Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that from the equalities, $\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$ follows $x^yy^x=y^zz^y=z^xx^z$. Problem :
Prove that from the equalities, $$\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$$ follows $$x^yy^x=y^zz^y=z^xx^z$$.
My approach :
$$\frac{x(y+z-... | $$x^yy^x=y^zz^y=z^xx^z$$
$$\Rightarrow \left\{\begin{array}{ll}
x^{y-z}=(\frac{z}{y})^x \\
y^{x-z}=(\frac{z}{x})^y \\
z^{y-x}=(\frac{x}{y})^z
\end{array} \right. $$
Take the log of both sides
$$\Rightarrow \left\{\begin{array}{ll}
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Strange trigonometric proof. I was trying to find out how to prove
$$ \sin(A-\arcsin(0.3 \ \sin \ A)) \ \cdot \ \sin(A+\arcsin(0.3 \ \sin \ A)) \ = \ 0.91 \ \sin^2 \ A \ \ . $$
When I put this equation into my calculator both sides appear to be exactly the same, but I have no idea how to prove it.
| $$\sin { \left( A-\arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) } \cdot \sin { \left( A+\arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) =0.91\sin ^{ 2 }{ \left( A \right) } }$$
Solution
:
$$\left( \sin { A\cos { \left( \arcsin { \left( 0.3\sin { \left( A \right) } \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate $\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$ I've been studying limits on Rudin, Principles of Mathematical Analysis for a while, but the author doesn't exactly explain how to calculate limits...so, can you give me a hint on how to solve this? $$\lim_{x\to0}\frac{... | Using the basic limit $$\lim_{t \to a}\frac{t^{n} - a^{n}}{t - a} = na^{n - 1}\tag{1}$$ we can see by putting $n = 1/2, a = 1, t = 1 + x$ that $$\lim_{x \to 0}\frac{\sqrt{1 + x} - 1}{x} = \frac{1}{2}\tag{2}$$ Replacing $x$ by $-x$ we get $$\frac{1 - \sqrt{1 - x}}{x} = \frac{1}{2}\tag{3}$$ From the equation $(2), (3)$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmat... | Note that
$$A=P\cdot\begin{bmatrix} 2i&0\\0&-2i\end{bmatrix}\cdot P^{-1}$$
With $P=\begin{bmatrix} -1&-1\\-2i&2i\end{bmatrix}$. We have
$$e^A=P\cdot e^{D}\cdot P^{-1}$$
With $D$ the diagonal matrix above
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Prove that the group $G$ is abelian if $a^2 b^2 = b^2 a^2$ and $a^3 b^3 = b^3 a^3$
In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$. Prove that the group $G$ is abelian.
My approach was the following:
Let $a,b\in G$
Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds.
Now, $$\begin{align}
a^... | Hint: We can show that
$$a^{6}b=ba^{6}$$
for all $a, b\in G$. With assumptions, we have
$$a^{2}=b^{-2}a^{2}b^{2}\ \ \text{and}\ \ a^{3}=b^{-3}a^{3}b^{3},$$
and we get
$$a^{6}=b^{-2}a^{6}b^{2}\ \ \text{and}\ \ a^{6}=b^{-3}a^{6}b^{3},$$
thus
$$a^{6}=b^{-2}a^{6}b^{2}=b^{-3}a^{6}b^{3}$$
and the above relation implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve trigonometric inequality $ \sin x+2 \cos x<2$ $$ \sin x+2 \cos x<2$$
$$ \dfrac{2t}{1+t^2}+2\dfrac{1-t^2}{1+t^2}<2$$
$$ 4t^2-2t>0$$
$$ 2t(2t-1)>0$$
$$ t(2t-1)>0$$
$$ (t>0 \wedge t>\dfrac{1}{2}) \vee (t<0 \wedge t<\dfrac{1}{2})$$
From this, I can only find $x<2\pi+2k\pi$, and, $x<2k\pi$, these are good (I think), ... | $\sin x+2\cos x=2(\frac{1}{2}\sin x + \cos x)$
Multiply and divide by $\frac{\sqrt{5}}{2}$
You will get: $$\sqrt{5}\big(\frac{1}{\sqrt{5}}\sin x + \frac{2}{\sqrt{5}}\cos x\big)$$
which can be expressed in the form of $a(\cos\theta\sin x+\sin\theta\cos x)=a\sin(x+\theta)$
You can express it as $$\sqrt{5}\sin \big(x+\arc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ then $\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$ Suppose that $\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ . Then , prove that $\displaystyle\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$
Attempt :
F... | Hint:
$$ (ab + bc + ca)(a + b + c) = (a + b)(b + c)(c + a) + abc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the sum of the n terms of the series $2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$ Find the sum of the n terms of the series:
$2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$
I don't know how to proceed. Please explain the process and comment on technique to solve questions of similar type.
Source: Barnard and Child Higher Algeb... | This lookslike a double sum. Try rewriting it the following way.
$$2\cdot 2^0+3\cdot 2^1+4\cdot 2^2+...+(n+2)\cdot 2^n =$$
$$=2\cdot 2^0+2\cdot 2^1+2\cdot 2^2+...+2\cdot 2^n$$
$$+1\cdot2^1+1\cdot 2^2+...+1\cdot 2^n$$
$$+1\cdot 2^2+...+1\cdot 2^n$$
$$ \cdots $$
The Terms are now a simple geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
I tried to solve this question but no luck.
My try:
$$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx=\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx\\
\int x^4(x^8+x^4+1)x^2(2+3x^{-4}+6x^{-8})^{1/4}dx$$
Now i got stuck,please help me reach the ... | Let $$\displaystyle I = \int (x^{12}+x^{8}+x^{4})\cdot (2x^8+3x^4+6)^{\frac{1}{4}}dx = \int (x^{11}+x^{7}+x^{3})\cdot (2x^{12}+3x^{8}+6x^{4})^{\frac{1}{4}}dx$$
Now Put $(2x^{12}+3x^{8}+6x^{4}) = t^4\;,$ Then $\displaystyle (x^{11}+x^{7}+x^{3})dx = \frac{t^3}{6}dt$
So Integral $$\displaystyle I = \frac{1}{6}\int t^{4}dt... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
How do I prove the above expression using mathematical induction? So f... | $$P(k)=\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2}$$
Therefore
$$P(k+1)=\frac{(-1)^{k} \cdot (k+1) \cdot (k + 2)}{2}$$
The way to go:
Write $P(k+1)$ as something containing $P(k)$.
Since you seem are asking only about some starting help, this should suffice. =)
After reading your comment:
The general way to go is start... | {
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"url": "https://math.stackexchange.com/questions/1433103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
... | hint: Let $x = t^6$, and simplify to a nicer expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Taking n steps forward and m steps back: probability of returning where you started if n and m are determined by random dice rolls. You've all heard the phrase "Three steps forward, two steps back."
I am trying to figure out the probability of returning to my starting point if the number of steps forward is determined ... | The difficulty in this problem lies in the calculation of the number of different ways a sum of n throws of a die can be determined. For one throw the probability that the 'sum' of the second die equals the first is $\frac{1}{6}$, which is calculated by adding the probabilities of rolling the same number on two dice i.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Here is what I have:
$a+b =b+a$ since $R$ is commutative.
Now,
$$(b+a) \cdot b... | Your proof is fine.
One way to discover it is by computing freely:
$$
\frac{1}{\dfrac{1}{a}+\dfrac{1}{b}}
=
\frac{1}{\dfrac{a+b}{ab}}
=
\frac{ab}{a+b}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show using the definition of limit that $\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$ can you help me with this excercise.
Show using the definition of limit that
$$\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$$
Definition of limit:
$\lim_{(x,y)\to(a,b)} f(x,y) =L$ if and only if f... | We need only use the inequality $\sin x<x$ for $x>0$ along with the trigonometric identity $\sin^2 x=\frac{1-\cos 2x}{2}$.
Then, we can write
$$\begin{align}
|1-\cos xy|&=|2\sin^2(xy/2)|\\\\
&\le\frac12(xy)^2\\\\
&\le \frac14(x^2+y^2)^2
\end{align}$$
along with
$$\begin{align}
|\sin y| &\le |y|\\\\
&\le(x^2+y^2)^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solutions to the Laplace Equation $\Delta u =0$, where $u= \log p$ Find all real solutions to the two dimensional Laplace equation $U_{xx} + U_{yy} =0$ of the form $u=\log p(x,y)$, where $p$ is a quadratic polynomial.
