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Integrate $\int\frac{\cos^2x}{1+\tan x}dx$ Integrate $$I=\int\frac{\cos^2x}{1+\tan x}dx$$
$$I=\int\frac{\cos^3xdx}{\cos x+\sin x}=\int\frac{\cos^3x(\cos x-\sin x)dx}{\cos^2x-\sin^2x}=\int\frac{\cos^4xdx}{1-2\sin^2x}-\int\frac{\cos^3x\sin xdx}{2\cos^2x-1}$$
Let $t=\sin x,u=\cos x,dt=\cos xdx,du=-\sin xdx$
$$I=\underbrace{\int\frac{-u^4du}{(2u^2-1)\sqrt{1-u^2}}}_{I_1}+\underbrace{\int\frac{u^3du}{2u^2-1}}_{I_2}$$
I have found(using long division): $$I_2=\frac{u^2}2+\frac18\ln|2u^2-1|+c=\frac12\cos^2x+\frac18\ln|\cos2x|$$
I have converted $I_1$ into this:
$$I_1=\frac12\left(\int(-2)\sqrt{1-u^2}du+\int\frac{du}{\sqrt{1-u^2}}\right)+\frac14\underbrace{\int\frac{du}{(2u^2-1)\sqrt{1-u^2}}}_{I_3}$$
Now I have took $v=1/u$ in $I_3$ so that $du=-(1/v^2)dv$:
$$I_3=\int\frac{vdv}{(v^2-2)\sqrt{v^2-1}}$$
Now I took $w^2=v^2-1$ or $wdw=vdv$ to get:
$$I_3=\int\frac{dw}{w^2-1}=\frac12\ln\left|\frac{w-1}{w+1}\right|$$
I have not yet formulated the entire thing;
*
*Is this correct?
*This is very long, do you have any "shorter" method?
|
Decompose the integrand as follows\begin{align}
\frac{\cos^2x}{1+\tan x}&= \frac{\cos^3 x \ (\cos x +\sin x)}{(\cos x+\sin x)^2}\\
&=\frac{(1+\cos2x)\ (\cos 2x + 1+ \sin2x)}{4(1+\sin 2x)}\\
&=\frac14\left( 1+\cos2x + \frac{\cos^2 2x+\cos2x}{1+\sin2x}\right)\\
&=\frac14\left( 2+\cos2x -\sin 2x +\frac{\cos2x}{1+\sin2x}\right)
\end{align}
Then
$$\int\frac{\cos^2x}{1+\tan x}dx
=\frac12x+\frac18\left[\sin 2x+ \cos 2x+\ln (1+\sin2x) \right]
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Variational problem concerning variances Let $\phi$ be the family consisting of all random variables $X$ such that $P(X\in [0,1])=1$, $EX=\frac{1}{3}$, $P(X<\frac{1}{4})<\frac{1}{2}$, $P(X>\frac{1}{4})\geq\frac{1}{2}$. Calculate $\sup \{Var(X):X \in \phi\} - \inf \{Var(X):X \in \phi\} .
$
|
Let us put $I=\inf({\sf Var}(X)|X\in\phi)$ and $S=\sup({\sf Var}(X)|X\in\phi)$. We will
compute $I$ and $S$.
Trivially $I=0$ (take $X$ constant equal to $\frac{1}{3}$).
Let us now compute $S$. To ease notation, we introduce the
characteristic functions $\alpha=1_{X\in[0,\frac{1}{4})}$ and $\beta=1_{X\in[\frac{1}{4},1]}$,
so that $\alpha+\beta=1$, and we shall abuse notation slightly by putting
$E_{\alpha}(\ldots)$ (or $E_{\beta}(\ldots))$ instead of $E(\alpha\ldots)$
( or $E(\beta\ldots))$. We then have a formal identity
$E=E_{\alpha}+E_{\beta}$. For $X\in\phi$, one has
$$
\begin{array}{lcl}
E(X^2) &= & E_{\alpha}(X^2)+E_{\beta}(X^2) \\
&\leq & E_{\alpha}\bigg(\frac{1}{4}X \bigg) +E_{\beta}\bigg(X^2+(1-X)(X-\frac{1}{4}) \bigg) \\
&= & \frac{1}{4}E_{\alpha}\bigg(X \bigg) +E_{\beta}\bigg(\frac{5}{4}X-\frac{1}{4} \bigg) \\
&= & \frac{1}{4}E_{\alpha}\bigg(X \bigg) +E\bigg(\frac{5}{4}X-\frac{1}{4} \bigg)-
E_{\alpha}\bigg(\frac{5}{4}X-\frac{1}{4} \bigg) \\
&= & \frac{1}{4}E_{\alpha}\bigg(X \bigg) +\frac{5}{4}\times\frac{1}{3}-\frac{1}{4} -
\frac{5}{4}E_{\alpha}\bigg(X\bigg)+\frac{1}{4}P(X<\frac{1}{4}) \\
&= & \frac{1}{6}-E_{\alpha}\bigg(X \bigg)+\frac{1}{4}P(X<\frac{1}{4}) \\
& \leq & \frac{1}{6}+\frac{1}{4}P(X<\frac{1}{4}) \leq \frac{1}{6}+\frac{1}{4}\times
\frac{1}{2} =\frac{7}{24} \\
\end{array}
$$
The inequality $E(X^2) \leq \frac{7}{24}$ cannot become an inequality, because
it would entail $P(X<\frac{1}{4})=\frac{1}{2}$ contradicting $X\in\phi$. The bound
$\frac{7}{24}$ is best possible, however. Indeed, for $n\geq 10$ consider $f_n:[0,1]\to {\mathbb R}^{+}$ defined
by
$$
f_n(x)=\left\lbrace\begin{array}{lcl}
2n-\frac{1}{4}+\frac{1}{8(n-1)}, & \text{if} & x\in[0,\frac{1}{4n}] \\
\frac{1}{8(n-1)}, & \text{if} & x\in[\frac{1}{4}-\frac{1}{4n},\frac{1}{4}] \\
\frac{8n}{27}+\frac{5}{27}+\frac{31}{216(n-1)}, & \text{if} & x\in[\frac{1}{4},\frac{1}{4}+\frac{3}{4n}] \\
\frac{10n}{27}-\frac{11}{108}-\frac{31}{216(n-1)}, & \text{if} & x\in[1-\frac{3}{4n},1] \\
0, & \text{otherwise} &
\end{array}\right.
$$
and let $X_n$ have density $f_n$. Then $E(X_n)=\frac{1}{3},P(X_n<\frac{1}{4})=\frac{1}{2}-\frac{1}{16n}$
so $X_n\in\phi$, and $E(X_n^2)=\frac{7}{24}-\frac{45n^2 - 17n - 2}{192n^3}$. So $\frac{7}{24}$ is best possible,
and hence $S=\frac{7}{24}-\bigg(\frac{1}{3}\bigg)^2=\frac{13}{72}$.
In the end, $S-I=\frac{13}{72}$.
|
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|
PEMDAS question: $F(x) = 3x^2 - x+2$. Find $[f(a)]^2$ How should I go about doing this? $(3a^2-a+2)^2$?
Thus, $9a^4-a^2+4$
|
Hint For any real numbers $x,y$ we have $$\color{green}{(x-y)^2 = x^2-2xy+y^2}, \quad \color{blue}{(x+y)^2 = x^2+2xy+y^2},$$
and note that with $\color{green}{x = (2-a)}$ and $y = 3a^2$ we get
$$(3a^2-a+2)^2=\color{blue}{(x+y)^2} = \ldots$$
Can you continue from here?
|
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|
Weighted Coin Toss Probablity Suppose two weighted coins are tossed. The first is weighted so that it comes up heads with probability $\frac{1}{3}$. The second is weighted so that it comes up heads with probability $\frac{1}{4}$. What is the probability that when both coins are tossed, one comes up heads and the other comes up tails?
|
*
*$P(HH)=\dfrac{1}{3}\cdot\dfrac{1}{4}=\dfrac{1}{12}$
*$P(HT)=\dfrac{1}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}$
*$P(TH)=\dfrac{2}{3}\cdot\dfrac{1}{4}=\dfrac{1}{6}$
*$P(TT)=\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{2}$
So the answer to your question is $P(HT)+P(TH)=\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{5}{12}$
|
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|
Modular calculus and square I want to prove that $4m^2+1$ and $4m^2+5m+4$ are coprimes and also $4m^2+1$ and $4k^2+1$ when $k\neq{m}$ and $4m^2+5m+4$ and $4k^2+5k+4$ when $k\neq{m}$. Firstly : Let $d|4m^2+1$ and $d|4m^2+5m+4$ then $d|4m^2+5m+4-(4m^2+1)=5m+3$ and $d|5m^2+3m$ thus $d|5m^2+3m-(4m^2+5m+4)=m^2-2m-4$ and $d|4m^2-8m-16$ or $d|4m^2+5m+4-(4m^2-8m-16)=13m+20$ but $d|15m+9$ thus $d|15m+9-(13m+20)=2m-11$ but $d|13m+20-(5m+3)=8m+17$ and $d|8m-88$ therefore $d|8m+17-(8m-88)=105=5.3.7$. So $4m^2+1=105k$ or $4m^2=104+105k'$ but $k'=4k''$ then $m^2=26+105k''$ or $4m^2+1=21k$ or $4m^2+1=35k$ or $4m^2+1=15k$ or $4m^2+1=5k$ or 3k or 7k. Are there solutions to these equations, please, or can I conclude that the two numbers are coprimes ?
|
You've already found $d$ divides $5m+3$
Again, $d$ divides $4m(5m+3)-5(4m^2+1)=12m-5$
$\implies d$ divides $12(5m+3)-5(12m-5)=61$
Now, $4m^2+1\equiv0\pmod{61}\iff m\equiv\pm36\pmod{61}$
|
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|
Without using Taylor expansion or L'Hospital rule evaluate the limit:
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$
I give the two alternate ways so that no one gives them again in their answer:
Using L'Hospital:
$$L=\lim_{x\to0}\frac{e^x-x-1}{3x^2}=\lim_{x\to0}\frac{e^x-1}{6x}=\lim_{x\to0}\frac{e^x}{6}=\frac16$$
Using Taylor:
$$L=\lim_{x\to0}\frac{\color{red}{(1+x+x^2/2+x^3/6+O(x^4))}-1-x-x^2/2}{x^3}=\frac16$$
Using nothing(not infact):
$$L=\lim_{x\to0}\frac{e^x-1-x^2/2}{x^3}\tag{sorry I omitted the x}\\
=\lim_{x\to0}\frac{e^{2x}-1-2x^2}{8x^3}\\
8L=\lim_{x\to0}\frac{e^{2x}-1-2x^2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{2x}-e^x-3x^2/2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{8x^3}\\
56L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{x^3}\\
49L=\lim_{x\to0}\frac{e^{4x}-2e^{2x}+e^x-9x^2/2}{x^3}=?$$
|
I will deal with the case $x \to 0^{+}$ and leave the $x \to 0^{-}$ case for OP and other readers. Let us use the formula $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ We then have $$\begin{aligned}L &= \lim_{x \to 0^{+}}\dfrac{e^{x} - 1 - x - \dfrac{x^{2}}{2}}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\lim_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} - 1 - x - \dfrac{x^{2}}{2}}{x^{3}}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}\dfrac{\left(1 + \dfrac{x}{n}\right)^{n} - 1 - x - \dfrac{x^{2}}{2}}{x^{3}}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}\frac{1}{x^{3}}\left\{-\frac{x^{2}}{2n} + \dfrac{\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)}{3!}x^{3} + \cdots\right\}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}\left\{-\frac{1}{2nx} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)\left(1 - \dfrac{3}{n}\right)}{4!}x + \cdots\right\}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}f(x, n) - \frac{1}{2nx}\end{aligned}$$ where $f(x, n)$ is a finite sum with number of terms dependent on $n$. By using Monotone convergence theorem we can show that the sum $f(x, n)$ tends to a limit (as $n\to\infty$) dependent on $x$ say $f(x)$ and we can also show that $f(x) = (1/6) + o(x)$. The term $-1/2nx$ tends to $0$. It follows that $L = \lim_{x \to 0^{+}}f(x) = 1/6$.
|
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|
Derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition
When not using the derivative definition I get $\cos (1/x) + 2x \sin(1/x)$,
which WolframAlpha agrees to.
However when I try solving it using the derivative definition:
$$\lim_ {h\to 0} = \frac{f(x+h) - f(x)}{h} $$
I get:
$$2x \sin \left(\frac{1}{x+h} \right ) + h \sin \left(\frac{1}{x+h}\right)$$
which in return results in, as $h \to 0$:
$$2x \sin (1/x)$$
So what am I doing wrong when using the def of derivatives?
|
You're doing wrongly the computation:
\begin{align}
f(x+h)-f(x)&=(x+h)^2\sin\frac{1}{x+h}-x^2\sin\frac{1}{x}\\
&=x^2\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)+2hx\sin\frac{1}{x+h}+h^2\sin\frac{1}{x+h}
\end{align}
When you divide by $h$ you get
$$
\frac{f(x+h)-f(x)}{h}=
\frac{x^2}{h}\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)+2x\sin\frac{1}{x+h}+h\sin\frac{1}{x+h}
$$
and you're left with computing
$$
\lim_{h\to0}\frac{1}{h}\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)
$$
because the second summand above tends to $2x\sin(1/x)$ and the third summand tends to $0$. You'll reinsert $x^2$ later.
Now you can use the identity
$$
\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}
$$
Setting $\alpha=1/(x+h)$ and $\beta=1/x$, we have
$$
\frac{\alpha+\beta}{2}=\frac{1}{2}\left(\frac{1}{x+h}+\frac{1}{x}\right)=
\frac{2x+h}{2x(x+h)}
$$
while
$$
\frac{\alpha-\beta}{2}=\frac{1}{2}\left(\frac{1}{x+h}-\frac{1}{x}\right)=
\frac{-h}{2x(x+h)}
$$
so your limit is
$$
\lim_{h\to0}\frac{2}{h}\cos\frac{2x+h}{2x(x+h)}\sin\frac{-h}{2x(x+h)}
$$
But
$$
\lim_{h\to0}2\cos\frac{2x+h}{2x(x+h)}=2\cos\frac{1}{x}
$$
so we just need to compute
$$
\lim_{h\to0}\frac{1}{h}\sin\frac{-h}{2x(x+h)}
$$
Set $k=-h/(2x(x+h))$, so
$$
2x^2k+2xhk=-h
$$
or $h(2xk+1)=-2x^2k$ that lends
$$
h=-\frac{2x^2k}{2xk+1}
$$
So this transformation is bijective (and bicontinuous) and $h\to0$ implies $k\to0$ so your limit is
$$
\lim_{h\to0}\frac{1}{h}\sin\frac{-h}{2x(x+h)}=
\lim_{k\to0}-\frac{2xk+1}{2x^2k}\sin k=
\lim_{k\to0}-\frac{2xk+1}{2x^2}\frac{\sin k}{k}=-\frac{1}{2x^2}
$$
In the end we get
$$
\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=
x^2\left(-\frac{1}{2x^2}\right)\cos\frac{1}{x}+2x\sin\frac{1}{x}=
-\cos\frac{1}{x}+2x\sin\frac{1}{x}
$$
This assumes, of course, $x\ne0$. If your function is defined by
$$
f(x)=\begin{cases}
x^2\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex]
0&\text{for $x=0$}
\end{cases}
$$
we are left with the derivative at $0$, that is
$$
\lim_{h\to0}\frac{f(h)-f(0)}{h}=
\lim_{h\to0}\frac{1}{h}h^2\sin\frac{1}{h}=
\lim_{h\to0}h\sin\frac{1}{h}=0
$$
with an easy application of the squeeze theorem.
Of course using the chain rule is much simpler.
Now that you know that
$$
f'(x)=\begin{cases}
-\cos\dfrac{1}{x}+2x\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex]
0&\text{for $x=0$}
\end{cases}
$$
it should be easy to verify whether the derivative is continuous at $0$.
|
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|
Calculus I - Simple Difference Quotient Question The problem is to calculate the difference quotient of $f(x) = \sqrt{x^2 +2x+1}$. But $\sqrt{x^2+2x+1}= \sqrt{x+1}^2 = x+1$ so can I just take the difference quotient of $x+1$? If not, how do I simplify $\frac{\sqrt{x^2+2xh+h^2+2x+2h+1} - \sqrt{x^2+2x+1}}{h}$ ?
|
Not observing the fact that $\sqrt{x^2+2x+1}=x+1$, that lonely $h$ will factor out of the denominator in your other form if you multiply the entire mess by
$$\frac{\sqrt{x^2+2xh+h^2+2x+2h+1} + \sqrt{x^2+2x+1}}{\sqrt{x^2+2xh+h^2+2x+2h+1} + \sqrt{x^2+2x+1}}.$$
Things will suddenly start to cancel in the numerator significantly. You will then factor an $h$ completely out of what remains in the numerator, and with that cancel the $h$ out of the denominator. Proceed cautiously, this is error prone.
Observing your simplification fact is a good idea, much more straightforward, and mathematically acceptable. The results either way will be identical.
|
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|
manipulations with Taylor expansions for log and sinh How could we derive the equality
$$ \frac14 \sum_{m=1}^\infty \frac1m \frac{1}{\sinh^2 \frac{m\alpha}2} = - \sum_{n=1}^\infty n\log (1-q^n)$$
where $q=e^{-\alpha}$ ?
|
$$ \begin{align} \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{\sinh^{2} \frac{m \alpha}{2}} &= \sum_{m=1}^{\infty} \frac{1}{m} \frac{4}{(e^{m \alpha /2}-e^{- m \alpha /2})^{2}} \\ &= 4 \sum_{m=1}^{\infty} \frac{1}{m} \frac{e^{-m \alpha }}{(1-e^{-m \alpha})^{2}} \\ &=4 \sum_{m=1}^{\infty} \frac{1}{m}\sum_{n=1}^{\infty} n(e^{-m \alpha})^{n} \tag{1} \\ &=4 \sum_{n=1}^{\infty} n \sum_{m=1}^{\infty} \frac{(e^{-\alpha n })^{m}}{m} \\ &= -4 \sum_{n=1}^{\infty} n \log(1-e^{- \alpha n}) \end{align}$$
$(1)$ If $|z| <1$, $ \displaystyle \sum_{n=1}^{\infty} nz^{n} = \frac{z}{(1-z)^{2}}$.
|
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|
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162
x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928
x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750
x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900
x^8+366120 x^6+87829920 x^4+13038019200
x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$
|
Are you sure about the $564x^6$ term? All polynomial coeffs are divisible by $k$ with the exception of that one.
|
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|
Using a given identity to solve for the value of an expression This problem caught my eye in the book yesterday. Till now I still get stuck. Here it is:
If $$\frac{x}{x^2+1}=\frac{1}{3},$$ what is the value of $$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}?$$
The denominator is a cyclotomic polynomial which can be expressed as $$\frac{x^7-1}{x-1}$$ but I have no idea if this even helps.
|
It can also solved by this equation $x^2+1=3x$.
$$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}=\frac{x^3}{3x^5+x^5+x^3+3x+x}=
\frac{x^3}{4x^5+4x^3+4x-3x^3}=\frac{x^3}{12x^4+4-3x^3}= \cdots $$
The steps is natural..., I omit; however the answer given by Jack is better.
|
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|
Help with dy/dx of natural log
Could someone explain how I would work this problem with steps?
|
It is easier to simplify first:
$$
y = \ln x - \ln(1+x^2).
$$
Then
\begin{align*}
\frac{dy}{dx} & = \frac{d}{dx}\ln x - \frac{d}{dx} \ln(1+x^2).
\end{align*}
The first derivative is $1/x$. For the second derivative, we need to use the chain rule:
\begin{align*}
\frac{d}{dx} \ln(1+x^2) & = \frac{d\ln(1+x^2)}{d(1+x^2)} \cdot \frac{d(1+x^2)}{dx} \\
& = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}.
\end{align*}
Hence
$$
\frac{dy}{dx} = \frac{1}{x}-\frac{2x}{1+x^2}.
$$
|
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"url": "https://math.stackexchange.com/questions/945201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Trigonometric Identity involving $\sin^2$ and $\cos^2$ I was trying to prove this trigonometric identity, it looks like using the elementary relations should be enough, but I still can't find how:
$$\frac{1}{2}\sin^2 a \ \sin^2 b + \cos^2a \ \cos^2 b = \frac{1}{3} + \frac{2}{3} \biggl( \frac{3}{2}\cos^2 a - \frac{1}{2} \biggr)\biggl( \frac{3}{2}\cos^2 b - \frac{1}{2}\biggr)$$
Thank you!
(taken from Celestial Mechanics)
|
The left hand side is
$$\begin{align}\frac 12\sin^2 a\sin^2b+\cos^2 a\cos^2 b&=\frac 12(1-\cos^2 a)(1-\cos^2b)+\cos^2a\cos ^2b\\&=\frac 12-\frac 12\cos^2a-\frac12\cos ^2b+\frac 32\cos^2a\cos^2b.\end{align}$$
The right hand side is
$$\frac 13+\frac 23\left(\frac 32\cos^2a-\frac 12\right)\left(\frac 32\cos^2b-\frac 12\right)$$$$=\frac 13+\frac 23\left(\frac94\cos^2\cos^2b-\frac{3}{4}\cos^2a-\frac 34\cos^2b+\frac 14\right)$$
$$=\frac 13+\frac 32\cos^2a\cos^2b-\frac 12\cos^2a-\frac12\cos^2b+\frac 16$$
$$=\frac 12-\frac 12\cos^2a-\frac12\cos ^2b+\frac 32\cos^2a\cos^2b.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
A triangle determinant that is always zero How do we prove, without actually expanding, that
$$\begin{vmatrix}
\sin {2A}& \sin {C}& \sin {B}\\
\sin{C}& \sin{2B}& \sin {A}\\
\sin{B}& \sin{A}& \sin{2C}
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried adding and subtracting from the rows and columns and I even tried using the sine rule, but to no avail.
|
$$\begin{pmatrix}
\sin(2A) & \sin C & \sin B \\
\sin C & \sin (2B) & \sin A \\
\sin B & \sin A & \sin (2C) \\
\end{pmatrix} =
\begin{pmatrix}
\sin(2A) & \sin (\pi-A-B) & \sin (\pi-A-C) \\
\sin (\pi-A-B) & \sin (2B) & \sin (\pi-B-C) \\
\sin (\pi-A-C) & \sin (\pi-B-C) & \sin (2C) \\
\end{pmatrix} $$
$$=
\begin{pmatrix}
\sin A \cos A + \cos A \sin A & \sin A \cos B+\cos A \sin B & \sin A \cos C + \sin C \cos A \\
\sin B \cos A + \cos B \sin A & \sin B \cos B + \cos B \sin B & \sin B \cos C + \cos B \sin C \\
\sin A \cos C + \sin C \cos A & \sin B \cos C + \cos B \sin C & \sin C \cos C + \cos C \sin C
\end{pmatrix}
$$
$$=\begin{pmatrix}
\sin A & \cos A \\ \sin B & \cos B \\ \sin C & \cos C
\end{pmatrix}
\begin{pmatrix}
\cos A & \cos B & \cos C \\ \sin A & \sin B & \sin C
\end{pmatrix}$$
My other post has a moral: exponentials are easier than trig functions. I can't see a moral in this one.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I verify the solution for this problem? (Advanced Calculus) I'm working on a problem in Rudin (Chapter 1, Exercise 1.19)...
Question:
Suppose $\mathbf a\in R^k, \mathbf b\in R^k.$ Find $\mathbf c \in R^k$ and $r>0$ such that
$|\mathbf{x-a}| = 2|\mathbf{x-b}|$
if and only if $|\mathbf{x-c}| = r$
Solution: $3 \mathbf c = 4\mathbf b - \mathbf a, 3r = 2|\mathbf b-\mathbf a|$
The solution is here: http://minds.wisconsin.edu/handle/1793/67009
Can somebody please explain to me why they squared both sides and the rest of the solution? I am confused on which equations they are squaring, how they squared it, and so on.
