Show that $\log(1+y) \approx y- \frac {y^2}2 + \cdots$ without Taylor Series

Question

For small $y$, prove that $\log(1+y)\approx y -\frac {y^2}2 + \cdots $

I have no idea to solve it.

Answer 1

I'm sure you know the geometric series

$$\sum_{n=0}^{\infty}(-y)^n=1-y+y^2-y^3+\ldots=\frac{1}{1+y},\qquad |y|<1\tag{1}$$

Now integrate both sides of $(1)$, and the result follows. For small $y$, the higher power terms become negligible.

Answer 2

An alternative is to assume that $\ln (1+y) \approx a + by + cy^2$ and then differentiate twice and use $y=0$ to get $\ln (1+y) \approx y - \frac{y^2}{2}$.

Answer 3

One approach directly based on a definition of $\log x$ is as follows. We use the following definition of $\log x$ $$\log x = \lim_{n \to \infty}n(x^{1/n} - 1)$$ Replacing $x$ by $(1 + x)$ and for $|x| < 1$ using binomial theorem we get \begin{align} \log (1 + x) &= \lim_{n \to \infty}n((1 + x)^{1/n} - 1)\notag\\ &= \lim_{n \to \infty}n\left(\frac{x}{1!}\frac{1}{n} - \frac{x^{2}}{2!}\frac{1}{n}\left(1 - \frac{1}{n}\right) + \frac{x^{3}}{3!}\frac{1}{n}\left(1 - \frac{1}{n}\right)\left(2 - \frac{1}{n}\right) + \cdots\right) \notag\\ &= \lim_{n \to \infty}\left(x - \frac{x^{2}}{2!}\left(1 - \frac{1}{n}\right) + \frac{x^{3}}{3!}\left(1 - \frac{1}{n}\right)\left(2 - \frac{1}{n}\right) + \cdots\right)\notag\\ &= x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots\notag \end{align} The last step where I took limit of an infinite series term by term requires some justification which might not be suitable at your stage of learning (as guessed from the content of question and comments to it). Moreover since you need only an approximation it is better to just take the terms till $x^{2}$ in the binomial expansion and then take limit.

Another approach is available here.

Answer 4

For any $y\in\mathbb{R}$ such that $|y|<1$ we have: $$ \log(1+y) = \int_{0}^{y}\frac{dx}{1+x} \tag{1}$$ but if $|x|<1$ we also have: $$ \frac{1}{1+x} = 1-x+\frac{x^2}{1+x}\tag{2} $$ and: $$ \log(1+y) = \color{red}{y-\frac{y^2}{2}}+\color{blue}{\int_{0}^{y}\frac{x^2}{1+x}\,dx} \tag{3}$$ where, by the Cauchy-Schwarz inequality: $$\left|\color{blue}{\int_{0}^{y}\frac{x^2}{1+x}\,dx}\right|^2 \leq \int_{0}^{y}\frac{dx}{(1+x)^2}\int_{0}^{y}x^4\,dx = \frac{y^6}{5(1+y)^2}.\tag{4} $$

Truth to be told, $(3)$ is just an instance of Taylor's formula with integral remainder: $$ \forall |y|<1,\qquad \log(1+y)=\color{red}{\sum_{k=0}^{n}\frac{(-1)^k}{k}\,y^k}+\color{blue}{(-1)^{n+1}\int_{0}^{y}\frac{x^{n+1}}{x+1}\,dx}.\tag{5} $$