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# How to perform Monte-Carlo simulations to price Asian options?
If I wish to price a fixed-strike Asian Call option via Monte-Carlo (This has no early-exercise), are my following steps correct?:
1) Simulate random asset prices. (Milstein)
$\ d S(t) = \ rS(t)dt + \sigma S(t) d B(t)$
$\ S_{t+dt} = S_t + r S_tdt + \sigma S_t \sqrt{ dt}Z + \frac{1}{2}\sigma^2dt(Z^2-1)$
2) Average the asset prices for each simulation.
$\ A[i]$ is the average for each simulation.
I'll be using both Geometric and Arithmetic averages
3) Calculate each payoff and discount it. Find the average of these payoffs
$\text{Payoff}[i]= \exp[-r(T-t)] * \max[A[i]-K,0]$
$\text{Average} = \frac{1}{N}\sum_{i=1}^N \text{Payoff}[i]$
I'm aware that there are some approximation formulae, Finite-Difference methods and closed-form solutions but I'm trying to focus on Monte-Carlo simulations for now.
• Are you sure of the last term in your Milstein scheme? $S_t$ seems to be missing IMHo. The rest looks fine to me although what you call payoff is actually the discounted payoff. – Quantuple Sep 29 '16 at 16:58
• This is interesting. I actually agree with you but looking at two different papers, frouah.com/finance%20notes/… and vlebb.leeds.ac.uk/bbcswebdav/orgs/SCH_Computing/FYProj/reports/…. They seemed to both not include the $/ S_t$. – mathnoob Sep 29 '16 at 17:10
• Take equation (16) of your paper and substitute the drift and diffusion terms as per the first lines of section 2.1. what do you get? Or simply analyse the dimensions of each term, or observe how the model should be homogeneous in space. – Quantuple Sep 29 '16 at 17:15
• The final term doesn't match. I get $0.5 \sigma_t^2 S_t dt(Z^2-1)$ – mathnoob Sep 29 '16 at 17:26
• Thanks for pointing that out. I'd give you a thumbs up but I don't really know how to do it for comments. Still figuring out how to use this website. – mathnoob Sep 29 '16 at 17:32 | {
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1. Instead of simulating the spot price, simulate its logarithm since the latter can be simulated exactly for any time step. $$\ln S_{t + \Delta t} = \ln S_t + \left( r - \frac{1}{2} \sigma^2 \right) \Delta t + \sigma Z,$$ where $Z \sim \mathcal{N}(0, \Delta t)$. You then just simply take the exponential of the simulated logarithmic price at each time step.
2. OK
3. OK
4. Calculate the standard error of your estimate. This is just as important as computing the estimate itself and for example allows you to construct confidence intervals. Let $$\text{Variance} = \frac{1}{N - 1} \sum_{i = 1}^N \left( \text{Payoff}_i - \text{Average} \right)^2$$ be your estimator of the variance. Then the standard error is
$$\text{Standard Error} = \sqrt{\frac{\text{Variance}}{N}}.$$
Looks good to me, although idk why you have (T-t) in the discounting... isn't big T the total time to maturity? What is little t in the equation? Shouldn't it just be exp[-rT] because you discount from the time of payoff which is the expiration of the option.
Kemna and Vorst (1990) [ download ] is a classic in Monte Carlo method for Asian option. Geometric mean, which can be analytically computed, is used as a control variate to reduce MC noise. | {
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# Asymptotic development of a recurrent sequence
Let $u_0 = 1$ and $u_{n+1} = \frac{u_n}{1+u_n^2}$ for all $n \in \mathbb{N}$.
I can show that $u_n \sim \frac{1}{\sqrt{2n}}$, but I would like one more term in the asymptotic development, something like $u_n = \frac{1}{\sqrt{2n}}+\frac{\alpha}{n\sqrt{n}} + o\bigl(\frac{1}{n^{3/2}}\bigr)$.
Here is the outline of my proof of $u_n \sim \frac{1}{\sqrt{2n}}$:
• $(u_n)$ is decreasing, and bounded from below by $0$, hence converges.
• The limit $\ell$ satisifies $\ell = \frac{\ell}{1+\ell^2}$, hence $\ell = 0$.
• A computation gives $v_n = u_{n+1}^{-2} - u_n^{-2} \to 2$.
• Using Cesàro lemma, $\frac{1}{n} \sum_{k=0}^{n-1} v_k \to 2$.
• Hence $\frac{1}{n} u_n^{-2} \to 2$.
Let $a_n=\dfrac1{u_n^2}$. Then $$a_{n+1}-a_n = \left(\frac{1+u_n^2}{u_n}\right)^2 - \frac1{u_n^2} = 2+u_n^2 = 2+\frac1{a_n}.$$ From this and $a_2=4$ we can find a successive sequence of estimates for $n\ge2$: \begin{align*} a_n &\ge 2n; \\ a_n &\le 2n + \sum_{k=2}^{n-1}\frac1{2k} < 2n+\frac12\log n; \\ a_n &\ge 2n + \sum_{k=2}^{n-1}\frac1{2k+\frac12\log k} > 2n+\int_2^n \frac{dx}{2x+\frac12\log x} \\ &> 2n+\int_2^n \frac{dx}{2x} - \int_2^n \frac{\frac12\log x}{2x(2x+\frac12\log x)}dx = 2n+\frac12\log n - \mathcal{O}(1). \end{align*}
Then $$u_n = a_n^{-1/2} = \frac1{\sqrt{2n}} \left(1+\frac{\log n}{4n}+\mathcal{O}\bigg(\frac1n\bigg)\right)^{-1/2} \\ = \frac1{\sqrt{2n}} \left(1-\frac12 \cdot \frac{\log n}{4n} + \mathcal{O}\bigg(\frac1n\bigg) \right) = \frac1{\sqrt{2n}} - \frac1{8\sqrt2}\cdot \frac{\log n}{n^{3/2}} + \mathcal{O}\left(\frac1{n^{3/2}}\right).$$
• Ooops, I started the sequence with $u_1=1$. But it changes only the constant in the error term. – G. Kós Dec 6 '14 at 18:47
• I wonder if this method can be iterated to yield abitrary deep asymptotic expansions. – Ewan Delanoy Dec 9 '14 at 19:00 | {
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Note: This answer does not contain a proof, but I'm fairly sure it's right. Posted in the hope that perhaps the answer will guide you towards a proof thereof.
Looking at the graph you can figure out that in fact the next correction looks to be $n^{-3/2} \log n$ dominating the next $n^{-3/2}$ term. Given this, one can do a plausible self-consistency check using the recurrence relation to find that the answer.
Specifically, suppose
$$u_n \sim \frac \alpha {\sqrt{2n}} + \frac {\beta \log n}{n^{3/2}}+\text{a sufficiently nice smaller series in particular }\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$$
Then $$u_{n+1}=\frac{u_n}{1+u_n^2}$$ implies $$u_{n+1}-\frac{u_n}{1+u_n^2} \sim \frac 1 2 (\alpha - 2 \alpha^3) \frac 1 {n^{3/2}}+\cdots$$ so that $\alpha=0,-1/\sqrt 2,1/\sqrt 2$ are the options. You have checked that $\alpha=1/\sqrt{2}$.
Substituting this back in gives $$u_{n+1}-\frac{u_n}{1+u_n^2} \sim \left(-\frac 1 {8\sqrt 2} - \beta\right) \frac 1 {n^{5/2}}+\cdots$$ and hence it seems the only consistent result has $$\boxed{\displaystyle u_{n} \sim \frac{1}{\sqrt{2n}}-\frac{\log n}{8\sqrt 2 n^{3/2}}}$$
Numerically, I find the error $$\epsilon = \frac{u_n - \frac{1}{\sqrt{2n}}}{-\frac{\log n}{8\sqrt 2 n^{3/2}}} - 1\approx \frac{3.4}{\log n} \to 0$$ suggesting that indeed this is correct with the next term in the series being the expected $n^{-3/2}$ term.
However, the coefficient of the $n^{-3/2}$ term cannot be determined by this self-consistency procedure, reflecting the fact that one can check this actually depends on the initial datum $u_0$.
Left to do: (Not that I think I'll come back to this myself, but others can!)
• Rigorously show the above logarithmic correction.
• Figure out the dependence of the next term on the initial data.
Mathematica code for my verification: | {
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Mathematica code for my verification:
min = 2;
max = 5000000;
dat = RecurrenceTable[{u[n + 1] == u[n]/(N[1, 50] + u[n]^2), u[0] == 1}, u, {n, min, max}];
errordat =
Table[{M, (dat[[M - min + 1]] -
1/Sqrt[2 M])/(-Log[M]/(8 Sqrt[2] M^(3/2))) - 1}, {M, min, max, 100000}];
nextconst = ((dat[[max - min + 1]] -
1/Sqrt[2 max])/(-Log[max]/(8 Sqrt[2] max^(3/2))) - 1)*Log[max]
Show[ListPlot[errordat, AxesOrigin -> {0, 0}, PlotStyle -> PointSize[Large]], Plot[nextconst/Log[x], {x, min, max}, PlotStyle -> Red, PlotRange -> All]]
Output: Estimate of $3.446...$ for next constant, and a plot of the $\epsilon$ against $n$ with a fitted curve based on a logarithmic next correction: Obviously the decay to 0 here is slow, which is to be expected given the predicted logarithmic correction. Feel free to go to larger ranges.
• Wouldn't it be better to plot the difference between the asymptotic (1st term) and numerical results and see if the result brings about the correction term? Showing that the error goes to zero is not good enough. – Ron Gordon Dec 6 '14 at 19:08
• I'm not entirely sure what you mean. The requirement that $\epsilon \to 0$ is precisely what it means for the boxed expression to be the first two terms of an asymptotic series, by definition. – Sharkos Dec 7 '14 at 11:58 | {
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# Is [0,1] closed covering of interval [0,1]?
I know it's true in case of open interval and open covering.
Also, in this case what will be Lebesgue number ?
$I = [0, 1]$ is closed as a subset of itself, making $\{ I \}$ a closed covering of $I$.
$B_{\delta}(x) \cap I \subset I$ for each $x \in I$ and each $\delta \in \mathbb{R}_{> 0}$, so that every such $\delta$ is a Lebesgue number.
• Yes but won't $\delta$ tends to zero ? As we take x near to 0 or 1, we have to take $\delta$ smaller correspondingly. – MeetR Apr 18 '17 at 2:52
• The open balls in $I$ as a metric space are all of the form $B_{\delta}(x_0) \cap I = \{ x \in I : |x - x_0| < \delta \}$ for $x_0 \in I$. – Dean Young Apr 18 '17 at 2:57 | {
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# Prove $\sum_{n=1}^\infty(e-\sum_{k=0}^n\frac1{k!})=1$
This comes from the comments section of this question here, credits Lucian.
The statement is
$$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)=1$$
This looks really interesting, so I was wondering if anyone has any ideas/suggestions or even better the whole proof?
Disclaimer: I added that link specifically because this might look like homework, leading to "What have you tried?". So the disclaimer is, I actually haven't tried anything yet, because yes, asking the community is just easier. It would be great if someone could answer this though since I, personally at least, think that this equation is very interesting and adds value to any math enthusiast stumbling upon it here someday. :) I will myself be trying it meanwhile too though(it really is interesting) , and add and accept my solution if I am successful and there are no answers yet.
I tried wolfram alpha for $\sum_{k=0}^n\frac1{k!}$ and it yields
$$\sum_{k=0}^n\frac1{k!} = \frac{e\Gamma (n+1,1)}{\Gamma(n+1)}$$
where $\Gamma(a,r)$ is the incomplete gamma function and $\Gamma(a)$ is the euler gamma function. Documentation here. Might be of help, might not. Just thought it'd be worth adding.
The first thought that comes into my head is to write $$e - \sum_{k=0}^n \frac{1}{k!} = \sum_{k=n+1}^\infty \frac{1}{k!},$$ so that the given sum is equivalent to a double sum: \begin{align*} \sum_{n=1}^\infty \sum_{k=n+1}^\infty \frac{1}{k!} &= \sum_{k=2}^\infty \sum_{n=2}^k \frac{1}{k!} \\ &= \sum_{k=2}^\infty \frac{1}{(k-2)!k} \\ &= \sum_{k=2}^\infty \frac{1}{(k-1)!} - \frac{1}{k!} = 1. \end{align*}. Of course, a somewhat more rigorous argument is needed to justify the interchange of the order of summation. A minor modification to consider the original series' partial sums, and taking the limit to get the double infinite sum, is sufficient and is left as an exercise. | {
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• Oh that was fast. Thanks. – Guy Mar 7 '14 at 18:11
• All the terms are positive, so it is okay to exchange sums. – abnry Mar 7 '14 at 18:12
• I am over my daily vote limit. I'll upvote this tomorrow. – Guy Mar 7 '14 at 18:17
• @nayrb of course, you are right. ^_^ – heropup Mar 7 '14 at 18:17
$$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right) =\sum_{n=1}^\infty \int_0^1 \exp(u) \frac{(1-u)^{n}}{n!} du$$as everything is positive: $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)= \int_0^1 \exp(u) \sum_{n=1}^\infty \frac{(1-u)^{n}}{n!} du \\= \int_0^1 \exp(u)(\exp(1-u) - 1) du = e - \int^1_0 \exp u du = 1$$
• Might be missing something obvious, but $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right) =\sum_{n=1}^\infty \int_0^1 \exp(u) \frac{(1-u)^{n}}{n!} du$$ how? – Guy Mar 7 '14 at 18:23
• this is the Taylor formula, obtained via $n$ integrations by parts. – mookid Mar 7 '14 at 18:23
• Ah, I am unfamiliar with this. I'll try to prove this myself though. Rest of the solution is great. (+1)(after my vote limit time restriction is lifted that is) – Guy Mar 7 '14 at 18:25
• you obtain it via $\exp 1 - 1 = \int_0^1 1 \exp u du = \ldots$ integrating the polynomial and derivating the $\exp$. – mookid Mar 7 '14 at 18:27
• Whoa dude, spoiler alert. :p – Guy Mar 7 '14 at 18:30 | {
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# Does introducing new open sets in topology improves it's “T”-axiom properties?
This question might be silly, however I'd like to be $100\%$ sure.
Suppose we have some topology $\tau$ on a set $X$ which gives us the $T_i$ space for some $i<6$. It's totally possible to improve topological properties (here I'd like to improve separation axioms) - simply consider basic anti-discrete topology, which consists only of $\emptyset$ and $X$.
My question is about the opposite situation. Is it possible to have topology $\tau_1$ on set $X$ and a bigger topology $\tau_2 \supset \tau_1$, such that $(X,\tau_1)$ is $T_i$ space while $(X,\tau_2)$ does not fulfill $T_i$ space axiom.
I'm pretty sure such situation is not possible, but if it is - please provide me with a proper counterexample.
For $i\leqslant 2$ I'm sure it's not the case, I wonder if making topology larger (which at the same time introduces new closed sets) can prevent $(X,\tau)$ from being $T_i$ space.
• Which cases do you speak about? If it's about $T_i$ for which $i<3$, then I've already pointed that out in my post. If you do speak about $i\geqslant 3$ please point out why it's so trivial, because I don't really see it. – I_Really_Want_To_Heal_Myself Jun 16 '17 at 20:05
• did you read my answer? – Jorge Fernández Hidalgo Jun 16 '17 at 21:13
• The point is, for $T_3$ and higher we have to consider closed sets as well, not just points. Adding open sets also adds closed sets, so more conditions to fulfill. – Henno Brandsma Jun 16 '17 at 22:06
• It's a very good Q. – DanielWainfleet Jun 18 '17 at 2:39
If a Hausdorf space is given a finer topology,
it's still Hausdorf. Proof is very simple.
The set of real numbers is regular and normal.
Add to the usual topolopy the set of rationals and
it is neither regular nor normal, yet still Hausdorf.
Pick a set $X$ along with an equivalence relation $\sim$. Consider the topology $\tau$ on $X$ in which the open sets are the sets that are unions of equivalence classes. | {
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In this space the closed and open sets coincide. This space is clearly $T_3$ .
Consider the refinement on $X$ which consists of taking one of the equivalence classes $C$ and refining the topology on $C$ into a non-$T_3$ topology.
• I think this approach works for the other ones also (Except $T_0,T_1,T_2$ and $T_{2.5}$) – Jorge Fernández Hidalgo Jun 16 '17 at 20:14
An example where $\tau_1$ is $T_6$ but $\tau_2$ is NOT $T_6.$
Let $I=[0,1].$ Let $\tau_1$ be the usual topology on $I^2.$ Let $<_L$ be the lexicographic order on $I^2.$ That is $(x,y)<_L(x',y')\iff (x<x'\lor [x=x'\land y<y']).$ The topology $\tau_2$ induced on $I^2$ by $<_L$ is stronger than $\tau_1.$ ( If $A,B$ are open real intervals then $(A\times B)\cap I^2$ is a union of $<_L$-open intervals.)
Then $\tau_1$ and $\tau_2$ are $T_5$ topologies. (For $\tau_2,$ all linear spaces are $T_5$).
But $\tau_1$ is $T_6$ and $\tau_2$ is NOT $T_6.$
(I). $\tau_1$ is $T_6$ because it is metrizable.
(II). Notation: ]a,b[ and [a,b[ are real intervals. (a,b) is an ordered pair.
Let $A=I\times \{0,1\}.$ Then $A$ is $\tau_2$-closed but $A$ is not a $G_{\delta}$ set in the $\tau_2$ topology.
