Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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prove that : $\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3$
For $a^2+b^2+c^2 =3$, with $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3.$$
Can any one help me with this problem?
| Multiplying both LHS and RHS by $a+b+c = s > 0$, you have:
$$\sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot (a+b) \right)+ \sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot c \right) \ge 3(a+b+c) $$
$$\iff 6+ \sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot c \right) \ge 3s $$
Now note that $a^2+b^2 \ge \frac12(a+b)^2$ using AM-GM, so it is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer? I'm trying solve: $~a^3 + b^3 = c^3~$ has no nonzero integer solutions.
If $~(c−b)=1~$ then $~c^3-b^3=3c^2-3c+1=a^3,~$
from Wolframalpha get:
$$
c = \dfrac{3- \sqrt{3}\sqrt{4a^{3}-1}}{6} \\
c = \dfrac{\sqrt{3}\sqrt{4a^{3}-1}+3}{6}
$$
How to prove $~\sqrt{3}\sq... | You have to prove that
$$ 12 a^3 = b^2+3 \tag{1}$$
has no integer solutions. Since $3\mid$LHS, we must have $b=3c$, hence:
$$ 4 a^3 = 3c^2+1,\tag{2}$$
and $c$ must be odd, hence $c=2d+1$ and:
$$ a^3 = 3 d^2 + 3 d + 1 = (d+1)^3-d^3.\tag{3}$$
So you just have to prove that the difference between two consecutive cubes is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$a,b,c \geq 0$,prove that $a^2+b^2+c^2+abc+5 \geq3(a+b+c) $
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^2+b^2+c^2+abc+5 \geq3(a+b+c).$$
I'm certain that this problem could be solved by using dirchlet's theory.but I do not know how to apply it exactly.
| Since $(a_1)^2(b-1)^2(c-1)^2\geq0$, we can assume that $(a-1)(b-1)\geq0$ or
$c(a-1)(b-1)\geq0$ or $abc\geq ac+bc-c$.
Thus, it remains to prove that
$$a^2+b^2+c^2+ac+bc-c+5\geq3(a+b+c)$$ or
$$c^2+(a+b-4)c+a^2+b^2-3a-3b+5\geq0,$$
for which it's enough to prove that
$$(a+b-4)^2-4(a^2+b^2-3a-3b+5)\leq0$$ or
$$(a-b)^2+2(a-1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How show that $ABC$ is equilateral? Let $D$, $E$ and $F$ be three points on sides $BC$,$AC$ and $AB$ of triangle $ABC$ such that lines $AD$, $BE$ and $CF$ concur at point $M$. If three trianles $MDB$, $MCE$ and $MAF$ have equal areas and equal perimeters. How show that $ABC$ is equilateral?
| This problem gave me a really hard time, lots of steps are involved.
Step 1. If $ABC$ is equilateral and $MAF,MBD,MCE$ have the same area, then $M$ is the center of $ABC$.
Given that $[x,y,z]$ are the trilinear coordinates of $M$, the areas of our triangles are proportional to: $$\frac{xy}{x+z},\frac{yz}{x+y},\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that (4-2/1)(4-2/2)...(4-2/n) in an integer. We have to prove that $(4-2/1)(4-2/2)...(4-2/n)$ is an integer for $n\in\mathbb{N}$. Can we do this by induction?
We prove for $n = 1$, which is trivial as $(4-2/1) = 2$ which is clearly an integer.
next we assume that the statement is true for some integer $n$
i.e. ... | [Cannot seem to add a comment (re request above on the induction method), so posting a new answer in response]
Assume result is true for $n$ terms, i.e. product is
$$2n\choose n$$
which is an integer.
Then, for $n+1$ terms, product becomes
$$\begin{align}
{2n\choose n}\left( 4-\frac 2{n+1}\right) &=\frac{4n+2}{n+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate the integral $\int e^{2x} \sin^2x\ dx$ How can I evaluate the following integral?
$$\int e^{2x} \sin^2x\ dx$$
| Using operator D and its properties, we have
$$
\int e^{2x}\sin^2xdx = \dfrac{1}{2D}[e^{2x} - e^{2x}\cos(2x)] = \dfrac{1}{4}e^{2x} - \dfrac{e^{2x}}{2}\dfrac{1}{D+2}\cos(2x)
$$
$$
= \dfrac{1}{4}e^{2x} - \dfrac{e^{2x}}{2}\dfrac{D - 2}{D^2 - 4}\cos(2x) = \dfrac{1}{4}e^{2x} + \dfrac{e^{2x}}{16}[-2\sin(2x) - 2\cos(2x)]
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Convergence of infinite $ \sum (\frac{n}{n+1})^{n^2} $? I have being trying to solve this convergence but with no success. Using the ratio test I have reached here:
$$
a_n = \left( \frac{n}{n+1} \right)^{n^2}
$$
And
$$
\frac{1}{a_n} = \left( \frac{n+1}{n} \right)^{n^2}
$$
Also
$$
a_{n+1} = \left(\frac{n+1}{n+2}\righ... | Instead of the ratio test, try the root test:
$\displaystyle\lim_{n \to \infty}|a_n|^{1/n} = \lim_{n \to \infty}\left|\left(\dfrac{n}{n+1}\right)^{n^2}\right|^{1/n} = \lim_{n \to \infty}\left(\dfrac{n}{n+1}\right)^{n} = \lim_{n \to \infty}\left(1-\dfrac{1}{n+1}\right)^{n}$
Do you recongnize this limit?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solution of equation $\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$
Solution of equation $\displaystyle \frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$
$\bf{My\; Try::}$ Clearly Here $x>0$, Now Using $\bf{A.M\geq G.M}$
So Here $\displaystyle x\cdot 2014^{\frac{1}{x}}>0$ and $... | Applying AM-GM probably gives the most elegant solution (and indeed the presentation is quite suggestive of AM-GM), but this can also be solved by using an idea behind AM-GM. Noting first that $x\ge0$, we have:
$$\begin{align}
&\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014
\\\implies&x2014^{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\log _{x^{2}-3}(x^{2}+6x)<\log _{x}(x+2)$? How to solve the following inequality $$\log _{x^{2}-3}\left(x^{2}+6x\right)<\log _{x}(x+2)\ ?$$
| Note first that $x \in A \cup B$ where $A = (\sqrt3, 2), \; B = (2,\infty)$, for the question to make sense.
Now if $x \in A$, note that $0< x^2-3 < 1 < x < x+2 < x^2+6x$, so LHS $< 0$, while $RHS > 0$ so this is a solution.
For $x \in B$, we have $ 1 < x < x+2 $ and $1 < x^2-3 < x^2+6x$, so both LHS and RHS are posi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Matrix Multiplication Understanding I have to multiply two 3x3 matrices. I don't understand the answer.
$$A=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
$$B=\begin{pmatrix}1&0&-1\\ 0&1&-1\\ 0&0&1 \end{pmatrix}$$
The answer given online is $$AB=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
I don't unders... | $\text{You have the right answer.}$
$$AB=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
Number of elements of order $2$ in $S_n$
How many elements of order $2$ are there in $S_n$?
Using combinatorics I arrived at this:
For $n$ even ($n=2k$) there are ${n\choose2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2} {n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}... | If you accept a confluent hypergeometric function as a closed form, then a Computer Algebra System like Mathematica will give you
for even $n = 2w$:
$\left(-\frac{1}{2}\right)^{-w} U\left(-w,\frac{1}{2},-\frac{1}{2}\right)-1$
and for odd $n = 2w-1$:
$\left(-\frac{1}{2}\right)^{1-w} U\left(1-w,\frac{3}{2},-\frac{1}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
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Sphere equation given 4 points Find the equation of the Sphere give the 4 points (3,2,1), (1,-2,-3), (2,1,3) and (-1,1,2).
The *failed* solution I tried is kinda straigh forward:
We need to find the center of the sphere.
Having the points:
$$
p_{1}(3,2,1),\,
p_{2}(1,-2,-3),\,
p_{3}(2,1,3),\,
p_{4}(-1,1,2)
$$
2 Triang... | I would cite the beautiful method from W.H.Beyer to find the center and radius of the sphere $(x-a)^2+(y-b^2)+(y-c)^2=R^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/894794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
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Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$ Let $a,b,c>0$, prove that
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$
I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.
Thanks.
| We need to prove that
$$\frac{a^2+b^2+c^2}{ab+ac+bc}-1\geq1-\frac{8abc}{(a+b)(a+c)(b+c)}$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{2(ab+ac+bc)}\geq\frac{\sum\limits_{cyc}c(a-b)^2}{(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc}(a-b)^2(a^2b+a^2c+b^2a+b^2c-c^2a-c^2b)\geq0.$$
Let $a\geq b\geq c$.
