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Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$ Is there any closed form for the infinite product $\prod_{l=1}^\infty \cos\dfrac{x}{3^l}$? I think it is convergent for any $x\in\mathbb{R}$.
I think there might be one because there is a closed form for $\prod_{l=1}^\infty\cos\dfrac{x}{2^l}$ if I'm not wrong.
|
Let $\displaystyle P=\prod_{i=1}^n\cos\left(\frac x{2^i}\right)$.
$\displaystyle P=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)\sin\left(\frac x{2^n}\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac12\,\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^{n-1}}\right)\sin\left(\frac x{2^{n-1}}\right)$
...
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac1{2^{n-1}}\,\cos\left(\frac x{2}\right)\sin\left(\frac x2\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac1{2^{n}}\,\sin(x)$
$\displaystyle P=\frac{\sin(x)}{2^{n}\sin\left(\frac x{2^n}\right)}$
$\displaystyle \lim_{n\to\infty}P=\lim_{n\to\infty}\frac{\sin(x)}x\frac{2^{-n}x}{\sin\left( 2^{-n}x \right)}=\frac{\sin(x)}x$
However, I am not able to solve the one with $3$ in the denominator.
|
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"url": "https://math.stackexchange.com/questions/1838157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Derivatives with different rules I'm having trouble with this one problem that just deals with deriving. I can't seem to figure out how they got their answer. Any help would be appreciated! Thanks!
$
\frac{(x+1)^2}{(x^2+1)^3}
$
The answer is:
$
\frac{-2(x+1)(2x^2-3x-1)}{(x^2+1)^4}
$
I seem to get the wrong answer when using the quotient rule.
|
Let's use quotient rule:
$$\frac{(x^2+1)^3\frac{d}{dx}(x+1)^2-(x+1)^2\frac{d}{dx}(x^2+1)^3}{{(x^2+1)^3}^2}$$
Do the derivatives out and simplify the denominator:
$$\frac{(x^2+1)^3\cdot 2(x+1)-(x+1)^2\cdot 2x\cdot 3(x^2+1)^2}{(x^2+1)^6}$$
Factor out $2(x+1)(x^2+1)^2$ from the numerator:
$$\frac{2(x+1)(x^2+1)^2((x^2+1)-(x+1)\cdot 3x)}{(x^2+1)^6}$$
Reduce the fraction and simplify the numerator:
$$\frac{2(x+1)(-2x^2-3x+1)}{(x^2+1)^4}$$
Factor out a $-1$ from $-2x^2-3x+1$:
$$\frac{-2(x+1)(2x^2+3x-1)}{(x^2+1)^4}$$
Thus, it seems that your book's answer is wrong and it should be $+3x$, not $-3x$. Wolfram Alpha also agrees with me on this.
|
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|
A convergent series: $\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$ I would like to find the value of:
$$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$
I could only see that the ratio of two consecutive terms is $\dfrac{1}{27\cos(2\theta)}$.
|
$$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$
By $\sin3\theta =3\sin \theta-4\sin^3\theta$
$$4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\sin\left(\frac{3\pi}{3^{n+1}}\right)-3\sin\left(\frac{\pi}{3^{n+1}}\right)$$
Then
$$\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\cdot4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\left[3\sin\left(\frac{\pi}{3^{n+1}}\right)-\sin\left(\frac{3\pi}{3^{n+1}}\right)\right]$$
$$=\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\left[3\sin\left(\frac{\pi}{3^{n+1}}\right)-\sin\left(\frac{3\pi}{3^{n+1}}\right)\right]$$
$$=\frac{1}{4}\sum_{n=0}^\infty \left[3^n\sin\left(\frac{\pi}{3^{n+1}}\right)-3^{n-1}\sin\left(\frac{3\pi}{3^{n}}\right)\right]$$
Now let $$\frac{3^n}{4}\sin\left(\frac{\pi}{3^{n+1}}\right) =f(n)$$
Then $$\frac{3^{n-1}}{4}\sin\left(\frac{\pi}{3^{n}}\right) =f(n-1)$$
$$\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\cdot4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\sum_{n=0}^m\left[ f(n)-f(n-1)\right]$$
$$=\sum_{n=0}^m\left[ f(n)-f(n-1)\right]=f(m)-f(-1)$$
$$=\frac{3^m}{4}\sin\left(\frac{\pi}{3^{m+1}}\right)-\frac{3^{-1}}{4}\sin\left(\frac{\pi}{3^{-1+1}}\right)$$
$$=\frac{3^m}{4}\sin\left(\frac{\pi}{3^{m+1}}\right)-0$$
Now see $$=\frac{3^m}{4}\sin\left(\frac{\pi}{3^{m+1}}\right)= \frac{\pi}{12} \cdot \left(\frac{\sin \frac{\pi}{3^{m+1}}}{\frac{\pi}{3^{m+1}}}\right)$$
Now take the limit $m \to \infty$.
|
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|
Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ In order to compute, in an elementary way,
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$)
i need to show, in a simple way, that:
$\displaystyle \int_0^1 \dfrac{\arctan x \log x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$
$G$ is the Catalan's constant.
|
Different approach:
start with applying integration by parts
$$I=\int_0^1\frac{\tan^{-1}(x)\ln(x)}{1+x}dx\\=\left|(\operatorname{Li}_2(-x)+\ln(x)\ln(1+x))\tan^{-1}(x)\right|_0^1-\int_0^1\frac{\operatorname{Li}_2(-x)+\ln(x)\ln(1+x)}{1+x^2}dx$$
$$=-\frac{\pi^3}{48}-\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx-\color{blue}{\int_0^1\frac{\ln(x)\ln(1+x)}{1+x^2}dx}\tag1$$
From $$\operatorname{Li}_2(x)=-\int_0^1\frac{x\ln(y)}{1-xy}dy$$
it follows that
$$\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx=\int_0^1\frac1{1+x^2}\left(\int_0^1\frac{x\ln(y)}{1+xy}dy\right)dx$$
$$=\int_0^1\ln(y)\left(\int_0^1\frac{x}{(1+x^2)(1+yx)}dx\right)dy$$
$$=\int_0^1\ln(y)\left(\frac{\pi}{4}\frac{y}{1+y^2}-\frac{\ln(1+y)}{1+y^2}+\frac{\ln(2)}{2(1+y^2)}\right)dy$$
$$=-\frac{\pi^3}{192}-\color{blue}{\int_0^1\frac{\ln(y)\ln(1+y)}{1+y^2}dy}-\frac12\ln(2)\ G\tag2$$
By plugging $(2)$ in $(1)$, the blue integral magically cancels out and we get $I=\frac12G\ln2-\frac{\pi^3}{64}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing that $c_{i}\equiv 0\pmod{p}$ Let the numbers $c_{i}$ be defined by the power series identity $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}= 1+c_{1}x+c_{2}x^{2}+\ldots$$ Show that $c_{i}\equiv 0\pmod{p}$ for all $i\geq 1$.
$\textbf{Solution.}$ First, from the Binomial Theorem we have $$(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!} x^{2} + \frac{n(n+1)(n+2)}{3!}x^{3} + \cdots$$
\begin{align*}
(1-x)^{-(p-1)} &= 1 + (p-1)x + \frac{(p-1)p}{2!} x^{2} + \frac{(p-1)p(p+1)}{3!}x^{3} + \cdots \\
&= 1 + \binom{p-1}{1} x + \binom{p}{2}x^{2} + \binom{p+1}{3}x^{3} + \binom{p+2}{4}x^{4} + \cdots \\
&= 1+\binom{p-1}{1}x + \binom{p}{2}x^{2} + \left[\binom{p}{2}+\binom{p}{3}\right]x^{3}+\left[\binom{p+1}{3}+\binom{p+1}{4}\right]x^{4} +...\\
&=1+\binom{p-1}{1}x + \binom{p}{2}x^{2} + \left[\binom{p}{2}+\binom{p}{3}\right]x^{3}+\left[\binom{p}{2}+2\binom{p}{3}+\binom{p}{4}\right]x^{4} +...
\end{align*}
where $\binom{p+1}{3}x^{3}=\Bigl\{\binom{p}{2}+\binom{p}{3}\Bigr\}x^{3}$ and $\binom{p}{4}x^{4}=\Bigl\{\binom{p}{2}+2\binom{p}{3}+\binom{p}{4}\Bigr\}x^{4}$ follows from using $\textit{Pascal's Identity.}$ Now we have $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}=(1+x+x^{2}+\ldots)\cdot (1-x)^{-(p-1)} =1+c_{1}x+c_{2}x^{2}+\ldots$$ Using the binomial expansion of $(1-x)^{-(p-1)}$ and comparing the coefficients, we get
\begin{align*}
c_{1} &= 1 + \binom{p-1}{1} =p\\
c_{2} &= \binom{p}{2} + \binom{p-1}{1} + 1 = \binom{p}{2} + c_{1}\\
c_{3} &= \binom{p+1}{3} + \binom{p}{2} + \binom{p-1}{1} + 1 =\binom{p}{2}+\binom{p}{3}+c_{2}\\
c_{4} &=\binom{p+2}{4}+c_{3} = \binom{p}{2}+2\binom{p}{3}+\binom{p}{4}+c_{3}
\end{align*}
In general $c_{i}=\binom{p+i-2}{i}+c_{i-1}$. Now by we know that $\binom{p}{i}\equiv 0\pmod{p}$ whenever $0<i<p$.
*
*So I am able to observe that $c_{1},c_{2},c_{3},c_{4}$ are $0\pmod{p}$. But still my argument is not correct. Since $c_{i}\neq\binom{p+i-2}{p}+c_{i-1}$ if $i\geq p$. A nice solution would be helpful.
|
In $R:=\mathbb{F}_p[\![x]\!]$, $(1-x)^p=1-x^p$. Because $1-x$ is an invertible element of $R$ (with inverse $1+x+x^2+\ldots$), $$\frac{1-x^p}{(1-x)^p}=1$$ in $R$. Therefore, [...]. The rest is up to you.
|
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Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image?
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is:
$90^{\circ}+ 720^{\circ}k$ where $k=1,2,3,...$
Intuitively, I understand the answer, but that's not the point, I would like to understand the correct mathematical solution. Thank you very much!
|
Firstly, denote $x=\frac{\theta}{2}$. Let $t=\tan \frac{x}{2}$. The equation $\sin x + \cos x = \sqrt 2$ became $$\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}=\sqrt 2.$$
Therefore $2t + 1 -t^2 = \sqrt 2(1+t^2) \Rightarrow (\sqrt 2+1)t^2 -2t + \sqrt 2 - 1 = 0$. This equation has the solutions $t_1=t_2=\sqrt 2 - 1$. So
$$\frac{x}{2}\in \{\arctan(\sqrt 2-1)+k\pi\;|\;k\in\mathbb Z\}.$$
Thus, since $\theta = 4x$, we get
$$\theta \in \{4\arctan(\sqrt 2 - 1)+4k\pi\;|\;k\in\mathbb Z\}.$$
We have $$\tan(2\arctan(\sqrt 2-1))=\frac{2(\sqrt 2-1)}{(\sqrt 2-1)^2-1}=1,$$
hence $2\arctan(\sqrt 2-1)=\frac{\pi}{4} \Rightarrow 4\arctan(\sqrt 2-1)=\frac{\pi}{2}$. In conclusion,
$$\theta \in \{\frac{\pi}{2}+4k\pi\;|\;k\in\mathbb Z\}.$$
|
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Which of the numbers has the largest number of divisors?
Which of the numbers $1,2,\ldots,1983$ has the largest number of divisors?
Firstly notice that each number less than or equal to $1983$ has at most $4$ different prime divisors since $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 > 1983$. We then analyze each of the cases of the amount of prime divisors.
Case 1: $1$ prime divisor
In this case there are only two divisors.
Case 2: Here we can have $2^a \cdot 3^b$. The set of all such numbers is $\{2^1 \cdot 3^1,2^1 \cdot 3^2, 2^1 \cdot 3^3, 2^1 \cdot 3^4, 2^1 \cdot 3^5, 2^1 \cdot 3^6, 2^2 \cdot 3^1, 2^2 \cdot 3^2, 2^2 \cdot 3^3, 2^2 \cdot 3^4, 2^2 \cdot 3^5, 2^3 \cdot 3^1, 2^3 \cdot 3^2, 2^3 \cdot 3^3, 2^3 \cdot 3^4, 2^3 \cdot 3^5, 2^4 \cdot 3^1, 2^4 \cdot 3^2, 2^4 \cdot 3^3, 2^4 \cdot 3^4, 2^5 \cdot 3^1, 2^5 \cdot 3^2, 2^5 \cdot 3^3, 2^6 \cdot 3^1, 2^6 \cdot 3^2, 2^6 \cdot 3^3, 2^7 \cdot 3^1, 2^7 \cdot 3^2, 2^8 \cdot 3^1, 2^9 \cdot 3^1\}$. Thus the maximum number of divisors in this case occurs for $2^6 \cdot 3^3$ which is $28$.
The other cases are similar but there are many numbers in the other sets. Is there a more efficient way of going about solving this?
|
If $n=p_1^{e_1}\cdots p_r^{e_r}$, then the number of divisors $d(n)$ of $n$ is given by
$$
d(n)=(e_1+1)(e_2+1)\ldots (e_r+1).
$$
We can have at most four different prime divisors, as you said,because otherwise $n>2\cdot 3\cdot5\cdot 7\cdot 11=2310$. Also, we may assume that we have the smallest primes, i.e., $n=2^{e_1}3^{e_2}5^{e_3}7^{e_4}$.
So your method works well, and we obtain, after only cheking a few cases,
$$
d(1680)=d(2^4\cdot 3\cdot 5\cdot 7)=5\cdot 2\cdot 2\cdot 2=40
$$
as the maximum.
|
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Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
|
If
$$I_n=\int\frac{1}{(ax^2+b)^n}dx\quad,\quad n\ge 2$$
$$I_1=\frac{\sqrt{\frac{b}{a}}}{b}\tan^{-1}\left(\sqrt{\frac{a}{b}}x\right)$$
then
$$I_n=\frac{2n-3}{2b(n-1)}I_{n-1}+\frac{x}{2b(n-1)(ax^2+b)^{n-1}}$$
set $a=1$ and $b=2$ apply
$$\frac{x^2-2}{(x^2+2)^3}=\frac{1}{(x^2+2)^2}-\frac{4}{(x^2+2)^3}$$
we have
$$\int \frac{x^2-2}{(x^2+2)^3}dx=-\left(\frac{\sqrt{2}}{16}\tan^{-1}\left(\frac{\sqrt{2}x}{2}\right)+\frac{x}{8x^2+16}+\frac{x}{2(x^2+2)^2}\right)+c$$
|
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Prove that $\max\{a_i \mid i \in \{0,1,\ldots,1984\}\} = a_{992}$
Consider the expansion
$$\left(1 + x + x^{2} + x^{3} + x^{4}\right)^{496} =
a_{0} + a_{1}x + \cdots + a_{1984}\,\,x^{1984}
$$
Prove that
$\max\left\{a_{i} \mid i \in \left\{0,1,\ldots,1984\right\}\right\} =
a_{992}\ $.
Since $1 + x + x^{2} + x^{3} + x^{4}$ is a palindromic polynomial, so is $\left(1 + x + x^{2} + x^{3} + x^{4}\right)^{496}$. Now we want to show that $a_{1},a_{2},\ldots,a_{1984}$ has exactly one maximum and we are done.
|
As you have noticed, for $(1 + x + x^2 + x^3 + x^4)^n = a_0 + a_1x + \cdots + a_{4n}x^{4n}$, we have
$$
a_i = a_{4n-i}\quad\text{for } 0 \leq i \leq 4n
$$
So it is sufficient to prove that $a_0 < a_1 < \cdots < a_{2n}$. We prove by induction below.
Basis. For $n = 2$, we have
$$
(1 + x + \cdots + x^4)^2 = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 4x^5 + 3x^6 + 2x^7 + x^8
$$
Thus $a_0 < a_1 < \cdots < a_4$.
Induction. Suppose for $n \geq 2$ we have
$$
(1 + x + \cdots + x^4)^n = a_0 + a_1x + \cdots + a_{4n}x^{4n}
$$
with $a_0 < a_1 < a_2 < \cdots < a_{2n}$ and
$$
(1 + x + \cdots + x^4)^{n+1} = b_0 + b_1x + \cdots + b_{4n+4}x^{4n+4}
$$
We aim to prove that $b_0 < b_1 < \cdots < b_{2n+2}$. Note that
$$
(1 + x + \cdots + x^4)^{n+1} = (1+x+\cdots + x^4)^n\cdot(1+x+\cdots +x^4)
$$
Therefore,
\begin{align}
b_i &= a_i + a_{i-1} + a_{i-2} + a_{i-3} + a_{i-4}\quad \text{for}\quad 4\leq i \leq 2n + 2\\
b_3 &= a_3 + a_2 + a_1 + a_0 \\
b_2 &= a_2 + a_1 + a_0 \\
b_1 &= a_1 + a_0 \\
b_0 &= a_0
\end{align}
Easy to observe that $b_0 < b_1 < b_2 < b_3 < b_4$. For $4 < i \leq 2n$, we have
$$
b_{i} - b_{i-1} = a_i - a_{i-5} > 0
$$
by our induction assumption. Moreover, we have
$$
b_{2n + 1} - b_{2n} = a_{2n+1} - a_{2n-4} = a_{2n-1} - a_{2n-4} > 0
$$
and
$$
b_{2n + 2} - b_{2n+1} = a_{2n + 2} - a_{2n - 3} = a_{2n - 2} - a_{2n - 3} > 0
$$
Therefore,
$$
b_0 < b_1 < \cdots < b_{2n + 2}
$$
When $n = 496$, by our proof, $a_{992}$ is maximum.
|
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On finding the complex number satisfying the given conditions Question:- Find all the complex numbers $z$ for which $\arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4}$ and $|z-3+i|=3$
My solution:-
$\begin{equation}
\arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4} \implies \arg\left(\dfrac{z-2-i}{z-4-3i}\right)=\dfrac{\pi}{4}
\end{equation}\tag{1}$
Now, what this suggests is that $z$ lies on a segment with the end points of the segment being $(2,1)$ and $(4,3)$. Now to find the equation of the circle we find the center and radius of the circle which comes out to be $(4,1)$ and $2$, respectively. So, the equation of the circle comes out to be
$$(x-4)^2+(y-1)^2=4$$
Now all we need to do is find the point of intersection of the circles $|z-4-i|=2$ and $|z-3+i|=3$ to find the desired $z$, which comes out to be
$$\boxed {z=\left(4 \mp \dfrac{4}{\sqrt{5}}\right) + i\left(1 \pm \dfrac{2}{\sqrt{5}} \right)}$$.
Now just to check which one is the solution I substituted both these $z$ back into $(1)$ but I found out that neither satisfy the condition as the argument comes out to be $\dfrac{\pm\sqrt{5}+5}{2}$, instead of 1, so according to my solution neither are the answer.
My problem with the question:-
But as it turns out as per the book I am solving, the answers are both the complex numbers $z$ that I found out. Also, I think only one of these is the solution because on some rough plotting I found out that the circle $|z-3+i|=3$ intersects the segment $|z-4-i|=2$ with end points as mentioned in the solution only once at $z=\left(4+\dfrac{4}{\sqrt{5}}\right) + i\left(1-\dfrac{2}{\sqrt{5}} \right)$.
So, is it me that is wrong or the book or maybe both. If there is anything wrong in my solution please point it out.
|
From the following where $\circ=\pi/4$,
$\qquad\qquad\qquad $
the circle $$(x-4)^2+(y-1)^2=4$$
you've found is correct, but note that we have only the part below the line $y=x-1$ passing through $(2,1)$ and $(4,3)$.
This is because the angle (as measured in counter-clock wise direction) from $\vec{zz_1}$ to $\vec{zz_2}$ is $\pi/4$ where $z_1=4+3i,z_2=2+i$.
$\qquad\qquad\qquad$
Thus, solving
$$(x-4)^2+(y-1)^2=4$$
$$y\lt x-1$$
$$(x-3)^2+(y+1)^2=3^2$$
gives only one solution $$x=4+\frac{4}{\sqrt 5},\quad y=1-\frac{2}{\sqrt 5}$$
(and this satisfies $(1)$.)
|
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|
Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$ Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$
Solutions are $x=2$ or $x=-1$.
But $x=2$ does not satisfy $\sqrt{x+2}+x=0$.Because $\sqrt{4}+2 \neq0$
So does it mean that they are different ? Why ?
|
Ofcourse there lies a difference,
Starting with $ x^2-x-2=0 \implies x^2=x+2 $ , Now follow from here that ,
$ x = \pm \sqrt{x+2} $ , When you take the negative value you get the above result as the solutions are $ x=2 $ or $x=-1 $ , so the negative value satisfies when you have a equation created with the negative sign taken.
