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How to know that $a^3+b^3 = (a+b)(a^2-ab+b^2)$ Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?
| So I guess the answer to this question is to expand the right-hand side of the equation, although it might not be too clear on how to do this.
It might be clearer if you make the substitution $u=a+b$. Then the right hand side is:
\begin{align*}
(a+b)(a^2-ab+b^2) & = u(a^2-ab+b^2) \\
& = ua^2-uab+ub^2 \\
& = (a+b)a^2-(a+b)ab+(a+b)b^2 \\
& = a^3+ba^2-a^2b-ab^2+ab^2+b^3 \\
& = a^3+b^3
\end{align*}
using the distributive law (and commutivity).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/20301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off) $\begin{align*}
\lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}
&=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}}\\
&=\lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}\\
&=\lim_{x\to 0}\frac{2 \times 2 + 2\sqrt{4-x}-2\sqrt{4-x}-((\sqrt{4-x})(\sqrt{4-x})) }{2 \times 2x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
&=\lim_{x\to 0}\frac{4-4+x}{4x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
&=\lim_{x\to 0}\frac{x}{x(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
&=\lim_{x\to 0}\frac{1}{(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
&=\frac{1}{(4\sqrt{4+0} + 2\sqrt{4+0}\sqrt{4-0})}\\
&=\frac{1}{16}
\end{align*}$
wolframalpha says it's negative. What am I doing wrong?
| In between the fourth and fifth steps, you go from
$$\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}} \text{ to } \lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}$$
which is not correct. It should be
$$\lim_{x\to 0}\frac{(2-\sqrt{4+x})(2+\sqrt{4+x})}{(2x\sqrt{4+x})(2+\sqrt{4+x})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/22704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prime factorization of square numbers Let n be a natural number with unique prime factorization $p^m$... $q^k$ . Show that n can be written as a square if and only if all (m, ...k) are even
| If $$n = \prod_{i = 1}^\infty {p_i}^{\alpha_i} \in \mathbb Z,$$ where $p_i$ are the primes in order by $i$ and $\alpha_i$ are the corresponding exponents which may be $0$ as needed, we then have $$\sqrt n = \sqrt{\prod_{i = 1}^\infty {p_i}^{\alpha_i}} = \prod_{i = 1}^\infty {p_i}^{\frac{\alpha_i}{2}}.$$ But if any of the $\alpha_i$ are odd, then $\frac{\alpha_i}{2}$ is not an integer and neither is ${p_i}^{\frac{\alpha_i}{2}}$.
For example, consider $129600 = 2^6 \times 3^4 \times 5^2 \times 7^0 \times \ldots$. The square root is found thus: $\sqrt{129600} = 2^3 \times 3^2 \times 5^1 \times 7^0 \times \ldots$
Now compare $648000 = 2^6 \times 3^4 \times 5^3 \times 7^0 \times \ldots$. The square root is found thus: $$\begin{align}\sqrt{648000} & = 2^3 \times 3^2 \times 5^{\frac{3}{2}} \times 7^0 \times \ldots \\ & = 2^3 \times 3^2 \times 5 \sqrt 5 \\ & = 360 \sqrt 5 \\ & \approx 804.98447\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/23360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Why determinant of a 2 by 2 matrix is the area of a parallelogram? Let $A=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$.
How could we show that $ad-bc$ is the area of a parallelogram with vertices $(0, 0),\ (a, b),\ (c, d),\ (a+b, c+d)$?
Are the areas of the following parallelograms the same?
$(1)$ parallelogram with vertices $(0, 0),\ (a, b),\ (c, d),\ (a+c, b+d)$.
$(2)$ parallelogram with vertices $(0, 0),\ (a, c),\ (b, d),\ (a+b, c+d)$.
$(3)$ parallelogram with vertices $(0, 0),\ (a, b),\ (c, d),\ (a+d, b+c)$.
$(4)$ parallelogram with vertices $(0, 0),\ (a, c),\ (b, d),\ (a+d, b+c)$.
Thank you very much.
| For the matrix $\left[\begin{array}{cc}
a & c \\
b & d \\
\end{array}\right]$ let
$$A = \left[\begin{array}{c}
a \\
b \\
\end{array}\right]
\;\text{and}\; B = \left[\begin{array}{c}
c \\
d \\
\end{array}\right]$$
as shown in the following figure.
Then the height of the parallelogram is
$$\text{height} = |B|\sin\alpha = |B|\cos\beta.$$
If we rotate $A$ by 90 degrees in the CCW direction as follows:
$$R_{90ΒΊ}A =
\left[\begin{array}{cc}
0 &-1 \\
1 &0 \\
\end{array}\right]
\left[\begin{array}{c}
a \\
b \\
\end{array}\right] =
\left[\begin{array}{c}
-b \\
a \\
\end{array}\right],$$
maintaining the magnitude of the base as
$$\text{base} = |A| = |R_{90ΒΊ}A|,$$
then it is clear that the area of the parallelogram is therefore
$$
\text{base}\times\text{height}=(|A|)(|B|\sin\alpha) = |R_{90ΒΊ}A|\;|B|\cos\beta = (R_{90ΒΊ}A)\cdot B =
\left[\begin{array}{c}
-b \\
a \\
\end{array}\right]
\cdot
\left[\begin{array}{c}
c \\
d \\
\end{array}\right]
=
ad-bc.
$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/29128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "80",
"answer_count": 11,
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How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? How can I evaluate
$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$?
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.
In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
| I first encountered this sum with the following problem:
Evaluate
$$\bigg(\frac{1}{2}\bigg)^\dfrac{1}{3}\bigg(\frac{1}{4}\bigg)^\dfrac{1}{9}\bigg(\frac{1}{8}\bigg)^\dfrac{1}{27}\bigg(\frac{1}{16}\bigg)^\dfrac{1}{81}\dots$$
Which , of course simplified to
$$\bigg(\frac{1}{2}\bigg)^{\dfrac{1}{3^1}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+\dots}=\bigg(\frac{1}{2}\bigg)^S$$
Getting back to your problem, now
$$\sum_{n=1}^\infty \frac{2n}{3^{n+1}}=\frac{2}{3}\sum_{n=1}^\infty \frac{n}{3^n}=\frac{2}{3}S$$
Using a method similar to deriving geometric series suppose that
$$S_k = \sum_{n=1}^k \frac{n}{3^{n}}$$
Then we have
$$\begin{array}{lll}
3S_k-S_k &=& 1+\frac{2}{3^1}+\frac{3}{3^2}+\frac{4}{3^3}+\dots+\frac{k}{3^{k-1}}\\
&&-\frac{1}{3^1}-\frac{2}{3^2}-\frac{3}{3^3}\dots-\frac{k-1}{3^{k-1}}-\frac{k}{3^k}\\
2S_k&=&\bigg(1+\frac{2-1}{3^1}+\frac{3-2}{3^2}+\frac{4-3}{3^3}+\dots+\frac{k-(k-1)}{3^{k-1}}\bigg)-\frac{k}{3^k}\\
&=&\frac{1-(\frac{1}{3})^{k}}{1-\frac{1}{3}} - \frac{k}{3^k}\\
2S&=&\lim_{k\to\infty} \frac{1-(\frac{1}{3})^{k}}{1-\frac{1}{3}} - \frac{k}{3^k}\\
2S&=&\frac{1}{1-\frac{1}{3}}=\frac{3}{2}\\
\frac{2}{3}S&=&\frac{1}{2}\\
\end{array}$$
by a similar method one can show, that if the series converges, that
$$\sum_{n=0}^\infty (n+1)x^n = \frac{1}{(1-x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/30732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "438",
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All natural solutions of $2x^2-1=y^{15}$ How can I find all positive integers $x$ and $y$ such that $2x^2-1=y^{15}$?
PS. See here.
| $(1,1)$ is available by inspection. Otherwise, if you write $2x^2=y^{15}+1=(y+1)(y^2-y+1)(y^4-y^3+y^2-y+1)(y^8+y^7-y^5-y^4-y^3+y+1)$ (Thanks, Alpha) you can look at where the factors of $2x^2$ can come from. As the last three terms are odd, $y+1$ must have an odd numbers of factors of $2$. $y^4-y^3+y^2-y+1=(y^3-2y^2+3y-4)(y+1)+5$, so these two can share a 5, but no other prime. $y^2-y+1=(y-2)(y+1)+3$ so these can share a factor $3$ but no other prime. No other pairs can have common factors. Aside from $2, 3, 5$, all primes must divide each factor an even number of times and can only divide one of them. This gives you a lot of information on how the factors of $2x^2$ can be distributed. I think it is pretty unlikely there are any more solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/30935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Complete induction of $10^n \equiv (-1)^n \pmod{11}$ To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.
It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$
For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$
$n\Rightarrow n+1$
$$\begin{align*}
(-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.
| If $a \equiv b \left(\bmod m \right)$, then $a^n \equiv b^n \left(\bmod m \right)$ where $n \in \mathbb{N}$ and $a,b,m \in \mathbb{Z}$.
The reason being $a^n - b^n$ factors as $(a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + b^{n-1})$ and hence if $m | (a-b)$, $m | (a^n-b^n)$
If you want to prove by induction, first note that
if $a \equiv b \left(\bmod m \right)$ and $c \equiv d \left(\bmod m \right)$, then $a \times c \equiv b \times d \left(\bmod m \right)$.
This is true since $ac-bd = a(c-d) + d(a-b)$
Hence at the inductive step, if you have that $10^k \equiv (-1)^k \left(\bmod m \right)$, and we have that $10 \equiv (-1) \left(\bmod m \right)$ and from the above result we have
$$10^{k+1} \equiv (-1)^{k+1} \left(\bmod m \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/39882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 3
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Primitive polynomials of finite fields there are two primitive polynomials which I can use to construct $GF(2^3)=GF(8)$:
$p_1(x) = x^3+x+1$
$p_2(x) = x^3+x^2+1$
$GF(8)$ created with $p_1(x)$:
0
1
$\alpha$
$\alpha^2$
$\alpha^3 = \alpha + 1$
$\alpha^4 = \alpha^3 \cdot \alpha=(\alpha+1) \cdot \alpha=\alpha^2+\alpha$
$\alpha^5 = \alpha^4 \cdot \alpha = (\alpha^2+\alpha) \cdot \alpha=\alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1$
$\alpha^6 = \alpha^5 \cdot \alpha=(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha+1+\alpha^2+\alpha=\alpha^2+1$
$GF(8)$ created with $p_2(x)$:
0
1
$\alpha$
$\alpha^2$
$\alpha^3=\alpha^2+1$
$\alpha^4=\alpha \cdot \alpha^3=\alpha \cdot (\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1$
$\alpha^5=\alpha \cdot \alpha^4=\alpha \cdot(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1$
$\alpha^6=\alpha \cdot (\alpha+1)=\alpha^2+\alpha$
So now let's say I want to add $\alpha^2 + \alpha^3$ in both fields. In field 1 I get $\alpha^2 + \alpha + 1$ and in field 2 I get $1$. Multiplication is the same in both fields ($\alpha^i \cdot \alpha^j = \alpha^{i+j\bmod(q-1)}$. So does it work so, that when some $GF(q)$ is constructed with different primitive polynomials then addition tables will vary and multiplication tables will be the same? Or maybe one of presented polynomials ($p_1(x), p_2(x)$) is not valid to construct field (altough both are primitive)?
| The situation is not so different in a simpler context, the field of 5 elements, also known as the integers modulo 5. Whether $\alpha$ is $2$ or $3$, the field is $0,1,\alpha,\alpha^2,\alpha^3$, but whether $\alpha+\alpha+1=0$ depends on which $\alpha$ you choose.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/40326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Compound angle formula confusion I'm working through my book, on the section about compound angle formulae. I've been made aware of the identity $\sin(A + B) \equiv \sin A\cos B + \cos A\sin B$. Next task was to replace B with -B to show $\sin(A - B) \equiv \sin A\cos B - \cos A \sin B$ which was fairly easy. I'm struggling with the following though:
"In the identity $\sin(A - B) \equiv \sin A\cos B - \cos A\sin B$, replace A by $(\frac{1}{2}\pi - A)$ to show that $\cos(A + B) \equiv \cos A\cos B - \sin A\sin B$."
I've got $\sin((\frac{\pi}{2} - A) - B) \equiv \cos A\cos B - \sin A\sin B$ by replacing $\sin(\frac{\pi}{2} - A)$ with $\cos A$ and $\cos(\frac{\pi}{2} - A)$ with $\sin A$ on the RHS of the identity. It's just the LHS I'm stuck with and don't know how to manipulate to make it $\cos(A + B)$.
P.S. I know I'm asking assistance on extremely trivial stuff, but I've been staring at this for a while and don't have a tutor so hope someone will help!
| Update: Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$, you have
$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B).$$
Replacing $A$ by $\frac{\pi }{2}-A$ in
$$\sin (A-B)\equiv \sin A\cos B-\cos A\sin B,$$
gives
$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left( \frac{\pi }{2}-A\right)
\cos B-\cos \left( \frac{\pi }{2}-A\right) \sin B.$$
Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$ and $\cos (\frac{\pi }{2}-x)\equiv \sin x$, you have
$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B),$$
$$\sin \left( \frac{\pi }{2}-A\right) \equiv \cos A,$$
and
$$\cos \left( \frac{\pi }{2}-A\right) \equiv \sin A.$$
Hence
$$\cos (A+B)\equiv \cos A\cos B-\sin A\sin B.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/41133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Constructing idempotent matrices Is there a general method for constructing an idempotent matrix if we are given the values of the diagonal entries?
| The following should work "generically" (given a list of diagonal elements $d_i, i=1\ldots n$ where $\sum_{i=1}^n d_i = r$ is an integer, $1 \le r \le n-1$). Start with a diagonal matrix with $r$ $1$'s and $n-r$ $0$'s on the diagonal. First conjugate with a random matrix.
Then for $k$ from 1 to $n-1$, conjugate with a matrix that differs from the identity only in position $(k,k+1)$, chosen to make the $(k,k)$ element $d_k$.
For example, suppose we want a $3 \times 3$ idempotent matrix with diagonal elements $2,1,-1$. Start with $\left(\matrix{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 0\cr}\right)$ and conjugate with $\left(\matrix{12 & -2 & 6\cr 15 & -7 & 3\cr 3 & 1 & -3\cr}\right)$ to get $M_0 = \left(\matrix {2/3 & -1/9 & 1/3\cr -1/2 & 5/6 & 1/2\cr 1/2 & 1/6 & 1/2\cr}\right)$. The conjugate of $M_0$ by $\left(\matrix{1 & b & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr}\right)$ has $(1,1)$ element $2/3 + b/2$, which is $2$ for $b = 8/3$. Using this, we get $M_1 = \left(\matrix{2 & 3 & -1\cr
-1/2 & -1/2 & 1/2 \cr 1/2 & 3/2 & 1/2\cr}\right)$. The conjugate of $M_1$ by $\left(\matrix{1 & 0 & 0\cr 0 & 1 & b\cr 0 & 0 & 1\cr}\right)$ has $(2,2)$ element $-1/2 - 3b/2$, which is $1$ for $b = -1$. Using this, we obtain
$M_2 = \left( \matrix{2 & 3 & -4\cr 0 & 1 & 0\cr 1/2 & 3/2 & -1\cr} \right)$, which satisfies the requirements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/42283",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$ How do I show that:
$$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$$
This is actually problem B $4371$ given at this link. Looks like a very interesting problem.
My attempts: Well, I have been thinking about this for the whole day, and I have got some insights. I don't believe my insights will lead me to a $\text{complete}$ solution.
*
*First, I wrote $\sin\frac{5\pi}{14}$ as $\sin\frac{9 \pi}{14}$ so that if I put $A = \frac{\pi}{14}$ so that the given equation becomes, $$\frac{1}{\sin^{2}{A}} + \frac{1}{\sin^{2}{3A}} + \frac{1}{\sin^{2}{9A}} =24$$ Then I tried working with this by taking $\text{lcm}$ and multiplying and doing something, which appeared futile.
*Next, I actually didn't work it out, but I think we have to look for a equation which has roots as $\sin$ and then use $\text{sum of roots}$ formulas to get $24$. I think I haven't explained this clearly.
*
*$\text{Thirdly, is there a trick proving such type of identities using Gauss sums ?}$ One post related to this is: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ I don't know how this will help as I haven't studied anything yet regarding Gauss sums.
| This may be a $3$ year old question, but I would like to add to the list an answer that relies on the sum of $\tan^2$ identity.
Let $$\begin{align}S &= \frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}}\\
&=\frac{1}{\cos^{2}\frac{3\pi}{7}} + \frac{1}{\sin^{2}\frac{2\pi}{7}} + \frac{1}{\sin^{2}\frac{\pi}{7}}\\
&= \sec^2{\frac{3\pi}{7}} + \sec^2{\frac{2\pi}{7}} + \sec^2{\frac{\pi}{7}}\\
&= \tan^2\frac{\pi}{7} + \tan^2\frac{2\pi}{7} + \tan^2\frac{3\pi}{7} + 3\\
&= \sum_{k = 1}^3 \tan^2\frac{k\pi}{7} + 3\end{align}$$
From the sum of $\tan^2$ identity discussed over here, we have
$$
\sum_{k=1}^n\tan^2\frac{k\pi}{2n+1} = 2n^2+n,\quad n\in\mathbb{N}^+.
$$
In our case, set $n = 3$. Then,
$$\begin{align}S &= 2(3)^2 + 3 + 3 \\&=24 \end{align}$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/45144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
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linear algebra - equations
Determine real number(s)for $a,b$ such that the system has no solution, has a unique solution, and has more than one solution:
$$\begin{align*}
x-2y+az-t&=1\\
-x+y-z+t&=-1\\
(a+1)y-a^2z+at&=0\\
(b+1)y-abz-a^2t&=b
\end{align*}$$
I could not transform the matrix into row reduced form.So that I could not find $a$ and $b$. I will be glad if someone could solve it.
| Your matrix will have $a$ and $b$ indicated. You then have to be careful with them.
(I hope I didn't make any silly arithmetic or algebra mistakes; but even if I did, you should be able to see how one proceeds: doesn't matter what $a$ and $b$ are, so long as you are careful not to divide by an expression containing them, since you won't know ahead of time that what you are dividing by is not zero).
The augmented matrix for the system is:
$$\left(\begin{array}{rrrr|r}
1 & -2 & a & -1 & 1\\
-1 & 1 & -1 & 1 & -1\\
0 & a+1 & -a^2 & a & 0\\
0 & b+1 & -ab & -a^2 & b
\end{array}\right).$$
Proceed as usual. First, we add the first row to the second row:
$$\left(\begin{array}{rrrr|r}
1 & -2 & a & -1 & 1\\
0 & -1 & a-1 & 0 & 0\\
0 & a+1 & -a^2 & a & 0\\
0 & b+1 & -ab & -a^2 & b
\end{array}\right).$$
Then, multiply the second row by $-1$:
$$\left(\begin{array}{rrrr|r}
1 & -2 & a & -1 & 1\\
0 & 1 & 1-a & 0 & 0\\
0 & a+1 & -a^2 & a & 0\\
0 & b+1 & -ab & -a^2 & b
\end{array}\right).$$
Now, subtract $(a+1)$-times the second row from the third row; multiplying the second row by $-(a+1)$ gives $(0\quad -(a+1)\quad (a^2-1)\quad 0\>|\>0)$, so we have:
$$\left(\begin{array}{rrrr|r}
1 & -2 & a & -1 & 1\\
0 & 1 & 1-a & 0 & 0\\
0 & 0 & -1 & a & 0\\
0 & b+1 & -ab & -a^2 & b
\end{array}\right).$$
Multiply the third row by $-1$:
$$\left(\begin{array}{rrrr|r}
1 & -2 & a & -1 & 1\\
0 & 1 & 1-a & 0 & 0\\
0 & 0 & 1 & -a & 0\\
0 & b+1 & -ab & -a^2 & b
\end{array}\right).$$
Now subtract $(b+1)$-times the second row from the fourth row. Multiplying the second row by $-(b+1)$ gives $(0\quad (b+1)\quad (ab+a-b-1)\quad 0\>|\>0)$, so we get:
$$\left(\begin{array}{rrrr|r}
1 & -2 & a & -1 & 1\\
0 & 1 & 1-a & 0 & 0\\
0 & 0 & 1 & -a & 0\\
0 & 0 & a-b-1 & -a^2 & b
\end{array}\right).$$
Now, subtract $(a-b-1)$-times the third row from the fourth row. Multiplying the third row by $-(a-b-1)$ gives $(0\quad 0\quad (1+b-a)\quad (a^2-ab-a)\>|\>0)$, so we get:
$$\left(\begin{array}{rrrr|r}
1 & -2 & a & -1 & 1\\
0 & 1 & 1-a & 0 & 0\\
0 & 0 & 1 & -a & 0\\
0 & 0 & 0 & -(ab+a) & b
\end{array}\right).$$
Now we can figure out the rank easily enough.
If $-(ab+a) = -a(b+1)\neq 0$, then the coefficient matrix has rank $4$ and the augmented matrix has rank $4$, so the system has exactly one solution.
If, however, $-a(b+1)=0$ and $b\neq 0$, then the coefficient matrix would have rank $3$ and the augmented matrix would have rank $4$, so that would be no solutions.
And, finally, if $-a(b+1)=0$ and $b=0$, then both the coefficient and augmented matrices have rank $3$, so the system would have an infinite number of solutions.
So now, it's just a matter of determining what are the conditions on $a$ and $b$ that give you those three possibilities. Suggestion: Deal with them in the inverse order I listed them: first the case where $-a(b+1)=0$ and $b=0$, because you'll be able to say exactly what $a$ and $b$ are; then consider the middle case; and finally the first mentioned case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/46157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Exponents in the denominator? I'm having trouble understanding exponents in the denominator.
For example: I have the expression: $\displaystyle 1 - \frac{1}{3^n} + \frac{2}{3^{n+1}}$.
I know that this simplifies to $\displaystyle 1 - \frac{1}{3^{n+1}}$, but how/why? Can someone please list the steps?
My understanding is that the exponent $(n+1)$ in the expression $x^{n+1}$ means that $x^{n+1} = x x^n$, but how does this fit with the above problem?
|
My understanding is that the exponent $n+1$ means that $x^{n+1}=x\cdot x^n$
Your understanding is correct, and you can apply it to the problem at hand by remembering that when an exponent, say $n$, of an expression is in the denominator of a fraction, it can be written as an expression raised to the $-n$ power. Then you can manipulate exponents by the respective rules to which you are accustomed. $$\frac{1}{x^{n+1}} = x^{-(n+1)} = x^{-n-1} = x^{-1}x^{-n} = \frac {1}{x\cdot x^n} < \frac{1}{x^n} = x^{-n}$$(Note: the inequality holds, as stated, provided $ n > 1$.)
That is, so we have that the "least common denominator" is $x\cdot x^n$, with $$\frac{1}{x^n} = x^{-n} = x^1\cdot x^{-n - 1} = \frac{x}{x\cdot x^n} = \frac {x}{x^{n+1}}$$
In your problem, then $$1-\frac{1}{3^n}+\frac{2}{3^{n+1}} = 1 - 3^{-n} + 2\cdot 3^{-(n+1)} = 1 - 3\cdot 3^{-n - 1} + 2\cdot 3^{-n - 1} $$ $$= 1 + (-3 + 2)\cdot 3^{-n-1} = 1 - 3^{-(n+1)} = 1 - \frac{1}{3^{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/47180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Deriving the rest of trigonometric identities from the formulas for $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, and $\cos (A-B)$ I am trying to study for a test and the teacher suggest we memorize $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, $\cos (A-B)$, and then be able to derive the rest out of those. I have no idea how to get any of the other ones out of these, it seems almost impossible. I know the $\sin^2\theta + \cos^2\theta = 1$ stuff pretty well though. For example just knowing the above how do I express $\cot(2a)$ in terms of $\cot a$? That is one of my problems and I seem to get stuck half way through.
| I'm not certain, but I think perhaps you need to revisit the basic trigonometric definitions.
$$\begin{array}{lll}
\sin\theta = \frac{opposite}{hypotenuse}&\csc\theta=\frac{hypotenuse}{opposite}\\
\cos\theta = \frac{adjacent}{hypotenuse}&\sec\theta=\frac{hypotenuse}{adjacent}\\
\tan\theta = \frac{opposite}{adjacent}&\cot\theta=\frac{adjacent}{opposite}
\end{array}$$
Having done that, we can now manipulate the definitions. For example
$$\cot\theta = \frac{adjacent}{opposite}\cdot\frac{\frac{1}{hypotemnuse}}{\frac{1}{hypotemnuse}}=\frac{\frac{adjacent}{hypotenuse}}{\frac{opposite}{hypotenuse}}=\frac{\cos\theta}{\sin\theta}$$
or alternatively
$$\cot\theta = \frac{adjacent}{opposite}=\frac{\frac{1}{opposite}}{\frac{1}{adjacent}}=\frac{\frac{1}{opposite}}{\frac{1}{adjacent}}\cdot\frac{hypotenuse}{hypotenuse}=\frac{\frac{hypotenuse}{opposite}}{\frac{hypotenuse}{adjacent}}=\frac{\csc\theta}{\sec\theta}$$
After that, move on to making 3 triangles that are similar to the adjacent-opposite-hypotenuse triangle. These triangles will rotate which side is equal to 1. This is where we get the familiar Pythagorean identities.
