Percentage error

coconut64

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

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thanks

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Reply 1

username2769500

Too many décimal places in your table?

Reply 2

coconut64

Original post by Anfanny

Too many décimal places in your table?

But I thought it has to be as accurate as possible? Thanks

Reply 3

username2769500

Did It say that in your notes?

Reply 4

coconut64

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

Reply 5

NotNotBatman

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 not found

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.

Reply 6

coconut64

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..

Reply 7

NotNotBatman

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?

Reply 8

coconut64

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

Reply 9

NotNotBatman

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.

Reply 10

coconut64

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...

Reply 11

NotNotBatman

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}).

(edited 7 years ago)

Reply 12

h3rmit

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

Reply 13

NotNotBatman

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.

Reply 14

h3rmit

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

Reply 15

coconut64

So which method is right ?

Reply 16

NotNotBatman

Original post by h3rmit

Oh, you're right, I'll edit it in

Original post by coconut64

So which method is right ?

Reply 17

coconut64

Original post by NotNotBatman

Oh, you're right, I'll edit it in

Could you explain again simply please, I am lost. Thanks

Reply 18

NotNotBatman

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.