Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How prove that $ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$? Let $ a,b,c>0$ be such that $ a+b+c=1$. How prove that
$ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$?
| There is no obvious symmetry in this inequality, so let's just try to simplify this with brute force. From $a + b + c = 1$:
*
*$1 - 3c = 3(1 - c) - 2 = 3(a + b) - 2$
*$1 - 4c = 4(1 - c) - 3 = 4(a + b) - 3$
*$c^2 = 1 + a^2 + b^2 -2a - 2b + 2ab$
After replacing these, our goal is now to show:
$3(a^3 + b^3 + a + b) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that for all positive real numbers $a,b$, not both of $a(1-b),b(1-a)$ are greater than $\frac{1}{4}$ Question:
Show that for all positive real numbers $a,b$, not both of $a(1-b),b(1-a)$ are greater than $\frac{1}{4}$
Attempt:
I have attempted several things with this problem. I will note what I feel is the most... | Suppose for some positive real numbers $a,b$ we have $a(1-b) > \dfrac{1}{4}$ and $b(1-a) > \dfrac{1}{4}$.
Then, $a(1-b) \cdot b(1-a) > \dfrac{1}{4} \cdot \dfrac{1}{4}$, which is equivalant to $a(1-a) \cdot b(1-b) > \dfrac{1}{4} \cdot \dfrac{1}{4}$.
Is this possible?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/894658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Co... | Since $y(1)=1$ then by your answer we get
$$\frac A3+\frac B2+c+d=1\iff d=1-\frac A3-\frac B2-c$$
and then the solution is
$$y=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+1-\frac A3-\frac B2-c\\=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How to prove this inequality without using Muirhead's inequality? I ran into a following problem in The Cauchy-Schwarz Master Class:
Let $x, y, z \geq 0$ and $xyz = 1$.
Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.
The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.
The proo... | Another approach, not using the general power inequalities.
For convenience, we say $a=x^{\frac 13}$ and similar for $b$ and $c$.
Just apply the inequality arithmetic mean $>$ geometric mean in the following way (and cyclic permutations):
\begin{align}
\frac 79 a^9+\frac 19 b^9+\frac 19 c^9&=\frac{\underbrace{a^9+a^9+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Sum of the $11^\mathrm{th}$ power of the roots of the equation $x^5+5x+1=0$
Find the sum of the $11^\mathrm{th}$ power of the all roots of the equation
$$
x^5+5x+1=0
$$
My Attempt:
Let $R=\{\alpha,\beta,\gamma,\delta,\mu\}$ be the set of all roots of the equation ${x^5+5x+1=0}$, and let $x\in R$. Then we have
$$
\beg... | If $r$ is a root, then $r^5 = -5r - 1$, hence $r^{10} = (5r + 1)^2$ and $r^{11} = r(5r + 1)^2$. This greatly simplifies subsequent calculations. We can also note $r^3 = -5r^{-1} - r^{-2}$, so we get $$\begin{align*} r^{11} &= 25r^3 + 10r^2 + r \\ &= 25(-5r^{-1} - r^{-2}) + 10r^2 + r \\ &= r - 125r^{-1} + 10r^2 - 25r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Sum of an unorthodox infinite series $ \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots $
This is a pretty unorthodox problem, and I'm not quite sure how to simplify it. Could I get a solution? Thanks.
| $$\begin{align}
\sum_{k=1}^\infty \frac {2k-1}{2^k} & =\frac {1}{2}+\frac {3}{2^2}+\frac {5}{2^3}+\frac {7}{2^4}+\frac {9}{2^5}+\cdots+\frac {2k-1}{2^k}+\cdots
\\[1ex]
2\sum_{k=1}^\infty \frac {2k-1}{2^k} & = \sum_{k=1}^\infty \frac {2k-1}{2^{k-1}}
\\ &= \sum_{j=0}^\infty \frac {2j+1}{2^j}
\\ &= 1 + \sum_{j=1}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Geometry problem involving infinite number of circles What is the sum of the areas of the grey circles? I have not made any progress so far.
| This is an alternate approach to derive the area of the circles
using Descartes four circle theorem.
WOLOG, assume $R = 1$. Let us call
*
*the outer circle (with radius $r_a = 1$) as $C_a$.
*the inner green circle (with radius $r_b = \frac12$) as $C_b$.
*the $1^{st}$ gray circle (the largest one at bottom) as $C_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 2,
"answer_id": 1
} |
What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \... | Perhaps a more unorthodox approach: From the AM-GM inequality,
$$\sqrt[3]{\frac{1}{2}\sin^2x\cdot\frac{1}{2}\sin^2x\cdot\cos^2x} \le \frac{\frac{1}{2}\sin^2x\cdot + \frac{1}{2}\sin^2x\cdot + \cos^2x}{3}$$
$$\sqrt[3]{\frac{1}{4}\sin^4x\cos^2x} \le \frac{\sin^2x + \cos^2x}{3}$$
But it is well known that $\sin^2x + \cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value let $x,y,z\in R$,and such $x>y>z$,and such
$$(x-y)(y-z)(x-z)=16$$
find this follow minimum of the value
$$I=\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$$
My idea: since
$$\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}=\dfrac{x-z}{(x-y)(y-z)}... | You have reached
$$I = \frac{(x-z)^2}{16}+\frac1{x-z} = \frac{(x-z)^2}{16}+\frac4{x-z}+\frac4{x-z}-\frac7{x-z} \ge 3-\frac7{x-z} \ge \frac54$$
The last inequality is because $$x-z = \frac{16}{(x-y)(y-z)} \ge \frac{64}{(x-z)^2} \implies x-z \ge 4$$
P.S. Equality is when $x-y = y - z = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Primitive of $\frac{3x^4-1}{(x^4+x+1)^2}$ How to find primitive of:
$$\frac{3x^4-1}{(x^4+x+1)^2}$$
I am having a faint idea of a type which may or maynot be in the primitve, i.e.:
$$\frac{p(x)}{x^4+x+1}$$
The problem is I am not getting an idea of a substitution to solve this problem.
I might show my work but it is tot... | $\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4-1}{x^2\cdot \left(x^3+1+x^{-1}\right)^2}dx = \int\frac{(3x^2-x^{-2})}{(x^3+1+x^{-1})^2}dx$$
Now Let $$(x^3+1+x^{-1}) = t\;,$$ Then $$(3x^2-x^{-2})dx = dt$$
So $$\displaystyle I = \int\frac{1}{t^2}dt = -\frac{1}{t}+\mathbb{C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the sum $1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + \dots + n\cdot 10^n$ How can I calculate
$$1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + 4\cdot 10^4+\dots + n\cdot 10^n$$
as a expression, with a proof so I could actually understand it if possible?
| Your expression is
$$S = \sum_{k=1}^n k 10^k$$
We can pull out a factor of $10$ to get
$$S = 10 \sum_{k=1}^n k 10^{k-1}$$
Now, consider the function
$$f(x) = \sum_{k=1}^{n}kx^{k-1}$$
So $S = 10f(10)$. For $x \neq 1$, we can rewrite the sum as follows:
$$f(x) = \begin{align}
\frac{d}{dx}\sum_{k=1}^{n} x^k
&= \frac{d}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Simplify $x^3 - 4x^2 + 10x - 125$ I've been trying to simplify $x^3 - 4x^2 + 10x - 125$ for a while now, and I don't seem to progress.
I know that the factors of $125$ are $1$, $5$, $25$ and $125$, but none of these seem to help here. So far I have managed to get $x(x^2 - 4x + 10) - 125$.
