Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Apostol's method of exhaustion to find area under x^2 I'm a high school student currently going through Apostol's calculus. I'm not that familiar with proofs, but I learned Calculus in school up to partial fractions, but we focused more on problems instead of the concept/proof, so please bear with me. I'm stuck in the part where he used the method of exhaustion to prove that the area of $b^2$ is $\frac{b^3}{3}$. After some inequalities, we find that there are 3 possibilities for the area: $A>\frac{b^3}{3}$, $A<\frac{b^3}{3}$, and $A=\frac{b^3}{3}$
To proof that $A=\frac{b^3}{3}$, we can do this by contradiction. I can prove by contradiction that $A>\frac{b^3}{3}$ is not possible, namely through the following method (please correct me if this is wrong):
$$A<\frac{b^3}{3}+\frac{b^3}{n}$$ for all $n>=1$
$$A-\frac{b^3}{3}<\frac{b^3}{n}$$
Since we assume that $A>\frac{b^3}{3}$, then $A-\frac{b^3}{3}$ >0, so we can divide from both sides and multiply both sides by n
$$n<\frac{b^3}{A-\frac{b^3}{3}}$$
$$\frac{b^3}{A-\frac{b^3}{3}}>0$$
$$\frac{b^3}{A-\frac{b^3}{3}}+1>1$$
Since $\frac{b^3}{A-\frac{b^3}{3}}+1$ is more than one it could be a value of n since $n>=1$, therefore it contradicts $n<\frac{b^3}{A-\frac{b^3}{3}}$. I tried to use the same method for the other possibility, $A<\frac{b^3}{3}$, but I can't get it to work. I'm also a bit confused, for instance, which inequality should I use, from?
$$\frac{b^3}{3}-\frac{b^3}{n}<A<\frac{b^3}{3}+\frac{b^3}{n}$$
Apostol chose the latter, but I'm not sure why. I tried to use both to contradict the second inequality, but I fail to contradict it. Please help.
|
You should use the left-hand inequality
$$
\frac{b^3}3-\frac{b^3}n<A\qquad\text{for all $n\ge1$}\tag1
$$
to prove that $A\ge\frac{b^3}3$. The argument by contradiction follows the same lines as the proof you've given: If $A<\frac{b^3}3$, then you can use algebra to rearrange (1) into the statement
$$
n<\frac{b^3}{\frac{b^3}3-A}\qquad\text{for all $n\ge1$}\tag2
$$
which is an impossibility since (2) is violated for any $n$ that exceeds $\frac{b^3}{\frac{b^3}3-A}$.
Why did we choose the left-hand inequality (1) to prove that $A\ge\frac{b^3}3$? Inequality (1) is saying that $A$ is larger than $\frac{b^3}3$ minus a small positive quantity (namely $\frac{b^3}n$). As $n$ gets larger, the quantity $\frac{b^3}n$ gets smaller and smaller and the expression $\frac{b^3}3-\frac{b^3}n$ gets closer and closer to $\frac{b^3}3$ from below. But we know $A$ is larger than the expression $\frac{b^3}3-\frac{b^3}n$. Thinking this way, your intuition should be that it's impossible for $A$ to then be strictly less than $\frac{b^3}3$, i.e. you should be able to contradict the assertion that $A<\frac{b^3}3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3593829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that for every $n\in\mathbb{N}$, $n^2$ is divisible by 3 or has a form $3k+1$? I tried to do this by induction, but it doesn't make any sense: $n^2=3k$ or $n^2=3k+1$
*
*option: $(n+1)^2= 3k+2n+1$
*option: $(n+1)^2= 3k+2n+2$
Is there any other way on proving this problem?
|
For all $n$ natural number, $n = 3k$ or $n=3k+1$ or $n=3k+2$
When $n=3k$, $n^2=9k^2=3(3k^2)$, thus is divisible by 3
When $n=3k+1$, $n^2=9k^2+6k+1=3(3k^2+2k)+1$, thus is in the form of $3n+1$
When $n=3k+2$, $n^2=9k^2+12k+4=3(3k^2+4k+1)+1$, thus is in the form of $3n+1$
Thus, covering all the cases and finishing the proof
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3599291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
}
|
Show that the line $x+y=q$ intersects the ellipse $x^2-2x+2y^2=3$ at two different points if $q^2<2q+5$ What should be the proper way to answer this question?
My working is as follows, but is it the right way to answer the question, since it seems like I am only showing that $q^2<2q+5$?
$x+y=q \tag{1}$
$x^2-2x+2y^2=3 \tag{2}$
From $(1)$, $y=q-x\tag{3}$
Substitute $(3)$ into $(2)$,
$\begin{align}x^2-2x+2(q-x)^2&=3\\x^2-2x+2(q^2-2qx+x^2)&=3\\3x^2-(2+4q)x+2q^2-3&=0\end{align}$
Since the line intersects the ellipse at two different points, therefore $b^2-4ac>0$
$(-2-4q)^2-4(3)(2q^2-3)>0$
$4+16q+16q^2-24q^2+36>0$
$8q^2-16q-40<0$
$q^2-2q-5<0$
$q^2<2q+5$
Does my working answer the quesiton, or are there more accurate approaches to solve this question?
|
If $y=x-q$ intersects the ellipse $$x^2-2x-2y^2=3 \implies x^2-2x+2(x-q)^2=3$$
Then the roots of this quadratic have to be real. So we demand $B^2 >4AC$ in the simplified quadratic,
$$3x^2-(4q+2)x+2q^2-3=9 \implies q^2 <2q+5$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3601570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$ Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$
My Attempt:
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}{2}\log_2 (x^2+7))=-2$$
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+1+\log_{1}{2}(\frac {1}{2} \log_2(x^2+7))=-1$$
$$\log_\frac {3}{4} (\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {3}{4}} (\dfrac {3}{4})=-(1+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-(\log_{\frac {1}{2}} (\frac {1}{2})+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-\log_{\frac {1}{2}} (\frac {1}{4} \log_2 (x^2+7))$$
|
After your first line you may set
$$y= \log_2(x^2+7)$$
for better readability and continue to calculate with $\log_2$.
You get
$$\log_{\frac 34}(\frac 13 y) + \log_{\frac 12}\frac 12 y=-2$$
Now, using $\log_b a =\frac{\log_2 a}{\log_2 b}$ you can isolate $\log_2 y$ (I leave the intermediate steps to you):
$$\log_2 y = 2\Leftrightarrow x^2+7 = 16 \Leftrightarrow x=\pm 3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3602742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y)dy$. Find $f(x)$ $f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$. Find $f(x)$
I've tried taking $\int_{0}^{1} (xy^2 + x^2y) f(y) dy$ to be $k(x)$ since it comes out to be a function of $x$. That transforms our equation to $f(x) = x + k(x)$.
$f(y) = y + k(y) \implies (xy^2 + x^2y)f(y) = (xy^2 + x^2y)y + (xy^2 + x^2y)k(y)$
Now I tried integrating on both sides against $dy$ from $0$ to $1$ in an attempt to find $k(x)$
$\int_{0}^{1} (xy^2 + x^2y)f(y).dy = k(x) = \int_{0}^{1} ((xy^2 + x^2y)y + (xy^2 + x^2y)k(y)).dy$
But I got stuck trying to integrate the right hand side. Any solutions or ideas are appreciated.
|
Let $A=\int_0^{1} y^{2}f(y)dy$ and $B=\int_0^{1} yf(y)dy$. Then $f(x)=x+xA+x^{2}B$. Multiply by $x$ and integrate to get $B=\frac 1 3 +\frac A 3 +\frac 1 4 B$. Similarly multiply by $x^{2}$ and integrate to get $A=\frac 1 4 +\frac A 4 +\frac 1 5 B$. Solve these two equations for $A$ and $B$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3607585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Is it possible to apply L'Hospital to $ \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8}))) $? I have a question about how to calculate this limit; can I apply L'Hospital?
$$
\lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8})))
$$
Is it possible to make a change of variable such as $$ t^2 = \frac{1}{x^4}\;?$$
|
$$\lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8})))$$
$$ \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )
= \lim_{x \to\infty}\left(\dfrac{1}{e^{x^2}}+\dfrac{2}{x} \right)
= 0$$
$$ \lim_{x \to\infty}\sin^2\left(\frac{2}{x^4}\right)x^8
= \left[ 2\lim_{x \to\infty}
\dfrac{\sin\left(\frac{2}{x^4}\right)}{\frac{2}{x^4}}\right]^2
= 4$$
$$
\lim_{x \to\infty} \cos \left(\frac{3}{x^8}\right) = 1
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3608845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Question regarding floor function subtraction Let:
*
*$a,b,c$ be integers with $a \ge b$ and $c > 0$
There exists integers $s,t$ such that:
$$\left\lfloor\frac{a}{c}\right\rfloor = \frac{a - s}{c}\text{ },\text{ }\left\lfloor\frac{b}{c}\right\rfloor = \frac{b-t}{c}$$
Does it follow that:
$$\left\lfloor\frac{a}{c}\right\rfloor - \left\lfloor\frac{b}{c}\right\rfloor = \left\lfloor\frac{a-b}{c}\right\rfloor \iff s \ge t$$
Here's my thinking:
If $s \ge t$, then $c > (s-t) \ge 0$ and:
$$\left\lfloor\frac{a}{c}\right\rfloor - \left\lfloor\frac{b}{c}\right\rfloor = \frac{a - b - (s-t)}{c} = \left\lfloor\frac{a-b}{c}\right\rfloor$$
If $s < t$, then $0 > (s-t) > -c$ and:
$$\frac{a - b - (s-t)}{c} = \frac{a - b - (s-t)}{c} = \left\lfloor\frac{a-b}{c}\right\rfloor + 1$$
Did I make a mistake? If I did not make a mistake, is there a simpler argument?
|
What you've done looks correct. Here's a somewhat different way to show it, although not really much simpler, but it does include a few points you don't show explicitly. Note you can use a similar proof in the case $c \lt 0$ to confirm your proposition, but you would need to change it to use $s \le t$ instead.
You have by Euclidean division that for some unique integer $m$, there is an integer $r$ such that
$$a = mc + r, \; 0 \le r \lt c \tag{1}\label{eq1A}$$
Thus, you then have
$$m = \left\lfloor\frac{a}{c}\right\rfloor = \frac{a - s}{c} \implies a = mc + s \tag{2}\label{eq2A}$$
This means that $s$ is uniquely the value of $r$ in \eqref{eq1A}. Similarly, this means for some unique integer $n$
$$\left\lfloor\frac{b}{c}\right\rfloor = \frac{b-t}{c} = n \implies b = cn + t, \; 0 \le t \lt c \tag{3}\label{eq3A}$$
As for your proposition of
$$\left\lfloor\frac{a}{c}\right\rfloor - \left\lfloor\frac{b}{c}\right\rfloor = \left\lfloor\frac{a-b}{c}\right\rfloor \iff s \ge t \tag{4}\label{eq4A}$$
note you have
$$m - n = \left\lfloor\frac{(m-n)c + s - t}{c}\right\rfloor = m - n + \left\lfloor\frac{s - t}{c}\right\rfloor \iff \left\lfloor\frac{s - t}{c}\right\rfloor = 0 \iff s \ge t \tag{5}\label{eq5A}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3609002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Complex numbers algebra problem involving cyclic summation
Let $a_1$, $a_2$, $a_3\in \mathbb{C}$ and $|a_1|=|a_2|=|a_3|=1$.
If $\sum\frac{a_1^{2}}{a_2 a_3}=-1$, find $|a_1 + a_2 + a_3|$
What I have done till now:
First, I tried to attack the required sum directly.
Let $\alpha=|a_1 + a_2 + a_3|$ , then squaring both sides we get ,
$$\alpha^{2}=(a_1 + a_2 + a_3)\left(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)$$ since $|a_1|=|a_2|=|a_3|=1$ and $|z|^{2}= z\overline{z}$, but it did not yield much as i perceived.
Then in the given sum, $\sum\frac{a_1^{2}}{a_2 a_3}=-1$, i tried taking LCM on left side and on solving I got, $a_1^{3} + a_2^{3} + a_3^{3} = -a_1 a_2 a_3$. On manipulations, we get
$$(a_1 + a_2 + a_3)(a_1^{2} + a_2^{2} + a_3^{2} - a_1 a_2 - a_2 a_3 - a_3 a_1) = 2a_1 a_2 a_3.$$
Here I am facing a dead end. I even tried taking conjugate of $\sum\frac{a_1^{2}}{a_2 a_3}=-1$, and add the 2 equations but it does not seem to be helping much .
Please help me with this problem.
|
Note that each term in the sum has length $1$. If we add $1$ as a fourth complex number to this sum, we get $0$. In this way, we get a (possibly degenerate) quadrilateral with four sides of equal length, producing a rhombus. In particular this means that pairs of these terms (including $1$) must be negatives of each other. In other words, out of the numbers $\frac{a_1^2}{a_2 a_3}, \frac{a_2^2}{a_1 a_3}, \frac{a_3^2}{a_1 a_2}$, two must sum to $0$, and the other must be $-1$.
Without loss of generality, assume $\frac{a_1^2}{a_2 a_3} = -1 \implies a_1^2 = -a_2a_3$. Then,
$$0 = \frac{a_2^2}{a_1 a_3} + \frac{a_3^2}{a_1 a_2} = a_1(a_2^3 + a_3^3) = a_2^3 + a_3^3 = (a_2 + a_3)(a_2^2 - a_2a_3 + a_3^2).$$
Suppose $a_2 + a_3 = 0$. Then $a_1^2 = a_2^2 \implies a_1 = \pm a_2$. So, $a_1 = -a_2$ or $a_1 = -a_3$, so in either case, $|a_1 + a_2 + a_3| = 1$.
Otherwise, we have $a_2^2 - a_2a_3 + a_3^2 = 0$. Note that
$$(a_2 - a_3)^2 = a_2^2 - a_2 a_3 + a_3^2 - a_2 a_3 = a_1^2.$$
Thus $a_1 - a_2 + a_3 = 0$ or $a_1 + a_2 - a_3 = 0$. In the former case, $a_1 + a_2 + a_3 = 2a_2$, and hence is of length $2$. Similarly, in the latter case, the length is still $2$.
So, in conclusion, the only possible values of $|a_1 + a_2 + a_3|$ are $1$ or $2$.
Let's finish by proving sharpness. If $a_1 = 1$, $a_2 = 1$, and $a_3 = -1$, then the cyclic sum comes to $-1$, and $|a_1 + a_2 + a_3| = 1$. On the other hand, let $a_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$, $a_2 = \overline{a_1} = a_1^{-1}$, and $a_3 = 1$. Then,
$$\frac{a_1^2}{a_2 a_3} + \frac{a_2^2}{a_1 a_3} + \frac{a_3^2}{a_1 a_2} = \frac{a_1^2}{a_2} + \frac{a_2^2}{a_1} + \frac{1}{1} = a_1^3 + a_2^3 + 1 = -1 + -1 + 1 = -1.$$
In this case, $|a_1 + a_2 + a_3| = 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3610470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Computing the binomial sum $\sum_{0\le i
Find the sum
$$\sum_{0\le i<j\le n}j\binom ni.$$
My 1st attempt: replacing $j$ with $n-j$. So the expression turns out to be
$$S=\sum n\binom ni-\sum j\binom ni$$
So adding the original and the final expression we get simply $S=n2^{n-1}$.
My second attempt: considering 3 parts:
Part 1: $i=j$, $\sum_{i=j}i\binom ni= n2^{n-1}$
Part 2: $i<j$ and $i>j$ they are equivalent, so we get $2S$
Part 3 : taking $i\in[0..n]$ and $j=[0..n]$ which gives $\frac{n(n+1)}22^n$
Combining all the parts I get $n^22^{n-2}$.
But none of my answers match with the given answer.
|
\begin{align}
\sum_{0 \le i<j\le n} j \binom{n}{i}
&= \sum_{i=0}^n \sum_{j=i+1}^n j \binom{n}{i}\\
&= \sum_{i=0}^n \binom{n}{i} \sum_{j=i+1}^n j \\
&= \sum_{i=0}^n \binom{n}{i} \frac{(n+i+1)(n-i)}{2} \\
&= \frac{1}{2}\sum_i \binom{n}{i} (n^2+n-i(i-1)-2i) \\
&= \frac{1}{2}(n^2+n)\sum_i \binom{n}{i} -\frac{1}{2}\sum_i \binom{n}{i} i(i-1)-\sum_i \binom{n}{i} i \\
&= \frac{1}{2}(n^2+n)\sum_i \binom{n}{i} -\frac{1}{2}n(n-1)\sum_i \binom{n-2}{i-2}-n\sum_i \binom{n-1}{i-1} \\
&= \frac{1}{2}(n^2+n)2^n -\frac{1}{2}n(n-1)2^{n-2}-n 2^{n-1} \\
&= (3n^2+n)2^{n-3}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3611850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Given: $\frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y},$ prove: a) $x^x\times y^y\times z^z =1,$ b) $x\times y \times z =1.$ Given: $ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y} $
Prove: $ a)\ x^x\times y^y\times z^z =1 \\ b) \ x\times y \times z =1 $
Here is a few step I tried to solve: $$ \log_{10}z^zy^yx^x => z\log_{10}z + y\log_{10}y + x\log_{10}x $$ and find each $\log_{10}z, \log_{10}y \ and \log_{10}x$ in the form of one given the equation by equalizing the denominator:
$(x-z)(y-x)\log_{10}z = (y-x)(z-y)\log_{10}y = (y-x)(x-z)log_{10}x $
and calculating $ \frac{\log_{10}z }{\log_{10}y}, \frac{\log_{10}z }{\log_{10}x} $
Somehow I couldn't figure out the how to get $ x^x\times y^y\times z^z =1... $
Can you help me to prove the equation? Thank you.
|
Here the common base is 10, then let all tharatios b equal to $l$, then
$$\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=k \implies x=10^{k(y-z)}, y=10^{k(z-x)}, z=10^{k(x-y)}.$$ $$F=x^x. y^y. z^z=10^{k(xy-xz)}.10^{k(yz-yx)}. 10^{(zx-zy)}=10^0=1$$
$$G=x.y.z=10^{k(y-z)}.10^{k(z-x)}.10^{k(x-y)}=10^0=1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3612751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators - Part II This post is inspired by this earlier question:
On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators.
I quote:
An example of a splitting to Odd Egyptian fraction is given below:
$$\frac{1}{3}= \frac{1}{5}+\frac{1}{9}+\frac{1}{45}$$
$$\frac{1}{5}= \frac{1}{9}+\frac{1}{15}+\frac{1}{45}$$
$$\frac{1}{7}= \frac{1}{15}+\frac{1}{21}+\frac{1}{35}$$
$$\frac{1}{7}= \frac{1}{9}+\frac{1}{45}+\frac{1}{105}$$
$$\frac{1}{9}= \frac{1}{15}+\frac{1}{35}+\frac{1}{63}$$
$$\frac{1}{11}= \frac{1}{21}+\frac{1}{33}+\frac{1}{77}$$
Notice that
$$3 \mid 45$$
$$5 \mid 45$$
$$7 \mid 35$$
$$7 \mid 105$$
$$9 \mid 63$$
$$11 \mid 77$$
My question is:
If
$$\frac{1}{y}=\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}$$
where all of $y$, $x_1$, $x_2$, and $x_3$ are distinct positive integers, and $x_1 < x_2 < x_3$, does it follow that $y \mid x_3$?
MY ATTEMPT
In the accepted answer, we have the following:
A general solution is for every positive integer $\ n\ $ :
*
*If $n$ is odd , then $$\frac{1}{3n+2}+\frac{1}{6n+3}+\frac{1}{18n^2+21n+6}=\frac{1}{2n+1}$$ is a solution with odd denominators
*If $n$ is even , then $$\frac{1}{3n+3}+\frac{1}{6n+3}+\frac{1}{6n^2+9n+3}=\frac{1}{2n+1}$$ is a solution with odd denominators
So, for every odd $\ k\ge 3\ $ we can write $\ \frac 1k\ $ with $\ 3\ $ distinct fractions with odd denominators.
Notice that
$$18n^2 + 21n + 6 = 3(2n + 1)(3n + 2)$$
$$6n^2 + 9n + 3 = 3(2n + 1)(n + 1).$$
Follow-Up Questions
If the answer to my first question is YES, how can such a claim be proved? Lastly, if the answer to my first question is NO, what is/are some of the smallest counterexample(s)?
The link below is useful for further details about Egyptian fractions: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#section9.5
|
There are lots of counterexamples . . .
A counterexample with $y=2$:
$$\frac{1}{2}=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}$$
A counterexample with all odd denominators:
$$\frac{1}{9}=\frac{1}{11}+\frac{1}{63}+\frac{1}{231}$$
A counterexample such that none of $y,x_1,x_2,x_3$ is a multiple of one of the others:
$$\frac{1}{10}=\frac{1}{15}+\frac{1}{55}+\frac{1}{66}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3619843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Number of solution in cubic residue I need to find the number of non zero solutions in $\mathbf Z_p $ to the equation $x+y=1,$ where $x\in \Gamma$ and $y\in g\Gamma$ where $ g $ is primitive root modulo an odd prime $p$ and $\Gamma=\langle g^3\rangle$. I could do it when both $x$ and $y$ are in $\Gamma$.
Clearly, it is equivalent to solving $g^{3k_1}+g^{3k_2+1}=1$ for $k_1$ and $k_2$. I have tried a few ways but nothing promising . Kindly help.
|
Let's count the solutions with $x \in \Gamma$ and $y \in g^n \cdot \Gamma$ as $n$ varies; it should only depend on $n \mod 3$.
Naturally $p \equiv 1 \mod 3$, so write $p-1=3m$ with $m$ even, so $(-1)^m = 1$.
We have:
$$\frac{1}{|\Gamma|} \sum_{x \in \Gamma} x^k = \begin{cases} 1, & k \equiv 0 \mod m, \\
0, & \text{otherwise}, \end{cases}$$
$$\frac{1}{|\Gamma|} \sum_{y \in g^n \Gamma} y^k = \begin{cases} g^{nk}, & k \equiv 0 \mod m, \\
0, & \text{otherwise}. \end{cases} $$
Moreover, by Fermat's Little Theorem, $1 - (x+y-1)^{p-1} = 0$ if $x+y=1$ and $0$ otherwise. Of course, $1/|\Gamma| = 3/(p-1) \equiv -3 \mod p$.
Let $S = S_n$ denote the set of solutions. Then
$$\begin{aligned}
9|S| = & \ 9 \sum_{S} 1 \\
\equiv & \ \frac{1}{|\Gamma|^2} \sum_{S} 1 \mod p \\
= & \ \frac{1}{|\Gamma|^2} \sum_{x \in \Gamma,y \in g \Gamma} 1 - (x+y-1)^{p-1} \\
= & \ \frac{1}{|\Gamma|^2} \sum_{x \in \Gamma,y \in g \Gamma} 1 - \sum_{a+b+c=p-1}
x^a y^b (-1)^c \binom{p}{a,b,c} \\
= & \ \frac{-1}{|\Gamma|^2} \sum_{x \in \Gamma,y \in g \Gamma} \sum_{a+b+c=p-1}^{(a,b,c) \ne (0,0,3m)}
x^a y^b (-1)^c \binom{p}{a,b,c} \\
= & \ - \sum_{a+b+c=p-1}^{(a,b,c) \ne (0,0,3m)} \frac{1}{|\Gamma|^2} \sum_{x \in \Gamma,y \in g \Gamma}
x^a y^b (-1)^c \binom{p}{a,b,c} \\
\end{aligned}$$
The inner sum is zero unless $a \equiv b
\equiv c \equiv 0 \mod m$, or equivalently if
$$(a,b,c) \in (3m,0,0), (2m,m,0), (2m,0,m), (m,2m,0), (m,m,m), (m,0,2m), (0,3m,0), (0,2m,m), (0,m,2m).$$
If we collect the $6$ terms of the form $[0,m,2m]$ up to re-ordering, we get
$$\binom{3m}{0,m,2m}(1 + 1 + g^{mn} + g^{mn} + g^{2mn} + g^{2mn}).$$
If $(n,3) = 1$, then $g^{mn}$ is a non-trivial $3$rd root of unity $\omega \mod p$, so this vanishes,
since $1 + \omega + \omega^2 = 0$.
I $3|n$, then $g^{mn} = 1$, so this is
$$6 \frac{(3m)!}{m! (2m)!} \equiv 6 \mod p,$$
since $m! = 1 \cdot 2 \cdot \ldots \cdot m \equiv (-1)^m (3m)(3m-1)\ldots (2m+1)$,
so $m! (2m)! \equiv (-1)^m (3m)! \equiv (3m)! \mod p$.
If we collect the two remaining terms of the form $[0,0,3m]$ we get $-1-1=-2$. Putting this together, we get
$$9|S_n| \equiv
\begin{cases} \displaystyle{ - 2 - \frac{(3m)!}{m!^3} g^{mn}}, & (3,n) = 1, \\
\displaystyle{-8 - \frac{(3m)!}{m!^3}}, & n| 3, \end{cases} \mod p$$
As a test case, if we add up the three possibilities, we get
$$9 (|S_1| + |S_2| + |S_3|) \equiv - 12 \mod p.$$
On the other hand, the left hand side can be computed; for any $x \in \Gamma$,
there is a $y = 1- x$ which is either in $\Gamma$, $g \Gamma$, or $g^2 \Gamma$ unless $x = 1$.
