Online Software for Polynomial Multiplication

Question

I have been working with Lagrange polynomials and I have some very complicated calculations such as $$(x-1)(x-2)(x-3)(x-4)(x-5)(x-22)$$ Can you suggest an (possibly give url) online application which will give out the answer. Also I have Wolfram Mathematica 8, but I do not know how to use it.

Answer 1

The website Wolfram Alpha will do it for you. Just type, e.g.

$$\tt expand (x-1)*(x-2)*(x-3)*(x-4); $$

I typed that in, and here's what I got. Your expression will, of course, be a bit longer.

If you have a product, say

$$\prod_{n=1}^{10} \left(x-\frac{1}{n}\right) = \left(x-1\right)\left(x-\frac{1}{2}\right)\left(x-\frac{1}{3}\right)\cdots \left(x-\frac{1}{10}\right) $$

it'll even do that for you. Just type:

$$\tt expand \ \ (product \ (x-1/n) \ , \ n=1..10)$$

I did that, and here's what I got.

Answer 2

It's an interesting problem. You can either expand it with any application (online or desktop) but if you want to know how to write it, you should check what is called the "Guelf expansion". It states that:

$$\displaystyle\prod_{i=1}^n(x-\lambda_i)=x^n+\sum_{k=1}^n\left((-1)^k x^{n-k}C_k(\lambda_1,\ldots,\lambda_n)\right)$$

Where $C_k$ is the sum of the products of all possible $k$-tuples from the set $\{\lambda_1,\ldots,\lambda_n\}$. There is exactly ${n \choose k}$ possible tuples for each $C_k$.

Example for $n=4$: $$P(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)$$

The expanded polynomial is:

$$P(x)=x^4-C_1x^3+C_2x^2-C_3x+C_4$$

And the coefficients are given by:

$$\cases{C_1=a_1+a_2+a_3+a_4 \\ C_2=a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4 \\ C_3=a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4 \\ C_4=a_1a_2a_3a_4}$$

In your case you get:

$$P(x)=x^6-C_1x^5+C_2x^4-C_3x^3+C_4x^2-C_5x+C_6$$

$$\mbox{With}\space\cases{C_1=1+2+3+4+5+22 \\ C_2=1\cdot2+1\cdot3+1\cdot4+1\cdot5+1\cdot22+2\cdot3+\cdots \\ C_3=1\cdot2\cdot3+1\cdot2\cdot4+1\cdot2\cdot5+1\cdot2\cdot22+1\cdot3\cdot4+\cdots \\C_4=1\cdot2\cdot3\cdot4+1\cdot2\cdot3\cdot5+1\cdot2\cdot3\cdot22+\cdots \\ C_5=1\cdot2\cdot3\cdot4\cdot5+1\cdot2\cdot3\cdot4\cdot22+1\cdot3\cdot4\cdot5\cdot22+\cdots \\ C_6=1\cdot2\cdot3\cdot4\cdot5\cdot22}$$

Which gives you:

$$\cases{C_1=37 \\ C_2=415 \\ C_3=2095 \\C_4=5224 \\ C_5=6148 \\ C_6=2640}$$

And you indeed get:

$$P(x)=x^6-37 x^5+415 x^4-2095 x^3+5224 x^2-6148 x+2640$$

I agree this is quite a long procedure for big $n$ but up to $n=5$ or even $n=6$ it can be good to know. And especially if you don't have access to a computer (during exams for instance) ;)