Series Problem

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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.

[math]1[/math][math]2-x[/math][math]3-2x+x^2[/math][math]4-3x+2x^2-x^3[/math][math]5-4x+3x^2-2x^3+x^4[/math]...........................
and so on,
Where, [imath]1 >x>-1[/imath] 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.

[math]1[/math][math]2-x[/math][math]3-2x+x^2[/math][math]4-3x+2x^2-x^3[/math][math]5-4x+3x^2-2x^3+x^4[/math]...........................
and so on,
Where, [imath]1 >x>-1[/imath] 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 [imath]n[/imath]?
 
Thank you Pedja and Blamocur,

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

I need to know is there any limit, if [imath]'n'[/imath] grows to infinity? Where [math]1>x>-1[/math]
If so, would you please solve it.
Click to expand...
If, for example, you set [imath]x=0[/imath] then the [imath]n[/imath]th term is equal to [imath]n[/imath], i.e., it grows to infinity. I.e., there is no limit.
Pedja said:
Your polynomials satisfy the following recurrence: [imath]P_n=-xP_{n-1}+n \text{ with } P_0=0[/imath]
Click to expand...
And thus [imath]P_n(x) = \sum_{k=0}^n k x^{n-k}[/imath]
 
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:
[imath]P_n = -xP_(n-1) +n[/imath] with [imath]P_0 = 0[/imath]
[imath]P_0 = 0[/imath]
[imath]P_1 = 1[/imath]
[imath]P_2 = -x+2[/imath]
[imath]P_3 = +x^2-2x+3[/imath]
[imath]P_4 = -x^3+2x^2-3x+4[/imath]

[imath]P_1-P_0 = 1[/imath]
[imath]P_2-P_1 = 1-x[/imath]
[imath]P_3-P_2 = 1-x+x^2[/imath]
[imath]P_4-P_3 = 1-x+x^2-x^3[/imath]
[imath]P_(n)-P_(n-1) = 1-x+x^2-x^3+... [/imath] = [imath]A[/imath]
[imath]P_(n-1) = P_(n) -A[/imath]

Since we had:
[imath]P_n = -xP_(n-1) +n[/imath]
[imath]P_n = -x(P_n-A)+n[/imath]
[imath]P_n=-xP_n+xA+n[/imath]
[imath]P_n+xP_n=xA+n[/imath]
[imath]P_n=\frac{xA+n}{x+1}[/imath]
If [imath]'n'[/imath] grows to infinity, then,
[math]P_n=\frac{x+(x+1)n}{(x+1)^2}[/math]And so [imath]P_n[/imath] 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
 
Morteza said:
If 'n' grows to infinity, then,
[imath]A=\frac{1}{x+1}[/imath]
So we get,
[math]P_n=\frac{x+(x+1)n}{(x+1)^2}[/math]
Click to expand...
Correct.

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