LaGrange Remainder Infinite Series

In summary: I'm not sure. If you could help me figure out where I went wrong, that would be much appreciated. In summary, the maximum error for f(1) = 10,000cos(1) is 2.45 x 10-5.
  • #1
carlodelmundo
133
0

Homework Statement



Let f be a function whose seventh derivative is f7(x) = 10,000cos x. If x = 1 is in the interval of convergence of the power series for this function, then the Taylor polynomial of degree six centered at x = 0 will approximate f(1) with an error of not more than

a.) 2.45 x 10-5

b.) 1.98 x 10-4

c.) 3.21 x 10-2

d.) 0.248

e.) 1.984

Homework Equations



The Lagrange Remainder Formula, it states that the biggest error is only as large as the next sum in the series.

The formula is:

Rn [tex]\leq[/tex][tex]\frac{f^{7}(z)(x-c)^7}{(n+1)!}[/tex]

The Attempt at a Solution



The maximum error is the next term in the sequence. Looking at the lagrange formula, we're looking for the maximum error, or the f7(z) term. Since we are given f7(x) = 10,000cos x and given that x = 1 is in the interval of convergence... I assumed that f7(1) = 10,000cos(1) is the maximum error.

Plugging it back in the LaGrange Formula I get the following:

R6 [tex]\leq[/tex] [tex]\frac{f^{7}(z)x^7}{7!}[/tex] = [tex]\frac{10000cos 1}{7!}[/tex]

I get a number that is not in the multiple choice answers. Any tips/ideas? Thanks
 
Physics news on Phys.org
  • #2
carlodelmundo said:

Homework Statement



Let f be a function whose seventh derivative is f7(x) = 10,000cos x. If x = 1 is in the interval of convergence of the power series for this function, then the Taylor polynomial of degree six centered at x = 0 will approximate f(1) with an error of not more than

a.) 2.45 x 10-5

b.) 1.98 x 10-4

c.) 3.21 x 10-2

d.) 0.248

e.) 1.984

Homework Equations



The Lagrange Remainder Formula, it states that the biggest error is only as large as the next sum in the series.

The formula is:

Rn [tex]\leq[/tex][tex]\frac{f^{7}(z)(x-c)^7}{(n+1)!}[/tex]

The Attempt at a Solution



The maximum error is the next term in the sequence. Looking at the lagrange formula, we're looking for the maximum error, or the f7(z) term. Since we are given f7(x) = 10,000cos x and given that x = 1 is in the interval of convergence... I assumed that f7(1) = 10,000cos(1) is the maximum error.
Why would you assume that? Cosine is a decreasing function between 0 and 1< [itex]\pi/2[/itex]. It takes its largest value at the lowest value of x in the interval, not the highest.

Plugging it back in the LaGrange Formula I get the following:

R6 [tex]\leq[/tex] [tex]\frac{f^{7}(z)x^7}{7!}[/tex] = [tex]\frac{10000cos 1}{7!}[/tex]

I get a number that is not in the multiple choice answers. Any tips/ideas? Thanks
 
  • #3
I see my error. What threw me off is at the beginning of the problem how "1" is in the interval of convergence. I immediately thought that f7 (1) was the maximum value.

We're just looking at the maximum value of f(x) = 10,000cos(x) correct? and cosine is between -1 and 1... so the maximum value is cos(0).

The lagrange remainder would be 10,000 / 7!.

Is this correct?
 

FAQ: LaGrange Remainder Infinite Series

1. What is the LaGrange Remainder in an Infinite Series?

The LaGrange Remainder in an infinite series is a mathematical concept that represents the difference between the sum of a partial infinite series and the actual value of the infinite series. It is used to estimate the accuracy of the partial sum in approximating the exact value of the infinite series.

2. How is the LaGrange Remainder calculated?

The LaGrange Remainder is calculated using the formula R_n = f^{(n+1)}(c)\frac{(x-a)^{n+1}}{(n+1)!}, where R_n is the remainder, f is the function, c is a point between x and a, x is the upper limit of the interval, and a is the lower limit of the interval.

3. What is the significance of the LaGrange Remainder in infinite series?

The LaGrange Remainder is significant because it provides a way to estimate the accuracy of the partial sum in approximating the exact value of the infinite series. It also helps in determining the convergence or divergence of an infinite series.

4. Can the LaGrange Remainder be used for all types of infinite series?

No, the LaGrange Remainder can only be used for certain types of infinite series, such as power series and Taylor series. It cannot be used for series with alternating signs or series with non-polynomial terms.

5. How is the LaGrange Remainder used in real-world applications?

The LaGrange Remainder has many applications in various fields of science and engineering, such as in calculating error bounds in numerical analysis, estimating errors in scientific measurements, and approximating solutions to differential equations.

Similar threads

Back
Top