1. Suppose f is continuous on R and f satisfies f(x) + f(2x) = 0 for all x E R. Prove that f = 0 on R.
1. Suppose f is continuous on R and f satisfies f(x) + f(2x) = 0 for all x E R. Prove that f = 0 on R.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Mathematical Analysis Problem:
1. **Problem Statement**:
Suppose \( f \) is continuous on \(\mathbb{R}\) and \( f \) satisfies
\[
f(x) + f(2x) = 0
\]
for all \( x \in \mathbb{R} \). Prove that \( f = 0 \) on \(\mathbb{R}\).
### Solution:
To solve this problem, we need to utilize the given functional equation along with the property of continuity of \( f \). Here's a step-by-step approach to derive the conclusion.
1. **Given Information**:
- \( f \) is continuous on \(\mathbb{R}\).
- \( f(x) + f(2x) = 0 \) for all \( x \in \mathbb{R} \).
2. **Initial Setup**:
- Evaluate the functional equation for specific values and investigate how \( f \) behaves.
- Particularly, choosing \( x = 0 \) may reveal some initial insights.
3. **Evaluation at Specific Points**:
- Substitute \( x = 0 \) into the equation:
\[
f(0) + f(0) = 0 \implies 2f(0) = 0 \implies f(0) = 0
\]
- This provides us with the initial value: \( f(0) = 0 \).
4. **Extend to Other Points**:
- Substitute \( x = \frac{1}{2} \):
\[
f\left(\frac{1}{2}\right) + f(1) = 0 \implies f(1) = -f\left(\frac{1}{2}\right)
\]
- Similarly, substituting \( x = 1 \):
\[
f(1) + f(2) = 0 \implies f(2) = -f(1) = f\left(\frac{1}{2}\right)
\]
- For \( x = 2 \):
\[
f(2) + f(4) = 0 \implies f(4) = -f(2) = -f\left(\frac{1}{2}\right)
\]
- Sub](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F833bf7b1-3e6b-4749-8e88-54090320a3f5%2Fe2848f7b-64c4-49aa-81e1-b4122e944f38%2F62spgrb_processed.png&w=3840&q=75)
Transcribed Image Text:### Mathematical Analysis Problem:
1. **Problem Statement**:
Suppose \( f \) is continuous on \(\mathbb{R}\) and \( f \) satisfies
\[
f(x) + f(2x) = 0
\]
for all \( x \in \mathbb{R} \). Prove that \( f = 0 \) on \(\mathbb{R}\).
### Solution:
To solve this problem, we need to utilize the given functional equation along with the property of continuity of \( f \). Here's a step-by-step approach to derive the conclusion.
1. **Given Information**:
- \( f \) is continuous on \(\mathbb{R}\).
- \( f(x) + f(2x) = 0 \) for all \( x \in \mathbb{R} \).
2. **Initial Setup**:
- Evaluate the functional equation for specific values and investigate how \( f \) behaves.
- Particularly, choosing \( x = 0 \) may reveal some initial insights.
3. **Evaluation at Specific Points**:
- Substitute \( x = 0 \) into the equation:
\[
f(0) + f(0) = 0 \implies 2f(0) = 0 \implies f(0) = 0
\]
- This provides us with the initial value: \( f(0) = 0 \).
4. **Extend to Other Points**:
- Substitute \( x = \frac{1}{2} \):
\[
f\left(\frac{1}{2}\right) + f(1) = 0 \implies f(1) = -f\left(\frac{1}{2}\right)
\]
- Similarly, substituting \( x = 1 \):
\[
f(1) + f(2) = 0 \implies f(2) = -f(1) = f\left(\frac{1}{2}\right)
\]
- For \( x = 2 \):
\[
f(2) + f(4) = 0 \implies f(4) = -f(2) = -f\left(\frac{1}{2}\right)
\]
- Sub
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