Thank you.
|
Working with the first equation, notice that:
\begin{align*}
&~|\mathbf{x-a}| = 2|\mathbf{x-b}| \\
&\iff |\mathbf{x-a}|^2 = 4|\mathbf{x-b}|^2 \qquad\text{since norms are nonnegative} \\
&\iff (\mathbf x - \mathbf a) \cdot (\mathbf x - \mathbf a) = 4[(\mathbf x - \mathbf b) \cdot (\mathbf x - \mathbf b)] \\
&\iff \mathbf x \cdot \mathbf x - \mathbf x \cdot \mathbf a - \mathbf a \cdot \mathbf x + \mathbf a \cdot \mathbf a = 4[\mathbf x \cdot \mathbf x - \mathbf x \cdot \mathbf b - \mathbf b \cdot \mathbf x + \mathbf b \cdot \mathbf b] \\
&\iff \mathbf{|x|}^2 - 2\mathbf{a \cdot x + |a|}^2 = 4[\mathbf{|x|}^2 - 2\mathbf{b \cdot x + |b|}^2]\\
&\iff \mathbf{|x|}^2 - 2\mathbf{a \cdot x + |a|}^2 = 4\mathbf{|x|}^2 - 8\mathbf{b \cdot x }+ 4\mathbf{|b|}^2 \\
&\iff 0 = 3\mathbf{|x|}^2 + 2\mathbf{a \cdot x} - 8\mathbf{b \cdot x} - |\mathbf a|^2+ {4|\mathbf b|}^2 \\
\end{align*}
Basically, we square both sides so that we can get rid of the norms and work with dot products instead.
|
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|
Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^2x$ we have:
$$\int_0^{\pi /4}\frac{\sec^2x}{\sec^2x-3\tan^2x}dx=\int_0^{\pi /4}\frac{\sec^2x}{1-2\tan^2x}dx=\int _0^1 \frac{dt}{1-2t^2}=\int _0^1 \frac{1}{2}\frac{dt}{\frac{1}{2}-t^2}=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}t}{1+\sqrt{2}t}\right|_0^1=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}}{1+\sqrt{2}}\right|$$
But when we do the same integration by dividing the initial term by $\sec^4x$ and solving it yields an answer $$\frac{\pi }{2}$$
Am I wrong somewhere?
|
When you divide the numerator and the denominator by $\cos^2 x$, you should get $$\displaystyle \int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3\sin^2 x}$$ instead of $$\displaystyle \int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3 \tan^2 x}$$
because
$$\begin{array}{rcll}
\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - 3\sin^2 x \cos^2 x} &=& \displaystyle\int^{\pi/4}_0 \dfrac{\frac{dx}{\cos^2 x}}{\frac{1 - 3\sin^2 x\cos^2 x}{\cos^2 x}}\\
&=&\displaystyle\int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\frac{1}{\cos^2 x} - \frac{3\sin^2 x\cos^2 x}{\cos^2 x}}\\
&=& \displaystyle\int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3 \sin^2 x}
\end{array}$$
Denote the original integral as $I$.
For me, at this point:$$\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - 3\sin^2 x \cos^2 x}$$
I would use double angle formula to write $3 \sin^2 x \cos^2 x$ in terms of $2x$.
From $\sin 2x = 2 \sin x \cos x$, square both sides:$$\sin^2 2x = 4\sin^2 x \cos^2 x$$Multiply by $\dfrac{3}{4}$:$$\dfrac{3}{4}\sin^2 2x = 3\sin^2 x\cos^2 x$$
The integral becomes $$\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - \frac{3}{4}\sin^2 2x}$$
Next, let $u = 2x$,$\;du = 2dx$.$$\begin{array}{rcll}
I&=&\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - \frac{3}{4}\sin^2 2x}\\ &=& \displaystyle \int^{\pi/2}_0 \dfrac{\frac{du}{2}}{1 - \frac{3}{4}\sin^2 u}\\
&=&2 \displaystyle \int^{\pi/2}_0 \dfrac{du}{4 - 3\sin^2 u}
\end{array}$$
Using the fact that $\sin^2 u = \dfrac{1 - \cos 2u}{2}$,
$$\begin{array}{rcll}
I&=&2\displaystyle\int^{\pi/2}_0 \dfrac{du}{4 - 3\sin^2 u}
\\&=& 2\displaystyle \int^{\pi/2}_0 \dfrac{du}{4 - 3 \left(\frac{1 - \cos 2u}{2}\right)}
\\&=& 4\displaystyle \int^{\pi/2}_0 \dfrac{du}{8 - 3(1 - \cos 2u)}
\\&=& 4\displaystyle \int^{\pi/2}_0 \dfrac{du}{5 + 3\cos 2u}
\end{array}$$
Next, noticing that $\cos 2u = \dfrac{1 - \tan^2 u}{1 + \tan^2 u}$,
$$\begin{array}{rcll}
I&=& 4\displaystyle \int^{\pi/2}_0 \dfrac{du}{5 + 3\cos 2u}
\\&=& 4 \displaystyle \int^{\pi/2}_0 \dfrac{du}{5 + 3\left(\frac{1 - \tan^2 u}{1 + \tan^2 u}\right)}
\\&=& 4 \displaystyle \int^{\pi/2}_0 \dfrac{\sec^2 u du}{5(1 + \tan^2 u) + 3\left(1 - \tan^2 u\right)}
\\&=& 4 \displaystyle \int^{\pi/2}_0 \dfrac{\sec^2 u du}{8 + 2 \tan^2 u}
\\&=& 2 \displaystyle \int^{\pi/2}_0 \dfrac{d\left(\tan u\right)}{4 + \tan^2 u}
\end{array}$$
Using the substitution $t = \tan u$,
$$\begin{array}{rcll}
I&=& 2 \displaystyle \int^{\pi/2}_0 \dfrac{d\left(\tan u\right)}{4 + \tan^2 u}
\\&=& 2 \displaystyle \lim_{\varepsilon \to 0^{+}}\int^{1/\varepsilon}_0 \dfrac{dt}{4 + t^2}
\\&=& 2 \displaystyle \lim_{\varepsilon \to 0^{+}}\left[\dfrac{1}{2} \arctan\left(\dfrac{t}{2}\right)\right]^{1/\varepsilon}_0
\\&=& 2 \displaystyle \lim_{\varepsilon \to 0^{+}} \left(\dfrac{1}{2}\arctan\left(\dfrac{1}{2\varepsilon}\right)\right)
\\&=& \displaystyle \lim_{\varepsilon \to 0^{+}} \arctan\left(\dfrac{1}{2\varepsilon}\right)
\\&=& \color{red}{\boxed{ \dfrac{\pi}{2} }}
\end{array}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function :
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
|
Your function
$$f(x) =\sqrt{x-1} + 2\sqrt{3-x}$$
is defined for all $x \in [1, 3]$, otherwise the square root are not defined for real number.
We can check what happens on the border of the segment $[1, 3]$:
*
*$f(1) = 2\sqrt{2}$;
*$f(3) = \sqrt{2}$.
Also, we can find the local maxima and minima of $x$ in the segment $[1, 3]$. In order to get them, let's evaluate the first derivative of $f(x)$:
$$\frac{\text{d} f(x)}{\text{d} x} = -\frac{1}{2\sqrt{x-1}} + \frac{1}{\sqrt{3-x}}$$
and find the values of $x$ such that $\frac{\text{d} f(x)}{\text{d} x} = 0$. Thus:
$$-\frac{1}{2\sqrt{x-1}} + \frac{1}{\sqrt{3-x}} = 0 \Rightarrow 2\sqrt{x-1}=\sqrt{3-x}.$$
Since both $x-1$ and $3-x$ are non-negative in the segment $[1, 3]$, we can elevate to the power of $2$ both members without particular problems:
$$4(x-1)=3-x \Rightarrow 4x+x = 3 + 4 \Rightarrow x = \frac{7}{5} \in [1, 3].$$
$x = \frac{7}{5}$ is a local maximum or minimum and $f\left(\frac{7}{5}\right) = \sqrt{10}.$
Summarizing:
*
*At the begining of the segment $[1, 3]$ the function assumes value $2\sqrt{2}$;
*at the end, the values is $2\sqrt{2}$;
*somewhere inside the segment (i.e. when $x = \frac{7}{5}$) the function has value $\sqrt{10}$.
It's clear now that the minimum value of $f(x)$ in $[1,3]$ is $\sqrt{2}$ while the maximum is $\sqrt{10}$.
We can conclude that the range of $f(x)$ is $[\sqrt{2}, \sqrt{10}]$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Series question,related to telescopic series, 1/2*4+ 1*3/2*4*6+ 1*3*5/2*4*6*8 ...infinity The series is $$\frac{1}{2*4}+ \frac{1*3}{2*4*6}+ \frac{1*3*5}{2*4*6*8}+....$$
It continues to infinity.I tried multiplying with $2$ and dividing each term by$(3-1)$,$(5-3)$ etc,starting from the second term which gives me $$\frac{1}{8} -\frac{1}{4*6} -\frac{1}{4*6*8}-\frac{1}{4*6*8*10}$$.
Also if anybody is wondering the answer 0.5
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We have
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)}&=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2(2n+2)}\\\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2 \cdot (2n+2)}\\\\
& =\frac{(2n)!}{2^{2n} (n!)^2 \cdot (2n+2)}\\\\
&=\frac{1}{(2n+2)\,4^n}\binom{2n}{n}\\\\&=\frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1}.
\end{align}
$$
The series is then telescopic:
$$\sum_{n=1}^{N}\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)}=\frac{1}{2}-\frac{1}{4^{N+1}}\binom{2N+2}{N+1}.
$$As $N$ is great, one may prove that
$$\frac{1}{4^{N+1}}\binom{2N+2}{N+1}=O\left(\frac{1}{\sqrt{N}}\right).
$$
The series is thus equal to $\dfrac{1}{2}$:
$$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)}=\frac{1}{2}.
$$
|
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Problem with Smith normal form over a PID that is not an Euclidean domain This is an homework exercise of the Algebra lecture.
I need to evaluate the Smith normal form of the following matrix
$$A:=\begin{pmatrix}1 & -\xi & \xi-1\\2 \xi&8&8\xi+7\\\xi& 4 & 3\xi +2 \end{pmatrix} \in M(3\times 3;\Bbb{Z}[\xi]),$$
where $\xi := \frac{1+\sqrt{-19}}{2}$.
We have seen in the lecture, that similarly as for rings, there exists the Smith Normal Form also for PID's. To solve the problem (we don't need to find $S, T \in M(3 \times 3; \Bbb{Z}[\xi])$, such that $SAT$ is in Smith Normal Form) we applied row, and column operations, paying particular attention to not multiply rows and columns by nonunits, since this is not allowed. (Also according to this answer). The steps that we performed are:
\begin{align*}\begin{pmatrix}1 & -\xi & \xi-1\\2 \xi&8&8\xi+7\\\xi& 4 & 3\xi +2 \end{pmatrix} &\overset{2\text{column}+\xi \text{ first column}}{\leadsto}
\begin{pmatrix}1 & 0 & \xi-1\\2 \xi&8+2\xi^2&8\xi+7\\\xi& 4+\xi^2 & 3\xi +2 \end{pmatrix}\\ \overset{3\text{column}-(\xi-1)
\text{ first column}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\2 \xi&8+2\xi^2&-2\xi^2+10\xi+7\\\xi& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}
&\overset{2\text{row}-(2\xi)
\text{ first row}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&8+2\xi^2&-2\xi^2+10\xi+7\\\xi& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}\\
\overset{3\text{row}-(\xi)
\text{ first row}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&8+2\xi^2&-2\xi^2+10\xi+7\\0& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}&
\overset{2\text{row}-2
\text{ third row}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&0&2\xi+3\\0& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}\\
\overset{\text{swap columns}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& -\xi^2+4\xi+2 & 4+\xi^2 \end{pmatrix} &
\\
\overset{\text{second column $+$ third column}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& 4 \xi +6 & 4+\xi^2 \end{pmatrix}
& \overset{\text{third row } - 2 \text{ second row}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& 0 & 4+\xi^2 \end{pmatrix}=: B
\end{align*}
Unfortunately the term $b_{33}$ is not divisible by the term $b_{22}$ as it should be. We have done the steps many many times, but we can't find the error (I'm doing this homework exercise with my classmates). Do you maybe find it? Thank you in advance for any help.
|
$\xi^2=\xi-5$, and therefore $4+\xi^2=\xi-1$ (which is a prime element). On the other side, $2\xi+3=2(\xi-1)+5=2(\xi-1)+\xi-\xi^2=-(\xi-1)(\xi-2)$, so $b_{33}\mid b_{22}$.
The SNF of your matrix is $$\begin{pmatrix}1 & 0 & 0\\0&\xi-1&0\\0& 0 & (\xi-1)(\xi-2) \end{pmatrix}.$$
|
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|
Evaluation of $\int \frac{x\sin(\sin x)}{x+5} \ dx$ How do we find $$\int \frac{x\sin(\sin x)}{x+5} \ dx\ ,$$ is there any way to take that $\sin x$ out from parent $\sin(\cdot)$ ?
|
Let $u=x+5$ ,
Then $x=u-5$
$dx=du$
$\therefore\int\dfrac{x\sin(\sin x)}{x+5}dx$
$=\int\dfrac{(u-5)\sin(\sin(u-5))}{u}du$
$=\int\sin(\sin(u-5))~du-\int\dfrac{5\sin(\sin(u-5))}{u}du$
For $\int\sin(\sin(u-5))~du$ ,
$\int\sin(\sin(u-5))~du$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}(u-5)}{(2n+1)!}du$
$=-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}(u-5)}{(2n+1)!}d(\cos(u-5))$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2(u-5))^n}{(2n+1)!}d(\cos(u-5))$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}(u-5)}{(2n+1)!}d(\cos(u-5))$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}(u-5)}{(2n+1)!k!(n-k)!}d(\cos(u-5))$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}(u-5)}{(2n+1)!k!(n-k)!(2k+1)}+C$
|
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|
Write it as an element of this ring? Since the degree of the irreducible polynomial $x^3+2x+2$ over $\mathbb{Q}[x]$ is odd, it has a real solution , let $a$. I am asked to express $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$.
I have done the following:
Since $a$ is a real solution of $x^3+2x+2$, we have that $a^3+2a+2=0$.
$\frac{1}{1-a}=y \Rightarrow a=\frac{y-1}{y}$
$$a^3+2a+2=0 \\ \Rightarrow \frac{(y-1)^3}{y^3}+2\frac{y-1}{y}+2=0 \\ \Rightarrow (y-1)^3+2(y-1)y^2+2y^3=0 \\ \Rightarrow 5y^3-5y^2+3y-1=0 \\ \Rightarrow 5\left ( \frac{a-1}{a} \right )^3-5\left ( \frac{a-1}{a} \right )^2+3\left ( \frac{a-1}{a} \right )-1=0 $$
That what I have found is not a polynomial of $a$, with coefficients at $\mathbb{Q}$, right??
Is there an other way to write $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$ ??
|
Without any trick: you know that $(1-a)^{-1}=\alpha+\beta a+\gamma a^2$ for some $\alpha,\beta,\gamma\in\mathbb{Q}$, that is
$$
(1-a)(\alpha+\beta a+\gamma a^2)=1
$$
This becomes
$$
\alpha+\beta a+\gamma a^2-\alpha a-\beta a^2-\gamma a^3=1
$$
Since $a^3=-2a-2$ this becomes
$$
(\alpha+2\gamma)+(\beta-\alpha+2\gamma)a+(\gamma-\beta)a^2=1
$$
or
\begin{cases}
\alpha+2\gamma=1\\
-\alpha+\beta+2\gamma=0\\
-\beta+\gamma=0
\end{cases}
This should be easy to solve.
|
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|
Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$ Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$.
My solution:
multiplying by: $\displaystyle\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}$
Which gives us: $\displaystyle\frac{x-1}{\sqrt{x^2+x-1}-x}$
dividing by $\sqrt{x^2}$ gives:
$\displaystyle \frac{1}{\sqrt{1}-1}$
which equals $1/0$
However, I double checked my answers, and this does not seem to be correct, am I making a mistake (perhaps when I take the $\sqrt{1}$ in the denominator of the last step?
|
Your mistake at this step:
dividing by $\sqrt{x^2}$ gives: $\displaystyle \frac{1}{\sqrt{1}-1}$
Since $x\to -\infty$, then $\sqrt{x^2}=-x$. So $\frac{x-1}{\sqrt{x^2+x-1}-x}=\frac{-1+\frac 1x}{\sqrt{1+\frac 1x-\frac{1}{x^2}}+1}.$
Take careful!
|
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Inequality: $ \frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq\frac{1}{2} $ I was trying to solve this inequality, but I wasn't able to do so:
Find the maximum number $k\in\mathbb R$ such that:
$$ \frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq\frac{1}{2} $$
Holds for all $a,b,c\ge0$ with $a+b+c=1$
Any help is highly appreciated.
|
Plugging, $a=1/2,b=1/2,c=0 \implies k \le 4$.
I will show that $k=4$.
$$ \frac{a}{1+9bc+4(b-c)^2}+\frac{b}{1+9ca+4(c-a)^2}+\frac{c}{1+9ab+4(a-b)^2}\ge \frac{1}{2} \\ \Longleftrightarrow \frac{a^2}{a+9abc+4a(b-c)^2}+\frac{b^2}{b+9abc+4b(c-a)^2}+\frac{c^2}{c+9abc+4c(a-b)^2} \ge \frac{1}{2}$$
By cauchy it suffice to show that, $$ \frac{(a+b+c)^2}{a+b+c+3abc+4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)} \ge \frac{1}{2} \\ \Longleftrightarrow \frac{1}{1+3abc+4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)} \ge \frac{1}{2} \\ \Longleftrightarrow 1 \ge 3abc+4a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \\ \Longleftrightarrow (a+b+c)^3 \ge 3abc+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \\ \Longleftrightarrow a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b) \ge 0 $$
Last inequality is true by Schur. $\Box$
|
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|
Question on an algebraic expression Why this is true?
$$
\frac{\frac{1}{\sqrt{x}}-1}{x-1} = -\frac{1}{\sqrt{x}+x}
$$
|
When you have complicated fractions, it is always a good idea to try making common denominators where applicable. The numerator is a bit complicated in this case, so let's work with that:
$$\frac{1}{\sqrt{x}} - 1 = \frac{1}{\sqrt{x}} - \frac{\sqrt{x}}{\sqrt{x}} = \frac{1-\sqrt{x}}{\sqrt{x}}.$$
Now let's replace $\frac{1}{\sqrt{x}} - 1$ with $\frac{1-\sqrt{x}}{\sqrt{x}}$ in your expression.
$$\frac{\frac{1}{\sqrt{x}}-1}{x-1} = \frac{\frac{1-\sqrt{x}}{\sqrt{x}}}{x-1}.$$
Moving the $\sqrt{x}$ to the denominator, we have
$$\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}.$$
In the numerator on the right hand side of your expression, you have $1$ but we have $1-\sqrt{x}$. What we should do now is rationalize the numerator by multiplying by $\frac{1+\sqrt{x}}{1+\sqrt{x}}$. Note that this does not change our expression any since it is really just a clever form of $1$. Doing so, we have
$$\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}\frac{1+\sqrt{x}}{1+\sqrt{x}} = \frac{(1-\sqrt{x})(1+\sqrt{x})}{\sqrt{x}(x-1)(1+\sqrt{x})} = \frac{1-x}{\sqrt{x}(x-1)(1+\sqrt{x})}.$$
Cancelling the common term of $1-x$ from the numerator and denominator gives us
$$-\frac{1}{\sqrt{x}(1+\sqrt{x})} = -\frac{1}{\sqrt{x}+x}$$
as desired.
|
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|
$\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers of $x$ and $y$ Give a convincing argument that $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers $x$ and $y$. Could someone please explain how to prove this? I attempted to say that the largest values that could be added to $x$ and $y$ is $0.99$ and that doing so still made $\lfloor x\rfloor + \lfloor y\rfloor < \lfloor x+y\rfloor$. However, my answer was not accepted.
|
Let $x = a + m$ and $y = b + n$ where $a, b$ are integers and $m,n$ are real numbers such that $0 \le m,n < 1$. Then,
$$\lfloor x \rfloor + \lfloor y \rfloor = a + b$$
while
$$\begin{align}\lfloor x + y \rfloor &= \lfloor a + b + m + n \rfloor\\
&= a + b + \lfloor m + n \rfloor\\
& \ge a + b\\
&= \lfloor x \rfloor + \lfloor y \rfloor\end{align}$$
Equality occurs when the fractional part of $x$ and $y$ sum up to be less than $1$, in which case $\lfloor m + n \rfloor = 0$. For example, $x = 0.5$ and $y = 1.1$.
|
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|
study of two sequences I need to study wether thos two sequences converge or not.
1) $u_n=n\sum_{k=1}^{2n+1} \frac{1}{n^2+k}$
2) $v_n=\frac{1}{n}\sum_{k=0}^{n-1} \cos(\frac{1}{\sqrt{n+k}})$
For the first i get it converges to $0$ by just expressing the sum as a fraction (not very smart imo).
|
How can $u_n$ converge to zero?
$$ u_n = n\sum_{k=1}^{2n+1}\frac{1}{n^2+k}\geq n\cdot\frac{2n+1}{(n+1)^2} \geq 2-\frac{3n+2}{(n+1)^2}.$$
On the other hand,
$$ u_n \leq n\cdot\frac{2n+1}{n^2+1} \leq 2+\frac{n-2}{n^2+1},$$
hence:
$$\lim_{n\to +\infty} u_n = 2.$$
For the second sum, since $\cos x \geq 1-\frac{x^2}{2}$ over $[0,1]$,
$$ v_n \geq \frac{1}{n}\left(n-\frac{1}{2}\sum_{k=0}^{n-1}\frac{1}{n+k}\right)\geq \frac{1}{n}\left(n-\frac{1}{2}\right),$$
but since $\cos x\leq 1-\frac{4x^2}{\pi^2}$ holds over the same interval,
$$ v_n \leq \frac{1}{n}\left(n-\frac{4}{\pi^2}\sum_{k=0}^{n-1}\frac{1}{n+k}\right)\leq \frac{1}{n}\left(n-\frac{4n}{\pi^2(2n-1)}\right),$$
so:
$$\lim_{n\to +\infty} v_n = 1.$$
|
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|
Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$ Given that $P$ is any point on the incircle of a regular $n$-sided polygon with edge length $l$ and vertices $V_1,V_2...V_n$, how do we prove that $$\displaystyle\sum_{i=1}^{n} \overline{PV_i}^2=\dfrac{nl^2}{4} \left(1+2\cot^2 \dfrac{\pi}{n}\right)$$
?
Here is a figure of what I understand
I don't even know where to begin with this one.. We want to prove that $\displaystyle\sum_{i=1}^{n} \overline{PV_i}^2$ is independent of the chosen point $P$..
Hints in the right direction and answers appreciated.
Please avoid using vector algebra or complex numbers to solve this problem.
|
Let $W_i$ be the midpoint of the segment $\overline{V_iV_{i+1}}$ (where $V_{n+1}=V_1$). Apply the Apollonius' theorem on the triangle $\triangle PV_iV_{i+1}$ with the midpoint $W_i$. We have
$$\overline{PV_i}^2+\overline{PV_{i+1}}^2=2(\overline{PW_i}^2+\overline{V_iW_i}^2)=2\overline{PW_i}^2+\frac{l^2}{2}$$
Summing this up with all $1\le i\le n$ gives
$$2\sum_{i=1}^n\overline{PV_i}^2=2\sum_{i=1}^n\overline{PW_i}^2+\frac{n}{2}l^2\tag{1}$$
Now we will show that if $T_1T_2\cdots T_n$ is a regular polygon with a circumcircle of radius $r$, we have $\displaystyle\sum_{i=1}^n\overline{PT_i}^2=2nr^2$ for every point $P$ on the circle.
