Taylor´s Inequality ln(1.25) (Calculus II)?
Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the value of x to be less than 0.001.
I have that the Absolute value of R sub n (1.25)= ( 1/z^(n+1) )*( (1.25)^(n+1) / (n+1) )
which should be <0.001. But because 0<z<1.25 I don`t get what value should I pick and I don`t know if the value x=1.25 really maximizes
If I were to pick some number z that would give me the maximum on [0,1.25], the first thing that comes to my mind is that I could pick z=0.1,.0001 and keep decreasing the values and the term 1/z will get bigger and bigger. Any ideas? Thanks in Advance and Thank You very Much.!
I think that 1/z^(n+1) can be written as (1/z)^(n+1) am I wrong?
abs()= absolute value.
Re: Taylor´s Inequality ln(1.25) (Calculus II)?
for an alternating series, error magnitude < |first omitted term|