Doubt in finding general term of the given sequence

Question

The following image has both the problem and its solution. I have a doubt in the solution, the details of which I have included below the image.

(Assume the terms in given series are generated from a polynomial)

Here, the author has assumed $T_n$ as an arbitrary cubic equation (Step indicated by the RED box).

My doubt is, why has he assumed it as a cubic equation and not a quadratic or biquadratic equation or any other degree equation?

Please do not use Newton's forward interpolation rule as I don't know such rules. Please explain in simple terms and properties.

Kindly clarify my doubt.

This is not a Mathematically valid question. The entire so-called solution makes no sense. — Kavi Rama Murthy, Aug 17 '19 at 06:17
@KaviRamaMurthy, if the so-called solution makes no sense, kindly give me a valid solution. This is from a book for JEE Mains and Advanced. — , Aug 17 '19 at 06:38
THIS A MATHEMATICS WEBSITE AND THE QUESTION DOES NOT MAKE SENSE IN MATHEMATICS. — Kavi Rama Murthy, Aug 17 '19 at 06:41
@KaviRamaMurthy, But the question is regarding Mathematics only, and its from the chapter Sequences and Series. — , Aug 17 '19 at 06:43
The author is essentially using the following fact: A sequence of real numbers ${ a_n}$ has constant (and non-zero) $d$-th differences if and only if $a_n$ is a $d$-th degree polynomial function of $n$. — Minus One-Twelfth, Aug 17 '19 at 06:45
Even 10 billion terms of sequence do not uniquely determine the next term. — Kavi Rama Murthy, Aug 17 '19 at 06:46
May I request the person who upvoted my comments to kindly vote for closure of this question? — Kavi Rama Murthy, Aug 17 '19 at 06:48
@KaviRamaMurthy, Understood that. But here we must assume they follow the property determined by the given set of numbers. — , Aug 17 '19 at 06:48
No property of the sequence given in the question. Only the first few terms are given. — Kavi Rama Murthy, Aug 17 '19 at 06:49
I agree with @kavi here. This problem is ill-posed. Maybe, as the answer states, you need the extra assumption that the sequence is generated by some polynomial. Jee kills formal statement and rigor many a times, especially, inadvertently. — Someone, Aug 17 '19 at 07:31
Possible duplicate of Doubt in finding the general term and sum of $n$ terms of the series $1+5+19+49+101+181+295+\dots+T_n$? — Michael Hoppe, Aug 17 '19 at 09:55
This is not a duplicate. Here the OP is looking for an alternative solution. — Toby Mak, Aug 17 '19 at 10:06
@MichaelHoppe, Yes. I actually copied most of the question from that. But my question is to look for another solution as I didn't understand any of the answers in that question. I wanted to look for an alternative solution without using Newton's forward interpolation rule. That's why I posted with such specification. — , Aug 17 '19 at 10:10
@MichaelHoppe, A similar question has been asked by Intellex as a follow up question in the comment of the accepted answer and is not answered by the author of the answer. But now, I got some useful links from the answers obtained from my question, that cleared my doubts. — , Aug 17 '19 at 10:27

John Omielan · Accepted Answer · 2019-08-17T07:58:34.390

3

The author is assuming the original terms of the series are values from a polynomial at increasing integral values. Thus, when a sequence of differences causes all of the terms to be the same at some level $m$, the sequence $T_n$ represents a polynomial of degree $m$. For more information, see Theorem $1$ in Difference Tables of Sequences. In your case, taking the sequence of the third order differences based on those second order differences causes all of the values to be $6$. This means the original sequence represents a cubic equation.

Also, although the author doesn't use this, and the linked page only hints at it, the sequence of constant values are $a_n(n!)$ where $a_n$ is the coefficient of the $x^n$ term. Thus, in this case, $6 = a_3(3!) \implies a_3 = 1$. The author could have used this to determine that $a = 1$ more directly.

edited Aug 17 '19 at 07:58

answered Aug 17 '19 at 06:18

John Omielan

47,976

Thank you for your answer. So, as the third difference here is same that is $6$, the terms are represented by a cubic equation. Will this hold good for higher orders? For example, if the 7th differences are same it will be represented by a polynomial of degree 7? Please clarify this doubt. – Aug 17 '19 at 06:21
3

@Sumathi Yes, that is true in general. However, note it's only true, of course, if all of the values are the same. If you're not given that the original sequence is from a polynomial, technically, no matter how many terms you calculate, you can never be sure of what the degree is (apart from it being at least that of where you got the constant set of values), or even if it's a polynomial. However, that would be a reasonable assumption in most cases. – John Omielan Aug 17 '19 at 06:24
Thank you very much for your kind help. – Aug 17 '19 at 06:26
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@Sumathi You're welcome. I added a link to a Web page that states about the various orders of differences and the degrees of the associated polynomials as their Theorem $1$, along with a proof. I hope that helps to explain why that works in more detail. – John Omielan Aug 17 '19 at 06:35
Thank you for sharing the link. I will go through it and ask if I have any doubts. – Aug 17 '19 at 06:39
2

@Sumathi You're welcome for the link. I also remembered you can determine the coefficient of the $x^3$ term from the constant sequence of values, as I explain in my updated answer. – John Omielan Aug 17 '19 at 06:51

score -1 · Answer 2 · answered Aug 17 '19 at 06:23

(Assuming the terms in given series are generated from a polynomial)

When you take the difference, you reduce the degree by $1$.

As an example, for a quadratic $T_n = n^2$, the first order differences would be in A.P: $$(n+1)^2 - n^2 = 2n+1$$

And for a cubic $T_n = n^3$, the first order differences gives you a quadratic: $$(n+1)^3-n^3=3n^2+3n+1$$

Now if you take the difference on above quadratic, you get a sequence in A.P.

Doubt in finding general term of the given sequence

2 Answers2

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