Advanced Math/Generating


QUESTION: Hello Scott

I have been given a standard formula which generates a "S" curve. The formula is;
Please note the x stands for the letter x not the multiplication symbol.
The formula is used for showing the cumulative spend on a project over time.
The total cost of the project is always fixed.
The time period of the project is always fixed.
The formula generates "Y" the first payment when "x"=1/12 if 1 is the first payment month of a 12 month project. You repeat the formula changing "x" to 2/12, 3/12 4/12 etc... to generate (in this case)12 "y"s or 12 cumulative payments which when plotted on a graph give a "s" curve.
The formula comes with a set of data for the values of S, C & K -here it is;
C   K   Project cost
-0.439   5.464   10,000
-0.37   4.880   50,000
-0.295   4.360   100,000
-0.22   3.941   200,000
-0.145   3.595   500,000
0.01   4.000   2,000,000
0.11   3.980   3,500,000
0.159   3.780   5,500,000
0.056   3.323   7,500,000
-0.028   3.090   9,000,000

So if the total cost of the project is 2,000,000 then that's the value of "S" and you use the corresponding values of 0.01 for "C" and 4.000 for "K"

My question is....

How do I tweak the formula or how do I tweak the data table to produce a slightly steeper "S" curve shape or a slightly gentler "S" curve shape ?

Thanking you in advance


ANSWER: The symbols and , on my end, are '' for squared and '' for cubed.

It looks like the formula should really be written with parentheses in
the denominator around the [K(6x-9x+3x)], so it would be
Y=S(x+Cx-Cx-1/([K(6x-9x+3x)]), for it looks like (6x-9x+3x)
is suppose to be in the denominator.

If not, the way to write it would be Y=S(x+Cx2-Cx-(6x-9x+3x)/K),
and in that case, the terms could be combined.

Once the correction has been made, it can then be rewritten as
Y = S(x + Cx2 - Cx - 1/([K(6x - 9x + 3x)]) - is that correct?

Assuming it is, let us make another function, W(X),
such that W(x) = K(6x - 9x + 3x).
From this, what we have is Y = S[x + Cx2 - Cx - 1/W(x)].

If so, the next thing to do is use calculus to take the derivative,
for the derivative of an equation is the slope.

This give Y' = S[1 + 2Cx - C + W'(x)/W(x)].

Given that W(x) = K(6x - 9x + 3x), it can be seen that W'(x) = K(18x - 18x + 3).

Putting this back in our formula for Y' gives
Y' =  S[1 + 2Cx - C + {K(18x - 18x + 3)}/{K(6x - 9x + 3x)}].

Now that we have that, next note that K cancels since it occurs once in the numerator and once in the denominator of that fraction.  That tells us that changing K does nothing to the slope.

Doing so gives us Y' =  S[1 + 2Cx - C + (18x - 18x + 3)/(6x - 9x + 3x)].

Looking at this formula, the only way to change the value of Y' (which is the slope)
is to change C.  Since x is in money, it is never negative.  If the formula were for
the amount of profit, it might be, but I believe this formula is for the amount of cash.

To see how C changes the slope, isolate the terms with C in them.

That is, we can rewrite the derivative as
Y' =  S[1 + T(C) + (18x - 18x + 3)/(6x - 9x + 3x)].
where T(C) = 2Cx - C.

From here, we can see that the slope of the equation is affected by the change of C alone.
The affect of changing C gives a linear result.  This means that the if we change C by twice as much, the value if changed by twice as much.

Note that if x were ever negative, it would have the opposite affect on the slope.

---------- FOLLOW-UP ----------

QUESTION: Your symbol assumptions are correct.

In regard to placing the denominator. The formula can be expressed as follows;


Does you answer still hold true ?



Since the real Y has stepped forward as Y=S{x+Cx2-Cx-[1/K](6x3-9x2+3x)},
the outcome changes.

I will write it as Y = S[x + Cx - Cx - (6x - 9x + 3x)/K].

The derivative is them Y'(x) = S(1 + 2Cx - C - (18x - 18x + 3)/K).

That can be rewritten as Y'(x) = S([1 - C - 3/K] + [2C + 18/K]x - [18/K]x).

The values of x are assumed to be positive.
If x is small (far less than 1), the term to look at is Sx - 3x/K.
If x is big (far greater than 1), the term to look at is -(6/K)x.

This will determine a rough idea of the slope of the curve at that point.

If the value of the derivative is positive, the function is increasing.

If the value is the derivative is negative, the function is decreasing.

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Scott A Wilson


I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?


Experience in the area; I have tutored people in the above areas of mathematics for over two years in I have tutored people here and there in mathematics since before I received a BS degree back in 1984. In just two more years, I received an MS degree as well, but more on that later. I tutored at OSU in the math center for all six years I was there. Most students offering assistance were juniors, seniors, or graduate students. I was allowed to tutor as a freshman. I tutored at Mathnasium for well over a year. I worked at The Boeing Company for over 5 years. I received an MS degreee in Mathematics from Oregon State Univeristy. The classes I took were over 100 hours of upper division credits in mathematical courses such as calculus, statistics, probabilty, linear algrebra, powers, linear regression, matrices, and more. I graduated with honors in both my BS and MS degrees. Past/Present Clients: College Students at Oregon State University, various math people since college, over 7,500 people on the PC from the US and rest the world.

My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.

Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

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I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.

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My clients have been students at OSU, people who live nearby, friends with math questions, and several people every day on the PC. I would guess that you are probably going to be one more.

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