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How to find number of real roots of a transcendental equation? The number of real roots of the equation
$$2\cos\left(\frac{x^2+x}6\right)=2^x+2^{-x}$$
Another question is...
can we use descartes rule of sign in here or in any transcendental equation ?
|
Notice that
$$2\cos\left(\frac{x^2+x}6\right) \le 2$$
$$2^x+2^{-x} \ge 2$$
First is true since $\cos\left(\frac{x^2+x}6\right) \le 1$ and the second is equivalent to $\left(2^{\frac{x}{2}} - \frac{1}{2^{\frac{x}{2}}} \right)^2 \ge 0$ which is always true. So in order to have an equality the second inequality also has to be an equality and happens only when $2^{\frac{x}{2}} = \frac{1}{2^{\frac{x}{2}}} \leftrightarrow2^x=1\leftrightarrow x = 0$.
|
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|
Use partial fractions to find the integral. Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{align*}
|
Hint: Write out the fraction given as $\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$ and equate the numerator to $2x^2+5x+3$. And the ensuing integral should be easy.
|
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|
How do I simplify this exponential expression?: $ 2(-3x^{-2}y^3)^{-1} \cdot (-3x^{-3}\cdot y)^2 $ How do I simplify this expression?
Simplify:
$$ 2(-3x^{-2}y^3)^{-1} \cdot (-3x^{-3}\cdot y)^2 $$
I tried and didn't get the answer.
|
Hint: Note that $$ \left( ax^p \cdot bx^q \right)^e = \left( ab \cdot x^{p+q} \right)^e = \left( ab \right)^e \cdot x^{e\cdot(p+q)}. $$These are just usages of exponent properties.
Solution, here: $$ \begin {align*} 2 \cdot \left( - 3 x^{-2} y^3 \right)^{-1} \cdot \left( -3 x^{-3} y \right)^2 &= 2 \cdot \left( - \dfrac {3y^3}{x^2} \right)^{-1} \cdot \left( - \dfrac {3y}{x^3} \right)^2 \\&= 2 \cdot \left( - \dfrac {x^2}{3y^3} \right) \cdot \left( \dfrac {9y^2}{x^6} \right) \\&= -6 \cdot x^{-4} \cdot y^{-1} = \boxed {-\dfrac {6}{x^4y}}. \end {align*} $$
|
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Proving a trigonometric identity: $\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x$ I really need some help with this question. I need to prove this identity:
$$\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x.$$
|
$\displaystyle\frac{2\sin^3x}{1-\cos x}=2\sin^2x\cdot\frac{\sin x}{1-\cos x}=2\sin^2x\cdot\frac{1+\cos x}{\sin x}$(using this)
$\displaystyle=2\sin x(1+\cos x)=2\sin x+2\sin x\cos x=\cdots$
|
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Help With Series (Apostol, Calculus, Volume I, Section 10.9 #9) I am looking for help finding the sum of a particular series from Apostol's Calculus (Volume I, Section 10.9, Problem 9). The trouble is that I can find the correct answer, but only using methods that aren't available at this point in the text, or with too much trouble. I would like some guidance on reducing the following series to a linear combination of convergent geometric or telescoping series, the methods by means of which I'm supposed to solve the problem. The series in question is $$\sum_{n = 1}^{\infty} \frac{(-1)^{n - 1}(2n + 1)}{n(n + 1)}$$ Decomposing the series using partial fractions and observing the series term-by-term, we see that
$$\begin{align*}
\sum_{n = 1}^{\infty} \frac{(-1)^{n - 1}(2n + 1)}{n(n + 1)} & ~ = ~ \sum_{n = 1}^{\infty} (-1)^{n - 1}\left(\frac{1}{n} + \frac{1}{n + 1}\right) \\
& ~ = ~ \sum_{n = 1}^{\infty}\frac{1}{n} - \frac{1}{n + 1} \\
& ~ = ~ 1 - \lim_{n \rightarrow \infty} \frac{1}{n} \\
& ~ = ~ 1
\end{align*}$$
The only problem is justifying the second line. The intuitive justification for this step is as follows
$$\begin{align*}
& \left(\frac{1}{1} + \frac{1}{2}\right) - \left(\frac{1}{2} + \frac{1}{3}\right) + \left(\frac{1}{3} + \frac{1}{4}\right) - \left(\frac{1}{4} + \frac{1}{5}\right) + \left(\frac{1}{5} + \frac{1}{6}\right) - \left(\frac{1}{6} + \frac{1}{7}\right) + \cdots \\
~ = ~ & \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{7}\right) + \cdots
\end{align*}$$
The problem is that I don't have any theorems about rearrangement of series at my disposal. Thus, justifying the second line devolves into demonstrating equivalence of the corresponding sequences. I'm using the test for equivalence of Cauchy sequences, while hoping it applies in this case
$$ \forall \varepsilon > 0 \exists N > 0 \forall n \geq N ~.~ \left| (-1)^{n - 1}\left(\frac{1}{n} + \frac{1}{n + 1}\right) - \left(\frac{1}{n} - \frac{1}{n + 1}\right)\right| \leq \varepsilon $$
Now this isn't particularly difficult, just tedious. However, given that no prior examples involved proving such equivalences, I assume that I'm missing something obvious that would let me transform the given series into a telescoping series. Either that, or I've messed up somewhere and need some correction. Any help would be appreciated.
|
Take a look at the sum of
the first $m$ terms.
$\begin{align}
\sum_{n = 1}^{m} \frac{(-1)^{n - 1}(2n + 1)}{n(n + 1)}
&=\sum_{n = 1}^{m} (-1)^{n - 1}\left(\frac{1}{n} + \frac{1}{n + 1}\right)\\
&=\sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n} + \sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n + 1}\\
&=\sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n} + \sum_{n = 2}^{m+1} (-1)^{n}\frac{1}{n}\\
&=1+\sum_{n = 2}^{m} (-1)^{n - 1}\frac{1}{n} + (-1)^{m+1}\frac{1}{m+1}+\sum_{n = 2}^{m} (-1)^{n}\frac{1}{n}\\
&=1-\sum_{n = 2}^{m} (-1)^{n}\frac{1}{n} + (-1)^{m+1}\frac{1}{m+1}+\sum_{n = 2}^{m} (-1)^{n}\frac{1}{n}\\
&=1+ (-1)^{m+1}\frac{1}{m+1}\\
\end{align}
$
This clearly converges to $1$.
|
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|
Simplification of $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}}$ I'm having trouble understanding how this expression:
$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}} \cdot \left(\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}\right)=$$
got to this one:
$$\frac2{\sqrt2\cdot\sqrt{2+\sqrt2}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\dots}$$
|
Now that I am reading this correctly I think I can make some sense of the claim.
Let $x=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$ and $ y=\sqrt{2-\sqrt{2+\sqrt{2+\cdots}}}$ The "equation" you ask about is $$y=y\frac{x}{x}=\frac{2}{\sqrt{2}\cdot\sqrt{2+\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\ \cdots} $$ Since the righthand side has a finite numerator and denominator an infinite product (of factors at least $\sqrt{2}$) this says $$y=0.$$ That does make sense, at least informally, since $x$ (if it makes any sense) is positive and satisfies $x=\sqrt{2+x}$ meaning $x^2=2+x$ and $x=2.$ Then, indeed, $$y=\sqrt{2-x}=\sqrt{0}=0. $$ But why that particular way of writing $0?$
The presumed meaning of the expression for $x$ is
$x=\lim_{n \to \infty}x_n$ where $x_0=0$, $x_1=\sqrt{2}$, $ x_2=\sqrt{2+\sqrt{2}}$ and in general $x_{n+1}=\sqrt{2+x_n}.$
Then we might also define
$y=\lim_{n \to \infty}y_n$ where $y_1=\sqrt{2}$, $y_2=\sqrt{2-\sqrt{2}}$, $y_3=\sqrt{2-\sqrt{2+\sqrt{2}}}$ and in general $y_{n+1}=\sqrt{2-x_n}.$
So the thing to be explained is $$y=y\frac{x}{x}\stackrel{?}{=}\frac{2}{\prod_{j=1}^{\infty}x_j} .\tag{*}$$
The sense I make of this is as the limit (in a somewhat informal sense) of $$y_n=y_n\frac{x_n}{x_n}=\frac{2}{\prod_{j=1}^nx_j}. \tag{**}$$ This can be shown by induction.
The base case is $y_1=\frac{2}{x_1}$ i.e. $$\sqrt{2}=\sqrt{2}\frac{x_1}{x_1}=\sqrt{2}\frac{\sqrt{2}}{x_1}=\frac{2}{x_1} $$
then the needed induction step is $y_{n+1}=\frac{y_{n}}{x_{n+1}}$ and indeed:
$$y_{n+1}=y_{n+1}\frac{x_{n+1}}{x_{n+1}}=\sqrt{2-x_{n}}\frac{\sqrt{2+x_{n}}}{x_{n+1}}=\frac{\sqrt{4-x_{n}^2}}{x_{n+1}}=\frac{y_{n}}{x_{n+1}}. $$
The last step makes use of $x_{n}^2=2+x_{n-1}$ so $\sqrt{4-x_{n}^2}=\sqrt{2-x_{n-1}}=y_{n}$
So, given $(**)$, I can make sense of $$y=\lim_{n \to \infty}y_n=\lim_{n \to \infty}y_n\frac{x_n}{x_n}=\lim_{n \to \infty}\frac{2}{\prod_{j=1}^nx_j}.$$ But, since my proof of $(**)$ is by induction, I do not see how one can, in one step, say $$y\frac{x}{x}=\frac{2}{\prod_{j=1}^{\infty}x_j} .$$
One way to more formally justify the value $x=2$ is to show $2-\frac{1}{2^{n-1}} \le x_n \lt 2$.
Finally, it does not directly relate to the specific question you ask, but see this question for a proof that $$y_n=2\sin{\frac{\pi}{2^{n+1}}}$$ so that
$$ \frac{\pi}{2^n}-\frac{\left(\pi/2^n\right)^3}{24} \lt y_n \lt \frac{\pi}{2^n} $$
|
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|
Proving an inequality by induction and figuring out intermediate inductive steps? I'm working on proving the following statement using induction:
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
Fair enough. I'll start with the basis step:
Basis Step: (n=1)
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
$$ \frac{1}{1^2} \le \frac{2}{1+1} $$
$$ 1 \le 1 \checkmark $$
Inductive Step:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
$$ \sum_{r=1}^n \frac{1}{r^2} + \frac{1}{(n+1)^2} \le \frac{2n}{n+1} + \frac{1}{(n+1)^2} $$
$$ \sum_{r=1}^{k+1} \le \frac{2n}{n+1} + \frac{1}{(n+1)^2} $$
My goal is to prove $\forall_{n\ge1} s(n) \implies s(n+1) $ or that this inequality holds true for all $n\ge1$.
I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
However, I'm not quite sure what needs to done to obtain this after attempting a few times.
|
For the inductive step you have to show that $ \sum_{r=1}^{n} \frac{1}{r^2} \le \frac{2n}{n+1} $ implies $ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n+2}{n+2} $.
Start with $ \sum_{r=1}^{n+1} \frac{1}{r^2}$ and show that is less or equal than $\frac{2n}{n+1}+\frac{1}{(n+1)^2}$ (use the inductive hypothesis). Then, show that the last quantity is less or equal than $\frac{2n+2}{n+2}$.
|
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Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx=\int_0^{\frac{\pi}{4}}\int_0^{b\sec(\theta)}r\sqrt{a^2-r^2}dr\,d\theta$$
$$=\int_0^{\frac{\pi}{4}}\frac{-1}{3}\left[(a^2-r^2)\sqrt{a^2-r^2}\right]_0^{b\sec(\theta)}d\theta$$
$$=\frac{1}{3}\int_0^{\frac{\pi}{4}}\left[a^3-(a^2-b^2\sec^2(\theta))\sqrt{a^2-b^2\sec^2(\theta)}\right]d\theta$$
but evaluating above integral is very difficult and antiderivative is very complexity! see here.
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$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}~dy~dx$
$=\int_0^b\left[\dfrac{y\sqrt{a^2-x^2-y^2}}{2}+\dfrac{a^2-x^2}{2}\sin^{-1}\dfrac{y}{\sqrt{a^2-x^2}}\right]_0^x~dx$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions)
$=\int_0^b\dfrac{x\sqrt{a^2-2x^2}}{2}dx+\int_0^b\dfrac{a^2-x^2}{2}\sin^{-1}\dfrac{x}{\sqrt{a^2-x^2}}dx$
Can you take it from here?
|
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Boolean Algebra simplify minterms I have this equation
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
and need to simplify it. I have got as far as I can and spent a good 2 hours at it. I've realized I now need to use De Morgan's law to continue however I am baffled as to which rule to use. If someone could send me in the right direction that would be great!
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot\bar{C} + B\cdot\bar{C} + B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B(\bar{C} + C)) $$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B) $$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(C + B) $$
Thanks
|
If you use a karnaugh map:
$$
\begin{array}{c|c|c|c|c}
C, AB & 00 & 01 & 11 & 10 \\ \hline
0 & 1 & & 1 & \\ \hline
1 & & & 1 & 1\\ \hline
\end{array}
\equiv
\bar{A}\bar{B}\bar{C} + A\bar{B} C + AB\bar{C} + ABC
$$
Which suggests Xor of the 2 groups:
$$
\begin{array}{c|c|c|c|c}
C, AB & 00 & 01 & 11 & 10 \\ \hline
0 & & & 1 & 1\\ \hline
1 & & & 1 & 1\\ \hline
\end{array}
\equiv A
$$
$$
\begin{array}{c|c|c|c|c}
C, AB & 00 & 01 & 11 & 10 \\ \hline
0 & 1 & & & 1\\ \hline
1 & & & & \\ \hline
\end{array}
\equiv \overline B \cdot \overline C
$$
Which gives:
$$A \oplus (\overline B \cdot \overline C)$$
|
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For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$ For any real numbers $a,b,c$ show that:
$$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \leq b \leq c$ because of the symmetry.
$a \leq b \leq c$ implies that $0 \leq b-a \leq c-a$. Since both sides are positive and $y=x^2$ is an increasing function for positive numbers we conclude that $(b-a)^2 \leq (c-a)^2$ and since $(a-b)^2=(b-a)^2$ we obtain $(a-b)^2 \leq (c-a)^2$.
Therefore:
$\min\{(a-b)^2,(b-c)^2,(c-a)^2\}=$ $$\min\{(b-c)^2,\min\{(a-b)^2,(c-a)^2\}\}=
\min\{(a-b)^2,(b-c)^2\}.$$
So, we have to prove the following inequality instead:
$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
Now two cases can happen:
*
*$$(a-b)^2 \leq (b-c)^2 \implies |a-b| \leq |b-c|=c-b \implies b-c \leq a-b \leq c-b \implies 2b-c \leq a \implies b \leq \frac{a+c}{2}.$$
*$$(b-c)^2 \leq (a-b)^2 \implies |c-b| \leq |a-b|=b-a \implies a-b \leq c-b \leq b-a \implies c \leq 2b-a \implies b \geq \frac{a+c}{2}.$$
So, this all boils down to whether $b$ is greater than the arithmetic mean of $a$ and $b$ are not.
Now I'm stuck. If $a,b,c$ had been assumed to be positive real numbers it would've been a lot easier to go forward from this step. But since we have made no assumptions on the signs of $a,b$ and $c$ I have no idea what I should do next.
Maybe I shouldn't care about what $\displaystyle \min\{(a-b)^2,(b-c)^2\}$ is equal to and I should continue my argument by dealing with $(a-b)^2$ and $(b-c)^2$ instead. I doubt that using the formula $\displaystyle \min\{x,y\}=\frac{x+y - |x-y|}{2}$ would simplify this any further. Any ideas on how to go further are appreciated.
|
Note that LHS does not change if you replace $(a,b,c)$ by $(a-t,b-t,c-t)$. Thus we can first minimize RHS with respect to $t$,after we replace $(a,b,c)$ by $(a-t,b-t,c-t)$. It turns out that RHS is minimized when $\displaystyle t = \frac{a+b+c}{3}$, with minimum value being
$$\frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2)$$
So it suffices to show that $$\min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2)$$
Without loss of generality, assume that $$a \leq b \leq c$$Then clearly, $\min\{(c-b),(b-a), (c-a)\} = \min\{(c-b),(b-a)\}$, so
$$\min\{(c-b)^2,(b-a)^2, (c-a)^2\} = \min\{(c-b)^2,(b-a)^2\}$$
Moreover,
$$(c-a) = (c-b)+(b-a) \ge 2 \min\{(c-b),(b-a)\}$$
and
$$\begin{eqnarray}
\frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2) &\ge& \frac{1}{6} (\min\{(c-b),(b-a)\})^2 + (\min\{(c-b),(b-a)\})^2 + (2\min\{(c-b),(b-a)\})^2 \\
&=& \min\{(c-b)^2,(b-a)^2\} \\
&=& \min\{(c-b)^2,(b-a)^2, (c-a)^2\}
\end{eqnarray}$$
|
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|
Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.
|
So, consider the following:
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \int_0^{1} dx \frac{\log{x}}{(1+x^2)^2} + \int_1^{\infty} dx \frac{\log{x}}{(1+x^2)^2}$$
Sub $x \mapsto 1/x$ in the latter integral on the RHS and get that
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \int_0^{1} dx \frac{1-x^2}{(1+x^2)^2} \log{x}$$
Note that
$$\frac{1-x^2}{(1+x^2)^2} = \sum_{m=0}^{\infty} (-1)^m (2 m+1) x^{2 m}$$
So we get
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \sum_{m=0}^{\infty} (-1)^m (2 m+1) \int_0^1 dx \, x^{2 m} \log{x}$$
Using the fact that
$$\int_0^1 dx \, x^{2 m} \log{x} = -\frac{1}{(2 m+1)^2}$$
we get that
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = -\sum_{m=0}^{\infty} \frac{(-1)^m}{2 m+1} = -\frac{\pi}{4}$$
|
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How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
Case $3$: $6$ games: Team A wins $4$ games, team B wins $2$ = $\binom{6}{4}\binom{6}{2}-2$...minus $2$ for the possibility of team A winning the first four games; and the middle four (games $2,3,4,5$), in which case there would be no game $6$.
Case $4$: $7$ games: Team A wins $4$ games, team B wins $3$ = $\binom{7}{4}\binom{7}{3}-3$...minus $3$ for the possibility of team A winning the first four games; games $2,3,4,5$; and games $3,4,5,6$.
Total = sum of the $4$ cases multiplied by $2$ since the question is asking for $2$ teams.
Is this correct?
|
Let the best of $n$ series be decided after $k$ games. This will happen if in the preceding $k-1$ games $A$ also wins $\lfloor n/2 \rfloor$ and wins the $k^{th}$ game. Hence,
$$C(k) = \dbinom{k-1}{\lfloor n/2 \rfloor}$$
Hence, the total number of ways is
$$\sum_{k=\lfloor n/2 \rfloor+1}^n C(k)$$
In your case, set $n=7$ to get the answer.
|
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|
Evaluating $\iint_D \sqrt{4x^2-y^2}\;\ \mathrm dx \ \mathrm dy$ I have to evaluate $\displaystyle\iint_Df(x,y)\;dxdy$ for $f(x,y) = \sqrt{4x^2-y^2}$ with $D = \{(x,y)\in\mathbb{R}^2: 0\leq x \leq 1, 0\leq y \leq x\}$.
It seems that i can't solve for $\displaystyle\int_0^1 \displaystyle\int_0 ^x\sqrt{4x^2-y^2} dydx$ but working with iterated integrals I could solve for $\displaystyle\int_0^1 \displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\sqrt{4x^2-y^2} dxdy$.
A small sketch of what i did until now:
$\displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\sqrt{4x^2-y^2} dx = $
$\displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\displaystyle\frac{4x^2}{\sqrt{4x^2-y^2}}dx - \displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\displaystyle\frac{y^2}{\sqrt{4x^2-y^2}}dx = $
$
4\left(\displaystyle\frac{x^2}{\sqrt{4x^2-y^2}} -\displaystyle\frac{1}{4}\displaystyle\int \sqrt{4x^2-y^2}dx\right)-y^2\displaystyle\frac{y}{2|y|}\arcsin\left(\displaystyle\frac{2}{y}x \right) $
$\implies \displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\sqrt{4x^2-y^2} dx = $
$\displaystyle\frac{1}{2}\left(\displaystyle\frac{4x^2}{\sqrt{4x^2-y^2}}-y^2\displaystyle\frac{y}{2|y|}\arcsin\left(\displaystyle\frac{2}{y}x \right) \right)$.
[wrong]
And it seems that after I evaluate the last one i get $\displaystyle\frac{y^2}{\sqrt{3y^2}} = \displaystyle\frac{y^2}{\sqrt{3}|y|}$
Then if I solve the integral of the expresion above for x I get $\displaystyle\frac{1}{2\sqrt{3}}$ if $y>0$ and $\displaystyle\frac{-1}{2\sqrt{3}}$ if $y<0$.[/wrong]
I'm almost sure i made a mistake somewhere. Can someone find any errors?
|
I am posting this about finding the proper limits not to solve the definite integrals. We have $$D = \{(x,y)\in\mathbb{R}^2: 0\leq x \leq 1, 0\leq y \leq x\}$$ and it seems you want to change the order of the integrals. OK! As you might already do, the following plot makes the region of $D$ clear.
