Proving f'''(x) = 0 using Taylor's Theorem

[stukbv]

Homework Statement

f is 3 times continuously differentiable,
f(a+h) = f(a) + f'(a+1/2*h)h whenever a is a real number and h>=0 .
By applying taylors theorem to f and f' show that the third derivative f''' of f is identically 0 ...

 

[micromass]

What have you tried already?

 

[stukbv]

I don't understand how to apply taylors theorem to this??!

 

[micromass]

Apply it anyway, we'll see where we end up. Can you start by finding the Taylor expansion of f and f' around a? Then apply it to f(a+h) and f(a+h/2)...

 

[stukbv]

errmm f(x) = f(a) + f'(a)(x-a)?

 

[micromass]

errmm f(x) = f(a) + f'(a)(x-a)?

Take a term more! And don't gorget the remainder!

And what's the expansion of f' around a (two terms are enough now) and don't forget the remainder!

 

[stukbv]

ok so first one f(x)= f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2 + R3(x)
and f'(x)=f'(a) +f''(a)(x-a) + R2 ? - not sure never seen it applied to f' before?

 

[micromass]

ok so first one f(x)= f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2 + R3(x)
and f'(x)=f'(a) +f''(a)(x-a) + R2 ?

OK! You'll probably have to write out the remainders later on (that's where the f''' comes in)

Now, subsitute the f and f' in the equation f(a+h) = f(a) + f'(a+1/2*h)h by their Taylor expansions...

- not sure never seen it applied to f' before?

Well, it applies to all continuous differentiable functions, so certainly to f', no?

 

[stukbv]

ok done that and i get that R3 = r2 ?

 

[micromass]

You should start by writing out the remainders...

 

[stukbv]

Ok now I've written out the remainders and i get that
f(a+h) = f(a) + f'(a)h + f''(a)h^2/2 + f'''(c)h^3/3!

f'(a+h/2) = f'(a) + f''(a)h/2 + f'''(a)h^2/4

so substituting them in i get
f(a) + f'(a)h + f''(a)h^2/2 + f'''(c)h^3/3! = f(a) +[ f'(a) + f''(a)h/2 + f'''(a)h^2/4]*h
^3=3f
when i cancel things i then get 2f'''(c)h^3=3f'''(c)h^3
Now what? i don't get it ?

 

[micromass]

Ok now I've written out the remainders and i get that
f(a+h) = f(a) + f'(a)h + f''(a)h^2/2 + f'''(c)h^3/3!

f'(a+h/2) = f'(a) + f''(a)h/2 + f'''(a)h^2/4

That should be f'''(d)h^2/4, with $d\in [a,a+h/2]$
[/QUOTE]

Thus

so substituting them in i get
f(a) + f'(a)h + f''(a)h^2/2 + f'''(c)h^3/3! = f(a) +[ f'(a) + f''(a)h/2 + f'''(a)h^2/4]*h
^3=3f
when i cancel things i then get 2f'''(c)h^3=3f'''(c)h^3

You should get (after cancelling the h): 2f'''(c)=3f'''(d), with $c\in [a,a+h]$ and $d\in [a,a+h/2]$.

Now, what happens if $h\rightarrow 0$?

 

[stukbv]

Ah i see so we take different values to put into the remainders, i was unsure about that.
so do we get c and d are both a as h->0?

 

[micromass]

Ah i see so we take different values to put into the remainders, i was unsure about that.
so do we get c and d are both a as h->0?

Indeed, so in the limit, we get 2f'''(a)=3f'''(a), thus...

 

[stukbv]

it = 0! so obvious now, thanks a lot !