# Asymptotic like expansion of log based on its Taylor series [closed]

I am trying to derive a general result on moments generating functions and need to prove that, for any $$K \ge 1$$, the following function:

$$f_K(x) = \frac{\log(1 + x) - \sum_{i = 1}^K \frac{(-1)^{i + 1}}{i} x^i}{(-1)^{K} x^{K + 1}}$$

is decreasing in $$\mathbb{R}^+$$. Simulations show that this function is indeed decreasing but manipulating its derivative is a mess.. any ideas on how to show this result?

• Do you really need it to be increasing on the whole of $(0,\infty)$ or just on $(0,1)$? On $(0,1)$ it equals $\frac{1}{K+1}-\frac{1}{K+2}x+\frac{1}{K+3}x^2-\cdots$ suggesting it is not even increasing there May 31 at 13:54
• Empirically, it decreases for any $K$. (plotting a few graphs). The claim should be false May 31 at 13:57
• Absolutely! I forgot a minus sign.. Or we can prove that it is decreasing in $\mathbb{R}^+$. I'll change the statement! May 31 at 14:00
• The result holds for all $\mathbb{R}^+$.. I would love to have a proof for $(0, \infty)$ May 31 at 14:01
• I learned a lot! Thank you everyone for your help, it is the first time I see this approach! May 31 at 14:17

Consider the numerator $$g(x) = \log(1 + x) - \sum_{i = 1}^K \frac{(-1)^{i + 1}}{i} x^i$$ first. We have $$g'(x) = \frac{1}{1+x} - \sum_{i = 0}^{k-1} (-x)^i = \frac{1}{1+x} - \frac{1-(-x)^K}{1+x} = \frac{(-1)^K x^K}{1+x}$$ and therefore $$\tag{*} f_K(x) = \frac{1}{x^{K+1}} \int_0^x \frac{t^K}{1+t} \, dt \, .$$

Inspired by Oliver Díaz' answer we now can substitute $$t = x u$$ and get $$\boxed{f_K(x) = \int_0^1 \frac{u^K}{1+xu} \, du}$$ which is clearly decreasing in $$x$$.

Alternatively we can compute the derivative of $$f_K$$ in $$(*)$$: $$f_K'(x) = \frac{1}{x(1+x)} - \frac{K+1}{x^{K+2}} \int_0^x \frac{t^K}{1+t} \, dt \, .$$ The right-hand side becomes larger if $$t$$ in the denominator of the integral is replaced by $$x$$, so that $$f_K'(x) < \frac{1}{x(1+x)} - \frac{K+1}{x^{K+2}} \int_0^x \frac{t^K}{1+x} \, dt = 0 \, .$$ This proves that $$f_K$$ is strictly decreasing on $$(0, \infty)$$.

• Beautiful derivation, thank you! May 31 at 14:18

Here is a slightly different solution based on the integral form of the Taylor approximation of function.

Theorem: Suppose $$f$$ is a function defined in an open interval $$I$$ containing a point $$a$$, and that$$f$$ has continuous derivative of order $$n+1$$. Then, for any $$x\in I$$ $$f(x) = \sum^n_{k=0}\frac{f^{(k)}(a)}{k!}(x-a)^k+ E_n(x)$$ where $$E_n(x)=\frac{1}{n!}\int^x_a(x-t)^nf^{(n+1)}(t)\,dt$$

This can be shown by induction on $$n$$; many textbooks on Calculus and real analysis discuss it; also, there are a few postings about it in MSE.

The change of variable $$t=t(u)=x+u(a-x)$$ yields another expression for $$E_n(x)$$ which is more useful for the OP's problem:

$$E_n(x)=\frac{(x-a)^{n+1}}{n!}\int^1_0 u^n f^{(n+1)}(x+u(a-x))\,du$$

In the case of the OP, $$f(x)=\log(1+x)$$ and $$a=0$$. Notice that the expression of interest is \begin{align} R_K(x):=\frac{E_K(x)}{(-1)^K x^{K+1}}&=\frac{1}{(-1)^K x^{K+1}}\frac{x^{K+1}}{K!}\int^1_0u^K \frac{(-1)^K K!}{(1+x(1-u))^{K+1}}\,du\\ &=\int^1_0\frac{u^K}{(1+x(1-u))^{K+1}}\,du \end{align} From this it is clear that $$R_K$$ is a decreasing function of $$x>0$$, since the integrand in the last expression above is positive and a decreasing function in $$x$$.

If one must, differentiation yields \begin{align} R'_K(x)&=\frac{\partial}{\partial x}\int^1_0\frac{u^K}{(1+x(1-u))^{K+1}}\,du\\ &=\int^1_0\frac{\partial}{\partial x}\frac{u^K}{(1+x(1-u))^{K+1}}\,du\\ &=-(K+1)\int^1_0\frac{u^K(1-u)}{(1+x(1-u))^{K+2}}\,du<0 \end{align}

Edit: to reconcile the reminder obtained by the the theorem above and the one obtained by @MartinR's using first principles, notice the substitution $$v=\frac{x-t}{1+t}=\frac{x+1}{1+t}-1$$ in $$E_n(x)$$, equivalently $$t=\frac{x+1}{v+1}-1$$, yields $$dt=-\frac{x+1}{(v+1)^2}dv$$ and $$E_n(x)=(-1)^n\int^x_0 v^n\frac{v+1}{x+1}\frac{x+1}{(1+v)^2}\,dv=(-1)^n\int^x_0\frac{v^n}{1+v}\,dv$$

• That substitution is elegant! If I apply that to my derivation of $f_k$ then I get $$f_K(x) = \frac{1}{x^{K+1}} \int_0^x \frac{t^K}{1+t} \, dt = \int_0^1 \frac{u^K}{1+xu} \, du$$ and now I will try to figure out what that is different from your $$R_K(x) = \int^1_0\frac{u^K}{(1+x(1-u))^{K+1}}\,du \, .$$ May 31 at 17:04
• I have verified it for $K \le 6$ with a symbolic calculator (Maxima), so it is definitely correct :) – Do you mind if I use your substitution and add $f_K(x) = \int_0^1 \frac{u^K}{1+xu} \, du$ to my answer? May 31 at 18:10
• @MartinR: not at all, If you think it will make your solution even better, go for it! May 31 at 18:11
• @MartinR: I as mentioned earlier, the substitution $v=\frac{x-t}{t+1}$ in my expression for $E_n$ yields the expression for the remainder that you obtained by first principles. May 31 at 19:23