/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 We return to the example begun i... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 6

We return to the example begun in Preview Activity 8.1 .1 to see how to derive the formula for the amount of money in an account at a given time. We do this in a general setting. Suppose you invest \(P\) dollars (called the principal) in an account paying \(r \%\) interest compounded monthly. In the first month you will receive \(\frac{r}{12}\) (here \(r\) is in decimal form; e.g., if we have \(8 \%\) interest, we write \(\frac{0.08}{12}\) ) of the principal \(P\) in interest, so you earn $$ P\left(\frac{r}{12}\right) $$ dollars in interest. Assume that you reinvest all interest. Then at the end of the first month your account will contain the original principal \(P\) plus the interest, or a total of $$ P_{1}=P+P\left(\frac{r}{12}\right)=P\left(1+\frac{r}{12}\right) $$ dollars. a. Given that your principal is now \(P_{1}\) dollars, how much interest will you earn in the second month? If \(P_{2}\) is the total amount of money in your account at the end of the second month, explain why $$ P_{2}=P_{1}\left(1+\frac{r}{12}\right)=P\left(1+\frac{r}{12}\right)^{2} $$ b. Find a formula for \(P_{3}\), the total amount of money in the account at the end of the third month in terms of the original investment \(P\). c. There is a pattern to these calculations. Let \(P_{n}\) the total amount of money in the account at the end of the third month in terms of the original investment \(P\). Find a formula for \(P_{n}\)

Short Answer

Expert verified
The amount in the account after \(n\) months is given by \(P_n = P \left( 1 + \frac{r}{12} \right)^n\).

Step by step solution

01

Calculate Interest for Second Month

After the first month, the principal has grown to \(P_1 = P \left( 1 + \frac{r}{12} \right)\). The interest for the second month is calculated based on this new principal amount. Thus, the interest for the second month is \(P_1 \left( \frac{r}{12} \right)\).
02

Calculate Total Amount after Second Month

The total amount at the end of the second month, \(P_2\), will be the principal from the end of the first month plus the interest earned in the second month. This can be expressed as: \(P_2 = P_1 + P_1 \left( \frac{r}{12} \right)\) Implementing the value of \(P_1\), we get: \(P_2 = P \left( 1 + \frac{r}{12} \right) \left( 1 + \frac{r}{12} \right) = P \left( 1 + \frac{r}{12} \right)^2\).
03

Calculate Total Amount after Third Month

Similarly, the principal after the third month, \(P_3\), can be calculated by applying the same process: \(P_3 = P_2 \left( 1 + \frac{r}{12} \right)\) Using the value of \(P_2\), we have: \(P_3 = P \left( 1 + \frac{r}{12} \right)^2 \left( 1 + \frac{r}{12} \right) = P \left( 1 + \frac{r}{12} \right)^3\).
04

Generalize the Pattern

We observe a pattern where the principal after \(n\) months can be generalized as: \(P_n = P \left( 1 + \frac{r}{12} \right)^n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Investment Growth
Investment growth is the process of how your invested money increases over time. This growth occurs due to the interest earned on the original investment, also known as the principal. The more interest you earn, the more your investment grows. For instance, if you start with an initial principal of \(P\) dollars, over time, your investment will grow as you continue to earn interest on the total amount in your account. The key to understanding investment growth is recognizing that your principal amount continuously increases due to the earned interest being reinvested.
Monthly Compounding
Monthly compounding refers to how interest is calculated and added to your account each month. With monthly compounding, the interest earned during one month gets added to the principal for the calculation of the next month鈥檚 interest. This means that each month, you鈥檒l earn interest on a slightly larger principal amount than the month before. For example, if your annual interest rate is \(r\) (in decimal form), the monthly interest rate would be \(\frac{r}{12}\). As a result, after the first month, your new principal would be \(P_1 = P \left( 1 + \frac{r}{12} \right)\). Each subsequent month follows the same pattern.
Mathematical Formula Derivation
Deriving the formula for the total amount of money in your account involves understanding the repeated application of interest calculation each month. We start with our original principal \(P\).
After one month, the interest added to the principal is given by \(P \left( \frac{r}{12} \right)\). Thus, the new principal after one month becomes $$ P_1 = P + P \left( \frac{r}{12} \right) = P \left( 1 + \frac{r}{12} \right) $$.
For the second month, the same process is applied to \(P_1\). Therefore, the new principal becomes $$ P_2 = P_1 \left( 1 + \frac{r}{12} \right) = P \left( 1 + \frac{r}{12} \right) \,^2 $$.
Extending this pattern, after the third month, we get $$ P_3 = P_2 \left( 1 + \frac{r}{12} \right) = P \left( 1 + \frac{r}{12} \right) \,^3 $$.
Thus, the general formula after \(n\) months is $$ P_n = P \left( 1 + \frac{r}{12} \right) \,^n $$.
Interest Rate Calculations
Interest rate calculations help determine how much interest you will earn over a specific period. To calculate monthly interest, we divide the annual interest rate \(r\) by 12 (since there are 12 months in a year). This results in the monthly interest rate of \(\frac{r}{12}\).
The interest for each month is then calculated based on the current principal. For example, if the principal is \(P_1\) at the start of a month, the interest for that month will be \(P_1 \times \frac{r}{12}\).
By understanding how to calculate interest each month, you can effectively predict and track the growth of your investment over time. This also forms the basis for creating the general formula for the total amount \(P_n\) in the account after \(n\) months.

