Truncated Exponential Series Inequality

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.