Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ How to find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ in the formal way? Numerically its value is $\approx 0.0217326$ and the partial sum formula contains the first derivative of the gamma function (by WolframAlpha).
| There is an alternate method and is as follows.
Notice that
$$ \frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3) $$
where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes
\begin{align}
S &= \sum_{n=2}^{\infty} \frac{1}{n \, (n+1)^{2} \, (n+2)} \\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Greatest of the numbers given To find out the greatest among the number given below:
$3^{1/3}, 2^{1/2}, 6^{1/6}, 1, 7^{1/7}$
I have plotted the following graph using graph plotter which is shown below:
It can be concluded that $3^{1/3}$ is the greatest.
I want to know that is there any other method to find greatest am... | Take lcm$(3,2,6,7)=42$
We need to check for $3^{1/3},2^{1/2},6^{1/6},7^{1/7}$
equivalently taking $42$nd power of each $3^{14},2^{21},6^7,7^6$
Now $2^3<3^2\iff(2^3)^7<(3^2)^7$
Again, $3^{14}-6^7=3^7(3^7-2^7)>0$
and finally $3^7>3^6=729>343=7^3\implies(3^7)^2>(7^3)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Badly formed question? $\|x\|=1=\|y\|$ and $\|x+y\|=\|x\|+\|y\|$, there is a line segment in the unit sphere Show that if a normed linear space $X$ contains linearly independent vectors $x$ and $y$ such that $\|x\|=1$ and $\|y\|=1$ with $\|x+y\|=\|x\|+\|y\|$, then there is a line segment contained in the unit sphere of... | HINT:
You have $\|\frac{1}{2}x + \frac{1}{2}y\| = 1$. Let's show for instance that
$\|\frac{1}{3}x + \frac{2}{3}y\| = 1$. Note that
$$\left\|\frac{1}{3}x + \frac{2}{3}y\right\| \le \frac{1}{3}\|x\| + \frac{2}{3}\|y\|=1$$
If we had $\|\frac{1}{3}x + \frac{2}{3}y\| <1$ then
$$\left\|\frac{1}{2}x + \frac{1}{2}y\,\right\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}... | For such form of the side lengths, the most convenient
would be a variation of the
Heron's formula
for the area:
\begin{align}
S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
\\
S&=\tfrac14\sqrt{
4(x^2+y^2)(y^2+z^2)-
(x^2+y^2+y^2+z^2-z^2-x^2)^2
}
\\
&=\tfrac14\sqrt{4(x^2+y^2)(y^2+z^2)-4y^4}
\\
&=\tfrac12\sqrt{x^2 y^2+y^2 z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Two numbers that cannot both be squares I was wondering where to start with the following question:
Show for $a,b \in \mathbb{N}$ that $a+b^2$ and $a^2+b$ cannot be both squares.
Here $\mathbb{N}$ is the positive integers ($0$ not included).
| Consider $a < b$:
Clearly, $b^2 < a+b^2$. Further, we see that
$$
a+b^2 < b + b^2 = b(b+1) < (b+1)^2
$$
Hence, $b^2 < a+b^2 < (b+1)^2$. Thus, $a+b^2$ is not a square. On the other hand, $a^2+b$ may be a square, depending on the choice of $a$ and $b$.
If we have $a=b$, then neither $a^2+b$ nor $a+b^2$ are squares, as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(... | First note that the factors $\sqrt {2 (2-x)}$ and $\sqrt {1-x}$ tend to $2$ and $1$ respectively, therefore they don't raise any problem. It remains to compute the limit of $\frac {1 - \sqrt {1-x^2}} {2 - \sqrt {4-x^2}}$. Note that this fraction can be rewritten as $\frac {1 - (1-x^2)} {1 + \sqrt {1-x^2}} \frac {2 + \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ contains $N$ integers. Find the value of $10N$.
I tried to find the mini... | Your simplified equation is correct only for $0 \leq x \leq \pi$. During your derivation be sure to consider both positive and negative square roots in the denominator. The result is an alternate version of your simplified equation
$f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{5-4\sin^2\left(\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
find a horizontal speed $\frac{dx}{dt}$ A girl slides down a slide in the shape of the parabola $y=(x−1)^2$ for $0≤x≤1$. Her vertical speed is $\frac{dy}{dt}=−y(1−y)$. Find her horizontal speed $\frac{dx}{dt}$ when $y=\frac 12$.
I have found that $$\frac{dy}{dt}=2(x+1)\frac{dx}{dt}$$, since $\frac{dy}{dt}=−y(1−y)$ and ... | Use the chain rule:
$$
\frac { dx }{ dt } \quad =\quad \frac { dx }{ dy } \frac { dy }{ dt } \\ \qquad \quad =\quad \frac { 1 }{ \frac { dy }{ dx } } y(y-1)\\ \qquad \quad =\quad \frac { y(y-1) }{ 2(x-1) } \\ \frac { dx }{ dt } \quad =\quad \frac { y(y-1) }{ -2\sqrt { y } }
$$
And now all you have to do is substitut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve this trigonometric equation $\cos3x\cos x=\sin 3x$ Solve this equation $$ \cos 3x\cos x=\sin 3x$$
I tried converting product into sum but with no results. I think they forgot to add $\sin x$ after $\sin 3x$.
| We have$$\begin{align}\\&\cos 3x\cos x=\sin 3x\\&\iff (4\cos^3x-3\cos x)\cos x=3\sin x-4\sin^3x\\&\iff 4\cos^4x-3\cos^2x=3\sin x-4\sin^3x\\&\iff 4(1-\sin^2x)^2-3(1-\sin^2x)=3\sin x-4\sin^3x\end{align}$$
Here, let $t=\sin x$.
$$\begin{align}\\& 4(1-t^2)^2-3(1-t^2)=3t-4t^3\\&\iff 4(1-2t^2+t^4)-3+3t^2-3t+4t^3=0\\&\iff 4-8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why does the elliptic curve for $a+b+c = abc = 6$ involve a solvable nonic? The curve discussed in this OP's post,
$$\color{brown}{-24a+36a^2-12a^3+a^4}=z^2\tag1$$
is birationally equivalent to an elliptic curve. Following E. Delanoy's post, let $G$ be the set of rational numbers that solve $(1)$. Courtesy of Aretino's... | Тhat is a beautiful solvable nonic! Here is its smallest real root:
With[{α = Sin[ArcSin[(1 + 3 × 3^(2/3))/8] / 3],
β = 3 + 9 × 3^(1/3) + 7 × 3^(2/3),
γ = 18 + 9 × 3^(1/3) + 2 × 3^(2/3)},
With[{
ρ = -51 - 33 × 3^(1/3) + 121 × 3^(2/3) - 8 α β + 4 α^2 (β - 60 × 3^(2/3)),
σ = 12 - 9 × 3^(1/3) - 7 × 3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
LU decomposition for cyclic tridiagonal matrices It is known that a tridiagonal matrix
$$
A = \begin{pmatrix}
b_1 & c_1 & 0 & 0 & \dots & 0\\
a_2 & b_2 & c_2 & 0 & \dots & 0\\
0 & a_3 & b_3 & c_3 & \dots & 0\\
0 & 0 & \ddots &\ddots &\ddots & 0\\
0 & \dots & 0 & a_{n-1} & b_{n-1} & c_{n-1}\\
0 & \dots & 0 & 0 & a_{n} &... | Actually, I was mistaken. There is a way to perform a symbolical decomposition of $B = A - \frac{a_1 c_n}{u_n} ZZ^\top$ with $u_n$ being a variable effectively in $O(n)$ operations.
One can note, that the first diagonal element of the matrix $B$ that is $b_1 - \frac{a_1 c_n}{u_n}$ has the following form
$$\frac{b_1 u_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$
My attempt
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\... | It is done much simpler in the following manner with just one application of LHR.
\begin{align}
L &= \lim_{x \to 0}\left(\frac{1}{x^{2}} - \frac{1}{\tan^{2}x}\right)\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{2}\tan^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot\frac{x^{2}}{\tan^{2}x}\not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
real values of $x$ in $\sqrt{5-x} = 5-x^2$.
