Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Irreducibility and Splitting Fields Show that over any field $F$, the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors.
Edited:
This is my attempt: Let $f(x)=x^3-3x+1$. Let $a_1,a_2,a_3$ be the roots of $f$. Suppose char $F\neq 2,3$. Suppose also that $f$ is neither irreducible nor splits in... | Here is one way to do it using almost no theory, just playing with algebra.
Suppose the polynomial has a root $a$ in $F$. If you divide $x^3 - 3x + 1$ by $x-a$ the quotient is the polynomial
$$
x^2 + ax + a^2 - 3
$$
in $F(a)[x]$. From the quadratic formula (assuming characteristic $\neq 2$ here, for the moment, I g... | {
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"url": "https://math.stackexchange.com/questions/26477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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solving quadratic equations by completing the square
Possible Duplicate:
How to “Re-write completing the square”: $x^2+x+1$
$b^2-4b+4=16$ can someone help me solve this equation. All the steps please.
| if $ax^2+bx^2+c=0$ then
$$
0=a\Big(x^2+\frac{b}{a}x\Big)+c
=a\Big(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\Big)+c
=a\Big(x+\frac{b}{2a}\Big)^2+c-\frac{b^2}{4a}
$$
hence
$$
\Big(x+\frac{b}{2a}\Big)^2=\frac{b^2-4ac}{4a^2}
$$
and
$$
x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{-b\pm\sqrt{b^2-4ac}}{... | {
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"timestamp": "2023-03-29T00:00:00",
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$x^3+48=y^4$ does not have integer (?) solutions How does one find all positive integer solutions to the equation $x^3+48=y^4$?
| I'm posting this as a community wiki "answer" because I'm not sure it works and the explanation wouldn't show up very well in the comments section I guess. If we factor the equation in $\mathbb{Z}[\sqrt{3}]$ we get $(y^2 - 4\sqrt{3})(y^2 + 4\sqrt{3}) = x^3$. Thus if $y^2 - 4 \sqrt{3} = (a + b\sqrt{3})^3$ for some $a + ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Help to find cubics with square discriminant If the discriminant $b^2-4c$ of the quadratic $x^2 + bx + c$ is a square then it factors. For every discriminant $d^2$ we have can parametrize them all $(b,c) = (d + 2 h,h(d+h))$. edit I realized now that the quadratic case is trivial because it the discriminant is a square ... | I would not focus on parametrization as such. Instead, note that $4 a^3 = d^2 + 27 b^2.$ This gives an easily stated restriction on the prime factorization of $a.$ It is necessary and sufficient that $a \geq 0$ and, whenever any prime $p | a$ and $p \equiv 2 \pmod 3,$ then the exponent of $p$ must be even. So $a = 2$ o... | {
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How do I prove this inequality?: $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ where $a,b,c>0$ and $abc=1.$ I have been thinking over this problem for a couple of days, but I have no idea how to solve it in a simple way.
I am interested if there is a way only using elementary me... | With $abc=1$, we can set :$a=\frac{y}{x}; b=\frac{z}{y}; c=\frac{x}{z}$. We have:
$\frac{x}{z}+\frac{y}{z}+\frac{z}{y}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\le 3+\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}$
$\Leftrightarrow 3xyz+x^3+y^3+z^3\ge xy(x+y)+yz(y+z)+zx(z+x)$.
It's true by Schur's inequality. Equality occurs if... | {
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at nex... | To say that $3n^3+6n$ is divisible by $9$ is equivalent to saying that $n^3+2n=n(n^2+2)$ is divisible by $3$. Now split to three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$ (explain why this is true.) Now you have to check just 3 cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of inequality I have problems with proving inequality :
$${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$$
where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$.
Thanks.
| We need to prove that
$$\frac{(a+b+c)(a^2+b^2+c^2)}{10}+\frac{2}{5}abc<\frac{(a+b+c)^3}{20}$$ or
$$(a+b+c)^3>8abc+2(a+b+c)(a^2+b^2+c^2)$$ or
$$\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)>8abc+2\sum_{cyc}(a^3+a^2b+a^2c)$$ or
$$\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)>0$$ or
$$(a+b-c)(a+c-b)(b+c-a)>0$$
Done!
| {
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Linear Algebra: solving for x I was wondering why in this exercise there isn't a minus $(x^2-2x-3)$. I thought that for this type of exercise you must use $(ad-bc)$.
Problem 50 (p 131) Solve for $x$.
$\begin{vmatrix} x-2& -1\\ -3 & x \end{vmatrix}=0$
$(x-2)\cdot x + (-3)\cdot(-1)=0$
$x^2-2x+3=0$
$x-3=0$ or $x... | This is a mistake. To calculate the determinant of a $2\times 2$ matrix you use $ad-bc$ which gives you $(x-2)x-(-1)(-3)=x^2-2x-3=(x+1)(x-3)=0$. Hence $x=3$ or $x=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$
How would you evaluate the following limit:
$$\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$$
I tried to use this formula: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, It didn't work.
Any hints?
| $$
\begin{aligned}
\lim _{x\to \infty }\left(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5}\right)
& = \lim _{t\to 0}\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}-\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right)
\\& = \lim _{t\to 0}\left(\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}-\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right)\cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/45970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to solve for $x$ in the equation, $4\sqrt{x-3} - \sqrt{6x-17} = 3$, with two square root terms? So guys, my girlfriend is taking a college algebra class this summer and I figured I would help her study for her upcoming final because I am an engineering major and this kind of math would be easy for me. As we were do... | There is a trick for problems like this that I really love to do.
\begin{align}
4\sqrt{x-3} - \sqrt{6x-17} &= 3 &\text{(multiply)} \\
\text{Let } 4\sqrt{x-3} + \sqrt{6x-17} &= y \\
\hline
16(x-3)-(6x-17) &= 3y \\
16x-48 - 6x + 17 &= 3y \\
y &= \dfrac{10x -31}{3} \\
\hline
4\sqrt{x-3} + \sqrt{6x-17} &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/52090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 8
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Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$? Given a series, how does one calculate that limit below? I noticed the numerator is an arithmetic progression and the denominator is a geometric progression — if that's of any relevance —, but I still don't know how to solve it.
$$\lim_{n\to\infty}... | Firstly, what you have is a limit of a finite sum, and the limit is a series:
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}:=\sum_{k=1}^\infty \frac{k+1}{3^k}$$
Now, to the question you asked: there are a few tricks you can use:
1) You know that $\sum_{k=1}^nx^k=\frac{1-x^{n+1}}{1-x}-1$ (geometric progression).
2) T... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 7
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Please help me to find Taylor expansion (or approximation) for $f(x)=\frac{1}{x^2(x-1)}$ around $a=2$ First, sorry if my translations is bad. I need help for this exercise, more precisely , I need to know if my result which I've found is good.
The exercise: Find Taylor expansion (or approximation) for $f(x)=\frac{1}{x^... | The general procedure is correct. And the details look pretty good too. I could now check the details, it would not be difficult.
Instead, I will do the problem in a somewhat different way, which in general is more efficient. It will turn out that a sign error crept into your calculation.
We have
$$f(x)=-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/57574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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differential equations A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (half life $1.28 \times 10^9$) and that one of every 9 potassium atom disintegrations yields an argon atom. What is the age of the rock me... | After Rahul's comment above, I understand the question now.
After $t$ years, exactly $\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$ of the potassium will remain, and exactly $\frac{1}{9}$ of the decays will have produced an argon atom. Thus, we have that
$$\text{argon}=\frac{1}{9}\cdot\text{ decayed potassium}$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/58137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trouble setting up double integrals I am studying for an entrance exam and have troubles to set up double integrals appropriately. Actually my biggest problem is that I don't get the notation.
$$ A = \{(x,y) | 0 \leq x + y \leq 1, 0 \leq x-y \leq \pi\} $$
$$ \iint_A e^{x+y} \sin(x-y)\mathrm dx\mathrm dy $$
and for an... | For the first example. The double inequality $0\leq x+y\leq 1$ means that
$$
\left\{
\begin{array}{c}
0\leq x+y \\
x+y\leq 1
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
y\geq -x \\
y\leq 1-x
\end{array}
\right.
$$
and $0\leq x-y\leq \pi $ means that
$$
\left\{
\begin{array}{c}
0\leq x-y \\
x-y\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/61982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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partial fraction question
$ \frac{125x^{2}+x+3}{x^{2}(x-5)} =
> \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-5)} | * x^{2}(x-5)$
$125x^{2}+x+3 = Ax(x-5) + B(x-5) + C (x^{2})$
$125x^{2}+x+3 = A x^{2} - 5Ax + Bx -5B +Cx^{2}$
$125x^{2}+x+3 = x^{2}(A+C) -x(A+B)-5B$
$3 = -5B \Rightarrow B = \frac{-3}{5}$
$-1 = A+B \Rightarrow ... | HINT $\ $ It's simpler to use the Heaviside cover-up method. First, evaluating your $\rm\:E_2 = 2$nd equation at $\rm\:x = 0\:$ yields $\rm\:3 = -5\:b\:.\:$ Next, differentiating $\rm\:E_2\:$ and evaluating at $\rm\:x = 0\:$ yields $\rm\: 1 = b - 5\:a\:.$ Solve those for $\rm\:a,b\:$. Finally evaluating $\rm\:E_2\:$ at... | {
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"timestamp": "2023-03-29T00:00:00",
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How to know what is the degree of the remainder The example question is
Find the remainder when $8x^4+3x-1$ is divided by $2x^2+1$
The answer did something like
$$8x^4+3x-1=(2x^2+1)(Ax^2+Bx+C)+(Dx+E)$$
Where $(Ax^2+Bx+C)$ is the Quotient and $(Dx+E)$ the remainder. I believe the degree of Quotient is derived from de... | EDIT to add these short answers.