Without the loss of generality, assume $P$ lie on the arc $T_1T_n$. From the law of cosines, $$\overline{PT_i}^2+\overline{PT_{i+1}}^2=2\overline{PT_i}\cdot\overline{PT_{i+1}}\cos\frac\pi n+(2r\sin\frac\pi n)^2=4\cot\frac\pi n|\triangle PT_iT_{i+1}|+(2r\sin\frac\pi n)^2$$ for all $1\le i\le n-1$. For $i=n$,\begin{align}\overline{PT_1}^2+\overline{PT_{n}}^2&=2\overline{PT_1}\cdot\overline{PT_{n}}\cos\frac{(n-1)\pi }n+(2r\sin\frac{(n-1)\pi} n)^2\\&=-4\cot\frac\pi n|\triangle PT_1T_n|+(2r\sin\frac\pi n)^2\end{align}
Sum this up by $i$ and get \begin{align}2\sum_{i=1}^n\overline{PT_i}^2&=4\cot\frac\pi n|T_1T_2T_3\cdots T_n|+4nr^2\sin^2\frac\pi n\\&=4\cot\frac\pi n\sum_{i=1}^n|\triangle OT_iT_{i+1}|+4nr^2\sin^2\frac\pi n\\&=4\cot\frac\pi n\sum_{i=1}^nr\sin\frac\pi nr\cos\frac\pi n+4nr^2\sin^2\frac\pi n\\&=4nr^2\end{align} Therefore $\displaystyle\sum_{i=1}^n\overline{PT_i}^2=2nr^2$
Now back to the original problem. The $W_1W_2\cdots W_n$ is a regular n-polygon lying on a circle with radius $r=\frac l2\cot\frac\pi n$. So $$\sum_{i=1}^n\overline{PW_i}^2=2nr^2=\frac{nl^2}{2}\cot^2\frac\pi n$$ Hence from $(1)$, we get $$\sum_{i=1}^n\overline{PV_i}^2=\frac{nl^2}4(1+2\cot^2\frac\pi n)$$ and we are done.
|
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|
Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$ I'm trying to prove that
$$
\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}.
$$
Let $\alpha = 2 \arcsin x$ and $\beta = \arccos x$; meaning $\sin\alpha = \frac{x}{2}, \cos\beta = x$. We know that:
$$
\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.
$$
Finding $\cos\alpha$ and $\sin\beta$:
$$\begin{align}
\cos\alpha & = \frac{\sqrt{4-x^2}}{2}, \\[0.1in]
\sin\beta & = \sqrt{1-x^2}.
\end{align}$$
So plugging everything in:
$$\begin{align}
\sin(2\arcsin x + \arccos x) & = \frac{x}{2} \cdot x + \frac{\sqrt{4-x^2}}{2} \cdot \sqrt{1-x^2}\\[0.1in]
& = \frac{x^2 + \sqrt{4-x^2}\sqrt{1-x^2}}{2}.
\end{align}$$
But this doesn't seem to lead to the right side. Is my method incorrect?
|
$$\arcsin x+\arccos x=\frac\pi2$$
So, we have $$\sin\left(\frac\pi2+\arcsin x\right)=\cos(\arcsin x)$$
Now $\cos(\arcsin(x)) = \sqrt{1 - x^2}$. How?
|
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|
Justify: $R$ is commutative
Let $R$ be a ring. If $(ab)^n=a^nb^n$ holds for all $n$ and all $a, b\in R$, then justify that $R$ is commutative ring.
No idea how to prove it. But also fail to get one counter example. Please help
|
By assumption
$$
(a(b+1))^2=a^2(b+1)^2=a^2b^2+2a^2b+a^2
$$
and by distrubutive law
$$
(a(b+1))^2=(ab+a)^2=(ab)^2+aba+a^2b+a^2
$$
From this it follows that
$$
a^2b=aba
$$
and this holds for all $a,b \in R$. Especially for $a+1$ and $b$
$$
(a+1)^2b=(a+1)b(a+1)=(ab+b)(a+1)=aba+ab+ba+b
$$
but compute the left side to get
$$
(a+1)^2b=(a^2+2a+1)b=a^2b+2ab+b=aba+ab+ab+b
$$
now compare
|
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|
Find $\int \frac{x^2}{2x-1}\mathrm{d}x$ Use the substitution $2x-1$ to find $\int \frac{x^2}{2x-1}\mathrm{d}x.$
I started out by using the substitution to find $\mathrm{d}x$ in terms of $\mathrm{d}u$.
$$
\frac {\mathrm{d}u}{\mathrm{d}x} = 2
$$
$$
\mathrm{d}x = \frac {1}{2}\mathrm{d}u
$$
Now the question looks like this:
$$
\int \! \frac{x^2}{u} \, \cdot \frac {1}{2}\mathrm{d}u
$$
I then found $x^2$ in terms of $u$:
$$
2x-1 = u
$$
$$
x = \frac {u+1}{2}
$$
$$
x^2 = \left(\frac {u+1}{2}\right)^2
$$
I then substituted this back into the equation.
$$
\int \! \frac {(\frac {u+1}{2})^2}{u} \cdot \frac {1}{2}\, \mathrm{d}u
$$
$$
\int \! \frac {(\frac {u+1}{2})^2}{2u}\, \mathrm{d}u
$$
From here I have tried expanding out the brackets, and using a substitution to put $2u$ on the top of the fraction, but neither method got the correct answer, and I have spent a long time on the question. Am I on the right track here, or should I be approaching this any differently? Thanks
|
HINT
Here it is another way to solve it: consider the following algebraic manipulation
\begin{align*}
\frac{x^{2}}{2x-1} & = \frac{1}{4}\times\frac{4x^{2}}{2x-1} = \frac{1}{4}\times\frac{(4x^{2}-1) + 1}{2x-1} = \frac{1}{4}\times\left[\frac{4x^{2}-1}{2x-1} + \frac{1}{2x-1}\right]\\\\
& = \frac{1}{4}\times\left[2x+1 + \frac{1}{2x-1}\right]
\end{align*}
Can you take it from here?
|
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|
Modular Arithmetic: using congruence to find remainder How do I use the fact that if $a = b \pmod n$ and $c = d\pmod n$ then $ac = bd\pmod n$ to find the remainder when $3^{11}$ is divided by $7$?
|
We know that:
$3\equiv3\pmod7$. Multiplying both sides by 3, we get $3^2\equiv9\pmod7$, which simplifies to:
$3^2\equiv2\pmod7$. Multiplying both sides by 3 again (and simplifying):
$3^3\equiv6\pmod7$
$3^4\equiv4\pmod7$
$3^5\equiv5\pmod7$
$3^6\equiv1\pmod7$
$3^7\equiv3\pmod7$
$3^8\equiv2\pmod7$
$3^9\equiv6\pmod7$
$3^{10}\equiv4\pmod7$
$3^{11}\equiv5\pmod7$
There are two ways to speed this up. One is to use Fermat's Little (not to be confused with Last) Theorem, which states that $a^{p-1}\equiv1\pmod p$ when $a\not\equiv0$ and $p$ is prime. So, we have:
$3^{11}\equiv3^63^5\equiv(1)3^5\equiv3^5\pmod7$.
To find $3^5\pmod7$, we only need to use the above method 5 times instead of 11.
Another is to go like this:
$3\equiv3\pmod7$. Squaring both sides:
$3^2\equiv2\pmod7$. Squaring:
$3^4\equiv4\pmod7$. Squaring:
$3^8\equiv2\pmod7$.
Since $11=8+2+1$, we have $3^{11}\equiv3^83^23^1\equiv(2)(2)(3)\equiv12\equiv5\pmod7$.
|
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|
A double Summation involving 7th roots of unity Is there possibly a closed form for $$\sum_{m=1}^{\infty} \left(\sum_{k=1}^{6} \dfrac{1}{m-\alpha^k}\right)^2$$ where $\alpha=e^{2\pi i/7}$ ?
I'm having problems evaluating the first sum, let alone the second. I tried some algebraic manipulation, but I could only find that $\displaystyle\sum_{k=1}^{6} \dfrac{1}{1-\alpha^k}=3$ using the fact that $\dfrac{1}{1-\alpha^k}+\dfrac{1}{1-\alpha^{7-k}}=1$
Hints and answers appreciated.
|
The six numbers $m-\alpha^k$ are the roots of $f(X) = \frac{(m-X)^7 - 1}{m-X-1} = X^6 + a_5 X^5 + \cdots + a_1 X + a_0$, so the sum of their reciprocals is $-\frac{a_1}{a_0}$.
Letting $X=0$, we have $a_0 = \frac{m^7-1}{m-1}$. Taking a derivative, we have $a_1 = -\frac{6m^7 -7m^6+1}{(m-1)^2}$, therefore:
$$s_m = \sum_{k=1}^6 \frac{1}{m-\alpha^6} = \frac{(6m^7 - 7m^6 + 1)}{(m-1)(m^7-1)} = \frac{6m^5 + 5m^4 + 4m^3 + 3m^2 + 2m + 1}{m^6 + m^5 + m^4 + m^3 + m^2 + m + 1}$$ $$= \frac{7m^6}{m^7-1} - \frac{1}{m-1}$$
(the last form comes from taking the logarithmic derivative, which ccorn suggested)
You can check that this gives $3$ for $m=1$. You can also quickly verify that $s_m \leq 21/m$, so $\sum_{m=1}^\infty s_m^2$ converges, and is less than $\frac{147}{2}\pi^2$. This is a very crude bound; after a few initial terms we should have $s_m \approx 6/m$, so the sum is closer to $6\pi^2$.
As for the possible existence of a closed form... well, good luck with that.
Addendum: I think this should not even be attempted until there is a solution for the same problem with third roots of unity, as this already seems quite hard, yet the corresponding problem for second roots of unity is quite tractable. In other words, is there a closed form for the following sum?:
$$\sum_{m=1}^\infty \left( \frac{2m+1}{m^2 + m + 1} \right)^2 = 1 + \sum_{m=2}^\infty \left( \frac{3m^2}{m^3-1} - \frac{1}{m-1} \right)^2$$
There are closed forms involving the logarithmic derivative of the gamma function... but I generally don't accept something as a closed form if it's both conceptually more difficult and harder to evaluate than the original formula.
Another observation:
$$\frac{(2m+1)^2}{(m^2 + m + 1)^2} = 4\cdot \frac{1}{m^2 + m + 1} - 3\cdot \frac{1}{(m^2 + m + 1)^2}$$
The second sum actually does have a closed form, though it's a doozy:
$$\sum_{m=0}^\infty \frac{1}{m^2 + m + 1} = \frac{2\pi}{\sqrt{3}} \left( \coth(\pi \sqrt{3}) -\frac{1}{2}\coth(\frac{\pi\sqrt{3}}{2}) \right)$$
(this is derived from the identity $\coth x = \sum_{m=-\infty}^\infty \frac{x}{\pi^2m^2 + x^2}$, which comes from Fourier something something)
What this means is that there is a closed form for $\sum_{m=1}^\infty \frac{(2m+1)^2}{(m^2 + m + 1)^2}$ if and only if there is a closed form for $\sum_{m=1}^\infty \frac{1}{(m^2 + m + 1)^2}$. This seems like a useful transformation, since the latter looks more standard and has probably been studied.
This seems to come down to finding a closed form for the real part of $\sum_{m=1}^\infty \frac{1}{(m-\omega)^2}$, where $\omega$ is a third root of unity. This looks promising, but there might still be a negative answer.
|
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|
Evaluate and find the principal value of $(-1+i)^ {2-i}$ Can anyone please help me evaluate and find the principal value of
$(-1+i)^{2-i}$
I got up to
$=e^{2-i}(ln(-1+i))$
$=e^{(2-i)(1/2 ln(2)+i(3pi/4))}$
|
\begin{align}
(-1+i)^{2-i} &= \left[ e^{\pi i} \, \sqrt{2} \, e^{- \pi i/4} \right]^{2 - i} \\
&= e^{2 \pi i + \pi} \, e^{(2-i) \ln(\sqrt{2})} \, e^{- \pi i/2 - \pi/4} \\
&= 2 \, e^{ 3 \pi/4} \left( e^{i(3 \pi + \ln(2))/2} \right)\\
&= 2 \, e^{ 3 \pi/4} \left[ \cos\left( \frac{3 \pi + \ln(2)}{2} \right) + i \,
\sin\left( \frac{3 \pi + \ln(2)}{2} \right) \right]
\end{align}
The principle value is
\begin{align}
2 \, e^{ 3 \pi/4} \, \cos\left( \frac{3 \pi + \ln(2)}{2} \right)
\end{align}
and the argument is
\begin{align}
2 \, e^{ 3 \pi/4} \, \sin\left( \frac{3 \pi + \ln(2)}{2} \right)
\end{align}
|
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|
Using $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ evaluate the limit Use $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$
to evaluate the limit
$$\lim_{x\to 0} \frac{x\tan (x^2)}{\cos(3x)\sin^3(2x)}$$
I'm really not sure how to go about this apart from trig identities and using the limit $=1$ to simplify it. Can anyone even get me started on it please?
|
$$\lim_{x\to0}\frac{x\tan(x^2)}{\cos(3x)\sin^3(2x)}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{x^2}\cdot\frac{\tan(x^2)}{\sin^3(2x)}=$$
$$\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{\sin^3(2x)}\cdot\frac{\tan(x^2)}{x^2}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot[\frac{x}{\sin(2x)}]^3\cdot\frac{\tan(x^2)}{x^2}$$
Now you have that:
$$\lim_{x\to0}\frac{x}{\sin(2x)}=\lim_{x\to0}\frac{1}{2}\frac{2x}{\sin(2x)}=\frac{1}{2}$$
and, setting $\ x^2=t:$
$$\lim_{x\to0}\frac{\tan(x^2)}{x^2}=\lim_{t\to0}\frac{\tan(t)}{t}=\lim_{t\to0}\frac{1}{\cos(t)}\frac{\sin(t)}{t}=1\cdot1=1$$
so the final result should be: $\ [\frac{1}{2}]^3=\frac{1}{8}$
|
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Finding all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors. Find all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors.
I've tried checking the first 6-7 $n$'s on wolframalpha, but I don't see any patterns for even nor odd $n$'s. At first I thought for all odd $n$'s it was divisible by $2,3,7$, but $n=5$ doesn't work ($n=1,3,7$ do work). How are these $n$ found?
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$2^n + 2^{2n}+2^{3n} = 2^n(1+2^n + 2^{2n}) = 2^n p^a$ for some prime $p$ and positive integer $a$.
Then, $p^a = 1+2^n + 2^{2n}$.
Suppose $a>1$.
If $a$ is even, $a = 2b$, meaning $p^a - 1 = (p^b - 1)(p^b+1) = 2^n(1+2^n)$, which is impossible (you can see that easily). Then $a$ must be odd.
Then, $2^n + 2^{2n}= (p-1)(p^{a-1} + \dots +1)$. $(p^{a-1} + \dots +1)$ must be odd (it is the sum of an odd number of odd numbers), then $2^n | p-1$. But since $p-1 < p^2 + p +1$, you must have $2^n = p-1$, which means $p= 2^n +1= (p^{a-1} + \dots +1)$, an absurd.
So, $a=1$.
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Is the derivative of the characteristic polynomial equal to the sum of characteristic polynomial of principal submatrices? Let $A$ by an $n \times n$ matrix over the complex numbers and let $\phi(A,x) = \det(xI-A)$ be the characteristic polynomial of $A$. Let $B_i$ be the principal submatrix of $A$ formed by deleting the $i$-th row and column of $A$. Let $\phi(B_i,x)$ be the characteristic polynomial of $B_i$.
Is it always true that the derivative of $\phi(A,x)$ is the sum of $\phi(B_i,x)$ as $i$ ranges across the rows/columns of $A$?
That is, is the following an identity?
$$ \frac{d\ \phi(A,x)}{dx} = \sum_i \phi(B_i,x) $$
If not, what is a small counterexample? (The equation works for all $2 \times 2$ matrices and all adjacency matrices of graphs....)
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It is true.
Proof:
$$
\phi(A,x) =
\begin{pmatrix}
x-a_{1,1} & -a_{1,2} & \cdots & -a_{1,n} \\
-a_{2,1} & x-a_{2,2} & \cdots & -a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
-a_{m,1} & -a_{m,2} & \cdots & x-a_{m,n}
\end{pmatrix}
$$Let us write the derivative using multilinearity:
$$
\phi'(A,x) =
\begin{pmatrix}
1 &0 & \cdots & 0 \\
-a_{2,1} & x-a_{2,2} & \cdots & -a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
-a_{m,1} & -a_{m,2} & \cdots & x-a_{m,n}
\end{pmatrix} +
\begin{pmatrix}
x-a_{1,1} & -a_{1,2} & \cdots & -a_{1,n} \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
-a_{m,1} & -a_{m,2} & \cdots & x-a_{m,n}
\end{pmatrix} +\dots +
\begin{pmatrix}
x-a_{1,1} & -a_{1,2} & \cdots & -a_{1,n} \\
-a_{2,1} & x-a_{2,2} & \cdots & -a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{pmatrix}
= \sum _{i=1}^n \phi(B_i,x)
$$
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|
Proof by induction for $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$ for $k > 4$ I was given this proof for hw.
Prove that $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
So, far I've gotten this
Basis:
$k = 5$, $2^{5 + 1} - 1 > 2\cdot5^2 + 2\cdot5 + 1$ => $63 > 61$ (So, the basis holds true)
Hypothesis:
for all $k > 4$, $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
Inductive step:
LHS
$2^{k + 1 + 1} - 1
= 2^{k + 2}
= 2 \cdot 2^{k + 1}$
$2[ 2^{k + 1} - 1 + 1] - 1 \geqslant 2\cdot(2k^2 + 2k + 1 + 1) - 1$
RHS (Attempt to prove that the RHS is less than the LHS )
$2(2k^2 + 2k + 1 + 1) - 1 \geqslant 2(k + 1)^2 + 2(k + 1) + 1$
Next step
$2(k + 1)^2 + 2(k + 1) + 1
= 2k^2 + 4k + 2 + 2k + 1 + 1
= (2k^2 + 2k + 1) + 4k + 3$
Now I'm stuck at how to prove that
$2(2k^2 + 2k + 1 + 1) - 1 \geqslant (2k^2 + 2k + 1) + 4k + 3$
So, what can I do?
Thanks
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Hypothesis:
$$\begin{align} \forall k > 4,&\quad\; 2^{k + 1} - 1 > 2(k)^2 + 2(k) + 1 \\ &\iff 2^{k+1} \gt 2k^2 + 2k + 2 \end{align}$$
Inductive step:
$$\begin{align} 2^{k + 1 + 1}& = 2\cdot 2^{k+1}\\
& \gt 2\cdot (2k^2 + 2k+2) \\
& = 4k^2 + 4k + 4\\
& = (2k^2 + 4k + 2) + (2k^2 + 2)\\
& = 2(k+1)^2 + 2k^2 + 2 \\
& \geq 2(k+1)^2 + 2(k + 1) + 2\tag{for $k \geq 2$}
\end{align}$$
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Show that inequality holds How would you show that the following inequality holds? Could you please write your reasoning by solving this problem too?
$a^2 + b^2 + c^2 \ge ab + bc + ca$ for all positive integers a, b, c
I have tried:
$a^2 + b^2 + c^2 - ab - ab + ab \ge bc + ca $
$(a^2 -2ab + b^2) + c^2 + ab \ge bc + ca $
$(a-b)^2 + c^2 + ab - bc - ca \ge 0 $
$(a-b)^2 + c(c - a) - b (c - a) \ge 0 $
$(a-b)^2 + (c - a)(c - b) \ge 0 $
Well, $(a-b)^2$ is positive. But how do we know if $(c-a)(c-b)$ is positive as well?
PS: The problem is from the book "How to think like a mathematician" by Kevin Housten.
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Hint
$$0\leq (a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$$
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|
Find the tangen to $\cos(\pi \cdot x)$ I have the following assignment.
Find the tangent to $y=f(x)=\cos(\pi \cdot x)$ at $x=\displaystyle\frac{1}{6}$.
First step would be to take the derivative of $f(x)$
$f'(x)= -\pi \sin(\pi \cdot x)$
Then I put the $x$-value into $f'(x)$ to find the slope
$f'(\displaystyle\frac{1}{6})= -\pi \sin(\pi \cdot \frac{1}{6})= \frac{-\pi}{2}$
And I put the $x$-value into $f(x)$ to find the $y$-value
$f(\displaystyle\frac{1}{6})=\cos(\pi \cdot \frac{1}{6})=\frac{\sqrt{3}}{2}$
Now I use the formula $y=m(x-x_0)+y_0$
$y= \displaystyle\frac{-\pi}{2}(x-\frac{1}{6})+\frac{\sqrt{3}}{2}$ = $\displaystyle\frac{-\pi \cdot x}{2} + \frac{\sqrt{3}}{2} + \frac{\pi}{12}$
Am I correct? Because my book got an different answer like $6 \pi x +12y =6\sqrt{3}+ \pi$
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Start with your answer: $y = -\dfrac{\pi x}{2} + \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{12}$
Multiply both sides by $12$: $12y = -6\pi x + 6\sqrt{3} + \pi$
Add $6 \pi x$ to both sides: $6\pi x + 12y = 6\sqrt{3} + \pi$
Therefore your answer is equivalant to the book's answer.
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|
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd)
+ (ad+bc)\sqrt{5}$
My work:
Since $\mathbb Q$ is a field, addition and multiplication defined by $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd) + (ad+bc)\sqrt{5}$ will produce elements in $F$ therefore $F$ is closed under multiplication and addition.
Because $F$ is a subset of $\mathbb R$ the operation on $F$ correspond to the usual operations on $\mathbb R$ so the associative, commutative and distributive conditions are inherited from $\mathbb R$
The additive identity is $0+0\sqrt{5}$ because $(a + b\sqrt{5})+(0+0\sqrt{5}) = (a + b\sqrt{5})$
The multiplicative identity is $1+0\sqrt{5}$ because $(a + b\sqrt{5})(1+0\sqrt{5}) = (a + b\sqrt{5})$
The additive inverse is $((-a) + (-b)\sqrt{5})$ because $(a + b\sqrt{5})+((-a) + (-b)\sqrt{5}) = 0$
The multiplicative inverse is where I went wrong, and not quite sure what to do, I think I have to find something such that $(a + b\sqrt{5})\times? = 1 = 1+0\sqrt{5}$ I thought of just putting $$\frac{1+0\sqrt{5}}{a+b\sqrt(5)}$$ but I was told this wasn't right so not sure what else to do.
If anyone can help me with checking the multiplicative inverse condition that would be really appreciated.
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We just need to simplify that expression so that it has the form:
$$
\alpha + \beta\sqrt{5}
$$
for some $\alpha,\beta \in \mathbb Q$. Indeed, we can do this via conjugates:
\begin{align*}
\frac{1}{a + b\sqrt 5}
&= \frac{1}{a + b\sqrt 5} \cdot \frac{a - b\sqrt 5}{a - b\sqrt 5} \\
&= \frac{a - b\sqrt 5}{a^2 - 5b^2} \\
&= \underbrace{\frac{a}{a^2 - 5b^2}}_{\in ~ \mathbb Q} + \underbrace{\frac{-b}{a^2 - 5b^2}}_{\in ~ \mathbb Q}\sqrt 5
\end{align*}
To finish this off, can you see why the denominator $a^2 - 5b^2$ is never zero if $a$ and
$b$ are not both zero?
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|
Find $D$ in RSA cryptosystem For the following encryption key $(n, E)$ in the RSA cryptosystem, compute $D$.
$(n, E)= (451, 231)$
So I know $n=11*41$, so $m=400$. Now $D=$ inverse of $231 \ (mod \ 400)$. However I am not sure how to find the inverse of $231$ in $mod\ 400$. Help with this would be great.
