Inequalities, Products and Sums of Numbers

By Anonymous on Thursday, December 9, 1999 - 03:16 pm :

How do you show that for p greater or equal to 1, |(x+y)/2|^p is less than or equal to (|x|^p + |y|^p )/2 for all real x, y?

What is the product of all terms [f(k)-1]/f(k) where f(k) = 2^k and the product is taken over all natural numbers k? If it's impossible to find the product, then is it possible to determine whether the product is 0 or some number in the interval (0,1)?

Is the sum of the odd terms in the harmonic series convergent or divergent?

By Amanda Turner (Agt24) on Thursday, December 9, 1999 - 07:14 pm :

For the first part of your question you need something known as Jensen's inequality. This uses the notion of convex and concave functions. A function

f

is convex if for all

x

y

in the domain of

f

$f (t x + (1 - t) y) \leq t f (x) + (1 - t) f (y)$

for all $t$ in the interval [0,1] and is concave if the inequality sign is the other way around. Geometrically you can think of this as the line drawn between two points on the graph of a function being respectively above or below the curve representing the function. If you know some calculus, then if a function has a second derivative, the function is convex if and only if its second derivative is positve and concave if and only if its second derivative is negative. Can you see why this is true?

Now Jensen's inequality states:

Suppose $f$ is a convex function. Then for any $x_{1}$ , $x_{2}$ , ..., $x_{n}$ we have

$f ((x_{1} + x_{\dots} + x_{n}) / n) \leq (f (x_{1}) + f (x_{2}) + \dots + f (x_{n})) / n$ .

If $f$ is concave, the inequality sign is the other way around.

You may like to try and prove Jensen's inequality using induction (if you do not know what induction is, please ask me). Under what conditions is the inequality strict i.e. when can you replace the inequality sign by an equality sign?

By Amanda Turner (Agt24) on Friday, December 10, 1999 - 03:28 pm :

There is a classical proof that the harmonic series diverges which works by considering intervals between adjacent powers of two:

Sum between 1/2^n-1 and 1/2ⁿ	Lower bound	Value
1/2	=1/2	=1/2
1/3+1/4	> 1/4+1/4	=1/2
1/5+1/6+1/7+1/8	> 1/8+1/8+1/8+1/8	=1/2
...	...	...

Try and extend this table. What does this tell you about the sum 1 + 1/2 + 1/3 + ... + 1/2ⁿ and hence the harmonic series? Try and adapt the above proof to deal with only the odd terms. Other variations you may want to think about are the series
*1+1/2^a +1/3^a +... for various values of a
*The sum of the reciprocals only of those numbers with no 9's in there decimal representation. So our series starts
1+1/2+1/3+...+1/8+1/10+...+1/18+1/20+...
* The reciprocals of the primes. So our series starts
1/2+1/3+1/5+1/7+1/11+...
What other variations can you think of?
I hope this is helpful. if you have any other questions or don't understand anything, please feel free to ask.

By Amanda Turner (Agt24) on Friday, December 24, 1999 - 01:29 pm :

For the second part of your question, by calculating values, the product seems to converge to 0.28878809508660.... Putting this number into the inverse symbolic calculator, which can be found here , only found the product itself, so it probably doesn't have a closed form. I am told, though, that your product is in a fact a modular form (called the eta-function) at the point q=1/2. (Something you will probably only learn about in 3rd or 4th year at university).

It is however possible to prove that the product is greater than 1/4. Let pk=(1-1/2)(1-1/4)...(1-2^-k ) and then prove by induction that pk > = 14 + 2^-(k+1) for all k. Can you improve this lower bound? How about an upper bound (much easier!)? What if you change all the 2's in your product to 3's say? What other ways of generalising your product can you think of?

I hope you have fun filling in the details. If you have any difficulties, please ask me.

Merry Christmas!
Amanda