Can I go any further than this... | If you have access to any computer algebra system, you can solve for the roots of a cubic equation. (If you don't have access to a computer algebra system then writing everything out this fully is a waste of time!)
$$x^3 - 4x^2 + 10x - 125=(x-x_0)(x-x_1)(x-x_2)=\left(x-\frac{1}{3} 7^{2/3} \left(1+i \sqrt{3}\right) \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$ Please help me to find a closed form for this integral:
$$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$
I suspect it might exist because there are similar integrals having closed forms:
$$\begin{align}\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx&=12\zeta... | An alternate form of the answers given by @Cleo and @Tunk-Fey as sum of $1$ and $1/2$ argumented polylogarithm-products with rational coefficients:
$$I = \frac{99}{16}\operatorname{Li}_5(1)-12\operatorname{Li}_5\left(\frac{1}{2}\right) + 15\operatorname{Li}_1\left( \frac{1}{2} \right)\operatorname{Li}_4(1) - 12\operato... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "131",
"answer_count": 5,
"answer_id": 1
} |
Prove $a^3+b^3+c^3\ge a^2+b^2+c^2$ if $ab+bc+ca\le 3abc$
if $a,b,c$ are positive real numbers and $ab+bc+ca\le 3abc$ Prove:$$a^3+b^3+c^3\ge a^2+b^2+c^2$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: We are given $a... | We only need to show that
$$(bc+ca+ab)(a^3+b^3+c^3)\ge3abc(a^2+b^2+c^2)$$
which is equivalent to
$$\sum_\text{cyc}a(b^4+c^4-b^3c-bc^3)\ge0$$
or
$$\sum_\text{cyc}a(b-c)^2(b^2+bc+c^2)\ge0$$
where $\sum_\text{cyc}$ is a summing operator $\sum_\text{cyc}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$
If $x$, $y$ and $z$ are positive numbers,prove:
$$(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}.$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so... | Because by AM-GM we obtain:
$$\prod_{cyc}(x+y)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\geq\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the integral of $(x^2+4x)/\sqrt{x^2+2x+2}$ Can somebody explain me how to calculate this integral?
$$\int \frac{\left(x^2+4x\right)}{\sqrt{x^2+2x+2}}dx$$
| Let $x+1=\tan\theta$, so $x=\tan\theta-1$, $dx=\sec^{2}\theta d\theta$, and $\sqrt{x^2+2x+2}=\sqrt{(x+1)^2+1}=\sec\theta$.
Then $\displaystyle\int\frac{x^2+4x}{\sqrt{x^2+2x+2}}dx=\int\frac{(\tan\theta-1)^2+4(\tan\theta-1)}{\sec\theta}\sec^{2}\theta d\theta$
$\displaystyle=\int(\tan^{2}\theta+2\tan\theta-3)\sec\theta d\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Are there other methods to evaluate $\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots}$? Are there other methods to evaluate the following series?
$$\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots}$$
My attempt is as follows,
\begin{align}
\frac{1^{-4}+2^{-4}+3^... | Your proof is fine.
The sigma notation can make some things cleaner. The below is the same argument, just written with different notation: $$x=\sum_{n=1}^\infty \frac{1}{n^4}\\y=\sum_{n=1}^\infty \frac{1}{(2n-1)^4}$$
Then $$x=\sum_{n=1}^{\infty} \left(\frac{1}{(2n-1)^4}+\frac{1}{(2n)^4}\right)=y+\frac{1}{2^4}x$$
and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving Example 1.1.15 of secrets in inequalities
if $a,b,c,d$ are positive real numbers,Prove:$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2\ge\frac{1}{a^2}+\frac{4}{a^2+b^2}+\frac{9}{a^2+b^2+c^2}+\frac{16}{a^2+b^2+c^2+d^2}$$
I was reading the solution of it from book and something was not understa... | For the first:
*
*$\displaystyle \frac1{ac+bc}\ge\frac1{a^2+b^2+c^2}$ or $a^2+b^2+c^2\ge ac+bc$
*
*By AM_GM:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2}=ab,\frac{b^2+c^2}2\ge bc,\frac{c^2+a^2}2\ge ca$$
*Since $ab\ge0$, Adding all:
$$a^2+b^2+c^2\ge ab+bc+ca\ge ac+bc$$
For the second:
*
*$\displaystyle \frac{4}{b^2+c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Taking Calculus in a few days and I still don't know how to factorize quadratics Taking Calculus in a few days and I still don't know how to factorize quadratics with a coefficient in front of the 'x' term. I just don't understand any explanation. My teacher gave up and said just use the formula to find the roots or so... | If we multiply two monomials
$$(x-a)(x-b)=x^2-(a+b)x+ab$$
Think of this in terms of the arithmetic mean of the roots $\mu=(a+b)/2$ and the geometric mean $\gamma=\sqrt{ab}$:
$$x^2-2\mu+\gamma^2$$
Then
$$\begin{align}\mu^2-\gamma^2& =\left(\frac{a+b}{2}\right)^2-ab\\
& =\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}\\
& =\frac{a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 7
} |
Finding a differential equation. Find a homogenous linear differential equation of order two of which $$y=e^x$$ and $$y=x+x^2$$ are two independent solutions. All I can think of is trial and error, how do I go backwards from these two equations to the DE?
| The equation is of the form $y''+ay'+by=0$.
Plugging $e^x$, $e^x+ae^x+be^x=0$, i.e. $1+a+b=0$.
Plugging $x+x^2$, $2+a(1+2x)+b(x+x^2)=0$.
Solving for $a$ and $b$,
$$a=\frac{-x^2-x+2}{x^2-x-1}\\b=\frac{2x-1}{x^2-x-1}.$$
Answer:
$$(x^2-x-1)y''+(-x^2-x+2)y'+(2x-1)y=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/918093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Selecting at least one ball of each color An urn contains five red, six white and seven blue balls. Five balls are selected without replacement. Find the probability that at least one ball of each color is selected.
Answer (attempt):
Getting red balls = $5 \choose 1$
Getting white balls = $6 \choose 1$
Getting blue bal... | The total number of combinations is:
$$\dbinom{5+6+7}{5}=\dfrac{18!}{5!\cdot13!}=8568$$
The number of combinations with no red balls is:
$$\dbinom{6+7}{5}=\dfrac{13!}{5!\cdot8!}=1287$$
The number of combinations with no white balls is:
$$\dbinom{5+7}{5}=\dfrac{12!}{5!\cdot7!}=792$$
The number of combinations with no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/920351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How do I prove that any unit fraction can be represented as the sum of two other distinct unit fractions? A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.
Noting that
$\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$
and
$\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$,
find a ... | $\frac{1}{n}$ - $\frac{1}{n + k}$ = $\frac{k}{n(n+k)}$
$\frac{k}{n(n+k)}$ = $\frac{1}{\frac{n(n+k)}{k}}$
So, as long as $\frac{n(n+k)}{k}$ is a whole number, we can say that $\frac{1}{\frac{n(n+k)}{k}}$ is a unit fraction and since
$\frac{1}{n}$ = $\frac{k}{n(n+k)}$ + $\frac{1}{n + k}$
$\frac{1}{n}$ can be expressed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Divergence of $u_{n+1}=1+\frac{n}{u_n}$ Let $u_n$ be defined by $u_0=1$ and $u_{n+1}=1+\frac{n}{u_n}$.
It can be shown easily that if it has a limit, then it must be $+\infty$.
Does $u_n$ diverge to $+\infty$ ?
What I have tried :
Let $x=u_n$
$U_{n+1}-u_n=1+\frac{n}x-x=f(x)$
$f'(x)=\frac{-n^2}{x^2}-1$
$f$ is sctrictly... | The first terms of the sequence are $u_0 = 1, u_1 = 1, u_2=2, u_3=2, \dots$
I prove by induction that
$$\forall n \geq 2,\qquad \ \sqrt{n} < u_n < \sqrt{n}+1.$$
From this follows that $\lim_n u_n = + \infty$.
For $n=2$ we are ok because $\sqrt{2} <2 < \sqrt{2} +1$.
For the inductive step we have
$$u_{n+1} = 1+ \frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seem... | A common trick: multiply by $1$.
That is: multiply $x(\sqrt{x^2+a^2}-\sqrt{x^2+b^2})$ by $\frac{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate the integral $e^{-(c\ln(\frac{1}{x}))^s} dx$? Can anyone help me evaluate
$$\int_{\alpha}^1 \exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}} dx$$,
Where $0 \leq \alpha \leq 1$ and $s \in \mathbb{R}$.
