Sum of Powers of 2Date: 08/28/2001 at 01:09:39 From: Amir Milbes Subject: Sums of powers Hi! I read and understood the method on how to derive a general formula for the sum of consecutive integers. (<> 1 + 2 + 3 + ... + n = S(n) = [n(n+1)]/2 Basically I did the following : 1 + 2 + 3 + ...+ n = S(n) n + (n-1) + (n-2) + ... + 3 + 2 + 1 = S(n) -------------------------------------------- (n+1)+(n+1)+ ... + (n+1) = 2S(n) n(n+1) = 2S(n) [n(n+1)]/2 = S(n). Then I wanted to derive a formula for the sum of powers of 2: 1^2 + 2^2 + 3^2 + ... + n^2 = S(n) This is were I get stuck. I tried the same method that I had used on consecutive integers but it did not seem to work, at least it got me lost. I don't know where to find a book that shows the DERIVATION of such formulas. I do have a precalculus book that shows the formulas for powers of 2, 3, 4, and even 5, but it does not show how they arrived at the formula. I want to see how the formula was derived. Thanks. Sincerely, Amir Milbes Date: 08/28/2001 at 10:19:50 From: Doctor Rob Subject: Re: sums of powers Thanks for writing to Ask Dr. Math, Amir. This is done in the following way. Define, recursively, f(x,0) = 1, f(x,k+1) = f(x,k)*(x-k), k >= 0. Then one has the equation f(x,k) = x*(x-1)*(x-2)*...*(x-k+1). This is called the "k-th falling factorial of x." Notice that f(x+1,k+1) - f(x,k+1) = (x+1)*f(x,k) - f(x,k)*(x-k), = ([x+1]-[x-k])*f(x,k). = (k+1)*f(x,k), f(x,k) = [f(x+1,k+1)-f(x,k+1)]/[k+1]. Now write the powers of x in terms of f(x,k) for various k's: m x^m = SUM S2(m,k)*f(x,k), k=0 for some constants S2(m,k). In particular, x^1 = 1*f(x,1), x^2 = 1*f(x,2) + 1*f(x,1), x^3 = 1*f(x,3) + 3*f(x,2) + 1*f(x,1), x^4 = 1*f(x,4) + 7*f(x,3) + 6*f(x,2) + 1*f(x,1), and so on. These constant coefficients S2(m,k) are called the Stirling Numbers of the Second Kind. They are always positive integers, and satisfy the following recursion relations: for m >= 1, S2(m,0) = 0, S2(m,1) = 1, S2(m,k) = 0 for all all k > m, S2(m+1,k) = S2(m,k-1) + k*S2(m,k). You can prove these formulas by using Mathematical Induction. They allow one to compute the Stirling Numbers of the Second Kind in a similar way to computing Binomial Coefficients using Pascal's Triangle. m\k| 1 2 3 4 5 6 ------------------------------- 1 | 1 2 | 1 1 3 | 1 3 1 4 | 1 7 6 1 5 | 1 15 25 10 1 5 | 1 31 90 65 15 1 For example, 90 = 15 + 3*25, and 65 = 25 + 4*10. (There are also Stirling Numbers of the First Kind S1(m,k), which are the coefficients you get when expressing f(x,m) in terms of the powers x^k for k = 1, 2, ..., m.) Now we can use the material above to compute the sum n T(m,n) = SUM x^m, x=1 n m = SUM SUM S2(m,k)*f(x,k), x=1 k=1 m n = SUM S2(m,k)*SUM f(x,k), k=1 x=1 m n = SUM S2(m,k)*SUM [f(x+1,k+1)-f(x,k+1)]/[k+1], k=1 x=1 m n = SUM S2(m,k)/(k+1)*SUM f(x+1,k+1) - f(x,k+1). k=1 x=1 Now the inner sum is a collapsing or telescoping sum, and f(1,k+1) = 0 for all k >= 1, so m T(m,n) = SUM S2(m,k)/(k+1)*[f(n+1,k+1)-f(1,k+1)], k=1 m T(m,n) = SUM S2(m,k)*f(n+1,k+1)/(k+1). k=1 This is the formula which you can use to find the sum of the first m n-th powers. For example, when m = 2, you get T(2,n) = S2(2,1)*f(n+1,2)/2 + S2(2,2)*f(n+1,3)/3, = 1*(n+1)*n/2 + 1*(n+1)*n*(n-1)/3, = n*(n+1)*(1/2 + [n-1]/3), = n*(n+1)*(1/2 + n/3 - 1/3), = n*(n+1)*(2*n+1)/6. As another example, when m = 4, you get T(4,n) = S2(4,1)*f(n+1,2)/2 + S2(4,2)*f(n+1,3)/3 + S2(4,3)*f(n+1,4)/4 + S2(4,4)*f(n+1,5)/5, = 1*(n+1)*n/2 + 7*(n+1)*n*(n-1)/3 + 6*(n+1)*n*(n-1)*(n-2)/4 + 1*(n+1)*n*(n-1)*(n-2)*(n-3)/5, = n*(n+1)*(1/2 + 7*[n-1]/3 + 3*[n-1]*[n-2]/2 + [n-1]*[n-2]*[n-3]/5), = n*(n+1)*(1/2 + 7*[n-1]/3 + 3*[n^2-3*n+2]/2 + [n^3-6*n^2+11*n-6]/5), = n*(n+1)*(1/2 + 7*n/3 - 7/3 + 3*n^2/2 - 9*n/2 + 3 + n^3/5 - 6*n^2/5 + 11*n/5 - 6/5), = n*(n+1)*(n^3/5 + 3*n^2/10 + n/30 - 1/30), = n*(n+1)*(6*n^3+9*n^2+n-1)/30, = n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30. These should agree with the formulas in your book. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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