Hence, $(a-c)^2\geq (b-c)^2$, $$a^2b+a^2c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894954",
"timestamp": "2023-03-29T00:00:00",
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Product rule for simplex numbers The $n$th triangular number is defined as $T_2(n) = n(n+1)/2$, and there is an interesting product rule for triangular numbers:
$$T_2(mn) = T_2(m)\,T_2(n) + T_2(m-1)\,T_2(n-1).$$
The tetrahedral numbers $T_3(n) = n(n+1)(n+2)/6$ satisfy a similar product rule:
$$T_3(mn) = T_3(m)\,T_3(n)... | By denoting with $(n)_k$ the falling Pochhammer symbol
$$(n)_k = n\cdot(n-1)\cdot\ldots\cdot(n-k+1)$$
we have:
$$ (mn+1)_2 = \frac{1}{2}\left((m+1)_2(n+1)_2+(m)_2(n)_2\right)\tag{1}$$
that leads to the first identity. Since:
$$ (mn+2) = \alpha(n+2)(m+2)+\beta(n-1)(m-1) $$
is solved by $\alpha=\frac{1}{3}$ and $\beta=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins I encountered the following problem in Herber Wilf's book Generatingfunctionology:
Prove that, in country that has $1$-cent, $2$-cent, and $3$-cent coins
only, the number of ways of changing $n$ cents is exactly the integer
nearest to $ \frac{(n+3)^2}{12}... | The following method could be explained to any student that understands equations for lines and knows how to sum an arithmetic sequence. However, you may find it at least as tedious as the method you already have!
The Idea
Think of points $(x,y)$ in the plane as representing "$x$ 2-cent coins and $y$ 3-cent coins".
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $ \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$
If $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$.
$\bf{My\; Try::}$ Given $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\Rightarrow 13!\left(1+\f... | Imagine bringing the left side to the common denominator $13!$. The numerator $x$ is then a sum of terms of the form $\frac{13!}{r}$. All these terms are congruent to $0$ modulo $11$ (divisible by $11$) except $\frac{13!}{11}$.
So the required remainder is the same as the remainder when $(10!)(12)(13)$ is divided by $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where did I go wrong in completing the square? $$2x^{ 2 }+8x+1=0$$
Move 1 to the other side of the equation:
$$2x^{ 2 }+8x\quad =-1$$
Divide both sides by 2 to get 1 as the leading coefficient:
$$x^{ 2 }+4x\quad =-\frac { 1 }{ 2 } $$
$$(\frac { 4 }{ 2 } )^{ 2 }$$
$$x^{ 2 }+4x+4\quad =-\frac { 1 }{ 2 } +4$$
$$(x+2)^{ 2... | Your calculations are right..
$$(x+2)^2=\frac{7}{2} \Rightarrow x+2=\pm \sqrt{\frac{7}{2}} \Rightarrow x=-2 \pm \sqrt{\frac{7}{2}}=-2 \pm \frac{\sqrt{14}}{2}$$
So:
$$x=-2 \pm \frac{\sqrt{14}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$ if $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
My try :
$2pp' = (b^2-a^2)\sin 2\theta$
$p'^2 + pp'' = (b^2-a^2)\cos 2\theta$
Thats it ! it doesn't simplify no matter what I try. Any help ... | We need to establish $$p+p''=\frac{a^2b^2}{p^3}$$
Multiplying by $2p'(\ne0),$ $$(p+p'')2p'=a^2b^2\cdot\frac{2p'}{p^3}$$
Integrating we get,
$$p^2+p'^2=-\frac{a^2b^2}{p^2}+K\text{, where }K\text{ is an arbitrary constant} $$
Multiplying by $p^2(\ne0),$
$$p^4+p'^2p^2=-a^2b^2 +p^2\cdot K\ \ \ \ (1)$$
Now, $$p^4+p'^2p^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integral of $\arcsin$ of a rational function, using integration by parts I'm a class 12 student and this a question from my textbook:
$$I=\int{\arcsin{2x\over 1+x^2}}\mathrm{d}x$$
I did it using integration by parts like this:
$$I=\arcsin{\left(2x\over 1+x^2\right)}\cdot\int1\cdot\mathrm{d}x-\int(\frac{\mathrm{d}}{\ma... | HINT
Your calculation of $ \frac{\mathrm{d}}{\mathrm{d}x} \sin^{-1}\left (\frac{2x}{1+x^2} \right) $ is wrong. It will be:
$$\begin{align}
\dfrac{\mathrm{d}}{\mathrm{d}x} \sin^{-1}\left (\dfrac{2x}{1+x^2} \right) &= {\dfrac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}} \cdot { 2(1+x^2) - (2x)(2x)\over(1+x^2)^2} \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of this series May I know how I should go about finding the sum of this series?
$\displaystyle\sum_{n=1}^\infty$ $\dfrac{n}{2^{n-1}}$
I am really stuck. Thanks!
| $$s=\sum_{n=1}^\infty \dfrac{n}{2^{n-1}}=1+2/2+3/2^2+4/2^3+5/2^4+....(1)$$
By dividing whole equation by $2,$
$$\dfrac{s}{2}=\dfrac{1}{2}(\sum_{n=1}^\infty \dfrac{n}{2^{n}})=1/2+2/2^2+3/2^3+4/2^4+5/2^5+....(2)$$
Now$(1)-(2)$ gives us,
$$\dfrac{s}{2}=1+1/2+1/2^2+1/2^3+1/2^4+....=1+1$$
Hence we can obtain $s=4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
| For the base case simply show that the premise is true for the smallest positive integer: $\mathsf P(1)$.
For the iterative case, you have to show that if the premise holds for $n$, then it is implied to hold for $n+1$. $\mathsf P(n)\implies\mathsf P(n+1)$
Do that by substituting $n+1$ for $n$ and demonstrate that th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find $\int_0^\pi \sin(x)\,dx$ explicitly A book asks me to prove that:
$$\int_0^{\pi}\sin(x)\,dx = 2$$
Using the identity:
$$\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right) = \frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\le... | Notice that we can rewrite the limit as:
$$
\left[\lim_{n\to\infty} \frac{\frac{\pi}{2n}}{\sin\left(\frac{\pi}{2n}\right)} \right]\left[\cos\left(\lim_{n\to\infty}\frac{\pi}{2n}\right)-\cos\left(\lim_{n\to\infty}\frac{(2n+1)\pi}{2n}\right) \right]
= 1 \cdot [\cos 0 - \cos \pi] = 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertices Triangle $ABC$ is an equilateral triangle and $P$ is an arbitrary point inside it. The distance from $P$ to $A$ is $4$ and the distance from $P$ to $B$ is $6$ and the distance from $P$ to $C$ is $5$. How to find the s... | If you call the distances from $P$ to $A, B, C$, say, $a, b, c$, respectively, then by the Law of Cosines the side length $s$ of the equilateral triangle is given by
$s^2 = a^2 + b^2 - 2 a b \widetilde{C} = b^2 + c^2 - 2 b c \widetilde{A} = c^2 + a^2 - 2 c a \widetilde{B}$ $(\ast)$,
where $\widetilde{A} = \cos \angle B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Proving determinant using properties of determinants $$\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a^3 & b^3 & c^3
\end{vmatrix} =
(a-b)(b-c)(c-a)(a+b+c)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hopi... | Using $C_2'=C_2-C_1, C_3'=C_3-C_1$
$$\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a^3 & b^3 & c^3
\end{vmatrix}$$
$$=\begin{vmatrix}
1 & 0 & 0\\
a & b-a & c-a\\
a^3 & b^3-a^3 & c^3-a^3
\end{vmatrix}$$
$$=-(a-b)(c-a)\begin{vmatrix}
1 & 0 & 0\\
a & 1 & 1\\
a^3 & b^2+ab+a^2 & c^2+ca+a^2
\end{vmatrix}$$
Use $C_2'=C_2-C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of a class of continued fractions Is there a closed-form way of writing the continued fraction:
$$
1 + \frac{2}{3+ \frac{4}{5 + \frac{6}{7 + ...}}}
$$
EDIT: Since the above has been determined as $\frac{1}{\sqrt{e}-1}$, is there a similar expression for:
$$
2 + \frac{3}{4+ \frac{5}{6 + \frac{7}{8 + ...}}}
$$... | It's $1/(e^{1/2}-1)$. You should be able to derive this by doing a term-by-term transformation on an appropriate infinite series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Checking a solution to an indefinite integral I recently did some work to try to find $\int{\frac{dx}{Ax^3 - B}}$, but I'm always paranoid that my solution has some minor trivial error in the middle of the process that screwed up the end result entirely, so could someone please help check my solution?
