So, I have to expand $f(x)=\sqrt{x}$ into a series with terms $(x-1)$ in it and then using that series I have to find how many of the serie's terms I have to use so I can calculate $\sqrt{0.9999999995}$ with a precision of 10 digits.
I have solved the first part, though I am stuck in the second one.
Let, $y=x-1$. Then: $$\sqrt{x}=x^{1/2}=(y+1)^{1/2}=\sum^\infty_{n=0}{{1/2}\choose n}*y^n=\sum^\infty_{n=0}{{1/2}\choose n}*(x-1)^n$$
This occured from the Taylor's expansion in $f(x)=(y+1)^{1/2}$.
Now, I can't figure out what the next step should be. Any hint would be appreciated. Thanks in advance!
We have $0.9999999995=1-5\cdot 10^{-10}=1+y$. The Maclaurin series with Lagrange's remainder is $$ \sqrt{0.9999999995}=(1+y)^{1/2}=1+\frac12 y-\frac18(1+\theta y)^{-3/2}y^2,\quad 0\le \theta\le 1. $$ Estimate $$ \left|-\frac18(1+\theta y)^{-3/2}y^2\right|\le\frac18\frac{25\cdot 10^{-20}}{(1-5\cdot 10^{-10})^{3/2}}\le\frac18\frac{25\cdot 10^{-20}}{(1-\frac34)^{3/2}}=25\cdot 10^{-20} $$ to see that it is small enough, then approximate $$ \sqrt{0.9999999995}\approx 1-\frac5210^{-10} $$ and take $10$ digits.