Changing the order of integrals makes $y$ to be the outer variable and $x$ to be the inner one. As you see $D$, I made a line parallel to $x$ -axe intersecting the region. Now if we read the points (follow the arrows), the line enters in with $x=y$ and enters out with $x=1$, so we have $$x|_{y}^1,~~y|_{0}^1$$ It seems switching the limits for $y$ is needed in your second integrals.
|
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|
How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$? Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this:
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.
|
Your idea is good, write
$$\frac{4x+4}{x^4+x^3+2x^2}=4\frac{x+1}{x^2 (x^2+x+2)}$$
Then, partial fraction will be of the form
$$\frac{x+1}{x^2 (x^2+x+2)}=\frac{a}{x} + \frac{b}{x^2} + \frac{cx+d}{x^2+x+2}$$
There are several methods to identify coefficients from this. The simplest IMHO is to rewrite the right hand side with the same denominator:
$$\frac{ax+b}{x^2} + \frac{cx+d}{x^2+x+2} = \frac{(ax+b)(x^2+x+2) + (cx+d)x^2}{x^2(x^2+x+2)}$$
Then develop, and write that the numerator must be $x+1$, thus you get equations by writing that the coefficients of $x^3$ and $x^2$ are $0$, and the coefficient of $x$ and the constant are $1$. Four linear equations, fours unknowns.
Is it enough for you to complete the process?
|
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|
limit problem (with roots) Is it possible to evaluate this limit without graphing or guessing
(ie to replace it by a simpler function)
$$\lim_{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I tried normalizing by multiplying by the conjugate (both denominator and numerator) didn't work.
|
$$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{(\sqrt{6-x}-2)\times (\sqrt{6-x}+2)\times(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)\times(\sqrt{3-x}+1)\times (\sqrt{6-x}+2)}$$
$$=\frac{\left((\sqrt{6-x})^2-4\right)\times(\sqrt{3-x}+1)}{\left((\sqrt{3-x})^2-1\right)\times (\sqrt{6-x}+2)}=\frac{\overbrace{\left((6-x)-4\right)}^{(2-x)}\times(\sqrt{3-x}+1)}{\underbrace{\left((3-x)-1\right)}_{(2-x)}\times (\sqrt{6-x}+2)}$$
$$=\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2},~~x\ne 2$$
|
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|
Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$
Has someone a different approach to solve this limit?
|
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$$
$$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^2\tan^2x}\right)$$
$$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)\times \lim_{x\to 0}\left(\frac{x}{\tan x}\right)^2=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)$$
Now, using L' Hospital Rule successively for $\frac{0}{0}$ form ,
$$=\lim_{x\to 0}\left(\frac{2\tan x\sec^2x-2x}{4x^3}\right)=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sec^2x \tan x-x}{x^3}\right)$$
$$=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sec^2x\sec^2x+2\tan^2x\sec^2 x-1}{3x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(\frac{\sec^4x-1+2\tan^2x\sec^2 x}{x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(\frac{(\sec^2x-1)(\sec^2x+1)+2\tan^2x\sec^2 x}{x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(\frac{\tan^2x(\sec^2x+1)+2\tan^2x\sec^2 x}{x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(3\sec^2x+1\right) \times \lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{4}{6}\times 1=\frac{2}{3}$$
|
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|
Show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3\cdot4^{n+2} +5\right)$ using congruences Using congruence theory, show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3 \cdot 4^{n+2} +5\right)$. The proof is quite simple with induction, but how can it be proved with congruences?
|
HINT:
As $10\equiv1\pmod 9, 10^n \equiv1^n\equiv1$
$\displaystyle 4^{n+2}=(1+3)^{n+2}$
$\displaystyle=1+\binom{n+2}13+\binom{n+2}23^2+\cdots+3^{n+2}\equiv1+(n+2)3\pmod 9\equiv3n+7$
Can you take it from here?
Another way
$$10^n+3\cdot4^{n+2}+5=(10^n-1^n)+3\left(4^{n+2}-1^{n+2}\right)+1+3+5$$
Using congruence,
$\displaystyle4\equiv1\pmod 3\implies 4^{n+2}\equiv1^{n+2}\pmod3\equiv1$
and $10^n\equiv1\pmod9$
|
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|
Fibonacci sequence proof Prove the following:
$$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$
For $n \ge 2$
Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$
So, here are my first steps:
$$ \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = \\
& = \frac 12(f_{3k+2}-1) + f_{3k+1} + f_{3k} + f_{3k+1} \\
& = \frac 12(f_{3k+2}-1) + f_{3k+1} + \frac 12(f_{3k+2}-1) + f_{3k+1} \\
& = f_{3k+2}-1 + 2 \cdot f_{3k+1}
\end{align} $$
And the fun stops around here. I don't see how to get to the conclusion: $\frac 12(f_{3k+5}-1) \\ $. Any help from this point would be great.
|
One way to derive/prove these identities is to start from
$$\begin{pmatrix} f_{n-1} & f_n \\ f_n &f_{n+1} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^n$$
Note that
\begin{align}
& \begin{pmatrix} 0 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} f_2+f_5+ \ldots +f_{3n-1} & f_3+f_6+ \ldots +f_{3n} \\ f_3+f_6+\ldots +f_{3n} & f_4+f_7+\ldots+f_{3n+1} \end{pmatrix} \\
& =\left( \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^3 - I \right)\left(\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^3+\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^6+\ldots +\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{3n}\right) \\
& =\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{3n+3}-\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^3 \\
&=\begin{pmatrix} f_{3n+2} & f_{3n+3} \\ f_{3n+3} & f_{3n+4} \end{pmatrix}-\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}
\end{align}
|
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|
Prove that $x_1^n+x_2^n$ is an integer and is not divisible by $5$ If $x_1$ and $x_2$ are the roots of the polynomial $x^2-6x+1$ then , for every non-negative integer, prove that $x_1^n+x_2^n$ is an integer and is not divisible by $5$ .
My trying:
$ x_1 = 3+2\sqrt{2}$ and $ x_1 = 3-2\sqrt{2}$
So $ x_1^n +x_2^n = (3+2\sqrt{2})^n + (3-2\sqrt{2})^n$ = $ \dfrac{(3+2\sqrt{2})^{2n}+1}{(3+2\sqrt{2})^n}$
Then what should I do to solve this problem ?
|
We have $x_1 + x_2 = \text{ Sum of roots }=6$. From the equation, we have
$$x_1^2 -6x_1 + 1 = 0 \text{ and }x_2^2 -6x_2 + 1 = 0$$
Adding both we get
$$x_1^2 + x_2^2 = 6\underbrace{(x_1+x_2)}_{\text{Integer}} - 2 = \text{Integer}$$
Now use strong induction and make use of the fact that
$$x_1^{n+2} -6x_1^{n+1} + x_1^n = 0 \text{ and }x_2^{n+2} -6x_2^{n+1} + x_2^n = 0$$
i.e.,
$$x_1^{n+2} + x_2^{n+2} = 6(x_1^{n+1}+x_2^{n+1}) - (x_1^n + x_2^n)$$
Use the same idea to show that
$$x_1^n + x_2^n \equiv \begin{cases} 1 \pmod5 & n\equiv 1 \pmod{4}\\
4 \pmod5 & n\equiv 0, 2 \pmod{4}\\ 3 \pmod5 & n\equiv 3 \pmod{4}\\\end{cases}$$
|
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|
$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
|
$$x = \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
$$x = \sqrt{7x}$$
$$x^2 - 7x = 0$$
$$x(x - 7) =0 \implies x = 7$$
Because $\sqrt{7} > 0$ we reject the $x=0$ solution.
|
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|
How prove this inequality $\sin{\sin{\sin{\sin{x}}}}\le\frac{4}{5}\cos{\cos{\cos{\cos{x}}}}$ Nice Question:
let $x\in [0,2\pi]$, show that:
$$\sin{\sin{\sin{\sin{x}}}}\le\dfrac{4}{5}\cos{\cos{\cos{\cos{x}}}}?$$
I know this follow famous problem(1995 Russia Mathematical olympiad)
$$\sin{\sin{\sin{\sin{x}}}}<\cos{\cos{\cos{\cos{x}}}}$$
This problem solution can see :http://iask.games.sina.com.cn/b/19776980.html
and everywhere have solution in china BBS
I post this problem solution
case1: if $x\in[\pi,2\pi]$,then
$$\cos{\cos{\cos{\cos{x}}}}>0,\sin{\sin{\sin{\sin{x}}}}\le 0$$
so
$$\cos{\cos{\cos{\cos{x}}}}>\sin{\sin{\sin{\sin{x}}}}$$
case2: if $x\in[0,\dfrac{\pi}{2}]$,then we have
$$\cos{x}+\sin{x}\le\sqrt{2}<\dfrac{\pi}{2}\Longrightarrow 0\le \cos{x}<\dfrac{\pi}{2}-\sin{x}$$
so
$$\cos{\cos{x}}>\cos{\left(\dfrac{\pi}{2}-\sin{x}\right)}=\sin{\sin{x}}$$
$$\sin{\cos{x}}<\sin{\left(\dfrac{\pi}{2}-\sin{x}\right)}=\cos{\sin{x}}$$
then
$$\cos{\cos{\cos{x}}}<\cos{\sin{\sin{x}}}$$
so
$$\cos{\cos{\cos{x}}}+\sin{\sin{\sin{x}}}<\cos{\sin{\sin{x}}}+\sin{\sin{\sin{x}}}<\dfrac{\pi}{2}$$
so
$$\cos{\cos{\cos{x}}}<\dfrac{\pi}{2}-\sin{\sin{\sin{x}}}$$
then
$$\cos{\cos{\cos{\cos{x}}}}>\cos{\left(\dfrac{\pi}{2}-\sin{\sin{\sin{x}}}\right)}=\sin{\sin{\sin{\sin{x}}}}$$
case3: if $x\in (\dfrac{\pi}{2},\pi)$,then let
$y=x-\dfrac{\pi}{2}$,so
$$\cos{\cos{\cos{\sin{y}}}}>\sin{\sin{\cos{\sin{y}}}}$$
and since
$f(t)=\sin{\sin{t}}$ is increasing,then
$$f(\cos{\sin{y}})>f(\sin{\cos{y}})\Longrightarrow \sin{\sin{\cos{\sin{y}}}}>\sin{\sin{\sin{\cos{y}}}}$$
so
$$\cos{\cos{\cos{\sin{y}}}}>\sin{\sin{\sin{\cos{y}}}}$$
so
$$\cos{\cos{\cos{\cos{x}}}}>\sin{\sin{\sin{\sin{x}}}}$$
But I found this $\dfrac{4}{5}$ maybe is strong,
so if $x\in[\pi,2\pi]$,then we have
$$\dfrac{4}{5}\cos{\cos{\cos{\cos{x}}}}\ge 0>\sin{\sin{\sin{\sin{x}}}}$$
But for the case $x\in [0,\pi]$, I can't prove this
$$4\cos{\cos{\cos{\cos{x}}}}\ge 5\sin{\sin{\sin{\sin{x}}}}$$
Thank you very much！
|
We still start from the original Russian Olympiad Problem:
$\cos \cos \cos \cos x> \sin \sin \sin \sin x$. It could have another numerical proof simply by doing in a calculator:
We have $-1\leq \cos x \leq 1, \text{that is }\cos 1\leq \cos \cos x\leq 1,
\text{that is }\cos 1 \leq \cos \cos \cos x \leq \cos \cos 1$.
Finally, we have,
$$
~0.6542 \simeq \cos \cos \cos 1\leq \cos \cos \cos \cos x \leq \cos \cos 1.
$$
Similarly, we have
$$
-\sin \sin \sin \sin 1 \leq \sin \sin \sin \sin x \leq \sin \sin \sin 1 \simeq0.6784...
$$
If the equation has the solution, that is
$$
\sin x\geq \sin^{-1} \sin^{-1}\sin ^{-1} \cos \cos \cos 1 \simeq 0.6086...
$$
Thus, we have
$$
|\cos x|\leq 0.7835...
$$
Therefore, $\cos \cos \geq 0.7013 \to \cos \cos \cos x \leq 0.7639 \to \cos \cos \cos \cos \cos x \geq 0.7221...$
Thus, it is not possible to have $\sin \sin \sin \sin \sin \leq 0.6784.$ The inequality holds.
So, the $\frac{4}{5}$ is still not a strong constant, and inequality proof as similar.
|
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|
Why is $\sum_{k=0}^{n} f(n,k) = F_{n+2}$? If $f(n,k)$ is the number of $k$ size subsets of $[ n ] = { 1 , \ldots , n }$ which do not contain a pair of consecutive numbers, how can I show that $\sum_{k=0}^{n} f(n,k) = F_{n+2}$?
($F_{n}$ is the nth Fibonacci number: $F_{0} = 0, F_{1} = 1, F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 2$.)
Thanks!
|
Here is a solution that uses generating functions. Suppose the subsets
are ordered with the smallest element first. Choosing this element
corresponds to the generating function
$$\frac{z}{1-z}.$$
The remaining elements of the subset are chosen by adding a series of
$k-1$ gap values $\ge 2$ consecutively starting with the first element,
giving a contribution of
$$\left(\frac{z^2}{1-z}\right)^{k-1}.$$
Finally we need to collect all subsets with largest element at most
$n$, giving a factor of
$$\frac{1}{1-z}.$$
We thus have $f(n, 0)=1$ and for $k\ge 1,$
$$f(n, k) = [z^n] \frac{1}{1-z} \times \frac{z}{1-z} \times
\left(\frac{z^2}{1-z}\right)^{k-1}
= [z^n] \frac{z^{2k-1}}{(1-z)^{k+1}}.$$
This implies for the sum that
$$\sum_{k=0}^n f(n, k) =
1 + [z^n] \sum_{k=1}^n \frac{z^{2k-1}}{(1-z)^{k+1}}
= 1 + [z^n] \frac{z}{(1-z)^2}
\sum_{k=1}^n \frac{z^{2(k-1)}}{(1-z)^{k-1}}.$$
Putting the sum into closed form we obtain
$$1 + [z^n] \frac{z}{(1-z)^2}
\frac{1-(z^2/(1-z))^n}{1-z^2/(1-z)}
= 1 + [z^n] z \times
\frac{1-(z^2/(1-z))^n}{(1-z)^2-z^2\times (1-z)}.$$
This finally yields
$$1 + [z^n]\frac{z}{1-2z+z^3}
\left(1-(z^2/(1-z))^n \right)$$
which is
$$1 + [z^n]
\left(-\frac{1}{1-z} + \frac{1+z}{1-z-z^2}\right)
\left(1-(z^2/(1-z))^n \right).$$
We proceed to coefficient extraction and observe that the $n$th power
term starts at $z^{2n}$ and hence no multiple of it can contribute to
$[z^n]$, leaving just two terms for coefficient extraction, the second
of which is the OGF of the Fibonacci numbers, for a final answer of
$$1 + (-1) + F_{n+1} + F_n = F_{n+2}.$$
|
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|
Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $\lfloor n+f \rfloor = n$ gives us
$$
\begin{align}
\lfloor (n+f)^2 \rfloor -3 \lfloor n+f \rfloor +2 &= 0\\
n^2+\lfloor f^2+2nf \rfloor -3n+2 &= 0
\end{align}
$$
How can I solve the equation from here?
|
The way I would approach it is to first observe that the zeroes of $y^2-3 y+2$ are at $y=1$ and $y=2$. Because these zeroes are integers, then $x=1$ and $x=2$ are part of the solution set.
To get the rest of the solution set, we must find all other $x$ near these zeroes that do not change the values of either $\lfloor x \rfloor$ or $\lfloor x^2 \rfloor$. Near $x=1$, this is when $x \in \left (1,\sqrt{2}\right)$; near $x=2$, we have $x \in \left (2,\sqrt{5} \right )$.
|
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|
Integer $a$ , If $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots If the equation $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots. Then the integer value of $a$ is
$\bf{My\; Try}::$ Let $\alpha,\beta\in \mathbb{Z}$ be the roots of the equation . Then $\alpha+\beta = (6-a)$ and $\alpha\cdot \beta = a$
Now $D = (a-6)^2-4a = $ perfect square. So $a^2+36-12a-4a = k^2$. where $k\in \mathbb{Z}$
$a^2-16a+36=k^2\Rightarrow (a-8)^2-28=k^2\Rightarrow (a-8)^2-k^2 = 28$
Now How can i solve after that
Help Required
Thanks
|
Another way:
We can set $a^2-16a+36$ to $(a-n)^2$ where $n$ is some integer
$\displaystyle \implies n^2=36-16a+2an$ which is even $\implies 2|n^2\iff 2|n$ as $2$ is prime.
So, we can set $n=2m$ where $m$ is some integer
$\displaystyle \implies 4m^2=36-16a+4am\implies a=\frac{m^2-9}{m-4}=m+4+\frac7{m-4}$
$\displaystyle\implies m-4$ must divide $7$ and $m-4\ne0$ as $a$ is finite
|
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|
Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
|
First way:
$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})$
set a=2, b=1
Edit: Second way.
Set $X=2^{(n-1)} + 2^{(n-2)} + … + 2 + 1$
$2X=2^n+2^{(n-1)} + 2^{(n-2)} + … + 2$
$2X=2^n+2^{(n-1)} + 2^{(n-2)} + … + 2 +1 -1$
$2X=2^n+X-1$
$X=2^n-1$
|
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|
Divisibility induction proof: $8\mid 7^n+3^n-2$ I'm stuck on the following proof by induction: $$8\mid3^n +7^n -2$$
And this is how far I've gotten:
$$\begin{aligned}3&\cdot3^n+7\cdot7^n-2\\3&(3^n+7^n-2)+7^n(7-3)-2\end{aligned}$$
Any help on where to go after this would be great!
|
Note that by hypothesis $$8|3^n+7^n-2$$ then $8|3^{n+1}+7^{n+1}-2$ if and only if also $$8|(3^{n+1}+7^{n+1}-2)-(3^{n}+7^{n}-2)=3^n(3-1)+7^n(7-1)=\underbrace{2\cdot 3^n+2\cdot 7^n-4}_{\mathrm{divisible\ by\ 8}}+\underbrace{4+4\cdot 7^n}_{\mathrm{divisible\ by\ 8}}.$$
|
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How do I find the Jordan normal form of a matrix with complex eigenvalues? I'm trying to obtain the Jordan normal form and the transformation matrix for the following matrix:
$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$
I've calculated its characteristic and minimum polynomials as $(λ - 1)^2(λ^2 + λ + 1)$, and thus the eigenvalues are $λ = 1$ (with an algebraic multiplicity of $2$) and $λ = \frac{-1 \pm i\sqrt{3}}{2}$.
An eigenvector for $λ = 1$ is $\begin{pmatrix} 0 \\\ 1 \\\ 1 \\\ 1 \end{pmatrix}$.
Since the minimum polynomial contains two identical factors, there must be at least a $2 x 2$ Jordan block associated with the eigenvalue $λ = 1$, and so the Jordan normal form must look something like the following:
$A = \begin{pmatrix} 1 & 1 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 + i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 - i\sqrt{3}}{2} \end{pmatrix}$
However, I don't know how to derive a transformation matrix $P$ such that $PJ = AP$. How would I go about solving for $P$?
|
We are given:
$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$$
We find that characteristic polynomial by solving $|A - \lambda I| = 0$, yielding:
$$(\lambda -1)^2 (\lambda^2 + \lambda +1) = 0$$
This yields a double and a complex conjugate pair of eigenvalues:
$$\lambda_{1,2} = 1, \lambda_{3,4} = -\dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2}$$
To find the eigenvectors, we solve $[A -\lambda_i I]v_i = 0$, so for $\lambda_1 = 1$, we get:
$$v_1 = (0,1,1,1)$$
This only gives us a single linearly independent eigenvector, so to find a generalized one, we set up and solve $[A -\lambda_1 I]v_2 = v_1$, yielding (using RREF):
$$v_2 = (3,2,1,0)$$
Next, we have a complex eigenvalue and follow the same procedure and what is nice is that the eigenvector will give us both of them since they too will be complex conjugates. We set up and solve $[A - \lambda_3 I]v_3 = 0$, where $\lambda_3 = -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$, which yields:
$$v_3 = (0, \dfrac{1}{2}(-1 + i \sqrt{3}), \dfrac{1}{2}(-1 - i \sqrt{3}), 1)$$
We can now write the last eigenvector as the complex conjugate, yielding:
$$v_4 = (0, \dfrac{1}{2}(-1 - i \sqrt{3}), \dfrac{1}{2}(-1 + i \sqrt{3}), 1)$$
Now, we have $P$ as a linear combination of these column eigenvectors:
$$P = [~v_1 ~| ~v_2 ~| ~v_3 ~| ~v_4 ~]$$
$$J = P^{-1} A P$$
However, we can figure out the JNF as:
$$J = \begin{pmatrix} 1 & 1 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 - i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 + i\sqrt{3}}{2} \end{pmatrix}$$
You could have also calculated it from what I wrote above $J = P^{-1}AP$.
