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# Calculus/Taylor Series

Question

WA
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 web-sites and books I have looked through so far have only managed to confused me further.

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,

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 may take a while to internalize, so be patient.

Hope it helps. Peace.

Calculus

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