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You are not logged in. #51 20070428 17:25:57
Re: Jane’s exercises#14 #52 20070429 00:47:02
Re: Jane’s exercises#13 I surrender.... Numbers are the essence of the Universe #53 20070602 00:14:54
Re: Jane’s exercises#15 This is an easy one. Last edited by JaneFairfax (20070602 00:15:24) #54 20070602 05:39:14
Re: Jane’s exercisesThis one is easier Last edited by Stanley_Marsh (20070602 05:39:28) Numbers are the essence of the Universe #55 20070602 06:24:51#56 20070602 18:01:07
Re: Jane’s exercisesYah , it comes from mutiplication law of mod Numbers are the essence of the Universe #57 20070624 00:18:53#58 20070804 23:22:16
Re: Jane’s exercises#17 Here is another exercise I made up myself. #59 20070808 07:54:03
Re: Jane’s exercises#18 Last edited by JaneFairfax (20070808 08:48:00) #60 20070811 20:14:03
Re: Jane’s exercisesThis result is important because I used it in proving Tony’s question: http://www.mathsisfun.com/forum/viewtopic.php?id=7807 Last edited by JaneFairfax (20070812 01:10:04) #61 20070817 18:38:09#62 20070914 09:24:41
Re: Jane’s exercises#20 I just made this problem up myself; hopefully I’ve got the maths correct. Last edited by JaneFairfax (20070914 09:26:02) #63 20071027 20:28:52
Re: Jane’s exercises#21 Prove that #64 20080223 02:44:32
Re: Jane’s exercisesa more powerful result of 15: which is two consecutive even integers, thus one can be written on the form 4k, and one on the form 4k+2, and thus k^{2n}1 is divisible by 8, ie k^{2n} gives rest 1 when divided by 8. #65 20090414 22:08:05
Re: Jane’s exercisesHi Jane; this has a characteristic equation of x^3x^2x+1=0 which has roots of {1,1,1} This then is the general form of the solution: we need to determine c1,c2, and c3 from the initial conditions So the general solution is: This agrees with your general solution at the bottom of problem 21 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. #66 20090414 23:15:29
Re: Jane’s exercisesbobbym. looks like you’ve got the answer. Well done! Then and are AP’s with common difference 7 and first terms 2 and 5 respectively. Thus Note that can be written as: The result follows. Last edited by JaneFairfax (20090414 23:22:51) #67 20090414 23:30:59
Re: Jane’s exercisesHi Jane; 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. #68 20090415 02:36:48
Re: Jane’s exercisesWell, I’m not that familiar with techniques on solving difference equations, but your solution looked right to me so I gave you full credit for it. #69 20090415 09:33:25
Re: Jane’s exercisesHi Jane; Last edited by bobbym (20090415 09:48:04) 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. #70 20090415 10:21:54
Re: Jane’s exercisesAw, thanks. Well, I thought up that problem myself, so for this particular problem I happen to have have my own solution handy. #71 20090415 23:13:54
Re: Jane’s exercisesHi Jane; Last edited by bobbym (20090420 16:24:25) 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. 