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Find the generating function of the given series $3+5x+9{{x}^{2}}+15{{x}^{3}}+23{{x}^{4}}+33{{x}^{5}}+...........$

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Last updated date: 20th Jun 2024
Total views: 441k
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Answer
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Hint: First consider the sequence as S then multiply it by x both S and its sequence then subtract it from S to get the form S (1-x). After that take out the function from S and take it as P then repeat the same process and get the function of P (1-x) and then take it as Q, again use the same process and find Q (1-x). Then find the function of Q in terms of x and from that get the function of P and finally S.

Complete step-by-step answer:
Infinite power series is simply a series with infinite number of terms in it, it is of the form of ${{c}_{n}}{{x}^{n}}$ where ${{c}_{n}}$ is some constant. So we might write a power series like this
$\sum\limits_{k=0}^{\infty }{{{c}_{k}}{{x}^{k}}}$
Or expand like this,
${{c}_{0}}+{{c}_{1}}x+{{c}_{2}}{{x}^{2}}+{{c}_{3}}{{x}^{2}}+{{c}_{4}}{{x}^{4}}............$
When viewed in the context of generating functions, we call such power series as generating series. The generating series generates the sequence of coefficients of the infinite polynomial.
Let the sequence $3+5x+9{{x}^{2}}+15{{x}^{3}}+23{{x}^{4}}+33{{x}^{5}}+...........$ be considered as ‘S’ so we can write ‘S’ as,
$S=3+5x+9{{x}^{2}}+15{{x}^{3}}+23{{x}^{4}}+33{{x}^{5}}+...........$
Now multiplying ‘x’ on both sides and then subtracting the new series from the former series,
$xS=3x+5{{x}^{2}}+9{{x}^{3}}+15{{x}^{4}}+.............$
Now on subtracting the above series with ‘S’, we get
$S\left( 1-x \right)=3+2x+4{{x}^{2}}+6{{x}^{3}}+8{{x}^{4}}.........-----(i)$
Let’s take the function $2x+4{{x}^{2}}+6{{x}^{3}}+8{{x}^{4}}+.........$ as P. Then we can write as,
$P=2x+4{{x}^{2}}+6{{x}^{3}}+8{{x}^{4}}+.........--(ii)$
Now, multiplying ‘x’ on both side and then subtracting the new series from the former series, we get
$\begin{align}
  & P=2x+4{{x}^{2}}+6{{x}^{3}}+8{{x}^{4}}+......... \\
 & xP=2{{x}^{2}}+4{{x}^{3}}+6{{x}^{4}}+........ \\
\end{align}$
Now on subtracting we get,
$P\left( 1-x \right)=2x+2{{x}^{2}}+2{{x}^{3}}+2{{x}^{4}}..........---(iii)$
Now let’s take $2x+2{{x}^{2}}+2{{x}^{3}}+2{{x}^{4}}..........$ as Q. Then we can write it as,
$Q=2x+2{{x}^{2}}+2{{x}^{3}}+2{{x}^{4}}.........----(iv)$
Now multiplying by x on both sides and then subtracting the new series from the former series.
$\begin{align}
  & Q=2x+2{{x}^{2}}+2{{x}^{3}}+2{{x}^{4}}........... \\
 & xQ=2{{x}^{2}}+2{{x}^{3}}+2{{x}^{4}}............ \\
\end{align}$
Now, on subtracting we get,
$\begin{align}
  & Q\left( 1-x \right)=2x \\
 & \Rightarrow Q=\dfrac{2x}{1-x} \\
\end{align}$
Now from equation (iii) and (iv), we see that P (1-x) = Q, substituting the value of ‘Q’, we get
$\begin{align}
  & P\left( 1-x \right)=\dfrac{2x}{1-x} \\
 & \Rightarrow P=\dfrac{2x}{{{\left( 1-x \right)}^{2}}}..........(v) \\
\end{align}$
Similarly, from equation (i) and (ii), we see that
S (1-x) = 3+P
Substituting the value of ‘P’ from equation (v), we get
$\begin{align}
  & S\left( 1-x \right)=3+\dfrac{2x}{{{\left( 1-x \right)}^{2}}} \\
 & \Rightarrow S=\dfrac{3}{1-x}+\dfrac{2x}{{{\left( 1-x \right)}^{3}}} \\
\end{align}$
So, the required generating function is $\dfrac{3}{1-x}+\dfrac{2x}{{{\left( 1-x \right)}^{3}}}$

Note: After getting the function of sequence of Q as $2x+2{{x}^{2}}+2{{x}^{3}}+.........$ we can do find the function another way by just using the formula of infinite sum which is $\dfrac{a}{1-r}$ where a is the first term and r is the common ratio.