In particular , $ (x+\sqrt{x+2})(x-\sqrt{x+2}) = x^2-x-2=0 $ here, so they doesn't represent the same equation.
|
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|
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff & x^3+y^3\geq \frac{1}{4}(x+y)^3
\end{split}$$
is the proof valid? is there a shorter way?
|
Write $x=m+z$ and $y=m-z$, so that $x+y=2m\gt0$. Then
$$x^3+y^3=(m+z)^3+(m-z)^3=2m^3+6mz^2\ge2m^3={1\over4}(2m)^3={1\over4}(x+y)^3$$
Note that we only need for the midpoint $m$ between $x$ and $y$ to be positive in order for the inequality to hold.
|
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|
Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{1}{2}\int\sin x\cos x\sin4x\,dx$
After this step, I tried evaluating the integral by using the $\sin a\sin b$ property.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$
|
Here is the most straight way to solve this problem:
Notice that $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x $
So $\sin^2 x \cos 4x = \sin^2 (1-2\sin^2 2x) = \sin^2 [1-2(1-2\sin^2 x)^2] = -8\sin^6 x +8\sin^4 x -\sin^2 x$
By Reduction Formula: The integral can be easily solved.
|
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|
Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for the second and third throw. Again the probability of getting a $6$ is fourth roll is $\frac{1}{6}$. So the probability of ending the game in the fifth roll is $\frac{5^3}{6^3}\times\frac{1}{6^2}=\frac{125}{6^5}$.
But the answer is not correct. Where is my mistake? Help please.
|
Let $X$ denote any value between $1-5$, then the optional sequences are:
*
*$XXX66$
*$X6X66$
*$6XX66$
Calculate the probability of each sequence:
*
*$P(XXX66)=\frac56\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^3}{6^5}$
*$P(X6X66)=\frac56\cdot\frac16\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}$
*$P(6XX66)=\frac16\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}$
Add up the above probabilities:
$$\frac{5^3+5^2+5^2}{6^5}\approx2.25\%$$
|
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|
$f(x)$ is a quadratic polynomial with $f(0)\neq 0$ and $f(f(x)+x)=f(x)(x^2+4x-7)$
$f(x)$ is a quadratic polynomial with $f(0) \neq 0$ and $$f(f(x)+x)=f(x)(x^2+4x-7)$$
It is given that the remainder when $f(x)$ is divided by $(x-1)$ is $3$.
Find the remainder when $f(x)$ is divided by$(x-3)$.
My Attempt:
Let $f(x)=ax^2+bx+c$ and $a,c \neq 0 $
I got $a+b+c=3$
And by the functional equation
$a[ax^2+(b+1)x+c]^2+b[ax^2+(b+1)x+c]+c= [ax^2+bx+c][x^2+4x-7]$
Then by putting $x=0$ , $ac^2 +bc+c=-7c$
Since $c \neq 0 $ , we have $ac+b+8=0$
Then comparing the coeffiecient of $x^4$ , we get $a^3=a$
Since $a \neq 0$ , $a=-1 $ or $a =1 $
Then how to proceed with two values of $a$ ?
or Is there a polynomial satisfying these conditions?
|
My answer to the question : Quadratic Functional equations.
Should work in the exact same manner, the extra conditions may or may not contradict the solution obtained, but in the case of a contradiction, then such a function does not exist.
|
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|
A closed form for $1^{2}-2^{2}+3^{2}-4^{2}+ \cdots + (-1)^{n-1}n^{2}$ Please look at this expression:
$$1^{2}-2^{2}+3^{2}-4^{2} + \cdots + (-1)^{n-1} n^{2}$$
I found this expression in a math book. It asks us to find a general formula for calculate it with $n$.
The formula that book suggests is this:
$$-\frac{1}{2}\times (-1)^{n} \times n(n+1)$$
Would you mind explaining to me how we get this formula?
|
Let $S(n)=1^2+2^2+\dots+n^2$ and $T(n)=1^2-2^2+3^2-4^2+\dots+(-1)^{n-1}n^2$.
Suppose first $n=2k$ is even; then
$$
T(n)=T(2k)=
S(n)-2\bigl(2^2+4^2+\dots+(2k)^2\bigr)=
S(n)-8S(k)
$$
Since
$$
S(n)=\frac{1}{3}n\left(n+\frac{1}{2}\right)(n+1)=\frac{n(2n+1)(n+1)}{6}
$$
We have
$$
T(2k)=\frac{2k(4k+1)(2k+1)}{6}-8\frac{k(2k+1)(k+1)}{6}=
-\frac{2k(2k+1)}{2}
$$
If $n=2k+1$ is odd, then
$$
T(n)=T(2k+1)=S(n)-8S(k)=
\frac{(2k+1)(4k+3)(2k+2)}{6}-8\frac{k(2k+1)(k+1)}{6}
$$
and an easy computation gives
$$
T(2k+1)=\frac{(2k+1)((2k+1)+1)}{2}
$$
So
$$
T(n)=(-1)^{n-1}\frac{n(n+1)}{2}
$$
|
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|
Solve the equation $x^{x+y}=y^{y-x}$ over natural numbers
Solve the equation $x^{x+y}=y^{y-x} \tag 1$ where $x,y \in \mathbb{N}$
$x = 1,y = 1$ is a solution, now suppose $x \ne 1, y \ne 1$.
Obviously $x, y \ne 0$ and $x, y$ have same prime divisors.
Because $x+y \gt y-x$ it follows that $x \mid y$ therefore $y = kx$ and (1) becomes: $x^{(k+1)x}=(kx)^{(k-1)x} \tag2$ and, dividing by $x^{k-1}$ we get
$x^{2x}=k^{(k-1)x} \tag 3$ therefore $x^2=k^{k-1} \tag 4$ and here I've got stuck. Any help is appreciated.
UPDATE
From (4) we have $k \ge 3$ therefore $2 \le k-1$ and $ k \mid x$. Let $x=km$ and (4) becomes $k^2m^2=k^{k-1} \tag5$ and $m^2=k^{k-3} \tag6$
UPDATE 2
$x=t^{t^2-1}, y=t^{t^2+1}$ is solution $\forall t \in \mathbb{N}-\{0\}$
|
From
$x^2=k^{k-1}$,
$k$ must be odd.
Let $k = 2j+1$,
so
$x^2 = (2j+1)^{2j}$
or
$x = (2j+1)^j$.
Then
$y = kx
=(2j+1)(2j+1)^j
= (2j+1)^{j+1}
$.
Check:
$x^{x+y}?y^{y-x}$
$\begin{array}\\
x^{x+y}
&=((2j+1)^j)^{(2j+1)^j+(2j+1)^{j+1}}\\
&=(2j+1)^{j(2j+1)^j(1+(2j+1))}\\
&=(2j+1)^{j(2j+1)^j(2j+2)}\\
&=(2j+1)^{2j(j+1)(2j+1)^j}\\
\end{array}
$
$\begin{array}\\
y^{y-x}
&=((2j+1)^{j+1})^{(2j+1)^{j+1}-(2j+1)^{j}}\\
&=(2j+1)^{(j+1)((2j+1)^{j+1}-(2j+1)^{j})}\\
&=(2j+1)^{(j+1)(2j+1)^{j}((2j+1)-1)}\\
&=(2j+1)^{(j+1)(2j+1)^{j}(2j)}\\
&=(2j+1)^{2j(j+1)(2j+1)^{j}}\\
\end{array}
$
And they match!
|
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|
Doubt in finding number of integral solutions Problem :
writing $5$ as a sum of at least $2$ positive integers.
Approach :
I am trying to find the coefficient of $x^5$ in the expansion of $(x+x^2+x^3\cdots)^2\cdot(1+x+x^2+x^3+\cdots)^3$ .
which reduces to coefficient of $x^3$ in expansion of $(1-x)^{-5}$ ,which is $${7\choose3}= 35$$
but we can count the cases and say that answer must be $6$ :
$4 + 1 $
$3 + 2 $
$3 + 1 + 1$
$2 + 2 + 1 $
$2 + 1 + 1 + 1$
$1 + 1 + 1 + 1 + 1$
At which stage am I making a mistake ? Thanks
|
Actually you count all the possible permutations that gives $x^5$ in
$\displaystyle \color{red}{\frac{x^2}{(1-x)^2}} \times
\color{blue}{\frac{1}{(1-x)^3}}$,
$\color{red}{(4+1)}+\color{blue}{(0+0+0)}$ counts
$\color{red}{2} \times \color{blue}{1}$ possibilities.
$\color{red}{(3+2)}+\color{blue}{(0+0+0)}$ counts
$\color{red}{2} \times \color{blue}{1}$ possibilities.
$\color{red}{(3+1)}+\color{blue}{(1+0+0)}$ counts
$\color{red}{2} \times \color{blue}{3}$ possibilities.
$\color{red}{(2+2)}+\color{blue}{(1+0+0)}$ counts
$\color{red}{1} \times \color{blue}{3}$ possibilities.
$\color{red}{(2+1)}+\color{blue}{(2+0+0)}$ counts
$\color{red}{2} \times \color{blue}{3}$ possibilities.
$\color{red}{(2+1)}+\color{blue}{(1+1+0)}$ counts
$\color{red}{2} \times \color{blue}{3}$ possibilities.
$\color{red}{(1+1)}+\color{blue}{(3+0+0)}$ counts
$\color{red}{1} \times \color{blue}{3}$ possibilities.
$\color{red}{(1+1)}+\color{blue}{(2+1+0)}$ counts
$\color{red}{1} \times \color{blue}{6}$ possibilities.
$\color{red}{(1+1)}+\color{blue}{(1+1+1)}$ counts
$\color{red}{1} \times \color{blue}{1}$ possibility.
Totally $35$ possibilities.
|
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|
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-sinxdx$
$$-\int \frac{(1-u^2 )u^2}{2-u^2}du=\int \frac{u^4-u^2 }{2-u^2}du$$
Polynomial division give us
$$\int-u^2-1+\frac{2}{u^2-2}du=-\frac{u^3}{3}-u=2\int \frac{1}{u^2-2}du$$
Using partial fraction we get to:
$$\frac{1}{u^2-2}=\frac{1}{(u+\sqrt{2})(u^2-\sqrt{2})}=\frac{A}{u^2+\sqrt{2}}+\frac{B}{u^2-\sqrt{2}}$$
So we get to:
$$1=(A+B)u^2+\sqrt{2}(A-B)$$
So $A=B$ but $0*\sqrt{2}\neq 1$, where did it go wrong?
|
$$\int\frac{\sin^3(x)\cos^2(x)}{1+\sin^2(x)}\space\text{d}x=$$
Use $\sin^2(x)=1-\cos^2(x)$:
$$-\int\frac{\sin(x)(\cos^2(x)-\cos^4(x))}{\cos^2(x)-2}\space\text{d}x=$$
Substitute $u=\cos(x)$ and $\text{d}u=-\sin(x)\space\text{d}x$:
$$\int\frac{u^2-u^4}{u^2-2}\space\text{d}u=$$
Use long division:
$$-\int\left[u^2+1+\frac{2}{u^2-2}\right]\space\text{d}u=$$
$$-\left[\int u^2\space\text{d}u+\int1\space\text{d}u+2\int\frac{1}{u^2-2}\space\text{d}u\right]=$$
$$-\left[\frac{u^3}{3}+u-\int\frac{1}{1-\frac{u^2}{2}}\space\text{d}u\right]=$$
Substitute $s=\frac{u}{\sqrt{2}}$ and $\text{d}s=\frac{1}{\sqrt{2}}\space\text{d}u$:
$$-\left[\frac{u^3}{3}+u-\sqrt{2}\int\frac{1}{1-s^2}\space\text{d}s\right]=$$
$$-\left[\frac{u^3}{3}+u-\sqrt{2}\text{arctanh}(s)\right]+\text{C}=$$
$$-\left[\frac{\cos^3(x)}{3}+\cos(x)-\sqrt{2}\text{arctanh}\left(\frac{\cos(x)}{\sqrt{2}}\right)\right]+\text{C}$$
|
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|
Integer solutions to $5m^2-6mn+7n^2 = 1985$
Are there integers $m$ and $n$ such that $$5m^2-6mn+7n^2 = 1985?$$
Taking the equation modulo $3$ gives $n^2-m^2 \equiv 2 \pmod{3}$. Thus, $3 \mid n$ but $3 \nmid m$. How can I use this to find a contradiction?
|
Taking modulo $5$ we have
$$n(2n-m)\equiv 0\pmod 5$$
This gives us two possibilities:
*
*If $n$ is a multiple of $5$, write $n=5u$ and $175u^2-30mu+5m^2=1985$, or $35u^2-6mu+m^2=397$. Then
$$m=\frac{6u\pm\sqrt{1588-104u^2}}{2}=3u\pm\sqrt{397-26u^2}$$
*
*If $2n\equiv m\pmod 5$, then $m=5u+2n$ and (after some easy calculations) $m^2-14mu-29u^2+397=0$. Now,
$$m=\frac{14u\pm\sqrt{312u^2-1588}}2=7u\pm\sqrt{78u^2-397}$$
Both possibilities require $-397$ to be a square modulo $13$, but $-397\equiv -7\pmod{13}$, and $-7$ is not a square modulo $13$.
Thus, there is no solution.
|
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|
Prove that $a_n$ is a perfect square
*
*Let $\,\,\,\left(a_{n}\right)_{\ n\ \in\ \mathbb{N}\,\,\,}$ be the sequence of integers defined recursively by
$$
a_{1} = a_{2} = 1\,,\qquad\quad a_{n + 2} = 7a_{n + 1} -a_{n} - 2\quad
\mbox{for}\quad n \geq 1
$$
*Prove that $a_{n}$ is a perfect square for every $n$.
We have $a_{3} = 4, a_{4} = 25, a_{5} = 169,\ldots$
Is there a way we can simplify the recursion or get its closed form in order to get that it is a perfect square ?.
|
For first, we may translate our sequence in order to have some sequence fulfilling $b_{n+2}=7b_{n+1}-b_n$. For such a purpose, it is enough to set $a_n=b_{n}+\frac{2}{5}$, leading to $b_0=b_1=\frac{3}{5}$. Now the characteristic polynomial of the sequence $\{b_n\}_{n\geq 0}$ is
$$ p(x)=x^2-7x+1 $$
with roots given by $\frac{7\pm 3\sqrt{5}}{2}$, so the explicit form of $b_n$ is given by:
$$ b_n = A\left(\frac{7+3\sqrt{5}}{2}\right)^2+B\left(\frac{7-3\sqrt{5}}{2}\right)^n $$
where the values of the constants $A,B$ are fixed by $b_0$ and $b_1$:
$$ b_n = \frac{-\sqrt{5}+3}{10}\left(\frac{7+3\sqrt{5}}{2}\right)^2+\frac{\sqrt{5}+3}{10}\left(\frac{7-3\sqrt{5}}{2}\right)^n. $$
Now the identity:
$$\boxed{ a_n = b_n+\frac{2}{5} = F_{2n-1}^2 }$$
is a simple consequence of the closed formula for Fibonacci numbers.
|
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|
Converting repeating decimal in base b to a fraction the same base The repeating decimal .36666... in base 8 can be written in a fraction in base 8.
I understand simple patterns such as 1/9 in base 10 is .1111.... so 1/7 in base 8 is .1111.
But I'm not too sure how to convert this decimal in this base to the fraction in the same base.
|
\begin{align}
0.3\bar{6}_8
&= \frac{3}{8} + 6\left(\frac{1}{8^2}+\frac{1}{8^3} + \cdots\right)\\
&= \frac{3}{8} + \frac{6}{8^2}\left(1+\frac{1}{8}+ \frac{1}{8^2} +\cdots\right)\\
&= \frac{3}{8} + \frac{6}{8^2}\frac{1}{1-(1/8)} & \text{geometric series}\\
&= \frac{3}{8} + \frac{3}{28}\\
&= \frac{27}{56}\\
&= \frac{33_8}{70_8}.
\end{align}
|
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|
How would I solve $n(a) := 3 1 7 2 a 6 3 7 5$ with $9|n(a)$? $n(a) := 3 1 7 2 a 6 3 7 5$
What value needs a to be so that $9|n(a)$?
Is there a way to solve this fast? My initial idea was to write it as a sum like
$5 + 7 * 10 + 3 * 100 + 6 * 1000 + a * 10000 ... $
Because $a + c = b + d (\mod m )$
But this seems a bit time intensive and my professor just said the answer is trivial, it is 2. So I assume that I can somehow directly know that it is 2? How would I do that?
update:
$34 + a = 0(\mod 9)$
$34 \mod 9 = 7$
$b = a \mod 9$
The rests need to add up to 9, therefore
$7 + b = 0 (\mod 9)$
$b = 2$
Therefore a needs also to be 2.
|
A number $n$ is divisible by $9$ if and only if the sum of the digits of $n$ is also divisible by $9$
(if and only if the sum of the digits of the sum of the digits of $n$ is also divisible by $9$) (if and only if the sum of the digits of the sum of the digits of the sum of the digits...)
E.g. $18324$ has sum of digits $1+8+3+2+4=18$ which is divisible by $9$ so $18324$ is also divisible by $9$.
As we desire $n(a)$ to be divisible by $9$, this will occur if and only if the sum of the digits is divisible by $9$.
$\iff$ $3+1+7+2+a+6+3+7+5\equiv 0\pmod{9}$
$\iff 34+a\equiv 0\pmod{9}$
$\iff a\equiv -34\pmod{9}$
$\iff a\equiv 2\pmod{9}$
As $2$ is the only digit which is equivalent to $2\pmod{9}$ it must be that $a=2$
Note: The final line is necessary as some other cases may yield multiple answers. If the number was instead $31\color{red}{9}2a6375$ we could have had $a=0$ or $a=9$
In a much more informal proof, one could "case out nines" very quickly. The $3$ and the $6$ will cancel, the $2$ and the $7$ will cancel. The $1,3,5$ together cancel, leaving you with two digits still needing to properly cancel one another: a $7$ and the $a$. The only digit which cancels the $7$ will be a $2$. This can be done in your head within a few seconds given proper mental arithmetic.
|
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|
Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$ If we have to find the value of the following (1)
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)
$$
I know that
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2 \right)
$$
I tried it lot and got a result but then stuck! (2)
(1) http://i.stack.imgur.com/26hA4.jpg
(2) http://i.stack.imgur.com/g2vBb.jpg
|
One may observe that summing
$$
u_{r+1}-u_{r-1}
$$ may be simplified as a telescoping sum:
$$
\sum_{r=1}^N\left(u_{r+1}-u_{r-1}\right)=\sum_{r=1}^N\left(u_{r+1}-u_{r}\right)+\sum_{r=1}^N\left(u_{r}-u_{r-1}\right)=u_{N+1}+u_N-u_1-u_0.
$$
From the identity
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2 \right)
$$ by telescoping you then obtain
$$
\begin{align}
\sum_{r=1}^N\arctan \left(\frac{4}{r^2+3} \right)&=\arctan \left(\frac{N+1}2 \right)+\arctan \left(\frac{N}2 \right)
\\\\&-\arctan \left(\frac12 \right)-\arctan \left(\frac{1-1}2 \right),
\end{align}
$$ giving, as $N \to \infty$ ,
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)=2\arctan \left(\infty \right)-\arctan \left(\frac12 \right)=\frac{\pi}2+\arctan 2.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x + \csc x + \sec x}=\frac{t(1- t)}{1+ t^2}$$
$$\implies \int \frac{\tan {\frac{x}{2}}(1-\tan {\frac{x}{2}})}{1+\tan^2 {\frac{x}{2}}}{dx}$$
I think, I have made the things more difficult. How can I proceed further? Is there any better substitution for it?
|
$$\begin{aligned}\int \frac{d x}{\tan x+\cot x+\csc x+\sec x}&=\int \frac{\cos x \sin x}{1+\cos x+\sin x} d x \\&= -\frac{1}{2} \int \frac{1-(\cos x+\sin x)^2}{1+\cos x+\sin x} d x \\& = -\frac{1}{2} \int(1-\cos x-\sin x) d x\\&= \frac{1}{2}(-x+\sin x-\cos x)+C\end{aligned} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving that $\lim_{x\to 2}\frac{x^2-5x}{x^2+2}=-1$ using the $\epsilon$-$\delta$ definition of a limit My attempt:
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|<\left|\frac{x^2-5x}{x^2+2}\right|<
\left|\frac{x^2-5x}{x^2}\right|<\frac{1}{x^2}|x^2-5x|,$$
using the restriction $|x-2|<2$, so $0<x<4$, thus
$$\frac{1}{x^2}|x^2-5x|=\frac{1}{x^2}|x^2-6x+4-4+x|=\frac{1}{x^2}|(x-2)^2+x-4|<\frac{(x-2)^2}{x^2}<\epsilon,$$
hence, due to the fact that $|x-2|<\delta\le2$ we get
$$
\delta^2<\epsilon x^2 < \epsilon16.
$$
Finally, for any $\epsilon >0$ the corresponding $\delta$ is $\min\{\epsilon, 4\epsilon^{1/2}\}$.