Proof: If $A\subset U\in \tau$ and $x\in [0,1[$ there is a $<_T$-open interval $J$ with $(x,1)\in J\subset U.$ And since $(x,1)$ is in the $\tau_2$-closure of $\{(x',y')\in I^2: (x,1)<_L (x',y')\},$ there is $(x',y')\in J$ with $x'>x.$ Since $J$ is a $<_T-$interval, therefore $]x,x'[\times I\subset J\subset U.$
So let $B(U)=\cup \{]x,x'[ \;: \;0\leq x<1\;\land \;]x,x'[\times I\subset U\}.$ With respect to the usual topology on $I,$ the set $B(U)$ is open and dense in $I.$ (If $0\leq x<y\leq 1$ there is $x'\in ]x,y[\cap B(U)$.)
So if $A\subset U_n\in \tau_2$ for $n\in \mathbb N$ then $$\cap_{n\in \mathbb N}U_n\supset (\cap_{n\in \mathbb N}B(U_n))\times I.$$ By the Baire Category Theorem applied to the usual topology on $I$ we have $\cap_{n\in \mathbb N}B(U_n)\ne \phi.$
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Therefore $\cap_{n\in \mathbb N}U_n\ne A.$
Remark: One-member subsets of $I^2$ are closed $G_{\delta}$ sets in $\tau_2$ but the closed set $A$ is not $G_{\delta}$ in $\tau_2$.
• I do like this one, although the notation makes it quite hard to see at the first glance. – I_Really_Want_To_Heal_Myself Jun 18 '17 at 21:31
• I don't usually use $]a,b[$ for open interval but I also have ordered pairs involved. And I made it very detailed to make sure I got it right. The lex-order on $I^2$ is suggested as a hint in General Topology by Engelking in an exercise asking for a space with the property in my "Remark". – DanielWainfleet Jun 19 '17 at 16:14
A $T_6$ topology $\tau$ and a finer topology $\tau'$ that is not even $T_4:$
Let $\tau$ be the usual (standard) topology on $\mathbb R^2.$
Let $S$ be the Sorgenfrey line (a.k.a. the lower limit topology on the set $\mathbb R$), which has a base of all real intervals $[x,y).$ Let $\tau'$ be the product topology on $S^2.$
A Q that has appeared repeatedly on this site is to prove that $S^2$ is not $T_4.$ The solution is to apply Jones' Lemma: If $X$ is a separable space with a closed discrete subspace $Y$ where the cardinal of $Y$ is $2^{\aleph_0}$ then $X$ is not normal. For the case $X=S^2$ let $Y=\{(x,-x): x\in \mathbb R\}.$ | {
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# Alternative for calculating the nth of quadratic sequence
Given the quadratic sequence $$f(n)=1, 7, 19, 37, \cdots$$
To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$
We start with the general quadratic function, then sub in for $n:=1,2$ and $3$
$$f(1)=a+b+c$$ $$f(2)=4a+2b+c$$ $$f(3)=9a+3b+c$$
Now solve the simultaneous equations
$$a+b+c=1\tag1$$ $$4a+2b+c=7\tag2$$ $$9a+3b+c=19\tag3$$
$(2)-(1)$ and $(3)-(2)$
$$3a+b=6\tag4$$ $$5a+b=12\tag5$$
$(5)-(4)$ $$a=3$$ $$b=-3$$ $$c=1$$
$$f(n)=3n^2-3n+1$$
This method is very long. Is there another easy of calculating the $f(n)$?
• Yes: just do what you did, but don't substitute the numbers; instead, keep $f(1), f(2), f(3)$ throughout. Then you end up with a fully general formula. – Patrick Stevens Mar 28 '18 at 11:13
Another standard way is to calculate a difference scheme and then to work backwards: $$\begin{matrix} 0 & & 1 & & 2 & & 3 & & 4 \\ & & 1 && 7 && 19 && 37 \\ && &6& & 12 && 18 & \\ &&&& 6 && 6 && \\ \end{matrix} \Rightarrow \begin{matrix} & 0 & & 1 & & 2 & & 3 & & 4 \\ \color{blue}{c}= &\color{blue}{1} & & 1 && 7 && 19 && 37 \\ \color{blue}{a+b}= & &\color{blue}{0}&&6& & 12 && 18 & \\ \color{blue}{2a}= &&& \color{blue}{6}&& 6 && 6 && \\ \end{matrix}$$ $$\Rightarrow a = 3, \; b= -3, \; c = 1 \Rightarrow f(n) = 3n^2-3n+1$$
Yes!
You know $f(1)= 1 \implies f(1) - 1 = 0 \implies f(x) - 1$ has a root at x=1
Now,
$f(x) - 1 = a(x-1)(x-b)$
Put $x = 2$, $a(2-b) = 6$
Put x = 3, $a*2*(3-b) = 18$
Divide both equation, we get $b = 0$ and a = 3
$f(x) - 1 = 3*(x-1)x \implies f(x) = 3x^2 - 3x + 1$
Take a better basis. Namely, $\{(n-1)(n-2),(n-1)(n-3),(n-2)(n-3)\}$. If $$f(n) = \alpha(n-1)(n-2) + \beta(n-1)(n-3) + \gamma(n-2)(n-3),$$ then: $$f(1) = 0 + 0 + \gamma(1-2)(1-3),$$ $$f(2) = 0 + \beta(2-1)(2-3) + 0,$$ $$f(3) = \alpha(3-1)(3-2) + 0 + 0.$$
A better explained version of @trancelocation's answer:
$$\begin{array}{} 1 & & 7 & & 19 & & 37 \\ & 6 & & 12& &18 & \\ & & 6 & & 6 & & \\ \end{array}$$ | {
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$$\begin{array}{} 1 & & 7 & & 19 & & 37 \\ & 6 & & 12& &18 & \\ & & 6 & & 6 & & \\ \end{array}$$
The second row has equation $6n$. Therefore, $\big(a(n+1)^2+b(n+1)+c \big) - \big(an^2+bn+c \big)$ $= 6n$, and so:
$$\big(a(n^2+2n+1)+b(n+1)+c \big) - \big(an^2+bn+c \big) = 6n$$ $$(2n+1)a + b = 6n$$
$a=3$ gives $6n+3$, and so $b=-3$. 'Plugging' $n=1$ into $3n^2-3n$ gives $0$, $c=1$, which gives $3n^2-3n+1$. | {
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# We have two coins, A and B. For each toss of coin A, the probability of getting head is 1/2…
We have two coins, A and B. For each toss of coin A, the probability of getting head is 1/2 and for each toss of coin B, the probability of getting Heads is 1/3. All tosses of the same coin are independent. We select a coin at random and toss it till we get a head. The probability of selecting coin A is ¼ and coin B is 3/4. What is the expected number of tosses to get the first heads?
The above problem is taken from the website https://www.analyticsvidhya.com/blog/2017/04/40-questions-on-probability-for-all-aspiring-data-scientists/ question 11
My solution is either 1/(1/4*1/2 + 3/4*1/3)=8/3, including the success toss, or 5/3, not including the success toss. My understanding is that it is geometric distribution question.
But the solution provided is 2.75, with the following explanation: "If coin A is selected then the number of times the coin would be tossed for a guaranteed Heads is 2, similarly, for coin B it is 3. Thus the number of times would be Tosses = 2 * (1/4)[probability of selecting coin A] + 3*(3/4)[probability of selecting coin B] = 2.75"
Is the solution provided incorrect? Or am I missing something?
It is just the application of the Law of total Expectation.
$$E(X)=\sum_{i} E(X|A_i)\cdot P(A_i)$$
In your case the definitions are: $X:=$random variable for the number of heads, $A_1$: Coin with head probabilty equal to $\frac12$ is selected, $A_2$: Coin with head probabilty equal to $\frac13$ is selected.
Due to the geometric distribution (as you mentioned) $E(X|A_1)=\frac{1}{p_1}=2$ and $E(X|A_2)=\frac{1}{p_2}=3$. And $P(A_1)=\frac14,P(A_2)=\frac34$. Consequently we have
$$E(X)=E(X|A_1)\cdot P(A_1)+E(X|A_2)\cdot P(A_2)=\frac14\cdot 2+\frac34\cdot 3=\frac{11}{4}$$
I think you are already familiar to the Law of total probability. This has a similar structure as the Law of total expectation, but for probabilities: | {
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$$P(A)=\sum_{i} P(A|B_i)\cdot P(B_i)$$
One may say that $E(X)$ is the weighted mean of the conditional expectations.
The provided answer of $2.75$ is correct. I wouldn't say that the explanation is great. The use of the word 'guaranteed' is misleading. It makes it sound like if the fair coin is selected there is a guarantee of getting a heads in $2$ tosses, which is, of course, not the case.
Your proposed solution tries to make a weighted average of geometric distributions into a geometric distribution, which it is not.
Here is an explanation for the correct answer of $2.75$:
If you were to repeat the experiment many times, about $\frac14$ of the time (when you got the fair coin), it would take an average of $2$ flips for heads; and $\frac34$ of the time it would take an average of $3$ flips. So in $n$ repetitions of the experiment, it would take about $\frac14 \cdot n \cdot 2 + \frac34\cdot n\cdot 3$ total flips. This works out to about $2.75$ flips per trial on average to get heads. | {
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# Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable?
Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?
Having some trouble with this one. I tried using the fact that $\lambda$ is an eigenvalue of $A$ iff there exist non-zero solutions to $Ax=\lambda x$. Well, clearly the $1\times n$ non-zero vector $x$ with all entries equal is an eigenvector of $A$, with corresponding eigenvalue $\lambda=n$, since $Ax=nx$. But I can't seem to get much further than that in terms of finding eigenvalues for $A$, any hints/suggestions?
• What is the rank of $A$? What does the rank-nullity theorem tell you about the nullity of $A$? What does that imply about the eigenvalue zero? What does the trace say about the eigenvalues? – JMoravitz Mar 14 '16 at 0:03
• – JMoravitz Mar 14 '16 at 0:06
• So the idea is that in general for a matrix with a $1$ in every entry, $p(\lambda)=(-1)^n\lambda^{n}+(-1)^{n-1}n\lambda^{n-1}$, so the only eigenvalues are $\lambda =0,n$? – Borat Sagdiyev Mar 14 '16 at 0:17
Observe:
• $Rank(A)=1$ since all columns are identical and nonzero (rank=number of linearly independent columns)
• $nullity(A)=n-1$ by the rank-nullity theorem (rank+nullity=n)
• $trace(A)=n$ by direct calculation (trace=sum of diagonal entries)
Since $nullity(A)=n-1=nullity(A-0I)$ is the dimension of the eigenspace associated with the eigenvalue zero, we know that we have zero as an eigenvalue of algebraic multiplicity at least $n-1$. This implies that zero is either of algebraic multiplicity $n-1$ or it is of algebraic multiplicity $n$.
Since $n=trace(A)=\sum \lambda\cdot algmu(\lambda)$, we know that zero is not of multiplicity $n$ and that the remaining eigenvalue must be $n$ itself. | {
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As for the question of if it is diagonalizable, as mentioned earlier, we know that $geomu(0)=n-1$ and that $n$ is an eigenvalue (and thus must have geometric multiplicity at least one like every eigenvalue) and has geometric multiplicity exactly one (as it can't exceed its algebraic multiplicity). Since the sum of the geometric multiplicities is $(n-1)+1=n$, it is seen to be diagonalizable.
• $A$ is diagonalizable since $dim(E_0)+dim(E_n)=n$, $dim(E_0)=n-1=mult_0$, and $dim(E_n)=1=mult_n$, right? But how would you solve $(A-nI)x=0$ to find $E_n$? – Borat Sagdiyev Mar 14 '16 at 3:05
• @Borat by inspection is a valid method which you already did in your original post. If you were to do it manually, consider adding all rows together in the row reduction process and see where that gets you. – JMoravitz Mar 14 '16 at 3:13
• Oh, I see. So $E_n=span\{(1,...,1)\}$? – Borat Sagdiyev Mar 14 '16 at 3:18
Note that $A$ is real and symmetric, so it is diagonalizable. We have $A^2=nA$, so all eigenvalues of $A$ satisfy $\lambda^2=n\lambda$: this tells us that the only eigenvalues are $0$ and $n$. Since the trace of $A$ (i.e., the sum of the eigenvalues) is $n$, we deduce that the multiplicity of $n$ is one and that the multiplicity of $0$ is $n-1$.
As for eigenspaces, if $e$ is the vector with all entries equal to one, we have $Ae=ne$, so $e$ spans the eigenspace of $n$. The eigenspace for $0$ consists of the kernel of $A$: $0=Ax=(x_1+\cdots+x_n,\ldots,x_1+\cdots+x_n)^t$. We can make a basis of this subspace by taking the vectors $$\begin{bmatrix}1\\-1\\0\\0\\0\\ \vdots\\ 0\end{bmatrix},\ \ \begin{bmatrix}0\\1\\-1\\0\\0\\ \vdots\\ 0\end{bmatrix},\ \ \begin{bmatrix}0\\0\\1\\-1\\0\\ \vdots\\ 0\end{bmatrix},\ \ \ldots\ \ ,\ \ \begin{bmatrix}0\\0\\0\\ \vdots \\0\\1\\-1\end{bmatrix}.$$ | {
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Here is a matrix $P$ that I made up some time ago. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal and therefore independent. The columns are eigenvectors for your matrix, in the 10 by 10 case. Multiply your matrix of all ones (10 by 10) on the left, this matrix on the right. What happens? $$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right).$$
If $\mathbf{1}$ denotes the column vector of all ones, what is $\mathbf{1} \mathbf{1}^T$?
Now, match this to the outer product form of the eigendecomposition to get the eigenvectors and eigenvalues.
Consider the matrix $B = A - \lambda I$, where $\lambda \in \text{Spec}(A)$. Then $\lambda_{1} = n - \lambda$ is an eigenvalue of $B$ with an associated eigenvector $v_{1} = \begin{pmatrix} 1 \\ 1 \\ 1\\ \cdot \\ \cdot \\ \cdot \\ 1 \\1 \\1\\ \end{pmatrix}$.
The remaining eigenvalues of $B$ are $\lambda_{2} = \lambda_{3} =...=\lambda_{n} = -\lambda$ and the corresponding eigenvectors are:
$v_{2} = \begin{pmatrix} 1 \\ -1 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, $v_{3} = \begin{pmatrix} 0 \\ 1 \\ -1\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, ..., $v_{n} = \begin{pmatrix} 0 \\ 0 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\1 \\-1\\ \end{pmatrix}$. | {
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Now, it is easier to see that $v_{1}, ..., v_{n}$ are linearly independent and they are eigenvectors of $A$ with $v_{1}$ associated with the eigenvalue $\lambda =n$ and $v_{j}$ for $j \geq 2$ associated with $\lambda = 0$. From the linearly independence of these eigenvectors we see $A$ is diagonalizable. | {
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×
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Note by Iq 131
4 years, 4 months ago
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Yes. Since the probability of choosing a goat in the first try is 2/3 , since the other goat will be revealed, behind the other door should be the car. This famous problem as far as i know is The Monty Hall Problem.
- 4 years, 4 months ago
- 4 years, 4 months ago
Where the same article also says that the question as posed is incomplete (what if the host knows Door 2 has a goat and hence offering the option; never offering the option if the player originally picked a goat?). So there is no satisfying answer for this problem as stated. Only after clarifying further that the problem can be solved.
- 4 years, 4 months ago | {
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- 4 years, 4 months ago
Probability of finding a car in door 2 is 2/3 whereas in door 1 is 1/2 . Apply the theory of probability whenever options seems to be close
- 4 years, 4 months ago | {
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# Moment of inertia of a bar with two spheres
1. Jul 24, 2013
### duplaimp
1. The problem statement, all variables and given/known data
A dumbbell has two spheres of radius 0.10m and mass 10kg (each). The two spheres are connected using a bar with length 1.0m and mass 12kg.
What is the moment of inertia of the dumbbell about an axis perpendicular to an axis at the bar's center of mass?
Moment of inertia of a sphere through its center of mass: $\frac{2}{5}MR^{2}$
Moment of inertia of a bar through its center of mass: $\frac{2}{12}ML^{2}$
3. The attempt at a solution
I tried the following:
(sphere) $Ip = Icm + Md^{2}$ <=> Ip = $\frac{2}{5} MR^{2} + M(\frac{L}{2} + R)^2$ <=> Ip = $\frac{7}{5} M R^{2} + M\frac{L^{2}}{4}$
(bar) I = $\frac{2}{12} ML^{2}$
Then I multiplied Ip by 2 and sum I (of the bar). Itotal = 2Ip + I
But the value I am getting is wrong. What I am doing wrong?
Last edited: Jul 24, 2013
2. Jul 24, 2013
### TSny
Hello.
I'm not sure I understand the phrase "about an axis perpendicular to an axis at the bar's center".
But I did note a couple of things.
$(\frac{L}{2} + R)^2 \neq \frac{L^2}{4}+R^2$
Are you sure the moment of inertia of a rod has a factor of 2/12?
3. Jul 24, 2013
### SteamKing
Staff Emeritus
Instead of writing equations off the cuff as it were, set up the problem in an organized manner.
Code (Text):
Item Mass C.O.M. Moment M*Dist^2 MMOI
Dist.
Sphere 1 10 0.55 5.5 3.03 (2/5)MR^2
Sphere 2 10 -0.55 -5.5 3.03 (2/5)MR^2
Bar 12 0.0 0.0 0.0 ML^2/12
----------------------------------------------------------
Totals
Add up the totals for each column and use the Parallel Axis Theorem
To find the MMOI for the bar bell.
4. Jul 24, 2013
### duplaimp
Oops.. The factor is 1/12 and I was thinking in $(L*R)^{2}$ instead of $(L+R)^{2}$
So, trying again I've got the right result but am I doing it right?
