Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way find the sum?
| The answer is $S_n = \frac{1}{5}n^5 + \frac{3}{2}n^4 + 4n^3 + \frac{9}{2}n^2 + \frac{54}{30}n $
$\sum n^4 $ is not really a problem.
You can do it using this method here
https://www.coursera.org/learn/calculus1/lecture/uR7YK/what-is-the-sum-of-n-4-for-n-1-to-n-k
I don't know if this is helpful but this is how i got it
$S_n = \sum a_n$
$a_n = n(2+n)(1+n)^2 = n^4+4n^3+5n^2+2n$
therefore : $ S_n = \sum {(n^4+4n^3+5n^2+2n)} $
$S_n = \sum n^4 + 4\sum n^3 + 5\sum n^2 + 2\sum n $
$S_n = \frac{1}{30}( 6n^5 +15n^4 + 10n^3- n) +4(\frac{1}{4}(n^4+2n^3+n^2) + 5(\frac{1}{6}(2n^3 + 3n^2 +n ) ) + 2(\frac{1}{2}(n^2 +n ) ) $
hence, $S_n = \frac{1}{5}n^5 + \frac{3}{2}n^4 + 4n^3 + \frac{9}{2}n^2 + \frac{54}{30}n $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Faster way to get a rational integral $\int_2^3 \frac{x^{100}+1}{x^3+1} \, dx$ So my problem is calculating this integral:
$$
\int_{2}^{3} \frac{x^{100}+1}{x^3+1} \, dx
$$
I know it can be done by polynomial division but that is really tedious I would have to divide $\approx 30$ times.
And i know that Mathematica would give me the answer in the blink of an eye.
Is there a clever way to do it pen&paper?
| We have
$$ J=\int_{2}^{3}\frac{x-1}{x^3+1}\,dx = \frac{1}{3}\log\left(\frac{21}{16}\right)\tag{1}$$
by partial fraction decomposition, and that is the only annoying part, since
$$ I = \int_{2}^{3}\frac{x^{100}+1}{x^3+1}\,dx = -J+\int_{2}^{3}x\cdot\frac{x^{99}+1}{x^3+1}\,dx \tag{2}$$
and
$$ \frac{x^{99}+1}{x^3+1} = 1-x^3+x^6-\ldots+x^{96} \tag{3} $$
so:
$$ \boxed{\int_{2}^{3}\frac{x^{100}+1}{x^3+1}\,dx} =-J+ \int_{2}^{3}\sum_{k=0}^{32}(-1)^k x^{3k+1}dx=\boxed{-\frac{1}{3}\log\left(\frac{21}{16}\right)+\sum_{k=0}^{32}(-1)^k\frac{3^{3k+2}-2^{3k+2}}{3k+2}}\tag{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
if the sum of the digits of $a+b$, $b+c$ and $a+c$ is $<5$, then is the sum of the digits of $a+b+c$ less or equal than 50? This is a problem from a math olympiad:
Let $S(x)$ denote the sum of the digits of a natural number $x$ in its decimal representation. Do there exist three natural numbers $a$, $b$ and $c$ with $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$ and $S(a+b+c)>50$?
My attempt:
NOTATION: Given a real number $x$, $[x]$ is the integer part of $x$.
REMARK 1: Let $x$ be a natural number with $2$ digits. Then the sum of its two digits is
$$ \left[\frac{x}{10}\right]+x-\left[\frac{x}{10}\right]\cdot 10=x-9\cdot \left[\frac{x}{10}\right]. $$
REMARK 2: If $x$ is a real number with $\left[ x\right]\geq1$, then $\left[ 2x\right]\leq 3\left[ x\right]$.
Let $a$, $b$ and $c$ be any natural numbers. Write
$$ a=\sum_{i=0}^{n-1}a_i 10^i,\quad b=\sum_{i=0}^{n-1}b_i 10^i, \quad c=\sum_{i=0}^{n-1}c_i 10^i, $$
where $a_i,b_i,c_i\in\{0,\ldots,9\}$ and $a_n\neq0$ or $b_n\neq0$ or $c_n\neq0$. By Remark 1,
$$ S(a+b+c)=\sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right), $$
$$ S(a+b)=\sum_{i=0}^{n-1}\left(a_i+b_i-9\cdot \left[ \frac{a_i+b_i}{10}\right]\right), $$
$$ S(b+c)=\sum_{i=0}^{n-1}\left(b_i+c_i-9\cdot \left[ \frac{b_i+c_i}{10}\right]\right), $$
$$ S(a+c)=\sum_{i=0}^{n-1}\left(a_i+c_i-9\cdot \left[ \frac{a_i+c_i}{10}\right]\right). $$
By Remark 2, if $\left[ (a_i+b_i+c_i)/10\right]\geq1$, then $$3\left[(a_i+b_i+c_i)/10\right]\geq \left[ 2(a_i+b_i+c_i)/10\right]\geq \left[ (a_i+b_i)/10\right]+\left[ (b_i+c_i)/10\right]+\left[ (a_i+c_i)/10\right].$$ Thus,
$$ \begin{align*}
S(a+b+c)= {} & \sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right) \\\\
= {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)\\
&\qquad-9\sum_{\substack{{0\leq i\leq n-1}\\{(a_i+b_i+c_i)/10\geq1}}}\left[ \frac{a_i+b_i+c_i}{10}\right] \\\\
\leq {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)-3\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad-3\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]-3\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
= & \frac12\sum_{i=0}^{n-1}\left(a_i+b_i-6\left[ \frac{a_i+b_i}{10}\right]\right)+\frac12\sum_{i=0}^{n-1}\left(b_i+c_i-6\left[ \frac{b_i+c_i}{10}\right]\right)\\
&\qquad+\frac12\sum_{i=0}^{n-1}\left(a_i+c_i-6\left[ \frac{a_i+c_i}{10}\right]\right) \\\\
= {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
\leq {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}S(a+b)+\frac{3}{2}S(b+c)+\frac{3}{2}S(a+c) \\\\
= {} & 2(S(a+b)+S(b+c)+S(a+c))\;.
\end{align*}
$$ Then, if $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$, we have $S(a+b+c)<2(5+5+5)=30$.
| Counterexample (the sum is 51):
a = 555 444 555 5
b = 555 555 444 5
c = 444 555 555 5
a+b = 1111 000 000 0
a+c = 1000 000 111 0
b+c = 1000 111 000 0
a+b+c = 1555 555 555 5
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Matrix information that can be extracted from the characteristic polynomial
Let $A\in M_{3}(\mathbb{R})$ with $f_{\lambda}(x)=x^3+4x^2+5x+2$
Find the eigenvalues, the trace and the determinant of $A$
$f_{\lambda}(x)=x^3+4x^2+5x+2$ the dividers of $2$ are $-1,-2$ we get that
$(-1)^3+4(-1)^2+5(-1)+2=0$ dividing the polynomial we get $f_{\lambda}(x)=x^3+4x^2+5x+2=(x+2)(x+1)^2$ so $
\lambda_{1}=-2$ and $\lambda_{2}=-1$
Now the trace is minus the coefficient of the element of the the $n-1$ power, in this case $-4$
The determinant is the value of the free variable which is $2$
Is this correct? where can I find more information about the coefficients of the characteristic polynomial?
| One has for an $n\times n$ matrix:
$$ f_\lambda(x) = \sum_{k=0}^n x^{n-k} (-1)^{k} \ {\rm tr } \wedge^k A
= \sum_{k=0}^n x^{n-k} (-1)^k \sum_{i_1<...<i_k} \lambda_{i_1} ... \lambda_{i_k} $$
where $\lambda_1,...,\lambda_n$ are the $n$ eigenvalues of $A$.
For the traces one has in terms of elements of $A$:
$$ {\rm tr} \wedge^k A= \sum_{i_1<i_2<...<i_k}
\left| \begin{matrix} a_{i_1 i_1} & ... & a_{i_1 i_k} \\
.. & ... & .. \\
a_{i_k i_1} & ... & a_{i_k i_k } \end{matrix} \right|$$
so e.g. in the case $n=3$
$$ {\rm tr} \wedge^2 A=
\left| \begin{matrix} a_{1 1} & a_{1 2} \\
a_{2 1 } & a_{2 2 } \end{matrix} \right| +
\left| \begin{matrix} a_{1 1} & a_{1 3} \\
a_{3 1 } & a_{3 3 } \end{matrix} \right|+
\left| \begin{matrix} a_{2 2} & a_{2 3} \\
a_{3 2 } & a_{3 3 } \end{matrix} \right|
$$
One has ${\rm tr } \wedge^1 A = {\rm tr } A$ (usual trace) and
${\rm tr } \wedge^n A = {\rm det\ }A$. In your case $\det A=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $9|2^{2n+1}+2^{4n+1}-4$
Show that $9|2^{2n+1}+2^{4n+1}-4$ ( n is a positive integer):
1-Using induction
2-Don't use induction
I have noticed that $2^m \equiv 2^{m+6} \pmod 9 $ but couldn't use it to solve the problem!
| $2^{2n+1}+2^{4n+1}-4 = f(2^{2n})$ for $f(x)=2x^2+2x-4=2(x-1)(x+2)$.
Now $2^{2n} \bmod 9$ is $1,4,7,1,4,7,\dots$ and
$f(1) \equiv 0 \bmod 9$
$f(4) = 36 \equiv 0 \bmod 9$
$f(7) \equiv 0 \bmod 9$
So $f(2^{2n}) \equiv 0 \bmod 9$ for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Does this constitute a valid proof that $\frac{x^2}{1+x^4} \leq \frac{1}{2}$?
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$
First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expression as below;
\begin{align}
\frac{x^2}{1+x^4} - \frac{1}{2} &\leq 0\\\\
\impliedby\frac{2x^2 - (1+x^4)}{2(1+x^4)} &\leq 0\\\\
\impliedby -\frac{(x^4 - 2x^2 + 1)}{2(1+x^4)}&\leq 0\\\\
\impliedby \frac{x^4 - 2x^2 + 1}{2(1+x^4)} &\geq 0\\\\
\end{align}
We then make the observation that $x^4 - 2x^2 + 1 = (x^2 - 1)^2 \geq 0$ for all $x \in \Bbb R$ and that $1 + x^4 \geq 1$ for all $x \in \Bbb R$, and so the final inequality holds for all $x \in \Bbb R. \\\hspace{92pt}\square$
Is it okay to prove it in this way? I have no solutions so it would be nice if I could have some validation!
| Start from
$$x^4-2x^2+1=(x^2-1)^2\ge0$$
which is true for all $x\in\Bbb R$. Then rearrange:
$$2x^2\le1+x^4$$
$$\frac{x^2}{1+x^4}\le\frac12$$
where the last step is done because 2 and $1+x^4$ are always positive for $x\in\Bbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
| In order to prove that $5$ is a divisor of $P(n)$ for even $n$, consider the following cases:
*
*$n\equiv\color\red0\pmod{10} \implies n+5\equiv\color\red0+5\equiv5\pmod{10} \implies 5|P(n)$
*$n\equiv\color\red2\pmod{10} \implies n+3\equiv\color\red2+3\equiv5\pmod{10} \implies 5|P(n)$
*$n\equiv\color\red4\pmod{10} \implies n+1\equiv\color\red4+1\equiv5\pmod{10} \implies 5|P(n)$
*$n\equiv\color\red6\pmod{10} \implies n+9\equiv\color\red6+9\equiv15\equiv5\pmod{10} \implies 5|P(n)$
*$n\equiv\color\red8\pmod{10} \implies n+7\equiv\color\red8+7\equiv15\equiv5\pmod{10} \implies 5|P(n)$
That said, I'm not sure how to prove that $5$ is indeed the largest divisor of $P(n)$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
If $a$ and $b$ are consecutive integers, prove that $a^2 + b^2 + a^2b^2$ is a perfect square. Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the same expression can be written as $$a^2 + (a^2 + 1)(a+1)^2$$ which would imply that there is no real root. Would really appreciate some help, thanks.
| Note that $a^2+b^2+a^2b^2 = (a^2+1)(b^2+1)-1$
Then this becomes a special case of the Brahmagupta-Fibonacci identity:
$
(x^2+y^2)(z^2+w^2) = (xz+yw)^2+(xw-yz)^2
$
Take $x = a, y = 1, z=b, w = 1$ and you get:
$
(a^2+1)(b^2+1) = (ab+1)^2+(a-b)^2 = (a^2+a+1)^2+1
$
since $b = a+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.
| We have $1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ and thus $$3(ab+bc+ca)=3/2-3/2(a^2+b^2+c^2)$$.
Using this and the fact you found, we find that an equivalent inequality is
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}-\frac{3}{2}a^2-\frac{3}{2}b^2-\frac{3}{2}a^2\geq1$$
Now $\frac{a}{1-a}\geq a(9/4a+3/4) $ (since $1/(1-a)\geq (9/4)a+3/4$) so we can conclude that
$$
\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}-\frac{3}{2}a^2-\frac{3}{2}b^2-\frac{3}{2}a^2\geq $$
$$\frac{3}{4}(a^2+b^2+c^2)+\frac{3}{4}(a+b+c)\geq$$
$$\frac{3}{4}+\frac{1}{4}=$$
$$1$$
And we're done.
Intuition:
I tried to go for an inequality of the form
$$\frac{1}{1-a}\geq ax+y$$
in an effort to 'homogenize' the found inequality. Due to the fact that $1/(1-a)$ is convex, we have that any line tangent to its graph on $x\in [0,1)$ would be below it on that region. The reason I chose $$\frac{1}{1-a}\geq \frac{9}{4}a+\frac{3}{4}$$
was that I noticed that the first inequality had equality when $a=b=c=1/3$ and thus I wanted my middle-man inequality to also have equality when $a=1/3$ and thus chose the line tangent to $1/(1-a)$ at $a=1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Partial sum of $k>2$ consecutive positive integers has no solution in prime Progress:
Let $a_1,...,a_k$ a sequence of $k$ consecutive positive integers. If $k>2$, show that no partial sum of the sequence $a_1+...+a_k$ has a solution in prime.
I made a heuristic attempt and found a pattern in the progression below:
$$x$$ $$x+(x+1)=2x+1$$ $$x+(x+1)+(x+2)=3x+3$$ $$x+(x+1)+(x+2)+(x+3)=4x+6$$ $$x+(x+1)+(x+2)+(x+3)+(x+4)=5x+10$$ $$x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=6x+15$$ $$x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=7x+21$$ $$...$$
Which clearly shows that after $2$nd line each right-hand side value tend to have a common factor greater than $1$ and implies they cannot be prime. Although, technically it's not a proof, so I would appreciate any elementary proof on this matter.
Regards
| The sum $a+(a+1)+\dots + (a+k-1)$ is equal to $\frac{k(2a+k-1)}{2}$.
If $k>2$ then both $k$ and $(2a+k-1)>2$. So your number is the product of two numbers that are larger than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If we know the GCD and LCM, how to find all unordered pairs? If we know that $\gcd(a,b)=12$ and $\operatorname{lcm}(a,b)=360$, how can we find all unordered pairs $a,b$?
The answer in the back of the book I got this from gives $(12,360), (24,180), (36,120), (60,72)$ as the solutions.
| When solving such questions, it is best to think in terms of prime factorizations instead of integers. So we start by prime-factorizing $360$ and $12$:
\begin{align*}
360 &= 2^3 \cdot 3^2 \cdot 5^1 \\
12 &= 2^2 \cdot 3^1
\end{align*}
$a$ and $b$ must divide $360$, since $360$ is the LCM, so they cannot have any prime factors other than $2, 3, 5$. So
\begin{align*}
a &= 2^x 3^y 5^z \\
b &= 2^{x'} 3^{y'} 5^{z'}
\end{align*}
Now from LCM and GCD being $360$ and $12$, we get
\begin{align*}
\text{max}(x,x') = 3, \text{min}(x,x') = 2 \implies \{x,x'\} = \{2,3\} \\
\text{max}(y,y') = 2, \text{min}(x,x') = 1 \implies \{y,y'\} = \{1,2\} \\
\text{max}(z,z') = 1, \text{min}(z,z') = 0 \implies \{z,z'\} = \{0,1\} \\
\end{align*}
So to write out all possibilities, we just have to assign $2$ and $3$ to $x$ and $x'$, $1$ and $2$ to $y$ and $y'$, and $0$ and $1$ to $z$ and $z'$.
Since we only want unordered possibilities, we can start out taking $z = 0$, $z' = 1$. So all unordered possibilities for $(a,b)$ are, as you wrote:
\begin{align*}
(2^2 3^1, 2^3 3^2 5^1) &= (12, 360) \\
(2^3 3^1, 2^2 3^2 5^1) &= (24, 180) \\
(2^2 3^2, 2^3 3^1 5^1) &= (36, 120) \\
(2^3 3^2, 2^2 3^1 5^1) &= (72, 60). \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Proof of an inequality involving different exponents I was asked the following interesting question but I can not come up with a proof.
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{a^3}{b^3 + c^3} + \frac{b^3}{a^3 + c^3} + \frac{c^3}{a^3 + b^3} \geq \frac{a^2}{b^2 + c^2} + \frac{b^2}{a^2 + c^2} + \frac{c^2}{a^2 + b^2} \tag{$\clubsuit$}
$$
One way to prove $(\clubsuit)$ may be to consider the following function
$$
f(x) = \frac{a^x}{b^x + c^x} + \frac{b^x}{a^x + c^x} + \frac{c^x}{a^x + b^x}
$$
and prove that $f(x)$ is a non-decreasing function on $[2, 3]$. When we take its derivative, the formula looks quite tedious and we just can not continue on. So is there any better solution?
| I think the solution with derivative is very smooth:
$$f'(x)=\sum\limits_{cyc}\frac{a^x\ln a(b^x+c^x)-a^x(b^x\ln b+c^x\ln c)}{(b^x+c^x)^2}=$$
$$=\sum\limits_{cyc}\frac{a^x(b^x(\ln a-\ln b)-c^x(\ln c-\ln a))}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}(\ln a-\ln b)\left(\frac{a^xb^x}{(b^x+c^x)^2}-\frac{a^xb^x}{(a^x+c^x)^2}\right)=$$
$$=\sum\limits_{cyc}\frac{a^xb^x(\ln a-\ln b)(a^x-b^x)(a^x+b^x+2c^x)}{(a^x+c^x)^2(b^x+c^x)^2}\geq0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find a trajectory $\sigma(t)$ to represent the ellipse $\{(x,y):4x^{2}+9y^{2}=36\}$? Given the ellipse $\{(x,y):4x^{2}+9y^{2}=36\}$, find a trajectory $\sigma(t)$ which represent it.
So far, I have this:
The standard equation for an ellipse is:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
So,
$4x^{2}+9y^{2}=36\}$
$=\left \langle \text{Arithmetic: Divide by}\ 36 \right \rangle$
$\frac{4x^{2}}{36}+\frac{9y^{2}}{36}=\frac{36}{36}$
$=\left \langle \text{Arithmetic} \right \rangle$
$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
I don't know how to continue developing the solution, but I think the trajectory could be represented as $(a\cos(t),b\sin(t))$
Some ideas, suggestions to solve this?
| For a circle $$x^2+y^2 = r^2$$
we can try
$$x=r\cos(t)$$
$$y=r\sin(t)$$
because
$$x^2+y^2 = r^2\cos^2(t)+r^2\sin^2(t) = r^2$$
So for an ellipse
$$a^2x^2+b^2y^2 = c^2$$
we can try
$$x=\frac{c}{a}\cos(t)$$
$$y=\frac{c}{b}\sin(t)$$
In our case we have
$$2^2x^2+3^2y^2 = 6^2$$
so let us try try
$$x=\frac{6}{2}\cos(t)$$
$$y=\frac{6}{3}\sin(t)$$
Trying:
$$2^2x^2+3^2y^2 = 2^2(\frac{6}{2}\cos(t))^2+3^2(\frac{6}{3}\sin(t))^2$$
$$ = 6^2\cos^2(t) + 6^2\sin^2(t) = 6^2 = 36$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Posch Teller potential analytical classical solution I am trying to obtain the classical solution for a 1D Schroedinger equation with a symmetric Posch-Teller potential, the equation is given by,
$ \frac{d^2 x}{d t^2} - V \frac{\sinh(x)}{\cosh^3(x)} = 0 $
Where $V>0$. The boundary conditions are given by $x(t_0) = x_0$ and $x(t_1) = x_1$ (with $t_0 < t_1$).
Does it exist an analytical solution for this equations? If not, which should be the best perturbative method in this situation?
The analytical solution is essential for the rest of the analysis as the quantum part depends on the classical path given the non-linearity of the Posch-Teller potential.
Thanks
Alvaro
| I think there is an analytic solution, I am not sure how convenient its form is. We have the equation
$$\frac{d^2 x}{dt^2} = \frac{V\sinh(x)}{\cosh^3(x)}.$$
Multiply both sides of the equation by the first derivative to obtain
$$\frac{d^2 x}{dt^2} \, \frac{d x}{dt} = \left(\frac{V\sinh(x)}{\cosh^3(x)} \right)\frac{d x}{dt}$$ which then can be written as
\begin{align}\frac{1}{2}\frac{d}{dt}\left(\left(\frac{d x}{dt}\right)^2 \right) &= \left(\frac{V}{\cosh^3(x)} \right)\frac{d}{dt}\big(\cosh(x)\big) = \frac{d}{dt}\left((-1) \frac{V}{2\,\cosh^2(x)}\right)
\end{align}
So if we integrate both sides with respect to $t$ we obtain the conservation of energy property of the system
$$\frac{1}{2}\left(\frac{d x}{dt}\right)^2 = E_0 \, - \, \frac{V}{2\,\cosh^2(x)} $$ or maybe in a more familiar form
$$\frac{1}{2}\left(\frac{d x}{dt}\right)^2 + \frac{V}{2\,\cosh^2(x)} = E_0 $$ which as you see says that the sum of the kinetic energy $\frac{1}{2}\left(\frac{d x}{dt}\right)^2$ and the potential energy $\frac{V}{2\,\cosh^2(x)}$ is a constant $E_0$ with respect to time $t$ for any solution $x=x(t)$ of the system. Since we are in the one degree of freedom case, we simply look at $$\frac{1}{2}\left(\frac{d x}{dt}\right)^2 = E_0 \, - \, \frac{V}{2\,\cosh^2(x)} $$ and take a square root and write the equation as
$$\frac{d x}{dt} = \epsilon\sqrt{2 E_0 \, - \, \frac{V}{\cosh^2(x)}} $$ where $\epsilon$ is either $+1$ or $-1$ depending on whether the derivative is negative or positive. It is locally constant (roughly speaking).