I would work this forward and backward to get your hands around it.
|
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|
Using the method of induction to show How can I use the method of induction to show for any real number $r$ does not equal $1$ and any positive integer $n$
show that
$$1+r+r^2+\cdots+r^n=\frac{1r^{n+1}-1}{r-1}$$
for $n=1$ it seems to work
$$1+r+\cdots+r^n=(1+r)$$
then $\dfrac{r^2-1}{r-1}$ for the right side
$$\frac{(r-1)(r+1)}{r-1}=1+r$$
Thus the formula is true for $n=1$ then assume the formula is true $n=k$ ,$k$ is an integer greater than $1$.
$$1+r+\cdots+r^k=\frac{r^{k+1}-1}{r-1}$$
then
$$1+r+\cdots+r^k+r^{k+1}$$
then
$$\frac{1r^{k+1}-1}{r-1} + \frac{r^{k+1}}{1}$$
$$\frac{r^{k+1}+r-1}{r-1}=1+r+\cdots+r^k+r^{k+1}$$
would this be correct method of induction?
|
Assuming $r\ne1$
Let $\displaystyle F(n): 1+r+r^2+....+r^n=\frac{r^{n+1}-1}{r-1}$ holds true for $n=m $
$\displaystyle\implies 1+r+r^2+....+r^m=\frac{r^{m+1}-1}{r-1}$
$\displaystyle \implies 1+r+r^2+....+r^m+r^{m+1}=\frac{r^{m+1}-1}{r-1}+r^{m+1}=\frac{r^{m+1}-1+r^{m+2}-r^{m+1}}{r-1}$
$\displaystyle \implies 1+r+r^2+....+r^m+r^{m+1}=\frac{r^{(m+1)+1}-1}{r-1}$
So, $f(n)$ will hold true for $n=m+1$ if it holds true for $n=m$
Establish the base case $n=1$
|
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|
combination problem. A small company employs $3$ men and $5$ women. If a team of $4$ employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly $2$ women?
So, first, how many possible combinations are there. Well there are $8$ people and a group of $4$ people are going to be selected. We want to know how many of these teams can be selected with $2$ women. It seems like order does not matter and so its a combination problem. MWMW is the same as MMWW here.
Anyway, how many total combinations are there are 8 ppl in 4 teams? Well its:
$_8C_4 = \frac{8!}{4!4!} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}=
\frac{2 \cdot 7 \cdot 2 \cdot 5}{1 \cdot 1 \cdot 2 \cdot 1}= \frac{140}{2} = 70$.
So there are $70$ total groups. So that goes in the denominator for the answer. What's our numerator?
How do I proceed from here?
|
There are $8$ employees. $4$ are chosen from the $8$ to organize the retreat: there are ${8 \choose 4} = 70 $ ways to do this. How many of these ways have just $2$ women? Well, remember that we have $5$ women. There are ${5 \choose 2} = 10$ ways to pick them. Then we have $3$ men. If in a team of $4$ we have $2$ women, then we must have $2$ men. There are ${3 \choose 2} = 3$ ways to pick them. By the product rule, we have $10 \cdot 3 = 30$ different ways of picking exactly $2$ men and $2$ women. So there are $30$ teams with exactly $2$ women in them. So the probability of picking such a team is $\frac{30}{70}=\frac37$.
|
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|
Induction: $n^{n+1} > (n+1)^n$ and $(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$ How do I prove this by induction:
$$\displaystyle n^{n+1} > (n+1)^n,\; \mbox{ for } n\geq 3$$
Thanks.
What I'm doing is bunch of these induction problems for my first year math studies.
I tried using Bernoulli's inequality at some point, but no success. Also, tried $(n+1)^{n+2}=(n+1)^{n+1}(n+1)$, then expanding $(n+1)^{n+1}$ by binomial formula to get the $n^{n+1}$ member to apply the induction hypothesis, still no success.
Here's another one I've been struggling with:
$$(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$$
EDIT: Finally solved the second one!
What I needed was the AM-GM inequality.
Therefore,
$$\frac{(n + 1)(2n + 1)}{6} = \frac{1}{n} \sum_{i=1}^{n} i^2 \geq \sqrt[n]{1^2 \cdot 2^2 \cdots n^2}$$
Thus,
$$\left(\frac{(n + 1)(2n + 1)}{6}\right)^n = \left(\frac{1}{n} \sum_{i=1}^{n} i^2\right)^n \geq 1^2 \cdot 2^2 \cdots n^2 = (n!)^2$$
Done.
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Supposed for $n\geq 3$ you have that $n^{n+1}>(n+1)^n$. WTS $(n+1)^{n+2}>(n+2)^{n+1}$. Since $n^{n+1}>(n+1)^n$ you get that $n^{n+1} \cdot \frac{(n+1)^{n+2}}{n^{n+1}}>(n+1)^n\cdot \frac{(n+1)^{n+2}}{n^{n+1}}=\frac{(n+1)^{2n+2}}{n^{n+1}}>(n+2)^{n+1}$, where the last inequality follows since $(n+1)^2>n(n+2)$. Hence $(n+1)^{n+2}>(n+2)^{n+1}$.
|
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|
Calculate how many ways to get change of 78 I been asked to calculate how many ways there are to get change of 78 cents with the coins of 25,10,5,1.
I been able to write this down:
$25a + 10b + 5c + d = 78$
But I do not know how to continue. Can you help me please?
|
Consider the product in alex.jordan's answer. The product of the last two factors is
$$1 + x^{10} + x^{20} + x^{25} + x^{30} + x^{35} + x^{40} + x^{45} + 2 x^{50} + x^{55}
+ 2 x^{60} + x^{65} + 2 x^{70} + 2 x^{75} + \ldots)$$
The product of the first two is
$$ 1 + \ldots + x^4 + 2 (x^5 + \ldots + x^9) + 3 (x^{10}+\ldots+x^{14}) + \ldots
+ 16 (x^{75}+\ldots+x^{79})+\ldots$$
So the coefficient of $x^{78}$ is
$$ \eqalign{ 16 & \text{ for } 1 \times 16 x^{78}\cr
+ 14 & \text{ for } x^{10} \times 14 x^{68}\cr
+ 12 & \text{ for } x^{20} \times 12 x^{58}\cr
+ \ldots\cr
+ 2 & \text{ for } 2x^{75} \times x^3}$$
|
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Ellipse problem : Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r. Problem :
Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r.
Few concepts about Ellipse :
Equation of Tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point $(x_1,y_1)$ is given by $T = \frac{xx_1}{a^2} +\frac{yy_1}{b^2}-1$
Also point of contact where line $y =mx +c $ touched the ellipse .
The line is y =mx +c touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ when $c =\pm \sqrt{a^2m^2+b^2}$
$(x_1,y_1) =(\pm \frac{a^2m}{\sqrt{a^2m^2 +b^2}}, \pm \frac{b^2}{\sqrt{a^2m^2+b^2}})$
Now, we know that equation of any tangent to the given ellipse is $y =mx \pm \sqrt{a^2m^2+b^2}$
Now if it touches $x^2+y^2=r^2$
Then
$ \sqrt{a^2m^2+b^2} = r\sqrt{1+m^2}$ I am unable to understand this condition ..please guide for this particular condition only... thanks.
|
You have done sufficient hard work.
As $m$ is the slope of the common tangent,
we have $\displaystyle \pm \sqrt{a^2m^2+b^2} = \pm r\sqrt{1+m^2}$
Squaring we get $\displaystyle a^2m^2+b^2=r^2(1+m^2)$
What is $m$, compare with my other answer?
|
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|
What is $\lim_{x\to0} \frac{(\cos x + \cos 2x + \dots+ \cos nx - n)}{\sin x^2}$? What is the limit of $$\lim_{x\to0} \frac{\cos x + \cos 2x + \dots+ \cos nx - n}{\sin x^2}$$
|
$$
\begin{align}
& \lim_{x\to0}\frac{\cos x+\cos 2x+\cdots+\cos nx-n}{\sin x^2} \\[10pt]
& =\lim_{x\to0} \frac{\cos x -1+\cos 2x -1+\cdots+\cos nx-1}{x^2} \cdot\frac{x^2}{\sin x^2} \\[10pt]
& = -\left(\frac{1}{2} + \frac{4}{2}+\frac{9}{2}+\cdots+\frac{n^2}{2}\right)\cdot1=-\frac{n(n+1)(2n+1)}{12}.
\end{align}
$$
We applied
$$\lim_{x\to0}\frac{1-\cos nx}{x^2}=\frac{n^2}{2} $$
|
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|
Probability that committee chosen from 8 men and 7 women has more men
A board of trustees of a university consists of 8 men and 7 women. A committee of 3 must be selected at random and without replacement. The role
of the committee is to select a new president for the university. Calculate the
probability that the number of men selected exceeds the number of women
selected.
My try:
Given that the number of men should be greater, so I'll find the probability that 2 out of the 3 are men.
Probability that the first three selected are men : $\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}$
Probability that the first two selected are men with the third a woman: $\frac{8}{15}\times\frac{7}{14}$
Probability that the first selected is a woman and the other two are men : $\frac{8}{14}\times\frac{7}{13}$
Total Probability: $\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}+\frac{8}{15}\times\frac{7}{14}+\frac{8}{14}\times\frac{7}{13} = .697$
The correct answer is: $\frac{36}{65}=0.5538$, and thanks in advance
|
In the line "Probability that the first two selected are men with the third a woman," the probability of choosing a woman third, after choosing two men, is 7/13. You need to multiply by that as well. Similarly, the probability of choosing the woman first is missing from the following line. Finally, you also need to add in the additional case where a man is chosen first, a woman second, and another man third.
Another approach:
Number of committees consisting of two men and one woman: $\binom{8}{2}\binom{7}{1}$.
Number of committees consisting of only men: $\binom{8}{3}$.
The number of all possible committees is $\binom{15}{3}$. I think you can take it from there :)
|
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|
Find $f(x)$ where $ f(x)+f\left(\frac{1-x}x\right)=x$ What function satisfies $ f(x)+f\left(\frac{1-x}x\right)=x$ ?
|
$f(x)+f\left(\dfrac{1-x}{x}\right)=x$
$f(x)+f\left(\dfrac{1}{x}-1\right)=x$
$\because$ The general solution of $T(x+1)=\dfrac{1}{T(x)}-1$ is $T(x)=\dfrac{(\sqrt5-1)^{x+1}+\Theta(x)(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2\Theta(x)(-\sqrt5-1)^x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
$\therefore f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}-1\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{(\sqrt5-1)^x(3-\sqrt5)+(-\sqrt5-1)^x(3+\sqrt5)}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{\dfrac{(\sqrt5-1)^x(\sqrt5-1)^2}{2}+\dfrac{(-\sqrt5-1)^x(\sqrt5+1)^2}{2}}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{(\sqrt5-1)^{x+2}+(-\sqrt5-1)^{x+2}}{2(\sqrt5-1)^{x+1}+2(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)=\Theta(x)(-1)^x-\sum\limits_x\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
|
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|
System of quadratic Diophantine equations Is there a method for determining if a system of quadratic diophantine equations has any solutions?
My specific example (which comes from this question) is:
$$\frac{4}{3}x^2 + \frac{4}{3}x + 1 = y^2$$
$$\frac{8}{3}x^2 + \frac{8}{3}x + 1 = z^2$$
I want to know if there are any positive integer triples $(x,y,z)$ which satisfy both equations.
|
Your specific example can be formalized in the following way:
$$
\begin{eqnarray}
4x^2+4x+3=3y^2\\
8x^2+8x+3=3z^2\\
4x^2+4x=3(z^2-y^2)\\
(z^2-y^2)=4x(x+1)/3
\end{eqnarray}
$$
let $x+1=3n$
$$
\begin{eqnarray}
(z^2-y^2)=4(3n-1)n
\end{eqnarray}
$$
Let $z=4n-1$ and $y=2n-1$ (more generally if $3n^2-n=uv$, then $z=u+v$ and $y=v-u$). So I would have thought there are tons of integer solutions, but there's some other constraint on the go. Let's punch $x+1=3n$ into the original expressions:
$$
\begin{eqnarray}
4x(x+1)+3&=&12n(3n-1)+3&=&3y^2\\
y^2&=&12n^2-4n+1&&\\
8x(x+1)+3&=&24n(3n-1)+3&=&3z^2\\
z^2&=&24n^2-8n+1&&\\
z^2-y^2&=&12n^2-4n&=&y^2-1\\
(z+y)(z-y)&=&(y+1)(y-1)&&
\end{eqnarray}
$$
The only solution I can come up with here is $z=y$, $y=1$, $x=0$. One other approach works too:
let $x=3n$
$$
\begin{eqnarray}
(z^2-y^2)=4(3n+1)n
\end{eqnarray}
$$
Let $z=4n+1$ and $y=2n+1$. Let's punch $x=3n$ into the original expressions:
$$
\begin{eqnarray}
4x(x+1)+3&=&12n(3n+1)+3&=&3y^2\\
y^2&=&12n^2+4n+1&&\\
8x(x+1)+3&=&24n(3n+1)+3&=&3z^2\\
z^2&=&24n^2+8n+1&&\\
z^2-y^2&=&12n^2+4n&=&y^2-1\\
(z+y)(z-y)&=&(y+1)(y-1)&&
\end{eqnarray}
$$
The end result is the same.
|
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|
calculation of $\int_{0}^{1}\tan^{-1}(1-x+x^2)dx$
Compute the definite integral
$$
\int_{0}^{1}\tan^{-1}(1-x+x^2)\,dx
$$
Failed Attempt:
Let $1-x+x^2=t$. Then
$$
\begin{align}
(2x-1)\,dx &= dt\\
dx &= \frac{1}{(2x-1)}dt
\end{align}
$$
Changing the limits of integration, we get
$$\int_{1}^{1}\tan^{-1}(t)\cdot \frac{1}{(2x-1)}dt = \int_{1}^{1}\tan^{-1}(t)\cdot f(t)dt = 0
$$
where $f(t)=\frac{1}{(2x-1)}$.
Is it true that $\int_{a}^{a}f(x)dx = 0$? If not, then where have I made a mistake in my attempted solution?
|
HINT:
As $\displaystyle 1-x+x^2=\frac{(2x-1)^2+3}4>0$ for real $x,$
using this, $\displaystyle \tan^{-1}(1-x+x^2)=\cot^{-1}\left(\frac1{1-x+x^2}\right)$
Now, we know $\displaystyle\cot^{-1}\left(\frac1{1-x+x^2}\right)=\frac\pi2-\tan^{-1}\left(\frac1{1-x+x^2}\right)$
Again, $\displaystyle\tan^{-1}\left(\frac1{1-x+x^2}\right)=\tan^{-1}\left(\frac{x+1-x}{1-x(1-x)}\right)=\tan^{-1}x+\tan^{-1}(1-x)$
From this, the last identity holds true if $\displaystyle x(1-x)\le1$ which is true as $\displaystyle x(1-x)=\frac{1-(2x-1)^2}4\le \frac14$ for real $x$
Finally, as $\displaystyle \int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$
$\displaystyle \int_0^1\tan^{-1}(1-x)dx=\int_0^1\tan^{-1}\{1-(\underbrace{1+0-x})\}dx=\int_0^1\tan^{-1}xdx$
|
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|
Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
|
Use Riemann sums to show that $S\in\Big[\tfrac{\ln2}2,\tfrac12\Big]$ :
$$\sum_{k=1}^n\frac k{k^2+n^2}<\sum_{k=1}^n\frac k{k+n^2}<\sum_{k=1}^n\frac k{n^2}\quad\iff\quad\int_0^1\frac x{1+x^2}dx<S<\int_0^1xdx$$
|
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|
Need help simplifiying a rational expression There's a math question on an online test which asks the following
Multiply the following expression, and simplify:
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I can't seem to figure out how to simplify it enough to get it right.
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
equals, $\frac{x^4 + 4x^3y - 16x^2y^2 - 64xy^3}{x^2-4xy}$. I then factored x out of the numerator and denominator to get $\frac{x(x^3 + 4x^2y - 16xy^2 - 64y^3)}{x(x-4y)}$ and cancelled out the factored x's to get $\frac{x^3 + 4x^2y - 16xy^2 - 64y^3}{x-4y}$. I don't know what to do from here though.
I've managed to get enough marks to be able to pass it but since it's a readiness test I want to understand all of the material going in.
|
As mentioned in the comments, by factoring out $x$ in the second fraction, $x^2+4xy$ becomes $x(x+4y)$. Which leaves you with:
$$\frac{x^2+16y^2}{x} \cdot \frac{x(x+4y)}{x-4y}=\frac{(x^2+16y^2)(x+4y)}{x-4y}$$
Alternately if you had not seen the cancellations and had ended up with $x^3+16xy^2+4x^2y+64y^3$ as your numerator. Grouping $x^3+4x^2y$ and $16xy^2+64y^3$ together, notice that $x^2$ and $16y^2$ are common (respectively). Factor them out to get,
$$x^2(x+4y) + 16y^2(x+4y) = (x^2+16y^2)(x+4y)$$
|
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|
How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$? How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$, where $\lfloor \cdot \rfloor$ denotes the greatest integer function?
Please help. I have no idea about this.
|
First, notice that the domain of $x$ is
$$2-x^2\ge0\iff-\sqrt2\le x\le \sqrt 2.$$
And the floor function $\lfloor x\rfloor$ is defined as
$$\lfloor x\rfloor=t\iff t\le x\lt t+1.$$
So, in your question, we have
$$\lfloor \sqrt{2-x^2}\rfloor =t\iff t\le\sqrt{2-x^2}\lt t+1$$
Now note that $t\ge 0 \in\mathbb Z$ (This is because $\sqrt{2-x^2}\ge0$). So we have
$$t^2\le 2-x^2\lt (t+1)^2\iff 2-(t+1)^2\lt x^2\le 2-t^2.$$
The fact that $\lfloor \sqrt{2-0^2}\rfloor=1, \lfloor \sqrt{2-({\sqrt 2})^2}\rfloor=0$ tells us that $t=0,1$.
Hence, we know
$$2-(0+1)^2\lt x^2\le 2-0^2\Rightarrow t=0,$$
$$2-(1+1)^2\lt x^2\le 2-1^2\Rightarrow t=1.$$
|
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|
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I wondering how to show easily this identity ? As a matter of the fact, it's a beautiful identity.
$$-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}$$
The one way I think about this I'll let here :
$$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b)(b+c)(c+a)+(b-c)(c+a)(a+b)+(c-a)(c+b)(b+a)}{(a+b)(b+c)(c+a)}$$
Now, I'll take one of the term of RHS
$(a −b)(b +c)(c + a)\\ =(a −b)(b−c + 2c)(a −c + 2c)\\ =(a −b)(b −c)(a −c)+ 2c(a −b)(b−c + a −c + 2c) \\=(a −b)(b −c)(a −c)+ 2c(a −b)(a +b)$
Similarly, We'll do that with the remainder
$(b−c)(a+b)(c + a)+(c − a)(a +b)(b+ c)\\=(a +b)[(b −c)(c + a)+(c −a)(b +c)]\\=(a +b)(bc +ba −c^{2} −ca +cb+c^{2} −ab −ac)\\= (a +b)(2bc − 2ac)\\= −2c(a −b)(a +b)$
and we get :
$$\boxed{\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}}$$
So it's too many work to show this identity. I just want to know if there's a simple way to show that or I don't know. I'm questing that because I didn't find anything on internet.
|
Let $x=\frac{a-b}{a+b}$ and so on. We then have: $(1-x)(1-y)(1-z)=(1+x)(1+y)(1+z)$
The terms with an even absolute degree will cancel, therefore: $x+y+z=-xyz$
|
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|
find general solution to the Differential equation Find the general solution to the differential equation
\begin{equation}
\frac{dy}{dx}= 3x^2 y^2 - y^2
\end{equation}
I get
\begin{equation}
y=6xy^2 + 6x^2 y\frac{dy}{dx} - 2y\frac{dy}{dx}
\end{equation}
rearrange the equation
\begin{equation}
\frac{dy}{dx} = \frac{y-6xy^2}{6x^2 y - 2y}
\end{equation}
simplified the equation
\begin{equation}
\frac{dy}{dx} = \frac{1-6xy}{6x^2 - 2}
\end{equation}
how can I solve this equation to get a general solution of $y=ax+c$
|
HINT :
$$\frac{1}{y^2}dy=(3x^2-1)dx$$
$$-\frac 1y=x^3-x+C$$
|
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|
If $a,b,c$ are positive integers and $a^2+b^2=c^2$ and $a$ is a prime, what can we conclude about primeness of b and c? Let $a,b,c$ be positive integers and they satisfy $a^2+b^2=c^2$, and if $a$ is prime, can we conclude whether $b$ and $c$, are both prime, composite or neither? If yes, why, if not why not?
I can conclude that $b$ and $c$ have to be one odd and the other one even using $a^2 = c^2-b^2=(c-b)(c+b)$. But I couldn't conclude anything about their primeness. Can anyone show me some ideas or its reasoning and maybe any useful and related theorems?
