY+c$ so that a graph of $Z$ against $X$ or $Y$ can be plotted.
X,Y,Z are variables while a,b,c are constants
| EDIT: Thanks to dylan37 for pointing out that the this is probably not the sort of "linearization" that curiousfurious (the OP) intended.
Maybe you could use the trick that's employed in deriving the Quadratic Formula.
Starting from $ X^{2}-Y^{2}=aZ+bZ^{2}$, first divide both sides by $b$ to get
$\frac{1}{b}X^{2}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the kernel of linear transformation Linear transformation $l:\mathbb{R}^3 \mapsto \mathbb{R}$ is determined as follows: $l(1,0,0)=1$; $l(1,4,0)=-1$; $l(0,0,1)=1$.
I need to find $\text{Ker}(l)$.
Answer should be $\text{Span}\{(0,2,1),(1,2,0)\}$.
But my answer is $\text{Span}\{(1,4,1),(2,4,0)\}$.
I tried to solve t... | Both answers are correct. That is $\mathcal{A}=((0,2,1)^T,(1,2,0)^T)$ and $\mathcal{B}=((1,4,1)^T,(2,4,0)^T)$ are both basis of the Kernel. Indeed for the basis $\mathcal{A}$ we have
$$\begin{pmatrix}0\\2\\1\end{pmatrix} = -\frac12\begin{pmatrix}1\\0\\0\end{pmatrix}+\frac12\begin{pmatrix}1\\4\\0\end{pmatrix}+\begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Number of solutions of $x_1 + x_2 + x_3 + x_4 = 14$ such that $x_i \le 6$
Let $x_1, x_2, x_3, x_4$ be nonnegative integers.
(a) Find the number of solutions to the following equation:
$$ x_1 + x_2 + x_3 + x_4 = 14 $$
I got $17 \choose 3$ for this. Is that correct?
(b) Find the number of solutions if... | Use generating functions. The ways to pick each $x_i$ is represented by:
$$
1 + z + z^2 + z^3 + z^4 + z^5 + z^6 = \frac{1 - z^7}{1 - z}
$$
Now you want four of those, and have them add up to $n$:
$\begin{align}
[z^n] \frac{(1 - z^7)^4}{(1 - z)^4}
&= [z^n] (1 - 4 z^7 + 6 z^{14} - 4 z^{21} + z^{28})
\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of the multiples of $3$ and $5$ below $709$? I just cant figure this question out:
Find the sum of the multiples of $3$ or $5$ under $709$
For example, if we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3$, $5$, $6$ and $9$. The sum of these multiples is $23$.
| If inc/exclusion is unknown then, note $\,3,5\mid n\!+\color{blue}{\!15k}\iff 3,5\mid n,\,$ so the multiples of $\,3,5\,$ have periodicity $15,\,$ so we can split the sum into chunks from each period as follows
$$\begin{eqnarray}
\color{blue}{0}+\overbrace{\{0,3,5,6,9,10,12\}}^{\large \rm sum\, =\, \color{#c00}{45}}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Double Integral $\int_{0}^{4} \int_{\sqrt{x}}^{2} \frac{1}{1+y^3} \mathrm{d}y\;\mathrm{d}x$ I am having trouble computing the double integral:
$$
\int_{0}^{4} \int_{\sqrt{x}}^{2} \frac{1}{1+y^3} \mathrm{d}y\,\mathrm{d}x
$$
I computed the inner integral:
$$
\left [ \frac{1}{3}\ln|y + 1| - \frac{1}{6}\ln(y^2 - y + 1) + \... | If you draw the region of integration, the integral you have is equal to this integral:
$$\int_0^2\int_0^{y^2} \frac{1}{1+y^3} \operatorname{d}x\operatorname{d}y.$$
This is a bit easier to integrate. *After integrating with respect to $x$, you can use a $u$ substitution, $u=1+y^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$
$(1)$ Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{a\sin x+b\cos x}{\sin \left(x+\frac{\pi}{4}\right)}dx$
$(2)$ Evaluation of $\displaystyle \int_{-1}^{1}\ln\left(\frac{1+x}{1-x}\right)\cdot \frac{x^3}{\sqrt{1-x... | (1)
We have:
$$\begin{eqnarray*}\int_{0}^{\pi/4}\frac{\sin x}{\sin(x+\pi/4)}\,dx &=& \int_{\pi/4}^{\pi/2}\frac{\sin(x-\pi/4)}{\sin x}\,dx = \frac{1}{\sqrt{2}}\int_{\pi/4}^{\pi/2}\left(1-\cot x\right)\,dx\\&=&\frac{1}{4\sqrt{2}}\left(\pi-\log 4\right),\end{eqnarray*}$$
$$\begin{eqnarray*}\int_{0}^{\pi/4}\frac{\cos x}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/( x^3+2x^2+2)$ What is the independent coefficient in the inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/(x^3+2x^2+2)$ ?
