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June 17

Arithmetic

shanu 07:42, 17 June 2008 (UTC)Suppose than x,y,z are real numbers not equal to zero. a & b are two negative real nos. Is it always possible to find another real no. c such that x^a + y^b = z^c . For example consider 2^-1 + 3^-1 = 1.2^-1 . I am thinking of this because fermat's last theorem doesen't allows it for natural nos. greater than 2.

For any given x, y, a and b values the left side of your equation is some constant value: v = x^a + y^b. Then the equation v = z^c with given z has a solution defined by logarithm: c = log_zv = log v / log z. --CiaPan (talk) 07:59, 17 June 2008 (UTC)[reply]

Cube Root

How can we find cube root of any no. ,like 2 without using calculator? —Preceding unsigned comment added by Rohit max (talk • contribs) 07:59, 17 June 2008 (UTC)[reply]

We can use a paper and a pencil. Possibly Tables of logarithms, too... --CiaPan (talk) 08:02, 17 June 2008 (UTC)[reply]

Without a log table it's a little harder. A good way to do it would be to start at some estimate

x_{0}

and repeatedly iterate

x_{n+1}={\frac {2}{3}}x_{n}+{\frac {a}{3x_{n}^{2}}}

. -- Meni Rosenfeld (talk) 08:54, 17 June 2008 (UTC)[reply]

Perhaps this will also be interesting. Not very effective for pen and paper, though. -- Meni Rosenfeld (talk) 09:10, 17 June 2008 (UTC)[reply]

Apparently, there are a couple of "paper and pencil" algorithms for finding cube roots. This is one example (haven't checked it out, to be honest). Pallida Mors 01:41, 18 June 2008 (UTC)[reply]

The simplest way is by the bisection method. Of course, if you actually mean *any* number, you'd have to consider calculating sines and cosines as well...--Fangz (talk) 01:55, 18 June 2008 (UTC)[reply]

-3^2 = ?

I brain farted with the initial question, asking 9^9. This is why Meni answered 'wrongly' - it's my fault -- 88.217.28.51 (talk) 08:54, 17 June 2008 (UTC)[reply]

Hi all,

-3^2 = 9

vs

-3^2 = -9

Which is correct? We can't agree, so I am asking here.. -- 88.217.28.51 (talk) 08:43, 17 June 2008 (UTC)[reply]

See Order of operations. Exponentiation takes precedence over negation. Thus to evaluate -9^2, we first replace 9^2 with 81, giving -81. Thus -9^2 = -81. If we wanted to do the negation first, we would use parentheses: (-9)^2 = (-9)*(-9) = 81. -- Meni Rosenfeld (talk) 08:49, 17 June 2008 (UTC)[reply]

^ vs **

Now, they claim that ^ (not the XOR kind) and ** are different operators. Opinions on that one? -- 88.217.28.51 (talk) 10:38, 17 June 2008 (UTC)[reply]

What is the context? In what way do "they" claim the operators are different? ** is not used in mathematical writing, only in some programming languages. Some languages use ^ to mean exponentiation, and some use **. I do not know of any other difference. -- Meni Rosenfeld (talk) 10:47, 17 June 2008 (UTC)[reply]

My thoughts exactly. Thanks -- 88.217.28.51 (talk) 10:52, 17 June 2008 (UTC)[reply]

The only language I know of that distinguishes between them is Haskell. ^ raises any number to an non-negative integer power (thus the result can have the same type as the base even in the case of integers, and it can be computed by repeated multiplication). ** raises numbers to floating-point powers and gives a floating-point result (in this case we can make no guarantee about whether the result is an integer or even rational, and we have to use a different algorithm to calculate the value). --Taejo|대조 11:51, 17 June 2008 (UTC)[reply]

Python is another example. ** is exponentiation and ^ is XOR. Dragons flight (talk) 16:04, 18 June 2008 (UTC)[reply]

OP said "not the XOR kind" --Taejo|대조 17:44, 22 June 2008 (UTC)[reply]