I tried changing variables, integration by parts etc. and got nowhere. Any clues o... | Consider:
$$
\begin{align*}
e^{-(c\ln\frac{1}{x})^s} &= e^{-(\ln(\frac{1}{x})^c)^s} \\
&= e^{-(\ln(\frac{1}{x^c}))^s} \\
&= \frac{1}{e^{(\ln(\frac{1}{x^c}))^s}} \\
&= \frac{1}{e^{\ln(\frac{1}{x^c})} \cdot e^{\ln(\frac{1}{x^c})} \ldots \text{s times} } \\
&= \frac{1}{\frac{1}{x^c} \cdot \frac{1}{x^c} \ldots \text{s time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Find the basis of the Ker(T) and Im(T) for $T: P_3 \to P_3$, $T(ax^3+bx^2+cx+d)=dx^3+cx^2+bx+a$ For the linear application, $T: P_3 \to P_3$, $T(ax^3+bx^2+cx+d)=dx^3+cx^2+bx+a$, find
*
*A matrix representation with respect to a basis of your choice
*A basis of Ker(T)
*A basis of Im(T)
What I have done so far...
... | You can try this way: choose basis $\displaystyle \{1,x,x^2,x^3\}$ of $P_3$. Then check:
$$T[x^3]=1$$
$$T[x^2]=x$$
$$T[x]=x^2$$
$$T[1]=x^3$$
So first element of basis is transofmed to fourth by $T$, second to third, third to second, fourth to first. So the matrix representation is:
$$T=\begin{bmatrix}0 && 0 && 0 && 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let$\ n$ be an odd abundant number. What is the best upper bound to $\ \frac{\sigma(n)}{n}$? NOTE: Of course the bound of the product of primes below is incorrect. I knew very little of this stuff and it seems was too lazy and gaga (I still am gaga) to check that W|A was acting stupid.
In order to obtain my upper bound... | Ummm...$\frac{\sigma(n)}{n}$ is unbounded above for odd $n$.
Lemma: For all $n\in\mathbb{N}$, we have $\frac{\sigma(n)}{n}=\sum_{d\mid n} \frac{1}{d}$.
Proof: Let $d_1<d_2<\dots<d_t$ be the divisors of $n$. Then $\frac{\sigma(n)}{n}=\frac{d_1}{n}+\cdots+\frac{d_t}{n}=\frac{1}{d_t}+\cdots+\frac{1}{d_1}$. QED.
Now if w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
on a recursive sequence (exercise 8.14 Apostol). The exercise asks to prove convergence and find the limit of the sequence:$$a_n= \frac{b_{n+1}}{b_n},\text{ where } b_1=b_2 =1, b_{n+2} = b_{n} + b_{n+1}. $$
It also gives a hint: Show that $ \ b_{n+2}b_n - b_{n+1}^2 = (-1)^{n+1}$ by induction.
I am having problems prov... | $$b_{n+3}b_{n+1}-b_{n+2}^2= (b_{n+2}+b_{n+1})b_{n+1}-b_{n+2}^2 = (b_{n+1}-b_{n+2})b_{n+2} + b^2_{n+1} = (b_{n+1}-b_{n+2})(b_{n+1} + b_n) + b^2_{n+1}= 2b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n -b_{n+2}b_n = (b^2_{n+1}-b_{n+2}b_n)+b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n$$
Now note that $$b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n = b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to simplify this complex integral? How to approximate this integral as a function of a and b?
$$\int_0^\pi\int_0^{2\pi}\sqrt{(a-b\sin\varphi\cos\theta)^2+(b\cos\varphi)^2+(b\sin\varphi\sin\theta)^2}d\theta d\varphi$$
where a and b are two variables.
| As was mentioned: the integrand simplifies to: $\sqrt{a^2 + b^2 - 2 a b \cos\theta \sin\phi}$
Factor out $\sqrt{a^2 +b^2}$ and use the binomial theorem to expand, treating $\cos\theta \sin \phi$ as small. This gives:
$$\sqrt{a^2 +b^2} \left(1-\frac{a b \cos (\theta ) \sin (\phi )}{a^2+b^2}-\frac{a^2 b^2 \cos ^2(\theta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
PEMDAS question: $F(x) = 3x^2 - x+2$. Find $[f(a)]^2$ How should I go about doing this? $(3a^2-a+2)^2$?
Thus, $9a^4-a^2+4$
| $(3a^2-a+2)^2$ is correct but your expansion of this is incorrect. Remember that:$$(3a^2-a+2)^2=(3a^2-a+2)(3a^2-a+2)=3a^2(3a^2-a+2)-a(3a^2-a+2)+2(3a^2-a+2)$$
You might find this useful: expanding or removing brackets
Here is an example to illustrate the point:$$(3-1+2)^2=(4)^2=16$$If we did this using your result, we w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Number theory problem from 11th Iberoamerican olympiads Given a number $n \in \mathbb{N}$, such that $n>1$, let us consider all the fractions of the form $1 \over{ab}$, where $a$ and $b$ are coprime natural numbers such that $0<a<b \leq n$ and $a+b>n$. Prove that the sum of all such fractions is equal to $1 \over{2}$.
| Induction on $n$. Let $S_n$ be your sum. The base case is easy:
$$S_2=\frac{1}{1\cdot 2}$$
In general,
$$S_{n+1} = S_n-
\left(
\sum
_{\stackrel{a=1}{(a,n+1-a)=1}}
^{\left\lfloor \frac n2 \right\rfloor}
\frac{1}{a(n+1-a)}\right)
+ \left(
\sum_{\stackrel{a=1}{(a,n+1)=1}}^{n}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding determinant of a 3x3 matrix Assuming y is a nonzero real number, I need to find the determinant of this matrix:
$$ \left[
\begin{array}{cc}
1 & y & y^2 \\
y & y^2 & y^3 \\
y^2 & y^3 & y^4
\end{array}
\right] $$
Can anyone help me get started? I know the answer is 0 from using Mathematica. Thanks
Edit
I... | Hint: Pick two rows (or two columns). Are they linearly independent?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Modular calculus and square I want to prove that $4m^2+1$ and $4m^2+5m+4$ are coprimes and also $4m^2+1$ and $4k^2+1$ when $k\neq{m}$ and $4m^2+5m+4$ and $4k^2+5k+4$ when $k\neq{m}$. Firstly : Let $d|4m^2+1$ and $d|4m^2+5m+4$ then $d|4m^2+5m+4-(4m^2+1)=5m+3$ and $d|5m^2+3m$ thus $d|5m^2+3m-(4m^2+5m+4)=m^2-2m-4$ and $d|... | you could also follow this method :-
let $4m^2+1 $ and $4m^2+5m+4 $ are coprimes then $GCD(4m^2+1,4m^2+5m+4)=1$
which means there exist x,y such that ${\color{Magenta} x}(4m^2+1)+{\color{Magenta} y }(4m^2+5m+4)={\color{Red} 1}$ , so lets try to find x,y that might makes it true.
by division algorithm(assume $m=2s$ or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the PDE $xu_x-2yu_y+u=e^x,$ with the side condition $u(1,y)=y^2$ 1b. $xu_x-2yu_y+u=e^x,$ side condition $u(1,y)=y^2$
My attempt: This has been a super endurance and I hope I got the whole thing right. So anyway, here it goes ...oh and one more thing... can someone please show me how to solve the side condition st... | Without looking at your attempt in details, a method of solving is shown below, so that you could compare with your calculus :
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the first derivative $y=\sqrt\frac{1+\cosθ}{1-\cosθ}$ $$y=\sqrt\frac{1+\cosθ}{1-\cosθ}$$ my professor said that the answer is $$y'=\frac{1}{\cosθ-1}$$ she said use half angle formula but I just end up with $\frac{(-2\sinθ)\sqrt{(1-\cosθ)(1+\cosθ)}}{2(1-\cosθ)^2(1+\cosθ)}$ I used the quotient rule. I also know that... | $$ y^2=\frac {1+\cos\theta}{1-\cos\theta}.$$
Differentiating gives
$$\frac {d}{d\theta} y^2=\frac {d}{d\theta} \frac {1+\cos\theta}{1-\cos\theta} = -\frac{2\sin\theta}{(1-\cos\theta)^2}. $$
Using the chain rule gives for the left hand side
$$2y\frac {dy}{d\theta}. $$ Hence, $$\frac {dy}{d\theta}=-\frac{\sqrt{1-\cos\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Having problems proving $ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (n+2)2^{n-1} $ I'm trying to prove the following
$$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (n+2)2^{n-1} $$
using the identity $$ \sum_{r=0}^{n} \binom{n}{r} = 2^{n} $$
but I'm not able to. This is what I did so far,
$$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (1) \bin... | Starting with the binomials expansion
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} t^{r} = (1+t)^{n}
\end{align}
it can be seen by differentiation that
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} r t^{r} = n t (1+t)^{n-1} = n \left[ (1+t)^{n} -
(1+t)^{n-1} \right].
