Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a series of numbers, say 2, 3, 7 that sum 12. I want to increase these numbers so that they sum 62. I could multiply each number by 62/12 to achieve that. However, all numbers would be scaled equally linearly. I want instead that 7 be scaled more than 2- i.e. bigger numbers to be scaled more.

One way to achieve that would be to square each number. $2^2+3^2+7^2=4+9+49=62$

In general, given $a_i$ and an $n$ I want to find a $k$ so that:

$$\sum a_i^k=n$$

I haven't been able to isolate the $k$. Can it be done or should I resort to finding it numerically (which sounds easy). I've tried using exponentiation...

$$e^{\sum a_i^k}=e^n$$

$$\prod e^{a_i^k} = e^n$$

But I can't seem to find any useful manipulation with $a^{b^c}$.

Thoughts?

share|improve this question
    
The bad news: I don't see how you will get an algebraic solution. The good news: because exponentiation is such a powerful operation, you can get a good starting value by just considering the largest $a_i$ when the others are rather smaller. In your example, the exponent would be $\frac{\ln 62}{\ln 7}\approx 2.12$, not far from (and obviously larger than) the answer $2$. So a root finder will work easily-your function is monotonic, has a single root, and you have the root bracketed. –  Ross Millikan Oct 8 '11 at 3:02
    
That's what I thought. If you post as an answer and no one finds another answer, I'll have to accept "there's no answer" as the valid answer. Also, if someone finds an alternative scaling method, that'd be cool too... –  alex Oct 8 '11 at 10:23

1 Answer 1

up vote 1 down vote accepted

The bad news: I don't see how you will get an algebraic solution. The good news: because exponentiation is such a powerful operation, you can get a good starting value by just considering the largest $a_i$ when the others are rather smaller. In your example, the exponent would be $\frac{\ln62}{\ln7}\approx 2.12$, not far from (and obviously larger than) the answer $2$. So a root finder will work easily-your function is monotonic, has a single root, and you have the root bracketed.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.