The Question

derivative of f(x) $$ 0 = (1-x)^n -n(1-x)^{n-1}x $$

$$ $$

finding the x in terms of n

divide f '(x) by $$ (1-x)^{n-1}$$ $$ $$

$$ 0 = (1-x) - xn $$ $$ x = \frac{1} {1 + n} $$

i'm stuck at how to find a-n and to find the limit

$$ \lim_{x \to \infty} \ (n+1)an$$ = $$ \lim_{x \to \infty} \ \frac{1}{x}an$$ = $$ \lim_{x \to \infty} \ lnx \ an$$

since I don't know to find a-n, I cannot find the lim, and cannot proceed to 3)

can you help me please?


For $a_n$, it's the maximum value of $f(x)$ on interval $I=[0,1]$.

Knowing that $f'(\frac{1}{n+1})=0$ (the only critical point), $f(0)=0, f(1)=0$, $\ f''(\frac{1}{n+1})<0$, you can conclude $f(\frac{1}{n+1})$ is the only local maximum on I.

Thus, $$lim_{n\rightarrow\infty}(n+1)\ f(\frac{1}{n+1})=\frac{1}{e}$$ is found after algebraic manipulation. (Try it by yourself :D ).

For the third question involving a definite integral, I suggest using integration by parts (Using the tabular method makes things easier). So $$\int_{0}^1=-\frac{x}{n+1}(1-x)^{n+1}+\frac{1}{(n+1)(n+2)}(1-x)^{n+2}|_{0}^{1}=\frac{1}{(n+1)(n+2)}$$

If you have any problem, please leave a comment!

  • $\begingroup$ Why there's negative sign at the first step of integration? $\endgroup$
    – hely123
    Jan 10 '18 at 4:44
  • $\begingroup$ @hely123 Because the integrand includes $(1-x)$ but not $(x-1)$ $\endgroup$
    – Macrophage
    Jan 10 '18 at 4:47
  • $\begingroup$ @hely123 If you differentiate and take out that negative sign, it will be positive again. $\endgroup$
    – Macrophage
    Jan 10 '18 at 4:48

From your derivative $$f'(x) = (1-x)^n -n(1-x)^{n-1}x$$ we find that the function is $$ f(x) = x(1-x)^n$$ $f'(x)=0$ implies $$ (1-x)^{n-1}(1-x-nx)=0$$If $n=1$ the only solution is $x=\frac {1}{2}$ and if $n>1$ we have also $x=1$ as a critical point.

For $x=1$ we get $ f(x) =0$ and for $x=\frac {1}{n+1}$ we get $f(x)=\frac {1}{n+1} (1-\frac {1}{n+1})^n.$

For odd values of $ n$, the function attains its maximum at $x= \frac {1}{n+1}$ and the maximum value is $\frac {1}{n+1} (1-\frac {1}{n+1})^n.$

For even values of $n$ the function $f(x) =x(1-x)^n$ does not have an absolute maximum.


Point 1

$$f'(x) = (1-x)^n -n(1-x)^{n-1}x=0\iff(1-x)^{n-1}(1-x-nx)=0\iff$$

$$\iff x=1\not\in(0,1) \quad \lor \quad x = \frac{1} {1 + n}\in(0,1)$$

Point 2

$$f(x) = x(1-x)^n \implies f(0)=f(1)=0\implies a_n=f_{max}=\frac{1} {1 + n}\left(1-\frac{1} {1 + n}\right)^n\\\implies a_n=\frac{1} {1 + n}\left(\frac{n} {1 + n}\right)^n$$


$$\lim_{n\to \infty}(n+1)a_n=\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n=\lim_{n\to \infty}\left(\frac{1}{\frac{n+1}{n}}\right)^n=\lim_{n\to \infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac1e$$

Point 3

By substitution $y=1-x$ and $dy=-dx$

$$\int_{0}^1 x(1-x)^n dx=-\int_{1}^0 (1-y)y^n dy=\int_{0}^1 y^n-y^{n+1} dy=\left[\frac{y^{n+1}}{n+1}-\frac{y^{n+2}}{n+2}\right]_0^1=$$ $$=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}$$


You did the first one correct as $x$ is really $\frac{1}{1+n}$.

The second one is covered pretty much by other answers but I would like to add that when you return the maximum $a_n=\frac{1}{n+1}\cdot\left(1-\frac{1}{n+1}\right)$ to the limit, you get the limit:
$$\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n=\lim_{n\to \infty}\left(\frac{1}{\frac{n+1}{n}}\right)^n=\lim_{n\to \infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}$$
then you substitute $\frac{1}{n}$ with t and get
For the third one I would recommend checking this answer on a similar question here on math.se.


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