\end{align}
Now by adding the original series to this resu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition
When not using the derivative definition I get $\cos (1/x) + 2x \sin(1/x)$,
which WolframAlpha agrees to.
However when I try solving it using the derivative definition:
$$\lim_ {h\to... | Problem:
$\frac{d}{dx}\left(x^2\sin\left(\frac{1}{x}\right)\right)$
Use the product rule, $\frac{d}{dx}\left(uv\right)=v\frac{du}{dx}+u\frac{dv}{dx}$, where $u=x^2$ and $v=\sin\left(\frac{1}{x}\right)$:
$=x^2\left(\frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right)\right)+\left(\frac{d}{dx}\left(x^2\right)\right)\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}dx$ How can I approach this integral? ($0<a \in \mathbb{R}$ and $n \in \mathbb{N}$)
$$\large\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx$$
Integration by parts doesn't seem to make it any simpler.
Hints please? :)
| Since the integrand is even then
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx
\end{align}
Using substitution $t=x^2$ we get
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx&=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left... | The denominators are fairly straightforward. I assume the next two are: (x^2 + 192)^9 and (x^2 + 243)^10? The recursion is: Given denominator (x^2 + k)^n, then the next denominator is (x^2 + k*(n/(n-1))^2)^(n+1). (Yes, this is same as the previous post, but I like to think in terms of recursions because it sometimes he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Using a given identity to solve for the value of an expression This problem caught my eye in the book yesterday. Till now I still get stuck. Here it is:
If $$\frac{x}{x^2+1}=\frac{1}{3},$$ what is the value of $$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}?$$
The denominator is a cyclotomic polynomial which can be expressed as... | The first identity gives:
$$ x+\frac{1}{x}=3, $$
hence:
$$ x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2 - 2 = 7, $$
$$ x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)= 18 $$
and
$$ x^3+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3} = 18+7+3+1 = 29 $$
so the answer to your question is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Difficulty in "expressing radical as square" I have to get from this expression:
$(4+2\sqrt3)(\sqrt{2-\sqrt3})$
To this expression:
$\sqrt2+\sqrt6$
I tried to square $(4+2\sqrt3)$ and put it inside the radical, so:
$\sqrt{(16+12+16\sqrt3)(2-\sqrt3)}$
but eventually I get to: $\sqrt{52+32\sqrt3-28\sqrt3-48}$ and therefo... | The idea is that you can compute square roots of numbers of the form $\sqrt{a+b\sqrt{d}}$. Indeed,
$$(x+y\sqrt{d})^2 = x^2 + dy^2 + 2xy\sqrt{d}. $$
We want $a = x^2 + dy^2$ and $b = 2xy$, so using $y = b/(2x)$, we get $a = x^2 + db^2/(4x^2)$, or $4x^4 - 4ax^2 + db^2 = 0$, which is a quadratic equation in $x^2$ whose so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/943491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How find this limit $\lim_{x\to 0}\frac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\frac{x^3}{3}+\frac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$ Find this limit
$$I=\lim_{x\to 0}\dfrac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$$
I think
$$I=6\lim_{x\to 0}\dfrac{\int_{0}^{x}(\sin{t}\ln{(1+t)}-t... | The factors of the integrand and the denominator have power series expansions that converge near $0$. If you work out the first few terms, you get the following.
$\begin{align}
I& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\sin{t}\ln{(1+t)}\,dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}\\
\\
& =\lim_{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving a recurrence equation that yields polynomials I am trying to solve the following recurrence equation:
$$
T(n) = kT(n - 1) + nd
$$
I have expanded the first 4 values ($n = 1$ was given):
$$\begin{align}
T(1) & = 1 \\
T(2) & = kT(2-1) + 2d = k + 2d \\
T(3) & = kT(3-1) + 3d = k(k + 2d) + 3d = k^2 + 2kd + 3d \\
T(... | Using
\begin{align}
\sum_{j=0}^{m} t^{j} = \frac{1-t^{m+1}}{1-t}
\end{align}
then
\begin{align}
\sum_{j=0}^{m} j \, t^{j} = \frac{t - (m+1) t^{m+1} + m t^{m+2}}{(1-t)^{2}}
\end{align}
for which
\begin{align}
\sum_{j=0}^{m} (j+1) \, t^{j} &= \frac{1}{(1-t)^{2}} \left[ (1-t)(1-t^{m+1}) + t - (m+1) t^{m+1} + m t^{m+2} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $({\sqrt{2}\!+\!1})^{1/n} \!+ ({\sqrt{2}\!-\!1})^{1/n}\!\not\in\mathbb Q$ How could we prove that for every positive integer $n$, the number
$$({\sqrt{2}+1})^{1/n} + ({\sqrt{2}-1})^{1/n}$$ is irrational?
I think it could be done inductively from a more general expression, but I don't know how.
I made an eff... | Use the fact that for every $n$ there exists a polynomial $P_n[x]$ monic of degree $n$ with integral coefficients so that we have the identity:
\begin{eqnarray*}
t^n + \frac{1}{t^n} = P_n(t+ \frac{1}{t})
\end{eqnarray*}
The $P_n$'s are connected with the Chebyshev polynomial of first kind since for $t = e^{i \theta}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^... | Multiplying both numerator and denominator by $\sec^4x$ yields
$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{4}} \frac{d x}{1-3 \cos ^{2} x \sin ^{2} x} \\
=& \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{4} x}{\sec ^{4} x-3 \tan ^{2} x} d x \\
=& \int_{0}^{1} \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}-3 t^{2}} d t \quad \textrm{ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\pi > 24\small{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}}$ Prove that $\pi > 24\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$.
I tried using trig but I couldn't solve it. A hint I was given is to use half angle identities. This should be easy for someone who is experienced in inequalities, but I'm looking for a "low power"... | $$\sqrt{3}=2\sin \frac{\pi}{3}$$
$$2+\sqrt{3}=2\left(1+\sin \frac{\pi}{3}\right)$$
$$2+\sqrt{3}=2\left(1+\cos \frac{\pi}{6}\right)$$
$$2+\sqrt{3}=2\left(2\cos^2 \frac{\pi}{12}\right)$$
$$\sqrt{2+\sqrt{3}}=2\cos \frac{\pi}{12}$$
$$2+\sqrt{2+\sqrt{3}}=2(1+\cos \frac{\pi}{12})$$
$$2+\sqrt{2+\sqrt{3}}=2\left(2\cos^2 \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$ I'm trying to prove $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$. Is the following a correct proof?
For all $n$ we have $0 \leq \left|\sqrt{n^2 + 1} - n\right| \leq \left|\sqrt{n^2+1} - 1 \right|$. For any $\epsilon > 0$ take $ N = \sqrt{(\epsilon+1)^... | Another way:
$\sqrt{n^2+1}
=n \sqrt{1+1/n^2}
$.
For $0 < x $,
$(1+x)^2
=1+2x+x^2
> 1+2x
$,
so
$\sqrt{1+2x} < 1+x$
or
$\sqrt{1+x} < 1+x/2$.
(You can directly prove this
by squaring
both sides.)
Putting
$x = 1/n^2$,
$\sqrt{1+1/n^2}
< 1+1/(2n^2)
$.
Therefore,
$\sqrt{n^2+1}
=n \sqrt{1+1/n^2}
< n(1+1/(2n^2))
= n+1/(2n)
$,
s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function :
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
| Let $y = f(x)$ be a function with domain $D$.
Let $\left[a,b\right] \le D$, then the global extrema are the extreme (largest and smallest) values in $\left[a,b\right]$.