Is it correct?
|
Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $\left|\frac{x^2-5x}{x^2+2}+1\right|<\epsilon$ whenever $|x-2|<\delta$. But, as Andre notes,
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|=\left|\frac{2x^2-5x+2}{x^2+2}\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|.
$$
We find a positive constant $C$ such that $\left|\frac{2x-1}{x^2+1}\right|<C\Rightarrow |x-2|\left|\frac{2x-1}{x^2+1}\right|<C|x-2|$, and we can make $C|x-2|<\epsilon$ by taking $|x-2|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-2|<1$ and note the following:
\begin{align}
|x-2|<1&\implies 1<x<3\\[1em]
&\implies 1>\frac{1}{x}>\frac{1}{3}\\[1em]
&\implies 1>\frac{1}{x^2}>\frac{1}{9}\\[1em]
&\implies \frac{1}{3}>\frac{1}{x^2+2}>\frac{1}{11}\\[1em]
&\implies \frac{5}{3}>\frac{2x-1}{x^2+2}>\frac{1}{11}\\[1em]
&\implies C=\frac{5}{3}.
\end{align}
Thus, we should choose $\delta=\min\left\{1,\frac{3\epsilon}{5}\right\}$. To see that this choice of $\delta$ works, consider the following:
Given $\epsilon>0$, we let $\delta=\min\left\{1,\frac{3\epsilon}{5}\right\}$. If $|x-2|<1$, then $\left|\frac{2x-1}{x^2+2}\right|<\frac{5}{3}$. Also, $|x-2|<\frac{3\epsilon}{5}$. Hence,
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|<\frac{3\epsilon}{5}\cdot\frac{5}{3}=\epsilon,
$$
as desired. $\blacksquare$
|
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|
Show that $a_n$ is decreasing
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.
First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_n) - 1}{3 - a_n} = \frac{8 - 3a_{n}}{3 - a_n} > 0$, which means showing $8 > 3a_{n}$, but I'm having trouble showing it.
|
Let us go for the overkill, i.e. to find an explicit formula for $a_n$. We may set $a_n=\frac{p_n}{q_n}$, with $p_1=2,q_1=1$, then deduce from $a_{n+1}=\frac{1}{3-a_n}$ the recurrence relation:
$$ w_{n+1}=\begin{pmatrix}p_{n+1} \\ q_{n+1} \end{pmatrix} = \begin{pmatrix}0 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix}p_{n} \\ q_{n} \end{pmatrix}=M w_m\tag{1}$$
The characteristic polynomial of the matrix $M$ is $x^2-3x+1$, having roots $\frac{3\pm\sqrt{5}}{2}=\left(\frac{1\pm\sqrt{5}}{2}\right)^2$.
By the Cayley-Hamilton theorem it follows that $x^2-3x+1$ is also the characteristic polynomial of the sequences $\{p_n\}_{n\geq 1}$ and $\{q_n\}_{n\geq 1}$, hence:
$$ p_n = p_{+}\left(\frac{1+\sqrt{5}}{2}\right)^{2n}+p_{-}\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\\ q_n = q_{+}\left(\frac{1+\sqrt{5}}{2}\right)^{2n}+q_{-}\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\tag{2} $$
and both $p_n$ and $q_n$ are linear combinations of the Fibonacci and Lucas numbers $F_{2n}$ and $L_{2n}$.
Interpolating through the values of $p_0,p_1,q_0,q_1$ we get:
$$\boxed{ a_n = \color{red}{\frac{F_{2n-5}}{F_{2n-3}}}}\tag{3} $$
and $\{a_n\}_{n\geq 1}$ is a decreasing sequence since $\{F_n\}_{n\geq 2}$ is an increasing sequence. Moreover, from $(3)$ it follows that:
$$ \lim_{n\to +\infty}a_n = \frac{1}{\varphi^2} = \frac{3-\sqrt{5}}{2} \tag{4} $$
as expected, since if $\{a_n\}_{n\geq 1}$ is converging, its limit has to fulfill $L=\frac{1}{3-L}$, i.e. has to be a root of the previous characteristic polynomial.
|
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|
Quadratic Inequality in terms of variable $x$
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help required, Thanks
|
Rewrite the inequality as $$(2a+x+6)^2< (6-x)(3x+6)$$
The right hand side is positive for $1<x<2$, so we get
$$-\sqrt{(6-x)(3x+6)}<2a+x+6<\sqrt{(6-x)(3x+6)}$$
and
$$-x-6-\sqrt{(6-x)(3x+6)}<2a<-x-6+\sqrt{(6-x)(3x+6)}$$
$$\frac{-x-6-\sqrt{(6-x)(3x+6)}}{2}<a<\frac{-x-6+\sqrt{(6-x)(3x+6)}}{2}$$
The maximum of the right hand side is $-4+2 \sqrt{3}$ (let $x=2$) and the minimum of the left side is $-\frac{7}{2}-\frac{3 \sqrt{5}}{2}$ (let $x=1$)
so $$-\frac{7}{2}-\frac{3 \sqrt{5}}{2}<a<-4+2 \sqrt{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
|
Set $f(x)=\left(1-\frac{1}{x+3}\right)^x=\left(\frac{x+2}{x+3}\right)^x$ where $x\ge 6$. We have
$$f'(x)=\left(\frac{x+2}{x+3}\right)^x\left(\ln\left(\frac{x+2}{x+3}\right)+\frac{x}{x^2+5x+6}\right)$$
$f'(x)<0$ for $x\ge 6$. Hence $f(x)\le f(6)=\left(\frac 89\right)^6<\frac 12$, i.e
$$\left(1-\frac{1}{x+3}\right)^x< \frac 12\tag 1$$
Note
$$\sum\limits_{k=1}^{n}{{{(k+2)}^{n}}}={{(n+3)}^{n}}$$
thus
$$\sum\limits_{k=1}^{n}{{{\left( \frac{k+2}{n+3} \right)}^{n}}}=1$$
in other words
$$\sum\limits_{k=1}^{n}{{{\left( 1-\frac{n+1-k}{n+3} \right)}^{n}}}=1$$
or
$$\sum\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+3} \right)}^{n}}}=1$$
Set
$$I=\sum\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+3} \right)}^{n}}}\tag 2$$
By application of Bernoulli's inequality, we have
$$\left( 1-\frac{k}{n+3} \right)\le {{\left( 1-\frac{1}{n+3} \right)}^{k}}$$
therefore as $n\ge 6$
$${{\left( 1-\frac{k}{n+3} \right)}^{n}}\le {{\left( 1-\frac{1}{n+3} \right)}^{nk}}< \left(\frac12 \right)^k\tag 3$$
$(2)$ and $(3)$
$$\color{red}{I=\sum\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+3}
\right)}^{n}}}<\sum_{k=1}^{n}\left(\frac12 \right)^k=1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1873787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Can anyone tell me the formula for this vector field? 3Blue1Brown's video on curl
$$V=\bigg[ \begin{array}{ccc}
P\left(x,y\right) \\
Q\left(x,y\right) \end{array} \bigg] $$
*
*$P$ gives you the x component at all points in space.
*$Q$ gives you the y component at any point in space.
What is the explicit formula for the above vector field?
*
*vortexes at $(6,0)$, $(-6,0)$, $(0,6)$, $(0,-6)$
*and a source and a sink at the origin.
*symmetrical
Hints, links welcome
|
Working through Andrew D. Hwang's answer above:
$$F\left(x,y\right)=\left(\frac{2(-y)}{1+\sqrt{x^2+y^2}},\frac{2(x)}{1+\sqrt{x^2+y^2}}\right)$$
$$G\left(x,y\right)=\left(\frac{-2(y)}{1+\sqrt{(x-6)^2+(y)^2}}+\frac{-2(y)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(y-6)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(y+6)}{1+\sqrt{(x)^2+(y+6)^2}},\\
\\
\\
\frac{-2(x-6)}{1+\sqrt{(x-6)^2+(y)^2}}\\
\\
+\frac{-2(x+6)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y+6)^2}}\right)$$
*
*first four terms give the x component
*the second four terms compute the y component
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration by means of partial fraction decomposition I'm trying to solve this indefinite integral by means of partial fraction decomposition:
$\int\dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$.
The denominator has complex (but not real) roots because $\Delta<0$; so, according with my calculus book, i try to decompose the integrand function in this form:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax+B}{\left(x^2+4x+5\right)}+\dfrac{Cx+D}{\left(x^2+4x+5\right)^2}$.
I get:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{\left(Ax+B\right)\left(x^2+4x+5\right)+Cx+D}{\left(x^2+4x+5\right)^2}$.
Multiplying the right term:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax^3+4Ax^2+5Ax+Bx^2+4Bx+5B+Cx+D}{\left(x^2+4x+5\right)^2}$.
Now i collect the terms with the same pwer of $x$:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D+ 5B}{\left(x^2+4x+5\right)^2}$.
Now i equate the two numerators:
$x+1=Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D$
and equate term by term: i get:
$A=0$, $B=0$, $C=1$, $D=1$.
With these values i get a correct identity:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{x+1}{\left(x^2+4x+5\right)^2}$
but this is unuseful in order to solve the integral.
Where is my mistake ?
|
Here is an easy way
$$\int { \frac { x+1 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\frac { 1 }{ 2 } \int { \frac { 2x+4-2 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\\ =\frac { 1 }{ 2 } \int { \frac { 2x+4 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } dx-\int { \frac { dx }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } } =\\ =\frac { 1 }{ 2 } \underset { { I }_{ 1 } }{ \underbrace { \int { \frac { d\left( x^{ 2 }+4x+5 \right) }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } } } -\underset { { I }_{ 2 } }{ \underbrace { \int { \frac { d\left( x+2 \right) }{ { \left( { \left( x+2 \right) }^{ 2 }+1 \right) }^{ 2 } } } } } =$$
Obviously,$$\\ { I }_{ 1 }=-\frac { 1 }{ 2\left( x^{ 2 }+4x+5 \right) } ,$$
now,let calculate ${ I }_{ 2 }$,substitute here $x+2=\tan { z } $,so that $${ I }_{ 2 }=\int { \frac { d\tan { z } }{ { \left( { \tan ^{ 2 }{ z } }+1 \right) }^{ 2 } } } =\int { \frac { \cos ^{ 4 }{ z } }{ \cos ^{ 2 }{ z } } } dz=\int { \cos ^{ 2 }{ z } =\frac { 1 }{ 2 } \int { \left( 1+\cos { 2z } \right
) dz } } =\frac { 1 }{ 2 } \left( z+\frac { \sin { 2z } }{ 2 } \right) $$
finally,
$$\int { \frac { x+1 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\frac { 1 }{ 2 } \left( -\frac { 1 }{ \left( x^{ 2 }+4x+5 \right) } -\arctan { \left( x+2 \right) -\frac { \sin { 2\left( \arctan { \left( x+2 \right) } \right) } }{ 2 } } \right) +C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that the matrices' linear transformations are similar. Let be $T: M_2(\mathbb{C}) \longrightarrow M_2(\mathbb{C})$ the following linear transformation $$ T\left( \begin{array}{cc}
x & y\\
z & w\\
\end{array} \right) = \left( \begin{array}{cc}
0 & x\\
z-w & 0\\
\end{array} \right) $$
a) Determine the matrix $T$ concerning to canonical basis.
b) Determine the matrix $T$ concerning to the following basis
$$ B = \left( \begin{array}{cc}
1 & 0\\
0 & 1 \end{array} \right), \left( \begin{array}{cc}
0 & 1\\
1 & 0 \end{array} \right), \left( \begin{array}{cc}
1 & 0\\
1 & 1 \end{array} \right), \left( \begin{array}{cc}
0 & 1\\
0 & 1 \end{array} \right) $$
of $M_2(\mathbb{R})$.
c) Show the matrix $M$ such that $[T]_B = M^{-1}[T]_{can}M$.
My attempt:
a) $[T]_{can} = \left( \begin{array}{rrrr}
0 & 0 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{array} \right)$
b) $[T]_B = \left( \begin{array}{rrrr}
2 & -1 & 1 & 1\\
1 & 0 & 1 & 0\\
-2 & 1 & -1 & -1\\
0 & 0 & 0 & 0\\
\end{array} \right)$
c) $[T]_{B,can} = \left( \begin{array}{rrrr}
1 & 0 & 0 & 1\\
0 & 1 & 1 & 0\\
1 & 0 & 1 & 1\\
0 & 1 & 0 & 1\\
\end{array} \right)$
and $[T]_{B,can}^{-1} = \left( \begin{array}{rrrr}
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
0 & 1 & 1 & 0\\
1 & 0 & 1 & 1\\
\end{array} \right)$, so $M = [T]_{B,can}$.
My doubt is how find the matrix $M$ without know about eigenvalues and eigenvectors and if my answers in $a$ and in $b$ are correct, thanks in advance!
EDIT: I edited my attempt, I think it's correct now.
|
Your answer for the first one is correct. I didn't check the second one fully but I think you made some mistakes. For instance, writing $B = (b_1,\ldots,b_4)$, we have $$T(b_1) = T\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}0&1\\-1&0\end{pmatrix} \neq \begin{pmatrix}1&0\\0&1\end{pmatrix} - \begin{pmatrix}1&0\\1&1\end{pmatrix} = b_1 - b_3,$$ so the first column can't possibly be $(1,0,-1,0)$.
For the last one, you are looking for the change of basis matrix from $B$ to $\mathit{can}$. The columns of this matrix simply consist of the coordinates (in the canonical basis) of the basis vectors in $B$. For instance, since $b_1 = e_1 + e_4$, the first column is $(1,0,0,1)$. (Here $\mathit{can} = (e_1,\ldots,e_4)$.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac{\pi}{12} $
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{3} -\sqrt{2}}{1+ \sqrt{6}}\right)$$
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) -\tan^{-1}{\sqrt3} + \tan^{-1} {\sqrt2} $$
$$\implies\frac{\pi}{2} -\frac{\pi}{3}=\frac{\pi}{6}$$
Another possibility is
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) +\tan^{-1}{\sqrt3} - \tan^{-1} {\sqrt2} $$
How to solve this ?
|
Hint the second bracket is $\tan(\frac{\pi}{3}-\tan^{-1}\sqrt{2})$ now can you proceed further
|
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|
how can I prove this $\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$ How can I prove the following equation?:
$$s=\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
Simplifying both terms of the equation:
$$\sum_{k=1}^{n}\frac{1}{4k^{2}-2k} = \sum_{k=1}^{n}\frac{1}{(2k-1)2k} =(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k})$$
$$\sum_{k=n+1}^{2n}\frac{1}{k}=(\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k})=(\sum_{k=1}^{n}\frac{1}{2k-1}+\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{k})$$
Now we have:
$$s=(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k})=(\sum_{k=1}^{n}\frac{1}{2k-1}\color{blue}{+}\sum_{k=1}^{n}\frac{1}{2k}\color{blue}{-\sum_{k=1}^{n}\frac{1}{k}})$$
How can I continue?
|
Let's do a cute induction on $n$:
For $n = 1$, we have $\frac{1}{4 - 2} = \frac{1}{2}$, so the claim holds.
Assume $n > 1$ and the claim holds for $n - 1$. Then $$\sum_{k = 1}^n \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = 1}^{n - 1} \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = n}^{2(n -1)} \frac{1}{k}$$
by the induction hypothesis. So we only need to show that $\frac{1}{2n} + \frac{1}{2n - 1} - \frac{1}{n} = \frac{1}{4n^2 - 2n}$, which is just a direct computation.
|
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|
$\tan^{-1}x$, $\tan^{-1}y$, $\tan^{-1}z$ are in arithmetic progression, as are $x$, $y$, $z$. Show ...
$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z $ are in arithmetic progression, as are $x$, $y$, $z$. (We assume $y \ne 0,1,-1$.) Show:
*
*$x$, $y$, $z$ are in geometric progression
*$x$, $y$, $z$ are in harmonic progression.
*$x=y=z$
*$(x-y)^2 +(y-z)^2+(z-x)^2 =0$
My attempt:
$$\text{A}=\tan^{-1}x \qquad \text{B}=\tan^{-1}y \qquad \text{C}=\tan^{-1}z$$
$$x=\tan A \qquad y=\tan B \qquad z=\tan C $$
$$x+z=2y$$
$$A+C=2B$$
$$\tan(A + B + C)=\frac{\tan A +\tan B +\tan C - \tan A\tan B\tan C }{1-\tan A\tan B -\tan B\tan C -\tan C\tan A}$$
$$\tan(3B)=\frac{x +y +z - xyz }{1-xy -yz -zx}$$
How do I continue from here?
|
(Not an answer so much as an elaborate comment ...)
For fun, I made a "trigonograph" to see why we might expect the elements to be equal (item (3)). Here, the arithmetic progressions are $\alpha$, $\gamma$, $\beta$ and $\tan\alpha$, $\tan\delta$, $\tan\beta$. (Also, I take all values to be positive, and assume $\alpha + \beta < 180^\circ$.)
$$\begin{align}
2 \gamma &\;=\; \alpha + \beta \\[4pt]
2 \tan\delta &\;=\; \tan\alpha + \tan\beta
\end{align}\qquad\implies\qquad
\begin{array}{c}
\gamma \;\leq\; \delta \\
\text{with equality when and only when} \\
\alpha = \beta \;\left(\; = \gamma = \delta\;\right)
\end{array}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1876639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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|
Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}[\ln(\tan \ x)^{\frac{2}{3}}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}-\ln(1)^\frac{2}{3}]=\frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}] = \bf \color{red}{\ln(\sqrt{3})}$$
However, the answer given is not this value, but instead $= \frac{3\sqrt[3]{3}-3}{2} \approx 0.663... $ while $\ln(\sqrt{3}) \approx 0.549...$
|
$$
\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx =
\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$
This step is wrong since
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x dx$$
Then $$\frac{f'(x)}{f(x)} = \frac{2}{3} \frac{ \sec^2x}{\tan x} \neq \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx dx$$
Try to use trigonometric identities to simplificate it first and then use substitute:
Let $u = \sqrt[3] {\tan{x}}$, then $u^3 = \tan x $, $2u^2 du = \sec^2 x dx$.
So
$$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{1}^{ 3^{1/6}}\frac{2}{3}\frac{2u^2}{u} du$$
|
{
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"url": "https://math.stackexchange.com/questions/1878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
|
When I have a rational function I usually start by making a part of the denominator "appear" in the numerator, and since we're dealing with trigonometric function, the following identity is quite handy:
$$\frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}=\frac{x^2\cos^2x+x^2\sin^2x}{(x \cos x - \sin x )(x \sin x + \cos x)} $$
$$\frac{x^2\cos^2x-x\sin x\cos x+x\sin x\cos x+x^2\sin^2x}{(x \cos x - \sin x )(x \sin x + \cos x)}=\frac{x\cos x(x\cos x-\sin x)+x\sin x(\cos x+x\sin x)}{(x \cos x - \sin x )(x \sin x + \cos x)} $$
$$ = \frac{x \cos x}{ x \sin x + \cos x } + \frac{x \sin x}{x \cos x - \sin x} $$
Let's treat each part separately:
$$I_1=\int \frac{x \cos x}{ x \sin x + \cos x }dx$$
Let $u=x \sin x + \cos x$ then $du=x \cos xdx$ then
$$I_1\int \frac {du}u=\ln(u)+c_1$$
Acting similarly on the second term, yields:
$$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx=\ln( x \sin x + \cos x)-\ln(x \cos x - \sin x)+c$$
|
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|
Inequality $a+b+c > abc$, prove $a^2+b^2+c^2 > abc$ Given, $$a+b+c > abc$$
Prove $$a^2+b^2+c^2 > abc$$
I tried to square the first one but it wouldn't work.
|
$$a^2+b^2+c^2\geq ab+bc+ac$$ holds, since it is equivalent with $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0.$$ Also $$ab+bc+ac\geq\sqrt{3abc(a+b+c)}.$$ Previous inequality holds, since it is equivalent with $$a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c),$$
which is equivalent with $$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$ for $x=bc$, $y=ac$, $z=ab$.
Finally, $$a^2+b^2+c^2\geq ab+bc+ac\geq\sqrt{3abc(a+b+c)}\geq\sqrt{3}abc>abc$$
|
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|
Find the maximum of the value $x+y+z$ if such condition $n(x+y+z)=xyz$ Let $n$ be give postive intger, and $x\ge y\ge z$ are postive integers,such
$$n(x+y+z)=xyz$$
Find the $(x+y+z)_{\max}$
I have see this problem only answer is $(n+1)(n+2)$,iff $x=n(n+2),y=n+1,z=1$
and How do it?
|
SUMMARY: it really is true that
$$(x+y+z)_{\max} = n^2 + 3n + 2,$$
this occurring only when $z=1, y=n+1, x=n^2 + 2n $
This works. I will leave $n=1,2,3$ as exercises for the reader. We take $n \geq 4.$ We are given $x \geq y \geq z \geq 1,$ with
$$ xyz = nx + ny + nz, $$
$$ x y z^2 = nxz + nyz + n z^2, $$
$$ z^2 xy - n zx - nzy = n z^2, $$
$$ z^2 xy - nzx - nzy + n^2 = n^2 + n z^2, $$
$$ (zx - n)(zy-n) = n(n+z^2). $$
First, if $z \geq n,$ then $x,y,z \geq n.$ We find $(zx-n)(zy-n) \geq (z^2 -n)^2.$ Therefore
$$ (zx - n)(zy-n) - n(n+z^2) \geq z^4 - 3 n z^2 = z^2(z^2 - 3n). $$
The factor $z^2 - 3n$ is positive for $ z \geq n \geq 4.$ ADDENDUM When we include $n=1,2,3,$ we still get the conclusion $z \leq n$ from the same calculation. If we assume $z \geq n+1,$ we get a contradiction because $ z^2 - 3n \geq n^2 + 2 n + 1 - 3n = n^2 - n + 1 \geq \frac{3}{4} > 0.$ Therefore, one may finish $n=1$ with $z=1,$ $n=2$ with $z=1,2,$ and $n=3$ with $z=1,2,3.$
We continue with $n \geq 4$ and $z < n.$ To get to the punchline, the largest possible value with such fixed $n,z$ is when $z=1,$
as suggested by the OP and my computer run last night.
