# Simple taylor expansion? (Complex)

#### scorpion007

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.

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#### tonio

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

• scorpion007

#### Prove It

MHF Helper
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

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

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

$$\displaystyle 3 > z - 2 > -3$$

$$\displaystyle -3 < z - 2 < 3$$

$$\displaystyle -1 < z < 5$$.

• scorpion007

#### scorpion007

Thanks a lot guys! Tonio, did you mean
$$\displaystyle \sum^\infty_{n=0}(-1)^{n}z^n$$?

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

#### tonio

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

#### dave0147

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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