So the LHS is
$$9 (|\Gamma| - 1) = 9 \cdot \frac{(p-4)}{3} = 3p - 12 \equiv -12 \mod p.$$
At this point we have computed $|S_n| \mod p$. But we can use this to compute $|S_n|$ as well,
using the Weil conjectures.That is because $9 |S_n|$ is more or less the number of points on the
curve
$$X_n:=x^3 + g^n y^3 = 1.$$
That is because each element of $|S_n|$ when $p \equiv 1 \mod 3$ gives $9$ points on this curve
(multiplying either $x$ or $y$ by $\omega$), and conversely an orbit of points gives an element of $|S_n|$
unless $xy =0$, which is at most $6$ points. So we have a formula for $9 |S_n| \mod p$, and then we have from
the Weil conjectures
$$|X_n(\F_p) - p - 1| \le 2 \sqrt{p}, \qquad 0 \le X_n(\F_p) - 9 |S_n| \le 6.$$
Equivalently, $9|S_n|$ is given modulo $p$, but then it is the integer that is this value modulo $p$ and is closest to $p$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3620169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Where Is my wrong The random variables given for the PDF distribution function:
$$
p_X(x) =
\begin{cases}
\frac{a}{7}x, & 2 <x\leq 3 \\
0, & x\notin(2,3] \\
\end{cases}
$$
$a)$ Find value of parametric $a$, and $F_X(x)$ (CDF).
$b)$ find the probability that the random variables is part of interval $(-\frac{5}{2},\frac{5}{2})\$
My attemp is: First part
$a)$ $$\int_2^3\frac{a}{7}xdx=1\Rightarrow a=\frac{14}{5}$$
Now we have:
$$
F_X(x) =
\begin{cases}
\frac{14}{35}x, & 2 <x\leq 3 \\
0, & x\notin(2,3] \\
\end{cases}
$$
Case 1: If $x<2$ we have $F_X(x)=0$
Case 2: If $2<x\leq 3$ we have: $F_X(x)=\int_2^x\frac{14}{35}udu\Rightarrow F_X(x)=\frac{7}{35}x^2-\frac{28}{35}$
Case 3: If $x>3$ we have $F_X(x)=1$
Now for the second part
$b)$ If $x<2$ for the function $F_X(x)$ we have $F_X(x)=0$ so the values of the interval remain $(\2,\frac{5}{2})\$. From here
$$P(X\in(2,\frac{5}{2}))=F(\frac{5}{2})-F(2)=\frac{7}{35}\frac{25}{4}-\frac{28}{35}-\frac{7}{35}4-\frac{28}{35}=-\frac{161}{120}????$$
i don't know where i went wrong. Please help me. Thanks
|
You're wrong at the last line by substracting $\frac{28}{35}$ twice, since you forgot to flip the sign the second time.
$$F_X(\frac{5}{2}) - F_X(2) = \frac{1}{5}\frac{25}{4} - \frac{4}{5} - \big(\frac{1}{5}4 - \frac{4}{5}\big) = \frac{1}{5}\big(\frac{25}{4} - 4\big)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3623746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
AIME I 2007 Problem $10$. Spot the overcounting?
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
My solution:
We can choose the $3$ squares from the first column in $\binom{6}{3}=20$ ways. I will fix $3$ squares from row $1,2,3$ which are in column $1$ and multiply them later by $20$. Also we will only find arrangements for the first $3$ columns since the last one is determined by them.
Case $1$: The second column has no common squares with the first column
There is $1$ way to choose the $3$ squares. Next, we can choose any $3$ squares from column $3$, therefore there are $1\cdot \binom{6}{3} =20$ ways to do this.
Case $2$: The second column has exactly $1$ square common with the first columm
There are $3$ ways to choose the common square and $3$ ways to choose $2$ squares from the remaining $3$. Next, there are 5 valid squares (since the row with common square has $2$ shadings) and we can choose $3$.Therefore there are $3\cdot 3\cdot \binom{5}{3}=90$ ways to do this.
Case $3$: The second column has $2$ common squares.
There are $3$ ways to choose the $2$ common squares, then there are $3$ ways to choose the $1$remaining square from the $3$ other squares. Next we choose $3$ squares from $4$ squares from the third column($2$ common squares hence their rows are filled). Therefore, there are $3\cdot 3 \cdot \binom{4}{3}=36$ ways to do this.
Case $4$: Second column has all $3$ squares common with the first column.
There is just $1$ way to choose the squares for the second column. Next there is just $1$ way for the third column since $3$ common squares have their rows filled. Therefore, there is just $1$ way to do this.
Adding up all the cases,
$$20+90+36+1=147$$
Finally multiplying this by $20$,
$$147×20=2940$$.
However the answer is $1860$ for the number of arrangements. I want to know where I overcounted.
Thanks
|
You took into account that some rows already have two shaded squares, but you didn’t take into account that some rows don’t have any shaded squares at all yet, and you need to shade them in both of the last two columns. So in case $2$ you can only choose $2$ of $4$ squares, with the third one fixed, and in case $3$ you can only choose $1$ of $2$ squares, with the other two fixed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3624563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Point out my fallacy in the three-of-kind problem, please How many hands of three-of-kind are there from a standard 52-card deck?
$$
13 \binom{4}{3} \binom{12}{2}\binom{4}{1}^2
$$
is the correct answer:
*
*$13$ : #ways to choose the denomination of the three-of-a-a-kind
*$\binom{4}{3}$ : #ways to choose their suits (three-of-a-a-kind);
*$\binom{12}{2}$ : #ways to choose two distinct denominations for the remaining cards
*$\binom{4}{1}^2$ : #ways to choose the suits of these latter two.
One could argue that the following is also correct
$$
52 \binom{3}{2} \binom{12}{2} \binom{4}{1}^2
$$
where
*
*$52$ : #ways to choose both denomination and suit of the first card in the three-of-a-kind
*$\binom{3}{2}$ : #ways to choose the suit of the remaining two cards (in the three-of-a-kind)
*$\binom{12}{2} \binom{4}{1}^2$ as before
But
$$
13 \binom{4}{3} < 52 \binom{3}{2}
$$
This means that there is an overcounting in the #ways to select the three-of-a-kind.
Can anyone kindly indicate me what I am overcounting? Thank you.
|
The $52 {3 \choose 2}$ counts each three of a kind three times. It counts each three of a kind once for having each of the three cards first. And in fact $52 {3 \choose 2}=3 \cdot \left(13 {4 \choose 3}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3625561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove $\lim_{t\rightarrow 0}\left[-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t\right] = \frac{4}{45}$ The author shows the following limit being taken
$\lim_{t\rightarrow 0}\left[-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t\right] = \frac{4}{45}$
I don't see how you could get anything but $\infty$...? The first term is
$\lim_{t\rightarrow 0}-t^{-4} = \infty$...
Or, finding a common denominator:
$-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t = \frac{t^5 + t^4(1+\frac{t^2}{3})\tan^{-1}t}{t^9} = \frac{3t^5 + (3t^4 + t^6)\tan^{-1}t}{3t^9}$
which doesn't illuminate anything
|
This is an approach which mirrors the use of Taylor series.
Let's observe that $$\arctan t =\int_{0}^{t}\frac{dx}{1+x^2}\tag{1}$$ and the expression under limit can be written as $$\left(1+\frac{t^2}{3}\right)\cdot\frac{1}{t^4}\left(\frac{\arctan t} {t} - \frac{1}{1+t^2/3}\right)$$ and hence the limit in question equals the limit of $$\frac{1}{t^4}\left(\frac{\arctan t} {t} - 1+\frac{t^2}{3}+1-\frac{t^2}{3}-\frac{1}{1+t^2/3}\right)$$ The above can be simplified as $$\frac{1}{t^4}\left(\frac{\arctan t} {t} - 1+\frac{t^2}{3}\right) - \frac{1}{9(1+t^2/3)}\tag{2}$$ The first term above can be written as $$\frac{1}{t^5}\int_{0}^{t}\left(\frac{1}{1+x^2}-1+x^2\right)\,dx$$ which is same as $$\frac{1}{t^5}\int_{0}^{t}\frac{x^4}{1+x^2}\,dx$$ Using substitution $x=u^{1/5}$ we can rewrite the above expression as $$\frac{1}{5t^5}\int_{0}^{t^5}\frac{du}{1+u^{2/5}}$$ and this tends to $(1/5)\cdot 1/(1+0^{2/5})=1/5$ via Fundamental Theorem of Calculus.
By equation $(2)$ it is now clear that the desired limit is $1/5-1/9=4/45$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3632114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Solve the equation $p^3x-p^2y-1=0$ where $p=\frac {dy}{dx}$ Solve the equation $p^3x-p^2y-1=0$ where $p=\frac {dy}{dx}$
My Attempt:
$$p^3x-p^2y-1=0$$
$$p^2y=p^3x-1$$
$$y=px-\frac {1}{p^2}$$
This is solvable for $y$ so differentiating both sides w.r.t $x$
$$\frac {dy}{dx} = p + x \cdot \frac {dp}{dx} - (-2) p^{-3} \cdot \frac {dp}{dx}$$
$$p=p+x\cdot \frac {dp}{dx} + \frac {2}{p^3} \cdot \frac {dp}{dx}$$
$$x\cdot \frac {dp}{dx} + \frac {2}{p^3} \cdot \frac {dp}{dx}=0$$
$$(x+\frac {2}{p^3}) \cdot \frac {dp}{dx}=0$$
How do I solve further?
|
The initial equation is
$$y'x-y-\frac1{y'^2}=0$$
and by differentiating,
$$y''x+2\frac{y''}{y'^3}=0$$
We split in two cases:
$$y''=0\to y=ax+b$$
and
$$y'=-\sqrt[3]{\frac2x}\to y=-\frac32\sqrt[3]{2x^2}+c.$$
Plugging back in the equation
$$p=a\to a^3x-a^2(ax+b)-1=0$$ implies $b=-\dfrac1{a^2}$, and
$$p=-\sqrt[3]{\frac2x}\to-\frac2xx+\sqrt[3]{\frac4{x^2}}\left(\frac32\sqrt[3]{2x^2}+c\right)-1=0,$$
requires $c=0.$
Finally
$$y=ax-\frac1{a^2}$$ or $$y=-\frac32\sqrt[3]{2x^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3632757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Understanding why a binary operator is associative. ( On a property of "fractional part of a sum" operation) Define the binary operation which gives the fractional part of a real number $x$ like so:
$$x +_1 y = \{x + y\}$$
where
$$\{x\}=x-\lfloor x \rfloor$$
Now we want to show that this operator is associative on the interval $[0, 1)$.
The proof I found in my book is like this:
We make use of the fact that $\lfloor x \rfloor$ is an integer:
\begin{align*}
x +_1 y &= \{x + y\} \\
&= x + y - \lfloor x + y \rfloor
\end{align*}
So $x +_1 y$ is equal to $x + y$ minus some integer.
Then for $x, y, z \in [0, 1)$:
\begin{align*}
(x +_1 y) +_1 z &= (x + y - p) +_1 z \\
&= (x + y - p) + z - q \\
&= (x + y + z) - (p + q)
\end{align*}
And:
\begin{align*}
x +_1 (y +_1 z) &= x +_1 (y + z - r) \\
&= x + (y + z - r) - s \\
&= (x + y + z) - (r + s)
\end{align*}
Where $p, q, r, s \in \mathbb{Z}$.
Since $(x +_1 y) +_1 z$ and $x +_1 (y +_1 z)$ both lie in the same interval,
the integers $(p + q)$ and $(r + s)$ must be equal and hence the following holds:
\begin{equation*}
(x +_1 y) +_1 z = x +_1 (y +_1 z)
\end{equation*}
I don't quite follow the bolded deduction.
*
*They say that both "$(x +_1 y) +_1 z$ and $x +_1 (y +_1 z)$ lie in the same interval", ok, that makes sense, because of the definition of the $+_1$ operator.
*But why does this imply that $(p + q)$ and $(r + s)$ must be equal?
|
We have that $(x +_1 y) +_1 z = (x+y+z) - (p+q)$ and that $x +_1 (y +_1 z) = (x+y+z) - (r+s)$. If we take the floor of both of these expressions both of the left hand sides will be equal to zero so we get.
$$0 = \left \lfloor (x+y+z) - (p+q) \right \rfloor = \left \lfloor (x+y+z) - (r+s) \right \rfloor$$
Because $p+q$ and $r+s$ are integers, they can be taken out of the floor function to give
$$0 = \left \lfloor x+y+z \right \rfloor - (p+q) = \left \lfloor x+y+z \right \rfloor - (r+s).$$
Ignoring the zero at the start and just focusing on the second part of the equality we can subtract $\left \lfloor x+y+z \right \rfloor$. This gives us
$$ -(p+q) = -(r+s) $$
from which the result easily follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3634684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Show that there is a $C$ such that $\frac{1}{n^{1+\alpha}} \leq C(\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha})$
1.(a) Show that if $\alpha>0$, then there is a constant $C$ such that for any $n\in \mathbb N$,
$$\frac{1}{n^{1+\alpha}} \leq C(\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha})$$
[Suggestion: Write
$$\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha}=\frac{1}{n^\alpha}[\frac{1-\frac{1}{(1+\frac{1}{n})^\alpha}}{\frac{1}{n}}],$$
and estimate the expression in square brackets as converging to a derivative as $n\to\infty$]
In this equation, I am wondering if I set $C$ to be infinity then I can confirm that there is surely a $C$ that exist? Or if not, can anyone tell me how should I interpret this $C$?
|
We choose $\displaystyle C=1+\dfrac{1}{\alpha}$ if $\alpha\ge 1$ and $C=1+\dfrac{2}{\alpha(\alpha+1)}$ otherwise. The inequality $\displaystyle \dfrac{1}{n^{1+\alpha}}\le C\left(\dfrac{1}{n^\alpha}-\dfrac{1}{(n+1)^\alpha}\right)$
is equivalent to $$\left(1+\dfrac{1}{n}\right)^\alpha\ge\dfrac{C}{C-\frac{1}{n}}=1+\dfrac{1}{Cn-1}$$
and the left-handed side above, when $\alpha\ge 1$, is $\ge 1+\dfrac{\alpha}{n}$, thus, it suffices to show that $$1+\dfrac{\alpha}{n}\ge 1+\dfrac{1}{Cn-1}\Longleftrightarrow n\left(C-\dfrac{1}{\alpha}\right)\ge 1$$ which is true for $C$ we had chosen.
For the case $0< \alpha <1$ the left-handed side of the above inequality, according to Taylor expansion, is $1+\dfrac{\alpha}{n}+\dfrac{\alpha(\alpha-1)}{2}\cdot\dfrac{1}{n^2}+\left(\left(\dfrac{\alpha(\alpha-1)(\alpha-2)}{3!}\cdot\dfrac{1}{n^3}\right)\left(\dfrac{4n+(\alpha-3)}{4n}\right)\dots\right)$ which in the last bracket is greater than or equal to zero each term, for $\alpha$ in this range.
We suffice to prove that $\displaystyle \dfrac{\alpha}{n}\left(1-\dfrac{1-\alpha}{2n}\right)=\dfrac{\alpha}{n}\left(1+\dfrac{\alpha-1}{2n}\right)\ge \dfrac{1}{Cn-1}$ for the chosen $C$
we have $\displaystyle 1-\dfrac{1-\alpha}{2n}\ge\dfrac{1+\alpha}{2}$ and plug it in, the inequality follows $\Box$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3637559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).
I’ve tried put $x=0,1$ and got \begin{align*}
f(0)+2f(2)&=0\\
f(2)+2f(0)&=-6
\end{align*}
which gives me $f(0)=-4$, $f(2)=2$.
Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.
But the above doesn’t seem to help for other values.
Thank you very much for helping.
|
Hint.
As $x^2-3x+2 = (x-2)^2+(x-2)$ calling $F(x) = f(x^2+x)$ we have
$$
F(x)+2F(x-2)=3x(3x-5)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3644316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
}
|
Find $y_1(x)$ and $y_2(x)$ from the recurrence relation of $r_2$ only, by using the Frobenius method The given original equation is:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
The series I have is:
$$x^r\left[\left(r^2-r+r+x^2-\frac{1}{4}\right)C_0+\left(r^2+r+1+r+x^2-\frac{1}{4} \right)C_1x\right]+\sum^\infty_{n=2}\left[(n+r)(n+r-1)+(n+r)+\frac{3}{4}\right]C_nx^n$$
The $r$ values I have are $$r_1=\frac{1}{2}, \space r_2=-\frac{1}{2}$$ Meaning that this problem is Case $2$, when $r_1-r_2$= a positive integer. The general formula for $y_1(x)$ and $y_2(x)$ in Case $2$ are:
$$y_1(x)=\sum^\infty_{n=0}C_nx^{n+r_1}$$
$$y_2(x)=Cy_1(x)\ln(x)+\sum^\infty_{n=0}b_nx^{n+r_2}$$
So when $r=-\frac{1}{2}$ the series should be
$$x^{-\frac{1}{2}}\left[(r^2+x^2-\frac{1}{4})C_0+(r^2+2r+x^2+\frac{3}{4})C_1x \right]+x\sum^\infty_{k=2}\left[((k+r)(k+r-1)+(k+r)+\frac{3}{4})C_kx^{n}\right]$$
The amount of terms and rigorous syntax required for these kinds of problems leads me to believe that I couldn't have gone this far without making a mistake, nor can I proceed as my professor is unreachable for help at the moment. My question is am I right so far and where do I go from here?
|
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
1) This is Bessel's equation of order $\dfrac 12$
2) Indicial equation:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
Substitute $y=x^a$
$$a(a-1)x^a+ax^a+x^{a+2}-\frac{1}{4})x^a=0$$
Take the coefficient of the lowest power of $x$:
$$P(a)=a^2-\dfrac 14$$
The indicial equation:
$$(a-\frac 12)(a+\frac 12)=0 \implies S_a=\{\frac 12,-\frac 12\}$$
So you have done this correctly.
3) Case 2: as you noted for the roots of the indicial polynomial. The formula for the second solution you have to use is correct too.
4) Recurrence formula:
$$ x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
$$(n+r)(n+r-1)a_n+(n+r)a_n-\dfrac 14a_n+a_{n-2}=0$$
$$(n+r)^2a_n-\dfrac 14a_n =-a_{n-2}$$
$$a_n =-\dfrac {a_{n-2}}{(n+r)^2-\frac 14}$$
For $r=\frac 12$ the recurrence formula becomes:
$$a_n =-\dfrac {a_{n-2}}{n(n+1)}$$
Calculate some coefficients and deduce the pattern:
$$a_{2n}= \dfrac {(-1)^na_0}{(2n+1)!}$$
Therefore one of the solution to the DE is:
$$y_1(x)=\sqrt x\sum_{n=0}^\infty \dfrac {a_0(-1)^n x^{2n}}{(2n+1)!}=\dfrac {a_0 } {\sqrt x}\sum_{n=0}^\infty \dfrac {(-1)^n x^{2n+1}}{(2n+1)!}$$
$$ \boxed {y_1(x)=\dfrac {a_0} {\sqrt x}\sin x }$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3646395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evaluate it directly. And the LHS equals to
\begin{align}
\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}
& = \frac{q}{1+q^2}+\frac{q^2}{1+q^4}\cdot\frac{q^3}{q^3}+\frac{q^3}{1+q^6}\cdot\frac{q}{q} \\
& = \frac{q}{1+q^2}+\frac{q^5}{1+q^3}+\frac{q^4}{1+q} \\
& = q\cdot\frac{(1+q)(1+q^3)+q^4(1+q)(1+q^2)+q^3(1+q^2)(1+q^3)}{(1+q)(1+q^2)(1+q^3)} \\
& = q\cdot\frac{1+q+q^3+q^4+q^4+q^5+q^6+1+q^3+q^5+q^6+q}{(1+q)(1+q^2)(1+q^3)} \\
& = \frac{-2q^3}{(1+q)(1+q^2)(1+q^3)} \\
\end{align}
And
$$(x-q)(x-q^2)(x-q^3)(x-q^4)(x-q^5)(x-q^6)=x^6+x^5+x^4+x^3+x^2+x+1$$
Let $x=-1$ I get that
$$(1+q)(1+q^2)(1+q^3)(1+q^4)(1+q^5)(1+q^6)=1$$
and
$$(1+q)(1+q^2)(1+q^3)\cdot q^4(q^3+1)\cdot q^5(q^2+1)\cdot q^6(q+1)=1$$
therefore
$$\left[(1+q)(1+q^2)(1+q^3)\right]^2=\frac{1}{q^{15}}=\frac{1}{q}$$
hence
$$\left[\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}\right]^2=\frac{q}{1}\cdot 4q^6=4$$
$$\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}=\pm 2$$
And I try for a solution as a polar-form method$.\\$Suppose $q=\cos\frac{2j\pi}{7}+i\sin\frac{2j\pi}{7}$
$$\frac{q^k}{1+q^{2k}}=\frac{\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}}{2\cos\frac{2jk\pi}{7}\left(\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}\right)}=\frac{1}{2\cos\frac{2jk\pi}{7}}$$
Am I going to the right direction? How I finish it? And please help to figure out what's wrong with my calculation at the first part. I appreciate for your help.
|
We have that $q$ is a root of
$$
X^6 +X^5+\dots +X+1=X^3(X^3+X^{-3} +X^{2}+X^{-2}+ X+X^{-1}+1).
$$
Hence $q+q^{-1}$ is a root of
$$
Y^3 - 3Y +Y^2 -2 +Y +1= Y^3+Y^2-2Y-1.
$$
Hence $\frac{1}{q+q^{-1}}$ is a root of
$$
Z^3+2Z-Z-1.
$$
Exactly the same is true for the roots $q^2, q^4$, so we get that the sum of the three roots
$$\frac{1}{q+q^{-1}}+\frac{1}{q^2+q^{-2}}+\frac{1}{q^4+q^{-4}}=-2.$$
The left hand side is equal to the expression we are to evaluate.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3649434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$ Question -
Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that
$$
a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}
$$
My try -
i tried putting $a+2 = x, b+2=y , c+2=z$
then we get $x+y+z=9$ and after simplification we have to prove that
$3>x/y + y/z + z/x$ which i am not able to prove...
i also tried C-S,Chebyshev,rearrangement..etc but none of them working
any hints ???
thankyou
|
Let $c=\min\{a,b,c\}$.
Thus, we need to prove that:
$$a^2+b^2+c^2-\frac{(a+b+c)^2}{3}\geq\frac{2+a}{2+b}+\frac{2+b}{2+a}-2+\frac{2+b}{2+c}-\frac{2+b}{2+a}+\frac{2+c}{2+a}-1$$ or
$$\frac{2}{3}((a-b)^2+(c-a)(c-b))\geq\frac{(a-b)^2}{(2+a)(2+b)}+\frac{(c-a)(c-b)}{(2+a)(2+c)},$$ which is true because
$$\frac{2}{3}>\frac{1}{4}>\frac{1}{(2+a)(2+b)}$$ and
$$\frac{2}{3}>\frac{1}{(2+a)(2+c)}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3652992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
If $f(x)$ is a common factor of $g(x)$ and $h(x)$ find $f(x)$ Given that $f(x)$ is a common factor of $g(x)=x^4-3x^3+2x^2-3x+1$ and $h(x)=3x^4-9x^3+2x^2+3x-1$ find $f(x)$
I tried to factorised $g(x)$ but it doesn't have any rational roots as I've already tried 1 and -1.
So how do I solve this?
|
$$x^4-3x^3+2x^2-3x+1=x^4-3x^3+x^2+x^2-3x+1=$$
$$=(x^2+1)(x^2-3x+1).$$
$$3x^4-9x^3+2x^2+3x-1=3x^4-9x^3+3x^2-x^2+3x-1=$$
$$=(x^2-3x+1)(3x^2-1).$$
Can you end it now?
The first factorization we can get by the following way:
$$x^4-3x^3+2x^2-3x+1=x^2\left(x^2+\frac{1}{x^2}+2-3\left(x+\frac{1}{x}\right)\right)=$$
$$=x^2\left(\left(x+\frac{1}{x}\right)^2-3\left(x+\frac{1}{x}\right)\right)=(x^2-3x+1)(x^2+1).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3655490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
If $x+y+z=1$ prove $ \sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3} $ Question -
Let $x, y, z$ be non-negative real numbers which satisfies $x+y+z=1$. Prove that
$$
\sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3}
$$
My work -
First, I tried to simplify this and then now I need to prove that $\sqrt{12x+(y-z)^2}+\sqrt{12y+(z-x)^2}+\sqrt{12z+(x-y)^2} < 6 $
which I can't prove.