Solution:
Let $p(x,y) = Ax^2 + By^2 +Cxy + D$ be a quadratic polynomial such that $A, B \not= 0$. The... | This is an heavy method, but you can continue.
$$U_{xx} + U_{yy} = \frac{2A ln(10)(Ax^2 + By^2 +Cxy + D) - ln(10)(2Ax + Cy)^2}{ln(10) (Ax^2 + By^2 +Cxy + D)^2} + \frac{2B ln(10)(Ax^2 + By^2 +Cxy + D) - ln(10)(2By + Cx)^2}{ln(10) (Ax^2 + By^2 +Cxy + D)^2} = 0$$
After simplification :
$$2A(Ax^2 + By^2 +Cxy + D)-(2Ax + Cy... | {
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"url": "https://math.stackexchange.com/questions/1435906",
"timestamp": "2023-03-29T00:00:00",
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Plane of all points equidistant from two other points? Find the equation of the plane that contains all the points that are equidistant from the given points
$(-9, 3, 3), (6, -2, 4)$
I think the plane described lies in the midpoint of these points, and it is perpendicular to the line connecting the two points. This mea... | Notice, there is another easy method to find the equation of the plane
Let the parametric point be $(x, y)$ on the plane which is equidistant from the given points $(-9, 3, 3)$ & $(6, -2, 4)$ hence, we have $$\sqrt{(x-(-9))^2+(y-3)^2+(z-3)^2}=\sqrt{(x-6)^2+(y-(-2))^2+(z-4)^2}$$
$$(x+9)^2+(y-3)^2+(z-3)^2=(x-6)^2+(y+2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a Quadratic Equation with roots of... I was given this problem at school to look at home as a challenge, after spending a good 2 hours on this I can't seem to get further than the last part of the equation. I'd love to see the way to get through 2) before tomorrow's lesson as a head start.
So the problem is as fol... | Given $\displaystyle \alpha,\beta$ are the roots of $2x^2+8x+4=0.$
So$\displaystyle \alpha+\beta = -\frac{8}{2}=-4$ and $\displaystyle \alpha\cdot \beta = \frac{1}{2}.$
Now for Second part, Using $\bullet\; \bf{x^2-(sum \; of \; roots)x+(product\; of \; roots) =0}$
So here $\displaystyle \bf{sum\; of \; roots } = 2\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this Taylor series correct taken correctly? Confused reasoning I have $dx/dy=-ay, x(0)=1$ initial value problem.
Then $x(y)=\frac{x(0)}{0!}y^0+ \frac{x'(0)}{1!}y+\frac{x''(0)}{2!}y^2=1+(-a)y+a^2y^2...$
| Given
$$
x'(y) = \frac{dx}{dy} = -ay.
$$
The second derivative can be found with
$$
x''(y) = \frac{d^2x}{dy^2} = \frac{d}{dy} \frac{dx}{dy} = -a,
$$
and thus the third derivative will be equal to
$$
x^{(3)}(y) = \frac{d^3x}{dy^3} = \frac{d}{dy} \frac{d^2x}{dy^2} = 0,
$$
any higher derivatives will also be zero.
So the ... | {
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"url": "https://math.stackexchange.com/questions/1439939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving equations with binomial coefficients
Let $n$ be a positive integer and let $x$ be a non zero real number. Prove the following.
*
*$\sum_{k=0}^n \dbinom{n}{k}2^{n-k} (1+x)(x^{-1}+x)^k = \frac{1}{x^n} (1+x)^{2n+1}$
*$\sum_{k=0}^n \dbinom{n}{k}2^{n-k} \dbinom{k}{\left\lfloor\frac{k}{2}\right\rfloor}= ... | Here is a proof that leaves some work for you to do in completing the
details.
Suppose we seek to verify that
$$\sum_{k=0}^n {n\choose k} 2^{n-k} {k\choose \lfloor k/2 \rfloor}
= {2n+1\choose n}.$$
This is
$$\sum_{q=0}^n {n\choose 2q} 2^{n-2q} {2q\choose q}
+ \sum_{q=0}^n {n\choose 2q+1} 2^{n-2q-1} {2q+1\choose q}.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k}$ I'm trying to prove the following equation above. So far I have:
\begin{align}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} & \text{(By the Binomial Theorem)}
\end{align}
I know I have to use the following identity somehow... | $$\sum_{k=0}^{n}\binom{2n+1}{k}=\frac{1}{2}\sum_{k=0}^{2n+1}\binom{2n+1}{k}=\frac{2^{2n+1}}{2}=2^{2n}$$
Using the relation $$\binom{2n+1}{k}=\binom{2n+1}{2n+1-k}$$ for $0 \le k \le n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Prove $\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} $ For postive integer $a, b, c $ prove the following inequality $$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} > \sqrt{2}$$
How we can p... | $$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} > \sqrt{2}$$
By$ AM-GM$ $$\sum_{cyc}\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}}=\sum_{cyc}\frac{\sqrt2}{2\cdot\frac{1}{\sqrt2}\sqrt{b+\frac{1}{a}+\frac{1}{2}}}\geq\sum_{cyc}\frac{\sqrt2}... | {
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"url": "https://math.stackexchange.com/questions/1445051",
"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\frac{1}{\sin^2 x+\sin x+1}dx$
Evaluation of $\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$
$\bf{My\; Try::}$ Using $$\; \bullet\; x^2+x+1 = (x-\omega)\cdot (x-\omega^2)\;,$$ where $\omega,\omega^2$ are cube root of unity
So we can write Integal $$\displaystyle I = \int\frac{1}{(\sin x-\omega)\... | By using Weierstrass substitution $x=2\arctan t$ the problem boils down to computing
$$ \int\frac{1+t^2}{1+2t+6t^2+2t^3+t^4}\,dt $$
through partial fraction decomposition. The roots of that palyndromic $4$th-degree polynomial are located at $t=-\frac{1}{2}\pm \frac{i \sqrt{3}}{2}-\sqrt{\frac{1}{2} \left(-3\pm i \sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating th... | We have:
$$
\sqrt{c}-\sqrt{c-1}≥\sqrt{c+1}-\sqrt{c}\iff\\
\\
\left(\sqrt{c}-\sqrt{c-1}\right)\frac{\sqrt{c}+\sqrt{c-1}}{\sqrt{c}+\sqrt{c-1}}≥\left(\sqrt{c+1}-\sqrt{c}\right)\frac{\sqrt{c+1}+\sqrt{c}}{\sqrt{c+1}+\sqrt{c}}\\
\\
\frac{1}{\sqrt{c}+\sqrt{c-1}}≥\frac{1}{\sqrt{c+1}+\sqrt{c}}\iff\\
\\
\sqrt{c+1}+\sqrt{c}≥\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Is there another way to solve this quadratic equation? $$\frac { 4 }{ x^{ 2 }-2x+1 } +\frac { 7 }{ x^{ 2 }-2x+4 } =2$$
Steps I took:
$$\frac { 4(x^{ 2 }-2x+4) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } +\frac { 7(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } =\frac { 2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+... | Setting $t=x^2-2x+1$ gives
$$\frac{4}{t}+\frac{7}{t+3}=2$$
$$4(t+3)+7t=2t(t+3)$$
$$2t^2+6t-4t-7t-12=0$$
$$2t^2-5t-12=0$$
$$(2t+3)(t-4)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
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How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3... | Partial fractions save the day:
$$\frac{2}{x^3-1} = \frac{a}{x-1} + \frac{bx+c}{x^2+x+1}$$
You get $a=\frac{2}{3}$, so:
$$\frac{2}{x^3-1} =\frac{2}{3}\frac{1}{x-1} + \frac{bx+c}{x^2+x+1}$$
Now $$\lim_{x\to 1}\frac{bx+c}{x^2+x+1} = \frac{b+c}{3}.$$
So you only need to compute $$\lim_{x\to 1} \left(\frac{1}{x-1}-\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the integral $\int \frac{dx}{\left(\sqrt{x^2+1}+x\right)^2}$ $$\int \frac{dx}{\left(\sqrt{x^2+1}+x\right)^2}$$
I think I need to use replace, but not sure:
$$x=\frac{u^2-1}{2\cdot u}$$
$$dx=\left(1-\frac{u^2-1}{2u^2}\right)$$
| Here, it might be a good idea to first simplify the integrand. We write
$$
\begin{aligned}
\frac{1}{\bigl(\sqrt{1+x^2}+x\bigr)^2}&=\frac{\bigl(\sqrt{1+x^2}-x\bigr)^2}{\bigl(\sqrt{1+x^2}+x\bigr)^2\bigl(\sqrt{1+x^2}-x\bigr)^2}\\
&=\bigl(\sqrt{1+x^2}-x\bigr)^2\\
&=1+2x^2-2x\sqrt{1+x^2}.
\end{aligned}
$$
Here, we have mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to get expected value from a probability mass function? A certain biased coin is flipped until it shows heads for the first time. If the probability of getting heads on a given flip is $5/11$ and $X$ is a random variable corresponding to the number of flips it will take to get heads for the first time, the expected... | The expectation is not a geometric series (at least not when you write
it directly), but its resemblance to a geometric series is a good observation.
