Deriving Quadratic Equation from Taylor Expansion: An Exercise

[kel]

Homework Statement

I have the following question to answer:
Show that
(X^2/h^2)*((1/2*y1) - y2 + (1/2*y3)) + (X/h)*((-1/2 y1)+(1/2 y3))+y2 (sorry about the format)

is equal to (taylor expansion):
y = y2+(x(dy/dx)¦0 + (x^2/2*((d^2)y)/(dx^2))¦0

Homework Equations

also given in dy/dx¦0 is value of dy/dx when x = 0

and

dy/dx¦0 (approximately) = 1/2h (y(+h)-y(-h))

The Attempt at a Solution

I know how to derive the quadratic, but I never really got the hang of taylor expansions so I'm lost.

I can see that there is a double derivative term, but as this i the derivative of the dy/dx¦0 term with no x values in it, does it then become zero? Also, all I have ended up with for the taylor expansion is:

y= y2 + x(1/2h(y(+h)-y(-h))) + x^2/2 ((double derivative term = 0)
giving
y= y2 + x(1/2h(y(+h)-y(-h))) and then I've tried to multiply out the brackets.

I can see that I've gone wrong as I can't get an answer.

Can anyone point me in the right direction please??

Cheers
Kel

 

[HallsofIvy]

I'm afraid this doesn't make much since to me. Is "X" in the first formula the same as "x" in the second? Are y1, y2, y3 arbitrary values of y? You use y2 in the second formula but not y1 or y3. What happened to them? There is an "h" in the first formula that is not present in the second.

I don't see how the two formulas could be the same since one contains much more information (h, y1, y3) than the other.

IF you are assuming that y1= y(0), y2= y(h), and y3= y(2h) please say so!

 

[kel]

Ok, here's a copy of the problem sheet (I'm on question 3)

and a copy of the lecturers notes - the relevant part is on page and 7, hopefully this will explain it a bit better than I can.

Cheers

 

[kel]

pages 6 & 7 I mean

 

[HallsofIvy]

Wouldn't it have been simpler just to say y2= y(X), y1= y(X-h), and y2= y(X+h)?

Notice an important difference: in deriving the Taylor's expansion (2nd degree Taylor polynomial), you must assume that you KNOW y(x), that is, y as a function of x. Then show that, whatever y(x) is, you can arrive at the interpolation polynomial.