The next step is to derive $\cos (\alpha+\beta)$ and $\sin (\alpha+\beta)$ by setting the lengths (I would use the squares of the lengths though) of the green lines in the diagram equal to each other. It's easy to do, since the endpoints are well known.
Finally, lets derive $\cot(A+B)$
$$\begin{array}{lll}
\cot(A+B)&=&\frac{\sin(A+B)}{\cos(A+B)}\\
&=&\frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\\
&=&\frac{\frac{\sin A\cos B + \cos A\sin B}{\sin A\sin B}}{\frac{\cos A\cos B - \sin A\sin B}{\sin A\sin B}}\\
&=&\frac{\frac{\cos B}{\sin B}+\frac{\cos A}{\sin A}}{\frac{\cos A\cos B}{\sin A\sin B}-1}\\
&=&\frac{\cot B+\cot A}{\cot A\cot B-1}
\end{array}$$
Note that $\cot 2A$ is just $\cot(A+B)$ where $A=B$, so
$$\cot 2A = \cot(A+A)=\frac{\cot A+\cot A}{\cot A\cot A-1}=\frac{2\cot A}{\cot^2 A-1}$$
Of course this list isn't comprehensive, for example we didn't derive the sine and cosines laws, nor did we consider the double angle and half angle formulas. But I hope that what little I did present here was helpful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/48938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
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} |
Why can ALL quadratic equations be solved by the quadratic formula? In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.
Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?
| I'll just let the algebra speak for itself.
For $ a \ \ne \ 0 $
If
$ax^2 = bx + c $
Then
$ x = \frac{b \ \pm \sqrt{b^2 \ + \ 4ac}}{2a} $
PROOF:
$ 4aax^2 = 4abx + 4ac $
$(2ax)^2 + b^2 = 4abx + 4ac + b^2 $
$(2ax)^2 - 4abx + b^2 = b^2 + 4ac $
$ (2ax - b)^2 = b^2 + 4ac $
$ 2ax - b = \pm \sqrt{b^2 + 4ac} $
$2ax = b \pm \sqrt{b^2 + 4ac} $
$ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $
Q.E.D.
SHORTCUT
Using the standard full definition , replace b with -b and c with -c to get the formula I derived in this post.
For $ a \ne 0 $
If
$ ax^2 + bx + c = 0 $
Then
$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Make the replacements in the 'if'
$ ax^2 + (-b)x + (-c) = 0 $
$ ax^2 = bx + c $
Make the replacements in the 'then'
$x = \frac{-(-b) \pm \sqrt{(-b)^2 - 4a(-c)}}{2a} $
$ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/49229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "294",
"answer_count": 22,
"answer_id": 2
} |
Property of an ellipse I need proof for the following question. Also, I want to know, can we apply the same for other conics. If yes, where and when... Please explain.
Show that there exists a point K on the major axis of E , having the property that for any chord $\overline{PQ}$ passing through K, $\dfrac{1}{PK^2} + \dfrac{1}{QK^2}$ is a constant. Also Show that $\lim_{e \to \infty}K = (2a, 0)$
| We use a brute force approach. Let our ellipse have equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
where $a \ge b >0$.
Let $K=(p,0)$ be a point (if there is one) such that $1/(PK)^2+1/(QK)^2$ is the same for all chords $PQ$ through $K$.
We will, regrettably, cheat, by (sort of) looking up the answer. Suppose that there is such a magic point $K=(p,0)$. Consider the horizontal and vertical chords through $K$. For the horizontal chord, the two relevant squares of distances are
$(a+p)^2$ and $(a-p)^2$. For the vertical chord, we calculate, by substituting $x=p$ in the equation of the ellipse. The two relevant squares of distances turn out to be each $(b^2/a^2)(a^2-p^2)$.
In order to satisfy our inverse square condition, we need
$$\frac{1}{(a+p)^2}+\frac{1}{(a-p)^2}=\frac{2a^2}{b^2(a^2-p^2)}.$$
Solving for $p^2$ turns out to be easy:
$$p^2=\frac{a^2(a^2-b^2)}{a^2+b^2}.$$
We have found the only possible candidates for $p$. Now we need to verify that that we get constant sum of inverse square distances for all chords through $(p,0)$.
The generic line passing through $(p,0)$ has equation of the form $y=m(x-p)$. It is somewhat more convenient to rewrite this as
$ny=x-p$, where $n$ is the reciprocal of the slope. One line through $(p,0)$ gets missed. We hope it doesn't feel bad.
Let $(x_1,y_1)$ and $(x_2,y_2)$ be the two points where the line $ny=x-p$ meets the ellipse. We want, with $p$ as determined above, to show that
$$\frac{1}{(x_1-p)^2+y_1^2}+ \frac{1}{(x_2-p)^2+y_2^2}$$
does not depend on $n$.
Since $(x_i-p)^2=n^2y_i^2$, our condition reduces to showing that
$$\frac{1}{n^2+1}\left(\frac{1}{y_1^2}+\frac{1}{y_2^2}\right)$$
is independent of $n$. Equivalently, we want to show that
$$\frac{1}{y_1^2}+\frac{1}{y_2^2}\qquad\text{or equivalently}\qquad \frac{(y_1+y_2)^2-2y_1y_2}{y_1^2y_2^2}$$
is a constant times $n^2+1$.
So now we want (sort of) to find $y_1$ and $y_2$. The ellipse has equation $b^2x^2+a^2y^2=a^2b^2$. But since $ny=x-p$, by substituting, we arrive at the equation
$$b^2(ny+p)^2 +a^2y^2=a^2b^2.$$
Temporarily, rewrite this as $Ay^2+By+C$. Since $y_1+y_2=-B/A$, and $y_1y_2=C/A$, we find that
$$\frac{(y_1+y_2)^2-2y_1y_2}{y_1^2y_2^2} =\frac{B^2-2AC}{C^2}.$$
Now we need to compute. We have
$$\frac{B^2-2AC}{C^2}=\frac{4b^4n^2p^2-2(b^2n^2+a^2)(b^2p^2-a^2b^2)}{(b^2p^2-a^2b^2)^2}.$$
The expression $a^2-p^2$ that occurs a couple of times in the above equation simplifies to $2a^2b^2/(a^2+b^2)$. The denominator is constant. After not much work, the numerator simplifies to $(n^2+1)(4b^4a^4)/(a^2+b^2)$, a constant times $n^2+1$, and we are finished.
At least temporarily. The parabola has obviously only one point of the above type, namely the vertex. (For fans of points at infinity, there is another one.)
It would be interesting to look for a synthetic proof. There may be a simple one, since the expression for $p^2$ has obvious geometric significance.
Added: What happens when eccentricity becomes large.
Recall that
$$p^2=\frac{a^2(a^2-b^2)}{a^2+b^2}.$$
Now fix $a$, and let the eccentricity get large. As the eccentricity gets large, $b^2$ approaches $0$.
Note that
$$p^2=a^2\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}$$
(we have divided top and bottom by $a^2$.)
It follows that
$$\lim_{b^2\to 0} p^2=a^2.$$
Thus if we start with the standard ellipse, it is not true that $K$ approaches $(2a,0)$. All we can say is that $p$ is then awfully close to $(a,0)$ or $(-a,0)$, either end of the major axis. If the point $K$ moves smoothly as $b^2$ decreases, we can assert a bit more, that $K$ approaches $(a,0)$ or $(-a,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/52359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How $x^2 + 5x + 6 \ge 0$ implies $β3 \ge x \ge β2$? How $x^2 + 5x + 6 \ge 0$ implies $β3 \ge x \ge β2$ ?
I represented $x^2 + 5x + 6 \ge 0$ as $(x+3)(x+2) \ge 0$,which means that either $(x+3) \ge 0$ and $(x+5) \ge 0$ which giving $x \ge -3$ or $x \ge -2$ but how does this gives $β3 \ge x \ge β2$?
ADDED: The above expression could also mean that $(x+3) \le 0$ and $(x+2) \le 0$ which will give $x \le -3$ and $x \le -2$,now should combine this two to get that result? but in that case we would get both $β3 \le x \le β2$ and $β2 \le x \le β3$ (if I am not wrong) but then the first one should be takes as the answer is taken as the second one makes no sense,am I right?
Source of the problem.
Check the solution given there which causes the confusion.
| $x^{2}+5x+6=0$, Solutions are $x=-3,x=-2.$
Since $x^{2}+5x+6=(x+3)(x+2)$, we have three cases:
*
*If $x\geq -2$, then $x+3\geq 1$, $x+2\geq 0$, so $x^{2}+5x+6\geq 0.$
*If $x\leq -3$, then $x+3\leq 0$, $x+2\leq -1$, so $x^{2}+5x+6\geq 0.$
*If $-3<x<-2$, then $x+3>0$, $x+2<0$, so $x^{2}+5x+6<0.$
Remark: instead of "$x^2 + 5x + 6 \ge 0$ implies $β3 \ge x \ge β2$", we should say that "$x^2 + 5x + 6 \le 0$ is equivalent to $β3 \le x \le β2$". The double inequality $β3 \ge x \ge β2$ is a compact form of $β3 \ge x$ and $x \ge β2$, which is not satisfied by any real $x$.
| {
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"url": "https://math.stackexchange.com/questions/53396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Construction of generating function from identity I am trying to solve identity involving binomials and Fibonacci numbers by using generating functions:
$$\sum_{k=0}^n{n \choose k}{n+k\choose k}f_{k+1}=\sum_{k=0}^n{n \choose k}{n+k\choose k}(-1)^{n-k}f_{2k+1}$$
My computational approach by Mathematica lead me to derive this generating function:
$$\frac{\sqrt{3x^2-2x+3+2\sqrt{x^4-8x^3-2x^2-8x+1}}}{\sqrt{5}\sqrt{x^4-8x^3-2x^2-8x+1}}$$
Can someone show how to transform both or any of the identity sides to obtain (coefficiens of) this generating function.
| Here is a proof using complex variables. We seek to show that
$$\sum_{k=0}^n {n\choose k} {n+k\choose k} F_{k+1}
=\sum_{k=0}^n {n\choose k} {n+k\choose k} (-1)^{n-k} F_{2k+1}.$$
Start from
$${n+k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=1} \frac{1}{z^{k+1}} (1+z)^{n+k} \; dz.$$
This yields the following expression for the sum on the LHS
$$\frac{1}{2\pi i}
\int_{|z|=1}
\sum_{k=0}^n {n\choose k} \frac{1}{z^{k+1}} (1+z)^{n+k}
\frac{\varphi^{k+1} - (-1/\varphi)^{k+1}}{\sqrt{5}} \; dz$$
This simplifies to
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|z|=1} \frac{(1+z)^n}{z}
\sum_{k=0}^n {n\choose k}
\left(\varphi\left(\varphi\frac{1+z}{z}\right)^k
+\frac{1}{\varphi}\left(-\frac{1}{\varphi}\frac{1+z}{z}\right)^k
\right)\; dz$$
This finally yields
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|z|=1} \frac{(1+z)^n}{z}
\left(
\varphi\left(1+\varphi\frac{1+z}{z}\right)^n
+\frac{1}{\varphi}\left(1-\frac{1}{\varphi}\frac{1+z}{z}\right)^n
\right) \; dz$$
or
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|z|=1} \frac{(1+z)^n}{z^{n+1}}
\left(
\varphi\left(z+\varphi(1+z)\right)^n
+\frac{1}{\varphi}\left(z-\frac{1}{\varphi}(1+z)\right)^n
\right) \; dz$$
Continuing we have the following expression for the sum on the RHS
$$\frac{1}{2\pi i}
\int_{|z|=1}
\sum_{k=0}^n {n\choose k} (-1)^{n-k} \frac{1}{z^{k+1}} (1+z)^{n+k}
\frac{\varphi^{2k+1} - (-1/\varphi)^{2k+1}}{\sqrt{5}} \; dz$$
This simplifies to
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|z|=1} \frac{(1+z)^n}{z}
\\ \times \sum_{k=0}^n {n\choose k} (-1)^{n-k}
\left(\varphi\left(\varphi^2\frac{1+z}{z}\right)^{k}
+\frac{1}{\varphi}\left(\frac{1}{\varphi^2}\frac{1+z}{z}\right)^{k}
\right)\; dz$$
This finally yields
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|z|=1} \frac{(1+z)^n}{z}
\left(
\varphi\left(-1+\varphi^2\frac{1+z}{z}\right)^n
+\frac{1}{\varphi}\left(-1+\frac{1}{\varphi^2}\frac{1+z}{z}\right)^n
\right) \; dz$$
or
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|z|=1} \frac{(1+z)^n}{z^{n+1}}
\left(
\varphi\left(-z+\varphi^2(1+z)\right)^n
+\frac{1}{\varphi}\left(-z+\frac{1}{\varphi^2}(1+z)\right)^n
\right) \; dz$$
Apply the substitution $z=1/w$ to this integral to obtain
(the sign to correct the reverse orientation of the circle is
canceled by the minus on the derivative)
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|w|=1} \left(1+\frac{1}{w}\right)^n w^{n+1}
\\ \times \left(
\varphi\left(-\frac{1}{w}+\varphi^2(1+\frac{1}{w})\right)^n
+\frac{1}{\varphi}
\left(-\frac{1}{w}+\frac{1}{\varphi^2}(1+\frac{1}{w})\right)^n
\right) \frac{1}{w^2} \; dw$$
which is
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|w|=1} \left(1+\frac{1}{w}\right)^n \frac{1}{w}
\\ \times \left(
\varphi\left(-1+\varphi^2(w+1)\right)^n
+\frac{1}{\varphi}
\left(-1+\frac{1}{\varphi^2}(w+1)\right)^n
\right) \; dw$$
which finally yields
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i}
\int_{|w|=1} \frac{(1+w)^n}{w^{n+1}}
\\ \times \left(
\varphi\left(-1+\varphi^2(w+1)\right)^n
+\frac{1}{\varphi}
\left(-1+\frac{1}{\varphi^2}(w+1)\right)^n
\right) \; dw$$
This shows that the LHS is the same as the RHS
because
$$-1 + \varphi^2(w+1) = -1 + (1+\varphi)(w+1)
= w + \varphi(w+1)$$
and
$$-1+\frac{1}{\varphi^2}(w+1)
= -1 + (1-\frac{1}{\varphi}) (w+1)
\\ = -1 + (w+1) - \frac{1}{\varphi} (w+1)
= w - \frac{1}{\varphi} (w+1).$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
| {
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"url": "https://math.stackexchange.com/questions/53830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
| Here's the equations typed out if you want save time with writing it yourself.
(x/7)^2*SQRT(ABS(ABS(x)-3)/(ABS(x)-3))+(y/3)^2\*SQRT(ABS(y+3*SQRT(33)/7)/(y+3*SQRT(33)/7))-1=0
ABS(x/2)-((3*SQRT(33)-7)/112)*x^2-3+SQRT(1-(ABS(ABS(x)-2)-1)^2)-y=0
9*SQRT(ABS((ABS(x)-1)*(ABS(x)-0.75))/((1-ABS(x))*(ABS(x)-0.75)))-8*ABS(x)-y=0
3*ABS(x)+0.75*SQRT(ABS((ABS(x)-0.75)*(ABS(x)-0.5))/((0.75-ABS(x))*(ABS(x)-0.5)))-y=0
2.25*SQRT(ABS((x-0.5)*(x+0.5))/((0.5-x)*(0.5+x)))-y=0
(6*SQRT(10))/7+(1.5-0.5*ABS(x))*SQRT(ABS(ABS(x)-1)/(ABS(x)-1))-((6*SQRT(10))/14)*SQRT(4-(ABS(x)-1)^2)-y=0
Also: http://pastebin.com/x9T3DSDp
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "466",
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Rule for multiplying rational numbers The rule for multiplying rational numbers is this:
$\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$
Can the rule be proven or is it meant to be taken as a given?
Edit: Where $b\neq 0$ and $d\neq 0$.
| An approach to make it more clear might be to seperate each rational into a product of its parts ie. $\frac {a}{b}= \frac{a}{1} \cdot \frac{1}{b}$ and $\frac {c}{d}= \frac{c}{1} \cdot \frac{1}{d}$ then use the commutative property to group the "numerator" fractions and the "denominator" fractions seperately: $\frac{a}{b}\cdot \frac{c}{d}=(\frac{a}{1} \cdot \frac{c}{1})\cdot (\frac{1}{b} \cdot \frac{1}{d})=\frac{ac}{1}\cdot\frac{1}{bd}=\frac{ac}{bd}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many triplets of real numbers $(x, y, z)$ which satisfy these $3$ restriction: How many triplets of real numbers $(x, y, z)$ which satisfy :
$$(x + y)^3 = z$$
$$(y + z)^3 = x$$
$$(z + x)^3 = y$$
I need some approaches for solving this problem.
| Look first for solutions of the shape $x=y=z=a$. Our equations all reduce to
$(2a)^3=a$, which has the solutions $a=0$ and $a=\pm 2^{-3/2}$.
Now look for solutions where not all the variables are equal. For definiteness, look for solutions with $x \lt z$.
If $x \lt z$, then $x+y \lt y+z$ and therefore $(x+y)^3 \lt (y+z)^3$. From the first two equations, it follows that $z \lt x$, which is impossible.
Thus there are $3$ triples that satisfy the system of equations.
One can write up the same idea by starting from $x \le z$ and concluding that $z\le x$, which shows that $x=z$. So for a solution, we must have $x=y=z$.
Other approaches: We sketch a more "algebraic" approach which happens to be more work. From the first two equations, we obtain
$$(x+y)^3-(y+z)^3=z-x.$$
Let $Z=x+y$ and $X=y+z$. Factoring the expression $Z^3-X^3$ on the left, we obtain
$$(Z-X)(Z^2+ZX+X^2)= (x-z)(Z^2+ZX+X^2)=z-x.$$
If $z \ne x$, this forces $Z^2+ZX+X^2=-1$. But this last equation does not have real solutions. There are many ways to see this, such as the Quadratic Formula. A cuter way is to note that
$$4(Z^2 +ZX+X^2)=(2Z+X)^2+3X^2 \ge 0.$$
So we must have $z=x$. Similarly, $z=y$, and we end up looking for solutions of $(2a)^3=a$.
| {
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| $$n^3=S_n-S_{n-1}=\left(\frac{n(n+1)}2\right)^2-\left(\frac{n(n-1)}2\right)^2=n^2\left(\frac{n+1}2-\frac{n-1}2\right)\left(\frac{n+1}2+\frac{n-1}2\right)\\
=n^2\cdot1\cdot n.$$
This is the inductive step. The rest is easy.
| {
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Is $\frac{1}{(\textrm{can}-a-b\cdot i)^2}-\frac{1}{(\textrm{can}+\textrm{cod}-a-b \cdot i)^2}$ in any way simplifiable with Maple? So I wonder if Maple can reduce such lines like $$\frac{1}{(\textrm{can}-a-b\cdot i)^2}-\frac{1}{(\textrm{can}+\textrm{cod}-a-b \cdot i)^2}$$ (assuming all variables but $i$ are real)?
| Writing $x = can - a - i\cdot b$ and $y = cod$, your expression is equivalent to
$$\frac{1}{x^2} - \frac{1}{(x + y)^2}$$
You could try reducing it to a single fraction:
$$\frac{1}{x^2} - \frac{1}{(x + y)^2} = \frac{(x + y)^2 - x^2}{x^2 (x + y)^2} = \frac{2xy + y^2}{x^2 (x + y)^2} = \frac{y(2x + y)}{x^2 (x + y)^2}$$
I guess you'll have to decide which one looks cleanest or works best for your purposes, but it will not get much better than what you had originally.
| {
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Product of $ |z^k - 1| $ Problem: Prove the following identity about the product involving the nth roots of unity:
$$
\prod_{k=1}^{N-1}|z^k-1| = N
$$
where $ z^k $ is the primitive nth root of unity.
Attempt:
$$
\begin{align}
\prod_{k=1}^{N-1}|z^k-1| &= \prod_{k=1}^{N-1}\left|(\cos(\frac{2\pi k}{N})-1)+i\sin(\frac{2\pi k}{N})\right| \\
&=\prod_{k=1}^{N-1}\sqrt{\cos^2(\frac{2\pi k}{N})-2\cos(\frac{2\pi k}{N})+1+\sin^2(\frac{2\pi k}{N})} \\
&=\prod_{k=1}^{N-1}\sqrt{2-2\cos(\frac{2\pi k}{N})} \\
&=\prod_{k=1}^{N-1}2\sqrt{\frac{1}{2}-\frac{1}{2}\cos(\frac{2\pi k}{N}))} \\
&=2^{N-1}\prod_{k=1}^{N-1}\sin(\frac{k\pi}{N})
\end{align}
$$
I found on Wikipedia that there is an identity for the last product: $ \prod_{k=1}^{N-1}\sin(\frac{k\pi}{N}) = N/2^{N-1} $. However I do not know how to prove it.
Could someone help me prove the last identity or perhaps suggest a different approach to the problem?
| Aha, I just solved it:
First consider the polynomial $$ \prod_{k=0}^{N-1}(x-z^k) $$
The roots of the polynomial are the nth roots of unity, which are precisely the roots of the polynomial $ x^n-1 $ and so the two are equal.
Dividing both sides by $ x-1 $, we get
$$ \prod_{k=1}^{N-1}(x-z^k) = 1+x+\dots+x^{N-1} $$
Substituting $ x=1 $, we get that the product equals $ N $.
The product of the magnitudes is simply the magnitude of the product, so we get the desired result $$ \prod_{k=1}^{N-1}|1-z^k| = N $$
| {
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Find x to keep the equality $\sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}$ $$\mbox{ Find }x \in \mathbb{Q} \mbox{ to keep the equality: } \sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}$$
I tried to write the roots using powers:
\begin{align*}\sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}&\Rightarrow
[(-2x)^{\frac{3x}{11-4x}}]^{\frac{1}{2x+1}}=(7x+2)^{\frac{1}{3x-1}}\\
&\Rightarrow (-2x)^{\frac{\frac{3x}{11-4x}}{2x+1}}=(7x+2)^{\frac{1}{3x-1}}\\
&\Rightarrow (-2x)^{\frac{3x}{(11-4x)(2x+1)}}=(7x+2)^{\frac{1}{3x-1}}\\
&\Rightarrow (-2x)^{\frac{3x}{-8x^2+18x+11}}=(7x+2)^{\frac{1}{3x-1}}
\end{align*}
I hope I did it right until this point. But I've stuck here. Can someone help me?
Thanks.
| I agree with Peter Taylor.
You get:
\begin{equation}
\frac{3x}{(2x+1)(11-4x)}\ln(-2x) = \frac{1}{3x-1}\ln(7x+2)
\end{equation}
Graphing gives about -0.15649: http://bit.ly/q5Rn4l
I don't think you can solve it exactly. A rational solution is possible but unlikely in general. Maybe you're meant to just try a whole bunch of rational numbers...
| {
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Show that $n + 2$ and $n^2 + n + 1$ cannot both be perfect cubes
Question: If $n$ is a nonnegative integer, prove that $n + 2$ and $n^2 + n + 1$ cannot both be perfect cubes.
Possible solution: Suppose $n+2$ and $n^2 + n + 1$ are perfect cubes, their product $(n+2)(n^2 + n + 1)$ must also be a perfect cube.
However, note that $(n+2)(n^2 + n + 1)=n^3 + 3n^2 + 3n + 2 = (n + 1)^3 + 1^3$
By Fermat's Last Theorem, $a^n + b^n \neq c^n $ if $a,b,c,n$ are positive integers and $n>2$, therefore $a^3 + b^3 \neq c^3$ and $(n + 1)^3 + 1^3$ cannot be a perfect cube (can't be expressed in the form $c^3$ where $c$ is a positive integer)
I'm looking for alternative methods of solution, and some verification that the above proof is correct.
| Except for $-1$, $0$ and $1$, the distance between consecutive perfect cubes is always greater than one. This is enough to conclude that $(n+1)^3+1$ is not a perfect cube when $n$ is nonnegative. (No need to invoke Fermat.)
| {
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Why is the math for negative exponents so? This is what we are taught:
$$5^{-2} = \left({\frac{1}{5}}\right)^{2}$$
but I don't understand why we take the inverse of the base when we have a negative exponent. Can anyone explain why?
| $$
\begin{align}
a\cdot 3^{10} & = a\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3 \\ \\
a\cdot 3^6 & = a\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3
\end{align}
$$
To go from the second line to the first, multiply by 3 four times.