I believe the degree of Quotient is derived from degree of $8x^4+3x-1$
- degree of divisor.
That's right.
But for remainder?
The degree of the remainder is less than the degree of the divisor, by definition of polynomial division.
*
*Let me start with this specific case. To fi... | {
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"source": "stackexchange",
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given $a^2+b^2=28ab$ what's $\log_{3}\left (\frac{(a+b)^2}{ab}\right)$? given $a^2+b^2=28ab$ what's $\log_{3} \left(\dfrac{(a+b)^2}{ab}\right)$?
$\log_{3} \left(\dfrac{(a+b)^2}{ab}\right)$
$\log_{3} \left(\dfrac{a^2+b^2+2ab}{ab}\right)$
$\log_{3} \left(\dfrac{a^2+b^2}{ab}+\dfrac{2ab}{ab}\right)$
$\log_{3} \left(\dfrac... | The unwritten core of the problem is that it is possible to determine
$C = (a+b)^2/ab$ (and hence also its logarithm to the base 3) given the first equation on $a$ and $b$. This would not be true if the condition had been modified to
$a^2 + b^2 = 28ab + 5, \quad$ or
$a^3 + b^3 = 28ab, \quad$ or
$\sqrt{a^2+b^2}=28a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/66702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit of a function $$\lim_{x\to 3}\frac{\sqrt{3x} - 3}{\sqrt{2x-4} - \sqrt{2}}.$$
Letting
$$F(x) = \frac{\sqrt{3x} - 3}{\sqrt{2x-4}-\sqrt{2}},$$
we have
$$F(x) = \frac{\sqrt{3}(\sqrt{x} - \sqrt{3})}{\sqrt{2}(\sqrt{x-2}-1)}.$$
Multiplying numerator and denominator by $\sqrt{x-2} + 1$,
$$F(x) = \frac{
(3)^{1/... | Hint: $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$.
Applying the hint twice,
$$
F(x)=\sqrt{\frac32}\frac{\sqrt{x}-\sqrt3}{\sqrt{x-2}-1}=\sqrt{\frac32}\frac{x-3}{\sqrt{x}+\sqrt3}\,\frac{\sqrt{x-2}+1}{(x-2)-1}=\sqrt{\frac32}\frac{\sqrt{x-2}+1}{\sqrt{x}+\sqrt3},
$$
hence
$$
\lim\limits_{x\to3}\,F(x)=\sqrt{\frac32}... | {
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"timestamp": "2023-03-29T00:00:00",
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Complete the square of a 4th degree expression $k^4 + 2k^3 -k^2 - 2k + 1$ I was reading the solution of a problem and I can't figure out how to go from here:
$k^4 + 2k^3 -k^2 - 2k + 1$
to:
$((k^2 + k) -1)^2$
I guess it's using the complete-the-square method but I've only used that with quadratic expressions.
| $k^4+2k^3−k^2−2k + 1 = k^4 + (k^2 - 2k^2) + 2k^3 - 2k + 1 \quad \Rightarrow$
$k^4+2k^3−k^2−2k + 1 = (k^4 - 2k^2 + 1) + 2k(k^2 - 1) + k^2$
$= (k^2 - 1)^2 + 2(k^2 - 1)k + k^2 = (k^2 - 1 + k)^2 = ((k^2 + k) - 1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/73076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Beta Function -- finding a lower bound based on parameters I would like to show that
$$ 1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq \frac{1}{c+1}.$$
for all $c \geq 2$.
I have plotted it out for $c$ up through 200, and it seems to hold.
Does anyone have any tips on if this can be formally shown? I imagine an e... | I was composing the following when the original answer by @MikeSpivey appeared; his edited answer provides considerably more information. The short calculation below arrives at the desired result without requiring integration by parts but does need knowledge of the relationship between the Beta and Gamma functions.
For... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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$n! \leq \left( \frac{n+1}{2} \right)^n$ via induction I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.
This is where I am stuck:
$$\left( \frac{n+2}{2} \right)^{n+1}
\geq \dots \geq
=2 \left( \frac{n+1}{2} \right)^{n+1}
= \left( \frac{n+1}{2} \right)^n(n+1)
\geq n!(n+1)
= (n+1)! $$
I appro... | Mine is a tad less elegant but arguably a bit clearer and assumes only the complete basics.
0. The task
Prove that for $ \forall n \in \mathbb{N}$ it is true that $n! \leq (\frac{n+1}{2})^n $ :
I. Base steps
(I need four base steps because my final inequality works for $k \ge 4$.)
$$n=0: 1 \leq (\frac{1}{2})^0 = 1 $$
$... | {
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"url": "https://math.stackexchange.com/questions/76130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
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When do addition and inversion of matrices commute? Can you guys help me out with the following problem:
Problem: Find conditions that the matrices $A$ and $B$ have to satisfy in order for the following to be valid: $(A+B)^{-1} = A^{-1} + B^{-1}$.
My solution. The first condition is that $A$ and $B$ have to have the sa... | For third condition you have that:
$(A+B)(A+B)^{-1}=(A+B)(A^{-1}+B^{-1})$
$(A+B)^{-1}(A+B)=(A^{-1}+B^{-1})(A+B)$
$\Rightarrow (A+B)(A^{-1}+B^{-1})=(A^{-1}+B^{-1})(A+B)\Rightarrow $
$\Rightarrow AA^{-1}+AB^{-1}+BA^{-1}+BB^{-1}=A^{-1}A+A^{-1}B+B^{-1}A+B^{-1}B \Rightarrow$
$\Rightarrow AB^{-1}+BA^{-1} =A^{-1}B+B^{-1}A$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Can you explain this please $T(n) = (n-1)+(n-2)+…1= \frac{(n-1)n}{2}$
Possible Duplicate:
Proof for formula for sum of sequence 1+2+3+…+n?
Can you explain this please
$$T(n) = (n-1)+(n-2)+…1= \frac{(n-1)n}{2}$$
I am really bad at maths but need to understand this for software engineering.
| For example, if we wish to add the numbers
$$
1+2+3+4+5+6
$$
you can pair them as such
$$
(1+6)+(2+5)+(3+4)=7+7+7=21.
$$
But, here $n=7$ and there are $\frac{n-1}{2}=3$ pairs of $n=7$. Thus we have that the sum is $\frac{n-1}{2}\cdot n=3\cdot 7=21$. In general, you can pair the sum
$$
1+2+3+\cdots +(n-2)+(n-1)
... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Limit of geometric mean of N radii of an ellipse Is this equation correct?
$$\lim_{N \to \infty} \prod_{n=1}^N (a^2\cos^2 (2\pi n/N)+b^2\sin^2(2\pi n/N))^{1/N}=ab$$
If so, why?
| The answer has been substantially revised. In particular, the previous revision quoted a wrong value for the integral $I(a,b)$, which resulted in an overall incorrect answer. I hope that this revision is correct. :-)
The limit is $\Big(\frac{a+b}{2} \Big)^2$ and is not $ab$ (as claimed). I break down the solution into... | {
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Prove that $\sum\limits_{k=1}^n \frac{1}{k^2+3k+1}$ is bounded above by $\frac{13}{20}$ I want ask a question about a sum. The exercise is as follows:
Prove the following inequality for every $n \geq 1$:
$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$
| $$
\begin{align}
\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} & \leq \sum\limits_{k=1}^n \frac1{k(k+3)}\\
& = \sum\limits_{k=1}^n \frac13 \left(\frac1k - \frac1{k+3} \right)\\
& = \frac13 \left( 1 + \frac12 + \frac13 - \frac1{n+1} - \frac1{n+2} - \frac1{n+3} \right)\\ & \leq \frac13 \frac{11}{6}\\
& = \frac{11}{18}
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/85316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Hard elementary combinatorics problem How does one compute (without brute force) the smallest integer $n$ such that
$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?