Sketch of proof for fixed $1 \leq z < n.$ We have $$ (zx - n)(zy-n) = n(n+z^2). $$ If both factors on the left hand side are non-positive, that means
$ z(x+y) < 2n, $ or $x+y \leq 2n,$ whence $x+y+z \leq 3n.$ This is small, we can do better. When both factors are positive, in particular we have $zy > n.$ Let
$$ n \equiv \delta \pmod z, $$
with $$ 0 \leq \delta < z. $$
Then $$ n + (z - \delta) \equiv 0 \pmod z. $$
AUDIENCE REQUEST: if I have positive real numbers $AB=C,$ with fixed $C$ and lower bound $A\geq B \geq \epsilon > 0,$ the largest value of $A+B$ occurs when $B = \epsilon.$ This is calculus or Lagrange multipliers. To maximize $x+y,$ we are going to maximize $(zx - n)+(zy-n).$ These two summands have a fixed product $n(n+z^2),$ so the biggest sum occurs when $zy-n$ is as small as possible, that is $y$ is as small as possible.
To minimize $y$ (check with Lagrange multipliers) we can take
$$ zy = n + (z - \delta) \leq n+z. $$ As a result,
$$ 1 \leq zy - n \leq z. $$ With
$$ (zx - n)(zy-n) = n(n+z^2), $$
$$ (zx - n) \leq n(n+z^2). $$
We have
$$ z^2 \leq z^2, $$
$$ zy \leq n + z,$$
$$ zx \leq n^2 + (z^2+1)n,$$
$$ z(x+y+z) \leq n^2 + (z^2+2)n + (z^2 + z) , $$
$$ x+y+z \leq \frac{n^2 + (z^2+2)n + (z^2 + z) }{z} = \frac{(n+1)z^2 + z + n^2 + 2n }{z} = (n+1)z + 1 + \frac{ n^2 + 2n }{z} $$
$$ x+y+z \leq (n+1)z + 1 + \frac{ n^2 + 2n }{z} $$
The second derivative (in $z$) of the right hand side is positive, the first derivative of the right hand side is negative for small $z$ such as $1.$ The next value of $z$ for which the bound is as large as its value at $z=1$ is
$$ z = n + 1 - \frac{1}{n+1} > n. $$
This means that with $z < n,$ the best value is when $z=1.$ Then $y = n+1$ and $x = n^2 + 2n.$
|
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|
Evaluate $\int_{3}^{\infty}\frac{dx}{x^2-x-2}$ $$\int_{3}^{\infty}\frac{dx}{x^2-x-2}=\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}$$
$$\frac{1}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$
$$1=Ax+A+Bx-2B$$
$$1=(A+B)x+A-2B$$
$A+B=0\iff A=-B$
$-3B=1$
$B=-\frac{1}{3}$, $A=\frac{1}{3}$
$$\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}=\frac{1}{3}\int_{3}^{\infty}\frac{dx}{x-2}-\frac{1}{3}\int_{3}^{\infty}\frac{dx}{(x+1)}=|\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1)|_{3}^{\infty}$$
$$=lim_{t\to \infty}(\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1))-\frac{1}{3}ln(1)+\frac{1}{3}ln(4)=\infty-\infty-0+\frac{ln(4)}{3}$$
But the answer is $\frac{2ln(2)}{3}$, What have I done wrong?
|
$\frac{\ln(4)}{3} = \frac{\ln(2^2)}{3} = \frac{2\ln(2)}{3}$
And to handle the first two terms, combine the logs:
$\frac{1}{3}\ln(t-2) - \frac{1}{3}\ln(t+1) = \frac{1}{3}\ln(\frac{t-2}{t-1})$
Then taking the limit as $t \rightarrow \infty$ will make it $\frac{1}{3}\ln(1) = 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: Corrected the final integral and the limit from $45$ to $\pi/4$
|
You can take the long way using partial fractions, which gives you:
$$\underbrace{\frac{1}{3(x^2+1)}}_{f_1(x)}+\underbrace{\frac{\sqrt 3 x+2}{6(x^2+\sqrt3 x+1)}}_{f_2(x)}+\underbrace{\frac{3\sqrt3x-6}{18(-x^2+\sqrt3x-1)}}_{f_3(x)}$$
$$\int_0^1f_1(x)dx=\frac 13\int_0^1\frac 1{1+x^2}dx=\left[\arctan(x)\right]_0^1$$
$$\int_0^1f_2(x)dx=\int_0^1\frac{\sqrt 3 x+2}{6(x^2+\sqrt3 x+1)}dx=\frac 16\int_0^1\frac{\sqrt3x+1}{x^2+\sqrt3 x+1}+\frac{1}{x^2+\sqrt3 x+1}$$
From here on out it's pretty straightforward. The answer will involve logarithms and $\arctan$'s. It's similar for $f_3(x)$...
|
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|
Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$ How does one find the first four terms of the Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$? My approach was this:
$(z^2+1) = f(z)\sin(z) = \left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\sin(z)=\left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\left(z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040} \right)$.
How do we proceed further?
|
$$\frac{z^2+1}{\sin z}=\biggl( z+\frac 1z\biggr)\frac z{\sin z},$$
so all you have to find is the expansion of
$$\frac z{\sin z}=\frac 1{1-\cfrac{z^2}6+\cfrac{z^4}{120}+\cfrac{z^6}{5040}+o(z^7)}$$
It can be obtained with a division by increasing powers of $x$:
$$\begin{array}{rrrrr}
&&1&{}+\dfrac{z^2}6&{}+\dfrac{7z^4}{360}&{}+\dfrac{31z^6}{15120}\\
1-\dfrac{z^2}6+\dfrac{z^4}{120}-\dfrac{z^6}{5040}&\Big(&1\\%
&&-1&{}+\dfrac{z^2}6&{}-\dfrac{z^4}{120}&{}+\dfrac{z^6}{5040}\\
\hline
&&&\dfrac{z^2}6&{}-\dfrac{z^4}{120}&{}+\dfrac{z^6}{5040}\\
&&&-\dfrac{z^2}6&{}+\dfrac{z^4}{36}&{}-\dfrac{z^6}{360}\\
\hline
%&&&&{}+\dfrac{7z^4}{360}&{}-\dfrac{z^6}{840}\\
%&&&&{}-\dfrac{7z^4}{360}&{}\dfrac{7z^6}{2160}\\
%\hline
&&&&&\dfrac{31z^6}{15120}\end{array}$$
whence the Laurent series after multiplication by $z+\dfrac1z$:
\begin{alignat*}{6}
\frac{z^2+1}{\sin z}=\biggl(z+\frac 1z\biggr)\frac z{\sin z}&=&&z+\dfrac{z^3}6&&+\dfrac{7z^5}{360}+\dfrac{31z^7}{15120}+o(z^8)
\\
&=\quad \frac1z+&&\dfrac{z}6+\dfrac{7z^3}{360}&&+\dfrac{31z^5}{15120}+o(z^6)\\
& =
\quad \color{red}{\frac1z+}&&\color{red}{\dfrac{7z}6+\dfrac{67z^3}{360}}&&\color{red}{{}+\dfrac{65z^5}{3024}+o(z^6)}.\end{alignat*}
Added: Division by increasing power order
It's like Euclidean division of polynomials $A(x)$ (dividend) and $B(x)$ (divisor), except it is defined when $B(0)\neq 0$, i.e. the divisor has a non-zero constant term, and at each step, one divides the lowest term of the dividend by the constant term of the divisor in order to eliminate the lowest term of the dividend.
The result we use is the following:
Let $A(x)$, $B(x)$ be polynomials. For each natural number $n$, there exists a unique pair of polynomials $Q_n(x)$ and $R_n(x)$ such that
$$A(x)=Q_n(x)B(x)+x^{n+1}R_n(x),\qquad\deg Q_n(x)\le n.$$
From unicity, we see that, for $m\le n$, $Q_m(x)$ is just the polynomial $Q_n(x)$, truncated at degree $m$.
This division can be used for the expansion in power series of the quotient $f(x)/g(x)$ of two analytic functions, and to find quickly the decomposition into partial fractions in the case of multiple poles.
|
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|
General formula for two sequences Please help me to find general 'analytical' formula fot these two sequences $\{x_n\}$, $n=1,2,3,\ldots$.
I. $0, 1, 2, 2, 3, 6, 6, 7, 14, 14, \ldots$
This meens the following.
$x_{1}=0$, $x_{2}=x_{1}+1$, $x_{3}=2x_{2}$, $x_{4}=x_{3}$, $\ldots$, $x_{3k-1}=x_{3k-2}+1$, $x_{3k}=2x_{3k-1}$, $x_{3k+1}=x_{3k}$.
II. $1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, \ldots$
That is, $x_{n}=1$ if $n=6k-5$, $k=1,2,3,\ldots$; $x_{n}=0$ otherwise.
I tried to use something like
$$
x_{n}=\frac{(-1)^{n-1}+1}{2},
$$
so, we get $1, 0, 1, 0, 1, 0, \ldots$ but I need 5 zeroes after 1.
|
For case II), shift the index to start from $0$ i.e. put $n=m+1$.
Then you have a function of period $6$:
$$
f\left( m \right) = \left\{ \begin{gathered}
1\quad \left| {\;\bmod \left( {m,6} \right) = 0} \right. \hfill \\
0\quad \left| {\;\bmod \left( {m,6} \right) \ne 0} \right. \hfill \\
\end{gathered} \right.
$$
There are plenty of ways to render it. Some are:
a) Floor
Delta of step functions
$$
f\left( x \right) = \left\lfloor {\frac{x}
{6}} \right\rfloor - \left\lfloor {\frac{{x - 1}}
{6}} \right\rfloor = 1 + \left\lfloor {\frac{x}
{6}} \right\rfloor - \left\lceil {\frac{x}
{6}} \right\rceil \quad \Rightarrow \quad x_n = \left\lfloor {\frac{{n - 1}}
{6}} \right\rfloor - \left\lfloor {\frac{{n - 2}}
{6}} \right\rfloor
$$
b) Discrete Fourier Transform
Taking the real part of the DFT gives:
$$
f\left( x \right) = \frac{1}
{6}\sum\limits_{0\, \leqslant \,k\, \leqslant \,5} {\cos \left( {\frac{{2\pi \,k\,x}}
{6}} \right)}
$$
c) Discrete Cosine Tranform
Applying the DCT you get instead:
$$
\begin{gathered}
f\left( x \right) = \frac{1}
{6}\left( {1 + \cos \left( {\pi \,x} \right) + 2\cos \left( {\frac{{\pi \,x}}
{3}} \right) + 2\cos \left( {\frac{{2\pi \,x}}
{3}} \right)} \right) = \hfill \\
= \frac{1}
{6}\left( {4\cos ^2 \left( {\frac{{\pi \,x}}
{3}} \right) + 2\cos \left( {\frac{{\pi \,x}}
{3}} \right) + \cos \left( {\pi \,x} \right) - 1} \right) \hfill \\
\end{gathered}
$$
|
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|
How to find a characteristic polynomial of 5x5 matrix From an example that I am looking at, the characteristic polynomial of
$\begin{bmatrix}
-1&4&0&0&0 \\
0&3&0&0&0 \\
0&-4&-1&0&0 \\
3&-8&-4&2&1 \\
1&5&4&1&4
\end{bmatrix}$
is $(x-3)^3(x+1)^2$. I understand how to find the characteristic polynomial of 2x2 and 3x3 matrices but anything nxn beyond that I'm not sure what to do. Could someone walk me through the steps of the calculation to find the characteristic polynomial for this matrix?
|
Directly using elementary operations that don't change the value of the determinant. Develop by the second row twice:
$$\begin{vmatrix}
x+1&-4&0&0&0 \\
0&x-3&0&0&0 \\
0&4&x+1&0&0 \\
-3&8&4&x-2&-1 \\
-1&-5&-4&-1&x-4
\end{vmatrix}=(x-3)\begin{vmatrix}
x+1&0&0&0 \\
0&x+1&0&0 \\
-3&4&x-2&-1 \\
-1&-4&-1&x-4
\end{vmatrix}=$$$${}$$
$$=(x-3)(x+1)\begin{vmatrix}
x+1&0&0 \\
-3&x-2&-1 \\
-1&-1&x-4
\end{vmatrix}\stackrel{\text{1st row}}
=(x-3)(x+1)^2\begin{vmatrix}x-2&-1\\-1&x-4\end{vmatrix}=$$$${}$$
$$=(x-3)(x+1)^2(x^2-6x+7)$$
As you can see, the above isn't $\;(x-3)^3(x+1)^2\;$, so at least one of you or I is wrong...but I already checked mine with this program
|
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|
On convergence of $\cos(2\pi \alpha n!)$
Let $x\in \mathbb R$
Prove that there exists some $\alpha_x$ such that the sequence $\cos(2\pi \alpha_x n!)$ converges to $\cos(x)$
I've been stumped with this problem for a while. I've been looking for some $\alpha$ such that $$\forall n, 2\pi \alpha n! = x + 2\pi k_n + \epsilon_n$$
where $k_n$ is an integer and $\epsilon_n$ a sequence that goes to $0$.
Can someone help me build the $\alpha$ ?
|
For any $x \in \mathbb{R}$, take a $y \in [0,\pi]$ such that $\cos x = \cos y$.
Define
$$\alpha = \sum_{n=1}^\infty \frac{1}{n!}\left\lfloor \frac{y}{2\pi} n\right\rfloor$$
For any $N > 0$, we have
$$\alpha N! =
\underbrace{\sum_{n=1}^N \frac{N!}{n!}\left\lfloor \frac{y}{2\pi} n\right\rfloor}_{\in \mathbb{Z}}
+ \frac{1}{N+1}
\left\lfloor \frac{y}{2\pi} (N+1)\right\rfloor
+ \sum_{k=2}^\infty \frac{1}{\prod_{j=1}^{k-1} (N+j)}\left(\frac{1}{N+k}\left\lfloor \frac{y}{2\pi}(N+k)\right\rfloor\right)
$$
For the $2^{nd}$ term, it is a number near $\frac{y}{2\pi}$. More precisely,
$$0 \le \frac{y}{2\pi} - \frac{1}{N+1}\left\lfloor \frac{y}{2\pi} (N+1)\right\rfloor < \frac{1}{N+1} < \frac{1}{N}$$
For the $3^{rd}$ term, we have the bound
$$0 \le \sum_{k=2}^\infty \frac{1}{\prod_{j=1}^{k-1} (N+j)}\left(\frac{1}{N+k}\left\lfloor \frac{y}{2\pi}(N+k)\right\rfloor\right)
\le \sum_{k=2}^\infty \frac{1}{(N+1)^{k-1}}\frac{y}{2\pi}
= \frac{y}{2\pi N} < \frac{1}{2N}
$$
Combine these two bounds, we can conclude
$$\alpha N! = \beta_N + \frac{y}{2\pi} + \delta_N
\quad\text{ where }\quad \beta_N \in \mathbb{Z}\;\text{ and }\; -\frac{1}{N} < \delta_N < \frac{1}{2N}$$
Apply MVT to $\cos(t)$ for $t$ between $2\pi\alpha N!$ and $2\pi\beta_N + y$, we find
$$|\cos(2\pi\alpha N!) - \cos(y)| \le 2\pi|\delta_N| < \frac{2\pi}{N}$$
As a result,
$$\lim_{N\to\infty} \cos(2\pi\alpha N!) = \cos y = \cos x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How would you show that the series $\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$ diverges? How would you show that the series $$\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$$ diverges? Wolfram Alpha says it diverges "by comparison", but I'd like to know to what you would compare it? I've tried some basic things to no avail: The ratio test is inconclusive, comparing with something "smaller" which I know diverges is proving difficult. If possible, I'd like to show this without using anything fancy (such as Stirling's approximation).
Can you do this with a "basic" comparison?
|
Here is my attempt:
$a_n=\dfrac{(2n)!}{4^n(n!)^2}$
First i expanded factorial terms:
$a_n=\dfrac{2n(2n-1)(2n-2)\cdots 3\cdot 2\cdot 1}{4^n n(n-1)(n-2)\cdots 2\cdot 1\cdot n(n-1)(n-2)\cdots 2\cdot 1}$
At the numerator, rearrange the $n$ even factors and the $n$ odd factors (not necessary but useful for my description)
$a_n=\dfrac{2n(2n-2)(2n-4)\cdots 4\cdot 2\cdot (2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{4^n n^2(n-1)^2(n-2)^2\cdots 2^2\cdot 1^2}$
From the $n$ even factors collect $2$:
$a_n=\dfrac{2^n n(n-1)(n-2)\cdots 2\cdot 1 \cdot (2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{4^n n^2(n-1)^2(n-2)^2\cdots 2^2\cdot 1^2}$
Now you can simplify the square factors at the denominator with the terms $n(n-1)(n-2)\cdots 2\cdot 1$ obtained from collecting $2$; you have:
$a_n=\dfrac{2^n (2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{4^n n(n-1)(n-2)\cdots 2\cdot 1}$
Obviously you can simplify $\dfrac{4^n}{2^n}$; so:
$a_n=\dfrac{(2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{2^n n(n-1)(n-2)\cdots 2\cdot 1}$
Now consider the factors at the numerator:
$(2n-1)\geq (2n-2)$
$(2n-3)\geq (2n-4)$
$(2n-5)\geq (2n-6)$
and so on.
You have:
$a_n\geq \dfrac{(2n-2)(2n-4)(2n-6)\cdot 4\cdot 2}{2^n n!}$
Again collect a factor $2$ at the numerator:
$a_n\geq\dfrac{2^n (n-1)(n-2)\cdots 4\cdot 2}{2^n n!}$
Simplify the terms $2^n$ and note that the numerator is $(n-1)!$; so:
$a_n\geq\dfrac{(n-1)!}{n!}=\dfrac{1}{n}$
We know that $\sum_{n=1}^\infty\dfrac{1}{n}$ diverges; so by comparison our series:
$\sum_{n=1}^{\infty}a_n$ diverges.
Hope this help.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
|
This is probably a too complex answer.
Considering $$J_n=\int x^n \tan^{-1}(x)\,dx=\frac{x^{n+1} \left((n+2) \tan ^{-1}(x)-x \,
_2F_1\left(1,\frac{n}{2}+1;\frac{n}{2}+2;-x^2\right)\right)}{(n+1) (n+2)}$$ where appears the hypergeometric function, $$I_n=\int_0^1 x^n \tan^{-1}(x)\,dx=\frac{H_{\frac{n-2}{4}}-H_{\frac{n}{4}}+\pi }{4 (n+1)}$$ where appears the generalized harmonic numbers. So, $$A_n=n \left( (n+1)I_n-\frac{\pi }{4}\right)=\frac{1}{4} n \left(H_{\frac{n-2}{4}}-H_{\frac{n}{4}}\right)$$ Now, using the asymptotics of harmonic numbers $$A_n=-\frac{1}{2}+\frac{1}{2 n}+O\left(\frac{1}{n^3}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Gauss circle problem : a simple asymptotic estimation.
Find the number of integer points lying in or inside a circle of radius $n\in \mathbb N$ centered at the origin.
The problem asks for all $(a,b)\in \mathbb Z^2$ such that $a^2+b^2\leq n^2$. Looking at such points lying strictly in the North-East quadrant, pick an $a\in \{1,\ldots,n\}$ and let $b$ run through $\{1,\ldots,\lfloor \sqrt{n^2-a^2}\rfloor\}$ : there are exactly $\displaystyle \sum_{k=1}^n \lfloor \sqrt{n^2-k^2} \rfloor$ integer points strictly in the quadrant.
Multiply that by $4$ to account for all $4$ quadrants, add the origin and points on the $x$ and $y$ axis and the total number is $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor$$
I'm interested in asymptotics of this sum. A trivial estimate is the following:$$\begin{align}\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor &= \sum_{k=0}^n \sqrt{n^2-k^2} + O(n)\\ &= n^2\cdot \left(\frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}}\right) + O(n) \\ &=n^2 \left( \int_0^1 \sqrt{1-t^2}dt + o(1)\right) + O(n) \\ &= \frac \pi 4n^2 + o(n^2)\end{align} $$
Hence $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + o(n^2)$$
Is there a way to refine the estimate above and get $\pi n^2 + O(n)$ instead (or something better) ?
|
Pick a square with unit side length centered at every lattice point inside the region $x^2+y^2=n^2$, i.e. the circle with radius $n$ centered at the origin. Those squares entirely cover the circle with radius $n-\frac{1}{\sqrt{2}}$, but any square lies inside a circle with radius $n+\frac{1}{\sqrt{2}}$, hence the number of lattice points is between $\pi\left( n-\frac{1}{\sqrt{2}}\right)^2$ and $\pi\left(n+\frac{1}{\sqrt{2}}\right)^2$, so it is $\pi n^2+O(n)$.