There are two global extrema; global minima (least value) and global maxima (largest value)
Conceptually speaking,
$$f'(x) = 0\iff f(x) \text{ `makes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Series question,related to telescopic series, 1/2*4+ 1*3/2*4*6+ 1*3*5/2*4*6*8 ...infinity The series is $$\frac{1}{2*4}+ \frac{1*3}{2*4*6}+ \frac{1*3*5}{2*4*6*8}+....$$
It continues to infinity.I tried multiplying with $2$ and dividing each term by$(3-1)$,$(5-3)$ etc,starting from the second term which gives me $$\fr... | The $r$th term $\displaystyle\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$ where integer $r\ge1$
As the number of terms in the numerator $=$ the number of terms in the denominator $-1,$
If $\displaystyle S=\sum_{n=1}^\infty\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)},$
Multiplying the num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/954922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 +... | Subtract the two integrals in question and get
$$\int_0^{\pi} dx \frac{\cos{2 x}}{1+\frac12 \sin{2 x}} = \frac12 \int_0^{2 \pi} du \frac{\cos{u}}{1+\frac12 \sin{u}}$$
This may be shown to be equal to the complex integral
$$-\frac{i}{4} \oint_{|z|=1} \frac{dz}{z} \frac{z+z^{-1}}{1+\frac{1}{4 i} (z-z^{-1})} = \oint_{|z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
Limit at negative infinity Evaluate $$\lim_{x\to -\infty}\frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}}$$
I think the strategy is to divide the numerator and denominator by x^2. Help please. The textbook answer is $$\frac{1}{4+\sqrt{3}}$$
| Instead of dividing the numerator and denominator by $x^2$, it may be easier to factor out the appropriate factors from the radical expressions:
\begin{align}
\frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}} & = \frac{\sqrt[3]{x^6\big(1+\frac{8}{x^6}\big)}}{4x^2+\sqrt{x^4\big(3+\frac{1}{x^4}\big)}} \\ & = \frac{\sqrt[3]{x^6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Zeros of a function of degree 4 I'm trying to show that the following function has no zeros
$$
60x^4-44x^3-25x^2-44x+60=0.
$$
I already tried using Eisenstein's criterium, but since the first and the last coefficient are both $60$, there is no prime which holds.
| In this answer I showed that the polynomial
$$
p(x) = x^4 + ax^3 + bx^2 + ax + 1
$$
has no real zeros if $|a| \leq 4$ and $b > 2 |a| - 2$.
We have
$$
60x^4-44x^3-25x^2-44x+60 = 60 \left(x^4 - \frac{44}{60}x^3 - \frac{25}{60}x^2 - \frac{44}{60}x+1\right),
$$
so in this case $a = -44/60$ and $b = -25/60$. Indeed, $|a| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
prove that $ \binom n 2 + \binom {n-2} 2 + \binom {n-4} 2 + \dots + \binom 3 2 = \frac 1 {24} (n-1)(n+1) (2n+3) $ $$ \binom n 2 + \binom {n-2} 2 + \binom {n-4} 2 + \dots + \binom 3 2 = \frac 1 {24} (n-1)(n+1) (2n+3) $$ where n is odd.
Plesase help mi with that equation.
| Let $n=2m+1\iff m=(n-1)/2$. Then, we have
$$\begin{align}\sum_{k=1}^{m}\binom{2k+1}{2}&=\sum_{k=1}^{m}\frac{(2k+1)(2k)}{2}\\&=\sum_{k=1}^{m}(2k^2+k)\\&=2\cdot\frac{m(m+1)(2m+1)}{6}+\frac{m(m+1)}{2}\\&=\frac{m(m+1)(4m+5)}{6}\\&=\frac 16\times\frac{n-1}{2}\times \left(\frac{n-1}{2}+1\right)\times \left(4\cdot \frac{n-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/961298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Tricky sequences and series problem For a positive integer $n$, let $a_{n}=\sum\limits_{i=1}^{n}\frac{1}{2^{i}-1}$. Then are the following true:
$a_{100} > 200$ and
$a_{200} > 100$?
Any help would be thoroughly appreciated. This is a very difficult problem for me. :(
| $$\begin{gather}a_n = 1 + \underbrace {\frac{1}{2} + \frac{1}{3}}_{2\text{ terms}} + \underbrace {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}}_{4\text{ terms}} + \ldots + \underbrace { \frac{1}{2^{n -1}} + \frac{1}{2^{n -1}+1} + \ldots + \frac{1}{2^n -1}}_{2^{n-1}\text{ terms}}\\
> 1 + 2\cdot\frac{1}{4} + 4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Does the series $\sum_{n=1}^{\infty}\sin\left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$ converge? Let $\alpha$ be such that $0\leq \alpha \leq 1$. Since $\sin n$ has no limit as $n$ tends to $\infty$, I'm having trouble with finding if the series $$\sum_{n=1}^{\infty}\sin \left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\r... | This series is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin \left( 2\pi \sqrt{n^2+\alpha^2 \sin n+(-1)^n}\right)\\\\
&=\sin \left( 2\pi n \:\sqrt{1+\frac{\alpha^2\sin n}{n^2}+\frac{(-1)^n}{n^2}}\right)\\\\
&=\sin \left( 2\pi n \:\left(1+\frac{\alpha^2\sin n}{2n^2}+\frac{(-1)^n}{2n^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/964433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$ I'm trying to prove that
$$
\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}.
$$
Let $\alpha = 2 \arcsin x$ and $\beta = \arccos x$; meaning $\sin\alpha = \frac{x}{2}, \cos\beta = x$. We know that:
$$
\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.
$... | $\alpha=2\arcsin x$ means that $x=\sin(\frac\alpha2)$, not that $\frac x2=\sin\alpha$
$\frac\alpha2=\arcsin x$
$x=\sin(\frac\alpha2)$
an alternate and faster way is to use
$\arcsin x+\arccos x=\frac\pi2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Modular Arithmetic: using congruence to find remainder How do I use the fact that if $a = b \pmod n$ and $c = d\pmod n$ then $ac = bd\pmod n$ to find the remainder when $3^{11}$ is divided by $7$?
| $3^{11} \equiv (3^2)^5\cdot 3 \equiv 2^5 \cdot 3 \equiv 96 \equiv 5 \pmod 7$
Of course, it's actually easier to employ Fermat's Little Theorem (if you're permitted) like so:
$3^{11} \equiv 3^6 \cdot 3^5 \equiv 1 \cdot (3^2)^2 \cdot 3 \equiv 2^2 \cdot 3 \equiv 12 \equiv 5 \pmod 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How calculate the shaded area in this picture? Let the centers of four circles with the radius $R=a$ be on 4 vertexs a square with edge size $a$. How calculate the shaded area in this picture?
| Let $(APD)$ be the area of the figure $APD$.
And let $x,y,z$ be $(KEPM),(PAD),(MPD)$ respectively.
First, we have
$$(\text{square}\ ABCD)=a^2=x+4y+4z.\tag1$$
Second, we have
$$(\text{sector}\ BDA)=\frac{\pi a^2}{4}=x+2y+3z.\tag2$$
Third, note that $KA=KD=a$ and that $(\triangle KAD)=\frac{\sqrt 3}{4}a^2$ since $\triang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question on simplification of $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$? I am having trouble seeing how $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$ equals $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. I can see $\sum_{n=1}^{\infty}\frac{1}{2n+1}+\frac{1}{2n+3}$ but not $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1... | Since $2 = (2n+3) - (2n+1)$ then
\begin{align}
\frac{2}{(2n+1)(2n+3)} &= \frac{(2n+3) - (2n+1)}{(2n+1)(2n+3)} = \frac{1}{2n+1} - \frac{1}{2n+3}.
\end{align}
The series then becomes
\begin{align}
\sum_{n=1}^{\infty} \frac{2}{(2n+1)(2n+3)} &= \sum_{n=1}^{\infty} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right) \\
&= \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit of a Sequence involving 3rd root I'm not finding any way to simplify and solve the following limit:
$$
\lim_{n \to \infty} \sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1}
$$
I've tried multiplying by the conjugate, but this give a more complex limit.
| $$\begin{align*}&\sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1}=\\\\
&=\sqrt{n^2+n+1}-n +\left(n-\sqrt[3]{n^3+n^2+n+1}\right)=\\ \\
&=\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n } +\dfrac{n^3-n^3-n^2-n-1}{n^2+n(n^3+n^2+n+1)^{\frac{1}{3}}+(n^3+n^2+n+1)^{\frac{2}{3}}}= \\\\
&=\dfrac{n+1}{\sqrt{n^2+n+1}+n } +\dfrac{-(n^2+n+1)}{n^2+n(n^3+n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the tangen to $\cos(\pi \cdot x)$ I have the following assignment.