I have been calculating some combinations, but I don't know how I can calculate the inverse.
| This is in no way a standard solution, but I just want to include it for fun.
Note that $x^3=-(2x^2+2)$, we will find $x^{-3}$. Now
$$x^3-x^2=-2=1$$
implies
$$x(x^2-x)=x^2(x-1)=1.$$
Thus
$$x^{-1}=x(x-1)\text{ and }x^{-2}=x-1.$$
Multiplying the two we get
$$x^{-3} = x(x-1)^2=x(x^2+x+1)=x^3+x^2+x.$$
Using $x^3=x^2+1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$ I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.
Could you help me with it?
| By first finding the following integral by using the algebraic identity
$a^2b=\frac{1}{6}\left(a+b\right)^3-\frac{1}{6}\left(a-b\right)^3-\frac{1}{3}b^3$ one can easily prove avoiding Euler sums that:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=-\frac{1}{4}\zeta \left(4\right)+2\ln \left(2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
"answer_count": 4,
"answer_id": 3
} |
What is the greatest common divisor of $3^{3^{333}}+1$ and $3^{3^{334}}+1$? What is the greatest common divisor of $3^{3^{333}}+1$ and $3^{3^{334}}+1$ ? Could somebody please help me ?
| Write $T_n=3^{3^n}+1$. Then
$$\eqalign{T_n^3
&=(3^{3^n})^3+3(3^{3^n})^2+3(3^{3^n})+1\cr
&=3^{3^{n+1}}+1+3(3^{3^n})(3^{3^n}+1)\cr
&=T_{n+1}+3^{3^n+1}T_n\ .\cr}$$
From this you should be able to see an impotant relationship between $T_n$ and $T_{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/909483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $ab+bc+ca=0$ then the value of $1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is If $ab+bc+ca=0$
then the value of
$1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is...
| $$\frac{1}{a^2+ab+ca}+\frac{1}{b^2+ab+bc}+\frac{1}{c^2+ca+bc}$$
$$=\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)}+\frac{1}{c(a+b+c)}=\frac{ab+bc+ca}{abc(a+b+c)}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then $\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$ If $\displaystyle\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then prove that $\displaystyle\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$
I tried to prove this in many ways. First, I tri... | If we set
$$k=\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a},$$
then we have
$$x+y=(3a-b)k,\ \ y+z=(3b-c)k,\ \ z+x=(3c-a)k.$$
Hence, adding these three gives us
$$2(x+y+z)=(3a-a+3b-b+3c-c)k\iff \frac{x+y+z}{a+b+c}=k\tag 1$$
On the other hand, we have
$$a(x+y)=a(3a-b)k,\ \ b(y+z)=b(3b-c)k,\ \ c(z+x)=c(3c-a)k$$
$$\Ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
The length of the smallest possible ladder to change the bulb.