The first step to... | Everything looks fine until your partial fraction decomposition. Indeed $P=\frac{1}{3}$, but to find $Q$ and $R$, start with
$$1 = \frac{1}{3}(u^2+u+1) + (Qu+R)(u-1)$$
and choose $u=0$ to get
$$1 = \frac{1}{3} - R,$$
so that $R = -\frac{2}{3}$. You then get $Q = -\frac{1}{3}$. So the decomposition is
$$\frac{1}{u^3-1}... | {
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"url": "https://math.stackexchange.com/questions/915414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Get $5$ by doing any operations with four $7$s How can one combine four sevens with elementary operations to get $5$? For example $$\dfrac{(7+7)\times7}{7}$$ (though that does not equal $5$). I am not able to do this. Can you solve it or prove that it's impossible?
| How about:
$$7 - \frac{7+7}{7} = 5$$
$$7 - \log_7 (7·7) = 5$$
$$7 - \frac{\ln (7·7)}{\ln 7} = 5$$
$$\left\lfloor \sqrt{\frac{7^7}{7!}} - 7\right\rfloor = 5$$
$$\lfloor 7\sin 777^\circ\rfloor = 5$$
$$\lfloor 7\cos 7^\circ\rfloor - \frac{7}{7} = 5$$
$$\lfloor 7\cos 7\rfloor = 5 \text{ using radians}$$
You can also use ba... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 10,
"answer_id": 0
} |
Prove $a^2+b^2+c^2=\frac{6}{5}$ if $a+b+c=0$ and $a^3+b^3+c^3=a^5+b^5+c^5$
if $a,b,c$ are real numbers that $a\neq0,b\neq0,c\neq0$ and $a+b+c=0$ and $$a^3+b^3+c^3=a^5+b^5+c^5$$
Prove that $a^2+b^2+c^2=\frac{6}{5}$.
Things I have done: $a+b+c=0$ So $$a^3+b^3+c^3=a^5+b^5+c^5=3abc$$
also $$a^2+b^2+c^2=-2ab-2bc-2ca$$
I t... | Note that $(a^3+b^3+c^3)-(a+b+c)(a^2+b^2+c^2)+(ab+bc+ca)(a+b+c)-3abc = 0$.
Using the fact that $a+b+c = 0$, this reduces to $a^3+b^3+c^3 = 3abc$. Thus, $a^5+b^5+c^5 = 3abc$.
Now, note that $(a^5+b^5+c^5)-(a+b+c)(a^4+b^4+c^4)+(ab+bc+ca)(a^3+b^3+c^3)-abc(a^2+b^2+c^2) = 0$.
Using the facts that $a+b+c = 0$ and $a^5+b^5+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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How to prove the following limit: $\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0$? How to prove the following limit:
$$\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0?$$
| Starting the same as Kelenner's answer, but using the Beta function, we get
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}\log(n+k)
&=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\frac{n\,\mathrm{d}x}{n+kx}\\
&=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\int_0^1n\,t^{n-1+x+(k-1)x}\,\mathrm{d}t\,\mathrm{d}x\\
&=-\int_0^1\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
lebesgue's identity In the Lebesgue's identity $$ a^2+b^2+c^2=d^2$$ where: $$a=m^2+n^2-p^2-q^2$$ $$b=2(mp+nq) $$ $$c=2(mq-np) $$ $$d=m^2+n^2+p^2+q^2 $$how can we write $(m,n,p,q)$ as a function of the integers $(a,b,c,d)$? I have been trying and failing. Please help.
| Lebesgue's Identity is complete only if you add a scaling factor t. Hence,
$$\begin{align}
a &= (m^2+n^2-p^2-q^2)t\\
b &= 2(mp+nq)t\\
c &= 2(mq-np)t\\
d &= (m^2+n^2+p^2+q^2)t
\end{align}\tag{1}$$
As a function of the integers $a,b,c,d$, then the variables $m,n,p,q,t$ are simply,
$$\begin{align}
m &= \frac{-b-c}{a-d}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Basic partial fractions issue I've been starting to get the hang on partial fractions, whilst I've been able to do most of the basic ones, this kept causing some issues so I assumed:
*
*I'm using the wrong method
*I'm converting values when I shouldn't be
Before going on I'll post the question:
(a) Express $\fra... | $$x^2+2x+2=0 \Rightarrow \Delta=4-4 \cdot 2=4-8=-4<0$$
So it has no real roots.
Therefore, to express $\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}$ in partial fractions we do the following:
$$\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}=\frac{A}{x-3}+\frac{Bx+c}{x^2+2x+2}$$
(The polynomial at the numerator has to be one d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the derivative of the function $f(x)=(5-x^2)(\sqrt{x})$ I am stuck on this question, not sure what I am doing wrong:
\begin{align*}
f(x) &=(5-x^2)(\sqrt{x}) \\
\\
f'(x) &=(5-x^2)(x)^{1/2} \\
&=(-2x)(x)^{1/2}+(-x^2+5)(\dfrac{1}{2})(x)^{-1/2}\\
&=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\\
&=\dfrac{1}{2}x^{-1/2}\... | Note that
$$-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\not =\dfrac{1}{2}x^{-1/2}\bigg(x^2-2x^2+10\bigg).$$
We have
$$f'(x)=\cdots=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})=\frac 12x^{-\frac 12}(-4x^2-x^2+5)=\frac{5(1-x^2)}{2\sqrt x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Indefinite integral of a rational function: $\int\frac{6x+4}{x^2+4}\,dx$ Find $\displaystyle\int\frac{6x+4}{x^2+4}\,dx$
The question asks to find integral of the expression so I divided them into two parts:
$$
\int\frac{6x}{x^2+4}\,dx
$$
and
$$\int\frac{4}{x^2+4}.
$$
So, for the first integral, I set $u=x^2+4$ and got... | Since you've nailed the first of the split integral, I'll address only the second:
If you make the substitution $\,x = 2\tan\theta,\,$ then $\,\theta = \arctan\left(\frac x2\right)\,$ and $\,dx = 2\sec^2 \theta.$
We also make use of the identity: $$\tan^2 \theta + 1 = \sec^2\theta\tag{1}$$
That gives you $$\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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$6$ women and $4$ men wait in line. If their order in line is random, find the probability that all of the women are adjacent to one another. My thoughts on the problem are that the number of ways the women can be adjacent to each other is $5!$ and the total number of arrangements for all the people is $10!$. Is this c... | Total number of arrangements was correctly determined as $10!$.
The number of ways the $6$ women can all stand next to each other are easily seen through the following:
$W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __ __ __
__ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __ __
__ __ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/924102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
What is $\cdots ((((1/2)/(3/4))/((5/6)/(7/8)))/(((9/10)/(11/12))/((13/14)/(15/16))))/\cdots$? What does this number equal if it goes on forever?
$$\frac{\frac{\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac56}{\frac78}}}{\frac{\frac{\frac{9}{10}}{\frac{11}{12}}}{\frac{\frac{13}{14}}{\frac{15}{16}}}}}{\frac{\frac{\fr... | HINT:
let $u_n=\dfrac{n}{n+1}$ and $v_n=\dfrac{u_n}{u_{n+1}}$ where $n=1,2,3,..$
I think here you are asking about the$ \lim_{ n \to \infty} v_n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/924601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 6,
"answer_id": 5
} |
derive limit to make a function continuous Here's the problem:
Choose the value of k that makes the following
function continuous at $x = 1$:
$f(x)=\begin{cases}
\frac{-8x^2 + 48x - 40}{x - 1} & x < 1\\
-2x + k &x \geq 1
\end{cases}$
My steps:
$\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow 1} f(x)$
$\frac{-8x^2 + 48x - ... | To address your aside: this is likely a typo.
To address your note: the $-2x$ term contains a variable $x$ whose scope is unrelated to the other variable on the other side of the equation. It's a bit unfortunate that we reuse the same variable, as this might be confusing. To emphasize the fact that the variables on eac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solve the trigonometric equation/inequality $(1).\quad\cos^2(2x) + \sin^4(x) = 2$ $($solve the equation$)$
$(2).\quad2\cos^2(3x) + 5\cos(3x) - 3 < 0$.
For this question, I tried letting $t=\cos(3x)$.
Thus, $2t^2 + 5t - 3< 0$, but that doesn't factor properly.
I'm really stuck... It would be nice if someone co... | For the equation $$\cos^2(2x) + \sin^4x = 2$$ you can use one of the trigonometric formulas for $\cos(2x)$. They are
\begin{align*}
\cos^2(2x) & = \cos^2x - \sin^2x\\
& = 2\cos^2x - 1\\
& = 1 - 2\sin^2x
\end{align*}
If we use the last one, then we will have a polynomial in $\sin x$.
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Calculus Question: Improper integral $\int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}\text dx$ How to evaluate integral $$\int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}\text dx?$$ I tried substitution $x=u^3$ and I got $3\displaystyle\int_{0}^{\infty}u \cos(2u^3+1)\text du$. After that I tried to use integration by pa... | $$\color{blue}{\mathcal{I}=\frac{\Gamma(\frac{2}{3})\cos(1+\frac{\pi}{3})}{2^{2/3}}\approx-0.391190966503539\cdots}$$
\begin{align}
\int^\infty_0\frac{\cos(2x+1)}{x^{1/3}}{\rm d}x
&=\int^\infty_0\frac{\cos(2x+1)}{\Gamma(\frac{1}{3})}\int^\infty_0t^{-2/3}e^{-xt} \ {\rm d}t \ {\rm d}x\tag1\\
&=\frac{1}{\Gamma(\frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
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Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$ I am doing some math repetition and am a bit stuck on this exercise:
Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$.
Now, this is a geometric sum on both the $LHS$ and $RHS$, which I guess is something that I should use to solve the equation...
