MA 124 Fall-08, Exam 3 Answers
- 1) Applying the ratio test gives 0 |x| . Since 0 |x| < 1 for all
values of x the radius of convergence is infinity and the interval of
convergence is (- infinity, infinity). (Thus there are NO endpoints to
check !)
- 2) a) C + \sum_{n=1}^infinity (-1)^{n+1} 1/((2n+1)! (2n-1)) x^{2n-1}
This is obtained by starting with the power series for sin x (which was given on
a separate sheet of formulas allowed to use), then noting
that since this series begins with - x + x^3/3! one gets (x - sin x)/x^3 =
1/6 + \sum_{n=2}^infinity (-1)^{n+1} 1/(2n+1)! x^{2n-2} . Now integrate
this series to get C + (1/6)x + \sum_{n=2}^\infinity (-1)^{n+1} 1/((2n+1)!
(2n-1)) x^{2n-1} . Noting that n=1 gives (1/6)x we get the answer above.
- 2) b) Evaluating the series in a) gives:
\sum_{n=1}^\infinity (-1)^{n+1} 2^{2n-1}/((2n+1)!(2n-1))
- 2 c) This series converges since the series we started with, the
Maclaurin series for sin x, converges for all x and the operations we did to
get the above power series do not change the radius of convergence. So the
radius of convergence is still infinity, thus substituting x = 2 gives a
convergent series.
NOTE: Applying ratio test directly to the series in b) to determine the
radius of convergence is quite difficult.
- 3) a) We can start with the series for tan^{-1} x (which was given on
a separate sheet of formulas allowed to use) and multiply it by x
to obtain our answer:
\sum_{n=0}^infinity (-1)^n 1/(2n+1) x^{2n+2}
- 3) b) From the formula for Maclaurin series we know the coefficient of
x^8 is f^(8)(0)/8! . We note that x^8 occurs for n = 3 in the above
expression. The coefficient when n = 3 is -1/7. So we have f^(8)(0)/8! =
-1/7 or f^(8)(0) = -8!/7 = -8 x 6 x 5 x 4 x 3 x 2 = -5760.
- 4) a) First rewrite f(x) = (1/2) (1 + (-x/8))^{-1/3} . Then use the
binomial series for k = -1/3 . (The formula for the general binomail series
(1+x)^k was given on a separate sheet of formulas one is allowed to use). We
substitute -x/8 for x , and then multiply this series by (1/2) to get:
\sum_{n=0}^\infinity (-1)^n 1/(2 x 8^n) ((-1/3) \above n) x^n
where ((-1/3) \above n) is the binomial coefficient with (-1/3) on top and
n on the bottom.
- 4) b) When we substitute -x/8 for x we must also substitute it into |x|
< 1 which is where the original binomail series converges. Solving for |x|
we get that our series converges for |x| < 8.
- 4) c) Form the formula for the series the coefficient of x^2 is:
1/(2 x 8^2) ((-1/3) \above 2)
We must compute the binomial coefficient ((-1/3) \above 2) = (-1/3)(-4/3)/2
= 2/9. Substituting this into the above and simplifying we get that the
coefficient of x^2 is 1/576.
- 5) a) We can use the series for cos x (given on separate sheet of
formulas one allowed to use). We can substitute x^2 for x to get the
Maclauring series for cos(x^2) . We want the terms up to and including
degree 4 in x, so expanding it out we get cos(x^2) = 1 - (1/2) x^4 + .... .
So T_4(x) = 1 - (1/2)x^4 .
- 5) b1) If one graphs f^(5)(x) on a graphing calculator ne will see that
the largest value occurs for x = \pi/4 . Evaluating f^(5)(\pi/4) gives
112.2 ... , so rounding UP we can conclude f^(5)(x) < 113 . Thus we can
take M = 113.
One can get a value of M without graphing by first noting f^(5) (x) <
expression obtained by dropping - 32x^5 sin(x^2) (since this is
negative). Then use cos(x^2) and sin(x^2) are both less than or equal to 1 ,
hence f^(5)(x) is less than or equal to 160(\pi/4)^3 + 120(\pi/4) which is
171.7 ... . Hence rounding UP we get f^(5)(x) < 172. So with this approach
we take M = 172.
NOTE: M is NOT unique. It depends on how one gets an upper bound for the
inequality.
- 5) b2) Using M = 113 one gets |R_4(x)| < 113 |\pi/4|^5 / 5! , which
rounding UP to two significant decimals gives |R_4(x)| < .29 .
Using M = 172 one gets |R_4(x)| < 172 |\pi/4|^5 / 5! , which rounding UP to
two significant decimals gives |R_4(x)| < .43 .
- 5 c) I CANNOT be sure that the error would be < .25 since both of the
values obtained in b) are > .25 .
IF YOU HAVE FURTHER QUESTIONS ABOUT HOW TO DO THESE PROBLEMS I WILL BE HAPPY
TO DISCUSS THEM WITH YOU DURING MY OFFICE HOURS OR AT MATH HELP.