Question
Consider the complex function $f(z) = \cos z + e^{z^2}$. The coefficient of $z^5$ in the Taylor series expansion of $f(z)$ about the origin is ________ (rounded off to 1 decimal place).
To determine the coefficient of \( z^5 \) in the Taylor series expansion of the function \( f(z) = \cos z + e^{z^2} \) around the origin, we first expand each component separately and then combine.
1. Taylor Series Expansion of \( \cos z \):
The Taylor series for \( \cos z \) about the origin is:
\( \cos z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} \)
Only terms with even powers of \( z \) appear. The term with \( z^5 \) is zero.
\( \cos z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} \)
Only terms with even powers of \( z \) appear. The term with \( z^5 \) is zero.
2. Taylor Series Expansion of \( e^{z^2} \):
The Taylor series for \( e^{z^2} \) about the origin is:
\( e^{z^2} = \sum_{n=0}^{\infty} \frac{(z^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{z^{2n}}{n!} \)
\( e^{z^2} = \sum_{n=0}^{\infty} \frac{(z^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{z^{2n}}{n!} \)
Here, the power of \( z \) must be 5. We note that \( 2n = 5 \) gives \( n = 2.5 \), which is not an integer. Hence, from \( e^{z^2} \), no \( z^5 \) term is produced.
3. Combination and Result:
Since neither \( \cos z \) nor \( e^{z^2} \) contains a term with \( z^5 \) in their expansions, the coefficient of \( z^5 \) in \( f(z) = \cos z + e^{z^2} \) is 0.
Verification: Within the range 0,0 specified, the obtained coefficient of \( z^5 \) is indeed 0, confirming the solution is correct.
Final Answer: The coefficient of \( z^5 \) is 0.0.
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