First let's get that factor of $\frac{5}{11}$ out of the way,
because it will become annoying at some point if we keep it inside the
summation.
$$E[x] = \sum_{x=1}^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Constructing Multiplication Table for Multiplication Modulo I have this question
Consider G={1,5,7,11,13,17} under Multiplication Modulo 18.Construct Multiplication Table for G.I have constructed the following
Im i correct ?
| When just multiplying in the reals I get:
$$\begin{array}{c|cccccc} \times & 1 & 5 & 7 & 11 & 13 & 17\\ \hline1 & 1 & 5 & 7 & 11 & 13 & 17\\ 5 & 5 & 25 & 35 & 55 & 65 & 85\\ 7 & 7 & 35 & 49 & 77 & 91 & 119\\ 11 & 11 & 55 & 77 & 121 & 143 & 187\\ 13 & 13 & 65 & 91 & 143 & 169 & 221\\ 17 & 17 & 85 & 119 & 187 & 221 & 289... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Another beautiful arctan integral $\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx$ Do you think we can express the closed form of the integral below in a very nice and short way?
As you already know, your opinions weighs much to me, so I need them!
Calculate in closed-form
$$\int_{1/2}^... | Exploiting integration by parts, the problem boils down to computing:
$$ \int_{1/2}^{1}\left(\log(1+x)-\log(1-x)\right)\left(\frac{3}{5x^2+8x+5}-\frac{4}{5x^2+6x+5}\right)\,dx $$
and by partial fraction decomposition that is equivalent to computing:
$$ I_{\pm}(\zeta)=\int_{1/2}^{1}\frac{\log(1\pm x)}{x-\zeta}\,dx $$
wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Solving without L'Hopital
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$
That's
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}\right)-\lim_{x\to-\infty}\left(\sqrt{4x^2+x}\right)$$
I have been taught to get the highest exponents, so...
$$... | Set $-1/x=y\implies y\to0^+, |y|=+y$
and $4x^2-6=\dfrac{4-6y^2}{y^2},\sqrt{4x^2-6}=\dfrac{\sqrt{4-6y^2}}{|y|}=\dfrac{\sqrt{4-6y^2}}y$
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)=\lim_{y\to0^+}\dfrac{\sqrt{4-6y^2}-\sqrt{4-y}}y$$
$$=\lim_{y\to0^+}\dfrac{(4-6y^2)-(4-y)}{y(\sqrt{4-6y^2}+\sqrt{4-y})}$$
$$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Path Integral Problem Could anyone help understand why my answer disagrees with the given answer?
let the exact differential $dz = 3x^2(x^2+y^2)dx + 2y(x^3+y^4)dy$
Find $\int\limits_c dz$ from (1,2) to (2,1).
Here's my answer:
$\frac{\partial z}{\partial x}=3x^2(x^2+y^2)$
$\therefore$
$z = \int(3x^4+3y^2x^2) dx$
$\ther... | I write this only to confirm that I get the same $z$ as you (one could add an arbitrary constant, but that does not really change anything), and the same final answer. It looks like the book has a misprint.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of the inequality $e^x\le e^{x^2} + x$ The question is to prove the inequality $e^x\le e^{x^2} + x$. I tried the Taylor expansion like ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ...$ and $x + {e^{{x^2}}} = 1 + x + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + ...$ but cannot see anything usefu... | If $x \geq 1$, then $x^2 \geq x$, we then have $e^x \leq e^{x^2}$, which in turn means that $e^x \leq e^{x^2}+x$.
For $0 \leq x \leq 1$, then $x^2 \leq x$, we then have $e^x$, we then have
\begin{align}
e^x & = \sum_{k=0}^{\infty} \dfrac{x^k}{k!} = 1 + x + \sum_{k=2}^{\infty} \dfrac{x^k}{k!} \leq 1 + x + \sum_{k=2}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Calculus 1: limits using squeeze or sandwich theorem, verification. I have the next problem:
Suppose that a function $g$ satisfies that $-1\leq g(x) \leq 1$ for all $x\geq 0$.
Calculate $\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}$
So: $$\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}=\lim_{x\to\infty}\dfrac{\frac{x+g(2x)}... | Get rid of $g(x)$ entirely. $-1\le g(x) \le 1$. So $x - 1 \le x + g(2x) \le x + 1$.
Likewise. $4x - 1 \le g(3x) + 4x \le 4x + 1$.
So $\dfrac{x - 1}{4x + 1} \le \dfrac{x + g(2x)}{g(3x) + 4x} \le \dfrac{x + 1}{4x - 1}$
So $1/4 = \lim\dfrac{x-1}{4x+1} \le \lim \text{wholeness} \le \lim \dfrac{x + 1}{4x - 1} = \dfrac14$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
| Clearly, $x\ne0$ in $x^2-3x+1=0$
Dividing both sides by $x,$
$$\dfrac{x^2-3x+1}x=0\implies x+\dfrac1x=3$$
$$x^{2n}+\dfrac1{x^{2n}}=\left(x^n+\dfrac1{x^n}\right)^2-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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What is the value of the following determinant? \begin{vmatrix}
1 & n & n & \dots & n \\
n & 2 & n & \dots & n \\
n & n & 3 & \dots & n \\
\vdots & \vdots& \vdots & \ddots & \vdots\\
n & n & n & & n\\
\end{vmatrix}
I've the feeling that I should to eliminate with the last row the others ove and after that should I m... | By Gaussian elimination,
$$\begin{vmatrix}
1 & n & n & \dots & n \\
n & 2 & n & \dots & n \\
n & n & 3 & \dots & n \\
\vdots & \vdots& \vdots & \ddots & \vdots\\
n & n & n & \dots & n\\
\end{vmatrix}=\begin{vmatrix}
1 & n & n & \dots & n \\
n-1 & 2-n & 0 & \dots & 0 \\
n-1 & 0 & 3-n & \dots & 0 \\
\vdots & \vdots& \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$ I am struggling to show that $$\dfrac \pi{2\sqrt3}=1-\dfrac 15+\dfrac 17-\dfrac 1{11}+\dfrac 1{13}-\dfrac 1{17}+\dfrac 1{19}-\cdots$$ by using the Fourier series $$\frac \pi2-\frac x2=\sum_1^\infty \dfrac {\sin(nx)}{n}.$$
Can somebody give me any hint?
| In the figure below, in which the trigonometric circle has been divided into six equal parts, we can see clearly that for $x= \frac {\pi}{3}$ the following equalities are verified:
$\sin( \frac{n\pi}{3})=\frac{\sqrt3}{2}$ for $n=1,2,7,8,……, 1+6n,2+6n,…..$
$\sin( \frac{n\pi}{3})=\frac{-\sqrt3}{2}$ for $n=4,5,10,11,........ | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462803",
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"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
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If $2x+3y \propto \sqrt{xy}$ ,then prove that $9x^2+4y^2 \propto xy$. I am stuck on the following problem which one of my friends gave me:
If $2x+3y \propto \sqrt{xy}$ ,then prove that $9x^2+4y^2 \propto xy$.
The problem could have been easier if I had to prove $4x^2+9y^2 \propto xy$ but this problem has stumped me... | $$\text{A) }2x+3y \propto \sqrt{xy}\qquad \text{B) }9x^2+4y^2 \propto xy$$
From A we get
$$4 x^2+12 x y+9 y^2 \propto xy$$
Combining A and B we get
$$4 x^2+12 x y+9 y^2 = k(9x^2+4y^2)$$
$$\Rightarrow 4 x^2+12 x y+9 y^2 = 9kx^2+4ky^2$$
From this we see that no $k$ will give an $xy$ term, so the statement is false. If we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What... | By virtue of solving for $x$, the original equation $$y = \tan^{-1} (x - \sqrt{1+x^2})$$ implies $$x = \frac{1}{2}(\tan y - \cot y) = \frac{\sin^2 y - \cos^2 y}{2 \sin y \cos y} = -\cot 2y,$$ hence $$y = \frac{1}{2}\cot^{-1} (-x) = -\frac{1}{2} \cot^{-1} x.$$ Now recalling that $$\cot^{-1} x + \tan^{-1} x = \frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 1
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separating equation $$
\begin{align}
\frac {dy}{dx} &= \frac{(y-1)(x+2)}{(y+1)(x-3)}\\
\frac{y+1}{y-1}dy &= \frac{x+2}{x-3}dx
\end{align}
$$
Integrate both sides :
$$
\begin{align}
\int\frac{y+1}{y-1}dy & =\int\frac{x+2}{x-3}dx\\
\int 1+\frac{2}{y-1}dy &=\int 1+\frac{5}{x-3}dx\\
y+2\ln|y-1|&=x+5\ln|x-3|+C\\
\end{align}... | You are not trapped and the solution cannot be expressed in terms of elementary functions.
If fact, there is a solution in terms of Lambert function you will learn at a time and the solution would be $$y=1+2 W\left(\pm\frac{c \sqrt{(x-3)^5 e^{x}}}{2 \sqrt{e}}\right)$$ where $c$ is the integration constant you missed.
Y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Maximisation and minimisation of sum of squares, if sum is equal to 15 Find the numbers whose sum is $15$ and sum of squares is minimum
My answer:
Let the numbers be $x$ and $15-x$. Then
$$y=x^2+625-30x+x^2$$
$$=2x^2-30x+625$$
$$dy/dx=0$$
$$4x-30=0$$
$$X=7.5$$
Did I go wrong?
| Approach without using calculus:
Geometric viewpoint. This is the same as finding the point on the line $x+y=15$ which is closest to the origin/farthest from the origin. If you draw the picture, you can see that the point with coordinates $x=y=15/2$ is closest to the origin. (It lies on the perpendicular line $x=y$.) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the inequation for $x$ Solve for $x$ :
$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $
I tried cases like : $ x-1 \le 0 $ , $ x \le 0 $ , $ x+1 \le 0 $
| Well, let's see if we can simplify your expression first:
$$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $$
$$(x^2 - 1)^{2005}x^{2006}(x+1)^{2} \le 0$$
Think, first, what qualifies $x$ to be $\le 0$? If $x$ is $0$ or a negative number, correct?