Then i tried jensen for $f(x)=\sqrt{x}$ and taking weights as $x,y,z$ but it also does not work.
any hints?
thankyou
|
Since $\frac{1}{12}<\frac{1}{8},$ it's enough to prove that:
$$\sum_{cyc}\sqrt{x+\frac{(y-z)^2}{8}}\leq\sqrt3.$$
Now, by C-S
$$\left(\sum_{cyc}\sqrt{x+\frac{(y-z)^2}{8}}\right)^2\leq\sum_{cyc}\frac{x+\frac{(y-z)^2}{8}}{2x+y+z}\sum_{cyc}(2x+y+z)=\sum_{cyc}\frac{4x+\frac{(y-z)^2}{2}}{2x+y+z}$$ and it's enough to prove that
$$\sum_{cyc}\frac{4x+\frac{(y-z)^2}{2}}{2x+y+z}\leq3$$ or
$$\sum_{cyc}\frac{8x(x+y+z)+(y-z)^2}{2x+y+z}\leq6(x+y+z)$$ or
$$\sum_{cyc}\left(6x-\frac{8x(x+y+z)+(y-z)^2}{2x+y+z}\right)\geq0$$ or
$$\sum_{cyc}\frac{4x^2-y^2-z^2-2xy-2xz+2yz}{2x+y+z}\geq0$$ or
$$\sum_{cyc}\frac{(x-y)(2x+y-z)-(z-x)(2x+z-y)}{2x+y+z}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{2x+y-z}{2x+y+z}-\frac{2y+x-z}{2y+x+z}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2z(x+y)(2z+x+y)\geq0$$ and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3659899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Simplifying $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$
The expression $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$, where $0<x<1$, is equal to $x$ or $\sqrt{(1+x^2)}$ or $\frac1{\sqrt{(1+x^2)}}$ or $\frac x{\sqrt{(1+x^2)}}$? (one of these 4 is correct).
My attempt: $$0<x<1\implies 0<\arctan x<\frac{\pi}{4}\implies\frac{1}{\sqrt2}<\cos(\arctan x)<1$$ and $$0<x\sin(\arctan x)<\frac{x}{\sqrt2}$$$$\implies\frac{1}{\sqrt2}<\cos(\arctan x)+x\sin(\arctan x)<1+\frac{x}{\sqrt2}$$$$\implies\frac{1}{2}<(\cos(\arctan x)+x\sin(\arctan x))^2<(1+\frac{x}{\sqrt2})^2$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{(1+\frac{x}{\sqrt2})^2-1}$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{\frac{x^2}{2}-\sqrt2x}$$ Not sure if I have reached anywhere.
|
Let $\arctan x=y;x=\tan y, -\dfrac\pi2 <y<\dfrac\pi2$
As $\cos y>0,$
$$\cos y=+\dfrac1{\sqrt{1+x^2}};\sin y=\tan y\cdot\cos y=\dfrac x{\sqrt{1+x^2}}$$
$$\cos(\arctan x)+x\sin(\arctan x)=\sqrt{1+x^2}$$
$\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}=\sqrt{{(\sqrt{1+x^2}})^2-1}=\sqrt{x^2}=|x|$ for real $x$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3661245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Simplify the algebraic fraction $\frac{4x^2-8x+3}{4x^2+2x-12}$. Find the quotients as algebraic fractions of the following division:
$$\frac{4x^2-8x+3}{4x^2+2x-12}$$
The answer is
$$\frac{2x-1}{2(x+2)}$$
The method that I am applying to solve:
Write what $2$ numbers multiplied give me the far most right number in the numerator, and add, give me the number left to the far most right number in the numerator as the numerator for the new fraction and doing the same for the denominator then cancelling out whichever $2$ numbers i can.
Ex.
*
*$\frac{x^2 -2x -3}{x^2 -1}$
*$\frac{(x+1)(x-3)}{(x+1)(x-1)}$
*after I cancel out $(x+1)$, I’m left with $\frac{x-3}{x-1}$
I can't find any numbers that both multiply together to get $3$ and add together to get $-8$. i tried simplifying the original fraction then applying the method but i still didn't get the right answer. This method should work because it worked for part a) and b) in the question
|
$4x^2 - 8x + 3$
When you have a coefficent on the $x^2$ term, it gets a little bit more complicated to factor. There are two main techniques. Since 3 is prime, we can start with the guess
$(Ax - 3)(Bx - 1)$
Now we look for $A,B$ such that $AB = 4$ and $-A - 3B = -8$
Our candidates are $\{(1,4),(2,2), (4,1)\}$ and we try these until something hits.
The other way students are taught, is more generalizable, but in this case more work.
$4x^2 - 8x + 3$
We take the first coefficent and the last and multiply them together.
$4\cdot 3 = 12$
Now we look for two numbers that when added together make $-8$ and multiplied together make $12.$ This is exactly as you describe above.
$(-2),(-6)$ work.
Now we say:
$4x^2 - 2x - 6x + 3$
group the terms:
$(4x^2 - 2x) + (-6x + 3)$
The expression on the left has a common factor of $2x$ and the expression on the right has a common factor of $3.$
$2x (2x - 1) + 3(-2x + 1)$
Let's factor our a $-1$ from the expression on the right to flip some signs and make the common factor between the two stand out.
$2x (2x - 1) - 3(2x - 1)$
And factor out the common factor
$(2x-3)(2x - 1)$
In the denominator, before you start, notice that $2$ is a common factor to all three terms.
$4x^2 +2x - 12 = 2(2x^2 + x - 6)$ and then proceed as above.
We might guess that $(2x - 1)$ or $(2x - 3)$ will be a factor, in order to have something that canceles. While this isn't "mathematically rigorous" logic, thinking like a teacher, test designer or textbook publisher often points to short-cuts.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3662734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Cartesian equation for the curvature of a superellipse? I have an application where I need to determine the curvature of a superellipse at various points along the curve, where the Cartesian form of the superellipse is defined as:
$$\frac{x^n}{a^n}+\frac{y^n}{b^n}=1$$
For an ellipse, the Cartesian equation for the curvature $\kappa$ is easy to find:
$$\kappa= \frac{1}{a^2b^2}\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}\right)^{-{\frac32}}$$
but I have searched and searched and cannot find the Cartesian equation for superellipse curvature.
Could someone point me to a site that has this information?
|
Using Hessian matrix for implicit function $F(x,y)=0$,
\begin{align}
F(x,y) &= \frac{x^n}{a^n}+\frac{y^n}{b^n}-1 \\
F_x &= \frac{nx^{n-1}}{a^n} \\
F_y &= \frac{ny^{n-1}}{b^n} \\
\nabla F &=
\begin{pmatrix}
F_{x} \\
F_{y}
\end{pmatrix} \\
F_{xx} &= \frac{n(n-1)x^{n-2}}{a^n} \\
F_{yy} &= \frac{n(n-1)y^{n-2}}{b^n} \\
F_{xy} &= 0 \\
\mathbb{H}(F) &=
\begin{pmatrix}
F_{xx} & F_{xy} \\
F_{xy} & F_{yy}
\end{pmatrix} \\
\kappa &=
\frac{
\begin{vmatrix}
\mathbb{H}(F) & \nabla F \\
(\nabla F)^T & 0
\end{vmatrix}}{|\nabla F|^3} \\
&=
\frac{-n^3(n-1) \dfrac{x^{n-2} y^{n-2}}{a^n b^n}
\left( \dfrac{x^n}{a^n}+\dfrac{y^n}{b^n} \right)}
{n^3\left( \dfrac{x^{2n-2}}{a^{2n}}+\dfrac{y^{2n-2}}{b^{2n} }\right)^{3/2}} \\
&= -\frac{(n-1) x^{n-2} y^{n-2}}
{a^n b^n\left( \dfrac{x^{2n-2}}{a^{2n}}+\dfrac{y^{2n-2}}{b^{2n} }\right)^{3/2}} \\
\end{align}
For $n>1$, $\kappa<0$ implies the centre of curvature being on another side of the outward normal.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3666925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
When $ab/(a+b)$ is an integer, where $a,b$ are positive integers. When $ab/(a+b)$ is an integer, where $a,b$ are positive integers?
clear;
maxn:=30;
for a in [1..maxn] do
for b in [a..maxn] do
q1:=a*b; q2:=a+b;
if q1 mod q2 eq 0 then
print a,b,q1 div q2;
end if;
end for;
end for;
the Magma code given above outputs the following.
2 2 1
3 6 2
4 4 2
4 12 3
5 20 4
6 6 3
6 12 4
6 30 5
8 8 4
8 24 6
9 18 6
10 10 5
10 15 6
12 12 6
12 24 8
14 14 7
15 30 10
16 16 8
18 18 9
20 20 10
20 30 12
21 28 12
22 22 11
24 24 12
26 26 13
28 28 14
30 30 15
So I conjectrue that $\frac{ab}{a+b}$ is an integer if and only if $\frac{a}{d}+\frac{b}{d} \mid d$, where $d=\gcd(a,b)$. It is true if $a=b$.
If $\frac{a}{d}+\frac{b}{d} \mid d$, then
$$\frac{ab}{a+b}=\frac{a}{d} \cdot \frac{b}{d}\cdot \frac{d}{\frac{a}{d}+\frac{b}{d}}$$
is a product of three positive integers and hence is an integer.
Conversely, i.e., $\frac{ab}{a+b}$ is an integer. Let $b=\frac{p}{q}\cdot a$ with $\gcd(p,q)=1$ where $p=\frac{b}{d}$, $q=\frac{a}{d}$ and $d=\gcd(a,b)$. Then
$$\frac{ab}{a+b}=\frac{a\cdot \frac{p}{q}\cdot a }{(1+\frac{p}{q})a}=\frac{ap}{p+q}$$
Since $\gcd(p,p+q)=\gcd(p,q)=1$, it has $p+q \mid a$. Similarly, it has $p+q \mid b$. As a result, $p+q \mid \gcd(a,b)$, i.e. $\frac{a}{d}+\frac{b}{d} \mid d$.
It completes the proof.
|
The conjecture is true - just cancel the gcd $(A,B) =:c\,$ to reduce to the simpler coprime case.
$\begin{align}{\bf Theorem}\ \ \ A\!+\!B\mid AB &\iff\, a+b\mid c,\ \ \ \ a = A/c,\, b = B/c \\[.4em]
&\ \smash[t]{\overset{\times\ c}\iff}\ \:\! A\!+\!B\mid (A,B)^2\end{align}$
Proof $\,\ \dfrac{AB}{A+B} = \dfrac{acbc}{ac+bc} = \dfrac{abc}{a+b}\ $ is an integer
$\ \iff\, a\!+\!b\mid \color{#0a0}a\color{#c00}bc\iff a\!+\!b\mid c,\, $ by $\,(a\!+\!b,\color{#0a0}a)={\underbrace{(a,b)}_{\large 1}} = (a\!+\!b,\color{#c00}b)\,$ by Euclid
Remark $ $ Generally, as explained here, problems like this are usually simplified by reducing to coprime case by cancelling the gcd throughout (using gcd & lcm distributive laws).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3668069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
The tangent line to the curve of intersection of the surface $x^2+y^2=z$ and the plane $x+z=3$ at the point $(1,1,2)$ passes through The tangent line to the curve of intersection of the surface $x^2+y^2=z$ and the plane $x+z=3$ at the point $(1,1,2)$ passes through
(A)$(-1,-2,4)$
(B)$(-1,4,4)$
(C)$(3,4,4)$
(D)$(-1,4,0)$
I can find the equation of the tangent line, If I can find the suitable parametrization($\vec{R(t)}$) using the equation $w(\lambda)=\vec{R(t)}|_{(1,1,2)}+ R'(t)|_{(1,1,2)}\lambda$
I made a parametrization by choosing $x=t,z=3-t,y^2=t-t^2.$ But it is not differnetiable at that point. How do I find the suitable parametrization?
|
Two surfaces, paraboloid $z=x^2+y^2$ and plane $z=3-x$, intersect on ellipse:
$$x^2+y^2=3-x \iff x^2+x+y^2=3 \iff (x+\frac{1}{2})^2-\frac{1}{4}+y^2 = 3 \iff (x+\frac{1}{2})^2+y^2=\frac{13}{4} \\$$
$$\iff \frac{(x+\frac{1}{2})^2}{\frac{13}{4}}+\frac{y^2}{\frac{13}{4}}=1$$
Now the tangent on the intersection curve at the point $P=(1,1,2)$ is easy to find, as well as it is easy to check which one of 4 points lie on this tangent...
Figure.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3669062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Verify Stokes' Theorem for Hemisphere I am trying to answer the following question, but am having difficulty getting the same result for each side of Stokes' theorem.
Question:
Verify Stokes' Theorem for the hemisphere $D: x^2 + y^2 + z^2 = 9, z\geq0$ its bounding circle $C: x^2 + y^2 = 9, z=0$ and the vector field $\overrightarrow{A} = y\overrightarrow{i} - x\overrightarrow{j}$.
LHS:
$\oint_{C} \overrightarrow{A} \cdot d\overrightarrow{r} = \oint_{C} ydx - xdy = -9\int_{0}^{2\pi}sin^2\theta + cos^2\theta \quad d\theta = \quad ...\quad = -18\pi$
I'm not sure how correct this is so apologies if I have made a mistake here.
RHS:
For the RHS whichever way I work it I keep getting -18 as the solution and I'm not sure where I'm going wrong. This is the kind of working I have done, I'm definitely doing something wrong here:
$\int_{S} curl(\overrightarrow{A}) \cdot \overrightarrow{n} dS = \cdots = -2 \int_{R}dxdy = \cdots = -18\int_{0}^{2\pi} cos\theta sin\theta - sin\theta d\theta = \cdots $
|
Let's clarify our setting first. Note that a normal vector to the sphere of radius $3$ at $\vec{x}$ is $\vec{n}=\frac{\vec{x}}{\|\vec{x}\|}=\frac{\vec{x}}{3}$. Our hemisphere is given by
$$
S=\{(x,y,\varphi(x,y)): x^2+y^2\le 9\},
$$
with $\varphi(x,y)=\sqrt{9-x^2-y^2}$. Therefore the surface measure is
$$
dS(\vec{x})=\sqrt{1+\|\nabla \varphi(\vec{x})\|^2}=\sqrt{1+\frac{x^2+y^2}{9-x^2+y^2}}=\frac{3}{\sqrt{9-x^2+y^2}}
$$
and the integration against $dS$ is given for $f:\mathbb{R}^3\rightarrow \mathbb{R}$ by
$$
\int_S f dS= \int_{\{x^2+y^2\le 9\}} f\left(x,y,\sqrt{9-x^2+y^2}\right)\frac{3}{\sqrt{9-x^2+y^2}}\,dx\,dy
$$
Finally, the curl of the vector field $\vec{A}$ is $(0,0,-2)$.
Therefore we have
\begin{align*}
\int_{S} curl(\vec{A}) \cdot \vec{n}\, dS&=\int_S (0,0,-2)\cdot \frac{\vec{x}}{3}\,dS (\vec{x})\\
&=\frac{-2}3 \int_{\{x^2+y^2\le 9\}} \sqrt{9-x^2+y^2} \frac{3}{\sqrt{9-x^2+y^2}}\,dx\,dy
\\
&=-2 |\{(x,y)\in\mathbb{R}^2: x^2+y^2\le 9\}|=-18\pi.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3670217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate $\sum_{n,k} \binom{n}{k}^{-1} $
Evaluate $$\sum_{n,k} \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1<k<n-1$.
Thouhgts:
We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$
We may try to find a closed form of the inner summation which is of the form :
$$ \frac{1}{ {n \choose 2} } + \frac{1}{{n \choose 3} }+ \dotsb + \frac{1}{{n \choose n-2} }. $$
Notice that we may write $\frac{1}{ {n \choose 2} } = \frac{2! }{n(n-1)}$ and if keep doing the same for the other terms we obtain the following:
$$ \frac{ (n-2)! + (n-3)! (n-(n-2)) + (n-4)!(n-(n-2))(n-(n-3)) + \dotsb + 2! (n-3)! }{n!}, $$
which equals
$$ \frac{ (n-2)! + 2!(n-3)! + 3! (n-4)! + \dotsb + (n-3)! 2! }{n!} $$
and so this equals:
$$ \frac{1}{n(n-1)} + \frac{2}{n(n-1)(n-2)} + \dfrac{6}{n(n-1)(n-2)} + \dotsb + \dfrac{2}{n(n-1)(n-2) }. $$
But half of this terms are identical. Therefore, we are trying to sum up series of the form
$$\sum_{n \geq k} \frac{1}{(n-1)(n-2)(n-3)\dotso(n-k)} ,$$
which can be done by a telescoping trick, but it seems very formidable. Am I approaching this problem the right way? Any hints/suggestions?
|
Generally similar sums can be evaluated using the Beta function:
$$
B(x+1,y+1)=\int_0^1 t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}=
\frac{x!y!}{(x+y+1)!}=\frac{1}{x+y+1}\binom{x+y}x^{-1}.
$$
Applying this in your case ($x=n-k,y=k$) one has:
$$
\binom{n}{k}^{-1}=(n+1)\int_0^1 t^{n-k}(1-t)^{k}dt
$$
or
$$\begin{align}
I_n=\frac1{n+1}\sum_{k=2}^{n-2}\binom{n}{k}^{-1}
&=\sum_{k=2}^{n-2}\int_0^1 t^{n-k}(1-t)^kdt\\
&=\int_0^1\left[ t^n \sum_{k=2}^{n-2}\left(\frac{1-t}t\right)^k\right] dt\\
&=\int_0^1 t^n\left(\frac{1-t}t\right)^2
\frac{1-\left(\frac{1-t}t\right)^{n-3}}{1-\frac{1-t}t}dt\\
\end{align}$$
and, finally,
$$\begin{align}
\sum_{n=4}^\infty (n+1)I_n&
=\int_0^1\left[\frac{(1-t)^2}{2t-1}\sum_{n=4}^\infty (n+1)\left(t^{n-1}-t^2(1-t)^{n-3}\right)\right]dt\\
&=\int_0^1(1+3t-3t^2)dt=\frac32.
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3677292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
}
|
Does $a^2 + b^2 = 2 c^2$ have any integer solution?
Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?
I think not, because of these equations for pythagorean triplets:
$$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$
$$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$
I think I would need to multiply $x$ and $y$ by $\sqrt{2}$ and they would never be rational numbers.
|
It is equivalent to solving the circle equation $x^2 + y^2 = 2$ in rational numbers.
Take one point $(1,1)$ on this curve and consider the line with slope $t$ passing through that line: $y = t x - t + 1$, substitute this in $x^2 + (t x - t + 1)^2 - 2 = 0$ and divide out $(x-1)$ to find the second point of intersection between this line and the circle, $(t^2 + 1)x + (-t^2 + 2t + 1) = 0$.
We have parametrized all nontrivial solutions $$x = a/c = \frac{-t^2 + 2t + 1}{t^2 + 1}$$
For example $t = 22/7$ gives us $a = 127$, $c = 533$ and $127^2 + 743^2 = 2\cdot 533^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3683854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
}
|
Proving $\cos(A + B)\cos(A - B) = -(\sin A + \cos B)(\sin A - \cos B)$. Could you give me some guidance on proving the following trig identity?
$$\cos(A + B)\cos(A - B) = -(\sin A + \cos B)(\sin A - \cos B).$$
|
Consider $\cos(x\mp y)=\cos x\cos y\pm\sin x\sin y$, add them together, you'll get $$\cos(x+y)+\cos(x-y)=2\cos x\cos y.$$
Now let $x=A+B,\,y=A-B$, so $2\cos(A+B)\cos(A-B)=\cos(2A)+\cos(2B)=2\cos^2A-1+1-2\sin^2B$,
hence we get the desired result as $\cos^2A-\sin^2B=(\cos A+\sin B)(\cos A-\sin B)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3684809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Prove by induction that for all $n\in\mathbb N, (\sqrt3+i)^n+(\sqrt3-i)^n=2^{n+1}\cos(\frac{n\pi}6)$ I want to prove by induction that for all $n \in \mathbb{N}$, $$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\left(\frac{n\pi}{6} \right)$$
I can prove the identity using direct complex number manipulation and de Movire (which gets it for all integers), but I'm getting stuck when trying induction out of interest instead.
For the inductive step, I have $$(\sqrt{3} + i)^{k+1} + (\sqrt{3} - i)^{k+1} $$ but what can I do with that? If I separate the terms into $(\sqrt{3} + i)^{k+1} = (\sqrt{3} + i)^{k} (\sqrt{3} + i)$ etc, the minus sign from the second term is causing me troubles.
|
My try is not that 'pure' in the sense of using the trig. equation.
Let $ A = \sqrt 3 + i$, $B = \sqrt 3 - i$. With some initial step verified, we can write
\begin{align*}
A^{n+1} + B^{n+1}&= (A^n + B^n)( A+ B) - AB (A^{n-1} + B^{n-1}) \\
& = 2^{n+2}\left( 2 \cos(\frac{n\pi}{6})\cos(\frac{\pi}{6}) - \cos \frac{(n-1)\pi}{6} \right) \\
& = 2^{n+2} \cos \frac{(n+1)\pi}{6}
\end{align*}
with $AB = (\sqrt3)^2 - i^2 =4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3685150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
$f(m^2 + f(n)) = f(m)^2 + n$ for all natural numbers Find all functions such that $$ f(m^2 + f(n)) = f(m)^2 + n$$ for all natural numbers (1,2,3,4,..)
I have been stuck on this problem.
I have tried to set f(1) = k, however, did not make any progress.
|
Let $P(m, n)$ denote $f(m^2+f(n)) = f(m)^2 + n$. Note that
\begin{align*}
P(1, y) \implies f(1+f(y)) &= f(1)^2 + y\\
\implies P(f(x), 1+f(y)) \implies f(f(x)^2 + f(1)^2 + y) &= f(f(x))^2 + 1 + f(y) \qquad (1)\\
P(x, y) \implies f(x^2+f(y)) &= f(x)^2 + y\\
P(f(1), x^2+f(y)) \implies f(f(x)^2 + f(1)^2 + y) &= f(f(1))^2 + x^2 + f(y) \qquad (2)\\
(1), (2)\implies f(f(x))^2 &= x^2 + c
\end{align*}
where $c = f(f(1))^2 - 1$. Hence, $x^2 + c$ is a perfect square for any positive integer $x$, so $c=0$ (otherwise, you can choose $x > c$ such that $x^2 + c < x^2 + x < (x+1)^2$, leading to a contradiction).
Hence, $f(f(x)) = x$. Therefore
$$P(1, f(x)) \implies f(x+1) = f(x) + f(1)^2$$
so $f(x) = f(1)^2x + (f(1) - f(1)^2)$. Since
$$x = f(f(x)) = f(1)^4x + f(1)^2(f(1)-f(1)^2) + (f(1) - f(1)^2)$$
we must have $f(1) = 1$, so $f(x) = x$.
We can verify this to check it works, so we're done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3689239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
On $\sum_{x=0}^\infty x^n r^x$ I am looking for either a closed form or recursive expression for $$\sum_\limits{x=0}^\infty x^n r^x\ $$ that does not include differential operators, where $n \in \mathbb N_0$. It is clear to me that $$\sum_\limits{x=0}^\infty x^n r^x = r \cfrac{\text d}{\text dr}\sum_\limits{x=0}^\infty x^{n-1} r^x \ $$ and from this I found some patterns in the first few expressions:
$$\sum_{x=0}^\infty xr^x = \cfrac r {(1-r)^2} \\ \sum_{x=0}^\infty x^2 r^x = \cfrac {r+r^2} {(1-r)^3} \\ \sum_{x=0}^\infty x^3 r^x = \cfrac {r^3+4r^2+r} {(1-r)^4} \\ \sum_{x=0}^\infty x^4 r^x = \cfrac {r^4+11r^3+11r^2+r} {(1-r)^5} \\ \sum_{x=0}^\infty x^5 r^x = \cfrac {r^5+26r^4+66r^3+26r^2+r} {(1-r)^6} \\ \sum_{x=0}^\infty x^6 r^x = \cfrac {r^6 + 57r^5+302r^4+302r^3+57r^2+r} {(1-r)^7} \\ \sum_{x=0}^\infty x^7 r^x = \cfrac {r^7+120r^6 + 1191 r^5 + 2416r^4+1191r^3+120r^2+r} {(1-r)^8}$$
It's interesting that the denominator increases by a factor of $1-r$ each time, and another interesting thing is that the coefficients of row $n$ add up to $n!$. I haven't been able, though, to leverage this information in a way that spits out the individual coefficients. Is my goal of finding a closed form/recursive formula for these possible? Thanks.
|
Terminology ... polylogarithm
$$
\operatorname{Li}_s(z) = \sum_{k=0}^\infty \frac{z^k}{k^s}
$$
So your question asks about the polylogarithm $\operatorname{Li}_{-n}(r)$.
That page shows $\operatorname{Li}_s(z)$ for $s=-1,-2,-3,-4$, the first four
examples you computed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3691489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ . What is the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ ?
I have been stuck on this problem with no direction. I have tried multiplying the sequence with $x$ and trying out $S-Sx$ but have gotten nowhere. Any help?
Thanks.
|
Hint:
$F'(x)=S=1\cdot2x+2\cdot3x^2+3\cdot4x^3+\cdots$
$F(x)=x^2+2x^3+3x^4+\cdots=x^2(1+2x+3x^2+\cdots)=\dfrac{x^2}{(1-x)^2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3691770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Sum $\sum \frac{1}{(4k-3)(4k-2)(4k-1)(4k)}$ I am stuck on this problem for quite a while now, and I don't seem any closer to the solution. So, here it is:
$S = 1/4! + 4!/8! + 8!/12! + 12!/16! + ......$
I crossed out the factorials first, and it could be easily represented by the general term,
$T = \frac{1}{(4n-3)(4n-2)(4n-1)(4n)}$
It looked like it could be expressed as the difference of two expressions which could come useful to find the sum, what we call the 'diagonal cancellation',
$T = \frac{1}{3}(\frac{1}{(4n-1)(4n-2)(4n-3)} - \frac{1}{4n(4n-1)(4n-2)})$
but unfortunately it doesn't work. I even unintentionally split it further into subtractions of two more expressions for both terms and so on, which finally leads to a harmonic series, which is certainly not expressible in "closed-form'.