Rewrite equation as
$$\cosh(x) \, \frac{d x}{dt} = \epsilon\sqrt{2 E_0 \, \cosh^2(x) \, - \, V} $$ which becomes
$$ \frac{d}{dt} \big(\sinh(x)\big) = \epsilon\sqrt{2 E_0 \, \Big(\sinh^2(x) + 1 \Big) \, - \, V} $$
Now it is easy to change the variables for convenience as $y = \sinh(x)$. The equation turns into
\begin{align} \frac{dy}{dt} &= \epsilon\sqrt{2 E_0 \, \big(y^2 + 1 \big) \, - \, V}\\
&= \epsilon\sqrt{2 E_0 \, y^2 + 2E_0 - V} \\
&= \epsilon \sqrt{2E_0} \, \sqrt{ y^2 + \frac{2E_0 - V}{2 E_0}}
\end{align}
Thus, we have the first order scalar time homogeneous equation
\begin{align} \frac{dy}{dt} &= \epsilon \sqrt{2E_0} \, \sqrt{ y^2 + \frac{2E_0 - V}{2 E_0}}
\end{align} which is integrable and solvable, but there are three different cases, based on the level of total energy $E_0$ and its value with respect to the peak maximum of the potential energy, which depends on $V$. In the case, $E_0 < \frac{V}{2}$ you have solutions occupying only the right side of a given value $x_0>0$, so moving only on the right of $x_0$ (when $x_0 < 0$ it occupies only the left side of $x_0$ i.e. it moves only on the left of $x_0$). In the case when $E_0 > \frac{V}{2}$ the solution occupies the whole real line, i.e. goes to plus infinity and to minus infinity. When $E_0 = \frac{V}{2}$ then we have special solutions, one is an equilibrium, the rest are go either to the left of $0$ or to the right of $0$, never to reach $0$.
So let's $E_0 < \frac{V}{2}$. Let us set $$a_0 = \sqrt{\frac{V -2E_0}{2E_0}} = \sqrt{\left|\frac{V -2E_0}{2E_0}\right|}$$ then the equation becomes
\begin{align} \frac{dy}{dt} &= \epsilon \sqrt{2E_0} \, \sqrt{ y^2 - a_0^2}
\end{align} so write it as
\begin{align} \frac{dy}{ \epsilon \sqrt{2E_0} \, \sqrt{ y^2 - a_0^2}} &= dt
\end{align} so integrating both sides we obtain
\begin{align} \int \, \frac{dy}{ \epsilon \sqrt{2E_0} \, \sqrt{ y^2 - a_0^2}} &= t + C
\end{align}
\begin{align} \frac{\epsilon}{\sqrt{2 E_0}}\int \, \frac{dy}{\sqrt{ y^2 - a_0^2}} &= t + C
\end{align}
\begin{align} \frac{\epsilon}{\sqrt{2 E_0}} \, \cosh^{-1}\left(\frac{y}{a_0}\right) &= t + C
\end{align}
\begin{align} \cosh^{-1}\left(\frac{y}{a_0}\right) &= \epsilon \sqrt{2 E_0} \,\, t + C_0
\end{align}
\begin{align} y(t) &= a_0 \cosh{\left(\epsilon \sqrt{2 E_0} \,\, t + C_0\right)}
\end{align}
So finally, returning to the original variable $x$ we obtain
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \cosh{\left(\epsilon \sqrt{2 E_0} \,\, t + C_0\right)}\Big)
\end{align}
If we impose the initial value problem $x(0) = x_0 > 0$ for instance then
$$C_0 = \cosh^{-1}\left(\frac{1}{a_0}\sinh(x_0)\right).$$ If we have initial conditions such that the first derivative of $x(t)$ at the point $x_0$ is positive,, i.e. $\frac{dx}{dt}(0) > 0$, then the solution is
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \cosh{\left(C_0 + \sqrt{2 E_0} \,\, t \right)}\Big)
\end{align}
for all $t \geq -\frac{C_0}{\sqrt{2E_0}}$ and
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \cosh{\left(C_0 - \sqrt{2 E_0} \,\, t\right)}\Big)
\end{align}
for all $t <-\frac{C_0}{\sqrt{2E_0}}$.
When $E_0 > \frac{V}{2}$ then the equation looks like
\begin{align} \frac{dy}{dt} = { \epsilon \sqrt{2E_0} \, \sqrt{ y^2 + a_0^2}}
\end{align} and so the solution is
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \sinh{\left(C_0 + \epsilon\sqrt{2 E_0} \,\, t \right)}\Big)
\end{align}
When $E_0 = \frac{V}{2}$ then $a_0=0$ and the equation simplifies significantly, allowing you to obtain an exponent of $t$ as a solution. This are the so called separatrix solutions.
You have to double check some of the calculations and especially the signs $\epsilon$. It is also instructive to draw the phase portrait of this system (the dynamics in the $x, \frac{dx}{dt}$-plane, known as the phase space). Sorry I cannot draw it here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all $x$ so $\frac{1^2+2^2+\cdots+x^2}{x}$ is a perfect square... Find all $x$ so $\dfrac{1^2+2^2+\cdots+x^2}{x}$ is a perfect square...
Clearly 1 is solution, but I have to show that there is an infinity... I see that this happens when the numbers $1,2,...,x$ have a whole quadratic root. Am I on the right road?
| The main result is:
$$1^2+2^2+3^2+\dots + x^2=\frac{x(x+1)(2x+1)}{6}$$
So you need $\frac{(x+1)(2x+1)}{6}=y^2$ for some $y$.
Or:
$$2x^2+3x+1 = 6y^2$$
From here, we are essentially going to "complete the square" on the left side.
Multiply both sides by $8$ and you get:
$$16x^2+24x+8=(4x+3)^2-1=48y^2$$
So you need infinitely many solutions to:
$$(4x+3)^2-48y^2=1$$
So you need to know how to solve Pell's equations from here.
The generally positive solution to the equation $u^2-48y^2=1$ can be written as:
$$u+y\sqrt{48}=(7+\sqrt{48})^k$$
For any positive integer $k$.
You should be able to prove with induction that infinitely many $u$ have $u\equiv 3\pmod 4$.
The recursive sequence of values of $u$ is $u_0=1,u_1=7$ and $u_{n+1}=14u_n-u_{n-1}$. You can show that $u_k\equiv 3\pmod 4$ when $k$ is odd. The next answers after $x=1$ are $x=337$ and $x=65521$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the differential equation $(x+5)y'+5y=\frac{3\sin x}{(x+5)^4}, x>-5$ What does it mean to solve the differential equation? For example,
Solve the differential equation
$$
(x+5)y'+5y=\frac{3\sin x}{(x+5)^4}, x>-5
$$
I believe I am to put this in the form of $y' + P(x)y = Q(x)$. Then integrate using $$v(x)=e^{\int Pxdx}$$.
If this is the technique I am to use when solving these problems, I seem to be having trouble identifying what is $P(x)$ and $Q(x)$. Any assistance is appreciated. Thank you.
| $$y'(x)(5+x)+5y(x)=\frac{3\sin(x)}{(5+x)^4}\Longleftrightarrow$$
Divide both sides by $5+x$:
$$y'(x)+\frac{5y(x)}{5+x}=\frac{3\sin(x)}{(5+x)^5}\Longleftrightarrow$$
Let $v(x)=\exp\left[\int\frac{5}{5+x}\space\text{d}x\right]=(5+x)^5$
Multiply both sides by $v(x)$:
$$y'(x)(5+x)^5+5y(x)(5+x)^4=3\sin(x)\Longleftrightarrow$$
Set $5(5+x)^4=\frac{\text{d}}{\text{d}x}\left((5+x)^5\right)$:
$$y'(x)(5+x)^5+y(x)\cdot\frac{\text{d}}{\text{d}x}\left((5+x)^5\right)=3\sin(x)\Longleftrightarrow$$
Apply the reverse product rule:
$$\frac{\text{d}}{\text{d}x}\left(y(x)(5+x)^5\right)=3\sin(x)\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}x}\left(y(x)(5+x)^5\right)\space\text{d}x=\int3\sin(x)\space\text{d}x\Longleftrightarrow$$
$$y(x)(5+x)^5=\int3\sin(x)\space\text{d}x\Longleftrightarrow$$
$$y(x)=\frac{\int3\sin(x)\space\text{d}x}{(5+x)^5}=\frac{\text{C}-3\cos(x)}{(5+x)^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.
| This is a very intriguing problem. First of all, $x\geq 0$ and $a\geq\sqrt{a+x}$ must hold for real $x$.
Squaring both sides of the equation we get $x^2=a-\sqrt{a+x}$.
$(x^2-a)^2=a+x$
$\Rightarrow x^4-2ax^2-x+a^2-a=0$
$\Rightarrow (x^2-x-a)(x^2+x+(1-a))=0$
$\therefore x= \frac12\pm\sqrt{a+\frac14},-\frac12\pm\sqrt{a-\frac34}$
All four of these solutions may not work because some of them could be extraneous roots due to squaring. Two of the solutions give negative values for $x$, so it must be
$x= \frac12+\sqrt{a+\frac14},-\frac12+\sqrt{a-\frac34}$
But the 1st one again does not satisfy the given equation.
Hence the unique solution to the equation is $x=-\frac12+\sqrt{a-\frac34}$, and that solution only exists if $a\geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Number of roots of $x^4 + x^3 \sin(x) + x^2 \cos(x) = 0$ How many roots do the following polynomial have?
$$x^4 + x^3 \sin(x) + x^2 \cos(x)$$
Obviously $0$ is a root. Dividing by $x^2$, I am left with
$$x^2 + x\sin(x) + \cos(x)$$
How do I know this polynomial has no root?
| Another way is as follows: the function $$f(x)= x^4 + x^3 \sin(x) + x^2 \cos(x)$$ is an even function (since the cosinus is even and the sinus is odd), hence it is enough to see at $x\ge 0$. The derivative $$f’(x)=(4+\cos(x))x^3+2x^2\sin(x) +2x\cos(x)$$ is clearly positive for $x$ positive. Consequently $f(x)$ is increasing for $x\gt 0$. Thus the only root is the obvious $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why $4a^3+27b^2<0 \iff x^3+ax+b=0$ has exactly three solutions? According to Exc. 8 Sec. 4.3 of the book Advanced Calculus by Fitzpatrick,
For numbers $a$ and $b$, prove that the following equation has exactly three solutions if and only $4a^3+27b^2<0$: $$ x^3+ax+b=0, \ \ \ \ \ x \ \text{in} \ \mathbb R.$$
Long time struggling, I can't prove neither directions. The book is pretty rudimentary and discussions like here are beyond the scope of the book.
Let $f(x) = x^3 + ax + b$. So $f'(x) = 3x^2 + a$. Set $3x^2 + a = 0$, so that at the points $\sqrt{-a/3}$ and $-\sqrt{-a/3}$, $f'(x) = 0$. There exists exactly two maximum/minimum due to the existence of $3$ solutions. These $2$ max/min must be at $x = \sqrt{-a/3}$ and $x = -\sqrt{-a/3}$. Note that must $a<0$. Also must $f(-\sqrt{-a/3}) > 0$ and $f(\sqrt{-a/3})<0$. Putting $\pm \sqrt{-a/3}$ into $f(x)$, I fail to conclude $4a^3 + 27b^2 < 0$. As, $$-(-a/3)^{3/2}-a(-a/3)^{1/2}+b>0 \\ (-a/3)^{3/2}+a(-a/3)^{1/2}+b<0$$ results in $a<0$ and no more thing!
Please help!
| Let $\alpha$, $\beta$ and $\gamma$ be the roots, then $\alpha+\beta+\gamma=0$.
The discriminant:
\begin{align*}
\Delta &= (\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2 \\
&=
-27\alpha^2 \beta^2 \gamma^2-
4(\beta \gamma+\gamma \alpha+\alpha \beta)^3 \\
&= -27b^2-4a^3
\end{align*}
For real coefficients $a$, $b$ but with complex roots, there're two roots being conjugate.
Say $\beta=\bar{\gamma}=u+vi$, then
\begin{align*}
\Delta &= (\alpha-u-vi)^2(\alpha-u+vi)^2(2vi)^2 \\
&= -4v^2[(\alpha-u)^2+v^2]^2 \\
&< 0 \\
4a^3+27b^2 &> 0 \\
\end{align*}
For three real roots,
$$\Delta \ge 0 \implies 4a^3+27b^2 \le 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
From Complex number to polar I have to calculate $(3 + 3i)^{829}$
What I did:
$$|3 + 3i| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9}=\sqrt{18}$$
The argument of $3 + 3i$ is $45º$.
| The key is in the title: you need to write the number in polar form:
$$x+\mathrm i y = r(\cos\theta+\mathrm i \sin\theta)$$
where $r$ is the modulus of $x+\mathrm i y$ and $\theta$ is the argument.
In your case, you have $x + \mathrm i y = 3 + 3\mathrm i$.
The modulus is given by $\left|3+3\mathrm i\right|=\sqrt{3^2+3^2}=3\sqrt{2}$.
Since $3+3\mathrm i$ lies on the line $y=x$, and is in the upper-right quadrant. The argument is given by $\arg(3+3\mathrm i)=\frac{1}{4}\pi$.
That means that $r=3\sqrt{2}$ and $\theta=\frac{1}{4}\pi$, and hence
$$3+3\mathrm i = 3\sqrt{2}\left(\cos(\tfrac{1}{4}\pi)+\mathrm i \sin(\tfrac{1}{4}\pi)\right)$$
De Moivre's Theorem tells is that $(\cos \theta + \mathrm i \sin \theta)^n = \cos (n\theta) + \mathrm i \sin (n\theta)$.
It follows that
\begin{eqnarray*}
(3+3\mathrm i)^{829} &=& (3\sqrt 2)^{829}\left(\cos(\tfrac{829}{4}\pi)+\mathrm i \sin(\tfrac{829}{4}\pi)\right)
\end{eqnarray*}
The sine and cosine functions are $2\pi$-periodic, i.e. they repeat every $2\pi$ and so $\sin(x+2\pi)\equiv \sin x$ just like $\cos(x+2\pi) \equiv \cos x$ for all real $x$. We can subtract multiples of $2\pi=\frac{8}{4}\pi$ from $\frac{829}{4}\pi$ without changing the values of sine and cosine.
Since $829 = 4 \times 206 + 5$ it follows that $\frac{829}{4} = 206 + \frac{5}{4}$ and hence $\frac{829}{4}\pi = 206\pi + \frac{5}{4}\pi$.
Given that sine and cosine are $2\pi$ periodic:
\begin{eqnarray*}
\cos(\tfrac{829}{4}\pi) &=& \cos(206\pi + \tfrac{5}{4}\pi) \\ \\
&=& \cos(\tfrac{5}{4}\pi) \\ \\
\sin(\tfrac{829}{4}\pi) &=& \sin(206\pi + \tfrac{5}{4}\pi) \\ \\
&=& \sin(\tfrac{5}{4}\pi)
\end{eqnarray*}
It follows that
\begin{eqnarray*}
(3\sqrt 2)^{829}\left(\cos(\tfrac{829}{4}\pi)+\mathrm i \sin(\tfrac{829}{4}\pi)\right) &=& (3\sqrt 2)^{829}\left(\cos(\tfrac{5}{4}\pi)+\mathrm i \sin(\tfrac{5}{4}\pi)\right) \\ \\
&=& (3\sqrt 2)^{829}\left(-\tfrac{1}{\sqrt 2}-\tfrac{1}{\sqrt 2}\,\mathrm i \right) \\ \\
&=& -3^{829}\times 2^{414} \times (1+\mathrm i)
\end{eqnarray*}
This number is massive and shouldn't be written down.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$ If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$
Should I manipulate the equation? I'm stuck on this problem. Answers are greatly appreciated.
| Trig-free solution: If $x^2+y^2=1$ then
$$
(x+y)^2 + (x-y)^2 = 2(x^2+y^2)=2
$$
so the largest value of $(x+y)^2$ is $2$, attained when $x=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the smallest number $n > 1$ such that $n = a^2 = b^3 = c^3$ Find the smallest number $n > 1$ such that
\begin{equation}
n = a^2 = b^3 = c^5
\end{equation}
for three integers $a, b$ and $c$.
I thought about finding $\mathrm{lcm}(a^2,b^3)$ and $\mathrm{lcm}(b^3, c^5)$ by using the formula
\begin{equation}
\mathrm{lcm}(x,y) = \frac{x\cdot y}{\gcd(x,y)}
\end{equation}
and compare them to find $n$, but I'm struggling to find an expression for $\gcd(a^2,b^3)$ and $\gcd(b^3, c^5)$ that makes any sense.
I've tried to make expressions of $a$, $b$ and $c$, only to come up with this
\begin{align*}
a = b\sqrt{b} \quad &\Rightarrow \gcd(a, b) = b &\Rightarrow \mathrm{lcm}(a,b) = a\\
a = c^2 \sqrt{c}\quad &\Rightarrow \gcd(a,c) = c &\Rightarrow \mathrm{lcm}(a,c) = a\\
b = c^2 \frac{\sqrt{c}}{\sqrt{b}} \quad &\Rightarrow \gcd(b,c) = c &\Rightarrow \mathrm{lcm}(b,c) = b
\end{align*}
and
\begin{align*}
\gcd(a^2, b^3) = b^3 &\Rightarrow \mathrm{lcm}(a^2,b^3) = a^2 \\
\gcd(a^2, c^5) = c^5 &\Rightarrow \mathrm{lcm}(a^2,c^3) = a^2 \\
\gcd(b^3, c^5) = c^5 &\Rightarrow \mathrm{lcm}(b^3, c^5) = b^5 \\
\end{align*}
which doesn't get me anywhere.
Any suggestions to how I should tackle this problem?
| Let $n=p_1^{q_1}p_2^{q_2}...p_n^{q_n}$ be the prime factorization of $n$. Then we see that for $n$ to be a square, a cube, and a fifth power, we must have $2|{q_1}, 2|{q_2}, ... 2|{q_n}$, $3|{q_1}, 3|{q_2}, ... 3|{q_n}$, and $5|{q_1}, 5|{q_2}, ... 5|{q_n}$. To find the smallest $n$, we can let $n$ be of the form $p_1^{q_1}$ and choose $p_1=2$ as the smallest prime. Now we need to solve for $q_1$. Since we have $2|q_1$, $3|q_1$, and $5|q_1$, we see that $q_1=lcm(2,3,5)=30$. Then we have that $n=2^{30}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\sum\limits_{k=0}^{n-1} \sin^t(\pi k/2n)$? How to evaluate $\displaystyle\sum_{k=0}^{n-1} \sin^t\left(\frac{\pi k}{2n}\right)$?
$t,n$ are constants $\in \Bbb{Z}$.
My try:
$$\begin{align}
\zeta:=e^{i\pi/2n} \implies & \sum_{k=0}^{n-1}\sin^t\left(\frac{\pi k}{2n}\right) \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^k- \zeta^{-k}\right)^{t} \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^{-kt}\right)\left(\zeta^{2k}-1\right)^{t}\\
\end{align}$$
How to proceed from here?
If a closed-form solution is not possible, can we still work out cases such as $t = n, t = 2n, $ etc?
EDIT: Sangchul Lee provided an answer for even t. Still looking for a solution for odd t as well.
| Here is an answer when $t = 2s$ is even.
Let $n, s$ be non-negative integers. Then with $\zeta = \mathrm{e}^{i\pi/n}$ and $\omega = \zeta^2 = \mathrm{e}^{2i\pi/n}$, we have
\begin{align*}
2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right)
&= \sum_{k=0}^{n-1} \left( \frac{\zeta^k + \zeta^{n-k}}{i} \right)^{2s} \\
&= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} \zeta^{(2s-l)k+l(n-k)} \\
&= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} (-1)^l \omega^{(s-l)k}.
\end{align*}
Interchanging the order of summation, we have
\begin{align*}
2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right)
&= \sum_{l=0}^{2s} (-1)^{l-s} \binom{2s}{l} \left( \sum_{k=0}^{n-1} (\omega^{s-l})^k \right) \\
&= \sum_{l=0}^{2s} (-1)^{l-s} \binom{2s}{l} \left( n \cdot \mathbf{1}_{\{ l \equiv s \ (\mathrm{mod} \ n)\}} \right) \\
&= n \sum_{j} (-1)^{nj} \binom{2s}{s+nj},
\end{align*}
where the last summation runs over all integers $j$ such that $-\frac{s}{n} \leq j \leq \frac{s}{n}$.
Examples. As special cases, plugging $s = n$ yields
$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{2n}} \left[ \binom{2n}{n} + (-1)^n 2 \right], $$
and similarly, replacing $n$ by $2n$ and plugging $s = n$ yields
$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{2n} \right) = \frac{n}{2^{2n}} \binom{2n}{n} - \frac{1}{2}. $$
Addendum. Here is an intuition on why we expect the sum to be simplified. If $t = 2s$ is even, then we can expand $\sin^{2s} x$ into a linear combination of $1, \cos 2x, \cos 4x, \cdots, \cos 2sx$. So in our case, we may write
$$ \sin^{2s} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \cos \left(\frac{2\pi j k}{n}\right). $$
Now, as you sum this over $k = 0, \cdots, n-1$, all the cosine terms will cancel out except when $j$ is a multiple of $n$. This means that
$$ \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n}\right) = n (a_0 + a_n + a_{2n} + \cdots). $$
So it is enough to identify $a_j$ for $j$'s multiple of $n$. This can be done by expanding $\sin^{2s} x$ using complex exponential and the binomial theorem. This is essentially what we did in the computation above.
On the other hand, this trick does not work for odd $t$. Indeed, when $t = 2s+1$ is odd, we can write
$$ \sin^{2s+1} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \sin \left(\frac{\pi (2j+1)k}{n}\right) $$
for some constants $a_0, \cdots, a_n$. (In fact, $a_j = (-1)^j 2^{-2s} \binom{2s+1}{s-j}$.) Now summing both sides for $k = 0, \cdots, n-1$,
\begin{align*}
\sum_{k=0}^{n-1}\sin^{2s+1} \left(\frac{\pi k}{n}\right)
&= \sum_{j=0}^{s} a_j \sum_{k=0}^{n-1} \sin \left(\frac{\pi (2j+1)k}{n}\right) \\
&= \frac{1}{2^{2s}} \sum_{j=0}^{s} (-1)^j \binom{2s+1}{s-j} \cot \left(\frac{\pi(2j+1)}{2n}\right).
\end{align*}
It is hard for me to believe that this will ever simplify except for some nice $s$.
Addendum 2 (Just for fun). Although no longer simpler than the original sum, one can also prove that for complex $t$ with $\Re(t) > 0$ and for positive integer $n$ the following formula holds:
$$\sum_{k=0}^{n-1} \sin^{t} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{t}} \sum_{j=-\infty}^{\infty} (-1)^{nj} \binom{t}{\frac{t}{2}+nj}. $$
Here, $\binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$ is the extended binomial coefficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
Prove that G is laying on K and find the ratio of AH to HB. We are given $Δ$ $ABC$ with $∠C = 90°$ and $AC > BC$. The circle K has a diameter CH (where CH is a height in $Δ$ $ABC$) and intersects with $AC$ in point $P$ and with $BC$ in point $Q$. If $PQ$ bisects $CG$, where $G$ is the centroid of $Δ$ $ABC$, prove that $G$ is laying on $K$ and find the ratio of $AH$ to $HB$.
You can find the diagram here
|
We have that $PQ$ is another diameter of the circle having $CH$ as a diameter and $CM\perp PQ$.
