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I came across this nice answer on stack:

enter image description here

However, two things still bother me. First, why do we assume a function that equals its own derivative will take the form of $a_nx^n$? Why should it be an exponential function? I can understand the reason for the sigma a little, it implies a function immune to derivatives should work regardless of the input changing from $0$ to infinity (why not negative infinity as well?)

Second, the condition $f(0) = 1$ also bothers me. So assuming we determine the function will be exponential in some way, I'm guessing this is a necessary condition as all exponential will be $1$ when using $0$ as input. Or is there something more to it?

YiFan Tey
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SzL
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  • We're assuming that there exists some function that admits a power series. – Sean Roberson Oct 14 '19 at 00:58
  • Thanks! How do we make that assumption however? Could there be another function that could equal its own derivative for example, which is not exponential? – SzL Oct 14 '19 at 01:03
  • There could be another function, but that's not part of the assumption. – Sean Roberson Oct 14 '19 at 01:16
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    We don't assume it takes the form of $A_nx^n$ which not an exponential function. (An exponential function is $b^x$, not $x^n$). We assume that $f(x)$ takes a form of an infinite* sum of power values. In other words a power series. We can assume this, because if it is its own derivative it is infinitely differentiable and that is just it's Taylor series. – fleablood Oct 14 '19 at 01:26
  • Thanks that's much more clear. – SzL Oct 14 '19 at 01:59
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    If it's a stackexchange answer you are citing, you should link it rather than (or as well as) screen-shotting it. – spaceisdarkgreen Oct 14 '19 at 02:13
  • @space, probably https://math.stackexchange.com/questions/1054175/why-does-the-sum-of-the-reciprocals-of-factorials-converge-to-e/1758228#1758228 – Gerry Myerson Oct 14 '19 at 02:54

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