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January 28

How can I solve this lengthy problem in a simple way?

If n = 2 + 2^1/3 + 2^2/3, then find the value of n³ -6n² + 6n. (Not a homework problem) 106.215.118.66 (talk) 06:38, 28 January 2014 (UTC)[reply]

The simple solution is to use a calculator (and arrive at 2). The other way is a pain of algebra, but quite simple. To simplify writing, put x for 2^1/3, then you have

(x³ + x² + x)(x²(x² + x + 1)² - 6x(x² + x + 1) + 6), after substitutions. As you multiply through, you can simplify at each stage using the relations: xⁿ⁺³ = 2xⁿ (for example x⁶ = 2x³ = 4x⁰ = 4 and 5x⁴ = 10x, etc.). Essentially, after each multiplication, reduce all cubic and higher powers down, eventually, you arrive at 2.Phoenixia1177 (talk) 07:24, 28 January 2014 (UTC)[reply]

If you take the log of both sides, you get ln(n) = 2ln(2). Let f(n) = n³ -6n² + 6n. Then find the log of f(n) and reduce. This equation will be in terms of ln(n), so you can plug in the first equation and solve. It's not entirely clean, but you can simplify it down to a few log calculations. Just a thought! OldTimeNESter (talk) 20:05, 29 January 2014 (UTC)[reply]

How would you find ln(f(n)) in terms of ln(n), though? *edit: you're doing ln(x + y) = ln(x) + ln(y), it should be ln(xy) = ln(x) + ln(y).Phoenixia1177 (talk) 07:22, 30 January 2014 (UTC)[reply]

ln(n)=1.57843 and 2ln(2)=1.38629. Not quite equal. Bo Jacoby (talk) 10:39, 30 January 2014 (UTC).[reply]

A more algebraic approach: let z be a cube root of 2. Then

n-1=z^{2}+z+1={\frac {z^{3}-1}{z-1}}={\frac {1}{z-1}}

\Rightarrow n={\frac {1}{z-1}}+1={\frac {z}{z-1}}

\Rightarrow n^{3}-6n^{2}+6n=2(n-1)^{3}-n^{3}+2={\frac {2-z^{3}}{(z-1)^{3}}}+2=2

Gandalf61 (talk) 13:28, 30 January 2014 (UTC)[reply]

Interval of numbers, worded with "between"

In what way do you use "between" in context with number intervals? Is a number "between 1 and 10" the same as "from 1 to 10", or different in the way that it does exclude 1 and 10? Or is this expression used ambiguously? --KnightMove (talk) 11:29, 28 January 2014 (UTC)[reply]

Unless someone says "inclusive"/"exclusive", there's definitely some ambiguity, but context can usually help with this. I doubt that there are any hard rules pertaining to either case, I would say that if you have doubt, ask.Phoenixia1177 (talk) 11:35, 28 January 2014 (UTC)[reply]

Yes, ask if you are presented with the ambiguity, but write and speak to avoid the potential for confusion :) SemanticMantis (talk) 20:52, 28 January 2014 (UTC)[reply]

Yep, it's technically ambiguous. But, fortunately, careful writing/speaking can always resolve the issue. Consider "X is strictly between 1 and 10", "among 1 to 10, lies a number X", "A number X, one or greater, which is less than 10" (Until relatively recently, a lot of math was written this way!). For reference, see fence post error, Interval_(mathematics)#Notations_for_intervals, and maybe clopen, if you're feeling adventurous. SemanticMantis (talk) 20:50, 28 January 2014 (UTC)[reply]

Also see open interval/closed interval terminology. StuRat (talk) 04:42, 29 January 2014 (UTC)[reply]

Expressions like "between/and" and "from/to" are kind of informal for mathematical writing. What I'd expect to see is either an open/closed interval notation like (1, 10] or an explicit predicate like 1 < x ≤ 10. --50.100.193.107 (talk) 08:16, 29 January 2014 (UTC)[reply]

It would have been better to explain the background: I'm well informed about the formal ways to write in mathematics. The reason for my question was a mathematical puzzle for laymen which, however, was not carefully worded, with many ambiguities, and depending on those the puzzle had many different interpretations. I just wanted to check this specific aspect. Thanks for your answers. --KnightMove (talk) 08:56, 29 January 2014 (UTC)[reply]

Pascal's triangle problem

Any Internet sites where I can see pictures of the following triangles:

Pascal's triangle, with each number replaced by a colored cell where the color depends on the remainder the number leaves when divided by n. Examples:

For n=2, this means take Pascal's triangle and color all even numbers black and all odd numbers red.
For n=3, this means color all multiples of 3 black, all numbers one more than a multiple of 3 red, and all numbers 2 more than a multiple of 3 blue.
For n=4, this means color all numbers depending on remainder left when divided by 4 (black means 0, red means 1, blue means 2, green means 3.)
For n=5, this means color all numbers depending on remainder left when divided by 5 (black=0, red=1, blue=2, green=3, yellow=4.)
For n=6, this means color all numbers depending on remainder left when divided by 6 (black=0, red=1, blue=2, green=3, yellow=4, pink=5.)

And so on. Georgia guy (talk) 21:05, 28 January 2014 (UTC)[reply]

Try putting "pascal's traingle mod 3" into Wolfram Alpha [1]. --RDBury (talk) 02:54, 29 January 2014 (UTC)[reply]