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====== Differences ====== This shows you the differences between two versions of the page.

--- gibson:teaching:fall-2012:math445:lab3 [2012/09/05 19:38]
gibson
+++ gibson:teaching:fall-2012:math445:lab3 [2012/12/05 09:11] (current)
gibson [Problem F]
@@ Line 10: / Line 10: @@
 labeling different lines in the same plot via the ''legend'' command.
-A. Write a function ''mysort.m'' that will take in a vector of any size and sort
+==== Problem A====
+Write a function ''mysort.m'' that will take in a vector of any size and sort
 it from least to greatest. Use the strategy of comparing neighbors and swapping
 them if necessary. If this process is done throughout the entire vector
@@ Line 17: / Line 18: @@
 then by switching the outer loop to use a ''while'' command.
-B. Suppose you opened a savings account that promised a 100% interest rate
+==== Problem B====
+ Suppose you opened a savings account that promised a 100% interest rate
 (typical rates are closer to 2.25% today). We are going to try and calculate
 how much money that account would have after one year depending on how the
@@ Line 32: / Line 34: @@
 six months the other half of the interest is applied but this time to //B2// so
 that the ending balance is given by
 <latex>
 (1 + r/2)B2 = (1 + r/2)(1 + r/2)B = ((1 + r/2)^2 )B.
 </latex>
 Similarly if the interest was compounded quarterly the ending balance would be
 <latex>
 ((1 + r/4)^4 )B,
 </latex>
-and the balance at the end of //j//th quarter would be give by
+and the balance at the end of //j//th quarter would be given by
 <latex>
 ((1 + r/4)^j )B.
-<latex>
+</latex>
 Before you begin, enter the command ''format bank''. What did that command do?
 Compute the ﬁnal balance after one year of the initial investment of //$10,000//
@@ Line 48: / Line 56: @@
 quartely, monthly, and bi-weekly. Hint: one of your answers should be 26130.35.
-C. Now we will compound the interest weekly. Let's use a ''for'' loop to
+==== Problem C ====
+ Now we will compound the interest weekly. Let's use a ''for'' loop to
 compute not only how much money we will have in our account at the end of
 the year but each week as well. Use the command
 ''B=zeros(53,1)''
-to create a column vector of zeros values where we will store the account
+to create a column vector of zeros where we will store the account balances. Let
-balances. Let
 ''B(1)=10000''
 to set the initial balance. Next we need to compute the interest after each
 of the 52 weeks. We can use either of the following formulas,
 <latex>
 B_{j+1} = (1 + r/52)^j B_1
 </latex>
 or
 <latex>
 B_{j+1} = (1 + r/52) B_j
@@ Line 66: / Line 78: @@
 We can use the following code to evaluate the former formula
 <code>
 for j=1:52
   B(j+1)=(1+1.00/52)ˆj * B(1);
 end
-<code>
+</code>
-Now we can plot the results using the ''plot'' command:
-''plot(B)''
+Now we can plot the results using the ''plot'' command: ''plot(B)''
 Turn in a copy of your plot. Repeat the calculation using the equation
 <latex>
 B_{j+1} = (1 + r/52) B_j .
 </latex>
 making the appropriate adjustments to code given above. Either way you should
-get the exact same plot and a ﬁnal value of 26925.97 for B53 .
+get the exact same plot and a ﬁnal value of 26925.97 for B(53) .
-Bonus: Can you think of a way to calculate the same vector ''B'' in one line,
+**Bonus:** Can you think of a way to calculate the same vector ''B'' in one line,
 without using a loop?
-D. As the number of compounds increases, the ﬁnal amount appears to be get
+==== Problem D ====
-closer to some ﬁnal value. In order to check this, lets compound the interest
+As the number of compounds increases, the ﬁnal amount appears to be get
+closer to some ﬁnal value. In order to check this, let's compound the interest
 every second. Compute the result of compounding the interest every second on
-$10,000 at //r = 1.00// interest and check your answer versus the command
+''$''10,000 at //r = 1.00// interest and check your answer versus the command
 ''exp(1)*10000''
 How close are these two numbers? Note: the command
@@ Line 95: / Line 111: @@
 commonly in exponential and natural logarithms.