| Let's replace an explicit number $-3$ with a symbol $z$, so that we consider:
$$
\sum_{k=1}^n z^{k-1} \binom{2n}{2k-1} = \sum_{m=0}^{2n} \left( \frac{1-(-1)^m}{2} \right)z^{(m-1)/2} \binom{2n}{m} = \frac{\left(1+\sqrt{z}\right)^{2n} - \left(1-\sqrt{z}\right)^{2n}}{2 \sqrt{z}}
$$
Now substitute $z=-3$. Notice that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Volume of a cut sphere The sphere $x^2 + y^2 + z^2 = 4$ is cut by the plane $z = 1/2$. How do you calculate the volume of two parts of the sphere using integrals? Thank you!
| The standard setup is
$$
\begin{align}
\int_\frac{1}{2}^2\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}\int_{-\sqrt{4-z^2-y^2}}^\sqrt{4-z^2-y^2}\;\mathrm{d}x\;\mathrm{d}y\;\mathrm{d}z
&=\int_\frac{1}{2}^2\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}2\sqrt{4-z^2-y^2}\;\mathrm{d}y\;\mathrm{d}z\\
&=\int_\frac{1}{2}^2\pi(4-z^2)\;\mathrm{d}z\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/88302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Use induction to prove a product of sums of squares is a sum of squares For any natural number $n\ge 1$, given pairs $(a_1,b_1),(a_2,b_2),...,(a_n,b_n)$ of integer numbers, there exist integer number $c$ and $d$ such that
$$\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$$
My initial approach is
Base Case: $(a_1^2+b_1^2) = a_1... | The following identity is useful:
$$(a^2+b^2)(x^2+y^2)=(ax\pm by)^2 + (ay \mp bx)^2.\qquad\qquad(\ast)$$
This identity is connected with the multiplication of complex numbers. Note that
$$(a+ib)(x+iy)=ax-by+i(ay+bx).$$
The identity $(\ast)$ can be thought of as asserting that the product of the norms of $a+ib$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Using power series, compute: $\sum\limits_{n=1}^{\infty}\frac{n}{2^n}$
Possible Duplicate:
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?
I need to compute the sum of $$\sum_{n=1}^{\infty}\frac{n}{2^n}$$
using power series.
Any hints of how should I do that?
| So you want to compute $$S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots$$
Now consider $$ f(x)= \frac{x}{2} + \frac{x^2}{2^2} + \frac{x^3}{2^3} + \cdots = \displaystyle\frac{\frac{x}{2}}{1 - \frac{x}{2}}$$ From here evaluate the value of $f'(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Odd numbers expressed as : $x^2-y^2$ How to prove following statement :
Conjecture:
An odd number $n$ , $(n>1)$ can be uniquely expressed as : $n= x^2-y^2$ ; $x,y \in \mathbb{Z}^{*}$
if and only if $n$ is a prime number .
If $x-y=m$ , where $m>1$ then $m \mid n$
Proof :
$n=x^2-y^2=(y+m)^2-y^2=y^2+2\cdot y\cdot m +m^2... | HINT $\ $ (Non) uniqueness follows from the composition law for differences of squares
$\rm\qquad\qquad\ (a^2-b^2)\ (A^2-B^2)\ =\ (a\:A+b\:B)^2-(a\:B+A\:b)^2$
$\rm\qquad\qquad\ \phantom{(a^2-b^2)\ (A^2-B^2)}\ =\ (a\:A-b\:B)^2-(a\:B-A\:b)^2$
E.g. composing $\rm\ 7 = 4^2 - 3^2\ $ with $\ 11 = 6^2 - 5^2\ $ yields for $\rm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/97893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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$Tx = \frac{x}{2} +\frac{1}{x}$ is a contraction on $M = [1,\infty)$ in $(\mathbb{R},|\cdot|)$? I cannot seem to find a contraction factor such that $$Tx = \frac{x}{2}+\frac{1}{x}$$ is a contraction on the whole set $[1,\infty)$ in the complete normed space $(\mathbb{R}, |\cdot|)$.
My argument for $x,y\in [1,\infty)$:
... | When you wrote
$$\left|\frac{x-y}{2}+\frac{y-x}{xy} \right|
\leq\left|\frac{x-y}{2}\right| +\left|\frac{x-y}{xy} \right|,$$
you didn't take advantage of the fact that $x-y$ and $y-x$ have opposite signs, so there is cancellation that will give you a better estimate. Instead you could note that
$$\left|\frac{x-y}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/100061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there a quick way to compute the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$? Is there a way to find the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$?
Just analyzing the roots shows that the polynomial is separable, so I suppose the splitting field would be a Galois extension. Howev... | Just some culture, in case of curiosity. If you take any prime $q$ that has an integral expression as
$$ q = 2 x^2 + x y + 4 y^2, $$ such as $2, \; 5, \; 7\; 19,$ then
$$ z^3 + z + 1 \equiv 0 \pmod q $$ has no roots in integral $z.$
For the special case of $31,$ we have
$$ z^3 + z + 1 \equiv (z -3)(z - 14)^2 \pmod {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/101196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$ Q. Prove
$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$$
My try: expand and got:
$$\frac{5^{2x-2}+2(5^{x^2-1})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}$$
Now what? I find my pre-calculus skills esp with Indices, Logarithms & Trigo lacking ...... | Let $u=5^x$. Then $u^2=25^x$ and
$$
\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{(u/5+5u)^2}{u^2/25+25u^2}=\frac{u^2(1/5+5)^2}{u^2(1/25+25)}=\frac{(1/5+5)^2}{(1/25+25)}=\frac{338}{313}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/102260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
incremental computation of standard deviation How can I compute the standard deviation in an incremental way (using the new value and the last computed mean and/or std deviation) ?
for the non incremental way, I just do something like:
$$S_N=\sqrt{\frac1N\sum_{i=1}^N(x_i-\overline{x})^2}.$$
mean = Mean(list)
for i = 0 ... | In case someone has to "decrement" and not only "increment" the standard deviation $\sigma$ (for example, when a result $x_i$ in the set is incorrect and needs to be removed or recalculated), you can use this formula:
$
\sigma_{\text{without } x_i} = \sqrt{\frac{n}{n -1} \left[ \sigma_n^2 - \frac{(\bar{x}_n-x_i)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/102978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "79",
"answer_count": 8,
"answer_id": 1
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Simplest proof that $\zeta(s) \to \infty$ as $s \to 1$? For homework I had to prove the divergence of the series $1/(k\log^p k)$ for all real $p$ (it is simple to do so via integration.) However a more elegant means would be to appeal to the behavior of the zeta function $\zeta(s)=\sum\limits_{k=1}^\infty 1/k^s$ on th... | It seems that the most elegant proof is Euler's proof via comparison
\begin{align*}
\zeta (1) &= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{3} + \frac{1}{4}} \right) + \left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right) + \cdots \\
&> 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{4} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/103684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplify _Elementary Calculus_ section 1.6 problem 25 Let's try this again. We're still on problem 25 in section 1.6 of Elementary Calculus.
$$\frac{3-\sqrt{c+2}}{c-7}$$
My first thought is (again) to multiply by $3+\sqrt{c+2}$:
$$=\frac{(3-\sqrt{c+2})(3+\sqrt{c+2})}{(c-7)(3+\sqrt{c+2})}$$
$$=\frac{9-(c+2)}{(c-7)(3+\sq... | From this step:
$$=\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})}$$
you have
$$=\frac{-(c-7)}{(c-7)(3+\sqrt{c+2})}$$
Cancel the common factor and the rest should follow straightforwardly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim _{(x,y) \to (0,0)}$ $\frac{(x^3-y^2)^2}{x^4+y^4}$- two methods different conclusion. I tried to use couple of inequalities in this way:
$0\leq\left|\frac{(x^3-y^2)^2}{x^4+y^4}\right| \leq \frac{(x^3)^2}{x^4}=x^2$, so now I can use the squeeze theorem and conclude that this limit is 0, but when I choose a route s... | The second result is the correct one. To see this, write $\frac{(x^3-y^2)^2}{x^4+y^4}=\frac{x^6}{x^4+y^4}-2x\frac{x^2y^2}{x^4+y^4}+\frac{y^4}{x^4+y^4}$. Since
$$\left|\frac{x^6}{x^4+y^4}-2x\frac{x^2y^2}{x^4+y^4}\right|\leq x^2+|x|\frac{x^4+y^4}{x^4+y^4}=x^2+|x|,$$
we just have to study $\lim_{(x,y)\to (0,0)}\frac{y^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to analyze convergence and sum of a telescopic series? I can't find a generic form Hello I would like to know what is the generic form to calculate a telescopic series.