In the drawing, the point P, a bit located under the bulb, has coordinates (a, b), where a and b are two parameters. You want to change the bulb, and for this it is necessary to install a ladder such that
*
*It relies on the ground (the $Ox$) axis at... | Let $x$ be the base of the triangle and let $y$ be the height.
To minimize the length $L$ of the ladder, we can minimize $L^2=x^2+y^2$.
By similar triangles, $\displaystyle\frac{y}{x}=\frac{b}{x-a}$, so $\displaystyle y=\frac{bx}{x-a}$ and $\displaystyle f(x)=L^2=x^2+\frac{b^2x^2}{(x-a)^2}$.
Then $f^{\prime}(x)=2x+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4... | Here is yet another way, using (weighted) AM-GM. I would still prefer Holder (the solution posted earlier) for its simplicity, this is just to illustrate it can be done - especially if you know the point of equality.
Let $a = \sqrt{\frac2{15}}$. Then using weighted AM-GM, we can write the following inequalities. Not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How prove this inequality $a^3b+b^3c+c^3a+a^3b^3+b^3c^3+c^3a^3\le 6$ let $a,b,c>0$, and such
$$a^2+b^2+c^2=3$$
show that
$$a^3b+b^3c+c^3a+a^3b^3+b^3c^3+c^3a^3\le 6\tag{2}$$
I know this famous inequality( creat by valsie)
$$(a^2+b^2+c^2)^2\ge 3(a^3b+b^3c+c^3a)$$ this inequality if and only if
$$a=b=c, or,a:b:c=\sin^2{\... | both side multiply $a^2+b^2+c^2=3$, we have:
$(a^2+b^2+c^2)(a^3b+b^3c+c^3a)+3(a^3b^3+b^3c^3+c^3a^3)\le \dfrac{2}{3}(a^2+b^2+c^2)^3 \iff 2(a^2+b^2+c^2)^3-3((a^2+b^2+c^2)(a^3b+b^3c+c^3a)+3(a^3b^3+b^3c^3+c^3a^3)) \ge 0 $
let $x= $min{$a,b,c$}, the other two are $x+u,x+v$, we will always have
$Ax^4+Bx^3+Cx^2+Dx+F \ge 0 $
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equat... | $n=14k+10=7\cdot2\cdot k+7+3=(2k+1)7+3$, the remainder is $3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Trig Substitution Problem - Integration Suppose I want to integrate this (I chose an easy one):
$$\int \frac {dx}{\sqrt{x^2-4}}$$
Method 1: (Trig Substitution) $x=2\sec(\theta)$
$$\int \frac {dx}{\sqrt{x^2-4}}=\int \frac{\sec(\theta)\tan{(\theta)}}{|\tan(\theta)|}d\theta$$
Now removing the absolute value I find is toug... | Notice that the integrand $\dfrac{1}{\sqrt{x^2-4}}$ is only defined on $(-\infty,-2)\cup(2,\infty)$.
Hence, we can consider the antiderivative on $x > 2$ and on $x < -2$ seperately.
If $x > 2$, then we substitute $x = 2\sec \theta$ where $0 < \theta < \tfrac{\pi}{2}$.
Over this range, $\tan \theta > 0$. So, $\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$
$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?
Motivation: We already know that $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\l... | Using the relation between 3 trilogarithms given here:
\begin{equation}
\operatorname{Li}_3(z)=-\operatorname{Li}_3(\frac{z}{z-1})-\operatorname{Li}_3(1-z)+\frac{1}{6}\log^3(1-z)-\frac{1}{2}\log(z)\log^2(1-z)+\frac{\pi^2}{6}\log(1-z)+\zeta(3)
\end{equation}
with $z=(1+i)/2$, we have $z/(z-1)=-i$ and $1-z=(1-i)/2$. Wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/918680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 7,
"answer_id": 2
} |
Find $ y''$ and $ y'''$ if $b^2x^2+a^2y^2=a^2b^2$ answers :- $$y''=-\frac{b^4}{a^2y^3}$$ and $$ y'''=-\frac{3b^6x}{a^4y^5}$$i have tried solve in this manner:-$$b^2x^2+a^2y^2=a^2b^2 -1.)$$diff wrt x :$$2b^2x+2a^2y'=0$$ $$a^2y'=-b^2x$$$$\frac{dy}{dx}=\frac{-b^2x}{a^2y}$$on substituting eq -1 we get :-$$\frac{dy}{dx}=\... | The question is not very clear but it seems that you are writing $y_1$ for the derivative of $y$ and $y_2$ for the second derivative.
