Mathematics desk
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January 21

Converting fahrenheit to celsius

Ok so say this object is 70 celsius... What do I need to do mathematically to find the answer in fahrenheit? Bond Extreme 04:31, 21 January 2007 (UTC)[reply]

what you guys gave never answered the question... the correct formula from Celcius to Fahrenheit is as follows.

$(9/5)*C+32$

Divide 9 by 5. Multiple by the tempature in Celcius then add 32... you got your answer

random walk

I'm trying to understand a random walk in one dimension - such that for n steps each step can either be +1 or -1, the relative probability of ending at distance a after n steps (starting at n) seems to be of the form n!/(n-2a)! (2a)! times a constant. Could someone point me in the direction of the article I need and confirm I'm approximately right so far. I'd also like to be able to work out the case when n becomes infinite. Thanks87.102.44.44 12:36, 21 January 2007 (UTC)[reply]

This is definitely the place to start. –King Bee ^{(T • C)} 14:05, 21 January 2007 (UTC)[reply]

I would have thought that too but there is no mention of the probability of having travelled a given distance in that article that I can find. Nor does Bernoulli process or markov chain help me.87.102.44.44 14:26, 21 January 2007 (UTC)[reply]

When you say random, I assume you mean that the probability of going to the right or to the left is equal; if that's the case, then this ought to have what you need. There is a table below with some sample probabilities, and you should be able to work out the details. –King Bee ^{(T • C)} 15:01, 21 January 2007 (UTC)[reply]

Since you need to travel a more steps to the right than the left to get to a position a, that means that

b-(n-b)=a

, where b is your total number of rightward steps. Then

b={\frac {a+n}{2}}

, and I get that your relative probability should be

{n \choose b}={\frac {n!}{\left({\frac {a+n}{2}}\right)!\left({\frac {n-a}{2}}\right)!}}

, noting that there is no probability of ending up an odd distance away after an even number of steps (or vice versa). (The normalizing factor, if you don't have it, is

2^{-n}

.) Let me get back to you (if I have any useful thoughts) about large n. --Tardis 15:04, 21 January 2007 (UTC)[reply]

Yes thanks - that was what I was thinking about - it's a bell shaped curve (obviously) - I was wondering if it was the gaussian or something else.. But I was unable to make any progress for when n tends to infinity - if anyone can give help with the infinite steps case (and step length tending to zero) that's exactly what I was looking for. Thanks.87.102.44.44 15:11, 21 January 2007 (UTC)[reply]

Suppose we normalise the process so that if we have taken n steps we make each step length 1/n, so that if we start from 0, we never go outside of the interval [-1,1]. The average distance away from the origin after n steps is of the order of

{\sqrt {n}}

steps. If each step is length 1/n then the average distance from the origin after n steps is of the order of

n^{-{\frac {1}{2}}}

. This tends to 0 as n tends to infinity. In other words, the shape of the distribution of final locations after n steps of length 1/n becomes an ever narrower and higher peak around 0 as n increases. Gandalf61 16:28, 21 January 2007 (UTC)[reply]

English word for property describing surface area to volume ratio

I'm looking for an acceptable English word (preferably one word) that describes an object's property in terms of surface area to volume ratio. Assuming the word was "surfaceness" (which I hope it isn't), a sphere would have low surfaceness and a fern leaf would have high surfaceness. Or it could be the other way around. The word can be a scientific term or a more commonly used word. Thank you in advance. ---Sluzzelin 16:19, 21 January 2007 (UTC)[reply]

(I'd guess you'd get an answer on the maths desk - so I'll copy your question over..87.102.44.44 16:34, 21 January 2007 (UTC))[reply]

Someone here will no doubt know this - original question at Wikipedia:Reference_desk/language# English word for property describing surface area to volume ratio 87.102.44.44 16:34, 21 January 2007 (UTC)[reply]

I don't believe there is a common word for this. The Wiki article uses sa/v, but if I was going to use the term several times in an article I'd use SAVR, defined at the start. If it was referred to in speech, pronouncing it as "saver" would be easy.86.132.162.27 18:20, 21 January 2007 (UTC)[reply]

Thanks, everyone. ---Sluzzelin 20:19, 21 January 2007 (UTC)[reply]

My transfer book calls it the specific surface, not sure how widespread that is though. Apparently there's a stub on it here.

