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You are not logged in. #1 20121226 00:42:08
Calculation with a given accuracyHello, guys! Plzz help me out here.. #2 20121226 00:57:30
Re: Calculation with a given accuracyHi Serj; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20121226 01:16:49
Re: Calculation with a given accuracybobbym, thank you for your reply! #4 20121226 01:23:35
Re: Calculation with a given accuracyOkay I have identified the function and it is In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #6 20121226 01:41:40
Re: Calculation with a given accuracyI have the answer that you require it will take some time to post and to check. Please hold on. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #8 20121226 01:56:29
Re: Calculation with a given accuracyHi;
They obviously want a table. And they only want you to run the series for the first 10 terms. The problem is as x gets larger you will need more terms to meet the error. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #10 20121226 02:24:50
Re: Calculation with a given accuracyAhh! The other variant is correct! That is the bound you need to do what is expected. Basically without this constraint you would have x, ε, and the number of summands exceeding 10. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #12 20121226 02:32:54
Re: Calculation with a given accuracyHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #14 20121226 02:53:29
Re: Calculation with a given accuracyHi; What this means is you need 2 terms for an error < .001 For .0001 and .00001, 3 terms will suffice. For .000001, 4 terms are needed. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #15 20130211 19:57:39
Re: Calculation with a given accuracypls. don't judge me harshly, but the problem has not been completely solved yet (but the deadline is coming... uggg) Last edited by Serj (20130211 20:08:50) #16 20130211 22:29:01
Re: Calculation with a given accuracyHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #18 20130211 22:44:03
Re: Calculation with a given accuracyYes, epsilon is the estimate of the error between the truncated sum and the actual answer. Do you want to know how to calculate that? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #20 20130211 23:06:30
Re: Calculation with a given accuracyHi; is an alternating series. There is a nifty way to get the error of those types of series. Take a look at this table: Last edited by bobbym (20130212 00:11:05) In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #22 20130212 01:21:36
Re: Calculation with a given accuracy
Sometimes calculating the error can be quite difficult or perhaps impossible. In this case it is easy.
You do not need a formula for this particular problem we use the alternating series rule. If you look at the table you might see what it is. If not I will show you. Last edited by bobbym (20130212 02:22:16) In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #25 20130212 03:42:29
Re: Calculation with a given accuracyThat is right. The actual error is always less then the first neglected term of an alternating series. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 