Ano... | Note that the left side of the equation can be written as
$2^0+2^x+2^{2x}+2^{3x}...2^{5x}$
This is a geometric series, with
a=1, n=6, and $r=2^x$
We use the formula:
$S_n = \frac {a(1-r^n)}{1-r}$
Substitute the values, you get
$S=\frac{(1-2^{6x})}{1-2^x}$
We do the same for the right side
$S=3[\frac{1-2^{3x}}{1-2^x}]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding the limit by factoring the denominator and canceling I have the problem
$$\lim_{x\to10} \frac{x-3}{x^2+7x-30}$$
If I factor it to $\dfrac{x-3}{(x+10)(x-3)}$ then $x-3$ cancels and I'm left with $0$.
I know the real answer is $1/20$, but why is zero wrong?
| Your cancellation method needs an adjustment:
$$\frac{a}{a\cdot b} = \frac{1}{b}, \quad \textbf{not} \frac{0}{b}$$
To see why, note that $a=a\cdot 1$, so
$$\frac{a}{a\cdot b} = \frac{a\cdot 1}{a\cdot b}= \frac{1}{b}$$
You may also want to experiment with $\dfrac{3}{2\cdot 3}$, for a concrete example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/929301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partition of a Matrix In Linear Algebra, we have been taught that the partition of a matrix $A$ consists of matrices,or blocks. In other words, its elements are matrices. This same, partitioned matrix, however is said to be equal to the original matrix. But their elements are different, as one contains scalars and the ... | You have to concatenate the matrices. Here's an example:
$$A= \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, B = \begin{pmatrix} 10 \\ 20 \end{pmatrix}, C = (100,200), D = 1000,$$
then
$$\begin{pmatrix}A & B \\ C & D \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} & \begin{pmatrix} 10 \\ 20 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve $ x_{n+1} - x_n = 2n + 3$ Solve $$ x_{n+1} - x_n = 2n + 3, x_0 = 1, n \ge 0$$
I would try to find a homogen solution and used $$ r^2 - r = 0$$ and got $$x^h_n = A1^n$$ but this seems wrong and I'm stuck on how to continue.
=== Edit ===
Homogen solution:
$r^2 - r = 0 $ has the solutions $r=0, r=1 $ So we have $$... | $$x_{n+1} - x_n = 2n + 3 \tag A$$
$$n = \frac 12 x_{n+1} - \frac 12 x_n - \frac 32 \tag B$$
$$n-1 = \frac 12 x_{n} - \frac 12 x_{n-1} - \frac 32 \tag C$$
Subtract previous two:
$$1 = \frac 12 x_{n+1} + \left(-\frac 12 - \frac 12\right)x_n + \frac 12 x_{n-1} \tag D$$
$$x_{n+1} = x_{n} - x_{n-1} + 2 \tag E$$
Now it is a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
problem solving logarithmic equation and reaching an equivalence ok so i've had a problem trying to simplify the $\ln\left[ \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right]$ and this is supposed to be equal to : $\ln [ \sqrt{a^2+u^2} + u ]$
how is this posible ?? i've tried to solve this for more than 2 hours and could... | $\ln x$ is injective so, if $\ln x=\ln y\implies x=y$. So, the following should hold true:\begin{align}
\sqrt{1+\frac{u^2}{a^2}}+\frac{u}{a}=\sqrt{a^2+u^2}+u\end{align}
However, simply plugging in $u=1, a=2$ gives us
\begin{align*}
&\sqrt{1+\frac{1}{4}}+\frac{1}{2}=\sqrt{4+1}+1\\
&\implies \sqrt{\frac{5}{4}}+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove a sum formula by induction I am to prove through induction that $$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$
And well, my method seems to be working, but I get stuck when I'm nearly done. First I prove the formula work for $n = 1$.
$$\sum_{k=1}^1 (2k-1)^2 = 1$$
$$f(1)=\frac{(2*1-1)(2*1+1)}{3} = 1$$
I then ... | $$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$
*
*$$\sum_{k=1}^1 (2k-1)^2=1^2 = \frac{1(2*1-1)(2*1+1)}{3}=1$$ True
*Assuming it is true for $k=n$:
$$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$ Assumption of Truth
*Now finding for $k=n+1$:
$$\begin{align}\sum_{k=1}^{n+1} (2k-1)^2&=\sum_{k=1}^n (2k-1)^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$ 2.Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$
I know that under that condition $y = -x$ which is the reflection of the $x$ graph. I have problems wi... | I don't know if anyone will read this after 8 years, but I think there is another mistake higher up concerning the following two lines:
$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$
In the second line it should instead read:
$2V_w-2V_w+2V_z -4(v) = e^{\frac{w+z}{2}+z}$
You f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/931921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
symetric inequality for a rational function of three variables $a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that: $$\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2} \geqslant \dfrac{1}{2} $$ I have tried with Cauchy-Schwarz inequality in Engel form...
| We have $\frac{ab^2}{a^2 +b^2}\leq \frac{b}{2} ,\frac{bc^2}{c^2 +b^2}\leq \frac{c}{2} ,\frac{ca^2}{c^2 +a^2}\leq \frac{a}{2}$ hence $$\frac{a^3}{a^2 +b^2} +\frac{b^3}{c^2 +b^2} +\frac{c^3}{a^2 +c^2} =a+b+c - \left(\frac{ab^2}{a^2 +b^2}+\frac{bc^2}{c^2 +b^2}+\frac{ca^2}{c^2 +a^2}\right) \geq \frac{a+b+c}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \fr... | No the most rigorous, but fast. Define $n=\frac 1x$, and assume it integer (keep in mind the dependency on $x$).
$$\frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}=\frac{(1 + \sin x + \sin^2 x)^n - (1 + \sin x)^n}{x}.$$
Factoring the numerator as a binomial of the form $a^n-b^n$, you will get two factors:
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
What is $\frac{2x}{1-x^2}$ when $x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$? If
$$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
Find
$$\frac{2x}{1-x^2}$$
I got till here by simplification by taking the previous value of x, ie,
$$x={\frac{\sqrt{1-\cos\theta}}{1+\cos\theta}}$$
$$\frac{2\tan\theta\sqrt{1+\cos\theta}}{\cos... | $$x=\sqrt\frac{1-cos(\theta)}{1+cos(\theta)}.$$ Suppose to the given equation, let it be defined such that $$y = \frac{2x}{1-x^2}$$ Just substitute x in the equation y will result $$ 1-x^2 = 1 - \frac{1-cos(\theta)}{1+cos(\theta)} \implies \frac{2 cos(\theta)}{1+cos(\theta)}.$$ similarly $$x=\sqrt\frac{1-cos(\theta)}{1... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solving an equation with exponents by using logarithms Solve the equation
$$0.25^5 = 4^{(5x-3)/3} \cdot (0.125)^{6x}$$
So would I just bring down the exponents by taking the log of each constant?
| Logarithms here are actually unnecessary. Notice that each base is a power of $2$, so we get:
\begin{align*}
0.25^5 &= 4^{\frac{1}{3}(5x - 3)} \cdot (0.125)^{6x} \\
(2^{-2})^5 &= (2^2)^{\frac{1}{3}(5x - 3)} \cdot (2^{-3})^{6x} \\
2^{-10} &= 2^{\frac{2}{3}(5x - 3)} \cdot 2^{-18x} \\
2^{-10} &= 2^{\frac{2}{3}(5x - 3) - 1... | {
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"source": "stackexchange",
"question_score": "1",
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Solve a PDE: $ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$ Solve the PDE $$ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$$
where, $ p=\displaystyle \frac{\partial z}{\partial x}$ and $ q=\displaystyle \frac{\partial z}{\partial y}$
My attempt:
I start with Lagrange's auxiliary equation
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{z(x^2-... | since
$$\dfrac{\dfrac{dx}{x}}{y^2+z}=\dfrac{\dfrac{dy}{y}}{-x^2-z}=\dfrac{dz}{x^2-y^2}$$
$$\Longrightarrow \dfrac{\dfrac{dx}{x}+\dfrac{dy}{y}}{y^2-x^2}=\dfrac{dz}{x^2-y^2}$$
$$\Longrightarrow \dfrac{dx}{x}+\dfrac{dy}{y}=-dz$$
$$\Longrightarrow z=-C\ln|xy|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$ Does the following series or integral have a closed-form
\begin{equation}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx
\end{equation}
where $\Psi_3(x)$ is the polygam... | Computation of $\displaystyle U=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1-x}\,dx$
\begin{align*}
U&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x\frac{\ln^3 t}{1-t}\,dt\right)\,dx\\
&=-6\zeta(4)\ln 2+\int_0^1\int_0^1 \left(\frac{\ln^3(tx)}{(1+t)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 5,
"answer_id": 3
} |
Solve the following integration $$\int \sqrt{cot\theta} d\theta$$
I tried to set $t=\sqrt{cot\theta},t^2=cot\theta$ and substitute into the original integration and get$$-\int\frac{2t^2}{1+t^4}dt$$, but then what can I do?
| Consider the integral
\begin{align}
I = \int \sqrt{\cot(\theta)} \, d\theta.
