What is the Linear Approximation Range for Sin(f) Within 10% Error?

Giuseppe
Messages
42
Reaction score
0

Homework Statement


For g=Hf = sin (f), use a Taylor expansion to determine the range of input for which the operator is approximately linear within 10 %


Homework Equations


The taylor series from 0 to 1 , the linearization, is the most appropriate equation

The Attempt at a Solution



g(f) = sin(0) + f*cos(0) = f
g1(f) = sin(f) g2 (f) =f ( at f=0, g1=g2(f) )

g1(f) = g2(f) + error

sin(f) = f + (1/10) * sin (f)

(9/10)* sin (f) =f

the value I keep getting is when f is equal to 0. I really don't think I am doing this correctly. Any advice?
 
Physics news on Phys.org
Giuseppe said:

Homework Statement


For g=Hf = sin (f), use a Taylor expansion to determine the range of input for which the operator is approximately linear within 10 %


Homework Equations


The taylor series from 0 to 1 , the linearization, is the most appropriate equation

The Attempt at a Solution



g(f) = sin(0) + f*cos(0) = f
g1(f) = sin(f) g2 (f) =f ( at f=0, g1=g2(f) )

g1(f) = g2(f) + error

sin(f) = f + (1/10) * sin (f)

(9/10)* sin (f) =f

the value I keep getting is when f is equal to 0. I really don't think I am doing this correctly. Any advice?
The only place sin(f)- f= (1/10)sin(f) is at f= 0 but if one is positive and the other negative you can still compare them. Your "error" is |sin(f)- f| and you want that less than |(1/10)sin(f)|. Where are those equal?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top