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If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2}{x+y}=4\Rightarrow x+y-\frac{2xy}{x+y}=4\Rightarrow x+y=\frac{2xy}{x+y}+4$
$x-y=\sqrt{(\frac{2xy}{x+y}+4)^2-4xy}$, but I am not able to proceed. I am stuck here. Is my method wrong?
|
Given $$\displaystyle \frac{x^2+y^2}{x+y} = 4\Rightarrow x^2+y^2 = 4x+4y$$
So we get $$x^2-4x+4+y^2-4y+4 = 8\Rightarrow (x-2)^2+(y-2)^2 = 8$$
Now we can write $$x-y = (x-2)-(y-2) = \left[(x-2)+(2-y)\right]$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
$$\displaystyle \left[(x-2)^2+(2-y)^2\right]\cdot [1^2+1^2]\geq \left[x-2+2-y\right]^2$$
So $$8\times 2 \geq (x-y)^2$$
So $$(x-y)^2\leq 4^2$$
So $$-4 \leq (x-y)\leq 4\Rightarrow x-y\in \left[-4,4\right]$$
|
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|
How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
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Let $x = n + \epsilon$ where $n\in \mathbb Z$ and $0 \le \epsilon \lt 1$.
Then
$x^2 = (n + \epsilon)^2 = 19n - 88$
Since $0 \le \epsilon \lt 1$, then $n^2 \le 19n - 88 \lt (n+1)^2$
The line $y = 19x - 88$ intersects the parabola $y = x^2$ at $x = 8$ and at $x = 11$. Since the parabola $y = x^2$ is concave up, this implies that
$n \in \{8, 9, 10, 11 \}$ and $x = \sqrt{19n - 88}$
$n = 8 \implies x = 8$
$n = 9 \implies x = \sqrt{83}$
$n = 10 \implies x = \sqrt{102}$
$n = 11 \implies x = 11$
|
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|
$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
I was given this question by my senior.I tried to solve it but could not reach the answer.
Let $I= \int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}} $
$I=\int\frac{dx}{\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}$
Then after repeated attempts, i could not solve further.
I think this function is not integrable.Am i correct?If not,how should i move ahead.Please help.
|
Edit: with some reasoning, and without controversial part
I see no way of calculating this primitive using human tools like integration by parts, substitution and so on. However, inspired by the example here, we could try a function in the form
$$
a\log\bigl[p(x)\sqrt{x^4+4x^3-6x^2+4x+1}+q(x)\bigr],
$$
where $a$ is a constant and $p$ and $q$ are polynomials. And indeed, after a painful differentiation and comparison, it turns out that the function
$$
\begin{aligned}
-\frac{1}{6}\log\Bigl[&\bigl(x^4+10x^3+30x^2+22x-11\bigr)\sqrt{x^4+4x^3-6x^2+4x+1}\\
&\qquad-x^6-12x^5-45x^4-44x^3+33x^2-43\Bigr]+C
\end{aligned}
$$
does the job.
|
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Show that the sequence $(a_n)$ where $a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$ is of Cauchy. Show that the sequence $(a_n)$ where
$$a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$$
is of Cauchy.
Comments:
I will use the definition
$$\forall \epsilon > 0, \exists n_0 \in \mathbb{N}; n,m > n_0 \Rightarrow |a_n - a_m|< \epsilon.$$
If $n>m$
$$|a_n - a_m| = \left|\frac{1}{(n+1)!} + \ldots + \frac{1}{n!}\right| \leq \left|\frac{1}{(n+1)!}\right| + \ldots + \left|\frac{1}{n!}\right|$$
I do not know how to limit the $\epsilon$ and define $n_0$ without it depends of $n$ and $m$.
|
If $m>n>1$, then
\begin{align}
|a_m-a_n|&=\left|\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots+\frac{1}{m!}\right|\\&<\left|\frac{1}{(n+1)!}+\frac{1}{2}\frac{1}{(n+1)!}+\ldots+\frac{1}{2^{m-n-1}}\frac{1}{(n+1)!}\right|\\
&=\frac{1}{(n+1)!}\sum_{k=0}^{m-n-1}\frac{1}{2^k}\\
&<\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{2^k}\\
&=\frac{2}{(n+1)!}\\
&<\frac{1}{n!}
\end{align}
Then, given $\varepsilon>0$, by taking $n_0$ such that $n_0!>\max\left(\frac{1}{\varepsilon},1\right)$, we have
$$m>n>n_0\quad\implies\quad|a_m-a_n|<\varepsilon$$
|
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|
a conjectured continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$ Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)$ is true
$$\begin{split}\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)&=\frac{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}\\&=\cfrac{2z}{2z+n+\cfrac{(n)(4z+n)} {3(2z+n)+\cfrac{(2z+2n)(6z+2n)}{5(2z+n)+\cfrac{(4z+3n)(8z+3n)}{7(2z+n)+\ddots}}}}\end{split}$$
Corollaries:
By taking the limit(which follows after abel's theorem)
$$
\begin{aligned}\lim_{z\to0}\frac{\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)}{2z}=\frac{\pi}{4}\end{aligned},
$$
we recover the well known continued fraction for $\pi$
$$\begin{aligned}\cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}}=\pi\end{aligned}$$
If we let $z=1$ and $n=2$,then we have the square root of $2$ $$\begin{aligned}{1+\cfrac{1}{2+\cfrac{1} {2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}}}=\sqrt{2}\end{aligned}$$
Q: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $x\gt0$?
Update:I initially defined the continued fraction $\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.
|
The proposed continued fraction
\begin{equation}
\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\cfrac{2z}{2z+n+\cfrac{(n)(4z+n)} {3(2z+n)+\cfrac{(2z+2n)(6z+2n)}{5(2z+n)+\cfrac{(4z+3n)(8z+3n)}{7(2z+n)+\ddots}}}}
\end{equation}
can be written as
\begin{equation}
\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\cfrac{2z/\left( 2z+n \right)}{1+\cfrac{(n)/\left( 2z+n \right)\cdot(4z+n)/\left( 2z+n \right)} {3+\cfrac{(2z+2n)/\left( 2z+n \right)\cdot(6z+2n)/\left( 2z+n \right)}{5+\cfrac{(4z+3n)/\left( 2z+n \right)\cdot(8z+3n)/\left( 2z+n \right)}{7+\ddots}}}}
\end{equation}
Denoting $u=\cfrac{z}{4z+2n}$, the factors of the numerators are
\begin{equation}
\frac{n}{2z+n}=1-4u\,;\quad\frac{4z+n}{2z+n}=1+4u\,;\quad\frac{2z+2n}{2z+n}=2-4u\,;\quad\frac{6z+2n}{2z+n}=2+4u\,;\cdots
\end{equation}
Then, the fraction can be simplified as
\begin{equation}
\displaystyle\tan\left(\pi u\right)=\cfrac{4u}{1+\cfrac{\cfrac{1-16u^2}{1\cdot3}} {1+\cfrac{\cfrac{4-16u^2}{3\cdot5}}{1+\cfrac{\cfrac{9-16u^2}{5\cdot7}}{1+\ddots}}}}
\end{equation}
It is thus a special case of the continued fraction found in this answer:
\begin{equation}
\tan\left(\alpha\tan^{-1}z\right)=\cfrac{\alpha z}{1+\cfrac{\frac{(1^2-\alpha^2)z^2}{1\cdot 3}} {1+\cfrac{\frac{(2^2-\alpha^2)z^2}{3\cdot 5}}{1+\cfrac{\frac{(3^2-\alpha^2)z^2}{5\cdot 7}}{1+\ddots}}}}
\end{equation}
here $z=1$ and $\alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.
|
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Does $\lfloor x\lceil y\rceil\rfloor = \lceil x\lfloor y\rfloor\rceil$ have non-integer solutions? Does $\lfloor x\lceil y\rceil\rfloor = \lceil x\lfloor y\rfloor\rceil$ have any non-integer solutions? If so, how do you find them?
|
I'm assuming your $x$ and $y$ are real numbers. Now, we can write $x= \lfloor x \rfloor + \{x \} $ and $ y=\lfloor y \rfloor + \{y\} $ where the $\{ \}$ here denotes the fractional part. We have that $\{x\} \neq 0$ and $\{y\} \neq 0$ since $x$ and $y$ are not integers. Your equation then reads
$$ \lfloor x \lceil y \rceil \rfloor = \lceil x \lfloor y \rfloor \rceil $$
$$ \lfloor (\lfloor x \rfloor +\{x\})(\lfloor y \rfloor +1) \rfloor = \lceil (\lfloor x \rfloor +\{x\}) \lfloor y \rfloor \rceil $$ which gives
$$ \lfloor x \rfloor( \lfloor y \rfloor +1) + \lfloor \{x\}(\lfloor y \rfloor+1) \rfloor = \lfloor x \rfloor \lfloor y \rfloor + \lceil \{x\}\lfloor y \rfloor \rceil $$ which after simplification and rearranging becomes
$$ \lfloor x \rfloor = \lceil \{x\}\lfloor y \rfloor \rceil - \lfloor \{x\}(\lfloor y \rfloor+1) \rfloor $$
Now, $$ \{x\}\lfloor y \rfloor < \{x\}(\lfloor y \rfloor+1) < \{x\}\lfloor y \rfloor +1 $$ which means that after rounding the small one up and the big one down they either become equal or the one rounded up is one more than the other. Hence, we see that $ \lfloor x \rfloor = 0,1$. If $ \lfloor x \rfloor = 0$ we have that $x$ is between 0 and 1 and that $\{x\}=x$ and the rounded up and down are equal, so there is an integer between $x\lfloor y \rfloor$ and $x\lfloor y \rfloor + x$. Call it $n$. Then $$ x\lfloor y \rfloor \leq n \leq x\lfloor y \rfloor + x $$ giving $$ \frac{n}{x} -1 \leq \lfloor y \rfloor \leq \frac{n}{x} $$ What does this mean? It means, choose an $x$ between 0 and 1, and an integer $n$. Compute $ \frac{n}{x}$ and $\frac{n}{x} -1 $ and find an integer between them. In general, there will only be one integer between them, unless $\frac{n}{x}$ is an integer ( so $x$ is rational ), in that case both the upper and lower bounds are integers and there are two possibilities. This integer you find is $\lfloor y \rfloor$ - but what is $y$? It doesn't matter, as long as $ \lfloor y \rfloor$ is the right value - so add any fractional part to $y$ that you wish. As for the case $\lfloor x \rfloor = 1$ ... how about I leave that as an exercise for you.
|
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|
Show that $e^{x+y}=e^xe^y$ using $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$. I was looking for a proof of $e^{x+y}=e^xe^y$ using the fact that $$e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$
So I have that $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1}{n^k}\sum_{i=0}^k\binom{k}{i}x^iy^{k-i}=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\left(\frac{x}{n}\right)^i\left(\frac{y}{n}\right)^{k-i}$$
But I can't get $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\sum_{i=0}^n\binom{n}{i}\left(\frac{y}{n}\right)^i.$$
Any idea ?
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\begin{align*}
e^x e^y &= \lim_{n\rightarrow \infty}(1+\frac{x}{n})^n\lim_{n\rightarrow \infty}(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x}{n})^n(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n}+\frac{xy}{n^2})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \frac{(1+\frac{x+y}{n}+\frac{xy}{n^2})^n}{(1+\frac{x+y}{n})^n}\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^{\frac{n^2(1+\frac{x+y}{n})}{xy}\frac{xy}{n^2(1+\frac{x+y}{n})}n}\\
&= \lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\\
&=e^{x+y}.
\end{align*}
|
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higher college level combinatorics Show the following equality:
$$\sum_{n_1,\ldots,n_k\geq 0} \min(n_1,\ldots,n_k)x_1^{n_1} \cdots x_k^{n_k}=\frac{x_1\cdots x_k}{(1-x_1)\dots(1-x_k)(1-x_1x_2\cdots x_k)}.$$
This is a problem from Stanley's Enumerative Combinatorics, and I am trying to prove the equality. I tried to plug in some some numbers and this equality stays true, so I guess induction would work~ but I would also like to some other ways of showing this equality.
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Looking at the RHS, you should know that $\frac{1}{(1-x_i)} = (1+x_i+x_i^2+x_i^3+\dots)$
Let us simplify notation a little by letting $x_1x_2x_3\dots x_k=X$
and since the RHS is a product of such terms you have:
$=X(1+x_1+x_1^2+x_1^3+\dots)(1+x_2+x_2^2+x_2^3+\dots)\dots(1+x_k+x_k^2+x_k^3+\dots)(1+X+X^2+X^3+\dots)$
Distributing the $x_1x_2\dots x_k=X$ into its respective parenthesis, this is then
$=(x_1+x_1^2+x_1^3+\dots)(x_2+x_2^2+x_2^3+\dots)\dots(x_k+x_k^2+x_k^3+\dots)(1+X+X^2+X^3+\dots)$
Can you figure out how many ways there are by looking at the RHS to get a specific outcome such as $x_1^3x_2^2x_3^4\dots x_k^4$?
Looking at a smaller example, let $k=3$ and we ask what the coefficient is for $x_1^4x_2^3x_3^5$
$(x_1+x_1^2+x_1^3+x_1^4+\dots)(x_2+x_2^2+x_2^3+x_2^4+\dots)(x_3+x_3^2+x_3^3+x_3^4+\dots)(1+X+X^2+X^3+\dots)$
We try to pick a term from each parenthesis and count how many ways we could have made such a selection. We could have for example taken:
$(x_1+x_1^2+x_1^3+\color{red}{x_1^4}+\dots)(x_2+x_2^2+\color{red}{x_2^3}+x_2^4+\dots)(x_3+x_3^2+x_3^3+x_3^4+\color{red}{x_3^5}+\dots)(\color{red}{1}+X+X^2+X^3+\dots)$
Or we could have taken $(x_1+x_1^2+\color{red}{x_1^3}+x_1^4+\dots)(x_2+\color{red}{x_2^2}+x_2^3+x_2^4+\dots)(x_3+x_3^2+x_3^3+\color{red}{x_3^4}+\dots)(1+\color{red}{X}+X^2+X^3+\dots)$
In general, if we first pick from $(1+X+X^2+X^3+\dots)$, given our specific choice there will be exactly one choice (or zero choices) from each of the other parenthesis to reach the target product. The question now is how many different choices from $(1+X+X^2+\dots)$ are legal and would allow us to reach our desired product. There are in fact $\min(n_1,n_2,\dots,n_k)$ many such choices, which yields the identity asked.
Explicitly, the choices being $1,X,\dots,X^{m-1}$ where $m=\min(n_1,\dots,n_k)$. If you had chosen a higher power of $X$ than that, then the resulting product would have too large of a power on one of the terms. Having picked your power of $X$, then you are forced to pick $x_i^{n_i-m+1}$ as the choice from the parenthesis $(x_i+x_i^2+x_i^3+\dots)$
|
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Infinitely rational points in $y^2 = x^3 - 4$? If the $x$-coordinate of a rational point $P$ of $y^2 = x^3 - 4$ is given by $m/n$, the $x$-coordinate of $2P$ is given by$${{(m^3 + 32n^3)m}\over{4(m^3 - 4n^3)n}}.$$Using this fact, how do I show that$$144 \cdot H(x\text{-coordinate of }2P) \ge H(x\text{-coordinate of }P)^4.$$Using this fact, can we conclude that there exist infinitely many rational points in $y^2 = x^3 - 4$?
EDIT: Here, let $a$ be a rational number and write $a = m/n$ as a fraction in lowest terms. Define the height $H(a)$ to be $\max(|m|, |n|)$.
|
Let $m$ and $n$ be relatively prime integers and let$$A = |(m^3 + 32n^3)m|,\text{ }B = |4(m^3 - 4n^3)n|.$$Denote by $D$ the greatest common divisor of $A$ and $B$. In order to show the inequality in question, it suffices to show that $D$ is a divisor of $144$. For, if that is the case and if the $x$-coordinate of $P$ is given by $m/n$ ($n \neq 0$) in lowest terms, then we have$$H(x\text{-coordinate of }2P) = H\left({A\over{B}}\right) = {1\over{D}}\max(A, B)$$$$\ge {1\over{D}}\max(m, n)^4 = {1\over{D}}H(x\text{-coordinate of }P)^4.$$Let $p$ be a prime number. We have$$\text{ord}_p(D) = \min(\text{ord}_p(A), \text{ord}_p(B)),$$since $\text{ord}_p$ indicates how many times $p$ divides the number. If $p$ is a prime factor of $D$, then $p$ does not divide $n$, since if it does, $p$ does not divide $m$, and thus $p$ does not divide $m^3 + 32n^3$ and $A$. If $p$ is a prime factor of $D$ and $p \neq 2$, then $p$ does not divide $m$, since if $p \neq 2$ and $p$ divides $m$, then $p$ does not divide $B$. Thus, if $p$ is a prime factor of $D$ and $p \neq 2$, then we have$$\text{ord}_p(D) = \min(\text{ord}_p(m^3 32n^3),\text{ord}_p(m^3 - 4n^3))$$$$\le \text{ord}_p((m^3 + 32n^3) - (m^3 - 4n^3))$$$$=\text{ord}_p(36n^3) = \text{ord}_p(36).$$Hence, we have $p = 3$ and $\text{ord}_3(D) \le 2$.
Next, we consider $\text{ord}_2(D)$. If $m$ is odd, then $\text{ord}_2(A) = 0$. If $m$ is even, then $\text{ord}_2(m^3 - 4n^3) = 2$ since $n$ is even. Hence, we have $\text{ord}_2(B) = 4$. Therefore, $D$ is a divisor of $2^4 \times 3^2 = 144$, and thus the inequality in question is proved.
If $r \ge 6$, then we have the inequality $r^4/144 > r$. If a rational point $P$ on the elliptic curve in equation satisfies$$H(x\text{-coordinate of }P) \ge 6,$$then we have$$H(x\text{-coordinate of }P) > H(x\text{-coordinate of }2P).$$Then the height of the $x$-coordinate of$$P,\text{ }2P,\text{ }4P,\text{ }8P,\text{ }16P, \dots$$are all different. Thus, these points are all distinct. This means there are infinitely many rational points on this elliptic curve.
To be more precise, we can show the following. If $m$ and $n$ satisfy $m \not\equiv 0 \text{ (mod }3\text{)}$ or $n \not\equiv 0 \text{ (mod }3\text{)}$, then$$m^3 - 4n^3 \not\equiv 0 \text{ (mod }9\text{)}.$$This can be done by checking all the possibilities of $0 \le m \le 8$, $0 \le n \le 8$. Thus we can see that $D$ is a divisor of $2^4 \times 3 = 48$ and that$$48 \cdot H(x\text{-coordinate of }2P) \ge H(x\text{-coordinate of }P)^4.$$If $r \ge 4$, then $r^4/48 > r$. Thus, if $P = (5, 11)$, then the $x$-coordinates of$$P,\text{ }2P,\text{ }4P,\text{ }8P, \dots$$all have different heights. This implies that we see the existence of infinitely many rational points as soon as we find one rational point $(5, 11)$.
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What is the coefficient of $x^3 y^4$ in the expansion of $ (2x-y+5)^8$ I was thinking of doing $\binom{8}{4}$ but not sure if right.
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Pre-expansion, there are $8$ factors of $2x - y + 5$.
From those $8$ factors, choose the $3$ that contribute to the $x^3$, from the remaining $5$ factors, choose the $4$ that will contribute to $y^4$. There is only 1 factor left so it chooses itself.
Post-expansion, that's ${8 \choose 3}{5 \choose 4}{1 \choose 1}$ terms with $x^3y^4$, and taking into account the initial coefficients of $\{2, -1, 5\}$, the final coefficient is:
$${8 \choose 3}{5 \choose 4}2^3(-1)^45^1 = 11200$$
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Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$ Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$
Attempt.
I'm preparing for a trigonometry examination tomorrow. So I'm solving as many questions as possible. This is confusing. I've tried solving it but its not proofing. I've tried replacing with trigonometric identities and simplifying. Please someone should help even if it is how to start?