Please help by suggesting a simple math solution, this is merely a class notes illustration and I am not an advanced maths student.
|
\begin{align*}
& \frac{1}{(4n-3)(4n-2)(4n-1)(4n)} \\
&= \frac{1}{3}\left(\frac{1}{(4n-1)(4n-2)(4n-3)} - \frac{1}{4n(4n-1)(4n-2)}\right) \\
& = \frac{1}{3\cdot 2}\left( \frac{1}{(4n-2)(4n-3)} - \frac{2}{(4n-1)(4n-2)} + \frac{1}{4n(4n-1)}\right) \\
& = \frac{1}{3\cdot 2\cdot 1}\left( \frac{1}{4n-3} - \frac{3}{4n-2} + \frac{3}{4n-1} - \frac{1}{4n}\right) \\
\end{align*}
So the sum is \begin{align*}
&\frac{1}{6}\sum_{n=1}^{\infty}\left( \frac{1}{4n-3} - \frac{3}{4n-2} + \frac{3}{4n-1} - \frac{1}{4n}\right) =\\
& =\frac{1}{6}\sum_{n=1}^{\infty}\left( \int_0^1 x^{4n-4} dx -3 \int_0^1 x^{4n-3} dx+ 3\int_0^1 x^{4n-2} dx - \int_0^1 x^{4n-1} dx\right) = \\
& = \frac{1}{6}\sum_{n=1}^{\infty}\left( \int_0^1 x^{4n-4} - 3 x^{4n-3} +3x^{4n-2} - x^{4n-1} dx\right) = \\
& \stackrel{*}{=} \frac{1}{6} \int_0^1 \sum_{n=1}^{\infty} \left(x^{4n-4} - 3 x^{4n-3} +3x^{4n-2} - x^{4n-1}\right) dx = \\
& = \frac{1}{6} \int_0^1 \frac{1 - 3x + 3x^2 - x^3}{1-x^4} dx = \\
& = \frac{1}{6} \int_0^1 \frac{(1-x)^2}{(1+x)(1+x^2)} dx = \cdots = \frac{1}{24}(6\ln2 - \pi)\\
\end{align*}
Here the step $\stackrel{*}{=}$ should be justified; the integrand is of the form $x^{4n-4}(1-x)^3$ and this is nonnegative on $[0,1]$ so Fubini-Tonelli is appliable.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3692272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to find the intergral $I_{A}=\int_{0}^{2\pi}\frac{\sin^2{x}}{(1+A\cos{x})^2}dx$ Let $A\in (0,1)$be give real number ,find the closed form intergral
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx$$
This integral comes from a physical problem，following is my try:
since
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx=I_{1}+I_{2}$$
Where $$I_{1}=\int_{0}^{\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx,I_{2}=\int_{\pi}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx$$
For $I_{2}$ Let $x=\pi+t$,then we have
$$I_{2}=\int_{0}^{\pi}\dfrac{\sin^2{x}}{(1-A\cos{x})^2}dx$$
so
$$I_{A}=I_{1}+I_{2}=2\int_{0}^{\pi}\dfrac{\sin^2{x}(1+A^2\cos^2{x})}{(1-A^2\cos^2{x})^2}dx=4\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^2{x}(1+A^2\cos^2{x})}{(1-A^2\cos^2{x})^2}dx$$
Then I fell ugly, so how to prove it? Thank you
|
Use integration by parts
$$\int \frac{\sin^2 x\ dx}{(1+A\cos x)^2}$$
$$=\int \sin x\cdot \frac{\sin x}{(1+A\cos x)^2}\ dx$$
$$=\sin x\cdot \frac{1}{A(1+A\cos x)}-\int \frac{\cos x}{A(1+A\cos x)}\ dx$$
$$=\frac{\sin x}{A(1+A\cos x)}-\frac{1}{A^2}\int \frac{(1+A\cos x)-1}{1+A\cos x}\ dx$$
$$=\frac{\sin x}{A(1+A\cos x)}-\frac{x}{A^2}+\frac{1}{A^2}\int \frac{dx}{1+A\cos x}$$
$$=\frac{\sin x}{A(1+A\cos x)}-\frac{x}{A^2}+\frac{1}{A^2}\int \frac{dx}{1+A\frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}}$$
$$=\frac{\sin x}{A(1+A\cos x)}-\frac{x}{A^2}+\frac{2}{A^2}\int \frac{\frac 12\sec^2\frac x2dx}{(1-A)\left(\frac{1+A}{1-A}+\tan^2\frac x2\right)}$$
$$=\frac{\sin x}{A(1+A\cos x)}-\frac{x}{A^2}+\frac{2}{A^2(1-A)}\int \frac{d\left( \tan\frac x2\right)}{\left(\tan\frac x2\right)^2+\left(\sqrt{\frac{1+A}{1-A}}\right)^2}$$
$$=\frac{\sin x}{A(1+A\cos x)}-\frac{x}{A^2}+\frac{2}{A^2\sqrt{1-A^2}}\tan^{-1}\left(\tan\frac{x}{2}\sqrt{\frac{1-A}{1+A}}\right)$$
$$\therefore \int_0^{2\pi} \frac{\sin^2 x\ dx}{(1+A\cos x)^2}=2\int_0^{\pi} \frac{\sin^2 x\ dx}{(1+A\cos x)^2}=\color{blue}{\frac{2\pi}{A^2}\left(\frac{1}{\sqrt{1-A^2}}-1\right)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3694900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Angular integrals used in QED I am reading a research paper
and am stuck at a point where the author uses angular integrals. I don't have any idea about it and would like help.
The angular integral is:
$$I_k (y)=\int_0^\pi {\sin^2(\theta)\cos^{2k}(\theta) \over \beta^2(y)-\cos^2 (\theta)} d\theta$$
The author directly give results to this integral without proof.
The results are:
$$I_0 (y)=\pi \left(1-\sqrt{1-{1\over\beta^2 (y)} } \right),$$
$$I_1 (y)={-\pi \over 2}+\beta^2 (y)I_0 (y),$$
I don't know how he got these results. Can anyone please help me?
|
An interesting approach to these integrals is the use of complex integration. In this case, we can use the fact that
with
\begin{equation}
f(\sin\theta,\cos\theta) = \frac{\sin^2\theta \cos^{2k} \theta}{\beta^2-\cos^2\theta}
\end{equation}
we have
\begin{eqnarray}
\int_0^\pi \mathrm{d}\theta \, f(\sin\theta,\cos\theta) &=&
\frac{1}{2} \int_0^{2\pi} \mathrm{d}\theta \,f(\sin\theta,\cos\theta)
\\
&=& \frac{1}{2} \int_{|z|=1} \frac{\mathrm{d} z}{i z} \, f\left(\frac{z-z^{-1}}{2i},\frac{z+z^{-1}}{2}\right)
\\
&=& - \frac{i}{2^{2k+1}} \,\int_{|z|=1} \mathrm{d} z\, \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} (z^4-4 \beta^2 z^2+2z^2+1)}
\\
&=& - \frac{i}{2^{2k+1}} \,\int_{|z|=1} \mathrm{d} z\, \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} \Pi_{i=1}^4 (z-z_i)} \, ,
\end{eqnarray}
with $z_i$ the simple poles
\begin{eqnarray}
z_1 &=& -\beta - \sqrt{\beta^2 - 1}
\\
z_2 &=& \beta - \sqrt{\beta^2 - 1}
\\
z_3 &=& -\beta + \sqrt{\beta^2 - 1}
\\
z_4 &=& \beta + \sqrt{\beta^2 - 1} \,.
\end{eqnarray}
By looking at the result, seems like the author assumes the range of $\beta$ is such that only $z_2$ and $z_3$ lay inside the unit circle (besides the $(2k+1)$-order pole at the origin). Add the contributions from those 3 poles and you can actually find a formula for arbitrary $k$ with relative ease.
More precisely, if we write
\begin{equation}
g = \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} \Pi_{i=1}^4 (z-z_i)} \,
\end{equation}
then we find
\begin{eqnarray}
\int_0^\pi \mathrm{d}\theta \, f(\sin\theta,\cos\theta) = && \frac{\pi}{2^{2k}} \left\{ \frac{1}{(2k)!} \lim_{z\rightarrow 0} \frac{d^{2k}}{dz^{2k}} \left[z^{2k+1} g\right] + \lim_{z\rightarrow z_2} \left[(z-z_2)g\right] + \lim_{z\rightarrow z_3} \left[(z-z_3)g\right] \right\} \, .
\end{eqnarray}
Replacing and simplifying gives us
\begin{equation}
I_k(y) = \frac{\pi}{2^{2k}} \frac{1}{(2k)!} \lim_{z\rightarrow 0} \frac{d^{2k}}{dz^{2k}} \left[ \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^4+z^2\left[2-4\beta^2(y)\right] + 1} \right] - \pi \beta^{2k-1}(y) \sqrt{\beta^2(y) - 1} \, .
\end{equation}
Finally, when expanding the first term, it turns out this can be further simplified, producing
\begin{equation}
I_k(y) = - \pi \sum_{m=0}^k \frac{(2m-3)!!}{(2m)!!} \beta^{2k-2m}(y) - \pi \beta^{2k-1}(y) \sqrt{\beta^2(y) - 1} \, ,
\end{equation}
where we take $(-3)!! = -1$, $(-1)!! = 1$ and $0!! = 1$. This also allows us to generalize the relation between successive values of $I_k(y)$, where we have
\begin{equation}
I_k(y) = - \frac{(2k-3)!!}{(2k)!!} \pi + \beta^{2}(y) I_{k-1}(y) \,.
\end{equation}
In hindsight, this result could have been deduced in a couple of lines, by noticing that the difference between the integrands of $I_k(y)$ and $\beta^2(y) I_{k-1}(y)$ simplifies to
\begin{equation}
\frac{\sin^2\theta \cos^{2k} \theta}{\beta^2(y)-\cos^2\theta} - \beta^2(y) \frac{\sin^2\theta \cos^{2k-2} \theta}{\beta^2(y)-\cos^2\theta} = \cos^{2k} \theta - \cos^{2k-2} \theta \,.
\end{equation}
Hence
\begin{eqnarray}
I_k(y) - \beta^2(y) I_{k-1}(y) &=& \int_0^\pi \mathrm{d}\theta \left[ \cos^{2k} \theta - \cos^{2k-2} \theta \right]
\\
&=& \pi \left[ \frac{(2k-1)!!}{(2k)!!} - \frac{(2k-3)!!}{(2k-2)!!} \right]
\\
&=& - \frac{(2k-3)!!}{(2k)!!} \pi \, ,
\end{eqnarray}
for $k \geq 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3698554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
What are the steps to factor $x^2 - 1$ into $(x+1)(x-1)$? Does $(x+1)(x-1) = x^2+1x-1x-1$? If so where are the $+1x$ and the $-1x$ when it is being factored from $x^2-1$ into $(x+1)(x-1)$?
What exactly are we dividing $x^2-1$ by to get $(x+1)(x-1)$ and how did you know what to divide it by?
|
Method 0:
$+1x - 1x = 0$
So $(x+1)(x-1) = x^2 + 1x - 1x -1 =$
$x^2 +(1x-1x) -1 = $
$x^2 + 0 -1 =$
$x^2 -1 $.
Method 1:
$x^2 - 1 = $
$x^2 + x - x - 1 = $
$(x^2 +x) - (x+1) = $
$(x\cdot x + x\cdot 1) + (-1)\cdot(x+1) =$
$x\cdot(x + 1) + (-1)\cdot (x+1) =$
$\color{blue}x\cdot\color{red}{(x + 1)} + \color{blue}{(-1)}\color{red}{(x + 1)} = $
$(\color{blue}x + \color{blue}{(-1)})\color{red}{(x+1)} = $
$(\color{blue}x-\color{blue}1)\color{red}{(x+1)}$
Method 2:
$(x^2 - 1)\div (x+1) = ?????$
We need to find a firs term, $a$ so that when we multiply $a$ times $x+1$ and get $ax+a$ that $ax = x^2$. What can be $a$? Well $ax=x^2$ and so (if we assume $x$ isn't always $0$) then $a= x$.
So $x(x+1) = x^2 + x$.
But $x^2 -1 \ne x^2 + x$ we must find a "remainder".
$(x^2 - 1)- (x^2 + x) = (x^2-x^2) + (-1-x) = -x - 1$.
So $x^2 - 1= (x^2 + x) - x-1 = x(x+1) - x-1$.
And $(x^2 - 1)\div (x+1) = x + \frac {-x-1}{x+1}$
Now we must divide $x+1$ into $-x-1$.
We must find term $b$ so that when we multiply $b$ by $x+1$ and get $bx + b$ the first term $bx = -x$. What can that $b$ be. Clearly it is $b=-1$.
So $-1(x+1) = -x-1$.
And $-x-1$ DOES equal $-1(x+1)$ so we have no remainder.
So $(x^2 -1) = x(x+1) + (-1)(x+1)$.
And $(x^2 - 1)\div (x+1) = x + (-1) = x-1$
So $(x^2 - 1) \div (x+1) = x-1$ so that mean $(x^2-1)= (x-1)(x+1)$.
Method 3:
$(x^2 -1) = (x+a) (x+b)$
$= x^2 +ax +bx + ab =$
$x^2 + (a+b)x + ab$
And that is supposed to be $x^2 -1 = x^2 + 0*x + (-1)$.
So we need $a,b$ so that $a*b = -1$ and $a+b = 0$.
Okay so $ab = -1$ so $a = -\frac 1b$
And $-\frac 1b + b = 0$ so
$b = \frac 1b$ so
$b^2 = 1$.
So $b = \pm 1$.
And $a +b = a\pm 1 = 0$ so $a =\mp 1$.
So one of them is $1$ and the other is $-1$.
So $x^2 -1 = (x+a)(x+b) = (x+1)(x-1)$.
Method 4:
If $ax^2 + bx + c = 0$ has two solutions $x= m$ and $x = n$
Then $ax^2 + bx + c = a(x-m)(x+n)$.
So what are the two solutions to $x^2 -1 = 0$
They are:
$x^2 - 1 =0$ so
$x^2 = 1$ so
$x = \pm 1$
So $m = 1$ and $n=-1$ are the two solutions and
so $x^2 -1 = (x- 1)(x-(-1)) = (x-1)(x+1)$.
........
But in ALL of these methods the $+1x -1x =0$ and the "cancel out".
You can cancel them to $0$ or pop them out of nowhere from $0$.
The this is you just have to think of it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3699703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
What mistake am I making while deriving the expansion of $\cos(\alpha + \beta)$ I was trying to derive the formula for expansion of $\cos (\alpha + \beta)$ by equating the ratio of lengths of two specific chords to the ratio of angles opposite to them but I'm not getting the correct results. Here's how I'm doing it :
In the above diagram,$\angle AOB = \alpha$, $\angle BOC = \beta$, $\angle AOC = (\alpha + \beta)$, $a = \cos{\alpha}$, $b = \sin{\alpha}$, $x = \cos{(\alpha + \beta)}$ and $y = \sin {(\alpha + \beta)}$
And as $a$, $b$, $x$ and $y$ are sines and cosines of $\alpha$ and $(\alpha+\beta)$ respectively, so : $a^2+b^2=x^2+y^2=1$
Now, using the distance formula for coordinate geometry, which states that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ on the Cartesian Plane is : $\sqrt{(x_1-x_2)^2 - (y_1-y_2)^2}$ units, we obtain :
$$AB = \sqrt{(a-1)^2+(b-0)^2}=\sqrt{a^2+1-2a+b^2}=\sqrt{(a^2+b^2)+1-2a}=\sqrt{1+1-2a}$$
$$\therefore AB = \sqrt{2-2a}$$
$$AC = \sqrt{(x-1)^2+(y-0)^2}=\sqrt{a^2+1-2x+y^2}=\sqrt{(x^2+y^2)+1-2x}=\sqrt{1+1-2x}$$
$$\therefore AC = \sqrt{2-2x}$$
Now, the ratio of lengths of $AB$ and $AC$ would be equal to the ratio of angles opposite to them, that are $\alpha$ and $(\alpha + \beta)$ respectively (this is the part where I think I might be wrong but don't see how).
So, according to me,
$$\dfrac{AB}{AC}=\dfrac{\alpha}{\alpha + \beta} \implies \dfrac{AC}{AB} = \dfrac{\alpha + \beta}{\alpha} = 1 + \dfrac{\beta}{\alpha}$$
$$\implies \dfrac{\sqrt{2-2x}}{\sqrt{2-2a}} = 1 + \dfrac{\beta}{\alpha} \implies \dfrac{2-2x}{2-2a} = \Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$
$$\implies \dfrac{1-x}{1-a} = \Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2 \implies 1-x = (1-a)\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$
This leads us to the conclusion that :
$$\cos(\alpha + \beta) = x = 1-(1-a)\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2 = 1-(1-\cos{\alpha})\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$
which is not true...
So, where am I going wrong in this?
Thanks!
PS : I am really grateful to those people who are giving alternative methods of derivation but what I really want to know is the mistake in my derivation. Thanks!
|
You make a mistake exactly where you suspected. Note the ratio of the lengths of the chords is not equal to the ratio of the angles subtended by them. You could see this by applying the law of cosines on $\triangle AOB$ and $\triangle AOC$: $$AB^2=OB^2 + OA^2 -2OA\cdot OB\cos\alpha \\ =1+1-2\cos\alpha\\=2(1-\cos\alpha)\\ \implies AB=2\sin\frac \alpha 2$$ Similarly, $$AC= 2\sin\frac{\alpha+\beta}{2}$$ and $$\frac{AB}{AC} \ne \frac{\alpha}{\alpha+\beta}$$ As for your question in the comments, recall that the arc length is really $r\theta$ and the chord length is $2r\sin\frac{\theta}{2}$. Two quantities will be directly proportional iff their ratio is a constant. But $$\frac{2r\sin\frac{\theta}{2}}{r\theta}=\frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}}$$ which is clearly not constant, except maybe for the case when $\theta\approx 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3702726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the saddle point of $F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2+(-x_1+x_2)y_3$ Finding the saddle points of $F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2$+$(-x_1+x_2)y_3$ subject to the constraints $x_1+x_2+x_3=1, y_1+y_2+y_3=1$. Show that the saddle point is $x=(\frac{1/3}{1/3},\frac{1/3}{1/3},\frac{1/3}{1/3}),y=(\frac{2}{7},\frac{1}{7},\frac{4}{7})$.
I know how to find the saddle point of a function with two variable by using $\Delta=(f_{12})^2-f_{11}f_{22}$, then when $\Delta>0$ the point is a saddle point.
But for this question, by using the constraints, we can reduce the variables from 6 to 4, but still can not use the $\Delta$ formula, and this question requires not to use lagrange multiplier. Thanks.
|
At a saddle point, all of the partial derivatives are 0. This leads to a system of equations:
\begin{align}
x_1 -2x_2 +x_3=0,\tag{1}\\
2x_1 -2x_3=0,\tag{2}\\
x_2-x_1=0,\tag{3}\\
y_1+2y_2-y_3=0, \tag{4}\\
-2y_1 +y_3=0,\tag{5}\\
y_1-2y_2=0. \tag{6}
\end{align}
Adding 1 to both sides of equation (4) leads to
\begin{equation}
2 y_1 = 1-3 y_2. \tag{7}
\end{equation}
Solving for $y_1$ from equation (6) and substituting in (7) leads to $y_2= \frac{1}{7}$ and consequently $y_1=2y_2=\frac{2}{7}$ and $y_3=2y_1=\frac{4}{7}$.
Similarly, we can subtract 1 from both sides of equation (1) and use the constraint for x, which leads to $x_2=\frac{1}{3}$ and consequently from (3), $x_1=x_2=\frac{1}{3}$. Finally, from (3), we have $x_3=x_1=\frac{1}{3}$.
EDIT:
Computing the Hessian matrix yields:
\begin{equation}
H =
\begin{pmatrix}
0 & 0 & 0 & 1 & 2 & -1 \\
0 & 0 & 0 & -2 & 0 & 1\\
0 & 0 & 0 & 1 & 0 & 0\\
1 & -2 & 1 & 0 & 0 & 0\\
2 & 0 & 0 & 0 & 0 & 0\\
-1 & 1 & 0 & 0 & 0 & 0
\end{pmatrix}.
\end{equation}
Diagonalizing gives us the characteristic polynomial
\begin{equation}
\lambda^6-12\lambda^4 +27\lambda^2 -4=0.
\end{equation}
This polynomial has a $\lambda \rightarrow -\lambda$ symmetry. So if we can find one non-zero real solution, we automatically show that the Hessian is indefinite (There are both positive and negative solutions). Therefore the critical point above is indeed a saddle point.
EDIT 2:
In case you find it difficult to actually find the roots of the polynomial, substitute $\rho=\lambda^2$, this gives
\begin{equation}
\rho^3-12\rho^2 +27\rho -4=0.
\end{equation}
Now, using the fact that complex roots come in pairs and that we can see that zero is not a solution, we can deduce that there is at least one real solution of the cubic $\rho_0$ and therefore two opposing sign solutions $\lambda_{+/-} = \pm \sqrt{|\rho_0|}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3704400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Strategy to calculate $ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right) $. I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$
I simplify this a little bit, by moving the constant multiplicator out of the derivative:
$$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(\frac{x^2-6x-9}{x^2(x+3)^2}\right) $$
But, using the quotient-rule, the resulting expressions really get unwieldy:
$$ \frac{1}{2} \frac{(2x-6)(x^2(x+3)^2) -(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} $$
I came up with two approaches (3 maybe):
*
*Split the terms up like this: $$ \frac{1}{2}\left( \frac{(2x-6)(x^2(x+3)^2)}{(x^2(x+3)^2)^2} - \frac{(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} \right) $$
so that I can simplify the left term to $$ \frac{2x-6}{x^2(x+3)^2}. $$
Taking this approach the right term still doesn't simplify nicely, and I struggle to combine the two terms into one fraction at the end.
*The brute-force-method. Just expand all the expressions in numerator and denominator, and add/subtract monomials of the same order. This definitely works, but i feel like a stupid robot doing this.
*The unofficial third-method. Grab a calculator, or computer-algebra-program and let it do the hard work.
Is there any strategy apart from my mentioned ones? Am I missing something in my first approach which would make the process go more smoothly?
I am looking for general tips to tackle polynomial fractions such as this one, not a plain answer to this specific problem.
|
Note that $x^2-6x-9 = (x-3)^2 - 18$. So after pulling out the factor of $\frac 12$, it suffices to compute
$$\frac{d}{dx} \left(\frac{x-3}{x(x+3)}\right)^2$$
and
$$\frac{d}{dx} \left(\frac{1}{x(x+3)}\right)^2.$$
These obviously only require finding the derivative of what's inside, since the derivative of $(f(x))^2$ is $2f(x)f'(x)$.
For a final simplification, note that
$$\frac{1}{x(x+3)} = \frac{1}{3} \left(\frac 1x - \frac{1}{x+3}\right),$$
so you'll only ever need to take derivatives of $\frac 1x$ and $\frac {1}{x+3}$ to finish, since the $x-3$ in the numerator of the first fraction will simplify with these to give an integer plus multiples of these terms.
As a general rule, partial fractions will greatly simplify the work required in similar problems.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3707227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
|
Observe that $(x+1)$ divides all quadratics: the original equation is
$$\sqrt{(x+1)(x+7)} + \sqrt{(x+1)(x+2)} = \sqrt{(x+1)(6x+13)},$$
and rearranging we obtain the following:
$$(\sqrt{x+1})(\sqrt{x+7}+\sqrt{x+2}-\sqrt{6x+13})=0.$$
We are doing some trickery with allowing square roots to venture into $\mathbb C$ here, but note that it is all still correct: for the original equation to have solutions in $\mathbb R$, then either $x \geq -1$ (and all the square roots stay safely within $\mathbb R$) or $x \leq -7$ (in which case all of the square roots are of negative values, so the extra factors of $i$ safely distribute out, assuming we use the principal square root).
The first factor yields a solution of $-1$. The second factor gives solutions when
$$\sqrt{6x+13}= \sqrt{x+7}+\sqrt{x+2},$$
and upon squaring both sides gives
$$6x+13=2x+9+2\sqrt{(x+7)(x+2)},$$
which rearranges to
$$\sqrt{(x+7)(x+2)}=2x+2.$$
Square again, solve the quadratic, and test your solutions to finish.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3707727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Prove that if $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, then $V = W_{1}\oplus W_{2}$. (a) Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$ such that $V = W_{1}\oplus W_{2}$. If $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are bases for $W_{1}$ and $W_{2}$, respectively, show that $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$ and $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$.
(b) Conversely, let $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ be disjoint bases for subspaces $W_{1}$ and $W_{2}$, respectively, of a vector space $V$. Prove that if $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, then $V = W_{1}\oplus W_{2}$.
MY ATTEMPT
(a) Let $\mathcal{B}_{1} = \{\alpha_{1},\alpha_{2},\ldots,\alpha_{m}\}$ and $\mathcal{B}_{2} = \{\beta_{1},\beta_{2},\ldots,\beta_{n}\}$ where $\dim W_{1} = m$ and $\dim W_{2} = n$.
