第六题怎么做啊 10
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∫(x->2x) f(t) dt = e^x - 1
2f(2x) - f(x) = e^x
x=0, =>f(0) = e^0 =1
4f'(2x) - f'(x) = e^x
x=0, =>f'(0) = (1/3)e^0 =1/3
...
2^(n+1).f^(n)(x) -f^(n)(x) = e^x
x=0, =>f^(n)(0) = {1/[2^(n+1) -1] }. e^0 = 1/[2^(n+1) -1]
f(x)
=f(0) + [f'(0)/1!]x+...+ [f^(n)(0)/n!]x^n +...
=1 + (1/3)x+...+ {1/(n![2^(n+1) -1]) } x^n+.....
2f(2x) - f(x) = e^x
x=0, =>f(0) = e^0 =1
4f'(2x) - f'(x) = e^x
x=0, =>f'(0) = (1/3)e^0 =1/3
...
2^(n+1).f^(n)(x) -f^(n)(x) = e^x
x=0, =>f^(n)(0) = {1/[2^(n+1) -1] }. e^0 = 1/[2^(n+1) -1]
f(x)
=f(0) + [f'(0)/1!]x+...+ [f^(n)(0)/n!]x^n +...
=1 + (1/3)x+...+ {1/(n![2^(n+1) -1]) } x^n+.....
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