Numerical Methods Backward Difference

Hi guys. I was trying to differentiate that problem. I can solve it by hand but I'm really bad on matlab. I didn't see like this type of example on matlab. I don't have any guide for this type of questions. Problem is this:

Assuming y (x) is a smooth function defined on the interval [0; 1] ; obtain a second order of

accuracy approximation formula for y''' (1) (third order derivatives).

I know we have to use backward formula. We have 2+3= 5 unknows.

Backward formula is this: v(t) = ((fx(t)) - (fx(t)-h))/h;

The question is 'You can make use of MATLAB software to find

the unknown coefficients'.

Thank you for your answers.

[Dana]

I don't understand the question. What does it mean to have a "second order of accuracy approximation formula for y'''(1)"? I understand what it means to find, say, a second-order Taylor approximation around x=1, but in that case you want

What does this have to do with y'''(1)?

Also, do you have a particular functional form for y(x)? Or is it an arbitrary function? And what is fx(t)?

Spending a little more time trying to make your question clear and precise will improve your chances of getting a helpful reply.

[Hasan Dikka]

OKey forget the upper part. I will explain clearly. I have a example about forward formula with solitions in pictures. I written forward and backward formula in pictures. Am ı clear? And also the question is this.

Assuming y (x) is a smooth function dened on the interval [0; 1] ; obtain a second order of

accuracy approximation formula for y'''(1) : You can make use of MATLAB software to find

the unknown coefficients. Write down the formula as a result.

Answer 1

[John D'Errico]

I won't directly do your homework for you. I'll show you how you might solve for the coefficients of a formula you should already know though. And there is probably a simpler way to do it.

syms x h y0 yp ypp yppp ypppp
syms a0 a1

yhat = y0 + yp*x + ypp*x^2/2 + yppp*x^3/2 + ypppp*x^4/24
yhat =
(ypppp*x^4)/24 + (yppp*x^3)/2 + (ypp*x^2)/2 + yp*x + y0

Now, can I solve for a simple backwards finite difference formula for the first derivative of y, at x == 0?

Consider the general backwards finite difference, with unknown coefficients a0 and a1.

findiff = expand(a0*subs(yhat,x,0) + a1*subs(yhat,x,-h))
findiff =
(a1*ypppp*h^4)/24 - (a1*yppp*h^3)/2 + (a1*ypp*h^2)/2 - a1*yp*h + a0*y0 + a1*y0

Now, the simple way to extract the coefficient of some derivative is to just differentiate with respect to that parameter.

A = solve(diff(findiff,y0) == 0,diff(findiff,yp) == 1,a0,a1)
A = 
  struct with fields:

    a0: [1×1 sym]
    a1: [1×1 sym]
    
subs(findiff,[a0,a1],[A.a0,A.a1])
ans =
- (ypppp*h^3)/24 + (yppp*h^2)/2 - (ypp*h)/2 + yp

A.a0
ans =
1/h

A.a1
ans =
-1/h

As you can see, this is the simple backwards finite difference formula to estimate the first dervative of y at x==0. How about the error term? That is the lowest order term in h.

Can you do the same thing for a finite difference approximation to yield the third derivative? I hope so, since I pretty much did your homework here, at least if you think about it.

And, again, I am sure I could have done this more directly.