TITLE: Taylor Polynomial residue term
QUESTION [2 upvotes]: A function $f:\mathbb R \to \mathbb R$ given as $f(x)=\sqrt{2x-1}.$
I have found that the third degree taylor polynomial is 
$P_3(x) = x-\frac12(x-1)^2+\frac12(x-1)^3$ in point $0.$
I have found out that the residue term for this polynomial is 
$$R_3(x)=-\frac58\cdot \frac{1}{(2z-1)^{\frac72}}\cdot (x-1)^4 $$ for $z$ betwwen 1 and $x.$ 
Then I am trying to show by taking the absolute value of $R_3(x)$ that the wrror which occurs by using $P_3(\frac32)$ instead of $f(\frac32)$ is less than or equal to $\frac5{2^7}.$
What I have done so far is: 
$$|R_3(x)|=|-\frac58\cdot \frac{1}{(2z-1)^{\frac72}}\cdot (x-1)^4 |=\frac58\cdot \frac{1}{(2z-1)^{\frac72}}\cdot (x-1)^4 $$ Then I am having trouble. What is $(x-1)^4 $ less than ? what about $\frac{1}{(2z-1)^{\frac72}}$?

REPLY [1 votes]: Hint
$2>|2z-1|>1$ implies
$$
|R_3(x)| < \frac{5}{8} \frac{1}{1^{7/2}}(3/2-1)^{4} = \frac{5}{2^7}
$$
{"set_name": "stack_exchange", "score": 2, "question_id": 1931219}