-E. At this point I am going to make a guess that
+====Problem E====
+At this point I am going to make a guess that
 <latex>
-(1 + 1.00/n)^n ≈ e.
+(1 + 1.00/n)^n \approx  e.
 </latex>
@@ Line 104: / Line 122: @@
 Mathematically, that is to say that limit as //n// goes to $\infty$ of
 //(1 + 1.00/n)^n// is //e//, or
 <latex>
 \lim_{n \rightarrow \infty} (1 + 1.00/n)^n = e.
 </latex>
-In fact the above statement is the deﬁnition of e. Lets test this idea of a
+In fact the above statement is the deﬁnition of e. Let's test this idea of a
-limit though. Using the model for ''for'' loops above create a vector of 20
+limit, though. Using the model for ''for'' loops above create a vector of 20
 values for ''n'' where $n_j = 2^j$ with //j = 1, ..., 20//. Then for each of
 the values of n, again using a ''for'' loop, calculate
-$a_j = (1 + 1.00/n_j )^{n_j}$ for ''j = 1, ..., 20''. Plot the values of //a//.
+$a_j = (1 + 1.00/n_j )^{n_j}$ for //j = 1, ..., 20//. Plot the values of //a//.
 Next plot //e − a//.  You should see a graph that is not very informative
 since the values quickly go to zero. Instead we will plot the graph on a log
@@ Line 121: / Line 140: @@
 Turn in all plots.
-Bonus: What happens if you let ''j = 1, ..., 60''? What happens if when you
+**Bonus:** What happens if you let //j = 1, ..., 60//? What happens if when you
-let $n_j = 10^j$ for ''j = 1,...,16''? Can you make a reasonable guess as to
+let $n_j = 10^j$ for //j = 1,...,16//? Can you make a reasonable guess as to
 what's happening here?
-F. Using the same procedure as in the previous problem, conﬁrm that
+==== Problem F ====
+Using the same procedure as in the previous problem, conﬁrm that
 <latex>
-\lim_{n\rightarrow \infty} (1 + 0.754/n)^nn = e^{0.754}
+\lim_{n\rightarrow \infty} (1 + 0.754/n)^n= e^{0.754}
 </latex>
 by calculating approximations and storing them in a vector //a// using the
 same values for //n// and //j// as used above. Plot //e^0.754 − a// on a log
@@ Line 134: / Line 156: @@
 the command ''exp(0.754)''.
-G. We have used several MATLAB functions so far. Now we are going to write our
+==== Problem G ====
+ We have used several MATLAB functions so far. Now we are going to write our
 own function. In the main MATLAB window click File → New → Script
 (or File → New → m-ﬁle depending on your version of MATLAB). This will open a
@@ Line 149: / Line 173: @@
 value for the variable ''Bal'' which will be returned by the function.
 Mathematically this is what the code needs to return
 <latex>
 Bal = (1 + Rate ∗ Yrs/Ncomp)^{Ncomp} Investment
 </latex>
 where ''Bal'' will be the balance you would have after ''Yrs'' years if you
 invested ''Investment'' amount of money at an interest rate of ''Rate'' that
@@ Line 168: / Line 194: @@
 the last line. Turn in the code for your function.
-H. Enter the command ''format long'' then verify the following properties of
+==== Problem H ====
+ Enter the command ''format long'' then verify the following properties of
 exponentials and logarithms by testing the appropriate MATLAB functions with
 the parameters indicated:
 <latex>
 e^a e^b = e^{a+b} \text{ for } a= 0.3, b= 0.4,
@@ Line 180: / Line 209: @@
 <latex>
-log(ab) = log(a) + log(b), \text{ for } a = 0.1, b = 5
+\log(ab) = \log(a) + \log(b), \text{ for } a = 0.1, b = 5
 </latex>
 <latex>
-log(a^b) = b log(a), \text{ for } a = 3, b = 3
+\log(a^b) = b \log(a), \text{ for } a = 3, b = 3
 </latex>
 Note the ''log'' function in matlab is the natural logarithm. How would you
-calculate $\log_{10}$ , $log_2$ , or $log_5$?
+calculate $\log_{10}$ , $\log_2$ , or $\log_5$?

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