Suppose you make a deposit of P into a savings account that...

Suppose you make a deposit of $\$ P$ into a savings account that earns interest at a rate of 100 $\%$ per year. a. Show that if interest is compounded once per year, then the balance after $t$ years is $B(t)=P(1+r)^{t}$ b. If interest is compounded $m$ times per year, then the balance after $t$ years is $B(t)=P(1+r / m)^{m t} .$ For example, $m=12$ corresponds to monthly compounding, and the interest rate for each month is $r / 12 .$ In the limit $m \rightarrow \infty,$ the compounding is said to be continuous. Show that with continuous compounding, the balance after $t$ years is $B(t)=\bar{P} e^{n}$

Suppose you make a deposit of $\$ P$ into a savings account that earns interest at a rate of 100 $\%$ per year.
a. Show that if interest is compounded once per year, then the balance after $t$ years is $B(t)=P(1+r)^{t}$
b. If interest is compounded $m$ times per year, then the balance after $t$ years is $B(t)=P(1+r / m)^{m t} .$ For example, $m=12$ corresponds to monthly compounding, and the interest rate for each month is $r / 12 .$ In the limit $m \rightarrow \infty,$ the compounding is said to be continuous. Show that with continuous compounding, the balance after $t$ years is $B(t)=\bar{P} e^{n}$

Key Concepts

Compound Interest

Compound Interest is the process by which an investment grows not only on the original principal but also on the accumulated interest from previous periods. This means that interest is earned on both the initial deposit and on the interest that is added over time, leading to exponential growth of the investment.

Discrete Compounding

Discrete Compounding refers to the calculation of interest at fixed, regular intervals, such as annually, monthly, or quarterly. In each period, interest is applied to the balance, and the resulting amount becomes the new principal for the next period. This leads to the formula B(t)=P(1+r/m)^(m*t), where m is the number of compounding periods per year.

Limit Process

The Limit Process involves analyzing the behavior of a function as one of its parameters approaches a particular value. In the context of interest compounding, it is used to transition from discrete to continuous compounding by letting the number of compounding periods per year tend to infinity. This mathematical concept simplifies the discrete compounding formula into one that describes continuous growth.

Continuous Compounding

Continuous Compounding is the scenario where interest is added to the principal incessantly, resulting in the balance growing according to an exponential function. Mathematically, it is derived as the limit of the discrete compounding formula when the compounding frequency becomes infinite, leading to the formula B(t)=P e^(r*t), where e is the base of the natural logarithm.

Transcript

00:01 So here, for part a, if we are going to find the interest or find the balance after one year, that's going to be our principle times our 1 plus r.

00:14 Right.

00:15 Now, if we do b of 2, we're going to take our balance from year 1, and then we're going to multiply that by our interest rate.

00:26 So this is going to be, if we substitute this in now, it's going to be p times 1.

00:32 Plus r times 1 plus r which is going to be equal to p times 1 plus r squared then if we try b3 remember b3 is going to be equal to b2 times 1 plus r and then remember b2 is equal to 1 r squared times 1 plus r so b3 is going to be equal to p times 1 plus r cubed and then we'll continue this pattern for any b of t we'll see that b of t is going to be 1 plus r to the t now for a compounded interest so b of t is going to be equal to so we'll have our initial p times 1 plus r divided by m times and then to the or sorry to the m t so we're taking the limit of this as m goes to infinity of and then instead of just of this whole thing, i'm just going to take the limit of this, because the limit of that times this later.

01:44 So we'll do 1 plus r divided by m to the mt.

01:49 Now, if we try a direct substitution, if we plug in infinity here, we'll get 1 plus 0, which is going to be 1, and then we'll get 1 to the infinity, which is indeterminate.

01:58 So i'm going to set that first equal to l.

02:01 Then i'm going to take the natural log of both sides.

02:04 So that's going to bring down the mt here.

02:07 So i get limit as m goes to infinity of, and i'm gonna go ahead and take ln of 1 plus r over m and then divided by, now this mt, which went out front here, i'm going to bring down into the numerator as 1 divided by mt, or 1 over m times 1 over t like so...

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