If I am correct in this, then
$$\eqalign{b^2x^2+a^2y^2=a^2b^2\quad
&\Rightarrow\quad 2b^2x+2a^2yy'=0\cr
&\Rightarrow\quad 2b^2+2a^2(yy''+(y')^2)=0\qquad(*)\cr
&\Rightarrow\quad y''... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/918742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generating functions to solve number of integer solution problem If I have $x_1 + x_2 + x_3 =10$ with $1\leq x_1 \leq 5, \; 2 \leq x_2 \leq 6, \;3 \leq x_3 \leq 9$
I know that I compute $(t^1+\dots + t^5)(t^2 +\dots + t^6)(t^3+\dots +t^9)$ and look at the coefficient of $t^{10}$ to find number of integer solutions.
But... | You are looking for ($\dotsb$ at the end of a factor covers terms that don't affect the result):
$\begin{align}
[z^{10}] (z + \dotsb + z^5) &(z^2 + \dotsb + z^6) (z^3 + \dotsb + z^9) \\
&= [z^{10 - 1 - 2 - 3}] (1 + \dotsb + z^4) (1 + \dotsb + z^4) (1 + \dotsb + z^6) \\
&= [z^6] \frac{(1 - z^5)^2 (1 - z^7)}{(1 - z)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+C$ Wolfram gives this nice result:
$$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$
I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting ver... | $$\int\dfrac{\cos xdx}{\cos^{3/2}2x}=\int\dfrac{\cos xdx}{(\cos^2x-\sin^2x)^{3/2}}=$$
$$\dfrac{\cos xdx}{(1-\tan^2x)^{3/2}\cos^3x}=\int\dfrac{\sec^2xdx}{(1-\tan^2x)^{3/2}}$$
$$u=\tan x,du=\sec^2xdx$$
$$\int\frac{du}{(1-u^2)^{3/2}}$$
$$u=\sin\theta,du=\cos\theta d\theta$$
$$\int\dfrac{\cos\theta d\theta}{\cos^3\theta}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show that two expressions are equivalent I am trying to prove a hyperbolic trigonometric identity and I ran into the following expression:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} \quad.$$
This expression is supposed to be equivalent to
$$\sqrt{x^2+1} \quad.$$
I tried to algebrai... | For $b\ne 0$, $\frac{a}{b}=c\Leftrightarrow bc=a$.
So a strategy would be to try to show that
$$(\sqrt{x^2+1}+x)^2+1=2(\sqrt{x^2+1}+x)\cdot \sqrt{x^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove of Nesbitt's inequality in 6 variables I was just reading "Pham kim hung secrets in inequalities,Volume 1" book and there was an interesting problem on it's Cauchy-Schwarz and Holder section that caught my eye.
Prove that for all positive real numbers $a,b,c,d,e,f$,we always have $$\frac{a}{b+c}+\frac{b}{c+d}+\f... | $$\sum\limits_{i=1}^{6}\frac{x_i}{x_{i+1}+x_{i+2}} \ge 3$$ where, the indices are taken cyclically. Wlog, assume $\sum\limits_{i=1}^{6} x_i = 1$.
