# 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*to argue that

^{ n }*f*is positive or negative depending on whether

_{ n }*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.

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