Re: Analysis with continuous, sum.

In article
<9b0b689a-5a2c-41e8-a669-f2f992778402@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Achava Nakhash, the Loving Snake" <achava@xxxxxxxxxxx> wrote:

On Oct 20, 10:56 am, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:
In article <gdi4br$c8...@xxxxxxxxxxxxxxxxxxxxx>,

 "mina_world" <mina_wo...@xxxxxxxxxxx> wrote:
Hello teacher~

f(x) = sum{n=1 to oo} (x^n)(1-x) / n

Show that f(x) is continuous on [0, 1].

The maximum value of (x^n)(1-x)/n on [0,1] is (n/(n+1))^n *
[1-(n/(n+1)]/n <= 1/n^2. Because sum 1/n^2 < oo, your series converges
uniformly on [0,1] by the Weierstrass M test. The partial sums are all
continuous on [0,1], hence so is f.



--------------------------------------------------
Hm...maybe,
I must show that f(x) converges uniformly on [0, 1] ?

Hm...
For n >= m > N,
|f_n(x) - f_m(x)| =
|{(x^(m+1))(1-x) / (m+1)} + ... + {(x^n)(1-x) / n}|
<= (1-x).|{x^(m+1) / (m+1)} + ... + {(x^n) / (m+1)}|
<= (n-m)(x^(m+1)) / (m+1)
<= (n-m) / (m+1)
??

Maybe, I need your rebuke.- Hide quoted text -

- Show quoted text -

I get (simply setting the derivative of (x^n)*(1-x)/n to 0 and solving
to get either x = 0 or x = n/(n+1). To see that the value of x = n/(n
+1) is a maximum, calculating the second derivative, which is
( (n-1)n*(x^(n-2)) - n(n+1)*(x^(n-1) )/n = (x^(n-1))*( (n-1)x - (n
+1)). The exponenent terms is always positive. Plugging in x = n/(n
+1) into the linear term gives (1 - 3n)/(n+1) which is negative so
indeed x = n/(n+1) is a maximum.

You don't need to do that. (x^n)*(1-x)/n > 0 in (0,1), = 0 at the end
points. So the max occurs inside where the derivative = 0, i.e., at x
= n/(n+1).

Plugging this value of x back into
the function to be maximized, we end up with (n/(n+1))^n, which is a
convergent series because it is a geometric progression with constant
multiple strictly less than 1.

No, the max is what I wrote: (n/(n+1))^n * [1-(n/(n+1)]/n <= 1/n^2.
Also, sum (n/(n+1))^n is not a geometic series - not even close.

Now use the M test. I like this
better than adding another inequality and then relying on the OP to
know a proof about the convergence of the sums of reciprocal squares
which I am not sure is necessarily true at this stage of a real
analysis class.

Regards,
Achava
.

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