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Math Help - The Remainder Estimation Theorem Problem

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    The Remainder Estimation Theorem Problem

    Use the Remainder Estimation Theorem to find an interval containing x=0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval.

    f(x)=cosx p(x)=1 - x^2/2! + x^4/4!
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    Quote Originally Posted by soleilion View Post
    Use the Remainder Estimation Theorem to find an interval containing x=0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval.

    f(x)=cosx p(x)=1 - x^2/2! + x^4/4!
    This is an Alternating series so the error in less than next term so in this case we have


    |\frac{x^6}{6!}|<.0001

    |\frac{x^6}{720}|<.0001

    \frac{x^6}{720} < .0001 \iff < x^6 < 0.0720 \iff x < .650
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    Quote Originally Posted by TheEmptySet View Post
    This is an Alternating series so the error in less than next term so in this case we have


    |\frac{x^6}{6!}|<.0001

    |\frac{x^6}{720}|<.0001

    \frac{x^6}{720} < .0001 \iff < x^6 < 0.0720 \iff x < .650

    I think you just use p(x), how about f(x), it does not matter?

    By the way, I think the number should be less than 0.0005, not 0.0001
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    Quote Originally Posted by soleilion View Post
    I think you just use p(x), how about f(x), it does not matter?

    By the way, I think the number should be less than 0.0005, not 0.0001
    Not when using this theorem.

    Theorem:

    if S_n=\sum (-1)^nb_n is a sum of an alternating series
    that satisfies

    0 \le b_{n+1} \le b_n and \lim_{n \to \infty}b_n =0

    Then the remainder |R_n| \le b_{n+1}

    The question said
    to find an interval containing x=0
    It is not unique.
    we could use any decimal that doesn't change in the thousandths column
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    Quote Originally Posted by TheEmptySet View Post
    Not when using this theorem.

    Theorem:

    if S_n=\sum (-1)^nb_n is a sum of an alternating series
    that satisfies

    0 \le b_{n+1} \le b_n and \lim_{n \to \infty}b_n =0

    Then the remainder |R_n| \le b_{n+1}

    The question said

    It is not unique.
    we could use any decimal that doesn't change in the thousandths column
    so I have to prove f(x) is converges, then I use
    then, I can get x
    right ?
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    The Macclaurin Series for cosine converges for all real x

    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}

    So by the theorm above we use the forth term (n=3)

    b_3=-\frac{x^6}{6!} we set this less than the error

    |b_3| < .0001 and solve for x.

    I hope this clears it up.

    Good luck.
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    Quote Originally Posted by TheEmptySet View Post
    The Macclaurin Series for cosine converges for all real x

    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}

    So by the theorm above we use the forth term (n=3)

    b_3=-\frac{x^6}{6!} we set this less than the error

    |b_3| < .0001 and solve for x.

    I hope this clears it up.

    Good luck.
    yes!!!
    I understood this
    Thnak you
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    Quote Originally Posted by TheEmptySet View Post
    The Macclaurin Series for cosine converges for all real x

    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}

    So by the theorm above we use the forth term (n=3)

    b_3=-\frac{x^6}{6!} we set this less than the error

    |b_3| < .0001 and solve for x.

    I hope this clears it up.

    Good luck.

    sorry, I still have a question
    How do you know b3<=0.0001
    why not b2 or b4?
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    Quote Originally Posted by soleilion View Post
    sorry, I still have a question
    How do you know b3<=0.0001
    why not b2 or b4?

    You were using the fist three terms of the series n=0,1,2

    P(x)=\underbrace{1}_{n=0} - \underbrace{\frac{x^2}{2}}_{n=1} + \underbrace{\frac{x^4}{24}}_{n=2}

    so the next term would be when n=3.

    Good luck.
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    Quote Originally Posted by TheEmptySet View Post
    You were using the fist three terms of the series n=0,1,2

    P(x)=\underbrace{1}_{n=0} - \underbrace{\frac{x^2}{2}}_{n=1} + \underbrace{\frac{x^4}{24}}_{n=2}

    so the next term would be when n=3.

    Good luck.

    I know this
    I mean how do you know b3 less than 0.0001
    because b3 includes x, you do not know what x is
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    Quote Originally Posted by soleilion View Post
    I know this
    I mean how do you know b3 less than 0.0001
    because b3 includes x, you do not know what x is
    We set the term b_3 less then the error (x and all)

    and solve the inequality for x. This gives us the distance x can be from the center of the series and get the accuracy we want. The Series is centered at zero.

    since x<.65 we get the set (-.65,.65)

    lets do a quick check

    p(.65)=.796187

    cos(.65)=.79608

    The error is within what we wan't yeah!!
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  12. #12
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    Quote Originally Posted by TheEmptySet View Post
    We set the term b_3 less then the error (x and all)

    and solve the inequality for x. This gives us the distance x can be from the center of the series and get the accuracy we want. The Series is centered at zero.

    since x<.65 we get the set (-.65,.65)

    lets do a quick check

    p(.65)=.796187

    cos(.65)=.79608

    The error is within what we wan't yeah!!
    Sorry, I did not understand
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  13. #13
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    Quote Originally Posted by TheEmptySet View Post
    We set the term b_3 less then the error (x and all)

    and solve the inequality for x. This gives us the distance x can be from the center of the series and get the accuracy we want. The Series is centered at zero.

    since x<.65 we get the set (-.65,.65)

    lets do a quick check

    p(.65)=.796187

    cos(.65)=.79608

    The error is within what we wan't yeah!!
    why did I miss type??????????????????????

    I type it again

    can you see this the #2 post

    I mean we can set x^8/8! less than .0001

    also we can set x^4/4! less than .0001

    we can get two different answer

    How can we know x^6/6! is the right one?
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  14. #14
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    Quote Originally Posted by soleilion View Post
    why did I miss type??????????????????????

    I type it again

    can you see this the #2 post

    I mean we can set x^8/8! less than .0001

    also we can set x^4/4! less than .0001

    we can get two different answer

    How can we know x^6/6! is the right one?
    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=\underbrace{1-\frac{x^2}{2}+\frac{x^4}{4!}}_{p(x)}-\frac{x^6}{6!}+\frac{x^8}{8!}-...

    according to the AST remainder theorem we use the next term in the series

    -\frac{x^6}{6!} this is when n=3 in the sum we set that term less than the error.
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  15. #15
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    Quote Originally Posted by TheEmptySet View Post
    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=\underbrace{1-\frac{x^2}{2}+\frac{x^4}{4!}}_{p(x)}-\frac{x^6}{6!}+\frac{x^8}{8!}-...

    according to the AST remainder theorem we use the next term in the series

    -\frac{x^6}{6!} this is when n=3 in the sum we set that term less than the error.
    How about f(x)=1/(1+x^2) p(x)=1-x^2+x^4

    the next term is |x^6| less than 0.0005

    I have the answer x=.281

    the correct answer is 0.31
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