Calculate the real solutions $x\in\mathbb{R}$ to
$$
\sqrt{5-x} = 5-x^2
$$
My Attempt:
We know that $5-x\geq 0$ and thus $x\leq 5$ and
$$
\begin{align}
5-x^2&\geq 0\\
x^2-\left(\sqrt{5}\right)^2&\leq 0
\end{align}
$$
which implies that $-\sqrt{5}\leq x \leq \sqrt{5}$. Now le... | You can see the solutions as the abscissa of the intersection points of the two curves $y=5-x^2$ (a parabola) and $y=\sqrt{5-x}$ (upper half of a parabola). These twoo parabolas intersect at 4 points, but only two of them lie in the $y>0$ half-plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
| You can use the identity
$$(a+b+c)^2=3(a^2+b^2+c^2)-(a-b)^2-(b-c)^2-(c-a)^2$$
The right-hand side is clearly maximised when $a=b=c$ and then $a+b+c=12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful:
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$
Thanks in advance!
| Let $x = u^6$.
$$\lim _{ x\to 1 }\dfrac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } = \lim_{u\to 1}\dfrac{u^3 + u^2 - 1 - 1}{u^6-1}$$
$$
= \lim_{u\to 1}\dfrac{u^3 - 1}{(u^2)^3-1^3} + \lim_{u\to 1}\dfrac{u^2 - 1}{(u^2)^3-1^3}
$$
$$
= \lim_{u\to 1}\dfrac{(u - 1)(u^2+u+1)}{(u-1)(u+1)(u^4 + u^2 +1)} + \lim_{u\to 1}\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that neither $A$ nor $B$ is divisible by $5$ Let the sum
$$ {1+ \frac12 + \frac13 + \frac 14+ \dots +\frac1{99} + \frac 1{100}}$$
be written as $\frac AB$, where $A$ and $B$ are positive integers with no common factors. Show that neither $A$ nor $B$ is divisible by $5$.
Using Mathematica, I found the sum is $$\... | Firstly $\frac{1}{25} + \frac{1}{50} + \frac{1}{75} + \frac{1}{100} = \frac{1}{25} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) = \frac{1}{12}$.
Also $\frac{1}{25k+5} + \frac{1}{25k+10} + \frac{1}{25k+15} + \frac{1}{25k+20} = \frac{1}{5} \left( \frac{(5k+4)+(5k+1)}{(5k+1)(5k+4)} + \frac{(5k+3)+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{1}{x^2+x+1}$ and$\frac{1}{x^2-x+1}$ How to integrate two very similar integrals. I am looking for the simplest approach to that, it cannot be sophisticated too much as level of the textbook this was taken from is not very high. $$\int \frac{1}{x^2+x+1} dx$$ and$$\int \frac{1}{x^2-x+1} dx$$
| $$\int { \frac { dx }{ { x }^{ 2 }+x+1 } =\int { \frac { dx }{ { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } } } =\int { \frac { d\left( x+\frac { 1 }{ 2 } \right) }{ { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } } =\frac { 2 }{ \sqrt { 3 } } \arctan { \left( \frac { 2 }{ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Solve the equation in the questions Solve the equation $$(x+3)(\sqrt{2x^2+6x+2}-2x)=\sqrt[3]{x^2+1}+(x^2-7)\sqrt{x+3}$$
It's little complicated, any help will be appreciated.
| $$ (x+3)\sqrt{2x^2+6x+2}-(2x^2+6x+2)=\sqrt[3]{x^2+1}-2+(x^2-7)\sqrt{x+3}$$
$$\Rightarrow \sqrt{2x^2+6x+2}\frac{7-x^2}{x+3+\sqrt{2x^2+6x+2}}=\frac{x^2-7}{\sqrt[3]{(x^2+1)^2}+2\sqrt[3]{x^2+1}+4}+(x^2-7)\sqrt{x+3}$$
And so the solutions $${x\in\{-\sqrt 7,+\sqrt 7\}}$$
And $$-\frac{\sqrt{2x^2+6x+2}}{x+3+\sqrt{2x^2+6x+2}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $f$ is a decreasing function It's given that $f(x)=\frac{1}{x^3}-x^3$ for $x>0$ show that $f$ is a decreasing function.
My attempt $f'(x)=-3x^{-4}-3x^2$
$x^6=-1$
How to continue by my attempt ?
| Notice, we have $$f(x)=\frac{1}{x^3}-x^3$$
$$\frac{d}{dx}(f(x))=\frac{d}{dx}\left(\frac{1}{x^3}-x^3\right)$$
$$f'(x)=\frac{-3}{x^4}-3x^2$$
$$f'(x)=-3\left(\frac{1}{x^4}+x^2\right)$$
$$f'(x)=-3\left(\left(\frac{1}{x^2}\right)^2+x^2\right)\tag 1$$
We know that $$\left(\frac{1}{x^2}\right)^2> 0\ \ \ \ \forall \ x>0 \tag 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
I tried to solve it.
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac... | Considering $$I=\int\frac{1+2\cos (x)}{(2+\cos (x))^2}\,dx$$ you can do several things.
First, considering the squared denominator, you could suppose that the result of integration would be something like $$\frac {a+b\sin(x)+c\cos(x)}{2+\cos(x)}$$ which, one differentiated, would give $$\frac{(a-2 c) \sin (x)+2 b \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove this inequality $(a_{1}a_{2}\cdots a_{n})\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}<\sqrt{n}$ Assmue that $a_{i}\in (0,1),i=1,2,3,\cdots,n$,show that
$$(a_{1}a_{2}\cdots a_{n})\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}<\sqrt{n},$$
I've tried many things but all have failed.
| Let $C_n:=a_1\cdots a_n$. Note, that for $n=1$, the inequality is true:
$$
a_1\cdot\sqrt{1-a_{n+1}}<1\cdot\sqrt{1-0}=1
$$
Now we have the following for $n>1$:
$$
0≤\sqrt{n-1}\left(\sqrt{1-a_{n+1}}-\frac{C_n}{2\sqrt{n-1}}\right)^2 \iff \\
0≤\sqrt{n-1}\cdot (1-a_{n+1})-C_n\sqrt{1-a_{n+1}}+\frac{C_n^2}{4\sqrt{n-1}}\iff\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Multivariable limit of $\sqrt{x^2+y^2} \sin (\frac{1}{x^2+y^2})$ Find Multivariable limit of $$\lim \limits_{(x,y) \rightarrow (0,0) }\sqrt{x^2+y^2} \sin \left(\frac{1}{x^2+y^2}\right)$$
Limit is obviously zero, but my question is how to simplify it a bit more to make it more obvious?
| Notice that
$$
0 \leq \sqrt{x^2+y^2} \left| \sin \left( \frac{1}{x^2+y^2} \right) \right| \leq \sqrt{x^2+y^2}
$$
and by squeezing...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no... | Let $f(x)= 1+x+\frac{x^2}{2!}+\ldots +\frac{x^6}{6!}$
Differentiating $f(x)$:
$$f'(x)= 1+x+\frac{x^2}{2!}+\ldots+\frac{x^5}{5!}$$
Consider $a$ as one real root for $f'(x)=0$, i.e. $f'(a)=0$. Obviously $a\ne 0$.
Now, $f(a)= 0+ \frac{a^6}{6!}>0$. This means that all the potential local minima of $f(x)$ are already positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 9
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Proving a ratio that has a relation with the Perpendicular bisectors and circumcircle $ABC$ is a triangle, $D$ is a point on the side $BC$ of $\triangle ABC$, $R_b$ is circumradius of $\triangle ABD$ , and $R_c$ is the circumradius of $\triangle ACD$.
Prove that
$$ {Rb\over Rc} ={AB\over AC}$$
Thanks for your help
| Applying sine rule in $\triangle ABD$ as follows $$\frac{\sin\angle ABD}{AD}=\frac{\sin\angle ADB}{AB} $$ $$\implies \color{red}{\sin \angle ABD=\frac{AD}{AB}\sin\angle ADB}$$ Now, the circumscribed radius $R_b$ of $\triangle ABD$ is given as $$R_b=\frac{\text{side of triangle }}{2\times\sin \text{(opposite angle)}}$$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| As already pointed out, the roots of the polynomial are complex.
The analytic solving is a boring task. Only the main intermediate results are reported below :
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Upper and lower bounds for $S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.$
For a positive integer $n$ let $S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.$
Then which of the following are true.
*
*(a) $S(100)\leq 100$.
... | Group the terms
exactly as you have,
but get a lower bound
that can be easily manipulated.
