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why does the quotient become messy when we switch the terms of the divisor when using long division? If we were to use long divison to divide $x^2 - y^2$ by $x + y$, the long division easily gives $x-y$. But if I were to use long division to divide $x^2 - y^2$ by $y + x$ instead, the quotient becomes lengthy and fruitless. Forgive me if I'm fundamentally flawed. I don't seem to understand this at all.
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Dividing $x^2-y^2$ by $y+x$ gives:$$\frac{x^2}{y}-y-\frac{x^3}{y^2}+x+\frac{x^4}{y^3}-\frac{x^2}{y}-\frac{x^5}{y^4}+\frac{x^3}{y^2}+\frac{x^6}{y^5}-\frac{x^4}{y^3}-\frac{x^7}{y^6}+\frac{x^5}{y^4}+\frac{x^8}{y^7}-\frac{x^6}{y^5}-\cdots$$ The pattern of signs is + - - +, and every fractional term cancels with the fifth term following, leaving $-y + x$. So going at it this way is lengthy but not fruitless. It seems the order does not affect the result, only the ease of attaining it, at least in this example.
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form for $\prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$ Question: I am hoping to analytically continue the function:
$F(n) = \prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$
to $n=1/2$. My understanding is that means I will need a closed form for this product. What is a closed form of the above product? (Or are there other methods for performing the analytic continuation without having a closed form?)
Attempt: A similar product satisfies the identity
$\prod_{k=1}^{n-1} \sin(\frac{\pi k}{n}) = n 2^{1-n}$.
A proof of the above identity is given in another post (Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$). I have gone through the proofs given there and tried to extend the techniques used to my case, but no luck so far.
Any help is appreciated - a full solution, or just suggestions of theorems or related identities which may be of use.
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We break up the product in a different way (analogous to changing the order of products):
$$\begin{align}
F(n) &= \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \sin{\left(\frac{3\pi}{n}\right)} \\
&\quad \quad \vdots \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}.
\end{align}$$
(Note there are $n-1$ of the first term, $n-2$ of the second, down to one of the last). Now, consider $F(n)^2$: it is sufficient to do this and square root afterwards since every term in the finite product is positive, so the whole thing must be positive. Again, we reorder the terms:
$$\begin{align}
F(n)^2 &= \sin{\left(\frac{(n-1)\pi}{n}\right)} \sin{\left(\frac{(n-2)\pi}{n}\right)} \dotsm \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \times \sin{\left(\frac{(n-2)\pi}{n}\right)} \dotsm \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \quad \vdots \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \times \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}.
\end{align}$$
(so basically I've taken the triangle of terms, flipped it over and made a rectangle).
Now, $\sin{\left(\frac{(n-k)\pi}{n}\right)} = \sin{\left(\frac{k\pi}{n}\right)}$, so we can change the terms in the upper-right triangle as follows:
$$\begin{align}
F(n)^2 &= \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-1)\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \times \sin{\left(\frac{2\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-1)\pi}{n}\right)} \\
&\quad \quad \vdots \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \times \sin{\left(\frac{(n-1)\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}.
\end{align}$$
But of course this is just $n-1+1=n$ copies of $\prod_{k=1}^{n-1} \sin{\left( \frac{k\pi}{n} \right)} = n/2^{n-1}$, so we find
$$ F(n) = \sqrt{\frac{n^n}{2^{n(n-1)}}} = \frac{n^{n/2}}{2^{n(n-1)/2}}, $$
as conjectured by Claude Leibovici.
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"timestamp": "2023-03-29T00:00:00",
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|
exponential equation: $6^x+8^x+15^x=9^x+10^x+12^x$ What are the solutions of this equation? Or at least in which interval are they? $$6^x+8^x+15^x=9^x+10^x+12^x$$ I tried to find an increasing function, or use some inequalities but I got nothing out of it...
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Let $a=3^x$ and $b=2^x$ and $c=5^x$.
Then we have that
$$ab+b^3+ac=a^2+bc+ab^2$$
$$ab+b^3+ac-a^2-bc-ab^2=0$$
$$a(b-a)+b^2(b-a)-c(b-a)=0$$
$$(b-a)(a+b^2-c)=0$$
Now $a=b \implies x=0$ and $a+b^2-c=0 \implies 3^x+4^x=5^x$ which gives solution for $x=2$ only and not for any higher integer $x$ by Fermat's Last Theorem.
So these are the $2$ solutions.
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write the given matrix as a product of elementary matrices Please explain this in detail because i simply cannot understand previous explanations I have read for this type of problem.
By the way this is from elementary linear algebra 10th edition section 1.5 exercise #29. There is a copy online if you want to check the problem out.
Write the given matrix as a product of elementary matrices.
\begin{bmatrix}-3&1\\2&2\end{bmatrix}
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It took me a good 20 minutes to type this, so I'm gonna be pissed af if you don't read it.
Take the matrix $\begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix}$ and add $2/3$ times the first row to the second. You get $\begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$. Let $E_1$ be the elementary row matrix corresponding to the row operation you just did:
$$E_1 = \begin{pmatrix}1 & 0 \\ \frac{2}{3} & 1 \end{pmatrix}$$
Notice that $E_1 \begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$.
Next, take the matrix $\begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$ and add $-\frac{3}{8}$ times the second row to the first. You get $\begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$. Let $E_2$ be the the elementary row matrix corresponding to the row operation you just did:
$$E_2 = \begin{pmatrix} 1 & -\frac{3}{8} \\ 0 & 1 \end{pmatrix} $$
Notice that $E_2 \begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$.
Next, take $\begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$ and multiply the first row by $-\frac{1}{3}$. You get $\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$. Let
$$E_3 = \begin{pmatrix} -\frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}$$
Notice $E_3 \begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3}\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$
Now take $\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$ and multiply the bottom row by $-\frac{3}{8}$. You get $I =\begin{pmatrix} 1 &0\\0 & 1 \end{pmatrix}$. Let
$$E_4 = \begin{pmatrix} 1 & 0 \\ 0 & -\frac{3}{8} \end{pmatrix}$$
And notice $E_4\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix} = I$.
Now
$$I = E_4\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix} = E_4 E_3\begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3}\end{pmatrix} = E_4E_3E_2\begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$$
$$ = E_4E_3E_2E_1 \begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix}$$
and so
$$ \begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix} = E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}$$
where $E_1^{-1} = \begin{pmatrix} 1 & 0 \\ -\frac{2}{3} & 1 \end{pmatrix}, E_2^{-1} = \begin{pmatrix} 1 & \frac{3}{8} \\ 0 & 1 \end{pmatrix}, E_3^{-1} = \begin{pmatrix} -3 & 0 \\ 0 & 1 \end{pmatrix}, E_4^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & -\frac{8}{3} \end{pmatrix}$. Those are the elementary matrices you want.
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|
Geometric series and sequences $t_n$ is the $n^{th}$ term of an infinite sequence and $s_n$ is the partial sum of the first $n$ terms. Given that $t_1 = \frac{2}{4}$, $s_2 = \frac{4}{7}$, $t_3 = \frac{1}{35}$, $s_4 = \frac{8}{13}$. Determine a formula for $s_n$ and calculate its limit.
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$s_2 = t_1 + t_2, s_4 = s_2 + t_3 + t_4$ this is enough information to find $t_2, t_4$
$t_1,t_2,t_3, t_4 =$$ \frac 12, \frac 1{14}, \frac 1{35}, \frac 1{65}\\
\frac {1}{2\cdot 1}, \frac {1}{2\cdot 7}, \frac {1}{5\cdot 7},\frac {1}{5\cdot 13}$
Just a hunch, but it appears that the first factor in the denominator increases by $3$ every other term, and the second factor increases by $6$ every second term.
$t_{2k-1} = \frac {1}{(3k-1)(6k-5)}\\
t_{2k} = \frac {1}{(3k-1)(6k+1)}$
$s_{2k} = $$s_{2k-2} + \frac {1}{(3k-1)(6k-5)} + \frac {1}{(3k-1)(6k+1)}\\
s_{2k-2} + \frac {12k-4}{(3k-1)(6k-5)(6k+1)}\\ s_{2k-2} + \frac {4}{(6k-5)(6k+1)}\\ s_{k-2} + \frac {2}{3(6k-5)} - \frac {2}{3(6k+1)}\\
\sum_\limits{i=1}^{k} \left(\frac {2}{3(6i-5)} - \frac {2}{3(6i+1)}\right)$
We have a telescoping series.
$s_{2k} = \frac 23 - \frac {2}{3(6k+1)}$
as k goes to infinity
$s = \frac 23$
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"timestamp": "2023-03-29T00:00:00",
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|
Prove an identity involving factorials inside summations Prove that:
\begin{align*}\sum _{k=0}^N\left(k\cdot \frac{N!}{k!\left(N-k\right)!}\cdot \frac{\left(X+k-1\right)!}{\left(X-1\right)!}\cdot \frac{\left(L-X+N-k-1\right)!}{\left(L-X-1\right)!}\frac{\left(L-1\right)!}{\left(L+N-1\right)!}\right)&= \frac{XN}{L}
\end{align*}
for $[0<X<L]\in \mathbb{N}$
My attempt using induction: (I'll leave out the verification of the base cases, but they do hold). Factoring the non-iterates out and moving them to the other side of the equality of the induction hypothesis:
\begin{align*}
\frac{N!}{\left(X-1\right)!\left(L-X-1\right)!}\sum _{k=0}^N\frac{k}{k!\left(N-k\right)!}\cdot \left(X+k-1\right)!\cdot \left(L-X+N-k-1\right)!&=\frac{XN}{L!}\left(N+L-1\right)!
\end{align*}
From the domain of $X$ and $L, \quad X-1 \geq 0$ and $L-X-1 \geq 0$. The $(X+k-1)!$ term can be written as $(k+X-1)(k+X-2)...(k+1)k!$ -- similarly, the $\left(L-X+N-k-1\right)!$ term is expressible as $(N-k+L-X-1)(N-k+L-X-2)...(N-k+1)(N-k)!$ We can cancel the factorial terms in the denominator of the first term of the summation and express the rising factorials using product notation. This returns the induction hypothesis that will be used.
\begin{align*}
\frac{N!}{(X-1)!(L-X-1)!}\sum_{k=0}^N\left(k\prod_{i=1}^{X-1}(k+i)\prod_{i=1}^{L-X-1}(N-k+i)\right) &= \frac{XN}{L!}\left(N+L-1\right)!
\end{align*}
Inductive step:
First we note the required form: $\frac{X(N+1)}{L!}(L+N)!$.
Through some algebraic manipulation, we arrive at the following.
\begin{align*}
&(N+1)\left(\frac{N!}{(X-1)!(L-X-1)!}\sum_{k=0}^{N}\left(k\prod_{i=1}^{X-1}(k+i)\prod_{i=1}^{L-X-1}(N+1-k+i)\right)+\frac{(X+N)!}{(X-1)!}\right)
\end{align*}
Because of the $N+1$ nested in the second product of the summation, there does not seem to be a way to insert the induction hypothesis. If one eliminates the $+1$ via altering the limits of the product, a new $k$ term appears:
Inductive step:
\begin{align*}
&(N+1)\left(\frac{N!}{(X-1)!(L-X-1)!}\sum_{k=0}^{N}\left(k\prod_{i=1}^{X-1}(k+i)\prod_{i=2}^{L-X}(N-k+i)\right)+\frac{(X+N)!}{(X-1)!}\right)
\end{align*}
Compared to the induction hypothesis:
\begin{align*}
\frac{N!}{\left(X-1\right)!\left(L-X-1\right)!}\left(\sum _{k=0}^N\left(k\left(\prod _{i=1}^{X-1}\left(k+i\right)\right)\left(\prod _{i=2}^{L-X}\left(N-k+i\right)\right)\left(\frac{\left(N-k+1\right)}{N-k+L-X}\right)\right)\right)
\end{align*}
Is there any way to resolve this, or is a proof by induction not possible?
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Here is an approach without using induction. We start with the left-hand side, do some simplifications to finally obtain the right-hand side.
We obtain
\begin{align*}
\sum _{k=0}^N& k\cdot \frac{N!}{k!\left(N-k\right)!}\cdot \frac{\left(X+k-1\right)!}{\left(X-1\right)!}\cdot \frac{\left(L-X+N-k-1\right)!}{\left(L-X-1\right)!}\frac{\left(L-1\right)!}{\left(L+N-1\right)!}\\
&=\frac{N!(L-1)!}{(L+N-1)!}\sum_{k=1}^N\frac{(X+k-1)!}{(k-1)!(X-1)!}\cdot\frac{(L-X+N-k-1)!}{(N-k)!(L-X-1)!}\tag{1}\\
&=\binom{L+N-1}{N}^{-1}\sum_{k=1}^N X\binom{X+k-1}{k-1}\binom{L-X+N-k-1}{N-k}\tag{2}\\
&=X\binom{L+N-1}{N}^{-1}\sum_{k=0}^{N-1}\binom{X+k}{k}\binom{L-X+N-k-2}{N-k-1}\tag{3}\\
&=X\binom{L+N-1}{N}^{-1}\sum_{k=0}^{N-1}\binom{-X-1}{k}(-1)^k\binom{-L+X}{N-1-k}(-1)^{N-1-k}\tag{4}\\
&=(-1)^{N-1}X\binom{L+N-1}{N}^{-1}\binom{-L-1}{N-1}\tag{5}\\
&=X\binom{L+N-1}{N}^{-1}\binom{L+N-1}{N-1}\tag{6}\\
&=\frac{XN}{L}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we do some rearrangements, factor out terms independent of the index variable $k$ and start with index $k=1$ since the left-hand expression contains the factor $k=0$.
*In (2) we rewrite the expression using binomial coefficients.
*In (3) we shift the index by one to start with $k=0$.
*In (4) we use the rule $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (5) we apply Vandermonde's identity.
*In (6) we apply again the rule from (4).
Note: A proof by induction is also possible. But, if a direct transformation is within reach, this often provides more insight. The main aspect here which connects the left-hand side and the right-hand side is an application of Vandermonde's identity. We would probably not see this, when proving the identity using induction.
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For $a, b, c$ is the length of three sides of a triangle. Prove that $\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$ For $a, b, c$ is the length of three sides of a triangle. Prove that $$\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$$
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HINT: use that we get for the sum $$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b) (a-c) (b-c)}{(a+b) (a+c) (b+c)}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Show $|-\ln(\cos (x))-\frac{x^2}2| \le \frac23|x^3| \quad\forall x \in[-\frac{\pi}{4},\frac{\pi}{4}] $
Let
$$f:(-\frac{\pi}{2},\frac{\pi}{2}) \to \Bbb R$$
$$f(x)=-\ln(\cos (x))$$
Show that
$$\left| f(x)-\frac{x^2}2 \right| \le \frac23\left| x^3 \right| \qquad x
\in[-\frac{\pi}{4},\frac{\pi}{4}]$$
My attempt:
We have $\left| -\ln(\cos (x))-\frac{x^2}2 \right| \le \frac23\left| x^3 \right| \iff |\ln(\cos
(x))+\frac{x^2}2\left| \le \frac23 \right|x^3| $
The first thing that comes to my mind is to show that $\cos(\alpha)\ge
\frac{1}{\sqrt2}$ for $\alpha \in[-\frac{\pi}{4},\frac{\pi}{4}]$ (i)
After that, I would have to go on to show that $0 \ge \ln(\beta)$ for
$\beta \in[\frac{1}{\sqrt2},1]$ (ii) which I also don't know how to
prove.
Once that is done we have
$$\left| \frac{x^2}2 \right| \le \frac23\left| x^3 \right|$$ and we are done.
So my question is: how can I go on about (i) and (ii)?
Thanks in advance.
|
Applying Taylor's theorem with remainder to
$$ \begin{aligned}
f(x) &= -\ln(\cos(x)) & f(0) &= 0 \\
f'(x) &= \tan(x) & f'(0) &= 0\\
f''(x) &= 1 + \tan^2(x) & f''(0) &= 1 \\
f'''(x) &= 2 \tan(x) (1+\tan^2(x))
\end{aligned}$$
gives
$$
f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(\xi)}{3!} x^3
= \frac{x^2}{2} + \frac{f'''(\xi)}{6} x^3
$$
for some $\xi$ between $0$ and $x$. The assertion follows because
for $|\xi| \le |x| \le \frac{\pi}{4}$
$$
|\tan(\xi)| \le 1 \Longrightarrow
|f'''(\xi)| \le 4
$$
and therefore
$$
\left\lvert f(x) - \frac{x^2}{2} \right\rvert =
\frac{|f'''(\xi)|}{6} |x|^3 \le \frac 23 |x|^3
$$
Remark: Your approach cannot work because
$$
\left| \frac{x^2}2 \right| \le \frac23\left| x^3 \right|
$$
does not hold for $x$ close to zero.
|
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|
For $x,y,z>0$. Minimize $P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$ For $x,y,z>0$ and $xy+yz+xz=1$, minimize $$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$$
My try: Let $xy=a; yz=b;zx=c \Rightarrow a+b+c=1$
$\Rightarrow x^2=\frac{ac}{b};y^2=\frac{ab}{c};z^2=\frac{bc}{a}$
Hence $$P=\sum \frac{1}{\frac{4ac}{b}-b+2}=\sum \frac{1}{\frac{4ac}{b}-b+2(a+b+c)}=\sum \frac{1}{\frac{4ac}{b}+2a+b+2c}=\sum \frac{b}{4ac+2ab+b^2+2bc}$$
$$P=\sum \frac{b}{(2a+b)(2c+b)} \geq \sum \frac{4b}{(2a+2b+2c)^2}=\sum \frac{b}{(a+b+c)^2}=1$$
And i need new method ?
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If $x=y=z=\frac{1}{\sqrt3}$ we get $P=1$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{4x^2-yz+2}\geq1$$ or
$$\sum_{cyc}\left(\frac{1}{4x^2-yz+2}-\frac{1}{3}\right)\geq0$$ or
$$\sum_{cyc}\frac{1-4x^2+yz}{4x^2-yz+2}\geq0$$ or
$$\sum_{cyc}\frac{xy+xz+2yz-4x^2}{4x^2-yz+2}\geq0$$ or
$$\sum_{cyc}\frac{(z-x)(2x+y)-(x-y)(2x+z)}{4x^2-yz+2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{2y+z}{4y^2-xz+2}-\frac{2x+z}{4x^2-yz+2}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(z^2+2xy+2)(4z^2-xy+2)\geq0$$
Done!
|
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|
If $\frac{1}{n}A^n = \frac{1}{n+1}A^{n+1} = \frac{1}{n+2}A^{n+2}$ then any eigenvalue of $A$ is $0$, please check my proof I'd like to show that for a square matrix A with the property that there exists a positive integer n such that $\dfrac{1}{n}A^n = \dfrac{1}{n+1}A^{n+1} = \dfrac{1}{n+2}A^{n+2}$, any eigenvalue of A must be 0.
I proceed as follows:
Let $\lambda$ be an eigenvalue of A, and $v$ be an eigenvector of A, such that $v \neq 0$.
Recall: $A^nv=\lambda^nv$, then since
\begin{align}
\dfrac{1}{n}A^n &=\dfrac{1}{n+1}A^{n+1} \\
\Rightarrow \dfrac{n+1}{n}A^nv-A^{n+1}v &=0 \\
\Rightarrow \lambda^n\left(\dfrac{n+1}{n}-\lambda\right)v &= 0 \\
\end{align}
$\Rightarrow \lambda = 0$ or $\lambda = \dfrac{n+1}{n}$
Similarly:
\begin{align}
\dfrac{1}{n+1}A^{n+1} &=\dfrac{1}{n+2}A^{n+2} \\
\Rightarrow \dfrac{n+2}{n+1}A^{n+1}v-A^{n+2}v &=0 \\
\Rightarrow \lambda^{n+1}\left(\dfrac{n+2}{n+1}-\lambda\right)v &= 0 \\
\end{align}
$\Rightarrow \lambda = 0$ or $\lambda = \dfrac{n+2}{n+1}$
My conclusion is since it's mathematically impossible that $\lambda = \dfrac{n+1}{n} = \dfrac{n+2}{n+1}$, the only eigenvalue of A must be 0.
Is my demonstration incomplete or incorrect?
|
Essentially correct. You should stick to mathematical english or to $\Rightarrow$ language, though and not mix them up as you do in the first block of the proof.
|
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|
Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ , $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\times h$ where $r_1,r_2,r_3$ in $C[x,y,z]$ but then i realised that i cant possibly do that by hand.This was an exercise on my first computational algebra course so we havent really learnt anything much yet. I would like a hint if possible.
|
$$2(xy+yz+zx) = (x+y+z)^2 - (x^2+y^2+z^2) = 4$$
$$-3xyz = (x+y+z)^3 - (x^3+y^3+z^3) - 3(xy+yz+zx)(x+y+z) = 27 - 25 = 2$$
Thus $x,y,z$ are the roots of
$$t^3 -3t^2 + 2t + \frac{2}{3} = 0$$
Now
\begin{align*}
t^4 = 3t^3 -2t^2 -\frac{2}{3}t
\end{align*}
Putting $t=x,y,z$ in the above and adding, we get
$$x^4 + y^4 + z^4 = 3 \sum x^3 - 2\sum x^2 - \frac{2}{3} \sum x = 21 - 10 -2 = 9$$
Also,
$$t^5 = 3t^4 - 2t^3 - \frac{2}{3} t^2$$ and hence
$$\sum x^5 = 3\sum x^4 - 2 \sum x^3 - \frac{2}{3} \sum x^2 = 27 - 14 - 10/3 = 29/3$$
|
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|
Find the value of a given integral sequence: $I_n=\int_{0}^{1}(1-x^2)^ndx$ Let $(I_n)_{n \geq 1}$ be a sequence such that:
$$I_n = \int_0^1 (1-x^2)^n dx$$
Find the value of $I_n$ (The solution is $\frac{2}{3} \cdot \frac{4}{5} \cdot ... \cdot \frac{2n}{2n+1}$).
I've tried using the binomial expansion of $(1-x^2)^n$ but I couldn't get the proper answer.
Thank you!
|
Here is an approach that relies on the Beta function $\operatorname{B}(m,n)$, namely
$$\operatorname{B} (m,n) = \int_0^1 x^{m - 1} (1 - x)^{m - 1} \, dx.$$
In the integral for $I_n$ we begin by enforcing a substitution of $x \mapsto \sqrt{x}$. Thus
\begin{align}
I_n &= \frac{1}{2} \int_0^1 x^{-1/2} (1 - x)^n \, dx\\
&= \frac{1}{2} \int_0^1 x^{\frac{1}{2} - 1} (1 - x)^{(n + 1) - 1} \, dx\\
&= \frac{1}{2} \operatorname{B} \left (\frac{1}{2}, n + 1 \right )\\
&= \frac{1}{2} \frac{\Gamma (\frac{1}{2}) \Gamma (n + 1)}{\Gamma (n + \frac{3}{2} )}\\
&= \frac{\sqrt{\pi}}{2} \frac{\Gamma (n + 1)}{(n + \frac{1}{2}) \Gamma (n + \frac{1}{2})}.
\end{align}
Since $n$ is a positive integer, we have
$$\Gamma (n + 1) = n! \quad \text{and} \quad \Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)!}{2^{2n} n!} \sqrt{\pi},$$
then
$$I_n = \frac{2^{2n}}{2n + 1} \frac{(n!)^2}{(2n)!} = \frac{2^{2n}}{(2n + 1) \binom{2n}{n}}.$$
Here $\binom{2n}{n}$ denotes the central binomial coefficient.
|
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|
Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
|
Squaring both sides:
$x^{2} - x +\frac{2}{x} - 1 = 2\sqrt{(x-\frac{1}{x})(1-\frac{1}{x})}$
Squaring again we get:
$x^{4} - 2x^{3} - x^{2} + \frac{4}{x^{2}} + 6x -\frac{4}{x} - 3 = 4(\frac{1}{x^{2}} + x - \frac{1}{x} - 1)$
We can reduce this to:
$x^{4} -2x^{3} -x^{2} +2x + 1 = 0$
Notice that this factorises into $(x^{2} - x -1)^{2} = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{5}}{2}$
Note this is the 'golden ratio' $\phi = \frac{1 + \sqrt{5}}{2}$ with the property that: $\frac{1}{\phi} = \phi - 1$
Since we have squared our initial equation thus introducing extra solutions, we must test each of these in the original equation.