Well, our expression can be $0$ if $x = -1, 1$
Thus, now lets look when our ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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module of sum is less than... Somehow need to prove:
$$|x_1 + x_2 + ... + x_n| \le \sqrt{n(x_1^2 + x_2 ^ 2 + ... +x_n^2)}$$
$x_i$ is a real number; $i = 1,...,n$
Here's mentioned that mathematical induction should help. So I tried to use it but right after squaring, cause I had no idea how to do it without squa... | Your induction hypothesis is $|x_1 + \cdots + x_n| \le \sqrt{n(x_1^2 + \cdots + x_n^2)}$, which becomes $(x_1 + \cdots + x_n)^2 \le n(x_1^2 + \cdots + x_n^2)$ by squaring. Write
\begin{align}(x_1 + \cdots + x_n + x_{n+1})^2 &= (x_1 + \cdots + x_n)^2 + x_{n+1}^2 + 2(x_1 + \cdots + x_n)x_{n+1}\\
&= (x_1 + \cdots + x_n)^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\r... | Hint: You can use the fact that
$$
\sqrt{4-5t} = 2\sqrt{1-\frac{5}{4}t} = 2\left(1-\frac{5}{8}t + o(t)\right)
$$
when $t\to 0$. (This is the Taylor approximation of $\sqrt{1+t}$ around $0$.) Note that when $x\to-\infty$, $t=\frac{1}{x}\to 0$.
Additionally, you have a few issues in your derivation. For instance, $x\to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\lim... | HINT:
Double angle formula:
$$\sin(m+1)x=2\sin\dfrac{(m+1)x}2\cos\dfrac{(m+1)x}2$$
Werner Formula: $$2\sin\dfrac x2\cos\dfrac{(m+1)x}2=\sin\dfrac{(m+2)x}2-\sin\dfrac{mx}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Integral $\int \sqrt{x+\sqrt{x^2+2}}dx$ $$\int \sqrt{x+\sqrt{x^2+2}}\ dx$$
I tried various solving methods but I am not coming forward. I unformed the term also to ${x+\sqrt{x^2+2} \over \sqrt{x+\sqrt{x^2+2}}}$ and even multiplied with ${\sqrt{x-\sqrt{x^2+2}} \over \sqrt{x-\sqrt{x^2+2}}}$.
Trigonometric and hyperboli... | Notice, let $$x+\sqrt{x^2+2}=t^2\implies x=\frac{t^4-2}{2t^2}$$$$ \left(1+\frac{x}{\sqrt{x^2+2}}\right)dx=2tdt\iff \left(\frac{x+\sqrt{x^2+2}}{\sqrt{x^2+2}}\right)dx=2tdt$$$$\implies dx=\frac{t^4+2}{t^3}dt$$
Now, we get
$$\int\sqrt{x+\sqrt{x^2+2}}dx=\int t\frac{t^4+2}{t^3}dt$$
$$=\int \frac{t^4+2}{t^2}dt=\int\left(t^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How is $\frac{ds}{dt}$ related to $\frac{dx}{dt}$? The problem states: Let $x$ and $y$ be differentiable functions of $t$, and let $s = \sqrt{4x^2+6y^2}$ be a function of $x$ and $y$. How is $\frac{ds}{dt}$ related to $\frac{dx}{dt}$ if $y$ is constant?
My attempt:
\begin{align}
\frac{ds}{dt} & = \frac{d}{dt} \left( \s... | This is an example of the usefulness of Leibniz notation in dealing with these sorts of problems.
$$\frac{ds}{dt} = \frac{ds}{dx} \cdot \frac{dx}{dt}$$
This, intuitively, is because the $dx$'s "cancel out".
All to do here is to find $\frac{ds}{dx}$.
Since $y$ is not a function of $x$, we can treat it as constant. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$? Can anyone tell me how to solve this?
$x^6-x^5+x^4-x^3+x^2-x+1=0$
What I got to was $x^7+1=0$.
Thanks in advance.
| $$x^6-x^5+x^4-x^3+x^2-x+1=0 \Longleftrightarrow$$
$$(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0 \Longleftrightarrow$$
$$x^7+1=0 \Longleftrightarrow$$
This introduces the extraneous root of $x=-1$, so from now on we assume that $x\ne -1$:
$$x^7+1=0 \Longleftrightarrow$$
$$x^7=-1 \Longleftrightarrow$$
$$x^7=e^{\pi i} \Longleftrig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Calculate equilibrium points of a 2nd order system The following 2nd order system is given:
$$
\frac{d\textbf{x}}{dt}=\begin{bmatrix} -6 & -\frac{2}{\pi} \\ 0 & \frac{1}{2} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}+\begin{bmatrix}\sin(\frac{\pi}{2}-x_2) \\ x_1x_2\end{bmatrix}
$$
The task is to calculate all... | What you have found is correct. But $x_2 = \pi$ and $x_2 = 2 \pi$ are also solutions in the second equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ordinary differential equation with polynomial terms The original equation I had was:
$$-y''+(x^2+2x^4-2\alpha)y=0$$
Where $\alpha$ is a real parameter $\geq 0$ and we require the solutions to go to $0$ at infinity. With the substitution $y=p(x)e^{-\frac{x^2}{2}}$ I got the equation:
$$-p''+2xp'+p(1+2x^4-2\alpha)=0$$
I... | Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{du}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+2\dfrac{dy}{du}=2x\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+2\dfrac{dy}{du}=2x\dfrac{d^2y}{du^2}2x+2\dfrac{dy}{du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to Calculate $x^6+x^3y^3+y^6$ Given that $x,y$ real numbers such that :
$x^2+xy+y^2=4$
And
$x^4+x^2y^2+y^4=8$
How can one calculate :
$x^6+x^3y^3+y^6$
Can someone give me hint .
| Note that
$$(x^2+xy+y^2)^2-2xy(x^2+xy+y^2)=x^4+x^2y^2+y^4 $$
so that
$$ xy = 1.$$
Then
$$\begin{align}x^6+x^3y^3+y^6&=(x^4+x^2y^2+y^4)(x+xy+y^2)-xy^5-2x^2y^4-2x^4y^2-yx^5\\
&=8\cdot 4-xy(x^4-2x^3y-2xy^3-y^4)\\
&=32-1\cdot ((x^4+x^2y^2+y^4)-xy(2x^2-xy-2y^2))\\
&=24+1\cdot(2(x^2+xy+y^2)-3xy)\\
&=29
\end{align}
$$
(I s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:
Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .
Solution: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3}... | Good question: that is a big leap in the argument in my opinion.
If we let $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$, then $\sqrt{2} + \sqrt{3}$, $\sqrt{2} - \sqrt{3}$, $-\sqrt{2} + \sqrt{3}$, and $-\sqrt{2} - \sqrt{3}$ all lie in $K$.
In order to prove that they are all roots of $p(x)$, the minimal polynomial of $\sqrt{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What line is determined by the following complex equation? $$\left | z+1-i \right |= \frac{\Re z-\Im z}{\sqrt{2}}$$
This leads to the following equation, if we write $z=x+iy$ :
$$\sqrt{\left ( x+1 \right )^{2}+\left( y-1 \right)^2}=\frac{x-y}{\sqrt{2}}$$
which according to desmos, defines a set in $\mathbb{C}$ which is... | Suppose that there exist $(x,y)$ such that
$$\sqrt{(x+1)^2+(y-1)^2}=\frac{x-y}{\sqrt 2}$$
(note here that $x-y\ge 0$)
Then we have
$$\begin{align}(x+1)^2+(y-1)^2=\frac{(x-y)^2}{2}&\Rightarrow x^2+y^2+2x-2y+2=\frac{x^2}{2}+\frac{y^2}{2}-xy\\&\Rightarrow \frac{x^2}{2}+\frac{y^2}{2}+xy+2x-2y+2=0\\&\Rightarrow x^2+y^2+2xy+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to solve this limit without Hopital / Taylor / derivatives: $\lim\limits_{x \to 0} \frac{x-\sin(x)}{x^3} = \frac{1}{6}$? It's simple to prove with Hopital that
$$ \lim_{x \to 0} \frac{x-\sin(x)}{x^3} = \frac{1}{6}$$
Is it possible to solve this limit without Hopital or Taylor (without derivatives)?
| Let us assume $$\displaystyle \lim_{x\rightarrow 0}\frac{x-\sin x}{x^3} = L$$ (A finite quantity).
