I am guessing this goes here

[RPGer]

I apologize if this is not the correct section but it has been awhile since I have been in a class.

(X x Y)Days repeated + Current number of Days = Days
X and Y are variables but I am using an example below, and these would not change from day to day. So for the example I am using 10 units of X and 2 units of Y every day.

(10 x 2)Day 1 + Current total of days = since this would be the first day so 20
(10 x 2)Day 2 + 20 = 60
(10 x 2)Day 3 + 60 = 120
(10 x 2)Day 4 + 120 = 200
(10 x 2)Day 5 + 200 = 300

My question is how would reverse the process. For instance, how would I figure out how many days would it take me to get 15000 Days? I am hoping there is a formula.

Please let me know if I need to clarify anything up. I am feeling scrambled just proofreading this.

 

[Cubist]

it has been awhile since I have been in a class

I am feeling scrambled just proofreading this.

I think I've understood this conundrum! It was difficult with all the "day" words. It started to make more sense to me when I thought of using the word "events", and these "events" then have a number of days between them.

So, I'm pretty sure that we have an "arithmetic progression" here. That is, the number of days between events increases by the same amount event after event. There is an initial gap of \(a_1=x \times y\), and the so-called "common difference" is \(d=x \times y\)

The sum of the above progression is...

[math]s=\frac{x \times y \times(n+n^2)}{2} [/math]
where s is the total days, and n is the event number. When we use n=1,2,3,4,5,6,7,etc in this equation will give your sequence s=20, 60, 120, 200, 300, 420, 560...

So all that needs to be done is to make "n" the subject, ie write the above as n=<something in terms of the s>. Is this something that you could do? It's a little bit tricky. I'll be offline for a few hours now, but if you have a go then other helpers will probably be willing to help you out.

 

[RPGer]

If I understand you correctly (admittedly this is a struggle for me), this may not work as "sequence s=20, 60, 120, 200, 300, 420, 560... " will change as x and y can change. The only constant is "n=1,2,3,4,5,6,7". It may make more sense to me if I can see the formula with numbers from my example, if possible.

 

[Deleted member 4993]

I apologize if this is not the correct section but it has been awhile since I have been in a class.

(X x Y)Days repeated + Current number of Days = Days
X and Y are variables but I am using an example below, and these would not change from day to day. So for the example I am using 10 units of X and 2 units of Y every day.

(10 x 2)Day 1 + Current total of days = since this would be the first day so 20
(10 x 2)Day 2 + 20 = 60
(10 x 2)Day 3 + 60 = 120
(10 x 2)Day 4 + 120 = 200
(10 x 2)Day 5 + 200 = 300

My question is how would reverse the process. For instance, how would I figure out how many days would it take me to get 15000 Days? I am hoping there is a formula.

Please let me know if I need to clarify anything up. I am feeling scrambled just proofreading this.

Is this problem assigned to you from a class? Can you post a picture of the ACTUAL problem? What is the subject matter and the context of this problem?

 

[Cubist]

It may make more sense to me if I can see the formula with numbers from my example, if possible.

Consider this line from your example:-

(10 x 2)Day 4 + 120 = 200

Here x=10, y=2, n=4. therefore

[math]s=\frac{x \times y \times(n+n^2)}{2} [/math]
[math]=\frac{10 \times 2 \times(4+4^2)}{2} [/math]
[math]=\frac{20 \times(4+16)}{2} [/math]
[math]=\frac{20 \times(20)}{2}[/math]
[math]= 200[/math]
this is the same result that you obtained, but the big advantage of this equation is that you don't need to know the previous (n=3) result.

Is this problem assigned to you from a class? Can you post a picture of the ACTUAL problem? What is the subject matter and the context of this problem?

I too am curious!

 

[RPGer]

I was really hoping to avoid showing what I am working... but here ya go:

This is part of a rough draft for a magic system I intend to use in the [URL deleted]

To preface a bit as this may be unfamiliar territory.

A character's fatigue points are determined by their strength stat. So a character with a 12 Str, has 12 fatigue points to spend on casting spells.

This is the stuff I am making up:
The magic user casts their spells by manipulating Spheres of magic:

Water
Air
Earth
Fire
Light
Life

Up to 4 Spheres in each Sphere type.

Here is what I am trying to get your help with:

"
Enchanting

To hold a spell for days, years, decades, etc. the mage will need a ritual. This is simply infusing the spell with fatigue repeatedly, until the desired lifespan of the spell is reached.