This leads us to reject $x = \frac{1 - \sqrt{5}}{2}$
Therefore $x= {\frac{1 +\sqrt{5}}{2}}$
|
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|
Regarding complex roots of a polynomial I am having difficulty in finding the roots of the following polynomial when it is given that all the roots are complex:-
$$f(x)= x^4+4x^3+8x^2+8x+4$$
How can I factorize the polynomial to get its roots?
|
Hint:
$$
\begin{align}
f(x)= x^4+4x^3+8x^2+8x+4 & = x^2\left(x^2+\frac{4}{x^2} + 4\left(x + \frac{2}{x}\right) +8\right) \\
& = x^2\left(\left(x+\frac{2}{x}\right)^2 + 4\left(x + \frac{2}{x}\right) +4\right) \\
& = x^2\left(x+\frac{2}{x} + 2\right)^2 = \;\cdots
\end{align}
$$
[ EDIT ] The above is based on the observation that the ratio between "symmetric" coefficients of the polynomial $\;\color{blue}1\cdot x^4+\color{red}4\cdot x^3+8\cdot x^2+\color{red}8\cdot x+\color{blue}4 \cdot x^0\;$ follows the powers of $\frac{1}{2}$ from center outwards: $\;\cfrac{8}{8} = \cfrac{1}{2^0}\,, \;\color{red}{\cfrac{4}{8}}=\cfrac{1}{2^1}\,, \;\color{blue}{\cfrac{1}{4}}=\cfrac{1}{2^2}\,$.
This indicates that the polynomial is amenable to a reciprocal polynomial with the substitution $x=\sqrt{2}\,y\,$, and indeed $\color{blue}4\,y^4+\color{red}{8 \sqrt{2}}\,y^3 + 16\,y^2+\color{red}{8\sqrt{2}}\,y+\color{blue}4\,$ is reciprocal.
Lastly, a reciprocal polynomial of even degree $2n$ can be factored as $y^n$ times a polynomial of degree $n$ in $y+\cfrac{1}{y}=\cfrac{1}{\sqrt{2}}\left(x+\cfrac{2}{x}\right)\,$.
This explains why the posted answer started by factoring out $x^2$ then focused on the $x+\cfrac{2}{x}$ terms.
|
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|
Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\sqrt 2 t+1}dt+\frac{1}{2}\int\frac{1+\sqrt 2t}{t^2+\sqrt 2t+1}dt$$
Using substitutions $$u=t^2-\sqrt 2t+1$$ and $$v=t^2+\sqrt 2t+1$$
we get $$\int\frac{1-t^2}{1+t^4}dt=-\frac{\sqrt 2}{4}\ln|u|+\frac{\sqrt 2}{4}\ln|v|=-\frac{\sqrt 2}{4}\ln|t^2-\sqrt 2t+1|+\frac{\sqrt 2}{4}\ln|t^2+\sqrt 2t+1|$$
From $$x=\frac{1+t^2}{1-t^2}\Rightarrow t=\sqrt{\frac{x-1}{x+1}}\Rightarrow$$
$$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx=$$$$-\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}-\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}+\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+c$$
Is there another, quicker method to solve this integral, rather than Euler substitution.
|
Setting $$x=\sinh(t)$$ then we have $$dx=\cosh(t)dt$$ and our integral will be $$\int \frac{dt}{1+\sinh^2(t)}$$
then use $$\sinh(t)=-2\,{\frac {\tanh \left( t/2 \right) }{ \left( \tanh \left( t/2
\right) \right) ^{2}-1}}
$$
note that $$x^2-1=\sinh^2(t)-1=\cosh^2(t)$$
|
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|
Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum_{n=0}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}(n+1)x^{2n+1}\int_0^1t^{2n}dt=\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)x^{2n+1}t^{2n}\right)dt$$
$$=x\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}\right)dt$$
$$\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}=\sum_{n=1}^{+\infty}n(xt)^{2n}+\sum_{n=1}^{+\infty}(xt)^{2n}$$
$$\sum_{n=1}^{+\infty}(xt)^{2n}=\frac{(xt)^2}{1-(xt)^2}$$
How to find the sum of $$\sum_{n=1}^{+\infty}n(xt)^{2n}?$$
EDIT:
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x^2\sum_{n=1}^{+\infty}\frac{x^{2n}}{2n+1}$$
$$=\frac{1}{2}x^2\sum_{n=1}^{+\infty}x^{2n}\int_0^1t^{2n}dt=\frac{1}{2}x^2\int_0^1\left(\sum_{n=1}^{+\infty}(xt)^{2n}\right)dt$$
$$=\frac{1}{2}x^2\int_0^1\frac{(xt)^2}{1-(xt)^2}=\frac{1}{2}x^4\int_0^1\frac{t^2}{1-(xt)^2}dt$$
$$=\frac{1}{2}x^4\cdot\frac{1}{x^3}(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|)\Rightarrow$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x\left(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|\right)\Rightarrow$$
$$\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)\Rightarrow$$
Finally,
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)$$
Question: Is this correct?
|
Hint:
$$f(x)=\sum_{n=0}^\infty\frac{n+1}{2n+1}x^{2n+1}$$
$$\implies f'(x)=\sum_{n=0}^\infty(n+1)x^{2n}$$
And now use the geometric series
$$\frac1{1-r}=\sum_{n=0}^\infty r^n$$
$$\frac d{dr}\frac1{1-r}=\sum_{n=0}^\infty(n+1)r^n$$
|
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|
How to find limit of $\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$ if $f(x)=-\sqrt{25-x^2}$ I have a question , if then find $$\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$$
I got $f(1)=-\sqrt {24}$. Should I get limit = $\frac{\sqrt{24} - \sqrt{25-x^2}}{x-1}$ but answer is $\frac{1}{\sqrt{24}}$ . Should I use $f'(x)$? If I use $f'(x)$ then I got $-\frac{1}{\sqrt{24}}$ .I should not get minus in solution.
Please help, Thanks in advance.
|
Method $1$. One may recall that, for any differentiable function $f$ near $a$, one has
$$
\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a),
$$
Here we have
$$
f(x)=-\sqrt{25-x^2},\qquad f'(x)=\frac{x}{\sqrt{25-x^2}},\qquad a=1.
$$
Can you finish it?
Method $2$. (without derivatives) One has, a $x \to 1$,
$$
\begin{align}
\frac{f(x) - f(1)}{x-1}&=\frac{\sqrt{25-1}-\sqrt{25-x^2}}{x-1}
\\\\&=\frac{(\sqrt{25-1}-\sqrt{25-x^2})(\sqrt{25-1}+\sqrt{25-x^2})}{(x-1)(\sqrt{25-1}+\sqrt{25-x^2})}
\\\\&=\frac{((25-1)-(25-x^2))}{(x-1)(\sqrt{25-1}+\sqrt{25-x^2})}
\\\\&=\frac{x^2-1}{(x-1)(\sqrt{25-1}+\sqrt{25-x^2})}.
\end{align}
$$Can you finish it?
|
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|
Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
$$f(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
After derivation:
$$f'(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}$$
The problem here is that:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}=x^2-x^{29}+x^{104}-...$$
Is it possible to find the closed form for the last series?
2) Using partial sums:
$$S_n=\sum_{k=0}^{n}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
Now, using the formula:
$$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}\Rightarrow$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
$$S_n=\frac{1}{3}-\frac{1}{30}+...+(-1)^{n}\frac{1}{(n+1)(4(n+1)^2-1)}$$
$$\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}=T_n=-\frac{1}{30}+...+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
$$T_n=S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Going back to the formula $$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
we have that $S_n$ cancels, so we can't determine partial sums using this method?
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+T_n$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Question: How to find the sum of this series?
|
An alternative approach: since $\frac{1}{k(4k^2-1)} = \frac{1}{2k-1}-\frac{2}{2k}+\frac{1}{2k+1}$ is the integral over $(0,1)$ of $x^{2k-2}-2x^{2k-1}+x^{2k} = x^{2k-2}(1-x)^2$, we have:
$$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{k(4k^2-1)}=\int_{0}^{1}(1-x)^2\sum_{k\geq 1}(-1)^{k+1} x^{2k-2}\,dx = \int_{0}^{1}\frac{(1-x)^2}{1+x^2}\,dx $$
and the last integral can be re-written as
$$ 1-\int_{0}^{1}\frac{2x}{1+x^2}\,dx = 1-\log(1+x^2)_{x=1} = \color{red}{1-\log 2}.$$
|
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|
Ackermann function $A(x,y)$: Prove that $A(4,y) = \underbrace{2^{2^{.^{\cdot^2}}}}_\textrm{y+3 times} -3$. Prove that $A(4,y) = \underbrace{2^{2^{.^{\cdot^2}}}}_\textrm{y+3 times} -3$.
I'm having trouble with proving that there are indeed $y+3$ exponentiations:
\begin{align*}
A(4,y) &= A(3, A(4,y-1))\\
&= 2^{A(4,y-1)+3}-3\\
&= 2^{2^{A(4,y-2)}-3+3}-3\\
&= 2^{2^{A(4,y-2)}}-3\\
&= 2^{2^{.^{.^{2^{A(4,0)}}}}}-3
\end{align*}
So far, according to my understanding, we have $y$ exponentiations. Now by considering that $A(4,0) = A(3,1) = 2^4-3 = 2^{2^2}-3$, we get additionally $2$ exponentiations for a a total of $b+2$.
Where is the missing exponent?
|
Prove by induction that
$$A(1,n)=n+2$$
Use this to prove by induction that
$$A(2,n)=2(n+3)-3$$
Use this to prove by induction that
$$A(3,n)=2^{n+3}-3$$
And use this to prove by induction that
$$A(4,n)=^{n+3}2-3$$
where we used tetration notation.
$$\begin{align}A(4,n+1)&=A(3,A(4,n))\\&=2^{(^{n+3}2-3)+3}-3\\&=2^{(^{n+3}2)}-3\\&=^{n+4}2-3\end{align}$$
Thus, it holds by induction.
As a bonus, see if you can prove by induction that
$$A(k,n)=2\uparrow^{k-2}(n+3)-3$$
where we use Knuth's up-arrow notation.
|
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|
Count number of exact matching sequences Consider all pairs of binary strings $P$ and $T$. Let the length of $P$ be $n$ and the length of $T$ be $2n-1$. For each such pair, we can check if $P$ is exactly equal to each of the $n$ substrings of $T$ in order from left to right and output a sequence representing these results. For example:
$$P = 11011, T = 111011011$$
gives an output sequence:
$$01001$$
where $1$ represents an exact matching and $0$ represents a mismatch.
If we iterate over all possible pairs $P$, $T$ we can count how many distinct sequences we get from the outputs. For $n = 2, \dots, 9$ we get
*
*$n = 2$ gives 4
*$n = 3$ gives 8
*$n = 4$ gives 14
*$n = 5$ gives 23
*$n = 6$ gives 34
*$n = 7$ gives 49
*$n = 8$ gives 66
*$n = 9$ gives 87
Is it possible to give a formula for this count or alternatively to give an asymptotic approximation?
|
Lemma. Let $R = R(P, T)$ be an output sequence for given strings $P$ and $T$ of length $n$ and $2n - 1$ correspondingly. Then
$$R = \underbrace{00\ldots\ldots\ldots\ldots\ldots\ldots0}_{\text{non-negative number of zeros}}\underbrace{1\underbrace{0\ldots0}_{\text{$k - 1$ zeros}}1\underbrace{0\ldots0}_{\text{$k - 1$ zeros}}1\ldots 1\underbrace{0\ldots0}_{\text{$k - 1$ zeros}}1}_{\text{non-negative number of ones}}\underbrace{00\ldots\ldots\ldots\ldots\ldots\ldots0}_{\text{non-negative number of zeros}}$$
for some positive integer $k$. (In other words, the distance between any two neighbouring ones in $R$ is the same.)
Proof. Suppose
$$R = \underbrace{\ldots\ldots\ldots}_{\text{whatever}}1\underbrace{00\ldots0}_{\text{$k - 1$ zeros}}1\underbrace{00\ldots0}_{\text{$\ell - 1$ zeros}}1\underbrace{\ldots\ldots\ldots}_{\text{whatever}}$$
for some positive integers $k \ne \ell$.
Let $k < \ell$, otherwise we can reverse all $P$, $T$ and $R$.
Let $P = a_0a_1\ldots a_{n - 1}$ where $a_i$ are symbols. Let $d = \gcd\{\,k, \ell\,\}$. Then
$$T = T_1P_0P_1\ldots P_mT_2,$$
where $T_i$ and $P_i$ are strings such that $P = P_0 P_1 \ldots P_{x - 1}P'_x$, $|P_0| = |P_1| = \ldots = |P_{m - 1}| = d$, $0 \le |P_m| < d$ and $P'_x = P_m$ is a prefix of $P_x$.
(Here $AB$ means concatenation of strings $A$ and $B$.)
Let $K = \frac{k}{d}$ and $L = \frac{\ell}{d}$. Then $P_i = P_{i + K}$ for $0 \le i < m - K$ and
$P_i = P_{i + L}$ for $0 \le i < m - L$. Also $\gcd\{\,K, L\,\} = 1$.
So $P_0 = P_{K} = P_{2K} = \ldots = P_{yK}$ for $(y - 1)K < L \le yK$.
If $K \mid L$ then it is easy to see that $P$ is prefix of $P_0P_1\ldots P_x = P_1P_2 \ldots P_{x+1}$, therefore $R$ misses at least one 1.
Then $(y - 1)K < L < yK$ and $P_0 = P_{yK - L} = P_{yK - L + K} = \ldots$. Iterating this process we get $P_0 = P_1 = \ldots = P_{m - 1}$ and $P_m$ is a prefix of $P_0.$ Therefore $R$ misses at least one 1, and this contradiction proves lemma. $\square$
It is easy to see that every $R$ described in the condition of lemma is achievable, so lemma describes all $R$ possible.
To compute the number of such sequences it is better to count sequences of all 0's and sequences with one 1 and then for all $k$ find the nubmer of sequences with at least two 1's:
$$1 + n + \sum_{k = 1}^{n - 1} \sum_{r = 0}^{k - 1}\binom{\left\lceil\frac{n - r}{k}\right\rceil}{2}.$$
Here $k$ is the distance between ones, $r$ is a remainder of (zero-based) position number of the first 1 modulo $k$ and $\binom{\left\lceil\frac{n - r}{k}\right\rceil}{2}$ is the number of ways to choose the first and the last 1's.
P. S. It is possible to show that asymptotics of this functions is $\frac12n^2\ln n$.
Let $f(n)$ be the desired number of sequences. Using inequality $x \le \lceil x \rceil < x + 1$ we get
$$1 + n + \sum_{k = 1}^{n - 1} \sum_{r = 0}^{k - 1}\binom{\frac{n - r}{k}}{2} \le f(n) < 1 + n + \sum_{k = 1}^{n - 1} \sum_{r = 0}^{k - 1}\binom{\frac{n - r}{k} + 1}{2}\\
\sum_{k = 1}^{n - 1} \sum_{r = 0}^{k - 1}\frac12\cdot\frac{n - r}{k}\left(\frac{n - r}{k} - 1\right) \le f(n) - 1 - n < \sum_{k = 1}^{n - 1} \sum_{r = 0}^{k - 1}\frac12\cdot\frac{n - r}{k}\left(\frac{n - r}{k} + 1\right)\\
\sum_{k = 1}^{n - 1} \sum_{r = 0}^{k - 1}\frac{1}{k^2}(n - r - k)(n - r) \le 2(f(n) - 1 - n) < \sum_{k = 1}^{n - 1} \sum_{r = 0}^{k - 1}\frac{1}{k^2}(n - r)(n - r + k)\\
\sum_{k = 1}^{n - 1} \frac{1}{k^2}\sum_{r = 0}^{k - 1}(n^2 + O(kn)) \le 2(f(n) - 1 - n) < \sum_{k = 1}^{n - 1} \frac{1}{k^2}\sum_{r = 0}^{k - 1}(n^2 + O(kn))\\
\sum_{k = 1}^{n - 1} \frac{1}{k^2}(kn^2 + O(k^2n)) \le 2(f(n) - 1 - n) < \sum_{k = 1}^{n - 1} \frac{1}{k^2}(kn^2 + O(k^2n))\\
n^2\sum_{k = 1}^{n - 1} \left(\frac{1}{k} + O\left(\frac1n\right)\right) \le 2(f(n) - 1 - n) < n^2\sum_{k = 1}^{n - 1} \left(\frac{1}{k} + O\left(\frac1n\right)\right)\\
n^2\ln n \sim n^2(H_{n - 1} + O(1)) \le 2(f(n) - 1 - n) < n^2(H_{n - 1} + O(1)) \sim n^2\ln n.
$$
Thus $f(n) \sim \frac12n^2\ln n$.
|
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|
Solve a system of two congruences with modules not pairwise coprime I have the following system of congruences:
$\left \{ \begin{array}{rcl} x & \equiv & 225 & \mbox{ (mod 250) } \\ x & \equiv & 150 & \mbox{ (mod 1225) } \end{array} \right .$
Since $\gcd(250, 1225) \ne 1$ the Chinese Remainder Theorem is not applicable.
I have tried to calculate the prime factors of both $250$ and $1225$ and I rewritten the system like this and try to obtain modules pairwise coprime:
$250 = 2 \cdot 5 \cdot 5 \cdot 5 \\ 1225 = 5 \cdot 5 \cdot 7 \cdot 7$
$\left \{ \begin{array}{rcl} x & \equiv & 225 & \mbox{ (mod 2) } \\ x & \equiv & 225 & \mbox{ (mod 5) } \\ x & \equiv & 225 & \mbox{ (mod 5) } \\ x & \equiv & 225 & \mbox{ (mod 5) } \\ x & \equiv & 150 & \mbox{ (mod 5) } \\ x & \equiv & 150 & \mbox{ (mod 5) } \\ x & \equiv & 150 & \mbox{ (mod 7) } \\ x & \equiv & 150 & \mbox{ (mod 7) } \end{array} \right .$
since
$x \equiv 225 \mbox{ (mod 5) } \iff x \equiv 0 \mbox{ (mod 5) }$ and the same for
$x \equiv 150 \mbox{ (mod 5) } \iff x \equiv 0 \mbox{ (mod 5) }$
I have reduced the system as the following:
$\left \{ \begin{array}{rcl} x & \equiv & 1 & \mbox{ (mod 2) } \\ x & \equiv & 3 & \mbox{ (mod 7) } \end{array} \right .$
and I have applied the Chinese Remainder Theorem:
$M = 2 \cdot 7 = 14 \\ M_1 = \frac{14}{2} = 7 \\ M_2 = \frac{14}{7} = 2$
the auxiliary system is:
$\left \{ \begin{array}{rcl} 7x & \equiv & 1 & \mbox{ (mod 2) } \\ 2x & \equiv & 1 & \mbox{ (mod 7) } \end{array} \right . \iff \left \{ \begin{array}{rcl} x & \equiv & 1 & \mbox{ (mod 2) } \\ x & \equiv & 4 & \mbox{ (mod 7) } \end{array} \right .$
$x = 7 \cdot 1 \cdot 1 + 2 \cdot 4 \cdot 3 = 31 \equiv 3 \mbox{ (mod 14) }$
Another way I tried is to consider this equation, but I don't know how to proceed:
$225 + 250h = 150 + 1225k$
Please, can you give me any suggestions? Many thanks!
|
$x \equiv 225 \pmod {250}$ so $x=25(9+10j)$ for some $j \in \mathbb{Z} $.
$x \equiv 150 \pmod {1225}$ so $x=25(6+49k)$ for some $k \in \mathbb{Z} $. So we need to find $y$ satisfying
\begin{eqnarray*}
y \equiv 9 \pmod {10} \\
y \equiv 6 \pmod {49}
\end{eqnarray*}
$10$ and $49$ are coprime so we can use the Chinese remainder theorem. We have $y=349$ and $x=8725$.
|
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|
If $a+b+c=x+y+z=1$ then $\frac{a}{y+z}+\frac{b}{x+z}+\frac{c}{x+y}\geq1+\frac{27}{2}abc$
Let $a$, $b$, $c$, $x$, $y$ and $z$ be positives such that $a+b+c=x+y+z=1$. Prove that:
$$\frac{a}{y+z}+\frac{b}{x+z}+\frac{c}{x+y}\geq1+\frac{27}{2}abc$$
I tried to make the homogenization:
We need to prove that
$$(x+y+z)\left(\frac{a}{y+z}+\frac{b}{x+z}+\frac{c}{x+y}\right)\geq1+\frac{27}{2}abc$$ or
$$\frac{ax}{y+z}+\frac{by}{x+z}+\frac{cz}{x+y}\geq\frac{27abc}{2(a+b+c)^2}$$
and what is the rest?
Maybe now we need to get rid of $x$, $y$ and $z$, but I don't see how we can do it.
Thank you!
|
I used Lagrange Multipliers here. I'd love to see a way that doesn't rely on them, because I feel there should be a much snappier solution. I'm a bit rusty with contest math, but I think I've checked my own work well enough. That's no excuse for holes, but I offer it as a caveat. It's likely there can be improvements in the style, etc., even though I've checked several times for holes. If people have improvements, or point out outright errors, I'm of course open to that.
Let's look at the left hand side of the second equation you offer and try to see what values of x/y/z maximize it.
$\frac{ax}{1-x} + \frac{by}{1-y} + \frac{cz}{1-z} \ge \frac{27abc}{2(a+b+c)^2}$ (a)
The left hand side can be expressed as
$\frac{a}{1-x} + \frac{b}{1-y} + \frac{c}{1-z} - 1$
(adding a, b and c to the respective fractions)
and thus has a Lagrange function of
$\frac{a}{1-x} + \frac{b}{1-y} + \frac{c}{1-z} - 1 - \lambda(x+y+z)$
If we differentiate with respect to X, we get
$\lambda = \frac{a}{x^2}$
With respect to y, $\lambda = \frac{b}{y^2}$ and with z, $\lambda = \frac{c}{z^2}$.
That means $\frac{x}{\sqrt{a}} = \frac{y}{\sqrt{b}}$ or $y = x\sqrt{b/a}$. Similarly $z = x\sqrt{c/a}$.
Since x+y+z=1, then, $x(\sqrt{a}+\sqrt{b}+\sqrt{c})/\sqrt{a} = 1$, or $x = \sqrt{a}/(\sqrt{a}+\sqrt{b}+\sqrt{c})$. Similarly $y = \sqrt{b}/(\sqrt{a}+\sqrt{b}+\sqrt{c})$ and $z = \sqrt{c}/(\sqrt{a}+\sqrt{b}+\sqrt{c})$.
Now we can express (a) as follows, noting $\frac{x}{1-x} = \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}$.
$\frac{a\sqrt{a}}{\sqrt{b}+\sqrt{c}} + \frac{b\sqrt{b}}{\sqrt{a}+\sqrt{c}} + \frac{c\sqrt{c}}{\sqrt{b}+\sqrt{a}} \ge \frac{27abc}{(a+b+c)^2}$
Now here's where I think there must be a better solution, but I multiplied everything out by $((a+b+c)^2)(\sqrt{b}+\sqrt{a})(\sqrt{a}+\sqrt{c})(\sqrt{b}+\sqrt{c})$ and compared the polynomial coefficients it worked okay. The strategy here is to look at the coefficients $\sqrt{abc}$ on the left and right side with the AM-GM inequality, then to deal with the symmetrical $\sqrt{a}+$\sqrt{b}+$\sqrt{c}$.
Let's look at the right hand side, since it's simpler to expand.
$(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})(\sqrt{b}+\sqrt{c})$ becomes $a\sqrt{b} + a\sqrt{c} + b\sqrt{a} + b\sqrt{c} + c\sqrt{a} + c\sqrt{b} + 2\sqrt{abc}$ and that means:
The right side's $\sqrt{abc}$ term is $27abc$, and its $\sqrt{a}$ term is $27abc(b+c)/2$.
The left side is, of course, trickier.
