Question
This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.
A power series in $x$ is defined as a function of the form $f(x)=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$ where the $a_i \in \mathbb{R}$.
It can be considered as an infinite polynomial.
This is an example of a power series, but is only a finite power series, since only a finite number of the $a_i$ are non-zero.
We will now attempt to generalise further.
Suppose $(1+x)^q, q \in \mathbb{Q}$ can be written as the power series $a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$
a. Expand $(1+x)^5$ using the Binomial Theorem.
b. Consider the power series $1-x+x^2-x^3+x^4-\ldots$
By considering the ratio of consecutive terms, explain why this series is equal to $(1+x)^{-1}$ and state the values of $x$ for which this equality is true.
c. Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for $(1+x)^{-2}$.
d. Repeat this process to find the first four terms in a power series for $(1+x)^{-3}$.
e. Hence, by recognising the pattern, deduce the first four terms in a power series for $(1+x)^{-n}, n \in \mathbb{Z}^{+}$.
f. By substituting $x=0$, find the value of $a_0$.
g. By differentiating both sides of the expression and then substituting $x=0$, find the value of $a_1$.
h. Repeat this procedure to find $a_2$ and $a_3$.
i. Hence, write down the first four terms in what is called the Extended Binomial Theorem for $(1+x)^q, q \in \mathbb{Q}$.
j. Write down the power series for $\frac{1}{1+x^2}$.
k. Hence, using integration, find the power series for $\arctan x$, giving the first four non-zero terms.
▶️Answer/Explanation
a. $1+5 x+10 x^2+10 x^3+5 x^4+x^5$
M1A1
[2 marks]
b. It is an infinite GP with $a=1, r=-x \quad$ R1A1
$S_{\infty}=\frac{1}{1-(-x)}=\frac{1}{1+x}=(1+x)^{-1} \quad$ M1A1AG
[4 marks]
c. $(1+x)^{-1}=1-x+x^2-x^3+x^4-\ldots$
$-1(1+x)^{-2}=-1+2 x-3 x^2+4 x^3-\ldots$
A1
$(1+x)^{-2}=1-2 x+3 x^2-4 x^3+\ldots$
A1
[2 marks]
d. $-2(1+x)^{-3}=-2+6 x-12 x^2+20 x^3$.
A1
$(1+x)^{-3}=1-3 x+6 x^2-10 x^3 \ldots$
A1
[2 marks]
e. $(1+x)^{-n}=1-n x+\frac{n(n+1)}{2 !} x^2-\frac{n(n+1)(n+2)}{3 !} x^3 \ldots \quad$ A1A1A1
[3 marks]
f. $1^q=a_0 \Rightarrow a_0=1$
A1
[1 mark]
g. $q(1+x)^{q-1}=a_1+2 a_2 x+3 a_3 x^2+\ldots \quad$ A1
$a_1=q$
A1
[2 marks]
h. $q(q-1)(1+x)^{q-2}=1 \times 2 a_2+2 \times 3 a_3 x+\ldots$
A1
$a_2=\frac{q(q-1)}{2 !}$
A1
$q(q-1)(q-2)(1+x)^{q-3}=1 \times 2 \times 3 a_3+\ldots$
A1
$a_3=\frac{q(q-1)(q-2)}{3 !}$
A1
[4 marks]
i. $\quad(1+x)^q=1+q x+\frac{q(q-1)}{2 !} x^2+\frac{q(q-1)(q-2)}{3 !} x^3 \ldots \quad$ A1
[1 mark]
j. $\frac{1}{1+x^2}=1-x^2+x^4-x^6+\ldots \quad$ M1A1
[2 marks]
k. $\arctan x+c=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots \quad$ M1A1
Putting $x=0 \Rightarrow c=0 \quad$ R1
So $\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$
A1
[4 marks]