What is taylor e^x,x=1 ?

The answer to taylor e^x,x=1 is sum from n=0 to infinity of (e(x-1)^n)/(n!)

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taylor e^x,x=1

taylor

e^{x}, x = 1

n = 0 \sum \infty \frac{e ( x - 1 ) ^{n}}{n !}

Solution steps

Apply the Taylor Formula

= e + \frac{\frac{d}{d x} ( e ^{x} ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{x} ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{x} ) ( 1 )}{3 !} (x - 1)^{3} + \dots

Evaluate Derivatives

= e + \frac{e}{1 !} (x - 1) + \frac{e}{2 !} (x - 1)^{2} + \frac{e}{3 !} (x - 1)^{3} + \frac{e}{4 !} (x - 1)^{4} + \dots

The series sum representation

= n = 0 \sum \infty \frac{e ( x - 1 ) ^{n}}{n !}

x \to 0 lim (x^{3} + 5)

\frac{3}{( s ^{2} + 4 s + 13 )}

\frac{d y}{d x} - (sin (x)) y = 2 sin (x)

f (x) = cos^{2} (x) - 2 sin (x)

What is taylor e^x,x=1 ?
The answer to taylor e^x,x=1 is sum from n=0 to infinity of (e(x-1)^n)/(n!)