1. And the general term is…
Part (a)
f(x) = e-x2
We’ve memorized that ex = 1 + x + x2 + x3 + … 2! 3!
Therefore, e-x2 = 1 + (-x2 ) + (-x2 )2 + 2! 3!
(-x2 )3
+ …
So the 1st four non-zero terms are…
1 – x2 + ½ x4 – x6 1 6
And the general term is… n!
(-x2 )n

2. Part (b) f(x) = e-x2 From Part (a), f(x) = 1 – x2 + ½ x4 – x6 + …
lim
x0
- ½ x4 + x x8 + … 1 6 x4 24
lim
x0
1 – x2 - (1 – x2 + ½ x4 – x6 + … ) 1 6 x4
lim
x0
1 – x x2 - ½ x4 + x6 - … ) 1 6 x4

3. -½ Part (b) f(x) = e-x2 - ½ x4 + x6 - x8 + … lim x41 6 x4 24
(Re-write by splitting the denominator.)
lim
x0
- ½ x4 + x x8 + … 1 6 x4 24 -½
(-½ x x4 + … )
lim
x0 1 6 24

4. Once again, from Part (a),
Part (c)
Once again, from Part (a),
1 – x2 + ½ x4 – x6 + … 1 6
f(x) = e-x2 =
(1 – t2 + ½ t4 – t6 + … ) dt 1 6
e-t2 dt = x
= [ t – 1/3 t3 + 1/10 t5 – 1/42 t7 + … ] x
x – 1/3 x3 + 1/10 x5 – 1/42 x7
1st 4 terms:

5. Part (c) x – 1/3 x3 + 1/10 x5 – 1/42 x7 1st 4 terms: [ x – 1/3 x3 ]
= [(1/2 – 1/24) – (0 - 0)] =
11/24

6. Part (d) x – 1/3 x3 + 1/10 x5 – 1/42 x7 1st 4 terms: [ 1/10 x5 ] 1/320
According to Taylor’s Theorem, the Lagrange error is less than or equal to the absolute value of the first ignored term.
In this case, that’s 1/10 x5.
x – 1/3 x3 + 1/10 x5 – 1/42 x7
1st 4 terms:
In other words, our error will be less than or equal to
[ 1/10 x5 ]
1/320
(which is less than 1/200)
(1/10)(1/32) =
(Since we’re dealing with an alternating series whose terms are approaching zero, our reasoning is solid and will hold up!)