$(a+b+c)^2 (a\sqrt{a}(\sqrt{a}+\sqrt{c})(\sqrt{a}+\sqrt{b})+b\sqrt{b}(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b})+c\sqrt{c}(\sqrt{a}+\sqrt{c})(\sqrt{c}+\sqrt{b}))$
Let's isolate the abc-term, without expanding $(a+b+c)^2$. Each of the three products has one way to get a term with $\sqrt{abc}$ so that gives
$(a+b+c)^2 (a\sqrt{abc}+b\sqrt{abc}+c\sqrt{abc})$ which simplifies to
$(a+b+c)^3\sqrt{abc}$ but $a+b+c/3 > \sqrt[3]{abc}$ by the AM GM inequality, and cubing both sides of that equation gives
$(a+b+c)^3 > 27abc$. In other words, the $\sqrt{abc}$ term on the left of the equation is greater than or equal to the $\sqrt{abc}$ term on the right.
Now let's look at the left, ignoring the $\sqrt{abc}$ coefficients. Let's get rid of the $(a+b+c)^2$ for the moment.
$a\sqrt{a}(a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}) + b\sqrt{b}(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}) + c\sqrt{c}(c+\sqrt{bc}+\sqrt{ac}+\sqrt{ab})$
$(a^2+b^2+c^2)\sqrt{a} + (a^2+b^2+c^2)\sqrt{b} + (a^2+b^2+c^2)\sqrt{c}$
So the left's remaining coefficients are $(a+b+c)^2(a^2+b^2+c^2)(\sqrt{a}+\sqrt{b}+\sqrt{c})$.
Let's look at the right. It expands to
$27abc/2 * (a(\sqrt{b}+\sqrt{c}) + b(\sqrt{a}+\sqrt{c}) + c(\sqrt{b}+\sqrt{a}))$
Let's use the rearrangement inequality here. The sum in parentheses is the smallest for $x_1=a,x_2=b,x_3=c$ and $y_1=\sqrt{b}+\sqrt{c}, y_2=\sqrt{a}+\sqrt{c},y_3=\sqrt{b}+\sqrt{a})$.
If we let $z_1 = y_1 - (\sqrt{a}+\sqrt{b}+\sqrt{c})$ then we notice y_j decreases if we increase x_j, so we have the case where we have the rearrangement of the lowest sum. Thus, it is less than or equal to the average of all the possible sums, in other words,
$(a(\sqrt{b}+\sqrt{c}) + b(\sqrt{a}+\sqrt{c}) + c(\sqrt{b}+\sqrt{a})) < (a+b+c) * (\sqrt{a}+\sqrt{b}+\sqrt{c})*2/3$
In other words, the right is less than $9abc * (a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})$
That puts us in the home stretch. We need to show
$(a+b+c)^2(a^2+b^2+c^2)(\sqrt{a}+\sqrt{b}+\sqrt{c}) > 9abc(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})$
But a lot cancels out.
$(a+b+c)(a^2+b^2+c^2) > 9abc$
However, $a+b+c>3\sqrt[3]{abc}$ and $(a^2+b^2+c^2) > 3\sqrt[3]{abc}^2$ both by the AM GM inequalities.
|
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|
How to solve the integral $\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx$ .
$$\large \int \frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}~dx$$
In this I tried to substitute $x^2 +1$ by $t$.
after that got stuck.
|
HINT:
$$x^2-1=x^2\left(1-\dfrac1{x^2}\right)$$
As $\displaystyle\int\left(1-\dfrac1{x^2}\right)dx=x+\dfrac1x$
write
$$x^2+1=x\left(x+\dfrac1x\right)$$
$$x^4+1=x^2\left(x^2+\dfrac1{x^2}\right)\implies\sqrt{x^4+1}=|x|\sqrt{x^2+\dfrac1{x^2}}$$
Finally $ x^2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2-2$
|
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|
Characterize Numbers that can be written as the Sum Of Three Consecutive Positive Integers I am looking to characterize the numbers that can be characterized as the sum of 3 consecutive integers, but that can only be written in one way
so for example $12 = 3 + 4 + 5$
but not $15 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.$
So in general I'm looking for numbers
of the form $a + (a+1) + (a+2) = 3a+3.$
Where would I go from here? Would I show what even numbers & what odd numbers can be written?
|
-----------This answer is incomplete now due to the recent clarification in the edit specifying looking only for those numbers which can be written as the sum of 3 consecutive integers but not by any amount larger -----------
Original answer for when interpretation of question was just about whether or not it could be written as sum of 3 integers with no other constraints
Instead of characterizing them by the smallest number, characterize them by the middle number
$(b-1)+b+(b+1)=k$
You see then that these are the numbers of the form $3b$ greater than $6$ (* if you require all of the integers in the sum to be positive since the smallest sum of three consecutive positive integers is $1+2+3=6$, otherwise any multiple of three positive or negative will work*)
For completeness, we can show an if and only if condition here. The argument above shows that if a number can be written as the sum of three consecutive numbers that it must be a multiple of three.
Supposing a number $k$ is a multiple of three, then there is a $b$ such that $k=3b$ in which case $k=(b-1)+b+(b+1)$ is a way of writing it as the sum of three consecutive integers.
Attempt at a fix: The sum of $2a+1$ consecutive numbers with middle term $b$ will be $(2a+1)b$. If this is equal to our target number $k$ then that implies that $(2a+1)\mid k$.
Given that we want our number to be representable by the sum of three numbers that implies that $3\mid k$. Given that we do not want our number to be representable by the sum of any other number of odd integers implies that $(2a+1)\not\mid k$ for all $a\geq 2$. This in particular implies that no prime larger than $3$ and no power of three other than three itself divides $k$ implying then that our numbers must be $3$ times a power of two, i.e. of the form $2^n\cdot 3$ for some $n$.
Still left to complete: Characterize the numbers which can simultaneously be written as the sum of three consecutive numbers and an even number of consecutive numbers which cannot be written in any other way as the sum of an odd number of consecutive numbers
Characterizing the sum of an even number of consecutive based on the term just under the middle, let there be $2a$ numbers total and let $b$ be the term just under the average, we have $(b-a+1)+(b-a+2)+\dots+(b-1)+b+(b+1)+\dots+(b+a-1)+(b+a) = 2ab + a$
Supposing our number could be written as the sum of three consecutive integers and in no other way as the sum of an odd number of integers and as the sum of $2a$ numbers, that implies that $2ab+a=3\cdot 2^n$ implying that $a\mid 3\cdot 2^n$ and that $(2b+1)\mid 3\cdot 2^n$.
As $2b+1$ is odd that implies that $2b+1=3$ or $1$ as these are the only odd factors of $3\cdot 2^n$. This implies the middle number is either zero or the middle number is $1$. Indeed, $6=1+2+3$ can also be represented as $6=0+1+2+3$.
In fact, any number can be written as a sum of an even number of consecutive terms if you allow negatives can be made using $b=0$. Take $n$ for example:
$n=(1-n)+(2-n)+\dots+(-1)+0+1+\dots+(n-2)+(n-1)+n$
Assuming you are referring only to representing numbers as the sum of positive numbers, this shows that any number of the form $2^n3$ cannot be written as the sum of an even number of consecutive positive integers and thus any number is expressable only as the sum of three consecutive positive integers if and only if it is of the form $2^n3$.
|
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Prove that $ \sqrt{5} $ is irrational geometrically Hardy and Wright I am reading Hardy and Wright's Intro to Number Theory chapter about irrational number and there is a section proving that $ \sqrt{5} $ is irrational geometrically, but the proof is confusing me for some parts. The proof, extracted verbatim from the book, is presented below for convenience.
$ \textbf{Proof} $: We argue in term of $ \displaystyle x = \frac{\sqrt{5} - 1}{2} $. Then $ x^2 = 1 - x $. Geometrically, if $ AB = 1, AC = x $, then $ AC^2 = AB.BC $ and $ AB $ is divided in golden section by $ C $. (For the geometric figure, see Fig 4 here.)
If we divide $ 1 $ by $ x $, taking the largest integral quotient, viz. $ 1 $, the remainder is $ 1 - x = x^2 $. If we divide $ x $ by $ x^2 $, the quotient is again $ 1 $ and the remainder is $ x - x^2 = x^3 $. We next divide $ x^2 $ by $ x^3 $, and continue the process indefinitely; $ \textbf{at each stage the ratios of the number divided,} $
$ \textbf{the divisor, and the remainder are the same.} $ I am confused about this bold sentence. I work out a few examples and I got:
$ 1 = 1.x + (1 - x) = 1.x + x^2 $
$ x = 1.x^2 + (x - x^2) = 1.x^2 + x^3 $
$ x^2 = 1.x^3 + (x^2 - x^3) = 1.x^3 + x^4 $
and so on, so the ratios of the number divided, the divisor, and the remainder are all $ \displaystyle \frac{1}{x} $, is that what they are implying here?
Continue the proof:
Geometrically, if we take $ CC_{1} $ equal and opposite to $ BC $, $ AC $ is divided at $ C_{1} $ in the same ratio as $ AB $ at $ C $, i.e. in golden section; if we take $ C_{1}C_{2} $ equal and opposite to $ C_{1}A $, then $ C_{1}C $ is divided in golden section at $ C_{2} $; and so on. Since we are dealing at each stage with a segment divided in the same ratio, the process can never end.
I currently stuck on this paragraph, what do they mean by "$ AC $ is divided at $ C_{1} $ in the same ratio as $ AB $ at $ C $?" At first I thought this means $ \displaystyle \frac{AB}{AC} = \frac{AC}{AC_{1}} $, but this is not true since $ \displaystyle \frac{AB}{AC} = \frac{1}{x} $ while $ \displaystyle \frac{AC}{AC_{1}} = \frac{x}{2x - 1} $.
Continue the proof:
If $ x $ is rational, then $ AB $ and $ AC $ are integral multiples of the same length $ \delta $, and the same is true for $ C_{1}C, C_{1}C_{2}, \dots $, i.e. all the segments in the figure. Hence, we can construct an infinite sequence of descending integral multiples of $ \delta $, and this is plainly impossible.
If $ x = AC $ is rational, then $ \displaystyle AC = \frac{a}{b} $ and $ AB = 1 $, then how can "$ AB $ and $ AC $ are integral multiples of the same length $ \delta $?"
Any insight is appreciated.
|
$ \textbf{at each stage the ratios of the number divided,} $
$ \textbf{the divisor, and the remainder are the same.} $ I am confused about this bold sentence.
They mean that
$$\color{red}{\text{red}}:\color{blue}{\text{blue}}:\color{green}{\text{green}}=1:x:x^2$$
always holds in
$$\color{red}{1}=1\times \color{blue}{x}+\color{green}{x^2}$$
$$\color{red}{x}=1\times \color{blue}{x^2}+\color{green}{x^3}$$
$$\color{red}{x^2}=1\times \color{blue}{x^3}+\color{green}{x^4}$$
$$\vdots$$
I currently stuck on this paragraph, what do they mean by "$ AC $ is divided at $ C_{1} $ in the same ratio as $ AB $ at $ C $?"
$\qquad\qquad$
In the paragraph, they mean that
$$BC:CA=AC_1:C_1C=CC_2:C_2C_1=C_1C_3:C_3C_2=\cdots$$
where $BC=C_1C=x^2,AC_1=C_2C_1=x^3,CC_2=C_3C_2=x^4,\cdots$.
If $ x = AC $ is rational, then $ \displaystyle AC = \frac{a}{b} $ and $ AB = 1 $, then how can "$ AB $ and $ AC $ are integral multiples of the same length $ \delta $?"
$AB=b\times\frac 1b$ and $AC=a\times\frac 1b$ are integral multiples of $\delta=\frac 1b$.
|
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If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$ If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$.
My Attempt:
.$$\sin (\theta+\alpha)=a$$
$$\sin \theta. \cos \alpha+\cos \theta.\sin \alpha=a$$
Multiplying both sides by $2$
$$2\sin \theta.\cos \alpha + 2\cos \theta.\sin \alpha=2a$$
Squaring both sides,
$$4\sin^2 \theta.\cos^2 \alpha + 6\sin^2 \theta. \cos^2 \alpha + 4\cos^2 \theta.\sin^2 \alpha=4a^2$$.
How should I do further?
|
Set $\alpha-\beta=\gamma$, $\varphi=\theta+\alpha$, $\psi=\theta+\beta$; then $\gamma=(\theta+\alpha)-(\theta+\beta)=\varphi-\psi$, so
$$
\cos\gamma=
\cos\varphi\cos\psi+
\sin\varphi\sin\psi=
ab+\cos\varphi\cos\psi
$$
Also
$$
\cos2\gamma=2\cos^2\gamma-1=
2(a^2b^2+2ab\cos\varphi\cos\psi+\cos^2\varphi\cos^2\psi)-1
$$
so that
$$
\cos2\gamma-4ab\cos\gamma=
-2a^2b^2+2\cos^2\varphi\cos^2\psi-1
$$
Now $\cos^2\varphi=1-a^2$ and $\cos^2\psi=1-b^2$.
|
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Determine all pairs $(x,y)$ of integers
Determine all pairs $(x,y)$ of integers such that
$$1+2^{x}+2^{2x+1}=y^2.$$
I tried:
$$1+2^x+2^{2x+1}=y^2$$
$$2^x+2^{2x+1}=y^2-1$$
$$2^x+2^x2^x2=(y+1)(y-1)$$
$$2^x(1+ 2^{x+1})=(y+1)(y-1).$$
|
Completing the square gives
\begin{equation}
\left(2^{x+2}+1\right)^2=8y^2-7
\end{equation}
so $8y^2-7$ must be a perfect square of a number which is one more than four times a power of two. So two solutions are $(0,2)$ and $(0,-2)$.
Two more solutions are $(4,\pm23)$.
The steps to completing the square are as follows:
\begin{eqnarray}
2\cdot2^{2x}+2^x&=&y^2-1\\
16\cdot2^{2x}+8\cdot2^x&=&8y^2-8\\
16\cdot2^{2x}+8\cdot2^x+1&=&8y^2-7\\
\left(4\cdot2^x+1\right)^2&=&8y^2-7\\
\left(2^{x+2}+1\right)^2&=&8y^2-7
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find condition for $ a $ so that $ x^3 - 3ax + 1$ is separable Let $ K $ be an algebraically closed field of characteristic $ 0 $. What is a polynomial condition for $ a \in K $ such that $ f(x) = x^3 - 3ax + 1$ has distinct roots?
Here is my attempt: note that $ Df(x) = 3x^2 - 3a $. If $ a = 0 $, then $ (x^3 + 1, 3x^2 ) = 1 $, hence $ f $ is separable. Otherwise, it suffices to consider $ (x^3 - 3ax +1, x^2 - a) $. Using polynomial division, we have $$ (x^3 - 3ax +1, x^2 - a) = (x^2 - a, 2ax - 1) = (x-1/(2a), a - 1/(4a^2)). $$
Hence $ f $ is separable if $ a - 1/(4a^2) \neq 0 $, i.e. $ 4a^3 \neq 1 $. So the polynomial condition for $ a $ is that $ a $ does not satisfy $ 4x^3 - 1 = 0 $. Is this correct?
|
Your computation of the greatest common divisor is wrong.
The division of $x^3-3ax+1$ by $x^2-a$ has remainder $-2ax+1$.
Now we can divide $4a(x^2-a)$ by $2ax-1$, which has remainder $1-4a^3$.
(The case $a=0$ is trivial, so multiplying by a constant doesn't change the greatest common divisor, which is determined up to a multiplicative constant.)
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found the modulus to be $2\cos{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$ units and that the argument would be $\operatorname{arg}(z+i)=\frac{\pi}{4}+\frac{\theta}{2}$.
Now, the next step that I took was that I replaced every theta with $\frac{3\pi}{8}$ in the polar form of the complex number $z+i$. So now it would look like this:
$$z+i=\left[2\cos{\left(\frac{\pi}{8}\right)}\right]\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$$
Then, I expanded the $\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$ part to become $\cos{\left(\frac{3\pi}{8}\right)}+i\sin{\left({\frac{3\pi}{8}}\right)}$. So now I've got the $\cos\left({\frac{3\pi}{8}}\right)$ part but I don't really know what to do next. I've tried to split the angle up so that there would be two angles so I can use an identity, however, it would end up with a difficult fraction instead. So if the rest of the answer or a hint would be given to finish the question, that would be great!!
Thanks!!
|
Problem statement
$$
\cos \left( \frac{3\pi}{8} \right) = \cos \left( \frac{\pi}{4} - \frac{\pi}{8} \right)
$$
Basic formulas
Use the $\color{blue}{angle \ addition}$ formula
$$
\cos \left( \alpha + \beta \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,
$$
and the $\color{blue}{half\ angle}$ formula
$$
\cos 2 \theta = \sqrt{\frac{1\color{red}{+}\cos \theta}{2}}, \qquad
\sin 2 \theta = \sqrt{\frac{1\color{red}{-}\cos \theta}{2}}
$$
And unit circle
$$
\cos \left( \frac{\pi}{4} \right) =
\sin \left( \frac{\pi}{4} \right) =
\frac{1}{\sqrt{2}}
$$
Solution
$$
\begin{align}
\cos \left( \frac{3\pi}{8} \right)
&= \cos \left( \frac{\pi}{4} - \frac{\pi}{8} \right) \\[3pt]
%%
&= \cos \left( \frac{\pi}{4} \right)\cos \left(\frac{\pi}{8} \right)
-
\sin \left( \frac{\pi}{4} \right)\sin \left(\frac{\pi}{8} \right)\\[3pt]
%%
&= \cos \left( \frac{\pi}{4} \right) \sqrt{\frac{1\color{red}{+}\cos \frac{\pi}{4}}{2}}
-
\sin \left( \frac{\pi}{4} \right)\sqrt{\frac{1\color{red}{-}\cos \frac{\pi}{4}}{2}} \\[4pt]
%%
&= \frac{1}{\sqrt{2}} \frac{\sqrt{\sqrt{2}+2}}{2}
-
\frac{1}{\sqrt{2}} \frac{\sqrt{\sqrt{2}-2}}{2} \\[5pt]
%%
&= \frac{\sqrt{2 - \sqrt{2}}} {2} \\[3pt]
%%
&= \frac{1}{\sqrt{4+2 \sqrt{2}}}
%%
\end{align}
$$
|
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|
$a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$. Find $a_n$. Given $\{a_n\}$: $a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$, $a_1=1$.
Find $a_n$ like a function of $n$.
My trying.
Let $a_n=\tan\alpha_n$, where $\alpha_n\in\left(0,\frac{\pi}{2}\right)$.
Hence, $a_{n+1}=\sin\alpha_n$ and what is the rest?
Thank you!
|
Note that $${ a }_{ 2 }=\frac { 1 }{ \sqrt { 2 } } \\ { a }_{ 3 }=\frac { \frac { 1 }{ \sqrt { 2 } } }{ \sqrt { \frac { 3 }{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \\ { a }_{ 4 }=\frac { \frac { 1 }{ \sqrt { 3 } } }{ \sqrt { \frac { 4 }{ 3 } } } =\frac { 1 }{ \sqrt { 4 } }
$$
so $$\\ { a }_{ n }=\frac { 1 }{ \sqrt { n } } \\ $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove this condition if three vectors are colinear? So I was given this problem:
Let $\vec{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\vec{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \mathbf{0}.$$
I tried plugging it in and it looks like it is true:
Let $\vec{a} = \begin{pmatrix}2\\4\\6\end{pmatrix}$, $\vec{b} = \begin{pmatrix}4\\8\\12\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}8\\16\\24\end{pmatrix}$.
$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}
=\begin{pmatrix}0\\0\\0\end{pmatrix}
+\begin{pmatrix}0\\0\\0\end{pmatrix}
+\begin{pmatrix}0\\0\\0\end{pmatrix}\\
=
\begin{pmatrix}0\\0\\0\end{pmatrix}\\
$$
How could I prove this to be true? I am not entirely sure where to begin.
|
Hint: Collinear vectors have the same direction vector.
|
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|
Find answer of a system of equations Let $a,b,c\in \mathbb{R}$ that $a^{2}+b^{2}+c^{2}=1$. We want to find $x,y,z,w$ in the following equations:
$$\begin{cases}
\begin{align}
x^{2}+y^{2}&=\frac{1}{2}(1+a) \tag{1}\\
w^{2}+z^{2}&= \frac{1}{2}(1-a) \tag{2}\\
xw+yz&= \frac{1}{2} b \tag{3}\\
yw-xz&= \frac{1}{2} c.\tag{4}
\end{align}
\end{cases}$$
|
Hint (assuming $x,y,z,w$ are reals): let $u=y+ix$ and $v=w+iz$ then the system becomes:
$$
\begin{cases}
\begin{align}
|u|^2 &= \frac{1}{2}(1+a) \\
|v|^2 &= \frac{1}{2}(1-a) \\
uv & = \frac{1}{2}(c+ib)
\end{align}
\end{cases}
$$
From the first two equations $|u|^2|v|^2= \frac{1}{4}(1-a^2)=\frac{1}{4}(b^2+c^2)\,$. From the third equation $|uv|^2=\frac{1}{4}(c^2+b^2)\,$ as well, so the modulus part of the 3rd equation is redundant.
Then pick $u$ to be any complex number of modulus $\sqrt{\frac{1+a}{2}}\,$, and let $v=\frac{c+ib}{2u}$.
|
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|
How to prove that $|w_{n+1}-w_n| \leq (2/3)^2 |w_n - w_{n-1}|$? If a sequence is given by $S_{n+1}=S_n + S_{n-1}$ where $S_1=1$ and $S_2=2$ and we let $a_n=S_n/S_{n-1}$ for all $n \geq 2$. For each $n \geq 2$, prove that $|a_{n+1}-a_n| \leq (2/3)^2 |a_n - a_{n-1}|$?
This is what I'm thinking...
Let n=3. Then $|a_{n+1}-a_n| \leq (2/3)^{2}|a_n - a_{n-1}|$ is true for $n=3$ since
$|a_{4}-a_3| \leq (2/3)^{2}|a_3 - a_2|$
$|\frac{5}{3}-\frac{3}{2}| \leq (2/3)^{2}|\frac{3}{2}-2|$
$\frac{1}{6} \leq (2/3)^{2}(1/2$
$\frac{1}{6} \leq 2/9$
Now let $n=k$ where $k \in \mathbb{R}$ such that $ k \geq 2$. Then
$|a_{k+1}-a_k| \leq (2/3)^{2}|a_k - a_{k-1}|$
$|\frac{S_{k+1}}{S_k}-\frac{S_k}{S_{k-1}}| \leq (2/3)^{2}|\frac{S_{k}}{S_{k-1}}-\frac{S_{k-1}}{S_{k-2}}|$
$\frac{1}{S_kS_{k-1}} \leq (2/3)^{2}\frac{S_{k}S_{k-2}-S^2_{k-1}}{S_{k-1}S_{k-2}}$
Thus, it holds for $n=k+1$
|
Like user dezdichado already mentioned, this is the fibonacci sequence, but we can approach this independently without necessarily using known facts about the sequence.
