Expressing with "product" notation

[shamieh]

1]express j! in ∏ notation

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

 

[MarkFL]

No, they want you to take:

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

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

 

[shamieh]

No I'm not familiar.

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$$\prod^n_{i=1} j!$$ something like this?

 

[MarkFL]

It works like this:

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

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.

$$\prod^j_{k=1} n!$$

 

[MarkFL]

f is always multiplied by k.

$$\prod^j_{k=1} n!$$

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

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

Compare the two expressions I gave:

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

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

Do you see that in this case we want:

$$f(k)=k$$

Hence:

$$\prod_{k=1}^j\left[k\right]=j!$$

 

[shamieh]

OH I see what you're saying..

 

[MarkFL]

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

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