Thanks very much!
|
$5^2+12^2=13^2$ and $7^2+24^2=25^2$, so I don't think there is anything interesting about primality of $b$ and $c$...
What you may say is that, except for trivial cases, $c+b\neq c-b$, hence we must have $c-b=1$ and $c+b=2b+1=a^2$, so $b=\frac{a^2-1}2$ and $c=\frac{a^2+1}2$. Hence, you find that exists exactly one such triple $(a,b,c)$ for every odd prime number.
|
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|
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it is a correct identity, although I can't prove it.
I've tried to use the formula
$$ \cos(z) = \frac{e^ {iz} - e^ {iz}}{2} $$ but without any satisfying result.
This exercise is from chapter on series.
EDIT: I corrected a mistake in the formula I wanted to use.
|
Let $\zeta_n = e^{2\pi i/n} = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$ be a primitive $n^{\rm th}$ root of unity, so in particular, the roots of $z^5 - 1$ are $\zeta_5^k$ for $k = 0, 1, 2, 3, 4$. Since the sum of the roots of a polynomial of degree $n$ is equal to the negative of the coefficient of the degree $n-1$ term, it follows that $$\sum_{k=0}^4 \zeta_5^k = 1 + \zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4 = 0.$$ Taking the real part of both sides of the equation immediately gives the desired identity.
|
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|
Calculation of $ \int_{0}^{\sqrt{n}}\lfloor t^2 \rfloor dt\;$, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$. Calculation of $\displaystyle \int_{0}^{\sqrt{n}}\lfloor t^2 \rfloor dt\;$, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$.
$\bf{My\; Try}::$ Let $t^2 = u$ and $\displaystyle dt = \frac{du}{2\sqrt{u}}$ and changing Limits, we get $\displaystyle \frac{1}{2}\int_{0}^{n}\frac{\lfloor u \rfloor }{\sqrt{u}}du$
Now Let
$\displaystyle I = \int_{0}^{n}\lfloor u \rfloor \cdot u^{-\frac{1}{2}}du = \int_{0}^{1}0\cdot u^{-\frac{1}{2}}du+\int_{1}^{2}1\cdot u^{-\frac{1}{2}}du+\int_{2}^{3}2\cdot u^{-\frac{1}{2}}du+.........+\int_{n-2}^{n-1}(n-2)\cdot u^{-\frac{1}{2}}du+\int_{n-1}^{n}(n-1)\cdot u^{-\frac{1}{2}}du$
$\displaystyle =-2\left\{1\cdot \left(\sqrt{2}-1\right)+2\cdot \left(\sqrt{3}-\sqrt{2}\right)+3\cdot \left(\sqrt{4}-\sqrt{3}\right)+.......+(n-2)\cdot \left(\sqrt{n-1}-\sqrt{n-2}\right)+(n-1)\cdot \left(\sqrt{n}-\sqrt{n-1}\right)\right\}$
Now how can i calculate given sum in closed form
please Help me
Thanks
|
You can obtain $$\frac{1}{2}\int_0^n\frac{\lfloor u\rfloor}{u^{1/2}}\text{d}u =\frac{1}{2}\int_0^n\frac{u-\{u\}}{u^{1/2}}\text{d}u = \frac{1}{2}\int_0^n u^{1/2}\text{d}u - \frac{1}{2}\int_0^n\frac{\{u\}}{u^{1/2}}\text{d}u,$$ which gives
$$\frac{1}{2}\int_0^n\frac{\lfloor u\rfloor}{u^{1/2}}\text{d}u =\frac{1}{2}\left[\frac{2}{3}u^{3/2}\right]_0^n-\frac{1}{2}\int_0^n\frac{\{u\}}{u^{1/2}}\text{d}u=\frac{1}{3}n^{3/2}-\frac{1}{2}\int_0^n\frac{\{u\}}{u^{1/2}}.$$
Not sure if you can do much with the last integral, but for a "similar" integral notice that
$$\lim_{n\longrightarrow\infty}\int_0^n\frac{\{u\}}{u^{s+1}}\text{d}u = -\frac{\zeta(s)}{s},$$ where $\zeta(s)$ is the Riemann zeta function and $0<s<1$.
|
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|
Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$ Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$
Solution :
$4(2y-x-3)^2 = 4x^2-16xy+24x+16y^2-48y+36$
and $9(2x+y-1)^2 = 36x^2+36xy-36x+9y^2-18y+9$
$\therefore 4(2y-x-3)^2 -9(2x+y-1)^2 = 7y^2+60x -52xy-32x^2-30y+27 =80$
Can we have other option available so that we will be able to find the solution more quicker way,
Since this is a conic then the given lines let $L_1 =2y-x-3=0 $ and $L_2 = 2x+y-1=0$ are perpendicular to each other .. Can we use this somehow please suggest... thanks..
|
With a translation of the origin in the center of the conic (which is the point of intersection of the two lines, you get the equation in the form
$$
4(-X+2Y)^2-9(2X+Y)^2=80
$$
Now the lines are indeed orthogonal: since $1^2+2^2=5$, you can write the equation as
$$
4\cdot5\left(-\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\right)^2
-9\cdot5\left(\frac{2}{\sqrt{5}}X+\frac{1}{\sqrt{5}}Y\right)^2=80
$$
The transformation
\begin{cases}
\xi=-\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\\
\eta=\frac{2}{\sqrt{5}}X+\frac{1}{\sqrt{5}}Y
\end{cases}
is a rotation, so you get the canonical form
$$
4\xi^2-9\eta^2=16
$$
or
$$
\frac{\xi^2}{4}-\frac{\eta^2}{16/9}=1
$$
Thus you have
$$a^2=4,\quad b^2=\frac{16}{9}$$
|
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|
compute limit (no l'Hospital rule) I need to compute
$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}.$$
I can not use the l'Hospital rule.
|
$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{\cos x} -1+1- \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{1+\cos x-1} -1- (\sqrt[3]{1+\cos x-1}-1)}{\cos x-1}\cdot\frac{\cos x-1}{\sin^2x}=\lim_{x\to 0}[\frac{(1+\cos x-1)^\frac{1}{2}-1}{\cos x-1}-\frac{(1+\cos x-1)^\frac{1}{3}-1}{\cos x-1}]\cdot\frac{-1}{2}=(\frac{1}{2}-\frac{1}{3})\cdot\frac{-1}{2}=-\frac{1}{12}$$
We applied:$$ \lim_{t\to 0}\frac{(1+t)^r-1}{t} =r$$
|
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|
How to prove that $\sum_{i=0}^{a}\frac{i\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=\frac{ac(a+b)}{b(b+1)}$ let $$b\ge c,a,b,c\in N^{+}$$
Show that
$$\sum_{i=0}^{a}\dfrac{i\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=\dfrac{ac(a+b)}{b(b+1)}$$
This sum is similar to Hypergeometric distribution,
but different.
I know this
$$\sum_{i=0}^{a}\binom{a}{i}\binom{b}{c-i}=\binom{a+b}{c}$$
so
$$\sum_{i=0}^{a}\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}=\sum_{i=0}^{a}\binom{a+b-c-i}{b-c}\binom{c+i-1}{c-1}=\binom{a+b-1}{a}？$$
so
$$\sum_{i=0}^{a}\dfrac{\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=1$$
use this
$$k\binom{n}{k}=n\binom{n-1}{k-1}$$
$$i\binom{c+i-1}{i}\binom{a+b-c-i}{a-i}=(c+i-1)\binom{c+i-2}{i-1}\binom{a+b-c-i}{a-i}$$
Then I can't .Thank you
|
Your result follows from some standard results about binomial coefficients
(as found in [1], for example).
I'm not sure of the minimal conditions needed to make the argument work,
but I will assume that $a\geq 0$ and $b>c\geq 1$. I first rewrite your sum, without
the factor $i$ as
$$\sum_{i\geq 0}{a+b-c-i\choose b-c}{c-1+i\choose c-1}={a+b\choose b}.\tag1$$
This follows from an application of (5.26) from [1].
We will now exploit the fact that the sum in (1) is valid for all $a\geq 0$.
Let's re-introduce the factor of $i$ and define
$$f(a)=\sum_{i\geq 0}{a+b-c-i\choose b-c} i {c-1+i\choose c-1}.$$
Then
\begin{eqnarray*}
f(a)
&=& \sum_{i\geq 1}{a+b-c-i\choose b-c} i {c-1+i\choose i}\\[5pt]
&=& \sum_{i\geq 1}{a+b-c-i\choose b-c} [c+(i-1)] {c+i-2\choose i-1}\\[5pt]
&=& \sum_{j\geq 0}{a-1+b-c-j\choose b-c} [c+j] {c+j-1\choose j}\\[5pt]
&=& c{a-1+b\choose b}+ f(a-1).
\end{eqnarray*}
Since $f(0)=0$, we deduce that
$$f(a)=c\sum_{k=0}^{a-1}{k+b\choose b}=c{a+b\choose a-1},$$
where the last equation comes from page 174 of Concrete Mathematics
under "parallel summation".
It is not hard to check that
$$c{a+b\choose a-1}={\binom{a+b-1}{a}}\dfrac{ac(a+b)}{b(b+1)}$$
so this gives the desired result.
Reference [1] Concrete Mathematics (2e) by Graham, Knuth, and Patashnik.
|
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|
If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares. If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.
My work:
$3n+1=x^2$
$3n+3=x^2+2$
$3(n+1)=x^2+2$
$(n+1)=\dfrac{x^2+2}{3}$
I have no clue what to do next. Please help!
|
If $\displaystyle 3n+1=a^2, (a,3)=1\implies a$ can be written as $\displaystyle3b\pm1$ where $b$ is an integer
So we have $\displaystyle 3n+1=(3b\pm1)^2\implies n=3b^2\pm2b$
$\displaystyle n+1=3b^2\pm2b+1=b^2+b^2+b^2\pm2b+1=b^2+b^2+(b\pm1)^2$
|
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|
How find this postive integer $a$ such $a(x^2+y^2)=x^2y^2$ always have roots Find all postive integer numbers of $a$,such this equation
$$a(x^2+y^2)=x^2y^2,xy\neq0$$
always have integer roots $(x,y)$
my try: since
$$\dfrac{x^2y^2}{x^2+y^2}\in N$$
and I can't
Thank you
|
If $(x,y)=1$, then $(x^2+y^2,x^2y^2)=1$. Suppose $(x,y)=d$, then $\left(\frac xd,\frac yd\right)=1$ and therefore
$$
\left(\frac{x^2+y^2}{d^2},\frac{x^2y^2}{d^4}\right)=1\tag{1}
$$
which implies that both
$$
\left(\frac{x^2+y^2}{d^2},\frac{x^2}{d^2}\right)=1\quad\text{and}\quad\left(\frac{x^2+y^2}{d^2},\frac{y^2}{d^2}\right)=1\tag{2}
$$
Suppose that
$$
\frac{x^2y^2}{x^2+y^2}=a\in\mathbb{Z}\tag{3}
$$
then
$$
\frac{x^2\frac{y^2}{d^2}}{\frac{x^2}{d^2}+\frac{y^2}{d^2}}
=\frac{\frac{x^2}{d^2}y^2}{\frac{x^2}{d^2}+\frac{y^2}{d^2}}
=a\tag{4}
$$
Thus, $(2)$ and $(4)$ require
$$
\left.\frac{x^2}{d^2}+\frac{y^2}{d^2}\middle|\,x^2\right.\quad\text{and}\quad\left.\frac{x^2}{d^2}+\frac{y^2}{d^2}\middle|\,y^2\right.\tag{5}
$$
which in turn require
$$
\left.\frac{x^2}{d^2}+\frac{y^2}{d^2}\middle|\,d^2\right.\tag{6}
$$
Let $u=\frac xd$ and $v=\frac yd$. Then, $(u,v)=1$ and $(6)$ guarantees that
$$
u^2+v^2\mid d^2\tag{7}
$$
and $(3)$ becomes
$$
\frac{d^2}{u^2+v^2}u^2v^2=a\tag{8}
$$
So pick any $u,v$ so that $(u,v)=1$, and let $u^2+v^2=b^2c$ where $c$ is square-free, then by setting $d$ to be any multiple of $bc$, we get that $a$ can be any square multiple of $cu^2v^2$.
|
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|
How to prove that $n^7\equiv n^3\mod40,\forall n\in\mathbb{Z}$ I have a problem when I try to induction on $\mathbb{Z}$. I don't know how to solve when I try $n-1$ or $n+1$.
|
$$n^7-n^3=n^3(n^4-1)=n^2(n^5-n)$$
Using Fermat's Little Theorem, $5|(n^5-n)$
If $n$ is even $8|n^3$
Else $\displaystyle n$ is odd, $=2m+1,$(say),
Both $n+1,n-1$ are even, one is divisible by $4$ and the other is by $2$
Algebraically, $\displaystyle(2m+1)^2=4m^2+4m+1=8\frac{m(m+1)}2+1\implies 8|(n^2-1)$ if $n$ is odd
Alternatively,
$$F=n^7-n^3=n^3(n^4-1)=n^3(n^2-1)(n^2+1)$$
$$=n(n^2-4+4)(n^2-1)(n^2+1)$$
$$=n(n^2-4)(n^2-1)(n^2+1)+4\cdot n(n^2-1)(n^2+1)$$
$$=n(n^2-4)(n^2-1)\cdot (n^2+1)+4\cdot n(n^2-1)(n^2-4+5)$$
$$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{The product of } 5 \text{ consecutive integers}}\left[n^2+1+4\right]+20\underbrace{(n-1)n(n+1)}_{\text{The product of } 3 \text{ consecutive integers}}$$
Now utilize The product of n consecutive integers is divisible by n factorial or The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients) to find that $F$ is actually divisible by $20\cdot 3!=120$
|
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|
Calculate GCD$(x^4+x+1,x^3+x^2)$ and a Bezout Identity in $\mathbb{F_2}$ A really short task:
Calculate GCD$(x^4+x+1,x^3+x^2)$ and a Bezout Identity in $\mathbb{F_2}.$
I've tried it but my GCD is $1$ and I cannot see where my mistake is.
$x^4+x+1= x \cdot (x^3+x^2) + x^3 +x + 1$
$x^3+x^2 = 1 \cdot (x^3 + x + 1) + x^2 + x + 1$
$x^3+x+1 = x \cdot (x^2 + 1 + x) + x^2 + 1$
$x^2+x+1 = 1 \cdot (x^2+1) + x$
$x^2 + 1 = x \cdot x + 1$
$x = 1 \cdot x + 0$
|
This is easy when using the augmented-matrix form of the extended Euclidean algorithm.
$\begin{eqnarray}
(1)&& &&x^4\!+x+1 \,&=&\, \left<\,\color{#c00}1,\color{#0a0}0\,\right>\ \ \ {\rm i.e.}\,\ \ x^4\!+x+1 = \color{#c00}1\cdot (x^4\!+x+1) + \color{#0a0}0\cdot(x^3\!+x^2)\\
(2)&& && x^3\!+x^2 \,&=&\, \left<\,\color{#c00}0,\color{#0a0}1\,\right>\ \ \ {\rm i.e.}\,\quad\ \ \ x^3\!+x^2 = \color{#c00}0\cdot (x^4\!+x+1) + \color{#0a0}1\cdot(x^3\!+x^2)\\
(3)&=&(1)-x\cdot (2)\quad && x^3\!+x+1 \,&=&\, \left<\,1,\,x\,\right>\\
(4)&=&(2)-(3)\quad && x^2\!+x+1 \,&=&\, \left<\,1,\,x+1\,\right>\\
(5)&=&(2)-x\cdot(4)\quad &&\qquad\quad\ \ \ x \,&=&\, \left<x,\,x^2+x+1\right>\\
(6)&=&(4)-(x\!+\!1)\,(5)\!\!\!\! &&\qquad\quad\ \ \ 1 \,&=&\, \left<\color{#c00}{x^2+x+1},\ \color{#0a0}{x^3+x}\right>
\end{eqnarray}$
The Bezout Identity is $\ 1\, =\, (\color{#c00}{x^2\!+x+1})(x^4\!+x+1) + (\color{#0a0}{x^3\!+x})(x^3\!+x^2)$
|
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|
Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs
and I perform my steps I get what Wolfram Alpha shows as an alternate solution.
Any help is greatly appreciated
The problem is the following:
Show that
$$\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$$
What I have:
$P(1)$:
$$\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$$
Replace n with 1
$$\frac{1}{1} \le 2 - \frac{1}{1}$$
Conclusion
$$1 \le 1$$
Prove:
$$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
P(K) Assume
$$\sum_{i=1}^k \frac{1}{i^2} \le 2 - \frac{1}{k}$$
$P(K) \implies P(k + 1)$
Performed Steps:
Working the LHS to match RHS
$$2 - \frac{1}{k} + \frac{1}{(k+1)^2}$$
Edit: Fixed error on regrouping
$$2 - \left[\frac{1}{k} - \frac{1}{(k+1)^2}\right]$$
Work the fractions
$$2 - \left[\frac{1}{k} \frac{(k+1)^2}{(k+1)^2} - \frac{1}{(k+1)^2} \frac{k}{k} \right]$$
$$2 - \left[\frac{(k+1)^2 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + 2k + 1 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + k + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1) + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1)}{k(k+1)^2} + \frac{1}{k(k+1)^2} \right]$$
$$2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2}$$
EDIT: I fixed my mistake of my regrouping and signs; had completely missed the regrouping.
This is the final step I got to. I am hung on where to go from here. The answers given have been really helpful and I'm happy with them. I'd just like to know the mistake I made or next step I should take.
$$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
Thanks for the help
|
$$
\sum_{i=1}^n\frac1{i^2}=1+\sum_{i=2}^n\frac1{i^2}\leqslant1+\sum_{i=2}^n\frac1{i(i-1)}=1+\sum_{i=2}^n\left(\frac1{i-1}-\frac1i\right)=\ldots
$$
Edit: (About the Edit to the question 2014-01-14 21:25:21)
I'd just like to know the mistake I made or next step I should take.
None, neither mistake nor next step. Actually, what you did yields the result since you proved that
$$
\sum_{i=1}^{k+1}\frac1{i^2}\leqslant
2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2},
$$
which implies $P(k+1)$ since
$$
2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2}\leqslant2 - \frac{1}{(k+1)}.$$ Well done.
|
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|
Fourier Series of $f(x) = 0$ from $(-\pi, 0)$, $x$ from $(0,\pi)$ I need to determine the fourier series of the following function, (using trig method, not complex)
$$ f(x) = \begin{cases} 0 & \text{if } -\pi < x < 0, \\
x & \text{if } 0 < x < \pi \end{cases} $$
and then use it to show that $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{\pi^2}{8}. $$
|
The Fourier coefficients are
$$
\begin{align}
a_{0} & = \frac{1}{\pi}\int_{0}^{\pi}xdx = \frac{1}{\pi}\frac{\pi^{2}}{2}=\frac{\pi}{2},\\
a_{n} & = \frac{1}{\pi}\int_{0}^{\pi}x\cos(nx)dx = \frac{1}{\pi}\left[\left.\frac{1}{n}\sin(nx)x\right|_{x=0}^{\pi}-\frac{1}{n}\int_{0}^{\pi}\sin nx\,dx\right] \\
& = \frac{1}{\pi}\left[\left.\frac{1}{n^{2}}\cos nx \right|_{x=0}^{\pi}\right]
= \frac{\cos(n\pi)-1}{\pi n^{2}}=\frac{(-1)^{n}-1}{\pi n^{2}} \\
b_{n} & = \frac{1}{\pi}\int_{0}^{\pi}x\sin(nx) dx=
\frac{1}{\pi}\left[\left.-\frac{1}{n}\cos(nx)x\right|_{x=0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\cos(nx)dx\right] \\
& = \frac{1}{\pi}\left[-\frac{\pi}{n}\cos(n\pi)+\left.\frac{1}{n^{2}}\sin(nx)\right|_{x=0}^{\pi}\right]=-\frac{(-1)^{n}}{n}
\end{align}
$$
So the Fourier Series for $f$ converges pointwise to
$$
\frac{f(x+0)+f(x-0)}{2}=\frac{\pi}{4}+\sum_{n=1}^{\infty}
\left(\frac{(-1)^{n}-1}{\pi n^{2}}\cos(nx)-\frac{(-1)^{n}}{n}\sin(nx)\right)
$$
At the endpoints of $[-\pi,\pi]$, the left and right limits at $x$ are interpreted in terms of the periodic extension of $f$, which gives $\{f(-\pi+0)+f(\pi-0)\}/2=\pi/2$. Evaluate at $x=\pi$:
$$
\frac{\pi}{2}=\frac{f(\pi-0)+f(-\pi+0)}{2}=\frac{\pi}{4}+\sum_{n=1}^{\infty}\frac{(-1)^{n}-1}{\pi n^{2}}(-1)^{n}=\frac{\pi}{4}+\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}.
$$
Solving for the sum,
$$
\frac{\pi^{2}}{8} = \sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}
$$
|
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|
conditions on $\{a_n\}$ that imply convergence of $\sum_{n=1}^{\infty} a_n$ (NBHM 2011) Question is :
For a sequence $\{a_n\}$ of positive terms, Pick out the cases which imply convergence of $\sum_{n=1}^{\infty} a_n$.
