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# Percentage error watch

1. 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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2. Too many décimal places in your table?
3. (Original post by Anfanny)
Too many décimal places in your table?
But I thought it has to be as accurate as possible? Thanks
4. Did It say that in your notes?
5. (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
6. (Original post by coconut64)
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 , but round it to the appropriate number of s.f.
7. (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 , 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..
8. (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?
9. (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
10. (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.
11. (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...
12. (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}).
13. (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
14. (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.
15. (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
16. So which method is right ?
17. (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 ?
18. (Original post by NotNotBatman)
Oh, you're right, I'll edit it in
Could you explain again simply please, I am lost. Thanks
19. (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.
20. still not getting the right answer...
(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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