In Boas it's stated that $$ln(p+y\sqrt{p})$$ can be expanded as $$ln(p+y\sqrt{p}) = ln(p)+ln\left(1+\frac{y}{\sqrt{p}}\right) = ln(p)+\frac{y}{\sqrt{p}}-\frac{y^2}{2p}$$
At first I tried to expand it like $$z(x,y) = z(x_0,y_0) + \frac{\partial z}{\partial x}\Big|_{(x_0,y_0)}(x-x_0) + \frac{\partial z}{\partial y}\Big|_{(x_0,y_0)}(y-y_0)$$
but I got $$ln(p) + \frac{1}{p+y\sqrt{p}}(y-y_0) + \frac{2\sqrt{p}+y}{2p^{3/2}+2yp}\left(p-p_0\right)$$
So my two questions are,
What did I do wrong when expanding? Not include enough terms? Not simplify enough (I tried this for awhile)
I can understand that boas used a series for ln(1+x) from the 2nd to 3rd step, but how did they go from the first expression to the second?
Any help would be greatly appreciated!
The transisition from your step 2 to 3 is the logarithm functional equation $\ln(xy) = \ln x + \ln y,$ valid for $x,y>0.$ $$\ln(p+y\sqrt{p})=\ln(p+\frac{yp}{\sqrt{p}}) = \ln\left(p\left(1+\frac{y}{\sqrt{p}}\right)\right) = \ln(p)+\ln\left(1+\frac{y}{\sqrt{p}}\right)$$ $$ = \ln(p)+\frac{y}{\sqrt{p}}-\frac{y^2}{2p} + O\left(\left(\frac{y}{\sqrt{p}}\right)^3\right)$$