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Math Help - Weird problems

  1. #1
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    Weird problems

    I'm in an analysis class currently discussing differentiation. I'm pretty sure I need to use Taylor's Theorem on these but I'm not sure. Could someone work me through them?

    First, if x is greater than zero, show that (1+x)^(1/3)-(1+(1/3)x-(1/9)x^2) is less than or equal to (5/81)x^3.

    Second, if abs(x) is less than or equal to one, show that abs([sin(x)-(x-(x^3)/6+(x^5)/120]) is less than (1/5040).

    Sorry. I know it's quite complicated, but I have no idea. Anyways, thank you for any help you can give.
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  2. #2
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    Hello, 12MED34!

    You're right . . . you need the Taylor Series.
    Here's the first one . . .


    If x < 0, show that: . (1+x)^{\frac{1}{3}} -1 + \frac{1}{3}x - \frac{1}{9}x^2\;\leq\;\frac{5}{81}x^3

    First, we'll crank out the Taylor Series for (1 + x)^{\frac{1}{3}}

    Formula: . f(x) \;=\;f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \hdots

    We have: . \begin{array}{ccccc}f(x)\;= \\ f'(x)\:= \\ f''(x)\,= \\ f'''(x) = \\ \vdots\end{array}<br />
\begin{array}{ccccc}(1+x)^{\frac{1}{3}} \\ \frac{1}{3}(1+x)^{\text{-}\frac{2}{3}} \\ \text{-}\frac{2}{9}(1+x)^{\text{-}\frac{5}{3}} \\ \frac{10}{27}(1+x)^{\text{-}\frac{8}{3}} \\ \vdots\end{array}

    Then: . \begin{array}{ccccc}f(0)\;= \\ f'(0)\:= \\ f''(0)\,= \\ f'''(0)= \\ \vdots\end{array}<br />
\begin{array}{ccccc}1 \\ \frac{1}{3} \\ \text{-}\frac{2}{9} \\ \frac{10}{27} \\ \vdots\end{array}


    We have: . (1+x)^{\frac{1}{3}}\;=\;1 + \frac{1}{3}x -\frac{\frac{2}{9}}{2!}x^2 + \frac{\frac{10}{27}}{3!}x^3 - \hdots

    Hence: . (1 + x)^{\frac{1}{3}} \;\leq \;1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^2

    Therefore: . (1 + x)^{\frac{1}{3}} - 1 + \frac{1}{3}x - \frac{1}{9}x^2\;\leq \;\frac{5}{81}x^3 . . . . There!

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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    Hello, 12MED34!

    You're right . . . you need the Taylor Series.
    Here's the first one . . .


    First, we'll crank out the Taylor Series for (1 + x)^{\frac{1}{3}}

    Formula: . f(x) \;=\;f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \hdots

    We have: . \begin{array}{ccccc}f(x)\;= \\ f'(x)\:= \\ f''(x)\,= \\ f'''(x) = \\ \vdots\end{array} \begin{array}{ccccc}(1+x)^{\frac{1}{3}} \\ \frac{1}{3}(1+x)^{\text{-}\frac{2}{3}} \\ \text{-}\frac{2}{9}(1+x)^{\text{-}\frac{5}{3}} \\ \frac{10}{27}(1+x)^{\text{-}\frac{8}{3}} \\ \vdots\end{array}" alt="
    \begin{array}{ccccc}(1+x)^{\frac{1}{3}} \\ \frac{1}{3}(1+x)^{\text{-}\frac{2}{3}} \\ \text{-}\frac{2}{9}(1+x)^{\text{-}\frac{5}{3}} \\ \frac{10}{27}(1+x)^{\text{-}\frac{8}{3}} \\ \vdots\end{array}" />

    Then: . \begin{array}{ccccc}f(0)\;= \\ f'(0)\:= \\ f''(0)\,= \\ f'''(0)= \\ \vdots\end{array} \begin{array}{ccccc}1 \\ \frac{1}{3} \\ \text{-}\frac{2}{9} \\ \frac{10}{27} \\ \vdots\end{array}" alt="
    \begin{array}{ccccc}1 \\ \frac{1}{3} \\ \text{-}\frac{2}{9} \\ \frac{10}{27} \\ \vdots\end{array}" />


    We have: . (1+x)^{\frac{1}{3}}\;=\;1 + \frac{1}{3}x -\frac{\frac{2}{9}}{2!}x^2 + \frac{\frac{10}{27}}{3!}x^3 - \hdots

    Hence: . (1 + x)^{\frac{1}{3}} \;\leq \;1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^2

    Therefore: . (1 + x)^{\frac{1}{3}} - 1 + \frac{1}{3}x - \frac{1}{9}x^2\;\leq \;\frac{5}{81}x^3 . . . . There!
    What follows is a bit pedantic, but the OP does say that he is in an analysis
    class, so I think I will assume he needs to treat the remainder in a Taylor series
    explicitly.

    In fact this is not a proof, you are making an assumption about the remainder
    for a truncated Taylor series, expanding about a :

    <br />
f(x)=f(a)+f'(a)(x-a)+...+\frac{f^{(n-1)}(a)(x-a)^{n-1}}{(n-1)!} + R_n<br />

    where:

    <br />
R_n=\frac{f^{(n)}(\xi)(x-a)^n}{n!},\ \ \xi \mbox{ is between a and x}<br />

    In this case a=0, x>0, and f'''(x) is a decreasing function of x, which allows one to obtain a bound on the remainder:

    <br />
R_3 < \frac{f'''(0) x^3}{3!}<br />

    and the conclusion follows.

    RonL
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  4. #4
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    I really appreciate the help.

    For the second one, what would the derivative of abs(sin x) be?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by 12MED34 View Post
    I really appreciate the help.

    For the second one, what would the derivative of abs(sin x) be?
    You don't need the derivative of |\sin(x)|, what you are asked to show is that the absolute value of the difference between \sin(x), and its truncated Taylor series truncated at the x^5 term, about 0 is less than 1/5040 when |x| \le 1.

    So the series you are interested in is that of \sin(x) about 0.

    RonL
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  6. #6
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    Yeah I end up with at the end that

    abs([sin(x)-(x-(x^3)/6+(x^5)/120]) is less than or equal to abs((x^7)/5040).
    How would I show that this is just greater than the expression on the left (and not equal)?
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by 12MED34 View Post
    Yeah I end up with at the end that

    abs([sin(x)-(x-(x^3)/6+(x^5)/120]) is less than or equal to abs((x^7)/5040).
    How would I show that this is just greater than the expression on the left (and not equal)?
    As x^7<=1 for all x such that |x|<=1, so you have abs([sin(x)-(x-(x^3)/6+(x^5)/120])<=1/5040.

    Equality can only occur when |x|=1, as if |x|<1, then the bound can be made tighter. So you need only check that equality does not apply at x=+/-1.

    RonL
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