Problem
Pick any exercise in 8.8 of the calculus textbook. Solve and post your solution. If you have issues doing so, discuss them.
Solution
The problems I will be picking up are: Exercises 8.1, problems #9, #10, and #11.
In Exercises 9 – 12, determine the \(n-th\) term of the given sequence.
Please note that I will be applying some series of procedures and equations that I’ve been working on but I haven’t formalized.
Disclaimer: If someone has seeing something like the steps and algorithms that I will apply please point it out and let me know, that way I will stop taking credit since this has been a great discovery from long time ago. These steps really help in order to avoid guessing such formulas but I never ended up formalizing these findings. I have tested it with many complex polynomials of diverse degrees and some exponential functions and it works in a beautiful way!
Basically I will employ a series of steps in order NOT to guess the succession formula but determining it using mathematical procedures. I will call this in the mean time “Duubar’s Procedures in order to find the generating succesion formula from a series of given points”.
Problem # 9
Determine the \(n-th\) term of the given sequence of 4, 7, 10, 13, 16, . . .
Procedure:
- Assign paring counters for such values, I usually like to start with 0 but this could be any starting value and the final function will auto adjust in order to provide the correct values.
| i | series |
|---|---|
| 0 | 4 |
| 1 | 7 |
| 2 | 10 |
| 3 | 13 |
| 4 | 16 |
| … | … |
- Subtract the values from the \(Subs_n = series_{n+1} - series_{n}\); until we can not reduce it.
| i | series | subs1 | subs1Results |
|---|---|---|---|
| 0 | 4 | 7-4 | 3 |
| 1 | 7 | 10-7 | 3 |
| 2 | 10 | 13-10 | 3 |
| 3 | 13 | 16-13 | 3 |
| 4 | 16 | … | … |
| … | … |
- Repeat the same procedure until we reach to zero,
| i | series | subs1 | subs1Results | subs2 | subs2Results |
|---|---|---|---|---|---|
| 0 | 4 | 7-4 | 3 | 3-3 | 0 |
| 1 | 7 | 10-7 | 3 | 3-3 | 0 |
| 2 | 10 | 13-10 | 3 | 3-3 | 0 |
| 3 | 13 | 16-13 | 3 | … | … |
| 4 | 16 | … | … | … | … |
| … | … |
- Now, that we have those results, we count how many subtractions we had to perform in order to obtain the zero column; in this case we had to perform two subtractions; I will say that we then need to solve a system of equations of degree \(0 + 1 = 1\) degrees.
Since this is a linear equation, and we know is in the form
\(y = mx + b\)
We can take any two different ordinate pairs in the form \((x_i, y_i)\) and solve a systems of equations as follows:
I will take \(i = 0\) and \(i = 1\).
\(series_0 = m \cdot i_0 + b\)
\(series_1 = m \cdot i_1 + b\)
We could use row reduction if the polynomial function is bigger or any other form; I will just replace with values for simplicity.
\(4 = 0 \cdot m + b \quad \implies b=4\)
\(7 = 1 \cdot m + b \quad \implies 7 = m + 4 \quad \implies m = 3\)
\(\therefore a_n = 3n + 4\)
You can test the above formula by writing your own test.
Problem # 10
Determine the \(n-th\) term of the given sequence of 3, \(\frac{-3}{2}\), \(\frac{3}{4}\), \(\frac{-3}{8}\), . . .
Procedure:
For this I will separate the numerator from the denominator and I will treat them as two different series.
- Assign paring counters for such values, I usually like to start with 0 but this could be any starting value and the final function will auto adjust in order to provide the correct values.
| i | NumSeries | DenSeries | Comb |
|---|---|---|---|
| 0 | 3 | 1 | 3/1 |
| 1 | -3 | 2 | -3/2 |
| 2 | 3 | 4 | 3/4 |
| 3 | -3 | 8 | -3/8 |
| … | … | … | …/… |
1.a) Let’s focus on the numerator first.
As you can see there’s an alternating sign and a constant; so the only thing to do for the alternating sign is to define \((-1)^n\); as you can see this will reproduce the signs in the order we need and we will keep the \(3\) since it does not vary.
| i | NumSeries | DenSeries | Comb | NumRes |
|---|---|---|---|---|
| 0 | 3 | 1 | 3/1 | 3 x (-1)^0 = 3 |
| 1 | -3 | 2 | -3/2 | 3 x (-1)^1 = -3 |
| 2 | 3 | 4 | 3/4 | 3 x (-1)^2 = 3 |
| 3 | -3 | 8 | -3/8 | 3 x (-1)^3 = -3 |
| … | … | … | …/… | … |
1.b) Now, I will focus same as the previous example but with the denominator values only.