To go from the first to the second, multiply by 3 minus four times.
$10 = 6 + 4$
$6 = 10 + (-4)$
$$
\begin{align}
a\cdot 3^{10} & = a\cdot 3^{6+4} \\ \\
a\cdot 3^6 & = a\cdot 3^{10 + (-4)}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Where do I fail with this integral (coordinates on a simplex)? Let us regard the standard simplex in two dimensions
$S = \{ x_1 + x_2 \leq 1, x_1 \geq 0, x_2 \geq 0 \}$
Where does the following calculation fail:
$ \int_S x_1^2 = \int_0^1 x_1^2 \int_0^{1-x_1} dx_2 dx_1
= \int x_1^2 ( 1 - x_1 ) dx_1 = \int_0^1 x_1^2 - x_1^3 dx_1
= \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
while
$ \int_S x_1^2 = \int_0^1 \int_0^{1-x_2} x_1^2 dx_1 dx_2
= \int_0^1 \dfrac{(1-x_2)^3}{3} dx_2 = \frac{1}{3}\sum^3_{i=0}\int_0^1 x_2^i dx_2 = \frac{1}{3}\sum^3_{i=0} \dfrac{1}{i+1} dx_2$
$ = \frac{1}{3}( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} ) = \frac{7}{36}$
I suppose the error is the error is in the first equality signs (except I have done some very dumb mistake that I am blind to.
| There is an oversight in the second calculation.
You wanted
$$\int \frac{(1-x_2)^3}{3}\,dx_2,$$
and decided to expand $(1-x_2)^3$ and integrate term by term.
Note that by the Binomial Theorem, or otherwise,
$$(1+t)^3=1+3t+3t^2+t^3,$$
so $(1-x_2)^3=1-3x_2 +3x_2^2-x_2^3$.
Expanding is in any case an inefficient way to evaluate the integral. Instead, make the substitution $u=1-x_2$. Then our integral becomes
$$\int_1^0 -\frac{u^3}{3}\,du.$$
| {
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Which MΓΆbius transformations send 0, 1 and infinity to 0, 1 and infinity
Possible Duplicate:
How do I find out the symmetry of a function?
Let $f:\mathbf{P}^1 \longrightarrow \mathbf{P}^1$ be a MΓΆbius transformation $z\mapsto (az+b)/(cz+d)$ sending $\{0,1,\infty\}$ to $\{0,1,\infty\}$ with $ad-bc = 1$.
I suspect there are only a finite number of such MΓΆbius transformations. What are these?
A non-trivial example is $f(z) = 1/(1-z)$. It sends $0$ to $1$, $1$ to $\infty$ and $\infty$ to $0$. Note that $f(z) = -z+1$ is not an example, because $ad-bc = -1$ in this case.
| If $0\mapsto 0$, then $b=0$. This means that $d=\frac{1}{a}$.
*
*If $f(1)=1$ and $f(\infty)=\infty$, then we must have $a=c+d$ and $c=0$, so $ad=1$, $a=1$. Hence $a=d=\pm 1$, $b=c=0$. The only transformation is the identity.
*If $f(1)=\infty$ and $f(\infty)=1$, then we must have $a=c$ and $c+d=0$; since $d=\frac{1}{a}= -c = -a$, we have $-a^2=1$, so $a=c=\pm i$, $d=\mp i$.
If $0\mapsto 1$, then $b=d$, so $(a-c)d=1$.
*
*If $1\mapsto 0$ and $\infty\mapsto\infty$, then $a=-b$ to get $1\mapsto 0$, and $c=0$ to get $\infty\mapsto \infty$. Then $(a-c)d = -d^2=1$, so $b=d=\pm i$, $a=\mp i$. (corrected; apologies to the anonymous user who suggested the edit)
*If $1\mapsto\infty$ and $\infty\mapsto 0$ then we must have $a=0$, and $c+d=0$. So $1 = (a-c)d = -cd = d^2$. Hence $d=b=\pm 1$, $c=\mp 1$, $a=0$.
If $0\mapsto\infty$, then $d=0$, so $bc=-1$.
*
*If $1\mapsto 0$ and $\infty\mapsto 1$, then $a=c$ and $a+b=0$. So $-1 = bc = -ac = -c^2$, hence $a=c=\pm 1$, $b=\mp 1$.
*If $1\mapsto 1$ and $\infty\mapsto 0$, then $a=0$ and $b=c$. Hence $-1=bc=b^2$, so $b=c=\pm i$, $a=d=0$.
| {
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How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? How can one prove the statement
$$\lim_{x\to 0}\frac{\sin x}x=1$$
without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.
This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.
| Let $\sin(x)$ is defined as solution of $\frac{d^2}{dx^2}\textrm{f}(x)=-\textrm{f}(x)$ with $\mathrm f(0)=0,\,\frac{d}{dx}\mathrm f(0)=C$ initial conditions, so exact solution is $\mathrm f(x)=C\cdot\sin(x)$.
Define second derivative as
$$
\begin{align*}
\frac{d^2}{dx^2}\textrm{f}(x)=\lim_{\Delta x\to 0}{\frac{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}-\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}{\Delta x}}&=\\=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-2\cdot \mathrm f(x-\Delta x)+\mathrm f(x-2\cdot\Delta x)}{\Delta x^2}}
\end{align*}
$$
we can easy check this limit for any (?) functions. Similarly, we can define the first derivative for right, middle and left points:
$$
\frac{d}{dx}\textrm{f}(x)\left\{
\begin{aligned}
&=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}}
\\
&=\lim_{\Delta x\to 0}{\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}\\
&=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-2\cdot\Delta x)}{2\cdot\Delta x}}
\end{aligned}
\right.
$$
Let's use the finite elements method assuming $Td=\Delta x,\,y_n=\mathrm f(x),\,y_{n-1}=\mathrm f(x-\Delta x),\,y_{n-2}=\mathrm f(x-2\cdot \Delta x)$
Override differential equation as
$$
\frac{y_n-2\cdot y_{n-1}+y_{n-2}}{Td^2}=-y_n
$$
Now solve this implicit equation for $y_n$ to obtain explicit recurrence relation:
$$
y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
$$
Using arbitrarily small but non-zero quantity Td we can plot exponentially decaying sampled sine function (because the poles are inside the unit circle of the transfer function corresponding to the given recurrence relation).
Similarly we write three systems for the initial conditions:
$$
\left\{
\begin{aligned}
&y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
\\
&C=\frac{y_n-y_{n-1}}{Td}
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
\\
&C=\frac{y_{n-1}-y_{n-2}}{Td}
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
\\
&C=\frac{y_n-y_{n-2}}{2\cdot Td}
\end{aligned}\right.
$$
Solve this sequence of equations for $y_{n-1}$ and $y_{n-2}$:
$$
\left\{
\begin{aligned}
&y_{n-1} = -C\cdot Td + y_{n}\\
&y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1-Td^2\right)\
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\\
&y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+\frac{Td^2}{2}\right)\\
&y_{n-2}=-2\cdot C\cdot Td + y_{n}\
\end{aligned}\right.
$$
At zero point $y_n=\mathrm f(0)=0$ and we can see linear dependence:
$$
\begin{aligned}
&y_{n-1} = -C\cdot Td\\
&y_{n-2}=-2\cdot C\cdot Td
\end{aligned}
$$
for all three solutions. Replace back:
$$
\begin{array}{l}
\mathrm f(0)&=0\\
\mathrm f(0-\Delta x) &= -C\cdot \Delta x\\
\mathrm f(0-2\cdot \Delta x) &= -2\cdot C\cdot \Delta x
\end{array}
$$
So all three $\frac{d}{dx}\mathrm f(0)$ limits is equal to $C$ at $x=0$ and in accordance with $\mathrm f(x)=C\cdot\sin(x)$ by definition we can write
$$
\lim_{\Delta x\to 0}{\frac{\mathrm f(0)-\mathrm f(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{0-(-C \cdot \Delta x)}{\Delta x}}=C
$$
Thus
$$
\lim_{\Delta x\to 0}{\frac{\sin(0)-C\cdot\sin(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{C\cdot\sin(\Delta x)}{\Delta x}}=C\cdot\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=C
$$
and $\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=1$
| {
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Interscholastic Mathematic League Senior B Division #11 The roots of the equation 3x^3-38x^2+cx-192=0 form a geometric progression. Compute c.
| We solve the problem using too much machinery for a contest question that has to be solved quickly!
The following result is useful. Let $r_1$, $r_2$, and $r_3$ be the roots of the cubic equation $x^3+Ax^2+Bx+C=0$. Then (i) $r_1+r_2+r_3=-A$; (ii) $r_1r_2+_2r_3+r_3r_1=B$; and (iii) $r_1r_2r_3=-C$.
The roots of our cubic form a three-term geometric progression. Say they are $a$, $ar$, and $ar^2$. We divide the coefficients of our cubic by $3$, to make the coefficient of $x^3$ equal to $1$. So the sum of the roots is $38/3$, and the product of the roots is $192/3=64$.
Note that the product of the roots is $(a)(ar)(ar^2)$, which is $a^3r^3$. Thus $(ar)^3=64$. We conclude that $ar=4$. (This assumes that we are dealing with a
geometric progression of real numbers, which is surely what is intended. If $c$ is real, one can prove that $ar$ cannot be one of the complex roots of $z^3=64$.)
The sum of the roots is $38/3$. Thus $a+ar+ar^2=38/3$. Since $ar=4$, we find that $a+ar^2=26/3$. We have $a=4/r$. Substitute in the equation $a+ar^2=26/3$. We obtain $\frac{4}{r} +4r=\frac{26}{3}$.
Divide through by $2$, multiply by $3$, and by $r$, and rearrange. We get
the equation $6r^2-13r+6=0$. By factoring, or otherwise, we find that $r=2/3$ or $r=3/2$. We work with $r=3/2$. (The other choice gives us the same geometric progression, written backwards.) Our geometric progression is therefore $8/3, 4, 6$.
It follows that $r_1r_2+r_2r_3+r_3r_1=152/3$, and therefore $c=152$.
Commenst: To prove the result that connects the coefficients to the roots, note that if the roots of $x^3+Ax^2+Bx+C=0$ are $r_1$, $r_2$, and $r_3$, then
$x^3+Ax^2+Bx+C$ is the polynomial $(x-r_1)(x-r_2)(x-r_3)$. Expand this product, and compare coefficients.
Much more frequently useful is the fact that if $x^2+Ax+B=0$ has the roots $r_1$ and $r_2$, then $r_1+r_2=-A$ and $r_1r_2=B$.
There are analogous results for higher degree polynomials.
Added: There is an easy way to solve the problem in a contest situation. Guess that the roots will be rational. Then by the Rational Roots Theorem, the rational roots are of the form $a/b$, where $a$ and $b$ are integers, $a$ divides $192$, and $b$ is a positive divisor of $3$. The roots cannot all be integers. A little fiddling with possibilities fairly quickly zooms in on the roots. We left this out originally because with a small numerical change, the roots will no longer be rational. The procedure that we used in the main part works in this more general setting, and the ideas are worth knowing.
| {
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Find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$ I want to find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$.
My thoughts so far: I want to find $p$ such that $ \left( \frac{15}{p} \right) = 1$. By multiplicativity of the Legendre symbol, this is equivalent to $ \left( \frac{5}{p} \right) \left( \frac{3}{p} \right) = 1 $. Using the Law of Quadratic Reciprocity, this is equivalent to finding $p$ such that $ - \left( \frac{p}{5} \right) \left( \frac{p}{3} \right) = 1$.
So there are two cases:
*
*$ \left( \frac{p}{5} \right) = -1, \left( \frac{p}{3} \right) = 1$.
*$ \left( \frac{p}{5} \right) = 1, \left( \frac{p}{3} \right) = -1 $.
For case (1.), the quadratic residues modulo $5$ are $1$ and $4$, so for $ \left( \frac{p}{5} \right ) = -1$, we must have that $p$ is $2$ or $3$ modulo $5$. We must also have that $p$ is $1$ modulo $3$ from the other condition. One of these pairs is incompatible, and we can solve to give $p$ is $13$ modulo $15$.
Similarly for case (2.)
Is this the correct approach? I'm unsure if each step in my working is an "if and only if". If $p$ is $13$ modulo $15$, is $15$ necessarily a quadratic residue modulo $p$?
Thanks!
| You must also consider your primes mod 4, as that alters the behavior of reciprocity. So, in the order I found them, all values are
$$ 1,49; \; 17, 53; \; 11, 59; \; 7, 43 \pmod {60}.$$ In ordinary numerical order,
$$1, 7,11,17,43,49, 53,59 \pmod {60}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How could we find the largest number in the sequence $ \sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$? How to find the largest number in the sequence$$ \sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$$
I am interested in a "calculus-free" approach.
Thanks,
| The $n$-th term in the sequence is $n\sqrt{51-n}=\sqrt{n^2(51-n)}$. So the question is: for which $n$ ($1\le n\le 50$), does $n^2(51-n)$ become the largest?
If you want to avoid calculus, you could use the AM-GM inequality:
if $x,\,y,\,z\ge 0$, then $$\frac{x+y+z}{3}\ge\sqrt[3]{xyz},$$
with equality if and only if $x=y=z$.
If we set $x=y=n/2$ and $z=51-n$, we obtain: $$\frac{51}{3}\ge \sqrt[3]{\frac{n}{2}\cdot\frac{n}{2}\cdot (51-n)},$$
with equality if and only if $n/2=51-n$ or $n=34$.
It follows that $n^2(51-n)\le 4\cdot 17^3$, or $\sqrt{n^2(51-n)}\le 2\cdot 17^{3/2}$, where equality holds for $n=34$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/80148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$? How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$?
| Here is a much more elementary proof:
$$a^{3a}b^{3b}=a^3a^3 \cdot... a^3 b^3b^3 \cdot ....b^3 \,.$$
Using the AM-GM inequality with $x_1=...=x_a=a^3$ and $x_{a+1}=...=x_{a+b}=b$ Yields
$$\sqrt[a+b]{a^3a^3 \cdot ... a^3 b^3b^3 \cdot ....b^3} \leq \frac{aa^3+bb^3}{a+b} \,.$$
Thus
$$a^{3a}b^{3b} \leq \left( \frac{a^4+b^4}{a+b} \right)^{a+b} \,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/80550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
If $f(x)$ is a polynomial satisfying $ f(x)f(\frac 1x) = f(x)+f(\frac 1x)$ and $f(3)=28$, then how could we find $f(4)$?
If $f(x)$ is a polynomial satisfying $ f(x)f(\frac 1x) = f(x)+f(\frac
1x)$ and $f(3)=28$, then how could we find $f(4)$ ?
| Solving the functional equation for $f\left(\frac{1}{x} \right) = \frac{f(x)}{f(x)-1}$. This means that $f(x)-1$ must be a monomial. Let $f(x) = 1 + c x^d$. Then
$$
c \left( \frac{1}{x} \right)^d +1 = \frac{1}{c} \left( \left( \frac{1}{x} \right)^d + c \right)
$$
This, implies $c^2 = 1$. Now use $f(3) = 28$ to determined $c$ and $d$. Since $28 = 1 + 1 \times 3^3$, we conclude $c=1$ and $d=3$.
Thus $f(4) = 1 + 4^3 = 65$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Continued fraction: Show $\sqrt{n^2+2n}=[n; \overline{1,2n}]$ I have to show the following identity ($n \in \mathbb{N}$):
$$\sqrt{n^2+2n}=[n; \overline{1,2n}]$$
I had a look about the procedure for $\sqrt{n}$ on Wiki, but I don't know how to transform it to $\sqrt{n^2-2n}$.
Any help is appreciated.
EDIT:
I tried the following:
$\sqrt{n^2+2n}>n$, so we get $\sqrt{n^2+2n}=n+\frac{n}{x}$, and $\sqrt{n^2+2n}-n=\frac{n}{x}$ and further $x=\frac{n}{\sqrt{n^2+2n}}$.
So we get $x=\frac{n}{\sqrt{n^2+2n}-n}=\frac{n(\sqrt{n^2+2n}+n)}{(\sqrt{n^2+2n}-n)(\sqrt{n^2+2n}+n}$ I don't know if it's right and how to go on.
| Because $n^2+2n=(n+1)^2-1$, the leading term in the continued fraction expansion will be $n$. Multiplying by a form of $1$, we have
$$\sqrt{n^2+2n}-n=\frac{2n}{n+\sqrt{n^2+2n}}$$
and $1<\frac{n+\sqrt{n^2+2n}}{2n}<2,$ so the next term will be $1$. Since $\frac{n+\sqrt{n^2+2n}}{2n}-1=\frac{\sqrt{n^2+2n}-n}{2n}$ and we can again rationalize the denominator to simplify $\frac{2n}{\sqrt{n^2+2n}-n}=n+\sqrt{n^2+2n}$. Because $n<\sqrt{n^2+2n}<n+1$, the next term in the continued fraction expansion is $2n$, and subtracting 2n yields $\sqrt{n^2+2n}-n$, at which point the pattern continues.
In the end, it all boils down to the useful trick that $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{a}+\sqrt{b}}{a-b}$, which can be viewed as a application of $a^2-b^2=(a+b)(a-b)$. Judging from your edit, you are aware of this idea, but you just didn't carry out the necessary algebraic simplifications to see that it actually leads somewhere.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Discriminant of derivative of cubic equation being a perfect square Is it possible for the discriminant of the first derivative of a cubic polynomial $(x+a)(x+b)(x+c)$, where $a, b$ and $c$ are distinct non-zero integers (i.e. Discriminant $[d((x+a)(x+b)(x+c))/dx]$ in Wolfram|Alpha/Mathematica) to be a perfect square?
If so, what are the smallest absolute values of $a, b$ and $c$? Thanks!
| Let $p(x) = x^3 + (a + b + c)x^2 + ((b + c)a + bc)x + abc$. Then,
$$p'(x) = 3x^2 + 2(a+b+c)x + ((b+c)a+bc).$$
The discriminant of $p'(x)$ is
$$\Delta = 4(a+b+c)^2 - 4\cdot 3\cdot ((b+c)a+bc)=4(a^2 - (b + c)a + (b^2 - bc + c^2)).$$
So you need to find values of $a,b,c$ that make $\Delta$ a square... Notice that of course $\Delta$ is symmetric in $a,b,c$:
$$\Delta = 4(a^2+b^2+c^2-ab-bc-ac).$$
The solution $a=1$, $b=4$, $c=-4$ seems to be the "smallest", with $a,b,c$ all distinct, and non-zero (because of the symmetry of $\Delta$, you can permute $a,b,c$ and get the same solution). This gives
$$p(x)=(x+1)(x+4)(x-4)$$
and
$$p'(x)=3x^2 + 2x - 16,$$
whose discriminant is $4-4\cdot 3\cdot (-16) = 196$.
Notice that the very first expression we found for the discriminant:
$$\Delta = 4((a+b+c)^2 - 3\cdot ((b+c)a+bc)),$$
gives an infinity family of rational solutions to your problem when $(b+c)a+bc=0$, that is, when $b\neq -c$ and
$$a=-\frac{bc}{b+c}.$$
This is an integer if $(b+c)$ divides $bc$, but $bck=b+c$ implies that $b$ divides $c$, so $c=bf$ and $bc/(b+c)=b^2f/(b(1+f))=bf/(1+f)$. The numbers $f$ and $1+f$ are relatively prime, so we need $1+f$ to divide $b$, say $b=(1+f)g$. Then
$$b=(1+f)g, \quad c=bf=(1+f)fg$$
and
$$\frac{bc}{b+c} = \frac{(1+f)^2g^2f}{(1+f)g+(1+f)fg}=\frac{(1+f)fg}{(1+f)}=fg.$$
Thus, we get an infinite family of integer solutions:
$$a=-fg, \quad b=(1+f)g,\quad c=(1+f)fg,$$
for any $f,g\in\mathbb{Z}$. For instance, $f=2$, $g=-1$, yields
$$a=2,\quad b=-3,\quad c=-6$$
so
$$p(x)=(x+2)(x-3)(x-6)$$
with
$$p'(x)= 3x^2 - 14x$$
and the discriminant is $14^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solving quadratic equation $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
Rearranging it I get: $1-x^2=x-x^2$, and so $x=1$. But the question Im doing says to find 2 solutions. How would I find the 2nd solution?
Thanks.
| $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
$$\frac{1}{x^2} = \frac{1}{x} $$
multiply by x suppose that $x \ne0$
$$x^2-x=0$$
$$x(x-1)=0$$
The unique solution is $x=1$ solution $x=0$ is excluded.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$? Find the limit:
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$
I am not able to find it because I don't know how to prove or disprove $0$ is the answer.
| METHOD I
Firstly, notice that the expression under the limit is an odd function and consider that $\sin(x)<x$. Then we have that:
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\frac{\sin x - x}{x\sin x}\le\lim_{x \rightarrow 0}\frac{\sin x - x}{x^2}\le\lim_{x \rightarrow 0}\frac{\tan x - x}{x^2}=0$$
As regards the last limit you wanna see my proof here.
Q.E.D.
METHOD II
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\frac{\sin x - x}{x\sin x}\le\lim_{x \rightarrow 0}\frac{\sin x - x}{x^2}=\lim_{x \rightarrow 0}x\cdot\frac{\sin x - x}{x^3}=0\cdot-\frac{1}{6}=0$$
Let's solve now the auxiliary limit I used (elementarily):
$$L=\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}=\lim_{x \rightarrow 0}\frac{\sin 2x - 2x}{8x^3}=\lim_{x \rightarrow 0}\frac{\sin x \cos x - x}{4x^3}=\lim_{x \rightarrow 0}\frac{\sin x \cos x -x\cos x + x\cos x- x}{4x^3}=\lim_{x \rightarrow 0}\frac{\cos x(\sin x \ -x) }{4x^3}-\lim_{x \rightarrow 0}\frac{(1 - \cos x) }{4x^2}=$$
$$\lim_{x \rightarrow 0} \cos x \cdot\frac{L}{4} -\frac{1}{8}=\frac{L}{4}-\frac{1}{8}$$
$$L=\frac{L}{4}-\frac{1}{8}$$
$$L=-\frac{1}{6}.$$
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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A Limit problem : No-existence I find difficulty proving the no existence of this limit
I show my process
$$ \lim_{x\to 0} \biggl(1 + x e^{- \frac{1}{x^2}}+\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}$$
We begin with rewriting the limit as follows:
$$ \lim_{x\to 0} \biggl(1 + x e^{- \frac{1}{x^2}}+\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}=\lim_{x\to 0} \biggl( 1 + x \frac{1}{e^{\frac{1}{x^2}}} +\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}$$
and analyze the various addends and the exponent of the limit:
$$\begin{align*}
&x \frac{1}{e^{\frac{1}{x^2}}}\to 0\\
&\sin \frac{1}{x^4}\to \not \exists\\
&e^{\frac{1}{x^2}}\to+\infty.\\
\end{align*}$$
The problem here lies in the fact that we have an addendum that there is no limit, let's consider:
$$\sin{a_n}\quad\text{e}\quad\sin{b_n}\quad\text{con }\quad n\to+\infty$$
where the two sequences are:
$$a_n=\frac{\pi}{2}+2n\pi\quad\text{e}\quad b_n=2n\pi$$
Then, the function values ββcalculated in the sequence $ a_n $, with $ k $ positive integer, tends to $ 1 $, calculated values ββof the sequence $b_n$ tends to $0$,:
$$\lim_ {n \to\infty}\sin{a_n}=1 \quad\text{and}\quad \lim_{n\to \infty}\sin {b_n} = 0$$
and therefore, as we know, the limit of $\sin x$ ($x \to \infty$) not exists.
Now, to prove that the given limit does not exist,i continued in this way
$t= \frac{1}{x^2},$ (if $x\to0 \rightarrow t\to+\infty$) :
$$\lim_{x\to 0} \biggl( 1 + x \frac{1}{e^{\frac{1}{x^2}}} +\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}=\lim_{t\to +\infty} \biggl( 1 + \frac{ \sqrt{t}}{t}\cdot \frac{1}{e^t} +\sin{t^2}\biggr)^{e^t}$$
$\frac{ \sqrt{t}}{t}\cdot \frac{1}{e^t}\to 0$
i consider
$$\begin{align*}
&\lim_{t\to +\infty} \biggl(1 + \sin{({a_n})^2}\biggr)^{e^t}=\biggl(1+1\biggr)^{e^t}=+\infty\\
&\lim_{t\to +\infty} \biggl( 1 + \sin{({b_n})^2}\biggr)^{e^t}=e^{e^t\ln\biggl( 1 + \sin{({b_n})^2}\biggr)}=e^{+\infty\ln( 1 + 0)}=???
