Approximating sin(1/2) with Taylor Inequality

In summary, the conversation discusses the use of Taylor's inequality to approximate sin(1/2). The fifth Taylor polynomial is used, and the upper bound for the sixth derivative of sin(x) is 1. The conversation also mentions the possibility of using a better upper bound through an iterative process.
  • #1
vande060
186
0

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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  • #2


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..)
 
  • #3


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
 

What is Taylor Inequality?

Taylor Inequality is a mathematical concept used to approximate the value of a function at a specific point using information about the function and its derivatives at that point.

Why is Taylor Inequality useful for approximating sin(1/2)?

Sin(1/2) is a trigonometric function that cannot be easily calculated by hand. Using Taylor Inequality, we can approximate the value of sin(1/2) by using a finite number of terms in the Taylor series expansion.

What is the formula for approximating sin(1/2) with Taylor Inequality?

The formula for approximating sin(1/2) with Taylor Inequality is: sin(1/2) ≈ Tn(1/2) + Rn(1/2), where Tn(1/2) is the nth degree Taylor polynomial for sin(x) centered at x = 0 and Rn(1/2) is the remainder term.

How accurate is the approximation of sin(1/2) with Taylor Inequality?

The accuracy of the approximation depends on the number of terms used in the Taylor series. The more terms used, the closer the approximation will be to the actual value of sin(1/2).

Are there any limitations to using Taylor Inequality to approximate sin(1/2)?

Yes, Taylor Inequality is only accurate for values of x close to the point of approximation. As the value of x moves further away from the point of approximation, the accuracy of the approximation decreases.

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