\end{align}
Make the substitution $t = \sqrt{\cot(\theta)}$ to obtain the integral
\begin{align}
I = -2 \int \frac{t^{2} \, dt}{1+t^{4}}.
\end{align}
Now it can be seen that
\begin{align}
\frac{t^{2}}{1+t^{4}} &= \frac{t^{2}}{(1+i t^{2})(1- i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding the indefinite integral of a root function I'm stuck on a particular integral problem. The problem is stated as:
$$\int x \sqrt{2x+1} dx$$
My working thus far:
$$\int x \sqrt{2x+1} dx = \frac{1}{2}x^2\frac{2}{3}(2x+1)^\frac{3}{2}$$
$$\frac{1}{3}x^2(2x+1)\frac{3}{2}$$
...
I think I am totally wrong here and com... | It looks like to integrate the product, you found the product of the integration of each factor. We can't do that!
If $\int f(x) \,dx = F(x)$ and $\int g(x)\,dx = G(x)$
$$\int f(x)\cdot g(x) \,dx \neq F(x)G(x) + C$$
Let's start over.
Note that since the integrand is defined only for $2x+1\geq0$, we put
$$\underbrac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}d... | Here is an alternative way.
\begin{align}\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}&=\frac{\sin(x-a)}{\sqrt{\sin(x-a)\sin(x+a)}}=\frac{\sin x\cos a-\cos x\sin a}{\sqrt{\frac12(\cos 2a-\cos 2x)}}\\&=\frac{\cos a\sin x}{\sqrt{\cos^2a-\cos^2x}}-\frac{\sin a\cos x}{\sqrt{\sin^2 x-\sin^2 a}}\end{align}
Now for the first one, substi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 3
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Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle.
Expr... | Here is a coordinate geometry solution:
The diagonals of a rhombus are perpendicular bisectors of each other. Suppose that the diagonals intersect at the origin of the coordinate plane, that the vertices at the ends of the short diagonal are $A(-3, 0)$ and $C(3, 0)$, that the vertices at the ends of the long diagonal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Clearing the confusion of the area In the example 1:
They are finding the area as $8$ $\sqrt[2]{30}$. But, I am finding $8$ $\sqrt[2]{15}$.
I have done $\sqrt[2]{16\times5\times8\times3}$
Then,
Breaking it up,
$2\times2\times2\times\sqrt[2]{15}$
Please help me clear my confusion.
| $16 = 2^4$
$8 = 2^3$
$\sqrt{(16)(8)(5)(3)} = \sqrt{(2^4)(2^3)(5)(3)} = \sqrt{(2^{3+4})(5)(3)} = \sqrt{(2^6)(2)(5)(3)}$
$= \sqrt{(8^2)(2)(5)(3)} = 8\sqrt{(2)(5)(3)} = 8 \sqrt{30}$
You honestly just took out an extra two. Happens to the best of us.
:)
| {
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What algebraic manipulation is used to express the solution to this integral? According to WolframAlpha, $$
\int \frac{2}{\sqrt{x^2 + 4}} dx = 2 \sinh^{-1}\left(\frac{x}{2}\right) + c.$$
I'm wondering how this was obtained, as I got the following:
Let $x = 2\tan\theta$. Then $dx = 2\sec^2\theta$, so \begin{align*}\int\... | Well by definition, $\sinh^{-1}(z) = \ln\left(z + \sqrt{1+z^2} \right)$. Now, let $z = \frac{x}{2}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to prove this inequality using am gm? Is it true that
$$\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}>\frac{1}{2k+1}+\frac{1}{2k+2}$$
for all natural k? Is bashing going to work?
| You can compare the RHS and the LHS (line 3 in Jack d'Auruzio's answer) this way:
$\frac{1}{6k+2} \ $ (inside LHS), is bigger than: $\frac{1}{6k+3}\ $ (inside RHS)
$\frac{1}{6k+4} > \frac{1}{6k+6}$ aswell.
We will now prove that what's left still "respects" the original (weaker) inequality:
$\frac{1}{6k+2} + \frac{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Derivative of $f(x)=-6\sin^4 x$
$f(x)=-6\sin^4x$
$f(x)=-6\sin x^4$
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$
What am I doing wrong? Please show the steps.
| As others have mentioned, $\sin^4x$ and $\sin x^4$ are completely different.
By convention, $\sin^4x = (\sin x)^4$, the fourth power of $\sin x$.
Also by convention, $\sin x^4 = \sin (x^4)$, the sine of the fourth power of $x$.
But here are derivatives for both $f(x) = -6 \sin^4 x$ and $g(x) = -6 \sin x^4$:
$$f'(x) = -... | {
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"url": "https://math.stackexchange.com/questions/952037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Epsilon-N method - Proof verification
Prove (using the epsilon-N method) that the sequence of numbers $\dfrac{5n^3-2}{n^3}$ converges. Calculate the limit first.
First we calculate the limit: $\lim_{n \to \infty} \dfrac{5n^3-2}{n^3} = \lim_{n \to \infty} 5- \dfrac{2}{n^3} = 5$. So we are required to show, for a giv... | The proof looks correct. The only thing I noticed was your use of the ceiling function $\lceil$ and $\rceil$.
Unless your professor taught limits in this manner, traditionally one might invoke the Archimedean Principle to show that there exists $N \in \mathbb{N}$ such that $N > a$ for some real number $a$. In fact, th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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finding the $\lim_{x \rightarrow -2 }\frac{x+2}{x^3+8}$ How do you solve $\lim_{x \rightarrow -2} \frac{x+2}{x^3+8}$ to get 1/12? I tried factoring out the denominator into $(x+2)(x+2)(x+2)$ and cancelling it out with the top but when you plug in 0 for x the denominator is still 0.
| Here are the steps
$$ \lim_{x\to -2} \frac{x+2}{x^3+8}=\lim_{x\to -2} \frac{x+2}{(x+2)(x^2-2x+4)} $$
$$ =\lim_{x\to -2} \frac{1}{x^2-2x+4}= \frac{1}{(-2)^2-2(-2)+4}=\frac{1}{4+4+4}=\frac{1}{12} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/953262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$ This is a repost of a question i had written incorrectly earlier.
How do I integrate this without substitutions ?
$$
\frac{x^2+3}{x^6\left(x^2+1\right)}
$$
I got:
$$
\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)},
$$
but wasn't able to eliminat... | If you don't want to go the formal partial fractions route, you can systematically chip away at the denominator as follows:
$$\begin{align}
{x^2+3\over x^6(x^2+1)}&={3(x^2+1)-2x^2\over x^6(x^2+1)}\\
&={3\over x^6}-{2\over x^4(x^2+1)}\\
&={3\over x^6}-{2(x^2+1)-2x^2\over x^4(x^2+1)}\\
&={3\over x^6}-{2\over x^4}+{2\over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/954101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Irrational solutions to some equations in two variables The next statement is a conjecture of mine, so I dont know if it's true (though quite sure):
Let $x,y$ be irrational numbers such that $x^4+y^4=1$. Prove (or disprove) that $x^5+y^5$ is irrational or $x^6+y^6$ is irrational (or both of them are irrational).
Edit:... | It turns out that $K(x^4+y^4,x^5+y^5,x^6+y^6) = K(x+y,xy)$.
Letting $a = x^4+y^4, b = x^5+y^5, c = x^6+y^6$, we have the identities
$$x+y = - \frac{4c^{6}b-10c^{4}ba^{3}+80c^{3}b^{3}a^{2}-100c^{2}b^{5}a+14c^{2}ba^{6}+16cb^{7}-68cb^{3}a^{5}+70b^{5}a^{4}-6ba^{9}}{2c^{6}a-4c^{5}b^{2}-5c^{4}a^{4}+70c^{3}b^{2}a^{3}-170c^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/954726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Deriving $\frac{8}{\sqrt{x-2}}$ I'm not sure how to derive this:
$$\frac{8}{\sqrt{x-2}}$$
I tried
$$8 \cdot \frac{1}{\sqrt{x-2}}$$
$$8 \cdot (\sqrt{x-2})^{-1}$$
Differentiating w.r.t. $x$,
$$8 \cdot -1 \cdot (\sqrt{x-2})^{-2}$$
$$8 \cdot -1 \cdot \frac{1}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}... | Since $(\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$, then $$\frac{8}{\sqrt{x-2}}=\frac{0-8(\sqrt{x-2})'}{x-2}=\frac{-4\frac{1}{\sqrt{x-2}}}{x-2}=-\frac{4}{(x-2)\sqrt{x-2}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For how many $n$, $x^6+n$ factors? $\textbf{Question}.$
i) For how many integers $n$ with |$n$|$<500$, can the polynomial $p_n(x)=x^6+n$ be written as a product of two non-constant polynomials with integer coefficients?
ii) How will number of solutions depend on $k>0$ if we replace the condition |$n$|$<500$ by |$n$|$<... | First, consider two cases: $(A, B, k > 0, k \in \mathbb{N})$
Case 1: k is odd.