137.99.174.5 01:11, 28 January 2007 (UTC)[reply]

"knobbliness" or "bumpiness" ? Anything knobbly or bumpy has a high sa/v compared to a smooth object of similar volume. Rupert Clayton 09:39, 30 January 2007 (UTC)[reply]

R^n is a hausdorff space.

hey, i have to know how to prove that R^n (how do you type that properly?) is a hausdorff space. certainly it seems intuitively true but no book seems to spell it out in the way thats being asked of me. any ideas would be much appreciated! (my exam's on tuesday afternoon UK time) thanks! —The preceding unsigned comment was added by 130.88.84.19 (talk) 16:42, 21 January 2007 (UTC).[reply]

R is a Hausdorff space, and the product of two Hausdorff spaces is also Hausdorff. Can you see where to take it from there ? Gandalf61 17:13, 21 January 2007 (UTC)[reply]

Alternatively, if you know metric spaces, look at the connection of metric spaces (what

R^{n}

of course is) and Hausdorff spaces. 80.130.139.206 18:50, 21 January 2007 (UTC)[reply]

ok, thanks guys- thats really helpful. we havent really done much on metric spaces but the "R is hausdorff argument" is brilliant -thanks!.

So you learn general topology first and then specialize into metric spaces. R^n with the topology consisting of the empty set and the whole space is NOT Hausdorff. You should mention the USUAL topology. Twma 01:23, 22 January 2007 (UTC)[reply]

divisibility by three -proof

hey, on a lighter note from the topology nightmare of my last post, i was told (by a french teacher no less) that a number is divisible by three if its component sumbers add up to a multiple of three. certainly this seems to work (123,2001,193475939283 etc) but does anyone know how you would prove it? (is there even notation for describing what i mean?) also, does anyone know if its a "if and only if statement" or just "one way" so to speak. thanks! —The preceding unsigned comment was added by 130.88.84.19 (talk) 16:49, 21 January 2007 (UTC).[reply]

See our article on divisibility rules, which includes this rule (yes, it is an "if and only if" rule) and gives an explanation if you go far enough down the article. You need to know a bit about modular arithmetic to understand why it works - basically, because any power of 10 leaves a remainder of 1 when divided by 3, the remainder that is left after you divide any number n by 3 is the same as the remainder that is left after you divide the sum of its digits by 3. If this remainder is 0, then n is a multiple of 3. A similar rule can be used to test for divisibility by 9. Gandalf61 17:23, 21 January 2007 (UTC)[reply]

thanks again gandolf.i do indeed know about modular arithmetic so the article ,made perfect sense! thanks!

I have 2 points of an exponential curve - how do I find the rest of the line

I have 2 points of an exponential curve - (0, 27) and (1, 24). How do I find more points and the rest of the line? Basically, I need to put this on graph paper. Thanks a lot! NIRVANA2764 17:49, 21 January 2007 (UTC)[reply]

I'm not sure this is enough information to determine the line. Taking the most general case, with, say, f(x)=A*e^(cx+d) + B, that's not enough points. It would help to know any of these constants - e.g. what value does the graph tend to as x tends to infinity - but I think you need more points. Rawling4851 18:17, 21 January 2007 (UTC)[reply]

If you want a line of the form y=Ae^x, then you have enough points. Just take the logarithm of your y values, work out the line of the form y'=mx+c that passes through both points. Then exponentiate each side to get y=e^mx+c and simplify it to convert it into the form y=Ae^x RupertMillard (Talk) 18:26, 21 January 2007 (UTC)[reply]

y=Ae^x has only one unknown, so it's y=Ae^bx which should be found. Note that the value of A will come immediately from your first point where x=0, so only one equation need be solved to get the value of the other unknown b.86.132.162.27 18:41, 21 January 2007 (UTC)[reply]

???

7⁵⁴

7⁵⁴ = 4.318114567×10⁴⁵. Next time consider using a calculator (it might be helpful). NIRVANA2764 18:57, 21 January 2007 (UTC)[reply]

4318114567396436564035293097707728087552248849 to be exact. — Kieff 21:27, 21 January 2007 (UTC)[reply]

What internet site, calculator program, or computer program did you use for that? X [Mac Davis] (DESK|How's my driving?) 23:28, 22 January 2007 (UTC)[reply]

I just wrote "echo bcpow("7","54");" on a php file I was working on at the time. — Kieff | Talk 15:50, 23 January 2007 (UTC)[reply]

I just bring up "bc" in an xterm and type "7^54". --Carnildo 10:15, 23 January 2007 (UTC)[reply]

or Google :) - grubber 16:34, 30 January 2007 (UTC)[reply]

Wikipedia:Reference desk/Archives/Mathematics/2007 January 21

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