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        riddles >> medium >> interesting problem -approximation
        
(Message started by: puzzlecracker on Jan 14^th, 2005, 4:30pm)

Title: interesting problem -approximation
Post by puzzlecracker on Jan 14^th, 2005, 4:30pm

Where are all the GURUS?

I was asked very strange (possibly trivial) problem, can't really recall it completely or accurately, but iI'll try. If someone encountered a comparable one, that would be great for clarifying my ideas.

OK here it is:
The real question to to approximate 1/(8^(12), give a good floating point number -- NO CALCULATOR -SUGGESTIONS, PLEASE

In the beginning, I was asked to approximate x in this equation 10^x= 1/(8^(12) so I said something like -7 or -8, let's call it [alpha], Then, he says: suppose your approximation is equal give 10^[alpha] = 1/(2^20), using this information approximate the original 1/(8^(12), more accurately!

any ideas?

Title: Re: interesting problem -approximation
Post by Icarus on Jan 14^th, 2005, 4:51pm

on 01/14/05 at 16:30:39, puzzlecracker wrote:

...that would be great for clarifying my ideas.

What would really be great is if you would clarify your communication!

Quote:

The real question to to approximate 1/(8^(12), give a good floating point number

Easy: 243.0 (you never gave any criteria on how good of an approximation you wanted! ::))

Seriously, I assume you mean 8^-12. (What's with the 2nd opening parenthesis? you repeated it every time you gave this expression.)

8^-12 = 2^-36 = 2⁴*2^-40 = 16*(1024)^-4 [approx] 16 x 10^-12 = 1.6 x 10^-11.

Quote:

suppose your approximation is equal give 10^[alpha] = 1/(2^20), using this information approximate the original 1/(8^(12), more accurately!

I know people from your country have trouble with the English language. ;) But perhaps you could find someone proficient in it to help you give a more sensible translation?

Title: Re: interesting problem -approximation
Post by puzzlecracker on Jan 14^th, 2005, 5:24pm

Listen,

In my country we speak American. But I, on the other, hand, was very irritated about this problem, thereby inadequately trying to describe it at ultra-speed typing (if it can be called so). You, on the other hand, shouldn’t get that offensive, for it was somewhat clear to infer the description.

Your explanation is sublime! Really, the problem is trivial, but was so confused, demn it.

Thanks.

...cracker

Title: Re: interesting problem -approximation
Post by Icarus on Jan 14^th, 2005, 6:02pm

No, it wasn't clear. And it still isn't. I cannot make heads or tails of what you were trying to say in that last paragraph.

I find it frustrating when people post, but make little attempt to make the post understandable. I tried to make a joke of it, and sorry that it offended you. But, please try to say things more clearly. It makes people much more willing to respond if the biggest puzzle isn't trying to guess what was meant.

Title: Re: interesting problem -approximation
Post by SWF on Jan 14^th, 2005, 7:23pm

im srue you don't mind if I type ultrafast too. Normally I'd try to make the equation more readble. This should give some improvment in accuracy over Icarus, but with mroe complxty
16(1024)^(-4)=16(1000+24)^(-4) approx=16(1000^(-4)-4*24*1000^(-5)) approx= 16e-12-16e-11= 1.44e-11
Used first two term of binomial and 4*24 approx= 100