The necessary sine is (0.164375- (0.164375)3/3!=9.993513*10(-7))0.000001.
The series ∑∞n=1bnsin(nπLt) is known as the sine series of f(t) and the series a02+∑∞n=1ancos(nπLt) is known as the cosine series of f(t).
The development series of sin(x) near 0 is, according to
Sin(x) = x/1, x/3, and x/5!...............(1) (1)
Using the above estimate, S(x)=x-x3/3! ................(2)
Consequently, the incorrect phrase for the guess is
error=(2)- (1)
= -(x^5/5! - x^7/7! + x^9/9!)
According to the elective series hypothesis, we only want to make sure that the aggregate will be below as much as feasible and that the outright value of the initial term dropped (x5/5!) is not exactly as far as possible.
This declares what
|x^5/5!| < 0.000001
Speaking for x:
x^5=5!*0.000001=0.000120
x=(0.000120)^(1/5)=0.164375
In light of this, the estimation of sin(x) is x-x3/3! has a blatant error for the range [-0.164375,+0.164375] below 0.000001.
Thus,
sin(0.164375)- (0.164375)^3/3!=9.993513*10^(- 7) < 0.000001
Visit: to learn more about sine series, visit brainly.com/question/15518595
#SPJ4
