91Ó°ÊÓ

The Taylor polynomial expansion of a function of the form \(f(x) = \ln(1+x)\) is given by: \[P_n(x) = f(0) + f'(0)(x - 0) + \frac{f''(0)}{2!} (x - 0)^2 + ... + \frac{f^{(n)}(0)}{n!} (x - 0)^n\] Now we need to find the first- and second- order derivatives of the function \(f(x) = \ln(1+x)\): First derivative: \(f'(x) = \frac{1}{1+x}\), so \(f'(0) = 1\), Second derivative: \(f''(x) = -\frac{1}{(1+x)^2}\), so \(f''(0) = -1\).

Using the Taylor polynomial expansion up to the second-order term, we have: \[P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = 0 + 1\cdot x - \frac{1}{2}x^2\] Now we can substitute \(x = 10^{-20}\) and compare with the given inequality: \(P_2(10^{-20}) = 10^{-20} - \frac{1}{2}(10^{-20})^2\).

Now we can show that the inequality holds: 1. Statement: \(\frac{1}{10^{20}+1} < \ln(1+10^{-20})\) Proof: Since \(P_2(10^{-20}) = 10^{-20} - \frac{1}{2}(10^{-20})^2\) is smaller than the true value of \(\ln(1+10^{-20})\), and \(\frac{1}{10^{20}+1} < 10^{-20}-\frac{1}{2}(10^{-20})^2\), it follows that \(\frac{1}{10^{20}+1} < \ln(1+10^{-20})\). 2. Statement: \(\ln(1+10^{-20}) < \frac{1}{10^{20}}\) Proof: Since \(P_2(10^{-20}) = 10^{-20} - \frac{1}{2}(10^{-20})^2\) is smaller than the true value of \(\ln(1+10^{-20})\), and \(10^{-20} - \frac{1}{2}(10^{-20})^2 < \frac{1}{10^{20}}\), it follows that \(\ln(1+10^{-20}) < \frac{1}{10^{20}}\). Hence, we have shown that: \(\frac{1}{10^{20}+1} < \ln(1+10^{-20}) < \frac{1}{10^{20}}\)