Recent content by Persimmon

So if I fix the 1800 ls into 900 could I apply the same logic? Find out t' given t=0, x = +/- 900ls, then find out how long it would take light to travel to Alice from the ends using non relativistic relations, since I know the speed with be +/- c and the distance based on proper length. Then...

Ok, maybe not. So for the flash at the front, I got t' = -2.4 *10 ^3 sec. I used simple velocity relations, and got that it would take 3000 seconds in Alice's frame for the light of the event to reach her, all good since it's 3000 light seconds away. So t' for when the light from the front...

That is the contracted length for half of the train... I believe. Since Alice was originally 3000 ls from either end, she should now be 1800ls from either end, if I'm understanding this correctly. Thanks for the tip, I'm new at this. Right so at t = 0, the position of the back pulse is x...

Can you please expand on why part a) is incorrect? I tried to do length contraction, so that proper length is 9 * 10^11 m to one end, or 3000 light seconds. Gamma is 0.6, which works outs to 1800 Ls as mentioned, so the contracted length is 1800 Ls or 5.4 * 10^11 m. Why would it be less than...

Homework Statement A train is moving by a stationary observer, Bob, at 0.8c. An observer standing still in the middle of the train, Alice, measures the distance to each end of train as 9*10^11 m. Two lightning bolts hit each end of the train simultaneously, as seen by the observer at the...

Thanks, it seems really obvious to me now. D'oh!

Homework Statement Provide an example of a function such that f(x) has two and only two real roots and f'(x) has two and only two real roots, where f is defined for all real numbers and differentiable everywhere on its domain. Homework Equations The Attempt at a Solution I know that if a...

It's confusing me, because that's the book answer and the wolfram answer! There must be some arithmetic trick.

Yes, of course that makess sense regarding the minus sign. Oops. The reason why I specified that particular answer, ie ± √(x-6) is because that is what WolframAlpha gives as the result to the inverse of the original function. I can't think of how to arrive to that result.

Could you please elaborate? How is the function not one to one? f(x) = (abs(x))*x +6 results in a graph that essentially looks like a cubic function due to the absolute value sign.

Homework Statement Given the function f(x) = (abs(x))*x +6, find f^-1(x) Homework Equations The Attempt at a Solution for x≥ 0, f(x) = x^2 + 6 y=x^2 +6 x = √(y-6) for y≥6 → f^-1(x) = √(x-6) for x≥6 for x< 0, f(x) = -x^2 + 6 y= -x^2 +6 x = √(6-y) for y<6 → f^-1(x) = √(6-x) for x<6 But...

Sorry for the unclear notation. I was trying to make it clearer so there wouldn't be endless brackets but I guess it's even more confusing that way. Thank you for your help!

Homework Statement find the following derivative d/dx [g(x + 1)(√(2+ (x + 8)^(1/3))/(cos(tan(sin(tan(sin x))))] at x = 0 Homework Equations The Attempt at a Solution I split the big long derivative into 3 functions: a(x) = g(x + 1) b(x) = √(2+ (x + 8)^(1/3)) c(x) =...

Got it, thanks! I was thinking it must be more complex than it really is.

So can I simply say sin(x)^2 + cos(x)^2 = 1 is true for all x, so sin(nx)^2 + cos(nx)^2 = 1? I'm a bit confused about why I would put y into S(y)^2 + C(y)^2... why wouldn't I put y into the sin and cos identity? ie: sin(y)^2 + cos(y)^2 = 1 for all y, take y = nx, therefore S(x)^2 + C(x)^2 = 1