Mathematics desk
< September 23	<< Aug \| September \| Oct >>	September 25 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

September 24

My mathematical ability.

I have a mathematical ability and wonder if there is an actual name for it? Putting a number to every letter of the alphabet e.g. A is 26 to Z is one i can give the sum of a word faster than than people can work it out on paper or on a computer. My average for a ten letter word would be under five seconds and people have been quite shocked when i have demonstrated this ability.So, can you tell me please is there a name i can put to this? Thanks, Harry Remfrey.

I doubt there is a name for it. It is just mental arithmetic. You have most likely memorised (perhaps intentionally, perhaps not) the numbers for each letter (oddly, you have them backwards from the way most people do such associations - A is usually 1 and Z is 26). Being good at mental arithmetic usually involves learning lots of tricks to make calculations quicker (among the most common are things like adding one to a number and then minusing the appropriate amount at the end to compensate, eg. 5x9=5x10-5=45, but there are more complicated tricks). You may also have memorised (again, intentionally or otherwise) the totals for certain combinations - 10 letter words are usually built up from smaller words with prefixes and suffixes, so if you already know what "ing" adds up to, you can speed up a lot of long words. --Tango (talk) 01:14, 24 September 2009 (UTC)[reply]

Furthermore being able to associate two different symbols by rote memorization is no indication of mathematical ability; although you seem to be able to calculate long sums quickly, it is not the case that you could match a computer program in terms of raw speed. On top of this, being able to calculate sums and work with integers is obviously beneficial, but being a mathematician is about problem solving, hard work and having an artistic streak too. Do you know how to prove things? That is real math. 94.171.225.236 (talk) 12:16, 24 September 2009 (UTC)[reply]

I'm not suggesting that you suffer from this, but you might be interested in our article on Savant syndrome. Many mathematicians (such as Carol Vorderman?) and other "normal" people such as magicians have taught themselves similar tricks. Dbfirs 13:42, 24 September 2009 (UTC)[reply]

My average arithmetic speed is 50% that of a normal person, and I always make mistakes. Does this mean I'm suffering from something? And yes I agree with 94.171.225.236 proving things is a totally different matter from arithmetics Breath of the Dying (talk) 16:35, 24 September 2009 (UTC)[reply]

You probably just haven't practised enough and learnt all the tricks. You may also be misjudging other people's abilities. --Tango (talk) 17:35, 24 September 2009 (UTC)[reply]

Moment Generating Function

Alright, now I need to show the moment generating function for that same function as my last post,

f(x)={\frac {1}{{\sqrt {2\pi }}x}}e^{-(\log x)^{2}/2},\qquad 0\leq x<\infty ,

does not exist. I think I have done it but I just want to make sure this makes sense. I don't have any of my undergrad analysis books here.

I calculated the derivative of $e^{tx}f(x)$ . It is positive for x such that $tx-1-\log x$ is positive. Depending on t, I know this will eventually be positive for some x and also every x bigger than that x since x grows faster than $\log x$ . So, the derivative is eventually positive which means the function is eventually increasing. And, the function itself is always positive. So, it can not tend toward 0 so the integral must be divergent. Does this make sense? Thanks. StatisticsMan (talk) 02:09, 24 September 2009 (UTC)[reply]

Yes. (Detail: to be precise, I wouldn't just say it can not tend toward 0, but rather it is bounded away from 0 for large x, that also follows from your argument. Because a function may be integrable over R even if it doesn't tend to 0 at infinity). You may also check that the moments

m_{r}

grow too fast:

m_{r}/r!

are not the coefficients of a convergent power series. To this end you may use the Stirling formula for

r!

, but in fact it is sufficient to know

m_{r}=\exp(r^{2}/2)

and

r!\leq r^{r}

so

(m_{r}/r!)^{1/r}\geq r^{-1}\exp(r/2)

, that diverges as r goes to infinity, so the Cauchy test gives a null radius of convergence. --pma (talk) 07:18, 24 September 2009 (UTC)[reply]

Easy question

Hi I have a easy question that I need someone to confirm I'm right. Let X be a set, and F a collection of subsets of X satisfying the finite intersection property. By Hausdorff's maximality principle, we can show there exists a maximal collection E of subsets of X satisfying the f.i.p and is a superset of F. Now take the intersection of all elements of E, does this intersection contain at most one element? I say yes because if it's not empty and contains x then {x} is in the collection E since E is maximal. Am I right? This stuff is really confusing Breath of the Dying (talk) 12:29, 24 September 2009 (UTC)[reply]

You are not asking if an ultrafilter E (containing the filter generated by F, which is not relevant for the question) is necessarily a principal ultrafilter, that is, the collection of all supersets of a singleton {x}. The answer to the question you didn't ask is no (otherwise the theory of filters would be totally trivial): just note that if the intersection of all members of F is empty, the same is true a fortiori for the intersection of all members of E. So take for instance F:=all co-finite subsets of N.--pma (talk) 15:40, 24 September 2009 (UTC)[reply]