*
*$\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$
*$\sum_{n=1}^{\infty} n^2a_n^2<\infty$
*$\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$
For the first case $\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$ :
Suppose $a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n\in \mathbb{N}$ then we would have :
$n^{\frac{3}{2}}a_n > \frac{3}{2}.n^{\frac{3}{2}} (\frac{1}{n})^{\frac{3}{2}}=\frac{3}{2}$ but then $n^{\frac{3}{2}}a_n$ would not converge to $\frac{3}{2}$
So, for large $n$ we should have $$a_n \leq\frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}\Rightarrow \sum_{n=1}^{\infty}a_n \leq \sum_{n=1}^{\infty} \frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}=\frac{3}{2}\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$$
Right hand side converges and by comparison test $\sum_{n=1}^{\infty}a_n$ should converge.
For the second case $\sum_{n=1}^{\infty} n^2a_n^2<\infty$ :
Suppose that $na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ then we would have $$n^2a_n^2> \dfrac{1}{n}$$ but then it is given that $\sum_{n=1}^{\infty} n^2a_n^2$ which would imply that $\sum_{n=1}^{\infty}\dfrac{1}{n}$ converges which is a contradiction.
Thus we should have $na_n\leq \dfrac{1}{\sqrt{n}}$ i.e., $a_n\leq \frac{1}{n\sqrt{n}}=\frac{1}{n^{\frac{3}{2}}}$ for all $n$ large.
So, We have $$\sum_{n=1}^{\infty}a_n\leq\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$$
right hand side is convergent so by comparison test $\sum_{n=1}^{\infty}a_n$ converges.
For third case $\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$ :
we would have $\lim_{n\rightarrow \infty} \dfrac{a_{n+1}}{a_n} <\lim_{n\rightarrow \infty} (\frac{n}{n+1})^2=1$
Thus, by ration test, $\sum_{n=1}^{\infty} a_n$ Converges absolutely and so is convergent.
My choice of bounds :
*
*$a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n$ in first case
*$na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ in second case
All these came to me just by choice and i would like to know if there is some sense behind this choice.
Please help me to clear this and make this solution a bit more clear.
Please do not give an alternative solution
until this problem is fully verified.
Thank you.
|
Comment on your first case:
$n^{1.5}a_n$ can converge to 1.5 while always staying above 1.5. But it must eventually go below (say) 3 so that eventually $a_n \leq 3/n^{1.5}$.
Comment on your third case:
You wrote since $a_n \geq 1/n^2$ (which isn't true anyway), comparison test applies which is also false. What you really get is $a_n \leq a_1/n^2$ (easily verified) which is enough to give convergence.
|
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|
Prove that $f(x)=x$ if the following holds true Let $f\colon\mathbb R \to \mathbb R$ be a continuous odd function such that
1) $f(1+x)=1+f(x)$
2) $x^2f(1/x)=f(x)$ for $x\ne0$.
Prove that $f(x)=x$.
|
If we do it in a more "functional equation"-ish approach, we notice that
$$\left(\frac{x+1}{x}\right)^2f\left(\frac{x}{x+1}\right) = f\left(\frac{x+1}{x}\right)$$
On the other hand,
$$\begin{align}f\left(\frac{x}{x+1}\right) &= f\left(1 - \frac{1}{x + 1}\right)\\
&= 1 + \left(-\frac{1}{x+1}\right)\\
&= 1 - \left(\frac{1}{x+1}\right)\\
&= 1 - \frac{f(x+1)}{(x+1)^2}
\end{align}$$
Substituting back:
$$\left(\frac{x+1}{x}\right)^2\left(1 - \frac{f(x+1)}{(x+1)^2}\right) = f\left(\frac{x+1}{x}\right)$$
$$\begin{align}\frac{1}{x^2}\left((x+1)^2 - f(x+1)\right) &= f\left(1 + \frac{1}{x}\right)\\
&= 1 + f\left(\frac{1}{x}\right) \\
&= 1 + \frac{f(x)}{x^2}\end{align}$$
Dividing throughout by $\frac{1}{x^2}$ (valid since $x \neq 0$):
$$\begin{align}(x+1)^2 - f(x+1) &= x^2 + f(x)\\
x^2 + 2x + 1 - x^2 &= f(x) + f(x + 1)\\
2x + 1 &= 1 + 2f(x)\end{align}$$
$$2x = 2f(x)$$
Which proves the desired statement:
$$f(x) = x$$
This is a rather long winded approach and I'm pretty sure it can be simplified a bit.
|
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|
Coefficient of $x^4$ in multinomial expansion What is the coefficient of $x^4$ in $(1 + x - 2x^2)^7$? What is a quick way to solve this problem using the binomial theorem (I have not learned multinomial theorem)?
|
An alternative approach, if you absolutely, positively have to use the binomial theorem, would be to let $a=1+x$ and $b=-2x^2$. Note that in the expansion of $(a+b)^7$ only the terms $a^7$, $a^6b$ and $a^5b^2$ will, when expanded again, eventually contain $x^4$. But how many? $a^7 = (x+1)^7$ should be easy, $a^6b=-2x^2(x+1)^6$,so you will need the coefficient of $x^2$ in $(x+1)^6$, and $a^5b^2=4x^4(1+x)^5$ should be easier again.
If you can do without the binomial theorem, fewer calculations will be required, however.
$(1 + x - 2x^2)^7=(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)$
The full product is the sum of all products of each combination of one summand from each expression in parentheses (not sure I put that in the best possible way; I mean: simply apply distributive law). Which of them contain $x^4$, and how many are there?
*
*$1*1*1*x*x*x*x$ -- $\binom{7}{4}$ possibilities
*$1*1*1*1*(-2x^2)*x*x$ -- $7*\binom{6}{2}$ possibilities (or $\binom{7}{2}*5$)
*$1*1*1*1*1*1*(-2x^2)*(-2x^2)$ -- $\binom{7}{2}$ possibilities
Now calculate $\binom{7}{4}-2*7*\binom{6}{2}+(-2)(-2)\binom{7}{2}$ (note: This is also the expression you get with the former approach, and the result will be the one given in another answer already ;-)). I am not sure factorization saves much time here.
|
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|
Proving $~\sum_\text{cyclic}\left(\frac{1}{y^{2}+z^{2}}+\frac{1}{1-yz}\right)\geq 9$ $a$,$b$,$c$ are non-negative real numbers such that $~x^{2}+y^{2}+z^{2}=1$
show that $~\displaystyle\sum_\text{cyclic}\left(\dfrac{1}{y^{2}+z^{2}}+\dfrac{1}{1-yz}\right)\geq 9$
|
Letting $a = x^2, b = y^2, c = z^2$,
it suffices to prove that, for all $a, b, c > 0$ with $a + b + c = 1$,
$$\sum_{\mathrm{cyc}} \left(\frac{1}{b + c} + \frac{1}{1 - \sqrt{bc}}\right) \ge 9.$$
We have
$$\frac{1}{1 - \sqrt{bc}} = \frac{1 + \sqrt{bc}}{1 - bc}
\ge \frac{1 + \frac{2bc}{b + c}}{1 - bc} = \frac{b + c + 2bc}{(b+c)(1-bc)}.$$
It suffices to prove that
$$\sum_{\mathrm{cyc}} \left(\frac{1}{b + c} + \frac{b + c + 2bc}{(b+c)(1-bc)}\right) \ge 9$$
or
$$\sum_{\mathrm{cyc}} \frac{(1+b)(1+c)}{(b+c)(1-bc)} \ge 9.$$
The last inequality is killed by BW (Buffalo Way).
This proof is not nice. Hope to see a nice proof.
|
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|
Given $x^2 + y^2 + z^2 = 3$ prove that $x/\sqrt{x^2+y+z} + y/\sqrt{y^2+x+z} + z/\sqrt{z^2+x+z} \le \sqrt3$ Given $x^2 + y^2 + z^2 = 3$
Then prove that $${x\over\sqrt{x^2+y+z}} + {y\over\sqrt{y^2+x+z}} + {z\over\sqrt{z^2+x+y}} \le \sqrt 3$$
I tried using the Cauchy-Schwarz inequality but the inequality is coming in opposite direction.
|
Use $(a+b+c)^2 \leq 3(a^2+b^2+c^2)$. Then multiply out the denominators and solve.
Alternatively, maximize the left hand side subject to the constraint.
|
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|
Relative error machine numbers Suppose we are working with a machine that does arithmetical calculations with a relative accuracy of $\xi, |\xi| \leq \xi '$. We want to calculate $a^2 - b^2$ in the following two ways; $(A);$ $a^2 - b^2 = (a - b)(a+b)$ and $(B)$; $a^2 - b^2 = a \times a - b \times b$. Suppose $a$ and $b$ are machine numbers. Show that the relative error of the result using method $A$ is at most $3\xi'$ and $\frac{1 + (a^2 + b^2)}{|a^2 - b^2|}\xi'$ using method $B$.
The first part (method $(A)$) is not so difficult but I am making a mistake in the second part. Let's start by calculating $a \times a$. This should be equal to $a^2(1+\xi) = a^2 + a^2\xi$. Similarly, $b \times b = b^2 + b^2\xi$. Substracting these two terms gives $(a^2 + a^2\xi - b^2 + b^2\xi)(1+\xi)$ which is equal to $a^2 + a^2\xi - b^2 - b^2\xi + a^2\xi + a^2\xi^2 - b^2\xi - b^2\xi^2$. If we ignore the terms involving $\xi^2$ then we get $a^2 - b^2 + 2\xi(a^2 - b^2)$ and so the relative error is at most $2\xi'$. Obviously this is not even close to the answer I am supposed to be getting but I am not sure where I went wrong.
For the second part of the question, we assume that $a$ and $b$ are not machine numbers. Show that the relative errors are given by $\frac{3 + 2(a^2 + b^2)}{|a^2 - b^2|}\xi$ and $\frac{1 + 3(a^2 + b^2)}{|a^2 - b^2|}$ respectively. I am not really sure what happens when $a$ and $b$ are not machine numbers, besides the fact that we have some sort of extra approximation error that we need to calculate, however I can't find any info on how to do this.
|
You have to use different $\xi$ with absolute value bounded by the same $\xi'$.
I use the symbols $\oplus$ and $\odot$ for the machine operations. We get
$$(a \odot a )\ominus (b \odot b) = \\
(a^2(1+\xi_1) - b^2(1+\xi_2)) (1+\xi_3) = \\
a^2-b^2+a^2(\xi_1+\xi_3+\xi_1\xi_3)-b^2(\xi_1+\xi_2+\xi_1\xi_2)
$$
so the relative error is
$$\left|\frac{a^2(\xi_1+\xi_3+\xi_1\xi_3)-b^2(\xi_2+\xi_3+\xi_2\xi_3)}{a^2-b^2}\right|\le \\
\frac{a^2(|\xi_1|+|\xi_1||\xi_3|)+b^2(|\xi_2|+|\xi_2||\xi_3|)+|a^2-b^2||\xi_3|}{|a^2-b^2|} \le\\
\frac{a^2(\xi'+\xi'^2)+b^2(\xi'+\xi'^2)+|a^2-b^2|\xi'}{|a^2-b^2|} \approx \\
\left(\frac{(a^2+b^2)}{|a^2-b^2|}+1\right)\xi'
$$ if one ignores the quadratic terms.
If $a$ and $b$ are not machine numbers you have rounding errors. You have to calculate
$$(\Box a \ominus \Box b)\otimes (\Box a \oplus \Box b)$$
or
$$(\Box a \odot \Box a )\ominus (\Box b \odot \Box b) $$
where the function $\Box$ is defined as $\Box a=a(1+\xi_a)$ and $\Box b= b(1+\xi_b)$
|
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|
How to solve for matrix $A$ in $AB = I$ Given $B$ = $\begin{bmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
1 & 1 & 1
\end{bmatrix}$
I know that $B$ is equal to inverse of $A$, how can I go backwards to solve for $A$ in $AB = I$?
|
Using Gauss-Jordan to invert the given matrix $B$ you get
$$\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
1&1&0&0&1&0\\
1&1&1&0&0&1
\end{array}\right)
\leadsto
\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&-1&1&0\\
0&1&1&-1&0&1
\end{array}\right)
\leadsto
\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&-1&1&0\\
0&0&1&0&-1&1
\end{array}\right)
$$
Where $\left(\begin{array}{ccc}1&0&0\\-1&1&0\\0&-1&1\end{array}\right)$ is the inverse of $B$. Now you have
$$AB=I \Longleftrightarrow ABB^{-1} = IB^{-1} \Longleftrightarrow A = B^{-1}$$
Thus the previous found inverse of your matrix $B$ is the same as the matrix $A$ you are searching for.
|
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|
Show that it is a solution for all values of $t$ Show that $(1-t, 2+3t, 3-2t)$ is a solution for all values of $t$ to following linear system:
$a + b + c = 6$
$a - b - 2c = -7$
$5a + b - c = 4$
I have found that these rows are linearly dependent but I do not know how to continue. Can you help me? Here is after Gaussian Elimination:
$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & -\dfrac{1}{2} \\0 & 1 & \dfrac{3}{2} & \dfrac{13}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
|
Hint: One approach is to use Gaussian Elimination, you end up with:
$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & -\dfrac{1}{2} \\0 & 1 & \dfrac{3}{2} & \dfrac{13}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
Can you take it from here?
Update
We have:
*
*$a = -\dfrac{1}{2} +\dfrac{1}{2}c$
*$b = \dfrac{13}{2} -\dfrac{3}{2}c$
So, $c$ is a free variable and from the problem statement, they chose $c = 3-2t$.
Now, what are $a$ and $b$? Also, can you write this in another form?
|
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|
How prove this $x^2+y^2+z^2+3\ge 2(xy+yz+xz)$ let $x,y,z$ be real numbers, and such $$x+y+z+xyz=4$$ show that
$$x^2+y^2+z^2+3\ge 2(xy+yz+xz)$$
My try: let
$$x+y+z=p,xy+yz+xz=q,xyz=r$$
then
$$p+r=4$$
then
$$\Longleftrightarrow p^2+3\ge 4q$$
But I can't.Thank you
|
Note that it is sufficient to consider $x, y, z \ge 0$. Then the condition yields $4 = x+y+z+xyz \ge 3\sqrt[3]{xyz}+xyz \implies xyz \le 1$.
Also note that among $x-1, y-1, z-1$, at least two have the same sign. WLOG let $(y-1)(z-1) \ge 0$. Then we have:
$$(x-1)^2+(y-z)^2+2x(y-1)(z-1) \ge 0$$
$$\implies x^2+y^2+z^2+2xyz + 1 \ge 2(xy+yz+zx)$$
which along with $xyz \le 1$ gives you the result.
|
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|
Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$,
$$4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$$
I've tried expressing it as a sum of squares, but haven't got anywhere.
Hints are also welcome.
|
WLOG, let $x$ be the smallest, $y= a+x$ and $z = b+x$ for $a, b \ge 0$. Then we have
$$4(x + y + z)^3 - 27(x^2y + y^2z + z^2x) = 27x^3+27(a+b)x^2+9(a+b)^2x + (a-2b)^2 (4a+b)>0$$
|
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|
Question about writing integrals as a sum of integrals? So the question that is given is as follows:
Write the integral as a sum of integrals without absolute values and evaluate
$$ \int_{-2}^3 |x^3| \, dx$$
Can someone help me figure this out because I keep getting $20.25$ or $\dfrac{81}{4}$ but the answer is supposedly $24.25$ or $\dfrac{97}{4}$.
|
Note that $|x^3| = -x^3$ when $x \le 0$. We can therefore split the integral into two as follows:
$$\int_{-2}^3 |x^3| dx = \int_{-2}^0 |x^3| dx + \int_0^3 |x^3| dx = -\int_{-2}^0 x^3 dx + \int_0^3 x^3 dx = -\frac{x^4}{4}\bigg|_{-2}^0 + \frac{x^4}{4}\bigg|_0^3$$
$$= -\left(\frac{0}{4}-\frac{16}{4}\right) + \left(\frac{81}{4}-\frac{0}{4}\right) = \frac{16}{4}+\frac{81}{4} = \frac{97}{4}.$$
|
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|
Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdot3^n$ is divisible by $4$.
After that I am not sure what to do.
|
$$6\cdot7^{n+1}-2\cdot3^{n+1}=7(6\cdot7^n-2\cdot3^n)+14 \cdot 3^n-2 \cdot 3^{n+1}=7(6\cdot7^n-2\cdot3^n)+8 \cdot 3^n$$
The first term is divisible by 4 by inductive assumption and $8= 4 \cdot 2$
|
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|
$n$th derivative of $e^x \sin x$ Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:
Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.
Taking the derivatives: $f^{(n)}(x) = \dfrac{1}{2i} \cdot ((1+i)^n e^{(1+i)x} - (1-i)^n e^{(1-i)x})$
Now, I use that: $$ (1+i)^n = {\sqrt{2}}^n \cdot \left(\cos\dfrac{n \pi}{4} + i \sin\dfrac{n \pi}{4}\right) \\(1 - i)^n = \sqrt{2}^n \cdot \left( \cos \dfrac{-n \pi}{4} + i \sin \dfrac{-n \pi}{4} \right)$$
Plugging that mess, I get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\left(\cos \dfrac{n \pi}{4} + i \sin \dfrac{n \pi}{4}\right) e^{ix} - \left(\cos \dfrac{-n \pi}{4} + i \sin \dfrac{- n \pi}{4}\right) e^{-ix} \right)$$
But, $e^{ix} = \cos x + i \sin x$, and using Moivre's theorem, that makes: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\cos \left( x + \frac{n \pi}{4}\right) + i \sin \left( x + \frac{n \pi}{4}\right) - \left(\cos \left( - x - \frac{n \pi}{4}\right) + i \sin \left( -x -\frac{ n \pi}{4}\right)\right)\right)$$
and since $\cos$ is an even function, and $\sin$ is odd, we get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot 2i \sin \left(x + \dfrac{n \pi}{4}\right)$$
Simplifying, the answer would be $f^{(n)}(x) = e^x \cdot \sqrt{2}^n \cdot \sin\left(x + \dfrac{n \pi}{4}\right)$. I'm almost positive that this is it, but I just want to be sure. Thank you in advance!
|
Yes. Your derivation is correct. Here is a distillation that might make it simpler to see what is going on:
$$
\begin{align}
\frac{\mathrm{d}^n}{\mathrm{d}x^n}e^x\sin(x)
&=\frac{\mathrm{d}^n}{\mathrm{d}x^n}\frac1{2i}\left(e^{(1+i)x}-e^{(1-i)x}\right)\\
&=\frac1{2i}\left((1+i)^ne^{(1+i)x}-(1-i)^ne^{(1-i)x}\right)\\
&=\frac1{2i}2^{n/2}\left(e^{in\pi/4}e^{(1+i)x}-e^{-in\pi/4}e^{(1-i)x}\right)\\[4pt]
&=2^{n/2}e^x\sin(x+n\pi/4)
\end{align}
$$
|
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|
Field of order 8, $a^2+ab+b^2=0$ implies $a=0$ and $b=0$. I was able to come up with a proof for this problem however, it seems like my argument can work for any field of even order and not just odd powers of 2 so I'm convinced there is something wrong here. Can someone verify or see where the error in reasoning is?
Problem: Let $F$ be a field with $2^n$ elements, with $n$ odd. Show that for $a,b \in F$ that $a^2+ab+b^2=0$ implies that $a=0$ and $b=0$.
Proof:
Suppose $a,b \in F$ and $a^2+ab+b^2=0$.
$\implies a^2+2ab+b^2 = ab$
$\implies \frac{2^n}{2}(a^2+2ab+b^2) = \frac{2^n}{2}ab$
$\implies \frac{2^n}{2}a^2+ 2^nab+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$
$\implies \frac{2^n}{2}a^2+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$ (since F is a group under addition, then every element to the $|F|$ multiple is the identity thus $2^n(ab) = 0$)
$\implies \frac{2^n}{2}(a^2+b^2) = \frac{2^n}{n}ab$
$\implies a^2+b^2 = ab$
$\implies a^2-ab+b^2 = 0 = a^2+ab+b^2$
$\implies -ab = ab \implies 2ab=0 \implies ab=0$.
Thus, $a=0$ or $b=0$. However, if just one of them is zero, then so is the other ($a=0 \implies a^2+ab+b^2 = 0 \implies b^2 = 0 \implies b=0$). Thus, $a=0$ and $b=0$.
QED
Anyways, if there is something wrong with this proof, could someone give me a subtle hint perhaps? I've been stuck on this seemingly simple problem for awhile now.
|
The mistake you make is:
$$\frac{2^n}{2}(a^2+b^2) = \frac{2^n}{2}ab \Rightarrow a^2+b^2 =ab $$
Note that your field has characteristic $2$, which means that $\frac{2^n}{2}=0$!
You divide again by $0$ in the last line.
Hint
$$a^3-b^3=(a-b)(a^2+ab+b^2)=0$$
Thus $a^3=b^3$, and you also know what $a^7, b^7$ are....
