What is the second-order Taylor expansion of log(1 + x) about x = 0?

## The general recipe A Taylor expansion approximates a function near a chosen base point by a polynomial that matches the function's value and its first few derivatives there. Around `x = 0` (the Maclaurin case), the second-order expansion is ``` f(x) ~ f(0) + f'(0)*x + (1/2)*f''(0)*x^2 ``` The logic is simple: the constant term makes the polynomial pass through the right height, the linear term makes it leave at the right slope, and the quadratic term makes it bend at the right rate. Each extra term buys one more derivative of agreement. ## Applying it to log(1 + x) Take `f(x) = log(1 + x)`, the natural logarithm. - `f(0) = log(1) = 0` - `f'(x) = 1/(1 + x)`, so `f'(0) = 1` - `f''(x) = -1/(1 + x)^2`, so `f''(0) = -1` Substituting: ``` log(1 + x) ~ 0 + 1*x + (1/2)*(-1)*x^2 = x - x^2/2 ``` The full series continues `x - x^2/2 + x^3/3 - x^4/4 + ...`, converging for `-1 < x <= 1`. Truncating after the quadratic term leaves a remainder whose leading piece is `+x^3/3`. ## How good is it? Numbers make the error scale concrete. At `x = 0.1`, `log(1.1) = 0.0953102...`; the linear model gives `0.1` (error about `0.0047`) and the quadratic model gives `0.095` (error about `0.00031`). The quadratic term removed roughly 94% of the error. Now halve the displacement: at `x = 0.05`, `log(1.05) = 0.0487902...` and the quadratic model gives `0.048750`, an error near `0.00004` — about one eighth of the error at `x = 0.1`. That eightfold shrink is the signature of a **cubic** leading error: halve `x` and `x^3` falls by `2^3 = 8`. This is the single most useful thing to take away. A first-order model has error of order `x^2` (halving `x` quarters the error); a second-order model has error of order `x^3` (halving `x` cuts it eightfold). Bigger displacements degrade a second-order model fast in the other direction too: at `x = 0.5` the quadratic model gives `0.375` against a true `0.405`, an error of `0.03` — two orders of magnitude worse than at `x = 0.05`. Taylor accuracy is a *local* promise. ## The companion expansion for e^x The exponential is the other expansion worth knowing cold. With `f(x) = e^x`, every derivative is `e^x`, and every derivative at `0` equals `1`, so ``` e^x ~ 1 + x + x^2/2 ``` with the series continuing `+ x^3/6 + x^4/24 + ...`. Note the constant term: unlike `log(1 + x)`, which vanishes at `0`, the exponential starts at `1`. A frequent slip is to write `e^x ~ x + x^2/2`, dropping the constant, or to give `log(1 + x)` a leading `1` it does not have. The two are near-inverses of each other around zero, which is a quick sanity check: substituting the quadratic model for `x` in the other and keeping terms up to `x^2` returns `x` in both directions. ## Why this appears in interviews Three reasons. First, it is a check that you can actually differentiate and assemble a Taylor polynomial rather than recite one. Second, `log(1 + x) ~ x` is the workhorse behind small-change approximations — small log-returns approximating percentage changes, small log-likelihood differences, small-`x` reasoning in derivations — and knowing the next term tells you when the shortcut starts to lie. Third, the error-scaling argument generalises directly: the same reasoning explains why any second-order model of a multivariable surface is trustworthy only within a small displacement of its base point, with error growing like the cube of that displacement. ## Things to state carefully Say *about which point* the expansion is taken; `x - x^2/2` is specific to a base point of `0` and says nothing about the behaviour near, say, `x = 3`. Say which logarithm: this is the natural log, and a base-10 version would carry an extra factor `1/log(10)`. And note the domain: `log(1 + x)` is undefined for `x <= -1`, and the series only converges on `-1 < x <= 1`, so the expansion is not an approximation you can push arbitrarily far in the negative direction.