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Taylor and Einstein

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data
    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).

    3. 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.
  2. jcsd
  3. Oct 7, 2007 #2


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    The error will depend upon the specific value of v for which the values are calculated- the integral is from 0 to v.
  4. Oct 7, 2007 #3
    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?
  5. Oct 7, 2007 #4
    I mean, doesn't the restterm become negative?

    I mean: 0-E''(0)*v + 0-E'(0)*v + E(v)-E(0)
  6. Oct 7, 2007 #5
    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.
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