*
*$A^k+B^k=(A+B)(A^{k-1}+A^{k-2}B+\dots+AB^{k-2}+B^{k-1})$
*$A^k-B^k=(A-B)(A^{k-1}-A^{k-2}B+\dots-AB^{k-2}+B^{k-1})$
Case 2: k is even.
*
*$A^k+B^k$ cannot be factorized this way
*$A^k-B^k=(A-B)(A^{k-1}-A^{k-2}B+\dots-AB^{k-2}+B^{... | {
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Probability interview question Suppose we have three positive integers $A, B, C$. We randomly choose an integer $a$ in the range $[0,A]$ and an integer $b$ in the range $[0,B]$. Find the probability that $a + b\leq C$.
I am unable to proceed in this question. Any hints .
| Let $S_C$ be the number of couples $(a,b)$, $a\in [0,A]$ and $b\in [0,B]$ such that $a+b \leq C$. It is clear that $S_0=1$ because the only couple that can sum up to zero is $(0,0)$. Moreover if $C>A+B$ then $S_C=S_{A+B}$ since all couples that are counted in $S_C$ were already counted in $S_{A+B}$. Without loss in gen... | {
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"url": "https://math.stackexchange.com/questions/956647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the function and domain for $ (f\circ f)$ when $ f(x) = x+ \frac {1} {x} \ $ Find the function and domain for $ (f\circ f)$
My answer is $ \frac {x^4+3x^2+1x} {(x^2+1)(x)}?$
However, the program I am using states I am wrong. What have I done incorrectly?
| $$\ (f\circ f)=x+\frac{1}{x}+\frac{1}{x+\frac{1}{x}}=\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{x^4+x^2+x^2+1+x^2}{x(x^2+1)}=\frac{x^4+3x^2+1}{x(x^2+1)}$$
The domain is right, it's $\ x≠0$, but in the numerator you have 1 and not $\ 1x$
| {
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Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$ How can I prove that
$\displaystyle \log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log2$
using $\displaystyle \frac{x}{1+x}<\log(1+x)<x$
| Riemann sums do this pretty quickly. $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ is a
lower Riemann sum for $\int_n^{2n} {1 \over x}\,dx = \log 2$, and an upper Riemann sum for
$\int_{n+1}^{2n + 1} {1 \over x}\,dx = \log {2n +1 \over n + 1}$, so one has
$$\log {2n +1 \over n + 1} < \frac{1}{n+1}+\frac{1}{n+2}+...+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/957977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \... | You can always do a brute force method. Check n=0,1,2,3,4,5,6,7
$$
n=0,
n(n+1)(n+2)(n+3)=0(1)(2)(3)=0 (mod 8)
$$
$$
n=1,
1(2)(3)(4)=24=0 (mod8)
$$
$$....$$
$$n=7,
7(8)(9)(10)=7(0)(1)(2)=0 (mod8)
$$
This is sufficient since if $$n=8m+r, r=0,1,2,3,4,5,6,7, m\in Z $$ then
$$
n(n+1)(n+2)(n+3)=(8m+r)(8m+r+1)(8m+r+2)(8m+r+3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Number of ways to choose a sequence of three letters from the letters of MISSISSIPPI How many ways can a sequence of three letters be chosen from the letters of MISSISSIPPI?
I'm just a little confused how to go about this since so many letters repeat
I=4
S=4
P=2
So ultimately there are 4 different letters.
| For a generating function solution, use exponential generating functions.
Let's take an example that's a bit simpler than yours. How many three-letter words can be made from the letters in AABBB? If you were simply forming a multiset of three letters, but not arranging them in order, you could let $r$ be the number... | {
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Prove the given condition from given two quadratic equation
Question: If the quadratic equations $x^2+bx+c=0$ and $bx^2+cx+1=0$ have a
common root then prove that either $b + c + 1 = 0$ or $b^2 + c^2 + 1 =bc + b + c$
Till yet,
I had figured the common root of the given two quadratic equation. i.e.
Multiplying first... | Let $y$ be the common root then, consider $y^2+by+c=0$ and $by^2+cy+1=0.$ Clearly $y\not=0.$ Therefore $$y^3+by^2+cy=0$$
$$y^3-1=0$$
$$(y-1)(y^2+y+1)=0.$$
If $y=1,$ then clearly $b+c+1=0.$
If $y^2+y+1=0,$ then note that $by+c=y+1$ and $cy+1=b(y+1).$ Solve this two simple linear equations and equate the value of $y.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Convergence and estimation of the rests of $\sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}} $ For $x\in ]-1,1[$ is required to evaluate the expression
$$
S(x) = \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}
$$
with absolute... | Using cancellations yields smaller errors, since $$\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}=\frac{-2x}{\sqrt{k^6-x^2}\left(\sqrt{k^3+x}+\sqrt{k^3-x}\right)}\in\Theta\left(\frac1{k^{9/2}}\right),$$ which shows that, for every $K\geqslant1$, $$\sum_{k=K+1}^\infty\left(\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/961887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Solving congruence equations Solve: $7x^6\equiv 11 \pmod{23}$ and $5^x\equiv 19 \pmod{23}$
I can solve simple congruence equations but how do I go about solving these?
| $$2^2=4,2^3=8,2^4=16\equiv-7,2^5\equiv9,2^6\equiv18\equiv-5,2^{11}\equiv9(-5)\equiv1$$
$5^2\equiv2\pmod{23},5^5=5(5^2)^2\equiv5\cdot2^2\equiv20\equiv-3,5^6\equiv2^3\equiv8,$
$5^{11}=5^5\cdot5^6\equiv-3\cdot8\equiv-1$
So, $5$ is a primitive root $\pmod{23}$ unlike $2$
$7\equiv-16\equiv5^{11}(5^2)^4\equiv5^{19}\implies$i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simplifying a square root fraction Simplify the following
$$\frac{\sqrt{3}}{\sqrt{2}(\sqrt{6} - \sqrt{3})}$$
Apparently the answer is $\frac{1}{2} (2 + \sqrt{2})$ but can't for the life of me see how to get it. Any help is massively appreciated.
Thanks
| $\dfrac{\sqrt 3}{\sqrt 2 (\sqrt 6 - \sqrt 3)} = \dfrac{\sqrt 3}{\sqrt 2 \sqrt 3(\sqrt2 - 1)} = \dfrac{1}{\sqrt 2 (\sqrt 2 - 1)} = \dfrac{1}{(2 - \sqrt 2)} = \dfrac{(2 + \sqrt 2)}{(2 - \sqrt 2)(2 + \sqrt 2)} = \dfrac{2 + \sqrt 2}{2}$
The reason we write it this way is that it is bad practice to have roots at the denomin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate $ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos (2x)}\right) \mathrm{d}x $ Evaluate the following definite integral:
$$ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos(2x)}\right) \mathrm{d}x, $$ where $k \in \mathbb{N}_{>0}$... | Let
$$I_k = \int_0^\pi \left( \frac{2 + 2\cos x - \cos(k-1)x - 2\cos kx - \cos(k+1)x}{1 - \cos2x}\right) \mathrm dx.$$
Then, we have:
*
*$I_0 = 0$;
*$I_1 = \pi$.
For any $ k \in \mathbb{N}^*$, we have:
$$ \begin{align}
I_{k+1} - 2I_k + I_{k-1}
&= \int_0^\pi \left( \frac{\cos(-2+k)x - 2\cos kx + \cos (2 + k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
How can I prove this inequality by M.I. or otherwise? This question is from the past examination:
$$f(n)=\frac{2}{3}(n^{3/2}-1)+\sqrt{n}$$
$$g(n)=1+\frac{2}{3}((n+1)^{3/2}-2^{3/2})$$
My task to prove $f(n)≥g(n)$ for all $n≥1$.
I have tried M.I here.
But the problem is that I do not know how to expand the term $(n+2)^{3... | Since $f(n)\geq g(n)$ holds for $n=1$, it is sufficient to show that:
$$\begin{eqnarray*} f(n+1)-f(n) &=& \frac{2}{3}((n+1)^{3/2}-n^{3/2})+(n+1)^{1/2}-n^{1/2}\\&\geq& \frac{2}{3}((n+2)^{3/2}-(n+1)^{3/2})=g(n+1)-g(n)\end{eqnarray*}$$
that is equivalent to:
$$\frac{2}{3}\left((n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}\right)\leq (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Ray Sphere Intersection I'm reading a book on raytracing and there's a part where the author is working out the equation for ray-sphere intersection. There's a part I don't understand that I'm hoping I can get help with.
The equation starts with:
$(\mathbf{p} - \mathbf{c}) \bullet (\mathbf{p} - \mathbf{c}) - r^2 = 0$
B... | It's a matter of rearranging the terms inside and applying FOIL:
$$(\mathbf{o} + t \mathbf{d} - \mathbf{c}) \cdot (\mathbf{o} + t \mathbf{d} - \mathbf{c}) \\ = [t \mathbf{d} + (\mathbf{o} - \mathbf{c})] \cdot [t \mathbf{d} + (\mathbf{o} - \mathbf{c})] \\ = t^2 \mathbf{d} \cdot \mathbf{d} + 2t (\mathbf{o} - \mathbf{c}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
differentiation of the following equation 3 i already done the differentiation, just wanna confirm either i got it right or wrong. Can someone verify this for me.