I'll copy, paste, and edit
your equations to show what I mean.
$$\displaystyle S(n) = 1+\underbrace{\frac{1}{2}+\frac{1}{3}}+\underbrace{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}+....+\underbrace{\frac{1}{2^{n-1}}+..+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Maximize $xy^2$ on the ellipse $x^2+4y^2=4$ I was using Lagrange multiplier, any steps gone wrong?
$$f(x,y)=xy^2$$
$$c(x,y)=x^2+4y^2$$
Partial Derivatives
$$\frac {\partial f}{\partial x} = y^2 $$
$$\frac {\partial f}{\partial y} = 2xy $$
$$\frac {\partial c}{\partial x} = 2x $$
$$\frac {\partial c}{\partial y} = 8y $$... | You can suppose $x,y>0$ otherwise $xy^2\leq 0$. Then you can use that $\frac{x_1+x_2+x_3}{3}\geq \sqrt[3]{x_1x_2x_3}$ and the equality holds iff $x_1=x_2=x_3$. Note that $(x^2+2y^2+2y^2)^3\geq 27x^2y^4$. So $64\geq 27(xy^2)^2$ the eqaulity holds iff $x=\sqrt{2}y$ or $x=-\sqrt{2}y$. Therefore $xy^2\leq \frac{8}{3\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\lim \frac{3^n + 2^n}{5\cdot3^n + 7\cdot2^n} = \frac{1}{5}$ Prove using limit definition.
$$\lim \frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} = \frac{1}{5}
$$
My try:
$$\left| {\frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} - \frac{1}{5}} \right| < \varepsilon
$$
$$ \Leftrightarrow \left| \frac{3^n + 2^n - 5\cdot 3^n - ... | HINT
$$
\frac{ 3^n + 2^n }{ 5 \times 3^n + 7 \times 2^n }
= \frac{ \color{red}{1} + (2/3)^n }{ \color{red}{5} + 7 \times (2/3)^n }
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$ Could someone please show me the proof that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$
I have no idea where to begin with this one.
Thanks.
| Notice, we have $$\lim_{x\to \infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}$$
$$=\lim_{x\to \infty}\frac{2\cdot 2^{x}+x^2+2x+1}{2^x+x^2}$$
$$=\lim_{x\to \infty}\frac{(2^{x}+x^2)+(2^x+2x+1)}{2^x+x^2}$$
$$=1+\lim_{x\to \infty}\frac{2^x+2x+1}{2^x+x^2}$$
Using L-Hospital's rule 3 times:
$$=1+\lim_{x\to \infty}\frac{2^x\ln 2+2}{2^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\lim_{n\to\infty} (n - \sqrt {{n^2} - n} )$ Calculate limit:
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n})$$
My try:
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n} ) = \lim_{n\to\infty} \left(n - \sqrt {{n^2}(1 - \frac{1}{n}} )\right) = \lim_{n\to\infty} \left(n - n\sqrt {(1 - \frac{1}{n}})\right)$$
$$\sqrt {(1 -... | Given $\displaystyle \lim_{n\rightarrow \infty}\left[n-\sqrt{n^2-n}\right] = \lim_{n\rightarrow \infty}\left[n-n\cdot \left(1-\frac{1}{n}\right)^{\frac{1}{2}}\right]$
Now Using $\displaystyle (1+x)^{n} = 1+nx+\frac{n(n-1)}{2}x^2+.....$
So we get $\displaystyle \lim_{n\rightarrow \infty}\left[n-n\left(1-\frac{n}{2n}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can the trigonometric equation be proven? This question :
https://math.stackexchange.com/questions/1411700/whats-the-size-of-the-x-angle
has the answer $10°$. This follows from the equation
$$2\sin(80°)=\frac{\sin(60°)}{\sin(100°)}\times \frac{\sin(50°)}{\sin(20°)}$$
which is indeed true , which I checked with Wolf... | $$\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}=2\sin 80^\circ\sin 20^\circ\cos 50^\circ=2\sin 80^\circ\sin 20^\circ\sin40^\circ$$
as $\cos 50^\circ=\sin40^\circ$
Now by Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Trying to solve $\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$ The equation is
$$\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$$
We get the system
$$
\begin{cases}
7-4\sqrt 2 \sin(x)=4\cos^2(x)-2\sqrt2\cos(x)\tan(x)+2\tan^2(x) \\
2\cos(x)-\sqrt2 \tan(x)\ge 0
\end{cases}
$$
I transformed the equation thus:
$$7(\sin... | $$2\cos x-\sqrt2\tan x=\dfrac{2-2\sin^2x-\sqrt2\sin x}{\cos x}=-\sqrt2\cdot\dfrac{\sqrt2\sin^2x+\sin x-\sqrt2}{\cos x}$$
Now $\sqrt2\sin^2x+\sin x-\sqrt2=(\sqrt2\sin x-1)(\sin x+\sqrt2)$ and $\sin x+\sqrt2\ge\sqrt2-1>0$
So we need $\dfrac{\sqrt2\sin x-1}{\cos x}\le0$
If $\cos x>0, \sqrt2\sin x-1\le0\iff\sin x\le\dfrac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ Problem :
Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ where $x \in [0,2\pi]$
My approach :
$5.(\frac{1}{25})^{\sin^2x}+4.5^{1-2\sin^2x}=25^{\frac{\sin2x}{2}}$... | $$5\cdot (\frac { 1 }{ 25 } )^{ sin^{ 2 }x }+4\cdot { \left( 5 \right) }^{ cos2x }=\left( 25 \right) ^{ \frac { sin2x }{ 2 } }\\ { 5 }^{ -2\sin ^{ 2 }{ x+1 } }+4\cdot { 5 }^{ 1-2\sin ^{ 2 }{ x } }={ 5^{ sin2x } }\\ 5\cdot { 5 }^{ 1-2\sin ^{ 2 }{ x } }={ 5^{ sin2x } }\\ { 5 }^{ 2-2\sin ^{ 2 }{ x } }=5^{ sin2x }\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$
| I would start by proving each side of the equation using induction.
First prove that $1^2+2^2+..+n^2 = \sum\limits_{i=1}^n i^2 \geq \frac{n^3}{3}$
Induction says that we must first check that the relationship holds for $n=1$.
$\sum\limits_{i=1}^1 i^2 = 1 \geq \frac{1^3}{3}=\frac{1}{3}$
So it holds for $n=1$. Now we a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How do I solve $x^4-3x^2+2=0$? How do I solve $x^4-3x^2+2=0$ ?
I would appreciate some kind of hint here. I have no clue how to start this problem.
| Consider the product $(y-2)(y-1)$ for which the expansion is $y^2 - 3 y +2$. Now let $y = x^2$ to obtain $x^4 - 3 x^2 + 2 = (x^2 -2)(x^2 -1) = 0$. This leads to $x^2 = 2$ or $x^2 =1$. These two equations yields $x \in \{ - 1, 1, - \sqrt{2}, \sqrt{2} \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solve logarithmic equation $\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$ Find $x$ from logarithmic equation:
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0 $$
This is how I tried:
$$x^2-8x+16>0$$
$$ (x-4)^2>0 \implies x \not = 4$$
then
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq \log_{\frac{x}{5}}(\frac{x}{5})^0 $$
because of base $\frac{x}{... | $$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$$
$$=\frac{\log{(x^2-8x+16)}}{\log\frac{x}{5}}\geq0$$
So $x$ is non-positive. And now raising both sides on the base $10$.
$$x^2-8x+16\geq1$$
$$x^2-8x+15\geq0$$
$$(x-5)(x-3)\geq0$$
$$x\geq5; \ x\geq3$$
But $x$ cannot be $5$. So answer is $[3,5)\cup(5,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Pythagorean identities. I was remembering about Pythagorean trigonometric identities so I grabbed some of my old notebooks and I was having so much fun, solving some equalities, until I found these four equalities that I just can't find the way to solve them, probably they are easier than I thought but I really need so... | Lets start from the LHS, then multiply by $\frac{RHS}{RHS}$
$$\begin{array}{lll}
\displaystyle\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}&=&\displaystyle\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}\cdot\frac{\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}}{\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Double u-substitution Problem I have to find $$\int\limits_{0}^{1}x(x+a)^{1/p}\text{ d}x$$ and I'm told to use two substitutions, $u=(x+a)^{1/p}$ and $u=x+a$.
However, I can't find the derivative of $u$, $\text{d}u$, to replace once I've done the substitution - could someone offer a bit of insight into what I'm missin... | As the others have suggested, we need to use the anti-power rule:
$$\int{x^a}dx=\frac{1}{a+1}x^{a+1}$$
Also as the comment has suggested, we may use $x=(x+a)-a$ to substitute the first $x$:
$$\int_{0}^{1}x(x+a)^{\frac{1}{p}}dx = \int_{0}^{1}[(x+a)-a](x+a)^{\frac{1}{p}}dx$$
This works because $(x+a)-a = x+a-a = x$. Let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int_1^{\frac{5}{2}}\frac{|x|}{[2x-5]}dx$ Find $\int_1^{\frac{5}{2}}\frac{|x|}{[2x-5]}dx$, where $[x]$ is the greatest integer function.
First, $|x|=x$ in the interval $[1,5/2]$. Next we divide subintervals, so that
for $[1,3/2), [2x-5]=-3$, $[3/2,2), [2x-5]=-2,$ and for $[2,5/2) [2x-5]=-1$.
Hence $$\int_1^{\fra... | Your answer is correct.