Assuming $BC=a<b=AC$, we have $CH=PQ=\frac{ab}{\sqrt{a^2+b^2}}$ and $$CG=\frac{2}{3}CM=\frac{2}{3}AM=\frac{\sqrt{a^2+b^2}}{3}.$$
Moreover, since $CPQ$ is similar to $CBA$,
$$ CP = \frac{a}{\sqrt{a^2+b^2}}PQ = \frac{a^2 b}{a^2+b^2},\qquad CQ=\frac{ab^2}{a^2+b^2}.$$
If $G$ lies on the circumcircle of $CPQ$, by Ptolemy's theorem
$$ CG\cdot PQ = 2\,CP\cdot CQ $$
has to hold, hence
$$ \frac{\sqrt{a^2+b^2}}{3}\cdot \frac{ab}{\sqrt{a^2+b^2}} = \frac{2a^3b^3}{(a^2+b^2)^2} $$
or
$$ (a^2+b^2)^2 = 6a^2 b^2 $$
that is a second degree equation in $r=\frac{b}{a}$, leading to:
$$ r=\frac{b}{a}=\color{red}{\sqrt{2+\sqrt{3}}}. $$
The remaining part is easy: $\frac{HA}{HB}=\frac{b^2}{a^2}=r^2=\color{red}{2+\sqrt{3}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof of Euler's theorem in complex numbers I am an engineer and not a mathematician, but I have used the formula stated below many number of times.$$e^{i\theta} = \cos{\theta} + i\sin{\theta}$$ Just out of curiosity, what is the proof to this theorem? I know that adding the Taylor series of the sines and the cosines and comparing it to the Taylor series of $e^{i\theta}$ is a proof, but is there some other way to go about this?
| It can also be seen with series. Robert mentioned this but here are the details. It needs to be noted that the series expansions for $e^z$, $\sin x$, and $\cos x$, with $z \in \Bbb C$ and $x \in \Bbb R$, each have an infinite radius of convergence. Also, these series all converge absolutely. Absolute convergence makes it safe to add the sine and cosine series like we do in the end.
Recall:
\begin{align*}
e^x &= \sum_{n=0}^{+\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}2 + \frac{x^3}{3!} + \cdots\\[0.3cm]
\cos x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}2 + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\\[0.3cm]
\sin x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
\end{align*}
Therefore:
\begin{align*}
e^{ix} &= 1 + ix + \frac{(ix)^2}2 + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots\\[0.3cm]
&= 1 + ix - \frac{x^2}2 - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \frac{x^6}{6!} - i\frac{x^7}{7!} + \cdots\\[0.3cm]
i\sin x &= ix - i\frac{x^3}{3!} + i\frac{x^5}{5!} - i\frac{x^7}{7!} + \cdots
\end{align*}
And so:
\begin{align*}
\cos x + i\sin x &= \left(1 - \frac{x^2}2 + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) + \left(ix - i\frac{x^3}{3!} + i\frac{x^5}{5!} - i\frac{x^7}{7!} + \cdots\right)\\[0.3cm]
&= 1 + ix - \frac{x^2}2 - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \frac{x^6}{6!} - i\frac{x^7}{7!} + \cdots\\[0.3cm]
&= e^{ix}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
What is the Taylor series expansion of $\frac{2z+1}{(z+1)^2}$ at $z=i$
What is the Taylor series expansion of $\frac{2z+1}{(z+1)^2}$ at $z=i$?
I try to use that $a_n=\frac{f^{(n)}(i)}{n!}$, but the formula is too tedious. How should I proceed?
| We are looking for a representation
\begin{align*}
\frac{2z+1}{(z+1)^2}=\sum_{n=0}^\infty a_n(z-i)^n\qquad\qquad a_n\in\mathbb{C}
\end{align*}
We obtain
\begin{align*}
\frac{2z+1}{(z+1)^2}&=\frac{2(z-i)+(1+2i)}{((z-i)+(1+i))^2}\\
&=\frac{1}{(1+i)^2}\cdot\frac{2(z-i)+(1+2i)}{\left(1+\frac{z-i}{1+i}\right)^2}\\
&=-\frac{i}{2}\left(2(z-i)+(1+2i)\right)\sum_{n=0}^\infty \binom{-2}{n}\left(\frac{z-i}{1+i}\right)^n\tag{1}\\
&=\left(-i(z-i)+\left(1-\frac{1}{2}i\right)\right)
\sum_{n=0}^\infty (n+1)(-1)^n\left(\frac{z-i}{1+i}\right)^n\tag{2}\\
&=-i\sum_{n=0}^\infty\frac{(n+1)(-1)^n}{(1+i)^n}(z-i)^{n+1}\\
&\qquad +\left(1-\frac{1}{2}i\right)\sum_{n=0}^\infty\frac{(n+1)(-1)^n}{(1+i)^n}(z-i)^n\tag{3}\\
&=-i\sum_{n=1}^\infty\frac{n(-1)^{n-1}}{(1+i)^{n-1}}(z-i)^{n}\\
&\qquad + \left(1-\frac{1}{2}i\right)\sum_{n=0}^\infty\frac{(n+1)(-1)^n}{(1+i)^n}(z-i)^n\tag{4}\\
&=\sum_{n=1}^\infty\frac{1+\frac{1}{2}(n-1)i}{(1+i)^n}(-1)^n(z-i)^n+1-\frac{1}{2}i\tag{5}\\
&=\sum_{n=0}^\infty\frac{1}{2^n}(-1+i)^n\left(1+\frac{1}{2}(n-1)i\right)(z-i)^n
\qquad\qquad\quad |z-i|<\sqrt{2}\tag{6}
\end{align*}
Comment:
*
*In (1) we expand the binomial series at $z=i$ with radius of convergence $\left|\frac{z-i}{1+i}\right|<1$, i.e. $|z-i|<\sqrt{2}$.
*In (2) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (3) we rearrange the series according to powers of $z-i$.
*In (4) shift the index of the left series by one to obtain powers $(z-i)^n$.
*In (5) we collect the terms with equal powers.
*In (6) we do some final simplifications.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_0^{\infty}\frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2$ with real method I came across the following integral:
$$\large{\int_0^\infty \frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2}$$
I know it could be solved with resuide method, and I want to know if there are some real methods can sove it?
Meanwhile,I remember a similar integral:
$$\large{\int_0^\infty \frac{1}{(x^2+1)(x^a+1)}\ dx=\frac{\pi}{4}}$$
And I want to know the following one:
$${\color{red}{\large{\int_0^\infty \frac{\ln x}{(x^2+1)(x^a+1)}\ dx = \huge{?}}}}$$
Using the Mathematica I got the follow result.
Could you suggest some ideas how to prove this? Any hints will be appreciated.
| Here is a variant of @tired's solution. In this solution, you are only required to know:
*
*How to expand a rational function into partial fraction decomposition,
*How to compute complex logarithm, and
*The following claim:
Claim. For any $z \in \Bbb{C}\setminus(-\infty, 0]$ we have
$$ I(z) := \int_{0}^{\infty} \left( \frac{\log x}{x+z} - \frac{\log x}{x+1} \right) \, dx = -\frac{1}{2}\log^2 z \tag{1}$$
where $\log$ is the principal value of the complex logarithm.
Computation. Before proving the claim, let us see how $\text{(1)}$ allows us to compute the integral in question. Using the partial fraction decomposition, we get
\begin{align*}
\frac{1}{(1+x^2)(1+x^3)}
&= \frac{1-i}{4} \cdot\frac{1}{x-i} + \frac{1+i}{4}\cdot\frac{1}{x+i} \\
&\qquad \qquad + \frac{1}{6}\cdot \frac{1}{x+1} - \frac{1}{3}\cdot \frac{1}{x - e^{\pi i/3}} - \frac{1}{3}\cdot \frac{1}{x - e^{-\pi i/3}}.
\end{align*}
(Obviously this is a disguise of @tired's complex-analytic solution, since this decomposition often comes from residue computation. On the other hand, residue is no longer a necessity and thus one can work purely with algebra.)
Also notice that the sum of 'coefficients' are zero: $ \frac{1-i}{4} + \frac{1+i}{4} + \frac{1}{6} - \frac{1}{3} - \frac{1}{3} = 0$. Therefore it follows from the claim that
\begin{align*}
\int_{0}^{\infty} \frac{\log x}{(1+x^2)(1+x^3)} \, dx
&= \frac{1-i}{4} I(-i) + \frac{1+i}{4}I(i)\\
&\qquad \qquad + \frac{1}{6}I(1) - \frac{1}{3}I(-e^{\pi i/3}) - \frac{1}{3}I(-e^{-\pi i/3}) \\
&= -\frac{1}{2} \bigg[ \frac{1-i}{4} \left( -\frac{i\pi}{2} \right)^2 + \frac{1+i}{4}I\left( \frac{i\pi}{2} \right)^2 \\
&\qquad \qquad + 0 - \frac{1}{3}\left( -\frac{2i\pi}{3} \right)^2 - \frac{1}{3}\left( \frac{2i\pi}{3} \right)^2 \bigg] \\
&= -\frac{37}{432}\pi^2.
\end{align*}
Proof of Claim. Now it remains to compute the integral $\text{(1)}$ This is easily done by differentiating $I(z)$:
$$ I'(z) = - \int_{0}^{\infty} \frac{\log x}{(x+z)^2} \, dx. $$
In order to compute this integral, we first replace the lower limit by $\epsilon > 0$ to obtain
$$ - \int_{\epsilon}^{\infty} \frac{\log x}{(x+z)^2} \, dx = \frac{\epsilon \log \epsilon}{z(z+\epsilon)} - \frac{1}{z}\log (z+\epsilon). $$
Taking limit as $\epsilon \to 0^+$, we get
$$ I'(z) = -\frac{1}{z}\log z, \qquad I(1) = 0. $$
This is enough to prove the claim. ////
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
If $20x=\pi$, what is $\frac{\cos 4x - \cos 8x}{\cos 4x\cdot \cos 8x}$? If $20x=\pi$,
what is $$\frac{\cos 4x - \cos 8x}{\cos 4x\cdot \cos 8x}?$$
I've tried using the factor formula on the numerator but I haven't managed to get anywhere with it...
This is a multiple choice question with options $4$, $2$, $1$, $-1$, and $-2$.
Thanks in advance!
| $$\\ 20x=\pi \\ 10x=\frac { \pi }{ 2 } \\ \frac { \cos { \left( 10x-6x \right) -\cos { \left( 10x-2x \right) } } }{ \cos { \left( 10x-6x \right) \cos { \left( 10x-2x \right) } } } =\frac { \sin { \left( 6x \right) -\sin { \left( 2x \right) } } }{ \sin { \left( 6x \right) \sin { \left( 2x \right) } } } =\frac { 2\sin { \left( 2x \right) \cos { \left( 4x \right) } } }{ \sin { \left( 6x \right) \sin { \left( 2x \right) } } } =2\frac { \cos { \left( 4x \right) } }{ \sin { \left( 6x \right) } } =2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Write the following as a fraction $1+\dfrac{1}{6x}+\dfrac{1}{x^2+3x}$
I keep getting answer like $\dfrac{6x^2 + 19x + 9}{6x(x+3)}$ but I think
it's wrong because I get different answers. Please help me
Greetings from Russia!
| \begin{align}
1 + \frac{1}{6x} + \frac{1}{x^2+3x} &= 1 + \frac{1}{6 x} + \frac{1}{x(x+3)} \\
&= 1 + \frac{x + 3}{6x(x+3)} + \frac{6}{6x(x+3)} = 1 + \frac{x + 9}{6x(x+3)} \\
&= \frac{6x^2 + 19x + 9}{6x(x+3)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.
| Dumping my answer here to a different but similar question which was labeled as a duplicate and would no longer accept the answer that I had already typed.
$\lim\limits_{t\to1}\frac{5}{1-t^5}-\frac{2}{1-t^2}=$
$\lim\limits_{t\to1}\frac{5}{(1-t)(1+t+t^2+t^3+t^4)}-\frac{2}{(1-t)(1+t)}=$
$\lim\limits_{t\to1}\frac{5+5t-2-2t-2t^2-2t^3-2t^4}{(1-t)(1+t+t^2+t^3+t^4)(1+t)}$
Clearly the numerator evaluates to $0$ when $t=1$ so $t-1$ must be a factor.
$\lim\limits_{t\to1}\frac{5+5t-2-2t-2t^2-2t^3-2t^4}{(1-t)(1+t+t^2+t^3+t^4)(1+t)}=$
$\lim\limits_{t\to1}\frac{2t^4+2t^3+2t^2-3t-3}{(t-1)(1+t+t^2+t^3+t^4)(1+t)}=$
$\lim\limits_{t\to1}\frac{2t^3+4t^2+6t+3}{(1+t+t^2+t^3+t^4)(1+t)}=$
$\frac{2+4+6+3}{(1+1+1+1+1)(1+1)}=\frac{15}{(5)(2)}=\frac32$
(Long division was used to obtain $\frac{2t^4+2t^3+2t^2-3t-3}{t-1}=2t^3+4t^2+6t+3$ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 10
} |
rational solutions of $x^4+x^2y^2+y^4=x^2$ I am trying to find rational solutions of $x^4+x^2y^2+y^4=x^2$, except for $(\pm 1,0), (0,0) $. I guess these three solutions are the only ones, but I failed to prove it. It seems that $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2) $ may be useful, but I do not know how to proceed.
| Let $a=x^2,b=y^2$. Then the equation simplifies to $a^2 + ab+b^2 = a$, which on grouping becomes $a^2 + a(b-1) + b^2=0$. Solving for $a$,
$$
a= \frac{1-b + \sqrt{1+b^2-2b-4b^2}}{2} = \frac{1-b + \sqrt{1-2b-3b^2}}{2}
$$
(The other root would be negative, but a is positive).
Therefore, we would like $1-2b-3b^2$ to be a perfect square, however, this is clearly only possible when $b=0$, because the expression is negative for $b \geq 1$.
Therefore, $y^2=0 \implies y=0 \implies x^4=x^2$, and :
$$
x^4-x^2=0 \implies x^2(x-1)(x+1)=0
$$
Hence there are three possible solutions, namely $(x,y)=(1,0),(0,0),(-1,0)$.
For rational non-zero $b$, suppose that $b = \frac{p^2}{q^2}$ in simplest form, where $p,q\neq 0$, so that $\gcd(p,q) = 1$. We are basically asking whether $1 - 2b - 3b^2$ can be a perfect square. First of all, note that : $$
(3b^2-2b-1) =(3b+1)(b-1) = \left(\frac{3p^2}{q^2}+1\right)\left(\frac{p^2}{q^2} -1\right) = \left(\frac{(3p^2+q^2)(p^2-q^2)}{q^2}\right)
$$
We claim that $\gcd(3p^2 + q^2,p^2 - q^2) = 1$ or $4$. To see this,let $p^2 = a$ and $q^2 = b$. Then, if $t | 3a+b$ and $t | a-b$, by summing $t | 4a$ and by taking $t | 3a-3b$ and subtracting we get $t | 4b$. Therefore, $t | \gcd(4a,4b) = 4$ as $a,b$ are coprime. However, $3a+b \equiv a-b \mod 4$, and in our case $a-b = p^2- q^2$ cannot take the value $2$ mod $4$, therefore the $\gcd$ cannot be two, since $p^2-q^2$ cannot divide two without dividing $4$, in which case the other must too.
$(3b+1)(b-1)$ is a perfect square if and only if $(3p^2 + q^2)(p^2 - q^2)$ is. If the two terms are coprime, then they are individually squares. If they are not, then $\frac{3p^2+q^2}{4},\frac{p^2-q^2}{4}$ are coprime whose product is a perfect square, and now $4$ is a perfect square so these terms are perfect squares implies that $3p^2+q^2$ and $p^2-q^2$ are both perfect squares.
That is, for co-prime $p$ and $q$, we want $p^2-q^2=c^2$ and $3p^2+q^2=d^2$ to both be squares, which are co prime.
However, $c^2 + d^2 = 4p^2$. Since $c^2$ and $d^2$ cannot both divide $4$, $c^2+d^2$ cannot be a multiple of $4$ from what we know about squares modulo $4$.
Therefore, no non-zero rational value of $b$ produces a square value of the expression. Hence, we have only $b= 0$ to consider, which of course forms part of the integral possibilities we discussed earlier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Let $a, b \in \mathbb{R}^+$. Prove that $\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$.
Let $$a, b \in \mathbb{R}^+$$ Prove that $$\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$$
We can make the denominators common on the LHS by AM-GM but the problem is there are 2 terms and the power is 3, so I am unable to simplify it. It gives
$$5(a^2+b^2)(a+b) \le 8 \cdot 6^{3/2} \cdot (ab)^{3/2}$$
I am not able to proceed. Please help.
Thanks.
| Consider
$$\frac{2a^2+3b^2}{2a^3+3b^3}(a+b) = 1 + a b \frac{2a+3b}{2a^3+3b^3}$$
and from Holder, $(2a^3+3b^3)(2+3)^2 \geqslant (2a+3b)^3$, so we may write
$$LHS \times(a+b) \leqslant 2 + 25ab\left(\frac1{(2a+3b)^2} + \frac1{(2b+3a)^2}\right)$$
so it is enough to show that
$$25ab[(2b+3a)^2+(2a+3b)^2] \leqslant 2(2a+3b)^2(2b+3a)^2$$
$$\iff 59(a^2-b^2)^2 + 13(a^4+b^4-a^3b-ab^3) \geqslant 0$$
which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $m$ and $n$ so that the given function has the range $[-3, 5]$ Find $m$ and $n$ real numbers so that $f(x) = \frac{3x^2 + mx + n}{x^2 + 1}$ takes all and only the values from the interval $[-3, 5]$.
I started by solving the following double inequality:
$$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$$
From the left inequality I got $6x^2 + mx + n + 3 \geq 0$. The coefficient of $x^2$ is positive so the discriminant $\delta$ has to be less than or equal to $0$.
$\delta = m^2 - 24n - 72 \leq 0$
From the right inequality I got $2x^2 -mx + 5 - n \geq 0$. By using the same technique as the one used to solve the left inequality I got:
$\delta = m^2 - 40 + 8n \leq 0$
By combining these two resulted inequalities I got $m \leq 48$, that is $m \in [-4\sqrt3, 4\sqrt3]$.
Now I need to find the values of $m$ and $n$ so that $f$ is surjective, because it must take all the values in the given interval.
I don't know yet how to proceed, so I would appreciate any help from you guys!
Thank you!
| $\frac{3x^2 + mx + n}{x^2 + 1}$ reaching a maximum value of $5$ implies that $5(x^2+1)-3x^2+mx+n = 2x^2-mx-(n-5)$ has a double root. Similarly, $\frac{3x^2 + mx + n}{x^2 + 1}$ reaching a minimum value of $3$ implies that $3x^2+mx+n-(-3)(x^2+1) = 6x^2+mx+(n+3)$ has a double root. Quadratic expressions have a double root when their determinant $b^2-4ac = 0$, so we obtain the simultaneous equations
$$
m^2+8(n-5) = 0
$$
and
$$
m^2-24(n+3) = 0
$$
which yields $n = -1$ and $m = \pm 4\sqrt{3}$. From Desmos:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How can the equation of a hyperbola be $xy=1$? We know that the standard form of a hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which is not coherent with $xy=1$. So, how can we say that $xy=1$ is a hyperbola?
| Actually equations in the form of $Ax^2+2Bxy+C^2+Dx+Ey=F$ can be simplified to an equation with variables $u,v$ depends on $x,y$ such that $\{u,v\}$ is an orthogonal basis for $\Bbb R^2$, i.e. one can rotate the graph in $uv$ coordinate into $xy$ plane.
Let $Ax^2+2Bxy+C^2+Dx+Ey=F$.
Then this equation can be written as
$$q(x,y)=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}A&B\\B&C\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}D&E\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=F$$
Find the eigenvalues and eigenvectors of $\begin{pmatrix}A&B\\B&C\end{pmatrix}$,namely $v_1,\lambda_1,v_2,\lambda_2$ .
Then a rotation matrix$R$ can be defined as
$$R=\begin{pmatrix}\frac{v_1}{|v_1|}&\frac{v_2}{|v_2|}\end{pmatrix}\text{ if }\det\begin{pmatrix}\frac{v_1}{|v_1|}&\frac{v_2}{|v_2|}\end{pmatrix}=1$$
$$R=\begin{pmatrix}\frac{v_1}{|v_1|}&-\frac{v_2}{|v_2|}\end{pmatrix}\text{ if }\det\begin{pmatrix}\frac{v_1}{|v_1|}&\frac{v_2}{|v_2|}\end{pmatrix}=-1$$
In such way $\begin{pmatrix}x\\y\end{pmatrix}=R\begin{pmatrix}u\\v\end{pmatrix}$. One can verify that the $xy$ term in the original given equation would be eliminated. Draw the graph with variables $u,v$ on the $uv$ plane. Then draw $xy$ plane with $u,v$ axis on it, then just draw the graph with respect to $u,v$ axis.
For practice, try graph $x^2+2xy+y^2-3x+y=0$, which is a parabola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$
If $a,b,c$ are sides of a triangle, prove that ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .
| Variational Approach
Without loss of generality, assume that $a+b+c=1$. Then
$$
\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\tag{1}
$$
has an interior extreme when
$$
\frac{\delta a}{(1-a)^2}+\frac{\delta b}{(1-b)^2}+\frac{\delta c}{(1-c)^2}=0\tag{2}
$$
for any variation so that
$$
\delta a +\delta b+\delta c=0\tag{3}
$$
Linearity says that we need $a=b=c=\frac13$, so that
$$
\bbox[5px,border:2px solid #C0A000]{\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac32}\tag{4}
$$
Due to symmetry and the triangle inequality, an edge extreme occurs only when $c=a+b=\frac12$ or when $c=0$, in which case $a+b=1$. In either case, we are finding an extreme of
$$
\frac{a}{1-a}+\frac{b}{1-b}\tag{5}
$$
An interior extreme of $(5)$ occurs when
$$
\frac{\delta a}{(1-a)^2}+\frac{\delta b}{(1-b)^2}=0\tag{6}
$$
for any variation so that
$$
\delta a +\delta b=0\tag{7}
$$
Linearity says that we need $a=b$, so that if $c=\frac12$, then $a=b=\frac14$ and
$$
\bbox[5px,border:2px solid #C0A000]{\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac53}\tag{8}
$$
or if $c=0$, then $a=b=\frac12$ and
$$
\bbox[5px,border:2px solid #C0A000]{\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=2}\tag{9}
$$
We don't need to worry about the edge edge case where $a=0$ or $b=0$ since those cases are covered by $(9)$. Therefore, the possible extremes are given in $(4)$, $(8)$, and $(9)$. Thus,
$$
\bbox[5px,border:2px solid #C0A000]{\frac32\le\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\le2}\tag{10}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $A+B+C=\pi$, prove: $\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$
If $A+B+C=\pi$, prove that:
$$\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$$
My Attempt:
Here, $A+B+C=\pi$
Now,
$$\begin{align}
LHS &=\cos(B+2C)+\cos(C+2A)+\cos(A+2B) \\
&=\cos(B+C+C)+\cos(C+A+A)+\cos(A+B+B) \\
&=\cos(\pi-(A-C))+\cos(\pi-(B-A))+\cos(\pi-(C-B)) \\
&=-\cos(A-C)-\cos(B-A)-\cos(C-B)
\end{align}$$
Please help to continue from here.
| For the RHS, you can use a known formula for the product of three cosines (see eg. here):
$\cos x \; \cos y \; \cos z = \frac{1}{4} \Big(\cos (x+y-z) + \cos (y+z-x) + \cos (z+x-y) + \cos (x+y+z)\Big)$
Considering your RHS, you set $x = \frac{B-C}{2}$, $y = \frac{C-A}{2}$, $z = \frac{A-B}{2}$, and plug it into the formula. Then, for example,
$x+y-z = B-A$, likewise the other arguments.
