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

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August 22

Simplify a sum involving binomial coefficients

Perhaps the sum $\sum _{i=n}^{m}{\frac {\binom {i}{n}}{2^{i}}}$ (where 0≤n≤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

$2-{\binom {m+1}{n}}2^{-1-m}{}_{2}F_{1}(1,m+2;m-n+2;1/2).$

There are various identities one can try to play with. This is also equal to

$2-{\binom {m+1}{n}}2^{-m}{}_{2}F_{1}(1,-n;m-n+2;-1).$

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 {{{1}}}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⁻¹ (according to the muzzle velocity article) and gravity acts to decelerate it at 9.81 ms⁻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 $v_{0}=gt$, then plug that time into $h=v_{0}t-1/2gt^{2}$. 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? ΣΑΠΦ (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. ΣΑΠΦ (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. ΣΑΠΦ (Sapph)^Talk 11:26, 23 August 2011 (UTC)

Using Iverson brackets: f(x)=[5≤x]+[14≤x]+[24≤x]+[33≤x]+[42≤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. ΣΑΠΦ (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)