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Taylor series?

  • #1

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...
 

Answers and Replies

  • #2
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?
 
  • #3
24
0
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:
  • #4
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!
 
  • #5
I agree, Ive done it with n = 6 but the principles the same.
You guys done the 3rd question?
And if so, how?
 

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