Find the tangent to $y=f(x)=\cos(\pi \cdot x)$ at $x=\displaystyle\frac{1}{6}$.
First step would be to take the derivative of $f(x)$
$f'(x)= -\pi \sin(\pi \cdot x)$
Then I put the $x$-value into $f'(x)$ to find the slope
$f'(\displaystyle\frac{1}{6... | multiplying your equation
$y=-\frac{\pi}{2}x+\frac{\sqrt{3}}{2}+\frac{\pi}{12}$ by $12$ you will get
$12y+6\pi x=6\sqrt{3}+\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a... | You can use a similar trick as in complex numbers. Note that $(a+b\sqrt5)(a-b\sqrt5)=a^2-5b^2$, so $\frac{1}{a+b\sqrt5}=\frac{a-b\sqrt5}{a^2-5b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Interesting functional equations problem? $f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x$
Find all functions $f:\mathbb R \to \mathbb R$ that satisfy $f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x$.
How would we solve this? I noticed that if you plug in $\frac{x-1}{x}$ in for $x$, and then again, we can solve $f(x)$ a... | Sorry, this wouldn't fit into a comment, so I had to post this in as an answer, but when I plugged $f$ back into the original equation, I got:
\begin{align*}
f(x) + 3 f\left(\frac{x-1}{x}\right) &= \frac{1}{28}\left(x - 3 - \frac{3}{x} - \frac{9}{x-1}\right) + 3\left(\frac{1}{28}\right)\left(\frac{x-1}{x} - 3 - \frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Area between curves $y=x^3$ and $y=x$ I've tried to done one of my homework problems for several times, but the answer doesn't make sense to me. The question asks to find the area between $y=x^3$ and $y=x$. Those are odd functions, and I'm pretty sure that the area between their graphs should be 0. However, I keep gett... | The area between two curves is always positive. See the below graph.
The area in green and orange is the area you are finding. It is always going to be positive because it exists. When you subtract the two curves, you are finding the area between the curves, regardless of their position relative to the x axis.
Also, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}$? From numerical evidence it appears that whenever the integral converges, $$J_a :=\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}.$$
For $a \in \mathbb{N}$, I was able to prove this using induction (see below). How can we prove it for non-in... | $$\begin{align*}
\int_{0}^{1}\frac{1+x^{a}}{\left(1+x\right)^{a+2}}dx
&=\int_{0}^{1}\frac{1}{\left(1+x\right)^{a+2}}dx+\int_{0}^{1}\frac{x^{a}}{\left(1+x\right)^{a+2}}dx \\
&=\left.-\frac{1+x}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1}+\left.\frac{\left(1+x\right)x^{a+1}}{\left(a+1\right)\left(1+x\right)^{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that complex numbers are vertices of equilateral triangle 1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.
I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but ... | Here is a simple way. Let $z_{1}=cos\theta_{1}+isin\theta_{1}$,$z_{2}=cos\theta_{2}+isin\theta_{2}$, $z_{3}=cos\theta_{3}+isin\theta_{3}.$
By $z_{1}+z_{2}+z_{3}=0$ we get the sum of the cosines are zero and the sum of the sines is zero.
Squaring the equations
$cos\theta_{1}+cos\theta_{2}=-cos\theta_{3}$ and the
$sin\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Area bounded by the graphs $y = x^2 – 2x – 6$ and $y = 6 – x^2$? Find the area bounded by the graphs of the following functions
$$y = x^2 – 2x – 6$$ and $$y = 6 – x^2$$
Are the points $-3$ and $2$? Where to go form there?
| The points of intersection can be found by solving
$$x^2-2x-6=6-x^2 \Longleftrightarrow x^2-x-6=0 \Longleftrightarrow (x-3)(x+2)=0$$
Thus $x=\color{red}-2$ and $x=\color{red}+3$ are the points of intersection.
In the interval $[-2,3]$, $6-x^2>x^2-2x-6$, therefore the area of the bounded region is
$$\int_{-2}^36-x^2-(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find $\omega^7$ and $\omega^6$ from $\omega^5+1=0$
I did the first parts :
$$\omega= (\cos \pi + i \sin \pi)^\frac{1}{5} \implies \omega^5 = \cos \pi + i \sin \pi=-1$$
$\omega=-1$ is a root so :
$$\omega^5-1= (\omega+1) (\omega^4-\omega^3+\omega^2-\omega+1)=0$$
$$\omega^4-\omega^3+\omega^2-\omega=-1$$
$$\o... | Since $\omega^5+1=0$, you have that $\omega^5=-1$, so $\omega^6=\omega^5\cdot\omega=-\omega$ etc.
For the quadratic equation, use that $(x-a)(x-b)=x^2-(a+b)x+ab$, let $a=\frac{1}{2}(\omega-\omega^4)=\cos(\pi/5)$ and $b=\frac{1}{2}(\omega^3-\omega^2)$. You calculated values for $a+b$ and $ab$ just beforehand, so you hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
find critical point $f(x,y) = 8 + 2y^3+2x^3 - 3xy$ $f(x,y) = 8 + 2y^3+2x^3 - 3xy$
$f_x = 6x^2 - 3y$
$f_{xx} = 12x$, $x= 0$, $y=0$
$f_y = 6y^2 -3x$
$f_{yy}=12y$
$f_{xy}=-3$
at $(0,0)$, $D = -9 < 0$
Is it one saddle point $(0,0)$?
| You need to first find ALL the points where $f_x = f_y = 0$.
$f_x = 6x^2 - 3y = 0$ gives us $y = 2x^2$ (1)
$f_y = 6y^2 - 3x = 0$ gives us $x = 2y^2$ (2)
Substitute (1) into (2) to get $x = 2(2x^2) = 8x^4$, i.e. $8x^4-x = 0$.
This has another real root besides $x = 0$, which will give you another critical point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The dilogarithm function. Question on an identity of it Upon reading a journal article about manipulating series using the dilogarithm function, I have a few questions. But before I ask them, let me give the information the article provides.
Consider the series $\displaystyle\sum \frac{1}{k^2} = 1 + \frac{1}{4} + \frac... | Extending the hint in the comment section, one has
$$\frac{d}{dx} Li_2\left(-\frac{1}{x}\right)=\frac{\ln\left(1+\frac{1}{x}\right)}{x}=\frac{\ln(1+x)-\ln(x)}{x},$$
and
$$\frac{d}{dx} Li_2\left(-x\right)= -\frac{\ln\left(1+x\right)}{x}. $$
This implies
$$\frac{d}{dx} Li_2\left(-\frac{1}{x}\right)+ \frac{d}{dx} Li_2\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ im... | Hint: $$\frac{|x|}{(1+x^2)(1+a^2)} \leq \frac{|x|}{1+x^2} < 1$$
and
$$\frac{|a|}{(1+x^2)(1+a^2)} \leq \frac{|a|}{1+a^2} < 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 0
} |
Factoring $x^4 + 4x^2 + 16$ I was putting together some factoring exercises for my students, and came across one that I am unsure of how to factor.
I factored $x^6 - 64$ as a difference of squares, and then tried it as a difference of cubes, but was left with $(x^2 - 4)(x^4 + 4x^2 + 16)$ is there a general method for f... | Here is another method that might generalize to more situations.
Using DeMoivre's Theorem, the roots of $x^6-64 = 0$ are $x = 2e^{ik\pi/3}$ for $k = 0,1,2,3,4,5$.
So, the polynomial is the product of the following complex linear factors:
$(x-2)(x+2)(x-2e^{i\pi/3})(x-2e^{i5\pi/3})(x-2e^{i2\pi/3})(x-2e^{i4\pi/3})$
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}... | By Hölder's inequality
$$\left(\frac a{\sqrt{1+a}}+\frac b{\sqrt{1+b}}+\frac c{\sqrt{1+c}}\right)^2\Big(a(1+a)+b(1+b)+c(1+c)\Big)\ge(a+b+c)^3,$$
so we only need to prove
\begin{align*}(a+b+c)^3&\ge\frac92\Big(a(1+a)+b(1+b)+c(1+c)\Big)\\(a+b+c)(2(a+b+c)^2-9)&\ge9(a^2+b^2+c^2)\\\end{align*}
AM-GM tells us that $(a+b+c)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
Prove that for any positive integer $n$, $2\sqrt{n+1}-2\le 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}}\le 2\sqrt{n}-1$ Prove that for any positive integer n,$$ 2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1 $$
Could anyone give me a h... | Hints:
$$2(\sqrt{k+1}-\sqrt{k})=\frac{2}{\sqrt{k}+\sqrt{k+1}} < \frac{1}{\sqrt{k}}< \frac{2}{\sqrt{k-1}+\sqrt{k}}=2(\sqrt{k}-\sqrt{k-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$ by induction Prove that $$\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$$ for all $n\in \mathbb{N}$ where $n\geq2$. I've already proven the base case for $n=2$, but I don't know how to make the next step.
