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Question: How to find the Taylor series for the function \(\dfrac{1}{{1 + z + {z^2}}}\) about the point \(z = ...

How to find the Taylor series for the function 11+z+z2\dfrac{1}{{1 + z + {z^2}}} about the point z=0z = 0 ?

Explanation

Solution

Taylor series states that any function can be written as a sum of infinite number of terms. Each term in the series is expressed in terms of the function’s derivatives at a single point and the function must be infinitely differentiable at that point. The Taylor series is given as,
f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+...  (upto  ∞  terms)f(x) = f(a) + \dfrac{{f'(a)}}{{1!}}\left( {x - a} \right) + \dfrac{{f''(a)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{f'''(a)}}{{3!}}{\left( {x - a} \right)^3} + ...\;\left( {upto\;\infty \;terms} \right)

Complete step by step solution:
We have been given a function 11+z+z2\dfrac{1}{{1 + z + {z^2}}}. We have to find the Taylor series for this function about the point z=0z = 0.
The Taylor series for a function at any point aa is given as,
f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+...  (upto  ∞  terms)f(x) = f(a) + \dfrac{{f'(a)}}{{1!}}\left( {x - a} \right) + \dfrac{{f''(a)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{f'''(a)}}{{3!}}{\left( {x - a} \right)^3} + ...\;\left( {upto\;\infty \;terms} \right)
We can observe that the terms contain derivatives of the function at the given point.
We have the given function f(z)=11+z+z2=(1+z+z2)−1f(z) = \dfrac{1}{{1 + z + {z^2}}} = {\left( {1 + z + {z^2}} \right)^{ - 1}}. Taylor series for this function at point z=0z = 0 can be written as,
f(z)=f(0)+f′(0)1!z+f′′(0)2!z2+f′′′(0)3!z3+...  (upto  ∞  terms)f(z) = f(0) + \dfrac{{f'(0)}}{{1!}}z + \dfrac{{f''(0)}}{{2!}}{z^2} + \dfrac{{f'''(0)}}{{3!}}{z^3} + ...\;\left( {upto\;\infty \;terms} \right)
At point z=0z = 0, f(0)=(1+0+0)−1=1f(0) = {\left( {1 + 0 + 0} \right)^{ - 1}} = 1
Now we find derivatives of the given function. We will use the formula d(xn)dx=nxn−1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}} and the chain rule to find the derivatives.
The first derivative can be found as,
f′(z)=d(1+z+z2)−1dz=−1(1+z+z2)−1−1.(d(1)dz+d(z)dz+d(z2)dz)=−(1+z+z2)−2.(1+2z)f'(z) = \dfrac{{d{{\left( {1 + z + {z^2}} \right)}^{ - 1}}}}{{dz}} = - 1{\left( {1 + z + {z^2}} \right)^{ - 1 - 1}}.\left( {\dfrac{{d\left( 1 \right)}}{{dz}} + \dfrac{{d\left( z \right)}}{{dz}} + \dfrac{{d\left( {{z^2}} \right)}}{{dz}}} \right) = - {\left( {1 + z + {z^2}} \right)^{ - 2}}.\left( {1 + 2z} \right)
Similarly we can find the second derivative as,

f′′(z)=d[−(1+z+z2)−2.(1+2z)]dz=(1+2z)d[−(1+z+z2)−2]dz+(−(1+z+z2)−2)d[(1+2z)]dz =[2(1+z+z2)−3.(1+2z)2]−2(1+z+z2)−2  f''(z) = \dfrac{{d\left[ { - {{\left( {1 + z + {z^2}} \right)}^{ - 2}}.\left( {1 + 2z} \right)} \right]}}{{dz}} = \left( {1 + 2z} \right)\dfrac{{d\left[ { - {{\left( {1 + z + {z^2}} \right)}^{ - 2}}} \right]}}{{dz}} + \left( { - {{\left( {1 + z + {z^2}} \right)}^{ - 2}}} \right)\dfrac{{d\left[ {\left( {1 + 2z} \right)} \right]}}{{dz}} \\\ = \left[ {2{{\left( {1 + z + {z^2}} \right)}^{ - 3}}.{{\left( {1 + 2z} \right)}^2}} \right] - 2{\left( {1 + z + {z^2}} \right)^{ - 2}} \\\

