Simplify me

[Bazzy]

√16 = ±4
4² = (-4)² = 16
It was already stated above
The sign of 4 doesn't matter since it's removed by the even exponent: f ( x² ) = f ( (-x)² ) ∀ x ∈ ℝ, ∀ f : ℝ→ℝ

[gcampton]

Wolfram alpha is bad.

type in: November 23, 1982

results:

Time difference from today (Saturday, October 9, 2010):
27 years 10 months 16 days ago

[hamsterman]

Wolfram does indeed fail with sqrt(16),
but what's wrong with the dates?

  1982 11 23
+   27 10 16
= 2009 21 39
= 2010 10 9

[chrisname]

@hamsterman,
How does it fail? sqrt(16) = 4, sqrt(-16) = 4i. Those are both correct.

[hamsterman]

Bazzy wrote:

√16 = ±4
4² = (-4)² = 16
It was already stated above

OK. Maybe that's a problem of sqrt, not wolfram. It's either not a function opposite of x² or not a function at all.

[S6k9GNh0]

Wolfram|Alpha is excellent, it can do so much. Try "bacon as mass of solar system" ( http://www.wolframalpha.com/input/?i=bacon+as+mass+of+solar+system ).

Why is my regular diet on a C++ forum?

[chrisname]

Uhhh... two questions.
1. How often are you eating THAT MUCH bacon?
2. Are you greater in mass than the solar system, and if so, by how much?

[tition]

But √((x)²) = |x|

:

Actually, it depends on your definition of sqrt() function, so the answer might be no. The key words you should google are "branch of square root".

The most conceptually correct way of defining square root function is using the complex numbers. One of the clean way: set z^{1/2}= e^{1/2 ln(z)}, where ln stands for natural logarithm. The reason this definition is clean is that

e^w = 1 + w/1! + w^2/2! + w^3/3! + ...

is a well defined power series, easy to handle. So is the natural logarithm:

ln(1+z) = z/1 - z^2/2 + z^3/3 - z^4/4 + ...

If you plug in w= 1/2 ln(1+(z-1)) in e^w and uncover the infinite brackets you will get one power series expansion of z^{1/2} (for z close enough to 1).

The problem with the sqrt() function: in the above expression,

ln(1+z) = z/1 - z^2/2 + z^3/3 - z^4/4 + ... 
ln(1+z) = 2i\pi+ z/1 - z^2/2 + z^3/3 - z^4/4 + ... 
ln(1+z) = 4i\pi+ z/1 - z^2/2 + z^3/3 - z^4/4 + ... 
ln(1+z) = 6i\pi+ z/1 - z^2/2 + z^3/3 - z^4/4 + ... 

are all equally good definitions of the logarithm function (because e^{2i\pi}=1). The even multiples of 2i\pi will give you the positive square root, and the odd multiples - the negative.