This gives immediately the form you arrived at,
$-\cos(A-C)-\cos(B-A)-\cos(C-B)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\int \frac{\sqrt{9-x^2}}{x}$ I don't understand how the integral
$$\int \frac{\sqrt{9-x^2}}{x}=\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$$
I keep getting $-3/x +c$ as the answer.
| Differentiate:
$$\left(\sqrt{9-x^2}-3\log(\sqrt{9-x^2}+3)+3\log x+C\right)'=$$
$$=-\frac x{\sqrt{9-x^2}}+\frac{3x}{\sqrt{9-x^2}}\cdot\frac1{\sqrt{9-x^2}+3}+\frac3x=$$
$$=\frac x{\sqrt{9-x^2}}\left(-1+\frac3{\sqrt{9-x^2}+3}\right)+\frac3x=$$
$$=\require{cancel}\frac x{\cancel{\sqrt{9-x^2}}}\frac{-\cancel{\sqrt{9-x^2}}}{\sqrt{9-x^2}+3}+\frac3x=\frac{9-x^2+3\sqrt{9-x^2}}{x(\sqrt{9-x^2}+3)}=$$
$$\frac{\sqrt{9-x^2}\left(\sqrt{9-x^2}+3\right)}{x(\sqrt{9-x^2}+3)}=\frac{\sqrt{9-x^2}}x$$
so they are right, you are wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the sum $\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \cdots$ Suppose $|x| < 1$. Can you give any ideas on how to find the following sum?
$$
\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \frac{8x^8}{x^8+1} + \cdots
$$
| Prove first that for any $x\in(-1,1)$ the identity
$$ \frac{1}{1-x}=\prod_{k\geq 0}\left(1+x^{2^k}\right) \tag{1}$$
follows from a telescoping product or the fact that every $n\in\mathbb{N}^*$ has a unique representation in base-$2$. By considering $\frac{d}{dx}\log(\cdot)$ of both sides of $(1)$, we get:
$$ \frac{1}{1-x} = \sum_{k\geq 0}\frac{2^k x^{2^k-1}}{1+x^{2^k}}\tag{2}$$
from which:
$$\sum_{k\geq 0}\frac{2^k x^{2^k}}{1+x^{2^k}}=\color{red}{\frac{x}{1-x}}.\tag{3}$$
Essentially the same "trick" allows us to derive Euler pentagonal number theorem from Jacobi's triple product, for instance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Using elementary row or column operations to compute a determinant How do you use elementary row or column operations to find the determinant of the following matrix?
$$\begin{bmatrix} 1 & 7 & -3\\
1 &3 & 1\\
4&8&1\end{bmatrix}$$
| The key idea in using row operations to evaluate the determinant of a matrix is the fact that a triangular matrix (one with all zeros below the main diagonal) has a determinant equal to the product of the numbers on the main diagonal. Therefore one would like to use row operations to 'reduce' the matrix to triangular form.
However, the effect of using the three row operations on a determinant are a bit different than when they are used to reduce a system of linear equations.
(1) Swapping two rows changes the sign of the determinant
(2) When dividing a row by a constant, the constant becomes a factor written in front of the determinant.
(3) Adding a multiple of one row to another does not change the value of the determinant.
Let's apply these operations to your matrix to find its determinant.
First we want to produce two zeros in rows $2$ and $3$ of column $1$. (Remember our goal is to produce all zeros below the main diagonal, and we do this one column at a time beginning with column $1$.)
The two row operations $-R_1+R_2\to R_2$ and $-4R_1+R_3\to R_3$ will accomplish this goal, and will not change the value of the determinant.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}
\end{eqnarray}
Now all that remains is to obtain a $0$ in row $3$ column $2$. We see that adding $-5$ times row $2$ to row $3$ will accomplish this. That is, $-5R_2+R_3\to R_3$.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}\\
&=& \begin{vmatrix}
1 & 7 & -3\\0 & -4 & 4\\0 & 0 & -7
\end{vmatrix}
\end{eqnarray}
Since we only had to use the third row operation, the one which does not change the value of the determinant and since we now have a triangular matrix, we find the determinant by multiplying the numbers on the main diagonal.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}\\
&=& \begin{vmatrix}
1 & 7 & -3\\0 & -4 & 4\\0 & 0 & -7
\end{vmatrix}\\
&=&28
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
To find the locus of a point
A point moves such that the sum of its distances from the coordinate axes is equal to its distance from the circle $x^2+y^2=4$.
I tried and got this equation for the locus:
$$x+y=\sqrt{x^2+y^2}-2$$
Should I replace $x+y$ with $|x|+|y|$ because distances are absolute? In that case, how will I represent $y$ as an explicit function of $x$?
Thanks in advance.
| Not only should you replace $x$ and $y$ with $|x|$ and $|y|$, you should also take the absolute value of the right-hand side because it represents a distance too:
$$|x|+|y|=\left\vert\sqrt{x^2+y^2}-2\right\vert$$
To solve this for $y$, we can first restrict ourselves to non-negative $x$ and $y$, since substituting $x\to-x$ or $y\to-y$ in a valid solution yields another valid solution. With this restriction we can remove the modulus signs around $x$ and $y$. The triangle inequality implies that $x+y>\sqrt{x^2+y^2}-2$, so when removing the modulus signs in the RHS we must have
$$x+y=2-\sqrt{x^2+y^2}$$
which leads to
$$2-(x+y)=\sqrt{x^2+y^2}$$
$$4-4(x+y)+x^2+2xy+y^2=x^2+y^2$$
$$4-4(x+y)+2xy=0$$
$$2-2(x+y)+xy=0$$
$$(x-2)y+2(1-x)=0$$
$$y=\frac{2(x-1)}{x-2},x\le1$$
Mirroring this across the $x$- and $y$-axes we get the final equation for $y$ in terms of $x$:
$$y=\pm\frac{2(|x|-1)}{|x|-2},|x|\le1$$
Here is a plot of the locus asked for by the question, with $x^2+y^2=4$ added as a reference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{\frac{x^2-1}{x^2}}=\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2}\cdot\frac{x^2}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}$$
As you can see I'm going in circles. Can anyone give me a hint on how to start on this problem?
| hint:
factor the denominator as follows
$x^2-1=(x-1)(x+1)$.
the numerator $x$ goes to $1$.
in the denominator, $(x+1)$ goes to $2$ but $(x-1)$ tends to $0$.
so, the denominator goes to $0$.
but $x$ is less than $1$, so the denominator goes to $0$ by negatives values, let's say it tends to $0^-$.
thus the final limit is $"\frac{1}{0^-}"$
which means $-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find the $n$th derivative of $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2$ In this case $n=16$ and the point is $x=1$.
I know $\sqrt{x}$ is not differentiable in zero, but this function actually has left derivatives.
Also, I know that $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2=x^2+x-2+2\sqrt{(x^2-1)(x-1)}$, so it is enogth to compute the derivative of $2\sqrt{(x^2-1)(x-1)}$.
Also, I tried to find the $n$th derivative of $\sqrt{g(x)}$ but its expresion was too complicated.
| HINT:
Since $x\ge 1$, we may write $\sqrt{(x^2-1)(x-1)}=(x-1)\sqrt{x+1}$.
Then, use the General Leibniz Rule for product differentiation
$$\frac{d^n}{dx^n}\left((x-1)\sqrt{x+1}\right)=\sum_{k=0}^n\binom{n}{k}\frac{d^k}{dx^k}(x-1)\frac{d^{n-k}}{dx^{n-k}}(x+1)^{1/2}$$
Finish by recognizing that the only surviving terms occur at $k=0$ and $k=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$ Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$
given that $a_0 = 0$, and $a_1 = 3$
My Attempt:
Get the characteristic equation and solve it.
For homogeneous equation
$x^2 -4x + 4 = 0$
$x = 2 $ or $ x = 2$
Hence, $a_n^h = (A+Bn)\cdot2^n $
Guess a particular solution: $n^22^nC$
$n^22^nC - 4(n-1)^22^{n-1}C + 4(n-2)^22^{n-2}C = 2^n$
$n^2C - 2(n-1)^2C + (n-2)^2C = 1$
$C= \frac12$
Hence, $a^p_n = \frac12n^22^n$
$a_n = a^p_n + a^h_n$
$a_n = (A+Bn)\cdot2^n + \frac12n^22^n$
$a_0 = 0 = A$
$a_1 = 3 = 2B + 1$
$B = 1$
Therefore, $a_n = (n+ \frac12n^2)2^n$
| $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{a_{n} - 4a_{n - 1} + 4a_{n - 2} = 2^{n}\,;\qquad
a_{0} = 0\,,\quad a_{1} = 3}$.
\begin{align}
&a_{n} - 4a_{n - 1} + 4a_{n - 2} = 2^{n} \implies
1 = {a_{n} \over 2^{n}} - 2\,{a_{n - 1} \over 2^{n - 1}} +
{a_{n - 2} \over 2^{n - 2} }
\end{align}
Moreover,
\begin{align}
\sum_{n = 2}^{\infty}z^{n} & =
\sum_{n = 2}^{\infty}{a_{n} \over 2^{n}}\,z^{n} -
2\sum_{n = 2}^{\infty}{a_{n - 1} \over 2^{n - 1}}\,z^{n} +
\sum_{n = 2}^{\infty}{a_{n - 2} \over 2^{n - 2}}\,z^{n}
\\[5mm]
{z^{2} \over 1 - z} & =
\sum_{n = 2}^{\infty}{a_{n} \over 2^{n}}\,z^{n} -
2\sum_{n = 1}^{\infty}{a_{n} \over 2^{n}}\,z^{n + 1} +
\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n + 2}
\\[5mm] & =
\pars{\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n} - {3 \over 2}\,z} -
2z\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n} +
z^{2}\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n}
\\[5mm] & =
\pars{1 - z}^{2}\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n} - {3 \over 2}\,z
\end{align}
Then,
\begin{align}
\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n} & =
-\,{z^{2} - 3z \over 2\pars{1 - z}^{3}} =
-\,{1 \over 2}\sum_{n = 0}^{\infty}{-3 \choose n}\pars{-1}^{n}
\pars{z^{n + 2} - 3z^{n + 1}}
\\[5mm] & =
-\,{1 \over 4}\sum_{n = 0}^{\infty}\pars{n + 2}\pars{n + 1}
\pars{z^{n + 2} - 3z^{n + 1}}
\\[5mm] & =
-\,{1 \over 4}\sum_{n = 2}^{\infty}n\pars{n - 1}z^{n} +
{3 \over 4}\sum_{n = 1}^{\infty}\pars{n + 1}n\,z^{n}
\\[5mm] & =
{3 \over 2}\,z +
\sum_{n = 2}^{\infty}
\bracks{-\,{1 \over 4}\,n\pars{n - 1} + {3 \over 4}\,\pars{n + 1}n}\,z^{n} =
{3 \over 2}\,z +
\sum_{n = 2}^{\infty}{\pars{n + 2}n \over 2}\,z^{n}
\\[5mm] & =
\sum_{n = 0}^{\infty}{\pars{n + 2}n \over 2}\,z^{n} \implies
{a_{n} \over 2^{n}} = {\pars{n + 2}n \over 2} \implies
\bbox[10px,#ffe,border:0.1em groove navy]{\color{#f00}{a_{n}} =
\color{#f00}{\pars{n + 2}n\,2^{n - 1}}\,,\quad \forall\ n \geq 0}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Incorrect method to find a tilted asymptote Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.
Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.
I would like to find out, though, what is wrong with the following incorrect way of finding the asymptote:
$\displaystyle y=\frac{x^2+x-6}{x+2}=\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}\approx\frac{x+1}{1}=x+1$, so $y=x+1$ is the slanted asymptote.
| The problem with your second approach is that you've kept more precision than your approximation actually has — you can say $y \approx x$, but you don't have enough precision to clarify that more specifically to $y \approx x+1$ (or any other translate).
In more detail,
$$ \frac{1}{1 + \frac{2}{x}} \approx 1 - \frac{2}{x} $$
and consequently,
$$ \frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}}
\approx
\left( x+1-\frac{6}{x} \right) \left(1 - \frac{2}{x} \right)
\approx
x \cdot 1 + 1 \cdot 1 - x \cdot \frac{2}{x}$$
By neglecting the $\frac{2}{x}$ term of the denominator, you neglect the $x \cdot \frac{2}{x}$ term of this approximation — but that term is $-2$, so you're neglecting a nonnegligible quantity!
Keeping the $\frac{2}{x}$ term around, the above approximation gives $x-1$, as desired.
For more rigor, you can use big O notation:
$$\frac{1}{1 + \frac{2}{x}} = 1 - \frac{2}{x} + O(x^{-2}) $$
$$
\frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}}
= \left( x+1+O(x^{-1}) \right) \left(1 - \frac{2}{x} + O(x^{-2}) \right)
= x - 1 + O(x^{-1})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
What is $\sin(nx)$ iteration in terms of $\sin A$ and $\cos A$? I want to use sum angle formulas, $\sin(A+B)=\sin A\cos B+\cos A\sin B$
to get for any angles, $\sin(nA)$ in terms of powers of $\sin A$ and $\cos A$.
I know there are other ways, but I want to use trigonometry and iteration on that. The goal is to carry out iteration in terms of $\sin(nA)$?
| The angle addition formulas for sine and cosine are nicely captured by this matrix formula
$$
\left( \begin{array}{c} s_{n} \\ c_{n}\end{array} \right) =
\left( \begin{array}{r} c & s \\ -s & c\end{array} \right)
\left( \begin{array}{c} s_{n-1} \\ c_{n-1}\end{array} \right)
$$
where $s_k = \sin(kx)$, $c_k = \cos(kx)$, $s= s_1$ and $c=c_1$.
Playing this recursion backwards, we arrive at the formula
$$
\left( \begin{array}{c} s_{n} \\ c_{n}\end{array} \right) =
\left( \begin{array}{r} c & s \\ -s & c\end{array} \right)^n
\left( \begin{array}{c} s_{0} \\ c_{0}\end{array} \right)
$$
Note that $s_0=0$ and $c_0=1$.
It's worth pointing out that that
$$
\left( \begin{array}{r} c & s \\ -s & c\end{array} \right) =
c \left( \begin{array}{r} 1 & 0 \\ 0 & 1\end{array} \right) +
s \left( \begin{array}{r} 0 & 1 \\ -1 & 0\end{array} \right)
$$
and that
$$
\left( \begin{array}{r} 0 & 1 \\ -1 & 0 \end{array}\right)^2 =
\left( \begin{array}{r} -1 & 0 \\ 0 & -1\end{array} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove $4p-3$ is a square knowing that $n\mid p-1$ and $p\mid n^3-1$, $p$ prime I really need some help at this problem:
Let $p$ be a prime number and $n$ a natural number, $n\ge2$ such that $n \mid p-1$ and $p \mid n^3-1$. Prove that $4p-3$ is a square.
So $p \mid (n-1)(n^2+n+1)$
What if $p \mid n-1$?
Treating the cases wasn't too efficient.
I was thinking about Fermat's theorem but it didn't helped really much.
A hint would be really apreciated.
Thanks!
| Given that $p\mid(n^3-1)$, thus we have $p\mid(n-1)$ or $p\mid(n^2+n+1)$. We also have $n\mid(p-1)$. Let us make cases.
Case 1: $p\mid(n-1)$ and $n\mid(p-1)$. Since $p-1\ge 1$ and $n>1$. Thus we can conclude that $p\mid(n-1)\implies p\le n-1$ and $n\mid(p-1)\implies n\le p-1$. Thus we have $n\le p-1\le n-2\implies n\le n-2\implies 0\le-2$, which is a clear contradiction. Hence $p\not\mid(n-1)$.
Case 2: $p\mid(n^2+n+1)$ and $n\mid(p-1)$. Thus, $\exists k\in\mathbb{Z},$ such that $pk=n^2+n+1$ and $\exists q'\in\mathbb{Z}$, such that $p=nq'+1$. Now $n^2+n+1\equiv 1\pmod n$ and $p\equiv 1\pmod n\implies pk\equiv k\pmod n\implies k\equiv pk=n^2+n+1\equiv 1\pmod n\implies k\equiv 1\pmod n.$
Thus $\exists q\in\mathbb{Z}$, such that $k=nq+1$. Hence we have $$p=nq'+1 \text{ and } k=nq+1\\ \implies pk=n^2qq'+n(q+q')+1=n^2+n+1\\\implies n(1-qq')=q+q'-1.$$
Now $p=nq'+1\implies nq'=p-1$ and $p-1\ge 1$. Thus $nq'\ge 1$. Now since $n>1$, hence $q'\ge 1$. This again implies that $nq'>1\implies p>2\implies p\ge 3.$
Now $pk=n^2+n+1>3\implies pk>3.$ Now $p\ge 3\implies k\ge 1.$ Now again we have $k=nq+1$. Since, $k\ge 1\implies nq+1\ge 1\implies nq\ge 0\implies q\ge 0$ $(\because n>1).$
Thus $q\ge 0,q'\ge 1\implies qq'\ge 0.$ Also $q+q'\ge 1\implies q+q'-1\ge 0.$ This implies that $n(1-qq')=q+q'-1\ge 0\implies n(1-qq')\ge 0\implies 1-qq'\ge 0 \implies qq'\le 1.$
Therefore, we simultaneously have $qq'\ge 0$ and $qq'\le 1$, which implies that $qq'=0$ or $qq'=1$.
Let us make cases.
Case 1: $qq'=1\implies q=1$ and $q'=1$. Thus $q+q'-1=1$ and $n(1-qq')=0\implies n(1-qq')\neq q+q'-1$, which is a contradiction to the fact that $n(1-qq')=q+q'-1$. Hence $qq'\neq 1$.
Case 2: $qq'=0\implies q=0$. Thus $n=q'-1\implies q'=n+1$, which in turn implies that $p=n(n+1)+1.$ Thus $4p-3=4n(n+1)+1=4n^2+4n+1=(2n+1)^2$.
Hence after analyzing all the cases we can conclude that $4p-3$ is a perfect square and is equal to $(2n+1)^2$, and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the following Definite Integral $\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx$, where $\{.\}$ is the fractional part of $x$. Integrate: $$\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx$$
My attempt: We know that within the interval $[0,1)$, the following is true:
$$\{x+1\}=x$$
$$\{x^2+2\}=x^2$$
$$\{x^2+3\}=x^2$$
$$\{x^3+4\}=x^3$$
Therefore $$I_1=\int_{0}^{1}x^3+x^5=5/12.$$
Similarily on the interval $[-1,0)$ we can write:
$$\{x+1\}=x+1$$
$$\{x^2+2\}=x^2$$
$$\{x^2+3\}=x^2$$
$$\{x^3+4\}=x^3+1$$
Therefore, $$I_2=\int_{-1}^{0}x^3+2x^2+x^5=2/3-5/12.$$ Hence $$\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx=2/3.$$ However the answer is $-2/3.$ I think I've made a mistake in the second transformations. Regardless, please explain where am I going wrong.
| When you say $\{x\}=x+1$ for $x\in [-1,0) $ you are saying that the fractional part of $-0.3$ is $0.7$. Was that your intended meaning?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that the sequence $\frac{\sin(k)+k^2}{\root\of{1+3k^4}}$ converges using the $\epsilon$ criteria. Propose that the limit $L = \frac{1}{\root\of3}$.
Given $\epsilon \gt 0 \exists k_\epsilon$ such that for all $k \gt k_\epsilon$ we have $\lvert\frac{\sin(k)+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert \lt \epsilon$. We have that $\lvert\frac{\sin(k)+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert\leq \lvert\frac{1+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert$ as $\sin$ is bounded above by $1$.
$\lvert\frac{1+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert \lt 1+k^2 = \epsilon$.
I'm pretty sure the last inequality is wrong but I'm not certain could somebody state whether it is wrong and how I should proceed if I am wrong.
| First, noting that $|\sin(k)|\le 1$, then given $\epsilon>0$ we have
$$\left|\frac{\sin(k)}{\sqrt{1+3k^4}}\right|\le \frac{1}{k^2}<\frac{\epsilon}{2}$$
whenever, $k>K_1=\sqrt{\frac{2}{\epsilon}}$.
Next, we have
$$\left|\frac{k^2}{\sqrt{1+3k^4}}-\frac1{\sqrt3}\right|=\left|\frac{1}{\left(\sqrt3k^2+\sqrt{1+3k^4}\right)\sqrt{3}\sqrt{1+3k^4}}\right|\le \frac{1}{k^4}<\frac{\epsilon}{2}$$
whenever $k>K_2=\sqrt[4]{\frac{2}{\epsilon}}$.
Finally, using the triangle inequality, we see that given $\epsilon>0$,
$$\left|\frac{\sin(k)+k^2}{\sqrt{1+3k^4}}-\frac1{\sqrt3}\right|\le \left|\frac{\sin(k)}{\sqrt{1+3k^4}}\right|+\left|\frac{k^2}{\sqrt{1+3k^4}}-\frac1{\sqrt3}\right|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$
whenever $k>\max(K_1,K_2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculating the integral $\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$ I wanted to calculate $$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$
So I solved the indefinite integral first (by substitution):
$$\int\frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{b^2}\int\frac{d \theta}{\cos^2\theta \left(\frac{a^2}{b^2} \tan^2\theta+1 \right)} =\left[u=\frac{a}{b}\tan\theta, du=\frac{a}{b\cos^2\theta} d\theta \right ]\\=\frac{1}{b^2}\int\frac{b}{a\left(u^2+1 \right)}du=\frac{1}{ab}\int\frac{du}{u^2+1}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )+C$$
Then:
$$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan (2\pi) \right )-\frac{1}{ab} \arctan \left(\frac{a}{b}\tan 0 \right )=0$$
Which is incorrect (the answer should be $2\pi/ab$ for $a>0,b>0$).
On the one hand, the substitution is correct, as well as the indefinite integral itself (according to Wolfram it is indeed $\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )$ ), but on the other hand I can see that had I put the limits during the substitution I'd get $\int\limits_{0}^{0} \dots = 0$ because for $\theta = 0 \to u=0$ and for $\theta = 2\pi \to u=0$.
Why is there a problem and how can I get the correct answer?
Edit: Here is Wolfram's answer:
Wolfram is correct because $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$ is the area of an ellipse (defined by $x=a\cos t , y=b\sin t$), that is $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\pi ab$$
| The substitution is incorrect : the tangent is not bijective on the interval $[0,2\pi]$. First, you need to restrict yourself to an interval on which the tangent behaves better. Using the $\pi$-periodicity of the function you want to integrate, you can show that:
$$\int_0^{2 \pi} \frac{1}{a \sin^2 (\theta)+b \cos^2 (\theta)} d \theta = 2 \int_{-\pi/2}^{\pi/2} \frac{1}{a \sin^2 (\theta)+b \cos^2 (\theta)} d \theta,$$
and go from there.
