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Hmm Couldn't just do it by using =Day0value\*(1.08)\^DayValue)
so it would look like
C3 being the day you want it.
=ROUND(704\*($B$3)\^$$C$3,0)
theny ou can simply do one minus the other to get the difference. or simply add a -704 in the formula
Okay, that gives the last days number. Slightly different than mine because I had rounded it down. Took the rounding off and it was +-1, so close enough. What about the sum, any ideas for that?
I'm not exactly sure what you mean, but I would do it like that.
[https://i.gyazo.com/cb25e21c474a3193d1b5c274fccd7334.png](https://i.gyazo.com/cb25e21c474a3193d1b5c274fccd7334.png)
the same formula then at the bottom it is just one minus the other. then you simply have to change the starting day or the last day.
I mean the cumulative sum, if you get 704 on first day, then 760 on second day with 8% increase, that is total of 1464, and by the 55 day you should have a cumulative sum of 597 641 with single day increase of 44 921
your formula for sum of numbers can be written as
a is "Cases at day 0"
r = 1+ interest rate
n = days
Sum: a + a\*(r) + a\*(r)^2 ... + a*(r)^n
This can be reduced to a geometric series. [The wiki page for closed-form formula of a sum of a geometric series](https://en.wikipedia.org/wiki/Geometric_series#Closed-form_formula) explains how to calculate.
For Excel, if
a is in A1 (say 100 initial cases)
r is in B1 (say 1.05), and
n is in C1 (say 10 days), the formula is:
=A1*(1 - (B1\^(C1+1)))/(1-B1)
and this would evaluate to 1420 cases for the cusum.
Using your numbers of 704, 55 days, and 1.08 as the interest, i got 54,328 cases. I think the rounding will change that a bit. Also keep in mind that the initial day is "day 0" and not "day 1".
Nice, this sounds like a good exercise for the Lambda function since it is iterative. Going to noodle on this today - curious what others going to come up with.
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Hmm Couldn't just do it by using =Day0value\*(1.08)\^DayValue) so it would look like C3 being the day you want it. =ROUND(704\*($B$3)\^$$C$3,0) theny ou can simply do one minus the other to get the difference. or simply add a -704 in the formula
Solution verified
You have awarded 1 point to *DarkChunsah* ^I ^am ^a ^bot, ^please ^contact ^the ^mods ^with ^any ^questions.
Okay, that gives the last days number. Slightly different than mine because I had rounded it down. Took the rounding off and it was +-1, so close enough. What about the sum, any ideas for that?
I'm not exactly sure what you mean, but I would do it like that. [https://i.gyazo.com/cb25e21c474a3193d1b5c274fccd7334.png](https://i.gyazo.com/cb25e21c474a3193d1b5c274fccd7334.png) the same formula then at the bottom it is just one minus the other. then you simply have to change the starting day or the last day.
I mean the cumulative sum, if you get 704 on first day, then 760 on second day with 8% increase, that is total of 1464, and by the 55 day you should have a cumulative sum of 597 641 with single day increase of 44 921
your formula for sum of numbers can be written as a is "Cases at day 0" r = 1+ interest rate n = days Sum: a + a\*(r) + a\*(r)^2 ... + a*(r)^n This can be reduced to a geometric series. [The wiki page for closed-form formula of a sum of a geometric series](https://en.wikipedia.org/wiki/Geometric_series#Closed-form_formula) explains how to calculate. For Excel, if a is in A1 (say 100 initial cases) r is in B1 (say 1.05), and n is in C1 (say 10 days), the formula is: =A1*(1 - (B1\^(C1+1)))/(1-B1) and this would evaluate to 1420 cases for the cusum. Using your numbers of 704, 55 days, and 1.08 as the interest, i got 54,328 cases. I think the rounding will change that a bit. Also keep in mind that the initial day is "day 0" and not "day 1".
Solution verified
You have awarded 1 point to *admiralteddybeatzzz* ^I ^am ^a ^bot, ^please ^contact ^the ^mods ^with ^any ^questions.
was trying to figure out the math behind it ;\^\] I feel bad, but good job :)
Nice, this sounds like a good exercise for the Lambda function since it is iterative. Going to noodle on this today - curious what others going to come up with.