- Subtract the values from the \(Subs_n = series_{n+1} - series_{n}\); until we can not reduce it.
| i | NumSeries | DenSeries | Comb | DenSubs1 | DenSubs1Res |
|---|---|---|---|---|---|
| 0 | 3 | 1 | 3/1 | 2-1 | 1 |
| 1 | -3 | 2 | -3/2 | 4-2 | 2 |
| 2 | 3 | 4 | 3/4 | 8-4 | 4 |
| 3 | -3 | 8 | -3/8 | … | … |
| … | … | … | …/… |
Repeat the same procedure.
| i | NumSeries | DenSeries | Comb | DenSubs2 | DenSubs2Res |
|---|---|---|---|---|---|
| 0 | 3 | 1 | 3/1 | 2-1 | 1 |
| 1 | -3 | 2 | -3/2 | 4-2 | 2 |
| 2 | 3 | 4 | 3/4 | … | … |
| 3 | -3 | 8 | -3/8 | … | … |
| … | … | … | …/… |
Repeat the same procedure
| i | NumSeries | DenSeries | Comb | DenSubs3 | DenSubs3Res |
|---|---|---|---|---|---|
| 0 | 3 | 1 | 3/1 | 2-1 | 1 |
| 1 | -3 | 2 | -3/2 | … | … |
| 2 | 3 | 4 | 3/4 | … | … |
| 3 | -3 | 8 | -3/8 | … | … |
| … | … | … | …/… |
In this case, for the denominator, we have identified that it is an exponential sequence with the following form:
\(den_n = a \cdot 2^n\)
in this case our value \(a = 1\) obtained from the last column.
\(den_n = 2^n\)
- Putting all together.
\[a_n = \frac{formula \: numerator}{formula \: denominator}\]
\[\therefore a_n = (-1)^n\frac{3}{2^n}\]
Problem # 11
Determine the \(n-th\) term of the given sequence of 10, 20, 40, 80, 160, . . .
Procedure:
- Assign paring counters for such values, I usually like to start with 0 but this could be any starting value and the final function will auto adjust in order to provide the correct values.
| i | series |
|---|---|
| 0 | 10 |
| 1 | 20 |
| 2 | 40 |
| 3 | 80 |
| 4 | 160 |
| … | … |
- Subtract the values from the \(Subs_n = series_{n+1} - series_{n}\); until we can not reduce it.
| i | series | subs1 | subs1Results |
|---|---|---|---|
| 0 | 10 | 20-10 | 10 |
| 1 | 20 | 40-20 | 20 |
| 2 | 40 | 80-40 | 40 |
| 3 | 80 | 160-80 | 80 |
| 4 | 160 | … | … |
| … | … |
In this case, I will repeat the process until we can reduce since we don’t have a lot of points
| i | series | subs1 | subs1Results | subs2 | subs2Results |
|---|---|---|---|---|---|
| 0 | 10 | 20-10 | 10 | 20-10 | 10 |
| 1 | 20 | 40-20 | 20 | 40-20 | 20 |
| 2 | 40 | 80-40 | 40 | 80-40 | 40 |
| 3 | 80 | 160-80 | 80 | … | … |
| 4 | 160 | … | … | … | … |
| … | … |
Repeat one more time
| i | series | subs1 | subs1Results | subs2 | subs2Results | subs3 | subs3Results |
|---|---|---|---|---|---|---|---|
| 0 | 10 | 20-10 | 10 | 20-10 | 10 | 20-10 | 10 |
| 1 | 20 | 40-20 | 20 | 40-20 | 20 | 40-20 | 20 |
| 2 | 40 | 80-40 | 40 | 80-40 | 40 | … | … |
| 3 | 80 | 160-80 | 80 | … | … | … | … |
| 4 | 160 | … | … | … | … | … | … |
| … | … |
Repeat one last time
| i | series | subs1 | subs1Results | subs2 | subs2Results | subs3 | subs3Results | subs4 | subs4Results |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 10 | 20-10 | 10 | 20-10 | 10 | 20-10 | 10 | 20-10 | 10 |
| 1 | 20 | 40-20 | 20 | 40-20 | 20 | 40-20 | 20 | … | … |
| 2 | 40 | 80-40 | 40 | 80-40 | 40 | … | … | … | … |
| 3 | 80 | 160-80 | 80 | … | … | … | … | … | … |
| 4 | 160 | … | … | … | … | … | … | … | … |
| … | … |
In this case we, can quickly identify that it is very similar to our previous case, making it an exponential formula with the form
\(a_n = a \cdot 2^n\) in this case \(a = 10\)
- Formula
\(\therefore a_n = 10 \cdot 2^n\)
Please let me know your input!
END