I did this: | {
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So, trying again I've got the right result but am I doing it right?
I did this:
(sphere) $Ip = Icm + Md^{2}$ <=> Ip = $\frac{2}{5} MR^{2} + M(\frac{L}{2} + R)^2$ <=> Ip = $\frac{2}{5} M R^{2} + M(\frac{L^{2}}{4}+2\frac{LR}{2}+R^{2})$
(bar) I = $\frac{1}{12} ML^{2}$
Then plugged in the values of each, multiplied I of sphere by two and did a sum of I of bar and I of sphere. Is this correct?
5. Jul 24, 2013
### duplaimp
Oops.. The factor is 1/12 and I was thinking in $(L*R)^{2}$ instead of $(L+R)^{2}$
So, trying again I've got the right result but am I doing it right?
I did this:
(sphere) $Ip = Icm + Md^{2}$ <=> Ip = $\frac{2}{5} MR^{2} + M(\frac{L}{2} + R)^2$ <=> Ip = $\frac{2}{5} M R^{2} + M(\frac{L^{2}}{4}+2\frac{LR}{2}+R^{2})$
(bar) I = $\frac{1}{12} ML^{2}$
Then plugged in the values of each, multiplied I of sphere by two and did a sum of I of bar and I of sphere. Is this correct?
6. Jul 24, 2013
### TSny
Yes, that looks correct assuming the axis of rotation is as shown.
I would probably not bother to expand out the $(\frac{L}{2}+R)^2$, but just leave it as $I_p = \frac{2}{5} MR^{2} + M(\frac{L}{2} + R)^2$. The expression is more compact that way and it will be easier to evaluate numerically.
#### Attached Files:
• ###### Fig 1.png
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7. Jul 25, 2013
### duplaimp
The axis of rotation is as you shown. Thank you! | {
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# How many lines tangent to the graph of $f$ are parallel to the line
I'm getting stuck with this problem:
How many lines tangent to the graph of $f(x)=x^3+3x$ are parallel to the line $y=6x+1$?
What I've done is this:
Since we're looking for lines parallel to $y=6x+1$ then obviously we're looking for lines of the form $y=6x+b$, so the real question is finding the values of $b$ that satisfy the conditions.
Since the lines are tangent to $f$ we need the derivative of $f$, that is $f'(x)=3x^2+3$.
Since the derivative of $f$ represents the slope of line tangent to $f$, we need $f'(x)=3x^2+3=6$ and we get the solutions $x=1$ and $x=-1$.
And this is where I'm stuck. There's some insight that I'm not having, I guess.
*Edit
There is a follow-up question asking me to produce the equations of those two lines. I didn't want to share because I'm aware of people abusing MSE for their homework. This question was created by the Math department of my university and the official answer is $$y = 6x + 2$$ and $$y = 6x - 2$$
I asked my tutor, and he arrived at the official answers this way:
$$x³ + 3x = 6x + b => x³ - 3x = b$$
Then you plug in the $+/- 1$ found earlier to get $b = +/-2$ and finally arrive at the official answer for the equations of the lines:
$$y = 6x +/- 2$$
• As @amd said, you're not stuck -- you're done! :-) – zipirovich Apr 5 '18 at 18:51
You’ve already solved the problem: there are two such tangent lines. The function is well-behaved and has a tangent at each point, and you’ve determined that there are two points at which the derivative has the right value. There’s no need to go any further with this unless there’s a follow-up question that asks you to produce the equations of those lines. | {
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• Yes, that should have made sense to me... there are two solutions to f'(x) = 6 so there are 2 lines tangent to f(x) and parallel to y=6x+1. There is indeed a follow-up question that asks to produce the equations of those lines. I just didn't want to over rely on MSE because I'm well aware MSE has a problem with people coming here asking other people to do their homework. – Victor S. Apr 5 '18 at 19:35
Note: The question asks only for how many so the answer is immeediately two. For the actual equations of the tangents:
The derivative is $$3x^2+3=6\implies x=\pm1$$ The points are therefore $(-1,-4),(1,4)$ - substitute into $y=x^3+2x$.
Let's consider the first point. For a tangent line parallel to $y=6x+1$, the slope must be $6$. So we have that $$y=mx+c\implies-4=6(-1)+c\implies c=2\implies \boxed{y=6x+2}$$
Can you do the other one?
• f(x) = x³ + 3x, not x³ - 3x, sorry. – Victor S. Apr 5 '18 at 15:53
• No problem, edited! – TheSimpliFire Apr 5 '18 at 18:42
• The equation of the tangents is the follow-up question but I didn't want to have MSE do my homework as i'm well aware this is a big problem here... Check my edit on the OP. – Victor S. Apr 5 '18 at 19:37
Plot of the original function and the equation of the lines tangent to f with slope = 6 | {
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# Reason why these two probabilities are equal (Picking Balls in exact order without Replacement)
I realize the probability of the following two events are equal. I am curious: is there a reason, besides coincidence, that the probabilities are equal?
Suppose there are five balls in a bucket. 3 of the balls are labelled A, and 2 of the balls are labelled B. There is no way to distinguish between balls labelled A. There is no way to distinguish between balls labelled B.
Suppose I draw balls at random, without replacement. The event $\{AAABB\}$ means I pick 3 balls labelled A, then 2 balls labelled B (in that exact order). Then,
$$P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 1 = 0.1$$
Also,
$$P(\{AABAB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} \times 1 = 0.1$$
As you can see, the probability of the events $\{AAABB\}$ and $\{AABAB\}$ are exactly the same. I have seen the claim that any possible order of 3 A's and 2 B's is the same. Why is this true (if it indeed true)? If the claim is true, then I don't have to multiply out individually for every conceivable event.
Thanks.
-
Intuitively there is a degeneracy between "sublevels" of the same ordering (i.e there are 12 different ways to pick AAABB but they are equivalent). Thus there is nothing special about one ordering versus another, therefore you would expect results that only differ in ordering to have the same probability. – crasic Jun 4 '11 at 18:00
The claim indeed seems true.
For a way to see this.
Consider the first probability
$P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 1 = 0.1$
This can actually be written as
$P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} = 0.1$
Now notice the denominators: This will always be $5 \times 4 \times 3 \times 2 \times 1$.
Now the numerator will just be $3 \times 2 \times 1 \times 2 \times 1$, in some order. | {
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Now the numerator will just be $3 \times 2 \times 1 \times 2 \times 1$, in some order.
The $3$ will appear when you pick the first A, one of the $2$ will appear when you pick the second A or first B etc.
Thus the probability is $\frac{12}{120} = 0.1$.
-
I will describe a problem that has the same flavour as yours, since it may throw some additional light on your observation.
A standard deck of cards is thoroughly shuffled. Someone lifts up the top $5$ cards, without looking at them, and looks at the sixth card. What is the probability that this sixth card is a Queen? It is probably intuitively obvious that this probability is $4/52$, since all orderings are equally likely.
Now change the problem a tiny bit, we deal out the cards one by one, looking at each one. What is the probability that the sixth card dealt is a Queen? (We are allowing one or more of the first five cards to be a Queen.)
Whether we looked or not obviously does not change the probability, so the probability is $4/52$.
But we are accustomed to solving such problems by a "tree diagram" procedure, so we might want to trace out all ways in which we can get a Queen on the sixth draw. One of them, for example, could be QQNNNQ, another could be NNNNNQ. Calculate the probabilities for each path (they are not all the same), and add up. Not too hard, but it involves some work. After the smoke clears, the mess will simplify to $1/13$.
After performing the computation and noting that the answer is (equivalent to) $4/52$, which is the same as the probability that the first card drawn is a Queen, one might ask a question very similar to the one you asked.
The answer to the cards question has already been given in the first few paragraphs of this answer. | {
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The answer to the cards question has already been given in the first few paragraphs of this answer.
Now to your problem. Let's change your description a little. We have $5$ cards, the $10$, Jack, Queen, King and Ace of spades. Let's label the first three A, with invisible ink. Let's label the last two B, again with invisible ink. Shuffle the $5$ cards. Note that all orders of the cards are equally likely.
Now reveal the labels, using a magic light. It is I hope obvious that the order AAABB has the same probability as (say) the order ABABA.
Sometimes, as in this case, even when we are told that certain objects are identical, one gets a clearer analysis by imagining them distinct, with any identicalness temporarily "secret."
-
If we deal 5 cards, the probability that the sixth card is a queen does depend what the first 5 cards were (if we looked). If we didn't look, it wouldn't matter what the first 5 cards were. For example, if there were 4 queens in the first 5 cards, the probability of the 5th card being a queen is 0. – jrand Jun 4 '11 at 19:27
@jrand: I said we looked. Did not say what we saw. Did not say we used the information to update the probability estimate. – André Nicolas Jun 4 '11 at 19:47 | {
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# Partial Fraction Decomposition with Complex Number $\frac{1}{z^2 - 2i}$.
How do I decompose the fraction
$$\dfrac{1}{z^2 - 2i}$$
into partial fractions? I understand how to do partial fraction decomposition with real numbers, but I am unsure of how to do it with complex numbers. I attempted to find examples online, but all examples are with real numbers -- not complex.
I would greatly appreciate it if people could please take the time to demonstrate this.
• Factoring $\,z^2-2i\,$ would be a good starting point. – dxiv Aug 30 '17 at 22:02
• @dxiv Yes...that is precisely what I am asking... – The Pointer Aug 30 '17 at 22:03
• You do it in the same way over any field. In the case of complex numbers, it is simpler because you have only linear irreducible factors. – Bernard Aug 30 '17 at 22:04
• $z^2-2i=(z+\sqrt{2i})(z-\sqrt{2i})=(z+1+i)(z-1-i)$. From here the process is identical to the decomposition of linear terms with real numbers – Alex Aug 30 '17 at 22:05
• @ThePointer For the first equality, use difference of squares, $a^2-b^2=(a+b)(a-b)$. Next note that $\sqrt{2i}=\sqrt{2}(e^{i\pi/2})^{1/2}=\sqrt{2}e^{i\pi/4}=\sqrt{2} \frac{1}{\sqrt{2}}(1+i)=1+i$ – Alex Aug 30 '17 at 22:37
## 4 Answers | {
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## 4 Answers
Note that $z^2-2i=(z+\sqrt{2i})(z-\sqrt{2i})$ and $\sqrt{2i}=\sqrt{2}e^{i\pi/4}=1+i$. To simplify, let $b=1+i$, then $$\frac{1}{z^2-2i}=\frac{1}{(z+b)(z-b)}$$ From here it actually doesn't matter if you regard $b$ as real or complex, the process to find the partial fractions is the same as long as the terms are linear in $z$. So we let $$\frac{1}{(z+b)(z-b)}=\frac{A}{z+b}+\frac{B}{z-b}$$ for some $A,B\in \mathbb C$. Adding the two fractions on the right hand side we get that $$A(z-b)+B(z+b)=1$$ and so $$A+B=0$$ $$-bA+bB=1$$ which has solution $$A=-\frac{1}{2b}$$ $$B=\frac 1{2b}$$ Plugging in the original $b=1+i$ we have that $$\frac{1}{2b}=\frac12\frac 1{(1+i)}\frac{(1-i)}{(1-i)}=\frac 14(1-i)$$ Therefore $$\frac{1}{z^2-2i}=-\frac{\frac 14(1-i)}{z+1+i}+\frac{\frac 14(1-i)}{z-1-i}.$$ As you can see the process for computing the partial fraction coefficients with complex rationals is equivalent to that of real numbers.
• Thanks for the complete response. I did the calculations myself (with help from your post and those of others) and got $\dfrac{1}{(z - 1 - i)(z + 1 + i)} = \dfrac{1}{(z - 1 - i)(2 + 2i)} - \dfrac{1}{(z + 1 + i)(2 + 2i)}$. – The Pointer Aug 30 '17 at 23:30
• For whatever reason, this answer absolutely captivates me! Stellar! – gen-z ready to perish Aug 30 '17 at 23:49
the roots of $z^2 - 2i$ are $1+i$ and $-(1+i)$
$$(1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i$$
Hint: either will work to find the square roots of $\,2i\,$:
• $\;\; 2i=(a+ib)^2 = a^2-b^2+2abi \implies a^2-b^2=0\,,\;2ab=2\,$
• $\;\; 2 e^{i \pi/2} = \left(r e^{i \phi}\right)^2 \implies r^2 = 2\,,\;2\phi = \pi/2 + 2k\pi$
I suppose one difficulty some will encounter here is how to factor $z^2-2i.$
You can observe that $$2i = 2(\cos90^\circ + i\sin90^\circ),$$ and therefore $$\pm\sqrt{2i\,} = \pm\sqrt 2(\cos45^\circ + i\sin45^\circ) = \pm \sqrt 2\left( \frac 1 {\sqrt 2} + i \frac 1 {\sqrt 2} \right) = \pm( 1+i).$$ | {
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Therefore $z^2 - 2i = \Big(z - (1+i)\Big)\Big( z+ ( 1+i)\Big).$
For the rest, if you know how to do other sorts of arithmetic with complex numbers, just proceed as with real numbers. | {
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# Line integral of a conservative field over a squre
I am trying to evaluate $$\oint _C \frac{-ydx+xdy}{x^2+y^2}$$
clockwise around the square with vertices (−1,−1), (−1,1), (1,1), and (1,−1).
So from the question, $$\vec{F}=<\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}>$$
I first conducted the gradient test $\frac{\partial F_2}{\partial x}=\frac{\partial F_1}{\partial y}$ to see whether the field is conservative. And indeed, I found out that
$$\frac{\partial F_2}{\partial x}=\frac{\partial F_1}{\partial y}=\frac{y^2-x^2}{(x^2+y^2)^2}$$
Thus, $$\vec\nabla \times\vec F=0$$
In this case, since the domain(the square) is simply connected, I thought that the answer was: $$\oint _C \frac{-ydx+xdy}{x^2+y^2}=0$$
However, apparently this is wrong and the solution is
$$\oint _C \frac{-ydx+xdy}{x^2+y^2}=2\pi$$
• What about $x=y=0$? Think ${1 \over z}$ in complex terms. Revisit the simply connected part... – copper.hat May 21 '15 at 15:17
• The domain where the integrand is defined is the square without the origin! – Alex Fok May 21 '15 at 15:18
• Ah i see! what a stupid mistake.. Then does that mean the only way to solve this problem is to compute all of the four line integrals and sum them up?? – user223022 May 21 '15 at 15:20
• @user223022 I provided 3 distinct approaches. Please let me know how I can improve my answer. I just want to give you the best answer I can give you. – Mark Viola May 21 '15 at 18:23
The reason that Stokes' Theorem does not apply directly here is that $F$ is not continuous (and thus, not differentiable) throughout the region bounded by $C$. The singularity at the origin is the source the issue.
METHOD 1: Brute Force
The integration is comprised of the sum of integrals over the 4 line segments. From symmetry considerations, each contribution is identical and we therefore only need to carry out the integration over one of the four segments and multiply by $4$.
Thus, | {
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Thus,
\begin{align} \oint_C \ \frac{-ydx+xdy}{x^2+y^2}&=4\int_{-1}^{1} \frac{(-1)dx+x(0)}{x^2+(1)^2}\\\\ &=\left. -4\arctan(x)\right|_{-1}^{1}\\\\ &=-2\pi \end{align}
METHOD 2: Change Contour
Since $\nabla \times \vec F =0$ for $(x,y)\ne (0,0)$, we can deform the contour to any shape that encloses the origin. Let's choose a unit circle as the new contour $C'$. Then, we have
\begin{align} \oint_C \ \frac{-ydx+xdy}{x^2+y^2}&=\oint_{C'} \ \frac{-ydx+xdy}{x^2+y^2}\\\\ &=-\int_{-\pi}^{\pi} \frac{-\sin \phi (-\sin \phi) d\phi+\cos \phi (\cos \phi)d\phi}{\cos^2 \phi +\sin^2 \phi}\\\\ &=-2\pi \end{align}
METHOD 3: Complex Plane Analysis Interestingly, this problem is identical to the imaginary part of the complex plane contour integration of $1/z$
\begin{align} \text{Im}\left(\oint_C \frac{dz}{z}\right)&=\text{Im}\left(\oint_C \bar z \frac{dz}{|z|^2}\right)\\\\ &=\text{Im}\left(\oint_C \frac{xdx+ydy}{x^2+y^2}+i\oint_C \frac{-ydx+xdy}{x^2+y^2}\right)\\\\ &=\oint_C \frac{-ydx+xdy}{x^2+y^2}\\\\ &=-2\pi \end{align}
from the residue theorem recalling that $C$ is traversed clockwise.
Put a circle of radius $r$ centered about the origin, so that the circle fits inside the rectangle. Then you know by Green's Thm that the integral over the rectangle is the same as the integral over the circle, which when you switch to polar coordinates, will give you $-2\pi$.
Since $\nabla \times \vec{F}=0$ is valid in $\mathbb{R}\backslash{(0,0)}$, you can reshape your curve to a circle centered at origin and use polar coordinates to evaluate it.
Your field $\vec F$ is nothing else but $\nabla\arg$. | {
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Your field $\vec F$ is nothing else but $\nabla\arg$.
The argument (polar angle) function $$\arg: \>\dot{\mathbb R}^2\to{\mathbb R}/(2\pi{\mathbb Z})$$ is locally a nice real function, but is globally only defined up to multiples of $2\pi$. Therefore on the one hand its gradient is well defined. For $x>0$ it can be computed from $$\arg(x,y)=\arctan{y\over x}\qquad({\rm mod}\ 2\pi)\ ,$$ and the result is your $\vec F$, valid in all of $\dot{\mathbb R}^2$. It follows that ${\rm curl}(\vec F)\equiv0$.
