# Math Help - Prove the following Power Series is monotonous

1. ## Prove the following Power Series is monotonous

Let f(x)=sum_(n=1)^infinity((-1)^n x^n)/n^(3/2)

Prove f(x) is strictly monotonous (where f is defined) and that there exists one solution to
f(x)=1.5
and f(x)=-0.5

2. ## Re: Prove the following Power Series is monotonous

Originally Posted by GIPC
Let f(x)=sum_(n=1)^infinity((-1)^n x^n)/n^(3/2)

Prove f(x) is strictly monotonous (where f is defined) and that there exists one solution to
f(x)=1.5
and f(x)=-0.5
What have you tried so far?

Where is f defined? It's a power series, so you need to find its radius of convergence R. The function will be defined in the interval of convergence (–R,R) (and maybe also at the endpoints of that interval).

Is it monotonic? You can differentiate a power series term by term within its radius if convergence. So do that, and see whether the derivative always has the same sign.

If it is strictly monotonic, then as x goes along the interval of convergence the function will take each value exactly once. The extreme values will be taken at the ends of the interval of convergence. Is the greatest value of the function greater than 1.5, and is the least value less than –0.5? If so, then it must take each of those two values exactly once.

Linguistic note: monotonous means boring and dreary; monotonic means always going the same way (in other words, always increasing or always decreasing).

3. ## Re: Prove the following Power Series is monotonous

Originally Posted by GIPC
Let f(x)=sum_(n=1)^infinity((-1)^n x^n)/n^(3/2)

Prove f(x) is strictly monotonous (where f is defined) and that there exists one solution to
f(x)=1.5
and f(x)=-0.5
a) the series converges for $-1 \le x \le 1$...

b) is $f(0)=0$

c) f(*) can be written as $f(x)= f_{e}(x) + f_{o}(x)$ where...

$f_{e}(x) = \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)^{\frac{3}{2}}}$

$f_{o}(x) = - \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)^{\frac{3}{2}}}$

d) as consequence of a), b) and c) in [-1,1] is $|f_{o}(x)| \ge |f_{e}(x)|$

e) f'(*) can be written as $f'(x)=f'_{e}(x) + f'_{o}(x)$ where...

$f'_{e}(x)= \frac{d}{dx} f_{o}(x)$

$f'_{o}(x)= \frac{d}{dx} f_{e}(x)$

f) as in d) in [-1,1] is $|f_{e}(x)| \ge |f_{o}(x)|$

Consequrence of a), b), etc... is that in [-1,1] f(x) is 'single value'...

Kind regards

$\chi$ $\sigma$

4. ## Re: Prove the following Power Series is monotonous

I'm sorry but I'm not sure I follow. Steps f and d.
How did you show that $|f_{e}(x)| \ge |f_{o}(x)|$ and $|f_{o}(x)| \ge |f_{e}(x)|$ ?

And the last sentence I don't understand

How did you get that f(x) is 'single value' from your steps and how does that show f is monotonic?

I would much appreciate if you could elaborate a little. I am trying to study the subject so some points are not trivial to me.

Maybe we should split f into x>-0 and x<0 instead of odd and even and take the derivative there?

5. ## Re: Prove the following Power Series is monotonous

Originally Posted by GIPC
Maybe we should split f into x>-0 and x<0 instead of odd and even and take the derivative there?
Yes, that is what you need to do.

If $f(x)= \sum_{n=1}^\infty\frac{(-1)^n x^n}{n^{3/2}}$ then (differentiating term by term within the radius of convergence)

$f'(x) = \sum_{n=1}^\infty\frac{(-1)^n x^{n-1}}{n^{1/2}} = -1 + \frac x{\sqrt2} - \frac{x^2}{\sqrt3} + \frac{x^3}{\sqrt 4} - \frac{x^4}{\sqrt5} + \ldots .$

If $-1\leqslant x <0$ then every term in that series is negative and therefore $f'(x)<0.$ If $0< x\leqslant1$ then the terms alternate in sign but decrease (strictly) in size. In such an alternating series, the partial sums are alternately less than and greater than the true sum. Therefore the sum must lie between $-1$ and $-1+\tfrac x{\sqrt2}$ and is thus again negative.

Hence $f'(x)$ is always negative and therefore f is strictly decreasing.

6. ## Re: Prove the following Power Series is monotonous

Hmm, thanks. Interesting, the former poster suggested splitting f for odd and even and you split it for x<0 and x>0

On what do you base now that f is single value? What is the formal explanation?

And you've noted that
"The extreme values will be taken at the ends of the interval of convergence. Is the greatest value of the function greater than 1.5, and is the least value less than –0.5? If so, then it must take each of those two values exactly once."

On what do you base this, what theorem? Why does f take each of those values once?

7. ## Re: Prove the following Power Series is monotonous

Originally Posted by GIPC
Hmm, thanks. Interesting, the former poster suggested splitting f for odd and even and you split it for x<0 and x>0

On what do you base now that f is single value? What is the formal explanation?

And you've noted that
"The extreme values will be taken at the ends of the interval of convergence. Is the greatest value of the function greater than 1.5, and is the least value less than –0.5? If so, then it must take each of those two values exactly once."

On what do you base this, what theorem? Why does f take each of those values once?
Two theorems:

1. Intermediate value theorem: If a continuous function takes two values, then it also takes every intermediate value. So if you know that f(x) is continuous on the interval [–1,1], then it must take every value between f(–1) and f(1).

2. Rolle's theorem: If f is differentiable on the interval [a,b] and f(a)=f(b), then f'(x)=0 at some point between a and b. Conversely, if you know that f'(x) is never equal to 0 (for example if f'(x) is always negative) then f(a) cannot be equal to f(b) (in other words, f can take each of its values only once).