Now replace $x\rightarrow 3y$, then we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-\sin 3y}{27y^3} = L$$
Now, using the formula $$\sin 3y = 3\sin y-4\sin^3 y$$
we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-3\... | {
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"source": "stackexchange",
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How to find the value of $\sum_{i = 1}^{\infty } (\frac{1}{i} - \frac{1}{2i + 1} - \frac{1}{2i - 1})$ Is there any way to solve this summation?
$$\sum_{i = 1}^{\infty } (\frac{1}{i} - \frac{1}{2i + 1} - \frac{1}{2i - 1})$$
The value is 1 - log4... but I'm not able to proove it.
| @Olivier Oloa, obviously, there is something wrong in your solution.
For i>0, $\frac{1}{i}-\frac{1}{2i+1}-\frac{1}{2i-1}=\frac{1}{2i}-\frac{1}{2i+1}+\frac{1}{2i}-\frac{1}{2i-1}<\frac{1}{2i}-\frac{1}{2i+1}$, but using your method, $\sum_{i=1}^{\infty}\frac{1}{2i}-\frac{1}{2i+1}=\sum\int dx{x^{2i-1}-x^{2i}}=\int dx\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\arctan n$ always equal to $\arccos\sqrt{\frac{1}{n^2+1}}$? $$\arccos\sqrt\frac{1}{2}=\arctan 1$$
$$\arccos\sqrt\frac{1}{5}=\arctan 2$$
$$\arccos\sqrt\frac{1}{10}=\arctan 3$$
$$\arccos\sqrt\frac{1}{17}=\arctan 4$$
$$\arccos\sqrt\frac{1}{26}=\arctan 5$$
$$\arccos\sqrt\frac{1}{37}=\arctan 6$$
$$\arccos\sqrt\frac{1}{5... | Let $\arctan n = a$, then $\tan(a)=n$ and...
$$1 + \tan^2 (a) = 1/(\cos^2 (a) )$$
$$1 + n^2 = 1/ (\cos^2 (a))$$
$$\cos^2 (a) = 1/(1+n^2)$$
$$\cos(a) = (1/(1+n^2) )^{1/2}$$
$$\arccos (1/(1+n^2) )^{1/2} = a = \arctan(n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For the partial differential equation $u_{xx}-u_{yy}=-1$ find $u(1/2,1/2)$
Let , $u(x,y)$ be a solution of $u_{xx}-u_{yy}=-1$ , $x\in \mathbb R$ , $y>0$ with $u(x,0)=0$ , $u_y(x,0)=0$ , $x\in \mathbb R$. Then , find $u\left(\frac{1}{2},\frac{1}{2}\right)$.
Attempt :
Solution is $u(x,y)=f_1(y+x)+f_2(y-x)-\frac{x^2}{2}... | $$
u(x, y) = f_1(y + x) + f_2(y-x) - \frac {x^2}2 \implies u_y = f_1' + f_2'
$$
Now, substitute ICs
\begin{align}
u(x, 0) &= f_1(x) + f_2(-x) - \frac {x^2}2= 0 \tag 1 \\
u_y(x, 0) &= f_1'(x) + f_2'(-x) = 0 \tag 2
\end{align}
Differentiate $(1)$
$$
f_1'(x) - f_2'(-x) -x = 0 \tag 3
$$
And solve it in conjunction with th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the quaternion that performs a rotation I managed to find this answer here where Christian Rau says "axis/angle rotation (a,x,y,z) is equal to quaternion (cos(a/2),xsin(a/2),ysin(a/2),z*sin(a/2))"
Assuming I know what rotation I need to perform, how would I represent it?
eg, finding the quaternion that rotates... | To rotate about the $z$ axis (yaw) by $\alpha$ you need the following quaternion
$\begin{aligned}q = \begin{bmatrix}\cos(\tfrac{\alpha}{2})\\0\\0\\\sin(\tfrac{\alpha}{2})\end{bmatrix}\end{aligned}\tag{1},$
to rotate about the $x$ (pitch) axis you need
$\begin{aligned}q = \begin{bmatrix}\cos(\tfrac{\alpha}{2})\\0\\ \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Legendre symbol $(-3/p)$ where $p = 1 \mod 3$ Suppose $p = 1 \bmod 3$, prove the following statements:
*
*prove that $x^2 + x + 1 = 0 \mod p$ has a solution
*Prove that $\left(\frac{-3}{p}\right) = 1$ if $p = 1\mod 3$
*Determine the discriminant of $x^2 + x + 1$
*Prove using 2,3 that $\left(\frac{-3}{p}\right) = ... | *
*You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p \equiv 1 \mod 3$ then there is an $x \ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way,... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Show $\big\lvert\frac{1}{2} - \frac{1}{2}e^{-i2\pi f}\big\vert^{2} = \sin^2(\pi f)$? $\big\lvert\frac{1}{2} - \frac{1}{2}e^{-i2\pi f}\big\vert^{2} = \sin^2(\pi f)$?
Any help?
| HINT:
Assuming $f\in\mathbb{R}$:
$$\left|\frac{1}{2} - \frac{1}{2}e^{-2i\pi f}\right|^{2} = \sin^2(\pi f)\Longleftrightarrow$$
$$\left|\frac{1}{2}\left(1 - \frac{1}{2}e^{-2i\pi f}\right)\right|^{2} = \sin^2(\pi f)\Longleftrightarrow$$
$$\frac{1}{4}\left|1 - \frac{1}{2}e^{-2i\pi f}\right|^{2} = \sin^2(\pi f)\Longleftrig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that the sum of fractions has an odd numerator and even denominator. I'm struggling to show that, for all $n>1$
$$
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \frac{k}{m}
$$
where $k$ is an odd number and $m$ is an even number.
Proof: The proof is by induction on $n$.
Base Case: $1 + \frac{1}{2} = \f... | To complete your proof, first observe that
$$
\frac{k}{m} + \frac{1}{n+1} = \frac{k(n+1)+m}{m(n+1)}
$$
now $m$ is even so let $m=2^\alpha a$,where $\alpha$ is the biggest power of $2$ in $m$, so $a$ is odd.
if $n+1$ is odd obviously $k(n+1)+m$ is odd and ${m(n+1)}$ is even, so we are done.
So consider the case when $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Proving that $\lim \limits_{x \to \infty}\frac{x^3}{x^2 +10}=\infty$ using $\epsilon-\delta$ definition $\lim \limits_{x \to \infty}\frac{x^3}{x^2 +10}=\infty$
Therefore $\forall M>0$ we need to find a $\delta=\delta(M)$ such that $f(x)>M$ or $x<\delta(M)$
so $\frac{x^2}{x^2+10}>M$
$x^2>M(x^2+10)$
$x^2-Mx^2>10M$
$x^2(1... | The limit
$$
\lim_{x\to \infty} \frac{x^2}{x^2+10} = 1
$$
and not equal to $\infty$. You can see this from
$$
\lim_{x\to \infty} \frac{x^2 / x^2}{(x^2+10)/x^2} = \lim_{x\to \infty} \frac{1}{1 + 10/x^2} = 1.
$$
If you are considering the limit
$$
\lim_{x\to \infty} \frac{x^3}{x^2+10} = \lim_{x\to \infty} \frac{x}{1 + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503425",
"timestamp": "2023-03-29T00:00:00",
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roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4... | You want to see when the discriminant is non negative, that is,
$$
(4abcd)^2-4(a^4+b^4)(c^4+d^4)\ge0
$$
that can be simplified into
$$
4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)\ge0
$$
Consider the left-hand side:
\begin{align}
4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)
&=4a^2b^2c^2d^2-a^4c^4-b^4c^4-a^4d^4-b^4d^4\\
&=(-a^4c^4+2a^2b^2c^2d^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the irrationality of $\sqrt{5} = x \sqrt{7} + y$ I need to prove that there are no rational numbers $x, y$ that
$$\sqrt{5} = x \sqrt{7} + y$$
We know that square root of prime is irrational so $y = 5 - 7x$ so the only number for it to be rational is if $y = 0$ so $x = \frac{\sqrt{7}} {\sqrt {5}}$ but that is ir... | Assume we have rationals $x,y$ such that $\sqrt{5} = x\sqrt{7} + y$.
Square both sides, we get $5 = 7x^2 + y^2 + 2xy\sqrt{7}$.
So $\sqrt{7} = {{5 - 7x^2 - y^2} \over {2xy}}$ which is rational, contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\ri... | Perhaps a way to 'see the answer' is to use the MVT: for the numerator, write $f( x) = x^n$, $a = 1 + 1/n -1/n^2$, $b = 1 + 1/n + 1/n^2$. Then
$$ f( b) - f(a) = f'(c)(b-a),$$
for some $c \in (a,b)$.
Therefore the numerator is
$$ n\left(1 + 1/n + o(1/n^2)\right)^{n-1} 2/n^2.$$
Similarly, in the denominator, using the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\... | We can make use of the fact that $\gcd(a, b) = \gcd(a, b-ka)$ for any integer $k$. Note that
$\gcd(m^2 - n^2, m^2 + n^2) = \gcd(m^2 - n^2, 2n^2).$ You noted that one of $m, n$ is odd and one is even, so $2 \nmid m^2 -n ^2$. So $$\gcd(m^2 - n^2, 2n^2) = \gcd(m^2 - n^2, n^2) (\text{why?}) = \gcd(m^2, n^2).$$
From there,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$ Let $b, c, d, e$ be real numbers such that the following equation
$$x^5-20x^4+bx^3+cx^2+dx+e=0$$
has real roots only. Find the largest possibe value of $b$.
