Show that for all $x>0$ we have $\ln(1+x)>x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$.
I know it has to do with taylor expansion but somehow I cannot prove it rigorously.
Show that for all $x>0$ we have $\ln(1+x)>x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$.
I know it has to do with taylor expansion but somehow I cannot prove it rigorously.
Take the derivative of the LHS minus the RHS to get
$\frac{1}{1+x} - 1 + x - x^2 + x^3 = \frac{x^4}{x+1}$ which is clearly nonnegative.
So the difference is monotone increasing. Since both sides are equal at $0$, the LHS is always $\ge$ the RHS.
Hint: the two sides are equal at $x=0$. What about the derivatives?
Another way of showing that, having in mind that $\ln(x)$ is differentiable infinite many times on $(0,\infty)$, is looking at the remainder. First - $\ln(1+x)=\sum_{k=1}^n (-1)^{k+1} \frac{x^k}{k!} +R(c,n)$ Where $R(c,n)=\frac{[\ln(1+x)]^{(n+1)}\big|_{x=c}}{(n+1)!}\cdot x^{n+1}$ for $c \in [0,x]$
When plugging $n=4$ we get that $R(n,c)>0$, as desired.