On the other hand the integral of $\vec F$ along any arc $\gamma$ in the punctured $(x,y)$-plane is equal to the total increment of $\arg$ along this arc, and the integral of $\vec F$ along the chain $C:=\partial Q$ is equal to the totel increment of $\arg$ along $C$, which is obviouly $=2\pi$. | {
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24 May 2010, 22:35
Ok so if 5 people are to sit at a round table how many ways can they be seated. Why is the answer not 5!
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25 May 2010, 00:07
Total No. of ways in which n no. of persons could be arranged on a round table is given by : ( n - 1 ) ! and not n !
Therefore the ans shd be 4! and not 5!
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Yes but why n-1 ! And not n! Can you give some more color so I can understand
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Yes but why n-1 ! And not n! Can you give some more color so I can understand
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The number of arrangements of n distinct objects in a row is given by $$n!$$.
The number of arrangements of n distinct objects in a circle is given by $$(n-1)!$$. | {
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From Gmat Club Math Book (combinatorics chapter):
"The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:
$$R = \frac{n!}{n} = (n-1)!$$"
$$(n-1)!=(5-1)!=24$$
Check Combinatorics chapter of Math Book for more (link in my signature).
Hope it helps.
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26 Feb 2016, 17:21
In a circle there are n-1 ways to arrange a group
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28 Feb 2016, 22:32
Expert's post
One way to think about it, and these problems in general, is to start by counting the possibilities 'naively'. It seems logical that there should be 5! ways to arrange 5 people around a round table, so start with 5!. Then, account for any special circumstances by figuring out whether you actually counted any of the possibilities more than once. In this case, you actually counted each separate possibility five times by using 5!. For instance, you counted these five arrangements as being different, but they're actually the same (since they're just rotations around the table): | {
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(A B C D E)
(B C D E A)
(C D E A B)
(D E A B C)
(E A B C D)
In order to correct for the overcounting, you'll divide by 5. 5!/5 = 4!, or 24.
This works for a wide range of counting problems. Suppose you wanted to know how many ways a class of eight people could be split into two groups of four. Naively, there are 8*7*6*5 ways to select the first group of four (after which the second group is determined). But you've overcounted by doing that, since you actually counted these groups as being different:
A B C D
B C D A
B A C D
... etc.
That is, you counted each different group 4! times. So, the actual answer is (8*7*6*5)/4!, which is equivalent to what you'd get from the combinatorics formula.
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# Math Help - Integration of Zero where d^2z/dt^2 = 0
1. ## Integration of Zero where d^2z/dt^2 = 0
Suppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial
velocity ˆ−j +2ˆk and moves under a constant force F = −i +2ˆj. Find
the position and velocity for all time.
The Equation of motion ma = F implies that,
a = −i +2ˆj
which gives the three scalar equations:
d^2x/dt^2= −1,
d^2y/dt^2= 2,
d^2z/dt^2= 0
integrating for x yields:
x = −1/2t^2+1
integrating for y yields:
y = t^2− t
I am unable find z, where:
d^2z/dt^2 = 0
I need to integrate this equation twice to obtain:
z = 2t
But as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.
Any help or advice would be greatly appreciated as I am finding this area of Applied math confusing.
2. Originally Posted by GrahamJamesK
Suppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial
velocity ˆ−j +2ˆk and moves under a constant force F = −i +2ˆj. Find
the position and velocity for all time.
The Equation of motion ma = F implies that,
a = −i +2ˆj
which gives the three scalar equations:
d^2x/dt^2= −1,
d^2y/dt^2= 2,
d^2z/dt^2= 0
integrating for x yields:
x = −1/2t^2+1
integrating for y yields:
y = t^2− t
I am unable find z, where:
d^2z/dt^2 = 0
I need to integrate this equation twice to obtain:
z = 2t
But as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.
Any help or advice would be greatly appreciated as I am finding this area of Applied math confusing.
$\displaystyle\int 0 \ dt=c\Rightarrow\int c \ dt=ct+n \ \ \ c,n\in\mathbb{R}$
If c = 2 and n = 0, you have your z = 2t
3. Originally Posted by dwsmith
$\displaystyle\int 0 \ dt=c\Rightarrow\int c \ dt=ct+n \ \ \ c,n\in\mathbb{R}$ | {
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If c = 2 and n = 0, you have your z = 2t
Thank you for your reply. I understand this, but how can I arrive at 2 if I assume that "z=2t" is unknown.
4. Originally Posted by GrahamJamesK
Thank you for your reply. I understand this, but how can I arrive at 2 if I assume that "z=2t" is unknown.
Why are you assuming it is 2t? If you know that, you just set them equal.
$ct+n=2t\Rightarrow c=2 \ \ n=0\Rightarrow z=2t$
5. Ok that makes sense, thank you.
6. From $\frac{d^2z}{dt^2}= 0$, you get $\frac{dz}{dt}= C$, a constant. But you are told that the velocity at t= 0 is -j+ 2k so that the z component of velocity is 2: the constant is 2. From $\frac{dz}{dt}= 2$, integrating again gives $z(t)= 2t+ D$. Since you wer told that the initial position was (1, 0, 0), z(0)= 2(0)+ D= 0 and so D= 0. z= 2t.
7. Your explanation was very instructive, I was getting confused by the vector labels and the axis labels. Kudos! | {
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# Intuition behind logarithm change of base
I try to understand the actual intuition behind the logarithm properties and came across a post on this site that explains the multiplication and thereby also the division properties very nicely:
Suppose you have a table of powers of 2, which looks like this: (after revision)
$$\begin{array}{rrrrrrrrrr} 0&1&2&3&4&5&6&7&8&9&10\\ 1&2&4&8&16&32&64&128&256&512&1024 \end{array}$$
Each column says how many twos you have to multiply to get the number in that column. For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.
Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$. Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.
I found that very helpful to understand the actual proofs for this property.
I still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:
$$\log_a x = y \implies a^y = x$$ $$\log_b a^y = \log_b x$$ $$y \cdot \log_b a = \log_b x$$ $$y = \frac{\log_b x}{\log_b a}$$
But can somehow provide a explanation in the style of the quoted answer why this actually works? | {
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But can somehow provide a explanation in the style of the quoted answer why this actually works?
• Small nit-pick, $a$ must be positive, otherwise you can get the false statement $\log_b((-1)^2)=2\log_b(-1)$. – Simply Beautiful Art Dec 28 '17 at 2:38
• One idea: use the same table to rationalize why $\log_4(x) = \frac{\log_2(x) }{ \log_2(4) } = \frac{1}{2} \log_2(x)\,$. – dxiv Dec 28 '17 at 2:42
• One interesting thing about reading all these answers is that I'd always thought of the change of base formula in the divided form, $\log_a b = \frac{\log_c b}{\log_c a}$, and hadn't tried to develop any intuition behind it, just applying it rotely. But now the multiplicative form, $\log_a x = \log_a b \times \log_b x$, has a lot more meaning to me. Of course the two are mathematically equivalent, but it's weird how intuition can latch onto one more easily than the other. – JonathanZ Dec 28 '17 at 18:19
• Many excellent answers, I also didn't expect that the question gets this much attention. Hard to pick any answer specifically, as many of them contributed to my understanding. – Max Dec 29 '17 at 4:01
• @SimplyBeautifulArt: And $b = 1$ disproves Max's claim that it actually works. – user21820 Dec 29 '17 at 6:43
Here is one way of looking at it. (I'll assume that the numbers $a,b,x \in \mathbb R$ satisfy $a > 1, b > 1$, and $x > 0$.)
I dislike the name "logarithm" and I think a more descriptive name for $\log_b(x)$ is "the exponent from $b$ to $x$". We could also use the notation $[b \to x]$ instead of $\log_b(x)$. The change of base rule then tells us that the exponent from $b$ to $x$ is equal to the exponent from $b$ to $a$ times the exponent from $a$ to $x$: $$\tag{\spadesuit}[b \to x] = [b \to a][a \to x]$$ or equivalently $$[a \to x] = [b \to x]/[b \to a].$$ In standard notation, this formula states that $$\log_a(x) = \frac{\log_b(x)}{\log_b(a)}.$$ | {
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Note that equation $(\spadesuit)$ is obvious, because \begin{align} b^{[b \to a][a \to x]} &=(b^{[b\to a]})^{[a \to x]} \\ &= a^{[a \to x]} \\ &= x. \end{align}
• I agree with your first sentence. Sometime it helps me to read 'logarithm' as 'what's the exponent?'. – JonathanZ Dec 28 '17 at 3:25
• Please be more precise, otherwise your claims are false. Try $b=1$. – user21820 Dec 28 '17 at 11:31
• @user21820 I probably should have stated that $a$ and $b$ are greater than $1$ and $x$ is positive, but the question was only asking for intuition and I wanted to get to the point as fast as possible. – littleO Dec 28 '17 at 11:45
• @user21820 Just updated the answer as you suggested. – littleO Dec 29 '17 at 20:46
Intuition is always tricky to get across, but I can try.
$\log_bx$, as you noted, tells you how many $b$s you need to multiply together to get $x$. Now if you need $\log_ba$ number of $b$s to multiply to get $a$, and you need $\log_ax$ number of $a$s to multiply to get $x$, we can "expand" each of those $a$s into a number of $b$s. There will be $\log_ba$ number of $b$s for each $a$, so the total number of $b$s will be $\log_ax \log_ba$. These $b$s multiply to $x$, so $\log_ax \log_ba = \log_bx$.
For example, take $b=2, a=8, x=64$. We start with $\log_ax = 2$, which tell us we need two $8$s to get $64$:
$$8 \cdot 8 = 64$$
We use $\log_ba = 3$, i.e., $2 \cdot 2 \cdot 2 = 8$, to expand each $8$:
$$(2 \cdot 2 \cdot 2) \cdot (2 \cdot 2\cdot 2) = 64$$
Now the total number of $2$s we are multiplying is $2 \cdot 3 = 6$, so $log_2 64 = 6$
It might help to crank though some easy examples
$4096 = 2^{12} = (2^2)^6 = (2^3)^{4}$
$\log_2 4096 = 12\\ \log_4 4096 = \frac {12}{2} = \frac {\log_2 4096}{\log_2 4}\\ \log_8 4096 = \frac {12}{3} = \frac {\log_2 4096}{\log_2 8}\\ \log_{16} 4096 = \frac {\log_2 4096}{\log_2 16} = 3 \implies (2^4)^3 = 2^{12}$ | {
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Motivated by @dxiv's comment, we can adapt your table to understand the change of base rule, although I'm just going to illustrate changing base from $2$ to $4$ and prove the formula in the (equivalent) form $$\log_2 x = log_2 4 \times \log_4 x.$$ Let's add a row to your table:
$$\begin{array}{rrrrrrrrrr} \text{exponent}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{power of 2}&1& 2& 4& 8& 16& 32& 64& 128& 256& 512& 1024\\ \text{power of 4}&1& 4& 16& 64& 256& 1024& 4096& 16384& 65536& 262144& 1048576 \end{array}$$
The top row, labeled exponents, is where you read off the logarithms, and they can be to the base $2$ or the base $4$ depending on which lower row you are looking at.
Imagine that there are counters in the bottom two rows, and they both start in the first column, on their $1$. Start moving the bottom counter forward, one column at a time. Now suppose the 'powers of 2' counter starts moving forward and it tries to stay on the same number as the 'powers of 4' counter is on. So when 'powers of 4' jumps to '4', 'powers of 2' has to move two columns to get to its 4. When 'powers of 4' then moves to 16, 'powers of 2' again has to move two columns to get to its 16. I.e. the powers of 2 counter has to move twice as fast as the powers of 4 counter. That means when you read off the logarithms from the top row $$\log_2 x = 2 \times \log_4x.$$
If you added a 'powers of 8' row you could easily see that $\log_2 = 3 \times \log_8x$, and then generalize it to compare $\log_a$ and $\log_{a^n}$.
• This is a wonderfully intuitive answer that used the OP's framework to address the question. Thank you and +1 from me! – Slecker Oct 20 '18 at 21:47
• Thank you. I really liked the dynamic image of $\log_2$ traveling faster than the $\log_4$. Glad you did too. – JonathanZ Oct 20 '18 at 23:48
Will it help to write the formula in a different way:
$$\log_a c=\log_a b\times\log_b c$$
? | {
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Will it help to write the formula in a different way:
$$\log_a c=\log_a b\times\log_b c$$
?
This is motivated as follows: assume you start with $a$ and you want to "reach" $c$ by exponentiation.
• You can first raise $a$ to some exponent $x=\log_a b$ to get $b=a^x$, and then you can raise $b$ to some exponent $y=\log_b c$ to get $c=b^y=(a^x)^y=a^{xy}$...
• ... Or, you can directly raise $a$ to the exponent, well, $xy=\log_a b\times\log_b c$ to get $c$. The thing is, we call this last exponent $\log_a c$.
Thus, $\log_a c=\log_a b\times\log_b c$.
Take a slide rule. There is a linear scale on it (on this photo it's labeled as lgX). It is deliberately created to be linear with respect to length because this allows multiplying numbers by adding corresponding line segments. This ability is the main advantage of a slide rule.
Now imagine there are two such scales. In the example below they are the first two:
0-------1-------2 $\log_4 x$
0---1---2---3---4 $\log_2 x$
1---2---4---8--16 $x$
The trick is: whatever (sane) bases you take, if the first scale is linear, so is the second. More general picture:
0----1------p---... $\log_a x$ (linear)
0----q------1---... $\log_b x$ (linear)
1----a------b---... $x$ (non-linear)
The two "upper" scales are linear, they have $0$ at the same place, so there is a simple proportion behind them. For every (sane) $x$ we have:
$$\frac {\log_a x} {\log_b x} = \frac 1 q = \frac p 1$$
where obviously $p=\log_a b$ and $q=\log_b a$.
I understand the change of base formula as a graph transformation hinted by the comment by dxiv. The change of base formula encapsulates the idea that all logarithms are equivalent up to vertical dilations (and reflections). More precisely, if you have two positive numbers $a$ and $b$, then there is a constant $C$ such that $$\log_a(x)=C\cdot\log_b(x)$$ for every positive $x$. The change of base formula tells you that $C=1/\log_b(a)$. | {
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This can be paralleled with exponentials in a similar way, for there is a "change of base formula" for exponentials. But it doesn't get the same limelight as it does for logarithms. All exponentials are equivalent up to horizontal dilations (and reflections). More precisely, if $a$ and $b$ are positive numbers again, then there is a constant $C$ such that $$a^x=b^{C\cdot x}$$ for every real $x$. This time $C=\log_b(a)$. Interestingly, the $C$'s in this discussion are inverses of each other (which shouldn't be a surprise, but I never noticed before).
Similar to @JonathanZ's table, try this
\begin{array}{c|rrrrrrrrr} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2^n & 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ 8^{(n/3)} & 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ \end{array}
Hence $\log_8(x)= \dfrac{\log_2(x)}{3} = \dfrac{\log_2(x)}{\log_2(8)}$ | {
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# What is the probability that Camilla and Cameron are paired with each other?
Textbook problem: A teacher with a math class of 20 students randomly pairs the students to take a test. What is the probability that Camilla and Cameron, two students in the class, are paired with each other?
My answer: 1/190. There is only one such pair out of the 20-choose-2 possible pairings.
Their answer: 1/19. Camilla can be paired with 19 students and only one such pairing is with Cameron.
Question: What principle am I missing in my reasoning that would help me to see why their answer is right and mine wrong? It's like I follow their line of reasoning too but don't see what underlying assumption differentiates the answers to see how I can frame the problem correctly on my own.
Textbook: Chapter 26 from The Art of Problem Solving (Volume I) by Rusczyk and Lehoczky.
• But there are 10 pairs of students. That means Cameron and Camilla get 10 chances for their pair to be chosen. So the probability is definitely higher than 1/190. May 25 at 20:46
You're missing that the teacher will pair up all the students, making 10 pairs. Camilla and Cameron could be any of these 10. | {
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• I want to make the argument 10 pairs / 190 possible pairs = 1/19, but this doesn't work because not all choices of 10 pairs have no repeat students. Is there a variation of this argument that does work? May 25 at 20:55
• @eyeballfrog Yes: suppose the pairs are numbered (perhaps by the desk they are sent to). Then the probability Camilla and Cameron are both picked for desk $1$ is $\frac1{190}$ as you found. Similarly desk 2, or each of the other desks. Then use linearity of expectation May 25 at 21:16
• There are 190 total possible pairs, so 190-choose-10 total possible testing classrooms that could be set up by the teacher. Out of all those testing classrooms there are 189-choose-9 testing classrooms you could choose that contain the Camilla-Cameron pair; hence the probability is 189-choose-9 / 190-choose-10 which equals 1/19. May 26 at 1:19
The 190 comes from counting all the possible pairs of students, including those that are not Cameron or Camilla. You need to zero in on only Cameron or Camilla pairs. Camilla will be paired with someone, and that is a certainty. She cannot be paired with herself, so that leaves only 19 other students she can be paired with.
If instead of pairing up everyone, you only picked two students at random to make one pair out of the 20, then the odds would be 1/190.
$$\mathbf{\text{For another approach:}}$$
Let assume that the boxes represent groups such that two adjacent boxes are in the same group. $$\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad$$ | {
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Now , let Camilla select one of the position , she can do it $$\frac{20}{20}=1$$ ways.After that Cameron must select an empty space , but we want that they are in the same group , so if we assume that Camilla selected the position $$\color{red}{1}$$ , then Cameran must select position $$2$$ among $$19$$ possible empty position. As you see , the probability that Cameron select the position $$2$$ is $$1/19$$ , then the the probability that they are in same group is $$1/19$$ $$\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{\color{red}{1}}\fbox{2} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad$$ | {
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# The number of ways to divide 5 people into three groups
How many ways can 5 people be divided into three teams where each team must have at least one member?
