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## 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 ...

- Thread starter stukbv
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- #1

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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 ...

- #2

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What have you tried already?

- #3

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I dont understand how to apply taylors theorem to this??!

- #4

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- #5

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

- #6

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Take a term more! And don't gorget the remainder!errmm f(x) = f(a) + f'(a)(x-a)?

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

- #7

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and f'(x)=f'(a) +f''(a)(x-a) + R2 ? - not sure never seen it applied to f' before?

- #8

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

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

Well, it applies to all continuous differentiable functions, so certainly to f', no?- not sure never seen it applied to f' before?

- #9

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ok done that and i get that R3 = r2 ?

- #10

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You should start by writing out the remainders...

- #11

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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 dont get it ?

- #12

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That should be f'''(d)h^2/4, with [itex]d\in [a,a+h/2][/itex]Ok now ive 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

[/QUOTE]

Thus

You should get (after cancelling the h): 2f'''(c)=3f'''(d), with [itex]c\in [a,a+h][/itex] and [itex]d\in [a,a+h/2][/itex].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 happens if [itex]h\rightarrow 0[/itex]?

- #13

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so do we get c and d are both a as h->0?

- #14

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

so do we get c and d are both a as h->0?

- #15

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

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