# For small $x$, one has $\ln(1+x)=x$? [duplicate]

What does it mean that for small $x$, one has $\ln(1+x)=x$? How can you explain this thing ? Thanks in advance for your reply.

• Look at the first order Taylor approximation. – user137731 Oct 15 '14 at 20:39

Since no one else said it clearly in words. One does not have that $$\ln(1 + x) = x$$ for small $x$. One does, however, have that for small values of $x$, $\ln(1+x)$ can be approximated by $x$. As the other answers have already pointed out, this you see from the Taylor expansion $$\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \dots.$$ Now if $x$ is a small enough number, then $x^2, x^3, \dots$ are all insignificant. And so for small $x$ you can approximate $\ln(1+x)$ by $x$.

If you write its Taylor's expansion then you have: $\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{k}$. For small values of $x$, the values of $x^2, x^3,...$ are small in comparing with $x$(note that positive numbers which are less than 1, will decrees as we multiply themselves), hence we can ignore the terms with degree larger than $1$, and estimate $\ln{(1+x)}$ as its first degree part which is $x$.

Take the tangent line at of $f(x) = \ln(1+x)$ in $x = 0$.

\begin{align*} f(x) & \approx f(0) + f'(0) (x - 0) \\ & = \ln(1+0) + \left[\frac{d}{dx} \ln(1+x)\right]_{x = 0} (x-0) \\ & = 0 + 1 x \\ & = x \end{align*}

Look at the equivalent $\text{exp}(x) = 1 + x$ ($x$ teensy) and compare it to the Taylor series.

Hint:
One can prove [1] that $$f(x) \approx f(a) + f'(a)(x - a).$$ for $x$ near $a$.

First, let me parrot Thomas' answer: it is not true that $$\log(1+x) = x$$ for small $$x$$. This equation holds only when $$x=0$$. However, it is true that if $$x$$ is small, then $$\log(1+x)$$ is "well-approximated" by $$x$$. Slightly more formally, the function $$x \mapsto x$$ is the best linear approximation of the function $$x \mapsto \log(1+x)$$ for values of $$x$$ which are "near zero".

### Derivatives and Linear Approximation

The essential idea is that if a function $$f$$ is differentiable, then the best linear approximation of that function near a point $$a$$ is the line with slope $$f'(a)$$ which passes through the point $$(a, f(a))$$. An equation for this line is given by $$f(x) \approx f(a) + f'(a)(x-a) =: \ell(x).$$ What this means is that if $$x$$ is "near" $$a$$, then the value of $$f(x)$$ is "near" the value of the function $$\ell(x)$$, where the the graph of $$\ell$$ is a line. All of this can be made rigorous by careful examination of the definition of the derivative.

As others have pointed out, taking $$f(x) = \log(1+x)$$ and $$a=0$$, this becomes \begin{align} \log(1+x) &\approx \log(1+0) + \left[ \frac{\mathrm{d}}{\mathrm{d}x} \log(1+x)\right]_{x=0} (x-0) \\ &= \log(1) + \left[\frac{1}{1+x}\right]_{x=0} x \\ &= 0 + \frac{1}{1+0} x \\ &= x. \end{align} Hence when $$x$$ is "near" $$0$$, then $$\log(1+x)$$ is "near" $$x$$.

As others have noted, a better approximation for the value of a function "near" some point can be obtained by evaluating the Taylor series expansion of the function near that point—this is the approach of Thomas' answer.

### From the Integral Definition of $$\log$$

The definition of $$\log(x)$$ which I find most compelling is the following: for all $$x \in (0,\infty)$$, define $$\log(x) := \int_{1}^{x} \frac{1}{t}\,\mathrm{d}t.$$ Note that we can use this definition to obtain all of the other properties we might like (e.g. $$\log$$ is the inverse of $$\exp$$; $$\log$$ has a particular Taylor series expansion; etc). I want to use this definition to show that when $$x$$ is "small", then the error $$|\log(1+x)-x|$$ is also "small". For any $$x \in [0,\infty)$$, we have \begin{align} |\log(1+x) - x| &= | x - \log(1+x) | \\ &= \left| x - \int_{1}^{1+x} \frac{1}{t} \,\mathrm{d}t \right| \\ &= \left| \int_{1}^{1+x} 1 \,\mathrm{d}t - \int_{1}^{1+x} \frac{1}{t} \,\mathrm{d}t \right| \\ &= \int_{1}^{1+x} \frac{t-1}{t}\,\mathrm{d}t && \text{(t \ge 1 \implies \tfrac{t-1}{t} \ge 0, thus |\cdot| is redundant)}\\ &\le \int_{1}^{1+x} t-1 \,\mathrm{d}t && \text{(t \ge 1 \implies \tfrac{t-1}{t} \le t-1)} \\ &= \left[ \frac{1}{2} t^2 \right]_{t=1}^{1+x} - x \\ &= \frac{1}{2}(1+x)^2 - \frac{1}{2} - x \\ &= \frac{1}{2} + x + \frac{1}{2} x^2 - \frac{1}{2} - x \\ &= \frac{1}{2} x^2. \end{align} This means that if $$x \ge 0$$, then the error between $$\log(1+x)$$ and the approximation $$x$$ will be smaller than $$\frac{1}{2}x^2$$. If $$x$$ is close to zero (say, $$x < 1$$), then $$\frac{1}{2}x^2$$ will be smaller than $$x$$, so the approximation here looks pretty good.

For $$x \in (-1, 0)$$, the situation is a little more complicated, as the integral "blows up" as $$x \to -1$$. Hence we need to constrain $$x$$ a little more. Since the goal is to show that if $$x$$ is "near" $$0$$, then $$\log(1+x)$$ is "near" $$x$$, we might as well assume that $$x \ge -\frac{1}{2}$$. Under this assumption, we have \begin{align} |\log(1+x) - x| &= \int_{1}^{1+x} \frac{t-1}{t} \,\mathrm{d}t && \text{(as above)} \\ &\le 2 \int_{1}^{1+x} t-1\,\mathrm{d}t && \text{(t \ge 1+x \ge \tfrac{1}{2} \implies \tfrac{t-1}{t} \le 2(t-1))} \\ &= x^2. \end{align} This says that if $$x \in (-\frac{1}{2}, 0)$$, then the error between $$\log(1+x)$$ and $$x$$ is smaller than $$x^2$$. Again, when $$x$$ is small, $$x^2$$ is even smaller.

Summarizing the above, we have the following relatively "nice" statement:

If $$|x| \le \frac{1}{2}$$, then the error between $$\log(1+x)$$ and $$x$$ is smaller than $$x^2$$. That is, $$\operatorname{error}(x) = |\log(1+x) - x| \le x^2.$$ The closer $$x$$ is to zero, the smaller this error will be.