\end{align*}$$
| Given an integer $k$, let $\frac{1}{x^4}=2\pi k + \frac{\pi}{2}$, or $x = (2\pi k + \frac{\pi}{2})^{-\frac{1}{4}}$. Then $\sin {\frac{1}{x^4}} = 1$. Let $y=e^{\frac{1}{x^2}}$. Then your expression is:
$$(2+\frac{x}{y})^y$$.
Now, if $|x|<1$ then $y>e$, so $\frac{x}{y}> \frac{-1}{2}$. So this expression is at least as big as $(\frac{3}2)^y$.
Picking a large enough $k$, we can make $x$ arbitrarily small, so $\frac{1}{x^2}$ arbitrarily large, so $y=e^{\frac{1}{x^2}}$ arbitrarily large. So, we see that we can make your expression bigger than $(\frac{3}2)^Y$ for arbitrarily large $Y$, and hence it has no limit.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do we calculate the area of a region bounded by four different curves? Calculate the area(express both respectively in integral with one variable) bounded by the following curves (i.e. the shape with one side corresponding to one curve):
$$xy=1, \quad xy^2=3,\quad x^2-y^2=26,\quad x^2-y^3=11$$
This problem is created by myself, but it is beyond my knowledge to solve it.
| As long as you're only asking for an expression as an integral, and not an actual number, we can calculate the area as follows:
Let
*
*$a$ be the positive real solution of $x^5-11x^3-1=0$
*$b$ be the positive real solution of $x^{7/2}-11x^{3/2}-3\sqrt{3}=0$
*$c$ be the positive real solution of $x^4-26x^2-1=0$
*$d$ be the positive real solution of $x^3-26x-3=0$
We have $a<b<c<d$, and the dashed lines in the picture below indicate their positions. The curves are colored as follows:
$$\color{red}{xy=1},\quad \color{green}{xy^2=3},\quad \color{blue}{x^2-y^2=26},\quad \color{black}{x^2-y^3=11}$$
As you can see, the equations for $a,b,c,d$ were obtained by solving for the $x$-coordinate of the relevant intersections of the curves.
In the upper right quadrant, we can re-express our four curves as
$$\color{red}{y=\tfrac{1}{x}},\quad \color{green}{y=\sqrt{\tfrac{3}{x}}},\quad \color{blue}{y=\sqrt{x^2-26}},\quad \color{black}{y=(x^2-11)^{1/3}}$$
The area below the black curve and above the red curve, from $a$ to $b$, is
$$\int_a^b\left((x^2-11)^{1/3}-\tfrac{1}{x}\right)dx$$
The area below the green curve and above the red curve, from $b$ to $c$, is
$$\int_b^c\left(\sqrt{\tfrac{3}{x}}-\tfrac{1}{x}\right)dx$$
The area below the green curve and above the blue curve, from $c$ to $d$, is
$$\int_c^d\left(\sqrt{\tfrac{3}{x}}-\sqrt{x^2-26}\right)dx$$
Thus the area of the upper region is
$$\int_a^b\left((x^2-11)^{1/3}-\tfrac{1}{x}\right)dx+\int_b^c\left(\sqrt{\tfrac{3}{x}}-\tfrac{1}{x}\right)dx+\int_c^d\left(\sqrt{\tfrac{3}{x}}-\sqrt{x^2-26}\right)dx$$
We can do a similar computation for the lower region.
Mathematica code:
NSolve[x^5 - 11x^3 - 1 == 0, x]
NSolve[x^(7/2) - 11x^(3/2) - 3*Sqrt[3] == 0, x]
NSolve[x^4 - 26x^2 - 1 == 0, x]
NSolve[x^3 - 26x - 3 == 0, x]
a = 3.320739129529704
b = 3.437347103656831
c = 5.102784025451723
d = 5.155761179910075
ContourPlot[{x*y == 1, x*y^2 == 3, x^2 - y^2 == 26, x^2 - y^3 == 11,
x == a, x == b, x == c, x == d}, {x, 2.5, 6}, {y, -2, 2},
ContourStyle -> {{Red, Thick}, {Green, Thick}, {Blue, Thick}, {Black,
Thick}, {Black, Dashed}, {Black, Dashed}, {Black,
Dashed}, {Black, Dashed}}]
| {
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"url": "https://math.stackexchange.com/questions/95104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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difficult sequence Prove that the sequence:
$$a_n= \frac{1}{n}\left(e\cdot\sqrt e \cdots\sqrt[3]e\cdot\sqrt[n]e\right)$$
is decreasing to a finite limit. After having shown that the sequence:
$$b_n=\left(\sum_{k=1}^n\frac{1}{k}\right)-\log n$$
converges to a positive real number $b,$ say who is the limit of $ a_n $
| $$\eqalign{
& {a_n} = \frac{1}{n}\prod\limits_{k = 1}^n {{e^{\frac{1}{k}}}} \cr
& \log {a_n} = - \log n + \sum\limits_{k = 1}^n {\frac{1}{k}} \cr
& \log {a_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \log n \cr
& \mathop {\lim }\limits_{n \to \infty } \log {a_n} = \gamma \cr} $$
You can prove $0 < \gamma < 1$ since we can replace $\log n$ by $\log (n+1)$ and put
$$\eqalign{
& {b_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \sum\limits_{k = 1}^n {\log \frac{{k + 1}}{k}} \cr
& {b_n} = \sum\limits_{k = 1}^n {\left( {\frac{1}{k} - \log \frac{{k + 1}}{k}} \right)} \cr} $$
Since we know
$$1 - \frac{1}{x} \leqslant \log x \leqslant x - 1$$
We have
$$\frac{1}{{k + 1}} \leq \log \left( {1 + \frac{1}{k}} \right) \leq \frac{1}{k}$$
We can prove both bounds with this.
$$\eqalign{
& \frac{1}{k} - \frac{1}{{k + 1}} \geqslant \frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0 \cr
& \sum\limits_{k = 1}^n {\frac{1}{k} - \frac{1}{{k + 1}}} \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)} \geqslant 0 \cr
& 1 \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)} \geqslant 0 \cr} $$
The first inequality also proves each term is positive or zero (though the last is discarded by a simple look at the image), i.e
$$\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Uniqueness of polynomial Let $n$ be a nonnegative integer. Can you help me prove the following ?
There exists a unique polynomial $P_{n}$ such that for all $t \in [0,\frac{\pi}{2}]$,
$P_{n}(\operatorname{cotan}^2t)=\frac{\sin((2n+1)t}{(\sin t)^{2n+1}}$
with $\operatorname{cotan}(x)=\frac{\cos x}{\sin x}$.
| There may be an easier way but this is what I came up with:
Start with Euler's formula:
\begin{align}
e^{i(2n+1)t} &= \cos((2n+1)t) + i \sin((2n+1)t) = (\cos t + i \sin t)^{2n+1} \\
&= \sum_{k=0}^{2n+1} \binom{2n+1}{k}i^k \sin^k t \cos^{2n+1-k}t \\
\end{align}
Now we separate the sum for even and odd values of $k$:
\begin{align}
&\cos((2n+1)t) + i \sin((2n+1)t) \\
&= \sum_{l=0}^{n} \binom{2n+1}{2l} i^{2l} \sin^{2l} t \cos^{2n+1-2l}t + \sum_{l=0}^{n} \binom{2n+1}{2l+1} i^{2l+1} \sin^{2l+1} t \cos^{2n+1-2l-1}t \\
&= \sum_{l=0}^{n} \binom{2n+1}{2l} (-1)^l \sin^{2l} t \cos^{2n+1-2l}t+ i \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \sin^{2l+1} t (\cos^2 t)^{n-l} \\
\end{align}
Since the imaginary parts must coincide, we get:
\begin{align}
\sin((2n+1)t) &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \sin^{2l+1} t\ (\cos^2 t)^{n-l} \\
&= \sin^{2n+1} t \quad \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l (\sin^2 t)^{l-n} (\cos^2 t)^{n-l} \\
&= \sin^{2n+1} t \quad\sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \Big(\frac{\cos^2t}{\sin^{2} t}\Big)^{n-l} \\
\end{align}
Hence
$$\frac{\sin((2n+1)t)}{\sin^{2n+1}t} = \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \Big(\frac{\cos^2t}{\sin^{2} t}\Big)^{n-l} = P_n(\cot^2 t),
$$
where $P_n$ is the (unique) $n$-degree polynomial given by $P_n(x) = \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l x^{n-l}$.
| {
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Finding the limit of $\frac{1}{t\sqrt{1+t}} - \frac{1}{t}$ as $t$ tends to $0$ $$\lim_{t\rightarrow 0}\left(\frac{1}{t\sqrt{1+t}} - \frac{1}{t}\right)$$
I attemped to combine the two fraction and multiply by the conjugate and I ended up with:
$$\frac{t^2-t^2\sqrt{1+t}}{t^3+{t\sqrt{1+t}({t\sqrt1+t})}}$$
I couldn't really work it out in my head on what to do with the last term $t\sqrt{1+t}({t\sqrt{1+t}})$ so I left it like that because I think it works anyways. Everything is mathematically correct up to this point but does not give the answer the book wants yet. What did I do wrong?
| Perhaps you were trying something like
$\dfrac{1}{t\sqrt{1+t}} - \dfrac{1}{t} = \dfrac{1-\sqrt{1+t}}{t\sqrt{1+t}} = \dfrac{1-(1+t)}{t\sqrt{1+t}(1+\sqrt{1+t})} = \dfrac{-1}{\sqrt{1+t}(1+\sqrt{1+t})} $
which has a limit of $\dfrac{-1}{1 \times (1+1)} = -\dfrac{1}{2}$ as $t$ tends to $0$.
Added: If you are unhappy with the first step, try instead $\dfrac{1}{t\sqrt{1+t}} - \dfrac{1}{t} = \dfrac{t-t\sqrt{1+t}}{t^2\sqrt{1+t}} = \dfrac{t^2-t^2(1+t)}{t^3\sqrt{1+t}(1+\sqrt{1+t})} = \dfrac{-t^3}{t^3\sqrt{1+t}(1+\sqrt{1+t})} $ $= \dfrac{-1}{\sqrt{1+t}(1+\sqrt{1+t})}$ to get the same result
| {
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Solutions of some Diophantine equations Respected Mathematicians,
The Diophantine equation
$$2^x + 5^y = z^2$$
has solutions $$x = 3, y = 0, z = 3$$ and $$x = 2, y = 1, z = 3$$ I got these solutions by trial and error method. To be honest, these solutions are below the number $5$. So, I easily verified them by trial and error method. I would like to know the method which will give the solutions of the above equation, as well as the solutions of equations below.
a)
$$4^x + 7^y = z^2$$
b)
$$4^x + 11^y = z^2$$
Looking forward to your solution and support.
baba
| Assuming x, y, z are non-negative integers, there are no solutions to $4^x + 7^y = z^2$ and $4^x + 11^y = z^2$. This can be shown through modular arithmetic.
Taking the equation a) mod 3 you'll find the following:
*
*$4^x \equiv 1 \pmod{3}$
*$7^y \equiv 1 \pmod{3}$
*$z^2 \equiv 0$ or $1 \pmod{3}$
Since $1 + 1 = 2$ and there are no squares equivalent to $2$ (mod 3), there are no solutions to $4^x + 7^y = z^2$.
Now taking equation b) mod 3:
*
*$4^x \equiv 1 \pmod{3}$
*$11^y \equiv (-1)^y \pmod{3}$
*$z^2 \equiv 0$ or $1 \pmod{3}$
Since, once again, there are no squares equivalent to $2$ (mod 3), $y$ must be odd. However taking the equation mod 4 you'll get the following:
*
*$4^x \equiv 0 \pmod{4}$
*$11^y \equiv (-1)^y \pmod{4}$
*$z^2 = 0$ or $1 \pmod{4}$
$y$ cannot be odd since this would imply that $11^y$ is equivalent to $-1 \equiv 3$ (mod 4) and there are no squares equivalent to 3 mod 4. This is a contradiction and therefore there are no solutions to equation b) either.
EDIT: Actually I just realised that I neglected the case where $x = 0$ and thus $4^0 \equiv 1 \pmod{4}$. But taking the resulting equation $1 + 11^y = z^2$ mod 5 proves that there's no solution anyway, as the quadratic residues for modulo 5 are 0, 1 and 4.
| {
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evaluate $\int \frac{x^2\cdot\log(x)}{x+1} dx$ I am trying to evaluate integral:
$$\int \frac{x^2\log(x)}{x+1}dx$$
But I have some problems with it. If I use Wolfram Alpha like this
I get a result, but I need evaluate it by hand. Which method should I use?
If we represent it as:
$$\int(x^2\log(x)\cdot d(\ln(x))dx$$
then it is unclear what to dot here, even we can represent as power, take $x^2$ in power of $\log(x)$, not hint yet. Please help me to evaluate it.
| We have
\begin{align*}
\int\frac{x^2\log x}{x+1}dx&=\int x\log x\frac{x+1-1}{x+1}dx\\
&=\int x\log x-\int \frac{x+1-1}{x+1}\log xdx\\
&=\frac{x^2}2\log x-\int\frac{x^2}2\frac 1xdx-\int \log x+\int\frac{\log x}{x+1}dx\\
&=\frac{x^2}2\log x-\frac{x^2}4-x\log x+x+\log x\log(x+1)-\int\frac{\log(x+1)}xdx\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)-\int\frac{\log(x+1)}xdx\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)-\sum_{k=1}^{+\infty}\int \frac{(-1)^{k+1}x^k}{kx}dx\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)+\sum_{k=1}^{+\infty}(-1)^k \frac{x^k}{k^2}\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)+\operatorname{Li}_2(-x),
\end{align*}
which is the result given by Wolfram Alpha.
| {
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Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!
| one way is partial fractions on
$$
\frac{x^4+1}{x^6+1}=\frac{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}{(x-e^{\pi i/6})(x-e^{3\pi i/6})(x-e^{5\pi i/6})(x-e^{7\pi i/6})(x-e^{9\pi i/6})(x-e^{11\pi i/6})}
$$
$$
=\frac{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}{(x^2+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)}
$$
| {
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"source": "stackexchange",
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Pythagorean triplets Respected Mathematicians,
For Pythagorean triplets $(a,b,c)$, if $c$ is odd then any one of $a$ and $b$ is odd. Here $(a, b, c)$ is a Pythagorean triplet with $c^2 = a^2 + b^2$.
Now, I will consider $c = b + \Omega$. The reason for considering $c = b + \Omega$ is, $c$ is a hypotenuse side of right triangle and it is obviously larger than the other side $b$.
Now,
$$a^2 + b^2 = (b + \Omega)^2 = b^2 + 2b \Omega + \Omega ^2\qquad\qquad(1)$$ which is same as
$$b = [a^2 - \Omega ^2] \div 2\Omega.$$
Which implies that $\Omega$ divides $a^2$ for $a^2- \Omega ^2) \gt 0$ or $(a - \Omega) (a + \Omega) \gt 0$, which implies that
$$a \gt \Omega\qquad\qquad(2).$$
Now, I will consider $a = 2^m$; then $\Omega$ is also even. Otherwise, if $a = 2^m + 1$ then obviously $\Omega$ is odd.
Now, I will consider both $a$ and $\Omega$ is an even numbers such that, $a = 2^m$ and $\Omega = 2^r$ for some $m$ and $r$. By (2), we have $m \gt r$ and by (1), we have
$$(2^m)^2 = 2^r (2b + 2^r)$$
or
$$b = \frac{2^r}{2}((4^m \div 4^r) - 1))\qquad\qquad(3)$$
As I said earlier, $a$ and $\Omega$ is an even, then $b$ should be an odd number. i.e., $r = 1$
Therefore, the required triplets for even numbers in powers of $2$ are
$(2^m, (4^m \div 4) - 1, (4^m \div 4 ) + 1))$
Now my question is, how one can generalize the same for following?
Case 1: if we take odd numbers for powers of some prime
Case 2: if we take even numbers with prime powers.
Thanking you,
| It is unclear to me what you are going for; for one thing, not every even number is a power of $2$. For another, even if $a$ is a power of $2$, you have no warrant to assert that $\Omega$ will necessarily be a power of $2$ as well; at least, no warrant that you have given. Of course, you may assume that's the case, but then you are dealing with a rather restrictive subset of pythagorean triples.
For another, it is false that if $a$ is even then $b$ is necessarily odd, though it is true for primitive pythagorean triples (ones where $a$, $b$, and $c$ are pairwise relatively prime). But you never make that assumption.
And finally, a complete description of primitive pythagorean triples is well known. You can deduce your formula from them rather easily.
Explicitly, suppose that $(a,b,c)$ is a primitive Pythagorean triple, and let us assume that $a$ is even; we have
$$(c+b)(c-b) = c^2-b^2 = a^2;$$
since $a,b,c$ are pairwise coprime, and $a$ is even, then $b$ and $c$ are both odd, so $c+b$ and $c-b$ are both even. Any common divisor of $c+b$ and $c-b$ will divide $(c-b)+(c+b) = 2c$, and also $(c+b)-(c-b) = 2b$; since $\gcd(2c,2b) = 2\gcd(b,c) = 2$, the gcd of $c-b$ and $c+b$ is $2$. Dividing through by $4$ (which we can do since $a$ is even, so $a^2$ is a multiple of $4$, we get
$$\left(\frac{c+b}{2}\right)\left(\frac{c-b}{2}\right) = \left(\frac{a}{2}\right)^2.$$
Since $\frac{c+b}{2}$ and $\frac{c-b}{2}$ are relatively prime, and their product is a square, they are each a square. So we can write $\frac{c+b}{2} = s^2$, $\frac{c-b}{2}=r^2$, $\frac{a}{2} = rs$, with $r$ and $s$ positive, $s\gt r$, relatively prime, and of opposite parity (since $c$ is odd).
This gives
$$\begin{align*}
c &= \frac{c+b}{2} + \frac{c-b}{2} = s^2+r^2.\\
b &= \frac{c+b}{2} - \frac{c-b}{2} = s^2 - r^2.\\
a &= 2rs.
\end{align*}$$
In your case, you want $a=2^m$; since $s$ and $r$ are of opposite parity, one must be equal to $1$; since $s\gt r\gt 0$, we will have $r=1$, $s=2^{m-1}$, and this gives
$$(a,b,c) = (2^m, 4^{m-1}-1, 4^{m-1}+1),$$
which was your result.
Note, however, that there are other (non-primitive) Pythagorean triples that have $a$ a power of $2$: namely,
$$(2^{m+k}, 2^{2m+k-2}-2^k, 2^{2m+k-2}+2^k)$$
is a Pythagorean triple for all $k\geq 0$.
If $a$ is odd, then just exchange the roles of $a$ and $b$ above.
| {
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Prove $\frac{\cos^3{x}-\sin^3{x}}{\cos{x}-\sin{x}} =1+\frac{1}{2} \sin{2x}$ Prove
$$\frac{\cos^3{x}-\sin^3{x}}{\cos{x}-\sin{x}} =1+\frac{1}{2} \sin{2x}$$
How do I start :( which identity do I use?
| Hint: use the identity:
$(a^3-b^3)=(a-b)(a^2+ab+b^2)$.
Solution follows:
We have
$\cos^3 x -\sin^3 x =(\cos x-\sin x)(\cos^2 x +\cos x\sin x +\sin^2 x )$.
So:
$$\eqalign{{
\cos^3 x -\sin^3 x\over \cos x-\sin x}&=
{(\cos x-\sin x)(\cos^2 x +\cos x\sin x +\sin^2 x )\over (\cos x-\sin x)}\cr
&=1+\cos x\sin x\cr
&=1+\textstyle{1\over2}\sin 2x.}
$$
(The last equality used the trigonometric identity $\sin(2x)=2\sin x\cos x$.)
As @Dilip Sarwate points out in the comment below, the above does not hold when $\cos x=\sin x$. In this case, ${
\cos^3 x -\sin^3 x\over \cos x-\sin x}$ is not defined. Of course, whenever ${
\cos^3 x -\sin^3 x\over \cos x-\sin x}$ is defined, it is equal to $1+{1\over2}\sin 2x$.
| {
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Integral without using Euler substitution Help me please with integral:
$$\int \frac{2x-\sqrt{4x^{2}-x+1}}{x-1}\;dx$$
I must solve it without using Euler substitution.
Thanks!
| Hint: writing
$$\frac{2x-\sqrt{4x^2-x+1}}{x-1}=\frac{2x-\sqrt{4x^2-x+1}}{x-1}\frac{2x+\sqrt{4x^2-x+1}}{2x+\sqrt{4x^2-x+1}},$$
we find $$\frac{2x-\sqrt{4x^2-x+1}}{x-1}=\frac{4x^2-(4x^2-x+1)}{(x-1)(2x+\sqrt{4x^2-x+1})}=\frac 1{2x+\sqrt{4x^2-x+1}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to define the sign of a function $$y=\arctan\frac{x+1}{x-3} + \frac{x}{4}$$
I know that is necessary to put the function $>$ than $0$, but then?
$$\arctan\frac{x+1}{x-3} + \frac{x}{4}>0$$
It's a sum, so I can't set up a "false system" putting the two factors $>0$. In this case which is the rule to study the sign of this function?
Note: These are not homeworks.
| Let $$f(x)=\arctan\frac{x+1}{x-3}+\frac{x}4\;.$$
You know that the algebraic sign of $\arctan u$ is the same as the sign of $u$, so
$$
\arctan\frac{x+1}{x-3}\text{ is }\begin{cases}
\text{positive}&\text{when }x>3\\
\text{negative}&\text{when }-1<x<3\\
0&\text{when }x=-1\\
\text{positive}&\text{when }x<-1\;.
\end{cases}$$
Without further work we can be sure that $f(x)>0$ when $x>3$, and $f(x)<0$ when $-1\le x\le 0$; itβs only the intervals $(\leftarrow,-1)$ and $(0,3)$ that we need to investigate in detail.
Now
$$\begin{align*}
f\,'(x)&=\frac14-\frac4{(x-3)^2+(x+1)^2}\\
&=\frac{x^2-2x-3}{2\big((x-3)^2+(x+1)^2\big)}\\
&=\frac{(x-3)(x+1)}{2\big((x-3)^2+(x+1)^2\big)}
\end{align*}$$
is negative throughout the interval $0<x<3$, so $f$ is decreasing on that interval; since $f(0)<0$, it must be the case that $f(x)<0$ for all $x\in(0,3)$. Similarly, $f\,'(x)>0$ when $x<-1$, so $f$ is increasing on the interval $(\leftarrow,-1)$, and $f(-1)=0$, so $f(x)<0$ for all $x<-1$. To summarize,
$$f(x)\text{ is }\begin{cases}
\text{positive}&\text{when }x>3\\
\text{negative}&\text{when }x<3\;.
\end{cases}$$
I donβt at the moment see any simple way to avoid using some calculus here. You can rewrite
$$\frac{x+1}{x-3}=1+\frac4{x-3}$$
to see easily that $\frac{x+1}{x-3}<1$ when $x<3$, so $0<\arctan\frac{x+1}{x-3}<\frac{\pi}4$, and therefore $f(x)<0$ when $x\le-\pi$, but that still leaves the intervals $(-\pi,-1)$ and $(0,3)$ to be dealt with.
| {
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about Pythagorean quadruples Respected Mathematicians,
I would like to prepare a function, which will generate Pythagorean quadruples (a, b, c, d) = $d^2$ = $a^2$ + $b^2$ + $c^2$...-> (1). How far I am correct I don't know. For some set of integers a and b, consider $a^2$ + $b^2$ = m and d = c + Ξ΅. Now by (1), c = (m - $Ξ΅^2$)/2Ξ΅)....>(2) Now I will discuss the generation of such quadruples.
For a is even and b is odd or b is even and a is odd;
(1) Let us take a = 12 and b = 15 then m = 369 = $3^2$ X 41
then Ξ΅ = 1, $3^2$ but Ξ΅ = 41 is not possible as $Ξ΅^2$ > m.
Now for c = 184 and d = 185 for Ξ΅ = 1; for c = 16 and d = 25 when $3^2$ = Ξ΅;
So the primitive quadruples for a = 12 and b = 15 are (12, 15, 184, 185) and also (12, 15, 16, 25).
(2) For a = 2 X 3 X 5 X 7 = 210 and b = $3^3$ X 5 = 135 then m = $3^2$ X $5^2$ X 277 = 62325 then Ξ΅ = 1, $3^2$, $5^2$, $3^2$ X $5^2$. Other combination of Ξ΅ are not possible due to $Ξ΅^2$ > m.
If I am correct with the cited examples, let me know the generalization of the Pythagorean quadruples.
Note: I considered a and b have common factors $p_1$, ...$p_n$.
Thanking you.
BABA
| You can assume that $a, b, c, d$ are relatively prime (no prime dividing them all), which in this case need not be the same as coprime (no prime dividing any pair).
