Series Problem

M

Morteza

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Jul 20, 2021
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I am working on a physics problem on my own research and I got the following series.

112−x2-x3−2x+x23-2x+x^24−3x+2x2−x34-3x+2x^2-x^35−4x+3x2−2x3+x45-4x+3x^2-2x^3+x^4...........................
and so on,
Where, 1>x>−11 >x>-1 and the number of sentences of series go to infinity.
Since I am stuck from the beginning and I couldn’t do anything with that, I can’t show my work in order to solve this problem. First, is there any limit for this series and second, if there is a limit, what is that?

Thanks for your help
 
Morteza said:
I am working on a physics problem on my own research and I got the following series.

112−x2-x3−2x+x23-2x+x^24−3x+2x2−x34-3x+2x^2-x^35−4x+3x2−2x3+x45-4x+3x^2-2x^3+x^4...........................
and so on,
Where, 1>x>−11 >x>-1 and the number of sentences of series go to infinity.
Since I am stuck from the beginning and I couldn’t do anything with that, I can’t show my work in order to solve this problem. First, is there any limit for this series and second, if there is a limit, what is that?

Thanks for your help
Click to expand...
What kind of limit can you expect if, according to Pedja's comment, your constant grows as nn?
 
Thank you Pedja and Blamocur,

I need to know is there any limit, if ′n′'n' grows to infinity? Where 1>x>−11>x>-1
If so, would you please solve it.
 
Morteza said:
Thank you Pedja and Blamocur,

I need to know is there any limit, if ′n′'n' grows to infinity? Where 1>x>−11>x>-1
If so, would you please solve it.
Click to expand...
If, for example, you set x=0x=0 then the nnth term is equal to nn, i.e., it grows to infinity. I.e., there is no limit.
Pedja said:
Your polynomials satisfy the following recurrence: Pn=−xPn−1+n with P0=0P_n=-xP_{n-1}+n \text{ with } P_0=0
Click to expand...
And thus Pn(x)=∑k=0nkxn−kP_n(x) = \sum_{k=0}^n k x^{n-k}
 
No, no, no. You need to try to solve this recurrence. If you need help doing so, then you'll receive it. This is what is meant by a math help forum. Maybe you should read/reread the posting guidelines for this forum.
 
So, the recurrence is:
Pn=−xP(n−1)+nP_n = -xP_(n-1) +n with P0=0P_0 = 0
P0=0P_0 = 0
P1=1P_1 = 1
P2=−x+2P_2 = -x+2
P3=+x2−2x+3P_3 = +x^2-2x+3
P4=−x3+2x2−3x+4P_4 = -x^3+2x^2-3x+4

P1−P0=1P_1-P_0 = 1
P2−P1=1−xP_2-P_1 = 1-x
P3−P2=1−x+x2P_3-P_2 = 1-x+x^2
P4−P3=1−x+x2−x3P_4-P_3 = 1-x+x^2-x^3
P(n)−P(n−1)=1−x+x2−x3+...P_(n)-P_(n-1) = 1-x+x^2-x^3+... = AA
P(n−1)=P(n)−AP_(n-1) = P_(n) -A

Since we had:
Pn=−xP(n−1)+nP_n = -xP_(n-1) +n
Pn=−x(Pn−A)+nP_n = -x(P_n-A)+n
Pn=−xPn+xA+nP_n=-xP_n+xA+n
Pn+xPn=xA+nP_n+xP_n=xA+n
Pn=xA+nx+1P_n=\frac{xA+n}{x+1}
If ′n′'n' grows to infinity, then,
Pn=x+(x+1)n(x+1)2P_n=\frac{x+(x+1)n}{(x+1)^2}And so PnP_n grows to infinity.

But, I’m not sure that I got the correct answer. Any help, appreciated.
 
Hi everyone, I need your comment for my solution to see if I’m correct.

Thanks
 
Morteza said:
If ′n′'n'′n′ grows to infinity, then,
Pn=x+(x+1)n(x+1)2P_n=\frac{x+(x+1)n}{(x+1)^2}Pn=(x+1)2x+(x+1)n
Click to expand...
I do not "get" that step (the last equation you wrote) - can you please show your intermediate steps
 
Subhotosh Khan said:
I do not "get" that step (the last equation you wrote) - can you please show your intermediate steps
Click to expand...
If 'n' grows to infinity, then,
A=1x+1A=\frac{1}{x+1}
So we get,
Pn=x+(x+1)n(x+1)2P_n=\frac{x+(x+1)n}{(x+1)^2}
 
Morteza said:
If 'n' grows to infinity, then,
A=1x+1A=\frac{1}{x+1}
So we get,
Pn=x+(x+1)n(x+1)2P_n=\frac{x+(x+1)n}{(x+1)^2}
Click to expand...
Correct.

In my opinion you should restate "since |x| < 1 " to be complete.
 
Morteza said:
If 'n' grows to infinity, then,
A=1x+1A=\frac{1}{x+1}
So we get,
Pn=x+(x+1)n(x+1)2P_n=\frac{x+(x+1)n}{(x+1)^2}
Click to expand...
Not sure I agree with this statement: you are using the same AA for different nn's. I.e., when you write that Pn−Pn−1=AP_n - P_{n-1} = A you are assuming that the left hand side does not depend on nn, which is not true. A cleaner way would be Pn−Pn−1=AnP_n - P_{n-1} = A_n, and lim⁡n→∞An=1x+1\lim_{n\rightarrow\infty} A_n = \frac{1}{x+1}. But your final expression for PnP_n seems incorrect too.
 
blamocur said:
Not sure I agree with this statement: you are using the same AA for different nn's. I.e., when you write that Pn−Pn−1=AP_n - P_{n-1} = A you are assuming that the left hand side does not depend on nn, which is not true. A cleaner way would be Pn−Pn−1=AnP_n - P_{n-1} = A_n, and lim⁡n→∞An=1x+1\lim_{n\rightarrow\infty} A_n = \frac{1}{x+1}. But your final expression for PnP_n seems incorrect too.
Click to expand...
I noted "If 'n' grows to infinity, then," A=1x+1A=\frac{1}{x+1}. So, it means that A is related to 'n'
And please explain why my final expression is incorrect?
 
Not sure how to interpret "A is realated to 'n'", but 1x+1\frac{1}{x+1} does not depend on nn as far as I can tell.
And your final expression for PnP_n does not match those of the your original post. E.g., in you original post P1=2−xP_1 = 2-x, but in the final expression P1=2x+1(x+1)2P_1 = \frac{2x+1}{(x+1)^2}
 
'A' is the limit of a series when 'n' grows to infinity.
The final expression 'Pn is true, when 'n' grows to infinity.
 
"PnP_n is true when 'n' grows to infinity" -- I don't know how to interpret this either. If you are talking about the limit of PnP_n when 'n' grows to infinity then you should put the limits on both sides of the '=' sign, but in this case the left hand side is obviously infinity and you would still have to prove it.
But I agree with you that lim⁡n→∞(Pn−Pn−1)=1x+1\lim_{n\rightarrow \infty} (P_n-P_{n-1}) = \frac{1}{x+1}. Since the right hand side is positive and constant for all x>−1x > -1 this is enough to prove that lim⁡n→∞Pn=∞\lim_{n\rightarrow \infty} P_n = \infty for all such xx's.
 
I didn't write it down professionally as you did. But the result is correct.
 
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