Another way to perform the calculation is to first observe as you did that $|x| = x$ for $x > 0$. Then the function $$f(x) = \frac{|x|}{\lfloor 2x-5 \rfloor} = \begin{cases} -x/3, & 1 \le x < 3/2 \\ -x/2, & 3/2 \le x < 2 \\ -x, & 2 \le x < 5/2 \end{cases}$$ on the interval of interest, and more... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A Floor Function based limit $\large{ \lim_{n \to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n-1} \ k \cdot \left \lfloor x + \dfrac{n-k-1}{n} \right \rfloor = \ ? }$
Find the value of the above limit upto 3 decimal places when $(x = \sqrt{2015})$.
Any suggestions on how to handle the floor function?
| By my comment above, some of the floor functions will evaluate to $\lfloor x\rfloor$, and some to $\lfloor x + 1\rfloor = \lfloor x\rfloor +1$. If we move that $+1$ out to a separate sum, we get
$$
\frac{1}{n^2} \sum_{k=1}^{n-1} \ k \cdot \left \lfloor x + \dfrac{n-k-1}{n} \right \rfloor = \frac{1}{n^2}\left( \sum_{k=1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding positive integral solutions of $3^a=b^2+2$ I am trying to find integral solutions of the equation $3^a=b^2+2$. I could get that $a$ and $b$ have to be odd. But I am unable to get any further. I believe that $(a,b)=(1,1);(3,5)$ are the only solutions but I am unable to show that. If any one has any hints, it wou... | The quadratic field $\mathbb Q(\sqrt {-2})$ we will use below in solving this question, is a very good field: its ring of integers is $\mathbb Z[\sqrt {-2}]$ (because $-2\equiv 2 (mod\space 4)) $; its only units are $\pm1$ (because $-2<0$ and distinct of $-1$ and$-3$); it is a registered unique factorization domain and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of... | Second line: You can get 2 of a type from A,B,C or D so $4\choose 1$. Then ${4\choose3}$ to select 3 different letters from the 4 types not yet selected, and finally, apply multiplication principle.
Second last line: 4 of a type only from A or B, so ${2\choose1}$, 1 more selection from the remaining 4 types, ${4\choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431903",
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If $a,b,c$ and $d$ are positive reals such that $a+b+c+d=1$.Find the minimum value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ If $a,b,c$ and $d$ are positive reals such that $a+b+c+d=1$.
Find the minimum value of $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$.
I have proved it by using Lagran... | Given $a+b+c+d = 1$ and We have to calculate $\bf{Minimum}$ of $\displaystyle \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$
where $a,b,c,d>0$
Using $\bf{A.M\geq H.M}$ Inequality
$$\displaystyle \frac{a+b+c+d}{4}\geq \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\Rightarrow \left(\frac{1}{a}+\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$2\sqrt n-2<1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....+\frac{1}{\sqrt n}<2\sqrt n-1$ If $n\in N$,then prove that $2\sqrt n-2<1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....+\frac{1}{\sqrt n}<2\sqrt n-1$
How should i prove this inequality,neither AM-HM nor any other inequality theorem is working here.
| This inequality comes in hand: $$ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}} < \sqrt{n} - \sqrt{n-1} $$
Part 1: $ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}} $
$$ \sqrt{n+1} - \sqrt{n} = (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$$
Since $\sqrt{... | {
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"timestamp": "2023-03-29T00:00:00",
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Can I show $\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$ by induction? It is known that $a_{n+1} > a_n$. I tried to prove
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$$
via induction.
Base case: $a_0 = \frac{1}{0+1} = 1 \leq 1$
Inductive step: assume $a_n \leq 1$, prove $a_{n+1} \leq ... | The result is "obvious." There are $n$ terms in the sum, each $\le \frac{1}{n+1}$. So the sum is $\le \frac{n}{n+1}$, which is $\lt 1$.
If we, unreasonably, want to prove the result by induction, we run into the problem mentioned in the OP. The solution is to strengthen the induction hypothesis. We prove by induction ... | {
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"timestamp": "2023-03-29T00:00:00",
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Euler's constant greater than 0 for all values of n? If Euler's constant is described as the limit as n approaches infinity of the following:
$$t_n = 1 + \frac 12 + \frac 13 \cdots + \frac1n -\ln(n)$$
How can one prove that $t_n$ is greater than $0$ for all values of $n$?
Thanks!
| First, put $s_n = 1+\dfrac{1}{2}+\cdots + \dfrac{1}{n}$, then
$1+\dfrac{1}{2}+\cdots + \dfrac{1}{n} - \ln n > 0$ because:
$1 = \displaystyle \int_{1}^2 1dx \geq \displaystyle \int_{1}^2 \dfrac{1}{x}dx$,
$\dfrac{1}{2} = \displaystyle \int_{2}^3 \dfrac{1}{2}dx \geq \displaystyle \int_{2}^3 \dfrac{1}{x}dx$, and
$.....$
$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Which one is greater $\left(5/2\right)^{2/5}$ or $\left(7/2\right)^{2/7}$? I have a question here:
Which of $\left(5/2\right)^{2/5}$ and $\left(7/2\right)^{2/7}$ is greater?
I tried comparison by the function $y=x^{1/x}$ and found the derivative as follows
$$\frac{\partial y}{\partial x}=\frac{1}{x}(x^{\frac{1}{x}-1}... | It is enough to notice that:
$$ f(x) = \log\left(x^{\frac{1}{x}}\right) = \frac{\log x}{x} \tag{1}$$
fulfills:
$$ f'(x) = \frac{1-\log x}{x^2},\qquad f''(x) = \frac{2\log x-3}{x^3} \tag{2}$$
hence $f(x)$ is a concave function over the interval $I=\left[\frac{5}{2},\frac{7}{2}\right]$, increasing from $\frac{5}{2}$ to $... | {
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"url": "https://math.stackexchange.com/questions/1437774",
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"question_score": "3",
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Unique Solution to a System of Equivalences
Show that $(2, 3, 7)$ is the only solution set to the following system of equivalences:
\begin{align}
ab = -1 & \mod c \\
ac = -1 & \mod b \\
bc = -1 & \mod a \\
\end{align}
I already attempted systems of equations like $ab + cm = -1$, $ac + bn = -1 \ldots$ but each time ... | I suppose you don't count solutions like $(1,1,1)$ although they are technically solutions to the system. So for now assume w.l.o.g. $1<a<b<c$.
The equations are equivalent to $c \mid ab+1,b \mid ac+1,a \mid bc+1$.
Since (as you stated) $a,b,c$ clearly must be pairwise relatively prime, we conclude that
$$abc \mid (ab+... | {
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Solving the nonlinear Diophantine equation $x^2-3x=2y^2$ How can I solve (find all the solutions) the nonlinear Diophantine equation $x^2-3x=2y^2$?
I included here what I had done so far. Thanks for your help.
Note: The equation above can be rewritten into $x^2-3x-2y^2=0$ which is quadratic in $x$. By quadratic formul... | Multiply through by $4$ and complete the square, $(2x-3)^2 - 8 y^2 = 9,$ so that $(2x-3)^2 + y^2 \equiv 0 \pmod 3.$ It follows that both $(2x-3)$ and $y$ are divisible by $3,$ therefore $x$ as well. Write $x=3t, y=3v.$ So far $(2t-1)^2 - 8 v^2 = 1.$ Let $u=2t-1,$ so $u^2 - 8 v^2 = 1.$
...if we write $$ x = \frac{3(1+u... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Troubles with solving $\sqrt{2x+3}-\sqrt{x-10}=4$ I have been trying to solve the problem $\sqrt{2x+3}-\sqrt{x-10}=4$ and I have had tons problems of with it and have been unable to solve it. Here is what I have tried-$$\sqrt{2x+3}-\sqrt{x-10}=4$$ is the same as $$\sqrt{2x+3}=4+\sqrt{x-10}$$ from here I would square bo... | Alternatively, take the reciprocal to get:
$$\frac{1}{\sqrt{2x + 3} - \sqrt{x - 10}} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{(2x+3)-(x-10)} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{x+13}$$
which we know is equal to $\frac 1 4$.
Thus we have the following:
$$\sqrt{2x+3} - \sqrt{x+10} = 4 \tag{1}$$
$$\sqrt{2x + 3} + \sqrt{x... | {
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"source": "stackexchange",
"question_score": "4",
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Polynomial Division by product of two polynomials f(x) is a polynomial, when it is divided by (x-3) it leaves remainder 15. when f(x) is divided by square of(x-1) it leaves remainder 2x+1. Find the remainder when f(x) is divided by product of two above divisors.
| HINT:
Let $f(x)=g(x)(x-3)(x-1)^2+Ax^2+Bx+C$
$15=f(3)=0+3^2A+3B+C$
$2x+1=A\{x^2-(x-1)^2\}+Bx+C$
$\iff2x+1=x(B+2A)+C-A$
Compare the constants & the coefficients of $x$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A$ is a unitary ring with $xy=1$ implies $yx=1$, prove that if $a^2=3b^2$ and $3ab=1+4ba$, then $ab=ba$ Let $(A,+, \cdot)$ be a unitary ring with the property that if $xy=1$ for $x,y \in A$, then $yx=1$. Let $a,b \in A$ with $a^2=3b^2$ and $3ab=1+4ba$. Prove that $ab=ba$.
