How do I find the minimum value of f(k) with summations and exponents?

[onako]

To determine the value of the unknown for which the function is minimized, we take the derivative and equate with 0. This is relatively simple with linear functions, but I have problem with a function involving
summations and exponents. What would be the way to express the value of k for which f(k) is minimized:
<br /> f(k)=\sum a_i^{2k} - 2 \sum a_i^kb_i<br />
Suppose equal number of terms a_i and b_i in the summation.

 

[HallsofIvy]

Assuming the sums are finite, you can differentiate term by term.

f&#039;(k)= 2\sum a_i^{2k}ln(a_i)- 2\sum a_i^k b_i ln(a_i)= 0

f&#039;(k)= 2\sum (ln(a_i))(a_i^{2k}- a)_i^k b_i)= 0[/quote]<br /> <br /> f&amp;#039;(k)= 2\sum ln(a_i)a_i^k(a_i^k- 1)b_i)= 0[/quote]&lt;br /&gt; &lt;br /&gt; Solving that equation may be quite difficult.

 

[onako]

Assuming the sums are finite, you can differentiate term by term.

f&#039;(k)= 2\sum a_i^{2k}ln(a_i)- 2\sum a_i^k b_i ln(a_i)= 0

f&#039;(k)= 2\sum (ln(a_i))(a_i^{2k}- a_i^k) b_i= 0

f&#039;(k)= 2\sum ln(a_i)a_i^k(a_i^k- 1)b_i= 0

Solving that equation may be quite difficult.

The last equation means that for k=0 the term (a_i^k-1) is 0, leading to complete 0 summation.
But, what would be the way to discover other values of k for which f(k) is min?
Is there a closed form solution, or would I have to employ some other procedure approaching the correct value gradually?
(I have some problems with viewing tex portions and the edit does not seem to work)
I really appreciate your help.