Let $v\in V = W_{1}\oplus W_{2}$. Then $v = w_{1} + w_{2}$ where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$.
Consequently, there are scalars $a_{1},a_{2},\ldots,a_{m}$ and $b_{1},b_{2},\ldots,b_{n}$ such that
\begin{align*}
\begin{cases}
w_{1} = a_{1}\alpha_{1} + a_{2}\alpha_{2} + \ldots + a_{m}\alpha_{m}\\\\
w_{2} = b_{1}\beta_{1} + b_{2}\beta_{2} + \ldots + b_{n}\beta_{n}
\end{cases}
\end{align*}
Thence we conclude that
\begin{align*}
v = w_{1} + w_{2} = a_{1}\alpha_{1} + a_{2}\alpha_{2} + \ldots + a_{m}\alpha_{m} + b_{1}\beta_{1} + b_{2}\beta_{2} + \ldots + b_{n}\beta_{n}
\end{align*}
Thus $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ spans $V$. Besides that, $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$. Indeed, if it were not the case, we would have $b\in\mathcal{B}_{1}\cap\mathcal{B}_{2}\subseteq W_{1}\cap W_{2}$ such that $b\neq 0$, which contradicts the fact that $W_{1}\cap W_{2} = \{0\}$.
Finally, let us prove that $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is LI. Indeed, if
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} + d_{1}\beta_{1} + d_{2}\beta_{2} + \ldots + d_{n}\beta_{n} = 0
\end{align*}
then we should have
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} = -d_{1}\beta_{1} - d_{2}\beta_{2} - \ldots - d_{n}\beta_{n}
\end{align*}
which implies that
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m}\in W_{1}\cap W_{2} = \{0\}
\end{align*}
whence we conclude that $c_{1} = c_{2} = \ldots = c_{m} = 0$. Similar reasoning shows that $d_{1} = d_{2} = \ldots = d_{n} = 0$, and the result holds.
(b) Based on the same notation as previously established, let $v\in V$. According to the given assumptions, there are scalars $a_{1},a_{2},\ldots,a_{m}$ as well as $b_{1},b_{2},\ldots,b_{n}$ such that
\begin{align*}
v = a_{1}\alpha_{1} + a_{2}\alpha_{2} + \ldots + a_{m}\alpha_{m} + b_{1}\beta_{1} + b_{2}\beta_{2} + \ldots + b_{n}\beta_{n} = w_{1} + w_{2}
\end{align*}
where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$. Thus $V = W_{1}+W_{2}$. It remains to prove that $W_{1}\cap W_{2} = \{0\}$.
Let us assume that $w\in W_{1}\cap W_{2}$. Then we conclude that
\begin{align*}
w = c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} = d_{1}\beta_{1} + d_{2}\beta_{2} + \ldots + d_{n}\beta_{n}
\end{align*}
Rearranging this relation, it results that
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} - d_{1}\beta_{1} - d_{2}\beta_{2} - \ldots - d_{n}\beta_{n} = 0
\end{align*}
thence $a_{1} = a_{2} = \ldots = a_{m} = b_{1} = b_{2} = \ldots = b_{n} = 0$, and we are done.
Are the provided proofs correct? Is there a neater way to rephrase my arguments? Any comments are appreciated.
|
Another approach without appealing dimensionality.
*
*Suppose that $\mathcal B_1$ is a basis for $W_1$ and that $\mathcal B_2$ is a basis for $W_2$. Since $V = W_1 \oplus W_2$, any vector in $V$ can be uniquely written as a sum of a vector in $W_1$ and a vector in $W_2$; but also, at the same time, every vector in $W_i$ can be uniquely written as a linear combination of vectors in $\mathcal B_i$. In conclusion, every vector in $V$ can be expressed in only one way as a linear combination of vectors in $\mathcal B_1 \cup \mathcal B_2$, and hence, $\mathcal B_1 \cup \mathcal B_2$ is a basis for $V$. Your explanation of why $\mathcal B_1 \cap \mathcal B_2 = \varnothing$ is fine.
*Now, suppose that $\mathcal B_1$ and $\mathcal B_2$ are two disjoint basis, the first one for $W_1$ and the second one for $W_2$. If $\mathcal B_1 \cup \mathcal B_2$ is a basis for $V$, the same argument above works to show that $V = W_1 \oplus W_2$, just be careful.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3708857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Sum $\sum_{n = 1}^{\infty}\left[\frac1n\sin nx + \frac{1}{n^2}\cos nx\right]$ I want to find the following sum by using the complex methods for series ($z = \cos nx + i \sin nx$).
$$
\sum_{n = 1}^{\infty}\left[\frac1n\sin nx + \frac{1}{n^2}\cos nx\right]
$$
Here is my attempt:
$$
S_N = \sum_{n = 1}^{N}\frac1n\sin nx + \sum_{n = 1}^{N}\frac{1}{n^2}\cos nx
$$
I'm done with the first sum:
$$
\sum_{n = 1}^{N}\frac1n\sin nx = \Im \sum_{n = 1}^{N}\frac{z^n}{n} \Rightarrow \lim_{N \to \infty} \sum_{n = 1}^{N}\frac1n\sin nx = -\Im \ln(1 - z)
$$
But I'm stuck with the second one:
$$
\sum_{n = 1}^{N}\frac{1}{n^2}\cos nx = \Re \sum_{n = 1}^{N}\frac{z^n}{n^2}
$$
|
Continue with
\begin{align}
\sum_{n = 1}^{\infty}\frac{\sin nx }n
= -\Im \ln(1 - z)
= \frac i2 [\ln( 1-z) -\ln (1-\bar z)]
=\frac i2\ln(-z)= \frac{\pi-x}2
\end{align}
and use the result to evaluate
$$
\sum_{n = 1}^{\infty}\frac{1- \cos nx }{n^2}=
\sum_{n = 1}^{\infty}\int_0^x \frac{\sin nt }{n}dt
= \int_0^x \frac{\pi-t}{2}dt= \frac{\pi}2x-\frac14x^2
$$
Thus
\begin{align}
\sum_{n = 1}^{\infty}\left(\frac1n\sin nx + \frac{1}{n^2}\cos nx\right)
&= \frac{\pi-x}2 - \frac{\pi}2x+\frac14x^2
+ \sum_{n = 1}^{\infty}\frac{1}{n^2} \\
&=\frac{x^2}4 - \frac{(\pi+1)x-\pi}2 +\frac{\pi^2}6
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3711768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$ I was training for upcoming Olympiads, working on inequalities, and the following inequality came up:
$$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$$ with the obvious delimitations $3b\geq a;\: 3a\geq b.$
I've been pondering the question for quite a while, and tried the using CS among others but didn't find the solution, which, given the level of sophistication the problem should have, surprises me.
Any help would be appreciated.
|
As hinted by Quasi's solution, repeated squaring works for this problem. This approach should definitely be in your bag, especially since it's so easy to get rid of square roots.
If you want to simplify it some, then consider a change of variables: $ x = 3a -b , y = 3b-a$.
This gives us $ a = \frac{3x+y}{8}, b = \frac{ x + 3y } { 8}$, $ a + b = \frac{ x+y} { 2}$ and $ab = \frac{ 3x^2 + 10xy + 3y^2 } { 64}$.
So, we WTS
$\sqrt{2} \sqrt{ x+y } ( \sqrt{x} + \sqrt{y}) \leq \sqrt{ 3x^2 + 10 xy + 3y^2}$
$ \Leftrightarrow 2(x+y) ( x+y + 2 \sqrt{xy} ) \leq 3x^2 + 10xy + 3y^2$
$\Leftrightarrow 4(x+y)\sqrt{xy} \leq x^2 + 6xy + y^2 $
$\Leftrightarrow 0 \leq (\sqrt{x} - \sqrt{y} )^4 $
We have equality iff $ x = y$, or that $ a = b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3715265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed{n\cdot2^{n-2}}.$ Am I on the right track?
|
Let $S$ be the sum in question.
Let $f(x)=(1+x)^n+(1-x)^n$.
Then $f'(1)=2S$.
Now $f'(x)=n(1+x)^{n-1}-n(1-x)^{n-1}$ and so $f'(1)=n 2^{n-1}$.
Therefore, $S=n 2^{n-2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
}
|
Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?
|
$$\text{You wrote: } x=\frac{y+2}{y^2+2y+1} $$
$$x(y^2+2y+1)=y+2$$
$$
xy^2 + ( 2x-1 )y +(x-2)=0 \tag 1
$$
$$
ay^2 + by + c = 0, \quad\text{where } a=x,\, b=(2x-1), \, c = x-2 \tag 2
$$
$$
y = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{1-2x\pm \sqrt{(2x-1)^2 -4x(x-2)}}{2x} \tag 3
$$
except that $(2)$ can be a valid solution of $(1)$ only when $a=x\ne 0.$ When $x=0,$ equation $(1)$ becomes
$$
y+2=0, \text{ so } y=-2.
$$
Thus the domain of the function defined on line $(3)$ excludes only those values of $x$ for which the expression under the radical is negative. So we want
$$
(2x-1)^2 - 4x(x-2)\ge0.
$$
$$
-4x + 1+8x\ge0
$$
$$
x \ge \frac {-1} 4.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
}
|
Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$
I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \sin ^{2} x+\cos ^{2} x=1$
$t^{2}+2-2 t^{2}+3 t \sqrt{1-t^{2}}=0$
$\left(t^{2}+2\right)^{2}=9 t^{2}\left(1-t^{2}\right)$
$t^{4}+4 t^{2}+h=9 t^{2}-9 t^{4}$
$10 t^{4}-5 t^{2}+4=0$
So the number of solution must be 4
P.s- Any other approach will be greatly appreciated!
correct me if I am wrong
|
Well, in a case very unusual for these type of questions the equation can be factored.
$\sin^2 x + 2\cos^2x + 3\sin x \cos x = (\sin x + \cos x)(\sin x + 2\cos x)=0$.
so
we have solutions when $\sin x = -\cos x$ or $\sin x=-2\cos x$. Looking at the unit circle it is clear that these can only occur in the 2nd and 4th quadrants where $\sin$ and $\cos$ are opposite signs. Furthermore in the quadrants the absolute values of one trig function is increasing from $0$ to $1$ and the other is decreasing from $1$ to $0$ so there will be exactly one solution to each equation in each quadrant.
so there are four solutions:
....
We could do what you did. In fact that would be my preferred way of doing it.
!!BUT!!!
*
*When you squared both sides you risked adding extraneous solutions.
*Declaring a fourth degree polynomial has four roots doesn't take into account that maybe there are multiple roots or that there might not be real roots.
*$\cos x \ne \sqrt{1 - \sin^2 x}$. $\cos x=\pm \sqrt {1-\sin^2 x}$
And most importantly.
4)If $t = t_1,t_2,t_3,t_4$ are the four solutions then $\sin x = t_i$ do not have one solution only. If $|t_1| > 1$ then $\sin x=t_i$ has no solution. And if $|t_i| < 1$ then $\sin x = t_1$ has two solutions.
and least importantly
*$-4 - 9 = -13$ not $-5$.
Let's redo your work
$\sin^2 + \cos^2 = 1$ so
$\sin^2 x+ 2cos^2 x + 3\sin\cos x =$
$1 + \cos^2 x + 3\sin \cos x$.
Not sure why you chose $t = \sin x$ rather than $t=\cos x$ but it shouldn't matter.
$1 + (1 - t^2) \pm 3t\sqrt{1-t^2}=0$
$2-t^2 = \pm 3t\sqrt{1-t^2}$ If we square both sides we need to take note that $1-t^2 \ge 0$ or $t^2 \le 1$. On the other hand we don't need to worry about the extraneous solution that the we lose the sign of the RHS because we don't know the sign of the RHS.
$4 - 4t^2 + t^4 = 9t^2(1-t^2)$
$10t^4 - 13t^2 + 4=0$
$t^2 = \frac {13 \pm {169-160}}{20} = \frac {13\pm 3}20= \frac 12, \frac 45$.
So four solutions to $t$: $\pm \frac{\sqrt 2}2; \pm \frac {2\sqrt 5}5$.
But for each $\sin x = t$ there are two values that $x$ could be. So $8$ values?
But no....we have to bear in mind that $\sin^2 x + 2\cos^2 x \ge 0$ so $\sin x\cos x \le 0$ so $\sin x $ and $\cos x$ are opposite signs. (That slipped me by!)
So when we had $2-t^2 = \pm 3t\sqrt{1-t^2}$ we DO know that that the RHS is positive and that we did get extraneous solutions.
Now we have $\sin x = \pm \frac{\sqrt 2}2; \pm \frac {2\sqrt 5}5$ AND $\cos x$ is the opposite sign. There is only one solution for each value.
.....
Moral. Be careful!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3718209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Find the range of $f:x \mapsto a+b\cos x$ The function $f:x \mapsto a+b\cos x$, is defined for $0 \le x\le 2\pi$. Given that $f(0) = 10$ and $f\left(\frac{2}{3}\pi\right) = 1$, find the values of $a$ and $b$, the range of $f$, and the exact value of $f\left(\frac{5}{6}\pi\right)$.
I was able to get the value of a and b given that $f(0) = 10$ and $f\left(\frac{2}{3}\pi\right) = 1$ which gives $a + b\cos (0) = 10$ and $a + b\cos\left(\frac{2}{3}\pi\right) = 1$. Therefore $a = 4$ and $b = 6$.
For the exact value of $f\left(\frac{5}{6}\pi\right)$; Since $a = 4$ and $b = 6$, $f(x) = a + b\cos x \Rightarrow 4 + 6\cos x$, Therefore $f\left(\frac{5}{6}\pi\right) = 4 + 6\cos\left(\frac{5}{6}\pi\right) = 4 + 6(-\frac{\sqrt3}{2})$ = $4-3\sqrt{3}$ or $-1.1962$.
Getting the range of f is what I don't know how to go about, can anyone here offer an explanation or answer? Thanks
|
Your calculation for $a,b$ is correct, and so is the calculation of $f\left(\frac56\pi\right)$.
For calculating the range of $f$, consider the following hint:
*
*For all $x$, you $-1\leq \cos x \leq 1$.
*The bounds above are strict, i.e., there exist values of $x$ for which $\cos x$ reaches values of $-1$ and $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3720457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Find all possible values $m$ such that $f(x)=x^2-5mx+10m-4$ has 2 roots and one of them is twice as the other Consider the polynomial $f(x)=x^2-5mx+10m-4$, find $m$ such that there exist a number $a$ that satisfies $f(a)=f(2a)=0$. This was my attempt:
$f(a)=f(2a)$
$a^2-5ma+10m-4=4a^2-10ma+10m-4$
$a^2-5ma=4a^2-10ma$
$-3a^2+5ma=0$
$a*(-3a+5m)=0$
If $a=0$ then $f(a)=10m-4=0$ and $m=\frac{2}{5}$, else if $-3+5m=0$ then I can't solve.
Is my first answer correct? And if yes, how do I solve for the second one
|
The given equation can be written as
$\begin{align}
f(x)& =x^2-4-5mx+10m\\
&=(x-2) (x+2) -5m(x-2) \\
&=(x-2) (x+2-5m)
\end{align}$
Now, we have two possibilities: $\alpha=2\beta$ or $\beta=2\alpha$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3723019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Value of $\alpha$ for which $x^5+5\lambda x^4-x^3+(\lambda\alpha-4)x^2-(8\lambda+3)x+\lambda\alpha-2=0$ has roots independent of $\lambda$
Consider the equation $$x^5 + 5\lambda x^4 -x^3 + (\lambda \alpha -4)x^2 - (8\lambda +3)x + \lambda\alpha - 2 = 0$$ The value of $\alpha$ for which the roots of the equation are independent of $\lambda$ is _______
My approach: The equation can be rewritten as:
$$\underbrace{(x-2)(x^4 + 2x^3 + 3x^2 + 2x + 1)}_{f(x)} + \lambda\underbrace{(5x^4 + \alpha x^2 -8x + \alpha)}_{g(x)} = 0$$
For this equation to be valid independent of $\lambda$, $f(x) = g(x) = 0$. $f(x)$ has $2$ as one of it's roots. Solving $g(2) = 0$, the value of $\alpha$ comes out to be
$$\alpha = -\frac{64}{5}$$
which is unfortunately not the correct answer. Where is my approach breaking down?
|
The question may be phrased incorrectly, as it is not possible to make the set of all roots independent of $\lambda$. The question I will answer is: For what value of $\alpha$ does the equation have some roots which are independent of $\lambda$?
As demonstrated in the question, $\alpha=-\frac{64}{5}$ is one possibility, which gives the root $x=2$, independent of $\lambda$.
But there is one other possibility that we can find by further factoring: $f(x) = (x-2)(x^2+x+1)^2$. Is there a value of $\alpha$ for which $g(x)$ shares a root with $x^2+x+1$? Setting $x=\omega$ with $\omega^3=1$ and $g(\omega)=0$ gives us $\alpha \omega^2 - 3\omega + \alpha = 0$. Reducing further using $\omega^2+\omega+1 = 0$ gives $-a\omega - 3\omega = 0$ or $a=-3$.
Indeed, we can verify that if $a=-3$, the original equation is divisible by $x^2 + x+1$ regardless of the value of $\lambda$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3723526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Why is $|x|'=\frac{x}{|x|}$? Why is $|x|'=\frac{x}{|x|}$ ?
May someone explain this clearly? If $x>0$ then it should be 1, and if $x<0$ then it's $-1$...
|
You can derive this equation by writing the absolute value as $|x| =\sqrt{x^2}$. From here we see that
$$
\frac{d}{dx} |x| = \frac{d}{dx} \sqrt{x^2} = \frac{1}{2 \sqrt{x^2} } \left( \frac{d}{dx} x^2 \right) = \frac{2x}{2 \sqrt{x^2} } = \frac{x}{\sqrt{x^2} } = \frac{x}{|x|}
$$
where we use the chain rule. Notice that the function $|x|$ is differentiable on $\mathbb{R}\setminus\{0\}$, which means that in our derivation above $x \neq 0$, which is why we can have an $x$ in the denominator.
You can see that this is consistent with the equation
$$
|x|'=\begin{cases}1 & \text{ if } x > 0\\ -1 & \text{ if } x <0\end{cases}
$$
Since if $x>0$, then $|x| =x$, this means that
$$
\frac{x}{|x|} = \frac{x}{x} = 1
$$
And similarly, for $x<0$ we have $|x| = -x$, so we get
$$
\frac{x}{|x|} = \frac{x}{-x} = -\frac{x}{x} = -1
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3726935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
|
Proving that this function all over the positive integer gives us this sequence? Firstly, we have this sequence : $1,1,2,1,2,3,1,2,3,4,...$ which is the sequence of integers $1$ to $k$ followed by integers $1$ to $k+1$. We could say a fractal sequence.
Secondly, we have this formula : $$a_n=\frac{1}{2}(2n+\lfloor\sqrt{2n}+\frac{1}{2}\rfloor-\lfloor\sqrt {2n}+\frac{1}{2}\rfloor^2)$$ where $n\ge1$
$a_1=1$ ; $a_2=1$ ; $a_3=2$ ; $a_4=1$ ; $a_5=2$ ; $a_6=3$ ; $a_7=1$
I don't know for sure but i think this formula gives us this sequence.
How to prove this ?
|
$\newcommand{\bb}[1]{\left( #1 \right)}$
$\newcommand{\f}[1]{\left\lfloor #1 \right\rfloor}$
Key observation: Given $m \in \Bbb{Z}^+$, if:
$$
\sum_{k=1}^m k = \frac{m(m+1)}{2} < n \leq \frac{(m+1)(m+2)}{2} = \sum_{k=1}^{m+1} k
$$
then:
$$
a_n = n - \frac{m(m+1)}{2}
$$
Now rewrite your formula in the following manner:
\begin{align*}
a_n &= n - \frac{1}{2}\bb{\f{\sqrt{2n} + \frac{1}{2}}^2 - \f{\sqrt{2n} + \frac{1}{2}}} \\
&= n - \frac{1}{2}\f{\sqrt{2n} + \frac{1}{2}}\bb{\f{\sqrt{2n} + \frac{1}{2}} - 1} \\
&= n - \frac{1}{2}\bb{\f{\sqrt{2n} - \frac{1}{2}} + 1}\f{\sqrt{2n} - \frac{1}{2}} \\
\end{align*}
Observe that if $\f{\sqrt{2n} - \frac{1}{2}} = m$, then we're done. Thus, it suffices to show that this indeed holds if $\frac{m(m+1)}{2} < n \leq \frac{(m+1)(m+2)}{2}$ for some $m \in \Bbb{Z}^+$. This is because:
\begin{align*}
\f{\sqrt{2n} - \frac{1}{2}} = m &\iff m \leq \sqrt{2n} - \frac{1}{2} < m + 1 \\
&\iff m + \frac{1}{2} \leq \sqrt{2n} < m + \frac{3}{2} \\
&\iff \bb{m + \frac{1}{2}}^2 \leq 2n < \bb{m + \frac{3}{2}}^2 \\
&\iff \frac{1}{2}\bb{m^2 + m + \frac{1}{4}} \leq n < \frac{1}{2}\bb{m^2 + 3m + \frac{9}{4}} \\
&\iff \frac{m(m+1)}{2} + \frac{1}{8} \leq n < \frac{(m+1)(m+2)}{2} + \frac{1}{8} \\
&\iff \frac{m(m+1)}{2} < n \leq \frac{(m+1)(m+2)}{2}
\end{align*}
where the last $\iff$ holds because $\frac{m(m+1)}{2},n,\frac{(m+1)(m+2)}{2}$ are all integers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3727685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Proof of $d^k + (a-d)^k = a[d^{k-1}-d^{k-2}(a-d)+\dots+(a-d)^{k-1}]$ It is stated to me as part of a larger problem that for odd $a > 2$, positive integer $n$, $k=a^n$, and positive integer $d$, we have
\begin{equation}
d^k + (a-d)^k = a[d^{k-1}-d^{k-2}(a-d)+\dots+(a-d)^{k-1}].
\end{equation}
I have tried expanding but I'm still unsure of how the coefficients on the RHS is equivalent to the binomial expansion on the LHS.
|
This is just a different form of the factorization for positive odd $n$ $$x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \cdots + y^{n-1}) = (x+y) \sum_{k=0}^{n-1} (-1)^k x^{n-k-1}y^{k}.$$ This is because
$$\begin{align*}
(x+y) \sum_{k=0}^{n-1} (-1)^k x^{n-k-1} y^k
&= \sum_{k=0}^{n-1} (-1)^k x^{n-k} y^k + \sum_{k=0}^{n-1} (-1)^k x^{n-k-1}y^{k+1} \\
&= x^n + \sum_{k=1}^{n-1} (-1)^k x^{n-k} y^k + \sum_{k=1}^n (-1)^{k-1} x^{n-k} y^k \\
&= x^n + \sum_{k=1}^{n-1} \left( (-1)^k + (-1)^{k-1} \right) x^{n-k} y^k + y^n \\
&= x^n + y^n + \sum_{k=1}^{n-1} 0 \\
&= x^n + y^n.
\end{align*}$$
Your identity follows by setting $n = k$, $x = d$, $y = a-d$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3728148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Geometry Proof to Find Maximum area of $\triangle PIE$
Circle $\omega$ is inscribed in unit square $PLUM,$ and points $I$ and $E$ lie on $\omega$ such that
$U, I,$ and $E$ are collinear. Find, with proof, the greatest possible area for $\triangle PIE.$
I'm not sure if there is a solution possible without trigonometry.
Also, for my diagram in my solution, I'm not sure how to center it. Sorry about that.
|
I would provide a solution without trigonometry, as asked in the OP. Consider the circle as centered in point $(1/2,1/2)$ of a Cartesian plane, so that the square corners are $L(0,0)$, $U(0,1)$, $M(1,1)$, $P(1,0)$. The circle equation is $(x-1/2)^2+(y-1/2)^2=1/4$. Let us call $k$ the slope of a line passing through $U$ and intersecting the circle. Its equation is then $y=kx+1$. By construction, the line intersects the circle only if $k\leq 0$.
