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Topi
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This is not a question, but I'm answering myself to show how powerseries and differentiation works.
 one year ago
 one year ago
Topi Group Title
This is not a question, but I'm answering myself to show how powerseries and differentiation works.
 one year ago
 one year ago

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Topi Group TitleBest ResponseYou've already chosen the best response.0
If we assume that every differentiable function can be expressed in a power series:\[\sum_{n=0}^{\infty}a_{n}x^{n}\]and we know that\[\frac{ dy }{ dx }x^n=nx ^{n1}\]Then if we try to find the function f(x) that folowes the rule that f'(x)=f(x) and f(0) =1 we can show that for the power series expansion for that function:\[\frac{ d }{ dx }\sum_{n=0}^{\infty}a _{n}x ^{n}=\sum_{n=0}^{\infty}na _{n}x ^{n1}\]And as these should be equal then\[a _{n1}x ^{n1}= na _{n}x ^{n1} \rightarrow a _{n1} =n a _{n}\rightarrow a _{n}=\frac{ a _{n1} }{ n }\]So because a(0) must be 1 so that f(0) can be 1 we get\[a_{1}=\frac{ 1 }{ 1 }, a _{2}=\frac{ 1 }{ 1*2 }, a _{3}=\frac{ 1 }{ 1*2*3 },...a _{n}=\frac{ 1 }{ 1*2*3,....,*n }\] So the function were after is\[f(x) =\sum_{n=0}^{\infty}\frac{ x^n }{ n! }\]Unfortunately this does not tell you what that funcion is, but it still tells you it's the function that is it's own derivative. So if you find a function that is it's own derivative then this is it's powerseries expansion.
 one year ago
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