What value do you get when you convert 811to decimal? You get 0.0123456790123456790....
What value do you get when you convert 98011 to decimal? You get 0.000102030405060708091011...9697990001.
What value do you get when you convert 9980011 to decimal? You get 0.000001002003...100101102...996997999000....
These decimals list every n digit numbers (81 is 1, 9801 is 2, 998001 is 3, etc.) apart from the second last number. There is a pattern to find one of these fractions.
Can you see something special about the denominators? 81 is 92, 9801 is 992, 998001 is 9992.
This means that if you did 999800011 you would get 0.0000000100020003...9996999799990000....
Can you find a fraction that, when converted to decimal, lists every n digit number?
#Fractions
#CosinesGroup
#RecurringDecimals
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Watch the numberphile video which explains this very well at :http://www.youtube.com/watch?v=daro6K6mym8
This is a result of generating functions, and plugging x=101 into the function. For example, the generating function for the Fibonacci sequence is 1−x−x2x. If we plug in x=101, we get 1−(1/10)−(1/10)21/10=100−10−110=8910=0.11235….
To answer your question, sure you can. If you want to list every n-digit number, you'll want to have the sum ∑i=1∞i×10−in−n. Recall that the generating function for i is (1−x)21, so we'll have (1−10−n)210−n.
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Though, for the Fibonacci sequence, note that with x=101, you 'add' the tens digit to the preceding units digit, so you don't get the sequence of 0.112358132134…, but instead 8910=0.1123595…. Ah, if only patterns were verified by checking the first 5 terms.
Here's a spinoff question.
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You see the effects of carrying the digit, yes, but it seemed more magical to post the first 5 digits.
Irrational. I would provide a proof,but then I would be guilty of stealing your answer from MSE.