Assumably they can either be put in one group of 3 people then two groups with 1 person, or two groups with 2 people then one group of 1 person. Hence my answer was $$^5C_3 +\, (^5C_2) \cdot (^3C_2)$$ However the provided answer was $$^5C_3 +\, (^5C_2) \cdot (^3C_2) \cdot (1/2)$$ Where did the 1/2 come from?
Thanks.
• Because the groupings $(A),(BC),(DE)$ and $(A),(DE),(BC)$ are the same (for example). – lulu Aug 11 '17 at 12:42
• Is there a way of knowing how many will be repeated without listing them out? – Casper C. Aug 11 '17 at 12:46
• Sure, there's a symmetry. the two $2-$member teams can be interchanged. You have the same problem with the other term, though you short-circuited it. You could have written the first term as $\binom 53\times \binom 21 \times \binom 11$ but then you'd have had to divide by $2$ again to cancel the symmetry between the $1-$ member teams. – lulu Aug 11 '17 at 12:48
• Thanks, but how can you tell the amount of symmetry there is? – Casper C. Aug 11 '17 at 12:56
• Not following. the symmetry is that every pair of $2$ member teams is counted twice (switching the order). There are situations in which this can get tricky...for example, in counting rolls of a pair of dice you have to distinguish the cases in which you get two of a kind. But there is no problem here. – lulu Aug 11 '17 at 13:03
• The factor $\frac{1}{2!}$ occurs in fact twice in your example, since we have \begin{align*} &(^5C_3) (^2C_\color{blue}{1})(^1C_{\color{blue}{1}})\color{blue}{\frac{1}{2!}} +\, (^5C_\color{blue}{2})(^3C_\color{blue}{2})(^1C_1)\color{blue}{\frac{1}{2!}}\\ &\quad=10\cdot2\cdot1\cdot\frac{1}{2}+10\cdot 3\cdot 1\cdot \frac{1}{2}+\\ &\quad=25 \end{align*} | {
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• We can reformulate the problem and ask for the number of ways to partition a set consisting of $5$ elements into $3$ non-empty subsets. These numbers are known as Stirling numbers of the second kind ${n\brace k}$.
Here we are looking for \begin{align*} {5\brace 3}=25 \end{align*}
When you pick the team as two players, another two, and then just the final one is left over, consider the following: if the five players are named $A, B, C, D, E$, then you might start by picking the duo $\{A, B\}$ and then $\{C, D\}$, which results in the breakdown of $\{\{A, B\}, \{C, D\}, \{E\}\}$ or you might have first picked the duo $\{C, D\}$ and then picked $\{A, B\}$, which would result in the same grouping of $\{\{A, B\}, \{C, D\}, \{E\}\}$.
To avoid this double-counting, you divide in this scenario by $2$. (Or, equivalently, multiply by $1/2$.)
• Thanks, but how do I tell how much repeat-counting I've done without listing out the combinations? – Casper C. Aug 11 '17 at 12:57
• @CasperC. Yes, this is the difficult part of combinatorics. You may wish to return to the definition of, e.g., $^5 C_3$ to see how and why it was defined; its definition, too, involves a bit of division to avoid overcounting, and the scenario here involves a similar bit of reasoning. – Benjamin Dickman Aug 11 '17 at 13:04
• May I rephase that? How do we know there are only two repetitions of $\{\{A, B\}, \{C, D\}, \{E\}\}$ ? How would we know how many repetitions of groups there would be if there were 7 people instead of 5? Thanks for helping. – Casper C. Aug 11 '17 at 13:45
• @CasperC. This relies on the observation that $n$ objects can be arranged in $n!$ ways. Here, you have the $2$ groups overcounting by a factor of $2! = 2$. – Benjamin Dickman Aug 11 '17 at 13:50 | {
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# Need help figure out a Fibonacci related math trick
My math teacher used to do a trick where he would have a student write $$2$$ numbers on the board then add the first to the second to create the third then add the second to the third and so on until there were $$10$$ numbers. He would then turn around and add them up in $$2$$ seconds. How did he do this?
• Have you tried a few simple cases to see the pattern? – rtybase Sep 5 '19 at 20:04
• have you heard of induction? – ggg Sep 5 '19 at 20:06
• Check my answer, I think it is easy; the seventh number in the list multiplied by 11. – Hussain-Alqatari Sep 5 '19 at 21:55
Multiplying any natural number by $$11$$ is so easy, check here.
Now the solution for your problem is to multiply the $$7^\text{th}$$ number in the list by $$11$$
Have this example: our first two numbers are $$16$$ and $$21$$
So the list is:
$$16$$
$$21$$
$$37$$
$$58$$
$$95$$
$$153$$
$$248$$
$$401$$
$$649$$
$$1050$$
The sum of those numbers is just $$248$$ (which is the $$7^\text{th}$$ number) $$\times 11=2728$$.
The rule is: $$\boxed{7^\text{th}\text{ number }\times 11}$$
Try it algebraically starting with $$a$$ and $$b$$ $$\begin{eqnarray*} a,b,a+b,a+2b,2a+3b,3a+5b,5a+8b, \\ 8a+13b,13a+21b,21a+34b. \end{eqnarray*}$$ Now add these together and we get $$55a+88b=11 (5a+8b)$$.
So I guess your teacher took the first value multiplied by $$5$$ and added it to the second value multiplied by $$8$$ and then multiplied by $$11$$. Your teacher would have had plenty of time to do this calculation while then values were being added.
• There's no need for the teacher to calculate $5a+8b$ - it's already written on the board! – Carmeister Sep 6 '19 at 4:27
• Actually the teacher can calculate $5a+8b$ essentially faster than it will be written on the board, and have additional time to perform multiplication by $11$ (and check the answer twice). – Oleg567 Sep 7 '19 at 9:14
Hint: | {
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Hint:
$$\begin{array}{rl} F(1) &= \color{blue}{F(3)}-F(2)\\ F(2)&= F(4)\color{blue}{-F(3)}\\ F(3)&=\color{red}{F(5)}-F(4)\\F(4)&=F(6)\color{red}{-F(5)}\\ \vdots\end{array}$$
$$F(1)+F(2)+\dots+F(n) = F(n+2)-F(2)$$
That is because Fibonacci numbers have a number of properties, one of them being:
$$\sum_{i=0}^nF_i = F_{n+2} - 1 = 2F_n + F_{n-1} - 1$$
Proof is by induction
Hence, if the numbers are $$0,1,1,2,3,5,8,13$$, the sum will be $$13*2 + 8 - 1 = 33$$
Well, to answer the question as to how he did it: If the first number is $$x$$ and the second number is $$y$$ then every other number and the sum of all ten numbers will a combination of $$x$$ and $$y$$. As you do the same thing every time the final sum will be the same combination. Your teacher merely memorized that the final sum would be $$55x + 88y$$.
As to how we would know the final number is $$55x+88y$$ we can
1) Simply do it. The ten numbers are $$x,y,x+y, x+2y, 2x+3y, 3x+5y,5x+8y,8x+13y,13x+21y, 21x+34y$$ and the sum is $$55x+88y$$.
2) Try to find a way to generalize this without doing each sum.
We notice the number of $$x$$s involved are $$1,0,1,1, 2,etc.$$. After a slow start once we have $$1,1$$ this has to follow the Fibonacci sequence. So if the $$k$$th number is $$a_k x + b_k y$$ we know $$a_k= F_{k-2}$$, the $$k$$th fibonacci number.
We notice the number of $$y$$s involved are $$0,1,1,2, etc.$$ and those are the Fibonacci numbers too but with a quicker start. $$b_k = F_{k-1}$$.
So the $$k$$th number is $$F_{k-2}x + F_{k-1}y$$.
The final total after ten numbers is therefore $$(1 +\sum_{k=1}^8 F_k)x + (\sum_{k=1}^9 F_k) y$$.
There's an interesting formula that
$$\sum_{k=1}^n F_k = F_{n+2} - 1$$.
So the sum is $$F_{10}x + (F_{11}-1)y$$
======
Another answer states that the answer is that the sum is the $$7$$th number times $$11$$.
So does $$11(F_{5}x + F_{6}y) = F_{10}x + (F_{11} -1)y$$?
Well, $$11(F_5x + F_6y) = 11(5x + 8y)=55x + 88y = 55x + (89-1)y$$. | {
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Well, $$11(F_5x + F_6y) = 11(5x + 8y)=55x + 88y = 55x + (89-1)y$$.
The Fibonacci series is $$1,1,2,3,5,8,13,21,34,55,89$$ so, yes, indeed this is true.
So that's actually how the teacher did it so quickly. You wrote down the $$7$$th term and he multiplied it by $$11$$ in his head.
• Thanks everyone! – PotatoHeadz35 Sep 6 '19 at 11:45 | {
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Cyclic quadrilateral is a special kind of quadrilateral inscribed in a circle. Inscribe means that the vertices of quadrilateral are on the circumference of the circle. A square, rectangle, isosceles trapezoid are some of cyclic quadrilaterals. Below are the some facts about this special figure.
1. The opposite angle of cyclic quadrilateral is supplementary.
$\angle{A}+\angle{C}=180$°
$\angle{B}+\angle{D}=180$°
2. Ptolemy’s Theorem. This theorem is a special case of Ptolemy’s Inequality. He formulated the following equation relating the sides and diagonals.
$d_1d_2=ac+bd$
3. Diagonals as Cords. Using the basic theorem in geometry, we can relate the diagonals with respect to their intersections as follows.
$AO(OC)=OD(OB)$
4. Brahmagupta’s Formula. Brahmagupta of India derived a formula to find the area of cyclic quadrilateral.
$A=\sqrt{s(s-a)(s-b)(s-c)(s-d)}$
Where S is the semiperimeter
$s=\displaystyle\frac{a+b+c+d}{2}$
Use the facts above to solve for the following problems.
Problem 1: In a cyclic quadrilateral $ABCD$, If $\angle{A}=105$°. Find the measure of $\angle{C}$
Solution:
Using the first fact, opposite angles are supplementary.
$\angle{A}+\angle{C}=180$°
$105$°$+\angle{C}=180$°
$\angle{C}=(180-105)$°
$\angle{C}=75$°
Problem 2: In cyclic quadrilateral $ABCD$, $\angle{C}=\angle{A}=90$°. If $\overline{BD}=5$, and two opposite sides have length 2 and 3. What is the area of quadrilateral?
Solution:
Since $\Delta BCD$ and $\Delta ABD$ are right triangle and $\overline{BD}$ is the hypotenuse of both triangles, we can find sides $\overline{AD}$ and $\overline{BC}$ by Pythagorean Theorem.
$\overline{BD}^2=\overline{BC}^2+\overline{CD}^2$
$5^2=\overline{BC}^2+3^2$
$\overline{BC}^2=25-9=16$
$\overline{BC}=4$
$\overline{BD}^2=\overline{AB}^2+\overline{AD}^2$
$5^2=2^2+\overline{AD}^2$
$\overline{AD}^2=25-4=21$
$\overline{AD}=\sqrt{21}$ | {
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$5^2=2^2+\overline{AD}^2$
$\overline{AD}^2=25-4=21$
$\overline{AD}=\sqrt{21}$
The area can be found using Brahmagupta’s formula or using the fact that both triangles are right triangle.
$Area_{quadrilateral}=Area_{\Delta BCD}+Area_{\Delta ABD}$
$Area_{quadrilateral}=\displaystyle\frac{\overline{BC}\cdot \overline{CD}}{2}+\displaystyle\frac{\overline{AB}\cdot \overline{AD}}{2}$
$Area_{quadrilateral}=\displaystyle\frac{1}{2}(3\cdot 4+2\cdot \sqrt{21})$
$Area_{quadrilateral}=6+\sqrt{21}$ sq. units
### Dan
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# Open or Closed Intervals? It Depends
#### (Archive Question of the Week)
Students commonly expect that textbooks all say the same thing (in fact, some think they can ask us about “Theorem 6.2” and we’ll know what they’re talking about!). The reality is that they can even give conflicting definitions, depending on the perspective from which they approach a topic. Here, I want to show how and why they differ in talking about intervals on which a function is increasing or decreasing. Let’s see if we can resolve the “fight”.
## In this corner … it must be open
This one page in our archive actually contains two of the three answers to the question in our archive, the second being a challenge to the first. Let’s start with the 2009 question from Bizhan, answered by Doctor Minter:
Endpoints of Intervals Where a Function is Increasing or Decreasing
Why do some calculus books include the ends when determining the intervals in which the graph of a function increases or decreases while others do not? I feel that the intervals should be open and the ends should not be included as they may be, for example, stationary points where a horizontal tangent can be drawn.
I noticed that the AP Central always include the ends in the formal solutions of such problems (one can see many examples there), but an author like Howard Anton never does. Could you clarify this for me please? Can both versions be correct?
For example, for the function $$y = -x^3+12x$$,
we see that it is increasing from -2 to +2. But should we describe that as the open interval $$\left(-2,2\right)$$, or as the closed interval $$\left[-2,2\right]$$? Different textbooks give different answers.
Doctor Minter began:
You pose an excellent question, and I agree that the discrepancy among the various textbooks is quite misleading. | {
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I completely agree with your claim that the intervals should be open (that is, should not include the endpoints). Let me attempt to give comprehensive reasoning as to why this should be.
We use derivatives to decide whether a function is increasing and/or decreasing on a given interval. Intervals where the derivative is positive suggest that the function is increasing on that interval, and intervals where the derivative is negative suggest that the function is decreasing on that interval.
He goes on to explain that in order for the derivative to be positive at a point, the derivative must exist there, so the function must be defined, and increasing, and differentiable, in some interval around that point. So he concludes,
In summary, for a function to be increasing (all of these concepts are similar for decreasing intervals as well), we have to be able to show that the function is greater for larger values of "x," and less for smaller values of "x" in a small neighborhood around each point in the interval. An endpoint cannot have both of these properties.
It is important to note that both the question and the answer come from the perspective of calculus, and depend on defining “increasing” in terms of the derivative.
## In that corner … it can be closed
But that is not the only way to define “increasing on an interval”!
Back in 1997, Doctor Jerry had answered a similar question, pertaining to the rules for AP calculus:
Brackets or Parentheses?
We have been discussing a problem in my Advanced Placement Calculus class. It concerns increasing/decreasing functions as well as concave up/concave down. ...
When expressing these answers as an interval, should I use a bracket, symbolizing that the endpoint is included, or a parenthesis, symbolizing that the endpoint is not included?
Doctor Jerry replied (in part): | {
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Doctor Jerry replied (in part):
Different books, teachers, and mathematicians use slightly different definitions of increasing functions, but this is not a matter of much consequence as long as one is consistent. Suppose your definition of an increasing function is: f is increasing on an interval I if for each pair of points p and q in I, if p < q, then f(p) < f(q). Note that I may be open (a,b), half-open [a,b) or (a,b], or closed [a,b].
Consider f(x) = x^2, defined on R. The usual tool for deciding if f is increasing on an interval I is to calculate f'(x) = 2x. We use the theorem: if f is differentiable on an open interval J and if f'(x) > 0 for all x in J, then f is increasing on J.
Okay, let's apply this to f(x) = x^2. Certainly f is increasing on (0,oo) and decreasing on (-oo,0). What about [0,oo)? The theorem, as stated, is silent. However, one can go back to the definition of increasing. To show that f is increasing on I = [0,oo), let u and v be in I and u < v. If 0 < u, then the theorem applies. Otherwise, 0 = u < v and we see that f(u) = 0 < f(v) = v^2.
Many instructors, books, and even AP exams often skip consideration of endpoints. If you want to be ultra-safe, then you can do the above kind of analysis. It just takes a few extra steps, usually easy, past the standard test.
Note that here, the question and answer are still in the context of calculus, but Doctor Jerry defines “increasing” without calculus, and then applies a theorem that relates this definition to the derivative: If the derivative is positive on an interval, then the function is increasing on that interval. This theorem doesn’t tell us anything about intervals in which the derivative is sometimes zero, so we have to fall back on the definition, not the theorem. And he concludes that his function is increasing on the half-closed interval $$\left[0,\infty\right)$$.
## Shake hands and come out fighting | {
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## Shake hands and come out fighting
There seems to be a difference of opinion here! We have to bring them together and compare them.
Doctor Minter gave an argument for why endpoints should *not* be included when determining intervals where a function is increasing or decreasing.
Implicit in your answer is that "increasing at a point" means "has a positive derivative in a neighborhood of that point." I wonder if it makes sense to define increasing at a point. I also wonder about another definition that I came up with (which, I acknowledge, doesn't work for a point).
We could define increasing for an interval [a, b] as: whenever x and y are in [a, b] then f(x) < f(y).
This makes no reference to derivatives, so you could still talk about a function being increasing even if it fails to be differentiable some places (e.g., we could say x^(1/3) is increasing everywhere). I'm also thinking about piecewise functions with jumps; again, it seems like we should be able to say they're increasing even if the derivative doesn't exist everywhere.
With this second definition, it seems to me that if a function is increasing on (a, b) and continuous at a and b, then it would be guaranteed to be increasing on [a, b].
What do you think? Do we need to have a definition of "increasing at a point" for some reason? Is there any way to reconcile these two definitions?
