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.

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

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.

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3