Approximating sin(1/2) with Taylor Inequality

[vande060]

Homework Statement

I have to approximate sin(1/2) with the taylor inequality

Homework Equations

taylors inequality |R_n(x)| ≤ M/(n+1)! | x-a|ⁿ⁺¹

The Attempt at a Solution

Im not really sure what the significance of this is, but ill do the derivatives

f(x) = sin(x)
f'(x) = cos(x)
f''(x) = -sin(x)
f'''(x) = -cos(x)
f''''(x) = sin(x)

then ill plug in n and a for taylors inequality

|R₅(x)| ≤ M/(6)! | x|⁶

not sure what to do next, although I have to incorporate the 1/2 somewhere,maybe for x, but then what is M? and I am not sure what the derivatives are used for

 

[HallsofIvy]

Well, as to "what the deratives are used for" I am sure you are aware that the 5th Taylor Polynomial, around x= 0 is given by
f(0)+ f'(0)x+ (f''(0)/2)x^2+ (f'''(0)/3!)x^3+ (f^{IV}(x)/4!)x^4+ (f^V(x))/5!)x^5

Also, when you learned that error formula:
R_5\le (M_6/6!)|x^6|
You should have learned that M_6 is a an upper bound on the sixth derivative of f, f^{VI}(x), between 0 and x. 1 is an obvious upper bound on sin(x).

(Since 1/2< \pi/2, and sine is increasing between 0 and \pi/2 is an increasing function, sin(1/2) is a better upper bound but, of course, you don't know what sin(1/2) is until you do this approximation. What you could do is use the upper bound 1 in that to get an upperbound on sin(1/2), then use that to get a better upper bound, then use that to get a still better upper bound, etc..)

 

[vande060]

Well, as to "what the deratives are used for" I am sure you are aware that the 5th Taylor Polynomial, around x= 0 is given by
f(0)+ f'(0)x+ (f''(0)/2)x^2+ (f'''(0)/3!)x^3+ (f^{IV}(x)/4!)x^4+ (f^V(x))/5!)x^5

Also, when you learned that error formula:
R_5\le (M_6/6!)|x^6|
You should have learned that M_6 is a an upper bound on the sixth derivative of f, f^{VI}(x), between 0 and x. 1 is an obvious upper bound on sin(x).

(Since 1/2< \pi/2, and sine is increasing between 0 and \pi/2 is an increasing function, sin(1/2) is a better upper bound but, of course, you don't know what sin(1/2) is until you do this approximation. What you could do is use the upper bound 1 in that to get an upperbound on sin(1/2), then use that to get a better upper bound, then use that to get a still better upper bound, etc..)

THe problem is solved, thanks for the hint