There doesn't seem to be consensus here. It's a basic calculus concept and there seems to be two (very convincing) ways of looking at it that are in conflict. | {
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This was a very perceptive question. There are really two different concepts here, just as Kevin suggested. The concept of increasing on an interval does not require calculus, and applies to functions that are not differentiable. The concept of increasing at a point requires calculus, and is often what the authors of calculus books are really talking about; Doctor Minter took “increasing on an interval” to mean “increasing at every point in the interval” in this sense. [Doctor Fenton, in an unarchived 2007 answer, mentioned that “increasing at a point” can be defined instead as “f(a) is larger than any f(x) for x to the left of a, and f(a) is less than any f(x) when x is larger than a“. This makes it still independent of the derivative.]
I started by discussing the definition Kevin gave, which was (when clarified) identical with Doctor Jerry’s:
This is the proper definition of increasing on an interval, which applies to any function, and is found, for example, here:
http://mathworld.wolfram.com/IncreasingFunction.html
A function f(x) increases on an interval I if
f(b) ≥ f(a) for all b > a, where a, b ∈ I.
If f(b) > f(a) for all b > a, the function is
said to be strictly increasing.
...
If the derivative f'(x) of a continuous function
f(x) satisfies f'(x) > 0 on an open interval
(a, b), then f(x) is increasing on (a, b).
However, a function may increase on an interval
without having a derivative defined at all
points. For example, the function x^(1/3) is
increasing everywhere, including the origin
x = 0, despite the fact that the derivative is
not defined at that point.
They state the theorem Doctor Jerry used, and give the same example Kevin gave, which shows that the issue applies not only to endpoints, but to any isolated point where the derivative is zero or undefined. For reference, here is the graph of $$y = x^{1/3}$$, the cube root: | {
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Note that although its tangent is vertical at the origin, so it is not differentiable there, it is clearly increasing everywhere. Similarly, the cube function, $$y = x^{3}$$, is increasing everywhere, although at the origin it is (momentarily) horizontal:
## A split decision
I also commented on Kevin’s final paragraph about consensus:
There are actually two different concepts: a precalculus concept, applicable to any function; and a calculus concept, applicable to differentiable functions. I would prefer not to confuse them. Much as we distinguish uniform vs. pointwise continuity, these notions of increasing could be better distinguished like this:
* The function f is increasing on the interval [0, 1], meaning that comparing any two points, the one on the right is higher.
* The function f is increasing at every point on the interval (0, 1), meaning that the derivative is positive everywhere.
Then I quoted from a 2013 conversation I’d had in which this topic had arisen, in which I referred to both Doctor Minter’s and Doctor Jerry’s answers. The question there (in the midst of a long discussion with Aakarsh) made the conflict even stronger:
I have y = f(x). On the x-axis there are points a, b, and c. When x = a, y = 0; when x = b, y = 4; when x = c, y = 1.
I realise the function increases on [a, b].
I also realise the function decreases on [b, c].
But why is the b in brackets? I know that they indicate closed intervals; that's no problem. If the graph increases from point a to point b, that is [a, b], but then the graph *MUST* decrease on (b, c].
If it increases TO "b," it decreases FROM "b" ... EXCEPT "b"?
Whoa! Can the function be both increasing and decreasing at the same point? If not, how would you decide which? (Notice that Aakarsh was taught to use closed intervals, and accepted that.) I first clarified the question:
I'll suppose that you were given a graph like this: | {
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I'll suppose that you were given a graph like this:
I think what you are asking is why they include the endpoints in the intervals. That's strange, because we've had other questions about why the endpoints are NEVER included in the interval of increase or decrease!
Different texts have different policies on this. How does your text DEFINE "increasing on an interval"? Can you show me the first example they give?
See these pages, which emphasize this variability among texts:
Brackets or Parentheses?
http://mathforum.org/library/drmath/view/53566.html
Endpoints of Intervals Where a Function is Increasing
or Decreasing
http://mathforum.org/library/drmath/view/73202.html
I'm inclined to agree more with the first of these than the second; but I think in pre-calculus, it's a good idea to ignore this detail and either always use open intervals or always use closed intervals. It's not really an important issue; but your concern that a function can't be increasing AND decreasing at the same point would tilt me in the direction of using open intervals just to avoid confusing students like you!
Really, however, you need to notice that your definition probably is only about increasing or decreasing IN AN INTERVAL, not AT A POINT. That is, they are not saying the function is increasing at b -- only that it is increasing in the interval [a, b]. So no claim is being made that the function is both increasing and decreasing at b!
Once I see your book's definition, I can be more clear on that.
As I continued writing to Kevin,
This student never replied with his text's definition, so I didn't get to explore the details with him. One such detail would have been the distinction between saying that a function is increasing on an interval and saying that that is a maximal interval, in the sense that there is no containing interval (open or closed) on which it is increasing. | {
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In my experience, texts leave a lot unstated. While these omitted details might keep things simple for the less mature student, they would be worth exploring with a curious, capable one like you!
As I see it, when a textbook (particularly at the precalculus level) asks for “the interval on which f is increasing”, it is not asking for any such interval, but for the largest. Ideally it would say that; but often they will rely on our instinct to give the “best” answer possible – just as when we are asked what shape a square is, we don’t say “a rectangle”, but give the most precise answer that fits. If they have defined “increasing” as I have, then the closed interval is the correct answer. But there is room for disagreement, especially among students who are thinking more informally.
Some texts, to avoid confusion about endpoints, will specify that they are asking for the largest open interval on which the function is increasing. I think this is the best solution at that level. At a higher level, where precision of definitions is important, just state your definition and act on it.
## Handling differences
It happens that, two months before Kevin wrote, someone else had written about the same issue, suggesting that we should add to Doctor Minter’s answer some information about the reasons for different answers. Kevin’s question gave the occasion to do just that. In corresponding with Ken about his suggestion, I said the following: | {
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I agree with you on your point about different definitions. That's something I often emphasize in my answers; and it's also an explanation for our having answers on our site that don't agree. I like to refer "patients" to past answers, in part to show them that the same problem can be looked at from different perspectives. I also recognize that each answer has its own context, answering a particular student's question either in the light of that student's level or, perhaps, that Math Doctor's personal context. Our goal is not, generally, to give a comprehensive survey of a topic covering all possible variations, but to show a variety of individual interactions. No one answer will cover everything I wish it did (even if it's one I wrote myself a couple years ago). So I just write another and link to the old one(s).
On Doctor Minter’s answer in particular, I said this:
I think Dr. Minter was probably assuming the discussion is about maximal intervals; he also seems to be assuming the function is differentiable, which is not necessary to be an increasing function. Actually, that's my own main objection to his answer: he hasn't actually defined what he means by increasing on an interval, or what context he is assuming. The answer to many questions depends heavily on the context, and we seldom get enough context to be able to give a perfectly appropriate answer. I personally try to state my assumptions up front (if I don't just ask for the definitions and context), so it can be clear what we are talking about -- especially if I am thinking of having my answer archived.
### 3 thoughts on “Open or Closed Intervals? It Depends”
1. Hi, Samrath. This is clearly a complicated issue. If you have specific questions about it, the best thing to do is to ask us directly via the Ask a Question link. There we can interact to find out your specific context and concerns.
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How is irrational exponent defined?
I am trying to understand the most significant jewel in mathematics - the Euler's formula. But first I try to re-catch my understanding of exponent function.
At the very beginning, exponent is used as a shorthand notion of multiplying several identical number together. For example, $5*5*5$ is noted as $5^3$. In this context, the exponent can only be $N$.
Then the exponent extends naturally to $0$, negative number, and fractions. These are easy to understand with just a little bit of reasoning. Thus the exponent extends to $Q$
Then it came to irrational number. I don't quite understand what an irrational exponent means? For example, how do we calculate the $5^{\sqrt{2}}$? Do we first get an approximate value of $\sqrt{2}$, say $1.414$. Then convert it to $\frac{1414}{1000}$. And then multiply 5 for 1414 times and then get the $1000^{th}$ root of the result?
Thanks to the replies so far.
In the thread recommended by several comments, a function definition is mentioned as below:
$$ln(x) = \int_1^x \frac{1}{t}\,\mathrm{d}t$$
And its inverse function is intentionally written like this:
$$exp(x)$$
And it implies this is the logarithms function because it abides by the laws of logarithms.
I guess by the laws of logarithms that thread means something like this:
$$f(x_1*x_2)=f(x_1)+f(x_2)$$
But that doesn't necessarily mean the function $f$ is the logarithms function. I can think of several function definitions satisfying the above law.
So what if we don't explicitly name the function as $ln(x)$ but write it like this:
$$g(x) = \int_1^x \frac{1}{t}\,\mathrm{d}t$$
And its reverse as this:
$$g^{-1}(x)$$
How can we tell they are still the logarithm/exponent function as we know them? | {
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$$g^{-1}(x)$$
How can we tell they are still the logarithm/exponent function as we know them?
• Check this mse question – user378947 Oct 31 '16 at 2:24
• You should check for similar questions before asking a new one. – user378947 Oct 31 '16 at 2:25
• @mathbeing Thanks. But that thread may not fully solve my confusion. I updated my question. – smwikipedia Oct 31 '16 at 3:14
• Well, much to Pytagoras' shock two millennia ago, there is no rational number whose square is 2. So again, its definition is through successive approximation like 1.4, 1.41, etc. Such a sequence is called Cauchy Sequence. – Momo Oct 31 '16 at 3:33
• If you have time, I recommend you to study the axioms of real numbers and constructions of reals from rationals of Cauchy, Dedekind, Weierstrass. It's not trivial; that is why it has not been formalized earlier. If you don't have time to waste, you may think at $5^\sqrt{2}$ as something mysterious between $5^{1.41}$ and $5^{1.42}$, that you can get with as many decimals as you need, but never exactly :) – Momo Oct 31 '16 at 3:49
Yes, you can approximate the result by approximating the irrational exponent with a rational number and proceed with computing integer powers and integer roots. But this does not give you much insight into what an irrational exponent might mean, and I think this is what you mostly care about.
The best insightful explanation I've seen comes from Khalid at BetterExplained.com.
The short summary is that we have to stop seeing exponents as repeated multiplication and start seeing them as continuous growth functions, where $e$ plays a central role.
So $5^{\sqrt2}$ can be written as $(e^{ln(5)})^\sqrt2 = e^{\sqrt2\cdot ln(5)}$. This can be interpreted as continuous growth for $1$ unit of time at a rate of $\sqrt2\cdot ln(5)$, or continuous growth for $\sqrt2$ units of time at a rate of $ln(5)$, or continuous growth for $ln(5)$ units of time at a rate of $\sqrt2$. They are all equivalent. | {
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Check out these links for a much more detailed explanation:
• The continuous growth explanation of $e$ is unbelievably amazing! I have never viewed $e$ like this! Thanks. Hope it can help me solve my confusion. – smwikipedia Oct 31 '16 at 8:01
• And a useful link about various growth functions. betterexplained.com/articles/… – smwikipedia Oct 31 '16 at 8:02
• Regarding the continuous compound growth, I have another question posted here: math.stackexchange.com/questions/1992743/… – smwikipedia Oct 31 '16 at 8:38
One way of defining the real numbers is as equivalence classes of "the collection of all Cauchy sequences (or, equivalently, the collection of all increasing sequences with upper bound) of rational numbers" with "$\{a_n\}$ equivalent to $\{b_n\}$ if and only if $\{a_n- b_n\}$ converges to 0. The essentially says that, for example, $\pi$ is "represented" by the infinite decimal 3.1415926.... From that definition, if a is an irrational number then there exist a sequence of rational numbers $r_1, r_2, r_3, ...$ that converges to a. We then define $x^a$ to be the limit of the sequence $x^{r_1}, x^{r_2}, x^{r_3}, ...$. Using the same example as before, $2^\pi$ is defined as the limit of the sequence $2^3, 2^{3.1}, 2^{3.14}, 2^{3.141}, 2^{3.1415}, 2^{3.14159}, 2^{3.141592}, 2^{31415926}, ...$.
• One key point is that the limit of $x^{r_{\large n}}$ does not depend on which sequence $r_n$ converging to $a$ is chosen. – anon Nov 5 '17 at 21:41
My two cents:
The definition has to do with the fact that $\mathbb R$ is by definition the completion of $\mathbb Q$:
$5^\sqrt{2}=\sup\{5^a:a\in \mathbb Q \wedge a<\sqrt 2\}$
You may show that the set on the RHS is bounded and nonempty, so $\sup$ exists.
So your approach is right (calculate $5^a$ for successive approximations of $\sqrt 2$ like 1.4, 1.41 and so on, and it will converge to the result). | {
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I've seen the theory developed the other way around (first study the series, then define elementary functions as series), but this approach seems very "unnatural" to me, as it is not how this functions were developed historically.
The exponential function is an order-preserving bijection over the rationals. Filling in the holes gives an exponential function over the reals that is an order-preserving bijection. This is done by letting a^i be the supremum of {a^(p/q) | (p/q)
• If you used the proper TeX style formatting, your answer would likely be a bit clearer. – rajb245 Oct 31 '16 at 2:49
• @rajb245 yeah i typed it clear but the damn compiler or whatever was written by a faschist. – Jacob Wakem Oct 31 '16 at 2:55
• You just need to put dollar signs to make math Jax read it. You will also need to use \ to escape your curly braces for set notation. – Ian Oct 31 '16 at 2:56
• @Alephnull: please use polite words, and don't call names. – P Vanchinathan Oct 31 '16 at 3:49
• Especially if you don't know how to spell the names! – user247327 Nov 5 '17 at 21:38 | {
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# Maths
The sixth term of GP is 16 and the third term is 2.Find the first term and common ratio. 6th term =16 3rd term=2 So ar^5=16 and ar^2=2, then divide. So r^3=8, and r^3 =2^3. Therefore r =2 and a=1/2
1. 👍
2. 👎
3. 👁
1. Looks good to me.
1. 👍
2. 👎
2. The formula of the GP
1. 👍
2. 👎
3. Thanks 😊
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2. 👎
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How to determine the eigen values of $A^tA$ where $x^tA A^tx=\alpha x^tx ~~~~ \forall x \in \Bbb{R}^m$ holds for some $\alpha$
Problem. A be an $$m\times n$$ matrix of rank $$m$$ with $$n>m$$. If for some nonzero real $$\alpha$$ we have $$x^tA A^tx=\alpha x^tx ~~~~ \forall x \in \Bbb{R}^m$$ then $$A^tA$$ has
1. exactly two distinct eigen values
2. $$0$$ as an eigen value with multiplicity $$n-m$$
3. $$\alpha$$ is a non zero eigen value.
4. exactly two non zero eigen values.
My attempt.
We know, $$rank(A^tA)=rank(A)=m$$. So, $$Nullity(A^tA)=n-m>0$$ So, $$0$$ is an eigen value of $$A^tA$$ with geometric multiplicity $$n-m$$. But $$A^tA$$ is symmetric so is diagonalizable i.e. $$0$$ is regular. Hence $$0$$ is an eigen value of (algebraic) multiplicity $$n-m$$ i.e. Option 2 is correct. (without using the given equation)
Now we have to conclude about the rest. I think for this we have to use the given equation.
I start by checking for the non zero eigen values of $$A^tA$$ (In fact, $$A^tA$$ should have $$m$$ non zero real eigen values)
Let $$\lambda$$ be a non zero eigen value of $$A^tA$$. Then there exists a non-null vector $$v \in \Bbb{R}^n$$ such that $$A^tAv=\lambda v$$. Now to use the given equation I think we need a vactor $$x \in \Bbb{R}^m$$. So, I try to choose $$x=Av$$ in that equation.....But I can't get any conclusion from there...!!
Any help is appreciated. Thank you.
N.B: This question has an answer here. But I cannot find out how to use $$AA^t=\alpha I$$ further.
EDIT: Well, I have found a way to proceed further:
Let $$\lambda$$ be a non-zero eigen value of $$A^tA$$. Then it can be proved it is also an eigen value of $$AA^t$$ (see here). Therefore, there exits a non-zero vector $$v \in \Bbb{R}^m$$ such that $$AA^tv=\lambda v$$. Again from the given equation we have, $$v^tAA^tv=\alpha v^tv$$, i.e., $$v^t(AA^tv)=\alpha v^tv$$, or, $$\lambda v^tv=\alpha v^tv$$ this implies, $$\lambda = \alpha$$. | {
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Hence the only non-zero eigen value of $$A^tA$$ is $$\alpha$$.
So, finally Correct options are $$1,2$$ and $$3$$.
Your logic looks good to me, though it can be done more simply than you have it. You were on the right track with setting $$x = Av$$. Then $$A^tx = A^tAv = \lambda v\\AA^tx = \lambda Av = \lambda x\\x^tAA^tx = \lambda x^tx\\\alpha x^tx = \lambda x^tx$$ Since $$\|x\| \ne 0$$, we have $$\lambda = \alpha$$. | {
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# Want to show $\sum_{n=2}^{\infty} \frac{1}{2^{n}*n}$ converges
Want to show $\sum_{n=2}^{\infty} \frac{1}{2^{n}*n}$ converges. I am trying to show this by showing that the partial sums are bounded. I have tried doing this by induction but am not seeing how to pass the inductive assumption part. Do I need to instead look for a closed form? thanks
-
The sum is less than $\sum_{n=2}^{\infty} \frac{1}{2^n} = \frac{1}{2}$.
-
Hint All the terms are postive and $$\frac{1}{n2^n}\leq \frac 1 {2^n}$$
Think about the monotone convergence, or the so called comparison test.
Spoiler
The partial sums are bounded above, and are monotone increasing, thus the monotone convergence theorem implies convergence.