This is because by homogeneity any other solution is a multiple of one that is a relatively prime. (This assumption can't be made, where the degrees are non-homogenous, such as $x^2 + y^3 = z^5$.)
Then considering the equation mod 8, noting that $x^2 \equiv$ 1 mod(8) if $x$ is odd, one concludes that exactly two of $a, b, c$ must be even, say $a, b$, so that $c, d$ are both odd and $d - c$ and $d + c$ are both even. So each bracketed term in the following is an integer:
$$\left(\dfrac{a}{2}\right)^2 + \left(\dfrac{b}{2}\right)^2 = \left(\dfrac{d - c}{2}\right) \left(\dfrac{d + c}{2}\right)$$
Now any prime $\equiv$ 3 mod(4) dividing either of the terms on the RHS must divide $a, b$ and hence must divide exactly one of the terms on the RHS to even multiplicity (or else it would divide $a, b, c, d$, contrary to hypothesis).
This, combined with the fact that 2 and every prime $\equiv$ 1 mod(4) can be expressed as the sum of two squares of integers, and their product composed into the same form by repeated use of:
$$(p^2 + q^2) (r^2 + s^2) = (p q - r s)^2 + (p r + q s)^2$$
means that each of
$$\frac{d - c}{2}\quad\text{and}\quad\frac{d + c}{2}$$
must be the sum of two squares of integers, say $v^2 + w^2$, $t^2 + u^2$ respectively
You can think of a sum of two squares as the norm of a so-called Gaussian integer, of the form $x + y \sqrt{-1}$, the norm being the product of this and its conjugate $x - y \sqrt{-1}$ to give $x^2 + y^2$.
The nice thing about Gaussian integers, also referred to as the ring $Z[\sqrt{-1}]$, is that factorisation into "primes" of the same form is unique, just like the rational ("ordinary") integers.
So treating the first equation above as expressing a Gaussian integer as the product of two others, i.e.
$$\dfrac{a}{2} + \dfrac{b}{2} \sqrt{-1} = (v + w \sqrt{-1}) (t + u \sqrt{-1})$$
where the norms of the bracketed terms on the RHS correspond to $\dfrac{d - c}{2}$ and $\dfrac{d + c}{2}$ respectively, one can conclude by multiplying out the RHS and equating coefficients (of $\sqrt{-1}$ and "1") that there must be some set of integers $t, u, v, w$ for which:
$$\begin{align*}
\dfrac{a}{2} &= t v - u w\\
\dfrac{b}{2} &= u v + t w
\end{align*}$$
and taking norms gives:
$$\begin{align*}
\dfrac{d - c}{2} &= v^2 + w^2\\
\dfrac{d + c}{2} &= t^2 + u^2\\
\end{align*}$$
i.e.
$$\begin{align*}
c &= t^2 + u^2 - v^2 - w^2\\
d &= t^2 + u^2 + v^2 + w^2
\end{align*}$$
I imagine this is very much like the solutions given in the earlier questions cited; but it's useful revision working it out again ;-)
| {
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Use mathematical induction to prove that for all integers $n \ge 2$, $2^{3n}-1$ is not prime I had a homework due yesterday with this problem.
The TA did the problem last week in discussion but I didn't understand it.
She pulled out a $7k$ almost immediately, and I have no idea from where.
It was like, it wasn't prime if $2^{3n}-1$ was 7 times some constant. I understand that that would make $2^{3n}-1$ not prime, but I don't understand how she just used "7".
Where did she get that from? I was thinking, maybe it was like $2^3 \cdot 2^n-1$... which is $8 \cdot 2^n-1$... but you can't just do $8-1$. How?
| Iβm sure that she used the following factorization
$$\begin{align*}
2^{3n}-1&=(2^3)^n-1\\
&=8^n-1^n\\
&=(8-1)(8^{n-1}\cdot 1^0+8^{n-2}\cdot 1^1+\dots+8^2\cdot 1^{n-3}+8^1\cdot 1^{n-2}+8^0\cdot 1^{n-1}\\
&=7(8^{n-1}+8^{n-2}+\dots+8^2+8+1)\\
&=7\sum_{k=0}^{n-1}8^k\;,
\end{align*}$$
which as Dylan Moreland pointed out is a special case of $$\begin{align*}a^n-b^n&=(a-b)(a^{n-1}b^0+a^{n-1}b^1+\dots+a^1b^{n-2}+a^0b^{n-1})\\
&=(a-b)\sum_{k=0}^{n-1}a^kb^{n-1-k}\end{align*}\;.$$
This is really just the formula for the sum of a finite geometric series in disguise:
$$\begin{align*}
\sum_{k=0}^{n-1}a^kb^{n-1-k}&=b^{n-1}\sum_{k=0}^{n-1}\left(\frac{a}b\right)^k\\
&=b^{n-1}\frac{\left(\frac{a}b\right)^n-1}{\frac{a}b-1}\\
&=\frac{\frac{a^n}b-b^{n-1}}{\frac{a-b}b}\\
&=\frac{a^n-b^n}{a-b}\;,
\end{align*}$$
and multiplying through by $a-b$ yields the factorization formula.
| {
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Computing $\sum_{1}^{\infty}\frac{1}{2^{2n+1}n} $ with a power series- What did I do wrong? The requested sum: $\sum_{1}^{\infty}\frac{1}{2^{2n+1}n}=
\frac{1}{2}\sum_{0}^{\infty}\frac{1}{4^nn}$
My helper- this power series: $
\sum_{0}^{\infty}\frac{x^{n}}{4^n}=\frac{1}{1-\frac{4}{x}}
$
Integration due to uniform convergence: $ \int \sum_{0}^{\infty}\frac{x^{n}}{4^n}=\int \frac{1}{1-\frac{4}{x}}$
This is what I get:$\sum_{0}^{\infty}\frac{x^{n+1}}{4^{n+1}(n+1)}=-4\ln(4-x)$
Or: $
\sum_{1}^{\infty}\frac{x^{n}}{4^{n}n}=-4\ln(4-x)
$
Finally what we need is: $\frac{1}{2}\sum_{1}^{\infty}\frac{x^{n}}{4^{n}n}=-2\ln(4-x)$
Now I want to plug-in $x=1$ and get the requested result, but what bothers me is that this is a positive series and I get a negative number in the RHS, obivously something's wrong.
Please tell me what's wrong with the steps described above.
Thanks! :)
| The problem with your method is that you're using primitives and not definite integrals:
$$\sum\limits_{n = 0}^\infty {\frac{1}{{n + 1}}{{\left( {\frac{x}{4}} \right)}^{n + 1}}} = - \log \left( {1 - \frac{x}{4}} \right) = - \log \left( {4 - x} \right) + \log 4 = \int\limits_0^x {\frac{1}{{4 - t}}dt} $$
Since you know
$$-\log \left( {1 - x} \right) = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{n}} $$
You simply plug in $1/4$. You get
$$\eqalign{
& - \frac{1}{2}\log \left( {1 - \frac{1}{4}} \right) = \frac{1}{2} \sum\limits_{n = 1}^\infty {\frac{1}{{n{4^n}}}} \cr
& \frac{1}{2}\log \left( {\frac{4}{3}} \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{1}{{n{4^n}}}} \cr} $$
You should be thinking about differentiating, not integrating. You have
$$\frac{1}{2}\sum\limits_{n = 1}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}} $$
So you might want to find
$$\frac{1}{2}\sum\limits_{n = 1}^\infty {n{x^n}} = f\left( x \right)$$
Adn the plug in $1/4$. Use
$$\eqalign{
& F\left( x \right) = \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} \cr
& x\frac{d}{{dx}}F\left( x \right) = x\frac{d}{{dx}}\frac{1}{{1 - x}} = x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty {{x^n}} \cr
& xF'\left( x \right) = \frac{x}{{{{\left( {1 - x} \right)}^2}}} = \sum\limits_{n = 1}^\infty {n{x^n}} \cr
& \frac{1}{2}xF'\left( x \right) = \frac{1}{2}\frac{x}{{{{\left( {1 - x} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty {n{x^n}} \cr} $$
Now plug in $1/4$ to get
$$\frac{1}{2}\frac{{\frac{1}{4}}}{{{{\left( {1 - \frac{1}{4}} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}} $$
$$\frac{2}{9} = \frac{1}{2}\sum\limits_{n = 0}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/108925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere.
Can anybody provide a hint as to how I should be looking at this problem?
| Let $d=\gcd(n^3+1,n^2+2)$ then we have:
$d|n^3+2n\Rightarrow\ d|2n-1\Rightarrow\ d|2n^2-n\Rightarrow\ d|2n^2+4\Rightarrow\ d|n+4\Rightarrow\ d|2n+8\Rightarrow\ d|9\Rightarrow\ d=1,3,9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
} |
Combining Taylor expansions How do you taylor expand the function $F(x)={x\over \ln(x+1)}$ using standard results? (I know that WA offers the answer, but I want to know how to get it myself.) I know that $\ln(x+1)=x-{x^2\over 2}+{x^3\over 3}+β¦$ But I don't know how to take the reciprocal. In general, given a function $g(x)$ with a known Taylor series, how might I find $(g(x))^n$, for some $n\in \mathbb Q$?
Also, how might I evaluate expressions like $\ln(1+g(x))$ where I know the Taylor expansion of $g(x)$ (and $\ln x$). How do I combine them?
Thank you.
| Concerning the multiplicative inverse of $\ln(1+x)$ you may use the classical division (here $x$ is divided by $\ln(1+x)$ ) :
$\begin{array} {r|l}
x +0 x^2+0 x^3 & x-\frac{x^2}2+\frac{x^3}3+\cdots \\
-x+\frac{x^2}2-\frac{x^3}3 & 1+\frac x2-\frac {x^2}{12}\\
\frac{x^2}2-\frac{x^3}3 & \\
-\frac{x^2}2+\frac{x^3}4 & \\
-\frac{x^3}{12} & \\
\end{array}
$
So that $\displaystyle \frac x{\ln(1+x)}=1+\frac x2-\frac {x^2}{12}$
Concerning integer powers ($f(x)^n$ with $n\in \mathbb{N}$) just apply distributivity of multiplication $n-1$ times ($f(x)=\ln(1+x)$ here) :
$\ln(1+x)^2= \left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)\cdot\left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)$
$=x\cdot\left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)-\frac{x^2}2\cdot\left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)+\frac{x^3}3\cdots$
For higher powers use :
$\ln(1+x)^3=\ln(1+x)^2\cdot \ln(1+x)$ and so on...
Sometimes it will be more convenient to rewrite the exponent in powers of $2$.
For example $\displaystyle f^9= f^{1+2^3}=f\cdot(((f^2)^2)^2$.
To compute $(1+f(x))^p$ you may use following expansion :
$(1+f(x))^p= 1+ p f(x)+ \frac{p(p-1)}{2!} f(x)^2 + \frac{p(p-1)(p-2)}{3!} f(x)^3+\cdots \binom{p}{k} f(x)^k+\cdots$
Note that you'll sometimes have to divide or multiply by a power of $x$ to apply
these methods (in the division we replaced $\frac1{\ln(1+x)}$ by $\frac x{\ln(1+x)}$, here we have $1+f(x)$ so that when computing for example $(x+f(x))^p$ we may rewrite this as $x^p\cdot(1+\frac{f(x)}x)^p$).
For any power $p$ (or when $\ln(1+f(x))$ is simple) you may use :
$\displaystyle (1+f(x))^p=e^{p\cdot \ln(1+f(x))}=e^{p\cdot \left(f(x)-\frac{f(x)^2}2+\cdots\right)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Using double angle formulas in integration, trouble following an example. I have just started looking at integration and I am having trouble understanding what has been done in one of the examples in the book I am working through.
It involves using the double angle formula for $\sin(2\theta)$ to provide a rearrangement for which an indefinite integral can then be found.
The double angle formula provided is $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and the example is as follows:
$$\int\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)dx=\int\frac{1}{2}\sin\left(x\right)dx$$
$$=-\frac{1}{2}\cos\left(x\right)+c$$
The part of this example I am specifically stuck with is the first line where $\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)$ is rewritten as $\frac{1}{2}\sin\left(x\right)$ using the previously stated double angle formula.
| You know that $$\sin(2 \theta)=2\sin(\theta)\cdot\cos(\theta)$$ Now, put $\theta=\dfrac{1}{2}x$ to see that, $$\begin{align}\sin\left(2 \cdot\dfrac{1}{2}x\right)&=2\sin\left(\dfrac{1}{2}x\right)\cdot\cos\left(\dfrac{1}{2}x\right)\\\sin(x)&=2\sin\left(\dfrac{1}{2}x\right)\cdot\cos\left(\dfrac{1}{2}x\right)\\\sin\left(\dfrac{1}{2}x\right)\cdot\cos\left(\dfrac{1}{2}x\right)&=\dfrac{1}{2}\cdot\sin(x)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Absolute Convergence of an Infinite Product based on Weierstrass's Factor Theorem I am trying to show that $\left\{ \left( 1-\dfrac {z} {\pi }\right) e^{\left( \dfrac {z} {\pi }\right) }\right\} \left\{ \left( 1+\dfrac {z} {\pi }\right) e^{\left( -\dfrac {z} {\pi }\right) }\right\} \left\{ \left( 1-\dfrac {z} {2\pi }\right) e^{\left( \dfrac {z} {2\pi }\right) }\right\} \left\{ \left( 1+\dfrac {z} {2\pi }\right) e^{\left( -\dfrac {z} {2\pi }\right) }\right\}\ldots $ Converges absolutely.
So the general term is of the form $\left( 1\pm \dfrac {z} {m\pi }\right) e^{\pm \dfrac {z} {m\pi }}$ I think it's the $e^{\pm \dfrac {z} {m\pi }}$ which is giving me grief. Any help would be much appreciated.
Edit:
Is multiplying the $2m -1 $ and the $2m$ terms and canceling out the $e^{\pm \dfrac {z} {m\pi }}$ parts and $\left( 1^2\ - (\dfrac {z} {m\pi })^2\right)$ terms be allowed or would that run the risk of altering the value of the product ?
| This product is quite a famous one:
We have that the Gamma function defined as:
$$\Gamma(z) = \int\limits_0^\infty e^{-t} t^{z}\frac{dt}{t}$$
is convergent for $\Re(z)>0$
It is not too difficult to prove that, fo $n\in \mathbb{N}$,
$$\lim\limits_{n \to \infty} \frac{\Gamma({z+n})}{ n^{z-1}n!}=1$$
This with the functional equation of $\Gamma$ gives the identity:
$$\Gamma(z) = \lim\limits_{n \to \infty} \frac{n^{z} n!}{\prod\limits_{v=0}^{n} (z+v)}$$
Let's write
$$\tag{1} \frac{1}{\Gamma_n(z)} = \frac{\prod\limits_{v=0}^{n} (z+v)}{ n^{z} n!}$$
We know that
$$\lim\limits_{n \to \infty} \frac{1}{\Gamma_n(z)} =\frac{1}{\Gamma(z)} $$
We can write $(1)$ differently, as
$$ \frac{1}{\Gamma_n(z)} = \frac{z}{n^z}{\prod\limits_{v=1}^{n} \left(\frac{z}{v}+1 \right)}$$
Now,
$${n^z} = {e^{z\log n}} = {e^{ - z\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n} \right)}}{e^{z\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}} \right)}}$$
so we can write
$$\frac{1}{{{\Gamma _n}(z)}} = z{e^{z\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n} \right)}}\prod\limits_{v = 1}^n {\left( {\frac{z}{v} + 1} \right)} {e^{ - \frac{z}{v}}}$$
Letting $n\to \infty$ gives:
$$\frac{1}{{\Gamma (z)}} = z{e^{z\gamma }}\prod\limits_{v = 1}^\infty {\left( {\frac{z}{v} + 1} \right)} {e^{ - \frac{z}{v}}}$$
where $\gamma\approx 0.577\dots$ is Euler's constant.
This means that
$$\frac{1}{{\Gamma ( - z)}} = - z{e^{ - z\gamma }}\prod\limits_{v = 1}^\infty {\left( {1 - \frac{z}{v}} \right)} {e^{\frac{z}{v}}}$$
or that
$$\eqalign{
& \frac{1}{{\Gamma ( - z)}}\frac{1}{{\Gamma (z)}} = - z{e^{ - z\gamma }}z{e^{z\gamma }}\prod\limits_{v = 1}^\infty {\left( {1 - \frac{z}{v}} \right)} \left( {1 + \frac{z}{v}} \right){e^{ - \frac{z}{v}}}{e^{\frac{z}{v}}} \cr
& \frac{1}{{\Gamma ( - z)}}\frac{1}{{\Gamma (z)}} = - {z^2}\prod\limits_{v = 1}^\infty {\left( {1 - \frac{z}{v}} \right)} \left( {1 + \frac{z}{v}} \right) \cr} $$
But $$\frac{1}{{ - z\Gamma ( - z)}} = \frac{1}{{\Gamma (1 - z)}}$$ so
$$\frac{1}{{\Gamma (1 - z)}}\frac{1}{{\Gamma (z)}} = z\prod\limits_{v = 1}^\infty {\left( {1 - \frac{{{z^2}}}{{{v^2}}}} \right)} $$
Now put $z=\dfrac{s}{\pi}$, and your expression will appear. Note that the exponentials cancel out, and the infinite product converges everywhere.
| {
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"url": "https://math.stackexchange.com/questions/116417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Solve $ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $ $$ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $$
I tried squaring and/or using $1-\cos x=2\sin^2{\frac{x}2}$, but no luck.
| We go some distance to a solution. Might as well get rid of the silly extra $2$'s, and solve
$$\sqrt{1-\cos x}+\sqrt{5-3\cos x}=\sqrt{5-3\cos 2x}=\sqrt{8-6\cos^2 x}.$$
From now on write $w$ instead of $\cos x$. So we are solving
$$\sqrt{1-w}+\sqrt{5-3w}=\sqrt{8-6w^2}.$$
Square both sides, simplify a bit. We get
$$\sqrt{1-w}\sqrt{5-3w}=-(3w^2-2w-1)=(1-w)(3w+1).$$
Note that each side has a $1-w$. Cancel, remembering the root $w=1$. Then square both sides again. We get
$$5-3w=(1-w)(3w+1)^2.$$
A cubic! But not a terrible cubic, it is $9w^3-3w^2-8w+4=0$, which happens to have $-1$ as a root. So divide by $w+1$, we get a quadratic. Solve, and throw
away anything that has snuck its way in as a result of the squaring process.
Remark: Things could have turned ugly. There was the happy "accident" that allowed us to partially get rid of a $1-w$ factor. And then there was the other happy accident of an obvious root $w=-1$. A small perturbation of the numbers could make things difficult.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Mathematical Analysis: Mean Value Theorem I'm trying to learn how the mean value theorem works by attempting to answer the following but i just dont understand it at all.
Consider the function $f(x)$ $=$ $\sqrt x$ for $x > 0$
i. Show that $f'(x)$ is a decreasing function on $(0, \infty)$
ii. Apply the Mean Value Therorem to $f$ on the interval $[100,102]$ to prove
$10 + \frac{1}{11} \leq \sqrt {102} \leq 10 + \frac{1}{10}$
Any help would be appreciated.
| We are given that $f(x) = \sqrt{x}$.
Hence, the first and second derivatives of $f$ are$f'(x) = \frac1{2\sqrt{x}}$ and $f''(x) = -\frac1{4x^{3/2}}$. Note that $f''(x) < 0$ for all $x \in \left(0, \infty\right)$. If the first derivative of a function is always negative, then the function is a decreasing function.
In our case, $f''(x)$ is the first derivative of $f'(x)$ and since $f''(x)<0, \forall x \in \left(0, \infty \right)$, we get that $f'(x)$ is a decreasing function.
For the second part, by mean value theorem, we have that if a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a < b$, and differentiable on the open interval $(a, b)$, then there exists a point $c \in (a, b)$ such that
$$f'(c) = \frac{f(b)-f(a)}{b-a}$$ i.e. $f(b) = f(a) + (b-a) \times f'(c)$. To prove the second part, choose $f(x) = \sqrt{x}$, $a = 100$ and $b=102$. Clearly $f$ satisfies the assumptions in the mean value theorem. Hence, $\exists c \in (a,b)$ such that $\sqrt{b} = \sqrt{a} + (b-a) f'(c) = \sqrt{a} + (b-a) \frac1{2 \sqrt{c}}$. Plugging in the values for $a$ and $b$ we get that $$\sqrt{102} = \sqrt{100} + \frac{2}{2 \sqrt{c}} = 10 + \frac1{\sqrt{c}}$$
Since $c \in (100,102)$, we have that $10 < \sqrt{c} <\sqrt{102} < \sqrt{121} = 11$. Hence, $\frac1{11} < \frac1{\sqrt{c}} < \frac1{10}$. Hence, $10 + \frac1{11} < 10 + \frac1{\sqrt{c}} < 10 + \frac1{10}$ i.e. $10 + \frac1{11} < \sqrt{102} < 10 + \frac1{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For complex $z$, find the roots $z^2 - 3z + (3 - i) = 0$
Find the roots of:
$z^2 - 3z + (3 - i) = 0$
$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$
$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$
So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I proceed?
| Use $(z-a)(z-b)=z^2-(a+b)+ab=0$ to get
$$
(a+b)=3 \tag{1}
$$
$$
ab=(3-i) \tag{2}
$$
From (1) you get $\Im(a)=-\Im(b)$. So $(a_r+ia_i)(b_r-ia_i)=(a_rb_r+a_i^2)+i(-a_r+b_r)a_i=(3-i)$, thus $a_rb_r+a_i^2=3$ and $(b_r-a_r)a_i=-1$. Now you can try a few values like ...
$a_i=1$ and figure out that $a_r=2$ and $b_r=1$.
So you finally rewrite it as $(z-(2+i))(z-(1-i))$.
You could also just use the standard way (as proposed by anon) to get:
$$
\frac{3\pm\sqrt{9-4(3-i)}}{2}=\frac{3\pm\sqrt{-3+4i}}{2}.
$$
Now note that $(i(2-i))^2=-3+4i$, so
$$ \frac{3\pm i(2-i)}{2}=\frac{3\pm (2i+1)}{2}=\frac{3\pm 1 }{2} \pm i $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all integer solutions to $\displaystyle{2^k = 9^m+7^n}$ I can only find the trivial ones
$(m,n,k)$ as $(0,0,1), (1,1,4)$
Cannot find any more. Are there any more values? More imporantly how to show those are the only ones.
| Suppose $m>0$.
If $2^k = 9^m + 7^n$, then $2^k \equiv 1 \pmod 3$, thus $k \equiv 0 \pmod 2$.
Put $k = 2l$ : $7^n = 4^l - 9^m = (2^l - 3^m)(2^l + 3^m)$.
If $(2^l - 3^m) > 1$, then both factors are multiples of $7$, and thus so is $3^m$, which is impossible.
Thus $2^l - 3^m = 1$. The only solution to this is $l=2$ and $m=1$ which gives the $(1,1,4)$ solution.
If $m=0$ then we need $2^k - 7^n = 1$, and the only solutions to this are $k=1,n=0$ and $k=3,n=1$ which give the solutions $(0,0,1)$ and $(0,1,3)$.
