Alternative argument to show that function diverges everywhereConsider the function:f(x)=x12+12x−12+23x32−14x−32+115x52+⋯which is constructed such that f is (or at least looks like) a solution to f′=f.Then, f(x)=Aex for some constant A. But f diverges at x=0, so A=±∞" and f diverges everywhere.

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Alternative argument to show that function diverges everywhereConsider the function:f(x)=x12+12x−12+23x32−14x−32+115x52+⋯which is constructed such that f is (or at least looks like) a solution to f′=f.Then, f(x)=Aex for some constant A. But f diverges at x=0, so A=±∞&quot; and f diverges everywhere.

Deven Paul · Accepted Answer

Answer: Clearly,&amp;nbsp;f(x)=∑n=0∞4nn!(2n+1)!x(2n+1)/2+∑n=0∞(2n-1)!4n(n-1)!x-(2n+1)/2The first term converges when&amp;nbsp;0&amp;lt;x&amp;nbsp;while the radius of convergence of the second term is&amp;nbsp;R=0Hence the function is nowhere definable

Alternative argument to show that function diverges everywhere Consider

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