Note that this is a good warning about using Wolfram (or any formal computation system) : the formula for the indefinite integral is good, but it holds only on each interval $(k\pi -\pi/2, k\pi+\pi/2)$, which the program does not tell you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Why $(x-2)^2$ cannot be divided by $(x-2)(x-3)$ I know the question may seem ridiculous. My question is:
Why can't $(x-2)^2$ be divided by $(x-2)(x-3)$?
I know the answer is obvious if we do the division by hand. However, we also know that if remainder of $\frac{P(x)}{Q(x)}$ is zero, then $Q(x)$ is a divisor of $P(x)$.
Notice that:
$$(x-2)^2 = (x-3)(x-2)Q(x) + R(x).$$ If x =2 then
$$(2-2)^2 = (2-3)(2-2)Q(x) + R(x) \to R(x) = 0,$$ so $(x-2)^2$ can be divided by $(x-2)(x-3)$.
This happens because $R(x) = ax +b$, and it isn't a single number, like $b$. Is there is another reason for it?
Thanks.
| If you divide $(x-2)^2$ with $(x-2)(x-3)$ you get $\frac{(x-2)}{(x-3)}$. The polynomial $q(x)$ divides the polynomial $p(x)$ if there exist a polynomial $g(x)$ such that $p(x) = g(x)\times q(x)$. You have that $\frac{(x-2)}{(x-3)} = \frac{1}{x-3} + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Number of polynomials of degree less than 4 satisfying 5 points Let polynomial $P(x)$ have the property that $P(1),$ $P(2),$ $P(3),$ $P(4)$ and $P(5)$ are equal to $1$, $2$, $3$, $4$, $5$ in some order. How many possibilities are there for the polynomial $P,$ given that the degree of $P$ is strictly less than $4$?
| By Lagrange interpolation, given the values of $P(1)$, $P(2)$, $P(3)$, $P(4)$, and $P(5)$, there is a unique polynomial of degree $\leq 4$ taking those values. So the question is, for which permutations of the numbers $1$ through $5$ will the resulting $P(x)$ actually have degree $<4$? To determine this, we can look at the actual explicit formula for Lagrange interpolation and see what the coefficient of $x^4$ will be in terms of our five values. That formula is
$$\begin{align*}P(x)=P(1)\frac{(x-2)(x-3)(x-4)(x-5)}{(1-2)(1-3)(1-4)(1-5)}&+P(2)\frac{(x-1)(x-3)(x-4)(x-5)}{(2-1)(2-3)(2-4)(2-5)}\\
&+P(3)\frac{(x-1)(x-2)(x-4)(x-5)}{(3-1)(3-2)(3-4)(3-5)}\\
&+P(4)\frac{(x-1)(x-2)(x-3)(x-5)}{(4-1)(4-2)(4-3)(4-5)}\\
&+P(5)\frac{(x-1)(x-2)(x-3)(x-4)}{(5-1)(5-2)(5-3)(5-4)},
\end{align*}$$
so the coefficient of $x^4$ will be $$\frac{P(1)}{24}-\frac{P(2)}{6}+\frac{P(3)}{4}-\frac{P(4)}{6}+\frac{P(5)}{24}=\frac{P(1)-4P(2)+6P(3)-4P(4)+P(5)}{24}.$$ So we get a polynomial of degree $<4$ iff $$P(1)+6P(3)+P(5)=4(P(2)+P(4)).$$
Now we just have some casework to consider. If $P(3)=1$, the RHS will be at least $4(2+3)=20$ and the LHS is at most $5+6+4=15$, so there are no solutions. Since the problem is symmetric with respect to conjugating by $x\mapsto 6-x$, there are no solutions with $P(3)= 5$ either.
If $P(3)=2$, then $P(1)+P(5)$ is divisible by $4$, which means $P(1)$ and $P(5)$ must be $1$ and $3$ (in some order) or $3$ and $5$ (in some order). The equation will hold in the second case but not the first case. This gives four different solutions, since you can swap the two values of $P(1)$ and $P(5)$, and also the two values of $P(2)$ and $P(4)$. Again, by symmetry there are four more solutions if $P(3)=4$.
Finally, suppose $P(3)=3$. Then $P(1)+P(5)$ must be $2$ mod $4$, so $P(1)$ and $P(5)$ must be $1$ and $5$ (in some order) or $2$ and $4$ (in some order). Both cases work, and each gives four solutions. Thus there are eight solutions total if $P(3)=3$.
Taking all the cases together, then, we find there are sixteen different solutions.
Some closing remarks: It's not a coincidence that the coefficients of the values of $P$ in the equation we got are binomial coefficients; you can see this by noticing that the denominator in the Lagrange interpolation term with $P(n)$ is $\pm(n-1)!(5-n)!$ (grouping the positive and negative factors together), and this generalizes if you replace $5$ by another positive integer. Still, it would be nice to have a more conceptual explanation for why we're getting binomial coefficients. It would also be nice to have a more conceptual explanation for how to count the solutions (in particular, something that would generalize if you replaced $5$ by any positive integer).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$ When doing induction should you always try to put your final answer as the "desired " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$
I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this:
Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.
For $n = 1$,
$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$
and
$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$
Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$ $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$ $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.
| First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}(k+2)(k+4)=\frac{2\cdot1^3+21\cdot1^2+67\cdot1}{6}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}(k+2)(k+4)=\frac{2n^3+21n^2+67n}{6}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}(k+2)(k+4)=$
$\color\red{\sum\limits_{k=1}^{n}(k+2)(k+4)}+(n+1+2)(n+1+4)=$
$\color\red{\frac{2n^3+21n^2+67n}{6}}+(n+1+2)(n+1+4)=$
$\frac{2(n+1)^3+21(n+1)^2+67(n+1)}{6}$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Laplace transform of a gaussian function? How do I find the Laplace transform of this Gaussian function $f(t)=\sqrt{t}e^{-\frac{1}{t}}$.
I have from a table that $L\Big[{\frac{1}{\sqrt{t}} e^{-\frac{a^2}{4t}}}\Big ]=\sqrt{\frac{\pi}{s}}e^{-a\sqrt{s}}$ for a Gaussian Normal Distribution.
If I set $a=2$ then the formula are alike apart for $\frac{1}{\sqrt{t}}$. How do I deal with that?
| The LT you seek may be written as
$$\begin{align} \int_0^{\infty} dt \, \sqrt{t} \, e^{-1/t} e^{-s t} &= 2 \int_{0}^{\infty} du \, u^2 \, e^{-(s u^2+1/u^2)} \\ &= 2 e^{2 \sqrt{s}} \int_0^{\infty} du \, u^2 \, e^{-\left (\sqrt{s} u + 1/u \right )^2} \\ \end{align}$$
Sub $v=\sqrt{s} u+1/u$, which implies two branches of the integration region:
$$u = \frac1{2 \sqrt{s}} \left (v \pm \sqrt{v^2-4 \sqrt{s}} \right ) $$
$$du = \frac1{2 \sqrt{s}} \left (1 \pm \frac{v}{\sqrt{v^2-4 \sqrt{s}}} \right ) dv$$
Thus, the LT we seek may be written as
$$e^{2 \sqrt{s}} \int_{\infty}^{2 s^{1/4}} dv \frac1{\sqrt{s}} \left (1 - \frac{v}{\sqrt{v^2-4 \sqrt{s}}} \right ) \left (\frac{v - \sqrt{v^2-4 \sqrt{s}}}{2 \sqrt{s}} \right )^2 e^{-v^2} \\ + e^{2 \sqrt{s}} \int_{2 s^{1/4}}^{\infty} dv \frac1{\sqrt{s}} \left (1 + \frac{v}{\sqrt{v^2-4 \sqrt{s}}} \right ) \left (\frac{v + \sqrt{v^2-4 \sqrt{s}}}{2 \sqrt{s}} \right )^2 e^{-v^2} $$
This looks horrendous but it simplifies considerably. Further, if we sub $v=2 s^{1/4} \cosh{w}$ and perform a little algebra, we get for the LT:
$$2 s^{-3/4} e^{-2 \sqrt{s}} \int_0^{\infty} dw \, \cosh{w} (1+4 \sinh^2{w}) e^{-4 \sqrt{s} \sinh^2{w}}$$
which again simplifies considerably to
$$2 s^{-3/4} e^{-2 \sqrt{s}} \int_0^{\infty} dp \, (1+4 p^2) e^{-4 \sqrt{s} p^2} $$
And now we are at a standard integral for which no further elaboration should be needed. The result we seek is
$$\int_0^{\infty} dt \, \sqrt{t} \, e^{-1/t} \, e^{-s t} = \frac12 \sqrt{\pi} e^{-2 \sqrt{s}} \left (s^{-3/2} +2 s^{-1} \right ) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Why doesn't this limit exist? I'm trying to compute this limit:
$\lim_{x\to +\infty}x^2-x^2\sqrt{1+\frac{(-1)^x}{x^2}}$
which, according to the results, is not defined; I initially thought about a $\infty - \infty$ form but it looks like I'm missing something...
| Note first that there's a potential issue with $(-1)^x$ being ambiguous or undefined for non-integer values of $x$. That's not a fundamental problem, however; we can just consider the question as asking about a sequence.
With that in mind, consider we can multiply by the conjugate root:
\begin{align*}
x^2 - x^2\sqrt{1 + \frac{(-1)^x}{x^2}}
&= \frac{x^4 - x^4\left(1 + \frac{(-1)^x}{x^2}\right)}{x^2 + x^2\sqrt{1 + \frac{(-1)^x}{x^2}}} \\
&= \frac{-x^2(-1)^x}{x^2 + x^2\sqrt{1 + \frac{(-1)^x}{x^2}}} \\
&= \frac{-(-1)^x}{1 + \sqrt{1 + \frac{(-1)^x}{x^2}}} \\
&= (-1)^{x+1} \frac{1}{1 + \sqrt{1+\frac{(-1)^x}{x^2}}}
\end{align*}
Notice that $\frac{1}{1 + \sqrt{1+\frac{(-1)^x}{x^2}}}$ is always positive and does not approach 0. Therefore, since the $(-1)^{x+1}$ term out front makes the whole expression alternate signs, the limit does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a real number $t$ such that $\alpha$ is represented with respect to the basis Doing some self study as this topic really stumps me, every time I see it I cringe. I believe this question requires knowledge of change of basis and using $\phi_{BD}(\alpha)$.
Question: Let $\alpha\in End(\Bbb R^3)$ be represented with respect to the canonical basis by the matrix $\begin{bmatrix}
4 & 4 & 1 \\
2 & 2 & 3 \\
2 & 2 & 3 \\
\end{bmatrix}$.
Find a real number $t$ such that $\alpha$ is represented with respect to the basis $B=\begin{Bmatrix}\begin{bmatrix} t \\-2\\0 \end{bmatrix} \begin{bmatrix} -1 \\t \\2 \\\end{bmatrix}\begin{bmatrix} 2\\2 \\t \\ \end{bmatrix}\end{Bmatrix}$ by the matrix $\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 4 \\
\end{bmatrix}$.
Can anyone explain this question? I know what the canonical basis is but im not seeing how to connect everything in this problem. Do you work backwards from the first matrix to find $\alpha$ then use alpha to find t?
I greatly appreciate someone explaining this to me.
Edited to add correct matrices: (I am leaving the original so the comments and answer given still make sense)
Let $\alpha\in End(\Bbb R^3)$ be represented with respect to the canonical basis by the matrix $\begin{bmatrix}
2 & 2 & 0 \\
1 & 1 & 2 \\
1 & 1 & 2 \\
\end{bmatrix}$.
Find a real number $t$ such that $\alpha$ is represented with respect to the basis $B=\begin{Bmatrix}\begin{bmatrix} t \\-1\\0 \end{bmatrix} \begin{bmatrix} -2 \\t \\1 \\\end{bmatrix}\begin{bmatrix} 1\\1 \\t \\ \end{bmatrix}\end{Bmatrix}$ by the matrix $\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 4 \\
\end{bmatrix}$.
| Because the matrix is diagonal, the form is very simple. If $B=\{b_1,b_2,b_3\}$, the matrix $$\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 4 \\
\end{bmatrix}$$ means that $$\tag{*} \alpha b_1=0,\ \ \alpha b_2=b_2,\ \ \alpha b_3=4b_3.$$ So
$$
\begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}
=\alpha b_1=\begin{bmatrix}
4 & 4 & 1 \\
2 & 2 & 3 \\
2 & 2 & 3 \\
\end{bmatrix}
\begin{bmatrix}
t \\
-2\\
0\\
\end{bmatrix}
=\begin{bmatrix}
4t-8 \\
2t-4 \\
2t-4 \\
\end{bmatrix}.
$$
So you need $t=2$. Now you still need to check that this $t$ also works for $b_2$ and $b_3$, which doesn't in this case.
In general, that $A $ is the matrix for the basis $b_1,b_2,b_3$ means that
$$\alpha b_1=A_{11}b_1+A_{12}b_2+A_{13}b_3, $$ and so on.
Edit: with the new data, the equations in $(*)$ now are
$$
\begin{bmatrix}
2 & 2 & 0 \\
1 & 1 & 2 \\
1 & 1 & 2 \\
\end{bmatrix}\begin{bmatrix} t \\-1\\0 \end{bmatrix}
=\begin{bmatrix} 2t-2 \\t-1\\t-1 \end{bmatrix},
$$
which forces $t=1$. Now, for the other two vectors,
$$
\begin{bmatrix}
2 & 2 & 0 \\
1 & 1 & 2 \\
1 & 1 & 2 \\
\end{bmatrix}
\begin{bmatrix}
-2 \\
1 \\
1 \\
\end{bmatrix}
=
\begin{bmatrix}
-2 \\
1 \\
1 \\
\end{bmatrix}
$$
and
$$
\begin{bmatrix}
2 & 2 & 0 \\
1 & 1 & 2 \\
1 & 1 & 2 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
=
\begin{bmatrix}
4 \\
4 \\
4 \\
\end{bmatrix}=4\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability of taking at least one ball of specific color, no reposition, order doesn't matter
A bag contains 9 balls. 4 are red, 3 are white and 2 are yellow. Three
balls are extracted in a succession, without reposition.
What is the probability of extracting at least 1 yellow ball?
I did:
*
*probability of 1 yellow ball:
(A is yellow and B is other color)
$$p(ABB) = \frac{2}{9}$$
$$p(BAB) = \frac{2}{8}$$
$$p(BBA) = \frac{2}{7}$$
*
*total = $\frac{2}{9}+\frac{2}{8}+\frac{2}{7}=\frac{191}{252}$
$$\\$$
*
*probability of getting 2 yellow balls:
$$p(AAB)=\frac{2}{9}*\frac{1}{8} = \frac{1}{36}$$
$$p(ABA)=\frac{2}{9}*\frac{1}{7} = \frac{2}{63}$$
$$p(BAA)=\frac{2}{8}*\frac{1}{7} = \frac{1}{28}$$
*total = $\frac{1}{36}+\frac{2}{63}+\frac{1}{28}=\frac{2}{21}$
So the probability of getting at least 1 yellow ball is $$\frac{2}{21}+\frac{191}{252} = \frac{215}{252} = .85%$$
But my book says the solution is about 58.3%.
What did I do wrong? How do I solve this?
| There are $\binom{7}{3}$ ways of extracting 3 balls that are not yellow, and $\binom{9}{3}$ possibile extractions. So the probability of extracting 3 non-yellow balls is
$$
p=\frac{\binom{7}{3}}{\binom{9}{3}}.
$$
Then the probability of extracting at least one yellow ball is
$$
\overline{p}=1-p=1-\frac{\binom{7}{3}}{\binom{9}{3}}=1-\frac{5}{12}=\frac{7}{12}\approx 58.3\%.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $a, b$ and $c$ from this system of linear equations? Find $a$, $b$ and $c$ from:
$$
\begin{align}
a+c &= 3 \\
b+a &= 2 \\
c+b &= -1 \\
\end{align}
$$
I tried the following way, the answer comes wrong:
$$(a+c) + (b+a) = 3+2$$
| $$a+c=3.....(1)$$
$$b+a=2.....(2)$$
$$c+b=-1.....(3)$$
Try your way and solve like this-
$$(a+c)+(b+a)=3+2$$
$$2a+c+b=5.......(4)$$
Now, subtract equation(3) from (4)
$$ \begin{array}\ 2a+c+b=5 \\\;\;\;\;\;-c-b=1\\ \hline2a\;\;\;\;\;\;\;\;\;\;\;=6
\end{array}$$
$$2a=6\;\implies a=3$$
Now, Put the value value of $a$ in equations(1) and (2)-
$$3+c=3\;\implies c=0$$
$$b+3=2\;\implies b=-1$$
I hope it'll help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx$? Suppose $g>0$, $g$ decreasing on $[a,b]$ where $0<a<b<1$. Is it true or false that
$$\displaystyle\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx ?$$
| The inequality does not hold. Counterexample: $g(x)=1$ and $0 \lt a=\frac{1}{4} \lt b=\frac{3}{4} \lt 1$ then:
$$\int_{1/4}^{1/2}\frac{dx}{(1-x)^2} = \frac{1}{1-x} \bigg|_{1/4}^{1/2} = \;\frac{2}{3} \quad \lt \quad 2\; = \frac{1}{1-x} \bigg|_{1/2}^{3/4} = \int_{1/2}^{3/4}\frac{dx}{(1-x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Arithmetic sequence common difference involving cube or square!
Evaluate $$\sum_{k = -3}^2 k^3$$
If I evaluate it using the expression above by putting the lower and upper limits it produces $-27$ for an answer, but when evaluated using the closed form expression of the geometric sequence where I put first term $a = -27$, $n = 3$ and common ratio $r = 3$, it produces the different result.
I believe my common ratio $r = 3$ is incorrect. Can anyone tell me is $r$ value wrong? If yes, then how to calculate common ratio involving the cube or square etc.
EDIT: Earlier I thought it was geometric sequence, but it is arithmetic sequence so I'm looking for d (common difference).
EDIT I: As @Brian M. Scott pointed out, it is neither an arithmetic sequence nor geometric sequence.
| \begin{align}
S
&= \sum_{k=-3}^2 k^3 \\
&= (-3)^3 + (-2)^3 + (-1)^3+ 0^3 + 1^3 + 2^3 \\
&= (-3)^3 - (2)^3 + 2^3 - 1 + 1 \\
&= -27
\end{align}
In general you could transform into a combination of sums of $j^k$, which is the topic of Faulhaber's formula.
\begin{align}
S
&= \sum_{k=-3}^2 k^3 \\
&= \sum_{j=1}^6 (j-4)^3 \\
&= \sum_{j=1}^6 (j-4)(j^2 - 8j + 16) \\
&= \sum_{j=1}^6 (j^3 - 12j^2 + 48j-64) \\
&= \sum_{j=1}^6 j^3 - 12 \sum_{j=1}^6 j^2 + 48 \sum_{j=1}^6 j- 64 \sum_{j=1}^6 1 \\
&= \left. \left( \frac{n(n+1)}{2} \right)^2\right\vert_{n=6}
-12 \left. \frac{n(n+1)(2n+1)}{6} \right\vert_{n=6}
+48 \left. \frac{n(n+1)}{2}\right\vert_{n=6} - 64 \cdot 6 \\
&= 441 - 1092 + 1008 - 384 \\
&= -27
\end{align}
which is overkill manually, but nice for a computer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Compute limit if it exist of $a_1 = 2$, $a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$ $$a_1 = 2$$
$$a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$$
I tried to use induction but it looks as if the sequence isn't monotonic.
| Let $\displaystyle\;b_n = \frac{a_n + 1}{a_n - 1}$, we have
$$
b_{n+1} = \frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n + \frac{1}{a_n} + 2}{a_n + \frac{1}{a_n} - 2} = \left(\frac{a_n + 1}{a_n - 1}\right)^2 = b_n^2$$
Together with
$\displaystyle\;b_1 = \frac{a_1-1}{a_1+1} = \frac{2+1}{2-1} = 3$, we get
$$b_n = b_1^{2^{n-1}} = 3^{2^{n-1}}
\quad\implies\quad
a_n = \frac{b_n + 1}{b_n - 1} = \frac{3^{2^{n-1}} + 1}{3^{2^{n-1}} - 1}
\to 1 \quad\text{ as }\quad n \to \infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Prove $n^2 \leq 2^n + 1$ by induction Suppose $n \in \mathbb{N}$, prove the given inequality.
So far I have this:
Show that the above is true for $n=1$.
$$ 1 \leq 3 $$
Suppose that it holds for all $n$. Show that it holds for $n+1$
$$ (n+1)^2 \leq 2^{n+1} + 1 $$
$$ \color{red}{n^2} + 2n + 1 \leq 2^n + \color{red}{2^n + 1} $$
So I have to use induction again. But here, when I try the case where $n=1$ it doesn't hold ... ?
i.e.
$$ 2n+1 \leq 2^n $$
Is not true for $n=1$. What do I do?
| You can first look for what you need.
Assume $n^2\le 2^n+1$. Then
$$
(2^{n+1}+1)-(n+1)^2=2\cdot2^n-n^2-2n=2\bigl((2^n+1)-n^2\bigr)+(n^2-2n-2)
$$
which is surely $\ge0$ so long as $n^2-2n-2\ge0$.
This happens when $n\ge1+\sqrt{3}$, so for $n\ge3$.
Now you can give the proof.
*
*Case $n=0$: $0^2\le 2^0+1$, true
*Case $n=1$: $1^2\le 2^1+1$, true
*Case $n=2$: $2^2\le 2^2+1$, true
*Case $n=3$: $3^2\le 2^3+1$, true
Suppose $n\ge3$; then $n^2-2n-2\ge0$. If $n^2\le 2^n+1$, we have
$$
(2^{n+1}+1)-(n+1)^2=2\cdot2^n-n^2-2n=2\bigl((2^n+1)-n^2\bigr)+(n^2-2n-2)\ge0
$$
and therefore $(n+1)^2\le 2^{n+1}+1$.
Another strategy. Assume $n^2\le 2^n+1$. Then
$$
(2^{n+1}+1)-(n+1)^2=2\cdot 2^n-n^2-2n=(2^n-n^2)+(2^n-2n)
$$
which is $\ge0$ so long as $2^n-2n\ge0$. Well, this can be proved by induction as well. The inequality $2n\le2^n$ holds for $n=0$ and $n=1$. If $n\ge1$ and $2n\le 2^n$, then
$$
2(n+1)=2n+2\le2^n+2\le2^n+2^n=2^{n+1}
$$
So we can give the proof.
Lemma. For all $n$, $2n\le 2^n$.
Proof. (Repeat the above)
Proposition. For all $n$, $n^2\le 2^n+1$.
