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November 27th, 2015, 12:19 AM   #1
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Question Need a formula to match equation set. Math already done, only syntax issue

Need a formula to match equation set. Math already done, only syntax issue.

What am I trying to figure out in order to solve a certain problem is the best formula to describe a specific system:

1. ) This system works only by integers of the input x, starting with x = 1, and can only go up one integer at a time, it cannot go down to 0 or to negative integers (x=2, x=3, x=4, etc). The output, however, f(x), can be any positive decimal number as the tally grows and is not restricted to integers.

2. ) This system attempts to take x and keep a running tally of a series of values which changes by factors of 5, meaning that from x = 1 to x = 5 the system counts this tally by a specific linear function, but then from x = 6 to x = 10 it counts it by a slightly different linear function, and so on, every 5 values of x, determined by the algorithm of f(x).

3. ) At 1, f(x) = 5, so the system starts with x = 1, f(x) = 5.

4. ) Every integer added after that up to x = 5 increments f(x) by 1 more, such that:

f(1) = 5
f(2) = 6
f(3) = 7
f(4) = 8
f(5) = 9

4. ) After x = 5, however, starting with x = 6, the tally grows by only 2/3 (0.6666) for each increment of x, up until x = 10, such that:

f(6) = 9.6666
f(7) = 10.3333
f(8 ) = 11.0
f(9) = 11.6666
f(10) = 12.3333

Basically the point of the system is to decrease the rate at which the tally rises as x grows.

5. ) After x = 10, however, starting with x = 11, the tally grows by only 2/4 (0.5) for each increment of x, up until x = 15, such that:

f(11) = 12.8333
f(12) = 13.3333
f(13) = 13.8333
f(14) = 14.3333
f(15) = 14.8333

As you can see, every time x grows by a factor of 5, the tally f(x) grows only by a fraction 2/b for each subsequent integer value of x, where b increases by one after every factor of 5.

Here is the equation I came up with to describe this behavior (note the floor function in special "L" shaped brackets for certain parts):

(4+((x-5floor[((x-1)/5)])((2/(floor[((x-1)/5)]+2)))))

( I know that's not pretty or easy to read, I tried my best to import ready-made images, but this forum tech seems pretty primitive, it wouldn't let me. Maybe try to plug that in to a math program to see what the formula actually looks like (such as wolframalpha.com). )

The interesting problem here is that this seems to get the right values for each factor of 5 worth of x values, but it doesn't describe the whole system, since the formula must be done at every factor of 5 again, while including the tally from the very last integer belonging to the preceding "set". For example, using this formula anywhere in the range of 1-5, gets the right value for any x value between 1 and 5. But then I can't just use this formula to get the correct answer when x = 6 since then the formula essentially counts every integer value of x being processed the same exact way from x = 1 all the way to x = 6, but we know this can't be true from the description of the system above.

Instead, the formula must be applied to x = 5, that number stored, then the formula reapplied to x = 10, that number added to the previous number to create a growing tally, then the formula reapplied to x = 15, that number added to the tally, and so forth, ad-infinitum, in order to gain the correct tally.

In that case, a better description of this system might look like this:

For f(5), ( f(x) = (4 + ((x-5floor[((x-1)/5)])((2/(floor[((x-1)/5)]+2))))) ) = a

For f(10), ( f(x) = a + (4 + ((x-5floor[((x-1)/5)])((2/(floor[((x-1)/5)]+2))))) ) = b

For f(15), ( f(x) = b + (4 + ((x-5floor[((x-1)/5)])((2/(floor[((x-1)/5)]+2))))) ) = c

ad-infinitum...

The above sequence comes up with the right results, but it is not neat, nor correctly written out.

Maybe now you can appreciate my problem? I understand the fundamental math going on to drive this system, but my actual math skills and knowledge are still insufficient to be able to correctly annotate the above behavior of this system into the proper formula which I know is capable of encapsulating all of that much more neatly, and much more correctly according to formal mathematical notation and symbolism. Also, my level of math might be too low, such that I understand the logic, but i just simply don't know how to express it properly. Is this an infinite series problem? Also, if one were to plot all of these points on an (x,y) graph, is there a curve of some type that more smoothly incorporates the same behavior but using another formula entirely (whose naturally curving line hits very close to most of the points on my system, and which therefore might be much simpler and superior to this "system")?

Thank you for your help, any assistance or attempt to shed further light on this problem is welcome.
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November 28th, 2015, 11:54 PM   #2
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I did make a mistake in my equations I just saw...

I added the 4 the first time, which is correct, but then all the subsequent uses of that formula do NOT add 4, just the first time, if done that way as I described in the procedure toward the end there. I was lazy and didn't double check myself until now.

But otherwise the formula seems to work, if performed iteratively as I described. Even so, despite that minor correction, I am still in need of the same exact help as before...

Nobody can help? Thanks.

Last edited by skipjack; November 29th, 2015 at 04:03 AM.
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November 29th, 2015, 04:28 AM   #3
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You can produce a nicely formatted formula here.

You type, for example, \$\left\lfloor\dfrac{x-1}{5}\right\rfloor\$, and the result is $\left\lfloor\dfrac{x-1}{5}\right\rfloor$. You could also simplify your working by temporarily using a letter of the alphabet to represent this particular expression.

I suggest you devise a formula that works for x = 5 or 10 or 15, etc., then spot how to modify it so that it works for 1 or 6 or 11, etc., as well as the original values of x.
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December 2nd, 2015, 09:23 PM   #4
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Good tips, btw could you point me to the page (legend) on this site that reveals all the math-imaging code and syntax? That way I'm sure I could hereafter be able to image any math formula I wanted to, thanks

Regarding the problem, though, I thought about it a bit recently and decided I'm going at it from a different approach...

I thought about where the points of the above clunky system might roughly lie on a graph, and the function (squareroot x) came to mind. I then fooled around on wolfram for a couple minutes and came up with:

floor((squareroot(2x-2))+5))

Honestly, this formula works even better than the system above, and gives me a very nice continually decelerating and practical return value as x grows, works very well for that specific need/problem.

I'm glad there's such a lively math community here though! I will have many more math problems to come as the needs arise, and even as my own math skills get better I will still undoubtedly come back here for advice, this site is a valuable resource, thanks

Last edited by CuriousSphere; December 2nd, 2015 at 09:34 PM.
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