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Here is the question A sum of 25000 is compounded annually at 6.5 percent per annum for 3 years.Calculate the amount received after 3 years?

When I solved it using this formula C.A :-Final Amount P:-Principle amount R:- Rate T:- Period

C.A=P(1+R/100)^T

C.A.=3019.87

While When I used the differential equation

So:-Initial Amount S:- Final Amount R:- Rate T:- Time Period

S=So * e^(RT)*

S=30382.77

Why are the two answers different shouldn't they should be equal?? Thank you.

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The second answer represents the result of compounding continuously, while the first answer represents the result of compounding only once per year.

The more frequently compounding occurs at a fixed rate, the more interest is earned. Continuous compounding is more frequent than any periodic compounding (in the limit, the time between subsequent compoundings drops to zero).

The continuous compounding rate $\delta$ that gives the same result as the annual compounding rate $6.5\%$ can be determined by solving

$$e^{\delta} = 1.065$$ $$\delta = \ln 1.065 \approx 0.062975 = 6.2975\%$$

In other words, continuous compounding at $6.2975\%$ is equivalent to annual compounding at $6.5\%$.

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  • $\begingroup$ I got your point but not completely.During the derivation of differential equation we never assume that t are in terms of days or years. So why adding years or days makes a difference for variable t.Thank you $\endgroup$ Jul 17 '20 at 14:14
  • $\begingroup$ A differential equation can only model continuous compounding. For periodic compounding, the amount $A(t)$ in the account is constant and equal to $A(0)$ for the whole year until the moment compounding occurs, at which time the interest for the year is deposited and $A(1) = A(0) + 0.065\cdot A(0)$. Then it remains constantly equal to this value for another year, until the next interest deposit occurs, etc. $\endgroup$
    – MPW
    Jul 17 '20 at 14:22
  • $\begingroup$ (continued)... Now because the value is mostly constant, $A'(t)=0$ at all times $t$ except for the instant the interest is deposited at the end of each year; at those instants ($A(1), A(2), A(3), \cdots$) the function $A$ has a jump discontinuity and so is not differentiable. Long story short, a differential equation can't model a series of discrete deposits. The value $A(t)$ satisfies $\frac{dA(t)}{dt}=0$ except at the end of each year! $\endgroup$
    – MPW
    Jul 17 '20 at 14:22
  • $\begingroup$ If I got it right .You are implying that the differential equation calculates the compounds each second.While the formula only once per year right?Thank you $\endgroup$ Jul 17 '20 at 14:32
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    $\begingroup$ Not just each second, even faster! Every single instant! But yes, that's the idea. The interest flows into the account like water trickling into a tank. It doesn't get dropped in a quart at a time, or even a teaspoon at a time. It constantly flows in. That's continuous compounding. Transferring water into a tank by repeatedly scooping in some amount at discrete moments is like periodic compounding. Turning on the hose is like continuous compounding $\endgroup$
    – MPW
    Jul 17 '20 at 14:35
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The answers are different because the formulae are for different circumstances.

For the first formula, the interest is assumed to be paid yearly. So, the interest of the first day, starts earning interest only in the second year.

For the second formula, the interest of the first day starts earning further interest on the second day and so on.Thus, the final amount is larger.

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  • $\begingroup$ I got your point but not completely.During the derivation of differential equation we never assume that t are in terms of days or years. So why adding years or days makes a difference for variable t.Thank you $\endgroup$ Jul 17 '20 at 14:14
  • $\begingroup$ During the derivation using differential equation, we assume time (and hence, the extra money) elapses instantaneously, not yearly, monthly or even secondly. $\endgroup$
    – Beetel
    Aug 11 '20 at 11:14

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