Problem 1 is done by the book as an example of a pretty sophisticated discussion. These hints are out of order. Scroll down.
Use "ln" instead of "log", but in fact Maple treats these as the same.
Use symbolic limits to get an equation for a that can be solved exactly. Then graph the function for that value of a and make sure from studying the left and right side formulas that they are continuous everywhere else on the stated domain of the function. While you are looking at the graph, is the function differentiable at the joining point of the piecewise definition?
Here we have a family of functions with the added complication that only for special values of the power a is the power function defined for negative values of x, like a rational value (in lowest terms) a = p/q with q odd and p either even or odd. [Don't fail to note that a 0 or negative value of a surely makes infinite one-sided limits since the log already goes to negative infinity and the negative power has division by zero, so goes to plus or minus infinity.] Back up your symbolic commands with some representative plots of typical members of the family. While you are studying this, why not explore when the function is continuous, and if continuous, when differentiable.
Test your procedure for all cases, and then just think why there is an obvious relation between the 3 functions and explain why.
Simplify the exact ratio of successive terms by hand and show that it equals the ratio of the lhs to the rhs of the inequality given in the book hint. and then use the book hints to show why it must be a proper fraction. Why can this ratio be compared to the b sequence given there? Then why is its b sequence monotonically decreasing? I still have not cracked this one. (What is an obvious lower bound for this sequence of positive terms?) Explore how fast it converges by plotting or looking at enough terms... where enough means really big... To check limit itself, follow book hint to substitute the Stirling approximation for n! take the log and then use the expansion for ln(1+x) for small x as indicated.
Look at an analogous problem:
../../students/bob/2-5-6-hint.mw
Adapt the discussion of the similar problem to the present one.
The hint in the back enables one to look instead at a quadratic equation
condition for the limit of the squared term sequence, where the quadratic
formula finds two roots, one of which is negative and must be excluded.
The first inequality in the book hint is due to the binomial expansion,
dropping all terms after the second one and since they are positive, you get
their first inequality. Why is their second inequality true? Use Maple to
simplify it. This can be easily done with L'Hopital's rule to check that Maple
is right.
More explicit help:
The ratio of successive terms is of the form an+1/an =
n/(n+1) Xn where X = (n/(n-1))
/ ((n+1)/n) . Use Maple to show that this is equivalent to
X = (1 + 1/(n2-1) ). Let x = 1/(n2-1)
.
The power term can then be expanded with the binomial theorem
(1+x) n = 1 + n x +
½ n(n-1) x2 +
... > 1 + n x
where the inequality follows since all terms dropped in the expansion are positive
if x > 0 . Thus
an+1/an >
n/(n+1) ( 1 + n x) .
Simplify this and expand the numerator and denominator each to a
third degree polynomial expression. Apply first normal to the entire
expression to put it into factored rational function form, then use numer
and denom to get the numerator and denominator, and then expand each one
separately.
Compare the numerator and denominator and
the result is obvious that the numerator is always bigger than the denominator.
Why is the sequence bounded? We are raising a proper fraction to a positive power, so the result must always be less than 1, so 1 is an upper bound to this monotonically increasing sequence and hence it must have a limit. Maple easily evaluates this limit. Examining many terms in the sequence you see that it converges very slowly, with only 2 decimal place accuracy after 100 terms.
The lowest order terms in the taylor series:
ln(1+x) = x, exp(x)-1 = x, if |x| << 1.
Show that the series can be represented as:
a_n = (2k^(1/n)-1)^n = (1+2b_n)^n, where b_n = k^(1/n)-1 = exp(ln(k)/n) - 1.
Show that b_n -> 0 for any k>0 (this is a determinate limit, think about
the cases 0<k<1, k=1 and k>1)
Then take the ln:
ln(a_n) = n ln(1+2b_n).
Use the taylor expansion on the ln function, then the exp taylor
expansion on b_n in that expression.
Simplify.
Exponentiate limit(ln(a_n)) to get back the limit for a_n itself.
To go further you might keep the quadratic terms in both taylor
expansions to see what the next term in the taylor series of ln (a_n)
would be (proportional to 1/n) to see an error estimate.
Problem 2.5.4 replaces k by n in the formulas, making the expression
grow faster, in fact like n^2, so that one must divide by n^2 to get a
finite limit.
but a bit more involved application of the same ideas will crack it.
Sorry that it took me so long to decipher the hints of the author. This
is not typical of the problems in the book. These limits are tricky.
see above [note misprint in book refers correctly to page 61 where problem 11 is found, but to some worksheet problem which is irrelevant], and:
> k:=n->(1+1/n)^(n^2)/n^(n+1/2)*n!; > seq(evalf(k(n)),n=1..20); Seems to be a decreasing sequence coming to a limit. > stirling:=n->(n/exp(1))^n*sqrt(2*Pi*n); Substituting the stirling approximation for factorial and simplifying by gathering powers leads to: > subs(n!=stirling(n),k(n)); > simplify(%,symbolic); This can be expressed in the form: > (1+1/n)^(n^2)/exp(1)^n*sqrt(2*Pi); This can be regrouped as: > ( (1+1/n)^n/exp(1) )^n*sqrt(2*Pi); The numerator inside the nth power is just the limit definition expression for e =exp(1), (calc book!) so the quotient approaches e/e = 1 so we have a 1^infinity type indeterminate limit, for which one then takes the ln and converts to a quotient to apply L'Hopital's rule. Try this by hand. If we take the logarithm of this expression, we can then expand the factor ln(1+1/n) in its Taylor series expansion valid for small x = 1/n (look in your calc book) > ln(1+x)=taylor(ln(1+x),x,4); If you then simplify the entire expression with this in place, the leading term that comes from this expansion when substituted into the log of the sequence expression (leading term = the constant term, the rest are inverse powers of n) apart from the log of the constant factor of sqrt(2*Pi) is -1/2 = -ln(sqrt(exp(1))), which then can be combined together (see book hint).