What I have done is:
Let $x_1, x_2, x_3, x_4, x_5$ be the 5 real roots of the equation. Then we have
$$x_1+x_2+... | The answer is correct and you are wong.
The mistake in your solution happens when you apply the Cauchy Inequality where you missed to square the right side.
$(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)(1+1+1+1+1)\ge(x_1+x_2+x_3+x_4+x_5)^2=400$
So $x_1^2+x^2_2+x_3^2+x_4^2+x_5^2\ge\frac{400}{5}=80$
and $b_{max}=\frac{1}{2}[20^2-80]... | {
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Show that $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 }$ So I'm trying to show that:
$\dfrac{ -1 }{ 2 } \le \dfrac{ x }{ 1+x^2 } \le \dfrac{ 1 }{ 2 }$
for every value of x.
I know I have to use mean value theorem so I tried to show it with cases. First I tried showing that $\dfrac{-1}{2} \le \dfrac{x}{1... | Let $u=|x|$, then $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 } \iff \frac{ u }{ 1+u^2 } \le \frac{ 1 }{ 2 }$.
You are using $AM-GM$ here and not Mean Value Theorem since $1+u^2\geq2u$ by $AM-GM$.
To do the calculus way, just find the derivative of $x\over {1+x^2}$.
$df\over dx$$={1+x^2-2x^2\over{(1+x^... | {
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Equation of locus Point P$(x, y)$ moves in such a way that its distance from the point $(3, 5)$ is proportional to its distance from the point $(-2, 4)$. Find the locus of P if the origin is a point on the locus.
Answer:
$$(x-3)^2 + (y-5)^2 = (x+2)^2 + (y-4)^2$$
or, $$10x+2y-14=0$$
or, $$5x+y-7=0$$
but answer given is... | $$(x-3)^2 + (y-5)^2 = k\left[ (x+2)^2 + (y-4)^2 \right]$$ where $k$ is a constant
Now $(0,0)$ lies on the locus.
Therefore $$9+25=k(4+16) \Rightarrow k=\frac{34}{20} = \frac{17}{10}$$
Using this value of $k$ in the equation, we get
$$(x-3)^2 + (y-5)^2 = \frac{17}{10}\left[ (x+2)^2 + (y-4)^2 \right]$$
$$10\left[(x-3)^2... | {
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"answer_id": 1
} |
Sum of the rows of Pascal's Triangle. I've discovered that the sum of each row in Pascal's triangle is $2^n$, where $n$ number of rows. I'm interested why this is so. Rewriting the triangle in terms of C would give us
$0C0$ in first row.
$1C0$ and $1C1$ in the second, and so on and so forth. However, I s... | There are various different ways to look at this. Here's one:
Two adjacent numbers in a row get added to get the number in the row below it:
$$
\begin{array}{cccccccccc}
& & 1 & & & & & 8 & & & & 28 & & & & 56 & & & & 70 & & \cdots \\
& & & & & & & & \searrow & & \swarrow \\
1 & & & & & 9 & & & & 36 & & & & 84 & & & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$ $$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$
I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?
| Trigonometric substitution There is another simpler method to solve the problem
Let $2x=3\sec\theta \implies dx=\frac{3}{2}\sec\theta \tan\theta \ d\theta$
$$\int \frac{\sqrt{4x^2-9}}{x^2}\ dx=\int \frac{\sqrt{9\sec^2\theta-9}}{\frac{9}{4}\sec^2\theta}\ \frac{3}{2}\sec\theta \tan\theta \ d\theta$$
taking positive valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding the shortest distance between an arbitrary point and a parabola I'm attempting to find the shortest distance between a point and a parabola. The point in question is (0,b), for any b, and the parabola that we are given is$\ y = x^2 $.
How would you approach the problem and find the shortest distance for any giv... | 2D
We consider the distance between a query point $Q=(0,b)$ and some point $P(x) = (x, x^2)$ on the graph of the function.
This leads to
$$
d(x) = d(Q, P(x)) = \sqrt{x^2 + (x^2 - b)^2}
$$
The distance to the graph is the minimum of those distances:
\begin{align}
d &= \min_{x \in \mathbb{R}} d(x) \\
&= \min_{x \in \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Proving a complex equality. How to prove the following equality:
$$2i\tan^{-1} \left(\frac{ia}{b}\right) = \log \left|\frac{a-b}{a+b}\right|$$
where $i = \sqrt{-1}$ and $a,b \in \mathbb{R},a \neq \pm b$
| Hint:
$$\int_0^x \frac{1}{x^2-a^2} \mathrm{d}x = \frac{1}{2a}\int_0^x \frac{1}{x-a} - \frac{1}{x+a} \mathrm{d}x = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right|$$
$$\int_0^x \frac{1}{x^2-a^2} \mathrm{d}x = \int_0^x \frac{1}{x^2+(ia)^2} \mathrm{d}x = \frac{i}{a}\tan^{-1} \left(\frac{ix}{a}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Rewrite an expression as a sum of even powers - proving injectivity without calculus I teach algebra to undergraduates - nothing too fancy. But we recently covered injective functions, and I was a little disappointed that I couldn't give rigorous justifications why certain functions were injective. For the handful of f... | There is a simple recursive reduction, at least for $7$.
Write:
$$\begin{align}
x^6+x^5+x^4+x^3+x^2+x+1 &= \left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}x^4+x^3+x^2+x+1\\
&=\left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}\left(x^2+\frac{2}{3}x\right)^2 + \frac{2}{3}x^2+x+1\\
&=\left(x^3+\frac{1}{2}x^2\right)^2 + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int_0^{2\pi}\frac{1}{a\cos \theta+b\sin\theta+d}d\theta$ where $a,b,d\in\mathbb{R}$ and $a^2+b^2$\int_0^{2\pi}\frac{1}{a\cos \theta+b\sin\theta+d}d\theta$ where $a,b,d\in\mathbb{R}$ and $a^2+b^2<d^2$
Here, I solve it by Residue Theory. By substituting $d\theta=dz/iz,\cos \theta=(1/2)(z+1/z),\sin\theta=(1/2i)(z-1/z),$... | $I
=\int\frac{1}{a\cos \theta+b\sin\theta+d}d\theta
$
Letting
$\tan(\theta/2)
=t
$,
$d\theta
=\frac{2dt}{1+t^2}
$,
$\sin \theta
=\frac{2t}{1+t^2}
$,
and
$\cos \theta
=\frac{1-t^2}{1+t^2}
$.
Therefore
$\begin{array}\\
I
&=\int\frac{1}{a\cos \theta+b\sin\theta+d}d\theta\\
&=\int\frac{1}{a\frac{1-t^2}{1+t^2}+b\frac{2t}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1524329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series
$$\sum\limits_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^2}$$
Second derivative is
$$\left(\frac{1}{1-x^2}\right)^{''}=\frac{1}{(1-x^2)^4}\cdot x(1+8x-2x^2-8x^3+x^4)$$
This gives
$$\frac{1}{(1-x^2)^4}=\sum\limits_{n=0}^... | $$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
$$\frac{d^3}{dx^3}(\frac{1}{1-x})=\frac{6}{(1-x)^4}=\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$
$$\frac{1}{(1-x)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$
now replace $x$ by $x^2$
$$\frac{1}{(1-x^2)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{2n-6}=1+4x^2+10x^4+20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the best way to solve an equation of the form $(f(x))^2-a(f(x))+b=x$? On a math contest I was told to solve the equation $$(x^2-3x+1)^2-3(x^2-3x+1)+1=x$$
For this particular problem I simplified by letting $$a\equiv x^2-3x+1$$
Then I continued with $$a^2-3a+1-x=0$$
$$a=\frac{3\pm\sqrt{9-4\left(1-x\right)}}{... | Let $f(x) = x^2-3x+1$.
Idea
The important observation here is following: If you have a solution $a$ that satisfies $f(a) = a$ then obviously
$a$ also solves the original equation because $f(f(a)) = f(a) = a$.
