# O Level maths needs Degree maths Knowledge

• May 18th 2012, 01:42 PM
SMQ
O Level maths needs Degree maths Knowledge
Hi everone, I'm sorry to invade your e-space as I am nowhere near the intellectual level of you guys when it comes to maths!

However I do have a question and I thought as u guys are the Canines Testicles at this sort of thing, one of u would know the answer and be able to communicate it in a way I can grasp (hopefully) but if that is not possible maybe u can provide the formula which returns the value I am querying...any help would be truly appreciated.

Heres the info,

I have a period of time - 1 to 36 Months.
In each of those Months (1 to 36) I have a single value which represents the loss in value due to depreciation over the period.
This is represented by incresing currency values (£). For Example

Month 1= 500
Month 2 = 750
Month 3 = 800....etc etc

A cummulative monthly curve over 36 months based on the overall distribution of these values adding up to 100% produces a factor of 0.6. If you adjust this value it adjusts the curve.

My problem is that I do not know why any changes are represented by a decimal number - i.e. what does 0.6 mean when applied to a curve? is it the deviation from 1? and if so does 1 represent a true average?

I can provide an excel version of what I mean if it helps

As I said O Levle maths is my limit so if any of this seems idiotic I apologise and bow to your superior knowledge but could really use the help anyhow...

Cheers

Steve
• May 18th 2012, 02:06 PM
emakarov
Re: O Level maths needs Degree maths Knowledge
Quote:

Originally Posted by SMQ
A cummulative monthly curve over 36 months based on the overall distribution of these values adding up to 100% produces a factor of 0.6.

So, you are not sure in what sense the function "produces" the value 0.6 and you would like to find this out? Yes, maybe the complete table would help...
• May 19th 2012, 02:42 AM
SMQ
Re: O Level maths needs Degree maths Knowledge
Thanks for replying so quickly the picture file is attached.

Cheers

SMQAttachment 23903
• May 19th 2012, 11:17 AM
emakarov
Re: O Level maths needs Degree maths Knowledge
I am not sure how 0.6 is related to this function, but it seems that the difference between consecutive values in the third column are approximately equal. This means the third column is a linear function. Since the values in the third column are themselves differences of consecutive values in the second column, this means that the derivative of the function in the second column is linear and the function itself is a quadratic polynomial. If you'd like, I can explain this using arithmetic progressions instead of derivatives.

Let's try to find the function f(x) in the second column assuming it has the form $ax^2 + bx + c$ for some constants a, b and c. I chose to substitute x = 0, x = 18 and x = 36, which gives the equations

c = 0
$a\cdot18^2+b\cdot18+c = 35.04$
$a\cdot36^2+b\cdot36+c = 100$

WolframAlpha says the solution is $a\approx0.0461728$ and $b\approx1.11556$. Indeed, $f(x) = 0.0461728x^2 + 1.11556x$ seems to approximate the second column pretty well.
• May 20th 2012, 12:58 AM
SMQ
Re: O Level maths needs Degree maths Knowledge
Hi thanks again for your response but I'm not sure I have explained it properly in the first place let me try again and if the answers the same thats fine.

Scenario
I have £36,000 that needs to be distributed (or "earned") over 36 months. When you apply a factor of 0.6 it "earns" or distributes the £36,000 over the 36 months in the %'s shown in column 3. If you apply a cummulative sum of these values i.e. Month 1 + Month 1 = Value 1, Month 1 + Month 2 = Value 2, Month 1 + Month 2 + Month 3 = Value 3 etc you get the values in column 2.

When you plot these (column 2") vales on a graph over the 36 months you get the curve. If you adjust the factor of 0.6 to lets say 0.25 the values in each month will change and therefore the curve will change. so what I need to know is how does the factor of 0.6 (or whatever other factor I use) when used against £36,000 over 36 Months produce the distribution % values for each individual month in Column 3?

In simpe terms if the factor is 0.0 the distribution will be £36,0000 / 36 Months giving £1000 in each of the months or 2.77% in each month producing a straight line when plotted on a graph if you adjust the factor the distribution changes - if you adjust the factor above zero it produces an inverse curve like the one shown. On the other hand if you were to apply a negative factor of -0.6 it would produce a convex curve in the mirror image of the positive 0.6 factor curve.

I have a spreadsheet where all I need to do is input the values for the amount of money (36,000 in this case), the number of months over which it is to be distributed (36) and the Factor (0.6). The spreadsheet then distributes the money over the 36 months. The problem is that I don't have access to the formula which is being used. Although I can physically see what effect changing the factor has on this distribution I do not know how the interaction is working.

Sorry to be a pain!.....

Steve
• May 20th 2012, 07:15 AM
emakarov
Re: O Level maths needs Degree maths Knowledge
Quote:

Originally Posted by SMQ
In simpe terms if the factor is 0.0 the distribution will be £36,0000 / 36 Months giving £1000 in each of the months or 2.77% in each month producing a straight line when plotted on a graph if you adjust the factor the distribution changes - if you adjust the factor above zero it produces an inverse curve like the one shown.

Why do you call the curve corresponding to 0.6 "inverse"?

Quote:

Originally Posted by SMQ
On the other hand if you were to apply a negative factor of -0.6 it would produce a convex curve in the mirror image of the positive 0.6 factor curve.

The curve corresponding to 0.6 is already convex in the mathematical sense. What do you mean by "convex"? Also, the curve corresponding to -0.6 is the mirror image of the curve corresponding to 0.6 with respect to which straight line?