How do we find Maclaurin Series of $e^{t^2/2}$? I see one way on one book to do it is applying the fact that $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$, then $$e^{t^2/2} = \sum_{n=0}^{\infty}\frac{(\frac{1}{2}t^2)^n}{n!}$$ But I cannot really understand why can we directly subsitute $\frac{1}{2}t^2$ into $e^x$ equation since it is not a constant. May someone explain to me? I am really confused.
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1It is due to the definition of $e^x$. Let $x=\frac{1}{2} t^2$ then substitute into the definition equation and we are done. – Szeto Feb 15 '18 at 01:17
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I'm not sure why it bothers you that $1/t^2$ is not a constant. For any value of $t,$ it's just some number. You substitute that into the formula for $e^x.$ That gives you a formula in terms of $t$. – saulspatz Feb 15 '18 at 01:19
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Consider the what you do with function composition with more simple minded examples ... $f(x) = x^2 + 2. f(e^x) = (e^x)^2 + 2$ You are doing the same thing into a function that is defined as an infinite series. – Doug M Feb 15 '18 at 01:19
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If you choose a value of $t$ it becomes a constant :D – Chickenmancer Feb 15 '18 at 01:20
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Oh, I see. Thanks a lot for the help. – PassingBy Feb 15 '18 at 01:23