What is the Taylor series expansion of [math]e^{-x}[/math] about zero?

first, you should know what is Taylor series expansion and what is the meaning of about zero

Taylor series- if you have any function, the function is differentiable then we differentiate continuously and about of any point is called the Taylor series

if any function is differentiated about the origin or x=0 then this series is called as Maclaurin series

so I move the question

question the what is the Taylor series expansion of exp(-x) about zero point

according to Taylor series expansion

y= f(x)

where x=a

and a is the point

so that

f(x)= f(a)+ (x-a)f’(a)+(x-a)^2/ 2! +…………………………..infinte

at point a

in question

f(x) = exp(-x)

x=0

first term = f(x) = exp(-x)

put the point f(0)= exp(0) =1

second term

f’(x) = -exp(-x)

(x-a)f’(a)= (x-0) f’(0)= x(-1)=-x

third term

f’’(x)= exp(0)

(x-a)^2.f’’(a)/2!= (x-0)^2*(1)/2! = x^2/2

and you will find the other terms fourth, fifth terms and so on according to your question

now finally the series of exp(-x)

exp(-x)= 1 - x + x^2 /2! - x^3/3!+ ……………………….