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Given that $f(0.8)=2.2255, f(1.0)=2.7183, \int_{0.8}^{1.0}f(x)e^{-x}dx=0.2$,

I need to find an approximation of $f'(1.0)$ using a method with error of degree $~2~$.

I'm studying for a test and I've seen many questions like this one. My guessing is that I should use Central Difference as this is the only method of error of degree $~2~$ that I studied, though I don't know how to to proceed really.

I'd like to get some intuition on how to approach such questions.

nmasanta
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Tamir Shalev
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  • Yes, only $f(x+h)$ at the numerator, not the derivative. So how do you suggest to approach this? – Tamir Shalev Jul 11 '19 at 11:44
  • @Sorry, I screwed up, I meant $\frac{f(x+h)-f(x-h)}{2h}$. I am afraid you cannot do it that way. The value of the integral has to be used somewhere. What results do you know regarding that? (I don't know any myself) – David Jul 11 '19 at 11:46

1 Answers1

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You have 3 pieces of data, thus you can compute an approximating function with 3 parameters, for ease of application take the product of a quadratic polynomial with the exponential function, $p(x)=e^x(ax^2+bx+c)$. Then you get the 3 equations \begin{align} e^{0.8}(a0.8^2+b0.8+c)&=2.2255\\ e^1(a+b+c)&=2.7183\\ \frac{a}3(1-0.8^3)+\frac{b}2(1-0.8^2)+c(1-0.8)&=0.2 \end{align} or \begin{align} 0.64a+0.8b+c&=1\\ a+b+c&=1\\ 0.81333a+0.9b+c&=1 \end{align}

Solve this and then compute $p'(1)=e^1(3a+2b+c)$.

Lutz Lehmann
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  • Thank you for responding. Could you explain though why this is an approximating function? And why is the error of degree 2? – Tamir Shalev Jul 11 '19 at 13:52
  • You probably can't with any certainty. The function $f$ could be any of $e^x[1+(x-0.8)(x-0.9)(1-x)g((x-0.8)(1-x))]$, one would have to assume that $f$ is nice, slowly changing in all derivatives, to restrict the size of $g$. After that, the factor before $g$ leads to a quadratic term in the derivative. – Lutz Lehmann Jul 11 '19 at 14:17