We easily get that $(a-2b)(2a+3b)=1$, so, by h... | Counterexample: $A = \mathbb{F}_7^{2\times 2}$ (the $2\times 2$ matrix ring over $\mathbb{F}_7$), $a = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$, $b = \left(\begin{matrix} 0 & 0 \\ 5 & 0 \end{matrix}\right)$.
| {
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"question_score": "7",
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(Beginner) Intuition to solve a functional equation and steps for this particular- Please can someone please tell me the intuition behind solving a functional equation?
For example, $$f(x)+f(y)=2f\left(\frac{x+y}2\right)f'\left(\frac{x-y}2\right)$$
Now at the first look, it seems like $f(x)=\sin x$, but how to solve i... | Taking $x=y$ gives
$$ 2f(x) = 2f(x)f'(0), \tag{1} $$
so either $f(x) \equiv 0$ or $f'(0)=1$. Suppose it is the latter (else we have found $f$). Setting $x=-y$ gives
$$ f(x)+f(-x) = 2f(0)f'(x). \tag{2} $$
Differentiating the original equation with respect to $x$ gives
$$ f'(x) = f'\left( \frac{x+y}{2} \right)f'\left( \f... | {
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"question_score": "4",
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GCSE level trigonometry question In the UK we do GCSE exams and this is one of the GCSE questions I can't solve at this time of night lol. First part is easy but I can't see a way around part B. Any help would be much appreciated.
http://s17.postimg.org/8me1qdbkf/Screen_Shot_2015_09_21_at_21_28_51.png
http://postimg.... | $\angle DBC = 45^\circ$ and $\angle BCD = \arctan \dfrac{18} 6 $. Therefore $\angle BDC = 180^\circ - 45^\circ - \arctan\dfrac{18} 6$.
So then you have the three angles of $\triangle BDC$ and you have the length of one of the sides, so you can use the law of sines.
$$\sin x = \sin \angle BCD = \frac{\text{opposite}}{\... | {
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"source": "stackexchange",
"question_score": "2",
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How many pairs of positive integers $(x,y)$ satisfy the equation $x^2 - 10! = y^2$? I have a question about number theory:
How many pairs of positive integers $(x,y)$ satisfy the following equation? $$x^2 - 10! = y^2$$
My attempt:
Move the $y^2$ from right to the left and 10! From left to the right such that:
$$x^2-y... | Write
$10!
= 10.9.8.7.6.5.4.3.2
=2^83^45^27
$.
You want
$(x+y)(x-y)=10!=ab$
with
$x+y=a$ and $x-y=b$.
Then
$x = (a+b)/2$ and
$y = (a-b)/2$.
So $a$ and $b$
have to have the same parity.
Then choose $2^a3^b5^c7^d$
with all possible choices
satisfying the above.
Since they have to
have the same parity,
$a$ must have
$2^n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}... | If you let $t=4-3x,\;$ so $x=\frac{1}{3}(4-t)$, you get
$\displaystyle\lim_{x\to 1}\frac{1-\sqrt[3]{4-3x}}{x-1}=\lim_{t\to1}\frac{1-\sqrt[3]{t}}{\frac{1}{3}(1-t)}=\lim_{t\to1}\frac{3(1-t^{1/3})}{(1-t^{1/3})(1+t^{1/3}+t^{2/3})}=\lim_{t\to1}\frac{3}{1+t^{1/3}+t^{2/3}}=1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the determinant of $n\times n$ matrix Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
-x & a_2&a_3&a_4&\cdots &a_n\\
a_{1} & -x & a_3&a_4&\cdots &a_n\\
a_1&a_{2} & -x &a_4&\cdots &a_n\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
a_1&a_{2} & a_3&a_4&\cdots & -x\\
\end{array}... | Subtract the first row from each of the other rows. Most of the terms are now zero, and you can expand across the first row. Each product misses one of the $(x+a_i)$ factors, replaced by $a_i$. So the determinant is
$$(-1)^n\prod_i(x+a_i)\left[\frac x{x+a_1}-\frac {a_2}{x+a_2}-\frac {a_3}{x+a_3}...\right]\\
=(-1)^n... | {
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Applic. Of derivatives The straight line $x\cos \alpha + y \sin \alpha = p$ touches the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Prove that $a^{2}\cos 2\alpha+b^{2} \sin 2\alpha=p^{2}$
| Notice, solving the given equations by setting $y=\frac{p-x\cos \alpha}{\sin\alpha}$ in the equation of the curve: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get $$\frac{x^2}{a^2}+\frac{\left(\frac{p-x\cos \alpha}{\sin\alpha}\right)^2}{b^2}=1$$ $$b^2x^2\sin^2\alpha+a^2(p-x\cos \alpha)^2=a^2b^2\sin^2\alpha $$
$$(a^2\cos^2\... | {
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Is it possible to describe the Collatz function in one formula? This is related to Collatz sequence, which is that
$$C(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
Is it possible to describe the Collatz function in one formula? (without modular conditions)... | Let f be the iteration function for every even number and g for every uneven number.
\begin{align}
f(n) = \frac{n}{2}
\end{align}
\begin{align}
g(n) = 3n + 1
\end{align}
It is possible to use the right equation for the right number by using a sine function with a period of 2. This way, even numbers get iterated by f an... | {
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"source": "stackexchange",
"question_score": "39",
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"answer_id": 3
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Partial Fraction Decomposition trouble $\frac{1}{(1+x^2)(1+x^{2015})}$ I wish to decompose $\frac{1}{(1+x^2)(1+x^{2015})}$
I have $1 = (Ax+B)(1+x^{2015}) + (Cx+D)(1+x^2) = Ax^{2016} + Bx^{2015} + Cx^{3} + Dx^{2} + x(A+C) + B+D$
Doesn't this imply that $B+D = 1$, but we also have that $B=D = 0$ as well.
What did I do wr... | There is a trick:
$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}&=&\int_{0}^{1}\frac{dx}{(1+x^2)(1+x^m)}+\int_{1}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}\\&=&\int_{0}^{1}\frac{dx}{1+x^2}\left(\frac{1}{1+x^m}+\frac{1}{1+x^{-m}}\right)\\&=&\int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\color{red}{\frac{\pi}{4}}\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $2^n+(-1)^{n+1}$ is divisible by 3. Prove that $2^n+(-1)^{n+1}$ is divisible by 3 for $n\in\mathbb{N}$.
My attempt:
For $n=1$:
$2^1+(-1)^2 = 2 + 1 = 3, 3 |3$
We assume that $3|(2^n+(-1)^{n+1})$
Then for $n+1$:
$2^{n+1} + (-1)^{n+2} = 2\cdot 2^n -(-1)^{n+1}$
I am trying to show that for $n+1$, the expression ... | If $n=1$, then $2^{n} + (-1)^{n+1} = 3$, which is divisible by $3$; if $n \geq 1$ such that $2^{n} + (-1)^{n+1} = 3q$ for some integer $q \geq 1$, then
$$
2^{n+1} + (-1)^{n+2} = 2[ 3q - (-1)^{n+1} ] + (-1)^{n+2} = 6q + 2(-1)^{n+2} + (-1)^{n+2} = 6q + 3(-1)^{n+2},
$$
which is again divislble by $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot... | HINT: $$\frac{1-\sqrt{\cos(x)}}{x^2}=\frac{1-\cos(x)}{x^2(1+\sqrt{\cos(x)})}=\frac{1-\cos(x)^2}{x^2(1+\sqrt{\cos(x)})(1+\cos(x))}$$
| {
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"source": "stackexchange",
"question_score": "4",
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Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$
Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.
