Use the Error Bound to find a value of n for which the given...

Key Concepts

Taylor Series Expansion

A Taylor series expansion expresses a function as an infinite sum of terms calculated from the function's derivatives at a particular point. This expansion approximates the function locally with a polynomial, using the derivatives at that point to capture the behavior of the function, which allows for approximations and analytic insights.

Lagrange Error Bound

The Lagrange error bound provides an upper limit on the error made when approximating a function with a Taylor polynomial. It is based on the (n+1)th derivative of the function evaluated at some point within the interval of interest, ensuring that the approximation error is within a predictable tolerance, which is crucial when determining how many terms to include for a desired accuracy.

Remainder Term in Taylor Series

In the context of Taylor series, the remainder term quantifies the difference between the actual function value and its Taylor polynomial approximation. Expressed in Lagrange form, this remainder involves the (n+1)th derivative, the factorial of the order, and a power of the difference between the point of interest and the expansion point, which collectively provide a valuable estimate on the magnitude of the neglected terms.

Logarithmic Function Approximation

Approximating the natural logarithm using Taylor series is a practical application of these ideas. By expanding ln(x) about a convenient point (such as x = 1), one can use a finite series to closely estimate the value of ln(x). The error bound techniques ensure that the polynomial approximation reaches a desired level of precision, guiding the choice of the number of terms required.

Transcript

00:01 Welcome to this lesson and this lesson i use the error band to find a value of n for which the given inequality is satisfied.

00:10 Alright, so here you are looking for the error, what is equal to f, the derivative n plus 1 of z of z then times x minus a.

00:31 It has been given us one divided by n plus one factor right so this inactive head should be less than 10 to the power negative four so we are looking for the particular end all right so now let's have f of x equals to the lane of x we'll look at the first few derivatives so here we have f prime of x equals you the then the derivative of lean of x is equals 1 over x.

01:25 All right.

01:26 Then we have the second derivative, which is equals to negative 1 over x.

01:35 We take the third derivative.

01:38 We have 2 over x to the power 3.

01:43 We take the 4 derivative.

01:45 We have negative 6 over x to the power 4.

01:50 All right, then we take the fifth derivative.

01:57 We have 24 over x to the power 5.

02:01 So let's sit a table to look for the least value of n.

02:07 So here let's have the n plus 1 derivative, evaluated at 1, all right.

02:18 Then we can have n the nf or n plus 1th derivative evaluate that 1 okay multiplying 1 minus well let's have it that's 1 .3 minus 1 okay which is 0 .3 so he has 0 .3 to the power n plus 1 all over n plus 1 all right and now let's see the results so we start with if we evaluate at this we have us one over one right when n is equals you but here we remember that we are not just taking the m factor you are taking sorry the nf derivative we are taking n plus 1 derivative okay so when n is equals you you actually looking at the second derivative all right so this we have us negative one all right and when n is close to two this we have asked when n is equals to three here we have as negative six and when n is equals to four we have us 24 then this one is one we have 0 .3 to the power n plus 1 so 0 .3 to the power 2 we have this as negative 1 then 0 .09 all over it will have the 2 factor which is 2 all right and that evalu is to negative 0 .045 all right then we now look at when n is equal 2 so here we have the 2 all right multiplying 0 .0.

04:36 3 to the power 3 which is 0 .07 and we have the 3 factorial which is 6.

04:49 Okay so this evaluates to 0 .003 which is still no up to what we are looking for.

04:58 We are looking for 10 to the part negative 4...

Please give Ace some feedback