Question

Find Taylor series of the function :

f(z) = 4 (z^3)+ 2 (z^2)-z-5 center Zo = -1
f(z) = cube-root of [ Exp ^(z-2)] center Zo=i
f(z)=e^z . Cosh z center Zo=0

Answer 1

Answer to Question # 49958

Find Taylor series of the function:

1) $f(z) = 4z^3 + 2z^2 - z - 5$, $z_0 = -1$

Solution:

$$f(z) = \sum_{k=0}^{n} \frac{f^{(k)}(z_0)}{k!}(z - z_0) + \overline{\partial}((z - z_0)^n) = \sum_{k=0}^{3} \frac{f^{(k)}(-1)}{k!}(z + 1)^k$$

At first we need to compute the derivatives:

$$f^{(0)}(z) = f(z) = 4z^3 + 2z^2 - z - 5$$ $$f^{(1)}(z) = f'(z) = 12z^2 + 4z - 1$$ $$f^{(2)}(z) = f''(z) = 24z + 4$$ $$f^{(3)}(z) = f'''(z) = 24$$

So the solution is:

$$f(z) = \sum_{k=0}^{3} \frac{f^{(k)}(-1)}{k!}(z + 1)^k = [4 \cdot (-1)^3 + 2 \cdot (-1)^2 + 1 - 5] + [12 \cdot (-1)^2 + 4 \cdot (-1) - 1](z + 1) + \\ + [24 \cdot (-1) + 4] \frac{(z + 1)^2}{2} + 24 \cdot \frac{(z + 1)^3}{6} = 4(z + 1)^3 - 10(z + 1)^2 + 7(z + 1) - 6$$

Answer: $4(z + 1)^3 - 10(z + 1)^2 + 7(z + 1) - 6$

2) $f(z) = \sqrt[3]{e^{z - 2}}$, $z_0 = i$

Solution:

$$f(z) = \{new\_var iable\_t = z - i\} = \sqrt[3]{e^{t + i - 2}} = e^{\frac{t + i - 2}{3}} = e^{\frac{i - 2}{3}} \cdot e^{\frac{t}{3}} = e^{\frac{i - 2}{3}} \cdot \sum_{n=0}^{\infty} \frac{\left(\frac{t}{3}\right)^n}{n!} = e^{\frac{i - 2}{3}} \cdot \sum_{n=0}^{\infty} \frac{\left(\frac{z - i}{3}\right)^n}{n!} = e^{\frac{i}{3}} \cdot \sqrt[3]{e^2} \cdot \sum_{n=0}^{\infty} \frac{\left(\frac{z - i}{3}\right)^n}{n!}$$

Answer: $\frac{e^{\frac{i}{3}}}{\sqrt[3]{e^2}} \cdot \sum_{n=0}^{\infty} \frac{\left(\frac{z - i}{3}\right)^n}{n!}$

3) $f(z) = e^{z} Cosh(z)$, $z_0 = 0$

**Solution:**

we know that

$$Cosh(z) = \frac{e^{z} + e^{-z}}{2}$$

so we obtain:

$$f(z) = e^{z} \cdot \frac{e^{z} + e^{-z}}{2} = \frac{e^{2z}}{2} + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{(2z)^n}{n!}$$

**Answer:**

$$\frac{1}{2} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{(2z)^n}{n!}$$

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