5 ms·

Care to elaborate? Why does 0.987654 simplify to 80/81 and 0.123456 to 10/81?

by Stolpe 11mo ago

madcaptenor 11mo ago
.123456... = x + 2 x^2 + 3 x^3 + ... with x = 1/10. Then you have (x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...) + (x^3 + x^4 + x^5 + x^6 + ...) (count the number of occurrences of each power of x^n on the right-hand side) and from the sum of a geometric series the RHS is x/(1-x) + x^2/(1-x) + x^3/(1-x) + ..., which itself is a geometric series and works out to x/(1-x)^2. Then put in x = 1/10 to get 10/81. Now 0.987654... = 1 - 0.012345... = 1 - (1/10) (10/81) = 1 - 1/81 = 80/81.
- gus_massa 11mo ago
  I don't know who downvoted this, but it's correct. The use of series is a little "sloppy", but x + 2 x^2 + 3 x^3 + ... has absolute uniform convergence when |x|<r<1, even more importantly that it's true even for complex numbers |z|<r<1. The super nice property of complex analysis is that you can be almost ridiculously "sloppy" inside that open circle and the Conway book will tell you everything is ok. [I'll post a similar proof, but mine use -1/10 and rounding, so mine is probably worse.]
- gowld 11mo ago
  Don't need the clutter of infinite series and polynomials: 1/9 = 0.1111... 1/81 = 1/9 * 1/9 = 0.111... * 0.111... = Sum of: 0.0111... 0.00111... 0.000111... ... = 0.012345...
  - madcaptenor 11mo ago
    This is better than my answer, at least if you can get your brain to interpret it in base b. In that case the first two lines would become 1/(b-1) = 0.1111... 1/((b-1)^2) = 1/b * 1/b = 0.111... * 0.111... =
  - GuB-42 11mo ago
    Isn't it essentially the same thing, but less formal 0.1111... is just a notation for (x + x^2 + x^3 + x^4 + ...) with x = 1/10 1/9 = 0.1111... is a direct application of the x/(1-x) formula The sum of 0.0111... + 0.00111... ... = 0.012345... part is the same as the "(x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...)" part (but divided by 10) And 1/81 = 1/9 * 1/9 ... part is the x/(1-x)^2 result
alyxya 11mo ago
If you set x = 0.123456..., then multiplying it by (10 - 1) gives 9x = 1.111111..., and multiplying it by (10 - 1) again gives 81x = 10, or x = 10/81. I’m not writing things formally here but that’s the rough idea, and you can do the same procedure with 0.987654... to get 80/81.

Care to elaborate? Why does 0.987654 simplify to 80/81 and 0.123456 to 10/81?

by Stolpe 11mo ago

Care to elaborate? Why does 0.987654 simplify to 80/81 and 0.123456 to 10/81?

.123456... = x + 2 x^2 + 3 x^3 + ... with x = 1/10. Then you have (x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...) + (x^3 + x^4 + x^5 + x^6 + ...) (count the number of occurrences of each power of x^n on the right-hand side) and from the sum of a geometric series the RHS is x/(1-x) + x^2/(1-x) + x^3/(1-x) + ..., which itself is a geometric series and works out to x/(1-x)^2. Then put in x = 1/10 to get 10/81. Now 0.987654... = 1 - 0.012345... = 1 - (1/10) (10/81) = 1 - 1/81 = 80/81.

gus_massa 11mo ago
I don't know who downvoted this, but it's correct. The use of series is a little "sloppy", but x + 2 x^2 + 3 x^3 + ... has absolute uniform convergence when |x|<r<1, even more importantly that it's true even for complex numbers |z|<r<1. The super nice property of complex analysis is that you can be almost ridiculously "sloppy" inside that open circle and the Conway book will tell you everything is ok. [I'll post a similar proof, but mine use -1/10 and rounding, so mine is probably worse.]
gowld 11mo ago
Don't need the clutter of infinite series and polynomials: 1/9 = 0.1111... 1/81 = 1/9 * 1/9 = 0.111... * 0.111... = Sum of: 0.0111... 0.00111... 0.000111... ... = 0.012345...
- madcaptenor 11mo ago
  This is better than my answer, at least if you can get your brain to interpret it in base b. In that case the first two lines would become 1/(b-1) = 0.1111... 1/((b-1)^2) = 1/b * 1/b = 0.111... * 0.111... =
- GuB-42 11mo ago
  Isn't it essentially the same thing, but less formal 0.1111... is just a notation for (x + x^2 + x^3 + x^4 + ...) with x = 1/10 1/9 = 0.1111... is a direct application of the x/(1-x) formula The sum of 0.0111... + 0.00111... ... = 0.012345... part is the same as the "(x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...)" part (but divided by 10) And 1/81 = 1/9 * 1/9 ... part is the x/(1-x)^2 result

alyxya 11mo ago

If you set x = 0.123456..., then multiplying it by (10 - 1) gives 9x = 1.111111..., and multiplying it by (10 - 1) again gives 81x = 10, or x = 10/81. I’m not writing things formally here but that’s the rough idea, and you can do the same procedure with 0.987654... to get 80/81.