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Size:  390.0 KB for this question, I don't quite get c) I got 0.653% instead of 0.687% which should be the right answer.

    Attachment 613146613148 thanks
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    Too many décimal places in your table?
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    (Original post by Anfanny)
    Too many décimal places in your table?
    But I thought it has to be as accurate as possible? Thanks
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    Did It say that in your notes?
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    (Original post by Anfanny)
    Did It say that in your notes?
    What do you mean? I think if you round it, the answer will be even more inaccurate right? Especially when the question does not specify that you should round the answers... Thanks
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    (Original post by coconut64)
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Size:  390.0 KB for this question, I don't quite get c) I got 0.653% instead of 0.687% which should be the right answer.

    Attachment 613146613148 thanks
    Use exact natural logs for the exact integration and in your table, you normally wouldn't put the exact value of \frac{2\sqrt{3}}{3}, but round it to the appropriate number of s.f.
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    (Original post by NotNotBatman)
    Use exact natural logs for the exact integration and in your table, you normally wouldn't put the exact value of \frac{2\sqrt{3}}{3}, but round it to the appropriate number of s.f.
    Thanks for the help. What do you mean by 'Use exact natural logs'. If I sub a certain value into secx, this wouldn't give me a log number, it is just decimal number..
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    (Original post by coconut64)
    Thanks for the help. What do you mean by 'Use exact natural logs'. If I sub a certain value into secx, this wouldn't give me a log number, it is just decimal number..
    Have you integrated secx?
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    (Original post by NotNotBatman)
    Have you integrated secx?
    Yes, and it is a value involving ln. So do you suggest I should sub the values straight into that value? Thanks
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    (Original post by coconut64)
    Yes, and it is a value involving ln. So do you suggest I should sub the values straight into that value? Thanks
    Yes for the exact value, then your percentage difference will hopefully be correct.
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    (Original post by NotNotBatman)
    Yes for the exact value, then your percentage difference will hopefully be correct.
    I have just checked and saw that the answer for this integral is ln 2+root 3. How do I sub values in as there is no x...
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    (Original post by coconut64)
    I have just checked and saw that the answer for this integral is ln 2+root 3. How do I sub values in as there is no x...
    No, that is the exact value, found after substituting limits 0 and pi/3, use this in finding the percentage error. Also, for the approximation you have to find use y values to 4.s.f, so no exact values, which should continue into your calculations.

    So you would do percentage error = [(approximation - exact value)/exact value]*100% , where the exact value is ln(2+sqrt{3}).
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    (Original post by NotNotBatman)
    No, that is the exact value, found after substituting limits 0 and pi/3, use this in finding the percentage error. Also, for the approximation you have to find use y values to 4.s.f, so no exact values, which should continue into your calculations.

    So you would do percentage error = [(actual value - approximation)/approximation]*100% , where the actual value is ln(2+sqrt{3}).
    I think you should divide by the actual value rather than the approximation
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    (Original post by h3rmit)
    I think you should divide by the actual value rather than the approximation
    You could swap it around as percentage error = [(approximation - actual value)/actual value]*100%
    If you take the modulus either way the numerator is the error, so the answer would be the same.
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    (Original post by NotNotBatman)
    You could swap it around as percentage error = [(approximation - actual value)/actual value]*100%
    If you take the modulus either way the numerator is the error, so the answer would be the same.
    But the denominator would be different: you'd be finding the approximation as a proportion of the actual value, then taking away 1 when dividing by the actual. When dividing by the approximation, you'll find the actual value as a proportion of the approximation and do 1 - that, so you'll get different answers
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    So which method is right ?
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    (Original post by h3rmit)
    But the denominator would be different: you'd be finding the approximation as a proportion of the actual value, then taking away 1 when dividing by the actual. When dividing by the approximation, you'll find the actual value as a proportion of the approximation and do 1 - that, so you'll get different answers
    Oh, you're right, I'll edit it in

    (Original post by coconut64)
    So which method is right ?
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    (Original post by NotNotBatman)
    Oh, you're right, I'll edit it in
    Could you explain again simply please, I am lost. Thanks
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    (Original post by coconut64)
    Could you explain again simply please, I am lost. Thanks
    Percentage error = [(approximate value - exact value)/exact value]*100% ; remember this. The exact value is what you find in terms of natural logs and the approximate value is what you find from doing the table. Plug these into a calculator and that's your answer.
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    (Original post by NotNotBatman)
    Percentage error = [(approximate value - exact value)/exact value]*100% ; remember this. The exact value is what you find in terms of natural logs and the approximate value is what you find from doing the table. Plug these into a calculator and that's your answer.
 
 
 
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