The intersection points $I$ and $E$ between the line and the circle are given by the solutions of the system composed by the equations of the line and the circle. These solutions are
$$I\left(\frac{1-k- \sqrt{-2k}}{2(k^2 + 1)}, \frac{k^2+k+2-k\sqrt{-2k} }{2(k^2 + 1)}\right) $$
$$E\left(\frac{1-k+\sqrt{-2k}}{2(k^2 + 1)}, \frac{k^2+k+2+k\sqrt{-2k} }{2(k^2 + 1)}\right) $$
By the standard formula for the distance between points, after some calculations and simplifications we obtain
$$IE=\frac{\sqrt{-2k}}{\sqrt{k^2 + 1}}$$
Now let us consider the perpendicular to $IE$ drawn from $P(1,0)$. This line must have angular coefficient $-1/k$ and has to satisfy $0=-1/k+r$, which implies $r=1/k$. The line has then equation $y=-x/k+1/k$. The coordinates of $X$ are the solutions of the system composed by the two lines. Solving the system, it follows
$$X\left( \frac{1-k}{k^2+1}, \frac{1+k}{k^2+1}\right)$$
and using again the formula for the distance between two points we get
$$PX= \frac{|k+1|}{\sqrt{k^2+1}}$$
Therefore, the area of $\triangle{PIE}$ is
$$ A(\triangle{PIE})\\= \frac{IE \cdot PX}{2}= \frac{1}{2} \cdot \frac{\sqrt{-2k}}{\sqrt{k^2 + 1}} \cdot \frac{|k+1|}{\sqrt{k^2+1}}\\
= \frac{|k+1| \sqrt{-2k}}{2(k^2 + 1)} $$
Taking the derivative we have
$$\frac{(k^3 + 3 k^2 - 3 k - 1)}{(2 \sqrt{-2k} (k^2 + 1)^2)} \,\text{for} \,\,k>-1$$
$$-\frac{(k^3 + 3 k^2 - 3 k - 1)}{(2 \sqrt{-2k} (k^2 + 1)^2)} \,\text{for} \,\,k<-1$$
As expected by the symmetry of the problem with respect to the case $k=-1$, setting these equations equal to zero we obtain that the area function has two maxima in $-2+\sqrt{3}$ and $-2-\sqrt{3}$. Substituting these values in the formula of $\triangle{PIE}$ area we conclude that the maximal area is
$$A_{max}(\triangle{PIE})= \frac{(\sqrt{3} -1) (\sqrt{ 4-2\sqrt{3}} )}{4 (4-2\sqrt{3} )}\\=\frac{1}{4}\,\,\, \left( \text{for}\,\, k=
-2+\sqrt{3}\right) $$
and
$$A_{max}(\triangle{PIE})= \frac{(\sqrt{3} +1) (\sqrt{ 4+2\sqrt{3}} )}{4 (4+2\sqrt{3} )}\\=\frac{1}{4}\,\,\, \left( \text{for}\,\, k=
-2-\sqrt{3}\right) $$
Here is a graph of the area as a function of the slope $k$, as provided by the formula above:
Note that for $k=0$, i.e. the points $I$ and $E$ coincide in $(1/2,1)$ with the midpoint of the upper side of the square, as expected the area formula gives zero. As $k$ decreases, the area increases, achieves its first maximal value of $1/4$ in $k=-2+\sqrt{3}$, and again decreases to zero in $k=-1$ (this is the case where the points $I$, $E$, and $P$ are aligned on the diagonal $UP$). As $k$ further decreases, the area increases again, achieves its second maximal value of $1/4$ in $k=-2-\sqrt{3}$, and progressively decreases tending to zero for $k\rightarrow -\infty$ (this is the case in which the points $I$ and $E$ coincide in $(0,1/2)$ with the midpoint of the left side of the square).
Lastly, note that, if we call $\alpha$ the angle $\angle{MUI}$, the slopes of $-2+\sqrt{3}$ and $-2+\sqrt{3}$ correspond to the the values $\alpha=\pi/12=15°$ and $\alpha=5\pi/12=75°$ easily obtained by the trigonometric approach.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3730695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:$ If $a+b+c \neq 0$ then $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c} = \frac {\left [\{xb+(1-x)c\} + \{xc + (1-x)a\} + \{xa+(1-x)b\} \right]} {a+b+c} =1.$$ Therefore $$x = \frac {a-c}{b-c} = \frac {b-a} {c-a} = \frac {c-b} {a-b}.$$
Comparing the first two expressions of $x$ and simplifying we get \begin{align*} a^2+b^2+c^2 - ab - bc -ca & = 0 \\ \implies (a-b)^2+(b-c)^2 +(c-a)^2 & = 0. \end{align*} Therefore we have $a=b=c.$ But then $x$ would be an indeterminant form. Does it imply that $a+b+c = 0$? How to proceed further? Any help will be highly appreciated. Thank you very much.
|
if $a=b=c$ is not true one can multiply three expressions of $x$ in from OP's thisrd step of $x$ to get $x^3=-1$. So $a+b+c \ne 0$ from OP's second equatiom. Hence either $x^3=-1$ or $a=b=c$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3733834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
Second system of equations I've solved that system and recieved $y=\frac{1}{4}$, $x = -\frac{4}{5}$ but there are one more pair of $(x, y)$ in book and I don't know what I have to do to find its.
\begin{cases} 2x - 3xy + 4y=0 \\ x + 3xy -3x = 1 \end{cases}
In book there are two answers and the second answer is $(1, -2)$.
I've reached $y=\frac{1}{4}$, $x = -\frac{4}{5}$ through that way:
\begin{array}{lcl}2x - 3xy + 4y = 0\\2x - 3xy = -4y\\x(2 - 3y) = -4y\end{array}
2)
\begin{array}{lcl}x + 3xy - 3x = 1\\-2x + 3xy = 1\\-x(2 - 3y) = 1\end{array}
Get new system:
\begin{cases} x(2 - 3y) = -4y \\ -x(2 - 3y) = 1 \end{cases}
Then I divide first to second and get
\begin{array}{lcl}-1 = -4y\\y = \frac{1}{4}\end{array}
Next solve one of equation and get \begin{array}{lcl} x = -\frac{4}{5}\end{array}
But how to get $x = 1, y = -2$?
|
In case you made a typo and the 2nd equation is actually $x+3xy-3y=1$, that equation can be factored:
$$
x-1 + 3y(x-1) = 0
$$
$$(x-1)(3y+1)=0$$
Which means that either $x=1$ or $y=-\frac{1}{3}$
Substitute that into the first equation to get the 2 pairs.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3735764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Find the cardinality of $\big\{(x,y,z)\mid x^2+y^2+z^2= 2^{2018}, xyz\in\mathbb{Z} \big\}$.
What is the cardinality of set $\big\{(x,y,z)\mid x^2+y^2+z^2= 2^{2018}, xyz\in\mathbb{Z} \big\}$?
Since I have very limited knowledge in number theory, I tried using logarithms and then manipulating the equation so that we get $$10^{2018}+2=x^2+y^2+z^2.$$
Then setting one of $x,y,z$ equal to $\sqrt{2}$ we find all values of $x$ and $y$ where $$2x^2+y^2=10^{2018}.$$
Finally we use combinatorics to get the required answer.
However this led to no-where.
What is the correct way to solve this problem?
|
For $n \in \mathbb N$, consider the equation
$$ x^2 + y^2 + z^2 = 2^n $$
where $x,y,z$ are integers. Since $x \mapsto -x$, $y \mapsto -y$, $z \mapsto -z$ does not change the equation, we may assume $x,y,z \ge 0$. We may henceforth suppose $x \ge y \ge z$.
Note that there is no solution when $n=1$.
Suppose $n \ge 2$. Since $x^2+y^2+z^2$ is even, exactly one of $x,y,z$ is even, or all three are even. The first of these cases is ruled out since $a^2 \equiv 0\pmod{4}$ if $a$ is even and $a^2 \equiv 1\pmod{4}$ when $a$ is odd. Therefore, $x,y,z$ are all even.
Writing $x=2x_1$, $y=2y_1$, $z=2z_1$ gives
$$ x_1^2 + y_1^2 + z_1^2 = 2^{n-2}. $$
If $n-2=1$, there is no solution. If $n-2 \ge 2$, we repeat the above argument to arrive at the equation
$$ x_m^2 + y_m^2 + z_m^2 = 2^e, $$
where $e=0\:\text{or}\:1$.
The only solution in the case $e=0$ is $x_m=1$, $y_m=z_m=0$. There is no solution in the case $e=1$. From $x=2x_1=2^2x_2=\ldots=2^mx_m$, etc., we get $x=2^m$ when $n=2m$ is even, and $y=z=0$. There is no solution when $n$ is odd.
We conclude that the equation $x^2+y^2+z^2=2^n$ has no solution when $n$ is odd, and that the only solutions when $n$ is even are $(x,y,z)=\pm(2^{n/2},0,0)$, and its permutations, giving a total of six solutions. $\blacksquare$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3738579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove by mathematical induction, that $81\times 3^{2n} - 2^{2n}$ is divisible by $5$ where $n \in \mathbb{Z}^+$
For all $k$, the equation I came up with is $3^{4+2k} = 5m + 2^{2k}$ where $m$ is a positive integer.
For all $k+1$, the expression is $3^{6+2k} - 2^{2k+2}$.
I tried to plug in the first equation to reach an expression that can be expressed with the integer $5$ taken common, but I am unable to figure out the required manipulation of the expressions to reach the result.
|
$$81\cdot3^{2n}-2^{2n}=5k\implies
\\\begin{align}81\cdot3^{2(n+1)}-2^{2(n+1)}=81\cdot9\cdot3^{2n}-4\cdot2^{2n}&=9\cdot(81\cdot3^{2n}-2^{2n})+5\cdot2^{2n}
\\&=9\cdot5k+5\cdot2^{2n}
\\&=5k'.\end{align}$$
Shorter:
$$81\cdot3^{2n}-2^{2n}=9^n-4^n\mod 5\implies 9^{n+1}-4^{n+1}=9\cdot9^n-4\cdot4^n=4(9^n-4^n)\mod5.$$
Yet shorter:
$$81\cdot3^{2n}-2^{2n}=9^n-4^n=4^n-4^n\mod 5.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3740090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$
My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{2 x}{3 y}$
$2 x^{2}+3 y^{2}=1$
What to do next? Any suggestion or Hint would be greatly appreciated!
|
You can use the method of Lagrange multipliers. If $f(x,y)=x+y$, then, for every $(x,y)\in\Bbb R^2$, $\nabla f(x,y)=(1,1)\ne(0,0)$. Therefore, the maximum of $f$ can only be attained at the boundary of the disk, that is, when $2x^2+3y^2=1$. So, solve the system$$\left\{\begin{array}{l}1=4\lambda x\\1=6\lambda y\\2x^2+3y^2=0.\end{array}\right.$$The only solutions are $(x,y)=\pm\left(\sqrt{\frac3{10}},\sqrt{\frac2{15}}\right)$. So, the maximum is attained at $\left(\sqrt{\frac3{10}},\sqrt{\frac2{15}}\right)$ and that maximum is $\sqrt{\frac3{10}}+\sqrt{\frac2{15}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3740226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
}
|
Proof for the general formula for $a^n+b^n$. Based on the following observations. That is
$$a+b = (a+b)^1 \\ a^2+b^2 = (a+b)^2-2ab \\ a^3+b^3 = (a+b)^3-3ab(a+b) \\ a^4+b^4= (a+b)^4-4ab(a+b)^2+2(ab)^2\\ a^5+b^5
= (a+b)^5 -5ab(a+b)^3+5(ab)^2(a+b)\\\vdots$$
I came to make the following conjecture as general formula.
$$ a^n +b^n =\sum_{k=0}^{n-1}(-1)^k \frac{n\Gamma(n-k)}{\Gamma(k+1)\Gamma(n-2k+1)}(a+b)^{n-2k}(ab)^k $$ where $\Gamma(.) $ is gamma function.
I tried up proving the result using binomial theorem $\displaystyle (a+b)^n=\sum_{r=0}^n a^{n-r}b^r$ for positive integers $a,b$ however, I didn't find any elegance in the work. So in the expect of some beautiful proofs, I wish to share general formula here.
Thank you
|
$a$ and $b$ are roots of $x^2=(a+b)x-ab$. Therefore, $a^{n+2}=(a+b)a^{n+1}-(ab)a^n$ and analogously for $b$.
Let $p_n=a^n+b^n$. Then $p_{n+2}=(a+b)p_{n+1}-(ab)p_n$ is a simple recurrence. The initial values are of course $p_0=2$ and $p_1=a+b$.
This recurrence is a special case of Newton's identities.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3741313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Definite integral $\int_{-\infty}^\infty \frac{\log(x^2+a^2)}{(x-ib)^2} dx$ How can I evaluate the integral
$$\int_{-\infty}^\infty \frac{\log(x^2+a^2)}{(x-ib)^2} dx?$$
Here $a, b$ are positive real constants. When I plug this expression in MATLAB, I obtain the answer as
$$ - \frac{\mathrm{log}\!\left(x - a\, \mathrm{i}\right)\, \mathrm{i}}{a - b} - \frac{\mathrm{log}\!\left(a^2 + x^2\right)\, \mathrm{i}}{b + x\, \mathrm{i}} + \frac{\mathrm{log}\!\left(x + a\, \mathrm{i}\right)\, \mathrm{i}}{a + b} + \frac{b\, \mathrm{log}\!\left(x - b\, \mathrm{i}\right)\, 2\, \mathrm{i}}{a^2 - b^2}$$
for the indefinite integral. However I have a problematic complex logarithm, which is ambiguous depending on the branch cut. Furthermore, the MATLAB does not give an answer of the definite integral for the integration range $(-\infty, \infty)$.
This integral is motivated from physics, especially when computing a Feynman diagram.
|
Let $f \colon (0,\infty)^2 \to \mathbb{C}, \, f(a,b) = \int_{-\infty}^\infty \frac{\log(a^2+x^2)}{(x-\mathrm{i} b)^2} \, \mathrm{d} x$. We can get rid of $a$ by letting $x = a t$ and using the result $\int_{-\infty}^\infty \frac{\mathrm{d} t}{(t-\mathrm{i} c)^2} = 0$ for $c > 0$:
$$ f(a,b) = \frac{1}{a} \int \limits_{-\infty}^\infty \frac{2 \log(a) + \log(1+t^2)}{\left(t - \mathrm{i} \frac{b}{a} \right)^2} \, \mathrm{d} t = \frac{1}{a} \int \limits_{-\infty}^\infty \frac{\log(1+t^2)}{\left(t - \mathrm{i} \frac{b}{a} \right)^2} \, \mathrm{d} t \, . $$
Integration by parts then yields
\begin{align}
f(a,b) &= \frac{2}{a} \int \limits_{-\infty}^\infty \frac{t}{(1+t^2)\left(t - \mathrm{i} \frac{b}{a}\right)} \, \mathrm{d} t = \frac{2}{a} \int \limits_{-\infty}^\infty \frac{t \left(t + \mathrm{i} \frac{b}{a}\right)}{(1+t^2)\left(\frac{b^2}{a^2} + t^2\right)} \, \mathrm{d} t \\
&= \frac{2}{a} \int \limits_{-\infty}^\infty \frac{t^2}{(1+t^2)\left(\frac{b^2}{a^2} + t^2\right)} \, \mathrm{d} t \, ,
\end{align}
since the imaginary part of the integrand is an odd function. For $a \neq b$ we can now use partial fractions to compute the remaining integral, while for $a = b$ integrating by parts once more does the trick. The final result in either case is
$$ f(a,b) = \frac{2}{a} \frac{\pi}{1 + \frac{b}{a}} = \frac{2\pi}{a+b} \, . $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3742239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Locus of the circumcenter of triangle formed by the axes and tangent to a given circle.
A circle centered at $(2,2)$ touches the coordinate axes and a straight variable line $AB$ in the first quadrant, such that $A$ lies the $Y-$ axis, $B$ lies of the $X-$ axis and the circle lies between the origin and the line $AB$. Find the locus of the circumcenter of the triangle $OAB$, where $O$ denotes the origin.
Answer: $xy=x+y+\sqrt{x^2+y^2}$
I was able to solve this question with a very lengthy approach outlined below:
Since $\Delta OAB$ is always right angled at $O$, it's circumcenter will be the midpoint of the segment $AB$.
Also, the equation of the given circle would be
$$(x-2)^2+(y-2)^2=4$$
Then I considered the question of the line $AB$ to be
$y+mx=c$, where $m$ is positive number.
Then I used the fact that this line is tangent to the given circle, i.e. the perpendicular distance of the line from the point $(2,2)$ is $2$ units to obtain a quadratic equation in $c$:
$$c^2-4(1+m)c+8m=0$$
Now, this yielded two values for $c$ of which one has to be rejected because in that case the line was trapped between the circle and the origin.
Thus, $c=2+2m+2\sqrt{1+m^2}$.
From here, the coordinates of the circumcenter required are:
$\left(\frac{1+m+\sqrt{1+m^2}}{m},1+m+\sqrt{1+m^2}\right)$.
Luckily, this was on a MCQ test and all the options had the terms $xy$, $x+y$ and $\sqrt{x^2+y^2}$ present. So I could evaluate these values and then see which option is correct.
But the above method is far too lengthy, and I'm looking for a shorter method, if it exists.
Please note that the average time for a question in the test was nearly two to three minutes.
Thanks a bunch!
|
The first thing to note is that $\triangle OAB$ is always a right triangle, and its incenter is always $(2,2)$. Thus, hypotenuse $AB$ is always a diameter of the circumcircle, and the circumcenter is therefore the midpoint of $AB$. If the line passing through $AB$ has equation $$\frac{x}{a} + \frac{y}{b} = 1,$$ then the $x$-intercept is $(a,0)$, and the $y$-intercept is $(0,b)$. The area enclosed by the triangle may be calculated in two ways: $$|\triangle OAB| = \frac{ab}{2} = rs$$ where $r = 2$ is the inradius and $$s = \frac{1}{2}\left(a + b + \sqrt{a^2 + b^2}\right)$$ is the semiperimeter. Consequently, $$\frac{ab}{2} = a+b+\sqrt{a^2+b^2}.$$ Since $(x,y) = (a/2, b/2)$ is the circumcenter, we obtain the relationship $$2xy = 2x + 2y + \sqrt{(2x)^2+(2y)^2},$$ or $$xy = x + y + \sqrt{x^2 + y^2},$$ as claimed.
The implicit formula for the locus admits a natural parametrization using the angle $\theta$ formed by the ray from the origin to $(x,y)$ and the positive $x$-axis: $$(x,y) = \left(1 + \tan \left(\frac{\theta}{2} + \frac{\pi}{4}\right), 1 + \cot \frac{\theta}{2} \right), \quad 0 < \theta < \frac{\pi}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3744657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$
For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$
My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}$$
Thus $$\sum \frac{a+c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
So it suffices to show that $$6\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow 3\sqrt[3]{abc}\ge a+b+c$$
Which is clearly wrong. :"(
Thank you very much.
|
Due to homogeneity, assume that $abc = 1$.
Let $p = a+b+c, q = ab+bc+ca, r=abc=1$.
We need to prove that $p\cdot \frac{q}{r} + 3 \ge 4 \cdot \frac{p}{\sqrt[3]{r}}$
or $pq + 3 \ge 4 p$.
Since $q^2 \ge 3pr$, it suffices to prove that
$p \cdot \sqrt{3p} + 3 \ge 4p$ or $\frac{1}{3}(\sqrt{3p} - 3)(3p - \sqrt{3p} - 3)\ge 0$
which is true since $p\ge 3$. We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3745912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Ambiguity with $\sin 2x$ and $\cos 2x$ squaring So basically I was learning about trigonometric functions of multiples of $x$, and they were defined as
$$\sin 2x= 2\sin x\cos x$$
and
$$\cos 2x= \cos^2 x- \sin^2 x$$
Now, I wanted to work from $\sin 2x= 2\sin x\cos x$ and, using $\sin^2 2x+ \cos^2 2x=1$, get the formula for $\cos 2x$
Then, I have
$$\sin^2 2x= 4\sin^2 x\cos^2 x$$
$$1-\sin^2 2x= 1-4\sin^2 x\cos^2 x$$
$$\cos^2 2x=1-4\sin^2 x(1-\sin^2 x)$$
$$\cos^2 2x=1-4\sin^2 x+4\sin^4 x$$
$$\cos^2 2x= {(2\sin^2 x-1)}^2$$
$$\pm\cos 2x=2\sin^2 x-1$$
Why do I get this ambiguity of $\pm$?? This is obviously wrong
This further confuses me because, if we continue from the formula of $\cos 2x$, we get
$$\cos 2x=\cos^2 x-\sin^2 x$$
$$\cos 2x=2\cos^2 x-1$$
$$1+\cos 2x= 2\cos^2 x$$
$$\frac{1+\cos 2x}{2}=\cos^2 x$$
$$\cos x=\pm {(\frac{1+\cos 2x}{2})}^{1/2}$$
And here, all of a sudden, its correct to have a $\pm$??
PS: I know I can derive $\cos 2x$ by $\sin(\pi/2+2x)$, I was just curious why this method of squaring doesnt work.
|
Unnecessarily squaring produces ambiguities. For example, if $x=1$, then starting with the squared equation as follows produces ambiguity
\begin{align*}
x^2&=1^2=1\Rightarrow x=\pm 1
\end{align*}
Unnecessarily squaring $\cos 2x=\cos(x+x)=\cos^2x-\sin^2x=1-\sin^2x-\sin^2x=1-2\sin^2x$ results similarly.
As for $\cos x=\pm \sqrt{\frac{1+\cos 2x}{2}}$, there is no ambiguity. For $x=\frac{\pi}3$, $\cos x=\pm \sqrt{\frac{1+\cos 2x}{2}}=+\frac 12$, whereas for $x=\frac{2\pi}3$, $\cos x=\pm \sqrt{\frac{1+\cos 2x}{2}}=-\frac 12$ since $\cos^2 x$ appearing in $\cos 2x=2\cos^2 x-1$ is $\frac12$ for both $x=\frac{\pi}3,\ \frac{2\pi}3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3748275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
total differentiability of $\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$ if the denominator equals $0$ Let $f:\mathbb{R}^3\to\mathbb{R}$ be given by $f(x,y,z)=\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$ if the denominator is not equal to $0$ and otherwise by $f(x,y,z)=0$.
Find all the points where $f$ is differentiable.
$f$ is surely differentiable when the denominator is not equal to zero so we only need to consider the case that $(x^2+y^2)(y^2+z^2)=0$.
However, even if we can show that the partial derivatives of $f$ are not continuous at those points, that still doesn't imply that $f$ is not differentiable there. How do I continue from here?
|
Choose $(x,y,z) \neq 0$ and look at $f((tx,ty,tz)) = t^3\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$. From this, if $f$ is differentiable, we would expect that
$Df(0) = 0$.
Note that ${x^2 \over x^2+y^2} \le 1$ and ${z^2 \over z^2+y^2} \le 1$ so
$|f((x,y,z))| \le |xz^2|$.
Note that $|xz| \le {1 \over 2} (x^2+z^2)$.
Suppose $|z| \le 2$, then we have $|f((x,y,z))| \le x^2+z^2 \le x^2+y^2+z^2= \|(x,y,z)\|^2$.
Hence $f$ is (Frechet) differentiable at $0$ with derivative $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3748721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
find the coefficient of $x^5$ in $(1+2x-3x^2)^6$ I know that I should first factor the expression within the brackets to $-(3x+1)^6(x-1)^6$. However, after that I do not know what to do.
|
We can also apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ of an expression.
We obtain
\begin{align*}
\color{blue}{[x^5]}&\color{blue}{(1+2x-3x^2)^{6}}\\
&=[x^5]\sum_{j=0}^{6}\binom{6}{j}(-3x^2)^j(1+2x)^{6-j}\tag{1}\\
&=\sum_{j=0}^2\binom{6}{j}(-3)^j[x^{5-2j}]\sum_{k=0}^{6-j}\binom{6-j}{k}(2x)^k\tag{2}\\
&=\sum_{j=0}^2\binom{6}{j}(-3)^j\binom{6-j}{5-2j}2^{5-2j}\tag{3}\\
&=\binom{6}{0}(-3)^0\binom{6}{5}2^5+\binom{6}{1}(-3)^1\binom{5}{3}2^3+\binom{6}{2}(-3)^2\binom{4}{1}2^1\\
&=192-1\,440+1\,080\\
&\,\,\color{blue}{=-168}
\end{align*}
Comment:
*
*In (1) we apply the binomial theorem once to $((1+2x)-3x^2))^6$.
*In (2) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$ and we set the upper index of the outer sum to $2$, since other terms do not contribute. We also apply the binomial theorem once again.