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Expanding on the comment, the Euclidean algorithm gives:
\begin{align*}
400 &= 231 \times 1 + 169.\\
231 &= 169 \times 1 + 62.\\
169 &= 62 \times 2 + 45.\\
62 &= 45 \times 1 + 17.\\
45 &= 17 \times 2 + 11.\\
17 &= 11 \times 1 + 6.\\
11 &= 6 \times 1 + 5.\\
6 &= 5 \times 1 + 1.
\end{align*}
Then, we do this in reverse...
\begin{align*}
1 &= 6 - 5\\
&= 6 - (11 - 6)\\
&= 2 \times 6 - 11\\
&= 2 \times (17 - 11) - 11\\
&= ...\\
&= 71 \times 231 - 41 \times 400.
\end{align*}
(Check the details.) Viewing the last equation modulo $400$ gives
$$
231^{-1} \equiv 71 \mod 400.
$$
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How to find $α^2(β^4 +γ^4 +δ^4)+β^2(γ^4 +δ^4 +α^4)+γ^2(δ^4 +α^4 +β^4)+δ^2(α^4 +β^4 +γ^4)$
How to do the part (iv) . Please help.
Here are my answers to the first parts:
(i)
α a root of given equation $\implies \alpha^4-5 \alpha^2 + 2 \alpha -1 = 0$
$\implies \alpha^{n+4} - 5 \alpha^{n+2} + 2\alpha^{n+1} -\alpha^n=0$
Summing over $α, \beta , \gamma , \sigma$, leads to
$S_{n+4}– 5S_{n + 2} + 2S_{n +1}- S_n=0$
(ii)
$S_2=10$
$S_4 = 5S_2 – 2S_1 + 4 = 50 – 0 + 4 = 54$
(iii)
$S_{-1} = 2$ from $y^4 – 2y^3+5y^2-1=0$
$S_3 = 5S_1 – 2S_0 + S_{–1} = –6$
$S_6 = 5S_4 – 2S_3 + S_2 = 292$
I have no clue how to (iv)
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Note that $\beta^4 + \gamma^4 + \delta^4 = S_4 - \alpha^4$, and similarly for the other quantities in parentheses. Substitute this in your desired expression, and expand it. You can then write the entire expression in terms of various $S$'s.
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Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I used the result to find its closed-form. The possible candidate closed-form from Wolfram Alpha is
$$\pi\sqrt{\frac{1+\sqrt{2}}{2}}-\pi$$
Is this true? If so, how to prove it?
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Here I present a completely elementary proof using only high school level techniques.
With the obvious substitution $\displaystyle \large t = \tan{x}$
We proceed as follows:
Using the Euler Substitution of the third kind, the following substitution is obtained:
$\displaystyle \large \sqrt{t-t^2} = tz \Rightarrow t = \frac{1}{1+z^2}$
$\displaystyle \large \text{d}t = \frac{-2z \, \text{d}z}{(1+z^2)^2}$
The borders change from $\displaystyle \large (0 \to 1)_t$ to $\displaystyle \large (\infty \to 0)_z$, and note the negative from the differential element flips this back into more "sensible" borders
The computation of the transform is tedious algebra and I will omit said details.
$\displaystyle \large \int_0^{\frac{\pi}{4}} \sqrt{\tan{x}}\sqrt{1-\tan{x}} \text{d}x = 2 \int_0^\infty \frac{z^2 \, \text{d}z}{(z^4+2z^2+2)(z^2+1)}$
It is easy to show that:
$\displaystyle \large \frac{z^2}{(z^4+2z^2+2)(1+z^2)} \equiv \frac{z^2+2}{z^4+2z^2+2} - \frac{1}{1+z^2}$
Simplifying:
$\displaystyle \large \int_0^{\frac{\pi}{4}} \sqrt{\tan{x}}\sqrt{1-\tan{x}} \text{d}x = 2 \int_0^\infty \frac{(z^2 + 2) \, \text{d}z}{z^4+2z^2+2} - \pi$
With the simultaneous substitutions $\displaystyle \large z = \frac{\sqrt[4]{2}}{u}$ and $\displaystyle \large z = \sqrt[4]{2}u$, and averaging, the integral is reduced to:
$\displaystyle \large \frac{1+\sqrt{2}}{\sqrt[4]{2}} \int_0^\infty \frac{(u^2+1) \, \text{d}u}{u^4+\sqrt{2}u^2+1}$
The substitution $\displaystyle \large v = u - \frac{1}{u}$ bijects the interval $\displaystyle \large (0,\infty)_u \mapsto (-\infty, \infty)_v$ and the integral is reduced to:
$\displaystyle \large \frac{1+\sqrt{2}}{\sqrt[4]{2}} \int_{-\infty}^\infty \frac{\text{d}v}{v^2+2+\sqrt{2}} - \pi = \pi \left(\frac{1+\sqrt{2}}{\sqrt{2(1+\sqrt{2})}} - 1 \right) = \pi \left(\sqrt{\frac{1+\sqrt{2}}{2}} - 1\right)$
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Algorithm - iteration method
$$\begin{align} &T(n)=n+T(n-1)= \\
&= n + (n-1)+T(n-2)= \\
&= n + (n-1)+(n-2)+T(n-3)= \\
&= n + (n-1)+(n-2)+...+2+T(1)= \\
&= \frac{n(n+1)}{2}-1+T(1) \\
& \mathrm{Hence,} \ T(n)=\frac{n^2+n}{2}-1+T(1)=\Theta(n^2)
\end{align} $$
What is $n(n+1)$?
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$T(n) = n + (n-1) + (n-2) + ... + 2 + T(1)$
$\quad\quad\; = n + (n-1) + (n-2) + ... + 2 + 1 - 1 + T(1)$
$\quad\quad\; = (\sum_{i=1}^{n} i ) - 1 + T(1)$
$\quad\quad\; = \frac{n(n+1)}{2} - 1 + T(1)$
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If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?
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It's the coefficient of $x^{13}$ in the product $(x+x^2 + x^3 + x^4+ x^5 + x^6)^3$.
To see this, note that to compute that coefficient we have to identify all ways we can form $x^{13}$ by picking one term from each of the three terms $(x + x^2 + x^3 + x^4 + x^5 + x^6)$ we have; we could have $x$ from the first, $x^6$ from the second and the third and this would correspond to throwing $(1,6,6)$ with the three different dice (which we imagine to have different colours to distinguish them). This choice gives us one way to get $x^{13}$ in the final gathering of terms, and all other choices (so pairs $(a,b,c)$ with $a + b + c = 13, 1 \le a,b,c \le 6$) give us one extra power of $x^{13}$. So the final coefficient just counts all those triples.
E.g. try this with two dice: $$(x+x^2 + x^3 + x^4+ x^5 + x^6)^2 = x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + \\ 3x^{10} + 2x^{11} + x^{12}$$ and we see that the coefficient of $x^n$ is just the number of ways we can throw $n$ with two dice.
Write this as $(x(1+x+x^2+x^3+x^4+x^4))^3 = x^3(1+ x + x^2 + x^3 + x^4 + x^5)^3$, so we are looking for the coefficient of $x^{10}$ in $(1+ x + x^2 + x^3 + x^4 + x^5)^3$.
The fancy way to do this is to write $(1+x+x^2+\ldots+x^5) = \frac{1-x^6}{1-x}$ (standard geometric series) and so $(1+ x + x^2 + x^3 + x^4 + x^5)^3 = (1-x^6)^3 (1-x)^{-3}$.
The first term on the right can be evaluated using the binomial formula as $1 - 3x^6 + 3x^{12} - x^{18}$.
The second term on the right can be evaluated by the generalised binomial formula as $\sum_{k=0}^{\infty} {k+2 \choose k} x^k$.
So to get $x^{10}$ from the product of these, we get the $1$ from the first times the ${12 \choose 10}x^{10}$ from the second and the $-3x^6$ from the first times ${6 \choose 4}x^4$ of the second. Other terms have too high powers of $x$.
So the answer is ${12 \choose 10} - 3{6 \choose 4} = 21$.
You can use Wolfram alpha to expand the original polynomial $(1+x+\ldots+x^5)^3$ and we get $$x^{15}+3 x^{14}+6 x^{13}+10 x^{12}+15 x^{11}+21 x^{10}+\\25 x^9+27 x^8+27 x^7+25 x^6+21 x^5+15 x^4+10 x^3+6 x^2+3 x+1$$
The alternative is simple enumeration. But I like the complicated ways, as they generalise to more dice and higher sums. E.g. in the final expansion we see that there are $25$ ways to throw $9+3 = 12$ with three dice, etc. We get all the probabilities for all the sums at the same time.
|
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How to find $\omega^7$ and $\omega^6$ from $\omega^5+1=0$
I did the first parts :
$$\omega= (\cos \pi + i \sin \pi)^\frac{1}{5} \implies \omega^5 = \cos \pi + i \sin \pi=-1$$
$\omega=-1$ is a root so :
$$\omega^5-1= (\omega+1) (\omega^4-\omega^3+\omega^2-\omega+1)=0$$
$$\omega^4-\omega^3+\omega^2-\omega=-1$$
$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (\cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5})$$
$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (-\cos \frac{\pi}{5} + i \sin \frac{\pi}{5})$$
$$\omega-\omega^4= 2 \cos \frac{\pi}{5}$$
Did similar method for $\omega^3-\omega^2$ and got the desired result.
For this part I only know the first part which is:
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}$$
$$2\cos {\pi}{5} + 2\cos \frac{3\pi}{5}=\omega-\omega^4+\omega^3-\omega^2=1$$
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}=\frac{1}{2}$$
$$\cos \frac{\pi}{5} \cos \frac{3\pi}{5} = \frac{1}{4} (\omega-\omega^4)(\omega^3-\omega^2)$$
$$\frac{1}{4} (\omega^4-\omega^3-\omega^7+\omega^6)$$
How do I find $\omega^7$ and $\omega^6$ ???? please help.
Also I would love if I receive support for the last part too, I don't know how to make a quadratic equation
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For the quadratic equation part, note that if $(x-a)(x-b)=0$ the roots of the equation are $a$ and $b$.
Expand the brackets and you get the equation $x^2-(a+b)x+ab=0$ with roots $a$ and $b$.
This means that if you know the sum of the roots $a+b$ and the product of the roots $ab$, you can write down a quadratic equation which has the roots $a$ and $b$. This is what your question is leading you through.
|
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Partial Fractions Decomposition I am failing to understand partial fraction decomposition in cases like the following:
Provide the partial fraction decomposition of the following:
$$\frac{x+2}{(x-4)^3(x^2 + 4x + 16)}$$
I see this and I think of
$$\frac{A}{x-4} + \frac{Bx+C}{(x-4)^2} + \frac{Dx^2 + Ex + F}{(x-4)^3} + \frac{Gx+H}{x^2 + 4x + 16}$$
But I am told that the correct answer is
$$\frac{A}{x-4} + \frac{B}{(x-4)^2} + \frac{C}{(x-4)^3} + \frac{Dx+E}{x^2 + 4x + 16}$$
What exactly is the numerator of each fraction based on?
|
You have the correct understanding that the numerator is one degree lower than the denominator. For example:
$$\frac{A}{x+1}$$
$$\frac{Ax+B}{x^2+3x+2}$$
$$\frac{Ax^2+Bx+C}{x^3+2x^2+5x+1}$$
However, repeated terms are an exception. The numerator is one degree less than the repeated polynomial in the denominator. You have as many fractions as the power on the repeated term. With each fraction's denominator with a different power such that one fraction's denominator is raised to the first power, and so on raising each denominator to a higher power, until the power on the original denominator is reached. Let's look at an example:
$$\frac{x}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}$$
$$\frac{x}{(x+1)^3}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}$$
$$\frac{x}{(x^2+x+1)^2}=\frac{Ax+B}{(x^2+x+1)}+\frac{Cx+D}{(x^2+x+1)^2}$$
|
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|
Test for convergence for $\ln \frac{n^2}{n^2-1}$ I've tried to figure out if this converges using the comparison test, and the ratio test, but with no luck: $\sum^\infty_{n=2} \ln(n^2/(n^2-1))$.
I'd appreciate any help
|
Here is a way
to show convergence
without finding the sum.
If $x > 0$,
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
$.
Therefore
(the $1+x$ in the denominator
of the next integral
is intentional)
$\int_1^{1+x} \frac{dt}{1+x}
< \ln(1+x)
< \int_1^{1+x} \frac{dt}{1}
$
or
$\frac{x}{1+x}
< \ln(1+x)
< x
$.
Since
$\frac{n^2}{n^2-1}
=\frac{n^2-1+1}{n^2-1}
=1+\frac{1}{n^2-1}
$,
$\ln \frac{n^2}{n^2-1}
<\frac{1}{n^2-1}
$.
Since
$\sum^\infty_{n=2} \frac{1}{n^2-1}
$ converges,
so does
$\sum^\infty_{n=2} \ln(n^2/(n^2-1))$.
For a lower bound,
$\ln \frac{n^2}{n^2-1}
>\frac{\frac{1}{n^2-1}}{1+\frac{1}{n^2-1}}
=\frac{1}{n^2}
$.
Note that,
in general,
if $m > 1$,
$\frac{1}{m}
<\ln \frac{m}{m-1}
<\frac{1}{m-1}
$.
|
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|
Is the Pattern in the Number of Digits in the Bernoulli Numbers Showing Something Significant For the first couple of powers of $10$, the number of digits in these show a certain pattern, is this a coincidence or is their a reasonable explanation. Specifically if we look at
$$ \lfloor \log_{10}\left|B\left(10^n\right)\right| \rfloor $$
Where $ B(k) $ is the kth Bernoulli number and n is a positive integer.
For values of n between $2$ and $8$, the number of digits are as follows:
$$ 078,\ 1769,\ 27677,\ 376755,\ 4767529,\ 57674260,\ 676752569 $$
For each of the number of digits, the pattern seems to be start with a leading digit of $ n-2$ and then carry on a pattern of adding a digit which the next element will repeat, and then add a new one. This leads to a number that seems to begin $76752(5/6)6...$ For $n=7$ this seems to break up a bit and give a digit of $6$ when $n=8$ implies that it should be $5$. MY question is three parted; does this pattern continue, if so, what number is it creating as $n \to\infty$, and is there any reason for why this pattern is what it is?
|
The reason is the asymptotic equality(1)
$$\lvert B(2k)\rvert \sim \frac{2\cdot(2k)!}{(2\pi)^{2k}}.$$
Thus we obtain
$$\log_{10} \lvert B(2k)\rvert = \log_{10} \left((2k)!\right) + \log_{10} 2 - 2k\log_{10}(2\pi) + o(1),$$
and using Stirling's approximation
$$\log (n!) = \left(n+\tfrac{1}{2}\right)\log n - n + \tfrac{1}{2}\log (2\pi) + O(n^{-1})$$
that becomes
$$\begin{aligned}
\log_{10}\lvert B(2k)\rvert &= \left(2k+\tfrac{1}{2}\right)\log_{10} (2k) - \frac{2k}{\log 10} + \tfrac{1}{2} \log_{10} (2\pi) + \log_{10} 2 - 2k\log_{10}(2\pi) + o(1)\\
&= \left(2k+\tfrac{1}{2}\right)\log_{10} (2k) - 2k\left(\log_{10} (2\pi) + \frac{1}{\log 10}\right) + \tfrac{1}{2} \log_{10}(8\pi) + o(1).
\end{aligned}$$
Now inserting $2k = 10^n$, that becomes
\begin{align}
\log_{10} \lvert B(10^n)\rvert &= \left(10^n + \tfrac{1}{2}\right)n - 10^n\left(\log_{10} (2\pi) + \frac{1}{\log 10}\right) + \tfrac{1}{2} \log (8\pi) + o(1)\\
&= 10^n\left(n - \log_{10}(2\pi) - \frac{1}{\log 10}\right) + \frac{n}{2} + \frac{1}{2} \log (8\pi) + o(1).
\end{align}
Since
$$\log_{10} (2\pi) + \frac{1}{\log 10} \approx 1.2324743502613666,$$
we expect indeed the first digits to form $n-2$, then followed by $76752564973...$ until the $\frac{n}{2}$ intervenes, which happens later and later for growing $n$.
(1) Which can be derived from the partial fraction decomposition of the cotangent, for example.
|
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|
Find all values of $\alpha$ so that all solutions approach $0$ as $x \to \infty$ Given the equation
$x^2y′′+\alpha xy′+4y=0$
find all values of α so that all solutions approach zero as $x \to \infty$.
Anyone have advice for this question?
So I get $y = c_1 x^{\frac{1}{2}\sqrt{a^2 -2a -15}-a+1} + c_1 x^{\frac{1}{2}\sqrt{a^2 -2a -15}-a+1}$
I tried solving $\sqrt{a^2 -2a -15}-a+1 < 0$ and got $[5, \infty)$. However, this itnerval is apparently not the answer. What did I do wrong?
|
As Dr. Sonnhard Graubner said:
In
$x^2y′′+\alpha xy′+4y=0$,
set
$y = x^r
$.
Then,
writing $a$ for $\alpha$
because I am lazy,
$y'
=rx^{r-1}
$
and
$y''
=r(r-1)x^{r-2}
$
so
$0
=r(r-1)x^{r}+arx^{r}+4x^r
= x^r(r(r-1)+ar+4)
= x^r(r^2+(a-1)r+4)
$
or
$r^2+(a-1)r+4=0$.
Therefore,
$r
=\dfrac{-(a-1)\pm\sqrt{(a-1)^2-16}}{2}
=\dfrac{1-a\pm\sqrt{a^2-2a-15}}{2}
$.
Call the two roots
$r_1
=\dfrac{1-a+\sqrt{a^2-2a-15}}{2}
$
and
$r_2
=\dfrac{1-a-\sqrt{a^2-2a-15}}{2}
$.
The solutions are therefore
$y
=c_1x^{r_1}
+c_2x^{r_2}
$.
For all solutions
to approach
$\infty$,
both roots
must be positive.
For all solutions
to approach
zero,
both roots
must be negative.
From these,
you can work out
conditions on $a$.
Note that
the roots must be real;
if not,
the solutions will oscillate.
This imposes another condition on $a$.
Also note that
if
$(a-1)^2 = 16$,
the roots are equal,
so you get another type of solution,
which has to be considered.
|
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|
If $|z +\frac{1}{z}|=a$ find extreme values of $|z|$ Let $a> 0$. Knowing that $z$ is complex number with $|z +\frac{1}{z}|=a$ find extreme values of $|z|$.
My partial solution:
$$|z +\frac{1}{z}|=a \iff (z +\frac{1}{z})(\overline{z} +\frac{1}{\overline{z}})=a^2\iff |z|^2+\frac{1}{|z|^2} + (\frac{z}{\overline{z}}+\frac{\overline{z}}{z})=a^2 \iff (\frac{z}{\overline{z}}+\frac{\overline{z}}{z})=a^2 - |z|^2+\frac{1}{|z|^2}$$
It can be shown easily that $(\frac{z}{\overline{z}}+\frac{\overline{z}}{z}) \in [-2, 2].$
Using inequalities $-2\leq a^2 - |z|^2+\frac{1}{|z|^2}\leq 2$ should obtain the required result but we came to a very complicated.
Does anyone have a simple idea?
I found a solution using trigonometric form of complex numbers. This solution is further exposed, may interest someone.
$$|z +\frac{1}{z}|=a \iff |r(\cos t+i\sin t)+\frac{1}{r(\cos t+i\sin t)}|=a$$
$$\iff |r(\cos t+i\sin t)+r(\cos t-i\sin t)|=a
\iff |(r+\frac{1}{r})\cos t+i(r-\frac{1}{r})\sin t|=a$$
$$\iff (r+\frac{1}{r})^2\cos^2 t+(r-\frac{1}{r})^2\sin^2 t = a^2\iff r^2+\frac{1}{r^2}+2\cos {2t}=a^2$$
$$\iff r^4-(a^2-2\cos {2t})r^2+1=0$$
Denote $r^2=u$. The roots of the equation $u^2-(a^2-2\cos {2t})u+1=0$ are $u_{1,2}=\frac{a^2-2\cos {2t}+-\sqrt{a^4-4a^2\cos {2t}-4\sin^2{2t} }}{2}= \cdots = \frac{a^2+4\sin^2 t-2+-\sqrt{(a^2-4\sin^2 t-4)(a^2-4\sin^2 t)}}{2}$
Using the notation $a^2+4\sin^2 t-2=x$ we have $u_1= \frac{x+\sqrt{x^2-4}}{2} $.
It is clear that this is maximum for $x$ max ie for $ \sin^2 t =1$ and consequently
$r_{max}=\sqrt{\frac{a^2+2+\sqrt{a^4+4a^2}}{2}}= \frac{a+\sqrt{a^2+4}}{2}$ and
$r_{min}=\frac{1}{r_{max}}=\frac{\sqrt{a^2+4}-a}{2}$.
Extreme values are obtained for $z_{max}=+-\frac{a+\sqrt{a^2+4}}{2}i$ and $z_{min}=+-\frac{\sqrt{a^2+4}-a}{2}i.$
(For simple writing of $r_{max}$ we used the formula
$\sqrt{A+\sqrt{B}}= \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A+C}{2}}$, where $C=\sqrt{A^2-B}$ ).
|
You can simply use the fact that:
$|z|+\frac{1}{|z|} \geq |z+\frac1z| \geq ||z|-\frac{1}{|z|}|$
Substituting $|z+\frac1z|=a$, we get two inequalities:
$|z|+\frac{1}{|z|} \geq a$ and $a\geq ||z|-\frac{1}{|z|}|$
Taking $|z| = t$$\geq1$, we get:
$t^2 - at + 1\geq0$ and $t^2 - at - 1\leq0$
For 1st inequality let $t_1$ and $t_2$ be the roots such that $t_2 > t_1$ and similarly for 2nd inequality let $t_3$ and $t_4$ be the roots such that $t_2 > t_1$, which can be figured out using the quadratic formula (trust me, they are ugly to type).
Next, the inequalities simplify to :
$(t-t_1)(t-t_2) \geq0$ and $(t-t_3)(t-t_4) \leq0$
Getting first set as $t\in$R$-(t_1,t_2)$ and second as $t\in [t_3,t_4]$
Plotting these two on the number line along with $t>1$,
We get one extremity (maximum) as $t_4$; either repeating for $t<1$ or replacing $t$ by $\frac1t$, we get the second extremity (minimum) as $\frac1t_4 $ (whose denominator can be rationalized).
|
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|
Evaluation of $\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}}\;\;,$ Where $i\neq j \neq k\neq l$
Evaluation of following Infinite Geometric series.
$(a)\;\; \displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^{i+j}}\;\;,$ Where $i\neq j\;\;\;\;\;\;\;\;\;\; (b)\;\; \displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{3^{i+j+k}}\;\;,$ Where $i\neq j \neq k$
$\displaystyle (c)\;\; \displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}}\;\;,$ Where $i\neq j \neq k\neq l.$
$\bf{My\; Try\; for }$ First one $(a)::$ Given $\displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^{i+j}} = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^i}\cdot \frac{1}{3^j}\;\;,$ Where $i\neq j$
So we can write $\displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^i}\cdot \frac{1}{3^j}=\sum_{i=0}^{\infty}\frac{1}{3^i}\cdot \sum_{j=0}^{\infty}\frac{1}{3^j}-\sum_{i=0}^{\infty}\frac{1}{3^{2i}} = \frac{1}{1-\frac{1}{3}}\times \frac{1}{1-\frac{1}{3}}-\frac{1}{1-\frac{1}{3^2}}=\frac{9}{8}$
Actually i have used the fact $\displaystyle \sum(i\neq j) = \sum(\bf{no\; condition})-\sum(i=j)$.
