[General boards] [Winter 2019 courses] [Fall 2018 courses] [Summer 2018 courses] [Older or newer terms] # Can't find an upper bound for an error

#1

2017 Spring Exam Question 4

From what I have got in question (d) and (e),
p(x) = x + 7/6x(x-1)
0 < ksi < 1
Error formula = (15/8 ksi(^-1/2))/(6) x(x-1)(x-1/4)

With the above information, I’ve concluded that we can’t find the bound in question (f) as ksi = 0 will lead to the bound being undefined. Can someone verify this is a valid conclusion and if not,please explain under what condition we won’t be able to find the upper bound. Thanks!

#2

You cannot use the polynomial error interpolation formula in this case,
since the assumptions of that theorem are not satisfied.
You need f to belong to C^3 in [0,1] to apply the theorem, and f is x^(5/2),
so it is not C^3.
Of course, if you try to apply the formula, with ksi -> 0, you will get infinity.

#3

If we can’t use that theorem, what should we use for the error formula in (e)?

#4

I meant to say that you can’t use the error formula to calculate a bound for the error.
You can still try to write the formula, but when ksi->0 the f^{(3)} is not defined,
so the formula says the error could grow to infinity (unbounded).