Attempt:
I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.
| Sine addition identity: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$
Cosine addition identity: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$$
From the above, we obtain the tangent addition identity: $$\begin{align*} \tan (\alpha + \beta) &= \frac{\sin (\alp... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_{c}^{c+2L} cos(\frac{2\pi nx}{L})$ Wolfram Alpha gives me this result but when doing I dont get it $\int_{c}^{c+2L} cos(\frac{2\pi nx}{L}) = \frac{L sin(2\pi n) cos(\frac{2 \pi n(c+L)}{L})}{\pi n}$ dx with $n = 0,1,2,...$
I put this integral in Wolfram Alpha and gives me this result, but when I try to do it at h... | Upon integrating, we have for $n\ne 0$, the term
$$\begin{align}
\sin \left(\frac{2\pi n(c+2L)}{L}\right)-\sin \left(\frac{2\pi nc}{L}\right)&=\sin \left(\frac{2\pi nc}{L}\right)\cos(4\pi n)\\\\
&+\cos \left(\frac{2\pi nc}{L}\right)\sin(4\pi n)\\\\
&-\sin \left(\frac{2\pi nc}{L}\right)\\\\
&=\sin \left(\frac{2\pi nc}{L... | {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer? $x,y,z \in \Bbb Z$, if $xy+yz+xz=1$ then prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer
$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+(xy)^2+(yz)^2+(xz)^2+(xyz)^2$
$(xy)^2+(yz)^2+(xz)^2+2xyz(x+y+z)=1$
| Let me try. $$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = x^2 + y^2 + z^2 + 2.$$
So, $$(1+x^2)(1+y^2)(1+z^2) = 1 + (x^2+y^2+z^2) + (x^2y^2+y^2z^2 + z^2x^2) +(xyz)^2 = 1 + (x+y+z)^2-2 + 1 - 2xyz(x+y+z) + (xyz)^2 = (x+y+z-xyz)^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Where am I going wrong in solving this exponential inequality? $$(3-x)^{ 3+x }<(3-x)^{ 5x-6 }$$
Steps I took:
$$(3-x)^{ 3 }\cdot (3-x)^{ x }<(3-x)^{ -6 }\cdot (3-x)^{ 5x }$$
$$\frac { (3-x)^{ 3 }\cdot (3-x)^{ x } }{ (3-x)^{ 3 }\cdot (3-x)^{ x } } <\frac { \frac { 1 }{ (3-x)^{ 6 } } \cdot (3-x)^{ 5x } }{ (3-x)^{ 3 }\cdo... | $$(3-x)^{ 3+x }<(3-x)^{ 5x-6 }$$
For $3>x$
$$(3+x)\log_{3-x}(3-x)<(5x-6 )\log_{3-x}(3-x)$$
$$3-x<5x-6$$
$$x>1.5$$
Thus, $$1.5<x<3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to evaluate $ \int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx $ Evaluate $$ \int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx $$ $$(a>0) $$
How can I use contour appropriately?
What is the meaning of this integral?
(additionally posted)
I tried to solve this problem.
First, I take a branch $$ \Omega=\mathbb C - \{z|\text{... | Real Method
We first evaluate the integral
$$
J=\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x
$$
Putting $x=\tan \theta$ yields
$$
\begin{aligned}
J& =\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\ln (a \tan \theta)}{a^2 \sec ^2 \theta} \cdot a \sec ^2 \theta d \theta \\
& =\frac{1}{a} \int_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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On solving the equation $\sin ^{2}\left ( z \right )= i\pi$ One obvious approach is to rewrite $\sin\left ( z \right )$ as $\frac{e^{iz}-e^{-iz}}{2i}$ and then proceed applying an appropriate substitution. However that leads to a quartic complex equation, which gets a bit(actually a lot) messy. Is there another(maybe o... | $$
sin(z) = \pm \sqrt{i \pi} \\
e^{iz} - e^{-iz} = \pm 2i \sqrt{i \pi}
$$
Let $e^{iz} = x$, then $x^2 \pm 2i \sqrt{i \pi} x - 1 = 0$.
Hence:
$e^{iz} = x = \frac{\pm 2i \sqrt{i \pi} \pm \sqrt{-4i \pi +4}}{2} = \pm i \sqrt{i \pi} \pm \sqrt{1-i \pi}$
Thus, $z = -i \log(\pm i \sqrt{i \pi} \pm \sqrt{1-i \pi})$.
Or:
$$
z_1... | {
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Evaluating $\sum_{n=1}^\infty \frac{n^3}{e^{2\pi n}-1}$ using inverse Mellin transform inspiration on the post Evaluating $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform
it is possible to calculate in close form
$$\sum _{k=1}^{\infty } -\frac{k^3}{e^{2 \pi k}-1}=\frac{3840 \pi ^4 \psi... | This is an alternative approach which is too long for comment. If we put $q = e^{-\pi}$ then the desired sum is $$-\sum_{n = 1}^{\infty}\frac{n^{3}}{q^{-2n} - 1} = -\sum_{n = 1}^{\infty}\frac{n^{3}q^{2n}}{1 - q^{2n}} = \frac{1 - Q(q^{2})}{240}$$ and we know that $$Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + ... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of integer solutions for $x_1+x_2+x_3 = 15$ under some constraints by IEP. For equation
$$
x_1+x_2+x_3 = 15
$$
Find number of positive integer solutions on conditions:
$$
x_1<6, x_2 > 6
$$
Let: $y_1 = x_1, y_2 = x_2 - 6, y_3 = x_3$
than, to solve the problem, equation $y_1+y_2 +y_3 = 9$ where $y_1 < 6,0... | You solved the problem in the nonnegative integers rather than the positive integers.
You wish to determine the number of solutions of the equation
$$x_1 + x_2 + x_3 = 15 \tag{1}$$
in the positive integers subject to the constraints $x_1 < 6$ and $x_2 > 6$.
Let's deal with the constraint $x_2 > 6$ first. Let $y_2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Olympiad inequality problem with $a+b+c+abc=4$ If $a,b,c \in \mathbb R_{> 0}$ and $a+b+c+abc=4$, prove that
$$({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}})^2(ab+bc+ca) \ge {\frac 12}(4-abc)^3$$
This can be solved by AM-GM-HM or the Cauchy-Schwarz inequality. I'd tried for some hours but couldn'... | Consider using the following Hölder's inequality:
If $a_1,\ldots, a_n$ and $b_1,\ldots, b_n$ are positive real numbers, then $$\sum_{k = 1}^n a_kb_k \le \left(\sum_{k = 1}^n a_k^p\right)^{1/p}\left(\sum_{k = 1}^n b_k^q\right)^{1/q}$$ where $p$ and $q$ are real numbers such that $p,q \ge 1$ and $\frac{1}{p} + \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the partial sums of $4+44+444+\cdots$
Find the sum to $n$ terms of $4+44+444+\cdots $
My attempts:
*
*Used successive difference method
*Used $4+(4+40)+(4+40+400)\dots$ method
Failed to get a formula for partial sum in both ways. What am I missing?
| We can Write $\displaystyle 4 = 4(1) = \frac{4}{9}\left[9\right] = \frac{4}{9}\left[10-1\right]$
Similarly $\displaystyle 44 = 4(11)=\frac{4}{9}\left[99\right] = \frac{4}{9}\left[10^2-1\right]$
Similarly $\displaystyle 444 = 4(111)=\frac{4}{9}\left[999\right] = \frac{4}{9}\left[10^3-1\right]$
Now you can proceed from h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove by induction: $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
For $n=1$ equality is true.
For $n=m$
$m-{m\choose 2}\frac{1}{2}+...+(-1)^{m+1}\frac{1}{m}=1+\frac{1}{... | Define
$$
f(n)=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac1k\tag1
$$
Then, $f(0)=0$ and for $n\gt0$,
$$
\begin{align}
f(n)
&=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac1k\tag2\\
&=\sum_{k=1}^n(-1)^{k-1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\frac1k\tag3\\
&=\sum_{k=1}^{n-1}(-1)^{k-1}\binom{n-1}{k}\frac1k+\sum_{k=1}^n(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $x^2+4y^2=4.$ Then find range of $ x^2+y^2-xy$
If $x^2+4y^2=4.$ Then find range of $ x^2+y^2-xy$
$\bf{My\; Try::}$ Given $$x^2+4y^2 = 4\Rightarrow \frac{x^2}{4}+\frac{y^2}{1} = 1$$, So parametric Coordinate for Ellipse are
$x = 2\cos \phi$ and $y = \sin \phi$.
Now Let $$f\left(x,y\right) = x^2+y^2-xy = 4-4y^2+y^2-... | Let's solve $\max_{(x,y)}x^2+y^2-xy$ s.t. $x^2+4y^2=4$.
*
*Let first $x=-\sqrt{4-4y^2}$ to have $$f(y)=-3 y^2+\sqrt{4-4 y^2} y+4$$ here we have $\frac{\sqrt{1-y^2}}{2}f'(y)=-2 y^2-3 \sqrt{1-y^2} y+1$. Solving this equation we get that $y_1=-\sqrt{\frac{1}{2}+\frac{3}{2 \sqrt{13}}}$ and $y_2=\sqrt{\frac12-\frac{3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the area of the triangle There are two points $N$ and $M$ on the sides $AB$ and $BC$ of the triangle $ABC$ respectively. The lines $AM$ and $CN$ intersect at point $P$. Find the area of the triangle $ABC$, if areas of triangles $ANP, CMP, CPA$ are $6,8,7$ respectively.
| Simple Geometry Approach
Note that, with the same altitude, ratio of areas of two triangles is equal to the ratio of their bases.
