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Approximating f(x - 1) in Terms of Its First and Second Derivatives

Date: 08/15/2012 at 11:31:57
From: Ajith
Subject: Approximation of taylor's series

When I was going through a proof of a theorem, a standard text made use of
a fact that I couldn't verify:

   f(x - 1) = f(x) - f'(x) + (1/2)f''(e)
Here, e is not the exponential constant. It is just a constant that lies
between x - 1 and x.

I know that any function f(x) can be written as 

   f(a) + (x - a)f'(a) + (((x - a)^2)/2)f''(e)

Here, a < e < x -- assuming the function converges, of course. The
Lagrangian form of the remainder is used.

Substituting x - 1 for a gives us 

   f(x - 1) = f(x) - f'(x - 1) - (1/2)f''(e)

   x - 1 < e < x

Where did I go wrong? The text evidently used Taylor's series to get to
that approximation, but it didn't work for me. First of all, can I
substitute x - 1 in the place of a?

Date: 08/15/2012 at 16:50:46
From: Doctor Schwa
Subject: Re: Approximation of taylor's series

Hi Ajith,

Maybe things would be more clear if you renamed your function.

For example, let g(x) = f(x - 1). Then f(x) = g(x + 1).

Using Taylor's expansion for g(x), we get ...

   g(x) = g(a) + g'(a) (x - a) 
               + 1/2 g''(e) (x - a)^2

... for some e between a and x.

Put all this back in terms of f:

   f(x - 1) = f(a - 1) + f'(a - 1) (x - a)
                       + 1/2 f''(e - 1) (x - a)^2

Now can you see what you want to set a equal to, and how to finish?

- Doctor Schwa, The Math Forum 

Date: 08/15/2012 at 17:15:33
From: Ajith
Subject: Thank you (Approximation of taylor's series)

Yeah. I got it. Thanks a lot :)
Associated Topics:
High School Functions

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