Question #2eeda

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Answer 1 · 2017-01-24T10:36:32

$e^x = e sum_(n=0)^oo (x-1)^n/(n!)$

We have that:

$d/(dx) e^x = e^x$

and therefore for higher orders:

$d^n/(dx^n) e^x = e^x$

and for $x=1$

$[d^n/(dx^n) e^x]_(x=1) = e$ for $n=0,1,2,...$

The expression for the Taylor series is then:

$e^x = sum_(n=0)^oo e (x-1)^n/(n!) = e sum_(n=0)^oo (x-1)^n/(n!)$

We can note that in general we have that for every point $x_0$ :

$e^x = e^(x-x_0+x_0)= e^(x-x_0)e^(x_0)$

and so if we start from the MacLaurin series:

$e^x = sum_(n=0)^oo x^n/(n!)$

the Taylor series around a point $x_0$ always have the form:

$e^x = sum_(n=0)^oo x^n/(n!) = e^(x_0) sum_(n=0)^oo (x-x_0)^n/(n!)$