Finding the $a$ that solve $f(a) = a$ is easy, that is just a quadratic equation. This means you can divide $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
integrations by parts. Show $\int_0^\infty 2y^2e^{-y^2} dy = .886$
$2\int_0^\infty y^2e^{-y^2} dy$
Then by parts;
$f=y^2$ $dg=e^{-y^2}$
$df=2y$ $g=-2ye^{-y^2}$
$2\int_0^\infty y^2e^{-y^2} dy$ = $-2y^3e^{-y^2} -\int_0^\infty -4y^2e^{-y^2}$=$-2y^3e^{-y^2} +4\int_0^\infty y^2e^{-y^2}$
equivalently; $-2\int_0^\inf... | $$\int y^2e^{-y^2}dy=\int \underbrace{y}_{=u}\cdot\underbrace{ ye^{-y^2}}_{=v'}dy=[-\frac{1}{2}ye^{-y^2}]+\frac{1}{2}\int_0^\infty e^{-y^2}dy$$
The integral $\int_0^\infty e^{-y^2}dy$ is not that obvious. If you set $I=\int_0^\infty e^{-y^2}dy$, you can compute $I^2$ using polar coordinate. You can also know that $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_{0}^\infty \mathrm{d}y\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}x=\int_{0}^\infty \mathrm{d}x\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}y=\pi/4$ How can we prove that
\begin{aligned}
&\int_{0}^\infty \mathrm{d}y\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}x\\
=&\int_{0}^\infty \mathrm{d}x\int_{0}^\infty \sin(x^2+y^2)\ma... | I do not know if you are supposed to know this. So, if I am off-topic, please forgive me.
All the problem is around Fresnel integrals. So, using the basic definitions,$$\int_{0}^t \sin(x^2+y^2)dx=\sqrt{\frac{\pi }{2}} \left(C\left(\sqrt{\frac{2}{\pi }} t\right) \sin
\left(y^2\right)+S\left(\sqrt{\frac{2}{\pi }} t\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$ for nonnegative random variables $X,Y$ and $p\ge0$ Suppose $X \geq 0$ and $Y \geq 0$ are random variables and that $p\geq 0$
*
*Prove
$$E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$$
Proof
Since $(X+Y)^p \leq (2 \> \max\{X,Y\})^p=2^p \> \max \{X^p,Y^p\}\leq 2^p(X^p+Y^p)$ $ \implies ... | If $p>1$, then by Holder inequality,
\begin{align*}
X+Y &\le (X^p+Y^p)^{\frac{1}{p}} 2^{1-\frac{1}{p}}.
\end{align*}
That is,
\begin{align*}
(X+Y)^p \le 2^{p-1} (X^p+Y^p).
\end{align*}
For $0 \le p \le 1$, we note that
\begin{align*}
\left(\frac{X}{X+Y} \right)^p + \left(\frac{Y}{X+Y} \right)^p \ge \frac{X}{X+Y}+ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
An Algebraic Proof that $|y^3 - x^3| \ge |(y - x)|^3/4 $ I can prove this using calculus, but not by simple algebra: can anyone help ?
Calculus Proof:
Fix the separation of $x$ and $y$ so that $y = x + d$ with $d>0$ ($ \implies y > x \implies y^3 > x^3$) and now consider $ f(x) = y^3 - x^3 = (x+d)^3 - x^3$.
$ f(x) = (... | $ |y^3-x^3| \geq |y-x|^3/4 \\
\iff |y-x|(x^2+xy+y^2) \geq |y-x|(y-x)^2/4 \\
\iff 4(x^2+xy+y^2) \geq (x^2-2xy+y^2)$
(Assuming that $y\neq x$)
$\iff 3x^2+6xy+3y^2 \geq 0\\
\iff (x+y)^2 \geq 0$
EDIT
$x^2+xy+y^2 = (x+y/2)^2 + 3y^2/4 \geq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Minimizing $f(x)=A^{\frac{tx-1}{x-1}} \left( c^x \frac{\Gamma(0.5+x)}{\sqrt{\pi}} \right)^{\frac{1-t}{x-1}}$ subject to the constraint Let $f(r)$ be a function defined as follows
\begin{align}
f(x)=A^{\frac{tx-1}{x-1}} \left( c^x \frac{\Gamma(0.5+x)}{\sqrt{\pi}} \right)^{\frac{1-t}{x-1}}
\end{align}
where $0 < A,c$ an... | Hopefully the following can somehow help or give some idea.
Rewrite
\begin{align}
h(x)=\ln \left( \frac{ \sqrt{\pi} A}{c\Gamma(x+0.5)} \right)+ (x-1)\psi^{(0)}(x+0.5)
\end{align}
as
\begin{align}
h(x)= (x-1)\psi^{(0)}(x+0.5)-
\ln \left( \Gamma(x+0.5) \right) +\alpha
\end{align}
where $\alpha =\ln \left( \frac{ \sqrt{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the following limit 0/0 $$\lim_{t\to 0} \frac{\left(\sqrt{t+9}-3 \sqrt[3]{2 t+1}\right)}{\sqrt[3]{t+8}-2 \sqrt[3]{3 t+1}}$$
I tried to multiply the nominator by conjugation but got even bigger expression.
| Let $a=\sqrt{t+9}$, $b=3\sqrt[3]{2t+1}$, $c=\sqrt[3]{t+8}$ and $d=2\sqrt[3]{3t+1}$.
We have
$$ \lim_{t\to 0} \frac{a-b}{c-d} = \lim_{t\to 0}\left(\frac{a^6 -b^6}{c^3-d^3}\right)\left( \frac{c^2+cd + d^2}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}\right)
= \lim_{t\to 0}\left(\frac{(t+9)^3 -(3^6(2t+1)^2)}{t+8 - 8(3t+1)}\right)\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Compute the limit of the sequence of functions of $\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}$ Compute the limit of the sequence of functions of
$$
\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}.
$$
Attempt.
$$\lim_{n\to \infty} f_n(x) = \lim_{n\to \infty} \frac{x^2}{x^2+(1-nx)^2} = \lim_{n\to \infty} ... | For $x=0$ and all $n\in \mathbf{N}$, we see that $$f_{n}(0)=\frac{0}{0+1}=0$$ For $x\neq 0$, then we note that $$\lim_{n\rightarrow \infty}(1-nx)$$ is $+\infty$ for $x<0$ and is $-\infty$ for $x>0$. Thus, $$\lim_{n\rightarrow \infty}(1-nx)^{2}=\left(\lim_{n\rightarrow \infty}(1-nx)\right)^{2}=+\infty$$ and $$\lim_{n\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove by induction $ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $
Prove by induction
$$ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $$
What I have for now:
$$ \frac {\sin (\frac {n + 1} {... | You can also see that
\begin{align}
\sum_{k=0}^{n}\sin kx & =\Im \sum_{k=0}^{n}\mathrm{e}^{\mathrm{i}kx}\\
& = \Im \sum_{k=0}^{n}\left(\mathrm{e}^{\mathrm{i}x}\right)^k\\
& = \Im\frac{\left(\mathrm{e}^{\mathrm{i}x}\right)^{n+1}-1}{\mathrm{e}^{\mathrm{i}x}-1}\\
& = \Im\frac{\mathrm{e}^{\mathrm{i}(n+1)x}-1}{\mathrm{e}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$ The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the ... | There is a much simpler and efficient way to solve this problem.
First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.
Now since the triangle is isosceles and right-angled it should be obvious that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $\frac{1+4n^2}{2+2n^2}$ is a cauchy sequence? I know that the following sequence is a
$$\frac{1 + 4n^2}{2+2n^2}$$
How can I show that it is a cauchy sequence - well it has to cause its convergent but I want to understand it with the cauchy definition.
I know that i have to get the $|a_m - a_n | < \var... | For any $p > 1$, we have
\begin{align}
& \left|\frac{1 + 4(n + p)^2}{2 + 2(n + p)^2} - \frac{1 + 4n^2}{2 + 2n^2} \right|\\
= & \frac{2 + 2n^2 + 8(n + p)^2 + 8n^2(n + p)^2 - 2 - 2(n + p)^2- 8n^2 - 8n^2(n + p)^2}{4(1 + n^2)(1 + (n + p)^2)} \\
= & \frac{8p(2n + p) - 2p(2n + p)}{4(1 + n^2)(1 + (n + p)^2)} \\
= & \frac{3p(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Matrix multiplication of $4 \times 1$ matrix with $4 \times 4$ matrix I'm new to matrix multiplication and just wondered how I would evaluate the following:
$$
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1& 0 & 0 \\
-1& 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c \\
d \\
\end{pmatrix}
$$
and even:
$$... | There is a formula to compute the $(i, j)$th entry of $C = AB$ (provided the sizes of $A$ and $B$ match): assume $A$ is $n \times p$ and $B$ is $p \times m$, then $C$ is $n \times m$, and
$$c_{i, j} = \sum_{k = 1}^p a_{\color{red}{i}k}b_{k\color{red}{j}}.$$
Take your first question for example, the result should be a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Upper bound of $a_{n+1}=a_n + \frac{1}{a_n}$ How can I prove:
If $$a_0=\alpha>0\quad and\quad a_{n+1}=a_n + \frac{1}{a_n}$$, then $$a_n^2<\alpha^2+2n+\frac{1}{\alpha^2}+\frac{1}{2}ln\left ( \frac{2n}{\alpha^2}+1 \right )$$
?
I'll really appreciate your help. Thanks.
| We have $\displaystyle a_{n+1}^2=a_n^2+\frac{1}{a_n^2}+2$. Hence for $n\geq 1$
$$a_n^2=a_0^2+2n+\sum_{k=0}^{n-1}\frac{1}{a_k^2}$$
This imply that $\displaystyle a_k^2\geq a_0^2+2k$ for $k\geq 1$. Hence
$$a_n^2\leq a_0^2+2n+\frac{1}{a_0^2}+\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}$$
Now $$\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}\leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Taking the PMF to find the CDF, probability. Question:We toss a fair coin three times. Let X be the number of heads minus the number of tails obtained.