Is the base case right?
for $n=2$
$$\sum^{2}_{k=1} \frac{1}{\sqr... | Assume $\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$.
Then
$$\sum^{n}_{k=1} \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{n+1}}>\sqrt{n} + \frac{1}{\sqrt{n+1}} = \frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}}.$$
Now since $n\ge 2$, we have $$\frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}} > \frac{\sqrt{n}\sqrt{n}+1}{\sqrt{n+1}} = \frac{n+1}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why is $x(x+2)(x-3)$ not $x^2+2x(x-3)$? How would you explain the principle why $x(x+2)(x-3)$ is not $x^2+2x(x-3)$ but $(x^2+2x)(x-3)$? This may involve the fundamentals of eliminating parenthesis.
| This is the distributive property for multiplication: $(a+b)(c+d)=a(c+d)+b(c+d)$. In your case:
$$k(a+b)(c+d)=(ka+kb)(c+d)=ka(c+d)+kb(c+d).$$
This comes from:
$$k(a+b) = \overbrace{(a+b)+(a+b)+\dotsc+(a+b)}^{k\text{-times}} = a+b+a+b+\dotsc+a+b=\overbrace{a+a+\dotsc+a}^{k\text{-times}}+\overbrace{b+b+\dotsc+b}^{k\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to write a fourier series using periodic boundary conditions Would writing
$$
f(x) = x^2
$$
as a Fourier series using periodic boundary conditions on $-L < x < L$ with a basis of
$$
e^{\frac{i\pi nx}{L}}
$$
be just
\begin{align}\bigl\langle e^{\frac{i\pi nx}{L}},x^{2}\bigr\rangle & = \Bigl\langle e^{\frac{i\pi ... | You should write
\begin{align}
c_n &= \langle x^2, e^{-in\pi x/L}\rangle\\
&= \frac{1}{2\pi}\int_{-L}^Lx^2\bigg(\cos\Big(\frac{n\pi x}{L}\Big) - \sin\Big(\frac{n\pi x}{L}\Big)\bigg)dx
\end{align}
Since $x^2\sin$ is an odd term, the sine terms integrate to zero so we are left with
$$
c_n = \frac{1}{\pi}\int_0^Lx^2\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$
$a,b,c$ are positive reals with $abc = 1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$
I try to use AM $\ge$ HM.
$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{... | As above write $x=a^{-1}$, $y=b^{-1}$, $c=z^{-1}$, so we are required to prove
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2}, \qquad xyz=1.$$
Consider now that $\frac{x}{y+z}$, $\frac{y}{z+x}$, $\frac{z}{x+y}$ are in the same order as $x,y,z$. Then by the Rearrangement Inequality
$$ \frac{x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Limit of $\prod\limits_{k=2}^n\frac{k^3-1}{k^3+1}$ Calculate $$\lim_{n \to \infty} \frac{2^3-1}{2^3+1}\times \frac{3^3-1}{3^3+1}\times \cdots \times\frac{n^3-1}{n^3+1}$$
No idea how to even start.
| Using the suggested factorizations,
and using
$\begin{array}\\
k^2-k+1
&=k(k-1)+1\\
&=(k-1+1)(k-1)+1\\
&=(k-1)^2+(k-1)+1\\
\end{array}
$
(this is really the key),
$\begin{array}\\
\prod_{k=2}^n \dfrac{k^3-1}{k^3+1}
&=\prod_{k=2}^n \dfrac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}\\
&=\dfrac{\prod_{k=2}^n (k-1)}{\prod_{k=2}^n (k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to find the value of this function If $f:\mathbb{R} \rightarrow\mathbb{R}$ is a function which satisfies $$f(x)+f(y)=\frac{f(x-y)}{2}\cdot \frac{f(x+y)}{2}$$ for all $x\in\mathbb{R}$ and $f(1)=3$, what is $f(6)$?
| The function you described does not exist because no value of $f(0)$ makes sense:
*
*$f(0) + f(0) = \frac{f(0 - 0)}{2} \cdot \frac{f(0+0)}{2} = \frac{f(0)^2}4$
implies that $8f(0) = f(0)^2$ which means $f(0)$ can either be $0$ or $8$.
*$3+f(0) = f(1) + f(0) = \frac{f(1-0)}{2}\cdot \frac{f(1+0)}{2} = \frac{f(1)^2}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $... | HINT:
Write $8-x = 2 a$, $\ x-6=2b$. Then $\ a+b=1$ and $a^4+b^4=1$. Draw some level curves.
$\bf{added:}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \s... | If you are allowed to use Taylor series, you could start with $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ Now, use $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Replacing $y$ by the previous developments, you then have $$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Triangle in Triangle I have the lengths of three sides of an acute triangle ABC as shown below. Assume a point P on the side AB such that, if Q is the projection of P onto BC, R is the projection of Q onto CA, P becomes the projection of R onto AB. How can I Find the length PB.
| I start with coordinates, choosing one edge length equal to $1$ w.l.o.g.:
$$B=(0,0) \qquad A=(x,y) \qquad C=(1,0) \qquad P=\lambda A+(1-\lambda)B$$
with $y\neq 0$ and some $\lambda\in[0,1]$ we'll need to determine.
Then using some computation (which I did using projective geometry, but there are other ways as well) you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Triple integration question involving sketching
Sketch the region $R=\{(x,y,z):0\le z\le 1-|x|-|y|\}$ and evaluate the integral
$$\iiint\limits_R(xy+z^2)dV$$
I REALLY need someone to confirm this!!!!!! This is what I did:
I used wolfram alpha to calculate the triple integrals, http://www.thestudentroom.co.uk/att... | Take a region of the 3D graph, and plot points on it. I would recommend the range $x \in \mathbb{R}:[-2,2]$ and $ y \in \mathbb{R}:[-2,2]$. This gives you some positive, and some negative. When you do this, you make a note of the symmetry of the graph, as well as the points of intersection with the xy-plane, since that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?
| If $k=z$, there are no solution by Fermat's Last Theorem.
When $k=1$ a solution can be $x=1$, $y=2$, $z = 3$, because $1^{3} + 2^{3} = 3^{2}$
Another $k=38$ have a solution can be $x =3$, $y=5$, $z=2$ and there are lots of $k$ can be
solution of that equation, but now I don't know any technique for count how many or w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Solve for constants: Derivatives using first principles
*
*Question
Find the values of the constants $a$ and $b$ such that $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{5}{12}$$
*
*My approach
*
*Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
*I view limit as... | It is fine. However, there is a short proof if you know this
$(1+x)^\frac1n=1+\frac1nx+o(x)$, where $\lim_{x\to 0}\frac{o(x)}{x}=0.$
So $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \lim_{x \to 0}\frac{b^\frac13 \sqrt[3]{\frac a bx + 1}-2}{x}=\lim_{x \to 0}\frac{(b^\frac13-2)+ \frac13 b^\frac13 \frac abx+b^\frac13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Calculating the value of $\frac{a-d}{b-c}$ If $\frac{a-b}{c-d}=2$ and $\frac{a-c}{b-d} = 3$ then determine the value of:
$$\frac{a-d}{b-c}$$
Where $a,b,c,d$ are real numbers.
Can someone please help me with this and give me a hint? I tried substitutions and solving them simultaneously but I couldn't determine this val... | This might not be the most mathletic solution, but here it is anyway.
