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syms x f(x) p

f1=taylor(f(x),x,'order',3)

(D(D(f))(0)=p; D(f)(0)=1; f(0)=0;

%%% I want to put initial conditios (D(D(f))(0)=p; D(f)(0)=1; f(0)=0; in f1 to

find f1=x+p*x^2/2 in symbolic form. Guide me please

John D'Errico
on 5 Jan 2020

Edited: John D'Errico
on 5 Jan 2020

Why not try it! ??? Make an effort. For example, this seems the obvious thing to try. So what does this do?

syms x f(x) p

f1=taylor(f(x),x,'order',3)

f1 =

(D(D(f))(0)*x^2)/2 + D(f)(0)*x + f(0)

subs(f1,f(0),0)

ans =

(D(D(f))(0)*x^2)/2 + D(f)(0)*x

Can you finish the next two steps on your own?

John D'Errico
on 5 Jan 2020

Yes, but you cannot do what you want.

syms x f(x) p g(x) a q h(x) r

f1 = taylor([f(x) g(x) h(x)],x,'order',[3 3 2]);

Error using sym/taylor (line 99)

The value of 'Order' is invalid. It must satisfy the function: isPositiveInteger.

The taylor function appears not to be vectorized in the sense that you can expand each term ot the vector of functions to a different order.

So this next is valid:

f1 = taylor([f(x) g(x) h(x)],x,'order',3);

And then you should be able to proceed further.

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