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My book is demanding that I show that the Maclaurin polynomial for $\ln(1+x)$ is

$$T_n (x) = x - \frac{x^2}{2} + \frac{x^3}{3}+ \cdots + (-1)^{n-1} \frac{x^n}{n}$$

I don't think this is true at all actually. Following the given formula for finding this

$$f(a) + \frac{f'(a)}{1!}(x)+\cdots$$

well just those two terms fail the test.

$a$ is zero so $\ln0$ is undefined. For the next part I get 1 which is also not on here. What is going on here? My book is suggesting that this is correct but I don't see that.

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$f(x) \ne \ln(x)$, but rather $f(x) = \ln(x+1)$. Plugging in $a=0$ to $f(x)$ does not return $\ln(0)$. – apnorton Jul 21 '13 at 21:35
up vote 3 down vote accepted

Although $a$ is zero, that doesn't mean that $f(a)$ isn't defined. Recall that $f(x) := ln(1 + x)$, so that $f(0) = ln(1) = 0$.

Similarly, $f'(x)) = \frac{1}{x + 1}$, so $f'(0) = 1$.


Edit: For clarification, the formula for a Maclaurin series states that

$$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + ...$$

In this case, $f(x)$ was defined as

$f(x) = ln(1 + x)$

Hence, we see that $f(0) = 0$, $f'(0) = 1$, and $f''(0) = -1$. Thus,

$$f(x) = 0 + \frac{1}{1} x + \frac{-1}{2} x^2 + ... = x - \frac{x^2}{2} + ...$$

as was desired.

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ln1 is similarly bad since I need to get an x not a one, there is no 1 in the thing I need. – Paul the Pirate Jul 21 '13 at 21:34
The computation ln(1) = 0 implies that the constant term in the series is zero. Which is exactly what you want. – user61527 Jul 21 '13 at 21:37
But there is no 1 in the series. – Paul the Pirate Jul 21 '13 at 21:38
@Paul Note that x = 1x. Read the formula for the coefficients again: the coeff of x^k is just $f^k(0) / k!$, so the coefficient of $x = x^1$ is $f'(0)/1! = 1/1 = 1$. – user61527 Jul 21 '13 at 21:40
But isn't x defined as 0? – Paul the Pirate Jul 21 '13 at 21:41

Your book is correct. $f(0) = \ln{(1+0)} = \ln{1} = 0$. As for the other coefficients,

$$f'(0) = \frac{1}{1+x}|_{x=0} = 1$$

$$f''(0) = -\frac{1}{(1+x)^2}|_{x=0} = -1$$

etc. You can easily bulid this up and verify your book's formula.

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Apparently not, but thanks for thinking so. – Paul the Pirate Jul 21 '13 at 21:35
@PaulthePirate: sure you can. Just apply the formula $$\frac{d}{dx} (1+x)^{-n} = -\frac{n}{(1+x)^{-n-1}}$$ – Ron Gordon Jul 21 '13 at 21:36
I really don't understand this, it is too confusing since I have x and a and f(a). f(a) gives x in the fucntion it's value and a points to zero as defined by the Maclaurin function right? – Paul the Pirate Jul 21 '13 at 21:40

Useing the geometric series formula,


Under the substitution $x:-t$, and then integrating with respect to $t$ from $0$ to $x$ we get, $$\ln(x+1)=\sum_{k=1}^n(-1)^{k-1}\frac{x^k}{k}+(-1)^n\int_{0}^x\frac{t^n}{t+1} dt$$

But, $$(-1)^n\int_{0}^x\frac{t^n}{t+1}dt<\int_{0}^xt^{n-1}=\frac{x^n}{n}$$

So we have, $$|\ln(x+1)-T_n(x)|<\frac{x^n}{n}$$

And if $|x|<1$, $\lim_{n\to\infty}\frac{x^n}{n}=0$

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Is it wrong to use the Taylor Polynomial formula with a = 0? – Paul the Pirate Jul 21 '13 at 21:48

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