https://file.aiquickdraw.com/mathbot/file/17257834701810tuprbs6.jpg

The image contains a multi-part mathematical problem. I will break down and solve the problem based on the information visible:

---

### (a) Given the function \( f(z) = e^{z^2}, z \in \mathbb{C}, s > 0 \), you are asked to:
1. **Show** that for every \( n \geq 2 \), the \( n \)-th derivative \( f^{(n)}(z) \) is of the form:
   \[
   f^{(n)}(z) = P_n(z) e^{z^2}
   \]
   where \( P_n(z) \) is some polynomial.

### Solution for (a):
To prove this, we can use induction on \( n \).

- Base case \( n = 0 \): \( f(z) = e^{z^2} \), so clearly, \( f^{(0)}(z) = P_0(z)e^{z^2} \), where \( P_0(z) = 1 \).
  
- First derivative \( n = 1 \):
  \[
  f^{(1)}(z) = \frac{d}{dz} \left( e^{z^2} \right) = 2z e^{z^2}.
  \]
  Here, \( P_1(z) = 2z \), which is a polynomial of degree 1.

- Now, assume that for \( n = k \), \( f^{(k)}(z) = P_k(z)e^{z^2} \), where \( P_k(z) \) is a polynomial.

- For \( n = k + 1 \), the derivative of \( f^{(k)}(z) \) is:
  \[
  f^{(k+1)}(z) = \frac{d}{dz} \left( P_k(z) e^{z^2} \right) = P_k'(z)e^{z^2} + 2zP_k(z)e^{z^2} = \left( P_k'(z) + 2zP_k(z) \right) e^{z^2}.
  \]
  Hence, \( P_{k+1}(z) = P_k'(z) + 2zP_k(z) \), which is also a polynomial since the sum and product of polynomials are polynomials.

By induction, the result holds for all \( n \geq 0 \).

---

### (b) Define:
\[
 g(z) = \begin{cases} 
      e^{-\frac{1}{z^2}} & \text{if } z \neq 0, \\
      0 & \text{if } z = 0.
   \end{cases}
\]
You are asked to prove that \( g(z) \) is **infinitely differentiable** and that \( g^{(n)}(0) = 0 \) for all \( n \geq 0 \).

### Solution for (b):
- For \( z \neq 0 \), the function \( g(z) = e^{-\frac{1}{z^2}} \) is smooth (infinitely differentiable) because exponentials are smooth functions.
  
- At \( z = 0 \), the value of the function is 0. We need to check the behavior of derivatives at \( z = 0 \).

- The derivatives of \( g(z) \) for \( z \neq 0 \) involve terms that decrease very rapidly to 0 as \( z \to 0 \). Intuitively, as \( z \to 0 \), each successive derivative involves multiplying by additional powers of \( \frac{1}{z^2} \), which causes the function and all its derivatives to approach 0 at \( z = 0 \). Therefore:
  \[
  g^{(n)}(0) = 0 \quad \text{for all } n \geq 0.
  \]
This makes \( g(z) \) a smooth function with all derivatives zero at \( z = 0 \), as required.

---

### (c) Consider a function that is not equal to its Taylor series at any point.

The problem likely refers to the fact that the function \( g(z) \) defined above is an example of a **smooth, non-analytic** function. Although \( g(z) \) is infinitely differentiable at \( z = 0 \), its Taylor series at \( z = 0 \) is identically zero (since all derivatives are zero at \( 0 \)), but \( g(z) \) is not identically zero. This means that \( g(z) \) is not equal to its Taylor series in any neighborhood of \( z = 0 \).

---

Would you like further details on any part of the solution, or additional questions on this topic? Here are 5 related questions for practice:

1. What is the general form of the \( n \)-th derivative of \( e^{z^2} \)?
2. Why is \( g(z) \) considered a smooth function even though it is not analytic?
3. How can we construct other functions that are smooth but not analytic?
4. What is the difference between a smooth function and an analytic function?
5. Can a function be analytic at some points and non-analytic at others?

**Tip:** Always check the behavior of derivatives at specific points to understand the nature of the function (whether it is analytic or smooth).

(a) Given the function f(z) = e^{z^2}, z ∈ ℂ, s > 0. Show that for every n ≥ 2, the nth derivative f^(n)(z) is of the form P_n(z) e^{z^2} for some polynomial P_n (depending on n).
(b) Define g(z) as a piecewise function:
  g(z) = { e^{-1/z^2} if z ≠ 0; 0 if z = 0 }
  Prove that g(z) is infinitely differentiable and that g^(n)(0) = 0 for all n ≥ 0.
(c) Consider a function that is not equal to the sum of its Maclaurin series.

This problem explores the nth derivative of the function f(z) = e^{z^2} and the infinitely differentiable nature of the piecewise function g(z). Learn how to prove that f^(n)(z) can be expressed as P_n(z) e^{z^2} and why g(z) is a smooth, non-analytic function.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation