## Ok, a simple problem

Posted by ragknot on September 23, 2011 – 9:30 pm

Did anyone learn my solution method for the Colebrook Equation?

I am talking about the simplicity of the method.

Here’s an easy one. Find X where ** X=100*Log(X).**

If you learned my method, you should laugh at how easy this is.

September 24th, 2011 at 8:59 am

x = 1.02386 or 237.581 (both are approx)

September 24th, 2011 at 10:00 am

For the 237.58120875934262 use: X

_{n+1}= 100 Log10(X_{n})For the 1.0238552281443278 use: X

_{n+1}= 10^(X_{n}/100)September 24th, 2011 at 11:08 am

Almost all numbers are “approximate”, right?

For example, how many “real” numbers are more than zero but less than 10. How many “approximate” numbers are in that limit?

I suppose to math majors there might be a difference in real vs approximate, but to smart engineer most all computations are approximate.

How tall are you? a real number? (I suppose that height changed since I asked.)

Are your approximate answers are right? Of course they are.

Are they right to a math major? I wouldn’t guess.

Thanks for your specific reply(s).

I began my reply before I read you second one.

September 24th, 2011 at 3:46 pm

Hi ragknot. I think you’re confusing “real” with “exact”. If by “approximate” you mean e.g. 10 decimal places representations of numbers that need more than 10 DP to be represented (more) accurately, then there are a finite number of approximate numbers and an infinite number of real numbers (the cardinality of which is the same as that of the continuum [some say aleph-1, but I'm not convinced about that]) in the range 1 to 10.

There’s a mini-challenge, how many approximate numbers are there in the range 1.0000000000 to 10.0000000000?

September 25th, 2011 at 8:08 pm

as to Chris’s mini-challenge…89999991

September 27th, 2011 at 5:11 am

Hi Knightmare. I’m not sure how you got that.

Preliminary: (a-1)(a

^{n}+ a^{n-1}+ … + a^{2}+ a +1)= (a

^{n+1}+ a^{n}+ … + a) – (a^{n}+ a^{n-1}+ … + a + 1) = a^{n+1}– 1.So a

^{n}+ a^{n-1}+ … + a^{2}+ a +1 = (a^{n+1}– 1)/(a-1).We have a = 10 (as there are 10 possible digits at each decimal place), and n = 10 (as we have 10 decimal places) => (10

^{11}-1)/9 = 11 111 111 111We have 9 initial, full ranging numbers, i.e. 1.xxx, 2.xxx,…, 9.xxx, so they give 99 999 999 999, then add 11 more for 10, 10.0, 10.00,…, 10.000 000 000 0. Altogether that’s 100 000 000 010 numbers. I’ve taken e.g. 1.23 to be distinct from 1.230 etc.

I hope I haven’t goofed that.