You asked for a subtle hint, I didn't include more details, let me know if it is helpful, or you want more details.
|
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|
Area of a quadrilateral inside right angled triangle $ABC$ is right angle triangle. $AB=24 cm$, $BC=10 cm$, $AC=26 cm$.
Point $D$ on $AC$ (hypotenuse) bisects $AC$ and connects point $E$ on side $AB$ such that $ED$ is perpendicular to $AC$. Side $AC$ is folded into half so that angle $A$ falls on angle $C$, creating line $ED$ perpendicular to hypotenuse $AC$ and bisecting side $AC$. What is the area of quadrilateral $BEDC$?
|
Let us look at $\triangle AED$ and $\triangle CED$:
*
*$\angle EDA = \angle EDC = 90\,^{\circ}$
*$DC= DA = 13$ cm
*$ED$ is common
So, by $SAS$ congruence, those two triangles are congruent, and hence $EC = AE$. Now let $BE$ be $x$. So, by Pythagoras Theorem, $(EC)^2=x^2+100$. But, $(EC)^2= (AE)^2 = (24-x)^2$. Therefore, we get an equation, $(24-x)^2= 576 + x^2 - 48x = x^2+100$. Which upon solving, we get, $x = \frac{119}{12}$ .
Back to $\triangle AED$, we see that $(ED)^2 = (24-x)^2 - 169$. So, we know $x$ and we can get $ED$ as well. The final touch, $$Area(BEDC) = Area(ABC) - Area(EDA) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2}$$
Edit: Details of calculation:
$(ED)^2 = (24 - x)^2 - 169 = (24 - \frac{119}{12})^2 - 169 = (\frac{169}{12})^2 - 169 = 169(\frac{169}{144}- 1)= \frac{169 \cdot 25 }{144}$
Therefore, $ED = \sqrt{\frac{169 \cdot 25 }{144}} = \frac{13 \cdot 5 }{12}$.
So, $$Area(BEDC) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2} = 120 - \frac{13 \cdot 13 \cdot 5}{24} = 120- \frac{845}{24} = \frac{2035}{24} = 84.791666...$$
|
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|
Evaluate of number represented by the infinite series $\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$. Evaluate of number represented by the infinite series
$$\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$$
|
HINT:
Let $a = \sqrt{1/3 + ...}$
Then $a = \sqrt{1/3 + a}$
|
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|
Express the following in terms of y. If $y = 2^{2x}$, express the following in terms of y.
$2^{2x-1} - 4^{2x+1} + 16^{x-1}$
I began it in this way:
$2^{2x}\cdot 2^{-1} - (2^2)^{2x+1} + (2^4)^{x-1}$
$2^{2x}\cdot \frac{1}{2} - 2^{4x}\cdot2^2 + 2^{4x}\cdot2^{-1}$
$2^{2x}\cdot\frac{1}{2}-2^{2x}\cdot2^{2x}\cdot2^2+2^{2x}\cdot2^{2x}\cdot\frac{1}{2}$
$\frac{y}{2} - 4y^2 + \frac{y^2}{2}$
Is this right?
|
Almost. Note that $$(2^4)^{x-1}=2^{4(x-1)}=2^{4x-4}=2^{4x}\cdot2^{-4}\ne2^{4x}\cdot 2^{-1}.$$ Also note that you can gather like terms (specifically, the $y^2$ terms).
|
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|
Polynomial $P(x)$ such that $P(3k)=2$, $P(3k+1)=1$, $P(3k+2)=0$ for $k=0,1,2,\ldots,n-1$, $P(3n)=2$, and $P(3n+1)=730$ Let $n$ be a positive integer such that there exists a polynomial $P(x)$ over $\mathbb{Q}$ of degree $3n$ satisfying the conditions below:
$$P(0) = P(3) = \ldots = P(3n) = 2\,,$$
$$P(1) = P(4) = \ldots= P(3n - 2) = 1\,,$$
$$P(2) = P(5) = \ldots = P(3n - 1) = 0\,,$$
and
$$P(3n + 1) = 730\,.$$
Determine the value of $n$.
|
Let $\omega:=\exp\left(\frac{2\pi\text{i}}{3}\right)=\frac{-1+\sqrt{-3}}{2}$. Define $$\tau(z):=\left(\frac{1-\omega^2}{3}\right)z^2+\left(\frac{1-\omega}{3}\right)z+1\text{ for all }z\in\mathbb{C}\,.$$
Note that, for each $k\in\mathbb{Z}$, we have
$$\tau(\omega^k)=\left\{
\begin{array}{ll}
2&\text{if }k\equiv0\pmod{3}\,,\\
1&\text{if }k\equiv1\pmod{3}\,,\\
0&\text{if }k\equiv2\pmod{3}\,.
\end{array}
\right.$$
It is well known (see also here) that, if $f(x)\in \mathbb{K}[x]$ is a polynomial over a field $\mathbb{K}$ of degree $d$, then
$$\sum_{r=0}^{d+1}\,(-1)^r\,\binom{d+1}{r}\,f(x+r)\equiv 0\,.$$
From the result above, we have that
$$\sum_{r=0}^{3n+1}\,(-1)^r\,\binom{3n+1}{r}\,P(r)=0\,.$$
Recall that $P(r)=\tau(\omega^r)$ for $r=0,1,2,\ldots,3n$, and $P(3n+1)=3^6+1=3^6+\tau(\omega^{3n+1})$. That is,
$$\sum_{r=0}^{3n+1}\,(-1)^r\,\binom{3n+1}{r}\,\tau(\omega^r)=(-1)^{3n}\,3^6\,.$$
Ergo,
$$\left(\frac{1-\omega^2}{3}\right)\left(1-\omega^2\right)^{3n+1}+\left(\frac{1-\omega}{3}\right)\left(1-\omega\right)^{3n+1}+1(1-1)^{3n+1}=(-1)^{3n}3^6\,.$$
Thus,
$$2\,\text{Re}\left((1-\omega)^{3n+2}\right)=\left(1-\omega^2\right)^{3n+2}+\left(1-\omega\right)^{3n+2}=(-1)^{3n}3^7\,.$$
Since $1-\omega=\sqrt{-3}\,\omega^2$, we see that
$$(1-\omega)^{3n+2}=\sqrt{-3}^{3n+2}\omega^{6n+4}=\sqrt{-3}^{3n+2}\omega\,.$$
Therefore,
$$3^{\frac{3n+2}{2}}\,\text{Re}\left(\sqrt{-1}^{3n+2}\omega\right)=\text{Re}\left(\sqrt{-3}^{3n+2}\omega\right)=(-1)^{3n}\frac{3^7}{2}\,.$$
Since $\frac{1}{2}\leq \Big|\text{Re}\left(\sqrt{-1}^{3n+2}\omega\right)\Big|\leq\frac{\sqrt{3}}{2}$, we have that
$$\frac{3^{\frac{3n+2}{2}}}{2}\leq \frac{3^7}{2} \leq \frac{3^{\frac{3n+3}{2}}}{2}\,.$$
Consequently,
$$3n+2\leq 14\leq 3n+3\,.$$
This proves that $n=4$ is the only possibility. It is not difficult to see that $n=4$ indeed works.
|
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|
Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s etc.
Still I have absolutely no idea how to generalize it for $n$ first terms...
|
The way I approached this problem is dividing the sum into two halves. Where the first half contains all the numbers starting from $1$ and reaching upto $m$ such that $\sqrt{m}$ is greatest integer $<\lfloor\sqrt{n}\rfloor$, and second half contains the leftover numbers. Intuitively $m$ can be calculated by using simple formula $\lfloor\sqrt{n}\rfloor - 1$Mathematically:
$\displaystyle\sum_{k=1}^n\lfloor\sqrt{k}\rfloor = \underbrace{\displaystyle\sum_{p=1}^{\lfloor\sqrt{n}\rfloor-1}p\times(2p + 1)}^{\text{first part}} + \underbrace{(n - {\lfloor\sqrt{n}\rfloor}^2 + 1)\times{\lfloor\sqrt{n}\rfloor}}^{\text{second part}}$
The same can be resolved using the following identities.
$\displaystyle\sum_{x = 1}^n x = \frac{x\times(x+1)}{2}$ and $\displaystyle\sum_{x = 1}^n x^2 = \frac{x\times(x+1)\times(2 \times x+1)}{6}$
I didn't tested this solution with every value, but the solution seems convincing.
I am taking an example to demonstrate it, taking $n = 5$
The answer should contain $\underbrace{1+1+1+2+2}+2+2+2+3+... = 7$
The $\lfloor\sqrt{5}\rfloor-1 = 1$(upper limit for the first part)So, our first part is calculated as $1\times (2+1) = 3$
Similarly second part is calculated as $(5-{\lfloor\sqrt{5}\rfloor}^2)\times \lfloor\sqrt{5}\rfloor = (5 - 2^2+1)\times 2 = (2)\times 2 = 4$
So we get $4 + 3 = 7$ as our answer, which is correct.
The same intuition can be used to derive formula for $\displaystyle\sum_{k=1}^n\lfloor\sqrt[x]{k}\rfloor$.
i.e.
$\displaystyle\sum_{k=1}^n\lfloor\sqrt[x]{k}\rfloor = \displaystyle\sum_{p=1}^{\lfloor\sqrt[x]{p}\rfloor}p\times ((p+1)^x-p^x)+(n - (\rfloor\sqrt[x]{n}\rfloor)^x + 1)\times\lfloor\sqrt[x]{n}\rfloor$
P.S. This is just an intuition, if anything wrong, feel free to comment.
|
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|
Write your answer in simplest Simplify: sqrt(2009*2011*2015*2017+36)+10. Write your answer in simplest form
|
Let $\displaystyle\frac{2009+2011+2015+2017}4=2013=a$
$\displaystyle\implies 2009\cdot2011\cdot2015\cdot2017=(a-4)(a-2)(a+2)(a+4)$
$\displaystyle=(a^2-16)(a^2-4)=a^4-20a^2+64$
Now, $\displaystyle 2009\cdot2011\cdot2015\cdot2017+36$
$\displaystyle=a^4-20a^2+64+36=(a^2)^2+10^2-2\cdot a^2\cdot10=(a^2-10)^2$
|
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|
Number of factors of summation Let $a(n)$ be the number of $1$'s in the binary expansion of $n$. If $n$ is a positive integer, show that
$$\Bigg|\sum_{k=0}^{2^n-1}(-1)^{a(k)}\times 2^k\Bigg|$$
has at least $n!$ divisors.
I think this can be solved using induction, but I'm not sure. The first three values of the sum are $1$, $3$, and $45$, which indeed have $1$, $2$, and $6$ factors, repectively.
|
If we set
$$S(n) = \sum_{k=0}^{2^n-1} (-1)^{a(k)}\cdot 2^k,$$
then we note that $a(2^n + k) = a(k) + 1$, and hence
$$S(n+1) = \sum_{k=0}^{2^{n+1}-1} (-1)^{a(k)}\cdot 2^k = \sum_{k=0}^{2^n-1} (-1)^{a(k)}\cdot \left(2^k - 2^{2^n+k}\right) = -\left(2^{2^n}-1\right)\cdot S(n).$$
Thus
$$\lvert S(n)\rvert = \prod_{k=0}^{n-1} \left(2^{2^k}-1\right).$$
Now
$$\begin{align}
2^{2^n} - 1 &= \left(2^{2^{n-1}}-1\right)\left(2^{2^{n-1}}+1\right)\\
&= \underbrace{\left(2^{2^0}-1\right)}_{=1}\prod_{k=0}^{n-1} \left(2^{2^k}+1\right),
\end{align}$$
and hence
$$\lvert S(n)\rvert = \prod_{k=0}^{n-2}\left(2^{2^k}+1\right)^{n-1-k}.$$
Since the $F_k = 2^{2^k}+1$ are pairwise coprime - we have $F_n - 2= \prod\limits_{k=0}^{n-1} F_k$ by the above, so a common divisor of $F_m$ and $F_n$ for $m < n$ must divide $2 = F_n - \prod\limits_{k=0}^{n-1} F_k$, but the $F_k$ are all odd, so the $\gcd$ is $1$ -, and $F_k^m$ has at least $m+1$ divisors, the number of divisors of $\lvert S(n)\rvert$ is at least
$$\prod_{k=0}^{n-2} (n-1-k+1) = n!$$
Since $F_k$ is prime for $0 \leqslant k \leqslant 4$, we have equality for $n \leqslant 6$. $F_5 = 641 \cdot 6700417$ is composite, so for $n \geqslant 7$, $\lvert S(n)\rvert$ has strictly more divisors than $n!$. The Fermat numbers $F_k$ for $5 \leqslant k \leqslant 32$ are all known to be composite, it is an open problem whether any further Fermat primes beyond $F_4$ exist.
|
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|
A problem about power series and big-O The problem is:
Prove: There exist constants $a$, $b$ such that $\frac{z^3-5z^2+3z}{(z+2)^3}=1+\frac{a}{z}+\frac{b}{z^2}+O(\frac{1}{z^3})$ as $z\rightarrow \infty$ and find an explicit values for $a$ and $b$.
My thought: we know the power series of $\frac{1}{1-z}$ so by taking derivative we can get the power series of $\frac{1}{(1-z)^3}=\sum_{n=0}^{\infty }\frac{(n+1)(n+2)z^n}{2}$. Can we adapt this to get a power series for $(2+z)^{-3}$? Or: I have tried to divide $z^3$ both at the numerator and denominator of $\frac{z^3-5z^2+3z}{(z+2)^2}$, and get $\frac{1-\frac{5}{z}+\frac{3}{z^3}}{(1+\frac{2}{z})^3}$, so we can use the power series for $(1+\frac{2}{z})^{-3}$, which is $\sum_{n=0}^{\infty }\frac{(n+1)(n+2)(-2z)^n}{2}$, but how to continue?
Thanks in advance.
|
Take the series for $(1+\frac{2}{z})^{-3}$, write out the terms up to $(1/z)^3$, multiply each of the terms by $1-5/z+3/z^3$ and collect like terms (for the terms after $(1/z)^3$, note that if you hit them with a 3rd degree polynomial in $1/z$, their contributions are at most $(1/z)^4$ (more precisely, if you hit $(1/z)^n$ with $1-5/z+3/z^3$, you get a $(1/z)^{n+3}, (1/z)^{n+2},(1/z)^{n)}$ term, so for $n\geq 4$, the largest term is $(1/z)^n$ which is $O(1/z^3)$), so you only need to worry about coefficients up to $(1/z)^3$ for doing the multiplication (the rest will be captured by the $O$)).
|
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|
How do I determine the sign of $\sin\theta$ in this question.
$\alpha$ and $\beta$ are two different values of $\theta$ lying between $0$ and $2\pi$ which satisfy the equation $\begin{equation}6\cos\theta + 8\sin\theta = 9\end{equation}$. Find $\sin(\alpha+\beta).$
I solved and got $\cos(\alpha+\beta) = \frac{-28}{100}$.
So $\sin(\alpha+\beta) = \pm\frac{24}{25}$.
I think the sign of $\sin(\alpha+\beta)$ could be either of '+' and '-', however my book says it must be '+'. How do I determine the sign?
|
Your book is correct. We don't need to find either $\alpha$ or $\beta$, i.e. to solve the given linear equation in $\cos\theta$ and $\sin\theta$, which could be transformed into a quadratic equation in $\tan\frac{\theta}{2}$.
[EDIT. Since you've commented that your course was a beginner's one, I edited this answer to add the derivation of all trigonomatric identities I used, from the addition and subtraction formulas.]
Equating the equations satisfied by $\alpha$ and $\beta$
\begin{eqnarray*}
6\cos \alpha +8\sin \alpha &=&9\tag{1} \\
6\cos \beta +8\sin \beta &=&9,\tag{2}
\end{eqnarray*}
and then simplyfying and rearranging, we get
\begin{eqnarray*}
3\left( \cos \alpha -\cos \beta \right) &=&4\left( \sin \beta -\sin \alpha
\right) .\tag{3}
\end{eqnarray*}
Applying the sum-to-product identities
\begin{eqnarray*}
\cos \alpha -\cos \beta &=&-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha
-\beta }{2} \tag{4}\\
\sin \beta -\sin \alpha &=&-2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha
+\beta }{2},\tag{5}
\end{eqnarray*}
assuming$^1$ that $\sin \frac{\alpha-\beta }{2}\ne 0 $ we have that
\begin{eqnarray*}
-3\times 2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}
&=&-4\times 2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}\tag{6} \\
3\sin \frac{\alpha +\beta }{2} &=&4\cos \frac{\alpha +\beta }{2}\tag{7} \\
\tan \frac{\alpha +\beta }{2} &=&\frac{4}{3}.\tag{8}
\end{eqnarray*}
Finally expressing $\sin \left( \alpha +\beta \right) $ in terms of $\tan
\frac{\alpha +\beta }{2}$, using the double-angle identity we find
\begin{equation*}
\sin \left( \alpha +\beta \right) =\frac{2\tan \frac{\alpha +\beta }{2}}{
1+\tan ^{2}\frac{\alpha +\beta }{2}}=\frac{2\left( \frac{4}{3}\right) }{
1+\left( \frac{4}{3}\right) ^{2}}=\frac{24}{25},\tag{9}
\end{equation*}
which agrees with the $\cos \left( \alpha +\beta \right) $ value you found:
\begin{equation*}
\cos \left( \alpha +\beta \right) =\frac{1-\tan ^{2}\frac{\alpha +\beta }{2}
}{1+\tan ^{2}\frac{\alpha +\beta }{2}}=\frac{1-\left( \frac{4}{3}\right) ^{2}
}{1+\left( \frac{4}{3}\right) ^{2}}=-\frac{7}{25}.\tag{10}
\end{equation*}
EDIT. Justification of:
*
*Identity $(4)$
\begin{eqnarray*}
\cos \left( a\pm b\right) &=&\cos a\cos b\mp \sin a\sin b \\
&&\text{If }\left\{
\begin{array}{c}
a+b=\alpha \\
a-b=\beta ,
\end{array}
\right. \text{ then }\left\{
\begin{array}{c}
a=\frac{\alpha +\beta }{2} \\
b=\frac{\alpha -\beta }{2}
\end{array}
\right. \text{ and} \\
\cos \alpha &=&\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
-\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \\
\cos \beta &=&\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
+\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \\
\cos \alpha -\cos \beta &=&-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha
-\beta }{2}.
\end{eqnarray*}
*Identity $(5)$
\begin{eqnarray*}
\sin \left( a\pm b\right) &=&\sin a\cos b\pm \sin b\cos a \\
&&\text{If }\left\{
\begin{array}{c}
a+b=\alpha \\
a-b=\beta ,
\end{array}
\right. \text{ then }\left\{
\begin{array}{c}
a=\frac{\alpha +\beta }{2} \\
b=\frac{\alpha -\beta }{2}
\end{array}
\right. \text{ and} \\
\sin \alpha &=&\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
+\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\
\sin \beta &=&\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
-\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\
\sin \beta -\sin \alpha &=&-2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha
+\beta }{2}.
\end{eqnarray*}
*identities $(9)$ and $(10)$
\begin{eqnarray*}
\sin \left( \alpha +\beta \right) &=&2\sin \frac{\alpha +\beta }{2}\cos
\frac{\alpha +\beta }{2}=\frac{2\sin \frac{\alpha +\beta }{2}\cos \frac{
\alpha +\beta }{2}}{\cos ^{2}\frac{\alpha +\beta }{2}+\sin ^{2}\frac{\alpha
+\beta }{2}} \\
&=&\frac{2\tan \frac{\alpha +\beta }{2}}{1+\tan ^{2}\frac{\alpha +\beta }{2}}
\\
&& \\
\cos \left( \alpha +\beta \right) &=&\cos ^{2}\frac{\alpha +\beta }{2}-\sin
^{2}\frac{\alpha +\beta }{2}=\frac{\cos ^{2}\frac{\alpha +\beta }{2}-\sin
^{2}\frac{\alpha +\beta }{2}}{\cos ^{2}\frac{\alpha +\beta }{2}+\sin ^{2}
\frac{\alpha +\beta }{2}} \\
&=&\frac{1-\tan ^{2}\frac{\alpha +\beta }{2}}{1+\tan ^{2}\frac{\alpha +\beta
}{2}}
\end{eqnarray*}
--
$^1$ The assumption $\sin \frac{\alpha -\beta }{2}\neq 0$ is valid because if $
0\neq \frac{\alpha -\beta }{2}=\pi $, then without loss of generality we might
assume that $0=\beta <\alpha =2\pi $, but we would get a contradition:
\begin{eqnarray*}
6\cos \left( 2\pi \right) +8\sin \left( 2\pi \right) &\neq &9 \\
6\cos \left( 0\right) +8\sin \left( 0\right) &\neq &9.
\end{eqnarray*}
|
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|
Double integrals in polar coordinates Determine the domain of
$$\!\!\!\!\!\!\!{\small
D \equiv
\left\{\left(x,y\right) \in \mathbb{R}^{2}\
{\large\mid}\
x \in \left[\,{-\,\frac{1}{\,\sqrt{\,{2}\,}\,},
\frac{1}{\,\sqrt{\,{2}\,}\,}}\,\right],\
y \in \left[\,{\left\vert\,{x}\,\right\vert,
\,\sqrt{\,{1 - x^{2}}\,}\,}\,\right]\right\}}
$$
in polar coordinates and draw it.
