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How can a nested power series be shown as a single power series?


Let sum(a[n],k) denote the infinite sum from k to infinity of summands a[n]. Then let f(x) = sum(a[i]x^i,0) and g(x) = sum(b[j]x^j,0). Then h(x) = f(g(x)) = sum(c[n]x^n,0) is the nested sum. Assuming convergence, a[i] = f^{i}(0)/i!, b[j] = g^{j}(0)/j! and c[n] = h^{n}(0)/n!, where f^{i}(0) denotes the ith derivative of f evaluated at 0 and similarly for g and h. Note that the mth derivative of sum(d[k]x^k,0) = sum((k+1)(k+2)...(k+m)a[k]x^k,0) c[0] = h^{0}(0)/0! = f(g(0)) = f(b[0]) = sum(a[i]b[0]^i,0). Since h^{1}(x) = f^{1}(g(x))g^{1}(x), c[1] = h^{1}(0)/1! = f^{1}(g(0))g^{1}(0) = sum((i+1)a[i]g(0)^i,0)b[0]. Since h^{2}(x) = f^{2}(g(x)(g^{1}(x))^2 + f^{1}(g(x))g^{2}(x), c[2] = h^{2}(0)/2! = (1/2){f^{2}(g(0)(g^{1}(0))^2 + f^{1}(g(0))g^{2}(0)} = (1/2){sum((i+1)(1+2)a[i]b[0]^i,0)b[1]^2+ sum((i+1)a[i]g(0)^i,0)2b[2]}. Using this approach, you can compute as many c[n]s as you want. For example, lets compute the second Taylor polynomial of sin(e^x) of degree 2. (This example is only to check the validity of the method and is *not* the best way to actually compute the Taylor polynomial.) f(x) = sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ... and g(x) = 1 + (1/1!)x^1 + (1/2!)x^2 + .... So, a[i] = 0, if i is even and (-1)^k/(2k+1)!x^(2k+1), if i = 2k+1. b[j] = 1/j!. So, c[0] = sum(a[i]b[0]^i,0) = sin(b[0]) = sin(1). Moreover, c[1] = sum((i+1)a[i]g(0)^i,0)b[0]. g(0) = e^0= 1 and b[0] = 1 also. So, c[1] = sum((i+1)a[i]g(0)^i,0)b[0] = sum((i+1)a[i]1^i,0) = sin^{1}(1) = cos(1). Also, c[2] = (1/2){sum((i+1)(1+2)a[i]b[0]^i,0)b[1]^2+ sum((i+1)a[i]g(0)^i,0)2b[2]} = (1/2){sum((i+1)(1+2)a[i]1^i,0)1^2+ sum((i+1)a[i]1^i,0)2(1/2!)} = (1/2){-sin(1)+cos(1)}. Hence, the second Taylor polynomial of sin(e^x) is sin(1) + cos(1)x + (cos(1)-sin(1))x^2/2, which can be checked by direct computation.)

R. Paul Beem

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