# Thread: Approximation Error and Tough Integral

1. ## Approximation Error and Tough Integral

Hello Everyone! Today We are asked to find $\displaystyle \int_{1}^{2} e^\frac{1}{x}dx$. We are given two option. First is by evaluating the definite integral itself or we can do numerical approximation using both Trapezoidal and Mid point Rules with sub interval n valued at 10. Since I've figured out that integrating such integral would be a , I just gone on and used the Numerical Approximation way. Everything is well until I am asked to find the error for both $\displaystyle T_{10}$ and $\displaystyle M_{10}$. I know the formula and the value of for the second derivative of the function $\displaystyle e^\frac{1}{x}$ to use it as a rule to find constant $\displaystyle K$ from the formula $\displaystyle |E_T|\leq\frac{K(b-a)^3}{12n^2}$ and $\displaystyle |E_M|\leq\frac{K(b-a)^3}{24n^2}$. Now I am stuck at how to find K when $\displaystyle K\geq|\frac{e^\frac{1}{x}(2x+1)}{x^4}|$ at interval $\displaystyle [1,2]$. Anyone got any idea to find $\displaystyle K$? Thanks y'all in advance!

Edit: I have an idea on how to solve this one. As we can see, x is bounded in [1,2], and that 1/x would only max out at x=1 since anything more than 1 would make x less so we can conclude that $\displaystyle \frac{1}{x}\leq 1$. Since K is the max value of the second derivative, then max value is acquired when 1/x is evaluated at x=1. So that $\displaystyle K\geq|(e^\frac{1}{x}(\frac{2}{\frac{1}{x}}+1)) (\frac{1}{x})^4|$ so if we substitute 1 to all 1/x, we get: $\displaystyle K=3e$. Is that the right way to do it?Hehe it is late in the morning and I am still pumped to do this problem!

2. ## Re: Approximation Error and Tough Integral

I think you need to find an upper bound for K, not a lower bound.

But you can't get a bound of any kind by setting x to the same value when you have the product of a decreasing and an increasing function. For example, if you have y=x(1-x) on [0,1], if you set x=0 across the board, you get a bound of zero, but y goes above zero. The correct procedure in this simplistic case is to recognize that x is always less than or equal to 1, and 1-x is always less than or equal to 1. So x(1-x) is always less than or equal to 1 times 1, which is 1. In this case we know the function has a maximum of $\displaystyle \frac{1}{4}$, so it is indeed always less than or equal to 1.

- Hollywood

3. ## Re: Approximation Error and Tough Integral

Hi Hollywood! Well I think from the definitions of Error bounds, K is just a constant in which the second derivative of integrand, f(x), maxes out. And if we look at the second derivative of f(x), we can see that we have product of 3 kinds of functions-exponential(decreasing), linear(increasing), and rational(decreasing). If we take the limit of $\displaystyle f"(x)$ as x goes to infinity, the value would go to zero. That means that for any increase in x in the bounds [1,2], the second derivative of f(x) only decreases, meaning, in the bounds of the integral [n-k,n], the second derivative maxes out on the smallest value of n-k which is 1. So to find the max value for K, we have to plug in 1 for any x in the second derivative. If we do that correctly, we should have$\displaystyle K=3e$. And since the Error is only bounded by the formula that involves max K, then we can just plug in K to all variable K in the formula. Is that reasoning good? Not really sure.

4. ## Re: Approximation Error and Tough Integral

Showing the limit of f''(x) as x goes to infinity is not good enough to do what you're doing. You would have to show that it is a decreasing function (i.e. that f'''(x)<0). Since it is a decreasing function, your answer is correct.

Another way to look at it is to write |f''(x)| as:

$\displaystyle \left|e^\frac{1}{x}\left( \frac{2}{x^3} + \frac{1}{x^4} \right)\right|$

and all the functions are decreasing, so you can just plug in 1, the smallest value for x.

And yes, I also get 3e for the numerical answer.

- Hollywood

5. ## Re: Approximation Error and Tough Integral

Hey pal. As far as Im concern, K is the abs(f''(x))<=K

6. ## Re: Approximation Error and Tough Integral

Thanks both of you for your answers. From what I understand, since K is defined from the inequality $\displaystyle \mid f''(x) \mid \leq K$ then K must be the highest value of the second derivative of the function.