SOLUTIONS CHAPTER 8.3
MATH 549 AU00
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Proof. The first inequality is clear: for a fixed x we have
ex = 1 +
x
x2
xn
xn+1
x
x2
xn
+
+ ··· +
+
+ ··· > 1 + +
+ ··· +
1!
2!
n!
(n + 1)!
1!
2!
n!
As for the second part (again, fixed x):
ex − (1 +
=
x
x2
xn−1
xn
xn+1
+
+ ··· +
)=
+
+ ··· =
1!
2!
(n − 1)!
n!
(n + 1)!
xn
x
x2
xn
x
x2
(1 +
+
+ ...) ≤
(1 + +
+ ...) ≤
n!
n + 1 (n + 1)(n + 2)
n!
1!
2!
xn
a
a2
ea xn
≤
(1 + +
+ ...) =
n!
1!
2!
n!
4
Proof. We use the fact that
e=1+
1
1
+ + ...
1! 2!
so we have:
1
1
1
+ + ··· + )
1! 2!
n!
(a finite sum is less than the whole series, when the series has only positive
terms) so we get
1
1
1
0 = 0 ∗ n! < en! − (1 + + + · · · + )n!
1! 2!
n!
which solves the first inequality.
For the second we have:
1
1
1
1
e − (1 + + + · · · + ) <
+ ...
1! 2!
n!
(n + 1)!
1
1
1
<
(1 +
+
+ ...) <
(n + 1)!
n + 2 (n + 2)(n + 3)
1
1
1 e
1 1
<
(1 + + + . . . ) =
n! n + 1
1! 2!
n! n + 1
0 < e − (1 +
Date: 11/04/2000.
1
2
MATH 549 AU00
hence
1
1
1
e
+ + · · · + )n! <
1! 2!
n!
n+1
The above (strict) inequalities are actually true for all n.
For the third inequality:
1
1
1
1
3 − e = 3 − 1 − − − ··· = − − ...
1! 2!
2 3!
but we have the following obvious inequality for n ≥ 2:
1
1
1
1
1
≥
⇐⇒
−
>
n(n + 1)
(n + 1)!
n (n + 1)!
n+1
en! − (1 +
which gives us:
1
1
1
− ≥
2 3!
3
1
1
1
− ≥
3 4!
4
..
.
add all the above and we get that
1
1
1
1
1
− − − ··· −
> >0
2 3! 4!
n!
n
for all n, hence
3−e>0
hence
e
<1
n+1
for n ≥ 2.
Assume now that e is rational ⇒ e = m
n ⇒ en! = m(n−1)(n−2) . . . 1 ∈ Z;
1
but we have that (1 + 1 + 2!1 + . . . n!
)n! ∈ Z as well; their difference is also an
integer. But look at the inequality we just proved: it says that this particular
difference is less than 1! so we have an integer, which is positive, but less
than 1, impossible ... so the assumption that e was rational is false.
5
Proof. We have that:
(x + 1)(1 − x + x2 − x3 + · · · + (−x)n−1 ) =
= x + 1 − x2 − x + x3 + x2 − x4 − x3 + · · · + x(−x)n−1 + (−x)n−1 =
= 1 − (−x)n
and just rearrange the equation ... By Theorem 8.3.9 we have that the log
is the antiderivative of x1 ; use this and we have:
Z x
Z x
1
(−t)n
dt =
(1 − t + t2 − t3 + · · · + (−t)n−1 +
) dt ⇐⇒
1+t
0 t+1
0
SOLUTIONS CHAPTER 8.3
3
Z x
n
x2
(−t)n
n−1 x
log(x + 1) − log(1) = (x −
+ · · · + (−1)
)−0+
dt ⇐⇒
2
n
0 1+t
Z x
n
x2
(−t)n
n−1 x
log(1 + x) = x −
+ · · · + (−1)
+
dt
2
n
0 1+t
From here we get:
Z x
x2
xn
(−t)n
| log(1 + x) − (x −
+ · · · + (−1)n−1 )| = |
dt| ⇐⇒
2
n
0 1+t
Z x
Z x
n
x2
(−t)n
n−1 x
⇐⇒ | log(1+x)−(x− +· · ·+(−1)
)| ≤
|
| dt ≤
tn dt =
2
n
1+t
0
0
xn+1
−0
n+1
(since t + 1 > 1) and the inequality is proved.
=
8
Proof. Let g(x) =
f (x)
ex ; let’s compute the derivative of this
f 0 (x)ex − f (x)ex
f (x)ex − f (x)ex
g 0 (x) =
(ex )2
=
e2x
function:
=0
Hence the function g must be a constant, g(x) = K, ∀x ∈ R. So
K ⇐⇒ f (x) = Kex .
f (x)
ex
=
9
Proof. Actually the inequality 1 + x ≤ ex is true for all x: taking h(x) =
ex − x − 1 we have that h0 (x) = ex − 1 which is negative up to 0 and
then positive from 0, which means that 0 is a local (and actually a global)
minimum; but h(0) = e0 − 0 − 1 = 1 − 1 = 0 hence the lowest posible value
for h(x) is 0, so ex − x − 1 ≥ 0 ⇒ ex ≥ x + 1.
Computing all the values xk , some will be negative - but we just observed
that it’s OK to plug in even negative numbers into the inequality that we
have to use. What we get is:
ak
ak
ak
ak
− 1 + 1 ≤ e A −1 ⇐⇒
≤ e A −1
A
A
We also notice that after pluging the values xk in the inequality we actually
get all positive numbers - and it’s safe to multiply:
a1
a2
an
a1 a2
an
...
≤ e A −1 e A −1 . . . e A −1 ⇐⇒
A A
A
a1
a2
an
a1 a2 . . . an
⇐⇒
≤ e A + A +···+ A −n ⇐⇒
n
A
a1 +a2 +···+an
a1 +a2 +···+an
a1 +a2 +···+an −n
a1 a2 . . . an
−n
A
n
⇐⇒
⇐⇒
≤
e
=
e
n
A
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MATH 549 AU00
1
a1 a2 . . . an
1 −n
n
≤
e
= en−n = 1 ⇐⇒
n
A
1
1
⇐⇒ a1 a2 . . . an ≤ An ⇐⇒ (a1 a2 . . . an ) n ≤ (a1 + a2 + · · · + an )
n
Equality is obtained when all values xk provide us with equality when
pluged into the inequality (remember, 0 was a global minimum) - hence
ak
A − 1 = 0 ⇒ ak = A, ∀k; but then a1 = a2 = · · · = an = A. Done.
⇐⇒