http://www.dummies.com/how-to/content/how-to-analyze-a-telescoping-series.html
here uses a sum that startes at $n=1$, but what if it started at $n=... | If a telescoping sum starts at $n=m$, then
$$
\sum_{n=m}^{N}\left( a_{n}-a_{n+1}\right) =a_{m}-a_{N+1}
$$
and the telescoping series is thus
$$
\begin{eqnarray*}
\sum_{n=m}^{\infty }\left( a_{n}-a_{n+1}\right) &=&\lim_{N\rightarrow
\infty }\sum_{n=m}^{N}\left( a_{n}-a_{n+1}\right) \\
&=&a_{m}-\lim_{N\rightarrow \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Location of a root of a cubic polynomial For $\alpha\in(0,\frac12)$, $\beta\in(0,\infty)$, $N\in\mathbb N\backslash\{0\}$ and $n\in\{0,\ldots,N\}$, how can I prove that exactly one zero of the cubic polynomial
$$
(N+2\beta)x^3-(N+n+3\beta)x^2+(n+\beta+N\alpha-N\alpha^2)x+n\alpha^2-n\alpha
$$
lies in $[\alpha,1-\alpha... | Evaluating at $\alpha$, we have:
$$\begin{align*}
f(\alpha) &= (N+2\beta)\alpha^3 - (N+n+3\beta)\alpha^2 + (n+\beta+N\alpha-N\alpha^2)\alpha + n\alpha^2 - n\alpha\\
&= (N+2\beta - N)\alpha^3 + (-N-n-3\beta+N+n)\alpha^2 + (n+\beta-n)\alpha\\
&= 2\beta\alpha^3 - 3\beta\alpha^2 + \beta\alpha\\
&= \alpha\beta(2\alpha^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/107519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $ 1 + \sum_{i=0}^{n-2} [ \prod_{k=0}^{i} (n-k)/(r-k) ] $ Is there a way in which the expression
$$ 1 + \sum_{i=0}^{n-2} \prod_{k=0}^{i} \dfrac{n-k}{r-k} $$
may be simplified?
It would simplify some equations I am working with.
| First, using recurrence equation for Euler's $\Gamma$-function:
$$
\prod_{k=0}^m \frac{n-k}{r-k} = \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma(r+1)}
$$
Then, changing summation variable as $m \to n-2-m$:
$$
1 + \sum_{m=0}^{n-2} \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/111803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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What is the minimum possible value of $(a + b + c)$? $a$, $b$ and $c$ are real positive numbers satisfying
*
*$\dfrac{1}{3} \le ab + bc + ca \le 1$ and
*$abc \ge \dfrac{1}{27}$
What is the minimum possible value of $(a + b + c)$?
| By the AM-GM inequality, $\sqrt[3]{abc}\leq \frac{a+b+c}{3}$.
Since $abc \geq \frac{1}{27}$, this implies that $\sqrt[3]{abc}\geq \frac{1}{3}$.
So $\frac{1}{3} \leq \frac{a+b+c}{3}$ and hence $1 \leq a+b+c$.
This minimum value is achieved in the symmetric case where $a=b=c=\frac{1}{3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of a? Let $a$ and $b$ be two positive numbers such that $a\gt b$. Let $G$ be the geometric mean of $a$ and $b$ (that is, $G=\sqrt{ab}$), and $H$ be the Harmonic mean of $a$ and $b$, that is, $$H = \frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
If $4G = 5H$, what is the value of $a$?
| According to the given information we have,
$$4\sqrt{ab} = 5(\frac{2ab}{a+b})$$
$$(a+b) =\frac{5}{2}\sqrt{ab})$$
$$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}} = \frac{5}{2}$$
Let , $t$ $ =$ $\sqrt{\frac{a}{b}}$
$$[t+\frac{1}{t}=\frac{5}{2}].....................Eq(1)$$
A clever person will immediately infer that $t=\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3... | Note that this works
to find the derivative
of $x^{1/n}$
where $n$
is a positive integer.
We use
$a^n-b^n
=(a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k}
$.
$\begin{array}\\
\frac{(x+h)^{1/n}-x^{1/n}}{h}
&=\frac{(x+h)^{1/n}-x^{1/n}}{h}
\frac{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to prove that $\frac{\cos\alpha}{\cos\beta}=a/b$ If $\alpha \not= \beta$, and $$ a\tan \alpha+b\tan\beta=(a+b)\tan\frac{\alpha+\beta}{2}$$ then can we prove that $\frac{\cos\alpha}{\cos\beta}=\frac{a}{b}$?
Seems like I am stuck on this one.
| This is true if we assume $\displaystyle \sin\left(\frac{\alpha -\beta}{2}\right) \neq 0$ and $\displaystyle b \neq 0$.
Let $\displaystyle \frac{a}{b} = c$
Rewrite your equation as
$$c\tan \alpha + \tan \beta = (c+1)\tan(\frac{\alpha+\beta}{2})$$
$$ c\left(\tan \alpha - \tan\left(\frac{\alpha+\beta}{2}\right)\right) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limits: dividing by the highest power does not work in this case I believe to solve expressions like:
$$\lim_{x\to \infty} (2x^2 + 1)$$
we need to divide each term in the numerator and denominator by the highest power. In the previous case, the highest power is $x^2$, so we get
$$\lim_{x \to \infty} \left(2 + \frac{1}... | What you are saying in your calculations is $2x^2 +1 = (2 + 1/x^2)$, which is wrong. What you want is $2x^2 +1 = x^2(2 + 1/x^2)$, thus
$$\lim_{x \to \infty} 2x^2 +1 = \lim_{x \to \infty} x^2(2+ 1/x^2) = \lim_{x\to \infty} 2x^2 = \infty$$
for your second limit you are making the same mistake: $\sqrt{n} + 2 \neq 1+ 2/\sq... | {
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"url": "https://math.stackexchange.com/questions/116282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Which formula do I use to integrate $ \int {\sqrt{x^2 + 81} \over 2} \,dx $ I am having trouble with a question really need help please.
$$
\int {\sqrt{x^2 + 81} \over 2} \,dx
$$
I thought about taking the square root off and turning the question into $\frac 12 \int (x^2 +81)^{1/2}\, dx$ but then wondered if I could ... | $$ \int \sqrt{x^2+81} dx $$
Integration by parts formula
$$ \int u(x)v^{'}(x) dx = u(x)v(x) - \int v(x)u^{'}(x) dx$$
Therefore assume that $v^{'}(x) = 1$ in this case
Denote the integral
$$ I = \int \sqrt{x^2+81} dx$$
$$
\begin{align*}
I &= x \sqrt{x^2+81} - \int \frac{x^2}{\sqrt{x^2+81}} dx\\
&= x \sqrt{x^2+8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Help evaluating $\int \frac{dx}{(x^2 + a^2)^2}$ I have following integral and it should be simple, however whatever substitution I use and no matter how many times I integrate it by parts (or combine both) I never get the correct solution (or any alternative solution):
$$\int \frac{dx}{(x^2 + a^2)^2}$$
I'm looking for ... | $$\int\dfrac{1}{(x^2+a^2)^2}dx$$
Hint : Does there exist a function which when differentiated gives something similar to $1\over (x^2+1)$ ?
Put $\tan^{-1} \frac{x}{a} = t$. This gives $\dfrac{1}{1+(\dfrac{x}{a})^2}.\dfrac{1}{a}.dx = dt = \dfrac{a}{a^2+{x}^2}.dx$
$$\int\dfrac{1}{(x^2+a^2)^2}dx=\int \dfrac{1}{a.( (a\tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Why Doesn't This Integral $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$ Work? I am trying to solve this integral, which is incorrect compared to Wolfram|Alpha. Why doesn't my method work?
Find $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$
Side work:
$\sin{\theta} = \frac{3}{x}$
$x = \frac{3}{\sin{\theta}} = 3 \csc{\theta}$
$dx = ... | Let $I = \int \frac{\sqrt{x^2 – 9}}{x^2}$
Easiest is a hyperbolic substitution:
Put $x = 3\cosh{u}, u \le 0$, so $dx = 3\sinh(u)du$
Then \begin{align}
I &=\int\frac{3\sinh(u^2)}{9 \cosh^2(u)}du\\\\
&= \int\tanh^2(u)du\\\\
&= \int\left(1–\mathrm{sech}^2(u)\right)du\\\\
&= u–\tanh(u)+c\\\\
&= u-\frac{3\sinh(u)}{3\cosh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Is there a way to create a summation function for this? I have an algorithm that generates terms as follows:
$$\begin{align*}
\text{Term 1}: &2\times 3\times\left(\frac{1}{2}+\frac{1}{3}\right) + 4\\
\text{Term 2}: &2\times 3\times 4\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + 3\times 5\\
\text{Term 3}: ... | The last term is $$\frac{(n+3)!}{2\times (n+2)},$$
or
$$\frac{(n+1)!(n+3)}{2}$$
where $n$ is the number of the term you are in.