-
+1 for the spoiler! I hadn't saw this before. – 000 Dec 29 '12 at 0:51
*hadn't seen =P – Pedro Tamaroff Dec 29 '12 at 1:15
You might want to realize a general principle here.
If we have $$\sum_{a \le n \le b}\frac{1}{f(n)g(n)}$$ and $f(n)$ and $g(n)$ are positive on $[a,b]$, then we always have that $$\sum_{a \le n \le b}\frac{1}{f(n)g(n)}\le\sum_{a\le n\le b}\frac{1}{f(n)} \text{ and } \sum_{a \le n \le b}\frac{1}{f(n)g(n)}\le\sum_{a \le n \le b}\frac{1}{g(n)}$$ when the sums $$\sum_{a \le n \le b}\frac{1}{f(n)}$$ and $$\sum_{a \le n \le b}\frac{1}{g(n)}$$ both converge.
In our case, we have that only one converges. Thus, only one of the inequalities is true.
-
"positive" $\mapsto$ "$\ge 1$". Also, you are describing finite sums, so there is never any question of convergence (boundedness, sure). – Erick Wong Dec 29 '12 at 1:42
Nite approach +1 – Babak S. Dec 29 '12 at 7:16
Hint:
Try using the root test. ${}{}{}$ This will give you that the series converges.
You need only know that $\lim_{n\to \infty}\sqrt[\large n]{n} = 1.\quad$ Then | {
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You need only know that $\lim_{n\to \infty}\sqrt[\large n]{n} = 1.\quad$ Then
$$\lim_{n\to \infty} \sqrt[\large n]{\dfrac{1}{n2^n}} = \sqrt[\large n]{\frac{1}{n}}\cdot\sqrt[\large n]{\frac{1}{2^n}} = 1 \cdot \dfrac{1}{2} = \dfrac{1}{2} < 1 \implies \sum_{n=2}^{\infty} \frac{1}{n\cdot2^{n}}\quad \text{converges}.$$
Note also that you'd get the same result (that the series converges), by using the the ratio test. Both the root test and ratio test are very useful for establishing convergence and/or divergence of many series.
-
Since you mentioned induction:
Let $s_m = \sum_{n=2}^{m} \frac{1}{2^{n}*n}$. Then $s_m \leq 1-\frac{1}{m}$.
$P(2)$ is obvious, while $P(m) \Rightarrow P(m+1)$ reduces to
$$1-\frac{1}{m}+\frac{1}{2^{m+1}(m+1)} \leq 1- \frac{1}{m+1}$$
which is equivalent to:
$$m \leq 2^{m+1}$$
-
Or we can go this way $$\sum_{n=2}^{\infty} \frac{1}{n2^{n}}\le\sum_{n=2}^{\infty} \frac{1}{n^2}\le \sum_{n=2}^{\infty} \frac{1}{n(n-1)}=1$$ and the sum is precisely $\log 2-\frac{1}{2}$ (hint: use the Taylor expansion of $\ln(1-x), |x|<1)$
Chris.
-
Go back to the Cauchy test for convergence of a sequence.
Let $S_n = \sum_{k=2}^{n} \frac{1}{k2^k}$.
If $m < n$, $S_n-S_m = \sum_{k=m+1}^{n} \frac{1}{k2^k} < \sum_{k=m+1}^{n} \frac{1}{2^k} < \frac{1}{2^m}$ and this $\to 0$ as $m \to \infty$.
Therefore the sequence converges.
-
How do you know that the last inequality is true? – Jmaff Dec 29 '12 at 2:34
Is induction required on n and m to show that? – Jmaff Dec 29 '12 at 2:48
No, only the observation I make in my answer. – Pedro Tamaroff Dec 29 '12 at 5:02
sorry to get back to this late, but i mean the portion" $\sum_{k=m+1}^{n} \frac {1}{2^{k}} < \frac{1}{2^{m}}$" – Jmaff Dec 29 '12 at 21:23
You can show that $-\ln(1-x)=\sum_1^\infty {x^n \over n}$ for $-1\le x<1$. Hence the series in question converges to $\ln{2}$.
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# Math Help - fibonacci series - a slight variant
1. ## fibonacci series - a slight variant
Let me define a recurence like this
S(1) = 1
S(2) = 1
S(n) = S(n-1) + S(n-2) + a; where a is some number
Q1: What is the relation beween S(n) and F(n) [nth Fibonacci number]
I found out this is S(n) = F(n) + K(n)*a
Where K(n) is another series of the type S(n) where a = 1
Q2: Is there a way to find the generic term for S(n) ?
Any pointers will be of great help. Thanks
2. Well, you can always use the theory of difference equations (recurrence relations). You can find the nth term of the Fibonacci sequence this way, and I'm sure your S(n) sequence can be found the same way. The theory is very similar to that of differential equations.
3. Let $f(n)$ be a sequence that $f(n+2) = f(n+1) + f(n)$ , not necessarily be Fibonacci seq. , depends on the initial conditions .
Then we have $S(n) = f(n) - a$ because
$f(n+2) - a = f(n+1) + f(n) - a = (f(n+1) - a) + (f(n) - a ) + a$
$= S(n+1) + S(n) + a$
4. Originally Posted by aman_cc
Let me define a recurence like this
S(1) = 1
S(2) = 1
S(n) = S(n-1) + S(n-2) + a; where a is some number
Q1: What is the relation beween S(n) and F(n) [nth Fibonacci number]
I found out this is S(n) = F(n) + K(n)*a
Where K(n) is another series of the type S(n) where a = 1
Q2: Is there a way to find the generic term for S(n) ?
Any pointers will be of great help. Thanks
We have $S_3=2+a, \; S_4=3+2a$.
You can use induction to show $S_n=F_n+a\cdot (F_n-1)$.
Edit: Credit goes to Soroban's post below. I had made a mistake.
5. Hello, aman_cc!
Let me define a recurence like this:
. . $S(1) = 1$
. . $S(2) = 1$
. . $S(n) \:= \:S(n\!-\!1) + S(n\!-\!2) + a\:\text{ for some constant }a.$
What is the relation beween $S(n)$ and $F(n)$, the $n^{th}$ Fibonacci number?
Crank out the first few terms: | {
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Crank out the first few terms:
. $\begin{array}{|c||c|c|}n & S(n) & F(n)\\ \hline
1 & 1 & 1\\
2 & 1 & 1\\
3 & 2+a & 2 \\
4 & 3+2a & 3\\
5 & 5+4a & 5 \\
6 & 8 + 7a & 8 \\
7 & 13 + 12a & 13 \\
8 & 21 + 20a & 21 \\
\vdots & \vdots & \vdots
\end{array}$
And we see that: . $S(n) \;=\;F(n) + [F(n)-1]\cdot a$
6. Thanks for all the posts | {
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Association Mining - is buying Independent?
I have a problem. I can't not solve this exerciese. What is the best way to solve this exerciese? What are the approaches for this kind of exerciese?
The following table summarizes transactions in a supermarket where customers bought tomatoes and/or mozzarella cheese or neither.
Is buying mozzarella independent of buying tomatoes in the data given above? If they are not independent, explain whether they are positively or negatively correlated, i.e. does buying one of them increase or decrease the probability of buying the other?
As you can see I calculated the lift for lift(Moz => Tom) = 1,33 And I calculated lift(Moz => NoTom) = 0,5.
So I think they are not independent, and it is positively correlated
3 Answers
You are correct. They are not independent and they are positively correlated.
Let $$A$$ and $$B$$ (or $$X$$ and $$Y$$) be two events for stating general theorems and $$M$$ and $$T$$ be the events "customer purchases mozzarella" and "customer purchases tomatoes" in this specific example. We will use $$\wedge$$ to mean "and" so that $$M \wedge T$$ is the event "customer purchases mozzarella and tomatoes".
The simplest way to check independence is directly from the definition. $$A$$ and $$B$$ are independent if $$P(A \wedge B) = P(A)P(B)$$. From the data table we have $$P(M \wedge T) = 2000 / 5000 = 0,4$$ $$P(M) = 2500 / 5000 = 0,5$$ $$P(T) = 3000 / 5000 = 0,6$$ $$P(M)P(T) = 0,5 \times 0,6 = 0,3$$
Since 0.4 does not equal 0.3 we conclude that the events are not independent. | {
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Since 0.4 does not equal 0.3 we conclude that the events are not independent.
I have to say that I've never seen the term "lift" before. But its definition is perfectly reasonable and we could calculate independence in those terms. We can rearrange the definition of independence to give $$P(A \wedge B) = P(A)P(B)$$ $$\frac{P(A \wedge B)}{P(A)P(B)} = 1$$ $$\text{lift} = 1$$ to see that two events are independent if the associated lift is exactly 1. Your notes correctly show that lift can be rewritten as $$\frac{\frac{P(A \wedge B)}{P(A)}}{P(B)}$$
Here we have a lift of $$0,4 / (0,5 \times 0,6) = 1,33...$$ or $$0,8 / 0,6 = 1,33...$$ as you correctly calculated in your notes. Since the lift is not exactly 1 we conclude that the two events are not independent.
I would have approached the question of correlation via conditional probabilities, as follows. Two events, $$A$$ and $$B$$ are positively correlated if $$P(A | B) > P(A)$$. That is to say, two events $$A$$ and $$B$$ are correlated if the probability of $$A$$ given $$B$$ is greater than the unconditional probability of $$A$$. In other words, if we discover that event $$B$$ has occurred then the chances that event $$A$$ has occurred increase.
The conditional probability is defined as $$P(A | B) = \frac{P(A \wedge B)}{P(B)}$$ So in this example $$P(M | T) = P(M \wedge T) / P(T) = 0,4 / 0,6 = 0,66...$$ whereas $$P(M) = 0,5$$, so we conclude that $$M$$ and $$T$$ are positively correlated.
But hang on! Surely correlation is supposed to symmetric. So let's test whether $$P(T | M) > P(M)$$. We have $$P(T | M) = 0,4 / 0,5 = 0,8$$ and $$P(T) = 0,6$$ so, yes, they are correlated.
A lucky escape? We can be more smug than that. Let's rewrite our definition of correlation $$P(M | T) > P(M)$$ $$\frac{P(M | T)}{P(M)} > 1$$ $$\frac{\frac{P(M \wedge T)}{P(T)}}{P(M)} > 1$$ $$\frac{P(M \wedge T)}{P(T)P(M)} > 1$$
If we approach correlation the other way we get the same result | {
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If we approach correlation the other way we get the same result
$$P(T | M) > P(T)$$ $$\frac{P(T | M)}{P(T)} > 1$$ $$\frac{\frac{P(T \wedge M)}{P(M)}}{P(T)} > 1$$ $$\frac{P(T \wedge M)}{P(M)P(T)} > 1$$ which is the same as above because both the and operator and multiplication are commutative.
As I'm sure you've noticed, $$\frac{P(M \wedge T)}{P(T)P(M)} > 1$$ is simply asking whether lift is above 1. That's probably the approach that your instructor was expecting you to take.
A couple of additional points to fully close the loop with your notes. You correctly calculated lift(Moz => NoTom) = 0,5 but that information was not required to answer the question because you had already calculated lift(Moz => Tom). Support({X} -> {Y}) looks very much like a definition of $$P(X \wedge Y)$$ and Confidence({X} -> {Y}) looks like $$P(Y | X)$$.
I have two final notes to place this all in a broader context.
First, my definition of correlation fits well to your definition of correlation as "does buying one of them increase or decrease the probability of buying the other". But our definitions are not completely standard and the word "correlation" is usually associated with some specific correlation measure -- most often Pearson's correlation coefficient but also things like Kendall's rank correlation coefficient. Having said that, our definition is totally reasonable and I'm fairly confident that we could prove that, for example, Kendall's rank correlation coefficient is positive if and only if our definition of positive correlation above is positive. | {
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Second, inspired by the definition of lift in probabilistic terms in your notes, I've conflated historical frequencies given in the table with probabilities. This is standard practice in much of data science and certainly in exercises but it is not completely uncontroversial. Probabilities are about the future and you have data about the past. The extent to which you can infer future probabilities from past data is philosophically unresolved. But that is not something to worry about in this situation.
Market Basket Analysis is the way to go about it.
Market Basket Analysis is a technique which identifies the strength of association between pairs of products purchased together and identify patterns of co-occurrence. A co-occurrence is when two or more things take place together.
Market Basket Analysis creates If-Then scenario rules, for example, if item A is purchased then item B is likely to be purchased. The rules are probabilistic in nature or, in other words, they are derived from the frequencies of co-occurrence in the observations. Frequency is the proportion of baskets that contain the items of interest. The rules can be used in pricing strategies, product placement, and various types of cross-selling strategies.
Here is the calculation :
Support({Moz}--{Tom} = Transaction containing both moz and tom / total number of transaction
= 2000/ 5000 = 2/5 = 0.4
Confidence({Moz}--{Tom}) = Transaction containing both moz and tom / total number of transaction containing Moz
= 2000 / 2500 = 2/2.5 = 0.8
Lift({Moz}--{Tom}) = (Transaction containing both moz and tom / total number of transaction containing Moz ) / Fraction of transaction containing Y
= 0.8 / (3000/5000)
= 0.8/0.6 = 1.33
For your table this association mining rule is strong. But as the relation is not commutative please do same calculation from P(tom--> Moz) and if values of support, confidence and lift is smaller that does not indicate strong relationship | {
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Given you have two categorical variables and the associated contingency table, one option is to calculate the joint and marginal probabilities:
$$P = \frac{count}{total}$$
Probabilities Tomatoes No Tomatoes Row
Mozzarella 0.4 0.1 0.5
No Mozzarella 0.2 0.3 0.5
Column 0.6 0.4 0.1
The probabilities then can be used to answer questions about the data - If a person has bought tomatoes, is the person more likely to buy Mozzarella?
A person who has bought tomatoes is four times as likely to buy mozzarella compared to a person who has not bought tomatoes (i.e., 0.4 vs 0.1). There is a strong positive relationship between the two purchases.
There are many other possible statistical analyses that can be conducted. Examples include odds ratio, phi coefficient, tetrachoric correlation coefficient, and hypothesis testing with Chi-squared. | {
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# Is the inverse of a symmetric matrix also symmetric?
Let $A$ be a symmetric invertible matrix, $A^T=A$, $A^{-1}A = A A^{-1} = I$ Can it be shown that $A^{-1}$ is also symmetric?
I seem to remember a proof similar to this from my linear algebra class, but it has been a long time, and I can't find it in my text book.
You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more detailed and complete proof.
Given A is nonsingular and symmetric, show that $$A^{-1} = (A^{-1})^T$$.
Since $$A$$ is nonsingular, $$A^{-1}$$ exists. Since $$I = I^T$$ and $$AA^{-1} = I$$,
$$AA^{-1} = (AA^{-1})^T.$$
Since $$(AB)^T = B^TA^T$$,
$$AA^{-1} = (A^{-1})^TA^T.$$
Since $$AA^{-1} = A^{-1}A = I$$, we rearrange the left side to obtain
$$A^{-1}A = (A^{-1})^TA^T.$$
Since $$A$$ is symmetric, $$A = A^T$$, and we can substitute this into the right side to obtain
$$A^{-1}A = (A^{-1})^TA.$$
From here, we see that
$$A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$ $$A^{-1}I = (A^{-1})^TI$$ $$A^{-1} = (A^{-1})^T,$$
thus proving the claim.
In fact, $(A^T)^{-1}=(A^{-1})^T$. Indeed, $A^T(A^{-1})^T=(A^{-1}A)^T=I$.
Yes.
$$AB=BA=I\quad\Rightarrow\quad B^TA^T=A^TB^T=I\quad\Rightarrow\quad B^TA=AB^T=I$$
Another way to see that is to recall the formula $$A^{-1} = \frac{1}{\det(A)} \mathrm{Adj}(A)^T$$ and to note that the adjoint matrix of a symmetric matrix is by construction symmetric.
Yes. The inverse $$A^{-1}$$ of invertible symmetric matrix is also symmetric:
\begin{align} A & = A^T &&\text{(Assumption: A is symmetric)}\\ \\ A^{-1} & = (A^T)^{-1} &&\text{(A invertible \implies A^T = A invertible)}\\ \\ A^{-1} & = (A^{-1})^T &&\text{(Identity: (A^T)^{-1} = (A^{-1})^T)} \\ \\ {\large \therefore}\quad \rlap{\text{If A is symmetric and invertible, then A^{-1} is symmetric.}} \end{align} | {
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All the proofs here use algebraic manipulations. But I think it may be more illuminating to think of a symmetric matrix as representing an operator consisting of a rotation, an anisotropic scaling and a rotation back. This is provided by the Spectral theorem, which says that any symmetric matrix is diagonalizable by an orthogonal matrix. With this insight, it is easy to see that the inverse of the operator is a similar three-step sequence. Finally, we need to establish that any such three-step sequence produces a symmetric matrix: given any orthogonal matrix $$V$$ and diagonal matrix $$D$$, $$(V D V^T)^T = V D V^T$$. Hence the inverse of a symmetric matrix is also symmetric.
• Very nice intuitive answer. The reason this is getting fewer upvotes is that it’s posted much later than the others. That doesn’t stop this from being a good answer though. Oct 18, 2020 at 20:22
We have two properties to use: $A^{T}=A$ and $A^{-1}$exist, here we go!
$$A^{-1}A=I$$since $A^{-1}$exist $$(A^{-1}A)^{T}=A^{T}(A^{-1})^{T}=A(A^{-1})^{T}=I^{T}=I$$since $A^{T}=A$ $$A^{-1}A(A^{-1})^{T}=A^{-1}I$$left multiple by $A^{-1}$
Thus $$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$$
Inverse of $$A$$ can be expressed as a polynomial $$p(A)$$ of $$A$$ (from Cayley-Hamilton theorem).