Actually, the OP is asking whether the intersection of an ultrafilter contains at most one point, not at least. The answer is yes, and the given reasoning is correct. — Emil J. 15:59, 24 September 2009 (UTC)[reply]

ops, sorry I misread the question. Thanks Emil --pma (talk) 17:21, 24 September 2009 (UTC)[reply]

Funniest fix I've seen in a while :-) -- Meni Rosenfeld (talk) 20:05, 24 September 2009 (UTC)[reply]

you should see my Beetle ;-) --pma (talk) 21:32, 24 September 2009 (UTC)[reply]

Solving systems of equations in three variables

Resolved

By my teacher

7x+5y+ z=  0
-x+3y+2z= 16
 x-6y- z=-18

I got this correct

    z=4
7x+5y+4=0
 7(-x-3y+2(4))=16

  7x+5y+4=0
 -7x-21y+56=112

 -16y+60=112

   -16y=52

Y needs to be 2 (check by using calculator.) Accdude92 (talk) (sign) 13:38, 24 September 2009 (UTC)[reply]

OK. Do you have a question ? Rkr1991 ^{(Wanna chat?)} 14:04, 24 September 2009 (UTC)[reply]

As I recall sometime solutions of three equations with three variables don't exist, this looks like it might be a case of that.83.100.251.196 (talk) 15:53, 24 September 2009 (UTC)[reply]

Try solving for z=4 into the 3rd equation - see what happens.83.100.251.196 (talk) 15:55, 24 September 2009 (UTC)[reply]

The OP has indeed correctly solved the set of equations... I don't understand what more there is to say... Rkr1991 ^{(Wanna chat?)} 18:10, 24 September 2009 (UTC)[reply]

The OP wanted to know what was the mistake in the given calculation. Which is in the line

7(-x-3y+2(4))=7\cdot 16

, where the sign of

3y

was reversed. -- Meni Rosenfeld (talk) 19:58, 24 September 2009 (UTC)[reply]

I get x = − 2, y = 2 and z = 4. ~~ Dr Dec (Talk) ~~ 19:27, 24 September 2009 (UTC)[reply]

Complex number

OK so this is a homework question but I'm only asking to have my work checked.

"If a complex number z is defined by $z=z_{1}+z_{2}$ and $|z_{1}|=1,|z_{2}|=2$ then sketch the subset of the Argand diagram represented by z."

So, $z_{1}$ must be a circle of radius one centred at the origin and $z_{2}$ must be a circle of radius 2 centred at the origin. So does this mean that z is a circle of radius 3 centred at the origin? Thanks 92.4.52.65 (talk) 18:54, 24 September 2009 (UTC)[reply]

No it does not: for you only have an inequality :

|z|=|z_{1}+z_{2}|\leq |z_{1}|+|z_{2}|=1+2=3

. Actually you can obtain all z in the closed disk of radius 3, not just the circle. --84.221.209.230 (talk) 19:03, 24 September 2009 (UTC)[reply]

As this is homework, I'll say only this: 84's answer isn't right either, but the general logic is sound. --Tardis (talk) 19:21, 24 September 2009 (UTC)[reply]

Indeed: 84.221.209.230's solution isn't quite right. Here are a few tips:

We know that |z₁| = 1 and so z₁ = e^iθ for some 0 ≤ θ < 2π.
We know that |z₂| = 2 and so z₂ = 2e^iφ for some 0 ≤ φ < 2π.
SInce z₁ + z₂ = e^iθ + 2e^iφ, what can we say about |z₁ + z₂|?

You'll find that |z₁ + z₂| can only get so small. For example, can z₁ + z₂ = 0? ~~ Dr Dec (Talk) ~~ 19:45, 24 September 2009 (UTC)[reply]

I think there's an even easier solution. I'm pretty sure that the following inequality always holds in a normed vector space.
Please, correct me if I'm wrong (add any missing assumptions, etc)

||z_{1}|-|z_{2}||\leq |z_{1}+z_{2}|\leq |z_{1}|+|z_{2}|\ .

This shows why 84's solution was not quite right: he only had the right hand part of the inequality. ~~ Dr Dec (Talk) ~~ 20:54, 24 September 2009 (UTC)[reply]

It does always hold. The right inequality, the triangle inequality of course, is part of the definition of a normed linear space. The left inequality comes from using the triangle inequality twice. Here is once

|z_{1}|=|z_{1}+z_{2}-z_{2}|\leq |z_{1}+z_{2}|+|z_{2}|\Rightarrow |z_{1}|-|z_{2}|\leq |z_{1}+z_{2}|

Just do it again with

z_{1},z_{2}

switched and combine the two to get the left inequality. StatisticsMan (talk) 21:08, 24 September 2009 (UTC)[reply]

Good. OP: and don't forget the other inclusion (if z has 1 ≤ |z| ≤ 3, it can be written as z = z₁ + z₂). --pma (talk) 21:23, 24 September 2009 (UTC)[reply]

So is the defined area the area between a circle of radius one and a circle of radius 3? 92.4.52.65 (talk) 17:13, 25 September 2009 (UTC)[reply]

Yes. However above there's only the proof of one inclusion. --pma (talk) 23:30, 25 September 2009 (UTC)[reply]