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The first thing to remember is
$$
\sin2x=2\sin x\cos x
$$
that settles the $-\sqrt{3}\sin2x$ part. Thus you just need to massage
$$
5\cos^2x+3\sin^2x
$$
recalling $\cos2x=\cos^2x-\sin^2x$; so rewrite it as
$$
\cos^2x+4\cos^2x-\sin^2x+4\sin^2x=\cos2x+4(\cos^2x+\sin^2x)
$$
You can also equivalently prove that
$$
5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x - \cos 2x + \sqrt{3}\sin2x - 4=0
$$
that goes like
\begin{multline}
5\cos^2x+3\sin^2x-\cos^2x+\sin^2x-4-\sqrt{3}(2\sin x\cos x-\sin2x)\\
=4\cos^2x+4\sin^2x-4-\sqrt{3}(\sin2x-\sin2x)=0
\end{multline}
|
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|
Finishing my $\epsilon - \delta$ proof Hi fellas I have this limit: $$lim_{x\to-1}\dfrac{x^2-1}{x^2+x}=2$$
$Dem:$
Let $\epsilon >0$ random but fixed then:
$$|\dfrac{x^2-1}{x^2+x}-2|=|\dfrac{x^2-1-2x^2-x}{x^2+x}|=|\dfrac{-x^2-2x-1}{x^2+x}|=|\dfrac{(x+1)^2}{x(x+1)}|=|\dfrac{x+1}{x}|<M|x-1|<M(\dfrac{\epsilon}{M})=\epsilon$$
So I have to choose M such that $|\dfrac{1}{x}|<M$
Let's suppose that $|x+1|<c$ for some $c$. (My problem is that I don't know how to choose this $c$)
So $|x-1|<c \Rightarrow -c<x-1<c$ and so $-c-1<x<c-1$.
The problem is that for a little $c<1$, $-c-1$ and $c-1 $ are both negative and for a bigger c, $-c-1$ is always negative and so I can't find my beloved $M$.
|
What you want to show is that,
for any $\delta > 0$
there is an $\epsilon > 0$
such that
if
$|x+1| < \epsilon$
then
$|\dfrac{x^2-1}{x^2+x}
-2| < \delta
$.
You have worked out
very nicely that
$|\dfrac{x^2-1}{x^2+x}
-2|
=|\frac{x+1}{x}|
$.
Suppose
$|x+1| < \epsilon$.
To make sure that
the $x$ in the denominator
does not cause problems,
we want to choose $x$ so it is
not close to zero.
If we choose
$\epsilon < \frac12$,
then,
since
$-1-\epsilon < x
< -1+\epsilon
$,
then
$x < -\frac12$,
so
$|x| > \frac12$.
Therefore,
$|\frac{x+1}{x}|
<\frac{\epsilon}{\frac12}
=2\epsilon
$.
This shows that
if
$|x+1| < \epsilon$
and
$\epsilon < \frac12$,
then
$|\dfrac{x^2-1}{x^2+x}
-2| < 2\epsilon
$.
Therefore,
to make
$|\dfrac{x^2-1}{x^2+x}
-2| < \delta
$,
choose
$2\epsilon
< \delta$,
or
$\epsilon
< \min(\frac12,\frac{\delta}{2})
$.
|
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|
Proof that $3^c + 7^c - 2$ by induction I'm trying to prove the for every $c \in \mathbb{N}$, $3^c + 7^c - 2$ is a multiple of $8$. $\mathbb{N} = \{1,2,3,\ldots\}$
Base case: $c = 1$
$(3^1 + 7^1 - 2) = 8$ Base case is true.
Now assume this is true for $c=k$.
Now I prove this holds for $c=k+1$ $(3^{k+1}+7^{k+1}-2)$.
$(3^{k+1}+7^{k+1}-2)$
$(3^k*3+7^k*7-2)$
But now I'm stuck...
|
Notice
\begin{align}
3^k\cdot 3+7^k\cdot 7-2&=3^k\cdot(3+4)-3^k\cdot 4+7^k\cdot 7-2\cdot 7 +2\cdot 7-2\\
&=7(3^k+7^k-2)-3^k\cdot 4+2\cdot 7-2\\
&=7(3^k+7^k-2)-3^k\cdot 4+12\\
&=7(3^k+7^k-2)-12\cdot(3^{k-1}-1)\\
\end{align}
by hypothesis $8$ divides $3^k+7^k-2$, and for $k\ge 1$ the number $3^{k-1}-1$ is even, then $8$ divides $12\cdot(3^{k-1}-1)$. Therefore $8$ divides $3^k\cdot 3+7^k\cdot 7-2.$
|
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|
Proving that $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} =\tan \left ( \frac{\alpha+\beta}{2} \right )$ Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.
$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \frac{\alpha}{2}\right ) \cos \left ( \frac{\alpha}{2}\right ) + \sin \left ( \frac{\beta}{2}\right ) \cos \left ( \frac{\beta}{2}\right )}{\cos^2 \left ( \frac{\alpha}{2} \right) - \sin ^2 \left ( \frac{\beta}{2} \right )}$$
But I know that if $\alpha$ and $\beta$ are angles in a triangle, then this expression should simplify to
$$\tan \left ( \frac{\alpha + \beta}{2} \right )$$
I can see that the denominator becomes $$\cos \left ( \frac{\alpha + \beta}{2} \right ) $$
But I cannot see how the numerator becomes
$$\sin \left ( \frac{\alpha + \beta}{2} \right )$$
What have I done wrong here?
|
Assume that $$P=(\cos\ a,\sin\ b),\ Q=(\cos\ b,\sin\ b)$$ Then what is
slope of line passing through $-Q,\ P$ :
For convenience $0<a<b<\pi/2$, let $R=(0,1)$. Then $$\angle\
ROQ=\frac{\pi}{2}-b,\ O=(0,0)$$
and $$\angle\ Q(-Q)P = \frac{b-a}{2} $$
When $\angle\ PP'(-Q) = \pi/2$, then $\angle\ P(-Q)P'=
\frac{a+b}{2}$
Hence $$ \tan\ \frac{a+b}{2} = \frac{{\rm difference \ of}\ y-{\rm
coordinates\ of}\ P,\ -Q }{{\rm difference \ of}\ x-{\rm coordinates
\ of}\ P,\ -Q } $$
|
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|
Min of $\frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $ I got this problem I tried several time to solve it by many inequalities but I got stuk. My question is how I get the minimum value of $$ \frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $$ if you know that $a, b, c \in (0, \infty) $ and $ a+b+c=1$? Any hint?
|
First note that
$$(a^3-b^3)(a^7-b^7)\ge 0\tag{1}$$
so
$$2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3).\tag{2}$$
It follows that
$$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{c^{10}+a^{10}}{c^7+a^7}\ge a^3+b^3+c^3.$$
Note that the equality happens when $a=b=c$. Next, we show that $a^3+b^3+c^3$ attains its minimum also when $a=b=c$. This can be seen by showing
$$3(a^3+b^3+c^3)\ge (a+b+c)(a^2+b^2+c^2)\ge (a+b+c)\frac{(a+b+c)^2}{3}\tag{3},$$
using Rearrangement Inequality, which can be proved as in (1) $\implies$ (2).
Or we can use Cauchy-Schwarz inequality
$$(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2\ge \frac{(a+b+c)^4}{3^2}.$$
|
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|
Finding the Roots of Unity I have the following equation, $$z^5 = -16 + (16\sqrt 3)i$$I am asked to write down the 5th roots of unity and find all the roots for the above equation expressing each root in the form $re^{i\theta}$. I am just wondering if my solutions are correct. Here are my solutions,
5th roots of unity, $$z = 1, e^{\frac{2\pi}{5}i}, e^{\frac{4\pi}{5}i}, e^{-\frac{2\pi}{5}i}, e^{-\frac{4\pi}{5}i}$$
and for all the roots for the above equation, $$z = e^{\frac{2\pi k} {5}i}, 2e^{\frac{(6k+2)\pi}{15}i}$$
where k = 0, 1, 2, 3, 4.
Please correct me if there are any mistakes. I will leave my working at the answer section for future reference.
|
For the fifth roots of unity, the roots are $\frac{2\pi}{5}$ radian apart from one another. Since the principal value for arg(z) is (-$\pi$,$\pi$], hence, the fifth roots of unity are $$z = 1, e^{\frac{2\pi}{5}i}, e^{\frac{4\pi}{5}i}, e^{-\frac{2\pi}{5}i}, e^{-\frac{4\pi}{5}i}$$
For the all the roots of the above equation, one of the roots will always be $$z = e^{\frac{2k\pi}{n}i}$$ where n = the power of z and k = 0, 1, 2, ..., n-1
Hence, $$z = e^{\frac{2k\pi}{5}i}$$
where k = 0, 1, 2, 3, 4
For the other root, $$|z^5| = \sqrt {(-16)^2 + (16\sqrt3)^2}$$
$$|z| = \sqrt[5] {32} = 2$$ and since the coordinate lies in the 2nd quadrant, $$arg(z) = \pi - tan^{-1} (\frac{16\sqrt3}{16}) = \frac{2\pi}{3}$$
Therefore, $$\begin{align} z & = re^{i\theta}\\
& = 2e^{(\frac{2k\pi}{5}+\frac{1}{5}\cdot\frac{2\pi}{3})i}\\
& = 2e^{\frac{(6k+2)\pi}{15}i}\\
\end{align}$$
|
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|
A trigonometry question involving medians and a sum of cotangents The medians of a triangle $ABC$ make angles $\alpha$, $\beta$ and $\gamma$ with each other. I have to prove that:
$$\cot\alpha+ \cot\beta+ \cot\gamma +\cot(A)+\cot(B)+\cot(C)= 0.$$
Any ideas?
|
By the sine and cosine theorems,
$$\cot(\widehat{BGC}) = \frac{\cos\widehat{BGC}}{\sin\widehat{BGC}} = \frac{\frac{BG^2+CG^2-BC^2}{2\, BG\cdot CG}}{\frac{2[BGC]}{BG\cdot CG}}=\frac{1}{4}\cdot\frac{BG^2+CG^2-BC^2}{[BGC]}\tag{1}$$
but $BG=\frac{2}{3}m_a$ and Stewart's theorem gives $m_a^2 = \frac{2b^2+2c^2-a^2}{4}$, hence:
$$\cot(\widehat{BGC}) = \frac{3}{4}\cdot\frac{BG^2+CG^2-BC^2}{[ABC]} = \frac{b^2+c^2-5a^2}{12[ABC]}\tag{2}.$$
On the other hand,
$$ \cot(\widehat{BAC}) = \frac{\cos\widehat{BAC}}{\sin\widehat{BAC}}=\frac{\frac{AB^2+AC^2-BC^2}{2\,AB\cdot AC}}{\frac{2[ABC]}{AB\cdot AC}}=\frac{b^2+c^2-a^2}{4[ABC]}\tag{3}$$
hence:
$$ \cot A+\cot B+\cot C = \frac{a^2+b^2+c^2}{4[ABC]} = -\left(\cot\alpha+\cot\beta+\cot\gamma\right)\tag{4} $$
as wanted.
|
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|
Rationalise the denominator and simplify $\frac {3\sqrt 2-4}{3\sqrt2+4}$ Does someone have an idea how to work $\dfrac {3 \sqrt 2 - 4} {3 \sqrt 2 + 4}$ by rationalising the denominator method and simplifying?
|
For problems like this, you need to multiply numerator and denominator by the denominator's conjugate. A conjugate is found by flipping the sign from positive to negative or negative to positive in a binomial. So, the denominator's conjugate in the above's equation is $ 3\sqrt{2} -4 $. Multiplying an irrational binomial by it's conjugate will rationalize it, as seen below:
$$ \frac{3\sqrt{2}-4}{3\sqrt{2}+4}\times \frac{3\sqrt{2}-4}{3\sqrt{2}-4} $$
Multiply out the bottom and top
$$ \frac{9\times2-12\sqrt{2}-12\sqrt{2}+16}{9\times2 -16} $$
$$ = \frac{18-24\sqrt{2}+16}{18 -16} $$
$$ = \frac{34-24\sqrt{2}}{2} $$
$$ = 17-12\sqrt{2}$$
|
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|
If $1+x^2=\sqrt{3}x$ then $\sum _{n=1}^{24}\left(x^n+\frac{1}{x^n}\right)^2$ is equal to I tried this problem by different methods but i am not able to get the answer in easy way . First i found the roots of equation and then represented it in polar form of complex number . I got $\cos \left(30^{\circ}\right)\pm i\sin \left(30^{\circ}\right)$ .
Then i tried to open whole square of question and the only thing i could sum up was that the terms of $x^2$ and $1/x^2$ must have cancelled and got the answer as $-48$ . I have seen one question like this in a book where equation was $x^2 + x + 1 =0$ and the roots of this equation are $\omega$ and $\omega ^2$ . Then they solved by putting values of $\omega$ in different sums but what if sum is large like upto $27$ terms ?
Is there a specific way to approach these kind of problems ?
Thanks for the help
|
For $\quad x=\cos 30^{\circ}+i\sin 30^{\circ}\quad$ from De Moivrè's formula follows
\begin{align}
x^n&=\cos (30^{\circ}n)+i\sin (30^{\circ}n)\\
\frac{1}{x^n}=x^{-n}&=\cos (30^{\circ}n)-i\sin (30^{\circ}n)\\
\end{align}
Hence,
$$x^n+\frac{1}{x^n}=2\cos(30^{\circ}n)\tag{1}$$
On the other hand, $\cos (x+360^{\circ})=\cos x$, so
$$\sum_{n=1}^{24}\left(x^n+\frac{1}{x^n}\right)^2=\sum_{n=1}^{24}\left[2\cos(30^{\circ}n)\right]^2=4\sum_{n=1}^{24}\left[\cos(30^{\circ}n)\right]^2=2\cdot4 \sum_{n=1}^{12}\cos^2(30^{\circ}n)\tag{2}$$
As $\cos (x+180^{\circ})=-\cos x$ we get $\quad\cos^2 [30^{\circ}(k+6)]=\cos^2 (30^{\circ}k),\quad$ then ($2$) becomes
\begin{align}
\sum_{n=1}^{24}\left(x^n+\frac{1}{x^n}\right)^2&=2\cdot4\cdot 2 \sum_{n=1}^{6}\cos^2(30^{\circ}n)\\
&=16\left[\cos^2 30^{\circ}+\cos^2 60^{\circ}+\cos^2 90^{\circ}+\cos^2(60^{\circ})+\cos^2(30^{\circ})+\cos^2(180^{\circ})\right]\\
&=16\left[2\cdot\left(\frac{\sqrt{3}}{2}\right)^2+2\cdot\left(\frac{1}{2}\right)^2+0+1\right]\\
&=16(3)\\
&=48
\end{align}
Here we have used the fact that $\;\;\;\cos (180^{\circ}-x)=-\cos x\;\;\;$ to get $\;\;\;\cos^2 (180^{\circ}-x)=\cos^2 x$. The same is true for $\;\;\; x=\cos 30^{\circ}-i\sin 30^{\circ}\;\;$ cause ($1$) is still true in such case.
|
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|
Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first inequality
$$\frac{a^4+b^4+c^4}{3abc} > \sqrt[3]{abc}$$
second inequality:
$$\frac{a+b+c}{3} > \sqrt[3]{abc}$$
I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??
|
You have to apply it repeatedly on different levels.
First, $(a^4+b^4)/2\ge a^2b^2$. Also, $(a^4+c^4)/2\ge a^2c^2$. Now apply the AM-GM to these two.
$${(a^4+b^4)/2+(a^4+c^4)/2\over2}\ge\sqrt{a^2b^2\cdot a^2c^2}=a^2bc=abc\cdot a$$
Now write two other similar inequalities with $a,b,c$ in different order, and add them all together.
|
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|
Positive integer solutions to $5a^2 + 8a + 4 = 4b^2$ I'm convinced $5a^2 + 8a + 4 = 4b^2$ can be somehow turned into Pell's equation.
My first steps: Rewrite as $5(a + 4/5)^2 + 4/5 = 4b^2$. Rewrite $B = 2b, A = 5a + 4$ to get $A^2 + 4 = 5B^2$. I'm almost there, I just need to know how to solve this Pell-like equation.
|
user236182's answer: $a$ must be even, $a = 2k$. Rewrite as $5(4k^2) + 16k +4 = 4b^2$, then multiply by $5/4$ to get $25k^2 + 20k + 5 = 5 b^2 \iff (5k + 2)^2 - 5b^2 = -1$. This is Pell's equation in the form $x^2 - 5y^2 = -1$. This is sufficient for all positive integer solutions.
|
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|
Nonlinear first order ODE In trying to compute an integral curve for a certain fiber of the Hopf fibration in $\mathbb R^3$ under stereographic projection, I've come across the following IVP:
$$
2y' - y^2 - 1=0, \quad y(0) = 0.
$$
I'm at a loss on how to solve this simple looking thing. Any help would be appreciated!
|
This is Riccati equation
Rewrite $$y'-\frac{1}{2}y^2=\frac{1}{2}$$
Let $y=-2\frac{w'}{w}$
then $$y'=-2\frac{w''(x)}{w(x)}+2\frac{w'(x)^2}{w(x)^2}=-2\frac{w''(x)}{w(x)}+\frac{1}{2}y^2$$
or
$$y'-\frac{1}{2}y^2=-2\frac{w''(x)}{w(x)}$$
thus we got
$$-2\frac{w''(x)}{w(x)}=\frac{1}{2}$$ or
$${w''(x)}+\frac{1}{4}w(x)=0$$
which has a general solution $$w=a\sin\frac{1}{2}x+b\cos\frac{1}{2}x$$
therefore $$y=-2\frac{w'}{w}=-2\frac{a\frac{1}{2}\cos\frac{1}{2}x-b\frac{1}{2}\sin\frac{1}{2}x}{a\sin\frac{1}{2}x+b\cos\frac{1}{2}x}$$
Since $y(0)=0$ we got that $$a\frac{1}{2}\cos\frac{1}{2}\cdot0-b\frac{1}{2}\sin\frac{1}{2}\cdot0=a\frac{1}{2} + 1=0$$
that is $a=0$, and we arrived at $$y=\tan\frac{1}{2}x$$
Another option
Thanks to @Evgeny
This is separable DE
Rewrite as $$2y'=1+y^2$$
then
$$\frac{2y'}{1+y^2}=1$$
so we get
$$2\,\mathrm{arctan}\,y=x+C$$
using initial condition one get $C=0$ and therefore
$$y=\tan\frac{1}{2}x$$
|
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|
$\lim _{x\to 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$ Find the limit of $f(x)=\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$ when x approaches 4.
I have no idea where to start with this problem.
|
Without using L'Hospital's rule
Notice,
Rationalizing the numerator & denominator as follows
$$\lim_{x\to 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$$
$$=\lim_{x\to 4}\frac{(\sqrt{5-x}-1)(\sqrt{5-x}+1)(2+\sqrt{x})}{(2-\sqrt{x})(2+\sqrt{x})(\sqrt{5-x}+1)}$$
$$=\lim_{x\to 4}\frac{(5-x-1)(2+\sqrt{x})}{(4-x)(\sqrt{5-x}+1)}$$
$$=\lim_{x\to 4}\frac{(4-x)(2+\sqrt{x})}{(4-x)(\sqrt{5-x}+1)}$$
$$=\lim_{x\to 4}\frac{2+\sqrt{x}}{\sqrt{5-x}+1}$$
$$=\frac{2+\sqrt{4}}{\sqrt{5-4}+1}=\frac{4}{2}=\color{blue}{2}$$
|
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|
rank of product of full rank matrices Please help me with the following question:
Let $A$ be $m \times n$, $B$ be $n \times p$ matrices with $\text{rank}(A)=m$, $\text{rank}(B)=p$, where $p < m < n$.
What are conditions such that $\text{rank}(AB)=\text{rank}(B)$???
