Expressing with "product" notation

[shamieh]

1]express j! in ∏ notation

Are they just wanting something like \(\displaystyle j! + (j-1)! + (j-2)! +(j-3)!\)...?

 

[MarkFL]

No, they want you to take:

\(\displaystyle j!=1\cdot2\cdot3\cdots(j-2)(j-1)j\)

And rewrite the right side using \(\displaystyle \prod\) notation. Are you familiar with how to use this notation?

 

[shamieh]

No I'm not familiar.

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\(\displaystyle \prod^n_{i=1} j!\) something like this?

 

[MarkFL]

It works like this:

\(\displaystyle \prod_{k=1}^n\left[f(k)\right]\equiv f(1)\cdot f(2)\cdot f(3)\cdots f(n-2)\cdot f(n-1)\cdot f(n)\)

So, what do you suppose the relationship between $f$ and the index $k$ would be? What do you suppose should be the lower and upper limit for the index $k$?

 

[shamieh]

f is always multiplied by k.

\(\displaystyle \prod^j_{k=1} n!\)

 

[MarkFL]

f is always multiplied by k.

\(\displaystyle \prod^j_{k=1} n!\)

You have the limits correct, but what you have written is:

\(\displaystyle \prod^j_{k=1} n!=(n!)^j\)

Compare the two expressions I gave:

\(\displaystyle j!=1\cdot2\cdot3\cdots(j-2)(j-1)j\)

\(\displaystyle \prod_{k=1}^n\left[f(k)\right]\equiv f(1)\cdot f(2)\cdot f(3)\cdots f(n-2)\cdot f(n-1)\cdot f(n)\)

Do you see that in this case we want:

\(\displaystyle f(k)=k\)

Hence:

\(\displaystyle \prod_{k=1}^j\left[k\right]=j!\)

 

[shamieh]

OH I see what you're saying..

 

[MarkFL]

Because of the commutativity of multiplication, you could also write:

\(\displaystyle j!=\prod_{k=0}^{j-1}\left[j-k\right]\)