using Pascal's triangle for an investigation

Grandpa Bob

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if we let x= (y +1/y) , and call
yn+ 1/yn = Pn (the n should be suffices throughout)
so that P1 -y +1/y = x
then we can find P2, P3, etc and ideally derive a formula for Pn in terms of x?
for example (y +1/y)²= P2 => P2= x²=2
similarly P3= x³- 3y
I can get P4- P10 but can’t get to a formula
Can anyone help?
 
if we let x= (y +1/y) , and call
yn+ 1/yn = Pn (the n should be suffices throughout)
so that P1 -y +1/y = x
then we can find P2, P3, etc and ideally derive a formula for Pn in terms of x?
for example (y +1/y)²= P2 => P2= x²=2
similarly P3= x³- 3y
I can get P4- P10 but can’t get to a formula
Can anyone help?
I think a lot of this is garbled (including the word "suffices", which is probably meant to be "superscripts" or "exponents"). It will be good if you can show us an image of the source from which you got this idea, and/or of your work on paper, so we can be sure what you mean.

I think it may be something like this:

If we let x = y + 1/y, that is, `x = y + 1/y`,​
and P_n = y^n + 1/y^n, that is, `P_n = y^n + 1/y^n`,​
so that P_1 = y + 1/y = x, that is, `P_1 = y + 1/y = x`,​
then we can successively find P_2, P_3, and so on to derive a formula for P_n in terms of x.​
For example, x^2 = (y + 1/y)^2 = y^2 + 2 + 1/y^2 = P_2 + 2, that is, `x^2 = (y + 1/y)^2 = y^2 + 2 + 1/y^2 = P_2 + 2`​
so that P_2 = x^2 - 2, that is, `P_2 = x^2 - 2`.​

Part of what I am doing is teaching you a better notation for typing math; and everything that follows "that is" is the very same thing put between "back quotes", which on my keyboard is just below ESC. That provides an easy way to display math neatly.

Let me know if I am interpreting correctly, and then show us your work (perhaps as an image) so we can see whether you did it correctly, and then work on bringing Pascal's triangle into this and maybe finding a general formula.
 
I don't know formulas; I only know how to think. In particular, I don't know a simple formula for this, but could look it up if that were the goal. Did you make up this problem, so that you don't know if there is a formula, or is it an exercise somewhere that implies there is one?

As far as terminology goes, subscripts and superscripts, or indices and suffices, are not the same thing, which is why I commented on your "yn+ 1/yn = Pn (the n should be suffices throughout). I will admit that though I am familiar with "indices", I don't think I've seen "suffices".

As for "it is considered inappropriate to write mathematics work using ^ as a power/index/superscript as it is computing notation", that is not true online. It is standard for typing math, if you don't want to use the superscript button on the interface (x2) or the LaTeX markup language. See here, for example: https://www.purplemath.com/modules/mathtext.htm
 
I'd be interested if you'd answer @Dr.Peterson 's question (just out of curiosity)...
...Did you make up this problem, so that you don't know if there is a formula, or is it an exercise somewhere that implies there is one?

--

I had a go at finding a solution for odd number n (I didn't consider a formula for even n). Using the binomial theorem a recurrence relation was easy enough to find. I can't currently see a way to directly obtain a RHS involving only x.

[math]P_n=x^n - \sum_{k=1}^{(n-1)/2}{\binom{n}{k}P_{n-2k}} [/math]
P3 = x^3 - 3*P1
P5 = x^5 - 5*P3 - 10*P1
P7 = x^7 - 7*P5 - 21*P3 - 35*P1
P9 = x^9 - 9*P7 - 36*P5 - 84*P3 - 126*P1
etc

I wrote a computer program, using the above idea, to find the RHS in terms of x via recursion...
P3 = x^3 -3*x
P5 = x^5 -5*x^3 +5*x
P7 = x^7 -7*x^5 +14*x^3 -7*x
P9 = x^9 -9*x^7 +27*x^5 -30*x^3 +9*x
P11 = x^11 -11*x^9 +44*x^7 -77*x^5 +55*x^3 -11*x
P13 = x^13 -13*x^11 +65*x^9 -156*x^7 +182*x^5 -91*x^3 +13*x
P15 = x^15 -15*x^13 +90*x^11 -275*x^9 +450*x^7 -378*x^5 +140*x^3 -15*x
P17 = x^17 -17*x^15 +119*x^13 -442*x^11 +935*x^9 -1122*x^7 +714*x^5 -204*x^3 +17*x
P19 = x^19 -19*x^17 +152*x^15 -665*x^13 +1729*x^11 -2717*x^9 +2508*x^7 -1254*x^5 +285*x^3 -19*x
etc
 
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