Thanks!
|
No. In general that is wrong, consider the following example: $p= 1$, $m=2$, $n=3$, $\def\M#1#2{\operatorname{Mat}_{#1,#2}(\mathbf R)}$
$$ B = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \in \M 31,
D = \begin{pmatrix} 2 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1 \end{pmatrix} \in \M 33,
A =\begin{pmatrix} -1 & 2 & 0\\
0 & 0 & 1 \end{pmatrix} \in \M 23 $$
Then $A$, $B$ and $D$ have full rank, we have
$$ AB = \begin{pmatrix} 1\\ 0\end{pmatrix},
AD = \begin{pmatrix} -2 & 2 & 0\\ 0 & 0 & 1 \end{pmatrix}, ADB = \begin{pmatrix} 0 \\ 0\end{pmatrix}
$$
so $\def\r{\operatorname{rank}}\r AB = p = 1 > 0 = \r ADB$.
|
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|
Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.
|
Hint. By considering polar coordinates,
$$
x:=r\cos\theta,\quad y:=r\sin\theta,
$$ you get
$$
\begin{align}
f(r\cos\theta,r\sin\theta)&=\frac{r^5(\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta)}{r^4}\\\\
&=r(\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta)
\end{align}
$$ then observe that
$$
\left|\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta\right|\leq7
$$ gives
$$
\left|f(r\cos\theta,r\sin\theta)\right|\leq7r
$$
for all $r>0$ and $\theta$.
|
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|
how to find this limit : lim x to infinity how can I find this limit which is become infinite
$\lim _{x\to \infty }\left(x(\sqrt{x^2+1}-x)\right)$
can I use conjugate method ?
that what I'm doing until now
$= x\left(\frac{\sqrt{x^2+1}-x}{1}\cdot \:\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)\:$
$= x\left(\frac{1}{\sqrt{x^2+1}+x}\right)$
$= \left(\frac{x}{\left(\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}+\frac{x}{x}\right)}\right)\:$
$= \:\left(\frac{x}{\left(\sqrt{1+\frac{1}{x^2}}+1\right)}\right)\:$
|
Yes Using Conjugate.
$$\displaystyle \lim_{x\rightarrow \infty}x\left(\sqrt{x^2+1}-x\right) = \lim_{x\rightarrow \infty}x\left[\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x}\times \sqrt{x^2+1}+x\right]$$
So $$\displaystyle \lim_{x\rightarrow \infty}x\left[\frac{1}{\sqrt{x^2+1}+x}\right] = \lim_{x\rightarrow \infty}\frac{x}{x\left[\sqrt{1+\frac{1}{x^2}}+1\right]} = \frac{1}{2}$$
|
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|
What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
|
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
$\frac{(x+1)(x-3)}{(x-1)(x+2)(x-3)}-\frac{(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{(x+1)(x-3)-(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$
the inequations are:
*
*$x>-2$
*$4x+3>0$ or $x>-\frac{4}{3}$
*$x>1$
*$x>3$
We need to have one or three inequations true to have the big inequality hold.
You will find that the answer is $(-2,-\frac{3}{4})\cup(1,3)$
|
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|
Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods would be welcome. I was solving a couple of AM GM inequality questions and I'm assuming this should be solved in a similar way. Correct me.
|
let $a= \frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then some manipulations would show that
$(a+b+c)^3= 3(a+b+c)((a^2+b^2+c^2)) - 2(a^3+b^3+c^3) + 6abc$
we already have $a+b+c = 7, abc=1$, now its left to find what $a^2+b^2+c^2$ is.
$(\frac{x}{y} + \frac{y}{z}+ \frac{z}{x})^2 = (\frac{x}{y})^2 + (\frac{y}{z})^2+ (\frac{z}{x})^2 + 2(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})$
$a^2+b^2+c^2=(\frac{x}{y})^2 + (\frac{y}{z})^2+ (\frac{z}{x})^2 = (\frac{x}{y} + \frac{y}{z}+ \frac{z}{x})^2 - 2(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}) = 7^2-2(9)=37$
so we have
$7^3 = 3(7)(37)-2(a^3+b^3+c^3)+6$
$((\frac{x}{y})^3+(\frac{y}{z})^3+(\frac{z}{x})^3) = 157$
|
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|
Distributing multiplication of rational functions I am having trouble distributing with fractions. This
$$ \left(\frac{1}{(x + 3)} + \frac{(x + 3)}{(x - 3)}\right)\, (9 - x^2) = -\frac{(x^2-3)}{9-x^2}(9-x^2) $$
has the answer $\left\{\left[x = - \frac{9}{7}\right]\right\}$.
I start solving by cancelling $(9-x^2)$ on the right hand side, and multiplying $(9-x^2)$ with the terms on the left hand side by foiling (multiplying term by term). So the first step is
$$9\times\frac{1}{x+3}+\frac{9 (x+3)}{x-3}-x^2\times\frac{1}{x+3}-\frac{x^2 (x+3) }{x-3}=-x^2-3$$
This is apparently wrong because it has another answer, $\left\{\left[x = - \frac{3}{7}\right]\right\}$ What am I doing wrong?
|
Cancelling $(9-x^2)$ on the RHS leaves you with $-(x^2-3)$. Next I'd recommend finding a common denominator and combining the fractions inside the parentheses on the LHS. You should get $$ \frac{1}{\left(x + 3\right)} + \frac{\left(x + 3\right)}{\left(x - 3\right)} = \frac{x-3+(x+3)^2}{x^2 - 9}$$ which means the LHS will simplify to $-(x-3+(x+3)^2)$, leaving you with $$-(x-3+(x+3)^2) = -(x^2-3)$$ Can you take it from here? If I carry out the rest of this algebra I find that $x = \frac{-9}{7}$.
|
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|
showing global min point let's $F(x)= x'Ax + a'x $
where $x'Ax$ is a quadratic form and $a'$ is defined as a vector.
$$A:= \left[ \begin{matrix} 6 &1&1 \\ 1&2&0 \\ 1&0&4\end{matrix}\right] $$
Does there exist a global minimum point (absolute min) for this function $F(x)$?
note: I know that if the function is given as a linear structure, whether this is convex or concave is not important. we only exemine $x'Ax$.
but i cannot perfectly prove this question. thank you for helping.
|
Isn't minimum or maximum were the gradient is zero (vector)?
$$ \frac{{\rm d}F(x)}{{\rm d}x} = 2 A x + a' =0 $$
So the unique solution
$$ x = - \frac{1}{2} A^{-1} a' $$
if it exists must be a minimum or maximum.
Edit 1
So for your example suppose $a'=(a_1,a_2,a_3)$ and $x=(x_1,x_2,x_3)$ then
$$ \frac{{\rm d}F(x)}{{\rm d}x} = 2 \begin{pmatrix} 6 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix}+ \begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} = \begin{pmatrix} 12 x_1+2 x_2+2 x_3 + a_1 \\ 2 x_1+4 x_2 + a_2 \\ 2 x_1 + 8 x_4 + a_3\end{pmatrix}$$
To make this the zero vector you need
$$ \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix} =-\frac{1}{2} \begin{pmatrix} 6 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 4 \end{pmatrix}^{-1} \begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} = \begin{pmatrix} -\frac{4 a_1-2 a_2-a_3}{42} \\ \frac{4 a_1-23 a_3-a_3}{84} \\ \frac{2 a_1-a_2 -11 a_3}{84} \end{pmatrix} $$
This makes $${\rm extrema}(F(x)) = \frac{-8 a_1^2+8 a_1 a_2+4 a_1 a_3-23 a_2^2-2 a_2 a_3 -11 a_3}{168} $$
|
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|
What is the smallest possible value of $\lfloor (a+b+c)/d\rfloor+\lfloor (a+b+d)/c\rfloor+\lfloor (a+d+c)/b\rfloor+\lfloor (d+b+c)/a\rfloor$? What is the smallest possible value of
$$\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\frac{d+b+c}{a}\right\rfloor$$
where $a,b,c,d>0$?
|
Let $$g(a,b,c,d)=\dfrac{a+b+c}{d}+\dfrac{b+c+d}{a}+\dfrac{c+d+a}{b}+\dfrac{d+a+b}{c}$$
use Cauchy-Schwarz inequality we have
$$g(a,b,c,d)=(a+b+c+d)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)-4\ge 16-4=12$$
so
$$\lfloor\dfrac{a+b+c}{d}\rfloor+\lfloor\dfrac{b+c+d}{a}\rfloor+\lfloor\dfrac{c+d+a}{b}\rfloor+\lfloor\dfrac{d+a+b}{c}\rfloor>12-4=8$$
and other hand,when,$a=10,b=c=d=11$
$$\lfloor\dfrac{a+b+c}{d}\rfloor+\lfloor\dfrac{b+c+d}{a}\rfloor+\lfloor\dfrac{c+d+a}{b}\rfloor+\lfloor\dfrac{d+a+b}{c}\rfloor=9$$
|
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|
Prove that the lim f(x) = 1 using the $\epsilon-\delta$ definition
Show directly (from the $\epsilon-\delta$ definition ) that
$$ \lim_{x\to-1}\frac{x^{2}-3}{x-1} = 1$$
Attempt
Using
$$
\left\lvert \frac{x^{2}-3}{x-1} - 1 \right\rvert
= \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert
= \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert
$$
and
$|x-2|\leq |x - 1 - 1 | \leq |x-1|+|-1| = |x-1|+1 \leq \delta + 1 = 2$ (if $\delta \leq$ 1)
and
$|x-1| = |x+1-2| \geq |-2| - |x+1| > 2 - \delta \geq 1$ (if |x+1| < $\delta$)
hence
$$
\left\lvert \frac{x^{2}-3}{x-1} - 1 \right\rvert
= \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert
= \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert
< 2 \delta \leq \epsilon
$$
so $\delta = \frac{\epsilon}{2}$
and picking $\delta$ = min(1,$\frac{\epsilon}{2}$)
hence
$$
\forall \epsilon>0 \exists \delta
= \min(1,\frac{\epsilon}{2}) > 0 : \forall
0 < \lvert x--1\rvert < \delta
= \min(1,\frac{\epsilon}{2}) \implies
\left\lvert\frac{x^{2}-3}{x-1} - 1 \right\rvert
= \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert
= \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert
< 2 \delta \leq \epsilon$$
hence
$$ \lim_{x\to-1}(\frac{x^{2}-3}{x-1}) = 1$$
|
I find it easier to make simplifying assumptions first. You have an
expression $| { (x-2)(x+1) \over x-1 } |$ which you would like to bound near
$x=-1$. Start by bounding $\delta$ by, say, $1$ (that is $|x+1| < 1$), in which case you have
$|x-1| > 1$, and $|x-2| < 4$. Then you have
$| { (x-2)(x+1) \over x-1 } | \le 4 |x+1|$.
Now choose $\epsilon>0$ and let $\delta = \min (1, {1 \over 4} \epsilon)$ to
get the desired result.
|
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|
Find the angle of complex number Let
$$
z = \frac{a-jw}{a+jw}.
$$
Then the angle of $z$ is
$$
-\tan^{-1}z\left(\frac{w}{a}\right) -\tan^{-1}\left(\frac{w}{a}\right) = -2\tan^{-1}\left(\frac{w}{a}\right).
$$
How is that so?
|
Assuming $a,w\in\mathbb{R}$:
$$\arg\left(\frac{a-wi}{a+wi}\right)=$$
$$\arg\left(\frac{a-wi}{a+wi}\cdot\frac{a-wi}{a-wi}\right)=$$
$$\arg\left(\frac{(a-wi)^2}{a^2+w^2}\right)=$$
$$\arg\left(\frac{a^2-w^2-2awi}{a^2+w^2}\right)=$$
$$\arg\left(a^2-w^2-2awi\right)-\arg\left(a^2+w^2\right)=$$
$$\arg\left(a^2-w^2-2awi\right)-0=$$
$$\arg\left(a^2-w^2-2awi\right)=$$
$$\arg\left(a^2-w^2-2awi\right)=$$
$$\arg\left((a-wi)^2\right)=$$
$$\arg\left((a-wi)(a-wi)\right)=$$
$$\arg\left(a-wi\right)+\arg\left(a-wi\right)=$$
$$2\arg\left(a-wi\right)=$$
$$-2\tan^{-1}\left(\frac{w}{a}\right)$$
|
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|
Expanding brackets problem: $(z - 1)(1 + z + z^2 + z^3)$ I have:
$(z - 1)(1 + z + z^2 + z^3)$
As, I have tried my own methods and enlisted the help on online software, but as well as them not all arriving at the same solution, I can't follow their reasoning.
I tried to gather all the like terms:
$(z - 1)(z^6+1)$
And then thought it may offer a route to a difference or two cubes if I shifted powers across the brackets, but that's $z^7$ in total, isn't it? I can't just do $z^7-1$, can I?
If I try to expand, I see:
$(z - 1)(z^6 + 1) = z^7+z-z^6-1$
Well...
$z^7-z^6=z$
Add the other z leaves:
$z^2-1$?
I expect this is all wrong.
Will you help please?
On a side note, I have difficulty with understanding how I'm expected to tag this post accurately when I can't used "expand", "multiply", or use any of the subjects it covers as tags?
|
You seem to have some very unfortunate ideas about algebra! As David Mitra said, When you are adding powers of z you do not add the powers themselves. That is a property of multiplication: $z^n\cdot z^m= z^{n+ m}$.
To multiply $(z- 1)(1+ z+ z^2+ z^3)$ use the "distributive law" $a(b+ c)= ab+ ac$ and $(b+ c)a= ab+ ac$.
Think of $z- 1$ as $(b+ c)$ with $b= z$ and $c= -1$ and a as $(1+ z+ z^2+ z^3)$.
$(z- 1)(1+ z+ z^2+ z^3)= z(1+ z+ z^2+ z^3)- 1(1+ z+ z^2+ z^3)$.
Now, for each of those, use the distributive law again: $z(1+ z+ z^2+ z^3)= z(1)+ z(z)+ z(z^2)+ z(z^4)$.
NOW use the rule for adding exponents (with $1= z^0$ and $z= z^1$): $z(1)+ z(z)+ z(z^2)+ z(z^3)= z+ z^2+ z^3+ z^4$.
And, of course, $-1(1+ z+ z^2+ z^3)= -1- z- z^2- z^3$.
Putting those together, $(z- 1)(1+ z+ z^2+ z^3)= z+ z^2+ z^3+ z^4- 1- z- z^2- z^3= -1+ (z- z)+ (z^2- z^2)+ (z^3- z^3)+ z^4= z^4- 1$.
|
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|
Solve $[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ How can one solve the equation :
$[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ ?
Such that $[x]$ is the integer part of $x$.
By definition :
$[\frac{x^2-x+1}{2}]<=\frac{x^2-x+1}{2}<[\frac{x^2-x+1}{2}]+1$.
So :
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}<\frac{x-1}{3}+1$.
1)
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}$ <=> $3x^2-5x+5>0$.
2)
$\frac{x^2-x+1}{2}<\frac{x-3}{2}+1$ <=> $3x^2-5x-1<0$.
How can I solve the system ?
|
The key is to recognize that the left hand side is, by definition of the floor function, an integer, so the right hand side must also be an integer, and hence we must have $x=3k+1$ for some $k\in\mathbb{Z}$. This gives (on doing a little algebra)
$${x^2-x+1\over2}={3k(3k+1)\over2}+{1\over2}$$
Note that $2$ divides either $k$ or $3k+1$, so $3k(3k+1)/2$ in an integer for any $k$, hence
$$\big\lfloor{x^2-x+1\over2}\big\rfloor={3k(3k+1)\over2}$$
Thus the equation to solve (for integer $k$) is
$${3k(3k+1)\over2}=k$$
But the only integer solution is $k=0$, since $k\not=0$ implies $3(3k+1)=2$. So $x=1$ is the only real solution to the original equation.
|
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|
$ x+y+z = 3, \; \sum\limits_{cyc} \frac{x}{2x^2+x+1} \leq \frac{3}{4} $
For positive variables $ x+y+z=3 $, show that $ \displaystyle \sum_{cyc} \dfrac{x}{2x^2+x+1} \leq \dfrac{3}{4} $.
Apart from $ (n-1)$ EV, I could not prove this inequality. I've tried transforming it into a more generic problem - it looks fairly more interesting to me.
Consider a continuous function $ f(x) \geq 0, \; 0 < x < \alpha $, with a unique value of $ \gamma $ such that $ f'(\gamma) = 0 $ and a unique value of $ \theta $ such that $ f''(\theta) = 0 $, being $ \gamma < \theta $.
If $ \displaystyle \sum_{i=1}^{n} x_i = \alpha, \; f(0)=0$ and $ f \left ( \frac{\alpha}{n} \right ) = \frac{\beta}{n} $, is it always correct to conclude that $ \displaystyle \sum_{i=1}^{n} f (x_i) \leq \beta $, being $ f(\gamma) > \frac{\beta}{n} $? Provide an example other than $ f(x) = \frac{x}{2x^2+x+1} $ and a counter-example.
|
Also, the Tangent Line method helps.
Indeed, we need to prove that:
$$\sum_{cyc}\left(\frac{1}{4}-\frac{x}{2x^2+x+1}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(2x-1)}{2x^2+x+1}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(2x-1)}{2x^2+x+1}-\frac{x-1}{4}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(5-2x)}{2x^2+x+1}\geq0,$$
which says that our inequality is true for $\max\{x,y,z\}\leq\frac{5}{2}.$
Now, let $x>\frac{5}{2}.$
Thus, since by AM-GM $$\frac{y}{2y^2+y+1}=\frac{1}{2y+\frac{1}{y}+1}\leq\frac{1}{2\sqrt2+1},$$ we obtain:
$$\sum_{cyc}\frac{x}{2x^2+x+1}\leq\frac{2}{2\sqrt2+1}+\frac{2.5}{2\cdot2.5^2+2.5+1}<\frac{3}{4}.$$
|
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|
What is the remainder of $31^{2008}$ divided by $36$? What is the remainder of $31^{2008}$ divided by $36$?
Using Euler's theorem, we have:
$$
\begin{align*}
\gcd(31,36) = 1 &\implies 31^{35} \equiv 1 \pmod{36} \\
&\implies 31^{2008} \equiv 31^{35(57) + 13} \equiv (31^{35})^{57+13} \equiv 1^{57}*31^{13} \\
&\implies 31^{13} \equiv 31\pmod{36}
\end{align*}
$$
So the remainder is $31$, is this correct?
|
The line that goes $31^{35}\equiv 1 \pmod {36}$ is false. Euler's theorem says that $31^{\phi(n)}\equiv 1 \pmod {36}$, but $\phi(36) = 12$. So we would say $31^{2008} \equiv 31^{4} \pmod{36}$ using the same reduction method (taking away multiples of 12) to get $13 \pmod{36}$.
|
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Calculate the volume of the region trapped by $z^2=x^2+y^2, z=2(x^2+y^2)$ Task
Calculate the volume of the region trapped by $z^2=x^2+y^2, z=2(x^2+y^2)$ using a triple integral.
I'm kind of lost on this one, here's my (probably wrong) attempt:
Calculate the intersection, I get $z=0 \wedge z=\frac 12$. That means the region is: $E=\{(x,y,z): x^2+y^2=1/2, z=1/2\} \cup \{(x,y,z): x^2+y^2=0\}$.
That means I should integrate the constant function $1$ over $E$ to get the volume of $E$, however, I don't think I can get a triple integral this way.
How do I proceed?
|
Your definition of the region is incorrect (though your bounds for $z$ are correct). When $0\le z\le\frac{1}{2}$, the cone lies above the paraboloid; the circle of intersection at $z=\frac{1}{2}$ is $x^2+y^2 = \frac{1}{4}$. So one definition of the region is
\begin{equation*}
-\frac{1}{2}\le x\le \frac{1}{2},\quad
-\sqrt{\frac{1}{4}-x^2} \le y \le \sqrt{\frac{1}{4}-x^2},\quad
2(x^2+y^2) \le z \le \sqrt{x^2+y^2}.
\end{equation*}
If you integrate $1$ over this region, you should get the answer. However, you may wish to use a different set of coordinates to make the integral simpler.
|
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|
Prove inequality with binomial coefficient: $6 + \frac{4^n}{2 \sqrt{n}} \le \binom{2n}n$ I have to prove inequality, where $n \in N$
$$6 + \frac{4^n}{2 \sqrt{n}} \le {2n \choose n}$$
I have checked and it is true when $n\ge4$, however I have no idea how I should start. Can anyone give a hint?
|
$$\frac{1}{4^n}\binom{2n}{n} = \frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)$$
and by squaring both sides:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)$$
hence:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)^{-1}$$
and:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{\pi n}\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)^{-1}=\frac{1}{\pi n}\prod_{k>n}\left(1-\frac{1}{(2k-1)^2}\right)$$
but in a neighbourhood of the origin $(1+x)^{-1}\geq e^{-x}$, hence:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 \geq \frac{1}{\pi n}\exp\left(-\sum_{k>n}\frac{1}{4k(k-1)}\right)=\frac{e^{-\frac{1}{4n}}}{\pi n}$$
and we have:
$$ \frac{1}{4^n}\binom{2n}{n}\geq \frac{e^{-\frac{1}{8n}}}{\sqrt{\pi n}}\geq \frac{8n-1}{8n\sqrt{\pi n}}$$
that is stronger than the given inequality for every $n\geq 5$.
