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The error bound for Trapezoidal rule is: ${\left|E_{T}\right| \leq k \frac{(b-a)^{3}}{12 n^{2}}} $

I am trying to calculate the error for $x^x$ in the interval $[0,1]$ and let $n = 500$. The problem is I am not able to compute $ k = \max (f^{(2)}(x))$ because $\max (f^{(2)}(x))$is undefined. How do I go about calculating the error bound?

Graphs: https://www.desmos.com/calculator/qvxrytufm6

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The second derivative of $f(x)=x^x$ is $$f^{(2)}(x)=x^x((\ln(x)+1)^2+1/x)$$ which is continuous, positive and convex in $(0,1]$ . Hence for $0<a< 1$ $$\max_{[a,1]}|f^{(2)}(x)|=\max(f(a),f(1))=\max(f(a),2).$$ Unfortunately $$\lim_{a\to 0^+}f^{(2)}(a)=+\infty$$ therefore you should apply the Trapezoidal rule to a smaller interval $[a,1]$ with $0<a< 1$.

Note that $x^x$ is integrable in $[0,1]$ and for $0<a<1$ we can easily estimate the error on the missing part $[0,a]$: $$0<\int_0^a x^x \,dx< a$$ because $0<x^x\leq 1 $ in $[0,1]$.

P.S. The following identity (see Sophomore's dream) $$\int_0^1x^x\ dx=\sum_{n=1}^\infty(-1)^{n+1}n^{-n}\approx 0.7834305107.$$ allows you to approximate the given integral quite efficiently.

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  • $\begingroup$ Im so sorry. There was a typo in the question. I meant $[0,1]$ $\endgroup$ Nov 9 '19 at 8:27

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