Wikipedia:Reference desk/Archives/Mathematics/2011 August 22

= August 22 =

Simplify a sum involving binomial coefficients
Perhaps the sum $$\sum_{i=n}^{m} \frac{\binom{i}{n}}{2^{i}}$$ (where 0&le;n&le;m) can be simplified. Please tell me! The result would be very useful. (See here). Bo Jacoby (talk) 09:05, 22 August 2011 (UTC).


 * It's a special value of the hypergeometric function. I get
 * $$2-\binom{m+1}{n}2^{-1-m} {}_2F_1(1, m + 2; m -n + 2; 1/2).$$
 * There are various identities one can try to play with. This is also equal to
 * $$2-\binom{m+1}{n}2^{-m} {}_2F_1(1, -n; m - n +2; -1).$$
 * I can't find an easy closed form, though. Sławomir Biały  (talk) 14:18, 22 August 2011 (UTC)


 * Thank you very much Sławomir ! This will eventually help me speeding up my program for large values of n and m. Bo Jacoby (talk) 08:16, 23 August 2011 (UTC).


 * I can see it's very similar to the Binomial Theorem (bottom formula) with ${{math|x{=}}2^{-1}}}$, except that the binomial coefficient is upside-down. That might be a dead-end, but one could consider wandering through the proofs? SamuelRiv (talk) 21:08, 25 August 2011 (UTC)

Trajectories, gravity, acceleration, straightforward calculus
It's been too long since school when I last did any calculus. I was reading this article about the dangers of firing guns into the air earlier and it quotes a ballistics expert as saying bullets "go a long way up when they're fired". The question is, how high? If a bullet is accelerated by its firing to a speed of somewhere between 120 to 1700 ms&minus;1 (according to the muzzle velocity article) and gravity acts to decelerate it at 9.81 ms&minus;2, at what height does it stop going up? I believe this is a straightforward calculus problem, but I can't remember how to frame it! I can picture a graph of height against time as an inverse parabola intersecting the origin and a point on the x-axis corresponding to the time when it hits the ground (neglecting air resistance, because the bullet will reach terminal velocity on its way back down), and I know that I need the y value at the point of inflection, where the derivative (speed) is zero, but this is about as far as I get. Beorhtwulf (talk) 18:54, 22 August 2011 (UTC)
 * If I recall correctly, tehn we can think about it like this. If it starts at 120 to 1700m/s, and comes to instantaneous rest at the top, its average speed will be 60 to 850m/s. The time it spends doing this is the time it spends decelerating, or 120 to 1700 divided by 9.81 (assuming g is constant in both cases) which is between 12 and 173. So the total height reached is one value times the other, or 720 to 147,000m, which is quite a range if you did mean 120 to 1700m/s, I have no idea of muzzle velocity myself. Clearly this overly simplistic - 147km up is way more than the atmosphere, so clearly there's a problem modelling it like that. Grandiose (me, talk, contribs) 20:49, 22 August 2011 (UTC)

Basic approach: figure out the time at which the vertical velocity reaches zero using $$v_0 = g t$$, then plug that time into $$h = v_0 t - 1/2 g t^2$$. Note that for a muzzle velocity of 1700 m/s (= Mach 5), air resistance will be huge and the formula will be miles off the mark. Looie496 (talk) 21:33, 22 August 2011 (UTC)


 * Energy conservation tells you that hg=v2/2. Bo Jacoby (talk) 08:12, 23 August 2011 (UTC).
 * That would agree with my figures (thankfully) - however absurd they are. Grandiose (me, talk, contribs) 13:30, 23 August 2011 (UTC)


 * Unfortunately, I don't think you can neglect air resistance at any point in the bullet's motion - I think drag will be a significant force on the bullet at all points in its trajectory. There is some discussion of this in our external ballistics article; the bottom line seems to be that mathematical modelling of drag on bullets is complicated. Gandalf61 (talk) 13:47, 23 August 2011 (UTC)