Therefore, $\dfrac {x}{y} = \dfrac {NP}{PC} = \dfrac {u}{v + z}$ ….. (*)
Similarly, $\dfrac {z}{y} = \dfrac {v}{u + x}$ ….. (#)
After eliminating v from (*) and (#) and making u as subject,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ After reading this question, the most popular answer use the identity
$$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1},$$
or, what is equivalent,
$$\sum_{t=k}^n \binom{t}{k} = \binom{n+1}{k+1}.$$
What's the name of this ident... | First proof
Using stars and bars, the number of ways to put $n$ identical objects to $k$ bins(empty bin allowed) is $\binom{n+k-1}{k-1}$.
If we reduce the number of bins by one, the number of ways to put $n$ identical objects to $k-1$ bins is $\binom{n+k-2}{k-2}$.
We can also count the number of ways to put $n$ identic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "73",
"answer_count": 18,
"answer_id": 17
} |
Finding a triangle ratio.
In the triangle ABC, the point P is found on the side AB. AC = 6 cm, AP = 4 cm, AB = 9 cm. Calculate BC:CP.
For some reason, I cannot get this even though I tried for half an hour.
I got that, $BC/CP < 17/10 = 1.7$ by the triangle inequality. $AP/PB = 4/5$
But that does not help one but, I'm... | Notice, let $BC=x$ & $CP=y$ now, applying cosine rule in $\triangle ABC$ $$\cos\angle BAC=\frac{AB^2+AC^2-BC^2}{2(AB)(AC)}=\frac{9^2+6^2-y^2}{2(9)(6)}=\frac{117-y^2}{108}\tag 1$$
similarly, applying cosine rule in $\triangle APC$
$$\cos\angle BAC=\frac{AP^2+AC^2-PC^2}{2(AP)(AC)}=\frac{4^2+6^2-x^2}{2(4)(6)}=\frac{52-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$ Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 \ \ \ n \in \mathbb{N}$
So for $n=1$
$$ 1 < 2$$
For $n > 1$
Assumption: $$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$$
Hypothesis (inductive ste... | inequility： $\frac{2}{n^2}\leq \frac{2}{n^2-1}=\frac{1}{n-1}-\frac{1}{n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the coordinate C. A triangle $A$, $B$, $C$ has the coordinates:
$A = (-1, 3)$
$B = (3, 1)$
$C = (x, y)$
$BC$ is perpendicular to $AB$. Find the coordinates of $C$
My attempt:
Grad of $AB$ =
$$\frac{3-1}{-1-3} = -0.5$$
Grad of $BC = 2$ ($-0.5 \times 2 = -1$ because AB and BC are perpendicular).
Equation of $BC$... | Yes, the gradient of line AB is $-1/2$ so the gradient of any line perpendicular to $AB$ is $2$. The line with gradient $2$ through $B = (3, 1)$ is $y = 2(x- 3)+ 1$ or $y= 2x- 5$. All points on that line, $(x, 2x- 5)$, satisfy the conditions for C.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does $\lim\limits_{n \to \infty } ((1 + x)(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$? Let $\left| x \right| < 1$.
Why does $$\lim\limits_{n \to \infty } ((1 + x)\cdot(1 + {x^2})\cdot(1 + {x^4})\cdot\ldots\cdot(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$$
| Using $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{1-x}\left[\overbrace{(1-x)(1+x)}^{(1-x^{2^{1}})}(1+x^2)(1+x^4).......(1+x^{2^{n}})\right]$$
So we get $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{1-x}\left[\overbrace{(1-x^2)(1+x^2)}^{(1-x^{2^2})}(1+x^4)....(1+x^{2^{n}})\right]$$
In a similar way we get th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Multivariable Epsilon Delta Proof I am trying to prove that a function $\ f(x,y)=y^2x$ is continuous everywhere, using epsilon delta proofs.
I've set it up like:
$\ |x^2y-a^2b| < \epsilon $
and
$\ \sqrt{\left(x-a\right)^2+\left(y-b\right)^2} < \delta $
But other than that, have absolutely no clue about what operations... | Fix $(a,b)\in \Bbb R^2$. For all $(x,y)\in \Bbb R^2$,
\begin{align}|f(x,y) - f(a,b)| &= |y^2x - b^2 a|\\
& = |(y^2x - y^2 a) + (y^2a - b^2 a)|\\
& \le |y^2x - y^2a| + |y^2a - b^2a|\\
& = y^2|x - a| + |y^2 - b^2||a|.
\end{align}
Now if $(x - a)^2 + (y - b)^2 < 1$, then in particular $|y - b| < 1$. So by the triangle i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integration of $\ln(1+x^2)\arctan x$ We need to evaluate $$\int \ln(1+x^2)\arctan x \, \mathrm{d}x$$
My thoughts were to set $u = \arctan x \implies x = \tan u$ so that our integral is transformed to $$\int 2u \sec^2 u \ln \sec u \, \mathrm{d}u$$
Am I on the right path? How do I continue from here? Are there... | We integrate by parts. First we find $\int \arctan x\, dx$.
$$\int \arctan x\, dx = (\arctan x)(x)-\int \frac{x}{x^2+1}\, dx$$
$$=(\arctan x)(x)-\frac{1}{2}\int \frac{d\left(x^2+1\right)}{x^2+1}=(\arctan x)(x)-\frac{1}{2}\ln(x^2+1)+C$$
$$\int \left(\ln\left(1+x^2\right)\right)\left(\arctan x\, dx\right)$$
$$=\ln\left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Write on my own my first mathematical induction proof I am trying to understand how to write mathematical induction proofs. This is my first attempt.
Prove that the sum of cubic positive integers is equal to the formula
$$\frac{n^2 (n+1)^2}{4}.$$ I think this means that the sum of cubic positive integers is equal to a... | You should use the second one:
Suppose it holds for the first $k$ numbers. So their sum is equal to $\frac{k²(k+1)^2}{4}$. Then the first sum of the first $k+1$ is equal to $1^3 + 2^3 + 3^3 + ... + (k+1)^3=\frac{k²(k+1)^2}{4}+(k+1)^3=\frac{k²(k+1)^2}{4}+\frac{4(k+1)^3}{4}$ which is equal to $\frac{k²(k+1)^2+4(k+1)^3}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Prove: If $|z|<\frac{1}{2}\Rightarrow |(1+i)z^3+iz|<\frac{3}{4},z\in\mathbb{C}$ Prove: If $|z|<\frac{1}{2}\Rightarrow |(1+i)z^3+iz|<\frac{3}{4},z\in\mathbb{C}$
$|z|=\sqrt{x^2+y^2}<\frac{1}{2}\Rightarrow x^2+y^2 <\frac{\sqrt{2}}{2}$
$$|(1+i)z^3+iz|=|(x^3-3xy-3x^2y-y-y^3)+i(x^3-3xy+3x^2y+x-y^3)|=\sqrt{(x^3-3xy-3x^2y-y-y^... | HINT:
Assuming $z=ai$ with $a\in\mathbb{R}$:
$$|(1+i)(ai)^3+i(ai)|<\frac{3}{4}\Longleftrightarrow$$
$$|(1+i)(ai)^3-a|<\frac{3}{4}\Longleftrightarrow$$
$$|(1-i)a^3-a|<\frac{3}{4}\Longleftrightarrow$$
$$|a((1+i)a^2-i)|<\frac{3}{4}\Longleftrightarrow$$
$$|a||(1+i)a^2-i|<\frac{3}{4}\Longleftrightarrow$$
$$\sqrt{a^6+(a^3-a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Understanding telescoping series? The initial notation is:
$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$
I get to about here then I get confused.
$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$
How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1... | The correct way to analyze this is to write
$$\begin{align}
\sum_{n=5}^N\frac{2}{n^2-1}&=\sum_{n=5}^{N}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\\\\
&=\left(\frac14-\frac16\right)+\left(\frac15-\frac17\right)+\left(\frac16-\frac18\right)+\cdots \\\\
&+\left(\frac1{N-3}-\frac1{N-1}\right)+\left(\frac1{N-2}-\frac1N\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$
I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses?
| This is called Trigonometric Substitution.
$$\int\sin^2 x\cos xdx, \color{green}{\text{let }u=\sin x}\implies du=\cos xdx$$
$$\int\sin^2 x(\cos xdx)=\int u^2du =\frac{u^3}{3}+C =\frac{\sin^3 x}{3}+C$$
Whereas this is called Inverse Trigonometric Substitution
$$\int\sqrt{1-4x^2}dx = \int\sqrt{1-(2x)^2}dx, \color{green}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Associative operation on a set $S$ Let $S$ be a non-empty set and $n\in \mathbb{N}, n\geq 2$ a fixed integer. We consider an associative operation $"\cdot"$ on $S$ with the following properties:
*
*$x^{n+1}=x, \forall x\in S$
*$xy^nx=yx^ny, \forall x,y\in S$
Show that this operation is commutative. This problem h... | We have
$$ \tag1x^2 = (x^2)^{n+1}=x(x^2)^nx=x^2x^nx^2=xx^{n+1}x^2=x^4$$
and consequently
$$\tag2 x = x^{n+1}=x^2x^{n-1}\stackrel{(1)}=x^4x^{n-1}=x^2x^{n+1}=x^3$$
and
$$ \tag3x^n=x^{n-1}x\stackrel{(2)}=x^{n-1}x^3=x^{n+1}x=x^2$$
for all $x$.