We start with the fact that $f(s) = \frac{1}{1-s}$ is convex on the interval $[0,1)$. Applying Jensen Inequality, $$\sum\limits_{i=1}^{6} \frac{x_i}{1-(x_i + x_{i-1}+x_{i-2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Closed form for $1 + 3 + 5 + \cdots +(2n-1)$ What is the closed summation form for $1 + 3 + 5 + \cdots + (2n-1)$ ?
I know that the closed form for $1 + 2 + 3+\cdots + n = n(n+1)/2$ and I tried plugging in $(2n-1)$ for $n$ in that expression, but it didn't produce a correct result:
$(2n-1)((2n-1)+1)/2$
plug in 3
$(2n-... | Sum of Arithmetic Progression $=\frac{N}{2}(a_1+a_n)$, where N is the nunber of terms
$N = \frac{(a_n-a_1)}{d} + 1 = \frac{(2n-2)}{2} + 1 = n-1+1 = n$
Thus the sum is $ = \frac{n}{2}(1+2n-1) = n^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
The remainder of $1^1+2^2+3^3+\dots+98^{98}$ mod $4$ How can I solve this problem:
If the sum $S=(1^1+2^2+3^3+4^4+5^5+6^6...+98^{98})$ is divided by $4$ then what is the remainder?
I know that all the even terms I can ignore since $(2n)^{2n}=4^nn^{2n})$ which is divisible by $4$,but i dont know what to do next it. ... | In mod $4$, note that for any positive integer $m$,
$$(2m)^{2m}=(4m^2)^m\equiv 0$$$$(4m-1)^{4m-1}\equiv (-1)^{4m-1}\equiv -1$$$$(4m-3)^{4m-3}\equiv 1^{4m-3}\equiv 1$$
and that $97=4\cdot 25-3,\ 95=4\cdot 24-1$.
Hence, we have
$$S\equiv 24\cdot (-1)+25\cdot 1\equiv 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $x+3y=4y^3,y+3z=4z^3 ,z+3x=4x^3$ in reals
Find answers of this system of equations in reals$$
\left\{
\begin{array}{c}
x+3y=4y^3 \\
y+3z=4z^3 \\
z+3x=4x^3
\end{array}
\right.
$$
Things O have done: summing these 3 equations give $$4(x+y+z)=4(x^3+y^3+z^3)$$
$$x+y+z=x^3+y^3+z^3$$
I've also tried to show tha... | Consider $f(x) = 4x^3 - 3x$. We have: $f'(x) = 12x^2 - 3 = 0 \iff x = \pm \dfrac{1}{2}$. Thus:
If $x \in \left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ then $f'(x) < 0$, and $f$ decreases. This means:
If $x > y > z$, then: $x = f(y) < f(z) = y$, a contradiction. And we can obtain a contradiction for any other inequalities o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/924596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
$$\implies y^3-x^3=8x^2-6x+8$$
$$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the... | We have
$y^3 − (x + 1)^3 = x^3 + 8x^2 − 6x + 8 − (x^3 + 3x^2 + 3x + 1) = 5x^2 − 9x + 7$.
Consider the quadratic equation
$5x^2 − 9x + 7 = 0$.
The discriminant of this equation is
$D = 92 − 4 × 5 × 7 = −59 < 0$ and hence the expression $5x^2 − 9x + 7$ is positive for all real values of x.
We conclude that $(x + 1)^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor ... | This is a very general approach.
$$n^3+n^2 = n(n-1)(n-2) + 4n(n-1) + 2n$$
So $$(n^3+n^2)x^n = \left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)x^n$$
So $$\sum (n^3+n^2)x^n =\left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove that $2|ab| \leq a^2 + b^2$ and $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$ I am having issues solving these problems:
I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.
