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The formula for relativistic mass and its Taylor series. It explains the practical significance of the interval of convergence of the series. a math problem solution from the University of Michigan Department of Mathematics.
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m = m 0
v^2 c^2
where c is the speed of light and m 0 is the mass of the object when it is at rest.
(a) (8 points) Write the first four nonzero terms of the Taylor series for m in terms of v. (Hint: You may want to use the binomial series.)
Since m is given by
m = m 0
v^2 c^2
and 0 ≤ v < c, we may use the binomial series
(1 + x)p^ = 1 + px + p(p − 1) 2! x^2 + p(p − 1)(p − 2) 3! x^3 + · · · ,
with x = −v^2 /c^2 and p = − 1 /2, to obtain:
m = m 0
2 c^2 v^2 +
8 c^4 v^4 +
16 c^6 v^6 + · · ·
(b) (3 points) The series you derived in part (a) converges for v in the interval [0, c). Interpret the practical significance of this interval of convergence in the context of this problem (that is, as far as the relativistic mass of an object is concerned.)
When an object’s speed (v) is less than the speed of light (c) the relativistic mass (m) of the object is finite. In this case, the Taylor series for the relativistic mass (derived above) converges to the actual relativistic mass.
University of Michigan Department of Mathematics Fall, 2005 Math 116 Exam 2 Problem 9 (relativity) Solution