How to do taylor series expansion of #e^(-x^2"/2")#?

Answer 1
When you construct a Taylor series, you're going to have to take some #n#th order derivatives. To get a substantial way into the series, let's try going to #n = 4#.

The Taylor series can be written out as:

#sum_(n=0)^N (f^((n))(a))/(n!)(x-a)^n#
You didn't specify what #a# was, but I will just assume a general case of #a = a#.
#color(green)(f^((0))(x)) = f(x) = color(green)(e^(-x^2"/2"))#
#color(green)(f'(x)) = e^(-x^2"/2") * -x = color(green)(-xe^(-x^2"/2"))#
#color(green)(f''(x)) = (-x)[-xe^(-x^2"/2")] + e^(-x^2"/2")(-1)# #= x^2e^(-x^2"/2") - e^(-x^2"/2")# #= color(green)((x^2 - 1)e^(-x^2"/2"))#
#color(green)(f'''(x)) = (x^2 - 1)(-xe^(-x^2"/2")) + e^(-x^2"/2")(2x)# #= (-x^3 + x)(e^(-x^2"/2")) + 2xe^(-x^2"/2")# #= color(green)((-x^3 + 3x)e^(-x^2"/2"))#
#color(green)(f''''(x)) = (-x^3 + 3x)(-xe^(-x^2"/2")) + e^(-x^2"/2")(-3x^2 + 3)# #= (x^4 - 3x^2)e^(-x^2"/2") - 3x^2e^(-x^2"/2") + 3e^(-x^2"/2")# #= color(green)((x^4 - 6x^2 + 3)e^(-x^2"/2"))#
Then we can write this out, using #x->a# and #n = 4# terms:
#sum_(n=0)^4 (f^((n))(a))/(n!)(x-a)^n#
#= (f^((0))(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + ...#
#= e^(-a^2"/2") + (-ae^(-a^2"/2"))(x-a) + ((a^2 - 1)e^(-a^2"/2"))/(2)(x-a)^2 + ((-a^3 + 3a)e^(-a^2"/2"))/(6)(x-a)^3 + ((a^4 - 6a^2 + 3)e^(-a^2"/2"))/(4!)(x-a)^4 + ...#
#= color(blue)(e^(-a^2"/2") - ae^(-a^2"/2")(x-a) + ((a^2 - 1)e^(-a^2"/2"))/(2)(x-a)^2 + ((-a^3 + 3a)e^(-a^2"/2"))/(6)(x-a)^3 + ((a^4 - 6a^2 + 3)e^(-a^2"/2"))/(4!)(x-a)^4 + ...)#
If we have #a = 0#, this simplifies significantly:
#= color(blue)(1 - x^2/2 + x^4/8 - x^6/48 + ...)#
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Answer 2

# e^(-x^2/2)= 1 - x^2/2 + x^4/8 - x^6/48 + x^8/384 - x^10/3840 + ...#

or,

# e^(-x^2/2) = sum_(n=0)^oo (-1)^(n)x^(2n)/(2^n n!) #

Assuming a TS pivoted about #x=0# (ie a MacLaurin Series) we start with well known series:
# e^x= 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ...#
Replacing "#x#"with #-x^2/2# we get:
# e^(-x^2/2)= 1 + (-x^2/2) + ((-x^2/2)^2)/(2!) + ((-x^2/2)^3)/(3!) + ((-x^2/2)^4)/(4!) + ((-x^2/2)^5)/(5!) + ...#
# \ \ \ \ \ \ = 1 - x^2/2 + (x^4/2^2)/(2) - (x^6/2^3)/(6) + (x^8/2^4)/(24) - (x^10/2^5)/(120) + ...#
# \ \ \ \ \ \ = 1 - x^2/2 + (x^4/4)/(2) - (x^6/8)/(6) + (x^8/16)/(24) - (x^10/32)/(120) + ...#
# \ \ \ \ \ \ = 1 - x^2/2 + x^4/8 - x^6/48 + x^8/384 - x^10/3840 + ...#
And we note that the #n^(th)# term is given by:
# u_n = (-1)^(n)x^(2n)/(2^n n!) #
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Answer 3

To find the Taylor series expansion of (e^{-\frac{x^2}{2}}), we use the standard formula for the exponential function's Taylor series. The Taylor series expansion of (e^x) around (x = a) is given by:

[ e^x = \sum_{n=0}^{\infty} \frac{{f^{(n)}(a)}}{{n!}}(x - a)^n ]

Here, (f^{(n)}(a)) represents the (n)th derivative of (f(x)) evaluated at (x = a). For (e^{-\frac{x^2}{2}}), (a = 0).

To find the derivatives of (e^{-\frac{x^2}{2}}), use the chain rule. The first few derivatives are:

[ f(x) = e^{-\frac{x^2}{2}} ] [ f'(x) = -x e^{-\frac{x^2}{2}} ] [ f''(x) = (x^2 - 1) e^{-\frac{x^2}{2}} ] [ f'''(x) = (3x - x^3) e^{-\frac{x^2}{2}} ]

Substitute these derivatives into the Taylor series formula and evaluate them at (a = 0) to find the coefficients. The Taylor series expansion of (e^{-\frac{x^2}{2}}) around (x = 0) will be the sum of these terms.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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