Since the series is a geometric series, N = (1 - 10^4024)/(1 - 10). 10^4024 = 10000....000 implies, 1 - 10^4024 = - (10^4024 - 1) = - 999999.....999. Hence we get N = -999.....999/(-9) = 999....999/9 = 111.....111.roberthun said:Let \(\displaystyle N=1+10+10^2+\cdots+10^{4023}\). Find the 2013-th digit after the decimal comma of \(\displaystyle \sqrt{N}\).

I have no idea what to do about sqrt(N).

This is almost surely false. Maybe you meant that the 2013-th digit of N. but OP asked for sqrt(N).greg1313 said:

Aha! I will do arithmetic on the numbers of the form 11.....111 (agentredlum discovered such numbers in a very old topic : viewtopic.php?f=40&t=18441).

the agenterdlum number sqrt(111.....11) can be mapped in the reals by transforming sqrt(11...1) to sqrt(1.111.....) where the place of the decimal doesn't matter. Now, sqrt(1.111....) = sqrt(11(0)/99) where 11(0) means there can be uncountably/countably many zeros after 11 (i.e. 110, 1100,...). we will take 11(0) = 11 for our computational matters. sqrt(11/99) = sqrt(11)/3sqrt(11) = 1/3 = 0.333333.... which is in turn represents the agentredlum number 3333.....333. Hence, no matter where the decimal is situated, the next number after the decimal is 3, and so as the next and next and e.t.c. Hence, the 2013-th digit (irrelevant) after the decimal (irrelevant,too) is always 3.

NOTE : @agentredlum, seems like we find a practical application of your idea I must admit, you're just brilliant!!

Balarka

.

:evil: It's not false, it ismathbalarka said:This is almost surely false.

\(\displaystyle \sqrt{11}\,\approx\,3.31\) (accurate to 2 decimal places).roberthun said:Let \(\displaystyle N=1+10+10^2+\cdots+10^{4023}\). Find the 2013-th digit after the decimal comma of \(\displaystyle \sqrt{N}\).

\(\displaystyle \sqrt{1111}\,\approx\,33.331\) (accurate to 3 decimal places).

\(\displaystyle \sqrt{111111}\,\approx\,333.3331\) (accurate to 4 decimal places).

. . . and so on.

I agree that the digit 1 is in the decimal expansion but how would you guarantee that '1' is thegreg1313 said::evil: It's not false, it is1!

He gave an approximationgelatine1 said:\(\displaystyle sqrt(1111)=33.33166=33.332\)

and not \(\displaystyle 33.331\)

but even though it is 1.

\(\displaystyle sqrt(11(2times))= 3(1time).3(1time)16 =3.316

sqrt(11(4times))= 3(2times).3(2times)16=33.3316

sqrt(11(6times))= 3(3times).3(3times)16=333.33316\)

.

.

.

\(\displaystyle sqrt(11(4024times))= 3(2012times).3(2012times)16\)

so 2013th digit after comma is 1.

Similar Math Discussions | Math Forum | Date |
---|---|---|

Anyone have some interesting problems? | Real Analysis | |

Interesting math problem | Elementary Math | |

Interesting Mathematical Find. | General Math | |

Interesting triangle headache | Geometry |