Simple taylor expansion? (Complex)

Jul 2006
364
44
Using

\(\displaystyle \frac{1}{1+z} = \frac{1}{3+(z-2)}\)

expand \(\displaystyle \frac{1}{1+z}\) is a Taylor series about z=2.

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Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

EDIT: It also mentions I need the first four non-zero terms of the series.
 
Last edited:
Oct 2009
4,261
1,836
Using

\(\displaystyle \frac{1}{1+z} = \frac{1}{3+(z-2)}\)

expand \(\displaystyle \frac{1}{1+z}\) is a Taylor series about z=2.

---
Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

\(\displaystyle \frac{1}{1+z}=1-z+z^2-z^3+\ldots=\sum^\infty_{n=0}(-1)^{n+1}z^n\,,\,|z|<1\Longrightarrow \) \(\displaystyle \frac{1}{1+z}=\frac{1}{3}\,\frac{1}{1+\frac{z-2}{3}}=\) \(\displaystyle \frac{1}{3}\left(1-\frac{z-2}{3}+\frac{(z-2)^2}{3^2}-\ldots\right)=\) \(\displaystyle \sum^\infty_{n=0}\frac{(z-2)^n}{3^{n+1}}\) , \(\displaystyle \left|\frac{z-2}{3}\right|<1\iff |z-2|<3\) .

Tonio
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
Using

\(\displaystyle \frac{1}{1+z} = \frac{1}{3+(z-2)}\)

expand \(\displaystyle \frac{1}{1+z}\) is a Taylor series about z=2.

---
Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

Assume you can write \(\displaystyle \frac{1}{1 + z}\) as a Taylor polynomial.


So \(\displaystyle \frac{1}{3 + (z - 2)} = c_0 + c_1(z - 2) + c_2(z - 2)^2 + c_3(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_0 = \frac{1}{3}\).


Differentiate both sides

\(\displaystyle -\frac{1}{[3 + (z - 2)]^2} = c_1 + 2c_2(z - 2) + 3c_3(z - 2)^2 + 4c_4(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_1 = -\frac{1}{3^2}\).


Differentiate both sides

\(\displaystyle \frac{2}{[3 + (z - 2)]^3} = 2c_2 + 3\cdot 2c_3(z - 2) + 4\cdot 3c_4(z - 2)^2 + 5\cdot 4c_5(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_2 = \frac{1}{3^3}\).


Differentiate both sides

\(\displaystyle -\frac{3\cdot 2}{[3 + (z - 2)]^4} = 3\cdot 2c_3 + 4\cdot 3 \cdot 2c_4(z - 2) + 5\cdot 4 \cdot 3 c_5(z - 2)^2 + 6\cdot 5 \cdot 4 c_6(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_3 = -\frac{1}{3^4}\).


Differentiate both sides

\(\displaystyle \frac{4\cdot 3 \cdot 2}{[3 + (z - 2)]^5} = 4 \cdot 3 \cdot 2c_4 + 5 \cdot 4 \cdot 3 \cdot 2c_5(z - 2) + 6\cdot 5 \cdot 4 \cdot 3 c_6 (z - 2)^2 + 7\cdot 6 \cdot 5 \cdot 4 c_7(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_4 = \frac{1}{3^5}\).



I think you can see that the series is taking the form

\(\displaystyle \frac{1}{1 + z} = \frac{1}{3} - \frac{1}{3^2}(z - 2) + \frac{1}{3^3}(z - 2)^2 - \frac{1}{3^4}(z - 2)^3 + \frac{1}{3^5}(z - 2)^4 + \dots - \dots\)

\(\displaystyle \frac{1}{1 + z} = \sum_{k = 0}^{\infty}\frac{(-1)^k}{3^{k + 1}}(z - 2)^k\).


Now you need to check the values for which this series will converge.

Since this is a geometric series with common ratio \(\displaystyle r = -\frac{1}{3}(z - 2)\), it only converges for

\(\displaystyle \left|-\frac{1}{3}(z - 2)\right| < 1\)

\(\displaystyle -1 < -\frac{1}{3}(z - 2) < 1\)

\(\displaystyle 3 > z - 2 > -3\)

\(\displaystyle -3 < z - 2 < 3\)

\(\displaystyle -1 < z < 5\).
 
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Jul 2006
364
44
Thanks a lot guys! Tonio, did you mean
\(\displaystyle \sum^\infty_{n=0}(-1)^{n}z^n\)?

and
\(\displaystyle
\sum^\infty_{n=0}\frac{(-1)^n(z-2)^n}{3^{n+1}}
\)
 
Oct 2009
4,261
1,836
Thanks a lot guys! Tonio, did you mean
\(\displaystyle \sum^\infty_{n=0}(-1)^{n}z^n\)?

and
\(\displaystyle
\sum^\infty_{n=0}\frac{(-1)^n(z-2)^n}{3^{n+1}}
\)

Yes. I just forgot the sum begin at zero and not at 1...(Giggle)

Tonio
 
Oct 2012
11
0
United States
multiply top and bottom by 1/3 and re-express the function as 1/3*(-1/(2-z)/3) now substitute (2-z)/3 into the geometric expansion and the rest is algebra! God I wish I could express myself better symbolically. Much less messy than ProveIt's approach!
 
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