Exponential Maclaurin Series
Is anyone able to help with the attached question that I've copied down from my textbook? I can't see where I've gone wrong in my working but my textbook lists the answers as:
e^-1/2 = 0.60653
0.00011% error
e^-1/2 = 0.60653
0.00011% error
Original post by Interea
Your calculations are fine, you just may need to do a few more terms of the expansion to get the required degree of accuracy in the question.
I expanded to 4 terms and got e^-1/2 = 29/48, giving me an answer accurate to more than 5 dp which I rounded off?
Original post by beachpanda
I expanded to 4 terms and got e^-1/2 = 29/48, giving me an answer accurate to more than 5 dp which I rounded off?
No. You're adding terms like 1/48. That's only to 2 decimal places. You need 1/100,000 to get 5 dps.
This does seem a bit extreme, but factorial*exponential on the denominator grows quickly.
Original post by mqb2766
No. You're adding terms like 1/48. That's only to 2 decimal places. You need 1/100,000 to get 5 dps.
This does seem a bit extreme, but factorial*exponential on the denominator grows quickly.
This does seem a bit extreme, but factorial*exponential on the denominator grows quickly.
Sorry I don't understand at all - I put 29/48 in my calculator and it gives a long decimal which I rounded to 5 d.p?
Original post by beachpanda
Sorry I don't understand at all - I put 29/48 in my calculator and it gives a long decimal which I rounded to 5 d.p?
Yes but there was no reason for you to stop after those first 4 terms - the question wants you to get an answer which is identical to the exact answer to 5d.p., so you need to keep going with the expansion until you're adding terms that are too small to change those first 5 decimal places.
Original post by beachpanda
Sorry I don't understand at all - I put 29/48 in my calculator and it gives a long decimal which I rounded to 5 d.p?
Each extra term in the maclaurin expansion adds a bit extra to the approximation. The more terms you have, the more accurate it is. The last term you add is 1/48 which is ~0.02. So your solution is accurate to ~2 decimal places. You need to add extra terms (higher powers) until the increments are less than 0.00001 (5dps).
Original post by mqb2766
No. You're adding terms like 1/48. That's only to 2 decimal places. You need 1/100,000 to get 5 dps.
This does seem a bit extreme, but factorial*exponential on the denominator grows quickly.
This does seem a bit extreme, but factorial*exponential on the denominator grows quickly.
Original post by Interea
Yes but there was no reason for you to stop after those first 4 terms - the question wants you to get an answer which is identical to the exact answer to 5d.p., so you need to keep going with the expansion until you're adding terms that are too small to change those first 5 decimal places.
Ok I see (I wish textbooks would actually explain this stuff
). I'm just unsure about the procedure of doing a question like this then? Is it a case of adding a term, then evaluating it, then if it's not accurate enough adding another and repeating until you have an accurate enough next term (to 5 d.p in this case)?Original post by mqb2766
Each extra term in the maclaurin expansion adds a bit extra to the approximation. The more terms you have, the more accurate it is. The last term you add is 1/48 which is ~0.02. So your solution is accurate to ~2 decimal places. You need to add extra terms (higher powers) until the increments are less than 0.00001 (5dps).
Actually I just evaluated 1/48 on its own and it's = 0.0208333333.... so I don't get how this is only accurate to 2 dp?
Original post by beachpanda
Actually I just evaluated 1/48 on its own and it's = 0.0208333333.... so I don't get how this is only accurate to 2 dp?
It's a term which you're adding onto an approximation. It's irrelevant how many decimal places you expand the term to, what's relevant is how much the approximation changes. The expansion of 1//48 is done in a calculator,.
The approximation will change by ~0.02. Adding the next term on, the approximation change by about 0.003. So thats how accurate each approximation is. You know that if you add on the infinite number of correction terms you'll get the precise answer. The correction terms (generally) decrease in value as you increase the power of x.
(edited 3 years ago)
Original post by beachpanda
Ok I see (I wish textbooks would actually explain this stuff ). I'm just unsure about the procedure of doing a question like this then? Is it a case of adding a term, then evaluating it, then if it's not accurate enough adding another and repeating until you have an accurate enough next term (to 5 d.p in this case)?
). I'm just unsure about the procedure of doing a question like this then? Is it a case of adding a term, then evaluating it, then if it's not accurate enough adding another and repeating until you have an accurate enough next term (to 5 d.p in this case)?Original post by beachpanda
Actually I just evaluated 1/48 on its own and it's = 0.0208333333.... so I don't get how this is only accurate to 2 dp?
You seem to be seriously missing the point here. You want your answer to be accurate to 5d.p., not one individual term!
If you wanted an answer accurate to 1 d.p. and you stopped at a term that was 3/10 = 0.3 you couldn't say that the answer was 0.3 to 1 d.p. because the next term could be 0.2 or 0.1 (or possibly 0.05 which would still affect the rounding). You need to look at the terms beyond the one where you've stopped to see how rapidly they're decreasing.
The usefulness of the Maclaurin series is that the terms usually decrease fairly rapidly because of the factorial in the denominator, but it depends on the value of x - hence why you need to check
Original post by beachpanda
Actually I just evaluated 1/48 on its own and it's = 0.0208333333.... so I don't get how this is only accurate to 2 dp?
Just to give an extreme example of why "When I plug a value into my series approximation I get 5 dp, so why isn't the result accurate to 5 d.p"?:
The series for sin x is x - x^3/3! + x^5 - ...
If I try to find sin(pi) just using the first term, I'd get that sin(pi) = 3.14159265 (etc). So there are *lots* of digits when I use my approximation, but since the true value of sin(pi) is 0, my approximation is hopelessly poor regardless of how many digits I've calculated pi to.
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