# Using Taylor's Theorem to show that $\ln(1 + x^2) \leq x^2$

Can we show that if $\operatorname{abs}(x) \lt 1$, then $$\ln(1+x^2) \leq x^2\;,$$ using Taylor's Theorem?

I am thinking of expanding it about $x=0$ but I got something like $$f(x) = -x^2 + \frac{x^4}{2} - \dots$$

Is my approach correct? Could you give me some hints/guides here?

Thanks.

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$$\begin{array}{rcl} \ln(1 + x^2) & \le & x^2 \\ e^{\ln(1 + x^2)} & \le & e^{x^2} \\ 1 + x^2 & \le & e^{x^2} \\ 1 + x^2 & \le & 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \\ {} 0 & \le & \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \dots \\ {} \end{array}$$

Which is true for all real $x$.

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This is not a proof! Using this kind of argument you can prove that $\pi=e$: (i) $\pi=e$; (ii) $0\cdot\pi=0\cdot e$; (iii) $0=0$. Now (iii) is true, so (i) has to be true. – That this answer obtained 8 votes is beyond me. – Christian Blatter Apr 21 '13 at 13:16
@ChristianBlatter Thank you for stating your opinion. Can you please clearly tell me at what step I was wrong, and why? – hkBattousai Apr 21 '13 at 17:34
Your argument is of the form "$\ A\Rightarrow B\Rightarrow \ldots\Rightarrow Z\Rightarrow {\rm True}$; therefore $A$ has to be true". In reality you would need such a chain where the arrows point into the opposite direction. – Christian Blatter Apr 21 '13 at 17:46
I started with the given expression and did the same operations to the both sides. I can't find any invalid algebraic manipulation in my derivation. – hkBattousai Apr 21 '13 at 21:48
You correctly proved that (a) $\log(1+x^2)\leq x^2$ implies (b) ${x^4\over2!}+{x^6\over3!}+\ldots\ \geq0$. This is not a proof of (a). – That's my last word on this matter. – Christian Blatter Apr 22 '13 at 10:02

Yes, but it is a nuisance to use the Taylor expansion of $\log(1+x^2)$. Instead, let $t=x^2$, and look at what the Taylor expansion of $\log(1+t)$ tells you for non-negative $t$.

Use the Lagrange form of the remainder. The error when you truncate the expansion of $f(t)$ at the linear term is equal to $\frac{1}{2!}f''(\xi)t^2$, where $\xi$ is a number between $0$ and $t$. Since the second derivative in our case is negative, truncating at the $t$ term gives us a negative error term, meaning that the first term $t$ overestimates $\log(1+t)$.

Or else if you are familiar with alternating series, you can get the same result.

There are plenty of other ways to prove the inequality. For example, let $g(t)=\log(1+t)-t$. We have $g(0)=0$, and $g'(t)=\frac{1}{1+t}-1$. So for $t\gt 0$, the function $g(t)$ is decreasing, and therefore $g(t)\lt 0$ if $t\gt 0$. That shows $\log(1+t)\lt t$ if $t\gt 0$.

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Sure.

$$\ln(1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots$$

In particular, the terms alternate sign. If we subtract $x^2$, we are left with

$$\ln(1 + x^2) - x^2 = -\frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \ldots$$

The first term of the sum is negative, and since the sum is a strictly alternating sum whose terms go to $0$ (since $|x| < 1$), the error between a sum of the first $n$ terms and the actual value is bounded by the $n+1$st term. Here, we only need to note that the first term is negative and the second term is smaller in absolute value - thus we conclude that

$$\ln(1 + x^2) - x^2 \leq 0$$

with equality only when $x = 0$.

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As $1+t^2\ge 1$ $$\ln{\left(1+x^2\right)}=\int_{0}^{x}{\frac{2t}{1+t^2}dt}\le\int_{0}^{x}{\frac{2t}{1}dt}\le x^2$$

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