So the $n$th term is:
$$\begin{align*}
t_n &= (n+2)!\left(\frac{1}{2}+\cdots+\frac{1}{n+2}\right) + \frac{(n+3)!}{2\times(n+2)}\\
&= (n+2)!\left(\frac{1}{2}+\cdots+\frac{1}{n+2}\right) + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing point is the orthocenter Given a rectangle WXYZ, let R be a point on its circumscribed circle. Show that, out of the orthogonal projections of R onto WX, XY, YZ, and ZW; one out of these 4 points is the orthocenter of the triangle created by the other three.
| Continuing with the notation and analytic-geometric approach from my comment,
$$
\eqalign{
R_1R_2 \perp R_3R_4 \iff
&
\{y=r\sin\theta\} \perp
\{x=r\cos\theta\}
\\\\&\text{True (horizontal/vertical lines)}
\\\\
R_1R_3 \perp R_2R_4 \iff
&
\overline{(a,r\sin\theta)(r\cos\theta,b)} \perp
\overline{(-a,r\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to solve $\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2$? I have tried something to solve the series
$$\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2.$$
My approach is :
$$(1+x)^n=\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^... | First we address the overflow issue. Note that $10^9+7$ is relatively prime to all the numbers that come up in a naive calculation of $\binom{2n}{n}$. Indeed $10^9+7$ happens to be prime. So when we are calculating, we can always reduce modulo $10^9+7$ as often as necessary to prevent overflow.
Now to the identity. We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
Show by substitution that $\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \,\mathrm dx = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm dx$ How do you show
$$\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \,\mathrm{d}x = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm{d}x$$
without integrating by parts... | For any integral of the form
$$I = \int_0^{\pi} x f(\sin x) \mathrm{d}x $$
since $\sin(\pi-x) = \sin(x)$, using substitution $x = \pi-u$, or $u=\pi- x$, one may reduce this integral to the following using
when $x=0, u=\pi$, and when $x=\pi, u=0$, and also $ \mathrm{d}x = \mathrm{-d}u$
$$
\begin{align*}
I = \int_0^{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Showing the identity: $\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$ My knowledge of trigonometry is still insufficient to resolve this problem. Any help would be greatly appreciated.
$$\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$$
| Hoping this is not homework:
$$\cot x - 8\cot 8x = \cot x -8\cot 8x= \cot x-8\frac{\cot^24x -1}{2\cot 4x}$$
$$= \cot x-4\frac{\cot^24x -1}{\cot 4x}= \cot x -4\cot 4x +4\tan 4x$$
$$= \cot x -4 \frac{\cot^22x -1}{2\cot 2x}+4\tan 4x=\cot x -2\cot 2x + 2\tan 2 x + 4 \tan 4x$$
$$=\cot x -2 \frac{\cot^2x -1}{2\cot x}+2\tan 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
find the largest of the three angle of the triangles?
The sides of a triangle are given to be $$ x^2 +x +1 ,$$ $$2x+1,$$ $$x^2-1$$Then find the largest of the three angle of the triangles .
Source:ISI entrance exam sample question 2016
I tried by applying hero's formula and then equating to the formula
$$ 0.5 bc \s... | Note that :
$
\begin{cases}
x^2+x+1 > 2x+1 > x^2-1, & \text{if } x\in(1,1+\sqrt 3) \\
x^2+x+1 > x^2-1 > 2x+1, & \text{if } x\in(1+\sqrt 3,+\infty)
\end{cases}$
Hence , the largest angle of the triangle is angle opposite to the side $x^2+x+1$ . Let us denote that angle as $\alpha$ , then according to Cosine rule w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Please check my answer to $\sum\limits_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$ $$\sum_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$$
I have this answer, please let me know if there is a more beautiful proof.
My answer:
at first, we prove two inequalities:
*
*If $x\in (x,\pi)$ then $\sin x \leq x$
*When $x\i... | The best it is possible to state is:
$$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots$$
Call $f_N(x)=\sum_{n=1}^N \frac{\sin(nx)}{n}$: it is a $2\pi$-periodic function converging to $\frac{\pi-x}{2}$ in $L_2\left([0,2\pi]\right)$.
Since:
$$\frac{d f_N(x)}{dx}= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| The complex roots of $x^5-1$ are:
$$
\begin{align}
x_1&=1\\
x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\
x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\
x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\
x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5
\end{align}
$$
using Vieta's formulas you get
$$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 18,
"answer_id": 5
} |
Factorize polynomial over $GF(3)$ I want to factorize $x^{11}-1$ over $GF(3)$ but I'm stuck at
$(x-1)(x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1).$
I have tried to do it trial and error but failed. Is
$$
x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1
$$
already irreducible over $GF(3)$?
| The polynomial
$$
\phi_{11}(x):=\frac{x^{11}-1}{x-1}=\sum_{i=0}^{10}x^i\in GF(3)[x]
$$
is not irreducible. Because $11\mid 3^5-1=242$, there is a primitive eleventh root of unity in the field $GF(3^5)$. This is a quintic extension field of $GF(3)$, so the minimal polynomials of all the primitive eleventh roots of unity... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is $\cos \cos 1 - \sin \sin \sin 1$ positive? How can I show that
$$\cos \cos 1 - \sin \sin \sin 1$$
is positive?
This is motivated by this question. If
$$\begin{align}
f(x) &= \cos \cos \cos \cos(\pi/2 + ix) - \sin \sin \sin \sin(\pi/2+ix)\\
&= \cos \cos \cos \sinh x - \sin \sin \sin \cosh x,
\end{align}$$
then ... | Recall that
$$
1-\frac{1}{2}x^2 \;<\; \cos x \;<\; 1-\frac{1}{2}x^2 + \frac{1}{24}x^4
$$
for $x>0$. From the upper bound, we see that $\cos 1 < 13/24$. Since $\cos$ is decreasing, it follows that
$$
\cos(\cos 1) \;>\; \cos\left(\frac{13}{24}\right) \;>\; 1-\frac{1}{2}\left(\frac{13}{24}\right)^2 \;>\; 0.85.
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/132172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Positive integers expressed in the form : $\frac{a^b+c}{a+c}$ Note : This problem has no specific source .
Is it true that :
Every positive integer $n$ greater than $1$ can be expressed in the form :
$n=\frac{a^b+c}{a+c}$ , where $a,b>1$ , and $c \in \mathbb{Z} \backslash \{0\}$
I am able to prove following :
$\foral... | Let $a = n$, $b = 3$, $c = n^2$. This does it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum_{n=0}^{\infty} \left(\frac{1}{2^{n-1}+1 }+ \frac{1}{2^n+1}\right)$ Evaluating $$\sum_{n=0}^{\infty} \left(\frac{1}{2^{n-1}+1 }+ \frac{1}{2^n+1}\right)$$
Full Question
Provided Answer
But how do I get from
$$\sum \frac{1}{2^{k-1}+1} - \frac{1}{2^k+1} = \frac{1}{2^{\color{red}0-1}+1}...$$
Why is the... | It is a finite telescoping sum. For any sequence $\rm\{ x_k\}_{k=1}^\infty$ we have
$$\rm \begin{array}{c l} \sum_{k=a}^b (x_k-x_{k+1}) & \rm =(x_a-\color{Maroon}{x_{a+1}})+(\color{Maroon}{x_{a+1}}-x_{a+2})+\cdots+(x_{b-1}-\color{Purple}{x_b})+(\color{Purple}{x_b}-x_{b+1} ) \\ & \rm =x_a-x_{b+1}. \end{array}$$
Notice t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
inequality-positive semidefinite matrices Let $A$ and $B$ be positive semidefinite matrices of the same size. If the largest eigenvalues of $A$ and $B$ are less than equal to $1$. probe that: $$AB+BA\geq- \frac{1}{4}I$$
In the hint, it says use the fact: $0\leq (A+B-\frac{1}{2}I)^{2}$.