But I did not understand how can i used the logic in $(b)$ and $(c)$ part ,
plz explain it to me in detail
Thanks
|
We have:
$$(a)\;\; S_2 = \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{1}{3^{i+j}} (i \ne j) = 2\sum_{i< j}\frac{1}{3^{i+j}} = 2\;I_2$$
$$(b)\;\; S_3 = \sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty\frac{1}{3^{i+j+k}} (i \ne j \ne k) = 6\sum_{i<j<k}\frac{1}{3^{i+j+k}} = 6\;I_3$$
$$(c)\;\; S_4 = \sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty\sum_{l=0}^\infty\frac{1}{3^{i+j+k+l}} (i \ne j \ne k \ne l) = 24\sum_{i<j<k<l}\frac{1}{3^{i+j+k+l}} = 24\;I_4$$
And in general:
$$(d)\;\; S_n = \sum_{i_1=0}^\infty...\sum_{i_n=0}^\infty\frac{1}{3^{i_1+\cdots+i_n}} (i_1 \ne ... \ne i_n) = n!\sum_{i_1<...<i_n}\frac{1}{3^{i_1+\cdots+i_n}} = n!\;I_n$$
where
$$I_n = \sum_{i_1<...<i_n}\frac{1}{3^{i_1+\cdots+i_n}}$$
$I_1 = \sum_{i=0}^\infty 3^{-i} = \frac32$, and for $n \ge 1$:
$$\begin{align}
I_{n+1} & = \sum_{i_1<...<i_{n+1}}\frac{1}{3^{i_1+\cdots+i_{n+1}}} \\
& = \sum_{i_1=0}^\infty \frac{1}{3^{i_1}} \sum_{i_1<i_2<...<i_{n+1}}\frac{1}{3^{i_2+\cdots+i_{n+1}}} \\
& = \sum_{i_1=0}^\infty \frac{1}{3^{i_1}} \sum_{0\le i_2<...<i_{n+1}}\frac{1}{3^{(i_1+i_2+1)+\cdots+(i_1+i_{n+1}+1)}} \\
& = \sum_{i_1=0}^\infty \frac{1}{3^{i_1}} \sum_{0 \le i_2<...<i_{n+1}}\frac{1}{3^{n(i_1+1)}}\frac{1}{3^{i_2+\cdots+i_{n+1}}}\\
& = \frac{1}{3^n}\sum_{i_1=0}^\infty \frac{1}{3^{(n+1)i_1}} \sum_{0 \le i_2<...<i_{n+1}}\frac{1}{3^{i_2+\cdots+i_{n+1}}}\\
& = \frac{1}{3^n}\frac{3^{n+1}}{3^{n+1}-1} I_n \\
& = \frac{3}{3^{n+1}-1}I_n
\end{align}$$
Thus $S_{n+1} = (n+1)!\;I_{n+1} = \dfrac{3}{3^{n+1}-1}(n+1)!\;I_n = \dfrac{3(n+1)}{3^{n+1}-1}S_n$
Starting with $S_1 = I_1 = \frac32$, we get:
$$S_2 = \frac{6}{8}S_1 = \frac{9}{8}$$
$$S_3 = \frac{9}{26}S_2 = \frac{81}{208}$$
$$S_4 = \frac{12}{80}S_3 = \frac{243}{4160}$$
|
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|
Questions about the computation of a limit. By L'Hôpital's rule, it is easy to see that
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} = 1/3.
$$
But if we use the following procedure to compute the limit, we get a wrong answer. The procedure is in the following.
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} =
\lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^2 \sin x}. \quad (2)
$$
Since $\sin x \sim x$ ($\sin x$ and $x$ are equivalent infinitesimals), we replace $\sin x$ by $x$ in (2). Then we obtain
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} =
\lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^2 \sin x} \\
= \lim_{x \to 0} \frac{\frac{x}{\cos x} - x}{x^3} \\
= \lim_{x \to 0} \frac{\frac{1}{\cos x} - 1}{x^2} \\
= \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x} \\
= 1/2.
$$
I don't konw where is the problem. Thank you very much.
|
Another solution could be to double check the result using Taylor series $$f(x)=\frac{\tan x - x}{x^2 \sin x}$$ and use sucessively $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ So, the numerator is $$\tan(x)-x=\frac{x^3}{3}+O\left(x^4\right)$$ and denominetaor is $$x^2\sin(x)=x^3-\frac{x^5}{6}+O\left(x^6\right)$$ and then the limit of $\frac{1}{3}$.
Personally, what I enjoy with Taylor series is that they show how the limit is reached. In this case, close to $x=0$, we obtain $$f(x)=\frac{1}{3}+\frac{17 x^2}{90}+O\left(x^4\right)$$
|
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|
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
|
By the same way as Mr. Mike$,$ it's enough to prove $$2\left(\sum_{cyc}(x^2+3yz)\right)^3\geqslant 9\sum xy(x+y)(2z+x+y)^3$$
Or
$$\frac18\sum \left( 16\,{x}^{4}+100\,{z}^{4}+104\,{x}^{3}y+243\,{y}^{2}{z}^{2
}+330\,{z}^{3}x+416\,{y}^{2}zx+342\,{z}^{2}xy \right) \left( x-y
\right) ^{2}+$$
$$+\frac18\sum x{y}^{2} \left( 18\,y+41\,x \right) \left( z+x-2\,y \right) ^{2}\geqslant 0$$
|
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|
If $f$ is continuous and $\,f\big(\frac{1}2(x+y)\big) \le \frac{1}{2}\big(\,f(x)+f(y)\big)$, then $f$ is convex Let $\,\,f :\mathbb R \to \mathbb R$ be a continuous function such that $$
f\Big(\dfrac{x+y}2\Big) \le \dfrac{1}{2}\big(\,f(x)+f(y)\big) ,\,\, \text{for all}\,\, x,y \in \mathbb R,
$$
then
how do we prove that $f$ is convex that is
$$
f\big(tx+(1-t)y\big)\le tf(x)+(1-t)f(y) , \forall x,y \in \mathbb R , t\in (0,1)?
$$
I can prove it for dyadic rational's $t=\dfrac k {2^n}$ and then argue by continuity ; but I would like a more direct proof . Thanks in Advance
|
I would like to properly show by induction that if $ m\in\{0,1,2,\ldots,2^{n}\} $ then
$$
f\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)\le \frac{m}{2^n}f(x)
+\Big(1-\frac{m}{2^n}\Big)f(y).
$$
and then the result will follows by contuinity since $$ m=\lfloor2^nt\rfloor \in\{0,1,2,\ldots,2^{n}\} ~~~and~~\frac{\lfloor2^nt\rfloor}{2^n}\to t$$
The initial state $n = 1$ is trivial by hypothesis.
Now we assume that for every $k<n$, whenever $m\in\{0,1,2,\ldots,2^{k-1}\}$,we have
$$
f\left(\frac{m}{2^k}x+\Big(1-\frac{m}{2^k}\Big)y\right)\le \frac{m}{2^k}f(x)
+\Big(1-\frac{m}{2^k}\Big)f(y). \tag{I}\label{I}
$$
we want to prove \eqref{I} for $k=n$.
Let $m \in\{0,1,2,\ldots,2^{n}\}$ then the division by 2 yields $m =2p +r$ with $r\in \{0,1\}$
$$X:=\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)= \left(\frac{2p +r}{2^n}x+\Big(1-\frac{2p +r}{2^n}\Big)y\right) \\= \left(\frac{p }{2^n}x+\frac{p +r}{2^n}x+\Big(1-\frac{p }{2^n}y-\frac{p +r}{2^n}y\Big)y\right)
\\= \frac12 \left(\frac{p }{2^{n-1}}x+\frac{p +r}{2^{n-1}}x+\Big(2-\frac{p }{2^{n-1}}y-\frac{p +r}{2^{n-1}}y\Big)y\right)
\\=\frac12\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right) \\:=
\color{red}{\frac{1}{2}\left(X_1+X_2\right)}$$
On the other hand, since $r\in \{0,1\}$ and $m\in\{0,1,2,\ldots,2^{n}\}$ it is to check using parity that $p,p+1 \in \{0,1,2,\ldots,2^{n-1}\}$
that is $p,p+r \in \{0,1,2,\ldots,2^{n-1}\}$
By hypothesis of induction we obtain
\begin{align}f(X) &= f\left(\frac{1}{2}\left(X_1+X_2\right)\right)\le \frac12f(X_1) +\frac12f(X_2) \\&=
\frac12f\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12f\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right)
\\&\le\frac12\left(\frac{p }{2^{n-1}}f(x)+ \Big(1-\frac{p }{2^{n-1}}f(y)\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}f(x)+ \Big(1-\frac{p +r}{2^{n-1}}f(y)\Big)\right) \\&= \frac{2p+r}{2^n}f(x)
+\Big(1-\frac{2p+r}{2^n}\Big)f(y)\\&= \frac{m}{2^n}f(x)
+\Big(1-\frac{m}{2^n}\Big)f(y).\end{align}
|
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|
Prove that for any positive integer $n$, $2\sqrt{n+1}-2\le 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}}\le 2\sqrt{n}-1$ Prove that for any positive integer n,$$ 2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1 $$
Could anyone give me a hint on this question? Does it have something to do with Riemann Sums?
|
Hint. The double inequality
$$
2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + .. + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1
$$
can be proved inductively. For $n=1$ is clear. Assume that it holds for $n=k$. Then
$$
2\sqrt{k+1} - 2 +\frac{1}{\sqrt{k+1}}\le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + .. + \frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \le 2\sqrt{k} - 1 +\frac{1}{\sqrt{k+1}}.
$$
It remains to show the following two inequalities:
$$
2\sqrt{k+2} - 2\le 2\sqrt{k+1} - 2 +\frac{1}{\sqrt{k+1}},
$$
and
$$
2\sqrt{k} - 1 +\frac{1}{\sqrt{k+1}}\le 2\sqrt{k+1}-1.
$$
|
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|
Prove inequalities $2\sqrt{n+1}-2 \leq 1 + 1/\sqrt{2} + 1/\sqrt{3} + .... + 1/\sqrt{n}\leq 2\sqrt{n}-1$ Prove that for any positive integer $n$,
$$
2\sqrt{n+1}-2 \leq 1 + 1/\sqrt{2} + 1/\sqrt{3} + .... + 1/\sqrt{n}\leq 2\sqrt{n}-1
$$
the middle part can be expressed as a sum. of $1/\sqrt{x}$ from 1 to $n$, which leads to a $Hn^{0.5}$. Im not sure how that helps.
I tried differentiating both end functions and found they are both increasing.. But it doesn't seem to help much... help pls
|
Hint: $2\sqrt{k+1} - 2\sqrt{k}<\dfrac{1}{\sqrt{k}} = \dfrac{2}{2\sqrt{k}} < \dfrac{2}{\sqrt{k}+\sqrt{k-1}} = 2\sqrt{k} - 2\sqrt{k-1}$. Now let $k$ runs from $1$ to $n$, and add up.
|
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|
Describe how the number of solutions to an equation system depend on the variable $a$. Describe how the number of solutions to the system:
\begin{eqnarray*}
x+y+z &=&1 \\
2x+y+a^2z &=&a \\
x+3y+3z &=&3
\end{eqnarray*}
is dependant on the value of $a$.
My attempt at a solution:
First we write the the equation system in matrix form:
$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\2 & 1 & a^2 & a\\1 & 3 & 3 & 3\end{array}\right]$$
Now, after subtracting the two times first row from the second we get:
$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\0 & -1 & a^2-2 & a-2\\1 & 3 & 3 & 3\end{array}\right]$$
And subtract one time the first row from the third:
$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\0 & -1 & a^2-2 & a-2\\0 & 2 & 2 & 2\end{array}\right]$$
Then we subtract one half the third row from the first:
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & a^2-2 & a-2\\0 & 2 & 2 & 2\end{array}\right]$$
We continue with adding two times the second row to the third:
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & a^2-2 & a-2\\0 & 0 & 2+2(a^2-2) & 2+2(a-2)\end{array}\right]$$
We divide the third row by $2$ and simplify the expressions.
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & a^2-2 & a-2\\0 & 0 & a^2-1 & a-1\end{array}\right]$$
We subtract one time the third row from the second row:
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & -1 & -1\\0 & 0 & a^2-1 & a-1\end{array}\right]$$
This result gives me the following conclusion:
If $a = 1$ the system will have infinitely many solutions.
If $a = -1$ the system will not have any solutions,
For all other values of $a$ the system will have a unique solution.
Am I thinking correctly?
|
If $A$ denotes the $3\times 3$-matrix of your equation $Ax=b$, then $\det(A)=-2(a-1)(a+1)$. Hence if the field has not characteristic two, and $(a-1)(a+1)\neq 0$, then we have a unique solution. You have already solved this case correctly. Note that you have divided by $2$ here.
For $char(K)=2$ we always have $\det(A)=0$. The system then is given by
\begin{align*}
x+y+z& =1, \\
y+a^2z & = a. \\
\end{align*}
We can solve this by setting $x=1-y-z$ and $y=a(1-az)$. If $K$ has infinitely many elements, then we have infinitely many solutions, independent of the value of $a$.
|
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|
$10\sin(x)\cos(x) = 6\cos(x)$ In order to solve
$$10\sin x\cos x = 6\cos x$$
I can suppose: $\cos x\ne0$ and then:
$$10\sin x = 6\implies \sin x = \frac{6}{10}\implies x = \arcsin \frac{3}{5}$$
And then, for the case $\cos x = 0$, we have:
$$x = \frac{\pi}{2} + 2k\pi$$
Is my solution rigth? Because I'm getting another strange things at wolfram alpha
|
$$10\sin x\cos x=6\cos x\iff 10\sin x\cos x-6\cos x=0\iff 2(\cos x)(5\sin x -3)=0\stackrel{:2}{\iff} (\cos x)(5\sin x -3)=0\iff \cos x=0 \; \; \lor \;\; 5\sin x-3=0$$
Is it any more clear now? Also, that $\cos x=0$ in your solution has to be $x=\dfrac{\pi}{2}+k\pi\, ,k\in \Bbb Z$ and your $\sin x=0$ has to include solutions in the form $\dfrac{\pi}{2}-\sin^{-1}\left (\dfrac35\right )$
|
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|
Limit of $\left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- 1/ {n^4}} {4+ n^7/ {3^n}}$ as $n$ tends to infinity I already took some steps to simplifying the original question and im stuck at this point:
$$ \lim_{n \to \infty} \left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- \frac {1} {n^4}} {4+ \frac {n^7} {3^n}} $$
I know that $$\lim_{n \to \infty}\left(\frac {2}{3}\right)^n=0$$
since $$0<\frac{2}{3}<1$$
and that $$\lim_{n \to \infty}\frac {1} {n^4}=0$$
and that I have to use d'Alembert's criterion to solve what $$\frac {n^7} {3^n}$$ is
but what I get is $$\frac 13*\frac {(n+7)^7}{n^7}$$
I suppose the next step is just $$\frac 13 <1$$
"therefore the limit is zero" but my question is how do I get rid of the
$$\frac {(n+7)^7}{n^7}$$
The last step is just:
$$0*\frac {1-0}{4+0}=0$$
|
HINT:
We have $$\frac{2^n(n^4-1)}{4\cdot3^n+n^7}<\frac{2^n}{4\cdot3^n}<\frac14\left(\frac23\right)^n$$
|
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|
Bijection, and finding the inverse function I am new to discrete mathematics, and this was one of the question that the prof gave out. I am bit lost in this, since I never encountered discrete mathematics before. What do I need to do to prove that it is bijection, and find the inverse? Do I choose any number(integer) and put it in for the R and see if the corresponding question is bijection(both one-to-one and onto)?
Show that the function $f: \Bbb R \setminus \{-1\} \to \Bbb R \setminus \{2\}$ defined by
$$
f(x) = \frac{4x + 3}{2x + 2}
$$
is a bijection, and find the inverse function.
(Hint: Pay attention to the domain and codomain.)
|
A function is bijective if it is injective (one-to-one) and surjective (onto).
You can show $f$ is injective by showing that $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
You can show $f$ is surjective by showing that for each $y \in \mathbb{R} - \{2\}$, there exists $x \in \mathbb{R} - \{-1\}$ such that $f(x) = y$.
If $f(x_1) = f(x_2)$, then
\begin{align*}
\frac{4x_1 + 3}{2x_1 + 2} & = \frac{4x_2 + 3}{2x_2 + 3}\\
(4x_1 + 3)(2x_2 + 2) & = (2x_1 + 2)(4x_2 + 3)\\
8x_1x_2 + 8x_1 + 6x_2 + 6 & = 8x_1x_2 + 6x_1 + 8x_2 + 6\\
8x_1 + 6x_2 & = 6x_1 + 8x_2\\
2x_1 & = 2x_2\\
x_1 & = x_2
\end{align*}
Thus, $f$ is injective.
Let $y \in \mathbb{R} - \{2\}$. We must show that there exists $x \in \mathbb{R} - \{-1\}$ such that $y = f(x)$. Suppose
$$y = \frac{4x + 3}{2x + 2}$$
Solving for $x$ yields
\begin{align*}
(2x + 2)y & = 4x + 3\\
2xy + 2y & = 4x + 3\\
2xy - 4x & = 3 - 2y\\
(2y - 4)x & = 3 - 2y\\
x & = \frac{3 - 2y}{2y - 4}
\end{align*}
which is defined for each $y \in \mathbb{R} - \{2\}$. Moreover, $x \in \mathbb{R} - \{-1\}$. To see this, suppose that
$$-1 = \frac{3 - 2y}{2y - 4}$$
Then
\begin{align*}
-2y + 4 & = 3 - 2y\\
4 & = 3
\end{align*}
which is a contradiction.
The inverse function is found by interchanging the roles of $x$ and $y$. Hence, the inverse is
$$y = \frac{3 - 2x}{2x - 4}$$
To verify the function
$$g(x) = \frac{3 - 2x}{2x - 4}$$
is the inverse, you must demonstrate that
\begin{align*}
(g \circ f)(x) & = x && \text{for each $x \in \mathbb{R} - \{-1\}$}\\
(f \circ g)(x) & = x && \text{for each $x \in \mathbb{R} - \{2\}$}
\end{align*}
\begin{align*}
(g \circ f)(x) & = g\left(\frac{4x + 3}{2x + 2}\right)\\
& = \frac{3 - 2\left(\dfrac{4x + 3}{2x + 2}\right)}{2\left(\dfrac{4x + 3}{2x + 2}\right) - 4}\\
& = \frac{3(2x + 2) - 2(4x + 3)}{2(4x + 3) - 4(2x + 2)}\\
& = \frac{6x + 6 - 8x - 6}{8x + 6 - 8x - 8}\\
& = \frac{-2x}{-2}\\
& = x\\
(f \circ g)(x) & = f\left(\frac{3 - 2x}{2x - 4}\right)\\
& = \frac{4\left(\dfrac{3 - 2x}{2x - 4}\right) + 3}{2\left(\dfrac{3 - 2x}{2x - 4}\right) + 2}\\
& = \frac{4(3 - 2x) + 3(2x - 4)}{2(3 - 2x) + 2(2x - 4)}\\
& = \frac{12 - 8x + 6x - 12}{6 - 4x + 4x - 8}\\
& = \frac{-2x}{-2}\\
& = x
\end{align*}
Hence, $g = f^{-1}$, as claimed.
|
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|
Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$
$a,b,c$ are positive reals with $abc = 1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$
I try to use AM $\ge$ HM.
$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$
Then how I proceed.
|
Hint: Let $x=a^{-1},y=b^{-1},z=c^{-1}$. Rewrite $a^3=\dfrac{a^2}{bc}$, the inequality becomes
$$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac32,$$
where $xyz=1$. That should be easy by Cauchy-Schwarz.
|
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|
What is the limit for the radical $\sqrt{x^2+x+1}-x $? I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1.
But when i try to find the m-value for my oblique asymptote by taking the limit of:
$$
\lim_{x\to\infty}\sqrt{x^2+x+1}-x=m
$$
I'm stuck.
How do i simplify this expression to find the limit?
I've tried manipulating the radical by converting it to the denominator:
$$\lim_{x\to\infty}\frac{x^2+x+1}{\sqrt{x^2+x+1}}-x.$$
Or by multiplying both terms with the conjugate:
$$\lim_{x\to\infty}\frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}}$$
But in neither case do I know how to take the limit. Any help would be greatly appreciated.
|
I think there's a mistake in your last expression. We have
$$ \sqrt{x^2+x+1}-x = \frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x} = \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} $$
Therefore we have $$\lim(\sqrt{x^2+x+1}-x)= \lim\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} = \frac{1}{2} $$
|
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|
Limit of $\prod\limits_{k=2}^n\frac{k^3-1}{k^3+1}$ Calculate $$\lim_{n \to \infty} \frac{2^3-1}{2^3+1}\times \frac{3^3-1}{3^3+1}\times \cdots \times\frac{n^3-1}{n^3+1}$$
No idea how to even start.
|
An overkill. Since we have $$\prod_{k\geq2}\frac{k^{3}-1}{k^{3}+1}=\prod_{k\geq2}\frac{\left(k-1\right)\left(k^{2}+k+1\right)}{\left(k+1\right)\left(k^{2}-k+1\right)}=\prod_{k\geq2}\frac{\left(k-1\right)\left(k-\frac{-1+i\sqrt{3}}{2}\right)\left(k-\frac{-1-i\sqrt{3}}{2}\right)}{\left(k+1\right)\left(k-\frac{1+i\sqrt{3}}{2}\right)\left(k-\frac{1-i\sqrt{3}}{2}\right)}$$ $$=\prod_{k\geq0}\frac{\left(k+1\right)\left(k+2-\frac{-1+i\sqrt{3}}{2}\right)\left(k+2-\frac{-1-i\sqrt{3}}{2}\right)}{\left(k+3\right)\left(k+2-\frac{1+i\sqrt{3}}{2}\right)\left(k+2-\frac{1-i\sqrt{3}}{2}\right)}
$$ and recalling the identity $$\prod_{k\geq0}\frac{\left(k+a_{1}\right)\left(k+a_{2}\right)\left(k+a_{3}\right)}{\left(k+a_{4}\right)\left(k+a_{5}\right)\left(k+a_{6}\right)}=\frac{\Gamma\left(a_{4}\right)\Gamma\left(a_{5}\right)\Gamma\left(a_{6}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right)\Gamma\left(a_{3}\right)},\, a_{1}+a_{2}+a_{3}=a_{4}+a_{5}+a_{6}
$$ which follows from the Euler's definition of the Gamma function, we get $$P=\prod_{k\geq0}\frac{\left(k+1\right)\left(k+2-\frac{-1+i\sqrt{3}}{2}\right)\left(k+2-\frac{-1-i\sqrt{3}}{2}\right)}{\left(k+3\right)\left(k+2-\frac{1+i\sqrt{3}}{2}\right)\left(k+2-\frac{1-i\sqrt{3}}{2}\right)}=\frac{\Gamma\left(3\right)\Gamma\left(2-\frac{1+i\sqrt{3}}{2}\right)\Gamma\left(2-\frac{1-i\sqrt{3}}{2}\right)}{\Gamma\left(1\right)\Gamma\left(2-\frac{-1+i\sqrt{3}}{2}\right)\Gamma\left(2-\frac{-1-i\sqrt{3}}{2}\right)}
$$ and since $\Gamma\left(x+1\right)=x\Gamma\left(x\right)
$ we get $$P=\frac{2}{\left(2-\frac{1+i\sqrt{3}}{2}\right)\left(2-\frac{1-i\sqrt{3}}{2}\right)}=\color{red}{\frac{2}{3}}.$$
|
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|
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $f(x)>16$
$\bf{\bullet\; }$ If $6 <x<8\;,$ Then $f(x)<16$.
