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October 13th, 2017, 06:39 PM   #1
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Question Multivariable (maybe) Calculus Help

Hi all - I'm looking for help on how to write an equation. My somewhat blurry memory of high school and college math tells me that this is a calculus problem, but I'm not certain.

In (hopefully) simple english, I'm solving for a value equal to the sum of four functions where the variable is a number between one and five. Here's the function with x:

f(x) = 5^(x-1)

In this case x represents a result that can be the same or different for each instance of the function. For instance, x could equal 1 twice, 3 once and 4 once.

The following is as close as I've gotten to representing the entire equation:

x = 5^(a-1) + 5^(b-1) + 5^(c-1) + 5^(d-1), when a, b, c & d are all integers between 1 and 5.

Is there a way to write this elegantly? I'm using the equation to express the level of risk across four categories where each category gets a score of 1-5, but structured such that the risk value for four scores of 1 (a score of 1 in all four categories) will never exceed a score of 2 in a single category, and so on.

Let me know if I can add context or clarification.

Thank you!
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October 14th, 2017, 04:24 PM   #2
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The "and so on" does not really give much guidance on understanding the entire problem.

With respect to what you have said, what for example is to happen if you have have a and b equal 1 and c and d equal 3, giving a total of 8, versus all four equal to 2, again giving a total of 8. Do those two possibities generate equal results. If not, which is higher and by how much?

EDIT: As I understand you so far, you have 256 possibilities. Do you know which numerical result SHOULD BE associated with each possibility?
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Last edited by JeffM1; October 14th, 2017 at 04:33 PM.
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October 14th, 2017, 10:21 PM   #3
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Jeff - Thanks for taking a look at this. I'll address your questions in inverse order.

1. This equation is going to get run multiple times for different types of scenarios, each with up to four distinct types of risk, with the impact of each risk falling somewhere between 1 (low) and 5 (high). The numerical result is relatively unimportant as long as this primary rule is always true: a scenario with four risks of a lower level (on the 1-5 scale) should never come out greater than a scenario with a single risk of a higher level. I'm applying the exponent to "5" because this seemed like the simplest way to ensure this.

2.
Quote:
if you have have a and b equal 1 and c and d equal 3, giving a total of 8, versus all four equal to 2, again giving a total of 8. Do those two possibities generate equal results. If not, which is higher and by how much?
If one variable is equal to three, the result for that scenario is going to be at least 25, based on the f(x) = 5^(x-1) equation, yes? Even if a, b, c + d are two in the other scenario, the result will be capped at 20. This is what I was trying to explain above.

3. Does it make more sense now? I want to show that the result is equal the sum of the same equation executed four times, either as four separate variables (all 1-5) or the same variable, as long as it can change each time. I wasn't sure if the {set} symbol might work, or if I could use the [sum] symbol, or if there's something else I just don't know about.
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October 15th, 2017, 04:14 AM   #4
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If I understand what you want, what would be very simple to code and to express is

$g = max(a,\ b,\ c,\ d) \text { and } f(a,\ b,\ c,\ d) = 5^{(g - 1)}.$

The possible scores would be 1, 5, 25, and 125.
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October 15th, 2017, 05:04 AM   #5
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When you say "the variable is a number between 1 and 5", do you mean an integer between 1 and 5, inclusive?
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October 15th, 2017, 06:04 AM   #6
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A somewhat more sophisticated scheme would be:

$w_1 = a - 1,\ w_2 = b - 1,\ w_3 = c - 1,\ \text { and } w_4 = d - 1.$

$\displaystyle \text {Score } = \sum_{j=1}^4 4^{w_j}.$

The lowest possible score is 4. The highest is 1024.

I suggest that you work out the possible values to see whether you like the scoring that results
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October 16th, 2017, 09:44 AM   #7
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Jeff - Brilliant. Thank you so much.
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