In both cases, we use Catalan's (now proven) conjecture, stating that the only solution to $x^a - y^b = 1$ for $a,b > 1$ is $3^2=1+2^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show $ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$ Show $$ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$$
| First notice that the function $\displaystyle{\frac{1}{1+\cos^2 x}}$ is an even function, and therefore
$$ \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = 2 \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+\cos^2 x}$$
$$ \begin{align*}
2 \int_0^{\frac{\pi}{2}} \frac{1}{1+\cos^2 x} \,\mathrm{d}x &= 2 \int_0^{\frac{\pi}{2}} \frac{\frac{1}{\cos^2 x}}{\left(\frac{1}{\cos^2 x}+1\right)}\mathrm{d}x \\
&= 2 \int_0^{\frac{\pi}{2}} \frac{\sec^2 x}{\sec^2 x+1} \mathrm{d}x \\
&= 2 \int_0^{\frac{\pi}{2}} \frac{\sec^2 x}{(\tan^2 x+1)+1} \mathrm{d}x & (\text{because} \hspace{4pt} \sec^2 x=\tan^2 x+1)\\
&= \int_0^{\frac{\pi}{2}} \frac{{\small{2}} \sec^2 x}{\tan^2 x+2} \mathrm{d}x \\
&= \int_0^{\frac{\pi}{2}} \frac{\sec^2 x}{\left(\frac{\tan x}{\sqrt 2}\right)^2+1}\mathrm{d}x
\end{align*}
$$
Now substitute
$$\frac{\tan x}{\sqrt 2} = t \Longrightarrow \sec^2 x \; \mathrm{d}x = \sqrt 2 \; \mathrm{d}t \hspace{5pt}$$
applying the new limits, the integral gets simplfied to
$$
\begin{align*}
\sqrt 2 \int_0^\infty \frac{\mathrm{d}t}{t^2+1} &= \sqrt 2 \left( \left. \tan^{-1} t \right|_0^\infty \right) \\
&= \sqrt 2 \left(\frac{\pi}{2}\right) = \frac{\pi}{\sqrt{2}}
\end{align*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing $ \iint_{[0,1]^2} \frac{-x\ln(xy)}{1-xy} \mathrm dx \mathrm dy $ I would like to compute $$ \iint_{[0,1]^2} \frac{-x\ln(xy)}{1-xy} \mathrm dx \mathrm dy $$
Without going into detail, here is what I found:
$$ \int_{0}^{1}(\int_{0}^{1} \frac{-x\ln(xy)}{1-xy} \mathrm dx ) \mathrm dy=\int_{0}^{1}(-\sum_{n=0}^{\infty} \int_{0}^{1} x^{n+1}y^n\ln(xy) \mathrm dx)\mathrm dy $$
$$ \int_{0}^{1}\sum_{n=1}^{\infty}\frac{y^n}{(n+1)^2}\mathrm dy=\sum_{n=1}^{\infty} \frac{1}{(n+1)^3}\approx0.202 $$
However Wolfram gives: $$ \iint_{[0,1]^2} \frac{-x\ln(xy)}{1-xy} \mathrm dx \mathrm dy=1 $$
Where is the problem?
| \begin{align*}\int_0^1x^{n+1}y^n\ln(xy)dx&=y^n\int_0^1x^{n+1}\ln xdx+y^n\ln y\int_0^1x^{n+1}dx\\
&=y^n\left[\frac{x^{n+2}}{n+2}\ln x\right]_0^1-y^n\int_0^1\frac{x^{n+2}}{n+2}\frac 1xdx
+y^n\ln y\frac 1{n+2}\\
&=-\frac{y^n}{(n+2)^2}+y^n\frac{\ln y}{n+2}\\
&=\frac{y^n}{n+2}\left(\ln y-\frac 1{n+2}\right)
\end{align*}
and
\begin{align*}
\int_0^1\frac{-x\ln(xy)}{1-xy}&=-\sum_{n=0}^{+\infty}\int_0^1\int_0^1x^{n+1}y^n\ln(xy)dxdy\\
&=-\sum_{n=0}^{+\infty}\frac 1{n+2}\int_0^1y^n\left(-\frac{1}{n+2}+\ln y\right)dy\\
&=-\sum_{n=0}^{+\infty}\frac 1{(n+2)^2(n+1)}-\sum_{n=0}^{+\infty}\frac 1{n+2}\frac 1{n+1}\left(-\int_0^1y^{n+1}\frac 1ydy\right)\\
&=-\sum_{n=0}^{+\infty}\frac 1{(n+2)^2(n+1)}+\sum_{n=0}^{+\infty}\frac 1{(n+2)(n+1)^2}\\
&=-\sum_{n=0}^{+\infty}\frac 1{(n+2)(n+1)}\left(\frac 1{n+2}-\frac 1{n+1}\right)\\
&=\sum_{n=0}^{+\infty}\frac 1{(n+2)(n+1)}\\
&=\sum_{j=1}^{+\infty}\frac{j+1-j}{(j+1)j}\\
&=\sum_{j=1}^{+\infty}\frac 1j-\frac 1{j+1}=1
\end{align*}
so Wolfram is right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/129660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the expected value of a dice sum If fair dodecahedron is rolled until at least $k$($k$ is fixed between 2 and 12) is gotten, and $X$ is the sum of all numbers appeared until the last time, what is $E(X)$?
| The probability that any roll is greater than or equal to $k$ is
$$
\frac{13-k}{12}
$$
so the expected number of rolls until a roll of $k$ or greater is
$$
\frac{12}{13-k}.
$$
All but the last one of these rolls is less than $k$, so the sum of those rolls has an expected value of
$$
\left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2}.
$$
Add to this the expected value of the final roll
$$
\frac{k+12}{2}
$$
and so the expectation of the sum is
$$
\left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2} + \frac{k+12}{2} = \frac{78}{13-k}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Differentiation of $y = \tan^{-1} \Bigl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\Bigr\}$ How do i differentiate the following: $$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}$$
I know that $\text{derivative}$ of $\tan^{-1}{x}$ is $\frac{1}{1+x^{2}}$ but not sure as to how to do this.
| First the idea is to observe as to what would you substitute for $x$ in order to remove the square root. After some manipulations you find that the correct substitution is $x^{2} = \sin{2\theta}$. Once you have done this then you have $$y = \tan^{-1}\biggl\{ \frac{(\cos\theta + \sin\theta) - (\cos\theta - \sin\theta)}{(\cos\theta + \sin\theta)+(\cos\theta - \sin\theta)}\biggr\} = \theta = \frac{1}{2}\sin^{-1}{x^2}$$
So if $y = \frac{1}{2}\sin^{-1}{x^2}$ then $$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1-x^{4}}} \cdot 2x = \frac{x}{\sqrt{1-x^{4}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Finding lowest value measures between three locations I've made an image of what the geography of my problem looks like:
Essentially, there are two "chemical plants" (A and B) located on a road $12$ miles apart. The pollution from plant A is given by the following equation (for some constant $K$):
$$\frac{K}{x^2 + 10}$$
The pollution from plant B (at $x$ miles from plant B) is $1/4$ that of A.
There's also a third plant, C, which is located on the perpendicular road (the one branching off of Road A-B on the map). Plant C is $5$ miles from A and $10$ miles from B. The pollution from plant C is twice that of B.
I am trying to find the point on Road A-B where the pollution count from the three plants is minimal. I'm not really sure where to even begin with this.
The pollution from Plant B must be
$$\frac{K}{4x^2 + 40},$$
and from Plant C must be
$$\frac{K}{2x^2 + 20}.$$
Adding the three equations would yield the following:
$$\frac{K}{x^2 + 10} + \frac{K}{4x^2 + 40} + \frac{K}{2x^2 + 20} = \frac{K + 4K + 2K}{x^2 + 10} = \frac{7K}{x^2 + 10}$$
Finding the derivative of this would be? Not sure I'm doing this right.
$$\left(\frac{7K}{x^2 + 10}\right)' = -\frac{14K}{(x^2 + 10)^2}$$
| Choose a co-ordinate system with Plant A as origin.
Basic trigonometry to find the angles of the triangle ABC
You will get (in Radians)
$\angle A = 0.9582$
$\angle B = 0.4214$
$\angle C = 1.762$
Now the position of Plant C is straightforward $(5\cos A, -5\sin A)$
Assuming the origin is fixed at Plant A the amount of pollution at any arbitary point (x, y) is
$\frac{K}{x^2 + y^2 + 10}+ \frac{K}{4((x-12)^2 + y^2 + 10))} + \frac{K}{2((x-2.874968)^2+(y+4.0907895)^2+10))}$
On Road AB this reduces to $\frac{K}{x^2+10}+ \frac{K}{4((x-12)^2+10))} + \frac{K}{2((x-2.874968)^2+26.7345587))}$
Factoring out K and minimizing $\frac{1}{x^2+10}+ \frac{1}{4((x-12)^2+10))} + \frac{1}{2((x-2.874968)^2+26.7345587))}$
Minimum occurs at $x \approx 8.7758$ miles from Plant A on Road AB
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the $\sin$, $\cos$ and $\tan$ without using calculator?
Evaluate the $\sin$, $\cos$ and $\tan$ without using calculator?
$150$ degree
the right answer are $\frac{1}{2}$, $-\frac{\sqrt{3}}{2}$and $-\frac{1}{\sqrt{3}} $
$-315$ degree
the right answer are $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$ and $1$.
| It is very simple.
$\sin(150) = \sin(90+60) = \cos(60) = \frac{1}{2}$
$\cos(150) = \cos(90+60) = - \sin(60) = -\frac{\sqrt{3}}{2}$
$\tan(150) = \tan(90+60) = - \cot(60) = -\frac{1}{\sqrt{3}} $
similarly
$\sin(-315)=-\sin(270+45)=\cos(45) = \frac{1}{\sqrt{2}}$
$\cos(-315)=\cos(315)=\cos(270+45)=\sin(45) = \frac{1}{\sqrt{2}}$
$\tan(-315)=-\tan(270+45)=\cot(45) = 1$
Check this link for more info on converting trigo identities
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How to solve this Diophantine Equation. Step by Step What are all the solutions to the diophantine equation:
6x-6y-xy=0.
| The 18 integer solutions are :
x = -42, y = 7
x = -24, y = 8
x = -18, y = 9
x = -15, y = 10
x = -12, y = 12
x = -10, y = 15
x = -9, y = 18
x = -8, y = 24
x = -7, y = 42
x = -5, y = -30
x = -4, y = -12
x = -3, y = -6
x = -2, y = -3
x = 0, y = 0
x = 3, y = 2
x = 6, y = 3
x = 12, y = 4
x = 30, y = 5
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is this series called? $\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!}$ I remember learning about this series in Precalculus the other day but I neglected to get the name of it. It looks something like this:
$
\begin{align*}
\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!}
\end{align*}
$
All I remember is that it helps in the modeling of $\sin{x}$.
| Are you sure you didn't get the summands flipped?
The following series is called the Taylor series expansion of $\sin{x}$:
\begin{align*}
\sin{x} &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} \\[8pt]
\end{align*}
It's derived here.
| {
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"url": "https://math.stackexchange.com/questions/138470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Interesting Determinant Let $x_1,x_2,\ldots,x_n$ be $n$ real numbers that satisfy $x_1<x_2<\cdots<x_n$.
Define \begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} & \cdots & x_{n-1}-x_{1} & x_{n}-x_{1} \\
x_{2}-x_{1} & 0 & \cdots & x_{n-1}-x_{2} & x_{n}-x_{2} \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
x_{n-1}-x_{1} & x_{n-1}-x_{2} & \cdots & 0 & x_{n}-x_{n-1} \\
x_{n}-x_{1} & x_{n}-x_{2} & \cdots & x_{n}-x_{n-1} & 0%
\end{bmatrix}%
\end{equation*}
Could you determine the determinant of $A$ in term of $x_1,x_2,\ldots,x_n$?
I make a several Calculation:
For $n=2$, we get
\begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} \\
x_{2}-x_{1} & 0%
\end{bmatrix}%
\text{ and}\det (A)=-\left( x_{2}-x_{1}\right) ^{2}
\end{equation*}
For $n=3$, we get
\begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} & x_{3}-x_{1} \\
x_{2}-x_{1} & 0 & x_{3}-x_{2} \\
x_{3}-x_{1} & x_{3}-x_{2} & 0%
\end{bmatrix}%
\text{ and}\det (A)=2\left( x_{2}-x_{1}\right) \left( x_{3}-x_{2}\right)
\left( x_{3}-x_{1}\right)
\end{equation*}
For $n=4,$ we get
\begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} & x_{3}-x_{1} & x_{4}-x_{1} \\
x_{2}-x_{1} & 0 & x_{3}-x_{2} & x_{4}-x_{2} \\
x_{3}-x_{1} & x_{3}-x_{2} & 0 & x_{4}-x_{3} \\
x_{4}-x_{1} & x_{4}-x_{2} & x_{4}-x_{3} & 0%
\end{bmatrix} \\%
\text{ and} \\ \det (A)=-4\left( x_{4}-x_{1}\right) \left( x_{2}-x_{1}\right)
\left( x_{3}-x_{2}\right) \left( x_{4}-x_{3}\right)
\end{equation*}
Finally, I guess that the answer is $\det(A)=2^{n-2}\cdot (x_n-x_1)\cdot (x_2-x_1)\cdots (x_n-x_{n-1})$. But I don't know how to prove it.
| Clearly the determinant is $0$ if $x_i = x_{i+1}$ (because two adjacent rows are identical) or $x_1 = x_n$ (last row is $-$ first row). So the determinant must be a polynomial divisible by $(x_1 - x_2)(x_2 - x_3) \ldots (x_{n-1} - x_n)(x_n - x_1)$. But the determinant has degree $n$, so it is a constant times this product. To determine what the constant is,
you might try a special case: $x_i = i$.
EDIT: Thanks to J.M.'s remark, you can show that in that special case the inverse of your matrix $A_n$ looks like this:
$$ \pmatrix{ -\frac{1}{2}+\frac{1}{2n-2} & \frac{1}{2} & 0 & 0 & \ldots & 0 & \frac{1}{2n-2}\cr
\frac{1}{2} & -1 & \frac{1}{2} & 0 & \ldots & 0 & 0\cr
0 & \frac{1}{2} & -1 & \frac{1}{2} & \ldots & 0 & 0\cr
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \cr
0 & 0 & 0 & 0 & \ldots & -1 & \frac{1}{2}\cr
\frac{1}{2n-2} & 0 & 0 & 0 & \ldots & \frac{1}{2} & -\frac{1}{2} + \frac{1}{2n-2}\cr}$$
where the elements on the main diagonal are all $-1$ except for the first and last, those
just above and below the diagonal are all $1/2$, the top right and bottom left are $1/(2n-2)$, and everything else is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/144818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Probability question (CDF , PDF etc)
Possible Duplicate:
Please correct my answer (Probability)
I have to calculate the density function of the random variable
$Y= 1-X^2$, given that: $f(x) = \frac{1}{9}(x+1)^2$, where $-1 < x < 2$.
So I finally understood that the domain of Y is $-3 < Y < 1$.
But it seems that i cant continue further... My method is to find the distribution of
Y by using the density function of X , and finally ill find the density function of Y
by taking the derivative of Y's distribution func.
Its not working at all, i am taking wrong integral, wrong spaces, everything wrong...
Could someone please explain me what is the correct way?
Thanks
| It is easiest to start with CDF:
$$
F_Y(y) = \mathbb{P}(Y \leqslant y) = \mathbb{P}(1-X^2 \leqslant y) = \mathbb{P}(X^2 \geqslant 1 - y) = \mathbb{P}(X \geqslant \sqrt{1 - y}) + \mathbb{P}(X \leqslant -\sqrt{1 - y})
$$
Since $$F_X(x) = \begin{cases} 0 & x < -1 \\ 1 &x > 2 \\ \left(\frac{x+1}{3}\right)^3 & -1 \leqslant x \leqslant 2 \end{cases}$$
one easily arrives as $F_Y(y)$
$$
F_Y(y) = F_X(-\sqrt{1-y}) + 1 - F_X(\sqrt{1-y}) = \begin{cases} 1 & y \geqslant 1 \\ 0 & y \leqslant -3 \\ \frac{23 + 3 y}{27} + \frac{y-4}{27} \sqrt{1-y} & y \leqslant 0 \\
1 + 2 \sqrt{1-y} \frac{y-4}{27} & 0 < y < 1
\end{cases}
$$
Now, differentiating with respect to $y$:
$$
f_Y(y) = \begin{cases} \frac{\left(1 + \sqrt{1-y}\right)^2}{18 \sqrt{1-y}} & -3 < y \leqslant 0 \\ \frac{2-y}{9 \sqrt{1-y}} & 0 < y< 1 \\ 0 & \text{otherwise}\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/149012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Evaluating $\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$
| This is a nice little question. I am assuming that $p \in \mathbb{Z}^+$, though same could be said about it when $p \notin \mathbb{Z}^+$. Before getting to the answer lets experiment a bit for small positive integers $p$. To start off, you could try for some values $p$.
For $p=1$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n(n+1)}{2}}{n} - \frac{n}{1+1} \right) = \lim_{n \rightarrow \infty} \frac12 = \frac12$$
For $p=2$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n(n+1)(2n+1)}{6}}{n^2} - \frac{n}{2+1} \right) = \lim_{n \rightarrow \infty} \left(\frac{(n+1)(n+1/2)}{3n} - \frac{n}3 \right)\\ = \lim_{n \rightarrow \infty} \left(\frac{n}3 + \frac12 + \frac1{6n} - \frac{n}3 \right)= \frac12$$
For $p=3$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n^2(n+1)^2}{4}}{n^3} - \frac{n}{3+1} \right) = \lim_{n \rightarrow \infty} \left(\frac{n^2 + 2n + 1}{4n} - \frac{n}4 \right)\\ = \lim_{n \rightarrow \infty} \left(\frac{n}4 + \frac12 + \frac1{4n} - \frac{n}4 \right)= \frac12$$
Hence, we would guess that it is $\dfrac12$ independent of $p$. And this turns out to be right.
Let us denote $1^p + 2^p + \cdots n^p = P_p(n)$. This is a polynomial of degree $p+1$ and is given by
$$P_p(n) = \frac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{p+1-k}$$ where $B_k$ are the Bernoulli numbers. These polynomials are related to the Bernoulli polynomials and there are some really nice results on these polynomials and more can be found here.
Hence, $$\dfrac{P_p(n)}{n^{p}} = \dfrac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{1-k} = \dfrac1{p+1} \left(B_0 n + (p+1) B_1 + \mathcal{O} \left(\frac1n\right) \right)$$
where $B_0 = 1$ and $B_1 = \frac12$.
What you are looking for is $$\lim_{n \rightarrow \infty} \left(\dfrac{P_p(n)}{n^{p}} - \dfrac{n}{p+1} \right) = \lim_{n \rightarrow \infty} \left(\dfrac1{p+1} \left(n + (p+1) B_1 + \mathcal{O} \left(\frac1n\right) \right) - \dfrac{n}{p+1} \right)\\ = \lim_{n \rightarrow \infty} \left(B_1 + \mathcal{O} \left(\dfrac1n \right)\right)= B_1 = \frac12$$ independent of $p$.
Users Did and Ragib Zaman have provided excellent solutions. You might also want to look at EulerβMaclaurin formula which is of significance in this context.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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Something is wrong with this proof, limit $\lim\limits_{(x,y) \to (0,0)} \frac{xy^3}{x^4 + 3y^4}$ Could someone please tell me what is wrong with this proof?
Show that $\lim\limits_{(x,y) \to (0,0)} \dfrac{xy^3}{x^4 + 3y^4}$ does not have a limit or show that it does and find the limit.
I know it is wrong because the limit doesn't exist, but this proof is contradicting me
Proof
Case 1
Assume for $x,y > 0$, then $x^4 + 3y^4 > x^4 > x > 0$
$$\begin{align*}
0 < x < x^4 + 3y^4 &\iff 0 < \dfrac{x}{x^4 + 3y^4} < 1 \\
& \iff 0 < \dfrac{x|y^3|}{x^4 + 3y^4} < |y^3| \\
& \iff \lim\limits_{(x,y) \to (0,0)} 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < \lim\limits_{(x,y) \to (0,0)}|y^3|\\
&\iff 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < 0
\end{align*}$$
Case 2. WLOG Assume $x,y <0$ and combine both cases.
What's wrong the ppoof? I don't find the flaw
| The easiest way to see that a limit doesn't exist is to write
$$
\frac{xy^3}{x^4+y^4}=\frac{x/y}{(x/y)^4+1}\tag{1}
$$
Using $(1)$, it is easy to see that, over any circle around the origin, $\frac{xy^3}{x^4+y^4}$ can take on any value that $\frac{t}{t^4+1}$ can; that is, any value in $\left[{-}\sqrt[4]{\frac{27}{256}}\;,\;\sqrt[4]{\frac{27}{256}}\right]$. Therefore,
$$
\lim_{x,y\to0}\frac{xy^3}{x^4+y^4}\tag{2}
$$
does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/149509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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$\cos(x)$ and $\arccos(x)$ couple limit
Find the value of the following limit:
$$\lim_{n\to\infty} \frac {\cos 1 \cdot \arccos \frac{1}{n}+\cos\frac
{1}{2} \cdot \arccos \frac{1}{(n-1)}+ \cdots +\cos \frac{1}{n} \cdot
\arccos{1}}{n}$$
| A more convenient way to state the sequence is:
$$
\frac{\sum_{k=1}^n\cos\frac{1}{k}\arccos\frac{1}{n-k+1}}{n}
$$
Note that for $0\leq x\leq 1$ we have $1-x\leq\cos x\leq 1$ and $\frac{\pi}{2}-\frac{\pi}{2}x\leq\arccos x\leq \frac{\pi}{2}-x$. Therefore, we have
$$
\tfrac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})\leq\cos\frac{1}{k}\arccos\frac{1}{n-k+1}\leq\frac{\pi}{2}-\frac{1}{n-k+1}.
$$
Thus, a lower bound for the limit is given by the sequence
$$
\frac{\sum_{k=1}^n\frac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})}{n}=\frac{\pi}{2}-\frac{\sum_{k=0}^n\frac{(n-k+1)+k-1}{k(n-k+1)}}{n}=\frac{\pi}{2}-\sum_{k=1}^n\frac{1}{k(n-k+1)}
$$
The terms in the latter sum can be rewritten to
$$
\frac{1}{k(n+1)}+\frac{1}{(n-k+1)(n+1)}
$$
so we get the sums
$$
\frac{1}{n+1}\sum_{k=1}^n\frac{1}{k}\qquad\text{and}\qquad\frac{1}{n+1}\sum_{k=1}^n\frac{1}{n-k+1}.
$$
Sacha indicated a proof in the comments that these sums tend to zero. Similarly, for the upper bound we have
$$
\frac{\sum_{k=1}^n\frac{\pi}{2}-\tfrac{1}{n-k+1}}{n}=\frac{\pi}{2}-\frac{\sum_{k=1}^n\frac{1}{n-k+1}}{n}\to\frac{\pi}{2}
$$
This gives an upper bound of $\frac{\pi}{2}$. This finishes the proof that the limit is $\frac{\pi}{2}$.
Note: Another proof that $\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\to 0$. Using Cauchy-Schwarz, we see that
$$
\sum_{k=1}^n\frac{1}{k}\leq \sqrt{n}\sqrt{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}
$$
Therefore we get
$$
\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\leq\sqrt{\frac{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}{n}}
$$
In the square root on the right hand side, the numerator converges, so the whole tends to zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Another limit related to pi number Find the value of the limit:
$$\lim_{n\to\infty} \sum_{k=0}^n \frac{{k!}^{2} {2}^{k}}{(2k+1)!}$$
I'm trying to find out if this limit can be computed only by using high school
knowledge for solving limits. Thanks.
| Mimicking robjohn's solution to the series, and after proving convergence, we may proceed as follows:
$$\sum\limits_{k = 0}^\infty {\frac{{k!^2{2^k}}}{{\left( {2k + 1} \right)!}}} = \sum\limits_{k = 1}^\infty {\frac{{\left( {k - 1} \right)!^2{2^{k - 1}}}}{{\left( {2k - 1} \right)!}}} = \sum\limits_{k = 1}^\infty {\frac{{{\Gamma ^2}\left( k \right)}}{{\Gamma \left( {2k} \right)}}{2^{k - 1}}} = \sum\limits_{k = 1}^\infty {\operatorname{B} \left( {k,k} \right){2^{k - 1}}} $$
$$\sum\limits_{k = 1}^\infty {\operatorname{B} \left( {k,k} \right){2^{k - 1}}} = \sum\limits_{k = 1}^\infty {\int\limits_0^1 {{{\left[ {2t\left( {1 - t} \right)} \right]}^{k - 1}}dt} } = \int\limits_0^1 {\sum\limits_{k = 1}^\infty {{{\left[ {2t\left( {1 - t} \right)} \right]}^{k - 1}}} dt} $$
Then
$$=\int\limits_0^1 {\frac{{dt}}{{1 - 2t\left( {1 - t} \right)}}} = \int\limits_0^1 {\frac{{dt}}{{1 - 2t + 2{t^2}}}} = \frac{1}{2}\int\limits_0^1 {\frac{{dt}}{{{{\left( {t - \frac{1}{2}} \right)}^2} + \frac{1}{4}}}} $$
Now let $t-1/2=u$.
$$\frac{1}{2}\int\limits_{ - 1/2}^{1/2} {\frac{{du}}{{{u^2} + {{\left( {1/2} \right)}^2}}}} = \arctan 2\frac{1}{2} - \arctan 2\left( { - \frac{1}{2}} \right)$$
$$=2\arctan 1=2\frac{\pi}{4}=\frac{\pi}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Every number $2^N > 4$ can be expressed by the sum of 2 primes? For example the first cases are:
$2^3= 8 = 3+5$
$2^4= 16 = 3+13$
and so on ...
| 32=2^5.It is of the form 4k.we know that primes of the form 4a+1,4b-1. Now we write 32=4a+1+4b-1.then a+b=8. By solving a+b=8, we get the pairs (a,b)=(3,5),(7,1) for which 4a+1,4b-1 are primes. In this way we can find 2 primes for every power 2. For ex. 22 is of the form 4k+2. Now we write 22= 4a+1+4b+1where 4a+1 and 4b+1 are primes now we get a+b=5 For (a,b)=(1,4),we get 22=5+17.now we write 22=4a-1+4b-1 From this we get a+b=6For (a,b)=(1,5),(3,3) we can write 22=3+19=11+11.If N =4n=p+q where p, q are primes of the form 4a+1 and 4b-1 If N=4n+2=4a+1+4b+1=4p-1+4q-1 where 4a+1,4b+1,4p-1,4q-1 are primes. For given any even integer N, we can find 2 primes as the sum is N.This is an elementary approach to write for any given even integer as a sum of 2 odd primes.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating the surface area of sphere above a plane How do I calculate the surface area of the unit sphere above the plane $z=\frac12$?