You also have a typo on your final inequality which it should instead read as:
$$\vert \frac{S_{k+1}S_{k-1}-S_k^2}{S_kS_{k-1}} \vert \leq (2/3)^2\vert \frac{S_{k}S_{k-2}-S_{k-1}^2}{S_{k-1}S_{k-2}} \vert$$
From there we can use the following simple fact
$S_{k+1}S_{k-1}-S_k^2 = (S_{k}+S_{k-1})S_{k-1}-(S_{k-1}+S_{k-2})^2 = S_kS_{k-1} - 2S_{k-1}S_{k-2} - S_{k-2}^2 = $
$S_{k-1}(S_k - S_{k-2}) - S_{k-2}(S_{k-1} + S_{k-2}) = S_{k-1}^2 - S_kS_{k-2}$
We conclude that:
$$\vert S_{k+1}S_{k-1}-S_k^2 \vert = \vert S_{k-1}^2 - S_kS_{k-2} \vert = \vert S_kS_{k-2} - S_{k-1}^2 \vert$$
and thus your original simplifies to
$$\vert \frac{1}{S_k} \vert \leq (2/3)^2\vert \frac{1}{S_{k-2}} \vert \Rightarrow \frac{1}{S_k} \leq (2/3)^2 \frac{1}{S_{k-2}} \Rightarrow \frac{S_{k-2}}{S_k} \leq (2/3)^2 \Rightarrow \frac{S_{k}}{S_{k-2}} \geq (3/2)^2 = \frac{9}{4}$$
Doing a bit more work we get:
$\frac{S_{k}}{S_{k-2}} = \frac{S_{k-1}+S_{k-2}}{S_{k-2}} = 1 + \frac{S_{k-1}}{S_{k-2}}$. So it's enough to show that:
$$\frac{S_{k-1}}{S_{k-2}} \geq \frac{5}{4}$$
The last should be easy to verify using a simple inductive argument
|
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|
Solving equations with floor function and square roots? I was solving a problem to discover n and after I transformed the problem it gave me this equation:
\begin{equation*}
\left\lfloor{\frac{2}{3}\sqrt{10^{2n}-1}}\right\rfloor = \frac{2}{3}(10^{n}-1)
\end{equation*}
So I tried to simplify it by defining:
\begin{equation*}
k = 10^{n}-1
\end{equation*}
and was left with:
\begin{equation*}
\left\lfloor{\frac{2}{3}\sqrt{k(k+2)}}\right\rfloor = \frac{2}{3}k
\end{equation*}
But I can't get past that. Can anyone help me?
|
Hint: $10^n-1$ is a multiple of $9$, so $\frac{2}{3}k$ is an integer, then:
$$
\begin{align}
\left\lfloor{\frac{2}{3}\sqrt{k(k+2)}}\right\rfloor = \frac{2}{3}k \;\;&\iff\;\; \frac{2}{3}k \le \frac{2}{3}\sqrt{k(k+2)} \lt \frac{2}{3}k \,+\, 1 \;\; \\
&\iff\;\; \frac{4}{9}k^2 \le \frac{4}{9}(k^2+2k) \lt \frac{4}{9}k^2 + \frac{4}{3}k+1 \\
&\iff\;\; 0 \le \frac{8}{9} k \lt \frac{4}{3}k+1 \\
\end{align}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integration of rational functions How would you integrate a rational function like this:
$$\frac{2x+3}{x^2+2x+10}$$
|
Notice that:
$$\frac{2x + 3}{x^2 + 2x + 10} = \frac{2x + 2}{x^2 + 2x + 10} + \frac{ 1}{x^2 + 2x + 10} = \frac{2x + 2}{x^2 + 2x + 10} + \frac{ 1}{(x+1)^2 + 3^2}$$
Therefore,
$$\int \frac{2x + 3}{x^2 + 2x + 10} dx = \int\frac{2x + 2}{x^2 + 2x + 10}dx+ \int\frac{ 1}{(x+1)^2 + 3^2}dx$$
$$ = \ln|x^2 + 2x + 10| + \int\frac{ 1}{(x+1)^2 + 3^2}dx $$
$$= \ln|x^2 + 2x + 10| + \int\frac{1/9}{(\frac{x+1}{3})^2 + 1}dx $$
$$= \ln|x^2 + 2x + 10| + \int\frac{1/9}{(\frac{x+1}{3})^2 + 1}dx$$
$$= \ln|x^2 + 2x + 10| + \frac{1}{9}\int\frac{1}{(\frac{x+1}{3})^2 + 1}dx$$
$$= \ln|x^2 + 2x + 10| + \frac{1}{3} \arctan(\frac{x+1}{3}) + c$$
We can safely drop the absolute value, because $x^2 + 2x + 10 > 0 \quad \forall x \in \mathbb{R}$, obtaining:
$$\int \frac{2x + 3}{x^2 + 2x + 10} dx = \ln(x^2 + 2x + 10) + \frac{1}{3} \arctan(\frac{x+1}{3}) + c$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4 u} \, du\\
&= \frac1{16}\left(\tan u + {\tan^3 u\over 3} \right) + C\\
&= \frac1{16}\left(\tan (\sec^{-1} 2x) + {\tan^3 (\sec^{-1} 2x)\over 3} \right) + C.
\end{align*}$$
Given answer : $$\dfrac{\left(2x^2+1\right)\sqrt{4x^2-1}}{24}+C$$
Why is my answer incorrect ?
|
$$\tan\left(\sec^{-1}(2x)\right)=\sqrt{4x^2-1}$$
Make this substitution into your answer and you will get the result you expected.
|
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|
Continuity of some function related to the Hopf fibration of $S^3$ By the identification $\mathbb R^4$ with the field of quaternions $\mathbb H$ by $(x_0,x_1,x_2,x_3) \sim x_0+x_1 i+x_2 j+x_3k$ and $S^3$ with the set of unit quaternions and $S^2$ with the set of purely imaginary unit quaternions, let's consider the Hopf mapping $p: S^3 \rightarrow S^2$, $p(x)=xix^*=(x_0^2+x_1^2-x_2^2-x_3^2)i+2(x_1x_2+x_0x_3)j+2(x_1x_3-x_0x_2)k$ for $x=x_0+x_1i+x_2j+x_3k\in S^3$, where $x^*$ denotes the conjugate quaternion to $x$. For fixed $v=(v_1,v_2,v_3)\in S^2\setminus \{S\}$, where $S=(-1,0,0)\in S^2$, we have that the preimage of $v$ is equal $p^{-1}(\{v\})=\{ \sqrt{\frac{1+v_1}{2}}(\cos t, \sin t, \frac{v_2 \sin t-v_3 \cos t}{1+v_1},\frac{v_2 \cos t+v_3 \sin t}{1+v_1}): t\in [0,2\pi) \}$ (see here).
In order to know that the structure $(S^3, S^1, p, S^2)$ is a fiber bundle (see here), we have to know that the trivialization map $f_1: (S^2\setminus\{S\}) \times S^1 \rightarrow p^{-1}(S^2 \setminus \{S\})$, $f_1(v,(a_1,a_2))= \sqrt{\frac{1+v_1}{2}}(a_1, a_2, \frac{v_2 a_2-v_3 a_1}{1+v_1},\frac{v_2 a_1+v_3 a_2}{1+v_1})$ for $(a_1,a_2)\in S^1$, $v=(v_1,v_2,v_3)\in S^2 \setminus \{S\}$, is the homeomorphism satisfying condition $p \circ f_1=proj_1$ and similarly for $f_2: (S^2\setminus\{N\}) \times S^1 \rightarrow p^{-1}(S^2 \setminus \{N\})$, where $N=(1,0,0)\in S^2$.
My problem concerns continuity of the inverse function to $f_1$.
Now $p^{-1}(S^2 \setminus \{S\})=\{\sqrt{\frac{1+v_1}{2}}(\cos t, \sin t, \frac{v_2 \sin t-v_3 \cos t}{1+v_1},\frac{v_2 \cos t+v_3 \sin t}{1+v_1}): t\in [0,2\pi), (v_1,v_2,v_3)\in S^2\setminus \{S\} \}$
Why the function $f_1^{-1}: p^{-1}(S^2 \setminus \{S\}) \rightarrow (S^2\setminus\{S\}) \times S^1$, $f_1^{-1}(\sqrt{\frac{1+v_1}{2}}(\cos t, \sin t, \frac{v_2 \sin t-v_3 \cos t}{1+v_1},\frac{v_2 \cos t+v_3 \sin t}{a+v|_1}))=((v_1,v_2,v_3),(\cos t, \sin t))$
is contunuous?
|
We can get the explicite formula on $f_1^{-1}$. From the system of equations $(a,b,c,d)=\sqrt{\frac{1+v_1}{2}}(a_1, a_2, \frac{v_2 a_2-v_3 a_1}{1+v_1},\frac{v_2 a_1+v_3 a_2}{1+v_1})$,
we get
$v_1=2(a^2+b^2)-1$,
$v_2=2\sqrt{a^2+b^2}(c a_2+da_1)$,
$v_3=2\sqrt{a^2+b^2}(da_2-ca_1)$,
$a_1=\frac{a}{\sqrt{a^2+b^2}}$,
$a_2=\frac{b}{\sqrt{a^2+b^2}}$.
Hence we obtain
$
f_1^{-1}(a,b,c,d)=\left( \left( 2(a^2+b^2)-1, 2\sqrt{a^2+b^2}(c \frac{b}{\sqrt{a^2+b^2}}+d\frac{a}{\sqrt{a^2+b^2}}), 2\sqrt{a^2+b^2}(d\frac{b}{\sqrt{a^2+b^2}}-c\frac{a}{\sqrt{a^2+b^2}}\right), \left(\frac{a}{\sqrt{a^2+b^2}},\frac{b}{\sqrt{a^2+b^2}}\right) \right)=
$
$
\left( \left( 2(a^2+b^2)-1, 2(bc +ad),2(bd-ac)\right), \left(\frac{a}{\sqrt{a^2+b^2}},\frac{b}{\sqrt{a^2+b^2}}\right) \right).
$
Now observe that points of the form $(0,0,c,d)\in S^3$ are not in $p^{-1}(S^2\setminus \{S\})$ because $p(0,0,c,d)=(-1,0,0)=S$. Hence the formula above defines $f_1^{-1}$ on the whole set $p^{-1}(S^2\setminus \{S\})$ and proves the continuity of $f_1^{-1}$.
|
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|
Mathematical Olympiad Treasures Problem 1.11 Let $n$ be a positive integer, prove that:
$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1$
is not prime
The solution states:
Observe:
$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1 = a^{3} + b^{3} + c^{3} - 3abc$
for $a = 3^{3^{n-1}}$, $b = 9^{3^{n-1}}$, $c = -1$
After that main insight the solution follows quite simply.
How are you supposed to find what $a$ and $b$ are. It makes sense to guess that $c = -1$ but finding $a$ and $b$ seems much harder, how do you guess it?
|
$$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1\equiv 1\cdot 0 +1-1\equiv 0\bmod 2$$ hence it's not a prime
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$ Show that
$$\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$$
How many ways are there to prove it ?
Is there a standard way ?
I was thinking about making it a Riemann sum.
Or telescoping.
What is the easiest way ?
What is the shortest way ?
|
Note that
$$\sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \sum_{k=1}^n \frac{2k}{k^2+n^2+1} - \frac{2}{2+n^2} - \frac{4}{5+n^2}.$$
We can ignore the last two terms since they converge to $0$.
Consider
$$\sum_{k=1}^n \frac{2k}{k^2+n^2+1} = \frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/n^2)}. $$
This is almost a Riemann sum for $\int_0^1 2x/(1 +x^2) \, dx$ except for the annoying term $1/n^2$.
However, it is valid to take a limit of a double sequence
$$S_{mn} =\frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/m^2)}, $$
as
$$\lim_{n \to \infty}S_{nn} = \lim_{n \to \infty} \lim_{m \to \infty} S_{mn},$$
since the inner limit on the RHS exhibits uniform convergence.
Thus,
$$\lim_{n \to \infty}\sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \lim_{n \to \infty} \lim_{m \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/m^2)} \\ = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2} \\ = \int_0^1 \frac{2x}{1+ x^2} \, dx \\ = \log 2$$
|
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How to find all possible solutions to $a^2 + b^2 = 2^k$? I'm working on the following problem, this is for an introductory discrete mathematics class.
Find all possible solutions to the equation $a^2 + b^2 = 2^k, k\geq1$ and $a$ and $b$ positive integers.
I've observed that the following are answers:
$2^1 = 1^2 + 1^2$
$2^3 = 2^2 + 2^2$
$2^5 = 4^2 + 4^2$ or $2^5 = 2^4 + 2^4$
From that, there seems to be a pattern such as $2^{k+1}=2^k + 2^k$ for some even $k \geq 0$
I've also been able to observe that $k$ has to be odd. I've tried the following:
$k = 2l + 1$ for some $l$
$2^{2l+1} = a^2 + b^2$
$2^l.2 = a^2 + b^2$ Now, if I set $a=b$
$2.2^l = 2a^2$ <=> $2^l = a^2$
But I don't know how to prove that $a$ has to be equal to $b$
Also, in the case where $k$ is even:
$k=2l$ for some $l$
$2^{2l}$ = $a^2 + b^2$ Again, setting $a=b$
$2^{2l+1} = a^2$
$2^{(2l-1)/2} = a^2$ where $(2l-1)/2$ is not an integer.
But again, I don't know how to go about it when $a \neq b$.
Am I on the right path? Any tips or suggestions on how to go forward?
|
Your statement that $2^{k+1} = 2^k + 2^k$ is proven correct by the observation on the right-hand side that for any integer $c$, $c + c = 2c$. My first thought is that $a,b$ must have the same parity for this to ever work.
If both are even, let $a = 2A$ and $b = 2B$, so that $a^2 + b^2 = 4A^2 + 4B^2 = 4(A^2 + B^2)$. This factorization should help you.
If both are odd, let $a = 2k + 1$ and let $b = 2l + 1$, so that
$$a^2 + b^2 = (2k+1)^2 + (2l+1)^2 = ...$$
Keep going and you should get something interesting with this. Hope that helps!
|
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|
If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$? Let $a_n$ and $b_n$ be integers defined in the following way: $$a_n+b_n\sqrt{3}=(2+\sqrt{3})^n.$$ Compute $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}.$$
I tried expanding using binomial theorem:
$$ (2+\sqrt{3})^n=\binom{n}{0}2+\binom{n}{1}2^2\sqrt{3}+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}+\binom{n}{n}\sqrt{3}^{n}$$
and I think we have that: $$ a_n=\binom{n}{0}2+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n}\sqrt{3}^{n}$$
$$ b_n=\binom{n}{1}2^2\sqrt{3}+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}$$
But, frankly, I do not know what to do next.
|
Note that
$$a_n - b_n\sqrt{3} = (2-\sqrt{3})^n$$
Hence,
$$2a_n = (a_n - b_n\sqrt{3}) + (a_n + b_n\sqrt{3}) \Rightarrow a_n = \dfrac{(2+\sqrt{3})^n + (2-\sqrt{3})^n}{2}$$
Expressing $b_n$ we obtain:
$$b_n = \dfrac{(2+\sqrt{3})^n - a_n}{\sqrt{3}} = \dfrac{(2+\sqrt{3})^n - (2-\sqrt{3})^n}{2\sqrt{3}}$$
Now it should be easy to compute the limit of $\dfrac{a_n}{b_n}$. Since $(2-\sqrt{3})^n \to 0$, we can conclude that
$$\lim\limits_{n\to+\infty}\dfrac{a_n}{b_n} = \lim\limits_{n\to+\infty} \dfrac{(2+\sqrt{3})^n}{(2+\sqrt{3})^n/\sqrt{3}} = \sqrt{3}$$
|
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|
Show that the sequence {$a_n$}, such that $a_1 =4$ and $a_{n+1}=3-{{2}\over\ a_n}$ is convergent to 2. Show that the sequence {$a_n$}, such that
$a_1 =4$ and $a_{n+1}=3-{{2}\over\ a_n}$ is convergent to 2.
I show that the sequence is bounded but I cannot show that is monotone.
|
Let's solve the difference equation.
Find the first few terms:
$a_2=3-\frac{2}{4}=\frac{5}{2}=\left(2+\frac{1}{2}\right)$
$a_3=3-\frac{4}{5}=\frac{11}{5}=\left(2+\frac{1}{5}\right)$
$a_4=3-\frac{10}{11}=\frac{23}{11}=\left(2+\frac{1}{11}\right)$
$a_5=3-\frac{22}{23}=\frac{47}{23}=\left(2+\frac{1}{23}\right)$
...
$a_{n+1}=2+\frac{1}{3\cdot2^{n-1}-1}$.
Thus, $\lim \limits_{n\to\infty}a_{n+1}=2.$
Note: $3\cdot2^{n-1}-1$ is a solution of the difference equation: $b_{n+1}=2b_n+1,b_1=2.$
|
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|
How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far:
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\right)\\
\log_5{2x^2} &= \log_5\left(\frac{5}{4}\right)\\
5^{2x^2} &= 5^\frac{5}{4}\\
(2x^2)\log5 &= \frac{5}{4}\log5\\
2x^2 &= \frac{5}{4}\\
2x^2 &= 1.25\\
x^2 &= 0.625\\
\sqrt {x^2} &= \sqrt {0.625}\\
x &= \frac{\sqrt {10}}{4}\end{align}
Substituting the value for $x$ into the equation doesn't equate it to zero. I've tried several different ways but I have still not come up with a correct answer.
|
$\require{enclose}5^{2x}+4(5)^{x}-5=0 \enclose{updiagonalstrike}{\implies} 5^{2x}+5^{x}=\frac{5}{4}$
Instead:
Let $y=5^x\quad$ noting that $y\geq 0$, then we have:
$y^2+4y-5=0\implies (y+5)(y-1)=0$
$\implies y=-5\quad\text{reject since}\quad y\geq 0\quad \text{or}\quad y=1\implies 5^x=1\implies x=0$
|
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|
Maximization of $f(\theta)=\frac{1}{\sqrt{2\pi c\theta}}e^{-\frac{1}{2c\theta}(x-\theta)^2}$ with inequality constraints I want to maximize the following function:
$$f(\theta)=\frac{1}{\sqrt{2\pi c\theta}}e^{-\frac{1}{2c\theta}(x-\theta)^2}$$
with respect to $\theta\in[0,1]$. Assume the rest of the variables known and $c>0$. Therefore, I took the log:
\begin{eqnarray*}
\mathrm{ln}(f(\theta))&=&\mathrm{ln}(2\pi c\theta)^{-1/2}-\frac{1}{2c\theta}(x-\theta)^2\\
&=&-\frac{1}{2}\mathrm{ln}(2\pi c\theta)-\frac{x^2}{2c\theta}+\frac{x}{c}-\frac{\theta}{2c}
\end{eqnarray*}
and eventually
\begin{eqnarray*}
\frac{\vartheta\mathrm{ln}(f(\theta))}{\vartheta\theta}=0&\Rightarrow&-\frac{1}{2}\frac{2\pi c}{2\pi c\theta}+\frac{x^2}{2c}\frac{1}{\theta^2}-\frac{1}{2c}=0\\
&\Rightarrow&\frac{x^2-c\theta}{c\theta^2}=\frac{1}{c}\\
&\Rightarrow&\theta^2+c\theta-x^2=0
\end{eqnarray*}
Is this a correct approach? What should I do next? I also need to satisfy the constraint $\theta\in[0,1]$.
|
Hint. What you have done is right. Then solving the quadratic equation gives two potential solutions
$$
\theta_0=-\frac12 \left(\sqrt{c^2+4 x^2}+c\right), \qquad \theta_1=\frac12 \left(\sqrt{c^2+4 x^2}-c\right).
$$ Which $\theta_i$ satisfies $0\le\theta_i \le1$ ?
|
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|
Evaluate $\int \frac{\cos\pi z}{z^2-1}\, dz$ inside rectangle with vertices $2+i,2-i,-2+i,-2-i$ My attempt: The poles are $z=1,-1$, both lie inside the rectangle.
Residue at the poles are $-\frac12$ each, since residue at $$f(a)=\left[\frac{\cos \pi z}{\frac{d}{dz}(z^2+1)}\right]_{z=a}=\frac{\cos\pi a}{2a}.$$
So, by residue theorem,
$$\int \dfrac{\cos\pi z}{z^2-1}\, dz= 2\pi i \left[-\frac{1}{2}-\frac{1}{2}\right]=-2\pi i.$$
Is this correct?
|
The residue of $\frac{\cos(\pi z)}{z^2-1} =\frac{h(z)}{z-1}, \ h(z) = \frac{\cos(\pi z)}{z+1}$ at $z = 1$ is $h(1)=\frac{\cos(\pi)}{2} = -1/2$.
The residue of $\frac{\cos(\pi z)}{z^2-1} =\frac{H(z)}{z+1}, H(z) = \frac{\cos(\pi z)}{z-1}$ at $z = -1$ is $H(-1) = 1/2$.
The proof of the residue theorem in the case of finite contours and poles of order $1$ is not complicated :
$g(z) = f(z) - \frac{-1/2}{z-1}- \frac{1/2}{z+1}$ is holomorphic (on a simply connected open containing the contour) so the Cauchy integral theorem applies :
$\int_C g(z)dz =0$
and $$\int_C f(z)dz = \int_C (\frac{-1/2}{z-1}+ \frac{1/2}{z+1})dz = 2i \pi (-1/2+1/2) = 0$$
(for evaluating $\int_C \frac{1}{z-a}dz$, use that when choosing the correct branch for the logarithm : $\frac{d}{dz}\log(z-a) = \frac{1}{z-a}$)
|
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|
Roots of Quadratic lies in $(0, 1)$ Given that the Quadratic equation $f(x)=ax^2-bx+c=0$ has two roots in $(0 \: 1)$
if $a,b,c \in \mathbb{N}$ Find Minimum values of $a$ and $b$
Since $a$ is Natural number graph of parabola will be open upwards.
Now $f(0) \gt 0$ and $f(1) \gt 0$ so we get
$c \gt 0$ and $a+c \gt b$ and since roots are in $(0 \: 1)$ product of roots also lies in $(0 \: 1)$
we get
$$\frac{c}{a} \lt 1$$ or $$a \gt c$$
Also Discriminant $$b^2 \gt 4ac \gt 4c^2$$
now how can we find minimum values of $a$ and $b$
|
For $f(x) = ax^2 - bx + c$ , we get $f^{'}(x) = 2ax - b$.
Thus the minima is at $\displaystyle f^{'}(x) = 0 \iff 2ax-b = 0 \iff x = {b\over 2a}$.
Therefore a root must be in $[0, b/2a]$ and one in $[b/2a, 1]$.
And since $f(0) > 0$ therefore $f(b/2a) < 0$
therefore $f(b/2a) < f(0) \implies c < b^2/4a$
Similarly, $\displaystyle f(1) > f(b/2a) \implies a-b +c > {-b^2\over4a} + c \implies b < 2a $
Since there is no zero at $1$ and $f(1) > 0$ we get $a + c > b$.
So we need $3$ three numbers that satisfy,
*
*$c < b^2/4a$
*$b < 2a$
*$a + c > b$
The smallest number $z \in \mathbb{N}$ that satisfy $z > b$ is $b+1$ therefore on comparing this with $(3)$ we get $a = b$ and $c = 1$(*).
Plugging $a = b$ and $c = 1$ in $(1)$ we get $a = b = 5$ and $c = 1$.
(*) :- If $a = 1$ and $c = b$ then from $(1)$ we get $4 < b$ and from $(2)$ we will get $b < 2$ which is not possible, thus $a = b$ and $c = 1$.
|
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Given functions $f,g$ find the direct form for $f(x)=g(x)+f(x-1)$ Let $g(x)=1+x+\frac{x(x+1)}{2}$ and let $f(x)=g(x)+f(x-1)$. Find the closed form for $f(x)$ and $x\geq0$ given that $f(0)=0$.
So what we are asked are is a direct form for $$\sum_{n=1}^{k}g(n)$$
I want to show that $$\sum_{n=0}^{k}g(n)=\frac{1}{2}(\sum_{n=0}^{k+1}n+\sum_{n=0}^{k+1}{n^2})=\dfrac{(k+1)(k+2)(k+3)}{6}$$
edit 1: $f,g$ are defined on the integers.
|
Use generating functions.