1) f(x) = $ -3\over x^{5/2}$
f '(x)
= $ 3({ 5\over 2}x^{3/2})$ . $\frac{1}{x^5}$
= $ { 15\over 2}x^{(3/2)-5}$
= ${ 15\over 2}x^{-7/2}$
2) f(x) = $\frac{2x... | $$(-\frac3{x^{5/2}})'=-3x^{-5/2}=-3(-\frac52)x^{-\frac52-1}=\frac{15}{2x^{7/2}}$$
$$\left(\frac{2x^2+3}{(x^3-4)^3}\right)'=\frac{(4x)(x^3-4)^3-(2x^2+3)\big(3(3x^2)(x^3-4)^2\big)}{((x^3-4)^3)^2}=\frac{4x(x^3-4)-9(2x^2+3)x^2}{(x^3-4)^4}=-x\frac{14x^3+27x+16}{(x^3-4)^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Integration of $\sqrt{x+\sqrt{x^2+3x}}$ I faced the following indefinite integration problem:
$$\int \sqrt{x+\sqrt{x^2+3x}}dx$$
This result by WolframAlpha suggests that there is an elementary way to compute this integration. But I don't know how to start. Any hints would be appreciated.
| See the other answer. To find $\int \frac{27}{2(2t^2+3)^2}\, dt$, you can avoid using non-real numbers (notice that $2t^2+3=at^2+bt+c$, where $b^2-4ac<0$, has no real roots and no non-trivial factorization over the real numbers. We can't use partial fractions with real numbers).
See this link (link) (Wikipedia (link) a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
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Absolute convergence of $\sum_{n=1}^{\infty} (-1)^n (1-\cos1/n)$ I would like to see if $$\sum_{n=1}^{\infty} (-1)^n \left(1-\cos \frac{1}{n}\right)$$ converges absolutely.
| Donote your sum by $S$. Taylor series for $\cos$ gives us $\cos x = 1 - \frac{1}{2}x^2 + o(x^3)$, hence $1 - \cos x = \frac{1}{2}x^2 + o(x^3)$. Since $|(-1)^n (1-\cos\frac{1}{n})| = \cos\frac{1}{n}$, it follows that
$$S \leq \sum_{n=1}^{\infty}\left(\frac{1}{n^2} + |o(\frac{1}{n^3})|\right).$$
Since $\frac{o(x^3)}{x^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Outer Product of Two Matrices? How would I go about calculating the outer product of two matrices of 2 dimensions each? From what I can find, outer product seems to be the product of two vectors, $u$ and the transpose of another vector, $v^T$.
As an example, how would I calculate the outer product of $A$ and $B$, wh... | To extend davcha's answer, for your specific example, you would get:
$$A\otimes B=
\left(
\begin{array}{cc}
\left(
\begin{array}{ccc}
5 & 6 & 7 \\
8 & 9 & 10 \\
\end{array}
\right) & \left(
\begin{array}{ccc}
10 & 12 & 14 \\
16 & 18 & 20 \\
\end{array}
\right) \\
\left(
\begin{array}{ccc}
15 & 18 & 21 \\
24 & 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 1
} |
Solving the system $a^2-6=2\sqrt{2c+6}, \, b^2-6=2\sqrt{2a+6}, \, c^2-6=2\sqrt{2b+6}$ Question:
Solve the following system for $a,b,c\in \mathbb{R}$:
$$\begin{cases}
b^2-6=2\sqrt{2a+6}\\
c^2-6=2\sqrt{2b+6}\\
a^2-6=2\sqrt{2c+6}
\end{cases}$$
I found the following:$$ (b^2-6)^2=4(2a+6)$$
$$(c^2-6)^2=4(2b+6)$$
$$(a^2-6... | $$(b^2-6)^2-(c^2-6)^2=8(a-b)\Leftrightarrow (b^2-c^2)(b^2+c^2-12)=8(a-b)$$
$$(b^2-6)^2-(a^2-6)^2=8(a-c)\Leftrightarrow (b^2-a^2)(b^2+a^2-12)=8(a-c)$$
$$(c^2-6)^2-(a^2-6)^2=8(b-c)\Leftrightarrow (c^2-a^2)(c^2+a^2-12)=8(b-c)$$
Assume that $a\geq b\geq c\geq 0$ then
$$b^2+c^2\geq 12$$
on the other hand
$$b^2+a^2\leq12$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $X+Y = 10 $ ($X$ and $Y $ both are positive) then what is the maximum value of $(X^3)(Y^2)$? I can get to the result by trying different values of X and Y but that is of course time taking. I want to know if there is a better way to get to the result?
| $Y = 10 - X$,and $X^3Y^2 = X^3(10-X)^2 = X^3(X^2 - 20X + 100) = X^5 - 20X^4 + 100X^3 = f(X)$.
$f'(X) = 5X^4 - 80X^3 + 300X^2 = 0 \iff 5X^2(X^2 - 16X + 60) = 5X^2(X-6)(X-10) = 0 \iff X = 0, 6, 10$. Since $0 < X < 10$, $X = 6$, and $Y = 10 - 6 = 4$. From this the max value of $X^3Y^2$ is $6^34^2 = 3,456$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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If $\int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}.$ Then $f(x)$ is
If $\displaystyle \int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}.$ Then $f(x)$ is
$\bf{Options:}$ $(a)\;\displaystyle \frac{1}{a^2\sin^2 x+b^2 \cos^2 x} \;\;\;\;\;\; (b)\; \frac{1}{a^2\... | $\displaystyle\int\frac{f^{'}(x)}{(f(x))^2}dx=(a^2-b^2)\int\sin 2xdx\\
\Rightarrow \frac{1}{f(x)}=\frac{(a^2-b^2)}{2}\cos 2x+C\\
=\frac{(a^2-b^2)}{2}(\cos^{2}x-\sin^{2}x)+\frac{(a^2+b^2)}{2}(\cos^{2}x+\sin^{2}x)+C\\
=a^2\cos^{2}x+b^2\sin^{2}x+C$
So We get $\displaystyle f(x) = \frac{1}{a^2\cos^{2}x+b^2\sin^{2}x+C}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/976324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrate $\sin^n{x}$ How do you integrate:
$$\int(\sin^n{x}) dx$$
The link to WolframAlpha : (Integration Answer)
No definite limits...
What is that hypergeometric function in that answer. Please help!
Thanks
| Using binomial series (link), we have:
$$(1-\cos^2x)^{\frac{n-1}{2}} = 1 - \frac{n-1}{2}\cos^2x + \frac{n-1}{2}\frac{n-3}{4}\cos^4x-\ldots. $$
Then:
$$ \int\sin^nx dx = \int\sin x (1-\cos^2x)^{\frac{n-1}{2}} dx $$
$$ = \int\sin x \left(1 - \frac{n-1}{2}\cos^2x + \frac{n-1}{2}\frac{n-3}{4}\cos^4x-\ldots\right) dx $$
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/976909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all posi... | Let $x^2+2x+1=(x+1)^2$ but $(x+1)^2\equiv0,1\pmod4$
i.e., $x^2+2x+1=4a$ or $4a+1$ for any integer $a\ge0$
So, $x^2+2x+2=4a+1$ or $4a+2$
In either case, the floor by $\pmod4$ remains same even for $x^2+2x+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How to find the $n$th term of a repeating pattern What is the nth term of following sequence
$1,2,3,4,5,6,1,2,3,4,5,6,\ldots$
$(n, n+1, n+2, n+3,\ldots,$ $,n+p, n, n+1, n+2, n+3,$ $n+4, n+5,$ $\ldots,$ $n+p,$ $n,$ $n+1,$ $\ldots)$
Actually, I am trying to solve the following puzzle
$6$ girls pick a captain by formin... | The $n$-th term of sequence $1,2,3,4,5,6,1,2,3,4,5,6,\ldots$ is given by $a_n=1+(n-1 \bmod 6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Triplets of distinct integers > 1 that return integer values. If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?
UPDATE: I've done some more work and have come up with the fol... | $f(A,B,C)$ has no integral solutions for distinct integers $A,B,C$.
Let $f(A,B,C)=k$ as in the question. $k<C$ and $k=C$ have been shown above to lead to contradictions. $k>C$ has been shown to reduce to solving $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ $$C=\frac{AB}{A+B}$$ for positive integers as restricted above.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the missing angle of similar triangle Find the missing angle $\theta$ in the triangle below given that $R>r$, $l\geq R$, $0< \theta < \frac{\pi}{2}$.
Attempted Solution
I attempted to use similar triangles to find the angle $\theta$ but the resulting expression for $\theta$ is pretty ugly.
