Find the Value of h for Sin(x) Approximation

Click For Summary
SUMMARY

The discussion centers on determining the value of h for the approximation of sin(x) using the Taylor series expansion. Participants concluded that h can be calculated as h = (6!/10^4)^(1/6), ensuring that the remainder R in the approximation |R| < 10^(-4). The approximation used is sin(x) = x - x^3/6 + x^5/120 + R, with R defined as the difference between sin(x) and the Taylor polynomial T(x). The calculated value of h is approximately 0.64499.

PREREQUISITES
  • Understanding of Taylor series expansions
  • Familiarity with the concept of remainders in approximations
  • Knowledge of factorial notation and its application in series
  • Basic calculus, particularly derivatives and continuity of functions
NEXT STEPS
  • Explore the derivation of Taylor series for trigonometric functions
  • Learn about the error analysis in Taylor series approximations
  • Investigate the continuity and differentiability of sin(x) in various domains
  • Study the application of L'Hôpital's Rule in evaluating limits involving sin(x)
USEFUL FOR

Students studying calculus, particularly those focusing on series expansions and approximations, as well as educators looking for examples of Taylor series applications in trigonometric functions.

mathmathmad
Messages
50
Reaction score
0

Homework Statement


find a value of h such that for |x|<h implies sin(x)=x - x^3/6 + x^5/120 + R where |R|<10^(-4)


Homework Equations





The Attempt at a Solution


it's tedious to type out my working but I've got h= (6!/10^4)^1/6 but I'm not sure about this...
 
Physics news on Phys.org
That certainly works since the difference in sin(x) and your T(x)with your x value is 9.16x10^-6
How did you get this answer?
 
blackscorpion said:
That certainly works since the difference in sin(x) and your T(x)with your x value is 9.16x10^-6
How did you get this answer?

I got that (mathmathmad's) as well but I chose x such that I could evaluate without using a calculator. It's not too difficult to be honest. What concerns me is that do we need to prove that sin(x) and its derivatives of all orders are continuous in the given domain or should we take it as given.
 
Last edited:
sin(x)=x - x^3/6 + x^5/120 + R

erm, I take |R| = |sin x - x + x^3/6 - x^5/120 |

since R_n = f(x) - T_n (which the value of n I'm not sure of but I take n=5)

and there's a formula which states there exists c in (x,0) such that :
R_n = f^(n+1)(c)*x^(n+1)/(n+1)!

since I take n as 5 then I evaluate |R|<10^(-4) blablabla
get x^6 < 6!/10^4

can we just get the 6th root of 6!/10^4 to evaluate x?

so does h equal to (6!/10^4)^1/6 which is approximately 0.64499...?

please correct me!
 
I agree, I've done it with n = 6 but the principles the same.
You guys done the 3rd question?
And if so, how?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K