Try our Free Online Math Solver!

Math Problems
Problem 1:
Use power series to calculate
Hint: Expand numerator and denominator into a power
series, carrying as many terms as needed
to have some power of x whose coefficient is not 0. Then cancel as many factors
x as possible
from the fraction , and do a long division of power series (just the bare
beginning of long division
suffices).
How many times would you have needed to use l’Hôpital to
get the same result? (Think what the
differentiations needed in l’Hôpital would do to the power series of numerator
and denominator
respectively; then you can answer this question.) Try to imagine how much paper
you’d have
needed to carry out these derivatives. (I am convinced it would be far more than
the power
series need.)
Problem 2: Finding the perimeter of an ellipse is
an old problem, and it leads alas to an
integral for which you won’t find a neat antiderivative. Write down an integral
for the length
of the ellipse x ^{2} + 2y^{2} = 1. Substitute x = sin φ in
this integral and show that the length is
Now this is an integral for which I had warned you that
you won’t find an antiderivative. But
we’ll find a numerically good procedure with power series to evaluate the
integral
anyways, for ‘small t’ (t < 1 actually.) To this end,
expand the integrand in a power series in
the variable t . Its coefficients will depend on sin φ,
and we don’t convert these in power series.
Now integrate the power series term by term. To this end, you need to find
etc. Do you remember how to use integration by parts to obtain
(I have hidden a small part of the formula with
,
so
you must do the calculation for yourself .)
Finally you put the pieces together. Can you write down
the first 6 terms in the power series
I(t) =?+?t+?t^{2}+?t^{3}+?t^{4}+?t^{5}+. . .? Make sure that you do not evaluate
products like 1 · 3 · 5 · 7
because doing so would obliterate the pattern that displays the general
coefficient.
Now evaluate I(t) for t = 1/2 with an error of at most
0.5%. Use a simple comparison with a
geometric series to estimate the truncation error.
Problem 3: The general binomial formula says
You may or
may not be familiar with this version. It can be proved by induction (see
appendix of the book,
p 335), but this is not your job right now. I just wanted to state this result
for you.
Now your job is to multiply the power series for e^{x} and
e^{y} and prove that e^{x}e^{y} = e^{x+y}. The
punchline of this calculation is that it guarantees the validity of this formula
even for complex
numbers x and y. Of course you are familiar with this formula for real x and y ;
but with the
exponential function recently defined for complex numbers as well, using the
power series, we
should in principle make sure that the usual rules still hold in the new
situation.
Prev  Next 