During the ritual, the mage can rest to replenish the lost energy points but cannot use any Life Spheres. The Life Sphere(s) is devoted to the ritual.

The Ritual

(Energy x Life Spheres)Days Repeated + Current number of Days from last day’s total = Days enchanted.

The ritual will take one hour per two fatigue. The mage must rest, eat, and drink a full night between days, or suffer a -1 fatigue for the next day. The -1 is accumulative and the modifier remains until the mage eats and drinks well, and sleeps a full night.

The mage may use any other Spheres except Life. If he or she does, then the Enchanting ritual ends with the total number of days from the last completed day, added to the spell's duration. The ritual is also ended, if a day is skipped. The mage may then continue to add more days by starting the ritual again from day 1.

The ritual may be restarted as often as the mage wants, meaning he or she could do the ritual for a week, every other week.

Enchanting a magical lock for a mage with a Mag Apt +2 and a strength of 12 (12 fatigue), would look something like this:

If your fatigue drops to one, the character falls unconscious, so we are only going to spend 10 fatigue points.

On the first day:
(10 fatigue x 2 Life Spheres)day 1 + (no previous total)
(20)1+0 = 20

Day 2
(10 fatigue x 2 Life Spheres)day 2 + Yesterday’s total 20
(20)2 + 20 = 60

Day 3
(10 fatigue x 2 Life Spheres)day 3 + Yesterday’s total 60
(20)3 + 60 = 120

Day 4
(10 fatigue x 2 Life Spheres)day 4 + Yesterday’s total 120
(20)4 + 120 = 200

Day 5
(10 fatigue x 2 Life Spheres)day 5 + Yesterday’s total 200
(20)5 + 200 = 300

The magic lock spell will now last 300 days

Instead of spending game time to work out how many days the ritual needs to be done, to reach the desired days, use the formula below." <-- this is the formula I am hoping to get help with.

Nothing as extreme or serious as for a class or job, just a bit of fun. I apologize if this is an inappropriate use for this website.

 

[Cubist]

I was really hoping to avoid showing what I am working... but here ya go:

I personally didn't need that much detail, but thanks for the explanation!

I apologize if this is an inappropriate use for this website.

I don't mind although some helpers might prefer to only help students therefore it's good to let us know.

Ideally we want to teach a poster how they can solve a problem themselves. Then they will hopefully gain some maths knowledge and perhaps find it interesting! It's often hard to achieve this if we don't know about the full problem (and sometimes a bit of background about the person asking).

So back to the problem, did you understand post #5 (and if so, then did post #2 make more sense)?

 

[RPGer]

Sorry no. If I am using the below formula, then I would have to know what n is to solve it, correct? n is what I am trying to find. I am trying to figure out given that I am using Fatigue =10 and Spheres = 2, how many days would it take me to get to an enchantment length of 1500 days. I would also need to be able change how much fatigue and how many spheres used per spell, and I would decide how long I want the enchant to last. Feed all that into the formula and it would give me the number of days to repeat the ritual. Is this possible?

s=x×y×(n+n2)
2

I am trying to figure out given that I am using Fatigue =10 and Spheres = 2, how many days would it take me to get to an enchantment length of 1500 days. I would also need to be able change how much fatigue and how many spheres used per spell, and I would need to decide how long I want the enchant to last. Feed all that into the formula and it would give me the number of days to repeat the ritual. Is this possible?

Amount of fatigue, number of spheres, and how many days the enchant will last is what I would know, but they could also change each time.

I feel like I may be getting close, otherwise I would leave the thread so I am not wasting anyone's time. I dont want to start over if I do not have to.

 

[Cubist]

Sorry no. If I am using the below formula, then I would have to know what n is to solve it, correct? n is what I am trying to find.

Precisely. I fully understand that "n" is your final goal. The formula that I wrote in step 2 was a stepping stone towards achieving this...

So all that needs to be done [now] is to make "n" the subject, ie write the above as n=<something in terms of the s>. Is this something that you could do?

A further hint is:- doing this involves using the quadratic formula.

--

I'm actually going to give you the answer below as a "spoiler" because I'm picking up that you may not be interested in the details of how it's done. If I'm wrong then simply don't click on the spoiler, and hopefully I can guide you towards solving it yourself.

Spoiler: final answer

[math] n= \frac{\sqrt{1+8s \mathbin{/} (xy)}-1}{2} [/math]
Plugging in x=10, y=2, s=15000 gives n≈38.233

 

[RPGer]

Awesome! I read over your original post and it seemed to snap into place. I really appreciate it.