$\bf{\bullet \; }$ If $x=6\;,x=8\;,$ Then $f(x) = 16$
So Solutions of the above equation are $x=6$ and $x=8$
Can we solve it using Derivative or Algebraic way
Thanks
|
You can exploit the symmetry of the situation to simplify things if you set $y=7-x$ so that the equation becomes $(y-1)^4+(y+1)^4=16$ or $2y^4+12y^2+2=16$ or $y^4+6y^2-7=0$
This factorises as $(y^2-1)(y^2+7)=0$ with real solutions $y=\pm 1$ corresponding to $x=6, x=8$
Note this symmetric trick is bound to work, because it gives a symmetric function in $y$, which is inevitably a function of $y^2$ (no odd order terms) - and the order is such that it can't be worse than quadratic in $y^2$, which can therefore be solved.
|
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|
What can we say If lim U(n+1) - U(n) = 0? So i proved that $$\lim_{n\to \infty} U(n+1) - U(n) = 0$$ and we also have $a< U(n) < b$
What can we say?
Does this prove that the serie converge?
|
Consider the series
$$1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}- \frac{1}{6}-\frac{1}{6}-\cdots.$$
|
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|
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
factor out $\sin x$ in the numerator
$$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $
$$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.
I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?
|
$\begin{align}
\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} &= \lim_{x\to 0}\frac{1}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}\cdot\lim_{x\to 0}\frac{\sec x-1}{x^2}\\
&=\frac12 \cdot \lim_{x\to 0}\frac{\sec x-1}{x^2}
\end{align}$
This last limit should yield to L'Hopital's rule, or you could multiply by the conjugate of the numerator and apply a trig identity, plus the fact that $\frac{\tan x}{x}\to 0$ as $x\to 0$.
|
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"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Triangle in Triangle I have the lengths of three sides of an acute triangle ABC as shown below. Assume a point P on the side AB such that, if Q is the projection of P onto BC, R is the projection of Q onto CA, P becomes the projection of R onto AB. How can I Find the length PB.
|
Here is another approach, which works directly and systematically. Call the angles of the triangle $A, B, C$ and the sides opposite those angles $a, b, c$. We use the cosine rule $a^2=b^2+c^2-2bc \cos A$ so that $\cos A=\cfrac {a^2+b^2+c^2-2a^2}{2bc}$ and if I define $2d^2=a^2+b^2+c^2$ then $\cos A=\cfrac {d^2-a^2}{bc}, \cos B=\cfrac {d^2-b^2}{ac}, \cos C=\cfrac {d^2-c^2}{ab}$.
Let BP=p.
Then $$p=c-AP=c-AR\cos A= c-(b-CR)\cos A=c-b\cos A+CR\cos A=$$
$$=c-b\cos A+CQ\cos C\cos A=c-b\cos A+(a-BQ)\cos C\cos A=$$
$$=c-b\cos A+a\cos A\cos C-BQ\cos A\cos C=$$
$$=c-b\cos A+a\cos A\cos C-p\cos A\cos B\cos C$$
So that $$p\left(1+\cos A\cos B\cos C\right)=c-b\cos A+a\cos A\cos C=$$$$=c-\frac {d^2-a^2}c+\frac {(d^2-a^2)(d^2-c^2)}{b^2c}=\frac {b^2c^2-b^2d^2+b^2a^2+d^4-a^2d^2-c^2d^2+a^2c^2}{b^2c}$$
Now with $a^2+b^2+c^2=2d^2$ this is equal to $$\frac {a^2b^2+b^2c^2+a^2c^2-d^4}{b^2c}$$
We also have $$1+\cos A\cos B\cos C=\frac {a^2b^2c^2+d^6-(a^2+b^2+c^2)d^4+(a^2b^2+b^2c^2+a^2c^2)d^2-a^2b^2c^2}{a^2b^2c^2}=\frac {d^2(a^2b^2+b^2c^2+c^2a^2-d^4)}{a^2b^2c^2}$$
So that everything cancels nicely to give $$p=\frac {a^2c}{d^2}=\frac {2a^2c}{a^2+b^2+c^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1011699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Base for Vector Space I have problem with one assignment and I am pretty sure that it's not that hard. So, I have 4 vectors :
$$
v_{1}=\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix},
v_{2}=\begin{pmatrix}
0 \\ 3 \\ 1
\end{pmatrix},
v_{3}=\begin{pmatrix}
1 \\ -2 \\ 0
\end{pmatrix},
v_{4} =\begin{pmatrix}
-2 \\ 1 \\ -1
\end{pmatrix}
$$
and I have to find one base for the $span\{v_{1}, v_{2}, v_{3}, v_{4} \}$. Problem is that every combination of 3 vectors is linearly dependent (also all 4 are linearly dependent).
Have I misunderstood something or am I doing something wrong?
|
Here is one general method that involves matrix reduction (I assume you've covered that?)
Set up the matrix: $$\begin{bmatrix} 1 & 0 & 1 & -2 \\ 1 & 3 & -2 & 1 \\ 1 & 1 & 0 & -1 \end{bmatrix}$$
Then use column reduction:
$$\begin{bmatrix} 1 & 0 & 1 & -2 \\ 1 & 3 & -2 & 1 \\ 1 & 1 & 0 & -1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 3 & -3 & 3 \\ 1 & 1 & -1 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{bmatrix}$$
The nonzero columns will then span your space.
That is $\text{span}(v_1, v_2, v_3, v_4) = \text{span}\left(\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3\\ 1\end{bmatrix}\right)$. So one of the infinite number of bases of your space is $\left\{\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3\\ 1\end{bmatrix}\right\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1012119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding the sum of x of two power series. Could someone give me a hint on finding the sum of all $x$ for the following power series:
$$
\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{2n+1}}{2n+1}
$$
I am pretty sure we need to compare this with $$arctan (x) = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1} $$
I'm just not sure how. Would integrating help us?
The other series: $$
\sum_{n=2}^{\infty}\frac{2^n-n}{n+1}x^n
$$
|
This is not the full answer,this is work in progress anyway it might help author to get idea.
$$\sum_{n=2}^
\infty \frac{2^nx^n}{n+1}-\frac{nx^n}{n+1}=\frac{1}{x}\sum_{n=2}^\infty\frac{2^nx^{n+1}}{n+1}-\frac{1}{x}\sum_{n=2}^\infty\frac{nx^{n+1}}{n+1}=\frac{1}{x}(-x-x^2+\sum_{n=0}^\infty\frac{2^nx^{n+1}}{n+1})-\frac{1}{x}(-\frac{x^2}{2}+\int\sum_{n=0}^\infty nx^ndx)=\frac{1}{x}(-x-x^2+\sum_{n=0}^\infty\frac{2^nx^{n+1}}{n+1})-\frac{1}{x}(-\frac{x^2}{2}+\int\frac{x}{(1-x)^2})=-1-x+\frac{1}{2x}\sum_{n=1}^\infty\frac{2^{n}x^{n}}{n}+\frac{x}{2}-\frac{1}{x}(\frac{1}{1-x}+\log(1-x)) =\frac{}{}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $\sin A+\sin B =a,\cos A+\cos B=b$, find $\cos(A+B),\cos(A-B),\sin(A+B)$ If $\sin A+\sin B =a,\cos A+\cos B=b$,
*
*find $\cos(A+B),\cos(A-B),\sin(A+B)$
*Prove that $\tan A+\tan B= 8ab/((a^2+b^2)^2-4a^2)$
|
From the two given conditions you get $$\tan \left(\frac{A+B}{2}\right)=\frac{a}{b}$$
From that we get $$\cos (A+B)=\frac{b^2-a^2}{b^2+a^2}\\
\sin (A+B)=\frac{2ab}{a^2+b^2}$$
And if you square and add both sides of the two given equations from that you'll get $$\cos (A-B)=\frac{a^2+b^2-2}{2}$$
Now, we get $$\cos^2A+\cos^2B+2\cos A\cos B=b^2\Rightarrow 1+\cos(A+B)\cos(A-B)+2\cos A\cos B=b^2\Rightarrow \cos A\cos B=1/2\left(b^2-1-\frac{(b^2-a^2)(a^2+b^2-2)}{2(a^2+b^2)}\right)$$Hence $$\tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B}=\frac{8ab}{2(b^2-1)(a^2+b^2)-(b^2-a^2)(a^2+b^2-2)}=\\\frac{8ab}{(a^2+b^2)^2-4a^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1012928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Given a vector field $\mathbf{H}$, find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g). Let$\;\mathbf{H}(x,y,z) = x^2y\mathbf{i}+y^2z\mathbf{j}+z^2x\mathbf{k}$. Find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g).
I took divergence on both sides which gave me $2xy+2yz+2xz=∇^2g$. I took curl on both sides which gave me three horrible equations one of which I have written down below:
$$\frac{\partial^2f_1}{\partial y^2}-\frac{\partial^2f_2}{\partial x\partial y}-\frac{\partial^2f_3}{\partial x\partial z}+\frac{\partial^2f_1}{\partial z\partial y}=-y^2$$
where I have taken
$$\mathbf{F}=f_1\mathbf{i}+f_2\mathbf{j}+f_3\mathbf{k}$$
I tried to assign some values hoping to work out the rest but this gave me nothing. Please help.
|
Below a slightly different approach which first gives $\mathbf{F}$
and then $G$.
The problem at hand is quite similar to that of finding the vector and
scalar potentials in terms of the electric and magnetic fields in
electrodynamics. In that case there are standard solutions. But here, since $%
\mathbf{H}$ is not decaying for large $\mathbf{x}$, they do not apply. But
it seems that a suitable $\mathbf{F}$ can easily be found. Thus we have$$
\mathbf{H}=\left(
\begin{array}{c}
x_{1}^{2}x_{2} \\
x_{2}^{2}x_{3} \\
x_{3}^{2}x_{1}
\end{array}
\right) =\partial _{\mathbf{x}}\times \mathbf{F}+\partial _{\mathbf{x}}G..
$$
We can assume
$
\partial _{\mathbf{x}}\cdot \mathbf{F}=0.
$
Now
\begin{eqnarray*}
\partial _{\mathbf{x}}\times \mathbf{H} &=&\left(
\begin{array}{c}
-x_{2}^{2} \\
-x_{3}^{2} \\
-x_{1}^{2}
\end{array}
\right) =\partial _{\mathbf{x}}\times (\partial _{\mathbf{x}}\times \mathbf{F%
})=\mathbf{-\partial }_{\mathbf{x}}^{2}\mathbf{F}, \\
\partial _{\mathbf{x}}\cdot \mathbf{H} &\mathbf{=}%
&2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})=\mathbf{-\partial }_{\mathbf{x}}^{2}G.
\end{eqnarray*}
Thus, $\mathbf{-\partial }_{\mathbf{x}}^{2}F_{1}=-x_{2}^{2}$ and assuming $
F_{1}=F_{1}(x_{2})$ we find $F_{1}(x_{2})=a+bx_{2}+cx_{2}^{2}+dx_{2}^{3}+
\frac{1}{12}x_{2}^{4}$. Setting the coefficients $a,b,c,d$ equal to $0$, we
obtain
$$
\mathbf{F}(\mathbf{x})=\frac{1}{12}\left(
\begin{array}{c}
x_{2}^{4} \\
x_{3}^{4} \\
x_{1}^{4}
\end{array}
\right) ,
$$
which is divergence-free. Now
$$
\partial _{\mathbf{x}}\times \mathbf{F}=\left(
\begin{array}{c}
-\partial _{3}F_{2} \\
-\partial _{1}F_{3} \\
-\partial _{2}F_{1}
\end{array}
\right) =-\frac{1}{3}\left(
\begin{array}{c}
x_{3}^{3} \\
x_{1}^{3} \\
x_{2}^{3}
\end{array}
\right) ,
$$
and
$$
\mathbf{H}-\partial _{\mathbf{x}}\times \mathbf{F}=\left(
\begin{array}{c}
x_{1}^{2}x_{2} \\
x_{2}^{2}x_{3} \\
x_{3}^{2}x_{1}
\end{array}
\right) +\frac{1}{3}\left(
\begin{array}{c}
x_{3}^{3} \\
x_{1}^{3} \\
x_{2}^{3}
\end{array}
\right) =\left(
\begin{array}{c}
x_{1}^{2}x_{2}+\frac{1}{3}x_{3}^{3} \\
x_{2}^{2}x_{3}+\frac{1}{3}x_{1}^{3} \\
x_{3}^{2}x_{1}+\frac{1}{3}x_{2}^{3}
\end{array}
\right) =\partial _{\mathbf{x}}G.
$$
\begin{eqnarray*}
\partial _{1}G &=&x_{1}^{2}x_{2}+\frac{1}{3}x_{3}^{3}\Longrightarrow G=\frac{
1}{3}x_{1}^{3}x_{2}+\frac{1}{3}x_{3}^{3}x_{1}+M \\
\partial _{2}G &=&x_{2}^{2}x_{3}+\frac{1}{3}x_{1}^{3}=\frac{1}{3}
x_{1}^{3}+\partial _{2}M \\
\partial _{2}M &=&x_{2}^{2}x_{3}\Rightarrow M=\frac{1}{3}x_{2}^{3}x_{3}+N
\end{eqnarray*}
Setting $N=0$,
$$
G=\frac{1}{3}(x_{1}^{3}x_{2}+x_{3}^{3}x_{1}+x_{2}^{3}x_{3})$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1013503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Flipping coins probability of $6$ flips having more heads than $5$ flips. I have $6$ fair coins and you have $5$ fair coins. We both flip our own coins and observe the number of heads we each have. What is the probability that I have more heads than you?
Not sure how to start this, any help please?
|
An answer using generating functions:
$$G(x)=\left(\frac{1}{2}+\frac{x}{2}\right)^6 \left(\frac{1}{2}+\frac{1}{2 \
x}\right)^5$$
Expanding that I get:
$$\frac{231}{1024}+\frac{1}{2048 x^5}+\frac{11}{2048 \
x^4}+\frac{55}{2048 x^3}+\frac{165}{2048 x^2}+\frac{165}{1024 \
x}+\frac{231 x}{1024}+\frac{165 x^2}{1024}+\frac{165 \
x^3}{2048}+\frac{55 x^4}{2048}+\frac{11 x^5}{2048}+\frac{x^6}{2048}$$
Use only the terms that have positive powers of x:
$$\frac{231 x}{1024}+\frac{165 x^2}{1024}+\frac{165 x^3}{2048}+\frac{55 \
x^4}{2048}+\frac{11 x^5}{2048}+\frac{x^6}{2048}$$
Enter x = 1 to clear out the x variable and sum the terms.
$$ \frac{231}{1024}+\frac{165}{1024}+\frac{165}{2048}+\frac{55}{2048}+
\frac{11}{2048}+\frac{1}{2048}=\frac{1}{2} $$
The answer is $ \frac{1}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove that $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 Prove that a number $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 for every natural $n\ge2$. I am not sure how to start.
|
The base case is true, $T_0=4+2+1+64-8=63$ is a multiple of $7$.
Now let us relate $T_{n+1}$ to $T_n$:
$$T_{n+1}=2.2^n+2.2^{n-1}+2.2^{n-2}+8.8^n-8.8^{n-2}.$$
Subtracting $T_n$, we have
$$T_{n+1}-T_n=2^n+2^{n-1}+2^{n-2}+7.8^n-7.8^{n-2}.$$
We can discard the last two terms, obviously multiples of $7$. Remains to prove that $2^n+2^{n-1}+2^{n-2}$ is a multiple of $7$.
The base case is true, $U_0=4+2+1=7$.
Now let us relate $U_{n+1}$ to $U_n$:
$$U_{n+1}=2.2^n+2.2^{n-1}+2.2^{n-2}=2.U_n.$$
As $U_n$ is a multiple of $7$, so is $U_{n+1}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1015460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Need help in figuring out what I am doing wrong when solving for n.. Here is the expression that I am trying to solve for n:
$$ \frac{4}{16+n} = \frac{10}{16+n} \frac{10}{16+n}$$
I am doing the following:
\begin{align}
\frac{4}{16+n} & = \frac{100}{(16 + n)^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{16^2 + 32n + n^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{256 + 32n + n^2} \\[8pt]
\frac{1}{16+n} & = \frac{100}{4(256 + 32n + n^2)} \\[8pt]
\frac{1}{16+n} & = \frac{20}{256 + 32n + n^2} \\[8pt]
16+n & = \frac{256 + 32n + n^2}{20} \\[8pt]
n & = \frac{256 + 32n + n^2 - 320}{20} \\[8pt]
n & = \frac{-64 + 32n + n^2}{20} \\[8pt]
20n & = -64 + 32n + n^2 \\[8pt]
20n -32n - n^2 & = -64 \\[8pt]
12n - n^2 & = -64 \\[8pt]
n(12 - n) & = -64
\end{align}
Not sure what to do next now... Book says that n = 9
Cannot get 9...
Where am I wrong?
Thank you
|
You can cancel one of the $16 + n$ terms on each side, greatly simplifying the problem. Multiply both sides by $16 + n$. This gives
$$4 = \frac{10 \times 10}{16 + n}$$
Multiply both sides by $16 + n$ again. This gives
$$4(16 + n) = 100$$
Divide both sides by $4$:
$$16 + n = 25$$
Subtract $16$ from both sides:
$$n = 9$$
|
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"url": "https://math.stackexchange.com/questions/1016234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Solving $\int \frac{1}{\sqrt{x^2 - c}} dx$ I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$
How do I do this?
It looks like it is close to being an $\operatorname{arcsin}$?
I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$
But apparently not.
|
$$ \int \frac{1}{\sqrt{x^2-c}}dx = \int \frac{1}{\sqrt{c\left(\frac{x^2}{c}-1\right)}}dx=\frac{1}{\sqrt{c}} \int \frac{1}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2-1}}dx $$
Let $u=\frac{x}{\sqrt{c}}$, then $du=\frac{1}{\sqrt{c}} dx$. So now we have
$$ \int \frac{1}{\sqrt{u^2-1}}du =\mathrm{arcosh}(u)+C = \mathrm{arcosh}\left( \frac{x}{\sqrt{c}}\right)+C $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1016585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving a Triangle inequality If $ a^2 + b^2 > 5c^2 $ in a triangle ABC then show c is the smallest side.I tried to solve this by cosine rule but i was not able to find the answer
|
First, recall that in any triangle, the triangle inequality states that $a+b>c$, $b+c>a$, and $c+a>b$.
Combining this with your condition $a^2+b^2>5c^2$, we have
$$5c^2<a^2+b^2<(b+c)^2+b^2=2b^2+2bc+c^2,$$
hence $$2c^2<b^2+bc.$$
Now, suppose it were the case that $b\leq c$. Then $2c^2<b^2+bc\leq 2c^2$, a contradiction.
A similar argument holds when $a\leq c$, hence $c$ is indeed the smallest side.
|
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|
Upper bound for the widest matrix with no two subsets of columns with the same vector sum Over at PPCG there is an ongoing contest going on to find the largest matrix without a certain property, called property $X$. The description is as follows (copied from the question).
A circulant matrix is fully specified by its first row $r$. The remaining rows are each cyclic permutations of the row $r$ with offset equal to the row index. We will allow circulant matrices which are not square so that they are simply missing some of their last rows. We do however always assume that the number of rows is no more than the number of columns. For example, consider the following $3\times5$ circulant matrix.
$$\begin{pmatrix}1&0&1&1&1\\
1&1&0&1&1\\
1&1&1&0&1\end{pmatrix}$$
We say a matrix has property $X$ if it contains two non-empty sets of columns with non-identical indices which have the same (vector) sum. The vector sum of two columns is simply an element-wise summation of the two columns. That is the sum of two columns containing $x$ elements each is another column containing $x$ elements.
The matrix above trivially has property $X$ as the first and last columns are the same. The identity matrix never has property $X$.
If we just remove the last column of the matrix above then we get an example which does not have property $X$. The score of a matrix is defined to be the number columns divided by the number of rows. The following matrix therefore does not have property $X$ and gives a score of $4/3$.
\begin{pmatrix}1&0&1&1\\
1&1&0&1\\
1&1&1&0\end{pmatrix}$$
The task they were given is to find the highest scoring circulant matrix whose entries are all 0 or 1 and which does not have property $X$.
So far the numerical evidence points towards an upper bound of two.
Is there an upper bound of $2$ for this score?
The highest scoring matrix found so far has a score of 36/19 by Peter Taylor. This has 000001001010110001000101001111111111 as the first row.
|
I think any nxm circulant matrices (where m=n+1) of this form will not have property X, with all entries 0 or 1
$$\begin{pmatrix}1&0&1&1&\cdots&1\\
1&1&0&1&\cdots&1\\
1&1&1&0&\cdots&1\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&1&\cdots&0\end{pmatrix}$$
So the score of a matrix of this form is $$S=\frac{n+1}{n}$$
S is largest when n=1, thus the max possible score is $$S_{max}=\frac{2}{1}=2$$
However a 1x2 matrix is just a row vector $$\begin{pmatrix}1&0\end{pmatrix}$$ and not a circulant matrix
Thus the highest scoring required matrix is a 2x3 matrix of the form
$$\begin{pmatrix}1&0&1\\1&1&0\end{pmatrix}$$
with max score $$S=\frac{3}{2}=1.5$$
================
EDIT (Attempt to fix my answer using the advice pointed out by Lembik)
From the requirements of property X, it seems any circulant (0,1) matrix must not have any columns identical
In addition, any column cannot be the linear combination of any 2 other columns
It is pointed out by Lembik that the following matrix
$$\begin{pmatrix}1&0&1\\1&1&0\end{pmatrix}$$
has property X since
$$\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}$$
In order to construct a matrix B such that it does not have property X and has no empty columns, it must first have unique non empty columns like so (or permultation thereof):
$$\begin{pmatrix}1&0&1&1&\cdots&1\\
1&1&0&1&\cdots&1\\
1&1&1&0&\cdots&1\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&1&\cdots&0\end{pmatrix}$$
The score of a matrix of this form is $$S=\frac{m}{n}$$
S is largest when n=1
However a 1xm matrix is just a row vector e.g. for 1x2 matrix $$\begin{pmatrix}1&0\end{pmatrix}$$ and not a circulant matrix, and since 0+1=1 as pointed out by the PPCG link, it has property X
For n=2, the only way there could be unique columns that are not linear combination of any two other columns is when m=2, but this violates the constraint m>n
Thus nxm where m>n, n must be > 2
We also note that for the case where n=3, m has to be < 5 as otherwise the exist a column which will either be non unique or a linear combination of any 2 of the existing columns (since all entries are restricted to 0 or 1)
Thus for n=3, m must = 4
We knew that by finding the limit of S as $$n \rightarrow \infty$$ is zero.
Therefore the highest scoring matrix must have the smallest m and n
Therefore the set of highest scoring required circulant matrices are (there are 24)
$$B=\begin{pmatrix}1&0&1&1\\1&1&0&1\\1&1&1&0\end{pmatrix}, \begin{pmatrix}1&0&1&1\\1&1&1&0\\1&1&0&1\end{pmatrix},\begin{pmatrix}1&1&0&1\\1&0&1&1\\1&1&1&0\end{pmatrix},\begin{pmatrix}1&1&0&1\\1&1&1&0\\1&0&1&1\end{pmatrix},\begin{pmatrix}1&1&1&0\\1&0&1&1\\1&1&0&1\end{pmatrix},\begin{pmatrix}1&1&1&0\\1&1&0&1\\1&0&1&1\end{pmatrix}\\\begin{pmatrix}1&1&1&0\\1&1&0&1\\0&1&1&1\end{pmatrix} etc.$$
with max score $$S=\frac{4}{3}=1.3333...$$
|
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|
Find $t$ in $i = 50\sin\left(120\pi t -\frac{3\pi}{25}\right)$ where $i = 25$
An alternating current generator produces a current given by the equation
$$i = 50\sin\left(120\pi t - \frac{3\pi}{25}\right)$$
where $t$ is the $\text{time}$ in $\text{seconds}$.
(Q) Find the first two values of $t$ when $i = 25 \text{ amperes}$
[src]
Is my working correct? Kinda stuck with this problem, what do I do next?