In order for $ (A+B-\frac{1}{2}I)... | you have $0 \le (A + B - \frac 12 I)^2$ as the left hand side is a square:
\[ \textstyle
\langle (A + B - \frac 12 I)^2x, x\rangle = \langle (A+B-\frac 12I)x, (A+B-\frac 12I)^*x\rangle =\langle (A+B-\frac 12I)x, (A+B-\frac 12I)x\rangle = \|(A + B -\frac 12I)x\|^2 \ge 0
\]
And for the second part we have $A^2 \le A$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$
Prove that for all integers $n$, $n \geq 1$,
$$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$
How would I go about proving this?
| $$\begin{eqnarray*} (1 + 3 + 5 + ... + (2n - 1)) &=& (1 + 2 + 3 + ... + 2n) - (2 + 4 + ... + 2n)\\
&=& \sum_{i=1}^{2n}{i} - 2\sum_{i=1}^{n}{i}\\
&=& \frac{(2n)(2n+1)}{2} - 2\left(\frac{n(n+1)}{2}\right)\\
&=& n(2n+1) - n(n+1)\\
&=& n(2n + 1 - n - 1) = n^{2}\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 9,
"answer_id": 0
} |
Simplify $\int \frac{1}{\sqrt{2-x^2}}\, dx$ Somewhere in the provided answer:
$$\int \frac{1}{\sqrt{2-x^2}} dx = \sin^{-1}{\frac{x}{\sqrt{2}}}$$
How did they get that? What I have:
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2(1-\frac{x^2}{2})}} = \frac{1}{\sqrt{2} \sqrt{1-\frac{x^2}{2}}}$$
$$\frac{1}{\sqrt{2}} \int \f... | You made a mistake in the last step. To see why, let $u = \frac{x}{\sqrt{2}}$, $du = \frac{dx}{\sqrt{2}}$.
\begin{align*}
\frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^2}} &= \frac{1}{\sqrt{2}} \int \frac{\sqrt{2}du}{\sqrt{1-u^2}} \\
&= \int \frac{du}{\sqrt{1-u^2}} \\
&= \arcsin{u} + c \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Induction Proof and divisibility by $2^n$ I'm trying to use induction to prove that for every integer $n > 0$ there exists an $n$-digit integer A(n) that is divisible by $2^n$ and that consists entirely of digits “1” and “2”.
Does anyone have a clue where to start ?
I found examples that work on small numbers and the i... | Define the $n$-digit number as follows. Let $f(1)=2$. Suppose that we have defined $f(n)$, and $2^n$ divides $f(n)$. Then define $f(n+1)$ as follows.
If $2^{n+1}$ divides $f(n)$, then $f(n+1)$ is obtained by putting a $2$ in front of the decimal expansion of $f(n)$. Or, to put it another way, $f(n+1)=2\cdot 10^n+f(n)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Convergence of the series $\sum \frac{n}{(1+n)\left( 1 + n \sqrt{n}\right)}$ I need to determine whether the following series converges or diverges:
$$ \sum_{n=1}^\infty \frac{n}{(1+n)\left( 1 + n \sqrt{n}\right)} $$
My first idea was to show that $a_n$ tends to zero, and so it does.
My next plan was to rewrite the su... | For deciding convergence, Siminore's comparison test answers affirmatively.
If for real $x>0$ we wish to approximate the sum over $\mathbb{Z}^+$ of
$$
\eqalign{
f(x)
&=\frac{x}{(x+1)(x^{3/2}+1)}
=\left(1-\frac1{1+x}\right)\left(1+x^{3/2}\right)^{-1}\\
&=\frac1{1+x^{3/2}}
-\frac1{\left(1+x\right)\left(1+x^{3/2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Maclaurin Series
Integrate using first three terms of appropriate series...
$$\int_0^1 \sin x ~dx.$$
So I use
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}$$
for my three terms and if I integrate just that I get the answer which is $.3103$.
However the solution book is showing a negative is taken outside the integ... | Yes, the minus sign is a typo and should not be there.
Note as well that you don't actually have equalities. Since
$$\sin x\approx x - \frac{x^3}{3!} + \frac{x^5}{5!},$$
and we don't have actual equality, we have that
$$\int_0^1\sin(x^2)\,dx \approx \int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx.$$
L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is my Riemann Sum correct? For the integral: $$\int_{-1}^{5} \left( x^{2} -4 \right) dx$$
My calculations:
$$\begin{align*}\Delta x &= \frac6n\\\\
x_i &= -1 + \frac{6i}n\\\\
f(x_i) &= 1 + \frac{36i^2}{n^2} -4\\\\
A&=72
\end{align*}$$
I'm unsure if this is correct as it is my first attempt at doing this type of problem.... | Your $\Delta x$ and $x_i$ are fine, but you've gone astray after that: $$f(x_i)=x_i^2=\left(-1+\frac{6i}n\right)^2=1-\frac{12i}n+\frac{36i^2}{n^2}\;.$$ Your Riemann sum is then
$$\begin{align*}
\sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\right)\frac6n&=\frac6n\sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Derivative of $\ln(x\sqrt{x^2-1})$ I am trying to find the derivative of $\ln(x\sqrt{x^2-1})$ but I can not get what the book gets.
I get $$\frac{1}{x \sqrt{x^2-1}} \cdot \sqrt{x^2-1} + x\cdot\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x$$
which I reduce to
$$\begin{align}
&\frac{1}{x\sqrt{x^2-1}}\sqrt{x^2-1} + x^2(x^2-1)^\fr... | $\log (x\sqrt{x^{2}-1})=\log x+\log(\sqrt{x^{2}-1})=\log x+ \frac{1}{2}\log({x^{2}-1})$
Let $y=\log x+ \frac{1}{2}\log({x^{2}-1}).$
Now differentiate with respect to $x$, we get,
$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{2}\frac{1}{x^{2}-1}2x$
$\therefore \frac{dy}{dx}=\frac{1}{x}+\frac{x}{x^{2}-1}$=$\frac{x^{2}-1+x^{2}}{x(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions? We have the following evaluations:
$$\begin{aligned}
&\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\
&\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\
&\sum_{n=1}^\infty \... | The paper gives (second-to-last page, right-hand column)
$$\sum_{n\ge1}\frac{1}{n^5\binom{2n}{n}}=\frac{9\sqrt{3}\pi}{8}\color{Purple}{L\left(4,\left(-3\atop\circ\right)\right)}+\frac{\pi^2\zeta(3)}{9}-\frac{19\zeta(5)}{3}. \tag{1}$$
You seek a way to write the $L$-function in purple as a linear combination of Hurwitz ... | {
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"url": "https://math.stackexchange.com/questions/144985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 1,
"answer_id": 0
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How did they get this result? Please, explain these computations:
1) $-\left(\frac{1}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{3}}{2} = \cos x$
How did we get $\frac{\sqrt{3}}{2}$ from $-\left(\frac{1}{2}\right)^2 +1$?
2) $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{2}}{2} = \cos x$
How did we g... | You have:
*
*$$\cos^2 x = -(\frac{1}{2})^2 + 1$$
$$\cos^2 x = -\frac{1}{4} + 1=\frac{3}{4}$$
Which is:
$$\sqrt \cos^2 x = \sqrt \frac{3}{4}$$
$$\cos x = \pm \frac{\sqrt 3}{2}$$
Same works for the second one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Are there any integer solutions to $x^2 - (n^2 - 2)y^2 = -1$? I was just wondering if there are any integer solutions to the Diophantine equation:
$x^2 - (n^2 - 2)y^2 = -1 \ \ $ for $n > 2$
I don't think there are any but can't prove why.
| Note that if $d$ is divisible by a prime $p$ of the form $4k+3$, then the equation $x^2-dy^2\equiv -1$ cannot have a solution, for $x^2\equiv -1\pmod{p}$ does not have a solution.
If $n>2$ is odd, then $n^2-2\equiv -1\pmod{4}$, so $n^2-2$ is divisible by a prime of the form $4k+3$.
If $n$ is divisible by $4$, then agai... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
does $\int_0^\infty x/(1+x^2 \sin^2x) \mathrm dx$ converge or diverge? $$\int_0^\infty x/(1+x^2\sin^2x) \mathrm dx$$
I'd be very happy if someone could help me out and tell me, whether the given integral converges or not (and why?). Thanks a lot.
| Since $-1 \leq \sin{x} \leq 1$, you have $\sin^{2}{x} \leq 1$, and from this you can conclude that
\begin{align*}
1+x^{2}\sin^{2}{x} &\leq 1+2x^{2} \\\ \Longrightarrow \frac{1}{1+x^{2}\sin^{2}{x}} &\geq \frac{1}{1+2x^{2}} \\\ \Longrightarrow \frac{x}{1+x^{2}\sin^{2}{x}}&\geq \frac{x}{1+2x^{2}} \\\
\Longrightarrow \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/152424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Diophantine equation $x^y-y^x=11$ How can one find all integer solutions to $x^y-y^x=k$, for a given k?
