# Thread: Three Random Questions....

1. ## Three Random Questions....

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!!!

2. Originally Posted by osiron23
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
Overkill. This is an alternating series so the absolute value of the truncation error is less (or equal) the absolute value of the first neglected term. The third term is $\displaystyle <0.01$, so if we truncate at the second term we will have the required prescission.

$\displaystyle \sin(1)\approx 1-0.16666 \approx 0.833$

CB

3. Originally Posted by osiron23
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

Looks OK

CB

4. suh-weet!! THX!!!