For example, mathematicians recently discovered a way to expand it to these kinds of polynomials.

例，数学家最近发现了一个展开这种多项式的方法。

Charles Babbage proposed a new mechanical device called the Difference Engine, a much more complex machine that could approximate polynomials.

Charles Babbage提出了一种称为差分机的新型机械装置这是一个可以近似多项式的更加复杂的机器。

However, some of you might complain that the only way that this is useful is  if we can actually evaluate this wild expression on our polynomial.

然，你们中的一些人可能会抱怨，这种方法只有在我们能够实际计算出这个多项式的疯狂表达式的情况下才有用。

And if I had to teach a lesson on synthetic division on fourth order polynomials, you better bet I would prepare for it.

果我不得不教授关四次多项式的综合法程，那你最好相信我会为此做好准备。

The only roots of that polynomial are the imaginary numbers I and negative eye.

那个多项式的根只有虚数I和负眼。

And these don’t have to be straight lines; they can be polynomials or some other fancy mathematical function.

这些不一定是直线，可以是多项式或其他复杂的数学函数。

A formula which involves an expansion in separate terms, polynomial terms as we say, for certain functions.

一个涉及特定函数展开式的公式，我们称之为多项式项的独立项展开。

It turns out though, if you think about the group which permutes the roots of such a polynomial, there's something about the nature of this group that reveals that no quintic formula can exist.

事实证明， 果你考虑一下置换这种多项式根的群，这个群的性质揭示了不存在五次方程的求根公式。

Namely, taking a  discrete sequence that we want to understand and treating it as the coefficients on a polynomial, then evaluating that polynomial on complex values.

也就是说， 取一个我们想要理解的离散序列， 并将其视为多项式的系数，然后在复数值上评估该多项式。

There's even a quartic formula for degree 4 polynomial, which is an absolute mess.

甚至有解四次多项式的公式，那简直是一团糟。

Which in the back of your mind, you can just think of as treating them like they're the coefficients of a polynomial and evaluating it at a very specially selected set of points.

这在你的脑海中，可以简单地认为是将它们视为多项式的系数，并在一个非常特别选择的点集上进行评估。

So for example, if you want to find the complex roots of a polynomial, or some of those million dollar zeros of the Riemann zeta function, this algorithm will do it for you.

所以例， 果你想找到多项式的复数根，或者黎曼zeta函数中那些价值百万美元的零点，这个算法就能帮你找到。

And letting our equation solver continue in this same way, it eventually converges on lots of zeros of this polynomial.

让我们的方程求解器继续以这种方式运行，它最终会收敛到这个多项式的许多零点。

And then suppose I say, I'll just tell you what the outputs of this polynomial would be if you inputted various different inputs like 0, 1, 2, 3 on and on.

然后假设我说，我会告诉你这个多项式在输入不同的值比0, 1, 2, 3等等时的输出结果。

Just remember each pixel on the right  corresponds to a unique polynomial, as determined by this parameter lambda.

只需记住右侧的每个像素对应一个唯一的多项式，由参数 lambda 确定。

And, in this case, it's an infinite polynomial so  I should really be calling it a power series.

，在这种情况下，它是一个无穷多项式的级数，所以我应该称之为幂级数。

Another way to think about that is  that they are roots of the polynomial z to the fifth minus one.

另一种思考方式是，它们是多项式z的五次方减一的根。

In the case of the functions arising from Newton's method, by  design they have a fixed point at the roots of the relevant polynomial.

在牛顿法产生的函数情况下，设计上它们在相关多项式的根处有一个不动点。

Points in this parameter space are colored black  if and only if the Newton's method process for the corresponding polynomial produces an attracting  cycle.

在这个参数空间中，果仅果对应多项式的牛顿法过程产生一个吸引周期，则该点被着色为黑色。

So the rough algorithm outline then would be whenever you want to convolve two lists of numbers, you treat them like they're coefficients of two polynomials.

因此，粗略的算法概述是， 每当您想要对两个数字列表进行卷积时，您都将它们视为两个多项式的系数。