

EQUATION 












EXAMPLE IX.A.1: Suppose `f (x) =x^2 + x^3 + x^100`
and
we wish to find
`f (0.5)`.
Of course the exact answer is `.25 + .125 + (.5)^100`, but if we only need an estimate of the answer it is easy to settle for .375. The error we make in using this estimate is relatively small [namely `(.5)^100`]. In this example we should note two things.
Taylor's TheoryObjective and Key Ideas: The main concerns
of Taylor's theory for estimating function values are
Taylor's theory generalizes these two ideas that use the derivative to estimate.
IX.A. Estimating `e^x`^{ }with Polynomials: An Introduction to Taylor's Theory.
To focus our discussion more specifially, in this section we will
consider
only `f(x)=e^x`and begin by trying to estimate `f(1)=e` illu
strating the key ideas
of
Taylor's theory.
At `x = 0`, we find that `f(0) =1`, `f '(0) =1`, and since
`f
'(x)
= f(x)` we have that `f^{(n)}(0) =1` for all
natural numbers `n`. We now consider our first result, typical of
the Taylor's theory approach to estimation using polynomials and
derivative
information.
PROPOSITION IX.A.1: If `P_n(x)= 1 + x+ {x^2}/2 +
{x^3}/{2*3} + {x^4}/{4!}+...+{x^n}/{n!}`
then
`P_n(x)` is a polynomial of degree n
so that `P_n(0)= P_n'(0)=P_n''(0)=...=P_n^{(n)}(0)=1`
and `P_n(b)` is approximately equal to `e^b`.
In
fact, if we let `R_n = e^b  P_n(b)`, then for some `c` between
`0` and `b`, `R_n = e^c {b^{n+1}}/{(n+1)!}`
.
GeoGebra: Table, Graph, and Mapping Diagram of `P_n(x)` and `R_n(x)`
Estimating e: Before we proceed to justify this result,
we'll
apply
this result using `n=5` to estimate the value of `e [=e^1]`.
First, `P_5(x)= 1 + x+ {x^2}/2 + {x^3}/{2*3} +
{x^4}/{4!}+{x^5}/{5!} = 1 + x+ {x^2}/2 + {x^3}/{6} +
{x^4}/{24}+{x^5}/{120}` ,
so according to the proposition, `e` is approximately equal to
`P_5(1)=1 + 1+ {1^2}/2 + {1^3}/{6} + {1^4}/{24}+{1^5}/{120} ~~
2.716667` and
`R_5 = eP_5(1)` where `R_5 = e^c {1^6}/{6!}=e^c 1/720` for some `c`
between `0` and `1`. Thus, `e` is approximately `2.716667` and the
error
in using this estimate is `R_5`. Since `c` is between `0` and `1`,
`e^c`
is between `1` and `e`, so we can deduce that `R_5`, the difference
between e and the estimate, `P_5(1)`, is no greater than `e/720`. Using
estimates of `e` from our earlier work in Chapter VI we know
that `e< 3` so our error `R_5(1)` is no larger than `3/720 = 1/240
\approx 0.004167`.This compares roughly well with the error from the
GeoGebra applet that shows an error after setting the slider $n = 5$
to find `R_5(1) = 0.00162`.
A more accurate estimate can be obtained by using a larger value for `n`. Try using `n=6` and `7` to see the improvement. [This can be done progressively using the GeoGebra applet above or download a spreadsheet which you can examine now or later.]
Proof of Proposition: We'll begin our proof by finding `P_n'(x)`.
`P_n'(x)=1+ x+{x^2}/2 +{x^3}/{2*3}+{x^4}/{4!}+...+{nx^{n1}}/{n!}`
`= 1 + x+ {x^2}/2 + {x^3}/{2*3} +
{x^4}/{4!}+...+{x^{n1}}/{(n1)!}=P_{n1}(x)`
From this it follows easily that `P_n(0)= P_n'(0)=P_n''(0)=...=P_n^{(n)}(0)=1`.
Now suppose that `b` is not `0` and let `R_n = e^b