EDIT: I have been attempting things and I am thinking about parameterizing this... While I know that surface area is given by the double integral of the cross products of partial derivatives of the new parameters, I don't know what to set them to.. (sorry I'm not good with the fancy notation)
| Surface area is given by
$$
\iint_R \left| \vec r_u \times \vec r_v \right| \ dA
$$
where $\vec r(u,v)$ is the parametrization of the surface. We can rewrite this as (derivation shown here: http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx):
$$
\iint_D \sqrt{ \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1} \ dA
$$
for a function $z = f(x,y)$ where $D$ is the projection of the surface onto the xy-plane.
Since we are only concerned with the portion of the unit sphere above $z = 0$, we can write it as
$$
z = \sqrt{1-x^2-y^2}
$$
Computing the partial derivatives with respect to $x$ and $y$,
$$
\frac{\partial z}{\partial x} = \frac{-x}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial x}\right)^2 = \frac{x^2}{1-x^2-y^2}
$$
$$
\frac{\partial z}{\partial y} = \frac{-y}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial y}\right)^2 = \frac{y^2}{1-x^2-y^2}
$$
Substituting these into our expression for surface area,
$$
\iint_D \sqrt{ \frac{x^2}{1-x^2-y^2} + \frac{y^2}{1-x^2-y^2} + 1} \ dA
$$
which simplifies to (omitting a bit of algebra)
$$
\iint_D \frac{1}{\sqrt{1-x^2-y^2}} \ dA
$$
Observe that $D$ (the projection of our surface into the xy-plane) is given by
$$
z = \sqrt{1-x^2-y^2}
$$
$$
\frac{1}{2} = \sqrt{1-x^2-y^2}
$$
$$
\frac{1}{4} = 1-x^2-y^2
$$
$$
x^2+y^2 = \frac{3}{4}
$$
which is a circle of radius $\frac{\sqrt{3}}{2}$. The integral over $D$ is easiest done in polar coordinates. I'll assume you know how to do that and omit the computation.
$$
\int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \frac{1}{\sqrt{1-r^2}} \ r \ dr \ d\theta
$$
$$
= \pi
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\int \frac {dx}{\sqrt {x^2 + 16}}$ I can not get the correct answer.
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$
$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$
$$\int \sec \theta$$
$$\ln| \sec \theta + \tan \theta|$$
Then I solve for $\theta$:
$x = 4 \tan \theta$
$x/4 = \tan \theta$
$\arctan (\frac{x}{4}) = \theta$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$
This is wrong and I do not know why.
| What you have done is correct! Note that whenever you have inverse trigonometric expressions you can express your answer in more than one way! Your answer can be expressed in a different way (without the trigonometric and inverse trigonometric functions) as shown below.
We will prove that $$\sec \left( \arctan \left( \dfrac{x}4 \right) \right) = \sqrt{1 + \left(\dfrac{x}{4} \right)^2}$$
Hence, your answer $$\ln \left \lvert \dfrac{x}4 + \sec \left(\arctan \left( \dfrac{x} 4\right) \right) \right \rvert + c$$ can be rewritten as $$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$
Note that $$\theta = \arctan\left( \dfrac{x}4 \right) \implies \tan( \theta) = \dfrac{x}4 \implies \tan^2(\theta) = \dfrac{x^2}{16} \implies 1 + \tan^2(\theta) = 1+\dfrac{x^2}{16}$$
Hence, we get that $$\sec^2(\theta) = 1+ \left(\dfrac{x}{4} \right)^2 \implies \sec (\theta) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \implies \sec \left(\arctan\left( \dfrac{x}4 \right) \right) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2}$$
Hence, you can rewrite your answer as
$$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$
Also, you have been a bit sloppy with some notations in your argument.
For instance, when you substitute $x = 4 \tan (\theta)$,
$$\dfrac{dx}{\sqrt{x^2+16}} \text{ should immediately become }\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}} d \theta$$
Also, you need to carry the $d \theta$ throughout the answer under the integral.
Writing just $\displaystyle \int\sec(\theta)$ or $\displaystyle \int\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}}$ without the $d \theta$ is notationally incorrect.
Anyway, I am happy that you are slowly getting a hang of these!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Sum with binomial coefficients: $\sum_{k=1}^m \frac{1}{k}{m \choose k} $ I got this sum, in some work related to another question:
$$S_m=\sum_{k=1}^m \frac{1}{k}{m \choose k} $$
Are there any known results about this (bounds, asymptotics)?
| You know that $\displaystyle (x + 1)^m = \sum_{k=0}^m {m \choose k} x^k$. So
$$\int_0^1 \frac{(x + 1)^m - 1}{x} \, dx = \sum_{k=1}^m {m \choose k} \frac{1}{k}.$$
Letting $y = x + 1$ this is just
$$\int_1^2 \frac{y^m - 1}{y - 1} \, dy = \sum_{k=1}^m \frac{2^k - 1}{k}.$$
The contribution of the $-1$ terms is $-H_m \sim - \log m$, so let's concentrate on estimating
$$T_m = \sum_{k=1}^m \frac{2^k}{k}.$$
There is an obvious lower bound
$$T_m \ge \sum_{k=1}^m \frac{2^k}{m} = \frac{2^{m+1} - 2}{m}.$$
To get an upper bound, we'll split the sum as
$$\sum_{k=1}^{m-r} \frac{2^k}{k} + \sum_{k=m-r+1}^m \frac{2^k}{k}$$
for some $r$ (depending on $m$, although we do not specify the dependence for now). (Thanks to leonbloy for improvements in the part of the argument that follows!) Noting that $f(k) = \frac{2^k}{k}$ is an increasing function of $k$, the first part is bounded above by $(m-r) \frac{2^{m-r}}{m-r} = 2^{m-r}$ while the second is bounded above by
$$\sum_{k=m-r+1}^m \frac{2^k}{m-r+1} \le \frac{1}{m-r+1} \sum_{k=0}^m 2^k < \frac{2^{m+1}}{m-r+1}.$$
The first part gets smaller as $r$ increases while the second gets larger; to minimize their sum it is generally a good idea to make the two parts about as equal as possible. Thus we want $2^{m-r} \approx \frac{2^{m+1}}{m-r+1}$, or
$$(m-r+1) \approx 2^{r+1}.$$
This gives $r \approx \log_2 m$. Setting $r = (1 + \epsilon) \log_2 m$ for some $\epsilon > 0$ makes the first part negligible relative to the second without really increasing the second and together with the lower bound gives an asymptotic
$$T_m \sim \frac{2^{m+1}}{m}.$$
Edited by leonbloy: With respect to the upper bound - we have that, for any $r = 1..m$:
$$T_m \le 2^{m-r} + \frac{2^{m+1}}{m-r+1} = 2^m \left(2^{-r} + \frac{2}{m-r+1}\right)$$
The expression in parentheses, thought as a function of $r$ (continuous), has a global minimum at $ 2^{r+1} = (m+1-r)^2 \log(2)$, which for large $m$ would give $r\approx 2 \log_2(m)$. Inspired by this, we can choose $r=\lceil 2 \log_2(m) \rceil$,
and hence we can bound the two terms:
$$r \ge 2 \log_2(m) \Rightarrow 2^{-r} \le \frac{1}{m^2}$$
$$r \le 2 \log_2(m) +1 \Rightarrow \frac{2}{m-r+1} \le \frac{2}{m-2 \log_2(m) }$$
So, finally we have the bounds:
$$ \frac{2}{m}(1+ 2^{-m}) \le \frac{T_m}{2^{m}} \le \frac{2}{m-2 \log_2(m) } +\frac{1}{m^2} $$
which, agrees with the above asymptotic.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral of secant cubed $\int \sec^3 x\>dx$ I need to calculate the following indefinite integral:
$$I=\int \frac{1}{\cos^3(x)}dx$$
I know what the result is (from Mathematica):
$$I=\tanh^{-1}(\tan(x/2))+(1/2)\sec(x)\tan(x)$$
but I don't know how to integrate it myself. I have been trying some substitutions to no avail.
Equivalently, I need to know how to compute:
$$I=\int \sqrt{1+z^2}dz$$
which follows after making the change of variables $z=\tan x$.
| $\int \sqrt (1 + x^2) dx$
let $x = \tan \theta $
then $dx = \sec ^2\theta d\theta $
we have the integral is then:
$\int \sec ^3\theta d\theta $
recall:
$\tan ^2\theta + 1 = \sec ^2\theta $
and write as:
$\int \sec \theta (\sec ^2\theta )d\theta $
continue with integration by parts,
by letting:
$u = \sec \theta $
and
$dv = \sec ^2\theta d\theta $
we have:
$v = \tan \theta $
and
$du = \sec \theta \tan \theta d\theta $
and thus,
$uv - \int vdu$
is then:
$\sec \theta \tan \theta - \int \sec \theta \tan ^2\theta d\theta $
recall:
$\tan ^2\theta + 1 = \sec ^2\theta $
we have:
$\sec \theta \tan \theta - \int \sec \theta (\sec ^2\theta - 1) d\theta $
after distribution, altogether we have:
$\int \sec ^3\theta d\theta = \sec \theta \tan \theta - \int \sec ^3\theta d\theta - \int \sec \theta d\theta $
rearranging and collecting like-terms:
$2\int \sec ^3\theta d\theta = \sec \theta \tan \theta - \int \sec \theta d\theta $
$2\int \sec ^3\theta d\theta = \sec \theta \tan \theta - \ln |\sec \theta + \tan \theta |$
$\int \sec ^3\theta d\theta = (1/2)[\sec \theta \tan \theta - \ln |\sec \theta + \tan \theta |]$
+++++++++++++++++++++++++++
without u\sin g integration tables, we can derive
$\int \sec \theta d\theta $
..........
$=\int 1/\cos \theta d\theta $
$=\int \cos \theta /\cos ^2\theta d\theta $
$=\int \cos \theta /(1 - \sin ^2\theta ) d\theta $
let $u = \sin \theta $
then $du = \cos \theta d\theta $
we have:
$\int 1/(1 - u^2) du$
partial fractions:
$A/(1 - u) + B/(1 + u) = 1$
$A + B = 1$
$A - B = 0$
$A = 1/2$
$B = 1/2$
we have:
$-(1/2)\ln |1 - u| + (1/2)\ln |1 + u|$
by rules of logarithms:
$(1/2)\ln |(1 + u)/(1 - u)|$
$(1/2)\ln |(1 - u^2)/(1 - u)^2|$
by rules of logarithms:
$\ln |\sqrt (1 - u^2)/(1 - u)|$
recall:
$u = \sin \theta $
$\ln |\sqrt (1 - \sin ^2\theta )/(1 - \sin \theta )|$
$\ln |\sqrt \cos ^2\theta /(1 - \sin \theta )|$
$\ln |\cos \theta /(1 - \sin \theta )|$
$\ln |\cos \theta (1 + \sin \theta )/(1 - \sin ^2\theta )|$
$\ln |\cos \theta (1 + \sin \theta )/\cos ^2\theta |$
$\ln |1/\cos \theta + \sin \theta /\cos \theta |$
$\ln |\sec \theta + \tan \theta |$
++++++++++++++++++++++++++++
but the integral of $\int \sec ^3\theta d\theta $ is:
$\int \sec ^3\theta d\theta = (1/2)[\sec \theta \tan \theta - \ln |\sec \theta + \tan \theta |]$ + CONS\tan T
and \sin ce
$\tan \theta = x = x/1 =$ opposite/adjacent,
with the pythagorean theorem, we derive:
hypotenuse = $\sqrt (1 + x^2)$
and thus, $\sec \theta $ = hypotenuse/adjacent = $\sqrt (1 + x^2)$
$\int \sqrt (1 + x^2) dx$
$= (1/2)[x\sqrt (1 + x^2) - \ln |\sqrt (1 + x^2) + x|]$ + CONS\tan T
| {
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"source": "stackexchange",
"question_score": "20",
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how many 5-digit numbers satisfy the following conditions How many five-digit numbers divisible by 11 have the sum of their digits equal to 30?
I am able to get the 5-digit numbers divisible by 11
and
I am also able to get the five-digit numbers whose sum of their digits equal to 30.
But i am not able to get how i can get the count of 5 digit numbers satisying both the condition.
Thanks in advance.
Thanks in advance.
combinatorics permutations
| A number is divisible by $11$ if the sum of the digits in the tens and thousands place minus the sum of the digits in the ones, hundreds, and ten thousands place is divisible by 11.
So take a number:
$a b c d e$
$(b+d)-(a+c+e)$ is divisible by $11$ (it may be $0$).
$a+b+c+d+e=30$.
$a,b,c,d,e$ are all integers that may be $0$ through $9$. $a\not=0$.
Combining the two equations above, $2(b+d)-30$ is divisible by $11$, but is also even (since $b$ and $d$ are integers), so it may be $0$, $22$, or $-22$.
If it is $22$, then $(b+d)=26$, which is a contradiction.
If it is $0$, then $(b+d)=(a+c+e)=15$ (condition 1)
If it is $-22$, then $(b+d)=4$ and $(a+c+e)=26$ (condition 2)
Take all such $a,b,c,d,e$ that satisfy the above statements.
If condition 1 holds, then there are $4$ different ways to choose $b$ and $d$ and $4+5+6+7+8+9+9+8+7+6=69$ ways to choose $a,c,e$ that satisfy $a+c+e=15$. Thus, for condition 1, there are $276$ distinct numbers that satisfy condition 1.
If condition 2 holds, then there are $4$ different ways to choose $b$ and $d$ and $3$ ways to choose $a,c,e$. So there are $12$ distinct numbers that satisfy condition 2.
Thus, there are $288$ distinct 5-digit numbers that satisfy the desired condition.
| {
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Evaluating $\int \frac{dx}{x^2 - 2x} dx$ $$\int \frac{dx}{x^2 - 2x}$$
I know that I have to complete the square so the problem becomes.
$$\int \frac{dx}{(x - 1)^2 -1}dx$$
Then I set up my A B and C stuff
$$\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{-1}$$
With that I find $A = -1, B = -1$ and $C = 0$ which I know is wrong.
I must be setting up the $A, B, C$ thing wrong but I do not know why.
| $$\begin{align} & {} \quad \int \frac{dx}{x^2 - 2x}\\ &=\int \frac{dx}{(x - 1)^2 -1}dx\\ &=\int \frac{dx}{(x - 1-1) (x-1+1)} \\ &=\int \frac{dx}{x(x-2)}\end{align}$$
The rest is easy (partial fractions):
$$\frac 12 \left[\int \frac {dx}{x-2} - \int \frac {dx}x\right]=\frac 12 \ln|x-2| -\frac 12 \ln |x| +C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit exercise from Rudin: $\lim\limits_{n \to \infty} \sqrt{n^2+n} -n$ This is Chapter 3, Exercise 2 of Rudin's Principles.
Calculate $\lim\limits_{n \to \infty} \sqrt{n^2+n} -n$.
Hints will be appreciated.
| When you have a sequence that involves difference between two roots, it's often a good idea to try using the identity $(a-b)(a+b) = a^2-b^2$ to git rid of the root difference:
$$
\sqrt{a} - \sqrt{b} = \left(\sqrt{a} - \sqrt{b}\right)\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right)}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Area of a revolution of $x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}$ I think my biggest problem here is I can not find a good way to find the square root in this problem
$$x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2} \ \ \ \ 1 \le x \le 2$$
$$\int_1^2 2 \pi \cdot {\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}} \sqrt{1 + (1/3 (y^2 + 2)^{\frac{3}{2}})^2}$$
$$\int_1^2 2 \pi \cdot \frac{1}{3}\left(y^2+2\right)^\frac{3}{2} \sqrt{1 + y^3 + 2y}$$
Here is where I am stuck, I have tried u-substitution and trig and I can not make this work. I have two pages of notes but I expect that they would be useless to type up.
| My suggestion is: take it easy and take it step by step. Don't throw everything in one expression, because it is easy to make a slip:
$$x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}$$
$$x'=\frac{1}{2}\left(y^2+2\right)^\frac{1}{2}\cdot\underbrace{ 2y}_{\text{chain rule}}=y\left(y^2+2\right)^\frac{1}{2}$$
$$x'^2=y^2\left(y^2+2\right)$$
$$1+x'^2=1+y^2\left(y^2+2\right)=y^4+2y^2+1=\left(y^2+1\right)^2$$
Now calculating the differential it is useful to take it a step too far to simplify calculations in the end:
$$ds=2\pi x\sqrt{1+x'^2}=2\pi y\left(y^2+1\right)dy=\pi\left(y^2+1\right)\cdot 2ydy=\pi\left(y^2+1\right)d\left(y^2\right)=\pi\left(y^2+1\right)d\left(y^2+1\right)$$
Where I am "pushing expressions under $d$": $f'(x)dx=d\left(f(x)+C\right)$
$$s=\pi\int_1^2\left(y^2+1\right)d\left(y^2+1\right)=\left.\frac{\pi(y^2+1)^2}{2}\right|_1^2=12\pi$$
| {
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Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$ One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
| Here's a way which avoids derivatives and integrals.
Assume that we know that $\frac{\sin x}{x} \to 1$ as $x \to 0$.
Then we also know that
$\frac{1-\cos x}{x^2} = \frac12 \left( \frac{\sin(x/2)}{x/2} \right)^2 \to \frac12$.
Now,
$$
\frac{\tan x - x}{x^3} = \frac{1}{\cos x} \left( \frac{\sin x - x}{x^3} + \frac{1-\cos x}{x^2} \right),
$$
so we are done if we can compute $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac16$.
The reason that I rewrote it like this is that I was asked by a
colleague about ten years ago whether that limit could be done in an
elementary way. :-) I came up with the following:
Let
$$
f(x) = \frac{x - \sin x}{x^3} = \frac{1 - \frac{\sin x}{x}}{x^2}.
$$
(Here I've changed the sign so that the limit
will be positive.)
Since $f$ is an even function, it's enough to consider $x>0$.
Fix a positive integer $n$. To begin with, we have
$$
x = 2^n \frac{x}{2^n} > 2^n \sin \frac{x}{2^n}.
$$
(I'm assuming that we also know that $0 < \sin x < x < \tan x$ for $0 < x < \pi/2$.)
Multiply this inequality by $\prod_{k=1}^n \cos\frac{x}{2^k}$ and use the double angle formula repeatedly,
as follows (illustrated for the case $n=3$):
$$
\begin{split}
x \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2}
& > 2^3 \sin\frac{x}{8} \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2}
\\
& = 2^2 \sin\frac{x}{4} \cos\frac{x}{4} \cos\frac{x}{2}
\\
& = 2^1 \sin\frac{x}{2} \cos\frac{x}{2}
\\
& = \sin x.
\end{split}
$$
This implies (again for $n=3$, but the general pattern is hopefully clear)
$$
\begin{split}
1 - \frac{\sin x}{x}
& > 1 - \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2}
\\
& =
\left( 1 - \cos\frac{x}{8} \right)
+ \cos\frac{x}{8} \left( 1 - \cos\frac{x}{4} \right)
+ \cos\frac{x}{8} \cos\frac{x}{4} \left( 1 - \cos\frac{x}{2} \right).
\end{split}
$$
We know that $\frac{1 - \cos(x/2^k)}{x^2} = \frac{1 - \cos(x/2^k)}{2^{2k} (x/2^k)^2} \to \frac{1}{2^{2k+1}}$,
so after dividing this inequality by $x^2$ we find in the limit
(for general $n$) that
$$
\liminf_{x \to 0^+} f(x) \ge \sum_{k=1}^n \frac{1}{2^{2k+1}} = \frac16 \left( 1 - \frac{1}{4^n} \right).
$$
This holds for every $n$, hence
$$
\liminf_{x \to 0^+} f(x) \ge \frac16.
$$
The other direction is similar.
Start with
$$
x = 2^n \frac{x}{2^n} < 2^n \tan\frac{x}{2^n} = 2^n \frac{\sin(x/2^n)}{\cos(x/2^n)}.
$$
This leads to
$$
\begin{split}
1 - \frac{\sin x}{x}
& < 1 - \cos\frac{x}{2^n} \cdot (\text{same product of cosines as above})
\\
& = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (1 - (\text{that product}))
\\
& = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (\text{same expression as above}).
\end{split}
$$
Divide by $x^2$ and let $x \to 0^+$:
$$
\limsup_{x \to 0^+} f(x) \le \frac{1}{2^{2n+1}} + \frac16 \left( 1 - \frac{1}{4^n} \right).
$$
Let $n \to \infty$:
$$
\limsup_{x \to 0^+} f(x) \le \frac16.
$$
It follows that $\lim_{x \to 0^+} f(x) = \frac16$, and therefore by symmetry $\lim_{x \to 0} f(x) = \frac16$,which is what we wanted to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Combining a radical and simplifying? How would I combine and simplify the following radical:
$$\sqrt {\frac{A^2}{2}} - \sqrt \frac{A^2}{8}$$
| Recall the following facts.
$$\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} \text{ whenever }a,b>0$$
$$\sqrt{x^2} = \lvert x \rvert \text{, where $x \in \mathbb{R}$}$$
$$\sqrt{ab} = \sqrt{a} \sqrt{b} \text{ whenever }a,b>0$$
Using the above, we then get that
\begin{align}
\sqrt{\dfrac{A^2}{2}} - \sqrt{\dfrac{A^2}{8}} & = \sqrt{\dfrac{A^2}{2}} - \sqrt{\dfrac{A^2}{2^2 \times 2}} = \dfrac{\sqrt{A^2}}{\sqrt{2}} - \dfrac{\sqrt{A^2}}{\sqrt{2^2 \times2}}\\
& = \dfrac{\lvert A\rvert}{\sqrt{2}} - \dfrac{\lvert A \rvert}{\sqrt{2^2} \times \sqrt{2}} = \dfrac{\lvert A\rvert}{\sqrt{2}} - \dfrac{\lvert A \rvert}{2\sqrt{2}}\\
& =\dfrac{\lvert A \rvert}{2\sqrt{2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prime as sum of three numbers whose product is a cube Good evening!
I am very new to this site. I would like to put the following material from Prof. Gandhi's note book and my observations. Of course it is little long with more questions. But, with good belief on this site, I am sending for good solutions/answers.
If we take other than primes $2$, $5$ and $11$, every prime can be written as $x + y + z$, where $x$, $y$ and $z$ are some positive numbers. Interestingly, $x \times y \times z = c^3$, where $c$ is again some positive number.
Let us see the magic for primes $3,7,13,31,43,73$
$$
\begin{align}
3 = 1 + 1 + 1 &\Longrightarrow 1 \times 1 \times 1 = 1^3\\
7 = 1 + 2 + 4 &\Longrightarrow 1 \times 2 \times 4 = 2^3\\
13 = 1 + 3 + 9 &\Longrightarrow 1 \times 3 \times 9 = 3^3\\
31 = 1 + 5 + 25 &\Longrightarrow 1 \times 5 \times 25 = 5^3\\
43 = 1 + 6 + 36 &\Longrightarrow 1 \times 6 \times 36 = 6^3\\
73 = 1 + 8 + 64 &\Longrightarrow 1 \times 8 \times 64 = 8^3\\
\end{align}
$$
Can you justify the above pattern? How to generalize the above statement either mathematically or by computer?
But, I observed that it is true for primes less than $9500$. Can your provide a computational algorithm to describe this?
Also, prove that, we conjecture that except $1, 2, 3, 5, 6, 7, 11, 13, 14, 15, 17, 22, 23$, every positive number can be written as a sum of four positive numbers and the product is again can be expressible in 4th power. Now, can we generalize this? Also, I want to know that, is there any such numbers can be expressible as some of $n$-integers with their product is again in $n$-th power?
Thank you so much.
edit
Concerning this cubic property :
Notice that this can be extended to hold for almost all squarefree positive integers $> 2$, not just the primes.
for instance :
we know for the prime $7$ : $7=1+2+4$ so we also get $7A = 1A + 2A + 4A$ and $1A * 2A * 4A$ is simply equal to $8A^3$.
In fact this can be extended to all odd positive integers $>11$ if $25,121$ have a solution.
Hence I am intrested in this and I placed a bounty.
I edited the question because its to much for a comment and certainly not an answer.
Btw Im curious about this Ghandi person though info about that does not get the bounty naturally.
