5
$\begingroup$

So from what I've seen of sequences a common format would be something like $a_{n+1} = 2a_n + 5 $

I was doing some review for sequences and come across a format that looks something like $a_{n+1}=a_n^2-1$ given $a_1=1$

According to my solution manual, $a_1=1,a_2=0,a_3=-1,a_4=0$

I initially thought that maybe it was just a weird notation and I could write it as $a_n$ but that would just make is a continually decreasing sequences so $a_1=1,a_2=0,a_3=-1,a_4=-2,a_5=-3$ and so on. So this doesnt fit the answer.

I'd like to know how this $a_n^c$ where c is just some number actually works in case it comes up on a test. Any example at all or just an explanation would be great, I can't actually find this notation in the section before the question.

For reference, this is in Briggs, Cochran Calculus Early Transcendentals 2nd edition, 8.1 #20

Edit: Thanks to u/Rob Arthan for this

Since $a_n^2 = a_n \times a_n=(a_n)^2$ this makes the answers from above as

$a_2=(a_1)^2-1=1^2-1=0$

$a_3=(a_2)^2-1=0^2-1=-1$

$a_4=(a_3)^2-1=(-1)^2-1=1-1=0$

and so on and so forth

$\endgroup$
  • $\begingroup$ $a_n^c$ here means $a_n$ to the $c$-th power. So $a_n^2 = a_n\cdot a_n$, the square of $a_n$. (People do occasionally use superscripts as indexes rather than exponents but that is fairly rare.) $\endgroup$ – Rob Arthan Nov 19 '17 at 0:04
  • $\begingroup$ That makes a lot of sense and clears this up a lot. Thanks, i'll update my question accordingly and I think that officially makes this solved. Not really sure how to mark is solved without a submitted answer though $\endgroup$ – Vin Nov 19 '17 at 0:14
  • $\begingroup$ You can also post your own answer $\endgroup$ – Dylan Nov 19 '17 at 0:45
1
$\begingroup$

In the comments, Rob Arthan wrote that "People do occasionally use superscripts as indexes rather than exponents but that is fairly rare." In my opinion, that is reason enough to consider $a_n^c$ as very poor style when ${a_n}^c$ is meant. For extra clarity, you can use parentheses, like so: $(a_n)^c$.

Either way, that's two extra bytes, or four if you're using UTF-16. That's nothing, at least until Trump's swamp FCC destroys net neutrality. In terms of ink expenditure, another pair of parentheses is also negligible in a book of more than a thousand pages.

So yeah, $a_{n + 1} = a_n^2 - 1$ should have been written as $a_{n + 1} = {a_n}^2 - 1$ or better yet $a_{n + 1} = (a_n)^2 - 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.