Define $f(-n) = 0$ and say that for all $n \geq 1$ we have $$f(n) = 1+n+\frac{1}{2} n(n+1) + f(n-1)$$
Then $$F(y) := \sum_{n=1}^{\infty} f(n) y^n = \sum_{n=1}^{\infty} [1+n+\frac{1}{2}n(n+1) + f(n-1)] y^n \\
= \sum_{n=1}^{\infty} y^n + \sum_{n=1}^{\infty} n y^n + \sum_{n=1}^{\infty} \frac{n(n+1)}{2} y^n + y \underbrace{\sum_{n=1}^{\infty} f(n-1) y^{n-1}}_{\sum_{n=0}^{\infty} f(n) y^n \ = \ f(0) + F(y)} \\
= y (1-y)^{-1} + y (y-1)^{-2} + y (1-y)^{-3} + y F(y)$$
so $$F(y)(1-y) = y(1-y)^{-1} + y (y-1)^{-2} + y (1-y)^{-3}$$
whereupon $$F(y) = y (1-y)^{-2} + y (1-y)^{-3} + y (1-y)^{-4}$$
Finally, we can just work out the coefficient of $y^n$ in this expression: it's $$f(n) = \frac{1}{6} n (11 + 6 n + n^2)$$
|
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Taylor series expansion of $\frac{(1- \cos x)}{x^2} $ According to my calculations, the Taylor series expansion for $1- \cos x$ around $x=\pi$ is
$$2 - \frac{(x-\pi)^2}{2} + \frac{(x-\pi)^4}{24} - \frac{(x-\pi)^6}{720} + \frac{(x-\pi)^8}{40320} -\dotsb
$$
In solving the Taylor series expansion of $\frac{(1- \cos x)}{x^2}$ around $x= \pi$, would it be the same if I'll just multiply $\frac{1}{x^2}$ to the series written above? If that's not possible, how can I solve the series expansion of the quotient of two functions without doing ridiculously long sessions of product rule differentiation?
|
As $x \to \pi$, you may express $\dfrac1{x^2}$ as
$$
\begin{align}
\dfrac1{x^2}&=\dfrac1{(\pi+\color{red}{x-\pi})^2}
\\\\&=\dfrac1{\pi^2} \cdot \dfrac1{\left(1+\frac{\color{red}{x-\pi}}\pi\right)^2}
\\\\&=\dfrac1{\pi^2} \cdot \sum_{n=0}^\infty\frac{(-1)^n(n+1)}{\pi^n}\left({\color{red}{x-\pi}}\right)^n
\end{align}
$$ then multiplying by
$$
1-\cos x=2+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left({\color{red}{x-\pi}}\right)^{2n}
$$ one gets, as $x \to \pi$,
$$
\dfrac{1-\cos x}{x^2}=\frac{2}{\pi^2}-\frac{4 (x-\pi)}{\pi^3}+\left(\frac{6}{\pi^4}-\frac{1}{2\pi^2}\right) (x-\pi)^2-\left(\frac{8}{\pi^5}-\frac{1}{\pi^3}\right) (x-\pi)^3+\cdots.
$$
|
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Show that there is a unique polynomial $\int_{x}^{x+1} b_{n}{(t)} dt = x^n$ For each n = 1, 2, 3, . . . show that there is a unique polynomial $ b_{n}{(x)} $ that
satisfies the equation
$\int_{x}^{x+1} b_{n}{(t)} dt = x^n$
n = 1, I find $ b_{1}{(t)} = t + x_{0} $ and find $x_{0} = -\frac{1}{2}$
n = 2, $ b_{2}{(t)} = t^2 - t + \frac{1}{6} $
How can I prove that the polynomial is unique and always satisfy the equation?
|
First observe that
$$
\int_x^{x + 1} t^n \,dt = \frac{(x + 1)^{n + 1} - x^n}{n+1} = \frac{1}{n + 1} \sum_{k = 0}^n {n + 1 \choose k} x^k.
$$
Therefore
$$\int_x^{x+1} t^n \,dt = x^n + \text{lower order terms}.$$
Therefore the matrix which writes $\{\int_{x}^{x+1} t^n \,dt : n = 0, 1, 2, 3\dots \}$ in terms of $\{1,x,x^2,x^3,\dots\}$ is triangular with ones down the diagonal. In particular, it is invertible. Because $\{1,x,x^2,x^3,\dots\}$ is a basis for the ring of polynomials, this means that $\{\int_{x}^{x+1} t^n \,dt : n = 0, 1, 2, 3\dots \}$ is also a basis. Therefore every polynomial in $x$ can be written uniquely as a linear combination
$$
\begin{align*}
p(x) &= a_n \int_{x}^{x + 1} t^n \,dt + a_{n - 1}\int_x^{x+1} t^{n - 1} \,dt + \dots + a_1 \int_x^{x+1} t \,dt + a_0 \int_x^{x+1} 1\, dt \\
&= \int_x^{x+1} a_nt^n + a_{n-1}t^{n-1} + \dots + a_1t + a_0 \, dt.
\end{align*}
$$
|
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How to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without quadratic equation? After almost seven years I recently started again learning math and have few holes in my algebra knowledge, so I apologize for the beginner question.
My question is:
Is there any simple trick to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without using quadratic equation?
And if it's possible, what is step by step procedure?
|
Not sure what you mean by "without quadratic equation" but I'll assume it means without using the quadratic formula. So we can factorise it by completing the square:
Let $f(x) = 12x^2-8x+1$, then:
$f(x) = 12(x^2-\frac{8}{12}x+\frac{1}{12})=12(x^2-\frac{2}{3}x+\frac{1}{12})$
$f(x) = 12[(x-\frac{1}{2}\cdot\frac{2}{3})^2-(\frac{1}{3})^2+\frac{1}{12}]$
$f(x) = 12[(x-\frac{1}{3})^2-\frac{1}{9}+\frac{1}{12}]$
$f(x) = 12[(x-\frac{1}{3})^2+\frac{3-4}{36}]=12[(x-\frac{1}{3})^2-\frac{1}{36}]$
So then we set $f(x) = 0:$
$(x-\frac{1}{3})^2-\frac{1}{36}=0\implies x=\frac{1}{3}\pm\frac{1}{6}\implies x=\frac{1}{2}\quad\text{or}\quad x=\frac{1}{6}$
Therefore $f(x) = 12(x-\frac{1}{2})(x-\frac{1}{6})=2(x-\frac{1}{2})\cdot6(x-\frac{1}{6})=(2x-1)(6x-1)$
|
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|
Mapping of a horizontal line $y = c$ by $w = \frac{1}{z}$ onto which region? I was trying to solve this problem ,
so $y = c$. Suppose for ease let us take $c > 0$.
so $w = \frac{1}{z} = \frac{1}{x + iy} = \frac{x - iy}{x^2 + y^2} = \frac{x - ic}{x^2 + c^2} = \frac{x}{x^2 + c^2} - i( \frac{c}{x^2 + c^2})$.
Now as $w = u + iv$,
comparing we get $u = \frac{x}{x^2 + c^2}$ and $v = \frac{-c}{x^2 + c^2}$ and also $u^2 + v^2 = \frac{1}{x^2 + c^2} \leq \frac{1}{c^2}$ and I think this is a disc centered at $0$ and of maximum radius of $\frac{1}{c}$.
Next I am thinking that since $v = \frac{-c}{x^2 + c^2}$ and $c > 0$ that would imply that $v$ is always negative but now if this sis the case then what I have got the region as disc will be wrong as it contains some positive $v$ also the upper part of the disc and hence now I am getting the region as semicircular disc with arc in the negative $v$ axis.
Is this correct?.if not where I am making mistake.
Any help is great.
|
Hint: Consider the absolute value of $w+\frac i{2c}$.
Also look at Inversive Geometry.
|
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|
If matrix $A$ in $\mathbb{R}^3 $ such that, $A^3 = I$, $\det A = 1$. Is there a such matrix which is not orthogonal, rotation and identity? I tried to use Cayley–Hamilton theorem to learn something about the matrix.
Using the theorem we have:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right) A + \det A$.
Since $-A^3 + \det A \cdot I = 0$, then $A = \frac{\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right)}{\text{tr} A}I$. With this in mind I cannot obtain even rotation matrix to say nothing about something else. Any hints?
|
Here's a quick example:
$$
A = \pmatrix{\cos(2\pi/3) & -10\sin(2 \pi /3) & 0\\
\frac 1{10}\sin(2 \pi /3) & \cos (2 \pi /3) & 0\\0 & 0 & 1}
$$
Is like the usual rotation, but with the off-diagonal entries slightly altered. Show that $A^3 = I$, but $A$ is not orthogonal (or a rotation) since $A^TA \neq I$.
Another easy example:
$$
A = \pmatrix{\cos(2\pi/3) & \sin(2 \pi /3) & 2\\
\sin(2 \pi /3) & \cos (2 \pi /3) & 3\\0 & 0 & 1}
$$
the same argument applies.
|
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|
Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15. Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. But, apparently there are no primitive solutions to $mod 15.$ So what's another option?
|
$x^2\equiv 1 \pmod 3$ and $x^2\equiv 1 \pmod 5$. Now, $x\equiv 1 \text{ or } 2 \pmod 3$. There are $2$ solutions. Also $x\equiv 1 \text{ or } 4 \pmod 5$. There are $2$ solutions. By Chinese remainder theorem: $x^2\equiv 1 \pmod {15}$ has $2\cdot 2=4$ incongruent solutions.
|
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|
prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know how to start.
Do I need contradiction?
|
Start with the hint . . .
Let $a,b,c$ be given by
\begin{align*}
a &= 3x+2z+1\\[4pt]
b &= 3x+2z+2\\[4pt]
c &= 4x+3z+2
\end{align*}
where $x,z$ are unknown positive integers.
If we were lucky, the equation $a^2 + b^2 = c^2$ would hold identically, for all $x,z$.
Let's try . . .
\begin{align*}
a^2 + b^2 - c^2 &= (3x+2z+1)^2 + (3x+2z+2)^2 - (4x+3z+2)^2\\[4pt]
&= 2x^2 + 2x + 1 - z^2\qquad\text{[by expanding and then combining like terms]}\\[4pt]
\end{align*}
So unfortunately, we were not that lucky, since $a^2 + b^2 - c^2$ simplifies to $2x^2 + 2x + 1 - z^2$, which is not identically zero.
But what if it was the case that $2x^2 + 2x + 1 - z^2 = 0\,$ for some $x,z$?
With that idea, let's look more closely at the equation $2x^2 + 2x + 1 - z^2 = 0$, to see if we can find a way to make it happen.
The key observation is
\begin{align*}
&2x^2 + 2x + 1 - z^2 = 0\\[4pt]
\iff\; &2x^2 + 2x + 1 = z^2\\[4pt]
\iff\; &x^2 + (x^2 + 2x + 1) = z^2\\[4pt]
\iff\; &x^2 + (x+1)^2 = z^2\\[4pt]
\end{align*}
Given that all transitions above are of the form "if and only if", it follows that
$$(x,x+1,z)$$
$$\text{is a pythagorean triple}$$
$$\text{if and only if}$$
$$(3x+2z+1,3x+2z+2,4x+3z+2)$$
$$\text{is a pythagorean triple}$$
Thus, starting with any pythagorean triple of the form $(x,x+1,z)$ we can get a new one which is strictly larger (e.g., larger perimeter).
Note also that any pythagorean triple of the form $(x,x+1,z)$ is automatically a primitive pythagorean triple since $\gcd(x,x+1) = 1$.
Thus, all we need is one such triple to get started. Fortunately, that's easy$\,{\large{-}}\,$just start with the triple $(3,4,5)$.
So we weren't so unlucky after all!
|
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|
Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where a and b are integers. Find the greatest common factor of b and 81. Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where $a$ and $b$ are integers. Find the greatest common factor of $b$ and $81$.
How would one solve this question? I tried to use the binomial theorem but that approach did not work. How would one solve this problem if it were to appear on a Math Olympiad?
I know the answer is one of:
(A) 1 (B) 3 (C) 9 (D) 27 (E) 81.
According to one of the Stack Exchange members, the answer is 3. It was found using Python.
|
It is a bit annoying, but certainly feasible, to compute the answer by repeated squaring in $(\mathbb{Z}/81\mathbb{Z})[\sqrt{2}]$.
$$(1+\sqrt{2})^2 \equiv 3+2\sqrt{2}$$
$$(1+\sqrt{2})^4 \equiv 17+12\sqrt{2}$$
$$(1+\sqrt{2})^8 \equiv 10+3\sqrt{2}$$
$$(1+\sqrt{2})^{16} \equiv 37+60\sqrt{2}$$
$$(1+\sqrt{2})^{32} \equiv 64+66\sqrt{2}$$
$$(1+\sqrt{2})^{64} \equiv 10 + 24\sqrt{2}$$
$$(1+\sqrt{2})^{128} \equiv 37 + 75\sqrt{2}$$
$$(1+\sqrt{2})^{256} \equiv 64 + 42\sqrt{2}$$
$$(1+\sqrt{2})^{512} \equiv 10 + 30\sqrt{2}.$$
Now in binary, $1000 = 1111101000_2$, or in other words, $1000 = 512 + 256 + 128 + 64 + 32 + 8.$ Therefore for any $a$
$$a^{1000} = a^{8}a^{32}a^{64}a^{128}a^{256}a^{512},$$
and we can calculate this product for $a=(1+\sqrt{2})$ using the table above:
$$(1+\sqrt{2})^{40}\equiv 64+42\sqrt{2}$$
$$(1+\sqrt{2})^{104}\equiv 64+12\sqrt{2}$$
$$(1+\sqrt{2})^{232}\equiv 37+60\sqrt{2}$$
$$(1+\sqrt{2})^{488}\equiv 37+48\sqrt{2}$$
$$(1+\sqrt{2})^{1000}\equiv 10+51\sqrt{2}$$
and $(51,81)=3.$
|
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|
Least possible polynomial deegree of complex roots What is the least possible deegree of polynomial with real coefficients having roots $2\omega , 2+3\omega , 2+3\omega ^2 , 2-\omega -\omega ^2$
As there are four roots so the deegree should be four but the answer is given as five . how ?
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The least possible degree of polynomial with real coefficients having these roots is the product of $(x-\alpha)(x-\bar\alpha)$, where $\alpha$ are the given roots, unless $\alpha$ is real, in which case the second factor is omitted.
Since $\bar\omega=\omega^2$ and $1+\omega+\omega^2=0$, the polynomial is
$$
(x-2\omega)(x-2\bar\omega)(x-2-3\omega)(x-2-3\bar\omega)(x-3)
\\=
(x^2+2x+4)(x^2-x+7)(x-3)
\\=
x^5 - 2 x^4 + 6 x^3 - 17 x^2 - 2 x - 84
$$
|
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|
Finding $\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$ Finding $$\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$$
Attempt: $$\lim_{n\rightarrow \infty}\bigg[\frac{1}{n^n}+\frac{2^2}{n^n}+\frac{3^3}{n^n}+\cdots \cdots +\frac{n^n}{n^n}\bigg] = 1$$
because all terms are approaching to zero except last terms
but answer is not $1$ , could some help me to solve it , thanks
|
Bounding by a geometric series,
$$
\begin{align}
&\frac{n^n}{n^n}+\frac{(n-1)^{n-1}}{n^n}+\frac{(n-2)^{n-2}}{n^n}+\cdots+\frac{1^1}{n^n}\\
&\le1+\frac1n+\frac1{n^2}+\frac1{n^3}+\cdots\\
&=\frac{n}{n-1}
\end{align}
$$
Since the sum is obviously always $\ge1$, and $\le\frac{n}{n-1}$, the Squeeze Theorem says that the limit is $1$.
|
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|
Number of integer partitions $p(n,m)$ In "Integer partitions" by Andrews and Eriksson, the authors provide formulas to compute $p(n,m)$, i.e., the number of partitions of $n$ into parts less than or equal to $m$, for $m=1,2,3,4,5$. As discussed in this question, it seems that no formula for $m>5$ is known.
Howevere, the fact that $p(n,m)$ for $m\leq 5$ are upper-bounded by polynomials, made me naturally wonder whether this holds for $m>5$ or not.
This is also related to the result in Section 6.4, in which the authors show that $p(n)$ is sub-exponential.
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Via Flajolet-Sedgewick: $p_m(z)=\frac{1}{\left(1-z\right) \left(1-z^2\right)\text{...} \left(1-z^m\right)}$ is an ordinary generating function. For general k, the coefficients are "polynom + some small (O(1)) periodic function". Using Mathematica,
$$ p(n,5) = \frac{30 n \left(15 \left(3 (-1)^n+85\right)+n (n (n+30)+310)\right)+10125 (-1)^n+5400 \sin \left(\frac{\pi n}{2}\right)+5400 \cos \left(\frac{\pi n}{2}\right)+6400 \cos \left(\frac{2 \pi n}{3}\right)+6912 \cos \left(\frac{2 \pi n}{5}\right)+6912 \cos \left(\frac{4 \pi n}{5}\right)+50651}{86400}$$
and for k=6
$$ \frac{(2 n+21) (2 n (n+21) (3 n (n+21)+434)+28511)}{1036800}+\frac{1}{32} \sin \left(\frac{\pi n}{2}\right)+\frac{i (6 n (n+21)+581) \sin (\pi n)}{4608}+\frac{1}{25} \sqrt{1+\frac{2}{\sqrt{5}}} \sin \left(\frac{2 \pi n}{5}\right)+\frac{1}{25} \sqrt{1-\frac{2}{\sqrt{5}}} \sin \left(\frac{4 \pi n}{5}\right)+\frac{\sin \left(\frac{2 \pi n}{3}\right)}{81 \sqrt{3}}+\frac{1}{32} \cos \left(\frac{\pi n}{2}\right)+\frac{1}{18} \cos \left(\frac{\pi n}{3}\right)+\frac{1}{162} (2 n+21) \cos \left(\frac{2 \pi n}{3}\right)+\frac{1}{25} \cos \left(\frac{2 \pi n}{5}\right)+\frac{1}{25} \cos \left(\frac{4 \pi n}{5}\right)+\frac{(6 n (n+21)+581) \cos (\pi n)}{4608} $$
|
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|
Solve $2^{x+5} - 2^{x+2} = 7$ I have a logarithms related problem at hand which I need to know its related formulas.
$$2^{x+5} - 2^{x+2} = 7$$
I already know the answer which is: $-2$.
So far, I have reached this form:
$$x = \frac{\ln(7 + 2^x+2)}{\ln (2)} - 5.$$
Any tips or hints will be appreciated. Thanks in advance.
|
Note that $2^{x+5} = 2^x\cdot 2^5 = 32\cdot 2^x$ and $2^{x+2} = 2^x\cdot 2^2 = 4\cdot 2^x$ so your expression becomes
$$ 32\cdot 2^x - 4\cdot 2^x = 7.$$
Can you take it from here?
|
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|
How to solve $2^{33} \mod 4725$? Basically what I should do is:
$$2^{33} \equiv x \pmod{4725}.$$
I need to find $x$ which will give same result as $2^{33}$ when modulo $4725$.
I should find prime factors of $4725$ which are $4725 = 3^3 × 5^2 × 7^1$, and use them to calculate
\begin{align*}
2^{33} &\equiv x \pmod{3^3}\\
2^{33} &\equiv x \pmod{5^2}\\
2^{33} &\equiv x \pmod{7^1}
\end{align*}
so I could apply Chinese remainder theorem to calculate result, but I am stuck with calculating these. I don't know what to do with $2^{33}$, how to decompose it?
Any help is welcome.
|
I will compute $2^{33}\pmod {27}$ as the OP is ok with the rest of the calculation.
In what follows $\equiv $ denotes congruence $\pmod {27}$.
Method I (iterated squaring): This works unusually well in this case because $33=32+1$ is so near a power of $2$. We write:
$$2^2=4\implies 2^4=16\implies 2^8=16^2\equiv 13 \implies 2^{16}\equiv 13^2\equiv 7$$
$$\implies 2^{32}\equiv 7^2\equiv -5$$
It follows that $$2^{33}=2^{32}\times 2\equiv 17\pmod {27}$$
Method II (Euler): As $\varphi(27)=18$ we have $2^{18}\equiv 1$. Thus $2^{36}\equiv 1$ so $2^{33}\equiv 2^{-3}$. As the inverse of $2\pmod {27}$ is clearly $14$ we see that $$2^{33}\equiv 14^3\equiv 17\pmod {27}$$
Note: It's not especially harder to do the entire problem by iterated squaring. Working $\pmod {4725}$ we have $$2^8=256 \implies 2^{16}\equiv 4111 \implies 2^{32}\equiv {4111}^2 \equiv 3721$$ So $$2^{33}=2^{32}\times 2 \equiv 2717\pmod {4725}$$
|
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|
Equation with integrals and absolute complex log function problem (physics related) I was working on a problem that I described in a previous question. The problem is the following: I need to solve this for $\theta$:
$$\frac{\text{n}}{\text{A}}\int_0^{2\pi}\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x\tag1$$
Now, for the LHS:
$$\frac{\text{n}}{\text{A}}\int_0^{2\pi}\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\frac{\text{n}}{\text{A}}\cdot\frac{2\pi}{\sqrt{\left(1-\epsilon^2\right)^3}}\tag2$$
And the RHS:
$$\int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\frac{\epsilon}{\epsilon^2-1}\cdot\frac{\sin\left(\theta\right)}{1+\epsilon\cos\left(\theta\right)}-\frac{1}{\sqrt{\left(\epsilon^2-1\right)^3}}\cdot\ln\left|\frac{1+\frac{\epsilon-1}{\sqrt{\epsilon^2-1}}\cdot\tan\left(\frac{\theta}{2}\right)}{1-\frac{\epsilon-1}{\sqrt{\epsilon^2-1}}\cdot\tan\left(\frac{\theta}{2}\right)}\right|\tag3$$
Now, in the problem I have a value for $\epsilon$ that gives:
$$\epsilon^2-1<0\tag4$$
So, the value between the absolute value become a complex number and that results into:
$$\ln\left|\frac{1+\frac{\epsilon-1}{\sqrt{\epsilon^2-1}}\cdot\tan\left(\frac{\theta}{2}\right)}{1-\frac{\epsilon-1}{\sqrt{\epsilon^2-1}}\cdot\tan\left(\frac{\theta}{2}\right)}\right|=0\tag5$$
So, we get the equality:
$$\frac{\text{n}}{\text{A}}\cdot\frac{2\pi}{\sqrt{\left(1-\epsilon^2\right)^3}}=\frac{\epsilon}{\epsilon^2-1}\cdot\frac{\sin\left(\theta\right)}{1+\epsilon\cos\left(\theta\right)}\tag6$$
But the known variables are: $\epsilon\approx0.0167086$ and $\text{A}\approx365.25636$ and for $1\le\text{n}\le\text{A}$
But this gives a complex solution for $\theta$ and that is not a possible solution for my problem, so where am I wrong?
|
Using Wolfram :
$$\int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\frac{\epsilon}{\epsilon^2-1}\cdot\frac{\sin\left(\theta\right)}{1+\epsilon\cos\left(\theta\right)}-\frac{2}{\sqrt{\left(\epsilon^2-1\right)^3}}\cdot\tanh^{-1}\left({\frac{\epsilon-1}{\sqrt{\epsilon^2-1}}\cdot\tan\left(\frac{\theta}{2}\right)}\right)\tag3$$Now, to avoid complex values, recall that $$ i \tanh^{-1} x=\tan^{-1} ix$$ so:
$$\int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\frac{\epsilon}{\epsilon^2-1}\cdot\frac{\sin\left(\theta\right)}{1+\epsilon\cos\left(\theta\right)}+\frac{2}{\sqrt{\left(1-\epsilon^2\right)^3}}\cdot\tan^{-1}\left({\frac{1-\epsilon}{\sqrt{1-\epsilon^2}}\cdot\tan\left(\frac{\theta}{2}\right)}\right)$$
If I made no mistake, now all the values are real and you can find your angle.
|
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Limit of trigonometric function $\lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$ I want to compute this limit:
$\displaystyle \lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$
Using L'Hopital, is easy to get the result, which is $\sqrt{3}$
I tried using linear approximation (making $u = x - \displaystyle \frac{\pi}{3}$)
$\displaystyle \lim_{u\to 0} \frac{1 - 2\cos(u + \frac{\pi}{3})}{\sin(u)} = \lim_{u\to 0} \frac{1 - \cos(u) + \sqrt{3}\sin(u)}{u} \approx \lim_{u\to 0} \frac{1 - 1 + \sqrt{3}u}{u} = \sqrt{3}$
But it bothers me using that sort of linear approximation, I want to get the result in a more formal way. I have tried using double angle properties
$$\cos(2x) = \cos^2(x) - \sin^2(x)$$
$$\sin(2x) = 2\sin(x)\cos(x)$$
But I reach to a point of nowhere, I cannot come up with a way of simplifying expressions to get the results:
$\displaystyle\lim_{x\to\pi/3} 2\frac{3\sin^2(\frac{x}{2}) - \cos^2(\frac{x}{2})}{\sqrt{3}\sin(x) - \cos(x)}$
Is there a way of computing this limit without approximations and without L'Hopital?