For info about the Voronoi bound, please see this survey.
|
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|
How to solve this determinant
Question Statment:-
Show that
\begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\
\end{vmatrix}
=2abc(a+b+c)^3
\end{align*}
My Attempt:-
$$\begin{aligned}
&\begin{vmatrix}
\\(a+b)^2 & ca & bc \\
\\ca & (b+c)^2 & ab \\
\\bc & ab & (c+a)^2 \\\
\end{vmatrix}\\
=&\begin{vmatrix}
\\a^2+b^2+2ab & ca & bc \\
\\ca & b^2+c^2+2bc & ab \\
\\bc & ab & c^2+a^2+2ac \\\
\end{vmatrix}\\
=&\dfrac{1}{abc}\begin{vmatrix}
\\ca^2+cb^2+2abc & ca^2 & b^2c \\
\\ac^2 & ab^2+ac^2+2abc & ab^2 \\
\\bc^2 & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\\\\
&\qquad (C_1\rightarrow cC_1, C_2\rightarrow aC_2, C_3\rightarrow bC_3)\\\\
=&\dfrac{2}{abc}\times\begin{vmatrix}
\\ca^2+cb^2+abc & ca^2 & b^2c \\
\\ab^2+ac^2+abc & ab^2+ac^2+2abc & ab^2 \\
\\bc^2+a^2b+abc & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\\\\
&\qquad (C_1\rightarrow C_1+C_2+C_3)\\\\
=&\dfrac{2abc}{abc}\left(\begin{vmatrix}
\\a^2+b^2 & a^2 & b^2 \\
\\b^2+c^2 & b^2+c^2+2bc & b^2 \\
\\c^2+a^2 & a^2 & c^2+a^2+2ac \\\
\end{vmatrix}+
\begin{vmatrix}
\\1 & ca^2 & b^2c \\
\\1 & ab^2+ac^2+2abc & ab^2 \\
\\1 & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\right)
\end{aligned}$$
The second determinant in the last step can be simplified to
\begin{vmatrix}
\\1 & ca^2 & b^2c \\
\\0 & ab^2+ac^2+2abc-ca^2 & ab^2-b^2c \\
\\0 & a^2b-ca^2 & bc^2+a^2b+2abc-b^2c \\\
\end{vmatrix}
I couldn't proceed further with this, so your help will be appreciated and if any other simpler way is possible please do post it too.
|
One can use factor theorem to get a simpler solution. If we put $a=0$, we get
\begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\
\end{vmatrix}
=\begin{vmatrix}
b^2 & 0 & bc \\
0 & (b+c)^2 & 0 \\
bc & 0 & c^2 \\
\end{vmatrix} = 0
\end{align*}
Hence $a$ is a factor. Similarly $b, c$ are factors. Again, put $a+b+c=0$, we get
\begin{align*}
\begin{vmatrix}
c^2 & ca & bc \\
ca & a^2 & ab \\
bc & ab & b^2 \\
\end{vmatrix}
=abc\begin{vmatrix}
c & a & b \\
c & a & b \\
c& a & b \\
\end{vmatrix}
\end{align*}
Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$. Putting $a=b=c=1$, we obtain
\begin{align*}
27k = \begin{vmatrix}
4 & 1 & 1 \\
1 & 4 & 1 \\
1 & 1 & 4 \\
\end{vmatrix} =54
\end{align*}
and $k=2$. Thus the given determinant equals $2abc(a+b+c)^3$
|
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|
How can we prove this inequality? Let $a$ , $b$ and $c$ be positive real numbers and $a+b+c=1$ How can we show this inequality?
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
Thanks.
|
Set $a=x^2$ , $b=y^2$ and $c=z^2$ and define
$$f(x,y,z)=x^4+y^4+z^4+2\sqrt{3}xyz-1$$
and
$$g(x,y,z)=x^2+y^2+z^2-1=0$$
By application of Lagrange method, we have
$$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$$
thus
\begin{cases}
4x^3+2\sqrt{3}yz=2\lambda x\\
4y^3+2\sqrt{3}xz=2\lambda y\\
4z^3+2\sqrt{3}xy=2\lambda z\\
\end{cases}
so
\begin{cases}
2x^4+\sqrt{3}xyz=\lambda x^2\\
2y^4+2\sqrt{3}xyz=\lambda y^2\\
2z^3+2\sqrt{3}xyz=\lambda z^2\\
\end{cases}
it is equivalent by
\begin{cases}
2x^4-2y^4=\lambda (x^2-y^2)\\
2y^4-2z^4=\lambda (y^2-z^2)\\
2z^4-2x^4=\lambda (z^2-x^2)\\
\end{cases}
or
\begin{cases}
x^2+y^2=\frac 12\lambda \\
y^2+z^2=\frac 12\lambda\\
z^2+x^2=\frac 12\lambda \\
\end{cases}
indeed
$$x^2+y^2=y^2+z^2=z^2+x^2$$
as a result
$$x^2=y^2=z^2\tag 1$$
on the other hand
$$x^2+y^2+z^2-1=0\tag 2$$
$(1)$ and $(2)$
$$x=y=z=\frac{1}{\sqrt{3}}\tag 3$$
then
$$f(x,y,z)\le f\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)=0$$
finally
$$x^4+y^4+z^4+2\sqrt{3}xyz-1\le 0$$
or
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
|
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|
Inequalities in integration We are given the value of $S_n$ below and we have to find if $S_n$ is greater or smaller than $\dfrac{\pi}{3\sqrt{3}}$.
$$S_n=\sum\limits_{k=1}^n \frac{n}{n^2+kn+k^2}$$
$$S_n<\dfrac{\pi}{3\sqrt{3}} \quad \text{or}\quad S_n>\dfrac{\pi}{3\sqrt{3}}$$
I tried it as follows:
$$S_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\left(\frac{k}{n}\right)^2+\frac{k}{n}+1}$$
$$\rightarrow \frac{1}{x^2+x+1}$$
But at last I thought to draw the graph so that I can do something with area under the curve or can thought at extreme points i.e at $n$ is approaching to infinity.
I got stuck not able to do anything. You can see the image of my attempt here.
|
Look at this picture. The red curve is the function $\displaystyle f(x)=\frac{1}{x^2+x+1}$
The red curve is decreasing on $[0,1]$, since $\displaystyle f'(x)=-\frac{2x+1}{\left(x^2+x+1\right)^2}<0$ for $0\leq x \leq 1$.
The gray shaded area is the value of $\displaystyle\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\left(\frac{k}{n}\right)^2+\frac{k}{n}+1}$ when $n=6$.
Can you see why? The width of each box is $\displaystyle \frac{1}{n}$, and the height of the $k$th box is $\displaystyle f\left(\frac{k}{n}\right)=\frac{1}{\left(\frac{k}{n}\right)^2+\frac{k}{n}+1}$.
As you can see, as $n$ increases, the number of boxes will increase, but the shaded area will never be equal to or greater than the area under the red curve for the interval $[0,1]$. (By the way, these approximations of the area under the curve are called Riemann sums.)
Ergo, we conclude that $\displaystyle S_n<\int_{0}^{1}\frac{dx}{x^2+x+1}$
$$\begin{align}
\int_{0}^{1}\frac{dx}{x^2+x+1}&=\int_{0}^{1}\frac{dx}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}
\\
&=\frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{du}{u^2+1} \qquad\text{(Substitute $u=\frac{2x+1}{\sqrt{3}}$, $\frac{du}{dx}=\frac{2}{\sqrt{3}}$)}\\
&=\frac{2}{\sqrt{3}}\arctan(u)\bigg|_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\\
&=\frac{\pi}{3\sqrt{3}}
\end{align}$$
And thus we have our desired result.
(By the way, since I am bad at calculus, I used the following sites: http://www.integral-calculator.com/, http://www.derivative-calculator.net/)
|
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|
Smallest $n$ such that $H_{n,2^{-1}} \geq n$ where $H_{n,2^{-1}}$ are generalized Harmonic numbers Consider the generalized harmonic numbers evaluated at $2^{-1}$ $$H_{n,2^{-1}} = 1+ {1 \above 1.5pt \sqrt{2}}+ \ldots+ {1 \above 1.5pt \sqrt{n}}$$ The table below lists some initial values:
$$\begin{array}{nc|cccccc}
n &1&2&3&4&5&6&7 \\ \hline
H_{n,2^{-1}} & 1.00 & 1.71 &2.28&2.78&3.23&3.64&4.01\\
\end{array}$$
Let $$s =min\{n\text{ }|\text{ }H_{n,2^{-1}} \geq n \}$$ For example the smallest $n$ such that $H_{n,2^{-1}}\ge 1$ is $1$. Similarly the smallest $n$ such that $H_{n,2^{-1}}\ge 2$ is $3$. We have the following sequence for $s$
$$\mathfrak a(s) =1,3,5,7,10,14,18,22,\ldots$$ I am asking if the following claim is true -
$$\mathfrak a(s) = \sum_{n\leq s}\Bigg(\Bigg\lfloor{n+2 \above 1.5pt
4}\Bigg\rfloor+\Bigg\lfloor{n+1 \above 1.5pt 4}\Bigg\rfloor\Bigg)$$
Note that $\mathfrak a(s)$ is the sequence A054040.
|
Note that
$$H_{n,2^{-1}}=\int_1^{n+1}\frac{dt}{\sqrt{\lfloor t\rfloor}}$$
So
$$\int_1^{n+1}\frac{dt}{\sqrt{t}}\le H_{n,2^{-1}}\le1+\int_2^{n+1}\frac{dt}{\sqrt{t-1}}$$
That is,
$$2(\sqrt{n+1}-1)\le H_{n,2^{-1}}\le 1+2(\sqrt{n}-1)$$
For example, $1998 \le H_{999,999,\,2^{-1}} < 1999$. This means that $\mathfrak a(1998)\le 999,999 < \mathfrak a(1999)$.
But
$$\begin{align}\sum_{n\le 1999}&\left(\left\lfloor\frac{n+2}4\right\rfloor+\left\lfloor\frac{n+1}4\right\rfloor\right)\\
&=\sum_{n\le 1999}\frac{2n+3}4\\
&-\left(\frac12+\frac34\right)\cdot499\qquad\text{for $n=4k+1$}\\
&-\frac34\cdot499\qquad\text{for }n=4k+2\\
&-\frac14\cdot499\qquad\text{for }n=4k+3\\
&-\left(\frac12+\frac14\right)\cdot498\qquad\text{for }n=4k\\
&=999,413<999,999
\end{align}$$
|
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|
$5$ divides one of $x,y,z$ in $x^2+y^2=z^2$? 1) If $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, show that $3$$∣$$xy$.
2) Now again if $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, can a similar result to the one in (part 1) be said mod $5$? Not that $5$$∣$$xy$, but that $5$ divides one of $x,y,z$?
I believe I have 1) down:
We note that $3$ is a prime, so $3∣xy⟺3∣x$ or $3∣y$. Suppose to the contrary that $3∤x$, and $3∤y$, then:$x=1$, $2$ mod $3$ and $y=1$, $2$ mod $3$. Then:
$x^2+y^2=2$ mod $3$, and $z^2=0$, $1$ mod $3$. Contradiction, proving the claim.
But I am having trouble with 2). Can anyone show a solution for 2)?
|
For the second part since $5$ is prime, if $5 \mid xy \implies 5 \mid x$ or $5\mid y$. For the first part, we have: $x = 2mn, y = m^2-n^2=(m-n)(m+n)$. Thus if $3 \nmid x$, then $3 \nmid m, 3 \nmid n$. Thus if $m = n \pmod 3$, then $m-n = 0 \pmod 3$, otherwise say $m = 1\pmod 3, n = 2\pmod 3 \implies m+n = 0 \pmod 3$.
|
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|
Prove that $ABC$ is right-angled Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.
I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.
Thank you in advance for your answers!
|
As you found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$ note that $\sin^2 \theta + \cos^2 \theta =1$
Then $1-\cos^2 A + 1- \cos^2 B + 1 - \cos^2 C = 2$
$ 3 -(\cos^2 A + \cos^2 B + \cos^2 C)= 2$
$3 - 2 = \cos^2 A + \cos^2 B + \cos^2 C$
$\cos^2 A + \cos^2 B + \cos^2 C =1$
|
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|
Prove that if $a+b+c=1$ then $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}\le\frac{9\sqrt{2}}{2}$ Let $a,b,c>0$,and such $a+b+c=1$,prove or disprove
$$\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\le\dfrac{9\sqrt{2}}{2}\tag{1}$$
My try:since
$$\sqrt{a^2+b^2}\ge\dfrac{\sqrt{2}}{2}(a+b)$$
it suffices to prove that
$$\sum_{cyc}\dfrac{1}{a+b}\le\dfrac{9}{2}$$
But since Cauchy-Schwarz inequality we have
$$2\sum_{cyc}(a+b)\sum_{cyc}\dfrac{1}{a+b}\ge 9$$
or
$$\sum_{cyc}\dfrac{1}{a+b}\ge \dfrac{9}{2}$$
so this try can't works.so How to prove $(1)$
|
It's wrong. Try $a=b\rightarrow0^+$
|
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Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2}}$ which evaluates to $1<0$Do you know what to do here ?
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Prove by induction the stronger inequality: for any integer $n\geq 1$,
$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$$
The basic step is true: $1/2<1/\sqrt{3}$.
Then in the inductive step you will obtain the inequality
$$\frac{1}{\sqrt{2n+1}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+3}}$$
that is equivalent to
$$4n^2+8n+3=(2n+1)(2n+3)<(2n+2)^2=4n^2+8n+4$$
which holds.
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Find coefficients of polynomial, knowing its roots are consecutive integers In the function $f(x)= x^3-15x^2+ax+b$ the graph has $3$ consecutive points where it crosses the x-axis.
These $3$ points are consecutive integers. Find $a$ and $b$ for this is you know that $a$ and $b$ are real numbers.
How do I start to find the answer?
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The brute force method that doesn't require any formulas or theorems would be like so:
You know that there are $3$ consecutive zeroes $(n-1)$, $n$, $(n+1)$. It's important to pick them like so as opposed to $n$, $(n+1)$, $(n+2)$ because this will save you a lot of algebra.
$0=(n-1)^3-15(n-1)^2+a(n-1)+b $
$0=n^3-15n^2+an+b $
$0=(n+1)^3-15(n+1)^2+a(n+1)+b $
$\color{red}{\text {This is how to start it. Below is the answer, so if you just wanted a hint you can stop here.}}$
Get rid of the $b$ first because it's easy. To do this, subtract the middle equation from the other two, producing two equations with $a$ and $n$ as unknowns:
$$0-0=[(n-1)^3-15(n-1)^2+a(n-1)+b ]- (n^3-15n^2+an+b )$$
Which simplifies to
$$0=-a-3n^2+33n-16 $$
And
$$0-0 = [(n+1)^3-15(n+1)^2+a(n+1)+b ] - (n^3-15n^2+an+b )$$
Which simplifies to
$$0=a+3n^2-27n-14 $$
So we have the system
$$0=-a-3n^2+33n-16 $$
$$0=a+3n^2-27n-14 $$
Now add these $2$ equations together, and get
$$0=6n-30$$
Therefore
$$n=5$$
Now that you have $n$, you know that the other solutions are of the form $n \pm 1$, so the $3$ roots are $4, 5, 6$.
Now plug just two of these into your cubic (maybe the two smallest ones) and you will have two equations with $a$ and $b$ as unknowns.
$$0=4^3-15\cdot4^2+4a+b$$
$$0=5^3-15\cdot5^2+5a+b$$
Simplifying,
$$0=4a+b-176$$
$$0=5a+b-250$$
Solving this system, we get
$$a=74, \ b=-120$$
EDIT:
As noticed by mathguy in the comments below, we can cut down on the work required. Once we find $n$, we can use the equation
$$0=a+3n^2-27n-14 $$
to find $a$:
$$0=a+3 \cdot(5)^2-27 \cdot (5) -14$$
$\implies$ $a=74$
Now we can use one of the original $3$ equations (the one with $n-1$ for the easiest computation?) to find $b$:
$$0=(5-1)^3-15(5-1)^2+74(5-1)+b $$
$\implies$ $b = -120$
$$a=74, \ b=-120$$
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How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$ How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$?
E.g. I know that $6x^2 + 5x + 1$ will factor to $(3x + 1)(2x + 1)$, but is there a recipe or an algorithm that will allow me to factorize this?
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The roots of a quadratic polynomial $ax^2+bx+c$ are $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$. Then the factorization is just $a(x-x_1)(x-x_2)$.
In your example, $a=6,b=5,c=1$, so $x_1=\frac{-5+\sqrt{5^2-4\cdot 6\cdot1}}{2\cdot 6}=-\frac13$ and $x_2=\frac{-5-\sqrt{5^2-4\cdot 6\cdot1}}{2\cdot 6}=-\frac12$. The factorization is then $6(x+\frac13)(x+\frac12)$, or $(3x+1)(2x+1)$.
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Find all integer solutions to the equation $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$ I was going through one of my Mathematics books and I came to this problem:
Find all integer solutions to the equation: $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$
I tried a few things to begin with but none went in the right direction, any suggestions?
Another thing, if we have the left side of the equation to be equal to 9, there are supposed to be 16 solutions in integers. What way can they be found?
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Since obviously $xyz\ne 0$, you have $$(xy)^2+(xz)^2+(yz)^2=15xyz\iff (y^2+z^2)x^2-(15yz)x+(yz)^2=0$$ hence the discriminant must be non-negative.
$$(15yz)^2-4(y^2+z^2)(yz)^2\ge 0$$ It follows $$y^2+z^2\le 56$$ This involves only the set $\mathcal S$ of the first seven squares
$$\mathcal S=\{1,4,9,16,25,36,49\}$$ Searching for $x=y=z$ one has $3x^4=15x^3$ whose only solution is $x=5$ which can be done with $y=\pm 5$ and $z=\pm 5$.
We make then a table in which we discard the cases $x=y=z$ and also $y=z$ because in those cases the equation $2X^2-15X+1=0$ has no rational solution.
$$\begin{array}{|c|c|}\hline x & y^2+z^2 & yz\\\hline 1&5&4\\\hline1&10&9\\\hline 1&17&16\\\hline1&26&25\\\hline1&37&36\\\hline1&50&49\\\hline2&13&36\\\hline2&20&64\\\hline2&29&100\\\hline2&40&144\\\hline2&53&196\\\hline 3&25&144\\\hline3&34&225\\\hline 3&45&324\\\hline4&41&400\\\hline4&52&576\\\hline\end{array}$$
Because of the congruence $$x^2(y^2+z^2)+(yz)^2\equiv 0\pmod 5$$ the discarding of this sixteen it is straightforward. Thus there are only four solutions basically refered to the $(|x|,|y|,|z|)=(5,5,5)$ changing1 signes in two of the variables.
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How do I get the closed form of this recurrence using generating functions? Recurrence: $T(n) = n + nT(n-1)$ and $T(0) = 0$.
What I tried:
Let $G(x) = \sum_{n=0}^{\infty} T(n) x^n$ so that $xG'(x) = \sum_{n=0}^{\infty} nT(n)x^n$
Solving:
\begin{align}
G(x) &= \sum_{n=0}^{\infty} T(n) x^n
\\&= 0x^0 + \sum_{n=1}^{\infty} T(n) x^n
\\&= \sum_{n=1}^{\infty} n x^n + \sum_{n=1}^{\infty} nT(n-1) x^n
\\&= -0x^0 + \sum_{n=0}^{\infty} n x^n + x\sum_{n=1}^{\infty} nT(n-1) x^{n-1}
\\&= \frac{x}{(1-x)^2} + x\sum_{n=0}^{\infty} (n+1)T(n) x^{n}
\\&= \frac{x}{(1-x)^2} + x\sum_{n=0}^{\infty} nT(n) x^{n} + x\sum_{n=0}^{\infty} T(n) x^{n}
\\&= \frac{x}{(1-x)^2} + x^2G'(x) + xG(x)
\end{align}
And $G(x) = \frac{x}{(1-x)^2} + x^2G'(x) + xG(x)$ rearranges to $G(x) = \frac{x}{(1-x)^3} + \frac{x^2}{1-x}G'(x)$
No idea where to take it from here.
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$$G(x)-xG'(x)=\frac{x}{(1-x)^2}$$
Now use the fact that
$$\left(\frac{G(x)}{x} \right)'=\frac{G'(x) x -G(x)}{x^2}$$
to reduce your equation to
$$\left(\frac{G(x)}{x} \right)'=-\frac{1}{x (1-x)^2}$$
Integrate and you are done.
Edit To address the updated question:
$$ G'(x) -\frac{1-x}{x^2}G(x) =-\frac{1}{x(1-x)^2} $$
is a first order linear differential equation.