I would like to remind David Speyer's comment : Every prime that is $1 mod 3$ is of the form $a^ 2 +ab+b^ 2$ , so that covers half the primes immediately.
So that might be a line of attack.
| Here is an attempt that doesn't quite work. In this post, $\left( \frac{a}{p} \right)$ is the quadratic residue symbol.
If $\left( \frac{-3}{p} \right) = 1$, then $p$ is of the form $x^2+xy+y^2$.
If $\left( \frac{85}{p} \right) = 1$, then $p$ is either of the form $9 x^2 + 25 xy + 15 y^2$ or $3 x^2 + 25 xy + 45 y^2$.
If $\left( \frac{-255}{p} \right) = 1$, then $p$ is of one of the forms $x^2+xy+64 y^2$, $2 x^2 + xy + 32 y^2$, $4 x^2 + xy + 16 y^2$, $8 x^2 + xy + 8 y^2$, $3 x^2 + 3 xy + 22 y^2$, $5 x^2 + 5 xy + 14 y^2$, $6 x^2 + 3 xy + 11 y^2$ or $7 x^2 + 5 xy + 11 y^2$.
Now, one of the three quadratic residues must be $1$, since $\left( \frac{-255}{p} \right) = \left( \frac{-3}{p} \right)\left( \frac{85}{p} \right)$. And for a lot of the quadratic forms above, we win: $(x^2)(xy)(y^2)$, $(9 x^2)(25 xy)(15 y^2)$, $(3 x^2)(25 xy)(45 y^2)$, $(x^2)(xy)(64 y^2)$, $(2 x^2)(xy)(32 y^2)$, $(4 x^2)(xy)(16 y^2)$ and $(8 x^2)(xy)(8 y^2)$ are all cubes. Unfortunately, the last $4$ quadratic forms of discriminant $-255$ don't work when written in reduced form. I haven't yet done a serious search to see whether some change of basis might put them into the form $a x^2 + bxy + c y^2$, with $abc$ a cube.
Also, we don't know what the signs of $x$ and $y$ are, so this doesn't necessarily get us three positive integers whose product is a cube.
Still, I am optimistic that we might be able to find enough quadratic forms of the form $a x^2 + bxy + c y^2$, with $abc=1$, to cover all the primes, at least if we ignore the sign issue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
"answer_count": 5,
"answer_id": 1
} |
Proof about $z\cot z=1-2\sum_{k\ge1}z^2/(k^2\pi^2-z^2)$ In Concrete Mathematics, it is said that
$$z\cot z=1-2\sum_{k\ge1}\frac{z^2}{k^2\pi^2-z^2}\tag1$$
and proved in EXERCISE 6.73
$$z\cot z=\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}+\sum_{k=1}^{2^{n-1}-1}\frac z{2^n}\left(\cot\frac{z+k\pi}{2^n}+\cot\frac{z-k\pi}{2^n}\right)$$
The trigonmetric identity is not hard, but I cannot understand the rest:
It can be shown that term-by-term passage to the limit is justified, hence equation (1) is valid.
How can we conclude that? Thanks for help!
| NOTE: This is incomplete. A tighter bound should be produced. Anyone able to do so is free to edit and add it.
For $x$ near the origin, $\cot x \sim \dfrac{1}{x}$. Since $\dfrac{1}{2^n}\to 0 $ we can use this. More precisely,
$$\frac{1}{x}-1<\cot x <\frac{1}{x} $$
Namely, we can dissect
$$z\cot z=\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}+\sum_{k=1}^{2^{n-1}-1}\frac z{2^n}\left(\cot\frac{z+k\pi}{2^n}+\cot\frac{z-k\pi}{2^n}\right)$$
into
$${A_n} = \frac{z}{{{2^n}}}\cot \frac{z}{{{2^n}}} = \cos \frac{z}{{{2^n}}}\frac{{\frac{z}{{{2^n}}}}}{{\sin \frac{z}{{{2^n}}}}} = \cos u\frac{u}{{\sin u}} \to 1$$
$${B_n} = \frac{z}{{{2^n}}}\tan \frac{z}{{{2^n}}} = \frac{z}{{{2^n}}}\sin \frac{z}{{{2^n}}}\frac{1}{{\cos \frac{z}{{{2^n}}}}} = \frac{{u\sin u}}{{\cos u}} \to 0$$
where $u \to 0$.
Then we have
$$\eqalign{
& \frac{{{2^n}}}{{z - k\pi }} + \frac{{{2^n}}}{{z + k\pi }} - 2 < \left( {\cot \frac{{z + k\pi }}{{{2^n}}} + \cot \frac{{z - k\pi }}{{{2^n}}}} \right) < \frac{{{2^n}}}{{z - k\pi }} + \frac{{{2^n}}}{{z + k\pi }} \cr
& \frac{z}{{z - k\pi }} + \frac{z}{{z + k\pi }} - \frac{z}{{{2^{n - 1}}}} < \frac{z}{{{2^n}}}\left( {\cot \frac{{z + k\pi }}{{{2^n}}} + \cot \frac{{z - k\pi }}{{{2^n}}}} \right) < \frac{z}{{z - k\pi }} + \frac{z}{{z + k\pi }} \cr
& \frac{{2{z^2}}}{{{z^2} - {k^2}{\pi ^2}}} - \frac{z}{{{2^{n - 1}}}} < \frac{z}{{{2^n}}}\left( {\cot \frac{{z + k\pi }}{{{2^n}}} + \cot \frac{{z - k\pi }}{{{2^n}}}} \right) < \frac{{2{z^2}}}{{{z^2} - {k^2}{\pi ^2}}} \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Compute the limit of $\frac1{\sqrt{n}}\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)^{1/n^2}$ Compute the following limit:
$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}} $$
I'm interested in almost any approaching way for this limit. Thanks.
| Another way: by Stolz-Cesaro theorem, we have
$$\begin{align}
L&=\lim_{n\to\infty} \frac{1}{n^2} \left(\sum_{k=1}^{n} k\log (k) - \frac{n^2\log(n) }{2}\right)\\
&=\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}\\
&=\lim_{n\to\infty} \frac{1}{n^2-(n-1)^2} \left(n\log (n) - \frac{n^2\log(n) }{2}+\frac{(n-1)^2\log(n-1)}{2}\right)\\
&=\lim_{n\to\infty} \frac{1}{4n-2} \left(2n\log (n) -n^2\log(n)+(n-1)^2(\log(n)+\log(1-1/n))\right)\\
&=\lim_{n\to\infty} \frac{1}{4n-2} \left(-n+o(n)\right)=-\frac{1}{4}
\end{align}$$
and therefore
$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}}=e^{-1/4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
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Help Verifying Trigonometric Identity I could really use help, hint or otherwise, in proving a trigonometric identity:
We are only allowed to work on one side of the equation.
$$\dfrac{2\sin^2(x)-5\sin(x)+2}{\sin(x)-2} = 2\sin(x)-1$$
| HINT: Factorize the numerator and cancel terms arguing why the terms you are canceling are not zero.
Move your mouse over the gray area for the answer.
$$\dfrac{2 \sin^2(x) - 5 \sin(x) + 2}{\sin(x) - 2} = \dfrac{2 \sin^2(x) - 4 \sin(x) - \sin(x) + 2}{\sin(x) - 2}\\ = \dfrac{2 \sin(x) (\sin(x) - 2) - ( \sin(x) - 2)}{\sin(x) - 2}\\= \dfrac{(\sin(x) - 2) (2 \sin(x) - 1)}{\sin(x) - 2} = 2 \sin(x) - 1$$ We are allowed to cancel $\sin(x) - 2$ since $\sin(x) \neq 2$, $\forall x \in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding solutions to $(4x^2+1)(4y^2+1) = (4z^2+1)$ Consider the following equation with integral, nonzero $x,y,z$
$$(4x^2+1)(4y^2+1) = (4z^2+1)$$
What are some general strategies to find solutions to this Diophantine?
If it helps, this can also be rewritten as $z^2 = x^2(4y^2+1) + y^2$
I've already looked at On the equation $(a^2+1)(b^2+1)=c^2+1$
| Here is one general approach. Since the product of the sum of two squares is itself the sum of two squares, then,
$$\tag{1}(4x^2+1)(4y^2+1) = 4z^2+1$$
is equivalent to,
$$\tag{2}(2x+2y)^2+(4xy-1)^2 = 4z^2+1$$
The complete solution to the form,
$$\tag{3}x_1^2+x_2^2 = y_1^2+y_2^2$$
is given by the identity,
$$\tag{4}(ac+bd)^2 + (bc-ad)^2 = (ac-bd)^2+(bc+ad)^2$$
One can then equate the terms of (2) and (4), solve for {x, y, z}, with {a, b, c, d} chosen such that one term on the RHS is equal to unity.
EDITED MUCH LATER:
In response to your questions, let's have a simpler solution to (3) as,
$$\tag{5}(6n+2)^2+(6n^2+4n-1)^2=(6n^2+4n+2)^2+1$$
Equate the terms of (2) and (5) and we find that,
$$x = \frac{1}{2}\big(1+3n-\sqrt{3n^2+2n+1}\big)$$
$$y = \frac{1}{2}\big(1+3n+\sqrt{3n^2+2n+1}\big)$$
$$z = (6n^2+4n+2)/2$$
To get rid of the $\sqrt{N}$ and solve the form,
$$an^2+bn+c^2 = \square$$
one simply chooses,
$$n = \frac{-2cuv+bv^2}{u^2-av^2}$$
for arbitrary {u, v}. Of course, since you want integer n, you have to solve the denominator as the Pell equation $u^2-av^2 = \pm 1$.
In summary, and after simplification, an infinite number of integer solutions to,
$$(4x^2+1)(4y^2+1) = 4z^2+1$$
is given by the rather simple,
$$x = (u-3v)(u-v)$$
$$y = 2uv$$
$$z = (u^2-2uv+3v^2)^2$$
where,
$$u^2-3v^2=1$$
P.S. It is quite easy to find other solutions similar to (5), and appropriate ones would need other Pell equations.
| {
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"url": "https://math.stackexchange.com/questions/162862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Single-digit even natural number solutions to the equation $a+b+c+d = 24$ such that $a+b > c+d$
Possible Duplicate:
Two algebra questions
How to approach the below question:
How many single-digit even natural number solutions are there for the equation $a+b+c+d = 24$ such that $a+b > c+d$?
| Here is one approach you can use:
The even single-digit (presumably in base 10) natural numbers are precisely 2, 4, 6, 8.
Thus, if $y$ and $z$ are such numbers, then $4\leq y+z\leq 16$, and of course $y+z$ is even.
Break the problem up as follows: find the number of solutions $w$, $x$ to $w+x=24$ such that $w$ and $x$ are even numbers such that $4\leq x<w\leq 16$, and then for each choice of $w$ and $x$, find the number of solutions to $a+b=w$ and the number of solutions to $c+d=x$ such that $a,b,c,d\in\{2,4,6,8\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Singular points of ODE My friend and I have conflicting answers and since his phone is off, I can't get his full solution and I don't understand his argument.
Consider this ODE
$$(x+1)y''+\frac{1}{x}y' + (x+3)y= 0$$
Basically what I did was divide out that $(x+1)$ on $y''$ and got
$$y''+\frac{1}{x(x+1)}y' + \frac{(x+3)}{(x+1)}y= 0$$
The singularities are x = -1, and 0 (both are regular)
My friend said we had to get our ODE in the form of
$$a(x-x_0)^2y'' + b(x-x_0)y' + (x-x_0)y =0$$ otherwise we cant' do anything. and he got x = 3 as an irregular singular point which i have no idea how even got this. Sorry if this is too vague, but my first source was also as vague.
what was my friend doing and who's right?
| $(x+1)y''+\dfrac{1}{x}y'+(x+3)y=0$
$y''+\dfrac{1}{x(x+1)}y'+\dfrac{x+3}{x+1}y=0$
It should be noted that the positions of finite singular points are always appear at the positions that makes one of the coefficients diverge.
$\therefore$ the positions of finite singular points in this question are $x=0$ and $x=-1$ only, absolute not like your friend said $x=3$ also is the finite singular point.
The next step is to determine whether the finite singular points are regular or irregular.
$\lim\limits_{x\to 0}\left(x\times\dfrac{1}{x(x+1)}\right)=\lim\limits_{x\to 0}\dfrac{1}{x+1}=1$
$\lim\limits_{x\to 0}\left(x^2\times\dfrac{x+3}{x+1}\right)=\lim\limits_{x\to 0}\dfrac{x^2(x+3)}{x+1}=0$
$\therefore$ the finite singular point $x=0$ is regular.
$\lim\limits_{x\to-1}\left((x+1)\times\dfrac{1}{x(x+1)}\right)=\lim\limits_{x\to-1}\dfrac{1}{x}=-1$
$\lim\limits_{x\to-1}\left((x+1)^2\times\dfrac{x+3}{x+1}\right)=\lim\limits_{x\to-1}(x+1)(x+3)=0$
$\therefore$ the finite singular point $x=-1$ is regular.
That's not the end. We should also check the singularities at infinity. Because these also act as one of the key points of distinguishing the ODE type.
Let $u=\dfrac{1}{x}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{du}=-u^2\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-u^2\dfrac{dy}{du}\right)=\dfrac{d}{du}\left(-u^2\dfrac{dy}{du}\right)\dfrac{du}{dx}=\left(-u^2\dfrac{d^2y}{du^2}-2u\dfrac{dy}{du}\right)\left(-\dfrac{1}{x^2}\right)=\left(-u^2\dfrac{d^2y}{du^2}-2u\dfrac{dy}{du}\right)(-u^2)=u^4\dfrac{d^2y}{du^2}+2u^3\dfrac{dy}{du}$
$\therefore u^4\dfrac{d^2y}{du^2}+2u^3\dfrac{dy}{du}+\dfrac{1}{\dfrac{1}{u}\left(\dfrac{1}{u}+1\right)}\left(-u^2\dfrac{dy}{du}\right)+\dfrac{\dfrac{1}{u}+3}{\dfrac{1}{u}+1}y=0$
$u^4\dfrac{d^2y}{du^2}+2u^3\dfrac{dy}{du}-\dfrac{u^4}{u+1}\dfrac{dy}{du}+\dfrac{3u+1}{u+1}y=0$
$u^4\dfrac{d^2y}{du^2}+\dfrac{u^4+2u^3}{u+1}\dfrac{dy}{du}+\dfrac{3u+1}{u+1}y=0$
$\dfrac{d^2y}{du^2}+\dfrac{u+2}{u(u+1)}\dfrac{dy}{du}+\dfrac{3u+1}{u^4(u+1)}y=0$
$\lim\limits_{u\to 0}\left(u\times\dfrac{u+2}{u(u+1)}\right)=\lim\limits_{u\to 0}\dfrac{u+2}{u+1}=2$
$\lim\limits_{u\to 0}\left(u^2\times\dfrac{3u+1}{u^4(u+1)}\right)=\lim\limits_{u\to 0}\dfrac{3u+1}{u^2(u+1)}=\infty$
$\therefore$ the singularities at $x=\pm\infty$ are irregular.
So $(x+1)y''+\dfrac{1}{x}y'+(x+3)y=0$ belongs to Heun's Confluent type ODE. If you solve it in MATLAB, MATLAB will express its general solution in terms of Heun's Confluent function.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine whether $\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$ is convergent or divergent. If convergent, find the sum. $$\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$$
It's geometric, since the common ratio $r$ appears to be $\frac{-3}{4}$, but this is where I get stuck. I think I need to do this: let $f(x) = \frac{(-3)^{x-1}}{4^x}$.
$$\lim\limits_{x \to \infty}\frac{(-3)^{x-1}}{4^x}$$
Is this how I handle this exercise? I still cannot seem to get the answer $\frac{1}{7}$
| $By the given series have:
$a_{n}=\frac{(-3)^{n-1}}{4^{n}}$, $a_{n+1}=\frac{(-3)^{n}}{4^{n+1}}$
By the criterion of Dalamber have:
$A=\lim\frac{a_{n+1}}{a_{n}}=\frac{3}{4}<1$
Under this criterion we have that A<1 conclude that given series is convergent.
Since the given series is convergent exist sum of this series. Mark wis $S_{n}$ sum of this series, it is $S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$.
Hance we
$S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$
$S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty \frac{(-3)^{n}}{4^n}$
$\frac{(-3)^{n}}{4^n}$ is it a geometric series. Now find sum this series. Find the sum must assign a_{1} dhe q.
$a_{1}=-\frac{3}{4}$, $q=-\frac{3}{4}$.
Sum accounst
$S_{n}=\frac {a_{1}(1-q^{n})}{1-q}=-\frac{3}{7}{[1-\frac{3^{n}}{4^{n}}]}$
$\lim S_{n}=-\frac{3}{7}$
Theres definitely have
$S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty\frac{(-3)^{n}}{4^{n}}$
$S_{n}=(-\frac{1}{3})(-\frac{3}{7})$
$S_{n}=\frac{1}{7}$
Conlude the: Given series is the convergent and its sum $\frac{1}{7}$.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/166097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Complex series: $\frac{z}{(z-1)(z-3)} = -3 \sum\limits_{n=0}^\infty \frac {(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$ for $0 < |z-1| < 2$
Show that when $0 < |z-1| < 2$,
$$\frac{z}{(z-1)(z-3)} = -3 \sum_{n=0}^\infty \frac {(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$$
I thought to attack this using a partial fraction decomposition and then breaking the partial fractions into Maclaurin series. I got the $$-\frac{1}{2(z-1)}$$ from that, but the other part has me a bit stumped.
I have:
$$
\frac{3}{2(z-3)}$$
for the other partial fraction. And I factor out the 3/2 and then have:
$$\frac{1}{z-3}$$
So I get:
$$\frac{1}{1 - (-(z-2))}$$
but this does not give me what I need.
| $$\dfrac{z}{(z-1)(z-3)} = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}$$
$$\dfrac1{z-3} = \dfrac1{(z-1) -2} = -\dfrac12 \left( \dfrac1{1-(z-1)/2}\right) = -\dfrac12 \left( \sum_{k=0}^{\infty} \left( \dfrac{z-1}2\right)^k\right)$$
Hence, $$\dfrac3{(z-1)(z-3)} = - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)$$
Hence,
\begin{align}
\dfrac{z}{(z-1)(z-3)} & = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}\\
& = \dfrac1{z-1} - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)\\
& = - \dfrac12 \dfrac1{z-1}- 3 \left( \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+2}}\right)\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/166823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that :
$$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$
I find that
$$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$
By AM-GM
$$\ \frac{ab}{a^{2}+3b^{2}} \leq \frac{1}{2 \sqrt{3}}=\frac{\sqrt{3}}{6} $$
$$\ \sum_{cyc} \frac{ab}{a^{2}+3b^{2}} \leq \frac{\sqrt{3}}{2} $$
But this is obviously is not working .
| I have answered this question in a slightly different way.
Let us assume the following : $ \frac{a}{b}=x$ and $ \frac{b}{c}=y.$ This converts the above equation to a equation
with two variables. $$ f(x,y)=\frac{x}{3+x^2} + \frac{y}{3+y^2}+\frac{xy}{1+3(xy)^2}$$ Now to get a maxima or minima point of $f(x,y)$ we partially differentiate it with $x$ and $y$ and equate them to $0$. Hence we have $$ \frac{\partial f(x,y)}{\partial x}= \frac{3-x^2}{(x^2+3)^2}+\frac{y(1-3x^2y^2)}{(3x^2y^2+1)^2}=0 $$ $$\Rightarrow \frac{3-x^2}{y(x^2+3)^2}=\frac{(3x^2y^2-1)}{(3x^2y^2+1)^2}...........Eqn(1)$$ Since $f(x,y)$ is symmetric with $x$ and $y$ ,we also have for $\frac{\partial f(x,y)}{\partial y}=0$ $$\frac{3-y^2}{x(y^2+3)^2}=\frac{(3x^2y^2-1)}{(3x^2y^2+1)^2}...........Eqn(2)$$ Combining equation 1 and 2 we get $$x \cdot\frac{3-x^2}{(x^2+3)^2}=y \cdot\frac {3-y^2}{(y^2+3)^2}$$ This immediately shows that $x=y $ is a critical point (maxima or minima). Now this clearly shows that at the critical point $$x=y$$ $$\Rightarrow ac=b^2.........Eqn (3)$$ In a similar fashion assuming $\frac{b}{c}=x$ and $\frac{c}{a}=y$ we again the same set of equation $$ f(x,y)=\frac{x}{3+x^2} + \frac{y}{3+y^2}+ \frac{xy}{1+3(xy)^2}$$ Following the same steps we get $$x=y$$ $$\Rightarrow ab=c^2...........Eqn(4)$$ Equation 3 and 4 show that $$a=b=c$$ at the critical point. Hence $a=b=c$ gives us that $$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}=\frac{3}{4}$$ is a maxima or a minima. To check maxima we double differentiate and check $\frac{\partial^2f}{\partial x^2}$ and $\frac{\partial^2f}{\partial y^2}$ We get the following : $$\frac{\partial^2f}{\partial x^2}=\frac{\partial^2f}{\partial y^2}=\frac{2x^3-18x}{(3+x^2)2}+\frac{18x^4-18x^8}{(1+3x^4)^2}=-\frac{16}{64} $$ at $ x=y=1$. Both being negative we see that the function $f$ has a maxima at $a=b=c$ which is $\frac{3}{4}$ Hence $$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2} \le \frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
Polynomial-related manipulation My question is:
Factorize: $$x^{11} + x^{10} + x^9 + \cdots + x + 1$$
Any help to solve this question would be greatly appreciated.
| Since $x^{11}+x^{10}+\ldots + x+1 = \frac{x^{12}-1}{x-1}$ we may first factorize $x^{12}-1$ and then divide by the factor $x-1$:
\begin{align*}
x^{12}-1 &= (x^6-1)(x^6+1)\\
&= (x^3-1)(x^3+1)(x^6+1)\\
&=(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1),
\end{align*}
hence
$$x^{11}+x^{10}+\ldots +x+1 = (x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1).$$
It is an easy exercise to show that the factors are irreducible over $\mathbb Q$.
In fact, the factors are the cyclotomic polynomials of the divisors of 12 (except 1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $\lim\limits_{n\to\infty} \prod\limits_2^n \left(1-\frac1{k^3}\right)$ I've just worked out the limit $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^2}\right)$ that is simply solved, and the result is $\frac{1}{2}$. After that, I thought of calculating $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^3}\right)$, but I don't know how to do it. According to W|A, the result is pretty nice, but I don't see how W|A gets that. (See here.) Is there any easy way to get the answer?
| The last step of Andrew getting
\begin{align}\lim_{n\to \infty}\prod _{k=2}^n \left(1-\frac{1}{k^3}\right)= \frac{\cosh \frac{\sqrt{3} \pi }{2} \Gamma \left(n-\frac{i \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(n+\frac{i \sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi n^3 \Gamma^2 (n)}\end{align}
was a bit ambigous.
using another method, note that
\begin{align*}\Gamma(z)=\frac{1}{z e^{\gamma z}}\prod_{k=1}^{\infty}\frac{k e^{\frac{z}{k}}}{z+k}
\end{align*}
holds for all complex number $z$ except negative integer, we obtain
\begin{align}g(z)=\prod_{k=1}^{\infty} (1+\frac{z}{k})e^{\frac{-z}{k}}=\frac{1}{z\Gamma(z)e^{\gamma z}}\end{align}
Thus
\begin{align}
g(\omega)g(\omega^2)=\prod_{k=1}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}=\frac{1}{\Gamma(\omega)\Gamma(\omega^2) e^{\gamma}}=\frac{3}{e}\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}
\end{align}
where $-\omega$ is the root of $x^3=1$
From
\begin{align}
\prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\lim_{n\to \infty}\frac{1}{n} e^{\frac{1}{2}+\cdots+\frac{1}{n}}=e^{\gamma -1}
\end{align}
Thus
\begin{align}
\prod_{k=2}^{\infty}\left(1-\frac{1}{k^3}\right)=\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}\prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\frac{1}{3\Gamma(\omega)\Gamma(\omega^2)}
\end{align}
and hence the result
By the similar way we may get $\prod_{k=2}^{\infty}(1-\frac{1}{k^n})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/168740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 0
} |
$(x^n-x^m)a=(ax^m-4)y^2$ in positive integers How do I find all positive integers $(a,x,y,n,m)$ that satisfy $ a(x^{n}-x^{m}) = (ax^{m}-4) y^{2} $ and $ m\equiv n\pmod{2} $, with $ax$ odd?
| Brute force finds three solutions so far:
a=3, x=3, y=12, n=5, m=1
a=1, x=3, y=12, n=6, m=2
a=1, x=9, y=12, n=3, m=1
which are just different ways of saying that $729 - 9 = (9-4)\times12^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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