*In (3) we select the coefficient of $x^{5-2j}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3749351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Continuity of $a^x+b$ with $a, b \in \mathbb R$ Let $a,b \in \mathbb{R}$ with $a > 0$. find $a$, $b$ so the function would be continuous
$$
f(x) = \begin{cases} a^x + b, & |x|<1 \\
x, & |x| \geq 1 \end{cases}
$$
I got $b = -a^x+x$ as my answer, but I'm unsure.
|
Since $f(x) = a^x + b$ will be continuous on $|x| < 1$ for $a > 0$, we only need to match up this portion of $f$ with it's definition on $|x| \geq 1$ at the endpoints $x = \pm 1$. Evidently, $f(-1) = -1$ and $f(1) = 1$. So we need $a^{-1} + b = -1$ and $a + b = 1$. Using the latter gives $b = 1 -a$, so substitution yields:
$$
\frac{1}{a} + 1 - a = -1.
$$
This becomes:
$$
a^2 -2a -1 = 0.
$$
A quick application of the quadratic formula yields $a = 1 \pm \sqrt{2}$, and we can discard $a = 1 - \sqrt{2}$ since we require $a > 0$. Thus,
$$
a = 1 + \sqrt{2},
\; \; \; \; \; \;b = -\sqrt{2}.
$$
Indeed, the function:
$$
f(x) = \begin{cases} (1+\sqrt{2})^x - \sqrt{2} & |x| < 1, \\
x & |x| \geq 1, \end{cases}
$$
is continuous, as shown below:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3750136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=\frac1{2\sqrt2}$$
$$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac1{2\sqrt2}$$
$$\alpha-\frac{\pi}{4}=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)$$
I calculated the value of $\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)\approx 20.705^\circ$,
so I got $\alpha\approx 45^\circ+20.705^\circ=65.705^\circ$
I calculated
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{1}{\sin^365.705^\circ}-\frac{1}{\cos^3 65.705^\circ}\approx -13.0373576$$
My question: Can I find the value of above trigonometric expression without using calculator? Please help me solve it by simpler method without solving for $\alpha$. Thanks
|
$$\frac{1}{\sin^3\alpha} -\frac{1}{\cos^3 \alpha}=\frac{\cos^3\alpha-\sin^3\alpha}{\sin^3\alpha \cos^3 \alpha}= -\frac{(\sin\alpha-\cos\alpha)(1+\sin\alpha\cos \alpha)}{(\sin \alpha \cos\alpha)^3}$$ Now,$$ (\sin \alpha -\cos \alpha )^2 =\frac 14 \implies 1-2\sin\alpha\cos\alpha =\frac 14 \\\implies \sin\alpha \cos\alpha =\frac 38$$ Just plug in the values of $\sin \alpha-\cos \alpha$ and $\sin \alpha\cos \alpha$ to finish.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
}
|
Compute $ \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $ Find $\displaystyle \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $
I tried to do like this :
Let $A=\displaystyle \lim_{x \to\frac{\pi}{2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $
Then $\ln A=\lim_{x \to\frac{\pi} {2}} \cos^2 x \ln\{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}$
which implies
$$\ln A = \lim_{x \to\frac {\pi}{2}} \frac{1^{\sec^2 x} + 2^{\sec^2 x}+ \cdots + n^{\sec^2 x}}{\sec^2 x}$$
I'm stuck here. If I am on the right way please guide me to reach conclusion. Otherwise please describe the actual way. Thanks in advance.
|
Let $f_n(x)$ be the sequence given by
$$f_n(x)=\left(\sum_{k=1}^n k^{\sec^2(x)}\right)^{\cos^2(x)}$$
Then, we have
$$\begin{align}
f_n(x)&=\left(n^{\sec^2(x)}\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
&=n\left(\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
\end{align}$$
For any fixed $n\ge1$
$$\lim_{x\to\pi/2}(k/n)^{\sec^2(x)}=\begin{cases}0&1\le k<n\\\\1&,k=n\end{cases}$$
Therefore, we have
$$\lim_{x\to \pi/2}f_n(x)=n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3752948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
|
In fact, substitution makes it very easy. Starting with
$$\int \frac{u^3}{(u^2 + 1)^3}\ du$$
take $v = u^2 + 1$, then $dv = 2u\ du$ so $u\ du = \frac{1}{2} dv$. Then we get
$$\begin{align}
\int \frac{u^3}{(u^2 + 1)^3}\ du &= \int \frac{u^2 \cdot u}{(u^2 + 1)^3}\ du\\
&= \int \frac{u^2 \cdot \overbrace{(u\ du)}^{\frac{1}{2}\ dv}}{(u^2 + 1)^3}\\
&= \frac{1}{2} \int \frac{u^2\ dv}{v^3}\end{align}$$
Now since $v = u^2 + 1$, we have $u^2 = v - 1$ and
$$\begin{align}
\int \frac{u^3}{(u^2 + 1)^3}\ du &= \frac{1}{2} \int \frac{v - 1}{v^3}\ dv\end{align}$$
which is now easy.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 5
}
|
Prove that a polynomial ring is integrally closed
Let $V \subseteq {\mathbb{A}}^2_{\mathbb{C}}$ be the curve defined by $x^2-y^2+x^3=0$, and let $\mathbb{C}\left [ V \right ]$ the coordinate ring of $V$. Let $\Theta :=\bar{y}/\bar{x} \in \mathbb{C}\left ( V \right )$. I must show that the ring $B:=\mathbb{C}\left [ V \right ]\left [ \Theta \right ]$ is a UFD.
I could show that this ring is a FD, because it is noetherian, but I am not sure how to prove that the factorizations are unique. Any help would be appreciated. I also tried to show that $B$ is isomorphic to a UFD, but i am not sure what domain would be suitable for this argument.
(Note: I need to prove that $B$ is a UFD to say that $B$ is integrally closed.)
|
Let us show that the $\Bbb{C}[V][\Theta]$ is in fact generated by $\Theta$ as a $\Bbb{C}$-subalgebra of $\Bbb{C}(V).$
Define a morphism $\phi$ as follows:
\begin{align*}
\phi : \Bbb{C}[x,y]&\to\Bbb{C}[t]\\
x&\mapsto t^2 - 1,\\
y&\mapsto t^3 - t.
\end{align*}
It is not difficult to check that this factors through the quotient map $\Bbb{C}[x,y]\to\Bbb{C}[V].$
Now, clearly $\phi$ is a surjection onto $\Bbb{C}[t^2 - 1,t^3 - t],$ so that $\Bbb{C}[x,y]/\ker\phi\cong\Bbb{C}[t^2 - 1,t^3 - t].$ We need to prove that $\ker\phi = (x^3 + x^2 - y^2)$. To do so, note that both $\ker\phi$ and $(x^3 + x^2 - y^2)$ are prime. Since $\ker\phi$ is not maximal ($\Bbb{C}[x,y]/\ker\phi$ is visibly not a field), it is properly contained in some maximal ideal $\mathfrak{m}.$ This gives us a chain of prime ideals
$$
(0)\subsetneq (x^3 + x^2 - y^2)\subseteq \ker\phi\subsetneq\mathfrak{m}.
$$
But, $\dim\Bbb{C}[x,y] = 2,$ so that we must have $(x^3 + x^2 - y^2) = \ker\phi.$
Thus, we get an isomorphism
$$
\Bbb{C}[x,y]/(x^3 + x^2 - y^2)\cong\Bbb{C}[t^2 - 1,t^3 - t],
$$
and $y/x = \Theta$ in the fraction field of the left hand side corresponds to $t$ in the fraction field of the right hand side, because $x = \Theta^2 - 1$ and $y = \Theta^3 - \Theta.$ Now, it is easy to see the result, as $\Bbb{C}[V][\Theta]\cong\Bbb{C}[t^2 - 1,t^3 - t][t] = \Bbb{C}[t].$ This implies that $\Bbb{C}[V][\Theta] = \Bbb{C}[\Theta]$ (and that $\Theta$ satisfies no relations over $\Bbb{C}$) as claimed initially.
Edit: As user26857 notes, the initial solution I presented (below) is not totally rigorous -- we need some condition on $x$ and $y$ to guarantee that $R[\frac{y}{x}]\cong R[T]/(xT - y).$ In fact, it isn't true that $\Bbb{C}[V][T]/(xT-y)\cong\Bbb{C}[V][y/x]$: the ideal $(xT-y)$ should be $(xT -y, T^2 - x - 1)$ -- this second relation is implicitly assumed and explicitly used. What is below can be made rigorous, either by justifying that the kernel of $\Bbb{C}[V][T]\to\Bbb{C}[V][y/x]$ is precisely $(xT -y, T^2 - x - 1),$ or by writing $x$ and $y$ in terms of $\Theta$ and justifying that $\Theta$ satisfies no additional relations.
First, note that $\Bbb{C}[V]\cong\Bbb{C}[x,y]/(x^3 + x^2 - y^2)$ and observe that $\Theta^2 = \frac{y^2}{x^2} = \frac{x^3 + x^2}{x^2} = x + 1.$
Now, using the fact that $x = \Theta^2-1,$ we find
\begin{align*}
\Bbb{C}[V][\Theta]&\cong(\Bbb{C}[x,y]/(x^3 + x^2 - y^2))[\Theta]/(x\Theta - y)\\
&= \Bbb{C}[x,y,\Theta]/(x\Theta - y,x^3 + x^2 - y^2)\\
&= \Bbb{C}[y,\Theta]/((\Theta^2 - 1)\Theta - y,(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2)
\end{align*}
However, it is now clear that $y = \Theta^3 - \Theta,$ and hence that
\begin{align*}
(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2 &=(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - (\Theta^3 - \Theta)^2\\
&= (\Theta^2 - 1)^2(\Theta^2 - 1 + 1) - (\Theta^3 - \Theta)^2\\
&= \Theta^2(\Theta^2 - 1)^2 - (\Theta^3 - \Theta)^2\\
&= 0.
\end{align*}
As such, we find that $$((\Theta^2 - 1)\Theta - y,(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2) = ((\Theta^2 - 1)\Theta - y),$$ so that
\begin{align*}
\Bbb{C}[V][\Theta]&\cong \Bbb{C}[y,\Theta]/((\Theta^2 - 1)\Theta - y,(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2)\\
&=\Bbb{C}[y,\Theta]/((\Theta^2 - 1)\Theta - y)\\
&=\Bbb{C}[\Theta^3 - \Theta,\Theta]\\
&= \Bbb{C}[\Theta].
\end{align*}
A polynomial ring in one variable over a field is clearly a UFD.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3755115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
For the first question, I tried to work it out with algebra; I solved for x through the equation given, then multiplied it by y and I got the value of $\frac{1}{xy} = 2005\left(\frac{1}{y-2005y^2}\right) $. Then I tried proving that $\frac{1}{y-2005y^2} =\frac{1}{x} + \frac{1}{y} $ but I failed at this.
|
For the second part of your question,
For $y \neq x$,
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l = \left(\frac{y}{y-x} - \frac{x}{y-x}\right) - \frac{y-x}{y} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l= -(y-x)\left(\frac{1}{y} + \frac{1}{x}\right) + \frac{(y+x)(y-x)}{xy} + 3$$
$$\implies l = (y-x)\left(\frac{y+x}{xy} - \frac1y - \frac1x\right) + 3$$
$$\implies l =3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3756456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Extreme points of a function at domain ends Consider the following function: $f(x) = x\sqrt{9-x^2}$
$\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad $
$f'(x) = \frac{-2x^2+9}{\sqrt{9-x^2}}$ and $D(f) = [-3,3]$ therefore the critical points of the function are $x_{c_i} = \left\{ -3, -\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}, 3 \right\}$
Apparently the points $\{-\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}\}$ are global minumum and global maximum respectively.
But what about the domain ends $\{-3, 3\}$? Are they considered to be saddle points, local minimums, or local maximums and why?
|
If $D(f)=[-3,3]$, then $f( \pm 3)=0.$
Let $x \in (-3,0)$, then $f(x)<0=f(-3)$, henc $f$ has a local maximum at $x=-3.$
Let $x \in (0,3)$, then $f(x)>0=f(3)$, henc $f$ has a local minimum at $x=3.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3757621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
|
$$\int \frac{x^3}{\left(4x^2+9\right)^{\frac{3}{2}}}dx$$
if $4x^2+9=t^2$ as in the answer of the user @Anton Vrdoljak you have:
$$=\int \frac{t-9}{32t^{\frac{3}{2}}}\ dt=\frac{1}{32}\cdot \int \left(\frac{1}{t^{\frac{1}{2}}}-\frac{9}{t^{\frac{3}{2}}}\right)dt=\frac{1}{32}\left(2t^{\frac{1}{2}}-\left(-\frac{18}{t^{\frac{1}{2}}}\right)\right)+k, \quad k\in\Bbb R$$
And at the end:
$$=\frac{1}{32}\left(2\left(4x^2+9\right)^{\frac{1}{2}}-\left(-\frac{18}{\left(4x^2+9\right)^{\frac{1}{2}}}\right)\right)+k, \quad k\in\Bbb R$$
$$=\frac{2x^2+9}{8\left(4x^2+9\right)^{\frac{1}{2}}}+k, \quad k\in\Bbb R$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
}
|
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.
|
After multiplying the numerator and denominator of the expression in the limit by $$\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}$$, I get $$\lim_{x \to 0} \frac{1+x\sin(x)-\cos(x)}{x\tan(x) \left(\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}\right)}$$
It is clear that $$\lim_{x \to 0}\left( \sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)} \right) = \sqrt{1 + 0 \sin(0)} + \sqrt{\cos(0)} = 2$$
so the original limit simplifies to $$\frac{1}{2} \lim_{x \to 0} \frac{1 + x\sin(x) - \cos(x)}{x \tan(x)} = \frac{1}{2} \left( \lim_{x \to 0} \frac{x\sin(x)}{x\tan(x)} + \lim_{x \to 0} \frac{1-\cos(x)}{x \tan(x)} \right)$$
The first limit is simply $\cos(0) = 1$, so the limit becomes $$\frac{1}{2} \left(1 + \lim_{x \to 0}\frac{1 - \cos(x)}{x \tan(x)} \right) = \frac{1}{2} \left(1 + \lim_{x \to 0} \frac{\cos(x) (1 - \cos(x))}{x \sin(x)} \right) = \frac{1}{2} \left(1 + \lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)} \right)$$
Then using L'Hôpital's rule, I get that $$\lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x\cos(x) + \sin(x)}$$
Using L'Hôpital's rule once more: $$\lim_{x \to 0} \frac{\cos(x)}{2\cos(x) - x\sin(x)} = \frac{1}{2}$$
and therefore $$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} = \frac{1}{2} \left(1 + \frac{1}{2} \right) = \frac{3}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
Sum of squares and linear sum
For which positive integer $n$ can we write $n=a_1+a_2+\dots+a_k$ (for some unfixed $k$ and positive integers $a_1,a_2,\ldots,a_k$) such that $\sum_{i=1}^k a_i^2 = \sum_{i=1}^k a_i + 2\sum_{i<j}a_ia_j$?
When $k=1$, the equation is $a_1^2=a_1$, so $a_1=1$ and $n=1$ is the only possibility.
When $k=2$, we get $a_1^2+a_2^2 = a_1+a_2+2a_1a_2$, or $(a_1-a_2)^2 = a_1+a_2$, hence $n$ must be a perfect square. For perfect square $n=r^2$, we can solve $a_1+a_2=r^2$ and $a_1-a_2=r$, which must have a solution because $r^2\equiv r\pmod 2$. So all perfect squares work.
|
Partial answer.
If we write $m = \sum_i a_i^2$, then the identity $$\left(\sum_i a_i\right)^2 = \sum_i a_i^2 + 2\sum_{i < j} a_i a_j$$ allows us to translate our condition to $$ 2m = n^2 + n.$$ We know that $m \equiv n^2\mod 2$, hence a necessary condition is $4\mid n^2 - n$, which means $n\equiv 0, 1\mod 4$.
Now we can try the first several possibilities, as below ("-" means there is no solution):
1: 1
4: 3, 1
5: -
8: -
9: 6, 3
12: -
13: 9, 3, 1
16: 10, 6
17: 12, 2, 2, 1
20: 14, 3, 2, 1
21: 15, 1, 1, 1, 1, 1, 1
24: 17, 2, 2, 1, 1, 1
25: 15, 10
I did these calculations manually, so there are possibly errors.
My feeling is that for sufficiently large $n\equiv 0, 1\mod 4$, it is always possible to find a solution. However it seems hard to prove.
EDIT:
With a simple computer program, I verified that for all $n < 200$ with $n \equiv 0, 1\mod 4$, only $5, 8, 12$ cannot be obtained. I also corrected some errors in the calculations above.
Here is a brief idea of a proof. Firstly, we find the maximum integer $x$ such that $x^2 < \frac{n^2 + n}2$. For large $n$, this $x$ is about $\frac 1{\sqrt 2}n \sim 0.7 n$.
We take either $x$ or $x - 1$ (in case $x^2$ is too close to $\frac{n^2 + n}2$) as our first term $a_1$. This leaves us with a remaining sum of about $0.3n$ and a remaining sum of squares of about $1.4n$, which is the size of the difference between $x^2$ and $(x + 1)^2$.
Thus the problem becomes to find integers who sum up to $0.3n$ while their squares sum up to $1.4n$. It suffices to search for small values of $a_i$, say with $a_i\in [1, 6]$.
If we denote by $c_i$ the number of $a_i$ that are equal to $i$, then it becomes a group of two linear equations in $6$ variables:
\begin{eqnarray}
c_1 + 2c_2 + 3c_3 + 4c_4 + 5c_5 + 6c_6 &=& n - a_1 (\sim 0.3n)\\
c_1 + 4c_2 + 9c_3 + 16c_4 + 25c_5 + 36c_6 &=& \frac{n^2 + n}2 - a_1^2 (\sim 1.4n)\\
\end{eqnarray}
It should be possible to show that this system has non-negative integer solutions for sufficiently large $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3760391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
|
Let us ise Euler's transformation
$$I=\int \frac{dx}{(x^2-4x+13)^2}=\int \frac{dx}{(x-a)^2 (x-b)^2},~~a,b=2\pm 3i.$$
Let $$t=\frac{x-a}{x-b} \implies x=\frac{bt-a}{t-1} \implies dx=\frac{a-b}{(t-1)^2}.$$
Then $$I=(b-a)^{-3} \int \frac{dt}{t^2(t-1)^2}=(a-b)^{-3}\int \frac{u^2 du}{(u-1)^2}, u=1/t .$$
Next use $u=v+1$, then
$$I=(a-b)^3 \int [1-2/v+1/v^2] dv= (a-b)^{-3}[v-2 \ln v -1/v] $$
$$I=(a-b)^{-3}\left(\frac{a-b}{x-a}-2\ln \frac{a-b}{x-a}-\frac{x-a}{a-b}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 5
}
|
Find the value of $\lim _{a \to \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x $ Find the value of : $$
\lim _{a \rightarrow \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x
$$ I have tried to evaluate this integral by L'Hospitals rule, by separately diffrentiating the numerator and the denominator in the following steps: $$\int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} d x\cdot\dfrac{\pi}{2} (\text{by taking x=1/t and adding the integrands)}
$$ I end up converting the above integral in the form $(\text{by Leibnitz's integral rule})$ $$ \dfrac{\pi}{2}
\left(\int_{0}^{\infty} \frac{x}{1+x^{4}} d x\right)
$$ please provide an approach to this problem after this step.
|
We can rewrite the integral as
$$\lim_{a\to\infty} \frac{1}{a}\int_0^\infty \frac{x^2+1}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx + \int_0^\infty \frac{x}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx$$
which we are allowed to split up since both pieces are absolutely convergent. Then notice that
$$ \frac{x}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right) = \frac{1}{x^2}\cdot \frac{\frac{1}{x}}{1+\frac{1}{x^4}}\tan^{-1}\left(\frac{1}{x}\right) = \frac{d}{dx}\left[-\frac{1}{4}\arctan^2\left(\frac{1}{x}\right)\right]$$
therefore the integral evaluates to $\frac{\pi^2}{16}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3762470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
$a = \log_{40}100, b = \log_{10}20$.How can I express $b$ depending only on $a$? Let $a = \log_{40}{100}, b = \log_{10}{20}$. How can I express $b$ depending only on $a$? I tried using the formula to change the base from $40$ to $10$, but couldn't get it just depending on $a$.
I used the base change formula $\log_a b = \dfrac{\log_ c b}{\log_c a}$ and got that $\log_{40} 100 = \dfrac{\log_{10} 100}{\log_{10}{40}} = \dfrac{2}{\log_{10}{20} + \log_{10}{2}}$. But then how could I express $\log_{10}2$ depending on $\log_{10}20$? I think that's what it suffices to show.
|
Find $a=\log_{40}100$ only in terms of $b=\log_{10}20$.
\begin{align*}
2b&=2\log_{10}20=\log_{10}(20)^2=\log_{10}(40\cdot10)=\log_{10}40+1=2\log_{10^2}40+1=\frac2a+1
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3763274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Why is my variance negative? Let $Y_1 \sim Gamma(2,1)$, and $Y_2|Y_1 \sim f_{Y_2}(Y_2)$, where $y_1\geq y_2\geq 0$.
Where
$$f_{Y_2}(y_2) = \frac{1}{y_1}$$
$$f_{Y_1}(y_1) = y_1 \exp( -y_1)$$
Goal: Find standard deviation of $Y_2 - Y_1$
Start with
$$Var(Y_1) = 1, E(Y_1) = 1$$
$$E(Y_2|Y_1) = \int^{y_1}_0 \frac{y_2}{y_1}dy_2 = \frac{y_1}{2}$$
$$E(Y_2^2|Y_1)= \int^{y_1}_0 \frac{y_2^2}{y_1}dy_2 = \frac{y_1^2}{3} $$
$$Var(Y_2|Y_1) = E(Y^2_2|Y_1) - (E(Y_2|Y_1))^2 = \frac{y_1^2}{3} - \frac{y_1^2}{4} = \frac{y_1^2}{12}$$
\begin{align*}
Var(Y_2) = & E(Var(Y_2|Y_1))+ Var(E(Y_2|Y_1))\\
= & E\left (\frac{y_1^2}{12} \right ) + Var\left (\frac{y_1}{2}\right )\\
= & E\left (\frac{y_1^2}{12}\right ) + Var\left (\frac{y_1}{2}\right )\\
= & \frac{2}{12} + \frac{1}{4}\\
= & \frac{5}{12}
\end{align*}
The formula I am using is $Var(Y_1 - Y_2) = Var(Y_1) + Var(Y_2) - 2Cov(Y_1, Y_2)$
So now I need
\begin{align*}
Cov(Y_1, Y_2) = & E(Y_1 Y_2) - E(Y_1)E(Y_2)\\
= & E(Y_1 Y_2) - E(Y_1)E(E(Y_2|Y_1))\\
= & E(Y_1 Y_2) - (1)\frac{1}{2}
\end{align*}
Now find
\begin{align*}
E(Y_1Y_2) = &\int^{\infty}_0 \int^{y_1}_0 y_1 y_2 \exp(-y_1) dy_2 dy_1\\
= & 3
\end{align*}
Thus $Cov(Y_1, Y_2) = 3 - \frac{1}{2} = \frac{5}{2}$
Hence
\begin{align*}
Var(Y_1-Y_2) = & Var(Y_1) + Var(Y_2) - 2Cov(Y_1, Y_2) \\
= & 1 + \frac{5}{12} - 5
\end{align*}
Which clearly can't be true. I don't where I went wrong.
I tried another approach by transforming the random variables and finding the distribution of $Y_1 - Y_2$ which did produce a realistic answer. However, I would like to know what was wrong with this approach.
|
I found my mistake.
$$E(Y_1) = 2$$
$$Var(Y_1) = 2$$
Therefore
\begin{align*}
Var(Y_2) = & E\left (Var(Y_2|Y_1)\right) + Var\left ( E(Y_2|Y_1)\right ) \\
= & E \left ( \frac{y^2_1}{12}\right ) + Var \left( \frac{y_1}{2} \right ) \\
= & \frac{6}{12} + \frac{2}{4}\\
= &1
\end{align*}
In addition, the $Cov(Y_1, Y_2) = 3 - (2)\frac{2}{2} = 1$
\begin{align*}
Var(Y_1 - Y_2) = &Var(Y_1) + Var(Y_2) - 2 Cov(Y_1, Y_2)\\
= & 2 + 1 - 2(1)\\
= & 1
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3763411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Prove that the following sequence $(\sin\frac{\pi k}{3})_{k=1}^\infty$ diverges $$x_k := \sin \left(\frac{\pi k}{3} \right),\qquad x_{\infty} := \lim_{k \to \infty}{x_k}$$
I tried to prove by contradiction. Assume limit exists and is $L$. Fix $\epsilon=\frac{\sqrt{3}}{4}$. There exists $K\in\mathbb{Z}^{+}$ such that $k>K$ implies $$|\sin\frac{\pi k}{3}-L|<\epsilon$$ Notice that for $k=6l+1$ or $k=6l+2$ where $l\in\mathbb{Z}^{\geq 0}$, $\lvert \frac{\sqrt{3}}{2}-L\rvert<\frac{\sqrt{3}}{4}$. Also, for $k=6l+4$ or $k=6l+5$ where $l\in\mathbb{Z}^{\geq 0}$, $\lvert -\frac{\sqrt{3}}{2}-L\rvert<\frac{\sqrt{3}}{4}$. Now we know that $\sqrt{3}=\lvert\frac{\sqrt{3}}{2}-\big(-\frac{\sqrt{3}}{2}\big)\rvert=|\frac{\sqrt{3}}{2}+L-L-\big(-\frac{\sqrt{3}}{2}\big)|\leq|\frac{\sqrt{3}}{2}+L|+|-L-\big(-\frac{\sqrt{3}}{2}\big)|$ $=|-\frac{\sqrt{3}}{2}-L|+|\frac{\sqrt{3}}{2}-L|<\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\;\Rightarrow\;\sqrt{3}<\frac{\sqrt{3}}{2}$. Hence, we get a contradiction. Therefore, the sequence diverges.
Is this proof correct? If not, can you help me to prove using epsilon-delta definition?
|
Let's consider 3 sub sequences
$$k=6n \Rightarrow \sin\frac{\pi k}{3}=\sin 2\pi n=0 $$
$$k=6n+1 \Rightarrow \sin\frac{\pi k}{3}=\sin \left(2\pi n +\frac{\pi }{3} \right) = \frac{\sqrt{3} }{2}$$
$$k=9n+1 \Rightarrow \sin\frac{\pi k}{3}=\sin \left(2\pi n +\pi+\frac{\pi }{3} \right) = -\frac{\sqrt{3} }{2}$$
So we have 3 limit points.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3764271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Existence of limit for sequence $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right)$ with initial values $x_0=5,x_1=10$ Let $x_0=5,x_1=10,$ and for all integers $n\ge2$ let $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right).$ By induction, we have $\forall m\in\mathbb Z_{\ge0}\enspace x_m>0,$ so we can avoid division by $0$ and the sequence is well-defined.