Show that for $a,b\in (-\infty,+\infty)$... | (a) $0 \leq (|a|-|b|)^2 = |a|^2 -2|a||b|+|b|^2 = a^2 - 2|ab|+ b^2\Rightarrow 2|ab| \leq a^2+b^2 $
(b) by (a) we have $(|a| + |b|)^2 = a^2 + 2|ab| + b^2 \leq a^2+b^2+a^2 +b^2 = 2(a^2+b^2) \Rightarrow |a|+|b| \leq \sqrt {2(a^2+b^2)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/929199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to Solve $ \int \frac{dx}{x^3-1} $ I am having quite a difficult time integrating
$$
\int \frac{\mathrm{d}x}{x^3-1}
$$
My first approach was to apply a partial fraction decomposition
$$
\int \frac{\mathrm{d}x}{x^3-1} = \int \frac{\mathrm{d}x}{(x-1)(x^2+x+1)} = \frac{1}{3} \int \frac{\mathrm{d}x}{x-1} - \frac{1}{3}... | You can write $\displaystyle\int\frac{x+2}{x^2+x+1} dx=\frac{1}{2}\int\frac{2x+1}{x^2+x+1} dx + \frac{3}{2}\int\frac{1}{x^2+x+1} dx$
$\displaystyle=\frac{1}{2}\ln(x^2+x+1)+\frac{3}{2}\int\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx$, and now let $u=x+\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ a... | $$|x^2-4|<2 \\ \implies -2<x^2-4<2 \\ \implies 2<x^2<6 \\ \implies \sqrt{2}<\pm x<\sqrt{6} \\ \implies \sqrt{2}<x<\sqrt{6},\ \ -\sqrt{2}>x>-\sqrt{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the exact length of the curve $y=\frac 12 x^2- \frac 12 \ln(x)$
Find the exact length of the curve $y = \frac 12 x^2- \frac 12 \ln(x)$, for $2 \le x \le 4$.
My attempt:
\begin{align}
L&= \int_2^4 \sqrt{1+\left[x-\frac 1{2x} \right]^2} \, dx \\
&= \int_2^4 \sqrt{1+x^2+\frac 1{4x^2}-1} \, dx \\
&= \int_2^4 \sqrt{x... | It is possible that there is a typo in the question: often in arclength problems, the constants are chosen so that $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$ miraculously turns out to be the square of something nice. However, let us assume there is no typo.
We want to integrate $\frac{1}{2x}\sqrt{4x^4+1}$. Multiply top a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{... | In first you have:
$$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
It should be:
$$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}du=\frac{1}{4}\cdot 2 \sqrt{1+u}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Calculate the binomial sum $ I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i} $ I need any hint with calculating of the sum
$$
I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i}.
$$
Maple give the strange unsimplified result
$$
I_n={\frac {1/12\,i\sqrt {3} \left( - \left( \left( 1+i\sqrt {3} \right)
^{2\,{\it n}+2} \right) ^... | Start by restating the problem: we seek to evaluate
$$\sum_{q=0}^n (-1)^q {2n+1-q\choose q}
= \sum_{q=0}^n (-1)^q {2n+1-q\choose 2n+1-2q}.$$
Introduce the integral representation
$${2n+1-q\choose 2n+1-2q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1-q}}{z^{2n+2-2q}} \; dz.$$
This gives the following for the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
How to show this equals 1 without "calculations" We have $$ \sqrt[3]{2 +\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = 1 $$
Is there any way we can get this results through algebraic manipulations rather than just plugging it into a calculator?
Of course, $(2 +\sqrt{5}) + (2-\sqrt{5}) = 4 $, maybe this can in some way help?
| I'll expand on Guillermo's comment: let
$$
a=\sqrt[3]{2+\sqrt{5}},\quad b=\sqrt[3]{2-\sqrt{5}}.
$$
You want to show that $a+b=1$. Consider
$$
(a+b)^3=a^3+3a^2b+3ab^3+b^3=(a^3+b^3)+3ab(a+b)=4+3\times(-1)(a+b)
$$
so if you let $x=(a+b)$ , then $x$ is a real number satisfying
$$
0=x^3+3x-4=x^3-x^2+x^2-x+4x-4\\
=(x-1)(x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.