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Math Help - Three Random Questions....

  1. #1
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    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!!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by osiron23 View Post
    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 <0.01, so if we truncate at the second term we will have the required prescission.

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

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by osiron23 View Post
    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
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  4. #4
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    suh-weet!! THX!!!
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