Natasha Oslinger

Scott’s Macho March Task

Math 3, Period 7

12-7-15

Table:

Scott’s Macho March Task

Math 3, Period 7

12-7-15

Table:

Equation:

g(n) and d(n)= n(n+2)

f(n)= 2n+1

m(n)= n(n+1) (2n+7) /6

Explanation:

Dear Scott,

To start your problem, I looked for obvious patterns and tried to interpret the question into a way I could better understand it. The most easily noticeable pattern is the push ups done each day last year, which is a simple growth of two for each day. With that knowledge, I found g(n) by adding the total of each of the days together starting from the day I was looking to find. Since he did the same number of push ups this year as he did last year, I knew d(n) had the same data as g(n). Then, to find m(n), I used the same method I used to find g(n).

After finding the data, I used what I knew to find formulas that could predict the next data point. I started with f(n), knowing it would be the most simple. I knew that I needed to make a formula that would always be constant, and I knew generally the parts I would need to make the final problem. Work:

f(n) + f(n-1) + f(n-2)=

(2n+1) + 2 (n-1) +1=

2(n-2)+1

In the end, I found the simple formula f(n)=2n+1, with “n” being the amount of days. This formula is linear and is made by adding the slope to the problem each time. For example, if he was looking at how many push ups he did last day each year on day 3, plug in 3 for n, 2*3+1 and solve to get 7. By checking the chart I made earlier, I can see that this formula was correct and found the correct answer every time.

Next, I would try to find the formula for g(n) and d(n). I knew I needed to start with the base formula n(n-1)/2. Work:

2(1+2+3+4)

2*10

n+2 (n+1)/2

n^2 +n+n= n^2+2n

n(n+2)

For the final formula for g(n) and d(n) I found n(n+2) to be the simplest solution. Finding this formula was very easy, and I can prove it to be correct by plugging in data and solving. For example if n was 3 days, I could plug in 3(3+2) to get 15. Then, by looking at the chart I made earlier, I can see that is formula is correct.

The final formula is the most complicated and took me an unreasonably long amount of time to figure out. I figured out that m(n) was not a power function and that the ratios didn’t seem to make sense. I realized that the ration were getting smaller, and I wondered if there was a limit. I explored the idea of using summation, but I figured it was probably had a better quadratic or cubic fit. In the end, I figured the best method to find the formula would be to use graphing. When I graphed it, I could see if the problem would be linear, quadratic, or cubic ect.. Essentially, I used guess and check work:

4.5x^2 -6.7x +5.6

45x^2 -67x +56 /10

1/3n^3 +3/2n^2 +7/6n

2n^3+9n^2+7n /6 =cubic ?

m(n) = 2n^3 +9n^2 +7n /6 ← the difference between two quadratics

*the difference had to be n^2+2n

⅙ (n) (2n^2 +9n +7) ← easy quadratic

n/6 (2n+7) (n+1)

2n^2 +2n+7n+7

2n^2+9n+7

n(n+1)(2n+7)/6

I found this formula after a lot of trial and error. I know this formula is correct because I can plug in whatever day I need to find the number of push ups in for n and find m(n). For example, if n is 1, 3(3+1)(2*3+7) /6, n=26. By checking my graph, I can see that this formula is correct.

There are many recognizable patterns that helped me through this problem, and overall I found the first two equations mostly by figuring out patterns. You can use these three formulas to figure out how many push ups you did last year and how many to do this year. The first graph is of f(n) and g(n) and grows at a pretty rapid but constant pace. The points are evenly distributed and stay pretty close to the y-axis. The second graph is of d(n) and m(n) and grows at a very rapid and slightly irregular pace. The points grow farther and farther apart as they get larger and get farther away from the y-axis than the first graph. Overall, this problem was soloable but challenging, and I all of the work I have done is correct and can be backed up mathematically.

g(n) and d(n)= n(n+2)

f(n)= 2n+1

m(n)= n(n+1) (2n+7) /6

Explanation:

Dear Scott,

To start your problem, I looked for obvious patterns and tried to interpret the question into a way I could better understand it. The most easily noticeable pattern is the push ups done each day last year, which is a simple growth of two for each day. With that knowledge, I found g(n) by adding the total of each of the days together starting from the day I was looking to find. Since he did the same number of push ups this year as he did last year, I knew d(n) had the same data as g(n). Then, to find m(n), I used the same method I used to find g(n).

After finding the data, I used what I knew to find formulas that could predict the next data point. I started with f(n), knowing it would be the most simple. I knew that I needed to make a formula that would always be constant, and I knew generally the parts I would need to make the final problem. Work:

f(n) + f(n-1) + f(n-2)=

(2n+1) + 2 (n-1) +1=

2(n-2)+1

In the end, I found the simple formula f(n)=2n+1, with “n” being the amount of days. This formula is linear and is made by adding the slope to the problem each time. For example, if he was looking at how many push ups he did last day each year on day 3, plug in 3 for n, 2*3+1 and solve to get 7. By checking the chart I made earlier, I can see that this formula was correct and found the correct answer every time.

Next, I would try to find the formula for g(n) and d(n). I knew I needed to start with the base formula n(n-1)/2. Work:

2(1+2+3+4)

2*10

n+2 (n+1)/2

n^2 +n+n= n^2+2n

n(n+2)

For the final formula for g(n) and d(n) I found n(n+2) to be the simplest solution. Finding this formula was very easy, and I can prove it to be correct by plugging in data and solving. For example if n was 3 days, I could plug in 3(3+2) to get 15. Then, by looking at the chart I made earlier, I can see that is formula is correct.

The final formula is the most complicated and took me an unreasonably long amount of time to figure out. I figured out that m(n) was not a power function and that the ratios didn’t seem to make sense. I realized that the ration were getting smaller, and I wondered if there was a limit. I explored the idea of using summation, but I figured it was probably had a better quadratic or cubic fit. In the end, I figured the best method to find the formula would be to use graphing. When I graphed it, I could see if the problem would be linear, quadratic, or cubic ect.. Essentially, I used guess and check work:

4.5x^2 -6.7x +5.6

45x^2 -67x +56 /10

1/3n^3 +3/2n^2 +7/6n

2n^3+9n^2+7n /6 =cubic ?

m(n) = 2n^3 +9n^2 +7n /6 ← the difference between two quadratics

*the difference had to be n^2+2n

⅙ (n) (2n^2 +9n +7) ← easy quadratic

n/6 (2n+7) (n+1)

2n^2 +2n+7n+7

2n^2+9n+7

n(n+1)(2n+7)/6

I found this formula after a lot of trial and error. I know this formula is correct because I can plug in whatever day I need to find the number of push ups in for n and find m(n). For example, if n is 1, 3(3+1)(2*3+7) /6, n=26. By checking my graph, I can see that this formula is correct.

There are many recognizable patterns that helped me through this problem, and overall I found the first two equations mostly by figuring out patterns. You can use these three formulas to figure out how many push ups you did last year and how many to do this year. The first graph is of f(n) and g(n) and grows at a pretty rapid but constant pace. The points are evenly distributed and stay pretty close to the y-axis. The second graph is of d(n) and m(n) and grows at a very rapid and slightly irregular pace. The points grow farther and farther apart as they get larger and get farther away from the y-axis than the first graph. Overall, this problem was soloable but challenging, and I all of the work I have done is correct and can be backed up mathematically.