Recursive defintion of the product notation

[neutrino]

I'm asked to define recursively (definition by induction) $\prod_{k=1}^{n}a_k$

Well, I wrote down the following:

$$\prod_{k=1}^{1}a_k = a_1$$

Assuming $\prod_{k=1}^{n}a_k$ has been defined for some $n\geq1$,

$$\prod_{k=1}^{n+1}a_k = a_{n+1}\prod_{k=1}^{n}a_k$$

A similar method was used to define the summation notation in the text, so I used it here.

But the answer given at the back is

$$\prod_{k=1}^{0}a_k = 1; \prod_{k=1}^{n+1}a_k = a_{n+1}\prod_{k=1}^{n}a_k$$

I don't understand why the index goes from 1 to 0, and why they have defined it to be 1. Please clarify this.

 

[mathman]

A little background - in Quantum mechanics a massless particle is needed to carry a force, the photon carries the electric force and W and Z particles carry the forces that hold atoms together.

It is a somewhat pedantic point. When the upper limit goes from n=0 to n=1 (induction step), you need 1 for the n=0 product to get a₁ for the n=1 product.

 

[Moo Of Doom]

The reason they define the product starting from k=1 to 0 is to define the empty product to be 1. This is a useful definition, as it avoids many special cases. It is clearly an equivalent definition to yours for all upper bounds greater than 0, since their inductive step is the same and by their definition,

$$\prod_{k = 1}^1 a_k = a_1 \prod_{k = 1}^0 a_k = a_1 \cdot 1 = a_1.$$

 

[neutrino]

Thanks for the replies. Could you give me a simple example where this definition will be useful? Also, why was not the summation notation defined to be $\sum_{1}^{0} = 0$? [As implied in my first post, $a_1$ was defined as the sum from 1 to 1.]

 

[mathman]

In summation notation sum from 1 to 0 is considered 0.