Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
expand_more
expand_more
format_list_bulleted
Related questions
Question
![### 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](https://content.bartleby.com/qna-images/question/833bf7b1-3e6b-4749-8e88-54090320a3f5/e2848f7b-64c4-49aa-81e1-b4122e944f38/62spgrb_processed.png)
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
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by stepSolved in 3 steps with 2 images
Knowledge Booster
Similar questions
- Which of the following functions f: R → R is continuous at the point 0 € R? f(x) = x² f(x): {2²2 = f(x) = { √x 0 {²+2 f(x) = { f(x) = { 1 if x ≤ 0 if x > 0 if x ≤ 0 x + 2 if x > 0 -1 if x > 0 if x 0 Choose... Choose... Choose... Choose... Choose... ✓ Choose... ◄► ◄► O Continuous at 0 Not continuous at 0arrow_forwardSuppose f: IR→IR is a monotonic function. True or false: If f is differentiable, then f must be uniformly continuous. True Falsearrow_forwardWhich of the following functions f: R → R is continuous at the point 0 € R? f(x) = x² f(x): {2²2 = f(x) = { √x 0 {²+2 f(x) = { f(x) = { 1 if x ≤ 0 if x > 0 if x ≤ 0 x + 2 if x > 0 -1 if x > 0 if x 0 Choose... Choose... Choose... Choose... Choose... ✓ Choose... ◄► ◄► O Continuous at 0 Not continuous at 0arrow_forward
- [2, 7] → [2, it has a fixed point in [0, 7]. Show that if f: 7] is continuous thenarrow_forward* Suppose that f: R R is continuous. Show that, for any c E R, the function -> cf : R Ris also continuous. (cf is the function obtained by multiplying the output of f by c.)arrow_forward(5) Let f: RR be twice-differentiable f(x) ≤0 and Prove that f is constant on R. on R. Suppose that f"(x) ≥0 for all x = Rarrow_forward
- Use FTC I to prove that if |f '(x)| ≤ K for x ∈ [a, b], then |f (x) − f (a)| ≤ K|x − a| for x ∈ [a, b].arrow_forward3. Let f: ZZ be defined by f(x) = 2x² + x. Is f surjective? Is f injective? Justify your answers.arrow_forwardh5. Prove that f ( x ) = x ⋅ | x | is continuous at all points c in ℝ.arrow_forward
- Use the e-d definition to prove that f(x) = x³ + 3 is continuous at xo = 3.arrow_forwardProve f(x) = 1 is continuous function in R. nº +x* n=1arrow_forwardLet f be defined by f(q)=3q for q rational and f(t)=t2 for t irrational.. b) is f continuous at 3? prove your answer, remember to prove true you MUST prove it to be true, not use a specific example.arrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
- Advanced Engineering MathematicsAdvanced MathISBN:9780470458365Author:Erwin KreyszigPublisher:Wiley, John & Sons, Incorporated
Numerical Methods for EngineersAdvanced MathISBN:9780073397924Author:Steven C. Chapra Dr., Raymond P. CanalePublisher:McGraw-Hill Education
Introductory Mathematics for Engineering Applicat...Advanced MathISBN:9781118141809Author:Nathan KlingbeilPublisher:WILEY 
Mathematics For Machine TechnologyAdvanced MathISBN:9781337798310Author:Peterson, John.Publisher:Cengage Learning,
Advanced Engineering Mathematics
Advanced Math
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:9780073397924
Author:Steven C. Chapra Dr., Raymond P. Canale
Publisher:McGraw-Hill Education
Introductory Mathematics for Engineering Applicat...
Advanced Math
ISBN:9781118141809
Author:Nathan Klingbeil
Publisher:WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:9781337798310
Author:Peterson, John.
Publisher:Cengage Learning,