It just remains to check the first cases by hand.
|
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|
$A+B$ is non singular and $C=(A+B)^{-1}(A-B)$,then prove that $C^TAC=A$ If $A$ is a symmetric and $B$ is a skew symmetric matrix and $A+B$ is non singular and $C=(A+B)^{-1}(A-B)$,then prove that $C^TAC=A$.
My Attempt:
$C^T=((A+B)^{-1}(A-B))^T=(A-B)^T((A+B)^{-1})^T=(A^T-B^T)((A+B)^T)^{-1}$
$C^T=(A+B)(A-B)^{-1}$
$C^TAC=(A+B)(A-B)^{-1}A(A+B)^{-1}(A-B)$
But i am stuck here and could not solve further.Please help me.Thanks.
|
First of all, we show that $C^T(A+B)C=A+B.$ To do that:
$$\begin{align}(A+B)(A-B)^{-1}(A+B)(A+B)^{-1}(A-B) & \\ & \underbrace{=}_{(A+B)(A+B)^{-1}=I}(A+B)(A-B)^{-1}(A-B)\\ & \underbrace{=}_{(A-B)(A-B)^{-1}=I}A+B.\end{align}$$
In a similar way we can show that $C^T(A-B)C=A-B.$ Indeed:
$$\begin{align}(A+B)(A-B)^{-1}(A-B)(A+B)^{-1}(A-B) & \\ & \underbrace{=}_{(A-B)(A-B)^{-1}=I}(A+B)(A+B)^{-1}(A-B)\\ & \underbrace{=}_{(A+B)(A+B)^{-1}=I}A-B.\end{align}$$
Finally adding both equalities we get the desired result.
|
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|
To determine if a polynomial has real solution I have the following polynomial : $x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$
I must determine if this polynomial has at least 1 real solution and justify why. We have a theorem which says that all polynomials with real coefficients can be decomposed in a product of polynomials of real coefficients with degree 1 or 2.
So this means we have four scenarios :
Factors : 2+2+2+1 , 2+2+1+1+1, 2+1+1+1+1+1, 1+1+1+1+1+1+1
In all these cases, we have atleast one factor of degree 1, so there is atleast one real solution in each case. What do you think ?
|
Since
$$x^7+x^6+x^5+x^4+x^3+x^2+x+1=\frac{x^8-1}{x-1}$$
we can use the fact that $x^8-1$ has roots at $-1$ and $1$ to get the root of $x=-1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How many solutions exist for a non-linear system How many solutions exist to the following system:
$$
\begin{eqnarray}
xy+xz &=& 54+x^2 \\
yx+yz &=& 64+y^2 \\
xz+yz &=& 70+z^2
\end{eqnarray}
$$
I have guessed that the solutions, if exist, have to be integer, then we get:
$$
\begin{eqnarray}
x(y+z-x) &=& 2\cdot 3\cdot 3\cdot 3 \\
y(x+z-y) &=& 2\cdot2\cdot2\cdot2\cdot2\cdot2 \\
z(x+y-z) &=& 2\cdot5\cdot7
\end{eqnarray}
$$
Looking at the third equation, there are 8 possibilities for the value of $z$ and the situation is much worse for the other two equations.
My questions are:
*
*is my guess about the solutions is correct, and how one can prove it?
*how to proceed without exhaustive testing for each of the possibilities?
*is there a better way to approach this problem?
Thanks!
|
Firstly as you had factorized:
$x(y+z-x) = 54$.
$y(z+x-y) = 64$.
$z(x+y-z) = 70$.
Also the differences between equations are:
$(x-y)z = (x-y)(x+y) - 10$.
$(y-z)x = (y-z)(y+z) - 6$.
$(x-z)y = (x-z)(x+z) - 16$.
Which are equivalent to:
$(x-y)(x+y-z) = 10$.
$(y-z)(y+z-x) = 6$.
$(x-z)(z+x-y) = 16$.
And combining with the factorizations gives:
$\frac{x-y}{z} = \frac{10}{70}$.
$\frac{y-z}{x} = \frac{6}{54}$.
$\frac{x-z}{y} = \frac{16}{64}$.
You can already easily guess an answer at this point: $7,8,9$.
But continuing we can always scale $(x,y,z)$ so that $z = 7$ and so:
$x-y = 1$ and $9(y-7) = x$.
Whence you get the scaled solution and unscale to get the 'general' solution.
It turns out that only the two obvious integer triples satisfy the original.
|
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|
Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$ Prove that:
$$
\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}
$$
if $a,b$ are positive real numbers such that $a+b=1$.
I have tried expanding the squares and rewriting them such that $a+b$ is a term/part of a term but what I get is completely contradictory to what is asked to prove
|
For your revised question, another way is to note that $(x + \frac1x)^2$ is convex, so by Jensen's inequality:
$$\left(a + \frac1a\right)^2 + \left(b + \frac1b\right)^2 \ge 2\left(\frac{a+b}2 + \frac2{a+b}\right)^2=2\left(\frac12 + 2\right)^2=\frac{25}2$$
|
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|
How do I complete solving this homogeneous system of linear equations using Gaussian elimination? $${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=0\\ { x }_{ 1 }+{ x }_{ 2 }+2{ x }_{ 3 }-{ x }_{ 4 }=0\\ 2{ x }_{ 1 }+2{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }=0\\ -{ x }_{ 1 }-{ x }_{ 2 }+{ x }_{ 3 }-2{ x }_{ 4 }=0\\ $$
Steps I took:
$$\left[\begin{array}{rrrr|r}
1 & 1 & 1 & 0 & 0 \\
1 & 1 & 2 & -1 & 0 \\
2 & 2 & 1 & 1 & 0 \\
-1 & -1 & 1 & -2 & 0
\end{array}\right]$$
$${ R }_{ 2 }={ 2R }_{ 2 }-{ R }_{ 3 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
1 & 1 & 1 & 0 & 0 \\
0 & 0 & 3 & -3 & 0 \\
2 & 2 & 1 & 1 & 0 \\
-1 & -1 & 1 & -2 & 0
\end{array}\right]$$
$${ R }_{ 3 }={ R }_{ 3 }+{ 2R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
1 & 1 & 1 & 0 & 0 \\
0 & 0 & 3 & -3 & 0 \\
0 & 0 & 3 & -3 & 0 \\
-1 & -1 & 1 & -2 & 0
\end{array}\right]$$
$${ R }_{ 4 }={ R }_{ 1 }+{ R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
1 & 1 & 1 & 0 & 0 \\
0 & 0 & 3 & -3 & 0 \\
0 & 0 & 3 & -3 & 0 \\
0 & 0 & 2 & -2 & 0
\end{array}\right]$$
$${ R }_{ 3 }={ R }_{ 3 }-{ R }_{ 2 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
1 & 1 & 1 & 0 & 0 \\
0 & 0 & 3 & -3 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & -2 & 0
\end{array}\right]$$
$${ R }_{ 2 }=2{ R }_{ 2 }-3{ R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
1 & 1 & 1 & 0 & 0 \\
0 & 0 & 3 & -3 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]$$
At this point I am stuck. I am having a lot of trouble figuring out what I need to do once I complete all the necessary row operations on my matrix. I would like to be guided through the process as of this point.
Please explain it to me in layman terms. I am hoping to use what I learn from being guided through the final steps of this problem in order to understand how to solve ones without any hand holding.
I'm also really frustrated with the fact that I am doing these operations like a computer without actually understanding what is going on with the "independent" and "dependent" variables.
|
You end up with
$$\left[\begin{array}{rrrr|r}
1 & 1 & 0& 1& 0 \\
0 & 0 & 1 & -1& 0 \\
\end{array}\right].$$
All you have to do is paint in just straightening the tableau, i.e., create an upper triangle matrix whose diagonal elements are $\pm1$ in the following way. The first row of your reduced matrix has the first $1$ at position $1$ so that would be the first row, the second row's position of the leading $1$ is $3$ so that becomes the third row; just leave space for row $2$ and $4$:
$$\left[\begin{array}{rrrr|r}
1 & 1 & 0& 1 & 0 \\
\strut &&&&\\
0 & 0 & 1& -1& 0 \\
\strut &&&&
\end{array}\right].$$
Now fill the empty rows with zeroes except for the row number's position: you insert a $-1$: the second row consists of zeroes except for the second positon, you fill in a $-1$ here:
$$\left[\begin{array}{rrrr|r}
1 & 1 & 0& 1& 0 \\
0&-1&0&0&0\\
0 & 0 & 1& -1& 0 \\
\strut &&&&
\end{array}\right].$$
Row number four gets a $-1$ at position four:
$$\left[\begin{array}{rrrr|r}
1 & 1 & 0& 1& 0 \\
0&-1&0&0&0\\
0 & 0 & 1& -1& 0 \\
0&0&0&-1&0
\end{array}\right].$$
Now the columns with the numbers of the inserted rows, namely column two
$\left[\begin{array}{c}
1 \\-1\\ 0 \\0
\end{array}\right]$
and column four
$\left[\begin{array}{r}
1\\ 0\\ -1\\ -1
\end{array}\right]$
span the kernel of the matrix. Cute, isn't it?
|
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|
Show that $\sum_{n=1} ^{\infty} \frac {1}{(x+n)^2} \leq \frac{2}{x} $
Show that $\displaystyle \sum_{n=1} ^{\infty} \frac {1}{(x+n)^2} \leq \frac{2}{x} $
For any real number x $\geq 1$, I want to show the above
Very rusty on my analysis, I think I need to do a comparison test to show the series converges, but then not sure what after that to get the inequality.
|
Let $n\ge 1$, for $x> 0$ we have $$\frac{1}{(x+n)^2}\le\frac{1}{(x+t)^2}\le\frac{1}{(x+n-1)^2}\qquad\text{for }n-1\le t\le n$$
Then
\begin{align}
\int_{n-1}^{n}\frac{1}{(x+n)^2}\,dt&\le \int_{n-1}^{n}\frac{1}{(x+t)^2}\,dt\\[6pt]
\frac{1}{(x+n)^2}&\le\frac{1}{x+n-1}-\frac{1}{x+n}\\[6pt]
\sum_{n=1}^N\frac{1}{(x+n)^2}&\le\frac{1}{x}-\frac{1}{x+N}
\end{align}
Then, as $N\to\infty$ we get
$$\sum_{n=1}^{\infty}\frac{1}{(x+n)^2}\le\frac{1}{x}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
For which $n$ is $2^{2x+2} \equiv 2\pmod n \quad\text{and}\quad 2^{2x+2} \equiv 4\pmod {n-1} $ We begin with
\begin{align*}
2(2^p+n^2-1-2n) \equiv 0\pmod n \quad&\text{and}\quad 2(2^p+n^2-1-2n) \equiv 0\pmod {n-1} \\
\end{align*}
Which we reduce to
\begin{align*}
2^p\equiv1\pmod{\frac n{2}} \quad&\text{and}\quad 2^{p-1}\equiv1\pmod{\frac{n-1}{4}}.
\end{align*}
We are looking to prove that $p$ cannot be odd, considering that $n \ge 3$
So we let $p=2x+1$ (looking to prove by contradiction)
\begin{align*}
2^{2x+2} \equiv 2\pmod n \quad&\text{and}\quad 2^{2x+2} \equiv 4\pmod {n-1} \\
\end{align*}
I know that $a^b \equiv a\pmod b$, so I am assuming that I should go somewhere in that direction. And one more thing, $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ where $s$ is an integer and $s \ge 3$. I have reason to believe that the only time that $p$ can be odd is when $n=2, s=2^p-1$
Also all 3 variables must be positive integers ($s,n,p$)
EDIT: It seems that there are solutions even when $n$ does not equal $2$, but the question still remains, for which $n$ can it hold true
|
EDIT: For a counterexample, try $p=11$ and $n=23$ We have $2(2^p+n^2−1−2n)=5060=10 \cdot 23 \cdot 22$.
INCORRECT ANSWER:
I think I found a counterexample. Set $p = 3$ and $n = 7$. Then, $2(2^p+n^2−1−2n) = 2(8+49-1-14) = 2(42) = 84$. This is divisible by both $n = 7$ and $n-1 = 6$, but $p = 3$ is odd.
|
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|
Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$ How can one evaluate the below sum? Any help would be greatly appreciated.
$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$$
|
We can built the formula, bottom up.
Let $s_n = \sum_{k=1}^\infty k^n r^k $
$n-(n-1)=1 \Rightarrow (1-r)s_1 = s_0$
$n^2-(n-1)^2=2n-1 \Rightarrow (1-r)s_2 = 2 s_1 - s_0$
$s_0 = r + r^2 + r^3 + \cdots = \frac{r}{1-r}$
$s_1 = \frac{s_0}{1-r} = \frac{r}{(1-r)^2}$
$s_2 = \frac{2 s_1 - s_0}{1-r} = \frac{r(r+1)}{(1-r)^3}$
For $r=\frac{1}{15}, s_0=\frac{1}{14}, s_1=\frac{15}{196}, s_2=\frac{30}{343}$
$$ \sum_{k=1}^\infty \frac{k(k+2)}{15^k} = s_2 + 2 s_1 = \frac{165}{686} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent Let $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\cdot A=A$ and $B\cdot B=B$?
Steps I took:
*
*Let $A= \left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right]$ and let $B= \left[\begin{array}{rr}
w & x \\
y & z \\
\end{array}\right]$
*$\left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right] \cdot \left[\begin{array}{rr}
w & x \\
y & z \\
\end{array}\right] = \left[\begin{array}{rr}
aw+by & ax+bz \\
cw+dy & cx+dz \\
\end{array}\right] $ (Which is also equal to A)
*$\left[\begin{array}{rr}
w & x \\
y & z \\
\end{array}\right] \cdot \left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right] = \left[\begin{array}{rr}
wa+xc & wb+xd \\
ya+zc & yb+zd \\
\end{array}\right] $ (Which is also equal to B)
At this point I am stuck. I don't know how to proceed and I imagine that I started off on the wrong track to begin with. I'd like a hint to guide me in the right direction.
My proof after consultation with answerers below:
Proof:
1) Since $AB=A$, we can say that: $(AB)A=AA$ which is equal to $A^2$
2) Then, (by associativity of matrix multiplication), we can say that $A(BA) = AB$ (since $BA=B$)
3) Then, $AB=A$ (since $AB=A$ was given)
4) Therefore, $AA$ is equal to $A$
5) Since $BA=B$, we can also say that: $(BA)B=BB$ which is equal to $B^2$
6) Then, (by associativity of matrix multiplication) we can say that $B(AB)=BA$ (since $AB=A$)
7) Then, $BA=B$
8) Therefore, $BB$ is equal to $B$
9) Thus, $AA$ is equal to $A$ and $BB$ is equal to $B$
Q.E.D.
|
Since $AB=A$ and $BA=B$, to show that $A$ is an idempotent we have to show that $A=A^2$
Therefore $A^2=(AB)^2 = A^2B^{ } = AB = A(BA) = AA = A^2 = A$. Proven
|
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|
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &\neq \sqrt{x+2}\\
\end{align}
*
*Is my proof correct?
*I'm interested in more ways of proving it.
|
Your proof is correct, because each inequality you write is equivalent to the previous one (it should be noted, probably).
Changing all $\ne$ into $=$ would make it a proof by contradiction, that's however unnecessary.
In a different way, you could just swap terms and square, again changing inequalities into equivalent ones:
\begin{gather}
\sqrt{x+2}+\sqrt{x}\ne 2\sqrt{x+1}\\[10px]
x+2+x+2\sqrt{x(x+2)}\ne 4x+4\\[10px]
\sqrt{x(x+2)}\ne x+1\\[10px]
x^2+2x\ne x^2+2x+1\\[10px]
0\ne1
\end{gather}
|
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"timestamp": "2023-03-29T00:00:00",
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|
find the solution to recurrent relation Solving some math problem, I have faced this recurrent equation: $$S(n) = 3 S(n - 3) + 2 \sum\limits_{k = 2}^{n / 3} S(n - 3 k) \times k.$$
Here $n = 3 \alpha$, means, $n$ can be divided by $3$ ($\alpha$ is integer), and $S(0) = 1$.
Could anyone please help me to find $S(n)$ in a closed form if it's possible? Thanks!
UPD: Of course, $n >= 6$.
|
Call $S(3 n) = s_n$, write the recurrence with no subtractions in indices:
$\begin{align}
s_{n + 2}
&= 3 s_{n + 1} + 2 \sum_{2 \le k \le n + 2} k s_{n + 2 - k} \\
&= 3 s_{n + 1} + 2 \sum_{0 \le k \le n} (k + 2) s_{n - k} \\
&= 3 s_{n + 1} + 2 \sum_{0 \le k \le n} (n - k + 2) s_k
\end{align}$
Define the generating function $S(z) = \sum_{n \ge 0} s_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$:
$\begin{align}
\sum_{n \ge 0} s_{n + 2} z^n
&= 3 \sum_{n \ge 0} s_{n + 1} z^n
+ 2 \sum_{n \ge 0} z^n \sum_{0 \le k \le n} (n + 2 - k) s_k \\
\frac{S(z) - s_0 - s_1 z}{z^2}
&= 3 \frac{S(z) - s_0}{z}
+ 2 \sum_{n \ge 0} z^n \sum_{0 \le k \le n} (n -k) s_k
+ 2 \sum_{n \ge 0} z^n \sum_{0 \le k \le n} s_k \\
&= 3 \frac{S(z) - s_0}{z}
+ 2 \left( \sum_{n \ge 0} n z^n \right)
\left( \sum_{n \ge 0} s_n z^n \right)
+ 2 \frac{S(z)}{1 - z} \\
&= 3 \frac{S(z) - s_0}{z}
+ 2 \frac{z}{(1 - z)^2} S(z)
+ 2 \frac{S(z)}{1 - z}
\end{align}$
This results in:
$\begin{align}
S(z)
= \frac{(1 - z)^2 (s_0 + s_1 - 3 s_0 z)}
{1 - 5 z + 5 z^2 - 3 z^3}
\end{align}$
This doesn't factor easily, so I'll leve it at this. Next step would be splitting into partial fractions, and get the coefficients off the resulting simpler terms (geometric series, binomial theorem).
|
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|
Prove that $\lim\limits_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$ using the definition
Prove $$\lim_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$$ using the definition.
Let there be $\varepsilon>0$ we need to find $N<n$ such that $\Big|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}$$
\Big|<\varepsilon$
$$\left|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}
\right|=\left|\frac{3n^2+3n-3-3n^2-1}{9n^2+3}
\right|=\left|\frac{3n-4}{9n^2+3}\right|$$
What should I do next, can I say that because $n\to \infty$ we can neglect $-4$ and $3$ from the expression?
|
Hint: $\frac{n-3}{3n^2+1}\leq \frac{1}{3n}$
|
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|
System of recurrence relations with Taylor series expansion Find $a_n,b_n$ where $a_0=1,b_0=0$ for the following relations:
$a_{n+1}=2a_n+b_n$
$b_{n+1}=a_n+b_n$
Using generating functions, the system is:
$f(x)-a_0=2xf(x)+xg(x)$
$g(x)-b_0=xf(x)+xg(x)$
Solving for $f(x)$ and $g(x)$ gives:
$$f(x)=\frac{1-x}{x^2-3x+1}$$
$$g(x)=\frac{x}{x^2-3x+1}$$
Taylor series for $f(x)$ and $g(x)$,
$\sum\limits_{n=0}^{\infty}\frac{f^{(n)}(a_0)}{n!}(x-a_0)^n$
$\sum\limits_{n=0}^{\infty}\frac{g^{(n)}(b_0)}{n!}(x-b_0)^n$
$n^{th}$ derivatives of $f,g$ can be found using partial fractions:
$$f^{(n)}(x)=\frac{(-1)^n n!(\sqrt{5}-5)}{10\left(x-\frac{3-\sqrt{5}}{2}\right)^{n+1}}+\frac{(-1)^n n!(-\sqrt{5}-5)}{10\left(x-\frac{3+\sqrt{5}}{2}\right)^{n+1}}$$
$$g^{(n)}(x)=\frac{(-1)^n n!(3\sqrt{5}-5)}{10\left(x-\frac{3-\sqrt{5}}{2}\right)^{n+1}}+\frac{(-1)^n n!(-3\sqrt{5}-5)}{10\left(x-\frac{3+\sqrt{5}}{2}\right)^{n+1}}$$
$n^{th}$ derivatives at $a_0$ and $b_0$:
$$f^{(n)}(a_0)=\frac{(-1)^n n!(\sqrt{5}-5)}{10\left(\frac{\sqrt{5}-1}{2}\right)^{n+1}}+\frac{(-1)^n n!(-\sqrt{5}-5)}{10\left(\frac{-1-\sqrt{5}}{2}\right)^{n+1}}$$
$$g^{(n)}(b_0)=\frac{(-1)^n n!(3\sqrt{5}-5)}{10\left(-\frac{3-\sqrt{5}}{2}\right)^{n+1}}+\frac{(-1)^n n!(-3\sqrt{5}-5)}{10\left(-\frac{3+\sqrt{5}}{2}\right)^{n+1}}$$
How to derive $a_n$ and $b_n$ in Taylor series?
|
You seem to have done all the necessary work, but confused a couple things. The Taylor series should be evaluated about the point $x=0$, not about the first terms of the sequences $a_0=1$, $b_0=0$. In other words, the $n^{\rm th}$ derivatives of $f$ and $g$ that you computed with partial fractions should be evaluated at $0$, and then you can use Taylor's formula to obtain the terms of the sequences: $a_n=f^{(n)}(0)/n!$ and $b_n=g^{(n)}(0)/n!$. So, as long as your derivatives are correct,
$$a_n=\frac{5-\sqrt5}{10}\left(\frac2{3-\sqrt5}\right)^{n+1}+
\frac{5+\sqrt5}{10}\left(\frac2{3+\sqrt5}\right)^{n+1}$$
and
$$b_n=\frac{5-3\sqrt5}{10}\left(\frac2{3-\sqrt5}\right)^{n+1}+
\frac{5+3\sqrt5}{10}\left(\frac2{3+\sqrt5}\right)^{n+1}.$$
For that matter, it's possible to skip the differentiation altogether and find the terms of the sequences directly from the partial fractions decomposition of the functions.
|
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|
Helping Im stuck with Question of Integral of $\frac{e^{2x}}{6e^{x}+2}$ So for this question I set $u= e^x$ and $du= e^x dx$, if put it back in it will be integral of $\frac{u}{6u+2}$. How would I continue on? Do I have to break it down into parts if so how?