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Note that $\cos (\pi/3) = 1/2$ and hence we can write $$\lim_{x \to \pi/3}\frac{1 - 2\cos x}{\sin (x - \pi/3)} = -2\lim_{x \to \pi/3}\frac{\cos x - \cos (\pi/3)}{x - \pi/3}\cdot\frac{x - \pi/3}{\sin (x - \pi/3)} = -2(-\sin \pi/3)\cdot 1 = \sqrt{3}$$ here we have used the fact that $(\cos x)' = -\sin x$ so that $$\lim_{x \to a}\frac{\cos x - \cos a}{x - a} = -\sin a$$
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|
If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.
|
An alternative solution, not to disparage the other answers.
$$\begin{aligned}
\frac{a}{b}=\frac{c}{d}\quad&\Rightarrow\quad\frac{ad}{b}=c&\text{solve for $c$}\\
\frac{a+c}{b+d}&=\frac{a+\left(\frac{ad}{b}\right)}{b+d}&\text{substitute $c$}\\
&=a\cdot\frac{1+\left(\frac{d}{b}\right)}{b+d}&\text{factor $a$ from numerator}\\
&=a\cdot\frac{b+d}{b(b+d)}&\text{multiply by $\frac{b}{b}$}\\
&=\frac{a}{b}\quad\blacksquare&\text{cancel $(b+d)$}\\
\end{aligned}$$
|
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|
Expressing partial fractions in ascending powers of $x$ Let $f(x) = \frac {5x^2 + x + 6}{(3-2x)(x^2 + 4)}$
$i) $ Express $f(x)$ in partial fractions.
$ii)$ Hence obtain the expansion of $f(x)$ in ascending powers of $x$, up to and including the term in $x^2$.
I got part $i)$ but how do I do part $ii)$, I know how to the binomial expansion of each individual component, but I don't get the correct answer.
For $i)$ I got this, which is correct:
$f(x) = \frac {3}{3 - 2x} + \frac {-x -2}{x^2 + 4}$
|
Note that we can expand $\frac{3}{3-2x}$ as a geometric series given by
$$\begin{align}
\frac{3}{3-2x}&=\frac{1}{1-(2/3)x}\\\\
&=\sum_{n=0}^\infty \left(\frac23\right)^n\,x^n\\\\
&=1+\frac23x+\frac49x^2+O(x^3)
\end{align}$$
for $|x|<3/2$.
Analogously, we can expand $\frac{-x-2}{x^2+4}$ as a geometric series given by
$$\begin{align}
\frac{-(x+2)}{x^2+4}&=\frac{-(x+2)/4}{1+(x/2)^2}\\\\
&=-\frac14(x+2)\sum_{n=0}^\infty \left(\frac{-1}{4}\right)^n x^{2n}\\\\
&=-\frac12-\frac14x+\frac18x^2+O(x^3)
\end{align}$$
for $|x|<1$.
Therefore for $|x|<1$, we find that
$$\bbox[5px,border:2px solid #C0A000]{\frac{5x^2+x+6}{(3-2x)(x^2+4)}=1+\frac5{12}x+\frac{41}{72}x^2+O(x^3)}$$
|
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Polynomials $P(x)$ such that $P(x-1)$ $=$ $P(-x)$ Are there an infinite number of polynomials $P(x)$ of even degree such that
$P(x-1) = P(-x)$
$P(x) = x^2+x+1$ is a good example because $P(x-1) = (x-1)^2+(x-1)+1 = x^2-x+1 = P(-x)$.
$P(x) = x^4+2x^3+4x^2+3x+1$ is another example because $P(x-1) = (x-1)^4+2(x-1)^3+4(x-1)^2+3(x-1)+1 = x^4-2x^3+4x^2-3x+1 = P(-x)$.
Is there an easy way to generate such polynomials? Thanks for help.
|
One easy way to generate infinitely many such polynomials is to start with your example $P(x)=x^2+x+1$ and consider $P(x)^2,P(x)^3,P(x)^4,\dots$
|
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|
$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$
My attempt:I've tried it by considering the sum
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$
along with
$$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots +\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$ which gives $ 100$ as resultant but failed to separate the sum of
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots\\ \dots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$ at last.
I tried the next approach by using de Movire's theorem but failed to separate the real and imaginary part.
I've invested a great amount of time in the so it would be better if someone please come up with an answer.
|
$$\cos\left(\frac{k\pi}{101}\right)= \frac{1}{2} \left(e^{i\frac{k\pi}{101}}+e^{-i\frac{k\pi}{101}} \right) \\
\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right) \\
\sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \sum_{k=1}^{100}\left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right)
$$
Now,
$$1+\sum_{k=1}^{100}e^{2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{2i\frac{\pi}{101}})^{101}}{1-e^{2i\frac{\pi}{101}}}=0 \\
1+\sum_{k=1}^{100}e^{-2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{-2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{-2i\frac{\pi}{101}})^{101}}{1-e^{-2i\frac{\pi}{101}}}=0 $$
Therefore
$$\sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(-1-1+200\right) $$
|
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|
Prove $f(x) =\ \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1$ has only one root. We have to prove that the equation $\displaystyle \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1=0$ have exactly one real root .
My sir told me it is just an application of derivative .
But I could not understand what he mean by that . Can anybody please explain me .
|
With a little inspection you can note that you have a truncated exponential function, which is monotonically increasing, and as pointed out can only have one real root. However, letting $f(x) = \frac{x^5}{120} + \frac{x^4}{24} + \frac{x^3}{6} + \frac{x^2}{2} + x + 1$, and noting that
$$f'(x) = \frac{x^4}{24} + \frac{x^3}{6} + \frac{x^2}{2} + x + 1$$
Setting $f'(x) = 0$ we see that we have no real solutions. So there are no critical numbers, which shows the slope of the function does not change signs (either always increasing or decreasing), in this case always increasing. It suffices to show that the function $f(x)$ takes on a negative value and a positive value, and concluding by the intermediate value theorem that there exists a point in between those two values where it crosses the $x$-axis. We have that
$f(-3) = \dfrac{-13}{20}$ and $f(0) = 1$, so your real root lies between $-3$ and $0$.
|
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|
show that $\sum_{i = 1}^{2k}\frac{ (-1)^{i+1}}{i} = \sum_{i = k+1}^{2k} \frac{1}{i}$ I have a proof but it does not seem elegant. Is there a more elegant solution? Thanks.
Consider $X = \sum_{i = 1}^{2k}\frac{(-1)^{i+1}}{i} = X_1 + X_2$ where
$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$
$X_2 = \frac{1}{k+1} - \frac{1}{k+2} + ... + \frac{1}{2k-1} - \frac{1}{2k}$
$k$ is even.
Every negative term $d$ in $X_2$ is of the form $\frac{1}{2^my}$ where $y$ is odd and could be $1$. Also $y \le k$
For every such $d$, we will have the following terms
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y}$ in $X_1$.
Adding $d$ to the above, we get
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y} - \frac{1}{2^my} = \frac{1}{2^my}$
So every negative term $d$ in $X_2$ is now replaced with a term of the same magnitude but with a positive sign.
Hence $X = \frac{1}{k+1} + \frac{1}{k+2} + ... + \frac{1}{2k-1} + \frac{1}{2k}$
|
Let $S_k=\sum_{k=1}^{2k}(-1)^{i+1}/i$ and $T_k=\sum_{i+1}^{2i}1/i$. Then
$$T_k-T_{k-1}=\frac1{2k-1}+\frac1{2k}-\frac1k=\frac1{2k-1}-\frac1{2k}=
S_k-S_{k-1}.$$
Obviously, $S_0=T_0$ so by induction....
|
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|
How to find two vectors orthogonal to the gradient space of the feasible set? After finding the solution to the minimization of $f(x,y,z)=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2$ constrained by $h(x,y,z)=x^2+y^2+z^2-1=0$ by using the Lagrange multiplier method:
$$\nabla f(x,y,z) = 2 \begin{pmatrix} x-\frac{1}{2}\ \\ y-\frac{1}{2} \\ z-\frac{1}{2}\end{pmatrix}=\begin{pmatrix} 2x-1\ \\ 2y-1 \\ 2z-1\end{pmatrix}, \nabla h(x,y,z)=\begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$$
such that: $\begin{pmatrix} 2x-1\ \\ 2y-1 \\ 2z-1\end{pmatrix}=\lambda \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}, x=y=z=\frac{1}{2(1-\lambda)}$
solving for $\lambda$ using constraint: $3\left(\frac{1}{2(1-\lambda)}\right)^2=1 \Rightarrow 2(1-\lambda)= \pm \sqrt{3} \Rightarrow \lambda=1 \pm \frac{\sqrt{3}}{2}$
which yields: $x=y=z=\pm \frac{1}{\sqrt{3}}$
making the final solution: $(x,y,x,\lambda)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1 - \frac{\sqrt{3}}{2}\right)$
Using that solution, it was suggested to use Gramm-Schmidt orthogonalization to construct two independent vectors orthogonal to $\nabla h\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ which will be in the tangential space of the feasible set.
I understand that I could just take two solutions to $a\frac{1}{\sqrt{3}}+b\frac{1}{\sqrt{3}}+c\frac{1}{\sqrt{3}}=0$ where $\vec{v}=(a,b,c)$ are the orthogonal vectors, but I'm curious how to implement Gramm-Schmidt with just one vector because as soon as you get to the second iteration, it seems that you need another vector. Is the idea to construct another linearly independent vector first and then use Gramm-Schmidt?
|
The optical inspection method works. The target vector is of the form
$$
u =
\left[ \begin{array}{c}
1 \\ 1 \\ 1
\end{array} \right]
$$
Two orthogonal vectors are
$$
v_{1} =
\left[ \begin{array}{r}
1 \\ 0 \\ -1
\end{array} \right],
\quad
v_{2} =
\left[ \begin{array}{r}
1 \\ -1 \\ 0
\end{array} \right].
$$
The process of Gram and Schmidt will orthogonalize the pair.
Update: once again @amd has a more elegant solution. Form the third vector as a cross product
$$
v_{2} = u \times v_{1}
$$
Final solution
$$
%
u = \frac{1}{\sqrt{3}}
\left[ \begin{array}{c}
1 \\ 1 \\ 1
\end{array} \right], \quad
%
v_{1} = \frac{1}{\sqrt{2}}
\left[ \begin{array}{r}
1 \\ 0 \\ -1
\end{array} \right], \quad
%
v_{2} = \frac{1}{\sqrt{6}}
\left[ \begin{array}{r}
1 \\ -2 \\ 1
\end{array} \right]
%
$$
|
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|
X and Y intercept basic issue Finding the Intercepts of the Graph of an Equation
Given an equation involving x and y, we find the intercepts of the graph as follows
*
*x-intercepts have the form (x,0); set y = 0 in the equation and solve for x.
*y-intercepts have the form (0,y); set x = 0 in the equation and solve for y
\begin{align}
y & = 2\sqrt{{x + 4}}
-2
\end{align}
Find the X and Y intercepts by setting each variable to 0. I believe I have X right but not Y, please explain where I am messing this up, should be a basic algebra folly.
X intercept: (I believe I have this right, though I should have just subtracted out the -2 early on)
\begin{align}
y & = 2\sqrt{{x + 4}}
-2
\end{align}
\begin{align}
0 & = 2\sqrt{{x + 4}}
-2
\end{align}
\begin{align}
0^2 & = 2^2{{(x + 4)}}
-2^2
\end{align}
\begin{align}
0 & = 4{{x + 16}}
-4
\end{align}
\begin{align}
0 & = 4{{x + 12}}
\end{align}
\begin{align}
-12 & = 4{{x}}
\end{align}
\begin{align}
-3 & = {{x}}
\end{align}
Thus the X intercept is (-3, 0)
Y intercept: I know this is wrong, textbook lists the answer as 2
\begin{align}
y & = 2\sqrt{{x + 4}}
-2
\end{align}
\begin{align}
y & = 2\sqrt{{0 + 4}}
-2
\end{align}
\begin{align}
y & = 2\sqrt{{4}}
-2
\end{align}
(Square both sides to remove the radix?)
\begin{align}
y^2 & = 4*4
-4
\end{align}
\begin{align}
y^2 & = 12
\end{align}
\begin{align}
y & = \sqrt12
\end{align}
Actual answer is 0, 2 for y intercept. So what's my mistake?
|
For the $x$-intercept: You need to isolate the square root before you square both sides. So when you have
$$ 0 = 2\sqrt{x+4}-2,$$
you don't want to square both sides here because you will have to FOIL out the right side like
$$0^{2} = (2\sqrt{x+4}-2)^{2}$$
and you will end up with a mess.
If you get the square root by itself you will have
$$1 = \sqrt{x+4}$$
then you can square both sides to get
$$1^{2} = (\sqrt{x+4})^{2}$$
and you can solve from here. (Notice that you had the same answer in your attempt, but that was pure coincidence)
For the $y$-intercept: You don't need to square both sides, you have
$$y = 2\sqrt{4} -2$$
and you know that $\sqrt{4}=2$, so this is just
$$y=2\cdot2-2=4-2=2.$$
|
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|
Integrate $\int \frac {1}{(x+2)(x+3)} \textrm {dx}$ Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$
My Attempt:
$$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$
$$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$
$$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$
$$\textrm {log} (x+2) . \textrm {log} (x+3) + C$$
Is this correct? Or, How do I proceed the other way?
|
Nope! That's not how integration works.
You want to split the denominator using partial fractions.
$\displaystyle\int \frac{1}{(x+3)(x+2)}\,dx =\int \left(\frac{1}{x+2}-\frac{1}{x+3}\right)\,dx=\ln|x+2|-\ln|x+3|+C=\boxed{\ln \left|\frac{x+2}{x+3}\right|+C}$
|
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|
Finding all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ Find all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ for all $x>0$
Attempt: Let $$\sqrt{9-(x-a)^2}> \sqrt{16-x^2}$$
So $$x^2-(x-a)^2>7$$
So $$a(2x-a)>7\Rightarrow a^2-2ax+7<0$$
could some help me how to find range of $a,$ Thanks
|
For the square roots to be defined, we need $|x|\leq 4$ and $|x-a|\leq 3$.
To solve a quadratic inequality, it's useful to find its roots. We have
$$a^2-2ax+7=0\iff a=\frac{2x\pm\sqrt{4x^2-28}}{2}=x\pm\sqrt{x^2-7}$$
In particular, if $|x|\leq\sqrt{7}$ the polynomial is always nonnegative in $a$.
If $\sqrt{7}<|x|\leq 4$, then the solution set will be all $a$ such that $|x-a|\leq 3$ and
\begin{align}&x-\sqrt{x^2-7}<a<x+\sqrt{x^2-7}\\
\iff &-\sqrt{x^2-7}<a-x<\sqrt{x^2-7}\\
\iff &|a-x|=|x-a|<\sqrt{x^2-7}\end{align}
Now, observe that if $|x|\leq 4$, then $x^2-7\leq 9$, so that $\sqrt{x^2-7}\leq 3$. In the end, all we need is:
$$\sqrt{7}<|x|\leq 4\,\,\text{ and }\,\,|x-a|<\sqrt{x^2-7}.$$
|
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|
geometric series calculating the sum for first 45 terms I am able to calculate up to the step in blue, but I cant understand how do I simply the terms from -3^45 to 10^20 ?
Can anyone explain?
|
$$S_{45}=-\frac{1}{8}(1-(-3)^{45})=-\frac{1}{8}+\frac{1}{8}(-3)^{45}$$
$$=-\frac{1}{8}-\frac{1}{8}(3)^{45}=-\frac{1}{8}-\frac{1}{8}(3^5)^9$$
Since $ $ $3^5=3\times3\times3\times3\times3=243$, we have
$$-\frac{1}{8}-\frac{1}{8}(243)^9=-\frac{1}{8}-\frac{1}{8}(2.43\times10^2)^9=-\frac{1}{8}-\frac{1}{8}(2.43^9\times10^{18})$$
Since $ $ $2.43^9=2954.31270…$ (you can either do this manually by multiplying 2.43 by 2.43, systematically, 9 times (not recommended), or use a calculator), we have
$$-\frac{1}{8}-\frac{1}{8}(2954.31270…\times10^{18})$$
$$=-\frac{1}{8}-\frac{2954.31270…}{8}\times10^{18}$$
$$=-\frac{1}{8}-369.2890…\times10^{18}$$
$$=-\frac{1}{8}-3.692890…\times10^2\times10^{18}$$
$$=-\frac{1}{8}-3.692890…\times10^{20}$$
$$\simeq-3.69\times10^{20}$$
|
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|
Induction proof of divisibility I am trying to solve a proof by induction, which says
For all $n \in \mathbb{N}_0$,
$$ ((2n+1)^2 - 2^{n+1}(2n+1) + 2^{2n+1}) \mid ((2n+1)^4 + 4^{2n+1}). $$
Every try ended in a dead end. Does anybody have a hint?
|
Notice that we can expand the right hand side as
\begin{equation}
\begin{split}
(2n+1)^4+4^{2n+1} &= \left((2n+1)^2+2^{2n+1}\right)^2-2^{2n+2}\cdot(2n+1)^2 \\&= \left((2n+1)^2-2^{n+1}\cdot(2n+1)+2^{2n+1}\right)\left((2n+1)^2+2^{n+1}\cdot(2n+1)+2^{2n+1}\right)
\end{split}
\end{equation}
Hence $$\left((2n+1)^2-2^{n+1}\cdot(2n+1)+2^{2n+1}\right)| \left((2n+1)^4+4^{2n+1}\right) $$
as desired.
|
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|
How do you prove that the sum of 1 to k divided by k is a multiple of 1/2? Expressed mathematically: $\frac{1}{k}\sum_{i=1}^k n = \frac{x}{2}$ where x is an integer. I tested this series mathematically and I found this trend, but I don't know how to actually prove it. Thanks in advance.
|
Notice that $\frac 1k \sum\limits_{n=1}^k n$ is, by definition the average of $1,2,3,4......, .... k$. Convince yourself that the average of $1....n$ (equally distributed) is $\frac {k+1}2$.
.... or ...
Let $N = \sum\limits_{n=1}^k n = \sum\limits_{n=k;-1}^1 n = \sum\limits_{n=1}^k ((k+1) -n)$
So $2N = (\sum\limits_{n=1}^k n) + (\sum\limits_{n=1}^k ((k+1) -n))$
$= \sum\limits_{n=1}^k [n + (k=1) - n] = \sum\limits_{n=1}^k (k+1)$
$= k*(k+1)$
So $N = \frac {k(k+1)}{2}$
.... or ....
$N = 1+2+3..........+ (k-2) + (k-1) + k$
$N = k + (k-1) + (k-2) + .....+ 3 + 2 + 1$
$N+N = (1+k) + (2+k-1) + (3k-2 ) + ........ +(k-2+3) + (k-1+2) + (k+1)$
$2N = (k+1) + (k+1) + (k+1) + ..... + (k+1) + (k+1) + (k+1)$
$2N = k*(k+1)$.
So $N = \frac {k(k+1)}2$
..... or do induction.....
$\frac 1k \sum^k n = \frac {k+1}2$ is true for $k = 1$ as $\frac 11 = \frac 22$.
$\frac 1k \sum^k n = \frac {k+1}2$ then
$\sum^k n = \frac {k(k+1)}2$
$\sum^{k+1}n = \frac {k(k+1)}2 + (k+1)$
$= \frac {k(k+1)}2 +\frac {2(k+1)}2$
$= \frac {(k+2)(k+1)}2$
So $\frac {1}{k+1} \sum^{k+1}n = \frac {k+2}2$.
|
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|
$p$,$q$ and $r$ are roots of $x^3+2x^2+3x+3=0$ $p$,$q$ and $r$ are roots of $x^3+2x^2+3x+3=0$. Find value of
$$\sum_{cyc}\left(\frac{p}{p+1}\right)^3$$
I have modified the equations as $$(x+1)^3=x^2-2$$ so
$$ \frac{x^3}{(x+1)^3}=\frac{x^3}{x^2-2}=x+\frac{1}{x-\sqrt{2}}+\frac{1}{x+\sqrt{2}}$$ So
$$\sum_{cyc}\left(\frac{p}{p+1}\right)^3=\sum_{cyc}p+\sum_{cyc}\frac{1}{p-\sqrt{2}}+\sum_{cyc}\frac{1}{p+\sqrt{2}}$$
After this i have find equation whose roots are $p-\sqrt{2}$, $q-\sqrt{2}$ and $r-\sqrt{2}$ and similarly for other cyclic expression. but its too lengthy
|
There is an easier way if $y=\frac{1}{x}$ then $y$ is a root of
$$3y^3+3y^2+2y+1=0$$ if $z=y+1$ then $z$ is a root of
$$3z^3-6z^2-6z-2=0$$ and if $w=\frac{1}{z}$ then $w$ is a root of
$$2w^3+6w^2+6w-3=0$$ and in the highly unlikely event I have not made a mistake, you can use Newton formula to find the sum of the cubes of the roots of this last equation.
|
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|
Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?
|
We can restrict to $x\in[0,\pi/2]$, because of symmetries.
The maximum and minimum of $v$ are attained at the same values as the maximum and minimum of $v^2$:
$$
v^2=a^2+b^2+
2\sqrt{(a^2\cos^2(x)+b^2\sin^2(x))(b^2\cos^2(x)+a^2\sin^2(x))}
$$
We can also remove $a^2+b^2$, then the factor $2$ and square again, so we reduce to finding the maximum and minimum of
$$
f(x)=(a^2\cos^2(x)+b^2\sin^2(x))(b^2\cos^2(x)+a^2\sin^2(x))
$$
If we set $t=\cos^2x$, we want to look at
$$
g(t)=((a^2-b^2)t+b^2)((b^2-a^2)t+a^2)
$$
subject to $0\le t\le 1$.
If $a^2=b^2$, the function is constant, so it's not restrictive to assume $a^2\ne b^2$. The (global) function $g$ assumes its maximum at the middle point between its zeros:
$$
\frac{1}{2}\left(\frac{b^2}{b^2-a^2}+\frac{a^2}{a^2-b^2}\right)=\frac{1}{2}
$$
which is in $[0,1]$ so it's our required maximum.
The minimum is either at $0$ or $1$ (the graph of $g$ is a parabola): since $g(0)=a^2b^2=g(1)$, we can choose either one.
Thus we have a maximum for $t=1/2$, which corresponds to $x=\pi/4$, and a minimum at $x=0$ (or $x=\pi/2$).
We have
$$
v(\pi/4)=2\sqrt{\frac{1}{2}(a^2+b^2)}
$$
and
$$
v(0)=v(\pi/2)=|a|+|b|
$$
|
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|
Find all functions $f$ such that $f(x)+f(\frac{1}{1-x})=x$ I would like to find all functions $f:\mathbb{R}\backslash\{0,1\}\rightarrow\mathbb{R}$ such that
$$f(x)+f\left( \frac{1}{1-x}\right)=x.$$
I do not know how to solve the problem. Can someone explain how to solve it?
In one of my attempts I did the following, which is confusing to me: By the substitution $y=1-\frac{1}{x}$ one gets
$f(y)+f\left( \frac{1}{1-y}\right)=\frac{1}{1-y}$. So with $x=y$ it follows that $0=x-\frac{1}{1-x}$. So it would follow that there is no solution. Is that possible or is there a mistake?
Best regards
|
make $x:= \frac{1}{1-x}$ then
$$f\left( \frac{1}{1-x}\right)+f\left( \frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}\to
f\left( \frac{1}{1-x}\right)+f\left(1- \frac{1}{x}\right)=\frac{1}{1-x}\quad (1)$$
do it again in the last equation:
$$f\left( \frac{1}{1-\frac{1}{1-x}}\right)+f(x)=\frac{1}{1-\frac{1}{1-x}}\to f\left(1- \frac{1}{x}\right)+f(x)=1- \frac{1}{x}\quad (2)$$
now make $(1)-(2)$ and get:
$$f\left( \frac{1}{1-x}\right)-f(x)=\frac{1}{1-x}-1+\frac{1}{x}$$
Subtract the equation is the statement and this last one.
$$2f(x)=x-\frac{1}{1-x}+1-\frac{1}{x}\to f(x)=\frac12\left(x-\frac{1}{1-x}+1-\frac{1}{x}\right)$$
|
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|
Is there a possible explanation in plain English how to use the Chinese Reminder Theorem? For example, if it is the problem of
Find the smallest integer that leave a remainder of 3 when divided by
5, a remainder of 5 when divided by 7, and a remainder of 7 when
divided by 11
In some explanation such as in this article it started to use mod, and the whole explanation seemed to very difficult to understand. Is there a more plain English explanation of how this is solved?
|
It's definitely worthwhile learning a little bit about modular arithmetic for questions such as this.