P_n(b)`.
We need only justify the formula for evaluating `R_n`.
For convenience we'll write `R=R_n` and let
Furthermore `g'(t) = e^t P_n(bt)  e^t P_n'(bt)R{(n+1)(bt)^{n}}/{b^{n+1}}=e^t[P_n(bt)  P_n'(bt)]R{(n+1)(bt)^{n}}/{b^{n+1}}`
But `P_n(x)  P_n'(x) = {x^n}/{n!}`, so`g'(t) = e^t {(bt)^n}/{n!}R{(n+1)(bt)^{n}}/{b^{n+1}}`.
Now we apply the Mean Value (or Rolle's) Theorem to the function g,
we
can
say
that there is a number `c` between `0` and `b` where
`g'(c)=0`.
Thus
`0 = e^c {(bc)^n}/{n!}R{(n+1)(bc)^{n}}/{b^{n+1}}`and `e^c
{(bc)^n}/{n!}= R{(n+1)(bc)^{n}}/{b^{n+1}}`.
Solving this last equation for R gives
`R_n = R = e^c {b^{n+1}}/{(n+1)!}`.
Note: Since the exponential function has a positive
derivative for all `x`, the function is increasing for all
`x`.
(i) If `0<c<b` then `1< e^c < e^b < 3^b` and `0 <
{b^{n+1}}/{(n+1)!} < R_n = R = e^c {b^{n+1}}/{(n+1)!} < e^b
{b^{n+1}}/{(n+1)!}< 3^b{b^{n+1}}/{(n+1)!}`.
(iia) If `0>c>b` then `1> e^c >e^b > 2^b` and if `n`
is odd then `{b^{n+1}}/{(n+1)!` is positive and thus
`{b^{n+1}}/{(n+1)!} > R_n = R = e^c
{b^{n+1}}/{(n+1)!} > 2^b {b^{n+1}}/{(n+1)!} > 0`.
(iib) If `0>c>b` then `1> e^c >e^b > 2^b` and if `n`
is even then `{b^{n+1}}/{(n+1)!` is negative and thus
`{b^{n+1}}/{(n+1)!} < R_n = R = e^c
{b^{n+1}}/{(n+1)!} < 2^b {b^{n+1}}/{(n+1)!} < 0`.
Putting this information about the quality of `R_n` together we can see
the following:
For all `x > 0` and for any `n`, `e^x > P_n(x)`,
while for ` x< 0` the polynomials differ in relation to `e^x`:
When `n` is odd then `e^x > P_n(x)`;
when `n` is even then `P_n(x) > e^x`.
Estimating the value of e is not the only use for the previous proposition. Before turning to the more general Taylor's theory in the next section, here are two more examples of its application in estimating definite integrals related to e^{x}.
EXAMPLE IX.A.2: Estimate `int_0^1 e^x dx = e1`. Use this to estimate `e`.
Solution: Using Proposition IX.1 with `n=4` for each x
between `0` and `1` we have
`0<e^x =P_4(x) + R_4(x) = 1 + x+ {x^2}/2 + {x^3}/6 +
{x^4}/24+R_4(x)` where `R_4(x) = e^c {x^5}/120`
for some `c` with `0<c<x<1`.
Thus`0<R_4(x)<3 {x^5}/120 = {x^5}/40`.
[Remember `e^c<3`.] and so `P_4(x)<e^x<P_4(x) + {x^5}/40`
for all `x` between `0` and `1`. Now we use the monotone property of
the
definite integral [Ch. V . ** ] to obtain the estimate:
`int
_0^1P_4(x)dx<int_0^1e^x dx<int_0^1P_4(x) + {x^5}/40`,
so
Therefore `e` is approximately `1 + 1+ 1/2+1/6+1/24+1/120 ~~2.716667` as in the first estimate we saw in the note after the statement of the theorem, and the error in using this estimate is less than `1/240` as we also saw previously.
Substituting `t^2` for `x` ,we see that
Now since `0>R_4(t^2)>{(t^2)^5}/120 ={t^10}/120` we have `int_0^1 P_4(t^2)dt >int_0^1 e^{t^2} dt > int_0^1 P_4(t^2) {t^10}/120 dt`.
By evaluating these integrals we obtain
or `1 1/3 + 1/10 1/42+ 1/216 > int_0^1e^{t^2}dt > 1  1/3 + 1/10 1/42+ 1/216  1/1320`
Therefore, `int_0^1 e^{t^2}dt` is approximately equal to `1  1/3 + 1/10  1/42 + 1/216 ~~ 0.747486...`;
and this is an overestimate by no more than `1/1320`.
Comment: In both of these examples we have been able to
estimate
a definite integral involving the exponential function by using a
Taylor
polynomial of degree 4. It should be apparent that by using a higher
degree
for the estimating polynomial, the error term will become smaller and
we
will obtain a more precise estimate. The systematic pattern in these
polynomials
should allow you to find more precise estimates for the last example
without
much difficulty.