You can solve it by using the method listed in these notes:
Linear Differential Equations
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Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which yields:
$$\frac{1}{3} > \sqrt{x}\sqrt{1 -x}.$$
Squaring it again we get inequality:
$$9x^2 - 9x + 1 > 0,$$
solution to which is a domain $[0, \frac{3 - \sqrt{5}}{6}) \cup (\frac{3 + \sqrt{5}}{6}, 1]$. But, solution to original inequality is just $[0, \frac{3 - \sqrt{5}}{6})$. What am I omitting/where I'm doing mistakes?
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Well, there is this but I'm not sure how to generalize it.
$\sqrt{1-x} - \sqrt{x} > 0$ so $1-x > x$ so $x < \frac 12$ and $x \in [0,1/2)$.
When you square but sides of $\sqrt{1-x} - \sqrt{x} > 1/\sqrt{3}$ to get $\frac{1}{3} > \sqrt{x}\sqrt{1 -x}$ you are extraneously adding the possibility that $\sqrt{x} - \sqrt{1-x} > 1/\sqrt{3}$ (i.e that maybe $x \in (1/2, 1]$) which we know can not be true.
When we square a second time to get $9x^2 - 9x + 1 > 0$ we are extraneously adding the possiblities that $x < 0$ or $x > 1$. (You caught those).
The solution to $9x^2 - 9x + 1 > 0$ is $(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)$.
You caught it should be $[(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)]\cap [0,1]$ but you didn't catch it should be
$[(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)] \cap [0,1/2)$.
=====
I guess the way to generalize this to note every time you square an inequality make a note to combine with the known inequality of sign.
i.e.
$\sqrt{1-x} - \sqrt{x} > 1/\sqrt{3}$
Square both sides and conclude
$\sqrt{x}\sqrt{1-x} < 1/3$ AND $\sqrt{1-x} - \sqrt{x} \ge 0$.
So $\sqrt{1-x} \ge \sqrt{x}$.
Square those both side to get $1-x \ge x$ AND $x \ge 0$ and $x \le 1$
So $\sqrt{x}\sqrt{1-x} < 1/3$ And $x \in (-\infty, 1/2]) \cap [0,\infty) \cap (-\infty, 1] = [0,1/2]$.
Square both sides to get
$9x^2 - 9x + 1 > 0$ AND $\sqrt{x}\sqrt{1-x} \ge 0$ and $x \in [0,1/2]$
So $9x^2 - 9x + 1 > 0$ and $x \in [0,\infty) \cap (-\infty,1] \cap [0,1/2] = [0,1/2]$.
So $x \in (\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)\cap [0,1/2]= [0,\frac{3-\sqrt{5}}{6})$
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Let a, b, c, x and y be integers. Why are there no solutions for $\sqrt {a^2 + b^2 + c^2} = x$, where $x=2^y$ or $x=5 \times 2^y$ Edit - An addition condition: $a$, $b$, and $c$ do not equal $0$.
I'm really digging into 3D vectors and their properties. I've decided to look and see which combinations of three integers will, when plotted in 3D space, generate a hypotenuse whose length is also an integer. For example; $\sqrt {6^2 + 42^2 + 85^2} = 95$
What I've found is quite interesting. It seems that there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x=2^y$ or $x=5 \times 2^y$ and $y$ is an integer.
$$\begin{array}{c|c|c}
y & 2^y & 5 \times 2^y \\ \hline
0 & 1 & 5 \\ \hline
1 & 2 & 10 \\ \hline
2 & 4 & 20 \\ \hline
3 & 8 & 40 \\ \hline
4 & 16 & 80 \\ \hline
5 & 32 & 160 \\ \hline
6 & 64 & \text{see below} \\ \hline
7 & 128 & \text{see below} \\
\end{array}$$
So, there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x$ equals any value from the following series: $1,2,4,5,8,10,16,20,32,40,64,80,128,160...$
I wrote a program which brute-force checked this for values up to $250$ for $a, b, \text {and } c$ (which is why I show no solutions to $5 \times 2^x$ using $6$ and $7$ above, since they should be $320$ and $640$ respectively). The program found at least 1 solution for all other integers from $1 \text{ to }250$.
Here's the code I used (excel VBA): http://pastebin.com/jVA27jYp
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It appears you do not permit any of $a,b,c$ to be zero. You can write it as $a^2+b^2+c^2=x^2$. In that case, it comes from the fact that the squares $\bmod 8$ are $0,1,4$ To have $2^{2y}$ be the sum of three squares you would have to have all of the squares be $0 \bmod 8$ or one be $0 \bmod 8$ and two of them be $4 \bmod 8$. In either case you can divide $a,b,c$ by $2$ and get a solution for $2^{2y-2}$, but then you make the same argument and there is no soluton by infinite descent. Similarly, if $x^2=5^22^{2y}$ you can perform the same infinite descent to get to $x^2=25$ and you have checked the cases for that.
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What are some easy but beautiful patterns in Pascal's Triangle? Pascal's triangle has wide applications in Mathematics.I have seen that the most important applications relate to the binomial coefficients and combinatorics.
Are there some other beautiful interesting patterns in Pascal's triangle that can be found by selecting some other combination modes ?
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There is a nice hexagonal property in the Pascal triangle:
\begin{array}{ccccccccccccccccc}
&&&&&&&&1\\
&&&&&&&1&&1\\
&&&&&&1&&2&&1\\
&&&&&1&&3&&3&&1\\
&&&&1&&4&&6&&4&&1\\
&&&1&&\mathbf{\color{blue}{5}}&&\mathbf{\color{blue}{10}}&&10&&5&&1\\
&&1&&\mathbf{\color{blue}{6}}&&\mathbf{\color{red}{15}}&&\mathbf{\color{blue}{20}}&&15&&6&&1\\
&1&&7&&\mathbf{\color{blue}{21}}&&\mathbf{\color{blue}{35}}&&35&&21&&7&&1\\
1&&8&&28&&56&&70&&56&&21&&8&&1\\
\end{array}
The hexagon around $\binom{n}{k}$ fulfils
\begin{align*}
\binom{n-1}{k-1}\binom{n}{k+1}\binom{n+1}{k}&=\binom{n-1}{k}\binom{n}{k-1}\binom{n+1}{k+1}\\
\end{align*}
The example $\binom{n}{k}=\binom{6}{2}=\color{red}{15}$ shows
\begin{align*}
\binom{n-1}{k-1}\binom{n}{k+1}\binom{n+1}{k}&=\binom{5}{1}\binom{6}{3}\binom{7}{2}=\color{blue}{5}\cdot \color{blue}{20}\cdot \color{blue}{21}=2100\\
\binom{n-1}{k}\binom{n}{k-1}\binom{n+1}{k+1}&=\binom{5}{2}\binom{6}{1}\binom{7}{3}=\color{blue}{10}\cdot \color{blue}{6}\cdot \color{blue}{35}=2100
\end{align*}
In other words:
*
*A hexagon around a binomial coefficient $\binom{n}{k}$ is always a perfect square. :-)
See Generalized hidden hexagon squares by A.K. Gupta for a generalized version of this relationship.
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Indeterminate Limit using Second Fundamental Theorem of Calculus I am trying to find $$\lim_{x\to 0}\frac{\int_0^x(x-t)\sin(t^2) \, dt}{\ln(1+x^4)}$$It seems that I am meant to use the 2nd Fundamental Theorem of Calculus to solve this, but I have never used it on an integral that has $x$ in it. Do I approach it any differently? Or am I on the wrong track entirely?
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$$\frac{\int_{0}^{x}(x-t)\sin(t^2)\,dt}{\log(1+x^4)}$$ is of the form $\frac uv=\frac 00$ if $x=0$. So, let us use L'Hospital rule considering $\frac {u'}{v'}$.
$$u=\int_{0}^{x}(x-t)\sin(t^2)\,dt \implies u'=\frac d {dx} \int_{0}^{x}(x-t)\sin(t^2)\,dt$$ The fundamental theorem of calculus gives $$\frac d {dx} \int_{0}^{x}(x-t)f(t)\,dt=\int_0^x f(t) \, dt$$ So, for the considered case $$u'=\int_0^x \sin(t^2) \, dt=\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} x\right)$$ where appears the Fresnel sine integral.
On the other hand $$v=\log(1+x^4)\implies v'=\frac{4 x^3}{1+x^4}$$ All of that makes $$\frac {u'}{v'}=\frac{\sqrt{\frac{\pi }{2}} \left(1+x^4\right) S\left(\sqrt{\frac{2}{\pi }}
x\right)}{4 x^3}$$
In the Wikipedia page, you will find the expansion (that converges for all $z$)
$$S(z) =\int_0^z \sin(t^2)\,dt=\sum_{n=0}^{\infty}(-1)^n\frac{z^{4n+3}}{(2n+1)!(4n+3)}$$ Keeping the first term of the expansion, we then have $$S\left(\sqrt{\frac{2}{\pi }}
x\right)=\frac{1}{3} \sqrt{\frac{2}{\pi }} x^3+O\left(x^4\right)$$ which makes $$\frac {u'}{v'}=\frac{\sqrt{\frac{\pi }{2}} \left(1+x^4\right) S\left(\sqrt{\frac{2}{\pi }}
x\right)}{4 x^3} \sim \frac{\sqrt{\frac{\pi }{2}} \left(1+x^4\right) \frac{1}{3} \sqrt{\frac{2}{\pi }} x^3}{4 x^3}=\frac{1}{12} \left(1+x^4\right) $$ and, if $x \to 0$, the limit is $\frac{1}{12}$.
Added just for your curiosity
We could have done more since a simple integration gives $$u=\int_{0}^{x}(x-t)\sin(t^2)\,dt=\frac{1}{2} \left(\sqrt{2 \pi } x S\left(\sqrt{\frac{2}{\pi }} x\right)-\sin
\left(x^2\right)\right)$$ So, using Taylor series around $x=0$ $$\frac uv=\frac{\frac{x^4}{12}-\frac{x^8}{336}+O\left(x^9\right)} {x^4-\frac{x^8}{2}+O\left(x^9\right)}=\frac{1}{12}+\frac{13 x^4}{336}+O\left(x^5\right)$$ which shows the same limit but also how it is approached.
Update
As @Vim commented, a pure Taylor approach could be used and this is much simpler than all the above.
Around $t=0$, we have $$(x-t)\sin(t^2)=t^2 x-t^3-\frac{t^6 x}{6}+\frac{t^7}{6}+O\left(t^9\right)$$ $$u=\frac{x^4}{12}-\frac{x^8}{336}+O\left(x^{10}\right)$$
$$\frac uv=\frac{\frac{x^4}{12}-\frac{x^8}{336}+O\left(x^{10}\right)} {x^4-\frac{x^8}{2}+O\left(x^9\right)}=\frac{1}{12}+\frac{13 x^4}{336}+O\left(x^6\right)$$
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simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ simplify $$\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$$
1.$\frac{3}{2}$
2.$\frac{\sqrt[3]{65}}{4}$
3.$\sqrt[3]{2}$
4.$1$
I equal it to $\sqrt[3]{a}+\sqrt[3]{b}$ but I cant find $a$ and $b$
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Let $X=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ From $(A+B)^3=A^3+B^3+3AB(A+B)$ we get the equation $$X^3-3ABX-(A^3+B^3)=0$$
We have $(AB)^3=({5+2\sqrt{13}})({5-2\sqrt{13}})=-27\Rightarrow {AB}=-3$ and $A^3+B^3=10$. Hence our equation $$X^3+9X-10=0\iff (X-1)(X^2+X+10)=0$$
Thus $$X=1$$ It is the $4$ the asked simplification.
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Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square
Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square.
We can observe that $49=7^2, 4489=67^2, 444889=667^2, \ldots$
I have tried expanding terms of the sequence, and to express it as a whole square. But it was too tough for me.
Any help will be appreciated.
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We can prove it as follows:
\begin{align*}
\underbrace{666\ldots6}_{n}\,7^2
&= \left[7 + 6(10^1 + 10^2 + \cdots + 10^n) \right]^2 \\
&= \left[7 + 6 \frac{10^{n+1} - 10}{9}\right]^2 \\
&= \left[ \frac{2 \cdot 10^{n+1} + 1}{3}\right]^2 \\
&= \frac{4 \cdot 10^{2n+2} + 4 \cdot 10^{n+1} + 1}{9} \\
&= 4 \cdot \frac{10^{2n+2} - 1}{9} + 4 \cdot \frac{10^{n+1} - 1}{9} + 1 \\
&= 4 \cdot \underbrace{111 \ldots 1}_{2n+2} + 4 \cdot \underbrace{111 \ldots 1}_{n+1} + 1 \\
&= \underbrace{444 \ldots 4}_{2n+2} + \underbrace{444 \ldots 4}_{n+1} + 1 \\
&= \underbrace{444 \ldots 4}_{n+1}\,\underbrace{888 \ldots 8}_{n}\,9.
\end{align*}
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If the chord $x+y=b$ of the curve ... If the chord $x+y=b$ of the curve $x^2+y^2-2ax-4a^2=0$ subtends a right angle at the origin, prove that: $b(b-a)=4a^2$
My Approach.
Given,
Equation of the chord,
$$x+y=b$$
$$\frac {x+y}{b}=1$$
Now,
Equation of the curve,
$$x^2+y^2-2ax-4a^2=0$$
$$x^2+y^2-2ax=4a^2$$
$$(b-y)^2+(b-x)^2-2ax=4a^2$$
I got stuck at here. Please help me to complete it.
|
The combined equation of the lines joining the origin to the end points of the chord can be obtained by "homogenising" the equation of the curve. This is
\begin{align*}
x^2+y^2 - 2ax\left(\frac{x+y}{b}\right) - 4a^2\left(\frac{x+y}{b}\right)^2 = 0
\end{align*}
These lines are perpendicular if the sum of the coefficients of $x^2$ and $y^2$ is zero. Thus the required condition is
\begin{align*}
1+1-\frac{2a}{b}-\frac{4a^2}{b^2}(1+1) &= 0\\
b^2 - ab -4a^2 &= 0\\
b(b-a) &= 4a^2
\end{align*}
|
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|
Find derivative of $\frac{x}{\sqrt{x^2 +2}}$ using definition I recognise that the derivative of $\frac{x}{\sqrt{x^2 +2}}$ is given by the expression $\lim_{h\to 0}\frac{\frac{x+h}{\sqrt{(x+h)^2}+2}-\frac{x}{\sqrt{x^2 +2}}}{h}$. But I have trouble proceeding from here. Any kind soul, please help!
|
we have $$\frac{1}{h}\frac{(x+h)\sqrt{x^2+2}-x\sqrt{(x+h)^2+2}}{\sqrt{(x+h)^2+2}\,\sqrt{x^2+2}}$$
and now multiply numerator and denominator by $$(x+h)\sqrt{(x^2+2)}+x\sqrt{((x+h)^2+2)}$$
can you proceed?
|
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|
Find the smallest positive integer that satisfies the system of congruences $N \equiv 2 \pmod{11}, N \equiv 3 \pmod{17}. $ Find the smallest positive integer that satisfies the system of congruences
\begin{align*}
N &\equiv 2 \pmod{11}, \\
N &\equiv 3 \pmod{17}.
\end{align*}
The only way I know to solve this problem is by listing it all out, and so far, it's not working. Is there a faster way? Thanks for posting a solution!
|
How it can be done using 'Euclid's algorithm': $N\equiv 3$ mod 17 can be written as N= 17i+ 3. Since 17= 11+ 6 is equivalent to 6 mod 11, we can write $N\equiv 2$ (mod 11) as $N\equiv 17i+ 3\equiv 6i+ 3= 2$ (mod 11) which is the sae as $6i= 2- 3= -1= 10$ (mod 11) or, dividing by 2, $3i\equiv 5$ (mod 11).
That is the same as 3i= 5+ 11j or 3i- 11j= 5. Euclids algorithm: 3 divides into 11 three times with remainder 2: 11- 3(3)= 2. 2 divides into 3 once with remainder 1: 3- 2= 1. Replace that "2" with 11- 3(3): 3- (11- 3(3))= 4(3)- 1(11)= 1. Multiplying both sides by 5, 20(3)- 5(11)= 5.
So one solution to 3i- 11j= 5 is i= 20, j= 5. Since we had before N= 17ii+ 3, N= 17(20)+ 3= 340+ 3= 343.
To check: 11 divides into 343 31 times with remainder 2: 343= 2 mod 11. 17 divides into 343 20 times with remainder 3: 343= 3 mod 17.
|
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|
Symmetric polynomials with Vieta's and Newton's theorems
Let $ x_{1}, x_{2}, x_{3}$ be the solutions of the equation $ x^3 -3x^2 + x - 1 = 0.$
Determine the values of $$\frac{1}{{x_{1}x_{2}}} + \frac{1}{{x_{2}x_{3}}} + \frac{1}{{x_{3}x_{1}}}$$
and also
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$
|
Given $(x-x_1)(x-x_2)(x-x_3)=x^3-3x^2+x-1$, the elementary symmetric polynomials of $x_1,x_2,x_3$ are given by Vieta's formulas:
$$ e_1=x_1+x_2+x_3=3,\quad e_2=x_1 x_2+x_1 x_3+x_2 x_3 = 1,\quad e_3=x_1 x_2 x_3 = 1 \tag{1}$$
hence
$$ \frac{1}{x_1 x_2}+\frac{1}{x_1 x_3}+\frac{1}{x_2 x_3} = \frac{e_1}{e_3} = \color{red}{3}.\tag{2}$$
Given the elementary symmetric polynomials, the power sums are given by Newton's identities:
$$ p_1=x_1+x_2+x_3 = 3,\quad p_2=x_1^2+x_2^2+x_3^2=e_1^2-2e_2 = 7,\tag{3} $$
and since $x_j^3 = 3x_j^2-x_j+1$ for every $j\in\{1,2,3\}$,
$$ p_3 = 3p_2-p_1+3 = 21-3+3 = \color{red}{21}.\tag{4}$$
|
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|
Does $A^TMA = MA^2$ where $A$ is symmetric and $M$ is not necesarily symmetric Let $A, M \in \mathbb{R}^{2\times 2}$, $A$ is symmetric
Does $A^TMA = MA^2$ hold regardless of whether $M$ is symmetric?
I have tested some simple examples and have not found a violation. Does a counter example exist?
|
Another counterexample:
$$
\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}
=\begin{pmatrix}4a & 2b \\ 2c & d\end{pmatrix},
$$
but
$$
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}
\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}
=
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
\begin{pmatrix}4 & 0 \\ 0 & 1\end{pmatrix}
=\begin{pmatrix}4a & b \\ 4c & d\end{pmatrix}.
$$
|
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|
Help on a very interesting integral: $\int_0^1\frac{x^5(1-x)^4}{1+x^3}\,dx$ I've been starting to work on problems with difficult integrals, and I came to
one of my problems where I end up having to integrate
$$\int_0^1 \frac{x^5(1-x)^4}{1+x^3} dx.$$
Wolfram seems to time out on integrals like these, so I decided to tackle this
integral by myself.
It seems like to me that there should be a meaningful integration by parts
that occurs here, in particular a relationship that could be developed between
$x^5$ and $\frac{(1-x)^4}{1+x^3}.$ The problem is, I'm having some difficulties
with the difficulty of building a meaningful integral of $dv = \frac{(1-x)^4}{
1+x^3} dx.$ any recommendations on how to go about starting with direction on
this problem?
|
Using the binomial theorem, expand $(1-x)^4$ in the numerator:
$$
I = \int_0^1 \frac{x^5(1-x)^4}{1+x^3}\,dx = \int_0^1 \frac{x^5(1 - 4x + 6x^2 - 4x^3 + x^4)}{1+x^3}\,dx \\
= \int_0^1 \frac{x^5 - 4x^6 + 6x^7 - 4x^8 + x^9}{1+x^3}\,dx.
$$
Using polynomial long division,
$$
I = \int_0^1 \left(x^6 - 4x^5 + 6x^4 - 5x^3 + 5x^2 - 6x + 5 + \frac{-5x^2 + 6x - 5}{x^3 + 1}\right)\,dx.
$$
At this point let's focus on the term $\frac{-5x^2 + 6x - 5}{x^3 + 1}$. Notice that $x^3 + 1 = (x+1)(x^2-x+1),$ and perform partial fraction decomposition:
$$
\int_0^1\frac{-5x^2 + 6x - 5}{x^3 + 1}\,dx = \int_0^1\left(\frac{x+1}{3(x^2-x+1)} - \frac{16}{3(x+1)}\right)\,dx.
$$
Put $\alpha = 3/2$, $\beta^2 = \alpha^2/3$, $u = x-1/2$, and let's focus on the term $\frac{x+1}{3(x^2-x+1)}$. Completing the square in the denominator:
$$
\int_0^1\frac{(x-\tfrac{1}{2}) + \tfrac{3}{2}}{3(x-\tfrac{1}{2})^ 2 + \tfrac{9}{4}}\,dx = \int_{-1/2}^{1/2}\frac{u + \alpha}{3u^2 + \alpha^2}\,du = \frac{1}{3}\int_{-1/2}^{1/2}\frac{\alpha}{u^2 + \tfrac{\alpha^2}{3}}\,du \\= \frac{2\alpha}{3}\int_{0}^{1/2}\frac{1}{u^2 + \beta^2}\,du = \frac{1}{\beta}\,\arctan{\frac{u}{\beta}}\,\Big|_{0}^{1/2} = \frac{2}{\sqrt{3}}\,\arctan{\frac{1}{\sqrt{3}}} = \frac{2}{\sqrt{3}}\frac{\pi}{6} = \frac{\pi}{3\sqrt{3}},
$$
where we made use of the facts that $2\alpha/3 = 1$, $f(u) = u/(3u^2 + \alpha^2)$ is odd, $g(u) = 1/(3u^2+\alpha^2)$ is even, and our region of integration is symmetric about $u = 0$.