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Most popular questions from this chapter

For the following alternating series, \(\sum_{n=1}^{\infty} a_{n}=1-\frac{1}{10}+\frac{1}{100}-\frac{1}{1000}+\ldots\) how many terms do you have to go for your approximation (your partial sum) to be within 1e-07 from the convergent value of that series?

Airy's equation \(^{2}\) $$ y^{\prime \prime}-x y=0 $$ can be used to model an undamped vibrating spring with spring constant \(x\) (note that \(y\) is an unknown function of \(x\) ). So the solution to this differential equation will tell us the behavior of a spring-mass system as the spring ages (like an automobile shock absorber). Assume that a solution \(y=f(x)\) has a Taylor series that can be written in the form $$ y=\sum_{k=0}^{\infty} a_{k} x^{k} $$ where the coefficients are undetermined. Our job is to find the coefficients. (a) Differentiate the series for \(y\) term by term to find the series for \(y^{\prime}\). Then repeat to find the series for \(y^{\prime \prime}\). (b) Substitute your results from part (a) into the Airy equation and show that we can write Equation \((8.6 .4)\) in the form$$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}-\sum_{k=0}^{\infty} a_{k} x^{k+1}=0 $$ (c) At this point, it would be convenient if we could combine the series on the left in \((8.6 .5)\), but one written with terms of the form \(x^{k-2}\) and the other with terms in the form \(x^{k+1}\). Explain why $$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}=\sum_{k=0}^{\infty}(k+1)(k+2) a_{k+2} x^{k} $$ (d) Now show that $$ \sum_{k=0}^{\infty} a_{k} x^{k+1}=\sum_{k=1}^{\infty} a_{k-1} x^{k} $$ (e) We can now substitute \((8.6 .6)\) and \((8.6 .7)\) into \((8.6 .5)\) to obtain $$ \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^{n}-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0 $$ Combine the like powers of \(x\) in the two series to show that our solution must satisfy $$ 2 a_{2}+\sum_{k=1}^{\infty}\left[(k+1)(k+2) a_{k+2}-a_{k-1}\right] x^{k}=0 $$ (f) Use equation (8.6.9) to show the following: i. \(a_{3 k+2}=0\) for every positive integer \(k\), ii. \(a_{3 k}=\frac{1}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)} a_{0}\) for \(k \geq 1\), iii. \(a_{3 k+1}=\frac{1}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)} a_{1}\) for \(k \geq 1\). (g) Use the previous part to conclude that the general solution to the Airy equation \((8.6 .4)\) is $$ \begin{aligned} y=& a_{0}\left(1+\sum_{k=1}^{\infty} \frac{x^{3 k}}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)}\right) \\ &+a_{1}\left(x+\sum_{k=1}^{\infty} \frac{x^{3 k+1}}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)}\right) \end{aligned} $$ Any values for \(a_{0}\) and \(a_{1}\) then determine a specific solution that we can approximate as closely as we like using this series solution.

For the following alternating series, \(\sum_{n=1}^{\infty} a_{n}=0.45-\frac{(0.45)^{3}}{3 !}+\frac{(0.45)^{5}}{5 !}-\frac{(0.45)^{7}}{7 !}+\ldots\) how many terms do you have to compute in order for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series?

Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to \(\infty\), and -infinity if it diverges to \(-\infty\). Otherwise, enter diverges. $$ \int_{1}^{\infty} \frac{3 d x}{x^{2}+1}= $$ Does the series \(\sum_{n=1}^{\infty} \frac{3}{n^{2}+1}\) converge or diverge? [Choose: converges | diverges to +infinity | diverges to -infinity | diverges]

Determine the sum of the following series. $$ \sum_{n=1}^{\infty}\left(\frac{3^{n}+5^{n}}{9^{n}}\right) $$
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