P.S: the answer should come out to $(\frac{e^x}{6})-(\frac{1}{18})ln|6e^x+2|+C$
|
If you do what Andre recommended. Let $u = 6e^x + 2$. In order to write the top, we can square $u$.
\begin{align}
u^2 = (6e^x + 2)^2 &= 36e^{2x} + 24e^x +4
\end{align}
So
\begin{equation}
e^{2x} = u^2 - 35e^{2x} - 24e^{x} - 4
\end{equation}
Also $\frac{du}{6e^x} = dx$. Now we can solve the integral(remember $6e^x = u - 2$)
\begin{align}
\int \frac{e^{2x}}{6e^x + 2} dx &= \int \frac{u^2 - 35e^{2x} - 24e^x - 4}{u(u - 2)} du\\
&= \int \frac{u}{u-2} - \frac{24e^x}{u(6e^x)} - \frac{4}{u(u - 2)} du - \int \frac{35e^{2x}}{u} \frac{du}{6e^x}\\
&= u - 2\ln|u| + C - 35\int \frac{e^{2x}}{6e^x + 2} dx\\
36 \int \frac{e^{2x}}{6e^x + 2} dx &= 6e^x + 2 + C - 2\ln|6e^x + 2|\\
\int \frac{e^{2x}}{6e^x + 2} dx &=\frac{e^x}{6} -\frac{1}{18}\ln|6e^x + 2| + C
\end{align}
|
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|
Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so far: $\int_{0}^{a}\sqrt{1+\left(\left(\sqrt{x}\right)'\right)^{2}}dx=\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\left[\begin{array}{cc}
t^{2}=1+\frac{1}{4x} & \sqrt{1+\frac{1}{4N}},\sqrt{1+\frac{1}{4a}}\\
2tdt=-\frac{1}{8x^{2}}dx & x=\frac{1}{4\left(t^{2}-1\right)}
\end{array}\right]=\\\lim_{N\to0}-\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\frac{t^{2}dt}{\left(t^{2}-1\right)^{2}}=\lim_{N\to0}\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\left(\frac{1}{4\left(t+1\right)}-\frac{1}{4\left(t+1\right)^{2}}-\frac{1}{4\left(t-1\right)}-\frac{1}{4\left(t-1\right)^{2}}\right)dt=\\=\lim_{N\to0}\frac{1}{4}\left[\ln\left(t+1\right)+\frac{1}{t+1}-\ln\left(t-1\right)+\frac{1}{t-1}\right]_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4a}}}{1+\frac{1}{4a}-1}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4N}}}{1+\frac{1}{4N}-1}\right]=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-8N\sqrt{1+\frac{1}{4N}}\right]=\\=\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}-0+0=\ln\left(4a\left(\sqrt{\frac{1}{a}+4}+2\right)+1\right)-8a\sqrt{1+\frac{1}{4a}}$
But it doesn't seem right... any Ideas what went wrong?
|
If $u^2=1+\frac1{4x}$ then $x=\frac1{4(u^2-1)}$ and $\mathrm{d}x=-\frac{u}{2(u^2-1)^2}\,\mathrm{d}u$
$$
\begin{align}
\int_0^a\sqrt{1+\frac1{4x}}\,\mathrm{d}x
&=\frac12\int_{\sqrt{1+\frac1{4a}}}^\infty\frac{u^2}{(u^2-1)^2}\,\mathrm{d}u\\
&=\frac18\int_{\sqrt{1+\frac1{4a}}}^\infty\left(\frac1{(u-1)^2}+\frac1{u-1}+\frac1{(u+1)^2}-\frac1{u+1}\right)\mathrm{d}u\\
&=\frac18\left[-\frac1{u-1}+\log(u-1)-\frac1{u+1}-\log(u+1)\right]_{\sqrt{1+\frac1{4a}}}^\infty\\
&=\frac18\left[\frac{2u}{u^2-1}+\log\left(\frac{(u+1)^2}{u^2-1}\right)\right]_{u=\sqrt{1+\frac1{4a}}}\\
&=\frac14\left(\sqrt{4a(4a+1)}+\log\left(\sqrt{4a}+\sqrt{4a+1}\right)\right)
\end{align}
$$
|
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|
Rationality of the Lemniscate. This question is exercise 2 of Chapter 4 in Kunz' textbook of algebraic curves.
Let $f$ be the lemniscate with equation
$$(X^2 + Y^2 )^2 = α(X^2 − Y^2) \;\; (\alpha \in K^\times )$$
and let $x, y \in K[f]$ be the associated coordinate functions. Prove the
rationality of $f$ by showing that $R(f) = K(t)$ with $t := \frac{x^2+y^2}{x-y}.$
Kunz defines a function as rational if $\mathcal{R}(f)\cong K(t)$ where $\mathcal{R}(f)$ is the ring of rational functions on $f$, and $t$ is transcendental over $K$.
So far I have not got far. I have tried parametrizing in $t$, however I ended up with trig functions. I have also tried the map $K[t]\to \mathcal{R}(f)$ which maps $t\mapsto\frac{x^2+y^2}{x-y}$ but I couldn't show that this extends to a map on $k(t)$. Thanks for any advice.
|
In the case $K=\mathbb{R}$ (where we can graph the lemniscate) the way to parametrize the lemniscate is by using the circles $x^2+y^2=t(x-y)$. You can check that:
$$
\begin{split} &&x^2+y^2=t(x-y)\\ &\implies& \left(x-\frac{t}{2}\right)^2+\left(y+\frac{t}{2}\right)^2=\frac{t^2}{2} \end{split}
$$
so that the parametrization is by circles centered at $(t/2,-t/2)$ with radius $t/\sqrt{2}$. Graphing some of these circles, you can see that they intersect the lemniscate at the origin (with multiplicity 2) and also at one other point. It is this other point that will give us the parametrization. We first check where the lemniscate intersects these circles using the substitution $x^2+y^2=t(x-y)$ in the equation for the lemniscate:
$$
\begin{split} && t^2(x-y)^2=a^2(x^2-y^2) \\ &\implies& t^2(x-y)^2=a^2(x-y)(x+y) \\ &\implies& t^2(x-y)=a^2(x+y) \\ &\implies& (t^2-a^2)x=(t^2+a^2)y \\ &\implies& y=\frac{t^2-a^2}{t^2+a^2}x\end{split}
$$
Where we have divided by $x-y$ since the only solution when $x=y$ is the $(0,0)$ point which we already know about (and is not the point we are interested in). This last equation defines a line so all we need to do is check where this line intersects the original circle (other than the origin, $x=0$). Given $ y=\frac{t^2-a^2}{t^2+a^2}x$, the equation for the circle becomes:
$$
\begin{split} && x^2+y^2=t(x-y) \\ &\implies& x^2+\left(\frac{t^2-a^2}{t^2+a^2}x\right)^2 =t\left(x-\frac{t^2-a^2}{t^2+a^2}x\right) \\ &\implies& \frac{(t^2+a^2)^2+(t^2-a^2)^2}{(t^2+a^2)^2}x^2=t\left(\frac{t^2x+a^2x-t^2x+a^2x}{t^2+a^2}\right) \\ &\implies& \frac{(t^2+a^2)^2+(t^2-a^2)^2}{t^2+a^2}x =2a^2t \\ &\implies& x=\frac{2a^2t(t^2+a^2)}{t^4+2a^2t^2+a^4+t^4-2a^2t^2+a^4} \\ &\implies& x=\frac{a^2t(t^2+a^2)}{t^4+a^4} \end{split}
$$
Substituting this in to get $y$ yields:
$$
y=\frac{t^2-a^2}{t^2+a^2}x=\frac{a^2t(t^2-a^2)}{t^4+a^4}
$$
This gives the required parametrization of the lemniscate in terms of $t$.
|
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|
Finding a sequence satisfying this recurrence relation? I just don't even know where to start with this,
Find a sequence $(x_n)$ satisfying the recurrence relation:
$2x_n$$_+$$_2$ = $3x_n$$_+$$_1$ + $8x_n$ + $3x_n$$_-$$_1$ Where n is a natural number and
$x_0$ = -1,
$x_1$ = 3 and
$x_2$ = 3
Thanks in advance!
|
A general way to solve such is using generating functions. Define:
$\begin{align*}
g(z)
&= \sum_{n \ge 0} x_n z^n
\end{align*}$
Write your recurrence shifted (subtraction in indices gets messy), multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
2 x_{n + 3}
&= 3 x_{n + 2} + 8 x_{n + 1} + 3 x_n \\
2 \sum_{n \ge 0} x_{n + 3} z^n
&= 3 \sum_{n \ge 0} x_{n + 2} z^n
+ 8 \sum_{n \ge 0} x_{n + 1} z^n
+ 3 \sum_{n \ge 0} x_n z^n \\
2 \frac{g(z) - x_0 - x_1 z - x_2 z^2}{z^3}
&= 3 \frac{g(z) - x_0 - x_1 z}{z^2}
+ 8 \frac{g(z) - x_0}{z}
+ 3 g(z)
\end{align*}$
Solve this for $g(z)$ (substitute the initial values), express as partial fractions:
$\begin{align*}
g(z)
&= \frac{1 - 5 z}{1 - 2 z - 3 z^2} \\
&= \frac{1 - 5 z}{(1 + z) (1 - 3 z)} \\
&= - \frac{3}{2 (1 + z)}
+ \frac{1}{2 (1 - 3 z)}
\end{align*}$
We want the coefficient of $z^n$ of this. But this is just two geometric series:
$\begin{align*}
[z^n] g(z)
&= - \frac{3}{2} (-1)^n + \frac{1}{2} \cdot 3^n \\
&= \frac{3^n - (-1)^n}{2}
\end{align*}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
computing $\lim\limits_{(x,y) \to(0,0)} \frac{e^{x^2-y^2} -1}{ x - y}$ without Taylor series how can i compute this limit without using Taylor series?
$\lim\limits_{(x,y) \to(0,0)} \frac{e^{x^2-y^2} -1}{ x - y}$, while $x\neq y$.
i tried to put $r^2=x^2-y^2$ and transfer to polar but it didn't work for me.
Thank you for your help.
|
By making the substitution $z = e^{x^2-y^2}-1$, we obtain that
\begin{align*}
\lim_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} &= \lim_{z\rightarrow 0}\frac{z}{\ln(1+z)}\\
&= \lim_{z\rightarrow 0}\frac{1}{\ln\left((1+z)^{1/z}\right)}\\
&=1.
\end{align*}
Then,
\begin{align*}
\lim_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x-y} &= \lim_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} \frac{x^2-y^2}{x-y}\\
&=\lim_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} (x+y)\\
&=\lim_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} \lim_{(x, y)\rightarrow (0, 0)}(x+y)\\
&=0.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determinant of an $n\times n$ Toeplitz matrix Let $A = (a_{ij}) \in R^{n\times n}$. Find the determinant if:
$$a_{ij}= |i-j|$$
So we have the symmetric matrix
\begin{bmatrix}
0 & 1 & 2 & 3 & 4 & \dots & n-1 \\
1 & 0 & 1 & 2 & 3 & \dots & n-2 \\
2 & 1 & 0 & 1 & 2 & \dots & n-3 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
n-1 & n-2 & n-3 & \dots & \dots & \dots & 0
\end{bmatrix}
But i can't find a way to diagonalize the matrix nor find the determinant by Laplace expansion.
Any ideas ??
|
If we do row reduction, i.e. $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_2-R_1$, and then again do $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_3-R_2$ and finally $R_2+R_1$, then we get
\begin{align}
&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-2 \\
2 & 1 & 0 & 1 & 2 & \cdots & n-3 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
n-1 & n-2 & n-3 & \cdots & \cdots & \dots & 0
\end{vmatrix}\\ \ \\
=&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & -1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & -1 & -1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1&1&1&\cdots&\cdots&-1&-1 \\
1 & 1 & 1 & \cdots & \cdots & 1 & -1
\end{vmatrix}\\ \ \\
=&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\
0 & 2 & 0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0&\cdots&2&0&0 \\
0 & 0 & 0 & \cdots & \cdots & 2 & 0
\end{vmatrix}\\
\end{align}
Now we calculate the determinant and its minors all by the last row:
\begin{align}
=&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\
0 & 2 & 0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0&\cdots&2&0&0 \\
0 & 0 & 0 & \cdots & \cdots & 2 & 0
\end{vmatrix}\\ \ \\
=&-2\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\
0 & 2 & 0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0&\cdots&\cdots&2&0 \\
\end{vmatrix}\\ \ \\
&=(-1)^{n-2}2^{n-2}\begin{vmatrix}0&n-1\\1&n-1 \end{vmatrix}\\ \ \\
&=(-1)^{n-1}2^{n-2}(n-1).
\end{align}
|
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|
if $a ≡ b\pmod {2n}$ then prove $a^2 ≡ b^2 \pmod {4n}$
Let $n$ be positive number, if $a \equiv b \pmod{2n}$, prove that
$a^2 \equiv b^2 \pmod{4n}$.
By the congruence in hypothesis, we have $a-b = 2nk$ where $k$ is an integer.
Then $a = b+2nk$ and $a^2 = b^2+4n^2k^2+4knb$. From this we get $a^2-b^2 = 4kn(kn+b)$.
Now I have a question here. On the right hand side, we only know that $k$ and $4$ are integers because question says nothing about $n$ and $b$. So can we go and say that $\frac{a^2-b^2}{4n} = k(kn+b)$ is an integer?
|
Here's an alternative proof. Assume that $2n|(a-b)$. Then $a$ and $b$ are either both even or both odd, so $2|(a+b)$. Since $a^2-b^2 = (a-b)(a+b)$, we have that $4n|(a^2-b^2)$.
|
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|
Determining matrix $A$ and $B$, rectangular matrix Let $A$ be a $3\times 2$ matrix and $B$ be a $2\times 3$ be matrices satisfying $$AB=\begin{pmatrix} 8 & 2 & -2\\ 2 & 5 & 4\\ -2 & 4 & 5\end{pmatrix}$$
Calculate $BA$. How would you go for this problem? Do we start by noticing the matrix is symmetric? Any hints/ideas? Thanks
|
Using the suggestion of @WillJagy and the direct approach suggested by @Dac0, we have
$$
A =
\begin{bmatrix}
a & b\\
c & d\\
e & f\\
\end{bmatrix}\qquad
B =
\begin{bmatrix}
a & c & e\\
b & d & f\\
\end{bmatrix}
$$
and from $AB$ we get the following set of equations
\begin{align}
a^2 + b^2 &= 8\\
c^2 + d^2 &= 5\\
e^2 + f^2 &= 5\\
ac + bd &= 2\\
ae + fb &= -2\\
ec + fd &= 4\\
\end{align}
I solved this in WolframAlpha and tested with this particular solution(out of infinite solutions):
$$
A =
\begin{bmatrix}
-2 & -2\\
-2 & 1\\
-1 & 2\\
\end{bmatrix}\qquad
B =
\begin{bmatrix}
-2 & -2 & -1\\
-2 & 1 & 2\\
\end{bmatrix}
$$
Interestingly enough,
$$BA =
\begin{bmatrix}
9 & 0\\
0 & 9\\
\end{bmatrix}
$$
which corresponds to the solution proposed by @AndreasCap. It's also interesting to note that the columns of $A$ are orthogonal and their squared length are equal to $9$, the non-zero eigenvalue of $AB$ with multiplicity $2$.
I gave importance to @WillJagy suggestion because somehow it remembered me of the covariance matrix. I am not saying is the same, but there are nice similarities.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
My work:
We consider the congruences $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$, $x \equiv 5 \pmod 6$. We can reduce this further to $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$. We have
$N_1 = 4 \cdot 5 = 20 \implies 20 x_1 \equiv 1 \pmod{3} \implies 2x_1 \equiv 1 \pmod{3} \implies x_1 \equiv 2 \pmod {3}$
$N_2 = 3 \cdot 5 = 15 \implies 15x_2 \equiv 1 \pmod{4} \implies -x_2 \equiv 1 \pmod{4} \implies x_2 \equiv 3 \pmod {4}$
$N_3 = 3 \cdot 4 = 12 \implies 12 x_3 \equiv 1 \pmod{5} \implies 2x_3 \equiv 1 \pmod{5} \implies x_3 \equiv 3 \pmod {5}$
Now,
\begin{align*}
\bar x &= a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 \\
&= 3 \cdot 20 \cdot 2 + 4 \cdot 15 \cdot 3 + 5 \cdot 12 \cdot 3 \\
&= 480 \equiv 0 \pmod {3 \cdot 4 \cdot 5}
\end{align*}
Is this correct, or is something wrong in my work? I don't like how I have $0$ remainder.
|
We have
$$x \equiv 2 \pmod {3}$$
$$x \equiv 3 \pmod {4}$$
$$x \equiv 4 \pmod {5}$$
in the form $x\equiv a_i\pmod{m_i}$ and $M=m_1m_2m_3=3\cdot4\cdot5 = 60$
Then using chinese remainder theorem we have to find $b_i$ such that $b_i\dfrac{M}{m_i} = 1\pmod{m_i}$
Then
$$b_1 \cdot\frac{60}{3}\equiv 1 \pmod {3} \implies b_1 \equiv 20^{-1} \pmod {3} \implies b_1 = 2$$
$$b_2 \cdot\frac{60}{4}\equiv 1 \pmod {4} \implies b_2 \equiv 15^{-1} \pmod {4}\implies b_2 = 3$$
$$b_3 \cdot\frac{60}{5}\equiv 1 \pmod {5} \implies b_3 \equiv 12^{-1} \pmod {5}\implies b_3 = 3$$
Now
$$x\equiv a_1b_1\frac{M}{m_1}\pmod {M}+a_2b_2\frac{M}{m_2}\pmod {M}+a_3b_3\frac{M}{m_3}\pmod {M}$$
$$\implies x\equiv 359\pmod {60}\implies x\equiv 59\pmod {60}$$
smallest such $x$ is $x=59$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Improper integral $\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$ I have to determine the convergence of
$\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$
Can I say that
$(1-x^{2}\sin(\frac{1}{x^{2}}))\leq1+x^2$
and since $\int_0^3 1+x^2 dx$
is not even improper, the integral
$\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$ is convergent?
|
$$\int_0^3 \left(1-x^{2}\sin\left(\frac{1}{x^2}\right)\right)\ dx$$
$$=\int_0^3\ dx-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$
$$=3-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$
Since
$$\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx \leq \int_0^3 x^2\ dx$$
And
$$\lim\limits_{t\to 0^+}\int_t^3 x^2\ dx=\frac13\lim\limits_{t\to 0^+} \left(3^3-t^3\right)=9$$
Therefore by the integral comparison test,
$$\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx=\mbox{convergent}$$
Which implies that
$$\int_0^3 \left(1-x^{2}\sin\left(\frac{1}{x^2}\right)\right)\ dx=\mbox{convergent}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to Evaluate $\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$? How to find this limit without using L'Hospital rule
$$\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$
|
Use, binomial expansion of $(1-x)^{1/3}$ & Taylor's series expansion of $4^x$ & $3^x$ as follows $$\lim_{x\to 0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$
$$=\lim_{x\to 0}\frac{\left(1+\frac{\frac{1}{3}}{1!}(-x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(-x)^2+\ldots\right)-1}{\left(1+x\ln 4+\frac{x^2}{2!}(\ln 4)^2+\ldots\right)-\left(1+x\ln 3+\frac{x^2}{2!}(\ln 3)^2+\ldots\right)}$$
$$=\lim_{x\to 0}\frac{\frac{\frac{1}{3}}{1!}(-x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(-x)^2+\ldots}{\left(x\ln 4+\frac{x^2}{2!}(\ln 4)^2+\ldots\right)-\left(x\ln 3+\frac{x^2}{2!}(\ln 3)^2+\ldots\right)}$$
Dividing numerator & denominator by $x$,
$$=\lim_{x\to 0}\frac{-\frac{1}{3}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(x)+\ldots}{\left(\ln 4+\frac{x}{2!}(\ln 4)^2+\ldots\right)-\left(\ln 3+\frac{x}{2!}(\ln 3)^2+\ldots\right)}$$
$$=\frac{-\frac{1}{3}+0}{\left(\ln 4+0\right)-\left(\ln 3+0\right)}$$
$$=\color{red}{-\frac{1}{3\ln\left(\frac{4}{3}\right)}}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Another way to express $\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}$? I believe that the sum $$\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}$$ converges and it is about $1.85193$. Is there another way that this number can be expressed without summation notation. If not, perhaps there is a good approximation?
|
$\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)
=\sin\left(\frac{2\pi n}{2m+2}\right)
=\sin\left(\frac{\pi n}{m+1}\right)
$
so
$\frac1{n}\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)
=\frac1{n}\sin\left(\frac{\pi n}{m+1}\right)
=\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)
=\frac{\pi}{m+1}\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)
$.
