How to obtain the taylor series for this function?

Hi, 

We're currently learning about Taylor series and I'm having a lot of trouble understanding the concept and the math involved in it. I typed into Wolfram Alpha the function e^-(x^2) at point 0 and order 4 and have attached the output:

My understanding is you start by differentiating the function 4 times but I don't understand how to get from there to the equation Wolfram Alpha has given me. I would really appreciate some help with this as the websites and books I have looked through so far have only managed to confused me further.

Thanks in advance.

Student X

The Taylor series is explained by 

f(x) = f(a) + f'(a)*(x-a) +  f" (a)*( x-a)^2 /2 ! +  f''' (a)*( x-a)^3 / 3! +........

When the function to be represented is about x=0 as mentioned in your problem, the taylor series is reduced to that of a Maclaurin's Series, 

ie f(x)= f(0) + f'(0) *x +  f" (0) *x^2/ 2! + f'''(0) *x^3/3! +........

Let's work things out to obtain a clearer picture. 

Assuming y =f(x)= e^-(x^2) , and therefore implying  f'(x)=dy/dx, f"(x)= d^2 y/dx^2  so on and so forth, 

Differentiating this once wrt x on both sides gives 

dy/dx = -2x * e^-(x^2) = -2xy        (Note: The e^-(x^2) component is replaced simply by y)

Differentiating a second time wrt x on both sides 

d^2 y/dx^2 = -2 [ x*dy/dx +y] = -2x*dy/dx -2y

Differentiating a third time wrt x on both sides, 

d^3 y/dx^3 = -2[ x*(d^2 y/dx^2) +dy/dx] -2 dy/dx  

               = -2x*(d^2 y/dx^2) -4 dy/dx

Differentiating one final time wrt x on both sides, 

d^4 y/ dx^4  = -2 [ x*(d^3 y/dx^3) + (d^2 y/dx^2)] -4 (d^2 y/dx^2)

Now, we seek to find the values of the various orders of derivatives when x=0, ie

f(0)= value of y when x=0,

f'(0) = value of dy/dx when x=0,

f"(0)= value of (d^2 y/ dx^2) when x=0, so on and so forth 

By making a series of substitutions into the above block of differential equations herein:

When x=0,  

y=1, dy/dx= 0, d^2 y/dx^2 =-2 ,  d^3 y/dx^3 = 0  and  d^4 y/dx^4 = 12 

Hence, the expansion of the series is given by 

f(x)= 1 + 0 * x + (-2)*x^2/2! + 0*x^3/ 3! + (12)*x^4/4! +....... 

    = 1- x^2 + 0.5*x^4 -.........

This whole affair may take a while to internalize, so do be patient.

Hope it helps. Peace. 

Best Regards,

Mr Koh