According to a Math GRE practice problem, the limit exists. How can we prove that? Note that, if we assume the limit exists, then we can show it equals $\sqrt8,$ but finding the value of the limit is not my goal here.
My work: We can compute $x_2=5.8,x_3=3.3,$ which are strictly between $4/3$ and $6,$ and then, assuming an inductive hypothesis, for all integers $n\ge4$ we have $4/3<x_{n-1}<6$ and $4/3<8/x_{n-2}<6,$ so that $4/3<x_n<6.$ We can probably compute more values of $x_n$ to get tighter bounds, but I don't see how to actually show convergence.
|
Let $$x_{n+1}=\tfrac{1}{2}(x_n+\frac{a}{x_{n-1}})$$
Then for $d_n=x_n-\sqrt{a}$, \begin{align}
x_{n+1}-\sqrt{a}&=\tfrac{1}{2}(x_n-\sqrt{a})+\frac{a}{2}\left(\frac{1}{x_{n-1}}-\frac{1}{\sqrt{a}}\right)\\
d_{n+1}&=\tfrac{1}{2}d_n-\frac{\sqrt{a}}{2}\frac{d_{n-1}}{d_{n-1}+\sqrt{a}}=\tfrac{1}{2}d_n-\frac{1}{2}\frac{d_{n-1}}{\frac{d_{n-1}}{\sqrt{a}}+1}\\
\end{align}
So if $|d_{n-1}|<\sqrt{a}/3$, $$|d_{n+1}|\le \begin{cases}\tfrac{1}{2}|d_n|,&d_{n-1}d_n>0\\
\frac{1}{2}|d_n|+\frac{3}{4}|d_{n-1}|,&d_{n-1}d_n<0\end{cases}$$
Since the worst case cannot happen twice in succession, we must have $$|d_{n+2}|\le\tfrac{1}{4}|d_n|+\tfrac{3}{8}|d_{n-1}|$$
This recurrence inequality can be solved, $|d_n|\le A|r_1|^n+B|r_2|^n+C|r_3|^n\to0$ since $r_1\approx0.84$, $|r_2|=|r_3|\approx0.67$.
Hence, as long as some $d_k$ comes close enough to $\sqrt{a}$, $x_n\to\sqrt{a}$. (In fact, the sequence may converge to $-\sqrt{a}$, e.g. $x_0=x_1=-1$ for $a=8$. )
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3767713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game stops.
What is the probability that the game ends on an even turn when $A$ rolls first?
Now the book gives the answer as $\frac{4}{7}$, however, when try to calculate I end up with $\frac{2}{11}$.
Below is my work:
To calculate this probability, we decompose the event into two disjoint events, (a) the event where $A$ wins on an even roll, and (b) the event where $B$ wins on an even roll.
(a) Now, the probability $A$ wins can be calculated as follows
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{1}{3}\\
= \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{4}{81}\biggr)^k = \frac{2}{27}\cdot \frac{1}{1- \frac{4}{81}} = \frac{6}{77}.
\end{align*}
(b) Similarly we calculate the probability $B$ wins on an even roll as
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot \frac{2}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot\frac{2}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{4}{9}\\
= \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{4}{81}\biggr)^k = \frac{8}{81}\cdot \frac{1}{1- \frac{4}{81}} = \frac{8}{77}.
\end{align*}
Therefore, it follows that the probability of the game ending on an even number of rolls is
\begin{equation*}
\frac{6}{77} + \frac{8}{77} = \frac{2}{11}.
\end{equation*}
Am I missing something?
|
Working under the assumption that the intended interpretation of the question was merely asking the probability that $B$ wins (i.e. distinguishing between the term "rounds" as iterating whenever A has a turn and "turns" iterating whenever either A or B has a turn) two other approaches have already been written. Here I will include yet another approach:
Consider the final round, that is a roll of $A$ followed by a roll of $B$, where we allow $B$ to roll even in the event that $A$ has already won despite the roll not influencing the final result of the game.
Ordinarily there are $6\times 6 = 36$ equally likely results for a round. Here, we condition on the fact that it is the last round, implying that it was not the case that both players missed their respective targets. This gives $6\times 6 - 4\times 2 = 28$ equally likely possible final rounds.
Of these, $4\times 4 = 16$ of them end with $A$ missing their target and $B$ hitting theirs.
The probability of $B$ winning the game is then: $$\dfrac{16}{28} = \dfrac{4}{7}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3768049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$ The original question is:
Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$
But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$$
At first, I assumed $x^2+y^2=z^2$. But, it is so complicated. Then, I assumed $x=r\cos\theta$ and $y=r\sin\theta$. But, I can't calculate the limits. Solving the equations I got three values of $\theta$ i.e. $\theta=0$, $\theta=\frac{\pi}{4}$ and $\theta=\frac{\pi}{2}$. I am just confused. Please help.
|
\begin{align}
\text{Let } & y = x\tan \theta, \\[8pt]
\text{so that }& dy = x\sec^2\theta\,d\theta \\[8pt]
\text{and } & x^2 + y^2= x^2\sec^2\theta,
\end{align}
and as $y$ goes from $0$ to $1$ then $\theta$ goes from $0$ to $\arctan(1/x)$.
Then
\begin{align*}
& \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \\[8pt]
= {} & \int_0^{\arctan(1/x)} \frac{x^2 - x^2 \tan^2\theta}{(x^2 + x^2\tan^2\theta)^2} \big( x\sec^2\theta\,d\theta\big) \\[8pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \frac{1-\tan^2 \theta}{\sec^2 \theta} \, d\theta \\[8pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} (\cos^2\theta-\sin^2\theta) \, d\theta \\[8pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \cos(2\theta) \, d\theta
= \frac 1 {2x} \sin\left(2\arctan \frac 1 x\right) \\[8pt]
= {} & \frac 1 x \sin\left(\arctan \frac 1 x \right) \cos\left( \arctan \frac 1 x \right) \\[8pt]
= {} & \frac 1 x \cdot \frac 1 {\sqrt{1+x^2}} \cdot \frac x {\sqrt{1+x^2}} = \frac 1 {1+x^2}. \\[5pt]
\text{And then}
& \int_0^1 \frac{dx}{1+x^2} = \frac \pi 4.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3768872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Find $a,b,c$ if ${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$ If $${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$$
and $$(a+b+c)=21, a\gt6$$
We have to find $a,b,c$
My attempt
I use a little fact that the sum of first $n$ odd numbers is $n^2$
From that i get $$a^2+b^2=c^2$$
Which means that the solutions are pythagorean triples, and found no such soultion.
While if i try to do this by hand, $$a=7,b=5,c=9$$ satisfies the question but not the pythaoarean equation that i set up above
Why is this so?
|
If $a=2n-1$ so $$1^2+3^2+...+a^2=n^2=\left(\frac{a+1}{2}\right)^2.$$
Id est, $$(a+1)^2+(b+1)^2=(c+1)^2$$ or
$$a^2+b^2+2a+2b+2=(22-a-b)^2$$ or
$$ab-23(a+b)=-241$$ or
$$(a-23)(b-23)=288.$$
Also we have $a\geq7$, $b+c\leq14$, $c>b$ and $c>a$.
Can you end it now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3769123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
System of congurences and the Chinese Remainder Theorem I have the following system of congruences:
\begin{align*}
x &\equiv 1 \pmod{3} \\
x &\equiv 4 \pmod{5} \\
x &\equiv 6 \pmod{7}
\end{align*}
I tried solving this using the Chinese remainder theorem as follows:
We have that $N = 3 \cdot 5 \cdot 7 = 105$ and $N_1=35, N_2=21, N_3=15$.
From this, we get the following
\begin{align*}
35x_1 &\equiv 1 \pmod{3} \\
21x_2 &\equiv 1 \pmod{5} \\
15x_3 &\equiv 1 \pmod{7}
\end{align*}
and this will result in
\begin{align*}
2x_1 &\equiv 1 \pmod{3} \\
x_2 &\equiv 1 \pmod{5} \\
x_3 &\equiv 1 \pmod{7}
\end{align*}
so from CRT $x =x_1N_1b_1 + x_2N_2b_2 + x_3N_3b_3 = 2 \cdot 35 \cdot3 + 1 \cdot 21 \cdot 5 + 1 \cdot 15 \cdot7 = 420 $.
However $420$ doesn't seem to satisfy the given system, what would be the problem here?
|
From the Chinese remainder theorem:
$$x = x_1N_1b_1 + x_2N_2b_2 + x_3N_3b_3$$
$$= 2 \cdot 35 \cdot \color{red}{1} + 1 \cdot 21 \cdot \color{red}{4} + 1 \cdot 15 \cdot \color{red}{6} = 244$$
The general solution is when $x \equiv 244 \pmod {\text{lcm}(3,5,7)} \Rightarrow x \equiv 244 \pmod {105} \equiv 34 \pmod {105}$, or when $x = 105k + 34$ when $k$ is an integer.
This problem is the exact problem that appears in the Brilliant article on the Chinese remainder theorem. Their method involves rewriting the largest congruence, $x \equiv 6 \pmod 7$ in the form $7j+6$, then substituting the expression into the next largest congruence, so that $7j + 6 \equiv 4 \pmod 5$. Solving this congruence gives $j \equiv 4 \pmod 5$. Repeating this process gives $j = 5k +4$, and substituting into the equation for $j$ gives $x = 7(5k + 4) + 6 = 35k + 34$. $35k + 34 \equiv 1 \pmod 3$ results in $k \equiv 0 \pmod 3$. Therefore $k = 3l$ and $x = 35(3l) + 34$, so $x$ is in the form $105k + 34$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3769723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $ I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[n]{\left(1+\dfrac{1}{t}\right)\left(2+\dfrac{1}{t}\right)...\left(n+\dfrac{1}{t}\right)}-\dfrac{1}{t}\\
&=\lim\limits_{t \to 0} \sqrt[n]{\left(\dfrac{t+1}{t}\right)\left(\dfrac{2t+1}{t}\right)...\left(\dfrac{nt+1}{t}\right)}-\dfrac{1}{t} \\
&=\lim\limits_{t \to 0} \dfrac{\sqrt[n]{(t+1)(2t+1)...(nt+1)}-1}{t}
\end{align*}
My idea is to use $\lim\limits_{x\to0}\dfrac{(ax+1)^{\beta}-1}{x} =a\beta .$ But after some steps (above), now I'm stuck.
Thanks for any helps.
|
The limit can also be shown using HM-GM-AM.
Setting $u = x^2$ and considering $u\to +\infty$ we have
$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u \leq \sqrt[n]{\prod_{k=1}^n (k+u)} - u \leq \frac{\sum_{k=1}^n(k+u)}n - u = \frac{n+1}{2}$$
For the LHS we have
$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u = \frac{n - \sum_{k=1}^n\frac u{k+u}}{\frac 1u\sum_{k=1}^n\frac 1{\frac ku+1}} $$ $$= \frac{u \sum_{k=1}^n\frac k{k+u}}{\sum_{k=1}^n\frac 1{\frac ku+1}}= \frac{\sum_{k=1}^n\frac k{\frac ku+1}}{\sum_{k=1}^n\frac 1{\frac ku+1}}\stackrel{u\to+\infty}{\longrightarrow}\frac{\sum_{k=1}^n k}{n} = \frac{n+1}{2}$$
Now, squeezing gives the limit $ \frac{n+1}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3770941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
}
|
A problem involving the roots of quartic polynomial $x^4+px^3+qx^2+rx+1$ Let $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ be the roots of the following polynomial
$$P(x)=x^4+px^3+qx^2+rx+1$$
Show that
$$(1+{\alpha_1}^4)(1+{\alpha_2}^4)(1+{\alpha_3}^4)(1+{\alpha_4}^4)=(p^2+r^2)^2+q^4-4pq^2r.$$
I came across this problem on a facebook page as an challenge.The only way that Strike me is multiply the terms in LHS and put the corresponding values but this method would be very long as polynomial is of 4th degree.
If anybody know any other method then Please tell me. Thank you for your help!
|
Squaring both sides
$$(x^4+qx^2+1)^2=x^2(px^2+r)^2$$
Let $x^2=y$
$$(y^2+qy+1)^2=y(py+r)^2$$
$$\iff y^4+y^2A+1=y^3B+yC$$
Let $z=1+y^2\implies y=\sqrt{z-1}$
$$\implies(z-1)^2+(z-1)A+1=\pm\sqrt{z-1}((z-1)B+C)$$
$$ z^2+zE+F=\pm\sqrt{z-1}(zG+H)$$
Squaring both sides
$$z^4+\cdots+F^2=(z-1)(\cdots+H^2)$$
$$z^4+\cdots+F^2+H^2=0$$
$$\implies\prod_{r=1}(1+\alpha_r^4)=\prod_{r=1}z_r=\dfrac{F^2+H^2}1$$ using Vieta's formula
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3771973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?
|
Write $ax^9+bx^8+1=(x^2-x-1)\sum\limits_{k=0}^7c_kx^{7-k}$ which gives the initial conditions $c_7=-1$ and $-c_7-c_6=0\implies c_6=1$. Notice that $c_0=a$ and $c_1-c_0=b$ on equating coefficients.
Further, we have $c_i=c_{i-1}+c_{i-2}$ for $i>1$ which is the negative Fibonacci sequence shifted by one. In particular, $c_{7-i}=(-1)^{i+1}F_{i+1}$ so $c_0=F_8$ and $c_1=-F_7$. Hence $a=21$ and $b=-34$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
}
|
Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$ How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3}
\sec\theta\ \tan\theta d\theta $$
$$=\int \tan^2\theta \sec^4\theta(1-3\cos^2\theta)^{4/3} \sec\theta\ \tan\theta d\theta $$
$$=\int \tan^3\theta \sec^5\theta(1-3\cos^2\theta)^{4/3}\ d\theta $$
$$=\int\dfrac{ \sin^3\theta}{ \cos^8\theta}(1-3\cos^2\theta)^{4/3}\ d\theta $$
I can't see if this substitution will work or not. This has become so complicated.
Please help me solve this integral.
|
If you multiply and divide by $3$, you get
$$ \int (x^2 -1)(x^3 - 3x)^{4/3}dx = \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx $$
changing variable to $u = x^3 - 3x$ you have $du = (3x^2 - 3x)dx$ so
$$
\begin{split}
\int (x^2 -1)(x^3 - 3x)^{4/3}dx &= \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx\cr
&= \frac{1}{3} \int u^{4/3} du \cr
&= \frac{1}{3} \times \frac{3u^{7/3}}{7} + C \cr
&= \frac{1}{7} (x^3 - 3x)^{7/3} + C \cr
\end{split}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3773220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
}
|
Indefinite integral of $\sin^8(x)$ Suppose we have the following function:
$$\sin^8(x)$$
We have to find its anti-derivative
To find the indefinite integral of $\sin^4(x)$, I converted everything to $\cos(2x)$ and $\cos(4x)$ and then integrated. However this method wont be suitable to find the indefinite integral $\sin^8(x)$ since we have to expand a lot. Is there any other way I can evaluate it easily, and more efficiently?
|
By expanding
\begin{align}
\sin^8x
&=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)=\\
&=\frac{1}{128}\left(
\frac{e^{8ix}-e^{-8ix}}{2}
-8\frac{e^{6ix}-e^{-6ix}}{2}
+28\frac{e^{4ix}-e^{-4ix}}{2}
-56\frac{e^{2ix}-e^{-2ix}}{2}+35\right)=\\
&=\frac{1}{128}\left[\cos8x-8\cos6x+28\cos4x-56\cos2x+35\right]
\end{align}
or by using the identity
$$
(\sin x)^{2m}=\frac{2}{4^m}\left[\sum_{k=0}^{m-1}\binom{2m}{k}(-1)^{m-k}\cos[2(m-k)x]+\frac{1}{2}\binom{2m}{m}\right]
$$
that for $m=4$ provides
\begin{align}
\sin^8x
&=\frac{2}{4^4}\left[\sum_{k=0}^3\binom{8}{k}(-1)^k\cos[2(4-k)x]+\frac{1}{2}\binom{8}{4}\right]=\\
&=\frac{1}{128}\left[\cos8x-8\cos6x+28\cos4x-56\cos2x+35\right]
\end{align}
we have
\begin{align}
\int\sin^8xdx
&=\frac{1}{128}\left[\frac{1}{8}\sin8x-\frac{4}{3}\sin6x+7\sin4x-28\sin2x+35x\right]+C\\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3773595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Calculus of $ \lim_{(x,y)\to (0,0)} \frac{8 x^2 y^3 }{x^9+y^3} $ By Wolfram Alpha I know that the limit
$$
\lim_{(x,y)\to (0,0)} \dfrac{8 x^2 y^3 }{x^9+y^3}=0.
$$
I have tried to prove that this limit is $0$, by using polar coordinate, the AM–GM inequality and the change of variable $ x^9= r^2 \cos^2(t) $ and $y^3= r^2 \sin^2(t)$, but these attempts were unsatisfactory.
I also have reviewed the similar questions and their answers but there are difference between those functions and mine one, I think the principal difference is that the powers of the denominators are odd.
|
Actually, that function isn't even bounded near $(0,0)$ and therefore, the limit at that point doesn't exist. You can check that$$\frac{8x^2(-x^3+x^6)^3}{x^9+(-x^3+x^6)^3}=\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}$$and that$$\lim_{x\to0}\left|\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}\right|=\infty.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3774244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Finding a Mistake for a Particular Form of Inequality My book depicts that the following problem uses ${x^3\over (1+y)(1+z)}+{(1+y)\over 8}+{(1+z)\over 8} \ge {3x\over 4} $.
Let $x, y, z$ be positive real numbers such that $xyz = 1$. Prove that $$ {x^3\over (1+y)(1+z)}+{y^3\over (1+z)(1+x)}+{z^3\over (1+x)(1+y)}\geq{3\over 4} $$
and also points the mistake on using $ {x^3\over (1+y)(1+z)}+{(1+y)}+{(1+z)} \ge {3x} $ as the equality can not hold.
Why does the equality does not hold?
|
The inequality $$ {x^3\over (1+y)(1+z)}+{(1+y)}+{(1+z)} \ge {3x} $$ is indeed true by AM-GM with the equality for
$$ {x^3\over (1+y)(1+z)}=1+y=1+z,$$ which gives $y=z$ and $x^3=(1+y)^3$ or $x=1+y,$ which with the condition $xyz=1$ gives $$(1+y)y^2=1.$$
But in the original inequality the equality occurs for $x=y=z=1$ and we got that your using of AM-GM does not save an equality occurring.
By the way, this AM-GM $${x^3\over (1+y)(1+z)}+{1+y\over 8}+{1+z\over 8} \ge {3x\over 4} $$ saves it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3774372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum of $\sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$ and concluding that $f(x)=(x-1)\log(1 - 2 x) -2x$ or to replace $\log(1 - 2 x)$ for it's Taylor Expansion and resaulting in $ \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$. I've tried both ways, starting from the Taylor Expansion of $\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n}$ but I've failed both ways. Any hints on how to prove this? Thanks in advance.
Edit: What I did
$$\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n} = -2x + \sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} \\ \rightarrow (x-1)\log(1 - 2 x)=-2x^2 + x\sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} + 2x +\sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ \rightarrow (x-1)\log(1 - 2 x) -2x =-2x^2 + \sum_{n=2}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^{n} x^{n+1}}{n} + \sum_{n=1}^{\infty} \frac{2^{n+1} x^{n+1}}{n+1} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1} (n+1) + 2^{n+1} x^{n+1} n}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}[(n+1)-2n]}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}(1-n)}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = 0 + \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)}$$
Which is what I was trying to prove.
|
$$\log(1-2x)=-\sum_{n=1}^\infty\frac{(2x)^n}n\;,\;\;|2x|<1\iff |x|<\frac12\implies$$
$$(x-1)\log(1-2x)-2x=\sum_{n=1}^\infty\frac{2^nx^{n+1}-2^nx^n}n-2x$$
and
$$\sum_{n=1}^\infty\frac{2^n(n-1)x^{n+1}}{n^2+n}=\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n+1}}^{:=f(x)}-\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n^2+n}}^{:=g(x)}$$
and now
$$f'(x)=\sum_{n=1}^\infty2^nx^n\;,\;\;g'(x)=\sum_{n=1}^\infty\frac{2^nx^n}n\implies f'(x)-g'(x)=\frac1{1-2x}-1+\log(1-2x)$$
Now try to take it from here...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3777861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How would I simplify this function $\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{\dots}}}}$ How do I simplify $\rho(x)$ into simple terms?
$$\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x+\sqrt{x-\sqrt{\dots}}}}}}}}$$
where the subtracting and the adding follows the Thue–Morse sequence $$+,-,-,+,-,+,+,-,-,+,+,-,+,-,-,+,\dots$$
I tried doing it with $x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\dots}}}}}}}}$ and got a answer by myself and I did it with $x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{\dots}}}}}}}}$ and found a post here Simplify the radical $\sqrt{x-\sqrt{x+\sqrt{x-...}}}$ and I understood how it worked I would like to know how to solve a problem like this? where the adding and the subtracting never repeats.
|
I wouldn't call this an exact closed form, but a 'close' one indeed. I remember a result in the paper (Page $28$); A chronology of continued square roots and other continued compositions by Dixon J. Jones; he refers another problem reffered in $1899$ by Karl Bochow (Problem 1740. Zeitschrift f¨ur mathematischen und
naturwissenschaftlichen Unterricht) Which asks the reader:
Assuming $0<a<1/2$: $$2\sin(\pi
a)=l_0\sqrt{2+l_{1}\sqrt{2+l_{2}\sqrt{2+l_{3}\sqrt{2+l_{4}\sqrt{...}}}}}$$ For $l_n$ being either $-1$ or $+1$. Then;
$$a=\frac{l_{0}}{2^{2}}+\frac{l_{0}l_{1}}{2^{3}}+\frac{l_{0}l_{1}l_{2}}{2^{4}}+\frac{l_{0}l_{1}l_{2}l_{3}}{2^{5}}+...$$
In your posed problem, we have
$$\small{\begin{align}
ρ\left(x\right) & = x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x+...}}}}}} \\
& = x+\left(-1\right)^{0}\sqrt{x+\left(-1\right)^{1}\sqrt{x+\left(-1\right)^{1}\sqrt{x+\left(-1\right)^{0}\sqrt{x+\left(-1\right)^{1}\sqrt{x+\left(-1\right)^{0}\sqrt{x+...}}}}}} \\
& = x+\left(-1\right)^{m_{0}}\sqrt{x+\left(-1\right)^{m_{1}}\sqrt{x+\left(-1\right)^{m_{2}}\sqrt{x+\left(-1\right)^{m_{2}}\sqrt{x+\left(-1\right)^{m_{3}}\sqrt{x+\left(-1\right)^{m_{4}}\sqrt{x+...}}}}}}
\end{align}}$$
Where $m_n$ is the n-th Thue-Morse element (having $m_0,m_1,m_2... = 0,1,1,0..$). Now applying the first result:
$$\small{\begin{align}
ρ\left(2\right) & = 2+\left(-1\right)^{m_{0}}\sqrt{2+\left(-1\right)^{m_{1}}\sqrt{2+\left(-\right)^{m_{2}}\sqrt{2+\left(-1\right)^{m_{2}}\sqrt{2+\left(-\right)^{m_{3}}\sqrt{2+\left(-1\right)^{m_{4}}\sqrt{2+...}}}}}} \\
& = 2+2\sin\left\{\pi\left(\frac{\left(-1\right)^{m_{0}}}{2^{2}}+\frac{\left(-1\right)^{m_{0}+m_{1}}}{2^{3}}+\frac{\left(-1\right)^{m_{0}+m_{1}+m_{2}}}{2^{4}}+\frac{\left(-1\right)^{m_{0}+m_{1}+m_{2}+m_{3}}}{2^{5}}+...\right)\right\} \\
& = 2+2\sin\left\{\pi\left(\frac{\left(-1\right)^{Sm_{0}}}{2^{2}}+\frac{\left(-1\right)^{Sm_{1}}}{2^{3}}+\frac{\left(-1\right)^{Sm_{2}}}{2^{4}}+\frac{\left(-1\right)^{Sm_{3}}}{2^{5}}+...\right)\right\} \\
& = 2+2\sin\left\{\frac{\pi}{4}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{Sm_{n}}}{2^{n}}\right\}\tag{1}
\end{align}}$$
Where $Sm_n=\sum m_k=m_0+m_1+...+m_n$. $Sm_n$ is also called the partial sum of the Theu-Morse sequence (find the sequence of that here). Now I'm not sure whether that sum has a closed form or not. I've seen certain other infinite series using Thue-Morse that has closed forms. There's a similar series to the sum in $(1)$:
$$\sum_{n=0}^{\infty}\frac{\left(-1\right)^{m_{n}}}{2^{n}}=2\left(1-2\tau\right)$$
Where $\tau$ is called the Thue-Morse constant
EDIT: To compute the series in $(1)$, you can use the following asymptote
$$\sum_{n=0}^{\infty}\frac{\left(-1\right)^{Sm_{n}}}{2^{n}}\sim\sum_{n=0}^{4x}\frac{\left(-1\right)^{Sm_{n}}}{2^{n}}-\frac{3}{5}2^{-4x}$$
Which stems from the result that:
$$\lim_{n \rightarrow \infty}\frac{Sm_n}{n}=\frac{1}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3778048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.