Taylors theorem

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  • #1
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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 ...
 

Answers and Replies

  • #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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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)...
 
  • #5
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errmm f(x) = f(a) + f'(a)(x-a)?
 
  • #6
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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!
 
  • #7
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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?
 
  • #8
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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? :smile:
 
  • #9
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ok done that and i get that R3 = r2 ?
 
  • #10
22,089
3,293
You should start by writing out the remainders...
 
  • #11
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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

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
22,089
3,293
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
That should be f'''(d)h^2/4, with [itex]d\in [a,a+h/2][/itex]
[/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 [itex]c\in [a,a+h][/itex] and [itex]d\in [a,a+h/2][/itex].

Now, what happens if [itex]h\rightarrow 0[/itex]?
 
  • #13
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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?
 
  • #14
22,089
3,293
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...
 
  • #15
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it = 0! so obvious now, thanks a lot !!
 

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