What's the difference between 1000e^0.05t and 1000*1.05^t?

In summary, the difference between 1000e^0.05t and 1000*1.05^t is that the first expression represents exponential growth with a constant value of 1000 and a growth rate of 5%, while the second expression represents compound interest with a starting value of 1000 and a growth rate of 5% per time period. This means that the first expression will result in a higher value over time, as the growth rate is applied to the current value each time period, while the second expression will result in a slower growth rate as the growth is only applied to the initial value. Additionally, the first expression is continuous, while the second expression is discrete.
  • #1
Karagoz
52
5
We have a population of y = 1000 at year 1980 (call it year 0).

Every year the population growth rate is 5% per year.

y' shows the growth rate of the y (population).

Since the population grows by 5% every year, the growth rate is:
y' = 0.05y.

This is a simple differential equation.
When y(0) = 1000

Then using a math software, the formula for the population is:
y(t) = 1000*e^(0.05t)

OR

We have a population of z = 1000 at year (1980) (call it year 0)

The population growth rate 5% per year.

Since the population grows by 5% per year, we can say:
z(t) = 1000*(1+0.05)^t = 1000*1.05^t

Derivation of z(t):
z’(t) = 1000(ln1.05)*e^(t*ln1.05)

Written as differential equation:

z’(t)=(ln1.05)*z(t)

The formula similar to z(t) is used when describing the growth of a money (in a bank at a interest rate of 5%).

Both the formula y(t) and formula z(t) describes growth rate by 5% per year.

But it’s obvious that z(t) ≠ y(t)

What is the difference between y(t) = 1000*e^(0.05t) and z(t) = 1000*1.05^t when both describes a growth rate of 5% per year?

What does z(t) describe and what does y(t) describe, and what’s the difference between what each formula describe?
 
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  • #2
Karagoz said:
What is the difference between y(t) = 1000*e^(0.05t) and z(t) = 1000*1.05^t when both describes a growth rate of 5% per year?
The difference is that y(t) is the exact description and z(t) is an approximation. If you write ##y(t)=y_0e^{\lambda~t}## and do a series expansion for the exponential, you get ##y(t)=y_0 (1 +\lambda t+\lambda t^2/2+\lambda t^3/6 +...)=y_0 \sum_{k=0} ^{\infty} (\lambda t)^k/k!##
Here ##y_0=1000## and ##\lambda = 0.05##. Your expression for z(t) tosses out all terms higher than first order, therefore it is approximately correct and not equal to y(t).
 
  • #3
To clarify my question:
When calculating what the value of the money C will be after x years, when having an interest rate at n, we use the formula:

f(x) = C*(1+n)^x

E.g. z(t) = 1000*(1+0.05)^t = 1000*1.05^t

But why don't we use the other formula: g(x) = C*e^(0.05n)?

What would it mean if we used g(x) instead of f(x) when calculating how the value C in a bank account will after x years with an interest rate n?
 
  • #4
Karagoz said:
What would it mean if we used g(x) instead of f(x) when calculating how the value C in a bank account will after x years with an interest rate n?
The plot below shows the difference 1000*e0.05*t - 1000*1.05t for 0 < t < 50 years. At the end of 50 years, you will be able to withdraw $12182.49 according to the exponential calculation and $11467.40 according to the approximation, a shortfall of $715.09.
Karagoz said:
But why don't we use the other formula: g(x) = C*e^(0.05n)?
It's not that "we" don't use the exponential formula, it's that the banks don't use it. Any guesses why? If you choose a bank that compounds interest more often than yearly, your interest will be reinvested sooner so the bank keeps less of your money.
InterestDifference.png
 

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  • #5
So the formula f(x) means yearly interest of n is compounded once a year.

But if banks did use g(x), that would mean the yearly interest of n is compounded far more frequently?
 
  • #6
Karagoz said:
But if banks did use g(x), that would mean the yearlt interest of n was compounded every second?
Or millisecond, or microsecond, or ...

Edit: Some banks compound interest quarterly, some monthly and some daily.
 

Related to What's the difference between 1000e^0.05t and 1000*1.05^t?

1. What do the variables in the equations represent?

The variable "t" represents time in years, while the number "1000" represents an initial value or amount.

2. What is the significance of the "^" symbol in the equations?

The "^" symbol represents an exponent, meaning that the number that follows it is raised to a certain power.

3. How do the two equations differ?

The first equation, 1000e^0.05t, uses the mathematical constant "e" which is approximately equal to 2.71828. The second equation, 1000*1.05^t, uses a constant multiplier of 1.05. This results in a difference in the rate of growth or decay over time.

4. Which equation should be used in different situations?

The choice of which equation to use depends on the specific situation and the type of growth or decay being measured. For example, if the growth or decay is continuous, such as in a biological population, the first equation may be more appropriate. If the growth or decay is discrete, such as in a financial investment, the second equation may be more applicable.

5. Can these equations be used to predict future values?

Yes, these equations can be used to predict future values by plugging in a specific value for "t" and solving for the resulting value. However, the accuracy of the prediction may depend on the accuracy of the initial values and the assumptions made about the growth or decay process.

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