Therefore
$\begin{array}\\
\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}
&=\sum_{n=1}^m\frac{\pi}{m+1}\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)\\
&=\frac{\pi}{m+1}\sum_{n=1}^m\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)\\
&=\frac{\pi}{m+1}\sum_{n=1}^m\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)\\
&=\frac{\pi}{m+1}\sum_{n=1}^m\frac{\sin\left(\frac{\pi n}{m+1}\right)}{\frac{\pi n}{m+1}}\\
&\to \int_0^{\pi} \frac{\sin(t)dt}{t}
\qquad\text{as } m \to \infty\\
&=1.8519370519824...\\
\end{array}
$
according to Wolfy.
|
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|
Coordinate of $S(s,t)$ for Which Area of Quadrilateral is Maximum.
Let $P(-2,3)\;\;,Q(-1,1)\;\;,R(s,t)$ and $S(2,7)$ be $4$ points in order on the parabola
$y=ax^2+bx+c$. Then the coordinates of $R(s,t)$ such that the area of Quadrilateral
$PQRS$ is maximum.
$\bf{My\; Try::}$ Here $P(-2,3)\;,Q(-1,1)\;\;,S(2,7)$ be the points lie on the parabola $y=ax^2+bx+c$
So Put $(x,y) = (-2,3)$ in $y=ax^2+bx+c\;,$ We get $4a-2b+c=3.................(1)$
Similarly Put $(x,y) = (-1,1)$ in $y=ax^2+bx+c\;,$ We get $a-b+c=1...........(2)$
Similarly Put $(x,y) = (2,7)$ in $y=ax^2+bx+c\;,$ We get $4a+2b+c=7...........(2)$
Now Solving These three equation , We get $a=1\;,b=1\;,c=1$
So we get equation of parabola be $y=x^2+x+1$
Now $\bf{Area\; of\; Quadrilateral\; PQRS = Area\; of \; \triangle PQR+Area\; of \; \triangle PRS}$
So Area Of Quadrilateral $$A(s,t) = \begin{vmatrix}
-2& 3 & 1\\
-1& 1 & 1\\
s & t & 1
\end{vmatrix}+\begin{vmatrix}
-2& 3 & 1\\
s& t & 1\\
2 & 7 & 1
\end{vmatrix}$$
So $$A(s,t) = 2s+t+1+20+4s-4t = 6s-3t+21$$
Now $R(s,t)$ lie on $y=x^2+x+1.$ So we get $t=s^2+s+1$
So we get $$A(s) = 6s-3(s^2+s+1)+21 = -3s^2+3s+18$$
So using Derivative Test $A'(s)=-6s+3$ and $A''(s)=-6$
So for Max. or min., Put $\displaystyle -6s+3=0\Rightarrow s=\frac{1}{2}$ and $\displaystyle t=\frac{1}{4}+\frac{1}{2}+1 = \frac{7}{4}$
My Question is can we solve it any Shorter way, Means Maximise area using Inequality or Geometrically
If yes Then plz explain here.
Thanks
|
All that calculation was really unnecessary. Here's a fast way:
The area of the quadrilateral PQRS is equal to the sum of the areas of the triangles $\triangle RQS$ and $\triangle PQS$. And since the area of the triangle $\triangle PQS$ is fixed, we need only to maximise the area of the $\triangle RQS$
Additionally since the base QS of the $\triangle RQS$ is also fixed, we need to just maximise the height corresponding to it. To do so we find the point R on the parabola where the tangent is parallel to the base QS.
Doing so, we have $y'=2x+1=2\implies x=\frac 12$ and therefore $R\equiv(\frac 12,\frac 74)$
|
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|
Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx = \int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx
$$
We can also replace $1-x$ in the denominator with $\sqrt{(1-x)^2}$
$$
\int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx = \int \frac{\sqrt{(1-x)^2}}{\sqrt{(1-x)(2x+1)}}\,dx
$$
If we simplify this fraction we get:
$$
\int \frac{\sqrt{1-x}}{\sqrt{2x+1}}\,dx
$$
Next we apply the following substitutions
$$
u = -x
$$
so : $-du = dx$
We can rewrite the integral as following:
$$-\int \frac{\sqrt{1+u}}{\sqrt{1-2u}}\,du$$
Then we apply another substitution:
$\sqrt{1+u} = t $ so $ \frac{1}{2\sqrt{1+u}} = dt $
We rewrite: $ \sqrt{1+u} $ to $\frac{1}{2}t^2 \,dt $
We can also replace $\sqrt{1-2u} $ as following:
$$\sqrt{-2t^2+3}=\sqrt{-2(1+u)+3}=\sqrt{1-2u}$$
With al these substitutions the integral has now the following form:
$$-\frac{1}{2}\int \frac{t^2}{\sqrt{-2t^2+3}}\,dt$$
Next we try to ''clean'' up the numerator:
$$-\frac{1}{2} \int \frac{t^2}{\sqrt{\frac{1}{2}(6-t^2)}} \, dt$$
$$-\frac{\sqrt{2}}{2} \int \frac{t^2}{\sqrt{6-t^2}} \, dt$$
And that's where I got stuck. I can clearly see that an arcsin is showing up in the integral but don't know how to get rid of the $t^2$.
|
Let $u^2=\frac{1-x}{1+2x}$. Then $x=\frac{1-u^2}{1+2u^2}$ and $\mathrm{d}x=-\frac{6u}{(1+2u^2)^2}\,\mathrm{d}u$.
Let $\sqrt2u=\tan(\theta)$, then $\sqrt2\,\mathrm{d}u=\sec^2(\theta)\,\mathrm{d}\theta$.
$$
\begin{align}
\int\frac{1-x}{\sqrt{1+x-2x^2}}\,\mathrm{d}x
&=\int\frac{1-x}{\sqrt{(1-x)(1+2x)}}\,\mathrm{d}x\\
&=\int\sqrt{\frac{1-x}{1+2x}}\,\mathrm{d}x\\
&=-\int\frac{6u^2}{(1+2u^2)^2}\,\mathrm{d}u\\
&=-\frac3{\sqrt2}\int\frac{\tan^2(\theta)}{\sec^4(\theta)}\sec^2(\theta)\,\mathrm{d}\theta\\
&=-\frac3{\sqrt2}\int\sin^2(\theta)\,\mathrm{d}\theta\\
&=-\frac3{2\sqrt2}\int(1-\cos(2\theta))\,\mathrm{d}\theta\\
&=-\frac3{4\sqrt2}(2\theta-\sin(2\theta))+C\\
&=-\frac3{2\sqrt2}\left(\theta-\frac{\tan(\theta)}{1+\tan^2(\theta)}\right)+C\\
&=-\frac3{2\sqrt2}\left(\arctan\left(\sqrt2u\right)-\frac{\sqrt2u}{1+2u^2}\right)+C\\
&=\frac12\sqrt{1+x-2x^2}-\frac3{2\sqrt2}\arctan\left(\sqrt{\frac{2-2x}{1+2x}}\right)+C\\
\end{align}
$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
System of linear and quadratic congruences Solve the system:
$x^2+x+1\equiv 0\pmod {7}$
$2x-4\equiv 0\pmod {6}$
Completing the square for first equation gives:
$$(2x+1)^2\equiv 4\pmod {7}$$
$$y^2\equiv 4\pmod{7}\Rightarrow y_1=2,y_2=5$$
$$2x+1\equiv 2\pmod{7}\Rightarrow (2,7)=1|7\Rightarrow x=4$$
$$2x+1\equiv 5\pmod{7}\Rightarrow (2,7)=1|4\Rightarrow x=2$$
Second equation:
$$2x-4\equiv 0\pmod{6}\Rightarrow (2,6)=2|4\Rightarrow x=2,x=1$$
One particular solution of the system is $x=2$.
How to find general solution?
|
You solved the quadratic congruence correctly, but the solutions of $2x-4\equiv 0\pmod{6}$ are $x\equiv \{2,5\}\pmod{6}$.
Divide everything by $\gcd(2,4,6)=2$: $$\iff x-2\equiv 0\pmod{3}\iff x\equiv 2\pmod{3}$$
$$\iff x\equiv \{2,5\}\pmod{6}$$
The solutions of $x^2+x+1\equiv 0\pmod{7}$ are $x\equiv \{2,4\}\pmod{7}$ (as you've found).
There are $4$ solutions of the whole system: $\begin{cases}x\equiv 2\pmod{7}\\x\equiv 2\pmod{6}\end{cases}$, $\begin{cases}x\equiv 4\pmod{7}\\x\equiv 2\pmod{6}\end{cases}$, $\begin{cases}x\equiv 2\pmod{7}\\x\equiv 5\pmod{6}\end{cases}$, $\begin{cases}x\equiv 4\pmod{7}\\x\equiv 5\pmod{6}\end{cases}$. Use Chinese Remainder Theorem to find the $4$ solutions of your system.
|
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|
Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further from there.
|
Let $x=2 u^2$:
$$2 \sqrt{2} \int du \, \frac{u}{\sqrt{1+u^2}} \arctan{u} = 2\sqrt{2} \sqrt{1+u^2} \arctan{u} - 2 \sqrt{2} \int \frac{du}{\sqrt{1+u^2}}$$
The latter integral is easily done using the sub $u=\sinh{v}$, so we have as the integral
$$2 \sqrt{x+2} \arctan{\sqrt{\frac{x}{2}}} - 2 \sqrt{2} \log{\left (\sqrt{\frac{x}{2}}+\sqrt{1+\frac{x}{2}} \right ) }+C$$
|
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|
$c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$ $a=\sqrt{57+40\sqrt2}-\sqrt{57-40\sqrt2}$ and $b=\sqrt{25^{\frac{1}{\log_85}}+49^{\frac{1}{\log_67}}}$ and $c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$.
I tried to solve and simplify this problem but no luck.Please help me.Thanks.
|
$$a^2=57+40\sqrt2+57-40\sqrt2-2\sqrt{57^2-40^2\cdot2}=2\cdot57-2\sqrt49=100$$
$$a=10$$
$$b=\sqrt{5^{2\log_5 8}+7^{2\log_7 6}}=\sqrt{8^2+6^2}=\sqrt100=10$$
$$x=a-\frac1a,\quad a=(7+5\sqrt2)^{\frac13}$$
$$x^3+3x-14=\left(a^3-3a+\frac3a-\frac1{a^3}\right)+3\left(a-\frac1a\right)-14\\
=a^3-\frac1{a^3}-14=7+5\sqrt2-\frac1{7+5\sqrt2}-14=7+5\sqrt2-\frac{7-5\sqrt2}{-1}-14\\
=0=c$$
|
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|
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
|
Actually there are solid arguments why $0/0$ can be defined as $0$.
Here is a graphic of the function $f(x,y)=x/y$:
The function is odd against both $x$ and $y$ variables.
Along the axis $x=0$ it is constant zero.
Along the axis $y=0$ it is unsigned infinity, but its Cauchy principal value is constant zero.
Along the diagonal $y=x$ it is constant 1.
Along the diagonal $y=-x$ it is constant -1.
As all odd functions its behavior around zero does not have any preference to positive or negative values. As all odd analytic functions, this function in zero(if defined) can either be zero or unsigned infinity. But unsigned infinity is not a part of affinely extended real line or complex plane, which is usually used in analysis (it is part of projectively extended real line/complex plane). As such, the only variant that remains is $0$.
Note also that we can define a Cauchy principal value of a two-variable function in a point $(x_0,y_0)$ as a limit of the following integral:
$$\lim_{r\to 0}\frac 1{2\pi}\int_0^{2\pi}f(x_0+ r \sin (t),y_0+r \cos(t)) dt$$
This integrates the values of a function over a circle around a point. For analytic functions as long as the radius of the circle r comes to 0, the value of the integral comes to the value of the function in $(x_0,y_0)$. For our function, in $x_0=0, y_0=0$ the value of the integral is constant zero (because the function is odd) and does not depend on r. Consequently, the limit is also zero.
We can extend our analysis to complex numbers, but we again will get the same result, $0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the area of the region that lies inside the first curve and outside the second curve. $r = 10 \cos\theta,\ r = 5$ I am not sure of my answer. In the figure, $r=10 \cos\theta$ is a circle that doesn't look like a circle.
The area of $r=5$ is $\pi r^2 = 25 \pi$.
You remove the area from $-\pi/3$ to $\pi/3$ of $10 \cos\theta$ from $25\pi$.
That is remove $$\frac 12 \int (10 \cos\theta)^2\,d\theta = 74.0105$$
Required area = $25 \pi - 74.0105 = 4.5293$
|
Let $S$ be the area of the red part in the following figure :
$\qquad\qquad\qquad$
You want to find $2S$.
$S$ can be obtained by
$$S=\frac 12\times 5^2\pi-[\text{sector $OAB$}]-[\text{the gray part}]\tag1$$
Since $\triangle{OAB}$ is an equilateral triangle, we have $\angle{AOB}=\frac{\pi}{3}$. So, $$[\text{sector $OAB$}]=\frac 12\times 5^2\times\frac{\pi}{3}=\frac{25}{6}\pi\tag2$$
Also,
$$[\text{the gray part}]=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac 12(10\cos\theta)^2 d\theta=25\left[\frac 12\sin(2\theta)+\theta\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}=\frac{25}{6}\pi-\frac{25}{4}\sqrt 3\tag3$$
From $(1)(2)(3)$, we have
$$2S=5^2\pi-2\cdot\frac{25}{6}\pi-2\left(\frac{25}{6}\pi-\frac{25}{4}\sqrt 3\right)=\color{red}{\frac{25}{3}\pi+\frac{25}{2}\sqrt 3}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
2 problems related to the number 2015
*
*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$
*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, but no result. For 2, I know that $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, but the proofs i found are over the elementary level the problem is aimed for. I also proved that it's smaller than $\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{1024^2} < 1$, but that obviously is missing $\frac{1}{2015}$
|
For 1 you can use the quadratic equation:
$n^2+n-p=0 \rightarrow n=\frac{-1 \pm \sqrt{4 \cdots 4 8 \cdots 89}}{2} $.
Now show $\sqrt{4 \cdots 4 8 \cdots 89}=\underbrace{66\cdots67}_\text{2015}$ and take the positive root to get $n=\underbrace{33\cdots3}_\text{2015}$
|
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|
Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$ How can i solve something like that?
$$\int\frac{x+1}{(x^2+7x-3)^3}dx$$
How should I start? Should I try rewrite it in partial fractions?
|
HINT:
As $x^2+7x-3=\dfrac{(2x+7)^2-61}4$ start with $2x+7=\sqrt{61}\sec y$
Alternatively,
as the highest power of $x$ in $(x^2+7x-3)^2$ is $4,$
differentiate $\dfrac{ax^3+bx^2+cx+d}{(x^2+7x-3)^2}$ wrt $x$
and compare with $\dfrac{x+1}{(x^2+7x-3)^3}$
|
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|
Proving of this trigonometric identity $$\frac{\cot \beta}{\csc \beta - 1} + \frac{\cot \beta}{\csc \beta + 1} = 2 \sec \beta$$
What I've done:
$$\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} +1} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} -1}=\\
=\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1-\sin\beta}{\sin \beta}} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1+\sin\beta}{\sin \beta}}=\\
=\cos \beta\frac{(1 - \sin\beta)}{\sin^2\beta} + \cos \beta\frac{1 + \sin\beta}{\sin^2\beta}=\\
=\frac{ \cos \beta - \cos\beta \sin\beta + \cos\beta + \cos\beta \sin\beta}{\sin^2 \beta}=\\
=\frac{2\cos\beta}{\sin^2\beta}=\\
=\frac{2\cos\beta}{1-\cos^2\beta}$$
Right side:
$$2 \sec\beta=\frac{1}{2\cos\beta}$$
Could you tell me where I went wrong? I've tried using a proof program online (symbolab) though those steps are a bit hard for me to follow.
Note: I only want to use what I have on the left side to solve the left side.
Thank you very much.
(Might be some errors first time using math jax..)
|
You first mistake was listening to your teacher. This has turned you into a $\sin$ / $\cos$ robot, causing you go against your better intuition and find a common denominator. So let's do that now
$$\begin{array}{lll}
\frac{\cot\beta}{\csc\beta-1}+\frac{\cot\beta}{\csc\beta+1}&=&\frac{\cot\beta}{\csc\beta-1}\cdot\frac{\csc\beta+1}{\csc\beta+1}+\frac{\cot\beta}{\csc\beta+1}\cdot\frac{\csc\beta-1}{\csc\beta-1}\\
&=&\cot\beta\bigg(\frac{\csc\beta+1}{\csc^2\beta-1}+\frac{\csc\beta-1}{\csc^2\beta-1}\bigg)\\
&=&\cot\beta\bigg(\frac{\csc\beta+1+\csc\beta-1}{\csc^2\beta-1}\bigg)\\
&=&\cot\beta\bigg(\frac{2\csc\beta}{\csc^2\beta-1}\bigg)\\
&=&\cot\beta\bigg(\frac{2\csc\beta}{\cot^2\beta}\bigg)\\
&=&2\frac{\csc\beta}{\cot\beta}\\
&=&2\csc\beta\tan\beta\\
\end{array}$$
At this point feel free to do the $\sin$/$\cos$ thingy, use the following identity
$$\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{\sin\beta}{\cos\beta}\cdot \frac{\frac{1}{\sin \beta \cos \beta}}{\frac{1}{\sin \beta \cos \beta}}=\frac{\frac{1}{\cos\beta}}{\frac{1}{\sin\beta}}=\frac{\sec\beta}{\csc\beta}$$
Continuing we have
$$2\csc\beta\tan\beta=2\csc\beta\cdot\frac{\sec\beta}{\csc\beta}=2\sec\beta$$
without the mess.
|
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|
If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\;,$ We get
$$k\sin C\left[k\sin A+k\sin B\right]\cdot \cos \frac{B}{2} = k\sin B\left[k\sin A+k\sin C\right]\cdot \cos \frac{C}{2} $$
So we get $$\sin C\left[\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\sin \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
Now Using $A+B+C=\pi\;,$ We get $\displaystyle \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$ and $\displaystyle \frac{A+C}{2}=\frac{\pi}{2}-\frac{B}{2}$
So we get $$\sin C\left[\cos \frac{C}{2}\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\cos \frac{B}{2}\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
So we get $$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
Now if we put $B=C\;,$ Then these two are equal.
