What is the correct range for c in Taylor's Theorem error bound?

[member 508213]

For the error bound for taylor's theorem, for the n+1 derivative evaluated at some c which maximizes the derivative my textbook says c must be between a and x..but today my teacher said that c must be between absolute value x and negative absolute value x, which is different than I thought.

An example would be calculating the error of using a second degree taylor polynomial to estimate e^x at x=-1...the n+1 derivative would be e^x, so the question would be do I use 0 because 0 maximizes e^x on [-1,0] or do I use 1 because of absolute value x being 1 and 1 maximizes e^x on [-1,1].

Hopefully my question makes sense, just to reiterate I am wondering if c is between a and x (which is what textbook says and is what I thought in the past) or between absolute value x and negative absolute value x.

Additionally I already tried to talk to my teacher to clarify and he insisted it must be between absolute value x and negative absolute value x...but in the past I learned it was x and a which is confirmed by my book.

Any help on this is appreciated

 

[mfb]

How did you define a and the variable you call x (despite using it as free variable at the same time)?

 

[member 508213]

How did you define a and the variable you call x (despite using it as free variable at the same time)?

Sorry for not being more clear, a is where the polynomial is centered at and x is where it is being evaluated at... In my example above the polynomial is centered at 0 (forgot to say that sorry) and we are evaluating it at x=-1

 

[mfb]

Between a and x then.

If a=0, then "between -|x| and |x|" is a weaker statement. It is not wrong, but sometimes (like here) it leads to a weaker estimate.

 

[member 508213]

Between a and x then.

If a=0, then "between -|x| and |x|" is a weaker statement. It is not wrong, but sometimes (like here) it leads to a weaker estimate.

Yes thank you that's what I thought, is there ever a situation where you would need to use abs value x to negative abs value x? Because I am trying to figure out why my teacher said that when it seems like a to x works and is more accurate

 

[mfb]

Maybe if you want an upper limit for the whole range of the expansion.