Consider the taylor series expansion of about the point ,
which is
Now , I want to ask can we write this as ?
That is, we are compressing all the infinite terms into a single term, where the derivative is evaluated about any other point in the domain of .
In this case, for example I modified the third term, but it could be done for any finite term in the series.
Can we prove this that for any particular , we can simply write the summation into above form?
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There are many forms of the exact remainder for a Taylor’s series. The one closest to what you want is this....
For a sufficiently differentiable function, and for any a,x and n, there exists c between a and x such that f(x)=j=0∑n−1j!f(j)(a)(x−a)j+n!f(n)(c)(x−a)n
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Yes sir, this is what I wanted...but I wanted to know the proof of the claim or if you could write a proof of it...or you could provide a reference from where I can read more about it.
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Put F(t)=j=0∑n−1j!f(j)(t)(x−t)jG(t)=(x−t)n Then F(a)=j=0∑n−1j!f(j)(a)(x−a)jF(x)=f(x)F’(t)=(n−1)!f(n)(t)(x−t)n−1 While G(a)=(x−a)nG(x)=0G’(t)=−n(x−t)n−1 Now use the Generalised Mean Value Theorem, to find c between a and x such that (F(x)−F(a))G’(c)=(G(x)−G(a))F’(c) Most textbooks on real analysis should give proofs of Taylor’s Theorem with remainder...
@Mark Hennings, @Chew-Seong Cheong, @David Vreken , @Chris Lewis