My question is how can we prove it.
Help me, Thanks
|
with $\cos(\beta/2)=\sqrt{\frac{s(s-b)}{ac}}$ and $\cos(\gamma/2)=\sqrt{\frac{s(s-b)}{ab}}$ and $s=(a+b+c)/2$ we get after squaring
$$-1/2\,bc \left( b-c \right) \left( {a}^{3}+{a}^{2}b+{a}^{2}c+3\,abc+{
b}^{2}c+b{c}^{2} \right)
=0$$ thus $b=c$
|
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|
Solve $y^3-3y-\sqrt{2}=0$ using trigonometry This is a part of a larger question.
I had to show that for $4x^3-3x-\cos 3\alpha=0$ one of the solutions is $\cos \alpha$ and then find the other two solutions. Here they are:
$$4x^3-3x-\cos 3\alpha = (x-\cos \alpha)(2x+\cos \alpha + \sqrt{3} \sin \alpha)(2x+\cos \alpha - \sqrt{3} \sin \alpha)$$
I have to use the above and the following results: $\cos 15^{\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2\sqrt{2}}$ to find the solutions of the following:
$$y^3-3y-\sqrt{2}=0$$
I assumed that the constant term must be the equivalent of the cosine term and tried to find alpha so that I have one solution and then can derive the other. But since $\arccos {\sqrt{2}}$ is not trivially defined, this is not a correct approach. Or at least it is not correct the way I am doing it. Also I would have to do a bit more trigs since the second polynomial is not equivalent to the first one. There must be an easier, neater solution.
|
Thee trigonometric method relies on the trigonometric identity:
$$\cos 3\theta=4\cos^3\theta -3\cos\theta.$$
Set $\;y=A\cos\theta,\enspace A>0$. The equation becomes
$$A^3\cos^3\theta -3A\cos\theta=\sqrt2$$
We choose $A>0$ such that the left-hand side of the equation is proportional to the development of $\cos3\theta $, i.e.
$$\frac{A^3}4=\frac{3A}3\iff A^2=4.$$
Thus we set $y=2\cos\theta $, so that we get the trigonometric equation
$$2\cos3\theta =\sqrt2\iff \cos3\theta =\frac{\sqrt2}2=\cos\frac\pi4,$$
and the solutions are
$$3\theta \equiv\pm\frac\pi4\mod2\pi\iff\theta \equiv\pm\frac\pi{12}\mod\frac{2\pi}3$$
There are $3$ different values for $\cos\theta$, and the different values for $y$ are
$$2\cos\frac\pi{12},\quad2\cos\frac{3\pi}4=-\sqrt2,\quad2\cos\frac{-7\pi}{12}=-\sin\frac\pi{12}$$
Now, the addition formulae yield:
\begin{align*}
\cos\frac\pi{12}&=\cos\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6+\sqrt2}4,\\
\sin\frac\pi{12}&=\sin\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6-\sqrt2}4.
\end{align*}
|
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|
Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x^8\equiv3\pmod{13}$ has exactly $4$ incongruent solutions modulo $13$.
I was able to find on my calculator (using brute force) that these solutions are $4,6,7,$ and $9$ (i.e. $\pm4,\pm6$), but how would I go about finding them without using a calculator?
|
We can just list all the values $x^8$ takes for $x\in\Bbb{Z}/13\Bbb{Z}$:
$$\begin{array}{c|cccccccccccccc}
x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10& 11& 12\\
\hline
x^8 & 0 & 1 & 9 & 9 & 3 & 1 & 3 & 3 & 1 & 3 & 9 & 9 & 1 \\
\end{array}$$
If you use the fact that $(-x)^8=x^8$ and $(xy)^8=x^8y^8$, the only 'big' computations you need are
$$2^8\equiv256\equiv9\pmod{13}\qquad\text{ and }\qquad3^8\equiv6561\equiv9\pmod{13},$$
and even these can be broken down. Anyway, the solutions are clearly $4$, $6$, $7$ and $9$.
|
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|
How to solve $\sin{14x} - \sin{12x} + 8 \sin x - \cos{13x} = 4$? Find all solutions of $$\sin{14x} - \sin{12x} + 8 \sin x - \cos{13x} = 4$$ on an interval $(0^\circ, 360^\circ)$
|
Using the trigonometric identity $$\sin A-\sin B\equiv2\cos\frac{A+B}2\sin\frac{A-B}2,$$with $A=14x$ and $B=12x$, we can write the equation as $$(2\sin x-1)(\cos13x+4)=0.$$The general solution is therefore $$x=n\pi+(-1)^n\frac\pi6\quad(n\in\Bbb Z).$$For the range you specify, $x$ is $30^\circ$ or $150^\circ$.
|
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|
Prove that $\frac{\binom{n}{k}}{n^k} < \frac{\binom{n+1}{k}}{(n+1)^k}$ I have this math question that I'm kind of stuck on.
Prove that for all integers $1 < k \le n$, $$\frac{\binom{n}{k}}{n^k} <
\frac{\binom{n+1}{k}}{(n+1)^k}$$
I have to use mathematical induction on $k$ to prove this. So, I first check the base case $k=2$, I get $$f(2): \frac{\binom{n}{2}}{n^2} <
\frac{\binom{n+1}{2}}{(n+1)^2}$$$$\frac{n(n-1)}{2n^2} < \frac{n(n+1)}{2(n+1)^2}$$$$= \frac{n-1}{n} < \frac{n}{n+1}$$
Now I assume that $f(z): \frac{\binom{n}{z}}{n^z} <
\frac{\binom{n+1}{z}}{(n+1)^z}$ is true.
Check $f(z+1)$ is also true.
$$f(z+1): \frac{\binom{n}{z+1}}{n^{z+1}} <
\frac{\binom{n+1}{z+1}}{(n+1)^{z+1}}$$
$$= \frac{\frac{n!}{(z+1)!(n-z-1)!}}{n^{z+1}} < \frac{\frac{(n+1)!}{(z+1)!(n-z)!}}{(n+1)^{z+1}}$$$$= \frac{n!}{n^{z+1}(z+1)!(n-z-1)!} < \frac{(n+1)!}{(n+1)^{z+1}(z+1)!(n-z)!}$$
I'm not sure what to do from here. Thanks.
|
First,
I would just take a look
at their ratio.
$\begin{array}\\
\frac{\frac{\binom{n+1}{k}}{(n+1)^k}}
{\frac{\binom{n}{k}}{n^k}}
&=\frac{\frac{\frac{(n+1)!}{k!(n+1-k)!}}{(n+1)^k}}
{\frac{\frac{n!}{k!(n-k)!}}{n^k}}\\
&=\frac{\frac{(n+1)}{(n+1-k)}}{(1+1/n)^k}\\
&=\frac{\frac{(1+1/n)}{1+(1-k)/n}}{(1+1/n)^k}\\
&=\frac{1}{(1-(k-1)/n)(1+1/n)^{k-1}}\\
\end{array}
$
For $a > 2$,
$(1+1/n)^a
>1+a/n+a(a-1)/(2n^2)
$,
so
$\begin{array}\\
(1-a/n)(1+1/n)^a
&>(1-a/n)(1+a/n+a(a-1)/(2n^2))\\
&=(1+a/n+a(a-1)/(2n^2))-(a/n)(1+a/n+a(a-1)/(2n^2))\\
&=(1+a/n+a(a-1)/(2n^2))-(a/n)-(a^2/n^2)-a^2(a-1)/(2n^3))\\
&=1+((a(a-1)/2)-a^2)/(n^2))-a^2(a-1)/(2n^3))\\
&=1-(a(a+1)/2)/(n^2))-a^2(a-1)/(2n^3))\\
\end{array}
$
so it looks like
the denominator is less than $1$.
To show this,
let
$f(x)
=(1-ax)(1+x)^a
$.
Then
$f(0) = 1$
and
$\begin{array}\\
f'(x)
&=(1-ax)a(1+x)^{a-1}-a(1+x)^a\\
&=(1+x)^{a-1}(a(1-ax)-a(1+x))\\
&=(1+x)^{a-1}(a-a^2x-a-ax)\\
&=(1+x)^{a-1}(-a^2x-ax)\\
&=-ax(1+x)^{a-1}(a+1)\\
&<0
\end{array}
$
Therefore
$(1-(k-1)/n)(1+1/n)^{k-1}
< 1
$
so
$\frac{1}{(1-(k-1)/n)(1+1/n)^{k-1}}
> 1
$
and the terms increase.
|
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|
Number Theory: If $d\mid(4^n+1)$, show that $d$ is a sum of two squares I have a proof for the following problem, but I'm not sure if it's correct:
If $d\mid(4^n+1)$, show that $d$ is a sum of two squares.
Proof
$d\mid(4^n+1)\implies dm=4^n+1$, some $m\in\mathbb{Z}$.
Suppose that one of $d,m\equiv3\pmod4$. Then, $dm\equiv3\pmod4$ is not of the form $4^n+1$.
Next, suppose $d\equiv m\equiv3\pmod4$. Then $d=4u+3, m=4v+3$, some $u,v\in\mathbb{Z}\implies dm=16uv+4\cdot3u+4\cdot3v+9=4(4uv+3u+3v+2)+1$. But $4uv+3u+3v+2$ is not of the form $4^w$, hence $dm\neq4^n+1$.
Hence, we must have $d\equiv1\pmod4$, which means if a prime $p\mid d$ and $p\equiv3\pmod4$, then it occurs an even number of times in the prime factorization of $d$, so $d$ can be written as a sum of two integer squares.
Adjusted Proof
$d\mid(4^n+1)\implies dm=4^n+1$, some $m\in\mathbb{Z}$.
Let $d=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$, $p_i$ prime, $k_i\geq1$. Now suppose that some $p_i\mid d$ is congruent to $3$ mod $4$.
But then $p_i\mid4^n+1\implies 4^m=(2^m)^2\equiv-1\pmod{p_i}$, contradicting the fact that $x^2\equiv-1\pmod{p}$ has no solution for $p$ a prime, $p\equiv3\pmod4$.
Hence, we must have $d=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ with all $p_i\equiv1\pmod4$, so $d$ can be written as $d=a^2+b^2$, for some $a,b\in\mathbb{Z}$.
|
If $p$ is a prime divisor of $4^n+1$, then $p$ is odd and $(2^n)^2 \equiv -1\pmod{p}$. Therefore, $p \equiv 1\pmod{4}$. So if $d \mid (4^n+1)$ and $d>1$, then $d$ is a product of (not necessarily distinct) primes of the form $4k+1$. Each of these primes is a sum of two squares, and so $d$ is also a sum of two squares. On the other hand, $d=1=1^1+0^2$.
More generally, if $d \mid (a^2+b^2)$ with $\gcd(a,b)=1$, then we can again show that $d$ is a sum of two squares. For if $p$ is an odd prime divisor of $a^2+b^2$, then $b^2 \equiv -a^2\pmod{p}$ and $p \nmid ab$ (since $p \mid a$ $\Leftrightarrow$ $p \mid b$). But then $(ba^{-1})^2 \equiv -1\pmod{p}$, and again $p \equiv 1\pmod{4}$. The rest of the argument is the same as above, with the additional observation that $2=1^2+1^2$. $\blacksquare$
|
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|
Supremum and Infimum of set I just got the set and Ι tried to find the supremum and infimum and prove it. $$(x-2)\sqrt{\frac{x+1}{x-1}} \quad \text{ for } \quad 2< x\leq 54$$
I succeed to get to this set $\frac{x-2}{x-1}\sqrt{x^{2}-1}$ but I'm stuck.
What can I do now ?
Thanks.
|
On one hand, since $x>2$ we have
$$
(x-2)\sqrt{\frac{x+1}{x-1}}>0.
$$
On the other hand
$$
\lim_{x\rightarrow 2^+}(x-2)\sqrt{\frac{x+1}{x-1}}=0.
$$
Hence
$$
\inf_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=0.
$$
Since $2<x\leq 54$
$$
(x-2)\sqrt{\frac{x+1}{x-1}}=\frac{x-2}{x-1}\sqrt{x^2-1}=\left(1-\frac{1}{x-1}\right)\sqrt{x^2-1}\leq \left(1-\frac{1}{54-1}\right)\sqrt{54^2-1}.
$$
Hence
$$
\sup_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=\max_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=\left(1-\frac{1}{54-1}\right)\sqrt{54^2-1}.
$$
|
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|
Calculating square roots using the recurrence $x_{n+1} = \frac12 \left(x_n + \frac2{x_n}\right)$
Let $x_1 = 2$, and define $$x_{n+1} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right).$$ Show that $x_n^2$ is always greater than or equal to $2$, and then use this to prove that $x_n − x_{n+1} ≥ 0$. Conclude that $\lim x_n =
\sqrt2$.
My question:
So I know how to do this problem but I don't know how to prove that $x_n >0$ for all $n\in \mathbb{N}$.
|
From
$x_{n+1}
= \frac{1}{2} \left(x_n + \frac{2}{x_n}\right)
$
we get, squaring,
$x_{n+1}^2
= \frac{1}{4} \left(x_n^2 +4+ \frac{4}{x_n^2}\right)
$
so that
$\begin{array}\\
x_{n+1}^2 -2
&= \frac{1}{4} \left(x_n^2 -4+ \frac{4}{x_n^2}\right)\\
&= \frac{1}{4} \left(x_n - \frac{2}{x_n}\right)^2\\
&= \frac{1}{4} \left(\frac{x_n^2-2}{x_n}\right)^2\\
\end{array}
$
From this,
with the initial
$x_1$,
it is straightforward
to show that
$x_n \to \sqrt{2}$
and that the convergence is quadratic
(as it should be,
since this is Newton's iteration).
More generally,
using copy, paste, and edit:
From
$x_{n+1}
= \frac{1}{2} \left(x_n + \frac{a}{x_n}\right)
$
we get, squaring,
$x_{n+1}^2
= \frac{1}{4} \left(x_n^2 +2a+ \frac{a^2}{x_n^2}\right)
$
so that
$\begin{array}\\
x_{n+1}^2 -a
&= \frac{1}{4} \left(x_n^2 -2a+ \frac{a^2}{x_n^2}\right)\\
&= \frac{1}{4} \left(x_n - \frac{a}{x_n}\right)^2\\
&= \frac{1}{4} \left(\frac{x_n^2-a}{x_n}\right)^2\\
\end{array}
$
From this,
with the initial
$x_1$,
it is straightforward
to show that
$x_n \to \sqrt{a}$
and that the convergence is quadratic
(as it should be,
since this is Newton's iteration).
|
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|
Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
|
Let's grind it out. We want all ordered quadruples, but to get them it is enough to find all ordered triples $(a,b,c)$ where $a\le b\le c$. Then any triple of this kind we find can be permuted arbitrarily.
There cannot be triples $(a,b,c)$ with $a\le b\le c$ and $a\gt 2$, since for these we get that $3$ divides $a!+b!+c!$. So $a=1$ or $a=2$.
You don't seem to have considered $a=1$. So here goes. If $a=1$ then $b=c=1$ doesn't work. Let $b=1$. Then $c=2$ works, and we get the triple $(1,1,2)$. We also have the triple $(1,1,3)$. There is no triple of the shape $(1,1,k)$ for $k\gt 3$, for then $k!$ is divisible by $4$, so $1!+1!+k!$ cannot be a power of $2$.
Now look for triples of the shape $(1,b,c)$ where $b\ge 2$. There are none, for then $1!+b!+c!$ is odd.
Now look for triples of the form $(2,b,c)$. Try for $b=2$. Note that $c=2,3$ don't work, and $c\gt 3$ doesn't work, for in that case $8$ divides $c!$, so $2!+2!+c!$ cannot be a power of $2$.
So try for $b=3$. For $c$, calculate. We find, as you did, that $c=4$ and $c=5$ work. Nothing bigger than $5$ works, for if $c\gt 5$ then $c!$ is divisible by $16$, so $2!+3!+c!$ cannot be a power of $2$.
Now look at $b\ge 4$. This cannot work for any $c$, because then $2!+b!+c!$ has remainder $2$ on division by $4$.
|
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|
Question on Divergence Theorem Evaluate $\iint F.ds$
where $F= \frac{59}{3} x^3 \hat i + \frac{59}{3} y^3 \hat j + \frac{59}{3} z^3 \hat k$, and $S$ is the surface:
$S= \{(x,y,z) | x^2 + y^2 +z^2 =9\}$
Please explain in detail how to get the answer.
|
Simple application of the Divergence theorem:
$$\iint F\cdot ds = \iiint \nabla \cdot F dV$$
$$=\iiint \nabla \cdot (\frac{59}{3} x^3 \hat i + \frac{59}{3} y^3 \hat j + \frac{59}{3} z^3 \hat k) dV$$
$$=\iiint (59 x^2 \hat i + 59 y^2 \hat j + 59 z^2 \hat k) dV$$
$$=\iiint 59(x^2+y^2+z^2) dV$$
$$=59 \cdot \iiint r^2 \cdot r^2 \sin \theta \, dr \, d\theta \, d\phi$$
$$=59 \cdot \int_{0}^{3} r^4 \, dr \int_{0}^{\pi} \sin \theta \, d\theta \int_{0}^{2\pi} d\phi$$
$$=59 \cdot \frac{3^5}{5} \cdot 2 \cdot 2\pi $$
|
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|
How to find all pairs $(x,y)$ of integers such that $y^2 = x(x+1)(x+2)$? Here $y^2$ is divisible by $12.$ And satisfying all those conditions I think $y=0$ is the only solution. But I can't show it mathematically.
|
Setting $z=x+1$, we need $y^2 = z^3-z = z(z^2-1)$. Since $\gcd(z,z^2-1)=1$, we need $z=m^2$ and $z^2 - 1 = n^2$. This forces $z^2-n^2=1 \implies (z+n)(z-n) = 1 \implies z = \pm 1 \text{ and }n=0$. Apart from this, clearly $z=0$ is also a solution. Hence, the only solutions are $$(x,y) = (0,0);(-1,0);(-2,0)$$
|
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|
The functional equation $ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 $ I came across the functional equation
$$ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 \text . $$
So far I tried plugging $ x = f ( y ) $ and got $ f ( x ) = \frac { f ( 0 ) - x ^ 2 + 1 } 2 $ which holds for every $ x $ that is equal to $ f ( y ) $ for some $ y $. I suppose that $ f ( 0 ) = 1 $ and $ f ( x ) = 1 - \frac { x ^ 2 } 2 $, which would be true if I prove that $ f $ is surjective, which I still haven't proven. Could anybody help me?
|
You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is $ f ( x ) = 1 - \frac { x ^ 2 } 2 $. It's straightforward to check that this function indeed satisfies \eqref{0}. We prove the converse.
Let $ c = f ( 0 ) $ and put $ y = 0 $ in \eqref{0} to get
$$ f ( x - c ) - f ( x ) = c x + f ( c ) - 1 \text . \tag 1 \label 1 $$
In particular, setting $ x = 0 $ in \eqref{1}, we have $ f ( - c ) - c = f ( c ) - 1 $, which shows that $ c $ cannot be equal to $ 0 $. This mean that the right-hand side of \eqref{1} is a surjective function, and thus for any $ x \in \mathbb R $, there are $ a _ x , b _ x \in \mathbb R $ such that $ f ( a _ x ) - f ( b _ x ) = x $. Also, letting $ x = f ( y ) $ in \eqref{0} we have
$$ f \big( f ( y ) \big) = \frac { c + 1 - f ( y ) ^ 2 } 2 \text . \tag 2 \label 2 $$
Now, we can substitute $ f ( a _ x ) $ for $ x $ and $ b _ x $ for $ y $ in \eqref{0}, and use \eqref{2} to get
$$ f \big( f ( a _ x ) - f ( b _ x ) \big) = f \big( f ( a _ x ) \big) + f ( a _ x ) f ( b _ x ) + f \big( f ( b _ x ) \big) - 1 = \\
\frac { c + 1 - f ( a _ x ) ^ 2 } 2 + f ( a _ x ) f ( b _ x ) + \frac { c + 1 - f ( b _ x ) ^ 2 } 2 - 1 = c - \frac { \big( f ( a _ x ) - f ( b _ x ) \big) ^ 2 } 2 \text , $$
or equivalently
$$ f ( x ) = c - \frac { x ^ 2 } 2 \text . \tag 3 \label 3 $$
Finally, we can set $ x = f ( y ) $ in \eqref{3} and compare the result to \eqref{2} and see that $ c = 1 $, which implies $ f ( x ) = 1 - \frac { x ^ 2 } 2 $ for all $ x \in \mathbb R $.
|
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|
How to prove $\sin (x)+ \sin(y) = 2 \sin \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$ using addition theorems? I am stuck at one task, it's to proof the following equation using addition theorems.