$50\sin(120\pi t-3\pi/25)= 25$
$\sin(120\pi t-3\pi/25)= 25/50$
$\sin(120\pi t-3\pi/25)= 0.5$
$\text{Ref angle }= \sin^{-1}(0.5)$
Since $\sin(120\pi t-3\pi/25)= 0.5 > 0$, it is at $1^{\text{st}}$ or $2^{\text{nd}}$ quadrant
$(120\pi t-3\pi /25)=\pi/6$
|
Hint: Let $x=120\pi t-3\pi/25$. If $x=\theta $ is one
solution of the equation $y=\sin x$ then another solution is $x=\pi -\theta $.(*)
Complete solution using your work. If the hint isn't enough, hover your mouse over the grey to see
You have correctly found the first angle $\theta >0$ for which $\sin \theta =1/2$. It's $\arcsin 1/2=\pi /6\in \left[0,\pi /2\right] $.
The first value of $t$ when $i=25\, \textrm{A}$ (I denote $t_{1}$) satisfies your last equation.
\begin{equation*}
120\pi t-\frac{3\pi }{25}=\frac{\pi }{6}.
\end{equation*}
Simplify by dividing both sides by $\pi $\begin{equation*}120t-\frac{3}{25}=\frac{1}{6}\end{equation*}and solve it
\begin{equation*}
t=t_{1}=\frac{1/6+3/25}{120}=\frac{43}{18\,000}\text{ }\mathrm{s}\approx
2.389\text{ }\mathrm{ms}
\end{equation*}
Since in general $\sin x=\sin \left( \pi -x\right) $, we have
\begin{equation*}\sin (\pi /6)=\sin \left( \pi -\pi /6\right) =\sin (5\pi /6).\end{equation*}
Hence $\theta _{2}=5\pi /6\in \left[ \pi /2,\pi \right]$ is the second angle $\theta>0$ for which $\sin \theta =1/2$.
As a consequence, and similarly to $t_{1}$, the second value of $t$ when $i=251 \, \textrm{A}$ (I denote $t_{2}$), is the solution of the equation
\begin{equation*} 120\pi t-\frac{3\pi }{25}=\frac{5\pi }{6},\end{equation*}
which is
\begin{equation*}
t=t_{2}=\frac{5/6+3/25}{120}\text{ }\mathrm{s}=\frac{143}{18\,000}\approx
7.944\text{ }\mathrm{ms}
\end{equation*}
--
(*) The general form for the solutions $x$ of $y=\sin x$ combine both
solutions $x=\theta ,x=\pi -\theta $ added to any multiple of $2\pi $. So $
x=2k\pi +\theta ,k\in \mathbb{Z}
$ and $x=\pi -\theta +2k\pi =(2k+1)\pi -\theta ,k\in \mathbb{Z}$ are solutions, which can be written in a single expression as $x=n\pi
+(-1)^{n}\theta $, $\theta \in \left[ -\frac{1}{2},\frac{1}{2}\right] $ and $n\in
\mathbb{Z}.$
|
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|
When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?
|
$\qquad\qquad$ Too long for a comment: Positive coprime integers x and y less than $6,000$ :
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=1$
$\qquad\qquad\qquad\qquad$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=2$
$\qquad\qquad\qquad\qquad$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=3$
$\qquad\qquad\qquad\qquad$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=4$
$\qquad\qquad\qquad\qquad$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=5$
$\qquad\qquad\qquad\qquad$
|
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|
Evaluate the integrals in $\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$ Note that when $0\le x \le 1$ we have $$\frac 12 \le \frac 1 {1+x^2} \le 1.$$
Hence, $$\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$$
Evaluate the integrals above to show that $$\frac1{1260}\le\frac {22}7-\pi\le\frac1{630}.$$
All the integrands include the expression $x^4(1-x)^4$ but I'm not sure how to continue.
|
To evaluate the first and third integrals, simply expand:
$$x^4 (1 - x)^4 = x^4 - 4 x^5 + 6 x^6 - 4 x^7 + x^8.$$ Then, integrate term by term.
To evaluate the second integral, apply polynomial long division, which will give that the middle integrand has the form
$$\frac{x^4 (1 - x)^4}{1 + x^2} = p(x) - \frac{4}{1 + x^2},$$
where $p(x)$ is some polynomial.
|
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|
Find maximum/minimum for $\cos(2x) + \cos(y) + \cos(2x+y) $ I have not been able to find the critical points for $\cos(2x) + \cos(y) + \cos(2x+y) $
|
This might be a solution.
Set $X=2x$ and $\cos(X)=(1-t^2)/(1+t^2),\cos(y)=(1-s^2)/(1+s^2)$,$\sin(X)=(2t)/(1+t^2),\cos(y)=(2s)/(1+s^2)$, then the original expression becomes:
$$\cos(X) + \cos(y) + \cos(X+y)=f(s,t)=\frac{3 - s^2 - 4 s t - t^2 - s^2 t^2}{(1+s^2)(1+t^2)}$$
where
$$f(s,t)=3 - s^2 - 4 s t - t^2 - s^2 t^2$$
From $f(s,t)=0$, we can solve for $t$:
$$t=\frac{-2 s \pm \sqrt{3 + 6 s^2 - s^4}}{1 + s^2}$$
Notice that:
$$\sqrt{3 + 6 s^2 - s^4}=\sqrt{(s^2-(3-2\sqrt{3}))(3+2\sqrt{3}-s^2)}$$
Thus when $3+2\sqrt{3}\ge s^2\ge(3-2\sqrt{3})$ and $s\in \mathbb{R}$, we have real solutions for $t$. Thus we require $3+2\sqrt{3}\ge s^2\ge 0$.
|
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|
Evaluate $ \int_0^3 \frac{x^3}{1-x^4}\, dx. $ Evaluate
$$
\int_0^3 \frac{x^3}{1-x^4}\, dx.
$$
I evaluated the integral and got $\left[\dfrac{-\ln(1-x^4)}{4}\right]_0^3$ which ended up diverging. Any help is appreciated!
|
The integral gives up easily to direct attack, but with a little twist.
Use the obvious substitution $u=x^4$:
$$\int_0^3 \frac{x^3}{1-x^4}\, dx=\frac{1}{4} \int_0^{81} \frac{du}{1-u}=$$
Now it makes sense to separate the integral into two parts, so the denominator is positive for each part:
$$=\frac{1}{4} \int_0^1 \frac{du}{1-u}-\frac{1}{4} \int_1^{81} \frac{du}{u-1}=$$
Now in the first integral we use the substitution $t=1-u$ and in the second $p=u-1$:
$$=\frac{1}{4} \int_0^1 \frac{dt}{t}-\frac{1}{4} \int_0^{80} \frac{dp}{p}=\frac{1}{4} \int_0^1 \frac{dt}{t}-\frac{1}{4} \int_0^1 \frac{dp}{p}-\frac{1}{4} \int_1^{80} \frac{dp}{p}=$$
But the first two divergent integrals just cancel each other out (they are equal), so we are allowed to write:
$$\int_0^3 \frac{x^3}{1-x^4}\, dx=-\frac{1}{4} \int_1^{80} \frac{dp}{p}=-\frac{1}{4} \ln 80=-\frac{1}{4} (\ln 5+4 \ln 2)$$
This is the correct answer (formally). Although Wolfram Alpha writes that the original integral diverges, and the answer we obtained is just Cauchy principal value of the integral.
Edit. A little justification for the calcelling the divergent integrals. The problem is at $t \to 0$, thus:
$$\int_0^1 \frac{dt}{t}=\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dt}{t}$$
In this case the limit doesn't exist.
But for any positive real number $\epsilon>0$ we have formally:
$$\int_\epsilon^1 \frac{dt}{t} \equiv \int_\epsilon^1 \frac{dp}{p}$$
We just renamed the variable after all. Moreover, in our particular example we have $t=-p$. So setting:
$$\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dt}{t}-\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dp}{p}=\lim_{\epsilon \to 0} \left( \int_\epsilon^1 \frac{dt}{t}-\int_\epsilon^1 \frac{dp}{p} \right)=0$$
Makes perfect sense.
But the rigorous justification should involve contour integration in the complex plane.
|
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|
How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}$$
But I don't know how to evaluate this?
Many thanks for any help.
|
Using the identity $sin \ x =3sin(\frac{x}{3})-4 sin^{3}(\frac{x}{3})$, our limit $l=lim \frac{sin \ x -x +\frac{x^{3}}{6}}{x^{3}}$ becomes,
$l= \frac{1}{9} lim \frac{sin \ \frac{x}{3} -\frac{x}{3} +(\frac{x}{3})^{3}.\frac{1}{6}}{x^{3}}+ \frac{4}{27}- lim \ 4 \frac{sin^{3}(\frac{x}{3})}{x^{3}}$, which means $l=\frac{1}{9} l$, so $l=0$.
|
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|
Write an expression for $(\cos θ + i\sin θ)^4$ using De Moivre’s Theorem. Obtain another expression for $(\cos θ + i \sin θ)^4$ by direct multiplication (i.e., expand the bracket). Use the two expressions to show
$$
\cos 4\theta = 8 \cos^4 \theta − 8 \cos^2 \theta + 1,\\
\sin 4\theta = 8\cos^3\theta \sin\theta − 4 \cos \theta\sin \theta.
$$
You may use the well-known identity: $\sin^2 \theta + \cos^2 \theta = 1$, but do not use any multiple angle formula.
I got using DMT that $(\cos \theta + i\sin \theta)^4 = (cos 4\theta + i\sin 4\theta)$
and using direct multiplication, I got $\cos^4\theta + \sin^4\theta - 4\cos^3\theta*\sin\theta*i - 4\cos\theta \sin^3\theta i - 6\cos^2\theta\sin^2\theta$
|
By De Moivre’s theorem,
$$
(\cos\theta+i\sin\theta)^4=\cos 4\theta+i\sin 4\theta.
$$
On the other hand
$$
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4
$$
Expand the binomial and equate the real and imaginary parts. Where you find $\sin^2\theta$, substitute $1-\cos^2\theta$.
You have, almost correctly,
$$
\cos4\theta+i\sin4\theta=
\cos^4\theta+4i\cos^3\theta\sin\theta+6i^2\cos^2\theta\sin^2\theta
+4i^3\cos\theta\sin^3\theta+i^4\sin^4\theta
$$
Now $i^2=-1$, $i^3=-i$ and $i^4=1$, so, by equating the real and imaginary parts, we get
$$
\cos4\theta=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\
\sin4\theta=4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta
$$
Now it's just a matter of substituting $\sin^2\theta=1-\cos^2\theta$, so, for example, we have
\begin{align}
\sin4\theta&=4\sin\theta(\cos^3\theta-\cos\theta\sin^2\theta)\\
&=4\sin\theta(\cos^3\theta-\cos\theta+\cos^3\theta)\\
&=8\cos^3\theta\sin\theta-4\cos\theta\sin\theta.
\end{align}
Do similarly for $\cos4\theta$.
|
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|
If the number $x$ is algebraic, then $x^2$ is also algebraic Prove that if the number $x$ is algebraic, then $x^2$ is also algebraic. I understand that an algebraic number can be written as a polynomial that is equal to $0$. However, I'm baffled when showing how $x^2$ is also algebraic.
|
Basically $P(x)\cdot P(-x)= Q(x^2)$, a polynomial in $x^2$. A similar trick using $n$-th roots of $1$ allows us to get $P(x) \cdot P_1(x) = Q(x^n)$ for some $Q$ so an equation for $x^n$. To get an equation satisfied by a general $\phi(x)$ ( $\phi(x) = x^2$ in the above example) we can use the companion matrix as @Julian Rosen: wrote in his answer. Here is a concrete example: We know that $2 x^5 + x-2=0$. We want the equation satisfied by $\phi(x) = x^3 - x^2 -1$.
Consider the companion matrix (see http://en.wikipedia.org/wiki/Companion_matrix)
\begin{eqnarray}
A = \left(
\begin{array}{ccccc}
0 & 0 & 0 &0 &1 \\
1 & 0 & 0 &0 &-\frac{1}{2}\\
0 & 1 & 0 &0 &0\\
0 & 0 & 1 &0 &0\\
0 & 0 & 0 &1 &0\\
\end{array} \right)
\end{eqnarray}
Take $\phi(A) = A^3 - A^2 -I_5 = B$ where
\begin{eqnarray}
B = \left(
\begin{array}{ccccc}
-1 & 0 & 1 &-1 &0 \\
0 & -1 & -\frac{1}{2} &\frac{3}{2} &-1\\
-1 & 0 &-1 &-\frac{1}{2} &\frac{3}{2}\\
1 & -1 & 0 &-1 &-\frac{1}{2}\\
0 & 1 & -1 &0 &-1\\
\end{array} \right)
\end{eqnarray}
The characteristic polynomial of $B$ is
$$R(t) =t^5+5 t^4+16 t^3+\frac{61}{2} t^2+\frac{271}{8}t+\frac{127}{8}$$
and one checks that
$$R[x^3 - x^2 -1] =\frac{1}{8} (2 x^5+x-2) (4 x^10-20 x^9+40 x^8-40 x^7+18 x^6+10 x^5-16 x^4-12 x^3+23 x^2-x-2)$$
that is, $x^3 - x^2 -1$ is a root of $R(\cdot)$.
Or, with numerics, the equation $2 x^5 + x-2=0$ has a unique real solution $0.8890618537791..$. We get $x^3 - x^2 -1 = -1.0876889476...$ and $R(-1.0876889476...) \simeq 0$
$\bf{Added}$ This method can be used to find equations for polynomial expressions involving several algebraic numbers $x_1$, $x_2$, $\ldots$ each satisfying a given equation. In that case we work with Kronecker products. This is finiteness effective; also see @Amitai Yuval: answer.
|
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|
Find exact value of $\cos (\frac{2\pi}{5})$ using complex numbers.
Factorise $z^5-1$ over the real field.
Show that $\cos \frac{2\pi}{5}$ is a root of the equation $4x^2+2x-1=0$ and hence find its exact value.
I have worked out that $$ z^5-1=(z-1)(z^4+z^3+z^2+z+1)\\=(z-1)(z^2-2z\cos72^o+1) (z^2+2z\cos 72^o+1) $$
|
Find the polynomials that cancel $\cos\left(\frac{2\pi}{5}\right)$ :
We know that the solution of $z^5=1$ are given by $\alpha^k=e^{\frac{2ik\pi}{5}}$, $k=0,1,2,3,4$. We can observe that $\alpha^{4}=\alpha^{-1}$ and $\alpha^{3}=\alpha^{-2}$. Moreover,
$$(*):\quad 0=1+\alpha+\alpha^2+\alpha^3+\alpha^4=1+\alpha+\alpha^{-1}+\alpha^2+\alpha^{-2}$$
and $\alpha+\alpha^{-1}=2\cos\left(\frac{2\pi}{5}\right)$.
Let set $X=\alpha+\alpha^{-1}$, then
$$X^2=(\alpha+\alpha^{-1})^2=\alpha^2+\alpha^{-2}+2$$
and so
$$\alpha^2+\alpha^{-2}=X^2-2$$
if we replace in $(*)$ we obtain
$$0=X+X^2-2+1=X^2+X-1.$$
Now set $x=\frac{X}{2}=\cos\left(\frac{2\pi}{5}\right)$, therefore, $X=2x$ and so
$$0=(2x)^2+2x-1=4x^4+2x-1$$
and thus, $\cos\left(\frac{2\pi}{5}\right)$ is solution of
$$4x^2+2x-1=0$$
Find the value of $\cos\left(\frac{2\pi}{5}\right)$ :
$$4x^2+2x-1=0\iff x^2+\frac{1}{2}x-\frac{1}{4}=0\iff \left(x+\frac{1}{4}\right)^2-\frac{1}{16}-\frac{1}{4}=0\iff\left(x+\frac{1}{4}\right)^2=\frac{5}{16}\iff x+\frac{1}{4}=\pm\frac{\sqrt 5}{4}\iff x=\frac{-1\pm\sqrt 5}{4}$$
But $\cos\left(\frac{2\pi}{5}\right)>0$ because $\frac{2\pi}{5}<\frac{\pi}{2}$, then,
$$\cos\left(\frac{2\pi}{5}\right)=\frac{-1+\sqrt 5}{4}$$
|
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|
How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in excel the software cannot handle it (because of truncation of significant figures) and the results are nonsense.
I thought to approximate the result using the first few terms of a series in increasing powers of $k$. I can multiply out the first few terms and examine the patterns in the following pyramid...
$$(1+k)^1 = k +1 $$
$$(1+k)^2 = k^2 +2k +1 $$
$$(1+k)^3 = k^3 +3k^2 +3k +1 $$
$$(1+k)^4 =k^4+4 k^3+6 k^2+4 k+1$$
$$(1+k)^5 =k^5+5 k^4+10 k^3+10 k^2+5 k+1$$
$$(1+k)^6 =k^6+6 k^5+15 k^4+20 k^3+15 k^2+6 k+1$$
So for example, for $N=3$ we would obtain the sum
$$
S= k^3 +4k^2 +6k +1
$$
The results suggest a solution with a pattern of the form
$$
S = a + bk^1 +ck^2+dk^3...
$$
I can see that $a=N$. The other coefficients increase monotonously and it might be possible to determine a formula for the coefficients from the pattern. Although the general pattern is not convergent, it is possible that for certain restricted ranges of $N$ and $k$ a convergent formula could be obtained. If so then it is possible that a useful approximation of S can be obtained just by evaluating the first few terms in the series.
But is there a well-known general formula for the terms in this series or can one be derived algebraically from the original formula?
UPDATE
following on from the answer by User73985...
$$ S=\sum_{i=1}^N (1+k)^i = N + \sum_{j=2}^{N+1}\binom{N+1}{j}k^{j-1}$$
So
$$ S= N + \sum_{j=2}^{N+1}\frac{(N+1)!}{(N+1-j)!j!}k^{j-1}$$
then
$$ S= N
+ \frac{(N+1)!}{(N-1)!2!} k^{1}
+ \frac{(N+1)!}{(N-2)!3!} k^{2}
+ \frac{(N+1)!}{(N-3)!4!} k^{3} +...
$$
giving
$$ S= N
+ \frac{(N+1)(N)}{2!} k^{1}
+ \frac{(N+1)(N)(N-1)}{3!} k^{2}
+ \frac{(N+1)(N)(N-1)(N-2)}{4!} k^{3} +...
$$
thus
$$ S= N
+ \frac{N^2+N}{2} k^{1}
+ \frac{N^3-N}{6} k^{2}
+ \frac{N^4-2 N^3-N^2+2 N}{24} k^{3} +...
$$
For $N=1 to 10,000$ and $k= 2.40242 * 10^{-12}$ this formula can be truncated to
$$ S = N + \frac{N^2+N}{2} k^{1}
$$
and then gives results very close to those expected. Because $k$ is so small relative to $n$ the terms in higher powers of $k$ can be ignored. Note that the coefficient of $k^1$ is consistent with that found by examination of the coefficients in the "pyramid" presented above.
|
for $k<1$ we can use
$$
(1+k)^N \approx 1+ kN + \frac{N(N-1)}{2}k^2 + \frac{N(N-1)(N-2)}{3!}k^3
$$
thus
$$
\begin{align}
\sum (1+k)^i &\approx&\frac{(k+1)(1+ kN + \frac{N(N-1)}{2}k^2 + \frac{N(N-1)(N-2)}{3!}k^3-1)}{k}\\
&=&(k+1)\left(1+\frac{N-1}{2}k+\frac{(N-1)(N-2)}{3!}k^2\right)
\end{align}
$$
|
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|
Partial fraction of $\frac{2x^2-9x-9}{x^3-9x}$ I'm doing some questions from Anton, 8th edition, page 543, question 13. I've found a answer but it does not match with the answer given at the last pages.
Questions asks to solve $\int{\frac{2x^2-9x-9}{x^3-9x}}$
So:
$$\frac{2x^2-9x-9}{x(x^2-9)}$$
then:
$$\frac{2x^2-9x-9}{x(x-3)(x+3)}$$
(using $a(x-x')(x-x'')$)
Continuing:
$$2x^2-9x-9 = \frac{A}{x} + \frac{B}{(x-3)} + \frac{C}{(x+3)}$$
Solving the product:
$$A(x^2-9)+Bx(x+3)+Cx(x-3)$$
Ended up with a system, because: $(A+B+C)x^2+(3B-3C)x+(-9A)$
\begin{cases} A+B+C =2 \\ 3B-3C = -9 \\ -9A = -9\end{cases}
then:
$A = 1$
$B = -1$
$C = 2$
The answer I found:
$$ ln |x| - ln|x-3| + 2 ln|x+3| + C $$
answer given by author:
$$ ln |\frac{x(x+3)^2}{x-3}| + C$$
What am I missing?
|
You're answer is fine and is equivalent to the author's.
To see why, recall:
$$\ln a - \ln b = \ln\left(\frac ab\right)$$
$$\ln a + \ln b = \ln(ab)$$
$$2 \ln a = \ln(a^2)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The integral $\int_{|z|=2}\log\frac{z+1}{z-1}dz$ Let $\log$ be the branch of the logarithm that extends the usual real logarithm, and consider on $D=\Bbb C\smallsetminus [-1,1]$ the function $$f(z)=\log\frac{z+1}{z-1}$$
I have to find the integral of $f$ around the circle $|z|=2$. Now, as an example, consider the integral $$\int_{|z|=2}\frac{e^{z+z^{-1}}}{1-z^2}dz$$
Using the biholomorphic mapping $B(0,1)^\times\to D$ that sends $z\to \frac 1 2(z+z^{-1})$ I got the integral $$\frac 1 2\int_\gamma \frac{e^{2z}}{1-z^2}dz$$
and $\gamma$ is a closed path inside $B(0,1)^{\times}$. This means the integral vanishes. I am trying to do something similar here. So if I take $w=\frac{z+1}{z-1}$, I get that $\frac{(w-1)^2}2=\frac{2}{(z-1)^2}$ and $dw=-2dz/(z-1)^2$, so I ultimately want to look at $$-2\int_{\gamma^{-}}\frac{ \log w}{(w-1)^2}dw$$ where $\gamma$ is a circle that passes through $1/3,3,-3/5+4/5 i$. Using the computer I got the circle is $$(x-5/3)^2+(y-5/2)^2=\frac{7225}{900}$$
Add This circle's orientation is now reversed.
Even without this one sees $1$ is an interior point, so the integral should equal, by Cauchy, $4\pi i$ (not $-4\pi i$). Can anyone confirm this is correct, and/or suggest another approach?
|
As mike said in the comments above, I think the answer is $4 \pi i$.
Since $1+ \frac{1}{z}$ and $1- \frac{1}{z}$ both lie in the right half-plane for $|z| >1$,
$$- \pi < \text{Arg} \left(1 + \frac{1}{z} \right) - \text{Arg} \left(1- \frac{1}{z} \right) \le \pi \ \ \text{for} \ \ |z| >1.$$
Thus
$$ \begin{align} f(z) = \log \left(\frac{z+1}{z-1} \right) &= \log \left( \frac{1 + \frac{1}{z}}{1- \frac{1}{z}} \right) \\&= \log \left(1 + \frac{1}{z} \right) - \log \left(1 - \frac{1}{z} \right) \\ &= \frac{2}{z} + \text{O}(z^{-3}) \ \ \text{for} \ |z| >1. \end{align}$$
So $$ \int_{|z|=2} \log \left(\frac{z+1}{z-1} \right) \ dz = - 2 \pi i \ \text{Res} [f(z), \infty] = - 2 \pi i (-2) = 4 \pi i $$ since by definition the residue at infinity of $f(z)$ is the negative of the coefficient of the $\frac{1}{z}$ term in the Laurent expansion at $\infty$.
|
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|
Finding a line through 4 other lines! This one's probably easy, but I'm dreadfully stuck and can't seem to figure out a decent method.