Getting a Formula Given Points: Step-Like Function
I've got a series of points: for x of 1-4, y is 0. For x of 5-13 (9 points), y is 1. For x of 14-23 (10 points), y is 2. For x of 24-32 (9 points), y is 3. For x of 33-41 (9 points), y is 4. For x of 42-50 (9 points), y is 5. All further steps are 9 points wide. How would I go about making a formula that matches these points? Σ Α Π Φ (Sapph) Talk 20:38, 22 August 2011 (UTC)
 * Could you clarify where this problem comes from? If it is a homework problem we won't do it, although we may be able to offer suggestions for how to approach it. Looie496 (talk) 21:25, 22 August 2011 (UTC)
 * It comes from a diet program. Carbs are worth points, in the amounts listed above. Instead of using a lookup table, I want to have an automated widget on my desktop/homepage that can give me results.  And programming the lookup table directly is . . . inelegant.  Σ Α Π Φ (Sapph) Talk 11:26, 23 August 2011 (UTC)


 * See lagrange interpolation and trigonometric interpolation. Bo Jacoby (talk) 08:07, 23 August 2011 (UTC).
 * A "formula" is not the right tool for describing such a function (though you could get a decent fit with enough trigonometric terms). What's wrong with using a piecewise definition? -- Meni Rosenfeld (talk) 08:52, 23 August 2011 (UTC)
 * I should have been more specific - I'm not necessarily looking for a formula like y=5x. I'm looking for something I might plug into a spreadsheet or webpage or program to automate the calculation. So something like round(x/5+int(x/10)) or whatever would be just fine.   Σ Α Π Φ (Sapph) Talk 11:26, 23 August 2011 (UTC)
 * Using Iverson brackets: f(x)=[5&le;x]+[14&le;x]+[24&le;x]+[33&le;x]+[42&le;x]. Bo Jacoby (talk) 13:20, 23 August 2011 (UTC).


 * The points you've described are almost described by simply rounding x/9 to the nearest integer, except that x-values of 23 and above should have 1 subtracted from them first, before the division (because of that one interval of width 10). The formula round(x/9) will give you the right value most of the time. To accommodate that one strange interval, either put in a line of code beforehand that subtracts 1 from x if x is greater than or equal to 23, or try something like round((x-(x>=23))/9), which will work if the >= operator yields 1 for true and 0 for false. In Excel the formula =ROUND((x-(x>=23))/9,0) will work. —Bkell (talk) 15:14, 23 August 2011 (UTC)
 * Bkell, that worked great. As it turned out, I found another source that stated the ratio as 19/175 - which is pretty close to 1/9, but accounts for the discrepancy at 23.  Thanks again.  Σ Α Π Φ (Sapph) Talk 22:38, 23 August 2011 (UTC)
 * Ah, I see. So if the ratio is really 19/175, then it's not quite true that "all further steps are 9 points wide"—there is another width-10 interval somewhere down the road, and another one after that eventually, and so on. —Bkell (talk) 22:47, 23 August 2011 (UTC)

Derivation of Poiseuille's Law
Hello. I read the derivation of Poiseuille's law here. I would like to know the steps immediately prior to rearranging $$ \frac{1}{\eta} \frac{\Delta P}{\Delta x} = \frac{d^2 v}{dr^2} + \frac{1}{r} \frac{dv}{dr} $$ by using the chain rule to $$ \frac{1}{\eta} \frac{\Delta P}{\Delta x} = \frac{1}{r} \frac{d}{dr} r \frac{dv} {dr}$$. Links to the calculus techniques involved would be greatly appreciated. Thanks in advance. --Mayfare (talk) 23:55, 22 August 2011 (UTC)
 * From the product rule we have
 * $$\frac{1}{r} \frac{d}{dr} \left [ r \frac{dv} {dr} \right ] = \frac{1}{r} \left [ r \left ( \frac{dv}{dr} \right )' + (r)' \frac{dv}{dr} \right ] = \frac{1}{r} \left [ r \frac{d^2 v}{dr^2} + \frac{dv}{dr} \right ] = \frac{d^2 v}{dr^2} + \frac{1}{r} \frac{dv}{dr}.$$
 * Not sure where the chain rule comes in. I'm also assuming the position of the square brackets on the left-hand side; the notation in the article is a little ambiguous in my eyes. — Anonymous Dissident  Talk 07:46, 23 August 2011 (UTC)

How does the product rule come to mind without knowing the next step? --Mayfare (talk) 19:27, 23 August 2011 (UTC)
 * Because I noticed we're considering the derivative of a product of two functions f and g, where f(r) = r and g(r) = dv/dr. — Anonymous Dissident  Talk 21:33, 23 August 2011 (UTC)