1. Homework Statement [/b]
Let V be the space of all polynomials of degree less than or equal to 2 over the reals. Define the transformation, H, as a mapping from V to R[x] by (Hp)(x)=\int^x_{-1}p(t)dt\\. a) Find the monic generator, d, which generates the ideal, M, containing the range of H...
The amplitude is 3. Max of cos2x is 1 and min of cos2x is -1. So max of y = (3*1+1) & min of y = (3*-1+1). The other answer are posted above 8-14 posting.
8. Correct
9. Correct
10. Correct
11. Max of cosine is 1 and min is -1. Use that to answer this question.
12. When x=pi, it reaches it's max. Minimum of sine is -1. When is sine -1?
13. If the function was 2/3sin(theta), then it would be 1 cycle. If it was 2/3sin(2*theta), then it would be 2...
You didn't put an answer for 1. What is sin(3pi/2)? What is cos(3pi/2)? What is cos(0) and sin(0)? 2. Is wrong? The sine function max is 1 and min is -1. So 5+2*(-1)=3. 3. Not the y-axis. Sin(pi/2)=1;whereas sin(-pi/2)=-1. So symmetric with the origin. Being symmetric with the origin means...
It's an BVP since I have boundary conditions. x(-\pi)=0 and x(\pi)=0. I don't know anything about initial conditions. So I don't think I can do what you did.
Ok thanks. One question you said you got the solution by taking two l derivatives. Of what exactly. Did you guess at the solution, then checked to see if it worked by taking two derivatives. Or did you used a Fourier series expansion?
Hi,
I need help finding the solution to the homogenous BVP. Normally I could do this but I'm lost on this one. \frac{d^2}{dt^2}x(t)+\int^{\pi}_{-\pi}\sin(t-s)x(s)ds=0. I'm hoping the only solution is the zero solution. If not, I need to know a method to find all solutions. I thought about...