But if you want to know in very concrete terms how to solve questions such as this, here's an attempt.
Suppose you want a number $n$ (not necessarily the least, we'll handle that later) with remainder $3$ when divided by $5$ and $5$ when divided by $7$, which we write as $n \equiv 3 \mod 5$ and $n \equiv 5 \mod 7$. The Chinese remainder theorem tells us that this is possible, because $5$ and $7$ are
relatively prime, i.e. have no common factor greater than $1$.
We start out by finding integers $x$ and $y$ such that $5 x + 7 y = 1$.
If so, then $5 x = 1 - 7 y$ is an integer $u$ such that $u \equiv 0 \mod 5$ and $u \equiv 1 \mod 7$, while $7 y = 1 - 5 x$ is a number $v$ such that $v \equiv 1 \mod 5$ and $v \equiv 0 \mod 7$. Combining these, $n = 5 u + 3 v$ would have $n \equiv 5 \cdot 0 + 3\cdot 1 = 3 \mod 5$ and $n \equiv 5 \cdot 1 + 3 \cdot 0 = 5 \mod 7$.
The method for finding $x$ and $y$ is called the Euclidean algorithm. It goes like this. We want $5 x + 7 y = 1$. Write $7 = 5 + 2$, so
$5 x + 5 y + 2 y = 5 (x + y) + 2 y$. Thus with $x_1 = x + y$, we want to solve $5 x_1 + 2 y = 1$ (a similar equation to what we started with, but with smaller coefficients). If we can do that, then we can take $x = x_1 - y$. to solve our original equation.
Similarly, $5 = 2 \cdot 2 + 1$, so $5 x_1 + 2 y = x_1 + 2 (2 x_1 + y) = x_1 + 2 y_1$ where $y_1 = 2 x_1 + y$.
But the equation $x_1 + 2 y_1 = 1$ is easy: we can see at a glance that $x_1 = 1$, $y_1 = 0$ is a solution. Then working backwards, $y = y_1 - 2 x_1 = -2$
and $x = x_1 - y = 1 - (-2) = 3$. So we have found $x = 3$ and $y = -2$ such
that $5 x + 7 y = 1$. We then take $u = 5 x = 15$ and $v = 7y = -14$, and
$n = 5 u + 3 v = 33$, which has $n \equiv 3 \mod 5$ and $n \equiv 5 \mod 7$.
Now, $n = 33$ is not the only possible solution. In fact, we could get another solution by adding or subtracting any multiple of $5 \times 7 = 35$.
Since $5$ and $7$ are relatively prime, the theory tells us that those are all the other solutions: if $m$ is another solution, $n - m$ would have to be a multiple of $5$ and a multiple of $7$, and therefore a multiple of $35$.
Now in this particular case $33 - 35$ is negative, in fact $33$ is the smallest positive solution.
Now, what about the original problem with $n \equiv 7 \mod 11$ in addition to $n \equiv 3 \mod 5$ and $n \equiv 5 \mod 7$? We know now that from the first two conditions we need $n$ to differ from $33$ by a multiple of $35$, i.e.
$n \equiv 33 \mod 35$. So we go through a similar procedure, finding
$x$ and $y$ such that $35 x + 11 y = 1$, etc. I'll leave the details to you.
|
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|
Problem in Rationalization The Question is :
If $1 < a <2$ then
$\sqrt{a - 2 \sqrt{a - 1}} - \sqrt{a + 2 \sqrt{a - 1}}$ can be:
A) $2$
B)$-2 \sqrt{a-1}$
C) $0$
D) $\sqrt{a-1}$
I have tried solving it a number of times but failed every time.
I took $\sqrt{a-1}$ as b then tried to rationalise by multiplying with unity ($\sqrt{a-2b}$ with $\sqrt{a+2b}$ and vice versa). However, after a few steps I always arrived at the question itself not the answer.
Can someone please help me with this sum?
|
$$k=\sqrt{a-2\sqrt{a-1}}-\sqrt{a+2\sqrt{a-1}}$$
square both sides and get:
$$k^2=2a-2\sqrt{a^2-4(a-1)}=2a-2|a-2|$$
once $1<a<2$ then $|a-2|=2-a$, so
$$k^2=2a-2(2-a)=4a-4\to k=\pm2\sqrt{a-1}$$
but $k<0$ because $\sqrt{a-2\sqrt{a-1}}<\sqrt{a+2\sqrt{a-1}}$, so
$$k=-2\sqrt{a-1}$$
|
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|
How to find these singular matrices? Let $X$ and $Y$ be two matrices different from $I$ such that $XY=YX$ and $X^n-Y^n$ is invertible for some natural number $n$ . If $X^n-Y^n=X^{n+1}-Y^{n+1}=X^{n+2}-Y^{n+2}$, then:
A) $I-X$ is singular
B) $I-X$ is singular
C) $X+Y=XY+I$
D) $(I-X)(I-Y)$ is non-singular
Like everyone, I started off taking $n=1$.
Now $|X-Y|\ne 0$ (because it's invertible)
then $X-Y=(X-Y)(X+Y)=(X-Y)(X^2+Y^2+XY)$
making it $(X-Y)(X-Y-I)=O$ and $(X-Y)(X^2+Y^2+XY-I)=O$
and after this. I'm lost and require help.
|
Our assumptions are as follows:
$1.) \: X,Y \neq I;$
$2.) \:X \: \text{and} \:Y \:\text{commute;}$
$3.) \:\text{there is some }n \text{ such that }X^n - Y^n \:\text{is invertible;}$
$4.) \: \text{for that same }n \text{, we have that } X^n - Y^n = X^{n+1} - Y^{n+1} = X^{n+2}-Y^{n+2}.$
$\text{--} $
By 2.), it is easily verifiable that $X^jY^k = Y^kX^j $, for any $j,k \in \Bbb{N}. $
Then, for the $n$ in 3.), we have
$\begin{align}
\smash{(X^{n+1} - Y^{n+1})(X+Y)}&=X^{n+2} + X^{n+1}Y - Y^{n+1}X-Y^{n+2}\\
&=X^{n+1}Y - Y^{n+1}X + X^{n+2} - Y^{n+2}\\
&=X^nXY - Y^{n}XY + X^{n+2} - Y^{n+2}\\
&=(X^n - Y^n)XY + (X^{n+2} - Y^{n+2})\\
\end{align}$
By 4.), we replace in the above LHS and RHS to achieve
$(X^n - Y^n)(X +Y) = (X^n - Y^n)XY + (X^n - Y^n)$
Since $X^n - Y^n$ invertible, we can left-side multiply both sides by whatever that inverse is to get
$X+Y = XY + I $
Hence, C) is true.
$\text{--} $
From that, we now rearrange and have
$\begin{align}
\smash{0}&=X+Y - XY - I\\
&=X-XY + Y-I\\
&=X(I-Y)-(I-Y)\\
&=(X-I)(I-Y)\\
\end{align}$
Multiply both sides by $-1$, and we have
$ (I-X)(I-Y) = 0$,
which is not invertible. Hence, D) is false.
$\text{--} $
Since $Y \neq I $, then $ I-Y \neq 0$. This means that $I-Y$ doesn't send everything to $0$;
i.e. there is some $u,v \neq 0 $ such that
$ (I-Y)u=v$.
Left-side multiply each side by $ (I-X)$ and we get
$(I-X)(I-Y)u = (I-X)v$
But $(I-X)(I-Y)=0$, so $ (I-X)v = 0 $, meaning there is a non-zero $v$ such that
$(I-X)v = 0v$.
Thus, $ 0$ is an eigenvalue of $ (I-X)$, and since the determinant of an operator is the product of its eigenvalues, then
$ \det(I-X) = 0 \implies (I-X) $ is singular $\implies $A) is true.
$\text{--} $
I assume B) from your problems was meant to read "$ I-Y$ is singular", and by the commutativity of $X$ and $Y$, the above argument can be reworked to show just that; i.e. B) is true.
|
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|
Integrals with Taylor expansion? Can I use the Taylor series to find this integrals?
$$ \lim_{n\to \infty } \int _{0}^n \frac{\arctan x}{x^2 + x + 1}dx $$
$$\lim _{n\to \infty }n \int_{-1}^0(x + e^x)^{n}dx = \: ?$$
|
Here is a relatively simple standard approach to evaluating the first of your integrals.
Let
$$I = \int_0^\infty \frac{\arctan x}{x^2 + x + 1} \, dx.$$
Dividing up the interval of integration as follows
$$I = \int_0^1 \frac{\arctan x}{x^2 + x + 1} \, dx + \int_1^\infty \frac{\arctan x}{x^2 + x + 1} \, dx.$$
In the second of the integrals appearing to the right, enforcing a substitution of $x \mapsto 1/x$, one has
\begin{align}
I &= \int_0^1 \frac{\arctan x}{x^2 + x + 1} \, dx + \int_0^1 \frac{\arctan \left (\frac{1}{x} \right )}{x^2 + x + 1} \, dx\\
&= \int_0^1 \frac{\arctan (x) + \arctan \left (\frac{1}{x} \right )}{x^2 + x + 1} \, dx\\
&= \frac{\pi}{2} \int_0^1 \frac{dx}{x^2 + x + 1} \tag1\\
&= \frac{\pi}{2} \int_0^1 \frac{dx}{\left (x + \frac{1}{2} \right )^2 + \frac{3}{4}}\\
&= \frac{\pi}{\sqrt{3}} \left [\arctan \left (\frac{2x + 1}{\sqrt{3}} \right ) \right ]_0^1\\
&= \frac{\pi^2}{6 \sqrt{3}}
\end{align}
At step (1) the following well-known result for the inverse tangent function has been used:
$$\arctan (x) + \arctan \left (\frac{1}{x} \right ) = \frac{\pi}{2}, \qquad x > 0.$$
|
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|
What is the result of the following binomial sum? When computing the expectation of $S_n^{-1}I_{[S_n > 0]}$ with $S_n \sim \text{binomial}(n, p)$, I need to evaluate the sum:
\begin{align*}
\sum_{k = 1}^n \frac{1}{k}\binom{n}{k}p^{k}(1 - p)^{n - k}
\end{align*}
for $p \in (0, 1)$.
Is there any well-known binomial identity that can be directly applied to get the above sum? Or any other methods?
My attempt:
Denote $\sum_{k = 1}^{n - 1} \frac{1}{k}\binom{n}{k}p^{k}(1 - p)^{n - k}$ by $f(p), p \in (0, 1)$. Differentiating once with respect to $p$, we get
\begin{align*}
f'(p) = - \frac{n}{1 - p}f(p) + \frac{1}{p(1 - p)}(1 - p^n - (1 - p)^n)
\end{align*}
This nonhomogeneous linear ODE seems solvable but could be tedious...
|
Here is a transformation which simplifies the sum somewhat by extracting Harmonic numbers $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$.
We obtain for $n\geq 1$
\begin{align*}
\color{blue}{f_n}&\color{blue}{=\sum_{k = 1}^n \frac{1}{k}\binom{n}{k}p^{k}(1 - p)^{n - k}}\\
&=\sum_{k = 1}^n \frac{1}{k}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right)p^{k}(1 - p)^{n - k}\\
&=\sum_{k = 1}^n \frac{1}{k}\binom{n-1}{k}p^{k}(1 - p)^{n - k}
+\frac{1}{n}\sum_{k = 1}^n \binom{n}{k}p^{k}(1 - p)^{n - k}\\
&=(1-p)f_{n-1}+\frac{1}{n}\left((p+(1-p))^n-(1-p)^n\right)\\
&\color{blue}{=(1-p)f_{n-1}+\frac{1}{n}-(1-p)^n\frac{1}{n}}\\
\end{align*}
Iterating this recurrence relation we get with $f_1=p$
\begin{align*}
\color{blue}{f_n}&=(1-p)f_{n-1}+\frac{1}{n}-(1-p)^n\frac{1}{n}\\
&=(1-p)^2f_{n-2}+(1-p)\frac{1}{n-1}+\frac{1}{n}-(1-p)^n\left(\frac{1}{n-1}+\frac{1}{n}\right)\\
&=(1-p)^3f_{n-3}+(1-p)^2\frac{1}{n-2}+(1-p)\frac{1}{n-1}+\frac{1}{n}\\
&\qquad-(1-p)^n\left(\frac{1}{n-2}+\frac{1}{n-1}+\frac{1}{n}\right)\\
&=\cdots\\
&=(1-p)^{n-1}f_1+\sum_{k=0}^{n-2}(1-p)^k\frac{1}{n-k}-(1-p)^n\left(H_n-1\right)\\
&=(1-p)^{n}\left(\frac{p}{1-p}+1-H_n\right)+\sum_{k=0}^{n-2}(1-p)^{n-k-2}\frac{1}{k+2}\tag{1}\\
&\color{blue}{=(1-p)^{n}\left(-H_n+\sum_{k=1}^{n}(1-p)^{-k}\frac{1}{k}\right)}\tag{2}\\
\end{align*}
Comment:
*
*In (1) we change the order of summation $k\rightarrow n-2-k$, use $f_1=p$ and do some simplifications.
*In (2) we do some simplifications, shift the index and start with $k=1$.
|
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|
Factorial expression of $\frac{n!}{(n-r)!}$ If $n \gt r,$
Then how is the following valid?
$$n \cdot (n-1) \cdot (n-2) \cdots (n-(r-1))=\frac{n!}{(n-r)!}?$$
I thought it the list of multiplications would be equal to $\frac{n!}{(n-(r-1))!}?$
|
$\require{cancel}$Maybe a concrete example will make things clearer:
\begin{align}
& \frac{9!}{(9-4)!} = \frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1} = \frac{9\cdot8\cdot7\cdot6\cdot\cancel{5\cdot4\cdot3\cdot2\cdot1}} {\cancel{5\cdot4\cdot3\cdot2\cdot1}} = 9\cdot8\cdot7\cdot6 \\[12pt]
= {} & \underbrace{9\cdot8\cdot7\cdot 6}_{\text{4 factors since} \\ 9 \,-\, 4 \text{ rather} \\ \text{than } 9 \text{ minus} \\ \text{something else} \\
\text{appeared.}} \quad \text{and the last factor, 6, is } (9-4+1).
\end{align}
|
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|
$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y = 0$ by Frobenius, finding $s$ in $\sum a_n x^{n+s}$ I need to solve:
$$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y = 0, \ \ \ a\ge 0$$
by Frobenius method. So I did:
$$y = \sum_{n=0}^{\infty}a_nx^{n+s}$$
$$y' = \sum_{n=0}^{\infty}a_n(n+s)x^{n+s-1}$$
$$y'' = \sum_{n=0}^{\infty}a_n(n+s)(n+s-1)x^{n+s-2}$$
so
$$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y = \\ \sum_{n=0}^{\infty}a_n(n+s)(n+s-1)x^{n+s-2} + \sum_{n=0}^{\infty}a_n(n+s)x^{n+s-3} + \sum_{n=0}^{\infty}(-a^2)a_nx^{n+s-2}$$
I couldn't find a way to make $\sum_{n=0}^{\infty}a_n(n+s)x^{n+s-3}$ as a series with coefficient $x^{n+s-2}$ so I had to change the other terms to have $x^{n+s-3}$. I eneded up with:
$$\sum_{n=1}^{\infty}a_{n-1}(n-1+s)(n+s-2)x^{n+s-3} + \\ a_0\cdot s \cdot x^{s-3} + \sum_{n=1}^{\infty}a_n(n+s)x^{n+s-3} + \\ -a^2\cdot a_0\cdot x^{s-2} + \sum_{n=1}^{\infty} a_{n-1}x^{n+s-3}$$
For $x^{s-2}$ we have: $x^{s-2}(a_0s-a^2a_0) = 0 \implies s = a^2$. But the term $x^{s-3}$ says that $a_0$ must be $0$ or $s$ must be $0$. I guess there's some problem here. What's happening?
Also, I ended up with the following relation:
$$x^{n+s-3}(a_{n-1}(n-1+s)(n+s-2) + a_n(n+s) + a_{n-1}) = 0$$
Do you think this relation is right, even though I had a problem determining the possible $s$ variables that can be used?
|
The coefficient of $y'$ is $1/x^2$, so $x=0$ is not a regular singular point. The Frobenius method won't work here.
|
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|
Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$
Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$
frist I use $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$
and multiple two side by $x^2$
can get $$\sum_{n=1}^\infty x^{n+2} = \frac{x^3}{1-x}$$
then diff each side two times
we can obtain $$\sum_{n=1}^\infty (n+1)(n+2)x^n= \frac{2x^3-6x^2+6x}{(1-x)^3}$$
but not the solution $=\frac{1}{(1-x)^3}$
|
Starting from zero,
$\begin{array}\\
\sum\limits_{n=0}^\infty \frac{(n+2)!}{(2!)(n!)}x^n
&=\sum\limits_{n=0}^\infty \binom{n+2}{2}x^n\\
&=\frac1{x^2}(1-x)^{-3}\\
\end{array}
$
by the generalized binomial theorem
(since
$\binom{-n}{k}
=(-1)^n \binom{n+k-1}{k}
$).
|
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|
Error in Polynomial Factoring
From Ramanujan's Notebooks IV:
Let $\alpha,\beta$ and $\gamma$ be the roots of$$x^3-ax^2+bx-1=0\tag1$$Now, choose cube roots such that $(\alpha\beta\gamma)^{1/3}=1$ and then let$$z^3-\theta z^2+\varphi z-1=0\tag2$$Denote the cubic polynomial with roots $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$. Cubing both sides of the equality$$z^3-1=\theta z^2-\varphi z\tag3$$We find that$$(z^3-1)^3-\theta^3z^6+\varphi^3z^3+3\theta\varphi z^3(z^3-1)=0\tag4$$Since $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$ are roots of $(2)$, they are also the root of $(4)$. As a cubic polynomial in $z^3$, $(4)$ thus has the roots $\alpha,\beta,\gamma$. Comparing $(1)$ with $(4)$, we deduce$$a=\theta^3+3-3\theta\varphi\tag5$$$$b=\varphi^3+3-3\theta\varphi\tag6$$ If we define $t$ by$$\theta^3=a+6+3t\tag7$$Then, by $(2)$ and $(7)$, $$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=\theta=(a+6+3t)^{1/3}$$And similarly, $$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=\varphi=(b+6+3t)^{1/3}$$
While the proof is fairly elementary, there is one problem that I'm not sure how Bruce got there. From $z^3-1=\theta z^2-\varphi z$, to the next step, how'd they get the $3\theta\varphi z^3(z^3-1)$. When you expand $\theta z^2-\varphi z$, you get$$(\theta z^2-\varphi z)^3=\theta^3z^6-3\theta^2\varphi z^5+3\theta\varphi^2z^4-\varphi^3z^3$$So where did that term come from?
|
The equation (4) is correct. After both sides of equation (3) are cubed and the terms on the right get moved to the left there is only zero on the right. Now the cube of the right side of (4) is expanded by binomial theorem to four terms. Two of them are combined and factored with one factor being the right side of (3) which is immediately replaced with the left side of (3). A simple example will make this clear. $$a = b-c,$$ $$a^3 = b^3 - 3b^2c +3bc^2 -c^3,$$ $$a^3 -b^3 +c^3 = -3bc(b - c),$$ $$a^3 - b^3 + c^3 = -3abc.$$ The author just combined several steps. Notice that if we change the sign of $b$ we get the pretty result that if $a+b+c=0$ then $a^3+b^3+c^3=3abc$, which is a fact that Ramanujan was very familiar with.
|
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Maximum value for $M^4$
$$M=|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+4}|$$
Find the maximum value of $M^4$.
I think it could have a geometric solution. Because it looks like the difference between two points formula. Please help.
|
Assuming $x \in \mathbb{R}$,
$$M(x) = \left \lvert \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right \rvert$$
and
$$M(x)^4 = (M(x)^2)^2 = g(x)^2$$
where
$$g(x) = \left ( \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right )^2$$
As Jaideep Khare and Abishanka Saha already succintly answered, we can write
$$M(x) = \left \lvert \sqrt{(x - 2)^2 + (0 \pm 1)^2} - \sqrt{(x - 1)^2 + (0 \pm \sqrt{3})^2} \right \rvert$$
where we can choose each $\pm$ between $+$ or $-$ without affecting the result. If we choose arbitrarily
$$M(x) = \left \lvert \sqrt{(x - 2)^2 + (0 - 1)^2} - \sqrt{(x - 1)^2 + (0 - \sqrt{3})^2} \right \rvert$$
then we can immediately see, as aforementioned members already explained, that $M(x)$ is the difference in distances from $(x,0)$ to $(2,1)$ and $(x,0)$ and $(1,\sqrt{3})$, which allows us to describe a result based on geometric reasoning.
Personally, as a non-mathematician, I felt unsure which triangle inequality would apply here, and was too lazy to find out. We have a triangle, with $(x, 0)$ at one vertex, and $(2,\pm1)$ and $(1,\pm\sqrt{3})$ as the other two, arbitrarily choosing either $+$ or $-$ for each $\pm$ to get the maximum difference in the length of edges meeting at $(x,0)$.
I get unreasonably suspicious — paranoid — whenever I need to make arbitrary decisions without a clear rule as to exactly why an arbitrary decision is necessary here, as opposed to some other rule stating which choice should be taken. I'm a physicist, and the first thing I always examine is whether the model I applied makes sense; and here, I just could not tell offhand why it should.
So, I decided to tackle it as a non-geometric math problem, and find out.
We know that a real-valued function $g(x)$ with $x \in \mathbb{R}$, reaches its (local) extrema whenever its derivative is zero. The global maximum is one of those. For the derivative in this case, we need to apply the chain rule (fixed!),
$$\frac{d b(a(x))}{d x} = \left [ \frac{d b(y)}{d y}\right ]_{y = a(x)} \frac{d a(x)}{d x}$$
twice. The subscript of the bracket means that after derivation with respect to $y$, we substitute $y$ with $b(x)$. In this case, the outermost function is square, the inner function is a square root, and the innermost are the functions inside the square roots in $g(x)$. We get
$$\frac{d g(x)}{d x} = 2 \left ( \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right ) \left ( \frac{2 x + 4}{2 \sqrt{x^2 + 4 x + 5}} - \frac{2 x + 2}{2 \sqrt{x^2 + 2 x + 4}} \right )$$
$$= 2 \left ( \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right) \left (\frac{ (x + 2)\sqrt{x^2 + 2 x + 4} - (x + 1)\sqrt{x^2 + 4 x + 5} }{\sqrt{(x^2 + 4 x + 5)(x^2 + 2 x + 4)}}\right)$$
This reaches zero when either part in parentheses is zero; the latter part in parentheses is only zero if and only if the numerator is zero:
$$\frac{d g(x)}{d x} = 0 \; \iff \; \begin{cases}
\sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} = 0 & \text{or} \\
(x + 2)\sqrt{x^2 + 2 x + 4} - (x + 1)\sqrt{x^2 + 4 x + 5} = 0 \end{cases}$$
The upper one is zero when $x = -1/2$, and the lower one when $x = -5/2 - \sqrt{3}/2$. Since $g(-1/2) = 0$ (and $g(x) \ge 0$ for all $x \in \mathbb{R}$, since $(y)^2 \ge 0$ for all $y \in \mathbb{R}$), it is the (or a) minimum; and since we only have two extrema (and $g(x) \ne 0$ for at least some $x \in \mathbb{R}$), $g(-5/2-\sqrt{3}/2)$ must be the global maximum.
Therefore, the maximum value of $M(x)^4 = g(x)^2$ when $x = -5/2 - \sqrt{3}/2$.
It was interesting (and reassuring!) to note that $g(-5/2-\sqrt{3}/2)$ corresponds to the solution obtained via geometric interpretation, $\left(\sqrt{(2-1)^2 + (1 - \sqrt{3})^2}\right)^2$. If we were to examine why this is so, and explore other similar problems, we'd onclude that the reason is we're governed by the basic triangle inequality here — which also explains the hint in Jaideep Kahre's answer: it is the special equality case, maximum, in the triangle inequality rule.
