- #1
Akorys
- 23
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Homework Statement
I wanted to ask anyone willing to check my proofs to two problems, so that I may know if I am making any mistakes or assumptions.
Let the function f be continuous and strictly monotonic on the positive real axis and let g denote the inverse of f. If a_1\lt a_2\lt...\lt a_n are n given positive real numbers, we define their mean value with respect to f to be the number M_f defined as follows:
M_f=g(\frac{1}{n}\sum_{i=1} ^{n} f(a_i))
1. Show that a_1\lt M_f\lt a_n.
2. If h(x)=af(x)+b, show that M_h = M_f.
Homework Equations
As above, M_f=g(\frac{1}{n}\sum_{i=1} ^{n} f(a_i))
The Attempt at a Solution
Solution 1.
Since f is a strictly monotonic function, we must have have either
f(a_1) \lt f(a_2) \lt ... \lt f(a_n) \text{ or,}f(a_1) \gt f(a_2) \gt ... \gt f(a_n)
It follows that nf(a_1) \lt \sum_{i=1}^{n} f(a_i) \lt nf(a_n) \text{ or,}nf(a_1) \gt \sum_{i=1}^{n} f(a_i) \gt nf(a_n) for each set of inequalities, respectively. Further, these inequalities imply f(a_1) \lt \frac{1}{n} \sum_{i=1}^{n} f(a_i) \lt f(a_n) \text{ or,}f(a_1) \gt \frac{1}{n}\sum_{i=1}^{n} f(a_i) \gt f(a_n) Now we simply consider each case. First, assume that f strictly increases. Then g also strictly increases, and we may apply g to the first set of inequalities above to derive: a_1 \lt M_f \lt a_n
Now assume f strictly decreases. Then g strictly decreases as well. If we apply g to the second set of inequalities above, the inequality signs must reverse since the function is decreasing. Therefore, we derive: g[f(a_1)] \lt g[\frac{1}{n}\sum_{i=1}^{n} f(a_i)] \lt g[f(a_n)]a_1 \lt M_f \lt a_n
This completes the proof of problem 1.
Solution 2.
If h(x) = af(x) + b, then f(x) = \frac{h(x) - b}{a}. We may apply g to both sides to get x = g(\frac{h(x)-b}{a}) Let s(y)=x be the inverse function of h(x), such that y = h(x). Through this we attain: s(y) = g(\frac{y - b}{a}) Now we may consider M_h. M_h = s(\frac{1}{n}\sum_{i=1}^{n} h(a_i))M_h = g(\frac{\frac{1}{n}\sum_{i=1}^{n} h(a_i) - b}{a}) We may write b = \frac{1}{n}\sum_{i=1}^{n}b, which allows us to work with both terms together in the numerator. Combine the denominator into the sum to make the proof clearer. Through this, we may derive: M_h = g(\frac{1}{n}[\sum_{i=1}^{n}\frac{ h(a_i) - b}{a}]) Since f(x) = \frac{h(x) - b}{a} for each x = a_1, x = a_2, ..., x=a_n, the sum reduces to M_h = g(\frac{1}{n}\sum_{i=1}^{n} f(a_i) = M_f This is true by definition, and thus completes the proof.