Guest
Displaying Posts 1 - 9 of 9|
#1 Posted 10:50am 07-03-10 Own prob - Minimum AP SumDenote by S[ss]k[/ss] the sum to k terms of an AP. If for some two indices m and n it is known that S[ss]m[/ss]=S[ss]n[/ss], for what index i, does S[ss]i[/ss] attain a real extremum. |
|
#2 Posted 2:47pm 08-03-10 Re: Own prob - Minimum AP SumAssuming m≠n, the extremum is attained for the index [frac]m+n[/]2[/frac] if that's an integer which will be when both are either even or odd. If one of them is even and one odd then the extremum is taken twice at [frac]m+n[/]2[/frac] &ndash [frac]1[/]2[/frac] and [frac]m+n[/]2[/frac]+[frac]1[/]2[/frac] |
|
#3 Posted 6:42pm 08-03-10 Re: Own prob - Minimum AP SumRight! Solution please. |
|
#4 Posted 9:06pm 08-03-10 Re: Own prob - Minimum AP SumThe solution just involves quadratics. Since for the indices m and n, we have S[ss]m[/ss] = S[ss]n[/ss] so [im]http://codecogs.izyba.com/gif.latex?%5Cdfrac%7Bm%7D%7B2%7D%5Cleft%282a+%28m-1%29d%5Cright%29%3D%5Cdfrac%7Bn%7D%7B2%7D%5Cleft%282a+%28n-1%29d%5Cright%29[/im] which gives [im]http://codecogs.izyba.com/gif.latex?%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7Ba%7D%7Bd%7D%3D%5Cdfrac%7Bm+n%7D%7B2%7D[/im] ------ (1) assuming m≠n. Next, [im]http://codecogs.izyba.com/gif.latex?S_k%20%3D%20%5Cdfrac%7Bk%7D%7B2%7D%5Cleft%282a+%28k-1%29d%5Cright%29[/im] which can finally be written as [im]http://codecogs.izyba.com/gif.latex?S_k%20%3D%20%5Cdfrac%7Bd%7D%7B2%7D%5Cleft%28k-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7Ba%7D%7Bd%7D%5Cright%29%5Cright%29%5E2%20-%5Cdfrac%7Bd%7D%7B2%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7Ba%7D%7Bd%7D%5Cright%29%5E2[/im] which gives, using (1), [im]http://codecogs.izyba.com/gif.latex?S_k%20%3D%20%5Cdfrac%7Bd%7D%7B2%7D%5Cleft%28k-%5Cdfrac%7Bm+n%7D%7B2%7D%5Cright%29%5Cright%29%5E2%20-%5Cdfrac%7Bd%7D%7B2%7D%5Cleft%28%5Cdfrac%7Bm+n%7D%7B2%7D%5Cright%29%5E2[/im] From where its not difficult to get the conclusion I already stated.
Edited on 00:05am 10-03-10 |
|
#5 Posted 09:25am 09-03-10 Re: Own prob - Minimum AP SumThe motivation for the problem is the fact that the expression for the sum to n terms of an AP is a quadratic in n. We know that for a quadratic f(x), points y and z such that f(y) = f(z), y≠z, lie symmetrically about the extremum point x[ss]o[/ss] and so [im]http://codecogs.izyba.com/gif.latex?x_0%20%3D%20%5Cfrac%7By+z%7D%7B2%7D[/im] Hence if m and n are of the same parity, [im]http://codecogs.izyba.com/gif.latex?S_%7B%5Cfrac%7Bm+n%7D%7B2%7D%7D[/im] is our real extremum. Otherwise, [im]http://codecogs.izyba.com/gif.latex?S_%7B%5Cfrac%7Bm+n-1%7D%7B2%7D%7D%20%3D%20S_%7B%5Cfrac%7Bm+n+1%7D%7B2%7D%7D[/im] will be the real extremum. Cool, na?
Edited on 09:26am 09-03-10 |
|
#6 Posted 8:16pm 09-03-10 Re: Own prob - Minimum AP SumI couldnt stop posting a solution to this one.. The thing is summations works very much like integration.. The sum of a few terms is zero in between, which directly means that the area of the curve is zero... (I know this is not absolutely correct.. but am trying to give a perspective of my own) SO what it says is that plot of the terms of the AP will be a straight line cutting the x axis exactly at the mid point of m and n... So this would mean that the area will be maximum or minimum at the point where it cuts the x axis... As a "unanswered point" , which i expect answers from users other than kaymant sir, bipin and ofcourse prophet sir is that If that is the case, then why is Kaymant sir's answer not simply (m+n)/2 ??
-when spring comes, it melts the snow one flake at a time… |
|
#7 Posted 8:57pm 09-03-10 Re: Own prob - Minimum AP Sumbecause its an index and so it has to be a natural number. So when m and n are not integers [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7Bm+n+1%7D%7B2%7D[/im] and [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7Bm+n-1%7D%7B2%7D[/im] are the integers that are closest to m+n/2 and since they lie symmetrically on either side of this point the sums at these 2 points will be equal. There are quite a few AP summation problems in the textbooks that are very simply solved by directly assuming a quadratic of the form [im]http://codecogs.izyba.com/gif.latex?an%5E2+bn[/im] |
|
#8 Posted 9:29pm 09-03-10 Re: Own prob - Minimum AP SumKAYMANT SIR: ""[b]The solution just involves quadratics. Since for the indices m and n, we have Sm = Sn so which gives ------ (1) assuming m≠n. Next, which can finally be written as which gives, using (1), From where its not difficult to get the conclusion I already stated.[/b]"" the equations cannot be seen...[2][2][2][2][2]
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
|
#9 Posted 00:04am 10-03-10 Re: Own prob - Minimum AP SumIs it? But I can see it perfectly. Anyway, here are the equations without the latex. The solution just involves quadratics. Since for the indices m and n, we have S[ss]m[/ss] = S[ss]n[/ss] so [frac]m[/]2[/frac](2a+(m-1) d) = [frac]n[/]2[/frac] (2a + (n-1)d) which gives [frac]1[/]2[/frac] &ndash [frac]a[/]d[/frac] = [frac]m+n[/]2[/frac] ----- (1) assuming m≠n. Next, S[ss]k[/ss] = [frac]k[/]2[/frac] (2a +(k-1)d) which can finally be written as S[ss]k[/ss] = [frac]d[/]2[/frac] (k &ndash (1/2 &ndash a/d))[p]2[/p] &ndash [frac]d[/]2[/frac] (1/2 &ndash a/d)[p]2[/p] which gives, using (1), S[ss]k[/ss] = [frac]d[/]2[/frac] ( k &ndash [frac]m+n[/]2[/frac])[p]2[/p] &ndash [frac]d[/]2[/frac] ([frac]m+n[/]2[/frac])[p]2[/p] From where its not difficult to get the conclusion I already stated. |
Displaying Posts 1 - 9 of 9