(I like that: it ties the loose ends I was unsure about in this problem very nicely. I often check my math this way, too — although I usually use e.g. Maple to do the hard work for me. I just pick the models that I think apply to this case, and see if they agree. If they do not, I definitely erred in choosing the model, and must examine where my understanding failed. If they agree, I've examined the problem from at least two different viewpoints, and therefore have at least some confidence in my solution. This is also the reason I posted this answer, even though there are already two concise answers posted: I wanted to show how I would approach the stated problem, as a weak-math-fu non-mathematician.)
The actual maximum value is obtained with straightforward substitution:
$$M \left(\frac{-5}{2} - \frac{\sqrt{3}}{2}\right)^4 = g \left(\frac{-5}{2} - \frac{\sqrt{3}}{2}\right)^4 = \left[\sqrt{\left(-\frac{5}{2} - \frac{\sqrt{3}}{2}\right)^2 - 5 - 2 \sqrt{3}} - \sqrt{\left(-\frac{5}{2} - \frac{\sqrt{3}}{2}\right)^2 - 1 - \sqrt{3} }\right]^8$$
$$g \left(\frac{-5}{2} - \frac{\sqrt{3}}{2}\right) = \left( 5 - 2 \sqrt{3} \right)^4 = 2569 - 1480 \sqrt{3} \approx 5.564804$$
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|
Find all primes such that $a^2+b^2=9ab-13$.
Let $a,b$ be primes. Find all primes such that $a^2+b^2=9ab-13$.
Whatever I've done is from parity checking. But I can't proceed with the case when both $a,b$ are odd primes.
I tried with some modulo-chasing but couldn't complete.
|
Not a complete answer, just a few restrictions...
Proposition 1. For any prime $p>3 \Rightarrow 3 \mid p^2-1$ (from LFT) and $8 \mid p^2-1$ (from $(2k+1)^2 \equiv 1 \pmod{8}$). As a result $24 \mid p^2-1$.
For $a>3,b>3 \Rightarrow 24 \mid a^2-1$, $24 \mid b^2-1$
and $$24 \mid a^2 + b^2 -2=9ab-15$$
or
$$8 \mid 3ab-5 \tag{1}$$
But, any prime $p>3$ is of the $p=4k+1$ or $p=4k+3$ form. None of $a,b$ can be of the same form at the same time:
*
*$a=4k_a+1,b=4k_b+1 \Rightarrow 3ab-5=3(4k_a+1)(4k_b+1)-5=12Q-2$ is not divisible by 4 and thus not divisible by 8.
*$a=4k_a+3,b=4k_b+3 \Rightarrow 3ab-5=3(4k_a+3)(4k_b+3)-5=12Q-22$ is not divisible by 4 and thus not divisible by 8.
As a result, either
$$a=4k_a+1,b=4k_b+3 \color{red}{\text{ or }} a=4k_a+3,b=4k_b+1 \tag{2}$$
Going further:
$$3ab-5=3(4k_a+1)(4k_b+3)-5=3(16k_ak_b+12k_a+4k_b)+4$$
reveals that both $k_a,k_b$ can't be odd or even at the same time.
We can also assume $a\leq b$ and from $a^2+b^2-9ab=-13<0$, by checking $x^2-9x+1<0$ where $x=\frac{b}{a}$, we have
$$a\leq b < 9a \tag{3}$$
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Find integer part of the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$ for $x+y=3$
If $M$ is the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$. Subject to $x+y=3$. Find the value of $\lfloor M \rfloor$.
I think the maximum of $M$ occurred when $x=y=3/2$. And I just put the value of that in M and found 34. What is the original answer with good solution ?? Somebody please help me.
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Extended hint: let $s=x+y, \,p=xy\,$, then $\,x^2+y^2=s^2-2p, \;x^3+y^3=s^3-3sp\,$ so:
$$
\begin{align}
x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4 &= xy(x^3+x^2+x+1+y+y^2+y^3) \\
&= xy(1+x+y+x^2+y^2+x^3+y^3) \\
&= p(1+s+s^2-2p+s^3-3sp) \\
&= p(40-11p)
\end{align}
$$
The maximum value is attained for $p=\cfrac{20}{11}\,$ and equals $\cfrac{400}{11}\,$, corresponding to the $x,y$ which are the roots of $z^2 - s z + p = 0 \;\iff\;z^2 - 3z + \cfrac{20}{11} = 0\,$.
|
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|
How to find the range of unknowns from a logarithmic equation? If the equation $\log(ax)\cdot \log(bx) + 1 = 0$ with $a>0, b>0$ constants has a solution $x>0$, it follows that $$b/a \geq (?)$$ or $$(?)\geq b/a >(?)$$
Hints maybe (for both :p)?
EDIT:
Answer $$b/a \ge 100$$ $$1/100 \ge b/a \ > 0$$
|
$\begin{array}\\
-1
&=\log(ax)\cdot \log(bx) \\
&=(\log(a)+\log(x))(\log(b)+\log(x))\\
&=(A+X)(B+X)
\qquad\text{with } A=\log a, B=\log b, X = \log x\\
&=AB+X(A+B)+X^2\\
\end{array}
$
so
$X^2+X(A+B)+AB+1 = 0$.
Solving,
$\begin{array}\\
X
&=\dfrac{-A-B\pm\sqrt{A^2+2AB+B^2-4AB-4}}{2}\\
&=\dfrac{-A-B\pm\sqrt{A^2-2AB+B^2-4}}{2}\\
&=\dfrac{-A-B\pm\sqrt{(A-B)^2-4}}{2}\\
\end{array}
$
If this has a real solution,
then
$(A-B)^2 \ge 4$
so
$|A-B| \ge 2$
or
$2
\le |A-B|
= |\log(a)-\log(b)|
= |\log(a/b)|
$.
Therefore
$\log(a/b) \ge 2$
or
$\log(a/b) \le -2$.
In the first case,
$a/b \ge e^2$
(assuming natural logs);
in the second
$a/b \le e^{-2}$.
|
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|
$\frac{(10^4+324)(22^4+324)\cdots(58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$
From AIME 1987, compute $$\frac{(10^4+324)(22^4+324)\cdots (58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$$
So basically the way used to solve this is by Sophie Germain's Identity which is $a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$
My question is , how is possible a student to solve this question without knowing this identity? Is there another way to solve this?
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Noticing that $x^4+324=0$ implies $x=-3\pm3i$ and $x=3\pm 3i$ each root corresponds to one factor. Now lets try to generalize our problem with placing $x$ instead of $7$.
$$\frac{((x+3)^4+324)((x+15)^4+324)((x+27)^4+324)((x+39)^4+324)((x+51)^4+324)}{((x-3)^4+324)((x+9)^4+324)((x+21)^4+324)((x+33)^4+324)((x+45)^4+324)}$$
Now notice that for $(x+3)^4+324=0$ the roots are $$x+3=-3\pm3i\\x=-6\pm3i$$ and $$x+3=3\pm 3i\\x=\pm3i$$
In general the roots of $(x+3+12k)^4+324=0$ are $$x=-6-12k\pm 3i$$ and $$x=-12k\pm 3i$$
And the roots of $(x-3+12k)^4+324=0$
$$x=-12k\pm 3i\\x=6-12k\pm3i$$
You can see that almost all the factors of numerator and denominator are the same except for the factors that correspond to the roots $-54\pm3i$ in the numerator and to roots $6\pm 3i$ in the denominator.so we are left with
$$\frac{(x+54-3i)(x+54+3i)}{(x-6-3i)(x-6+3i)}=\frac{(x+54)^2+9}{(x-6)^2+9}$$
For $x=7$ we get
$$\frac{61^2+9}{10}=\frac{(60+1)^2+9}{10}=\frac{3600+121+9}{10}=373$$
|
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|
What are the integration limits for this volume? I had a test today and want to know whether I set the limits right for this volume computation. The solid is the one defined by $$z= 2-2x^2-y^2, x + y + z = 3, 2x^2 + y^2 = 2$$, and the limits I set are (in adapted cylindrical coordinates in which
$$x=r\cos t , y = \sqrt{2}r\sin t, z = z$$ and the Jacobian $J = \sqrt{2}r$:
$$ 0\le t \le 2\pi , 0\le r \le 1 , 2-2r^2 \le z \le 3-r\cos t - \sqrt{2}r\sin t$$
I got the volume $ V = 2\sqrt{2}\pi$
Can anyone verify this please?
Thank you so much!!
|
We can visualize the three surfaces as follows:
Looking at the image, we want the volume that is above the upside-down paraboloid, inside the elliptic cylinder, and below the plane.
What you are doing is not really converting to cylindrical coordinates, because the coordinate transformation is not:
$$ x = r \cos \theta \qquad y = r \sin \theta \qquad z = z $$
rather we define a new coordinate transformation:
$$ x = r \cos \theta \qquad y = \sqrt{2} r \sin \theta \qquad z = z$$
in which case, our volume of integration now corresponds to:
$$ 0 \leq r \leq 1 \qquad 0 \leq \theta \leq 2\pi \qquad 2 - 2r^2 \leq z \leq 3 - r \cos \theta - \sqrt{2} r \sin \theta $$
and the Jacobian is:
\begin{align*}
\frac{\partial(x,y,z)}{\partial(r,\theta,z)} &= \frac{\partial(x,y)}{\partial(r,\theta)} \\
&= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} \\
&= \begin{vmatrix} \cos \theta & -r \sin \theta \\ \sqrt{2} \sin \theta & \sqrt{2} r \cos \theta \end{vmatrix} \\
&= \sqrt{2}\begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} \\
&= \sqrt{2}r
\end{align*}
So
$$ dV = dx \, dy \, dz = \sqrt{2}r \, dr \, d\theta \, dz $$
So our volume integral becomes:
\begin{align*}
V &= \iiint \, dV \\
&= \int_0^{2\pi} \int_0^1 \int_{2-2r^2}^{3 - r \cos \theta - \sqrt{2}r \sin \theta} \sqrt{2}r \, dz \, dr \, d\theta \\
&= \sqrt{2}\int_0^{2\pi} \int_0^1 r\left(1 + 2r^2 - r\cos \theta - \sqrt{2} r\sin \theta \right) \, dr \, d\theta \\
&= \sqrt{2}\int_0^{2\pi} \int_0^1 (r+2r^3) \, dr \, d\theta - \sqrt{2}\int_0^{2\pi} \int_0^1 r^2\left(\cos \theta + \sqrt{2} \sin \theta \right) \, dr \, d\theta \\
&= 2\sqrt{2} \pi \int_0^1 (r+2r^3) \, dr - \sqrt{2}\int_0^{2\pi} \frac{1}{3}\left(\cos \theta + \sqrt{2} \sin \theta \right) \, d\theta \\
&= 2\sqrt{2} \pi - \frac{\sqrt{2}}{3}\int_0^{2\pi} \left(\cos \theta + \sqrt{2} \sin \theta \right) \, d\theta \\
&= 2\sqrt{2} \pi - 0 \\
&= 2\sqrt{2} \pi
\end{align*}
|
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|
Cubic equation with unknown coefficients given roots I was given this equation. $x^3 + 3px^2 + qx + r=0$. The roots are $1, -1$, and $3$.
Ive tried dividing the equation by $(x-1)$ to get a quadratic to make it easier for me. But that ended up really badly.
I also inputted the different roots into the equation to get different equations that I could solve. When I tried to prove my answer it turned out to be a flop.
And I also tried multiplying the three factors and comparing coefficients. It didnt seem right.
|
Subbing in these roots to the equation gives us three equations.
$1 + 3p + q + r = 0$
$-1 + 3p -q + r = 0$
$27 + 27p + 3q + r = 0$
This can become a matrix equation
$\begin{bmatrix}1 \\ -1 \\ 27 \end{bmatrix} + \begin{bmatrix}3 & 1 & 1\\3 & -1 & 1 \\27 & 3 & 1\end{bmatrix} \begin{bmatrix}p \\ q \\ r \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} $
$\begin{bmatrix}3 & 1 & 1\\3 & -1 & 1 \\27 & 3 & 1\end{bmatrix} \begin{bmatrix}p \\ q \\ r \end{bmatrix} = \begin{bmatrix}-1 \\ 1 \\ -27 \end{bmatrix} $
$ \begin{bmatrix}p \\ q \\ r \end{bmatrix} = \begin{bmatrix}3 & 1 & 1\\3 & -1 & 1 \\27 & 3 & 1\end{bmatrix} ^{-1} \begin{bmatrix}-1 \\ 1 \\ -27 \end{bmatrix} = \begin{bmatrix}-1 \\ -1 \\ 3 \end{bmatrix} $
|
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|
Find the minimum value of $xy+yz+xz$ Question:
Find the minimum value of $xy+yz+xz$, given that $x,y,z$ are real and $x^2+y^2+z^2=1$
My attempt,
Since $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$=1+2(xy+xz+yz)$
$(x+y+z)^2\geq0$
So that $1+2(xy+xz+yz)\geq0$
$(xy+xz+yz)\geq -\frac{1}{2}$
So the minimum value is $-\frac{1}{2}$.
My question is am I correct ? Because I don't have the solution. And is there another way to solve this? Thanks in advance.
|
You are almost done. You are right that, since $(x+y+z)^2 \geq 0$, we have $xy+xz+yz \geq -\frac{1}{2}$. Now, to find the actual minimum, turn the inequality sign into an equality sign:
$$x+y+z = 0 \Rightarrow xy+xz+yz=-\frac{1}{2}$$
Now, just find values of $x, y, z$ such that
$$\begin{cases} x^2+y^2+z^2=1 \\ x+y+z=0 \end{cases}$$
This is an equation of a circle (sphere intersected with a plane), and all it's points give you the minimum. To find one, you can simply take almost any values for $x,y,z$, then project this point on the plane, and then normalize the vector so that it lies on the sphere.
For example, take $x=0, y=0, z=-1$. One possible way of projecting on the plane $x+y+z=0$ is to take the sum of coordinates $s$, and decrease every coordinate by $\frac{s}{3}$. For the chosen vector we get $s=-1$, and the vector transforms into $$x=\frac{1}{3}, y=\frac{1}{3}, z=-\frac{2}{3}$$
Now, compute this vector's norm: $$\sqrt{x^2+y^2+z^2} = \sqrt{\frac{1^2+1^2+2^2}{3^2}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$$
Now, dividing the vector's coordinates by the norm (thus multiplying by $\sqrt{\frac{3}{2}}$), we get a vector on the sphere (and, since it remains being in the plane, it appears to be on the desired circle):
$$x = \frac{1}{3}\sqrt{\frac{3}{2}}=\frac{1}{\sqrt{6}}$$
$$y = \frac{1}{3}\sqrt{\frac{3}{2}}=\frac{1}{\sqrt{6}}$$
$$z = -\frac{2}{3}\sqrt{\frac{3}{2}}=-\frac{2}{\sqrt{6}}$$
Note that it is exactly the solution by CY Kwong in the comments.
|
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|
Find the smallest cosntant $k>0$ such that $\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b} \leq k(a+b+c)$ for every $a,b,c>0$. In the book 'Putnam and beyond', page $173$, it has the following problem:
Find the smallest cosntant $k>0$ such that
$$\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b} \leq k(a+b+c)$$ for every $a,b,c>0$.
In the solution, it states that:
Note that the inequality remains unchanged on replacing $a,b,c$ by $ta,tb,tc$ with $t>0$. Consequently, the smallest value of $k$ is the supremum of
$$f(a,b,c) = \frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b}$$ over the domain $\Delta= \{ (a,b,c) :a,b,c>0, a+b+c=1 \}.$
Question: Why the smallest value of $k$ is the supremum of the function above on the domain? In particular, I do not understand why they only care about the domain and not other domain, and why the fact 'remain unchanged ...' will lead to the above conclusion.
|
The inequality is equivalent to $f(a,b,c) \leq k$ where
$$
f(a,b,c)=\frac{\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b}}{a+b+c}
$$
Now, if we simplify $f(ta,tb,tc)$ we find that it is equal to $f(a,b,c)$ : it "remains
unchanged". For any $p_0=(a_0,b_0,c_0)$ with $a_0,b_0,c_0>0$, if we put $p_1=(a_1,b_1,c_1)=(t_1a_0,t_1b_0,t_1c_0)$ with $t_1=\frac{1}{a_0+b_0+c_0}$ then
$f(p_1)=f(p_0)$ and $p_1\in \Delta$.
|
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"source": "stackexchange",
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Infinite sum of rationals is irrational
a) Let $x$ be a number between $0$ and $1$. Let $a_1$ be the smallest positive integer such that $x_1=x-a_1^{-1}\geq 0$, let $a_2$ be the smallest positive integer such that $x_2=x_1-a_2^{-1}\geq 0$, etc. Show that this leads to a finite expansion $$x=\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}$$ (that is, that $x_{n+1}=0$ for some $n$) if and only if $x$ is rational.
b) Show that if the integers $1<b_1<b_2<\cdots$ increase so rapidly that $$\frac{1}{b_{k+1}}+\frac{1}{b_{k+2}}+\dots<\frac{1}{b_k-1}-\frac{1}{b_k}~~\text{for }k\geq 1,$$ then the number $\sum b_k^{-1}$ is irrational. Use this then to prove that $\sum\limits_{k=0}^\infty (2^{3^k}+1)^{-1}$ is irrational.
I solved (a), but (b) is a little difficult.
This is my way to (b)
suppose $\sum b_k^{-1}$ is rational and I want this is a contradiction.
using the result of (a), 1/b_k−1 can be expressed with finite rational expansion.
1/b_k-1>$\sum b_k^{-1}$ which is rational
so I put 1/b_k-1=$\sum b_k^{-1}$+a/b which b is not zero.
Now, What should I do?
how to solve (b)?
|
To solve (b), we can use (a). Let $x=\sum b_k^{-1}$, and for $n\ge 1$, let $x_n=\sum_{k>n} b_k^{-1}$.
We have to show $b_{n + 1}$ is the smallest integer s.t. $x_n - b_{n+1}^{-1}\ge 0$ for all $n$.
Let $n\in\mathbb{N}$. On the one hand, $x_n - b_{n+1}^{-1}=x_{n+1}$ is indeed positive. On the other hand
\begin{eqnarray}
x_{n} - \frac{1}{b_{n+1} - 1} & = & \left(\frac{1}{b_{n+1}} - \frac{1}{b_{n+1} - 1}\right) + \sum_{k>n+1}\frac{1}{b_k}\\
& < & \left(\frac{1}{b_{n+1}} - \frac{1}{b_{n+1} - 1}\right) + \left(\frac{1}{b_{n+1} - 1} - \frac{1}{b_{n+1}}\right)\\
& = & 0
\end{eqnarray}
thus, $b_{n + 1}$ is the smallest integer s.t. $x_n - b_{n+1}^{-1}\ge 0$. Hence, $x$ is of the form shown in (a), with all the terms $x_n >0$, thus it is irrational.
To prove $x=\sum_k (2^{3^k}+ 1)^{-1}$ is irrational, note that for all $k$,
$$
\frac{1}{2^{3^k}} - \frac{1}{2^{3^k} + 1} = \frac{1}{2^{3^k}}\left
(1 - \frac{2^{3^k}}{2^{3^k} + 1}\right) = \frac{1}{2^{3^k}}\left
(\frac{1}
{2^{3^k} + 1}\right) := m_k
$$
while, on the other hand,
$$
\sum_{n>k}\frac{1}{2^{3^n} + 1} < \frac{1}{2^{3^{k+1}} + 1} + \sum_{n\ge 3^{k+2}}\frac{1}{2^n} = \frac{1}{2^{3^{k+1}} + 1} + \frac{2}{2^{3^{k+2}}} := n_k
$$
We can show that for all $k, m_k>n_k$. Hence by part (b) we conclude $x$ is irrational.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2289412",
"timestamp": "2023-03-29T00:00:00",
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|
Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot)-\frac{1}{10}(1+\frac{1}{10}+\cdot\cdot\cdot)$$ using 3 5 7 terms and 2 4 6 terms to form two geometric series. then if we calculate sum for them we will have $$S_1 = \frac{10}{11}$$ and $$S_2 = \frac{10}{9}$$ and after substituting them we'll finally have sum of series $$S = \frac{88-9\sqrt{10}}{99}$$
Is everything right here?
But how to check it for convergence?
|
My idea: suppose that $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$ is a convergent serie. Let $S$ the sum of the serie $$S= 1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}\left(1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\cdot\cdot\cdot\right)=1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}S$$
The sum of the serie must satisfy the equation
$$S= 1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}S$$
i.e. $$S=\frac{10\sqrt{10}-90}{\sqrt{10}-100}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2289659",
"timestamp": "2023-03-29T00:00:00",
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|
$F(\alpha, \beta):F$ is a simple extension Let $K:F$ be a extension, $F$ characteristic other than $2$, $\alpha, \beta \in K-F$ with $\alpha \neq \beta$ and $\alpha \neq -\beta$ such that $\alpha^2, \beta^2 \in F$. I have to prove extension $F(\alpha, \beta):F$ is simple.
So as $F$ characteristic other than $n$, then $n(1)=0$ in $F$, $n \neq 2$.
We want to show that there exist $x$ in $F(\alpha, \beta)$ such that $F(\alpha, \beta)=F(x)$.
What can I do next?
Can somebody help me with a hint or solution ?
|
As with the hint of @Chickenmancer, consider $F(a+b)$. Then clearly $F(a+b) \subset F(a, b)$. We'll show we have the reverse inclusion and hence equality.
Since $(a+b) \in F(a+b)$ we must have that $(a+b)^2 = a^2 + 2ab + b^2 \in F(a+b)$.
Now, $a^2, b^2 \in F \subset F(a+b)$ gives us that $2ab \in F(a+b)$.
Now the characteristic isn't 2, so we can divide by the $2$ since we're in a field so $ab \in F(a+b)$.
Then, since $a \ne -b$ we have $a+b \ne 0$ and so we can divide by $(a+b)$ in $F(a+b)$.
Now we see that, $(a-b) = \frac{(a-b)(a+b)}{a+b}= \frac{a^2 - b^2}{a+b} \in F(a+b)$.
Lastly $(a-b) + (a+b) = 2a \in F(a+b)$. Again by characteristic not equal to $2$ we get that $a \in F(a+b)$. Similarly $b \in F(a+b)$. So $F(a, b) = F(a+b)$ and we're done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2289863",
"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}.$$
Note that:
$$\ln\left(2-\dfrac{x}{a} \right)\sim_{a}1-\dfrac{x}{2}.$$
Now we have $\dfrac{\pi x}{2a}\underset{x\to a}{\longrightarrow} \dfrac{\pi}{2}$, i.e.$\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \underset{x\to a}{\longrightarrow} 0$ and $\tan h \sim_{0}h.$
I'm stuck.
Update: here is another way :
\begin{aligned}
\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}&=\exp\left[{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}\right].\\
&=\exp\left[ \left(1-\dfrac{x}{a}\right)\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2}+\dfrac{\pi}{2} \right).\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{1-\dfrac{x}{a}}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(-\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ \dfrac{2}{\pi} \dfrac{\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
\end{aligned}
Thus
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left(\dfrac{\pi x}{2a}\right)} =e^{\dfrac{2}{\pi}} $$
*
*Am i right beside i'm intersted in way which use equivalents
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Hint:
$$\left(2-\frac xa\right)^{\tan(\pi x/2a)}=\left[\left(1+\left(1-\frac xa\right)\right)^{1/(1-x/a)}\right]^{(1-x/a)\tan(\pi x/2a)}\\\stackrel{x\to a}\longrightarrow e^{2/\pi}$$
Since,
$$e=\lim_{u\to0}(1+u)^{1/u}$$
$$\frac2\pi=\lim_{u\to1}(1-u)\tan(\pi u/2)$$
for those who want to make this rigorous, use squeeze theorem:
$$\underbrace{\left(2-\frac xa\right)^{\frac2\pi\left(\frac a{a-x}-1\right)}}_{\text{Laurent expansion of }\tan(z)\text{ at }z=\frac\pi2}\le\left(2-\frac xa\right)^{\tan(\pi x/2a)}\le\underbrace{e^{\left(1-\frac xa\right)\tan\left(\frac{\pi x}{2a}\right)}}_{\text{Maclaurin expansion of }e^x}$$
where the inequalities are true for $x<a$ and flipped for $x>a$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2294071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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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.