Math Problem Statement

Resoudre l'exercice etape par etape et faire tous xalcules demandé

Résoudre l'exercice étape par étape et faire tous les calculs demandés pour la série entière f(z) = ∑ (z^(n+1)) / (2^n).

Solution

The exercise involves analyzing the series given by:

f(z)=n=0zn+12nf(z) = \sum_{n=0}^{\infty} \frac{z^{n+1}}{2^n}

Let's go through each part of the exercise step-by-step.


Part (a): Déterminer le rayon de convergence de la série entière f(z)f(z)

To determine the radius of convergence RR of a power series, we can use the formula:

1R=lim supnann\frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{\left| a_n \right|}

For this series, the general term is: an=12na_n = \frac{1}{2^n}

Thus, the series can be rewritten as: f(z)=n=0(zn+12n)f(z) = \sum_{n=0}^{\infty} \left(\frac{z^{n+1}}{2^n}\right)

Alternatively, the ratio test can also help determine the radius of convergence: limnan+1an\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

Applying the ratio test for 12n\frac{1}{2^n}, we observe that it converges for all zz values such that: z2<1z<2\left| \frac{z}{2} \right| < 1 \Rightarrow |z| < 2

Therefore, the radius of convergence RR is 2.


Part (b): Quelle est la valeur du prolongement analytique de ff au point z=23iz = 2 - 3i?

Since the radius of convergence R=2R = 2, the point z=23iz = 2 - 3i lies outside this radius (23i=22+(3)2=133.61|2 - 3i| = \sqrt{2^2 + (-3)^2} = \sqrt{13} \approx 3.61, which is greater than 2).

This means that the series does not converge at z=23iz = 2 - 3i, so f(z)f(z) cannot be analytically continued to that point using this series.


Part (c): Représenter ce prolongement analytique par une série de Taylor convergente au point z=23iz = 2 - 3i

To represent the analytic continuation of f(z)f(z) around z=23iz = 2 - 3i, we would typically need to find another series expansion (such as a Laurent series) around this point, as the original Taylor series centered at z=0z = 0 does not converge beyond z=2|z| = 2.

However, constructing such an expansion requires more context about the function f(z)f(z) beyond the initial series. Therefore, additional techniques or transformations (like Möbius transformation or Laurent series) would be needed for an explicit representation around z=23iz = 2 - 3i.


Part (d): Que remarque-t-on ?

The primary observation here is that:

  1. The series f(z)f(z) has a limited radius of convergence (R=2R = 2), which restricts its convergence to points within this circle.
  2. For points outside z=2|z| = 2, such as z=23iz = 2 - 3i, the series does not converge, highlighting the necessity of alternative methods (like analytic continuation) for evaluating f(z)f(z) beyond this boundary.

Would you like further explanations on any of these steps or have additional questions?

Related Questions

  1. How would the radius of convergence change if the series terms were modified?
  2. What is the impact of complex points like z=23iz = 2 - 3i on convergence in general?
  3. How does the ratio test help in determining the radius of convergence?
  4. What is the difference between a Taylor series and a Laurent series for analytic continuation?
  5. Can f(z)f(z) have different convergence behavior if centered at z=23iz = 2 - 3i?

Tip

In problems involving series convergence, always start with the radius of convergence to understand the domain in which the series is valid before proceeding with specific evaluations.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Power Series
Radius of Convergence
Analytic Continuation

Formulas

Radius of convergence formula: 1/R = limsup_{n -> ∞} (|a_n|)^(1/n)
Ratio test: lim_{n -> ∞} |a_{n+1} / a_n|

Theorems

Radius of Convergence
Analytic Continuation

Suitable Grade Level

Undergraduate Level (Complex Analysis or Advanced Calculus)