Questions: Taylor Series for Complex Functions

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

The function f(x) = 1/(1 + x²) is smooth and well-defined for all real x, yet its Taylor series around x = 0 converges only for |x| < 1. What is the correct explanation?

AThe function grows too steeply beyond x = 1 for polynomial approximation to keep up
BThe Taylor series converges only where the function is increasing, and 1/(1+x²) decreases for x > 0
CIn the complex plane, f has singularities at z = ±i, which are distance 1 from the origin — those singularities govern the radius of convergence
DThe function's Taylor coefficients become large enough at n = 1 to cause divergence
Question 2 Multiple Choice

Two holomorphic functions f and g agree on a small open disk D. What must be true on their shared domain?

Af and g agree only within D; outside D their values may diverge
Bf and g agree everywhere on their entire shared domain
Cf and g agree provided they share the same singularities
Df and g agree only if D contains a zero of f − g
Question 3 True / False

In complex analysis, a holomorphic function equals its Taylor series exactly within the disk of convergence — not approximately, as in real analysis.

TTrue
FFalse
Question 4 True / False

The radius of convergence of the Taylor series of a complex function is determined by the behavior of the function near the real expansion point.

TTrue
FFalse
Question 5 Short Answer

Explain why the Taylor series of 1/(1 + z²) around z = 0 has radius of convergence exactly 1, even though the real function 1/(1+x²) is smooth and bounded for all real x.

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