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Universidad Carlos III de Madrid
Escuela Politécnica Superior
DEPARTMENT OF MATHEMATICS
CALCULUS I
First Year Degree in Biomedical Engineering Solutions to the Final Exam January 9, 2018
Problem 1
a) (2 points) Find the Taylor expansion up to 5th order in powers of x of the function f (x) = log( 2 − cos x). b) (2 points) Determine whether the series ∞
n= 1
(sin n) log
2 − cos( 1 /n)
converges absolutely, only conditionally, or does not converge at all.
SOLUTION:
a) On the one hand cos x = 1 −
x^2 2 +^
x^4 24 +^ o(x
(^5) ), (x → 0 );
on the other hand, f (x) = log( 1 + y) where y = 1 − cos x. Then
y =
x^2 2 −^
x^4 24 +^ o(x
Since log( 1 + y) = y − y
2 2
3 3
- o(y^3 ), (y → 0 ), then log( 2 − cos x) =
x^2 2 −^
x^4 24 +^ o(x
(^5) ) − x^4 8 , because the remaining powers are all higher than x^5. Thus
f (x) =
x^2 2 −^
x^4 6 +^ o(x
(^5) ), (x → 0 ).
b) The series can be written (^) ∞
n= 1
(sin n) f ( 1 /n).
To determine whether it converges absolutely we must decide on the convergence of ∞
n= 1
|(sin n) f ( 1 /n)|.
Since |(sin n) f ( 1 /n)| 6 | f ( 1 /n)|, if the series ∞
n= 1
| f ( 1 /n)|
converges so does the previous one. Now, from the result obtained in a) we know that f (x) ∼ x^2 / 2 as x → 0, therefore, as n → ∞, | f ( 1 /n)| ∼
2 n^2. So the series (^) ∞
n= 1
| f ( 1 /n)| < ∞
and, by the comparison test, ∞
n= 1
|(sin n) f ( 1 /n)| < ∞.
Thus our series converges absolutely.
Problem 2 (1 point)
If a, b, x > 0, prove the inequality
a log x − bx 6 a log
( (^) a b
− a.
SOLUTION:
Take the function f (x) = a log x − bx and differentiate it:
f ′(x) = a x
− b = a^ −^ bx b
Then f ′(x) > 0 for 0 < x < a/b and f ′(x) < 0 for x > a/b. Hence f (x) reaches its maximum value at x = a/b, and therefore f (x) 6 f (a/b). Substituting
a log x − bx 6 a log
( (^) a b
− (^) b a b^
= a log
( (^) a b
− a.
Problem 3 (2 points)
Let f be a differentiable function such that ∫ (^) x 0
f (t) dt =
∫ (^1) x
t^2 f (t) dt + x
16 8
18 9
Find f (x) and the constant c.
SOLUTION: Differentiating the equation,
f (x) = −x^2 f (x) + 2 x^15 + 2 x^17 ⇒ ( 1 + x^2 ) f (x) = 2 x^15 ( 1 + x^2 ) ⇒ f (x) = 2 x^15.
Now substituting back into the equation and setting x = 1, ∫ (^1) 0
f (t) dt =
8 +^
9 +^ c^ ⇒^
t^16 8
1
0
8 +^
9 +^ c^ ⇒^
8 =^
8 +^
9 +^ c^ ⇒^ c^ =^ −^
