Truncated Exponential Series Inequality
Check out John Cook's explanation of truncated exponential series inequality.
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define t n to be the taylor series for exp( x ) truncated after n terms:
how does this function compare to its limit, exp( x )? we might want to know because it’s often useful to have polynomial upper or lower bounds on exp( x ).
for x > 0 it’s clear that exp( x ) is larger than t n ( x ) since the discarded terms in the power series for exp( x ) are all positive.
the case of x < 0 is more interesting. there exp( x ) > t n ( x ) if n is odd and exp( x ) < t n ( x ) if n is even.
define f n ( x ) = exp( x ) – t n ( x ). if x > 0 then f n ( x ) > 0.
we want to show that if x < 0 then f n ( x ) > 0 for odd n and f n ( x ) < 0 for even n .
for n = 1, note that f 1 and its derivative are both zero at 0. now suppose f 1 is zero at some point a < 0. then by rolle’s theorem, there is some point b with a < b < 0 where the derivative of f 1 is 0. since the derivative of f 1 is also zero at 0, there must be some point c with b < c < 0 where the second derivative of f 1 is 0, again by rolle’s theorem. but the second derivative of f 1 is exp( x ) which is never 0. so our assumption f 1 ( a ) = 0 leads to a contradiction.
now f 1 (0) = 0 and f 1 ( x ) ≠ 0 for x < 0. so f 1 ( x ) must be always positive or always negative. which is it? for negative x , exp( x ) is bounded and so
f 1 ( x ) = exp( x ) – 1 – x
is eventually dominated by the – x term, which is positive since x is negative.
the proof for n = 2 is similar. if f 2 ( x ) is zero at some point a < 0, then we can use rolle’s theorem to find a point b < 0 where the derivative of f 2 is zero, and a point c < 0 where the second derivative is zero, and a point d < 0 where the third derivative is zero. but the third derivative of f 2 is exp( x ) which is never zero.
as before the contradiction shows f 2 ( x ) ≠ 0 for x < 0. so is f 2 ( x ) always positive or always negative? this time we have
f 2 ( x ) = exp( x ) – 1 – x – x 2 /2
which is eventually dominated by the – x 2 term, which is negative.
for general n , we assume f n is zero for some point x < 0 and apply rolle’s theorem n +1 times to reach the contradiction that exp( x ) is zero somewhere. this tells us that f n ( x ) is never zero for negative x . we then look at the dominant term – x n to argue that f n is positive or negative depending on whether n is odd or even.
another way to show the sign of f n ( x ) for negative x would be to apply the alternating series theorem to x = -1.
Published at DZone with permission of John Cook, DZone MVB. See the original article here.
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