Regression test 回归测试，测试所有的情况。
Speed of convergence 收敛速度
Newton's method 牛顿法
The basic idea is, you take a guess and you find the tangent of that guess 简单的说，先设定一个初始猜测值guess，然后求得该值对应函数的切点斜率。
f(guess) = guess² - x
So let's say I guessed 3, I look for the tangent of the curve at 3. And then my next guess is going to be where the tangent crosses the x axis. So instead of dividing it in half, I'm using a different method to find the next guess. The utility of thie relies upon the observation that, most of the time, tangent line is a good approximation to the curve for values near the solution. And therefore, the x intercept of tangent will be closer to the right answer.
举个例子，先假定猜测值是3，找到3点处的切线。 然后将下一次猜测值guess设定在该切线与X轴相交处。 这里不再是一分为二的方法，而是使用了另一种方法来找到下一个猜测值guess。 这种方法使用依据于一个发现：绝大多数情况下，对于问题的解附近的值来说， 切线是曲线很好的近似。因此切线在X轴上的截距会比先前更接近正确的解。
So how do we find the intercept of the tangent, the x intercept? Well, this is where derivatives come in. What we know is that the slope of the tangent is given by the first derivative of the function f at the point of the guess. So the slope of the guess is the first derivative. Which is dy over dx, change in y divided by change in x.
我们如何得到切线在X轴上的截距。这里导数就派上用场了。 该点切线的斜率将会等于函数一阶导数在该点的值。斜率就是一阶导数。 即dy/dx，y的该变量除以x的该变量。
f'(guessi) = 2*guessi guessi+1 = guessi - f(guessi)/2guessi
以求开根号16为例，guess从3开始： f(3) = 3*3 - 16 = -7 guessi+1 = 3 - (-7 / 2*3) = 4.1666 错过了正解4，但会逐渐接近它。
牛顿法在解决复杂问题时更加出色，比如： 求squareRoot(2, 0.01)时，牛顿法循环了3次，二分法8次。 求squareRoot(2, 0.0001)时，牛顿法循环了4次，二分法14次。 求squareRoot(2, 0.000001)时，牛顿法循环了5次，二分法22次。 随着问题复杂度增加，好方法同差方法之间的差距就会越来越大。
Non-scalar type 非标量 Immutable: Tuples, Strings Mutable - List
Techs = [ 'MIT', 'Cal Tech' ] Ivys = [ 'Harvard', 'Yale', 'Brown' ]
append操作，产生list的list Univs = [] Univs.append(Techs) Univs.append(Ivys) [ [ 'MIT', 'Cal Tech' ], [ 'Harvard', 'Yale', 'Brown' ] ]
连接操作 Univs = Techs + Ivys [ 'MIT', 'Cal Tech', 'Harvard', 'Yale', 'Brown' ]
把各种类型元素保存到list L = [ 'L', 'MIT', 3.3, ['a'] ]
可变性，删除元素，跟split切片很不同 L.remove('MIT')