Also how would you integrate $$\int\int_D \frac{1}{1+x^2 + y^2}dA$$ which is i guess
$$\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{|x|}^{\sqrt{1-x^2}} \frac{1}{1+x^2 + y^2}dydx$$
I guess in the integral you can use the polar coordinates
$$\int\int_D \frac{1}{1+r^2\cos^2(\phi) + r^2\sin^2(\phi)}rdrd\phi$$
$$\int\int_D \frac{r}{1+r^2}drd\phi$$
$$\int_{\frac{1}{4\pi}}^{\frac{3}{4\pi}}\int_0^1 \frac{r}{1+r^2}drd\phi=\int_{\frac{1}{4\pi}}^{\frac{3}{4\pi}}\left(\frac 12 \ln(1+1^2)-\frac 12\ln(1+0^2) \right)d\phi$$
$$\int_{\frac{1}{4\pi}}^{\frac{3}{4\pi}}\frac{\ln{2}}{2}d\phi=\left(\frac{3}{4\pi}\frac{\ln{2}}{2}-\frac{1}{4\pi}\frac{\ln{2}}{2} \right)=\frac{\pi}{4}\ln{2}$$
Did I get it right?
|
Draw a picture. A simple plot reveals that the domain $D$ is simply the sector of the circle $r=1$ between two values of $\theta$. A little thought provides those values of $\theta$ (i.e., what purpose does the absolute value serve?).
The integrand you show is also wrong, as $1+r^2 \ne 2$.
The answer I get is $(\pi/4) \log{2}$.
|
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|
$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Again, we can have, $ab+bc+ca\le b(3-b)+\dfrac{1}{b}$
Now, I have to show that, $b(3-b)+\dfrac{1}{b}\le 3$
How can I prove this now? Please help.
|
I think you can check this inequality by Lagrange multiplier method. Consider the following Lagrange function:
\begin{equation}
L(a,b,c)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)+\lambda(a+b+c-3)
\end{equation}
For finding the extreme value, we have:
\begin{equation}
\frac{\partial L}{\partial a}=2ab^2c^2+2(3-2b)(3-2c)+\lambda=0\\
\frac{\partial L}{\partial b}=2a^2bc^2+2(3-2a)(3-2c)+\lambda=0\\
\frac{\partial L}{\partial c}=2a^2b^2c+2(3-2a)(3-2b)+\lambda=0\\
\frac{\partial L}{\partial \lambda}=a+b+c-3=0
\end{equation}
Because $a,b,c$ is symmetric in systems, I only show part of the solutions and the corresponding extreme values:
\begin{equation}
a=1,b=1,c=1,L=0\\
a=\frac{3}{2},b=\frac{3}{2},c=0,L=0\\
a=\sqrt[3]{2},b=\sqrt[3]{2},c=3-2\sqrt[3]{2},L\approx 0.1107011764
\end{equation}
So, we obtain the minimum of $L(a,b,c)$ is $0$. In summary, under the constrain $a+b+c=3$, we have $L(a,b,c)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)+\lambda(a+b+c-3)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)\geq0$.
This is the inequality you want to prove.
|
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|
Taylor series of $\frac 1 {1+x^2}$ I have to construct the Taylor series of
$$\frac 1 {1+x^2}$$
around $0$ and $1$ and analyze the convergence in both cases. Also (but this is a consequence of the previous series) I have to construct the Taylor series of
$$arctan(x)$$
What I have so far:
*
*I know that the Taylor polynomial around $0$ of $\frac 1 {1+x^2}$ is $$1-x^2+x^4-x^6+....+(-1)^nx^{2n}$$
*If I have the power series of a function f such as $F'=f$, I can construct the power series of $F$ with $F(x)=F(a)+\sum_{n=1}^{\infty}\frac {a_n} {n+1} (x-a)^{n+1}$
|
If I have the power series of a function f such as $F'=f$, I can construct the power series of $F$ with $F(x)=F(a)+\sum_{n=1}^{\infty}\frac {a_n} {n+1} (x-a)^{n+1}$
Have you tried this? Note that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$.
You have the power series for $\frac{1}{1+x^2}$ centered at $0$, for which
$$
a_n = \begin{cases}
(-1)^{n/2} & n \text{ is even}\\
0 & \text{otherwise}
\end{cases}
$$
In order to find the Taylor expansion of $\frac{1}{x^2 + 1}$ at $1$, note that
$$
\begin{align}
\frac{1}{x^2 + 1} &=
\frac{1}{1 + (1+(x-1))^2} =
\frac{1}{1 + 1 + 2(x-1) + (x-1)^2}
\\ &
= \frac{1}{2 + 2(x-1) + (x-1)^2}
= \frac{1}{2} \cdot \frac{1}{1 + \left[(x-1) + \frac{(x-1)^2}{2}\right]}
\\ &
= \frac 12 \left(1 - \left[(x-1) + \frac{(x-1)^2}{2}\right] + \left[(x-1) + \frac{(x-1)^2}{2}\right]^2 - \cdots \right)
\\ &
= \frac 12 - \frac 12 (x-1) + \frac 14(x-1)^2 \\
&\qquad- \frac 18 (x-1)^4 + \frac 18 (x-1)^5 -\frac 1{16} (x-1)^6\\
&\qquad+ \frac 1{32} (x-1)^8 - \frac 1{32} (x-1)^9 + \frac 1{64}(x-1)^{10} - \cdots
\end{align}
$$
|
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|
How to solve this trig problem? $\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$ Basic trig problem my brother ask me, but I don't know how to do it:
$$\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$$
|
$\sec\left(\sin^{-1}\left(\dfrac{-5}{13}\right)-\tan^{-1}\left(\dfrac{4}{3}\right)\right)$
Let $\alpha = \sin^{-1}\left(\dfrac{-5}{13}\right)$
Then $\alpha$ is the angle in the fourth quadrant that corresponds to the point
$(x,y) = (12,-5)$ with amplitude $r = 13.$
So $\cos(\alpha) = \dfrac xr = \dfrac{12}{13}$ and
$\sin(\alpha) = \dfrac yr = \dfrac{-5}{13}$
Let $\beta = \tan^{-1}\left(\dfrac{4}{3}\right)$
Then $\beta$ is the angle in the first quadrant that corresponds to the point $(x,y) = (3,4)$ with amplitude $r = 5.$
So $\cos(\beta) = \dfrac xr = \dfrac 35$ and
$\sin(\beta) = \dfrac yr = \dfrac 45$
So
\begin{align}
\sec\left(\sin^{-1}\left(\dfrac{-5}{13}\right)-
\tan^{-1}\left(\dfrac{4}{3}\right)\right)
&= \sec(\alpha - \beta)\\
&= \dfrac{1}{\cos(\alpha - \beta)}\\
&= \dfrac{1}{\cos \alpha \; \cos \beta + \sin \alpha \; \sin \beta}\\
&= \dfrac{1}{\dfrac{12}{13} \cdot \dfrac{3}{5} +
\dfrac{-5}{13} \cdot \dfrac 45}\\
&= \dfrac{65}{16}
\end{align}
|
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|
Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
|
By Holder
$$\sum_{cyc}\frac{a^3}{bc}\geq\frac{(a+b+c)^3}{3(ab+ac+bc)}=\frac{(a+b+c)\cdot(a+b+c)^2}{3(ab+ac+bc)}\geq a+b+c$$
|
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|
Finding intersection points of 2 functions. My method is incomplete. These are the 2 functions :
$y = x^{4}-2x^{2}+1$
$y = 1-x^{2} $
Here's how I solved It :
$x^{4}-2x^{2}+1 = 1-x^{2}$
$x^{4}-x^{2} = 0$
$x^2(x^2-1)=0$
$x^2-1=0$
$x=\pm \sqrt{1} $
Value of $y$ when $x=1$
$y=1-x^2\\y=1-1\\y=0$
Value of $y$ when $x=(-1)$
$y=1-x^2\\y=1-(-1)^2\\y=1-1\\y=0$
So the intersection points of the 2 functions are $(1,0)$ and $(-1,0)$.
The problem
The problem is when I used a graphing calculator to find the intersection points of the above 2 functions it gave me 3 results instead of 2. They were :
$(-1,0),(0,1),(1,0)$
So what am I missing ? Why don't I get 3 intersection points ?
Best Regards !
|
Functions:
a.) y = x^4 - 2x^2 + 1
b.) y = 1 - x^2
Solving:
x^4 - 2x^2 + 1 = 1 - x^2
x^4 - x^2 = 0
x^2 * ( x^2 - 1) = 0
Now set each part equal to 0
x^2 = 0
x = 0
x^2 - 1 = 0
x = +/-1
Now plug in each value of x to get the corresponding value of y
X = 0
y = 1 - (0)^2
y = 1
Intersection at (0,1)
X = 1
y = 1 - (1)^2
y = 1 - 1
y = 0
Intersection at (1,0)
X = -1
y = 1 - (-1)^2
y = 1 - 1
y = 0
Intersection at (-1,0)
Intersections for function a and b are:
(0,1)
(-1,0)
(1,0)
|
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|
$\pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z},\ n \neq 0} \frac{1}{z-n}+ \frac{1}{n}$ I'm reading the proof that $$\pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z},\ n \neq 0} \frac{1}{z-n}+ \frac{1}{n}$$
There is a function $$h(z) =\pi \cot (\pi z) -[ \frac{1}{z} + \sum_{n \in \mathbb{Z},\ n \neq 0} \frac{1}{z-n}+ \frac{1}{n} ]$$ and the proof states that $h(1) = 0$.
I don't understand this point, why $h(1) = 0$ ?
|
Consider the function
$$d(z) = \pi \cot \pi z - \frac{1}{z-1}$$
in a neighbourhood of $1$. Writing $z = 1+h$, we have
$$\begin{align}
d(1+h) &= \pi\frac{\cos \pi(1+h)}{\sin \pi(1+h)} - \frac{1}{h}\\
&= \pi \frac{-\cos \pi h}{-\sin \pi h} - \frac{1}{h}\\
&= \frac{1 - \frac{\pi^2h^2}{2} + O(h^4)}{h\left(1 - \frac{\pi^2h^2}{6}+O(h^4)\right)} - \frac{1}{h}\\
&= \frac{1}{h}\left(1- \frac{\pi^2h^2}{2}+O(h^4)\right)\left(1 + \frac{\pi^2h^2}{6}+O(h^4)\right) - \frac{1}{h}\\
&= \frac{1}{h}\left(1-\frac{\pi^2h^2}{3} + O(h^4)\right) - \frac{1}{h}\\
&= -\frac{\pi^2h}{3} + O(h^3),
\end{align}$$
so $d$ has a removable singularity in $1$, and after removing it, we have $d(1) = 0$.
Then it remains to see that
$$e(z) = \left(\frac{1}{z} + \sum_{n\neq 0} \frac{1}{z-n}+\frac{1}{n}\right) - \frac{1}{z-1} = \frac{1}{z} + 1 + \sum_{n\notin \{0,1\}} \frac{1}{z-n} + \frac{1}{n}$$
also has a zero in $1$. Having cancelled the $\frac{1}{z-1}$ term, we can compute $e(1)$ simply by substituting $1$ for $z$ in the last representation. The sums nicely telescope then and yield the expected result. Since $h(z) = d(z) - e(z)$, that concludes the proof.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How do i solve the cubic equation?
$$x^3 - 3x^2 - 3x +2 = 0$$
The rational root test does not work; there are no rational roots.
|
Here is a step by step procedure.
First shift $x$ so there is no quadratic term. To do this we shift by 1/3 the quadratic term (assuming leading coefficient is 1) This give
$$
x=y+1 \Rightarrow y^3-6\,y-3 =0
$$
We now reduce the cubic by the substitution $y = A(z+1/z)$
$$
y = A(z+1/z) \Rightarrow \frac{z^6\,A^3+3\,z^4\,A^3+3\,z^2\,A^3+A^3-6\,z^4\,A-6\,z^2\,A-3\,z^3}{z^3}=0
$$
We now pick $A \neq 0$ so that $Z^4$ term is zero. This requires
$$ A(A^2-2)=0$$
So pick $A=\sqrt{2}$ and the equation becomes
$$2^{{{3}\over{2}}}\,z^6-3\,z^3+2^{{{3}\over{2}}}=0$$
This is a quadratic in $z^3$. So set $z=u^{1/3}$ to get
$$
z=u^{1/3} \Rightarrow 2^{{{3}\over{2}}}\,u^2-3\,u+2^{{{3}\over{2}}}=0$$
Solve this quadratic for $u$. You will have two roots.
Use them to find $z$. Recall each number will have 3 cube roots, but you will have only 3 possible values and not 6.
Once you have $z$, calculate $y$ and then finally $x$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Functional equation of function defined over non-negative reals satisfying $f \big(xf(y)\big)f(y)=f(x+y)$, $f(2)=0$ and $f(x)\ne0$ for $x\in[0, 2)$
Find all $ f:[0,\infty)\to [0,\infty) $ such that $ f (2)=0 $, $ f (x)\ne 0 $ for $ x\in [0, 2) $ and
$$ f \bigl(xf (y)\bigr) f (y)=f (x+y) $$
for all $ x, y\ge 0 $.
I tried plugging in values of $ x $ and $ y $ but didn't succeed.
Do you know how to proceed?
|
Substitute in $x=y=0$, we get that $ f(0) f(0) = f(0)$. Since $f(0) \neq 0$, thus $f(0) = 1$.
Substitute in $x=x, y = 2$, $0 = f(x\cdot 0) \cdot 0 = f(x+2)$. Hence for $x \geq 2, f(x) = 0 $.
We now focus our attention to the region $x\leq 2$.
Substitute in $x=2-y, y = y < 2$. We get that $ f[ (2-y) f(y) ] f(y) = f(2) = 0$. Since $f(y) \neq 0$ thus $(2-y) f(y) \geq 2$, or that $f(y) \geq \frac{2}{2-y}$.
Suppose that there exists a value $y$ such that $f(y) > \frac{2}{2-y}$. Then, take the value $x$ such that $ \frac{2}{f(y)} < x < 2-y$. We get that
$$ 0 = f( x f(y) ) f(y) = f(x+y) \neq 0,$$
which is a contradiction. Hence, at best, $f(y) = \frac{2}{2-y}$.
It is easy to check that $ f(x) = \begin{cases} \frac{2}{2-x}& 0\leq x < 2 \\ 0 & 2 \leq x \\ \end{cases}$ is a solution to the functional equation. (The simplest approach is to condition on $x+y<2$ and $x+y \geq 2$). Hence, it is the only solution.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/681955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Problem related to DE Solve and comment on the solution behaviour at $|x|$ approaches infinity (bounded or unbounded as $x$ approaches $\pm\infty$.
$$\frac{1}{1+x^2} + \sin y + y'\left(x\cos y + \frac{y^2}{2}\right)=0,\quad y(1)=\Pi.$$
Can I have some hints on how to get started on this problem? Thanks
I cannot even separate it.
|
Deal with the LHS of the ODE and we can obtain the solution
\begin{equation}
\frac{1}{1+x^2}+(\sin(y)+y'x\cos(y))+y'\frac{y^2}{2}=0\\
\frac{1}{1+x^2}+(x\sin(y)+\frac{y^3}{6})'=0\\
\int(x\sin(y)+\frac{y^3}{6})'dx+\int\frac{1}{1+x^2}=0\\
x\sin(y)+\frac{y^3}{6}+\arctan(x)+C=0
\end{equation}
which $C$ is a constant given by the initial condition that $y(1)=\pi$
\begin{equation}
\frac{\pi^3}{6}+\frac{\pi}{4}+C=0\\
C=-(\frac{\pi^3}{6}+\frac{\pi}{4})
\end{equation}
Rewritten the solution by the following form:
\begin{equation}
\frac{y^3}{6}=-x\sin(y)-\arctan(x)-C
\end{equation}
If $|x|\rightarrow+\infty$, we found that the $\arctan(x)$ is bounded, and $x\sin(y)$ is unbounded. So we can claim that $y(x)$ is unbounded when $|x|$ going to infinity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/682918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
I cannot find the derivatives of these functions using the product rule? I need help finding the derivative of $y=x^3(3x+7)^2$ at $x=-2$
I tried to simplify the function to $y=x^3 (3x+7)(3x+7)$ and the simplify it into two terms and the derivatives of those terms using the product rule, but that doesn't work.
Since I don't understand how to approach this type of question (with the power on the outside), I couldn't do this one either: $y=(2x+1)^5(3x+2)^4$ at $x=-1$
I don't need help finding the derivative at the specific values of $x$, I just can't seem to get the general case. BTW, I can only use the product rule as my teacher did not teach the quotient and chain rule to us, so she doesn't allow us to use them.
the answers(aka the derivative):
*
*${(3x^2)(3x+7)^2}+{(x^36)(3x+7)}$
*$5(2x+1)^4 (2) (3x+2)^4 + (2x+1)^54(3x+2)^3(3)$
|
The product rule is: if $f(x) = g(x)h(x)$, then $f'(x) = g'(x)h(x)+g(x)h'(x)$. Extending it to three terms, we get:
$$f(x) = g(x)h(x)j(x) \implies f'(x) = g'(x)\left[h(x)j(x)\right]+g(x)\left[h(x)j(x)\right]' \\ = g'(x)h(x)j(x)+g(x)\left[h'(x)j(x)+h(x)j'(x)\right] \\
= g'(x)h(x)j(x)+g(x)h'(x)j'(x)$$
as we might expect.
So, solving via the product rule, $$f'(x) = (x^3)'(3x+7)(3x+7)+x^3(3x+7)'(3x+7)+x^3(3x+7)(3x+7)'.$$
Notice that the last two terms are the same.
$$f'(x) = 3x^2(3x+7)^2+2x^3(3x+7)'(3x+7) \\
= 3x^2(3x+7)^2+2x^3(3x+7)(3) \\
= 9x^3+21x^2+3(6x^4+14x^3) \\
= 18x^4+51x^3+21x^2.$$
This is not the easiest way to solve the problem. The easiest way is to use the product rule once, and the chain rule once:
$$f'(x) = 3x^2(3x+7)^2+x^3\cdot 2 \cdot(3x+7)\cdot (3x+7)' \\
= 9x^3+21x^2+2x^3(3x+7)(3) \\
= 18x^4+51x^3+21x^2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
I determined the coefficients of the Fourier series, which are
$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}$$
Then, I get
$$x^3 = \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(nx)$$
If $x = \pi$, then
$$\begin{aligned}
\pi^3 &= \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(n\pi)\\
\dfrac{3\pi^3}{4} &= \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}(-1)^n
\end{aligned}$$
I'm stuck. It's easy to compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$, using the Fourier series, but for this type of problem I'm stuck.
Any comments or suggestions? By the way, I know that
$$\sum\limits_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$
I need to know how to get there.
|
From your identity
\begin{equation*}
\frac{3\pi ^{3}}{4}=\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{
\pi n^{4}}(-1)^{n}
\end{equation*}
expanding the right hand side and using the result $\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi^2}{6}$, we get
\begin{eqnarray*}
\frac{3\pi ^{4}}{4} &=&\sum_{n=1}^{\infty }\frac{6(\pi
^{2}n^{2}-2)(-1)^{n}+12}{n^{4}}(-1)^{n} \\
&=&6\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}}-12\sum_{n=1}^{\infty }\frac{1
}{n^{4}}+12\sum_{n=1}^{\infty }\frac{(-1)^{n}}{n^{4}} \\
&=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-12\sum_{n=1}^{\infty }
\frac{(-1)^{n-1}}{n^{4}}.
\end{eqnarray*}
Now we need to express the alternating series $\sum_{n=1}^{\infty }\frac{
(-1)^{n-1}}{n^{4}}$ in terms of $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$, e.g.
as follows
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n-1}}{n^{4}} &=&\sum_{n=1}^{
\infty }\frac{1}{n^{4}}-2\sum_{n=1}^{\infty }\frac{1}{(2n)^{4}}
=\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{1}{2^{3}}\sum_{n=1}^{\infty }
\frac{1}{n^{4}}=\frac{7}{8}\sum_{n=1}^{\infty }\frac{1}{n^{4}}.
\end{eqnarray*}
Then
\begin{eqnarray*}
\frac{3\pi ^{4}}{4} &=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{
21}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}} \\
&=&\pi ^{4}-\frac{45}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}}.
\end{eqnarray*}
Solving for $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$ we finally obtain
\begin{equation*}
\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{2}{45}\left( \pi ^{4}-\frac{3\pi
^{4}}{4}\right) =\frac{\pi ^{4}}{90}.
\end{equation*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Prove that $a^ab^b>\left(\frac{a+b}{2}\right)^{a+b}$ where $a\ne b$
Prove that $a^a\cdot b^b>\left(\dfrac{a+b}{2}\right)^{a+b}$ where $a\ne b$.