Proof. The inequality holds for $n=0$. Suppose it holds for $n$; then
\begin{align}
(n+1)^2 &= n^2+2n+1 \\
&\le 2^n+2n+1 &&\text{induction hypothesis} \\
&\le 2^n+2^n+1 &&\text{by the lemma} \\
&= 2^{n+1}+1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Surface area of a cone intersecting a horizontal cylinder I'm trying to find the surface area of the cone $x^2 + y^2 = z^2$ above the $xy$-plane and under the cylinder $y^2 + z^2 = 16$. I thought of using spherical coordinates $x = \rho \sin \phi \cos \theta,\quad y = \rho \sin \phi \sin \theta,\quad z= \rho \cos \theta$. In this case $\phi=\pi/4$. I know $0\leq \theta < 2\pi$ and $\rho$ ranges from 0 but the upper bound depends on $\theta$. To solve for this upper bound I calculate
$$x^2+y^2=z^2,\quad y^2+z^2 = 16 \Rightarrow x^2 + 2y^2 = 16$$
$$\frac{1}{2}\rho^2\cos^2\theta + \rho^2\sin^2\theta = 16$$
$$\rho =\frac{4}{\sqrt{\frac{1}{2}\cos^2\theta +\sin^2\theta}}$$
Now I think I integrate
$$\int_0^{2\pi}\int_0^{\frac{4}{\sqrt{\frac{1}{2}\cos^2\theta+\sin^2\theta}}}\rho^2\sin\theta\, d\rho d\theta$$
But I'm not sure this is the right integral or even the right approach since that integral looks like hell.
| $x = \rho \cos\theta \sin\phi\\
y = \rho \sin\theta \sin\phi\\
z = \rho \cos\phi$
On the surface of the cone.
$\phi = \frac {\pi}{4}$
$x = \frac {\sqrt 2}{2}\rho \cos\theta\\
y = \frac {\sqrt 2}{2}\rho \sin\theta \\
z = \frac {\sqrt 2}2\rho$
We can drop the $\frac {\sqrt 2}2$ across terms and still have a good perametarizatinon for the surface.
Next what is dS?
$dS = \|(\frac{dx}{d\rho},\frac{dy}{d\rho},\frac{dz}{d\rho})\times(\frac{dx}{d\theta},\frac{dy}{d\theta},\frac{dz}{d\theta})\|$
$\begin{bmatrix}\cos\theta&\sin\theta&1\\
-\rho \sin\theta&\rho\cos\theta&0\end{bmatrix}$
$dS = \|(\rho\cos\theta, \rho \sin\theta, \rho)\| = 2\rho$
$\iint 2\rho \;d\rho\;d\theta$
Limits:
$y^2 + z^2 = 16\\
\rho^2\sin^2\theta + \rho^2 = 16\\
\rho = \sqrt {\frac{16}{1+\sin^2\theta}}$
$\int_{0}^{2\pi}\int_{0}^{\sqrt \frac{16}{1+\sin^2\theta}} 2\rho \;d\rho\;d\theta$
$\int_{0}^{2\pi}\rho^2|_0^{\sqrt \frac{16}{1+\sin^2\theta}} \;d\theta\\
\int_{0}^{2\pi}\frac{16}{1+\sin^2\theta} \;d\theta\\
\int_{0}^{2\pi}\frac{16\sec^2\theta}{\sec^2\theta+\tan^2\theta} \;d\theta\\
\int_{0}^{2\pi}\frac{16\sec^2\theta}{1+2\tan^2\theta} \;d\theta\\
8{\sqrt2}\tan^{-1}(\sqrt 2 \tan\theta)|_0^{2\pi}\\
16\sqrt 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the points on the curve $x^2+xy+y^2=7$ where tangent is parallel to (a) X axis (b) parallel to Y axis.
Find the points on the curve $x^2+xy+y^2=7$ where tangent is parallel
to (a) X axis (b) parallel to Y axis.
For part (a) if we differentiate w.r.t $x$ we get $y'=\large-\frac{2x+y}{x+2y}$, this when equated to zero (parallel to X axis means slope is zero) gives $x=-y/2$, this when substituted in the equation of the curve gives $\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right),\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)$
For part (b) we find $dx/dy$, equate it to zero and find that the points are $\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right),\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)$
$\underline{\text{My question}}$ is if we consider two points on the curve $(x_1, y_1),(x_2, y_2)$ such that $x_1 = x_2$, because for the line parallel to Y axis coordinates of x remain the same. Substituting these two points in the equation of the curve we get
$$x_1^2+x_1y_1+y_1^2=7$$ $$x_1^2+x_1y_2+y_2^2=7$$
solving them simultaneously we get $x_1=-(y_1+y_2)=x_2$, but this does not agree with the points that we calculated $\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right),\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)$ in part (b). $y_1+y_2=0\neq x_1 \hspace{10pt}\text{OR} \hspace{10pt} \neq x_2$
| If $x_1,x_2$ with $x_1=x_2$ are the roots of the equation $$x^2+xy+y^2-7=0$$
we need the discriminant$=0$ $$\implies y^2-4(y^2-7)=0\iff y^2=?$$ and consequently $$x_1=x_2=-\dfrac y2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Factorize expression $(4a+3)^2-a^4$ The question is to factorize $(4a+3)^2-a^4.$
The answer I ended up with was $(4a+3-a^2)(4a+3-a^2).$
I got this by difference of two squares $\cdots.$
I would like to know if this is correct.
| SIMPLE
The question was to factorize $(4a+3)^2-a^4.$
IF you use difference of two squares we will get
LET $4a+3$ be $X$ and $a^2$ be $Y$
the $X^2-Y^2$
WE GET
$(4a+3-a^2)(4a+3+a^2).$
UNDERSTOOD
Further each of the quadratic equation can be factorised
$(x-1)(x-3)(x-/7+2)(x+/7+2)$
SO THE FINAL AWSER IS HERE
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the transformation I am trying to find the transformation of $x = 2$ under $T(z) = \frac{z+1}{z-1}$
My approach was as follows,
I put $u +iv = \frac{z+1}{z-1} $ and then rationalized the RHS.
I get $u = \frac{3-y^2}{1+y^2}$ and $v = \frac{4y}{1+y^2}$
I was trying to substitute $y = tan(\theta)$ and prove its a circle.
Thanks
Anupam
| Note that if $\Re(z)=2$,
then $\forall y\in \mathbb{R}$,
$z=2+iy$
$$T(z)=\frac{2+iy+1}{2+iy-1}=\frac{3+iy}{1+iy}=\frac{(1-iy)(3+iy)}{1+y^2}=\frac{3+y^2-2iy}{1+y^2}=2+\frac{1-y^2-2iy}{1+y^2}$$
Now notice that:
$$|T(z)-2|^2=\frac {1}{(1+y^2)^2} ((1-y^2)^2+2^2y^2)=\frac{1+2y^2+y^4}{(1+y^2)^2}=1$$
Which means that the distance between $T(z)$ and the point $2$ doesn't depend on the value of $y$, so it is the circle centered at $2$ and has radius $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
I want to find a pair of integers $X$, $Y$ which satisfy $X^2 - 2Y^2=1$ such that $X > Y > 50$
Find a pair of integers $X, Y$, which satisfy $X^2 - 2Y^2 = 1$, such that $X > Y > 50$.
I have started by finding a pair of much smaller integers that work: $X(1) = 3$ and $Y(1) = 2$.
When I looked up a solution it was as follows:
$$\begin{array}
&X(2) = 3 × 3 + 4 × 2 = 17, & Y(2) = 2 × 3 + 3 × 2 = 12 \\
X(3) = 3 × 17 + 4 × 12 = 99, & Y(3) = 2 × 17 + 3 × 12 = 70
\end{array}$$
So $X = 99$ and $Y = 70$ is such a pair.
But what is the method used? I tried searching for similar questions, but didn't find an answer to this.
| Another (more naive) way of looking at this: we want all possible $k$ for which $(Y+k)^2-2Y^2=1$. Solving this for $Y$ (in terms of $k$) we get
$$
Y=k\pm\sqrt{2k^2-1}
$$
We will want to ignore the minus, since our solutions must be positive and we only want the solutions for when $\sqrt{2k^2-1}$ is an integer.
By inspection we get the first 4 solutions:
$$
\begin{array}{c|c|c|c}
k & \sqrt{2k^2-1}& Y=k+\sqrt{2k^2-1} & X=Y+k \\ \hline
1 & 1 & 2 & 3 \\
5 & 7 & 12 & 17 \\
29 & 41 & 70 & 99 \\
169 & 239 & 408 & 577 \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Method for finding sum of series $\sum_{r = 1}^{10} \binom{10}{10 - r}\binom{20}{r}$? I came across this in a problem:
$$\sum_{r = 1}^{10} \binom{10}{10 - r}\binom{20}{r}$$
Please help me solve this and give a general method to solve such sums.
| Using $$(1+x)^{20} = \binom{20}{0}+\binom{20}{1}x+\binom{20}{2}x^2+\cdots\cdots +\binom{20}{20}x^{20}.........(1)$$
Similarly $$(1+x)^{10} = \binom{10}{0}+\binom{10}{1}x^{10}+\binom{10}{2}x^{2}+\cdots \cdots \cdots+\binom{10}{10}x^{10}.....(2)$$
Now coeficients of $x^{10}$ in multiplication of $(1)$ and $(2)$
$$(1+x)^{20}\cdot (1+x)^{10} = \binom{20}{0}\cdot \binom{10}{10}+\binom{20}{1}\cdot \binom{10}{9}+\cdots \cdots +\binom{20}{10}\cdot \binom{10}{0}$$
So $$\binom{30}{10}-1 = \sum^{10}_{r=1}\binom{20}{r}\binom{10}{10-r}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximum value of a vector If a,b,c are unit vectors , then we have to know $|a-b|^2$$+|b-c|^2$$+|c-a|^2$ does not exceed .
I tried it a lot , but not able to get how to start it .
| Using $||\mathbf{a}||^2=\langle\mathbf{a},\mathbf{a}\rangle$ and the linearity properties of the inner product (if you don't want this you may just expand on components) you may prove:
$||\mathbf{a}+\mathbf{b}+\mathbf{c}||^2+||\mathbf{a}-\mathbf{b}||^2+||\mathbf{b}-\mathbf{c}||^2+||\mathbf{c}-\mathbf{a}||^2=3(||\mathbf{a}||^2+||\mathbf{b}||^2+||\mathbf{c}||^2)=9$
since you said that $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are on the unit circle.
So:
$||\mathbf{a}-\mathbf{b}||^2+||\mathbf{b}-\mathbf{c}||^2+||\mathbf{c}-\mathbf{a}||^2=9-||\mathbf{a}+\mathbf{b}+\mathbf{c}||^2\le9$
Equality is attained when $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$, that is when $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ form an equilateral triangle on the unit circle.
EDIT: to prove the last statement at OP's request:
If $\theta$ is the angle between two of the vectors (say $\mathbf{a}$ and $\mathbf{b}$) then:
$\cos\theta=\frac{\langle\mathbf{a},\mathbf{b}\rangle}{||\mathbf{a}||\cdot||\mathbf{b}||}=\langle\mathbf{a},\mathbf{b}\rangle$
since $||\mathbf{a}||=||\mathbf{b}||=1$
From $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$ we get $\mathbf{a}+\mathbf{b}=-\mathbf{c}$ so $||\mathbf{a}+\mathbf{b}||=||\mathbf{c}||=1$
Now:
$1=||\mathbf{a}+\mathbf{b}||^2=||\mathbf{a}||^2+||\mathbf{b}||^2+2\langle\mathbf{a},\mathbf{b}\rangle=2+2\langle\mathbf{a},\mathbf{b}\rangle$
$\cos\theta=\langle\mathbf{a},\mathbf{b}\rangle=-\frac{1}{2}$
so
$\theta=\frac{2\pi}{3}=120^{\circ}$
The angle between any two vectors is $120^{\circ}$, so the vectors form an equilateral triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Counting the number of ways three dice can sum to $n$ I want to count in how many ways three dice can sum to a given number without brute forcing it. In fact, I would like to do it using generating functions and without having to expand out the product.
To do this, I have thought of making a differential equation out of $y=(x+x^2+x^3+x^4+x^5+x^6)^3$ and then solving the equation through the use of power series, or getting the $n$th taylor coefficient from the equation.
What I attempted was:
$$y=(x+x^2+x^3+x^4+x^5+x^6)^3$$
$$y'=3y^{2/3}(1+2x+3x^2+4x^3+5x^4+6x^5)$$
Then, by setting $x=0$ we see $y(0)=0$ and $y'(0)=0$, which is what I want, however, when trying to obtain $y''(0)$ by differentiating both sides, I get division by $0$ on the RHS.
Am I making a mistake in the set up or is calculating the expression through the use of a differenrial equation impossible?
| Here is an alternate approach. We could at first transform $y(x)$ so that an expansion after that becomes less cumbersome. Maybe this variant is also useful for your needs.
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We also use Iverson brackets
\begin{align*}
[[P(x)]]=\begin{cases}
1&\qquad P(x) \ \text{ true}\\
0&\qquad P(x) \ \text{ false}
\end{cases}
\end{align*}
This way we can treat multiple cases in one expression.
We obtain
\begin{align*}
[x^n]y(x)&=[x^n](x+x^2+x^3+x^4+x^5+x^6)^3\\
&=[x^n]x^3(1+x+x^2+x^3+x^4+x^5)^3\\
&=[x^{n-3}][[n\geq 3]]\left(\frac{1-x^6}{1-x}\right)^3\tag{1}\\
&=[x^{n-3}][[n\geq 3]](1-3x^6+3x^{12}-x^{18})\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\tag{2}\\
&=\left([x^{n-3}][[n\geq 3]]-3[x^{n-9}][[n\geq 9]]\right.\\
&\qquad\quad\left.+3[x^{n-15}][[n\geq 15]-[x^{n-21}][[n\geq 21]\right)
\sum_{j=0}^\infty\binom{j+2}{2}x^j\tag{3}\\
&=\binom{n-1}{2}[[n\geq 3]]-3\binom{n-7}{2}[[n\geq 9]]\\
&\qquad\quad+3\binom{n-13}{2}[[n\geq 15]]-\binom{n-19}{2}[[n\geq 21]]\tag{4}
\end{align*}
Comment:
*
*In (1) we apply the coefficient of rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
and we use the formula for the finite geometric series. Since there is no contribution to the coefficient of $x^n$ if $n<3$ we respect this by using $[[n\geq 3]]$.
*In (2) we expand the binomial and we also expand $\frac{1}{(1-x)^3}$ using the binomial series expansion.
*In (3) we use the linearity of the coefficient of operator and we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q=\binom{p+q-1}{p-1}(-1)^q
\end{align*}
*In (4) we select the coefficient of $x^{n-k}, k\in\{3,9,15,21\}$.
Note: The usage of the Iverson brackets covers the general case. If we need to calculate a specific case only, the calculation becomes even more straight forward.
Example: $[x^{10}]y(x)$
We obtain
\begin{align*}
[x^{10}]y(x)&=[x^{10}](x+x^2+x^3+x^4+x^5+x^6)^3\\
&=[x^7]\left(\frac{1-x^6}{1-x}\right)^3\\\
&=[x^7](1-3x^6)\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\\
&=\left([x^7]-3[x]\right)\sum_{j=0}^\infty\binom{j+2}{2}x^j\\
&=\binom{9}{2}-3\binom{3}{2}\\
&=27
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of the value $\max\left(\frac{1}{ac}+b,\frac{1}{a}+bc,\frac{a}{b}+c\right)$ Let $a,b,c>0$ find the following minimum of the value
$$\max\left(\dfrac{1}{ac}+b,\dfrac{1}{a}+bc,\dfrac{a}{b}+c\right)$$
the book is hint:when $a=b=c=1$,then the minimum of the value is $2$
| Let $M =\max \left[ \frac{1}{ac}+b, \frac{1}{a}+bc, \frac{a}{b}+c \right]$
Case I: $c\ge 1$
\begin{align*}
M &=\max \left[ \frac{1}{a}+bc, \frac{a}{b}+c \right]
\tag{$\frac{1}{a}+bc \ge \frac{1}{ac}+b$} \\
& \ge \max \left[ 2\sqrt{\frac{bc}{a}}, 2\sqrt{\frac{ac}{b}} \right]
\tag{$AM \ge GM$} \\
& \ge \max \left[ 2\sqrt{\frac{b}{a}}, 2\sqrt{\frac{a}{b}} \right]
\tag{$c \ge 1$} \\
& \ge 2
\end{align*}
The lower bound is achievable when $a=b=c=1$
Case II: $c<1$
\begin{align*}
M &=\max \left[ \frac{1}{ac}+b, \frac{a}{b}+c \right]
\tag{$\frac{1}{ac}+b > \frac{1}{a}+bc$} \\
& \ge \max \left[ 2\sqrt{\frac{b}{ac}}, 2\sqrt{\frac{ac}{b}} \right]
\tag{$AM \ge GM$} \\
& \ge 2
\end{align*}
When $ac=b$,
\begin{align*}
\frac{b}{a} & < 1 \\
\frac{a}{b}+\frac{b}{a} &> 2 \\
\frac{1}{b}+b & \ge 2 \\
M &=\max \left[ \frac{1}{b}+b, \frac{a}{b}+\frac{b}{a} \right] \\
&> 2
\end{align*}
No values of $c\in (0,1)$ give $M=2$
The smallest possible $M$ is $2$ when $a=b=c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that the function f(x,y) is not differentiable I have the following problem:
Prove that the function:
$f(x,y)=
\ \begin{cases}
\frac{x^3-x\cdot y^2}{x^2+y^2} & (x,y)\neq (0,0) \\
\\0 & (x,y)=(0,0)
\end{cases}
\\$
is continuous on $R^2$ and has its first order partial derivatives.
everywhere on $R^2$, but $f$ is not differentiable at $(0,0)$
I know how to prove that it is continuous on $R^2$ and its partial derivatives exist at $(0,0)$ (I use limit definition of a derivative). But I do not know how to prove that this function is not differentiable.
| We have $${ f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } }{ x } } =1\\ \\ { f }_{ y }^{ \prime }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac { f\left( 0,y \right) -f\left( 0,0 \right) }{ y } } =\lim _{ y\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } }{ y } } =0\\ f\left( x,y \right) -f\left( 0,0 \right) =\frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } =x+\left( \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x \right) ={ f }_{ x }^{ \prime }\left( 0,0 \right) x+{ f }_{ y }^{ \prime }\left( 0,0 \right) y+\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$
where $\alpha \left( x,y \right) =\frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } =\frac { -2x{ y }^{ 2 } }{ \left( x^{ 2 }+y^{ 2 } \right) \sqrt { x^{ 2 }+y^{ 2 } } } $
however when $n\rightarrow \infty $ $$\alpha \left( \frac { 1 }{ n } ,\frac { 1 }{ n } \right) =\frac { -\frac { 1 }{ { n }^{ 3 } } }{ \frac { 1 }{ { n }^{ 3 } } \sqrt { 2 } } =-\frac { 1 }{ \sqrt { 2 } } $$
which shows $$\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \neq o\left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Decide whether or not a function is continuous Is the piecewise function
$$f(x)=\begin{cases}
\frac{x^2-x-6}{x-3} & x>3 \\
\frac{\sqrt{x}-3}{x-9} & x \leq 3
\end{cases}
$$
continuous?
I've done the following:
$\lim_{x\rightarrow3^+}\frac{\sqrt{x}-3}{x-9}=\frac{\sqrt{3}-3}{3^+ -9}=\frac{\sqrt{3}-3}{6}$
$\lim_{x\rightarrow3^-}\frac{\sqrt{x}-3}{x-9}=\lim_{x\rightarrow3^-}\frac{(x-3)(x+2)}{(x-3)}=\lim_{x\rightarrow3^-}(x+2)=3^- +2=5$
Since $\lim_{x\rightarrow3^+} f(x)\neq \lim_{x\rightarrow3^-} f(x)$, the function is not continuous.
Any thoughts whether this is correct?
| Let us find the value of $f(3)$ first (i.e. $x=3$).
$$f(3)=\frac{\sqrt{3}-3}{3-9}=-\frac{\sqrt{3}-3}{6}$$
Now, let us find the left-side limit as $x\to3$ (i.e. $x\lt3$).
$$\lim\limits_{x\to3^-}(f(x))=\frac{\sqrt{3}-3}{3-9}=-\frac{\sqrt{3}-3}{6}$$
The left-side limit as $x\to3$ is equal to $f(3)$. Let us find the right-side limit as $x\to3$ (i.e. $x\gt3$).
$$\lim\limits_{x\to3^+}(f(x))=\lim\limits_{x\to3^+}\frac{x^2-x-6}{x-3}=\lim\limits_{x\to3^+}\frac{(x-3)(x+2)}{(x-3)}=\lim\limits_{x\to3^+}(x+2)=5$$
The left-side and right-side limits as $x\to3$ are not equal. Thus, the ordinary limit does not exist and the function is not continuous. There exists a jump discontinuity at $x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the value of $\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$ How to find the exact value of the following limit?
$\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$
| HINT:
$$\dfrac{e^{x\cos x^2}-e^x}{x^5}=e^x\cdot\dfrac{e^{x(\cos x^2-1)}-1}{x^5}$$
$$\dfrac{e^{x(\cos x^2-1)}-1}{x^5}=\dfrac{e^{x(\cos x^2-1)}-1}{x(\cos x^2-1)}\cdot\dfrac{x(\cos x^2-1)}{x^5}$$
$$\dfrac{\cos x^2-1}{x^4}=-\left(\dfrac{\sin x^2}{x^2}\right)^2\cdot\dfrac1{(1+\cos x^2)}$$
Now use $\lim_{h\to0}\dfrac{e^h-1}h=1$ and $\lim_{y\to0}\dfrac{\sin y}y=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to disprove $8q^2+15=7m^2$ for any integral value of $q$ and $m$
Question Statement:-
Show that the quadratic equation $x^2+7x-14(q^2+1)=0$, where $q$ is an integer, has no integral roots.
My attempt at a solution:-
For the given quadratic equation having rational coefficients to have integral roots the discriminant $D$ of the equation needs to be a perfect square.
$$\therefore D=(7)^2+56(q^2+1)=56q^2+105=7(8q^2+15)$$
Now, from the expression for the discriminant we see that $(8q^2+15)$ is always odd so for the discriminant to be a perfect square, the following condition needs to be satisfied
$$8q^2+15=7(2n+1)^2,\qquad\qquad\text{where n $\in Z$}$$
So $$2(q^2+1)=7(n^2+n)$$
After this I was having trouble coming up with a way to disprove the above equality that I came up with, if you could help me out as to how to proceed from here on or maybe suggest a way that makes it easier for me to prove what the question asks.
| If $x^2+7x-14(q^2+1)=0$, where $x,q\in\mathbb Z$, then $$7\mid 7\left(2\left(q^2+1\right)-x\right)=x^2$$ so by Euclid's Lemma $7\mid x$, so $7^2\mid x^2$ and $7^2\mid 7x$, so $$7^2\mid x^2+7x=7\left(2\left(q^2+1\right)\right),$$
so $7\mid 2\left(q^2+1\right)$. Since $\gcd(7,2)=1$, by Euclid's Lemma $7\mid q^2+1$, i.e. $q^2\equiv -1\pmod{7}$, which is impossible, because $0^2\equiv 0$, $(\pm 1)^2\equiv 1$, $(\pm 2)^2\equiv 4$, $(\pm 3)\equiv 2$ mod $7$, so $-1$ isn't a quadratic residue mod $7$.
