Find the Value of h for Sin(x) Approximation

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Homework Help Overview

The discussion revolves around finding a value of h such that for |x|

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss different values of n for the Taylor series expansion and how to derive h from the remainder term R. There are questions about the continuity of sin(x) and its derivatives in the specified domain, as well as the validity of the chosen approach to evaluate h.

Discussion Status

Some participants have provided calculations and approximations for h, while others are seeking clarification on the assumptions made regarding continuity and the choice of n. There is an acknowledgment of different methods being explored, but no explicit consensus has been reached.

Contextual Notes

Participants are considering the implications of the remainder term in the Taylor series and whether certain properties of sin(x) can be assumed without proof. There is also a mention of evaluating without a calculator, which may influence the choice of x values.

mathmathmad
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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...
 
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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?
 

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