B What is the general formula for solving polynomial series?

[johann1301h]

Hi, I am trying to solve this series generally:

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

But this doesn't work.. Please help.

 

[PeroK]

Hi, I am trying to solve this series generally:

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

But this doesn't work.. Please help.

I would add $0$ at the beginning and see whether that helps.

 

[mfb]

Calculate how much the numbers increase in each step. That should lead to some pattern.

 

[OmCheeto]

Hi, I am trying to solve this series generally:

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

But this doesn't work.. Please help.

Did you try graphing it?
When I graphed your series of numbers, I got a curve.
That implied to me, that "4n", being a linear function, was not correct.

 

[Ssnow]

You must find $f(n)$ such that $a_{n+1}=f(n)+a_{n}$ with $a_{0}=3$, hint: think simple!
Ssnow

 

[Svein]

Most sequences can be resolve using difference (not differential) analysis.

 3 
 7    4  
12    5    1
18    6    1
25    7    1

 

[DShane_MU]

Just stumbled across this page. I'm doing independent math research and the series you are inquiring about is a smaller part of what I am doing. The formula you are looking for is F(n)= n(n+5) / 2 = {3,7,12,18,25,...}. It is similar to the formula for triangular numbers T(n) = n(n+1)/2 = {1,3,6,10,15,...}. Hope this helps!

Regards,
David

 

[maka89]

Difference between two numbers is (2+n), so you can write x(n)= (2+n) + x(n-1) when n > 1.

This is a recursive formula, that also can be written out as:
x(n) = (2+n) + (2+n-1) + (2+n-2) + (2+n-3) + ... + x(1)

Recognizing the pattern as:
x(n) = x(1) + \sum_{i=2}^n(2+i) = x(1) - 3 + \sum_{i=1}^n(2+i) = \sum_{i=1}^n2+ \sum_{i=1}^ni = 2n+ n(n+1)/2 = n(n+5)/2

So we find x(n) = n(n+5)/2.

\sum_{i=1}^nn = n(n+1)/2 can be found from a table or WolframAlpha(or could also derivated without too much trouble) :)

 

[Kumar Dev]

 

[etotheipi]

This is not necessarily a better method, since yours is actually probably faster in this case, however any series with a constant second difference is called a quadratic series and can be generalised as such:

$u_{n} = an^{2} + bn + c$

First term: $u_{1} = a + b + c$
First difference: $u_{n} - u_{n-1} = 2an + (b-a)$
Second difference: $(u_{n} - u_{n-1}) - (u_{n-1} - u_{n-2}) = 2a$

So in the case of the sequence $3,7,12,18,25...$ we can write down three relations

$a + b + c = 3$
$3a + b = 4$
$2a = 1$

Which gives $a = \frac{1}{2}, b = \frac{5}{2}, c= 0$, so

$u_{n} = \frac{1}{2}n^{2} + \frac{5}{2}n$

which is the answer you obtained.

 

[wle]

More generally it's easy to check by induction that $$\sum_{n=1}^{a} n (n + 1) (n + 2) \dotsm (n + p - 1) = \frac{a (a + 1) \dotsm (a + p)}{p + 1} \,.$$ The relation can be be seen as the discrete analogue of $$\int_{0}^{a} \mathrm{d}x \, x^{p} = \frac{a^{p+1}}{p + 1}$$ and it can be used to determine the partial sum of any polynomial. For example, $$\begin{eqnarray*}
\sum_{n=1}^{a} n^{3} &=& \sum_{n=1}^{a} \bigl( n - 3 n (n + 1) + n (n + 1) (n + 2) \bigr) \\
&=& \frac{1}{2} a (a + 1) - 3 \frac{1}{3} a (a + 1) (a + 2) + \frac{1}{4} a (a + 1) (a + 2) (a + 3) \\
&=&\frac{1}{4} a^{2} (a + 1)^{2} \,.
\end{eqnarray*}$$