Now
$$\tag4 xy\cdot yx=xy^2x\stackrel{(3)}=xy^nx=yx^ny\stackrel{(3)}=yx^2y=yx\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Epsilon and Delta proof for limit $\lim_{x\to-1} \frac{x+5}{2x+3}=4$ I have a function $f(x)=\frac{x+5}{2x+3}$ and want to show that the limit
$$\lim_{x\to -1} f(x) = 4$$
is true using $\epsilon$-$\delta$. I have a trouble finding $\delta$ so that $f(x)-4$ is less than every $\epsilon$.
Can anyone help?
| Given $\epsilon > 0$, for $x \in (-1.2, -0.8)$, $|2x + 3| > 0.6$. Take $\delta = \min(0.2, \epsilon/12)$, then for all $|x - (-1)| < \delta$, we have
\begin{align}
& \left|f(x) - 4\right| \\
= & \left|\frac{x + 5}{2x + 3} - 4\right| \\
= & \left|\frac{-7x - 7}{2x + 3}\right| \\
= & 7\left|\frac{x + 1}{2x + 3}\right| \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $ Got these two problems on an exam recently and was unsure if I managed to answer them correctly:
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
Second:
$$\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+... | For the first, start from $\sin^2x+\cos^2x=1$ and square both sides.
For the second, note that the second part is the reciprocal of the first. So you need to solve $p+\frac 1p=\frac 52$ first.
But what have you tried?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the probl... | Too long for a comment:
As long as $x\in\Bbb{R}$, you can consider $|\frac{x}{x+2}|$ as $(\frac{x}{x+2})\text{sign}(\frac{x}{x+2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Solving $z^6=64$
$$z^6=64$$
My attempt:
$$\stackrel{\text{Euler}}{=}[r(\cos(\theta)+i\sin(\theta))]^{6}=64$$
$$\stackrel{\text{De moivre}}{=}r^6(\cos(6\theta)+i\sin(6\theta))=64$$
$$\stackrel{\text{Euler}}{=}r^6e^{i6\theta}$$
I know that I should get $6$ roots..
| $$z^6=64\Longleftrightarrow$$
$$z^6=\left|64\right|e^{\arg\left(64\right)i}\Longleftrightarrow$$
$$z^6=64e^{oi}\Longleftrightarrow$$
$$z=\left(64e^{2\pi ki}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$z=2e^{\frac{1}{6}\cdot 2\pi ki}\Longleftrightarrow$$
$$z=2e^{\frac{1}{3}\cdot \pi ki}$$
With $k\in\mathbb{Z}$ and $k:0-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to compute this finite sum $\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$? I do not know how to find the value of this sum:
$$\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$$
(Yes, the last term is added twice).
Of course I've already plugged it to wolfram online, and the answer is $$2-\frac{1}{2^{n-1}}$$
But I do not kn... | A trick with generating functions:
$\begin{align}
\sum_{k \ge 0} 2^{-k} z^k
&= \frac{1}{1 - 2^{-1} z} \\
z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{2 (1 - 2^{-1} z)}
&= \frac{z}{2 (1 - 2^{-1} z)^2} \\
&= \sum_{k \ge 0} k 2^{-k} z^k \\
\frac{1}{1 - z} \cdot \frac{z}{(1 - 2^{-1} z)^2}
&= \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Find the limit? How to calculate this limit ?
$$\lim _{x\to 1}\frac{\sqrt{x^2-1}}{\sqrt[3]{x-1}}$$
An image for clarification.
| $$\lim _{ x\rightarrow 1 }{ \left( \frac { \sqrt { { x }^{ 2 }-1 } }{ \sqrt [ 3 ]{ x-1 } } \right) = } \lim _{ x\rightarrow 1 }{ \left( \sqrt [ 6 ]{ \frac { { \left( { x }^{ 2 }-1 \right) }^{ 3 } }{ { \left( x-1 \right) }^{ 2 } } } \right) = } \lim _{ x\rightarrow 1 }{ \left( \sqrt [ 6 ]{ \frac { { \left( { x }-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
Assuming that :
$\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
So
$(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$
$\sqrt{(x-1)(y-1)} \leq xy-x-y+2$
$ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$
Here I'm stuck !
| WLOG $\sqrt x=\sec A,\sqrt y=\sec B$ where $0<A,B<\dfrac\pi2$
$\sqrt{x-1}+\sqrt{y-1}=\tan A+\tan B=\dfrac{\sin(A+B)}{\cos A\cos B}\le\sqrt{\sec A\sec B}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What are roots of $x^{3}+3x^{2}+4x+1$? There are no divisor of 1 in this polynomial for which would be satisfied $x^{3}+3x^{2}+4x+1=0$.
How to find roots here?
| This can be done by hand with a bunch of substitutions.
Let $y=x+1$, then
$$1+4(y-1)+3(y-1)^2+(y-1)^3=0$$
If we expand all the terms we get
$$1+4y-4+3y^2-6y+3+y^3-3y^2+3y-1=0$$
which boils down to
$$y^3+y-1=0$$
Now we can do a change in coordinates with another substitution.
Let $y=u+\frac{\lambda}{u}$. We will find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Modified Leibnitz integral: $\lim\limits_{a \to\infty}\frac1a\int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx=?$ $\lim\limits_{a \to \infty} \frac{1}{a} \int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx $ ,where $a$ is a parameter.
ATTEMPT:- Let $I(a)=\frac{1}{a} \int _0^\infty \frac{(x^2+ax+1)\... | Hint. One may recall that
$$
\arctan x +\arctan \frac1x =\frac{\pi}2,\qquad x>0.
$$
Then by the change of variable $x \to \dfrac1x$, one gets that
$$
aI(a):=\int _0^\infty\frac{(x^2+ax+1)\arctan(\frac1x)}{1+x^4}dx=\int _0^\infty\frac{(x^2+ax+1)\arctan x}{1+x^4}dx
$$ Thus we obtain
$$
aI(a)=\frac{\pi}2\int_0^\infty\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
as a product of linear factors.
I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(... | This is maybe not the most elegant answer, but I think it's relatively efficient. By trial and error we find that $a=b$, $a=c$ and $b=c$ make the expression vanish. Hence we write it as
$$\begin{align}
(a-b)(a-c)(b-c)f&=(a^2b-a^2c-ab^2+ac^2+b^2c-bc^2)f\\
&=-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3
\end{align}$$
Now you could... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Solve $\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}$ without L'Hôpital's rule Find limit without L'Hôpital's rule: $\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}$
I'm learning limits and I'm stuck with this example. So far, I tried modify denominator but I have no luck with it.
| $$\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}=\lim\limits_{x \to 1} \frac{2(x-1)}{x^2(x^2-1)^2}=\lim\limits_{x \to 1} \frac{2(x-1)}{x^2(x-1)^2(x+1)^2}$$
Therefore $\lim\limits_{x \to 1^+} \frac{2}{x^2(x-1)(x+1)^2}=+\infty$
and $\lim\limits_{x \to 1^-} \frac{2}{x^2(x-1)(x+1)^2}=-\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the limit without using L'Hopitals $\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$ $$\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$$
I think I need to do in the numerator by multiplying the difference of cubes.
$\lim _{x\to 1}\left(\frac{\left(\sqrt[3]{x+7}-2\right)\cdot \left(\s... | $$
\begin{aligned}
\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)
& = \lim _{t\to 0}\left(\frac{\sqrt[3]{\left(t+1\right)+7}-2}{2\left(t+1\right)^2+3\left(t+1\right)-5\:}\right)
\\& = \lim _{t\to 0}\left(\frac{\sqrt[3]{t+8}-2}{2t^2+7t}\right)
\\& = \lim _{t\to 0}\left(\frac{2+\frac{1}{12}t-\frac{1}{288}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum of $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ for $00$.
Find the minimum of $\displaystyle (u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2$ for
$0<u<\sqrt{2}$ and $v>0$.
I think I have to use the Arithmetic and Geometric Means Inequalities. Or $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\fr... | $\bf{Using\; Cauchy\; Schwartz\; Inequality::}$
$$[1^2+1^2]\cdot \left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq \left[v+\frac{9}{v}-\left(u+\sqrt{2-u^2}\right)\right]^2$$
Now Using $\bf{A.M\geq G.M}\;,$ We get
$\displaystyle v+\frac{9}{v}\geq 6$ and equality hold when $v=3$ and $\displaystyle \left(u+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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