Write the cumulative distribution function of X.
What I already know: I know how to get the pmf from this problem.
$$
\ P(X=x) = \left\{
\begin{array}{l l}
\ {3 \... | The c.d.f. is not $x\mapsto \Pr(X=x)$; it is $x\mapsto\Pr(X\le x)$. So, for example, its value at $x=2$ is $\Pr(X\le 2) = \Pr(X=-3) + \Pr(X=-1) + \Pr(X=1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I find the maximum perimeter of a rectangle inscribed in an ellipse? The problem I've been stuck on is this:
A rectangle is inscribed in the ellipse $$\frac{x^2}{20} + \frac{y^2}{12} = 1$$ What is the maximum perimeter of the rectangle?
I don't even know if I'm taking the right approach. So far, I've been tryin... | One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$, If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$,then find the value of $|z_1^2+z_2^2|$.
$z_1^3-3z_1^2z_2=2$
$3z_1z_2^2-z_2^3=11$
Adding them,we get
$z_1^3-3z_1^2z_2+3z_1z_2^2-z_2^3=13$
$(z... | dividing both the equations we get
$$\frac{z_1^2}{z_2^2} \times \frac{z_1-3z_2}{3z_1-z_2}=\frac{2}{11}$$ $\implies$ assuming $\frac{z_1}{z_2}=p$ we get
$$\frac{p^2(p-3)}{3p-1}=\frac{2}{11}$$ i.e.,
$$11p^3-33p^2-6p+2=0$$ and this cubic equation has three real roots which means
$\frac{z_1}{z_2}$ is Real.
Now $$|z_1^2+z_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Can this approximation be made more formal? When considering oscillating systems in physics, we end up with some response function like $$F(\omega) = \frac{\omega^2}{(\omega_0^2 - \omega^2)^2 + (\omega/\tau)^2},$$ where $\omega_0$ and $\tau$ are characteristic properties of the system, and $\omega$ is the driving frequ... | Let $F(\omega)$ be given by
$$\begin{align}
F(\omega)&=\frac{\omega^2}{(\omega^2-\omega_0^2)^2+(\omega/\tau)^2}\\\\
&=\frac{1/4}{(\omega_0/2)^2(1-\omega/\omega_0)^2(1+\omega/\omega_0)^2+(1/2\tau)^2}
\end{align}$$
Now, let's denote $\omega/\omega_0=1+x$. Then, we have
$$\begin{align}
F(\omega)&=F(\omega_0(1+x))\\\\
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the $n^{\text{th}}$ term of $-1,-1,-1,-1,1,1,1,1,...$ as a repeating 8-block In my work I came across that sequence
$-1,-1,-1,-1,1,1,1,1,\dots$ and repeating this 8-block so on forever
Now I cant find an ( e.g. trigonometric/complex ) expression
$f(n)$ ( e.g. $f(n) =(-1)^g(n)$ ) which gives me the sequence star... | By symmetry, we can assume
$$
f(n) =
a_1 \sin\frac{\pi n}{4}
+a_3 \sin\frac{3\pi n}{4}
+b_1 \cos\frac{\pi n}{4}
+b_3 \cos\frac{3\pi n}{4}.
$$
Then
$$
\begin{aligned}
f(1) &= \frac{a_1}{\sqrt{2}} +\frac{a_3}{\sqrt{2}} + \frac{b_1}{\sqrt{2}} -\frac{b_3}{\sqrt{2}} = -1\\
f(2) &= a_1 - a_3 = -1\\
f(3) &= \frac{a_1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Why is there at least a $50$% chance that greatest common divisor $(n,x-y)$ is a factor of $n$ not equal to $0$, $1$, or $n$? Given integers $x$, $y$, and $n$ such that $x^2\equiv y^2 ($mod$ n)$, there is at least a $50$% chance that greatest common divisor $(n,x-y)$ is a factor of $n$ not equal to $1$, $0$, or $n$. W... | The given statement is correct for numbers $n$ with at least two distinct odd prime factors. (to be more precise, if there is no primitive root modulo $n$ , but the case of $n$ odd and $n$ has two distinct prime factors is the most important in practice)
Suppose, $n$ is divisible by the odd distinct primes $p$ and $q$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Volume between cone and sphere - First octant Find the volume between $z=\sqrt{x^2+y^2}$ and the sphere $x^2+y^2+z^2=1$ that lies in the first octant using cylindrical coordinates.
So I found the intersection and got $r=\frac{\sqrt2}{2}$.
I know theta has to be between $0$ and $\frac{\pi}{2}$ but not sure about z
| Using Cylindrical coordinates:
$\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r}^{\sqrt{1-r^2}}rdzdrd\theta=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{2}}}{r(\sqrt{1-r^2}-r)}drd\theta=\int_{0}^{\frac{\pi}{2}}[-\frac{1}{3}(1-r^2)^{\frac{3}{2}}-\frac{r^3}{3}]_{0}^{\frac{1}{\sqrt{2}}}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate Point Coordinates As you can see, In the image a rectangle gets translated to another position in the coordinates System.
The origin Coordinates are A1(8,2) B1(9,3) from the length 7 and the height 3 you can also guess the vertices of the rectangle.
Now the Rectangle gets moved.
Now A1 is at A2(16,9) and B1 ... | Your transformation contains translation (2 parameter), rotation (1 parameter) and stretching, which I hope means scaling (1 parameter).
This in general is no linear but an affine transform, except for the case that the origin gets mapped to the origin, which I doubt here.
The transform would be like this, using homoge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate determinant with induction I need to prove the following, with induction to every $1 \leq n$:
$$D(a_1,...,a_n) = \left| \begin{array}{ccc}
a_1+x& a_2 & a_3 & \cdots & a_n \\
a_1& a_2+x & a_3 & \cdots & a_n \\
a_1& a_2 & a_3+x & \cdots & a_n \\
\vdots & \vdots & \vdots & & \vdots \\
a_1& a_2 & a_3 & \cdots ... | Developing with respect to the last row, after performing those elementary row operations (that don't change the determinant), you get
$$
D(a_1,\dots,a_n,a_{n+1})=\\
xD(a_1,\dots,a_n)+(-1)^{(n+1)+1}(-x)\det\begin{bmatrix}
a_2 & a_3 & \dots & a_n & a_{n+1} \\
a_2+x & a_3 & \dots & a_n & a_{n+1} \\
a_2 & a_3+x & \dots & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If an integer $n$ is chosen at random from $1$ to $96$ inclusive ,what is the probability that $n(n+1)(n+2)$ is divisible by 8? In this one If I consider that n is even then probability that the number $n(n+1)(n+2)$ will be divisible by 8 will be 1/2 ,Now if n is odd then for n(n+1)(n+2) to be divisible by 8 ,n+1 shoul... | the total numbers are $96$. for an odd number to be multiple of 8 using expression $n(n+1)(n+2)$ the number $n$ should be a number preceding to a multiple of $8$ starting from $8$ itself so first number is $7$ and such is an AP whose last term is $95$. So total terms which are $1$ less than a multiple of $8$ ie odd are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Convergence rate of the sequence $a_{n+1} = a_n-a_n^2, a_0=1/2$. The sequence converges to zero at a rate that seems to be slightly faster than $1/n$.
What are the best known results on the convergence rate of this sequence?
| Here is a slightly better bound:
first, it is more convenient to look at $b_n = 1/a_n$. We have $$\frac{1}{b_{n+1}}=\frac{1}{{b_n}}-\frac{1}{{b_n^2}}=\frac{b_n-1}{b_n^2}$$ or, $$b_{n+1}=\frac{b_n^2}{b_n-1}=\frac{(b_n+1)(b_n-1)+1}{b_n-1}=b_n+1+\frac{1}{b_n-1}.$$
It is easy to see that $b_k\geq 2$ for all $k$, which impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Define $a$ and $b$ rational numbers so they satisfy equation Define $a, b \in \mathbb{Q}$ so that
$$\frac{a}{\sqrt{7 + 4\sqrt{3}}} + \frac{b}{\sqrt{7 - 4\sqrt{3}}} = \sqrt{4 + 2\sqrt{3}}$$
Using $\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$ I got $\frac{a}{2 + \sq... | Let's use the completion of square as @Kay K: did.
$\sqrt{ 4 + 2 \sqrt{3}} = \sqrt{ \sqrt {3} ^ 2 + 2 \sqrt{3} + 1} = \sqrt{ (\sqrt{3} + 1)^2} = \sqrt{3} + 1$
$\sqrt{ 7 + 4 \sqrt{3}}= \sqrt{ 2^2 + 2\cdot 2 \sqrt{3} + \sqrt{3}^2} = \sqrt{ (2+ \sqrt{3})^2}= 2+ \sqrt{3}$
$\sqrt{ 7 - 4 \sqrt{3}}= \sqrt{ 2^2 - 2\cdot 2 \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.