Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs

Dashboard
Notifications
Mark all as read
Q&A

What's wrong with evaluating $n(n-1) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$ at $k = 1$?

+0
−1

This snag arose out of this post, and these comments by r~~. In that post, I couldn't imagine how

By convention, $n(n-1) \dots {\color{red}{(n-k+1)}} = n$ for k = 1.

Thus I wrote out the LHS $= n(n-1)(n - 2) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$. Then I substituted $k=1$ into this expression, to deduce $LHS|_ {k = 1} = n(n-1)(n - 2) \dots (n+2)(n+1)n \quad \neq \quad n$.

But r~~ counseled that

you can't simply substitute a variable into that kind of informal, descriptive expression without thinking about what the notation as a whole represents. It's not an algebraic expression, so the rules of algebra don't apply. is a signal to look at the pattern being established to the left of the dots, and extend it until it meets what is right of the dots. But if the patterns can't meet, the expression doesn't mean anything.

I don't know why, but I'm unpersuaded. How does my substitution above differ from evaluating a function at an integer like the following? I can re-word my substitution above, if I define $LHS = n(n-1) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$ as $f(n,k)$. Then I just evaluated $f(k=1)$! I see no germane difference!

  • James Stewart's Calculus Early Transcendentals (2011 7e), p xxvii, Diagnostic Test C: Functions.

Image alt text

  • Ron Larson's Calculus (2018 11e), p 20.

Image alt text

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

1 answer

+1
−0

The informal expression stands for the formal expression $$\prod_{j=1}^k (n-j+1)$$ where, since it is a formal expression, you indeed can insert $k=1$, to get $$\prod_{j=1}^1 (n-j+1) = (n-1+1) = n$$ as the text states.

Now the author of the text obviously wanted to avoid the product notation, either to accomodate those who don't know about the product notation, or alternatively on the assumption that it is more easily understandable that way.

The formal definition of the product (ignoring the special case of the empty product, as it is not relevant here) is: \begin{align} \prod_{j=m}^m f(j) &= f(m)\\ \prod_{j=m}^{n+1} f(j) &= \left(\prod_{j=m}^n f(j)\right)f(n+1) \end{align}

Let's apply this to the product above: \begin{align} \prod_{j=1}^1 (n-j+1) &= (n-1+1) = n\\ \prod_{j=1}^2 (n-j+1) &= (n-1+1)(n-2+1) = n(n-1)\\ \prod_{j=1}^3 (n-j+1) &= (n-1+1)(n-2+1)(n-3+1) = n(n-1)(n-2) \end{align} and so on. As you see, the number of factors equals exactly $k$. Now how to write the right hand side for an unspecified $k$? Well, we can't write it down formally since the number of factors depends on $k$. But assuming $k$ is large enough, we know that the first two factors are $n(n-1)$, the last factor id $(n-k+1)$, and the factors are all of the same form, which can be easily derived from the last factor (just replace $k$ by $j$ for the $j$-th factor). Therefore we can informally write that product as $$n(n-1)\ldots (n-k+1)$$ where the missing terms are to be inferred by the reader, and crucially, it is also expected that the reader understands that if $k<3$, there are less terms than are explicitly written.

The author obviously expected the reader to understand this for the case $k=2$ (where the term after the dots equals the second term), but felt the need to explicitly point it out for $k=1$ where also the second term, the one immediately preceding the dots, goes away, so you are left with just the first factor, $n$, which is therefore the value of the full product.

Why does this post require moderator attention?
You might want to add some details to your flag.

1 comment thread

Thanks. Can you please recapitulate and underscore my mistake? I read your answer, but I still don't ... (1 comment)

Sign up to answer this question »

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!

Like what we're doing? Support us! Donate