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Prove that the Taylor series for f(x) = sin(x) centered at a = π/2 represents sin(x) for all x. In other words, show that limn→[infinity] Rn(x) = 0 for each x, where Rn(x) is the remainder between sin(x) and the nth degree Taylor polynomial for sin(x) centered at a = π/2.

Respuesta :

Answer:

Note that when you derive sin(x), you loop betweeen

cos(x)

-sin(x)

-cos(x)

and again sin(x)

Also, sin(π/2) = 1, -sin(π/2) = 0 and cos(π/2) = -cos(π/2) = 0

Therefore, Rn(x) will have an expression of this type

[tex]\frac{sin(\pi/2) (x- \pi/2)^n}{n!} + \frac{cos(\pi/2) (x- \pi/2)^{n+1}}{(n+1)!} - \frac{sin(\pi/2) (x- \pi/2)^{n+1}}{(n+2)!} - \frac{cos(\pi/2) (x- \pi/2)^{n+1}}{(n+3)!}\\+ ... = \frac{(x-\pi/2)^n}{n!} - \frac{(x-\pi/2)^{n+2}}{(n+2)!} + ......[/tex]

Note that this is the tail of an alternating series with the generic term being

(x-π/2)^{2k}/(2k)! (note that the series takes even entries thats why the 2k); which limit is equal to 0 for any fixed value of x because we have a factorial dividing. Then, for the Leibnitz criterion, the tail of the series tends to 0, and thus, we can conclude that the Taylor series represents sin(x) for all x.

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