I have answers for them. I'm just not sure if they're correct. They're probably not.
1. Derive the Taylor series for f(x) = cosx centered at π/4.
f(x) = cos(x)
f'(x) = -sin(x)
f''(x) = -cos(x)
f'''= sin(x)
f''''(x) = cos(x)
f(π/4) = sqrt(2)/2
f'(π/4) = -sqrt(2)/2
f''(π/4) = -sqrt(2)/2
f'''(π/4) = sqrt(2)/2
f''''(π/4) = sqrt(2)/2
the definition of the Taylor series expansion of f(x) centered at a is:
sum of(n = 0 to infinity): [(f^n)/n!)(x - a)^n
where: f^n denotes the nth derivative of f, and we take f^0 as being equal to f.
cos x = [(sqrt(2)/2)/0!](x - π/4)^0 + [(-sqrt(2)/2)/1!](x - π/4)] + [(-sqrt(2)/2)/2!](x - π/4)^2 + [(sqrt(2)/2)/3!](x - π/4)^3 + [(sqrt(2)/2)/4!](x - π/4)^4 + .....
2. Approximate sin 1 (1 is in radians) with an error of less than .01.
n x = x - ((x^3)/6) + ((x^5)/120) - ((x^7)/5040) +
sin 1 = 1 - (1/6) + (1/120) - (1/5040) +
= 1 - 0.1666666667 + 0.00833333333 - 0.000198412698 +
= 0.83333333333 + 0.00833333333 - 0.000198412698 +
= 0.841666666667 - 0.000198412698 +
=0.841468254 +
sin 1 ≈ 0.84
3. What range of values for x does the approximation sinx = x has an error of less than .1%?
sin x = x - (x^3)/3! +...
so approximating sin x by x has an error smaller than x^3/3!
so find x such that [(x^3)/3! ] / x = 0.001
(0.001 is the fraction that is 0.1%)
x^2 = 0.006
x = 0.077
Thanks for the help!!!