My work:
$$a^a\cdot b^b>\left(\frac{a+b}{2}\right)^a\cdot\left(\frac{a+b}{2}\right)^b\implies 1>\left(\frac{1+\frac{b}{a}}{2}\right)^a\cdot\left(\frac{1+\frac{a}{b}}{2}\right)^b$$
Now, I am stuck. Please help.
|
Take the function $f(x)=x\ln x,$ where $x>0$, then $f'(x)=1+\ln(x)$ and $f''(x)=\frac{1}{x}>0,$ so $f(x)$ is convex functon. Apply jensen-inequality in $f$, we have $$\dfrac{a\ln a + b\ln b}{2}\ge \dfrac{(a+b)}{2}\ln\bigg(\dfrac{a+b}{2}\bigg),$$
on simplification and taking anti-log on both sides it becomes, $a^ab^b\ge \bigg(\dfrac{a+b}{2}\bigg)^{a+b}$.
|
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"url": "https://math.stackexchange.com/questions/689667",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$ Assume I have a function $f(n) = \frac{4n+1}{n(2n-1)}$ with $n \in \mathbb{N} \setminus \left\{ 0 \right\}$. The objective is to find all $n$ for which $f(n)$ has a proper decimal fraction. I know that any given fractiononly has a proper decimal fraction whenever the denominator only has the prime factors $2$ and $5$.
I wrote a script in Mathematica that calculated that for $n < 100000$ this only applies to $n_1 = 2 \lor n_2 = 8$. However this doesn't give me any explanation as to why this is the case. Assuming the fraction is already completely reduced: For $n(2n-1)$ to have the prime factors $2$ and $5$ only, $2n-1$ needs to have these prime factors as well. From this and the assumption I was able to figure that $n = \frac{1}{2} \left(5^i + 1 \right)$, whose last digit always is a $3$, which in turn means it is uneven. Also $n$ needs to have the prime factors $2$ and $5$ only, but as it needs to be uneven as shown before it can only have the prime factors $5$, which would however mean the last digit would have to be a $5$. I conclude that for every $n$ for which $f(n)$ has a proper decimal fraction, $f(n)$ cannot be already completely reduced, which means there are additional prime factors in the denominator that are also in the numerator. This coincides with the results the Mathematica script gave me, but I have no idea how to go about finding these common prime factors and possibly eliminating them, etc.
I would greatly appreciate any help here!
Regards,
bk1ng
|
Note that $\gcd(4n+1,n)=1$ and $\gcd(4n+1,2n-1)=\gcd(3,2n-1)\in\{1,3\}$.
Thus $n$ must be of the form $n=2^a5^b$ and the odd number $2n-1$ must be of the form $2n-1=5^c$ or $2n-1=3\cdot 5^c$ with $a,b,c\in\mathbb N_0$. So from $1=2\cdot n-(2n-1)$ we find
$$\tag1 1 = 2^{a+1}5^b-5^c\qquad \text{or}\qquad 1=2^{a+1}5^b-3\cdot 5^c.$$In both cases $(b\ge1\land c\ge1)$ leads to a contradiction as the right hansd side then is a multiple of $5$, hence $b=0$ or $c=0$.
The case $c=0$ is easy: $1=2^{a+1}5^b-1$ leads to $a=b=0$, $\color{red}{n=1}$; and $1=2^{a+1}5^b-3$ leads to $a=1, b=0$, i.e. $\color{red}{n=2}$. Remains the case $b=0$ and $c>0$, i.e.
$$\tag2 1=2^{a+1}-(1\text{ or }3)\cdot5^c.$$
Since $c>0$, by taking remainders modulo $5$ we find in both cases
$$\tag3 1\equiv 2^{a+1}\pmod 5.$$
This implies that $a+1$ is a multiple of $4$, so $2^{a+1}=16^d$ for some $d$.
Rearranging $(2)$ we find
$$\tag4(1\text{ or }3)\cdot 5^c = 16^d-1=(4^d-1)(4^d+1).$$
At most one of the factor $4^d\pm1$ is a multiple of $5$, hence the other must be $\le3$. The only possibility is $d=1$, which leads to $a=3$, i.e. $\color{red}{n=8}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Reflection across the plane Let $T: \Bbb R^3 \rightarrow \Bbb R^3$ be the linear transformation given by reflecting across the plane $S=\{x:-x_1+x_2+x_3=0\}$ (...)
Then, $S={\rm gen}[(1,1,0),(1,0,1)].$
But how can I get the matrix $R_v$ such that reflects across $S$?
Thanks!
|
The vector $\vec{n}=\langle -1,1,1 \rangle$ is normal to the plane, thus $T$ maps $\vec{n}$ to $-\vec{n}=\langle 1,-1,-1 \rangle$. It also fixes the vectors $\vec{u}= \langle 1,1, 0 \rangle$ and $\vec{v} = \langle -1,1,-2 \rangle$ that are perpendicular to each other and to $\vec{n}$. More generaly, $\vec{u}$ could be any vector whose dot product with $\vec{n}$ is zero (such vectors are easy to find) and $\vec{v} = \vec{n}\times\vec{u}$.
In the basis determined by $\vec{n}$, $\vec{u}$, and $\vec{v}$, your matrix looks like
$$\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right).$$
The matrix $A$ whose columns are formed by $\vec{n}$, $\vec{u}$, and $\vec{v}$ form a similarity transformation that changes the diagonal matrix $E$ formed by the eigenvalues $-1$, $1$, and $1$ into the matrix you want. Thus, your matrix is
$$A \cdot E \cdot A^{-1} =
\left(
\begin{array}{ccc}
-1 & 1 & -1 \\
1 & 1 & 1 \\
1 & 0 & -2 \\
\end{array}
\right)
\cdot
\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
\cdot
\frac{1}{6}\left(
\begin{array}{ccc}
-2 & 2 & 2 \\
3 & 3 & 0 \\
-1 & 1 & -2 \\
\end{array}
\right)
=
\frac{1}{3}\left(
\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
2 & -2 & 1 \\
\end{array}
\right).$$
Of course, it's easy to check that this matrix behaves as advertised by multiplying it by the vectors above.
Addendum
This technique can be used to derive Bubba's answer, although I used Mathematica to do the heavy lifting for me.
A = Transpose[{{a, b, c}, {0, -c, b}, {b^2 + c^2, -a*b, -a*c}}];
FullSimplify[A.{{-1, 0, 0}, {0, 1, 0}, {0, 0, 1}}.Inverse[A]] /.
a^2 + b^2 + c^2 -> 1 // MatrixForm
$$
\left(
\begin{array}{ccc}
1-2 a^2 & -2 a b & -2 a c \\
-2 a b & 1-2 b^2 & -2 b c \\
-2 a c & -2 b c & 2 \left(a^2+b^2\right)-1 \\
\end{array}
\right)
$$
However, seeing Bubba's answer reminded me wonderful and simple Householder reflections are. Just define the function $f:{\mathbb R}^n \rightarrow {\mathbb R}^n$ by $f(x)=\vec{x}-2\vec{n}(\vec{n}^T\vec{x})$, where $\vec{x}$ is a column vector and $\vec{n}$ is a unit column vector denoting your normal. It's super simple to check that $f$ maps $\vec{n}$ to $-\vec{n}$:
$$f(\vec{n}) = \vec{n}-2\vec{n}(\vec{n}^T\vec{n}) = \vec{n}-2\vec{n}1=-\vec{n},$$
where $\vec{n}^T\vec{n}=1$, since $\vec{n}$ is a unit vector. It's also super simple to see that any vector $\vec{x}$ perpendicular to $\vec{n}$ is fixed by $f$, since
$$f(\vec{x}) = \vec{x}-2\vec{n}(\vec{n}^T\vec{x}) = \vec{x}-2\vec{n}0=\vec{x}.$$
Factoring out the $\vec{x}$ from the definition of $f$, we see the matrix you need:
$$I-2\vec{n}\,\vec{n}^T.$$
Applying this to your situation, we get the same matrix as above.
\begin{align}
I-\vec{n}\,\vec{n}^T &=
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
-
2\frac{1}{\sqrt{3}}\left(
\begin{array}{c}
-1 \\
1 \\
1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
-1 & 1 &
1 \\
\end{array}
\right)/\sqrt{3}\\
&= \left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
- 2\frac{1}{3}
\left(
\begin{array}{ccc}
1 & -1 & -1 \\
-1 & 1 & 1 \\
-1 & 1 & 1 \\
\end{array}
\right)
=
\frac{1}{3}
\left(
\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
2 & -2 & 1 \\
\end{array}
\right)
\end{align}
|
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|
Infinite series for $ \sqrt 2 $ What is infinite series for $ \sqrt 2 $? I don't mean continued fraction. That kind of series such as like for $e, \pi, $etc.
|
$$\sqrt{2}=\frac{1}{\left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{6^2}\right) \left(1-\frac{1}{10^2}\right) \left(1-\frac{1}{14^2}\right) \cdots}$$
$$\sqrt{2}=\left(1+\frac{1}{1}\right) \left(1-\frac{1}{3}\right) \left(1+\frac{1}{5}\right) \left(1-\frac{1}{7}\right) \cdots$$
$$\sqrt{2}=1+\frac{1}{2}-\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots$$
$$\sqrt{2}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}$$
https://en.wikipedia.org/wiki/Square_root_of_2#Series_and_product_representations
The first one is the true answer. It is in the format of the natural number $e = \left(1+\frac{1}{\infty}\right)^\infty$ except it's a minus sign instead and a reciprocal. Consider it's reduced form here:
$$\sqrt{2}=\frac{1}{\left(1-\frac{1}{4\cdot1^2}\right) \left(1-\frac{1}{4\cdot3^2}\right) \left(1-\frac{1}{4\cdot5^2}\right) \left(1-\frac{1}{4\cdot7^2}\right) \cdots}$$
This is the most reduced true form of the infinite series that is $\sqrt2$. An amazing property of $\sqrt2$ is that the reciprocal is equal to exactly $\frac{1}{2}$ of its value. So
$$\sqrt{2}=2 \left(1-\frac{1}{4\cdot1^2}\right) \left(1-\frac{1}{4\cdot3^2}\right) \left(1-\frac{1}{4\cdot5^2}\right) \left(1-\frac{1}{4\cdot7^2}\right) \cdots$$
|
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"url": "https://math.stackexchange.com/questions/694699",
"timestamp": "2023-03-29T00:00:00",
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|
Different methods of evaluating $\int\sqrt{a^2-x^2}dx$: Is there a simple and nice way to solve $\int\sqrt{a^2-x^2}dx$:
PS:I am not looking for a substitution like $x=a\sin p$,
|
Another way to attack such kinds of integrals is using Chebyshev's substitution. Consider the integral:
$$
I_{m,p} = \int x^m (\alpha x^n + \beta)^pdx
$$
Chebyshev has shown that $I_{m, p}$ is expressed in elementary functions if and only if one of the following cases is satisfied:
$$\begin{align*}
&1)\ p \in \Bbb Z \\
&2)\ {m+1 \over n} \in \Bbb Z \\
&3)\ {m+1 \over n} + {1\over p} \in \Bbb Z
\end{align*}
$$
For the first case substitute:
$$
x = t^N,\ \text{where}\ N\ \text{is common denominator of fractions}\ m\ \text{and}\ n
$$
Second case, substitute:
$$
\alpha x^n + \beta = t^M\ \text{where}\ M\ \text{is the denominator of}\ p
$$
Third case substitute:
$$
\alpha + \beta x^{-n} = t^M\ \text{where}\ M\ \text{is the denominator of}\ p
$$
If you check the third case is satisfied, hence the integral exists in elementary functions.
$$
(a^2 - x^2)^{1\over 2} = x^m(\alpha x^n + \beta)^p
$$
Therefore:
$$
p = {1\over 2}\\
m = 0\\
n = 2\\
\alpha = -1\\
\beta = a^2
$$
Let:
$$
\alpha + \beta x^{-n} = t^M
$$
So:
$$
t^2 = {a^2\over x^2} - 1\\
x^2 = {a^2\over t^2 + 1}\\
x^3 = \left({a^2\over t^2 + 1}\right)^{3\over 2}\\
tdt = -{a^2 \over x^3}dx \implies dx = -{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt
$$
The integral becomes:
$$\begin{align}
I &= -\int \sqrt{a^2 - {a^2\over t^2 + 1}}{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt \\
&= -\int \sqrt{a^2 t^2 \over t^2 + 1}{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt\\
&= -\int {|t|t\over a^2}\left({a^2\over t^2 + 1}\right)^2dt
\end{align}
$$
And this is a regular rational function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/694853",
"timestamp": "2023-03-29T00:00:00",
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|
issues with simple algebraic equations $ab + a + b = 250$
$bc + b + c = 300$
$ac + a + c = 216$
then find $a + b + c = ?$
MY APPROACH:
(i) * c , (ii) * a , (iii) * b then we get
$abc + ac + bc = 250c$
$abc + ab + ac = 300a$
$abc + ab + bc = 216b$
(iv)+(v)+(vi)
$3abc + 2ac +2ab + 2bc = 300a + 216b +250c$
now i cant solve it ?? how to solve??
|
\begin{cases}
ab+a+b=250 \\
bc+b+c=300 \\
ac+a+c=216 \\
\end{cases}
it seems of no difficulty: a system of 3 equations and 3 variables...
Let's start by adding 1 to every equations, we obtain:
\begin{cases}
ab+a+b+1=(a+1)(b+1)=251 \\
bc+b+c+1=(c+1)(b+1)=301 \\
ac+a+c+1=(a+1)(c+1)=217 \\
\end{cases}
Now let's substitute $a+1 = x, b+1 =y, c+1 = z$, so we can reduce the amount of calculus needed, in fact to solve
\begin{cases}
xy=251 \\
yz=301 \\
xz=217 \\
\end{cases}
you only need to find $x$ (or $y$) in the first equation, substitute it in the last (or second) equation, and then do another substitution. Then subtract 1, and you'll find the solutions:
$(a,b,c) = (-1-\sqrt{\frac{7781}{43}}, -1-\sqrt{\frac{10793}{31}}, -1-7\sqrt{\frac{1333}{251}})$
and
$(a,b,c) = (\sqrt{\frac{7781}{43}}-1, \sqrt{\frac{10793}{31}}-1, 7\sqrt{\frac{1333}{251}}-1)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplfying $\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$ I am trying to simplify the expression:
$\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$
I tried to square the expression but I can't do that because it is not an equation so I got stuck. Can someone please give me some pointers on how to proceed?
|
The general way to evaluate these kinds of expressions is to try to factor the contents of the square root into a perfect square.
One can try to find the factorization by letting $31 + 8\sqrt{15} = (a + b\sqrt{15})^2 = (a^2 + 15b^2) + 2ab\sqrt{15}$. We then test some possible values of $a$ and $b$. In this case, we see that $a = 4$ and $b = 1$ works. So, we can deduce that :
$$\begin{align}(4 + \sqrt{15})^2 &= 16 + 15 + 8 \sqrt{15} \\
&= 31 + 8 \sqrt{15}\end{align}$$
Similarly, we can deduce that
$$\begin{align}(4 - \sqrt{15})^2 &= 16 + 15 - 8 \sqrt{15} \\
&= 31 - 8 \sqrt{15}\end{align}$$
Hence,
$$\begin{align}\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}} &= \sqrt{(4 - \sqrt{15})^2} + \sqrt{(4 + \sqrt{15})^2} \\
&= 4 - \sqrt{15} + 4 + \sqrt{15} \\
&= 8\end{align}$$
Of course, you can alternatively go ahead to obtain the solution by squaring the original expression, but it'll be rather tedious if you are working with bigger numbers.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. Show that if $m,n$ are positive integers and $m$ is odd, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
(Hint: Let $s=1^m+2^m+\cdots+(n-2)^m+(n-1)^m$. Obviously $s=(n-1)^m+(n-2)^m+\cdots+2^m+1^m$.
Consider these relations as equivalent ${}\bmod n$ and add them.)
$$1^m+2^m+\cdots+(n-2)^m+(n-1)^m=((n-1)^m+(n-2)^m+\cdots+2^m+1^m)\bmod n$$
$$(n-1)^m+(n-2)^m+\cdots+2^m+1^m=(1^m+2^m+\cdots+(n-2)^m+(n-1)^m)\bmod n$$
By adding them we get:
$$(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)=((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))\bmod n$$
That means that:
$$n|[(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)]-[((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))]$$
Or not??
And that is equal to $$n\mid 0$$
Is this correct? How can I continue??
|
Hint: if $m$ is odd, then $(n-i)^m +i^m \equiv (-1)^mi^m+i^m\equiv -i^m+i^m\equiv 0 \bmod n.$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$ $$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$
I think to do :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$
but i dont get anything.
Or to divied by $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$
|
Hint:
$$\sqrt3\sin x-\cos x=\sqrt2\iff \sin\frac\pi3\sin x-\cos\frac\pi3\cos x=\frac{\sqrt2}2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/698964",
"timestamp": "2023-03-29T00:00:00",
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|
What is the average pathlength and probability to randomly traverse any given graph?
To get specific first off, it's about this graph:
I want to get from $A$ to $B$. Every edge has the same length (e. g. 1 m). The shortest walk from $A$ to $B$ is easily found ($A-2-5-B$ and $A-3-6-B$). But how can I calculate the average length of a walk and its probability, when I am navigating randomly through this graph (what is the length/probability of $n$ steps)? At every node, there is a chance of $p= \frac{1}{degree}$ to choose any next edge (also back, where I just came from).
For example, there would be a chance of walking in circles for a long time, but probability decreases - also to get straight through.
I wrote a computer simulation to find a statistic answer: After a megaattempt (1000000 attempts), it averaged at about 20.329 edges for this specific graph here.
I'd also like to know how to calculate such things in general, of course ;)
Path vs. Walk: Some authors (e.g. Bondy and Murty 1976) use the term "walk" for a path in which vertices or edges may be repeated, and reserve the term "path" for what is here called a simple path. - https://en.wikipedia.org/wiki/Path_(graph_theory)
|
It's difficult to claim an average because there exist many possible infinite length paths that never reach $B$. You may want instead to calculate the probability that you have reached $B$ after $n$ steps.
@draks ... is on to something, but if you're going to determine the probably that you have "seen" node $B$ after $n$ steps, then you definitely need to include the probability as weights in the adjacency matrix.
Question: For the given graph $G$, what is the probability of having seen node $B$ at the $n^{th}$ step?
Answer: Let us write the adjacency matrix of $G$ weighted with edge probabilities as
$$M = \begin{bmatrix}
0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 & 0 & 0 \\
\frac{1}{3} & 0 & \frac{1}{3} & 0 & \frac{1}{3} & 0 & 0 & 0 \\
\frac{1}{5} & \frac{1}{5} & 0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\
\frac{1}{3} & 0 & \frac{1}{3} & 0 & 0 & 0 & \frac{1}{3} & 0 \\
0 & \frac{1}{4} & \frac{1}{4} & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{4} \\
0 & 0 & \frac{1}{4} & 0 & \frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4} \\
0 & 0 & 0 & \frac{1}{3} & 0 & \frac{1}{3} & 0 & \frac{1}{3} \\
0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\
\end{bmatrix}$$
Let $N$ be the matrix $M$ with the last row and column removed. Then the probability of having seen $B$ at step $n$ is then equal to 1 minus the sum of the first row of $N^{n}$ (because that is the row corresponding to the node we started from $A$).
Here's a pretty picture of how the probability changes as $n$ increases:
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $(1-\frac{1}{2^2}\cdots 1-\frac{1}{9\,999^2})(1-\frac{1}{10\,000^2})=0.500\,05$ Prove that $\displaystyle\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{9\,999^2}\right)\left(1-\frac{1}{10\,000^2}\right)=0.500\,05$
Here are all my attempts to solve this problem:
So the first thing I thought about is to transform the expression of the form $$\left(1-\frac{1}{n^2}\right)$$
To the expression: $$\left(\frac{n^2-1}{n^2}\right)$$
But in evaluating the product $$\prod_{k=2}^n \frac{n^2-1}{n^2}$$ seems way complicated than what I'm capable of, I don't know how to evaluate products even if I know that notation.
The other idea is to transform again $$\left(1-\frac{1}{n^2}\right)\to \left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)$$
But this isn't helpful in any way.
My other attempts were to try to see what happens when I start evaluating this sum, and it turns out that a lot of things cancel out but again no result.
I'll be happy if someone could guide me to solve this problem. (I feel that there is some kind of symmetry that I should remark, a symmetry that would allow me to cancel things out and to have my final)
|
Note that you can factor $n^2 - 1 = (n - 1)(n + 1)$ to find something like
\begin{align*}
\left(1 - \frac 1 {2^2}\right)&\left(1 - \frac 1 {3^2}\right) \cdots \left(1 - \frac 1 {10000^2}\right) \\ & =\left(\frac 1 2 \cdot\frac 3 2\right) \left(\frac 2 3 \cdot\frac 4 3\right) \left(\frac 3 4 \cdot \frac 5 4\right) \left(\frac 4 5 \cdot \frac 6 5 \right)\cdots \left(\frac{10000 - 1}{10000} \cdot \frac{10000+1}{10000}\right)
\end{align*}
Convince yourself that everything cancels (i.e. the product telescopes), except for a factor of $1/2$, and the final factor of $10001/10000$.
|
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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.