Thread: Economic Mathematical Equation which is constantly decreasing.

1. Economic Mathematical Equation which is constantly decreasing.

I don't know where this goes so I'll just put this here. It's a question regarding a decreasing value over time.

If you have an equation B = (1 - R) A + I,

where A is the value of stock in one period, and B is the value of stock in a next period,
R is a depreciation rate of 2%, and I is an investment value added upon every period.

Would the value ever become constant?

E.G. the Initial value of capital stock is 200, and I = 10, R = 0.02

where you get the first period formula to be (1 - 0.02)*200 + 10 = 206
second period (1-0.02)*206 + 10 = 211.88
third period (1-0.02)*211.88 + 10 = 217.6424

The difference between each period becomes smaller just by inspecting that 211.88 - 206 > 217.6424 - 211.88.

However, since it is a depreciation rate, the value becomes continuously smaller. So, does the stock eventually become constant? I assumed it shouldn't, but the answers differ.

This is a macroeconomics class so I'm unsure if they take mathematical theories as seriously. The answers say that it does, and perhaps this is due to the disregard of dollar value if it reaches a point where it is too miniscule to consider.

2. Re: Economic Mathematical Equation which is constantly decreasing.

Just a wild guess - does the long term value settle at 500? Must be 500.

3. Re: Economic Mathematical Equation which is constantly decreasing.

Originally Posted by Drongoski
Just a wild guess - does the long term value settle at 500? Must be 500.
Yes.

Originally Posted by ProdigyInspired
I don't know where this goes so I'll just put this here. It's a question regarding a decreasing value over time.

If you have an equation B = (1 - R) A + I,

where A is the value of stock in one period, and B is the value of stock in a next period,
R is a depreciation rate of 2%, and I is an investment value added upon every period.

Would the value ever become constant?

E.G. the Initial value of capital stock is 200, and I = 10, R = 0.02

where you get the first period formula to be (1 - 0.02)*200 + 10 = 206
second period (1-0.02)*206 + 10 = 211.88
third period (1-0.02)*211.88 + 10 = 217.6424

The difference between each period becomes smaller just by inspecting that 211.88 - 206 > 217.6424 - 211.88.

However, since it is a depreciation rate, the value becomes continuously smaller. So, does the stock eventually become constant? I assumed it shouldn't, but the answers differ.

This is a macroeconomics class so I'm unsure if they take mathematical theories as seriously. The answers say that it does, and perhaps this is due to the disregard of dollar value if it reaches a point where it is too miniscule to consider.
"Would the value ever become constant?"

Yes, it asymptotically approaches a constant.

$\noindent We have X_{n+1} = cX_{n} + I, where I is constant and c \equiv 1 - R (replacing B with X_{n+1} and A with X_{n} to reflect evolution in time). This is an inhomogeneous first order difference equation with characteristic root c, and |c| < 1 with your value of R. This (that the characteristic root satisfies |c| < 1) implies that X_{n} approaches the particular solution'' part of this difference equation as n\to \infty. As this particular solution is a constant (because I is constant and c\neq 1), X_{n} will tend to a constant as n\to \infty. In fact you can show that X_{n} = \alpha c^{n} + \frac{I}{1-c} for some constant \alpha (that depends on the initial value X_{0}), and so as n\to \infty, we have that X_{n} converges to the constant value \frac{I}{1-c} = \frac{I}{R}, which in your case equals \frac{10}{0.02} = 500.$

$\noindent (By the way, this difference equation is structurally the same as the ones you would have got in the loan repayments questions in HSC 2U maths, if you remember how to solve those questions.)$

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