Science:Math Exam Resources/Courses/MATH104/December 2011/Question 06 (b)

MATH104 December 2011

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Question 06 (b)

In this problem, you will approximate $\ln(0.9)$ in two ways: one using linear approximation and the other using quadratic approximation. You may find it useful to draw the graph of $\ln(x)$ for $x$ near 1 to help you answer some parts of this question. b) Use the formula $|{\mbox{Error}}|\leq {\frac {M}{2}}(x-a)^{2},$ where $M$ > 0, to estimate an error bound for your approximation of $\ln(0.9)$ in part (a). It will be useful to remember how to find such an $M$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How does M relate to the second derivative of f(x)?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Here we are working with $x=0.9$ and $a=1$. We find that the error takes the following value $|{\text{Error}}|\leq {\frac {M}{2}}(0.9-1)^{2}={\frac {M}{200}}$ where $M$ is the maximum value of the absolute value of the second derivative on the interval [0.9, 1]. In other words, we would like to estimate how large the second derivative can be on this interval. Given that $|f''(x)|=\left|{\frac {-1}{x^{2}}}\right|={\frac {1}{x^{2}}}$ and since this is a decreasing function on the interval [0.9, 1] we can deduce that the maximum value it takes on the interval is at $x=0.9$ and hence $M={\frac {1}{0.9^{2}}}={\frac {100}{81}}$ This allows us to find the upper bound on the error: $|{\text{Error}}|\leq {\frac {\frac {100}{81}}{200}}={\frac {1}{162}}$

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