Approximating sin(1/2) with Taylor Inequality

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Homework Statement



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

Homework Equations



taylors inequality |Rn(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
 
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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
[itex]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[/itex]

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

(Since [itex]1/2< \pi/2[/itex], and sine is increasing between 0 and [itex]\pi/2[/itex] is an increasing function, [itex]sin(1/2)[/itex] 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..)
 


HallsofIvy said:
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
[itex]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[/itex]

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

(Since [itex]1/2< \pi/2[/itex], and sine is increasing between 0 and [itex]\pi/2[/itex] is an increasing function, [itex]sin(1/2)[/itex] 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