Show Taylor Formula Proves E > T_2_E for 0 to c

[Niles]

Homework Statement

I have E(v) = (m*c^2)/sqrt(1-v^2/c^2).

I also have a second-order Taylor-polynomial around v = 0, T_2_E, which is mc^2+½mv^2.

I have to use Taylors formula with restterm to show that E is bigger than T_2_E for all v in the interval [0,c).

The Attempt at a Solution

I have written an expression:

E(v) = T_2_E + 1/n! * int [E'''(t)*(v-t)^2] dt,

where n of course is 2, so it's 1/2 infront of my integral.

I am very uncertain whether my expression is correct or not - do I have to use the limits 0 to c, or 0 to v?

Thank you in advance.

 

[HallsofIvy]

The error will depend upon the specific value of v for which the values are calculated- the integral is from 0 to v.

 

[Niles]

I get that E(v) = T_2_E(v) + ½*[E''(t)*(v-t)^2] + [E'(t)*(v-t)] + [E(t)] where the limits are from 0 to v.

From this, I don't see how E(v) >= T_2_E(v)? The "(v-t)"-part will cancel out for t=v?

 

[Niles]

I mean, doesn't the restterm become negative?

I mean: 0-E''(0)*v + 0-E'(0)*v + E(v)-E(0)

 

[Niles]

Sorry guys, but this paper is due in 5 hours. I've been trying for the past 2-3 hours, but I have no more solutions.

First I thougth of just looking at E'''(t) - so I wouldn't have to integrate. Apparently E''''(t) (4 * ') only has complex roots, so no solutions in R. Then it must mean that E'''(t) (3 * ') is growing, so the integral must be positive or 0.

That just doesn't seem like a valid solution.