-1
$\begingroup$

Consider the product series:

$(x-a)$ $(x^{2} -b)$ .......$(x^{14} -n)$

I want to express the result in this form:

$x^{1+2+3+.....14}$ + (some constant)$x^{1+2+3+...14-1}$ +.......so on

My question is how to find the general expression for the coefficients for every powers of x(which are decreasing)?

I have no idea please help me

Is there any formula which can help me?

$\endgroup$
  • $\begingroup$ I would start with setting $a, b, \ldots,n$ to $1$ first and see what you get. Then set them all to $-1$, and see what you get. That's difficult enough, and at the same time ought to give you a good idea of how to attack the general problem. $\endgroup$ – Arthur Jan 31 '18 at 7:33
  • $\begingroup$ @Arthur Yes I completely understand but I need to express it in decreasing powers of x and all I want is some formula or generalization for it's respective coefficients $\endgroup$ – David Jan 31 '18 at 7:36
  • $\begingroup$ @Arthur I think setting them to 1 is an assumption sir. Can't we do it without any assumption? $\endgroup$ – David Jan 31 '18 at 7:41
  • $\begingroup$ You are entirely correct that it is an assumption. As I said, I thiunk you should do it with some restrictive assumptions first, just to see what happens. Then once you've done that, you can try to do the whole thing. $\endgroup$ – Arthur Jan 31 '18 at 7:46
  • $\begingroup$ @Arthur But sir in that case the whole question will become different. But I want to thank you for giving me the idea...Hope someone solves it $\endgroup$ – David Jan 31 '18 at 7:51
0
$\begingroup$

No, there isn't a general formula. You know that the leading term is $x^105$ because $\sum_{k=1}^{14}{k}=105.$ Now the next term must be $-ax^{104}$ because there's only one way to get $x^{104}.$ Similarly, the third term is $-bx^{103}.$ But when we get to $x^{102}$ things change. We might have multiplied all the variables but $x^3$ or all the variables but $x$ and $x^2$, so the coefficient is $ab-c$. Things will get more complicated as you go along. There are $2^{14}=16,384$ ways to pick which constants to multiply together, and only 106 terms among which to distribute the products, so it's going to get messy.

May I ask why you want to know the individual terms? It seems like you've got a pretty simple formula now.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sir, I want to know it because I require it for a problem in my exercise $\endgroup$ – David Jan 31 '18 at 7:48
  • $\begingroup$ I still believe there can be some way sir $\endgroup$ – David Jan 31 '18 at 7:54
  • $\begingroup$ David, there has to be something more to this problem. This would be far too arduous as an exercise. No teacher would make you compute 16,000 terms -- he wouldn't be able to grade it! What exactly is the problem statement? $\endgroup$ – saulspatz Jan 31 '18 at 7:55
  • $\begingroup$ It was given as a bounty problem. I guess we need to invent some formula for it if it is possible. The problem was the same written here. I just want to devise some formula not because of the reward but because of the love for mathematics $\endgroup$ – David Jan 31 '18 at 8:06
  • $\begingroup$ I wish you good luck, but I'm not optimistic. $\endgroup$ – saulspatz Jan 31 '18 at 19:17

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.