From a draft I understood that the equation is correct or true. But that's it.
What's a good way to explain it mathematically?
$$\sin (x)+ \sin(y) = 2 \sin \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$$
As usual, any help is upvoted immediately.
|
The relevant addition formulae tell you that:
$$
\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)\\
\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)
$$
Applying these to $2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$ yields:
$$
\begin{align*}
2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) & = 2\left[\sin(x/2)\cos(y/2) + \cos(x/2)\sin(y/2)\right]\left[\cos(x/2)\cos(y/2) + \sin(x/2)\sin(y/2)\right]\\
& = 2\sin(x/2)\cos(x/2)\cos^2(y/2) + 2\sin^2(x/2)\sin(y/2)\cos(y/2)\\
& \hspace{20pt}+ 2\cos^2(x/2)\sin(y/2)\cos(y/2) + 2\sin(x/2)\cos(x/2)\sin^2(y/2)\\
& = 2\sin(x/2)\cos(x/2)\left[\sin^2(y/2) + \cos^2(y/2)\right]\\
& \hspace{20pt}+ 2\sin(y/2)\cos(y/2)\left[\sin^2(x/2) + \cos^2(x/2)\right]\\
& = 2\sin(x/2)\cos(x/2) + 2\sin(y/2)\cos(y/2)
\end{align*}
$$
From here, can you make the final step?
|
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|
Find the value of $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})$ Find the value of $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})$
Let $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})=x$
$\arcsin\frac{\sqrt{63}}{8}=4\arcsin x$
$\arcsin\frac{\sqrt{63}}{8}=\arcsin(4x\sqrt{1-x^2})(2x^2-1)$
$\frac{\sqrt{63}}{8}=(4x\sqrt{1-x^2})(2x^2-1)$
$\frac{63}{64}=16x^2(1-x^2)(2x^2-1)^2$
Let $x^2=t$
$\frac{63}{64}=16t(1-t)(2t-1)^2$
$64t^4-128t^3+80t^2-16t+\frac{63}{64}=0$
Now solving this fourth degree equation is getting really difficult,even after using rational roots theorem.
Is there a better method possible?The answer given is $x=\frac{\sqrt2}{4}$.Please help me.
|
Let
$$x = \arcsin{\frac{\sqrt{63}}{8}} = \arccos{\frac18} $$
Use succession of half-angle formulae:
$$\sin{\frac{x}{4}} = \sqrt{\frac{1-\cos{(x/2)}}{2}} = \sqrt{\frac{1-\sqrt{\frac{1+\cos{(x)}}{2}} }{2}} $$
In this case, $\cos{x} = 1/8$ so that
$$\sin{\left (\frac14 \arcsin{\frac{\sqrt{63}}{8}} \right )} = \sqrt{\frac{1-\sqrt{\frac{9/8}{2}} }{2}} = \frac1{2 \sqrt{2}}$$
|
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|
I would like to calculate $\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$ I want to calculate the following limit: $$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$$
or prove that it does not exist. Now I know the result is $-3$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
|
You may just observe that, $\color{#3366cc}{\sin (\pi/6) =1/2}$ and that, as $x \to \pi/6$,
$$
\require{cancel}
\frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}=\frac{\cancel{\color{#3366cc}{(2 \sin{x}-1)}}(\sin{x}+1)}{\cancel{\color{#3366cc}{(2 \sin{x}-1)}}(\sin{x}-1)}=\frac{\sin{x}+1}{\sin{x}-1} \color{#cc0066}{\longrightarrow} \frac{1/2+1}{1/2-1}=\color{#3366cc}{-3}
$$
Some details. One has
$$
\begin{align}
2u^2+u-1 &=2\left(u^2+\frac{u}2-\frac12 \right) \\
& = 2\left[\left(u +\frac14\right)^2-\frac1{16}-\frac12 \right] \\
& = 2\left[\left(u +\frac14\right)^2-\frac9{16}\right] \\
& = 2\left[\left(u +\frac14-\frac34\right)\left(u +\frac14+\frac34\right)\right] \\
& = 2\left[\left(u -\frac12\right)\left(u +1\right)\right] \\
& = \left(2u -1\right)\left(u +1\right)
\end{align}
$$ giving, with $u:=\sin x$,
$$
2 \sin^2{x}+\sin{x}-1=\color{#3366cc}{(2 \sin{x}-1)}\left(\sin x +1\right).
$$
Similarly,
$$
2 \sin^2{x}-3\sin{x}+1=\color{#3366cc}{(2 \sin{x}-1)}\left(\sin x -1\right).
$$
|
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|
Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$ Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$
Using Taylor series:
$$\ln(1+4^x)=\frac{2\cdot 4^x-4^{2x}}{2}+O(4^{2x}),\ln(1+3^x)=\frac{2\cdot 3^x-3^{2x}}{2}+O(3^{2x})\Rightarrow$$
$$\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}=\lim\limits_{x\to \infty}\frac{2\cdot 4^x-4^{2x}}{2\cdot 3^x-3^{2x}}=\infty$$
The limit should be $0$. Could someone point out what is wrong?
|
Use equivalents:
*
*$\ln(1+4^x)\sim_\infty\ln 4^x=x\ln 4$,
*similarly $\ln(1+3^x)\sim_\infty\ln 3^x=x\ln 3$,
hence $$\frac{\ln(1+4^x)}{\ln(1+3^x)}\sim_\infty\frac{x\ln 4}{x\ln 3}=\frac{\ln 4}{\ln 3}.$$
|
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|
If two integer triples have the same sum of 6th powers, then their sums of squares agree $\bmod 9$ Given $$a^6 + b^6 + c^6 = x^6 + y^6 + z^6$$
prove that $$a^2 + b^2 + c^2 - x^2 - y^2 - z^2 \equiv 0 \bmod{9}$$
I was thinking of using $n^6 \pmod{27}$ and showing both sides have the same pattern but it's getting really confusing..
|
Write
$$
a^6+b^6+c^6-x^6-y^6-z^6 = (a^2-x^2)(a^4+a^2x^2+x^4) + (b^2-y^2)(b^4+b^2y^2+y^4) + (c^2-z^2)(c^4+c^2z^2+z^4) ;
$$
by Thomas's comment, we can also pair off in such a way that $a,x$ are either both $\equiv 0\bmod 3$ or both $\not\equiv 0\bmod 3$, and similarly for $b,y$ and $c,z$.
The non-zero squares $\bmod 9$ are $1,4,7$, and now some experimentation shows that $a^4+a^2x^2+x^4\equiv 3\bmod 9$ if $a,x\not\equiv 0\bmod 3$. Thus the expression from above becomes
$$
3(a^2-x^2+b^2-y^2+c^2-z^2)
$$
when considered $\bmod 9$; note that this is also correct when the other case ($a,x\equiv 0\bmod 3$) applies. Since the difference of the sums of squares is $\equiv 0\bmod 3$ (by Thomas's comment again, and since $u^2\equiv 1\bmod 3$ for $u\not\equiv 0\bmod 3$), the claim follows now.
|
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|
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series and seems to me of the form:
$-3-7-11-15\ldots $
I feel like its of the closed form:
$\sum(-4i+1)$
So how do I prove that the equality is right?
|
Proceed inductively. Verify that for $k=1$, $1 = 1$.
Now suppose the result holds for $n - 1$. Then
$$ 1^2 + \dots + (-1)^{n-1} n^2 = 1 + \dots + (-1)^{n-2} (n-1)^2 + (-1)^{n-1} n^2 = (-1)^n \frac{n(n-1)}{2} + (-1)^{n-1} n^2 = (-1)^n \left(\frac{n(n-1)-2n^2}{2} \right) = (-1)^n \left(\frac{-n^2-n}{2} \right) = (-1)^{n-1} \left(\frac{n(n+1)}{2}\right)$$
The induction hypothesis was applied at the second equality.
|
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|
Divisibility set of problems It may seem as a begginer question, but i was wondering if there exist a certain method to solve problems like the following :
$$1) \text{ if } n ∈ \Bbb N \\ n(n^2+5)\,⋮\,6$$
$$2) \text{ if } n ∈ \Bbb N \\n^4+6n^3+11n^2+6n\,⋮\,24$$
$$3)\text{ if } n=2k \text{ then } \frac{n}{12}+\frac{n^2}{8}+\frac{n^3}{24}∈ \Bbb Z $$
NOTE:I havent dont this types of problems in a long time and it would be useful to see a detailed solution.Thanks!a
|
Another possible method is to write these numbers in terms of binomial coefficients which are always integers :
1) $$\frac{n(n^2+5)}{6}=\frac{n^3+5n}{6}=n+\frac{n^3-n}{6}=n+\frac{n(n+1)(n-1)}{3!}=n+\binom{n+1}{3}$$ which is an integer so $6 \mid n(n^2+5n)$
2) $$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$$ so $$\frac{n^4+6n^3+11n^2+6n}{24}=\frac{n(n+1)(n+2)(n+3)}{4!}=\binom{n+3}{4}$$ which is an integer .
3) $$\frac{k}{6}+\frac{k^2}{2}+\frac{k^3}{3}=\frac{2k^3+3k^2+k}{6}=\frac{k(k+1)(2k+1)}{6}$$
This can be seen either by this method :
This number is actually :
$$2\binom{k+1}{3}+\binom{k+1}{2}$$
Or you can do this in an awesome way noticing it's actually :
$$1^2+2^2+3^2+\ldots+k^2$$ (if you know this formula ) .
|
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|
Remainder of the numerator of a harmonic sum modulo 13 Let $a$ be the integer determined by
$$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}.$$
Determine the remainder of $a$ when divided by 13.
Can anyone help me with this, or just give me any hint?
|
Note that
$$a = \frac{23!}{1} + \frac{23!}{2} + \dots + \frac{23!}{23}$$
Since each term is an integer, $a$ is an integer.
To find $a \pmod{13}$ (if that's what you're asking), note that due to the factorial term, each of these fraction has at least a factor of $13$, except $\frac{23!}{13}$.
As such, you'll have to determine
$$a \equiv 1\cdot2\cdot\dots12\cdot14\cdot15\cdot\dots23 \pmod{13}$$
which can be done quickly using Wilson's Theorem:
$$12!\equiv-1\pmod{13}$$
and
$$\begin{align}14\cdot15\cdot\dots23&\equiv 1\cdot2\cdot\dots\cdot10 \pmod{13}\\ &\equiv 12!\cdot12^{-1}\cdot{11}^{-1}\pmod{13}\\
&\equiv 12!\cdot12\cdot6\pmod{13}\\
&\equiv -1 \cdot -1\cdot6\\
&\equiv 6\end{align}$$
Multiplying both together, we get
$$\begin{align}a&\equiv -6 \pmod{13}\\
&\equiv 7 \pmod{13}\end{align}$$
|
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Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of solution given is $\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq2+2+2=6$
I don't know how to proceed from the question to the first step of solution. Can anyone explain?
|
More easier and obvious answer:
$$\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\geq\frac{3}{2}$$
$$\frac{2a}{b+c}+\frac{2c}{a+b}+\frac{2b}{a+c}\geq3$$
$$\frac{2a+b+c}{b+c}+\frac{b+2c+a}{a+b}+\frac{c+2b+a}{a+c}\geq6$$
$$\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq6$$
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"timestamp": "2023-03-29T00:00:00",
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|
Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1}$$
Now put $\left(x-\frac{1}{x}\right)=t,x\ne0$. Then we get $$f(t) = \frac{t}{t^2+1} = \frac{1}{2}\left(\frac{2t}{1+t^2}\right)$$
Now put $t=\tan \theta$. Then $$f(\theta) = \frac{1}{2}\frac{2\tan \theta}{1+\tan^2 \theta} = \frac{1}{2}\sin 2\theta$$
So we get $$-\frac{1}{2}\leq f(\theta)\leq \frac{1}{2}\Rightarrow f(\theta)\in \left[-\frac{1}{2}\;,\frac{1}{2}\right]$$
My question is: Is my solution right? If not, then how can we solve it?
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we have $x^4-x^2+1=(x^2-\frac 12)^2+\frac {3}{4}≥0$$$\frac{1}{2}-f(x)=1/2*{\frac { \left( {x}^{2}-x-1 \right) ^{2}}{{x}^{4}-{x}^{2}+1}}\geq 0$$
and $$f(x)+\frac{1}{2}=1/2*{\frac { \left( {x}^{2}+x-1 \right) ^{2}}{{x}^{4}-{x}^{2}+1}}\geq 0$$
thus $$|f(x)| \le \frac{1}{2}$$
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|
Find the number of solutions of the equation $G=G'$ in $[0,2\pi]$ Let $G$ be the sum of infinite geometric series whose first term is $\sin\theta$ and common ratio is $\cos\theta$,while $G'$ be the sum of a different infinite geometric series whose first term is $\cos\theta$ and common ratio is $\sin\theta$.Find the number of solutions of the equation $G=G'$ in $[0,2\pi]$
I found $G=\frac{\sin\theta}{1-\cos\theta}$ and $G'=\frac{\cos\theta}{1-\sin\theta}$
According to the question,
$\frac{\sin\theta}{1-\cos\theta}=\frac{\cos\theta}{1-\sin\theta}$
$\sin\theta-\sin^2\theta=\cos\theta-\cos^2\theta$
$(\sin\theta-\cos\theta)(1-\sin\theta-\cos\theta)=0$
Either $\sin\theta-\cos\theta=0$ or $\sin\theta+\cos\theta=1$
$\sin\theta+\cos\theta=1$ gives $\theta=0,\frac{\pi}{2},2\pi$ but these value are not allowed as they make denominators zero.
$\sin\theta-\cos\theta=0$ gives $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.So there are two solutions.
But the book gives answer as no solutions.I do not understand why?Please help me point my mistake.
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So we have $G=\frac{\sin\theta}{1-\cos\theta}$ and $G'=\frac{\cos\theta}{1-\sin\theta}$,
an important condition for the sum of the geometric series is $-1<r<1$ where r is the common ratio, thus
$$\frac{\sin\theta}{1-\cos\theta}=\frac{\cos\theta}{1-\sin\theta}$$
$${\sin\theta}*(1-\sin\theta)={\cos\theta}*(1-\cos\theta)$$
$${\sin\theta}-{\cos\theta}=({\sin\theta})^2-({\cos\theta})^2$$
$${\sin\theta}-{\cos\theta}=({\sin\theta}-{\cos\theta})*({\sin\theta}+{\cos\theta})$$
$$1=({\sin\theta}+{\cos\theta})$$ with $({\sin\theta}-{\cos\theta})$ not zero.
Now we see that this equation holds for $\theta=\pi/2$ or $\theta=0$, but those solutions are in contradiction with initial conditions about the common ratio.
The other possible solution is $({\sin\theta}-{\cos\theta})=0$ thus $\theta=\pi/4$. I don't see contradictions with the initial conditions here, so this is the solution.
If we substitute the solution in the first equation we get
$$1+\sqrt{2}=1+\sqrt{2}$$
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|
How to solve the integral $\int\frac{x-1}{\sqrt{ x^2-2x}}dx $ How to calculate $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx $$
I have no idea how to calculate it. Please help.
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Notice,
$$\int \frac{x-1}{\sqrt{x^2-2x}}\ dx=\frac 12\int \frac{2(x-1)}{\sqrt{x^2-2x}}\ dx $$
$$=\frac 12\int \frac{d(x^2-2x)}{\sqrt{x^2-2x}}$$
$$=\frac 12\int (x^2-2x)^{-1/2}d(x^2-2x)$$
$$=\frac 12\frac{(x^2-2x)^{-\frac 12+1}}{-\frac 12+1}+C$$
$$=\frac 12\frac{(x^2-2x)^{1/2}}{1/2}+C$$
$$=\color{red}{\sqrt{x^2-2x}+C}$$
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate expression with $x^6$ So I'm trying to integrate this expression, but I'm not figuring out what's the best substitution to do...
$ \int \frac {1}{x^6+1} dx $
I tried to take $x^6 +1 $ and write $ (x^2 + 1) (x^4 -x^2 + 1) $ and then do partial functions, so I reach to the sum of two expressions.
One of them is easy to integrate...
But the other one:
$ \frac {x^2-1} {x^4 - x^2 + 1} $
I'm having trouble integrating...
Is there any easier method or is there a way to integrate this right from the beginning or can someone give a hint about how to integrate this last expression?
Thanks!
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HINT: the partialfrac decomposition is given by $$1/3\, \left( {x}^{2}+1 \right) ^{-1}+1/3\,{\frac {-{x}^{2}+2}{{x}^{4}-
{x}^{2}+1}}
$$
and $$x^4-x^2+1=(x^2-1/2)^2+\frac{3}{4}$$
$$x^4-x^2+1=(x^2+1)^2-3x^2$$ is better (see above) since we get a product finally we obtain
$$1/3\,\arctan \left( x \right) -1/12\,\sqrt {3}\ln \left( {x}^{2}-
\sqrt {3}x+1 \right) +1/6\,\arctan \left( 2\,x-\sqrt {3} \right) +1/12
\,\sqrt {3}\ln \left( {x}^{2}+\sqrt {3}x+1 \right) +1/6\,\arctan
\left( 2\,x+\sqrt {3} \right)
$$ for the first integral and
$$1/6\,\sqrt {3}\ln \left( {x}^{2}-\sqrt {3}x+1 \right) -1/6\,\sqrt {3}
\ln \left( {x}^{2}+\sqrt {3}x+1 \right)
$$ for the second one
|
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|
Show $\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}$. Let $p$ be prime and $d \ge 2$. I want to show that
$$
\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}.
$$
I have a proof, but I think it is complicated, and the statement appears in a book as if it is very easy to see. So is there any easy argument to see it?
My proof uses
$$
\frac{p^n - 1}{p-1} = 1 + p + \ldots + p^{n-1}.
$$
So
$$
\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} = \frac{(1 + p + \ldots + p^{d-1})(1 + p + \ldots + p^{d-2})}{p+1}
$$
If $d - 2$ is odd, set $u := (1 + p + \ldots + p^{d-1})$ and then proceed
\begin{align*}
& \frac{u(1 + p + \ldots + p^{d-2})}{p+1} \\
& = \frac{u(1+p) + u(p^2 + \ldots + p^{d-4})}{p+1} \\
& = u + p^2\cdot \frac{u(1+p+\ldots + p^{d-4})}{p+1} \\
& = u + p^2\cdot \left( \frac{u(1+p) + u(p^2 + \ldots p^{d-4})}{p+1} \right) \\
& = u + p^2\cdot \left( u + p^2\cdot \frac{u(1+p+\ldots p^{d-6})}{p+1} \right) \\
& \quad \qquad\qquad \vdots \\
& = u + p^2\cdot \left( u + p^2 \left( u + p^2\left( u + \ldots + p^2 \frac{u(1+p)}{1+p} \right) \right) \right) \\
& = u + p^2\cdot ( u + p^2 \cdot ( u + p^2 ( u + \ldots + p^2 u ))) \\
& \equiv u \pmod{p} \\
& \equiv 1 \pmod{p}
\end{align*}
and similar if $d-1$ is odd then successively multiply $(1+p+\ldots + p^{d-1})$ out with $v := (1+p+\ldots p^{d-2})$. But as said, this seems to complicated for me, so is there another easy way to see this?
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There are even easier proofs than the answers others have supplied. If you look at the numerator, one of the terms is guaranteed to be divisible by $p^2-1$, whichever has an even exponent. This is because $(x^2-1)|(x^{2n}-1)$. The division clearly results in a polynomial with constant term $1$, as does the division of the other term by $p-1$. Then their product does as well, and so taking the expression mod p gives 1.
Alternatively, we observe that
$p^d=0$ (mod p) and then just be done. This means that the fraction trivially simplifies to $$\frac{(-1)^2}{(-1)^2}=1$$
|
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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.