Note that for the given information to make sense, we must have $c\neq d$ and $b\neq d$. From these two observations, we also see that $b\neq c$ since otherwise the two given equations are equal, i.e., $2=3$ which is nonsense. Let $k=\frac{a-d}{b-c}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$ What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
| Let $I = \int_0^1 x \tan(\pi x) \log(\sin(\pi x)) dx$. We have
$$I= \int_0^1 \dfrac{x \tan(\pi x)}2 \log(1-\cos^2(\pi x))dx$$
Hence,
$$2I = -\sum_{k=1}^{\infty}\dfrac1k\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx \,\,\,\,\,\,\,\, \spadesuit$$
\begin{align}
\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx & = \int_0^1 x\sin(\pi x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 0
} |
Proving this inequality $\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\sqrt[n]{(3x)^n+\cdots}}}< (x+\frac{1}{n-1})$ How can I prove this inequality $$\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\sqrt[n]{(3x)^n+\cdots}}}< \left(x+\frac{1}{n-1}\right)$$ if $n$ and $x$ are positive integer number $$x>=1$$ $$n>1$$
| As A.D. noted, the inequality is wrong. So what is the correct upper bound?
Let
$$ u_{km} = \dfrac{1}{kx}\sqrt[n]{(k x)^n + \sqrt[n]{((k+1) x)^n + \sqrt[n]{\ldots + \sqrt[n]{(mx)^n}}}}$$
with $x \ge 0$, $m \ge k \ge 1$, $n \ge 1 $, so the left side of your inequality is $\lim_{m \to \infty} x u_{1m}$.
We have
$$ u_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x... | You can continue, using $\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = 1-2\sin^2(\frac{x}{2}) \iff \cos x -1 = -2\sin^2(\frac{x}{2})$ after your last step so that $$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2} = \lim_{x\to0} \frac{1+2\Big(\frac{\cos x - 1}{x^2}\Big)}{6} = \lim_{x\to0} \frac{1+2\Big(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find the least square solution and optimal least square solution of the linear system Find the least square solution and optimal least square solution of the linear system:
$$x_1 + 2x_3 = 1$$
$$x_2 +3x_3 = 0$$
$$-x_1 + x_2 + x_3 =0$$
$$-x_2 -3x_3 =1$$
Letting $A$ be the matrix of coefficents, I get that $A^*A$ is
$$A^... | You want the least-squares solution of
$$
\left[\begin{array}{rrr}1 & 0 & 2 \\
0 & 1 & 3 \\
-1 & 1 & 1 \\
0 & -1 & -3 \end{array}\right]x
=\left[\begin{array}{r}1 \\ 0 \\ 0 \\ 1\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How many solutions does this have, modulo $p$? $p$ is a prime number, and $p<14$. Furthermore, $x,y\in\mathbb{Z}$ and $0\leq x<13$ and $0\leq y<13$. Let $a,b,c$ denote unknown variables. How many solutions does the following system of equations have, modulo $p$? (that is, two solutions are the same if they are congruen... | You have to compute the determinant:
$$\left|\begin{array}{ccc}
1&1&1\\
1&2&x+1\\
1&3&x^2+2x+2
\end{array}
\right|$$
It is
$$2x^2+4x+4-3x-3-x^2-2x-2+x+1+3-2=x^2+1$$
which is not $0$ for $p\in\{3,7,11\}$
For $p=2$, $p=5$ and $p=13$ the determinant can be $0$. Namely:
*
*If $p=2$, when $x$ is odd
*If $p=5$, when $x\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
First Order Ordinary Differential Equation by Any Method (1R-24) I just need for someone to check my work and suggest a better way to solve this if one exists. I can use any method but not numerical or any other iterative series approximation. The following documents my process of attempting to solve this problem. It d... | I think you could go faster to the solution rewriting $$\frac{dy}{dx} = \frac{y}{x + (y + 1)^2}$$ as $$\frac{dx}{dy} = \frac{x + (y + 1)^2}{y}$$ that is to say $$\frac{dx}{dy}-\frac{x}{y} =\frac{ (y + 1)^2}{y}=y+2+\frac{1}{y}$$ which gives for the homogeneous equation $x=Cy$. So, the equation write now $$y \frac{dC}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in... | Using
$$ \sum_1^N (1+k)^i = \frac{(k+1)\,((k+1)^N-1)}{k}$$
(which since this is a geometric series is not hard to prove)
we get
$$ \sum_1^N (1+k)^i = \frac{\left(\sum_{j=0}^{N+1}\binom{N+1}{j}k^j\right) - (k+1)}{k}$$
$$ = \frac{1 + (N+1)k + \left(\sum_{j=2}^{N+1}\binom{N+1}{j}k^j\right) - (k+1)}{k}$$
$$ = N + \sum_{j=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$\sum_{p \in \mathcal P} \frac1{p\ln p}$ converges or diverges? We will denote the set of prime numbers with $\mathcal P$.
We know that the sum $\sum_{n=1}^{\infty}\frac1n$ and $\sum_{n=2}^{\infty}\frac1{n\ln n}$ diverges. It is also known that $\sum_{p \in \mathcal P} \frac1p$ is also diverges, where the sum runs over... | We begin by proving a weaker version of Mertens' first theorem:
$$\left\lvert\sum_{p\leqslant n} \frac{\ln p}{p} - \ln n\right\rvert \leqslant 2\tag{1}$$
for all $n\geqslant 2$.
Although Mertens' first theorem isn't too hard to prove, a complete proof would be too long for this answer, so we only prove
$$\sum_{p\leqsla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 2
} |
Partial fraction of $\frac{2x^2-9x-9}{x^3-9x}$ I'm doing some questions from Anton, 8th edition, page 543, question 13. I've found a answer but it does not match with the answer given at the last pages.
Questions asks to solve $\int{\frac{2x^2-9x-9}{x^3-9x}}$
So:
$$\frac{2x^2-9x-9}{x(x^2-9)}$$
then:
$$\frac{2x^2-9x-9}{... | Using properties of logarithm ($\ln a + \ln b = \ln (ab)$) you have
$$\ln |x| - \ln |x-3| + 2 \ln |x+3| = \ln \frac{|x|(x+3)^2}{|x-3|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many zeros does $f(x)= 3x^4 + x + 2 $ have?
How many zeros does this function have?
$$f(x)= 3x^4 + x + 2 $$
| $f(x)$ has $4$ zeros by the fundamental theorem of algebra.
Here is an alternative way to find that none of them is real:
$$ \ \ \ \ \ \ \ \ 3x^4+x+2=0 \ \ \ \ \ (1)\\ 3x^4=-x-2,$$ which, as $0\ne-2$, is equivalent to $$3x^3=-1-\frac{2}{x}.$$
Letting $g(x)=3x^3$ and $\displaystyle h(x)=-1-\frac{2}{x}$ we have $$
\left\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/... | $14=2^{2^2}-2$and$18=2^{2^2}+2$$13=\frac{22}2+2$$24=\frac{(2^2)!}{\frac22}$$20=\sqrt{{22}^2}-2=(2^2)!-2^2$$22=\sqrt{\left[(2^2)!-2\right]^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 0
} |
The convergence of a recurrcively defined sequence. Let $a_1=\sqrt{2}$ and $a_n=\sqrt{2+a_{n-1}}$ determine the convergence of the sequence and find its limit.
I know the sequence converges to $2$ and i can show this informally. But I don't know how to prove that formally.
| Once proved the convergence by induction as suggested by Henno, you might be interested in seeing directly that the limit is $2$. So let's find $a_n$ (without the dependence from $a_{n-1}$) and then take the limit as $n\to \infty$. Here we have: $$ a_1=\sqrt{2}\\ a_2=\sqrt{2+\sqrt{2}}=\sqrt{\sqrt{2}(\sqrt{2}+1)}=\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving a recurrence with diagonalization? Considering the recurrence $F_n=F_{n-1}+3F_{n-2}-3F_{n-3}$ where $F_0=0$, $F_1=1$ and $F_2=2$. Use diagonalization to find a closed form expression for $F_n$.
So I first continued the recurrence to find $F_3=5$, $F_4=8$, $F_5=17$ ... etc
From this is I got a vector with three ... | Your matrix $A$ is not correct. It should represent the recurrence. You define $u_k = \begin{pmatrix} F_{k+2} \\ F_{k+1} \\ F_{k}\end{pmatrix}$ and the matrix $A$ is such that $u_{k+1}=Au_k$, so $A=\begin{pmatrix} 1 & 3 & -3 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$ where the first line is the recurrence and the next t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.