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    [Edexcel Heinemann P6]

    For small values of x : It seems that each time I use the approximation sinx = x, the text book uses sinx = x-(x³/6) and vice versa!

    What have I failed to see?

    Aitch
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    (Original post by Aitch)
    [Edexcel Heinemann P6]

    For small values of x : It seems that each time I use the approximation sinx = x, the text book uses sinx = x-(x³/6) and vice versa!

    What have I failed to see?

    Aitch
    I remember doing a question in one of the exercises using \sin x \approx x but then having to redo it with \sin x \approx x-\frac{x^{3}}{6}.
    I'd consider it preferable to use \sin x \approx x-\frac{x^{3}}{6} rather than \sin x \approx x as it's generally more accurate. Remember that you can always get rid of terms in x^{3} and higher after substituting in the approximation and simplifying the expression if necessary.
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    (Original post by Aitch)
    [Edexcel Heinemann P6]

    For small values of x : It seems that each time I use the approximation sinx = x, the text book uses sinx = x-(x³/6) and vice versa!

    What have I failed to see?

    Aitch
    It depends on the context of the question.

    Like to work out the limit of sinx/x as x->0 the first approx would do

    To work out (sinx-x)/x^3 as x->0 you'd need the latter

    you could think up examples where you'd need to go to the x^5 power
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    (Original post by Gaz031)
    I remember doing a question in one of the exercises using \sin x \approx x but then having to redo it with \sin x \approx x-\frac{x^{3}}{6}.
    I'd consider it preferable to use \sin x \approx x-\frac{x^{3}}{6} rather than \sin x \approx x as it's generally more accurate. Remember that you can always get rid of terms in x^{3} and higher after substituting in the approximation and simplifying the expression if necessary.
    I did wonder about this as an approach:

    Use the more accurate approximation, then omit the higher powers as required. I'm not sure if, in certain circumstances, this produces an over-elaborate solution which cannot be easily simplified...

    Aitch
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    Do you have some examples where you've been uncertain which to choose between?
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    (Original post by RichE)
    Do you have some examples where you've been uncertain which to choose between?
    Here's one:

    Evaluate (ln(1+x) -x)/(sin²x) [lim x=>0]

    My first instinct was to put sin²x = (x-(x³/6))² until I considered the overhead, then I wasn't sure...

    ...then decided on sinx =x, just because it might save an hour or two!

    I felt that the route was decided pragmatically, rather than truly logically...

    Aitch
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    (Original post by Aitch)
    Here's one:

    Evaluate (ln(1+x) -x)/(sin²x) [lim x=>0]

    My first instinct was to put sin²x = (x-(x³/6))² until I considered the overhead, then I wasn't sure...

    ...then decided on sinx =x, just because it might save an hour or two!

    Aitch
    Use sinx = x and ln(1+x) = x - x²/2 as then

    In (ln(1+x) -x)/(sin²x)

    has a denominator and numerator that are both non-zero

    Basically you need to go as far as you need to in order to ensure a non-zero vanishing first term in the denominator's and numerator's power series
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    (Original post by RichE)
    Use sinx = x and ln(1+x) = x - x²/2 as then

    In (ln(1+x) -x)/(sin²x)

    has a denominator and numerator that are both non-zero

    Basically you need to go as far as you need to in order to ensure a non-zero vanishing first term in the denominator's and numerator's power series
    Thanks!

    I think I may try the longer approximation (for sinx) tomorrow just to look at the results...

    I've got some time to think about this , since for those of us who are teaching ourselves, the academic year starts on 01/07, so we have a bit more time for exploration... :cool:

    Aitch
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    (Original post by Aitch)
    Thanks!

    I think I may try the longer approximation (for sinx) tomorrow just to look at the results...

    For those of us who are teaching ourselves, the academic year starts on 01/07, so we have a bit more time for exploration... :cool:

    Aitch
    Well let me give you one as practice. Find the limit as x tends to 0 of

    

\frac{(\sinh x - x)^6 (\sin x - x)^3}{(\cos x - 1 + x^2/2)^5 (\exp x - 1)^7}

    Eventually with practice you get to the point that you can just write the answer down as

    [(1/6)^6 (-1/6)^3]/[(1/24)^5 (1)^7]

    without having to do any real working.

    Just make sure to get the first non-zero power in each bracket.
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    (Original post by RichE)
    Well let me give you one as practice. Find the limit as x tends to 0 of

    

\frac{(\sinh x - x)^6 (\sin x - x)^3}{(\cos x - 1 + x^2/2)^5 (\exp x - 1)^7}

    Eventually with practice you get to the point that you can just write the answer down as

    [(1/6)^6 (-1/6)^3]/[(1/24)^5 (1)^7]

    without having to do any real working.

    Just make sure to get the first non-zero power in each bracket.
    ...provided that you're confident that the question setter got the power arithmetic right!

    Thanks for this - good example.


    The following is quite a good example of choosing the "form which gets the answer":

    [1B Q4] Given that x is small, show that

    (sinx - xcosx)/x³ ≈ (1/3)

    sinx ≈(6x - x³)/6 ... does the job.

    It would be clearer if we had
    x is small for sinx ≈ x - x³/3!
    and
    x is very small for sinx ≈ x

    Aitch
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    (Original post by Aitch)
    Here's one:
    Evaluate (ln(1+x) -x)/(sin²x) [lim x=>0]
    My first instinct was to put sin²x = (x-(x³/6))² until I considered the overhead, then I wasn't sure...
     \ln (1+x) \approx x-\frac{x^{2}}{2}+\frac{x^{3}}{3} \\

\\

\sin x \approx x -\frac{x^{3}}{6} \\

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \approx \frac{-\frac{x^{2}}{2}+\frac{x^{3}}{3}}  {x^{2}-\frac{1}{3}x^{4}+\frac{x^{6}}{36  }}

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \approx \frac{-\frac{1}{2}+\frac{x}{3}}{1-\frac{1}{3}x^{2}+\frac{x^{4}}{36  }}\\

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \rightarrow \frac{-\frac{1}{2}+0}{1-0+0} \\

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \rightarrow -\frac{1}{2} \textit{ as } x \rightarrow 0

    This shows it can be done but you just need to divide top and bottom by the lowest power of x in the denominator so as to ensure the denominator isn't zero.
    It may often be better to decide on the context and go from there though.
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    (Original post by Aitch)
    ...provided that you're confident that the question setter got the power arithmetic right!
    And that they didn't mean as x tends to 1 instead.
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    (Original post by Gaz031)
     \ln (1+x) \approx x-\frac{x^{2}}{2}+\frac{x^{3}}{3} \\

\\

\sin x \approx x -\frac{x^{3}}{6} \\

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \approx \frac{-\frac{x^{2}}{2}+\frac{x^{3}}{3}}  {x^{2}-\frac{1}{3}x^{4}+\frac{x^{6}}{36  }}

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \approx \frac{-\frac{1}{2}+\frac{x}{3}}{1-\frac{1}{3}x^{2}+\frac{x^{4}}{36  }}\\

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \rightarrow \frac{-\frac{1}{2}+0}{1-0+0} \\

\\

\frac{\ln (1+x) - x}{\sin ^{2}} \rightarrow -\frac{1}{2} \textit{ as } x \rightarrow 0

    This shows it can be done but you just need to divide top and bottom by the lowest power of x in the denominator so as to ensure the denominator isn't zero.
    It may often be better to decide on the context and go from there though.
    Thanks for this one, which I've worked through.

    I've done a few questions using both approximations, and am getting used to spotting the most useful form, I think...

    Aitch
 
 
 

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