I have the following lines:
$$a: \vec{x}(\lambda)= \left( \begin{array}{ccc}
4 \\
-2 \\
-2 \end{array} \right) + \lambda\left( \begin{array}{ccc}
1 \\
-1 \\
-1 \end{array} \right)
$$
$$b: \vec{x}(\mu)= \left( \begin{array}{ccc}
-1 \\
1 \\
-3 \end{array} \right) + \mu\left( \begin{array}{ccc}
1 \\
0 \\
2 \end{array} \right)
$$
$$c: \vec{x}(\nu)= \left( \begin{array}{ccc}
1 \\
0 \\
5 \end{array} \right) + \nu\left( \begin{array}{ccc}
0 \\
-2 \\
1 \end{array} \right)
$$
$$d: \vec{x}(\tau)= \left( \begin{array}{ccc}
3 \\
-2 \\
0 \end{array} \right) + \tau\left( \begin{array}{ccc}
-1 \\
1 \\
1 \end{array} \right)
$$
I have to find the line that intersects all four of these lines. How do I go about doing this?
Cheers!
|
Note that $a$ and $d$ are parallel, so whatever line it is, it needs to lie in the plane containing $a$ and $d$. Considering it also needs to intersect $b$ and $c$, try figuring out which two points those two lines intersect the $ad$-plane.
Edit: Here's a full answer.
$ad$-plane: The $ad$-plane's normal vector is orthogonal to $(-1, 1, 1)$ as well as $\vec{a}(0) - \vec{d}(0) = (1, 0, -2)$. We therefore have a normal vector given by
$$
(-1, 1, 1)\times (1, 0, -2) \\
= (1\cdot(-2) - 1\cdot 0, 1\cdot 1 - (-1)\cdot (-2), (-1)\cdot 0 -
1\cdot 1)\\
= (-2, -1, -1)
$$
I elect to choose the negative of this vector, for estethic reasons.
Inserting $\vec{d}(0)$ into the general equation for a plane, we have:
$$
2\cdot 3 + 1\cdot (-2) + 1 \cdot 0 = 4
$$
and therefore the $ad$-plane is given by $2x + y + z = 4$.
$b$-intersection: The $\mu$ for the point where the $b$-line intersects the $ad$-plane is given by
$$
2(-1 + \mu) + 1 -3 + 2\mu = 4\\
4\mu = 8\\
\mu = 2
$$
so the intersection point is $B = \vec{b}(2) = (1, 1, 1)$.
$c$-intersection: The $\nu$ for the point where the $c$-line intersects the $ad$-plane is given by
$$
2\cdot 1 -2\nu + 5 + \nu = 4\\
-\nu = -3\\
\nu = 3
$$
so the intersection point is $C = \vec{c}(3) = (1, -6, 8)$.
The line: We need the line $\vec{l}(\gamma)$ that goes from $B$ to $C$. It is given by
$$
\vec{l}'(\gamma) = B + \gamma(C - B)\\
= (1, 1, 1) + \gamma(0, -7, 7)
$$
which I would like to rewrite to:
$$
\vec{l}(\gamma) = (1, 1, 1) + \gamma(0, -1, 1)
$$
For reference, the four intersection points are:
*
*$al$: $\lambda = -3, \gamma = 0, (1, 1, 1)$
*$bl$: $\mu = 2, \gamma = 0, (1, 1, 1)$
*$cl$: $\nu = 3, \gamma = 7, (1, -6, 8)$
*$dl$: $\tau = 2, \gamma = 1, (1, 0, 2)$
|
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|
How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$
Evaluate
$$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$
I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta
$$
After this, I thought of using the Taylor Expansion of
$\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.
Please Help!
|
Let $I$ be our integral
Since $\displaystyle \small\int \ln\left(\frac{1+x^2}{2x^2}\right)\ dx=x\ln\left(\frac{1+x^2}{2x^2}\right)+2\arctan(x)$, so by integration by parts we have
$$I=2\left(\frac{\pi}{4}\right)^3-4\underbrace{\int_0^1\frac{\arctan^2(x)}{1+x^2}\ dx}_{\frac13\left(\frac{\pi}{4}\right)^3}-\underbrace{\int_0^1\frac{2x}{1+x^2}\ln\left(\frac{1+x^2}{2x^2}\right)\arctan(x)\ dx}_{IBP}$$
$$=\frac{\pi^3}{96}+\int_0^1\frac{\ln(1+x^2)}{1+x^2}\ln\left(\frac{1+x^2}{2x^2}\right)\ dx-\int_0^1\frac{2\ln(1+x^2)\arctan(x)}{x(1+x^2)}\ dx$$
$$=\frac{\pi^3}{96}+A-B\tag1$$
For $A$, use $x=\tan\theta$
$$A=2\ln2\int_0^{\pi/4}\ln(\cos\theta)\ d\theta+4\int_0^{\pi/4}\ln(\cos\theta)\ln(\sin\theta)\ d\theta$$
$$A=2\ln2\underbrace{\int_0^{\pi/4}\ln(\cos\theta)\ d\theta}_{\text{common integral}}+2\underbrace{\int_0^{\pi/2}\ln(\cos\theta)\ln(\sin\theta)\ d\theta}_{\text{beta function}}$$
$$=2\ln2\left(\frac{G}{2}-\frac{\pi}{2}\ln^22\right)+2\left(\frac{\pi}{2}\ln^22-\frac{\pi^3}{48}\right)$$
$$\boxed{A=G\ln2+\frac{\pi}{2}\ln^22-\frac{\pi^3}{24}}$$
For $B$, write $\frac{2}{x(1+x^2)}=\frac{2}{x}-\frac{2x}{1+x^2}$
$$B=2\int_0^1\frac{\ln(1+x^2)\arctan(x)}{x}\ dx-\int_0^1\frac{2x\ln(1+x^2)\arctan(x)}{1+x^2}\ dx$$
The first integral is already evaluated here
$$\int_0^1\frac{\ln(1+x^2)\arctan(x)}{x}\ dx=\frac{\pi^3}{16}+\frac{\pi}{8}\ln^22+G\ln2+2\Im\operatorname{Li}_3(1-i)$$
For the second one, apply integration by parts
$$\int_0^1\frac{2x\ln(1+x^2)\arctan(x)}{1+x^2}\ dx=\frac{\pi}{8}\ln^22-\frac12\int_0^1\frac{\ln^2(1+x^2)}{1+x^2}\ dx$$
$$=\frac{\pi}{8}\ln^22-2\int_0^{\pi/4}\ln^2(\cos\theta)\ d\theta$$
we proved here
$$\int_0^{\pi/4}\ln^2(\cos\theta)\ d\theta=\frac{7\pi^3}{192}+\frac{5\pi}{16}\ln^22-\frac{G}{2}\ln2+\Im\operatorname{Li_3}(1-i)$$
Collecting the results we have
$$\boxed{B=\frac{19\pi^3}{96}+\frac{3\pi}{4}\ln^22+G\ln2+6\Im\operatorname{Li_3}(1-i)}$$
Now plug the boxed results in $(1)$ we get
$$I=-\frac{11\pi^3}{48}-\frac{\pi}{4}\ln^22-6\Im\operatorname{Li_3}(1-i)$$
|
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|
Find the domain of $\frac{x}{\sqrt{6x^2+3x+3/4}+x}$. My attempt:
Let's assume that $\sqrt{6x^2+3x+\frac{3}{4}}+x$ $>$ 0$$
\rightarrow\sqrt{6x^2+3x+\frac{3}{4}} > -x
\rightarrow {6x^2+3x+\frac{3}{4}} > x^2\\
\rightarrow {5x^2+3x+\frac{3}{4}} > 0\\$$
If x $<$ 0 then $x^2$ $>$ 0, and ${5x^2+3x+\frac{3}{4}} > 0$
the domain is $(-\infty,+\infty)\\$
But, I'm not totally convinced. Any help is appreciated.
Thank you
|
All we need is $\sqrt{6x^2+3x+\dfrac34}+x\ne0$ and of course $6x^2+3x+\dfrac34\ge0$
$6x^2+3x+\dfrac34=6\left(x+\dfrac14\right)^2+\dfrac34-\dfrac6{16}>0$ for all real $x$
Now $\sqrt{6x^2+3x+\dfrac34}+x=0\iff x=-\sqrt{6x^2+3x+\dfrac34}\le0$
$\sqrt{6x^2+3x+\dfrac34}=-x$
Squaring we get $5x^2+3x+\dfrac34=0$ but $5x^2+3x+\dfrac34=5\left(x+\dfrac3{10}\right)^2+\dfrac34-5\cdot\dfrac9{100}>0$ for all real $x$
So, the denominator is non-zero real for all real $x$
|
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|
Which rational primes less than 50 are also Gaussian primes? Which rational primes less than 50 are also Gaussian primes?
My attempt: First we need to list all of the rational prime numbers that are less than $50$
$1,2,3,5,7,11,13,17,19,23,29,31,37,39,41,43,47$
A Gaussian prime is defined as a Gaussian integer that is not the product of Gaussian integers of similar norm.
The $Z[i]$ norm is $norm(a+bi)=\mid a+bi \mid^2 = a^2+b^2$ We have to look at the norms to determine which rational prime is a Gaussian prime.
For example $4+i$ is a Gaussian prime because our norm is $norm(4+i)=4^2+(i)^2=4^2+(-1)^2=16+1=17$. Since $17$ is also a prime in a set of all integers, $4+i$ is not the product of Gaussian integers of smaller norm because no such norms divide 17.
Also we need to separate the prime factors into two groups: any prime factors congruent to 3 mod 4 and any prime factors congruent to 1 mod 4
So all of the numbers congruent to 3 mod 4 which is also in $4k+3$ format are $3,7,11,19,23,31,39,43$ and $47$. These numbers can't be expressed as the sum of two squares.
All of the numbers congruent to 1 mod 4 which is also $4k+1$ are $1,5,13,17,29,37$, and $41,$ so these numbers can be expressed as the sum of squares and can be expressed as Gaussian Integers as well.
$1=1^2+0^2 = (1+0i)(1-0i)$
$5=2^2+1^2=(2+i)(2-i)$
$13 =2^2+3^2=(2+3i)(2-3i)$
$17=4^2+1^2=(4+i)(4-i)$
$29=5^2+2^2=(5+2i)(5-2i)$
$37=6^2+1^2=(6+i)(6-i)$
$41=5^2+4^2=(5+4i)(5-4i)$
Now my problem is...how do I really know that these numbers which are represented as a sum of squares and Gaussian Integers is really a Gaussian Prime. I know 17 counts as one, and for some reason 2 is not a Gaussian Prime because $norm(1+i)=1+(i^2)=1+((-1)^2)=1+1=2$ is smaller than $norm(2^2)=4$. So the norm has to be the same value in order to be considered as a Gaussian Prime?
|
What might really help you understand here is the concept of conjugate pairs. The conjugate of $a - bi$ is $a + bi$. And $(a - bi)(a + bi) = a^2 + b^2$, which should look very familiar because it's the norm function.
Given $p$ a "rational prime" (for lack of a better term) with nonzero real part but no imaginary part, its norm is $p^2$. But if $p = a^2 + b^2$, then it is composite in $\mathbb{Z}[i]$ and its factorization is $(a - bi)(a + bi)$. It is a very well known result that the only odd rational primes that are the sum of two squares are primes of the form $4k + 1$. For example, $13 = 3^2 + 2^2 = (3 - 2i)(3 + 2i)$, so $3 + 2i$ is a Gaussian prime but $13$ is not.
Thus, minding the special case $(1 - i)(1 + i) = 2$, to find the rational primes between $1$ and $50$ that are also Gaussian primes, we just take the rational primes in that range, remove $2$ and remove the rational primes of the form $4k + 1$. So we take $$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$$ and whittle it down to $$3, 7, 11, 19, 23, 31, 43, 47.$$
The reason we can be absolutely sure that these really are Gaussian primes is by verifying their divisors. For example, what are the divisors of $3$ in $\mathbb{Z}[i]$? We only need to look at Gaussian integers with a norm of $9$ or less to find: $1, i, -1, -i, 3, 3i, -3, -3i$. These are just units and $3$ multiplied by units. Gaussian primes have exactly $8$ divisors.
As for $39$, it can't be expressed as a sum of two squares, but it is a composite number. If it's composite in $\mathbb{Z}$, it sure as heck is also composite in $\mathbb{Z}[i]$. Its factorization is $39 = 3(3 - 2i)(3 + 2i)$.
|
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|
Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{1}{2}n^2-\frac{3}{4}n+\frac{9}{8}.$$
I am just wondering if someone could check my work to make sure I have the procedure correct.
|
Well you've made a mistake,I'm not sure where is your mistake anyway here's is a different solution
$$a_n=3a_{n-1}+n^2-3\\a_{n-1}=3a_{n-2}+n^2-2n-2\\a_n-a_{n-1}=3a_{n-1}+n^2-3-3a_{n-2}-n^2+2n+2\\a_n=4a_{n-1}-3a_{n-2}+2n-1\\a_{n-1}=4a_{n-2}-3a_{n-3}+2n-3\\a_n-a_{n-1}=4a_{n-1}-3a_{n-2}+2n-1-4a_{n-2}+3a_{n-3}-2n+3\\a_n=5a_{n-1}-7a_{n-2}+3a_{n-3}+2\\a_n=5a_{n-1}-7a_{n-2}+3a_{n-3}+2\\a_{n-1}=5a_{n-2}-7a_{n-3}+3a_{n-4}+2\\a_n-a_{n-1}=5a_{n-1}-7a_{n-2}+3a_{n-3}+2-5a_{n-2}+7a_{n-3}-3a_{n-4}-2\\a_n=6a_{n-1}-12a_{n-2}+10a_{n-3}-3a_{n-4}\\t^4-6t+12t-10t+3=0\\t_{1,2,3}=1\\t_4=3\\a_n=C_1+nC_2+n^2C_3+C_43^n\\1=C_1+C_4\\a_1=1,a_2=4,a_3=18\\a_1=C_1+C_2+C_3+3C_4\\a_2=C_1+2C_2+4C_3+9C_4\\a_3=C_1+3C_2+9C_3+27C_4\\C_1=0,C_2=-\frac{3}{2},C_3=-\frac{1}{2},C_4=1\\a_n=-\frac{3}{2}n-\frac{1}{2}n^2+3^n$$
|
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|
What is the maximum value of $f(\theta) = \sin\theta \cos\theta$ What is the maximum value of $f(\theta) = \sin\theta \cos\theta$ ?
|
Hint $1$:
$\sin \theta \cdot \cos \theta = \dfrac{\sin 2\theta}{2}$
Hint $2$:
$\sin \theta \cdot \cos \theta \leq \dfrac{\sin^2 \theta + \cos^2 \theta}{2} = \dfrac{1}{2}$.
Choose the hint you like most...
|
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|
Inequalities with arctan I don't understand how to solve inequalities with arctan, such as:
$$\arctan\left(\frac{1}{x^2-1}\right)\ge \frac{\pi}{4} $$
If someone could solve this and give me a very brief explanation of what they did, I'd be thankful.
|
Here are the steps
$$\arctan\left(\frac{1}{x^2-1}\right)\ge \frac{\pi}{4} $$
$$\frac{1}{x^2-1}\ge \tan\left(\frac{\pi}{4}\right)$$
$$\frac{1}{x^2-1}\ge 1$$
$$\frac{1}{x^2-1}-1\ge 0$$
$$\frac{1-x^2+1}{x^2-1}\ge 0$$
$$\frac{2-x^2}{x^2-1}\ge 0$$
$$\frac{\left(\sqrt{2}-x\right)\left(x+\sqrt{2}\right)}{(x+1)(x-1)}\ge 0$$
Can you take it from here?
|
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|
Solve $\lfloor \sqrt x \rfloor = \lfloor x/2 \rfloor$ for real $x$ I'm trying to solve $$\lfloor \sqrt x \rfloor = \left\lfloor \frac{x}{2} \right\rfloor$$
for real $x$.
Obviously this can't be true for any negative reals, since the root isn't defined for such.
My approach is the following: Let $x=:n+r$, $n \in \mathbb{N}_0, 0\leq r < 1$.
For the left hand side $\lfloor \sqrt {n+r} \rfloor = \lfloor \sqrt {\lfloor n+r\rfloor} \rfloor = \lfloor \sqrt n \rfloor$ holds (without further proof).
$$\left\lfloor \frac{x}{2} \right\rfloor = \left\lfloor \frac{n+r}{2} \right\rfloor = \left\{ \begin{array}{l l} \frac{n}{2} & \quad \text{for n even} \\ \frac{n-1}{2} & \quad \text{for n odd} \end{array}\right.$$
Now I don't really know if that'd lead me in the right direction, but I'll write my thoughts down anyways.
Let $\sqrt n =: n'+r'$, $n \in \mathbb{N}_0, 0\leq r < 1$.
Therefore $\lfloor n'+r' \rfloor = n'$. And $n = (n'+r')^2 = n'^2 +2n'r' +r'^2$
For which $n',r'$ holds $$ n' < n'^2 + 2n'r' + r'^2.$$
Well and now I'm stuck and don't know how to proceed.
I'd appreciate any help.
|
Clearly $x$ can’t be negative, and $\sqrt{x}$ and $\frac{x}2$ can’t be too far apart. They’re equal for $x=4$, and after that $\frac{x}2$ exceeds $\sqrt{x}$ by more and more as $x$ increases. Thus, we should focus initially on relatively small non-negative values of $x$. Points where one or the other floor increases are perfect squares and even integers, so we look at intervals that have such integers as endpoints.
For $x\in[0,1)$, we clearly have
$$\lfloor\sqrt{x}\rfloor=0=\left\lfloor\frac{x}2\right\rfloor\;.$$
For $x\in[1,2)$ we have $\lfloor\sqrt{x}\rfloor=1$, but $\left\lfloor\frac{x}2\right\rfloor=0$. So far, then, we have solutions $[0,1)\cup\{2\}$.
For $x\in[2,4)$ we have $$\lfloor\sqrt{x}\rfloor=1=\left\lfloor\frac{x}2\right\rfloor\;,$$
and for $x\in[4,6)$ we have
$$\lfloor\sqrt{x}\rfloor=2=\left\lfloor\frac{x}2\right\rfloor\;,$$
so the solution set has expanded to $[0,1)\cup[2,6)$. I claim that for $x\ge 6$, however, the separation between $\sqrt{x}$ and $\frac{x}2$ is too large, and we always have
$$\lfloor\sqrt{x}\rfloor<\left\lfloor\frac{x}2\right\rfloor\;.$$
First, for $x\in[6,9)$ we have $\lfloor\sqrt{x}\rfloor=2$, but $\left\lfloor\frac{x}2\right\rfloor\ge 3$. Now suppose that $n\in\Bbb Z^+$, $n\ge 3$, and $n^2\le x<(n+1)^2$. Then $\lfloor\sqrt{x}\rfloor=n$, and
$$\left\lfloor\frac{x}2\right\rfloor\ge\left\lfloor\frac{n^2}2\right\rfloor\;.$$
We’re done if we can show that $\left\lfloor\frac{n^2}2\right\rfloor\ge n+1$ for $n\ge 3$. And this is an easy proof by induction. The inequality certainly holds when $n=3$. Suppose that it holds for some $n\ge 3$. Then
$$\left\lfloor\frac{(n+1)^2}2\right\rfloor=\left\lfloor\frac{n^2+2n+1}2\right\rfloor=\left\lfloor\frac{n^2+1}2\right\rfloor+n\ge\left\lfloor\frac{n^2}2\right\rfloor+n\ge 2n+1>n+2\;,$$
and the induction step is complete.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1038396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
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Evaluate $\int^\pi_{-\pi}(1-a\cos\theta)^{-b-2}\log(1-a\cos\theta)d\theta$
$$ \int^\pi_{-\pi} \left(1-a\cos\theta\right)^{-b-2}
\log\left(1-a\cos\theta\right)d\theta$$
Under the condition $0<a<1$ and $b>0$.
Mathematica found the following form.
$$ 2\pi\left(a+1\right)^{-b-2}
\bigg({}_2F_1\left(\frac{1}{2},b+2;1;\frac{2a}{a+1}\right)
\left(\log\left(a+1\right)+\Psi\left(b+2\right)\right)
+\Gamma\left(b+2\right){}_3\tilde{F}_2^{\left(\{0,0,0\},
\{0,1\},0\right)}
\left(\frac{1}{2},b+2,b+2;1,b+2;\frac{2a}{a+1}\right)\bigg)$$
I need to use this in an algorithm, but calculation of $\displaystyle {}_3\tilde{F}_2^{\left(\{0,0,0\}, \{0,1\},0\right)}$ takes time. So I am looking for some simpler expression.
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Consider $\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}~d\theta$ ,
$\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_{-\pi}^0(1-a\cos\theta)^{-b-2}~d\theta+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_\pi^0(1-a\cos(-\theta))^{-b-2}~d(-\theta)+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=2\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=2\int_1^{-1}(1-ax)^{-b-2}~d(\cos^{-1}x)$
$=2\int_{-1}^1\dfrac{(1-ax)^{-b-2}}{\sqrt{1-x^2}}dx$
$=2\int_0^2\dfrac{(1-a(x-1))^{-b-2}}{\sqrt{1-(x-1)^2}}d(x-1)$
$=2\int_0^2x^{-\frac{1}{2}}(2-x)^{-\frac{1}{2}}(a+1-ax)^{-b-2}~dx$
$=2\int_0^1(2x)^{-\frac{1}{2}}(2-2x)^{-\frac{1}{2}}(a+1-2ax)^{-b-2}~d(2x)$
$=\dfrac{2}{(a+1)^{b+2}}\int_0^1x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}\left(1-\dfrac{2ax}{a+1}\right)^{-b-2}~dx$
$=\dfrac{2\pi}{(a+1)^{b+2}}~_2F_1\left(b+2,\dfrac{1}{2};1;\dfrac{2a}{a+1}\right)$
$\therefore$ $\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}\log(1-a\cos\theta)d\theta$
$=\int^\pi_{-\pi}\left(-\dfrac{d}{db}\left(1-a\cos\theta\right)^{-b-2}\right)d\theta$
$=-2\pi\dfrac{d}{db}\left(\dfrac{1}{(a+1)^{b+2}}~_2F_1\left(b+2,\dfrac{1}{2};1;\dfrac{2a}{a+1}\right)\right)$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1044120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Using Ratio test/Comparison test I have $\displaystyle\sum_{n=1}^{\infty}{\frac{(2n)!}{(4^n)(n!)^2(n^2)}}$ and need to show whether it diverges or converges.
I attempted to use the ratio test, but derived that the limit of $\dfrac{a_{n+1}}{a_n}=1$ and hence the test is inconclusive.
So I now must attempt to use the comparison test, but I am struggling to find bounds to compare it to to show either divergence or convergence.
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See below for my solution. Definition:
$$\lim_{x\rightarrow\infty} \frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty} \frac{\frac{(2n+2)!}{4^{n+1}\cdot((n+1)!)^2 \cdot (n+1)^2}}{\frac{(2n)!}{4^n\cdot (n!)^2 \cdot n^2}}$$
Expand the factorials, to prepare for removal due to ratio.
$$=\lim_{n\rightarrow\infty} \frac{\frac{(2n+2)(2n+1)\cdot(2n)!}{4\cdot 4^n\cdot (n+1)^2 \cdot (n!)^2 \cdot (n+1)^2}}{\frac{(2n)!}{4^n\cdot (n!)^2 \cdot n^2}}$$
Cancel out the factorial operations:
$$=\lim_{n\rightarrow\infty} \frac{\frac{(2n+2)(2n+1)}{4\cdot (n+1)^2 \cdot (n+1)^2}}{\frac{1}{n^2}}$$
Put into simple ratio form for final simplification:
$$=\lim_{n\rightarrow\infty} \frac{(2n+2)(2n+1)\cdot n^2}{4\cdot (n+1)^4}$$
Expand the powers of polynomials:
$$=\lim_{n\rightarrow\infty} \frac{4n^4+6n^3+2n^2}{4\cdot (n^4+4n^3+6n^2+4n+1)}$$
Approximation for simplicity. Throwing out the lower order positive terms in the denominator makes the fraction larger, so implied comparison test. Ditto for swallowing the numerator terms in the highest order.
$$\approx\lim_{n\rightarrow\infty} \frac{4n^4}{4n^4}\rightarrow 1$$
Therefore, the test is inconclusive.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1045156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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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.