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Math Help - Find the value of the a and b

  1. #1
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    Find the value of the a and b

    Question :
    Find the value of a and b in order that

    <br />
\lim_{x \to 0} \frac{x(1 + a cosx) - b sinx}{x^3} = 1.<br />
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question :
    Find the value of a and b in order that

    <br />
\lim_{x \to 0} \frac{x(1 + a cosx) - b sinx}{x^3} = 1.<br />
    The numerator behaves like x(1+a+b)-x^3(a/2 + b/6) for small x and we need this to be \sim x^3 so a=-5/2 and b=3/2.

    Note: you need to expand x(1+ a \cos(x) ) and b \sin(x) up to and including the terms in x^3 then the coefficient of x^3 in the numerator should be 1 and the coefficients for all lower power terms should be 0 .

    CB
    Last edited by CaptainBlack; December 7th 2009 at 03:21 AM. Reason: correct error (missed out a term)
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  3. #3
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    Could u please explain me this a little more

    Quote Originally Posted by CaptainBlack View Post
    The numerator behaves like x(1+a+b)-x^3(a/2 + b/6) for small x and we need this to be \sim x^3 so a=-5/2 and b=3/2.
    CB

    I am not that brilliant in this and need a little more explanation ....
    Last edited by CaptainBlack; December 7th 2009 at 03:13 AM.
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  4. #4
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    Quote Originally Posted by zorro View Post
    I am not that brilliant in this and need a little more explanation ....
    Replace sin and cos with their series expansions up to and including the third and second degree terms respectivly and simplify. Alternatively try L'Hopital's rule.

    CB
    Last edited by CaptainBlack; December 7th 2009 at 03:15 AM. Reason: correct omission
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  5. #5
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    Sir i am not that brilliant to replace sin and cos with series expansion

    Quote Originally Posted by CaptainBlack View Post
    Replace sin and cos with their series expansions up to and including the third and second degree terms respectivly and simplify. Alternatively try L'Hopital's rule.

    CB

    Sir i am not that brilliant to replace sin and cos with series expansion...If u could just guide me through the first few step i would be able to do it .....sorry for the language but i am very frustrated right now with this question
    Last edited by CaptainBlack; December 7th 2009 at 03:15 AM. Reason: correct error in quote
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  6. #6
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    Quote Originally Posted by zorro View Post
    Sir i am not that brilliant to replace sin and cos with series expansion...If u could just guide me through the first few step i would be able to do it .....sorry for the language but i am very frustrated right now with this question
    You will find the required series for sin x and cos x here: Taylor series - Wikipedia, the free encyclopedia

    Now apply the advice given in post #4.
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  7. #7
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    Is this correct

    Quote Originally Posted by mr fantastic View Post
    You will find the required series for sin x and cos x here: Taylor series - Wikipedia, the free encyclopedia

    Now apply the advice given in post #4.

    \lim_{x \to 0} \frac{x(1 - a cosx) - bsinx}{x^3} = 1

    sinx = \sum_{n = 0}^{ \infty } \frac{(-1)^n}{(2n + 1)} = x - \frac{x^3}{3!}

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

    Substituting in the equation we get

    \lim_{x \to 0} \frac{x \left[ 1 + a \left(1 - \frac{x^2}{2!} \right) \right] - b \left(x - \frac{x^3}{3!} \right)}{x^3}


    \lim_{x \to 0} \frac{x \left[ 1 + a \left( \frac{2 - x^2}{2} \right) \right] - b \left(\frac{6x - x^3}{6} \right)}{x^3}

    \lim_{x \to 0} \frac{f(x)}{f(y)} = \lim_{x \to 0} \frac{f'(x)}{f'(y)}

    \lim_{x \to 0} \frac{f'(x)}{f'(y)}= \lim_{x \to 0} \frac{ \frac{6x + 6ax - 3ax^3 - 6bx +bx^3}{6}}{3x^2}

    \lim_{x \to 0} \frac{(1 + a - b) - \frac{1}{2} x^2(3a - b)}{3x^2}

    I am stuck here !!!!
    What i have done is it correct or no?
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  8. #8
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    Quote Originally Posted by zorro View Post
    \lim_{x \to 0} \frac{x(1 - a cosx) - bsinx}{x^3} = 1

    sinx = \sum_{n = 0}^{ \infty } \frac{(-1)^n}{(2n + 1)} = x - \frac{x^3}{3!}

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

    Substituting in the equation we get

    \lim_{x \to 0} \frac{x \left[ 1 + a \left(1 - \frac{x^2}{2!} \right) \right] - b \left(x - \frac{x^3}{3!} \right)}{x^3}
    The original limit is equal to (since we are ignoring a term or order x in our truncation of the series):

    \lim_{x \to 0} \frac{x \left[ 1 + a \left( \frac{2 - x^2}{2} \right) \right] - b \left(\frac{6x - x^3}{6} \right)}{x^3} = <br />
\lim_{x \to 0} \frac{x(1+a-b)-x^3(a/2 +b/6)}{x^3}= <br />
\lim_{x \to 0} \frac{(1+a-b)}{x^2}- (a/2 +b/6)

    Now if this last limit is equal to 1, we must have 1+a-b=0 (as otherwise the limit would be infinite), and a/2+b/6=-1

    CB
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    I'll just add that the equation a/2+b/6=-1 can be re-written as 3a + b = -6, which may be easier to work with ....
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  10. #10
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    Is the answer correct ?