Another way here is noticing that $7$ is a prime of the form $4k+3$, $k\in\mathbb Z$, so by Quadratic Reciprocity (or a separate lemma, which is a lot easier to prove than quadratic reciprocity) $-1$ isn't a quadratic residue mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Solve the integral $ \int{\frac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx $ My attempt:
\begin{align} \int{\frac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx &\\ &=\int{\frac{x+\sqrt{x^2+5x+6}}{x+\sqrt{x^2+5x+6}}}dx+\int{\frac{-\sqrt{x^2+5x+6}-\sqrt{x^2+5x+6}}{x+\sqrt{x^2-5x+6}}}dx \\
&=x-\int{\frac{\sqrt{x^2+5x+6}}{x+\sqrt{x^2+5x+6}}}dx-\int{\frac{\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx \\
&=x-I_1-I_2
\end{align}
I know how to solve the integral $I_1$ using Euler substitution, but I don't know how to solve the integral $I_2$.
| Hint:
$\int\dfrac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}~dx$
$=\int\dfrac{x-\sqrt{(x-3)(x-2)}}{\sqrt{(x+3)(x+2)}+x}~dx$
$=\int\dfrac{\left(x-\sqrt{(x-3)(x-2)}\right)\left(\sqrt{(x+3)(x+2)}-x\right)}{\left(\sqrt{(x+3)(x+2)}+x\right)\left(\sqrt{(x+3)(x+2)}-x\right)}~dx$
$=\int\dfrac{x\sqrt{(x+3)(x+2)}}{5x+6}~dx+\int\dfrac{x\sqrt{(x-3)(x-2)}}{5x+6}~dx-\int\dfrac{x^2}{5x+6}~dx-\int\dfrac{\sqrt{(x+3)(x+2)}\sqrt{(x-3)(x-2)}}{5x+6}~dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Limit with sum of cube roots I am trying to evaluate
$$\lim \limits_{n \to \infty} \left(\sqrt[3]{{1 \over n^4}} + \sqrt[3]{{2 \over n^4}} + \sqrt[3]{{3 \over n^4}} + \dots + \sqrt[3]{{n \over n^4}} \right)$$
I simplify the expression to
$${\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \dots + \sqrt[3]{n} \over \sqrt[3]{n^4}}$$
$y_n=\sqrt[3]{n^4} \to \infty$, so we might try to use the Stolz theorem to get the limit.
$${x_{n+1}-x_n \over y_{n+1}-y_n} = {\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \sqrt[3]{4} + \dots + \sqrt[3]{n+1} - \left(\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \dots + \sqrt[3]{n}\right) \over \sqrt[3]{(n+1)^4}-\sqrt[3]{n^4}}={\sqrt[3]{n+1} \over \sqrt[3]{(n+1)^4} - \sqrt[3]{n^4}}$$
Whne I use the Stolz theorem, the limit seems to be $\infty$. Stolz works for $\infty$, so this would contradict the limit ${3 \over 4}$ obtained below by integration which I'm fairly confident is correct. Any suggestions on what I'm missing?
| Let $f(x)=x^\frac 13$.
$$S_n=\sum_{k=1}^n (\frac{k}{n^4})^{\frac 13}\;\;$$
is a Riemann sum and
$$\lim_{n\to+\infty}S_n=$$
$$\lim_{n\to+\infty}\frac{1-0}{n} \sum_{k=1}^n f(0+k\frac{1-0}{n})$$
$$=\int_0^1 x^\frac{1}{3}dx$$
$$=\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Diophantine equation $7b^2+7b+7=a^4$.
Solve in positive integers $(a,b)$ the equation
$$7b^2+7b+7=a^4.$$
As the left side is divisible by $7$ we must have $7|a^4\implies 7|a$. Let $a=7c$, then the given equation reduces to $$b^2+b+1=343c^4.$$
I'm not sure how to proceed. Any hints or solutions are welcome.
| If
$$
b^2+b+1=343c^4
$$
then $b^2+b+1-343c^4=0$
$$
b=\frac{-1-\sqrt{1-4(1-343c^4)}}{2}=\frac{-1-\sqrt{1372c^4-3}}{2}
$$
So there is an integral solution iff there exist $d$ such as $1372c^4-3=d^2$, that is
$$
1372c^4-d^2=3
$$
A solution for the last equation is $c=1$, $d=37$. Hence $b=18$ or $b=-19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$
I tried:
$$18^{n+1} = 2^{n+1} \cdot 27 \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18 \cdot 18^n - 54 \cdot 2^n = 0 \Leftrightarrow ???$$
What do I do next? Am I doing it right?
| $18^{n+1}=2^{n+1}⋅27$
$\implies (3^{2}⋅2)^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}⋅2^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}=3^{3}$
$\implies 2(n+1) = 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Using the Runge Kutta's Method to solve a 2nd derivative question Given that the motion of two bodies subject to a gravitational force of $\frac{d^2x}{dt^2}$ = $\frac{-x}{(x^2+y^2)^{3/2}}$, $x(0) = 0$, $\frac{dx}{dt}$$(0) = -0.5$ and $\frac{d^2y}{dt^2}$ = $\frac{-y}{(x^2+y^2)^{3/2}}$, $y(0) = 1$, $\frac{dy}{dt}(0) = -0.5$
It is also given that at those points given, the motion is periodic.
Find the largest and smallest $\sqrt{x^2+y^2}$(distance between the two bodies).
First thoughts: I am only experienced working with just first derivative so I'm not really sure if I am supposed to use the Runge Kutta method two times to find the original. I will also be computing later via matlab and not by hand as the computations can get extremely difficult. However, I am stuck on just where to begin. Or is creating a systems of solutions the proper way to go about this question?
EDIT: searching around the internet, I read that it might be possible to solve this using a system of equations? So I created two matrices:
//sorry I'm unaware of how to create matrices:
Matrix 1: X' = [$x_2$ $\frac{-x_1}{(x_1^2+y^2)^{3/2}}$] but this should be 1 columned and two rows.
Matrix 2: X(0) = $[0, -0.5]^T$
| There is an alternative method to my previous answer when the acceleration function is defined as a 2nd order diff. equation $$\ddot{\mathbf{Y}} = f(t,\mathbf{Y},\dot{\mathbf{Y}})$$
with the state vector $\mathbf{Y} = (x,y)$ in your case, holding the positions in the two coordinates, and $\mathbf{V} = \dot{\mathbf{Y}}= (\dot{x},\dot{y})$ holding the velocity vectors
$$ \begin{pmatrix}\ddot{x}\\ \ddot{y}\end{pmatrix} = f(t,\begin{pmatrix}x\\y\end{pmatrix},\begin{pmatrix}\dot{x}\\ \dot{y}\end{pmatrix}) = \begin{pmatrix} -\frac{x}{\sqrt{(x^2+y^2)^3}} \\ -\frac{y}{\sqrt{(x^2+y^2)^3}}\end{pmatrix} $$
Then you follow the following steps.
$$\begin{aligned}
\mathbf{C}_0 &= \mathbf{V} & \mathbf{K}_0 & = f(t, \mathbf{Y}, \mathbf{C}_0) \\
\mathbf{C}_1 &= \mathbf{V}+ \frac{h}{2} \mathbf{K}_0 & \mathbf{K}_1 & = f(t+\frac{h}{2}, \mathbf{Y} + \frac{h}{2} \mathbf{C}_0, \mathbf{C}_1)\\
\mathbf{C}_2 &= \mathbf{V}+ \frac{h}{2} \mathbf{K}_1 & \mathbf{K}_2 & = f(t+\frac{h}{2}, \mathbf{Y} + \frac{h}{2} \mathbf{C}_1, \mathbf{C}_2) \\
\mathbf{C}_3 &= \mathbf{V}+ h \mathbf{K}_2 & \mathbf{K}_3 & = f(t+h, \mathbf{Y} + h \mathbf{C}_2, \mathbf{C}_3)
\end{aligned} $$
and
$$\begin{aligned}
\Delta \mathbf{Y} & = \frac{h}{6} \left( \mathbf{C}_0 + 2 \mathbf{C}_1 + 2 \mathbf{C}_2 + \mathbf{C}_3 \right) \\
\Delta \mathbf{V} & = \frac{h}{6} \left( \mathbf{K}_0 + 2 \mathbf{K}_1 + 2 \mathbf{K}_2 + \mathbf{K}_3 \right)
\end{aligned}$$
Essentially your are running two Rk4 schemes in parallel, one for the positions and one for the velocities and you keep all cross term influences appropriately.
NOTE: The position step can be simplified to $\Delta \mathbf{Y} = h \left(\mathbf{V}+\frac{h}{6} \left( \mathbf{K}_0 + \mathbf{K}_1 + \mathbf{K}_2 \right)\right) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
maximum value of $xy+yz+zx.$ given $x+2y+z=4$ if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$
putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$
i wan,t be able to go further,could some help me with this
| $$xy+z(x+y)=xy+(4-x-2y)(x+y)=4x+4y-x^2-2y^2-2xy=k\text{(say)}$$
$$x^2+2x(y-2)+2y^2-4y-k=0$$
As $x$ is real, the discriminant $$4(y-2)^2-4(2y^2-4y-k)\ge0$$
The equality occurs if $x=-\dfrac{2(y-2)}2$
$$\iff k\le4-y^2\le4$$
The equality occurs if $y^2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
A limit without invoking L'Hopital: $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^2}$ The following limit
$$\ell=\lim_{x \rightarrow 0} \frac{x \cos x - \sin x}{x^2}$$
is a nice candidate for L'Hopital's Rule. This was given at a school before L'Hopital's Rule was covered. I wonder how we can skip the rule and use basic limits such as:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} \quad , \quad \lim_{x \rightarrow 0} \frac{\cos x -1}{x^2}$$
| We have,
$$\lim_{x \to 0} \dfrac{x\cos x - \sin x}{x^2} = \lim_{x \to 0} \dfrac{\cos x -1}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2} $$
$$ = -2\lim_{x \to 0} \dfrac{\sin^2 \left(\frac{x}{2}\right)}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2} $$
The first limit is zero since $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$, and,
$$ 0 \leq \lim_{x \to 0}\dfrac{x - \sin x}{x^2} \leq \lim_{x \to 0}\dfrac{\tan x - \sin x}{x^2}$$
But,
$$\lim_{x \to 0}\dfrac{\tan x - \sin x}{x^2} = \lim_{x \to 0} \ \left( \sin x \times \dfrac{1-\cos x}{x^2 \cos x} \right) = \lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0$$
Thus, by the Squeeze Theorem,
$$\lim_{x \to 0} \dfrac{x\cos x - \sin x}{x^2} =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$$
I have no idea on how to face this limit. Its value at $0$ seems to be $0$, but its limit equals $\frac{23}{24}$. I tried l'Hôpital but the new expressions were as confusing. I think maybe using squeeze theorem would be useful, but I don't know how to apply it in this case.
Any help would be appreciated, thanks in advance.
| Preliminary remark. I turned the previous hint into a full answer because it has been downvoted two times. It seems that the little-o notation is not so popular :-(
Use Taylor'expansion. Note that
$$\frac{x}{1-x^2}=x(1+x^2+o(x^2))=x+x^3+o(x^3),$$
and
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4).$$
Then
$$\cos\left(\frac{x}{1-x^2}\right)=
1-\frac{(x+x^3+o(x^3))^2}{2}+\frac{(x+x^3+o(x^3))^4}{4!}+o(x^4)\\
=1-\frac{x^2+2x^4}{2}+\frac{x^4}{24}+o(x^4)
=1-\frac{x^2}{2}-\frac{23x^4}{24}+o(x^4).
$$
Now it should be easy to verify that the desired limit is $\frac{23}{24}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculating $\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)$ I've managed to solve it by rewriting the expression as
$$\frac{1 - \frac{\sqrt{x^2 +x + 1}}{x e^{\frac{1}{x}}} }{ \frac{1}{x e^{\frac{1}{x}}} }$$
then applying L'Hospital's rule.
This took up one whole page and was very hairy, even after substituting $t = \sqrt{x^2+x+1}$. I'm wondering if there's a simpler way. A friend suggested substituting $e = (1+\frac{1}{x})^x$, but that's a bit suspicious.
In both cases, the answer is $\frac{1}{2}$, as confirmed by my computer.
| Multiplying by $\displaystyle\frac{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}$ we have
$$\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)=\lim_{x\to\infty}\frac{x^2 (e^{\frac{2}{x}} -1)-x-1 }{x e^{\frac{1}{x}} + \sqrt{x^2+x+1} }=\lim_{x\to\infty}\frac{x (e^{\frac{2}{x}} -1)-1-1/x }{ e^{\frac{1}{x}} + \sqrt{1+1/x+1/x^2} }.$$
Now $\displaystyle\lim_{x\to\infty}x(e^{\frac{2}{x}} -1)=\lim_{x\to\infty}\frac{e^{\frac{2}{x}} -1}{1/x}=\lim_{x\to\infty}\frac{e^{\frac{2}{x}}(-2/x^2)}{-1/x^2}=2$.
Thus the final limit is: $\frac{2-1}{1+\sqrt{1}}=1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
New Question: Find $x,y,z$ s.t. $x+y+z=1$ and $x^2+y^2+z^2=1$, and $x,y,z \in ( - 1,0) \cup (0,1)$ If in my earlier question [1], we relieve the constraints such that $x,y,z$ can also get values between $(−1,0)$, it seems that the system has a solution. Now, the question is how we can find this solution?
$$\begin{cases}&x + y + z = 1\\
&{x^2} + {y^2} + {z^2} = 1\\
&x \ne y \ne z, x \ne z\\
&x,y,z \in ( - 1,0) \cup (0,1)
\end{cases}$$
--
[1] Multivariate-Multi-objective Optimization Problem: $x+y+z = 1$ and $x^2+y^2+z^2 = 1$
| The set of solutions become a circle which intersect of next two spheres.
$$(x-1/3)^2+(y-1/3)^2+(z-1/3)^2=2/3$$
$$x^2+y^2+z^2=1$$ except $x=y,y=z,z=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
maximum value of the expression : $2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1$ I have to find out the maximum value of the expression :
$2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1.$
| Let $(x,y,z)=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) \in S^2$
\begin{align*}
2\cos \phi+3\sin \phi & \le \sqrt{2^2+3^2} \\
2x+3y &= \sin \theta (2\cos \phi+3\sin \phi) \\
& \le \sqrt{13} \, |\sin \theta| \\
2x+3y+z & \le \sqrt{13} \, |\sin \theta|+\cos \theta \\
& \le \sqrt{13+1} \\
&= \sqrt{14}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$n^{\text{th}}$ term of The Maclaurin Expansion of $\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}$? I am trying to find the coefficient of $n^{\text{th}}$ term of the Maclaurin series of $$\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}.$$
How can I find the coefficient of $n^{\text{th}}$ term of this function?
| As noted by @Daniel, it equals $1/(1-x)(1-x^2)(1-x^3)$. Another way to put that is $$\frac{(1+x+...+x^5)(1+x^2+x^4)(1+x^3)}{(1-x^6)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Note that $a = \frac{1+\sqrt{5}}{2}$ satisfies the equation $a^2 - a - 1 = 0$
Substituting $a=x+\frac{1}{x}$ gives:
$$0 = \left(x+\frac{1}{x}\right)^2 - \left(x+\frac{1}{x}\right) - 1 = x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2}$$
$$\iff \quad x^4 - x^3 + x^2 - x + 1 = 0$$
Multiplying by $x+1$ results in $x^5+1=0$, so $x$ is a complex $5^{th}$ root of $-1$ therefore $x^5 = -1$.
Then $x^{2000}+\frac{1}{x^{2000}} = \big(x^{5}\big)^{400} + \cfrac{1}{\big(x^{5}\big)^{400}} = (-1)^{400} + \cfrac{1}{(-1)^{400}} = 1 + 1 = 2$.
P.S. For a heavy-handed "solution" to the tune of "how to crack a nut with a sledgehammer", let Wolfram Alpha do all the work:
resultant[resultant[x^2 - a x + 1, a^2 - a - 1, a ], x^4000 - b x^2000 + 1, x] = (b-2)^4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
} |
What is the value of $\prod_{n=1}^\infty (1+\frac{1}{n^2})$? How would I go about finding $\prod_{n=1}^\infty (1+\frac{1}{n^2})$ ? Setting $a_k = \prod_{n=1}^k (1+\frac{1}{n^2})$, the first few partial products are $a_1 = 2$, $a_1 = 2.5$, $a_3 \approx 2.78$, $a_4 \approx 2.591$, $a_{20} \approx 3.50$, and they eventually appear to converge at around $3.639$. (These values were found with a quick python program. So much for my initial conjecture that the product converged to $e$.)
I tried writing $1+\frac{1}{n^2}$ as $\frac{n^2 + 1}{n^2}$, but it wasn't immediately obvious how to prove convergence / find the limit because the product doesn't really factor or telescope. Is there a way to calculate the limit?
| Six years too late
Similar to @Felix Marin's answer to this question but focusing on the partial product
$$P_k=\prod_{n=2}^{k} \left( 1+\frac{1}{n^2}\right)=\prod_{n=2}^{k} \left( \frac{n^2+1}{n^2}\right)=\frac{\prod_{n=2}^{k}(n+i)\,\prod_{n=2}^{k}(n-i) } {\prod_{n=2}^{k}n\,\prod_{n=2}^{k}n }$$ that is to say
$$P_k=\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{\Gamma (2-i)\, \Gamma (2+i)\, \Gamma (k+1)^2}$$
But we know by Euler's reflection formula that $\Gamma (2-i)\, \Gamma (2+i)=2 \pi \, \text{csch}(\pi )$. So
$$2 \pi \, \text{csch}(\pi ) P_k=\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{ \Gamma (k+1)^2}$$ Using Stirling approximation and continuing with Taylor series
$$\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{ \Gamma (k+1)^2}=1-\frac{1}{k}+\frac{1}{k^2}-\frac{2}{3 k^3}+\frac{1}{6
k^4}+O\left(\frac{1}{k^5}\right)$$ Taking more terms and transforming the expansion as a Padé approximant
$$\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{ \Gamma (k+1)^2}=\frac {84 k^4+126 k^3+186 k^2+86 k+27 }{84 k^4+210 k^3+312 k^2+244 k+85 }+O\left(\frac{1}{k^9}\right)$$
$$P_k\sim \frac{\sinh (\pi )}{2 \pi }\,\frac {84 k^4+126 k^3+186 k^2+86 k+27 }{84 k^4+210 k^3+312 k^2+244 k+85 }$$
Using it for $k=4$
$$\frac 5 4 \times \frac{10}{9} \times \frac{17}{16} =\frac{425}{288}=1.475694444$$ while
$$P_4=\frac{32915 \sinh (\pi )}{81994 \pi }=\color{red}{1.475694}617$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the lowest value possible of $f(x) = e^{1-0.5x}+e^{0.5x-1}$
Let $f$ be the function defined by: $$f(x) = e^{1-0.5x}+e^{0.5x-1}$$
Find the values of $k$ for which the equation $f(x) = k$ has two non
negative solutions, using the graphical calculator.
This problem could be solved very easily using the calculator, but I want to try solving solving it analitically.
If you look at the graph of this equation http://www.wolframalpha.com/input/?i=e%5E%7B1-0.5x%7D%2Be%5E%7B0.5x-1%7D you'll notice that the wanted values are in the set ]lowest value of $f(x)$; $f(0)$]
Finding $f(0)$ is easy:
$$ e^{1-0.5\cdot 0}+e^{0.5\cdot 0-1} = e + e^{-1}$$
I think there are 2 ways of solving it:
*
*Simplify this equation to the form $ax^2 +bx +c$, and so c would be the wanted value;
*Solve $e^{1-0.5\cdot x}+e^{0.5\cdot x-1} < e + e^{-1} $, or $f(x) < f(0)$
For the first option I tried:
$$e^{1-0.5\cdot x}+e^{0.5\cdot x-1} = e^{1-\frac{x}{2}}+e^{\frac{x}{2}-1} = \sqrt{e^{2-x}} + \sqrt{e^{x-2}} = \frac{e^2+e^x}{e^{\frac{x}{2}+1}} = ???$$
For the second one I tried:
$$e^{1-0.5\cdot x}+e^{0.5\cdot x-1} < e + e^{-1} \\ \Leftrightarrow \frac{e}{e^{\frac{x}{2}}} + \frac{e^{\frac{x}{2}}}{e} < e + e^{-1} \\ \Leftrightarrow \frac{1}{e^{\frac{x}{2}}} \cdot e + \frac{1}{e} \cdot e^{\frac{x}{2}} < e + e^{-1}$$
Then I did $y = e^{\frac{x}{2}}$
$$\frac{1}{y} \cdot e + y \cdot \frac{1}{e} < e + e^{-1} \Leftrightarrow ????$$
My questions:
*
*How do I solve this using method 1 (if it is possible)?
*How do I solve this using method 2?
*Can I factorize $\frac{1}{y} \cdot e + y \cdot \frac{1}{e}$? How?
Thanks
| For the second way:
\begin{equation}
\frac{1}{y} \cdot e + y \cdot \frac{1}{e} < e + e^{-1}\\\frac{1}{y} \cdot e + y \cdot \frac{1}{e} - e - e^{-1}<0\\\frac{e^2+y^2-e^2y-y}{ey}<0
\end{equation}
Now we know that the denominator is always positive, so we have to solve $e^2+y^2-e^2y-y<0$. This is a parabola thus it s easy:
\begin{equation}
e^2+y^2-e^2y-y=0\\y^2+y(-e^2-1)+e^2=0\\y=\frac{e^2+1\pm(e^2-1)}{2}
\end{equation}
hence we get $y=1$ and $y=e^2$. Thus the solution of the inequality is $1< y< e^2$ but $y=e^{x\over2}$ so $1< e^{x\over2}< e^2\longrightarrow 0<x<4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Sum of an infinite series of fractions I am taking a fun walk into Number Theory land, and I am conducting an investigation about an infinite sum of fractions. These fractions have to do with the amount of composites within an even number. All the fractions have to do with the reciprocal of a prime.
This is my sum:
$\frac{1}{3} + \frac{1}{5} - (\frac{1}{3}*\frac{1}{5}) + \frac{1}{7}-(\frac{1}{3} * \frac{1}{7} + \frac{1}{5}*\frac{1}{7}) + \frac{1}{11} - (\frac{1}{3}*\frac{1}{11} + \frac{1}{5} * \frac{1}{11} + \frac{1}{7} * \frac{1}{11}) + ... $
I don't quite know how to simplify the sequence, or even whether it is convergent or divergent.
Each term $ a_n $ in the series involves the next largest prime.
$ a_n = \frac{1}{p_n} - (\frac{1}{p_{n-1}}*\frac{1}{p_n} + \frac{1}{p_{n-2}}*\frac{1}{p_n} + \ \ ...)$
Therefore,
$\frac{1}{7}-(\frac{1}{3} * \frac{1}{7} + \frac{1}{5}*\frac{1}{7})$
is a single term, where 7 would be $p$ and 5 would be $p_{n-1}$ and 3 would be $p_{n-2}$.
EDIT:
I have been having trouble understanding some of the answers (I understand the math, just not the answer). The basis for the problem was the following: if 1/3 of all positive integers are divisible by 3, and 1/5 of all positive integers are divisible by 5, and there is an overlap here of 1/15, ($\frac{1}{3} * \frac{1}{5}$), then $\frac{1}{3} + \frac{1}{5} - \frac{1}{15}$ represents all numbers that are divisible by either 3 or 5. I expanded this to include all prime numbers in my series, and was hoping that someone would show that the series would converge to 1, allowing me to continue on my way with a new piece of knowledge.