The integral $\int_0^1 \frac{16}{3(x+1)}\,dx$ should be straightforward, and of course the integral of the polynomial portion is easy too.
|
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|
Existence of a quadrilateral with side lengths $a,b,c,d$ where $a+b+c>d$ and $d$ is largest length I tried to prove triangle inequality, given side lengths $a, b, c$ if $a + b \gt c$, then a triangle always exists, by placing the side length $c$ on the $x$-axis and proving that intersection points of two circles of radius $b$ and $c$ always exists.
Now, I wanted to do the same for any polygon with number of sides greater than $3$. I tried to prove for quadrilateral by using same method. But I can't progress.
Any idea on how prove if sides are $a, b, c, d$, where $d$ is the largest side and $a + b + c \gt d$, then quadrilateral exists.
|
Assume $a < b < c < d$.
Suppose that WLOG, $a+b > d$. Then , construct a triangle with lengths $a,b,d$, and at the vertex joining $b$ and $d$, make a separation and attach the length $c$ to make a quadrilateral. Similarly for the other pairwise sums. Hence, we will assume that each pairwise sum is less than $d$.
Let $\epsilon = a+b+c-d > 0$.
A triangle exists with length $a,b,$ and $ d-c + \frac{\epsilon}{2}$ , as $d-c + \frac{\epsilon}2 < d-c + \epsilon = a+b$, and $a < b$, and $b+c < d \implies b < d-c + \frac{\epsilon}{2}$.
Furthermore, a triangle exists with lengths $d-c + \frac{\epsilon}{2} , d, c$ as $c+d-c+\frac{\epsilon}{2} > d$ , $c< d$, and if $d-c + \frac{\epsilon}{2} > d +c \implies c < \frac{\epsilon}{4} \implies a+b+c < \epsilon \implies a+b+c < a+b+c-d$, which is a contradiction.
Now, you have two triangles having the side $d-c+\frac{\epsilon}2$ common. Glue them together to get a quadrilateral with sides $a,b,c,d$.
|
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|
Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$
and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$
So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure how to show that
|
Hint: use Cauchy-Schwarz in $\mathbb{R}^2$ on the vectors $(a,b)$ and $(c,d)$. This technique should provide a one-line proof of the desired result.
More directly, from what you've already computed, you can observe that
$$
a^2d^2 - 2abcd + b^2c^2 = (ad - bc)^2 \geq 0
\text{,}
$$
so $a^2d^2 + b^2c^2 \geq 2abcd$.
|
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|
In any triangle $ ABC $ $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $. In any triangle $ ABC $ prove that $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $.
Please help. Thanks in advance.
|
In the above configuration, $R \sin(B-C) = OJ_A = M_A H_A$ and $R\sin(B+C)=R\sin(A)= BM_A$. Moreover, by the Pythagorean theorem we have $H_A B^2-H_A C^2=AB^2-AC^2=c^2-b^2$, hence
$ b^2-c^2 = (H_A B+H_A C)(H_A B-H_A C) $ and
$ H_A B - H_A C = \frac{c^2-b^2}{a}.$
It follows that
$$ BH_A = \frac{1}{2}\left(\frac{c^2-b^2}{a}+a\right) = \frac{a^2+c^2-b^2}{2a} $$
and
$$ M_A H_A = \frac{a}{2}-BH_A = \frac{b^2-c^2}{2a} $$
so
$$ \boxed{\frac{\sin(B-C)}{\sin(B+C)}=\frac{OJ_A}{BM_A} = \frac{2 M_A H_A}{a} = \color{red}{\frac{b^2-c^2}{a^2}}.}$$
An alternative approach through the sine addition formula and the cosine theorem:
$$\begin{eqnarray*}\frac{\sin(B-C)}{\sin(B+C)}=\frac{2R \sin(B-C)}{2R\sin(A)}&=&\frac{2R\sin(B)\cos(C)-2R\sin(C)\cos(B)}{a}\\&=&\frac{2ab\cos(C)-2ac\cos(B)}{2a^2}\\&=&\frac{(a^2+b^2-c^2)-(a^2+c^2-b^2)}{2a^2}\\&=&\color{red}{\frac{b^2-c^2}{a^2}}.\end{eqnarray*}$$
|
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|
Skip a number in a summation $$\sum_{n=1}^{10} n^2$$
Returns:
$1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100$
and I would like it to return:
$1 + 9 + 25 + 49 + 81 + 121 + 169 + 225 + 289 + 361$
How will I go about this and, more importantly, how does it work?
|
You are looking for $1^2 + 3^2 +5^2 +7^2 +9^2 +11^2+13^2+15^2+17^2 +19^2$
so whats wrong with
$$\sum_{n=0}^9(2n+1)^2$$
or
$$\sum_{n=1}^{10}(2n-1)^2$$
It's $2n$ because we are going up in twos
|
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|
How to integrate $ \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} $? I am having a little problem with my maths homework. The problem is as follows:
\begin{equation}
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
\end{equation}
I tried to do the following but got stuck halfway:
Let $\ \ x \ = asin\theta, \ hence, \ dx = acos\theta \ d\theta $
$
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
$
$
= \int^\frac{\pi}{2}_0{\cfrac{acos\theta}{asin\theta \ + \ \sqrt{a^2 \ - \ a^2sin^2\theta}}}\ d\theta
$
$
= a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ \sqrt{a^2cos^2\theta}}}\ d\theta
$
$
= a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ acos\theta }}\ d\theta
$
$
= \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{sin\theta \ + \ cos\theta }}\ d\theta \\ \\
$
$
= \int^\frac{\pi}{2}_0{\left(1 \ + \ \cfrac{(cos\theta \ - \ sin\theta)}{sin\theta \ + \ cos\theta } - \cfrac{cos\theta}{sin\theta \ + \ cos\theta}\right)}\ d\theta
$
Could someone please advise me how to solve this problem?
|
Assuming $a>0$ and applying the substitutions $x=az$, $z=\sin\theta$:
$$ I(a)=\int_{0}^{a}\frac{dx}{x+\sqrt{a^2-x^2}}=\int_{0}^{1}\frac{dz}{z+\sqrt{1-z^2}}=\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta+\cos\theta}\,d\theta $$
but due to the substitution $\theta=\frac{\pi}{2}-\varphi$ we also have $I(a)=\int_{0}^{\pi/2}\frac{\sin\theta}{\sin\theta+\cos\theta}\,d\theta$, hence:
$$ 2\cdot I(a) = \int_{0}^{\pi/2}\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\,d\theta = \frac{\pi}{2}$$
and $\boxed{I(a)=\color{red}{\large\frac{\pi}{4}}}$ holds by symmetry.
|
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|
Prove that the function $(12-6x+x^2) e^x - (12+6x+x^2)$ is positive on $(0, \infty)$
Prove that the function $(12-6x+x^2) e^x - (12+6x+x^2)$ is positive on $(0, \infty)$.
I am unable to prove this.
|
Let $f(x)=x+\ln(x^2-6x+12)-\ln(x^2+6x+12)$.
Since $f'(x)=1+\frac{2x-6}{x^2-6x+12}-\frac{2x+6}{x^2+6x+12}=\frac{x^4}{(x^2-6x+12)(x^2+6x+12)}\geq0$,
we obtain that $f(x)\geq f(0)=0$.
Done!
|
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For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-3a^2+9$ a prime number?
Here is what I did: $$a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a)$$ To find $a$, I looked at the following situations:
$$a^4-3a^2+9=1$$
$$a^4-3a^2+9=3$$
$$a^4-3a^2+9=5$$
$$a^4-3a^2+9=7$$
$$a^4-3a^2+9=11$$
$$a^4-3a^2+9=13$$
$$....$$
|
Here is a fundamental fact about prime numbers: IF: a prime number $p$ factors as $p = xy$, where $x$ and $y$ are integers, THEN: $x = \pm 1$ or $y = \pm 1$.
You have already correctly noticed that
$$
a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a).
$$
Assuming that $a^4 - 3a^2 + 9$ is prime, it follows by the above result that there are two possibilities:
\begin{align*}
a^2 + 3 + 3a &= \pm 1 \\
a^2 + 3 - 3a &= \pm 1.
\end{align*}
This is really four possibilities in disguise:
\begin{align*}
a^2 + 3 + 3a &= 1 \\
a^2 + 3 + 3a &= -1 \\
a^2 + 3 - 3a &= 1 \\
a^2 + 3 - 3a &= -1.
\end{align*}
Now, your job is to solve for $a$ in each of these four cases. Then, check for each case whether the $a^4 - 3a^2 + 9$ is actually prime.
|
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|
Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do this?
My method:
$$
{f'(x)}=\left [\ln \left (\frac{x}{(x^2+1)^\frac{1}{2}} \right) \right ]'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right)\left (\frac{x}{(x^2+1)^\frac{1}{2}} \right)'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[(x^2+1)^\frac{1}{2}]'}{[(x^2+1)^\frac{1}{2}]^2} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(x^2+1)']}{\left | x^2+1 \right |} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)]}{x^2+1} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x^2(x^+1)^{-\frac{1}{2}}}{x^2+1} \right)=\frac{(x^2+1)^{(\frac{1}{2}+\frac{1}{2})}-x^2(x^2+1)^{\frac{1}{2}+-\frac{1}{2}{}}}{x(x^2+1)}=-\frac{x^2}{x}=-x
$$
and
$$
f''(x)=(-x)'=-1\
$$
This took me much more than 1.5 hours just to type into LaTex :'(
|
Hint
$$\ln(\frac{x}{\sqrt{x^2+1}})=\ln x-\frac{1}{2}\ln(x^2+1)$$
|
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|
$\lim_{n\rightarrow\infty} n^2C_n$ for $C_n=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}nx}{\sin^{-1}nx}dx$ equals? Let $$C_n=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}nx}{\sin^{-1}nx}dx$$ then $\lim_{n\rightarrow\infty} n^2C_n$ equals?
I am having trouble in finding the integral.Wolfram alpha too doesnt give any answer http://www.wolframalpha.com/input/?i=integrate+arctan(nx)%2Farcsin(nx).
Any idea?
|
By the MVT for integrals, there is $\xi_n\in[\frac{1}{n+1},\frac1n]$ such that
$$ \int_\frac{1}{n+1}^\frac1n\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx=\frac{\tan^{-1}(n\xi_n)}{\sin^{-1}(n\xi_n)}\frac{1}{n(n+1)}.$$
Note that both $\sin^{-1}x$ and $\tan^{-1}x$ are increasing in $[0,1]$ and hence
$$\frac{\tan^{-1}(n\cdot\frac1{n+1})}{\sin^{-1}(n\cdot\frac1{n})}\le\frac{\tan^{-1}(n\xi_n)}{\sin^{-1}(n\xi_n)}\le \frac{\tan^{-1}(n\cdot\frac1n)}{\sin^{-1}(n\cdot\frac1{n+1})}. $$
So
$$ n^2\frac{\tan^{-1}(n\cdot\frac1{n+1})}{\sin^{-1}(n\cdot\frac1{n})}\frac{1}{n(n+1)}\le n^2\int_\frac{1}{n+1}^\frac1n\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx\le n^2\frac{\tan^{-1}(n\cdot\frac1n)}{\sin^{-1}(n\cdot\frac1{n+1})}\frac{1}{n(n+1)} $$
which implies
$$ \lim_{n\to\infty}n^2\int_\frac{1}{n+1}^\frac1n\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx=\frac12.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$ Let $a,b,c>0; a+b+c=1$. Prove the inequality
$$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$
My work so far:
I tried AM-GM and used fact $a+b+c=1$.
|
An homogenization and a full expanding give:
$$\sum\limits_{cyc}(a^6b^2+2a^5b^3+a^5c^3+2a^4b^4+a^6bc-2a^5b^2c+4a^5c^2b+$$
$$+a^4b^3c+4a^4c^3b-7a^4b^2c^2-7a^3b^3c^2)\geq0$$
which is obvious by AM-GM.
|
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|
How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rightarrow \quad z_1 = -3+2i \quad z_2 = 3-2i
$$
My question is: can the following method (below) be used to solve the problem above? Motivation for this question: if I were to find the cubic roots of the number given, I couldn't use the first method.
I will use this "other method" it in a different problem.
Find the square roots of $ z = 2i $
The method:
Since $ \rho = 2 $ and $ \theta = \frac{\pi}{2} $ we have to find a complex number such that
\begin{align*}
z_k^2 &= 2i\\
z_k^2 &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\rho_k^2 \left( \cos 2\theta_k + i \sin 2 \theta_k \right) &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\end{align*}
\begin{cases}
\rho^2 &= 2\\
2\theta_k &= \frac{\pi}{2} + 2k\pi
\end{cases}
\begin{cases}
\rho &= \sqrt{2}\\
\theta_k &= \frac{\pi}{4} + 2k\pi
\end{cases}
\begin{align*}
z_0 &= \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 1+i\\
z_1 &= \sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = -1-i
\end{align*}
Using this method for the first example seems like a dead end, specially because of the fact that the angle of $z$ is not as straightforward as the angle of the second example.
Thank you.
|
Your method should work fine in both cases (though the trigonometry is a bit more complicated), but I point out that your first method can still work on the square roots of $2i$: We see that
$$
a^2-b^2 = 0
$$
so $a = \pm b$, and then, secondly, $2ab = 2$. This gives us $a = b = \pm 1$ as the solutions, so the square roots of $2i$ are $1+i$ and $-1-i$.
|
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|
Easier way to discover the area of a right triangle In the following right triangle: $y-x=5$ and the altitude to the hypotenuse is 12. Calculate its area.
I've managed to discover its area using the following method, but it ends up with a 4th degree equation to solve. Is there an easier way to solve the problem?
$ha=xy \Rightarrow 12 \sqrt{x^2+y^2} = xy$
Substitute $y=5+x$ and square both sides:
$144 (x^2 + (5+x)^2)=x^2 (5+x)^2 \Rightarrow x^4+10x^3-263x^2-1440x-3600=0$
Which only positive solution is $x=15$ and therefore $y=20$ and the area is $\frac{15 \cdot 20 }{2}=150$
Thanks in advance.
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Slightly easier: Write the unknown area as $A:=xy/2$. From $12 \sqrt{x^2+y^2} = xy$ deduce $$x^2+y^2=(A/6)^2.$$ Substitute this into $x^2-2xy+y^2=25$. This gets you a quadratic for $A$:
$$
\left(\frac A6\right)^2-4A-25=0
$$
|
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|
Prove that $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$ (suggestions for improvement on basic induction style/proof) Please see my proof to the following proposition below. From a standpoint of critiquing does this seem like a great execution or would someone feel that this proof was overly cumbersome? Specifically for the purposes of marks in a class how would this proof be viewed?
Prove that $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$ for $n\in\mathbb{N}$.
$Proof.$ We use mathematical induction.
Base Case. Let $n=1$. We see that $3^n=3^{(1)}=3$ and $\frac{3^{n+1}-3}{2}=\frac{3^{(1)+1}-3}{2}=\frac{6}{2}=3.$ Thus our equality holds for $n=1$.
Inductive Step. Let $n\ge1$. Assume $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$. Now observe that
$$\begin{align*}
3^1+3^2+3^3+3^4+\cdots+3^n+3^{n+1}&=\\
\frac{3^{n+1}-3}{2}+3^{n+1}&=\\
\frac{3^{n+1}-3+2(3^{n+1})}{2}&=\\
\frac{3^{n+1}+2(3^{n+1})-3}{2}&=\\
\frac{3^{n+1}(1+2)-3}{2}&=\\
\frac{3^{n+1}(3)-3}{2}&=\\
\frac{3^{n+2}-3}{2}&=\\
\frac{3^{(n+1)+1}-3}{2}&=\\
\end{align*}$$
It follows by induction that $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$ for $\forall n\in\mathbb{N}$
|
Your proof is good!
Here is another nice way to prove it:
$$S=3^1+3^2+3^3+...+3^n\\3S=3(3^1+3^2+3^3+...+3^n)\\3S=3^2+3^3+3^4+...+3^{n+1}\\3S-S=3^{n+1}[-3^n+3^n]-...-[3^2+3^2]-3^1\\2S=3^{n+1}-3\\S=\frac{3^{n+1}-3}{2}$$
|
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|
Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$.
Let $x,y$ be real numbers s.t.
$a + b = 6$,
$ax + by = 10$,
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$.
I am new to problem solving exercises like this and therefore appreciate diverse approaches to this problem as well as comments on how to tackle those types of exercises.
|
$$ \color{red}{3 \cdot 62 - 24 = 162}. $$
$$ 3 \cdot 10 - 6 = 24. $$
$$ 3 \cdot 24 - 10 = 62. $$
This is the observation of @Ross,
$$ (xy-1)(x-y)^2 = 0. $$ Ross points out in comment (below) that an ingredient was $44=abxy(x-y)^2,$ so that
$$ a \neq 0, b \neq 0, x \neq 0, y \neq 0, x \neq y. $$ Finally,
$$ xy = 1. $$
The simple fact I got was
$$ a (x^2 - 3x+1) + b ( y^2 - 3 y + 1) = 0. $$
We have $xy = 1.$ If $x^2 - 3 x + 1 \neq 0$ and $y^2 - 3 y + 1 \neq 0,$
$$ y^2 - 3 y + 1 = (x^2 - 3 x + 1) / x^2,$$
$$ x^2 ( y^2 - 3 y + 1 ) = x^2 - 3 x + 1. $$
$$ a x^2 (y^2 - 3 y + 1) + b (y^2 - 3 y + 1) = 0,$$
$$ (a x^2 + b)(y^2 - 3 y + 1) = 0. $$
Switch the letters, we get
$$ (a + b y^2)(x^2 - 3 x + 1) = 0. $$
Alright,
$$ a x^2 + b y^2 = 24, a x^2 + b = 0. $$
$$ b(y^2 - 1) = 24. $$ Also
$$ a ( x^2 - 1) = 24. $$
Add,
$$ a(x^2 - 1) + b ( y^2 - 1) = 48.$$ However, we know
$$ a x^2 + b y^2 - a - b = 24 - 6 = 18. $$ This contradicts the assumption
$ x^2 - 3 x + 1 \neq 0. $
$$ \color{red}{ x^2 - 3 x + 1 = 0} $$
$$ \color{red}{ y^2 - 3 y + 1 = 0} $$
|
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|
About the primitives of $\frac1{a^2 + (2x)^2}$ So I was to find the integral $\int^{\frac{3}{2}}_0\frac{1}{9+4x^2}dx$
I noticed that the denominator is equal to $3^2+(2x)^2$ and thought I could use the integral $\int\frac{1}{a^2+x^2}dx = \frac{1}{a}arctan(\frac{x}{a}) + C$.
Hence, $[\frac{1}{3}arctan(\frac{2z}{3})]^{3/2}_0 = \frac{\pi}{12} - 0 = \frac{\pi}{12}$
The textbooks answer was $\frac{\pi}{24}$. It also gives a solution by dividing the denominator and taking $\frac{1}{4}$ outside of the integral before integrating so the denominator is $\frac{9}{4} + x^2 = (\frac{3}{2})^2 + x^2$.
So my question is why does the integral for $\int\frac{1}{a^2+x^2}dx$ not work when instead of ${x^2}$ you have ${(kx)^2}$ for some constant k?
|
The same reason why $\sin (2x)$ is not an anti-derivative (= primitive function) of $\cos (2x)$, although $\sin (x)$ is an anti-derivative of $\cos (x)$.
To see what happens, you can use the substitution method for integrals (let $t = kx$ etc).
