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Solutions To Mathematics Textbooks/Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)/Chapter 5

< Solutions To Mathematics Textbooks‎ | Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)


Chapter 5Edit


We will compute the derivative and show that it is always  . The derivative of   is given by the quotient rule to be   The denominator is clearly always positive as we know it is the square of some real number  , so in order to show that the entire derivative is always positive we need to show that the numerator is positive.

However,   and we know that, by the mean value theorem there is some   such that   since  . Since   is monotonically increasing it must be that  .

Thus,   and so the derivative of   is positive, which implies that   is monotonically increasing.


Since the denominator of   goes to zero as   we can employ L'Hopital's rule. Using this we get   Since the denominator is again zero we apply L'Hopital's rule again    

An example of a function that is not differentiable, but where this limit exists is   The second derivative of this function does not exist at zero, as   however we can compute the above limit at zero by  


A function   is said to be convex if  

First we shall prove ( ) that   monotonically increasing implies   is convex.

Assume   is monotonically increasing. Let   for  . The mean value theorem implies that   such that  . Similarly   such that  . Thus, since   is assumed to be monotonically increasing we have   Because   we know that   for some  . Moreover, we have           We note here that the direction of the inequality is preserved since we have   and   and   (*)       reducing yields         but since the initial choice of   was arbitrary (up to ordering) this holds for any  . We must still address the case corresponding to  . This is easy though.

For let  . Then by what we have proven already   now let   so it follows from the above that    

Thus if   is monotonically increasing   for  ,  

Now we prove ( ) that   convex implies   is monotonically increasing.

The proof of this direction follows similarly to the previous. Let's assume   is convex.

Then   for   (Note: we must have   by (*)) and  ,  . Following the previous reasoning in reverse we see that   for  .

So in the limit   we have   and in the limit   we have   thus, combining these statements we have   so   is monotonically increasing as desired.

Now it only remains to prove that (assuming   exists for  )   is convex if and only if   for  .

  is convex     is monotonically increasing by the previous proof. Moreover,   is monotonically increasing     (proved in class).

So we have shown that (assuming   exists for  )   is convex if and only if   for   as desired.