Example case $x^y-y^x=11$
| The numbers $x^y - y^x$ blow up when $x$ and $y$ get large. Besides the trivial solutions of the form $(k+1, 1)$ the only values of $k < 1000$ giving solutions are:
$$\begin{align} 3^2 - 2^3 &= 1 \\ 2^5 - 5^2 &= 7 \\ 3^4 - 4^3 &= 17 \\ 2^6 - 6^2 &= 28 \\ 2^7 - 7^2 &= 79 \\ 3^5 - 5^3 &= 118 \\ 2^8 - 8^2 &= 192 \\ 4^5 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Multiple roots of a polynomial over a field of characteristic $p$ I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$
have a root of multiplicity $>1$ over the field of characteristic $p$.
$ p=2, 3, 5, 7 $
Please help.
For $F$ a field of characteristic $3$, $f(x)= x^4 + x = x(x^3+1)$ ... | Here's an elementary answer:
a multiple root of $x^4 + x + 6$ in $\mathbf F_p$ is the exact same thing as a root common to $x^4 + x + 6$ and its derivative $4x^3 + 1$. But, on $\mathbf Z$, we have the following formula:
$$4(x^4 + x + 6) = (4x^3 + 1) x + (3x + 24).$$
So, for $p \neq 2$, a root $\alpha$ of $x^4 + x + 6$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\ln(2)$ using Riemann sum.
Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?
Show that
$$\ln(2) = \lim_{n\rightarrow\infty}\left( \frac{1}{n + 1} + \frac{1}{n + 2} + ... +
\frac{1}{2n}\right)$$
by considering the lower Riemann sum of $f$ where $f(x) = \frac{1}{x... | METHOD I
We may recall the celebre limit that yields Euler-Mascheroni constant, namely:
$$\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$$ $\tag{$\gamma$ is Euler-Mascheroni constant}$
Then everything boils down to:
$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
| WLOG assume $a \leq b$, therefore, $a^2 \leq b^2$ and $\frac{1}{b - 1} \leq \frac{1}{a - 1}$. Therefore, by Rearrangement Inequality,
$$\frac{a^2}{b-1} + \frac{b^2}{a - 1} \geq \frac{a^2}{a-1} + \frac{b^2}{b - 1}$$
From here, one can simply follow Valentin's steps,
$$\frac{a^2}{a - 1} + \frac{b^2}{b - 1} = 4 + (a - 1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 5
} |
Limit of $ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}},m\rightarrow\infty$ I would like to show that $$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{2}$$
Using integrals:
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} \leq \sum_{n=m+1}^{\infty} \frac{m}{n\s... | Use this and think about
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} \leq \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}} \leq m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}}+\frac{m}{(m+1)\sqrt{2m+1}}$$
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} = \frac{\pi}{2}-\arctan \left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the limit I need to find the limit of this problem. I pretty much know you have to multiply by the conjugate but I get lost after I do that.
$$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$$
| $$\begin{eqnarray}
-\lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{1-x} &=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\\
&=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\frac{1/\sqrt{x}+1}{1/\sqrt{x}+1}\\
&=& \lim\limits_{x\to 1} \frac{1/x - 1}{(x-1)(1/\sqrt{x}+1)}\\
&=& \lim\limits_{x\to 1} \frac{1}{1/\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Calculate: $\sum_{k=0}^{n-2} 2^{k} \tan \left(\frac{\pi}{2^{n-k}}\right)$ Calculate the following sum for integers $n\ge2$:
$$\sum_{k=0}^{n-2} 2^{k} \tan \left(\frac{\pi}{2^{n-k}}\right)$$
I'm trying to obtain a closed form if that is possible.
| We have this nice identity $$\tan(\theta) = \cot(\theta)-2 \cot(2 \theta)$$
Making use of this, and denoting $\displaystyle \sum_{k=0}^{m} 2^k \tan(2^k \theta)$ as $S$, we get that \begin{align}S & = \tan(\theta) + 2 \tan(2 \theta) + 4 \tan(4 \theta) + \cdots + 2^m \tan(2^m \theta)\\
& = \cot(\theta) -2 \cot(2 \theta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Integral as a limit of a sum How do I get
$$
\int_a^b \frac{1}{x}dx = \ln\left(\frac{b}{a}\right)$$
as a limit of sum. The constant width partition of the interval $(a,b)$ doesn't seem to work.
| $$\int _a^b {f(t) dt}=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n f(\frac{k(b-a)}{n}+a)$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{1}\sum \limits_{k=1}^n \frac{1}{k(b-a)+na}$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Homogenous polynomials In section 3.1 (3rd paragraph on page 4) in this paper, I cannot understand why $Q$ and $R$ are homogeneous:
(Given $A$, $B$ are homogenous. Capital letters denote homogenous polynmials.)
| The trick is to realize that all terms of $q$ (or $r$) are homogeneous of the same degree BUT you have to introduce a grading with negative degrees in $K(y,z)$.
For example, if you want to divide $yz^2+x^3$ by $xz+y^2$ you first use the euclidean algorithm to obtain $$ yz^2+x^3= (\frac {y}{x}z -\frac {y^3}{x^2}) \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/163200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factoring by grouping: $x^4 - y^4 -4x^2 + 4$ Please help me factor $x^4 - y^4 -4x^2 + 4$ by grouping terms.
Thank you.
| $$x^4-y^4-4x^2+4=(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify $\frac{9}{2}(1 + \sqrt 5)\sqrt{10 - 2\sqrt 5} + 9\sqrt{5 + 2\sqrt 5}$ Simplify $\displaystyle{\frac{9}{2}(1 + \sqrt 5)\sqrt{10 - 2\sqrt 5} + 9\sqrt{5 + 2\sqrt 5}}$.
I get this when I was doing another Q,
but I don't know how to further simplify it.
Can anyone help me, please?
| Hint: Let's note $o:=\frac {1+\sqrt{5}}2$, $a:=\sqrt{10-2\sqrt{5}}$ and $b:=\sqrt{5+2\sqrt{5}}$
then $ab=\sqrt{30+10\sqrt{5}}=5+\sqrt{5}$
Compute $(o\cdot a+b)^2$ to conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/165214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Computing large powers $\!\bmod n$ when $n$ has multiple prime factors. How will the congruence modulo works for large exponents? What theorem/s may be used? For example to show that $7^{82}$ is congruent to $9 \pmod {40}$.
| Maybe just start calculating. Note that $7^2=49\equiv 9\pmod{40}$. So $7^4\equiv 81\equiv 1\pmod{40}$. We got lucky.
It follows that $7^{80}=(7^{16})^5\equiv 1\pmod{40}$. Thus
$7^{82}=7^{80}7^2\equiv 49\equiv 9\pmod{40}$.
In general, repeated squaring, and reduction with respect to our modulus (in this case $40$) g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/165670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What's $T\left(n\right)$? If $T\left( n \right) = 8T\left( n-1 \right) - 15T\left( n-2 \right); T\left(1\right) = 1; T\left( 2 \right) = 4$,
What's $T\left(n\right)$ ?
I use this method:
Let $c(T(n) - aT(n-1)) = T(n-1) - aT(n-2)$
from $T(n) = 8T(n-1) - 15T(n-2)$, we can get $\begin{cases}c = \frac{1}{3}\\a = 5\end{cas... | This is second order recurrence equation. Since $T(n)=8T(n-1)-15T(n-2)$, we first solve the quadratic equation: $x^2=8x-15$. Its roots are $3$ and $5$. Therefore, $T(n)$ must be in the form:
$$T(n)=a\cdot 3^n+b\cdot 5^n, n\geq 1$$for some constants $a$ and $b$.
To find $a$ and $b$, since $1=T(1)=3a+5b$ and $4=T(2)=9a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Recurrence relation $C_n = n + 1 + \dfrac{2}{n}\sum\limits_{k=0}^{n-1}C_k$. A Discrete Mathematics book from which I'm self-studying ("Discrete Mathematics and Its Applications", by Kenneth Rosen) asks me to do the following:
Given the following recurrence relation:
$$C_n = n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k$$
The ... | As other answers have shown, the approach that you took is not the easiest one, but it can be made to work. Note first that the new recurrence is equivalent (after division by $n$) to
$$C_n=\left(1+\frac1n\right)C_{n-1}+2\;;\tag{1}$$
this will be a little easier to work with. I’ll assume $(1)$ as an induction hypothe... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
| A solution for straightforward mathematicians. :-)
It seems the following.
