Using Lagrange Error Bound

[armolinasf]

Homework Statement

Im supposed to use the lagrange error bound to find a bound for the error when approximating ln(1.5) with a third degree taylor polynomial about x=0, where f(x)=ln(1+x)

Homework Equations

Lagrange error bound

m/(n+1)! abs(x-a)^n+1, m=f(n+1)(c)

The Attempt at a Solution

The error bound is basically the next term in the series (correct?) So if I am looking for a bound on a third degree taylor polynomial i would have 4 factorial in the denominator. And I would let m=ln(1.5) since that is the greatest possible value on the interval.

So I would have something that looks like this:

Error < ln(1.5)/4! * abs(.5)^4 =.001055

Answer in my book says it should be .0156. So where am I going wrong? Thanks for the help.

 

[mighty2000]

Dear student,

You are on the right track, but there are a few errors in your calculations. First, the Lagrange error bound formula is actually m/(n+1)! * abs(x-a)^n+1, where m = f(n+1)(c). In this case, we are looking at a third degree polynomial, so n = 3 and a = 0. This gives us:

Error < f(4)(c)/4! * abs(x-0)^4

= (1/(1+c)^4)/4! * abs(x)^4 (since f(n)(x) = 1/(1+x)^n)

= (1/(1+c)^4)/24 * abs(x)^4

= (1/(1+c)^4)/24 * (.5)^4

Now, we need to find the value of c that will give us the maximum error. Since ln(1.5) is the greatest possible value on the interval, we can let c = 1.5. This gives us:

Error < (1/(1+1.5)^4)/24 * (.5)^4

= (1/2.5^4)/24 * (.5)^4

= (1/625)/24 * (.5)^4

= 1/150000 * (.5)^4

= 1/60000 * (.5)^4

= 1/9600 * (.5)^4

= 1/38400

= 0.00002604

Therefore, the bound for the error when approximating ln(1.5) with a third degree Taylor polynomial about x = 0 is 0.00002604. This is much closer to the answer given in your book than your previous attempt. I hope this helps!