How do you find the Maclaurin series for #e^(x^2)# ?

The Maclaurin series is simply the Taylor series centered around #a = 0#.

The Taylor series formula is:

#sum_(n=0)^(N) (f^((n))(a))/(n!)(x-a)^n#

Hence, the Maclaurin series formula is:

#sum_(n=0)^(N) (f^((n))(0))/(n!)x^n#

Now, we need to take some derivatives. Let's go to #n = 4#.

#f^((0))(x) = color(green)(f(x) = e^(x^2))#

#color(green)(f'(x) = e^(x^2)*2x)#

#color(green)(f''(x)) = e^(x^2)*2 + 2x*e^(x^2)*2x = color(green)(e^(x^2)(4x^2 + 2))#

#color(green)(f'''(x)) = 2e^(x^2)*2x + e^(x^2)*8x + 4x^2*e^(x^2)*2x#

#= 4xe^(x^2) + 8xe^(x^2) + 8x^3e^(x^2)#

#= color(green)(e^(x^2)(8x^3 + 12x))#

#color(green)(f''''(x)) = (e^(x^2)*24x^2 + 8x^3*e^(x^2)*2x) + e^(x^2)*12 + 12x*e^(x^2)*2x#

#= color(green)(e^(x^2)[16x^4 + 48x^2 + 12])#

Now we can construct the Maclaurin series:

#sum_(n=0)^(N) (f^((n))(0))/(n!)x^n#

#= (f(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + (f''''(0))/(4!)x^4 + ...#

#= e^(0^2) + cancel([e^(0^2)*2(0)]x)^(0) + cancel([(e^(0^2)(4(0^2) + 2))/(2)])^(1) x^2 + cancel([(e^(0^2)(8(0)^3 + 12(0)))/(6)]x^3)^(0) + [(e^(0^2)(16(0)^4 + 48(0)^2 + 12))/(24)]x^4 + ...#

#= color(green)(0!)/color(highlight)(0!)*1/(0!) + color(green)(2!)/color(highlight)(1!)*x^2/(2!) + overbrace(color(green)(4!)/color(highlight)(2!))^(12)*x^4/underbrace(4!)_("x"^4"/24") + color(green)(6!)/color(highlight)(3!)*(x^6)/(6!) + ...#

See the pattern? It requires a bit of manipulation to figure out, but the constant happens to be #color(green)(n!)/color(highlight)((n/2)!)#, while the #n!# in the actual formula remains there. That cancels out to give #(n/2)!# in the denominator of the simplified answer.

#n = 0, 2, 4, 6, ...#

#color(blue)(sum_(n=0)^N (f^((n))(0))/(n!)x^n = 1 + x^2 + x^4/2 + x^6/6 + x^8/24 + ...)#

Given the well known series for #e^x#, we can write:

#e^y = 1 + y/(1!) + y^2/(2!) + y^3/(3!) +... = sum_(n=0)^oo y^n/(n!)#

Let #y = x^2# to find:

#e^(x^2) = 1 + x^2/(1!) + x^4/(2!) + x^6/(3!) +... = sum_(n=0)^oo x^(2n)/(n!)#