After multiplication of both sides of the inequality by a common denominator and simplification, we reduce the initial inequality to
$$1+ab^2+bc^2+ca^2\ge 4a^3b^3c^3.$$
Since $$ab^2+bc^2+ca^2\ge 3abc$$ it suffices to check that
$$1+3abc-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/168704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Computing a double series. How to compute following double series,
$$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \cfrac{(-1)^{(i+j)}}{i^2+2nij+j^2} $$
where $n\in \mathbb{N}$.
| This double series is not absolutely convergent. You might try a kind of Abel summation:
$\lim_{z \to -1+} f_n(z)$ where
$$f_n(z) = \sum_{i=1}^\infty \sum_{j=1}^\infty \dfrac{z^{i+j}}{i^2 + 2nij + j^2}$$
But I would find a closed form surprising.
It might be interesting to try this asymptotically as $n \to \infty$, u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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$(\sin\theta+\cos\theta)^2=1+\sin2\theta$ 49) $(\sin\theta+\cos\theta)^2=1+\sin2\theta$
Left Side:
\begin{align*}
(\sin\theta+\cos\theta)^2=\sin^2\theta+2c\cos\theta\sin\theta+cos^2\theta=1+2\cos\theta\sin\theta
\end{align*}
This can either be $1$ or I can power reduce it. I don't know.
Right Side:
\begin{align... | Open parentheses and use:
$$(1)\,\,\sin^2x+\cos^2x=1$$
$$(2)\,\,\sin 2x=2\sin x\cos x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to evaluate $\int 1/(1+x^{2n})\,dx$ for an arbitrary positive integer $n$? How to find
$$\int\dfrac{dx}{1+x^{2n}}$$
where $n \in \mathbb N$?
Remark
When $n=1$, the antiderivative is $\tan^{-1}x+C$. But already with $n=2$ this is something much more complicated. Is there a general method?
| If the integral is taken from $0$ to $\infty$, there is more than one way to evaluate this. One is
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}t}{1+t^{2n}}
&=\int_0^1\frac{\mathrm{d}t}{1+t^{2n}}+\int_0^1\frac{t^{2n-2}\,\mathrm{d}t}{1+t^{2n}}\\
&=\int_0^1(1-t^{2n}+t^{4n}-t^{6n}+\dots)\,\mathrm{d}t\\
&+\int_0^1(t^{2n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$
Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$
$$\cos^2x-\sin^2x=\sin$$
Edit
$$1-\sin^2x-\sin^2x=\sin x$$
$$2\sin^2 x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$x=-1,\dfrac{1}{2}$$
$$x=\sin^{-1}(.5)=30^{\circ}=\dfrac{\pi}{6}$$
$$... | The calculation is almost completely correct. You reached the two possibilities $\sin x=\frac{1}{2}$ and $\sin x=-1$.
We are interested in solutions in the interval $-\pi \lt x\le \pi$.
Certainly $x=\frac{\pi}{6}$ is a solution, since $\sin(\pi/6)=\frac{1}{2}$. But there is another $x$ in our interval whose sine is $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$ How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$
I have not been sucessful to do this.
| Hint:
$$ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} $$
and
$$ 1 - \frac12 + \frac13 - \frac14 + \dotsb = \ln 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving a number with $3^n$ equal digits is divisible by $3^n$ Prove a number with $3^n$ equal digits is divisible by $3^n$.
My thoughts about the problem are: a number with $3^n$ equal digits $d$ is equal to $d\frac {10^{3^n} - 1} {9}$. We use Lifting The Exponent lemma, or plain induction.
| This follows from the fact that $x^3-1=(x-1)(x^2+x+1)$ and that $10\equiv1\pmod{3}$. That is,
$$
10^{3^n}-1=\left(10^{3^{n-1}}-1\right)\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)\tag{1}
$$
and
$$
\begin{align}
\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)
&\equiv1+1+1\\
&\equiv0\pmod{3}\tag{2}
\end{align}
$$
so tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Help on differential equation $y''-2\sin y'+3y=\cos x$ $y''-2\sin y'+3y=\cos x$
I'm trying to solve it by power series, but I just can't find the way to get $\sin y'$. Is there any special way to find it?
| Maple finds the two fundamental solutions:
(with $y(0)=0$ and $y'(0)=b$)
$$\eqalign{ y \left( x \right) &=bx+ \left( \sin \left( b \right) +{\frac {1}{2}}
\right) {x}^{2}+ \left( \frac23\,\cos \left( b \right) \sin \left( b
\right) +\frac13\,\cos \left( b \right) -\frac12\,b \right) {x}^{3}\cr+& \left( \frac2
3\, \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution.
| $f(x)=1-x^2$ inserting that gives:
$$
\begin{eqnarray}
x^2 f(x) + f(1-x) &=&x^2(1-x^2) + (1 - (1-x)^2)\\
&=&x^2-x^4 +(1-(1-2x+x^2))\\
&=&x^2-x^4 +(1-1+2x-x^2)\\
&=&x^2-x^4+2x-x^2 \\
&=& 2x -x^4
\end{eqnarray}
$$
as required. But I admit: I had some help...
But you can also write it as
$$
x^4+f(x) x^2 -2x-f(1-x)=0
$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
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Prove $\cos 3x =4\cos^3x-3\cos x$ How would I solve the following double angle identity.
$$\cos 3x =4\cos^3x-3\cos x $$
I know $\,\cos 3x = \cos(2x+x)$
So know I have $\,\cos 2x +\cos x \,$ , Which is $\,(2\cos^2x-1)\cos x$
But I am not sure what to do next.
| I will attempt to answer my own question now.
$$\begin{eqnarray}
\cos(2x)(\cos x)-\sin(2x)(\sin x)& =& (2\cos^2 x-1)(\cos x)-2\sin x\cdot\cos x\cdot\sin x\\
&=&2\cos^3x-\cos x-2\sin^2 x\cos x\\
&=&2\cos^3x-\cos x-2(1-\cos^2 x)(\cos x)\\
&=&2\cos^3x-\cos x-2\cos x+2\cos^3 x\\
&=&4\cos^3 x-3\cos x
\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
A three variable binomial coefficient identity I found the following problem while working through Richard Stanley's Bijective Proof Problems (Page 5, Problem 16). It asks for a combinatorial proof of the following:
$$ \sum_{i+j+k=n} \binom{i+j}{i}\binom{j+k}{j}\binom{k+i}{k} = \sum_{r=0}^{n} \binom{2r}{r}$$
where $n \... | Restating your question, you are seeking to find the generating function of the left-hand-side:
$$
g(x) = \sum_{n=0}^\infty x^n \sum_{i+j+k=n}\binom{i+j}{i} \binom{j+k}{j} \binom{k+i}{k} = \sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty x^{i+j+k} \frac{(i+j)! (i+k)! (j+k)!}{i!^2 j!^2 k!^2}
$$
First, carry out ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 0
} |
Proving an estimate for this integral How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$
| In the interval from $1$ to $2$, $\frac{(1+x)^{1/3}}{x^2}\le \frac{3^{1/3}}{x^2}$.
Thus
$$\int_1^2 \frac{(1+x)^{1/3}}{x^2}\,dx\le \int_1^2\frac{3^{1/3}}{x^2}\,dx= \frac{1}{2}(3^{1/3}).$$
In the interval from $2$ to $\infty$, $(1+x)^{1/3}\le \left(\frac{x}{2}+x\right)^{1/3}$. So
$$\int_2^\infty \frac{(1+x)^{1/3}}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Proving $\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$ I would like to know how to prove the following assertion :
For every $n>0$: $$\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$$
| Or you could obfuscate the problem. Let $t=\sqrt{n+1}$. Then we have $t>1$ and $\sqrt{n}=\sqrt{t^2-1}$, so
$$
\begin{align*}
&\sqrt{t^2-1}+\frac{1}{t} \ge t\\
\text{iff}\qquad &\sqrt{t^2-1}\ge t-\frac{1}{t}\\
\text{iff}\qquad &t^2-1 \ge t^2-2+\frac{1}{t^2}\\
\text{iff}\qquad &1 \ge \frac{1}{t^2}\\
\text{iff}\qquad &t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$ using $\frac{dy}{dx}=\frac{2}{(x+3)^2}$ where $y=\frac{x+1}{x+3}$ Find derivative of $$y= \frac{ax+b}{cx+d}$$
I found it to be $$\frac{dy}{dx}=\frac{a}{cx+d}-\frac{c(ax+b)}{(cx+d)^2}$$
Use it to evaluate:
$$\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\fr... | You should have been careful when multiplying by $2$, I think that you should have multiplied by $\frac{1}{2}$ in your formal calculation.
Also, you may need to rewrite your answer a bit to get it right, such as computing $\frac{2}{9} - 1 = -\frac{7}{9}$, which is implicitly required from you by any textbook, I presume... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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.