So it is sufficient to prove that if $$A$$ is symmetric then power $$A^k$$ is symmetric, sum of symmetric matrices is symmetric and multiply by scalar is symmetric.
• I.e. for symmetric $A,B$ $(A^k)^T=A^k,(tA)^T=tA, (A+B)^T=A+B,$ Jun 15, 2019 at 10:56
$$(AA^{-1})^T=I^T=I=(A^{-1})^TA^T=(A^{-1})^TA=I\quad\Rightarrow\quad(A^{-1})^T=IA^{-1}\quad\Rightarrow\quad(A^{-1})^T=A^{-1}$$ By definition $$A=A^T$$ Used linear algebra equations $$(AB)^T=B^TA^T;AA^{-1}=I;I^T=I$$ | {
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# A manufacturer makes and sells 2 products, P and Q. The revenue from t
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A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink]
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A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is $20.00 and the revenue from the sale of each unit of Q is$17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?
A. $28.50 B.$27.00
C. $19.00 D.$18.50
E. $18.00 _________________ Md. Abdur Rakib Please Press +1 Kudos,If it helps Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges ##### Most Helpful Expert Reply CEO Joined: 11 Sep 2015 Posts: 3341 Location: Canada A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags Updated on: 05 Dec 2018, 16:06 1 Top Contributor 5 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50
B. $27.00 C.$19.00
D. $18.50 E.$18.00
Let's plug in some NICE VALUES that satisfy the given information...
Last year the manufacturer sold twice as many units of Q as P.
So, let's say the manufacturer sold 1 P and 2 Q's | {
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What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?
TOTAL revenue = 1($20.00) + 2($17.00)
= $20.00 +$34.00
= $54 There were 3 units sold. So, average price =$54/3 = $18 Answer: E RELATED VIDEO _________________ Test confidently with gmatprepnow.com Originally posted by GMATPrepNow on 17 Jun 2017, 07:31. Last edited by GMATPrepNow on 05 Dec 2018, 16:06, edited 1 time in total. ##### General Discussion Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4328 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 17 Jun 2017, 22:48 1 1 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50
B. $27.00 C.$19.00
D. $18.50 E.$18.00
Revenue last year from the sale of P is 20*1 = 20
Revenue last year from the sale of Q is 17*2 = 34
Total Revenue from the sale of P and Q last year is 54
Average Revenue from the sale of P and Q last year is $$\frac{54}{2+1} = 18$$
Hence, answer will be (E) 18
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Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink]
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02 Jul 2017, 20:49
Ratio of quantity of P and Q sold, P:Q = 1:2. | {
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### Show Tags
02 Jul 2017, 20:49
Ratio of quantity of P and Q sold, P:Q = 1:2.
Thus, average revenue per unit = (20*P + 17*Q)/(P+Q) = (20*1 + 17*2)/(2 + 1) = 54/3 = $18 Hence E answer Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4551 Location: United States (CA) Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 03 Aug 2018, 14:07 1 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50
B. $27.00 C.$19.00
D. $18.50 E.$18.00
We can compute a weighted average to solve. Let’s assume that 2 units of Q and 1 unit of P were produced last year. So the total revenue is 2 x 17 + 1 x 20 = $54, and thus the average revenue per unit is thus 54/3 =$18.
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Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t &nbs [#permalink] 03 Aug 2018, 14:07
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Which is the greatest possible natural number that divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?
Which is the greatest possible natural number that definitely divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?
This one is from my module, comes as a fill in the blanks with no answer. I have a feeling that there is something wrong with this question, since for $p=5$, one has $(p+3)(p-7)=-16$.
As mentioned, in one answer I tried substituting $p=5,7,9,11,13,17,19,...$ and spotted that the greatest number is $8$. I am just wondering .. is there any other way (probably using modulus) to directly find this number without actually substituting ?
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Why does this have a vote to close as "not constructive"? – Zev Chonoles Dec 3 '11 at 13:31
Playing around (substituting) is always a good idea. – André Nicolas Dec 3 '11 at 16:22
I agree with Zev Chonoles. It looks quite constructive to me and shows some thought about the problem. +1 from me. – Ross Millikan Dec 3 '11 at 16:24
I have a feeling that there is something wrong with this question
The "greater than $3$" condition is a bit strange, because the answer is unchanged if $p=3$ is allowed. And if they wanted to exclude negative values of $(p+3)(p-7)$, they should say "greater than 7". But the question as it stands is technically correct. It would perhaps be better like this:
Which is the greatest natural number that divides $(p+3)(p−7)$ for every odd prime number $p$?
If $p$ is odd, then $p = 2k+1$ for some $k$. So $(p+3)(p-7) = (2k+4)(2k-6) = 4(k+2)(k-3)$, which is divisible by $8$ since one of $k+2, k-3$ must be even. So $8$ divides $(p+3)(p−7)$ for every odd prime number $p$. | {
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That's half the question. The other half is to show that no natural number $n > 8$ divides $(p+3)(p−7)$ for every odd prime number $p$. Note that if $n$ has this property, then so does every factor of $n$. So it is enough to show that (i) $16$ doesn't have this property; and (ii) no odd prime $q$ has this property.
To show that $16$ doesn't have this property, just put $p=11$. So suppose now that $q$ is an odd prime. We need to produce another odd prime $p$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. Then we can be sure that $q$ doesn't divide $(p+3)(p-7)$. It seems a bit like overkill to me, but we can use Dirichlet's theorem on arithmetic progressions for this: for coprime $a$ and $q$, the arithmetic progression $a + rq$ $(r = 0, 1, 2,...)$ contains infinitely many primes. So we just have to choose $a$ with $1 \le a \le q-1$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. If $q=3$, we can choose $a=2$; and if $q \ne 3$, we can choose $a=3$.
And we're done.
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Put $p=5$. Anything $m$ that divides all the $(p+3)(p-7)$ must divide $16$. Put $p=11$ as you did. Thus $m$ must divide $56$. It follows that $m$ must divide $\gcd(16,56)$. We would need fancier machinery only if we wanted to show that if $\gcd$ is restricted to primes $>M$, it is still $8$. – André Nicolas Dec 3 '11 at 16:18
@AndréNicolas: Doesn't the paragraph starting "If $p$ is odd" prove that $8$ divides $(p+3)(p-7)$ for all odd $p$, prime or not? – Ross Millikan Dec 3 '11 at 16:22
@Ross Millikan: The answer above used Dirichlet's Theorem to show that no odd prime $q$ can divide all the $(p+3)(p-7)$. – André Nicolas Dec 3 '11 at 16:27
Well, at least I can say I was right about the overkill :-) André's comment $-$ and Bill Dubuque's answer $-$ provide a much simpler solution. – TonyK Dec 3 '11 at 17:27
There's nothing wrong with the question. | {
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There's nothing wrong with the question.
You're looking for a natural number. You know it divides $(p+3)(p-7)$. And you can put in some primes into that equation, as you have done already - you've got $-16$. What happens if you stick in $p=11$? Or $p=13$?
OK, now you've identified the natural number. Can you see why the product is always divisible by that number?
clue word: modulo
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HINT $\ \rm\:d\ |\ f_p = (p+3)\:(p-7)\ \Rightarrow\ d\ |\ f_5,\:f_7,f_{11}\: =\:\: -16,\:0,\:56\ \Rightarrow\ d\ |\ gcd(16,56) = 8\:.\:$ Conversely
$\$ mod $8$, since odd^2 $\equiv 1\:,\:$ we have $\rm\ f_p\equiv (p+3)(p+1)\: \equiv\: p^2+4\:p+3 \:\equiv\: 4\:(p+1) \:\equiv\: 0$
Generally for a sequence satisfying a monic $\rm\:n\:$th order linear recurrence with integer coefficients one easily proves by induction that the gcd of all terms is the gcd of the first $\rm\:n\:$ terms, since successive terms are integer linear combinations of the first $\rm\:n\:$ terms so they don't change the gcd.
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If $p$ is prime it can be written $p=4a+1$ or $p=4a+3.$
Substituting the first of these gives
$$(p+3)(p-7)=p^2-4p-21 = 16a^2+8a+1-16a-4-21 = 16a^2-8a-24 = 8(2a^2-a-3)$$
and the other gives $$(p+3)(p-7)=p^2-4p-21 = 16a^2+24a+9-16a-12-21 = 16a^2+8a-24 = 8(2a^2+a-3)$$
and so $(p+3)(p-7)$ is divisible by $8$ either way.
Edit
To show that 8 is the largest such divisor, we must show that $2a^2-a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\text{ and }5$ produces 25 and 42 respectively which have no prime factors in common.
And we must also show that $2a^2+a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\text{ and }5$ again produces 33 and 52 respectively which also have no prime factors in common.
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Strictly spoken, you have not proven yet that it 8 is the greatest number for which it holds. However, as MaX noted, trying the first few numbers shows that it is at most 8. – Sjoerd Dec 3 '11 at 15:44 | {
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Observe that $(5+3)(5-7) = -16$ and $(11+3)(11-7) = 56$; these have greatest common multiple 8. So the answer must be some factor of 8. If you keep trying more numbers you'll see it looks to be 8.
How to prove it? $p$ is odd. So $p+3$ and $p-7$ are even numbers which differ by 10. Let $p = 2n+7$; then we want to show that for any integer $n$, $2n(2n+10)$ is divisible by 8. But we can rewrite this as $4n(n+5)$, so we just need to show that $n(n+5)$ is always even; either $n$ is even or $n+5$ is.
A challenge for you, to help you understand what's going on here: prove in a similar way that for every integer $n$, $n^3-n$ is a multiple of 6 and $n^5-n$ is a multiple of 30.
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If p is odd number, p-7 is even =2k(say). Then (p+3)(p-7)=(2k+10)2k=8k(k+1)/2+16k, is clearly divisible by 8 for all integral value of k.
Now if f(k) = (2k+10)2k=4k(k+5), f(k+1)= 4(k+1)(k+6) (f(k),f(k+1))=(4k(k+5), 4(k+1)(k+6))= 4(k(k+5), (k+1)(k+6)) =4(k(k+5), (k+1)(k+6) - k(k+5)) as (a,b)=(a, b-a) =4(k(k+5), 2k+6) =8(k(k+5)/2, k+3) as k(k+5)/2 is integer for all integral value of k.
Now, 2 k(k+5)/2 - k(k+3)= 2k. We know ax+by=c is solvable iff (a,b)|c where a,b,c,x,y are integers. Here, x=2, a=k(k+5)/2, y=k, b=k(k+3), c=2k So, (k(k+5)/2, (k+3)) | 2k, =>2|(k+3) =>k is odd. For even k, 2∤(k(k+5)/2, (k+3)). In that case, no natural number>1, can divide f(k) for all integral value of k. So, 8 will be the greatest natural number that divides (p+3)(p-7) for all odd integral value of p.
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I would interpret the question as, "What is the greatest natural number that divides $(p+3)(p-7)$ for all primes $p>3$?" | {
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We can write $(p+3)(p-7)=4\cdot\frac{p+3}2 \cdot\frac{p-7}2$ but one of the latter two factors is even, since they differ by five, an odd number, so we can conclude that $(p+3)(p-7)$ is always divisible by eight, but $(5+3)(5-7)=-16$, $(9+3)(9-7)=24$, and $\gcd(-16,24)=8$. Therefore no natural number larger than eight divides $(p+3)(p-7)$ for all primes $p>3$, and eight is in fact the greatest natural number that divides $(p+3)(p-7)$ for all primes $p>3$.
-
since p is odd 2| (p+3)(p-7). so the answer should be \frac{1}{2} (p+3)(p-7)
-
You have not read the question carefully. We want the highest number that divides $(p+3)(p-7)$ for any odd $p$, not for a given odd $p$. – Ross Millikan Nov 6 '13 at 19:28
The question is sloppily phrased, because to me it reads more like "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$, where $p$ is some given prime greater than 3?", in which case the answer is trivially $(p+3)(p-7)$ !
But I imagine (as does everyone else by the look of it) the OP meant "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$ for every prime $p$ greater than 3?"
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# Concept check: Standard deviation
## Introduction
Unlike most questions on Khan Academy, some of these questions aren't graded by a computer. You'll learn the most if you try answering each question yourself before clicking "explain".
## The formula (for reference)
The formula for standard deviation (SD) is
$\Large\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
where sum means "sum of", x is a value in the data set, $\bar{x}$ is the mean of the data set, and n is the number of values in the data set.
## Part 1
Consider the simple data set left brace, 1, comma, 4, comma, start color redD, 7, end color redD, comma, 2, comma, 6, right brace.
How does the standard deviation change when start color redD, 7, end color redD is replaced with start color greenD, 12, end color greenD?
Please choose from one of the following options.
How can we see this in the formula for standard deviation?
### How does the standard deviation change?
The standard deviation increases because the data becomes more spread out:
### How can we see this in the formula?
We can see this in the formula
$\text{SD} = \sqrt{\dfrac{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{n}}$
because ${{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{}}$ is the sum of the squares of the distances from each data point to the mean. As the data, gets more spread out, the value of ${{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{}}$ increases.
### I'm curious, what are the actual standard deviations of the data sets?
The standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color redD, 7, end color redD, right brace is approximately 2, point, 28.
The standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color greenD, 12, end color greenD, right brace is approximately 3, point, 90.
## Part 2 | {
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## Part 2
Is it possible to create a data set with 4 data points that has a standard deviation of 0?
Please choose from one of the following options.
If it is possible, do it! Can you create two different data sets? How about three?
### Yes, it's possible!
In fact, there are an infinite number of possible data sets.
Here's one:
5, comma, 5, comma, 5, comma, 5
Here's another:
8, comma, 8, comma, 8, comma, 8
Any data set where all of the data points are the same has a standard deviation of 0 because the distance from each data point to the mean is 0.
### Show me the calculation for $\{ 5,5,5,5 \}$left brace, 5, comma, 5, comma, 5, comma, 5, right brace.
#### Step 1: Find the mean
$\bar{x} = \dfrac{5 + 5 + 5 + 5}{4} = \dfrac{20}{4} = \blueD5$
#### Step 2: Find the square of the distances from each of the data points to the mean
x$\lvert x - \bar{x} \rvert^2$
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
#### Step 3: Apply the formula
\begin{aligned} \text{SD} &= \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}} \\\\\\\\ &= \sqrt{\dfrac{0 + 0 + 0 + 0}{4}} \\\\\\\\ &= \sqrt{\dfrac{{0}}{4}}\\\\\\\\ &= \sqrt{{0}}\\\\\\\\ &= 0\end{aligned}
## Part 3
Can standard deviation be negative?
Please choose from one of the following options.
Why or why not? | {
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Why or why not?
### No, standard deviation cannot be negative!
To see why, think about the numerator and denominator inside the radical:
$\Large\text{SD} = \sqrt{\dfrac{\blueD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{\maroonD{n}}}$
Notice how start color maroonD, n, end color maroonD is always positive. It's the number of data points, and we can't have a negative number of data points.
Also notice that $\blueD{\sum\lvert x - \bar{x} \rvert^2}$ involves a quantity getting squared. Whenever we square something, we get a non-negative number.
Since both the denominator and numerator are positive, the entire expression must be positive too.
## Part 4
Standard deviation is a measure of spread of a data distribution.
What do you think deviation means?
In everyday language, deviation is how different something is from what might be considered normal.
In statistics, when discussing measures of spread, deviation is the amount by which a single measurement differs from the mean.
### Part 5
Here are the formulas for standard deviation (SD) and the formula for mean absolute deviation (MAD), both of which are measures of spread:
$\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
$\text{MAD} = {\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert}}}{n}}$
What are the similarities between the formulas? What are the differences?
### What are the similarities?
The formulas are very similar! They are both based on the distance from each data point to the mean $\lvert x - \bar{x} \rvert$, and they both include dividing by the number of data points n.
### What are the differences?
The difference between the two formulas is that when calculating standard deviation, we square the distance from each data point to the mean, and we take the square root as the last step of the formula.
### Which one is better?
Standard deviation is more complicated, but it has some nice properties that make it statisticians' preferred measure of spread.
### Part 6 | {
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### Part 6
Here's the formula that we've been using to calculate standard deviation:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
Here's the formula that statisticians actually use:
$\sqrt{\dfrac{\sum\limits_{}^{}{{( x-\bar{x})^2}}}{n}}$
Are the two formulas equivalent?
Please choose from one of the following options.
### What's the difference between the formulas?
In the formula that we've been using, we take the absolute value of $x - \bar{x}$:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\tealD{\lvert x-\bar{x}\rvert}^2}}}{n}}$
In the formula that statisticians use, they put parentheses around $x - \bar{x}$:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\purpleC{( x-\bar{x})^2}}}}{n}}$
### Are the formulas equivalent?
Yes, both formulas are equivalent!
Statisticians realize that squaring will make the distance positive, so they don't bother using absolute value signs and just use parentheses instead.
For example, let's evaluate $\tealD{\lvert x - \bar{x} \rvert^2}$ and $\purpleC{(x - \bar{x})^2}$ for x, equals, 2 and $\bar{x} = 5$:
$\tealD{\lvert x - \bar{x} \rvert^2} = \lvert 2 - 5 \rvert^2 = \lvert-3\rvert^2 = 3^2 = \greenD9$
$\purpleC{(x - \bar{x})^2} = (2 - 5)^2= (-3)^2 = \greenD9$
They're both positive! They're both start color greenD, 9, end color greenD! They're equivalent! | {
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mathematician Miloš Petrović. As the contest starts on Monday but at the very first day no one could answer correctly till the end of the week. | {
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