The proofs below show that this sum goes to negative infinity--that is, the amount of composite numbers expressed as some number between 0 and 1 goes to negative infinity. How is this possible? (please provide an explanation in your answer or update it) Maybe I have made some sort of logical or conceptual mistake that needs to be addressed, or maybe there is actually a good way of interpreting this answer. Thanks!
| The partial sums can be rearranged as
$$\sum_{k=2}^n a_k = \sum_{k=2}^n \frac{1}{p_k}\left(1 - \sum_{j=2}^{n-1} \frac{1}{p_j}\right)
= \sum_{k=2}^n \frac{1}{p_k} - \sum_{2\le j < k \le n}\frac{1}{p_jp_k}\\
= \sum_{k=2}^n \frac{1}{p_k} - \frac12\left[ \left(\sum_{k=2}^n \frac{1}{p_k}\right)^2 - \left(\sum_{k=2}^n \frac{1}{p_k^2}\right)\right]
$$
It is known that following limits exists
$$\lim_{N\to\infty}\left(\sum_{p_k \le N} \frac{1}{p_k} - \log\log N\right)
= \gamma + \sum_{k=1}^\infty\left[\log\left(1 - \frac{1}{p_k}\right) + \frac{1}{p_k} \right]$$
(known as Mertens constant).
Together with Prime number theorem which essentially say
$p_n \sim n \log n$ for large $n$ and following bound:
$$\sum_{k=2}^n \frac{1}{p_k^2} \le \sum_{k=2}^n \frac{1}{k^2} < \frac{\pi^2}{6} - 1 < \infty$$
We obtain
$$\sum_{k=2}^n \frac{1}{p_k} \sim O(\log\log(n\log n))
\quad\implies\quad
\sum_{k=2}^n a_k \sim O((\log\log(n\log n))^2)$$
The series at hand diverges to $-\infty$ very slowly.
Update
About the updated question why the answer can be negative.
Let's look at number $105 = 3\times 5 \times 7$ as an example,
*
*you counted it 3 times, in $\frac13$, $\frac15$ and $\frac17$.
*you cancelled it 3 times, in $\frac13\times\frac15$, $\frac13\times\frac17$ and $\frac15\times\frac17$.
This means you are not counting $105$ at all.
In general, for any number that contain $m \ge 3$ distinct prime factors, you
over cancel it $\frac{m(m-1)}{2} - (m - 1) = \frac{(m-1)(m-2)}{2}$ times.
Even though the fraction of numbers having a specific $m$ getting smaller and smaller as $m$ increases, it does not drop fast enough to suppress this $O(m^2)$ dependence.
At the end, you over cancel too much and the result diverges to $-\infty$.
To properly deal with this sort of over counting, you need to use
Inclusion-exclusion principle.
The expression you should use should be something like:
$$\sum_{2\le i_1\le n} \frac{1}{p_{k_1}}
+ \sum_{2\le i_1<i_2 \le n} \frac{(-1)}{p_{i_1} p_{i_2}}
+ \sum_{2\le i_1<i_2<i_3 \le n} \frac{(-1)^2}{p_{i_1} p_{i_2} p_{i_3}}
+ \cdots
+ \sum_{2\le i_1<\cdots<i_{n-1}\le n} \frac{(-1)^{n-2}}{p_{i_1} \cdots p_{i_{n-1}}}
$$
This can be rewritten as
$\displaystyle\;1 - \prod_{k=2}^n\left(1 - \frac{1}{p_k}\right)\;$
and it does converge to $1$ as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
dividing a group 0f 10 men and women into couples of same sex
There're 10 men and 10 women. We need to divide them into couples of same sex.
The answer is: $$\left( \frac{10!}{5!*2^5} \right)^2$$ I don't understand why.
I know that we can divide into couples where there's exactly 1 man and 1 woman in each couple in $10!$ ways (1). Now I would need to find the total possible number of couples regardless of sex $\left( \frac{20!}{10!*2^{10}} \right)$ (2). Then (2) - (1) should give the answer. But that doesn't give the same result.
| You are right about the total possible number of couples regardless of sex. The number $\frac{20!}{10!*2^{10}}$ includes every possibility: out of the 10 couples, they can be of the same gender, or some of them can be of 1 man and 1 woman. The number $10!$ represents the case in which none of the couples are of the same sex, so you can subtract $10!$ from that to get the answer.
We first divide them into two groups of 10 persons by sex. In each of these two groups,
we have $\binom{10}{2}$ ways to choose the first couple while leaving $8 \times 2$ persons behind, then we have $\binom{8}{2}$ ways to pick another couple while leaving $6 \times 2$ persons behind. We continue this process until everyone has its partner. Therefore, we have
$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$, but we need to divide this by $5!$ since we ask for the number of ways of dividing them into couples. This gives the desired result for the group of 10 persons. Squaring this will give the required answer.
\begin{align}
\frac{1}{5!}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} &= \frac{1}{5!} \frac{10\times9}{2}\,\frac{8\times7}{2}\,\dots\frac{2\times1}{2} \\
&= \frac{10!}{5!*2^{5}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place I am trying to evaluate
$$\sum_{n=1}^{\infty}{1\over n^5}$$
up to the second decimal place. While the series is convergent, I have no idea how to construct such a bound, preferably using basic properties of series and sequences. Any hints?
| Considering the telescoping sum
$$\sum_{n=2}^{\infty} \left[ \frac{1}{n^2(n-1)^2}-\frac{1}{n^2(n+1)^2} \right]=\frac{1}{4}$$
Also $$\frac{1}{n^2(n-1)^2}-\frac{1}{n^2(n+1)^2}=\frac{4}{n^{5}}+O\left( \frac{1}{n^7} \right)$$
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^5} &=
1+\frac{1}{16}+\sum_{n=2}^{\infty}
\left[
\frac{1}{n^5}-\frac{1}{4n^2(n-1)^2}+\frac{1}{4n^2(n+1)^2}
\right] \\
&= \frac{17}{16}-\sum_{n=2}^{\infty} \frac{2n^2-1}{n^5(n^2-1)^2} \\
&\approx \frac{17}{16}-\sum_{n=2}^{\color{red}{3}} \frac{2n^2-1}{n^5(n^2-1)^2} \\
&= 1.03710 \ldots \\
&= \zeta(5)-0.000173 \ldots
\end{align*}
Error bound
\begin{align*}
E(N) &=\sum_{n=N+1}^{\infty} \frac{2n^2-1}{n^5(n^2-1)^2} \\ &<
\sum_{n=N+1}^{\infty} \frac{2}{n^3(n^2-1)^2} \\
&< \sum_{n=N+1}^{\infty} \int_{n}^{n+1} \frac{2}{x^3(x^2-1)^2} \, dx \\
&= \frac{2N^2-1}{N^2(N^2-1)}-\ln \frac{N^4}{(N^2-1)^2}
\end{align*}
$$
\begin{array}{|c|c|}
\hline
N & E(N)< \\
\hline
2 & 0.007969 \ldots \\
3 & 0.000545 \ldots \\
4 & 0.000089 \ldots \\
5 & 0.000022 \ldots \\
\hline
\end{array}
$$
That's why $N=\color{red}{3}$ is enough for $2$ decimal places.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
find the maximum of sum of roots without calculus A question asks me to evaluate the maximum and minimum value of the following function:
$$y=\sqrt{x}+\sqrt{27+x}+\sqrt{13-x}, 0\le x\le13$$
I have tried to compute y using $$x=0,1,2,...,13$$ I believe such maximum occurs at x=9, and the minimum value lies on boundary points, but how can I prove that without calculus?
Thanks!
| Shifting the OP's function $9$ to the left we get a function of the form
$$g(x)=\sqrt{x+(n+1)^2}+\sqrt{x+\bigl(n(n+1)\bigr)^2}+\sqrt{n^2-x}.$$
Now it's easy to prove without calculus simply by squaring that $\sqrt{x+a^2}\leq \frac{1}{2a}x+a$ and $\sqrt{a^2-x}\leq-\frac{1}{2a}x+a$ for positive $a$ and $-a^2< x<a^2$. Equality occurs iff $x=0$. From here
$$\sqrt{x+(n+1)^2}\leq\frac{1}{2(n+1)}x+n+1,$$
$$\sqrt{x+\bigl(n(n+1)\bigr)^2}\leq\frac{1}{2n(n+1)}x+n(n+1),$$
$$\sqrt{n^2-x}\leq-\frac{1}{2n}x+n.$$
Now adding up yields $g(x)\leq n+1+n(n+1)+n$ with equality iff $x=0$.
Hence the OP's function (with $n=2$) has its global maximum in $x=9$ of value $11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How to find $\sum_{i=1}^n\left\lfloor i\sqrt{2}\right\rfloor$ A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)). A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).
If $n = 5$ then
$$\left\lfloor1\sqrt{2}\right\rfloor+ \left\lfloor2\sqrt{2}\right\rfloor + \left\lfloor3\sqrt{2}\right\rfloor +\left\lfloor4 \sqrt{2}\right\rfloor+ \left\lfloor5\sqrt{2}\right\rfloor
= 1+2+4+5+7 = 19$$
Sequence from $1$ to $20$ is:
$S=\{1,2,4,5,7,8,9,11,12,14,15,16,18,19,21,22,24,25,26,28\}$
I want to find answer for $n = 10^{100}$.
| It is clear that, because the sequence $\{<n\sqrt{2}>\}$(the fractional part) is equidistributed over the interval $[0,1)$, we have
$$\tag{1}\sum_{n=1}^N \lfloor n\sqrt{2} \rfloor = \frac{N(N+1)\sqrt{2}}{2}-\sum_{n=1}^N <n\sqrt{2}>\label{1}$$
and for the latter sum,
$$\tag{2}\frac{1}{N}\sum_{n=1}^N <n\sqrt{2}> \to \frac{1}{2}\label{2}$$
as $N \to \infty$.
In other words, we have
$$\tag{3}\sum_{n=1}^N \lfloor n\sqrt{2} \rfloor = \frac{N(N+1)\sqrt{2}}{2}-\frac{N}{2}+o(N)\label{3}$$
as $N \to \infty$.
So , in average, we have
$$\tag{4}\frac{1}{N}\sum_{n=1}^N \lfloor n\sqrt{2} \rfloor = \frac{(N+1)\sqrt{2}}{2}-\frac{1}{2}+o(1)\label{4}$$
and in fact the remainder term is smaller than $1/2$.
So we conclude that $\frac{1}{N}\sum_{n=1}^N \lfloor n\sqrt{2} \rfloor$(which is not an integer) is very close to the nearest integer to the number $\frac{N\sqrt{2}+\sqrt{2}-1}{2}$.
One interesting thing I observed is that we in fact have more nice decay of the error term, that is,
$$\tag{5}\frac{1}{N}\sum_{n=1}^N \lfloor n\sqrt{2} \rfloor = \frac{(N+1)\sqrt{2}}{2}-\frac{1}{2}+o(\frac{1}{N}),\label{5}$$
so, return to our original problem, we come up with
$$\tag{6}\sum_{n=1}^N \lfloor n\sqrt{2} \rfloor = \frac{N(N+1)\sqrt{2}}{2}-\frac{N}{2}+o(1)\label{6}$$
and in fact the error term is again smaller than 1/2. So the sum is the nearest integer to the number
$$\tag{7}\label{7}\frac{N(N+1)\sqrt{2}}{2}-\frac{N}{2}=\frac{N(N\sqrt{2}+\sqrt{2}-1)}{2}$$.
But the proof could possibly require more nice approximation than just the equidistribution of the sequence.
(And there seems even more faster decay of the error term!!)
++Added))
What always true in the above discussion is $\eqref{3}$ or the equivalent form $\eqref{4}$. So we can exactly figure out the average value of the Beatty sequence of $\sqrt{2}$, that is, the division of $\eqref{1}$ by $N$.
However, for the exact computation of the value of the sum $\eqref{1}$, we need more precise approximation on the error term like $\eqref{5}$ or $\eqref{6}$. Unfortunately, $\eqref{5}$ is not true and so is $\eqref{7}$.
I think the best we can do is this: For any irrational $\gamma$, let $L(\gamma)=1-|1-2<\gamma>|$. Then we have
$$\left\vert \sum_{n=1}^N \lfloor n\gamma \rfloor - \left(\frac{N(N+1)\gamma}{2}-\frac{N}{2}\right) \right\vert \leq \frac{c}{L(\gamma)}$$
with $c$ a constant irrelevant to $\gamma$ and $N$($c=2$ would actually work)
This essentially asserts that the randomness of the distribution of the sequence $\{<n\gamma>\}$ in $[0,1)$ depends on how close $<\gamma>$ is to $0$ or $1$(Note that $L(\gamma)/2$ is the minimum distance from $<\gamma>$ to $0$ and to $1$. Of course this is really a naive approximation, need to be adjusted in many ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 0
} |
The limit of a function with sum of two roots. I need to find the limit of the following function:
$$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$
I derived it to the form of:
$$\lim_{x\to-\infty}\frac{(x^2+2x)^3-(x^3+x^2)^2}{\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+x^2}\right)\left((x^2+2x)^2+(x^2+2x)\sqrt[3]{(x^3+x^2)^2}+\sqrt[3]{(x^3+x^2)^4}\right)},$$
hoping that I'd be able to simplify something but still have $\frac{0}{0}$ and don't see how to do that.
| For $x<0$,
$$f(x)=-x(1+\frac{2}{x})^{\frac{1}{2}}+x(1+\frac{1}{x})^\frac{1}{3}$$
$$=-x-1+x+\frac{1}{3}+\frac{1}{x}\epsilon(x).$$
the limit is $$-\frac{2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\int y \ln y \; \mathrm{d}y$ Can you please show me how to integrate this?
$$\int y \ln y \; \mathrm{d}y$$
I attempted integration by parts but it is getting more complicated.
| Do you know integration by parts?
$$\int fg' = fg - \int f' g.$$
Let's pose $f = \log(y)$ and $g' = y$. Then:
$$f' = \frac{1}{y}, g = \frac{1}{2}y^2.$$
Joining all:
$$\int y\log(y) dy = \frac{1}{2}y^2\log(y) - \int \frac{1}{y}\frac{1}{2}y^2 dy = \\
\frac{1}{2}y^2\log(y) - \frac{1}{4}y^2 + c. $$
If $h(y) = y\log(y)$, then let's call $H(y) = \frac{1}{2}y^2\log(y) - \frac{1}{4}y^2.$
Beware of $0$!!!
Indeed, the function $y \log(y)$ is not defined for $y=0$ (what happens to logarithm when $y$ goes to $0$?). Then:
$$\int_0^1 h(y)dy = H(1) - \lim_{y \to 0} H(y).$$
The first part is simple:
$$H(1) = \frac{1}{2}1^2 \cdot \log(1) - \frac{1}{4}1^2 = -\frac{1}{4}.$$
The second part is "harder":
$$\lim_{y \to 0} H(y) = \lim_{y \to 0}\left( \frac{1}{2}y^2\log(y) - \frac{1}{4}y^2\right) = \lim_{y \to 0}\left( \frac{1}{2}y^2\log(y)\right) - 0 = \\
= 0 - 0 = 0.$$
Then:
$$\int_0^1 h(y)dy = -\frac{1}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Definition of the binomial coefficient $a_{n}=\binom{-\frac{1}{2}}{n}$ Applying the definition of the binomial coefficient I can't figure out how to simplify the following expression:
$$a_{n}=\binom{-\frac{1}{2}}{n}$$
I want to find:
$$a_{n}=(-1)^n\frac{(2n-1)!}{(n-1)!2^{2n}n!}$$
I'm stuck at the beginning.
$$a_{n}=\frac{-\frac{1}{2}!}{n!\left ( n-\frac{1}{2} \right )!}$$
Thank you so much!
| Here we use the following definition of the binomial coefficient
\begin{align*}
\binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}
\qquad\qquad\alpha\in\mathbb{C},\quad n\in \mathbb{N}\tag{1}
\end{align*}
It is also convenient to use double factorials for a more compact notation
\begin{align*}
(2n)!! &= (2n)(2n-2)\cdots 4\cdot2\qquad\qquad\qquad n\in\mathbb{N}\\
(2n-1)!! &= (2n-1)(2n-3)\cdots 3\cdot1\\
\end{align*}
We obtain
\begin{align*}
\binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)
\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\tag{2}\\
&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)
\cdots\left(-\frac{2n-1}{2}\right)}{n!}\\
&=\frac{(-1)^n}{2^n}\cdot\frac{1\cdot 3\cdots \
\left(2n-1\right)}{n!}\tag{3}\\
&=\frac{(-1)^n}{2^n}\cdot\frac{(2n-1)!!}{n!}\tag{4}\\
&=\frac{(-1)^n}{2^n}\cdot\frac{(2n)!}{(2n)!!n!}\tag{5}\\
&=\frac{(-1)^n}{2^n}\cdot\frac{(2n)!}{2^nn!n!}\tag{6}\\
&=\frac{(-1)^n(2n)!}{2^{2n}n!n!}\tag{7}\\
&=\frac{(-1)^n(2n-1)!}{2^{2n-1}(n-1)!n!}\tag{8}\\
\end{align*}
Comment:
*
*In (2) we use the definition of the binomial coefficient with $\alpha=-\frac{1}{2}$
*In (3) we factor out $-\frac{1}{2}$ from each of the $n$ factors giving $(-1)^n\frac{1}{2^n}$
*In (4) we use double factorials $(2n-1)!!$
*In (5) we apply $$(2n)!=(2n)!!(2n-1)!!$$
*In (6) we use the formula from (5) and note that
$$(2n)!!=(2n)(2n-2)\cdots4\cdot 2=2^nn!$$
*
*In (7) we collect terms.
*In (8) we devide numerator and denominotor by $2n$.
Note there is a factor $2^{2n-1}$ in the denominator of $a_n$ which gives for $n=1$
\begin{align*}
\binom{-\frac{1}{2}}{1}=-\frac{1}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \frac {a^c}{a^a}=\frac{b^b}{b^c}$
$ \frac {a^c \cdot{b^c}}{a^a} =b^b\qquad \text{(i)}$
Similarly, taking the next two terms we obtain,
$b^b=\frac{b^a \cdot c^a}{c^c}\qquad \text{(ii)}$
I tried to solve the two equations obtained to get to the desired statement but I couldn't. Is the way adopted correct or is there another way to reach the desired answer. Please help me proceed with this question
| Let $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=t\ $ implies $a=e^{(b-c)t}$, $b=e^{(c-a)t}$ & $c=e^{(a-b)t}$
so now, using values of a, b, c: $$a^ab^bc^c=e^{a(b-c)t}e^{b(c-a)t}e^{c(a-b)t}=e^{(ab-ac+bc-ab+ac-bc)t}=e^0=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What is the contrast used in calculating the $2^p$ ANOVA table? ${{}}$ According to my statistics textbook, you need the contrast of each effect and interaction in order to compute the sum of squares (for ANOVA) of that effect/interaction. Unfortunately, it never explains what contrast is.
Here are how the contrasts are calculated for each effect/interaction in a $2^3$ experiment:
\begin{align}
\text{Contrast}_A & =-\overline{X}_{1} + \overline{X}_A - \overline{X}_{B} - \overline{X}_{C} + \overline{X}_{AB} + \overline{X}_{AC} - \overline{X}_{BC} + \overline{X}_{ABC}\\[6pt]
\text{Contrast}_B & =-\overline{X}_{1} - \overline{X}_A + \overline{X}_{B} - \overline{X}_{C} + \overline{X}_{AB} - \overline{X}_{AC} + \overline{X}_{BC} + \overline{X}_{ABC}\\[6pt]
\text{Contrast}_C & =-\overline{X}_{1} - \overline{X}_A - \overline{X}_{B} + \overline{X}_{C} - \overline{X}_{AB} + \overline{X}_{AC} + \overline{X}_{BC} + \overline{X}_{ABC}\\[6pt]
\text{Contrast}_{AB} & =\overline{X}_{1} - \overline{X}_A - \overline{X}_{B} + \overline{X}_{C} + \overline{X}_{AB} - \overline{X}_{AC} - \overline{X}_{BC} + \overline{X}_{ABC}\\[6pt]
\text{Contrast}_{AC} & =\overline{X}_{1} - \overline{X}_A + \overline{X}_{B} - \overline{X}_{C} - \overline{X}_{AB} + \overline{X}_{AC} - \overline{X}_{BC} + \overline{X}_{ABC}\\[6pt]
\text{Contrast}_{BC} & =\overline{X}_{1} + \overline{X}_A - \overline{X}_{B} - \overline{X}_{C} - \overline{X}_{AB} - \overline{X}_{AC} + \overline{X}_{BC} + \overline{X}_{ABC}\\[6pt]
\text{Contrast}_{ABC} & =-\overline{X}_{1} + \overline{X}_A + \overline{X}_{B} + \overline{X}_{C} - \overline{X}_{AB} - \overline{X}_{AC} - \overline{X}_{BC} + \overline{X}_{ABC}
\end{align}
The sum of squares is calculated with $\text{SS} = \dfrac{(\#\text{replicates per group}) \cdot \text{contrast}^2}{2^3}$
So what is contrast?
| What I think is the most frequently seen definition (but I haven't checked all the books, even among those I have at hand) is this:
A contrast is a linear combination in which the sum of the coefficients is $0$, but not all coefficients are $0$.
A bit more precisely, if you multiply all the coefficients by the same scalar, it's not really a different contrast, e.g. if the coefficients are $2, -1, -1$, that's really the same contrast as $4,-2,-2$.
Contrasts among averages of different populations or among estimates of those averages assess the ways in which they differ from each other. For example $2,-1,-1$ would measure the diffence between the first population and the average of the second and third.
Your example involves a three-factor anova. Your first contrast differs significantly from $0$ if there is a significant difference between the two levels of factor $A$. That is because all of the terms corresponding to one of the two levels of factor $A$ have plus signs and those corresponding to the other level have a minus sign. The one with the subscript $AB$ differs significantly from $0$ if there is significant evidence of interaction between those two factors. Etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the integral part of the product $\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdots \frac{2016}{2015}.$
Find the integral part of the following number $$T = \dfrac{2}{1} \cdot \dfrac{4}{3} \cdot \dfrac{6}{5} \cdot \dfrac{8}{7} \cdots \dfrac{2016}{2015}.$$
We can show that $T = 2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$, since $$\int_0^{\frac{\pi}{2}} \sin^{2n+1}(x) dx = \dfrac{2}{3} \cdot \dfrac{4}{5} \cdot \dfrac{6}{7} \cdots \dfrac{2n}{2n+1},$$ but how do we calculate the integral part of $2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$?
| Although @winther and @achillehui did this the smart way and used the standard approximations I took the "battering ram" approach by first rewriting the product in the form
\begin{equation}
P=\dfrac{2^{2016}(1008!)^2}{2016!}
\end{equation}
This gave
\begin{equation}
\ln P=2016\ln(2)+\sum_{k=1}^{1008}\ln\left(\dfrac{k}{k+1008}\right)\approx4.030351009
\end{equation}
which gives $P\approx56.28066279$
Rather than write a short program to calculate the sum I used a spreadsheet but still obtained a result in agreement with approximation formulas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.