    Quote Originally Posted by mr fantastic View Post
    I'll just add that the equation a/2+b/6=-1 can be re-written as 3a + b = -6, which may be easier to work with ....

    since (1 + a - b) = 0

    therefore, a = b - 1

    Substituting a = b - 1 in 3a + b = -6

    3(b - 1) + b = -6

    3b - 3 + b = -6

    4b = -6 + 3

    4b = -3

    b = - \frac{3}{4}

    Substiting b in 3a + b = -6 weget

    3a + \left( - \frac{3}{4} \right) = -6

    3a = -6  + \frac{3}{4}

    3a = \frac{-24 + 3}{4}

    3a = \frac{-21}{4}

    a = 3 \times \frac{-21}{4}

    a = - \frac{63}{4}

    Is this correct ????
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  11. #11
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    Quote Originally Posted by zorro View Post
    since (1 + a - b) = 0

    therefore, a = b - 1

    Substituting a = b - 1 in 3a + b = -6

    3(b - 1) + b = -6

    3b - 3 + b = -6

    4b = -6 + 3

    4b = -3

    b = - \frac{3}{4}

    Substiting b in 3a + b = -6 weget

    3a + \left( - \frac{3}{4} \right) = -6

    3a = -6 + \frac{3}{4}

    3a = \frac{-24 + 3}{4}

    3a = \frac{-21}{4}

    a = 3 \times \frac{-21}{4}

    a = - \frac{63}{4}

    Is this correct ????
    Does your answer satisfy

    1 + a - b = 0 .... (1)

    3a + b = -6 .... (2)

    You should know how to solve simultaneous equations at the level you're apparently studying at. Personally, I'd just add equations (1) and (2) and solve for a. Then I'd substitute my value for a into (1) and solve for b.
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  12. #12
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    Trying the L'Hospital rule

    Quote Originally Posted by CaptainBlack View Post
    Replace sin and cos with their series expansions up to and including the third and second degree terms respectivly and simplify. Alternatively try L'Hopital's rule.

    CB

    I am trying to solve the same question by using the L'Hospital rule , but i have got stuck at the following step


    \lim_{x \to 0} \ \frac{x(1+acosx)-bsinx}{x^3} = 1

    f'(x) \ = \ x(1+acosx) - bsinx

    = 1- ax (sinx) +  (cosx) - b(cosx)

    and g'(x) = 3 x^2

    \lim_{x \to 0} \frac{ 1- ax (sinx) +  (cosx) - b(cosx)}{3 x^2}

    \lim_{x \to 0} \frac{1}{3x^2} - \frac{a}{3} \frac{sinx}{x} + (a-b). \frac{cosx}{3x^2}

    \lim_{x \to 0} \frac{1}{3x^2} - \frac{a}{3} (1) + \lim_{x \to 0} (a-b). \frac{cosx}{3x^2}
    I am stuck here ............
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  13. #13
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    Quote Originally Posted by zorro View Post
    I am trying to solve the same question by using the L'Hospital rule , but i have got stuck at the following step


    \lim_{x \to 0} \ \frac{x(1+acosx)-bsinx}{x^3} = 1

    {\color{red}{f(x)}} \ = \ x(1+acosx) - bsinx

    \color{red}f'(x)=(1+a \cos(x))+x(-a\sin(x))+b \cos(x)

    ....... \color{red}= 1- ax \sin(x) +  a\cos(x) - b\cos(x)

    and g'(x) = 3 x^2

    \lim_{x \to 0} \frac{{\color{red} 1- ax \sin(x) +  a\cos(x) - b\cos(x) }}{3 x^2}
    Stop here. If this limit is finite the numerator must be equal to zero at x=0. Since otherwise the limit does not exist. So:

    1+a-b=0

    or:

    b=1+a

    use this to substitute for b in the last limit. Now another stage of L'Hopitals rule is needed

    CB
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  14. #14
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    I did get u

    Quote Originally Posted by CaptainBlack View Post
    Stop here. If this limit is finite the numerator must be equal to zero at x=0. Since otherwise the limit does not exist. So:

    1+a-b=0

    or:

    b=1+a

    use this to substitute for b in the last limit. Now another stage of L'Hopitals rule is needed

    CB

    Where should i substitute the value of b ???
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  15. #15
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    Quote Originally Posted by zorro View Post
    Where should i substitute the value of b ???
    Here:

    <br /> <br />
\lim_{x \to 0} \frac{{ 1- ax \sin(x) +  a\cos(x) - b\cos(x) }}{3 x^2}<br />
<br />
 =\lim_{x \to 0}\frac{ 1- ax \sin(x) +  a\cos(x) - (a+1)\cos(x)}{3x^2} <br />
 =\lim_{x \to 0}\frac{ 1- ax \sin(x) - \cos(x)}{3x^2}

    CB
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