Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (h)

MATH104 December 2012

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Question 01 (h)
Suppose you approximate $\cos(0.25)$ by $L(0.25)$, where $L(x)$ is the linear approximation of $\cos(x)$ at $x=0$. Provide an estimate for the absolute value of the error in your approximation.

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
Recall that an upper bound on the error in a linear approximation is given by $|R_{1}(x)|\leq {\frac {M}{2}}|x-0|^{2}$ where $\displaystyle R_{1}(x)$ is the error in a linear approximation at the point x and $\displaystyle M$ is an upper bound on the second derivative, that is $\displaystyle |f''(x)|\leq M$ for all values of x inside our interval $\displaystyle [0,0.25]$.

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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. Following the hint, we first find a suitable value for M. To do this, we need the second derivative. Computing gives ${\begin{aligned}f(x)&=\cos(x)\\f'(x)&=-\sin(x)\\f''(x)&=-\cos(x)\end{aligned}}$ and thus ${\begin{aligned}|f''(x)|=|-\cos(x)|\leq 1=M\end{aligned}}$ where the above bound holds for any value of x. Plugging into the upper bound on the error in a linear approximation as given by the hint, we have $|R_{1}(0.25)|\leq {\frac {M}{2}}|0.25-0|^{2}={\frac {1}{2}}|0.25-0|^{2}={\frac {1}{32}}$ completing the question.

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