Help needed to understand the following steps in example of series.

Question

Here's the original example

To prove that when $y$ is real and numerically $y<1$.
$$\log_e(1+y)=y-{1\over2}y^2+{1\over3}y^3-{1\over4}y^4+...$$ ad inf.

Here are the first 2 steps of the solution. I understand the first one but not how 2nd one came.

$(1+y)^x=1+x\log_e(1+y)+{\frac{x^2}{2!}}[\log_e(1+y)]^2+...$ this comes from expansion of $a^x$ which I know.

But, since $y$ is real and numerically $<1$, we have
$$\Rightarrow (1+y)^x=1+x.y+\frac {x(x-1)}{1.2}y^2+\frac{x(x-1)(x-2)}{1.2.3}y^3+...$$

How is this second line concluded? Please help me derive this statement not necessarily from first as it seems to independent derivation in the example.

Answer 1

The first line is the Taylor series expanded about $x=0.$ The second line is the Taylor series expanded about $y=0.$ It's just a change of independent variable.

Answer 2

I'll be providing a different approach. Although this does not completely answer your question, it does provide you a relatively easier way to prove the result. Both ultimately somehow are connected as indicated by the coefficients of your infinte sum.

From the Formula for Sum of a Geometric Series we have:

$$\dfrac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^k x^k$$ Now, Integrating term by term gives us $\ln(1+x)$ on the left hand side as follows: $$\ln(1+x)=\int\sum_{k=0}^{\infty}(-1)^kx^k\mathrm dx=\sum_{k=0}^{\infty}(-1)^k\int x^k\mathrm dx=\sum_{k=0}^{\infty}(-1)^k\dfrac{x^{k+1}}{k+1}=x-\dfrac{x^2}{2}+-\cdots$$