Or the other way around: because the chain rule for derivatives would produce an extra factor $2$ (in the case of $2x$) or generally $k$ (in the case of $kx$).
|
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|
definite integral of a trigonometric function with log $$I=\int_0^{\frac\pi 2} \log(1+\tan x)\,dx$$
I attempted it by substituting $(\frac\pi 2-x)$ as $x$ but we get $\log(1+\cot x)$ not sure how to proceed... and on adding $$2I=\log[(1+\tan x)(1+\cot x)]$$
|
Another solution using Clausen function
Start with $$ I=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( 1+\tan \left( x\right) \right) dx=\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \sin \left( x\right) +\cos \left( x\right) \right) dx} \limits^{I_{1}}-\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \cos \left( x\right) \right) } \limits^{I_{2}}$$
$$ I_{1}=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \sin \left( x\right) +\cos \left( x\right) \right) dx=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \sqrt{2} \left( \sin \left( \frac{\pi }{4} \right) \cos \left( x\right) +\cos \left( \frac{\pi }{4} \right) \sin \left( x\right) \right) \right) dx= \frac{-\pi }{4} \ln\left( 2\right) +\int \limits^{\frac{\pi }{2} }_{0}\ln\left( 2\sin \left( x+\frac{\pi }{4} \right) \right) dx=\frac{-\pi }{4} \ln\left( 2\right) +\frac{1}{2} \int \limits^{\frac{3\pi }{2} }_{\frac{\pi }{2} }\ln\left( 2\sin \left( \frac{t}{2} \right) \right) dt=\frac{-\pi }{4} \ln\left( 2\right) -\frac{1}{2} \text{Cl}_{2}\left( \frac{3\pi }{2} \right) +\frac{1}{2} \text{Cl}_{2}\left( \frac{\pi }{2} \right) =\boxed {\frac{-\pi }{4} \text{ln}\left( 2\right) +G}$$
$$ I_{2}=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \cos \left( x\right) \right) dx=\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( \sin \left( x\right) \right) dx $$
$$ 2I_{2}=\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( \sin \left( 2x\right) \right) dx-\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( 2\right) dx=\frac{1}{2} \int \limits^{\pi }_{0}\text{ln}\left( \sin \left( u\right) \right) du-\frac{\pi }{2} \text{ln}\left( 2\right) =\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( \sin \left( u\right) \right) du} \limits^{I_{2}}-\frac{\pi }{2} \text{ln}\left( 2\right) $$
$$ I_{2}=\boxed {\frac{-\pi }{2} \text{ln}\left( 2\right)}$$
We have $I=I_{1}-I_{2} $ to get the final result
$$ \color{purple}{\boxed {I= G+\frac{\pi }{4} \text{ln}\left( 2\right) } }$$
|
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|
Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let n=0. Then $\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$. Thus for $n=0$ our equation is satisfied.
Inductive Step. Let $n\ge0$. Assume $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$. Now observe that
$$\begin{align*}\sum_{k=0}^{n+1}k{m+k \choose m}&=\\
\sum_{k=0}^{n}k{m+k \choose m}+(n+1){m+n+1\choose m}&=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}+(n+1){m+n+1\choose m} \end{align*}$$
This is where I'm getting stuck... using Pascal's Identity to combine some of these terms seems ideal. However, the factors of $n$ and $n+1$ are making a clever manipulation difficult for me.
|
A combinatorial proof is also possible. I have $m+n+1$ white balls numbered $0$ through $m+n$. I’m going to paint $m+2$ of them red, then choose any of the red balls except the one with the highest number and put a gold star on it, and I want to know how many different outcomes are possible.
Suppose that the highest-numbered red ball is ball $m+k$. There are $m+k$ balls with smaller numbers (since I started the numbering at $0$), so there are $\binom{m+k}m$ ways to choose the other $m$ red balls that don’t have the gold star. Once they’ve been chosen, there are $k$ ways to pick one of the remaining balls numbered below $m+k$, paint it red, and slap a gold star on it. Thus, there are $k\binom{m+k}m$ outcomes in which ball $m+k$ is the highest-numbered red ball. Summing over $k$ gives the total number of possible outcomes: it’s
$$\sum_{k=0}^nk\binom{m+k}m\;.$$
But we can count these outcomes in another way. There are $\binom{m+n+1}{m+1}$ ways to choose $m+1$ balls to be the unstarred red balls, and we can then choose any of the remaining $n$ balls to be the starred red ball, so there are
$$n\binom{m+n+1}{m+1}\tag{1}$$
ways to choose a set of $m+2$ balls, paint them red, and put a gold star on one of the balls. However, this includes the outcomes in which the gold star is on the highest-numbered red ball, and we don’t want those. For each possible set of $m+2$ red balls there is exactly one unwanted outcome, the one in which we put the gold star on the highest-numbered red ball, and there are
$$\binom{m+n+1}{m+2}$$
possible sets of red balls, so we need to subtract this from $(1)$ to get the correct count of
$$n\binom{m+n+1}{m+1}-\binom{m+n+1}{m+2}\;.$$
This shows that
$$\sum_{k=0}^nk\binom{m+k}m=n\binom{m+n+1}{m+1}-\binom{m+n+1}{m+2}\;,$$
as desired.
Added: Let me suggest a way to proceed from the point at which you got stuck with your induction argument. First, it’s not too hard to notice that you have both $n\binom{m+n+1}{m+1}$ and $n\binom{m+n+1}m$, which can be combined using Pascal’s identity:
$$\begin{align*}
&\sum_{k=0}^nk\binom{m+k}m+(n+1)\binom{m+n+1}m\\
&\qquad=n\binom{m+n+1}{m+1}-\binom{m+n+1}{m+2}+(n+1)\binom{m+n+1}m\\
&\qquad=n\left(\binom{m+n+1}{m+1}+\binom{m+n+1}m\right)+\binom{m+n+1}m-\binom{m+n+1}{m+2}\\
&\qquad=n\binom{m+n+2}{m+1}+\binom{m+n+1}m-\binom{m+n+1}{m+2}\;.\tag{1}
\end{align*}$$
You want to show that this is equal to
$$(n+1)\binom{m+n+2}{m+1}-\binom{m+n+2}{m+2}\;.\tag{2}$$
You might try taking the difference and trying to show that it’s $0$. Subtracting $(1)$ from $(2)$, we get
$$\binom{m+n+2}{m+1}-\binom{m+n+2}{m+2}-\binom{m+n+1}m+\binom{m+n+1}{m+2}\;.\tag{3}$$
Pascal’s identity allows us to combine the second and fourth terms:
$$\binom{m+n+2}{m+2}=\binom{m+n+1}{m+2}+\binom{m+n+1}{m+1}\;,$$
so
$$\binom{m+n+1}{m+2}-\binom{m+n+2}{m+2}=-\binom{m+n+1}{m+1}\;,$$
and $(3)$ reduces to
$$\binom{m+n+2}{m+1}-\binom{m+n+1}{m+1}-\binom{m+n+1}m\;,$$
and one more application of Pascal’s identity verifies that this is indeed $0$.
|
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|
Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$ Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$.
How do i do this by using these identities.
$C^r_r + C^{r+1}_r + C^{r+2}_r + ....+C^{n}_r = C^{n+1}_{r+1}$
Or
$C^r_0 + C^{r+1}_1 + C^{r+2}_2 + ....+C^{r+k}_k = C^{r+k+1}_{k}$
|
Hint:
$$(1\times2)+(2\times3)+(3\times4)+\dots(n\times(n+1)) =\sum_{k=1}^nk(k+1) \\
\begin{align}
& =2\sum_{k=1}^n\frac{k(k+1)}2 \\
& =2\sum_{k=1}^nC_2^{k+1} \\
\end{align}$$
|
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If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold? PROBLEM STATEMENT
If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold?
$$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<\frac{B}{D}<\frac{B}{C}<\frac{C}{D}<1$$
$$1<\frac{D}{C}<\frac{C}{B}<\frac{D}{B}<\frac{B}{A}<\frac{C}{A}<\frac{D}{A}$$
MY ATTEMPT
Since $A, B, C, D \in \mathbb{N}$, we obtain
$$\frac{A}{D} < \frac{B}{D} < \frac{C}{D} < 1$$
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C}$$
$$\frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B}$$
$$1 < \frac{B}{A} < \frac{C}{A} < \frac{D}{A}$$
Summarizing the first and the second:
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C} < \frac{D}{B} < \frac{D}{A}$$
Summarizing the third and the fourth:
$$\frac{A}{D} < \frac{A}{C} < \frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B}$$
Summarizing the first and the third:
$$\frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B} < \frac{D}{A}$$
Summarizing the second and the fourth:
$$\frac{A}{D} < \frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C}$$
Summarizing the first and the fourth:
$$\frac{A}{D} < \frac{B}{D} < \frac{C}{D} < 1 < \frac{B}{A} < \frac{C}{A} < \frac{D}{A}$$
Summarizing the second and the third:
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{C}{B} < \frac{D}{B}.$$
Following a different approach, we also have
$$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<1$$
$$\frac{B}{D}<\frac{B}{C}<1<\frac{B}{A}$$
$$\frac{C}{D}<1<\frac{C}{B}<\frac{C}{A}$$
$$1<\frac{D}{C}<\frac{D}{B}<\frac{D}{A}.$$
CONCLUSION
We are therefore sure about the validity of the series of inequalities
$$\frac{A}{D} < \frac{A}{C} < \frac{B}{C} < 1 < \frac{C}{B} < \frac{D}{B} < \frac{D}{A}.$$
(Note that $\min(A,B,C,D)=A$ and $\max(A,B,C,D)=D$. Therefore, $A/D$ is the minimum possible fraction with numerator and denominator (distinct from the numerator) coming from the set $\{A,B,C,D\}$. Reciprocally, $D/A$ is the maximum possible fraction.)
However, we cannot conclusively "decide" which member of the (unordered) pairs
$$\left\{\frac{D}{C},\frac{C}{B}\right\}$$
and
$$\left\{\frac{D}{B},\frac{B}{A}\right\}$$
is larger (or equivalently, smaller).
QUESTION
Is my analysis of the problem correct? Or is the problem "decidable" using some other approach/in another context?
|
You are correct. Consider the following examples:
$$A=1,B=3,C=4,D=5$$
In which $$\frac{D}{C}<\frac{C}{B},\frac{D}{B}<\frac{B}{A}$$
Also:
$$A=2,B=3,C=4,D=6$$
yields $$\frac{D}{C}>\frac{C}{B}, \frac{D}{B}>\frac{B}{A}$$
|
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|
Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.
|
If it simplifies, then $7+5\sqrt 2$ is a cube $(a+b \sqrt 2)^3$, in the ring of integers of $\Bbb Q(\sqrt 2)$, which is $\Bbb Z[\sqrt 2]$, so $a$ and $b$ must be integers (sometimes you can only deduce that $2a,a+b,2b$ are integers but it's still very good)
Moreover, you have $2a = (7+5\sqrt 2)^\frac 13 + (7-5\sqrt 2)^\frac 13$
Since $5\sqrt 2$ is between $7$ and $8$,
The first term is between $2$ and $3$ the second term is between $-1$ and $0$, so the sum has to be $2$ if it's going to be an even integer. So we can bet on $a=1$.
Then writing $7+5\sqrt 3 = (1+b\sqrt 2)^3$ you get $7 = 1+6b^2$, thus $b^2=1$, and you also get $5 = 3b+2b^3 =b(3+2b^2) = b(3+2) = 5b$ so $b=1$.
Since it is compatible with $b^2=1$, it shows that $(1+\sqrt 2)^3 = 7+5\sqrt 2$
|
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|
Find $f$ if $f'(x)=\dfrac{x^2-1}{x}$ knowing that $f(1) = \dfrac{1}{2}$ and $f(-1) = 0$ I am asked the following problem:
Find $f$ if $$f'(x)=\frac{x^2-1}{x}$$
I am not sure about my solution, which I will describe below:
My solution:
The first thing that I've done is separate the terms of $f'(x)$
\begin{align*}
f'(x)&=\frac{x^2-1}{x}\\
&=x-\frac{1}{x}\\
\therefore \quad f(x)&=\frac{x^2}{2}-\ln|x|+c
\end{align*}
For ( x > 0 ):
\begin{align*}
f(x)&=\frac{x^2}{2}-\ln x+c\\
f(1)&=\frac{1^2}{2}-\ln 1+c=\frac{1}{2} \quad \Rightarrow \quad c=0\\
f(x)&=\frac{x^2}{2}-\ln |x|
\end{align*}
For ( x < 0 )
\begin{align*}
f(x)&=\frac{x^2}{2}-\ln (-x)+c\\
f(-1)&=\frac{(-1)^2}{2}-\ln [-(-1)]+c=0 \quad \Rightarrow \quad c=-\frac{1}{2}\\
f(x)&=\frac{x^2}{2}-\ln |x|-\frac{1}{2}
\end{align*}
Is my solution correct? Should I really find two different answers, one for $x > 0$ and another for $x < 0$?
Thank you.
|
Looks fine! However a little typo when calculate $f(-1)=0$ not $-1$
|
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|
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
|
First you need to factor before bring the terms into one side. $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$ Now you subtract the right hand side, where it becomes $$(x^2-3x+1)^2-(2x-3)^2=0$$ You can factor by using differences of squares where $a^2-b^2=(a+b)(a-b)$.
Now finish this off on your own:)
If you really want the full solution just mouse over this part
$$(x^2-3x+1)^2-(2x-3)^2=0 $$ $$(x^2-x-2)(x^2-5x+4)=0$$ $$(x-2)(x+1)(x-4)(x-1)=0$$ $$x=-1,1,2,4$$
|
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|
Product of digits of a 7 digit number is $2^43^4$ Find the number of seven digit numbers whose product of digits is $2^43^4$.
One method is to list out all possible sets of seven digits that give this product, and then find number of permutations for each case. But there are too many cases that way.
Is there a shorter method?
|
I propose to count these numbers with respect to the number of digits 6.
We have five cases: ZERO, ONE, TWO, THREE, FOUR. In the following formulas we place the $6$s then the powers of $2$ (i. e. $2,4,8$) and finally the powers of $3$ (i. e. $3,9$).
If there are ZERO digits 6 then we have
$$
\binom{7}{4}\left[\binom{3}{4}+3\binom{3}{3}+\binom{3}{2}\right]
+3\binom{7}{3}\left[\binom{4}{4}+3\binom{4}{3}+\binom{4}{2}\right]
\\+
3\binom{7}{2}\left[\binom{5}{4}+3\binom{5}{3}+\binom{5}{2}\right]
=5040.$$
If there is ONE digit 6 then we have
$$\binom{7}{1} \left[ \binom{6}{3}\left[\binom{3}{3}+2\binom{3}{2}\right]
+2\binom{6}{2}\left[\binom{4}{3}+2\binom{4}{2}\right]+
\binom{6}{1}\left[\binom{5}{3}+2\binom{5}{2}\right]
\right]
=5600.$$
If there are TWO digits 6 then we have
$$\binom{7}{2}\left[ \binom{5}{2}\left[\binom{3}{2}+\binom{3}{1}\right]
+\binom{5}{1}\left[\binom{4}{2}+\binom{4}{1}\right]
\right]=2310.$$
If there are THREE digits 6 then we have
$$\binom{7}{3}\binom{4}{1}\binom{3}{1}=420.$$
If there are FOUR digits 6 then we have
$$\binom{7}{4}=35.$$
Finally, the total number is
$$5040+ 5600+2310+420+35=13405.$$
|
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|
Find the global minimum of a function of two variables without derivatives If $f:(\mathbb{R}^{+*})^2\to\mathbb{R}$ is defined by $$f(x,y)=\sqrt{x+y}\cdot\Big(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\Big),$$
how would you prove that $f$ has a global minimum without using all the differential calculus tools ? I am asked to only use the inequality $x+y\geq 2\sqrt{xy},$ but steps as $f(x,y)\geq 2\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}\geq 2$ are too violent and do not seem to give a reachable minimum. Thank you for your help !
|
$$f(x,y) = \sqrt{x+y} \cdot \left( \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \right) = \sqrt{x+y} \frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$$
Squaring we get
$$f(x,y)^2 = (x+y) \frac{x + y + 2 \sqrt{xy}}{xy} = \frac{(x+y)^2}{xy} + 2 \frac{x+y}{\sqrt{xy}}$$
Now AM-GM implies that the first summand and second summand are both minimized at $x=y$.
|
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|
$1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p$, and $2\cdot4\cdot6\cdots(p - 1) = ( -1)^{m +k} \pmod p$. Prove that if $p$ is a prime having the form $4k + 3$, and if $m$ is the number of quadratic residues less than $\frac p2$, then we have
$$1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p, \text{ and } 2\cdot4\cdot6\cdots(p - 1) = ( -1)^{m +k} \pmod p.$$
I am stuck with the problem....Help Needed.
|
Let $\phi: \mathbb{Z}_p \to \{-1,1\}$, where $\phi(n) = 1$ is a homomorphism given by $1$ if $n$ is a quadratic residue modulo $p$ and $-1$ if not.
Now consider $1 \cdot 3 \cdots (p-2)$. Now switch every number $x$ bigger than $\frac p2$ to $(p-x)$. Eventually as you will make $\frac{p-3}{4} = k$ changes we have that:
$$1 \cdot 3 \cdots (p-2) = \left(\frac{p-1}{2}\right)! \cdot (-1)^k$$
It's fairly easy to notice that $\left(\frac{p-1}{2}\right)!$ is equal to either $1$ or $-1$ modulo $p$, so the sum is equal to $-1$ or $1$. Also let's note that $\phi(1) = 1$ and $\phi(-1) = -1$, so therefore it's enough to see whether $ \left(\frac{p-1}{2}\right)!$ is sent by $\phi$ to $1$ or $-1$. But by the definition of $\phi$ we have that $\phi\left(\frac{p-1}{2}\right)! = (-1)^t$, where $t$ is the number of non-equadratic residues modulo $p$ and less than $\frac{p}{2}$. As there are odd amount of numbers less than $\frac{p}{2}$ we have that $t$ and $m$ have a different parity, hence $(-1)^t = (-1)^{m+1}$ and:
$$1 \cdot 3 \cdots (p-2) = \left(\frac{p-1}{2}\right)! \cdot (-1)^k \equiv (-1)^{k+t} \equiv (-1)^{k+m+1} \pmod p$$
Similar reasoning yields the wanted proof for the second part.
|
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|
Evaluate : $\int _0^{\infty } \sin (ax^2)\cos \left(2bx\right)dx$ How can I solve
$$\int_{0}^{\infty} \sin \left (ax^2 \right) \cos \left (2bx \right)dx$$
Thanks!
|
Concerning the antiderivative, start writing$$\sin \left (ax^2 \right) \cos \left (2bx \right)=\frac12\left(\sin(ax^2+2bx)+\sin(ax^2-2bx)\right)$$ Now $$ax^2+2bx=\left(\sqrt a x +\frac{b}{\sqrt a}\right)^2-\frac{b^2}a$$ $$ax^2-2bx=\left(\sqrt a x -\frac{b}{\sqrt a}\right)^2-\frac{b^2}a$$ Now, consider $$I=\int \sin(ax^2+2bx)\,dx$$ and change variable accordingly for each sine. Expand the sine and you will arrive to a linear comination of sine and cosine Fresnel integrals. Back to $x$, you should get $$I=\frac{\sqrt{\frac{\pi }{2}} \left(\cos \left(\frac{b^2}{4 a}\right) S\left(\frac{b+2
a x}{\sqrt{a} \sqrt{2 \pi }}\right)-\sin \left(\frac{b^2}{4 a}\right)
C\left(\frac{b+2 a x}{\sqrt{a} \sqrt{2 \pi }}\right)\right)}{\sqrt{a}}$$ Similarly $$J=\int \sin(ax^2-2bx)\,dx$$ $$J=-\frac{\sqrt{\frac{\pi }{2}} \left(\sin \left(\frac{b^2}{4 a}\right) C\left(\frac{2 a
x-b}{\sqrt{a} \sqrt{2 \pi }}\right)-\cos \left(\frac{b^2}{4 a}\right)
S\left(\frac{2 a x-b}{\sqrt{a} \sqrt{2 \pi }}\right)\right)}{\sqrt{a}}$$ I suppose that you have all pieces to finish your problem.
|
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|
$\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$ by Fourier series of $x^2$
Prove that
$$\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$$ by the Fourier series of $x^2$.
By Parseval's identity, I can only show $\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}$. Could you please give me some hints?
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Considering $f(x)=x^3$ By Parseval identity we can prove that $\sum_{n=1}^{\infty}\frac{1}{n^6}=\frac{\pi^6}{945}$.
Then $$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=0,$$$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx=0$$ and \begin{align}b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\\&=(-1)^{n+1}\frac{2\pi^2}{n}+(-1)^{n}\frac{12\pi}{n^3}\end{align}
From the relation $$\frac{1}{\pi}\int_{-\pi}^{\pi}|f|^2dx=\frac{a_0}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2)$$we get \begin{align}\sum_{n=1}^\infty(\frac{144}{n^6}+\frac{4\pi^4}{n^2}-\frac{48\pi^2}{n^4})=\frac{2\pi^6}{7}\end{align}$$\sum_{n=1}^\infty\frac{144}{n^6}=\frac{16\pi^6}{105}$$ $$\sum_{n=1}^\infty\frac{1}{n^6}=\frac{\pi^6}{945}$$
|
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|
Proving $\frac{\sin x}{x} + \frac{x^2}{4} >1$, for $x \in [0,\frac{\pi}{2}]$
Prove that, for every $x \in \left(0,\frac{\pi}{2}\right)$,
$$\frac{\sin x}{x} + \frac{x^2}{4} >1$$
I have tried using differentiation to prove that the left-hand side is strictly increasing on the interval, but no success. Please, I need a hint.
|
Fix $x \in (0, \pi/2)$. Using Taylor's theorem
\begin{align}
\sin x = x-\frac{1}{3!}x^3+\frac{\cos(\xi)}{5!}x^5
\end{align}
where $\xi \in (0, x)$.
Hence it follows
\begin{align}
\frac{\sin x}{x}+\frac{1}{4}x^2-1 = \left(\frac{1}{4}-\frac{1}{6}\right)x^2+\frac{\cos(\xi)}{5!}x^4>0
\end{align}
since $\cos x$ is non-negative on $(0, \pi/2)$.
|
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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.