Further the third derivative can be found as,

f′′′(z)=d[[2(1+z+z2)−3.(1+2z)2]−2(1+z+z2)−2]dz =d[2(1+z+z2)−3.(1+2z)2]dz−d[2(1+z+z2)−2]dz =[(1+2z)2d[2(1+z+z2)−3]dz+(2(1+z+z2)−3)d[(1+2z)2]dz]−[−4(1+z+z2)−3.(1+2z)] =−6(1+z+z2)−4.(1+2z)3+8(1+z+z2)−3(1+2z)+4(1+z+z2)−3.(1+2z) =−6(1+z+z2)−4.(1+2z)3+12(1+z+z2)−3(1+2z)  f'''(z) = \dfrac{{d\left[ {\left[ {2{{\left( {1 + z + {z^2}} \right)}^{ - 3}}.{{\left( {1 + 2z} \right)}^2}} \right] - 2{{\left( {1 + z + {z^2}} \right)}^{ - 2}}} \right]}}{{dz}} \\\ = \dfrac{{d\left[ {2{{\left( {1 + z + {z^2}} \right)}^{ - 3}}.{{\left( {1 + 2z} \right)}^2}} \right]}}{{dz}} - \dfrac{{d\left[ {2{{\left( {1 + z + {z^2}} \right)}^{ - 2}}} \right]}}{{dz}} \\\ = \left[ {{{\left( {1 + 2z} \right)}^2}\dfrac{{d\left[ {2{{\left( {1 + z + {z^2}} \right)}^{ - 3}}} \right]}}{{dz}} + \left( {2{{\left( {1 + z + {z^2}} \right)}^{ - 3}}} \right)\dfrac{{d\left[ {{{\left( {1 + 2z} \right)}^2}} \right]}}{{dz}}} \right] - \left[ { - 4{{\left( {1 + z + {z^2}} \right)}^{ - 3}}.\left( {1 + 2z} \right)} \right] \\\ = - 6{\left( {1 + z + {z^2}} \right)^{ - 4}}.{\left( {1 + 2z} \right)^3} + 8{\left( {1 + z + {z^2}} \right)^{ - 3}}\left( {1 + 2z} \right) + 4{\left( {1 + z + {z^2}} \right)^{ - 3}}.\left( {1 + 2z} \right) \\\ = - 6{\left( {1 + z + {z^2}} \right)^{ - 4}}.{\left( {1 + 2z} \right)^3} + 12{\left( {1 + z + {z^2}} \right)^{ - 3}}\left( {1 + 2z} \right) \\\

From above found derivatives we can find the derivatives at the point z=0z = 0.

f′(0)=−(1+0+02)−2.(1+2×0)=−1 f′′(0)=[2(1+0+02)−3.(1+2×0)2]−2(1+0+02)−2=2−2=0 f′′′(0)=−6(1+0+02)−4.(1+2×0)3+12(1+0+02)−3(1+2×0)=−6+12=6  f'\left( 0 \right) = - {\left( {1 + 0 + {0^2}} \right)^{ - 2}}.\left( {1 + 2 \times 0} \right) = - 1 \\\ f''\left( 0 \right) = \left[ {2{{\left( {1 + 0 + {0^2}} \right)}^{ - 3}}.{{\left( {1 + 2 \times 0} \right)}^2}} \right] - 2{\left( {1 + 0 + {0^2}} \right)^{ - 2}} = 2 - 2 = 0 \\\ f'''\left( 0 \right) = - 6{\left( {1 + 0 + {0^2}} \right)^{ - 4}}.{\left( {1 + 2 \times 0} \right)^3} + 12{\left( {1 + 0 + {0^2}} \right)^{ - 3}}\left( {1 + 2 \times 0} \right) = - 6 + 12 = 6 \\\

Thus, we can put these values in the Taylor series to get the final result as,
f(z)=f(0)+f′(0)1!z+f′′(0)2!z2+f′′′(0)3!z3+... ⇒f(z)=1+−11!z+02!z2+63!z3+... ⇒f(z)=1−z+z3+...  f(z) = f(0) + \dfrac{{f'(0)}}{{1!}}z + \dfrac{{f''(0)}}{{2!}}{z^2} + \dfrac{{f'''(0)}}{{3!}}{z^3} + ... \\\ \Rightarrow f(z) = 1 + \dfrac{{ - 1}}{{1!}}z + \dfrac{0}{{2!}}{z^2} + \dfrac{6}{{3!}}{z^3} + ... \\\ \Rightarrow f(z) = 1 - z + {z^3} + ... \\\
Hence, we get the required Taylor series as 1−z+z3+...1 - z + {z^3} + ...

Note: We found the Taylor series of the function at a given point. For the Taylor series to exist at that point, the function must be infinitely differentiable at that point. The series contains an infinite number of terms, but for representation we can write only three or four terms. The calculation of derivatives can be messy after the second or third derivative, so utmost care should be taken while finding the derivatives.