$\frac{r}{R\cos\theta+\... | Hint
\begin{equation}
\begin{cases}
S = {S_1} + {S_2} = {1 \over 2}Rr\sin {\theta _1} + {1 \over 2}R{l_2}sin\theta = {1 \over 2}r{l_2} \\ \\
{\cos }{\theta _1} = {{{r^2} + {R^2} - l_1^2} \over {2Rr}} \\ \\
{\cos}\theta = {{l_2^2 + {R^2} - {l^2}} \over {2R{l_2}}} \\ \\
{\cos}{\theta _1} = {\sin}\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove that the gradient of the tangent to $xy=d^2$ is $-\frac{c^2}{4d^2}$ I have this question:
Given that $y=mx+c$ is a tangent to $xy=d^2$ prove that $m=-\frac{c^2}{4d^2}$.
I'm not sure what direction to take - I tried differentiating the hyperbola equation, but that gave me a gradient of $-\frac{y}{x}$. How do I g... | $xy = d^2 \Leftrightarrow y = \frac{d^2}{x}$.
Tangent line equation to $f(x)$ in point $x=a$ is
$g(x) = f(a) + f'(a)(x-a)$.
$f'(x) = -\frac{d^2}{x^2}$, so $g(x) = \frac{d^2}{a} - \frac{d^2}{a^2}(x-a) = -\frac{d^2}{a^2}x + \frac{2d^2}{a}$, hence $m = -\frac{d^2}{a^2}, c = \frac{2d^2}{a}$ and $m = -\frac{c^2}{4d^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using recursion tree to solve recurrence $T(n) = 3T(n/2)+n$ I am trying to solve the recurrence $T(n) = 3T(n/2)+n$ where $T(1) = 1$ and show its time complexity.
$n$ can be assumed to be a power of $2$.
So basically, I drew out the tree and found that:
the number of levels in the tree will be $h = \log_2 n+1$.
The num... | Our time formula is:
$$
T(n) = n + \frac{3}{2}n + \frac{3^2}{2^2}n + \cdots + \frac{3^{\log_2 n}}{2^{\log_2 n}}n
$$
[As an aside, note that the last term is just $n^{\log_2 3}$.] Now this is a finite geometric series with initial term $a = n$, common ratio $r = \frac{3}{2}$, and $N = \log_2 n + 1$ terms. By applying th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluating $(\frac{\cos x}{1-\sin x})^2$ $(\dfrac{\cos x}{1-\sin x})^2$
$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x+\sin^2x-\cos^2x}{(1-\sin x)^2})$
Does $\sin^2x-\cos^2=1$? or $-1$? Then it could factor with the bottom and the answer would be $\dfrac{2\cos x}{(1-\sin x)^2}$. Is this right?
| $\sin^2(x)-\cos^2(x)$ does not equal $-1$ or $1$.
$$\sin^2(x)+\cos^2(x)=1,$$
So $\sin^2(x)-\cos^2(x)$ is equal to:
$$\sin^2(x)-(1-\sin^2(x))= 2\sin^2(x)-1,$$ or alternatively,
$$1-\cos^2(x)-\cos^2(x)= 1-2\cos^2(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the co-ordinates of the point of intersection I have the function $$y=2x^2-3x$$ How do I find the co-ordinates of the point of intersection of the lines tangent to the curve at $y=-1$?
One point where $y = -1$ is when $x = \dfrac 12$.
I took the derivative of the given $y$: $\dfrac{dy}{dx} = 4x- 3$, then found s... | Solve for $x$ when $y = -1$: $$2x^2 - 3x = -1 \iff 2x^2 - 3x + 1 = 0 \iff (2x-1)(x-1) = 0 \implies x = \frac 12 \text{ or } x = 1$$
So, there are two points corresponding to $y = -1$: $\left(\frac 12, -1\right), (1, - 1)$.
Perhaps you need to find the point of intersection of the line tangent to the first point, with t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluation of $\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx$ and $\int_{0}^{2} \left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\right)dx$ Evaluation of Some Integrals::
$\displaystyle (a)\;\;\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx\;\;\;\;\;\;(b)\;\; \int_{0}^{2} \left(\sqrt{1+x^3}+\s... | Hint
For the first integral, use $\sin(3x)=3\sin(x)-4\sin^3(x)$ . So $$\frac{1+\sin 3x}{1+2\sin x}=\frac{-4 \sin(x)^3+3 \sin(x)+1}{2 \sin(x)+1}$$ Remark that equation $-4z^3+3z+1=0$ has an obvious solution $z=1$ and two other roots equal to $z=-\frac{1}{2}$ ... which is also the root of $2z+1=0$ !
Then,
$$\frac{1+\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to prove the determinant? We have to prove the following result without expanding
$\left|\begin{array}{lll}
a^3 & a^2 &1 \\
b^3 & b^2 &1\\ c^3 & c^2 &1
\end{array}
\right|=(ab+bc+ca)\left|\begin{array}{lll}
a^2 & a &1 \\
b^2 & b &1\\ c^2 & c &1
\end{array}
\right|$
Progress :
$\left|\... | $ \begin{align}\left|\begin{array}{lll}
a^2 & a &bc+ab+ac+a^2 \\
b^2 & b & ca+ba+b^2+bc\\ c^2 & c &ab+ca+cb+c^2
\end{array}
\right|\end{align} $ by $(a+b+c) \times C_2 + C_3\rightarrow C_3' $
Then $
=\left|\begin{array}{lll}
a^2 & a &bc+ab+ac \\
b^2 & b & ca+ba+bc\\ c^2 & c &ab+ca+cb
\end{arr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.
I begin with rewriting the complex equation to $a+bi$ format.
1 $z^3 = i = 0 + i$
2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$
3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.
4 The complex... | Use polar coordinates.
$z^{3}=i=e^{i(\frac{\pi}{2}+2k\pi)}$, $k\in \mathbb{Z}$
And from here it is much simpler
EDIT: what I mean by much simpler...
The OP kind of used polar form, but not really. You should stock with polar form until the very end. The OP introduces $\theta$, $cosinus$ and $sinus$ functions etc. You ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Find the distribution function for Y for the following density function I am to find the distribution function for Y given the following density function
$$f(y)=\begin{cases}
y,\quad 0<y<1\\
2-y, \quad 1 \leq y < 2\\
0, \quad\text{elsewhere}\\
\end{cases}$$
So since $f(y)$ is given by $\frac{dF(y)}{dy}$, then $F(y)$ sh... | Let us compute $F(x) = \int_{-\infty}^x f(y)dy$ the distribution function:
If $x<0$ then clearly $F(x)=0$.
If $0<x<1$ then
$$F(x) = \int_{-\infty}^x f(y)dy = \int_{0}^x f(y)dy = \int_0^x ydy = \frac{x^2}{2}.$$
If $1<x<2$ then
$$F(x) = \int_{-\infty}^{x} f(y) dy = \int_0^1 f(y)dy+ \int_1^x (2-y)dy = F(1)+ \bigg(2y - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differentiation of $f(x)=2x(2+3x^2)^3$ The question:
Differentiate $f(x)=2x(2+3x^2)^3$.
How do I approach this problem? Do I only have to use the product rule...? I have the answer but I don't know how to get there.
Here is my attempt to this problem:
$$\frac{df}{dx}=(2)(2+3x^2)^3 + (2x)(6x)(3)(2+3x^2)^2 = \\ =(2+3x... | You get the right answer:
$$(2+3x^2)^2 (42x^2+4)=(2+3x^2)^2 \cdot 2 \cdot (21x^2+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that : $a^n+b^n+c^n=x^n+y^n+z^n$; $\forall n\in \mathbb{N}$ $a;b;c;x;y;z \in \mathbb{R}$ such that : \begin{matrix}
a+b+c=x+y+z & \\
a^2+b^2+c^2=x^2+y^2+z^2 & \\
a^3+b^3+c^3=x^3+y^3+z^3 &
\end{matrix}
Prove that : $a^n+b^n+c^n=x^n+y^n+z^n$; $\forall n\in \mathbb{N}$
P/s : I don't have any ideas about this pr... | Hint:
Let $p_1=a+b+c, \; q_1 = ab+bc + ca,\; r_1 = abc, \; p_2 = x+y+z, \; q_2 = xy+yz+zx, \; r_2 = xyz$. Then it is easy to show that:
$$p_1 = p_2, \quad q_1 = q_2, \quad r_1 = r_2$$
Further, $a^n+b^n + c^n$ and $x^n + y^n + z^n$ are both symmetric functions and hence can be expressed in terms of $p_1, q_1, r_1$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$ I want to prove this limit by using $(\epsilon,\delta)$ definition
$$\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$$
Here is what I have done
$$|\frac{1}{x} - \frac{1}{2}|<\epsilon \Leftrightarrow -\epsilon < \frac{1}{x} - \frac{1}{2} < \epsilon$$
$$\Leftrightarrow -\epsil... | Let $\epsilon >0$. Then you have to bound $|\frac{1}{x}-\frac{1}{2}|=\frac{1}{2}\frac{|x-2|}{|x|}$. So, by choosing $ \delta=min\{1,2\epsilon\} $ you have $-1<x-2<1\Longrightarrow 1<x<3\Longrightarrow \frac{1}{x}<1$ and $|x-2|<2\epsilon$ therefore
$ |\frac{1}{x}-\frac{1}{2}|=\frac{1}{2}\frac{|x-2|}{|x|}<\epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"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.