Purchase Solution

Periodic function proof

Not what you're looking for?

Ask Custom Question

Please provide explanation to prove that f(x)=sin(x)+sin(x/sqrt(2)) is not periodic.

Purchase this Solution
Solution Summary

This proves a function is not periodic.

Solution Preview

The solution is on the attached file.
I assumed that the formulas for the derivatives of sin and cos are known and also that sin(y)= sin(y+2nPI) for any n.
In other words the sin function has period 2PI.

Solution

f(x)=sin(x)+sin(x/sqrt(2))

Suppose that f(x) is periodic with period p. Then f(x+p) - f(x)=0. Now let g(x) = f(x+p) - f(x)=0 .
Since g(x)=0 then g"(x) =0 (this is ...

Solution provided by:
Chris Busu, PhD (IP)
Education
  • BSc, University of Bucharest
  • MSc, Ovidius
  • MSc, Stony Brook
  • PhD (IP), Stony Brook
Recent Feedback
  • "Thank you "
  • "Thank You Chris this draft really helped me understand correlation."
  • "Thanks for the prompt return. Going into the last meeting tonight before submission. "
  • "Thank you for your promptness and great work. This will serve as a great guideline to assist with the completion of our project."
  • "Thanks for the product. It is an excellent guideline for the group. "
Purchase this Solution
Free BrainMass Quizzes
Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

Know Your Linear Equations

Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

Probability Quiz

Some questions on probability

Graphs and Functions

This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.

View More Free Quizzes
Related BrainMass Solutions

  • Laplace transforms

    limit sum value as: Hence we have: LF [f(t)]= or LF [f(t)]= This is a proof regarding the Laplace transformation of a given periodic function.

  • Periodic Functions : Bounded and Continuous

    Show that a continuous periodic function on R is bounded and uniformly continuous. Please see the attached file for the complete solution. Thanks for using BrainMass. Proof: First, we consider the closed interval .

  • Laplace transform

    t)} = e-sna e-st F(t) dt ---------------------------(6) Hence from (6) it follows that L { F(t)} = { e-st F(t) dt }/ (1- e-st) , t > 0 This problem is a proof regarding a periodic continuous function and its Laplace transform.

  • Integrating Periodic Functions

    A periodic function is integrated. The soution is well presented.

  • Wave Equations and Periodic Differentiable Functions

    Proof.

  • Proofs : Pairwise Real Numbers, Natural and Irrational Numbers

    (Note: Since sin(x) is a periodic function, we know that , so we can choose u large enough such that sin(u)=x ) Since y=sin(x) is continuous at u, given any positive real number , there exists so that, if , then , i.e.,

  • Fourier Mathcad

    The function cannot be periodic as x is not a periodic function and sinx is a periodic function and their multiplication is not a periodic function. Above is the graph of the function y=x. This is an odd function as it is symmetric in origin.

  • Periodic Functions via Convolution

    (b) Let p(x + c) = p(x) be a periodic function. Prove or disprove: p(x) is the convolution (i.e. f * g (x) E f(x ? t)g(t) dt) of a periodic train of Dirac delta functions with a non-periodic function, say g(x) in L2 (?co, oo). What is g(x)?

  • Sketch the graph of the following function.

    Sketch the graph of the function (include two full periods). (1) Solution. Since and y=sin(x) is a periodic function. Thus, is a periodic function with a period . We analyze for Since , We can sketch its graph for as follows.

View More Related Solutions