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Math Help - How to express / simplify f(x) = x + (x-1) + (x-2) ... (x - x)

  1. #1
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    How to express / simplify f(x) = x + (x-1) + (x-2) ... (x - x)

    I've forgotten my math...


    Is there an easier way to express the function in my title? Here's the context:

    In Simcity you can buy a coal plant for $10000. Every year you own it it costs $1 per month * number of years owned * 12 months to maintain. If I've owned the plant for 20 years, it costed me $240 that year, $228 the previous year, $216 the year before that, etc. If I buy a new plant, the maintenance costs go back down to $12 for the year again. So what's the most affordable way to keep using coal plants?

    I was thinking, let x = number of years between buying a new plant, and then I could come up with a function that tells me what the cost would be for every x years. It would be something like 10,000 + 12x + 12(x-1) + 12(x-2) and so on, but that's not so easy to evaluate for larger numbers.

    So is there a way to simplify that expression? (Or should I just go wind power)?

    Thanks in advance!
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  2. #2
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    You can use sigma notation to simplify the title function.
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  3. #3
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    I guess what I meant was that it would be tedious to evaluate by hand (or brain) for large numbers. I was hoping that was some easier way to express it using only a few terms.

    If there isn't, I'll accept that.
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  4. #4
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    Quote Originally Posted by supaju View Post
    I've forgotten my math...


    Is there an easier way to express the function in my title? Here's the context:

    In Simcity you can buy a coal plant for $10000. Every year you own it it costs $1 per month * number of years owned * 12 months to maintain. If I've owned the plant for 20 years, it costed me $240 that year, $228 the previous year, $216 the year before that, etc. If I buy a new plant, the maintenance costs go back down to $12 for the year again. So what's the most affordable way to keep using coal plants?

    I was thinking, let x = number of years between buying a new plant, and then I could come up with a function that tells me what the cost would be for every x years. It would be something like 10,000 + 12x + 12(x-1) + 12(x-2) and so on, but that's not so easy to evaluate for larger numbers.

    So is there a way to simplify that expression? (Or should I just go wind power)?

    Thanks in advance!
    1. As far as I understand your question you are going to calculate a sum with n+1 summands where n summands are terms in brackets. Right?

    2. If so I'm going to modify your notation a little bit:

    s(x) = x+(x-1)+(x-2)+(x-3)+(x-5)+(x-6)+...+(x-n)

    s(x)=\underbrace{x+x+x+x+...+x}_{\text{(n+1) summands}} - (\underbrace{1+2+3+4+5+6+...+n}_{\text{(n) summands}})

    3. The sum of the first n natural numbers is calculated by:

    1+2+3+...+n=\sum_{i=1}^n(i)=\dfrac n2(n+1)

    4. Thus your sum becomes:

    s(x)=(n+1) x - \dfrac n2(n+1)
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  5. #5
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    Yes!

    Yes, that's the one! Sum of natural numbers...

    Thank you earboth!
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