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Consider the piecewise constant interpolation of the function $f(x) = ln(x)$ , $10 ≤ x ≤ 11$ , at points $x_i = 10+ih$, where $h = 0.1$. Thus, our interpolant satisfies $v(x) = ln(10+ih)$, $10 + ih − 0.5h ≤ x < 10 + ih + 0.5h$, for $i = 0, 1, . . . , 11$.

Find a bound for the error in this interpolation, i.e. find a (reasonably small) constant M such that $|f(x) − v(x)| < M$ for any $10 ≤ x ≤ 11$.

I decided to first plug in all the i's so I can get my x's:

$x_0 = 10$

$x_1 = 10.1$

$x_2 = 10.2$

$x_3 = 10.3$

$x_4 = 10.4$

$x_5 = 10.5$

$x_6 = 10.6$

$x_7 = 10.7$

$x_8 = 10.8$

$x_9 = 10.9$

$x_10 = 11$

$x_11 = 11.1$

I also plugged in the the lowest i and highest i in $10 + ih − 0.5h ≤ x < 10 + ih + 0.5h$ and I got the interval of $9.95 ≤ x ≤ 11.15$, which all my x's fall in the interval of.

I'm not entirely sure how to start this (other than what I've done), if I could get a few pointers on what to do next would be great.

user129509
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  • No :/, I saw a section on it in my Numerical book but it really didn't explain it well and I couldn't really understand the example all that well – user129509 Mar 12 '15 at 02:50
  • Thanks so much! A First Course On Numerical Methods, what's been confusing me the most has been my prof. jumping around the chapters – user129509 Mar 12 '15 at 03:05
  • This is a practice problem from the professor and I wrote it as is given, but there is a similar problem in the book that he referenced for the problem. Here's the screencap of the book problem http://i.imgur.com/hk75ZUY.png – user129509 Mar 12 '15 at 03:14
  • Yeah it is odd, it's made it difficult for me to find other explanations on how to do it to. Thanks for the website though, I think I can use that to solve it – user129509 Mar 12 '15 at 03:30

1 Answers1

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$f(x)=\ln(x)$ is monotonically increasing, which means that the error is greatest at the jump points $$10+(i+0.5)·0.1,\qquad i=1,...,9.$$

The slope $f'(x)=1/x$ is falling which gives $x=10.05$ at the jump up from the left as the point of greatest error with value $$|\ln(10.05)-\ln(10)|=\ln(1.005)\approx 0.005$$

Lutz Lehmann
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